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https://mathoverflow.net/questions/414361 | 0 | Let $b\geq 2$ be an integer. A real $r \in [0,1]$ is said to be *normal with respect to $b$* if every finite string made from the elements $\{0,\ldots,b-1\}$ appears in the $b$-ary expansion of $r$.
Are there integers $b, b'\geq 2$ as well as a real number $r\in[0,1]$ such that $r$ is normal with respect to $b$, but not with respect to $b'$?
| https://mathoverflow.net/users/8628 | Does normalcy in one base imply normalcy in any other base? | A [1960 paper by Wolfgang Schmidt](https://msp.org/pjm/1960/10-2/p22.xhtml) states the following:
>
> We write $r \sim s$, if there exist integers $n$, $m$ with $r^n = s^m$. Otherwise, we put $r \not\sim s$.
>
>
> In this paper we solve the following problem. Under what conditions on $r$, $s$ is every number $\xi$ which is normal to base $r$ also normal
> to base $s$? The answer is given by
>
>
> ### THEOREM 1.
>
>
> A) Assume $r \sim s$. Then any number normal to base $r$ is normal to base $s$.
>
> B) If $r \not\sim s$, then the set of numbers $ξ$ which are normal to base
> $r$ but not even simply normal to base $s$ has the power of the continuum.
>
>
>
| 7 | https://mathoverflow.net/users/70594 | 414363 | 169,000 |
https://mathoverflow.net/questions/371951 | 4 | Let $\mathcal{C}$ be a braided monoidal category. We have a canonical functor $\mathcal{C} \to \mathcal{Z}(\mathcal{C})$ from $\mathcal{C}$ to the Drinfeld center $\mathcal{Z}(\mathcal{C})$ sending an object $V$ in $\mathcal{C}$ to $(V,c\_{V,\,\\_})$. Here, $c$ is the braiding in $\mathcal{C}$.
When does this functor admit left/right adjoints, and how do they look like? You are free to assume as much as you want on the category $\mathcal{C}$ (abelian, finite, factorizable, etc).
| https://mathoverflow.net/users/58211 | Does the functor $\mathcal{C} \to \mathcal{Z}(\mathcal{C})$ have adjoints? | **Short answer**: Yes, it can possibly have an adjoint.
**Longer answer**:
Assume that $\mathcal{C}$ is rigid, and that the coend $L = \int^{X \in \mathcal{C}} X^\* \otimes X$ exists.
It is a coalgebra.
Your assumptions on $\mathcal{C}$ were that it is braided, and in that case, it is well-known that $L$ is even a bialgebra.
Moreover, we know that ${}\_L\mathcal{C} = \mathcal{Z}(\mathcal{C})$, i.e. the center of $\mathcal{C}$ is isomorphic to the category of modules over $L$.
Under this isomorphism, your "free central object" $(V, c\_{V, -})$ is sent to the trivial $L$-module on $V$, i.e. the action is $\varepsilon \otimes V \colon L \otimes V \to V$, where $\varepsilon \colon L \to 1$ is the counit of $L$.
It is an algebra morphism.
Thus, walking everything through the isomorphisms, the inclusion functor can actually be interpreted as the pullback functor
\begin{align}
\varepsilon^\* \colon {}\_1\mathcal{C} = \mathcal{C} \to {}\_L\mathcal{C}
\ .
\end{align}
A sufficient condition for pullbacks along algebra morphisms to have adjoints was identified in [my answer to my own question over on M.SE](https://math.stackexchange.com/a/4362375/100833).
Translating to our situtation, $\varepsilon^\*$ has a left adjoint if $\mathcal{C}$ has coequalizers and $L$ is coflat (i.e. $L \otimes - $ preserves coequalizers).
Then the left adjoint sends an $L$-module $(V, r)$ to the coequalizer of
$$
r,\ \varepsilon \otimes id\_V \colon L \otimes V \to V
\ .
$$
So for a particular situation where it works:
take $\mathcal{C}$ to be a braided finite tensor category in the sense of EGNO.
Then in particular, $\mathcal{C}$ is abelian, so it has coequalizers, and the tensor product is exact, so every object is coflat.
Moreover, it's well-known that for these kinds of categories, the coend $L$ indeed does exist.
| 7 | https://mathoverflow.net/users/41706 | 414369 | 169,001 |
https://mathoverflow.net/questions/414305 | 2 | Let $L/K$ be an abelian extension of number fields with Galois group $G$ and let $\chi : G \to \{\pm 1\}$ denote a real linear character of $G$. Denote $L(\chi,s)$ the Artin L-function associated to $\chi$, $n=\mathrm{ord}\_{s=0} L(\chi,s)$ and
$$
L^\ast(\chi,0)=\lim\_{s\to 0} L(\chi,s)s^{-n}
$$
its special value at $s=0$. It is known this L-functions is also the L-function associated to a Hecke character. My question is: what are the conjectures and results regarding the sign of $L^\ast(\chi,0)$ ? My current work has gotten me to conjecture that it is $-1$ for the trivial character and $1$ otherwise, and I wonder if this is something known or conjectured.
| https://mathoverflow.net/users/138396 | Sign of the special value at s=0 of Hecke L-functions | If $\chi$ is real-valued, then the question makes sense. Using the functional equation, it reduces to computing the sign of the non-zero real number $L(\chi, 1)$ if $\chi$ is non-trivial, or the residue of $L(\chi, s) = \zeta\_K(s)$ at $s = 1$ if $\chi$ is trivial. In either case, $L(\chi, s)$ tends to $+1$ for $s$ large and real, and it cannot vanish on $Re(s) > 1$, so $L(\chi, 1+\epsilon) > 0$ for all positive $\epsilon$. This shows that $L(\chi, 1) > 0$ for $\chi \ne 1$, and that $Res\_{s=1} \zeta\_K(s) < 0$.
| 4 | https://mathoverflow.net/users/2481 | 414371 | 169,003 |
https://mathoverflow.net/questions/414375 | 2 | During my self study to the calculus of variations I come across this problem. Because of my search, I know what I wanted to do but I need some help to do them.
The function $f:[-1,1] \times \mathbb R \to \mathbb R$ is defined by
$$
f(x,\xi) = \left[ w(x) \xi - 2 x w(x) \sin\left(\frac{\pi}{x}\right) + \pi w(x) \cos\left(\frac{\pi}{x}\right) \right]^2,
$$
where
$$
w(x) =
\begin{cases}
e^{- \frac{1}{x^2}},& x\not = 0\\
0, & x=0
\end{cases}.
$$
In this problem we consider the regularity of solutions of the Euler-Lagrange equation corresponding to the one-dimensional minimization problem
$$
\inf\_{u \in X} I[u],
$$
where $X= \big\{ u \in \operatorname{Lip}[-1,1]: u(-1) = u(1)=0 \big\}$ and
$$
I[u] = \int\_{-1}^1 f(x, u'(x)) dx.
$$
I want to prove the so-called **regularity theorem**; Let $\overline{u} \in \operatorname{Lip}[a,b]$ satisfy the integral Euler equation for the functional $\displaystyle \int\_a^b f(x,u(x),u'(x)) dx$, where for every $x \in [a,b]$ the function $\xi \to f(x,\overline{u}(x),\xi)$ is strictly convex. Then $\overline{u}$ lies in $C^1 [a,b]$.
As far as reached, to prove the theorem we can solve the following sub-problems, to explain why the regularity theorem does not apply to this minimization problem.
**The steps are:**
* Showing that $f$ is infinitely differentiable, $\xi \mapsto f(x,\xi)$ is convex and $f\_{\xi \xi} (x,\xi) > 0$ holds for all $x$ except for $x=0$.
* Showing that the function
$$\overline{u}(x) = \begin{cases} x^2 \sin \frac \pi x,& x \not = 0\\ 0, & x=0\end{cases}$$
yields the minimum of the minimization problem. Also confirm that this function is Lipschitz continuous on $[-1,1]$.
* And we show that there is no other minimizer except for the function $\overline{u}$ above.
* And we need to show that $\overline{u}$ does not belong to $C^1 ([-1,1])$.
The main step is to explain why the regularity theorem does not apply
**My Proof:**
First we prove that the function $w(x)$ is infinitely differentiable. The function to be infinitely differentiable, one would need to show that in the limit the function $w(x)$, and all its derivatives, go to zero as $x$ goes to $0$. Since $x \mapsto \frac{1}{x}$ is smooth for $x \neq 0$ and $x \mapsto e^x$ is smooth, it is clear that $w$ is smooth for $x \neq 0$.
Suppose $x \ne 0$, then ${w}^{(k)}$ has the form ${w}^{(k)}(x) = e^{-{1 \over x^2}} p\_k({1 \over x})$ for some polynomial $p\_k$. This is clearly true for $k=0$, so suppose it is true for $k=0,...,n$. Then ${w}^{(n)}(x) = e^{-{1 \over x^2}} p\_n({1 \over x})$ and the chain rule gives
$$
\begin{split}
{w}^{(n+1))}(x) & = {w}^{(1)}(x) p\_n\left({1 \over x}\right) - {w}^{(0)}(x) p\_n'\left({1 \over x}\right) \left({1 \over x^2}\right) \\
& = e^{-{1 \over x^2}} \left[{2 \over x^3}p\_n\left({1 \over x}\right)-p\_n'\left({1 \over x}\right) \left({1 \over x^2}\right) \right].
\end{split}
$$
If $p\_{n+1}(y) = 2 y^3p\_n(y)-p\_n'(y) y^2 $, then
$$
{w}^{(n+1)}(x) = e^{-{1 \over x^2}}
p\_{n+1}\left( {1 \over x} \right),
$$ and so the result is true for all $n$. If $x \neq 0$, we have
$$
e^{-{1 \over x^2}} = {1 \over {e^{1 \over x^2}}}\quad \text{ and }\quad e^{1 \over x^2} \ge \sum\_{k=0}^n {1 \over k!} {1 \over x^{2k}}
$$
and thus
$$
e^{-{1 \over x^2}} \le {x^{2n} \over \sum\_{k=0}^n {1 \over k!} {x^{2(n-k)}}} \le {x^{2n} \over n!}.
$$
Suppose $p$ is a polynomial of degree $d$. Then for any $n$ we see that there is some constant $K$ such that $|e^{-{1 \over x^2}} p({1\over x})| \le K |x|^{2n-d}$ whenever $0 <|x| \le 1$. In particular, there is some $K$ such that
$|e^{-{1 \over x^2}} p({1\over x})| \le K x^2$ for all $0 < |x| \le 1$. We have ${w}^{(0)}(x) \le x^2$ for all $x$, and so ${w}$ is continuous at $x=0$. Since
$|{w}^{(0)}(x) - {w}^{(0)}(0) -0| \le x^2$, we see that ${w}^{(0)}$ is differentiable at $x=0$, and ${w}^{(1)}(0) = 0$.
Now suppose ${w}^{(k)}$ is differentiable at $x=0$ and ${w}^{(k)}(0) = 0$ for $k=0,...,n$. Then
$|{w}^{(n)}(x) - {w}^{(n)}(0) -0| \le K x^2$ for some $K$ and $|x| \le 1$. Hence
${w}^{(n)}$ is differentiable at $x=0$, and ${w}^{(n+1)}(0) = 0$. With a similar argument we find that the function
$$
\cos {\pi \over x} + i \sin \frac \pi x= \begin{cases}
e^{- \frac{\pi}{x}},& x\not = 0\\
0, & x=0
\end{cases}$$
is infinitely differentiable. Therefore, $\cos{\frac \pi x}$ and $\sin \frac{\pi}{x}$ are smooth functions. Since the summation and product of smooth functions is smooth function then $f$, defined above, is smooth.
Next step is showing $f(x,\xi)$ is convex w.r.t. $\xi$. For notational simplicity we put
$$g=2 x \sin \frac{\pi}{x} + \pi \cos\frac{\pi}{x}.,$$
then for fix $x \in [-1,1]$, we have $f(\xi)= w^2(\xi-g)^2$ is a convex function, for let $\lambda\in [0,1]$, then
$$
\begin{split}
f&(x,\lambda \xi\_1 + (1-\lambda) \xi\_2) - \lambda \xi\_1 f( \xi\_1) - (1-\lambda) f(\xi\_2) \\
& = w^2(\lambda \xi\_1 + (1-\lambda) \xi\_2 -g)^2 - \lambda w^2( \xi\_1-g)^2 - (1-\lambda) w^2(\xi\_2-g)^2 \\
& = w^2(\lambda (\xi\_1-g) + (1-\lambda) (\xi\_2 -g))^2 - \lambda w^2( \xi\_1-g)^2 - (1-\lambda) w^2(\xi\_2-g)^2 \\
&=w^2(\lambda^2 (\xi\_1-g)^2 +2\lambda(1-\lambda)(\xi\_1-g)(\xi\_2 -g) + (1-\lambda)^2 (\xi\_2 -g)^2 -\lambda( \xi\_1-g)^2 -(1-\lambda)(\xi\_2-g)^2 )\\
& = w^2( (\lambda^2-\lambda)(\xi\_1-g)^2 + 2\lambda(1-\lambda)(\xi\_1-g)(\xi\_2 -g) + ((1-\lambda)^2 - (1-\lambda) ) (\xi\_2 -g)^2)\\
& = w^2(\lambda(1-\lambda)(\xi\_1-g)^2 +2\lambda(1-\lambda)(\xi\_1-g)(\xi\_2 -g) + \lambda(1-\lambda) (\xi\_2 -g)^2)\\
& = \lambda(1-\lambda)w^2( (\xi\_1-g)^2 + 2)(\xi\_1-g)(\xi\_2 -g) + (\xi\_2 -g)^2) \\
&= \lambda(1-\lambda)w^2((\xi\_1-g) + (\xi\_2 -g))^2 \geq 0
\end{split}
$$
(The last inequality is as a product of two non-negative numbers and two square numbers is non-negative)
Could you please help with the other steps!
| https://mathoverflow.net/users/471464 | The regularity theorem, a non-regular minimizer problem | $\newcommand\ol\overline$The steps were:
1. Show that $f$ is infinitely differentiable, $\xi \mapsto f(x,\xi)$ is convex and $f\_{\xi \xi} (x,\xi) > 0$ holds for all $x$ except for $x=0$.
2. Show that the function
\begin{equation}
\overline{u}(x) = \begin{cases} x^2 \sin \frac \pi x,& x \not = 0\\ 0, & x=0\end{cases} \tag{0}
\end{equation}
yields the minimum of the minimization problem. Also confirm that this function is Lipschitz continuous on $[-1,1]$.
3. Show that there is no other minimizer except for the function $\overline{u}$ above.
4. Show that $\overline{u}$ does not belong to $C^1 ([-1,1])$.
**Step 1:** You have already checked that $f$ is infinitely differentiable.
Next,
\begin{equation\*}
f(x,\xi)=w(x)^2(\xi - g(x))^2, \tag{1}
\end{equation\*}
where
\begin{equation\*}
g(x):=2 x \sin\frac{\pi}{x} - \pi \cos\frac{\pi}{x}.
\end{equation\*}
Note that $f(x,\xi)$ is so far undefined at $x=0$ (since $g(x)$ is so far undefined at $x=0$). So, let $g(0):=0$ and $f(0,\xi):=0$, so that (1) holds even for $x=0$.
Then clearly $\xi \mapsto f(x,\xi)$ is convex and $f\_{\xi \xi} (x,\xi)=2w(x)^2 > 0$ for all $x\ne0$.
**Steps 2 and 3:** We have $I[\ol u]=0$, since
\begin{equation\*}
I[u] = \int\_{-1}^1 w(x)^2(u'(x) - g(x))^2\, dx
\end{equation\*}
and $\ol u'=g$. Also, if $u\in X\setminus\{\ol u\}$, then the Lebesgue measure of the set $\{x\colon u'(x)\ne g(x)\}=\{x\colon u'(x)\ne\ol u'(x)\}$ is $>0$ and hence $I[u]>0$.
So, $\ol u$ is the only minimizer of $I[u]$.
Also, $|\ol u'(x)|=|g(x)|\le2|x|+\pi\le2+\pi$ for $x\in[-1,1]\setminus\{0\}$ and $\ol u$ is continuous. So, $\ol u$ is Lipschitz continuous on $[-1,1]$.
**Step 4:** By the definition of the derivative and (0), $\ol u'(0)=0$. However, $\ol u'(x)$ does not converge as $x\to0$. Indeed, otherwise, $-\pi \cos\frac{\pi}{x}=g(x)-2 x \sin\frac{\pi}{x}=\ol u'(x)-2 x \sin\frac{\pi}{x}$ would converge as $x\to0$, which is clearly not so. So, $\ol u\notin C^1 ([-1,1])$.
This completes the steps.
| 1 | https://mathoverflow.net/users/36721 | 414377 | 169,005 |
https://mathoverflow.net/questions/414378 | 16 | I usually work in the field of differential geometry, but I have encountered the following problem in my research: Are there infinitely many positive integers $k,l,m\in\mathbb N^{>0}$ such that $$(3+3k+l)^2=m\,(k\,l-k^3-1)\,?$$ Obviously, taking $l=k^2$
and $m=-(3+3k+l)^2$
gives infinitely many integer solutions, but $m<0$ is negative. As a non-expert, I imagine that there is either a simple answer to this question, or the problem is not so simple to solve. Of course, I've played around with the equations a bit, but other than finding numerous examples, I haven't made any progress.
I would appreciate an existence or non-existence statement for infinitely many positive integer solutions, but also some hints that the problem is most likely hard to solve would help me.
Background: I am looking for certain integer representations of a surface group, and I can show that integer solutions to this diophantine equation actually give rise to integer representations. The condition that $k,l,m$ are positive is equivalent to the condition that the corresponding representation is contained in a higher Teichmüller component (which is important for my differential geometric application).
| https://mathoverflow.net/users/4572 | Are there infinitely many positive integer solutions to $(3+3k+l)^2=m\,(k\,l-k^3-1)$? | It does have infinitely many positive solutions. Here is just one such series.
Consider the following recurrence sequence:
$$u\_0=1,\ u\_1=2,\ u\_{n+1} = 23 u\_n - u\_{n-1} - 4\qquad (n\geq 1).$$
Let $t,k$ be any two consecutive terms of this sequence, then setting $l:=k^2+t$ produces the following equality:
$$(3+3k+l)(t+1) = (k+26)(kl-k^3-1),$$
which gives solution $m:=\frac{(k+26)(3+3k+l)}{t+1}$ (which is an integer) to the original equation.
---
In fact, integrality of $m$ follows from the identity:
$$(u\_{n+2}+1)(u\_n+1) = (u\_{n+1}+26)(u\_{n+1}+1),$$
which can be verified from the recurrence for $u\_n$.
In summary, the values $(k,l,m)$ in this solution series are given by
$$\begin{cases}
k = u\_{n+1}, \\
l = u\_{n+1}^2 + u\_n, \\
m = (u\_{n+2}+2)(u\_{n+1}+2) + 24.
\end{cases}\qquad (n\in\mathbb{Z}\_+)
$$
---
**ADDED.** I've added $u\_n$ to the OEIS as [sequence A350917](https://oeis.org/A350917). Together with 9 other similar recurrences it gives all solutions $k$ to $(tk-1)\mid (k+1)^4$, which are now listed in [sequence A350916](https://oeis.org/A350916).
| 23 | https://mathoverflow.net/users/7076 | 414380 | 169,007 |
https://mathoverflow.net/questions/414344 | 1 | Let $(X, \mathcal{T}\_X)$ and $(Y, \mathcal{T}\_Y)$ be topological spaces, $Z = X \times Y$, $\mathcal{T}\_Z$ be the product topology on $Z$, $f : Z \to X$ be defined by $f(x, y) = x$, and $C \subset Z$ be compact. Is $f \restriction C = f|(C \to f[C])$ a quotient map?
Background
----------
In what at first seems like a simple question, any counter-example or proof eludes me. This question was originally asked in [Math StackExchange](https://math.stackexchange.com/q/4350112/51161), but has gone unanswered. That question arose in turn in the context of [another question](https://math.stackexchange.com/questions/4348707/is-projection-of-locally-connected-compact-subset-locally-connected).
Partial results
---------------
### $f\restriction C$ is a quotient when $X$ is Hausdorff
Suppose $X$ is Hausdorff. Then $f\restriction C$ is a continuous map from a compact space to a Hausdorff space, [hence a closed map](https://math.stackexchange.com/questions/143988/how-to-prove-the-closed-map-lemma), [hence a quotient map](https://math.stackexchange.com/questions/3981375/a-surjective-continuous-and-closed-map-is-a-quotient-map).
### $f\restriction V$ is a quotient when $Y$ is compact and $V \subset Z$ is closed
Suppose $Y$ is compact. [Then $f$ is closed](https://math.stackexchange.com/questions/22697/projection-map-being-a-closed-map). Let $V \subset Z$ be closed. Then $f\restriction V$ is a closed map, hence a quotient map.
### $f\restriction U$ is a quotient when $U \in \mathcal{T}\_Z$
Suppose $U \in \mathcal{T}\_Z$. [It can be shown that $f$ is open](https://math.stackexchange.com/questions/247542/projection-is-an-open-map). Then $f\restriction U$ is an open map, hence a quotient map.
### $f\restriction C$ is a quotient when $C$ is closed
Let $\pi\_X : Z \to X$ be defined by $\pi\_X(x, y) = x$ and $\pi\_Y : Z \to Y$ be defined by $\pi\_Y(x, y) = y$. Let $C\_X = \pi\_X(C)$ and $C\_Y = \pi\_Y(C)$. By continuity, $C\_X$ and $C\_Y$ are compact. Therefore $D = C\_X \times C\_Y$ is compact. By a previous section, $\pi\_X \restriction D$ is closed. Since $C$ is closed in $D$, $\pi\_X \restriction C$ is closed. Therefore $f\restriction C$ is a quotient map.
### The previous strategy fails when $C$ is not closed
The previous proof does not generalize to the case when $C$ is not closed. Let $X = Y = \{0, 1\}$ and $\mathcal{T}\_X = \mathcal{T}\_Y = \{\emptyset, \{0\}, \{0, 1\}\}$. Then $\{(0, 0), (1, 0), (0, 1)\}$ is compact, but not closed in $X \times Y$.
### Compact slices are not sufficient to be a quotient
Let $X = Y = \mathbb{R}$, $Z' = \{(0, 1)\} \cup \{(1/n, 0) : n \in \mathbb{N}^{> 0}\}$, and $g = f \restriction Z'$. Then $(\{x\} \times Y) \cap Z'$ is compact for each $x \in X$ as a singular subset. Let $V = \{0\}$, and $U = g^{-1}(V) = \{(0, 1)\}$. Then $U \in \mathcal{T}\_Z|Z'$, and $V \not\in \mathcal{T}\_X|g(Z')$. Therefore $g$ is not a quotient map.
| https://mathoverflow.net/users/32487 | Is the restriction of a projection to a compact subset a quotient map? | A counterexample with finite topological spaces:
recall that on a set $X$, an "Alexandroff discrete" topology ($\mathrm T\_0$ where all intersections of open sets are open)
is the same thing as a partial order; the minimum open set containing $x$ is the principal order filter $Fx$ in the order.
The dual order gives the topology where open and closed sets are interchanged.
Take a finite $X$, and as $Y$ its dual as above; take as $C$ the principal diagonal of points $(x,x)$.
The minimum open set which contains $(x,x)$ is the direct product of the principal order filter $Fx$ and the principal ideal $Ix$.
The only $y$ with both inequalities $x\leq y$, $x\geq y$ is $y=x$, so the diagonal $C$ is discrete.
Essentially, one has the identity map on a finite set $X$, with the discrete topology in the domain and an arbitrary $\mathrm T\_0$ topology
in the codomain. No identifications but no homeomorphism either (except for the antichain poset order on $X$). No quotient.
Edit. Take any two distinct quasi compact topologies on a set $X$,
and the projections from the diagonal as above. At least one
of the two projections is not a quotient.
| 3 | https://mathoverflow.net/users/474159 | 414388 | 169,010 |
https://mathoverflow.net/questions/414394 | 5 | Given a (0,1)-matrix $A$, I'll denote by $\mu(A)$ the number of maximal monochromatic polyominoes in $A$ (i.e., the number of connected polyominoes contained in $A$ each of which is either all 0 or all 1 such that each polyomino is as large as possible - so they thus tile $A$). I do not know of any literature on $\mu(A)$, so I apologize if there is already some notation for the quantity that I am not using.
So for example if $A$ is the $n \times n$ identity matrix, $\mu(A) = n + 2$. For an $n \times n$ matrix $A$, we have $\mu(A) \leq n^2$, with equality if and only if $A$ is one of the "checkerboard matrices".
I would like to understand what $\mu(A)$ is on average, as we allow the size of the matrix $A$ to tend toward infinity. In particular, is it true that
$$
\lim\_{n \to \infty} \frac{\text{Average value of $\mu$ over all $n\times n$ (0,1)-matrices}}{n^2} = 1
$$
| https://mathoverflow.net/users/99414 | Average number of tiles of a (0,1)-matrix? | The limit is positive but well below $1$; I show that the limit is in $(0.09448, 0.2646)$, and outline how to approximate it much more closely than this.
The average of $\mu$, call it ${\rm E}(\mu)$,
is twice the average number of maximal all-0 polyominos;
so the limit of ${\rm E}(\mu) / n^2$ is twice the expected number of
maximal all-0 polyominos per unit area of a large square.
That limit is the special case $p=q=1/2$ of
the expected number, call it $\epsilon(p)$,
of maximal all-0 polyominos per unit area
when each entry is independently 0 or 1 with probability $p,q$ respectively
($p+q = 1$). We can approximate $\epsilon(p)$ by writing it as a sum
$\sum\_P \epsilon\_P(p)$ over all possible polyomino shapes $P$,
and calculating partial sums over all $P$ of size at most $k$.
The contribution $\epsilon\_P(p)$ is $p^{|P|} q^{|\delta(P)|}$,
where $|P|$ is the size of $P$, and $|\delta(P)|$ is the number of
cells at distance $1$ from $P$. (We do *not* identify different orientations;
for example, for $k=4$ there is one square $P$, two straight ones,
four S/Z shaped tetrominos, four T-shaped, and eight that are L-shaped.)
So we seek $\sum\_P p^{|P|} q^{|\delta(P)|}$. Note that
$\sum\_P |P| p^{|P|} q^{|\delta(P)|} = p$ because this is just
the expected number of 0 entries per unit area.
The analysis so far works in any dimension; for example, in dimension $1$
there is a unique $P$ of each size $k \geq 1$, and $|\delta(P)| = 2$ always,
so $\epsilon(p) = \sum\_{k=1}^\infty p^k q^2 = p q^2/(1-p) = p q$
(and indeed $\sum\_{k=1}^\infty k p^k q^2 = p q^2/(1-p)^2 = p$).
For example, $\epsilon(1/2) = 1/4$, and the expected number of
maximal all-0 or all-1 strings in a random bitstring of length $n$
is asymptotically $2\epsilon(1/2) n = n/2$.
(Exercise [noted in my comment]: in fact that expected number is
exactly $(n+1)/2$. What happens for arbitrary $p$?)
In dimension $2$, the sum of $\epsilon\_P(p)$ over $|P| \leq 4$ is
$$
p q^4 + 2 p^2 q^6 + p^3 (2 q^8 + 4 q^7)
+ p^4 (q^8 + 2 q^{10} + 4 q^8 + 4 q^8 + 8 q^9),
$$
and the sum of $|P| \epsilon\_p(P)$ up to $|P| \leq 4$
is obtained from by multiplying the $k$-th term by $k$ ($k=1,2,3,4$),
and comes to $p - 315 p^5 + 1824 p^6 - 4848 p^7 + \cdots$.
(The coefficient $315$ seems to come from the number of rooted 5-ominos,
see <https://oeis.org/A048664> .)
For $p=1/2$ these sums are only $387/2^{13}$ and $612/2^{13}$
respectively. It follows that the $|P| \geq 5$ terms must contribute
$1/2 - 612/2^{13}$ to the sum of $|P| \epsilon\_p(P)$,
and thus can contribute at most $1/5$ of that to the sum of $\epsilon\_p(P)$.
We conclude that
$$
\frac{387}{2^{13}} < \epsilon(1/2) <
\frac{387}{2^{13}} + \frac15 \left( \frac12 - \frac{612}{2^{13}} \right)
= \frac{5419}{40960};
$$
numerically these bounds are $0.04724+$ and just under $0.1323$.
Hence the answer to the OP's question is between $0.09448$ and $0.2646$,
as claimed.
Polyominos have been tabulated well beyond $k=4$.
It must be feasible to compute $\sum\_{|P| \leq k} \epsilon\_P(1/2)$ and
$\sum\_{|P| \leq k} |P| \epsilon\_P(1/2)$ for $k$ large enough to obtain
a reasonably close approximation to $\epsilon(1/2)$, and thus to
the value $2\epsilon(1/2)$ of the limit that the OP seeks.
| 8 | https://mathoverflow.net/users/14830 | 414405 | 169,013 |
https://mathoverflow.net/questions/414402 | 23 | On MSE this got 5 upvotes but no answers not even a comment so I figured it was time to cross-post it on MO:
Is the Moebius strip a linear group orbit? In other words:
**Does there exists a Lie group $ G $ a representation $ \pi: G \to \operatorname{Aut}(V) $ and a vector $ v \in V $ such that the orbit
$$
\mathcal{O}\_v=\{ \pi(g)v: g\in G \}
$$
is diffeomorphic to the Moebius strip?**
My thoughts so far:
The only two obstructions I know for being a linear group orbit is that the manifold (1) must be smooth homogeneous (shown below for the the group $ \operatorname{SE}\_2 $) and (2) must be a vector bundle over a compact Riemannian homogeneous manifold (here the base is the circle $ S^1 $).
The Moebius strip is homogeneous for the special Euclidean group of the plane
$$
\operatorname{SE}\_2= \left \{ \
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1
\end{bmatrix} : a^2+b^2=1 \right \}.
$$
There is a connected group $ V $ of translations up each vertical line
$$
V= \left \{ \
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & y \\
0 & 0 & 1
\end{bmatrix} : y \in \mathbb{R} \right \}.
$$
Now if we include the rotation by 180 degrees
$$
\tau:=\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
then $ \langle V, \tau \rangle$ has two connected components and
$$
\operatorname{SE}\_2/\langle V, \tau \rangle
$$
is the Moebius strip.
| https://mathoverflow.net/users/387190 | Is it possible to realize the Moebius strip as a linear group orbit? | Yes. Here is one way: Consider standard $\mathbb{R}^3$ endowed with the Lorentzian quadratic form $Q = x^2+y^2-z^2$, and let $G\simeq\mathrm{O}(2,1)\subset\mathrm{GL}(3,\mathbb{R})$ be the symmetry group of $Q$. Then $G$ preserves the hyperboloid $H$ of $1$-sheet given by the level set $Q=1$, which is diffeomorphic to a cylinder. Consider the quotient of $H$ by $\mathbb{Z}\_2$ defined by identifying $v\in H\subset\mathbb{R}^3$ with $-v$. This abstract quotient is a smooth Möbius strip.
This quotient can be identified as a linear group orbit as follows: Let $V = S^2(\mathbb{R}^3)\simeq \mathbb{R}^6$ and consider the smooth mapping $\sigma:\mathbb{R}^3\to V$ given by $\sigma(v) = v^2$ for $v\in\mathbb{R}^3$. Then $\sigma$ is a $2$-to-$1$ immersion except at the origin. The action of $G$ on $\mathbb{R}^3$ extends equivariantly to a representation $\rho:G\to \mathrm{Aut}(V)$ such that $\rho(g)(v^2) = \rho\bigl(\sigma(v)\bigr)=\sigma(g v)= (gv)^2$. It follows that $\sigma(H)\subset S^2(\mathbb{R}^3)\simeq\mathbb{R}^6$, which is a Möbius strip, is a linear group orbit under the representation $\rho$.
Note that the representation of $G$ on $S^2(\mathbb{R}^3)\simeq\mathbb{R}^6$ is actually reducible as the direct sum of a trivial $\mathbb{R}$ and an irreducible $\mathbb{R}^5$. Projecting everything into the $\mathbb{R}^5$ factor, one obtains a representation of $G$ on $\mathbb{R}^5$ that has a Möbius strip as a $G$-orbit.
| 28 | https://mathoverflow.net/users/13972 | 414425 | 169,022 |
https://mathoverflow.net/questions/414382 | 5 | Let $S$ be a minimal compact complex surface of general type with ample canonical class $K\_S$. In [**1**, Theorem 3] the following result is stated:
>
> **Theorem.** Every symmetric power $S^n \Omega\_S$ of the cotangent bundle $\Omega\_S$ is stable with respect to $K\_S$, unless $S$ is
> uniformized by the bi-disk.
>
>
>
I understand why the assumption about the universal covering is necessary: indeed, if for instance $S=C\_1\times C\_2$, with $C\_i$ smooth curve of genus $g \geq 2$, then $\Omega\_S$ is the direct sum of two line bundles; thus, every symmetric power is also reducible as a sum of line bundles, in particular, it is non-stable.
However, if I made the computations correctly, in this example all direct summands in $S^n \Omega\_S$ have the same slope with respect to $K\_S$, hence the symmetric powers of $\Omega\_S$ are $K\_S$-semistable.
I wonder if this is true in general; I asked some people and I was told that this should be in fact a result of Bogomolov, but I was not given a precise reference, and I was unable to locate one. So, let me ask the
>
> **Question.** If $S$ is a compact complex surface as above, is it true that every symmetric power $S^n\Omega\_S$ is $K\_S$-semistable? If so, what is a
> reference?
>
>
>
**References.**
**[1]** *Lu, Steven Shin-Yi*, On hyperbolicity and the Green-Griffiths conjecture for surfaces, Noguchi, J. (ed.) et al., Geometric complex analysis. Proceedings of the conference held at the 3rd International Research Institute of the Mathematical Society of Japan, Hayama, March 19-29, 1995. Singapore: World Scientific. 401-408 (1996). [ZBL0941.32024](https://zbmath.org/?q=an:0941.32024).
| https://mathoverflow.net/users/7460 | Semi-stability of $S^n\Omega_S$ with respect to $K_S$ | Ciao Francesco!
The answer to your question is yes, and the work of Bogomolov you are looking for about semistability of the tangent space (for minimal surfaces of general type indeed, no need of ampleness of the canonical bundle) is
"Holomorphic tensors and vector bundles on projective varieties", Math. USSR Izvestija 13/3 (1979) 499-555.
But let me give you a more general framework.
Given a Hermitian holomorphic vector bundle $(E,h)$ over a compact Kähler manifold $(X,\omega)$, if $(E,h)\to (X,\omega)$ is Hermite-Einstein then its dual as well as its symmetric powers (with the induced Hermitian structures) also are Hermite-Einstein.
Next, if $(E,h)\to (X,\omega)$ is Hermite-Einstein, then $E\to X$ is $[\omega]$-semistable (polystable, indeeed). This is the "easy" direction of the Kobayashi-Hitchin correspondence.
If a projective manifold $X$ has ample canonical bundle, then it admits a Kähler-Einstein metric $\omega$ of negative Einstein constant, *i.e.* a Kähler metric such that $\operatorname{Ric}(\omega)=-\omega$ (by the Aubin-Yau Theorem).
In particular $(T\_X,\omega)\to (X,\omega)$ is Hermite-Einstein, and then so does any symmetric power of the cotangent bundle (with the induced Hermitian structure). This makes these vector bundles $[\omega]$-semistable, as we saw above.
Finally, in this case, begin $[\omega]$-semistable means that these vector bundle are $K\_X$-semistable, since $[\omega]=[-\operatorname{Ric}(\omega)]=c\_1(K\_X)$.
You can see all this and much more on S. Kobayashi "Differential geometry of complex vector bundles".
P.S. For quite recent advances about semistability of the tangent sheaf of possibly singular varieties (even in the logarithmic setting), you might want to take a look to [this](https://hal.archives-ouvertes.fr/hal-01878999/document) paper by H. Guenancia.
| 3 | https://mathoverflow.net/users/9871 | 414427 | 169,023 |
https://mathoverflow.net/questions/414367 | 0 | Let $X$ be a compact metric space, $A,B\subset X$ be subsets and $f\colon X\times X\to \mathbb{R}$ a continuous function that is strictly positive on $A\times B$. Do there exist increasing sequences of subsets $A\_1\subseteq A\_2\subseteq \dots$ and $B\_1\subseteq B\_2\subseteq \dots$ such that:
* $A=\bigcup\_{n\in\mathbb{N}}A\_n$ and $B=\bigcup\_{n\in\mathbb{N}}B\_n$;
* for every $n\in\mathbb{N}$, the restriction of $f$ to $A\_n\times B\_n$ is bounded away from zero?
We can also rephrase the question as follows. Let $X,A,B$ as before and let $C\subset X\times X$ be a closed subset such that $(A\times B)\cap C= \emptyset$. Do there exist increasing sequences of compact sets $K^A\_1\subseteq K^A\_2\subseteq \dots$ and $K^B\_1\subseteq K^B\_2\subseteq \dots$ such that:
* $A\subseteq\bigcup\_{n\in\mathbb{N}}K^A\_n$ and $B\subseteq\bigcup\_{n\in\mathbb{N}}K^B\_n$;
* $(K^A\_n\times K^B\_n)\cap C = \emptyset$ for every $n\in\mathbb{N}$?
If true, this sounds like a useful lemma and I would not be surprised if it already appeared somewhere else. I need this fact (or something similar) to reduce some statement about operators on product spaces to the compact case, but I thus far failed to find a proof or a counterexample. Any suggestion?
| https://mathoverflow.net/users/54309 | Exhaustions of product subsets by smaller product subsets | Take $A,B$ complementary dense subsets of $X$ (say, rationals and irrationals in $[0,1]$).
$d(a,b)$ is strictly positive on $A\times B$.
Suppose $A$ increasing join of the $A\_n$, dually $B\_n$, with positive distance between $A\_n,B\_n$ i.e.
between their closures $A'\_n,B'\_n$.
Then Baire assures that for one $n$ (and then the successive ones), $A'\_n$ (or $B'\_n$) contains a open set.
By density, it contains a point $b$, which must be in some $B\_m$, and so for index $j$ greater than $n,m$
$A'\_j,B'\_j$ i.e. $A\_j,B\_j$ cannot have positive distance.
Corollary: in any partition as above, at least one set cannot be $\sigma$-compact.
| 1 | https://mathoverflow.net/users/474159 | 414439 | 169,027 |
https://mathoverflow.net/questions/414460 | 5 | Is the following a theorem of $\sf ZF(C)$?
**Countable reflection:** If $\phi$ is a sentence in which $W$ is not free, then:
$\phi \to \exists W: |W|= \omega \land \operatorname {Transitive}(W) \land \phi^W$
Where $\phi^W$ is the sentence obtained by merely bounding all quantifiers in $\phi$ by $ \in W$.
That is: every true sentence is reflected upon a countable transitive set.
So all true sentences are reflected upon some elements of $V\_{\omega\_1}$ (or $V\_{\omega\_2}$ in case of $\sf ZF$).
| https://mathoverflow.net/users/95347 | Does ZF(C) prove countable reflection? | As Monroe Eskew points out in his comment to the question, the positive answer is well-known for ZFC, thanks to the ZF reflection theorem and the Löwenheim-Skolem Theorem (in the form: every model in a countable language has a countable elementary submodel). See, e.g., part (iii) of Theorem 12.14 in Jech's canonical textbook *Set theory, third millenial edition.*
Note that ZFC can be weakened in the above to ZF+ DC (DC = dependent choice) since it is well-known that in the presence of ZF, DC implies (and indeed is equivalent to, see, e.g., [Asaf Karagila's note](https://karagila.org/wp-content/uploads/2012/10/Lowenheim-Skolem-and-Choice.pdf)) the statement that every structure in a countable language has a countable elementary submodel.
>
> The point of this answer is to point out that even DC can be eliminated from the proof of countable reflection.
>
>
>
**Theorem.** ZF *proves every instance of countable reflection.*
Proof: In light of the fact that the Mostowski collapses can be carried out in ZF, and the constructible universe L satsifies ZFC (and therefore L satisfies "every model in a countable language has a countable elementary submodel), the proof is complete once we note:
$(\*)$ For any set-theoretical sentence $\phi$, if $\phi$ has a transitive model, then the sentence "$\phi$ has a transtive model" holds in the constructible universe L.
The above statement, as far as I know, was first noted in a (famous) paper of Barwise and Fisher entitled "The Shoenfield Absoluteness Lemma" (Israel J. Math. 8 1970, pp.329-339) in which a fine-tuning of the Shoenfield Absoluteness Lemma is presented (in particular, it is shown that the Lemma can be proved without invoking DC).
More specifically, $(\*)$ follows from Theorem 1b (page 336) of the Barwise-Fisher paper, which states that if $\phi$ has a transitive model $A$ with $\rho(A)=\alpha$ (where $\rho$ is the usual ordinal-valued rank function on sets), then there is a transitive model of $\phi$ in the $\kappa^{+}(\alpha)$-th level $L\_{\kappa^{+}(\alpha)}$ of the constructible universe $L$, where $\kappa^{+}(\alpha)$ is the next admissible ordinal after the least admissible ordinal above $\alpha$.
| 18 | https://mathoverflow.net/users/9269 | 414474 | 169,045 |
https://mathoverflow.net/questions/414404 | 1 | If we add to the language of set theory a total one place function symbol $\mathcal P$ standing for powerset operator, and then add to ZF-Power the following axioms:
**Power:** if $\phi$ is a formula in which only the symbol $y$ occurs free, then: $$ X \subseteq A \land X=\{ y \mid \phi\} \to X\in \mathcal P(A) $$
**Countability:** $\forall X: X \text { is countable }$
>
> Is this theory interpretable in Kripke-Platek set theory (with Infinity)?
>
>
>
>
> If not, would it constitute a subsystem of second order arithmetic? If yes, what would be its proof theoretic ordinal?
>
>
>
| https://mathoverflow.net/users/95347 | Does this restriction on powersets in ZF have a proof theoretic ordinal? | Your theory interprets $\mathsf{ZF}^-$. In fact, $\mathsf{ZF}-$ (namely, $\mathsf{ZF}$ without Powerset) interprets $\mathsf{ZF}^-+(V=L)$.
It answers your questions negatively since the proof-theoretic strength of $\mathsf{ZF}^-+(V=L)$ is that of Full Second-order Arithmetic.
The reason is that we can construct $L$ from $\mathsf{ZF}-$.
