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https://mathoverflow.net/questions/290491 | 23 | In a comment to a [recent question](https://mathoverflow.net/questions/290459/is-it-consistent-with-zf-that-v-to-v-ast-ast-is-always-an-isomorphism), Jeremy Rickard asked whether it is consistent with ZF that the map $V \to V^{\*\*}$ from a vector space to its double dual is always surjective. We know that "always injective" is consistent (since that's what happens in ZFC) and Jeremy Rickard's argument shows that "always an isomorphism" is *not* consistent. But what about "always surjective"?
---
Gro-Tsen points out that Harry West has already showed this is impossible [elsewhere on MO](https://mathoverflow.net/a/401202/297). I am missing one step in West's answer. I thought I'd write up the issue here, and someone can explain to me what I am missing.
First of all, for any field $F$ and any set $X$, we can form the free vector space with basis $X$, call it $FX$, and the vector space of functions from $X$ to $F$, call it $F^X$. It is easy to see that $(FX)^{\ast} \cong F^X$ so, if $V \to V^{\ast \ast}$ is always surjective, then the obvious injection $FX \to (F^X)^{\ast}$ must always be an isomorphism. So we may and do assume:
**Key Consequence** For every set $X$, the obvious injection $FX \to (F^X)^{\ast}$ is an isomorphism.
Now, suppose that $\alpha$ is an ordinal with cofinality $>\omega$.
**Lemma 1** Let $X \subset \alpha$ have the property that $X \cap \beta$ is finite for every $\beta < \alpha$. Then $X$ is finite.
**Proof** If not, then $X$ is an infinite well-ordered set (by restricting the order from $\alpha$) so it contains a copy of $\omega$. By the hypothesis on $\alpha$, there is some $\beta\_0 < \alpha$ containing this copy of $\omega$. But then $X \cap \beta\_0$ is infinite. $\square$
Let $V$ be the subspace of $F^{\alpha}$ consisting of functions which are supported on $F^{\beta}$ for some $\beta< \alpha$.
**Lemma 2** $V^{\ast} = F \alpha$.
**Proof:** Let $\phi \in V^{\ast}$. We can restrict $\phi$ to $F^{\beta}$ for each $\beta < \alpha$ and, by the definition of $V$, the functional $\phi$ is determined by the list of these restrictions. On each $F^{\beta}$, by the Key Consequence, $\phi|\_{F\_{\beta}}$ coincides with some unique vector from $F \beta$. Let the support of that vector be $X\_{\beta}$ and let $X = \bigcup\_{\beta} X\_{\beta}$. Then $X \cap \beta= X\_{\beta}$ for each $\beta < \alpha$ so, by Lemma 1, $X$ is finite. The functional $\phi$ is then induced by a functional in $FX \subset F \alpha$. $\square$.
But then $V^{\ast \ast} = (F \alpha)^{\ast} = F^{\alpha}$, whereas $V$ is a proper subspace of $F^{\alpha}$. **QED**
---
My only issue is, in ZF, are we sure that there are ordinals of cofinality $>\omega$? In ZFC, one simply takes the first uncountable ordinal, $\omega\_1$. If we had a cofinal sequence $0=x\_0$, $x\_1$, $x\_2$, \dots, in $\omega\_1$, then $\omega\_1$ would be the union of the countable intervals $[x\_i, x\_{i+1})$, and would hence be countable.
But in ZF, a countable union of countable sets doesn't have to be countable. I tried some tricks to get around this and failed; please let me know what I missed.
| https://mathoverflow.net/users/3106 | Is it consistent with ZF that $V\to V^{\ast \ast}$ is always surjective? | $\def\fin{\text{finite}}\def\count{\text{countable}}$This is a CW answer to write up the proof which Harry West gave on [another answer](https://mathoverflow.net/a/401202/297), as pointed out by Gro-Tsen. Thanks to Prof. West for the clever solution and Gro-Tsen for pointing it out. I'm just rewriting it to record details which took me a while to follow. The answer is "no", there is always some vector space with $V^{\ast \ast}$ not spanned by the image of $V$.
Let $F$ be a field. For any set $X$, the vector space $F^X$ is the vector space of functions $X \to F$. It contains the subspace $F^X\_{\fin}$ of finitely supported functions, which is isomorphic to the free vector space with basis $X$. It is straightforward that $(F^X\_{\fin})^{\ast} = F^X$ so, if $(F^X)^{\ast} \supsetneq F^X\_{\fin}$, then we are done already. Thus, we may make the **key assumption** that $(F^X)^{\ast} = F^X\_{\fin}$ for every set $X$.
Now, let $X$ be a well-ordered uncountable set, such as the first uncountable ordinal $\omega\_1$. The only place that we will use the well-order on $X$ is to be sure that every infinite subset of $X$ contains a copy of $\omega$ (in other words, infinite subsets of $X$ can't be Dedekind finite).
Let $F^X\_{\count}$ be the subspace of $F^X$ consisting of countably supported functions.
**Claim:** $(F^X\_{\count})^{\ast} = F^X\_{\fin}$.
**Proof:** Let $\phi$ be any linear function $\phi : F^X\_{\count} \longrightarrow F$. For any countable subset $Y$ of $X$, we have $F^Y \subset F^X\_{\count}$. So the restriction of $\phi$ to $F^Y$ is a vector in $(F^Y)^{\ast}$ and, by the key assumption, $(F^Y)^{\ast} = F^Y\_{\fin}$. So, for each $Y$, there is some unique finitely supported vector $\phi\_Y$ in $F^Y$ such that $\phi(\psi)$ is the dot product $\psi \cdot \phi\_Y$ whenever $\psi$ is supported on $Y$. By the uniqueness of $\phi\_Y$, if $Y\_1 \subseteq Y\_2 \subset X$ are finite sets, then $\phi\_{Y\_1}$ is the restriction of $\phi\_{Y\_2}$ to $Y\_1$.
Let $S = \bigcup\_{X \subset Y,\ \count} \text{Support}(\phi\_Y)$. For every countable $Y$, we know that $S \cap Y$ is finite. We claim that this forces $S$ to be finite; if $S$ is infinite then (using the well order on $X$) there is a countably infinite subset $Z$ of $S$ and $S \cap Z= Z$ is infinite, contradicting that $S \cap Y$ is finite for every countable $Y$.
So $S$ is finite, and $\phi$ is given by dot product with a vector in $F^S$, so $\phi$ is given by dot product with a vector in $F^X\_{\fin}$. We have shown that $(F^X\_{\count})^{\ast} = F^X\_{\fin}$. $\square$
But then $(F^X\_{\count})^{\ast \ast} = (F^X\_{\fin})^{\ast} = F^X$, and $F^X\_{\count}$ is a proper subspace of $F^X$. This completes the proof. $\square$
As Prof. West observes in his write up, one can identify a specific vector space $V$ for which this proof shows $V$ does not surject onto $V^{\ast \ast}$: The vector space $F^{\omega}\_{\fin} \oplus F^{\omega\_1}\_{\count}$.
| 15 | https://mathoverflow.net/users/297 | 415774 | 169,477 |
https://mathoverflow.net/questions/415773 | 4 | Here is a link for the definition of Wick product
<https://encyclopediaofmath.org/wiki/Wick_product>, which defines the Wick product recursively. My question is where do these two equations come from? I mean the equations
$$
\left\langle: f\_{1}^{k\_{1}} \cdots f\_{n}^{k\_{n}}:\right\rangle=0
$$
and
$$
\frac{\partial}{\partial f\_{i}}\left(: f\_{1}^{k\_{1}} \cdots f\_{n}^{k\_{n}}:\right)=k\_{i}: f\_{1}^{k\_{1}} \cdots f\_{i}^{k\_{i}-1} \cdots f\_{n}^{k\_{n}}:
$$
What is the motivation or intuition behind two equations? Is there a good reference on it?
| https://mathoverflow.net/users/69279 | Motivation for the axioms in Wick product | The Wick product :$A\_1A\_2A\_3$: is a specific way to order noncommuting operators $A\_1,A\_2,A\_3$. The concept was introduced by Gian-Carlo Wick in 1950 to avoid "infinite expectation values" that arise from the zero-point-motion of harmonic oscillators. Basically you reorder the operators in such a way that the expectation value of the reordered product vanishes – that motivates the first of the two equations in the question. The second equation follows because the order of the operators no longer matters once we have reordered them, so we can take the derivative as if these are ordinary functions – rather than non-commuting operators.
A general prescription for Wick ordering in quantum field theory in a creation/annihilation operator formalism is: “Permute all the creation operators $a^\dagger$ and the annihilation operators $a$, treating them as if they commute, so that in the end all $a^\dagger$ are to the left of all $a$.”
An introduction which I found instructive is [Wick calculus](https://arxiv.org/abs/physics/0212061) by Alexander Wurm and Marcus Berg.
*Note:* Hida and Ikeda introduced a related but distinct concept in probability theory. This is discussed in [The Wick product](https://www.duo.uio.no/bitstream/handle/10852/43474/1/H-Gjessing.pdf) by H. Gjessing et al.
| 6 | https://mathoverflow.net/users/11260 | 415786 | 169,481 |
https://mathoverflow.net/questions/415743 | 0 | Given iid random variables $\xi\_1,\dots,\xi\_n,\dots$ with probability $P(\xi\_1=-\log 3)=\frac{1}{2}=P(\xi\_1=\log 5-\log 3)$. Let $S\_n=\sum\_{i=1}^n \xi\_i$. This is a random walk on $R$. If let a stopping time $T:=\inf\{n: S\_n\le -4\log 3\}$. I want to get $\mathbb{P}(\tau<\infty)$.
I try to follow the solution in <https://math.stackexchange.com/questions/1108134/unbounded-stopping-time>
1. The stopped process $M\_n := e^{\xi\_1+\ldots+\xi\_{n \wedge \tau}}$ is also a martingale; hence, $$\mathbb{E}M\_n = \mathbb{E}M\_1=E(e^{\xi\_1})=(1/3)\*(1/2)+(5/3)\*(1/2)=1.$$ Calculate $\mathbb{E}M\_1$.
2. By the strong law of large numbers, $$\frac{\xi\_1+\ldots+\xi\_n}{n} \to \mathbb{E}\xi\_1 <0.$$ Thus, $$\xi\_1+\ldots+\xi\_n \to - \infty \quad \text{almost surely as} \, \, n \to \infty.$$
3. Conclude that $$\begin{align\*} M\_n &= e^{\xi\_1+\ldots+\xi\_n} 1\_{\{ \tau=\infty\}} + e^{\xi\_1+\ldots+\xi\_{n \wedge \tau}} 1\_{\{\tau<\infty\}}\\ &\to 0 + e^{-4log 3} 1\_{\{\tau<\infty\}}. \end{align\*}$$
4. Deduce from step 1 and 3 that
$$
1=3^{-4}P(\tau<\infty)
$$
$$\mathbb{P}(\tau<\infty) = 3^4.$$
But this is impossible? Where am I wrong?
| https://mathoverflow.net/users/168083 | Expectation value of random walk on $\mathbb{R}$ | 1. In the [MSE setting](https://math.stackexchange.com/questions/1108134/unbounded-stopping-time), there could be no overshoot: there, $X\_1+ \dots+X\_\tau$ is exactly $1$ (on the event $\{\tau<\infty\}$). In contrast, in your setting there will be, with a nonzero probability, an undershoot: $P(S\_T<-4\ln3)>0$. So, the convergence you claimed in Step 3 does not actually take place.
