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https://mathoverflow.net/questions/446484 | 0 | Hermite polynomials $H\_k(x), x \in \mathbb{R}, k \in \mathbb{N}$ are defined by the formula
$$
H\_k(x)=(-1)^k e^{x^2} \frac{d^k}{d x^k}\left(e^{-x^2}\right) .
$$
Each $H\_k(x)$ is a polynomial of exact degree $k$. The Hermite polynomials are also given by the generating function
$$
e^{2 x w-w^2}=\sum\_{k=0}^{\infty} \frac{H\_k(x)}{k !} w^k
$$
Define the Hermite functions $\tilde{h}\_k(x)$ by
$$
\tilde{h}\_k(x)=H\_k(x) e^{-\frac{1}{2} x^2} .
$$
We have the Mehler's formula, for the Hermite functions $\tilde{h}\_k(x)$.
**Proposition** For $w \in \mathbb{C},|w|<1$ and $x, y \in \mathbb{R}$,
$$\sum\_{k=0}^{\infty} \frac{\tilde{h}\_k(x) \tilde{h}\_k(y)}{2^k k !} w^k=\left(1-w^2\right)^{-\frac{1}{2}} e^{-\frac{1}{2} \frac{1+w^2}{1-w^2}\left(x^2+y^2\right)+\frac{2 w}{1-w^2} x y}$$
One has $$\int\_\Bbb R\left(\tilde{h}\_k(x)\right)^2 d x=2^k k ! \sqrt{\pi} .$$
Thus we can normalise $\tilde{h}\_k(x)$ by defining
$$h\_k(x)=\left(2^k k ! \sqrt{\pi}\right)^{-\frac{1}{2}} \tilde{h}\_k(x)$$
This family $\{h\_k(x): k \in \mathbb{N}\}$ is an orthonormal system in $L^2(\mathbb{R})$. But we can say more.
**Theorem** The system $\{h\_k(x): k \in \mathbb{N}\}$ is an orthonormal basis for $L^2(\mathbb{R})$. Consequently, every $f \in L^2(\mathbb{R})$ has an expansion
$$f(x)=\sum\_{k=0}^{\infty}\left(f, h\_k\right) h\_k(x)$$
where the series converges to $f$ in the $L^2$ norm.
My question is there a close formula for this sum: $$\sum\_{k=0}^{\infty}\frac{1}{k+a}h\_k(x)h\_k(y)$$
| https://mathoverflow.net/users/172078 | Closed formula for Hermite polynomials | Up to some normalization, the harmonic oscillator $H$ is self-adjoint such that
$$
\langle Hu, u\rangle=\sum\_{k\ge 0}(\frac12+k) \vert u\_k\vert^2,
$$
and thus defining a self-adjoint $A$ by the equality
$$
\langle Au, u\rangle=\sum\_{k\ge 0}(a+k) \vert u\_k\vert^2,
\quad\text{implying}\ A=H+a-\frac12.
$$
As a result your sum is the kernel of the operator
$$
(H+a-\frac12)^{-1},
$$
which makes sense for $a>0$.
| 3 | https://mathoverflow.net/users/21907 | 446509 | 179,893 |
https://mathoverflow.net/questions/446437 | 10 | **I. Four quintics?**
The general quintic can be transformed in radicals to at least ***three*** one-parameter forms. For simplicity, assume this free parameter to be some generic "*alpha*". Hence,
$$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0\tag1$$
$$x^5-5\alpha x -\alpha = 0\tag2$$
$$x^5+5\sqrt{\alpha}\, x^2 -\sqrt{\alpha} = 0\tag3$$
which are the *Brioschi*, *Bring-Jerrard*, and *Bring-Euler* quintics, respectively. Naturally, these are $5T5$ with order $5!=120$. Their discriminants are,
\begin{align}
d\_1 &= 5^5\,(1-1728\alpha)^2\,\alpha^8\\
d\_2 &= 5^5\,(1-256\alpha)\,\alpha^4\\
d\_3 &= 5^5\,(1-108\alpha)\,\alpha^2
\end{align}
We can do a minor transformation to get their respective variants,
$$y^3(y^2+5y+40) = j\_1\tag4$$
$$y(y-5)^4 = j\_2\tag5$$
$$y^3(y-5)^2 = j\_3\tag6$$
with discriminants,
\begin{align}
D\_1 &= 5^5\,(j\_1-1728)^2\,{j\_1}^2\\
D\_2 &= 5^5\,(j\_2-256)\,{j\_2}^3\\
D\_3 &= 5^5\,(j\_3-108)\,{j\_3}^3
\end{align}
However, it seems we are missing one quintic with discriminant $D\_4$,
$$D\_4 = 5^5(j\_4-64)^a\,{j\_4}^b\quad$$
which has level $p=6,7,8$ versions discussed [in this MO post](https://mathoverflow.net/q/446263/12905). (The octics in that post, after tedious manipulation, can be reduced to their deg-$7$ resolvents.)
---
**II. Eta quotients and the Monster**
Given [*Dedekind eta function*](https://en.wikipedia.org/wiki/Dedekind_eta_function#Eta_quotients) $\eta(\tau)$, define the four eta quotients which in fact are the first four *McKay-Thompson* series 1A, 2A, 3A, 4A of the Monster,
\begin{align}
\quad j\_1 &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3 \\
\quad j\_{2} &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 \\
\quad j\_{3} &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 \\
\quad j\_{4} &=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}
\end{align}
where $j\_1$ is just the *j-function*. Let $\tau = \sqrt{-d}$ or $\tau =\frac12+ \sqrt{-d}$ such that the $j\_i$ are **radicals**.
---
**III. Question 1**
Let $j\_i(\tau)$ be the radicals defined as above. Then is it true that for the quintics,
$$y^3(y^2+5y+40) = j\_1\tag4$$
$$y(y-5)^4 = j\_2\tag5$$
$$y^3(y-5)^2 = j\_3\tag6$$
the Galois group is **now** solvable, and the $y$ are solvable in radicals? For example, let,
$$j\_2\left(\tfrac{\sqrt{-232}}4\right)=396^4$$
which appears in Ramanujan's pi formula in the title. So, then a change of variable to $z$,
$$y(y-5)^4 = 396^4$$
$$z(z^4-5) = 396$$
$$z^5-5z-396 = 0$$
which is a solvable Bring-Jerrard quintic. (In fact, it factors).
---
**IV. Question 2**
However, there is still $j\_4$. To recall, for level $p=7$ ([in this MO post](https://mathoverflow.net/q/446263/12905)), the one-parameter formulas are complete for **all four**.
**Q:** So does this imply the general quintic can be reduced to a ***fourth*** one-parameter form (still unknown) and analogous to the three above?
---
**V. Sextic version?**
The "missing" quintic may have a sextic version (also with order $5! = 120$) and is given by,
$$j\_4 =\frac{(x + 1)^5 (x + 5)}x\tag7$$
with expected discriminant,
$$\text{Discrim}\_4 = 5^5\,(j\_4-64)^2\,{j\_4}^4$$
So what we're looking for might be its quintic subextension. But I am uncertain how to generate the *correct* quintic from this sextic. Ironically, the octics were easier. (*Note*: The correct quintic must have order $120$ for general $j\_4$ but be solvable when $j\_4 = j\_4(\tau)$ as defined in Section II.)
| https://mathoverflow.net/users/12905 | On Ramanujan's pi formula $\frac 1\pi=\sum_{k=0}^\infty\frac {(4k)!}{k!^4}\frac {Ak+B}{396^{4k}}$ and the solvable quintic $z^5-5z-396 = 0$? | (*This addresses Question 2*.)
**I.** In general, it asks if we can reduce the general quintic to a one-parameter form with a *specified* discriminant $D\_i$ that is different from the other well-known forms. These $D\_i$ involve the integers $\color{blue}{1728, 256, 108, 64}$, numbers which appear per level in *Ramanujan's theory of elliptic functions to alternative bases*, and his *pi formulas* with radii of convergence $\color{blue}{1/1728, 1/256, 1/108, 1/64}.$ ("*[Rational analogues of Ramanujan's series for 1/Pi](https://www.researchgate.net/publication/232016510_Rational_analogues_of_Ramanujan%27s_series_for_1I)*", Cooper and Chan, p. 23).
**II.** As a further constraint, we require its single parameter $j$, if equated to the appropriate eta quotient in Section II, must render the quintic **solvable** in radicals.
**III.** For the 4th form, it turns out one way is to start with the Brioschi quintic,
$$w^5 - 10 c w^3 + 45c^2 w - c^2 = 0\tag1$$
then using a rational Tschirnhausen transformation $x = P(w)/Q(w)$,
$$\left(w^2-3c\right)x - \left(\frac{j\,w}{j+512}-24c\right)=0\tag2$$
$$c= \frac{j^2}{(j-64)(j+512)^2}\tag3$$
Eliminating $w$ between $(1)$ and $(2)$ using resultants, then factoring, one gets the new form,
$$x^5 - 5j\, x^2 - 15j\, x - (8j + j^2) = 0\tag4$$
which, for general $j$, has order $5! = 120$. However, it has the desired property that if $j = j\_4(\tau)$, then it becomes solvable in radicals. The two quintics have the discriminants,
$$d\_1 = 5^5\,(1-\color{blue}{1728}c)^2\,c^8$$
$$D\_4 = 5^5\,(j−\color{blue}{64})^2(j-1)^2j^4$$
respectively. Note that $(3)$ is a just a cubic in $j$, so the general quintic has been transformed to the one-parameter form $(4)$ in radicals.
**IV.** As an afterword, just like the Bring quintic, the parameter of the Brioschi has infinitely many *rational* $c$ such that it is solvable in radicals, namely,
$$ c = \frac{n}{(n^2 - 10n + 5)^3 + \color{blue}{1728}n}$$
for rational $n$. And since $c$ in $(3)$ is a cubic in $j$, then the new quintic $(4)$ is solvable for infinitely many cubic radicals $j$, in addition to when $j = j\_4(\tau)$ are eta quotients.
| 0 | https://mathoverflow.net/users/12905 | 446511 | 179,894 |
https://mathoverflow.net/questions/446486 | 3 | Let $X$ be a Riemann surface with analytic boundary. Assume that $X$ has negative Euler characteristic. Then there exists a conformal hyperbolic metric $X$ such that $\partial X$ consists of geodesics (conformal in the sense of being compatible with the given complex structure on $X$).
Does anyone has a reference (or even better, a quick proof) of this result?
Edit: Let me state a stronger version. Suppose $X$ is embedded in an open Riemann surface $Y$. Then there exists a conformal hyperbolic metric on a neighborhood of $X$ in $Y$ such that $\partial X$ consists of geodesics. Is this true?
| https://mathoverflow.net/users/90076 | Finding a hyperbolic metric with geodesic boundary on a given Riemann surface | A good reference is
W. Abikoff, The real analytic theory of Teichmuller space, Springer, 1980. (Chap. II section 1).
The idea is that you construct the double: it is the result of gluing of your surface with its mirror image. This is a compact surface, it has a hyperbolic metric, and the restriction of this metric on the original surface is the hyperbolic metric with geodesic boundary.
| 4 | https://mathoverflow.net/users/25510 | 446520 | 179,900 |
https://mathoverflow.net/questions/446154 | 7 | I am currently reading "On Subsets with Cardinalities of Intersections Divisible by a Fixed Integer" by P. Frankl And A. M. Odlyzko. They used the following result without citation:
For each number $l$, we have a decomposition $l=l\_1+\cdots+l\_q$ with $l\_i\geq\epsilon l$ for a fixed constant $\epsilon$, such that the Hadamard matrix of size $4l\_k$ exists.
Would anyone provide a reference or a short proof that I missed? THX
| https://mathoverflow.net/users/148253 | About a result on Hadamard matrix | This can be proved using the strategy of Fedor Petrov and a theorem from the following paper:
*Haselgrove, C. B.*, [**Some theorems in the analytic theory of numbers**](https://doi.org/10.1112/jlms/s1-26.4.273), J. Lond. Math. Soc. 26, 273-277 (1951). [ZBL0043.04704](https://zbmath.org/?q=an:0043.04704).
Let $63/64 < \theta < 1$.
According to Theorem A of Hasselgrove, if $m$ is a sufficiently large odd number, then $m$ is the sum of three primes $p\_1$, $p\_2$, $p\_3$ with $|p\_i-m/3| < m^{\theta}$. Putting $m = 2 \ell-3$, we obtain
$$4 \ell = 2(p\_1+1) + 2(p\_2+1) + 2(p\_3+1).$$
The [Paley construction](https://en.wikipedia.org/wiki/Paley_construction) gives a Hadamard matrix of size $2(p+1)$ for any odd prime $p$. Since $p\_i = (1/3) m + O(m^{\theta})$, we have $2(p\_i+1) = (4/3) \ell + O(\ell^{\theta})$.
The exponent $63/64$ has been improved on by many other authors; for example, [Matomakai, Maynard and Shao](https://arxiv.org/abs/1610.02017) push it down to $11/20$. But the OP just asked for each of the matrix sizes to be comparable to $\ell$, so I'll stop here.
| 5 | https://mathoverflow.net/users/297 | 446536 | 179,904 |
https://mathoverflow.net/questions/446530 | 1 | Let $\mathcal{X}$ be the input or feature space, let $\mathcal{B}$ be Borel $\sigma$-algebra on $\mathcal{X}$ and $P(\mathcal{X})$ denotes the set of all probability measures on $(\mathcal{X},\mathcal{B})$. Let $P\_1, P\_2 \in P(\mathcal{X})$, The total variation is defined by
$$\delta(P\_1, P\_2) = \sup\_{A\in \mathcal{B}}|P\_1(A)-P\_2(A)|$$
I would like to ask for the idea of proving the following lemma,
Let $P\_1, P\_2 \in P(\mathcal{X})$ and let $\mathcal{F}$ the unit ball in $L^{\infty}(\mathcal{X})$,
$$\mathcal{F}=\{f \in L^{\infty}(\mathcal{X}) | \; ||f||\_{\infty} \leq 1\}$$
then we have the following characterization for the total varation distance
$$\delta(P\_1, P\_2) = \sup\_{f\in\mathcal{F}}|\mathbb{E}\_{x\sim P\_1}f(x) - \mathbb{E}\_{x\sim P\_2}f(x)|$$
| https://mathoverflow.net/users/504474 | Total variation distance | $\newcommand{\X}{\mathcal X}\newcommand{\B}{\mathcal B}\newcommand{\F}{\mathcal F}\newcommand{\De}{\Delta}\newcommand{\de}{\delta}$Let $\mu:=P\_1-P\_2$, so that $\mu$ is a finite signed measure. By the [Hahn--Jordan decomposition](https://en.wikipedia.org/wiki/Hahn_decomposition_theorem), there exist sets $\X^\pm\in\B$ such that $\X^+\cap\X^-=\emptyset$, $\X^+\cup\X^-=\X$, the functions $\mu^\pm$ defined by the formulas $\mu^+(B):=\mu(B\cap\X^+)$ and $\mu^-(B):=-\mu(B\cap\X^-)$ for $B\in\B$ are (nonnegative) measures, and
\begin{equation}
\mu=\mu^+-\mu^-.
\end{equation}
So, for any $f\in\F$,
\begin{equation}
\begin{aligned}
\int\_\X f\,dP\_1-\int\_\X f\,dP\_2&=\int\_\X f\,d\mu \\
&=\int\_{\X^+} f\,d\mu+\int\_{\X^-} f\,d\mu \\
&=\int\_{\X^+} f\,d\mu^+ -\int\_{\X^-} f\,d\mu^- \\
&\le\int\_{\X^+} 1\,d\mu^+ -\int\_{\X^-} (-1)\,d\mu^- \\
&=\mu^+(\X)+\mu^-(\X).
\end{aligned}
\end{equation}
Similarly, $\int\_\X f\,dP\_2-\int\_\X f\,dP\_1\le\mu^+(\X)+\mu^-(\X)$ and hence
\begin{equation}
\Big|\int\_\X f\,dP\_1-\int\_\X f\,dP\_2\Big|\le\mu^+(\X)+\mu^-(\X),
\end{equation}
with the equality if $f=1\_{\X^+}-1\_{\X^-}$. So,
\begin{equation}
\begin{aligned}
\De(P\_1,P\_2)&:=\sup\_{f\in\F}|E\_{x\sim P\_1}f(x) - E\_{x\sim P\_2}f(x)| \\
&=\max\_{f\in\F}\Big|\int\_\X f\,dP\_1-\int\_\X f\,dP\_2\Big| \\
&= \mu^+(\X)+\mu^-(\X)=2\mu^+(\X);
\end{aligned}
\end{equation}
the latter equality holds because $0=P\_1(\X)-P\_2(\X)=\mu(\X)=\mu^+(\X)-\mu^-(\X)$, so that $\mu^+(\X)=\mu^-(\X)$.
On the other hand, for any $A\in\B$ we have $P\_1(A)-P\_2(A)=\mu(A)=\mu^+(A)-\mu^-(A)\le\mu^+(A)\le\mu^+(\X)$ and similarly $P\_2(A)-P\_1(A)=-\mu(A)=\mu^-(A)-\mu^+(A)\le\mu^-(A)\le\mu^-(\X)=\mu^+(\X)$, so that $|P\_1(A)-P\_2(A)|\le\mu^+(\X)$, with the equality if $A=\X^+$. So,
\begin{equation}
\de(P\_1,P\_2)=\mu^+(\X)=\frac12\,\De(P\_1,P\_2).
\end{equation}
(You had the factor $\frac12$ missing.)
---
*Remark:* Here $\B$ does not have to be a Borel $\sigma$-algebra; it can be any $\sigma$-algebra over any set $\X$.
| 3 | https://mathoverflow.net/users/36721 | 446537 | 179,905 |
https://mathoverflow.net/questions/446539 | 1 | Say we have 2 functions $f$ and $g$ such that:
$f(a)<f(b) \Leftrightarrow g(a)<g(b)\;\; \forall a,b \in \mathbb{R}^n$
Is there an accepted name for a couple of functions like these?
Is there a body of research or some known theorems on this kind of functions?
| https://mathoverflow.net/users/504503 | Pair of functions that vary in the same direction | $\newcommand\R{\mathbb R}$Such pairs of functions may be called comonotone -- cf. [comonotone approximation](https://mathworld.wolfram.com/ComonotoneApproximation.html), which, for $n=1$, is an approximation of a piecewise monotonic function by a polynomial with the same monotonicity.
If functions $f$ and $g$ are comonotone in this sense and are in $L^2(\mu)$ for some probability measure $\mu$ over (say) $\mathbb R^n$, then
$$\int fg\,d\mu\ge\int f\,d\mu\,\int g\,d\mu. \tag{1}\label{1}$$
This is Chebyshev's integral inequality -- cf. e.g. [this](https://en.wikipedia.org/wiki/Chebyshev%27s_inequality#Integral_Chebyshev_inequality) and [this](https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality).
---
For completeness, here is a proof of \eqref{1}:
We have $(f(a)-f(b))(g(a)-g(b))\ge0$ for all $a$ and $b$ in $\R^n$, and hence
$$0\le\iint\mu(da)\mu(db)(f(a)-f(b))(g(a)-g(b)) \\
=2\int fg\,d\mu-2\int f\,d\mu\,\int g\,d\mu. \quad\Box$$
---
More on such comonotonicity: Suppose again that $f$ and $g$ are comonotone, and also suppose that $f$ and $g$ are Borel measurable. Let $Z$ be any random vector in $\R^n$. Let $X:=f(Z)$ and $Y:=g(Z)$. Then it is easy to check that, for any real $x$ and $y$, one of the events $\{X\le x\}$ and $\{Y\le y\}$ is contained in the other one. So, for all real $x$ and $y$
$$P(X\le x,Y\le y)=\min(P(X\le x),P(Y\le y));$$
that is, the random vector $(X,Y)$ is comonotone in [this sense](https://en.wikipedia.org/wiki/Comonotonicity#Comonotonicity_of_Rn-valued_random_vectors).
| 1 | https://mathoverflow.net/users/36721 | 446543 | 179,907 |
https://mathoverflow.net/questions/446441 | 1 | Lions and Paul claim in their 1993 paper "Sur les mesures de Wigner" that the Hamiltonian flow
$$\dot{x} = \xi, \quad \dot{\xi} = - \nabla V(x) $$
of the Vlasov equation
$$\partial\_t f + \xi \cdot \nabla\_x f + \nabla\_x V \cdot \nabla\_{\xi} f = 0$$
is well defined if $V\in C^{1,1}(\mathbb{R}^d)$ and $\exists C>0$ such that
$$ V(x) \geq -C(1+|x|^2)$$
(Théorème IV.1)
Neither do they prove it nor do they give a reference for this claim. It is, however, not obvious to me.
| https://mathoverflow.net/users/146998 | Why is this Hamiltonian flow of the Vlasov equation well defined? | I have found an answer.
The condition $V\in C^{1,1}$ ensures that $\nabla V$ is Lipschitz which implies the existence of a global solution in the neighborhood of every point for the Cauchy problem of the differential system $\dot{x} = \xi, \dot{\xi} = -\nabla V(x)$.
We have the following inequality for the energy $E(t)$
$$E(t) := \frac12 |\xi(t)|^2 + V(q(t)) > \frac12 |\xi(t)|^2 -C(1+|x(t)|^2) \\ > \frac12 |\xi(t)|^2 - 2C - 2C|x(t)|^2 $$
Therefore,
$$ |\xi(t)|^2 < 2E +2C + 2|x(t)|^2.$$
Moreover,
$$\frac{d}{dt} |x(t)|^2 = 2\xi(t) x(t) \leq |\xi(t)|^2 +|x(t)|^2 \leq 2E+2C +(2C+1)|x(t)|^2.$$
Therefore, $|\xi(t)|+|x(t)|$ are exponentially bounded by Gronwall's inequality and hence do not blow up in finite time. Then we can conclude that the flow exists globally.
| 2 | https://mathoverflow.net/users/146998 | 446544 | 179,908 |
https://mathoverflow.net/questions/445491 | 3 | When underlying $4$ manifold is compact and hyperkähler, the philosophy of infinite dimensional moment map tells us that its instantons moduli space is also hyperkähler. I'm curious about the following two questions.
1. When our hyperkähler $4$ fold is no longer compact, is its instantons moduli space(maybe with some constriant about decaying condition) still hyperkähler?
2. For some noncompact kähler surface that are not hyperkähler, for example blowup of $\mathbb{C}^2$ at origin, does its framed instantons moduli space still admit a hyperkähler structure? I know this space can be constructed by a finite dimensional symplectic quotient, but the configuration space does not admit a natural hyperkähler structure in general.
| https://mathoverflow.net/users/494608 | Hyperkähler structure of framed instantons over $\smash{\overline{\mathbb{C}P}}^2$ |
>
> When our hyperkähler 4-fold is no longer compact, is its
> instantons moduli space (maybe with some constraint about
> decaying condition) still hyperkähler?
>
>
>
The "decay constraint" condition is called "gravitational instanton".
A gravitational instanton is a complete hyperkähler metric on
a 4-manifold with $L^2$-integrable curvature. There is a lot
of literature on gravitational instantons and on instanton
bundles on gravitational instantons. The gravitational instantons
are classified according to the asymptotic growth of a volume
of a ball: when it grows as a 4-th degree (like on Euclidean space),
it's ALE instanton, degree 3 is ALF, and the classification becomes
very detailed and intricate for smaller degrees. ALF and ALE spaces
can be constucted as hyperkähler quotients of flat hyperkähler
spaces, for lower degrees it's probably false.
The space of instantons on ALE space is hyperkähler:
P.B. Kronheimer, H. Nakajima, [Yang–Mills instantons
and ALE gravitational instantons](https://doi.org/10.1007/BF01444534). Math. Ann. 288 (1990),
263–307.
The same is true for instantons on Taub-NUT space,
which is ALF:
Cherkis, Sergey A.
[Moduli spaces of instantons on the Taub-NUT space](https://doi.org/10.1007/s00220-009-0863-8).
Comm. Math. Phys. 290 (2009), no. 2, 719–736.
and on arbitrary ALF space:
Cherkis, Sergey A.
[Instantons on gravitons](https://doi.org/10.1007/s00220-011-1293-y).
Comm. Math. Phys. 306 (2011), no. 2, 449–483.
There is no framing in these results, you need to consider
instantons with $L^2$-integrable curvature.
>
> For some noncompact kähler surface that are not
> hyperkähler, for example blowup of $\mathbb C^2$ at origin,
> does its framed instantons moduli space still admit a
> hyperkähler structure?
>
>
>
Mostly not, in fact, it can easily be odd-dimensional.
One could imagine results like that (for instance, $CP^2$
is not hyperkähler, but instantons on $CP^2$ are
hyperkähler, if the framing is chosen right), but
this would probably mean that the complement
to the curve where you have chosen framing
is hyperkähler.
| 5 | https://mathoverflow.net/users/3377 | 446548 | 179,909 |
https://mathoverflow.net/questions/446534 | 3 | A subgroup $H$ of a group $G$ is *malnormal* if $gHg^{-1}\cap H=\{e\}$ for all $g\in G$ with $g\notin H$. It is *almost malnormal* if we merely require $gHg^{-1}\cap H$ to be finite.
I am wondering whether $\mathrm{SL}(n,\mathbb{Z})$, or $\mathrm{PSL}(n,\mathbb{Z})$, for $n\geq 2$, have (almost) malnormal non-abelian free subgroups. And why or why not?
More generally, I'd be happy to know a bit more about when countably infinite groups contain malnormal non-abelian free subgroups. Thanks!
Edit: As pointed out in Moishe's answer, this really should be a question for $\mathrm{PSL}(n,\mathbb{Z})$, not $\mathrm{SL}(n,\mathbb{Z})$, to avoid the possibility of (trivially) conjugating with $-I$.
| https://mathoverflow.net/users/16107 | Does $\mathrm{SL}(n,\mathbb{Z})$ have an (almost) malnormal free subgroup? | First of all, you have to work with $G=PSL(2, {\mathbb Z})$ and not $SL(2, {\mathbb Z})$, for otherwise the claim is clearly false. Then $G$ is a nonelementary hyperbolic group with trivial maximal finite normal subgroup. According to Lemma 8 in
*Minasyan, Ashot; Olshanskii, Alexander Yu.; Sonkin, Dmitriy*, [**Periodic quotients of hyperbolic and large groups.**](https://arxiv.org/abs/0804.3328), Groups Geom. Dyn. 3, No. 3, 423-452 (2009). [ZBL1234.20051](https://zbmath.org/?q=an:1234.20051).
the group $G$ contains a malnormal subgroup which is also free of rank 2.
As for the group $PSL(n, {\mathbb Z})$, $n\ge 3$, the situation is unclear. I do know that it is impossible to find an almost malnormal Zariski dense subgroup because such subgroups always contain real-regular elements [2] and the latter have centralizers which are not virtually cyclic [1].
The references are
[1] *Prasad, Gopal; Raghunathan, M. S.*, [**Cartan subgroups and lattices in semi-simple groups**](https://doi.org/10.2307/1970790), Ann. Math. (2) 96, 296-317 (1972). [ZBL0245.22013](https://zbmath.org/?q=an:0245.22013).
[2] *Prasad, Gobal*, [**$\mathbb{R}$-regular elements in Zariski-dense subgroups**](https://doi.org/10.1093/qmath/45.4.542), Q. J. Math., Oxf. II. Ser. 45, No. 180, 541-545 (1994). [ZBL0828.22010](https://zbmath.org/?q=an:0828.22010).
| 5 | https://mathoverflow.net/users/39654 | 446556 | 179,910 |
https://mathoverflow.net/questions/446516 | 9 | In type theory, the dependent sum $\sum\_{x : A} T(x)$ and the dependent product $\prod\_{x:A} T(x)$ are defined by their introduction/elimination rules.
In category theory, we use a base-change functor. Given a morphism $f \colon b \to a$, we define the functor $f^\* \colon C/a \to C/b$ between slice categories using a pullback. The dependent sum $\sum\_f$ is then defined as the left adjoint, and the dependent product $\prod\_f$ as the right adjoint to $f^\*$.
My question is, what's the interpretation of $f$ ? It doesn't show in the type-theoretical definitions. It looks like one can get the type-theoretical definitions by picking $a$ to be the terminal object, with $f \colon b \to 1$ the unique terminal morphism $!$. Is there more to it?
| https://mathoverflow.net/users/34546 | Dependent sum/product and the base-change functor adjunctions | The morphism $f$ is invisible in type theory because it corresponds to weakening, which in type theory appears as context extension, rather than an explicitly applied substitution.
#### Type-theoretic explanation
More precisely, given a context $\Gamma$ and a type family $\Gamma \vdash A \; \mathsf{type}$
there is a substitution $\iota : (\Gamma, x {:} A) \to \Gamma$ which takes each variable $y \in \Gamma$ to itself (but is not identity, it's shifting in terms of de Bruijn indices). This is your $f$. It induces a weakening operation $\iota^{\*} : \mathsf{Type}(\Gamma) \to \mathsf{Type}(\Gamma, x{:}A)$ that takes a type family over $\Gamma$ to a type family over $\Gamma, x {:} A$. The dependent product and sum go in the opposite direction
$$\Pi, \Sigma : \mathsf{Type}(\Gamma, x{:}A) \to \mathsf{Type}(\Gamma)$$
and their rules state precisely that there are adjunctions $\Sigma \vdash \iota^{\*} \vdash \Pi$.
#### Set-theoretic explanation
Let us also explain what $\iota$ corresponds to set-theoretically.
There are two equivalent ways of giving a set-theoretic interpretation, namely families and display maps. You indicated in your question that you prefer the latter, so let us use that. (The families are more natural, please consult Section 5.1 of my [notes on realizability](https://github.com/andrejbauer/notes-on-realizability) for details.)
A context is a set. The empty context is the singleton set.
A type family $\Gamma \vdash A \; \mathsf{type}$ is an element of the slice $\mathsf{Set}/\Gamma$, which we write as $p\_A : \overline{A} \to \Gamma$.
A term of type $A$ in context $\Gamma$ is a section $t : \Gamma \to \overline{A}$ of $p\_A$, i.e., it has to satisfy $p\_A \circ t = \mathsf{id}$. In particular, this means that from $t(\gamma) \in \overline{A}$ we can reconstruct $\gamma = p\_A(t(\gamma))$, which means that $t(\gamma)$ should *not* be thought of as "one element of $A$", but rather as "one element of $A$ together with an environment $\gamma$".
Given a type family $p\_A : \overline{A} \to \Gamma$, the context extension $\Gamma, x {:} A$ is the set $\overline{A}$.
A substitution is a map $\sigma : \Gamma \to \Delta$. Specifically, the above substitution $\iota : (\Gamma, x {:} A) \to \Gamma$ is just the map $p\_A : \overline{A} \to \Gamma$.
The action of $\sigma : \Gamma \to \Delta$ is pullback
$$\sigma^\* : \mathsf{Set}/\Delta \to \mathsf{Set}/\Gamma.$$
When we plug in $p\_A$ we get the pullback
$$p\_A\* : \mathsf{Set}/\Gamma \to \mathsf{Set}/\overline{A}.$$
It has left and right adjoints $\Sigma\_A \dashv p\_A^\* \dashv \Pi\_A$. I can spell them out if you wish.
| 11 | https://mathoverflow.net/users/1176 | 446559 | 179,911 |
https://mathoverflow.net/questions/446540 | 1 | Let $N \geq 1$ be an integer and let $p$ be a prime not dividing $N$. For $r \geq 1$, let $M\_2(\Gamma\_0(Np^r))$ denote the space of weight $2$ modular forms of level $\Gamma\_0(Np^r)$. Let $$U\_p: M\_2(\Gamma\_0(Np^r)) \to M\_2(\Gamma\_0(Np^r))$$ denote the $p$-th Hecke operator, which acts on $q$-expansions by sending $\sum a\_n q^n$ to $\sum a\_{np} q^n$.
I did some numerical calculations on Sage, and it looks like $U\_p$ takes values in $M\_2(\Gamma\_0(Np))$, not merely in level $Np^r$. That is, for all $r \geq 1$, it looks like $U\_p$ is actually a map from $M\_2(\Gamma\_0(Np^r))$ to $M\_2(\Gamma\_0(Np))$.
Is this true? And if so, could anyone sketch a proof / point to a reference?
| https://mathoverflow.net/users/394740 | What is the image of the Hecke operator $U_p$? | The statement, as claimed, is false.
Let $p = 2, N = 11$, and let $f\_0$ be the unique normalised eigenform in $S\_2(\Gamma\_0(11))$; and set $f(\tau) = f\_0(8\tau)$. Then $f \in M\_2(\Gamma\_0(Np^3))$, but $U\_p(f) = f\_0(4\tau)$ is not in $M\_2(\Gamma\_0(Np))$.
However, $U\_p(f)$ is in $M\_2(\Gamma\_0(Np^2))$. The correct general statement is that for $r \ge 2$, $U\_p$ maps $M\_2(\Gamma\_0(Np^r))$ to $M\_2(\Gamma\_0(Np^{r-1}))$ (and likewise for forms of any weight $k$, not just weight $2$). But in general it doesn't go all the way from level $\Gamma\_0(Np^r)$ to $\Gamma\_0(Np)$.
PS. You asked for an outline of the proof. The idea is to check that (under the hypotheses of my last paragraph) the double coset $\Gamma\_0(Np^r) \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix} \Gamma\_0(Np^r)$ is actually invariant under right-translation by $\Gamma\_0(Np^{r-1})$, not merely by $\Gamma\_0(Np^r)$. (This kind of argument is very important in Hida theory.)
EDITED TO ADD. One can check that the composite of $U\_p$ with the twisted map $M\_2(\Gamma\_0(Np^{r-1})) \to M\_2(\Gamma\_0(Np^r))$, $f(\tau) \mapsto f(p\tau)$, is multiplication by a power of $p$. So $U\_p$ is *surjective* as a map $M\_2(\Gamma\_0(Np^r)) \to M\_2(\Gamma\_0(Np^{r-1}))$, answering the slightly more refined question in the title of Adithya's post.
| 4 | https://mathoverflow.net/users/2481 | 446563 | 179,913 |
https://mathoverflow.net/questions/446523 | 4 | I would like to know if there is a "moral" reason why in the definition of [triangulated categories](https://en.wikipedia.org/wiki/Triangulated_category#TR_2) the "rotation axiom" TR2 requires that we have to add a negative sign to an arrow when we rotate the triangles? What was (presumably Verdier's) initial motivation to pose this negative sign in the arrow $−u[1]$ as stated in Wikipedia or standard literature? Does it come only from a convention or is there a deeper reason for it?
| https://mathoverflow.net/users/501436 | Moral reason for negative sign in rotation axiom for triangulated categories | One deeper reason is that this is what you get (no convention) when you have a triangulated category that comes from a stable $\infty$-category.
Ultimately, this boils down to the fact that when you take a loop in some space and reverse its direction, you get minus that loop in $\pi\_1$. The connection to this becomes clear when you then examine what the rotation of triangles corresponds to in stable $\infty$-categories. It is worked out in detail in lemma 1.1.2.13 of Lurie's *Higher algebra*.
More historically, this comes from the example of the triangulated category of chain complexes, which was the motivating example. In this case, if I understand correctly, if you are careful about sign conventions, it is also forced on you. This is worked out in Lawson's note on sign conventions ([https://www-users.cse.umn.edu/~tlawson/papers/signs.pdf](https://www-users.cse.umn.edu/%7Etlawson/papers/signs.pdf)), in paragraph 11.
| 3 | https://mathoverflow.net/users/102343 | 446577 | 179,916 |
https://mathoverflow.net/questions/446451 | 11 | In 2011 when I was in school I created a formula to calculate square roots... For $x\in\mathbb{R}$ with $x>0$ the following holds:
$$\sqrt{x} = \sum\_{n=0}^{\infty}\frac{\left(\prod\_{k=1}^{n}\left(\frac{3}{2}-k\right)\right)\left(\frac{1}{4}+\frac{x}{2}+\frac{x^2}{4}\right)^{\frac{1}{2}-n}\left(-\frac{1}{4}+\frac{x}{2}-\frac{x^2}{4}\right)^{n}}{n!}$$
It was mostly a joke and the formula is not useful. Recently I stumbled upon this formula again, and decided to work on the ideas behind that formula, not on the formula itself, to create new formulas that are actually useful. With "useful" I mainly mean a good rate of convergence, whatever "good" exactly means in this context...
If you're curious about the specifics of how everything unfolded (and if you either understand German or are willing to use a translator): [The regarding thread on matheplanet.com](https://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=162777)
I've finished my work for now and, in my opinion, the results are quite nice. However, as I don't have a background in mathematics (I did not study mathematics), I don't know whether these things already exist, or whether they could potentially be useful. I haven't proven anything yet, but I am still confident that it's all correct.
I'd be grateful to get some information about my results from people who actually know mathematics, since I really have no idea what to think of my work.
Let $x,a \in \mathbb{R}\_+$ and $m \in \mathbb{N}\_0$ and let $\sqrt{x} \neq a > \sqrt{\frac{x}{2}} > 0$ hold
We define
$$S(x,a,m) := \sum\_{k=0}^{m}\left(\sum\_{n=k}^{m}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\binom{n}{k}\right)\left(-1\right)^{k+1}a^{1-2k}x^{k}\right)$$
$$S(x,a) := \lim\_{m\to\infty}S(x,a,m) = \sum\_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}a^{1-2k}x^{k}\right)\right)$$
It holds:
$$\forall m : S(x,a,m) \neq S(x,a) = \sqrt{x}$$
We define $\hat{a} := x+1$ and $z := S(x,\hat{a},m)$ which yields:
$$\sqrt{x} = S(x,z) = S(x,S(x,x+1,m))$$
Since I don't have any proofs, I used Mathematica at least... The [Mathematica output](https://i.stack.imgur.com/eBbiM.png) and the regarding [Mathematica notebook](https://matheplanet.com/matheplanet/nuke/html/dl.php?id=2472&1683559031).
Using [Nest](https://reference.wolfram.com/language/ref/Nest.html) I created a really quick & dirty (& (very) inefficient) iteration to make it converge fast.
Iterating $5$ times with $m=10$ and $x=2$ yields over $34000$ correct decimal places for $\sqrt{2}$.
$m=11$ yields over $51000$ correct decimal places.
It boils down to: The bigger $m$, the faster the iteration converges... $\text{¯\\_(ツ)\_/¯}$
In other words: You calculate $S(x,a,m)$ for a given $m$ once, the result is an easy function, see the [Mathematica output](https://i.stack.imgur.com/eBbiM.png) for $S(x,x+1,10)$ as an example. You can save that function, and use it for the iteration.
Furthermore $S(x,a,m)$ and consequently $S(x,x+1,m)$ are (hopefully) easy to compute.
Note that the outer sum runs from $k=0$ to $m$ and $a^{1-2k}x^{k}$ is nicely isolated at the end, so that everything in front of $a^{1-2k}x^{k}$ yields the coefficients.
For $S(x,a)$ everything is "rearranged", so that we do not get nested infinity due to $m\to\infty$.
To get nice $\sqrt{x}$ formulas we can e. g. do the following...
For $x\in\mathbb{R}$ and $x>0$ holds:
$$\sqrt{x} = \sum\_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}\left(x+1\right)^{1-2k}x^{k}\right)\right) = \sum\_{n=0}^{\infty}\left(\frac{\Gamma\left(n-\frac{1}{2}\right)}{2\sqrt{\pi}n!}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}\left(x+1\right)^{1-2k}x^{k}\right)\right)$$
Kind of by accident (I looked at $\sqrt{\pi}$) I also obtained the following two formulas:
$$\sqrt{\pi}=\int\_{-\infty}^{\infty}e^{-x^2}\,dx = \sum\_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}\pi^{1-k}\right)\right)$$
$$\pi = \sum\_{n=0}^{\infty}\left(\frac{\Gamma\left(n-\frac{1}{2}\right)}{2n!}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}\pi^{1-k}\right)\right)$$
The last one is just supposed to look cool... For $x \in \mathbb{R}$ and $x > 0$ and $m \in \mathbb{N}\_0$ holds:
$$\sqrt{x}=\sum\_{p=0}^{\infty}\left(-1\right)^{p-1}\frac{\left(\sum\_{q=0}^{m}\left(-1\right)^{q-1}\frac{\left(x+1\right)^{1-2q}}{q!}\frac{\Gamma\left(q-\frac{1}{2}\right)}{2\sqrt{\pi}}\left(x-(x+1)^2\right)^{q}\right)^{1-2p}}{p!}\frac{\Gamma\left(p-\frac{1}{2}\right)}{2\sqrt{\pi}}\left(x-\left(\sum\_{r=0}^{m}\left(-1\right)^{r-1}\frac{\left(x+1\right)^{1-2r}}{r!}\frac{\Gamma\left(r-\frac{1}{2}\right)}{2\sqrt{\pi}}\left(x-(x+1)^2\right)^{r}\right)^2\right)^{p}$$
**Edit**:
* I guess I should include, that I've never been able to speak to someone about my mathematics in my whole life, so I really have no idea...
* "Working in isolation" was mentioned in the comments. I'd like to clarify: Although I've worked in isolation only, that's not what I want. I simply didn't know who to speak to...
**Answer to [answer](https://mathoverflow.net/a/446587/504411)**:
While thinking about writing down what I did in regard to the Taylor series of $\sqrt{x}$ I noticed that I'm now able to ask specific questions.
However: I haven't developed abstract enough mathematics to express those questions, yet.
Consequence: I'll develop the needed mathematics, and when I've done so I'll edit this question (again) to ask those questions and I'll also explain what I did in regard to the Taylor series of $\sqrt{x}$ to give some context.
(Or should I open a new post, then? If so, please let me know)
| https://mathoverflow.net/users/504411 | New method to compute square roots | Let me "unclutter" the basic formula $S(x,a)=\sqrt{x}$, starting from the definition in the OP,
$$S(x,a) =\sum\_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}a^{1-2k}x^{k}\right)\right).$$
The finite sum over $k$ is the [binomial expansion](https://en.wikipedia.org/wiki/Binomial_theorem) of $-a(1-x/a^2)^n$, and with some further algebra we have
$$S(x,a)=-\frac{a}{2\sqrt{\pi}}\sum\_{n=0}^\infty\frac{\Gamma \left(n-\frac{1}{2}\right) }{ \Gamma (n+1)}\left(1-x/a^2\right)^n.$$
This is simply the Taylor series of $a\sqrt {x/a^2}=\sqrt{x}$ around $x=1$, independent of $a>0$.
So I am unsure whether $S(x,a)$ has more content than just this, a power series of $\sqrt x$ around $x=1$.
| 7 | https://mathoverflow.net/users/11260 | 446587 | 179,918 |
https://mathoverflow.net/questions/446512 | 3 | This is a refined version of a [question I have recently posted](https://mathoverflow.net/q/446461/9924).
For a prime $p$, let $\varphi\_p\colon\mathbb Z\to\mathbb Z/p\mathbb Z$ denote the canonical homomorphism from the integers onto the group of order $p$.
>
> Given an integer $n\ge 3$, what is the smallest $\varepsilon=\varepsilon(n)>0$ such that for any subset $A\subset\mathbb Z$ with $|A|=n$, not contained in an arithmetic progression with the difference greater than $1$, there exists a prime $p$ satisfying $(1-\varepsilon(n))n\le|\varphi\_p(A)|<n$?
>
>
>
To put it simply, I want a prime $p$ distinguishing between the elements of $A$ ``as much as possible", but not distinguishing between all of them - subject to the assumption that $A$ is not contained in a nontrivial arithmetic progression (see [this nice construction](https://mathoverflow.net/a/446470/9924) by Peter Mueller showing that the containment assumption is vital.).
As an example, $\varepsilon(3)=1/3$: for any pairwise distinct integers $a,b,c$ with $\gcd(b-a,c-b,a-c)=1$ there exists a prime $p$ dividing exactly one of $b-a$, $c-b$, and $a-c$.
| https://mathoverflow.net/users/9924 | Bins-and-primes (prime divisors of $\prod(a_i-a_j)$, II) | The answer is $\varepsilon(n)=1-\frac{2}{n}$. Clearly, $\lvert\varphi\_p(A)\rvert\ge2$ for all primes $p$. However,
for every $n\ge2$ there is a set $A$ of size $n$ such that $\lvert\varphi\_p(A)\rvert=2$ whenever $\varphi\_p$ is not injective on $A$:
Let $P=\{2,3,\ldots,p\_{n-1}\}$ be the set of the first $n-1$ primes and $\pi$ be their product. Set $$A=\{0\}\cup\{\frac{\pi}{p}\,|\,p\in P\}.$$
Then $\operatorname{gcd}(A)=1$, so $A$ is not contained in an arithmetic progression with difference $>1$.
For each $p\in P$, exactly one of the elements in $A$ is not divisible by $p$, so $\lvert\varphi\_p(A)\rvert=2$.
Now let $p<q$ be distinct elements from $P$. From $\frac{\pi}{p}-\frac{\pi}{q}=\frac{\pi}{pq}(q-p)$ and $0<q-p<p\_{n-1}$ we see that all the prime divisors of $\frac{\pi}{p}-\frac{\pi}{q}$ are in $P$. Thus $\lvert\varphi\_p(A)\rvert=2$ for $p\in P$, and $\lvert\varphi\_p(A)\rvert=n$ for each prime $p>p\_{n-1}$.
| 5 | https://mathoverflow.net/users/18739 | 446617 | 179,924 |
https://mathoverflow.net/questions/446611 | 2 | Let $A$ be a finite dimensional algebra over an algebraically closed field. I'm trying to better understand the difference between $A$ being representation infinite and $A$ being $\tau$-tilting infinite when $A$ is a string algebra. So, if $A$ is representation finite, then $A$ is also $\tau$-tilting finite.
I'm wondering what is an example of a string algebra $A$ that is representation infinite but $\tau$-tilting finite.
| https://mathoverflow.net/users/338456 | Rep infinite, but $\tau$-tilting finite | I think the group algebra of a dihedral $2$-group in characteristic two, mod its socle, is an example. The smallest of these is $k\langle x,y\rangle/(x^2,y^2,xy,yx)$. See Plamondon, "$\tau$-Tilting finite gentle algebras are representation finite."
| 3 | https://mathoverflow.net/users/460592 | 446618 | 179,925 |
https://mathoverflow.net/questions/446565 | 4 | While reading a paper [Hengang Li and Weiping Yan - Asymptotic stability and instability of explicit self-similar waves for a class of nonlinear shallow water equations](https://doi.org/10.1016/j.cnsns.2019.104928), I experienced that my calculation results kept differing from the author's calculation results.
The authors of the paper seem to believe that the following equation holds:
$$\int\_\mathbb{R} xf'(x) \Lambda^{2s}f dx= -\int\_\mathbb{R} (\Lambda^s f)^2 dx-\frac{1}{2}\int\_\mathbb{R}x[(\Lambda^s f)^2]' dx$$
where $s>3$ is a constant and $\Lambda^s f:=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}$, $\hat{f}(\xi):=\int\_\mathbb{R}f(x) e^{-2\pi i\xi x}dx$ (Fourier transform) and $\check{f}(x):=\int\_\mathbb{R} f(\xi)e^{2\pi i \xi x}dx$ (Inverse Fourier transform). $f$ is in Schwartz space $S(\mathbb{R})$.
However, my result is different with above. Following all lemmas and theorem are just my opinions.
Lemma 1. $\Lambda^s f=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}=[(1+\xi^2)^{\frac{1}{2}}\check{f}]^{\wedge}$.
