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https://mathoverflow.net/questions/438541 | 1 | We say that a bijection $\varphi:\omega\to\omega$ *parity-separates* $a\neq b\in \omega$ if $\varphi(a)$ is even and $\varphi(b)$ is odd, or vice versa.
Is there a finite set $\Phi$ of bijections such that for all $a\neq b\in\omega$ there is $\varphi\in\Phi$ such that $\varphi$ parity-separates $a,b$? If yes, how small can $|\Phi|$ be?
| https://mathoverflow.net/users/8628 | Can $\omega$ be parity-separated with finitely many bijections? | No, because if you have $n$ functions, then the number of possible parity patterns to be exhibited by a number with respect to them is $2^n$. So by the pigeon-hole principle there must be infinitely many numbers with the same pattern. So it doesn't separate those numbers.
| 2 | https://mathoverflow.net/users/1946 | 438542 | 177,135 |
https://mathoverflow.net/questions/438330 | 3 | Given symmetric monoidal $\infty$-categories $A, B, C$ and lax symmetric monoidal maps $F:A\to C$, $G:B\to C$, I am curious if the pullback (when I say pullback here I will really mean homotopy pullback) of $F$ along $G$ is necessarily lax symmetric monoidal in a natural way? The pullback is necessarily an $\infty$-operad still, but it doesn't seem immediately clear that one can still get cocartesian lifts of all morphisms, not just inert ones.
I am sort of curious if there is an $\infty$-cosmos (in the sense of Riehl-Verity) of symmetric monoidal $\infty$-categories with lax symmetric monoidal maps. If there were, one could just define it as cosmologically embedded in $\infty$-operads, full on the underlying simplicial categories. This would work if symmetric monoidal $\infty$-categories were stable under pullback along isofibrations in $\infty$-operads (which should be equivalent to the above question). I suspect this would be false, but I can't think of a counterexample.
| https://mathoverflow.net/users/489806 | Can there be a cospan of symmetric monoidal $\infty$-categories whose maps are lax symmetric monoidal but the pullback is not symmetric monoidal? | Oops, this is actually not hard, just using 1-categories. Explicitly, take two maps from the terminal category to Ab, one landing in $\mathbb{Z}$, one landing in 0, both are lax symmetric monoidal. The pullback (after replacing the first map by an equivalent isofibration but we can mostly ignore this technicality) is empty so not a symmetric monoidal $\infty$-category (where it would be an ordinary category anyways). So indeed, symmetric monoidal $\infty$-categories with lax symmetric monoidal functors do not form an $\infty$-cosmos.
Though, if we are pulling back $(G,F):C\to A\times B$ along a two-sided fibration $(p\_1,p\_0):D\to A\times B$ from $A$ to $B$ (which is to say, $p\_1$ is cocartesian, $p\_0$ is cartesian and lifts are compatible), where $p\_1,p\_0,$ and $F$ are symmetric monoidal, while $G$ is lax symmetric monoidal, this should work. This case can be used to show that cyclotomic spectra are symmetric monoidal, for example.
| 2 | https://mathoverflow.net/users/489806 | 438548 | 177,136 |
https://mathoverflow.net/questions/438522 | 0 | Recall that an operator $T:X\rightarrow Y$ is called absolutely summing if there exists a constant $C>0$ such that
$$\sum\_{i=1}^{n}\|Tx\_{i}\|\leq C \sup\_{x^{\*}\in B\_{X^{\*}}}\sum\_{i=1}^{n}|\langle x^{\*},x\_{i}\rangle|,$$ for all finite families $(x\_{i})\_{i=1}^{n}$ in $X$. The least $C$ for which the above inequality always holds is denoted by $\pi(T)$. The set of all absolutely summing operators from $X$ to $Y$ is denoted by $\Pi(X,Y)$.
An operator $T:X\rightarrow Y$ is said to be nuclear provided that there are sequences $(x^{\*}\_{n})\_{n}$ in $X^{\*}$ and $(y\_{n})\_{n}$ in $Y$ such that $\sum\limits\_{n}\|x^{\*}\_{n}\|\|y\_{n}\|<\infty$ and $Tx=\sum\limits\_{n}x^{\*}\_{n}(x)y\_{n}$ for $x\in X$. The nuclear norm $\nu(T)$ is defined to be the infimum of $\sum\limits\_{n}\|x^{\*}\_{n}\|\|y\_{n}\|<\infty$ over all possible such representations. The space of all nuclear operators from $X$ to $Y$ is denoted by $\mathcal{N}(X,Y)$. It is known that every nuclear operator is absolutely summing.
Question 1. For $1<p,q<\infty$, when is $\Pi(l\_{p},l\_{q})$ reflexive ?
Question 2. For $1<p,q<\infty$, $\Pi(l\_{p},l\_{q})=\mathcal{N}(l\_{p},l\_{q})$ ?
Question 3. For $1<p<q<\infty$, is the inclusion map $i\_{p,q}$ from $l\_{p}$ to $l\_{q}$ absolutely summing ?
Thank you !
| https://mathoverflow.net/users/41619 | Absolutely summing operators from $l_{p}$ to $l_{q}$ | As for Question 3, note that an absolutely summing operator is completely continuous, i.e., maps weakly null sequences to null sequences, a property not shared by $i\_{p,q}$ if $p>1$. Therefore this question has a negative answer.
| 3 | https://mathoverflow.net/users/127871 | 438550 | 177,137 |
https://mathoverflow.net/questions/438549 | 6 | The Langlands-Shahidi method says that the $L$-functions of automorphic representations appear in the constant terms of Eisenstein series. Since those Eisenstein series have analytic continuation and functional equation, the $L$-function also inherits those nice properties.
Most expositions of the Langlands-Shahidi method are phrased in the adelic language, which I am not too comfortable with. So I would like to understand it in a more classical language, without using adeles.
**My question is:** Suppose you have a weight $2$ eigenform $f \in S\_2(\Gamma\_0(N))$. Consider the $L$-function $L(f,s)$ attached to $f$. Is there a reference that explains the Langlands-Shahidi method for $L(f,s)$ without referring to adeles? Namely, *in classical language*, what is the Eisenstein series whose constant term contains the $L$-function $L(f,s)$? How can the nice analytic properties of $L(f,s)$ be read off from the analytic properties of the Eisenstein series?
| https://mathoverflow.net/users/394740 | Langlands-Shahidi method in classical language | Well, as part of a constant term of an Eisenstein series, it would appear in the constant term of an Eisenstein series attached to the 2,1 parabolic (or 1,2...) in GL3, with the cuspform on the GL2 factor of the Levi component. My relatively recent book(s) with Cambridge U Press... also available on-line... treat some aspects of this, in both classical and adelic terms.
But also you should look at Langlands' SLN 544, and Goldfeld/Goldfeld-Hundley's books on GLn...
| 7 | https://mathoverflow.net/users/15629 | 438557 | 177,139 |
https://mathoverflow.net/questions/438246 | 7 | In quantum information, much can be done with the averaging formula
$$
\int\_{\mathbb{C}P^{n-1}} (zz^\*)^{\otimes t} dz = {n + t -1 \choose t}^{-1} \operatorname{\Pi}\_{\mathrm{Sym}^t}$$
Here the integral gives an average over all unit vectors $z \in \mathbb{C}^n$ and $\operatorname{\Pi}\_{\mathrm{Sym}^t}$ denotes the orthogonal projector onto the totally symmetric subspace of $(\mathbb{C}^n)^{\otimes t}$. This result appears as Lemma 1 in [the following paper](https://arxiv.org/abs/quant-ph/0604049).
Does this result have an analog for the average over all unit vectors $x \in \mathbb{R}^n$?
$$
\int\_{\mathbb{R}P^{n-1}}
(xx^\top)^{\otimes t} dx = {?}
$$
By direct computations, I see that the first formula does not hold when $\mathbb{C}P$ is replaced by
$\mathbb{R}P$ for $t=2$.
Having a formula for
$\int\_{\mathbb{R}P^{n-1}} (xx^\top)^{\otimes t} dx$ (even for $t = 2$) would help with a problem I'm working on, but I have not been able to find a reference for this or discover a formula myself.
| https://mathoverflow.net/users/315386 | Average of polynomials over the real sphere | Here is a solution for the case $t=2$.
Let $$T= \int\_{\mathbb{R}P^{n-1}}
(xx^\top)\otimes (xx^\top) dx $$
(I'm gonna assume that the integral is normalized so that the measure of $\mathbb{R}P^{n-1}$ is $1$).
As in the complex case, it's easy to see that $T$ has image in the symmetric subspace $\text{Sym}^2 \mathbb{R}^n$ and vanishes on its complement $\bigwedge^2\mathbb{R}^n$. Hence it's enough to understand the action of $T$ on $\text{Sym}^2 \mathbb{R}^n$.
Let $\phi : \text{Sym}^2 \mathbb{R}^n \to \mathbb{R}$ be the linear map defined by $\phi(x\otimes y)=\langle x,y\rangle$ and let $W=\ker \phi$. As a representation of $SO(n)$, $\text{Sym}^2 \mathbb{R}^n$ decomposes into irreducibles as
$$\text{Sym}^2 \mathbb{R}^n = W \oplus \mathbb{R}\omega$$
where $\omega=\sum\_{i=1}^n e\_i\otimes e\_i \in \text{Sym}^2 \mathbb{R}^n$ for any orthogonal basis $(e\_i)\_{i=1}^n$ of $\mathbb{R}^n$ (and the two summands are still irreducible over $\mathbb{C}$; see <https://math.stackexchange.com/questions/4413836/decomposition-of-symmetric-powers-of-the-standard-representation-of-son>).
It's easy to see that $T$ commutes with the action of $SO(n)$ on $\text{Sym}^2 \mathbb{R}^n$, so by Schur's lemma it must act by a scalar $\lambda$ on $W$ and by another scalar $\mu$ on $\mathbb{R}\omega$.
To determine $\mu$, we can calculate explicitly
$$T\omega = \int\_{\mathbb{R}P^{n-1}}
\sum\_{i=1}^n \langle x, e\_i\rangle\>^2 x\otimes x \ dx = \int\_{\mathbb{R}P^{n-1}}
x\otimes x \ dx$$
which is sent under $\phi$ to
$$\int\_{\mathbb{R}P^{n-1}}
1 \ dx = 1$$
On the other hand, note that $\phi(\omega)=n$. Hence we must have $\mu =\frac{1}{n}$.
To determine $\lambda$, note that $xx^\top\otimes xx^\top$ has trace $1$ for every unit vector $x\in \mathbb{R}^n$, so $T$ itself has trace $1$. On the other hand the trace of $T$ is given by
$$\text{Tr}(T)=\lambda \cdot \dim(W) + \mu = \lambda \cdot \frac{n^2+n-2}{2} + \frac{1}{n}=\lambda\cdot\frac{(n-1)(n+2)}{2} + \frac{1}{n}.$$
Solving, for $\lambda$, we find that $\lambda = \frac{2}{n(n+2)}$.
This gives a complete description of the averaged operator $T$ in the case $t=2$. With more work, I think that the same ideas can be extended to arbitrary $t$, using the decomposition of $\text{Sym}^t \mathbb{R}^n$ into $SO(n)$-irreducibles described in the MSE question linked above.
| 4 | https://mathoverflow.net/users/160416 | 438563 | 177,141 |
https://mathoverflow.net/questions/438333 | 19 | Freyd's theorem in classical category theory says that any small category $\mathcal{C}$ admitting products indexed by the set $\mathcal{C}\_1$ of all its arrows is a preorder. I'm interested in whether this is also true for for precategories/univalent categories à lá HoTT Book (§II.9), provided we assume LEM.
I'll refer to the notation from Andrej Bauer's [very clear presentation](https://mathoverflow.net/a/336610/475802) of the proof. In particular, $f, g : a \to b$ are morphisms with $f \ne g$, and $z$ is the large product $\prod\_{\mathrm{Arr}(\mathcal{C})} b$.
In HoTT (and, potentially, [others where equality of objects is problematic](https://nforum.ncatlab.org/discussion/2449/complete-small-category/?Focus=25906#Comment_25906)), there is no general way of embedding $\mathcal{C}(a,b) \to \mathcal{C}\_1$, so while it is possible to construct the map the map $i : \{0,1\}^{\mathcal{C}\_1} \to \mathcal{C}\_1$ from the linked answer, showing this map is an embedding is impossible. If we instead consider the assignment $c \mapsto \langle r\_{c(i)}\rangle\_i $ from the linked answer to define a function $i : \{0,1\}^{\mathcal{C}\_1} \to \mathcal{C}(a, z)$ we *do* get an embedding.
An example of $\mathcal{C}, a, b$ for which the evident map $\mathcal{C}(a,b) \to \mathcal{C}\_1$ fails to be an embedding is obtained by taking $\mathcal{C} = \mathbf{Sets}$, and $a, b = \{0,1\}$. Then we have $(a, b, \mathrm{id}) = (a, b, \mathrm{not})$ in the groupoid of arrows, but the identity function is not the negation function. Of course, the category of sets fails to be large-complete for other reasons, but the point stands: we can not embed a $\mathrm{Hom}$-set into the space of all arrows unless we know *a priori* that $\mathcal{C}\_1$ is a set.
If $\mathcal{C}\_0$ admits a surjection $c : X \twoheadrightarrow \mathcal{C}\_0$ from a set, then we can consider a modified definition of $\mathcal{C}\_1$ which allows the traditional proof to be repaired: Rather than the sum $\sum\_{(a,b) : \mathcal{C}\_0 \times \mathcal{C}\_0} \mathcal{C}(a,b)$, we define $\mathcal{C}\_1'$ by summing over $X \times X$ instead, and use *this* type to construct $i$. Since we're aiming for a contradiction, we can assume we've been given $a', z' : X$ with identifications $p\_a : c(a') = a$ (resp. $p\_z : c(z') = z$); The composite
$$
\{0,1\}^{\mathcal{C}'\_1} \hookrightarrow \mathcal{C}(a,z) \simeq C(c(a'), c(z')) \hookrightarrow\mathcal{C}'\_1
$$
is verified to be an embedding. Replacing $\mathcal{C}\_0$ by $X$ comes in at the last step, where we're given $(a, b, i(f)) = (a, b, i(g)) : \mathcal{C}\_1'$ and want to conclude $i(f) = i(g)$ (thus $f = g$). Since $(a, b)$ now inhabits the set $X \times X$, this is allowed.
This modified proof establishes that Freyd's theorem holds in e.g. the simplicial sets model of HoTT, which satisfies excluded middle *and* sets cover. However, I'm a bit surprised that we had to assume sets cover! It's true that sometimes "small" means "small, strict" (e.g.: "the category of small categories is monadic over that of graphs"), but the statement of Freyd's theorem doesn't, at a glance, mention anything that disrespects equivalence: completeness, being a preorder, and the arrow category are all invariant.
1. Is Freyd's theorem *really* one of the cases where small means small *and strict*, or is there an alternative proof that does not assume $\mathcal{C}(a,z) \hookrightarrow \mathcal{C}\_1$? Other proofs of Freyd's theorem in the literature (e.g. in Shulman's [*Set theory for category theory*](https://arxiv.org/abs/0810.1279), §2.1; Freyd's original in *Abelian Categories*, §3.D) are all expressed in a set-theoretic framework, and follow the same strategy as Andrej's. The generalisation to the internal logic of a Grothendieck topos (e.g. [here](https://mathoverflow.net/questions/43433/small-complete-categories-in-a-grothendieck-topos?noredirect=1&lq=1)) doesn't help either, since they reduce to internal categories in $\mathbf{Sets}$: small + strict.
2. Assuming that the theorem isn't salvageable, can we have a complete small category that's not a preorder? This would be a very interesting gadget!
I think it can't be any of the usual suspects which become "large-complete" under "impredicative Set", since for any such $\mathcal{C}$ and endofunctor $F : \mathcal{C} \to \mathcal{C}$, the category of $F$-algebras inherits limits from $\mathcal{C}$, and so also has "limits its own size" --- in particular for the identity $\mathrm{Id} : F\mathrm{-Alg} \to F\mathrm{-Alg}$. By Lambek's theorem, this means every endofunctor on $\mathcal{C}$ has a fixed point.
| https://mathoverflow.net/users/475802 | Small complete categories in HoTT+LEM | Here is an alternative proof based on Russell's paradox rather than cardinality that doesn't require sets cover, although I do need to assume that hom sets are 0 truncated. The rough outline is to modify Gylterud's definition of the cumulative hierarchy by limiting it to sets that can be constructed from $\mathcal{C}$ to make it a small type, and then follow the usual proof of Russell's paradox.
The first step is the following lemma.
**Lemma** There is a small type $X : \mathcal{U}$ together with an embedding $\iota : X \hookrightarrow \mathcal{U}$ such that every type that can be expressed as a set quotient of a subtype of a type of the form $\mathcal{C}(a, b)$ for $a, b : \mathcal{C}\_0$, is equivalent to one in the image of $\iota$.
**Proof** For each $a, b : \mathcal{C}\_0$ we can express every subobject $Z \hookrightarrow \mathcal{C}(a, b)$ of $\mathcal{C}(a, b)$ as $\sum\_{f : \mathcal{C}(a, b)} \chi(f) = 1$ for some map $\chi: \mathcal{C}(a, b) \to 2$. Furthermore, every set quotient of $Z$ is the quotient of some equivalence relation $R : Z \times Z \to 2$. This gives a type $Y$ together with a map $Y \to \mathcal{U}$ whose image contains every set quotient of every subtype of $\mathcal{C}(a, b)$ for every $a, b : \mathcal{C}\_0$. By type theoretic replacement (Theorem 4.6 in Rijke, [*The join construction*](https://arxiv.org/abs/1701.07538)), we can take the image of this map to be small type, which is the required $X$.$\quad \square$
Next we construct a modified version of the Aczel cumulative hierarchy $M$ as the $W$-type with constructors $X$, and the arity of $x : X$ is just $\iota(x)$. Note that since $X$ is a small type, and all the arities are small, so is $M$, unlike the usual Aczel hierarchy. We then construct a retract of $M$ corresponding to the construction of the HIT cumulative hierarchy in Gylteud, [*From multisets to sets in homotopy type theory*](https://doi.org/10.1017/jsl.2017.84). Namely an element of $M$ of the form $\sup(x, \alpha)$ for $x : X$ and $\alpha : \iota(x) \to M$ is a *hereditary embedding* if $\alpha$ is an embedding and $\alpha(y)$ is a hereditary embedding for all $y : \iota(x)$. Write $V$ for the subtype of $M$ consisting of hereditary embeddings. Since $V$ is a subtype of $M$, it is also small. Following the same argument that Gylterud used for the cumulative hierarchy (section 5 of paper above), $V$ is necessarily 0-truncated. We define membership $\in$ the same as Gylterud does.
Now suppose that we are given $f \neq g : \mathcal{C}(a, b)$. Define $c$ to be the product $\prod\_V b$. We have injections $V \hookrightarrow 2^V \hookrightarrow \mathcal{C}(a, c)$. Since $V$ and $\mathcal{C}(a, c)$ are both sets, this gives us a surjection $s : \mathcal{C}(a, c) \twoheadrightarrow V$. We define a subtype $Z$ of $V$ as $Z := \sum\_{z : V} z \notin z$. Pulling back along $s$ gives us a subtype $Z'$ of $\mathcal{C}(a, c)$ together with a surjection $Z' \twoheadrightarrow Z$. Since $Z'$ and $Z$ are both sets, $Z$ must be a set quotient of $Z'$. Hence $Z$ is equivalent to $\iota(x)$ for some $x$. Hence we have $z\_0 : V$ such that for all $z \in z\_0 \;\simeq\; z \notin z$ for all $z : V$, which gives a contradiction by Russell's paradox.
| 6 | https://mathoverflow.net/users/30790 | 438568 | 177,143 |
https://mathoverflow.net/questions/438555 | 3 | Let $p(u,x)=\frac{1}{(4\pi u)^{q/2}}e^{-|x|^2/(4u)},u>0,x \in \mathbb{R}^q.$
1. Prove that for $r\geq 0,c>1$ there exists $C>0$ (depending on $r,c$) such that $$\forall x \in \mathbb{R}^q,u>0,\frac{|x|^{2r}}{u^r}p(u,x) \leq Cp(cu,x).$$
2. Deduce that for $n \in \mathbb{N}^q,k \in \mathbb{N},$ there exists $C'>0$ such that $|\partial\_x^n\partial\_u^k p(u,x)| \leq C' u^{-|n|/2-k}p(cu,x)$ where $\partial^n\_x=\partial x\_1^{n\_1}...\partial x\_q^{n\_q}$
1 follows from $\frac{|x|^{2}}{4ur}(1-\frac{1}{c}) \leq e^{\frac{|x|^2}{4ur}(1-1/c)}$ and we choose $C=c^{q/2} 4^r(r+1)^r/(1-1/c)^r$
How to deduce 2 (for simplicity we can take $q=1)$? *Is it possible to prove the inequality using induction*? If so, how?
| https://mathoverflow.net/users/138491 | Inequality: multivariate normal distribition | $\newcommand\p\partial$By induction,
$$\p\_u^k p(u,x)=\frac1{u^k}\,P\_k\Big(\frac{|x|^2}u\Big)p(u,x)
=\frac K{u^{k+q/2}}\,P\_k\Big(\frac{|x|^2}u\Big)e^{-|x|^2/(4u)}, \tag{1}\label{1}$$
where $K$ is a real number not depending on $u$ or $x$, and $P\_k(z)$ is a polynomial (with coefficients not depending on $u$ or $x$).
Next, for any polynomial $P$ (in $q$ variables) and any $j=1,\dots,q$, there is another polynomial $Q\_j$ (in $q$ variables) such that
$$\p\_{x\_j} \Big(P\Big(\frac{x}{u^{1/2}}\Big)e^{-|x|^2/(4u)}\Big)
=\frac1{u^{1/2}}\,Q\_j\Big(\frac{x}{u^{1/2}}\Big)e^{-|x|^2/(4u)}.$$
So, in view of \eqref{1},for some polynomial $R$ (in $q$ variables) and all real $u>0$,
$$|\p\_x^n\p\_u^k p(u,x)|
=\frac1{u^{|n|/2+k+q/2}}\,\Big|R\Big(\frac{x}{u^{1/2}}\Big)\Big|e^{-|x|^2/(4u)} \\
\le \frac C{u^{|n|/2+k+q/2}}\,e^{-|x|^2/(4cu)}
=\frac{C\_1}{u^{k+|n|/2}}\,p(cu,x),$$
where $C$ and $C\_1$ are positive real numbers not depending on $u$ or $x$. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 438569 | 177,144 |
https://mathoverflow.net/questions/438574 | -2 | A Dirichlet-$L$ function is typically defined by its series, and its Euler product is a consequence of the definition. Here my approach is the other way around.
I define the function
$$
L\_4^\*(s) = \prod\_{p\in P} \Bigg(1 - \frac{\chi(p)}{p^{s}}\Bigg)^{-1}
$$
with $P=\{p\_1,p\_2,\dots\}$ the set of prime numbers but ordered differently so that the sequence $\chi(p\_1),\chi(p\_2),\dots$ is perfectly alternating. So
if $p\_{k} \equiv 1 \bmod{4}$ then $p\_{k+1}\equiv -1 \bmod{4}$. In this case, $p\_{k+1}$ would be the smallest prime $> p\_k$, equal to $-1$ modulo $4$, and not already used previously in the product.
Let's pretend that $L\_4^\*(s)$ is different from the standard function $L\_4$ with the primes ordered by value in their natural order. After all no one knows, but there is considerable empirical evidence that they are identical. Is $L\_4^\*$ analytical? Even if different from $L\_4$, is it any easier to prove that the $L\_4^\*$ product converges if $\Re(s)>\frac{1}{2}$?
In the end, does the above product define a nice $L$-function worth studying? To put it differently, would proving convergence for $L\_4^\*$ be a major milestone, or a useless result because $L\_4^\*$ does not belong to the category of "interesting functions" under GRH. The reason I am asking is because I am interested in finding a Dirichlet-$L$ function easier to study (the convergence) than the classic ones, yet featuring nice properties.
| https://mathoverflow.net/users/140356 | Convergence of scrambled product for Dirichlet-$L$ function with modulo 4 character | The usual Euler product converges absolutely in $\{s:\Re s>1\}$, hence no matter how you permute its factors, you get the same function in $\{s:\Re s>1\}$. This function has a unique analytic continuation to $\mathbb{C}$, which is then $L(s,\chi)$. In short, the new function you are trying to define will either not be analytic in $\mathbb{C}$, or it will not be new.
| 7 | https://mathoverflow.net/users/11919 | 438578 | 177,146 |
https://mathoverflow.net/questions/438579 | 5 | Recall that
$\mathfrak{p}=\min\{|F|: F$ is a subfamily of $[\omega]^{\omega}$ with the *sfip* which has no infinite pseudo-intersection $\}$.
The cardinal $\mathfrak{q}\_0$ defined as the smallest cardinality of a subset of $\mathbf{R}$ which is not a $Q$-space.
**Q1.** Is it true that $\mathit{MA}(\omega\_1)$ iff $\omega\_1<\mathfrak{p}$ ?
**Q2.** Is there a model of set theory in which $\omega\_1=\mathfrak{q}\_0$ and a non-$\mathit{MA}(\omega\_1)$ ?
| https://mathoverflow.net/users/112417 | Is it true that $\mathit{MA}(\omega_1)$ iff $\omega_1<\mathfrak{p}$? | Q1: No, see [Between Martin's Axiom and Souslin's Hypothesis](https://eudml.org/doc/211011) by Kunen and Tall. Note: Bell proved in [The combinatorial principle $P(\mathfrak{c})$](https://eudml.org/doc/211293) that $\mathfrak{p}>\aleph\_1$ is equivalent to $\mathsf{MA}(\aleph\_1)$ for $\sigma$-centered partial orders.
Q2: $\mathsf{MA}(\aleph\_1)$, even $\mathfrak{p}>\aleph\_1$, implies $\mathfrak{q\_0}>\aleph\_1$, see page 162 of [Internal Cohen Extensions](https://doi.org/10.1016/0003-4843(70)90009-4) by Martin and Solovay. The principle $\mathsf{S}\_\aleph$ is proved on page 154, using a $\sigma$-centered partial order.
So, every model with $\mathfrak{q}\_0=\aleph\_1$ satisfies $\neg\mathsf{MA}(\aleph\_1)$ as well, for example every model of $\mathsf{CH}$ will do.
| 9 | https://mathoverflow.net/users/5903 | 438583 | 177,148 |
https://mathoverflow.net/questions/438577 | 4 | Let $U,V\in\mathbb{R}^{4\times n}$ such that $UU^T=VV^T=I$, and $A\in\mathbb{R}^{n\times n}$ be an Hermitian matrix.
Is it true that
$$\sqrt{\lambda\_{\text{max}}\left(\left(UAU^T-VAV^T\right)^2\right)}\leq \lambda\_{\text{max}}\left((U-V)A(U-V)^T\right)$$
?
| https://mathoverflow.net/users/103133 | largest eigenvalue of the difference between two quadratic forms | This is false for the following reason. Say that $n=2$. Choose
$$A=\begin{pmatrix} a & b \\ b & 0 \end{pmatrix},\quad U=I\_2,\quad V=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$
Then $(U-V)A(U-V)^T=0\_2$, while
$$UAU^T-VAV^T=\begin{pmatrix} 0 & b \\ b & 0 \end{pmatrix},$$
whose square is non-zero if $b\ne0$.
| 4 | https://mathoverflow.net/users/8799 | 438585 | 177,150 |
https://mathoverflow.net/questions/438536 | 2 | Let $A$ be a PSD random matrix, which has $0$ as one of its eigenvalues. The second smallest eigenvalue of the expectation of $A$ writes as $\lambda\_2(\mathbb{E}(A))>0$.
Why the following statement holds?
If $\|A-\mathbb{E}(A)\|<\lambda\_2(\mathbb{E}(A))$, then $\lambda\_2(A)>0.$
Note that $\|\cdot\|$ denotes the operator norm.
| https://mathoverflow.net/users/494410 | Deviation of random matrix from its expectation informs the positiveness of its second smallest eigenvalue | The random-matrix connection is a bit of a red herring: Since the weight of $A$ in the statistical average can be arbitrarily small, we might as well replace $\mathbb{E}(A)$ by an arbitrary PSD matrix $B$ with $\lambda\_2(B)>0$.
The statement in the OP
>
> If $\|A-B\|<\lambda\_2(B)$, then $\lambda\_2(A)>0,$
>
>
>
is a consequence of the "eigenvalue stability" inequality
$$|\lambda\_i(A)-\lambda\_i(B)|\leq \|A-B\|,$$
which is proven, for example, in [Tao's notes](https://terrytao.wordpress.com/2010/01/12/254a-notes-3a-eigenvalues-and-sums-of-hermitian-matrices/) (equation 13).
| 1 | https://mathoverflow.net/users/11260 | 438593 | 177,152 |
https://mathoverflow.net/questions/438573 | 0 | It is one of the concepts used in "ON THE REPRESENTATION OF CONTINUOUS FUNCTIONS OF SEVERAL VARIABLES AS SUPERPOSITIONS OF CONTINUOUS FUNCTIONS OF A SMALLER NUMBER OF VARIABLES", in the second paragraph, highlighted with black background,
[as shown in this image](https://i.stack.imgur.com/7vbus.png).
| https://mathoverflow.net/users/497862 | What does "a universal tree" mean? | In the beginning of Section 6 on p. 318 of Menger's Kurventheorie, referred to in [Kolmogorov's paper](https://cs.uwaterloo.ca/%7Ey328yu/classics/Kolmogorov56.pdf) on the page whose image you linked in your post, we find this:
>
> Wir bezeichen eine Baumkurve $B$ als einen Universalbaum bzw. als Universalbaum $n$-ter Ordnung, wenn $B$ zu jedem vorgelegten Baum bzw. zu jedem vorgelegten Baum $n$-ter Ordnung eine homöomorphe Menge als Teilmenge enthält.
>
>
>
which apparently means
>
> We refer to a tree curve $B$ as a universal tree, or a universal tree of the $n$th order, if $B$ contains as a subset a homeomorphic image of any given tree, or any given tree of the $n$th order.
>
>
>
| 3 | https://mathoverflow.net/users/36721 | 438597 | 177,153 |
https://mathoverflow.net/questions/438616 | 5 | $\newcommand{\g}{\mathfrak{g}}$Let $G$ be a reductive algebraic group over field $k = \overline{k}$ and consider the characteristic polynomial $\g \to \g/\!/G := \operatorname{Spec} (k[\g]^G)$ induced by the adjoint action. Assume $|W|$ is invertible in $k$, so Chevalley-Todd-Shepard implies the quotient is affine space.
I expect the following basic geometric properties to generalize from the $\operatorname{GL}\_n$ case.
1. The $k$-points of $\g/\!/G$ are in bijection with Zariski-closed $k$-orbits.
2. Every $k$-rational orbit has a unique closed orbit in its closure, and the map on $k$-points is this identification.
3. The locus of regular elements (those $X \in \g$ with $\dim G\_X = \operatorname{rank} G$, as small as possible) is exactly the smooth locus of the morphism.
4. The semisimple $X \in \g$ (in the sense of additive Jordan-Chevalley) are exactly those with closed $k$-orbits.
Do these hold in the setting above? If so, how much more general are they? The first two at least seem like general facts about actions on an affine $k$-scheme.
**Edit**: After some more reading and Ben's helpful answer, I was able to resolve most of this, though I haven't found anything that ensures **surjectivity of the quotient morphism** in the positive characteristic case: in characteristic zero the argument uses that for any $A^G$-algebra $S$, we have the identity $S = (S \otimes\_{A^G} A)^G$. This doesn't work in positive characteristic (and *GIT* never claims that $(\operatorname{Spec} A)/\!/G$ is a "universal categorical quotient" unless char 0).
I'll leave this open for a bit in case Ben or anyone else can chip in with a substitute argument or counterexample.
| https://mathoverflow.net/users/124185 | Geometric properties of the adjoint action of a reductive group | Yes, all these things are true, and 1. and 2. hold for any action of a reductive group $G$ on an affine variety $X$. The key point is that if I have two closed orbits $O\_1,O\_2$, then the functions on their disjoint union is a quotient of functions on $X$ (because $X$ is affine), so there's a function on $X$ that is 1 on $O\_1$ and 0 on $O\_2$. Since the group is reductive, you can assume this function is $G$ invariant by projecting it to the trivial rep.
| 8 | https://mathoverflow.net/users/66 | 438617 | 177,157 |
https://mathoverflow.net/questions/438621 | 2 | I have a sum of $n$ i.i.d random variables $X\_i$ such that $E[X\_i] = 0$,$\mathrm{E}[|X\_i|^{1 + \delta}]$ exists for some $0 < \delta < 1$ but $\mathrm{E}[|X\_i|^{1 + \delta+ \epsilon}]$ does not exists for any $\epsilon > 0$.
I would like to ask if it is possible to provide a lower bound for $P(|\sum\_{i=1}^n X\_i| > A\sqrt{n})$, showing that this term converges to $1$ when $n$ approaches infinity.
I am aware that when $X\_i$ obeys the tail balance condition, we would have $\frac{X\_1 + ... + X\_n}{a\_n}$ converges to a $\alpha-$stable distribution for proper scaling factor $a\_n$. So I am mostly interested in the case where the tail balance condition does not hold.
| https://mathoverflow.net/users/477986 | Lower bound on sum of independent heavy-tailed random variables | Certainly. All you need is $EX^2=+\infty$. Then the characteristic function $f\_X(t)$ satisfies $\lim\_{t\to 0}\frac{1-|f(t)|}{t^2}=+\infty$, so for every finite interval $I\subset \mathbb R$, we have $\lim\_{n\to\infty}\int\_I|f(t/\sqrt n)|^n\,dt=0$ and the usual Fourier trickery finishes the story. Of course, the efficient lower bound would require some quantitative information about the speed of divergence of the second moment: if you have essentially the uniform distribution on $[-1,1]$ combined with some large value dust of extremely low probability (which can still blow up any moment you like), then you'll have to wait quite long time before the spreading effect kicks in.
Edit: Just to explain the "Fourier trickery". The story is that if you have some \*smooth non-negative function $g$ with compact support that is identically $1$ on $[-A,A]$, then $P(|X|\le A)\le Eg(X)=c\int\_{\mathbb R}\widehat g(t)f\_X(-t)$, so if you know that $\int|f\_X|$ is small over a long interval where $\widehat g$ is still noticeable, then you can conclude that this probability is small and, therefore, $P(|X|>A)$ is close to $1$.
| 5 | https://mathoverflow.net/users/1131 | 438625 | 177,161 |
https://mathoverflow.net/questions/438623 | 7 | Let $\mathcal{M}\mathrm{fld}\_n$ denote the $\infty$-category of topological manifolds (without boundary) and embeddings; more precisely, it is the homotopy coherent nerve of the simplicial category whose objects are topological manifolds, and whose hom-spaces are given by $\operatorname{Sing}\operatorname{Emb}(M,N)$, where $\mathrm{Emb}(M,N)$ is equipped with the compact-open topology. In [Factorization homology of topological manifolds](https://arxiv.org/abs/1206.5522) by Ayala-Francis, they construct a **tangent classifier functor** $$\tau: \mathcal{M}\mathrm{fld}\_n\to\mathcal{S}\_{/B\mathrm{Top}(n)}$$
which maps a manifold $M$ to the classifying map $M\to B\mathrm{Top} (n)$ of the tangent microbundle of $M$. **I do not understand this construction well, and I need some clarification.**
Here's the detail. Let $B\mathrm{Top}(n)\subset \mathcal{M}\mathrm{fld}\_n$ for the full subcategory spanned by $\mathbb{R}^n$. (By the Kister-Mazur theorem, $B\mathrm{Top}(n)$ is a Kan complex which has the homotopy type of the classifying space of the simplicial group $\operatorname{Sing}\mathrm{Top}(n)=\operatorname{Sing}\operatorname{Homeo}(\mathbb{R}^n,\mathbb{R}^n),$ hence the notation.) The tangent classfier is defined as the composite
$$\mathcal{M}\mathrm{fld}\_n\to\mathcal{P}(B\mathrm{Top}(n))\simeq\mathcal{S}\_{/B\mathrm{Top}(n)},$$
where the first map is the restricted Yoneda embedding and the second is the straightening-unstraightening equivalence. It is claimed in Corollary 2.13 of Ayala-Francis's article cited above that for any $n$-manifold $M$, $\tau(M)$ may be regarded as the classifying map of the tangent microbundle of $M$. I do not follow their argument very well (due to my incompetence), but roughly, it can be summarized as follows:
1. Observe that the domain of $\tau(M)$ has the homotopy type of $\operatorname{Sing}(M)$. (This is proved in Proposition 6.1.6 of [Yonatan Harpaz's course note](https://www.math.univ-paris13.fr/%7Eharpaz/lecture_notes.pdf) also.)
2. Observe that $\tau(M)$ classifies the tangent bundle if $M=\mathbb{R}^n$, and conclude by using that the assignment $M\mapsto \pi\_M$ is functorial in $M$ by Step 1.
I do not understand Step 2 above; it is not clear to me how the functoriality in $M$ is applied there. (Besides, $\tau({\mathbb{R}^n})$ of course classifies the tangent microbundle, because its domain is contractible by Step 1.) **Can anyone explain why $\tau(M)$ classifies the tangent microbundle?** Thanks in advance.
| https://mathoverflow.net/users/144250 | Why does the tangent classifier classify the tangent (micro)bundle? | For an $\infty$-category $X$ let me explicitly write $\mathrm{Fun}(X,\mathcal{S})$ for $\mathcal{P}(X)$. The construction of the tangent classifier firsts associates to $M$ the presheaf $\mathrm{Emb}(-,M) \in \mathrm{Fun}(\mathrm{Top}(n),\mathcal{S})$. The result is literally the topological frame bundle $\mathrm{Emb}(\mathbb{R}^n,M)$ with the standard action of $\mathrm{Top}(n)$. It then passes through the equivalence $\mathrm{Fun}(\mathrm{Top}(n),\mathcal{S}) \simeq \mathcal{S}\_{/\mathrm{BTop}(n)}$ to arrive at a map $M \rightarrow \mathrm{BTop}(n)$.
This equivalence is not mysterious, at least on the $\pi\_0$ level. Taking the product $\mathrm{Top}(n)$, $- \times E \mathrm{Top}(n)$ and equipping it with the diagonal action shows that every $\mathrm{Top}(n)$ space, i.e. an element of $\mathrm{Fun}(\mathrm{Top}(n),\mathcal{S})$, is equivalent to a prinicipal $\mathrm{Top}(n)$ bundle in this category. Up to homotopy, this equivalence of $\infty$-categories is just the classical equivalence between isomorphism classes of principal $\mathrm{Top}(n)$-bundles and homotopy classes of maps into $\mathrm{Top}(n)$. But of course, the principal bundle associated to the topological frame bundle is the tangent microbundle as the tangent microbundle at a point $x$ is essentially given by the germ of the embeddings of $\mathbb{R}^n$ which send the origin to $x$.
| 7 | https://mathoverflow.net/users/134512 | 438628 | 177,162 |
https://mathoverflow.net/questions/438547 | 1 | A Verra fourfold is a Fano fourfold which is defined as double cover of $\mathbb{P}^2\times\mathbb{P}^2$ with branch divisor to be $(2,2)$-hypersurface of $\mathbb{P}^2\times\mathbb{P}^2$, which is an index two Fano fourfold of Picard rank two. The $(2,2)$-divisor is a Verra threefold.
But it is known from <https://www.fanography.info/2-6> that there are two types of Verra threefold. I call the $(2,2)$-divisor in $\mathbb{P}^2\times\mathbb{P}^2$ the type I Verra threefold and the one which is double cover of $(1,1)$-divisor of $\mathbb{P}^2\times\mathbb{P}^2$, branched over its anticanonical divisor the type II Verra threefold.
My question is besides the Verra fourfold I defined above, do we have another type of Verra fourfold? It seems that the hyperplane section of the Verra fourfold I defined above is exactly the type II Verra threefold. So I suspect there should be another type of Verra fourfold whose hyperplane section is type I Verra threefold.