2. More importantly, in the [MSE setting](https://math.stackexchange.com/questions/1108134/unbounded-stopping-time), the (positive) martingale $(M\_n)$ is bounded (by $3$), whereas your martingale $(M\_n)$ is unbounded. So, even if your claim in Step 3 were correct, you cannot use the dominated convergence theorem to go from Step 3 to Step 4.
Mainly for the latter reason, you get the absurd value, $3^4$, for $P(\tau<\infty)$ (you apparently use two different symbols, $T$ and $\tau$, to denote the same thing). In fact, in your setting, by the strong law of large numbers, $P(T<\infty)=1$.
| 2 | https://mathoverflow.net/users/36721 | 415795 | 169,482 |
https://mathoverflow.net/questions/415801 | 0 | An informal investigation of a sum.
Consider this sum:
$$S =\sum\_{k=2}^{\infty}(k^{1/k} -1)$$
Does this converge? How does it behave as it diverges, if it diverges? If $k$ equaled $1$ we would get $0$ so we start at $k=2$. Very generally we can find that: $k^{1/k}$ will always be $1$ + a remainder. After subtracting $1$ from each term, each term will be a value less than $1$. Not too illuminating so far.
Let's raise the second k to a power t so we can examine how tweaking that variable affects the output $S\_t = \sum\_{k=2}^{\infty}(k^{1/k^t} -1)$. When $t=1$ it is the original and the partial sums seem to rise and rise, but is unclear of the final behavior.
Breaking $t$ out in parts, lets set $t = (1 + \frac{1}{x})$,
$$S\_x=\sum\_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1).$$
Numerically, these sums seem to converge when $x$ is finite, but its difficult to calculate when x gets large. Calculating some values, we find a pattern at last! We find $S\_2 = 4$ plus a remainder less than $1$. Going further $S\_3 = 9$ plus a remainder less than $1$, and $S\_{100} = 10,000$ plus a remainder less than $1$.
The output always seems to be the square plus a remainder! It seems true that $$\sum\_{k=2}^{\infty}(k^{1/(k^{1+1/x})} -1) = \lfloor x^{2}\rfloor$$ when $x$ is the square root of any natural number greater than 1.
Here are some examples:
$S\_4 = 16.238932773$,
$S\_{12} = 144.5937831$,
$S\_{\sqrt{1729}} = 1729.84841$,
$S\_{50000} = 2500000000.988421705$.
Does that remainder ever get higher than one? Let's get another perspective on the sum to see if we can calculate that remainder at the limit as ${x \to {\infty}}$.
Time to transform the sum so we can get a handle on why its the square plus a remainder, and try to get a handle on that remainder. Remembering that $t = (1 + 1/x)$,
see that: $$k^{1/k^t}=e^{(\log k)/k^t} = \sum\_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$
Let's define $$S\_t = \sum\_{k=2}^{\infty}\left(\sum\_{n = 0}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}\right) -1$$ Notice when $n=0$ we get $1$, so it cancels the $-1$ term resulting in: $$\sum\_{k=2}^{\infty}\sum\_{n = 1}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}}$$
What is effectively adding column by column instead of row by row, we get:$$\sum\_{n=1}^{\infty}\sum\_{k=2}^{\infty} \frac{\left( \log k \right)^n}{n! k^{n t}} = \sum\_{n=1}^{\infty}\frac{1}{n!}\sum\_{k=2}^{\infty} \frac{\left( \log k \right)^n}{k^{n t}}$$
Using the definition $$\zeta^{(n)}(t) = e^{i \pi n}\sum\_{k=2}^{\infty}\frac{\log k ^{n}}{k^{t}}$$ valid for all $n\in\mathbb{R}, t\in\mathbb{C}$,
we get $$S\_t = \sum\_{n=1}^{\infty}\frac{(-1)^{n}}{n!}\zeta^{(n)}(nt)$$
Since we'll be looking at the Zeta function close to it's pole at $1$, let's expand $\zeta$ with the Laurent expansion of the Zeta function expanded around 1. $\gamma$ represents the Stieltjes constants.
$$\zeta(s) = \frac{1}{s-1} + \sum\_{n=0}^{\infty}(-1)^{n}\frac{\gamma\_n}{n!}(s-1)^{n}$$
Giving us $$\zeta^{'}(s) = -\frac1{(s-1)^2} -\gamma\_1 +\gamma\_2(s-1)-\frac12\gamma\_3(s-1)^2\cdots$$
$$\zeta^{''}(s) = \frac2{(s-1)^3} +\gamma\_2 -\gamma\_3(s-1)+\frac12\gamma\_4(s-1)^2\cdots$$
$$\zeta^{'''}(s) = -\frac6{(s-1)^4} -\gamma\_3 +\gamma\_4(s-1)-\frac12\gamma\_5(s-1)^2...$$
Using these interpretations and setting $s = n(1+\frac1x)$ and examining what occurs when $x \to \infty$ allows us to see why the output of $S\_x$ is always $x^2$ plus a constant less than 1. As before, we can add column by column instead of row by row for the summations of the Laurent series.
Examining the largest term when $n=1$, we have $$\frac1{((1+\frac1x)-1)^2} = x^2$$ $x^2$ obviously diverges, so already we can tell the original sum $S$ diverges. Let's keep going to see how the remainder term behaves as $x \to \infty$. When $n=2$ and greater, the terms such as $\frac{2}{(s-1)^3} = \frac2{(n-1)^3}$ turn into finite values. Summing all the terms in $S\_t$, setting $t=1+\frac1x$ and letting x zoom off, we arrive at the limiting behavior of the sum $S\_x$.
$$\lim\_{x \to \infty}\Big|\Big(\sum\_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum\_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum\_{n=1}^{\infty}\frac{\gamma\_n}{n!} + \sum\_{k=3}^{\infty}\sum\_{n=k}^{\infty}(-1)^k\frac{\gamma\_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} = C,$$
and $$\sum\_{n=1}^{\infty}\frac{\gamma\_n}{n!} = \frac12 - \gamma\_0,$$
and $$\lim\_{x \to \infty}\Big|\Big(\sum\_{k=2}^{\infty}(k^{1/k^{1+1/x}} -1)\Big)-x^2\Big| = \sum\_{n=2}^{\infty}\frac1{(n-1)^{n+1}} + \sum\_{k=3}^{\infty}\sum\_{n=k}^\infty(-1)^k\frac{\gamma\_n(n-k+1)^{k-2}}{(n-k+2)!(k-2)!} + \frac12 - \gamma\_0 = C$$ .
$\gamma\_n$ is bounded by $|\gamma\_n| < \frac{n!}{2^{n+1}}$, which can show convergence, and numerically it stabilizes within the realm of testing.
$$C = 0.988549601142268750644\ldots$$
So it seems $$\sum\_{k=2}^\infty (k^{1/(k^{1+1/x})} -1) = x^2 + C\_x,$$ where $-0.028501\ldots < C\_x < C$, $x\geq1$ , and $x \in \mathbb{R}$ , where $-0.028501\ldots$ is $C\_x$ when $x=1$. $C\_x$ is positive when roughly $x>~1.37$.
Edit: Hi everyone, I didn't know the strict format of this site. Please see the selected answer as a beautiful solution to my first question, and please consider the rest of this post as exposition of how the sum diverges.
| https://mathoverflow.net/users/nan | The exploration of the asymptotic behavior of a simple sum. $\sum_{k=1}^{\infty} (k^{1/k} - 1)$ | The OP asks the question whether the series $S =\sum\_{k=2}^{\infty}(k^{1/k} -1)$ diverges.
The answer is that the series diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\frac{\ln k}{k} > \frac{1}{k}$$
and $\sum\_{k=2}^\infty 1/k=\infty$.
| 10 | https://mathoverflow.net/users/11260 | 415802 | 169,486 |
https://mathoverflow.net/questions/415682 | 2 | Say I have $N$ random variables $X\_1,\cdots,X\_i,\cdots,X\_N$, with zero mean and finite variance. $X\_i$ and $X\_j$ are independent iif $|i-j|>m$, and positively correlated otherwise (say the covariance is of $\mathcal{O}(1)$). It is well-known that the sums of $N$ of these random variables are distributed as a normal distribution as $N \rightarrow \infty$, if $m$ is a finite number of $\mathcal{O}(1)$.
My question is, what if $m=\sqrt{aN}$, and we take a sum of $\sqrt{aN}$ of these random variables, randomly selected from the sequence $X\_1,\cdots,X\_i,\cdots,X\_N$. Would the sum become a normal distribution? Each time we take a sum, we resample $\sqrt{aN}$ random variables and take a sum of them.
My intuition says yes. The reason is if we randomly sample $\sqrt{aN}$ variables from the original sequence of length $N$ (and $m=\sqrt{aN}$), the expected value of the number of variables sampled from any given window of consecutive $\sqrt{aN}$ variables is $a$, based on the following calculation:
$$\text{number of samples}\times\text{probability of sampling from a given window}=\sqrt{aN} \frac{\sqrt{aN}}{N} = a$$
Now it is as if we are taking a sum of $\sqrt{aN}$ random variables that are independent if $|i-j|>a$, so the conventional '$m$-dependent CLT' still holds since our $a$ is of $\mathcal{O}(1)$. Is this intuition correct? How do I prove/disprove this more rigorously?
| https://mathoverflow.net/users/173974 | Weakly dependent central limit theorem | The answer is yes. Say we have a stationary process, but we observe samples at random times $\{t\_n\}$ which itself is a stochastic point process (e.g. Poisson process). The resulting sample is also a stationary process ([Ref](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.851.9334&rep=rep1&type=pdf)).
I assume that if the original sequence is m-dependent ($m=\mathcal{O(1)}$), the resulting autocovariance function (for the sampled sequence) decays fast enough that CLT holds for the sum of the sampled sequence.
| 0 | https://mathoverflow.net/users/173974 | 415804 | 169,487 |
https://mathoverflow.net/questions/415782 | 3 | Let $V$ be a $\mathbb{C}$-vector space of dimension $N$, let $d$ be a positive integer, let $l \leq N$ be a positive integer, and let $U \subseteq S^d(V)$ be a linear subspace of codimension $k=\binom{l+d-1}{d}$. Does there exist a linear subspace $W \subseteq V$ of dimension $l$ for which $S^d(V)=U \oplus S^d(W)$?
I tried to prove yes by showing that the dimension of the subvariety of $Gr(k,S^d(V))$,
consisting of subspaces of the form $S^d(W)$ for an $l$-dimensional linear subspace $W \subseteq V$, is greater than the dimension of the subvariety of subspaces that intersect $U$ non-trivially, but this is emphatically not the case. The former has dimension $l(N-l)$, while the latter has dimension $k(\binom{N+d-1}{d}-k)$. (See [here](https://mathoverflow.net/questions/415732/what-is-the-dimension-of-this-subvariety-of-the-grassmannian))
| https://mathoverflow.net/users/150898 | Given a subspace $U \subseteq S^d(V)$, does there always exist a complement of the form $S^d(W)$? | Without any further hypothesis on $U$, the answer is no. Take $V = \mathbb{C}^3$, $d=2$, $l=2$ and $U = S^2(W\_1)$ where $W\_1$ is a $\mathbb{C}^2$ inside $V$. Then for any other $W\_2 \subset V$ of dimension $2$, we have:
$$S^2(W\_1 \cap W\_2) \subset U \cap S^2(W\_2)$$
and obiously $\dim S^2(W\_1 \cap W\_2) \geq 1$. On the other hand, I think the answer should be *yes* if you assume $U$ to be generic (and perhaps some adequate numerical conditions). This should follow from a dimension count bounding the dimension of the subvariety of $\mathrm{Gr}\left( \binom{l+d-1}{l},S^d V \right)$ representing linear spaces which intersect $S^d(W)$ non-transversally for all $W \in \mathrm{Gr}(l,V)$.
| 6 | https://mathoverflow.net/users/37214 | 415807 | 169,488 |
https://mathoverflow.net/questions/415709 | 3 | Suppose $A$ is an abelian variety over a number field $K$ and call $M$ the maximal torsion free quotient of $A(\overline{K})$ equipped with its Galois action.