Proof) This holds because of change of variable $\xi \rightarrow -\xi$.
$$\Lambda^s f(x)=[(1+\xi^2)^{\frac{1}{2}}\hat{f}]^{\vee}=\int\_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\Bigl{(}\int\_\mathbb{R}f(y)e^{-2\pi i\xi y}dy\Bigr{)}d\xi$$
$$=\int\_\mathbb{R}(1+\xi^2)^{\frac{s}{2}}e^{-2\pi i\xi x}\Bigl{(}\int\_\mathbb{R}f(y)e^{2\pi\xi y}dy\Bigr{)} d\xi=[(1+\xi^2)^{\frac{1}{2}}\check{f}]^{\wedge}.\qquad\blacksquare$$
Lemma 2. $\int\_\mathbb{R}\Lambda^sg\Lambda^sfdx=\int\_\mathbb{R}g\Lambda^{2s}(f)dx$.
Proof) This results from weak Parseval's theorem i.e.
$$\forall f,g\in S(\mathbb{R})\ \ \ \ \int\_\mathbb{R}\hat{f}gdx=\int\_\mathbb{R}f\hat{g}dx.$$
If we use weak Parseval's theorem,
\begin{gather\*}
\int\_\mathbb{R}\Lambda^s{f}\Lambda^s{g}dx=\int\_\mathbb{R}[(1+\xi^2)^{\frac{s}{2}}\hat{f}]^{\vee}[(1+\xi^2)^{\frac{s}{2}}\check{g}]^{\wedge}dx=\int\_\mathbb{R}\hat{f}(1+\xi^2)^s\check{g}d\xi=\int\_\mathbb{R}[(1+\xi^2)^s\hat{f}]^{\vee}(\check{g})^{\wedge}dx \\
=\int\_\mathbb{R}g\Lambda^{2s}f dx.\qquad \blacksquare
\end{gather\*}
Lemma 3. $\frac{d}{dx}(\Lambda^sf)(x)=\Lambda^s\Big{(}\frac{df(x)}{dx}\Big{)}$. In other words, the linear operators $\Lambda^s$ and $\frac{d}{dx}$ commute.
Proof) In this proof, we use the well-known fact about Fourier transforms
$$2\pi i\xi \hat{f}(\xi)=(\frac{d}{dx}f)^{\wedge}(\xi),\ \ \ \frac{d}{d\xi}\hat{f}(\xi)=(-2\pi i) (xf(x))^\wedge(\xi).$$
If we use the above fact,
$$\Lambda^s\Bigl{(}\frac{df(x)}{dx}\Bigr{)}=[(1+\xi^2)^{\frac{s}{2}}(f')^\wedge]^{\vee}=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}.$$
Then,
$$\frac{d}{dx}(\Lambda^sf)(x)=\frac{d}{dx}\int\_{\mathbb{R}}(1+\xi^2)^{\frac{s}{2}}e^{2\pi i\xi x}\hat{f}(\xi)d\xi=2\pi i [(1+\xi^2)^{\frac{s}{2}}\xi\hat{f}]^{\vee}=\Lambda^s\Bigl{(}\frac{df(x)}{dx}\Bigr{)}.\qquad\blacksquare$$
**Theorem** $\int\_\mathbb{R}xf'\Lambda^{2s}fdx=-\frac{1}{2}\int\_\mathbb{R}x[(\Lambda^s f)^2]'dx+(s-1)\int\_\mathbb{R}(\Lambda^sf)^2dx
-s\int\_{\mathbb{R}}(\Lambda^{s-1}f)^2dx$ holds.
If we use integration by parts and Lemma 3 and Lemma 2
\begin{gather\*}
\int\_\mathbb{R}xf'(x)\Lambda^{2s} f dx=\Bigl{[}xf(x)\Lambda^{2s} f\Bigr{]}\_{-\infty}^\infty-\int\_\mathbb{R}xf(x)(\Lambda^{2s}f)'dx-\int\_\mathbb{R}f(x)\Lambda^{2s}fdx \\
=-\int\_\mathbb{R}\Lambda^s(xf)\Lambda^s(f')dx-\int\_\mathbb{R}(\Lambda^sf)^2dx.
\end{gather\*}
Meanwhile, if we use $(h')^{\vee}=-2\pi ix\check{h}$
\begin{gather\*}
x\Lambda^sf=x[(1+\xi^2)^{\frac{s}{2}}\hat{f}(\xi)]^{\vee}=\frac{i}{2\pi}[\frac{d}{d\xi}((1+\xi^2)^{\frac{s}{2}}\hat{f})]^{\vee} \\
=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}+(1+\xi^2)^{\frac{s}{2}}\hat{f}'(\xi)]^{\vee} \\
=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+[(1+\xi^2)^{\frac{s}{2}}(xf)^{\wedge}]^{\vee} \\
=\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}+\Lambda^s(xf).
\end{gather\*}
In other words,
$$\Lambda^s(xf)=x\Lambda^sf-\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}.$$
Therefore,
$$-\int\_\mathbb{R}\Lambda^s(xf)\Lambda^s(f')dx=-\int\_{\mathbb{R}}x\Lambda^sf\Lambda^s(f')dx
+\int\_\mathbb{R}\frac{i}{2\pi}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx.$$
If we focus on the second term,
$$\frac{i}{2\pi}\int\_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx
=\frac{i}{2\pi}\int\_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}[(1+\xi^2)^{\frac{s}{2}}(f')^{\vee}]^{\wedge}dx.$$
If we use weak Parseval theorem and $(h')^{\vee}=-2\pi ix\check{h}$,
\begin{gather\*}
=\int\_\mathbb{R}s\xi^2(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi
=\int\_\mathbb{R}s(1+\xi^2)^s\hat{f}\check{f}d\xi-\int\_\mathbb{R}s(1+\xi^2)^{s-1}\hat{f}\check{f}d\xi \\
=\int\_{\mathbb{R}}s(\Lambda^sf)^2dx-\int\_{\mathbb{R}}s(\Lambda^{s-1}f)^2dx.
\end{gather\*}
Therefore,
\begin{gather\*}
\int\_\mathbb{R}xf'(x)\Lambda^{2s}fdx=-\int\_\mathbb{R}(\Lambda^sf)^2dx-\int\_\mathbb{R}x\Lambda^sf\Lambda^s{f'}dx+
\frac{i}{2\pi}\int\_\mathbb{R}[s\xi(1+\xi^2)^{\frac{s-2}{2}}\hat{f}]^{\vee}\Lambda^s(f')dx
=-\frac{1}{2}\int\_\mathbb{R}x[(\Lambda^s f)^2]'dx+(s-1)\int\_\mathbb{R}(\Lambda^sf)^2dx
-s\int\_{\mathbb{R}}(\Lambda^{s-1}f)^2dx.\qquad \blacksquare
\end{gather\*}
I wonder if the calculations of the paper authors are correct or if my calculations are correct. If anyone has any ideas for the calculations, any help would be greatly appreciated.
| https://mathoverflow.net/users/113502 | Question about calculation in Schwartz space | The key computation is the commutator $$ \Lambda^s (xf) - x \Lambda^s f. $$ You can check this "classically" in the case $s = 4$ to find $$ (1 - \Delta)^2 (xf) = x (1-\Delta)^2 f - 4 (1-\Delta) f'$$
which differs from was found in the paper (their computation dropped the final commutator term).
In fact, your computations imply the following result
$$ \Lambda^s(xf) = x\Lambda^s f - s \Lambda^{s-2} f' $$
from which you'd get (returning to the author's original term)
\begin{gather\*}
\int \Lambda^s f \Lambda^s(xf') = \int \Lambda^s f\Lambda^s(xf)' - \Lambda^s f \Lambda^s f = - \int \Lambda^s f' \Lambda^s (xf) - \int (\Lambda^s f)^2 \\
= - \frac12 \int x [(\Lambda^s f)^2]' + s \int (\Lambda^{s-1} f')^2 - \int (\Lambda^s f)^2 = s \int (\Lambda^{s-1} f')^2 - \frac12 \int (\Lambda^s f)^2.
\end{gather\*}
(The fact that the extra term can be written as something with a sign means that this cannot just disappear due to some strange cancellations.)
This may have an impact on the main results of the paper; I haven't looked too carefully, but it appears that in deriving the Bernoulli type differential inequality after equation (3.21), the authors relied on this (and similar terms) to be signed to drop them from consideration. The corrected expression adds a term with the wrong sign. There may be a way to absorb this, but it is not immediately obvious.
| 5 | https://mathoverflow.net/users/3948 | 446619 | 179,926 |
https://mathoverflow.net/questions/446613 | 0 | Let $p\_n$ denote the $n$-th consecutive prime number and $g\_n=p\_{n+1}-p\_n$ a prime gap. There are many results about the upper bound for $g\_n$. Some of them still has astatus of conjecture, such as Firoozbakth conjecture (in a prime gap version):
$g\_n<p\_n\left( \sqrt[n]{p\_n}-1\right) $ , $\forall n\in N$, and its consequence $g\_n<\sqrt{n}$ , $n\geq 3645$ . Currently the best known proved result on upper bound is $g\_n\leq p\_n^{0.525}$ ( Baker-Harman-Pintz Theorem ). Heuristic and a few calculations that I made, suggest for upper bound:
**Conjecture**:
$$\frac{g\_n}{\log{g\_n}}<2\log{n}, n\geq 5 $$
If we combine Conjecture with B.K.P. Theorem we get $g\_n<1.05\log^2{n}$ , with Firoozbakth Conjecture we get $g\_n<\log^2{n}$, which seem to be in contradiction with A.Granville proposition :$\limsup\_{n\longrightarrow\infty}\frac{g\_n}{\log^2{p\_n}}\geq 2e^{-\gamma}\approx 1.1229$. There is also a simple consequence of Conjecture providing the lower bound for a prime gaps:
**Proposition**
$$g\_n>\left( \frac{p\_{n+1}}{p\_n}\right) ^{\frac{n}{2}} , n\geq 2$$
Proof:
$$\left( \frac{p\_{n+1}}{p\_n}\right) ^{\frac{n}{2}}=\left[ \left( 1+\frac{g\_n}{p\_n}\right) ^{\frac{p\_n}{g\_n}}\right] ^{\frac{n.g\_n}{2p\_n}}<e^{\frac{n.g\_n}{2p\_n}}<e^{\log g\_n}=g\_n$$
( follow from Conjecture and $\frac{p\_n}{n}>\log n$) .
**Question:** Is conjecture plausible? Can it be proved or disproved? If Conjecture is not true, can we prove the Proposition , now taken as Conjecture, with other means?
| https://mathoverflow.net/users/169583 | The lower bound for prime gaps | **1.** The Proposition in the post is almost equivalent to the Conjecture, namely it implies that
$$\frac{g\_n}{\log g\_n}\leq (2+o(1))\log n.$$
In particular, the Proposition (hence also the Conjecture) implies that
$$g\_n\ll\log n\,\log\log n.\tag{$\ast$}$$
This would contradict the common expectation that $g\_n\gg (\log n)^2$ holds for infinitely many $n$'s.
**2.** On the other hand, we don't know that $(\ast)$ is false. The best result in this direction is due to [Ford-Green-Konyagin-Maynard-Tao (2014)](https://arxiv.org/abs/1412.5029), and it states that
$$g\_n\gg\frac{\log n \,\log \log n\,\log\log\log\log n}{\log \log \log n}$$
holds for infinitely many $n$'s.
| 7 | https://mathoverflow.net/users/11919 | 446629 | 179,929 |
https://mathoverflow.net/questions/446631 | 4 | Let $X$ be a random variable and $Y=f(X)$ where $f$ is a deterministic function. Moreover, assume that there exists a deterministic function $g(.)$ such that the following probability is small.
\begin{align}
\mathbb{P}[g(Y)\neq X]\leq\delta.
\end{align}
Assume that $X'$ is another random variable which is close to $X$ in terms of total variation distance, i.e., $\|p\_X-p\_{X'}\|\_1\leq\epsilon$. What can be said about the following possibility?
\begin{align}
\mathbb{P}[g(f(X'))\neq X']\leq?.
\end{align}
| https://mathoverflow.net/users/68835 | Effect of small change in probability distribution on error probability | $\newcommand\de\delta\newcommand\ep\epsilon$Let $h:=g\circ f$, so that $g(Y)=h(X)$ and $g(f(X'))=h(X')$. Let $A:=\{x\colon h(x)\ne x)$. Then the condition $P(g(Y)\ne X)\le\de$ can be written as
$$\int\_A p\_X\le\de.$$
So,
$$P(g(f(X'))\ne X')=P(h(X')\ne X')=\int\_A p\_{X'} \\
=\int\_A p\_X+\int\_A (p\_{X'}-p\_X)
\le\int\_A p\_X+\|p\_X-p\_{X'}\|\_1\le\de+\ep.$$
Here we used the inequalities $\int\_A (p\_{X'}-p\_X)\le\int\_A |p\_{X'}-p\_X|\le\|p\_X-p\_{X'}\|\_1$.
---
Working slightly harder and letting $u\_+:=\max(0,u)$, we can write
$$\int\_A (p\_{X'}-p\_X)\le\int\_A (p\_{X'}-p\_X)\_+
\le\int (p\_{X'}-p\_X)\_+=\frac12\,\int|p\_X-p\_{X'}| \\
=\frac12\,\|p\_X-p\_{X'}\|\_1.$$
(The penultimate inequality above follows because $|p\_X-p\_{X'}|=(p\_{X'}-p\_X)\_+ + (p\_X-p\_{X'})\_+$, $p\_X-p\_{X'}=(p\_X-p\_{X'})\_+ - (p\_{X'}-p\_X)\_+$, and $\int(p\_X-p\_{X'})=0$, so that $\int(p\_X-p\_{X'})\_+=\int(p\_{X'}-p\_X)\_+=\frac12\,\int|p\_X-p\_{X'}|$.)
So, we get
$$P(g(f(X'))\ne X')\le\de+\ep/2.$$
| 4 | https://mathoverflow.net/users/36721 | 446635 | 179,932 |
https://mathoverflow.net/questions/446388 | 2 | I've been struggling a bit with a double sum that arose as the trace of an operator:
$$\sum\_{(j,k)\in Z^2 \setminus (0,0)} \frac{(j+k)^n}{(j^2+k^2)^n},$$
where $n$ is an even natural number. **Is there a closed form for this sum?**
Even in the case of $n=2$, I'm a bit miffed. If there were no numerator, I would use the Mellin transform and rewrite the sum in terms of theta functions, which is a standard approach in lattice sums (cf. MR3135109). However, there is a numerator, and I can't seem to be clever enough to make some change of variables work well.
I am also interested in the corresponding $d$-dimensional case.
Thanks for any thoughts/suggestions.
| https://mathoverflow.net/users/504359 | 2D lattice sum with numerator | The $n=4$ sum is accessible as follows. Expand $(j+k)^4$; by symmetry
the odd terms $4j^3k$ and $4jk^3$ cancel out so we need only sum
$(j^4 + 6j^2k^2 + k^4) / (j^2+k^2)^4$. You say you already know
the sum of $1 / (j^2+k^2)^2$, which is $(j^4 + 2j^2k^2 + k^4) / (j^2+k^2)^4$ [see also the "P.S." paragraph below];
so we only need one more sum $\sum^\prime (j^4 + c j^2 k^2 + k^4) / (j^2+k^2)^4$.
Well, we know the sum of $1 / (j+ik)^4$ from the theory of the Weierstrass
$\wp$-function; if I did this right it comes to
$\Gamma(1/4)^8 / (960 \pi^2) = 3.1512120+$. But
$$
\frac1{(j+ik)^4} = \frac{(j-ik)^4}{j^2+k^2)^4}
= \frac{(j^4 - 4i j^3 k - 6 j^2 k^2 + 4i j k^3 + k^4}{(j^2+k^2)^4},
$$
and the odd terms again cancel out, so this gives us the sum of
$(j^4 - 6j^2k^2 + k^4) / (j^2+k^2)^4$ — and we're done because
$$
2 (j^4 + 6j^2k^2 + k^4) = 3 (j^2+k^2)^2 - (j^4 - 6j^2k^2 + k^4).
$$
P.S. In general the sum of $1/(j^2+k^2)^n$ is $4 \zeta(n) L(n,\chi\_4)$
[because it's 4 times the value at $n$ of the zeta function for ${\bf Q}(i)$].
Thus when $n=2$ that's $2\pi^2 {\bf G} / 3$ where ${\bf G}$ is Catalan's constant
$$
L(2,\chi\_4) = 1 - \frac1{3^2} + \frac1{5^2} - \frac1{7^2} + - \cdots
= .915965594\ldots
$$
so $2\pi^2 {\bf G} / 3 = 6.026812\ldots$; no further simplification is known
or expected, though there are various ways known to compute the numerical
value of ${\bf G}$ efficiently to any desired accuracy.
| 4 | https://mathoverflow.net/users/14830 | 446640 | 179,934 |
https://mathoverflow.net/questions/446644 | 3 | Suppose that $K=[-N, N]$ is some compact subset of $\mathbb R$, for some $N>2.$
>
> Can we expect to choose $h$ such that $h=1$ on $K$ and the support of the Fourier transform of $\widehat{h}$ contained in $[-2/N, 2/N]$ and $\int\_{\mathbb R} h(x) dx \leq N$? if so how?
>
>
>
My attempts: Take $\widehat{f}= \chi\_{[-M, M]}^{\vee}g$ where the support of $g$ lies in $[-2/N, 2/N]$. Then by taking the inverse Fourier transform we get $f= \chi\_{[-M, M]}\ast g^{\vee}$. Now how to choose $g$ and $M$ so that $f=1$ on $[-N, N]$? (My approach could be wrong as well?)
| https://mathoverflow.net/users/173418 | How to choose some $h$ so its Fourier transform supported in some set? | No. It is already impossible for $h$ to be "band-limited" (i.e. with $\widehat h$ of compact support) and constant on an interval. Indeed by the Fourier inversion formula a band-limited function is analytic, so if $h=1$ on an interval then $h=1$ on all of ${\mathbf R}$, whence $h$ does not have a Fourier transform at all (and at any rate can't have $\int\_{\mathbf R} h < \infty$ even if you allow $\widehat 1$ to be a "delta function").
| 8 | https://mathoverflow.net/users/14830 | 446646 | 179,936 |
https://mathoverflow.net/questions/445814 | 3 | Let $a(n,k)$ be the sequence of $k$-Dowling numbers (for more information see [A007405](https://oeis.org/A007405) and its *CROSSREFS* section) with e.g.f.
$$\operatorname{exp}\left(x + \frac{\operatorname{exp}(kx) - 1}{k}\right)$$
Let
$$\ell(n)=\left\lfloor\log\_2 n\right\rfloor$$
$$f(n)=n-2^{\ell(n)}$$
$$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$
Here $f(n)$ is the same as $n$ without the most significant bit and $T(n,k)$ is the $(k+1)$-th bit from the right side in the binary representation of $n$.
Let $b(n,m,k)$ be an integer sequence such that
$$b(n,m,k)=mb(f(n),m,k)+\sum\limits\_{j=0}^{\ell(n)} k(1-T(n,j))b(f(n)+2^j(1-T(n,j)),m,k)$$
Here $b(n,1,1)$ is [A341392](https://oeis.org/A341392).
Let $s(n,m,k)$ be an integer sequence such that
$$s(n,m,k)=\sum\limits\_{j=0}^{2^n-1}b(j,m,k)$$
I conjecture that
$$s(n,1,k)=a(n,k)$$
I also conjecture that e.g.f. for $s(n,m,k)$ is
$$\operatorname{exp}(x + m\frac{\operatorname{exp}(kx) - 1}{k})$$
Here is the PARI prog to verify these conjectures:
```
s(n, m, k)=my(v, v1); v=vector(2^n, i, 0); v[1]=1; for(i=1, #v-1, my(L=logint(i, 2), A=i-2^L); v[i+1]=m*v[A+1] + sum(j=0, L, my(B=bittest(i, j)); k*(1-B)*v[A + 2^j*(1-B) + 1])); v1=[1]; for(i=1, n, v1=concat(v1, sum(j=0, 2^i-1, v[j+1]))); v1
[n, m, k]=[10, 1, 2]
x=s(n, m, k)
z='z+O('z^(n+1)); x1=Vec(serlaplace(exp(z + m*(exp(k*z) - 1)/k)))
test=x==x1
```
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Sequences that sum up to Dowling numbers | Cleaning up the notation a bit,
$$b\_{m,k}(n) = m\, b\_{m,k}(n-2^{\ell(n)}) + k \sum\_{j=0}^{\ell(n)-1} [n \,\&\, 2^j = 0] \,b\_{m,k}(n - 2^{\ell(n)} + 2^j)$$ where $\&$ is bitwise AND.
$$s\_{m,k}(n) = \sum\_{j=0}^{2^n-1} b\_{m,k}(j)$$
---
Let $\operatorname{wt}(n)$ be the Hamming weight of $n$, and for an arbitrary polynomial $p(z)$ define $$s\_{m,k}^p(n) = \sum\_{j=0}^{2^n-1} p(\operatorname{wt}(j)) b\_{m,k}(j)$$ This generalises $s\_{m,k} = s\_{m,k}^{z^0}$. The differences reduce nicely:
$$\begin{eqnarray\*}s\_{m,k}^p(n) - s\_{m,k}^p(n-1)
&=& \sum\_{j=2^{n-1}}^{2^n-1} p(\operatorname{wt}(j)) b\_{m,k}(j) \\
%&=& \sum\_{j=0}^{2^{n-1}-1} p(\operatorname{wt}(2^{n-1} + j)) b\_{m,k}(2^{n-1} + j) \\
&=& \sum\_{j=0}^{2^{n-1}-1} p(1 + \operatorname{wt}(j)) b\_{m,k}(2^{n-1} + j) \\
%&=& \sum\_{j=0}^{2^{n-1}-1} p(1 + \operatorname{wt}(j)) \left( m\, b\_{m,k}(j) + k \sum\_{i=0}^{n-2} [j \,\&\, 2^i = 0] \,b\_{m,k}(j + 2^i) \right) \\
&=& m \sum\_{j=0}^{2^{n-1}-1} p(1 + \operatorname{wt}(j)) b\_{m,k}(j) + k \sum\_{j=0}^{2^{n-1}-1} p(1 + \operatorname{wt}(j)) \sum\_{i=0}^{n-2} [j \,\&\, 2^i = 0] \,b\_{m,k}(j + 2^i) \\
%&=& m s\_{m,k}^{E\_z p}(n-1) + k \sum\_{j=0}^{2^{n-1}-1} \sum\_{i=0}^{n-2} [j \,\&\, 2^i = 0] p(\operatorname{wt}(j + 2^i)) b\_{m,k}(j + 2^i) \\
%&=& m s\_{m,k}^{E\_z p}(n-1) + k \sum\_{j=0}^{2^{n-1}-1} \operatorname{wt}(j) p(\operatorname{wt}(j)) b\_{m,k}(j) \\
&=& m s\_{m,k}^{E\_z p}(n-1) + k s\_{m,k}^{zp}(n-1) \\
\end{eqnarray\*}$$ where $E\_z$ is the raising operator $(E\_z p)(z) = p(z+1)$.
With the base case $s\_{m,k}^p(0) = p(0)$, we get $s\_{m,k}(n) = (1 + mE\_z + kz)^n z^0 \mid\_{z=0}$.
---
Two useful subresults towards the main proof:
**Theorem**: $(mE\_z + kz)^d z^0 \mid\_{z=0} = \sum\_{i=0}^d \genfrac{\lbrace}{\rbrace}{0pt}{}{d}{i} k^{d-i} m^i$ where $\genfrac{\lbrace}{\rbrace}{0pt}{}{d}{i}$ is a Stirling number of the second kind.
By induction.
* In the base case, $d=0$, we have $1 = \genfrac{\lbrace}{\rbrace}{0pt}{}{0}{0}$.
* Note that by repeated application of $E\_z z = (z+1)E\_z$ we can rewrite $(mE\_z + kz)^d$ as $\sum\_{i=0}^d k^{d-i} m^i q\_{k,m}(z) E\_z^m$ where the $q\_{k,m}$ are polynomials. Then $(mE\_z + kz)^d z^0 \mid\_{z=0} = \sum\_{i=0}^d \genfrac{\lbrace}{\rbrace}{0pt}{}{d}{i} k^{d-i} m^i$ is equivalent to $\forall i \in [0,d]: q\_{d-i,i}(0) = \genfrac{\lbrace}{\rbrace}{0pt}{}{d}{i}$. Now we multiply on the right to get $$\begin{eqnarray\*}(mE\_z + kz)^{d+1}
&=& \sum\_{i=0}^d k^{d-i} m^i q\_{k,m}(z) E\_z^m (mE\_z + kz) \\
&=& \sum\_{i=0}^d k^{d-i} m^{i+1} q\_{k,m}(z) E\_z^{m+1} + k^{d-i+1} m^i q\_{k,m}(z)(z+m) E\_z^m \\
\end{eqnarray\*}$$ so $q\_{k,m}(0) = q\_{k,m-1}(0) + k q\_{k-1,m}(0)$ $= \genfrac{\lbrace}{\rbrace}{0pt}{}{k+m-1}{m-1} + k \genfrac{\lbrace}{\rbrace}{0pt}{}{k+m-1}{m}$ $= \genfrac{\lbrace}{\rbrace}{0pt}{}{k+m}{m}$.
**Theorem**: $(\exp(z)-1)^i = \sum\_{n \ge i} \frac{i!}{n!} \genfrac{\lbrace}{\rbrace}{0pt}{}{n}{i} z^n$
Surely standard. By induction: $(\exp(z)-1)^0 = 1 = \sum\_{n \ge 0} \frac{0!}{n!} \genfrac{\lbrace}{\rbrace}{0pt}{}{n}{0} z^n$ checks out, and $$\begin{eqnarray\*}\left(\sum\_{n \ge i} \frac{i!}{n!} \genfrac{\lbrace}{\rbrace}{0pt}{}{n}{i} z^n\right)(\exp(z)-1)
&=& \left(\sum\_{n \ge i} \frac{i!}{n!} \genfrac{\lbrace}{\rbrace}{0pt}{}{n}{i} z^n\right)\left( \sum\_{j \ge 1} \frac{1}{j!} z^j\right) \\
&=& \sum\_{n \ge i+1} z^n \sum\_{j=1}^n \frac{i!}{(n-j)!j!} \genfrac{\lbrace}{\rbrace}{0pt}{}{n-j}{i} \\
&=& \sum\_{n \ge i+1} \frac{i!}{n!} z^n \sum\_{j=1}^n \binom{n}{j} \genfrac{\lbrace}{\rbrace}{0pt}{}{n-j}{i} \\
\end{eqnarray\*}$$
The inner sum counts pointed partitions of $n$ items into $i+1$ sets, so equals $(i+1) \genfrac{\lbrace}{\rbrace}{0pt}{}{n}{i+1}$, completing the proof.
---
**Theorem**: $\sum\_{n \ge 0} \frac{s\_{m,k}(n)x^n}{n!} =
\exp\left(x + m\frac{\exp(kx) - 1}{k}\right)$
For the LHS we have $$\begin{eqnarray\*}\sum\_{n \ge 0} \frac{s\_{m,k}(n)x^n}{n!}
&=& \sum\_{n \ge 0} \frac{x^n}{n!} \sum\_{d=0}^n \binom{n}{d} \sum\_{i=0}^d \genfrac{\lbrace}{\rbrace}{0pt}{}{d}{i} k^{d-i} m^i \\
&=& \sum\_{n,i,j \ge 0} \frac{1}{n!} \binom{n}{i+j} \genfrac{\lbrace}{\rbrace}{0pt}{}{i+j}{i} x^n m^i k^j
\end{eqnarray\*}$$
For the RHS we have \begin{eqnarray\*}\exp\left(x + m\frac{\exp(kx) - 1}{k}\right)
&=& \exp\left(x + mx\frac{\exp(kx) - 1}{kx}\right) \\
&=& \sum\_{u \ge 0} \frac{x^u}{u!} \left(1 + m\frac{\exp(kx) - 1}{kx}\right)^u \\
&=& \sum\_{u,i \ge 0} \frac{x^u}{u!} \binom{u}{i} \left(\frac{m}{kx} \right)^i (\exp(kx) - 1)^i \\
&=& \sum\_{u,i \ge 0} \frac{x^u}{u!} \binom{u}{i} \left(\frac{m}{kx} \right)^i \sum\_{v \ge i} \frac{i!}{v!} \genfrac{\lbrace}{\rbrace}{0pt}{}{v}{i} (kx)^v \\
&=& \sum\_{n,i,j \ge 0} \frac{1}{n!} \binom{n}{i+j} \genfrac{\lbrace}{\rbrace}{0pt}{}{i+j}{i} x^n m^i k^j
\end{eqnarray\*}
where the final line uses the substitutions $j = v-i$ and $n = u+j$.
---
Finally, note that the operator expression for $s\_{m,k}$ gives the elegant formulation of the main theorem as $$\exp(x(1 + mE\_z + kz)) z^0 \mid\_{z=0} = \exp\left(x + m\frac{\exp(kx) - 1}{k}\right)$$
| 1 | https://mathoverflow.net/users/46140 | 446656 | 179,937 |
https://mathoverflow.net/questions/446325 | 0 | For a given map $\phi :X\longrightarrow Y$, the mapping cylinder of $\phi$ is defined by $M\_{\phi}:=Y\cup\_{\phi} (X \times \{ 1\})$. Denote $\pi\_n (M\_{\phi},X \times \{ 1\} )$ by $\pi\_n (\phi)$. The map $\phi$ is called $n$-connected if $X$ and $Y$ are connected and $\pi\_i (\phi)=0$ for $1\leq i\leq n$.
Let $K$ be a CW-complex, $X$ have the homotopy type of one, and suppose that $\phi :K\to X$ to be $(n-1)$-connected. If $n\geq 3$, $\phi$ induces an isomorhphism of fundamental groups, so we can regard $\pi\_n (\phi)$ as a $\mathbb{Z}\pi\_1 (X)$-module. Select $\mathbb{Z}\pi\_1 (X)$-generators $\{ \alpha\_i \}$ for $\pi\_n (\phi)$. Then the $\partial \alpha\_i$ belong to $\pi\_{n-1} (K)$: use them to attach $n$-cells to $K$. Now use the $\alpha\_i$ themeselves to extend $\phi$ over these cells (recall that an element of $\pi\_n (\phi )$ is represented by a pair of maps $\alpha :S^{n-1}\longrightarrow X$ and $\beta :D^n \longrightarrow Y$ with $\beta\rvert\_{S^{n-1}}=\phi \circ \alpha$). If the resulting space is $L$ and map $\psi :L\to X$, then since the map $\alpha$ in the exact sequence $$\pi\_n (L,K)\overset{\alpha}{\to}\pi\_n (\phi)\to \pi\_n (\psi)\to \pi\_{n-1}(L,K)=0$$ is onto (for, if $n\geq 3$, $\pi\_n (L,K)\cong H\_n (\tilde{L},\tilde{K})=C\_n (\tilde{L})$, and the $\alpha\_i$ were chosen as generators of $\pi\_n (\phi)$), $\pi\_n (\psi)$ vanishes, and so $\psi$ is $n$-connected.
The above argument belongs to C.T.C. Wall's paper ``Finite conditions for CW-complexes" page 59. There are two things that I don't understand in the argument.
1. Why if $\phi$ induces an isomorhphism of fundamental groups, we can regard $\pi\_n (\phi)$ as a $\mathbb{Z}\pi\_1 (X)$-module?
2. I really appreciate if someone could explain me why such an exact sequence exists, $\pi\_{n-1}(L,K)=0$, $\pi\_n (L,K)\cong H\_n (\tilde{L},\tilde{K})=C\_n (\tilde{L})$, and $\alpha$ is onto in more detail. Clearly, $\pi\_{n-1}(L,K)=0$ and $\alpha$ is onto, then by the exact sequence we have $\pi\_n (\psi)=0$.
| https://mathoverflow.net/users/114476 | Explaining some detail in Wall's paper of CW-complexes | As to (1):
If we choose a basepoint in $K$, then $\phi$ can be viewed as a map of based spaces. Let $F$ be the homotopy fiber of $\phi$. Then there is
a well-defined action $\Omega K \times F \to F$ which induces a $\Bbb Z[\pi\_1(X)]$-module structure on $H\_\*(F)$.
As $F$ is $1$-connected (by the assumptions), the Hurewicz theorem says that $$\pi\_n(\phi) = \pi\_{n-1}(F) \cong H\_{n-1}(F).$$ So $\pi\_n(\phi)$ is indeed a $\Bbb Z[\pi\_1(X)]$-module.
As for (2):
Let $F'$ be the homotopy fiber of the map $\psi$, and let $F''$ be the homotopy fiber of the map $K\to L$. Then one has a commuting diagram:
$\require{AMScd}$
$$
\begin{CD}
F @>>> K @>\phi >> X \\
@VVV @VVV @VVV \\
F' @>>> L @>>\psi > X
\end{CD}
$$
where the rows are homotopy fiber sequences. Taking homotopy groups vertically gives a long exact sequence
$$
\cdots\to \pi\_\*(F'') \to \pi\_\*(F) \to \pi\_\*(F') \to \pi\_{\*-1}(F'') \to \cdots
$$
i.e.,
$$
\cdots\to \pi\_\*(L,K) \to \pi\_\*(\phi)) \to \pi\_\*(\psi) \to \pi\_{\*-1}(L,K) \to \cdots
$$
The given connectivity assumptions then say that this long exact sequence terminates on the right when $\* = n$.
| 3 | https://mathoverflow.net/users/8032 | 446669 | 179,941 |
https://mathoverflow.net/questions/446654 | 5 | *(For brevity, the level-6 functions have been migrated to [another post](https://mathoverflow.net/q/448777/12905).)*
**I. Level-10 functions**
Given the Dedekind eta function $\eta(\tau)$. To recall, for level-6,
$$j\_{6A} = \left(\sqrt{j\_{6B}} + \frac{\color{blue}{-1}}{\sqrt{j\_{6B}}}\right)^2 =\left(\sqrt{j\_{6C}} + \frac{\color{blue}8}{\sqrt{j\_{6C}}}\right)^2 = \left(\sqrt{j\_{6D}} + \frac{\color{blue}9}{\sqrt{j\_{6D}}}\right)^2-4$$
For level-10,
$$j\_{10A} = \left(\sqrt{j\_{10D}} + \frac{\color{blue}{-1}}{\sqrt{j\_{10D}}}\right)^2 = \left(\sqrt{j\_{10B}} + \frac{\color{blue}4}{\sqrt{j\_{10B}}}\right)^2 = \left(\sqrt{j\_{10C}} + \frac{\color{blue}5}{\sqrt{j\_{10C}}}\right)^2-4$$
where,
\begin{align}
j\_{10B}(\tau) &= \left(\frac{\eta(\tau)\,\eta(5\tau)}{\eta(2\tau)\,\eta(10\tau)}\right)^{4}\qquad \\
j\_{10C}(\tau) &= \left(\frac{\eta(\tau)\,\eta(2\tau)}{\eta(5\tau)\,\eta(10\tau)}\right)^{2} \\
j\_{10D}(\tau) &= \left(\frac{\eta(2\tau)\,\eta(5\tau)}{\eta(\tau)\,\eta(10\tau)}\right)^{6} \\
j\_{10E}(\tau) &= \left(\frac{\eta(2\tau)\,\eta^5(5\tau)}{\eta(\tau)\,\eta^5(10\tau)}\right)
\end{align}
Conway and Norton found these moonshine functions obey (or a version thereof),
$$j\_{10A}+2j\_{10E} = j\_{10B}+j\_{10C}+j\_{10D}+6$$
---
**II. Sequences**
Just like for level-6, we can use the relations above to get ***four*** sequences. In Cooper's paper, "*[Level 10 Analogues of Ramanujan's series for 1/Pi](https://www.researchgate.net/publication/266859001_Level_10_analogues_of_Ramanujan%27s_series_for_1p)*", he discussed $s\_{10}=s\_{10A}$ and two related sequences found by Zudilin (p.10), but **not** the other three below,
\begin{align}
s\_{10A}(k) &=\sum\_{m=0}^k \binom{k}{m}^4\\
s\_{10D}(j) &=\sum\_{k=0}^j (-u)^{j-k}\binom{j+k}{j-k}\,s\_{10A}(k)\\
s\_{10B}(j) &=\sum\_{k=0}^j (-v)^{j-k}\binom{j+k}{j-k}\,s\_{10A}(k)\\
s\_{10C}(n) &=\sum\_{j=0}^n\sum\_{k=0}^j (-w)^{n-j}\binom{n+j}{n-j}\binom{j}{k}\binom{2j}{j}\binom{2k}{k}^{-1}s\_{10A}(k)
\end{align}
where $u = \color{blue}{-1}$, $v = \color{blue}4$, $w = \color{blue}5$. Using the variable $h$ for uniformity, the first few terms are,
\begin{align}
s\_{10A}(h) &=1, 2, 18, 164, 1810, 21252,\ldots\\
s\_{10D}(h) &=1, 3, 25, 267, 3249, 42795, 594145,\ldots\\
s\_{10B}(h) &=1, -2, 10, -68, 514, -4100, 33940,\ldots\\
s\_{10C}(h) &=1, -1, 1, -1, 1, 23, -263, 1343, -2303,\ldots
\end{align}
such that all $s\_{10}(0) = 1.$ The sequences $(s\_{10A}, s\_{10D}, s\_{10B}, s\_{10C})$ have an $m$-term recurrence relation with $m=3,5,5,7$ (with the last one courtesy of G. Edgar's answer below).
---
**III. Pi formulas**
**A.** These four sequences can be used to generate new [*Ramanujan-Sato formulas*](https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series#Level_10) for $1/\pi$ of level 10. For example, let $\tau = \sqrt{-19/10}$, then,
\begin{align}
j\_{10A}(\tau) &= 76^2\\
j\_{10D}(\tau) &= (2+\sqrt5)^6\\
j\_{10B}(\tau) &= 4(3+\sqrt{10})^4\\
j\_{10C}(\tau) &= 5(1+\sqrt{2})^8\qquad
\end{align}
to get (the first one is known),
\begin{align}
\frac1{\pi} &= \frac{5}{\sqrt{95}}\,\sum\_{n=0}^\infty s\_{10A}(n)\,\frac{\;408n+47}{(76^2)^{n+1/2}}\\[4pt]
\frac1{\pi} &= \frac{2\sqrt{95}}{17\sqrt{5}}\sum\_{n=0}^\infty s\_{10D}(n)\,\frac{408n+47-\psi\_1}{\big((2+\sqrt5)^6\big)^{n+1/2}}\\[4pt]
\frac1{\pi} &= \frac{\sqrt{95}}{6\sqrt{10}}\sum\_{n=0}^\infty s\_{10B}(n)\,\frac{408n+47+\psi\_2\;}{\big(4(3+\sqrt{10})^4\big)^{n+1/2}}\\[4pt]
\frac1{\pi} &= \frac{1}{\sqrt{95}}\;\sum\_{n=0}^\infty s\_{10C}(n)\,\frac{An+B+\psi\_3}{\;\big(5(1+\sqrt{2})^8\big)^{n+1/2}}\\[4pt]
\end{align}
where $\psi\_1 = \frac{157}{38(2+\sqrt5)^3},$ and $\psi\_2 = \frac{157}{19(3+\sqrt{10})^2}.$ (The fourth to be added later.)
**B.** Furthermore, if within the radius of convergence, it seems that,
$$\sum\_{h=0}^\infty s\_{10A}(h)\,\frac{1}{\;\big(j\_{10A}\big)^{h+1/2}} = \sum\_{h=0}^\infty s\_{10B}(h)\,\frac{1}{\;\big(j\_{10B}\big)^{h+1/2}} = \\ \sum\_{h=0}^\infty s\_{10C}(h)\,\frac{1}{\;\big(j\_{10C}\big)^{h+1/2}} = \sum\_{h=0}^\infty s\_{10D}(h)\,\frac{1}{\;\big(j\_{10D}\big)^{h+1/2}}\;$$
---
**IV. Questions**
1. What is the recurrence relation for $s\_{10C}$?
2. Using the four given sequences of level $10$, is the last relation really true? And do their closed-forms have simpler versions, just like for level-6?
| https://mathoverflow.net/users/12905 | On level $10$ of the McKay-Thompson series of the Monster | For $s\_{10C}$, Maple finds this $7$-term recurrence:
`{(15625*n^3 + 46875*n^2 + 46875*n + 15625)*u(n) + (11250*n^3 + 61875*n^2 + 115625*n + 73125)*u(n + 1) + (4575*n^3 + 36600*n^2 + 99150*n + 90950)*u(n + 2) + (1116*n^3 + 11718*n^2 + 41434*n + 49322)*u(n + 3) + (183*n^3 + 2379*n^2 + 10371*n + 15157)*u(n + 4) + (18*n^3 + 279*n^2 + 1445*n + 2501)*u(n + 5) + (n^3 + 18*n^2 + 108*n + 216)*u(n + 6), u(0) = 1, u(1) = -1, u(2) = 1, u(3) = -1, u(4) = 1, u(5) = 23}`
\begin{align}
0 = &\left( 15625\,{n}^{3}+46875\,{n}^{2}+46875\,n+15625 \right)
u \left( n \right)
\\ &+ \left( 11250\,{n}^{3}+61875\,{n}^{2}+115625\,n+
73125 \right) u \left( n+1 \right)
\\ &+ \left( 4575\,{n}^{3}+36600\,{n}^{
2}+99150\,n+90950 \right) u \left( n+2 \right)
\\ &+ \left( 1116\,{n}^{3}+
11718\,{n}^{2}+41434\,n+49322 \right) u \left( n+3 \right)
\\ &+ \left(
183\,{n}^{3}+2379\,{n}^{2}+10371\,n+15157 \right) u \left( n+4
\right)
\\ &+ \left( 18\,{n}^{3}+279\,{n}^{2}+1445\,n+2501 \right) u
\left( n+5 \right)
\\ &+ \left( {n}^{3}+18\,{n}^{2}+108\,n+216 \right) u
\left( n+6 \right) ,
\\
&u \left( 0 \right) =1,u \left( 1 \right) =-1,u
\left( 2 \right) =1,u \left( 3 \right) =-1,u \left( 4 \right) =1,u
\left( 5 \right) =23
\end{align}
or
\begin{align}
0 = & \,5^6\, \left( k-2 \right) ^{3}u \left( k-3 \right)
\\ &+ 5^4 \left( 18\,k^3 - 63\,k^2 + 77\,k -33 \right) u \left( k-2 \right)
\\ &+ 5^2
\left( 183\,k^3 - 183\,k^2 + 123\,k - 25 \right) u \left( k-1
\right)
\\ &+2\, \left( 2\,k+1 \right) \left( 279\,{k}^{2}+279\,k+175
\right) u \left( k \right)
\\ &+ \left( 183\,{k}^{3}+732\,{k}^{2}+1038\,k
+514 \right) u \left( k+1 \right)
\\ &+ \left( 18\,{k}^{3}+117\,{k}^{2}+
257\,k+191 \right) u \left( k+2 \right)
\\ &+ \left( k+3 \right) ^{3}u
\left( k+3 \right)
\end{align}
| 3 | https://mathoverflow.net/users/454 | 446673 | 179,942 |
https://mathoverflow.net/questions/446663 | 3 | Let $u: D\_1 \to \mathbf{R}$ be a smooth function defined on the unit disk $D\_1 \subset \mathbf{R}^2$ which describes the minimal graph $G$. Suppose that at the origin $G$ is tangent to the horizontal plane $\{ x^{3} = 0 \}$, that is: $u(0) = 0$ and $Du(0) = 0$. Suppose that in a small disk $D\_r$ around the origin, the intersection of $G$ and the plane decomposes like
\begin{equation}
\{ u = 0 \} \cap D\_r \setminus \{ 0 \} = \gamma\_1 \cup \cdots \cup \gamma\_4.
\end{equation}
The $\gamma\_i$ are smooth curves meeting at right angles at the origin, their common endpoint. After rotation, we may assume that $\gamma\_1'(0) = e\_1$, $\gamma\_2'(0) = e\_2$, $\gamma\_3'(0) = -e\_1$ and $\gamma\_4'(0) = -e\_2$.
>
> Is it possible for $\gamma\_1 \cup \gamma\_2$ to form a graph over the $x^1$-axis, that is can there be $\varphi \in C^0(-r,r)$ so that $\gamma\_1 \cup \gamma\_2 = \{ (t,\varphi(t),0) \mid t \in (-r,r) \}$?
>
>
>
| https://mathoverflow.net/users/103792 | 'Degenerate' tangent point of a minimal graph | Yes, this is possible. Consider the intersection of the helicoid $z = \tan^{-1}(y/x)$ with its tangent plane $z = y$ at $(1,0,0)$. The projection of the intersection curves to $\{z = 0\}$ consists of the line $y = 0$ and the curve $x = y/\tan(y) \sim 1 - y^2/3$. After a rigid motion so that the tangent plane is horizontal and the tangent point is the origin, the union of $\gamma\_1$ and $\gamma\_2$ will be a $C^{1/2}$ graph $\{(t,\,\varphi(t),\,0)\}$ where $\varphi(t) = 0$ for $t \geq 0$ and $\varphi(t) \sim (-6t)^{1/2}$ for $t < 0$.
One can generate many more examples using Cauchy-Kovalevskaya, by choosing appropriate Cauchy data on a line segment through 0 and solving the minimal surface equation in a neighborhood of the origin. For example, taking $u = 0$ and $u\_y = x + x^2/2$ on $\{y = 0\}$ gives
$$u = xy + \frac{1}{2}x^2y - \frac{1}{6}y^3 + O((x^2+y^2)^2),$$
hence $\{u = 0\}$ locally resembles the line $y = 0$ and the curve $y^2 = 6x$.
| 3 | https://mathoverflow.net/users/16659 | 446681 | 179,945 |
https://mathoverflow.net/questions/446691 | 2 | I was trying to solve some integrals that appear in quantum electrodynamics but I was not able to do it on my own.
$$1/6\int\_0^1 \int\_0^1 { u^3 z^2(1-z^2/3) \over [u^2(1-z^2)+4(1-u)]}dudz $$
I know the answer should be $$(\pi^2 / 18) - (115 / 216)$$
Can anyone help me solve this or at least point me to a book that might cover those double integrals?
I would realy appreciate it. In attachment you have an image of the paper this came from. Of course I have problems with the other integrals too. But I chose to start with this one.
Thank you.
[Extract of paper by Soto, 1970 - Calculation of the Slope at q2=0 of the Dirac Form Factor for the Electron Vertex in Fourth Order][1]
| https://mathoverflow.net/users/504630 | Integral in the Lamb shift calculation – fourth order | The double integral in question is
$$I:=\int\_0^1 du\, J(u),$$
where
$$J(u):=\int\_0^1 dz\,{ u^3 z^2(1-z^2/3) \over {u^2(1-z^2)+4(1-u)}} \\
\text{[which is a standard integral, with a denominator of the integrand quadratic in $z$]} \\
=\frac{3 \left(u^3-6 u+4\right) \ln(1-u)+u \left(-5 u^2-12
u+12\right)}{9 u^2} \\
=J\_1(u)-J\_2(u),$$
$$J\_1(u):=\frac{3 \left(u^3-6 u+4\right)(-u)+u \left(-5 u^2-12
u+12\right)}{9 u^2}=\frac19\, (6 - 5 u - 3 u^2),$$
$$J\_2(u):=\sum\_{k=2}^\infty
\frac{3 \left(u^3-6 u+4\right) u^k/k}{9 u^2}; $$
here we used the Maclaurin series for $\ln(1-u)$.