Verra threefold is very similar to Gushel-Mukai threefold. We know that there are two types of Gushel-Mukai threefold, and there are also two types of Gushel-Mukai fourfold. The Verra fourfold I defined above is kind of similar to special Gushel-Mukai fourfold, since they are both double cover with branched divisor to be the corresponding type threefold. I was wondering if we have "ordinary" Verra fourfold.
| https://mathoverflow.net/users/41650 | There are only one type of Verra fourfold? | I am not aware of Verra fourfolds of second type, and I think the situation with Verra fourfolds is more similar to the situation with GM sixfolds, where also only one type is available.
| 2 | https://mathoverflow.net/users/4428 | 438630 | 177,163 |
https://mathoverflow.net/questions/438126 | 0 |
>
> **Question.** Is it possible to construct a right triangle with a given [cathetus](https://en.wikipedia.org/wiki/Cathetus) $a$ and a given opposite angle $\alpha$ using only compass and ruler in the absolute geometry (so without the axiom of parallels)?
>
>
>
Observe that such a triangle always exists (by the axiom of continuity), unlike the right triangle with a given angle and given cathetus (which needs not exist in the hyperbolic geometry, but if exists, then it can be constructed by ruler and compass).
**Remark.** In fact, I am interested in a stronger version of this question in which it is allowed to use the compass and ruler only to construct new points as intersecting points of two lines and intersection points of circles and their diameters.
| https://mathoverflow.net/users/61536 | Is a right triangle with the given cathetus and the opposite angle constructible in the absolute plane? | A right triangle with a given side and opposite angle can be constructed in the hyperbolic plane with the help of a starightedge and a compass. This is done in Construction 7.2.4 of [this Bachelor Thesis](https://studenttheses.uu.nl/bitstream/handle/20.500.12932/40052/thesis.pdf?sequence=1) of Ruben de Vries.
| 0 | https://mathoverflow.net/users/61536 | 438634 | 177,165 |
https://mathoverflow.net/questions/438632 | 1 | Consider a one dimensional random walk, in which the probability of moving left along a line is $q=1/2$ and the probability of moving right is $p=1/2$. The square root error $\langle d\_N \rangle$, which is the expectation of the absolute distance traveled after $N$ steps is known to be $\sqrt{2N/\pi}.$ I am interested in finding the square root error of a modified version of this problem, in which there are three distinct possibilities. You can move to the left, you can move to the right, or you can stay put on a given turn. There will now be three probabilities, the probability $p$ of moving left, the probability $q$ of moving right, and the probability $r$ of not moving at all. The particular problem of interest to me is when $p=q$. So we will have that $p+q+r=1$. Has this problem been studied, and what would be the reference; what would be the solution?
| https://mathoverflow.net/users/130113 | A question about the square root error of one dimensional random walks | Because of the central-limit-theorem, for large $N$ the absolute distance $d\_N$ converges in distribution as
$$P\_N(d\_N/\sqrt N)\to p(|X|),$$
where $X$ is a Gaussian random variable with mean zero and variance $2p=1-r$. Since $\mathbb{E}(|X|)=2\sqrt{p/\pi}$ we conclude that
$$\mathbb{E}(d\_N)\to \sqrt\frac{2(1-r)N}{\pi}.$$
So this "lazy" random walk differs from the simple random walk by a rescaling of the number of steps by a factor $1-r$.
| 1 | https://mathoverflow.net/users/11260 | 438640 | 177,167 |
https://mathoverflow.net/questions/438466 | 1 | I know several specific variations of the domino tiling problem has been determined to be decidable or undecidable, such as the seed domino problem. I have a variation which I have not been able to find, but as I am not familiar with this subject, I thought someone here might know better.
Given a Wang tile $T=(T\_N,T\_E,T\_S,T\_W)$ with colors $\mathcal{C}$, each entry corresponds to the color in the north, east, south or west sides. Given two Wang tiles $T^{(1)},T^{(2)}$ that agree on opposite sides, their fusion would be gluing them on the side on which they agree. For example, if $T^{(1)}\_N=T^{(2)}\_S$, their fusion along South-North side is a new Wang tile
$$T^{(1)} \oplus\_{SN} T^{(2)}=\Big( T^{(2)}\_N, (T^{(2)}\_E,T^{(1)}\_E), T^{(1)}\_S, (T^{(2)}\_W,T^{(1)}\_W) \Big),$$
over a different set of colors obviously. We can similarly define $T^{(1)}\oplus\_{WE} T^{(2)}$ when they satisfy $T^{(1)}\_E=T^{(2)}\_W$. If we take a Wang tile set $\mathcal{T}$ with colors $\mathcal{C}$, we can obtain new Wang tiles sets
$$ \mathcal{T\_1} = \big\{ T \oplus\_{SN} T' : T,T' \in \mathcal{T}, T\_N=T'\_S \big\} $$
and
$$ \mathcal{T\_2} = \big\{ T \oplus\_{WE} T' : T,T' \in \mathcal{T}\_1, T\_E=T'\_W \big\}. $$
$\mathcal{T}\_2$ will again be a collection of Wang tiles over the new color set $\mathcal{C}\times \mathcal{C}$. I can consider two related questions:
1. Does $\mathcal{T}\_2$ tile the plane periodically?
2. Does $\mathcal{T}\_1$ tile the plane periodically?
Is it known whether question (1) being *un\decidable* imply question (2) is *un\decidable*? I am assuming that $\mathcal{T}\_1$ tiles the plane.
This seems like a natural question to me, but I really have a bad intuition on the subject, so it may be obviously resolvable. I would be glad to find appropriate references, or even solutions.
| https://mathoverflow.net/users/143153 | A variation of domino tiling problem with fusions | Each tiling with tiles from $\mathcal{T}$ gives rise to exactly two tilings with tiles from $\mathcal{T}\_1$ and exactly two tilings with tiles from $\mathcal{T}\_2$, given by taking the two choices of fusing neighbouring pairs (either $2n\longleftrightarrow 2n+1$ or $2n-1 \longleftrightarrow 2n$). This process also preserves periodicity.
Up to translation, this correspondence is also reversible and again preserves periodicity. So $\mathcal{T}$ admits a periodic tiling if and only if $\mathcal{T}\_1$ admits a periodic tiling if and only if $\mathcal{T}\_2$ admits a periodic tiling.
The same argument works for non-periodic tilings of course.
| 1 | https://mathoverflow.net/users/21271 | 438647 | 177,169 |
https://mathoverflow.net/questions/438638 | 3 | Let $\chi\colon(\mathbb{Z}/N\mathbb{Z})^\times\rightarrow\{\pm1\}$ be a primitive quadratic Dirichlet character of conductor $N$. For any integer $m=1,2,\cdots,\infty$, consider the partial Dirichlet series
$$L\_m(\chi)=L\_m(1,\chi)=\sum\_{n=1}^m\frac{\chi(n)}{n}.$$
It is well-known that $L\_\infty(\chi)$ is positive, see chapter 6 of Apostol's 'Introduction to analytic number theory'. My question is about the partial sum $L\_N(\chi)$. Here the subscript $N$ is the conductor of $\chi$. I have done some verification in sagemath, and for $N\le1000$, $L\_N(\chi)$ is always positive. Is it always the case for any $N$? How to prove this? I asked this question in mathematics stack exchange([here](https://math.stackexchange.com/questions/4602090/positivity-of-partial-dirichlet-series-for-a-quadratic-character)) and got a partial answer. I would like to see if there are more relevant results in this direction.
Edit: note here that the $N$ in $L\_N(\chi)$ is the same as the conductor of the character $\chi$. I forgot to emphasise this point in the first version, which led to some misunderstandings.
| https://mathoverflow.net/users/207039 | Positivity of partial Dirichlet series for a quadratic character? | Theorem 6.18 in the same chapter of Apostol shows, by partial summation, that
$$L\_m(\chi) = L(1,\chi) + O\_{\chi}(m^{-1})$$
where the implied constant is effective and depends on $\chi$. This can be explicated by using the Pólya--Vinogradov inequality, which gives
$$L\_m(\chi) = L(1,\chi)+O( m^{-1} N^{1/2} \log N)$$
where $N$ is the conductor and the implied constant is effective and absolute (one reference is exercise 11.2.1.3 in p. 374 of Montgomery and Vaughan's book). Taking $m=N$ as you do gives
$$L\_N(\chi) = L(1,\chi)+O( N^{-1/2} \log N)$$
unconditionally with an absolute and effective constant. Explicit constants were worked out by various authors, see e.g. Pomerance's paper "Remarks on the Pólya-Vinogradov inequality", Integers 11, No. 4, 531-542, A19 (2011).
To conclude that this is positive, one needs a lower bound on $L(1,\chi)$. There are two lower bounds on $L(1,\chi)$: effective and non-effective.
We have $L(1,\chi)\gg N^{-1/2}$ with an effective constant, see Theorem 11.11 in the Montgomery--Vaughan book, following from a result of Page. This falls just short of being useful here, so I do not think your question can be verified unconditionally.
Siegel showed $$L(1,\chi) > C(\varepsilon)N^{-\varepsilon}$$ for any given $\varepsilon>0$. The positive constant $C(\varepsilon)$ is not effective (it is if one shows there are no Siegel zeros). See Theorem 11.14 of the same book. Taking, say, $\varepsilon=1/3$, we find $L\_N(1,\chi)$ is positive for $N$ sufficiently large. Assuming no Siegel zeros, you can make this effective. So conditionally, to verify your question one needs to check finitely many values. I wouldn't be surprised if the answer is positive.
On GRH, $L(1,\chi) \gg 1/\log \log N$ (Littlewood). The constant has been made effective by Lamzouri, Li and Soundararajan in "Conditional bounds for the least quadratic non-residue and related problems" (Math. Comput. 84, No. 295, 2391-2412 (2015); corrigendum ibid. 86, No. 307, 2551-2554 (2017)), see their Theorem 1.5. This should allow you to get a quick verification of your question under GRH.
| 5 | https://mathoverflow.net/users/31469 | 438649 | 177,170 |
https://mathoverflow.net/questions/438594 | 7 | I've always wondered if the *DeMoivre method* to generate an algebraic number $x\_p$,
$$x\_p = u\_1^{1/p}+u\_2^{1/p}$$
of degree $p$ using only quadratic roots $u\_i$ could be generalized using ***cubic*** roots $v\_i$,
$$x\_p = v\_1^{1/p}+v\_2^{1/p}+v\_3^{1/p}$$
I serendipitously found a method which works for prime $p=6m+1$, the clue being the *Klein quartic* for $p=7$. The surfaces I used starts with that,
$$a^3b+b^3c+c^3a\quad \text{(deg 7)}$$
$$\;a^4b+b^4c+c^4a\;\quad \text{(deg 13)}$$
$$a^5b^2+b^5c^2+c^5a^2\quad \text{(deg 19)}$$
**and so on**. First, given the generic cubic for $C\_3 = A\_3$,
$$x^3-nx^2+(n-3)x+1 = 0$$
with discriminant $D=-(n^2-3n+9)^2,$ hence all roots $a,b,c$ are real.
---
**I. Level 7**
Using the roots $a,b,c$ of the generic cubic, then,
$$x\_7 = (a^3b)^{1/7}+(b^3c)^{1/7}+(c^3a)^{1/7}$$
$$x\_7^{'} = (a^3c)^{1/7}+(b^3a)^{1/7}+(c^3b)^{1/7}$$
properly chosen, are the roots of,
\begin{align}
x^7 - 14x^4 + 7n x^3 - 14(n - 3)x^2 &= 28x - (n^2 + 5n + 9)\\
x^7 - 14x^4 + 7n x^3 - 14(n - 3)x^2 &= -7(n^2 - 3n + 5)x + (n^3 - 4n^2 + 4n - 9)
\end{align}
The method generates ***pairs*** of solvable equations that curiously differ only by 2 coefficients for $p=7$, by 4 coefficients for $p=13$, etc.
---
**II. Level 13**
Using the same $a,b,c$, then,
$$x\_{13} = (a^4b)^{1/13}+(b^4c)^{1/13}+(c^4a)^{1/13}$$
$$x\_{13}^{'} = (a^4c)^{1/13}+(b^4a)^{1/13}+(c^4b)^{1/13}$$
are the roots of,
$$x^{13} + 26x^{10} - 13n x^8 + 143x^7 - 65(3 - n)x^6 - 65n x^5 - 13 (39 - 13 n - 2 n^2) x^3 = A$$
$$x^{13} + 26x^{10} - 13n x^8 + 143x^7 - 65(3 - n)x^6 - 65n x^5 - 13 (39 - 13 n - 2 n^2) x^3 = B$$
where,
$$A = -442x^4 + 13 (9 + 15 n + n^2) x^2 - 13 (36 - 3 n + n^2) x + (189 - 72 n + 15 n^2 - n^3)$$
$$B = 26 (1 - 6 n + 2 n^2) x^4 + 13 (9 + 6 n + 4 n^2 - n^3) x^2 + 13 (-3 + 2 n)^2 x + (18 + n^2) (12 - 5 n + n^2)$$
hence differs only by 4 coefficients.
---
**III. Level 19**
Still using the same $a,b,c$, then,
$$x\_{19} = (a^5b^2)^{1/19}+(b^5c^2)^{1/19}+(c^5a^2)^{1/19}$$
$$x\_{19}^{'} = (a^5c^2)^{1/19}+(b^5a^2)^{1/19}+(c^5b^2)^{1/19}$$
are the roots of two $19$-deg equations that differ only by several coefficients. (A parametric example can be given, but will clutter up the post.)
---
**IV. Levels 31 and 37**
$$x\_{31} = (a^6b)^{1/31}+(b^6c)^{1/31}+(c^6a)^{1/31}$$
$$x\_{37} = (a^7b^3)^{1/37}+(b^7c^3)^{1/37}+(c^7a^3)^{1/37}$$
and so on.
---
**V. Questions**
1. Why, starting with the Klein quartic and the roots of the generic cubic, does the method work?
2. By trial and error, I've found the surfaces for $p=43,61,67,73$. If the method works, can we predict these in advance from first principles?
| https://mathoverflow.net/users/12905 | A method to generate solvable equations of degrees $p = 7, 13, 19, 31, 37,\dots$ using only cubics | The following can explain the shape of the expression and also why you always get pairs (why the minimal polynomials for the two elements in a pair differ by few coefficients, I don't know).
Let $p$ be a prime congruent to $1$ mod $6$, and consider the rational function $f(X):=g(X^p)$, where $g(X):=\frac{X^3-3X+1}{X-X^2}$ (note that setting $t:=g(X)$ and clearing denominators gives exactly your generic cubic, except that I like $t$ better than $n$ as a parameter). Then $X\mapsto f(X)$ yields a cover $\mathbb{P}^1\to \mathbb{P}^1$, and we can try to compute its monodromy group (i.e., the Galois group of $f(X)-t=0$ over $\mathbb{C}(t)$ - and, as a matter of fact, still over $\mathbb{Q}(\zeta\_p)(t)$). Since $Gal(g(X)-t/\mathbb{C}(t)) = C\_3$ and $Gal(X^p-t/\mathbb{C}(t))=C\_p$, this monodromy group will be a subgroup of $C\_p\wr C\_3 (=(C\_p\times C\_p\times C\_p)\rtimes C\_3) \le S\_{3p}$. We can be more explicit: IF $a,b,c$ denote the three roots of $g(X)-t=0$, and $\alpha:=a^{1/p}$, $\beta:=b^{1/p}$, $\gamma:=c^{1/p}$, then the roots of $f(X)-t=0$ fall into three blocks $\Delta\_1, \Delta\_2, \Delta\_3$, with $\Delta\_1:=\{\alpha, \zeta\alpha, \dots, \zeta^{p-1}\alpha\}$ (and analogously for $\Delta\_2,\Delta\_3$).
The inertia groups at the branch points of the cover $X\mapsto f(X)$ will now give more precise information about the Galois group. It's easy to see that $X\mapsto g(X)$ has exactly two (non-rational) branch points (namely the roots of your discriminant $D$), whereas $X\mapsto X^p$ has of course exactly the branch points $0$ and $\infty$, and these two get mapped to the same point $\infty$ under $X\mapsto g(X)$. So in total, the composition has exactly three branch points: the two irrational ones (whose inertia group generators are of order $3$ (permuting $a$, $b$, $c$ cyclically and thus permuting the three blocks $\Delta\_i$ in the same way), and $t=\infty$ at which the inertia group is a double-$p$-cycle inside the block kernel (since unramified under $X\mapsto g(X)$, but with two totally ramified preimages on the "upper" level). The permutation group generated by these elements can then be identified as $G:=K\rtimes C\_3$, where $K:=\{(x\_1,x\_2,x\_3)\mid x\_1,x\_2,x\_3\in \mathbb{F}\_p; \sum x\_i=0\}$ is the ``augmentation ideal". This group $G$ has two normal subgroups $N\_1$ and $N\_2$ isomorphic to $C\_p$, both with quotient group $G/N\_i\cong C\_p\rtimes C\_3$: namely, let $y\_1,y\_2$ be the two primitive third roots of unity in $\mathbb{F}\_p$ and let $N\_i:=\{(x,y\_i x, y\_i^2 x): x \in \mathbb{F}\_p\}\trianglelefteq K$.
We want to find an expression in $\alpha,\beta,\gamma$ which is fixed by $N\_i$, and claim that this can be done in the form $\alpha^j\beta^k$ for $j,k \in \{1,\dots, p-1\}$ suitable, so that the minimal polynomial of $\alpha^j\beta^k (=a^jb^k)^{1/p}$ over $\mathbb{C}(t)$ will have Galois group $G/N\_i\cong C\_p\rtimes C\_3$ (the remaining symmetrization $\alpha^j\beta^k + \beta^j\gamma^k + \gamma^j\alpha^k$ the only drops the degree by a factor $3$, reaching a degree-$p$ polynomial, but still with the same splitting field; the existence of such a polynomial is clear anyway, since $C\_p\rtimes C\_3 \le S\_p$, so from the theoretical point of view this last step is not so important anymore).
So let $\sigma\_i:=(1,y\_i,y\_i^2)\in N\_i$ be a generator of $N\_i$. This acts quite explicitly on the three blocks $\Delta\_1,\Delta\_2,\Delta\_3$ by
$\sigma\_i(\alpha) = \zeta\_p \alpha$, $\sigma\_i(\beta) = \zeta\_p^{y\_i}\beta$, $\sigma\_i(\gamma) = \zeta\_p^{y\_i^2}\gamma$.
Hence, $\sigma\_i(\alpha^j\beta^k) = \zeta\_p^{j+k\cdot y\_i} \alpha^j\beta\_k$, and we're laughing if we can get $j+k\cdot y\_i$ divisible by $p$. There's obviously some freedom here in choosing $j$. E.g., for $p=7$, the third roots of unity in $\mathbb{F}\_7$ are $y\_1=2$ and $y\_2=4$; now if you choose $j=3$ with $y\_2=4$, you get $k=1$, which after the symmetrization yields your first solution $x\_7$; whereas $j=1$ with $y\_1=2$ gives $k=3$, and that, after symmetrization is your second solution $x\_7'$.
Just to check one more case, for $p=37$, the third roots of unity are $10$ and $26$. Choosing $j=7$ with $y\_1=10$ gives $k=3$.
You could of course choose other exponents $j$ (whether choosing to increase $j$ by $1$ for each new prime is actually a natural choice, I don't know; it happens to give small $k$ for the first few terms, but if I made no mistake, choosing $j=8$ for $p=43$ should give you $k=13$; instead always choosing $j=1$ would give the rather systematically-looking $k=y\_i+1$!), although these would of course not generate new field extensions.
| 5 | https://mathoverflow.net/users/127660 | 438651 | 177,171 |
https://mathoverflow.net/questions/438653 | 3 | Theorem 7.5.2 states:
Let $v\_1, \dots, v\_n$ be vectors with $\|v\_i\| \leq 1.$ Let $\epsilon\_1, \dots, \epsilon\_n \in \{-1, 1\}$ be independent with uniform probability and let $X=\|\epsilon\_1 v\_1 + \dots + \epsilon\_n v\_n\|.$ Then $$\Pr[X - \mathbb{E}[X]] > \lambda \sqrt{n}] < e^{-\lambda^2/2},$$ $$\Pr[X - \mathbb{E}[X]] < \lambda \sqrt{n}] < e^{-\lambda^2/2}.$$
In the solution they state that the corresponding martingale $X\_0, \dots, X\_n$ obtained by exposing each edge one at a time satisfies $|X\_{i+1}-X\_i| \leq 2$. Why is this?
| https://mathoverflow.net/users/497926 | Probabilistic method Alon and Spencer Azuma's inequality | $\newcommand\ep\epsilon$The martingale $(X\_i)$ is given by the formula
$$X\_i:=E(X|\ep\_1,\dots\ep\_i).$$
By the independence of the $\ep\_i$'s,
$$X\_i=g\_i(\ep\_1,\dots,\ep\_i),$$
where
$$g\_i(t\_1,\dots,t\_i):=E\Big\|\sum\_{j=1}^i t\_jv\_j+\sum\_{j=i+1}^i \ep\_jv\_j\Big\|$$
for real $t\_1,\dots,t\_i$.
By the triangle/norm inequality,
$$|g\_{i+1}(t\_1,\dots,t\_i,t\_{i+1})-g\_i(t\_1,\dots,t\_i)|
\le E|t\_{j+1}-\ep\_{j+1}|\,\|v\_{j+1}\|\le2\times1=2$$
if $|t\_{j+1}|=1$.
So,
$$|X\_{i+1}-X\_i|=|g\_{i+1}(\ep\_1,\dots,\ep\_i,\ep\_{i+1})-g\_i(\ep\_1,\dots,\ep\_i)|\le2,$$
as desired.
| 3 | https://mathoverflow.net/users/36721 | 438655 | 177,174 |
https://mathoverflow.net/questions/438659 | 0 | Let $A$ be a real symmetric matrix in $M\_{2n}(\mathbb{R})$with $A^2=I\_{2n}$. Suppose that the Schur decomposition of $A$ is given by $A=\Lambda^t D \Lambda$. Let us consider the following matrix.
$$J=I\_n\oplus (-I\_n)$$
Q. By the eigen values/eigenvectors of $A$, can we find/make some eigenvalues/eigenvectors of the product $JA$?
p.s. We denote $I\_n$ by the $n\times n$ identity matrix and $-I\_n$ is its negative.
| https://mathoverflow.net/users/84390 | The spectrum of the product $JA$ where $J=I_n\oplus (-I_n)$ | **Original question:** *What information can be extracted concerning the eigenvalues/eigenvectors of the product $JA$?*
The matrix $JA$ is orthogonal,
$$JA(JA)^\top=JA^2J=J^2=I,$$
so its eigenvalues $\lambda\_p$ are complex conjugate pairs $e^{\pm i\phi\_p}$ on the unit circle. The eigenvectors are an orthonormal set.
| 1 | https://mathoverflow.net/users/11260 | 438665 | 177,180 |
https://mathoverflow.net/questions/438656 | 7 | The recurrence relations for division polynomials of elliptic curves are well known:
$$\Psi\_{2n} = \Psi\_n \left( \Psi\_{n+2} \Psi\_{n-1}^2 - \Psi\_{n-2} \Psi\_{n+1}^2 \right) / \ 2y$$
$$\Psi\_{2n+1} = \Psi\_{n+2} \Psi\_n^3 - \Psi\_{n+1}^3 \Psi\_{n-1}$$
Who originally discovered these relations?
~~For some reason, a number of works do not cite the origin of the relations, e.g. Silverman's 2nd edition of *The Arithmetic of Elliptic Curves*, exercise 3.7 just gives them plainly.~~ See edit:
SageMath cites Appendix A from [Mazur-Tate](https://projecteuclid.org/journals/duke-mathematical-journal/volume-62/issue-3/The-p-adic-sigma-function/10.1215/S0012-7094-91-06229-0.short) from 1991 and they take a few pages to describe the discovery in full. However, they also show up in [this paper](http://www.mat.uniroma2.it/%7Eschoof/ctpts.pdf) (p.485, pdf p.4) from Schoof from 1985, citing Lang's
*Elliptic Curves; Diophantine Analysis* from 1978. Indeed, on page 37 of Lang we find the above result as Theorem 1.3, but again without a reference.
EDIT: I was wrong about Silverman not giving citations, they are just in the back of the book (p. 461) and he cites Lang, and also the same work by Cassels as given in the answer by Jeremy Rouse.
| https://mathoverflow.net/users/110923 | Reference request for recurrence relation of division polynomials | These recurrences are stated explicitly in Weber's *Lehrbuch der Algebra* (published in 1908). See [volume 3, section 58, page 200](https://archive.org/details/lehrbuchderalgeb03webeuoft/page/200/mode/2up). Weber doesn't give a citation to them, so it's hard to know if they were worked out by him or not.
I found the passage in Weber via J.W.S. Cassels 1949 article ''[A note on the division values of $\wp(u)$](https://doi.org/10.1017/S0305004100024683)'' which I believe is the source for the exercise in Silverman's *Arithmetic of Elliptic Curves*. Cassels gives references for these recurrences to Fricke (1922), Weber (1908) and Jordan (1913).
| 11 | https://mathoverflow.net/users/48142 | 438675 | 177,182 |
https://mathoverflow.net/questions/438622 | 1 | Let $K$ be a Markov kernel from $\mathcal{X}$ to $\mathcal{Y}$, i.e., $K(\cdot|x)$ is a probability measure on $\mathcal{Y}$ for all $x\in \mathcal{X}$.
Let $\mu$ and $\nu$ be two probability measures on $\mathcal{X}$. Moreover, let $K\mu$ and $K\nu$ be the probability measures on $\mathcal Y$ induced by $K$ with input distributions $\mu$ and $\nu$, respectively. That is, $K\mu(A) = \int\_{\mathcal X} \mu(dx) K(A|x)$ for any measurable set $A\subset \mathcal Y$.
It is known that
$$\sup\_{\mu, \nu}\frac{\|K\mu-K\nu\|}{\|\mu-\nu\|} = \sup\_{x, x'\in \mathcal{X}} \|K(\cdot|x)- K(\cdot|x')\|,$$
where $\|\cdot\|$ is the $\ell\_1$ distance. This result is largely attributed to Dobrushin.
I'm trying to prove a similar result as follows.
Let $\kappa>1$ be given. I wish to show that
$$\sup\_{\mu, \nu}\frac{\|K\mu-\kappa K\nu\|}{\|\mu-\kappa \nu\|} \leq \sup\_{x, x'\in \mathcal{X}} \|K(\cdot|x)- K(\cdot|x')\|.$$
Any thoughts?
I suspect that this problem (or some versions of it) has already appeared in some literature.
| https://mathoverflow.net/users/41666 | An inequality relating $\ell_1$ distance of input and output of a Markov krnel | Let me assume that $X$ and $Y$ are topological spaces (satisfying the usual conditions), and denote by $M(X), M(Y)$ the respective Banach spaces of finite (signed) Borel measures endowed with the total variation norm. I will also need the closed codimension 1 subspaces $M\_0(X),M\_0(Y)$ that consist of the measures of total weight 0 (i.e., the ones for which the masses of the positive and the negative parts in the Hahn decomposition are equal). We are given a Markov operator $K$ from $X$ to $Y$ (as it has already been pointed out, the standard probabilistic notation for the action of a Markov operator on measures is $\mu\mapsto \mu K$).
First, about the norm of the restriction of $K$ to the subspace $M\_0(X)$. Since for the "elementary measure" $\mu=\delta\_x-\delta\_{x'}$ associated with any $x\neq x'\in X$ one has
$$
\frac{\| \mu K \|}{\|\mu \|} = \frac12 \|(\delta\_x - \delta\_{x'}) K \| \;,
$$
and any measure from $M\_0(X)$ can be decomposed into an integral of elementary ones, the norm of $K$ as an operator from $M\_0(X)$ to $M\_0(Y)\subset M(Y)$ is the supremum of the RHS's of the above formula over all $x\neq x'\in X$.
I don't see what is here to be attributed to Dobrushin - or to anyone else. By the way, when stating your claim you mistakenly assumed that $\|\delta\_x - \delta\_{x'}\|=1$ - this is how some probabilists (who apparently don't believe in numbers greater than 1) define the total variation, but you explicitly stated that your total variation is the $\ell^1$ one.
Now, when passing to the measures $\mu-\kappa\nu$ you leave the subspace $M\_0(X)$, so that (by elementary functional analysis) you are asking about the norm of the operator $K$ on the whole space $M(X)$ which is 1 for **any** Markov operator. For an explicit couterexample to your claim consider the situation when the $K$-image of $M(X)$ is one-dimensional, i.e., **all** measures $\delta\_x K$ are the same. In this case your RHS is 0 (as the whole subspace $M\_0(X)$ is in the kernel of $K$), whereas the LHS is 1.
| 1 | https://mathoverflow.net/users/8588 | 438679 | 177,184 |
https://mathoverflow.net/questions/438682 | 5 | Let G be a discrete or say for sake of simplicity a finite group. In Hatcher's book Algebraic Topology on p 89 the construction of universal bundle $EG$ carries structure of a $\Delta$-complex whose $n$-simplices are the ordered $(n + 1)$ tuples
$[g\_0, ... ,g\_n]$ of elements of $G$. As
quotient space the model of the classifing space $BG=EG/G$ inherits structure $\Delta$-complex where a $n$-simplex $BG$ can be written uniquely in the form $[g\_1 \vert g\_2\vert ... \vert g\_n]:=
G \cdot [e\_G, g\_1, g\_1g\_2,..., g\_1,.., g\_n]$. Note that by construction $EG$ is realized even as a simplicial complex, while $BG$ inherits only structure of honest $\Delta$-complex due to more subtle identificiations by passing to quotient.
My question is if there is a rather "canonical" way to choose a trivializing cover $\{U\_i\}\_{i \in I}$ of the universal principal $G$-bundle $EG \to BG$, ie a family of open subsets of $BG$ such that $EG \vert \_{U\_i} \cong U\_i \times G$ with relatively good controllable associated transition functions $h\_{ij}: U\_i \cap U\_j \to G$ dictating the patching structure $ U\_i \cap U\_j \times G \to U\_i \cap U\_j \times G, (u,g) \mapsto (u, h\_{ij}(u)g)$.
The question is closely cross posted from [MSE](https://math.stackexchange.com/questions/4608049/construct-a-trivializing-cover-of-universal-principal-g-bundle-eg-to-bg) where it not received any resonance.
| https://mathoverflow.net/users/108274 | Construct a 'nice' trivializing cover of universal principal $G$-bundle $EG \to BG$ | I think that this works.
EDIT: No, it doesn't. See John Rognes's comment.
Notation: For a point $x\in EG$ we may symbolically write $x=\sum\_{j=0}^nt\_jg\_j$, where $g\_0,\dots ,g\_n$ is an ordered tuple of distinct elements of $G$ and $t\_j\ge 0$ and $\sum\_jt\_j=1$. Here the face relations are accounted for by agreeing that if $t\_j=0$ for some $j$ then $g\_j$ can be omitted from the tuple and the term $t\_jg\_j$ can be omitted from the expression, without changing the point $x$.
Invariant open cover of $EG$: For $p\ge 0$ let $U\_p\subset EG$ consist of the points $x=\sum\_{j=0}^nt\_jg\_j$ such that there is a (necessarily unique) subset $S(x)\subset \{0,n\}$ of cardinality $p+1$ such that
(1) $t\_j>t\_k$ whenever $j\in S(x)$ and $k\notin S(x)$, and
(2) $t\_j>0$ whenever $j\in S(x)$.
Thus for each $n$-simplex $\sigma=[g\_0,\dots ,g\_n]$ the set $U\_p\cap\sigma$ is as follows. If $p\le n$ then it is the disjoint union of convex open subsets, one for each $p$-dimensional face $\tau\subset\sigma$, namely the open star, in the barycentric subdivision of $\sigma$, of the barycenter of $\tau$. It is empty if $p>n$.
This is well-defined (compatible with the face relations), and open. The union of the sets $U\_p$ is all of $EG$. For each $p$ the set $U\_p$ is invariant under the action of $G$.
Local trivializations: Now let's show that the bundle is trivial over the image of $U\_p$, by describing a cross-section. Let $V\_p\subset U\_p$ consist of those points $x=\sum\_jt\_jg\_j\in U\_p$ such that when $j\_0$ is the smallest element of the set $S(x)$ then $g\_j=e$. This is open in $U\_p$, and every point in $U\_p$ is uniquely expressible as $gx$ for some $g\in G$ and $x\in V\_p$.I think
| 5 | https://mathoverflow.net/users/6666 | 438688 | 177,186 |
https://mathoverflow.net/questions/394142 | 7 | $\newcommand\R{\mathbb R}\newcommand\la\lambda$It is well known and easy to see that the rotationally invariant
product of two probability measures on $\R$ has to be a Gaussian (or Dirac) measure; see e.g. [this answer](https://math.stackexchange.com/a/2599580/96609).
This appears to make the conjecture below somewhat plausible.
Let $\mu$ be any probability measure on $\R^2$ with a finite nonzero covariance matrix. Let $\mu\_1:=\mu$. For each natural $n$, consider the following three-step procedure:
1. let $$\la\_n:=\mu\_n^{(1)}\otimes\mu\_n^{(2)},$$
where $\mu\_n^{(1)}$ and $\mu\_n^{(2)}$ are the marginals of $\mu\_n$;
2. let $$\nu\_n:=\frac1{2\pi}\int\_0^{2\pi}\la\_n R\_t\,dt,$$
where $\la\_n R\_t$ is the pushforward measure obtained from $\la\_n$ by the rotation about the origin through angle $-t$;
3. let $\mu\_{n+1}$ be obtained by rescaling the probability measure $\nu\_n$ so that the covariance matrix of $\mu\_{n+1}$ be the unit matrix.
**Conjecture:** $\mu\_n$ converges weakly (as $n\to\infty$) to the standard Gaussian measure on $\R^2$.
Is this conjecture true?
*Comment:* Perhaps Step 3, the rescaling, is not essential. Of course, if we have the convergence to a (nondegenerate) Gaussian measure without Step 3, then we have such a convergence with Step 3 as well.
---
We can restate the problem (without rescaling) analytically as follows. Let $f\_n$ denote the characteristic function of $\nu\_n$, so that
$$f\_n(u,v)=\int\_{\R^2}\nu\_n(dx\times dy)e^{i(ux+uy)} \\
=\int\_{\R^2}\nu\_n(dx\times dy)\cos(ux+uy)\quad \text{(by symmetry)}$$
for all real $u$ and $v$. Then
$$f\_n(u,v)=g\_n\big(\sqrt{u^2+v^2}\big)$$
for some function $g\_n\colon[0,\infty)\to\mathbb R$ and all real $u$ and $v$. Then for all natural $n$ and all real $r\ge0$
$$g\_{n+1}(r)=\frac2\pi\int\_0^{\pi/2}
dt\,g\_n(r\cos t)g\_n(r\sin t). \tag{1}$$
We want to show that
$g\_n(r)\to e^{-c^2 r^2/2}$ for some $c\in(0,\infty)$ and all real $r\ge0$.
So, analytically, the problem may be viewed as one of stability of a (nonlinear) integral equation or as one of solving such an integral equation by iterations.
---
Using substitutions $g\_n(r)=h\_n(r^2)$ and $r^2=s$, we can rewrite (1) as
$$h\_{n+1}(s)=\frac1\pi\int\_0^s
du\,\frac{h\_n(s-u)}{\sqrt{s-u}}\frac{h\_n(u)}{\sqrt{u}}$$
and then as
$$\pi H\_{n+1}(s)\sqrt s=[(H\_n\*H\_n)(s)=]\int\_0^s
du\,H\_n(s-u)H\_n(u)$$
for all natural $n$ and all real $s\ge0$, where $H\_n(u):=h\_n(u)/\sqrt u$.
We want to show that $H\_n(u)\to e^{-c^2 u/2}/\sqrt u$ for some $c\in(0,\infty)$ and all real $u>0$.
| https://mathoverflow.net/users/36721 | Normal distribution by successive approximation? | In the rewritten form it doesn't require any ingenuity at all. For brevity, introduce the notation
$$
(Th)(s)=\frac 1\pi\int\_0^s\frac{h(s-u)h(u)}{\sqrt{u(s-u)}}\,du=\frac 1\pi\int\_0^1\frac{h(s(1-v))h(sv)}{\sqrt{v(1-v)}}\,dv
$$
Suppose that $h\_0$ is any real valued Lipschitz function on $[0,+\infty)$ (the finiteness of the second moment guarantees Lipschitzness for your $h\_0$ and, if you really want to discuss it, I can show how this particular condition can be dropped) such that $\|h\_0\|\_\infty=h\_0(0)=1$ and there exists $h\_0'(0)=-c< 0$ (the latter two conditions are essential). Then the iterations $h\_{n+1}=Th\_n$ converge to $e^{-cx}$ uniformly on compact subsets of $[0,+\infty)$.
The proof consists of several easy observations:
1. $\|Th\|\_\infty\le\|h\|\_\infty^2$ and, thereby, $\|h\_n\|\_\infty\le 1$ for all $n$.
2. If $h$ is bounded and $L$-Lipschitz, then $Th$ is $L\|h\|\_\infty$-Lipschitz.
Indeed,
$$
Th(s)-Th(S)=\frac1\pi\int\_0^1\frac{dv}{\sqrt{v(1-v)}}[h(s(1-v))h(sv)-h(S(1-v))h(Sv)]
$$
and
$$
|h(s(1-v))h(sv)-h(S(1-v))h(Sv)|
\\
\le\|h\|\_\infty[|h(s(1-v))-h(S(1-v))|+|h(sv)-h(Sv)|]
\\
\le\|h\|\_\infty L[|s-S|(1-v+v)]=\|h\|\_\infty L|s-S|\,.
$$
Thus all $h\_n$ are Lipschitz with the same Lipschitz constant $L$ as $h\_0$.
3. If $A\in\mathbb R$ and $h(s)\ge e^{-As}$ on $[0,s\_0]$, then $(Th)(s)\ge e^{-As}$ on $[0,s\_0]$.
3') If $a\in\mathbb R$ and $0\le h(s)\le e^{-as}$ on $[0,s\_0]$, then $(Th)(s)\le e^{-as}$ on $[0,s\_0]$.
(all exponential functions are fixed points of $T$ and we have monotonicity as long as $h$ stays non-negative)
4. Now choose any $0<b<a<c<A<B$ and choose $s\_0>0$ so that $e^{-As}\le h\_0(s)\le e^{-as}$ on $[0,s\_0]$ (by the derivative at zero condition such $s\_0$ exists).
Consider the largest $S\_n$ such that $e^{-Bs}\le h\_n(s)\le e^{-bs}$ on $[0,S\_n]$. We have $S\_0\ge s\_0$ and $S\_{n+1}\ge S\_n$. We want to improve the latter trivial inequality to some quantitative advance $S\_{n+1}\ge S\_n+\delta(S\_n)$ where $\delta>0$ is separated from $0$ on any compact subinterval of $[s\_0,+\infty)$. That is quite easy:
$$
h\_{n+1}(S\_n)=(Th\_n)(S\_n)=\frac1\pi\int\_0^{S\_n}\frac{h\_n(S\_n-u)h\_n(u)}{\sqrt{u(S\_n-u)}}\,du\ge
\\
\frac1\pi\int\_0^{S\_n}\frac{e^{-B(S\_n-u)}e^{-Bu}}{\sqrt{u(S\_n-u)}}\,du
+\frac1\pi\int\_0^{s\_0}\frac{e^{-B(S\_n-u)}[e^{-Au}-e^{-Bu}]}{\sqrt{u(S\_n-u)}}\,du
\\
\ge e^{-BS\_n}+e^{-BS\_n}S\_n^{-1}\int\_0^{s\_0}[e^{-Au}-e^{-Bu}]\,du=e^{-BS\_n}+\Delta(S\_n)
$$
Then, by the Lipschitz property of both $h\_{n+1}$ and $e^{-Bs}$, the inequality $h\_{n+1}(s)\ge e^{-Bs}$ persists on $[S\_n,S\_n+\delta(S\_n)]$ with $\delta(S\_n)=\frac{\Delta(S\_n)}{B+L}$, say. The extension of the upper bound is similar.