>
> For the first Galois cohomology of $M$, do we have $H^1(\text{Gal}\_K, M) = 0$? Is $H^1(\text{Gal}\_K, M)$ at least finite?
>
>
>
If $H^1(\text{Gal}\_K/U, M^U)$ is trivial for all open normal subgroups of $\text{Gal}\_K$, then so is $H^1(\text{Gal}\_K, M)$. If the vanishing is false, I'd be interested in a counterexample evev just for some $U$ and $H^1(\text{Gal}\_K/U, M^U)$.
| https://mathoverflow.net/users/nan | Galois cohomology of abelian varieties | I think that all Galois cohomology groups $\mathrm{H}^i(\mathrm{Gal}\_K,M)$ vanish for $i>0$. As you observed, it suffices to prove that
$\mathrm{H}^i(\mathrm{Gal}\_K/U,M^U)$ vanishes for all open normal subgroups $U$ of $\mathrm{Gal}\_K$. Since $\mathrm{Gal}\_K/U$ is a finite
group, the Galois cohomology group $\mathrm{H}^i(\mathrm{Gal}\_K/U,M^U)$ is torsion (via restriction/corestriction with the trivial subgroup,
see Serre: Corps locaux, p. 138). In fact, all of its elements have order a divisor of $n=|\mathrm{Gal}\_K/U|$.
It suffices then to show that the multiplication-by-$n$ map $[n]$ on $M^U$ is bijective.
The map $[n]\colon M^U\rightarrow M^U$ is injective since $M$ is torsion free. In order to show that it is surjective choose $x\in M^U$ and
consider the short exact sequence defining $M$
$$
0\rightarrow A(\bar K)\_{\mathrm{tor}}\rightarrow A(\bar K)\overset{\pi}{\rightarrow} M\rightarrow 0,
$$
where $A(\bar K)\_{\mathrm{tor}}$ is the torsion subgroup of $A(\bar K)$. Since $x\in M$ and $A(\bar K)$ is a divisible group, there is an element
$y\in A(\bar K)$ such that $\pi([n]y)=x$, where $[n]$ also denotes the multiplication-by-$n$ morphism on the abelian group $A(\bar K)$. Let us
prove that $\pi(y)\in M^U$. Choose $g\in U$. One has
$$
\pi([n]gy)=\pi(g[n]y)=g\pi([n]y)=gx=x=\pi([n]y)
$$
since $x\in M^U$. It follows that $[n](gy-y)\in\ker(\pi)$. Hence $[n](gy-y)$ is a torsion element of $A(\bar K)$. Then $gy-y$ is in $A(\bar
K)\_{\mathrm tor}$ as well. This means that
$$
g\pi(y)-\pi(y)=\pi(gy-y)=0,
$$
i.e., $g\pi(y)=\pi(y)$.
Hence $\pi(y)\in M^U$. Since $[n]\pi(y)=\pi([n]y)=x$, the morphism $[n]\colon M^U\rightarrow M^U$ is surjective indeed.
| 4 | https://mathoverflow.net/users/85592 | 415809 | 169,490 |
https://mathoverflow.net/questions/415805 | 4 | Suppose $L/F$ is a finite Galois extension of number fields. Let $E$ be an elliptic curve over $F$ and $E\_L$ its base change to $L$.
>
> Do we have that $\text{Sha}(E/F)$ is finite if and only if $\text{Sha}(E\_L/L)^{\text{Gal}(L/F)}$ is finite?
>
>
>
Analogously, write $\text{Sha}(E\_{\overline{F}}/\overline{F})$ for the direct limit of the system $\{\text{Sha}(E\_L/L)\}$ with $L$ running across the finite Galois extensions of $F$, with transition maps induced by functoriality.
>
> Same question with $\text{Sha}(E\_{\overline{F}}/\overline{F})^{\text{Gal}(\overline{F}/F)}$ replacing $\text{Sha}(E\_L/L)^{\text{Gal}(L/F)}$.
>
>
>
I was reading about the relation between the BSD conjecture for $E$ and that for $E\_L$ and the question came up.
**Remark**
I expect the inflation-restriction exact sequence in group cohomology to be key here. For example, in the first question the kernel of the restriction map $\text{Sha}(E/F)\to \text{Sha}(E\_{L}/L)^{\text{Gal}(L/F)}$ is contained in $H^1(\text{Gal}(L/F), E\_L(L))$. This implies that if $\text{Sha}(E\_{L}/L)^{\text{Gal}(L/F)}$ is finite, so is $\text{Sha}(E/F)$.
| https://mathoverflow.net/users/nan | Tate-Shafarevich groups under finite Galois field extensions | The remark added to the question shows that the kernel of $Ш(E/F) \to Ш(E/L)^G$ is finite where $G$ is the finite Galois group of $L/F$.
$\DeclareMathOperator{\coker}{coker}$
Here is an argument why the cokernel is finite. It may be too complicated.
Let $p$ be a prime and let us show that the cokernel on the $p$-primary part is finite (and that it is trivial for all $p$ that do not divide $\lvert G\rvert$). Let $\alpha\colon S(E/F) \to S(E/L)^G$ where $S$ stands for the $p$-primary Selmer group in $H^1\bigl(F, E[p^{\infty}]\bigr)$. Consider the exact sequence
$$ 0\to \Bigl( E(L) \otimes \mathbb{Q}\_p/\mathbb{Z}\_p \Bigr)^G \to S(E/L)^G \to Ш(E/L)[p^{\infty}]^G \to H^1\bigl( G, E(L) \otimes \mathbb{Q}\_p/\mathbb{Z}\_p\bigr) $$
Note that the last term is isomorphic to $H^2\bigl(G,E(L)\otimes \mathbb{Z}\_p\bigr)$, which is finite. Comparing that sequence with the sequence with $S(E/F)$ in the middle, shows that the cokernel on Ш lies in an exact sequence between $\coker(\alpha)$ and $H^2\bigl(G,E(L)\otimes \mathbb{Z}\_p\bigr)$.
Let $S$ be the finite set of places containing all places at $\infty$, all places of bad reduction and all places above $p$.
Let $G\_S(L)$ be the Galois group of the maximal extension of $L$ unramified outside $S$.
Next the exact sequence
$$
0\to S(E/L)^G \to H^1\bigl(G\_S(L), E[p^{\infty}]\bigr)^G \to \Bigl(\bigoplus\_{w \in S} H^1\bigl(L\_w, E\bigr)[p^{\infty}]\Bigr)^{G}
$$
Again we compare it with the similar sequence defining $S(E/F)$. This shows that there is a map $\beta\colon\coker(\alpha)\to H^2\bigl(G,E(L)[p^{\infty}]\bigr)$. The target of $\beta$ is finite; its kernel can be shown to be a subquotient of kernel of the restriction map on the right hand side:
$$\bigoplus\_{v \in S\_F} H^1\bigl( G\_w, E(L\_w)\bigr)[p^{\infty}]$$
where $S\_F$ is the set of places below $S$ and for each $v$ we choose one $w$ above $v$.
As this is a finite sum of finite groups, this is also finite.
For the second question. Since $Ш(E/L)\subset H^1\bigl(L,E)$, your limit is natrually a subgroup of the limit of these $H^1$, i.e of $H^1(\bar{L},E)=0$.
| 5 | https://mathoverflow.net/users/5015 | 415811 | 169,491 |
https://mathoverflow.net/questions/413714 | 0 | I would like to define standard Gram matrices, and use them to help me understand the symmetries of lattices.
I define "standard Gram matrix" as the Gram matrix g that minimizes the deviation from Toeplitz and that satisfies abs(inv(g)) = abs(g).
I have obtained Gram matrices for D4, E8, A15+ with small deviations from Toeplitz that satisfy abs(inv(g)) = abs(g) but I cannot be sure that I have obtained the global optimum. My method for solving the optimization problem specified by the definition of "standard Gram matrix" is ad hoc and I cannot be sure that it has found the global optimum.
QUESTIONS:
-- How do I uniquely determine a standard Gram matrix for a lattice? Are there any other definitions of "standard Gram matrix" for lattices? Do you have any literature references in which "standard Gram matrix" is defined for any reason?
I seem to be able to uniquely define standard Gram matrices for D4, E8, A15+ for instance.
But I may have missed the optimum for E8 and A15+.
**This is my ad-hoc method for determining the standard Gram matrix according to my definition:**
I find a Gram matrix g for which abs(inv(g)) - abs(g) is zero. g seems to be unique up to permutations and sign flips, but I am not sure. I choose g to have the most positive signs. inv(mat) is the matrix inverse of mat, and abs(mat) is the absolute values of the matrix elements.
Furthermore I permute the basis vectors to make the Gram matrix maximally Toeplitz.
That seems to uniquely define the standard Gram matrices for D4 and E8.
("maximally" refers to the L1 norm of the matrix elements, sum of absolute values of deviations of matrix elements)
For instance, for the 24-cell D4 = D4\* lattice I obtain a Toeplitz Gram matrix
$$ \mathrm{gd4} = \sqrt{\frac{1}{2}} \left( \begin{array}{cccc}
2 & 1 & 0 &-1 \\
1 & 2 & 1 & 0 \\
0 & 1 & 2 & 1 \\
-1& 0 & 1 & 2 \\
\end{array} \right)$$
The inverse is
$$ \mathrm{inv}(\mathrm{gd4}) = \sqrt{\frac{1}{2}} \left( \begin{array}{cccc}
2 &-1 & 0 & 1 \\
-1& 2 &-1 & 0 \\
0 &-1 & 2 &-1 \\
1 & 0 &-1 & 2 \\
\end{array} \right) $$
likewise E8 has gram matrix ge8, not quite Toepliz.
$$ \mathrm{ge8 =} \begin{array}{ccccccccc}
&& 2 && 0 && 1 && 0 && 0 && 0 && -1 && 1 \\
&& 0 && 2 && 1 && 1 && 0 && 0 && 0 && -1 \\
&& 1 && 1 && 2 && 0 && 1 && 0 && 0 && 0 \\
&& 0 && 1 && 0 && 2 && -1 && 1 && 0 && 0 \\
&& 0 && 0 && 1 && -1 && 2 && 0 && 1 && 0 \\
&& 0 && 0 && 0 && 1 && 0 && 2 && 1 && 1 \\
&& -1 && 0 && 0 && 0 && 1 && 1 && 2 && 0 \\
&& 1 && -1 && 0 && 0 && 0 && 1 && 0 && 2 \\
\end{array} $$
Each basis vector is perpendicular (90 degrees) to four other basis vectors, and makes angles of 60, 60, 120 degrees with the other three basis vectors.
inverse
$$ \mathrm{inv(ge8) =} \begin{array}{ccccccccc}
&& 2 && 0 && -1 && 0 && 0 && 0 && 1 && -1 \\
&& 0 && 2 && -1 && -1 && 0 && 0 && 0 && 1 \\
&& -1 && -1 && 2 && 0 && -1 && 0 && 0 && 0 \\
&& 0 && -1 && 0 && 2 && 1 && -1 && 0 && 0 \\
&& 0 && 0 && -1 && 1 && 2 && 0 && -1 && 0 \\
&& 0 && 0 && 0 && -1 && 0 && 2 && -1 && -1 \\
&& 1 && 0 && 0 && 0 && -1 && -1 && 2 && 0 \\
&& -1 && 1 && 0 && 0 && 0 && -1 && 0 && 2 \\
\end{array} $$
The "Dynkin diagrams" are a square for D4 and a cube for E8 as shown in the figure here: [Dynkin diagrams for D4 and E8](https://i.stack.imgur.com/xRSno.png)
In contrast the usual Dynkin diagram for D4 is a triangle: <https://commons.wikimedia.org/wiki/File:Dynkin_diagram_D4.png>
I wonder whether I have found the maximally Toeplitz Gram matrix for E8 and A15+.