The integral $\int\_0^1 du\, J\_1(u)$ is very easy, and
$$-\int\_0^1 du\, J\_2(u)=\sum\_{k=2}^\infty
\Big(\frac{2}{k^2}+\frac{7}{6 k}+\frac{1}{6 (k+2)}-\frac{4}{3 (k-1)}\Big).$$
Next, $\sum\_{k=2}^\infty \frac2{k^2}=2(\pi^2/6-1)$, and the sum
$$\sum\_{k=2}^\infty
\Big(\frac{7}{6 k}+\frac{1}{6 (k+2)}-\frac{4}{3 (k-1)}\Big)
=-\frac{53}{36}$$
can be found either by telescoping or using the fact that $\sum\_{k=1}^n\frac1k-\ln n$ converges as $n\to\infty$ to Euler's gamma constant $\gamma=0.577\ldots$ (which latter will get canceled, because $\frac76+\frac16-\frac43=0$).
Collecting the pieces, we get the result.
| 2 | https://mathoverflow.net/users/36721 | 446697 | 179,948 |
https://mathoverflow.net/questions/446378 | 16 | (cross-posted from [this math.SE question](https://math.stackexchange.com/questions/4679118/does-a-completely-metrizable-space-admit-a-compatible-metric-where-all-intersect))
It is well-known that given a metric space $(X,d)$, the metric is complete if and only if every intersection of nested (i.e. decreasing with respect to inclusion) closed balls of vanishing diameters have non-empty intersection. This is usually referred to as Cantor's Intersection Theorem.
*Is it true that every completely **metrizable** topological space **admits one compatible metric** such that every intersection of nested closed balls has a non-empty intersection?*
---
Some more details:
------------------
Let's call spherically complete a metric as above where every intersection of nested closed balls has non-empty intersection.
Not every complete metric is spherically complete (see e.g. [this question](https://math.stackexchange.com/questions/3314912/in-a-metric-space-the-intersection-of-nested-closed-balls-is-empty) and the list of its related questions).
However, this question is not about a fixed metric, but about the existence of one metric compatible with the topology.
We know there exist (classes of) completely metrizable spaces that admit compatible spherically complete metrics.
For example:
* All (complete) metrics on a compact space are spherically complete.
* The standard euclidean metric on $\mathbb{R}$ is spherically complete.
* All completely ultrametrizable spaces admit a compatible spherically complete metric.
So to be more precise my question is:
* Do all completely metrizable topological spaces admit a compatible spherically complete metric?
* If not, is it known what is the largest class of (completely metrizable) topological spaces that admit one?
I suspect that every locally compact completely metrizable space admits a spherically complete metric, and probably the same is true for Polish spaces (using that they are homeomorphic to $G\_\delta$ subsets of the Hilbert cube).
But I could not find any reference for this.
| https://mathoverflow.net/users/121875 | Does a completely metrizable space admit a compatible metric where all intersections of nested closed balls are non-empty? | Let us say that a topological space $X$ is *spherically completely metrizable* if the topology of $X$ is generated by a spherically complete metric.
>
> **Theorem.** Every closed subspace $X$ of the countable product of locally compact metrizable spaces is spherically completely metrizable.
>
>
>
*Proof.* We lose no generality assuming that $X$ is a closed subspace of the countable power $L^\omega$ of some locally compact metrizable space $L$. By the paracompactness, the locally compact metrizable space $L$ is a topological sum $\bigcup\_{\alpha\in\kappa}L\_\alpha$ of clopen $\sigma$-compact subspaces. Each space $L\_\alpha$ is locally compact and $\sigma$-compact, and hence its topology is generated by a metric $d\_\alpha$ whose closed balls are compact. On the space $L$ consider the metric $d$ defined by
$$d(x,y)=\begin{cases}1&\mbox{if $x\in L\_\alpha$ and $y\in L\_\beta$ for distinct $\alpha,\beta\in\kappa$};\\
\min\{d\_\alpha(x,y),1\}&\mbox{if $x,y\in L\_\alpha$ for some $\alpha\in\kappa$}.
\end{cases}
$$
It follows that for every $c\in L$ and every $r<1$ the closed ball $B\_L(c;r):=\{x\in L:d(x,c)\le r\}$ is compact and for every $r\ge 1$, $B\_L(c;r)=L$.
On the countable power $L^\omega$ consider the complete metric $\rho$ defined by $$\rho((x\_n)\_{n\in\omega},(y\_n)\_{n\in\omega})=\max\_{n\in\omega}\frac{d(x\_n,y\_n)}{2^n}.$$
We claim that the metric $\rho$ induces a spherically complete metric on the closed subspace $X\subseteq L^\omega$.
Let $(B\_n)\_{n\in\omega}$ be a sequence of nested closed balls in $X$. Let $(c\_n)\_{n\in\omega}$ and $(r\_n)\_{n\in\omega}$ be the sequences of the centers and radii of the balls $B\_n$. Since every sequence of positive real numbers contains a monotone subsequence, we lose no generality assuming that the sequence $(r\_n)\_{n\in\omega}$ is monotone.
If the sequence $(r\_n)\_{n\in\omega}$ is increasing (i.e., $r\_n\le r\_{n+1}$ for all $n$), then for every $n\in\omega$ we have $c\_n\in B\_n\subseteq B\_0$ and hence $\rho(c\_n,c\_0)\le r\_0\le r\_n$ and $c\_0\in B(c\_n;r\_n)=B\_n$.
So, we assume that $(r\_n)\_{n\in\omega}$ is strictly decreasing.
If $\inf\_{n\in\omega}r\_n=0$, then $\lim\_{n\to\infty}r\_n=0$ and the intersection $\bigcap\_{n\in\omega}B\_n$ is not empty by the completeness of the metric $\rho$.
So, we assume that $r:=\inf\_{ n\in\omega}r\_n>0$.
If $r\ge 1$, then every ball $B\_n$ coincides with $X$ and hence $\bigcap\_{n\in\omega}B\_n=X\ne\emptyset$.
So, we assume that $r<1$. Let $m\in\omega$ be the largest number such that $2^m r<1$.
Since $\lim\_{n\to\infty}r\_n=r$, we can replace the sequence $(B\_n)\_{n\in\omega}$ by a suitable subsequence, and assume that $2^mr\_0<1$. Then for every $k\in\omega$, the ball $B\_L(c\_0(k),2^mr\_0)$ in $L$ is compact. For every $n\in\omega$, the inclusion $c\_n\in B\_n\subseteq B\_0$ implies $\rho(c\_n,c\_0)\le r\_0$ and hence $d(c\_n(k),c\_0(k))\le 2^kr\_0$. Then for every $k\le m$, the sequence $(c\_n(k))\_{n\in\omega}$ is contained in the compact ball $B\_L(c\_0(k),2^mr\_0)$ and hence has a convergent subsequence.
Replacing $(B\_n)\_{n\in\omega}$ by a suitable subsequence, we can assume that for every $k\le m$ the sequence $(c\_n(k))\_{n\in\omega}$ is convergent in $L$, and moreover $d(c\_i(k),c\_j(k))<r$ for all $i,j\in\omega$.
We claim that $\rho(c\_0,c\_n)\le r\_n$ for every $n\in\omega$. This inequality will follow as soon as we check that $d(c\_0(k),c\_n(k))\le 2^kr\_n$ for all $k\in\omega$.
If $k>m$, then $d(c\_0(k),c\_n(k))\le 1\le 2^{m+1}r<2^kr\_n$ by the definition of the metric $d$.
If $k\le m$, then $d(c\_0(k),c\_n(k))<r\le 2^kr\_n$ by the choice of the (sub)sequence $(B\_i)\_{i\in\omega}$.
Therefore, $c\_0\in \bigcap\_{n\in\omega}B\_n$. $\quad\square$.
Since every Polish space is homeomorphic to a closed subspace of $\mathbb R^\omega$, Theorem implies
>
> **Corollary.** Every Polish space is spherically completely metrizable.
>
>
>
The Theorem suggests the following
>
> **Question.** Which metrizable spaces do embed into countable products of locally compact metrizable spaces?
>
>
>
**Remark 1.** The necessary condition of the embeddability of a topological space $X$ into the countable product of locally compact metrizable spaces is the separability of all quasicomponents of $X$. This condition implies that nonseparable connected metrizable spaces do not embed into countable products of locally compact metrizable spaces.
The following proposition answers the above Question.
>
> **Proposition.** A topological space $X$ is homeomorphic to a closed subspace of the countable product of locally compact metrizable spaces if and only if $X$ is homeomorphic to a closed subspace of $\mathbb R^\omega\times\kappa^\omega$ for some cardinal $\kappa$ endowed with the discrete topology.
>
>
>
*Proof.* The "if" part of this characterization is trivial. To prove the "only if" part, assume that $X$ is homeomorphic to a closed subspace of the product $\prod\_{n\in\omega}L\_n$ of locally compact metrizable spaces $L\_n$. By the paracompactness, every space $L\_\alpha$ is a topological sum of locally compact $\sigma$-compact metrizable spaces and hence is a topological sum of Polish spaces. Since every Polish space is homeomorphic to a closed subspace of the space $\mathbb R^\omega$, for every $n\in\omega$ the locally compact metrizable space $L\_n$ is homeomorphic to a closed subspace of $\mathbb R^\omega\times\kappa$ for some cardinal $\kappa$. Then $\prod\_{n\in\omega}L\_n$ is homeomorphic to a closed subspace of the space $(\mathbb R^\omega\times\kappa)^\omega$, which is homeomorphic to $\mathbb R^\omega\times\kappa^\omega$. $\quad\square$
Proposition and Theorem imply that every closed subspace of $\mathbb R^\omega\times\kappa^\omega$ is spherically completely metrizable.
>
> **Problem.** Let $\kappa$ be a cardinal. Is every closed metrizable subspace of the space $[0,1]^\kappa\times\kappa^\omega$ spherically completely metrizable?
>
>
>
**Remark 2.** For every cardinal $\kappa$, every closed metrizable subspace of the space $[0,1]^\kappa\times\kappa^\omega$ is completely metrizable. On the other hand, closed metrizable subspaces of the space $\mathbb R^\kappa$ are realcompact but needs not be completely metrizable (by Theorem 3.11.12 in Engelking's "General Topology", every Lindelof space is realcompact; in particluar, every metrizable separable space is realcompact).
| 10 | https://mathoverflow.net/users/61536 | 446702 | 179,949 |
https://mathoverflow.net/questions/446671 | 10 | I have discovered a pertinent solution to my problem in the article *On the Kinetic Theory of Rarefied Gases* by Harold Grad and the book *Thermodynamik und Statistik* by Arnold Sommerfeld, both of which present the same proof for the issue I was addressing. The proof assumes the continuity of the function $f$.
---
Suppose we have a function $f\_0:{\mathbb R}^3\rightarrow {\mathbb R}\_+$ that satisfies the following property
\begin{equation}
\begin{split}
&\mathbf{v}\_1^2 + \mathbf{v}\_2^2 = \mathbf{v}\_1'^2 + \mathbf{v}\_2'^2\newline
&\mathbf{v}\_1 + \mathbf{v}\_2 = \mathbf{v}\_1' + \mathbf{v}\_2'
\end{split}
\quad \Rightarrow \quad
f\_0(\mathbf{v}\_1)f\_0(\mathbf{v}\_2) = f\_0(\mathbf{v}\_1')f\_0(\mathbf{v}\_2')
\end{equation}
**Question**: Under what conditions, such as continuity, smoothness, or even analyticity as assumed by physicists, can $\log f\_0$ be written as a linear combination of $v^2$, the three components of $v$, and an arbitrary constant?
$f\_0$ originates from Boltzmann’s distribution of particles in the velocity space, which specifies the equilibrium state in the absence of external forces in classical statistical mechanics. The two equalities represent the conservation laws of kinetic energy and momentum, respectively, in a collision between two perfectly elastic spheres. I encountered $f\_0$ in the book *Mathematical Statistical Mechanics* by Colin J. Thompson.
Here is the original text from the book
>
> Taking logarithms of both sides of Equation 6.1 we have
> $$
> \log f\_0({\bf v\_1})+ \log f\_0({\bf v\_2}) = \log f\_0({\bf v\_1’}) + \log f\_0({\bf v\_2’})
> $$
> which has the form of a conservation law. Since for spinless molecules (e.g., hard spheres) the only conserved quantities are energy and momentum (and constants), it follows that must be a linear combination of $v^2$ and the three components of $v$, plus an arbitrary constant, i.e.,
> $$
> \log f\_0({\bf v})= \log A-B({\bf v}-{\bf v\_0})^2
> $$
>
>
>
I cannot see a rigorous proof provided to explain it, nor are any reference bibliographies given, despite the book being titled "Mathematical Statistical Mechanics"…
| https://mathoverflow.net/users/478784 | Proving the simple form of a function from statistical mechanics | We can indeed prove this for reasonable functions, $\log f\_0\in C^2$, say.
Let me write $F=\log f\_0$. By replacing $F$ by $F(v)-C-d\cdot v$, we can also assume that $F(0),\nabla F(0)=0$.
If $a,v$ are orthogonal, then, by assumption,
$$
F(v)+F(a)=F(v+a)+F(0)=F(v+a) ,
$$
and for small $a$, we have $F(v+a)\simeq F(v) +a\cdot\nabla F(v)$, $F(a)\simeq a\cdot\nabla F(0) =0$. This shows that $a\cdot \nabla F(v)=0$ for all $a$ with $a\cdot v=0$. In other words, $\nabla F(v)$ has the same direction as $v$. It follows that $F$ is radial since for a fixed sphere, the directional derivatives tangential to the sphere are zero. We have $F(v)=g(x^2+y^2+z^2)$, if $v=(x,y,z)$.
On the other hand, we can again decompose $v=xe\_1+(ye\_2+ze\_3)$, and our basic assumption gives $F(v)=g(x^2)+g(y^2+z^2)$. So $g$ satisfies [Cauchy's functional equation](https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation) $g(s+t)=g(s)+g(t)$, and this implies that $g(r^2)=cr^2$, as desired (for example, take the $s$ derivative at $s=0$ in the functional equation to conclude that $g'$ is constant).
| 8 | https://mathoverflow.net/users/48839 | 446732 | 179,955 |
https://mathoverflow.net/questions/446737 | 6 | Let $f\colon M\to N$ a smooth surjective map of compact oriented manifolds of the same dimension. Then there is a map $f\_!\colon H\_i(N)\to H\_i(M)$ obtained from the induced map on cohomology combined with Poincaré duality. This map has several names. I have seen it called the transfer, umkehr or wrong-way map. And it is treated in several question on mathoverflow and math.stackexchange:
[Pullback map in homology](https://mathoverflow.net/questions/115764/pullback-map-in-homology)
[Reference for push-pull formula in cohomology](https://mathoverflow.net/questions/244844/reference-for-push-pull-formula-in-cohomology?rq=1)
<https://math.stackexchange.com/questions/4683875/pullback-of-homology-via-trace-and-poincar%C3%A9-duality>
I am interested in a geometric interpretation of this map in the following situation. Assume that there is an open subset $U\subset N$ such that, letting $V=f^{-1}(U)$, the restriction $f|\_V\colon V\to U$ is an $r$-sheeted covering map ($r\in\mathbb{N}$) and $N'\subset U$ a compact oriented submanifold. If I understood correctly the answers and comments from the questions above, then we have
$f\_!([N'])=[f^{-1}(N')]$.
Is this correct? And is there a reference for this? Several books in the answers and comments to the questions above are mentioned but I have actually not found this statement.
I should also say that I want to use this statement in a paper (if true!) and neither me nor the audience of the paper are experts on algebraic topology. So I would very much appreciate a reference where this fact is explicitly stated or at least a statement which easily implies this fact.
Edit: One can construct $f\_!$ also in the case when $M$ and $N$ are oriented manifolds with boundary and $f^{-1}(\partial N)=\partial M$ (and all other assumptions the same). Now the map map $f\_!\colon H\_i(N)\to H\_i(M)$ is obtained from the induced map on cohomology relative to the boundaries combined with Lefschetz duality. Is the conclusion $f\_!([N'])=[f^{-1}(N')]$ still true? We still assume $f|\_V\colon V\to U$ being a $r$-sheeted covering, neither $U$ nor $V$ intersect the boundaries of $M$ and $N$, and $N'\subset U$ a compact oriented submanifold (without boundary).
| https://mathoverflow.net/users/36563 | Geometric interpretation of transfer map on homology | Let $K\subset N$ be a compact oriented smooth submanifold of codimension $k$ in an oriented smooth manifold $N$, $T\subset N$ a tubular neighbourhood of $K$ and $\tau$ a $k$-form on $N$ supported on $T$ such that the restriction of $\tau$ to the fibre of $T$ over any point of $K$ (which is of course diffeomorphic to $\mathbb R^k$) has compact support and integrates to 1. Then the class of $\tau$ in $H^k\_c(N)$ is Poincaré dual to $[K]$. Such a $\tau$ always exists; it is called a Thom form. This is discussed in Bott, Tu, Differential forms in algebraic topology, § 5--6.
If $M$ is an oriented smooth manifold, $f:M\to N$ a smooth map transversal to $K$ and $T$ a sufficiently small tubular neighbourhood of $K$, then $f^{-1}(T)$ [is a tubular neighbourhood](https://math.stackexchange.com/questions/1859972/preimage-of-tubular-neighborhood) of $f^{-1}(K)$; one can arrange trivializations in such a way that $f$ maps any fibre of $f^{-1}(T)$ diffeomorphically to a fibre of $T$. So the pullback via $f$ of a Thom form for $K$ will be a Thom form for $f^{-1}(K)$. This proves the desired statement $f\_!([K])=[f^{-1}(K)]$.
| 8 | https://mathoverflow.net/users/485324 | 446744 | 179,958 |
https://mathoverflow.net/questions/446742 | 8 | In the Author Commentary to the reprint of the paper paper *[Diagonal Arguments and Cartesian Closed Categories](http://www.tac.mta.ca/tac/reprints/articles/15/tr15abs.html)* in *Theory and Applications of Categories* Bill Lawvere wrote:
>
> Although the cartesian-closed view of function spaces and functionals was intuitively obvious in all but name to Volterra and Hurewicz (and implicitly to Bernoulli), it has counterexamples within the rigid framework advocated by Dieudonné and others. According to that framework the only acceptable fundamental structure for expressing the cohesiveness of space is a contravariant algebra of open sets or possibly of functions. Even though such algebras are of course extremely important invariants, their nature is better seen as a consequence of the covariant geometry of figures. Specific cases of this determining role of figures were obvious in the work of Kan and in the popularizations of Hurewicz’s k-spaces by Kelley, Brown, Spanier, and Steenrod, but in the present paper I made this role a matter of principle: the Yoneda embedding was shown to preserve cartesian closure, and naturality of functionals was shown to be equivalent to Bernoulli’s
> principle.
>
>
>
First of all, as if I understand correctly the mention of Bernoulli refers to [Johann Bernoulli](https://en.wikipedia.org/wiki/Johann_Bernoulli) (1667 – 1748) which is known for his contributions to infinitesimal calculus and educating Leonhard Euler in the pupil's youth. Am I correct thinking so?
Secondly, which principle exactly does the author mention in the passage? The only Bernulli principle which comes on my mind is [Daniel Bernoulli's principle](https://en.wikipedia.org/wiki/Bernoulli%27s_principle) in fluid dynamics.
| https://mathoverflow.net/users/73577 | Mention of Bernoulli principle by Bill Lawvere | Since the topic of Lawvere paper is differential geometry, the Bernoulli is likely [Jacob Bernoulli,](https://en.wikipedia.org/wiki/Jacob_Bernoulli) or his brother [Johann,](https://en.wikipedia.org/wiki/Johann_Bernoulli) and refers to their calculus of variations and the principle of virtual work. Further evidence is this [paper](http://denise.vella.chemla.free.fr/Lawvere-categ-3.pdf) by Lawvere on category theory, where he explicitly refers to Bernoulli in the context of the calculus of variations.
| 5 | https://mathoverflow.net/users/11260 | 446749 | 179,960 |
https://mathoverflow.net/questions/446658 | 2 | As far as I understand, model categories mainly provide tools for studying the "homotopy theories" (that is, $\infty$-categories) that are ubiquitous in mathematics. From this point of view, model categories (as an independent concept, not a tool) are of no interest and behave rather strangely (starting with the zigzags of Quillen equivalences). Simplicial model categories serve the same function, only they are much more convenient than ordinary model categories. But what is the function of model categories enriched in an arbitrary (good) monoidal model category?
>
> Are enriched model categories a representation of enriched $\infty$-categories?
>
>
>
If so, that would be the perfect answer to my question in the title. As far as I understand, the work [Rune Haugseng - Rectification of enriched infinity-categories](https://arxiv.org/abs/1312.3881v4) shows that categories enriched over a good monoidal model category represent $\infty$-categories enriched over the corresponding monoidal $\infty$-category. But this does not explain why a model structure on an enriched category is needed.
>
> If the answer to the first question is no, then maybe there is some other important invariant concept behind them? Otherwise, why do we study such a concept? What other areas (topics, issues) is the general theory of enriched model categories related to?
>
>
>
| https://mathoverflow.net/users/148161 | Why do we need enriched model categories? | To me, the interest in model categories stems from Quillen's observation that the tools of topology (e.g., CW approximation) can be applied in so many different settings, especially in algebra. But not all of those settings are simplicial model categories. Many of them are dg-model categories, i.e., enriched in the category of chain complexes over a commutative ring. There is a huge literature about dg categories, and they have real applications in representation theory. Using enriched model categories, people have been able to prove that the theory of dg-categories closely parallels the theory of spectral categories, i.e., categories enriched in (your favorite) monoidal category of spectra. This has been a fruitful way to use results in stable homotopy theory to prove things in representation theory, homological algebra, and triangulated categories. It also sets the stage for results about Fukaya categories, homological mirror symmetry, etc.
Results in higher category theory (by which I mean weak $n$-categories, not $(\infty,n)$-categories) also often require enriched categories. At its most basic level, an $n$-category is a category enriched in $(n-1)$-categories. So, many of the models for weak $n$-categories (e.g., invented by Bergner, Rezk, Barwick, Ara, Tamsamani, etc.) are cartesian model categories, so that the next level can be enriched in them. To do homotopy theory in this context, you need enriched model categories. I used them a lot in [a recent paper with Batanin](https://www.ams.org/journals/tran/2022-375-05/S0002-9947-2022-08600-1/) proving a strong version of the Baez-Dolan stabilization hypothesis.
Another important reason to care about enriched model categories is monoidal model categories, i.e., self-enriched model categories. Homotopy theory is full of situations where the monoidal structure has been crucial, e.g., in chromatic homotopy theory, K-theory, etc. You probably know there was a long (and eventually successful) search for a good monoidal category of spectra, and then tons of applications. You might be interested in the paper "[Presentably symmetric monoidal infinity-categories are represented by symmetric monoidal model categories](https://arxiv.org/abs/1506.01475)" by Thomas Nikolaus and Steffen Sagave, that proves every presentable monoidal $\infty$-category is modeled by a combinatorial monoidal model category. Perhaps you'd be interested to generalize this result with "monoidal" replaced by "enriched."
The MathOverflow community has already answered the question "[Why do we need model categories?](https://mathoverflow.net/questions/287091/why-do-we-need-model-categories)" and "[Do we still need model categories?](https://mathoverflow.net/questions/78400/do-we-still-need-model-categories?rq=1)" The same answers tell you that we still need monoidal model categories. And, even if you only want to work in the setting of monoidal $\infty$-categories, you will quickly find that you might actually need monoidal model categories to compute things, in much the same way as described in those answers.
| 9 | https://mathoverflow.net/users/11540 | 446755 | 179,964 |
https://mathoverflow.net/questions/446739 | 4 | Consider an $n$-dimensional convex set $K \subset \mathbb{R}^n$ and let $\mu$ denote the Gaussian measure with density
$$
\gamma(\mathbf{x}) = \frac{1}{(2\pi)^{n/2}} e^{-\lVert \mathbf{x} \rVert^2/2}.
$$
I am trying to figure out if, as I slide the convex body $K$ along a straight line, its Gaussian measure, viewed as a function of the line parameter, is a function that has a unique local and global maximum. In essence, I want to know if
$$
g \colon \ \mathbb{R} \to \mathbb{R}\_+, \ t \mapsto \mu(\mathbf{u} + t\mathbf{v} + K)
$$
is log-concave or quasi-concave.
| https://mathoverflow.net/users/100355 | Sliding a convex body over a Gaussian measure | $\newcommand\u{\mathbf u}\newcommand\v{\mathbf v}\newcommand\x{\mathbf x}\newcommand\R{\mathbb R}\newcommand\ga\gamma$We have
$$\mu(\u+t\v+K)=f(t):=\int\_{\R^n}d\x\,F(\x,t),$$
where
$$F(\x,t):=\ga(\x)\,1(\x\in\u+t\v+K).$$
The function $F\colon\R^n\times\R\to\R\_+$ is log concave, as the product of two log-concave functions. So, by the Prékopa–Leindler inequality (cf. e.g. [this](https://en.wikipedia.org/wiki/Pr%C3%A9kopa%E2%80%93Leindler_inequality#Applications_in_probability_and_statistics)), the function $f\colon\R\to\R\_+$ is log concave. $\quad\Box$
For related results, see e.g. [this paper](https://www.sciencedirect.com/science/article/pii/S0047259X13002534) and references therein.
| 7 | https://mathoverflow.net/users/36721 | 446756 | 179,965 |
https://mathoverflow.net/questions/446711 | 3 | I am looking for a sequence of topological spaces $X\_n$, $n\in\mathbb N$, with the following property. Let $\tilde{K}^0(X\_n)$ be the complex reduced $K$-theory group of $X\_n$ (with respect to some choice of base point). I would like, for each $n$, for there to be a class $\xi\_n\in\tilde{K}^0(X\_n)$ such that $\xi\_n$ cannot be represented as the difference of two vector bundles of rank at most $n$.
**Question:** How could one choose/construct $X\_n$ and $\xi\_n$ for all $n$?
*Comments:* I thought one might be able to take $X\_n=S^n$ (say for $n$ even), which has a Bott vector bundle $\beta\_n$ whose dimension increases with $n$. The class of $\beta\_n-1$ generates $\tilde{K}^0(S^n)$, and I suspect it cannot be represented by the difference of two lower-rank bundles, but I'm not sure if this is true or how to show it.
| https://mathoverflow.net/users/78729 | "High-dimensional" classes in topological $K$-theory | Let $X\_n = S^{2n+2}$.
Since $\operatorname{ch} : K(S^{2n+2})\otimes\_{\mathbb{Z}}\mathbb{Q} \to H^{\text{even}}(S^{2n+2}; \mathbb{Q})$ is an isomorphism, there is a complex vector bundle $E \to S^{2n+2}$ of rank $n + 1$ with $\operatorname{ch}\_{n+1}(E) \neq 0$, i.e. $c\_{n+1}(E) \neq 0$. Let
$$\xi\_n = E - \varepsilon^{n+1} \in \widetilde{K}(S^{2n+2}).$$
Note that $c(\xi\_n) = c(E)c(\varepsilon^{n+1})^{-1} = c(E) = 1 + c\_{n+1}(E)$. On the other hand, if $F \to S^{2n+2}$ and $G \to S^{2n+2}$ are vector bundles of rank at most $n$, then $c(F - G) = c(F)c(G)^{-1} = 1$. Since $c\_{n+1}(E) \neq 0$, we see that $c(\xi\_n) \neq 1$, so $\xi\_n$ cannot be represented as the difference of two vector bundles of rank at most $n$.
| 3 | https://mathoverflow.net/users/21564 | 446758 | 179,966 |
https://mathoverflow.net/questions/446775 | 4 | Let $A$ be a complete, Noetherian, local ring with finite residue field of characteristic $p$. If $F$ is a non-Archimedean local field, then we will denote the ring of integers of $F$ by $\mathcal{O}\_F$. Let $I$ denote the intersection of kernels of all (local) morphisms $A\to \mathcal{O}\_F$ where $F$ runs over all non-Archimedean local field of zero or $p$.
**Question:** Is $I=(0)$?
In other words, is a complete local ring determined by its values in local fields? Any comments and reference would be appreciated. The answer to the question is Yes when $A$ is flat over $\mathbb{Z}\_p$ and reduced, cf. Corollary 2.3 [Serre’s modularity conjecture (II)](https://link.springer.com/article/10.1007/s00222-009-0206-6).
Edit: As pointed out in the comment, I should assume that **$A$ is reduced**.
| https://mathoverflow.net/users/149460 | Is a complete local ring determined by its values in local fields? | The paper you cite itself cites Corollary 10.5.8 of EGA 4, part III. Corollary 10.5.9 says the points of dimension $1$ in $\operatorname{Spec} A$ are dense in $\operatorname{Spec} A - \mathfrak m$. Each point of dimension $1$ corresponds to a homomorphism to a complete local ring of dimension $1$, whose field of fractions is a local field, and the intersection of the kernels corresponds to an ideal whose vanishing set is dense, which if $A$ is reduced must be the zero ideal.
See also 10.5.10 which explains how to use these local rings to form a basis for the topology on $A$.
| 6 | https://mathoverflow.net/users/18060 | 446782 | 179,974 |
https://mathoverflow.net/questions/446788 | 5 | Let $L\_n(p)$ be the $2n+1$ dimensional Lens space
$$
S^{2n+1}/\mathbb{Z}\_p
$$
where the action is given as $z\_i\rightarrow e^{\frac{2\pi}{p}}z\_i$, $i=1,...,n+1$, with $z\_i$ the coordinates of $\mathbb{C}^{n+1}$ such that $S^{2n+1}$ is $|z\_1|^2+...+|z\_{n+1}|^2=1$. For $k\neq 0,2n+1$ the homology groups with coefficients in a commutative ring $R$ are
$$
H\_k(L\_n(p),R)=\left\{\begin{array}{cc}
R/pR & \text{if $k$ is odd}\\
T\_p(R) & \text{if $k$ is even }
\end{array}\right.
$$
where $T\_p(R)\subset R$ is the $p-$torsion subgroup:
$$
T\_p(R)=\left\{x\in R \ | \ px=0\right\} \ .
$$
Since $R$ is an abelian group, we can consider $R^{\vee}$, its Pontryagin. Fix an extension $\Gamma$ of $R$ by $R^{\vee}$:
$$
1\rightarrow R^{\vee}\rightarrow \Gamma \rightarrow R\rightarrow 1 \ .
$$
This induces a long exact sequence of homology and cohomology groups, with connecting homomorphisms given by the Bockstein maps. For instance in cohomology $\beta :H^k(L\_n(p),R)\rightarrow H^{k+1}(L\_n(p),R^{\vee})$.
Given $\Sigma \in H\_{2n}(L\_n(p),R)$ we consider its Poincare' dual cocycle $A=PD(\Sigma)\in H^1(L\_n(p),R)$, and its image under Bockstein $\beta(A)\in H^2(L\_n(p),R^{\vee})$. Since $R^{\vee}$ is also a ring, the cup product associated with the product $R^{\vee}\times R^{\vee}\rightarrow R^{\vee}$ allows to construct a class $\beta(A)^n\in H^{2n}(L\_n(p),R^{\vee})$, and the natural pairing $R\times R^{\vee}\rightarrow \mathbb{R}/\mathbb{Z}$ allows the construction of the self-linking invariant:
$$
lk(\Sigma)=\int \_{L\_n(p)}A\cup \beta(A)^n \in \mathbb{R}/\mathbb{Z}
$$
1. Preliminary question: how this self-linking invariant is related with the more standard linking form which only uses the sequence $\mathbb{Z}\rightarrow \mathbb{Q}\rightarrow \mathbb{Q}/\mathbb{Z}$?
2. Let us first fix $n=1$, $p=2$ so that $L\_1(p)=\mathbb{RP}^3$, and $R=\mathbb{Z}\_2$. Since $T\_2(\mathbb{Z}\_2)=\mathbb{Z}\_2$ there is precisely one 2-cycle $\Sigma \in H\_2(L\_1(2),\mathbb{Z}\_2)$, and one 1-cycle $\gamma\in H\_1(L\_1(2),\mathbb{Z}\_2)$. How can I compute
$$
lk(\Sigma)=\int \_{L\_1(2)}A\cup \beta(A)
$$
explicitely? Moreover I have the intuition that $\beta(A)$ should be the Poincare' dual of $\gamma$, but I don't know how to prove this.
3. For general $n$ and $p$ fix $R=\mathbb{Z}\_p$, so that $R/pR=T\_p(R)=\mathbb{Z}\_p$, and again choose one generator $\Sigma $ of $H\_{2n}(L\_n(p),\mathbb{Z}\_p)$. How can I compute $lk(\Sigma)$? By extending the intuition before I also guess that, fixing a generator $\gamma$ of $H\_1(L\_n(p),\mathbb{Z}\_p)$, $\beta(A)$ is the Poincare' dual of $\beta(A)^n$. Is this correct?
| https://mathoverflow.net/users/495347 | Computation of the linking invariant on Lens spaces | In my thesis, I gave the calculation of the linking form on homology, which is equivalent to the question you asked. I credited the calculation to de Rham ([Sur L'analysis situs des varietés a n dimensions](http://www.numdam.org/item/THESE_1931__129__1_0.pdf), J. Math. Pures et Appl., 10 (1931), 115-200.) See Proposition 4 in [Imbedding punctured lens spaces and connected sums](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwj6vZHruPX-AhUjF1kFHcolCOwQFnoECBwQAQ&url=https%3A%2F%2Fprojecteuclid.org%2Fjournals%2Fpacific-journal-of-mathematics%2Fvolume-113%2Fissue-2%2FImbedding-punctured-lens-spaces-and-connected-sums%2Fpjm%2F1102709207.pdf&usg=AOvVaw30d8npdEhbIfukeNw-zYcx). Pacific J. Math. 113 (1984), no. 2, 481–491.
The answer is given for a lens space $L(m;q\_1,\ldots,q\_{2k})$ of dimension $4k-1$; I imagine that it's similar for dimension $4k+1$. With respect to a certain generator $e$ of $H\_{2k-1}(L)$, de Rham showed $\lambda(e,e) = \frac1m q\_1\cdots q\_k \cdot r\_{k+1}\cdots r\_{2k}$ where $q\_jr\_j \equiv 1 \pmod{m}$. You'd have to look in de Rham's paper to find the details.
| 5 | https://mathoverflow.net/users/3460 | 446802 | 179,978 |
https://mathoverflow.net/questions/446790 | 3 | Let $(X, g)$ be a smooth, oriented, Riemannian 4-diemnsional manifold. Let $\Lambda^2$ denote the bundle of 2-forms on $X$. Then the Hodge-star operator decomposes $\Lambda^2$ into the space of self-dual and anti-self-dual 2-forms
$$\Lambda^2 = \Lambda^2\_+ \oplus \Lambda^2\_- .$$
Both $\Lambda^2\_+$ and $\Lambda^2\_-$ are $3$-dimensional bundles. We then have the projection
$$
P\_+: \Lambda^2\to \Lambda^2\_+
$$
and the operator $d\_+: \Lambda^1\to \Lambda^2\_+$ given by
$$
d\_+=P\_+\circ d
$$
where $d: \Lambda^1\to \Lambda^2$ is the de Rham operator.
The Riemannian curvature tensor could be considered as a map $\mathcal{R}: \Lambda^2\to \Lambda^2$ and under the decomposition $\Lambda^2 = \Lambda^2\_+ \oplus \Lambda^2\_- $, $\mathcal{R}$ could be decomposed as
$$
\mathcal{R}=\begin{bmatrix}
A & B
\\ B^\* & C
\end{bmatrix}.
$$
The Riemannian manifold $(X, g)$ is called *self-dual* if we have
$$
C-\frac{1}{3}tr(C)=0.
$$
Now let $\omega$ and $\theta$ be $1$-forms on $X$. Then $\omega\wedge \theta$ is a $2$-form and we can consider $P\_+(\omega\wedge \theta)$ and the $3$-form $d(P\_+(\omega\wedge \theta))$. On the other hand we have $3$-forms $(d\_+\omega)\wedge \theta$ and $\omega\wedge (d\_+\theta)$.
>
>
> >
> > My question is: do we have
> > $$
> > d(P\_+(\omega\wedge \theta))=(d\_+\omega)\wedge \theta-\omega\wedge (d\_+\theta)
> > $$
> > if $(X,g)$ is a self-dual manifold?
> >
> >
> >
>
>
>
| https://mathoverflow.net/users/24965 | Do we have $d(P_{+}(\omega\wedge \theta))=d_{+}\omega\wedge \theta-\omega\wedge d_{+}\theta$ on a self-dual manifold? | No. If this formula were true, then we would have
$$
\mathrm{d}\bigl(P\_+(\mathrm{d}f\wedge\mathrm{d}g)\bigr) = 0
$$
for all smooth functions $f$ and $g$, since $\mathrm{d}\_+(\mathrm{d}f) =P\_+\bigl(\mathrm{d}(\mathrm{d}f)\bigr) = 0$.
Now, consider $\mathbb{R}^4$ with its standard flat metric $g = (\mathrm{d}x\_1)^2+(\mathrm{d}x\_2)^2+(\mathrm{d}x\_3)^2+(\mathrm{d}x\_4)^2$ and orientation $\mathrm{d}x\_1\wedge\mathrm{d}x\_2\wedge\mathrm{d}x\_3\wedge\mathrm{d}x\_4>0$, which is clearly self-dual.
Let $f=f(x\_1)$ and $g = g(x\_2)$, then
$$
P\_+(\mathrm{d}f\wedge\mathrm{d}g) = \tfrac12\,f'(x\_1)g'(x\_2)\bigl(\mathrm{d}x\_1\wedge\mathrm{d}x\_2 + \mathrm{d}x\_3\wedge\mathrm{d}x\_4\bigr),
$$
so
$$
\mathrm{d}\bigl(P\_+(\mathrm{d}f\wedge\mathrm{d}g)\bigr)
= (f''(x\_1)g'(x\_2)\,\mathrm{d}x\_1 + f'(x\_1)g''(x\_2)\,\mathrm{d}x\_2)\wedge\mathrm{d}x\_3\wedge\mathrm{d}x\_4\,
$$
which will not usually be zero.
| 5 | https://mathoverflow.net/users/13972 | 446803 | 179,979 |
https://mathoverflow.net/questions/446701 | 2 | Let $\mathbb{S}^2\_+$ denote the closed upper hemisphere of the unit round sphere in $\mathbb{R}^3$. It is well known that the first positive eigenvalue of the Laplacian on the closed unit sphere is $2$, and the associated eigenfunctions are the coordinate functions $x,y,z$ restricted to the sphere. I wonder if the functions $x$ and $y$ generate the first eigenspace of the Laplacian the upper hemisphere with Neumann boundary condition. It is easy to see that $z$ does not satisfy Neumann boundary condition.
| https://mathoverflow.net/users/85934 | Are these the only first eigenfunctions on a hemisphere? | As Christian Remling already indicated in the comments: one can use reflection techniques and thus show: eigenfunctions of the Laplace-Beltrami operator on the $n$-dimensional hemisphere with Neumann boundary conditions are in 1-to-1 correspondence to eigenfunctions on the $n$-dimensional sphere that are invariant under reflection $(x\_1,x\_2,..,x\_n,z)\mapsto (x\_1,x\_2,...,x\_n,-z)$. The isomporphism is given by symmetric extension, thus its inverse is restriction. As the eigenfunctions on the sphere are explicitly known as restrictions of harmonic homogeneous polynomials on $\mathbb{R}^{n+1}$, one knows the eigenfunctions explicitly [see e.g. M. Berger, P. Gauduchon, E. Mazet, Le spectre d’une variété Riemannienne, Lecture Notes in Mathematics 194, Springer Verlag, Berlin, New
York, 1971]. E.g. the first eigenspace is generated by the constant function $1$, the second one by the restrictions of the $x\_i$. This holds in any dimension $n\geq 1$.
| 2 | https://mathoverflow.net/users/110127 | 446815 | 179,985 |
https://mathoverflow.net/questions/446830 | 2 | I am struggling to find a reference for the following statement, which I still believe to be true.
"Let $(\Omega\_1, \mathcal{A}\_1, \mu\_1), (\Omega\_2, \mathcal{A}\_2, \mu\_2)$ be finite measure spaces. Furthermore, let $(\Omega\_1\times\Omega\_2, \mathcal{A}\_1\otimes\mathcal{A}\_2, \mu\_1\otimes\mu\_2)$ the usual product measure space. Then, for every $A\in \mathcal{A}\_1\otimes\mathcal{A}\_2$ there are sequences $(B\_1^i)\_{i\in\mathbb{N}}\subset\mathcal{A}\_1, (B\_2^i)\_{i\in\mathbb{N}}\subset\mathcal{A}\_2$ such that $\bigcup\_{i=1}^n B\_1^i\times B\_2^i\subset A$ and $(\mu\_1\otimes\mu\_2)(A\backslash \bigcup\_{i=1}^n B\_1^i\times B\_2^i)\to 0$ for $n\to\infty$."
So, in $\mathbb{R}^2$ with the Borel-$\sigma$-algebra that is the classical picture that you can approach measurable sets by unions of rectangluar sets.
I would also be fine with a statement that approximates $A$ with bigger sets, so $A\subset \cup\_{i=1}^{n\_j} B\_1^{i,j}\times B\_2^{i,j}$ for any $j\in\mathbb{N}$ and for $j\to\infty$ I have that $(\mu\_1\otimes\mu\_2)((\cup\_{i=1}^{n\_j} B\_1^{i,j}\times B\_2^{i,j})\backslash A)$.
Thank you very much for every answer in advance.
EDIT: Sorry, I forgot one property, which I added. Also, if it is easier I would also be happy with an approximation with "bigger" sets.
| https://mathoverflow.net/users/504794 | Product sigma-algebra: approximating elements arbitrary good using the generating sets | $\newcommand\Om\Omega\newcommand\A{\mathcal A}\newcommand\I{\mathbb I}$Your original statement is trivial: Take $B\_1^i=\Omega\_1$ and $B\_2^i=\Omega\_2$ for all $i$.
If you additionally require that $B\_1^i\times B\_2^i\subseteq A$ for all $i$, then the statement will become false in general. For instance, for $j=1,2$ let $(\Om\_j,\A\_j,\mu\_j)$ be the standard Lebesgue measure space over $[0,1]$. Let $A=\{(x,y)\in[0,1]^2\colon x-y\in\I\}$, where $\I$ is the set of all irrational real numbers. Then (by, say, the Tonelli theorem), $(\mu\_1\otimes\mu\_2)(A)=1$. However, if $B\_1\times B\_2 \subseteq A$, then $(\mu\_1\otimes\mu\_2)(B\_1\times B\_2)=0$, because otherwise the set $B\_1-B\_2=\{x-y\colon x\in B\_1,y\in B\_2\}$ would contain a nonzero-length interval and therefore would contain a rational point as an element -- cf. e.g [this](https://www.researchgate.net/profile/Iosif-Pinelis/publication/280132389_150600537v2/data/55abc6d608aea3d0868593f2/150600537v2.pdf?origin=ResearchDetailAlternativeSimilarResearch&_rtd=eyJjb250ZW50SW50ZW50Ijoic2ltaWxhciJ9). So, for any $B\_j^i\in\A\_j$ such that $B\_1^i\times B\_2^i\subseteq A$ for all $i$, we will have $(\mu\_1\otimes\mu\_2)(A\setminus\bigcup\_{i=1}^n(B\_1\times B\_2))=1\not\to0$.
Similarly, in general you cannot approximate a set $A\in\A\_1\otimes\A\_2$ by bigger sets (containing $A$) belonging to the algebra of all finite unions of the products of sets in $\A\_1$ and $\A\_2$ -- just consider the complement of the set $A$ in the example above. (Do not confuse the product $\sigma$-algebra $\A\_1\otimes\A\_2$, generated by the mentioned algebra, with the product $\A\_1\times\A\_2$ of the $\sigma$-algebras $\A\_1$ and $\A\_2$.)
---
What is possible is for each natural $n$ to find a set $B\_n$ belonging to the mentioned algebra of all finite unions of of the products of sets in $\A\_1$ and $\A\_2$ such that $(\mu\_1\otimes\mu\_2)(A+B\_n)\to0$ as $n\to\infty$, where $+$ denotes the symmetric difference -- see e.g. Theorem 1.1 [here](https://link.springer.com/article/10.1007/s11117-017-0507-8) or [here](https://arxiv.org/abs/1702.01142).
| 0 | https://mathoverflow.net/users/36721 | 446837 | 179,994 |
https://mathoverflow.net/questions/446705 | 7 | Consider the quantum group $U\_q(\mathfrak{sl}\_2)$, with generators $E,F,K$ such that $[E,F]=\frac{K-K^{-1}}{q-q^{-1}}$. Write $[n]=\frac{q^n-q^{-n}}{q-q^{-1}}$, and $[n]!=[n][n-1]\dotsm[1]$.
In [Quantum deformations of certain simple modules over enveloping algebras](https://doi.org/10.1016/0001-8708(88)90056-4), Lusztig defined the following elements inside his integral form for $U\_q(\mathfrak{sl}\_2)$ (I am replacing $q$ by its square root in his notation):
$$ \newcommand\qbinom{\genfrac[]0{}}\qbinom{K;\ c}t=\frac{1}{[t]!}\prod\_{s=0}^{t-1}\frac{(q^{c-s}K-q^{s-c}K^{-1})}{(q-q^{-1})}.$$
These are $q$-deformations of the element $\binom{H+c}{t}$ in Kostant's $\mathbb{Z}$-form for $U(\mathfrak{sl}\_2)$.
In Kostant's $\mathbb{Z}$-form, we have identities such as $\binom{H}{1}^2=2\binom{H}{2}+\binom{H}{1}$. However, if one tries to quantize both sides, the result is
$$ \qbinom{K;\ 0}1^2=q^2[2]\qbinom{K;\ 0}2+qK^{-1}\qbinom{K;\ 0}1.$$
For various reasons, I am interested in identities whose coefficients do not involve powers of $K$. For instance, it turns out
$$ \qbinom{K;\ 0}1^2=\qbinom{K;\ 1}2+\qbinom{K;\ 0}2.$$
In general, I believe one may be able to apply iteratively a relation in Lusztig's paper, to prove that products between such elements are always $\mathbb{Z}(q)$-linear combinations of $\qbinom{K;c}t$'s. A possible caveat, is that these elements are not linearly independent over $\mathbb{Z}(q)$. For instance, one has the relation
$$ \qbinom{K;\ 0}2-[3]\qbinom{K;\ 1}2+[3]\qbinom{K;\ 2}2-\qbinom{K;\ 3}2=0.$$
[EDIT: Assuming a certain PBW-like result, I think one may be able to prove that Lusztig’s elements for $c\le t$ form a basis.]
This is such a well studied object, so I was hoping this ring structure would be worked out somewhere. In short:
>
> What is the product $\qbinom{K;\ a}t\qbinom{K;\ b}s$ in terms of Lusztig's elements?
>
>
>
| https://mathoverflow.net/users/138150 | What is the ring structure on Lusztig's integral form of quantum $\mathfrak{sl}(2)$? | I figured it out, but I would really appreciate a reference!