The outcome is that the double inequality $e^{-Bs}\le h\_n(s)\le e^{-bs}$ propagates from $[0,s\_0]$ to the entire real line. Since $0<b<a<c<A<B$ were arbitrary, we conclude that $h\_n$ tend to $e^{-cs}$ uniformly on compact intervals, finishing the story.
| 1 | https://mathoverflow.net/users/1131 | 438691 | 177,187 |
https://mathoverflow.net/questions/438390 | 4 | Consider bivariate copulas $C\_1$ and $C\_2$ with $\max\{C\_1(u,v), C\_2(u,v)\}< M\_2(u,v)$ for all $u,v \in(0,1)$, where $M\_2(u,v) := \min\{u,v\}$ is the Fréchet-Hoeffding upper bound.
Is there a copula $D$ with $\max\{C\_1(u,v), C\_2(u,v)\}\leq D(u,v) < M\_2(u,v)$ for all $u,v \in(0,1)$?
The problem here is that $\max\{C\_1(u,v), C\_2(u,v)\}$ is in general only a quasi-copula and I need a bound strictly below $M\_2$. I think such a bound should exist in full generality but I don't have a reference/proof for it.
| https://mathoverflow.net/users/497745 | Existence of copula bound pointwise strictly smaller than the Fréchet-Hoeffding upper bound | You can prove more. Let $F(u,v)$ be any $1$-Lipschitz function on $[0,1]^2$ such that $F(u,v)<\min(u,v)$ inside the square. Then there exists a copula $D(u,v)$ such that
$$
F(u,v)\le D(u,v)<\min(u,v)
$$
everywhere inside the square.
The second inequality will be immediate if we just construct the corresponding joint distribution measure $\mu$ on $[0,1]^2$ with full support. Indeed, then
$$
D(u,v)=\mu([0,u]\times [0,v])=\mu([0,u]\times [0,1])-\mu([0,u]\times(u,1])
\\
=u-\mu([0,u]\times(u,1])<u
$$
and similarly for $v$.
It remains to make the following two observations.
Observation 1: There is a copula $D\_0(u,v)>F(u,v)$ inside the unit square such that the associated $\mu$ has density $p\_0(u,v)\ge q\_0>0$ separated away from $0$ in some open neighborhood $\Omega$ of $\{(x,x):0<x<1\}$. To see it, just take a sufficiently fine countable partition of $(0,1)$ into disjoint intervals $I\_j$, consider the density $p=\sum\_j|I\_j|^{-1}\chi\_{I\_j\times I\_j}$ and mix two such distributions to take care of the corners. "Sufficiently fine" just means that the length of each $I\_j$ is much less than $u-F(u,u)$ for $u\in I\_j$. Now take any sequence $q\_0>q\_1>\dots \ge q\_0/2$ and
Observation 2: Suppose we have a copula $D\_n$ for which $D\_n(u,v)>F(u,v)$ inside the unit square and the density $p\_n$ of $\mu\_n$ is at least $q\_n$ in $\Omega$. Let $(u\_n,v\_n)\in(0,1)^2$ be any point outside the diagonal. Then, for every $\delta\_n\ge 0$, there exists a copula $D\_{n+1}$ with $D\_{n+1}>F$ inside the unit square, such that $p\_{n+1}\ge p\_n$ outside $\Omega$, $p\_{n+1}\ge q\_{n+1}$ in $\Omega$, and $(u\_n,v\_n)$ is in the support of $\mu\_{n+1}$. To do it, just choose two very short intervals $U,V$ of equal length containing $u\_n$ and $v\_n$ respectively and set
$$
p\_{n+1}=p\_n+t(-\chi\_{U\times U}-\chi\_{V\times V}+\chi\_{V\times U}+\chi\_{U\times V})
$$
with very small $t$. Notice that this changes $D\_n(u,v)$ only on a compact subset of the open unit square (say, $I\times I$ where $I$ is a closed interval such that $U\cup V\subset I\subset (0,1)$) and there the difference $D\_n-F$ is separated from $0$, so this surgery leaves it positive for small $t>0$.
Now just take any sequence $(u\_n,v\_n)\in (0,1)^2\setminus{\rm diag}$ dense in the square and run this recursion. Then take either the weak limit of $\mu\_n$, or the $L^1$-limit of $p\_n$, whichever you are more comfortable with, to get $D$.
| 3 | https://mathoverflow.net/users/1131 | 438715 | 177,192 |
https://mathoverflow.net/questions/438538 | 4 | In his paper "Hecke Algebras of type $A\_n$ (Inv. Math. 1988, [EUDML link](https://eudml.org/doc/143571)) and subfactors", in section 2 "Orthogonal representations...", Wenzl takes the usual third relation of a standard generator of Hecke algebras, namely $g\_i^2=(q-1)g\_i+q$, where $q$ is a complex number that is neither $0$ nor $-1$, and says that $g\_i$ can only have at most two eigenvalues. Understandable given that the minimal polynomial of $g\_i$ must divide $x^2-(q-1)x-q$. So the possible eigenvalues must be $q,-1$. Then proceeds to say that the spectral projection belonging to $-1$ is $e\_i= \frac{q-g\_i}{q+1}$. Can someone explain to me what he means by spectral projection and how he comes up with the formula for $e\_i$?
| https://mathoverflow.net/users/160919 | Spectral projection of an eigenvalue associated to a generator of Hecke algebras | Since $e\_i$ kills any eigenvectors with eigenvalue $q$ (clear) and squares to itself (due to the quadratic relation for $g\_i$ and normalisation), it is a projector onto the $-1$-eigenspace of $g\_i$. For the usual (reducible) representation of $H\_n(q)$ on $V^{\otimes n}$, where $g\_i$ becomes the simple transposition $s\_i$ as $q\to 1$, the element $e\_i$ acts as the antisymmetriser in the $i$ and $i+1$st factors of $V^{\otimes n}$. So one may think of $e\_i$ as a $q$-antisymmetriser.
For comparison, the spectral projector onto the $q$-eigenspace of $g\_i$ would be $-(1+g\_i)/(q+1)$. Note that these two projectors orthogonal (again due to the quadratic relation) and add to the identity. One can think of this as the corresponding $q$-symmetriser.
(If $g\_i$ acts on $V^{\otimes n}$ and $V \cong \mathbb{C}^2$, then the left- and right-hand sides of Wenzl's relation (H1) both vanish, and the $H\_n(q)$-representation on $V^{\otimes n}$ factors through a representation of the Temperley--Lieb algebra, with generators $e\_i$ up to normalisation depending on conventions.)
I hope this helps.
PS. Nowadays the Hecke generators are usually denoted by $T\_i$ rather than $g\_i$. The notation $e\_i$ for (something proportional to) $q-T\_i$ is standard.
| 3 | https://mathoverflow.net/users/45956 | 438717 | 177,193 |
https://mathoverflow.net/questions/438710 | 4 | I was trying to get a short, intuitive proof of Sierpinski’s theorem (gch implies axiom of choice) and I could but only by using the following gch2 for the generalized continuum hypothesis gch.
gch: Let S be an infinite set and S’ be the power set of S. If T is a subset of S’ such that |T| > |S|, i.e., T is not injectable into S, then |T| = |S’|, i.e., T and S’ are bijectable.
gch2($\alpha$): For ordinal $\alpha$ and V the cumulative hierarchy starting at the empty set, let S be a set where $\alpha$ is the least ordinal such that |S| $\leq$ |$V\_\alpha$| (i.e., S is injectable into $V\_\alpha$). Then |S| = |$V\_\alpha$|.
gch2: gch2($\alpha$) is true if $\alpha$ is an infinite ordinal (of course it is false on all finite ordinals except 0, 1,and 2).
gch2 is easily proven to be equivalent to gch in ZF using Sierpinski’s theorem, but I couldn’t see how to get gch2($\lambda$) in ZF from gch without using Sierpinski’s theorem for the case where $\lambda$ is a limit ordinal.
gch2 asserts that every infinite ordinal stage of the cumulative hierarchy presents exactly one new cardinality that wasn’t presented at lower levels. The axiom of choice is sufficient to easily prove this for limit ordinal stages (usually expressed as the continuity of the beth function), but is it necessary? That is essentially my question presented below.
Two possibilities that I am interested in are: (1) proving in ZF that if gch2($\lambda$) holds for all limit ordinals $\lambda$, then the axiom of choice holds. Thus the axiom of choice would be equivalent to gch2($\lambda$) holding on all limit ordinals $\lambda$. And at the other extreme (2) proving in ZF that gch2($\lambda$) holds on all limit ordinals $\lambda$ (unconditionally).
My question is can either one of these possibilities be settled? If (1) can be done then the axiom of choice follows from gch2 immediately since all limit ordinals are infinite ordinals. It also suggests that Sierpinski has more to do with unions than power sets.
Without (1), gch2 easily implies trichotomy of cardinals, hence axiom of choice, in a manner not as short as (1) provides but pretty short and intuitive:
Let S and T be infinite sets. Then S and T are in the cumulative hierarchy by axiom of foundation. Let $\alpha$ and $\beta$ be the least ordinals such that |S| $\leq$ |$V\_\alpha$| and |T| $\leq$ |$V\_\beta$|, respectively. Since S and T are infinite sets, $\alpha$ and $\beta$ are infinite ordinals. By trichotomy of ordinals either $\alpha$ = $\beta$ or one of the two is greater than the other (say $\alpha$ > $\beta$). By gch2, which applies since $\alpha$ and $\beta$ are infinite ordinals, |S| = |$V\_\alpha$| and |T| = |$V\_\beta$|.
Now if $\alpha$ = $\beta$ then |S| = |T|. If $\alpha$ > $\beta$ then $V\_\beta$ $\in$ $V\_\alpha$, so that |$V\_\beta$| < |$V\_\alpha$| and thus |T| < |S|. Therefore trichotomy of cardinals is easily established in ZF from gch2.
Please note that the following two conditions are easily provably equivalent in ZF to gch2($\alpha$) without using Sierpinski, and are easily provably equivalent in ZF to gch if Sierpinski is used.
gch3($\alpha$): For $\alpha$ an ordinal, if S is a set such that |S| < |$V\_\alpha$|, i.e., S is injectable into $V\_\alpha$ but not bijectable, then |S| = |x| for some x $\in$ $V\_\alpha$.
gch4($\alpha$): For $\alpha$ an ordinal, |$V\_\alpha$| is the sup of the cardinalities |$V\_\beta$| for $\beta$ < $\alpha$, in the sense that it is an upper bound with |$V\_\alpha$| > |$V\_\beta$|, for all $\beta$ < $\alpha$, and it is the least upper bound with |$V\_\alpha$| $\leq$ any other such upper bound |S| of those cardinalities.
Similarly, note that the usual aleph formulations of gch, such as: for all ordinals $\alpha$, the power set of $\aleph\_\alpha$ has cardinality $\aleph\_{\alpha+1}$, are easily proved in ZF without Sierpinski to imply gch2, and therefore gch2 also provides short proofs of Sierpinski for these formulations of gch. These implications producing gch2 are true in these aleph formulations because the desired properties on both the limit ordinal and infinite successor ordinal stages of the cumulative hierarchy are easily derived from these formulations using the similar properties of the alephs.
So in summary, the generalized continuum hypothesis can be viewed as failing in ZF on all but the smallest finite ordinal stages of cumulative hierarchy, succeeding in ZFC on all limit ordinal stages, and is up for grabs in ZFC at the infinite successor ordinal stages. This is surprising to someone like me with just a pedestrian knowledge of set theory. We pedestrians know that the generalized continuum hypothesis fails at most finite stages but are unaware that it succeeds at so many infinite stages. Could this succeeding be used as an argument for accepting that hypothesis as a good condition for distinguishing finite and infinite sets, for those looking for such an argument? Also, possibility (1) implies that the generalized continuum hypothesis is what happens to the axiom of choice when the latter progresses (as the gch2 condition) to hold on all infinite ordinals instead of just the limit ordinals. Does that make accepting the generalized continuum hypothesis more palatable for those who already accept the axiom of choice?
| https://mathoverflow.net/users/480831 | About the relationship between the generalized continuum hypothesis and the axiom of choice | Possibility (1) holds; i.e. ZF + "for all limit ordinals $\lambda$, GCH2($\lambda$) holds" implies choice. For it implies that for cofinally many ordinals $\lambda$, $V\_\lambda$ can be wellordered. For fix an ordinal $\alpha$; then we can easily find a limit ordinal $\lambda>\alpha$ such that $\lambda$ does not inject into $V\_\beta$ for any $\beta<\lambda$ (just define $\left<\lambda\_n\right>\_{n<\omega}$ by setting $\lambda\_0=\alpha$, and given $\lambda\_n$, $\lambda\_{n+1}$ is the least $\lambda'>\lambda\_n$ such that $\lambda'$ does not inject into $V\_{\lambda\_n}$; then set $\lambda=\sup\_{n<\omega}\lambda\_n$). But certainly $\lambda$ injects into $V\_\lambda$. So we can apply GCH2($\lambda$) with $S=\lambda$ (since $\lambda$ is the least ordinal $\beta$ such that $S$ embeds into $V\_\beta$). Therefore $|S|=|V\_\lambda|$, i.e. $|\lambda|=|V\_\lambda|$, i.e. there is a bijection between $\lambda$ and $V\_\lambda$, so $V\_\lambda$ is wellorderable, which suffices.
| 6 | https://mathoverflow.net/users/160347 | 438728 | 177,196 |
https://mathoverflow.net/questions/438727 | 17 | This is a question about the true number of constraints imposed by the Jacobi identity on the structure constants of a Lie algebra.
For an $n$-dimensional Lie algebra, there are $\frac{n^2(n-1)}{2}$ structure constants $f\_{ab}^c$, where I've accounted for antisymmetry but not the Jacobi identity. Accounting for obvious symmetries, the Jacobi identity $f\_{ad}^e f\_{bc}^d +f\_{bd}^e f\_{ca}^d + f\_{cd}^e f\_{ab}^d= 0$ seems to impose $\frac{n^2(n-2)(n-1)}{6}$ constraints.
For $n>5$, this exceeds the total degrees of freedom of the structure constants. The problem appears to be overspecified and there should (in all likelihood) be no nontrivial solutions. I.e. by such a naive counting of constraints there should be no finite-dimensional Lie algebras with $n>5$. Obviously, this is not the case.
It's certainly possible for an apparently overconstrained system of quadratic equations to have nontrivial solutions. However, it's natural to ask whether that's more than mere coincidence and there is some other mechanism at play — either a less obvious relationship between the equations or a reason they were formed in a manner which admits solutions.
I wrote a simple computer program to analyze how many of the Jacobi equations are trivial or redundant in a couple of simple cases ($n=3$ to $n=12$), and none are. In fact, they all are linearly independent (i.e. if we regard distinct quadratic terms as basis elements in some vector space, then the rank of the resulting $\big[\frac{n^2(n-2)(n-1)}{6}\big] \times \big[\frac{1}{2}\frac{n^2(n-1)}{2}\big(\frac{n^2(n-1)}{2}+1\big)\big]$ sparse matrix is the same as the number of Jacobi equations). The equations may be related in more complicated ways than mere linear independence, of course.
So am I missing something obvious? There must be some additional reduction in the number of independent equations that I'm missing. Somehow I must be massively overcounting the Jacobi identity equations.
I tried posting this on math.stackexchange a few years ago and got no responses, and now I'm revisiting the subject and realized it still puzzles me — so I figured I'd try posting it here.
Presumably, the set of solutions (modulo an overall scale factor) would provide an alternate approach to classifying Lie algebras of a given dimension over the relevant field.
Most likely, a similar issue arises for the structure constants of an associative algebra, though I haven't gone through the motions there.
Thanks in advance for your help!
| https://mathoverflow.net/users/30496 | Counting degrees of freedom in Lie algebra structure constants (aka why are there any nontrivial Lie algebras of dim >5?) | Linear independence does not really say much.
This algebraic variety is discussed in some detail in an old paper of Kirillov and Neretin: [The variety $A\_n$ of $n$-dimensional Lie algebra structures](https://www.mat.univie.ac.at/%7Eneretin/kiri87a.pdf).
The case of $n=4$ which already shows many interesting phenomena (but already is nontrivial from the computer algebra viewpoint) is analysed in detail in the recent preprint [Manivel, Sturmfels, and Sverrisdóttir - Four-Dimensional Lie Algebras Revisited](https://arxiv.org/abs/2208.14631) — you might find it enlightening.
Update: I checked the MathSciNet review of the paper of Kirillov and Neretin and found two other relevant references:
[Carles, Diakité - les variétés d'Algèbres de Lie de dimension $\leqslant 7$](https://www.sciencedirect.com/science/article/pii/002186938490125X)
[Gorbatsevich - Some properties of the space of n-dimensional Lie algebras](https://iopscience.iop.org/article/10.1070/SM2009v200n02ABEH003991) (where in particular your observation on linear independence is proved)
| 12 | https://mathoverflow.net/users/1306 | 438733 | 177,198 |
https://mathoverflow.net/questions/420921 | 6 | Consider the following proposition.
Proposition: let $\lambda$ be a limit ordinal and $V$ be the cumulative hierarchy starting with the null set, and $S$ be a set with
$\vert S\vert<\vert V\_\lambda\vert$. Then there exists an $x\in V\_\lambda$ with $\vert x\vert=\vert S\vert$.
The proposition is trivially true for $\lambda=\omega$, and is trivially true for all limit ordinals assuming the axiom of choice, but is it true for any other limit ordinal without assuming the axiom of choice? My uneducated guess is no.
I find this question interesting because what appears to be a gch condition on successor ordinals in the cumulative hierarchy is merely an axiom of choice condition on limit ordinals. If the above proposition holds then what appears to be a gch condition on successor ordinals is unconditionally true on all limit ordinals. This could be used as evidence (not decisive evidence but evidence none the less) for accepting gch.
I had to answer the below questions only by editing the above text because the software won’t let me comment. In my original post I forgot to indicate that I was asking about the case where we don’t have the axiom of choice. I edited the above appropriately. Sorry!
| https://mathoverflow.net/users/480831 | When does the cardinality of a set equal the cardinality of an element of $V_\lambda$ for $\lambda$ being a limit ordinal? | The proposition holds for the limit ordinal $\lambda$ iff $V\_\lambda$ is wellorderable.
For if $V\_\lambda$ is wellorderable where $\lambda$ is a limit, then it proposition easily follows at $\lambda$. Conversely, suppose the proposition holds at limit ordinal $\lambda$. (Without using AC) we can specify a certain ordinal $\eta$ and an injection $\pi:\eta\to V\_\lambda$, such that $\eta$ cannot be injected into any $V\_\alpha$ with $\alpha<\lambda$, but then applying the hypothesis with $S=\eta$, it follows that $|V\_\lambda|=\eta$.
To define $\eta,\pi$, first for each $\alpha<\lambda$, define $\eta\_\alpha$ and an injection $\pi\_\alpha:\eta\_\alpha\to V\_{\alpha+1}$, as follows. For ordinals $\beta$, let $W^\alpha\_\beta$ be the set of all binary relations $R\in V\_\alpha$ which are wellorders in ordertype $\beta$, and let $\eta\_\alpha$ be the least $\beta$ such that $W^\alpha\_\beta=\emptyset$. Then define $\pi\_\alpha:\eta\_\alpha\to V\_{\alpha+1}$ by $\pi\_\alpha(\beta)=(\alpha,W^\alpha\_\beta)$. Now let $D=\{(\alpha,\beta)\bigm|\alpha<\lambda\wedge\beta<\eta\_\alpha\}$, and define $\pi':D\to V\_\lambda$ as $\pi'(\alpha,\beta)=\pi\_\alpha(\beta)$. Note that $D$ is wellorderable, and let $\eta$ be an ordinal and $\sigma:\eta\to D$ a bijection. Then define $\pi:\eta\to V\_\lambda$ by $\pi(\xi)=\pi'(\sigma(\xi))$. Then $\pi$ is an injection, so $|\eta|\leq|V\_\lambda|$. I claim that $|\eta|=|V\_\lambda|$. For otherwise we have $|\eta|<|V\_\lambda|$. So applying the proposition at $\lambda$ with $S=\eta$, we get that $|\eta|<|V\_\alpha|$ for some $\alpha<\lambda$. But this easily contradicts the construction.
| 4 | https://mathoverflow.net/users/160347 | 438734 | 177,199 |
https://mathoverflow.net/questions/438737 | 2 | Let $G$ be a split reductive group and let $T$ be a split maximal torus whose rank is $l$. Is it possible to find a base $\gamma\_1,..., \gamma\_l$ of the weight lattice $X(T)$ such that the cone $C$ in the coweight lattice $Y(T)$ defined by
$$C=\{\psi\in Y(T)\vert (\psi, \gamma\_i)\geq 0\; i=1,...,l\}$$
forms a complete fan under the translation of the Weyl group in $Y(T)$?
If $G$ is an adjoint group, such a base is provided by simple roots and $C$ is simply the positive Weyl chamber. How about the general reductive group?
| https://mathoverflow.net/users/173314 | Find an analogue of Weyl chamber structure | Saying that the translates of C form a complete fan seems to imply that $C$ is a fundamental domain for the Weyl group: The translates cover space, and they form a fan, meaning the intersection of any two translates has lower dimension.
The Weyl group is generated by reflections. The interior of a fundamental domain can't intersect the fixed hyperplanes of these reflections, because then the intersection of the domain with its translate by that reflection must have full dimension. Since the fundamental domain is a cone, its interior is connected, and thus lies in a single connected component of the complement of these hyperplanes, i.e. a single Weyl chamber. Since its translates cover the space, it must make up a full Weyl chamber. So $C$ must be a Weyl chamber.
After translating, we can assume $C$ is the standard Weyl chamber. This happens only if the $\gamma\_i$s are positive real multiples of the simple roots.
So this is possible if and only if the least positive multiples of the simple roots that lie in the weight lattice in fact form a basis for the weight lattice. This fails, for example, for $SL\_n$.
| 3 | https://mathoverflow.net/users/18060 | 438740 | 177,201 |
https://mathoverflow.net/questions/438729 | 3 | Suppose $M$ and $N$ are complete metric spaces and $f, g: M \to N$ are uniformly continuous maps between them with common modulus of continuity $m$. Further suppose $f$ and $g$ are homotopy equivalent.
Must there be a a homotopy equivalence $\alpha\colon M \times [0, 1] \to N$ between $f$ and $g$ such that $\alpha$ is uniformly continuous?
If so, can we further require that $\alpha$ has a modulus of continuity $m^\*$ where $m^\*$ only depends on $m$? Can we require $m^\* = m$?
Similarly, if $M$ and $N$ are complete metric spaces which are homotopy equivalent, can we find a uniformly continuous homotopy between them? If so can we say anything about the modulus of continuity which is independent of the spaces (e.g. that they are Lipschitz)?
| https://mathoverflow.net/users/8106 | Uniformly continuous homotopy equivalence | I think not.
If $M$ is compact, then $\alpha$ must be uniformly continuous; but even then the modulus of continuity can be impossible to preserve; there is some interpretation on what is meant here, depending on the metric you use on the product $M\times [0,1]$, but we can have that there must exist some $t$ such that $\alpha\_t:=x\mapsto\alpha(x,t)$ does not have $m$ as modulus of continuity. For example, take $M$ the circle, $N$ an unduloid, and $f,g$ two short geodesics on two different necks of $N$. Then any homotopy must pass through a long curve at the fat part between the necks, which cannot have exactly the same modulus of continuity (here, $m^\*$ can be a multiple of $m$).
If $M$ is not compact even the equi-uniform continuity of $\alpha\_t$ can be prevented. Take $M=\mathbb{R}$, and for $N$ the plane with a sequence of hills centered at points $(0,n)$ where $n$ runs over $\mathbb{Z}$ and contained in a band $(\lvert y\rvert \le 1)$. Make the hills very steep and very high when $z$ is large. Let $f = x\mapsto (2,x)$ and $g = x\mapsto (-2,x)$. Then any homotopy from $f$ to $g$ has to incur high distortion to pass through the farther hills.
If $M$ and $N$ are homeomorphic, it might be impossible to get any modulus of continuity for an homeomorphism: a classical operation to see this is *snowflaking*: take $M=N$ but metric $d$ on $M$ and $d^q$ on $N$, where $q\in(0,1)$. Then they have different dimensions, ruling out a Lipschitz homeomorphism; and you can compose $d$ with an arbitrarily bad concave modulus of continuity to prevent any given modulus to be achieved, so nothing uniform over all metric spaces can be expected.
For homotopy equivalence, you can do this construction on a circle; a map of non-vanishing degree will incur the same distortion as a homeomorphism.
To prevent even uniformly continuous homotopies, one has to look for non-compact spaces again. I guess that using a loch-ness monster obtained by attaching a circle to the line at each integer point could do the trick. For $M$, make the metric give length $1$ to all circle. For $N$, make it give length $\lvert n\rvert$ to the circle at $n$. Any homotopy $M\to N$ must find a way to map a circle of $M$ to each of $N$, and this should prevent uniform continuity.
| 3 | https://mathoverflow.net/users/4961 | 438742 | 177,202 |
https://mathoverflow.net/questions/438746 | 1 | [This answer](https://math.stackexchange.com/a/3095770/897367) seems to imply that: for an ordinal $\alpha$, to be recursively inaccessible (i.e. $\alpha$ is admissible and limit of admissible) implies to be not locally countable (i.e. $L\_\alpha \models \exists \beta \ ``\beta \text{ is uncountable"}$). Here is the relevant excerpt:
>
> If there is some $r\in L\_\alpha$ with $\omega\_1^r=\alpha$, then $L\_\alpha$ will be *locally countable* (= $L\_\alpha\models$ "every set is countable"). But plenty of countable admissible $\alpha$s don't give rise to locally countable levels of $L$! In particular, if $\alpha$ is an admissible limit of admissibles (= "recursively inaccessible") then every real in $L\_\alpha$ is contained in some admissible $L\_\beta$ with $\beta<\alpha$.
>
>
>
However I believe that if by "admissible" we read here "$\Sigma\_1$-admissible", this statement does not hold as in this sense the first recursively inaccessible ordinal appears way before the first ordinal that is not locally countable (see resp. 2.3 and 2.21 in DA Madore's [zoo of ordinals](http://www.madore.org/%7Edavid/math/ordinal-zoo.pdf)). Whence my question:
>
> Does the statement hold if by "admissible" we read "$\Sigma\_n$-admissible for all $n$"? If no, can we describe the first ordinal which is not locally countable in term of higher recursive inaccessibility?
>
>
>
| https://mathoverflow.net/users/138089 | Recursively inaccessible ordinals and non locally countable ordinals | Suppose $L\_\alpha\models$"There is a largest cardinal". Then ($L\_\alpha$ is $\Sigma\_n$-admissible for all $n<\omega$) $\Leftrightarrow$ ($L\_\alpha\models$ ZF$^-$) $\Leftrightarrow$ ($L\_\alpha$ does not project ${<\alpha}$) $\Leftrightarrow$ ($\alpha$ is a cardinal in $L\_{\alpha+1}$). So the least $\beta$ such that $L\_\beta\models$"$\omega\_1$ exists" is $\beta=\alpha+1$ where $\alpha=\omega\_1^{L\_\beta}$ is the least ordinal such that $L\_\alpha$ is $\Sigma\_n$-admissible for all $n<\omega$, hence the least such that $L\_\alpha\models$ ZF$^-$.
However, if $\alpha$ is least such that ($L\_\alpha\models$ ZF$^-$ and $\alpha$ is a limit of $\beta$'s such that $L\_\beta\models$ ZF$^-$), then $L\_\alpha\models$"Every set is countable". (For each of those $\beta$'s, $L\_\beta$ is pointwise definable, so $\beta$ is countable in $L\_{\beta+2}$).
If we let $\alpha$ be least such that $L\_\alpha\models$" KP + $\omega\_1$ exists", then $\omega\_1^{L\_\alpha}$ is a limit of limits of ZF$^-$ levels, a limit of limits of limits of ZF$^-$ levels, etc.
| 3 | https://mathoverflow.net/users/160347 | 438752 | 177,205 |
https://mathoverflow.net/questions/438719 | 9 | The question is as in the title:
Is there a nonpolynomial $C^\infty$ function $f$ on $\mathbb{R}$ such that $\sup\_{x \in \mathbb{R}} \lvert f^{(q)}(x) \rvert \leq (\ln q)^{-q}$ for every natural number $q >1$?
Here, "nonpolynomial" excludes constant functions as well, of course.
I think nonpolynomial functions "very uniformly approximating" a constant function" might satisfy such a bound, but I cannot find such a nice example.
Could anyone please help me?
| https://mathoverflow.net/users/56524 | Is there a nonpolynomial $C^\infty$ function $f$ such that $\sup_{x \in \mathbb{R}} \lvert f^{(q)}(x) \rvert \leq (\ln q)^{-q}$ for every $q >1$? | The answer is no. From [Taylor's theorem with remainder](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder), we see that for any integer $q>2$, we have
$$ f^{(2)}(x) = \sum\_{j=0}^{q-3} \frac{f^{(2+j)}(0)}{j!} x^j + O( \frac{(\ln q)^{-q} |x|^{q-2}}{(q-2)!} )$$
and thus for $|x| \leq \frac{q \ln q}{10}$ (say) we have from Stirling's formula that
$$ \sum\_{j=0}^{q-3} \frac{f^{(2+j)}(0)}{j!} x^j = O( 1 ).$$
The key point here is that we have bounded control on a polynomial on an interval of length much wider than its degree (a situation in which one expects favorable estimates thanks to the uncertainty principle).
Making the trigonometric substitution $x = \frac{q \ln q}{10} \sin \theta$ we conclude
$$ \sum\_{j=0}^{q-3} \frac{f^{(2+j)}(0)}{j!} (\frac{q \ln q}{10})^j \sin^j \theta = O( 1 )$$
for all $\theta$. The LHS is a trigonometric polynomial of degree $O(q)$, so by the [Bernstein inequality](https://en.wikipedia.org/wiki/Bernstein%27s_theorem_(polynomials)#Bernstein%27s_inequality) we have
$$ \frac{d}{d\theta} \sum\_{j=0}^{q-3} \frac{f^{(2+j)}(0)}{j!} (\frac{q \ln q}{10})^j \sin^j \theta = O( q )$$
for all $\theta$. Evaluating this at $\theta=0$ we conclude that
$$ f^{(3)}(0) \frac{q \ln q}{10} = O(q).$$
Dividing by $q \ln q$ and sending $q \to \infty$ we conclude that $f^{(3)}(0)$ vanishes. Applying a translation to this argument, we see that $f^{(3)}$ vanishes everywhere, thus $f$ is a polynomial (in fact it is quadratic).
| 17 | https://mathoverflow.net/users/766 | 438755 | 177,208 |
https://mathoverflow.net/questions/438766 | 0 | $\mathcal{S}^{1/2}\_{1/2}(\mathbb{R})$ is defined to be the collection of $C^\infty$ functions $f$ on $\mathbb{R}$ such that
\begin{equation}
\sup\_{x \in \mathbb{R}} \lvert x^k f^{(q)}(x) \rvert \leq CA^kB^q k^{k/2} q^{q/2}
\end{equation}
where $k$ and $q$ are nonnegative integers and $A,B,C$ are positive constants depending on $f$.
In fact, I am aware that it is a special case of more general function spaces $\mathcal{S}^{\alpha}\_{\beta}(\mathbb{R})$ for nonnegative $\alpha, \beta$.
I am currently trying to find smooth functions satisfying some stronger bounds. For example, are there smooth functions $f$ satisfying the bounds
\begin{equation}
\sup\_{x \in \mathbb{R}} \lvert x^k f^{(q)}(x) \rvert \leq CB^q k^{1/8} q^{q/2}
\end{equation}
for positive constants $C, B$ depending on $f$? I suspect that Hermite functions might satisfy these bounds, but cannot prove rigorously.
Could anyone please provide any example?
| https://mathoverflow.net/users/56524 | Existence of smooth functions $f$ satisfying $\sup_{x \in \mathbb{R}} \lvert x^k f^{(q)}(x) \rvert \leq C B^q k^{1/8} q^{q/2}$ | Start from the estimate (with $q = K$ and $k = K+1$) (valid for $x > 0$)
$$ |f^{(K)}(x)| \leq C B^{K} (K+1)^{1/8} K^{K/2} x^{-K-1}$$
Integrate back from infinity $K$ times, you get
$$ |f(x)| \leq C \frac{B^K (K+1)^{1/8} K^{K/2}}{K!} \frac{1}{x} $$
Using Stirling's approximation
$$ |f(x)| \lesssim \frac{e^K B^K (K+1)^{1/8} K^{K/2}}{K^{1/2} K^K} \frac{1}{x} $$
with the implicit constant independent of $K$.
Take the limit $K \to \infty$ you see that the RHS converges to 0. This shows that the only function with the specified property is $f \equiv 0$.
| 4 | https://mathoverflow.net/users/3948 | 438768 | 177,212 |
https://mathoverflow.net/questions/438744 | 3 | I need the result that for all $t$,
$$\sum\_{i=0}^{\lfloor t/2 \rfloor} (-1)^{i+1} \binom{t-i}{i} C\_{t-i-1} = 0,$$
where $C\_{t-i-1}$ is the $(t-i-1)$-th Catalan number. I've checked for $t$ up to 1000 using Python and the result holds, but I don't really have an intuition for why it would be true. The terms of this sequence are on OEIS ([A068763](https://oeis.org/A068763)) but they're simply called a "generalized Catalan sequence".
Does anyone have a name for this sequence or a citation for this result?
| https://mathoverflow.net/users/497997 | Why is this alternating sum involving Catalan numbers $\sum_{i=0}^{\lfloor t/2 \rfloor} (-1)^{i+1} \binom{t-i}{i} C_{t-i-1} = 0$ for all $t$? | The formula (which holds for $t>1$ but not for $t=1$), is equivalent to
$$\sum\_{t=1}^\infty C\_{t-1}\bigl(x(1-x)\bigr)^t = x,$$
which follows directly from the generating function
$$\sum\_{t=1}^\infty C\_{t-1}x^t = \frac{1-\sqrt{1-4x}}{2}.$$
| 4 | https://mathoverflow.net/users/10744 | 438771 | 177,213 |
https://mathoverflow.net/questions/437803 | 2 | Let $A$ be a Cartan matrix, i.e. a $n\times n$ matrix with integer entries such that $A\_{ii}=2$ and $A\_{ij}\leq0$ for $i\neq j$. Then the corresponding Kac-Moody Lie algebra has a Cartan subalgebra $\mathfrak{h}$, and a root system $\Delta\subseteq\mathfrak{h}^\*$, along with a Weyl group $W$ acting on $\mathfrak{h}^\*$ and preserving $\Delta$.
Within $\Delta$ we have the simple roots $\Sigma$, which are a finite set. A root $\alpha\in\Delta$ is called real if it is in the $W$-orbit of a simple root.
Given a real root $\gamma$ and a simple root $\alpha$, we obtain an $\alpha$ root string via $\gamma,\gamma+\alpha,\dots,r\_{\alpha}(\gamma)$.
My question is: suppose that $A$ is $3\times 3$ with all entries nonzero. Under what conditions will there exist a real root $\beta$ such that it lies along an $\alpha$ root string of another real root $\gamma$ for some simple root $\alpha$, but with $\beta\neq\gamma,r\_{\alpha}\gamma$.
As an example: if $A=\begin{bmatrix}2 & -1 \\ -r & 2\end{bmatrix}$ for $r>1$, then if $\alpha\_1,\alpha\_2$ are the simple roots then we have $\alpha\_1+\alpha\_2$ is real and lies along the $\alpha\_1$-root string through $\alpha\_2$ given by $\alpha\_2,\alpha\_2+\alpha\_1,\dots,\alpha\_2+r\alpha\_1$. I believe for all other $2\times 2$ Cartan matrices this does not occur.
| https://mathoverflow.net/users/97652 | Real roots along root strings | The characterisation you are looking for is given by the following theorem of Jun Morita (*Root strings with three or four real roots in Kac-Moody root systems*, Tohoku Math. J. 40 (1988), p. 645-650):
**Theorem** (Morita): Let $A = (a\_{ij})$ be an $n\times n$ generalized Cartan matrix, $\Delta$ the associated root system, and $\Delta^{re}$ the set of real roots. Put $r(\alpha;\beta) = |\{\beta + k\alpha \ | \ k\in \mathbf{Z}\} \cap \Delta^{re}|$ for $(\alpha,\beta)\in\Delta^{re}\times\Delta$. Then the following two conditions are equivalent.
1. $r(\alpha;\beta)\geq 3$ for some $(\alpha,\beta)\in\Delta^{re}\times\Delta$.
2. $a\_{ij}=-1$ and $a\_{ji}<-1$ for some $i,j$ ($1\leq i,j\leq n$).
| 1 | https://mathoverflow.net/users/106751 | 438791 | 177,220 |
https://mathoverflow.net/questions/438666 | 5 | I have recently stumbled across the problem of describing the characteristic subgroups of a finite abelian group. With some discussions with some mathematicians in my lab, I managed to obtain a "visual" description of such, and wrote it down.
Nonetheless, I know that some of this work has already been clear by Kaplansky in "Infinite abelian groups". Combining both theorems 25 and 26, we obtain a description for every finite abelian $p$-group for $p\neq 2$, and then for every abelian finite group of odd order. I managed to obtain a (less aesthetical, but still understoodable) description for the finite abelian $2$-groups that Kaplansky mentions in the theorem 27 as problematic. In this book, I didn't manage to find some reference for such groups, but maybe some person here knows an article that has already tackles this question.
So my question is : has some text already given a reasonable parametrization of characteristic subgroups of finite abelian $2$-groups ?
| https://mathoverflow.net/users/242651 | Characteristic subgroups of a finite abelian $2$-group | It seems to me that there is the requisite description here Kerby, B. L., Rode, E. Characteristic Subgroups of Finite Abelian Groups. Communications in Algebra, 2011, 39:4, 1315-1343 (Section 2).
Theorem 2.9 describes characteristic subgroups of abelian 2-groups.
| 5 | https://mathoverflow.net/users/173068 | 438796 | 177,222 |
https://mathoverflow.net/questions/438793 | 4 | Let $n$ be a positive integer and $1\leq j\leq n$. Consider the following polynomial:
$$p\_{n,j}(x)=\frac{\prod\limits\_{i=1}^{n+1}\frac{x^{i}+1}{x+1}}{\prod\limits\_{i=1}^{j}\frac{x^{i}+1}{x+1}\prod\limits\_{i=1}^{n-j+1}\frac{x^{i}+1}{x+1}}\in\mathbb{F}[x]$$
This polynomial can be computed for each $j,n$ given. What I want to do is to understand in which cases $p(1)=0$ and in which cases $p(1)=1$ when working over a field $\mathbb{F}$ of characteristic $2$. I checked some cases by hand and this is what I got:
$$p\_{2,1}(1)=p\_{2,2}(1)=1$$
$$p\_{3,1}(1)=p\_{3,2}(1)=p\_{3,3}(1)=0$$
$$p\_{4,1}(1)=p\_{4,4}(1)=1~~p\_{4,2}(1)=p\_{4,3}(1)=0$$
$$p\_{5,2}(1)=p\_{5,4}(1)=1~~p\_{5,1}(1)=p\_{5,3}(1)=p\_{5,5}=0$$
$$p\_{6,1}(1)=p\_{6,2}(1)=p\_{6,3}(1)=p\_{6,4}(1)=p\_{6,5}(1)=p\_{6,6}(1)=1$$
So apparently there is no easy pattern. It is also easily expressible as a product of cyclotomic polynomials, but I'm not sure if this is of any help. I would like (if possible) to have a way to know if $p\_{n,j}(1)=0$ or $p\_{n,j}(1)=1$ just in terms of $n$ and $j$.