We need to check whether $\mathsf{ZF-}$ proves transfinite recursion, but you can check proofs in standard textbooks (like Jech or Kunen) work. Also, we can define $\operatorname{Def}(X)$ by using Replacement.
Work in $L$, we can see that $L$ has a natural rank function given by the $L$-hierarchy. Hence $L$ satisfies its own version of the reflection principle. We can see that $L$ satisfies $\mathsf{ZF}^-$ by using the reflection principle in $L$.
| 2 | https://mathoverflow.net/users/48041 | 414476 | 169,046 |
https://mathoverflow.net/questions/414487 | 0 | Title says it all. Sorry my previous question was wrong; I see now; very stupid of me. So this is what I meant to ask. I am looking for a field extension of $\mathbb Q$, let's call it $K$, s.t. $K$ is a proper subset of $\bar{\mathbb{Q}}$, and $X^n-a$ has a root in $K$ whenever $a$ is in $K$.
Or is this question unknown?
| https://mathoverflow.net/users/475698 | Does there exist a proper intermediate field between ℚ and ℚ̅ closed under taking nth roots? | Yes because the algebraic closure of $\mathbb{Q}$ contains all roots of all polynomials. But what we learn in elementary Galois theory is that there are polynomials that cannot be solved using any finite number of applications of the operations of taking $n$th roots for every $n$ in $\mathbb{N}$ and field operations. See for example this old post on Math Stack Exchange, where this questions might be more suited for <https://math.stackexchange.com/questions/286944/quintic-polynomial-with-galois-group-a-5>
| 4 | https://mathoverflow.net/users/471081 | 414490 | 169,050 |
https://mathoverflow.net/questions/414465 | 11 | Is every $ \mathbb{R}P^{2n} $ bundle over the circle trivial?
Are there exactly two $ \mathbb{R}P^{2n+1} $ bundles over the circle?
This is a cross-post of (part of) my MSE question
<https://math.stackexchange.com/questions/4349052/diffeomorphisms-of-spheres-and-real-projective-spaces>
which has been up for a couple weeks and got 8 upvotes and some nice comments but no answers.
My intuition for thinking both answer are yes is that there are exactly 2 sphere bundles over the circle. The trivial one and then the non-trivial (and non-orientable) one which can be realized as the mapping torus of an orientation reversing map of the sphere. So importing that intuition to projective spaces then the orientable $ \mathbb{R}P^{2n+1} $ should have a nontrivial (and non orientable) bundle over the circle while the non orientable $ \mathbb{R}P^{2n} $ should have only the trivial bundle. For $ n=1 $ this checks out since that projective space is orientable and thus we have exactly two bundles over the circle (the trivial one=the 2 torus and the nontrivial one=the Klein bottle).
| https://mathoverflow.net/users/387190 | $ \mathbb{R}P^n $ bundles over the circle | Your answer is correct if appropriately understood, but it's a little subtle. Here I should note that I'm interpreting your question as a purely homotopy theoretic one (in particular ignoring smooth structure), and that by "bundle" you mean "Serre bundle". If you care about smooth bundles, see Tom Goodwillie's answer.
The way I know to do this involves a little algebraic topology.
First, it's always the case that $X$-bundles over a circle (for $X$ some topological space) are classified by the group $\pi\_0(\mathrm{Aut}(X)),$ where $\mathrm{Aut}(X)$ is the group of maps $X\to X$ which are invertible up to homotopy. Thus a good technique to carry out this classification is to first find all self-maps $X\to X$ up to homotopy, then see which ones are invertible.
Now we want to specialize to the case $X = RP^n,$ for some $n$. However it turns out that it's much easier to classify maps not into $RP^n$ itself, but rather into $RP^\infty.$ Namely, maps from some space $X$ to $RP^\infty$ up to homotopy are always classified by $H^1(X, \mathbb{Z}/2)$ (equivalently, this is group homomorphisms from $\pi\_1(X)$ into $\mathbb{Z}/2$). Now the difference between $RP^n$ and $RP^\infty$ isn't actually too terrible. Namely, the CW approximation theorem tells us that any map $RP^n\to RP^\infty$ actually can be chosen up to homotopy such that it factorizes through $RP^n\subset RP^\infty.$ So we can compute
$$\pi\_0\mathrm{Maps}(RP^n, RP^\infty) \cong H^1(RP^n, \mathbb{Z}/2)\cong \mathbb{Z}/2.$$
The element $0\in \mathbb{Z}/2$ corresponds to the trivial map $$\*:RP^n\to RP^n$$ mapping everything to a basepoint in $RP^\infty$ and the nontrivial element $1\in \mathbb{Z}/2$ corresponds to the identity map, $$id:RP^n\to RP^n\subset RP^\infty.$$ And any other map $RP^n\to RP^n$ will be homotopy equivalent to one of these as an element of $\mathrm{Maps}(RP^n, RP^\infty)$.
However there's a catch: while any map $RP^n\to RP^n$ is homotopy equivalent to one of the maps $id, \*$ *as a map to $RP^\infty$,* the homotopy between the two maps might not live in $RP^n$. So a priori, there can be multiple homotopy classes of self-maps of $RP^n$ homotopic to one of these maps in $RP^\infty$ (and indeed, sometimes there are). To get a handle on how badly maps to $RP^\infty$ are undercounting, you can apply the CW approximation theorem again to see that any homotopy between two maps $RP^n\to RP^\infty$ will factorize, up to homotopy, through $$RP^{n+1}\subset RP^\infty.$$ The "error" of such a homotopy existing in $RP^n$ will be classified by a map to the quotient, $$RP^n\times [0,1]\to RP^{n+1}/RP^n\cong S^{n+1},$$ taking both $RP^n\times \{0\}, RP^n\times \{1\}$ to the basepoint, in other words, a based map from the space $$RP^n\_+\wedge S^1 = \Sigma(RP^n\_+)$$ (you can think of this as the ordinary suspension $\Sigma(RP^n)$ with the two suspension points identified) to $S^{n+1}.$ Let's write $$D: = \pi\_0\text{Maps}(RP^n\_+\wedge S^1, S^n)$$ for the set of possible such "defects" of a homotopy in $RP^{n+1}$ restricting to $RP^n$.
Based maps from a closed $n+1$-dimensional manifold to $S^{n+1}$ are classified by ordinary $H^{n+1},$ and so we have $$D \cong H^{n+1}(\Sigma(RP^n\_+)) \cong H^n(RP^n) \cong \begin{cases}
\mathbb{Z}/2, & n\text{ even}\\
\mathbb{Z}, & n\text{ odd}.
\end{cases}$$
Thus for each of the maps $\*, id: RP^n\to RP^n,$ there can be at worst $D$ worth of distinct other homotopy classes of self-maps $RP^n\to RP^n$ homotopic to it as maps to $CP^{\infty}.$ Since a map homotopic to $\*$ cannot be an automorphism (it would have to induce the trivial map on $H^1$ which cannot come from an automorphism), we can restrict our attention to maps homotopic in $CP^\infty$ to $id:RP^n\to RP^n.$ A priori, there could have been elements of $D$ which are not realizable as "defects" of homotopies between maps $RP^n\to RP^n$, but in this case we don't run into this problem: indeed, every element of $D$ occurs as the defect of some homotopy $RP^n\times [0,1]\to RP^{n+1}$ between the identity $id:RP^n\to RP^n$ and another map. Namely, recall that we have realized $D \cong H^n(RP^n)$ as a cyclic group, either $\mathbb{Z}$ or $\mathbb{Z}/2$ (depending on parity). Let $\alpha$ be a generator of this group. Then every element of $D$ can be written $k\alpha$ for some $k\in \mathbb{Z}$. By doing a calculation, you can see that each element $k\alpha\in D$ is realized as the defect of the homotopy $$[0,1]\cdot RP^n\to RP^{n+1}$$ induced by the map $[0,1]\times S^n\to S^{n+1}$ given by rotating $S^n$ in a circle around some ($n-1$-dimensional) axis inside $S^{n+1},$ by an angle of $$k\cdot \pi.$$ Now if $n$ is even, we see the resulting "new" map $RP^n\to RP^n$ is once again the identity. If $n$ is odd, the new map is induced from the "reflection" map given by
$$\sigma:(x\_1,x\_2,\dots, x\_n)\mapsto (-x\_1,x\_2, \dots, x\_n)$$
(in some coordinates). Thus from what we've done so far, there can be at most two homotopy invertible self-maps up to homotopy $$RP^n\to RP^n$$ for any $n$, namely $id$ and $\sigma.$ It remains to check whether the induced two self-maps $RP^n\to RP^n$ are homotopic to each other. When $n$ is odd, they cannot be homotopic to each other since $RP^n$ is orientable, and $\sigma$ reverses orientation (so $\sigma$ can be distinguished from $id$ by looking at action on $H^n$). But when $n$ is even, the map $\sigma$ is homotopy equivalent via sphere rotations to the map $(x\_1,x\_2\dots, x\_n)\mapsto (-x\_1,-x\_2\dots, -x\_n),$ and this induces a homotopy between $\sigma$ and $id$ as maps $RP^n\to RP^n.$ Thus we have $$\pi\_0(\mathrm{Aut}(RP^n)) \cong \begin{cases}
\{id\}, & n \text{ even}\\
\{id, \sigma\}, & n \text{ odd}.
\end{cases}
$$
As mentioned, $RP^n$-bundles on $S^1$ are classified by the same data, so your guess is correct.
| 8 | https://mathoverflow.net/users/7108 | 414496 | 169,052 |
https://mathoverflow.net/questions/413526 | 4 | Let $\left(\mathcal{M}^2,g\_\mathcal{M};X\right)$ and $\left(\mathcal{N}^2,g\_{\mathcal{N}};Y\right)$ be two smooth two-dimensional, simply connected Riemannian manifolds (with or without boundary), equipped with non-nonvanishing Killing fields $X$ and $Y$, respectively.
Does there exist a mapping $f:\mathcal{M}^2\rightarrow \mathcal{N}^2$ with constant, distinct singular values such that $\mathrm{d}f\left(X\right)=Y$ ?
In the affirmative case, how would one construct explicitly such a mapping, given the singular values ?
(It is well known that in general there are no local obstructions for mapping with constant singular values, but assuming that there exists a Killing field that is mapped to another Killing may set obstructions.)
| https://mathoverflow.net/users/171439 | Mappings between 2-manifolds with symmetries with fixed singular values | To understand the local conditions, it's convenient to establish canonically associated local coordinate expressions for the quantities involved. Thus, let $(M^2,g,X)$ and $(N^2,h,Y)$ be as described and suppose that we want to test whether, for a given $p\in M$ and $q\in N$, there exists an open $p$-neighborhood $U\subset M$ and a local diffeomorphism $f:U\to N$ satisfying $f(p) = q$ with the desired properties. To simplify notation a little bit, let us fix orientations on $M$ and $N$ and require $f$ to be orientation-preserving. (We'll see what comes of this choice later.)
Then it is easy to show that there exist oriented, $p$-centered coordinates $(r,\theta):V\to (-\epsilon, \epsilon)\times (-\epsilon,\epsilon)$ on an open $p$-neighborhood $V\subset M$ and a smooth function $A:(-\epsilon,\epsilon)\to\mathbb{R}$ with $A(0)=0$ and $A'>0$ such that, on $V$, we have $g = \mathrm{d}r^2 + A'(r)^2\,\mathrm{d}\theta^2$ and $X = \partial/\partial\theta$. Similarly, there exist oriented, $q$-centered coordinates $(s,\phi):W\to (-\delta, \delta)\times (-\delta,\delta)$ on an open $q$-neighborhood $W\subset M$ and a smooth function $B:(-\delta,\delta)\to\mathbb{R}$ with $B(0)=0$ and $B'>0$ such that, on $W$, we have $h = \mathrm{d}s^2 + B'(s)^2\,\mathrm{d}\phi^2$ and $Y = \partial/\partial\phi$. Such adapted coordinates are locally unique.
[Note that, since $M$ is simply-connected (and, let's assume, connected, though the OP didn't include that condition), the functions $r$ and $\theta$ extend globally to $M$ uniquely so that $g = \mathrm{d}r^2 + a^2\,\mathrm{d}\theta^2$ and $X=\partial/\partial\theta$ where $a$ is a positive smooth function on $M$ satisfying $\mathrm{d}r\wedge\mathrm{d}a = 0$. The smooth mapping $(r,\theta):M\to\mathbb{R}^2$ is an immersion, but there is no reason to believe, under the given hypotheses, that it is an embedding, nor is it necessarily true that $a = A'(r)$ for some function $A:r(M)\to\mathbb{R}$.
The situation with $(N,h,Y)$ is similar.]
Supposing that an $f$ exists with all the specified properties, we can, by shrinking $\epsilon$, assume that $f(V)\subset W$ and hence, using the fact that $f\_\ast(X) = Y$, conclude that
$$
f^\*(s) = s\circ f = R(r)\quad\text{and}\quad f^\*(\mathrm{d}\phi) = \mathrm{d}\theta + M(r)\,\mathrm{d}r.\tag1
$$
for some functions $R$ and $M$ on $(-\epsilon,\epsilon)$ with $R(0)=0$ and $R'>0$. This implies that, relative to the orthonormal coframings, we must have
$$
f^\*\begin{pmatrix}\mathrm{d}s\\ B'(s)\,\mathrm{d}\phi\end{pmatrix}
=
\begin{pmatrix}R'(r) & 0\\ B'(R(r))M(r) & B'(R(r))/A'(r)\end{pmatrix}
\begin{pmatrix}\mathrm{d}r\\ A'(r)\,\mathrm{d}\theta\end{pmatrix}\tag2
$$
Now, the constancy of the singular values implies that, in particular, the determinant of the above coefficient matrix must be constant, i.e., that there must be a constant $c\_2>0$ such that
$$
R'(r)B'(R(r))/A'(r) = c\_2\,.\tag3
$$
Since $B(0) = R(0) = A(0) = 0$, we then integrate to get $B(R(r)) = c\_2\,A(r)$. In particular, since $B$ is invertible, $R(r) = B^{-1}\bigl(c\_2\,A(r)\bigr)$ for some positive constant $c\_2$.
Now, the sum of the squares of the singular values of the coefficient matrix must be another constant $c\_1 > 2c\_2$ (so that the two constant singular values will be distinct) such that
$$
R'(r)^2 + B'(R(r))^2\,M(r)^2 + B'(R(r))^2/A'(r)^2 = c\_1\,.\tag4
$$
Using the above formula for $R'(0) = c\_2 A'(0)/B'(0)$, we see that, by taking $c\_1$ sufficiently large, we can guarantee that the above equation for $M(r)$ has (two) real solutions on a neighborhood of $r=0$.
Conversely, for $c\_2>0$ and $c\_1 > 2c\_2$ sufficiently large and $\epsilon>0$ sufficiently small, there will be functions $R(r)$ and $M(r)$ that satisfy the above equations (3) and (4) and the initial condition $R(0)=0$, and hence, via (1) and the initial condition $f(p)=q$, they will determine a unique mapping $f$ with the desired properties.
Thus, local mappings $f$ with constant, distinct singular values always exist carrying any desired point to any other. Moreover, it is clear that there is a 2-parameter family of such local mappings carrying any given point in the domain to any given point in the range.
The existence of a global such mapping $f:M\to N$ depends on the growth properties of the functions $A$ and $B$ and the validity of their domains. Little more can be said about this without more information or hypotheses about the functions $A$ and $B$.
| 4 | https://mathoverflow.net/users/13972 | 414499 | 169,053 |
https://mathoverflow.net/questions/414477 | 3 | Suppose that I want to send a message (consisting of bits) over a channel where from $n$ transferred bits as many as $n/2-\varepsilon n$ might be flipped, i.e., the distance of the code is $n-2\varepsilon n$. How does the number of encodable messages $m$ change as a function of $n$ and $\varepsilon$? I do not want to assume that $\varepsilon$ is a constant, so it can be for example $1/\log n$.
| https://mathoverflow.net/users/955 | Coding over very noise channel | For a binary code, as $n$ grows you cannot do better than the repetition code $$C=\{11\cdots1,00\cdots0\},$$ with two codewords as soon as the minimum distance required $d>n/2.$ See Theorem 4 of Venkat Guruswami's notes, for example, available [here][1]. The result for your case is
that if a binary code $C$ must satisfy $d>n/2$ then
$$
|C | = m \leq \frac{2d}{2d-n}.
$$
Letting $d=n(1-\frac{2}{\log n})$ gives
$$
m \leq \frac{2n(1-\frac{2}{\log n})}{2n(1-\frac{2}{\log n})-n}=
\frac{2n(1-\frac{2}{\log n})}{n(1-\frac{4}{\log n})}.
$$
For $d=n/2$ and below there is much more freedom: You have, for example, the first order Reed Muller Code which is formed by taking $n=2^k,$ and the rows of a $2^k\times 2^k$ Sylvester Hadamard Matrix as well as their translates obtained by adding the all 1 vector to obtain $m=2^{k+1}=2n.$
**Edit:** If $d=n/2-\varepsilon,$ then using the Gilbert-Varshamov lower bound the asymptotic relative rate $R(C)=m/n$ of the code obeys
$$
R(C)\geq 1-h\_2(\delta)-o(1),
$$
where $\delta=d/n$ is the relative distance. Here $\delta=1/2-\varepsilon$ yields
$$
R(C) \geq 1-h\_2(1/2-\varepsilon)-o(1)
$$
which can be estimated from below as
$$
R(C) \geq
1-\sqrt{1-4\varepsilon^2}\geq 1-(1-2 \varepsilon^2)=2 \varepsilon^2,
$$
by using $h\_2(q)\geq 2\sqrt{q(1-q)},$ for the binary entropy function.
[1]: [https://www.cs.cmu.edu/~venkatg/teaching/codingtheory/notes/notes4.pdf](https://www.cs.cmu.edu/%7Evenkatg/teaching/codingtheory/notes/notes4.pdf)
| 3 | https://mathoverflow.net/users/17773 | 414508 | 169,058 |
https://mathoverflow.net/questions/414322 | 6 | Recent references on the matter at hand include, a lecture slide [The Konvalinka-Amdeberhan conjecture
and plethystic inverses](https://people.brandeis.edu/%7Egessel/homepage/slides/K-A%20conjecture.pdf) and a preprint on [Counting tanglegrams with species](https://arxiv.org/pdf/1509.03867.pdf) by I. Gessel; the initial work [On the enumeration of tanglegrams and tangled chains](https://arxiv.org/pdf/1507.04976.pdf) by S. Billey et al.
For each prime $p$, let's consider the family of (integral) sequences
$$a\_p(n):=\sum\_{\lambda\vdash n}\frac{1}{z\_\lambda}\prod\_{i=2}^{\ell(\lambda)}(p\lambda\_i+\cdots+p\lambda\_{\ell(\lambda)}+1);$$
where the sum runs through all $p$-ary partitions $\lambda$, of $n$, and $\ell(\lambda)$ is the length of the partition and the numbers $z\_{\lambda}$ are well-known since the number of permutations in $\mathfrak{S}\_n$ with cycle type $\lambda$ is computed by $\frac{n!}{z\_{\lambda}}$.
On the other hand, Michael Somos proposed the functional equations $$
x\cdot A\_p(x)^p=\frac{x\cdot A\_p(x^p)}{1-p\cdot x\cdot A\_p(x^p)}
$$ and a couple of these are listed on OEIS as [A085748](https://oeis.org/A085748) and [A091190](https://oeis.org/A091190). However, there are not enough interpretations attached to $A\_p(x)$ or its coefficients on OEIS.
After some experimentation, I wish to ask:
>
> **QUESTION.** Can the following be justified or refuted?
> $$A\_p(x)=\sum\_{n\geq0}a\_p(n)x^n.$$
>
>
>
**Remark.** Observe that neither $A\_p(x)$ nor $a\_p(n)$ have been found to output integers, unless $p$ is a prime.
| https://mathoverflow.net/users/66131 | Tanglegrams and functional equations of M. Somos | In my lecture slides that Tewodros cites, I studied symmetric functions that I called $g\_m$, where $m$ is a positive integer, that have constant term 1 and satisfy
$$ -L\_m[g\_m] = p\_1,$$
where $L\_m$ is the Lyndon symmetric function given by
$$L\_m = \frac{1}{m} \sum\_{d\mid m} \mu(d) p\_d^{m/d}.$$
Here $\mu$ is the Möbius function, the $p\_i$ are power sum symmetric functions, and $L\_m[g\_m]$ is the plethysm of $L\_m$ and $g\_m$. I showed that $g\_m$ is an integral symmetric function; i.e., its coefficients in the underlying variables are integers, and that if $m$ is a power of the prime $q$ then for any $\alpha$,
\begin{multline\*}
\quad
g\_m^{-\alpha} = 1+
\sum\_{n=1}^\infty \sum\_{\lambda}\frac{p\_{\lambda}}{z\_\lambda}
\times
\alpha\prod\_{j=2}^{l(\lambda)} (m\lambda\_j+m\lambda\_{j+1}+\cdots +m\lambda\_{l(\lambda)}+\alpha),\quad\tag{1}
\end{multline\*}
where the sum on $\lambda$ is over all partitions of $n$ in which every part is a power of $q$. (I'm using $q$ here instead of $p$ to avoid confusion with the power sum symmetric functions.) This shows that if $\alpha$ is positive and $m$ is a power of a prime then all coefficients of $g\_m^{-\alpha}$ are positive. I don't know that this is true if $m$ is not a power of a prime, though I suspect that it is.
Now let us specialize the symmetric functions (in the variables $x\_1, x\_2,\dots$) by setting $x\_1=x$ and $x\_i=0$ for $i>0$, or equivalently, $p\_i =x^i$ for all $i$. Let $G\_m(x)$ be the image of $g\_m$ under this specialization. Then $G\_m(x)$ satisfies the functional equation
$$-\frac{1}{m}\sum\_{d\mid m}\mu(d) G\_m(x^d)^{m/d} = x.$$
If $m$ is a power of the prime $p$, this may be written
$$G\_m(x^p)^{m/p} -G\_m(x)^m = mx$$
and it's easy to check that if $m=p$ then $A\_p(x) =1/G\_p(x)$, where $A\_p(x)$ is as in the original question. Moreover, it follows from $(1)$ that if $m$ is a power of the prime $p$ then for any $\alpha$ we have
\begin{multline\*}
\quad
G\_m(x)^{-\alpha} = 1+
\sum\_{n=1}^\infty x^n
\sum\_{\lambda}\frac{\alpha}{z\_\lambda}\,
\times\prod\_{j=2}^{l(\lambda)} (m\lambda\_j+m\lambda\_{j+1}+\cdots +m\lambda\_{l(\lambda)}+\alpha),\quad\tag{2}
\end{multline\*}
where the sum on $\lambda$ is over all $p$-ary partitions of $n$. The case $m=p$, $\alpha=1$ answers the OP's questions in the affirmative.
In all cases $G\_m(x)$ has integer coefficients, but I can only prove that $G\_m(x)^{-1}$ has positive coefficients when $m$ is a prime power, when it follows from $(2)$, though this is likely true in all cases. The stronger statement that $1-G\_m(x)$ has positive coefficients also follows from $(2)$ when $m$ is prime power.
I am working on a paper with detailed proofs of these formulas, but it will take a while.
| 6 | https://mathoverflow.net/users/10744 | 414516 | 169,060 |
https://mathoverflow.net/questions/414514 | 5 | In proposition 2.7. of the condensed notes of professors Scholze and Clausen it is said that the category of extremally disconnected sets is a site, but in the definition of a site in the Stacks Project (<https://stacks.math.columbia.edu/tag/00VH>) it is necessary for a site to have fibre products sometimes (Axiom (3) in the definition) and extremally disconnected sets don't have all fibre products.
The category of extremally disconnected sets is a site using the definition from the Stacks Project?
| https://mathoverflow.net/users/130868 | The site of extremally disconnected sets | Usage varies. Let's at least stipulate that "site" is synonymous with "category equipped with a Grothendieck topology".
Some, but not all, authors, require a site to have pullbacks, because this assumption simplifies the definition a bit. But e.g. the [nlab](https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/Grothendieck+topology#definition_using_sieves) gives the definition which doesn't assume one has pullbacks. Apparently this version of the definition goes back at least to SGA 4.
| 8 | https://mathoverflow.net/users/2362 | 414520 | 169,062 |
https://mathoverflow.net/questions/413220 | 7 | I am working on $\mathbb{Z}/18\mathbb{Z}$ elliptic curves over cubic fields. The curves are created using the formulas on p. 584 of
>
> D. Jeon, C. H. Kim, Y. Lee, *Families of elliptic curves over cubic number fields with prescribed torsion subgroups*, Mathematics of Computation, V. 80, 273, January 2011, p. 579-591, JSTOR: [41104715](https://www.jstor.org/stable/41104715).
>
>
>
My [code snippet](https://mega.nz/file/eggGnRbb#adIzMWp1MIOg90fNILw4LwlHW4C9W45YJJJnNTzyLgM) with the saved output for [Magma Calculator online](http://magma.maths.usyd.edu.au/calc/) is available for download from MEGA.
I observe that the following triples of rational $t$-values produce curves with similar characteristics:
$$t\_1=t$$ $$t\_2=1-\frac{1}{t\_1}=1-\frac{1}{t}$$ $$t\_3=1-\frac{1}{t\_2}=\frac{1}{1-t}$$
For a triple $t\_1,t\_2,t\_3$, the three elliptic curves are different over three different cubic fields, with different discriminants and conductors.
But the ranks, $j$-invariants, and heights of all generators are the same (e.g., for rank $2$ there will be height $h\_1$ for generators $g\_{11}, g\_{12}, g\_{13}$ and height $h\_2$ for generators $g\_{21}, g\_{22}, g\_{23}$, where $g\_{ik}$ is the $i$-th generator for the curve created using $t\_k$).
It was also pretty straightforward to derive the formula for the $j$-invariant and see that it is always rational:
$$j=\frac{(t^3-3t^2+1)^3(t^9-9t^8+27t^7-48t^6+54t^5-45t^4+27t^3-9t^2+1)^3}{(t^3-6t^2+3t+1)(t^2-t+1)^3(-1+t)^9t^9}$$
Magma is unable to check whether the curves are isomorphic, as they are defined over different cubic fields:
```
>> IsIsomorphic(E1, E2);
^
Runtime error in 'IsIsomorphic': Curves must be defined over the same base ring
```
Isogeneity check is unavailable over number fields (only over rationals or finite fields):
```
>> IsIsogenous(E1, E2);
^
Runtime error in 'IsIsogenous': Bad argument types
Argument types given: CrvEll[FldNum[FldRat]], CrvEll[FldNum[FldRat]]
```
The curves seem to be **essentially the same (not distinct in any real sense)** to me, even though they might not be considered isomorphic and/or isogenous.
>
> **Question 1:** Does there exist a proper mathematical name for "essentially the same" used above, or the name for the observed connection between the curves?
>
>
>
>
> **Question 2:** Is it possible to map a generator discovered on one of the curves to the other two curves? The explicit expression for the map is not a priority yet.
>
>
>
>
> **Question 3:** If the height of the generator (but not the generator itself) is considered to be known, is it possible to speed up a search process for it? If so, how?
>
>
>
**Rationale for Questions 2 and 3:** It is very easy to determine both generators (with heights $1.798$ and $11.652$, default `Effort := 1` in $45$ seconds) for $t\_1=\frac{1}{5}$, harder to do so for $t\_2=-4$ (`Effort := 1000` helps, takes much longer), and very hard to recover the second generator for $t\_3=\frac{5}{4}$ (`Effort := 1600` fails).
| https://mathoverflow.net/users/95511 | ℤ/18ℤ elliptic curves over cubic fields | We already have an accepted answer, but since i had already started an answer and was at the half of the route beyond getting the essence of the structure, i completed it now, since it may be useful in similar contexts.
---
On the mathematical side the situation is as follows, recalled for the convenience of the reader from the literature.
Notations are as in the already cited paper:
[Families of Elliptic Curves over Cubic Number Fields with Prescribed Torsion Subgroups, Daeyeol Jeon, Chang Heon Kim, And Yoonjin Lee](https://www.ams.org/journals/mcom/2011-80-273/S0025-5718-10-02369-0/S0025-5718-10-02369-0.pdf)
A further reference that should not be omitted is:
[Markus Reichert, Explicit Determination of Nontrivial Torsion Structures of Elliptic Curves Over Quadratic Number Fields](https://www.ams.org/journals/mcom/1986-46-174/S0025-5718-1986-0829635-X/S0025-5718-1986-0829635-X.pdf)
In order to produce an example of a curve $E$ with torsion $\Bbb Z/18$, the Ansatz is to work with the Tate normal form, consider curves $E=E(b,c)$ parametrized by two
algebraic numbers $b,c$ from a cubic number field $K$,
$$
E = E(b, c)\ :\qquad y^2 + (1 − c)xy − by = x^3 − bx^2\ ,
$$
and arrange that the point
$P = (0, 0)$ has order $18$.
For this, pick two parameters $(U,V)$ satisfying the equation for $X\_1(18)$:
$$
\begin{aligned}
X\_1(18) \ :\qquad
g\_{18}(U,V) &= 0\ ,\qquad\text{ where}
\\
g\_{18}(U,V)
&:=(U-1)^2 V^2 - (U^3 - U + 1)V + U^2(U - 1) \\
&= U^3(1-V) + U^2 (V^2 -1) + U(V-2V^2) + (V^2-V)\\
&\sim\_{\Bbb Q(V)^\times}
U^3 - U^2(V + 1) + \frac{2V^2-V}{V-1} -V\ .
\ .
\end{aligned}
$$
Seen as a polynomial in $U$, it has degree $3$.
We set $V=t$ to be a "suitable" rational number,
and the polynomial $g\_{18}(U,t)$ defines a cubic field $K=\Bbb Q(\alpha\_t)$
generated by some $\alpha\_t$. Let me plot the connection to $X\_1(18)$ explicitly:
$$
g\_{18}(\alpha\_t,t)=0\ .
$$
Then the formulas for $b,c$ are given by one and the same
rational function in $(U,V)=(\alpha\_t,t)$. They are:
$$
\begin{aligned}
b(U,V) &=
-\frac
{V(U - V)(U^2 + V)(U^2 -UV + V)}
{(U^2 -V^2+V)(U^2 + UV -V^2 + V)^2}
\ ,
\\
c(U,V) &=
-\frac
{V(U - V)(U^2 -UV + V)}
{(U^2 -V^2+V)(U^2 + UV -V^2 + V)}
\ .
\end{aligned}
$$
---
*Warming up.* We proceed as follows in the given context from above.
We fix some $t$. To have a concrete example, $t$ may be specialized to $t=t\_1=1/5$, as the OP does it also. Let $t'$ be its cousin,
$$
t'=t\_3=\frac 1{ 1-t }\ .
$$
We build the corresponding field $K=\Bbb Q(\alpha)$, where $\alpha =\alpha\_t$ is a suitable root of the polynomial $g\_{18}(U,t)$, seen as a polynomial in $U$. Let $K'=\Bbb Q(\alpha')$ be the cousin field, where $\alpha'$ is a specific root for $g\_{18}(U, t')$.
**Question:** Are $K$ and $K'$ isomorphic (for some good choice of $\alpha'$)?
**Answer:** Yes, they are, take $\displaystyle \alpha'= 1-\frac 1\alpha$.
To illustrate the situation, we consider first the sample case $t=1/5$.
[Sage](https://www.sagemath.org) gives this information as follows:
```
def g18(U, V):
return (U^3*V - U^2*V^2 - U^3 + 2*U*V^2 + U^2 - U*V - V^2 + V)
R.<U> = PolynomialRing(QQ)
t1 = 1/5
a1 = g18(U, t1).roots(ring=QQbar, multiplicities=False)[0]
t3 = 1/(1 - t1)
a3 = 1 - 1/a1
print(f'g18(a3, t3) = {g18(a3, t3)}')
```
The sage interpreter gives after a copy+paste of the above code, together with one more line to be sure we get a clean zero:
```
g18(a3, t3) = 0.?e-17
sage: g18(a3, t3).minpoly()
x
```
Because of the rôle of $(\alpha,t)$ as a special value for $(U,V)$,
i will use below rather $(u,v)$ pairs instead. Now the whole context can be explained *structurally* as follows.
---
**Proposition:**
Let $F$ be a field (of characteristic $\ne 2,3$).
For two parameters $b,c\in F$, $b\ne 0$, let $E\_T(b,c)$ be the elliptic curve in Tate normal form
$$
E\_T(b, c)\ :\qquad y^2 +(1-c)xy -by = x^3 bx^2\ ,
$$
so that $P=(0,0)$ is a rational point on it.
For suitable ($\Delta(A,B)\ne 0$) parameters $A,B\in F$ let consider also the elliptic curve in short Weierstrass form
$$
E\_W(A,B)\ :\qquad y^2 = x^3 + Ax+B\ .
$$
Fix $u,v$ în $F$, $u\ne 1$, so that the pair
$(u,v)$ corresponds to a point on the moduli space $X\_1(18)$ parametrized
as mentioned above, i.e. it satisfies
$$
g\_{18}(u,v)=0\ ,\qquad\text{ where }\\
g\_{18}(U,V)=
(U-1)^2 V^2 -(U^3 -U + 1)V + U^2(U - 1)\ .
$$
Then the pair $(u',v')$ with components
$$
\begin{aligned}
u' &= \frac 1{1-u}\ ,\\
v' &= 1-\frac 1v\ ,
\end{aligned}
$$
is also defining a pointin the moduli space $X\_1(18)$,
i.e. $g\_{18}(u',v')=0$. Let $\underline A$, $\underline B$
be the rational functions given by
$$
\begin{aligned}
\underline A(b,c) &= -\frac 1{48}\Big(\ ((c-1)^2 - 4b)^2 - 24b(c - 1)\ \Big)\ ,\\
\underline B(b,c) &= \frac 1{864}\Big(\ ((c-1)^2 - 4b)^3 - 36b(c-1)^3 + 72b^2(2c + 1)\ \Big)\ .
\end{aligned}
$$
Consider with a slight abuse of notation $b,c\in F$ and $b',c'\in F$, then $A,B\in F$ and $A',B'\in F$ as follows
$$
\begin{aligned}
b &= b(u,v)\ ,\qquad &b' &= b(u',v')\ ,\\
c &= c(u,v)\ ,\qquad &c' &= c(u',v')\ ,\\[2mm]
A &= \underline A(b,c)\ ,\qquad &A' &=\underline A(b',c')\ ,\\
B &= \underline B(b,c)\ ,\qquad &B' &=\underline B(b',c')\ ,\\[2mm]
&\qquad\text{ and consider the elliptic curves}\\[2mm]
E\_T &= E(b, c)\ , \qquad &E'\_T &= E\_T(b', c')\\
E\_W &= E(A, B)\ , \qquad &E'\_W &= E\_W(A', B')\ .
\end{aligned}
$$
Then$$
\frac {A'}A = U^{12}\ ,\qquad
\frac {B'}B = U^{18}\ ,
$$
so the elliptic curves $E\_W$ and $E\_W'$ are canonically isomorphic via a map $\Phi$, as shown in the diagram below. The functions
$\underline A$, $\underline B$ were chosen to make
$E\_T(b,c)$ isomorphic $E\_W(A,B)$. Then the following diagram is commutative:
$\require{AMScd}$
$$
\begin{CD}
E\_T @>{\cong}>> E\_W\\
@A{\cong} AA @A\cong A\Phi A\\
E'\_T @>>\cong> E'\_W
\end{CD}
$$
So we can compare the rational points $P=(0,0)\in E\_T(F)$ and $P'=(0,0)\in E'\_T(F)$ in one or any of the common worlds, e.g. in $E\_W(F)$, and then $11P$ and $P'$
(or equivalently $P=5\cdot 11 P$ and $5P'$) correspond to one and the same torsion point of order (dividing) $18$. In a diagram:
$\require{AMScd}$
$$
\begin{CD}
P\_T @>{\cong}>> P\_W=5\Phi(P'\_W)=\Phi(5P'\_W)\\
@. @A\cong A\Phi A\\
5P'\_T @>>\cong> 5P'\_W
\end{CD}
$$
**Proof** by computer.
$\square$
---
Code for the proof. First let us define the needed functions, and needed objects.
```
def bmap(U, V):
return -(U^2 - U*V + V) * (U^2 + V) * (U - V) * V / (U^2 + U*V - V^2 + V)^2 / (U^2 - V^2 + V)
def cmap(U, V):
return -(U^2 - U*V + V) * (U - V) * V / (U^2 + U*V - V^2 + V) / (U^2 - V^2 + V)
def Amap(b, c):
return -1/48 * ( ((c-1)^2 - 4*b)^2 - 24*b*(c - 1) )
def Bmap(b, c):
return 1/864 * ( ((c-1)^2 - 4*b)^3 - 36*b*(c-1)^3 + 72*b^2*(2*c + 1) )
def f(U, V):
return (U^3*V - U^2*V^2 - U^3 + 2*U*V^2 + U^2 - U*V - V^2 + V)
R.<U,V> = PolynomialRing(QQ)
Q = R.quotient( f(U, V) )
FR = R.fraction_field()
FQ = Q.fraction_field()
u1, v1 = FQ(U), FQ(V)
u2, v2 = 1/(1 - u1), 1 - 1/v1
```
Now we can check:
```
print(f'Is f(u2, v2) zero? {bool( f(u2, v2) == 0 )}' )
b , c = bmap(U , V ), cmap(U , V )
b1, c1 = bmap(u1, v1), cmap(u1, v1)
b2, c2 = bmap(u2, v2), cmap(u2, v2)
A , B = Amap(b , c ), Bmap(b , c )
A1, B1 = Amap(b1, c1), Bmap(b1, c1)
A2, B2 = Amap(b2, c2), Bmap(b2, c2)
print(f'Is A2/A1 = u1^12? {bool( A2/A1 == u1^12 )}')
print(f'Is B2/B1 = u1^18? {bool( B2/B1 == u1^18 )}')
ET = EllipticCurve(FR, [1 - c, -b, -b, 0, 0])
EW = EllipticCurve(FR, [A, B])
phi = ET.isomorphism_to( EW )
PT = ET.point( (0, 0, 1) )
PW = phi( PT )
five_PW = 5*PW
x_PW , y_PW = PW.xy()
x_five_PW, y_five_PW = five_PW.xy()
x_PW.subs({U: u1, V:v1}) == x_five_PW.subs({U: u2, V: v2}) / u1^6
y_PW.subs({U: u1, V:v1}) == y_five_PW.subs({U: u2, V: v2}) / u1^9
```
This gives the needed confirmations:
```
Is f(u2, v2) zero? True
Is A2/A1 = u1^12? True
Is B2/B1 = u1^18? True
True
True
```
The last two `True` values confirm that the coordinates of $P=(0,0)=E\_T(F)$ and $5P'$ where $P'=(0,0)\in E\_T'$ are the same, when transported to $E\_W(F)$.
*Note:* Unfortunately, sage cannot build the needed curves over `FQ`.
| 5 | https://mathoverflow.net/users/122945 | 414529 | 169,066 |
https://mathoverflow.net/questions/414479 | 9 | *This question is a follow-up to another question of mine, with different language - see the link below.*
Say that an infinite regular cardinal $\kappa$ is **Fraissean** iff the logic $\mathcal{L}\_{\kappa,\omega}$ has the following property (called "SED" in the below-linked question):
>
> For every finite signature $\Sigma$ there is a larger signature $\Sigma'$ containing $\Sigma$ and two new unary predicate symbols $A,B$ - and possibly more symbols besides - and an $\mathcal{L}\_{\kappa,\omega}[\Sigma]$-sentence $\eta$ such that, for every pair of $\Sigma$-structures $\mathfrak{A},\mathfrak{B}$, we have $\mathfrak{A}\equiv\_{\kappa,\omega}\mathfrak{B}$ iff there is an $\mathfrak{M}\models\eta$ with $A^\mathfrak{M}\upharpoonright\Sigma\cong\mathfrak{A}$ and $B^\mathfrak{M}\upharpoonright\Sigma\cong\mathfrak{B}$.
>
>
>
*(That $\omega$ is Fraissean is an immediate consequence of Fraisse's characterization of elementary equivalence in terms of Ehrenfeucht-Fraisse games, hence the name.)*
[Farmer S showed](https://mathoverflow.net/a/414179/8133) that $\omega\_1$ is **not** Fraissean; that argument, however, does not seem to immediately generalize to higher cardinals, the issue being that $\mathcal{L}\_{\kappa,\omega}$-sentences are not generally coded by reals and so even in the presence of large cardinals we lose a necessary absoluteness result.
My question is: what can we say, in $\mathsf{ZFC}$ alone, about the situation re: $\omega\_2$? *Neither* possibility has an obvious consistency proof to me; I would tentatively hazard a guess that $L$ thinks $\omega\_2$ is not Fraissean and that the tree property at $\omega\_2$ implies that $\omega\_2$ is Fraissean, but both of these are essentially just free association.
| https://mathoverflow.net/users/8133 | Logics detecting their own equivalence notions, take two: $\mathcal{L}_{\omega_2,\omega}$ | (Working in ZFC.)
$\omega\_2$ is not Fraissean. In fact, it is not Fraissean with respect to $\Sigma$, where $\Sigma$ is the signature with a single binary relation $<$.
To see this we use a variant of the argument you linked in the question. Suppose otherwise, and let $\Sigma'$ and $\eta$ witness this. Let $\gamma$ be a large enough ordinal and let $\pi:M\to V\_\eta$ be elementary, with $M$ transitive and $M^{\omega\_1}\subseteq M$ and $\mathrm{crit}(\pi)=\kappa$ exists. Let $\mathfrak{A}=(\kappa,{{\in}\upharpoonright\kappa})$ and $\mathfrak{B}=(\pi(\kappa),{{\in}\upharpoonright\pi(\kappa)})$.
Since $M^{\omega\_1}\subseteq M$, $M$ is correct about $\mathcal{L}\_{\omega\_2,\omega}$-truth, and also by elementarity of $\pi$, therefore $\mathfrak{A}\equiv\_{\mathcal{L}\_{\omega\_2,\omega}}\mathfrak{B}$. So let $\mathfrak{M}$ witness the choice of $\Sigma',\eta$ with respect to $\mathfrak{A},\mathfrak{B}$.