The formula is quite nice. For $c,d\ge0$ such that $c\le t $ and $d\le s$, the following holds:
$$ \newcommand\qbinom{\genfrac[]0{}}\qbinom{K;\ c}t\qbinom{K;\ d}s=\sum\_{i\ge 0}\qbinom{t-c+d}{i-c}\qbinom{s-d+c}{i-d}\qbinom{K;\ i}{t+s}.$$
(Here, the binomial $\qbinom{a}{b}$ is set to zero whenever $b<0$ or $b>a$)
| 3 | https://mathoverflow.net/users/138150 | 446852 | 179,999 |
https://mathoverflow.net/questions/446738 | 2 | Recently I was thinking if there is a way to do the following: assuming I have some sampled points of distribution $\mathcal{X}$ and distribution $\mathcal Z$ (whose MGF I do not have in closed form) and I know that $\text{law}(\mathcal X)\cdot\text{law}(\mathcal Y) = \text{law}(\mathcal Z)$ can I obtain $\mathcal{Y}$? Are there any results/papers on this?
We can also that the support of all $3$ RVs is in $\mathbb{R\_{+}}$
| https://mathoverflow.net/users/504682 | Approximation to ratio distribution | $\newcommand\R{\mathbb R}\newcommand\Z{\mathbb Z}$After [James Martin's clarifying comment](https://mathoverflow.net/questions/446738/approximation-to-ratio-distribution?noredirect=1#comment1154351_446738), the question becomes as follows:
>
> Suppose that $Z=XY$, where $X$ and $Y$ are independent positive random variables (r.v.'s). The distributions $P\_X$ and $P\_Z$ of $X$ and $Z$ are known. Does this determine the distribution $P\_Y$ of $Y$?
>
>
>
The answer to this question is no. Indeed, write $X=e^U$, $Y=e^V$, and $Z=e^W$, where $U,V,W$ are real-valued r.v.'s such that $W=U+V$; $U$ and $V$ are independent; and the distributions $P\_U$ and $P\_W$ are known. The question can now be restated in terms of the characteristic functions (c.f.'s) $f\_U,f\_V,f\_W$ of $U,V,W$ as follows:
>
> Suppose that $f\_U\,f\_V=f\_W$ and suppose that $f\_U$ and $f\_W$ are known. Does this determine $f\_V$?
>
>
>
The answer to this equivalent question is of course still no. Indeed, note that (i) the function $f$ given by the formula $f(t)=\max(0,1-|t|/\pi)$ for real $t$ is a c.f. (of the absolutely continuous distribution with density $\R\ni x\mapsto\dfrac{1-\cos\pi x}{\pi^2 x^2}\,1(x\ne0)$ and (ii) the periodic function $g$ with period $2\pi$ such that $g=f$ on $[-\pi,\pi]$ is a c.f. (of the discrete distribution on $\Z$ with probability mass function
$\Z\ni x\mapsto\dfrac12\,1(x=0)+\dfrac{1-\cos\pi x}{\pi^2 x^2}\,1(x\ne0)$. Note that $fg=f^2$. So, if $f\_U=f$ and $f\_W=f^2$, then $f\_V$ can be either one of the two distinct c.f.'s: $f$ or $g$. So, $f\_U$ and $f\_W$ do not determine $f\_V$. $\quad\Box$.
This example is well known; see e.g. [this book](https://rads.stackoverflow.com/amzn/click/com/0852641702).
---
Such examples are of course an exception. Indeed, if the set $N\_U:=\{t\in\R\colon f\_U(t)=0\}$ is nowhere dense (as will usually be the case, apparently -- including the case when $f\_U$ is analytic in a neighborhood of $\R$), then we know the values of $f\_V(t)=f\_W(t)/f\_U(t)$ for all $t$ in the everywhere dense set $\R\setminus N\_U$. So, by the continuity of any c.f., we know $f\_V$ completely -- so that we know the distribution of $V$ and hence the distribution of $Y$.
---
Regarding concerns about statistical aspects of the problem, raised by [James Martin](https://mathoverflow.net/questions/446738/approximation-to-ratio-distribution/446854#comment1154360_446854), one can say the following.
(i) If we have an i.i.d. sample $(X\_1,Z\_1),\dots,(X\_n,Z\_n)$ from the **joint** distribution of the pair $(X,Y)$ for a large enough sample size $n$, we get the i.i.d. $Y$-sample $Y\_1:=Z\_1/X\_1,\dots,Y\_n:=Z\_n/X\_n$. So, we can use, say, the empirical c.d.f. based on $Y\_1,\dots,Y\_n$ to approximate in the standard manner the true c.d.f. of $Y$. The question as to whether the distribution of $Y$ is determined by the **individual** distributions of $X$ and and of $Z$ is completely irrelevant here, once the **joint** distribution of the pair $(X,Y)$ is known or available for sampling as described above.
(ii) If we **only** have an i.i.d. sample $X\_1,\dots,X\_n$ from the **individual distribution** of $X$ and an i.i.d. sample $Z\_1,\dots,Z\_n$ from the **individual** distribution of $Z$, then we only have partial knowledge of the **individual** distributions of $X$ and and of $Z$. But, as shown above, even complete knowledge of the **individual** distributions of $X$ and of $Z$ will in general **not** determine the distribution of $Y$. So, clearly, the partial knowledge of the **individual** distributions of $X$ and and of $Z$ cannot provide substantial partial knowledge of the distribution of $Y$ -- in particular, then we would even be unable to guess whether $Y$ is integer-valued or absolutely continuous.
Summarizing the statistical sampling aspects of the problem: depending on the kind of of sampling available -- (i) from the joint distribution of $(X,Z)$ or (ii) from the individual distributions of $X$ and $Z$, the sampling is either (i) irrelevant to the problem stated in the OP or (ii) not at all helpful to the problem.
| 2 | https://mathoverflow.net/users/36721 | 446854 | 180,000 |
https://mathoverflow.net/questions/446848 | 7 |
>
> Is there a cubic polynomial $c(x,y)$ with exactly 3 saddle point critical points?
>
>
>
In other words, can a cubic polynomial in two variables have three critical points, all of which are saddle points? Or could it have the maximum 4 critical points (as per Bézout's theorem), with only one of them being a local extremum?
A broader formulation of the question could be:
>
> What number of saddle points can a cubic polynomial in two variables have without having any local maximum?
>
>
>
Observations:
* By [Bézout's Theorem](https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem) a cubic polynomial (in two variables) has at most 4 critical points.
* It is [easy to see](https://math.stackexchange.com/q/4620638/1134951) that a cubic polynomial has at most one maximum / minimum. Thus every polynomial $c(x,y)$ with 4 critical points has at least 2 saddle points. For example, the polynomial $c(x,y) = x^3-3x + y^3-3y$ has four critical points: Local maximum at $(-1,-1)$; local minimum at $(1,1)$ and saddle points at $(1,-1)$ and $(-1,1)$.
* A quartic polynomial can have all 9 critical points without having any local maximum: [Can a real quartic polynomial in two variables have more than 4 isolated local minima?](https://mathoverflow.net/q/442736/497175)
* As argued in the answer to question <https://math.stackexchange.com/q/4620663/1134951>, there is an upper bound on the number of $M-s$, where $M$ is the number local extrema (local minima and maxima) and $s$ is the number of saddle points. Here we are interested in the lower bound.
| https://mathoverflow.net/users/497175 | Can a cubic polynomial in two real variables have three saddle points? | The cubic $x^3 - xy^2 - 2x^2 + x$ has critical points in $(1,0)$, $(0,-1)$, $(0,1)$ and $(1/3, 0)$. The determinant of the Hessian matrix is $-4(3x^2 + y^2 - 2x)$. It assumes the values $-4$, $-4$, $-4$ and $4/3$ in these four critical points. Thus the first three of them are saddle points.
**Added later:** A simpler example is the cubic $xy(x+y-1)$ with saddle points in $(0,0)$, $(1,0)$, $(0,1)$ (arising from setting $\alpha=\beta=0$, $\gamma=1$ in the comments below).
| 14 | https://mathoverflow.net/users/18739 | 446858 | 180,001 |
https://mathoverflow.net/questions/442076 | 0 | I have a 600-cell, whose coordinates are given by
$$\begin{array}{ccc}
\text{8 vertices} & \left(0,0,0,\pm1\right) & \text{all permutations,}\\
\text{16 vertices} & \frac{1}{2}\left(\pm1,\pm1,\pm1,\pm1\right),\\
\text{96 vertices} & \frac{1}{2}\left(\pm\tau,\pm1,\pm\tau^{-1},0\right) & \text{even permutations}.
\end{array}$$
Now I would like to isolate a tethraedral cell of the 600-cell and find the hyperplane that passes through it.Probably I should calculate the distances between the vertices to understand which of those are neighbours. Does anyone have a reference or a shorter path to do it?
| https://mathoverflow.net/users/83165 | How can I find the hyperplane passing through a 600-cell | You already could have considered your provided vertices within layers according to their last coord values. Within decreasing order you get:
$(0, 0, 0; 1)$: the single point `o3o5o` at the north pole
$\frac12(0,\pm\tau^{-1},\pm1; \tau)$: a full icosahedron `v3o5o` (with $\tau^{-1}$-sized edges)
$\frac12(\pm1, \pm1, \pm1; 1)$ and $\frac12(\pm\tau^{-1}, 0, \pm\tau; 1)$: a full dodecahedron `o3o5v` (with $\tau^{-1}$-sized edges)
$\frac12(\pm1, \pm\tau, 0; \tau^{-1})$: a full icosahedron `x3o5o` (with unit edges)
$(\pm1, 0, 0; 0)$ and $\frac12(\pm\tau, \pm1, \tau^{-1}; 0)$: a full icosidodecahedron `o3v5o` (with $\tau^{-1}$-sized edges) at the equator
The remainder then is the mirror image on the southern hemisphere (or rather -globe).
Next consider the $\tau^{-1}$-scaled icosahedral section in turn within layers according to its 3rd coordinate:
$\frac12(0,\pm\tau^{-1}; 1)$: top-most edge `v o` (of size $\tau^{-1}$)
$\frac12(\pm1, 0; \tau^{-1})$: pair of vertices at the polar circle `o x`
$\frac12(\pm\tau^{-1}, \pm1; 0)$: equatorial rectangle `x v`
Again the remainder would just be the mirror image on the southern hemisphere.
Therefrom one now derives e.g. the following set of 4 vertices belonging to a single tetrahedral facet of the hexacosichoron (= 600-cell), chosen here by means of a single triangle of that icosahedron at height $\frac12\tau$ plus the north pole vertex at height $1$ as its tip:
$(0, 0, 0, 1)$, $\frac12(0, \pm\tau^{-1}, 1, \tau)$, $\frac12(1, 0, \tau^{-1}, \tau)$.
--- rk
| 1 | https://mathoverflow.net/users/118679 | 446863 | 180,003 |
https://mathoverflow.net/questions/446851 | 3 | For integral homology groups there is the notion of linking form (<http://www.map.mpim-bonn.mpg.de/Linking_form>)
$$
Tor(H\_{l}(X,\mathbb{Z}))\times Tor(H\_{n-l-1}(X,\mathbb{Z}))\rightarrow \mathbb{Q}/\mathbb{Z}
$$
for the torsion part of the homology groups. This can be defined by using the Bockstein map associated with the sequence $\mathbb{Z}\rightarrow \mathbb{Q}\rightarrow \mathbb{Q}/\mathbb{Z}$.
Maybe this is a very trivial generalization, but is there an analogous notion for homology groups with coefficients in an arbitrary abelian group $A$? I am looking for something like
$$
Tor(H\_{l}(X,A))\times Tor(H\_{n-l-1}(X,A^{\vee}))\rightarrow \mathbb{Q}/\mathbb{Z}
$$
where $A^{\vee}$ is the dual group. If yes, what is the Bockstein map one should use?
| https://mathoverflow.net/users/495347 | Linking form for homology with general coefficients | Expanding on Ryan's comment, one way to get the torsion linking form you mention (at least for compact oriented manifolds) is through the sequence of isomorphisms $$Tor(H\_l(X;\mathbb{Z}))\cong Tor(H^{n-l}(X;\mathbb{Z})) \cong Ext(H\_{n-l-1}(X);\mathbb{Z})\cong Ext(Tor(H\_{n-l-1}(X));\mathbb{Z})\cong Hom(Tor(H\_{n-l-1}(X));\mathbb{Q}/\mathbb{Z}) .$$
The first isomorphism is Poincar'e duality, and the second is the universal coefficient theorem (using that the Ext summand corresponds to the torsion subgroup). The third is a basic property of Ext (write the homology group as a sum of a torsion group and a free group - this uses that the homology is finitely generated from the compactness assumption). The final can be seen to come from the short exact sequence you mention: take the Hom/Ext exact sequence and note that $Ext(-,\mathbb{Q})=0$ always and $Hom(A,\mathbb{Z})=0$ when $A$ is torsion.
So all this can be generalized in circumstances in which these properties all hold. For example, if $R$ is a Dedekind ring and $X$ is a compact $R$-orientable manifold, I believe everything should go through to give a pairing $Tor(H\_l(X;R))\otimes Tor(H\_{n-l-1}(X;R))\to Q(R)/R$, where $Q(R)$ is the field of quotients of $R$.
Offhand, I'm not sure about a version in which the coefficients are modules over $R$ (in your original question, modules over $\mathbb{Z}$). One would have to think through in what cases versions of these isomorphisms still hold.
\*Note: You don't mention manifolds explicitly, but the site you link to for your basic definitions does assume the context of closed, oriented manifolds.
\*\*It does take a little work to show that this definition of the linking form is equivalent to the one using the Bockstein.
| 3 | https://mathoverflow.net/users/6646 | 446873 | 180,004 |
https://mathoverflow.net/questions/446868 | 1 | Let $X$ be a real-valued standard normal variable. Then, for any differentiable function $f: \mathbb{R} \to \mathbb{R}$ such that $E[f(X)^2] < \infty$ and $E[\bigl( f'(X) \bigr)^2] < \infty$, it is well-known that
\begin{equation}
\text{Var}(f(X)) \leq E[\bigl(f'(X)\bigr)^2]
\end{equation}
and is called the Gaussian Poincare inequality.
I can see that this is the Poincare inequality with $p=2$ according to the Wikipedia article <https://en.wikipedia.org/wiki/Poincar%C3%A9_inequality>
Now, I wonder if the Gaussian Poincare inequality holds for $p=1$. That is, do we also have
\begin{equation}
E[\bigl \lvert f(X) - E[f(X)] \bigr \rvert ] \leq E[\bigl \lvert f'(X) \bigr \rvert ]
\end{equation}
in general?
In the link above, the generalization of the Poincare inequality to general measure spaces is considered as well. I searched for papers myself but was not able to find anything specialized to Gaussian measures.
Could anyone please help me?
| https://mathoverflow.net/users/56524 | Does the Gaussian Poincare inequality hold for $p=1$ as well as $p=2$? | Yes, Gaussians also satisfy a Poincaré inequality with $p = 1$ (such an inequality is equivalent to what is called a "Cheeger inequality"). More generally, E. Milman has shown that for log-concave measures, all $(p, q)$-Poincaré inequalities are equivalent:
Milman, E. On the role of convexity in isoperimetry, spectral gap and concentration. Invent. math. 177, 1–43 (2009). <https://doi.org/10.1007/s00222-009-0175-9>
| 1 | https://mathoverflow.net/users/37014 | 446874 | 180,005 |
https://mathoverflow.net/questions/446872 | 7 | It is well-known that for a finite group $G$ and field $k$ of characteristic 0, the linearization morphism $B(G) \to R\_k(G)$ has in most cases nontrivial kernel, and this can be used to find permutation modules which admit non-isomorphic permutation bases (as $G$-sets). I'm hoping to find a relatively easy-to-state example, but this time, for a permutation $\mathbb{Z} G$-module instead.
Unfortunately, all the examples I've cooked up over $\mathbb{Q}$ haven't extended to $\mathbb{Z}$. Given that Krull-Schmidt doesn't hold for $\mathbb{Z}G$ in general, I'm almost certain that examples of such permutation modules exist - does anyone have any examples of this phenomenon?
| https://mathoverflow.net/users/152544 | Examples of permutation $\mathbb{Z}G$-modules which admit non-isomorphic permutation bases? | A similar question was asked on [math.stackexchange](https://math.stackexchange.com/q/2730330) a few years ago, and I posted the following answer. I've just looked again at Conlon's paper, and I'm afraid it's a bit short on explicit examples.
*===========================================================*
There are fairly general examples due to Conlon in the paper:
*Conlon, S.B.*, [**Monomial representations under integral similarity**](http://dx.doi.org/10.1016/0021-8693(69)90111-2), J. Algebra 13, 496-508 (1969). [ZBL0185.06702](https://zbmath.org/?q=an:0185.06702).
There is even a transitive example, due to Scott:
*Scott, Leonard L.*, Integral equivalence of permutation representations, Sehgal, Surinder (ed.) et al., Group theory. Proceedings of the 21st biennial Ohio State-Denison mathematical conference, Granville, OH (USA), 14-16 May, 1992. Singapore: World Scientific. 262-274 (1993). [ZBL0828.20004](https://zbmath.org/?q=an:0828.20004).
In this example, $G$ is $\text{PSL}(2,29)$ and the permutation actions are on the cosets of two non-conjugate subgroups both isomorphic to the alternating group $A\_5$.
| 8 | https://mathoverflow.net/users/22989 | 446883 | 180,009 |
https://mathoverflow.net/questions/446877 | 5 | I asked this question some time ago in MSE but I didn't recieved any feedback.
<https://math.stackexchange.com/questions/4672664/diagonalization-of-symmetric-matrices-of-functions>
This problem arised to me when I was trying to find an analog to orthogonal reference frames for singular metric tensors.
Let $U\subseteq\mathbb{R}^m$ be an open subset such that $0\in U$ and $n\leq m$. Let
$G(p)=\begin{pmatrix}
g\_{11}(p) & g\_{12}(p) & \cdots & g\_{1n}(p)\\
g\_{12}(p) & g\_{22}(p) & \cdots & g\_{2n}(p)\\
\vdots & \vdots & \ddots & \vdots \\
g\_{1n}(p) & g\_{2n}(p) & \cdots & g\_{nn}(p)\\
\end{pmatrix}
$ be a symmetric matrix of infinitely differentiable functions $g\_{ij}:U\rightarrow\mathbb{R}$ such that
$G(0)=0\in\mathcal{M}\_{n\times n}(\mathbb{R})$, that is $g\_{ij}(0)=0$ for all $i,j$.
To diagonalize the matrix near $0\in\mathbb{R}^m$, the Gram-Schmidt process doesn't work because it will divide by $0\in\mathbb{R}$ in some steps. You can not even do an adaptation of this process beacause there is no initial orthonormal basis to start with or more acurately any basis will be orthogonal but not normal since the "norms" will be zero. I write norms between quotation marks because $G(p)$ isn't positive semidefinite so it doesn't define any norm and I don't want to put any condition about the "definitness" of the matrix.
Said that, the question is: **is it possible to find an open subset $V\subseteq U$ such that $0\in V$ where such a matrix is diagonalizable?** Maybe with a different process than Gram-Schmidt or even leaving the ring of infinitely differentiable functions. I just need to prove the existence theoretically, in other words, I don't need it to be computable.
Thank you in advance!
| https://mathoverflow.net/users/148711 | Diagonalization of symmetric matrices of functions | In general, this cannot be done. For example, in dimension $2$ in coordinates $(x,y)$, let
$$
G(x,y) = \left[\begin{matrix}x&y\\y&-x\end{matrix}\right].
$$
If $G$ could be diagonalized by a differentiable invertible matrix $A(x,y)$, i.e., if
$$
A^T G A = \left[\begin{matrix}\lambda\_1&0\\ 0&\lambda\_2\end{matrix}\right]
$$
where $\lambda\_1$ and $\lambda\_2$ were differentiable, then the $\lambda\_i$ would have to vanish at $x=y=0$. Taking determinants yields
$$
-(x^2+y^2)(\det A)^2 = \lambda\_1\lambda\_2\,.
$$
Then, looking at the lowest order terms on each side (the terms of order $2$), you'd have $x^2+y^2$ written as a product of two factors linear in $x$ and $y$, which is impossible.
For similar reasons, you cannot achieve
$$
G = A^T\left[\begin{matrix}\lambda\_1&0\\ 0&\lambda\_2\end{matrix}\right]A
$$
for a differentiable $A$ and $\lambda\_i$. The above argument shows that $A$ could not be invertible, so we would have to have $\det A$ vanishing at $x=y=0$.
Then $-(x^2+y^2) = (\det A)^2\lambda\_1\lambda\_2$ would imply that $\det A$ vanishes at most to order 1 at $x=y=0$ and that $\lambda\_1$ and $\lambda\_2$ do not vanish at $x=y=0$, which again gives a contradiction, since $x^2+y^2$ is not the square of a linear term.
In fact, one cannot have
$$
G = A^T\left[\begin{matrix}\lambda\_1&0\\ 0&\lambda\_2\end{matrix}\right]A
$$
with $A$ and $\lambda\_i$ being merely *continuous* on some disk $ x^2+y^2\le \epsilon^2$ for some $\epsilon>0$.
Here is why: The relation $-(x^2+y^2) = (\det A)^2\lambda\_1\lambda\_2$, shows that $\det A$, $\lambda\_1$ and $\lambda\_2$ must be nonzero away from $(x,y)=(0,0)$, so each $\lambda\_i$ cannot change sign and we must have $\lambda\_1\lambda\_2<0$ away from $(x,y)=(0,0)$. Without loss of generality, we can assume that $\lambda\_1 = \mu\_1^2$ and $\lambda\_2=-\mu\_2^2$, where the $\mu\_i$ are continuous, so by modifying $A$ in the obvious way, we can reduce to the case that $\lambda\_1 = -\lambda\_2 = 1$ and, moreover, that $\det A = \sqrt{x^2+y^2}>0$.
The mapping $f:S^1\to\mathrm{SL}(2,\mathbb{R})$ defined by
$$
f(\theta) = \frac1{\sqrt{\epsilon}} A(\epsilon\cos\theta,\epsilon\sin\theta)
$$
then satisfies
$$
f(\theta)^T\left[\begin{matrix}1&0\\ 0&-1\end{matrix}\right]f(\theta)
= \left[\begin{matrix}\cos\theta&\sin\theta\\ \sin\theta&-\cos\theta\end{matrix}\right].
$$
Now, consider the map $s:\mathrm{SL}(2,\mathbb{R})\to H$ (where $H$, a hyperboloid of one sheet, is the quadric surface in the symmetric $2$-by-$2$ matrices defined by setting the determinant equal to $-1$) defined by
$$
s(A) = A^T\left[\begin{matrix}1&0\\ 0&-1\end{matrix}\right]A.
$$
Both $\mathrm{SL}(2,\mathbb{R})$ and $H$ are homotopic to the circle and hence have $\pi\_1\simeq \mathbb{Z}$. The map $s$ carries the generator of $\pi\_1(\mathrm{SL}(2,\mathbb{R}))$ to twice a generator of $\pi\_1(H)$. However, the above formula for $f$ shows that $s\circ f:S^1\to H$ carries a generator of $\pi\_1(S^1)$ to a generator of $\pi\_1(H)$, which is impossible.
| 8 | https://mathoverflow.net/users/13972 | 446894 | 180,012 |
https://mathoverflow.net/questions/446843 | 1 | Let $B: C^{\infty}([0,1]^3)$ satisfy
$$B(t,t,x)=0 \quad \text{for all $t,x \in [0,1]$.}$$
Let $f \in C^{\infty}([0,1]^2)$ satisfy the following integral equation:
$$ \int\_0^1 f(t,x)\,dx + \int\_0^t\left(\int\_0^1 f(s,x)\,B(t,s,x)\,dx\right)\,ds =0, \quad \forall\, t\in (0,1).$$
Does it follow that
$$ \int\_0^1 f(t,x)\,dx=0$$
for all $t\in (0,1)$?
| https://mathoverflow.net/users/50438 | On an integral equation | The answer is no. A counterexample is
$$
f(t,x) = x - \frac{1}{2} -\frac{1}{24} t^2
$$
$$
B(t,s,x) = \left( x-\frac{1}{2} \right) (t-s)
$$
(Method: I obtained this by expanding $f$ and $B$ into power series in the arguments $t$ and $s$, considering the integral equation order by order in $t$, fiddling a bit with how few nonzero expansion coefficients I could get away with, and utilizing symmetry/antisymmetry around $x=1/2$).
| 3 | https://mathoverflow.net/users/134299 | 446912 | 180,020 |
https://mathoverflow.net/questions/446919 | -3 | The starting point of this question is the observation that in ${\sf (ZFC)}$, all ordinals $\alpha < \omega\_1$ can be [order-embedded in](https://math.stackexchange.com/questions/408300/countable-ordinals-are-embeddable-in-the-rationals-bbb-q-proofs-and-their) $\mathbb{R}$.
Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$. Given $f, g:\omega\to\omega$ we say $f\leq^\* g$ iff there is $N\in\omega$ such that $f(k) \leq g(k)$ for all $k\geq N$.
We say ${\cal D}\subseteq \omega^\omega$ is *dominating* if for all $f\in \omega^\omega$ there is $d\in {\cal D}$ such that $f\leq^\* d$, and we say ${\cal B}\subseteq \omega^\omega$ is *unbounded* if for all $f\in\omega^\omega$ there is $b\in {\cal B}$ such that $b\not\leq^\*f$. (A diagonalization argument shows that every unbounded (and therefore also every dominating) family must be uncountable, and there are interesting set-theoretical considerations in this context, see [here](https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum).)
We define ${\frak b}$ to be the smallest cardinality that an unbounded family ${\cal B}\subseteq\omega^\omega$ can have, and let ${\frak d}$ be the smallest cardinality that a dominating family ${\cal D}\subseteq\omega^\omega$ can have. It is consistent that $\omega\_1 < {\frak b} < {\frak d} \leq 2^{\aleph\_0}$.
**Question.** Is it provable in ${\sf (ZFC)}$ that ${\frak b}$ or even ${\frak d}$ can be order-embedded in $\mathbb{R}$? (The question could also be asked of other [cardinal characteristics of the continuum](https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum).)
| https://mathoverflow.net/users/8628 | Order-embeddability of ${\frak b}$ and ${\frak d}$ in $\mathbb{R}$ | Assume that $i : \alpha \to \mathbb{R}$ is an order-embedding of some ordinal $\alpha$ into $(\mathbb{R},<)$. We can modify $i$ to yield an order embedding $j : \alpha \to \mathbb{Q}$ by induction over $\beta < \alpha:
The interval $[i(\beta), i(\beta + 1)]$ is a non-degenerate interval in $\mathbb{R}$, which contains no points from the range of $i$ other than its endpoints. Pick rationals $p,q$ with $i(\beta) \leq p < q \leq i(\beta + 1)$, and set $j(\beta) = p$ (if $\beta$ is a limit ordinal, otherwise $j(\beta)$ is already defined) and $j(\beta + 1) = q$.
Thus, the ordinals that order-embed into $\mathbb{R}$ are precisely the countable ones.
| 1 | https://mathoverflow.net/users/15002 | 446920 | 180,023 |
https://mathoverflow.net/questions/446888 | 2 | Problem :
Show that :
$$\frac{1}{\zeta(3)}<2C-1$$
Where we can see the zeta function and the Catalan's constant .
After a bounty on Maths Stack Exchange there is no satisfying answer .
See <https://math.stackexchange.com/questions/4693000/show-that-frac1-zeta32c-1>
My last attempt was to use Cauchy-Schwarz inequality on a double integral as we have :
$$\int\_{0}^{1}\int\_{0}^{1}-\frac{\frac{1}{2}\ln\left(xz\right)}{1-xz}dxdz=\zeta(3),\int\_{0}^{1}\int\_{0}^{1}\left(\frac{2}{2-x^{2}-z^{2}}-1\right)dxdz=2C-1$$
We can introduce a continuous finite function $f(x)$ on $x\in(0,1)$ and we have :
$$\int\_{0}^{1}\int\_{0}^{1}-\frac{\frac{1}{2}\ln\left(xz\right)}{1-xz}+f(x)-f(z)dxdz=\zeta(3)$$
As we can invoke symmetry . So far I cannot find a good path .
One can show as intermediate inequality :
$$\frac{1}{\zeta(3)}<\left(\frac{\pi^{4}}{81}-\frac{1}{19\pi^{4}}\right)^{-1}<2C-1$$
We can also consider with Cauchy-Schwarz :
$$\int\_{0}^{1}\left(\left(\frac{2\arctan\left(x\right)}{x}-\left(1+u\right)x^{u}\right)\left(\frac{\left(d-x^{a}+b\right)^{c}}{\int\_{0}^{1}\left(d-t^{a}+b\right)^{c}dt}d\right)\right)^{\frac{1}{2}}dx=0.99988\cdots,u=1/2,d=\zeta(3),c=1.2,a=0.8,b=-1/10$$
We can improve the result we have with the notation above and here $b=-0.11r=1.2$:
$$\int\_{0}^{1}\left(\left(\frac{2\arctan\left(x\right)}{x}-\left(1+u\right)\left(x\right)^{u}+\frac{\ln\left(\tan\left(x\cdot\frac{\pi}{2}\right)\right)}{200}+\frac{\left(\cos\left(x\cdot\frac{\pi}{2}\right)-\sin\left(x\cdot\frac{\pi}{2}\right)\right)}{80}\right)\left(\frac{\left(r-vx^{a}+b\right)^{c}}{\int\_{0}^{1}\left(r-vt^{a}+b\right)^{c}dt}d\right)\right)^{\frac{1}{2}}dx=0.99992\cdots$$
If the following improvement of Cauchy Schwarz is true we have :
$$\left(1+\left(h\left(0.9978\right)\right)^{2}\right)\left(\int\_{0}^{1}\left(\left(\frac{2\arctan\left(x\right)}{x}-\left(1+u\right)\left(x\right)^{u}+\frac{\ln\left(\tan\left(x\cdot\frac{\pi}{2}\right)\right)}{200}+\frac{\left(\cos\left(x\cdot\frac{\pi}{2}\right)-\sin\left(x\cdot\frac{\pi}{2}\right)\right)}{p}\right)\left(\frac{\left(r-vx^{a}+b\right)^{c}}{\int\_{0}^{1}\left(r-vt^{a}+b\right)^{c}dt}d\right)\right)^{\frac{1}{2}}dx\right)^{2}>1$$
Where the target is the minima of $h(x)$ over $[0,1]$ define as :
$$h\left(x\right)=\left(\frac{2\arctan\left(x\right)}{x}-\frac{3}{2}\left(x\right)^{\frac{1}{2}}+\frac{\ln\left(\tan\left(x\cdot\frac{\pi}{2}\right)\right)}{200}+\frac{\left(\cos\left(x\cdot\frac{\pi}{2}\right)-\sin\left(x\cdot\frac{\pi}{2}\right)\right)}{p}\right)\left(\frac{\left(r-vx^{a}+b\right)^{c}}{\int\_{0}^{1}\left(r-vt^{a}+b\right)^{c}dt}d\right),p=90$$
Why not physics comes here ? see
<https://arxiv.org/abs/1703.09634>
>
>
> >
> > Question:
> >
> >
> >
>
>
>
How to show it without a calculator so by hand ?
---
Just few remarks on the use of Cauchy-Schwarz :
If we strenghening the condition here <https://math.stackexchange.com/questions/4701003/conjectured-improvement-of-cauchy-schwarz-for-integrals> we can find a $p$ as in the statement of the question such that :
$$\left(1+\left(h\left(0.9978\right)\right)^{2}\right)\left(\int\_{0}^{1}\left(\left(\frac{2\arctan\left(x\right)}{x}-\left(1+u\right)\left(x\right)^{u}+\frac{\ln\left(\tan\left(x\cdot\frac{\pi}{2}\right)\right)}{200}+\frac{\left(\cos\left(x\cdot\frac{\pi}{2}\right)-\sin\left(x\cdot\frac{\pi}{2}\right)\right)}{p}\right)\left(\frac{\left(r-vx^{a}+b\right)^{c}}{\int\_{0}^{1}\left(r-vt^{a}+b\right)^{c}dt}d\right)\right)^{\frac{1}{2}}dx\right)^{2}=1$$
Remains to show the existence .
In fact the conjecture above in the mse link is false see better <https://math.stackexchange.com/questions/4703716/conjecture-improvement-of-buniakowsky-integral-inequality-left-int-0ig-l>
Using this last conjecture we have assuming $g(0)=1$ :
$$I=\left(\frac{2}{\sqrt{7}}\int\_{0}^{\frac{\pi}{2}}\left(\left(-\ln\left(\tan\left(\frac{x}{2}\right)\right)-\frac{8}{\pi^{2}}\left(\frac{\pi}{2}-x\right)\right)\left(x\ln\left(\tan\left(x\right)+\sec\left(x\right)\right)\right)\right)^{\frac{1}{2}}dx\right)^{2}-\frac{4}{7}\int\_{0}^{\frac{\pi}{2}}x\ln\left(\tan\left(x\right)+\sec\left(x\right)\right)dx\int\_{0}^{\frac{\pi}{2}}\left(-\ln\left(\tan\left(\frac{x}{2}\right)\right)-\frac{8}{\pi^{2}}\left(\frac{\pi}{2}-x\right)\right)dx=I$$
then if true :
$$\int\_{0}^{-I}\left(-\ln\left(\tan\left(\frac{t}{2}\right)\right)-\frac{8}{\pi^{2}}\left(\frac{\pi}{2}-t\right)\right)^{\frac{1}{2}}dt+\left(\frac{2}{\sqrt{7}}\int\_{0}^{\frac{\pi}{2}}\left(\left(-\ln\left(\tan\left(\frac{x}{2}\right)\right)-\frac{8}{\pi^{2}}\left(\frac{\pi}{2}-x\right)\right)\left(x\ln\left(\tan\left(x\right)+\sec\left(x\right)\right)\right)\right)^{\frac{1}{2}}dx\right)^{2}<1$$
Unfortunetaly...
Fortunetaly using a stronger statement of the conjecture we have :
$$I=\left(\int\_{0}^{1}\left(\left(\frac{2\arctan\left(x\right)}{x}-\left(1+u\right)\left(x\right)^{u}+\frac{\ln\left(\tan\left(x\cdot\frac{\pi}{2}\right)\right)}{200}+\frac{\left(\cos\left(x\cdot\frac{\pi}{2}\right)-\sin\left(x\cdot\frac{\pi}{2}\right)\right)}{80}\right)\left(\frac{\left(r-vx^{a}+b\right)^{c}}{\int\_{0}^{1}\left(r-vt^{a}+b\right)^{c}dt}q\right)\right)^{\frac{1}{2}}dx\right)^{2}-\int\_{0}^{1}\left(\left(\frac{\left(r-vx^{a}+b\right)^{c}}{\int\_{0}^{1}\left(r-vt^{a}+b\right)^{c}dt}q\right)\right)dx\int\_{0}^{1}\left(\frac{2\arctan\left(x\right)}{x}-\left(1+u\right)\left(x\right)^{u}+\frac{\ln\left(\tan\left(x\cdot\frac{\pi}{2}\right)\right)}{200}+\frac{\left(\cos\left(x\cdot\frac{\pi}{2}\right)-\sin\left(x\cdot\frac{\pi}{2}\right)\right)}{80}\right)dx$$
$$\left(\int\_{0}^{1}\left(\left(\frac{2\arctan\left(x\right)}{x}-\left(1+u\right)\left(x\right)^{u}+\frac{\ln\left(\tan\left(x\cdot\frac{\pi}{2.01}\right)\right)}{200}+\frac{\left(\cos\left(x\cdot\frac{\pi}{2}\right)-\sin\left(x\cdot\frac{\pi}{2}\right)\right)}{80}\right)\left(\frac{\left(r-vx^{a}+b\right)^{c}}{\int\_{0}^{1}\left(r-vt^{a}+b\right)^{c}dt}q\right)\right)^{\frac{1}{2}}dx\right)^{2}+2.736\frac{\left(\int\_{0}^{-I}\left(\left(\frac{2\arctan\left(x\right)}{x}-\left(1+u\right)\left(x\right)^{u}+\frac{\ln\left(\tan\left(x\cdot\frac{\pi}{2.01}\right)\right)}{200}+\frac{\left(\cos\left(x\cdot\frac{\pi}{2}\right)-\sin\left(x\cdot\frac{\pi}{2}\right)\right)}{80}\right)\right)^{\frac{1}{2}}dy\right)}{1.9546}>1$$
---
I put here my progress :
Theorem (or equivalent form of continuous Callebaut inequality):
Let $f,g$ be continuous decreasing integrable on $[0,1]$ then let $p\in[0,1]$ we have :
$$I=\left(\int\_{0}^{1}\left(f\left(x\right)\right)^{2}dx\int\_{0}^{1}\left(g\left(x\right)\right)^{2}dx-\left(\int\_{0}^{1}f\left(x\right)g\left(x\right)dx\right)^{2}\right)$$
Then :
$$\frac{\left(\int\_{0}^{I}g\left(p\right)dy\right)}{g\left(0\right)}\leq I$$
So we have :
$$I=\left(\int\_{0}^{1}\left(f\left(x\right)\right)^{2}dx\int\_{0}^{1}\left(g\left(x\right)\right)^{2}dx-\left(\int\_{0}^{1}f\left(x\right)g\left(x\right)dx\right)^{2}\right),f\left(x\right)=\left(\frac{2\arctan\left(x\right)}{x}-\frac{3}{2}x^{\frac{1}{2}}\right)^{\frac{1}{2}},g\left(x\right)=\left(\frac{\left(\frac{6}{5}-x^{\frac{4}{5}}+-\frac{1}{10}\right)^{\frac{6}{5}}}{\int\_{0}^{1}\left(\frac{6}{5}-t^{\frac{4}{5}}+-\frac{1}{10}\right)^{\frac{6}{5}}dt}1.202056931\right)^{\frac{1}{2}}$$
$$k\left(x\right)=\left(\int\_{0}^{I}f\left(x\right)dy\right)+\left(\int\_{0}^{1}f\left(u\right)g\left(u\right)du\right)^{2}>1,x=1/9$$
Now if we fasely assume :
$$\int\_{0}^{1}\left(f\left(x\right)\right)^{2}dx\int\_{0}^{1}\left(g\left(x\right)\right)^{2}dx=1$$
We can use a kind of reductio ad absurdum to show that it's not equal for $x\in(0,1]$. It needs to evaluate the integral $\left(\int\_{0}^{1}f\left(x\right)g\left(x\right)dx\right)^{2}$ wich seems to be a monster so I stop here .
Perhaps someone can do better .
Last update :
Let's define :
$$r\left(x\right)=x-\frac{x^{3}}{3}+\frac{x^{5}}{6}-x^{7}\cdot\frac{17}{315}+\frac{x^{9}}{165},g\left(x\right)=\frac{r\left(x\right)}{r\left(1\right)}\cdot\frac{\pi}{4},h\left(x\right)=\arctan x,q\left(x\right)=\left(h\left(x\right)-g\left(x\right)\right)-\frac{1}{10}\left(\frac{1}{10}x^{4}\left(1-x\right)+\frac{1}{3}x^{6}\left(1-x\right)^{2}+x^{8}\left(1-x\right)^{3}\right)$$
Then we have $x\in[0,1]$:
$$ q(x)\geq 0$$
Remains to divide $q(x)$ by $x$ and integrate over the $[0,1]$.Obviously we can improve it again and again using Fourier series notably.
Let :
$$q\left(x\right)=\left(h\left(x\right)-g\left(x\right)\right)-\frac{1}{10}\left(\frac{1}{10}x^{4}\left(1-x\right)+\frac{1}{3}x^{6}\left(1-x\right)^{2}+x^{8}\left(1-x\right)^{3}+x^{10}\left(1-x\right)^{4}\right)$$
Then a graph of $q(x)$ suggest we can use Fourier series notably $10000q\left(x\right)$
Here we have used hyperbolic tangent and cheated a bit with Bernoulli's number .
For $\zeta(3)$ see the cross posted question where we have also Bernoulli's number .
If someone knows why there is Bernoulli's number feel free to comment .
| https://mathoverflow.net/users/147649 | An inequality related to Catalan's constant and $\zeta(3)$ | One has
\begin{equation}
C=\sum\_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}
=\sum\_{k=0}^\infty f\_1(k)=S\_{1,c}(f\_1)+S\_{2,c}(f\_1),
\end{equation}
where
\begin{equation}
S\_{1,c}(f):=\sum\_{k=0}^{c-1}f(k),\quad S\_{2,c}(f):=\sum\_{k=0}^\infty f(c+k),
\end{equation}
\begin{equation}
f\_1(x):=\frac{1}{(4 x+1)^2}-\frac{1}{(4 x+3)^2}=\frac{8 (2 x+1)}{(4 x+1)^2 (4 x+3)^2};
\end{equation}
note that all derivatives of $f\_1$ of even orders are decreasing on $[0,\infty)$.
To evaluate $S\_{2,c}(f)$, use the Euler--Maclarin formula
\begin{equation}
\sum\_{k=0}^\infty f(k)=A\_m(f)+R\_m(f)
\end{equation}
for $f(x)=f\_1(x+c)$, where $m$ is an integer $\ge2$,
\begin{equation}
A\_m(f):=\frac{f(0)}2-\sum\_{j=1}^{m-1}\frac{B\_{2j}}{(2j)!}\,f^{(2j-1)}(0),
\end{equation}
\begin{equation}
|R\_m(f)|\le\frac{2\zeta(2)}{(2\pi)^{2m-1}}f^{(2m-2)}(0)
=\frac{\pi^2/3}{(2\pi)^{2m-1}}f^{(2m-2)}(0);
\end{equation}
cf. e.g. formulas (2.1) and (2.4) [here](https://link.springer.com/article/10.1007/s00211-018-0978-y) or [here](https://arxiv.org/abs/1511.03247).
Choosing now $m=3$ and $c=3$, we get $|R\_m(f\_1)|<2/10^6$ and
\begin{equation}
S\_{1,c}(f\_1)+A\_m(f\_1)>\frac{915965}{1000000},
\end{equation}
so that
\begin{equation}
C>C\_\*:=\frac{915965}{1000000}-2/10^6.
\end{equation}
Similarly, with $f\_2(x):=\frac1{(x+1)^3}$ instead of $f\_1(x)$, $m=3$, and $c=4$, we get
\begin{equation}
\zeta(3)>\zeta\_\*(3):=\frac{1202056}{1000000} - 10^{-5}.
\end{equation}
So,
\begin{equation}
\zeta(3)(2C-1)>\zeta\_\*(3)(2C\_\*-1)=\frac{250003330149}{250000000000}>1.\quad\Box
\end{equation}
---
In this setting, the derivatives of $f\_1$ and $f\_2$ of all orders are very simple, and the required accuracy of calculations is low (to just $5$ or $6$ digits). Otherwise, much more effective is the alternative (to Euler--Maclarin) summation formula given in the linked paper.
| 6 | https://mathoverflow.net/users/36721 | 446926 | 180,025 |
https://mathoverflow.net/questions/441631 | 5 | Does every triangulable manifold have a vertex-transitive triangulation?
When I talk about a vertex-transitive triangulation of a manifold, I mean in the sense of realizing a manifold homeomorphically as a simplicial complex such that the automorphism group of the simplicial complex is transitive on the vertices or 0-simplexes of the complex.
This could also be formulated for other categories, like the category of simplicial sets, or the category of maps on surfaces (once you define G-actions suitably) but we'll stick with graphs and simplicial complexes here.
Two of the reasons I was interested in this problem is the work that's been done on local computability of characteristic classes of manifolds by Gelfand and others, as well as the conjecture of Kahn, Saks and Sturtevant on vertex transitive nonevasive complexes.
I'm surprised no one has really raised the question in print before, to my knowledge. Once, I thought I had an idea for a proof that some 2-manifolds can't be suitably triangulated, by focusing on surfaces which can't be realized suitably by regular maps (or flag transitive triangulations) and showing some of these surfaces can't be vertex transitively triangulated in any other way. I hope to return to this and complete the details sometime soon.
| https://mathoverflow.net/users/495429 | Does every triangulable manifold have a vertex-transitive triangulation? | There [exists](https://mathscinet.ams.org/mathscinet-getitem?mr=953960) many closed connected hyperbolic 3-manifolds $M$ with trivial symmetry group, and hence [trivial mapping class group](https://www.ams.org/journals/jams/1997-10-01/S0894-0347-97-00206-3/). $M$ cannot be homeomorphic to a simplicial complex $\tau$ which admits a vertex-transitive automorphism group $G$ (which must be non-trivial). Then the quotient $M/G = \tau/G$ is a 3-orbifold which by the [orbifold theorem](https://en.wikipedia.org/wiki/Orbifold#3-dimensional_orbifolds) must be hyperbolic. But this implies that the symmetry group was conjugate in the mapping class group to a group of isometries, a contradiction.
I imagine that this ought to be true for [symmetry-free hyperbolic manifolds](https://mathscinet.ams.org/mathscinet-getitem?mr=2132171) in any dimension >2, but I’m not quite sure how to prove that the automorphism group of the triangulation induces non-trivial outer automorphisms of the fundamental group (equivalently non-trivial isometries up to homotopy by Mostow rigidity).
For the 2-dimensional case, it does appear to be open in general which closed surfaces admit a vertex-transitive triangulation. [This paper](https://arxiv.org/abs/1001.2777) states that there are at most four exceptions with $\chi \geq -127$.
| 5 | https://mathoverflow.net/users/1345 | 446933 | 180,030 |
https://mathoverflow.net/questions/446911 | 11 | The eigenvalue map in question is
$\sigma: {\mathfrak gl}(\mathbb{C}, n) \to S\_n \backslash \mathbb{C}^n$,
from $n$ by $n$ complex matrices to $\mathbb{C}^n$ vectors modulo permutation of entries by $S\_n$, and is known to be *continuous*.
This map can be constructed by first considering the root map $\rho: \mathbb{C}^n \to S\_n \backslash \mathbb{C}^n$ of a monomial of degree $n$
$$
z^n + a\_{n-1}z^{n-1} + \ldots + a\_1 z + a\_0 = (z - \lambda\_1) \cdots (z - \lambda\_n)
$$
given by
$$
\rho(a\_0, a\_1, \ldots, a\_{n-1}) = [(\lambda\_1, \ldots, \lambda\_n)]
$$
which is known to be an *homemomorphism* with inverse given by Vieta's formulae ([see here, for example](https://www.jstor.org/stable/2045978)).
Now given $A \in {\mathfrak gl}(\mathbb{C}, n)$, the coefficients $(a\_0, a\_1, \ldots, a\_{n-1})$ of its characteristic polynomial $p\_A(z) = \det(A - zI)$ are *continuous* (indeed polynomial) functions on the entries of $A$. The eigenvalue map is then given by
$$
\sigma(A) = \rho(a\_0, a\_1, \ldots, a\_{n-1})
$$
so that it is continuous on $A$.
Is $\sigma$ an open map on $A$?
Besides being a natural question, a positive answer would have some nice applications. For example, when one proves that certain subset of matrices are open and dense when their eigenvalues satisfy some open and dense conditions (e.g. hyperbolic matrices, matrices with distinct eigenvalues, etc) one can use the continuity of $\sigma$ to prove openness but has to argue denseness in other ways, for example perturbing the matrices involved. If one could prove that $\sigma$ is an open map, one could use $\sigma$ all the way.
A negative answer by means of a counterexample would also be very instructive.
| https://mathoverflow.net/users/12170 | Is the eigenvalue map open? | Yes. Write $D(\lambda\_1, \ldots, \lambda\_n)$ as the diagonal matrix with diagonal entries $(\lambda\_1, \ldots, \lambda\_n)$.