If you are courious, these are the coefficients that has appeared in the computation of the homology groups of a certain family of Artin groups. I want to understand the behaviour of these polynomials over a field of characteristic $2$ in $x=1$ because this is the situation where I can assure that the homology groups are infinite dimensional.
Thanks for your help.
| https://mathoverflow.net/users/482329 | Vanishing of a product of cyclotomic polynomials in characteristic 2 | *Claim:* $p\_{n,j}(1)$ vanishes modulo $2$ if and only there is a carry when adding $j$ and $n+1-j$ in base-$2$.
*Proof:* Let us write $x^n+1 \equiv x^n-1= \prod\_{d \mid n} \Phi\_d(x)$ where $\Phi\_d$ is the $d$th cyclotomic polynomial. Then, since there are $\lfloor m/d\rfloor$ multiplies of $d$ in $\{1,2,\ldots,m\}$,
$$p\_{n,j}(x) = \prod\_{d=2}^{n} \Phi\_d(x)^{\lfloor \frac{n+1}{d} \rfloor- \lfloor \frac{j}{d}\rfloor - \lfloor \frac{n-j+1}{d}\rfloor }.$$
Note that $\lfloor a+b \rfloor \ge \lfloor a\rfloor + \lfloor b \rfloor$ so your expression is necessarily a polynomial (and not a strictly rational function).
The value of $\Phi\_d(x)$ at $x=1$ is well-understood. One can use the property $x^n-1= \prod\_{d \mid n} \Phi\_d(x)$ to show $\Phi\_d(1)$ is $1$ unless $d$ is a prime power $p^k$ in which case $\Phi\_d(1)=p$ (this is classical, see [here for proofs](https://math.stackexchange.com/questions/49956/value-of-cyclotomic-polynomial-evaluated-at-1)). This implies
$$p\_{n,j}(1) \equiv \prod\_{p^k \le n} p^{\lfloor \frac{n+1}{p^k} \rfloor- \lfloor \frac{j}{p^k}\rfloor - \lfloor \frac{n-j+1}{p^k}\rfloor }.$$
Since $2$ is the only even prime,
$$p\_{n,j}(1) \equiv \prod\_{2^k \le n} 2^{\lfloor \frac{n+1}{2^k} \rfloor- \lfloor \frac{j}{2^k}\rfloor - \lfloor \frac{n-j+1}{2^k}\rfloor }.$$
So your expression vanishes modulo $2$ if and only if
$$ \sum\_{2 \le 2^k \le n} \left\lfloor \frac{n+1}{2^k} \right\rfloor- \left\lfloor \frac{j}{2^k}\right\rfloor- \left\lfloor \frac{n-j+1}{2^k}\right\rfloor\ge 1.$$
Let us express this criterion using base-$2$ representation. If $j = \sum\_{i \ge 0} a\_i 2^i$, $n-j+1 = \sum\_{i \ge 0} b\_i 2^i$ and $n+1 = \sum\_{i \ge 0} c\_i 2^i$ then
$$ \left\lfloor \frac{n+1}{2^k} \right\rfloor- \left\lfloor \frac{j}{2^k}\right\rfloor- \left\lfloor \frac{n-j+1}{2^k}\right\rfloor =\sum\_{i \ge k} (c\_i-a\_i-b\_i) 2^{i-k} $$
so that
$$\begin{align}
\sum\_{2 \le 2^k \le n} \left\lfloor \frac{n+1}{2^k} \right\rfloor- \left\lfloor \frac{j}{2^k}\right\rfloor- \left\lfloor \frac{n-j+1}{2^k}\right\rfloor &= \sum\_{i \ge 0} (c\_i-a\_i-b\_i)(2^{i-1}+2^{i-2}+\ldots+1) \\
&= \sum\_{i \ge 0} (c\_i-a\_i-b\_i)(2^{i}-1)\\
&= n+1 - (j + (n+1-j)) - \sum\_{i \ge 0} (c\_i-a\_i-b\_i)\\
&= \sum\_{i \ge 0} (a\_i+b\_i-c\_i).
\end{align}$$
Let $s\_2(m)$ be the sum of digits of $m$ in base-$2$. We see that $p\_{n,j}(1)\equiv 0$ if and only if $s\_2(j) + s\_2(n+1-j) > s\_2(n+1)$.
In general, $s\_2(a)+s\_2(b)-s\_2(a+b)$ is the number of carries when adding $a$ and $b$ in base-$2$ (can be proved by induction, goes back to Kummer).
| 5 | https://mathoverflow.net/users/31469 | 438803 | 177,224 |
https://mathoverflow.net/questions/438785 | 1 | A positive irrational number $\alpha\in{\mathbb R}\setminus {\mathbb Q}$ is said to be a *[Brjuno number](https://en.wikipedia.org/wiki/Brjuno_number)* if $$\sum\_{i=1}^\infty\frac{\log q\_{i+1}}{q\_i} < \infty$$ where $q\_i>0$ is the denominator of the $i$th [convergent](https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents) $\frac{p\_i}{q\_i}$ of the [continued fraction](https://en.wikipedia.org/wiki/Continued_fraction) expansion of $\alpha$.
Is the set $B$ of Brjuno numbers measurable, and if yes, what is the measure of $B\cap[0,1]$?
| https://mathoverflow.net/users/8628 | Measurability of Brjuno numbers | I replace $\alpha$ by $x$ below.
Restricting ourselves to positive numbers is useless, since the property considered does not depend on $a\_0(x) = \lfloor x \rfloor$. Therefore, $B$ is the union of all images of $B \cap [0,1[$ by integer translations.
Now, call $G$ the Gauss-Kuzmin map from $[0,1[ \setminus \mathbb{Q}$ to itself: $G(x) = 1/x-\lfloor 1/x \rfloor$.
This map preserves the probability measure $\mu$ with density $x \mapsto (\ln 2)^{-1}(1+x)^{-1}$ on $[0,1[ \setminus \mathbb{Q}$ and is ergodic.
The partial quotients $(a\_n(x))\_{n \ge 1}$ of $\alpha$ are the integer parts of the iterates $(G^n(x))\_{n \ge 1}$. Hence $(a\_n)\_{n \ge 1}$ and $(q\_n)\_{n \ge 1}$ are Borel functions, so $B$ is a Borel set.
One checks that $\ln(a\_1+1)$ is integrable with regard to $\mu$. Birkhoff theorem applies, so when $n \to +\infty$
$$\frac{\ln q\_n}{n} \le \frac1n\sum\_{k=1}^n \ln(a\_k+1) \to C:= \int\ln(a\_1+1)\mathrm{d}\mu \quad \mu \text{-a.s..}$$
On the other hand, $(q\_n)\_{n \ge 1}$ is bounded below by Fibonacci sequence. Therefore, by comparison, the series $\sum\_n \frac{\ln q\_{n+1}}{q\_n} $ converges $\mu$-almost surely, and also almost surely for the Lebesgue measure.
| 1 | https://mathoverflow.net/users/169474 | 438804 | 177,225 |
https://mathoverflow.net/questions/438761 | 9 | Infinitesimal analysis (by which I mean that originating from topos theory---not the nonstandard analysis of Robinson) seeks to recover the pre-limit notions of calculus (which are sufficiently useful to persist in applications and intuition) in a formal way by throwing out the law of the excluded middle, which led to the paradoxes remedied in the 19th century.
My question relates not to the theory, which is certainly able to be constructed from scratch, and, rather, with how the models by which we're assured of the theory's consistency are founded.
As I understand it, from what little I've read of John Bell's 2008 monograph *par excellence,* these models are toposes into which it's possible to embed $\mathbf{Man}$ without introducing nonsmooth entities. However, if our goal is to completely supplant limit analysis (for, say, pedagogical or constructivist reasons; it's a noble crusade either way), it seems like this is a dependency loop: we need machinery to define $\mathbf{Man}$ as "the category of smooth manifolds and maps," but the entire utility of atlases is their reduction of computation on manifolds to flat analysis---which is what we're trying to replace.
Is this actually a problem? Is it known? Tractable?
| https://mathoverflow.net/users/160917 | Are the models of infinitesimal analysis (philosophically) circular? | It is not circular for us to prove the consistency of noneuclidean geometry by providing an interpretation of noneuclidean geometry within euclidean geometry, such as with the Poincaré disk model of hyperbolic geometry. Rather, these interpretations are important because they establish the basic coherence of the other theory—they give us the relative consistency result, which gives us the confidence that the other theory has its own basic integrity, at least as much as the standard alternative.
Similarly, it is not circular when we construct a model of ZF+$\neg$AC by forcing over a model of ZFC, or a model of ZFC+$\neg$CH by forcing over a model of ZFC+CH. Rather these arguments show the basic coherence and relative consistency of the other theory. The theories ZF+AC, ZF+$\neg$AC, ZFC+CH, ZFC+$\neg$CH are all equiconsistent with each other, equally safe from a consistency point of view.
It seems to be the same situation in your case. At issue historically was the worry whether the infinitesimal approach to calculus might simply be incoherent. Robinson's nonstandard analysis and the approach you mention show various (different) senses in which it is coherent.
| 21 | https://mathoverflow.net/users/1946 | 438808 | 177,227 |
https://mathoverflow.net/questions/438757 | 3 | Suppose $X\_n$ $n = 1, 2, \ldots$ are i.i.d random variables with $\mu := \mathbb{E}[X\_n]$ > 0. (although they are not necessarily non-negative). Then if $S\_n = \sum\_{k=1}^n X\_k$ and $\tau\_a$ = $\inf \{n \geq 1 : S\_n \geq a\}$ - so that $\tau$ is the first time that the random sum exceeds the value a. Does there exist some $b,c$ independent of $a$ such that $\mathbb{E}[\tau\_a] \leq b+c a$? (or is there even just some way to bound $\mathbb{E}[\tau\_a]$ in expectation?)
Additionally we may assume that all moments of $X\_n$ exist if needed.
I tried a naive bound using Chebyshev's inequality on $S\_n$ to say that it should be close to $n \mu$, and then $\mathbb{E}[\tau\_a] = \sum\_{k=1}^{\infty} \mathbb{P}[\tau\_a \geq k] \leq \sum\_{k=1}^{\infty}\mathbb{P}[S\_k \leq a]$ - however the details don't quite work out.
I also tried considering $M\_n = S\_n - n\mu$ as a martingale and trying to use optional stopping, but since $M\_{n \land \tau}$ isn't bounded from below I also can't quite see how to make it work.
Another idea is trying to use [Walds theorem](https://en.wikipedia.org/wiki/Wald%27s_equation#Example_with_dependent_terms), but assumption 3 in the wikipedia page gives me the exact same difficulties as trying to use optional stopping
I believe this should be possible, since if we view $S\_k$ as the martingale $M\_k$ and some drift, then we are in a similar case to Brownian Motion with drift,although our case is discreet. Then the hitting times $\rho\_a$ of Brownian motion with drift (which are now continuous) satisfy $\mathbb{E}[\rho\_a] \leq c a$, and an explicit pdf can be found.
| https://mathoverflow.net/users/497999 | First time random sum exceeds value | $\newcommand\al\alpha$Your desired bound is easy to get if we assume that $\alpha\_p:=E|X\_1-\mu|^p<\infty$ for some real $p\in(2,3)$.
Indeed,
$$E\tau\_a=E\sum\_{n=0}^{\tau\_a-1} 1=E\sum\_{n=0}^\infty 1(\tau\_a>n)=\sum\_{n=0}^\infty P(\tau\_a>n). \tag{1}\label{1}$$
Next, if $n\ge2a/\mu$, then
$$P(\tau\_a>n)\le P(S\_n<a)=P(S\_n-n\mu<a-n\mu)
\le\frac{E|S\_n-n\mu|^p}{|n\mu-a|^p}
\le2^p\frac{E|S\_n-n\mu|^p}{n^p\mu^p}
\le C\_1\frac{n\al\_p+n^{p/2}\al\_2^{p/2}}{n^p\mu^p}
\le\frac{C\_2}{n^{p/2}\mu^p};$$
here, $C\_1$ is a universal positive real constant, $C\_2$ is a positive real number depending only on $\al\_p$, and the penultimate inequality is an application of [Rosenthal's inequality](https://mathoverflow.net/a/298563/36721).
So, by \eqref{1},
$$E\tau\_a\le\sum\_{0\le n<2a/\mu}1+\sum\_{n\ge2a/\mu}\frac{C\_2}{n^{p/2}\mu^p}
\le\frac2\mu\,a+1+\frac{C\_3}{\mu^p}, \tag{2}\label{2}$$
where $C\_3$ is a positive real number depending only on $\varepsilon:=p/2-1>0$ and $\al\_p$. $\quad\Box$
One can replace the factor $\dfrac2\mu$ in the upper bound on $E\tau\_a$ in \eqref{2} by $\dfrac{1+\delta}\mu$, for any real $\delta>0$, but then $C\_3$ will have to depend on $\delta$ as well.
| 3 | https://mathoverflow.net/users/36721 | 438813 | 177,229 |
https://mathoverflow.net/questions/438811 | 1 | Let $\mathbf{V}$ be a $p\times p$ orthogonal matrix (i.e., $\mathbf{V}\mathbf{V}^\top = \mathbf{V}^\top \mathbf{V} = \mathbf{I}$) whose columns are
$$
\mathbf{V} = \begin{bmatrix} \mathbf{v}\_1 & \mathbf{v}\_2 & \cdots & \mathbf{v}\_p \end{bmatrix}.
$$
Denote its submatrix as
$$
\mathbf{V}\_r = \begin{bmatrix} \mathbf{v}\_r & \cdots & \mathbf{v}\_p \end{bmatrix}.
$$
for some $r>0$, so the $\mathbf{V}\_r$ is a $p\times (p-r+1)$ matrix.
Then the question is
>
> **Question.**
>
>
> Under what condition $p^{-1} \mathbf{1}^\top \mathbf{V}\_r \mathbf{V}\_r^\top \mathbf{1} \rightarrow C$ for some well-defined value $C$ as $p\rightarrow \infty$? (Here $r$ is fixed, and $\mathbf{1}$ is a vector of ones.)
>
>
>
I apologize that this question would be too naive, but I just want to find some intuition about under what special structure of $\mathbf{V}\_r$ makes $p^{-1} \mathbf{1}^\top \mathbf{V}\_r \mathbf{V}\_r^\top \mathbf{1}$ converges. This question has a quite important meaning in the high-dimensional inference of some factor model because under the factor model
$$
\mathbf{y}\_t = \mathbf{B} \mathbf{f}\_t + \mathbf{\varepsilon}\_t,
$$
where $\mathbf{f}\_t$ is a $r$-dimensional random factors, and its covariance
$$
\mathbf{\Sigma}\_y = \mathbf{B} \mathbf{\Sigma}\_f \mathbf{B}^\top + \mathbf{\Sigma}\_{\varepsilon}
= \mathbf{V} \mathbf{\Lambda} \mathbf{V}^\top, \qquad (\text{by the singular value decomposotion}.)
$$
its submatrix of eigenvectors $\mathbf{V}\_r$ corresponds to the idiosyncratic part $\mathbf{\Sigma}\_{\varepsilon}$. So $p^{-1} \mathbf{1}^\top \mathbf{V}\_r \mathbf{V}\_r^\top \mathbf{1}$ corresponds to the (normalized) variance of equally weighted portfolio due to the idiosyncratic risk. To estimate this value, we have to guarantee that it has a well-defined limit.
So it would be very appreciated if you let me know if there is any special structure for that (or is there any good material to study).
**Here what I have done.**
Notice that if $r=1$, then it is just
$$
p^{-1} \mathbf{1}^\top \mathbf{V}\_r \mathbf{V}\_r^\top \mathbf{1} = p^{-1} \mathbf{1}^\top \mathbf{V} \mathbf{V}^\top \mathbf{1} = 1,
$$
so it is meaningful to consider $r>1$.
It is easy to show that $p^{-1} \mathbf{1}^\top \mathbf{V}\_r \mathbf{V}\_r^\top \mathbf{1}$ is bounded:
$$
0 \le p^{-1} \mathbf{1}^\top \mathbf{V}\_r \mathbf{V}\_r^\top \mathbf{1} \le p^{-1} \| \mathbf{1}\|^2 \| \mathbf{V}\_r\|^2 = 1,
$$
where $\| \mathbf{V}\_r\|$ is a spectral norm of the matrix $\mathbf{V}\_r$.
So it has a convergent subsequence, and the problem is that showing every subsequence converges to the same limit $C$.
Also, notice that this value is invariant under any rotation $\mathbf{O}$ because
$$
\mathbf{1}^\top \mathbf{V}\_r \mathbf{O} \mathbf{O}^\top \mathbf{V}\_r^\top \mathbf{1} = \mathbf{1}^\top \mathbf{V}\_r \mathbf{V}\_r^\top \mathbf{1},
$$
so this value is the same for any $\mathbf{V}\_r'$ who has the same span of $\mathbf{V}\_r$. As an illustrative example of $\mathbf{V}\_r$, $\mathbf{v}\_i = \mathbf{e}\_i$ have a clearly well-defined limit of $p^{-1} \mathbf{1}^\top \mathbf{V}\_r \mathbf{V}\_r^\top \mathbf{1} = (p-r+1)/p \rightarrow 1$. Also, Paul (2007, [This paper](https://www3.stat.sinica.edu.tw/statistica/oldpdf/A17n418.pdf)) shows that eigenvector of covariance matrix of i.i.d. Gaussian normal $\mathcal{N}(\mathbf{0}, \mathbf{I})$ distributed uniformly on the surface of unit sphere $\mathbb{S}^{p-1}$. This implies that each element of the eigenvector distributed standard normal (see [this](https://mathoverflow.net/q/315221/159685) or [this](https://math.stackexchange.com/q/444700/498394)), and by the law of large number it has a clearly well-defined limit.
| https://mathoverflow.net/users/159685 | Convergent condition of the high-dimensional submatrix of some orthogonal matrix | For a given matrix $V$ this number $Z=p^{-1} \mathbf{1}^\top \mathbf{V}\_r \mathbf{V}\_r^\top \mathbf{1}$ will converge to $1$ if you take the limit $p\rightarrow\infty$ at fixed $r$. If you average over $V$ (with the Haar measure), the convergence to $(p-r+1)/p$ will apply also for finite ratio $p/r$.
At the left you see a plot of $Z$ for a single matrix $V$ (drawn from $O(p)$ with the Haar measure), as a function of $p$ for $r=5$. Notice the large statistical fluctuations, which are only damped out when $p/r\rightarrow\infty$. At the right the data has been averaged over 10 random choices of $V$, and the curve $(p-r+1)/p$ is followed more closely already for $p/r\gtrsim 5$.
Blue points are $(p-r+1)/p$, gold points are the number $Z$ for a single randomly chosen $V$.
| 1 | https://mathoverflow.net/users/11260 | 438817 | 177,230 |
https://mathoverflow.net/questions/438815 | 0 | Let $X$ and $Y$ be continuous random variables with finite first and second moments. Then, is it true that $Var[X]\geq Var[E(X|Y)]$?
| https://mathoverflow.net/users/490493 | Is the unconditional variance of a RV an upper bound for the variance of any conditional expectation of the RV? | It is enough to show
$$
Var[X] - Var[E[X|Y]] \ge 0.
$$
By the fact $Var[X] = E[X^2] - E[X]^2$,
$$
Var[X] - Var[E[X|Y]] = E[X^2] - E[X]^2 - E\big[ E[X|Y]^2 \big] + E \big[E[X|Y]\big]^2 \\
= E\big[ X^2 - E[X | Y]^2 \big] \\
= E \big[ E\big[X^2 - E[X|Y]^2 \big| Y\big] \big] = E[Var[X|Y]\big] \ge 0,
$$
where the second and the third equality is due to the law of total expectation.
| 2 | https://mathoverflow.net/users/159685 | 438819 | 177,232 |
https://mathoverflow.net/questions/438807 | 4 | Is $d^2+d+1$ prime for infinitely many $d\in \mathbb{Z}\_{>0}$?
This is expected by the [Bunyakovsky conjecture](https://en.wikipedia.org/wiki/Bunyakovsky_conjecture) which says that, under some conditions, given a polynomial $p(x) \in \mathbb{Z}[x]$ we have $p(d)$ prime for infinitely many $d\in \mathbb{Z}\_{>0}$. Is there some proof of this when $p(x) = x^2+x+1$?
| https://mathoverflow.net/users/100511 | Primes of the form $d^2+d+1$ | To answer the question and summarizing the comments: the answer is no, there is no known proof of this conjecture.
| 7 | https://mathoverflow.net/users/10898 | 438820 | 177,233 |
https://mathoverflow.net/questions/438798 | 2 | I am currently reading volume 2 of "Generalized Functions" by Gelfand and $\mathcal{S}^{2}\_0(\mathbb{R})$ is defined to be the collection of $C^\infty$ functions $f$ on $\mathbb{R}$ such that
\begin{equation}
\sup\_{x \in \mathbb{R}}\lvert x^k f^{(q)}(x) \rvert \leq CA^kB^q q^{2q}
\end{equation}
where $k$ and $q$ are arbitrary nonnegative integers and $A,B,C>0$ are constants depending on $f$.
I also figured out that any function in $\mathcal{S}^{2}\_0(\mathbb{R})$ must have a compact support.
**My question is this**: does there exist a certain $f$ for prescribed $A,B,C>0$? For example, if $A \in (0,1]$ is given, can we find some function $f \in \mathcal{S}^{2}\_0(\mathbb{R})$ satisfying the above bounds with this given $A$?
The function space of $\mathcal{S}$-type are quite new to me. Could anyone please help me understand them?
Add : Here, $2$ can be replaced by any bigger number. That is, could anyone provide a nontrivial element for some $\mathcal{S}^b\_0$ where $b$ can be as large as one wants?
| https://mathoverflow.net/users/56524 | Description of $\mathcal{S}^{2}_0(\mathbb{R})$ and certain class of functions inside it | $\newcommand\S{\mathcal S}\newcommand\R{\mathbb R}$For $x\in(0,1)$, let $g(x):=\frac1x$, and then let
$$f(x):=e^{-g(x)-g(1-x)}$$
for $x\in(0,1)$, with $f(x):=0$ for real $x\notin(0,1)$.
Then $f\in\S\_0^3(\R)$ -- see the details on this at the end of the answer.
As noted in [Willie Wong's comment](https://mathoverflow.net/questions/438798/description-of-mathcals2-0-mathbbr-and-certain-class-of-functions-in#comment1131557_438798), you can rescale $f$ vertically and horizontally to get desired $C$ and $A$ (and you said you do not need $B$ to be small).
**Details:** Everywhere here, $x\in(0,1)$. We have
$$f'(x)=P\_1(x)f(x),$$
where $P\_1(x):=\frac1{x^2}-\frac1{(1-x)^2}$. It is now easy to see by induction that for any $r=0,1,\dots$
$$f^{(r)}(x)=P\_r(x)f(x),$$
where $P\_r(x)$ is a polynomial in $\frac1x\,,\frac1{1-x}$ of total degree $\le2r$, containing $\le C^r$ monomials with coefficients $\le C^r r!$ in absolute value; here and elsewhere $C$ denotes various universal positive real constants. Also, for any nonnegative integers $m$ and $p$ such that $m+p\le2r\le2q$ we have
$$\dfrac1{x^m(1-x)^p}\,e^{-g(x)-g(1-x)}\le e^{-m-p}m^m p^p
\le q^{m+p}\le q^{2q}.$$
Thus,
$$|f^{(q)}(x)|\le C^q q^{3q}.$$
---
By a more accurate accounting on the coefficients of the monomials of $P\_r(x)$, one can get
\begin{equation\*}
|f^{(q)}(x)|\le C^q q^{2q} \tag{10}\label{10}
\end{equation\*}
and thus $f\in\S\_0^2(\R)$.
Indeed, for $r=0,1,\dots$,
\begin{equation\*}
f^{(r+1)}(x)=(P\_r(x)f(x))'=(P'\_r(x)+P\_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}))f(x),
\end{equation\*}
so that
\begin{equation\*}
P\_{r+1}(x)=P'\_r(x)+P\_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2}).
\end{equation\*}
Consider any monomial $m(x)=c(\frac1x)^i(\frac1{1-x})^{p-i}$ of the polynomial $P\_r(x)$
in $\frac1x\,,\frac1{1-x}$; this monomial is of total degree $p$ and with coefficient $c$. Then $(m(x)f(x))'/f(x)$ is the sum of four monomials of $P\_{r+1}(x)$. Two of these four monomials are monomials of $P'\_r(x)$ (say they are of the 1st kind) and the other two monomials are monomials of $P\_r(x)(\tfrac1{x^2}-\tfrac1{(1-x)^2})$ (say they are of the 2nd kind).
The total degree of each of the two monomials of the 1st kind is $p+1$ and the absolute value of its coefficients is $\le p|c|\le2r|c|$. The total degree of each of the two monomials of the 2nd kind is $p+2$ and the absolute value of its coefficient is $\le |c|$.
So, if a monomial of $P\_q(x)$ is obtained by $k$ steps of the 1st kind (and, hence, $q-k$ of the 2nd kind), then it is of the form
\begin{equation\*}
M(x)=c(\tfrac1x)^i(\tfrac1{1-x})^{p-i}
\end{equation\*}
with
\begin{equation\*}
p=k+2(q-k)\quad\text{and}\quad |c|\le (2q)^k\le2^q q^k.
\end{equation\*}
Recalling now the inequality $u^s e^{-u}\le s^s$ for $u>0$ and $s\ge0$ (with $0^0:=1$), we see that
\begin{equation\*}
|M(x)f(x)|\le2^q q^k i^i (p-i)^{p-i}\le2^q q^k p^p=2^q h(k),
\end{equation\*}
where $h(k):=q^k (k+2(q-k))^{k+2(q-k)}$. Note that $h$ is log convex, $h(0)=(2q)^{2q}$, and $h(q)=q^{2q}$. So, $h(k)\le(2q)^{2q}$ for all $k\in[0,q]$.
So, for any monomial $M(x)$ of $P\_q(x)$ we have
\begin{equation\*}
|M(x)f(x)|\le2^q (2q)^{2q}=8^q q^{2q}.
\end{equation\*}
Since $P\_q(x)$ contains $\le4^q$ monomials, we conclude that \eqref{10} holds. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 438823 | 177,235 |
https://mathoverflow.net/questions/438702 | 2 | In an arXiv preprint [[2108.05125v1]](https://export.arxiv.org/abs/2108.05125v1), the authors use the following vertical Fourier decomposition (page 7 therein).
Let $(M,g)$ be a Riemannian surface and $SM$ be its unit tangent bundle. Denote by $V$ the vertical vector field on $SM$, i.e. $V = \partial/\partial \theta$ generated by the angle coordinate, $(x,\theta) \in SM$. The authors then write $L^2(SM,\mathbb{C}^n) = \bigoplus\_{k \in \mathbb{Z}} H\_k$ as an orthogonal decomposition into the eigenspaces $H\_k$ of $-iV$ corresponding to eigenvalue $k$.
Now, suppose $A = \sum\_{k \in \mathbb{Z}} A\_k \in C^{\infty}(SM,\mathfrak{u}(n))$, where $\mathfrak{u}(n)$ is the space of skew-Hermitian matrices.
The authors claim on page 31 that $A\_{-k} = -A\_k^\*$ for all $k \in \mathbb{Z}$ due to the skew-Hermitian property.
I cannot verify this equality. Let me give an example that illustrates my problem.
Suppose $\omega$ is a $\mathfrak{u}(n)$-valued 1-form on $M$ and consider $A(x,v) = \omega\_x(v)$.
Then $A$ is linear in its second argument. As such, for any vector field of the form $Z(x,v) = (0,\lambda v)$ in the canonical splitting of $TSM$ into its horizontal and its vertical component, we have $Z(A)(x,v) = dA(Z)(x,v) = A\_x(\lambda v) = \lambda A\_x(v)$. In other words, $A$ is an eigenfunction of $Z$ for the eigenvalue $\lambda$. This eigenvalue depends only on $Z$.
So, regardless of whether $A$ is skew-Hermitian or not (we only used linearity of $A$ in $v$), I do not see how there could be 1-forms that are eigenfunctions of $V$ for the eigenvalue $1$ and also other 1-forms that are eigenfunctions of $V$ for the eigenvalue $-1$, which should be the case given $A\_{-1} = -A\_1^\*$.
Related to this problem, I am not sure how to understand $-iV$: while $V$ is a vector field on $SM$, it seems that $-iV$ lives in $TTM$ but not in $TSM$.
Any help would be greatly appreciated.
| https://mathoverflow.net/users/117619 | Vertical Fourier decomposition for skew-Hermitian 1-forms | If $SM$ is trivial (such that you can view it as $M\times \mathbb S^1$ and use the angle $\theta$ to describe the second variable), the $k$th Fourier mode of $\mathbb A\in C^\infty(SM,\mathbb C^{n\times n})$ is given by $$
\mathbb A\_k(x)=\int\_{0}^{2\pi}\mathbb A(x,\theta)e^{-ik\theta}d\theta.
$$
Take the complex conjugate (which flips the sign in the exponential) and transpose to obtain $(\mathbb A\_k)^\*=(\mathbb A^\*)\_{-k}$. If $\mathbb A$ is skew Hermitian, this means that $\mathbb A\_k^\*=-\mathbb A\_{-k}$.
I don't quite understand your computation with $Z$, especially the equality $dA(Z)(x,v)=A\_x(\lambda v)$ does not seem to make sense. Keep in mind that up to this point it should work for *any* smooth function $\mathbb A = A$ on $SM$, such that $\mathbb A(x,\lambda v)$ is not even well-defined unless $\vert \lambda \vert=1$. Say the vector field $Z$ actually equals $V$, such that its flow is given by $\psi\_t(x,v)=(x,e^{it}v)$. Then
$$
V\mathbb A=\frac d {dt}\big\vert\_{t=0} \psi\_t^\*\mathbb A=\frac d {dt}\big\vert\_{t=0} \mathbb A(x,e^{it}v)=?
$$
If $\mathbb A(x,v)=\omega\_x(v)$ for a $1$-form that is *complex linear* in $v$, you can continue the computation and obtain $i\mathbb A$, which is to say that $\mathbb A = \mathbb A\_1 \in H\_1$. Being complex linear means that $\omega$ is a $(1,0)$-form (so it looks like $f dz$ in a local holomorphic chart of the Riemann surface $M$). More generally, the splitting $\Omega^1(M)=\Omega^{1,0}\oplus \Omega^{0,1}$ corresponds to the splitting $H^{1}\oplus H^{-1}$ for its lift to $SM$.
Regarding the meaning of $-iV$: We frequently view vector fields as first order differential operators, that is, $V\colon C^\infty(SM,\mathbb C)\rightarrow C^\infty(SM,\mathbb C)$ acts by taking the vertical derivative. In this viewpoint it is clear how to multiply by $i$. Your confusion might arise because $i$ is sometimes also used to denote a $90$ degree rotation in $T\_xM$. If we rotate $V(x,v)$ inside the vertical subbundle in $T\_{(x,v)}(TM)$ we obtain a vector field $V\_\perp$, which indeed fails to be tangent to $SM$. But this vector field is of no relevance in the paper.
| 2 | https://mathoverflow.net/users/126651 | 438827 | 177,236 |
https://mathoverflow.net/questions/438814 | 0 | Does there exist a function which is holomorphic in $|z|<1,$ continuous in $|z|\leq1$ and such that the series $\sum |a\_n|$ is divergent, where $a\_n$'s coefficients in the Taylor series expansion of $f?$
| https://mathoverflow.net/users/143655 | A holomorphic function in the open unit disk satisfying certain properties | Yes there are lots of classical examples by Hardy and Littlewood like say $f(z)=\sum \_{n \ge 1}e^{in \log n}\frac{z^n}{n^{3/4}}$.
Using the second derivative test for exponential sums with $f(u)=(u\log u+u\theta)/(2\pi)$, so $f''(u) \sim 1/M$ when $u \sim M$, the exponential sum $\sum\_{n=M}^{2M}e^{in\log n}e^{in\theta} =O(\sqrt M)$ uniformly in $\theta$ and an easy dyadic argument shows that $\sum\_{n=1}^Ne^{in\log n}e^{in\theta}=O(\sqrt N)$ also uniformly in $\theta$, so it follows that $f$ converges uniformly on the unit circle and obviously satisfies $\sum |a\_n|=\infty$
| 3 | https://mathoverflow.net/users/133811 | 438829 | 177,237 |
https://mathoverflow.net/questions/438842 | 3 | Let $X\subset\mathbb{P}^N$ be a smooth scheme theoretical complete intersection, and $H\subset X$ a linear subspace. Denote by $N\_{H,X}$ the normal bundle of $H$ in $X$.
If $\dim(H) = 1$, that is $H$ is a line, then $N\_{H,X}$ splits as a sum of line bundles. Does there exists an example with $\dim(H) \geq 2$ in which $N\_{H,X}$ does not split as a sum of line bundles?
Thank you.
| https://mathoverflow.net/users/14514 | Normal bundle of a linear subspace | For instance, if $X$ is a smooth 4-dimensional quadric in $\mathbb{P}^5$ and $H = \mathbb{P}^2$, the normal bundle fits into the exact sequence
$$
0 \to N\_{H/X} \to \mathcal{O}(1)^{\oplus 3} \to \mathcal{O}(2) \to 0,
$$
which implies that
$$
N\_{H/X} \cong \Omega\_H(2).
$$
In particular, it does not split.
| 11 | https://mathoverflow.net/users/4428 | 438848 | 177,243 |
https://mathoverflow.net/questions/438288 | 2 | Let $\Delta:=\partial\_z\,\partial\_{\overline {z}} $ be the Laplacian operator. I look for a particular non-trivial solution $u$ of $$\Delta u=\frac{a}{1-|z|^2}u$$ where $u\in C^2(\mathbb{D})$ and $a\in\mathbb{C} $.
| https://mathoverflow.net/users/497676 | Linear elliptic equation | The equation is happily linear, so depending on the domain you may find separable analytical solutions to the Dirichlet problem thanks to Sturm–Liouville.
For instance, let's take the domain to be a unit circle and set $u(r=1,\theta)=f(\theta)$ as boundary value. Construct the solution as
\begin{align}
u(r,\theta)=\sum\_{n=0}^\infty\left[A\_n\cos n\theta+B\_n\sin n\theta\right]\frac{R^{(a)}\_n(r)}{R^{(a)}\_n(1)},
\end{align}
where
\begin{align}
A\_n=\frac{1}{\pi}\int\_0^{2\pi}d\theta\: f(\theta)\cos n\theta\:,\:\:\:B\_n=\frac{1}{\pi}\int\_0^{2\pi}d\theta\: f(\theta)\sin n\theta,
\end{align}
and $R^{(a)}\_n(r)$ satisfies
\begin{align}
\frac{d^2R^{(a)}\_n}{dr^2}+\frac{1}{r}\frac{dR^{(a)}\_n}{dr}-\left[\frac{a}{1-r^2}+n^2\right]R^{(a)}\_n=0.
\end{align}
This last equation has in general two linearly independent solutions, but only one of them is well behaved as $r\rightarrow 0 $.
For $n=0$ (which is the full solution if $f(\theta)$ is constant) one has, as rightly mentioned by [Zachary](https://mathoverflow.net/questions/438288/linear-elliptic-equation#comment1130075_438288), one solution that can be written in terms of Meijer G-functions, but has no $r\rightarrow 0 $ limit so we exclude it. The other one is
\begin{align}
R^{(a)}\_0=\, \_2F\_1\left(-\frac{i \sqrt{a}}{2},\frac{i \sqrt{a}}{2};1;r^2\right).
\end{align}
| 1 | https://mathoverflow.net/users/171439 | 438849 | 177,244 |
https://mathoverflow.net/questions/438741 | 17 | For real variable $x$, the function
\begin{equation}
f(x):=\sum\_{n=1}^\infty \frac{1}{n! n^n}(-x^2)^n
\end{equation}
clearly has infinite radius of convergence and defines a $C^\infty$ function on $\mathbb{R}$.
However, I wonder if this $f(x)$ is a bounded function on $\mathbb{R}$ as well. Also, is it possible to estimate the bound if it is indeed bounded?
I tried to apply summation by parts or the error bounds for an alternating series. But all I have found out is that the partial sums $\sum\_{n=1}^N \frac{1}{n!}(-x^2)^n$ are extremely slowly convergent to the Gaussian function..
Could anyone please help me with the above function $f(x)$?
| https://mathoverflow.net/users/56524 | Evaluating the sum $f(x):=\sum_{n=1}^\infty \frac{1}{n! n^n}(-x^2)^n$ and estimating bounds | As already stated by @Noam and @Alexandre in the comments above,
$$
f(x)= - x \int\_0^1 \mathrm d\tau \sqrt{-8\log \tau} \, J\_1\big(x \tau \sqrt{-8\log \tau}\big),\tag{1}
$$
with Bessel function $J\_1$. Here, I did one further substitution in order to simplify the expression. The function $f(x)$ oscillates at large $x$, and asymptotically seems to become
$$
f\_\infty(x)= \sqrt 2 \cos(2 e^{-1/2} x) + c + O(x^{-1/2}),\tag{2}
$$
with $c\approx -1.055$. The oscillation frequency stems from the oscillations of $J\_1(x \tau \sqrt{-8\log \tau})$ in the integrand at stationary maximal argument (see comment by @Noam), which is at $\tau=e^{-1/2}=0.60653\ldots$. The prefactor of $x^{-1/2}$ is around $-0.285$.
| 10 | https://mathoverflow.net/users/90413 | 438852 | 177,245 |
https://mathoverflow.net/questions/434233 | 6 | In 1-category theory a representation of a 1-functor $F:C\to Set$ is a 0-cell $X$ in $C$ together with a universal element $u\in FX$ such that the transformation $C(X,-)\to F$ is an isomorphism (=a 1-equality) in the 1-category of functors $Fun(C,Set)$. It is well known how unique a representation $(X,u)$ is: unique up to a unique compatible isomorphism.
To see the uniqueness of the representation better, one can unfold $F$ into its [category of elements](https://ncatlab.org/nlab/show/category+of+elements) $el(F)$. The representations of $F$ are then precisely the initial objects of $el(F)$. When we consider only the full subcategory $rep(F)$ of representations of $F$ in $el(F)$, and if we consider the unique functor $rep(F)\to \mathbf 1$, the one point category, then we see that this functor is an equivalence. This means that $rep(F)$ is contractible and the representation is [unique up to contractibility](https://youtu.be/X2kNt0ARVeI).
Can we mimic this in the (weak) 2-categorical setting?
All my $2$-categories, functors and transformations are per default weak (but not lax). A representation of a $2$-functor $F:C\to Cat$ is a pair $(X,u)$, where $X$ is a $0$-cell of $C$ and $u$ an object of $F X$, such that the transformation of 2-functors $$C(X,-)\to F$$
induced by the 2-Yoneda lemma is an equivalence in the $2$-category $2Cat(C,Cat)$. How unique is a representation of $F$?
Is it possible to unfold $F$ into a (weak) 2-category $el(F)$ such hat the representations are precisely (weak) 2-initial objects in that 2-category, and such that the full sub-2-category $rep(F)$ of representations is contractible? By this I mean that the 2-functor $rep(F)\to \mathbf 1$ into the one-point 2-category is a 2-equivalence.
| https://mathoverflow.net/users/219922 | Do the representations of a 2-functor naturally form a contractible 2-category? | All my 2-categories, 2-functors, 2-(co)limits etc. are per default weak (or pseudo- or bi- if you so like).
One can unfold a 2-functor $M: K\to Cat$ into the 2-category $el(M)$ of its elements, and any representation of $M$ will be a 2-initial object in that 2-category. Unfortunately, and in contrast to the 1-categorical case, a 2-initial object in $el(M)$ need not be a representation of $M$. Still, the full sub-2-category of representations in $el(M)$ will be 2-contractible (if a representation exists), so that a 2-representation is unique up to a 2-contractible choice.
**Definition 1.** The 2-category $el(M)$ has the following cells.
* The 0-cells are pairs $(C,m)$, where $m\in M(C)$.