Let $G$ be $V$-generic for $\mathbb{P}=\mathrm{Coll}(\omega\_1,\theta)$ where $\theta=\max(\pi(\kappa),\mathrm{card}(\mathfrak{M}))$ (collapsing $\theta$ to size $\aleph\_1$ with countable conditions). Then $V[G]\models$"There are structures $\mathfrak{A}',\mathfrak{B}',\mathfrak{M}'$, each having universe $\omega\_1$, such that $\mathfrak{A}',\mathfrak{B}'$ are in signature $\Sigma$, and $\mathfrak{M}'$ in signature $\Sigma'$, and $A^{\mathfrak{M}}\upharpoonright\Sigma\approx\mathfrak{A}'$ and $B^{\mathfrak{M}}\upharpoonright\Sigma\approx\mathfrak{B}'$, as witnessed by isomorphisms $\sigma,\tau$, and $\mathfrak{M}'\models\eta$ and there is a sentence $\varphi$ of $\mathcal{L}\_{\omega\_2^{V[G]},\omega}$, coded by a set $X\subseteq\omega\_1$, such that $\mathfrak{A}'\models\varphi$ but $\mathfrak{B}'\models\neg\varphi$" (consider the natural sentence specifying the ordertype of $\kappa$). Fix names $\dot{\mathfrak{A}}',\dot{\mathfrak{B}}',\dot{\mathfrak{M}}',\dot{\sigma},\dot{\tau},\dot{X}$ for such objects (we may assume the empty condition forces the above things to hold for these names). So these are all basically names for subsets of $\omega\_1$.
Now working in $V$, given an $\aleph\_1$-sized family $\mathscr{F}$ of dense subsets $D\subseteq\mathbb{P}$, we can build an $\mathscr{F}$-generic filter $G$, because $\mathbb{P}$ is countably closed. We claim that by picking $\mathscr{F}$ appropriately, and $G$ be $\mathscr{F}$-generic, then letting $\mathfrak{A}'$, etc, be the interpretations $\dot{\mathfrak{A}}'\_G$, and $\varphi$ the sentence in $\mathcal{L}\_{\omega\_2,\omega}$ the sentence coded by $X'$, then the sentence of the previous paragraph which held in the generic extension, holds in $V$ about these objects, which contradicts our assumptions.
To arrange $\mathscr{F}$: First, it is straightforward to arrange $\aleph\_1$-many dense sets which arrange that $\dot{\sigma}\_G$ and $\dot{\tau}\_G$ will truly be isomorphisms; the main thing is to arrange the that the domain and codomain are the right sets. To arrange that $\mathfrak{M}'\models\eta$, consider not just $\eta$, but the set $S$ of all subformulas thereof.
Now here that all formulas in $S$ have only finitely many distinct free variables (if $\psi$ has infinitely many distinct free variables, proceed by induction on the rank of formulas which have it as a subformula, to see that none of them are sentences).
Since $\eta$ is in $\mathcal{H}\_{\omega\_2}$, we can fix a surjection $\pi:\omega\_1\to S^+$, where $S^+$ is the set of pairs $(\psi,\vec{\alpha})$, where $\psi\in S$ and $\vec{\alpha}$ is an assignment of the (finitely many) free variables of $\psi$ to ordinals ${<\omega\_1}$.
Now the basic point is that if $(\psi,\vec{\alpha})\in S^+$ and $\psi$ is a disjunction $\bigvee\_{\gamma<\omega\_1}\psi\_\gamma$, and $p\in\mathbb{P}$ forces $\dot{\mathfrak{M}}'\models\psi(\vec{\alpha})$, then for each $q\leq p$ we can
pick some $\gamma<\omega\_1$ and $r\leq p$ such that $r$ forces that $\dot{\mathfrak{M}}'\models\psi\_\gamma(\vec{\alpha})$. Thus, we can include a dense set $D\_{\psi,\vec{\alpha}}$ consisting of those conditions $p$ such that either $p$ forces that $\dot{\mathfrak{M}}'\models\neg\psi(\vec{\alpha})$ or there is $\gamma<\omega\_1$ such that $p$ forces that $\dot{\mathfrak{M}}'\models\psi\_\gamma(\vec{\alpha})$. Similarly, if $\psi$ is of form $\exists x\varrho$, then we can include the dense set of conditions $p$ either forcing that $\dot{\mathfrak{M}}'\models\neg\psi(\vec{\alpha})$, or such that there is $\alpha<\omega\_1$ such that $p$ forces $\dot{\mathfrak{M}}'\models\varrho(\alpha,\vec{\alpha})$. And naturally, for each $(\psi,\vec{\alpha})\in S^+$, we include the dense set of $p$ deciding whether $\dot{\mathfrak{M}}'\models\psi(\vec{\alpha})$. There are also natural dense sets forcing the atomic formulas the correspond to the structure of $\dot{\mathfrak{M}}'$.
For $X'$ and $\varphi'$ it is similar, but we no longer have
$\varphi$ fixed in advance. But we can fix a name $\dot{S}\_{\dot{\varphi}}^+$ for the set of all pairs $(\psi,\vec{\alpha})$ such that $\psi$ is a subformula of $\varphi$ (note in an actual forcing extension, we could have $\psi\notin V$), and $\vec{\alpha}$ an assignment of the free variables of $\psi$ to elements ${\in\omega\_1}$. Fix a name for a surjection $\dot{f}:\omega\_1\to \dot{S}^+\_{\dot{\varphi}}$. Write $\dot{\psi}\_\gamma$ for the name for the first component of $\dot{f}(\gamma)$, and $\vec{\alpha}\_\gamma$ for the name for the second. Then for each $\gamma<\omega\_1$, we can use the dense set deciding what sort of formula $\dot{\psi}\_\gamma$ is (in particular, whether it is a disjunction, and whether it is existential), and the dense set of all conditions $p$ such that (i) there is $\vec{\alpha}$ such that $p$ forces "$\dot{\vec{\alpha}}\_\gamma=\vec{\alpha}$", (ii) for some $\beta<\omega\_1$, either
* $p$ forces that $\dot{\psi}\_\gamma$ is not a disjunction, or
* $p$ forces that $\dot{\mathfrak{A}}'\models\neg{\dot{\psi}}\_\gamma(\vec{\alpha})$, or
* $p$ forces "$\dot{\psi}\_\gamma$ is a disjunction,
$\dot{\psi}\_\beta$ is one of the disjuncts of $\dot{\psi}\_\gamma$, $\dot{\vec{\alpha}}\_\beta=\vec{\alpha}$ and $\dot{\mathfrak{A}}'\models\dot{\psi}\_\beta(\vec{\alpha})$",
and (iii) there are $\alpha,\beta<\omega\_1$ such that either
* $p$ forces that $\dot{\psi}\_\gamma$ is not existential, or
* $p$ forces that $\dot{\mathfrak{A}}'\models\neg\dot{\psi}\_\gamma$, or
* $p$ forces "letting $\varrho=\dot{\psi}\_\beta$, then $\dot{\psi}\_\gamma$ is the formula $\exists x\ \varrho$, and
$\vec{\alpha}\_\beta=\vec{\alpha}\frown(\alpha)$,
and $\dot{\mathfrak{A}}'\models\varrho(\vec{\alpha},\alpha)$".
Also include dense sets deciding whether $\dot{\mathfrak{A}}'\models\dot{\psi}\_\gamma(\dot{\vec{\alpha}}\_\gamma)$ holds, for each $\gamma$.
Likewise for $\dot{\mathfrak{B}}'$.
Also for each pair of ordinals $\alpha,\beta<\omega\_1$, include the dense set of conditions $p$ such that either
* $p$ forces that $\dot{\psi}\_\alpha$ is not a
disjunction/conjunction, or
* $p$ forces that $\dot{\psi}\_\alpha$ is a disjunction/conjunction,
and $\dot{\psi}\_\beta$ is one of the disjuncts/conjuncts, or
* $p$ forces that $\dot{\psi}\_\alpha$ is a disjunction/conjunction,
and $\dot{\psi}\_\beta$ is not one of the disjuncts/conjuncts.
Also for each $\alpha<\omega\_1$, the dense set of conditions deciding the number (finitely many) of free variables of $\dot{\psi}\_\alpha$,
and arranging that $\dot{f}\_G$ truly enumerates all of the relevant pairs $(\psi,\vec{\alpha})$.
(I think this is now about enough dense sets.)
Now let $G$ be $\mathscr{F}$-generic. Let $X'=\dot{X}'\_G$, etc. It is straightforward to see that $\mathfrak{M}'\models\eta$ and $\mathfrak{A}'\approx A^{\mathfrak{M}'}\upharpoonright\Sigma$ (as witnessed by $\sigma'$) and likewise for $\mathfrak{B}'$. So by our assumptions, $\mathfrak{A}'\equiv\_{\mathcal{L}\_{\omega\_2,\omega}}\mathfrak{B}'$. But $X'$ does really code a sentence $\varphi'$ of $\mathcal{L}\_{\omega\_2,\omega}$, because the ordertype along which it is built is wellfounded, because by the $\sigma$-closure of $\mathbb{P}$, otherwise we can find a condition $p$ which forces something to be an illfounded ordinal. And $\dot{S}^+{\dot{\varphi}}\_G$ is the set of pairs $(\psi,\vec{\alpha})$ such that $\psi$ is a subformula of $\varphi'$, and $\vec{\alpha}$ an assignment of its free variables, and this set is enumerated by $\dot{f}\_G$. And note that the names for the satisfaction relations amongst these formulas evaluate to the correct satisfaction relation. Thus, $\mathfrak{A}'\models\varphi'$ but $\mathfrak{B}'\models\neg\varphi'$, a contradiction.
| 12 | https://mathoverflow.net/users/160347 | 414533 | 169,067 |
https://mathoverflow.net/questions/414495 | 4 | Note: I asked this question a few months ago [here](https://math.stackexchange.com/questions/4303093/relation-between-two-permutation-metrics), but received no answer.
Consider the following two metrics on permutations of $\{1,2,\dots,n\}$:
$d\_\text{swap}(\sigma,\tau)$ is the minimum number of swaps of adjacent elements that are required to reach $\tau$ from $\sigma$ (or $\sigma$ from $\tau$). Alternatively, it is the number of *discordant pairs* for $\sigma$ and $\tau$. A pair of distinct elements $(x,y)$ is called a *discordant pair* for $\sigma$ and $\tau$ if $x$ and $y$ have different relative orderings in the two permutations. If I am not missing anything, $d\_\text{swap}$ is identical to the [Kendell tau distance](https://en.wikipedia.org/wiki/Kendall_tau_distance).
$d\_\text{sum}(\sigma, \tau)$ is given by $\sum\_{i=1}^n \left|\operatorname{pos}\_\sigma(i) - \operatorname{pos}\_\tau(i) \right|$, where $\operatorname{pos}\_\pi$ indicates the position of $i$ in the permutation $\pi$.
I need to understand the relationship between these two metrics for another problem I am working on. In particular, I would like to know whether the following two conjectures I have are true:
* $d\_\text{sum} \geq d\_\text{swap}$
* there exists a constant $C < 1$, such that $d\_\text{swap} \geq C \cdot d\_\text{sum}$
While I am sure that these distances have been well studied, I did not manage to find the answer to these questions. If you know the answer or even any relevant literature feel free to help me out :)
| https://mathoverflow.net/users/475708 | Relation between two permutation metrics | Both statistics are preserved under right multiplication by a permutation, so you can reduce to the case where $\sigma$ is the identity.
In that case, $d\_{\rm swap}$ is the inversion number or equivalently the Coxeter length of the permutation, and $d\_{\rm sum}$ is the total displacement, studied by Diaconis and Graham (and more recently by Petersen and Tenner).
Diaconis and Graham proved that total displacement is at least the sum of length and reflection length, so your first conjecture is true.
Your second conjecture is false. Take the family of permutations where $\tau=n2\cdots(n-1)1$ (and $\sigma$ is the identity). Then $d\_{\rm swap}=2n-3$ and $d\_{\rm sum}=2n-2$, and the limit of the ratios is 1.
EDIT: I misread the second conjecture. Diaconis and Graham prove that $C=1/2$ works (and Petersen and Tenner show the bound is realized precisely by the permutations avoiding 321).
EDITED TO ADD REFERENCES:
Persi Diaconis and Ronald L. Graham, *Spearman's Footrule as a Measure of Disarray*. J. R. Stat.Soc. Ser. B. Stat. Methodol. **39** (1977), 262--268.
T. Kyle Petersen and Bridget E. Tenner, *The depth of a permutation*. J. Comb. **6** (2015) 145--178.
| 5 | https://mathoverflow.net/users/3077 | 414537 | 169,068 |
https://mathoverflow.net/questions/414511 | 3 | Is there a family of continuous functions $(f\_n)\_{n \in \mathbb{N}}$ on $[0,1]$ whose span is dense in $L^1[0,1]$ for the $L^1$-norm, but not dense in $L^2[0,1]$ for the $L^2$-norm?
---
Some preliminary considerations: I suspect the answer is yes, since there exist a family of $L^2$ functions dense in $L^1$ but not in $L^2$. See my answer to [Completeness of $\{ f\_n : n \in \mathbb N \} \subset C[0,1]$ in $L^1[0,1]$](https://math.stackexchange.com/questions/4348499/completeness-of-f-n-n-in-mathbb-n-subset-c0-1-in-l10-1) that uses the general [Müntz-theorem in $L^p$ spaces (see [Borwein and Erdélyi - The full Müntz Theorem in $C[0,1]$ and $L^1[0,1]$](https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.434.9392)): for $p=1, 2$ and distinct $\lambda\_i > -\frac{1}{p}$,
$$\text{the space $\operatorname{Span}(1, x^{\lambda\_1}, x^{\lambda\_2}, \dotsc)$ is dense in $L^p[0,1]\ \ $ iff $\ \ \sum \limits\_{i=1}^{\infty} \frac{\lambda\_i + 1/p}{(\lambda\_i + 1/p)^2 + 1} = \infty$.}$$
We can take $\lambda\_i = \frac{-1}{2} + \frac{1}{i^2}$ to have the sum diverge for $p=1$ and converge for $p=2$.
For the continuous case however, the Müntz theorem means it's useless to look for a family $(f\_n)$ in the form of power functions. Also, any suitable candidate obviously cannot have a dense span in $\mathcal{C}[0,1]$, which rules out the usual suspects. I think any approach to construct dense families in $L^p[0,1]$ spaces without directly relying on the density of polynomials in $\mathcal{C}[0,1]$ could be useful to study, but I haven't found any.
| https://mathoverflow.net/users/105382 | Functions dense in $L^1[0,1]$ but not in $L^2[0,1]$ | $\newcommand{\ep}{\varepsilon}$User Z. M [provided](https://mathoverflow.net/questions/414511/functions-dense-in-l10-1-but-not-in-l20-1#comment1062924_414511) an elegant and brief (and yet complete) detailization of Fedor Petrov's [comment](https://mathoverflow.net/questions/414511/functions-dense-in-l10-1-but-not-in-l20-1#comment1062883_414511).
Here is another detailization, which is pedestrian but explicit.
Let $F:=\{f\_1,f\_2,\dotsc\}$ be any countable subset of $C[0,1]$, which is dense in $C[0,1]$ with respect (w.r.) to the norm $\lVert\cdot\rVert\_\infty$. For instance, $F$ can be the set of all polynomials with rational coefficients. Then $F$ is also dense in $C[0,1]$ w.r. to the norm $\lVert\cdot\rVert\_1$. So, $F$ is dense in $L^1[0,1]$ w.r. to the norm $\lVert\cdot\rVert\_1$.
Let $h(x):=x^{-1/3}$. Then $h\in L^2[0,1]$.
For each natural $n$, we can change the function $f\_n$ to a function $g\_n\in C[0,1]$ such that
$g\_n$ is orthogonal to $h$ and yet $\lVert g\_n-f\_n\rVert\_1\le1/n$, for all $n$ — so that $g\_n$ differs little from $f\_n$ in $L^1$ if $n$ is large. Then clearly the set $G:=\{g\_1,g\_2,\dotsc\}$ will be dense in $L^1[0,1]$, but even the span of $G$ will not be dense in $L^2[0,1]$.
Indeed, let
\begin{equation\*}
g\_n(x):=
\begin{cases}f\_n(x)&\text{ if }\ep\_n\le x\le1, \\
f\_n(\ep\_n)-c\_n(\ep\_n-x)&\text{ if }0\le x<\ep\_n,
\end{cases}
\tag{1}\label{-1}
\end{equation\*}
where
\begin{equation\*}
c\_n:=\frac{I\_n+f\_n(\ep\_n)\frac32\,\ep\_n^{2/3}}{J\_n}=\frac{10}9\,(I\_n\ep\_n^{-5/3}
+\tfrac32\,f\_n(\ep\_n)\ep\_n^{-1}),
\tag{2}\label{0}
\end{equation\*}
\begin{equation\*}
I\_n:=\int\_{\ep\_n}^1 f\_n h,
\end{equation\*}
\begin{equation\*}
J\_n=\int\_0^{\ep\_n}dx\,(\ep\_n-x)h(x)
=\frac9{10}\,\ep\_n^{5/3},
\end{equation\*}
and $\ep\_n\in(0,1)$ is small enough so that
\begin{equation\*}
\int\_0^{\ep\_n} |f\_n|\, +\, \|f\_n\|\_\infty\, \ep\_n
+\frac59\,(\|f\_n\|\_2\,\sqrt 3\,\ep\_n^{1/3}+\tfrac32\,\|f\_n\|\_\infty\,\ep\_n)
\le\frac1n. \tag{3}\label{1}
\end{equation\*}
Then $g\_n\in C[0,1]$,
\begin{equation\*}
\begin{aligned}
\lVert g\_n-f\_n\rVert\_1&\le\int\_0^{\ep\_n}dx\,(\lvert f\_n(x)\rvert+\lvert f\_n(\ep\_n)\rvert+\lvert c\_n\rvert(\ep\_n-x)) \\
&\le\int\_0^{\ep\_n} \lvert f\_n\rvert + \lVert f\_n\rVert\_\infty\, \ep\_n + \lvert c\_n\rvert \ep\_n^2/2.
\end{aligned}
\tag{4}\label{2}
\end{equation\*}
Next, $\lvert I\_n\rvert\le\lVert f\_n\rVert\_2\lVert h\rVert\_2=\lVert f\_n\rVert\_2\,\sqrt3$ and hence, by \eqref{0},
\begin{equation\*}
\lvert c\_n\rvert\ep\_n^2/2\le\frac59\,(\lVert f\_n\rVert\_2\,\sqrt3\,\ep\_n^{1/3}+\tfrac32\,\lVert f\_n\rVert\_\infty\,\ep\_n).
\end{equation\*}
Thus, by \eqref{2} and \eqref{1}, $\lVert g\_n-f\_n\rVert\_1\le1/n$.
Also, by \eqref{-1} and \eqref{0}, $\int\_0^1 g\_n h=0$,
as claimed.
| 3 | https://mathoverflow.net/users/36721 | 414538 | 169,069 |
https://mathoverflow.net/questions/414453 | 11 | Here's a mix of heuristic and precise questions as I try to grapple with topos theory.
I try to think of topoi as two notions of "$1$" being glued at the hip. One is the "building block" $1$, generating the naturals, the ordinals, the cardinals... with all its usual arithmetic properties. This generates the set theory of a topos.
The other is the "all-encompassing" $1$, the highest truth value, the top element of a Heyting/Boolean algebra, whose subobjects form the algebra of truth values in the topos. This generates the logic of a topos.
Hence the set-theoretical properties (existence of an NNO, choice, CH, large cardinals...) happen "above $1$" and the logical properties (being Boolean, two-valued, well-pointed...) happen "below $1$" (whatever this means, this is just my heuristic).
Now Grothendieck topoi are defined as sheaf topoi with values in a certain well-pointed topos, namely $\textbf{Set}$. I can construct topoi that are not Grothendieck by taking a $\textbf{Set}$-like topos that is not $\textbf{Set}$, for example $\textbf{FinSet}$, and then considering sheaves valued in it. My first question is: are all topoi generated like this? **Is every topos equivalent to a topos of sheaves on a site** (its logic) **with values a well-pointed topos** (its set theory)**?**
Taking the functor $(-)^1$ of global elements seems to suggest the affirmative, as it is a logical functor to its image $\mathcal{I}$, which is well-pointed. Then the topos is, I assume, equivalent to the $\mathcal{I}$-valued sheaves on its Heyting algebra.
My second suspicion was that the logic and the set theory of a topos are independent of each other, but upon reviewing forcing, this doesn't seem to be the case. For example, following Mac Lane and Moerdijk, I can start with $\textbf{Set}$, value presheaves on a forcing poset $P$ in it ($\widehat{P}$), make it Boolean ($\widehat{P}\_{\neg\neg}$) and mod out an ultrafilter ($\widehat{P}\_{\neg\neg}/\mathcal{U}$). But then I've changed the set theory of a topos by only tinkering with its logic. In this light, forcing seems really unexpected and counterintuitive, by creating an interplay between what's happening "above $1$" and "below $1$". [I suppose this is exactly what the forcing theorem in material set theory says, though.] So my second question is: **To what extent does forcing measure the interdependence of the logic and the set theory of a topos? Is there a theorem describing this?**
It also surprises me that the two constructions $(-)^1$ and $/\mathcal{U}$ to collapse a topos to a well-pointed one don't coincide. For example, $\textbf{Set} \cong (\widehat{P}\_{\neg\neg})^1$ and $\widehat{P}\_{\neg\neg}/\mathcal{U}$ generally differ. Is there a deep reason for this? Is it perhaps because the ultrafilter-quotient construction is not as well-behaved and purely logical as I think?
| https://mathoverflow.net/users/475672 | Is every topos a sheaf topos with values in a well-pointed one? | A solution to your first question is given in "Sketches of an Elephant" by Johnstone, Example A4.4.2(d). If a topos $\mathcal{E}$ is the topos of sheaves on some internal site in a topos $\mathcal{S}$, then there is a natural geometric morphism $p : \mathcal{E} \to \mathcal{S}$. It turns out that there are toposes that do not admit geometric morphisms to Boolean toposes whatsoever. Two examples given by Johnstone are the effective topos and the topos of triples $(A,B,f)$ where $A$ is a set, $B$ is a finite set, and $f : A \to B$ is a function.
So these toposes cannot be written as a topos of sheaves over a Boolean topos (so also not over a well-pointed topos, because well-pointed toposes are Boolean).
Below is the argument that a geometric morphism $p : \mathcal{E} \to \mathcal{S}$ does not exist, for $\mathcal{E}$ the topos of triples $(A,B,f)$ as above and $\mathcal{S}$ a Boolean topos. I follow the Elephant, Example A4.5.24. Because subtoposes of Boolean toposes are again Boolean, we can assume that $p$ is surjective, so $p^\*$ is faithful. Further, in $\mathcal{S}$ all objects are decidable, i.e. the diagonal embeddings $X \to X \times X$ have a complement. So the same holds for objects of the form $(A,B,f) = p^\*(X)$. This condition implies that $f$ is injective, and as a result both $A$ and $B$ are finite. We conclude that there are only finitely many morphisms $p^\*(X) \to p^\*(Y)$, and because $p^\*$ is faithful, it follows that there are only finitely many morphisms $X \to Y$, for $X,Y$ arbitrary. However, for $(A,1,f)$ with $A$ infinite, we get that there are infinitely many morphisms $1 \to p\_\*(A)$, so this gives a contradiction.
| 7 | https://mathoverflow.net/users/37368 | 414558 | 169,071 |
https://mathoverflow.net/questions/414547 | 3 | $\newcommand{\loc}{\mathrm{loc}}$Let $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n),\mu)$ denote the Euclidean space $\mathbb{R}^n$ with its Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^n)$ equipped with the Lebesgue measure $\mu$ and let $H$ be a separable Hilbert space. Let $L^1\_{\loc}(\mathbb{R}^n,H)$ denote the space of all (equivalence classes of) measurable vector-valued functions from $\mathbb{R}^n$ to $H$ such that, given any $x\in \mathbb{R}^n$ and any $\epsilon>0$
$$
\int I\_{B(x,\epsilon)}(u) \,\|f(u)\|\_Hd\mu(u)<\infty.
$$
We recall that $L^1\_{\loc}(\mathbb{R}^n,H)$ can be made into a Fréchet space when equipping it with the metric
$$
d(f,g):=\sum\_{n=1}^{\infty}\frac1{2^n}
\frac{\int I\_{B(x,n)}(u) \,\|f(u)-g(u)\|\_Hd\mu(u)}{
1+ \int I\_{B(x,n)}(u) \,\|f(u)-g(u)\|\_Hd\mu(u)}
.
$$
---
I expect that $L^1\_{\loc}(\mathbb{R}^n,H)$ admits a *Schauder basis* however, is it true that
$$
\left\{
\psi\_{i,j}\cdot h\_k
\right\}\_{i,j,k}
$$
is a Schauder basis of $L^1\_{\loc}(\mathbb{R}^n,H)$ where $\{\psi\_{i,j}\}\_{i,j=1}^{\infty}$ the [Haar-system](https://en.wikipedia.org/wiki/Haar_wavelet) *(defined by $\psi\_{i,j}(t)\triangleq 2^{i/2}\psi(2^it -j)$ and $\psi(t)=I\_{[0,1/2)}(t) -I\_{[1/2,1)}(t)$)* and where $\{h\_k\}\_{k=1}^{\infty}$ is a fixed orthonormal basis of $H$.
---
---
---
**Reasoning:**
My reasoning is the following.
The subset $L^1(\mathbb{R})\subset L^1\_{loc}(\mathbb{R})$ is dense and therefore, for each $f\in L^1\_{loc}(\mathbb{R})$ there exists a sequence $\{f\_n\}\_n\in L^1(\mathbb{R})$ satisfying
$$
\lim\limits\_{n\to\infty}\, d(f\_n,f)=0.
$$
Now, the Haar-system is a Schauder basis of $L^1(\mathbb{R})$, for its norm topology, and the norm topology is strictly stronger than the topology the subspace topology on the set $L^1\_{}(\mathbb{R})$ inherited from $L^1\_{loc}(\mathbb{R})$. Therefore, there exist unique $(F^{i,j})\_{i,j=1}^{\infty} \in L^1(\mathbb{R})'$ such that
$$
\lim\limits\_{s\to\infty}\,
\|f\_n-\sum\_{i',j'=1}^{s} F^{i,j}(f\_n)\psi\_{i',j'}\|\_{L^1}=0.
$$
Combining both expressions, we find that for every $f\in L^1\_{loc}(\mathbb{R})$ we have
$$
\lim\limits\_{n\to\infty} \,d(f,\sum\_{i',j'=1}^{n} F^{i,j}(f\_n)\psi\_{i',j'}) =0.
$$
*I suspect that there is an issue with the argument when discussing uniqueness, but I can't put my finger on it... am I missing something?*
| https://mathoverflow.net/users/36886 | Schauder basis of $L^1_{\mathrm{loc}}(\mathbb{R}^n,H)$ | Instead of the balls, you can equivalently look at convergence on cubes $Q\_k:=k+[0,1)^n$ for $k\in\mathbb{Z}^n$. More precisely: Convergence in your metric is equivalent to convergence to w.r.t. the set of seminorms
$$f\mapsto\int\_{B\_0(r)} \|f(x)\|\_H \,dx$$
for all $r\in\mathbb{N}$ which is equivalent to convergence w.r.t. the set of seminorms
$$f\mapsto\int\_{Q\_k} \|f(x)\|\_H \,dx$$
for all $k\in\mathbb{Z}^n$. Now since the $Q\_k$ are pairwise disjoint, it is sufficient to find a Schauerbasis for $L^1(Q\_k,H)=L^1([0,1]^n, H)$.
I think I can prove that if $(\psi\_m)\_m$ is a Schauder basis of $L^1([0,1]^n)$ and $h\_k$ an orthonormal basis of $H$, then there is Schauder basis consisting of products $\psi\_m\cdot h\_k$ of $L^1([0,1]^n,H)$.
I' being a bit vague, because $L^1$ does not have an unconditional Schauder basis, so the precise ordering matters and all relevant series will be conditionally convergent at most. I have to get back to work now, but tonight I'll write my proof down in detail.
| 1 | https://mathoverflow.net/users/3041 | 414560 | 169,072 |
https://mathoverflow.net/questions/414561 | 6 | Let $G$ be a free group of finite rank. Consider any commutative ring $R$ containing $\mathbb{Z}$. Consider the group ring $RG$.
Q) What can we say about the Hochschild cohomology groups of $RG$ with coefficients in $R$?
I can find the $HH^1(RG)$ in terms of outer derivations but again I don't know the complete description of derivations on a free group. Also, I could not find a single element of $HH^n(RG)$ for $n\geq 2$.
Any comment, reference, or suggestion will be extremely helpful. Thanks in advance.
| https://mathoverflow.net/users/9485 | Hochschild cohomology of group ring of a free group | Hochschild cohomology of group rings is reducible to ordinary group cohomology. (Not really if you're interested in multiplicative structure, but that's another story). Everyting works over any commutative base ring, you can assume it's $\Bbb Z$ if you'd like.
Namely, there's an adjunction between $kG-bimod$ and $kG-mod$ (I mean left modules): forgetful right adjoint functor $\rho$ sends a bimodule $A$ to the module $\rho A$ where action is conjugation, and left adjoint $\lambda$ is $M \mapsto \lambda M := M \otimes\_k kG$. Right action is multiplication in $kG$, and left action is $h \cdot \sum\_G m \otimes g = \sum\_G hm \otimes hg$.
So, now we can get a string of identifications
$$HH^\*(kG, A) = Ext\_{bimod}(kG, A) = Ext\_{bimod}(\lambda k, A) = Ext\_{mod}(k, \rho A) = H^\*(G, \rho A)$$
In particular, Hochschild cohomological dimension of a group ring is equal to comohomological dimension of a group itself, and $HH^\*(kG, k) = H^\*(G, k)$.
(so if your group is free, there's no second cohomology at all)
Everything above is pretty much copied verbatim from chapter X of Homological algebra by Cartan and Eilenberg.
I'd suggest looking at Dan Burghelea and Sarah Witherspoons papers for further reading, but cannot provide explicit references for a while (because searching old papers from a phone is somewhat difficult).
| 8 | https://mathoverflow.net/users/81055 | 414565 | 169,073 |
https://mathoverflow.net/questions/414542 | 2 | I would like to try my luck here for the following question after failing to elicit an answer to [it on math.stackexchange.com](https://math.stackexchange.com/q/4363044/64809).
---
For $r\ge -1$, the exponential of the negative Renyi entropy is defined as
$$M(p):=\Big(\sum\_i p\_i^{1+r}\Big)^{\frac1r},$$
for a probability measure as tuples $p:=(p\_i)\_i$
I would like to prove the convexity of $M(\cdot)$, or
$$M(ap+bq)\le aM(p)+bM(q),$$
$\forall\,a+b=1 \wedge a,b\ge0$, and two probability measures $p$ and $q$ with the same cardinalities.
For $r>0$, I can show the convexity
via the Minkowski inequality for $\big(\sum\_i x\_i^{1+r}\big)^{\frac1{1+r}}$ then the convexity of $f(x):=x^{1+\frac1r}$.
But how would one show the convexity for $-1<r<0$? The above technique does not work since the inequality signs from the two steps point in the opposite directions.
| https://mathoverflow.net/users/32660 | Convexity of the exponential of the negative Renyi entropy | Take any $r\in(-1,0)$, any vector $(p\_i)\_{i=1}^n$ with $p\_i>0$ for all $i$, and any vector $(h\_i)\_{i=1}^n\in\mathbb R^n$. For all real $t$ close enough to $0$, let
$$g(t):=M(p+th).$$
Then
$$g''(0)=(1+r)\Big(\sum\_{i=1}^n p\_i^{r-1}h\_i^2\,\sum\_{i=1}^n p\_i^{1+r}
+\frac{1-r^2}{r^2}\,\Big(\sum\_{i=1}^n p\_i^r h\_i\Big)^2\Big)
\Big(\sum\_{i=1}^n p\_i^{1+r}\Big)^{1/r-2},$$
which is manifestly $\ge0$.
It follows that $M$ is indeed convex (actually, convex on the entire positive orthant of $\mathbb R^n$).
| 1 | https://mathoverflow.net/users/36721 | 414566 | 169,074 |
https://mathoverflow.net/questions/414551 | 8 | In the first section of "A procedure for killing homotopy groups of differentiable manifolds", Milnor gives the surgery construction as follows. Let $W$ be an $n=p+q+1$ dimensional manifold. Given a smooth, orientation preserving embedding:
$$f:S^p\times D^{q+1}\to W$$
we may obtain a new manifold as the disjoint sum
$$(W-f(S^p\times 0))\cup (D^{p+1}\times S^q)$$
modulo some equivalence relation.
My question is: why in this construction is Milnor deleting $S^p\times 0$ from $W$ and not $S^p\times D^{q+1}$? I expected this disjoint sum to be
$$(W-f(S^p\times D^{q+1}))\cup (D^{p+1}\times S^q).$$
| https://mathoverflow.net/users/475761 | Regarding the surgery construction in "A procedure for killing homotopy groups of differentiable manifolds" by Milnor | There is no error in the paper as far as I can tell. One *thinks* of the surgery as taking out the interior of the image of $S^p\times D^{q+1}$, and glueing in a copy of $D^{p+1}\times S^q$ along the common boundary $S^p\times S^q$. However, that does only define a topological space and not an orientable differentiable manifold. That's why one takes $W'=W\setminus f(S^p\times\{0\})$ and glues in $D^{p+1}\times S^q$ on the overlap
$$
f(S^p\times (D^{q+1}\setminus\{0\}))\cong
(D^{p+1}\setminus\{0\})\times S^q
$$
the identification being $f(u,\theta v)\leftrightarrow (\theta u,v)$, for $(u,v)\in S^p\times S^q$ and $0<\theta\leq 1$.
| 8 | https://mathoverflow.net/users/85592 | 414570 | 169,075 |
https://mathoverflow.net/questions/414572 | 3 | The following is taken from [a post by Terence Tao on the Chowla conjecture and the Sarnak conjecture](https://terrytao.wordpress.com/2012/10/14/the-chowla-conjecture-and-the-sarnak-conjecture/):
Given a bounded sequence ${f: {\bf N} \rightarrow {\bf C}}$, define the topological entropy of the sequence to be the least exponent ${\sigma}$ with the property that for any fixed ${\varepsilon > 0}$, and for ${m}$ going to infinity the set ${\{ (f(n+1),\ldots,f(n+m)): n \in {\bf N} \} \subset {\bf C}^m}$ of ${f}$ can be covered by ${O( \exp( \sigma m + o(m) ) )}$ balls of radius ${\varepsilon}$ (in the ${\ell^\infty}$ metric). A sequence is called deterministic of the topological entropy defined as above is zero.
Based on Tao's post, the sequence $\{e^{2\pi i \alpha n^2}\}\_{n\in \bf N}$ is deterministic. His argument is:
>
> the ${m}$-blocks of such polynomials sequence have covering numbers that only grow polynomially in ${m}$, rather than exponentially, thus yielding the zero entropy.
>
>
>
But this argument is a little heuristic to me: It is hard for me to find a bound for the covering number. In fact, what we need to do is that for any $\sigma >0, \varepsilon >0 $ the sequence
$$S(m,f)=\{(e^{2 \pi i \alpha (n+1)^2}, \dots, e^{2 \pi i \alpha (n+m)^2)}): n\in \bf N\}$$
can be covered by ${O( \exp( \sigma m + o(m) ) )}$ balls of radius ${\varepsilon}$. I personally believe that the omega notations could involve $\varepsilon$, please let me know if I am wrong.
Based on Tao's argument, we can actually make $S(m,f)$ covered by $P(m,\varepsilon)$-many balls of radius $\varepsilon$ (in ${\ell^\infty}$ metric) where $P(m,\varepsilon)$ is a polynomial in $m$ whose coefficients might depend on $\varepsilon$. But how to find such a cover exactly?
| https://mathoverflow.net/users/64462 | An example of deterministic sequence from Terence Tao's blog | Let $S\_1$ be a subset of $[0,1]$ with consecutive elements separated by a distance of at most $\epsilon/2\pi $ and let $S\_2$ be a subset of $[0,1]$ with consecutive elements separated by at most $\epsilon / (2 \pi m)$.
We can take $|S\_1| < 2\pi\epsilon^{-1} +1$ and $|S\_2| < 2\pi m \epsilon^{-1} +1$.
Let $\gamma \in S\_1$ be the closest point to $\alpha n^2 \mod 1$ and let $\beta \in S\_2$ be the closest point to $2 \alpha n \mod 1$. Then for all $j$ from $1$ to $m$
$$ \left| e^{ 2\pi i \alpha(n+j)^2} -e^{ 2\pi i (\alpha j^2 + \beta j + \gamma) } \right| <2 \pi \left|\alpha(n+j)^2 -(\alpha j^2 + \beta j + \gamma)\right| = 2\pi \left| \alpha n^2- \gamma + 2 \alpha n j - \beta j + \alpha j^2 - \alpha aj^2 \right| \leq |\alpha n^2 - \gamma| + j |2\alpha n - \beta| \leq \frac{\epsilon}{2} + \frac{j \epsilon}{2m } \leq \epsilon $$
giving a $\delta$-covering of size
$$|S\_1| |S\_2| \leq( 2\pi\epsilon^{-1} +1 )( 2\pi m \epsilon^{-1} ). $$
This is indeed a polynomial in $m$ with coefficients depending on $\epsilon$ (and in fact is polynomial in $\epsilon$).
| 7 | https://mathoverflow.net/users/18060 | 414578 | 169,077 |
https://mathoverflow.net/questions/414563 | 12 | Let $\chi$ be a primitive Dirichlet character of conductor $q$. I want to compute numerically $$G(k)=\sum\_{n\bmod q}\chi(n)e^{2\pi i n(n-k)/(2q)}$$
for all $k$ with $0\le k<2q$ with $k\equiv q\pmod2$ (thanks to this last condition the sum $G(k)$ is well defined). I have two questions:
1. For now I compute these values in a naive way, so using $q\*q$ steps (of course precomputing the $2q$th roots of unity). Is there a better method (even reducing to $q^2/2$ steps would be nice)? Imagine $q=10^5$ or $q=10^6$.
2. When $q$ is not prime, $G(k)$ is often equal to $0$. For instance if $p$ is a prime congruent to $3$ mod 4 dividing $q$ and $k$, it seems that $G(k)=0$, at least for quadratic characters (but I am interested in all characters). My question is: give a necessary and sufficient condition for $G(k)=0$ (sufficient would already be nice).
| https://mathoverflow.net/users/81776 | Computation of modified Gauss sums | For the second question, if $p$ is an odd prime dividing $q$, $p$ divides $k$ with at least the multiplicity with which it divides $q$, and $\chi\_p(-1)=-1$, where $\chi\_p$ is the $p$-adic part of $\chi$, then $G(k)=0$.
This generalizes what you wrote in the $\chi$ quadratic case. In that case, the multiplicity which which $p$ divides $q$ is always one, and $\chi(-1)=-1$ if and only if $p$ is congruent to $3$ mod $4$.
Indeed, the Chinese remainder theorem gives
$$G(k)=\frac{1}{2} \sum\_{n\bmod 2q}\chi(n)e^{2\pi i n(n-k)/(2q)} = \prod\_{ p \mid 2q} \sum\_{n \bmod p^{v\_p(2q)}} \chi\_p (n) e^{ 2 \pi i \lambda\_p n (n-k)/ p^{v\_p(2q)}}$$
where $v\_p(2q)$ denotes the $p$-adic valuation of $2q$, $\lambda\_p$ is the inverse of $q/ p^{v\_p(2q)}$ modulo $p^{v\_p(2q)}$, and $\chi\_p$ is the $p$-adic part of $\chi$.
Fixing on an odd prime $p$, we see that if $k$ is a multiple of $p^{v\_p(2q)}=p^{v\_p(q)}$, the factor at $p$ simplifies to
$$ \sum\_{n \bmod p^{v\_p(2q)}} \chi\_p (n) e^{ 2 \pi i \lambda\_p n^2/ p^{v\_p(2q)}}$$
and if $\chi\_p(-1)= -1$ then the terms for $n$ and $-n$ in the sum always cancel, giving a value of $0$.
If $\chi(-1)= 1$ then we get a sum of two Gauss sums and I don't see any reason it should be zero.
In general, if $v\_p(2q)=1$, the local factor is a nice complete exponential sum. For fixed $k \neq 0$ mod $p$, Katz's equidistribution theory will tell us that the sum is nonzero for a density 1 subset of characters $\chi$. This probably can be proven also for fixed $\chi$ and a density one set of values of $k$, but might be harder to prove (at least by this method - there may be a clever congruence that proves nonvanishing).
| 5 | https://mathoverflow.net/users/18060 | 414580 | 169,078 |
https://mathoverflow.net/questions/414584 | 4 | Let $X = Q\_1\cap Q\_2$ be a complete intersection of two smooth quadrics, over a field $K$, in $\mathbb{P}^4$ with homogeneous coordinates $y\_0,y\_1,y\_2,y\_3,y\_4$.
Set $Q\_1 = \{F\_1 = 0\}$ and $Q\_2 = \{F\_2 = 0\}$ and assume that the monomial $y\_0^2$ does not appear in $F\_1$ so that $Q\_1$ is rational and that the monimial $y\_0y\_1$ does not appear in $F\_2$.
Under these hypotheses could we conclude anything about the unirationality over $K$ of $X$?
What if $X$ is a complete intersection of two quadrics in $\mathbb{P}^n$ with the same properties for $n\geq 5$?
| https://mathoverflow.net/users/14514 | Del Pezzo surfaces of degree four and complete intersections of two quadrics | No.
Let $Q\_1,Q\_2$ be arbitrary quadrics. Let $a$ be the coefficient of $y\_0^2$ in $F\_1$, $b$ the coefficient of $y\_0^2$ in $F\_2$, $c$ the coefficient of $y\_0 y\_1$ in $F\_1$, $d$ the coefficient of $y\_0 y\_1$ in $F\_2$.
Then the coefficient of $y\_0^2$ in $b F\_1 - a F\_2$ and the coefficient of $y\_0 y\_1$ in $d F\_1 - c F\_2$ both vanish. If $ad-bc \neq 0$ (the generic case), then $b F\_1 - a F\_2$ and $d F\_1 - c F\_2$ is $Q\_1 \cap Q\_2$.
So any intersection of two quadrics that satisfies a mild genericity property can be written in this form, and thus the unirationality problem is as hard as for a general intersection of two quadrics.
| 7 | https://mathoverflow.net/users/18060 | 414587 | 169,079 |
https://mathoverflow.net/questions/414590 | 5 | Fix a prime $p$. A p-adic field is a finite extension of $\mathbb{Q}\_p$.