Let $A$ be any complex $n \times n$ matrix. We can upper-triangularize $A$ as $A = S (D(\lambda\_1, \ldots, \lambda\_n)+N) S^{-1}$ where $N$ is upper triangular. Let $U$ be an open ball around $A$. We want to show that $\rho(U)$ contains an open ball $V$ around $(\lambda\_1, \ldots, \lambda\_n)$.
Indeed, we can choose $V$ small enough that $S (D(\mu\_1, \ldots, \mu\_n)+N) S^{-1}$ is in $U$ for all $(\mu\_1, \ldots, \mu\_n)$ in $V$. Then $\rho(U)$ contains $\rho(S (D(\mu\_1, \ldots, \mu\_n)+N) S^{-1}) = (\mu\_1, \ldots, \mu\_n)$ for all $(\mu\_1, \ldots, \mu\_n)$ in $V$, and $\rho(U)$ contains $V$, as desired.
| 13 | https://mathoverflow.net/users/297 | 446934 | 180,031 |
https://mathoverflow.net/questions/446932 | 4 | If $(P,\leq)$ is a partially ordered set and $a,b\in P$ we set $[a,b]:=\{x\in P: a\leq x\leq b\}$. We say that $P$ is *fractal* if whenever $a,b\in P$ and $[a,b]$ contains more than one element, then $[a,b]\cong P$.
So for instance, $[0,1]$ and $[0,1]\cap \mathbb{Q}$ are fractal with their usual linear orderings.
For $A,B\in{\cal P}(\omega)$ we say $A\subseteq^\* B$ if $A\setminus B$ is finite (that is, $A$ is "almost contained" in $B$). We write $A\simeq\_{\text{fin}} B$ if $A\subseteq^\* B$ and $B\subseteq^\* A$ (that is, the sets $A, B$ are "almost the same set" except for finitely many elements). It is easy to see that $\simeq\_{\text{fin}}$ is an equivalence relation on ${\cal P}(\omega)$.
We denote the collection of equivalence classes on ${\cal P}(\omega)$ with respect to $\simeq\_{\text{fin}}$ by ${\cal P}(\omega)/(\text{fin})$. Using $\subseteq^\*$ on representatives of equivalence classes, it is easy to see that we can make ${\cal P}(\omega)/(\text{fin})$ into a partially ordered set.
Is ${\cal P}(\omega)/(\text{fin})$ a fractal poset? If yes, is it true that if ${\cal B}$ is a fractal Boolean algebra on more than $2$ points, then ${\cal P}(\omega)/(\text{fin})$ can be order-embedded into ${\cal B}$?
| https://mathoverflow.net/users/8628 | Is ${\cal P}(\omega)/\text{(fin)}$ a fractal poset? | Yes, $P(\omega)/\text{fin}$ is fractal. If $A\subseteq^\* B$ but not equivalent, then the interval $[A,B]$ in $P(\omega)/\text{fin}$ consists of the sets that almost contain $A$ and are almost contained in $B$, and this is isomorphic to $P(B-A)/\text{fin}$, which is isomorphic to $P(\omega)/\text{fin}$, since $B-A$ will be infinite.
| 7 | https://mathoverflow.net/users/1946 | 446938 | 180,032 |
https://mathoverflow.net/questions/446937 | 1 | Let $0 < a < 1$ be an irrational number. Is it true that
$$\liminf\_{n \in \mathbb N, n \to \infty} n \{na\} = 0?$$
*Note: Here $\{\cdot\}$ denotes the fractional part.*
| https://mathoverflow.net/users/173490 | The liminf of an expression involving an irrational rotation | The answer is no, because for some real $c>0$ and all integers $q>0$ we have
$$q\{q\sqrt2\}=q(q\sqrt2-\lfloor q\sqrt2\rfloor)
=q|q\sqrt2-\lfloor q\sqrt2\rfloor| \\
\ge q\,\inf\_{p\in\mathbb Z}|q\sqrt 2-p|
=q^2\,\inf\_{p\in\mathbb Z}|\sqrt 2-p/q|\ge c.$$
So, $\liminf\limits\_{q\to\infty} q\{q\sqrt2\}\ge c>0$.
More specifically, letting $p:=p\_q:=\lfloor q\sqrt2\rfloor$ and $q\to\infty$, we have $p<q\sqrt2$, $p\sim q\sqrt2$, and hence
$$q\{q\sqrt2\}=q(q\sqrt2-p)
=q\frac{2q^2-p^2}{q\sqrt2+p}
\ge q\frac1{q\sqrt2+p}\sim\frac1{2\sqrt2}.$$
So,
$$\liminf\_{q\to\infty} q\{q\sqrt2\}\ge\frac1{2\sqrt2}>0.$$
| 6 | https://mathoverflow.net/users/36721 | 446941 | 180,034 |
https://mathoverflow.net/questions/446740 | 1 | I understand the concept of the 1 dimensional Brownian bridge with the form of:
$$dx\_t=\frac{-1}{1-t}x\_t \, dt + dw\_t$$
s.t. $x\_0=0$ and $x\_1=0$
where $dw\_t$ is a Wiener process.
I am thinking about the Brownian bridge in the phase space with arbitrary boundary conditions:
$$dx\_t=v\_t \, dt$$
$$dv\_t=a\_t\,dt+dw\_t$$
s.t. $x\_0=\phi\_0, v\_0=\psi\_0$ and $x\_1=\phi\_1, v\_1=\psi\_1$
Is there any clean close form of $a\_t$ that yields similar results to the Brownian bridge?
Any comments, or suggestions are welcome!
| https://mathoverflow.net/users/504685 | Phase space Brownian bridge | I use capital letters for random variables and small letters for possible values.
Let $W$ be a brownian motion, defined on the canonical space $\mathcal{C}(\mathbb{R}\_+)$ endowed with the Wiener measure $\mathbb{P}$ and with the canonical filtration $\mathcal{F}$. For $t>0$, call $p\_t$ the density of the random variable $(W\_t,\int\_0^t W\_s~ds)$, namely, the density of $\mathcal{N}(0,C\_t)$, where
$$C\_t = \left(\begin{array}{cc}
t & t^2/2 \\
t^2/2 & t^3/3
\end{array} \right),
\text{ so } C\_t^{-
1} = \frac{12}{t^4}\left(\begin{array}{cc}
t^3/3 & -t^2/2 \\
-t^2/2 & t
\end{array} \right)
= \left(\begin{array}{cc}
4/t & -6/t^2 \\
-6/t^2 & 12/t^3
\end{array} \right).$$
Thus $p\_t(x,y)$ is proportional to $\exp(2x^2/t - 6xy/t^2 + 6y^2/t^3)$.
Let $V,U$ the processes defined by
$$V\_t := v\_0+W\_t \text{ and } U\_t := u\_0+\int\_0^t W\_s~ds := u\_0+tv\_0+\int\_0^t W\_s~ds.$$
We want to condition $(V,U)$ by $(V\_1,U\_1) = (v,u)$.
The idea is to compute for each $t \in [0,1[$ the probability
$\mathbb{P}\big|\_{\mathcal{F}\_t} \big[\cdot \big|(V\_1,U\_1) = (v,u)\big]$ and to apply Girsanov Theorem.
Let $A \in \mathcal{F}\_t$ and $f : \mathbb{R^2} \to \mathbb{R}$ measurable and bounded. Observe that $V\_1 = V\_t + W\_1 - W\_t$ and
$$U\_1 = U\_t + (1-t)v\_0 + \int\_t^1 W\_s~ds = U\_t + (1-t)V\_t + \int\_t^1 (W\_s-W\_t)~ds.$$
Since $W\_{s+\cdot}-W\_s$ is a Brownian motion, independent of $\mathcal{F}\_s$, we get
\begin{eqnarray\*}
\mathbb{E}[\mathbb{1}\_A f(V\_1,U\_1)]
&=& \int\_{\mathbb{R^2}}
\mathbb{E}[\mathbb{1}\_A f(V\_t + x,U\_t + (1-t)V\_t + y) p\_{1-t}(x,y)] dxdy \\
&=& \mathbb{E}\Big[\int\_{\mathbb{R^2}}
\mathbb{1}\_A f(v,u) p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t) dvdu \Big]
\end{eqnarray\*}
Therefore
$$\mathbb{P}\big|\_{\mathcal{F}\_t}\big[\cdot\big|(V\_1,U\_1) = (v,u)\big]
= D\_t\mathbb{P}\big|\_{\mathcal{F}\_t}, \text{ where }
D\_t = \frac{p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)}{p\_1(v-v\_0,u-u\_0-v\_0)}.$$
The process $D$ thus defined on the time interval $[0,1[$ is a martingale.
Girsanov Theorem yields that under $\mathbb{P}\big[\cdot\big|(V\_1,U\_1) = (v,u)\big]$, the process
$$\hat{W} := W - \int\_0^\cdot \frac{d\langle D,W \rangle\_s}{D\_s}$$
is a local martingale, hence a Brownian motion, since it has the same quadratic variation as $W$.
Ito calculus yields
\begin{eqnarray\*}
dD\_t
&=& \frac{-1}{p\_1(v-v\_0,u-u\_0-v\_0)}
\Big( \partial\_1p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t) \\
& & \hspace 5 cm +(1-t)\partial\_2p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t) \Big) dW\_t \\
& & \quad + \textrm{ process with locally bounded variation.}
\end{eqnarray\*}
Thus
\begin{eqnarray\*}
\hat{W} &=& W + \int\_0^\cdot \frac{\partial\_1p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)}{p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)} dt \\
& & + (1-t)\int\_0^\cdot \frac{\partial\_2p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)}{p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)} dt.
\end{eqnarray\*}
Since the density $p\_{1-t}$ is well-known, this computation can be continued.
Hence, under $\mathbb{P}\big[\cdot\big|(V\_1,U\_1) = (v\_1,u\_1)\big]$, the process $W$ can be written as the sum of the Brownian motion $\hat{W}$ and an explicit drift.
| 0 | https://mathoverflow.net/users/169474 | 446953 | 180,038 |
https://mathoverflow.net/questions/446813 | 13 | I am looking for an example of a locally compact group $G$ and a continuous $G$ module $M$, which also is locally compact, such that the continuous cochain cohomology differs from group cohomology (ignoring the topologies). In the literature, I find many results, giving criteria when they agree, but I could not find an example, in which they don't.
| https://mathoverflow.net/users/473423 | Example of continuous cohomology vs cohomology | Consider the circle $G=\mathbb{R}/\mathbb{Z}$ and its trivial representation on $M=\mathbb{R}$.
Since $G$ is abelian, $H^1(G,M)\cong \text{Hom}(G,M)$ and $H^1\_c(G,M)\cong \text{Hom}\_c(G,M)$ (these are the continuous versions).
Clearly, $\text{Hom}\_c(G,M)=0$.
However, as abstract groups $\mathbb{R}\cong \mathbb{Q}\oplus\mathbb{R}/\mathbb{Q}$, thus $\mathbb{R}/\mathbb{Z}\cong \mathbb{Q}/\mathbb{Z}\oplus \mathbb{R}/\mathbb{Q}$, thus $\text{Hom}(G,M)\cong \text{Hom}(\mathbb{R}/\mathbb{Q},\mathbb{R})$, which is huge, as both arguments are high dimensional $\mathbb{Q}$-vector spaces.
| 8 | https://mathoverflow.net/users/89334 | 446956 | 180,039 |
https://mathoverflow.net/questions/446948 | 2 | (*Note*: This third method continues from [this post](https://mathoverflow.net/q/446886/12905).)
There are level-$7$ pi formulas based on the McKay-Thompson series $T\_{7A}$ and Cooper's $s\_7$ sequence in this [paper](https://www.researchgate.net/publication/257642843_Sporadic_sequences_modular_forms_and_new_series_for_1p). This third method, among other things, will enable us to use $T\_{7B}$.
---
**I. Method 3**
Given the *binomial coefficient* $\binom{n}{k}$, some free parameters $p, r,$ and a sequence $s\_1(n)$. Define a second sequence as,
$$s\_2(m) = \sum\_{n=0}^m r^{m-pn}\binom{m}{pn} s\_1(n)$$
Then we have the transformation,
$$\sum\_{n=0}^{\infty} s\_1(n)\,\frac{An+B}{C^n}=\left(\frac{C^{1/p}}{C^{1/p}+r}\right)^2\,\sum\_{m=0}^{\infty} s\_2(m)\,\frac{A/p\,m+ B-D\_3}{(C^{1/p}+r)^m}$$
where,
$$D\_3 = \frac{r\,(A/p-B)}{C^{1/p}}$$
---
**II. Examples**
Given the Dedekind eta function $\eta(\tau)$. First define the functions,
\begin{align}
j\_{7A}(\tau) &= \left(\sqrt{j\_{7B}(\tau)}+\frac{7}{\sqrt{j\_{7B}(\tau)}}\right)^2\\
j\_{7B}(\tau) &= \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4
\end{align}
Let $\tau = \frac{7+\sqrt{-427}}{14},$ note that $427 = 7\times61$, and we get,
\begin{align}
j\_{7A}(\tau) &= -22^3+1 = -(39\sqrt7)^2\\
j\_{7B}(\tau) &= -7\left(\frac{39+5\sqrt{61}}{2}\right)^2
\end{align}
where the latter involves the fundamental unit $U\_{61}$. We have Cooper's formula,
$$\frac{1}{\pi} = \frac{\sqrt7}{22^3}\sum\_{j=0}^\infty s\_7(j)\, \frac{11895j+1286}{(-22^3)^j}$$
However, we wish to find a sequence that uses the ***whole*** $j\_{7A}(\tau) = -22^3+1$ as this will lead to a second sequence that uses $j\_{7B}(\tau)$. Thus $r=1$, and applying Method 3, we get,
$$\frac{1}{\pi} = \frac{\sqrt7}{(-22^3+1)^2}\sum\_{k=0}^\infty t\_{7A}(k)\, \frac{22^3(11895k+1286)-(-22^3+39)}{(-22^3+1)^{k}}$$
Then using [Method 1](https://mathoverflow.net/q/446778/12905), we get,
$$\frac{1}{\pi} = \frac{1}{(-22^3+1)\sqrt{-\,j\_{7B}}}\sum\_{h=0}^\infty t\_{7B}(h)\, \frac{1272437 - 207636\sqrt{61}(1+2h)}{(j\_{7B})^{h}}$$
where $j\_{7B} = -7\left(\frac{39+5\sqrt{61}}{2}\right)^2$ as above.
---
**III. Sequences**
Starting with Cooper's sequence,
\begin{align}s\_7(j)
&= \sum\_{m=0}^j \binom{j}{m}^2\binom{2m}{j}\binom{j+m}{m}\\
&= 1, 4, 48, 760, 13840, 273504\dots
\end{align}
we derive,
\begin{align}t\_{7A}(k)
&= \sum\_{j=0}^k\binom{k}{j}\sum\_{m=0}^j \binom{j}{m}^2\binom{2m}{j}\binom{j+m}{m}\quad\\
&= 1, 5, 57, 917, 17185, 350805\dots\quad
\end{align}
\begin{align}
t\_{7B}(h)
&= \sum\_{k=0}^h(-7)^{h-k}\binom{h+k}{h-k}\sum\_{j=0}^k\binom{k}{j}\sum\_{m=0}^j \binom{j}{m}^2\binom{2m}{j}\binom{j+m}{m}\\
&= 1, -2, 1, 49, -602, 5257, -39095\dots
\end{align}
The advantage of Cooper's sequence $s\_{7}$ is it only has a 3-term recurrence relation. The recurrence status of $t\_{7A}$ and $t\_{7B}$ is unknown. However, we recover the nice relation,
$$\sum\_{n=0}^\infty t\_{7A}(n)\,\frac{1}{\;\big(j\_{7A}\big)^{n+1/2}} = \sum\_{n=0}^\infty t\_{7B}(n)\,\frac{1}{\;\big(j\_{7B}\big)^{n+1/2}}$$
with *closed-forms* for the sequences, so it is now found in levels $L = 1,2,3,4,6,7,8,10,$ (but not yet in $L=5,9$).
---
**IV. Questions**
1. Like the previous ones, why does Method 3 work, and how free are its parameters $p,r$?
2. Can the closed-forms of sequences $t\_{7A}$ and $t\_{7B}$ be simplified?
3. Lastly, what are their recurrence relations? (I've tested them, got nowhere, and I think it is an $m$-term relation with coefficients as polynomials of deg-$n$ where $m,n>4$.)
| https://mathoverflow.net/users/12905 | Finding a new level-$7$ pi formula using the relation $j_{7A}(\tau) = \left(\sqrt{j_{7B}(\tau)}+\frac{7}{\sqrt{j_{7B}(\tau)}}\right)^2$ | For Question $3$ about the recurrence relations, using my code from
[MMA question 285008](https://mathematica.stackexchange.com/q/285008/)
for $a\_n := T\_{7A}(n)$ I used
`findseqrecur[4, 4, Array[t7A, 33, 1], 1, "a", k, -1]` to get
$$ 0 = 14(n+1)(n+2)(2n+3) a\_n \\
-3(n+2)(19n^2+76n+80) a\_{n+1} \\
+ 5(2n+5)(3n^2+15n+19) a\_{n+2} \\
- (n+3)^3 a\_{n+3}. $$
For $b\_n := T\_{7B}(n)$ there are several recurrences.
For degree $4$ polynomials I used
`findseqrecur[6, 5, Array[t7B, 34, 1], 1, "b", k, -4]` to get
$$ 0 = -7^5(k-17)(k-2)^3 b\_{k-3}\quad \\
-7^3(19k^4-678k+2218k^2-2640k+1113)b\_{k-2}\;\; \\
-7(85k^4-8707k^3+9978k^2-7072k+2090)b\_{k-1} \\
+(85k^4+8707k^3+9978k^2+7072k+2090)b\_{k}\, \\
+(19k^4+678k+2218k^2+2640k+1113)b\_{k+1} \\
\quad +(k+17)(k+2)^3 b\_{k+2}. $$
For degree $3$ polynomials I used
`findseqrecur[8, 4, Array[t7B, 38, 1], 1, "b", k, -5]` to get
$$ 0 = 7^7(k-3)^3b\_{k-4}\; \\
+ 7^5(47k^3-300k^2+646k-470)b\_{k-3} \;\\
+ 2\cdot 7^3(480k^3-1830k^2+2483k-1206)b\_{k-2}\quad \\
+ 7^2 (1578k^3-2001k^2+1513k-433)b\_{k-1} \\
+ 7 (1578k^3+2001k^2+1513k+433)b\_{k}\;\; \\
\;\; + 2 (480k^3+1830k^2+2483k+1206)b\_{k+1} \\
\; + (47k^3+300k^2+646k+470)b\_{k+2} \\
\; + (k+3)^3 b\_{k+3}. $$
Notice the symmetry of these two recursions.
| 2 | https://mathoverflow.net/users/113409 | 446969 | 180,041 |
https://mathoverflow.net/questions/446962 | 3 | Consider the Segre embedding $$ \mathbb{P}^2 \times \mathbb{P}^2 \to \mathbb{P}^8.$$ Denote by $V$ the image of the Segre embedding and by $B$ the locus of triples $(H\_1, H\_2, H\_3)$ with $H\_i \in H^0(\mathcal{O}\_{\mathbb{P}^8}(1))$ a hyperplane section such that $V \cap H\_1 \cap H\_2 \cap H\_3$ is a proper intersection i.e., a curve. Clearly, $B$ is an open subset in $H^0(\mathcal{O}\_{\mathbb{P}^8}(1))^{\oplus 3}$. Denote by $B' \subset B$ the subloci consisting of triples $(H\_1, H\_2, H\_3)$ such that the intersection $V \cap H\_1 \cap H\_2 \cap H\_3$ is a singular curve. My question is: Is $B'$ necessarily of codimension $1$ in $B$ or can it be of higher codimension in $B$? Any hint/reference will be most welcome.
| https://mathoverflow.net/users/45397 | Segre embedding and intersections by hyperplanes | This is a standard projective duality argument.
Let $W = H^0(\mathcal{O}\_{\mathbb{P}^8}(1))$. Consider the variety $X$ of tuples
$$
(P,H\_1,H\_2,H\_3) \in V \times W^{\oplus 3}
$$
such that $V \cap H\_1 \cap H\_2 \cap H\_3$ is singular at $P$. Then it is easy to see that the projection
$$
X \to V,
\qquad
(P,H\_1,H\_2,H\_3) \mapsto P
$$
is flat and its fibers have dimension $22$ (indeed, there are 4 parameters for a 2-dimensional subspace in the tangent space $T\_PV$ and $3 \cdot 6 = 18$ parameters for the $H\_i$ passing through $P$ and containing this tangent space), hence $\dim(X) = 4 + 22 = 26$. It is also easy to see that $X$ is irreducible.
On the other hand, the general fiber of the map
$$
X \to W^{\oplus 3},
\qquad
(P,H\_1,H\_2,H\_3) \mapsto (H\_1,H\_2,H\_3)
$$
over its image is finite (to see this it is enough to find just one triple $(H\_1,H\_2,H\_3)$ such that $V \cap H\_1 \cap H\_2 \cap H\_3$ has a finite number of singular points), hence the dimension of the image of $X$ is 26, hence it is a divisor in $W^{\oplus 3}$.
| 5 | https://mathoverflow.net/users/4428 | 446974 | 180,043 |
https://mathoverflow.net/questions/446959 | 1 | Let $G$ be a finite abelian group and denote by $G^{\vee}=\mathrm{Hom}(G,U(1))$ its Pontryagin dual. For any positive integer $n$ one can define a homomorphism of abelian groups
$$
f:H^{n}(G,G^{\vee})\rightarrow H^{n+1}(G,U(1))
$$
Its action on the chains is
$$
f(c)(g\_1,...,g\_{n+1})=\langle c(g\_1,....,g\_n),g\_{n+1}\rangle
$$
where $\langle \alpha ,g\rangle=\alpha(g)$ denotes the natural pairing $G\times G^{\vee}\rightarrow U(1)$. One can check that this is a defines a map of cohomology groups and is a homomorphism (see this post <https://math.stackexchange.com/questions/4631490/a-map-in-group-cohomology-from-h2g-widehatg-to-h3g-u1> on MSE).
I would like to know if this is an isomorphism, or at least there if there are conditions on $G$ and $n$ which guarantee $f$ to be an isomorphism. For instance $H^2(\mathbb{Z}\_N,\mathbb{Z}\_N)=H^3(\mathbb{Z}\_N,U(1))=\mathbb{Z}\_N$ but i don't know if $f$ provides an isomorphism. Some hint for $G$ a product of cyclic groups will be also appreciated.
| https://mathoverflow.net/users/495347 | A map in group cohomology from $H^n(G,G^{\vee})$ to $H^{n+1}(G,U(1))$ | Let $I\_G$ be the augmentation ideal of $\mathbb{Z}G$. Given that $G$ is abelian, we have $I\_G/I\_G^2\cong G$ by the Hurewicz isomorphism sending the coset of $g-1$ to $g$. Then your map is induced by the surjection $I\_G\to I\_G/I\_G^2\cong G$ in the following sense. We can regard $H^n(G,G^\vee)$ as $\underline{Hom}(I\_G^{\otimes n},G^\vee)$ where $\underline{Hom}$ denotes $\mathbb{Z}G$-module homomorphisms modulo those that factor through a projective module. Then your map is $H^n(G,G^\vee)\cong \underline{Hom}(I\_G^{\otimes n},G^\vee) \cong \underline{Hom}(I\_G^{\otimes n} \otimes G,U(1)) \to \underline{Hom}(I\_G^{\otimes n+1},U(1))$. If $G=\langle g\rangle$ is cyclic of order $N$ then $I\_G$ is the ideal generated by $g-1$, and $I\_G^2$ is the ideal generated by $(g-1)^2$. So the map in cohomology induced by the inclusion $I\_G^2\to I\_G$ is multiplication by $g-1$, which is the zero map in positive degrees. Thus your map $H^n(G,G^\vee)\to H^{n+1}(G,U(1))$ is injective in this case, with cokernel $H^{n+1}(G,(I\_G^2)^\vee)$. If $n$ is even, this gives you an isomorphism, but not if $n$ is odd. For non-cyclic groups the situation is more complicated, but can be computed using this description.
| 5 | https://mathoverflow.net/users/460592 | 446976 | 180,045 |
https://mathoverflow.net/questions/446979 | 9 |
>
> By Bézout's theorem a quartic polynomial $p(x,y)$ can have at most 9 isolated critical points. Can all of them be saddle points?
>
>
>
In case of a cubic polynomial there is a mechanical way to answer this type of questions: One can find a general form of such polynomials with critical points at three given points, e.g. at $(0,0),(0,1),$ and $(1,0)$, and then play with the remaining free parameters until a polynomial with the desired properties if found, as there is only one remaining critical point whose position was not normalized by affine transformation [[1]](https://mathoverflow.net/q/446848/497175), [[2]](https://math.stackexchange.com/q/4701058/1134951).
However, the above approach won't work in case of a quartic polynomial, as then affine transformation normalizes position of only 3 out of the 9 potential critical points.
Note that a quartic polynomial in two real variables can have at most 5 minima out of its 9 potential critical points [[3]](https://math.stackexchange.com/q/4620663/1134951), [[4]](https://mathoverflow.net/q/442736/497175). Can it reversely have no extreme points, so that all 9 of the potential critical points would be saddle points?
| https://mathoverflow.net/users/497175 | How many saddle points can a quartic polynomial in two real variables have? All 9? | By (3.1) of [Counting Critical Points of Real Polynomials in Two Variables by Alan Durfee, Nathan Kronefeld, Heidi Munson, Jeff Roy, Ina Westby](https://www.jstor.org/stable/2324459?seq=7) a degree $d$ polynomial with only nondegenerate critical points can contain at most $d(d-1)/2$ saddle points. For $d=4$ this gives a max of $6$ saddle points.
This bound is attained by a product of $d$ linear polynomials, as long as each pair of lines intersects at a different point, as then the $d(d-1)/2$ intersection points are all saddle points.
(This paper was cited in a paper cited in [one of the answers you cite](https://math.stackexchange.com/questions/4620663/how-many-strict-local-minima-a-quartic-polynomial-in-two-variables-might-have).)
| 14 | https://mathoverflow.net/users/18060 | 446982 | 180,047 |
https://mathoverflow.net/questions/446978 | 36 | In a recent [talk](https://agenda.unige.ch/events/view/36495) at the University of Geneve, Dustin Clausen presented a "modified Hodge Conjecture". I found the abstract intriguing but couldn't find videos or notes available online.
If I'm reading the abstract right, his conjecture (which, as I understand from it, is part of joint work with Peter Scholze) is perhaps only similar in spirit to the classical Hodge Conjecture, but doesn't imply it/is not implied by it.
I couldn't deduce anything more precise from the abstract alone and I'd be curious to see a reference, or perhaps some seminar notes describing the statement and its relation to the classical Hodge Conjecture. Can anyone point me to any?
| https://mathoverflow.net/users/497064 | Clausen's modified Hodge Conjecture | It's a bit of a long story, but I can at least give the idea. Let $X$ be a smooth projective variety over $\mathbb{C}$. The Hodge conjecture says that for all $p\geq 0$, the cycle class map
$$Ch^p(X)\_{\mathbb{Q}} \to Hdg^p(X)\_{\mathbb{Q}}$$
from codimension p algebraic cycles to Hodge classes is surjective, with rational coefficients. First, a small modification: Grothendieck gave a Chern character isomorphism
$$K\_0(X)\_{\mathbb{Q}} \simeq \oplus\_p Ch^p(X)\_{\mathbb{Q}}$$
identifying rational Chow theory with rational algebraic K-theory, so the Hodge conjecture can be equivalently reformulated as saying that the map
$$K\_0(X)\_{\mathbb{Q}} \to \oplus\_p Hdg^p(X)\_{\mathbb{Q}}$$
is surjective. Now, the proposed modification involves modifying $K\_0(X)$ to a different kind of K-theory $K^{an}\_0(X)$, which takes into account the analytic nature of $\mathbb{C}$. I will describe more precisely what this analytic K-theory is below, but for now let me take it as a black box. The usual K-theory maps to the analytic K-theory, and the above map extends to a map
$$K^{an}\_0(X)\_{\mathbb{Q}} \to \oplus\_p Hdg^p(X)\_{\mathbb{Q}}.$$
Then the weak form of the modified Hodge conjecture is that this map is surjective. This modified form is obviously strictly weaker than the usual Hodge conjecture. But it admits a strengthening, in a different direction. Recall that the usual cycle class map from $Ch^p(X)$ to $Hdg^p(X)$ actually factors through a more refined target, the Deligne cohomology group $H^{2p}(X;\mathbb{Z}(p))$, which is an extension of $Hdg^p(X)$ by the Griffiths intermediate Jacobian. (This Deligne cohomology is essentially the "derived" version of the definition of Hodge classes.) Now, it is known that the Hodge conjecture fails on the level of Deligne cohomology, and in fact this obstructs some inductive methods for trying to prove the Hodge conjecture by induction on dimension and using hyperplane sections (see section 3 of <http://publications.ias.edu/sites/default/files/hodge.pdf>). However, the situation conjecturally improves when passing to analytic K-theory. Just as with the usual cycle class map to Hodge cycles, the refined cycle class map to Deligne cohomology also extends to a map from analytic K-theory
$$K^{an}\_0(X)\_{\mathbb{Q}} \to \oplus\_p H^{2p}(X;\mathbb{Q}(p)),$$
and we conjecture this map to be not just surjective, but an isomorphism. In fact, one can also make an analogous map out of $K^{an}\_i(X)\_{\mathbb{Q}}$ for any $i\in\mathbb{Z}$, with values in $\oplus\_p H^{2p-i}(X;\mathbb{Q}(p))$, and we even conjecture this to be an isomorphism for all $i$.
To finish I want to say something about what this analytic K-theory is, but first, a sobering remark: one of the reasons people really care about the Hodge conjecture is that it produces strong algebraic data (algebraic cycles) out of rather weak topological/analytic data (Hodge classes). The modified Hodge conjecture does not do this: it produces analytic data instead of algebraic data. But it maybe gives a new perspective on the Hodge conjecture: if the modified Hodge conjecture is true, then the usual Hodge conjecture would essentially amount to the statement that the comparison map $K\_0(X) \to K^{an}\_0(X)$ is surjective on connected components, i.e. any analytic K-theory class can be continuously deformed into an algebraic K-theory class. Moreover, the modified conjecture fits the spirit of the original Hodge conjecture, in the following sense: cycle classes, or algebraic $K\_0$ classes, can be viewed as a kind of universal source for cohomology classes, and the Hodge conjecture is saying that Hodge classes give universal classes. Similarly, the modified Hodge conjecture is saying is that, in some sense, Deligne cohomology is a universal cohomology theory for analytic spaces over the complex numbers. So we're not "missing" any cohomology theories; Hodge theory really has it all. This is actually why I'm interested in the statement, not because it's connected to the traditional Hodge conjecture. I just want to see if we're missing something. If we are missing something, then the modified Hodge conjecture will be wrong, and analytic K-theory will hopefully hint to us exactly what it is that we're missing.
Now, about the definition of analytic K-theory. It fits into the framework of Alexander Efimov's remarkable extension of the scope of algebraic K-theory. Usually, algebraic K-theory is defined in terms of small categories, like the category of vector bundles. Efimov says that this is a mistake, and we should think of it as an invariant of large categories, like the category of quasicoherent sheaves. (Actually, one needs to work with the derived analogs of these, namely perfect complexes and derived quasi-coherent sheaves, but let me slough over this point for the purposes of this already too lengthy explanation...) Now, this seems like an academic distinction, because the small and large categories determine each other, via passing to Ind-objects in one direction and compact objects in the other. But the point is that there are large categories which are *not* Ind-categories of small categories, but for which algebraic K-theory can still, magically, be defined, and with basically all the same formal properties as in the more restrictive classical context. These are the so-called *dualizable* categories studied by Lurie. It turns out these things are everywhere once you know to look for them, and I think it's important to get very comfortable with these non-compactly generated beasts...
Anyway, there is such a dualizable, non-compactly generated category of topological $\mathbb{C}$-vector spaces, let's call it $Mod^{an}(\mathbb{C})$ (formally it is a certain full subcategory of derived condensed $\mathbb{C}$-vector spaces), and we can build a theory of (derived) quasicoherent sheaves on $X$ where all the underlying vector spaces are objects of $Mod^{an}(\mathbb{C})$ instead of usual (smaller) category of abstract $\mathbb{C}$-vector spaces. We take Efimov's K-theory of this quasicoherent sheaf category and call it $K^{an}(X)$. The last thing to explain is what this category $Mod^{an}(\mathbb{C})$ is. It turns out that it can be described from a classical functional analysis perspective: its basic objects (which generate everything under colimits inside derived condensed $\mathbb{C}$-vector spaces) are the sequential colimits of (say) Hilbert spaces along injective transition maps which are compact operators, with singular values decaying rapidly. This is a somewhat obscure class of dual Frechet spaces, but it really is completely forced on us by the abstract considerations of Efimov's theory, and I don't believe there's any other choice than this precise one which should work.
To sum up, we replace usual vector bundles by some more nebulous analytic beasts described using functional analysis, and the hope is that this change makes K-theory match the most refined cohomology theory coming from Hodge theory.
| 59 | https://mathoverflow.net/users/3931 | 446992 | 180,049 |
https://mathoverflow.net/questions/446993 | 3 | Let $(\mathcal T, \otimes)$ be a tensor tringulated (tt-)category. Balmer defined a functor from the category of tt-categories to the category of locally ringed spaces, called the Balmer spectra or tt-spectra. He showed that the functor is not essentially injective if we set the target to be the category of topological spaces, but I am wondering if there is any known counter-example in our setting. In particular, when the Bamler spectra $X$ of $(\mathcal T, \otimes)$ is a qcqs scheme (or a smooth proper variety over a field if that makes a difference), is it true that we have $(\mathcal T, \otimes) \simeq (Perf X, \otimes\_{\mathcal O\_X}^{\mathbb L})$ as tt-categories? Thank you in advance.
| https://mathoverflow.net/users/177839 | "Essential injectivity" of Balmer spectra | No this is not true in general. In [Tensor Triangulated Categories in
Algebraic Geometry](https://www.math.uni-hamburg.de/home/sosna/diplom-online.pdf), Prop 4.0.9, Sosna shows that if X is a connected noetherian scheme then it is possible to put a tt-structure $\boxtimes$ on the derived category of $X\amalg X$ such that $Spc(D^{b}(X\amalg X),\boxtimes)\cong X$, yet $D^{b}(X\amalg X)\not\simeq D^{b}(X)$ so there cannot be an equivalence as tt-categories between $(D^{b}(X),\otimes\_{X}^{\mathbb{L}})$ and $D^{b}(X\amalg X,\boxtimes)$. This structure is just like a square zero extension for the $\otimes\_{X}^{\mathbb{L}}$ tt-structure on $D^{b}(X)$.
You need some more control on either the variety or the sort of tt-structure you want to put on the triangulated category. For example disregard spaces like $X\amalg X$ or fix some conditions on the unit.
| 5 | https://mathoverflow.net/users/44499 | 447011 | 180,056 |
https://mathoverflow.net/questions/447014 | 13 | Assuming the negation of CH, let $\omega\_1$ be the first uncountable ordinal, $\mathfrak{c}$ be the cardinality of the continuum. Does there exist a map $f: \omega\_1 \times [0, 1] \rightarrow \mathfrak{c}$ s.t. for all $t \in \mathfrak{c}$, we have $t \in f(\omega\_1 \times \{s\})$ for Lebesgue measure a.e. $s \in [0, 1]$? Obviously this would be true if we assume CH, and assuming the negation of CH this seems impossible intuitively. However, I couldn’t find any simple reason to show no such map can exist. (If it is impossible to prove the existence of such an $f$ without CH, is its existence at least consistent with the negation of CH?)
| https://mathoverflow.net/users/504602 | Almost everywhere “filling” of the continuum by the first uncountable cardinal without CH | The existence of such an $f$ is consistent with $\mathsf{ZFC+\neg CH}$. Suppose $\mathfrak{c}=\aleph\_2$ and every set of size $\le\aleph\_1$ is null *(this is consistent with $\mathsf{ZFC}$; it follows, for example, from [$\mathsf{2^{\aleph\_0}=\aleph\_2+MA}$](https://en.wikipedia.org/wiki/Martin%27s_axiom))*. Fix a bijection $b:[0,1]\rightarrow\omega\_2$, and let $f(\omega\_1\times\{s\})\supseteq b(s)$. Then for each $\alpha<\omega\_2$, there are only $\omega\_1$-many reals $s\in[0,1]$ such that $\alpha\not\in f(\omega\_1\times\{s\})$.
| 15 | https://mathoverflow.net/users/8133 | 447016 | 180,058 |
https://mathoverflow.net/questions/447024 | 6 | A poset $(P,\leq)$ is *homogeneous* if $P\cong [a,b]$ for all $a,b\in P$ with $a<b$ (where $[a,b] := \{x\in P: a\leq x\leq b\}$).
Examples of homogeneous posets include $[0,1]$, $[0,1]\cap \mathbb{Q}$, and [the Boolean algebra ${\cal P}(\omega)/(\text{fin})$](https://mathoverflow.net/questions/446932/is-cal-p-omega-textfin-a-fractal-poset).
Is every homogeneous poset a [lattice](https://en.wikipedia.org/wiki/Lattice_(order))?
| https://mathoverflow.net/users/8628 | Is every homogeneous poset a lattice? | **Counterexample.** Let
$$P=\{(x,i)\in\mathbb Q\times\{0,1\}:0\le x\le1,\ x\ne i\}$$
be ordered so that
$$(x,i)\lt(x',i')\iff x\lt x'.$$
| 11 | https://mathoverflow.net/users/43266 | 447025 | 180,059 |
https://mathoverflow.net/questions/447023 | 1 | Edges of the complete graph on $2n$ vertices can be colored with $2n-1$ colors such that only edges of different colors intersect.
*Can this always be done such that for every pair of different colors the set of edges of these two colors form a unique cycle?*
Reformulation: Are there $2n-1$ fixed-point-free involutions in the symmetric group $S\_{2n}$ such that the product of two (different) such involutions has always two cycles of length $n$?
(I suspect that the answer is well-known to specialists : This seems a fairly natural question.)
The answer is YES if $2n-1$ is a prime number. Indeed, dropping the condition that any maximal set of edges of two colors defines a hamiltonian cycle, there is an easy solution due to Soifer, see <https://en.wikipedia.org/wiki/Edge_coloring> which can be described as follows : Consider the set $S=\lbrace 0,1,\ldots,2n-2,\infty\rbrace$. For $a$ in $\lbrace 0,\ldots,2n-2\rbrace$ consider the
fixed point-free involution $\sigma\_a$ defined by $x\longmapsto 2a-x$ if $x\not\in \lbrace a,\infty\rbrace$ extended by $\sigma\_a(a)=\infty$, $\sigma\_a(\infty)=a$. A symmetry argument shows that it is enough to prove the claim for all pairs of colors $\lbrace \sigma\_0,\sigma\_a\rbrace$ for finite non-zero $a$. The composition of these two involutions is then essentially addition (or subtraction) of $2a$ modulo $2n-1$. We get therefore Hamiltonian cycles for all choices of $a$ if $2n-1$ is prime. We have however a problem if $a$ divides $2n-1$.
| https://mathoverflow.net/users/4556 | Hamiltonian edge colouring of complete graphs with even numbers of vertices | This is called a "perfect 1-factorisation". Existence is incompletely determined.
<https://core.ac.uk/download/pdf/82799957.pdf> is an older source.
<https://math.stackexchange.com/questions/3455298/perfect-1-factorization-of-k-2n> contains some more recent refs. Searching on the name will find many more.
| 3 | https://mathoverflow.net/users/9025 | 447026 | 180,060 |
https://mathoverflow.net/questions/447032 | -2 | I am looking for functions $f(x,y)$, real arguments, continuous,
with the following properties:
1. $f(m,n) = r$, where $r$ is integer $> 0$ if and only if $m,n$ are integers $> 0$.
2. $f(m,n) \le f(r,s)$ if and only if $m n \le r s$, with $m,n,r,s$ integers $> 0$.
I would like to know if functions with properties 1 alone, 2 alone, or both together exist, and if they have been studied and given a specific name (for further investigation). In particular, I would be curious to know if relatively simple closed-form functions satisfying both conditions exist, or can exist at all.
If useful to allow existence, the further constraints $m < n, r < s$ can be used.
| https://mathoverflow.net/users/116669 | Two-variable continuous function which results in an integer if and only if arguments are integer | Such functions as you require do not exist. Your requirements impose that $f(1,1) < f(1,2) < f(2,2)$ (for example, $f(1,1) \leq f(1,2)$ by (2), but $\neg (f(1,2) \leq f(1,1))$ also by (2), so we have $f(1,1) < f(1,2)$). Now consider the continuous function $x \mapsto f(x,x)$ on $[1,2]$: by the intermediate value theorem, there exists $x$ with $1<x<2$ such that $f(x,x) = f(1,2)$. Since $f(1,2)$ is an integer but $f(x,x)$ is not by your requirement (1), this is a contradiction.
| 8 | https://mathoverflow.net/users/17064 | 447033 | 180,062 |
https://mathoverflow.net/questions/446951 | 8 | Consider a rigid braided monoidal category, with braiding $\beta\_{x,y} : x \otimes y \cong y \otimes x$, and every object has a dual such that $\epsilon\_x : 1 \to a \otimes a^\*, \bar\epsilon\_x : a^\* \otimes a \to 1$ satisfying the zig-zag identities. Now we have the Reidemeister moves, e.g.
$$ (c \otimes \beta\_{a,b}) \circ \beta\_{a\otimes b, c} = \beta\_{b \otimes a, c} \circ (\beta\_{a,b}\otimes c) $$
saying that braiding the two string with another, and then braiding the two, is the same as first braiding the two and then with the other. This is Reidemeister III. Similarly $\beta\_{a\otimes a^\*, b} \circ (b \otimes \epsilon\_a) = \epsilon\_a \otimes b$ is Reidemeister II, and $(\bar\epsilon\_{a} \otimes a) \circ (a^\* \otimes \beta\_{a,a}) \circ (\epsilon\_a\otimes a) = \mathrm{id}\_a$ is Reidemeister I. This is rather like the [Morse link presentation](http://katlas.math.toronto.edu/wiki/MorseLink_Presentations) of knots and links.
Is there any reference that develops such an idea? How can the theory of quandles be integrated in this picture? I'm guessing that quandles either behave like modules over a knot category, or we have a joint generalization of quandles and knot categories.
| https://mathoverflow.net/users/136535 | Is there a notion of "knot category"? | To expand on my comment, this connection is indeed well-known and the key concept is that of ribbon category. A standard textbook reference is Turaev, Quantum Invariants of Knots and 3-Manifolds.
Braided and rigid is not enough to get links invariants, because RI will not hold in general (and in fact pretty much never). A nice exposition of that issue can be found in Selinger, A survey of graphical languages for monoidal categories (<https://arxiv.org/abs/0908.3347>) (in that reference autonomous means rigid and tortile means ribbon).
Any ribbon category provides an invariant of framed, oriented links for each object $X$. If $X$ is simple then the ribbon element $\theta\_X$ acts as a multiple of the identity so that you can renormalize to get an invariant of oriented links. The choice of an isomorphism $X\cong X^\*$, provided one exists, gets rid of the orientation.
If you want non trivial invariants of oriented links on the nose, you'd need that $\theta\_X=id\_X$ and $\theta\_{X\otimes X} \neq id\_{X\otimes X}$. While this isn't impossible (tautologically, the category of oriented tangles is a ribbon category which satisfies this for example) this is a pretty unnnatural condition.
| 3 | https://mathoverflow.net/users/13552 | 447039 | 180,064 |
https://mathoverflow.net/questions/447021 | 0 | Let $M$ be a von Neumann algebra and $\psi$ a normal faithful semifinite weight on $M$. Then one should be able to form the object
$$\iota \otimes \psi: (M\overline{\otimes} M)\_+ \to \widehat{M\_+}.$$
This is for example used in the theory of locally compact quantum groups (in the sense of Vaes-Kustermans). I have been told that this is an "operator valued weight". Takesaki's second book contains a section about operator-valued weights, but I cannot find the definition of tensor product of operator-valued weights in the book.
Concretely, my question is: how to define the tensor product of operator-valued weights? References are appreciated.
| https://mathoverflow.net/users/216007 | Tensor product of operator values weights (in the theory of locally compact quantum groups) | The approach I had in mind is the following. For a reference, see Section 4, Chapter IX of Takesaki 2. The extended positive part $\widehat{M\_+}$ is by definition the space of positive-homogeneous, additive, lower semi-continuous maps $M\_\*^+\rightarrow [0,\infty]$. Given $x\in (M\bar\otimes M)\_+$ define $(\iota\otimes\varphi)(x) \in \widehat{M\_+}$ to be the map
$$ \omega \mapsto \varphi\big( (\omega\otimes\iota)(x) \big). $$
For $\omega\in M\_\*^+$, as $x$ is positive, also $(\omega\otimes\iota)(x)$ is positive, and so we obtain a well-defined member of $[0,\infty]$. Clearly $(\iota\otimes\varphi)(x)$ is positive-homogeneous and additive. If $(\omega\_i)$ is a net in $M\_\*^+$ converging to $\omega$ in norm, then $(\omega\_i\otimes\iota)(x) \rightarrow (\omega\otimes\iota)(x)$ $\sigma$-weakly. As $\varphi$ is $\sigma$-weakly lower semi-continuous (see Theorem 1.11 in Chapter VII of Takesaki) it follows that $\lim\_i (\iota\otimes\varphi)(x)(\omega\_i) \geq (\iota\otimes\varphi)(x)(\omega)$. Thus $(\iota\otimes\varphi)(x)$ is lower semi-continuous.
Clearly the map $\iota\otimes\varphi$ is positive-homogeneous and additive. Given $a\in M, \omega\in M\_\*^+$,
$$ (\iota\otimes\varphi)((a\otimes 1)^\*x(a\otimes 1))(\omega)
= \varphi\big( (a \omega a^\* \otimes\iota)(x) \big)
= (\iota\otimes\varphi)(x)(a \omega a^\*)
= \big( a^\* (\iota\otimes\varphi)(x) a \big)(\omega). $$
Thus $(\iota\otimes\varphi)((a\otimes 1)^\*x(a\otimes 1)) = a^\* (\iota\otimes\varphi)(x) a$.
Finally, if $(x\_i)$ increases to $x$ in $(M\bar\otimes M)\_+$ then for each $\omega\in M\_\*^+$ we have that $(\omega\otimes\iota)(x\_i)$ increases to $(\omega\otimes\iota)(x)$ so by normality of $\varphi$, it follows that $(\iota\otimes\varphi)(x\_i)(\omega)$ increases to $(\iota\otimes\varphi)(x)(\omega)$. Hence $(\iota\otimes\varphi)(x\_i)$ increases to $(\iota\otimes\varphi)(x)$. So $\iota\otimes\varphi$ is normal. So $\iota\otimes\varphi$ is an operator-valued weight.
| 3 | https://mathoverflow.net/users/406 | 447042 | 180,065 |
https://mathoverflow.net/questions/447030 | 5 | Take $G=\operatorname{GL}\_3$, defined over the algebraic closure of a finite field $\mathbb{F}\_q$ and let $X$ be the set of Borel subgroups of $G$.
The Frobenius morphism $F:G\to G$ induces a map $F:X\to X$, as $G/B\cong X$ for any chosen Borel subset $B$.