* A 1-cell $(C,M)\to (D,n)$ consists of a 1-cell $f:C\to D$ in $K$ together with an isomorphism $f\_\ast m \cong n$. Here $f\_\ast m$ is an abbreviation for $(Mf)(m)$. I will not give extra notation the isomorphism $f\_\ast m\cong n$, although it is part of the data of the 1-cell.
* A 2-cell $\theta:f\to g$ in $el(M)$ is a $2$-cell $\theta:f\to g$ in $K$ such that $\theta\_\ast m$ is the composite $f\_\ast m\cong n=n\cong g\_\ast m$.
The composition of 2-cells is just that of $K$. Given 1-cells $f:(C,m)\to (D,n)$ and $g:(D,n)\to (E,l)$ in $el(M)$, we let their composite be the morphism $gf$ together with the isomorphism $$(gf)\_\ast m \cong g\_\ast(f\_\ast m)\cong g\_\ast n\cong l$$The coherence cells of $el(M)$ are those of $K$. I am skipping over a lot of details here, but it feels like it should work out (this is always a dangerous thing to say).
**Definition 2.** A 2-initial object in a 2-category $E$ is an object $I$ such that for each other $0$-cell $X$ the 1-category $E(I,X)$ is 1-contractible. More explicitly, this means that there is one and precisely one invertible $2$-cell between any two $1$-cells $I\to X$, and $E(I,X)$ is not empty.
**Claim 1.** If $(X,u)$ is a representation of $M$, then $(C,u)$ is 2-initial in $el(M)$.
*Proof.* Let $(A,m)$ be any other 0-cell in $el(M)$. The fact that "evaluation at $u$" is an equivalence $$K(X,A)\to M(A)$$ means that I can find some $f:X\to A$ together with an isomorphism $f\_\ast u \cong m$. This means that there is at least one 1-cell $(X,u)\to (A,m)$. Assume that $g$ is a second such 1-cell in $el(M)$. Then $g$ comes equipped with a choice of an isomorphism $g\_\ast u \cong m$. A 2-cell $f\to g$ in $el(M)$ consists of a 2-cell $\theta: f\to g$ in $K$ such that $\theta\_\ast m$ is the composite $f\_\ast u \cong m \cong g\_\ast u$.The fact that $K(X,A)\to M(A)$ is an equivalence of 1-categories tells me also that there is one and precisely one such $\theta$. $\square$
Let us denote by $rep(M)$ the full 2-subcategory of $el(M)$ with 0-cells the 2-representations of $M$.
**Claim 2.** The 2-category $rep(M)$ is 2-contractible. This means that the 2-functor $rep(M) \to pt$ is a 2-equivalence.
*Proof.* What this means in elementary terms is that each hom-1-category in $el(M)$ must be 1-contractible. That is, whenever there are two representations $(A,a)$ and $(B,b)$, then there must be at least one 1-cell $(A,a)\to (B,b)$, and whenever there is another one then the two 1-cells are comparable by a unique 2-cell which is also invertible. But we see immediately that this is the case, because we have already shown that all the representations are 2-initial objects. $\square$
**Claim 3.** Unfortunately, being 2-initial in $el(M)$ doesn't automatically mean that you are a 2-representation of $M$. So while $el(M)$ answers the question how unique representations are, it does not characterize them.
*Proof.* This is in length discussed in the paper [Bi-initial objects and bi-representations are not so different](https://arxiv.org/abs/2009.05545) by tslil clingman AND LYNE MOSER, which by accident asks precisely my question in the middle of page 2. :) I only found the text weeks after I have asked the question. It also contains some positive results. $\square$
| 2 | https://mathoverflow.net/users/219922 | 438854 | 177,246 |
https://mathoverflow.net/questions/438837 | 9 | $\DeclareMathOperator{\Tot}{Tot}$I have the following problem. Let $H$ and $G$ be groups such that $H$ acts on $G$, i.e., there exists a group homomorphism $H\to \mathrm{Aut}(G)$ and let $M$ be an abelian group treated as a trivial $G$- and $H$-module.
Further on, there is a double complex $C^{p,q}:= C^p(H,C^q(G,M))$ (I use here bar resolutions). To this double complex there are two spectral sequences associated, one of them is Lyndon--Hochschild--Serre spectral sequence. In particular, I know that both spectral sequences converge to $H^\ast(G\rtimes H, M)$. On the other hand, by the general construction of both spectral sequences, we know that they converge to $H^\ast(\Tot(C^{\ast \ast}))$.
My question is the following. If I have a specific cohomology class in $H^\ast(\Tot(C^{\ast \ast}))$, how can I decide to which class in $H^\ast(G\rtimes H, M)$ it corresponds? In particular, an answer to any of the following questions would do the job (I think):
1. Is there an explicit map $H^\ast(\Tot(C^{\ast \ast}))\to H^\ast(G\rtimes H, M)$ giving the isomorphism? Or is there a cochain level map?
2. Given a class in $H^2(\Tot(C^{\ast \ast}))$, how can I construct a group extension $1\to M \to E\to G\rtimes H\to 1$?
I am mostly interested in cohomology classes of degree 2 (as indicated by the second question). Also, reference would be enough.
| https://mathoverflow.net/users/123432 | Comparing cohomology of a total complex with the cohomology of semidirect product | Let me change the notation a little: let $\phi \colon H \to \operatorname{Aut}(N)$ be a group homomorphism, and consider $G = N \rtimes\_\phi H$ (we will drop the subscript $\phi$ in the rest of this post). As usual, write $^hn$ for $\phi(h)(n)$ if $h \in H$ and $n\in N$.
The point is that the double complex $C^{\bullet\bullet}$ actually comes from a natural bi-cosimplicial abelian group, as follows:
**Definition.** Define the bisimplical set $X\_{\bullet\bullet}$ via $X\_{p,q} = N^{q+1}\times H^{p+1}$, where a morphism $(f,g) \colon ([p],[q]) \to ([p'],[q'])$ in $\Delta \times \Delta$ acts as
\begin{align\*}
X\_{p',q'} &\to X\_{p,q}\\
\big(n\_0,\ldots,n\_{q'},h\_0,\ldots,h\_{p'}\big) &\mapsto \big(n\_{f(0)},\ldots,n\_{f(q)},h\_{g(0)},\ldots,h\_{g(p)}\big).
\end{align\*}
Endow each $X\_{p,q}$ with a $G$-action via
\begin{align\*}
G \times X\_{p,q} &\mapsto X\_{p,q}\\
\big((n,h),(n\_0,\ldots,n\_q,h\_0,\ldots,h\_p)\big) &\mapsto \big(n\ {^hn\_0},\ldots,n\ {^hn\_q},hh\_0,\ldots,hh\_p\big)\\
\big((n,h),\big(\underline n,\underline h\big)\big) &\mapsto \big(n\ ^h\underline n,h\underline h\big).
\end{align\*}
This is visibly functorial in $(p,q) \in \Delta^{\text{op}} \times \Delta^{\text{op}}$, so defines a bisimplicial $G$-set. Thus we can view $\mathbf Z[X\_{\bullet\bullet}]$ naturally as a bisimplicial $\mathbf Z[G]$-module, and I claim that
$$\operatorname{Hom}\_{\mathbf Z[G]}\big(\mathbf Z[X\_{\bullet\bullet}],M\big) \cong C^{\bullet\bullet}\label{1}\tag{1}$$
as double complexes, for any $G$-module $M$ (not necessarily trivial). Indeed, note that
\begin{align\*}
\operatorname{Hom}\_{\mathbf Z[G]}\big(\mathbf Z[X\_{p,q}],M\big) &\cong \operatorname{Map}\_G\big(X\_{p,q},M\big), \\
C^{p,q} &\cong \operatorname{Map}\_H\big(H^{p+1},\operatorname{Map}\_N(N^{q+1},M)\big).
\end{align\*}
We have a bijection $\operatorname{Map}(X\_{p,q},M) \stackrel\sim\to \operatorname{Map}(H^{p+1},\operatorname{Map}(N^{q+1},M))$, taking a function $\phi \colon X\_{p,q} \to M$ to $\psi \colon H^{p+1} \to \operatorname{Map}(N^{q+1},M)$ given by
$$\psi\big(h\_0,\ldots,h\_p\big)\big(n\_0,\ldots,n\_q\big) = \phi\big(n\_0,\ldots,n\_q,h\_0,\ldots,h\_p\big).$$
Now $\psi$ has image in $\operatorname{Map}\_N(N^{q+1},M)$ if and only if
$$\phi\big(n\underline n,\underline h\big) = n\phi\big(\underline n,\underline h\big) \qquad \text{for all } \big(\underline n, \underline h\big) \in X\_{p,q}, \text{ all } n \in N.\label{2}\tag{2}$$
The $G$-action on $\operatorname{Map}\_N(N^{q+1},M)$ is given by $(gf)(\underline n) = gf(g^{-1} \cdot \underline n)$, which is trivial on $N$ by definition of $\operatorname{Map}\_N$. As action of $H$, it can be written explicitly by $(hf)(\underline n) = hf(^{h^{-1}}\!\underline n)$. This means that $\psi \colon H^{p+1} \to \operatorname{Map}\_N(N^{q+1},M)$ is $H$-equivariant if and only if
$$\phi\big(\underline n,h\underline h\big) = h\phi\big({^{h^{-1}}}\!\underline n,\underline h\big) \qquad \text{for all } \big(\underline n,\underline h\big) \in X\_{p,q}, \text{ all } h \in H.\label{3}\tag{3}$$
Combining \eqref{2} and \eqref{3} shows that $\psi \in C^{p,q}$ if and only if $\phi(g \cdot (\underline n,\underline h)) = g \phi(\underline n,\underline h)$ for all $(\underline n,\underline h) \in X\_{p,q}$ and all $g$ in either $N$ or $H$. But this means exactly that $\phi \colon X\_{p,q} \to M$ is $G$-equivariant, so
$$\operatorname{Map}\_G\big(X\_{p,q},M\big) \stackrel\sim\to \operatorname{Map}\_H\big(H^{p+1},\operatorname{Map}\_N\big(N^{q+1},M\big)\big).$$
This identification is natural in $p$ and $q$, giving the isomorphism \eqref{1}. This also shows that $C^{\bullet\bullet}$ is naturally a bi-cosimplicial abelian group.
In particular, the Eilenberg–Zilber theorem ([nlab](https://ncatlab.org/nlab/show/Eilenberg-Zilber%20theorem#statement)) shows that the totalisation of $C^{\bullet\bullet}$ is naturally chain homotopic (!) to the diagonal $dC^\bullet$ (i.e. $dC^n = C^{n,n}$). But the diagonal of $X\_{p,q}$ is just the bar resolution of $G$, so we conclude that the inclusion $dX\_\bullet \hookrightarrow X\_{\bullet\bullet}$ induces a chain homotopy equivalence
$$\operatorname{Tot}C^{\bullet\bullet}\simeq C^\bullet(G,M).$$
Explicit descriptions of the map follow from sufficiently effective proofs of Eilenberg–Zilber. See for instance 8.5.4 in Weibel's *An introduction to homological algebra*. (For your direction, you need to use the Alexander–Whitney map on $\mathbf Z[X\_{\bullet\bullet}]$, because the direction flips when applying $\operatorname{Hom}\_{\mathbf Z[G]}(-,M)$.)
Concretely, the map $\operatorname{Tot}(C^{\bullet\bullet}) \to C^\bullet(G,M)$ takes a $(p,q)$-cocycle $f \colon X\_{p,q} \to M$ to the $(r,r)$-cocycle $X\_{r,r} \to M$ given by
$$\big(n\_0,\ldots,n\_r,h\_0,\ldots,h\_r\big) \mapsto f\big(n\_p,\ldots,n\_r,h\_0,\ldots,h\_q\big),$$
where $r = p+q$. (Here I am working with inhomogeneous cocycles, i.e. $G$-equivariant maps $G^{r+1} \to M$.)
| 3 | https://mathoverflow.net/users/82179 | 438859 | 177,248 |
https://mathoverflow.net/questions/438732 | 7 | *Throughout, work in $\mathsf{ZFC}$ + large cardinals (let's say a proper class of Woodin limits of Woodins but I'm happy to go higher if that would help).*
Let $\mathcal{R}=(\mathbb{R};+)$ be the additive group of real numbers. We have $\mathit{Aut}(\mathcal{R})^{L(\mathbb{R})}\not\cong\mathit{Aut}(\mathcal{R})^V$ due to large cardinals. However, I don't see that this means that $L(\mathbb{R})$ must not contain any copy of $\mathit{Aut}(\mathcal{R})$:
>
> Is there a group $G\in L(\mathbb{R})$ such that, in $V$, $G\cong \mathit{Aut}(\mathcal{R})$?
>
>
>
Unless I'm missing something, the set of all functions from reals to reals in ${L(\mathbb{R})}$ is too small to be a candidate for this. That said, I see no reason why a copy of $\mathit{Aut}(\mathcal{R})$ couldn't show up in $L(\mathbb{R})$ somewhere "far away" from $\mathcal{R}$ itself.
More generally, is there a structure $\mathcal{S}$ in $L(\mathbb{R})$ whose $V$-automorphism group does not have a copy $\mathcal{G}\in L(\mathbb{R})$? *(Note that - here and above - I **don't** demand that $L(\mathbb{R})$ in any way be able to connect elements of this strange copy $\mathcal{G}$ with $\mathcal{S}$; in particular, $L(\mathbb{R})$ need not have any interesting maps $\mathcal{G}\times\mathcal{S}\rightarrow\mathcal{S}$.)*
| https://mathoverflow.net/users/8133 | Does $\mathit{Aut}(\mathbb{R};+)$ have a copy in $L(\mathbb{R})$ granting large cardinals? | Assuming also CH, the answer to the more general question is yes,
there is a structure in $L(\mathbb{R})$ (in fact, just the set $\mathbb{R}$, with no additional structure), whose automorphism group in $V$ does not have an isomorphic copy in $L(\mathbb{R})$. (I haven't really thought about the original question, i.e. $(\mathbb{R},+)$.)
Assume ZFC + CH and $\mathbb{R}^{\#}$ exists
(i.e. there is an elementary embedding $j:L(\mathbb{R})\to L(\mathbb{R})$).
Consider the set $\mathbb{R}$ of reals
with no additional structure. So automorphisms
are just bijections $\pi:\mathbb{R}\to\mathbb{R}$. Let $G$ be the group of such bijections $\pi$ (under composition). I claim that $L(\mathbb{R})$ has no group which is (in $V$) isomorphic to $G$.
The proof will use the existence of $\mathbb{R}^{\#}$ to decompose the supposed copy $G'$ of $G$, with $G'\in L(\mathbb{R})$,
into the direct limit of elementary substructures $G'\_s$, each of which have cardinality $\leq$ the continuum (which is $\aleph\_1$ under the CH hypothesis), and such that there are only countably many isomorphism types of $G'\_s$ over some base substructure $G'\_{s\_0}$, modulo isomorphisms which fix $G'\_{s\_0}$ pointwise. Also, $G'\_{s\_0}$ will be rich enough that, transporting this over to $G$ and corresponding $G\_{s\_0}$ and $G\_s$, we can construct a bijection (an element of $G$) which couldn't appear in any of the $G\_s$'s, via a diagonalization.
For $x,y\in\mathbb{R}$
with $x\neq y$ let $\pi\_{xy}:\mathbb{R}\to\mathbb{R}$ be the bijection
such that $\pi\_{xy}(x)=y$ and $\pi\_{xy}(y)=x$ and $\pi\_{xy}(z)=z$ for all other $z$.
Lemma 1: Let $x\neq y$. Then the commutator $C\_{\pi\_{xy}}$ of $\pi\_{xy}$ is the set of all $\sigma\in G$ such that $\sigma(x)=y$ and $\sigma(y)=x$.
Lemma 2: Let $x\neq y$. Let $\pi\in G$
be such that the commutator $C\_\pi$ of $\pi$ is such that $C\_{\pi\_{xy}}\subsetneq C\_\pi$. Then $\pi=\mathrm{id}$, so $C\_\pi=G$.
So the collection of $\pi\_{xy}$'s
is definable over $G$ as those
group elements $\pi$ whose commutator
is maximal for being a proper subset of $G$.
In particular, any $G'\preccurlyeq G$
will contain infinitely many $\pi\_{xy}$'s.
Lemma 2.5: Let $x,y,z$ be pairwise distinct. Then
$\pi\_{xy}\pi\_{yz}\pi\_{xy}=\pi\_{xz}$.
Lemma 2.6:
Let $x\neq y$ and $x'\neq y'$,
such that $\pi\_{xy}\neq \pi\_{x'y'}$.
Then $\{x,y\}\cap\{x',y'\}=\emptyset$
iff $\pi\_{xy}\pi\_{x'y'}$ has order 2.
And $\{x,y\}\cap\{x',y'\}\neq\emptyset$
iff $\pi\_{xy}\pi\_{x'y'}$ has order 3.
Write $\mathscr{I}$ for the class of Silver indiscernibles for $L(\mathbb{R})$.
Let $s\in[\mathscr{I}]^{<\omega}$ with $s\neq\emptyset$. Then write $H\_s=\mathrm{Hull}\_{\Sigma\_1}^{L\_{\max(s)}(\mathbb{R})}(\mathbb{R}\cup\{s^-\})$
where $s^-=s\backslash\{\max(s)\})$.
Lemma 3: $H\_s$ is countably closed (in $V$).
Proof: Let $\left<x\_i\right>\_{i<\omega}\subseteq H\_s$. Let $\left<\varphi\_i,r\_i\right>\_{i<\omega}$
be such that $\varphi\_i$ is a $\Sigma\_1$ formula and $r\_i\in\mathbb{R}$ and $x\_i$ is the unique $x\in L\_{\max(s)}(\mathbb{R})$ such that $L\_{\max(s)}(\mathbb{R})\models\varphi\_i(x,r\_i,s^-)$.
Let $r=(\oplus\_{i<\omega}r\_i,\oplus\_{i<\omega}\varphi\_i)\in\mathbb{R}$. Then using the parameter $r$,
working in $L(\mathbb{R})$
we can identify $\left<x\_i\right>\_{i<\omega}$. In fact,
because $\max(s)$ is inaccessible in $L(\mathbb{R})$, there must be some $\gamma<\max(s)$ such that $L\_{\gamma}(\mathbb{R})\models\varphi\_i(x,r\_i,s^-)$
for each $i<\omega$. But then it easily follows that there is a $\Sigma\_1$ formula $\psi$ such that $\left<x\_i\right>\_{i<\omega}$ is the unique $y\in L\_{\max(s)}(\mathbb{R})$
such that $L\_{\max(s)}(\mathbb{R})\models\psi(y,r,s^-)$,
so $\left<x\_i\right>\_{i<\omega}\in H\_s$,
as desired.
Now suppose that $G'\in L(\mathbb{R})$
is a group isomorphic (in $V$)
to $G$.
So $G'\in H\_{s\_0}$
for some $s\_0\in[\mathscr{I}]^{<\omega}$
(but this doesn't say that $G'\subseteq H\_{s\_0}$; in fact it isn't,
because $G$ has cardinality $2^{2^{\aleph\_0}}$, but $H\_{s\_0}$
has cardinality $2^{\aleph\_0}$).
Let $G'\_{s\_0}=G'\cap H\_{s\_0}$.
We have $H\_{s\_0}\preccurlyeq\_{\Sigma\_1}^{L\_{\max(s\_0)}(\mathbb{R})}$,
and since $G'\in H\_{s\_0}$,
therefore $G'\_{s\_0}\preccurlyeq G'$
(fully elementary).
Say a group element $g\in G$
is \emph{good}, if $C\_g\subsetneq G$
and there is no $h\in G$ such that
$C\_g\subsetneq C\_h\subsetneq G$
(so equivalently, $g=\pi\_{xy}$
for some $x\neq y$).
Say a group element $g\in G'$ is \emph{good} under the same definition
(but we can't characterize this one as being a $\pi\_{xy}$;
it's important that the property
is definable just from the group structure). Given $g,h\in G$ which are both good, say that the pair $(g,h)$ is \emph{independent} if $gh$ has order 2 (cf. Lemma 2.6).
Lemma 4: (i) $G'\_{s\_0}$ contains uncountably many good elements. In fact,
(ii) there is an uncountable set $A\subseteq G'\_{s\_0}$ consisting of pairwise
independent good elements.
Proof: Part (i): $L\_{s\_0}(\mathbb{R})\models$``there are
uncountably many good group elements of $G'$'',
and because $H\_{s\_0}$ is countably closed
(in $V$), part (i) follows.
Part (ii): Otherwise, there would
be some countable set $A'$ of good elements such that for all good $g\in G'\_{s\_0}\backslash A'$, there is some $h\in A$
such that $(g,h)$ is dependent.
Let $f:G\to G'$ be an isomorphism.
Let $A=f^{-1}``A'$. Then
we can fix a countable set $C\subseteq\mathbb{R}$ such that for all $g\in A$, we have $g=\pi\_{xy}$ for some $x,y\in C$. Let $B'\subseteq G\_{s\_0}'$ be an uncountable set of good elements
such that $B'\cap A'=\emptyset$.
Let $B=f^{-1}``B'$.
Then we can find an uncountable $\bar{B}\subseteq B$ and an $x\in C$
such that for each $\pi\in\bar{B}$,
there is $y\in\mathbb{R}$ such that
$\pi=\pi\_{xy}$. But then taking $y\_1,y\_2$ such, by Lemma 2.5, we have $\pi\_{y\_1y\_2}=\pi\_{y\_1x}\pi\_{xy\_2}\pi\_{y\_1x}$.
So these generate an uncountable
set of pairwise independent $\pi$'s. Pulling these back under the isomorphism $f$, part (ii) is proven.
Now if $s,t\in\mathscr{I}^{<\omega}$
with $|s|=|t|$ then $H\_s\equiv H\_t$,
and the isomorhphism fixes $\mathbb{R}$.
Let $\kappa$ be the least limit Silver indiscernible with $\max(s)<\kappa$.
Then $L\_\kappa(\mathbb{R})=\bigcup\_{s\in[\mathscr{I}\cap\kappa]^{<\omega}}H\_s$,
and therefore letting $\mathscr{J}$ be the set of all $s\in[\mathscr{I}\cap\kappa]^{<\omega}$ such that $s\_0\subseteq s$, then $G'=\bigcup\_{s\in\mathscr{J}}G'\_s$.
Also, if $s,t\in\mathscr{J}$ with $s\subseteq t$ then $G'\_{s\_0}\preccurlyeq G'\_s\preccurlyeq G'\_t\preccurlyeq G'$,
each $G'\_s$ has cardinality continuum
(by Lemma 4
and since $H\_s$ has cardinality continuum and by CH).
Lemma 5: Let $s,s'\in\mathscr{J}^{<\omega}$ be such that $|s|=|s'|$
and the $s\_0$ sits in ordertype within $s$
just as $s\_0$ sits in $s'$. Then $G'\_s\cong G'\_{s'}$
via an isomorphism which fixes $G'\_{s\_0}$.
(Here by ``sits in... just as...'' I mean
that if $s=\{s^0,s^1,s^2,\ldots,s^k\}$
with $s^0<s^1<\ldots<s^k$,
and $s'=\{(s')^0,(s')^1,\ldots,(s')^k\}$
increasing likewise,
then for $i\in[0,k]$,
we have $s^i\in s\_0$ iff $(s')^i\in s\_0$.)
Proof of Lemma 5: Easy consequence of Silver indiscernibility.
Thus, there are only countably many isomorphism types for the subgroups $G'\_s$ over $G'\_{s\_0}$.
For $g\in G'$, let $E\_g$
be the set of equations satisfied by
elements in $G'\_{s\_0}\cup\{g\}$.
(E.g. if $h\_1,h\_2\in G'\_{s\_0}$,
we might have the equation ``$h\_1gh\_2=\mathrm{id}$'' in $E\_g$.)
Lemma 6: $\{E\_g\bigm|g\in G'\}$
has cardinality at most continuum $=\aleph\_1$
(in $V$).
Proof: If $s\in\mathscr{J}$ then $\{E\_g\bigm|g\in G'\_s\}$ has cardinality at most continuum, since $G'\_s$ has at most cardinality continuum. But if $s,s'$
are such that $G'\_s$ and $G'\_{s'}$
are isomorphic via an isormorphism which fixes $G'\_{s\_0}$,
then clearly $\{E\_g\bigm|g\in G'\_s\}=\{E\_g\bigm|g\in G'\_{s'}\}$. So
by Lemma 5, we get at most continuum
many such $E\_g$'s in total.
The structure established for $G'$
now shifts over to $G$ under the isomorphism $f:G'\to G$. Write $G\_{s\_0}$, $G\_s$ etc for the pointwise images of $G'\_{s\_0}$, $G'\_s$ etc.
The rest of the proof is a direct diagonalization, constructing some
$h\in G$ such that $E\_h$ is not
one of the $E\_g$'s above, which is a contradiction. The construction is just algebraic.
Let $\left<r\_\alpha\right>\_{\alpha<\aleph\_1}$ be an enumeration of $\mathbb{R}$
in ordertype $\aleph\_1$. We define by recursion on $\beta<\aleph\_1$,
some ordinals $\alpha\_\beta<\aleph\_1$,
and
$h\upharpoonright\left<r\_\alpha\right>\_{\alpha<\alpha\_\beta}$, with $h``\left\{r\_\alpha\right\}\_{\alpha<\alpha\_\beta}=\left\{r\_\alpha\right\}\_{\alpha<\alpha\_\beta}$. Let $\left<g\_\beta\right>\_{\beta<\aleph\_1}\subseteq G$ be such that all possible $E\_g$'s for $g\in G$ are produced by the $E\_{g\_\beta}$'s (by the lemmas, there are only continuum = $\aleph\_1$-many such $E\_g$'s).
Suppose we have $\beta<\aleph\_1$
and we have defined $\alpha\_\beta$
and $h\upharpoonright\{r\_\alpha\}\_{\alpha<\alpha\_\beta}$, as above.
Consider $g\_\beta$ and $E\_{g\_\beta}$.
We want to now make sure that $E\_h$
will not be $E\_{g\_\beta}$. We know
that $G\_{s\_0}$
has an uncountable subset of pairwise independent good elements. So let $\gamma<\delta<\aleph\_1$ be such that $\alpha\_\beta\leq\gamma,\delta$
and $\pi\_{r\_\gamma r\_\delta}\in G\_{s\_0}$.
We can now decide freely whether or not $h\pi\_{r\_\gamma r\_\delta}=\pi\_{r\_\gamma r\_\delta}h$. If $g\_\beta \pi\_{r\_\gamma r\_\delta}\neq\pi\_{r\_\gamma r\_\delta}g\_\beta$, then we will set the equation true, by defining
$h(r\_\gamma)=r\_\delta$
and $h(r\_\delta)=r\_\gamma$;
if otherwise, we will set the equation false, by
defining $h(r\_\gamma)=r\_{\delta+1}$
and $h(r\_{\delta+1})=r\_\gamma$;
in both cases set $h(r\_\xi)=r\_\xi$
for all other $\xi\in[\alpha\_\beta,\delta+1]$ where we haven't yet defined $h(r\_\xi)$. Then set $\alpha\_{\beta+1}=\delta+2$. Note that this ensures
that $E\_h\neq E\_{g\_\beta}$.
At limit $\beta$ define $\alpha\_\beta=\sup\_{\beta'<\beta}\alpha\_{\beta'}$.
This constructs some $h\in G$,
but $E\_h\neq E\_{g\_\beta}$ for all $\beta<\aleph\_1$, which is a contradiction.
| 4 | https://mathoverflow.net/users/160347 | 438860 | 177,249 |
https://mathoverflow.net/questions/437766 | 1 | Let $\Omega\subset \mathbb{C}^n$ be an open subset. Let $u\colon \Omega\to [-\infty,+\infty)$ be an upper semi-continuous function.
Recall that $u$ is called plurisubharmonic (psh) if its restriction to any complex line is subharmonic.
Any psh function $u$ satisfies the following property: for any point $x\in \Omega$ and for any $C^2$-smooth function $\phi$ defined near $x$ and such that $u\leq \phi$ and $u(x)=\phi(x)$ one has
$$(\Delta\_L (\phi|\_L))(x)\geq 0$$
for any complex line $L$ containing the point $x$. Here $\Delta\_L$ denotes the Laplacian on the line $L$.
**Is the converse true, i.e. if an upper semi-continuous function $u$ satisfies the above condition is it psh? A reference would be very helpful.**
This post is a continuation of [A possible characterization of subharmonic functions](https://mathoverflow.net/questions/437740/a-possible-characterization-of-subharmonic-functions/437750#437750)
| https://mathoverflow.net/users/16183 | A characterization of plurisubharmonic functions | You can consult [Harvey and Lawson](http://www.math.stonybrook.edu/%7Eblaine/MA8.pdf), sections 5 and 6, on that matter. Especially Lemma 5.5 and point (6) on p. 19 (note that for smooth $\phi$ condition you gave is equivalent to having complex hessian non negative).
| 2 | https://mathoverflow.net/users/85450 | 438866 | 177,252 |
https://mathoverflow.net/questions/438806 | 1 |
>
> Is it possible to iterate elementary embeddability and reflect on those stages that are elementary embeddable to themselves?
>
>
>
The following is a formal capture of that idea:
To the language of $\sf ZF$ (i.e., mono-sorted $\sf FOL(=,\in)$) add primitive partial unary functions $W$ and $j$.
To the axioms of $\sf ZF$, add the following axioms:
**Restriction:** $\forall \alpha: W\_\alpha \lor j\_\alpha \to \operatorname {ordinal}(\alpha)$
**Injectivity:** $W\_\alpha \land W\_\beta \land \alpha \neq \beta \to W\_\alpha \neq W\_\beta$
**Cumulation:** $\forall \operatorname {ordinal} \alpha \exists \lambda: W\_\alpha=V\_\lambda$
**Elementarity:** $\forall \operatorname {ordinal} \alpha \, (j\_\alpha: W\_\alpha \to W\_\alpha \land \\ \forall \vec{x} \in W\_\alpha [ \phi(\vec x) \leftrightarrow \phi(j\_\alpha[\vec x ])] \land \\ \exists x: j\_\alpha (x) \neq x) \\\text {where } \phi \text { is purely set theoretic }$
**Reflection:** $\forall \vec{x} \in W\_\alpha \, (\phi \to \phi^{W\_\alpha})$
if $\phi$ [in Reflection] is a formula of the language of set theory + $``j\_\alpha \!"$, meaning that $W$ doesn't occur in it and every occurence of $j$ must be subscripted with $\alpha$; also $``\alpha \!"$ only appears in $\phi$ as a subscript of $j$.
Where $V\_\lambda$ stands for the $\lambda^{th}$ stage of the cumulative hierarchy, defined in the customary manner. $\phi^X$ stands for relativising all quantifiers in $\phi$ to $``\in X\!"$.
>
> Is the above theory consistent relative to some large cardinal property? If so, Which one?
>
>
>
| https://mathoverflow.net/users/95347 | Can we iteratively reflect on self elementary embeddable stages of the cumulative hierarchy? | $\newcommand\Ord{\mathit{Ord}}$The edit to the question has changed it enough so I think it deserves its own answer.
I assume that in reflection we have $\forall \vec{x} \in W\_\alpha \, (\phi \to \phi^{W\_\alpha})$ holds for all ordinals $\alpha$.
The theory with this strong reflection is inconsistent in a strong sense.
Specifically it fails even when you only consider $W\_0$.
Indeed assume that $W\_0$ reflects all formulae in the language of $\{\in,j\_0\}$ then it reflects $\eta=\exists x(x\notin \operatorname{Dom}(j\_0))$, so let $p\in W\_0$ be witness of $\eta^{W\_0}$.
$W\_0$ also reflects $\psi(y)=\exists x\in \Ord (y\in V\_x)$, let $r$ be witness of $\psi(p)$ in $W\_0$.
Lastly reflect $\phi(y)=y\in \Ord\land V\_y\subseteq \operatorname{Dom}(j\_0)$ to get that $W\_0$ thinks that $\phi(r)$, but this is a contradiction to the choice of $r$, $p$ (note that I only chose 1 arbitrary element, $p$, so I didn't use any choice).
| 3 | https://mathoverflow.net/users/113405 | 438882 | 177,255 |
https://mathoverflow.net/questions/438881 | 6 | The ladder operator in quantum mechanics are the operators
$$a^\dagger \ = \ \frac{1}{\sqrt{2}} \left(-\frac{d}{dq} + q\right)$$
and
$$a \ \ = \ \frac{1}{\sqrt{2}} \left(\ \ \ \!\frac{d}{dq} + q\right).$$
They are differential operators on $\mathbb R.$ If one writes them in the Hermite basis, then
$$a^\dagger = \begin{pmatrix}
0 & 0 & 0 & 0 & \dots & 0 & \dots \\
\sqrt{1} & 0 & 0 & 0 & \dots & 0 & \dots \\
0 & \sqrt{2} & 0 & 0 & \dots & 0 & \dots \\
0 & 0 & \sqrt{3} & 0 & \dots & 0 & \dots \\
\vdots & \vdots & \vdots & \ddots & \ddots & \dots & \dots \\
0 & 0 & 0 & \dots & \sqrt{n} & 0 & \dots & \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots \end{pmatrix}$$ and
$$a =\begin{pmatrix}
0 & \sqrt{1} & 0 & 0 & \dots & 0 & \dots \\
0 & 0 & \sqrt{2} & 0 & \dots & 0 & \dots \\
0 & 0 & 0 & \sqrt{3} & \dots & 0 & \dots \\
0 & 0 & 0 & 0 & \ddots & \vdots & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots & \sqrt{n} & \dots \\
0 & 0 & 0 & 0 & \dots & 0 & \ddots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}.$$
Now assume I was interested in numerically computing the spectrum of
$$H = \begin{pmatrix} 0& a\\a^\* & 0\end{pmatrix}.$$
I absolutely know that this can be computed by hand, but I wonder about how to do this numerically.
A naive idea would be to truncate the above matrices at a large size $N$, but this leads to the wrong spectrum as both matrices then have a non-zero nullspace once they are truncated (it is clear since 0 is then an eigenvalue of geometric multiplicity $1$ for both matrices). Hence, the truncated numerics would predict that the Hamiltonian $H$ has an eigenvalue $0$ of multiplicity 2 rather than 1, which is correct.
Does anybody know how to numerically overcome this pseudospectral effect?
| https://mathoverflow.net/users/496243 | Spectrum of operator involving ladder operators | **Q:** *Does anybody know how to numerically overcome this pseudospectral effect?*
The key idea is "normal ordering". Rewrite the problem in such a way that annihilation operators $a$ appear to the right of creation operators $a^\ast$. In this particular case, first notice that $H$ has chiral symmetry, if $\lambda$ is an eigenvalue then also $-\lambda$ is an eigenvalue. We can thus reconstruct the full spectrum of $H$ without sign ambiguities from the spectrum of $H^2$,
$$H^2=\begin{pmatrix}
1+a^\ast a&0\\
0&a^\ast a
\end{pmatrix}.$$
We can now safely truncate each block to $N$ states. The spectrum of $H\_N^2$ contains 0 as eigenvalue with multiplicity 1.
| 6 | https://mathoverflow.net/users/11260 | 438883 | 177,256 |
https://mathoverflow.net/questions/438895 | 0 | I'm looking for a citable reference for the following identity involving the Stirling numbers of the second kind $S(n, k)$ stated in [Equation (27)](https://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html): For $n \geq 2$,
$$
\sum\_{m=1}^n S(n, m) (-1)^m (m-1)! = 0.
$$
Thank you.
| https://mathoverflow.net/users/479711 | Identity involving Stirling number of the second kind | The first formula in Section 24.1.4.I.B in Handbook of mathematical functions with formulas, graphs, and mathematical tables, Dover edition, 1965 (Library of Congress Catalog Card Number: 65.12253) by Abramowitz and Stegun is
$$
x^n=\sum\_{m=0}^n S(n,m) x(x-1)\cdots(x-m+1). \tag{1}\label{1}
$$
Assuming here $n>1$ (so that $S(n,0)=0$), dividing both sides by $x$, and then letting $x=0$, we get the desired identity
$$\sum\_{m=1}^n S(n,m)(-1)^m(m-1)!=0.$$
---
One may note that Riordan (An Introduction to Combinatorial Analysis, Wiley, 1958, formula (35) on p. 33) uses identity \eqref{1} to *define* the Stirling numbers $S(n,m)$, then deriving their combinatorial definition (as the number of ways of partitioning a set of $n$ elements into $m$ nonempty sets) -- cf. formula (38) on p. 33 and the third display from the bottom on p. 91 of the book.
| 4 | https://mathoverflow.net/users/36721 | 438902 | 177,260 |
https://mathoverflow.net/questions/438782 | 3 | Let $\ p\_n\ $ be the consecutive primes starting with
$\ p\_0:=2.\ $ Let $\ M(n)\ $ be the multiplicative monomial
generated by $\ \{p\_k:\ k=0\ldots n\}\ $ (of course $\ 1\in M(n)$).
Could you prove or disprove:
$$ \forall\_{n\in\mathbb Z\_0}\, \exists\_{K\ L\in M(n)}
\ \left( \prod\_{k=0}^n p\_k|K\cdot L\ \text{and}\ p\_{n+1}=L-K\right).$$
| https://mathoverflow.net/users/110389 | Generating prime $\ p_{n+1}\ $ (the complete version) | Assuming that the triple $(a,b,c)=(2,3^{10}\cdot 109,23^5)$ found by Eric Reyssat is the one with the highest quality $q=\log(c)/\log(\text{rad}(abc))=1.6299\ldots$ for the ABC conjecture, one quickly computes that there is indeed no solution for $p\_{n+1}=31$. Setting $Q=\text{rad}(KLp\_{n+1})=\prod\_{k=0}^{n+1}p\_k$, we get $\log(L)\le q\log(Q)$. Running through the possibilities for these $L$ with prime factors $\le29$, one checks that $K=L-p\_{n+1}$ either has prime factors $\ge31$, or $\prod\_{k=0}^{n}p\_k$ does not divide $K\cdot L$.
The (naive) pure python code which runs about 20 seconds on my machine for $p\_{n+1}=31$ (which is $q$ in the third line) is
```
from math import prod, floor, factorial
q = 31
a = [z for z in range(2, q) if factorial(z-1)%z == z-1] # primes < q
P = prod(a)
qual = 1.63
c = floor((q*P)**qual) # upper bound for L
def La(K, a): # check if all prime divisors of K are in a
for p in a:
while K%p == 0:
K //= p
return K == 1
def indent(i): # Create iterator for candidates of L
return ' '*(4*i)
s = 'def tmp():\n'
s += ' c0 = c\n'
s += ' x0 = 1\n'
for i in range(len(a)):
s += indent(i+1) + f'b{i} = 1\n'
s += indent(i+1) + f'while b{i} <= c{i}:\n'
s += indent(i+2) + f'x{i+1} = x{i}*b{i}\n'
s += indent(i+2) + f'c{i+1} = c{i}//b{i}\n'
s += indent(i+2) + f'b{i} *= a[{i}]\n'
s += indent(len(a)+1) + f'yield x{len(a)}\n'
exec(s)
f = tmp()
for L in f:
K = L-q
if (K*L)%P == 0:
if La(K, a):
print(K, L)
break
else:
print('no solution')
```
| 8 | https://mathoverflow.net/users/18739 | 438905 | 177,261 |
https://mathoverflow.net/questions/438694 | 0 | Let $R$ be a regular local ring and let $I \subset R$ be an almost complete intersection ideal, that is, $\mu(I)=\text{ht}(I)+1$ where $\mu(I)$ is the number of minimal generators of $I$ and $ht(I)=\text{dim}(R)-\text{dim}(R/I)$. Is it true that $\text{depth}(R/I)=\text{dim}(R/I)-1$? If not, is there any other relation between almost complete intersections and depth?
| https://mathoverflow.net/users/113200 | Depth of almost complete intersection rings | In general, depth can be zero. Let me give an example.