For A15+ the permutations are irrelevant. In this case I obtain a Gram matrix ga15plus satisfying
abs(ga15plus) = abs(inv(ga15plus))
$$ \mathrm{ga15plus \ =} \begin{array}{ccccccccccccccccc}
&& 2 && 1 && 1 && 1 && 1 && 1 && 1 && 4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 2 && 1 && 1 && 1 && 1 && 1 && 4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 2 && 1 && 1 && 1 && 1 && 4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 2 && 1 && 1 && 1 && 4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 2 && 1 && 1 && 4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 2 && 1 && 4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 2 && 4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 4 && 4 && 4 && 4 && 4 && 4 && 4 && 15 && 4 && 4 && 4 && 4 && 4 && 4 && 4 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && 4 && 2 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && 4 && 1 && 2 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && 4 && 1 && 1 && 2 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && 4 && 1 && 1 && 1 && 2 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && 4 && 1 && 1 && 1 && 1 && 2 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && 4 && 1 && 1 && 1 && 1 && 1 && 2 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && 4 && 1 && 1 && 1 && 1 && 1 && 1 && 2 \\
\end{array} $$
$$ \mathrm{inv(ga15plus) \ =} \begin{array}{ccccccccccccccccc}
&& 2 && 1 && 1 && 1 && 1 && 1 && 1 && -4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 2 && 1 && 1 && 1 && 1 && 1 && -4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 2 && 1 && 1 && 1 && 1 && -4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 2 && 1 && 1 && 1 && -4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 2 && 1 && 1 && -4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 2 && 1 && -4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 2 && -4 && 1 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& -4 && -4 && -4 && -4 && -4 && -4 && -4 && 15 && -4 && -4 && -4 && -4 && -4 && -4 && -4 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && -4 && 2 && 1 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && -4 && 1 && 2 && 1 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && -4 && 1 && 1 && 2 && 1 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && -4 && 1 && 1 && 1 && 2 && 1 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && -4 && 1 && 1 && 1 && 1 && 2 && 1 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && -4 && 1 && 1 && 1 && 1 && 1 && 2 && 1 \\
&& 1 && 1 && 1 && 1 && 1 && 1 && 1 && -4 && 1 && 1 && 1 && 1 && 1 && 1 && 2 \\
\end{array} $$
With this Gram matrix there is one unique basis vector, so for A15+, there are 15 equivalent permutations, 15 equivalent standard gram matrices according to my definition.
But perhaps a different ga15plus also satisfies the inverse equality and can be made more Toeplitz.
In summary with my draft definition of "standard Gram matrix",
I seem to obtain unique standard Gram matrices for D4 and E8 and 15 equivalent standard Gram matrices for A15+. But I am not sure that I have obtained the most Toeplitz Gram matrices for E8 and A15+.
Again my questions are
-- How do I uniquely determine a standard Gram matrix for a lattice? Are there any other definitions of "standard Gram matrix" for lattices? Do you have any literature references in which "standard Gram matrix" is defined for any reason?
| https://mathoverflow.net/users/385881 | Standard Gram matrices for lattices |
>
> How do I uniquely determine a standard Gram matrix for a lattice? Are there any other definitions of "standard Gram matrix" for lattices? Do you have any literature references in which "standard Gram matrix" is defined for any reason?
>
>
>
Yes, this has been done in [A Canonical Form for Positive Definite Matrices](https://arxiv.org/pdf/2004.14022.pdf). In particular, for positive definite $A$ (Gram matrices of positive definite lattices are positive definite), they
first define $A, B$ to be *arithmetically equivalent* if $A = U^tBU$ for uniomdular $U$.
They then define a mapping
$$A\mapsto Can(A)$$
such that
1. $Can(A)$ is equivalent to $A$, and
2. For any unimodular $U$, $Can(U^tAU)=Can(A)$.
Their (broad) strategy is to reduce to the case of graphs, for which there are recent quasi-polynomial time algorithms. This somewhat limits the maximum dimension they can handle (computationally) --- it appears they go up to dimension 40 at the most.
Note that this problem is thought to be hard though --- it would imply a solution to the *Lattice Isomorphism Problem*.
Efficient algorithms for this have been open for quite a while (there has been some interest for 10+ years), and recently [two](https://eprint.iacr.org/2021/1332.pdf) [cryptosystems](https://eprint.iacr.org/2021/1548) have been proposed assuming (variants of) this problem are hard, so now one can concretely say that efficient algorithms would break cryptographic proposals (so if you find one, there is a more obvious motivation to write it up).
| 2 | https://mathoverflow.net/users/101207 | 415814 | 169,492 |
https://mathoverflow.net/questions/415806 | 5 | I'm trying to calculate the weak limit of $\mathcal{E}\_N(x)=\sum\_{k=1}^{2^N}\delta\_{x-Z\_k}$ , with $Z\_k=X\_k-\max\_{k\leq 2^N}X\_k$, $\{X\_k\}$ being $2^N$ copies of i.i.d. Gaussians with mean zero and variance $N$(so it can be seen as discretization of Brownian motion).
I'm considering the Laplace transform, since it works well when $\max\_{k\leq 2^N}X\_k$ is replaced by the typical extrema $m\_N=\sqrt{2\log 2}N-\frac{1}{2\sqrt{2\log 2}}\log N$ in the definition of $Z\_k$. Which is to say, I want to calculate the limit of the following for every test function $\phi\in C\_0^+(\mathbb{R})$: $$\mathbb{E}\exp\left(-\int\phi(x)\mathcal{E}\_N(x)dx\right)=\mathbb{E}\exp\left(-\sum\_{k=1}^{2^N}\phi(X\_k-\max X\_k)\right).$$
The main difficulty is that these $\{X\_k-\max X\_k\}$ are now not independent, while the previous ones $\{X\_k-m\_N\}$ are independent. I don't know how to deal with this, even if I use some simpler special functions eg. $\phi(x)=1(x\geq-A)$ to simplify the transform to $$\mathbb{E}\exp\left(-\#\{X\_k:X\_k\geq\max X\_k-A\}\right).$$
How can I deal with the dependency here?
| https://mathoverflow.net/users/174600 | Limit of the extremal process of i.i.d. Gaussians see from the tip | $\newcommand\R{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\vpi}{\varphi}$For $\phi=c\,1\_{[-A,\infty)}$ with $c\ge0$ and $A\ge0$, the expectation in question converges to
\begin{equation\*}
\frac1{1+(e^c-1)e^{A\sqrt{\ln4}}}. \tag{1}\label{1}
\end{equation\*}
---
Indeed, let $n:=2^N\to\infty$. Let
\begin{equation\*}
h:=\phi=c\,1\_{[-A,\infty)} \tag{2}\label{2}
\end{equation\*}
for some real with $c\ge0$ and $A\ge0$.
Let $V\_i:=X\_i-\max\_{1\le j\le n}X\_j$. Let $g\colon\R^n\to\R$ be
any nonnegative Borel-measurable function that is symmetric (with respect to any permutation of its arguments). Let $f$ denote the pdf of each $X\_j$, so that
\begin{equation\*}
f(x)=\frac1s\,\vpi\Big(\frac xs\Big)
\end{equation\*}
for all real $x$, where $\vpi$ is the standard normal pdf and
\begin{equation\*}
s:=\sqrt N=\sqrt{\log\_2 n}=\sqrt{\frac{\ln n}{\ln2}}. \tag{3}\label{3}
\end{equation\*}
Then
\begin{equation\*}
\begin{aligned}
&Eg(V\_1,\dots,V\_n) \\
&=n\,Eg(X\_1-X\_1,X\_2-X\_1,\dots,X\_n-X\_1)\,1(X\_1>\max\_{2\le j\le n}X\_j) \\
&=n\,\int\_{\R^n}g(0,x\_2-x\_1,\dots,x\_n-x\_1)\,1(x\_1>\max\_{2\le j\le n}x\_j)
\prod\_{j=1}^n f(x\_j)dx\_j\, \\
&=n\,\int\_\R f(x\_1)\,dx\_1\,\int\_{(-\infty,0)^n}g(0,v\_2,\dots,v\_n)\,
\prod\_{j=2}^n f(x\_1+v\_j)dv\_j.
\end{aligned}
\end{equation\*}
Then the expectation in question is
\begin{equation\*}
L:=L\_n:=ne^{-c}I, \tag{3.5}\label{3.5}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
I:=I\_n&:=\int\_\R f(x\_1)\,dx\_1\,\Big(\int\_{-\infty}^0 e^{-h(v)} f(x\_1+v)\,dv\Big)^{n-1} \\
&=\int\_\R dz\,\vpi(z)\,H(z)^{n-1},
\end{aligned}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
H(z)&:=\int\_{-\infty}^z e^{-h(s(w-z))}\vpi(w)\,dw \\
&=e^{-c}\Phi(z)+(1-e^{-c})\Phi(z-A/s),
\end{aligned}
\end{equation\*}
where $\Phi$ is the standard normal cdf (and $h$ and $s$ are as defined in \eqref{2} and \eqref{3}).
Let
\begin{equation\*}
z\_\ep:=\sqrt{(2-\ep)\ln n},
\end{equation\*}
where $\ep\in(0,2)$.
Note that
\begin{equation\*}
I=J\_1+J\_2+J\_3, \tag{4}\label{4}
\end{equation\*}
where
\begin{equation\*}
J\_1:=\int\_{-\infty}^{z\_\ep} dz\,\vpi(z)\,H(z)^{n-1},\quad
J\_2:=\int\_{z\_\ep}^{z\_0} dz\,\vpi(z)\,H(z)^{n-1},\quad
J\_3:=\int\_{z\_0}^\infty dz\,\vpi(z)\,H(z)^{n-1}.
\end{equation\*}
Noting that $0<H\le\Phi<1$ and letting
\begin{equation\*}
G:=1-\Phi,
\end{equation\*}
we see that, for each $\ep\in(0,2)$,
\begin{equation\*}
nJ\_1\le\int\_{-\infty}^{z\_\ep} dz\,\vpi(z)\,n\Phi(z)^{n-1}
=\Phi(z\_\ep)^n \\
=(1-G(z\_\ep))^n\le\exp\{-nG(z\_\ep)\}=o(1/n), \tag{4.5}\label{4.5}
\end{equation\*}
since
\begin{equation\*}
nG(z\_\ep)=n\exp\Big\{-\frac{z\_\ep^2}{2+o(1)}\Big\}
=n\exp\Big\{-\frac{2-\ep}{2+o(1)}\,\ln n\Big\}=n^{\ep/(2+o(1))}.