The Weyl group $W$ of $G$ is isomorphic to the symmetric group $S\_3$, and we have a bijection between $W$ and the set of $G$-orbits on $X\times X$.
Fix a Borel subset $B^+\in X$.
For $w\in W$, define $\mathcal{O}(w)$ to be the orbit of $(B^+,\dot wB^+\dot w^{-1})$ in $X\times X$.
We say that $B\_1,B\_2\in X$ are in relative position $w$ if $(B\_1,B\_2)\in\mathcal{O}(w)$.
For any $w\in W$, we can define the Deligne–Lusztig variety associated to it as $X(w)=\{B\in X:B\text{ and }F(B)\text{ are in relative position }w\}$.
**My Question**
I want to compute $X((1\ 2))$.
I know that $X$ can be identified with the full flag variety $\mathcal{F}\_3$, but how I can interpret the condition of $B$ and $F(B)$ being in relative position $w$ in terms of flags?
Any help is appreciated.
| https://mathoverflow.net/users/504919 | An example of a Deligne–Lusztig variety for a general linear group | Let $\mathcal{F}\colon V\_1\subseteq\dotsb\subseteq V\_n$ and $\mathcal{F}'\colon U\_1\subseteq\dotsb\subseteq U\_n$ be two complete flags. Then $\mathcal{F}$ and $\mathcal{F}'$ are said to be in relative position $w$, where $w\in S\_n$ is a permutation, if
$$\dim V\_i\cap U\_j=\#(\{ 1,\dotsc,i \}\cap \{ w(1),\dots,w(j) \})$$
for any $i,j\in\{1,\dotsc,n\}$.
In particular, the variety $X((1\ 2))$ consists of the flags $V\_1\subseteq V\_2$ in which $V\_2$ is an $F$-stable plane and $V\_1$ is an $F$-unstable line. So it is
$$\operatorname{Gr}(1,3)^F\times \mathbb{P}^1\setminus\mathbb{P}^1(\mathbb{F}\_q),$$
a disjoint union of copies of the Drinfeld curve.
Computations like this are very useful for making explicit connections with combinatorics. For a reference, you may want to take a look at Steinberg's paper *[An occurrence of the Robinson–Schensted correspondence](https://doi.org/10.1016/0021-8693(88)90177-9)*.
| 9 | https://mathoverflow.net/users/56217 | 447051 | 180,068 |
https://mathoverflow.net/questions/447040 | 1 | This question is an extension of the one I posted on ME: <https://math.stackexchange.com/questions/4701500/if-alpha-nx-int-lvert-x-y-rvert-leq-1-n-lvert-x-y-rvert2-d-muy>
It might be elementary for here, but I would deeply appreciate any help.
Let for each $n \in \mathbb{N}$, let $r\_n : \mathbb{R}^m \to (0,\infty)$ be a sequence of smooth functions that converges to zero "pointwise" as $n \to \infty$.
Also let $\mu$ be a sufficiently regular Borel probability measure on $\mathbb{R}^m$. For concreteness, we can think of the stadard normal Gaussian measure.
Now, define
\begin{equation}
\alpha\_n(x):=\int\_{\lVert x-y \lVert \leq r\_n(x)} \lVert x-y \rVert^2 d\mu(y)
\end{equation}
for each $n$ and let $A\_n := \int\_{\mathbb{R}^m} \alpha\_n(x) d\mu(x)$.
Then, I suspect that for any bounded real-valued smooth function $F$ on $\mathbb{R}^m$, we have
\begin{equation}
\frac{1}{A\_n}\int\_{\mathbb{R}^m} F(x) \alpha\_n(x) d\mu(x) \to \int\_{\mathbb{R}^m} F(x) d\mu(x)
\end{equation}
as $n \to \infty$.
However, I cannot find a way to justify this guess rigorously. I tried to use convolution theorems but they do not fit in the above formula. Perhaps it is related to ergodicity?
Could anyone please help me?
| https://mathoverflow.net/users/56524 | A generalized form of the approximation to identity? | $\newcommand\al\alpha\newcommand\be\beta\newcommand\R{\mathbb R}$This is not true in general. E.g., suppose that $m=1$, $\mu=N(0,1)$, and $r\_n(x)=r:=1/n$ for all $n$ and $x$. Let $f$ be the standard normal pdf.
Then for each real $x$ (and $n\to\infty$) we have
$$\al\_n(x)\sim\int\_{x-r}^{x+r}(y-x)^2\,dy\,f(x)=\frac{2r^3}3\,f(x).$$
Next, letting $X$, $Y$, and $Z$ denote independent standard normal random variables, we get
$$A\_n=\int\al\_n f=E(X-Y)^2\,1(|X-Y|\le r)
=2EZ^2\,1(|Z|\le r/\sqrt2) \\
\sim 2f(0)\int\_{-r/\sqrt2}^{r/\sqrt2}dz\,z^2=f(0)\frac{r^3\sqrt2}3.$$
So, for $F:=f$ and
$$L\_n:=\frac1{A\_n}\int F\al\_n \,d\mu$$
we have
$$\liminf\_n L\_n\ge\int f\frac{\sqrt2}{f(0)}\,f \,f=L:=\frac{2}{\sqrt{3 \pi }},$$
whereas
$$R:=\int F\,d\mu=\int f^2=\frac{1}{2 \sqrt{\pi }}<L.$$
So, $L\_n\not\to R$. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 447054 | 180,070 |
https://mathoverflow.net/questions/447047 | 5 | The constructible universe $L$ has some nice properties:
1. $L$ has a $\mathit{\Delta}^1\_2$-good well-ordering of $\mathbb{R}$. (Gödel, Addison)
2. For any $\mathit{\Sigma}^1\_2$ formula $\varphi(x)$ and a real $r \in \mathbb{R}\cap L, \ \varphi(r) \iff \varphi(r)^L$. (Shoenfield)
3. For any $\mathit{\Sigma}^1\_2$ set $A$, either $A\subseteq L$ or $A$ contains a perfect set. (Mansfield, Solovay)
My questions are:
* Is there, in $\mathsf{ZFC}$, an inner model that satisfies all three statements above with $\mathit{\Delta}^1\_2$ and $\mathit{\Sigma}^1\_2$ replaced by $\mathit{\Delta}^1\_3$ and $\mathit{\Sigma}^1\_3$, respectively?
* What about $\mathit{\Delta}^1\_n$ and $\mathit{\Sigma}^1\_n$ with $n > 3$?
* What if instead we work in $\mathsf{ZFC}$ + large cardinals?
| https://mathoverflow.net/users/141146 | Inner model with a $\mathit{\Delta}^1_3$-good well-ordering of the reals | The situation is a bit more complicated than you might hope because of the periodicity phenomena in the projective hierarchy. For odd $n$, assuming $\mathbf{\Delta}^1\_{n-1}$-determinacy, the set $Q\_n$ of reals that are $\Delta^1\_n$ in a countable ordinal is $\Pi\_n^1$ definable, so its complement is a $\Sigma\_n^1$ set that has no element in any inner model with a good $\Delta^1\_n$ wellorder. So no such model is $\Sigma\_n^1$-correct.
The set $Q\_n$ is the set of reals in $M\_{n-2}$, the minimum inner model with $n-2$ Woodin cardinals, which does have a good $\Delta^1\_n$-wellorder of the reals. The union of all countable $\Sigma^1\_n$ sets is going to be the set of all $\Delta^1\_n$ reals by the effective perfect set theorem, so $M\_{n-2}$ will satisfy the version of the perfect set theorem you want. But really this is not quite the right theorem, I think, since $Q\_n$ is $\Pi^1\_n$ not $\Sigma^1\_n$. On the other hand, there is a largest countable $\Pi^1\_n$ set (analogous to the largest countable $\Pi^1\_1$ set, the mastercodes of $L$), but this is actually strictly larger than $Q\_n$.
On the other hand, if there is no inner model with a Woodin cardinal and every set has a sharp, then the core model $K$ is $\Sigma^1\_3$-correct and has a $\Delta^1\_3$-good wellorder. Well, I'm not sure if this has been proved in such generality, the version I know assumes there is a measurable cardinal. Maybe an expert can fill this in.
For the even levels, under large cardinal assumptions, the situation should be very similar to the one for $L$, so probably exactly what you want is true. I think this cannot be done without some large cardinal assumptions, or really determinacy/mouse existence assumptions. I'm getting a little tired of writing for now, hopefully I'll update later.
Some references: Steel's "Projectively wellordered inner models," Steel's "Ordinal definability in models of determinacy," Steel's *Core Model Iterability Problem* (for the correctness of $K$), Kechris-Martin-Solovay's "Introduction to Q-theory."
| 8 | https://mathoverflow.net/users/102684 | 447056 | 180,071 |
https://mathoverflow.net/questions/447059 | 0 | The motivation for my current question arises from [this MO post by R. Stanley](https://mathoverflow.net/questions/430741/number-of-coefficients-equal-to-k-in-certain-fibonacci-polynomials). *Caveat.* There's a slight alteration.
With the convention $F\_1=F\_2=1$ for the Fibonacci numbers, define the polynomials $f\_n(x)=\prod\_{i=1}^n(1+x^{F\_i})$.
For a given prime $p$, expand $f\_n(x)$ and compute the coefficients modulo $p$. Denote the resulting polynomial by $g\_{n,p}(x)$ and its number of non-zero coefficients by $N(g\_{n,p})$.
**Example.** $g\_{1,2}=1+x, g\_{2,2}=1+x^2, g\_{3,2}=1+x^4, g\_{4,2}=1+x^3+x^4+x^7$, we find that
$N(g\_{1,2})=2, N(g\_{2,2})=2, N(g\_{3,2})=2, g\_{4,2}=4$. I like to propose two sets of questions.
>
> **QUESTION 1.** Are any of these recurrences true?
> \begin{align}
> N(g\_{n,2})&=2N(g\_{n-1,2})-2N(g\_{n-2,2})+2N(g\_{n-3,2}), \\
> N(g\_{n,3})&=2N(g\_{n-1,3})-2N(g\_{n-2,3})+3N(g\_{n-3,3})-4N(g\_{n-4,3})+4N(g\_{n-5,3}), \\
> N(g\_{n,5})&=2N(g\_{n-1,5})-N(g\_{n-2,5})+N(g\_{n-3,5})-2N(g\_{n-4,5})+3N(g\_{n-5,5}) \\
> & \qquad \qquad \qquad-4N(g\_{n-6,5})+4N(g\_{n-7,3}).
> \end{align}
>
>
>
>
> **QUESTION 2.** For fixed prime $p$, does the sequence $\{N(g\_{n,p})\}\_{n\geq1}$ have a rational generating function?
>
>
>
| https://mathoverflow.net/users/66131 | Fibonacci and product polynomials | Question 2 follows from Theorem 6.1 of [arXiv:2101.02131](https://arxiv.org/pdf/2101.02131.pdf). (In this reference, I consider $\prod\_{i=1}^n(1+x^{F\_{i+1}})$ rather than $\prod\_{i=1}^n(1+x^{F\_i})$, but the proof still works.) The result holds for any positive integer $p$, not just primes. The proof is constructive, so in principle one can obtain actual recurrences.
| 4 | https://mathoverflow.net/users/2807 | 447063 | 180,073 |
https://mathoverflow.net/questions/447062 | 3 | I am reading the paper by Jardine and Goerss, Localization theories for simplicial presheaves and having troubles with understand an argument. In this paper, the two authors considered $\mathcal{C}$ to be a small Grothendieck site (so in particular the underlying class of objects is a set and morphisms between two objects form a set). They proposed a list of axiom, called localization theories, so that whenever the site $\mathcal{C}$ satisfies this list, the category $\mathbf{PreSh}(\mathcal{C},\Delta^{op}\mathbf{Sets})$ of simplicial presheavse is a model category. For simplicity, we can think of weak equivalences are sectionwise equivalences and cofibrations are monomorphisms.
One of the axioms is:
>
> Suppose that $\gamma$ is a limit ordinal and there is functor $X: \gamma \longrightarrow \mathbf{PreSh}(\mathcal{C},\Delta^{op}\mathbf{Sets})$ such that for each morphism $i \leq j$ in $\gamma$, $X(i) \longrightarrow X(j)$ is a trivial cofibration, then every canonical morphism
> $$X(i) \longrightarrow \operatorname{lim}\_{j \in \gamma}X(j)$$
> is a trivial cofibration.
>
>
>
As fas as I understand, this axiom was listed in order to make the small object argument work. Let me first pick a cardinal $\alpha$ bouding the set of morphisms of $\mathcal{C}$ and $\beta$ be some cardinal greater than $2^{\alpha}$ (this is where I do not understand the choice). Then the small object argument comes into the account: it factors any morphism $f: X \longrightarrow Y$ into
$$X \longrightarrow X\_{\beta} \longrightarrow Y$$
where $X\_{\beta}$ is defined by a transfinite induction argument (basically, it is like what we should do for simplicial sets, we should fill all the holes $\partial \Delta[n] \subset \Delta[n]$), i.e. $X\_{\beta}$ is some $\beta$-transfinite composition. What I do not understand is why $X\_{\beta} \longrightarrow Y$ has to be a fibration. The authors proved this by using the lifting property. They proved that $X\_{\beta} \longrightarrow Y$ has the right lifting property w.r.t all trivial cofibration $i:U \longrightarrow V$ with $V$ being $\alpha$-bounded (meaning that $\left|V\_n(A) \right| \leq \alpha$ for every $n\geq 0$ and $A \in \mathcal{C}$).
Their argument is:
A morphism $U \longrightarrow X\_{\beta}$ must factor through some stage $X\_{\gamma}$ for some $\gamma < \beta$ because otherwise $U$ ***would have too many subobjects***. Once the factorization is found, the lifting problem is solved.
I do not see why? Why $U \longrightarrow X\_{\beta}$ does not factor implies that $U$ would have too many subobjects? And precisely, how many is too many?
| https://mathoverflow.net/users/482398 | Injective model structure for simplicial presheaves | To answer the question as it is stated: $U$ is an object in a locally presentable category, therefore $U$ is a small object, hence the corepresentable functor of $U$ preserves $α$-filtered colimits for some regular cardinal $α$. More precisely, the image of $U$ in such a colimit has a cardinality bounded by the product of cardinalities of $U$ and $\cal C$. Therefore, it must factor through some fixed stage of an $α$-filtered colimit if $α$ is greater than this cardinality.
However, that being said, the result discussed in the main post has been superseded by a considerably more general and easier to use theorem of Smith: for any left proper combinatorial model category, the left Bousfield localization at any set of morphisms $S$, and, more generally, at any class of morphisms $S$ that forms an accessible subcategory of $C$, exists. The conditions of Goerss and Jardine (especially, condition E7) simply guarantee that there is a set $S$ of such morphisms. For a reference, see, for example, Theorem 4.7 in Barwick's [On left and right model categories and left and right Bousfield localizations](https://projecteuclid.org/journals/homology-homotopy-and-applications/volume-12/issue-2/On-left-and-right-model-categories-and-left-and-right/hha/1296223884.full).
| 1 | https://mathoverflow.net/users/402 | 447075 | 180,076 |
https://mathoverflow.net/questions/447067 | 3 | Let me describe the simplest non-trivial case of what I have in mind. Let $V$ be a 2-dimensional $\mathbb{R}$-vector space and fix an isomorphism $V \cong \mathbb{R}^2$, where $\mathbb{R}^2$ is equipped with the standard basis. This will be our starting frame of reference, so to speak. Let $S \subseteq V$ be an arbitrary subset. I am mostly interested in the cases where $S$ is a path, or a loop, or an open subset, but **for the purpose of this question let me give a formulation for the case when $S := U$ is an open subset of the plane.**
Let $I \in \operatorname{M}\_2(\mathbb{R}) \cong \operatorname{End}(V)$ denote the identity matrix, let $p\_I : U \twoheadrightarrow \mathbb{R}$ be the projection onto, say, the *first* coordinate with respect to the standard basis (hence the $I$-notation), and define
$$
U\_I := p\_I(U) \times \mathbb{R}.
$$
(It won't matter if we choose the first or the second projection.) Clearly, $U\_I \supseteq U$ is the best we can recover from only knowing $p\_I(U)$. Now change the basis (coordinate system) via $(A^T)^{-1} \in \operatorname{GL}\_2^+(\mathbb{R})$ and put similarly
$$
U\_A := p\_A(U) \times \mathbb{R},
$$
where $p\_A : U \twoheadrightarrow \mathbb{R}$ is the projection onto the *first* coordinate with respect to the new basis.
Define the *projection hull* (for a lack of a better term) of $U$ to be
$$
\tilde{U} = \bigcap\_{A \in \operatorname{GL}\_2^+(\mathbb{R})} U\_A
$$
Then $\tilde{U} \supseteq U$ is the best we can recover from knowing the 1-dimensional projection of $U$ in every coordinate system.
Of course, one can do a similar construction with $\operatorname{GL}\_2(\mathbb{R})$ and both projections, which produces the same result by rotation invariance of the problem, or one may consider more special types of coordinate changes, i.e. other subgroups of $\operatorname{GL}\_2(\mathbb{R})$.
Here is an explicit description of the set $U\_A$ with respect to the original frame of reference:
$$
U\_A = \bigg\{ \frac{1}{\det(A)} (a\_{11} a\_{22} x\_1 + a\_{12} a\_{22} x\_2 - a\_{12}y, -a\_{11} a\_{21} x\_1 - a\_{12} a\_{21} x\_2 + a\_{11} y) \mid (x\_1,x\_2) \in U, y \in \mathbb{R} \bigg\}
$$
where $a\_{11},a\_{12},a\_{21},a\_{22}$ are the coefficients of $A$. (I sincerely hope I haven't miscalculated or mistyped.)
Here are my questions:
**(Q1)** Is there an accepted name for $\tilde{U}$ in the literature?
**(Q2)** Assuming $U$ is a **bounded simply connected domain**, can we have $\tilde{U} \supsetneqq U$? If yes, is there a different, more tangible characterization of $\tilde{U}$?
*Please feel free to add more appropriate tags.*
| https://mathoverflow.net/users/1849 | Recovering a set from its projections in varying coordinate systems - a projection hull? | $\newcommand{\R}{\mathbb R}\newcommand{\tU}{\tilde U}$Suppose that $U$ is connected. Then all its projections are connected. So, all one-dimensional projections of $U$ will be convex. So, $\tilde U$ will be convex.
Now take any non-convex connected $U$. Then $\tilde U\ne U$ and hence $\tilde{U} \supsetneqq U$.
---
Suppose now that a connected subset $U$ of $\R^2$ is closed and bounded. Then
\begin{equation}
\tU=\bigcap\_{l\in(\R^2)'}l^{-1}([m\_l,M\_l]),
\end{equation}
where $(\R^2)'$ is the space of all linear functionals on $\R^2$, $m\_l:=\min l(U)$, and $M\_l:=\max l(U)$.
Let us show that
$$\tU=C,$$
the closed convex hull of $U$. (So, is then a new name for $\tU$ needed?)
Clearly $\tU$ contains $U$ and is closed and convex, as the intersection of closed convex sets. So, $C\subseteq\tU$.
To obtain a contradiction, assume that $\tU\not\subseteq C$. So, we have some $x\in\tU\setminus C$. Then $x$ can be separated from $C$ by some $l\_x\in(\R^2)'$ so that $l\_x(x)<\min l\_x(C)\le\min l\_x(U)=m\_{l\_x}$. Then $x\notin l\_x^{-1}([m\_{l\_x},M\_{l\_x}])$, which contradicts the condition $x\in\tU$. So, $\tU\subseteq C\subseteq\tU$. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 447077 | 180,078 |
https://mathoverflow.net/questions/447090 | 2 | A *Halin graph* is a graph constructed by embedding a tree with no vertex of degree
two in the plane and then adding a cycle to join the tree’s leaves.
We found a list of the number of Halin graphs within $14$ vertices on the website <https://oeis.org/A346779>.
```
n a(n)
1 0
2 0
3 0
4 1
5 1
6 2
7 2
8 4
9 6
10 13
11 22
12 50
13 106
14 252
```
It seems that there aren't many Halin graphs with at most 14 vertices (252 Halin graphs with 14 vertices is considered quite a small number).
Based on this, I estimate that the number of non-isomorphic Halin graphs within 20 vertices may also not be too large.
By [Wiki](https://en.wikipedia.org/wiki/Halin_graph), it is possible to test whether a given $n$-vertex graph is a Halin graph in linear time. To obtain Halin graphs with more vertices, such as a Halin graph with 20 vertices, one approach could be to generate 3-connected planar graphs with 20 vertices first and then filter them one by one. However, the problem is that the number of [3-connected planar graphs](https://oeis.org/A000944) with 20 vertices is astronomically large. Therefore, I wonder if there is an algorithm generating Halin graphs or existing graph data available (I haven't found it either).
| https://mathoverflow.net/users/171032 | Is there an algorithm to generate non-isomorphic Halin graphs? | There is a theorem due to Jordan which is useful for enumerating or generating trees: every tree has a centre or a bicentre. To enumerate/generate suitable bicentral trees, enumerate/generate suitable rooted trees grouping them by height and join two rooted trees of the same height with an edge between their roots. To enumerate/generate suitable central trees, enumerate/generate suitable rooted trees grouping them by height and join two or more to the centre by their roots, ensuring that the maximum height of the trees chosen is not unique.
This construction method also gives a natural way to generate the embeddings into the plane.
| 2 | https://mathoverflow.net/users/46140 | 447101 | 180,083 |
https://mathoverflow.net/questions/441734 | 1 | I am looking for a reference proving the existence of the minimal Steiner tree in the Euclidean Steiner tree problem:
Given N points in the d-dimensional Euclidean space, the goal is to connect them by lines of minimum total length in such a way that any two points may be interconnected by line segments either directly or via other points and line segments (it may be shown that the connecting line segments do not intersect each other except at the endpoints and form a tree called the Steiner minimal tree).
| https://mathoverflow.net/users/47256 | Existence of minimal Steiner tree | The existence theorem for a finite number of points is more-or-less obvious, so classical sources do not give it as a statement. But you can extract them from the very classical paper Gilbert E. N., Pollak H. O. Steiner minimal trees //SIAM Journal on Applied Mathematics. – 1968. – Т. 16. – №. 1. – С. 1-29, or from any paper with an algorithm.
If you need just a statement, then I know only the paper Paolini E., Stepanov E. Existence and regularity results for the Steiner problem //Calculus of Variations and Partial Differential Equations. – 2013. – Т. 46. – №. 3-4. – С. 837-860, which contains an existance theorem in a very general setting (infinite number of vertices, mild conditions on a metric space).
| 3 | https://mathoverflow.net/users/479618 | 447111 | 180,084 |
https://mathoverflow.net/questions/447114 | 7 | Let me begin by mumbling some abstract nonsense, and then attempt to be concrete.
The category of groups inherits the structure of a strict 2-category from the 2-category of small categories.
Explicitly, a 2-morphism between $\varphi$ and $\varphi'\colon H \to G$ is an element $g \in G$ such that for each $h \in H$, we have $\varphi(h) = g\varphi(h)g^{-1}$. I'm curious why this category has coinserters but not inserters.
So much for the mumbling.
Let $\varphi$ and $\varphi'\colon H \to G$ be homomorphisms of groups. Let $\psi\colon K \to H$ be a homomorphism and $g$ an element of $G$. We say that the pair $(\psi,g)$ *inserts* $\varphi$ and $\varphi'$ if we have that for all $k \in K$, the following equality holds in $G$
$$\varphi\psi(k) = g\varphi'\psi(k)g^{-1}.$$
The collection of pairs $(\psi,g)$ with domain $K$ that inserts $\varphi$ and $\varphi'$ forms a *category*: a morphism in this category from $(\psi,g)$ to $(\psi',g')$ is an element $h \in H$ such that for all $k$ in $K$, we have the equality
$$\psi(k) = h\psi'(k)h^{-1}$$
as well as the equality
$$\varphi(h)g' = g\varphi'(h).$$
A *strict inserter* of $\varphi$ and $\varphi'$ would be a group $\mathrm{Ins}(\varphi,\varphi')$ equipped with an inserting pair $(\Psi,g)$ which is *universal* for this property in the following sense:
If $(\psi',g')$ is an inserting pair with domain $K$, there exists a unique homomorphism $\eta\colon K \to \mathrm{Ins}(\varphi,\varphi')$ such that
$\psi' = \Psi\eta$ and such that $g = g'$. This latter requirement kills the possibility of strict inserters: even if $G$ is *abelian,* so that the condition of a pair $(\psi,g)$ inserting $\varphi$ and $\varphi'$ is just the condition that $\psi$ equalizes $\varphi$ and $\varphi'$, the free choice of $g$ makes it impossible to have a universal choice of $g$.
---
The dual notion is of a *coinserter*. This is a group $K$ equipped with a homomorphism $\Psi\colon G \to K$ and an element $t \in K$ such that $\Psi\varphi(h) = t\Psi\varphi'(h)t^{-1}$ for all $h \in H$. If a pair $(\psi',t')\colon G \to K'$ coinserts $\varphi$ and $\varphi'$, there must be a unique homomorphism $K \to K'$ satisfying the obviously dual condition to the above.
Now since $t$ and $t'$ live in *different groups,* suddenly we're able to construct coinserters, certainly at least in the case where $\varphi$ and $\varphi'$ are injective, but I imagine a variant of the construction produces a coinserter in all cases.
Explicitly the coinserter is the pair $(\iota, t)\colon G \to G\*\_H$,
where $G\*\_H$ denotes the *HNN extension* of $G$ with associated subgroups $\varphi(H)$ and $\varphi'(H)$, $\iota$ is the canonical inclusion of $G$ into $G\*\_H$ and $t$ is the "stable letter" for the HNN extension.
A presentation for $G\*\_H$, given that a presentation for $G$ is $G = \langle S \mid R \rangle$ is as follows:
$$\langle S, t \mid R, t\varphi(h)t^{-1} = \varphi'(h) \rangle$$
as $h \in H$ varies. It should be clear that if a pair $(\psi,t')\colon G \to K$ coinserts $\varphi$ and $\varphi'$, we can define a homomorphism $G\*\_H \to K$
taking $\iota(g)$ to $\psi(g)$ for all $g \in G$ and taking $t$ to $t'$.
---
Sorry for the ramble. Anyway: is there some deeper reason that the 2-category of groups should have (strict) coinserters but not (strict) inserters? After all, the 2-category of categories has both.
| https://mathoverflow.net/users/135175 | Why does the 2-category of groups have (some, strict) coinserters but not (strict) inserters? | One viewpoint goes as follows: the 2-categorical structure on groups can be seen as coming from inner automorphisms, so that a 2-cell is given by an inner automorphism that translates one map to the other. Now, inner automorphisms of an object can be defined in any category (see e.g. [this paper](https://doi.org/10.1016/j.entcs.2018.11.010)) using the notion of isotropy group, a particular functor from the starting category to groups.
Moreover, one can use these abstract inner automorphisms to promote any category into a 2-category (albeit not in a functorial way: a given functor might not become a 2-functor). Me and Pieter Hofstra have some results on how 2-categorical limits and colimits behave in the resulting 2-category. In particular, as soon as your starting category is finitely cocomplete, it has all coinserters iff the isotropy functor is representable. Moreover, all limits and connected colimits of the underlying category satisfy the required two-dimensional universal property as well. However, inserters, equifiers or 2-d coproducts exist iff the isotropy is trivial. For what it's worth, I've given a talk about this stuff [here](https://progetto-itaca.github.io/pages/fest22.html) but there's no publicly available writeup yet.
One can debate if this counts as a "deep reason", as ultimately the general proof that strict inserters do not exist boils down to the kind of situation you considered. That said, this does put the observation in context, so perhaps it still counts.
| 9 | https://mathoverflow.net/users/136562 | 447116 | 180,086 |
https://mathoverflow.net/questions/447099 | 9 | **I. Some functions**
As these will be used in the continued fraction evaluations below, recall the *Riemann zeta function* $\zeta(s),$ and *[Dirichlet beta function](https://mathworld.wolfram.com/DirichletBetaFunction.html)* $\beta(s),$
$$\beta(s) = \sum\_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^s}$$
and special cases of the *[Clausen function](https://mathworld.wolfram.com/ClausenFunction.html)* $\operatorname{Cl}\_s(x),$
$$\operatorname{Cl}\_2(x) = \sum\_{n=1}^\infty\frac{\sin(n\,x)}{n^2}$$
\begin{align}
\operatorname{Cl}\_2\left(\tfrac12\pi\right) &= K = \beta(2) \\
\operatorname{Cl}\_2\left(\tfrac13\pi\right) &= \kappa
\end{align}
with *Catalan's constant* $K$ and its cubic counterpart *[Gieseking's constant](https://mathworld.wolfram.com/GiesekingsConstant.html)* $\kappa$.
---
**II. Zagier's 6 sporadic sequences**
Inspired by Apery's result in proving the irrationality of $\zeta(3)$ using certain integer sequences, Zagier (via a computer) searched for sequences with recurrence relation and deg-$2$ coefficients of form,
$$(n+1)^2\,u\_{n+1} = (an^2+an+b)u\_k+ cn^2\,u\_{n-1}$$
that produced only integer values. Only six $(a,b,c)$ were found, namely,
$$(11,3,1),\quad (7,2,8) ,\quad (12,4,-32)$$
$$(-17,-6,-72),\quad (10,3,-9), \quad (-9,-3,-27)$$
It seems we can use **ALL** these coefficients to produce nice cfracs.
---
**III. Continued fractions**
Given a 3-term recurrence relation of form,
$$F\_1(n)\,u\_{n+1} = F\_2(n)\,u\_n + F\_3(n)\,u\_{n-1}$$
where $F\_i(n)$ are polynomials of degree $k$. Define two polynomial functions using the rules,
\begin{align}
p(n) &= F\_1(n-1)\, F\_3(n)\\
q(n) &= F\_2(n)
\end{align}
which implies $p(n)$ has degree ***twice*** that of $q(n)$. Define the [*continued fraction*](https://mathworld.wolfram.com/ContinuedFraction.html),
$$C =\cfrac{1}{q(0) + \cfrac{p(1)}{q(1) + \cfrac{p(2)}{q(2)+ \cfrac{p(3)}{q(3)+\ddots } }}}$$
More compactly,
$$C(m) = \frac1{q(0) + \large{\underset{n=1}{\overset{m}{\mathrm K}} ~ \frac{p(n)}{q(n)}}}$$
or in *Mathematica* notation,
$$C(m) = \frac1{q(0) + \text{ContinuedFractionK}[p(n),\;q(n),\, \text{{n, 1, m}}]}$$
It seems $C$ may have a nice closed-form based on the properties of the recurrence relation. Examples below.
---
**IV. Degree 2**
Recall Zagier's recurrence,
$$\color{blue}{(n+1)^2}\,u\_{n+1} = (\color{blue}{an^2+an+b})u\_k+\color{blue}{cn^2}\,u\_{n-1}$$
Define $p(n)$ and $q(n)$ according to the rules in the previous section,
\begin{align}
p(n) &= \color{blue}{n^2}\times \color{blue}{cn^2} = cn^4\\
q(n) &= \color{blue}{an^2+an+b}
\end{align}
Then define the cfrac,
$$C\_2(a,b,c) = \frac1{q(0) + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p(n)}{q(n)}}}$$
**Q:** Is it true that,
\begin{align}
C\_2(11,3,1) &= \frac15\,\zeta(2)\\
C\_2(-17,-6,-72) &=\color{green}{-\frac5{6\sqrt3}\operatorname{Cl}\_2\left(\tfrac13\pi\right) = -\frac5{6\sqrt3}\kappa}\\
C\_2(10,3,-9) &=\frac2{3\sqrt3}\operatorname{Cl}\_2\left(\tfrac13\pi\right) = \frac2{3\sqrt3}\kappa\\
C\_2(7,2,8) &= \frac14\,\zeta(2)\\
C\_2(12,4,-32) &= \frac12\operatorname{Cl}\_2\left(\tfrac12\pi\right) = \frac12\beta(2)=\frac12K\\
C\_2(-9,-3,-27) &=\;\color{red}{??}
\end{align}
where $K$ is *Catalan's constant* and $\kappa$ is *Gieseking's constant*, both of which **not** yet proven to be irrational.
*Note*: The first evaluation is valid since it was found by Apery, while the second (in $\color{green}{\text{green}}$) is courtesy of H. Cohen's answer. (**Update: May 22, 2023**) It turns out $C\_2(-9,-3,-27)$ has ***six limits***, one of which is divergent. See this [MO post](https://mathoverflow.net/q/447316/12905).
---
**V. Degree 3**
In [Cooper's paper](https://www.researchgate.net/publication/257642843_Sporadic_sequences_modular_forms_and_new_series_for_1p), we find the recurrence relation with deg-$3$ coefficients in $n$,
$$(n+1)^3\,v\_{n+1} = -(2n+1)(an^2+an+a-2b)v\_n +(-a^2-4c)n^3\,v\_{n-1}$$
and Zagier's $(a,b,c).$ Using the same rules, let,
\begin{align}
r(n) &= n^3\times(-a^2-4c)n^3 = -(a^2+4c)n^6\\
s(n) &= -(2n+1)(an^2+an+a-2b)
\end{align}
Define the cfrac,
$$C\_3(a,b,c) = \frac1{s(0) + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{r(n)}{s(n)}}}$$
**Q:** Is it true that,
\begin{align}
C\_3(11,3,1) &=\;\color{red}{??}\\
C\_3(-17,-6,-72) &= \frac16\,\zeta(3)\\
C\_3(10,3,-9) &= -\frac{7}{24}\,\zeta(3)\\
C\_3(7,2,8) &=\;\color{red}{??}\\
C\_3(12,4,-32) &= -\frac{7}{32}\,\zeta(3)\\
C\_3(-9,-3,-27) &= \frac{128}{243\sqrt3}\,\beta(3) = \frac{4\pi^3}{243\sqrt3}
\end{align}
where $d = a^2+4c =125, 1, 64, 81, 16, -27,$ respectively (and all powers of the smallest primes $2,3,5$).
*Note*: The second closed-form is valid since it was also found by Apery which he used (together with other methods) to prove the irrationality of $\zeta(3)$.
---
**VI. Degree 4 & 5**
Curiously, there is no known 3-term recurrence,
$$P\_1(n) v\_{n+1} = P\_2(n) v\_n + P\_3(n) v\_{n-1}$$
where $P\_i$ are polynomials of deg-$4$. ***Why?*** But Zudilin found,
$$Q\_1(n) v\_{n+1} = Q\_2(n) v\_n + Q\_3(n) v\_{n-1}$$
where $Q\_i$ are polynomials of deg-$5$ and used it in an analogous continued fraction for $\zeta(4).$ (*To be discussed in the [next post](https://mathoverflow.net/q/447166/12905).*)
---
**VII. Questions**
1. Are all cfracs with proposed closed-forms correct? (I know two of them are.)
2. What are the closed-forms of the others?
| https://mathoverflow.net/users/12905 | On 12 cfracs: for Catalan's $K$, Gieseking's $\kappa$, and $\pi^2$, $\pi^3$, plus three for $\zeta(3)$ using Zagier's "six sporadic sequences" | We have $$C\_2(-17,-6,-72)=-(5/8)L(\chi\_{-3},2)$$ and
$$C\_2(10,3,-9)=(1/2)L(\chi\_{-3},2)$$ so both are proportional to what you
call Gieseking's constant but which is simply the value at 2 of the
L function of the nontrivial character modulo 3, close analogue to Catalan's
constant which is the same with the nontrivial character modulo 4.
All the other "??" that you quote, both in degree 2 and in degree 3
are divergent cfracs (by the way, "degree" is more proper than "level").
Finally just a typo: $C\_3(12,4,-32)=-(7/32)\zeta(3)$ (minus sign omitted).
Two useful references:
O. Gorodetsky, New representations for all sporadic Ap'ery-like sequences, with applications to congruences,
arXiv:2102:2102.11839 (2021)
and
Y. Yang, Ap'ery limits and special values of $L$-functions,
J. Math. Anal. Appl. {\bf 343} (2008), 492--513.
| 12 | https://mathoverflow.net/users/81776 | 447123 | 180,088 |
https://mathoverflow.net/questions/447100 | 2 | The following question, which arose during my research, seems deceivingly simple to me, but I could not find any elegant and formal argument.
For a binomially distributed variable $X \sim \text{Bin} \left( n, \frac{1}{\sqrt[4]{n}} \right)$, I am looking for a preferably slick and short formal argument that: $$\text{Pr}[ X \geq \mathbb{E}[X]] = \Omega(1)$$ asymptotically, i.e. that the probability of $X$ lying above its expectation is lower bounded by a constant as $n$ grows large.
Any help is well appreciated. Thank you!
| https://mathoverflow.net/users/475708 | Simple anticoncentration bound for binomially distributed variable | Let $Z\sim N(0,1)$, $p\_n:=n^{-1/4}$, $q\_n:=1-p\_n$. By the [Berry--Esseen inequality](https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem#Identically_distributed_summands),
$$P(X\ge EX)\ge P(Z\ge0)-0.5\frac{n(p\_nq\_n^3+q\_np\_n^3)}{(np\_nq\_n)^{3/2}}=\frac12-o(1)$$
as $n\to\infty$. $\quad\Box$
More explicitly, (i) for $p\in(0,0.68]$ and $q=1-p$ we have $\dfrac{pq^3+qp^3}{pq^{3/2}}\le1$ and (ii) $p\_n<0.68$ for
$n\ge5$, so that
$$P(X\ge EX)\ge \frac12-0.4748\, n^{-1/8}$$
for $n\ge5$.
| 9 | https://mathoverflow.net/users/36721 | 447124 | 180,089 |
https://mathoverflow.net/questions/446908 | 3 | Let $[1]^n=\{0<1\}^n$ equipped with the product order. I consider the small category $\widehat{\square}$ of the category of partially ordered sets generated by the *coface maps* $\delta^\epsilon\_i:[1]^{n-1}\to [1]^n$ with $\epsilon=0,1$ defined by $$\delta^\epsilon\_i:(x\_1,\dots,x\_{n-1}) \mapsto (x\_1,\ldots,x\_{i-1},\epsilon,x\_i,\dots,x\_{n-1})$$ and by the strictly increasing maps $f:[1]^n\to [1]^n$.
The small category $\widehat{\square}$ contains the symmetry maps (the ones permuting the coordinates).
>
> Informally, I would like to remove the symmetry maps from
> $\widehat{\square}$, and only them, to obtain a subcategory of
> $\widehat{\square}$.
>
>
>
Every strictly increasing map $f=(f\_1,\dots,f\_n):[1]^n\to [1]^n$ gives rise to another strictly increasing map by permuting the coordinates. I need to find a way to make a choice among all permutations.
Every strictly increasing map $f=(f\_1,\dots,f\_n):[1]^n\to [1]^n$ satisfies the equalities $$f\_i(x\_1,\dots,x\_n) = \max\_{(\epsilon\_1,\dots,\epsilon\_n)\in
f\_i^{-1}(1)} \min \{x\_k\mid \epsilon\_k=1\}$$ for all $1\leq i\leq n$ (see <https://mathoverflow.net/a/429941/24563>).
>
> **Question:** In the formula above, is there a way to put a total order on the coordinates by using the syntax of the formula ?
>
>
>
**Motivation:** I work with the presheaves on $\widehat{\square}$ that I call *transverse sets*. They are a generalization of the category of precubical sets adapted for studying the directed homotopy for concurrency. And I would like to define the *non-symmetric transverse sets*. The two papers using transverse sets are [Combinatorics of labelling in higher dimensional automata](https://doi.org/10.1016/j.tcs.2009.11.013) and [Directed degeneracy maps for precubical sets](https://doi.org/10.48550/arXiv.2209.02667).
**EDIT (I add some details to give some intuition):** a (too) naive idea consists of defining this subcategory of $\widehat{\square}$ by using this lemma:
>
> **Fact:** Every map $f:[1]^m\to [1]^n$ of $\widehat{\square}$ factors uniquely as a composite $[1]^m\to [1]^m \to [1]^n$ where the right-hand $[1]^m\to [1]^n$ is a composite of coface maps.
>
>
>
And then to consider the subset of maps of $\widehat{\square}$ factorizing like $[1]^m\to [1]^m \to [1]^n$ such that the left-hand map is not one-to-one unless it is the identity of $[1]^m$ and such that the right-hand map is a composite of coface maps. Unfortunately, this subset of maps of $\widehat{\square}$ is not closed under composition. Here is a simple counterexample.
* $f:[1]^2\to [1]^4$ defined by $f(x\_1,x\_2)=(x\_1,x\_2,0,0)$
* $g:[1]^4\to [1]^4$ defined by $g(x\_1,x\_2,x\_3,x\_4) = (x\_2,x\_1,\max(x\_3,x\_4),\min(x\_3,x\_4))$.
$f$ is a composite of coface maps. $g$ is not one-to-one since $$g(x,x,1,0)=g(x,x,0,1)=(x,x,1,0)$$ for $x=0$ or $x=1$. However $$(g\circ f)(x\_1,x\_2)=(x\_2,x\_1,0,0)$$ which means that $g\circ f$ is the composite of a nontrivial permutation map $[1]^2\to[1]^2$ followed by a composite of coface maps.
There are four maps in $\widehat{\square}([1]^2,[1]^2)$:
1. the identity $f(x\_1,x\_2)=(x\_1,x\_2)$
2. the permutation $f(x\_1,x\_2)=(x\_2,x\_1)$
3. $\gamma\_1(x\_1,x\_2)=(\max(x\_1,x\_2),\min(x\_1,x\_2))$
4. $\gamma\_2(x\_1,x\_2)=(\min(x\_1,x\_2),\max(x\_1,x\_2))$.
The idea would be to keep from $\widehat{\square}([1]^2,[1]^2)$ the identity and one of the two maps crushing the square transversally $\gamma\_1$ or $\gamma\_2$, and to find a way to do the same thing for all sets $\widehat{\square}([1]^m,[1]^n)$ in such a way that we obtain a subcategory of $\widehat{\square}$.
| https://mathoverflow.net/users/24563 | Removing the symmetry maps from a small category of cubes | The naive idea has to be slightly modified. The point is not to sort out all terms (it is a wrong intuition), but only where the variables $x\_i$ are "alone". For example, the map $$(x\_1,x\_2,x\_3,x\_4)\mapsto (x\_2,x\_1,\max(x\_3,x\_4),\min(x\_3,x\_4))$$ is not kept in the subcategory because $x\_2$ which is alone is before $x\_1$ which is alone. On the contrary, the map $$(x\_1,x\_2,x\_3,x\_4)\mapsto (x\_1,x\_2,\max(x\_3,x\_4),\min(x\_3,x\_4))$$ will be kept.
This way, symmetry maps cannot show up by precomposing by coface maps, and the counterexample of the question cannot exist. Everything boils down to the following proposition ($\square$ is the box category, which is generated by the coface maps):
>
> **Proposition:** The set of maps $$\mathcal{A}=\{\phi:[1]^m\to[1]^n\in \widehat{\square}\mid \forall \delta:[1]^p\to [1]^m\in \square,
> \phi\delta \hbox{ one-to-one }\Rightarrow \phi\delta\in \square\}$$ is
> closed under composition and contains all identity maps. Moreover, the
> only one-to-one maps of $\mathcal{A}$ are the maps of $\square$.
>
>
>
**Proof:** Let $\phi\_1,\phi\_2\in \mathcal{A}$ such that $\phi\_1\phi\_2$ exists. Let $\delta\in \square$ such that $\phi\_1\phi\_2\delta$ exists and is one-to-one. Then $\phi\_2\delta$ is a one-to-one set map. Thus $\phi\_2\delta\in \square$, $\phi\_2$ belonging to $\mathcal{A}$. We deduce that $(\phi\_1\phi\_2)\delta =\phi\_1(\phi\_2\delta) \in \square$ since $\phi\_1\in \mathcal{A}$. This means that $\phi\_1\phi\_2 \in \mathcal{A}$ by definition of $\mathcal{A}$. $\mathcal{A}$ contains clearly the identity maps. Finally suppose that $f:[1]^m\to [1]^n\in \mathcal{A}$ is one-to-one. Then $f\mathrm{id}\_{[1]^m}$ is one-to-one, which implies that $f\in \square$.
| 0 | https://mathoverflow.net/users/24563 | 447128 | 180,090 |
https://mathoverflow.net/questions/447107 | 2 | Suppose that $e\_1, \cdots, e\_n$ are the standard vectors of the Euclidean space $\mathbb{R}^n$.
Let us consider the backward shift operator $T:\mathbb{R}^n\to \mathbb{R}^n$ given by $Te\_k=e\_{k-1}$ if $k\geq2$ and $Te\_1=e\_n$.
Let $A:\mathbb{R}^n\to \mathbb{R}^n$ be the operator given by $Ae\_3=Ae\_4=e\_1$ and $Ae\_k=0$ for any other $k$.
It is well-known that the columns of the discrete Fourier matrix are just the eigenvectors of $T$. What about the (analytical proof) eigenvectors of $T+A$?
| https://mathoverflow.net/users/84390 | The eigenvectors of adding a particular rank one matrix to the circulant matrix | $\newcommand{\la}{\lambda}\newcommand{\R}{\mathbb R}$For any $x=(x\_1,\dots,x\_n)\in\R^n$ we have $Tx=(x\_2,\dots,x\_n,x\_1)$ and $Ax=(x\_3+x\_4,0\dots,0)$, so that for $U:=T+A$ we have $Ux=(x\_2+x\_3+x\_4,x\_3,\dots,x\_n,x\_1)$.
Let now $(x\_1,\dots,x\_n)\in\R^n$ be an eigenvector of $U$ belonging to an eigenvalue $\la$ of $U$.
Then
\begin{equation}
\left\{
\begin{aligned}
&x\_k=\la x\_{k-1}\text{ for }k=3,\dots,n, \\
&x\_2+x\_3+x\_4=\la x\_1, \\
&x\_1=\la x\_n.
\end{aligned}
\right.
\end{equation}
So, $x\_k=\la^{k-2} x\_2$ for $k=3,\dots,n$, $x\_1=\la x\_n=\la^{n-1} x\_2$, $(1+\la+\la^2)x\_2=\la x\_1=\la^n x\_2$. So, $x\_2\ne0$ and without loss of generality (wlog) $x\_2=1$. So, $\la$ is one of the $n$ roots $\la\_1,\dots,\la\_n$ of the equation $1+\la+\la^2=\la^n$, and the eigenvector $(x\_{j,1},\dots,x\_{j,n})$ belonging to the eigenvalue $\la\_j$ is given by th formulas
\begin{equation}
x\_{j,2}=1,\quad x\_{j,1}=\la\_j^{n-1},\quad
x\_{j,k}=\la\_j^{k-2}\text{ for }k=3,\dots,n.
\end{equation}
---
It is now shown (see [this question and the answers to it](https://mathoverflow.net/q/447134/36721)) that the eigenvalues $\la\_1,\dots,\la\_n$ are pairwise distinct and hence the corresponding eigenvectors form a basis of $\R^n$.
| 3 | https://mathoverflow.net/users/36721 | 447130 | 180,091 |
https://mathoverflow.net/questions/446780 | 7 | Let $f(z):=\langle g(z),g(z)\rangle,$ where $z \mapsto g(z)$ is holomorphic and $\langle \bullet,\bullet\rangle$ is an inner-product on some function space, such as $L^2$, such that $\langle g(z),g(z)\rangle = \int\_X \vert g(z,x)\vert^2 d\mu(x).$
Then one has by a fairly explicit computation
$$ \partial\_z \partial\_{\bar z} \log(f(z)) = f(z)^{-2} ( f(z) \Vert \partial\_z g(z) \Vert^2 - \vert \langle \partial\_z g(z),g(z) \rangle \vert^2) \ge 0,$$
where positivity just follows from the Cauchy-Schwarz inequality.