Let $X\subset\mathbb{P}^n$ be a smooth variety of dimension $d$ which is subcanonical (this means $K\_X=O\_X(\*)$ for some integer $\*$) and NOT arithmetically normal (this means the map $H^0(O\_{\mathbb{P}^n}(l))\to H^0(O\_X(l))$ is not onto for some $l$).
Let $m=n-d$ and choose general $m$ hypersurfaces $f\_1,\ldots,f\_m$ vanishing on $X$ of large enough degree. Then, $V(f\_1,\ldots,f\_m)=X\cup Y$ where $Y$ is another $d$ dimensional variety and the $f\_i$s define a complete intersection. One has an exact sequence,
$$0\to K\_X(\*)\to O\_{X\cup Y}\to O\_Y\to 0,$$
for a suitable integer $\*$.
Let $R$ be the homogeneous co-ordinate ring of $\mathbb{P}^n$ and let $I$ the defining ideal of $X$. Taking global sections in the above sequence gives,
$$0\to H^0\_\*(K\_X)=A\to H^0\_\*(O\_{X\cup Y})=R/(f\_i)=S\to S/A\to 0.$$
One has a natural inclusion $R/I\to A$, by our assumption on $K\_X$ and this inclusion is not an isomorphism. The cokernel $T$ is non-zero of finite length. Composing this inclusion with the inclusion of $A\to S$, we get an inclusion, $R/I\to S$ and let the cokernel be $B$. Notice that $B$ is almost complete intersection and by snake lemma we have an exact sequence $0\to T\to B\to S/A\to 0$ and thus depth of $B$ is zero.
To get a regular local ring, we may localize at the irrelevant maximal ideal of $R$ without changing the final claim.
There are several ways of constructing $X$ as above. Let me describe one. Take a non-degenerate smooth complete intersection $X$ of dimension $d$ in $\mathbb{P}^{n+1}$ where $n>2d$. Then, take a generic projection to $\mathbb{P}^n$.
| 2 | https://mathoverflow.net/users/9502 | 438910 | 177,263 |
https://mathoverflow.net/questions/438873 | 8 | *I have asked [the same question on math.SE](https://math.stackexchange.com/questions/4620954/is-the-n-2-th-heat-kernel-coefficient-topological), without much success so I'm trying my luck here too.*
Let $M$ be an $n$-dimensional manifold, with $n$ *even* and consider the heat kernel of the Laplacian on $M$:
$$K\_M(t) := \mathrm{Tr}\Big(\mathrm{e}^{-t \Delta\_M}\Big) \equiv \sum\_{\lambda\in\mathrm{spec}(\Delta\_M)} \mathrm{e}^{-t \lambda}.$$
The heat kernel admits an asympotic, small-$t$ expansion as
$$K\_M(t) = \frac{1}{(4\pi t)^{n/2}}\sum\_{k=0}^\infty b\_k^{(M)} t^k,$$
where $b\_k^{(M)}$ are known as the *heat kernel coefficients* and are given by integrals of local geometric data of $M$.
I am particularly interested in what I'll call *middle-dimensional heat kernel coefficient*, namely
$$ b\_{n/2}^{(M)} = b\_{\mathrm{dim}M/2}^{(M)}.$$
This coefficient is interesting because it is very closely related to the value of the Minakshisundaram-Pleijel zeta function (i.e. the spectral zeta function for the Laplacian) at zero.
Is an explicit formula for the calculation of $b\_{n/2}^{(M)}$ known? In particular, is it a topological invariant? It seems to me that it wants to be related to the number of connected components of $M$ or to some (alternating?) sum of Betti numbers of $M$, but I haven't managed to show it.
If anyone feels brave, I'd be curious about how it generalizes to $p$-forms; namely what's the middle-dimensional heat kernel coefficient associated to the Laplacian, $\Delta\_p = \mathrm{d}\delta + \delta \mathrm{d}$, acting on $p$-forms on $M$?
| https://mathoverflow.net/users/498081 | Is the $n/2$-th heat kernel coefficient topological? | For $n=2$, the answer is given by $\frac{E}{6}$, where $E$ is the Euler characteristic of $M$, see McKean Jr, H. P., & Singer, I. M. (1967). Curvature and the eigenvalues of the Laplacian. Journal of Differential Geometry, 1(1-2), 43-69.
For $n>2$, if it is topological, then it is none of what you conjectured. Consider the case of hyperbolic manifolds: since those are locally isometric to the hyperbolic space, the coefficient $b^{M}\_{n/2}$ is proportional to the hyperbolic volume (Which, incidentally, is indeed topological by Mostow rigidity.)
**Update:** It is not topological for $n>2$. Let $M\_1=\mathbb{T}^2\times\mathbb{S}^2$ and $M\_2=(2\mathbb{T}^2)\times \mathbb{S}^2,$ where $2\mathbb{T}^2$ denotes the torus scaled by the factor of $2$. Then, $M\_1$ and $M\_2$ are homeomorphic and $b\_2^{M\_2}=4b\_2^{M\_1},$ because both spaces are homogeneous and locally isometric to each other. On the other hand, $$P^{M\_1}\_t(x,x)=P^{\mathbb{T}^2}\_t(x,x)P^{\mathbb{S}^2}\_t(x,x)=\left(\frac{1}{4\pi t}+o(t^{100})\right)P^{\mathbb{S}^2}\_t(x,x).$$ Therefore, we only need to show that $b^{\mathbb{S}^2}\_2$ does not vanish. According to the table on page 63 of the above reference, this is indeed the case.
| 6 | https://mathoverflow.net/users/56624 | 438923 | 177,268 |
https://mathoverflow.net/questions/438889 | 5 | This question came about when I was naively considering to what extent generating sets for finitely-generated groups are unique.
Let $G$ be a group and let $\phi\_1, \phi\_2 : F\_k \to G$ be two surjective homomorphisms from a free group on $k$ letters $F\_k$. Given a natural number $K \geq k$ let the surjective homomorphisms $\Phi\_1^K, \Phi\_2^K : F\_K \to G$ be the obtained by extending the corresponding maps $\phi\_1$ and $\phi\_2$ to send the $K-k$ other letters in $F\_K$ to the identity.
Does there exist some $K$ such that $\Phi\_1^K \circ \alpha = \Phi\_2^K$ for some automorphism $\alpha$ of $F\_K$?
| https://mathoverflow.net/users/419791 | Stable equivalence of generating sets of a finitely-generated group? | The answer is $K=2k$, which is an exercise. Here is a sketch: Let $f\colon \langle x\_i\rangle\to G$ be a surjective homomorphism, and $h\colon \langle x\_i\rangle\*\langle y\rangle\to G$ the map defined by $h(x\_i)=f(x\_i)$, $h(y)=e$. Let $g\in G$ be arbitrary, and choose $w\in\langle x\_i\rangle$ such that $f(w)=g$. Let $\alpha$ be the automorphism of $\langle x\_i\rangle\*\langle y\rangle$ which maps $y$ to $yw$ and leaves the other generators alone. Then $h(\alpha(y))=g$. Do this $k$ times, one for each element of your second generating tuple, then reverse the process, swapping the roles of the $x$ and $y$.
$K=2k-1$ doesn't suffice, by Kapovich and Weidmann, who in "Nielsen equivalence in a class of random groups" construct small-cancellation examples.
| 9 | https://mathoverflow.net/users/2225 | 438951 | 177,279 |
https://mathoverflow.net/questions/438878 | 5 | For a link $L\subset S^3$ and two *Seifert surfaces* (edit: a better name would be slice surfaces as the comments below [1](https://mathoverflow.net/questions/438878/are-two-slice-surfaces-with-minimal-genus-isotopic#comment1131877_438878) [2](https://mathoverflow.net/questions/438878/are-two-slice-surfaces-with-minimal-genus-isotopic#comment1131880_438878) point out) with minimal genus $S\_1,S\_2\subset B^4$, I have the following questions:
1. Is it true that the surfaces are isotopic?
2. And if the knots are algebraic?
3. I am interested in the Hopf link, so the surface is a cylinder, is it true if the surfaces have genus zero?
I know of the existence of the paper [Livingston - Surfaces bounding the unlink](https://doi.org/10.1307/mmj/1029002727) and [Using a 4th dimension to make Seifert surfaces isotopic?](https://mathoverflow.net/questions/376600/using-a-4th-dimension-to-make-seifert-surfaces-isotopic)
but it answers a particular case and I hope that there are updates since 1980 (does not seem so from the citations).
| https://mathoverflow.net/users/117639 | Are two slice surfaces with minimal genus isotopic? | The question is very loaded and the question would almost require a survey...
Anyway, the answer to your questions is mostly no. Let $S \subset S^4$ be a 2-knot (i.e. an embedded 2-sphere), $p \in S$ a point, and remove a small ball around $p$. The complement of $S$ in the complement of the ball is a slice disc $D$ for the unknot, and the fundamental group of the complement of $D$ is isomorphic to the complement of the fundamental group of $S^4\setminus S$, which can be non-trivial. (It is possible that $D$ is trivial if and only if $S$ is, but I don't want to make the claim without having properly checked). So, any $2$-knot with non-cyclic fundamental group gives you a slice disc for the unknot which is not isotopic to the standard one.
You can plant the construction into any minimal-genus Seifert surface, and probably you get the same construction for any knot. (Perhaps using the Alexander polynomial is a good way to show that you get different things, without worrying too much about Seifert--van Kampen.)
Now, you might ask: what about slice surfaces whose complement has cyclic fundamental group? Then the answer depends on the category. For locally-flat slice discs, Freedman showed that they are topologically isotopic. There's a [nice preprint](https://arxiv.org/abs/2205.12774) of Conway, Piccirillo, and Powell where they treat higher-genus surfaces and they give a complete classification in terms of algebro-topological data (Blanchfield forms). In the smooth category, things get pretty weird: Kyle Hayden has examples of slice discs that are topologically isotopic but not smoothly isotopic (see [here](https://arxiv.org/abs/2003.13681)) and of pushed-in Seifert surfaces that stay non-isotopic in $B^4$ (note that these automatically have cyclic fundamental group).
Back on the subject of pushed-in Seifert surfaces, Hayden, Kim, M. Miller, Park, and Sundberg proved that they can be topologically non-isotopic and topologically isotopic but not smoothly isotopic (see [here](https://arxiv.org/abs/2205.15283) and the Quanta Magazine [article](https://www.quantamagazine.org/special-surfaces-remain-distinct-in-four-dimensions-20220616/) on their result).
| 9 | https://mathoverflow.net/users/13119 | 438960 | 177,282 |
https://mathoverflow.net/questions/438701 | 3 | **I. Comparison**
It doesn't seem to be well-known that the generic cubic (prominent in this [MO post](https://mathoverflow.net/questions/438594/)) for $C\_3 = A\_3$,
$$x^3-nx^2+(n-3)x+1 = 0$$
has the nice property that its roots $a,b,c$, if in correct order, obey,
$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$
(I only noticed this after I asked an [MO question](https://mathoverflow.net/questions/438035/) about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)
Since the generic cubic is intimately connected to the roots of unity for prime $p\equiv 1\,\text{mod}\; 6$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the *Emma Lehmer quintic* which is for $p\equiv 1\,\text{mod}\; 10$, namely,
$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$
It turns out its roots $x\_k$ may have a similar property.
---
**II. Question:**
Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the **correct** ordering of its roots $a,b,c,d,e$?
***Update 1:*** Thanks to Peter Taylor in the comments, and using the fact that $abcde = -1$, we can get rid of the fifth roots and get the equivalent but more elegant form,
$$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$
Or more generally (for $\mu$ an integer),
$$\frac{\mu}{a}-\frac{\mu^{2}}{ab}+\frac{\mu^{3}}{abc}-\frac{\mu^{4}}{abcd}+\frac{\mu^{5}}{abcde} = 0$$
where $\mu^5$ is the constant term of the quintic and the Lehmer quintic the special case $\mu=1$.
***Update 2:*** I've already tested the [*Hashimoto septic*](https://mathoverflow.net/a/255424/12905) which fortunately has a seventh power $\mu^7$ as its constant term and the analogous relation,
$$\frac{\mu}{a}-\frac{\mu^{2}}{ab}+\frac{\mu^{3}}{abc}-\frac{\mu^{4}}{abcd}+\frac{\mu^{5}}{abcde}-\frac{\mu^{6}}{abcdef}+\frac{\mu^{7}}{abcdefg} = 0$$
and among the $(p-1)! = 720$ permutation of its roots, at least 9 works.
---
**III. Example**
Let $n=-1$. Then we get the quintic for $p=11$ and its roots,
$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$
$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$
with $k = 1, 4, 5, 2, 3$ as the correct order. One can then verify it obeys the relation in the question.
**P.S.** For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?
| https://mathoverflow.net/users/12905 | A similar relationship between the generic cubic and the Lehmer quintic? | The map $s(r) = \frac{n+2 + nr - r^2}{1 + (n+2)r}$ cyclically permutes the roots. This map is given in [2], and I found it through the reference in [1]. It turns out to give the correct order.
Explicitly, if we pick one root $x\_1$ we have
\begin{eqnarray\*}
D &=& n^3 + 5n^2 + 10n + 7 \\
Dx\_2 &=& (n^2 + 4n + 4)x\_1^4 + (n^4 + 4n^3 + 4n^2 - n - 2)x\_1^3 + \\&& (-2n^5 - 14n^4 - 43n^3 - 76n^2 - 80n - 39)x\_1^2 + \\&& (n^6 + 9n^5 + 37n^4 + 89n^3 + 131n^2 + 107n + 36)x\_1 + \\&& (n^4 + 8n^3 + 26n^2 + 39n + 22) \\
%
Dx\_3 &=& (-2n - 3)x\_1^4 + (-2n^3 - 4n^2 - 3n - 2)x\_1^3 + \\&& (3n^4 + 14n^3 + 31n^2 + 41n + 24)x\_1^2 + \\&& (-n^5 - 7n^4 - 21n^3 - 36n^2 - 29n - 6)x\_1 + \\&& (-2n^2 - 7n - 6) \\
%
Dx\_4 &=& (-n^2 - 3n - 2)x\_1^4 + (-n^4 - 3n^3 - 2n^2 + n + 1)x\_1^3 + \\&& (2n^5 + 12n^4 + 33n^3 + 54n^2 + 53n + 23)x\_1^2 + \\&& (-n^6 - 8n^5 - 29n^4 - 62n^3 - 81n^2 - 59n - 18)x\_1 + \\&& (-n^5 - 7n^4 - 24n^3 - 47n^2 - 52n - 25) \\
%
Dx\_5 &=& (n + 1)x\_1^4 + (n^3 + 2n^2 + 3n + 3)x\_1^3 + \\&& (-n^4 - 4n^3 - 9n^2 - 14n - 8)x\_1^2 + \\&& (-n^4 - 7n^3 - 19n^2 - 29n - 19)x\_1 + \\&& (n^4 + 6n^3 + 16n^2 + 20n + 9)
\end{eqnarray\*}
and it is pure computation to show that $$x\_1 x\_2 x\_3 x\_4 x\_5 - x\_2 x\_3 x\_4 x\_5 + x\_3 x\_4 x\_5 - x\_4 x\_5 + x\_5 = 0$$
[Online computation with sagecell.sagemath.org](https://sagecell.sagemath.org/?z=eJyFk01v2zAMhs8LkP9AeIdZTpxWsrXD0OzYc-sNuxQ1YNRyI8yRPFkd1A377yPleEmGoj0oovi8_JDoVJsr8xm2cGP7Z2P3uukrbR7T21u2XCwXrepgSAP7tFy8c8o_OQOhlrACU4ss1CXkkIrM1AW6PuIucOeXmYkbQwWB1KBwBfKg43wWyiiUpBOTjnj5Qh7a537G835SEpHeoCyHkAUGF5By8sZA9Me7VGJzFfx-eOG216558Nqaa636Nq0Y6qsC1Zy0GPfjyXqtjE_xLTADi_mCQIjNcDyFYrIF2eVkF2TLyS7Rdtb6EY93ga8hCFwFrhKXvKd87-Fb0-u28Qr8rvH4o54_OAVN30OrR6_Ng18uBqexj16ZdFQ-jTkZW0MyKyC6EvZWxiib090l9nsCepo1bLdwCaofFSRdo_sEOusggDZT0P0hN94W5xoKHFoocWJBsvo3v5B_6PkDJyaIFcTKmW2JIUE_ejEI1SiWR57z6KISX-wrmVavdECsIFYSk8T4KSuJSWKcmDhlkhgnJogVB0b9TN3XOV2gzgnWeaydx3T5v_r_3zE_O63OwOySjEo0poURB6nAWKPAdjS16eFBj_BLObuOrk53fhcBDI3zBDEiJ8FmnmvydaesU3sYnP2pTBxx4FnAP0aRhTI7dHY8UWdHMLvk6SdxgyU7oA9DtQn7C_6_KP4=&lang=sage&interacts=eJyLjgUAARUAuQ==)
However, this isn't a very satisfying proof because it doesn't address the question of necessary and sufficient properties of the original polynomial. Does a suitable permutation of roots exist for all irreducible polynomials of odd (or maybe odd prime) degree whose Galois group is cyclic? The equivalence between the forms
$$
(x\_1^4 x\_2^3 x\_3^2 x\_4)^{1/5} + (x\_2^4 x\_3^3 x\_4^2 x\_5)^{1/5} + (x\_3^4 x\_4^3 x\_5^2 x\_1)^{1/5} + (x\_4^4 x\_5^3 x\_1^2 x\_2)^{1/5} + (x\_5^4 x\_1^3 x\_2^2 x\_3)^{1/5} = 0 \\
\frac1{x\_1} - \frac1{x\_1 x\_2}+\frac1{x\_1 x\_2 x\_3} - \frac1{x\_1 x\_2 x\_3 x\_4} + \frac1{x\_1 x\_2 x\_3 x\_4 x\_5} = 0 \\
x\_1 x\_2 x\_3 x\_4 x\_5 - x\_2 x\_3 x\_4 x\_5 + x\_3 x\_4 x\_5 - x\_4 x\_5 + x\_5 = 0
$$
and a couple of other variants relies on the constant coefficient of Emma Lehmer's quintic being $1$. If there is a generalisation, which variant does it require? Or does generalisation require a cyclic Galois group and a unit constant coefficient?
---
[1] Henri Darmon, [*Note on a polynomial of Emma Lehmer*](https://www.math.mcgill.ca/darmon/pub/Articles/Research/03.Lehmer/paper.pdf), Math. Comp. vol. 56, no. 194, April 1991, pp. 795-800.
[2] Rene Schoof and Lawrence C. Washington, [*Quintic polynomials and
real cyclotomic fields with large class numbers*](https://www.ams.org/journals/mcom/1988-50-182/S0025-5718-1988-0929552-2/S0025-5718-1988-0929552-2.pdf), Math. Comp. vol. 50,
no. 182, April 1988, pp. 543-556.
| 3 | https://mathoverflow.net/users/46140 | 438962 | 177,284 |
https://mathoverflow.net/questions/438374 | 2 | Let $\mathfrak {g}$ be a non-degenerate triangular Lie bialgebra with the non-degenerate triangular structure $r \in \bigwedge^2 \mathfrak {g}.$ Then how does it induce $r^{-1} \in \bigwedge^2 \mathfrak {g}^{\ast}\ ?$
Since $r$ is a non-degenerate triangular structure on $\mathfrak {g}$ it induces an isomorphism $r^{\sharp} : \mathfrak {g}^{\ast} \longrightarrow \mathfrak {g}.$ So ${r^{\sharp}}^{-1} : \mathfrak {g} \longrightarrow \mathfrak {g}^{\ast}$ exists as an isomorphism of vector spaces. But how does it give rise to an element $r^{-1} \in \bigwedge^2 \mathfrak {g}^{\ast}\ $? Is it $\left ({r^{\sharp}}^{-1} \right )^{\otimes 2} (r)\ $?
Any help in this regard would be warmly appreciated. Thanks for your time.
| https://mathoverflow.net/users/491136 | How to define inverse of a non-degenerate triangular structure? | The answer to your last question is yes.
Note that the property of being triangular for $r$ reads as $[r,r]=0$, where $[-,-]$ is the Lie bracket of $\mathfrak{g}$ that one extends to $\wedge^\bullet\mathfrak{g}$ by the graded Leibniz rule.
This condition translates on $\omega:=r^{-1}$ in the following way: $d\_{CE}\omega=0$. Here $d\_{CE}$ is the differential for the Chevalley-Eilenberg cochain complex.
The whole story is parallel to how one identifies non-degenerate Poisson bivectors with symplectic forms (non-degenerate closed $2$-forms).
| 2 | https://mathoverflow.net/users/7031 | 439006 | 177,305 |
https://mathoverflow.net/questions/439005 | 8 | One can consider a variant of the Dedekind-Peano axioms
in which one replaces the assumption that every number
has exactly one successor by the assumption that every
number has **at most** one successor, leaving the other
axioms the same (and in particular retaining the
second-order induction axiom). The models of this theory
are then the initial segments of the natural numbers.
Call it a "size-neutral" version of the second-order
Dedekind-Peano theory. My question is, to what extent
can one prove analogues of standard theorems about the
natural numbers in this theory?
For instance, what would serve as a size-neutral analogue
of the proposition that the set of primes is infinite? Ideally it would be some proposition P with the property that if
you assume that P is true, *and* you assume that every
number has a successor, then the standard claim about
the set of primes being infinite would follow without
too much trouble (where "too much trouble" means "going back to the beginning and proving the infinitude of the primes in the usual way without invoking P at all").
I'm aware that even defining "prime" is problematic.
To begin with, in a size-neutral theory, one cannot
define addition or multiplication in the usual recursive
way, or rather, if one does, the operations are not
provably total; if one's model cuts off at n, then
sums and products that exceed this bound are undefined.
One could choose to remedy this with a definition that
basically says "If a number has no successor, define it
to be its own successor; now proceed using this new,
enlarged notion of succession instead of the old one",
and then sums and products that in ordinary arithmetic
exceed n would now equal n -- though I'm not sure this is
a good way to go.
This can't be a new line of thought, so I'd be happy
with answers of the form "Go read X".
| https://mathoverflow.net/users/3621 | Dedekind-Peano axioms, but numbers have at most one successor | Victoria Gitman and I recently looked at the first-order version of this theory, which we called fPA, to contrast it with the theory FPA, which Woodin has looked into, which is a possibly stronger version.
[**Update** As explained by Emil Jeřábek in the comments below and in Ali Enayat's answer, this theory has been well studied in models of arithmetic and is known as $\text{PA}^{\text{top}}$. The observations below about $I\Delta\_0$ are evidently due originally to Jeff Paris.]
So the language of fPA is the same as the ordinary language of arithmetic, except that there is possibly a largest number (without a successor), and the operations of + and $\cdot$ are merely partial, but defined as far as they can be according to the usual recursive definitions. Meanwhile, we have the first-order induction scheme.
The main observation we made is that if you have a model of fPA with a largest number $N$, then there is some largest number $m$ whose square exists $m^2\leq N$, and using this number, we can interpret a larger model of fPA. Simply use quadruples of numbers $abcd$ to represent larger numbers in base $m$. We know how to add and multiply such numbers in that representation, and this requires only arithmetic for numbers less than $m$, which works fine. Using this method, we can build a larger model of fPA in which $N^2$ exists. Basically, every model of fPA interprets a larger model of fPA, in which any two numbers from the original model now have a product.
**Answer to your question.** The main point I want to make is that the base $m$ method provides a means to express primality etc. also in your context. Since the interpreted model is definable in the original model using quadruples, we can express those notions entirely in the original model simply by using the base $m$ representation. Any concept that could be expressed in the context of the larger models can also be expressed in the original model.
**Iterating.** The further observation is that we can simply iterate the interpretation construction $\omega$ many times, interpreting to larger and larger models of fPA. The union model is a model of $I\Delta\_0$, since bounded induction amounts to unbounded induction in the individual models.
**Conclusion.** The models of fPA with a largest number all arise by chopping a model of $I\Delta\_0$ at some number $N$.
For this reason, questions about what is or what is not provable in fPA amount to the corresponding questions about $I\Delta\_0$, which is an intensely studied theory and there are many open questions.
**FPA.** Woodin had looked into the first-order theory FPA in which one adopts the pigeon-hole principle as a fundamental axiom scheme, rather than just the first-order induction scheme. PHP implies the induction scheme, but it is open whether they are equivalent.
| 14 | https://mathoverflow.net/users/1946 | 439012 | 177,311 |
https://mathoverflow.net/questions/439018 | 4 | I am wondering if there are some know results for the non-trivial roots at ${\rm Re}(s) = \frac{1}{2}$, even maybe a table of the first few roots with $t>0$. This sister function
$$
L\_4^\* (s,\chi\_4) = \prod\_{p} \Big(1 + \frac{\chi\_4(p)}{p^{s}}\Big)^{-1} = \left(1-\frac{1}{4^s}\right)
\frac{\zeta(2s)}{L\_4(s,\chi\_4)}
$$
seems easier to handle. I could not find any study or table regarding the zeros of
$L\_4(s,\chi\_4)$. The only interesting document I found is the following (below) but it focuses on computational aspects only all done in C. Looking for help with this.
<https://fredrikj.net/blog/2016/11/dirichlet-l-functions-in-arb/>
| https://mathoverflow.net/users/140356 | Zeros of Dirichlet function $L(s,\chi_4)$ | There is no need to use the subscript $4$ on the $L$-function: just write
$L(s,\chi\_4)$ and $L^\*(s,\chi\_4)$. The first nontrivial zero in the upper half of the critical strip is $1/2 + it$ where
$t \approx 6.020948i$.
Data for $L(s,\chi\_4)$ is available on the LMFDB: see <https://www.lmfdb.org/L/1/2e2/4.3/r1/0/0>. You can find imaginary parts of the initial zeros in the upper and lower half-plane towards the bottom of the page (the zeros are symmetric since $\chi\_4$ is quadratic).
In PARI, the command lfunzeros(L,T) computes imaginary parts $t$ of critical zeros with "$L$" being some $L$-function data (datum?) and $0 \leq t \leq T$. Being lazy, let's take L to be $x^2+1$, whose $L$-function is the zeta-function of $\mathbf Q(i)$, namely $\zeta(s)L(s,\chi\_4)$. Typing lfunzeros(x^2+1,20), I get the answer
>
> [6.0209489046975966549025115216120858689, 10.243770304166554552137757479109959025, 12.988098012312422507453109789562993765, 14.134725141734693790457251983562470271, 16.342607104587222194976861483456149939, 18.291993196123534838526004277590699428].
>
>
>
Only 14.1347... belongs to $\zeta(s)$; the rest are zeros of $L(s,\chi\_4)$.
See <https://pari.math.u-bordeaux.fr/pub/pari/manuals/2.13.3/refcard-lfun.pdf> for a printout listing PARI -function commands and <https://pari.math.u-bordeaux.fr/dochtml/html-stable/_L_minusfunctions.html> for a similar webpage.
| 11 | https://mathoverflow.net/users/3272 | 439019 | 177,313 |
https://mathoverflow.net/questions/438993 | 1 | Let $\mathcal{X}$ be some nice space, let $\Phi$ be some ordered space, and let $K :\mathcal{X} \times \mathcal{X} \times \Phi \to \mathbf{R}$ be a positive-semidefinite kernel indexed by a hyperparameter $\phi$, i.e. for any $\phi \in \Phi$, the mapping $K(\cdot, \cdot, \phi) :\mathcal{X} \times \mathcal{X} \to \mathbf{R}$ is a positive-semidefinite kernel in the sense of Mercer.
Now, fix a finite subset $\mathcal{X}\_N = \{ x\_1 , \ldots, x\_N \} \subseteq \mathcal{X}$ of size $N$, and consider the mapping $\mathbf{K}$ which takes a hyperparameter $\phi$ and maps it to the corresponding $N \times N$ kernel matrix, i.e.
\begin{align}
\mathbf{K} ( \phi) \in \mathrm{PSD}\_{N \times N}, \quad \\
\mathbf{K} ( \phi)\_{i, j} = K (x\_i, x\_j, \phi).
\end{align}
I'm interested in examples of parametric kernels $K$ for which this mapping is order-preserving (for any set $\mathcal{X}\_N$), in the sense that when $\phi\_1 \prec \phi\_2$ with respect to the ordering on $\Phi$, it follows that $\mathbf{K} ( \phi\_1 ) \prec \mathbf{K} ( \phi\_2)$ in the semidefinite ordering on matrices.
Some simple examples of which I'm already aware:
* Let $K\_1$, $K\_2$ be two PSD kernels, let $\Phi = \mathbf{R}\_+$, and let $K(x, y, \phi) = K\_1 (x, y) + \phi \cdot K\_2 (x, y)$.
* Let $\{ K\_i : i \in \mathbf{N} \}$ be a countable collection of PSD kernels, let $\Phi = \mathbf{N}$, and let $K(x, y, \phi) = \sum\_{i \leqslant \phi} K\_i (x, y)$.
I'd be particularly interested in cases where such an ordering holds as e.g. the length scale of a kernel varies, or similar.
| https://mathoverflow.net/users/121692 | Monotonicity of kernel matrices with respect to hyperparameters | This example may be a little bit ridiculous, but suppose we take $\mathcal{X}=\mathbb{R}$ and let $\Phi$ be any parametric subset of the set of PSD kernels itself. We define
$$ \mathbf{K}(\phi)\_{i,j} = \sum\_{i,j}\phi(x\_i, x\_j) $$
We will say that $\phi\_1 \preceq \phi\_2$ iff for any $N$, $x\_1,...,x\_N$, and $c\_1,...,c\_N\geq 0$, we have
$$\sum\_{i,j} c\_i c\_j (\phi\_2(x\_i,x\_j)-\phi\_1(x\_i,x\_j)) \geq 0$$
This gives a partial order on $\Phi$ that (by construction) respects the OP's property of interest.
Possibly the OP is less interested in the preceding construction and more interested in concrete parametric examples. If so, geodesics of the cone of PSD matrices provide some interesting examples. For example, take $K\_1, K\_2$ as two PSD kernels, and let $\Phi=[0,1]$. For a fixed $N$, let $A$ and $B$ be the two corresponding PSD matrices. Then define
$$K(x,y,\phi)=A^{1/2}\exp(\phi \log(A^{-1/2}BA^{-1/2}))A^{1/2}$$
More inspiration along that line might be found at Bonnabel and Sepulchre's ["Riemannian Metric and Geometric Mean for
Positive Semidefinite Matrices of Fixed Rank"](https://arxiv.org/pdf/0807.4462.pdf)
| 1 | https://mathoverflow.net/users/8938 | 439021 | 177,314 |
https://mathoverflow.net/questions/439028 | 8 | Let $M$ be the Mandelbrot set.
**Question:** Is the interior of $M$ dense in $M$?
My intuition is that this is true, and moreover that hyperbolic components of the interior are dense in $M$ as well, and moreover that this is known (as it is not very close to the Hyperbolicity Conjecture and thus not too hard). Is that right?
| https://mathoverflow.net/users/22930 | Does the Mandelbrot set have dense interior? | The answer is positive and this is not difficult (a normal families argument). The boundary
of the Mandelbrot set is the set of $J$-instability. Every point $c\_0$ of this set is a limit of $c\_n$ such that $z\mapsto z^2+c\_n$ has a superattracting cycle. So actually hyperbolic components accumulate
to every boundary point of $M$.
(MLC is related to a much harder statement, Fatou conjecture, that interior of $M$ consists of only hyperbolic components).
Refs: M. Lyubich, Some typical properties of the dynamics of rational mappings, English transl.: Russian Math surveys,
38, 1983, or
M. Lyubich, The dynamics of rational transforms: the topological picture,
Russ. Math. Surv. 41, No. 4, 43-117 (1986);
| 10 | https://mathoverflow.net/users/25510 | 439039 | 177,321 |
https://mathoverflow.net/questions/439030 | 7 | In some constructive systems, [every function from $\mathbb{R}\to\mathbb{R}$ is continuous](https://math.stackexchange.com/a/176285/24563) (roughly speaking from the classical fact that computable functions are continuous). More weakly, in Bishop's constructive approach [one cannot prove](https://mathoverflow.net/q/164694/36972) the existence of discontinuous functions.
I am wondering: in such a system, is the [Weierstrass function](https://en.wikipedia.org/wiki/Weierstrass_function)
$$W\_\alpha(x) = \sum\_{n=0}^\infty a^n \cos(b^n\pi x),\quad ab>1+3\pi/2$$
nowhere differentiable? (It is well-known classically to be continuous but nowhere differentiable.) Perhaps a broader, hence more interesting, question is whether one can construct a continuous but nowhere differentiable function in Bishop's mathematics.
| https://mathoverflow.net/users/36972 | Continuous nowhere differentiability and constructive mathematics | The usual proofs are either constructive or can be made constructive fairly easily, sometimes by a slight weakening of the theorem. For example, let us read through [this note](https://math.berkeley.edu/%7Ebrent/files/104_weierstrass.pdf) by Brent Nelson. (Please read the four page proof before reading the rest of the answer.)
The first part, establishing that the Weierstraß function is uniformly continuous, is constructive.
The second part proves non-differentiability by demonstrating that the limit $\lim\_{x \to x\_0} \frac{f(x) - f(x\_0)}{x - x\_0}$ does not exist. It is *not* an appeal proof by contradiction but a constructively valid [refutation by contradiction](https://en.wikipedia.org/wiki/Proof_by_contradiction#Refutation_by_contradiction).
There is just one sticky point, namely the proof tells us to select, for each $m \in \mathbb{N}$, an integer $\alpha\_m$ such that $$-1/2 < b^m x\_0 - \alpha\_m \leq 1/2.$$
These bounds are too tight, because constructively we cannot show that every real is within distance $1/2$ of an integer. (When $b^m x\_0$ is very close to $1/2$ we have to be able to decide whether it is actually smaller than $1/2$ or not in order to determine whether $\alpha\_m = 0$ or $\alpha\_m = 1$.)
So let us relax the condition for $\alpha\_m$ a bit and see if the rest of the proof goes through. Let $q > 0$ be a small "relaxation" rational (we shall impose additional conditions on it later as needed). Using countable choice, we can select integers $\alpha\_m$ such that
$$-1/2 - q < b^m x\_0 - \alpha\_m < 1/2 + q.$$
This is possible because $q$ gives us a bit of wiggle room.
Now the proof proceeds by showing that two partial sums, called $S\_1$ and $S\_2$, are respectively very small and very large. Smallness of $S\_1$ is unproblematic. The proof that $S\_2$ is large depends on bounding $x\_m = b^m x\_0 - \alpha\_m$ away from $1$ so that
$$
\frac{1 + \cos(b^n \pi x\_m)}{1 + x\_m}
$$
is seen to be positive. This is still the case with our $q$ thrown in as long as $q < 1/2$. But where the original proof gets the bound $2/3$ we obtain a slightly worse estimate
$$
\frac{1 + \cos(b^n \pi x\_m)}{1 + x\_m} \geq
\frac{1}{1 + \frac{1}{2} + q} = \frac{2}{3 + q}.
$$
We can still salvage the rest of the proof by picking a small enough $q$: so long as $a b > 1 + (3 + q) \pi / 2$, the estimates in the rest of the proof will work. Since we assumed $a b > 1 + 3 \pi / 2$, there is a rational $q > 0$ such that $a b > 1 + (3 + q) \pi / 2$, so we use one such. (This was *not* an application of choice, because "choosing one thing known to exist" is formally an elimination of $\exists$.)
One would likely have to work a bit harder to also eliminate countable choice.
| 8 | https://mathoverflow.net/users/1176 | 439047 | 177,324 |
https://mathoverflow.net/questions/439051 | 10 | [Baker's theorem](https://en.wikipedia.org/wiki/Baker%27s_theorem) in transcendental number theory states that
$$
\left|\beta\_0 + \sum\_{i=1}^n \beta\_i \log \alpha\_i\right| > H^{-C}
$$
where
* $\beta\_0, \ldots, \beta\_n$ are algebraic numbers, not all zero,
* $\alpha\_1, \ldots, \alpha\_n$ are multiplicatively independent algebraic numbers,
* $H$ is the maximum of the heights of $\beta\_i$, and
* $C$ is an effectively computable constant depending on $n$, $\alpha\_i$, and the maximum of the degrees of $\beta\_i$.
Is there a simple proof for a special case of Baker's theorem where $\beta\_0 = 0$, the $\beta\_i$ are rational integers, and the $\alpha\_i$ are positive rational integers?
| https://mathoverflow.net/users/4758 | Baker's theorem for integer combinations of logarithms of integers? | The main difficulty in proving Baker's theorem is in estimating $C$. If you don't care about those estimates, then the proof is not difficult. For example, Chapter 7 of Waldschmidt's *Diophantine approximation on linear groups* gives a simple proof for the homogeneous case $\beta\_0=0$. Some technicalities (like using Fel'dman integer-valued polynomials) are not necessary if you don't care about strong estimates.
The homogeneous case is slightly simpler because one can work with just $G\_m^n$ or (dually) with $G\_a^n$. However, the basic structure of the proof is the same:
1. Using group structure, consider several points (working with $G\_m^n$, the points are $(\alpha\_1^s,\dots,\alpha\_n^s)$).
2. construct an auxiliary function/polynomial either using Dirichlet principle (in the form of Siegel's lemma) or using interpolating determinant
3. Show that this polynomial with some derivatives is small (using either Schwarz's lemma or Hermite's formula) at those points.
4. Show that it can't vanish at all those points.
5. Show that if the polynomial or its derivative is small at one of those points, the value must be zero.
Assumptions, that $\beta\_i$ and $\alpha\_i$ are integers simplify the last step - instead of Liouville's inequalities, you are just using directly that if $x$ is an integer with $|x|<1$ then $x=0$. However, the other steps are the same.
Baker's theorem can also be proved using the Schnieder-Lang criterion, but I don’t know if that method will provide estimates sufficient for your needs.
| 14 | https://mathoverflow.net/users/1811 | 439054 | 177,328 |
https://mathoverflow.net/questions/439046 | 4 | The “second Kleene algebra” $\mathcal{K}\_2$ is defined, e.g. [here on nLab](https://ncatlab.org/nlab/show/Kleene%27s+second+algebra), or in section 1.4.3 of van Oosten's book *Realizability: an Introduction to its Categorical Side* (2008), or as example 3.4 of [the notes “Realizability” by Thomas Streicher (2017–2018)](https://www2.mathematik.tu-darmstadt.de/%7Estreicher/REAL/REAL.pdf), or (implicitly) in ¶1.9.12 of Troelstra's *Metamathematical Investigations of Intuitionistic Arithmetic and Analysis* (1973), or in various other places. Let me reproduce the essential part of the definition for readers' convenience:
>
> Let $\mathcal{B} = \mathbb{N}^{\mathbb{N}}$ denote Baire space, endowed with its usual product topology. Given $\alpha \in \mathcal{B}$, define $F\_\alpha$ a partial function from $\mathcal{B}$ to $\mathbb{N}$ by
> $$
> F\_\alpha(\beta) = n \text{ \; iff \; }\exists k\in\mathbb{N}.(\alpha(\bar\beta\upharpoonright k)=n+1 \land \forall \ell<k.(\alpha(\bar\beta\upharpoonright k)=0))
> $$
> (undefined if no such $n$ exists) where $\bar\beta \upharpoonright k$ denotes an integer encoding the finite sequence $\langle\beta(0),\ldots,\beta(k-1)\rangle$. This defines a continuous function $U \to \mathbb{N}$ with $U \subseteq \mathcal{B}$ open, and any such function is of the form in question.
>
>
> Now let
> $$
> \alpha\bullet\beta = (n \mapsto F\_\alpha(\langle n\rangle \smallfrown \beta))
> $$
> provided $F\_\alpha(\langle n\rangle \smallfrown \beta)$ is defined for every $n$, undefined otherwise, where $\langle n\rangle \smallfrown \beta$ denotes the function $0 \mapsto n$ and $k+1 \mapsto \beta(k)$. Then $\mathcal{K}\_2$ is $\mathcal{B}$ endowed with this partial operation $\bullet$.