\end{equation\*}
So, the conclusion
\begin{equation\*}
nJ\_1=o(1/n) \tag{5}\label{5}
\end{equation\*}
will hold if $\ep=\ep\_n$, for some sequence $\ep\_n\downarrow0$. In what follows, it is indeed assumed that $\ep=\ep\_n\downarrow0$.
Next,
\begin{equation\*}
J\_3\le\int\_{z\_0}^\infty dz\,\vpi(z)=G(z\_0)<\frac{\vpi(z\_0)}{z\_0}=o(\vpi(z\_0))=o(1/n).
\tag{6}\label{6}
\end{equation\*}
Further,
\begin{equation\*}
\begin{aligned}
H'(z)&=e^{-c}\vpi(z)+(1-e^{-c})\vpi(z-A/s) \\
&=\vpi(z)[e^{-c}+(1-e^{-c})e^{Az/s}e^{-A^2/(2s^2)}] \\
&\sim\vpi(z)[e^{-c}+(1-e^{-c})e^{Az/s}],
\end{aligned}
\end{equation\*}
since $s\to\infty$.
Also, for $z\in[z\_\ep,z\_0]$ we have $z\sim z\_0$ (since $\ep\downarrow0$) and hence
\begin{equation\*}
z/s\to z\_0/s=\sqrt{\ln4}.
\end{equation\*}
So,
\begin{equation\*}
\begin{aligned}
J\_2&=\int\_{z\_\ep}^{z\_0} dz\,\frac{\vpi(z)}{H'(z)}\,H'(z)H(z)^{n-1} \\
&\sim\frac1{e^{-c}+(1-e^{-c})e^{A\sqrt{\ln4}}}\,\int\_{z\_\ep}^{z\_0} dz\,H'(z)H(z)^{n-1}.
\end{aligned}
\tag{7}\label{7}
\end{equation\*}
Also,
\begin{equation\*}
\begin{aligned}
n\int\_{-\infty}^{z\_\ep} dz\,H'(z)H(z)^{n-1}= H(z\_\ep)^n\le\Phi(z\_\ep)^n =o(1),
\end{aligned}
\tag{8}\label{8}
\end{equation\*}
as shown in \eqref{4.5}.
Also,
\begin{equation\*}
\begin{aligned}
\int\_{z\_0}^\infty dz\,H'(z)H(z)^{n-1}
&\le\int\_{z\_0}^\infty dz\,H'(z) \\
&=e^{-c}G(z\_0)+(1-e^{-c})G(z\_0-A/s) \\
&\le G(z\_0-A/s) \\
&\le \frac{\vpi(z\_0-A/s)}{z\_0-A/s} \\
&=O\Big(\frac{\vpi(z\_0)}{z\_0-A/s}\Big)
=o(\vpi(z\_0))=o(1/n).
\end{aligned}
\tag{9}\label{9}
\end{equation\*}
Collecting \eqref{7}, \eqref{8}, and \eqref{9}, we get
\begin{equation\*}
\begin{aligned}
J\_2&\sim\frac1{e^{-c}+(1-e^{-c})e^{A\sqrt{\ln4}}}\, \\
&\times\Big(\int\_{-\infty}^\infty dz\,H'(z)H(z)^{n-1} \\
&\quad-\int\_{-\infty}^{z\_\ep} dz\,H'(z)H(z)^{n-1}-\int\_{z\_0}^\infty dz\,H'(z)H(z)^{n-1}\Big) \\
&=\frac1{e^{-c}+(1-e^{-c})e^{A\sqrt{\ln4}}}\,(1/n-o(1/n)-o(1/n)).
\end{aligned}
\tag{10}\label{10}
\end{equation\*}
Finally, collecting \eqref{3.5}, \eqref{4}, \eqref{5}, \eqref{6}, and \eqref{10}, we get the limit \eqref{1} for $L$.
---
It follows that $\sum\_{k=1}^n h(X\_k-\max\_{1\le i\le n} X\_i)$ converges in distribution to a geometrically distributed random variable $Y$ such that
\begin{equation}
P(Y=j)=\tfrac1B\,(1-\tfrac1B)^{j-1}\,1(j\in\{1,2,\dots\}),
\end{equation}
where
\begin{equation}
B:=e^{A\sqrt{\ln4}}>1.
\end{equation}
This result could probably be obtained directly, without using the Laplace transform.
| 6 | https://mathoverflow.net/users/36721 | 415825 | 169,496 |
https://mathoverflow.net/questions/414238 | 3 | It seems to be a commonplace in harmonic analysis that if some operator (say, Fourier multiplier) is bounded on $L^p(\mathbb{R}^n)$ then by transference the similar operator is also bounded on $L^p(\mathbb{T}^n)$ (the most common example of application of this general principle is Hilbert transform for $n=1$). There exist certain general theorems which allow us to transfer the results in the setting of $\mathbb{R}^n$ to the setting of $\mathbb{T}^n$, and vice versa (for example, such theorems may be found in Grafakos, "Classical Fourier Analysis").
Now, my question is -- can we do the same for the cyclic group $\mathbb{Z}\_n$ (with the standard Haar measure on it)? Say, for arbitrary number $m<n$ we may define the operator $\widehat{Tf}=\chi\_{[0,m]}\hat{f}$ on $L^p(\mathbb{Z}\_n)$. It seems to be not difficult to see directly that such operator is bounded and its norm does not depend on $n$ and $m$ but can we somehow derive it from the fact that Hilbert transform is bounded on $L^p(\mathbb{R})$ and $L^p(\mathbb{T})$? The intuition here is that for large $n$ the group $\mathbb{Z}\_n$ should resemble $\mathbb{Z}$ (or $\mathbb{T}$) but I do not know how to do a rigorous proof of such "transference".
After that, one may go further and prove the "Littlewood--Paley theorem for $\mathbb{Z}\_n$" (uniformly in $n$) and maybe certain multiplier theorems. Did anybody address such questions (with or without transference)? Or maybe these questions are trivial for some reason?
| https://mathoverflow.net/users/69086 | Fourier multipliers and transference on cyclic groups | Ok, I found the answer myself, so I'll post it here. It turned out that the results for multipliers should be transferred not from $L^p(\mathbb{T})$ but from $\ell^p(\mathbb{Z})$. That is, the boundedness of operators that figure in the question follows from the boundedness of the following operator on $\ell^p(\mathbb{Z})$: we take a sequence in $\ell^p(\mathbb{Z})$, consider its Fourier transform (which is a function on $\mathbb{T}$), multiply it by characteristic function of an arc and take the Fourier coefficients of the resulting function. The boundedness of such operator on $\ell^p(\mathbb{Z})$ is of course known --- it can be found in the book by Edwards and Gaudry "Littlewood--Paley and multiplier theory" (1977).
As for this transference, it is known, too (and not very difficult): it is written in an article "Transference methods in analysis" by Coifman and Weiss (in a more general context; to be more specific, Theorem 3.15 and Corollary 3.16 can be applied here).
| 1 | https://mathoverflow.net/users/69086 | 415827 | 169,498 |
https://mathoverflow.net/questions/415820 | 1 | Suppose $X$ is a Hausdorff (I'm happy to also assume "non compact") topological space that can be written as the topological direct limit
$$\varinjlim\_{a\in J} K\_a$$
for $J$ a directed set and $K\_a$ compact Hausdorff subspaces of $X$, where all bonding maps are continuous non-surjective immersions (I'm happy to assume, additionally, that they are not surjective).
Suppose that there is another system of compact Hausdorff subspaces of $X$ with bonding maps as in the previous system,
$$\{H\_a, a\in J'\}$$
for a directed set $J'$ with $J\subseteq J'$. Suppose that for every $a\in J$ we have
$$K\_a\subseteq H\_a.$$
Suppose that we still have
$$X = \varinjlim\_{a\in J'}H\_a$$
with the direct limit topology on the right side and $X$'s own topology on the left side.
>
> **Q1.** Can we conclude that for every $a\in J'$ there exists a $b\in J$ such that $H\_a\subseteq K\_b$?
>
>
>
>
> **Q2.** Can we conclude as in Q1 if in addition all the $K\_a$'s and $H\_a$'s are compact differentiable manifolds with differentiable immersions as bonding maps?
>
>
>
The [answer](https://mathoverflow.net/a/415828) by KP Hart answers Q1 in the negative.
| https://mathoverflow.net/users/nan | Approximations by compact sub-spaces | Q1. Consider the space of rationals $\mathbb{Q}$.
It is the direct limit of the family of finite unions of convergent sequences (including their limits), ordered by inclusion, as a set is closed iff its intersection with every convergent sequence is closed.
Now choose for every convergent sequence $\mathbf{q}=\langle q\_n:n\in\mathbb{N}\rangle$ a set $S\_\mathbf{q}$ of convergent sequences $\{\mathbf{r}\_n:n\in\mathbb{N}\}$, where $\mathbf{r}\_n$ converges to $q\_n$, and has diameter not larger than $1/n$.
If $K$ is a finite family of convergent sequences then let $H\_K=\bigcup\_{\mathbf{q}\in K}S\_\mathbf{q}$.
Then $K\subseteq H\_K$, and $H\_K$ is compact; and $\mathbb{Q}$ is also the direct limit of the system of $H\_K$s. But there are no $L$ and $K$ such that $H\_K\subseteq L$ ($L$ has finitely many accumulation points, $H\_K$ has infinitely many).
Addendum: this construction works in every metric space without isolated points.
There a set is closed iff its intersection with every convergent sequence is closed, and every point is the limit of a non-trivial convergent sequence of arbitrarily small diameter.
| 2 | https://mathoverflow.net/users/5903 | 415828 | 169,499 |
https://mathoverflow.net/questions/415832 | 0 | I'm looking for a closed solution or an approximation to $$\int x^{-a} \text{erf}\left( b - c x^{-d} \right) dx,$$
where $a, b, c, d > 0$.
| https://mathoverflow.net/users/103291 | Solution or approximation to $\int x^{-a} \text{erf}\left( b - c x^{-d} \right) dx$? | From the comments I understand that the OP seeks an approximation of
$$I=\int\_{x\_1}^{x\_2} x^{-a} \text{erf}\left( b - c x^{-d} \right)\, dx$$
for $x\_2\gg x\_1\gg 1$. A complication which will limit the accuracy of the approximation is that $d\ll 1$. If I ignore that for a moment, and assume all coefficients $a,b,c,d$ are of order unity, then a large-$x$ expansion of the integrand gives the approximation
$$I\_{\text{appr}}=\int\_{x\_1}^{x\_2}x^{-a}\left(\text{erf}(b)-\frac{2c e^{-b^2} }{x^{d}\sqrt{\pi }}\right)\,dx$$
$$\qquad={x\_2}^{-a} {x\_1}^{-a} \left(\frac{2 e^{-b^2} c {x\_2}^{-d} {x\_1}^{-d} \left({x\_2} {x\_1}^{a+d}-{x\_1} {x\_2}^{a+d}\right)}{\sqrt{\pi } (a+d-1)}+\frac{\text{erf}(b) \left({x\_1} {x\_2}^a-{x\_2} {x\_1}^a\right)}{a-1}\right).$$
The values of interest to the OP are $\{a,b,c,d,x\_1,x\_2\}=\{1.2743, 16.33, 18.7525, 0.0308, 284.959, 712090\}$ In this case $I=0.621072$ while $I\_{\text{appr}}=0.682988$, an error of 10%. If the parameter $d$ is increased slightly to $0.05$ the agreement improves to four decimal places.
| 3 | https://mathoverflow.net/users/11260 | 415843 | 169,504 |
https://mathoverflow.net/questions/415815 | 6 | Consider $\mathcal{X}$ a projective and flat scheme over $\text{Spec}(\mathcal{O}\_K)$, with $\mathcal{O}\_K$ the ring of integers of a number field $K$.