Question 1: Let $K$ be a $p$-adic field and fix $n$. Is there $m$ such that if $\alpha \in \mathbb{Q}\_p$ is an $m$th power in $K$ then $\alpha$ is an $n$th power in $\mathbb{Q}\_p$? $\quad$ (Probably $m$ is a multiple of $n$.)
I think that you'd approach this by decomposing $K/\mathbb{Q}\_p$ into a tower of extensions of some sort and then go up the tower. So we probably want to actually prove the following.
Question 2: Suppose that $K/L$ is an extension of $p$-adic fields and fix $n$. Is there $m$ such that if $\alpha \in K$ is an $m$th power in $L$ then $\alpha$ is an $n$th power in $K$?
I can prove Question 1 in the case when $K/\mathbb{Q}\_p$ is unramified, but I don't know about the general case. I think that one might be able to use class field theory, or maybe it's already well-known, or maybe I am just missing something elementary.
I have good motivation for this question, but it would take a bit of work to explain, the motivation comes from logic.
| https://mathoverflow.net/users/152899 | Powers in finite extensions of the p-adics | Yes.
I'll solve the more general question 2.
Let $v\_p(n)$ be the highest power of $p$ dividing $n$. Then if $\alpha \in L$ is congruent to $1$ modulo $p^{v\_p(n)+1}$, then $\alpha$ is an $n$th power in $L$.
Every $\alpha \in L$ can be written as $\pi^j u$ where $j\in \mathbb Z$ and $u$ is a unit in $L$. It suffices to find $m$ such that if $\pi^j u$ is an $m$th power in $K$ then $j$ is divisible by $n$ and $u$ is congruent to $1$ modulo $p^{v\_p(n)+1}$.
Let $e$ be the ramification degree of $K$ over $L$, i.e. the greatest $e$ such that there is an element of $K$ whose $e$'th power is $\pi$ times a unit and let $k$ be any natural number. Then if $\pi^j u$ is a $ke$'th power in $K$ then $j$ is divisible by $k$, and hence $u = \pi^j u / (\pi^{j/k})^k$ is also a $k$th power in $K$.
So it suffices to find $k$ such that if $u$ is a unit that is a $k$th power in $K$ then $u$ is congruent to $1$ mod $p^{v\_p(n)+1}$.
Let $q$ be the order of the residue field in $k$. If $u$ is a $k$th power, then it is a $k$th power of some unit $v$, and $v^{q-1}$ is congruent to $1$ modulo the uniformizer $\pi'$ of $K$. So $$ v^{ (q-1) p^r} = (1 + (v^{q-1}-1))^{p^r} = \sum\_{i=0}^{P^r} \binom{p^r}{i} (v^{q-1}-1)^i$$ where the $i$th term is divisible by $\binom{p^r}{i} \pi'^{i}$.
Taking $r$ sufficiently large, we can ensure that $\binom{p^r}{i} \pi'^{i}$ is divisible by $p^{ v\_p(n)+1}$ for all $i>0$, so $v^{ (q-1) p^r}$ is congruent to $1$ mod $p^{ v\_p(n)+1}$, so we may take $k= (q-1) p^r$.
| 4 | https://mathoverflow.net/users/18060 | 414592 | 169,081 |
https://mathoverflow.net/questions/414579 | 3 | Let $R$ be a unital ring. Let $\mathbf{A}\_\bullet$ and $\mathbf{C}\_\bullet$ be positive chain complexes of $R$-modules. If $\mathbf{A}\_\bullet$ consists of flat $R$-modules then there is *homology Künneth spectral sequence*
$$E^2\_{p,q}:=\bigoplus\_{s+t=q}\mathrm{Tor}\_p^R(H\_s(\mathbf{A}\_\bullet),H\_t(\mathbf{C}\_\bullet))\Rightarrow H\_{p+q}(\mathbf{A}\_\bullet\otimes\_R\mathbf{C}\_\bullet).$$
I am interested in a cohomological version, specifically, is the following true?
>
> Suppose $\mathbf{C}\_\bullet$ is a negative complex. If $\mathbf{A}\_\bullet$ consists of projective $R$-modules, then there is a *cohomology Künneth spectral sequence*
> $$E^2\_{p,q}:=\bigoplus\_{s+t=q}\mathrm{Ext}^p\_R(H\_s(\mathbf{A}\_\bullet),H\_t(\mathbf{C}\_\bullet))\Rightarrow H\_{p+q}(\mathrm{Hom}\_R(\mathbf{A}\_\bullet,\mathbf{C}\_\bullet)).$$
>
>
>
A version of this appears in Rotman's introduction to homological algebra (first edition) but it does not appear in the second edition and I do not know of another reference.
If the "theorem" is true, what is a reference for it?
| https://mathoverflow.net/users/121307 | Künneth spectral sequence for cohomology of chain complexes of $R$-modules | This is a special case of the hyper(co)homology spectral sequence from Chapter XVII, Section 2 of Cartan-Eilenberg (1953), for the functor $T(C,A) = \mathrm{Hom}\_R(A,C)$. The $E\_2$-term is given in equation (4) on page 368, essentially as
$$
E\_2^{p,q} = \prod\_{s+t=q} \mathrm{Ext}^p\_R(H\_s(\mathbf{A}\_\bullet), H^t(\mathbf{C}^\bullet))
\Longrightarrow H^{p+q}(\mathrm{Hom}\_R(\mathbf{A}\_\bullet, \mathbf{C}^\bullet)) \,,
$$
with $p$ as the filtration degree.
| 4 | https://mathoverflow.net/users/9684 | 414593 | 169,082 |
https://mathoverflow.net/questions/414526 | 1 | In the Levi-Civita field, are there elements such that the standard parts of their subsequent powers produce an arbitrary sequence?
Particularly, is there an element $w$ of the field such that the standard part (the zeroth element of the corresponding series) of $w^n$ is $B\_n$ (Bernoulli numbers)?
| https://mathoverflow.net/users/10059 | In the Levi-Civita field, are there elements such that the standard parts of their subsequent powers produce an arbitrary sequence? | Yes, it is true, and there even are such elements in the field of formal Laurent series. Specifically, let $a\_n,n\geq 1$ be any sequence of real numbers. We then take a Levi-Civita series
$$z=\varepsilon^{-1}+\sum\_{i=0}^\infty b\_i\varepsilon.$$
We want to show the $b\_i$ can be chosen so that the constant term of $z^n$ is $a\_n$ for $n\geq 1$. The key thing to note is that this constant term will only depend on coefficients $b\_i$ for $i<n$. This lets us define $b\_i$ recursively. Suppose we have constructed $b\_i$ for $i<n$ already. To determine $b\_n$, we consider $z^{n+1}$, and note that the condition imposed by its constant term being $a\_{n+1}$ involves a sum of various combinations of $b\_i,i<n$, but $b\_n$ only appears once, so we can always pick $b\_n$ to make the total sum equal to $a\_{n+1}$.
| 1 | https://mathoverflow.net/users/30186 | 414599 | 169,085 |
https://mathoverflow.net/questions/414352 | 2 | Let $(M^{2n},\omega)$ be a symplectic manifold of dimension $2n$. Let $L^k\_{\omega}:\Omega^q(M)\to \Omega^{q+2k}(M)$ be the map given by $L^k\_{\omega}(\alpha)=\alpha\wedge\omega^k$. Then is it true that $L^k\_{\omega}$ is injective for all $q\leq n-k$ and surjective for all $q\geq n-k$?
It seems that this is true, but I could not find a proof for it. Can anyone provide a proof for this, or atleast point out where can I get the proof for this.
| https://mathoverflow.net/users/153479 | $\omega$ is a symplectic form then $L^k_{\omega}:\Omega^q(M)\to \Omega^{q+2k}(M)$ is injective for all $q\leq n-k$ and surjective for all $q\geq n-k$ | The injectivity case is well-known and follows quite easily from the statement (usually attributed to Lefschetz) that $L^k\_\omega:\Omega^{n-k}(M)\to \Omega^{n+k}(M)$ is a isomorphism for $0\le k\le n$.
This is a purely linear algebra statement and only relies on $\omega$ being nondegenerate, i.e., that $\omega^n$ be nonvanishing; $\omega$ does not need to be closed for this isomorphism to hold.
N.B.: Just for clarity's sake, let me point out that I am assuming that the OP intends $\Omega^p(M)$ to mean the module of (smooth) $p$-forms on $M$, as is standard. I'm not sure why people are bringing up comments about cohomology and hard Lefschetz, as the question (as I understand it) really has nothing to do with that.
For a discussion of the linear algebra result, see [this MO question](https://mathoverflow.net/questions/316505/the-lefschetz-operator/316527#316527) and its answers.
| 5 | https://mathoverflow.net/users/13972 | 414605 | 169,088 |
https://mathoverflow.net/questions/414610 | 5 | Let $\operatorname{Col}(\omega,<\kappa)$ denote the Lévy collapse of an inaccessible cardinal $\kappa$. A variant of the Factor Lemma is as follows:
>
> **Lemma.** Suppose that $\kappa$ is an inaccessible cardinal and that $\mathbb{P}$ is a poset of size $<\kappa$. Let $G$ be $\operatorname{Col}(\omega,<\kappa$)-generic over $V$. If in $V[G]$ there is a filter $h \subseteq \mathbb{P}$ that is $\mathbb{P}$-generic over $V$, then there is $G^\* \in V[G]$ that is $\operatorname{Col}(\omega,<\kappa)$-generic over $V[h]$ and such that $V[h][G^\*] = V[G]$.
>
>
>
For those interested, I took this variant from the paper [*Happy and mad families in $L(\mathbb{R})$*](http://pi.math.cornell.edu/%7Ezbnorwood/files/hmlr.pdf), Lemma 17 of Page 10.
This lemma shows that, in particular, $h \in V[G]$. Is it true that there must exist some ordinal $\beta < \kappa$ such that $h \in V[G\upharpoonright\beta]$?
| https://mathoverflow.net/users/146831 | Locating generic filters in the Lévy collapse | Note that the lemma doesn't show that $h$ is in $V[G]$, it *assumes* this. But yes, if $h\in V[G]$ is a subset of a set $X\in V$ such that $|X| < \kappa$, then for some $\beta < \kappa$, $h\in V[G\restriction \beta]$. The argument to follow is by no means original, but I don't remember where I saw it. Fix a name $\dot h$ such that $h = \dot h\_G$, and for each $x\in X$, let $A\_x\subseteq \text{Col}(\omega,{<}\kappa)$ be a maximal antichain consisting of conditions that force either $\check{x}\in \dot h$ or its negation. Recall that $\text{Col}(\omega,{<}\kappa)$ has the $\kappa$-cc, so $|A\_x| < \kappa$. Let $A = \bigcup\_{x\in X} A\_x$. Since $|X| < \kappa$, $|A| < \kappa$, and therefore there is some $\beta < \kappa$ be such that $A\subseteq \text{Col}(\omega,{<}\beta)$. We have $h\in V[G\restriction \beta]$ since $h = \{x\in X : \exists q\in G\restriction \beta\, (q\Vdash \check{x} \in \dot h)\}$.
| 10 | https://mathoverflow.net/users/102684 | 414612 | 169,091 |
https://mathoverflow.net/questions/414596 | 6 | I posted this on MSE a couple months ago and it got three upvotes but no answers or even comments so I decided to cross-post it here:
For every pair $ a,b $ of real numbers define the operator $ U\_{a,b} $ on $ L^2(\mathbb{R}) $ sending $ \psi \in L^2(\mathbb{R}) $ to $ U\_{a,b}\psi $ defined by the equation
$$
[U\_{a,b}\psi](x)=e^{ibx}\psi(x+a)
$$
Consider the set of operators
$$
\mathcal{B}:=\{ U\_{a,b}:a,b \in \mathbb{R} \}
$$
Let $ V $ be the closure in the operator norm topology of the span of the set $ \mathcal{B} $. Does anyone have a good idea for a nice characterization of what sort of operators are and are not in $ V $? Does $ V $ include all trace class operators? All compact operators? All unitary operators?
This is a follow up question to my question:
<https://math.stackexchange.com/questions/4303824/is-this-a-basis-for-the-bounded-operators-on-l2-mathbbr>
| https://mathoverflow.net/users/387190 | Characterize this subspace of the bounded operators on $ L^2(\mathbb{R}) $ | Any linear combination $L$ of $U\_{a,b}$'s can be written $(L\psi)(x) = \sum\_{k=1}^n \alpha\_ke^{ib\_kx}\psi^{\to a\_k}(x)$, where $\psi^{\to a\_k}(x) = \psi(x + a\_k)$. Fix $L$.
Let $N \in \mathbb{N}$ be such that $Nb\_k$ is close to an integer multiple of $2\pi$, for all $k$. Then $$(L\psi^{\to N})(x) = \sum \alpha\_k e^{ib\_kx}\psi^{\to a\_k + N}(x) = (\sum \alpha\_k e^{ib\_k(x - N)}\psi^{\to a\_k}(x))^{\to N} = (\sum \alpha\_k'e^{ib\_kx}\psi^{\to a\_k}(x))^{\to N}$$ where each $\alpha\_k'$ is close to $\alpha\_k$. Okay, now find a sequence $N\_j \to \infty$ such that as $j \to \infty$ the multiples $N\_jb\_k$ get arbitrarily close to integer multiples of $2\pi$. Then $(L\psi^{\to N\_j})^{-N\_j} \to L\psi$ in $L^2(\mathbb{R})$, for every $\psi$.
But any compact operator $T$ satisfies $T\psi^{\to N\_j} \to 0$ in $L^2(\mathbb{R})$. Taking $\psi$ with $\|\psi\|\_2 = 1$ and $\|L\psi\|\_2$ close to $\|L\|$, we have $\|T\psi^{\to N\_j}\|\_2 \to 0$ but $\|L\psi^{\to N\_j}\|\_2 \to \|L\psi\|\_2 \cong \|L\|$. This shows you that $\|T - L\| \geq \|L\|$; that is, the distance from $L$ to the compact operators is $\|L\|$. Every element of $V$ will have the same property, so in particular $V$ contains no compact operators besides $0$.
Every operator is a linear combination of four unitaries, so if $V$ contained every unitary then it would be all of $B(L^2(\mathbb{R}))$, which we've just seen is not the case. On the other hand, the WOT closure of $V$ does equal $B(L^2(\mathbb{R}))$; this follows from the double commutant theorem, since any operator that commutes with $U\_{0,b}$ for all $b$ must be a multiplication operator and hence won't commute with $U\_{a,0}$ for all $a$ unless it is a scalar. That is, $V' = \mathbb{C}\cdot I$, so $V'' = B(L^2(\mathbb{R}))$.
| 7 | https://mathoverflow.net/users/23141 | 414613 | 169,092 |
https://mathoverflow.net/questions/323445 | 5 | Suppose that $X$ is a finite dimensional Hilbert space. Let $A\_{1},\dots,A\_{r}:X\rightarrow X$ be linear operators. Then define the multi-spectral radius of $(A\_{1},\dots,A\_{r})$ to be
$$\limsup\_{n\rightarrow\infty}(\sum\_{a\_{1},\dots,a\_{n}\in\{1,\dots,r\}}\|A\_{a\_{1}}\dots A\_{a\_{n}}\|)^{1/n}.$$
If $r=1$, then the multi-spectral radius of $(A\_{1},\dots,A\_{r})$ is simply the spectral radius of $A\_{1}$ which is equal to $\max(|\lambda\_{1}|,\dots,|\lambda\_{s}|)$ where $\lambda\_{1},\dots,\lambda\_{s}$ is an enumeration of the eigenvalues of $A\_{1}$. Is there a similar characterization of the multi-spectral radius in terms of eigenvalues or something similar?
Since this question can be formulated in terms of Banach algebras or other spaces, feel free to answer this question in a more general context.
If an answer in the general case is hard to obtain, then I would still be interested in the case where $A\_{1},\dots,A\_{r}$ are satisfy some condition (normal operators etc.), but I want the operators $A\_{1},\dots,A\_{r}$ to not commute with each other.
As we all expect, this question is motivated by some questions I have about large cardinals.
| https://mathoverflow.net/users/22277 | Spectral radius for multiple linear operators | Yes. There is a characterization of the multi-spectral radius in terms of the ordinary spectral radius.
Suppose that $A$ is a unital Banach algebra. If $a\_{1},\dots,a\_{r}\in A$, then define the multi-spectral radius of $a\_{1},\dots,a\_{r}$ to be
$\limsup\_{n\rightarrow\infty}(\sum\_{i\_{1},\dots,i\_{n}\in\{1,\dots,r\}}\|a\_{i\_{1}}\dots a\_{i\_{n}}\|)^{1/n}$.
I claim that the multi-spectral radius of $a\_{1},\dots,a\_{r}$ is the maximum value of a spectral radius $\rho(x\_{1}\iota(a\_{1})+\dots+x\_{n}\iota(a\_{r}))$ where $\iota:A\rightarrow B$ is an isometric embedding of unital Banach algebras and where $x\_{i}\iota(a\_{j})=\iota(a\_{j})x\_{i}$ for all $i$ and $j$ and where $\|x\_{i}\|\leq 1$ for all $i$.
Observe that if $\iota:A\rightarrow B$ is an isometric embedding of unital Banach algebras and where $x\_{i}\iota(a\_{j})=\iota(a\_{j})x\_{i}$ for all $i$ and $j$ and where $\|x\_{i}\|\leq 1$ for all $i$, then
$$\rho(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))$$
$$=\lim\_{n\rightarrow\infty}\|(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))^{n}\|^{1/n}
=\lim\_{n\rightarrow\infty}\|\sum\_{i\_{1},\dots,i\_{n}\in\{1,\dots,r\}}x\_{i\_{1}}\iota(a\_{i\_{1}})\dots x\_{i\_{n}}\iota(a\_{i\_{n}})\|^{1/n}$$
$$\leq\limsup\_{n\rightarrow\infty}[\sum\_{i\_{1},\dots,i\_{r}\in\{1,\dots,n\}}\|x\_{i\_{1}}\dots x\_{i\_{n}}\iota(a\_{i\_{1}})\dots\iota(a\_{i\_{n}})\|]^{1/n}$$
$$\leq\limsup\_{n\rightarrow\infty}[\sum\_{i\_{1},\dots,i\_{r}\in\{1,\dots,n\}}\|x\_{i\_{1}}\|\dots \|x\_{i\_{n}}\|\cdot\|\iota(a\_{i\_{1}})\dots\iota(a\_{i\_{n}})\|]^{1/n}$$
$$\leq\limsup\_{n\rightarrow\infty}[\sum\_{i\_{1},\dots,i\_{r}\in\{1,\dots,n\}}\|\iota(a\_{i\_{1}})\dots\iota(a\_{i\_{n}})\|]^{1/n}
\leq\limsup\_{n\rightarrow\infty}[\sum\_{i\_{1},\dots,i\_{r}\in\{1,\dots,n\}}\|a\_{i\_{1}}\dots a\_{i\_{n}}\|]^{1/n}.$$
From the following proposition, one can choose $x\_{1},\dots,x\_{r},B,\iota$ such that $\rho(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))$ is the multi-spectral radius of $A$.
Proposition: For each unital Banach algebra $A$ and cardinal $\lambda$, there is a Banach algebra $B$ and an isometric embedding $\iota:A\rightarrow B$ along with a subset $X\subseteq B$ where $|X|\geq\lambda$ such that $\|x\|\leq 1$ for each $x\in X$ and such that $$\|(\iota(a\_{1})x\_{1}+\dots+\iota(a\_{r})x\_{r})^{n}\|=\sum\_{i\_{1},\dots,i\_{n}\in\{1,\dots,r\}}\|a\_{i\_{1}}\dots a\_{i\_{n}}\|$$ whenever
$a\_{1},\dots,a\_{r}\in A$ and $x\_{1},\dots,x\_{r}$ are distinct elements in $X$.
Proof: Let $X$ be a set disjoint from $A$ with $|X|\geq\lambda$. Let $B$ be the collection of all sums of the form $\sum\_{w\in X^{\*}}w\cdot a\_{w}$ where $a\_{w}\in A$ for all $w\in X^{\*}$ and where $\sum\_{w\in X^{\*}}\|a\_{w}\|<\infty$. Define a norm on $B$ where we set $\|\sum\_{w\in X^{\*}}w\cdot a\_{w}\|=\sum\_{w\in X^{\*}}\|a\_{w}\|$.
$B$ is clearly a vector space. Define multiplication by letting
$$(\sum\_{u\in X^{\*}}u\cdot a\_{u})\cdot(\sum\_{v\in X^{\*}}v\cdot a\_{v})
=\sum\_{u,v\in X^{\*}}uv\cdot a\_{u}a\_{v}.$$
Observe that multiplication on $B$ satisfies the inequality $\|\mathbf{b}\mathbf{c}\|\leq\|\mathbf{b}\|\cdot\|\mathbf{c}\|$. One can now check that $B$ is a Banach space, but if you don't want to check that $B$ is a Banach space, then the proof still goes through when you replace $B$ with its completion.
The mapping $\iota:A\rightarrow B$ where $\iota(a)=\epsilon\cdot a$ and $\epsilon$ denoted the empty word is a Banach algebra embedding. All the conditions of the theorem are satisfied.
Q.E.D.
From the above proposition, we know that
$$\rho(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))=\lim\_{n\rightarrow\infty}\|(x\_{1}\iota(a\_{1})+\dots+x\_{r}\iota(a\_{r}))^{n}\|^{1/n}$$
$$=\lim\_{n\rightarrow\infty}(\sum\_{i\_{1},\dots,i\_{n}\in\{1,\dots,r\}}\|a\_{i\_{1}}\dots a\_{i\_{n}}\|)^{1/n},$$
so the limit superior in the definition of the multi-spectral radius is actually just a limit.
Here is a conjecture that I did not have too much time to think about.
Conjecture: Whenever $A\_{1},\dots,A\_{r}$ are complex matrices, then the multi-spectral radius of $A\_{1},\dots,A\_{r}$ is the supremum of the spectral radii of matrices of the form $A\_{1}\otimes U\_{1}+\dots+A\_{r}\otimes U\_{r}$ where there exists an $n$ where $U\_{1},\dots,U\_{r}$ are $n\times n$-unitary matrices .
| 0 | https://mathoverflow.net/users/22277 | 414615 | 169,093 |
https://mathoverflow.net/questions/414614 | 5 | Assume $V=L$.
Let $\alpha$ be the least ordinal such that there is a $\Diamond\_{\omega\_1}$-sequence in $L\_\alpha$.
It's obvious that $\omega\_1 < \alpha < \omega\_2$.
Do we have some better estimates of $\alpha$?
Also, what about $\Diamond^+$, squares and morasses?
| https://mathoverflow.net/users/170286 | Height of diamond | It depends on what diamond sequence you have.
$V=L$ proves there is a $\diamondsuit$-sequence of $L$-rank $\omega\_1+1$: recall the famous way of proving the existence of $\diamondsuit$-sequence. The definition goes as follows (for example, Theorem 13.21 of [Jech - Set theory (3rd edition)](https://doi.org/10.1007/3-540-44761-X)):
>
> $(S\_\alpha,C\_\alpha)$ is the $<\_L$-least pair $S\_\alpha\subseteq\alpha$, $C\_\alpha\subseteq\alpha$ a club, and $S\_\alpha\cap\xi\neq S\_\xi$ for all $\xi\in C\_\alpha$. Otherwise $S\_\alpha=C\_\alpha=\alpha$.
>
>
>
Observe that if $X\subseteq\alpha<\omega\_1$, then $(\alpha,X)\in L\_{\omega\_1}$. Furthermore, $<\_L$ is absolute between $L$ and $L\_{\omega\_1}$. Combining with that $L\_{\omega\_1}$ satisfies $\mathsf{ZFC}^-$, which implies $L\_{\omega\_1}$ is closed under recursively defined sequences over there, we can see that the above definition works over $L\_{\omega\_1}$. Hence the $\diamondsuit$-sequence defined as above is a member of $\operatorname{Def}(L\_{\omega\_1})=L\_{\omega\_1+1}$.
However, the $L$-rank of a $\diamondsuit$-sequence can be arbitrarily large for a silly reason. Observe that the $(\alpha+1)$-th component of $\diamondsuit$-sequence does not affect being $\diamondsuit$, and hence we may replace it arbitrarily.
Let $f:\omega\_1\to\omega\_1$ be the increasing enumeration of successor ordinals below $\omega\_1$. (Note that $f\in L\_{\omega\_1+1}$.) Recall the $\diamondsuit$-sequence $\langle A\_\alpha\mid \alpha<\omega\_1\rangle$ we previously defined (so of $L$-rank $\omega\_1+1$.) For any $B\subseteq \omega\_1$, define $A^B\_\alpha$ by
$$A^B\_{f(\alpha)}=\begin{cases} 0&\text{if }\alpha\in B \\ 1 & \text{if } \alpha\notin B\end{cases}$$
and $A^B\_\alpha=A\_\alpha$ for a limit $\alpha$. Then we can recover $B$ from $\langle A^B\_\alpha\mid\alpha<\omega\_1\rangle$ (possibly except for information about $0\in B$ and $1\in B$, but it does not matter.) Hence the $L$-rank of $\langle A^B\_\alpha\mid\alpha<\omega\_1\rangle$ is greater than or equal to that of $B$. (I suspect they are equal.)
*There is no reason to believe that the $L$-rank of a $\diamondsuit$-sequence is **equal** to a given ordinal less than $\omega\_2$. I think the question [A tree coding all of $L\_\alpha$ within $L\_\alpha\cap\mathcal{P}(\omega)$](https://math.stackexchange.com/q/4072513/53976) in MSE indicates, at least, the above construction does result in a $\diamondsuit$-sequence whose $L$-rank is equal to a given ordinal.*
I believe the same argument applies for $\diamondsuit^+$-sequences, but I have not checked in detail. I am not sure the same holds for $\square$-sequences or morasses.
| 8 | https://mathoverflow.net/users/48041 | 414625 | 169,095 |
https://mathoverflow.net/questions/414603 | 2 | Suppose that $E/\mathbf{Q}$ is an elliptic curve and $K$ is an imaginary quadratic field. Let $\mathbf{Q}\_{\infty}$ denote the cyclotomic $\mathbf{Z}\_p$ extension of $\mathbf{Q}$, and let $K\_{\infty}$ denote the cyclotomic $\mathbf{Z}\_p$ extension of $K$,
If we know that the $\mu$-invariant for the Selmer over $K\_{\infty}$ vanishes, does it follow that the $\mu$-invariant for the Selmer over $\mathbf{Q}\_{\infty}$ vanishes? That is, if we know $\mu=0$ for the bigger field $K\_{\infty}$, can we "propogate it down" to show that $\mu=0$ for the field $\mathbf{Q}\_{\infty}$?
| https://mathoverflow.net/users/394740 | Does $\mu=0$ for an imaginary quadratic field $K$ imply $\mu=0$ for $\mathbf{Q}$? | Under your setting
\begin{equation}
\operatorname{Sel}(E/K\_{\infty})\simeq\operatorname{Sel}(E/\mathbb Q\_{\infty})\oplus\operatorname{Sel}(E\otimes\chi/\mathbb Q\_{\infty})
\end{equation}
where $\chi$ is the quadratic character attached to $K/\mathbb Q$ and all these $\mathbb Z\_{p}[[X]]$-modules are torsion. Hence
\begin{equation}
\mu(\operatorname{Sel}(E/K\_{\infty}))=\mu(\operatorname{Sel}(E/\mathbb Q\_{\infty}))+\mu(\operatorname{Sel}(E\otimes\chi/\mathbb Q\_{\infty})).
\end{equation}
Indeed, if $\mu(\operatorname{Sel}(E/K\_{\infty}))=0$ then $\mu(\operatorname{Sel}(E/\mathbb Q\_{\infty}))=0$.
However, the very same argument suggests that it is probably very hard to show that $\mu(\operatorname{Sel}(E/K\_{\infty}))=0$ without showing that $\mu(\operatorname{Sel}(E/\mathbb Q\_{\infty}))=0$.
What has been known since the works Gillard, Schneps, Hida, Hsieh... is that it is sometimes possible to prove that $\mu$-invariant vanishes over the $\mathbb Z\_p^2$-extension $L\_\infty/K$ (the composite of the cyclotomic and anticyclotomic extension of $K$). Unfortunately, in that case, there is no obvious way (and indeed, at present no known way) to deduce from those results the vanishing of the $\mu$-invariant over $\mathbb Q\_\infty$, as the specialization of a power-series in $\mathbb Z\_p[[X,Y]]$ with vanishing $\mu$-invariant at $Y=0$ can very well have non-vanishing $\mu$-invariant.
| 3 | https://mathoverflow.net/users/2284 | 414630 | 169,096 |
https://mathoverflow.net/questions/414627 | 1 | $\DeclareMathOperator\Hom{Hom}$Let $G$ be a (pro-)cyclic group (topologically) generated by $\phi$, and let $V,W$ be two finite-dimensional (continuous) $k$-linear semisimple representations of $G$, where $k$ is some algebraically closed (complete) field.
Let $$P\_V(t) = \prod\_{i=1}^n (t - a\_i), P\_W(t) = \prod\_{j=1}^m (t - b\_j)$$
be the characteristic polynomials of $\rho\_V(\phi), \rho\_W(\phi)$ respectively.
I would like to know if
$$\dim\_k \Hom\_G(V, W) = \# \{ (i, j) : a\_i = b\_j \}$$
I know that $\dim\_k \Hom\_G(V, W) = \langle \chi\_V, \chi\_W \rangle$, and that $f \in \Hom\_G(V, W)$ iff $f \circ \rho\_V(\phi) = \rho\_W(\phi) \circ f$. But I don't know how to proceed from there.
(NB: already asked [here](https://math.stackexchange.com/questions/4355067/) two weeks ago, with no success).
| https://mathoverflow.net/users/196880 | Dimension of $\mathrm{Hom}_G(V, W)$ in terms of characteristic polynomial | If $k$ is of characteristic zero or characteristic $p$ and $G$ is a pro-$p'$-group, then yes. In that case, everything in sight is semisimple by Maschke's theorem, the eigenvalues $a\_i$ and $b\_j$ give you the decomposition into irreducibles, and every pair of equal eigenvalues contributes one degree of freedom by Schur's lemma.
Otherwise no, even in the simplest case, because of lack of semiplicity. If $k=\overline{\mathbb{F}\_p}$, $G=C\_p$, $\rho\_V(\phi) = \begin{pmatrix}1&1\\&1\end{pmatrix}$ and $\rho\_W(\phi)=\begin{pmatrix}1&\\&1\end{pmatrix}$, then $a\_1=a\_2=b\_1=b\_2$ so that your conjecture would predict $\dim\_k \operatorname{Hom}\_G(V,W) = 4$, but it is only 2, because $V$ is not semi simple, but $W$ is, and therefore every morphism must factor through $V/\operatorname{rad}(V)$.
| 7 | https://mathoverflow.net/users/3041 | 414635 | 169,097 |
https://mathoverflow.net/questions/414488 | 3 | Let $0< \alpha \ll 1$. I'm trying to minimize $\int\_0^\pi |f'|^2 dx$ over the functions $f \in W\_0^{1,2}([0,\pi])$ (or at least find "good" lower bound in terms of $\alpha$) such that satisfy the following constraints:
$$
\begin{cases}
\int\_0^\pi f^2 dx = 1, \\
\int\_{\pi/3}^{2\pi/3} f^2 dx \leq \alpha.
\end{cases}
$$
In other words, I'm trying to find a function that vanishes at $0$ and $\pi$, has most of its mass near the end points, and minimizes its Dirichlet energy.
I do have a trivial lower bound given by Poincare (Wirtinger) inequality which is not in terms of $\alpha$.
I don't know where to start studying this problem.
I have tried taking Fourier transform of $f$ but the second constraint gets complicated. To be honest, I started with the formulation in the Fourier space which is the following:
$$
\begin{cases}
\mbox{find sequence $(a\_n)\_n$ that minimizes} &\sum\_n na\_n^2, \\
\mbox{subject to:} &\sum\_n a\_n^2 = 1, \\
& \int\_{\pi/3}^{2\pi/3} (\sum\_n a\_n sin(n x))^2 dx \leq \alpha.
\end{cases}
$$
Does anyone know where to start or any related reference?
| https://mathoverflow.net/users/173610 | Non convex optimization problem in $W_0^{1,2}$ | You can treat this as a problem with two Lagrange multipliers. Then by standard methods, a minimizer $f$ has to exist (by convexity in $f'$) and has to be a weak solution to
$$-f'' + \lambda f + \mu f \chi\_{[\pi/3,2\pi/3]} = 0$$
with $\lambda, \mu \in \mathbb{R}$ and $\mu \leq 0$ because the constraint is one-sided.
Solving this equation on its sub-intervals and using the boundary condition gives you
$$ f(x) = \begin{cases} a\_1 \sin(\lambda x) &\text{ for } x \in [0,\pi/3] \\ b\_1 \sin((\lambda + \mu)x) +b\_2 \cos((\lambda + \mu)x) &\text{ for } x \in [\pi/3,2\pi/3] \\ a\_2 \sin(\lambda (x-\pi)) &\text{ for } x \in [2\pi/3,\pi]. \end{cases} $$
Using a bit of regularity theory on $f'' = - \lambda f - \mu f \chi\_{[\pi/3,2\pi/3]} \in L^2([0,\pi])$ gives you $f \in W^{2,2}([0,\pi])$, so since we are in 1d, $f'$ is absolutely continuous. This gives you two equalities at $\pi/3$ and $2\pi/3$ each.
You also can explicitly calculate $\int\_0^\pi f^2 = 1$ and have $\int\_{\pi/3}^{2\pi/3} f^2 = \alpha$ or $\mu = 0$.
This gives you a system of 6 (independent, if I am not mistaken) equations on 6 variables. I will not try to solve it here, but my guess would be that it is a tedious but doable task. You will likely end up with a countable family of solutions, but selecting the smallest should be easy.
| 1 | https://mathoverflow.net/users/51695 | 414638 | 169,098 |
https://mathoverflow.net/questions/414624 | 2 | Let's say we have two metric spaces, $(X,\rho)$ and $(Y,\tau)$. The continuity of $f:X\to Y$ is obvious and natural to define. What about semi-continuity? Without a natural ordering on $Y$, perhaps "upper" and "lower" won't make sense anymore. But are there other, weaker notions of continuity such that $f$ is continuous iff it is continuous in all of those weaker senses?
| https://mathoverflow.net/users/12518 | Metric analogue of upper/lower semicontinuity | A "weak" generalization of continuity for a function $f:X\rightarrow Y$ with values in a metric space:
for each fixed $y\in Y$, consider the real valued function $f\_y:x\mapsto d(f(x),y)$;
for $f$ to be continuous each $f\_y$ must be continuous
(i.e. both upper and lover semicontinuous).
"Weak" here means that it is exactly the idea of weak topologies in functional analysis,
but this is also the idea for an embedding or a metric space in a space of continuous functions
(as used in a exercise in Rudin's "principles")
I want to show now that this kind of idea is used in a wide spectrum of mathematical contexts
(corollary: the idea might be too general for your needs, but perhaps with specific needs in mind it can be concretized).
Suppose we have a class $A$ of structures [like metric spaces], and a class $B$ of "nicer" structures [the reals], and three notion of morphisms:
a general one between $A$-structures [continuous maps], another general one between $B$-structures [the identity], and a "nicer" one from a $A$-structure to a $B$-structure
[the maps of the form "distance from a fixed point" for a suitably compatible (pseudo)metric];
assume the obvious compatibility conditions (two categories that act, one on one side and one on the other, on the set of "nicer" maps).
Then one can ask how much of the structure of a $A$-object can be recovered by knowing only the "nice" morphisms originating from the object.
[One can dualize the situation and consider "nice" maps towards $A$-structures].
(Another obvious, and perhaps generally more useful question: is there a universal morphism among the nice ones?
This is a "adjoint functor question" or "existence of a nice reflection/completion of a $A$-structure". But it is *not* the question directly posed here.)
This setting is very general, but is also effectively used:
The compactification theory (of completely regular T$\_0$ spaces, ore general ones) uses continuous maps to compact T$\_2$ spaces.
One recovers the structure of "completely regular T$\_0$ space" (or the (co?)reflection into such category for a general space),
and even maps towards the real interval $[0,1]$ are enough (family of (co?)generators in a category, in this case a single one).
Dually (continuous maps from T$\_2$ compact spaces) one has the theory of $k$-spaces (a common generalization of sequential and locally compact T$\_2$ spaces);
A. Weil thought (see his commentaries in his collected papers) that a better setting for "Bourbaki like" measure theory, one the compact T$\_2$ case is known, is not a locally compact T$\_2$ space but
a set equipped with a collection of compact subsets (a slight variation of the above, and again expressible in this general setting).
A (real, infinitely or only continuously) differentiable structure is known when one knows the "nice" (differentiable) maps from the reals
(differentiable paths, even the ones with "stop points" where the differential is 0) or dually the "nice" functions from the manifold to the reals.
This (is equivalent to the usual way for finite dimensional manifolds, thanks to a theorem known since 1969, and) has be used to generalize the
notion of Banach manifold to obtain a theory that works better that the old attempts using locally convex spaces.
To obtain analytic manifolds or complex manifolds, everywhere defined "nice" maps are not enough. One can generalize the setting (sheaf of rings of "nice" functions);
a variation that is useful for settings other than "analytic manifolds" uses a filter of "dense" subsets
(open dense sets in a topological space, dense G$\_\delta$ sets in a Baire space, co-null sets in a (finitely or countably additive) positive measure space)
and "nice" maps defined on them, identified when they agree on a "nice dense" set. One obtains representation theorems for vector lattices (and even lattice ordered, or even directed, abelian groups)
for with suitable completeness assumptions (and then weaker representation theorems for such structures that have a completion, the "integrally closed" condition being the necessary and sufficient one for completability).
One uses such kinds of "nice dense subsets" also in algebra, in the theory of rings of quotients (huge generalization of the usual embeddings of a commutative domain in a field), and torsion theories for modules.
Also in algebra, specifically in ring theory, taking as "nice" the morphisms from a commutative associative ring, one sees that one recovers only the Jordan ring structure
($a\circ b=(ab+ba)/2$) of an associative ring with $1/2$ (in sufficiently indecomposable cases that means that one recovers the ring up to isomorphisms or anti-isomorphisms).
With morphisms from boolean algebras, seen as "observables", one can define quantum logics.
So, returning to your case, for any special kinds of (continuous) maps from a metric space to the reals, one can treat them as "nice" in the above setting,
and study the $f:X\rightarrow Y$ such that for any "nice" map on $Y$ the composition with $f$ gives a "nice" map on $X$, or only a weak form of it, like
a semicontinuous map. It might be that, when a specific application is in mind, a specific useful concept of "nice" naturally arises.
Edit: I now see that while I was writing my answer in the comments other different proposals had been made. These proposals are quite surely better for your specific needs. Sorry for the noise, if you want I delete this answer (which was really meant to be a comment, but it is impossible to decently format long comments).
| 6 | https://mathoverflow.net/users/474159 | 414641 | 169,099 |
https://mathoverflow.net/questions/414629 | 5 | Let $K$ be a knot in $S^3$. Let $N(K)$ be a tubular neighborhood of $K$, a solid torus. On $\partial N(K)$, we may specify a *preferred* longitude $\lambda$, i.e., a simple closed curve whose linking number with $K$ is $0$. Also, we can choose a canonical meridian $\mu$ whose linking number with $K$ is $1$.
A $p/q$-*surgery* on $S^3$ along $K$ is a closed oriented $3$-manifold given by $$S^3\_{p/q} (K) = (S^3 - \mathrm{int}(N(K))) \cup\_\varphi (D^2 \times S^1)$$ where $\varphi: S^1 \times S^1 \to S^1 \times S^1$ is a homeomorphism that sends $\partial D^2 \times \{ 0 \}$ to $p \mu + q \ell$.
I want to understand the generalization of this process to an arbitrary closed oriented $3$-manifold $M$.
1. The concept of integral surgery makes sense for any $M$? If yes, how?
2. Here, the choice of $\lambda$ is not obvious. How about the rational surgeries?
**P.S.** I checked several reference books about the knot theory. I couldn't find a precise approach for this generalization. Any reading advise will be appreciated.
| https://mathoverflow.net/users/475366 | Integral surgeries on $3$-manifolds | Two places in which this is discussed are Gompf and Stipsicz's book *4-manifolds and Kirby calculus* (Sections 5.2 and 5.3) and Ozbagci and Stipsicz's *Surgery on contact 3-manifolds and Stein surfaces* (Chapter 2). The reason this is not discussed in many knot theory references is that this is more of a 3.5-dimensional issue (that is, in between dimension 3 and dimension 4) than a knot-theoretical one.
1. Yes, this makes sense in arbitrary 3-manifolds, since the fact that $q=1$ is independent of your choice of longitude. The meridian is always well-defined, and any two longitudes differ by a multiple of the meridian, so having an expression of the form $p\mu + \lambda$ is independent of which $\lambda$ you choose. (The $p$ will vary, of course.) A different, arguably "better" perspective is that this corresponds to attaching a 4-dimensional 2-handle to $M \times I$ along $K$, which is what the two references I gave you are mostly about.
2. I'm not sure what you mean by "how about", so I'll at least tell you that the construction you have for knots in $S^3$ is completely general, and it goes by the name of "Dehn surgery". You can do it for any knot in any 3-manifold, and the datum you need is just the choice of (the homology class of) a simple closed curve in $\partial N(K)$, also called the slope (which is your $(p,q)$ or $p/q$). The only drawback is that if $K$ and $M$ are arbitrary you don't have a canonical way to translate this slope into a rational number, unless $K$ is null-homologous in $M$ (see below).
About 2., let me add that the choice of $\lambda$ is not only non-obvious, but also not possible in general. The only instance in which there is a canonical choice of $\lambda$ is when the knot $K$ is null-homologous in $M$, i.e. $[K] \in H\_1(M;\mathbb{Z})$ vanishes. Then you have a preferred longitude (still called the Seifert longitude), which is the only curve on $\partial N(K)$ (up to isotopy) whose homology class dies in $H\_1(M\setminus K; \mathbb{Z})$.
In general, whenever you have a knot $K \subset M$, you always have a rank-1 subgroup of $H\_1(\partial N(K); \mathbb{Z})$ which dies in $H\_1(M \setminus K; \mathbb{Z})$. This group needs not be generated by a primitive element, so there might be no simple closed curve on $\partial N(K)$ that bounds a surface in $M \setminus {\rm int}(N(K))$. It will be generated by a longitude if and only if $K$ is null-homologous in $M$.
| 5 | https://mathoverflow.net/users/13119 | 414642 | 169,100 |
https://mathoverflow.net/questions/414617 | 0 | **Summary**
I would like to pass from a sequence of probability measures whose "limit" satisfies a desired property to a new probability measure that satisfies this property.