>
>
>
This definition is not complicated, but it is… extremely opaque, and none of the sources I've cited bother to explain why we define things in this particular way ($F\_\alpha(\beta)$ searches for the first nonzero value among the values returned by $\alpha$ on the finite subsequences of $\beta$ and subtracts one to it: why do we want to do precisely this? rather than, say, return the index $k$ where this nonzero value was found), apart from the fact that “it works” (it gives a partial combinatory algebra).
Kleene's *first* algebra $\mathcal{K}\_1$ (namely $\mathbb{N}$ with the operation $e\bullet n = \varphi\_e(n)$ for some standard enumeration $\varphi$ of partial recursive functions) is easy to develop an intuition for: it's the set of computable functions, i.e., computer programs, and the operation is that of taking data and feeding it to a computer program. I understand that the idea behind $\mathcal{K}\_2$ is to do something similar for continuous functions on Baire space (and that $F\_\alpha$ must play a role analogous to $\varphi\_e$), but this doesn't really help me explain the particulars of the definition above (I feel like if in the case of $\mathcal{K}\_1$ I'd been described what a Turing machine was without any explanation as to how this definition was reached). So:
**Question:** What's the intuition behind the definition of $\mathcal{K}\_2$ and the reason for this particular choice of definition? Has anyone written an introduction to $\mathcal{K}\_2$ analogous to what a first course on computability theory would be on $\mathcal{K}\_1$?
| https://mathoverflow.net/users/17064 | Intuition behind Kleene's “second algebra” $\mathcal{K}_2$ | There's a bit of explanation in my [lecture notes on intuitionistic logic](https://awswan.github.io/teaching/intuitionisticlogic/index.html), in particular the [section on function realizability](https://awswan.github.io/teaching/intuitionisticlogic/lecturenotes/section14.pdf).
The idea is that you want to encode continuous partial functions $\mathbb{N}^\mathbb{N} \to \mathbb{N}^\mathbb{N}$ as elements of $\mathbb{N}^\mathbb{N}$ analogously to how computable partial functions $\mathbb{N} \to \mathbb{N}$ can be encoded as elements of $\mathbb{N}$. It turns out the main idea comes up just encoding partial continuous functions $\mathbb{N}^\mathbb{N} \to \mathbb{N}$, so I'll talk about that.
It's easiest to break it into two steps. Write $\mathbb{N}^{< \omega}$ for finite sequences of numbers. Then given a function $f : \mathbb{N}^{< \omega} \to \mathbb{N} + \{\bot\}$, we can view it as a continuous partial function $F : \mathbb{N}^\mathbb{N} \to \mathbb{N}$ as follows. To compute $F(g)$ we feed finite approximations of $g$, e.g. the length $k$ approximation $[g(0), g(1), g(2), \ldots, g(k)]$. $f([g(0), g(1), g(2), \ldots, g(k)])$ can either lie in the $\mathbb{N}$ component of $\mathbb{N} + \{\bot\}$ or it's equal to $\bot$. The former tells us that we can compute the value of $F(g)$ from this list, and it is equal to $f([g(0), \ldots, g(k)])$ (unless it was already computed at an earlier stage). The latter, $f([g(0), \ldots, g(k)]) = \bot$, tells us $f$ has not been given enough information and needs a longer list before it can return a value.
Finally note that $\mathbb{N}^{< \omega}$ and $\mathbb{N} + \{\bot\}$ are both countable, so we can view $f$ as a function $\mathbb{N} \to \mathbb{N}$. In particular checking for a number $> 0$ and then subtracting $1$ comes from the bijection $\mathbb{N} + \{\bot\} \cong \mathbb{N}$.
| 4 | https://mathoverflow.net/users/30790 | 439062 | 177,330 |
https://mathoverflow.net/questions/439063 | 4 |
>
> Can we have a sequence of transitive sets $\langle\mathcal V\_0, \mathcal V\_1, \mathcal V\_2,...\rangle$, all modeling $\sf ZF$, such that $\mathcal P(V\_n) \subset \mathcal V\_{n+1}$, and the cardinality of each is inaccessible from the cardinalities of the sets preceding it, and where choice fails inside each one of them, yet each successor set $\mathcal V\_{n+1}$ proves choice externally over its predecessor $\mathcal V\_n$?
>
>
>
Formal workup:
Add a primitive total unary function symbol $\mathcal V$ to the language of set theory; as a notation we shall write $\mathcal V(x)$ as $\mathcal V\_x$. To all axioms of $\sf ZF$ [in language $\{=,\in\}$], add the following axioms:
**Restriction:** $\forall x: x \not \in \omega \to \mathcal V\_x= \emptyset$
**Modeling:** if $\phi$ is an axiom of $\sf ZF\neg C$, then: $\forall n \in \omega: \phi^{\mathcal V\_n}$
**Transitivity:** $\forall n \in \omega: \mathcal V\_n \text { is transitive}$
**Power :** $\forall n \forall x: n \in \omega \land x \subseteq \mathcal V\_n \to x \in \mathcal V\_{n+1}$
**Inaccessibility:** $\forall n \in \omega: \operatorname {icc}( |\mathcal V\_n |)$
**Choice:** $\forall n \in \omega: \forall x \in \mathcal V\_n \exists f \in \mathcal V\_{n+1}: \\\operatorname {dom}(f)=x \setminus \{\emptyset\} \land \forall m \, (f(m) \in m ) $
Where for every $\Phi$ set of sentences, $[\Phi]^X$ is the relativization of all quantifiers in elements of $\Phi$ to $X$; and $``\operatorname {icc..}\!\!"$ stands for *".. is inaccessible"*, $``| \ |\!\!"$ stands for cardinality defined after Scott's. This means that the cardinality of each $\mathcal V\_n$ is not reachable by set unions and powering from below, i.e. it is strictly larger than the cardinality of any set union of a set of strictly smaller cardinality whose elements are of strictly smaller cardinalities, and it is strictly larger than the cardinality of the power set of any set of a strictly smaller cardinality.
| https://mathoverflow.net/users/95347 | Can we have this sequence where choice fails and returns? | Sure.
Start with countably many inaccessible cardinals, $\kappa\_n$, and now take the full support product adding $\kappa\_n^+$ subsets to each $\kappa\_n$. Then the $n$th model is the symmetric extension given by adding all the generics for the first $n-1$ coordinates, and violating choice in the $n$th one (e.g. a Cohen style model).
| 8 | https://mathoverflow.net/users/7206 | 439064 | 177,331 |
https://mathoverflow.net/questions/438850 | 1 | If $\lambda=(\lambda\_1\geq\lambda\_2\geq\dots\geq\lambda\_k)\vdash n$ is an integer partition of $n$ then $\lambda\_1$ is its largest part and $k$ is its length, $\ell(\lambda)$.
Define the statistic $c\_n(\lambda)=\max\{\lambda\_1,\ell(\lambda)\}$ for the above partition. Also consider the polynomial (in $t$),
$$Q\_n(t)=\sum\_{\lambda\vdash n}t^{c\_n(\lambda)}.$$
Here are some examples:
```
t
2
2 t
3 2
2 t + t
4 3 2
2 t + 2 t + t
5 4 3
2 t + 2 t + 3 t
6 5 4 3
2 t + 2 t + 4 t + 3 t
7 6 5 4 3
2 t + 2 t + 4 t + 5 t + 2 t
8 7 6 5 4 3
2 t + 2 t + 4 t + 6 t + 7 t + t
9 8 7 6 5 4 3
2 t + 2 t + 4 t + 6 t + 9 t + 6 t + t
10 9 8 7 6 5 4
2 t + 2 t + 4 t + 6 t + 10 t + 11 t + 7 t
```
**QUESTION 1.** It appears that the coefficients of $Q\_n(t)$, read from left to right, are twice the partition numbers $1,1,2,3,5,7,11,15,\dots$, up to (at least) the middle term. Is this true?
**QUESTION 2.** Is there a generating function for the polynomials $Q\_n(t)$?
| https://mathoverflow.net/users/66131 | Largest part and length of a partition in play | Your $c\_n(\lambda)$ is closely related to what [Billey, Konvalinka, Swanson 2020](https://arxiv.org/abs/1905.00975) call the aft of a partition:
$$ \text{aft}(\lambda) = | \lambda | - \max\{\lambda\_1, \lambda'\_1\}$$
where $\lambda'$ is the conjugate of $\lambda$, so $\lambda'\_1 = \ell(\lambda)$.
Following up on Martin Rubey's comment, your Question 1 is true: For $c\_n(\lambda) > n/2$, we have $\lambda \ne \lambda'$ since $\lambda\_1 = \lambda'\_1$ would make $2\lambda\_1 > n$, a contradiction. Therefore, $\lambda$ is not self-conjugate and each partition $\mu \vdash (n - \lambda\_1)$ is counted twice in the coefficient of $t^{\lambda\_1}$, once for the partition of $n$ consisting of $\lambda\_1$ followed by $\mu$ and once for the conjugate of that partition of $n$.
For your Question 2, a generating function in terms of Gaussian binomial coefficients is given in the related OEIS entry [A338621](https://oeis.org/A338621) submitted by Swanson, one of the article authors.
| 3 | https://mathoverflow.net/users/14807 | 439084 | 177,340 |
https://mathoverflow.net/questions/438875 | 2 | Let $n\geq m$ be non negative integers, and consider a list of $(n+m+1)$ distinct numbers (complex or real). I am interested in getting a closed form formula for the following determinant: $\det\left[\sum\_{k=i}^{n+i}s\_k^{m+1-j}e^{-s\_k}\left[\prod\_{l=i,l\neq k}^{n+i}(s\_k-s\_l)\right]^{-1}\right]\_{i,j=1\ldots m+1}$, i.e.
$$
\det\left(\begin{array}[ccc]
\,\sum\_{k=1}^{n+1}s\_k^me^{-s\_k}\left[\prod\_{l=1,l\neq k}^{n+1}(s\_k-s\_l)\right]^{-1}& \ldots &\sum\_{k=1}^{n+1}e^{-s\_k}\left[\prod\_{l=1,l\neq k}^{n+1}(s\_k-s\_l)\right]^{-1} \\
\vdots & \ddots & \vdots \\
\sum\_{k=m+1}^{n+m+1}s\_k^me^{-s\_k}\left[\prod\_{l=m+1,l\neq k}^{n+m+1}(s\_k-s\_l)\right]^{-1}& \ldots &\sum\_{k=m+1}^{n+m+1}e^{-s\_k}\left[\prod\_{l=m+1,l\neq k}^{n+m+1}(s\_k-s\_l)\right]^{-1}
\end{array}\right)
$$.
This determinant can be rewritten as the determinant of an alternant matrix $\left(F\_{j}(S\_i)\right)\_{i,j=1\ldots m+1}$, for suitable multivariate functions $F\_j:\mathbb{C}^{n+1}\rightarrow\mathbb{C}$.
If the exponential are not there, then this determinant should some king of Schur polynomial. I am wondering if this determinant can be interpreted in terms of representation theory.
A related post : [A class of matrix determinants between Wronskians and Vandermondes](https://mathoverflow.net/questions/180974/a-class-of-matrix-determinants-between-wronskians-and-vandermondes)
| https://mathoverflow.net/users/153070 | A Vandermonde like determinant with exponentials | After slight renumbering $(m,n)\mapsto (m,n)-1$ and horizontal reflection $j\mapsto m+1-j$, let's call the $m\times m$ matrix in the OP's determinant
$$
A=\left[\sum\_{k=i}^{i+n-1}s\_k^{j-1}e^{-s\_k}\prod\_{l=i,l\neq k}^{i+n-1}(s\_k-s\_l)^{-1}\right]\_{\,i,j=1}^{\,m},\tag{1}
$$
and define $r=m+n-1$. Note that this answer also holds for $n<m$.
$A$ can be factorized similar to [this posting](https://mathoverflow.net/questions/236323) using the $r \times m$ Vandermonde matrix
$$
V\_{r,m}=\left[ s\_\rho^{\mu-1}\right]\_{\rho,\mu=1}^{r,m},\tag{2}
$$
the diagonal matrix
$$
E\_r = \mathrm{diag}\,[e^{-s\_\rho}]\_{\rho=1}^r\tag{3}
$$
and the matrix of inverse derivatives
$$
D\_{m,r}^{(o)}=\left[ \frac{1}{P\_{\mu,o}'(s\_\rho)}
\begin{cases}
1 & \mu \leq \rho < m+\mu \\
0 & \text{else}
\end{cases}
\right]\_{\,\mu,\rho=1}^{\,m,r},\tag{4}
$$
of the polynomials
$$
P\_{l,o}(s)=\prod\_{\rho=l}^{l+o-1}(s-s\_\rho)\tag{5}
$$
according to
$$
A = D\_{m,r}^{(m)} E\_{r} \, V\_{r,m}.\tag{6}
$$
Note that $D$ is not diagonal in this case, e.g., for $m=3$, $n=3$, where $r=5$:
$$
D\_{3,5}^{(3)}=
\left(
\begin{array}{ccccc}
\frac{1}{\left(s\_1-s\_2\right) \left(s\_1-s\_3\right)} &
\frac{1}{\left(s\_2-s\_1\right) \left(s\_2-s\_3\right)} &
\frac{1}{\left(s\_3-s\_1\right) \left(s\_3-s\_2\right)} & 0 & 0 \\
0 & \frac{1}{\left(s\_2-s\_3\right) \left(s\_2-s\_4\right)} &
\frac{1}{\left(s\_3-s\_2\right) \left(s\_3-s\_4\right)} &
\frac{1}{\left(s\_4-s\_2\right) \left(s\_4-s\_3\right)} & 0 \\
0 & 0 & \frac{1}{\left(s\_3-s\_4\right) \left(s\_3-s\_5\right)} &
\frac{1}{\left(s\_4-s\_3\right) \left(s\_4-s\_5\right)} &
\frac{1}{\left(s\_5-s\_3\right) \left(s\_5-s\_4\right)} \\
\end{array}
\right).\tag{7}
$$
As the matrices $D$ and $V$ are not square, the determinant of $A$ does not factorize in a simple way. However, one could try to enlarge both matrices to $r \times r$ by inserting appropriate matrix elements $O(\epsilon)$, factorize and let $\epsilon\to 0$. Alternatively, one can naturally enlarge $D$ and $V$ and define the $r \times r$ matrix
$$
B=D\_{r,r}^{(m)} E\_{r} \, V\_{r,r},\tag{8}
$$
where $\det B$ factorizes. The matrix $A$ then is the upper left $m \times m$ submatrix of $B$.
| 1 | https://mathoverflow.net/users/90413 | 439088 | 177,343 |
https://mathoverflow.net/questions/436657 | 5 | The maximum symmetry metric on real projective space $ \mathbb{RP}^n $ is the round metric.
The maximum symmetry metric on complex projective space $ \mathbb{CP}^n $ is the Fubini–Study metric; see [Maximum symmetry metric on $ \mathbb{C}P^n $](https://mathoverflow.net/questions/433847/maximum-symmetry-metric-on-mathbbcpn).
Let $ M $ be a compact connected manifold. The degree of symmetry of $ M $, denoted $ N(M) $, is the maximum of the dimensions of the isometry groups of all possible Riemannian structures on $ M $. See for example [Ku, Mann, Sicks, and Su - Degree of symmetry of a product manifold](https://www.ams.org/journals/tran/1969-146-00/S0002-9947-1969-0250340-1).
Two metrics are considered to be equivalent if they are isometric up to a constant multiple.
I'm interested in manifolds $ M $ for which there is a unique up to equivalence metric with isometry group of dimension $ N(M) $.
Is the pushforward of the biinvariant metric of $ F\_4 $ onto the Cayley projective plane
$$
\mathbb{OP}^2 \cong F\_4/{\operatorname{Spin}(9)}
$$
a maximum symmetry metric in this sense?
| https://mathoverflow.net/users/387190 | Maximum symmetry metric on Cayley plane $ F_4/{\operatorname{Spin}(9)}$ | $\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}$A maximum symmetry metric on $M:=\mathbb{O}P^2$ must be equivalent to the one you described.
Here's one way to see it.
Suppose $G$ is the isometry group of some fixed Riemannian metric on $M$ of maximal symmetry. Then $G$ acts effectively on $M$. From Theorem 2.2 of
>
> Chang, Theodore, and Tor Skjelbred. “Lie Group Actions on a Cayley Projective Plane and a Note on Homogeneous Spaces of Prime Euler Characteristic.” American Journal of Mathematics, vol. 98, no. 3, 1976, pp. 655–78. JSTOR, <https://doi.org/10.2307/2373811>.
>
>
>
it follows that $G$ has rank at most $4$. Now a simple inspection shows that $F\_4$ has the largest dimension of all rank $4$ Lie groups. In fact, because we'll need it later, let me note that $\Sp(4)$ and $\Spin(9)$ are both 36 dimensional, and have the second largest dimension among at-most-rank-$4$ groups.
So, since $G$ is maximum symmetry, we conclude that the identity component of $G$ is , up to cover, $F\_4$. However, [$F\_4$ has trivial center](https://mathoverflow.net/questions/369887/non-isomorphic-complex-lie-groups-with-the-same-exceptional-lie-algebra-for-ma?_gl=1*1t5iech*_ga*Mjc3NDg0Mzc5LjE1NjgyMzcyOTc.*_ga_S812YQPLT2*MTY3NDM0NzE5NS40NDMuMS4xNjc0MzUxMzkyLjAuMC4w "Non-isomorphic complex Lie groups with the same exceptional Lie algebra for g2,f4,e6,e7,e8?"), so the identity component of $G$ is $F\_4$.
We still need to pin down the metric. To that end, select $p\in M$ and consider the the isotropy group $G\_p$. We have $\dim G - \dim G\_p = \dim (G\cdot p)\leq \dim M = 16$, so $\dim G\_p \geq 36$. As $G\_p\subseteq G$ has rank at most $4$, we thus conclude that either $\dim G\_p = 36$ (which implies the $G$ action is transitive) or that $G\_p = G$, so the $G$-action has a fixed point. The latter case cannot happen: if it does, then then by taking the differential, we obtain a $G$-action on $T\_p M$, so obtain a 16-dimensional representation of $G$. However, the smallest non-trivial representation of $F\_4$ is $26$-dimensional.
So we now know that $G$ acts transitively, which means a maximum symmetry metric on $M$ is determined once we know it at $p$. Moreover, the induced $G\_p$-action on $T\_p M$ must be by isometries.
To finish, we note that the transitive actions on $M$ are known: there is an essentially unique such action (see, e.g., Onishchik's Topology of Transitive Transformation Groups, Table 10 on pg. 265). In particular, $G\_p = \Spin(9)$, and the embedding of $\Spin(9)$ into $F\_4$ is the usual one. Thus, the isotropy action of $\Spin(9)$ is irreducible (in fact, it's transitive on the unit sphere), which implies that there is only one $G\_p$-invariant inner product on $T\_p M$. Thus, there is, up to scale, a unique maximum symmetry metric on $M$.
| 4 | https://mathoverflow.net/users/1708 | 439091 | 177,344 |
https://mathoverflow.net/questions/439095 | 3 | Let $C$ be a smooth genus two hyperelliptic curve and $\mathcal{M}\_C$ be a moduli space of stable rank two vector bundles of fixed degree(or fixed determinant). Then is $\mathrm{Aut}(\mathcal{M}\_C)\subset\mathrm{Aut}(C)?$. The motivation is following. Let $Y$ be an index two prime Fano threefold of degree $4$, one can show that such threefold is isomorphic to a moduli space of stable vector bundle of rank two fixed determinant over a genus two curve $C$ such that its degree is odd(say degree=1). Then, I am wondering if $\mathrm{Aut}(Y)\subset\mathrm{Aut}(C)?$. In the paper [Hilbert scheme of lines and conics and automorphism groups of Fano threefolds](https://arxiv.org/abs/1605.02010), the authors show that $\mathrm{Aut}(Y)\subset\mathrm{Aut}(J(C))$. But I think $\mathrm{Aut}(J(C))$ is not contained in $\mathrm{Aut}(C)$ since $J(C)$ has some translation which is not coming from $\mathrm{Aut}(C)$. On the other hand, for index one degree 12 prime Fano threefold $X$. The authors show that $\mathrm{Aut}(X)\subset\mathrm{Aut}(C)$, in this case $X$ is also purely determined by $C$.
| https://mathoverflow.net/users/41650 | Automorphism of moduli space of stable vector bundles over a curve | The moduli space of rank 2 vector bundles on $C$ with fixed determinant of odd degree is a smooth complete intersection of two quadrics in $\ \mathbb{P}^5\qquad$ (P. Newstead, Topology 7 (1968), 205-215); in an appropriate system of coordinates, it is given by $\sum X\_i^2=\sum \lambda \_iX\_i^2=0$, for $\lambda\_0,\ldots ,\lambda\_5\in\mathbb{C}$ distinct. It always admits a group $(\mathbb{Z}/2)^5$ of automorphisms (acting by changing the sign of the coordinates). On the other hand, $\operatorname{Aut}(C)=\mathbb{Z}/2 $ for $C$ general.
| 11 | https://mathoverflow.net/users/40297 | 439096 | 177,346 |
https://mathoverflow.net/questions/365879 | 3 | I have arrived at needing SPDEs and encountered a strange thing. In the literature, two approaches are mentioned: One where the equation is thought of as an SDE in an infinite dimensional space; an other where the solution is thought of a random field which changes over time (?). Now, I have read that these two approaches are not translatable one to one. Does that mean that I can have existence of a solution for one approach but not for the other? Same question for uniqueness? How do I chose which approach best suits my problem?
| https://mathoverflow.net/users/160649 | Two approaches two SPDEs not equivalent? | Indeed in the reference ["Stochastic integrals for spde’s: A comparison"](https://www.sciencedirect.com/science/article/pii/S0723086910000435),
they show that the two approaches
* Walsh random field approach
* the framework of the stochastic evolution in Hilbert spaces
are equivalent. However, as explained eg. ["Invariant measures for the nonlinear stochastic heat equation with no drift term"](https://arxiv.org/pdf/2209.04771.pdf) for example there can be differences in the theorems conditions that are preferable depending on the context:
>
> Here we emphasize that we study the invariant measure using the Walsh random field approach [27], whereas such studies are mostly carried out under the framework of the stochastic evolution in Hilbert spaces [13]. Even though both theories are equivalent (see [16]), the differences in many technical aspects are still substantial. As the random field approach often produces results that are more explicit, we try to use this approach to obtain more precise conditions for the existence of an invariant measure.
>
>
>
| 1 | https://mathoverflow.net/users/99863 | 439104 | 177,348 |
https://mathoverflow.net/questions/439105 | 39 | Is the number of "breakthroughs" in mathematics decreasing, as it is claimed to be in other sciences?
Background for the question:
1. Park, M., Leahey, E. & Funk, R.J. Papers and patents are becoming less disruptive over time. Nature 613, 138–144 (2023). <https://doi.org/10.1038/s41586-022-05543-x>
2. [What Happened to All of Science’s Big Breakthroughs?](https://www.nytimes.com/2023/01/17/science/science-breakthroughs-disruption.html)
>
> A new study finds a steady drop since 1945 in disruptive feats as a share of the world’s booming enterprise in scientific and technological advancement.
>
>
>
Has a similar analysis been conducted for the field of mathematics (or, say, pure mathematics)? If yes, how do the findings compare?
| https://mathoverflow.net/users/498238 | Is the number of "breakthroughs" in mathematics decreasing, as it is claimed to be in other sciences? | **Q:** *Is the number of "breakthroughs" in mathematics decreasing?*
To get some quantitative feel for the question I considered the [Timeline of mathematics](https://en.wikipedia.org/wiki/Timeline_of_mathematics) on Wikipedia. Not all entries are "breakthroughs", but most could be considered as such. Here is a plot of the cumulative number of entries since 1900. I do notice a kink in the slope around 1965, so based on this evidence on might conclude that, indeed, the rate of discovery has decreased somewhat. Or perhaps the 1960's was just an unusually productive decade.
Relative to the total output the discovery rate is obviously much smaller now than in the past.
| 19 | https://mathoverflow.net/users/11260 | 439118 | 177,353 |
https://mathoverflow.net/questions/297638 | 15 | I am looking for a reference to the bipartite version of the Schoenberg's criterion of embeddability into a Hilbert space. The Schoenberg criterion is formulated as Proposition 8.5(ii) of the book "[Geometric Nonlinear Functional Analysis" by Benyamini and Lindenstrauss](https://books.google.pl/books/about/Geometric_Nonlinear_Functional_Analysis.html?id=lXZ95EKwjYUC&redir_esc=y):
**Theorem (Schoenberg).** A metric space $(X,d)$ admits an isometric embedding into a Hilbert space if and only if the function $d^2$ is negative definite in the sense that $\sum d^2(x\_i,x\_j)c\_i\bar c\_j\le 0$ for all $x\_1,\dots,x\_n\in X$ and all complex scalars $c\_1,\dots,c\_n$ satisfying $\sum c\_j=0$.
This characterization has a bipartite version:
**Theorem (??).** A metric space $(X,d)$ admits an isometric embedding into a Hilbert space if and only if
$$\sum\_{i<j} d^2(x^+\_i,x^+\_j)+\sum\_{i<j} d^2(x^-\_i,x^-\_j)\le \sum\_{i,j=1}^n d^2(x^+\_i,x\_j^-)$$ for any points $x^+\_1,\dots,x^+\_n$ and $x^-\_1,\dots,x^-\_n$ in $X$.
>
> **Question.** I hope Theorem (??) is known. If yes, could you provide me with a suitable reference?
>
>
>
The reduction of Theorem (??) to Schoenberg's Theorem can be done in six steps:
**Lemma.** For a pseudometric $d$ on a set $X$ the following conditions are equivalent:
(1) $\sum d^2(x\_i,x\_j)c\_i c\_j\le 0$ for any $x\_1,\dots,x\_n\in X$ and any complex numbers $c\_1,\dots,c\_n$ with $\sum c\_j=0$.
(2) $\sum d^2(x\_i,x\_j)c\_i c\_j\le 0$ for any $x\_1,\dots,x\_n\in X$ and any real numbers $c\_1,\dots,c\_n$ with $\sum c\_j=0$.
(3) $\sum d^2(x\_i,x\_j)c\_i c\_j\le 0$ for any $x\_1,\dots,x\_n\in X$ and any rational numbers $c\_1,\dots,c\_n$ with $\sum c\_j=0$.
(4) $\sum d^2(x\_i,x\_j)c\_i c\_j\le 0$ for any $x\_1,\dots,x\_n\in X$ and any integer numbers $c\_1,\dots,c\_n$ with $\sum c\_j=0$.
(5) $\sum d^2(x\_i,x\_j)c\_i c\_j\le 0$ for any $x\_1,\dots,x\_n\in X$ and any integer numbers $c\_1,\dots,c\_n\in\{-1,1\}$ with $\sum c\_j=0$.
(6) $\sum\_{i,j=1}^m d^2(x^+\_i,x^+\_j)+\sum\_{i,j=1}^m d^2(x^-\_i,x^-\_j)\le \sum\_{i,j=1}^m (d^2(x^+\_i,x\_j^-)+d^2(x^-\_i,x^+\_j))$ for any points $x^+\_1,\dots,x^+\_m$ and $x^-\_1,\dots,x^-\_m$ in $X$.
*Proof.* The implications $(1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(4)\Rightarrow(5)$ are trivial.
$(2)\Rightarrow(1)$ Take any complex numbers $c\_1,\dots,c\_n$ with $\sum c\_k=0$. Write each complex number $c\_k$ as $c\_k=a\_k+ b\_ki$. The equality $\sum c\_k=0$ implies $\sum a\_k=0=\sum b\_k$. Consider the number $s=\sum d^2(x\_i,x\_j)c\_i\bar c\_j$ and observe that it is real: $\bar s=\sum d^2(x\_i,x\_j)\bar c\_ic\_j=\sum d^2(x\_j,x\_i)c\_j\bar c\_i=s$. Then
$$s=\sum d^2(x\_i,x\_j)c\_i\bar c\_j=\sum d^2(x\_ix\_j)(a\_ia\_j+b\_ib\_j)=\sum d^2(x\_ix\_j)a\_ia\_j+\sum d^2(x\_i,x\_j)b\_ib\_j\le 0$$by (2).
$(3)\Rightarrow(2)$ can be proved by a standard continuity argument.
$(4)\Rightarrow(3)$ can be proved by multiplying the rational numbers $c\_1,\dots,c\_n$ by their common denominator.
$(5)\Rightarrow(4)$ can be proved by repeating each point $x\_i$ $|c\_i|$ times.
$(5)\Leftrightarrow(6)$ The condition (6) is a rewritten condition (5) with $m=n/2$ and points $x^+\_1,\dots,x^+\_m$ corresponding to $c\_i=1$ and $x^-\_1,\dots,x^-\_m$ to $c\_i=-1$.
| https://mathoverflow.net/users/61536 | A reference to a characterization of metric spaces admitting an isometric embedding into a Hilbert space | In "[Geometry of cuts and metrics](https://www.researchgate.net/publication/242569526_Geometry_of_Cuts_and_Metrics)" by Deza and Laurent, your inequality is called *pure inequality of negative type*.
In Section 6.1.1, it is stated that pure inequalities of negative type imply all inequalities of negative type.
In Theorem 6.2.2 they formulate Schoenberg's criterion: a metric space admits an isometric embedding into a Hilbert space if and only it the square of the metric satisfies all the inequalities of negative type.
| 6 | https://mathoverflow.net/users/1441 | 439119 | 177,354 |
https://mathoverflow.net/questions/439110 | 1 | I want to draw a matrix $A\in \mathbb{R}^{n\times k}$ uniformally at random from the Stiefel manifold $\mathbb{V}\_k(\mathbb{R}^n)$, that is from the collection of all $n\times k$ matrices $A$ such that $A^TA=I\_{k\times k}$.
Is this true that generating a matrix $X\in\mathbb{R}^{n\times n}$ with iid standard normal entries $\mathcal{N}(0,1)$, performing a svd decomosition $X=U\Lambda V^H$ and taking the first $k$ vectors of $U$, correspond to sampling $\mathbb{V}\_k(\mathbb{R}^n)$ uniformally at random?
| https://mathoverflow.net/users/103133 | Is svd of a Gaussian iid matrix corresponds to Haar measure on the Stiefel manifold? | Yes, this is correct; the probability distribution of $X=U\Lambda V^\top$ is
$$P(X)\propto \exp\left(-\tfrac{1}{2}\,{\rm tr}\,XX^\top\right)=\exp\left(-\tfrac{1}{2}\,{\rm tr}\,\Lambda^2\right),$$
so it is independent of the orthogonal matrices $U$,$V$. These are therefore distributed uniformly in $O(n)$, and identifying $A$ with the first $k$ columns of $U$ will generate a uniformly distributed $A$ in $\mathbb{V}\_k(\mathbb{R}^n)$.
An alternative approach, which does not require you to perform a SVD, is to orthonormalize the first $k$ columns of $X$ and place these in $A$.
| 1 | https://mathoverflow.net/users/11260 | 439140 | 177,363 |
https://mathoverflow.net/questions/301577 | 15 | This is a followup question to [Does foundation/regularity have any categorical/structural consequences, in ZF?](https://mathoverflow.net/questions/300046/does-foundation-regularity-have-any-categorical-structural-consequences-in-zf)
As shown in answers to that question, the axiom of foundation (AF, aka regularity) has consequences over ZF–AF that can be formulated purely in structural terms (i.e. using sets only in ways that respect isomorphism).
In particular, it implies the statement “Every set can be embedded into a set equipped with a well-founded extensional relation”, which Mike Shulman has called the *axiom of well-founded materialisation*; and (by the arguments in that question) this axiom should have the same structural consequences as foundation.
This statement is a very mathematically reasonable one. However, I don’t think I have ever seen this or anything similar appearing in practice in ordinary mathematics — only in logical investigations like this one, where it occurs specifically because of its relation to AF. So my question here is:
**Does the axiom of foundation have any consequences (over ZF–AF) that have appeared in ordinary mathematics?**
Slightly more precisely:
* By “consequence of foundation over ZF–AF”, I of course mean principally a statement that’s provable in ZF, but not in ZF–AF. But I’d also be interested in e.g. a proof in ZF that doesn’t go through in ZF–AF and can’t be trivially modified to do so.
* By “ordinary mathematics”, I mean… this of course is subjective, and mostly I think “we know it when we see it”, and hope that most of us will find our judgement on it mostly agrees. But to articulate my own intent here a little: a sufficient (but not necessary) condition to would be that the statement has been considered in a field of mathematics other than logic; necessary conditions would be that it’s purely structural, and that it’s been considered outside work specifically investigating foundation (or related axioms) or modelling set theories that include foundation. To illustrate the non-necessity of the “outside logic” criterion — I would certainly be happy to consider consequences in e.g. general classical model theory.
See e.g. [Where in ordinary math do we need unbounded separation and replacement?](https://mathoverflow.net/questions/121406/where-in-ordinary-math-do-we-need-unbounded-separation-and-replacement) for a similar question, about different axioms, and relevant discussion of what should be considered “ordinary mathematics”.
| https://mathoverflow.net/users/2273 | Consequences of foundation/regularity in ordinary mathematics (over ZF–AF)? | I think Frucht's theorem, the statement "every group is isomorphic to the automorphism group of a graph", is a great example of what you're looking for; from the statement, it'd be hard to tell that any set theory is relevant in the proof.
Frucht's theorem is true in ZF (and ZFC-AF), but independent from ZF-AF. I'll sketch a proof below.
To construct a model of ZF-AF where Frucht's theorem fails, we start a permutation model of ZFA (ZF with atoms) where it fails. The basic Frankel model, where A is superamorphous, works fine ([see Jech's "The Axiom of Choice" book](https://gwern.net/docs/math/1973-jech-theaxiomofchoice.pdf) if you haven't seen this model before).
To see it fails in the Frankel model, consider the free abelian group $G$ generated by the atoms. Suppose $G$ were the automorphism group of a graph $\Gamma$. Since $\Gamma$ is symmetric, it has a finite support, $E$, and every permutation of $A$ which fixes $E$ must fix $\Gamma$. This means every element of $\mathrm{Perm}(A\setminus E)$ corresponds to an element of $\mathrm{Aut}(\Gamma)$, which is $G$ by assumption. In particular, let $\pi$ the a transposition swapping two elements $a, b\in A\setminus E$. $\pi$ has order 2, so the corresponding element of $G$ has order $\leq$2, but $G$ is free abelian, so we must have $\pi$ acts trivally on $\Gamma$. However, $\pi$ acts non-trivially on $G$; it swaps the generators $g\_a$ and $g\_b$ corresponding to $a$ and $b$. $g\_a$ and $g\_b$ must correspond to distinct automorphisms $\gamma\_a$ and $\gamma\_b$ of $\Gamma$. If $v$ is a vertex in $\Gamma$, then
$$\gamma\_a(v)=\pi(\gamma\_a(v))=\pi(\gamma\_a)(\pi(v)) = \gamma\_b(\pi(v)) = \gamma\_b(\pi(v)) = \gamma\_b(v)$$
So, $\gamma\_a = \gamma\_b$. Since $\gamma\_a$ and $\gamma\_b$ can't be distinct, $\mathrm{Aut}(\Gamma)$ can't be $G$.
Finally, to get a model of ZF-AF instead of ZFA (where atoms are all empty), you can replace each atom with a distinct Quine atom (a set so $x=\{x\}$). As long as you're careful to define free abelian group in a way that doesn't depend on $x$ not equaling the pair $\langle x,x\rangle$, there shouldn't be any difficulties.
As for why Frucht's theorem holds in ZF, it requires fairly trivial modifications to [Sabidussi's ZFC proof](https://link.springer.com/article/10.1007/BF01319053).
Given a group $G$, begin with a colored, directed Cayley graph of $G$. This has automorphism group $G$, but if we forget the edge colors and directions, the graph may have extra automorphisms. We can indicate directions by attaching a stick of length 1 and 2 in the middle of every edge in the Cayley graph, resulting in an undirected colored graph with the correct automorphism group. If there are countably many colors, a similar trick works for colors, but in general, we will need distinct rigid graphs for each generator of $G$ to label each color. Under AC, it suffices to build a unique rigid graph for each ordinal; however, Sabidussi's paper actually builds a unique rigid graph for every set, by induction on rank. The idea is, roughly, that given a set in $S\in V\_{\alpha+1}$, you begin with a complete graph on $V\_\alpha$, and attach to each vertex it's corresponding, inductively defined graph. The resulting graph is now rigid. Then you attach more things to this graph to indicate elements or non-elements of $S$.
I have a better write-up of this argument (with sorely needed pictures) in my thesis, but until that's somewhere I can link conveniently, Sabidussi's paper is probably the best reference for actual induction details.
| 12 | https://mathoverflow.net/users/498259 | 439141 | 177,364 |
https://mathoverflow.net/questions/439101 | 5 | Consider 6 people $p\_i$, $i=1,\dots 6$, standing on a sphere $S^2$. We label the positions of these people by $p\_i$ again. Suppose no pair of these points $p\_i$ are antipodal. At each point $p\_i$ each person divides the sphere in two 4 identical regions by two orthogonal planes passing through the center of the sphere. The four regions $R\_i^j$, $j=1,2,3,4$ can be uniquely ordered in counterclockwise order. Assume the person $p\_i$ sees everyone standing in regions $R\_i^{1}$ and $R\_i^{3}$ green and everyone standing in regions $R\_i^{2}$ and $R\_i^{4}$ red or vice versa. Indeed the odd number and even number regions are seen the same color. Also assume that no one is standing exactly on the plane that any other person has used to divide the sphere into 4 regions.
Can we show that there exists two people $p\_i, p\_j$ such that they both see each other the same color?
| https://mathoverflow.net/users/115905 | Six people standing on earth | It is possible for all pairs of these people to see each other in different colors.
Consider six people all on the arctic circle, with longitudes 0°, 61°, 122°, 183°, 244° and 305° east of the prime meridian. Suppose they all color the directions northwest and southeast of them red, and they all color the directions northeast and southwest of them blue.
The arcs connecting these people are geodesics which go somewhat north of the arctic circle, and not to the north pole. So when two of these people look at each other along these arcs, one sees the other to the east in blue, and one sees the other to the west in red.
| 1 | https://mathoverflow.net/users/nan | 439146 | 177,365 |
https://mathoverflow.net/questions/439103 | 4 | I posted this on Stackexchange but got no responses or comments.
Consider the following integral, for $\epsilon\ne 0:$
$$\displaystyle\frac{1}{(2\pi)^2\epsilon^4}\int\_{\Omega}yb\,e^{\frac{i}{\epsilon}[-ay+bx-yb]}\,dx\,da\,dy\,db\,,$$
where $\Omega$ is some compact neighborhood of the origin (you can assume the integral is over $\mathbb{R}^4$ and that there is a smooth cutoff function). The exponent $-ay+bx-yb$ has a nondegenerate and unique critical point at $(x,a,y,b)=(0,0,0,0)\,,$
and therefore the stationary phase approximation says that
$$\displaystyle\lim\limits\_{\epsilon\to 0}\frac{1}{(2\pi)^2\epsilon^2}\int\_{\Omega}yb\,e^{\frac{i}{\epsilon}[-ay+bx-yb]}\,dx\,da\,dy\,db=0\,.$$
My guess is that it's also true that
$$\lim\limits\_{\epsilon\to 0}\frac{1}{(2\pi)^2\epsilon^4}\int\_{\Omega}yb\,e^{\frac{i}{\epsilon}[-ay+bx-yb]}\,dx\,da\,dy\,db=0\,,$$
but I'm not sure how to show it. Does anyone have any suggestions?
| https://mathoverflow.net/users/40323 | How to compute the asymptotics of this oscillatory integral? | As Fedor Petrov suggested, you can diagonalize the quadratic form.