Let $F/K$ vary over all finite Galois number field extensions and define $\mathbf{A} := \mathbf{A}\_{\overline{K}}$ as the direct limit of the topological rings $\mathbf{A}\_F$ of the adèles of each $F$.
>
> **Question 1** Is there a good and intrinsic definition of $\mathcal{X}(\mathbf{A})$ and does it agree with the direct limit topological space $\varinjlim\_{F/K}\mathcal{X}(\mathbf{A}\_F)$? (or is this latter the definition usually given?)
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> **Question 2** Is $\mathcal{X}(\mathbf{A})$ compact?
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The ring $\mathbf{A}$ must probably be replaced with a restricted product of countably many copies of $\mathbf{C}$, $\mathbf{R}$, $(\mathbf{C}\_p,\mathcal{O}\_{\mathbf{C}\_p})$ for infinitely many $p$. Even so, $\mathbf{C}\_p$ is not locally compact, so I expect the answer to question 2 is "no", no matter how we put it.
| https://mathoverflow.net/users/nan | Adèlic points and algebraic closure | Since $\mathcal X$ is projective, a section is given by finitely many coordinates. If the $i$'th coordinate lies in $\mathbf A\_{F\_i}$ for some extension $F\_i$ of $K$, then all the coordinates lie in $\mathbf A\_{F}$ for $F$ the composition of the $F\_i$.
So $\mathcal X(\mathbf A\_{\overline {K}})$ agrees with the direct limit of $\mathcal X(\mathbf A\_{\overline {F}})$ - at least as sets: I didn't check the topology but I imagine it's fine.
I'm not sure what you mean by an intrinsic definition. Both these definitions seem fine to me.
$\mathcal X(\mathbf A\_{\overline {K}})$ is not compact. Choose a $p$-adic valuation on $\overline{K}$, We have maps $$\mathcal X ( \mathbf A\_{\overline{K}}) \to \mathcal X (\overline{K}\_p) = \mathcal X( \overline{\mathbb Q}\_p) = \mathcal X( \overline{\mathbb Z}\_p) \to \mathcal X ( \overline{\mathbb F}\_p)$$ and the composition is continuous with the discrete topology on $\mathcal X ( \overline{\mathbb F}\_p)$, but $\mathcal X ( \overline{\mathbb F}\_p)$ is infinite.
Restricted to each individual $\mathcal X( \mathbf A\_{F})$, we're restricting the valuation to $F$, mapping to the appropriate completion, clearing denominators (because we are working in a projective variety), and reducing modulo the uniformizer.
| 6 | https://mathoverflow.net/users/18060 | 415850 | 169,507 |
https://mathoverflow.net/questions/415846 | 2 | $\DeclareMathOperator\Vol{Vol}$In 1.7 on p.224 of the following paper, there is a rigidity result for compact manifolds whose sectional curvature is almost $-1$.
[Gromov, M.. Manifolds of negative curvature. J. Differential Geometry 13 (1978), no. 2, 223-230.](https://projecteuclid.org/journals/journal-of-differential-geometry/volume-13/issue-2/Manifolds-of-negative-curvature/10.4310/jdg/1214434487.full)
**Statement:** Given $n \geq 4$ and $C>0$, there is an $\epsilon\_0=\epsilon\_0(n, C)>0$ so that
for any $0<\epsilon<\epsilon\_0$, if a compact
Riemannian manifold $(M^n,g)$ satisfies $-1 \leq K \leq -1+\epsilon$ and the volume $\Vol(M,g) \leq C$, then it must admit a metric of constant negative sectional curvature.
The above result is mentioned as a standard application from the main "non-collapsing" result (in Sec 1.2 on p.223 of the above paper). Roughly speaking, under the above assumption the volume has a lower bound, and the diameter has an upper bound. However I could not figure out the proof of Statement.
Assume by contradiction, we have a sequence of $(M\_i, g\_i)$ whose curvature is $-1<K<-1+\epsilon\_i$ and $\Vol(M\_i, g\_i) \leq C$, but each of $M\_i$ does not admit a hyperbolic metric. then we may apply the $C^{1, \alpha}$ convergence theory of Cheeger-Gromov. Now we get a smooth manifold $N$ and diffeomorphisms $f\_i: N \rightarrow M\_i$ so that the pull back $f\_i^{\ast} g\_i$ converges to a $C^{1, \alpha}$ metric $h$ on $N$ under the $C^{1, \alpha}$ norm. Even if $\alpha$ could be $1$, the convergence is still weaker than $C^2$.
**Question:** Why is $(N, h)$ a hyperbolic metric?
| https://mathoverflow.net/users/12904 | On a closed manifold whose curvature is close to "hyperbolic" | I know two ways to complete the argument.
1. The lower curvature bound is preserved under GH convergence, and so is the upper curvature bound provided there is a lower bound on convexity radius. In the universal cover the convexity radius is infinite. So the universal cover converges to an Alexandrov space of curvature both $\ge -1$ and $\le -1$. Then there is a rigidity result saying that the metric must be Riemannian and hyperbolic. For the latter one usually refers to an old paper of Alexandrov [Uber eine Verallgemeinerung der Riemannschen Geometrie, Schriften Forschungsinst. Math (1957), no. 1, 33–84]. I suspect this is what Gromov had in mind.
2. One can also use smoothing (e.g. by Ricci flow run for short time) to get control of the covariant derivatives of the curvature tensor in the sequence without changing curvature bound very much. This will ensure that the convergence is $C^\infty$, and so the limit is hyperbolic.
EDIT: I take back method 2, but wish to offer a variation on 1. Instead of quoting Alexandrov's paper, (which I must confess I have not read because my German is very poor), one can argue that in the limit there is a weak curvature tensor that satisfies the Einstein equation of the hyperbolic metric, and hence by a standard regularity argument the limiting metric is $C^\infty$, and because it has curvature $\ge -1$ and $\le -1$ in the comparison sense, it is hyperbolic. The weak curvature tensor argument can be extracted from [Convergence of Riemannian manifolds; Ricci and
$L^{n/2}$-curvature pinching
L. Zhiyong Gao
J. Differential Geom. 32(2): 349-381 (1990).](https://projecteuclid.org/journals/journal-of-differential-geometry/volume-32/issue-2/Convergence-of-Riemannian-manifolds-Ricci-and-Lsp-n-2-curvature/10.4310/jdg/1214445311.full)
| 4 | https://mathoverflow.net/users/1573 | 415853 | 169,508 |
https://mathoverflow.net/questions/415794 | 8 | In the smooth setting, Whitney's approximation theorem says the following: If $M,N$ are smooth manifolds and $f,g:M\to N$ are smooth functions that are continuously homotopic (ie there is a continuous homotopy $H:M\times [0,1]\to N$ between them) then they are also smoothly homotopic (ie there exists a smooth homotopy $\tilde{H}$ between them).
Does something similar hold for Lipschitz functions between Lipschitz manifolds? More precisely, **if $M,N$ are two Lipschitz manifolds and $f,g:M\to N$ are Lipschitz functions that are continuously homotopic, is it true that they are also Lipschitz homotopic (ie there exists a Lipschitz homotopy between them)?**
| https://mathoverflow.net/users/351083 | Whitney's approximation theorem for Lipschitz manifolds | For compact Lipschitz manifolds this follows from the main result of the paper
by Liu, Luofei, Yu, Hanfu,
Liu, Ye
"[Converting uniform homotopies into Lipschitz homotopies via moduli of
continuity.](https://www.sciencedirect.com/science/article/abs/pii/S0166864120303205)" Topology Appl. 285 (2020)
But the above paper is surely an overkill. It gives quantitative bounds on the homotopy and qualitative result should be known much earlier.
I expect it's known that for a compact Lipschitz manifold $N$, for some Lipschitz embedding of $N$ into some $\mathbb R^n$ a small neighbourhood of $N$ Lipschitz retracts onto $N$. This would easily imply the result.
It's likely even true that $N$ is a Lipschitz ANR.
It's also clear that there is an appropriate version of this for noncompact manifolds too.
Edit: I found a reference to the fact that any Lipschitz manifold is a Lipschitz ANR. This immediately follows from Theorem 5.1 in [this paper](http://www.acadsci.fi/mathematica/Vol03/vol03pp085-122.pdf).
Therefore what I said above works. It is enough to show that any two sufficiently $C^0$ close Lipschitz maps $f,g:M\to N$ are Lipschitz homotopic.
Embed $N$ by a Lipschitz map into a Euclidean space $\mathbb R^n$, connect $f$ to $g$ by straight line homotopy in $\mathbb R^n$ and then use Lipschitz neighborhood retraction to push this homotopy into $N$.
| 5 | https://mathoverflow.net/users/18050 | 415856 | 169,509 |
https://mathoverflow.net/questions/415863 | 12 | Given a matrix $A\in \operatorname{SL}\_d(\mathbb{Z})$ ([the special linear group](https://en.wikipedia.org/wiki/Special_linear_group)) satisfying the two conditions: (1) no eigenvalue of $A$ is a root of unity, (2) the characteristic polynomial of $A$ is irreducible over $\mathbb{Q}$.
>
> **QUESTION.** Does it follow that at least one eigenvalue $\lambda$ of $A$ fulfills $\vert\lambda\vert>1$? Assuming true, it seems that there must be a theorem of a sort here, but I couldn't recall. It would also be nice if one can relax the conditions to gain the same conclusion, if possible.
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**UPDATE.** After exploring papers by Kronecker (thanks [Terry Tao](https://mathoverflow.net/questions/415863/is-there-an-eigenvalue-of-modulus-larger-than-1#comment1066826_415863)) and others, I realized that we don't quite need "irreducibility" but "monic" is enough.
| https://mathoverflow.net/users/66131 | Is there an eigenvalue of modulus larger than 1? | Since the characteristic polynomial is irreducible, eigenvalues are simple and hence the matrix is $\mathbf{C}$-diagonalizable. If all were on the unit cercle, it would follow that $\{ A^n:n\in\mathbf{Z}\}$ is bounded. But since it is contained in the set of integral points, this would force it to be finite, and hence all eigenvalues would be roots of unity, contradiction.
Hence one is not on the unit circle. Since the product of eigenvalues is $\pm 1$ (the determinant), it follows that some eigenvalue has modulus $>1$.