**Details**
We work on a probability space $(\Omega,\mathcal{F},P)$. Let $C\subseteq L^{\infty}$ be convex, contain $-L^{\infty}\_+$, and be closed in the weak\*-topology on $L^{\infty}$.
Suppose there exists a sequence of probability measures $(Q^n)\_{n\in\mathbb{N}}$, each of which is equivalent to $P$, such that for each $n$,
\begin{equation}
0\leq \sup\_{X\in C} E\_{Q^n}[X] \leq \frac{1}{n} \enspace (\*).
\end{equation}
The goal is to construct a probability measure $Q$, that is equivalent to $P$, such that
\begin{equation\*}
\sup\_{X\in C} E\_{Q}[X] =0.
\end{equation\*}
**Question**
The *intuitive idea* is to take the limit as $n\rightarrow \infty$ in $(\*)$. How can this be formalized? I.e. how can the probability measures $(Q^n)\_{n\in\mathbb{N}}$ be used to construct a new probability measure $Q$ with the desired properties? Are there any papers with a similar construction?
| https://mathoverflow.net/users/nan | Construction of a probability measure from a sequence of probability measures | $\newcommand\R{\mathbb R}$This is impossible to do in such generality.
E.g., suppose that $\Omega=\R$, $\mathcal{F}$ is the Borel $\sigma$-algebra over $\R$, $P$ is the standard normal distribution,
$$C=\{g\in L^\infty\colon g\le f\},$$
$f$ is the standard normal pdf, and $Q^n$ is the normal distribution with mean $n$ and variance $1$.
Then $Q^n$ is equivalent to $P$ for each $n$ and
\begin{equation}
0\le\sup\_{X\in C} E\_{Q^n}X= E\_{Q^n}f
=\int\_\R f(t)f(t-n)\,dt
=\frac1{\sqrt2}f\Big(\frac n{\sqrt2}\Big)\le\frac1n
\end{equation}
for each $n$.
However, for any probability measure $Q$ (even if not equivalent to $P$),
\begin{equation\*}
\sup\_{X\in C} E\_Q X=E\_Q f>0,
\end{equation\*}
since $f>0$.
| 2 | https://mathoverflow.net/users/36721 | 414649 | 169,103 |
https://mathoverflow.net/questions/414568 | 13 | Grothendieck topoi have cohomology: the abelian category of abelian group objects in a topos has enough injectives, hence one can consider the right derived functors of the global sections functor from abelian group objects to abelian groups. (For details see Chapter 8 in Johnstone's book *Topos theory*.)
Also, Grothendieck topoi have homotopy groups. (I think a reference is Artin and Mazur's *Etale homotopy*.)
**Question:** Can some of these algebraic invariants (or other "topological properties") of Grothendieck topoi be generalized to [pretopoi](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/pretopos)?
| https://mathoverflow.net/users/475784 | Do pretopoi have cohomology and homotopy groups? | There's a long story that can be told here but I will try to be brief.
In one sense, the answer is yes – you can certainly define cohomology and homotopy groups and so on for pretoposes and have them coincide with the classical definitions for Grothendieck toposes – but in another sense the answer is no – because you are essentially just embedding the pretopos into a suitable Grothendieck topos and reducing to that case.
Let $\mathcal{E}$ be a pretopos.
That means $\mathcal{E}$ is a category with a terminal object, pullbacks, finitary coproducts, and coequalisers of internal equivalence relations, such that finitary coproducts are disjoint and preserved by pullback, and coequalisers of internal equivalence relations are effective and preserved by pullback.
In short, $\mathcal{E}$ satisfies the exactness part of the Giraud axioms, with finitary coproducts eplacing infinitary coproducts.
That in itself should be a powerful reason to believe that any finitary constructions that can be carried out in a Grothendieck topos can also be carried out in $\mathcal{E}$ with the same results.
Indeed:
**Proposition.**
Assuming $\mathcal{E}$ is small, there is a fully faithful embedding of $\mathcal{E}$ into a Grothendieck topos where the embedding preserves finite limits, finitary coproducts, and coequalisers of internal equivalence relations.
*Proof.*
Regard $\mathcal{E}$ as a site where the covering sieves are those that contain a sieve generated by a finite family that is jointly strongly epimorphic, and take the topos of sheaves on this site. ◼
(If $\mathcal{E}$ is not small then go up to a universe where it is, or find a subpretopos that is small and contains the objects and morphisms you care about.)
Concretely, the category $\textbf{Ab} (\mathcal{E})$ of internal abelian groups in $\mathcal{E}$ is an abelian category (but not necessarily AB4 or AB4\*, let alone AB5).
So you can go on to define the category $\textbf{Ch} (\mathcal{E})$ of chain complexes in $\textbf{Ab} (\mathcal{E})$ and then the (unbounded) derived category $\mathbf{D} (\mathcal{E})$.
What you do not get is the existence of enough injectives in $\textbf{Ab} (\mathcal{E})$ itself.
Nonetheless, the definition of derived functors as (absolute, or at least pointwise) Kan extensions makes sense, and some derived functors can be constructed without injective resolutions.
For example, although $\mathbf{R} \textrm{Hom}\_{\textbf{Ch} (\mathcal{E})} (A, -) : \mathbf{D} (\mathcal{E}) \to \mathbf{D} (\textbf{Ab})$ itself does not have an obvious construction, $H\_0 \mathbf{R} \textrm{Hom}\_{\textbf{Ch} (\mathcal{E})} (A, -) : \mathbf{D} (\mathcal{E}) \to \textbf{Ab}$ always exists: you can directly check that $\textrm{Hom}\_{\mathbf{D} (\mathcal{E})} (A, -) : \mathbf{D} (\mathcal{E}) \to \textbf{Ab}$ works.
Also, if $\mathcal{E}$ is small, then any functor $\textbf{Ch} (\mathcal{E}) \to \textbf{Ab}$ whatsoever admits a pointwise left Kan extension along $\textbf{Ch} (\mathcal{E}) \to \mathbf{D} (\mathcal{E})$... but it is unclear to me whether this is consistent with what $(\infty, 1)$-category theory would give.
Similarly (in some sense...), the category $\textbf{Kan} (\mathcal{E})$ of internal Kan complexes in $\mathcal{E}$ is a category of fibrant objects (in the sense of Brown) where the fibrations are the internal Kan fibrations and the weak equivalences are the internal weak homotopy equivalences.
Thus the homotopy category $\mathbf{H} (\mathcal{E})$, obtained by localising $\textbf{Kan} (\mathcal{E})$ with respect to internal weak homotopy equivalences, is reasonable in the sense that there is a nice-ish formula for its hom-sets.
The category $\textbf{Set}\_\textrm{fin}$ of finite sets is the initial pretopos, so we get an induced functor $L : \mathbf{H} (\textbf{Set}\_\textrm{fin}) \to \mathbf{H} (\mathcal{E})$.
The homotopy type of $\mathcal{E}$ is "morally" a representing object for the functor $\textrm{Hom}\_{\mathbf{H} (\mathcal{E})} (1, L {-}) : \mathbf{H} (\textbf{Set}\_\textrm{fin}) \to \textbf{Set}$, where $1$ is the terminal object, but in practice this functor is rarely representable (even if $\mathcal{E}$ is a Grothendieck topos) so we are forced to make various tweaks like replacing $\textbf{Set}\_\textrm{fin}$ with $\textbf{Set}\_{< \kappa}$ (if $\mathcal{E}$ has coproducts of families of size $< \kappa$), or allowing more generalised notions of representability (e.g. pro-representability), or both.
(Actually, this phenomenon can already be seen at homotopy level 0 for Grothendieck toposes, so in some sense the difficulty does not (only) come from trying to work in higher homotopy levels or with general pretoposes instead of Grothendieck toposes.)
| 5 | https://mathoverflow.net/users/11640 | 414654 | 169,104 |
https://mathoverflow.net/questions/414552 | 5 | [Edit: added details on the Reed-Muller codes]
Are there explicit (non random) constructions of probability measures on $D\_N = \{0,1\}^N$ with support of size $O(N)$ and with all nontrivial Fourier coefficients less than $\frac 1 2$ is absolute value ?
I would expect that such questions are very well understood, but being from a different background I probably lack the common knowledge. Pointers to the relevent litterature are very much welcome.
Below are some explanations on the terminology and on the $O(N)$. I am happy with any answer with $\frac 1 2$ replaced by $c<1$ (but independent from $N$).
---
When I write that *$\mu$ has all nontrivial Fourier coefficients less than $\frac 1 2$ in absolute value*, I mean that $|\hat \mu(A)| \leq \frac 1 2$ for every nonempty subset $A$ of $\{1,\dots,N\}$, where $\hat \mu(A)$ is the Fourier (or Fourier-Walsh) coefficient:
$$ \hat \mu(A) = \int (-1)^{\sum\_{i \in A} \varepsilon\_i} d\mu(\varepsilon).$$
The trivial Fourier coefficient, corresponding to $A=\emptyset$, is always $1$.
The uniform probability measure on $D\_N$ has all non-trivial Fourier coefficients equal to $0$, but has full support of size $2^N$. So I am looking for a probability measure that looks like the uniform probability on the Fourier side, but is very far on the other side, having very small support.
---
Why the $O(N)$? Because it is the optimal rate for the mere existence of $\mu$.
Indeed, seeing $D\_N$ as a vector space of dimension $N$ over the field with two elements, the question can be translated in linear algebra terms. In particular, we see that for all nontrivial Fourier coefficients of $\mu$ to be $\neq 1$, we need that the support of $\mu$ contains at least a basis of $D\_N$ and therefore has to have cardinality at least $N$.
Conversely, taking for $\mu$ the uniform measure on a subset of cardinality $K N$ chosen uniformly at random, a basic application of union bound and large deviation (Hoeffding's inequality) arguments gives that $\mu$ works with high probability, provided that $K$ is large enough. What I am looking for is explicit non-random constructions.
---
Using [Reed-Muller codes](https://en.wikipedia.org/wiki/Reed%E2%80%93Muller_code), a back-of-the enveloppe-computation seems to provide an explicit $\mu$ with support $N^{K \log \log N}$.
[Added after request in comments]
Here is the construction. I do not know how standard it is ; I learnt it from [the MIP\*=RE paper](https://arxiv.org/abs/2001.04383).
There is an integer parameter $a$, and we using the code coming from multilinear functions over the field $\mathbf{F}\_q$ with $q=2^{a}$ elements in $m = 2^{a-1}$ variables.
Denote $\mathcal{P}(m)$ the set of all subsets of $\{1,\dots,m\}$, and define a map $\alpha:\mathbf{F}\_q^{m+1} \to \mathbf{F}\_q^{\mathcal{P}(m)}$ by setting
$$\alpha(u\_1,\dots,u\_m,a) = (a \prod\_{i \in A} u\_i \prod\_{i \notin A} (1-u\_i))\_{A \in \mathcal{P}(m)}.$$
We can identify $\mathbf{F}\_q^{\mathcal{P}(m)}$ as an additive group with $D\_N$ for $N=a 2^m = a 2^{2^{a-1}}$, and I claim that $\mu$, the image by $\alpha$ of the uniform measure on $\mathbf{F}\_q^{m+1}$ does the job.
$\mu$ clearly has support of size $q^{m+1} = 2^{a +a2^{a-1}} = 2^{O(\log N \log\log N)}$, so we are fine as far as the support is concerned.
Let me prove that $|\hat\mu(\chi)|\leq \frac 1 2$ for every nontrivial character $\chi$. Let us fix a nonzero character $\eta$ of $\mathbf{F\_q}$. Then by counting, every character of $\mathbf{F\_q}^{\mathcal P(m)}$ is of the form $\eta \circ T$ for a $\mathbf{F}\_q$-linear map $T:\mathbf{F}\_q^{\mathcal{P}(m)} \to \mathbf{F}\_q$. Moreover, for every $x \in \mathbf{F}\_q$, we have $\mathbf{E}\_{a \in \mathbf{F\_q}} \eta(ax)=1\_{x=0}$, so we can compute
\begin{align\*}\hat \mu(\chi) &= \mathbf{E}\_{u \in \mathbf{F}\_q^m} \mathbf{E}\_{a \in \mathbf{F}\_q} \eta(a T(\alpha(u,1)))\\
&= \mathbf{E}\_{u \in \mathbf{F}\_q^m} 1\_{T(\alpha(u,1))}=0\\
& = \mathbf{P}\_{u \in\mathbf{F}\_q^m}( T(\alpha(u,1))).
\end{align\*}
But if $T$ is a nonzero linear map, then $T(\alpha(u,1))$ is a nonzero polynomial in $m$ variables of individual degree $\leq 1$, so the last probability is $\leq\frac{m}{q} = \frac 1 2$. QED
| https://mathoverflow.net/users/10265 | Probability measure on the boolean cube with small support and small Fourier transform | Just to close the lid on my comment: the desired construction is known as an $\epsilon$-biased space, where here $\epsilon = 1/2$. The best known construction seems to be by Ta-Shma from 2017 (<https://eccc.weizmann.ac.il/report/2017/041/>), which yields a subset with support size $O(N/\epsilon^{2+o(1)})$. This is near-optimal, as it is known that $\Omega(N/\epsilon^2\log(1/\epsilon))$ is necessary. However, for the special case of $\epsilon = 1/2$ where the precise $\epsilon$-dependence is not relevant, it appears that earlier constructions based on well-known codes suffice to obtain $O(N)$.
| 7 | https://mathoverflow.net/users/170770 | 414657 | 169,105 |
https://mathoverflow.net/questions/409371 | 2 | I have been reading the paper *"Coherent orientations for periodic orbits problems in symplectic geometry"* by Floer and Hofer, trying to understand how we can orient the moduli spaces that appear in Floer Homology in a way that is coherent.
So to do this we first fix on the problem on considering the Fredholm operators of the form $L=\partial\_s+J\_0\partial\_t + S\_L(s,t)$ that are asymptotically stable and we try to see how the orientations of their determinant bundles induce an orientation on $L\#\_{\rho} K$ for a specific choice of $\rho$.
Now the part that I am having some trouble with is the associativity part. They say that to prove this it's analogous to what was done to proving that the induced orientation doesn't depend on the choice of constant asymptotically operator, provided that they have the same limit operators. This is done by using an homotopy argument. Now I am not sure how one can do something similar to this to prove associativity.
My idea was to just use the fact that $\text{Det}((L\#\_{\rho\_0} K)\#\_{\rho\_1} M) \cong \text{Det}(L\#\_{\rho\_0} K)\otimes \text{Det}(M)\cong (\text{Det}(L)\otimes \text{Det}(K))\otimes \text{Det}(M)$ and then use the associativity of the tensor product. However here the argument is not similar to the previous one, hence I am not sure this is correct. Also this would be a lot simpler, and hence I would assume that if this was correct it would be mentioned that this works.
Also in Schwarz's book *Morse homology*, he mentions how one can prove associativity, and it's using an homotopy argument, however I would like to understand why this method wouldn't work.
So I would like to understand if there is anything wrong with this argument. Any enlightenment is appreciated, thanks in advance.
| https://mathoverflow.net/users/155363 | Associativity of orientations of determinant bundles in Floer homology | The issue here are the isomorphisms in your suggested proof. There are choices and conventions involved. To give names to the isomorphisms clarifies what needs to be checked.
$$
\iota\_{K,L}:Det(K\sharp\_{\rho}L)\to Det(K)\otimes Det(L).
$$
You need to check that you've set things up so that:
$$
\iota\_{K\sharp\_{\rho\_0}L,M}\circ (Id\_{K} \otimes \iota\_{L,M})=\iota\_{K,L\sharp\_{\rho\_1}M}\circ (\iota\_{K,L}\otimes Id\_M).
$$
If you get your conventions wrong there is a sign lurking in here. As shameless promotion you can look at the proof of Lemma 20.2.1. on page 380 of **Monopoles and Three Manifolds** by Kronheimer and myself. The proof is actually on the pages before.
| 3 | https://mathoverflow.net/users/12605 | 414658 | 169,106 |
https://mathoverflow.net/questions/414661 | -1 | Let $(X, \lVert \cdot \rVert)$ be a separable normed space. Can we always guarantee that there is a nonempty compact set $K \subseteq B\_X$, where $B\_X$ is a closed unit ball in $X$ such that:
$$\forall \Lambda \in S\_{X^\*} \quad \sup\_{k \in K} \Lambda(k) > 0,$$
where $S\_{X^\*}$ denotes the unit sphere in the dual space $X^\*$? If not, is it possible to do so, if instead of $S\_{X^\*}$ we would take some weak\* dense subset of $S\_{X^\*}$?
In the case when $\mathrm{dim}X < \infty$ the answer is yes, since we can just take a closed unit ball for $K$, however, I'm not sure whether this result could be generalized to any separable normed space.
| https://mathoverflow.net/users/170491 | "Large" compact sets in separable normed space | Let $S=\{x\_n:n\in\mathbb N\}$ be a dense subset and choose
$\varepsilon\_n>0$ such that $y\_n=\varepsilon\_n x\_n\to 0$ (e.g., $\varepsilon\_n=1/n\|x\_n\|$). Then $K=\{ty\_n:n\in\mathbb N, |t|=1\}\cup\{0\}$ is compact and for every every $\Lambda\in X^\*\setminus \{0\}$ we have $\sup\{\Lambda(k):k\in K\}\ge |\Lambda(y\_n)|>0$ for some $n\in\mathbb N$ since otherwise $\Lambda$ would vanish on $S$ and hence on $X$.
| 2 | https://mathoverflow.net/users/21051 | 414664 | 169,107 |
https://mathoverflow.net/questions/414663 | 1 | Assume we are over $\mathbb C$. Let $C$ be a complete algebraic curve, and $E$ an algebraic vector bundle. Its Hilbert polynomial is
$$p(t)=rt+r(1-g)+d$$
where $r=\mathrm{rank}(E)$ and $d=\deg(E)$ and $g$ is the genus of $C$. So a condition on the Hilbert polynomial is a condition on the rank, degree and genus.
It is known that, if $r=1$ and $d<0$, then $E$ has no nonzero section. I wonder if there are similar conditions for general rank.
| https://mathoverflow.net/users/105537 | For a vector bundle over a curve, is there a condition on the Hilbert polynomial for no non-zero section? | If the bundle is semi-stable, then $H^0(E) \neq 0$ implies $\deg(E) \geq 0$. Indeed, a global section of $E$ induces a non-zero map of shaves $\mathcal{O}\_C \to E$, and semi-stability implies $$0= \mu (\mathcal{O}\_C) \leq \mu(E)=\frac{1}{r}\deg(E).$$ For strictly unstable bundles one cannot say anything. For instance, if $L$ is a line bundle of degree $-1$ on $C$, then $E=L^{\otimes n} \oplus \mathcal{O}\_C$ has degree $-n$ and $H^0(E)=1$, so for all $g \geq 0$ and for all $r \geq 2$ there exist unstable vector bundles on $C$ with non-zero global sections and arbitrarily negative degree.
| 2 | https://mathoverflow.net/users/7460 | 414665 | 169,108 |
https://mathoverflow.net/questions/414521 | 3 | Suppose that $f\in \mathbb{Z}[x\_1,\dots,x\_n]$ and $f$ is a homogenous polynomial of degree $d$. Can we always construct $f$ such that the hypersurface $S\_f=\{x \in \mathbb{Z}^n:f(x)=0\}$ exhibits the failure of the Hasse principle? In particular, I am trying to understand why the non-singularity condition in results of Birch (Forms in Many Variables), Browning-Prendiville etc. is vital. I am told that there in case of $d=4$, there is an example of Hasse failure, probably by Swinnerton-Dyer ($S\_f$ for $f(x\_1,x\_2,x\_3,x\_4)=7x\_1^4+8x\_2^4-9x\_3^4-14x\_4^4$) (not sure how it is a Hasse failure and how easy it is to prove that it is one, but at any rate, it is a non-singular hypersurface that demonstrates that 4 variables are not nearly enough). Can we find more examples of such hypersurfaces which fail to satisfy the hassy principle, for degree $d=4$ or otherwise? I was just hoping to find a large class of examples of such $f$, not just one.
| https://mathoverflow.net/users/392272 | Examples of non-singular hypersurfaces exhibiting Hasse principle failures | Daniel Loughran's suggestion to study the paper [Random Diophantine Equations by Poonen and Voloch](https://math.mit.edu/%7Epoonen/papers/random.pdf) was a good one.
Te conjecture 3.2 of the paper Daniel Loughran linked, together with Appendix A (both due to Colliot-Thélène), implies that the Hasse principle holds for nonsingular hypersurfaces of degree $4$ in $n$ variables as soon as $n>4$.
So, conjecturally, the condition $n>48$ in Browning-Prendiville is not at all necessary, and the $n>48$ condition is "merely" a restriction of the technique.
For $n=4$, there are more counterexamples. The paper [Counterexamples to the Hasse principle in families](https://doi.org/10.1017/S0004972721000885) contains two infinite families along the lines of Swinnerton-Dyer's example.
| 4 | https://mathoverflow.net/users/18060 | 414681 | 169,110 |
https://mathoverflow.net/questions/414639 | 2 | This question arose by reading the paper **[1]**, in particular, the remark at p. 737:
>
> As an example, consider a non-isotrivial smooth projective morphism $f \colon X \to Y$ from a smooth projective surface to a smooth
> projective curve. Suppose that there exists a $y \in Y$ such that the
> Kodaira-Spencer map $$\rho\_{f, \, y} \colon T\_{Y, \, y} \to H^1(X\_y, \,
> T\_{X\_y})$$ is zero. Then $\Omega\_X|\_{X\_y}=\mathcal{O}\_{X\_y} \oplus
> \Omega\_{X\_y}$, so $\Omega\_X|\_{X\_y}$ is not ample.
>
>
>
Smooth, non-isotrivial fibrations from a surface to a curve are called *Kodaira fibrations* (see **[2]** for a good introduction to the subject) and the usual way to construct them is taking suitable ramified covers of a product of two curves.
Now, I do not see how to perform the construction in such a way that the above condition on the Kodaira-Spencer map is satisfied. In the paper **[3]** Kodaira shows, by means of a local computation, that in all his original examples the Kodaira-Spencer map is everywhere non-vanishing (p. 212-213). On the other hand, I am not aware of any example in which it vanishes at some point (in fact, I am not aware of any other example of Kodaira fibration for which the Kodaira-Spencer map has been explicitly computed).
So let me ask the
>
> **Question.** What are examples of non-isotrivial smooth projective morphisms $f \colon X \to Y$ from a smooth projective surface to a
> smooth projective curve such that the Kodaira-Spencer map vanishes at
> some $y \in Y$?
>
>
>
**References.**
**[1]** *Jabbusch, Kelly*, [**Positivity of cotangent bundles**](http://dx.doi.org/10.1307/mmj/1260475697), Mich. Math. J. 58, No. 3, 723-744 (2009). [ZBL1186.14037](https://zbmath.org/?q=an:1186.14037).
**[2]** *Catanese, Fabrizio*, [**Kodaira fibrations and beyond: methods for moduli theory**](http://dx.doi.org/10.1007/s11537-017-1569-x), Jpn. J. Math. (3) 12, No. 2, 91-174 (2017). [ZBL1410.14010](https://zbmath.org/?q=an:1410.14010).
**[3]** *Kodaira, Kunihiko*, [**A certain type of irregular algebraic surfaces**](http://dx.doi.org/10.1007/BF02788717), J. Anal. Math. 19, 207-215 (1967). [ZBL0172.37901](https://zbmath.org/?q=an:0172.37901).
| https://mathoverflow.net/users/7460 | Smooth, non-isotrivial fibration with vanishing Kodaira-Spencer map at a point | Any ramified base change of a smooth, non-isotrivial fibration should give an example.
You can give a simple proof using one of the various characterizations of the Kodaira spencer map:
<https://en.wikipedia.org/wiki/Kodaira%E2%80%93Spencer_map#In_scheme_theory>
Namely, if $f:X\to Y$ is a smooth fibration over a curve $Y$, then the Kodaira spencer map at a point $\operatorname{Spec} k = y\in Y$ is obtained by restricting $f$ to a first order thickening $\overline{f}:\mathfrak{X}\_{y}\to \operatorname{Spec} k[\varepsilon]$ of $y$ and considering the associated sequence of differentials
$$0\to \overline{f}^\*\Omega\_{k[\varepsilon]/k}\to \Omega\_{\mathfrak{X}\_y}\to \Omega\_{\mathfrak{X}\_y/k[\varepsilon]}\to 0 $$
By smoothness of the morphism, this sequence is exact and the last nonzero term is locally free as an $\mathcal{O}\_{\mathfrak{X}\_y}$-module. In particular, the sequence remains exact when you restrict to the reduced fiber $X\_y$. By the base change properties of differentials, you get an exact sequence
$$0\to f^\*T\_{Y,y}^{\vee}\to \Omega\_{\mathfrak{X}\_y}|\_{X\_y} \to \Omega\_{X\_y}\to 0$$
The associated element in $T\_{Y,y}^{\vee}\otimes \operatorname{Ext}^1(\Omega\_{X\_y},\mathcal{O}\_X)$ is equivalent to the data of a map $T\_{Y,y}\to H^1(X\_y,T\_{X\_y})$ which is the Kodaira spencer map.
Now if $h:Y'\to Y $ is a morphism of smooth curves that is ramified at $y'\in Y'$ such that $h(y')=y$, we get a diagram
$$ \begin{array}{ccc} \mathfrak{X}\_{y'} & \rightarrow & \mathfrak{X}\_y \\
\downarrow & & \downarrow \\
\operatorname{Spec}k[\varepsilon] & \rightarrow & \operatorname{Spec}k[\varepsilon] \end{array}$$
in which the lower horizontal map is given by the dual map on algebras with $\varepsilon\mapsto 0$.
By functoriality of differentials, it follows that we have a morphism of exact sequences
$$ \begin{array}{ccccccccc} 0 & \rightarrow & f\_y^\*T\_{Y,y}^{\vee} & \rightarrow & \Omega\_{\mathfrak{X}\_y}|\_{X\_y}&\rightarrow &\Omega\_{X\_y}& \rightarrow & 0\\
& &\downarrow & & \downarrow& &\downarrow & & \\0 & \rightarrow & h^\*f\_{y'}^\*T\_{Y',y'}^{\vee} & \rightarrow & \Omega\_{\mathfrak{X}\_y}|\_{X\_y}&\rightarrow &\Omega\_{X\_y}& \rightarrow & 0 \end{array}$$
Here, the first vertical map is the differential of $h$, which vanishes when we have ramification, while the last vertical map is the identity.
Finally, it follows that we have a commutative diagram of Kodaira-Spencer maps
$$ \begin{array}{ccc} T\_{Y',y'} & \rightarrow & H^1(X\_{y'},T\_{X\_{y'}}) \\
\downarrow & & \downarrow \\T\_{Y,y} & \rightarrow & H^1(X\_y,T\_{X\_y})\end{array}$$
where the first vertical map is zero, and the last vertical map an isomorphism. This implies that the upper horizontal map is zero.
| 5 | https://mathoverflow.net/users/112142 | 414682 | 169,111 |
https://mathoverflow.net/questions/414522 | 19 | Does anybody know any biographical information about [М. М. Артюхов](http://www.mathnet.ru/php/person.phtml?&personid=34845&option_lang=eng) (e.g., first name, affiliation)?
It seems he discovered a criterion for primality equivalent to the Solovay–Strassen one in 1966, in this paper: [Некоторые критерии простоты чисел, связанные с малой теоремой Ферма](https://www.impan.pl/en/publishing-house/journals-and-series/acta-arithmetica/all/12/4/96314/nekotorye-kriterii-prostoty-chisel-svjazannye-s-maloj-teoremoj-ferma).
(Keith Conrad discusses the story a little bit here: [The Miller–Rabin test](https://kconrad.math.uconn.edu/blurbs/ugradnumthy/millerrabin.pdf).)
I'm just curious. :-)
| https://mathoverflow.net/users/658 | M. M. Artyukhov / М. М. Артюхов | In the books [1], pp. 41-42 and [2], p. 67, it is stated that his name (and patronymic) is Mikhail Mikhailovich, confirming [Anatoly Kochubei's comment](https://mathoverflow.net/questions/414522/m-m-artyukhov-%d0%9c-%d0%9c-%d0%90%d1%80%d1%82%d1%8e%d1%85%d0%be%d0%b2/414688#comment1063272_414525) to [Kostya\_I's answer](https://mathoverflow.net/a/414525/113756). References also state that he was born on the 26 of April 1910 in Moscow, laureated from (the then called) Leningrad University in 1936, become "aspirant" (aspirant in science) until 1939, then "kandidat fiz.-matem. nauk" (candidate in physics and mathematical sciences) and "docent" (professor) from 1945 onward. After 1954 he worked at the North Caucasus (Ordzhonikidze) Metallurgical Institute ([1], p. 41), thus Kostya\_I's answer is further confirmed. In [1], p. 42, there's a list of six of his works, the first being dated 1934 and the latest one being dated 1957: in [2], p. 67, the list is continued with four more works, dated from 1958 to 1963.
**References**
[1] Fomin, S. V.; Shilov, G. E., eds. (1970), Математика в СССР 1958–1967 [Mathematics in the USSR 1958–1967] (in Russian), Том второй: Биобиблиография выпуск второй М–Я, Москва: Издательство "Наука", p. 762, [MR 0250816](https://www.ams.org/mathscinet-getitem?mr=0250816), [Zbl 0199.28501](https://zbmath.org/?&q=an:0199.28501).
[2] Kurosh, A. G.; Vityushkov, V. I.; Boltyanskii, V. G.; Dynkin, E. B.; Shilov, G. E.; Yushkevich, A. P., eds. (1959), Математика в СССР за сорок лет 1917–1957 [Mathematics in the USSR 1917–1957] (in Russian), Том второй: Биобиблиография, Москва: Государственное Издательство Физико–Математическои Литературы, p. 819, [MR 0115874](https://www.ams.org/mathscinet-getitem?mr=0115874), [Zbl 0191.27501](https://zbmath.org/?q=an:0191.27501).
| 13 | https://mathoverflow.net/users/113756 | 414688 | 169,113 |
https://mathoverflow.net/questions/414594 | 9 | I'm reading [Amplitudes and the Riemann Zeta Function](https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.127.241602), which recently appeared in Physical Review Letters. It's received some [publicity](https://phys.org/news/2022-01-quantum-zeta-epiphany-physicist-approach.html), including [my own campus](https://www.news.ucsb.edu/2022/020520/quantum-zeta-epiphany)' PR operation. From the abstract (adapting the notation)
>
> "Physical properties of scattering amplitudes are mapped to the
> Riemann zeta function. Specifically, a closed-form amplitude is
> constructed, describing the tree-level exchange of a tower with masses
> $m\_n^2=\gamma\_n^2$, where $\zeta(1/2+i\gamma\_n)=0$. Requiring real
> masses corresponds to the Riemann hypothesis..."
>
>
>
NB: I'm skeptical that this will lead to any progress on the Riemann Hypothesis, and not interested in that aspect.
I'm trying to determine the new (if any) mathematical content. The author develops the identity
$$-4+\frac{\pi^2}{8}+G+\frac{\zeta^{\prime\prime}(1/2)}{2\zeta(1/2)}-\frac{1}{8}\left(C+\frac{\pi}{2}+\log8\pi\right)^2
=2\sum\_{n=1}^\infty\frac{1}{\gamma\_n^2}.$$
Here $G$ is the Catalan constant and $C$ denotes the Euler constant. (I can check this via the logarithmic derivative of the Hadamard product for $\Xi(\sqrt{s})$, and differentiating the functional equation for $\zeta(s)$ twice.)
More generally, with
$$\zeta\_n(s)=\frac{\zeta^{(n)}(s)}{\zeta(s)}, \qquad\zeta\_n^k=\zeta\_n(1/2)^k,$$
he has identities for $2\sum\_{n=1}^\infty 1/\gamma\_n^{2k}$ for odd $k$, for example
$$
2\sum\_{n=1}^\infty \frac{1}{\gamma\_n^6}=-128+\frac{1}{7680}\Psi^{(5)}(1/4)-\zeta\_1^6+3\zeta\_1^4\zeta\_2-\frac{9}{4}\zeta\_1^2\zeta\_2^2+\frac{1}{4}\zeta\_2^3-\zeta\_1^3\zeta\_3+\zeta\_1\zeta\_2\zeta\_3-\frac{1}{12}\zeta\_3^2+\frac{1}{4}\zeta\_1^2\zeta\_4-\frac{1}{8}\zeta\_2\zeta\_4-\frac{1}{20}\zeta\_1\zeta\_5+\frac{1}{120}\zeta\_6
$$
The author writes
>
> "[These] can be proven exactly albeit laboriously, without appeal to
> our amplitude, using repeated differentiation of the functional
> equation and the Hadamard product form of the zeta function, as well
> as various polygamma identities... what is remarkable is that our
> amplitude construction allows for much simpler, physical derivations
> of these identities."
>
>
>
Are such identities new? Are they interesting? Regarding the latter, the author in an [online talk](https://pirsa.org/21100004) describes such sums as moments of the $\{\gamma\_n\}$, and they look like moments to me.
| https://mathoverflow.net/users/6756 | Scattering amplitudes and the Riemann zeta function | Thanks to @reuns for the answer in the comments. I've asked him to post as an answer, and I will accept it if he does. Meanwhile, his comments encouraged me to look again, and here is another approach, quite easy (Lemma 4.11 in Equivalents of the Riemann Hypothesis vol. II)
Let $\Xi(s)=\xi(1/2+is)$, so $\Xi(-s)=\Xi(s)$. Let $\gamma\_n$ be the zeros with positive real part, so the Hadamard product looks like
$$
\Xi(s)=\xi(1/2)\prod\_{n=1}^\infty\left(1-\frac{s^2}{\gamma\_n^2}\right)
$$
Expanding $\Xi(s)$ as a power series about $s=0$
$$
\sum\_{k=0}^\infty \frac{(-1)^k\xi^{(2k)}(1/2)}{2k!}s^{2k}=\Xi(s)=\xi(1/2)\left(1-s^2\left(\sum\_n \gamma\_n^{-2}\right)+s^4\left(\sum\_{m,n}\gamma\_m^{-2}\gamma\_n^{-2}\right)-\ldots\right)
$$
Equating coefficients and the [Newton Identities](https://en.wikipedia.org/wiki/Newton%27s_identities) gives the moments as polynomials in the $\xi^{(2k)}(1/2)$, for example
$$
\sum\_n\frac{1}{\gamma\_n^6}=\frac{\xi^{(6)}(1/2)}{240\,\xi(1/2)}-\frac{\xi^{(2)}(1/2)\xi^{(4)}(1/2)}{16\,\xi(1/2)}+\frac{\xi^{(2)}(1/2)^3}{8\,\xi(1/2)}.
$$
| 9 | https://mathoverflow.net/users/6756 | 414689 | 169,114 |
https://mathoverflow.net/questions/414619 | 0 | Consider the following """easier""" conjectures:
**C1**. every sum of two semiprimes $n = pq + rs$, $p,q,r,s$ primes, can be expressed as $n = (a + b)/2$; with $a,b$ primes.
**C2**. every number $p(q + r)$, $p,q,r$ primes (sum of two squarefree semiprimes that share a factor), can be expressed as $p(q + r) = (a + b)/2$; with $a,b$ primes.
Do they have an official name/literature?
| https://mathoverflow.net/users/35419 | Name of conjectures similar to Goldbach conjecture | I don't believe either has a name in the literature.
Also, I don't know how much easier these conjectures would be. They don't seem to have any useful structure for the problem to take advantage of. C1 is almost surely equivalent to the Goldbach conjecture itself: probably every number greater than 33 (resp., 82) is the sum of two semiprimes (resp., squarefree semiprimes). See [A072931](https://oeis.org/A072931), [A072966](https://oeis.org/A072966), and [A329481](https://oeis.org/A329481).
| 3 | https://mathoverflow.net/users/6043 | 414703 | 169,119 |
https://mathoverflow.net/questions/414714 | 0 | Let $A\in \mathbb{R^{n\times n}}$ be a symmetric negative difinite matrix and
$D\in \mathbb{R}^{n\times n}$ be a diagonal matrix $D = \mathrm{diag}\{d\_i\}, (d\_i < 0)$.
From Weyl's inequality, the maximum eigenvalue of the sum of these matrices $S = A + D$ can be evaluated as follows.
$\sigma\_{1} \leq \alpha\_{1} + \delta\_{1}$,
where $\sigma\_1\geq \cdots \geq \sigma\_n$, $\alpha\_1\geq \cdots \geq \alpha\_n$ and $\delta\_1\geq \cdots \geq \delta\_n$, are eigenvalues of $S$, $A$ and $D$ respectively.
Weyl's inequality can be applied to the sum of two Hermitian matrices. However here we have a stricter condition for $A$ and $D$. Therefore, I am considering the possibility of obtaining a smaller upper bound of $\sigma\_{1}$.
I would appreciate it if you could give me some advice.
| https://mathoverflow.net/users/475899 | Change in the largest eigenvalue due to perturbation of diagonal components of a symmetric matrix | You don't really have any stricter conditions, and in fact it is simple to reduce the general case to yours: every matrix is diagonal in *some* basis, and every matrix is negative definite if you subtract a suitable multiple of the identity to it.
So no, there can be no better bounds than the general case.
| 2 | https://mathoverflow.net/users/1898 | 414716 | 169,123 |
https://mathoverflow.net/questions/414710 | 1 | Let $V$ be a TRO i.e. closed subspace of $B(H,K)$ such that $xy^\*z \in V$ for all $x,y,z \in V$. Let $C(V)$ and $D(V)$ denotes the $C^{\ast}$-algebra generated by $VV^{\ast}$ and $V^\*V$ respectively. We define $A(V)$, the linking $C^\*$-algebra of $V$ as follows:
$$A(V) = \begin{bmatrix}
C(V) & V\\
V^\* & D(V)
\end{bmatrix}$$
>
> Let $V$ be a TRO such that $A(V)= \mathbb{C}$, what can we say about $V$?
>
>
>
| https://mathoverflow.net/users/129638 | Let $V$ be a TRO such that $A(V)= \mathbb{C}$, what can we say about $V$? | So, I think when we write
$$ A(V) = \left[ \begin{matrix} C(V) & V \\ V^\* & D(V) \end{matrix} \right] $$
we implicitly mean taking the linear span. Thus $A(V) = \mathbb C$ means that $A(V)$ is spanned by a single matrix, which is necessarily a projection, as $A(V)$ is a $C^\*$-algebra. So $V$ must certainly be one-dimensional as well, say spanned by $v\in B(H,K)$. However, if $C(V)$ or $D(V)$ is non-zero, then because we *take linear spans*, $A(V)$ will still be more than one-dimensional. As $\|v^\*v\| = \|v\|^2$, we must have that $C(V)$ and $D(V)$ are non-zero. So I believe this is impossible.
---
If $V$ were spanned by a single [partial isometry](https://en.wikipedia.org/wiki/Partial_isometry) then $A(V)$ will be four dimensional.
| 1 | https://mathoverflow.net/users/406 | 414721 | 169,124 |
https://mathoverflow.net/questions/414268 | 0 | Are there any central simple algebras admitting a standard basis?
By a standard basis I mean a normal basis that has a cyclic property generalizing that of the familiar basis $1, i, j, k$ for Quaternion algebras $(a, b \mid F)$, satisfying relations $i^2=a, j^2=b, k=ij=-ji$ for $a, b \in F^\times$.
I saw [Cayley–Dickson construction](https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction) which yields $F$-algebras of dimension $2^n$, but these algebras are not associative so not central simple in general)
One way to construct quaternion algebras is to use abelian varieties:
Like in the case of elliptic curves, by using $\ell$-adic Tate modules we have the following:
>
> the endomorphism ring $\text{End}(A)$ is finitely generated $\mathbb{Z}$-module of rank at most $4g^2$.
>
>
>
In general, the endomorphism ring $\text{End(A)}$ of an abelian vairety $A$ is an order in a semi-simple algebra over $\mathbb{Q}$. When $B$ is a semi-simple algebra over $\mathbb{Q}$ admitting a positive definite anti-involution, such algebras have been classified by A. Albert and G. Scorza.
Type I: Totally real number field
Type II/III: Definite or indefinite quaternion algebra over totally real field
Type IV: Central simple algebra over a CM-field.
Central simple algebras of dimension $n^2$ are all quaternion algebras when $n=2$. I was wondering if other form of basis with cyclic property appears when $n$ is larger, but at least in the construction above only quaterinon algebras appear.
| https://mathoverflow.net/users/475480 | Are there any central simple algebras admitting a standard basis? | As suggested by [@Kimball](https://mathoverflow.net/questions/414268/are-there-any-central-simple-algebras-admitting-a-standard-basis/414750#comment1063039_414268) I develop my [comment](https://mathoverflow.net/questions/414268/are-there-any-central-simple-algebras-admitting-a-standard-basis/414750#comment1062278_414268). An important class of central simple algebras consists of the *cyclic algebras*: assume that the field $k$ contains a primitive $n$-th root of unity $\zeta $, with $n$ prime to $\operatorname{char}(k) $. Let $a,b\in k^\*$; the cyclic algebra $R\_{a,b}$ is the quotient of $k\langle x,y \rangle$ by the relations $x^n=a, y^n=b, yx=\zeta xy$. The monomials $x^{i}y^{j}$ for $0\leq i,j<n$ form a natural basis of $R\_{a,b}$ over $k$, generalizing the standard basis of the quaternions. Over a number field, every central simple algebra is
cyclic — this is a deep result of Albert–Hasse–Brauer–Noether. It is a classical conjecture that for $n$ prime this should hold over any field.
| 3 | https://mathoverflow.net/users/40297 | 414725 | 169,126 |
https://mathoverflow.net/questions/414729 | 2 | It is well-known that $[0,1]$ is not a nontrivial disjoint union of closed intervals -- e.g.:
<https://math.stackexchange.com/questions/1195179/the-interval-0-1-is-not-the-disjoint-countable-union-of-closed-intervals>
Using fat Cantor sets, I can construct disjoint unions of closed intervals in $[0,1]$ with arbitrarily large Lebesgue measure. Question: Can full Lebesgue measure be achieved?