Writing $A = a+y$, $B = (a-y)$, $C = (b+x-y)$ and $D = (b-x +y)$, up to some negligible constants (and restoring the smooth cut-off) your integral is
$$ \int \varphi \cdot (C+D)(B-A) e^{\frac{i}{\epsilon} (- A^2 + B^2 + C^2 - D^2)} $$
This you can write as (up to negligible constants)
$$ \epsilon^2 \int \varphi \cdot (\partial\_C - \partial\_D)(\partial\_A + \partial\_B) e^{\frac{i}{\epsilon} (- A^2 + B^2 + C^2 - D^2)}$$
After integration by parts, the standard argument shows that the leading order term is, up to a proportionality constant, given by
$$ \epsilon^4 (\partial\_C - \partial\_D)(\partial\_A + \partial\_B)\varphi(0) $$
If your cut-off function is really just a cut-off function (so it is constant in a neighborhood of the origin), then in fact your integral is effectively supported away from the stationary point of the phase, and so should actually decay faster than any polynomial.
| 4 | https://mathoverflow.net/users/3948 | 439154 | 177,369 |
https://mathoverflow.net/questions/439168 | 2 | Let $D$ be a diagonal matrix in $M\_{2n}(\mathbb{R})$ such that $D^2=I$ and Trace$(D)$=0
Suppose that $e\_k$s are the standard vectors in $\mathbb{R}^{2n}$, that is
$$e\_k=(0,\cdots 0,1,0,\cdots,0)^t$$ where $1$ is in the $k^{th}$ position.
Let us consider the following vectors for $k=1,\cdots,n$.
$$v\_k=e\_{2k}~~,~~~ v\_{n+k}=e\_{2k-1}$$
Let us consider the matrix $W$ whose $k^{th}$-coulmn $W\_k$ is given by $v\_k$
Q. How can we find the eigenvalues of $WD$?
| https://mathoverflow.net/users/84390 | The eigenvalues of the product $WD$ for some particular matrices | I noticed that the $n\times n$ matrix $M=WD$ is "periodic", $M^p=\pm M$ for some even integer $p\leq n$. This identifies the eigenvalues of $M$ as $p$-th roots of $\pm 1$.
For some $n$ I find $p=n$ and the eigenvalues are all distinct: $e^{2k\pi i/n}$ if $M^n=M$ or $e^{(2k+1)\pi i/n}$ if $M^n=-M$, with $k=0,1,2,\ldots n-1$. This is the case for $n=2,4,6,10,12$.
But for other $n$ I find $p<n$, and some eigenvalues have multiplicity greater than 1. This happens for $n=8$ (with $p=6$) and for $n=14$ (with $p=8$). I have not been able to find the rule that governs these cases.
| 2 | https://mathoverflow.net/users/11260 | 439175 | 177,373 |
https://mathoverflow.net/questions/439164 | 2 | The universal covering group $G$ of $\mathrm{SL}\_2({\mathbb R})$ has infinite center. Is there an irreducible unitary representation $\pi$ of $G$, whose central character is injective? Or does every $\pi$ factor through a finite cover of $\mathrm{SL}\_2({\mathbb R})$?
| https://mathoverflow.net/users/473423 | Unitary dual of universal cover | Yes.
(a) This is equivalent to ask about the existence of an extremal normalized positive-definite function $\phi$ on $G$ that is "faithful" on its infinite cyclic center $Z$, that is, $\phi^{-1}(\{1\})\cap Z=\{0\}$. This is just because these functions are precisely the functions of the form $x\mapsto\phi(x)=\langle \xi,\pi(x)\xi\rangle$ for unitary representations $\pi$ and unit vectors $\xi$. (Extremal means it lies in an extremal axis in the cone of positive-definite functions.)
(b) In turn, this is equivalent to ask about the existence of a nonzero unitary representation of $G$ for which $Z$ acts as a faithful character. Indeed if $\phi$ is as in (a) then the corresponding $\pi$ works. Conversely, if we have such $\pi$ and $\phi$ is any normalized positive-definite function associated to $\pi$, then $\phi$ in the convex hull of extremal positive-definite functions. Let $\chi$ be the restriction $\phi|\_Z$: this is by assumption a faithful character. Since $\phi$ is extremal in restriction to $Z$, we deduce that $\phi$ is in the convex hull of extremal positive-definite functions that equal $\chi$ on $Z$. Whence the existence of such a nonzero function.
(c) Now we need to construct $\pi$ as in (b). Fix a faithful unit character $\chi$ on $Z$ (i.e., an injective homomorphism into the unit circle in $\mathbf{C}$). Let $H\_0$ be the space of locally measurable functions $G\to\mathbf{C}$ (modulo equality on null subsets), and $H\_\chi$ its subspace of functions $f$ satisfying (1) $f(zx)=\chi(z)f(x)$ for every $z\in Z$ and almost every $x\in G$; (2) given (1), $|f|$ is $Z$-invariant, and we assume that the resulting function $|f|:G/Z\to\mathbf{C}$ is in $L^2$. This is a Hilbert space under the scalar product computed on a fundamental domain $X$ of the action of $Z$ on $G$, i.e., we identify $H\_\chi$ to $L^2(\chi)$ to make it a Hilbert space. The group $G$ acts on $H\_\chi$ by shifting: $g\cdot f(x)=f(g^{-1}x)$. This is a unitary representation, for which $Z$ acts as the scalar $\chi$, and $H\_\chi\neq\{0\}$.
---
Note that I just assumed, say, that $G$ is a second-countable locally compact group and $Z$ is a central discrete subgroup and proved that every character of $Z$ can be obtained by restricting some irreducible unitary representation of $G$, and hence the conclusion holds as soon as $Z$ possesses a faithful character.
| 2 | https://mathoverflow.net/users/14094 | 439180 | 177,374 |
https://mathoverflow.net/questions/437097 | 1 | It is well-known that the power-exponential function $x^x$ and its first few derivatives are often taught in calculus.
Does the general formula for the $n$th derivative of the power-exponential function $x^x$ exist somewhere?
What is the general formula for the $n$th derivative of the power-exponential function $x^x$?
| https://mathoverflow.net/users/147732 | Where or what is the general formula for the $n$th derivative of the power-exponential function $x^x$? | The Stirling numbers of the first kind $s(n,k)$ for $n\ge k\ge0$ can be analytically generated by
\begin{equation\*}%\label{1st-stirl-gen-funct}
\frac{[\ln(1+t)]^k}{k!}=\sum\_{n=k}^\infty s(n,k)\frac{t^n}{n!},\quad |t|<1,
\end{equation\*}
which can be rearranged as Maclaurin's power series expansion of the power function
\begin{equation}\label{1st-stirl-gen-funct-rewr}
\biggl[\frac{\ln(1+t)}{t}\biggr]^k=\sum\_{n=0}^\infty \frac{s(n+k,k)}{\binom{n+k}{k}}\frac{t^{n}}{n!}, \quad |t|<1, \quad k\ge0.
\end{equation}
Comtet's numbers $b(n,k)$ are generated by
\begin{equation}\label{b(n-k)-gen-eq}
\frac{[(1+t)\ln(1+t)]^k}{k!}=\sum\_{n=k}^{\infty}b(n,k)\frac{t^n}{n!},\quad k\ge0.
\end{equation}
For $n\in\mathbb{N}\_0=\{0,1,2,\dotsc\}$, we have
\begin{equation}\label{power-exp-deriv-eq}
(x^x)^{(n)}=n!x^{x-n}\sum\_{k=0}^{n} x^{k} \sum\_{j=0}^{k}\Biggl[\sum\_{q=0}^{n-k} \frac{s(q+j,j)}{(q+j)!} \binom{j}{n-k-q}\Biggr]\frac{(\ln x)^{k-j}}{(k-j)!}
\end{equation}
and
\begin{equation}\label{power-exp-deriv-b(n-k)}
(x^x)^{(n)}
=x^{x-n}\sum\_{k=0}^{n} x^{k} \sum\_{j=0}^{k}(-1)^{j}\binom{j-n-1}{j} b(n-j,k-j)(\ln x)^{j}.
\end{equation}
Consequently, Taylor's power series expansion around $x=1$ is
\begin{equation}\label{power-exp-taylor-ser}
x^x=\sum\_{n=0}^\infty\Biggl[\sum\_{k=0}^{n} \sum\_{q=k}^{n} \frac{s(q,k)}{q!} \binom{k}{n-q}\Biggr](x-1)^n, \quad |x-1|<1
\end{equation}
and
\begin{equation}\label{b(n-k)-ser=1-eq}
x^x=\sum\_{n=0}^\infty\Biggl[\sum\_{k=0}^{n} b(n,k)\Biggr]\frac{(x-1)^n}{n!}, \quad |x-1|<1.
\end{equation}
These texts are extracted from Theorem 1 of the following paper:
1. Jian Cao, Feng Qi, and Wei-Shih Du, *Closed-form formulas for the $n$th derivative of the power-exponential function $x^x$*, Symmetry **15** (2023), no. 2, Article 323, 13 pages; available online at <https://doi.org/10.3390/sym15020323>.
| 2 | https://mathoverflow.net/users/147732 | 439188 | 177,376 |
https://mathoverflow.net/questions/426487 | 1 | I am the author of the package for tomographic reconstruction <https://github.com/kulvait/KCT_cbct> I have implemented CGLS/CGNR , algorithm which applies conjugate gradients on normal equation
$$
A^\top A x = A^\top b
$$
I use CGLS/CGNR as a default algorithm for many tomographic problems. Let's say $x\_i$ is the solution after $i$-th iteration. I compute sequence of solutions. I report the norm of discrepancy
$$
d\_i = \|Ax\_i-b\|
$$
and the norm of residual
$$
r\_i = \|A^\top A x\_i -A^\top b\|
$$
to see the speed of convergence based on sequences $(r\_i)$ and (d\_i).
I note that the norm of residual $(r\_i)$ is not always decreasing monotonously but sometimes it jumps up. As conjugate gradients guarantee to minimize norm of error, this is expected behavior. What is interesting I have never seen for any problem the sequence of norm of discrepancy $(d\_i)$, which is non monotonous. Now the question is is that just coincidence?
So I would like to prove or to show an counterexample for the following **claim**.
For all positive definite $A^\top A$ the CGLS behaves such that
$$
\|Ax\_{i+1}-b\| \leq \|Ax\_{i}-b\|
$$
| https://mathoverflow.net/users/69346 | Does norm of discrepancy decrease monotonously in CGLS/CGNR | Let $(x\_i)$ be sequence of solutions to CGLS/CGNR and let $x^\*$ be a solution of
$$A^\top A x^\* = A^\top b$$
AS CG minimizes $A$ norm of error over growing Krylov subspace, CGLS/CGNR minimizes $A^\top A$ norm of error which is
$$\|x^\* - x\_i\|\_{A^\top A} = \|A x^\* - A x\_i\|$$
thus we have
$$\|A x^\* - A x\_{i+1}\| \leq \|A x^\* - A x\_i\|$$
Now from the properties of the least squares solution we can decompose
$$b = A x^\* + b^\perp, \mbox{where} $$
$b^\perp$ is orthogonal to the column space of $A$ and thus
$$\|b - A x\_{i}\| = \|b^\perp\| + \|A x^\* - A x\_{i}\|.$$
Therefore we conclude that the statement
$$\|A x\_{i+1} - b\| \leq \| A x\_{i} - b \|$$
is true.
Thanks to Zdeněk Strakoš for outlining this answer.
| 0 | https://mathoverflow.net/users/69346 | 439193 | 177,377 |
https://mathoverflow.net/questions/439158 | 5 | It is relatively easy to prove that if there exists a positive definite matrix $Q$ such that $Q - A^{H}QA$ is positive definite, where $A^{H}$ means the conjugate transpose of $A$, then the spectral radius of $A$ is less than $1$. Just look at every eigenpair $(v,\lambda)$. But as for the reverse problem, I am wondering how to construct a $Q$ such that $Q - A^{H}QA$ is positive definite, when the spectral radius of $A$ is less than $1$?
| https://mathoverflow.net/users/498280 | If the spectral radius of matrix $A$ is less than $1$, how to construct a positive definite $Q$ such that $Q - A^{H}QA$ is also positive definite? | I'll follow the suggestions in the comments and post my comment as an answer:
Take your favourite positive definite matrix $P$ and define $Q := \sum\_{k=0}^\infty (A^{\operatorname{H}})^k P A^k$; note that this series converges since, as the spectral radius of $A$ is $<1$, there exist numbers $\delta \in [0,1)$ and $M \ge 1$ such that $\|A^k\| \le M \delta^k$ for all integers $k \ge 0$.
The matrix $Q$ is positive definite and satisfies $Q−A^{\operatorname{H}}QA=P$.
*Remarks:*
* Both the result and the proof above are standard in systems and control theory; see for instance Section 3.3.5 in the book *Mathematical systems theory. I. Modelling, state space analysis, stability and robustness* (2005) by Hinrichsen and Pritchard ([link to zbMATH](https://zbmath.org/1074.93003)).
* The same argument also works for bounded linear operators on infinite dimensional Hilbert spaces.
| 3 | https://mathoverflow.net/users/102946 | 439200 | 177,382 |
https://mathoverflow.net/questions/439182 | 1 | I'm reading through Jannsen's paper [Motives, numerical equivalence, and semi-simplicity](https://epub.uni-regensburg.de/26642/1/jannsen10.pdf) and I'd like to pose two questions.
Suppose all motives are $F$-linear, for some characteristic zero field $F$, and $M$ is a **simple** numerical motive **of weight zero** over a fixed finite field $\mathbf{F}\_q$.
>
> If $\operatorname{End}(M)=F$, does $M$ have to be the trivial motive $1=(\operatorname{Spec}(\mathbf{F}\_q),\mathrm{identity},0)$?
>
>
>
Moreover, in the very definition of motives as in the incipit of the paper, say $X$ is a smooth projective $\mathbf{F}\_q$-variety, irreducible of dimension $d$:
>
> for a motive $M=(X,p,j)$ does the idempotent $p\in A^d(X\times X,F)$ need to be **central**?
>
>
>
I expect not, although if not then I find the $F$-algebra structure on $\operatorname{End}(M)=pA^d(X\times X, F)p$ confusing. How is it a sub-$F$-algebra of $A^d(X\times X, F)$? (the answer could well be: "it's not, and that's fine".) What is its identity?
**Note** that for the first question, since we are considering motives over $\mathbf{F}\_q$, the condition $\text{End}(M)=F$ means that the Frobenius element $\pi\_M$ acts as a rational number, and since $M$ has weight zero, $\pi\_M$ acts as the identity. It would be a consequence of the Tate conjecture that $M$ is isomorphic to $1$. However, what I'm asking assumes an a priori stronger condition than $\pi\_M=\text{id}$: I'm assuming that we have $\text{End}(M)=F$. For example, this condition implies that the center of $\text{End}(M)$ is $F[\pi\_M]$ (which is $F=\text{End}(M)$ itself), while if we merely assume $\pi\_M=\text{id}$ we would know nothing about the center of $\text{End}(M)$ without assuming the Tate conjecture. I'm hoping the stronger condition $\text{End}(M)=F$ is sufficient to prove $M=1$ unconditionally, and that's what the question is about.
As a final remark, since both $M$ and $1$ are indecomposable, it would be enough to find a nonzero morphism between $M$ and $1$, and it would have to be an isomorphism.
| https://mathoverflow.net/users/497064 | About simple motives | Your second question is easier to answer. The idempotent does not need to be central.
The subset $p A^d(X \times X, F) p $ of $A^d(X\times X, F)$ is closed under addition, additive inverses, and multiplication, and contains the additive identity, but not (usually) the multiplicative identity. However, $p1p=p$ is itself a multiplicative identity for $p A^d(X \times X, F) p $, so that $p A^d(X \times X, F) p $ is indeed an algebra (whether it is a subalgebra is a notational question, I guess).
Even if $p$ is central, $1$ is not contained in $p A^d(X \times X, F) p $ unless in fact $p=1$.
For the first, I think the answer is no because I am not sure I agree with your logic that Frobenius must act as a rational number. Why can't it act as an arbitrary algebraic number in $F$?
If $F= \mathbb Q$ I agree the answer should be yes by the Tate conjecture, but I don't think this should necessarily be any easier than the Tate conjecture itself (i.e. it may turn out to be easier in an eventual proof, but not at the current state of knowledge). Consider an algebraic variety $X$ where $H^{2i}(X, \mathbb Q\_\ell)$ has an $n$-dimensional subspace where Frobenius acts by $q^i$, and suppose that the Tate conjecture is false and $X$ only has an $n-1$-dimensional space of codimension-$i$ cycles up to $\ell$-adic homological equivalence. Suppose also enough standard conjectures that the Lefschetz pairing on these cycles is nondegenerate. Then isn't the Tate twist by $i$ of the orthogonal complement of the space generated by algebraic cycles inside the subspace of $H^{2i}(X)$ where Frobenius acts by $q^{i}$ a motive satisfying all your assumptions but not your conclusion?
I don't see why this case of the Tate conjecture should be easier than the general case.
| 1 | https://mathoverflow.net/users/18060 | 439201 | 177,383 |
https://mathoverflow.net/questions/439190 | 1 | **Statement.** To ensure the rank of $\operatorname{ddiag}(AQQ^T)-\sigma\Delta=n$, it is sufficient to require $\min\_i(\operatorname{diag}(AQQ^T))\_i>\sigma\lVert\Delta\rVert$.
**Note:** $Q\in\mathbb{R}\_{n\times 2}$, $\sigma$ is a scalar constant, $\Delta$ is $n\times n$ random matrix. The operator $\operatorname{ddiag}$ sets all off-diagonal entries of a matrix to zero. The operator $\operatorname{diag}$ takes the diagonal entries of a matrix.
I don't know why this holds. And also, could someone point out which theory or which direction this statement belongs to, so that next time I can be more clear about where I can find this kind of techniques. Is it just basic linear algebra or something more advanced?
| https://mathoverflow.net/users/494410 | Control the summation of a diagonal matrix and another matrix to be full rank | You must assume $\sigma > 0$. Then it's just the fact that the if $D$ is a diagonal $n \times n$ matrix with minimum diagonal entry $m > \|B\|$, where $B$ is an $n \times n$ matrix, $\|D^{-1} B\| \le \|D^{-1}\| \|B\| < 1$, so $I - D^{-1} B$ is invertible, and then $D - B = D (I - D^{-1} B)$ is invertible.
| 1 | https://mathoverflow.net/users/13650 | 439209 | 177,385 |
https://mathoverflow.net/questions/439137 | 8 | Let $S$ be a scheme. Let's define a Grothendick topology on $\mathrm{Sch}/S$ where a covering family $\{f\_i:Z\_i\rightarrow X\}\_{i\in I}$ on an $S$-scheme $X$ is a collection of **closed immersions** of $S$-schemes such that $\lvert X\rvert=\bigcup\_i f\_i(\lvert Z\_i\rvert)$ (notice that $\lvert Y\rvert$ denotes the underlying topological space of a scheme $Y$).
I am interested in the following inter-related questions:
Has sheaf theory on this topology been studied? If so, what do we know about it? In particular, can we compare it somehow to the sheaves on the Zariski site?
If this facilitates things, assume that we work on affine schemes, and our coverings are also given by affine morphisms in the same way, so that everything is reduced to rings and quotient of rings.
Maybe this is a very basic question and maybe it is fully settled in the literature, but I am not aware of a good reference. Any references/comments are highly appreciated!
| https://mathoverflow.net/users/35397 | The Grothendieck topology of closed immersions on schemes | The topology you mention is called the "closed topology" $J\_\mathrm{cl}$ in ["Points in algebraic geometry"](https://arxiv.org/abs/1407.5782) by Gabber–Kelly. See also the [comparison diagram](https://pbelmans.ncag.info/topologies-comparison/) by Pieter Belmans.
Note that Gabber and Kelly additionally ask that the covering families are finite. But since they ask for the base scheme to be Noetherian (and separable), I don't think there is a difference, because any covering family automatically has a finite refinement.
You can apply the [Comparison Lemma](https://ncatlab.org/nlab/show/comparison+lemma) to simplify the site $(\mathrm{Sch}/S,J\_\mathrm{cl})$ without changing the category of sheaves on it. To make the notation easier, I will consider $J\_\mathrm{cl}$ on $\mathrm{Sch}$, but the argument can be modified to work for $\mathrm{Sch}/S$ as well. The claim is that the category of sheaves $\mathbf{Sh}(\mathrm{Sch},J\_\mathrm{cl})$ is equivalent to the category of presheaves $\mathbf{PSh}(\mathrm{IntSch})$ on the category $\mathrm{IntSch}$ of integral schemes.
Any scheme $X$ can be written as a union
$$X = \bigcup\_{x \in |X|} \overline{\{x\}}$$
and each of the closed subsets $\overline{\{x\}}$ can be given the [reduced induced subscheme](https://stacks.math.columbia.edu/tag/01J4) structure. In this way, any scheme can be covered by integral (i.e. irreducible and reduced) schemes. The Comparison Lemma then says that the category of sheaves for your topology can equivalently be described as a category of sheaves on the full subcategory of integral schemes. The Grothendieck topology that we have to take on this category of integral schemes is the restriction of the one on schemes: we again look at covers by closed subschemes. But for an integral scheme $X$, if you have such a covering by closed subschemes, one of the closed subschemes will contain the generic point, and then this closed subscheme must be equal to $X$. In this way, you can see that the restricted Grothendieck topology is precisely the presheaf topology.
In the same way, you can also find $\mathbf{Sh}(\mathrm{Sch}/S,J\_\mathrm{cl})\simeq \mathbf{PSh}(\mathrm{IntSch}/S)$.
| 5 | https://mathoverflow.net/users/37368 | 439211 | 177,386 |
https://mathoverflow.net/questions/439124 | 11 | Let $T$ be a normed vector space, $K\subseteq T$ compact and convex and $O\subseteq K$ convex and open in $K$ (i.e. open w.r.t. the subspace topology of $K$ inherited by $T$).
Can we find an open set $O'\subseteq T$ such that $O' \cap K = O$? Can we also find such an $O'$ which is convex?
Edit: The convexity of $K$ is indeed necessary. There exists a counterexample in case of $K$ not being convex.
| https://mathoverflow.net/users/498220 | Existence of an open convex set | A convex $O'$ need not exist: a counterexample is given by setting $K=[-1,1]\times[0,2]\subseteq\mathbb{R}^2$ and $O=\{(x,y)\in K;y>x^3\}$. Indeed, any open $O'$ with $O'\cap K=O$ would contain some nhood of $(-\frac{1}{2},0)$, so it would contain some point $q=(-\frac{1}{2},-\varepsilon)$ with $\varepsilon>0$. Note that the tangent line from $q$ to the function $f:[0,\infty)\to\mathbb{R};f(x)=x^3$ passes below the point $(0,0)$. This easily implies that if $O'$ was convex, it would contain $(0,0)$, contradicting $O'\cap K=O$.
However, the conjecture is true for lots of convex sets $K$ (for example [strictly convex](https://en.wikipedia.org/wiki/Convex_set#strictly_convex) sets, which satisfy the lemma below). I will assume that the normed space is $\mathbb{R}^n$, with $n\geq2$ with norm $||\cdot,\cdot||\_2$, since the norm is irrelevant for the problem. First I prove the result for $\overline{\mathbb{B}^n}$, where it's easier to visualize, and then I generalize below.
**Proposition:** For each convex set $O\subseteq\overline{\mathbb{B}^n}$ open in $\overline{\mathbb{B}^n}$ we can find a convex open set $O'$ such that $O=O'\cap\overline{\mathbb{B}^n}$.
Proof: For each $v\in\mathbb{S}^{n-1}$ let $X\_v:=\{x\in\mathbb{R}^n;\langle x,v\rangle<1\}$. Let $B=\mathbb{S}^{n-1}\setminus O$, and let $A$ be the interior of $\bigcap\_{v\in B}X\_v$. Clearly $A$ is open and convex, and $O\subseteq A$. We will define $O'=O\cup(A\setminus\overline{\mathbb{B}^n})$.
By definition, $O=O'\cap\overline{\mathbb{B}^n}$, and $O'$ is open because $A$ is open, $O\setminus\mathbb{S}^{n-1}$ is open and any $p\in O\cap\mathbb{S}^{n-1}$ is in the interior of $A$, so it is in the interior of $O'$.
So we just need to deduce that $O'$ is convex: let $p,q\in O'$.
* If $p,q\in O$ then the segment $[p,q]$ is contained in $O\subseteq O'$.
* If $p\in A\setminus\overline{\mathbb{B}^n},q\in O$, then $[p,q]\subseteq A$. If $[p,q)$ doesn't intersect $\mathbb{S}^{n-1}$, then $[p,q)\subseteq A\setminus\overline{\mathbb{B}^n}\subseteq O'$ so we are done. If not, let $r$ be que unique intersection of $[p,q)$ with $\mathbb{S}^{n-1}$. As $r\in A\cap\mathbb{S}^{n-1}$, $r$ cannot be in $B$, so $r\in O$. Thus $[r,q]\subseteq O\subseteq O'$ and $[p,r)\subseteq A\setminus\overline{\mathbb{B}^n}\subseteq O'$, so as we wanted $[p,q]\subseteq O'$.
* A similar argument in the case $p,q\in A\setminus\overline{\mathbb{B}^n}$ (this time considering the up to $2$ points of intersection of $[p,q]$ with $\mathbb{S}^{n-1}$, which again will be in $O$) allows us to conclude that $O'$ is convex. $\square$
Now let $K$ be a general convex set with nonempty interior. We say an open half space $X$ is tangent to $K$ at $p\in\partial K$ if $K\subseteq\overline{X}$ and $p\in\partial X$. Let $\mathfrak{X}$ be the set of all half spaces tangent to $K$ at points of $\partial K\setminus O$ and let $A$ be the interior of $\bigcap\_{X\in\mathfrak{X}}X$. Then again, $A$ is open and convex, but it need not be true that $O\subseteq A$. If indeed $O\subseteq A$, we can still define $O'=O\cup(A\setminus K)$ and the rest of the proof works the same way (when proving convexity, we can define $r$ as the point of $[p,q)\cap\partial K$ closest to $p$).
A sufficient condition for $O\subseteq A$ is the following:
**Lemma:** Suppose that every hyperplane tangent to $K$ intersects $K$ at just one point. Then $O\subseteq A$.
Proof: Suppose some $p\in O$ is not in $A$. The statement of the lemma implies that $p\in X$ for every $X\in\mathfrak{X}$, so $p\in\bigcap\_{X\in\mathfrak{X}}X$. Thus $p$ is in the boundary of $A$, and there is a sequence of half spaces $X\_n\in\mathfrak{X}$ tangent to $K$ at some points $p\_n\in\partial K\setminus O$, such that $d(p,\partial X\_n)\to0$. Taking a subsequence if necessary, we can suppose that the vectors $v\_n$ normal to the hyperplanes $\partial X\_n$ converge to $v\_\infty$, and that the points $p\_n$ converge to $p\_\infty$ (note that $p\_\infty\not\in O$, so $p\_\infty\neq p$). So the half space $X$ based at $p\_\infty$ with normal vector $v\_\infty$ is tangent to $K$, and $p\in\partial X$. Which contradicts the statement of the lemma.$\square$
| 9 | https://mathoverflow.net/users/172802 | 439224 | 177,391 |
https://mathoverflow.net/questions/438331 | 7 | Consider two independent copies of IID random walk on ${\bf Z}$ starting from $0$, and let $N\_1(t)$ (resp. $N\_2(t)$) denote the number of times, up to time $t$, that the first (resp. second) walker has returned to $0$. We know that $N\_1(t),N\_2(t) \rightarrow \infty$ almost surely and that $N\_1(t)/t,N\_2(t)/t \rightarrow 0$ almost surely. Is it the case that $N\_1(t)/N\_2(t) \rightarrow 1$ almost surely? What if we replace random excursions from $0$ in ${\bf Z}$ by random excursions from $(0,0)$ in ${\bf Z}^2$?
| https://mathoverflow.net/users/3621 | Counting returns in null-recurrent random walk | You can easily derive everything from the first principles by the following back of envelope computation:
If we have a random walk starting *anywhere*, then after $t$ steps the expected number $EN(t)$ of returns is at most $\sum\_{m=1}^t \frac 1{\sqrt m}\approx \sqrt t$ and $E[N(t)^2]\le\sum\_{1\le m< M\le t} \frac 1{\sqrt m}\frac 1{\sqrt{M-m}}+\sum\_{1\le m\le t}\frac 1{\sqrt m}\approx t$ with approximate equality attained if we start at $0$. Note that it implies that $P(N(t)>c\sqrt t)>c$ for some fixed $t>0$ if we start at $0$. On the other hand, if we start above $C\sqrt t$ with large $C>0$, then the probability to ever reach $0$ is as small as we want, so the trivial Cauchy-Schwarz yields $EN(t)\le \sqrt{P(\exists \tau X(\tau)=0)E[N(t)^2]}<\frac{c^2}4\sqrt t$ if $C$ is large enough so $P(N(t)>\frac c2\sqrt t)\le c/2$.
Now just take $T>0$ and let $t$ be the first time after $T$ such that $X\_1(t)=0$. Then the probability that $|X\_2(t)|>C\sqrt t$ is some constant $p\_0$, say. From there and until the time $2t$ with probability $\ge c-\frac c2=\frac c2$, $X\_1$ acquires $>c\sqrt t$ returns but $X\_2$ less than $\frac c2\sqrt t$ returns. So, with fixed probability ($p\_0c/2$) they have noticeably different increments on the scale $\sqrt t$. Also, the probability that $N\_2(t)$ is much larger than $\sqrt t$ is small (we condition only on $X\_1$, so $X\_2$ still has the standard distribution). Thus, for every $M$, with constant probability, there exists $t>M$ such that $|N\_1(t)/N\_2(t)-1|+|N\_1(2t)/N\_2(2t)-1|>\delta>0$ for some fixed absolute $\delta$. That definitely rules out convergence to $1$ even in distribution, forget about a.s.
I tried to avoid any high-tech here and just answered the question as posed. I leave more advanced considerations to someone else.
In $\mathbb Z^2$, the situation is the same only with $\log t$ instead of the $\sqrt t$ everywhere.
I hope I haven't written any nonsense (lately I'm more prone to it than I should be).
| 0 | https://mathoverflow.net/users/1131 | 439225 | 177,392 |
https://mathoverflow.net/questions/439227 | 4 | Let $p$ be a prime, and let $K/\mathbb{Q}\_p$ be a tamely ramified finite extension of degree $n$. Let $q$ be a prime factor of $n$ with $q\neq p$. Must there exist an intermediate extension $L$ (with $\mathbb{Q}\_p\subset L\subset K$) of degree $q$ over $\mathbb{Q}\_p$?
---
Some partial progress, following Hasse's "Number Theory", Chapter 16:
We can decompose our field extension as $\mathbb{Q}\_p\subset K^{ur}\subset K$, where $K^{ur}$ is the maximal unramified subextension. Let $e=[K:K^{ur}]$ and $f=[K^{ur}:\mathbb{Q}\_p]$. If $q$ divides $f$, we are done, since $K^{ur}/\mathbb{Q}\_p$ is Galois with cyclic Galois group.
Otherwise, write $K^{ur}=\mathbb{Q}\_p(\zeta)$ and $K=\mathbb{Q}\_p(\zeta,(up\zeta^r)^{1/e})$, where $\zeta$ is a primitive $(p^f-1)st$ root of unity, $u$ is a unit in the $p$-adic integers, and $0\leq r<\gcd(p^f-1,e)$. If $\gcd(p^f-1,e)=1$, then $K$ contains $\mathbb{Q}\_p((up)^{1/e})$, and thus the intermediate field $\mathbb{Q}\_p((up)^{1/q})$, which has degree $q$ over $\mathbb{Q}\_p$.
We can also reduce to the case of $e=q$, by considering the subfield $\mathbb{Q}\_p(\zeta,(up\zeta^r)^{1/q})$. Since $q$ is prime and we dealt with the relatively prime case, we need only consider the case of $q$ dividing $p^f-1$. Note that if $q$ divides $p-1$, we have that $\mathbb{Q}\_p(\zeta,(up\zeta^r)^{1/q})$ is an abelian Galois extension of $\mathbb{Q}\_p$, also leading to the desired extension of degree $q$. That leaves the case of $q$ not dividing $p-1$ but still dividing $p^f-1$; in these cases, the extension is not Galois. Can we say anything about intermediate extensions of degree $q$ in this case?
| https://mathoverflow.net/users/127215 | Existence of intermediate field extensions for tamely ramified p-adic extensions | The answer is no. For example the group $S\_3\times C\_3$ occurs (as TransitiveGroup(6,5)) as the Galois group of an (automatically tame) extension of $\mathbb{Q}\_5$ (and more generally over $\mathbb{Q} \_p$ for all odd primes $p\equiv 2$ mod $3$, by taking the compositum of the unramified degree-$3$ extension with the splitting field of $x^3-p$), see <https://www.lmfdb.org/padicField/5.6.4.2> . The point stabilizer (namely, the diagonal subgroup $C\_3$) is contained in a maximal subgroup of index 2, but not one of index 3, meaning that there is a tame degree 6 extension not containing a degree 3 subextension.
| 2 | https://mathoverflow.net/users/127660 | 439231 | 177,395 |
https://mathoverflow.net/questions/438690 | 2 | Let $G$ be a locally profinite group and let $H$ be a closed normal subgroup. Let $\pi\_1$ and $\pi\_2$ be two smooth complex representations of $G$. Is there always a spectral sequence as follows?
$$E\_2^{p,q} = \operatorname H^p(G/H,\operatorname{Ext}\_H^q(\pi\_1,\pi\_2)) \implies \operatorname{Ext}\_G^{p+q}(\pi\_1,\pi\_2)$$
Here $\operatorname H^{\bullet}(G/H,\cdot) \simeq \operatorname{Ext}\_{G/H}^{\bullet}(\mathbb C,\cdot)$ denotes the group cohomology in the context of smooth representations.
I did find a paper [On the Continuous (Co) Homology of Locally Profinite Groups and the Künneth Theorem](https://doi.org/10.1006/jabr.1994.1028) by Hitta proving the existence of a Hochschild–Serre spectral sequence in the case $\pi\_1 \simeq \mathbb C$ (Theorem 2 loc. cit.). But what about arbitrary smooth $\pi\_1$?
| https://mathoverflow.net/users/125617 | For locally profinite groups $H\lhd G$, is there a spectral sequence $\newcommand\@[2]{{\rm Ext}_#1^{#2}(\pi_1,\pi_2)}H^p(G/H,\@Hq)\implies\@G{p+q}$? | I found a way to prove the existence of such a spectral sequence, at least when $H$ is open in $G$. In order not to leave this question unanswered, let me write things down.
If $G$ is a locally profinite group, let $\mathcal R(G)$ denote the category of smooth representations of $G$. If $V$ is an abstract representation of $G$, I will write $V^{G-\mathrm{sm}}$ for the subspace of smooth vectors. Thus $V^{G-\mathrm{sm}}\in \mathcal R(G)$.
Let $(\pi\_1,V) \in \mathcal R(G)$. The functor $\mathcal R(G) \to \mathcal R(G)$ mapping $(\pi\_2,W)$ to $\mathrm{Hom}(\pi\_1,\pi\_2)^{G-\mathrm{sm}}$ is exact and preserves injectives, see for instance the theorem in the section III.1.15 of the book by Renard and Schwartz. Here, the $\mathrm{Hom}$ space consists of $\mathbb C$-linear morphisms $V\rightarrow W$, and the $G$-action is given by
$$gf: v \mapsto g\cdot f(g^{-1}\cdot v).$$
If we compose this functor with taking the $G$-invariants, we obtain the functor $\mathcal R(G) \to \mathrm{Ab}$ given by $(\pi\_2,W) \mapsto \mathrm{Hom}\_G(\pi\_1,\pi\_2)$. Thus, we deduce that
$$\mathrm{Ext}^i\_G(\pi\_1,\pi\_2) = \mathrm H^i(G,\mathrm{Hom}(\pi\_1,\pi\_2)^{G-\mathrm{sm}}).$$
Now, by the paper of Hitta which I refer to in my question, we have a Hochschild-Serre spectral sequence computing the right-hand side.
$$E\_2^{i,j} = \mathrm H^i(G/H,\mathrm H^j(H,\mathrm{Hom}(\pi\_1,\pi\_2)^{G-\mathrm{sm}})) \implies \mathrm H^i(G,\mathrm{Hom}(\pi\_1,\pi\_2)^{G-\mathrm{sm}})$$
Inside the $E\_2^{i,j}$ term, one must be a litte bit careful. Namely, we want to replace $\mathrm H^j(H,\mathrm{Hom}(\pi\_1,\pi\_2)^{G-\mathrm{sm}})$, seen as a representation of $H$ by restriction, with $\mathrm{Ext}\_H^j(\pi\_{1|H},\pi\_{2|H}) \simeq \mathrm H^j(H,\mathrm{Hom}(\pi\_{1|H},\pi\_{2|H})^{H-\mathrm{sm}})$.
But $\mathrm{Hom}(\pi\_1,\pi\_2)^{G-\mathrm{sm}}\_{|H}$ consists of all linear morphisms $f:V\to W$ such that there exists $K \subset G$ an open compact subgroup with $f(k\cdot) = k\cdot f(\cdot)$ for all $k \in K$, and equipped with the action of $H$ restricted from $G$.
And $\mathrm{Hom}(\pi\_{1|H},\pi\_{2|H})^{H-\mathrm{sm}}$ consists of all linear morphisms $f:V\to W$ such that there exists $K' \subset H$ an open compact subgroup with $f(k'\cdot) = k'\cdot f(\cdot)$ for all $k' \in K'$, and equipped with the natural action of $H$.
Clearly the former is a subspace of the latter. If we assume that $H$ is open, then any $K'$ is also open compact in $G$, so that the two spaces are equal. Then, the Hochschild-Serre spectral sequence above is the desired result.
| 3 | https://mathoverflow.net/users/125617 | 439236 | 177,396 |
https://mathoverflow.net/questions/439223 | 2 | Is it possible to describe, up to isomorphism, all Lie groups $G$ whose underlying manifold is diffeomorphic to $\mathbb{R}^n$ (with its standard smooth structure)?
In fact, I haven't found any such group yet, other than the standard $(\mathbb{R}^n, +)$.
| https://mathoverflow.net/users/148161 | Classification of Lie group structures on $\mathbb{R}^n$ | YCor claims [here](https://math.stackexchange.com/questions/475385/under-what-conditions-is-the-exponential-map-on-a-lie-algebra-injective/1592257#1592257) that the contractible Lie groups (this is equivalent to being diffeomorphic to $\mathbb{R}^n$, since a connected Lie group is diffeomorphic to the product of a Euclidean space times its maximal compact) are exactly the Lie groups of the form $S \rtimes \widetilde{SL\_2(\mathbb{R})}^k$ where $S$ is a simply connected solvable Lie group and $\widetilde{SL\_2(\mathbb{R})}$ is the universal cover of $SL\_2(\mathbb{R})$.
So the classification reduces to the classification of solvable Lie algebras (then of semidirect products of these Lie algebras with $\mathfrak{sl}\_2(\mathbb{R})^k$), which I understand to be hopeless.
| 5 | https://mathoverflow.net/users/290 | 439240 | 177,398 |
https://mathoverflow.net/questions/420645 | 2 | Suppose the beth function is defined as follows:
beth[a]=|V[a]|, for all ordinals a. Here V[a] is the ath level of the cumulative hierarchy, and || is the cardinality function defined as in for example the Levy Basic Set Theory book.
The above definition of beth is equivalent to the usual one in the presence of the axiom of choice, and of course that beth function is known to be normal and hence continuous.
My question is whether in the absence of the axiom of choice the above beth function is continuous. Continuity here does not require the cardinalities to be ordinals but means |V[a]| is the lub of the set of |V[b]| such that b<a, for each limit ordinal a, and according to Levy’s definition of lub of cardinalities.
The non existence of Lubs for some sets of cardinals makes me doubtful but I’m too naive to know for sure.
| https://mathoverflow.net/users/480831 | Is the beth function continuous without the axiom of choice? | By the [answer](https://mathoverflow.net/a/438728) of Farmer S to my latest question, the answer to this question is no. Lub property on limit ordinal stages of cumulative hierarchy implies the axiom of choice. Thanks Farmer!
| 1 | https://mathoverflow.net/users/480831 | 439252 | 177,403 |
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