I wonder if one can generalize this formula to $f(z):= \operatorname{det}(\langle g\_i(z),g\_j(z) \rangle)\_{i,j}$, where $g\_i$ are holomorphic and if the log derivative still remains non-negative?
By a direct computation one has of course also in this case
$$ \partial\_z \partial\_{\bar z} \log(f(z)) = f(z)^{-2} ( f(z) \partial\_z \partial\_{\bar z} f(z) - \vert \partial\_z f(z) \vert^2),$$
but I do not quite see how to say anything more about this last expression, i.e. is $$f(z) \partial\_z \partial\_{\bar z} f(z) \ge \vert \partial\_z f(z) \vert^2?$$
Numerical experiments suggest that this holds, but it is hard for me to verify it.
| https://mathoverflow.net/users/496243 | Log-convexity of determinant | Your claim can be deduced from the case $n=1$.
Let $H$ be the Hilbert space you start with.
Let $H^{\otimes n}=H\otimes\dots\otimes H$ be the $n$-th tensor power. Equip it with the usual inner product and complete to get a Hilbert space $V$ again. Let $g=g\_1\wedge\dots\wedge g\_n$ be the function
$$
g=\sum\_\sigma\mathrm{sign}(\sigma)\ g\_{\sigma(1)}\cdots g\_{\sigma(n)},
$$
where the sum runs over all permutations.
Then, because of $\mathrm{sign}(\sigma)=\mathrm{sign}(\sigma^{-1})$ we get
$$
<g,g> =\sum\_\sigma\sum\_\tau\mathrm{sign}(\sigma^{-1}\tau)<g\_{\sigma(1)},g\_{\tau(1)}>\cdots <g\_{\sigma(n)},g\_{\tau(n)}>
$$
Replacing $\tau$ with $\sigma\tau$ and reordering the product we get
$$
<g,g> =\sum\_\sigma\sum\_\tau\mathrm{sign}(\tau)<g\_{1},g\_{\tau(1)}>\cdots <g\_{n},g\_{\tau(n)}>
$$
this equals $n!\ \det(<g\_i,g\_j>)$. You apply the $n=1$ case to this $g$ and get the claim.
| 5 | https://mathoverflow.net/users/473423 | 447132 | 180,092 |
https://mathoverflow.net/questions/446991 | 14 | $\newcommand{\A}{\mathcal A}\newcommand{\Tr}{\operatorname{tr}}$For $c$ and $C$ such that $0<c<C<\infty$, let $\A\_{d;c,C}$ denote the set of all symmetric positive-definite real $d\times d$ matrices $A$ such that
$$cI\le A\le CI,$$
where $I$ is the $d\times d$ identity matrix and $A\le B$ for $d\times d$ matrices $A$ and $B$ means that $B-A$ is positive semidefinite. Let $|A|$ denote the determinant of a square matrix $A$.
>
> **Proposition 1:** For any $A\_0$ and $A\_1$ in $\A\_{d;c,C}$
> \begin{equation}
> |A\_1|-|A\_0|\le L\|A\_1-A\_0\|\_F\le L\sqrt d\,\|A\_1-A\_0\|,
> \end{equation}
> where $L:=C^d\sqrt d/c$, $\|\cdot\|\_F$ is the Frobenius norm, and $\|\cdot\|$ is the spectral norm.
>
>
>
A proof of Proposition 1 will be given at the end of this post.
>
> **Question 1:** Is Proposition 1 known?
>
>
>
>
> **Question 2:** Can Proposition 1 be improved?
>
>
>
A correct and complete answer to either one of these questions will be considered a correct and complete answer to this entire post.
---
*Proof of Proposition 1:* In view of the inequality $\|B\|\_F\le\sqrt d\,\|B\|$ for any matrix $B$, it is enough to prove the first inequality in Proposition 1.
Note that the set $\A\_{d;c,C}$ is convex.
For $A\in\A\_{d;c,C}$,
\begin{equation}
f(A):=\sqrt{|A|}=(2\pi)^{-d/2}\int\_{\mathbb R^d}dx\,e^{-x^\top A^{-1}x/2}.
\end{equation}
Let $X\_A$ stand for any zero-mean Gaussian random vector in $\mathbb R^d$ with covariance matrix $A\in\A\_{d;c,C}$. Let $\Tr A$ denote the trace of a square matrix $A$.
Then the derivative of $f$ at $A$ applied to any $d\times d$ real matrix $D$ is
\begin{equation}
\begin{aligned}
f'(A)(D)&=(2\pi)^{-d/2}\int\_{\mathbb R^d}dx\,e^{-x^\top A^{-1}x/2}x^\top A^{-1}DA^{-1}x/2 \\
&=\frac{f(A)}2\,E(X\_A^\top A^{-1}DA^{-1}X\_A) \\
&=\frac{f(A)}2\,E\Tr(X\_AX\_A^\top A^{-1}DA^{-1}) \\
&=\frac{f(A)}2\,\Tr(EX\_AX\_A^\top A^{-1}DA^{-1}) \\
&=\frac{f(A)}2\,\Tr(DA^{-1}) \\
&\le\frac{f(A)}2\,\|D\|\_F \|A^{-1}\|\_F \\
&\le\frac{f(A)}2\,\|D\|\_F \sqrt d\,\|A^{-1}\| \\
&\le\frac{C^{d/2}}2\,\|D\|\_F \sqrt d\,/c=K\|D\|\_F,
\end{aligned}
\end{equation}
with $K:=\frac{C^{d/2}}2\,\sqrt d\,/c$. So, for any $A\_0$ and $A\_1$ in $\A\_{d;c,C}$
\begin{equation}
\sqrt{|A\_1|}-\sqrt{|A\_0|}=f(A\_1)-f(A\_0)
\le K\|A\_1-A\_0\|\_F,
\end{equation}
whence
\begin{equation}
|A\_1|-|A\_0|=(\sqrt{|A\_1|}+\sqrt{|A\_0|})(\sqrt{|A\_1|}-\sqrt{|A\_0|}) \\
\le 2C^{d/2}K\|A\_1-A\_0\|\_F=L\|A\_1-A\_0\|\_F,
\end{equation}
which proves the first inequality in Proposition 1. $\quad\Box$
| https://mathoverflow.net/users/36721 | Lipschitz property of the determinant | The best constant is $C^{d-1}\sqrt{d}$.
Write $D(A)=\det A$. We can rephrase the inequality as the claim that $\|D'\|\_F\le L$ for $c\le A\le C$. (It's perhaps best to think of the matrices as long column vectors and the Frobenius norm as the Euclidean norm, and then $D'$ can be viewed as the gradient of $D$.)
By [Jacobi's formula](https://en.wikipedia.org/wiki/Jacobi%27s_formula) for the derivative of a determinant, we have
$$
\frac{\partial D}{\partial a\_{jk}}= D(A)\;\textrm{tr}\: (A^{-1}E\_{jk}) = D(A) (A^{-1})\_{kj} ;
$$
here $E\_{jk}$ is the matrix with a $1$ in the $jk$ slot and zero entries otherwise.
So $\|D'(A)\|\_F= D(A) \|A^{-1}\|\_F$. (This immediately recovers the original inequality since $D\le C^d$, $\|A^{-1}\|^2\_F\le d/c^2$.) To find the optimal constant, we write
$$
D^2\|A^{-1}\|^2\_F= \prod \lambda\_j^2\cdot \sum \lambda\_j^{-2} = \lambda\_2^2\cdots\lambda\_d^2 +\lambda\_1^2\lambda\_3^2\cdots \lambda^2\_d +\ldots +\lambda\_1^2\cdots \lambda^2\_{d-1}
$$
in terms of the eigenvalues of $A$. Clearly this is maximized at $A=C$, with value $C^{d-1}\sqrt{d}$, so this constant works.
It is also optimal as we can confirm by simply taking $A=C$, $B=C-\epsilon$: then $\|A-B\|^2\_F=d\epsilon^2$, $D(A)-D(B)=C^d-(C-\epsilon)^d=dC^{d-1}\epsilon + O(\epsilon^2)$.
(The fact that the derivative does have this value $C^{d-1}\sqrt{d}$ does not immediately rule out still smaller constants since the derivative controls movements in arbitrary directions while we are dealing with positive definite matrices only. However, we can also observe, in more abstract style, that $D'(A)$ is symmetric, so we do stay inside positive definite matrices when following the direction of the largest change of $D(A)$.)
| 7 | https://mathoverflow.net/users/48839 | 447138 | 180,095 |
https://mathoverflow.net/questions/447106 | 7 | For elliptic curves over $\mathbb{Q}$ the Mumford-Tate group is either $\mathrm{GL}\_2$ or $\mathrm{Res}\_\mathbb{Q}^F (\mathbb{G}\_m)$ if it has CM with the imaginary quadratic field $F$. In this case the $\mathbb{Q}$-endomorphism algebra completely determines the Mumford-Tate group. On the automorphic side, having CM can be seen as the associated modular form $f$ satisfying $f=f\otimes \chi$ for some quadratic character $\chi$.
For a simple abelian surface $A/\mathbb{Q}$, again the endomorphism algebra determines the Mumford-Tate group. [Here](https://arxiv.org/pdf/math/9901113.pdf) one can find the classification of the Hodge group (special Mumford-Tate group) for low dimensional abelian varieties. In particular there are four cases for abelian surfaces. One expects to be able to associate an automorphic representation $\pi$ of $\mathrm{GSp}\_4$ to the abelian surface $A$.
My question is that if the four cases of Mumford-Tate groups (or the endomorphism algebra if you like) can be translated to properties (hopefully some kind of symmetries) of the automorphic representation $\pi$. For instance, I think the CM case (Type IV(2,1) [here](https://arxiv.org/pdf/math/9901113.pdf)) should be similar to the case of elliptic curves, namely $\pi=\pi \otimes \chi$ for some character.
My main concern is the two dimensional case but any comments on the higher dimensional cases where the Mumford-Tate group is not determined by the endomorphism algebra is also highly appreciated.
| https://mathoverflow.net/users/496065 | Automorphic classification of different types of abelian surfaces | Yes, knowing the endomorphism algebra of $A$ (conjecturally) translates to certain properties of an associated automorphic representation $\pi$.
First, you should look at the *Galois type*, which is labelled (A)-(F) in [FKRS's paper on Sato-Tate groups](https://arxiv.org/abs/1110.6638). (These labels also include non-simple $A$, which correspond to products of elliptic curves, and thus products of rational elliptic newforms.)
If the $\mathbb Q$-endomorphism algebra contains a real quadratic field $K$, i.e., $A$ has real multiplication, then $A$ is of GL(2) type as defined by Ribet. By Ribet + Serre's conjectures, this means $A$ corresponds to weight 2 newform of GL(2) with rationality field $K$. If further $A$ has quartic CM, then $f$ will have CM, and have inner twists.
Otherwise, then the (generalized) paramodular conjecture predicts $A$ corresponds to a weight 2 automorphic representation $\pi$ of GSp(4). See [Brumer and Kramer's formulation](https://arxiv.org/abs/1004.4699) and/or [BCGP's potential modularity paper](https://arxiv.org/abs/1812.09269). In the typical case (trivial endomorphisms), the Galois representations are strongly irreducible, but the other Galois types (B)-(F) correspond to representations that are reducible upon restriction to some finite extension $K/\mathbb Q$, with details depending on the type---see Section 9.2 of [BCGP]. Consequently this means that $\pi\_K$ is an isobaric sum, which is one way of stating the automorphic property corresponding to the endomorphism algebra of $A$.
There should be similar stories in higher dimensions. Certainly the GL(2)-type situation is understood in higher dimensions.
| 4 | https://mathoverflow.net/users/6518 | 447142 | 180,097 |
https://mathoverflow.net/questions/447118 | 3 | I am looking for further proofs, preferably in the literature, of the following result:
**Proposition:** Let $R$ be a unital commutative $\mathbb{Z}$-torsion free ring. If $f(x) \in R[x]$ with $f'(x) \in U(R[x])$, then there exists $g(x) \in R[x]$ such that $f(g(x)) = g(f(x)) = x$ in $R[x]$.
For example, $R = \mathbb{Z}$ is allowed (mixed characteristic) as well as $R = \mathbb{Z}[t]/(t^2)$, but $R = \mathbb{Z} \times \mathbb{Z} /(n)$ is **not** allowed.
I already have a proof of this fact which relies explicitly on a theorem of Comtet mentioned at least in [MO415617](https://mathoverflow.net/questions/415617/a-leibniz-like-formula-for-fx-fracddxn-fx), [MO337330](https://mathoverflow.net/questions/337330/%d0%a1losed-formula-for-g-partialn), and [MO80828](https://mathoverflow.net/questions/80828/differential-operator-power-coefficients). So, if there is a proof in the literature based on a Comtet's *type of result*, I would like to know about this reference, but I am mostly interested in "new" proofs that **do not** invoke this type of result.
The $1$-dimensional Jacobian Problem in for $\mathbb{Z}$-torsion free rings as stated above should be well-known to experts (as a non-problem). But one issue when searching through the literature is that people are understandably mostly focused on the complex case in higher dimensions. Indeed the above result is completely trivial for any ring with $\operatorname{nil}(R) = 0$, not just $\mathbb{C}$.
**EDIT:** By Remy van Dobben de Bruyn's comment below, the name $\mathbb{Z}$-torsion free is a much more standard name than what I called *strong characteristic 0* previously. Requiring only characteristic 0 is insufficient as shown by his example in the comments, but on the other hand requiring $R$ to be a $\mathbb{Q}$-algebra is a tad too strong as it excludes too many valid examples like the ones mentioned above. The point is that it suffices that multiplication by non-zero integers has a trivial kernel rather than non-zero integers being invertible in $R$.
| https://mathoverflow.net/users/1849 | The $1$-dimensional Jacobian Conjecture over $\mathbb{Z}$-torsion free rings | Here is a straightforward commutative algebra argument. We have to show the following:
**Lemma.** *Let $R$ be a $\mathbf Z$-torsion free ring and $f \colon R[x] \to R[x]$ an étale homomorphism of $R$-algebras. Then $f$ is an isomorphism.*
*Proof.* As noted, this is trivial if $R$ is reduced: then $f'(x) \in R[x]^\times = R^\times$ is constant, so $\deg f = 1$ as $R$ is $\mathbf Z$-torsion free and $\tfrac{\partial}{\partial x}\sum\_i a\_ix^i = \sum\_iia\_ix^{i-1}$. We obtain the general case by reduction to the reduced case. Firstly, note that $f$ is defined and étale over some finitely generated subring $A \subseteq R$ (namely the subring generated by the coefficients of $f(x)$ and of $f'(x)^{-1}$), and the result for $f\_A \colon A[x] \to A[x]$ implies the result for $f \colon R[x] \to R[x]$. Thus we may assume that $R$ is Noetherian. (Note that a subring of a $\mathbf Z$-torsion-free ring is still $\mathbf Z$-torsion-free.)
Then the radical $I = \operatorname{nil}(R)$ is finitely generated, so satisfies $I^n = 0$ for some $n \in \mathbf Z\_{>0}$. Note that $R/I$ is again $\mathbf Z$-torsion free: if $x \in R$ and $m \in \mathbf Z\_{>0}$ are such that $mx \in I$, then $m^n x^n = 0$ hence $x^n = 0$ since $R$ is $\mathbf Z$-torsion free. Write $\bar R$ for $R/I$, and $\bar f \colon \bar R[x] \to \bar R[x]$ for the reduction of $f$ modulo $I$.
Then $\bar f$ is still étale, hence an isomorphism since $R$ is reduced. By Nakayama's lemma [Tag [00DV](https://stacks.math.columbia.edu/tag/00DV)(11)], we see that $f \colon R[x] \to R[x]$ is surjective since $\bar f$ is. It is also flat and finitely presented, hence isomorphic to the localisation $R[x] \to R[x]\_e$ at an idempotent $e \in R[x]$ [Tag [00U8](https://stacks.math.columbia.edu/tag/00U8)]. Since $\bar f$ is an isomorphism, we get $e \equiv 1 \pmod I$, i.e. $1-e$ is nilpotent. But $1-e$ is also idempotent, so $1-e=0$ and $e=1$. $\square$
In general, for thickenings $R \twoheadrightarrow R/I$ there is an equivalence of categories between étale $R$-algebras and étale $R/I$-algebras; see [Tag [0BQB](https://stacks.math.columbia.edu/tag/0BQB)] for the case of finite étale algebras, and [Tag [04DZ](https://stacks.math.columbia.edu/tag/04DZ)] for the general statement (in an even more general setting: thickenings are examples of *universal homeomorphisms*). So for these types of results, you usually only need to think about the reduced case.
| 5 | https://mathoverflow.net/users/82179 | 447147 | 180,099 |
https://mathoverflow.net/questions/404849 | 2 | Let $\sigma(x)=\sigma\_1(x)$ be the *classical sum of divisors* of the positive integer $x$.
It is known that
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))}$$
if $q^k n^2$ is an odd perfect number with special prime $q$.
Hence, **if it is known that $n \mid \sigma(n^2)$**, then it follows that
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}$$
since we can compute
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{2n^2/\sigma(q^k)}=\frac{\sigma(q^k)}{2}\cdot\Bigg(\gcd\bigg(1,\frac{\sigma(n^2)}{n}\bigg)\Bigg)^2.$$
But since $n^2 \nmid \sigma(n^2)$, then $n \mid \sigma(n^2)$ implies that
$$n\cdot{\frac{\sigma(q^k)}{2}}=\gcd(n^2,\sigma(n^2))\cdot\gcd(\sigma(q^k),\sigma(n^2))=\Bigg(\gcd(n,\sigma(n^2))\Bigg)^2=n^2$$
from which we obtain
$$\frac{\sigma(q^k)}{2}=n=\frac{\sigma(n^2)}{q^k}.$$
---
**Edit: October 5, 2021 - 1:56 PM Manila time** I have just found a gap in the proof. If $n \mid \sigma(n^2)$, then it does not follow from $n^2 \nmid \sigma(n^2)$ that $\gcd(n^2, \sigma(n^2)) = n$. (In fact, since $\sigma(n^2) = cn$ for some $c > (8/5)n$, then $c$, ~~which is just a proper divisor of $n$~~, must be large.) We then have
$$\gcd(n^2, \sigma(n^2)) = cn$$
which contradicts
$$\gcd(n^2, \sigma(n^2)) = n$$
to several orders of magnitude.
---
(We can then derive the estimates
$$\frac{8n}{5} < q^k < 2n$$
by considering either the resulting abundancy index of $q^k$ or that of $n^2$.)
Note that we then have
$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}=n=\gcd(n^2,\sigma(n^2))$$
under the assumption that $n \mid \sigma(n^2)$.
Here is my question:
>
> Does $n \mid \sigma(n^2)$, if $q^k n^2$ is an odd perfect number?
>
>
>
**MY ATTEMPT**
I tried checking for examples of numbers $n$ satisfying the divisibility constraint
$$n \mid \sigma(n^2)$$
using a Pari-GP script, via [Sage Cell Server](https://sagecell.sagemath.org/?z=eJxLyy_SqLA10lEwNAADHYXMNA0N3_wUjeLM9NxEjYo4I02dCk0FW1sFA02dgqLMvBKNCp20xOQSkEZNIAAAZksS2Q==&lang=gp&interacts=eJyLjgUAARUAuQ==):
```
for(x=2, 1000000, if((Mod(sigma(x^2),x) == 0),print(x,factor(x))))
```
Here is the output:
```
39[3, 1; 13, 1]
793[13, 1; 61, 1]
2379[3, 1; 13, 1; 61, 1]
7137[3, 2; 13, 1; 61, 1]
13167[3, 2; 7, 1; 11, 1; 19, 1]
76921[13, 1; 61, 1; 97, 1]
78507[3, 2; 11, 1; 13, 1; 61, 1]
230763[3, 1; 13, 1; 61, 1; 97, 1]
238887[3, 2; 11, 1; 19, 1; 127, 1]
549549[3, 2; 7, 1; 11, 1; 13, 1; 61, 1]
692289[3, 2; 13, 1; 61, 1; 97, 1]
863577[3, 2; 11, 2; 13, 1; 61, 1]
```
Note that all of the known examples are **odd**.
**Alas, this is where I get stuck!**
| https://mathoverflow.net/users/10365 | Does $n \mid \sigma(n^2)$, if $q^k n^2$ is an odd perfect number? | Let $p^k m^2$ be an odd perfect number with special prime $p$.
It follows that
$$\frac{\sigma(m^2)}{p^k}\cdot\frac{\sigma(p^k)}{2}=m^2.$$
Let $t\_1 = \sigma(m^2)/p^k$, $t\_2 = \sigma(p^k)/2$. It follows that $m^2 = t\_1 t\_2$.
Now define the GCDs
\begin{align\*}
G&=\gcd(\sigma(p^k),\sigma(m^2))=\gcd(\sigma(p^k)/2,\sigma(m^2)/p^k) =\gcd(\sigma(p^k)/2,H) \\
H&=\gcd(m^2,\sigma(m^2))=\gcd(m^2,\sigma(m^2)/p^k)=\gcd(m^2,H) \\
I&=\gcd(m,\sigma(m^2))=\gcd(m,\sigma(m^2)/p^k))= \gcd(m,H).
\end{align\*}
It follows that
$$G = \gcd\left(\sigma(p^k)/2,H\right) = \gcd(t\_1, t\_2)$$
$$H = \gcd(m^2,H) = \gcd\left(t\_1 t\_2, p^k t\_1\right) = t\_1 \gcd(t\_2, p^k) = t\_1$$
$$I = \gcd(m,H) = \gcd\left(\sqrt{t\_1 t\_2}, p^k t\_1\right) = \sqrt{t\_1}\gcd\left(\sqrt{t\_2}, p^k \sqrt{t\_1}\right). \tag{\*}$$
These findings follow from Equation $(\*)$:
>
> **(1)** Since $t\_1$ is a square if and only if $t\_2$ is a square, if $t\_2 = \sigma(p^k)/2$ is a square, then we have
> $$I = \sqrt{H}\cdot\gcd\left(\sqrt{\sigma(p^k)/2},p^k \sqrt{H}\right)
> = \sqrt{H}\cdot\gcd\left(\sqrt{\sigma(p^k)/2}, \sqrt{H}\right)$$
> $$= \sqrt{H}\cdot\sqrt{\gcd(\left(\sigma(p^k)/2,H\right)} = \sqrt{HG},$$
> which checks with the expected result $G \times H = I^2$.
>
>
>
>
> **(2)** Note that neither $t\_1$ nor $t\_2$ could be squarefree in Equation $(\*)$, since it would make the $RHS$ irrational, while the $LHS$ must necessarily be an integer. (Another way to see it is via the following (lengthier) proof: The implication
> $$\left(t\_1 \text{ is squarefree}\right) \implies \left(t\_1 \mid I\right)$$
> holds. But $I \mid t\_1$ since $I=\gcd(m,\sigma(m^2))$ and $t\_1 = \gcd(m^2,\sigma(m^2))$. Hence, we have that $t\_1 \text{ is squarefree } \implies t\_1 = I$. Therefore, $t\_1=I=t\_2$. This implies that $J=\frac{t\_1}{t\_2}=1$. Consequently, by the contrapositive to [Theorem **4.2 (page $5$)** of this paper](https://arxiv.org/abs/2202.08116), $t\_2 = \sigma(p^k)/2$ is *not squarefree*.)
>
>
> **(3)** Lastly, note that if $t\_1$ and $t\_2$ are both not squares and not squarefree, then again $\sqrt{t\_1}$ and $\sqrt{t\_2}$ would be irrational, making the $RHS$ of Equation $(\*)$ irrational, contradicting the fact that the $LHS$ must necessarily be an integer.
>
>
>
I believe this rough sketch for a proof argument that $\sigma(p^k)/2$ (and therefore, $\sigma(m^2)/p^k$) must necessarily be squares holds water. (I thank Professor [Pace Nielsen](https://mathoverflow.net/users/3199) for giving the appropriate hints in this [MO answer](https://mathoverflow.net/a/446925/10365).) In particular, this appears to prove that the Descartes-Frenicle-Sorli Conjecture necessarily holds for an odd perfect number. (See [this paper](http://math.colgate.edu/%7Eintegers/n39/n39.pdf) for a proof of the implication
$$\sigma(p^k)/2 = \square \implies k = 1,$$
where $x = \square$ means that $x$ is a square.)
Note that
$$\gcd(\sigma(p^k),\sigma(m^2)) = G = t\_2 = \sigma(p^k)/2$$
**and, therefore**
$$I = \gcd(m,\sigma(m^2)) = m,$$
follows directly from
$$G \times H = I^2$$
and
$$(\sigma(p^k)/2)\cdot(\sigma(m^2)/p^k)=m^2,$$
since $H=\sigma(m^2)/p^k$.
| -1 | https://mathoverflow.net/users/10365 | 447149 | 180,100 |
https://mathoverflow.net/questions/446799 | 1 | A famous theorem of Beilinson gives a finite, locally free resolution of the diagonal for $\mathbf{CP}^{n}$ by exterior tensor products of locally free sheaves on $\mathbf{CP}^{n}$: for $1 \leq k \leq n$, the $k$th component of the resolution is given by $\mathcal{O}(-k) \boxtimes \Omega^{k}(k)$ where $\Omega^{k}(k) := \bigwedge^{k} \Omega^{1} \otimes \mathcal{O}(k)$ (e.g. see <https://johncalab.github.io/stuff/beilinson.pdf>).
For a projective hypersurface $Y$ (or more generally, for a projective scheme), when does there exist a finite resolution of the diagonal in $Y \times Y$ by (direct sums of) exterior tensor products of locally free sheaves on $Y$? Does such a resolution exist, for example, in the case $Y = V(y^{2}z - x^{3} + xz^{2}) \subset \mathbf{CP}^{2}$, and if so can one explicitly write it down?
| https://mathoverflow.net/users/504744 | Resolution of the diagonal for projective hypersurface | If there is a resolution of the diagonal of a smooth projective variety $Y$ with terms direct sums of $F'\_i \boxtimes F''\_i$, then it is easy to see that the Grothendieck group $K\_0(Y)$ is generated by the classes of $[F'\_i]$ (or of $F''\_i$). To see this just consider the Fourier--Mukai transform given by the structure sheaf of the diagonal. On the one hand, it is the identity functor. On the other hand, any object of the derived category in its image has a resolution by $F'\_i$. This means that any object has such a resolution, hence its class is a linear combination of $[F'\_i]$.
This observation rules out any variety $Y$ whose $K\_0(Y)$ is not of finite rank. For instance, it rules out any curve of positive genus, because for a curve
$$
K\_0(Y) = \mathbb{Z} \oplus \mathrm{Pic}(Y),
$$
and if $\mathrm{Pic}^0(Y) \ne 0$, it is not of finite rank.
| 3 | https://mathoverflow.net/users/4428 | 447152 | 180,102 |
https://mathoverflow.net/questions/447155 | 2 | Let $f: X \to Y$ be a surjective morphism of normal projective varieties with connected fibers (in my case, $X$ is $\mathbb{Q}$- factorial also). Let $E$ be an irreducible $f$-exceptional divisor (i.e. the codim of its image is at least $2$) and $W:= f(E)$. Suppose the irreducible components of $f^{-1}W$ are $E$ and $S$, where codim $S \geq 2$. Suppose, I remove $S$ from $X$ and then put back to $X \setminus S$ all the points of $E \cap S$ which were removed in the process. Let $X^{'}$ be the resulting variety and $f^{'}: X^{'} \to Y$ the restricted morphism.
Is $f^{'}$ still projective?
Apologies if this is elementary, but I could not find any useful reference for this. Thanks in advance!
| https://mathoverflow.net/users/150655 | Removing irreducible components from fibers of projective morphisms | No. If this happens then $X = X' \cup S$ where $X'$ and $S$ are closed subsets, because every projective subset of a projective variety is closed.
This means $X$ is not irreducible, meaning $X$ is not a variety (or, depending on your definitions, at least not normal).
| 3 | https://mathoverflow.net/users/18060 | 447159 | 180,105 |
https://mathoverflow.net/questions/444946 | 6 | Let $f:M\to Y$ be a continuous proper bijective map from a metrizable space $M$ onto a $T\_1$-space $Y$. The properness of $f$ means that for every compact subspace $K\subseteq Y$ the preimage $f^{-1}[K]$ is a compact subset of $M$.
Let $g:H\to Y$ be any continuous map from a compact Hausdorff space $H$ into $Y$.
>
> **Question.** Is the function $f^{-1}\circ g:H\to M$ continuous?
>
>
>
**Remark 1.** The map $f^{-1}\circ g:H\to M$ is sequentially continuous.
*Proof.* Assuming that the map $h:=f^{-1}\circ g:H\to M$ is not sequentially continuous, we can find a sequence $(x\_n)\_{n\in\omega}$ in $H$ that converges to some point $x\in H$ but no subsequence of the sequence $(h(x\_n))\_{n\in\omega}$ converges to $h(x)$. Then the set $\{n\in\omega:h(x\_n)=h(x)\}$ is finite and we can assume that $h(x\_n)\ne h(x)$ and hence $g(x\_n)\ne g(x)$ for all $n\in\omega$. Since $Y$ is a $T\_1$-space, we can replace $(x\_n)\_{n\in\omega}$ by a suitable subsequence and assume that $g(x\_n)\ne g(x\_m)$ for any distinct numbers $n,m$. Then the sequence $(h(x\_n))\_{n\in\omega}$ consists of pairwise distinct points. Since the space $M$ is metrizable, we can replace $(x\_n)\_{n\in\omega}$ by a suitable subsequence and assume that $\{h(x\_n):n\in\omega\}$ is a discrete subspace of $M$.
By the continuity of $g$, the set $K=\{g(x)\}\cup\{g(x\_n)\}\_{n\in\omega}$ is compact in $Y$ and by the properness of $f$, $f^{-1}[K]=\{h(x)\}\cup\{h(x\_n)\}\_{n\in\omega}$ is a compact subset of $M$. Then some subsequence $(h(x\_{n\_k}))\_{k\in\omega}$ converges to some point $z$ of the compact set $f^{-1}[K]$. The choice of the sequence $(x\_n)\_{n\in\omega}$ ensures that $z\ne h(z)$. Then $z=h(x\_m)$ for some $m\in\omega$, which also is not possible as the space $\{h(x\_n):n\in\omega\}$ is discrete. $\square$
**Remark 2.** Requirement on $Y$ to be a $T\_1$-space is essential as shown by the following
**Example.** Let $M$ be the doubleton $\{0,1\}$ with discrete topology and $Y$ be the doubletin $\{0,1\}$ with the $T\_0$-topology $\{\emptyset,\{1\},\{0,1\}\}$. It is clear that the identity map $i:M\to Y$ is continuous and proper. Consider the compact metrizable subspace $H=\{0\}\cup\{\frac1n:n\in\mathbb N\}$ of the real line and the continuous function $g:H\to Y$ defined by $g(0)=0$ and $g(\frac1n)=1$ for all $n\in\mathbb N$. It is easy to see that the function $i^{-1}\circ g:H\to M$ is discontinuous.
| https://mathoverflow.net/users/61536 | The continuity of certain maps on compact Hausdorff spaces | The answer to this question is affirmative.
>
> **Proposition.** Let $p:X\to Y$ be a proper bijective map from a Hausdorff topological space $X$ onto a $T\_1$-space $Y$. Then for every continuous map $f:K\to Y$ from a compact Hausdorff space $K$, the map $p^{-1}\circ f:K\to X$ is continuous.
>
>
>
*Proof.* By the $T\_1$-property of $Y$ and the continuity of $f$, for every $x\in X$ the set $f^{-1}(p(x))$ is closed in $K$. By the compactness of $K$, for every closed set $F\subseteq K$ the image $f[F]$ is compact in $Y$ and by the properness of $p$, the preimage $p^{-1}[f[F]]$ is compact in $X$ and closed in $X$, by the Hausdorff property of $X$. This means that the map $p^{-1}\circ f:K\to X$ is closed and has compact preimages of points. The following lemma implies that $p^{-1}\circ f$ is continuous. $\quad\square$
>
> **Lemma.** A function $f:X\to Y$ from a normal topological space $X$ to a topological space $Y$ is continuous if $f$ is closed, $f[X]$ is compact and for every $y\in Y$, the preimage $f^{-1}(y)$ is a closed subset of $X$.
>
>
>
*Proof.* To see that the function $f:X\to Y$ is continuous, take any point $x\in X$ and any neighborhood $O\_y$ of its image $y=f(x)$ in $Y$.
Let $\mathcal F$ be the family of all closed sets $F\subseteq X$ containing the set $f^{-1}(f(x))$ in its interior in $X$. Since the map $f$ is closed, for every $F\in \mathcal F$ its image $f[F]$ is a closed subset of $Y$. We claim that $\bigcap\_{F\in\mathcal F}f[F]=\{y\}$. Indeed, given any point $z\ne y$, we obtain that $f^{-1}(z)$ is a closed subset of $X$, disjoint with the closed set $f^{-1}(y)$. By the normality of the space $X$, there exists a set $F\in\mathcal F$ such that $F\cap f^{-1}(z)=\emptyset$ and hence $z\notin f[F]$. Since the space $f[X]$ is compact and $\bigcap\_{F\in\mathcal F}f[F]=\{y\}\subseteq O\_y$, we can apply Corollary 3.1.5. from Engelking's "General Topology", and conclude that $f[F]\subseteq O\_y$ for some $F\in\mathcal F$. Then $F$ is a neighborhood of $x$ with $f[F]\subseteq O\_y$, witnessing that the function $f$ is continuous. $\quad\square$
| 1 | https://mathoverflow.net/users/61536 | 447162 | 180,106 |
https://mathoverflow.net/questions/447160 | 1 | Let $G$ be a directed graph and let $P\_i$
be its vertices. Let $A$
be the corresponding adjacency matrix of $G$, i.e. $a\_{i,j}=1$
if and only if there is a directed edge from $P\_i$
to $P\_j$, ($a\_{i,j}=0$
otherwise).
Q. Any characterization for directed graphs whose adjacency matrix admits only 0 as the eigenvalue?
| https://mathoverflow.net/users/84390 | Directed graph whose adjacency matrix admits only 0 as eigenvalue | $0$ is the only eigenvalue of $A$ if and only if $A$ is nilpotent, which is equivalent to $A^n=0$. But $A^n=0$ expresses that for all $i$ and $j$, there is no path of length $n$ from $P\_i$ to $P\_j$. This in turn is equivalent to the graph containing no circles.
| 5 | https://mathoverflow.net/users/18739 | 447163 | 180,107 |
https://mathoverflow.net/questions/447164 | 3 | It is proved [here](https://mathoverflow.net/questions/447134/can-the-equation-1zz2-zn-have-multiple-complex-roots) that the equation $1+z+z^2=z^n$ have no multiple complex roots.
>
> Q. Let us consider the equation $1+z+z^q=z^n$ where $q$ and $n$ are natural numbers with $1<q<n$. Any characterization for the pair $(q,n)$ such that the equation $1+z+z^q=z^n$ have no multiple complex roots?
>
>
>
| https://mathoverflow.net/users/84390 | Can the equation $1+z+z^q=z^n$ have multiple complex roots $z$? | The roots are always simple, this follows without further reasoning from the old paper [On the irreducibility of certain trinomials and quadrinomials](https://doi.org/10.7146/math.scand.a-10593) by Ljunggren. Set $f(z)=z^n-1-z^p-z^q$. Then Ljunggren proves (Theorem 1) that $f(z)=Q(z)g(z)$, where the roots of $Q$ are roots of unity (so $Q=1$ if there are no such roots), and where $g(z)\in\mathbb Q[z]$ is irreducible none of its roots is a root of unity. As an irreducible polynomial, $g$ has no multiple roots. Furthermore (Theorem 2), Ljunggren shows that $Q$ has simple roots either, which settles the question.
Remark: In case Ljunggren's paper is not accessible, [this review at zbMATH](https://zbmath.org/scans/095/013.gif) contains the full statement of Ljunggren's results needed here.
| 7 | https://mathoverflow.net/users/18739 | 447168 | 180,108 |
https://mathoverflow.net/questions/445930 | 4 | Let $A$ be a Hopf algebra (over a field). Consider a unital subalgebra $B\subseteq A$ with $\Delta(B)\subseteq B\otimes A$. Put
$$B^+:= B\cap \ker(\epsilon).$$
It can be shown that $B^+$ is a two-sided coideal of $A$, so that $AB^+$ is also a two-sided coideal.
**Question**: If $x\in AB^+$, is it true that $S(x) \in B^+ A?$
Of course, it suffices to show that $S(B^+)\subseteq B^+ A$ but I am not sure how to show this.
Thanks in advance for your help/comments/suggestions!
| https://mathoverflow.net/users/216007 | Hopf algebra and coideal question | This is indeed true, and it forms part of the proof of a result known as ***Koppinen's Lemma***.
The argument is as follows: Let $\{x\_j\,|\, j \in J\}$ be a basis of $B^+$, and take an element $x \in B^+$. It follows from the counit axiom of a Hopf algebra, and linear independence of our basis, that we can write the coproduct of $x$ as
$$
\Delta(x) = x \otimes 1 + \sum\_{j} x\_j \otimes y\_j, \textrm{ for some } y\_j \in A^+.
$$
Applying $\epsilon \circ \eta = m \circ (S \otimes \mathrm{id})$ to this expression, we get the identity
$$
S(x) = - x\_jS(y\_j) \in B^+A.
$$
As I wrote above, this is a step in Koppinen's Lemma, which states, under the assumption of a bijective antipode, that
$$
S(AB^+) = B^+A.
$$
An elementary proof (from which I took the above argument) along with a reference to the original article of Koppinin, can be found in this [article](https://link.springer.com/article/10.1007/BF02810683). (An open access copy can also be found on Schneider's LMU Munich website.)
It is also interesting to note that if $B^+A = AB^+$, then Koppinen's Lemma implies that the Hopf algebra structure of $A$ descends to the quotient $A/B^+A$.
| 3 | https://mathoverflow.net/users/3072 | 447174 | 180,111 |
https://mathoverflow.net/questions/447202 | 4 | In the proof of Corollary 5.7 in the following link:
<https://arxiv.org/pdf/1610.05200.pdf>
the author claims that $E[\lVert X\rVert^2] \lesssim E[\lVert X\rVert ]^2$ for the standard normal distribution on $\mathbb{R}^n$.
I wonder if this result is still valid for cenetered Gaussian distributions on any infinite dimensional separable Banach space $V$ .
Moreover, the covariance for the measure is a symmetric nonnegative bilinear form on $V^\*$. If the above inequality is still valid on $v$, how would the covariance affect it?
Could anyone please provide any information or reference?
| https://mathoverflow.net/users/56524 | For centered Gaussian measures, is $E[\lVert X\rVert^2] \lesssim E[\lVert X\rVert ]^2$ true in infinite dimensions as well? | $\newcommand{\si}{\sigma}$Yes, we have
\begin{equation\*}
E\|X\|^2\le c(E\|X\|)^2 \tag{1}\label{1}
\end{equation\*}
for
\begin{equation\*}
c:=1+2\pi
\end{equation\*}
and any centered Gaussian random vector $X$ in any separable Banach space $V$.
This follows from the [Borell--Tsirelson--Ibragimov--Sudakov inequality](https://en.wikipedia.org/wiki/Borell%E2%80%93TIS_inequality):
\begin{equation\*}
P(\|f\|\_T\ge E\|f\|\_T+u)\le\exp-\frac{u^2}{2\si\_T^2}
\end{equation\*}
for real $u\ge0$, where $(f\_t)\_{t\in T}$ is a centered Gaussian process, $\|f\|\_T:=\sup\_{t\in T}|f\_t|$, and
\begin{equation\*}
\si\_T:=\sqrt{\sup\_{t\in T}Ef\_t^2}.
\end{equation\*}
Indeed, let $T=B^\*$, the unit ball in the dual $V^\*$ to $V$, with $f\_t:=t(X)$ for $t\in B^\*$, so that $\|f\|\_T=\|X\|$ and
\begin{equation\*}
\si\_T=\si:=\sqrt{\sup\_{t\in B^\*}Et(X)^2}=\sqrt{\frac\pi2\,\sup\_{t\in B^\*}(E|t(X)|)^2}
\le\sqrt{\frac\pi2}\,E\|X\|. \tag{2}\label{2}
\end{equation\*}
So, for real $u\ge0$,
\begin{equation\*}
P(\|X\|\ge E\|X\|+u)\le\exp-\frac{u^2}{2\si^2}.
\end{equation\*}
(More immediately, the latter inequality is inequality (4.4) in [Ledoux's book](https://link.springer.com/content/pdf/10.1007/BFb0095676.pdf).)
So,
\begin{equation\*}
\begin{aligned}
E\|X\|^2&=E\int\_0^\infty 2w\,dw\,1(\|X\|\ge w) \\
&=\int\_0^\infty 2w\,dw\,P(\|X\|\ge w) \\
&\le\int\_0^{E\|X\|} 2w\,dw+\int\_{E\|X\|}^\infty 2w\,dw\,\exp-\frac{(w-E\|X\|)^2}{2\si^2} \\
&=(E\|X\|)^2+2\si^2+\sqrt{2\pi}E\|X\|\si.
\end{aligned}
\end{equation\*}
Now \eqref{1} follows from \eqref{2}. $\quad\Box$
| 6 | https://mathoverflow.net/users/36721 | 447209 | 180,116 |
https://mathoverflow.net/questions/447213 | -1 | Let $X$ and $A$ be compact Polish spaces endowed with Borel $\sigma$-algebras. Let $\mathcal{A} = X\times \mathcal{B}(A)$ be the $\sigma$-algebra consisting of cylinders whose projections on $A$ are Borel sets. Let $u: X\times A\rightarrow\mathbb{R}$ be a continuous function. Let $\pi\in \Delta(X\times A)$ be a probability measure.
Let $\mathcal{F}\equiv \{f:A\rightarrow A|\;f \text{ is } \mathcal{A} \text{ measurable}\}$ denote the collection of all measurable functions from $A$ to $A$.
**Question:** Is it true that
$$\sup\_{f\in \mathcal{F}}\int\_{X\times A} E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] d\pi(x,a) = \int\_{X\times A}\; \sup\_{f\in \mathcal{F}}E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] d\pi(x,a)?$$
**An economic interpretation:** As Michael Greinecker pointed out in the comment, if we interpret $u$ as the a payoff function, $x\in X$ as an unknown state, $a\in A$ as an action, and $\pi$ as a system of stochastic action recommendations, then the claim I am trying to establish can be interpreted as saying that *choosing an optimal contingent plan ex ante leads to the same expected utility as maximizing for each recommended action at the interim stage*.
---
---
**My thoughts so far:**
My first instinct is to invoke the Measurable Selection Theorem, which would be similar to the arguments in Theorem 14.60 of Rockafella and Wets' "Variational Analysis". However, I do not know how to work with the conditional expectations in the expression above, which is itself a random variable that is only unique almost surely.
Specifically, to use the Measurable Selection Theorem as Rockafellar and Wets did, I would need to somehow establish that
$$
\Big\{k: E \big[u\big(x,k\big) \big| a\big] \ge c \Big\}
$$
is a closed set for each $a\in A$ and $c\in \mathbb{R}$, but I'm not sure why that would be true, especially since conditional expectation is only pinned down for **almost all** $a\in A$.
Any pointers would be greatly appreciated!
| https://mathoverflow.net/users/30374 | Conditional expectation: commuting integration and supremum | Note: This answer used to be a counterexample that missed the mark.
The way to get around he definitional issues with conditional expectations is to work with [regular conditional probabilities in product form](https://doi.org/10.1214/aop/1176993081), which guarantee that all conditional expectations fit together well. In particular, there exists a measurable function (or a transition probability, essentially the same thing) $\kappa:A\to X$ such that for $\pi\_A$ the $A$-marginal of $\pi$, we have
for every Borel set $E\subseteq X\times A$ that
$$\pi(E)=\int\int 1\_E(x,a)~\mathrm d\kappa\_a(x)~\mathrm d\pi\_A(a).$$
The function $\kappa$ is unique up to $\pi\_A$-null sets.
That way, one can show that
$$\max\_{f\in \mathcal{F}}\int\_{X\times A} u\big(x,f(a)\big)~\mathrm d\pi(x,a) =\max\_{f\in \mathcal{F}}\int\_A \int\_X u\big(x,f(a)\big)~\mathrm d\kappa\_a(x) ~\mathrm d\pi\_A(a)$$ $$= \int\_A \max\_{f\in \mathcal{F}} \int\_X u\big(x,f(a)\big)~\mathrm d\kappa\_a(x) ~\mathrm d\pi\_A(a).$$
The left side is trivially no larger than the right side. For the other direction, you show that the correspondence that associates to each $a$ the argmax of $\int\_X u\big(x,\cdot\big)~\mathrm d\kappa\_a(x)$ is measurable with nonempty compact values. So you can use the Kuratowski-Ryll-Nardzewski measurable selection theorem to turn a solution for the problem on the right to a solution of the problem on the left, which must, therefore, give the same value.
The argument does not require $X$ to be compact, any Polish space will do, and $u$ need not be continuous in $X$, any bounded (or integrably bounded) Carathéodory function will do.
| 2 | https://mathoverflow.net/users/35357 | 447222 | 180,121 |
https://mathoverflow.net/questions/445065 | 2 | There seems to be many valid ways of generalizing the notion of the spectral radius $\rho(A)$ of a complex matrix $A$ to spectral radii of multiple operators. I am wondering if there is an abstract theory of what it means to be a multi-spectral radius $\rho(A\_1,\dots,A\_r)$ of complex matrices $A\_1,\dots,A\_r$.
Example 0: Suppose that $A\_1,\dots,A\_r$ are complex matrices and $1\leq p<\infty$. Then define
$$\rho\_p(A\_1,\dots,A\_r)=\lim\_{n\rightarrow\infty}\big(\sum\_{i\_1,\dots,i\_n\in\{1,\dots,r\}}\|A\_{i\_1}\dots A\_{i\_r}\|^p\big)^{1/(pn)}.$$
We can call this notion of the spectral radius the $L\_p$-spectral radius.
Theorem: $\rho\_2(A\_1,\dots,A\_r)^2=\rho(A\_1\otimes\overline{A\_1}+\dots+A\_r\otimes\overline{A\_r})$. Here, $\overline{A}=(A^\*)^T=(A^T)^\*$. Alternatively, define the completely positive linear mapping $\Phi(A\_1,\dots,A\_r):M\_n(\mathbb{C})\rightarrow M\_n(\mathbb{C})$ by setting $\Phi(A\_1,\dots,A\_r)(X)=A\_1XA\_1^\*+\dots+A\_rXA\_r^\*$. Then $\rho\_2(A\_1,\dots,A\_r)^2=\rho(\Phi(A\_1,\dots,A\_r))$.
It seems like $\rho\_2(A\_1,\dots,A\_r)$ is the best way of generalizing the notion of a spectral radius to multiple operators if I had to choose one notion of a spectral radius.