Edit: Here is my precise question: Does there exist an infinite collection of disjoint, nonempty, closed intervals $I\_n=[a\_n,b\_n]$
such that $\mu([0,1]\setminus\bigcup\_{n\ge1}I\_n)=0$, where $\mu$ is the Lebesgue measure?
| https://mathoverflow.net/users/12518 | Disjoint union of closed sets | Yes, sure: If you have a finite union of closed intervals, the complement in [0,1] is a finite union of open (in [0,1]) intervals. Hence, you find in this complement a finite union of closed intervals such that, if you add these intervals to your previous collection, the measure of the complement is only half of the measure it was before. Proceeding by induction and taking the union over all intervals you have chosen, you obtain a countable union of disjoint closed intervals of full measure.
| 6 | https://mathoverflow.net/users/165275 | 414732 | 169,129 |
https://mathoverflow.net/questions/414604 | 4 | Let $\mathcal{A} \subset \mathbb{N}$ be an infinite sequence with positive density, in the sense that
$$
\tag{1}
\lim\_{x\to\infty} \frac{|\mathcal{A} \cap x|}{x} = c > 0,
$$
and define the remainder term $r\_p(x)$ by
$$
r\_p(x) = \sum\_{\substack{n\leq x\\ n\in \mathcal{A}\\ p\mid n}} 1 - \frac{|\mathcal{A}\cap x|}{p}.
$$
What can be said about $r\_p(x)$ on average, assuming only the hypothesis (1)? In particular, can one obtain an estimate of the form
$$
\tag{2}
\sum\_{p\leq Q} |r\_p(x)| \ll \frac{x}{\log x}
$$
in a large range of $Q$, potentially as large as $Q \leq x$? It seems to me that the positive density of the sequence $\mathcal{A}$ should preclude the possibility that $\mathcal{A}$ "misses" many primes, in the sense that $r\_p(x)$ is large for many $p$. However, perhaps there is a way to construct such a sequence via a clever use of the Chinese Remainder Theorem. Any comments/references are most appreciated.
**Edit.** The estimate (2) is false in general, as one need only consider the case when $\mathcal{A}$ is an arithmetic progression. As a concrete example, if $\mathcal{A} = \left\{4n+1: n\geq 0\right\}$ and $p=2$, then
$$
|r\_2(x)| = \frac{|\mathcal{A}|}{2} = \frac{x}{8}+O(1),
$$
since the sum is empty. For a general arithmetic progression $a\mod{q}$, this kind of thing happens for at most finitely many primes (the divisors of the modulus $q$). For the application I have in mind, I really only need something like
$$
\sum\_{y < p\leq Q} |r\_p(x)| \ll \frac{x}{\log x},
$$
where $y$ is a parameter that grows with $x$, and so a finite number of large $r\_p(x)$ is acceptable. This also allows for sets $\mathcal{A}$ where the density of integers divisible by a prime $p$ is only asymptotically $\frac{1}{p}$. For instance, if $\mathcal{A}$ is the set of squarefree numbers, then one expects
$$
\sum\_{\substack{n\leq x\\p\mid n}} \mu^2(n) \sim \frac{|\mathcal{A}\cap x|}{p+1}.
$$
| https://mathoverflow.net/users/307675 | Remainder terms of congruence sums in sets of positive density | Let $\mathcal A$ be the set of all $n\in \mathbb N$ with a prime factor $p>\sqrt{n}$. First, the number of elements $a\in \mathcal A$ with $a\leq x$ is $\gg x$. Indeed, this large prime factor is unique for any $a$, for a given $p\leq x$ there are $\left[\frac{\min(x,p^2)}{p}\right]$ integers $a\leq x$ with $p>\sqrt{a}$ and $p\mid a$. Therefore,
$$
|\mathcal A\cap [1,x]|=\sum\_{p\leq x}\left[\frac{\min(x,p^2)}{p}\right]=\sum\_{\sqrt{x}<p\leq x}\frac{x}{p}+\sum\_{p\leq \sqrt{x}}p+O(\pi(x))=
$$
$$
=x(\ln 2+o(1))\gg x
$$
On the other hand, for any $\sqrt{x}<p\leq x$ all the numbers between $\sqrt{x}$ and $x$, which are divisible by $p$, lie in $\mathcal A$. Therefore,
$$
r\_p(x)=\sum\_{p\mid n, n\in \mathcal A} 1-\frac{x(\ln 2+o(1))}{p}\geq
$$
$$
\geq \left[\frac{x}{p}\right]-\left[\frac{\sqrt{x}}{p}\right]-\frac{x(\ln 2+o(1))}{p}=
$$
$$
=\frac{x(1-\ln 2)+o(x)}{p}.
$$
Summing over all $\sqrt{x}\leq y<p\leq x$, we get
$$
\sum\_{y<p\leq x}|r\_p(x)|\geq x(1-\ln 2+o(1))\sum\_{y<p\leq x}\frac{1}{p}.
$$
For $y=x^{1-\varepsilon}$ we would get $\gg x$. Also, a trivial upper bound for $r\_p(x)$ is $\frac{x}{p}$, so for $y\geq \sqrt{x}$ there is no hope for an upper bound of non-trivial order.
| 4 | https://mathoverflow.net/users/101078 | 414734 | 169,130 |
https://mathoverflow.net/questions/414684 | 4 | Is the following true? For every $n \geq 1, k\geq 2$, there is a set $S \subseteq [n]^k$ of size $|S| = n^2$ such that every two $k$-tuples in $S$ have at most one common entry.
Does anyone know if this is true? Is there a reference?
| https://mathoverflow.net/users/141963 | k-partite design | Assuming that "at most common entry" means that any two tuples $(x\_1,\ldots,x\_k)$ and $(y\_1,\ldots,y\_k)$ *match in at most one position*, that is, there is at most one $i$ such that $x\_i=y\_i$.
The claim is false for $n=2$ and any $k \ge 4$. We would need $n^2=4$ tuples $x,y,z,w$. Then each of $y$ and $z$ must differ from $x$ in at least $k-1$ positions, which implies that $y$ and $z$ match in at least $k-2 \ge 2$ positions.
---
(If the meaning was that any two tuples contain at most one common *value*, regardless of position, the claim would fail already with $n=2$, $k=2$, because there are only four different tuples $(1,1),(1,2),(2,1),(2,2)$, and $(1,2),(2,1)$ have two common values $1$ and $2$.)
| 1 | https://mathoverflow.net/users/171662 | 414741 | 169,132 |
https://mathoverflow.net/questions/414730 | 10 | Can this number $\ln \omega$ be written in $\{L|R\}$ form? What's its birthday?
| https://mathoverflow.net/users/10059 | In surreal numbers, what is $\ln \omega$? | In general this is taken to mean the value of Gonshor's logarithm at $\omega$. This was defined in the tenth chapter of his 1986 book *An introduction to the theory of surreal numbers* where you can find a justification for my answer.
In an informal way, the function $\ln$ is the "simplest" function that is eventually smaller than each power function $x\mapsto x^r$ for $r \in \mathbb{R}^{>0}$ but eventually greater than any constant function. So if $\ln(\omega)$ could be the simplest number that is greater than each real number but smaller than each power $\omega^r$ for $r \in \mathbb{R}^{>0}$, that would be nice.
Indeed $\ln(\omega)=\{ \mathbb{R} \ | \ \omega^{r}: r \in \mathbb{R}^{>0}\}$. In Conway normal form, this is a monomial $\omega^{\omega^{-1}}$. You can also write $\ln(\omega)=\{ \mathbb{N} \ | \ \omega^{2^{-n}}: n \in \mathbb{N}\}$, then the difference is that the elements in brackets are simpler than $\ln(\omega)$ in the sense of the simplicity relation on surreal numbers.
re-edit: my past answer for the birth day was wrong. In fact each $\omega^{2^{-n}}$ has birth day $\omega+\omega^2.n$, so the birth day of $\ln(\omega)$ is actually $\omega^3$.
| 15 | https://mathoverflow.net/users/45005 | 414742 | 169,133 |
https://mathoverflow.net/questions/349352 | 4 | Let $X$ be a Banach space. Suppose $f:X^\*\to\mathbb R\cup\{\infty\}$ is convex, has weak\*-compact effective domain, and is weak\*-continuous on its effective domain. In particular, $f$ is weak\*-lower semicontinuous on $X^\*$.
Suppose I know $f$ is subdifferentiable at $x^\*\in \text{dom}(f)$, i.e. the subdifferential $\partial f(x^\*)\subseteq X^{\*\*}$ is nonempty. Does this necessarily imply that $X\cap \partial f(x^\*)$ is nonempty? If not, are there known sufficient conditions for $X\cap \partial f(x^\*)$ to be nonempty?
| https://mathoverflow.net/users/145424 | Subgradient in a predual under weak* continuity | Finally, I was able to cook up a counterexample. We choose $X = c\_0$ (zero sequences equipped with supremum norm). Thus, the dual spaces are (isometric to) $X^\* = \ell^1$ and $X^{\*\*} = \ell^\infty$.
We define
$$
C := \{ x \in \ell^1 \mid \forall n \in \mathbb N : |x\_n| \le 1/n^2 \}$$
and $f \colon \ell^1 \to \mathbb R \cup \{\infty\}$ via
$$
f(x) = \sum\_{n=1}^\infty x\_n \in \mathbb R
$$
for all $x \in C$ and $f(x) = \infty$ for all $x \in \ell^1 \setminus C$.
Let us check, that the assumptions are satisfied. The set $C$ is bounded due to $\sum\_{n = 1}^\infty 1/n^2 < \infty$ and weak-$\star$ closed since it is the intersection of the weak-$\star$ closed "stripes"
$$ \{x \in \ell^1 \mid |x\_n| \le 1/n^2\} \qquad\forall n \in \mathbb N.$$
Thus, it is weak-$\star$ compact.
The function $f$ is convex and it remains to check weak-$\star$ continuity on $C$. Let $x\_0 \in C$ be given and consider a net $(x\_i)\_{i\in I} \subset C$ with $x\_i \to x\_0$.
For an arbitrary $\varepsilon > 0$, there is $N \in \mathbb N$ with
$\sum\_{n = N+1}^\infty 1/n^2 < \varepsilon$.
Next, there is $i \in I$ with
$$
\left| \sum\_{n = 1}^N (x\_{j,n} - x\_{0,n}) \right| < \varepsilon
\qquad\forall j \ge i$$
since $y \mapsto \sum\_{n = 1}^N y\_n$ is weak-$\star$ continuous.
Thus,
$$
|f(x\_j) - f(x\_0)|
\le
\left| \sum\_{n = 1}^N (x\_{j,n} - x\_{0,n}) \right|
+
\sum\_{n = N+1}^\infty |x\_{j,n}|
+
\sum\_{n = N+1}^\infty |x\_{0,n}|
<
3 \varepsilon
\qquad\forall j \ge i.$$
Since $\varepsilon > 0$ was arbitrary, this shows weak-$\star$ continuity on $C$.
Finally, it is easy to check that $\partial f(0) = \{1\}$, but $1 \in \ell^\infty \setminus c\_0$.
| 3 | https://mathoverflow.net/users/32507 | 414752 | 169,136 |
https://mathoverflow.net/questions/414708 | 0 | Let $G(V,E)$ be a symmetric graph with $n$ vertices and $m$ edges that has a $2\text{-factor}$ with edge set $F$, i.e. $F$ are the edges of an undirected vertex-disjoint cycle cover of $G$.
**Question:**
given only $F$ represented as an unordered sequence $\big( (u\_1,v\_1),\,\dots,\,(u\_n,v\_n)\big)$, what is the complexity of determining a permutation of the $n$ vertices that has the same cycles as $F$, i.e. storing the vertices in an array $\boldsymbol{a}$ such that $(i,\boldsymbol{a}[i])\in F$
| https://mathoverflow.net/users/31310 | Reconstructing a 2-factor from its edge set | Construct an undirected graph on the edge set $F$. This graph has each vertex with even degree, i.e. it's an Eulerian graph. Construct an Eulerian cycle in each connected component, e.g. following the [Hierholzer's algorithm](https://en.wikipedia.org/wiki/Eulerian_path#Hierholzer%27s_algorithm), which takes linear time in $|F|$. Fix any orientation of these cycles, and impose it on the edges from $F$.
| 2 | https://mathoverflow.net/users/7076 | 414754 | 169,138 |
https://mathoverflow.net/questions/395980 | 2 | Let $H$ be a (commutative or non-commutative) monoid. We say that $H$ satisfies the ACCPL (ascending chain condition on principal left ideals) if there exists no infinite sequence of principal left ideals of $H$ that is strictly increasing with respect to inclusion, where a principal left ideal of $H$ is a set of the form $Ha := \{xa: x \in H\}$ (with $a \in H$). The ACCPR (ascending chain condition on principal *right* ideals) is defined in a left-right symmetric way.
Now, Proposition 0.9.3 in P.M. Cohn's
* *Free Ideal Rings and Localization in General Rings* (New Math. Monogr. 3, Cambridge Univ. Press, 2006)
reads as follows (an *atom* of $H$ is a non-unit $a \in H$ such that $a \ne xy$ for all non-units $x, y \in H$):
>
> **Theorem 1.** If $H$ is a *cancellative* monoid satisfying both the ACCPL and the ACCPR, then $H$ is *atomic*, meaning that every non-unit of $H$ factors as a product of atoms.
>
>
>
I'll refer to Theorem 1 as *Cohn's theorem on atomic factorizations in cancellative monoids*. I'm aware that the statement is, in its essence, much older than Cohn's book. In the section "Notes and comments on Chapter 0" (loc. cit.), Cohn himself writes, "The results of Section 0.9 are for the most part well known." In practice, I'd like to understand if it's possible to track the history of the theorem better than I've seen done so far. To my knowledge, one of the first occurrences of something of the "same" quality is Proposition 1.1 in
* P.M. Cohn, *Bezout rings and their subrings*, Math. Proc. Cambridge Phil. Soc. **64** (1968), No. 2, pp. 251-264,
where Cohn famously wrote:
>
> An element of an integral domain is called an *atom* if it is a non-unit which cannot
> be written as a product of two non-units. If every element of a ring $R$ which is not a
> unit or $0$ can be written as a product of atoms, $R$ is said to be atomic. The following result is easily verified:
>
>
> PROPOSITION 1.1. An integral domain is atomic if and only if it satisfies the maximum condition on principal ideals.
>
>
>
I say "famously" because the conclusion turned out to be not-so-easy-to-verify after all, as it was proved wrong by A. Grams in
* *Atomic rings and the ascending chain condition for principal ideals*, Math. Proc. Cambridge Phil. Soc. **75** (1974), No. 3, pp. 321-329.
In Cohn's paper, an integral domain is really a *commutative* domain: I'm stressing this point because, elsewhere in his work, Cohn is using the term "integral domain" to refer to both commutative and non-commutative domains. In particular, Proposition 2.5 in
* P.M. Cohn, *Free Rings and Their Relations*, Academic Press, 1985
reads as follows:
>
> **Theorem 2.** Any integral domain with left and right $ACC\_1$ is atomic.
>
>
>
Here, an "integral domain" is actually a commutative or non-commutative domain, and the left (resp., right) $ACC\_1$ is nothing else than the ACCPL (resp., ACCPR). So, I'll refer to Theorem 2 as *Cohn's theorem on atomic factorizations in domains*.
In my view, the standard proof of Theorems 1 and 2 in the commutative case is not really of the same difficulty as Cohn's proof of the result in the non-commutative case (although the latter is still an easy proof by any modern standards, at least in hindsight). Therefore, I'll focus on the *non-commutative* case and ask the following:
>
> **QUESTION.** Are you aware of any (published) results that predate Cohn's theorems on atomic factorizations (either in domains or in cancellative monoids)?
>
>
>
By the word "predate", I mean either a result of the form "If $H$ is any monoid (resp., domain) in a certain non-trivial class of *non-commutative* monoids (resp., domains) satisfying the ACCPL and the ACCPR, then $H$ is atomic". I count on your common sense for the actual meaning of "non-trivial class"; in particular, a group does not count as non-trivial, and neither does a monoid $H$ that comes by with a *length function*, that is, a function $\phi: H \to \mathbf N \cup \{\infty\}$ such that $\phi(x) < \phi(y)$ whenever $x$ divides $y$ (i.e., $y \in HxH$) but not the other way around.
**Edit #1.** I had a look at M.L. Dubreil-Jacotin's paper
* *Sur l'immersion d'un semi-groupe dans un groupe*, C. R. Acad. Sci. Paris **225** (1947), 787-788,
which was suggested by Benjamin Steinberg in the comments. Here is the main result:
>
> Let $S$ be a cancellative semigroup with no identity such that (1) if two elements $a, b \in S$ have a common right multiple (i.e., $aH \cap bH \ne \emptyset$), then one is a left divisor of the other (i.e., $a \in bH$ or $b \in aH$); (2) each element has only a finite number of left divisors. Then every element of $S$ can be *uniquely* written as a product of indecomposable elements of $S$ (i.e., elements having no left divisors).
>
>
>
Honestly, I don't see much of a resemblance to Proposition 0.9.3 in Cohn's 2006 book. To me, the result looks much closer to the Fundamental Theorem of Arithmetic.
To start with, there is no (implicit or explicit) reference to the ACC on principal left *and* on principal right ideals: Instead, there is a much stronger condition on the number of left divisors of an arbitrary element (i.e., there are only finitely many of them). Moreover, the conclusion is much stronger than the *existence* of an atomic factorization, because an "indecomposable factorization" à la Dubreil-Jacotin is unique (in the strongest possible sense). It follows that $S$ is a free semigroup and the "indecomposable elements" are "prime elements" (and again in a very strong sense). But this is not usually the case with atoms, let alone that atomic factorizations are, in general, all but unique in any sensible way.
| https://mathoverflow.net/users/16537 | Origins of a theorem on an atomic factorizations in domains and cancellative monoids satisfying the ACCPL and the ACCPR | A "close analogue" of (what I'm referring to as) *Cohn's theorem on atomic factorizations in cancellative monoids* (that is, Theorem 1 in the OP) is given by the unnumbered corollary on the bottom of p. 589 in P.M. Cohn's
* *Torsion modules over free ideal rings*, Proc. London Math. Soc. III. Ser. **17** (1967), 577-599.
After recalling (or introducing?) the notions of "atom" and "atomic ring" (the latter definition makes explicit reference to the monoid of regular elements, as can be seen from loc. cit., p. 587), Cohn proves that, if a ring $R$ is Morita-equivalent to a FIR, then the monoid of regular elements of $R$ is atomic: The proof relies on (i) Proposition 4.3 from the same paper (where Cohn shows that a FIR satisfies the ACCPR) and (ii) the observation that a FIR satisfies the ACCPR if and only if it satisfies the ACCPL.
The same idea (to combine the ACCPL and the ACCPR to prove atomicity) also appears in the proof of Theorem 2.8 from Cohn's *Free ideal rings*, which was published in the 1st issue of the 1st volume of J. Algebra (back in 1964). It seems that, at the time, Cohn had not yet coined the term "atom" and was rather using the term "prime" (with the same meaning), which was apparently common back then (cf. Cohn's 1963 TAMS paper on non-commutative unique factorization domains in Trans. AMS or R.E. Johnson's 1965 PAMS paper on unique factorization in principal right ideal domains).
It is perhaps worth noting that, in the 1964 paper (cited above), Cohn was not yet explicitly "thinking in monoids", to the contrary of the 1967 paper; and none of these papers includes any reference to the work of other people building on the idea that "ACCPL & ACCPR $\implies$ atomicity".
| 1 | https://mathoverflow.net/users/16537 | 414758 | 169,139 |
https://mathoverflow.net/questions/414718 | 2 | Consider a stochastic process $X$ defined by
$$X\_t:=1+\int\_0^t b(s,X\_s) \, ds+ W\_t,\quad \forall t\ge 0,$$
where $(W\_t)\_{t\ge 0}$ is a standard Brownian motion. Suppose that $b:\mathbb R\_+ \times \mathbb R \to \mathbb R$ is Lipschitz and of linear growth so that $X$ is uniquely defined. Under what kind of conditions one has
$$\mathbb P\big(X\_t>0 \text{ for all } t\ge 0\big)>0?$$
An obvious condition is $\inf\_{(t,x)}b(t,x)>0$. My question is whether we have more general conditions for the above inequality, especially for the case where $b$ changes sign?
Any answer, comments and references are highly appreciated.
PS : I'm looking for sufficient conditions, and we may consider the simple case $b\equiv b(t)$. Denote by $B(t):=\int\_0^tb(s) \, ds$ for $t\ge 0$. Then a necessary condition is $\limsup\_{t\to \infty} B(t)=\infty$. Can we impose some suitable condition on the growth of $B$ such that the desired inequality holds?
| https://mathoverflow.net/users/nan | Search for conditions of the positive probability that a stochastic process never hits zero | $\newcommand\ep\varepsilon$The case of interest when $b(t,x)=b(t)$ depends only on $t$ is comparatively simple.
Indeed, let
$$g(t):=\sqrt{(2t+1/2)\ln\ln(3+t)}$$
for real $t\ge0$.
By the law of the iterated logarithm,
$$\sup\_{s\in[t,\infty)}\frac{W\_s}{g(s)}\to1$$
as $t\to\infty$ almost surely and hence in probability.
So, for each real $\ep>0$ there is some real $t=t\_\ep>0$ such that
\begin{equation\*}
\tfrac\ep2\,g(t)>(1+\ep)g(0) \tag{1}\label{1}
\end{equation\*}
and
\begin{equation\*}
P(B)>0, \tag{2}\label{2}
\end{equation\*}
where
\begin{equation\*}
B:=\{W\_s<(1+\ep/2)g(s)\ \forall s\in[t,\infty)\}.
\end{equation\*}
Let
\begin{equation\*}
A:=\{W\_s<(1+\ep)g(s)\ \forall s\in[0,t]\}.
\end{equation\*}
Note that the function $g$ is concave. So, for all $s\in[0,t]$ we have $g(s)\ge g(0)+\frac st\,(g(t)-g(0))$ and hence for all real $u<(1+\ep/2)g(t)$
\begin{equation\*}
\begin{aligned}
&(1+\ep)g(s)-\tfrac st\,u \\
&\ge(1+\ep)g(0)+\tfrac st\,[(1+\ep)(g(t)-g(0))-u] \\
&\ge(1+\ep)g(0)+\tfrac st\,[(1+\ep)(g(t)-g(0))-(1+\ep/2)g(t)] \\
&=(1+\ep)g(0)+\tfrac st\,[\tfrac\ep2\,g(t)-(1+\ep)g(0)] \\
&\ge(1+\ep)g(0)>g(0),
\end{aligned}
\tag{3}\label{3}
\end{equation\*}
in view of \eqref{1}.
Note also that the Brownian bridge $W^{(t)}\_\cdot$ defined by the formula $W^{(t)}\_s:=W\_s-\tfrac st\,W\_t$ for $s\in[0,t]$ is independent of $W\_t$.
Recalling also the symmetry of $W\_\cdot$ and its Markov property, as well as \eqref{3}, we get
\begin{equation\*}
\begin{aligned}
&P(W\_s>-(1+\ep)g(s)\ \forall s\in[0,\infty)) \\
&=P(W\_s<(1+\ep)g(s)\ \forall s\in[0,\infty)) \\
&=P(A\cap B) \\
&=P(W\_s<(1+\ep)g(s)\ \forall s\in[0,t],W\_t<(1+\ep)g(t),B) \\
&=\int\_{-\infty}^{(1+\ep)g(t)} P(W\_t\in du,B)
P(W\_s-\tfrac st\,W\_t<(1+\ep)g(s)-\tfrac st\,u\ \forall s\in[0,t]) \\
&=\int\_{-\infty}^{(1+\ep)g(t)} P(W\_t\in du,B)
P(W^{(t)}\_s<(1+\ep)g(s)-\tfrac st\,u\ \forall s\in[0,t]) \\
&\ge\int\_{-\infty}^{(1+\ep)g(t)} P(W\_t\in du,B)
P(W^{(t)}\_s<g(0)\ \forall s\in[0,t]) \\
&=P(B)
P(W^{(t)}\_s<g(0)\ \forall s\in[0,t])>0,
\end{aligned}
\tag{4}\label{4}
\end{equation\*}
by \eqref{2} and because $g(0)>0$.
Note that $X\_s=1+B(s)+W\_s$. So,
\begin{equation\*}
\begin{aligned}
&P(X\_s>1+B(s)-(1+\ep)g(s)\ \forall s\in[0,\infty)) \\
&=P(W\_s>-(1+\ep)g(s)\ \forall s\in[0,\infty))>0,
\end{aligned}
\end{equation\*}
by \eqref{4}.
Thus, the condition that
\begin{equation\*}
B(s)>(1+\ep)g(s)-1 \tag{5}\label{5}
\end{equation\*}
for some real $\ep>0$ and all $s\in[0,\infty)$ is sufficient for
\begin{equation\*}
P(X\_s>0\ \forall s\in[0,\infty))>0.
\end{equation\*}
Note finally that for any $\ep\in(0,\frac1{g(0)}-1)=(0,3.61\ldots)$ there is a positive continuous function $b$ such that the function $B$ given by the formula $B(t)=\int\_0^t b(s)\,ds$ for all real $t\ge0$ satisfies condition \eqref{5} (however, of course, $b$ does not have to be everywhere positive or continuous in order for $B$ to satisfy condition \eqref{5}).
| 1 | https://mathoverflow.net/users/36721 | 414771 | 169,142 |
https://mathoverflow.net/questions/414705 | 1 | $\DeclareMathOperator\SU{SU}\DeclareMathOperator\SL{SL}$I'm working through some of the constructions in [*Introduction to Arithmetic Groups* by Dave Witte Morris](https://arxiv.org/abs/math/0106063), and I'm confused by the construction of example 6.3.1 on page 121. For reference, here's the setup:
Take $a,b \in \mathbb{Q}^+$, yielding the totally real number field $L=Q(\sqrt a) \subset \mathbb{R}$ with ring of integers $\mathcal{O}$ commensurable to $\mathbb{Z}[\sqrt a]$. Let $\tau$ be the nontrivial element of $\operatorname{Gal}(L/\mathbb{Q})$, and let $A=(\begin{smallmatrix} b & 0 \\ 0 & -1 \end{smallmatrix})$. Define a ''unitary group'' with entries in $\mathcal O$ as
$$G\_{\mathcal O} = \SU(A,\tau;\mathcal O) = \{g \in \SL(2,\mathcal O) \mid \tau(g^T)Ag=A \} \subset \SL(2,\mathbb{R}).$$
The statement is that $G\_\mathcal{O}$ is an arithmetic subgroup of $G=\SL(2,\mathbb{R})$.
There are a few points in the provided construction that are confusing me, but I think I might be able to work them out if someone could explain one thing to me:
#### Why do we expect $G\_{\mathcal O}$ to be an arithmetic subgroup of $G$?
From the avenue of restriction of scalars I would expect $G\_\mathcal{O}$ to be arithmetic in $G \times G$, one factor for each of the Galois conjugates of the number field. I don't think either of the conjugates lands in a compact factor since the form $A$ that's being preserved doesn't change under automorphisms of $L/\mathbb{Q}$.
| https://mathoverflow.net/users/151664 | Clarification on arithmetic groups example | YCor gave a brief explanation in the comments. Based on your comment about seeking intuition, I think it would be helpful to explain concretely the backstory to the calculation YCor did:
Given any kind of arithmetically-defined group $\Gamma$, you want to find a real algebraic group / Lie group $G$ in which $\Gamma$ is arithmetic. For this, a necessary condition is that $\Gamma$ is discrete in $G$, but it's not sufficient. If $\Gamma \subset H \subset G$, then $\Gamma$ is not arithmetic in $G$ unless $G/H$ is compact, so we want to take $G$ as small as possible containing $\Gamma$.
Making $\Gamma$ discrete is easy. You just have to express elements of $\Gamma$ by integer coordinates, since $\mathbb Z^n$ is discrete in $\mathbb R^n$, so any subset of $\mathbb Z^n$ is discrete in any subspace of $\mathbb R^n$ containing it.
For example, for $\Gamma = SL(2, \mathcal O)$ or $\Gamma$ any subgroup of $SL(2, \mathcal O)$ (in, for simplicity, the special case $\mathcal O = \mathbb Z[\sqrt{a}]$), we can write any matrix $g$ as $M + N \sqrt{a}$ where $M, N $ are $2 \times 2$ matrices over the integers. This embeds $SL(2,\mathcal O)$ into $\mathbb Z^8$. For an arbitrary group over an arbitrary ring of integers $\mathcal O$ you can do the same thing, you just need to pick a $\mathbb Z$-basis of $\mathcal O$. So $SL\_2 (\mathcal O)$ is discrete inside the manifold $\mathbb R^8$ parameterizing pairs $M ,N$ of $2\times 2$ matrices over the reals.
Of course we care about $\Gamma$ as a group and not just an abstract set. So you need to find a polynomial formula for the multiplication map. We have $$ (M\_1+ N\_1\sqrt{a} ) \cdot (M\_2+ N\_2\sqrt{a} ) = (M\_1 M\_2 + a N\_1 N\_2) + (M\_1 N\_2 + N\_1 M\_2) \sqrt{a}$$ so we ca write multiplication as $$(M\_1,N\_1) \cdot (M\_2, N\_2) = ( (M\_1 M\_2 + a N\_1 N\_2) , (M\_1 N\_2 + N\_1 M\_2) ).$$
Then $\Gamma$ is discrete inside the group of pairs of $2\times 2$ real matrices that have inverses under this multiplication map, i.e. those such that $M + N \sqrt{a}$ and $M - N \sqrt{a}$ are both invertible. This is isomorphic to $GL(2,\mathbb R) \times GL(2,\mathbb R)$.
But this is not the *minimal* group inside which $\Gamma$ is discrete. To find that, we need to take the *Zariski closure* of $\Gamma$, i.e. for all polynomial relations satisfied by $(M ,N) \in \Gamma$, we look at only those $M, N \in M\_2(\mathbb R)$. This defines an algebraic group $G$, which always contains $\Gamma$, and that's the group in which $\Gamma$ is arithmetic.
These include $\det (M + N \sqrt{a})=1$ and $\det (M - N \sqrt{a}) =1$, which cut the group down to $SL(2,\mathbb R) \times SL(2,\mathbb R)$, which if $\Gamma =SL(2,\mathcal O)$ is as far as we go, but for $\Gamma$ the unitary group, there is another polynomial relation:
$$ \tau( g^T) Ag = A$$
i.e.
$$ ( M^T- N^T \sqrt{a}) A ( M+ N \sqrt{a} ) = A $$
and
$$ ( M^T+ N^T \sqrt{a}) A ( M- N \sqrt{a} ) = A $$
These two relations cut you down to a smaller group. Whatever it is, it's clearly not $SL (2,\mathbb R) \times SL(2, \mathbb R)$, because these relations are not satisfied for every element in $SL(2, \mathbb R)$. In fact, it's $SL(2,\mathbb R)$, for the reason YCor gave: Essentially, either $M+N \sqrt{a}$ or $M- N \sqrt{a}$ is determined by the other one, so you only need one $2\times 2$ matrix to determine the element.
---
In summary, you always want to express your group with coordinates over the integers, and relations defined by polynomials (ideally over the integers, or else remember to write down all Galois conjugates of your polynomial relations). Then you consider the real solutions of the same set of polynomial equations, and that will give you the right algebraic group. Before tricky relations that involve the Galois group like the one appearing in the definition of the unitary group, this will split as a product over the real places of your number field, but these extra relations will relate the projections to different places, giving a smaller group.
| 4 | https://mathoverflow.net/users/18060 | 414775 | 169,144 |
https://mathoverflow.net/questions/414773 | 2 | I've found through evidence and have conjectured on a math publication that:
$$\Big\lfloor\int\_1^\infty (k^{1/(k^{1+1/\sqrt{x}})} - 1)dk\Big\rfloor = \Big\lfloor\sum\_{k=1}^{\infty}k^{1/(k^{1+1/\sqrt{x}})} -1\Big\rfloor = x $$
where $ x \in \mathbb{N}, x>1$.
It is very hard to compute these values. Repeated Shanks transformations and Richardson's Extrapolation will be required to compute, or using Pari GP techniques. Before you post a counter example below 10^7 for the sum, please check your precision.
Proving this has proved extremely difficult.
My question is, does anyone have any suggestions of how to prove this? The only information I have is that this is true from all tests for $x$ less than 10^7 and we're still running tests for the sums.
They aren't equal without the floor function, and each equal $x + C$, where $C$ is a constant less than 1, and $C$ is different for the integral and sum. As $x$ tends to infinity, $C$ tends to 1.
| https://mathoverflow.net/users/nan | A conjecture relating an integral and a sum, the floor function and squares | First of all,
$$k^{1/k^t}=e^{(\log k)/k^t} = \sum\_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$
and therefore
$$\intop\_{1}^{\infty} \left( k^{1/k^t} - 1 \right) d k = \sum\_{n = 1}^{\infty} \frac{1}{n!} \intop\_{1}^{\infty} \left( \log k \right)^n k^{- n t} d k =$$
$$\sum\_{n = 1}^{\infty} \frac{1}{n!} \intop\_{0}^{\infty} y^n e^{(1 - n t) y} d y = \sum\_{n = 1}^{\infty} \frac{1}{(n t - 1)^{n + 1}}.$$
If $t = 1 + 1/\sqrt{x}$ then the main contribution to this sum is the term $n = 1$ which gives a contribution of $x$, and so the sum is equal to
$$x + \sum\_{n = 2}^{\infty} \frac{1}{(n t - 1)^{n + 1}}.$$
In order for the floor of this value to be equal to $x$, then we must have $\sum\_{n = 2}^{\infty} \frac{1}{(n t - 1)^{n + 1}} < 1$ for each $t > 1$, or equivalently $\sum\_{n = 2}^{\infty} \frac{1}{(n - 1)^{n + 1}} \leq 1$. However, this is false: indeed, the second term alone is equal to $1$, so for large enough $x$ the floor of the integral is at least $x + 1$. What is true is that the integral is always strictly less than $x + 2$, since
$$\sum\_{n = 3}^{\infty} \frac{1}{(n - 1)^{n + 1}} < \sum\_{n = 3}^{\infty} \frac{1}{2^{n + 1}} = \frac{1}{8}.$$
---
Presumably, the sum can be treated in a similar manner, where the main term (which should be about $x$) would be $\sum\_{k = 1}^{\infty} \frac{\log k}{k^{1 + 1/\sqrt{x}}}$, and the other terms contributing at most a constant.
| 3 | https://mathoverflow.net/users/88679 | 414777 | 169,145 |
https://mathoverflow.net/questions/390655 | 4 | Suppose that $p\ge 5$ is a prime, $n$ a positive integer divisible by $p-1$,
and $L<\mathbb F\_p^n$ a subspace of dimension $d=n/(p-1)$. Do there exist
vectors $l\_1,\dotsc,l\_n\in L$ such that the matrix with $l\_1,\dotsc,l\_n$ as
its columns has a nonzero permanent? Clearly, the answer is negative if $L$
is contained in a coordinate hyperplane, or in a linear subspace like
$\{(x\_1,\dotsc,x\_n)\in\mathbb F\_p^n\colon x\_1=\dotsb=x\_p\}$; are there other
obstructions of this sort?
>
> Is it possible to classify those subspaces
> $L<\mathbb F\_p^n$ for which any square matrix of order $n$ with all its
> column vectors in $L$ has a vanishing permanent?
>
>
>
Notice that if the permanent vanishes for $l\_1,\dotsc,l\_n$ being the elements of some particular basis of $L$, with each element repeated $p-1$ times, then in fact it vanishes for any $l\_1,\dotsc, l\_n\in L$. (It is this property that depends critically on the assumption $d=n/(p-1)$.)
In the case where $d=1$, the requirement that $L$ is not contained in a coordinate hyperplane is easily seen to be also sufficient. The case $d=2$ does not look that easy to me.
| https://mathoverflow.net/users/9924 | Subspaces of vanishing permanent | Let $v\_1,\dots, v\_d$ be a basis for $L$. Then the permanent of the matrix obtained from $p-1$ repetitions each of $v\_1,\dots, v\_d$ is a polynomial function in the entries of $v\_1,\dots, v\_d$. Keeping all the entries but the last one in each $v\_i$ constant, we get a linear function. Since it is linear, among all $p^{nd}$ possible tuples of vectors, the number that satisfy it is at least $p^{nd-1}$.
Not all $d$-tuples of vectors in $\mathbb F\_p^n$ form a basis, but $1 - O( p^{d-n-1})$ do.
So the fraction of subspaces $L$ where this determinant vanishes is at least $$ \frac{1}{p} - O( p^{d-n-1}) $$
On the other hand, the fraction that satisfy one of your conditions is smaller than that for $d>1$ - just $\frac{ d (p-1)}{p^d}$ satisfy the coordinate condition and $\frac{ \binom{d (p-1) }{ p } } { p^{ (p-1) d }}$ satisfy the $p$-fold repetition condition.
So there certainly are other obstructions. I do not expect it is possible to classify them.
| 1 | https://mathoverflow.net/users/18060 | 414778 | 169,146 |
https://mathoverflow.net/questions/414733 | 5 | Are there situations in which the [polygamma](https://en.wikipedia.org/wiki/Polygamma_function) pops up naturally in a mathematical physics context? In particular: are there examples of potentials having some interest for which the dependence on the distance is expressed in terms of $\psi^{(n)}$?
**Update**: While Carlo Beenakker's answer is clearly useful, the references therein still don't contain exactly what I'm looking for in the second part of the question.
| https://mathoverflow.net/users/167834 | Polygamma function in mathematical physics | Abou-Salem, L. I., *A study on baryons spectroscopy using digamma-function as interacting potential*, <https://arxiv.org/abs/1311.6743> studies using the digamma function as an interaction potential (for quarks in baryons).
| 6 | https://mathoverflow.net/users/75890 | 414779 | 169,147 |
https://mathoverflow.net/questions/414623 | 4 | $\DeclareMathOperator\Sel{Sel}$Let $E$ be an elliptic curve defined over a number field $K$ with full $2$-torsion. The classical complete $2$-descent method tells that the $2$-Selmer group $\Sel\_2(E/K)$ can be identified with the set of locally solvable everywhere homogeneous spaces. More precisely, we consider a quadratic twist
$$E:y^2=x(x-na)(x+nb)$$
defined over $\mathbb Q$, where $n,a,b,\frac{a+b}{2}$ are positive odd integers. Then $\Sel\_2(E/\mathbb Q)$ can be identified with
$$\{\Lambda=(d\_1,d\_2)\in(\mathbb Q^\times/\mathbb Q^{\times2})^2: D\_\Lambda(\mathbb{A}\_K)\neq\emptyset\},$$
where
$$D\_\Lambda: d\_1u\_1^2-d\_2u\_2^2=na,\quad d\_1u\_1^2-d\_1d\_2u\_3^2=-nb.$$
Assume that $n$ is a positive square-free integer prime to $2ab(a+b)$ and $\Lambda=(d\_1,d\_2)$ where $d\_1,d\_2$ are positive square-free odd divisors of $nab(a+b)$.
Then we can show: if $D\_\Lambda$ is locally solvable everywhere except $v=2$, then it is also locally solvable at $v=2$.
This can be proved by the poduct formula on Hilbert symbols.
So my question is: when the following hold for any $\Lambda$? Does this relate the Selmer group or Shafarevich-Tate group of $E$?
>
> Let $D\_\Lambda$ be a homogeneous space in the form as above.
> If $D\_\Lambda$ is locally solvable everywhere except a place $v$, then it is also locally solvable at $v$.
>
>
>
| https://mathoverflow.net/users/475824 | An analogy of product formula for homogeneous space? | (**Edit**: I revise most of my question as my first answer overlooked that $d\_1$ and $d\_2$ are odd.)
$\DeclareMathOperator{\res}{res}$
Let $E$ be an elliptic curve over $\mathbb{Q}$ with $E[2]\subset E(\mathbb{Q})$. Let $S$ be the normal $2$-Selmer group and let $S'$ be the relaxed Selmer group, that is the subset of $H^1\bigl(\mathbb{Q},E[2]\bigr)\cong {}^{\mathbb{Q}^\times}\!/{}\_{\square} \times{}^{\mathbb{Q}^\times}\!/{}\_{\square}$ that satisfy all local conditions away from $2$, but maybe not at $2$.
Within $S'$ define $S''$ the subset of all $\xi=(d,e)$ where both $d$ and $e$ are odd. That is also a sort of a natural Selmer group, where the local condition at $2$ is that $\res\_2(\xi)$ lies in the kernel of the valuation ${}^{\mathbb{Q}\_2^\times}\!/{}\_{\square} \times{}^{\mathbb{Q}\_2^\times}\!/{}\_{\square}\to {}^{\mathbb{Z}}\!/\!{}\_{2\mathbb{Z}}\times {}^{\mathbb{Z}}\!/\!{}\_{2\mathbb{Z}}$. The question is: When is $S''\subset S$?
This is not a local question as the elements in the kernel of the valuation in $H^1\bigl(\mathbb{Q}\_2,E[2]\bigr)$ are not naturally compared with the image of the local Kummer map $\kappa\_2\colon {}^{E(\mathbb{Q}\_2)}\!/\!{}\_{2E(\mathbb{Q}\_2)} \to H^1\bigl(\mathbb{Q}\_2,E[2]\bigr)$. Here I give the answer in one particular case; the methods generalise to other similar cases and one could determine the answer in general I imagine. Since the actual calculations with Hilbert symbols in very simple, it may help to understand better the original proof alluded to in the question.
Assume $E(\mathbb{Q}\_2)[4]=E(\mathbb{Q}\_2)[2]$. That is no point of order $2$ is divisible by $2$ in $E(\mathbb{Q}\_2)$.
Also assume that $E$ is given by $y^2=x(x-a)(x-b)$ with $a$ and $b$ odd integers. (I don't think $a\equiv b\pmod{4}$ is needed here.)
First consider the following exact sequence coming from global duality:
$$
0\to S\to S' \xrightarrow{\alpha} H^1\bigl(\mathbb{Q}\_2, E\bigr)[2]\xrightarrow{\hat\beta} \hat{S}
$$
where $\alpha$ and $\beta$ are restriction maps.
First if $\beta\colon S\to {}^{E(\mathbb{Q}\_2)}\!/\!{}\_{2E(\mathbb{Q}\_2)}$ is surjective, then $S'=S$ and hence $S''\subset S$.
Therefore, suppose $\beta$ is not surjective.
My assumption above imposes that the three $2$-torsion points (which are global points and hence in the image of $\beta$) are all distinct in ${}^{E(\mathbb{Q}\_2)}\!/\!{}\_{2E(\mathbb{Q}\_2)}$. Therefore the image of $\beta$ is equal to group generated by the torsion points. It is $2$-dimensional in the $3$-dimensional target.