---
PS: Here's the straightforward way to extend this the general case (without irreducibility). Let $A\in\mathrm{M}\_d(\mathbf{Z})$ be a matrix whose eigenvalues are not only among $0$ and roots of unity. As any integral matrix, we can block-triangulate it (over $\mathbf{Z}$) so that all diagonal blocks are $\mathbf{Q}$-irreducible. Hence at least one diagonal block, with characteristic polynomial $P=P(t)$, also satisfies the assumption. So $P$ is $\mathbf{Q}$-irreducible, and not equal to $t$. If $|P(0)|\ge 2$ then clearly some root has modulus $>1$. Otherwise, $P(0)\in\{\pm 1\}$ and we are in the previous (main) case.
| 23 | https://mathoverflow.net/users/14094 | 415864 | 169,511 |
https://mathoverflow.net/questions/415871 | 2 | If $\{X\_i, i\in I\}$ is a directed system of abelian groups such that we have
$$\varinjlim\_{i\in I}X\_i = 0$$
is it true that for every $i$ and large enough $j\ge i$ the transition map $f\_{i,j} : X\_i\to X\_j$ is the zero map?
If the transition maps satisfy this, then of course the direct limit is zero. If the $X\_i$ are finitely generated abelian groups then the converse is also clearly true. The question is for general abelian groups $X\_i$ (I'm happy to assume the $X\_i$ are all countably generated and that $I=\mathbb{N}$).
| https://mathoverflow.net/users/nan | Transition maps in trivial direct limit | No. Take each $G\_n$ free abelian with basis $e\_k$ with $k\in \mathbb N$. Here $n$ runs over the natural number. Let the map from $G\_n$ to $G\_{n+1}$ kill $e\_0$ and send $e\_k$ to $e\_{k-1}$ for $k>0$. Then no map is zero but each element in each group eventually maps to $0$ far enough down. So the direct limit is 0.
| 3 | https://mathoverflow.net/users/15934 | 415872 | 169,513 |
https://mathoverflow.net/questions/415867 | 2 | Let $X$ and $Y$ be compact Hausdorff spaces and let $\varphi:X\to Y$ be continuous with a property that if $A$ is a nowhere dense zero-set in $Y$, then $\varphi^{-1}(A)$ is nowhere dense in $X$. Let $Z=\varphi(X)$.
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> Does $\varphi$ still have the analogous property as a map into $Z$?
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Note that the condition implies that the set $J\_{Z}$ of $f\in C(Y)$ which vanish on $Z$ is a $\sigma$-ideal in $C(Y)$, which means that it contains all existing supremums of countable sets in $J\_{Z}$. Such $Z$'s could be considered countable analogues of regular closed sets, because $Z$ is regular if and only if $J\_{Z}$ contains existing supremums of all sets.
| https://mathoverflow.net/users/53155 | Is a certain property of a continuous map preserved under "surjectification"? | The answer here is negative: for $Y$ take the remainder $\beta\omega\setminus\omega$ of the Stone-Cech remander of the discrete space $\omega$ of finite ordinals. In the space $Y$ take any countable discrete subspace $D$ and let $Z$ be the closure of $D$. Since $Y$ has no isolated points, the space $D$ is nowhere dense in $Y$ and so is its closure $Z$. Since $D$ is countable and discrete in the compact space $Z$, the remainder $R=Z\setminus D$ is a nonempty functionally closed nowhere dense set in $Z$. Consider the space $X=(Z\times\{0\})\cup (R\times\{1\})$ and the natural projection $\varphi:X\to Z\subseteq Y$.
Observe that the set $R$ is functionally closed and nowhere dense in $Z$ and its preimage $\varphi^{-1}[R]$ contains the nonempty clopen subset $R\times\{1\}$ of $X$.
On the other hand, each nonempty $G\_\delta$-subset of the space $Y=\beta\omega\setminus\omega$ has nonempty interior in $Y$. So, $Y$ contains no functionally closed nowhere dense subsets and hence the function $\varphi:X\to Y$ has the desired property: for every nowhere dense functionally closed set $A$ in $Y$ the preimage $\varphi^{-1}(A)$ has any desired property, in particular is nowhere dense in $X$.
| 2 | https://mathoverflow.net/users/61536 | 415873 | 169,514 |
https://mathoverflow.net/questions/415877 | 2 | Consider a number field $K$ and a finite Galois field extension $L/K$. Let $E$ be an elliptic curve over $K$ and consider the abelian group
$$E(L)\otimes L^{\times}.$$
Every element $g$ in $\text{Gal}(L/K)$ acts on it by $g(P\otimes \alpha) = g(P)\otimes g(\alpha)$ for $P\in E(L)$ and $\alpha\in L^{\times}$.
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> **Problem 1.1** Do we have $(E(L)\otimes L^{\times})^{\text{Gal}(L/K)} = E(K)\otimes K^{\times}$?
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(This question came up when others and I were trying our hand at the proof of the weak Mordell-Weil theorem. It has nothing to do with it, it just came up).
Also, what about $H^1$? That is
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> **Problem 1.2** Is there an analog of Hilbert's Theorem 90 so that $H^1(\text{Gal}(L/K), E(L)\otimes L^{\times})=0$?
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Now write $E(L)\_{tf}$ for the maximal torsion-free quotient of $E(L)$.
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> **Problem 2.1** Do we have $(E(L)\_{tf}\otimes L^{\times})^{\text{Gal}(L/K)} = E(K)\_{tf}\otimes K^{\times}$?
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> **Problem 2.2** Do we have $H^1(\text{Gal}(L/K),E(L)\_{tf}\otimes L^{\times})=0$?
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| https://mathoverflow.net/users/nan | Galois invariants and tensor products | Consider $E:y^2=x^3-3$ over $\mathbb Q$. Then $E(\mathbb Q)$ is trivial, hence so is $E(\mathbb Q)\otimes\mathbb Q^\times$. On the other hand, over $L=\mathbb Q(\sqrt{-3})$ we have a point $P=(0,\sqrt{-3})\in E(L)$ which is mapped to its opposite by the complex conjugation. The same is true of $\sqrt{-3}\in L$, therefore $P\otimes\sqrt{-3}\in(E(L)\otimes L^{\times})^{\text{Gal}(L/K)}$, and we can check this element is nontrivial.
An analogous example for torsion-free parts is given by the point $P=(0,\sqrt{-2})\in E(L)$ for $E:y^2=x^3+x-2$ and $L=\mathbb Q(\sqrt{-2})$. $E(\mathbb Q)\cong\mathbb Z/2$ and $P$ is a point of infinite order, so $P\otimes\sqrt{-2}$ will be nontrivial in $E(L)\_{tf}\otimes L^\times$.
I don't know any result for $H^1$ but I doubt there is anything as simple as Hilbert's 90. As a Galois representation, $E(L)$ can have quite nontrivial structure, even after tensoring with $\mathbb Q$, but I don't have an explicit counterexample at hand. (edit: actually see Chris Wuthrich's comment below though)
| 6 | https://mathoverflow.net/users/30186 | 415879 | 169,517 |
https://mathoverflow.net/questions/415885 | 3 | The generating function of the product of Legendre polynomials for the same $n$ is given by
\begin{aligned}
\sum\_{n=0}^{\infty} z^{n} \mathrm{P}\_{n}(\cos \alpha) \mathrm{P}\_{n}(\cos \beta)&=\frac{\mathrm{F}\left(\frac{1}{2} ,\frac{1}{2} ;1;\frac{4z\sin \alpha \sin \beta }{1-2z\cos (\alpha +\beta )+z^{2}}\right)}{\sqrt{1-2z\cos (\alpha +\beta )+z^{2}}} \\&=\frac{2K\left(\sqrt{\frac{4z\sin \alpha \sin \beta }{1-2z\cos (\alpha +\beta )+z^{2}}}\right)}{\pi \sqrt{1-2z\cos (\alpha +\beta )+z^{2}}}
\end{aligned}
where $P\_n$ is a Legendre polynomial, $F$ is a hypergeometric function and $K$ is a complete elliptic integral of the first kind.
<https://www.researchgate.net/publication/269015726_A_generating_function_for_the_product_of_two_Legendre_polynomials>
However, in the study of quantum physics, we need the similar result as above for following expression:
\begin{align}
\sum\_{n=0}^{\infty} z^{n} \mathrm{P}\_{n}(x) \mathrm{P}\_{n-1}(x)
\end{align}
Is there any result known? Any hints for this problem would be appreciated.
| https://mathoverflow.net/users/476902 | Generating function of the product of Legendre polynomials | Using the recursion relation
$$
P\_{n-1} (x) = x P\_n (x) - \frac{x^2 -1}{n} \frac{d}{dx} P\_n (x) \ ,
$$
you can reduce your expression to a sum of the generating function you quote and a combined derivative (the $d/dx$) and integral w.r.t. $z$ (to generate the $1/n$) thereof.
| 5 | https://mathoverflow.net/users/134299 | 415888 | 169,520 |
https://mathoverflow.net/questions/415886 | 2 | Consider the nonlinear map $F\_i:\mathbb R^2 \to \mathbb R$
$F\_i(x):=\varepsilon^2\langle x, A\_i x\rangle +\varepsilon\langle b\_i,x \rangle + x\_i,$
where $A\_i$ is some matrix and $b\_i$ some vector
Can we uniquely solve the equation $(F\_1(x),F\_2(x))=z\_0 \in \mathbb R^2$ for $\varepsilon>0$ small enough such that $x = \sum\_{n=0}^{\infty} \varepsilon^n z\_n$ where $z\_n$ are some coefficients?
| https://mathoverflow.net/users/150549 | Analytic solution of low-dimensional Riccati equation | This is a particular instance of the analytic implicit function theorem; see e.g. [this MO page](https://mathoverflow.net/questions/73388/analytic-implicit-function-theorem) and further references there.
| 2 | https://mathoverflow.net/users/36721 | 415898 | 169,522 |
https://mathoverflow.net/questions/415897 | 4 | Let $\mathbb{B}$ be an infinite $\sigma$-complete Boolean algebra. By $\mathbb{B}^\omega$ we denote the countable product of $\mathbb{B}$ with the coordinate-wise operations. Let us call a homomorphism $\varphi\colon\mathbb{B}^\omega\to\mathbb{B}$ *generalized limit* if for every sequence $(A\_n)\in\mathbb{B}^\omega$ we have:
$\bigvee\_{n\in\omega}\bigwedge\_{k\ge n} A\_k\le\varphi(A\_n)\le\bigwedge\_{n\in\omega}\bigvee\_{k\ge n}A\_k$.
(I guess it's immediately visible, why $\varphi$ is called *limit*.)
Are you aware of some standard way of constructing in ZFC such generalized limits? I am especially interested in the case of $\mathbb{B}$ being a measure algebra, e.g. $\mathbb{B}=\mathrm{Bor}([0,1])/\mathcal{N}$ (here $\mathcal{N}$ stands for the Lebesgue null ideal), but of course any general study will be useful.
I will appreciate any answer, hint, or reference to a research paper.
| https://mathoverflow.net/users/15860 | Generalized limits in Boolean algebras | If the Boolean algebra $B$ is complete, then I claim that there exists generalized limits. I am going to generalize this answer from limits indexed by $\omega$ to a limit of a net indexed by an arbitrary directed set.
Let $B$ be a complete Boolean algebra. Suppose that $D$ is a directed set. Then
let $I$ be the ideal on $B^{D}$ consisting of all $(b\_{d})\_{d\in D}$ where there exists some $e\in D$ where $b\_{d}=0$ for all $d\geq e.$
Proposition: Suppose that $\varphi:B^{D}\rightarrow B$ is a Boolean algebra homomorphism. Then the following are equivalent:
1. $$\varphi((b\_{d})\_{d\in D})\leq\bigwedge\_{d\in D}\bigvee\_{e\geq d}b\_{e}$$ for each tuple $(b\_{d})\_{d\in D}\in B^{D}$.