Example 1: Let $\mathfrak{k}=(k\_n)\_{n=0}^{\infty}$ be a sequence of numbers in the set $\{1,\dots,r\}$. Then define $\rho\_\mathfrak{k}(A\_1,\dots,A\_r)=\limsup\_{n\rightarrow\infty}\|A\_{k\_0}\dots A\_{k\_n}\|^{1/n}.$
Example 2: Define $\rho\_{2,1}(A\_1,\dots,A\_r)$ to be the largest value of $(|z\_1^2|+\dots+|z\_r^2|)^{-1/2}$ where $I-(z\_1A\_1+\dots+z\_rA\_r)$ is not invertible. Then $\rho\_{2,1}$ is another generalized notion of a spectral radius. More generally, if $\|\cdot\|$ is a complex norm on $\mathbb{C}^n$, then define $\rho\_{2,\|\cdot\|}(A\_1,\dots,A\_r)$ to be the largest value of
$\|(z\_1,\dots,z\_r)\|^{-1}$ where $I-(z\_1A\_1+\dots+z\_rA\_r)$ is not invertible.
If $\rho$ is a multi-spectral radius function, then it seems like $\rho$ should satisfy properties such as log-plurisubharmonicity, continuity, homogeneity of degree 1, invariance under joint-similarity, and a few mundane properties. For some particularly nice multi-spectral radius functions $\rho$, the value $\rho(A\_1,\dots,A\_r)$ only depends on the completely positive superoperator $\Phi(A\_1,\dots,A\_r)$. And if a multi-spectral radius function $\rho$ only depends on the superoperator $\Phi(A\_1,\dots,A\_r)$, then I would imagine that this multi-spectral radius easily generalizes to bounded operators between Hilbert spaces.
What would be a good axiomatization for multi-spectral radius functions? Is there an axiomatization for what is meant by a multi-spectrum of multiple operators? Clearly, one may consider
the spectrum of $\Phi(A\_1,\dots,A\_r)$ as a multi-spectrum of $A\_1,\dots,A\_r$. However, if $z\_1,\dots,z\_r$ are complex numbers with $|z\_1|^2+\dots+|z\_r|^2=1$ and where $\rho(z\_1A\_1+\dots+z\_rA\_r)$ is maximized, then the spectrum $\sigma(z\_1A\_1+\dots+z\_rA\_r)$ may be considered as another notion of a multi-spectrum.
I am interested in these generalized spectral radii since I have originally used a generalized notion of the spectral radius for cryptocurrency research and development, but such generalized spectral radii seem applicable for other machine learning applications.
| https://mathoverflow.net/users/22277 | Is there an abstract theory of multi-spectral radii? | I claim that there is a somewhat abstract notion of a multi-spectral radius and that there is probably an abstract theory behind this abstract notion. I will try to justify this abstract multi-spectral radius by showing that it captures the specific examples of multi-spectral radii that I have mentioned in the question and that the simplest examples of these multi-spectral radii are reasonable mathematical objects. With that being said, there are notions of a multi-spectral radius that I have not shown to fit within this framework, so more research on this topic is needed.
Suppose that $A$ is a complex Banach algebra. We say that a function
$\rho:A^r\rightarrow[0,\infty)$ is a multi-spectral radius if there is an isometric embedding $\iota:A\rightarrow B$ along with a bounded subset $\mathcal{C}\subseteq B^r$ where
1. $x\_j\iota(a)=\iota(a)x\_j$ whenever $a\in A,(x\_1,\dots,x\_r)\in\mathcal{C}$,
2. if $(x\_1,\dots,x\_r)\in\mathcal{C}$, then $(\lambda\_1x\_1,\dots,\lambda\_rx\_r)\in\mathcal{C}$ whenever $|\lambda\_j|=1$ for $1\leq j\leq r$, and
3. $$\rho(a\_1,\dots,a\_r)=\rho\_{\iota,\mathcal{C}}(a\_1,\dots,a\_r)=\sup\_{(x\_1,\dots,x\_r)\in\mathcal{C}}\rho(x\_1\iota(a\_1)+\dots+x\_r\iota(a\_r))$$ whenever $a\_1,\dots,a\_r\in A$.
One may also want to require that if $(x\_1,\dots,x\_r)\in\mathcal{C}$, then $(\lambda\_1x\_1,\dots,\lambda\_rx\_r)\in\mathcal{C}$ whenever $|\lambda\_j|=1$ for $1\leq j\leq r$, but this condition is not necessary. One can show that if $a,b\in B$, then the mapping from $\mathbb{C}$ to $[-\infty,\infty)$ defined by
$\lambda\mapsto \ln(\rho(a+\lambda b))$ is subharmonic, so by the maximum principle,
$\max\{\rho(\lambda\_1x\_1\iota(a\_1)+\dots+\lambda\_rx\_r\iota(a\_r)):|\lambda\_1|=\dots=|\lambda\_r|=1\}=\max\{\rho(\lambda\_1x\_1\iota(a\_1)+\dots+\lambda\_rx\_r\iota(a\_r)):|\lambda\_1|\leq 1,\dots,|\lambda\_r|\leq 1\}.$
The $L\_1$-spectral radius can be characterized in terms of our framework.
Theorem: $\rho\_1(a\_1,\dots,a\_r)$ is the maximum value of $\rho(x\_1\iota(a\_1)+\dots+x\_r\iota(a\_r))$ where $\iota:A\rightarrow B$ is an isometric embedding of Banach algebras, and $\|x\_j\|\leq 1$ for $1\leq j\leq r$.
The proof of the above result is not too hard, and I have given a proof of the above result in [this answer](https://mathoverflow.net/questions/323445/spectral-radius-for-multiple-linear-operators).
We say that a multi-spectral radius $\rho$ is unitary invariant if
$\rho(a\_1,\dots,a\_r)=\rho(b\_1,\dots,b\_r)$ whenever there is an $n\times n$-unitary matrix $(u\_{i,j})\_{i,j}$ where $b\_j=\sum\_{i=1}^ru\_{i,j}a\_i$ for $1\leq j\leq r$. The following lemma is a standard result from quantum information theory.
Lemma: Suppose that $A\_1,\dots,A\_r,B\_1,\dots,B\_r\in M\_n(\mathbb{C})$. Then $\Phi(A\_1,\dots,A\_r)=\Phi(B\_1,\dots,B\_r)$ if and only if there is an $r\times r$-unitary matrix $(u\_{i,j})\_{i,j}$ where $B\_j=\sum\_{i=1}^ru\_{i,j}A\_i$ for $1\leq j\leq r$.
Therefore, a multi-spectral radius $\rho:M\_n(\mathbb{C})^r\rightarrow[0,\infty)$ is unitary invariant if and only if $\rho(A\_1,\dots,A\_r)=\rho(B\_1,\dots,B\_r)$ whenever
$\Phi(A\_1,\dots,A\_r)=\Phi(B\_1,\dots,B\_r)$. The continuous unitary invariant multi-spectral radii are completely determined by the mapping $\Phi(A\_1,\dots,A\_r)\mapsto\rho(A\_1,\dots,A\_r)$ where $\Phi(A\_1,\dots,A\_r)$ is completely positive and trace preserving (a completely positive trace preserving map is known as a quantum channel).
The following easy lemmas show that how we can always upgrade a multi-spectral radius to a unitary invariant multi-spectral radius.
Lemma: Let $A$ be an algebra over a field $K$. Suppose that $(a\_1,\dots,a\_r),(b\_1,\dots,b\_r),(x\_1,\dots,x\_r),(y\_1,\dots,y\_r)\in A^r$. Let $(u\_{i,j})\_{i,j},(v\_{i,j})\_{i,j}\in M\_r(K)$ be inverse matrices. Suppose that
$a\_k=\sum\_{i=1}^ru\_{i,k}b\_i$ and $x\_k=\sum\_{j=1}^rv\_{k,j}y\_j$ for $1\leq k\leq r$. Then $$\sum\_{k=1}^ra\_kx\_k=\sum\_{i=1}^rb\_iy\_j.$$
Lemma: Suppose that $K$ is a field and $A$ is an algebra over $K$. Let $(u\_{i,k})\_{i,k}\in M\_r(K)$. Suppose furthermore that $a\_1,\dots,a\_r,b\_1,\dots,b\_r,x\_1,\dots,x\_r,y\_1,\dots,y\_r\in A$ and
$a\_k=\sum\_{i=1}^ru\_{i,k}b\_i$ for $1\leq k\leq r$ and $x\_i=\sum\_{k=1}^ru\_{i,k}y\_k$ for $1\leq k\leq r$. Then $\sum\_{k=1}^ra\_ky\_k=\sum\_{i=1}^rb\_ix\_i$.
Proposition: Let $A,B$ be Banach algebras. Let $\iota:A\rightarrow B$ be an isometric embedding. Suppose that $\mathcal{C}\subseteq B^r$ is a bounded subset with $x\_j\iota(a)=\iota(a)x\_j$ for $1\leq j\leq r$. Let $\mathcal{D}$ be the collection of all tuples $(y\_1,\dots,y\_r)$ where there is some $r\times r$-unitary matrix $(u\_{i,j})\_{i,j}$ and $(x\_1,\dots,x\_r)\in\mathcal{C}$ where
$y\_j=\sum\_{i=1}^ru\_{i,j}x\_i$ for $1\leq i\leq r$. Then
$$\rho\_{\iota,\mathcal{D}}(x\_1,\dots,x\_r)=\sup\{\rho\_{\iota,\mathcal{C}}(\sum\_ju\_{1,j}x\_j,\dots,\sum\_ju\_{r,j}x\_j)\mid (u\_{i,j})\_{i,j}\in U(r)\}.$$
By using the following version of Holder's inequality that can be proven using the classical Holder's inequality, we can show that the $L\_2$-spectral radius is a multi-spectral radius.
Theorem: $\rho(A\_1\otimes B\_1+\dots+A\_r\otimes B\_r)\leq
\rho\_p(A\_1,\dots,A\_r)\cdot\rho\_q(B\_1,\dots,B\_r)$ whenever $p,q\in(1,\infty)$ and $\frac{1}{p}+\frac{1}{q}=1$.
As a consequence, if $d\geq n$ and $A\_1,\dots,A\_r\in M\_n(\mathbb{C})$, then
$$\rho\_2(A\_1,\dots,A\_r)=\max\_{(X\_1,\dots,X\_r)\in M\_d(\mathbb{C})}\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho\_2(X\_1,\dots,X\_r)}.$$
We have another construction that allows us to show that the $L\_p$-spectral radius is a multi-spectral radius for $1\leq p<\infty$. Suppose now that $1\leq p<\infty$.
Now, let $A$ be a Banach algebra. Let $x\_1,\dots,x\_r$ be non-commutating variables. Let $B$ be the collection of all sums of the form
$\sum\_{k=0}^n\sum\_{i\_1,\dots,i\_k\in\{1,\dots,r\}}a\_{i\_1,\dots,i\_k}x\_{i\_1}\dots x\_{i\_k}$. We observe that for $p>1$ the Banach space $\ell^p$ indexed with the natural numbers cannot be endowed with a convolution operation since $(1/n)\_{n=1}^{\infty}\*(1/n)\_{n=1}^{\infty}=(+\infty)\_{n=1}^\infty$. We can give $B$ a norm that combines the $\ell^p$ and the $\ell^1$ norms that makes the completion of $B$ into a Banach algebra.
Then give $B$ the norm $$\|\sum\_{k=0}^n\sum\_{i\_1,\dots,i\_k\in\{1,\dots,r\}}a\_{i\_1,\dots,i\_k}x\_{i\_1}\dots x\_{i\_k}\|=\sum\_{k=0}^n\|(a\_{i\_1,\dots,i\_k})\_{i\_1,\dots,i\_k}\|\_p.$$
Give $B$ the multiplication defined by bilinearity along with the condition that $$(a\cdot x\_{i\_1}\dots x\_{i\_m})\cdot (b\cdot x\_{j\_1}\dots x\_{j\_n})=
ab\cdot x\_{i\_1}\dots x\_{i\_m}x\_{j\_1}\dots x\_{j\_n}.$$
In other words, each element in $A$ commutes with each variable $x\_j$, but we do not impose any other version of commutativity.
$B$ is submultiplicative: Let $$u=\sum\_{j=0}^\infty\sum\_{i\_1,\dots,i\_j\in\{1,\dots,r\}}a\_{i\_1,\dots,i\_j}x\_{i\_1}\dots x\_{i\_j}$$ and let $$v=\sum\_{j=0}^\infty\sum\_{i\_1,\dots,i\_j\in\{1,\dots,r\}}a\_{i\_1,\dots,i\_j}x\_{i\_1}\dots x\_{i\_j}$$ where only finitely many terms of these 'non-commutative polynomials' are non-zero.
Then
$$\|u\cdot v\|$$
$$=\|(\sum\_{k=0}^\infty\sum\_{i\_1,\dots,i\_k\in\{1,\dots,r\}}a\_{i\_1,\dots,i\_k}x\_{i\_1}\dots x\_{i\_k})\cdot (\sum\_{k=0}^\infty\sum\_{i\_1,\dots,i\_k\in\{1,\dots,r\}}b\_{i\_1,\dots,i\_k}x\_{i\_1}\dots x\_{i\_k})\|$$
$$=\|\sum\_{k=0}^{\infty}\sum\_{j=0}^k\sum\_{i\_1,\dots,i\_j\in\{1,\dots,r\}}\sum\_{i\_{j+1},\dots,i\_k}a\_{i\_1,\dots,i\_j}b\_{i\_{j+1},\dots,i\_k}x\_{i\_1}\dots x\_{i\_k}\|$$
$$\leq\sum\_{k=0}^{\infty}\sum\_{j=0}^k\|\sum\_{i\_1,\dots,i\_k\{1,\dots,r\}}a\_{i\_1,\dots,i\_j}b\_{i\_{j+1},\dots,i\_k}x\_{i\_1}\dots x\_{i\_k}\|$$
$$=\sum\_{k=0}^{\infty}\sum\_{j=0}^k\|(a\_{i\_1,\dots,i\_j}\cdot b\_{i\_{j+1},\dots, i\_k})\_{i\_1,\dots,i\_k\in\{1,\dots,r\}}\|\_p$$
$$\leq\sum\_{k=0}^{\infty}\sum\_{j=0}^k\|(a\_{i\_1,\dots,i\_j})\_{i\_1,\dots,i\_j\in\{1,\dots,r\}}\|\_p\cdot \|(b\_{i\_{j+1},\dots,i\_k})\_{i\_{j+1},\dots,i\_k\in\{1,\dots,r\}}\|\_p$$
$$=\sum\_{j=0}^\infty\|(a\_{i\_1,\dots,i\_j})\_{i\_1,\dots,i\_j\in\{1,\dots,r\}}\|\_p\cdot\sum\_{k=0}^\infty\|(b\_{i\_1,\dots,i\_k})\_{i\_1,\dots,i\_k\in\{1,\dots,r\}}\|\_p=\|u\|\cdot\|v\|.$$
Therefore, the completion $\overline{B}$ of $B$ is a Banach algebra, and the original Banach algebra $A$ embeds into $\overline{B}$. In this case, we simply have
$\rho\_p(a\_1,\dots,a\_r)=\rho(a\_1x\_1+\dots+a\_rx\_r)$.
One should be able to generalize the above construction to most sensible notions of a multi-spectral radius.
**Other examples:**
In order for our notion of a multi-spectral radius to be sensible, one would expect that the functions $\rho\_{\iota,\mathcal{C}}$ would be coherent and interesting for the simplest possible cases of $\iota,\mathcal{C}$. For example, if $\iota:A\rightarrow A$ is the identity function and $1\leq p\leq\infty$, and $\mathcal{C}$ is the unit ball in $\mathbb{C}^r$ with respect to the $p$-norm, then one should expect for $\rho\_{\iota,\mathcal{C}}$ to be about as reasonable of a function as the $L\_p$-spectral radii, and experimental computations indicate that this is indeed the case.
Define a mapping $F\_{\iota,\mathcal{C},a\_1,\dots,a\_r}:\mathcal{C}\rightarrow[0,\infty)$ by $F\_{\iota,\mathcal{C},a\_1,\dots,a\_r}(x\_1,\dots,x\_r)=
\rho(x\_1\iota(a\_1)+\dots+x\_r\iota(a\_r))$. Experimental computations suggest that the local maxima $(x\_1,\dots,x\_r)$ for the function $F\_{\iota,\mathcal{C},a\_1,\dots,a\_r}$ tend to resemble a sort of conjugate of $a\_1,\dots,a\_r$.
Let $\iota\_n:M\_n(\mathbb{C})\rightarrow M\_n(\mathbb{C})$ be the identity mapping, and let $\mathcal{L}\_{r;p}=\{(\lambda\_1,\dots,\lambda\_r)\in \mathbb{C}^r:\|(\lambda\_1,\dots,\lambda\_r)\|\_p=1\}$.
In some of my experiments with $A\_1,\dots,A\_r\in M\_n(\mathbb{R})$ and in all of my experiments with $A\_1,\dots,A\_r\in M\_n(\mathbb{C})$ that are Hermitian or real symmetric, when I computed $\lambda\_1,\dots,\lambda\_r$ locally maximizes $F\_{\iota\_n,S\_1^r,A\_1,\dots,A\_r}$, then one can find a $\lambda\in S\_1$ and $e\_1,\dots,e\_r\in\{-1,1\}$ where
$\lambda\_j=\lambda\cdot e\_j$ for $1\leq j\leq r$. A similar phenomenon holds when I locally maximized $F\_{\iota\_n,\mathcal{L}\_{r;p},A\_1,\dots,A\_r}$ for $1\leq p\leq\infty$ even though this phenomenon seems to break down as $p$ gets close to $1$ and it holds better for Hermitian matrices than it does for non-symmetric real matrices.
My computer experiments indicate that if we locally maximize $F\_{\iota,\mathcal{C},a\_1,\dots,a\_r}$, then as $\mathcal{C}$ better approximates $A$, the local maxima $(x\_1,\dots,x\_r)$ will become more and more similar to a conjugate version of $(a\_1,\dots,a\_r)$. On the other hand, if $\mathcal{C}$ is too complicated and has too much room to work with, then the local maxima $(x\_1,\dots,x\_r)$ will again poorly represent the elements in $A$. Therefore, in order to best represent the conjugates of the elements in $A$, it is best if $\mathcal{C}$ is a little bit simpler than $A$.
Let $\iota\_{r;n,d}:M\_n(\mathbb{C})\rightarrow M\_{n\times d}(\mathbb{C})$ be the algebra homomorphism defined by $\iota\_{r;n,d}(A)=A\otimes I\_d$. Let $\mathcal{C}\_{r;n,d}$ be the collection of all tuples $(I\_n\otimes X\_1,\dots,I\_n\otimes X\_r)$ where $\rho\_2(X\_1,\dots,X\_r)=1$.
Theorem: Suppose that $A\_1,\dots,A\_r,B\_1,\dots,B\_r$ are $n\times n$-complex matrices where $A\_1,\dots,A\_r$ do not have a common invariant subspace. Suppose furthermore that $\rho\_2(A\_1,\dots,A\_r)>0,\rho\_2(B\_1,\dots,B\_r)>0$. Then
$\rho(A\_1\otimes B\_1+\dots+A\_r\otimes B\_r)=\rho\_2(A\_1,\dots,A\_r)\rho\_2(B\_1,\dots,B\_r)$ if and only if there is some $\lambda$ and invertible $C$ where $B\_j=\overline{\lambda\cdot C\cdot A\_j\cdot C^{-1}}$ for $1\leq j\leq r$.
See [this answer](https://mathoverflow.net/a/424866/22277) or [this link](https://circcashcore.com/2022/07/24/lower-dimensional-approximations-to-the-l_2-spectral-radius/) for proofs that I gave of the above result.
From the above result, we see that if $I\_n\otimes\overline{X\_1},\dots,I\_n\otimes\overline{X\_r}\in M\_{n\times n}(\mathbb{C})$ globally maximizes
$F\_{\iota\_{r;n,n},\mathcal{C}\_{r;n,n},A\_1,\dots,A\_r}$ and $A\_1,\dots,A\_r$ have no common invariant subspace, then there are $C,\lambda$ where $X\_j=\lambda CA\_jC^{-1}$ for $1\leq j\leq r.$
If $A\_1,\dots,A\_r\in M\_n(\mathbb{C})$ does not have a common invariant subspace and $I\_n\otimes\overline{X\_1},\dots,I\_n\otimes\overline{X\_r}\in M\_{n\times d}(\mathbb{C})$ locally maximizes
$F\_{\iota\_{r;n,d},\mathcal{C}\_{r;n,d},A\_1,\dots,A\_r}$, then the matrices $X\_1,\dots,X\_r$ will (up-to-similarity and a constant factor) resemble $A\_1,\dots,A\_r$. For example, if $A\_1,\dots,A\_r$ are all real, complex symmetric, real symmetric, Hermitian, real positive semidefinite, complex positive semidefinite, quaternionic, rank $\leq k$, etc, and $I\_n\otimes\overline{X\_1},\dots,I\_n\otimes\overline{X\_r}$ locally maximizes $F\_{\iota\_{r;n,d},\mathcal{C}\_{r;n,d},A\_1,\dots,A\_r}$, then one will often be able to find a constant $\lambda$ and invertible matrix $C$ where
$Y\_j=\lambda CX\_rC^{-1}$ satisfy those properties respectively. Furthermore, one will often be able to find matrices $R,S$ where
$Y\_j=RA\_jS$ for $1\leq j\leq r$. In this case, $RS=I\_d$ and $P=SR$ will be a (non-orthogonal) projection matrix. Define linear operators $F,G:M\_n(\mathbb{C})\rightarrow M\_n(\mathbb{C})$ by setting $F(X)=\sum\_{k=1}^rA\_kX(PA\_kP)^\*$ and $G(X)=\sum\_{k=1}^rA\_k^\*XPA\_kP$ (here $F=G^\*$). Define $U\_0=I\_n,V\_0=I\_n$ and set $U\_{n+1}=F(U\_n)/\|F(U\_n)\|,V\_{n+1}=G(V\_n)/\|G(V\_n)\|$ for $n\geq 0$. Then $U\_n,V\_n$ experimentally converge to positive semidefinite matrices $U,V$, the dominant eigenvectors of $F$ and $G$. It seems like the strategy that the optimization algorithm chose for locally maximizing the spectral radius was to make the dominant eigenvalues of $F,G$ positive semidefinite matrices of rank $d$, but the best way to retain the positive semidefiniteness of the dominant eigenvectors of $F,G$ is to make the operators $PA\_kP$ closely related to the operators $A\_k$. It seems like the reason this strategy works is that in order for a spectral radius of a matrix $A$ to be large, the matrix $A$ should be designed to maximize a particular eigenvalue, and by making the operators $PA\_kP$ related to $A\_k$, we can maximize the spectral radius of $F,G$. Since the local maximum values of $F\_{\iota\_{r;n,d},\mathcal{C}\_{r;n,d},A\_1,\dots,A\_r}$ are closely related to the tuples $(A\_1,\dots,A\_r)$ themselves, I would regard the multi-spectral radius $\rho\_{\iota\_{r;n,d},\mathcal{C}\_{r;n,d}}$ as a legitimate generalization of the notion of the spectral radius to multiple operators which I call the $L\_{2,d}$-spectral radius $\rho\_{2,d}$.
Other multi-spectral radii $\rho\_{\iota,\mathcal{C}}$ are probably reasonably well-behaved, but more computer experiments are needed to verify whether other multi-spectral radii $\rho\_{\iota,\mathcal{C}}$ behave nearly as well as $\rho\_{\iota\_{r;n,d},\mathcal{C}\_{r;n,d}}$.
One can find more details on $\rho\_{\iota\_{r;n,d},\mathcal{C}\_{r;n,d}}$ at [my site here](https://circcashcore.com/2022/07/24/lower-dimensional-approximations-to-the-l_2-spectral-radius/), and [here is another page](https://circcashcore.com/2022/08/08/measuring-block-cipher-round-function-security-using-spectral-radii/) where I apply $\rho\_{\iota\_{r;n,d},\mathcal{C}\_{r;n,d}}$ to evaluate cryptographic algorithms. I also gave some experimental observations of $\rho\_{\iota\_{r;n,d},\mathcal{C}\_{r;n,d}}$ [right here](https://circcashcore.com/2023/02/28/empirical-observations-of-l_2d-spectral-radii-and-other-functions/).
**Multi-spectrum:**
There seems to be a somewhat reasonable definition of a multi-spectrum of a collection of operators.
Suppose that $\rho\_{\iota,\mathcal{C}}$ is a multi-spectral radius. If
$(x\_1,\dots,x\_r)\in\mathcal{C}$ and $\rho(x\_1\iota(a\_1)+\dots+x\_r\iota(a\_r))=\rho\_{\iota,\mathcal{C}}(a\_1,\dots,a\_r)$, then we say that the spectrum of $x\_1\iota(a\_1)+\dots+x\_r\iota(a\_r)$ is a multi-spectrum of $a\_1,\dots,a\_r$ with respect to the embedding $\iota$ and set $\mathcal{C}$.
Suppose that $(x\_1,\dots,x\_r)\in\mathcal{C}$ and for every neighborhood $U$ of $(x\_1,\dots,x\_r)$ with respect to the topology induced by the norm on $A$, then whenever $(y\_1,\dots,y\_r)\in U\cap\mathcal{C}$, we have
$\rho(x\_1\iota(a\_1)+\dots+x\_r\iota(a\_r))\geq\rho(y\_1\iota(a\_1)+\dots+y\_r\iota(a\_r))$; then we say that the spectrum of $x\_1\iota(a\_1)+\dots+x\_r\iota(a\_r)$ is a local multi-spectrum of $(a\_1,\dots,a\_r)$ with respect to the embedding $\iota$ and the set $\mathcal{C}$.
This notion of a multi-spectrum depends on the choice of $\iota,\mathcal{C}$ and not only on the multi-spectral radius $\rho\_{\iota,\mathcal{C}}$. For example, if $A\_1,\dots,A\_r$ are complex matrices with no common invariant subspace, then the multi-spectrum of $A\_1,\dots,A\_r$ with respect to $\iota\_{r;n,n}$ and $\mathcal{C}\_{r;n,n}$ is simply the spectrum of $A\_1\otimes\overline{A\_1}+\dots+A\_r\otimes\overline{A\_r}$. On the other hand, suppose that $x\_1,\dots,x\_r$ are the non-commuting variables in $\overline{B}$ where $\rho\_p(A\_1,\dots,A\_r)=\rho(x\_1\iota(A\_1)+\dots+x\_r\iota(A\_r))$ for all matrices $A\_1,\dots,A\_r$ and suppose that $\mathcal{C}=\{(\lambda\_1 x\_1,\dots,\lambda\_r x\_r)\mid \lambda\_1,\dots,\lambda\_r\in S\_1\}$. Then a multi-spectrum of $(A\_1,\dots,A\_r)$ with respect to $\iota$ and $\mathcal{C}$ is the spectrum of $x\_1A\_1+\dots+x\_rA\_r$. If $\lambda$ is complex number with $|\lambda|=1$, then there is an automorphism $\phi$ of the Banach algebra $B$ with $\phi(x\_1A\_1+\dots+x\_rA\_r)=\lambda(x\_1A\_1+\dots+x\_rA\_r)$. Therefore, since $x\_1A\_1+\dots+x\_rA\_r$ has the same spectrum as $\lambda(x\_1A\_1+\dots+x\_rA\_r)$, there is some compact set $C\subseteq[0,\infty)$ with $\sigma(x\_1A\_1+\dots+x\_rA\_r)=\{\lambda t:|\lambda|=1,t\in C\}$.
**Unresolved properties**
If the set $\mathcal{C}$ is compact, then the function $\rho\_{\iota,\mathcal{C}}$ is automatically upper-semicontinuous. If $\rho\_{\iota,\mathcal{C}}$ is not upper-semicontinuous, then we can just take the upper-semicontinuous regularization of $\rho\_{\iota,\mathcal{C}}$, but the possible lack of upper-semicontinuity is a potential problem with the abstract theory that I am proposing. I do not know if we should require $\mathcal{C}$ to always be compact in order to make $\rho\_{\iota,\mathcal{C}}$ always upper-semicontinuous.
So far, I have mainly experimental results about multi-spectral radii, but I would like for there to be more theorems about multi-spectral radii.
| 0 | https://mathoverflow.net/users/22277 | 447226 | 180,122 |
https://mathoverflow.net/questions/447220 | 2 | Is there a $\Pi^0\_2$ singleton that forms a minimal pair with $0''$? That is, is there a set $X$ such that $X$ is the unique solution to $\forall x \exists y \phi(X|\_y, x)$, $X$ and $0''$ are incomparable and if $Y \leq\_T 0'' \land Y \leq\_T X$ then $Y \leq\_T 0$.
For some motivation, note that the most common examples of $\Pi^0\_2$ singletons are the $\alpha$-REA sets. Because every $\alpha$-REA set is built up in uniform r.e. sets no $\alpha$-REA set forms a minimal pair with $0''$ (induction and use $0''$ at limit stages to convert r.e. indexes to recursive ones).
Even Harrington's construction of a $\Pi^0\_2$ singleton not of $\omega$-REA degree builds the $\Pi^0\_2$ singleton by stitching together a part that is built computably in $0'$ with parts built computably in $0''$, $0'''$ and so on.
So it's an attractive hypothesis to think that, in some sense, every $\Pi^0\_2$ singleton is somehow built up in pieces analagous to the $\omega$-REA situation. I suspect it's false but don't have a proof.
| https://mathoverflow.net/users/23648 | $\Pi^0_2$ singleton forming minimal pair with $0''$ | Maybe I should give a more detailed answer.
Harrington proved (or claimed) the following result in his handwritten draft.
>
> **Theorem** There is a $\Pi^0\_2$-singleton $x$ so that $\forall n<\omega (x^{(n)}\equiv\_T x\oplus \emptyset^{(n)}\wedge \forall m\geq n \forall z (z\leq\_T x^{(n)}\wedge z\leq\_T \emptyset^{(m)}\implies z\leq\_T \emptyset^{(n)}))$.
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>
An immediate conclusionis that there is a $\Pi^0\_2$-singleton which forms a minimal pair with $\emptyset^{(n)}$ forall $n<\omega$.
| 4 | https://mathoverflow.net/users/14340 | 447231 | 180,125 |
https://mathoverflow.net/questions/447225 | 2 | Since $\exp(\cdot)$ is locally Lipschitz, the following SDE has a strong solution:
$$
\mathrm{d}X\_s=\exp(X\_s) \, \mathrm{d}B\_s,\quad X\_0=1,
$$
where $B$ is a standard Brownian motion. I wonder if the following expression holds:
$$\mathbb{E}\int\_0^T\exp(2X\_s) \, \mathrm d s<\infty.$$
| https://mathoverflow.net/users/484728 | Existence of solution for a non-linear SDE | Let us show that
\begin{equation\*}
E\int\_0^T e^{2X\_t}\,dt=\infty \quad\text{for real }T\ge T\_\*:=e^{-2}/2. \tag{1}\label{1}
\end{equation\*}
Indeed, letting $Y\_t:=e^{2X\_t}$, we have
$Y\_0=e^2$ and, by [Itô's lemma](https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma#Mathematical_formulation_of_It%C3%B4%27s_lemma), for real $t\ge0$
\begin{equation\*}
dY\_t=2e^{4X\_t}dt+2e^{3X\_t}dB\_t= 2Y\_t^2dt+2e^{3X\_t}dB\_t.
\end{equation\*}
So, for real $t$ and $u$ such that $0\le t\le u$ and $m(t):=EY\_t$ we have $m(0)=e^2$ and
\begin{equation\*}
m(u)-m(t)=2\int\_t^u EY\_s^2ds\ge2\int\_t^u m(s)^2ds.
\end{equation\*}
So, on any interval $[0,T)$ where the function $m$ is finite, $m\ge m(0)=e^2>0$ and $m$ is a continuous increasing function. Moreover, then for any $t\in[0,T)$
\begin{equation\*}
m'\_+(t):=\liminf\_{u\downarrow t}\frac{m(u)-m(t)}{u-t}\ge 2m(t)^2.
\end{equation\*}
So, for $h:=-1/m$ and any $t\in[0,T)$
\begin{equation\*}
h'\_+(t)=\liminf\_{u\downarrow t}\frac{h(u)-h(t)}{u-t}
=\liminf\_{u\downarrow t}\frac{m(u)-m(t)}{u-t}\frac1{m(u)m(t)}\ge2.
\end{equation\*}
So,
\begin{equation\*}
0>h(T)\ge h(0)+2T=-e^{-2}+2T,
\end{equation\*}
whence $T<T\_\*$. So, if $T\ge T\_\*$, then the increasing function $m$ is not finite on $[0,T)$. So, $m(t)=EY\_t=Ee^{2X\_t}=\infty$ for all $t$ in a left neighborhood of $T$ if $T\ge T\_\*$. So, we get \eqref{1}.
$\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 447238 | 180,126 |
https://mathoverflow.net/questions/447191 | 1 | Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph). The *chromatic number* $\chi(H)$ is the smallest cardinal $\kappa \leq |V|$ such that there is a map $c:V\to \kappa$ such that for all $e\in E$ having more than one element, the restriction $c|\_e: e\to \kappa$ is not constant.
**Question.** If $H=(V,E)$ is a hypergraph with $V\neq \emptyset$ and $\kappa < \chi(H)$ is a non-empty cardinal, is there necessarily $E\_0\subseteq E$ such that $\chi(V, E\_0) = \kappa$?
| https://mathoverflow.net/users/8628 | Is the chromatic number of hypergraphs downward continuous? | Fred Galvin had conjectured that the answer is "yes" for *graphs* in [1] (conjecture 2), in his paper he showed that the variation of the problem to *induced* graphs is consistently false: Assume $2^{\aleph\_0}=2^{\aleph\_1}<2^{\aleph\_2}$ there exists a graph $(V,E)$ with a chromatic number of $\aleph\_2$ but for no subset $V'\subseteq V$ we have that the chromatic number of $(V', E\cap [V']^2)$ is $\aleph\_1$.
Later in [2] Péter Komjáth had shown the consistency of the failure of the above conjecture: he gave a model in which $2^{\aleph\_0}=2^{\aleph\_2}=\aleph\_3$ where there exists a graph with chromatic number of $\aleph\_2$ but no subgraph with chromatic number $\aleph\_1$.
Lastly, in [3] Shelah showed a partial positive direction: assume $V=L$, then for any $(V,E)$ and $\kappa$ such that $|V|<\kappa^{+\kappa}$ and $\chi(V,E)>\kappa$, then there exists a subgraph of chromatic number exactly $\kappa$. Further more, he had showed that under $V=L$ there is no counterexample of cardinality $\aleph\_2$: every graph with cardinality $\aleph\_2$ contains a subgraph in any smaller chromatic number.
>
> [1] *Galvin, F.*, [**Chromatic numbers of subgraphs**](https://doi.org/10.1007/BF02276099), Period. Math. Hung. 4, 117-119 (1973). [ZBL0278.05105](https://zbmath.org/?q=an:0278.05105).
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>
>
>
> [2] *Komjáth, Péter*, [**Consistency results on infinite graphs**](https://doi.org/10.1007/BF02772573), Israel J. Math. 61, No. 3, 285-294 (1988). [ZBL0668.05031](https://zbmath.org/?q=an:0668.05031).
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>
>
>
> [3] *Shelah, Saharon*, Incompactness for chromatic numbers of graphs, A tribute to Paul Erdős, 361-371 (1990). [ZBL0727.05025](https://zbmath.org/?q=an:0727.05025).
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>
| 2 | https://mathoverflow.net/users/113405 | 447242 | 180,128 |
https://mathoverflow.net/questions/447241 | 53 | String theory (and related areas of purely theoretical quantum gravity, like loop quantum gravity) has a unique position within the academic physics community. Many academic physicists don't really consider string theory to be physics at all (due to its disconnect with any experimental evidence) and think that it should be more properly considered a branch of philosophy or mathematics. Other academics strongly hold the opposite opinion.
Are there any branches of academic mathematics for which there is a similar dispute as to whether those branches constitute math at all, as opposed to philosophy or some other field?
Let me clarify the scope of this question:
1. It excludes the question of whether it's useful to separate pure math from applied math. Nor does it include the question of whether certain mathematical topics in *applied* math are so closely associated with an application field (e.g. computational biology) that they should be grouped within that topic (e.g. biology) rather than within mathematics. Instead, I'm focused on the boundary between *pure* math and (e.g.) philosophy.
2. It also excludes the question of whether any specific mathematical axioms (e.g. the axiom of choice) "should" be included in the set of axioms that are typically assumed, or the question of which is the "best" mathematical axiom system.
3. The actual question of whether string theory should be considered a branch of physics is out of scope. Similarly, the actual question of whether any given academic field of math *should* count as math is out of scope. Instead, I'm asking about whether there's *consensus within the academic community* that the field should count as math. This is a sociological question, which, while perhaps somewhat subjective regarding the term "consensus", is ultimately a factual question.
| https://mathoverflow.net/users/95043 | Are there any fields of academic mathematics whose epistemic status as math is controversial within the academic community? | There are some speculative mathematical concepts that come to mind, such as the [field of one element](https://en.wikipedia.org/wiki/Field_with_one_element) or [motives](https://en.wikipedia.org/wiki/Motive_(algebraic_geometry)), though perhaps these are more classifiable as "potential future mathematics" rather than "not mathematics at all", and certainly these speculative topics have at least *inspired* the creation of mainstream, commonly-accepted mathematics (rigorous theorems, applications to other fields of mathematics, precise conjectures, conceptual reworkings of existing theories, etc.). [And motives may be currently in transition from "potential future mathematics" to "actual mathematics"; I'll leave it to experts in the area to weigh in further on this.]
A more controversial example might be [inter-universal Teichmüller theory](https://en.wikipedia.org/wiki/Inter-universal_Teichm%C3%BCller_theory), where there is genuine debate as to whether this is "actual mathematics", "potential future mathematics", or "not mathematics at all".
If one turns from subfields of mathematics to modalities of mathematics, then in the recent past there were some debates as to whether [experimental mathematics](https://en.wikipedia.org/wiki/Experimental_mathematics) or [computer-assisted proofs](https://en.wikipedia.org/wiki/Computer-assisted_proof) counted as "real" mathematics, but I believe that the prevailing consensus nowadays (by which I mean in the last decade or so) is that these do broadly fall inside the realm of mathematics. (Though perhaps these debates may be re-ignited in coming years if AI-generated conjectures and/or AI-generated proofs of new mathematical theorems become commonplace.) Going back even further in time, we of course have some venerable debates about the use of non-constructive methods (cf. [Gordan's quote](https://academic.oup.com/princeton-scholarship-online/book/23470/chapter-abstract/184556786) on Hilbert's proof of his basis theorem being theology rather than mathematics), set-theoretic infinities, non-Euclidean geometry, complex numbers, etc., though again the modern consensus is very strongly in favor of classifying all of these methods and concepts as being part of the field of mathematics. (cf. Gordan's later quote - reported by Klein - on having convinced himself that theology has its advantages.)
Finally, in the 1990s, the topic of [Bible codes / Torah codes](https://en.wikipedia.org/wiki/Bible_code) did briefly attract some academic mathematical interest (and controversy), but it would be a stretch to consider it a "field of academic mathematics" currently.
EDIT: in the converse direction, there are certainly disciplines that are typically housed outside of academic mathematics departments that have a strong case of being considered to be primarily mathematical in nature. [Theoretical computer science](https://en.wikipedia.org/wiki/Theoretical_computer_science) is one example that comes to mind; there may well be others.
SECOND EDIT:
[Section 19 (Mathematical Education and Popularization of Mathematics) and Section 20 (History of Mathematics) of the (2022) International Congress of Mathematicians](https://www.mathunion.org/fileadmin/IMU/Report/SC/2019/structure_committee_final.pdf) are both devoted to fields which one could certainly argue do not have the epistemic status of mathematics, but are still perfectly valid fields of academic study, and which are the primary or secondary interests of a non-trivial number of faculty at mathematics departments. Whether they qualify as "fields of academic mathematics" depends on one's definitions, though.
THIRD EDIT: The [Online Encyclopedia of Integer Sequences (OEIS)](https://oeis.org/) is not, strictly speaking, a field, but it does have an active community of both professional and amateur mathematicians contributing to it, and is widely used within the academic mathematical community. One could pose the philosophical question of whether contributing to the OEIS is an activity that can be ascribed the epistemic status of "mathematics". Similar questions could be asked for the communities centered around developing mathematical software, such as [proof assistants](https://en.wikipedia.org/wiki/Proof_assistant). However, my personal view is to incline towards a "big tent" view of mathematics, and that excessive gatekeeping of what qualifies as "genuine" mathematics could be harmful towards achieving progress in the field.
| 59 | https://mathoverflow.net/users/766 | 447246 | 180,131 |
https://mathoverflow.net/questions/447019 | 12 | *This question is of course related to [this earlier MO question](https://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold), but I don't believe is answered by the posts there.*
My favorite proof of the Cantor-Schroeder-Bernstein theorem actually establishes something stronger: if $f:A\rightarrow B$ and $g:B\rightarrow A$ are injections, then there is a bijection $h:A\rightarrow B$ contained in the union $f\cup g^{-1}$. This is an extremely strong property. For example, let $A$ and $B$ each be copies of $\mathbb{N}$ as a linear order and consider the embeddings $f:A\rightarrow B:x\mapsto 2x$ and $g: B\rightarrow A: x\mapsto 3x$. Then $A\cong B$, but the unique isomorphism from $A$ to $B$ can't be pieced together from $f$ and $g$ in any reasonable way, let alone literally being $\subseteq f\cup g^{-1}$.
I'm interested in any situations where a strengthening of CSB along the above lines holds. To keep things reasonably narrow, the following potential CSB strengthening for vector spaces seems natural to consider but unlikely to hold:
>
> Suppose $f:V\rightarrow W, g:W\rightarrow V$ are linear embeddings between disjoint vector spaces over a field $k$. Let $E$ be the equivalence relation on $V\sqcup W$ generated by (the graphs of) $f$ and $g$. For $a\in V$ (resp. $b\in W$) let $\widehat{a}=\langle W\cap [a]\_E\rangle\_W$ (resp. $\widehat{b}=\langle V\cap [b]\_E\rangle\_V$), where "$\langle \cdot\rangle\_U$" means "span in $U$."
>
>
> Must there be an isomoprhism $h:V\cong W$ such that for all $a\in V$ and all $b\in W$ we have $h(a)\in \widehat{a}$ and $h^{-1}(b)\in\widehat{b}$?
>
>
>
| https://mathoverflow.net/users/8133 | The scope of a "strong Cantor-Bernstein" property | For infinite-dimensional vector spaces over any field, there is no such isomorphism. We will handle the countable-dimensional case, as the others are similar. Let $V$ be generated by the linearly independent vectors $v\_0,v\_1,\ldots$, and similarly let $W$ be generated by the linearly independent vectors $w\_0,w\_1,\ldots$. Let $f$ be the linear map uniquely determined by taking $v\_i\mapsto w\_{i+1}$, and similarly let $g$ be the linear map uniquely determined by taking $w\_i\mapsto v\_{i+1}$. (These might be termed "right-shift maps".) They are both injective.
Let $h\colon V\to W$ be any vector space isomorphism.
Now, $[v\_0]\_{E}=\{v\_0,w\_1,v\_2,w\_3,\ldots\}$ and $[v\_1]\_{E}=\{w\_0,v\_1,w\_2,v\_3,\ldots\}$. So, if $h$ is supposed to have the stated property, then $h(v\_0)\in {\rm Span}(w\_1,w\_3,\ldots)$, while $h^{-1}(w\_0)\in {\rm Span}(v\_1,v\_3,\ldots)$.
However, $[v\_0+h^{-1}(w\_0)]\_{E}$ does not have $w\_0$ in the support of any element. So, we cannot have $w\_0$ in the support of $$h(v\_0+h^{-1}(w\_0))=h(v\_0)+w\_0\in w\_0+{\rm Span}(w\_1,w\_3,\ldots),$$ which gives us the needed contradiction.
| 3 | https://mathoverflow.net/users/3199 | 447260 | 180,135 |
https://mathoverflow.net/questions/447050 | 4 | Let $M$ be a smooth $n$-manifold (which is not assumed to be orientable), and write $o(TM)\to M$ for its orientation bundle. Equivalently, it is the top exterior bundle $\Lambda^n(TM)\to M$. In any case, it is a real line bundle and is flat, where the locally constant transition functions are given by the sign of the Jacobian matrix of the transition functions of $TM\to M$. Let $\nabla^{o(TM)}$ be the corresponding flat connection.
My questions are:
1. Is the flat connection on $o(TM)\to M$ unique (in some/any sense)? I saw in somewhere the following sentence "Let $\nabla^{o(TM)}$ be the natural flat connection on $o(TM)\to M$". I don't understand what natural means there.
2. Is the flat connection $\nabla^{o(TM)}$ induced by some other connection? Let say I put a Riemannian metric on $M$, and denote by $\nabla^{TM}$ its Levi-Civita connection. What is the difference $\nabla^{o(TM)}-\nabla^{\Lambda^n(TM)}$?
Any reference related to the orientation bundle and its flat connection will be appreciated.
| https://mathoverflow.net/users/41686 | Orientation bundle and its flat connection | There is a different construction of orientation bundles. One considers the $\{\pm1\}$-principal bundle $o(TM)$ of fibrewise orientations of $TM$. The associated real line bundle $o(TM)\times\_{\pm 1}\mathbb R$ carries a flat connection which is natural under local diffeomorphisms.
This bundle can be described as having the signs of the determinants of Jacobi matrices as transition functions.
On the other hand, the bundles $\Lambda^{\max}TM$ and $\Lambda^{\max}T^\*M$ have as transition functions the actual determinant of the Jacobi matrix (or its inverse, respectively).
A choice of a volume density defines an isomorphism $\Lambda^{\max}T^\*M\cong o(TM)\times\_{\pm1}\mathbb R$. If the volume density comes from a Riemannian metric, this isomorphism will identify the connection above with the one induced from the Levi-Civita connection.
A different choice of volume density leads to a different identification, so in this sense, the connection on $\Lambda^{\max}T^\*M$ is only natural with respect to local diffeomorphisms that preserve the volume density.
| 3 | https://mathoverflow.net/users/70808 | 447272 | 180,138 |
https://mathoverflow.net/questions/447274 | 5 | I asked Chat GPT to suggest a number theoretic conjecture.
It came up with the following interesting conjecture:
**Conjecture (Chat GPT):** For each even natural number $n$, there is a prime $p$ such that $p+n^2$ is also prime.
(I am not sure it stated *even* but this is clearly necessary, e.g., take $n=5$.)
Equivalently, if $P$ is the set of primes, then the difference set $P-P$ contains all squares of even natural numbers.
Is this conjecture true or false?
What is known about variations of this conjecture, with some other function $f(n)$ instead of $n$?
I am not a number theorist, but every mathematician finds number theoretic questions interesting.
| https://mathoverflow.net/users/2415 | A number theoretic conjecture by Chat GPT | Ancient Greeks conjectured that there are infinitely many pairs of primes which differ by 2 (twin primes). A natural widely believed generalization is that 2 may be replaced by every even number. Moreover, it is expected that for every integer $T>0$ there are about $C n/\log^2 n$ numbers $p\leqslant n$ for which $p$ and $p+2T$ are both primes.
What is proved is that there are infinitely many pairs of primes with the same difference (Zhang and beyond). But it is not known for difference 2 or 4. If you (or chatgpt) need only one pair, this is certainly checked for all not too large integers, but I am afraid that still open for large enough integers.
| 10 | https://mathoverflow.net/users/4312 | 447276 | 180,139 |
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