Let now $\xi\in S''$ with $\res\_2(\xi) = (d,e)$. To say that $\xi\in S$ is equivalent to $\alpha(\xi)=0$. Since we know already that $\xi\in\ker(\hat\beta)$, all we need to check is that $\res\_2(\xi)\cup \kappa(P)=0$ for one point $P\in E(\mathbb{Q}\_2)$ which is not in the image of $\beta$. Here $\cup \colon H^1\bigl(\mathbb{Q}\_2,E[2]\bigr)\times H^1\bigl(\mathbb{Q}\_2,E[2]\bigr) \to {}^{\mathbb{Z}}\!/\!{}\_{2\mathbb{Z}}$ is the local duality pairing. Under our identification it corresponds to the pairing $(d,e)\cup (d',e') = (d,e')\_2 + (d',e)\_2$ on ${}^{\mathbb{Q}\_2^\times}\!/{}\_{\square} \times{}^{\mathbb{Q}\_2^\times}\!/{}\_{\square}$ where $(,)\_2$ is the Hilbert symbol with values in ${}^{\mathbb{Z}}\!/\!{}\_{2\mathbb{Z}}$.
In our concrete case, there is a point $P$ with $x$-coordinate equal to $\tfrac{1}{4}$ on $E(\mathbb{Q}\_2)$ because $a$ and $b$ are both odd. In fact the points $P$, $T\_1=(0,0)$ and $T\_2=(a,0)$ generate ${}^{E(\mathbb{Q}\_2)}\!/\!{}\_{2E(\mathbb{Q}\_2)}$ because $T\_2$ has bad reduction, $T\_1$ has good non-trivial reduction, and $P$ has trivial reduction. We have $\kappa(P)=(1,1-4a)=(1,-3)$ since $a$ is odd. Now
$$
(d,e)\cup(1,-3) = (d,-3)\_2 + (1,e)\_2 = 0
$$
because $d$ is odd and $-3\equiv 1 \pmod{4}$. Therefore $(d,e)\in S$.
Finally, I would like to point to Theorem 3 in Appendix 1 of Swinnerton-Dyer's [2-descent through the ages](https://cims.nyu.edu/%7Etschinke/princeton/papers/.miami/submitted/swd.pdf) where a question of that nature is discussed, but for places away from 2.
**Edit**: This is the original answer.
I actually have troubles believing the statement you make in your question. Maybe I am wrong or I misunderstood something, but since it is too long for a comment, I include it as a possible "answer".
Take $y^2 = x^3 - x$. So $n=a=b=1$. I claim: The relaxed Selmer group, imposing all conditions but the one at $2$, is $\mathbb{F}^3$, while the rank of the curve is $0$ so the usual Selmer group is of dimension $2$.
Concretely, the local conditions away from $2$ and $\infty$ impose that $d\_1$ and $d\_2$ both belong to the group generated by $-1$ and $2$ modulo squares, since the curve has good reduction away from $2$.
The condition at $\infty$ imposes that $d\_1$ and $d\_2$ have the same sign.
Now we are down to the group generated by $(1,2)$, $(-1,-1)$ and $(2,1)$.
The first two correspond to the torsion points $(1,0)$ and $(0,0)$ in $E(\mathbb{Q})[2]$. So the torsor
$$\begin{align\*}
2u^2 -v^2 & = 1 \\ 2u^2 -2w^2 &= -1
\end{align\*}$$
is locally soluble at all places, except for $2$.
The reason, I was suspicious initially is that your statement, using global duality, would be equivalent to the surjectivity of the map from the Selmer group $S\_2(E/\mathbb{Q})$ to $E(\mathbb{Q}\_2)/2E(\mathbb{Q}\_2)$. Your question would be equivalent to the surjectivity of the restriction from the Selmer group of some isogeny $\varphi$ to the group of points at this one place. And I don't see a reason why this should hold without conditions on $\varphi$.
| 1 | https://mathoverflow.net/users/5015 | 414781 | 169,149 |
https://mathoverflow.net/questions/414782 | 0 | Can all elements of the Levi-Civita field be represented as power series of a single element
$$p=\varepsilon^{-1}-\frac{\varepsilon }{24}+\frac{3 \varepsilon ^3}{640}-\frac{1525 \varepsilon ^5}{580608}+\dotsb$$
where the numerators of the terms are given in <https://oeis.org/A118050> and the denominators are in <https://oeis.org/A118051>?
How would look $\varepsilon$ and $\varepsilon^{-1}$ in this basis?
| https://mathoverflow.net/users/10059 | Levi-Civita field in unusual basis | Let $p$ be any Laurent series in $\varepsilon$ of the form $\varepsilon^{-1}+\sum\_{n=0}^\infty a\_n\varepsilon^n$, like the one in the question. Then infinite Laurent series in $p$ itself never converge (not in the ring of Laurent series, or Levi-Civita field, or Hahn series, or any related such field), because positive powers of $p$ do not converge to zero.
On the other hand, if we consider $p^{-1}=\varepsilon+\sum\_{n=2}^\infty b\_n\varepsilon^n$, then this element topologically generates $\mathbb R[[\varepsilon]]$ (as a ring) and $\mathbb R((\varepsilon))$ (as a field): proving this comes down to the usual method of showing that you can iteratively find coefficients of a power series in $p^{-1}$ to make it give an arbitrary element of $\mathbb R[[\varepsilon]]$.
On the other hand, there is no way such a Laurent series in $p^{-1}$ can give you anything outside the ring of formal Laurent series. There is no way to produce fractional powers of $\varepsilon$ through this procedure.
| 3 | https://mathoverflow.net/users/30186 | 414785 | 169,150 |
https://mathoverflow.net/questions/412741 | 1 | In great generality a Lie group mod its maximal compact subgroup is contractible (for example this is true for all connected Lie groups). Whenever this is true then the Lie group $ D $ is diffeomorphic to a cartesian product of its maximal compact $ K $ with a contractible piece $ D/K $. So (again assuming $ G $ is connected or semisimple or some other sufficient condition to guarantee $ D/K $ contractible) if $ D $ is dimension $ 2n $ and the maximal compact subgroup is dimension $ n $ then we have a diffeomorphism
$$
D \cong K \times (D/K)
$$
where $ D/K $ is a contractible piece of dimension equal to $ K $. So $ D $ is diffeomorphic to a trivial $ \dim(K) $ dimensional real vector bundle over $ K $. On the other hand, since $ K $ is a Lie group it must be parallelizable. That is, the tangent bundle of $ K $ is a trivial $ \dim(K) $ dimensional real vector bundle over $ K $. In other words, we have that the Lie group $ D $ is diffeomorphic to the tangent bundle of its maximal compact subgroup,
$$
D \cong T(K).
$$
A particular case of this is when $ G $ is a linear algebraic group whose real points $ G\_\mathbb{R} $ are compact. A compact group is always the maximal compact subgroup of its complexification (this follows from the fact that for subgroups of a complex linear algebraic group maximal compact is equivalent to compact plus Zariski dense). In other words, $ G\_\mathbb{R} $ is the ($ n $ dimensional) maximal compact subgroup of the ( $ 2n $ real dimensional) group $ G\_\mathbb{C} $. So, taking $ D=G\_\mathbb{C} $ and $ K=G\_\mathbb{R} $ in the argument above, we have that the complex points are diffeomorphic to the tangent bundle of the real points,
$$
G\_\mathbb{C} \cong T(G\_{\mathbb{R}}).
$$
Again this argument requires that the real points are compact. For example $ G\_\mathbb{R} $ compact implies $ G\_\mathbb{C} $ reductive group and so by Iwasawa decomposition for reductive groups we have that $ G\_\mathbb{C} $ mod its maximal compact is contractible.
**Let $ G $ be a linear algebraic group and $ H $ a linear algebraic subgroup. Suppose the real points $ G\_\mathbb{R}$, $H\_\mathbb{R} $ are compact. Consider the manifold of complex points
$$
M\_\mathbb{C}=G\_\mathbb{C}/H\_\mathbb{C}.
$$
Then is the tangent bundle of
$$
M\_\mathbb{R}=G\_\mathbb{R}/H\_\mathbb{R}
$$
diffeomorphic to $ M\_\mathbb{C} $?**
$\DeclareMathOperator\SO{SO}$Motivation:
$$
\SO\_{n+1}(\mathbb{C})/\SO\_{n}(\mathbb{C})
$$
is diffeomorphic to the tangent bundle of the $ n $ sphere
$$
S^n= \SO\_{n+1}(\mathbb{R})/\SO\_{n}(\mathbb{R}).
$$
More Motivation: When $ H $ is trivial the result follows from the discussion above.
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SU{SU}\DeclareMathOperator\SO{SO}$Comment on the non-compact case: Without the compactness assumption this is false. For example we can take $ H $ trivial and $ G=\SL\_n $ (and $ n\geq 2 $ ). Then $ G\_\mathbb{C}=\SL\_n(\mathbb{C}) $ is homotopy equivalent to $ \SU\_n $ while the tangent bundle of $ G\_\mathbb{R}=\SL\_n(\mathbb{R}) $ is homotopy equivalent to $ \SL\_n(\mathbb{R}) $ which is homotopy equivalent to $ \SO\_n(\mathbb{R}) $. But $ \SU\_n $ and $ \SO\_n(\mathbb{R}) $ are not homotopy equivalent. Thus $ \SL\_n(\mathbb{C}) $ and the tangent bundle $ T(\SL\_n(\mathbb{R})) $ are not homotopy equivalent so they certainly are not diffeomorphic.
Possible counter examples to consider next:
The Steifel manifold $ O\_4/O\_2 $ or the Grassmanian $ O\_4/(O\_2 \times O\_2) $. What do the tangent bundles look like? Are they diffeomorphic to the complex points?
| https://mathoverflow.net/users/387190 | Is the manifold of complex points of a quotient of compact groups just the tangent bundle? | This fact about the tangent space being the complexification is known to be true if $ G\_\mathbb{R}/H\_\mathbb{R} $ is a compact symmetric space. In other words, if $ H\_\mathbb{R} $ is the fixed points of an involution of $ G\_\mathbb{R} $. This includes, for example, all the spheres written as
$$
S^n \cong SO\_{n+1}(\mathbb{R})/SO\_n(\mathbb{R})
$$
This is taken directly from <https://mathoverflow.net/a/414174/387190>
| 0 | https://mathoverflow.net/users/387190 | 414792 | 169,153 |
https://mathoverflow.net/questions/414666 | 2 | I am interested in vector bundles over a nonsingular complete algebraic curve $C$ over $\mathbb C$. For a vector bundle $E$, its Harder-Narasimhan filtration is a filtration of subbundles
$$0=E\_0\subset E\_1\subset\cdots\subset E\_n=E$$
such that each $E\_i/E\_{i-1}$ is semitable and $\frac{\deg(E\_1/E\_0)}{\mathrm{rank}(E\_1/E\_0)}>\cdots>\frac{\deg(E\_n/E\_{n-1})}{\mathrm{rank}(E\_n/E\_{n-1})}$.
Now if we have two vector bundles $E,F$ and we know their HN filtation $\{E\_i\},\{F\_j\}$, can we get any information about the HN filtration of $E\otimes F$?
One information I am interested in is the upper bound of $\frac{\deg}{\mathrm{rank}}$ of subbundles of $E\otimes F$.
Thanks.
| https://mathoverflow.net/users/105537 | Do we know anything about Harder-Narasimhan filtrations of tensor products of vector bundles? | It is a result of Narasimhan and Seshadri that if $V$ is semistable and $W$ is semistable then $V \otimes W$ is semistable.
If $E$ has a filtration with associated graded $E\_i/ E\_{i-1}$, and $F$ has a filtration with associated graded $F\_j/ F\_{j-1}$, then $E \otimes F$ has a filtration with associated graded $(E\_i/ E\_{i-1}) \otimes (F\_j/ F\_{j-1})$. By the previous claim, the associated graded pieces of this filtration are semistable, and we can choose the filtration so that these pieces are in order of increasing slope. Hence it is the Harder-Narasimhan filtration.
Thus the slopes of $E \otimes F$ are the slopes of $E$ plus the slope of $F$, and in particular the maximal slope of $E \otimes F$ is the sum of the maximal slope of $E$ and the maximal slope of $F$.
| 4 | https://mathoverflow.net/users/18060 | 414793 | 169,154 |
https://mathoverflow.net/questions/414701 | 2 | I would like to prove that there are two sets $A,B\subset \mathbb{N}$ such that
* $A |\_T B$
* $\emptyset' \equiv\_T A\oplus B$
* for every $e$, if $\{e\}^{A\oplus B}=\emptyset'$ then the map sending $(B,n)$ to the prefix of $B$ used in the computation $\{e\}^{A\oplus B}(n)$ is not computable.
I believe the existence of $A,B$ as above can be proved with a not-so-hard (but maybe not entirely trivial) finite extension argument. However, since these arguments are often tedious to read, and even more to write, I was wondering whether the existence of $A,B$ follows more easily by some result in classical computability.
| https://mathoverflow.net/users/130978 | Computing the halting problem with no computable bound on the use function | The answer is yes.
Since Chaitin's $\Omega$ is $wtt$-reducible to $\emptyset'$, we may replace $\emptyset'$ with $\Omega$.
Now let $A<\_T \emptyset'$ be a $K$-trivial but promptly simple set. Then there is an incomplete c.e. set $B$ so that $\emptyset'\equiv\_T A \oplus B$.
Suppose that there is such an $e$. Since $B$ is c.e. but incomplete, it must be non-$DNC$. So there is a partial computable function $g$ so that
>
> **$ \exists^{\infty}n(g(n)\mbox{ is the prefix as required})$.**
>
>
>
(Note that $B$ can be low and so non-high. Then we may assume that $g$ is totally computable.)
Therefore for any such $n$, $K(\Omega \upharpoonright n)\leq K(A\upharpoonright |g(n)|)+K(g(n))\leq 2K(n)$ up to a constant. This is a contradiction.
| 1 | https://mathoverflow.net/users/14340 | 414795 | 169,155 |
https://mathoverflow.net/questions/414756 | 5 | Let $X$ be a Polish space and let $G\in\mathbf{\Sigma}^1\_1(X^2)$ be a graph on $X$, that is an irreflexive and symmetric relation on $X$.
Given a cardinal $\kappa$ we say that $G$ has chromatic number $\kappa$, in symbols $\chi(G)=\kappa$, if there is a function $\varphi\colon X\to Y$ for some Polish space $Y$ such that $|\varphi(X)|=\kappa$ and for every $y\in\varphi(X)$, we have that $\varphi^{-1}(y)$ is $G$-independent, meaning that $G\cap(\varphi^{-1}(y))^2=\varnothing$, and $\kappa$ is the least cardinal with this property. We say that $G$ has Borel chromatic number $\kappa$, in symbols $\chi\_B(G)=\kappa$ if we additionally require $\varphi$ to be Borel (of course the fact that $Y$ is Polish is completely irrelevant in the definition of $\chi(G)$ and I'm only phrasing it this way to define $\chi\_B(G)$ analogously).
**Question:** Does $\mathsf{ZFC}$ prove that $\chi(G)\in\{1,2,\ldots,\aleph\_0,2^{\aleph\_0}\}$? Of course the answer is positive under $\mathsf{CH}$ but that is hardly interesting.
**Remarks:**
* If $\mathsf{CH}$ fails and there are no regularity assumption on $G$
then there are trivial counterexamples, fix $A\subseteq\Bbb R$ with $|A|=\aleph\_1$ and let $G=A^2\setminus\Delta\_\Bbb R$. Then $\chi(G)=\aleph\_1$ but $G$ is not analytic.
* The same question for $\chi\_B(G)$ (or even for Baire measurable colourings) has a positive answer by the Kechris-Solecki-Todorcevic $G\_0$-dichotomy.
* $\chi(G)$ and $\chi\_B(G)$ can be wildly different, there are examples with $\chi(G)=2$ ($G$ can even be taken to be acyclic) and $\chi\_B(G)=2^{\aleph\_0}$.
| https://mathoverflow.net/users/49381 | On the chromatic number of an analytic graph | There is a counterexample due to Todorcevic. See Proposition 9.2 in Kechris, Solecki, Todorcevic, Borel Chromatic Numbers, Advances in Mathematics, Vol. 141, Issue 1, 1999, Pages 1-44
| 6 | https://mathoverflow.net/users/475361 | 414797 | 169,156 |
https://mathoverflow.net/questions/414789 | 14 | Given a metric space $(X, d)$, we can consider the set of all quasi-isometries $f: X \to X$, and quotient out by the equivalence relation identifying $f$ and $g$ if $\sup\_{x \in X}d(f(x), g(x))$ is finite. Doing so, we obtain a set of equivalence classes $\mathcal{QI}(X)$ that is a group under composition.
In the same spirit as the questions [Every group is a fundamental group](https://math.stackexchange.com/questions/939856/every-group-is-a-fundamental-group) and [Is every group the automorphism group of a group?](https://math.stackexchange.com/questions/253936/is-every-group-the-automorphism-group-of-a-group), we can ask: for which groups $G$ does there exist a metric space $X$ such that $\mathcal{QI}(X) \cong G?$
Surprisingly, someone told me today that basically nothing is known about this question. According to them, we do not even know how to construct a metric space $X$ such that $\mathcal{QI}(X)$ is a finite cyclic group.
Given this, my question is: what *do* we know about the quasi-isometry groups of metric spaces? For example, what are some metric spaces $X$ for which $\mathcal{QI}(X)$ has been computed? Do we know of any groups $G$ which are not isomorphic to $\mathcal{QI}(X)$ for any $X$?
| https://mathoverflow.net/users/474271 | Quasi-isometry groups of metric spaces | A first observation is that $\mathcal{QI}$ is quite complicated for most natural spaces. For instance, any two linear maps $x \mapsto \lambda x, x \mapsto \lambda' x$ are equal in $\mathcal{QI}(\mathbb{R})$ if and only if $\lambda = \lambda'$. A corollary of this is that $\mathcal{QI}(\mathbb{N})$ is uncountable.
In order to get small $\mathcal{QI}(X)$ to be small, one must spread apart points in $X$ to sabotage the flexibility of the quasi-isometry condition. A helpful building block and motivating example is $X\_0 = \{ n! \mid n \in \mathbb{N} \} \subset \mathbb{N}$. Any $(K,C)$-quasi isometry $f$ of $X\_0$ must have $f(n!) = n!$ for all $n$ large enough, for instance $n > \text{max}(K, C) + 1$, by considering $d(f(n!), f((n+1)!))$. So $\mathcal{QI}(X\_0)$ is trivial. If one allows metrics that obtain $\infty$ as a distance, one obtains a space with $\mathcal{QI}(X\_n) = S\_n$ (with $S\_n$ the symmetric group on $n$ letters) by taking $n$ copies of $X\_0$, with infinite distance between any two copies.
An addendum: this construction also gives direct products of symmetric groups by mixing growth rates in building blocks. For instance, let $Y\_0 = \{ (n!)! \, | n \in \mathbb{N}\}$. Then taking the disjoint union of $n$ copies of $X\_0$ and $m$ copies of $Y\_0$ as above gives a space with $\mathcal{QI}(X) = S\_n \times S\_m$. This is because for a $(K, C)$ quasi-isometry, one can not map sufficiently large elements of $X\_0$ into $Y\_0$ or vice-versa. One sees this by comparing the distances between $3$ consecutive elements in $X\_0$ or $Y\_0$.
A correction: as pointed out by Fedya below, a previous version of this answer incorrectly asserted that spaces $X\_k$ obtained as pinwheels of $k$ copies of $X\_0$ have $\mathcal{QI}(X\_k) \cong S\_k$. One can independently permute each of the $k$ copies of $n!$ while remaining a quasi-isometry, and finite-distance maps allow one to freely move around finitely many points freely. This yields the quasi-isometry group of $X\_k$ to be isomorphic to $(\prod\_{k=1}^{\infty}S\_k) / (\bigoplus\_{k=1}^\infty S\_k).$
It seems quite unclear how to build many other groups with explicit examples. In particular, one question I think is interesting but do not know how to answer is if there exist metric spaces (with only finite distances allowed) with finite and nontrivial $\mathcal{QI}$ group.
| 11 | https://mathoverflow.net/users/136267 | 414798 | 169,157 |
https://mathoverflow.net/questions/414737 | 1 | Let $G=(V\_G,E\_G)$ and $H=(V\_H,E\_H)$ be two undirected graphs. When considering the *two-dimensional Weisfeiler-Leman* ($2$-$\mathsf{WL}$) procedure (the one corresponding to the three-variable fragment of first-order logic with counting quantifiers [[1](https://link.springer.com/article/10.1007/BF01305232)]), one typically declares $G$ and $H$ to be *distinguishable* by $2$-$\mathsf{WL}$ if the multisets of $2$-$\mathsf{WL}$ colours of pairs of vertices in $G$ and $H$ differ.
Formally, let us denote by $\chi\_{G,\mathsf{2wl}}(v,v')$ the (stable) colour of the pair $(v,v')$ of vertices in $G$, as assigned by $2$-$\mathsf{WL}$. Then, $G$ is said to be distinguishable from $H$ by $2$-$\mathsf{WL}$ if
$$
\{\!\!\{ \chi\_{G,\mathsf{2wl}}(v,v')\mid v,v'\in V\_G\}\!\!\}\neq\{\!\!\{ \chi\_{H,\mathsf{2wl}}(w,w')\mid w,w'\in V\_H\}\!\!\},
$$
where $\{\!\!\{\,\}\!\!\}$ denotes a multiset.
Consider next the following weaker notion of distinguishability: $G$ is distinguishable from $H$ by $2$-$\mathsf{WL}'$ if
$$
\{\!\!\{ \chi\_{G,\mathsf{2wl}}(v,v)\mid v\in V\_G\}\!\!\}\neq\{\!\!\{ \chi\_{H,\mathsf{2wl}}(w,w)\mid w\in V\_H\}\!\!\},
$$
where one thus only considers the $2$-$\mathsf{WL}$ colours of *vertices* (represented by pairs $(v,v)$) in the graphs.
We note that, if $G$ and $H$ are distinguishable by $2$-$\mathsf{WL}'$ then they are also distinguishable by $2$-$\mathsf{WL}$, because the $2$-$\mathsf{WL}$ colouring encodes the equality type of pairs of vertices.
**Question:** One would expect $2$-$\mathsf{WL}$ distinguishability to be stronger than $2$-$\mathsf{WL}'$ distinguishability. Is this indeed the case? That is, do there exist two graphs $G$ and $H$ (of the same number of vertices) that are distinguishable by $2$-$\mathsf{WL}$ but not by $2$-$\mathsf{WL}'$? If so, I would be interested in those graphs.
[1](https://link.springer.com/article/10.1007/BF01305232) [Jin-Yi Cai, Martin Fürer, Neil Immerman. An optimal lower bound on the number of variables for graph identification, Combinatorica, 12, 389-410, 1992.](https://link.springer.com/article/10.1007/BF01305232)
| https://mathoverflow.net/users/9839 | Distinguishing graphs by 2-dimensional Weisfeiler-Leman vertex colourings | These notions of distinguishability are, in fact, equivalent. If 2-WL can distinguish $G$ from $H$, then their multisets of 2-WL labels are actually disjoint. And thus, in particular, there cannot be two vertices with the same 2-WL label. See Corollary 1 in "*On the Combinatorial Power of the
Weisfeiler-Lehman Algorithm*" by Martin Fürer, where this is stated in another way: it suffices to have the same 2-WL label for two pairs of vertices, one pair in $G$, the other pair in $H$, to ensure that the entire multisets of 2-WL labels in those graphs are the same.
| 2 | https://mathoverflow.net/users/475964 | 414802 | 169,158 |
https://mathoverflow.net/questions/414810 | 1 | Consider a family of smooth, atomless CDFs, $F\_x(\cdot)$, for each $x \in \mathbb R$. Suppose that $F\_x(\cdot)$ are FOSD ranked in $x$. That is, for any $x, x'$ such that $x \ge x'$, $F\_x(\cdot) \le F\_{x'}(\cdot)$.
Let $K > 0$ be a fixed scalar. I want to know whether there exists a function $g(\cdot)$ and a number $y^\*$ such that
\begin{align\*}
g(y) =& 0 & \text{ if } y < y^\*\\
g(y) >& 0 & \text{ if } y \ge y^\*\\
\mathbb{E}\_{y}[g(\cdot)] =& K & \forall y \ge y^\*\\
\mathbb{E}\_{y}[g(\cdot)] \le& K & \forall y < y^\*
\end{align\*}
where $\mathbb E\_y[g(\cdot)] = \int\_{\mathbb R} g(\cdot) d F\_y(\cdot)$.
I primarily want to know what is the right language to pose such a question. What should I be looking at? Is it optimal transport or something?
Thanks!
| https://mathoverflow.net/users/78761 | Expectation of a function according to a family of distributions | $\newcommand\R{\mathbb R}$This is true for some first order stochastic dominance (FOSD) families but not all of them.
Indeed, if $F\_x$ does not depend on $x$ and $c:=E\_x1\_{[y\_\*,\infty)}>0$, then just let $g:=\dfrac Kc\,1\_{[y\_\*,\infty)}$.
On the other hand, suppose that $F\_x$ is the cdf of $Z+x$, where $Z$ is a standard normal random variable. Then the family $(F\_x)\_{x\in\mathbb R}$ is FOSD. Suppose now that there exist some real $K,y\_\*$ and some Lebesgue-measurable function $g$ such that $g=0$ on $(-\infty,y\_\*)$ and $g>0$ on $[y\_\*,\infty)$ such that
$$h(y):=E\_y g=K$$
for all real $y\ge y\_\*$. Then
$$K=h(y)=Eg(Z+y)=\int\_\R g(z+y)f(z)\,dz=\int\_\R g(u)f(y-u)\,du$$
for all $y\ge y\_\*$, where $f$ is the standard normal pdf. Since
$$f(z+t)=e^{-t^2/2}f(z)e^{-tz}, \tag{1}$$
the derivatives $h^{(k)}$ of all natural orders $k$ of the function $h$ exist and and for all $y>y\_\*$
$$0= h^{(k)}(y)=\int\_\R g(u)f^{(k)}(y-u)\,du=(-1)^k\int\_\R g(u)H\_k(y-u)f(y-u)\,du,$$
where the $H\_k$'s are the ["probabilist's" Hermite polynomials](https://en.wikipedia.org/wiki/Hermite_polynomials#Definition). So, by the [completeness property](https://en.wikipedia.org/wiki/Hermite_polynomials#Completeness) of the Hermite polynomials, the function $g$ must be constant, which contradicts the conditions that $g=0$ on $(-\infty,y\_\*)$ and $g>0$ on $[y\_\*,\infty)$. Thus, in this case no function $g$ with the desired properties exists. (This conclusion can also be obtained from (1) and the fact that, if the moment generating function (mgf) of a distribution is everywhere finite, then the distribution is characterized by the values of the mgf on any nonempty open interval -- cf. e.g. [this post](https://stats.stackexchange.com/a/297243/86566).)
(The condition that $E\_y g\le K$ for $y<y\_\*$ is redundant: it follows from the condition that $E\_y g=K$ for $y\ge y\_\*$ and the FOSD condition.)
| 1 | https://mathoverflow.net/users/36721 | 414825 | 169,162 |
https://mathoverflow.net/questions/414818 | 1 | Over the past few decades, a vast research area in number theory is surrounded by the $p$-adic number field $\mathbb{Q}\_p$ and its extensions.
My question is on different perspective.
What are the lists of some specialized journals that publishes works in $p$-adic number theory ?
I have in mind the following journals:
* journal of number theory
* algebra and number theory
* acta arithmetica
* international journal of number theory
* Journal de Theorie des Nombres de Bordeaux
* $p$-adic numbers, ultrametric analysis and applications
May be my ranking is not correct exactly.
All of the above journals are SCI indexed or SCI-expanded indexed except the last one. The last one seems a quite new journal, probably consisting of $14$ volumes as of year $2021$. But it seems, the last journal is emerging well over the past years looking at its Scimago impact factor (placing it in quartile $Q\_2$ last year). It seems it is specially oriented for $p$-adic number field or more generally on nonarchimedian field.
As a PhD student in $p$-adic algebraic number theory, my question-
>
> Is it worthy to publish a research paper in any one of these journals ?
>
>
>
Any comments please
| https://mathoverflow.net/users/122445 | Some good journals in $p$-adic number theory | Your question is somewhat broad, since local and p-adic fields permeate number theory and other parts of mathematics. If you want to get an idea of which journals publish articles about aspects of p-adic fields, you can look on MathSciNet. For example, there's an entire category
* 11S Algebraic number theory: local and p-adic fields
with 14 subcategories, and there are subcategories in other sections that deal with local fields, including
* 11D88 Diophantine equations: p-adic and power series fields
* 11E95 Forms and linear algebraic groups: p-adic theory
* 11F85 p-adic theory, local fields: Discontinuous groups and automorphic forms
* 11G25 Arithmetic algebraic geometry: Varieties over finite and local
fields
* 11K41 Probabilistic theory: distribution modulo 1; metric
theory of algorithms: Continuous, p-adic and abstract analogues
So for example, if you're interested in Galois cohomology associated to local fields, you could search on 11S25 in the primary and secondary fields for the year 2021. I did that and found that there are such articles published in Trans AMS, RIMS, IJNT, Proc AMS, Selecta Math, Ann Inst Fourier, J Inst Math Jussieu, Mem Soc Math Fr, Acta Math Sin, and Acta Arith. So quite a variety of journals to choose from, and that's just 2021 articles in this particular part of $p$-adic number theory.
| 10 | https://mathoverflow.net/users/11926 | 414830 | 169,164 |
https://mathoverflow.net/questions/414713 | 11 | Let $G$ be a finite group, and $R(G)$ its representation ring over $\mathbb{C}$. We have the Adams operations $\psi^k:R(G)\rightarrow R(G)$, given on the level of characters by: $$\chi\_{\psi^k{V}}(g)=\chi\_V(g^k).$$
Since the power sums can be expressed as polynomials in the homogenous symmetric functions $h\_n$, these $\psi^kV$ correspond to virtual representations given by integer polynomials in the symmetric powers of $V$.
By using these same polynomials in the symmetric powers, we can define Adams operations on the level of Burnside rings $\psi^k:A(G)\rightarrow A(G)$, such that these operations intertwine the natural map $A(G)\rightarrow R(G)$ induced by taking the free (virtual) vector space on the (virtual) $G$ set.
Since $R(G)$ has a nondegenerate quadratic form, we can take the adjoint of the $\psi^k$ map, call it $\nu^k$. Since it is an adjoint, this map preserves characters/virtual representations. On characters, this map is given by:$$\chi\_{\nu^k V}(g)=\sum\_{h^k=g}\chi\_V(h).$$
My question is then, does a natural lift of $\nu^k$ to the Burnside ring exist?
It is natural condition for such a $\nu^k$ to commute with induction from subgroups, so it suffices to define $\nu^k(\ast)$ for the trivial $G$ set $\ast$. So to show that such a Burnside ring version of this map isn't possible, it would suffice to show that $\nu^k(\mathbb{1})$ isn't in the image of the natural map $A(G)\rightarrow R(G)$, but I'm struggling to find a counterexample to this claim.
| https://mathoverflow.net/users/128502 | A question about the adjoint of the Adams operations on representation rings | If I understand the question correctly, then $\nu^{2}(1)$ isn't in the image of the natural map $A(G) \to R(G)$ when $G = Q\_{8},$ the quaternion group of order $8$.
The number of square roots of the identity in $G$ is $2$, the number of square roots of the central involution $z$ in $G$ is $6$, and the number of square roots of each element of order $4$ is $0$.
Hence we have $\nu^{2}(1) = \lambda\_{1} + \lambda\_{2}+ \lambda\_{3} + \lambda\_{4} - \chi$, where the $\lambda\_{i}$ are the linear characters of $G$ and $\chi$ is the unique irreducible character of $G$ of degree $2$.
Now I claim that this virtual character is not a difference of permutation characters. Indeed, it is not even a difference of characters afforded by $\mathbb{R}G$-modules.
For the irreducible character $\chi$ has real Schur index $2$, so occurs with even multiplicity in any character of $G$ afforded by an $\mathbb{R}G$-module.
Hence $\chi$ occurs with even multiplicity in any difference of permutation characters, so that $\nu^{2}(1)$ is not expressible as a difference of permutation characters.
More generally, if $G$ is any finite group of even order which has an irreducible character $\chi$ with Frobenius-Schur indicator $-1$, then $\nu^{2}(1)$ is not expressible as a difference of permutation characters of $G$.
An induction theorem of G. Segal may be relevant to trying to characterize exceptions for which $\nu^{2}(1)$ is not a virtual permutation character of $G$.
| 6 | https://mathoverflow.net/users/14450 | 414833 | 169,165 |
https://mathoverflow.net/questions/414848 | 11 | In Besse's "Einstein manifolds", p. 177, he states that, until that moment, no general classification of homogeneous Einstein manifolds was know, even in the compact case. More specifically, he poses a problem: classify the compact simply connected homogeneous manifolds $M=G/K$ which admit a $G$-invariant Einstein metric.
Does that question remain open to this day?
| https://mathoverflow.net/users/101432 | Classification of homogeneous Einstein manifolds | Yes, the question is still open.
I suggest to read [this quite recent paper](https://arxiv.org/abs/2107.06609) by Kerr and Böhm. It reviews some of the most important advances in the problem and includes several open problems.
| 13 | https://mathoverflow.net/users/20052 | 414852 | 169,174 |
https://mathoverflow.net/questions/414767 | 3 | Let's define a coprime graph as a simple graph (undirected graph without any self-loops or multiple-edges) in which for all edges $(, )$, the property $\gcd(\mathrm{degree}\_u, \mathrm{degree}\_v) = 1$ is true (i.e., degrees of vertices $$ and $$ are coprime).
I conjecture that, such a graph with maximum number of edges ~~(is always)~~ can always be obtained as [a $k$-partite complete graph](https://en.wikipedia.org/wiki/Multipartite_graph).
It has been a while that I have investigated this problem and even proposed [an optimization task](https://lotfizadeh.org/wp-content/uploads/2021/12/Fuzzy-Graph.pdf) on that. You can also read [the solution](https://lotfizadeh.org/wp-content/uploads/2021/12/Fuzzy-Graph-Solution.pdf).
I would like to know if anyone has some proof or related work?
| https://mathoverflow.net/users/475929 | Maximum number of edges in a "coprime graph" | OK, I can now confirm that even your modified conjecture is false, although the first counterexample has $13$ vertices.
First we need to know which complete $k$-partite graphs on $13$ vertices are coprime and how many edges they have. Both of these things can be determined directly from the sizes of the partite-sets. So for each partition of $13$ with parts $p\_1,p\_2,\ldots,p\_k$ we need to have $n-p\_i$ coprime to $n-p\_j$ for all $i < j$. For each such partition the number of edges of the corresponding complete $k$-partite graph is $\sum\_{i<j}p\_ip\_j$
In this case, the winner is $K\_{3,4,6}$ which has 12+18+24 = 54 edges.
However, here is a nice $13$-vertex graph that is coprime but has $56$ edges.
```
0 [5, 6, 7, 9, 10, 11, 12] 7
1 [5, 6, 7, 9, 10, 11, 12] 7
2 [5, 6, 8, 9, 10, 11, 12] 7
3 [5, 7, 8, 9, 10, 11, 12] 7
4 [6, 7, 8, 9, 10, 11, 12] 7
5 [0, 1, 2, 3, 8, 9, 10, 11, 12] 9
6 [0, 1, 2, 4, 8, 9, 10, 11, 12] 9
7 [0, 1, 3, 4, 8, 9, 10, 11, 12] 9
8 [2, 3, 4, 5, 6, 7, 11, 12] 8
9 [0, 1, 2, 3, 4, 5, 6, 7, 11, 12] 10
10 [0, 1, 2, 3, 4, 5, 6, 7, 11, 12] 10
11 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 11
12 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 11
```
Each line contains a vertex name (from $0$-$12$), then a list a neighbours of that vertex, then the degree of that vertex.
The vertices of each fixed degree must be independent sets, and it is easy to check that $\{0,1,2,3,4\}$, $\{5,6,7\}$, $\{9,10\}$ and $\{11,12\}$ are indeed independent.
Any edge involving a vertex of degree $7$, $9$ and $11$ automatically has ends of coprime degree because $7$, $9$ and $11$ are coprime to every other degree. So it is only an edge connecting a vertex of degree $8$ to one of degree $10$ that can prevent this graph from being coprime. And no such edge exists.
You can put this directly into SageMath to construct the graph.
```
x = Graph("L?BvUo~~v}^~~}")
```
| 5 | https://mathoverflow.net/users/1492 | 414864 | 169,178 |
https://mathoverflow.net/questions/414869 | 1 | I am a mathematician studying the dynamics of the $N$-Body density matrix $\rho\_{N}(x;y)$ for $n$ particles, defined by
$$\rho\_{N, t}^{(n)} (x\_1,..,n\_n; y\_1,...,y\_n) =
\begin{cases}
\int \rho\_{N,t}(x\_1,...,x\_n,x\_{n+1},...,x\_N; y\_1,...,y\_n,x\_{n+1},...,x\_N) dx\_{n+1}...d\_{x\_N},\,\,\,\,\, 1 \leq n \le N\\
0,\,\,\,\,\, o.w.
\end{cases}$$
and the Wigner transform for one particle is:
$$w\_N(x;v):= \frac{1}{(3 \pi)^{3 N}} \int e^{- i v. y} \rho\_N \left( x + \frac{y}{2}, x- \frac{y}{2}\right) dy.$$
I cannot understand how to obtain that
$$\int w\_N(x,v) dx dv =1.$$
| https://mathoverflow.net/users/471464 | Integration of Wigner transform | I think your coefficients have typo's, but in any case, you first want to make sure that your reduced density matrix is properly normalized to unit trace. For one particle, that would mean that
$$\int\rho^{(1)}\_N(x,x)\,dx=1.$$
The Wigner transform can be then be defined as
$$w\_N^{(1)}(x,v)= \frac{1}{2\pi} \int e^{- i v y} \rho^{(1)}\_N \left( x + \frac{y}{2}, x- \frac{y}{2}\right)\, dy.$$
Now check that
$$\int w^{(1)}\_N(x,v) \,dx dv =\int \delta(y) \rho^{(1)}\_N \left( x + \frac{y}{2}, x- \frac{y}{2}\right)\, dxdy$$
$$\qquad=\int\rho^{(1)}\_N(x,x)\,dx=1,$$
where I used that $\int e^{-ivy}\,dy=2\pi\delta(y)$.
| 1 | https://mathoverflow.net/users/11260 | 414871 | 169,179 |
https://mathoverflow.net/questions/414868 | 4 | Is the following problem known? Suppose one is given some of the entries of an $n \times n$ matrix $A$ over $\mathbb{R}$, so that the given entries are symmetric. Can one assign values to the remaining entries so that the resulting matrix is positive definite?
Has this problem been studied from a theoretical or algorithmic point of view?
| https://mathoverflow.net/users/141963 | Completing partial matrix to a positive definite matrix | There is an extensive literature. Here are some entry points:
* [Positive definite completions of partial Hermitian matrices](https://www.sciencedirect.com/science/article/pii/0024379584902076)
* [The positive definite completion problem revisited](https://doi.org/10.1016/j.laa.2008.04.020)
* [Matrix Completion Problems](https://www.uibk.ac.at/mathematik/algebra/media/teaching/matrix-completion-problems_entner.pdf)
| 4 | https://mathoverflow.net/users/11260 | 414872 | 169,180 |
https://mathoverflow.net/questions/414870 | 0 | Let $(X,d)$ be a separable and connected metric space. My question is rather short and to the point: do there exist $\{x\_n\}\_{n=0}^{\infty}\subseteq X$ such that
$$
\left\{d(x\_n,\cdot)-d(x\_0,\cdot)\right\}\_{n=1}^{\infty},
$$
is a Schauder basis of $C\_b(X)$? If so, what is this "basis" called in the literature?
| https://mathoverflow.net/users/469470 | When does $C_b(X)$ admit a Schauder Basis? | Note that in order for $C\_b(X)$ to have a Schauder basis, $X$ has to be compact. Indeed, $C\_b(X)$ is naturally isomorphic to $C(\beta X)$ and the latter is non-separable (because $\beta X$ is non-metrisable) [as long as $X$ is a non-compact metric space](https://math.stackexchange.com/a/217422/17929).
Thus, you are in the realm of compact metric space to have a chance for a Schauder basis of some specific form since $C(X)$ always has a Schauder basis for a compact metric connected space. But do you have a basis of the required form already for $C[0,1]$?
| 2 | https://mathoverflow.net/users/15129 | 414873 | 169,181 |
https://mathoverflow.net/questions/414736 | 3 | Let $M$ be a complex manifold. Consider a connection $\nabla$ on the holomorphic tangent bundle $T^{1,0}M$. The *torsion* of $\nabla$ is defined as the torsion of the induced connection $D$ on the real tangent bundle, $$T\_\nabla(\alpha,\beta) = T\_D(\alpha,\beta) := D\_\alpha \beta - D\_\beta \alpha - [\alpha,\beta]$$ for smooth vector fields $\alpha$ and $\beta$. The connection $\nabla$ is *torsion-free* if $T\_\nabla = 0$.
Assume now that $\nabla$ is torsion-free. It is well-known that if $\nabla$ is also hermitian, i.e., compatible with a hermitian metric $h$ on $T^{1,0}M$, then it is the Chern connection of the metric, i.e., $\nabla$ is a $(1,0)$-connection in the sense that $\nabla^{0,1}=\bar\partial$, and $h$ provides a Kähler metric on $M$, see e.g., Huybrechts, Complex Geometry, Proposition 4.A.7.
I am interested in whether one can find torsion-free $(1,0)$-connections on $T^{1,0} M$ also on non Kähler manifolds? Such connections would thus necessarily not be hermitian by the above result. Or is there some obstruction?
| https://mathoverflow.net/users/49151 | Torsion free (1,0)-connections on the holomorphic tangent bundle? | I will write in terms of the holomorphic frame bundle, i.e. the bundle of choices of complex linear bases of tangent spaces, a holomorphic principal $\operatorname{GL}\_n$-bundle.
Any sum $\gamma=\sum h\_a \gamma\_a$ with $\sum h\_a=1$ of $(1,0)$-connection forms on the holomorphic frame bundle is a $(1,0)$-connection. If all $\gamma\_a$ are torsion-free, so is $\gamma$. To see this, take the soldering forms $\omega=(\omega^{\mu})$ on the frame bundle, and then the structure equations of a connection are $d\omega=-\gamma\wedge\omega+\frac{1}{2}a\omega\wedge\omega+b\omega\wedge\bar\omega+c\bar\omega\wedge\bar\omega$, with $a,b,c$ the components of the torsion. (Note that $c$ is the Nijenhuis tensor, so $c=0$ on a complex manifold.) If these vanish for all of the $\gamma\_a$, they still vanish for $\gamma$. Therefore if the $h\_a$ form a partition of unity, for local choices of torsion-free $(1,0)$-connections, then $\gamma$ is a torsion-free $(1,0)$-connection.
| 3 | https://mathoverflow.net/users/13268 | 414874 | 169,182 |
https://mathoverflow.net/questions/414746 | 8 | Exponential sums are a powerful tool in additive combinatorics and number theory. In my understanding, when it comes to estimate the cardinality of a certain set, exponential sums are (essentially) used in this way: (1) the indicator function of the set is replaced by an appropriate exponential sum; (2) the sum of the indicator function and the sum of the exponential sum are swapped; (3) the main term is extracted; 4) the error term is bounded.
However, recently I stumble upon some simple additive problems for which the exponential-sums approach seems to fail miserably. My (somehow philosophical) question is: Am I applying exponential sums in the wrong way? (And, in such a case, how should apply them to solve such problems?) Or are exponential sums are not good for these problems? (And, in such a case, what is the reason it is so?)
Below an extremely simple example in which exponential sums seem to fails. Of course, I do not really care about solving such problem with exponential sums, but it is just for the sake of example.
Let $1 \leq h \leq m$ be fixed integers. We want to estimate the number $C$ of $(x, y) \in \{0,\dots,h-1\}^2$ such that $x \equiv y \bmod m$. Obviously, $C = h$. Proceeding with exponential sums, we have:
$$C = \sum\_{0 \leq x < h} \sum\_{0 \leq y < h} \begin{cases} 1 & \text{ if } x \equiv y \bmod m \\ 0 & \text{ if not} \end{cases}$$
$$\stackrel{(1)}{=} \sum\_{0 \leq x < h} \sum\_{0 \leq y < h} \frac1{m} \sum\_{0 \leq k < m} \exp\Big(\frac{2 \pi i (x - y)k}{m}\Big)$$
$$\stackrel{(2)}{=} \frac1{m} \sum\_{0 \leq k < m} \sum\_{0 \leq x < h} \sum\_{0 \leq y < h} \exp\Big(\frac{2 \pi i (x - y)k}{m}\Big)$$
$$\stackrel{(3)}{=} ???$$
In step (3) comes the problem. It is not clear what the main term of $\sum\_{0 \leq k <m}$ is. The term $k = 0$ surely is not, since it equals $h^2 / m$. In fact, all the terms are positive an equal to $|\sum\_{0 \leq x < h} \exp\Big(\frac{2 \pi i x k}{m}\Big))|^2$.
One can compute explicitly all the geometric sums and get the result, but that is unwieldy.
| https://mathoverflow.net/users/475914 | Why are exponential sums so bad at solving this very easy problem? | The basic example of exponential sums in Number Theory is to count solutions to an equation such as $f(x\_1,\cdots,x\_n)\equiv 0$ $\mod p$ where $p$ is prime this is because $p$-adic solutions are a necessary condition for solutions in integers and also for getting methods like the [Hardy\_Littlewood circle method](https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_circle_method) to work. Essentially this is an example of the [Hasse principle](https://en.wikipedia.org/wiki/Hasse_principle) - local solutions give a global solution in certain circumstances.
In this case the main term from your exponential sum, given by setting $k=0$ in your problem, is the average number of solutions in the following sense. Pick values for the $x\_i$'s ranging over all values $\mod p$. This gives $p^n$ possible values for $f$. We now make the assumption that these values are approximately evenly distributed $\mod p$. It then makes sense that $p^n/p=p^{n-1}$ should be a good estimate for the total number of solutions to the congruence $f\equiv 0$.
The method of exponential sums then proceeds by showing that the error term given by the rest of the sum is small in comparison with the main term and this gives you solutions for large enough $p$. The classical example of this is for diagonal equations $a\_1x\_1^k+\cdots +a\_nx\_n^k \equiv 0$. (Andre Weil analysed these in his paper where he stated the famous [Weil Conjectures](https://en.wikipedia.org/wiki/Weil_conjectures))
You can also use similar ideas to solve equations over restricted sets of variables. In your case case, $f(x\_1,x\_2)=x\_1-x\_2$ solving $\mod m$ with a set $\{0,\cdots h-1\}$. The same idea gives you a main term of $h^2/m$. The problem is you don't have equi-distribution for the number of solutions of $x\_1-x\_2=r$ as $r$ varies unless $h$ is almost the same size as $m$ and in that case $h^2/m\approx h$ giving you an accurate estimate of the number of solution.
In this case of large $h$ the error term is also small so the standard approach of exponential sums does actually work.
Note that the reason why exponential sums breaks down is because the fourier coefficients of your set $\{0,\cdots h-1\}$ are not small if $h$ is small compared to $m$.
We can relate this to a classic example from additive combinatorics - Roth's Theorem for arithmetic progressions of length 3. we have $f(x\_1,x\_2,x\_3)=x\_1-2x\_2+x\_3$ and the exponential sum method allows us to count length 3 aps $\mod n$ where x\_i are taken from sets $A,B,C$ which are reasonably dense in $\{1,\cdots,n\}$. Then we can express the solution count in the same way with the main term being $|A|B||C|/n$ and the error term a sum of fourier coefficients of the characteristic functions, $N^2\sum\_{s} \widetilde{A}(s) \widetilde{B}(-2s) \widetilde{C}(s)$. If the fourier coefficients are small indicating that the sets are uniformly distributed in a certain sense, often called pseudo-random, then we can show that the error term is small and solutions exist.
Again, the above does not work if the sets are not sufficiently pseudo-random for the main term to approximate the number of solutions and in this case one has to use alternative ideas to show that a set with a large fourier coefficient is not uniformly distributed and hence concentrated in some arithmetic progression $\mod m$ which allows one to iterate. This is described very elegantly in Tim Gowers' papers on [Szemeredi's Theorem](https://en.wikipedia.org/wiki/Szemer%C3%A9di%27s_theorem).
Others know these ideas much better than I do! I just wanted to give some background and note the importance of pseudo-randomness in the standard number theoretic application of exponential sums but also that if this fails there are techniques that one can apply to deal with the case where the sets are more structured.
| 7 | https://mathoverflow.net/users/7113 | 414881 | 169,184 |
https://mathoverflow.net/questions/414858 | 3 | Consider some deterministic, monotonic, eroding binary cellular automata on some lattice $\mathbb{Z}^d$, and consider the set of initial states $I(L)$ in which all of the vertices are $0$ except for finitely many, and these finitely many $1$'s are within some hypercube of side length $L$ centered at the origin. I am interested on bounds depending on $L$ for the time it takes for this eroding cellular automata to evolve a state $J\in I(L)$ into the all $0$ state. Call this time $T(J)$.
I am interested in whether one could find an example where $T$ asymptotically dominates $L$. That is, can one find a monotonic, eroding binary cellular automata for which
$$\lim\_{L \to \infty} \sup\_{J \in I(L)} \frac{T(J)}{L} = \infty ? $$
My intuition says that the answer is no, as such a model would likely provide a heuristic counterexample to Toom's proof of stability of monotonic eroders under small amounts of probabilistic noise, but I would like to understand why monotonicity and erosion are sufficient to make the limit above non-infinite.
---
I am new to the field of cellular automata and still learning the language, but I hope my question is well-posed. Please let me know if any parts are unclear or if I should clarify definitions.
By monotonic and eroding, I mean the definitions from chapter one of [Lise Ponselet's thesis](https://arxiv.org/pdf/1312.3612.pdf).
Colloquially, I believe a monotonic update rule means that if I have some configuration $J$ and consider a configuration $K$ that looks like $J$ except some $0$'s are replaced with $1$'s, the update rule will leave $K$ with at least as many $1$'s as $J$ after a given update. Colloquially, I believe an eroding cellular automata takes a finite number of updates to remove a finite number of $1$'s from an otherwise all $0$ configuration.
| https://mathoverflow.net/users/153549 | Binary cellular automata: How slowly can an eroder remove $1$'s? | The answer is no.
According to [1] this is proved in [2] and in some paper of Toom from 1979 that does not seem to be listed in their bibliography. I cannot access [2] or any paper of Toom. I am a little unsure of myself here as I never had to write this stuff and don't have access to any of the classical references, but I think the following argument should be standard or wrong.
On page 12 of Ponselet's thesis it is stated that Toom proves that the precise criterion for erosion is the following erosion criterion.
>
> Definition (Erosion Criterion). A monotonic binary CA is said to satisfy the Erosion Criterion if $\bigcap\_{j=1}^J \mathrm{conv}(Z\_j) = \emptyset$ where $Z\_j$ are the *zero sets* meaning minimal subsets of the lattice such that if you have zeroes in those positions then you have zeroes at the origin in the image of the CA.
>
>
>
Here $\mathrm{conv}$ denotes the convex hull in $\mathbb{R}^d$. Note that the family of zero sets fully characterizes a monotone CA, and the zero sets are contained in its finite neighborhood as a CA.
>
> Theorem [Toom]. The Erosion Criterion is the erosion criterion for monotone CA over the binary alphabet.
>
>
>
Now if you satisfy the erosion criterion, then there is a simple geometric way to see that erosion is linear. Take an arbitrary cell $v$, and consider the area $(v+\mathrm{conv}(Z\_j)) \cap \mathbb{Z}^d$ on the previous time step. If all of those cells contain zero, then we have $0$ in $v$ at the present step. Consider then $v+2\mathrm{conv}(Z\_j) \cap \mathbb{Z}^d$; if all those cells contain zero then two steps later you have $0$ at $v$. Namely if $c|\_{v+2\mathrm{conv}(Z\_j) \cap \mathbb{Z}^d} \equiv 0$ then because
$v+Z\_j+Z\_j \subset v + \mathrm{conv}(Z\_j) + \mathrm{conv}(Z\_j) \cap \mathbb{Z}^d = v + 2\mathrm{conv}(Z\_j) \cap \mathbb{Z}^d$ we have $\phi(c)|\_{v + Z\_j} \equiv 0$ where $\phi$ is our CA, and then of course $\phi^2(c)\_v = 0$.
Continuing this logic, we see that if $t$ steps before we had only zeroes in $v + t\mathrm{conv}(Z\_j) \cap \mathbb{Z}^d$ then we have 0 at $v$ on the present step. But now if $\bigcap\_{j=1}^J \mathrm{conv}(Z\_j) = \emptyset$, then in linearly many steps $t = O(n)$, the $n^d$-hypercube does not even intersect one of the sets $v+t\mathrm{conv}(Z\_j)$ (no matter where this hypercube is located), since the distances between the blown-up convex hulls are blowing up proportionally to $t$. So we're done.
[1] *de Meneyes, Moisés Lima; Toom, André*, A nonlinear eroder in presence of one-sided noise, Braz. J. Probab. Stat. 20, No. 1, 1-12 (2006). [ZBL1272.60049](https://zbmath.org/?q=an:1272.60049).
[2] *Gal’perin, G. A.*, One-dimensional local monotone operators with memory, Sov. Math., Dokl. 17, 688-692 (1976); translation from Dokl. Akad. Nauk SSSR 228, 277-280 (1976). [ZBL0366.94059](https://zbmath.org/?q=an:0366.94059).
| 3 | https://mathoverflow.net/users/123634 | 414882 | 169,185 |
https://mathoverflow.net/questions/414875 | 9 | Let $X$ be a Banach space and let $\mathrm{Iso}(X)$ be its group of isometries, i.e., the set of surjective linear maps $T: X \to X$ with $\|Tx\| = \|x\|$.
**Q: Is $\mathrm{Iso}(X)$ a topological group under the strong topology?**
While it is easy to show that multiplication is continuous, it is not clear to me how to show that inversion is continuous. I did not find a reference in the literature for this statement.
If $X = H$ is a separable Hilbert space, then $\mathrm{Iso(X)} = \mathrm{U}(H)$, the unitary group of $H$, and the statement that this is a topological group is well-established. However, the proof (that I know) uses the fact that the weak and the strong operator topology agree on $\mathrm{U}(H)$ and that the inverse is just given by $u \mapsto u^\*$, which is continuous in the weak operator topology.
| https://mathoverflow.net/users/16702 | Group of isometries of Banach spaces a topological group? | That the inverse is continuous for the strong topology would actually be true for any *bounded* subgroup of $GL(X)$, the invertible operators on $X$.
Firstly, as translation is continuous, it suffices to consider continuity at the identity.
Now let $(T\_i)$ be a *bounded* net of invertibles converging strong to $I$, and with $(T\_i^{-1})$ also bounded, say a common bound of $K$. Then, for $x\in X$,
$$ \|T\_i^{-1}(x) - x\| = \|T\_i^{-1}(x - T\_i(x))\| \leq K \|x-T\_i(x)\| \rightarrow 0 $$
by assumption that $T\_i\rightarrow I$ strongly. Thus also $T\_i^{-1}\rightarrow I$ strongly.
| 10 | https://mathoverflow.net/users/406 | 414892 | 169,188 |
https://mathoverflow.net/questions/414582 | 3 | This has been on MSE for over a month with four upvotes but no answers or even comments so I'm cross-posting:
According to [Examples of two-dimensional Riemannian manifolds that can't be isometrically embedded into $\mathbb{R}^4$](https://mathoverflow.net/questions/231495/examples-of-two-dimensional-riemannian-manifolds-that-cant-be-isometrically-emb) there is no smooth isometric embedding of the round (=constant positive curvature) projective plane into $ \mathbb{R}^4 $.
**Is there some intuition for why there is a smooth isometric embedding into $ \mathbb{R}^5 $ and even into round $ S^4 $ but not into $ \mathbb{R}^4 $?**
Some things I already know:
* There is an isometric embedding of round $ \mathbb{R}P^n $ into $ \mathbb{R}^{N} $ where $ N=\frac{n(n+3)}{2} $
* This embedding is equivariant with respect to the isometry group $ O\_{n+1} $ at least for $ n=1,2 $. In fact the image of the embedding arises as the orbit of a vector with respect to an $ N $ real dimensional irreducible orthogonal representation of $ O\_{n+1} $.
* It is an open question if there exists a $ C^r $ embedding of $ \mathbb{R}P^2 $ into $ \mathbb{R}^4 $ for $ 1 < r <2 $
| https://mathoverflow.net/users/387190 | Intuition for no isometric embedding of round projective plane into $\mathbb{R}^4 $ | I'm not sure what you would accept as 'intuition' for this result. It's actually a simple consequence of two facts, which actually prove something much stronger:
The first fact is that, for any smooth surface $S\subset\mathbb{R}^n$ with positive Gauss curvature $K$, its mean curvature vector $H$ cannot vanish. This is because the Gauss equation says that for any smooth surface in flat space, we have $K = |H|^2 - |I\!I\_0|^2$, where $I\!I\_0$ is the trace-free part of the second fundamental form. In particular, if $K>0$ everywhere, then $H$ is a nowhere vanishing normal vector field on the surface.
The second fact is an old differential topology result of Whitney that says that, when $\mathbb{RP}^2$ is embedded smoothly in $\mathbb{R}^4$, its normal bundle has no nonvanishing (continuous) section. This is not obvious; Whitney proved it using his obstruction theory for sphere bundles. See his *On the topology of differentiable manifolds*, Lectures in Topology, University of Michigan Press, 1941.
Granted these facts, it is immediate that, for any smooth embdding of $\mathbb{RP}^2$ into $\mathbb{R}^4$, there must be a point of $\mathbb{RP}^2$ where the induced metric has non-positive Gauss curvature. *A fortiori*, a metric of *constant* positive Gauss curvature on $\mathbb{RP}^2$ cannot be isometrically embedded into $\mathbb{R}^4$. (This is essentially the argument that Gromov and Rohklin give in Appendix 4 of their 1970 article *Embeddings and immersions in Riemannian geometry* , though their argument for the first fact (from differential geometry) seems less immediate to me than just using the Gauss equation, as I did above. But, eye of the beholder, etc.)
As far as I know, it is still an open problem whether $\mathbb{RP}^2$ can be smoothly *immersed* into $\mathbb{R}^4$ so that the induced metric has positive Gauss curvature (let alone *constant* positive Gauss curvature).
| 14 | https://mathoverflow.net/users/13972 | 414894 | 169,189 |
https://mathoverflow.net/questions/342606 | 21 | It's well-known that the norm on a $C^\ast$-algebra is uniquely determined by the underlying $\ast$-algebra by the spectral radius formula. Therefore there should be a way to axiomatize $C^\ast$-algebras directly in terms of the $\ast$-algebra structure, without explicitly talking about a norm.
**Question 1:** How does one do this? That is, which $\ast$-algebras are $C^\ast$-algebras?
**Question 2:** How does one axiomatize those $\ast$-algebras which embed into a $C^\ast$-algebra (equivalently, embed into their $C^\ast$-enveloping algebra)?
Some possibilities:
* Perhaps a $\ast$-algebra is a $C^\ast$-algebra iff the spectral radius *is* a complete, submultiplicative norm?
* Perhaps a $\ast$-algebra embeds into a $C^\ast$-algebra iff every element has finite spectral radius?
If the first guess above (or something like it) is correct, it would still be nice to break it down into more manageable chunks.
**EDIT:** I'm currently fascinated by the following observation. Let that if $A$ be any algebra over $\mathbb C$, and $a \in A$. Let $B$ be the subalgebra of $A$ generated by $a$, and let $C$ be the subalgebra of $A$ obtained from $B$ by closing under those inverses which exist in $A$, so that $C \cong \mathbb C[a][\{(a-\lambda)^{-1} \mid \lambda \not \in Spec(a)\}]$. Writing a general element $c \in C$ as a rational function $c = \phi(a)$, we have $Spec(c) = \phi(Spec(a))$. It follows that he spectral radius is a homogenous, subadditive, submultiplicative, power-multiplicative function on $C$. If we assume that the spectral radius in $A$ of any nonzero element of $C$ is finite and nonzero, it follows that the spectral radius is in fact a submultiplicative, power-multiplicative norm on $C$. So it seems natural to stipulate that (if $A$ is a $\ast$-algebra, and maybe assuming that $a$ is normal?), every "Cauchy sequence" in $C$ should have a unique "limit" in $A$ with respect to the spectral radius. I wonder how far this condition is from guaranteeing that $A$ is a $C^\ast$-algebra?
| https://mathoverflow.net/users/2362 | Which $\ast$-algebras are $C^\ast$-algebras? | **TL;DR:** A $\*$-algebra embeds in a $C^\*$-algebra iff it is archimedean with no non-trivial infinitesimals. In this case the $\*$-algebra is a $C^\*$-algebra iff a certain norm is complete.
I do explicitly talk about a norm here, but it is explicitly constructed from a natural order structure associated with the $\*$-operation.
---
Let $A$ be a complex $\*$-algebra.
In what follows we assume $A$ is unital and regard accordingly $\mathbb{C}$ as subalgebra of $A$. This is not a real restriction, as one can always unitalize $A$.
We define
$$ A\_+=\left\{\sum\_{i=1}^n x\_i^\*x\_i\mid n\in\mathbb{N},~x\_1,\ldots,x\_n\in A\right\}$$
and note that it is a convex cone in $A$, thus it defines on it a partial order by
$$ x\leq y \quad \Longleftrightarrow \quad y-x\in A\_+. $$
Next we define for $x\in A$,
$$ \|x\|=\sqrt{\inf\{\alpha\in \mathbb{R}\_+\mid x^\*x\leq \alpha\}}\in [0,\infty] $$
(using the conventions $\inf\emptyset=\infty$ and $\sqrt{\infty}=\infty$).
**Claim:** $A$ is embeddable in a $C^\*$-algebra iff it is archimedean with no non-trivial infinitesimals iff $\|\cdot\|$ is a norm on $A$ and in this case, $A$ is a $C^\*$-algebra iff this norm is complete.
Let me elaborate.
If $-1\in A\_+$ then $\|\cdot\|=0$ identically, thus we assume from now on that $-1\notin A\_+$.
$A$ is said to be *archimedean* if $\|\cdot\|$ attains only finite values.
We define the set of *infinitesimal* elements in $A$ to be
$A\_i=\{x\mid \|x\|=0\}$.
The claim above follows from the following two facts.
For proofs see section 2.1 [here](https://arxiv.org/pdf/2001.10185.pdf).
**Fact 1:** $A$ has a unital $\*$-representation into a $C^\*$-algebra iff it is archimedean and $A\_i$ is a two-sided ideal which is in the kernel of every such a representation.
**Fact 2:** If $A$ is archimedean then
$\|\cdot\|$ is a seminorm on $A$ and the corresponding norm on $A/A\_i$ satisfies the $C^\*$-property.
In particular, $A/A\_i$ $\*$-embeds into the $C^\*$-algebra obtained by completion.
| 4 | https://mathoverflow.net/users/89334 | 414904 | 169,191 |
https://mathoverflow.net/questions/414898 | 5 | Let $M$ be an irreducible 3-manifold with incompressible boundary of genus > 1.
When is $M$ homotopy equivalent to an Eilenberg-MacLane space? Or it is never true?
| https://mathoverflow.net/users/17895 | Irreducible 3-manifold with boundary of genus greater than 1 | M is always aspherical, and hence a Eilenberg-Maclane space.
This is because $\pi\_1(\partial M)$ embeds to $\pi\_1(M)$, by the incompressibility condition. If the genus of the boundary is no less than 1, then $M$ has infinite fundamental group. This forces the universal cover $\tilde M$ to be a non-compact 3-manifold, so $H\_i(\tilde M)=0,\ i\geqslant 3$.
By sphere theorem, $M$ is irreducible implies that $\pi\_2(\tilde M)=\pi\_2(M)=\{e\}$. Note that $\pi\_1(\tilde M)=\{e\}$, then by Hurewicz theorem we have $H\_i(\tilde M)=0,\ i=1,2.$ Then $\tilde M$ has trivial integral homology and is simply connected, so $\tilde M$ is contractible by Whitehead theorem.
| 12 | https://mathoverflow.net/users/140203 | 414905 | 169,192 |
https://mathoverflow.net/questions/414877 | 10 | Find all integer solutions to the equation
$$
y(x^2+1)=z^2+1.
$$
There is, for example, an infinite family of solutions $x=u$, $y=(uv\pm1)^2+v^2$, $z=(u^2+1)v \pm u$, $u,v \in {\mathbb Z}$, but there are also solutions outside of this family, e.g. $(x,y,z)=(8,5,18)$ or $(x,y,z)=(12,2,17)$. The question is to describe all integer solutions. Any reasonable description is ok. An algorithm generating all solutions is also ok, provided that it does not involve any search by trial and error (otherwise there is a trivial algorithm that tries all triples $(x,y,z)$ in some order). Using parametric expressions, recurrence relations, or something like ``start with this solution and apply these operations in any order'' (like generating Markov numbers via Markov tree) would be ideal, but more complicated algorithms are also possible. For example, for equation $yz=x^3+1$, there is an obvious algorithm "let $x$ be an arbitrary integer, let $y$ be any divisor of $x^3+1$, and then let $z=(x^3+1)/y$", which I think is acceptable. The equation in question is one of the smallest/simplest ones for which I do not see any reasonable method/algorithm to describe all solutions, hence the question.
Remark: If we define, for any integer $u$, set $S(u)$ as a set of integers $0\leq r < u^2+1$ such that $\frac{r^2+1}{u^2+1}$ is an integer, then all the solutions to the equation are in the form $x=u$, $z=(u^2+1)v \pm r$ and $y=(u^2+1)v^2 \pm 2vr + \frac{r^2+1}{u^2+1}$ for $u,v \in {\mathbb Z}$ and $r \in S(u)$, but we need trial and error to construct set $S(u)$, so I do not think this is an acceptable answer.
| https://mathoverflow.net/users/89064 | Solve in integers: $y(x^2+1)=z^2+1$ | The equation says that $z^2 + 1 \equiv 0 \mod (x^2+1)$. For each positive integer $x$, you can enumerate the square roots of $-1$ in the integers mod $x^2+1$ as you remark in your last paragraph. This does not require "trial and error": if you can factor $x^2+1$, you can use the Tonnelli-Shanks algorithm
to find the square roots mod each prime power in the factorization, then put them together using the Chinese Remainder Theorem.
| 5 | https://mathoverflow.net/users/13650 | 414909 | 169,194 |
https://mathoverflow.net/questions/414911 | 10 | I'm interested in cohomology operations (in ordinary cohomology)
$$H^i(-, G)\rightarrow H^{i+1}(-, H)\;,$$
that is, elements of
$$H^{i+1}(K(G, i), H)\;.$$
I know that $K(G, 1)=BG$, so for $i=1$, those cohomology operations are in $H^2(BG, H)$, and therefore given by the Bocksteins of the corresponding central extensions of $G$ by $H$. Also, for $G=H=\mathbb{Z}\_p$, the **stable** cohomology operations are given by the Steenrod algebra, and the only degree-1 elements are Bocksteins.
However, I don't know how it is for unstable cohomology operations, or groups other than $\mathbb{Z}\_p$. I'm mostly interested in simple groups, such as finitely generated or $\mathbb{R}/\mathbb{Z}$.
| https://mathoverflow.net/users/115363 | Are all degree-1 cohomology operations Bocksteins? | Yes. For $i\ge1$ you can build $K(G,i)$ from the Moore space $M(G,i)$ by adding cells of dimension $\ge i+2$, so $H\_i(K(G,i); Z) = G$ and $H\_{i+1}(K(G,i); Z) = 0$. Hence $Ext(G, H) \cong H^{i+1}(K(G,i); H)$ by the UCT. The elements of $H^{i+1}(K(G,i); H)$ represent the cohomology operations $H^i(-;G) \to H^{i+1}(-;H)$, and the elements of $Ext(G,H)$ correspond to the Bockstein operations. The case $i=0$ may require special attention.
| 16 | https://mathoverflow.net/users/9684 | 414914 | 169,195 |
https://mathoverflow.net/questions/414902 | 3 | A hyperbolic 3-manifold has finite volume if and only if it is either closed or has toroidal boundary and it is not homeomorphic to $T^2\times I$.
This statement is from [3-Manifold Groups, page 18 (the link is editted)](https://www.uni-regensburg.de/Fakultaeten/nat_Fak_I/friedl/papers/3-manifold-groups-final-version-031115) by Matthias Aschenbrenner, Stefan Friedl and Henry Wilton, it seems that the three references in the book toward this statement only give partial results (when the boundary components are already cusps).
**Edits**: The precise statement in my opinion should be:
Let $M$ be a compact three dimensional manifold with incompressible toroidal boundary (possibly none). If the interior of $M$ admits a hyperbolic structure, then $M$ either has finite volume, or is homeomorphic to $T^2\times I$.
Thanks for any solutions or hints.
| https://mathoverflow.net/users/140203 | Volume of hyperbolic 3-manifolds with toroidal boundary | I think you are trying to ask the following question.
>
> Suppose that $M$ is a compact connected oriented three-manifold. Suppose that $M^\circ$, the interior of $M$, admits a hyperbolic metric. Then when must this hyperbolic metric have finite volume?
>
>
>
As Ryan points out, if $M$ is closed, then $M^\circ = M$ is compact and thus has finite volume. Also, as you note, the interior of $M = T^2 \times I$ admits (many) hyperbolic metrics, but all have infinite volume. There is another such manifold: namely the solid torus $D^2 \times S^1$.
I think that the place where you are confused (please correct me if I am wrong) is the case where $M$ has a boundary of higher genus. Here $M^\circ$ again always has infinite volume.
| 4 | https://mathoverflow.net/users/1650 | 414915 | 169,196 |
https://mathoverflow.net/questions/414244 | 1 | Given two distinct primes $P\_1$ and $P\_2$ picked randomly and uniformly in the interval $[T^2,2T^2]$ consider the set $\chi(P\_1,P\_2)$ of numbers of form $$xP\_1-yP\_2$$ where $x,y$ are in $[0,T^{1+\epsilon}]$ where $T>0$ and $\epsilon\geq0$ is small.
>
> Are there expected to be at least $\frac{T^\mu}{O(\log T)}$ many distinct primes in the set $\chi(P\_1,P\_2)$ which are of size $T^{1+\mu}$ for small $\mu\geq0$?
>
>
> If not what is the correct asymptotic at least under appropriate assumptions?
>
>
>
Assume $T^{\min(\mu, \epsilon)}=\omega(\mathsf{polylog}(T))$.
| https://mathoverflow.net/users/10035 | Expected number of primes of particular size and from a linear form | The set $\chi(P\_1,P\_2)$ has $T^{2+2\epsilon}$ distinct elements in $[-2T^{3+\epsilon},T^{3+\epsilon}]$ (i.e., density of order $T^{\epsilon-1}$),
so we would expect $\chi(P\_1,P\_2)$ to contain about $T^{\mu+\epsilon}$ elements in $[C\_1T^{1+\mu}, C\_2T^{1+\mu}]$ provided that $0<C\_1<C\_2$. Of these, a fraction of about $1/(\log T)$ should be primes, leading to a prediction of about
$$\frac{T^{\mu+\epsilon}}{\log T}$$ primes in $[C\_1T^{1+\mu}, C\_2T^{1+\mu}]$ .
| 0 | https://mathoverflow.net/users/7691 | 414925 | 169,199 |
https://mathoverflow.net/questions/414929 | 6 | According to the answers in the the following questions: [How to prove the spectrum of the Laplace operator?](https://math.stackexchange.com/questions/790401/how-to-prove-the-spectrum-of-the-laplace-operator?noredirect=1&lq=1) and [What is spectrum for Laplacian in $\mathbb{R}^n$](https://math.stackexchange.com/questions/766479/what-is-spectrum-for-laplacian-in-mathbbrn%5D) , the spectrum of the Laplace operator $\Delta :H^2(\mathbb{R}^2)\subset L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2)$ is in fact $\sigma(\Delta)=(-\infty,0].$ However, I was not able to find a discussion on the eigenvalues of $\Delta$.
The set of eigenvalues $\sigma\_p(\Delta)$ (also called point spectrum) is known to be contained in $\sigma(\Delta)$ and one can have $\sigma\_p(\Delta)\subsetneq \sigma(\Delta)$. Indeed, by taking the Fourier transform $\mathcal{F}:L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2)$ of the eigenvalue problem one has
$$\Delta u(x) = \lambda u(x),\;\;\forall x\in \Bbb R^2 \;\;\;\overset{\mathcal F}{\longrightarrow}\;\;\;\;-4\pi^2|\xi|^2\hat u(\xi) =\lambda \hat u(\xi), \;\;\;\forall\xi\in \Bbb R^2,$$ and this can only be satisfied by $\hat u=u=0$. ~~This means that the only eigenvalue-eigenvector pair in this setting is $(\lambda,u)=(0,0)$~~ . Also, the same argument applies when $\Delta$ is seen as $\Delta:W^{m,p}(\mathbb{R}^2)\subset L^p(\mathbb{R}^2)\to L^p(\mathbb{R}^2)$ with $p\in [1,2)$ and $\mathcal F:L^p(\mathbb{R}^2)\to L^{p^\*}(\mathbb{R}^2)$ with $1/p+1/p^\*=1$.
***Question 1***. *What happens when $p>2$ and the Fourier transform becomes distribution valued, so that the above elementary argument cannot be applied directly?*
It seems if $u\in C^2(\mathbb{R}^2)$ is in fact an eigenvalue of $\Delta$, then it cannot be in $L^p(\mathbb{R}^2)$ for any $p\in [1,2]$. Depending on the answer to Q1, this might also hold for $p>2$. In any case, it seems that the $L^p$ framework is not suitable for this problem.
***Question 2***. *On what space(s) could one define the domain of $\Delta$ to obtain non-trivial eigenvalues?*
---
*Edit.* The crossed out sentence should be replaced by: "The point spectrum $\sigma\_p(\Delta)$ is therefore empty."
| https://mathoverflow.net/users/105925 | Eigenvalues and eigenfunctions of the Laplace operator on entire plane | The point spectrum coincides with the spectrum minus 0 if $p>2n/(n-1)$ and it is empty in the remaining cases ($n$ is the dimension). This is proved in G. Talenti: "Spectrum of the Laplace operator acting in $L^p(R^n)$", Indam, Symposia Mathematica vol VII, Academic Press 1971.
| 8 | https://mathoverflow.net/users/150653 | 414932 | 169,202 |
https://mathoverflow.net/questions/414918 | 0 | Let $H$ and $K$ be Hilbert spaces and $D(T)$ a vector subspace of $H$. Let $T: D(T) \to K$ be a densely defined **antilinear** operator. Its adjoint $T^\*: D(T^\*)\to K$ is defined by the relation
$$\langle T^\*\eta, \xi\rangle = \langle T\xi,\eta\rangle$$
for all $\eta \in D(T^\*)$ and all $\xi \in D(T)$, where $D(T^\*)$ is the subspace of $K$ of all $\eta\in K$ such that $ D(T)\ni \xi \mapsto \langle \eta, T\xi\rangle$ is a bounded linear functional.
I want to prove that $T^\*$ has closed graph in a direct way.
Let me quickly revise the idea of the proof for unbounded **linear** operators. We introduce the unitary
$$V: H \oplus K \to K \oplus H: (\xi, \eta)\mapsto (\eta, -\xi).$$
Then, one shows the equality
$$G(T^\*) = V(G(T))^\perp$$
where $G(T)$ is the graph of $T$ and $G(T^\*)$ is the graph of $T^\*$. From this, it is clear that $G(T^\*)$ is a closed subset of $K \oplus H$.
Is there a way to repair this argument for antilinear operators? If $T$ is antilinear and densely defined, I can prove that $T^\*$ has closed graph using an argument with the 'adjoint' Hilbert space where the scalar multiplication is conjugated and reduce it to the linear case where I already know the result. However, I'm interested to see if we can save the above argument used to prove the linear case or do something similar and give a direct proof.
Thanks in advance for any ideas or suggestions.
| https://mathoverflow.net/users/216007 | Antilinear unbounded operator has closed graph | I'm not sure what you can expect here. Notice that if $T$ is *antilinear* then defining the "graph" as
$$ G(T) = \{ (T\xi, \xi) : \xi\in D(T) \} $$
does not give a subspace: it's not closed under (complex) scalar multiplication. The obvious way to fix this is to consider $T$ as a linear map $H\supseteq D(T)\rightarrow \overline K$, that is, use the conjugate Hilbert space construction.
However, the OP asks for other (not using the conjugate Hilbert space construction) options. If all you want to prove is that $T$ densely-defined implies $T^\*$ is closed, this can be argued directly. First, notice that $T$ being densely-defined is required to show that $T^\*$ is well-defined. By translating, we need only show that $T^\*$ is closed at $0$. That is, if $(\eta\_n)$ is a sequence in $D(T^\*)$ with $\eta\_n\rightarrow 0$ and $T^\*(\eta\_n)\rightarrow\alpha$, we wish to show that $\alpha=0$. However, then
$$ \langle \alpha,\xi \rangle = \lim\_n\langle T^\*\eta\_n,\xi\rangle=\lim\_n \langle T\xi, \eta\_n\rangle =0
\qquad (\xi\in D(T)). $$
As $D(T)$ is dense, this shows that $\alpha=0$.
---
In the comments, it's asked if we can show $D(T^\*)$ is densely-defined when $T$ is closed. I don't know how to do this if you want to avoid the conjugate Hilbert space. The usual proof has at its heart the fact that if $V\subseteq H$ is a subspace of a Hilbert space then $V^{\perp\perp}$ is the closure of $H$. The analogous result for (say, reflexive) Banach spaces uses Hahn-Banach. You want to apply this to $H\oplus\overline K$, and I'm not sure how to avoid this. (Or to do something terribly artificial, essentially sneaking in the conjugate Hilbert space.)
| 2 | https://mathoverflow.net/users/406 | 414939 | 169,204 |
https://mathoverflow.net/questions/414924 | 10 | In complex analysis, Hurwitz's theorem roughly states that, under certain conditions, if a sequence of holomorphic functions converges uniformly to a holomorphic function on compact sets, then after a while (above a certain rank ) those functions and the limit function have the same number of zeros in any open disk.
I read somewhere that the result is valid more generally on bounded convex sets , from what I remember, but no references were given.
Question. Where can I find this more general statement of Hurwitz's theorem, and where can I find some useful references in this direction? Thank you.
| https://mathoverflow.net/users/40906 | What is the most general form of Hurwitz's theorem in complex analysis? | This has nothing to do with convexity. The exact formulation is this:
Let $\Omega$ be an arbitrary region, and $f\_n\to f$ is a sequence of holomorphic functions converging uniformly on compacts in $\Omega$.
If $f\neq 0$ (this is an important condition!), and $D$ is a region,
such that $\overline{D}\subset \Omega$, and $f(z)\neq 0$ on $\partial D\_1$, then there exists $N$ (dependng on $D$) such that for $n\geq N$, $f$ and $f\_n$ have the same number of zeros
in $D$.
For the proof, you find an intermediate region $D\_1$ such that
$D\subset D\_1\subset\overline{D\_1}\subset\Omega$, and $\partial D\_1$ is piecewise smooth, and the number of zeros of $f$ in $\overline{D\_1}$ is the same as the number of zeros in $D$. Such region exists since $\overline{D}\subset\Omega$, and since the zeros
of $f$ are isolated and $f(z)\neq 0,\; z\in\partial D$.
Then the number
of zeros $f$ in $D$ is the same as in $D\_1$ and is equal to
$$\frac{1}{2\pi i}\int\_{\partial D\_1} \frac{df}{f},$$
and the number of zeros of $f\_n$ in $D\_1$ is
$$\frac{1}{2\pi i}\int\_{\partial D\_1} \frac{df\_n}{f\_n}.$$
Since $f\_n\to f$ uniformly on the compact set $K:=\overline{D\_1}\backslash D$, and $f(z)\neq 0$ on $K$, we conclude that for $n\geq N$, $f\_n$ has the same number of zeros
in $D$ and $D\_1$. Since the integrals for $f\_n^\prime/f\_n$ converge to the
integral for $f'/f$, and all these integrals are integers, we conclude that they are equal for $n\geq N$.
Refs. This is stated in full generality in the book
A. I. Markushevich, Theory of functions of a complex variable,
vol. I, Ch. IV, Sect 3 (p. 426 of the Russian original).
Special cases are in:
Ahlfors, Complex Analysis, p. 178, Theorem 2,
Titchmarsh, The theory of functions, sect. 3.45,
Marshall, Complex Analysis, Theorem 8.8,
But all these special cases have the same proof as the general theorem whose complete proof I wrote.
| 11 | https://mathoverflow.net/users/25510 | 414947 | 169,207 |
https://mathoverflow.net/questions/414922 | 3 | Let $X$ and $Y$ be Tychonoff (i.e. completely regular Hausdorff) topological spaces and let $\varphi:X\to Y$ be a continuous surjection that also has a property that $\operatorname{int}\overline{\varphi(U)}\ne\varnothing$, for any open nonempty $U\subset X$ (this property is called skeletal, weakly open, almost open etc in various sources; equivalently, the preimage of any nowhere set is nowhere dense).
Let us consider the topology $\tau$ on $Y$ which is the strongest Tychonoff topology with respect to which $\varphi$ is continuous (this property for $\varphi$ as a map into $(Y,\tau)$ is called $\mathbb{R}$-quotient, which may be viewed as a quotient map in the category of Tychonoff spaces).
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> Is $\varphi$ still skeletal as a map from $X$ into $(Y,\tau)$?
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This is true is $X$ is locally compact, so the counterexamples should be looked for outside of this class.
| https://mathoverflow.net/users/53155 | Is a certain property of a continuous map preserved under a modification of the topology on the target space? | The answer to this question is negative. A suitable counterexample can be constructed as follows.
Let $Y=\mathbb R$ be the real line with the standard Euclidean topology.
Let $\mathbb Q$ be the subspace of rational numbers in $\mathbb R$. Write $\mathbb R\setminus \mathbb Q$ as the union $\bigcup\_{q\in \mathbb Q}X\_q$ of pairwise disjoint dense sets in $\mathbb R$. Let $$X=\mathbb Q\oplus\bigoplus\_{q\in \mathbb Q}(\{q\}\cup X\_q)$$be the topological sum of the spaces $\mathbb Q$ and $\{q\}\cup X\_q$ for $q\in\mathbb Q$.
Let $\varphi:X\to Y$ be the natural projection. It is clear that the spaces $X,Y$ are separable, metrizable (and hence Tychonoff) and the function $\varphi:X\to Y$ is skeletal.
On the other hand, in the $\mathbb R$-quotient topology $\tau$ on $Y=\mathbb R$, the set $\mathbb Q$ is closed and nowhere dense in $(Y,\tau)$, witnessing that the function $\varphi:X\to(Y,\tau)$ is not skeletal.
To show that $\mathbb Q$ is closed in $(Y,\tau)$, choose any point $y\in\mathbb R\setminus \mathbb Q$. Find $q\in\mathbb Q$ such that $y\in X\_q$. Consider the function $f\_q:Y\to\mathbb R$ defined by
$f\_q(x)=|x-q|$ if $x\in X\_q$ and $f\_q(x)=0$, otherwise.
Observe that the composition $f\_q\circ \varphi:X\to\mathbb R$ is continuous, which implies that the set $X\_q=\{x\in Y:f\_q(x)>0\}$ is $\tau$-open, contain $y$, and does not intersect $\mathbb Q$.
This completes the proof of the closedness of $\mathbb Q$.
Assuming that the closed set $\mathbb Q$ is not nowhere dense in $(Y,\tau)$, we can find a nonempty open set $U\subseteq\mathbb Q$. Choose any point $q\in U$ and observe that the preimage $\varphi^{-1}(U)$ is an open set in $X$ containing the point $q\in \{q\}\cup X\_q$. Since $\{q\}$ is nowhere dense in $X\_q$, the set $\varphi^{-1}(U)$ has nonempty intersection with the set $X\_q$ and then $\emptyset \ne \varphi[X\_q\cap \varphi^{-1}(U)]\subseteq U\cap X\_q\subseteq U\setminus\mathbb Q$, which contradicts the choice of $U\subseteq\mathbb Q$.
| 2 | https://mathoverflow.net/users/61536 | 414954 | 169,210 |
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