2. $$\bigvee\_{d\in D}\bigwedge\_{e\geq d}b\_{e}\leq\varphi((b\_{d})\_{d\in D})$$ for each tuple $(b\_{d})\_{d\in D}\in B^{D}$.
3. $$\varphi(x)=0$$ whenever $x\in I$, and $\varphi((b)\_{d\in D})=b$.
4. There is a Boolean algebra homomorphism $f:B^{D}/I\rightarrow B$ such that $f((b)\_{d\in D}+I)=b$ for all $b\in B$ and where if $\pi\_{I}:B^{D}\rightarrow B^{D}/I$ is the quotient mapping, then
$\varphi=f\pi\_{I}.$
Proof outline: The equivalence between 1 and 2 follows by taking complementation. The equivalence between 3 and 4 follows from the first isomorphism theorem applied to Boolean algebras.
$(1\wedge 2)\rightarrow 3$. If $(b\_{d})\_{d\in D}\in I$, then there is some $d\in D$ where $b\_{e}=0$ for $e\geq d$. Therefore,
$$\varphi((b\_{d})\_{d\in D})\leq\bigvee\_{e\geq d}b\_{e}=0.$$
Furthermore, we have $$\varphi((b)\_{d\in D})\leq\bigwedge\_{d\in D}\bigvee\_{e\geq d}b=b\leq\bigvee\_{d\in D}\bigwedge\_{e\geq d}b\leq\varphi((b)\_{d\in D}),$$ so
$\varphi((b)\_{d\in D})=b.$
$(3\wedge 4)\rightarrow 1.$ Let $\varphi,\pi\_{I},f$ be the mappings stated in the proposition where $\varphi=f\pi\_{I}$.
Let $(b\_{d})\_{d\in D}\in B^{D}$, and suppose that $d\_{0}\in D$. Let $b=\bigvee\_{e\geq d\_{0}}b\_{e}$. Then
$\pi\_{I}(b\_{d})\_{d\in D}\leq\pi\_{I}((b)\_{d\in D})$. Therefore,
$$\varphi(b\_{d})\_{d\in D}=f\pi\_{I}(b\_{d})\_{d\in D}\leq f\pi\_{I}((b)\_{d\in D})
=\varphi((b)\_{d\in D})=b=\bigvee\_{e\geq d\_{0}}b\_{e}.$$
We conclude that
$$\varphi(b\_{d})\_{d\in D}\leq\bigwedge\_{d\_{0}\in D}\bigvee\_{e\geq d\_{0}}b\_{e}.$$
Q.E.D
The mappings $f:B^{D}/I\rightarrow B$ such that $f((b)\_{d\in D}+I)=b$ for all $b\in B$ can be obtained using Sikorski's extension theorem, so just let $\varphi=f\pi\_{I}$ to obtain your generalized limits.
| 2 | https://mathoverflow.net/users/22277 | 415920 | 169,532 |
https://mathoverflow.net/questions/415922 | 3 | Let $ X $ be an affine, normal, $ \mathbb{G}\_{a} $-variety with action $ \beta $ over a field $ k $ of characteristic zero. The action of $ \mathbb{G}\_{a} $ is obtained from $ \delta \in \operatorname{Der}\_{k}(\mathcal{O}(X)) $ via
\begin{align\*}
\beta^{\sharp}(f) =\exp(t\delta)(f)\\
= \sum\_{j=0}^{\infty} \frac{\delta^{j}(f)t^{j}}{j!}.
\end{align\*}
The plinth ideal is the ideal of $ \mathcal{O}(X)^{\mathbb{G}\_{a}} $ generated by all elements $ h $ such that there exists a $ g \in \mathcal{O}(X) $ such that $ \delta(g)=h $. The first paper I could find with this name usage was one by Dufresne and Kraft ([Invariants and Separating Morphisms for Algebraic Group Actions](https://doi.org/10.1007/s00209-015-1420-0)). Does anyone have any earlier references with this name usage or concept?
| https://mathoverflow.net/users/470753 | Does anyone know the first times the term plinth ideal of a $ \mathbb{G}_{a} $-representation was used? | G. Freudenburg, in [Algebraic theory of locally nilpotent derivations](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.470.10&rep=rep1&type=pdf) introduces the term "plinth ideal" in a way that suggests it was not used before, on page 10, with the footnote:
*The term plinth commonly refers to the base of a column or statue.*
S. Kuroda, in [arXiv:1412.1598](https://arxiv.org/pdf/1412.1598.pdf) writes that *"the notion of plinth ideal already appeared in Nakai (1978), although not called by this name."*
| 2 | https://mathoverflow.net/users/11260 | 415923 | 169,533 |
https://mathoverflow.net/questions/415890 | 1 | Given a simple complex Lie group $G$ (I might say upfront that I am mostly interested with exceptional Lie algebras) and a nilpotent orbit $\mathcal{O}\subset G$ I would like to describe the intersection of $\mathcal{O}$ with Levi subgroups (or probably, their derived groups) and Pseudo-Levi subgroups.
I guess that there is a way to take the weighted Dynkin diagram of $\mathcal{O}$ (or the nilpotent orbit associated to it in the Lie algebra) and construct the weighted Dynkin diagram of the intersection.
As an example to what I'm trying to figure out, I went to Birne Binegar's UMRK database (at <http://umrk.dynns.com:800/UMRK/UMRK.html>) and looked at the data for Nilpotent orbits of the Lie algebra of type $E\_7$.
For the orbit $A\_4+A\_1$ it says that the producing pseudo-Levi subalgebra is generated by the roots [0, 1, 2, 3, 5, 6, 7] (so the negative highest roots together with roots number 1,2,3,5,6,7 in the Dynkin diagram, enumeration as in Bourbaki I believe). The producing pseudo-Levi subalgebra means here (as far as I understand) a standard pseudo-Levi subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ such that the intersection of $A\_4+A\_1$ with $\mathfrak{h}$ is distinguished.
My question is, what orbit of $\mathfrak{h}$ is the intesection?
In particular, I wish to know if and how can I find a subalgebra such that the intersection is the principal orbit.
A closely related follow-up question is, can I then go about determining the stabilizer of the orbit from this data (say, the stabilizer of the intersection in $\mathfrak{h}$ and somethineg like $\mathrm{Aut}(\mathrm{Dyn}(\mathfrak{h}))$?
A bonus question, if someone happens to know: The UMRK attaches a further integer to the pseudo-Levi subalgebra on top of the list of roots (it's 4 in the case), what does this integer represents?
| https://mathoverflow.net/users/64702 | Unipotent orbits and intersection with Levi and pseudo-Levi subgroups | You can't always find a pseudo-Levi such that the intersection is a principal = regular orbit, but you can always find a **Levi** subalgebra such that the intersection is distinguished, i.e. the connected centraliser in the derived subgroup is unipotent. This is the Bala-Carter theorem.
Now, there is an improvement to the Bala-Carter theorem due to Sommers, see "A generalisation of the Bala-Carter theorem for nilpotent orbits" in IMRN in 1998 (can't put my hands on the precise reference at the moment). We consider simple groups of adjoint type. For any nilpotent element $e$ of ${\mathfrak g}$ let $A(e)=Z\_G(e) /Z\_G(e)^\circ$. Each coset in $A(e)$ has a semisimple representative $s$; $s$ generates $Z(L)/Z(L)^\circ$ where $L$ is the centraliser of $s$ (a pseudo-Levi subgroup); finally, this induces something close to a bijection between conjugacy classes in $A(e)$ and pseudo-Levi subalgebras of ${\mathfrak g}$ (up to conjugacy) which contain $e$ (perhaps modulo some issues with graph automorphisms). In particular, you can improve on "$e$ is distinguished in a Levi subalgebra" by instead saying "$e$ is semiregular in a pseudo-Levi subalgebra". Here semiregular means the centraliser is connected and unipotent (a stronger condition than distinguished).
For almost all cases, semiregular will imply regular, but e.g. I think from memory that there are three semiregular orbits in type $E\_6$.
| 1 | https://mathoverflow.net/users/26635 | 415926 | 169,535 |
https://mathoverflow.net/questions/415903 | 0 | Let $(X,d)$ be a metric space and $m$ be a Borel measure on $(X,d)$. The measure $m$ is called Ahlors regular if $m(B(x,r))\asymp r^q$ for some $q>0$ and each $x\in X$. Is there a name for measures satisfying the following relaxation:
There is some strictly monotone continuous increasing function $\omega:[0,\infty]\rightarrow [0,\infty]$ with $\omega(0)=0$ and $\omega(\infty)=\infty$ satisfying:
$$
m(B(x,r)) \leq \omega(r);
$$
what about in the simple case where $\omega(r)=r^q$ for some $q>0$?
| https://mathoverflow.net/users/36886 | Terminology "upper" Ahlfors regular measure | In the case $\omega(r) = Cr^q$:
* the terminology "upper Ahlfors ($q$-)regular" is certainly in use. See, e.g., <https://arxiv.org/pdf/1803.04819.pdf> .
* the terminology "($q$-)Frostman measure" is also used, referring to [Frostman's lemma](https://en.wikipedia.org/wiki/Frostman_lemma). See, e.g., [https://www.maths.ed.ac.uk/~jazzam/posts/2020-01-14-frostmann/](https://www.maths.ed.ac.uk/%7Ejazzam/posts/2020-01-14-frostmann/) .
In the case of a general $\omega$, I personally do not know of a name. I might call it some kind of "upper mass bound".
| 0 | https://mathoverflow.net/users/476947 | 415929 | 169,537 |
https://mathoverflow.net/questions/415904 | 2 | Let $A$ be a finite abelian group, and $\sigma$ a non-trivial action of $A$ on $\mathbb{C}$ by real algebra automorphisms. In particular, using $\sigma$, we can view $\mathbb{C}^{\times}$ as a module over $A$. Let $\xi$ be a 2-cocycle representing a class in $H^2(A;\mathbb{C}^{\times}\_{\sigma})$. We can consider the real algebra $\mathbb{C}^{\xi}\_{\sigma}\lbrack A\rbrack$ with multiplication given by $$z\_1a\_1\cdot z\_2a\_2 = (\xi(a\_1,a\_2)z\_1\sigma(a\_1)(z\_2)) a\_1a\_2.$$
How many real irreducible representations does this algebra have? Alternatively, one could try to count the quaternionic irreducible representations? (The center of this algebra is not too hard to describe explicitly.)
**EDIT:** By a real irreducible representation, I mean a representation $V$ of $\mathbb{C}^{\xi}\_{\sigma}\lbrack A\rbrack$ over $\mathbb{R}$ such that $End\_{\mathbb{C}^{\xi}\_{\sigma}\lbrack A\rbrack}(V)\cong \mathbb{R}$.
| https://mathoverflow.net/users/105094 | Real representations of twisted group algebras | The algebra $A:=\mathbb{C}\_\sigma^\xi[G]$ is the universal $\mathbb{R}$-algebra that
1. contains $\mathbb{C}$ and symbols $\{u\_g \mid g\in G\}$ such that
2. $u\_g u\_h = \xi(g,h) u\_{gh}$ and $u\_g z u\_g^{-1} = {^g z}$ holds, where I use ${^g z}$ as a shorthand for $\sigma(g)(z)$.
This is an example of a "crossed" $G$-graded $\mathbb{R}$-algebra, i.e. an algebra that is equipped with a decomposition $A=\bigoplus\_g A\_g$ such that $A\_g A\_h\subseteq A\_{gh}$ (that "$G$-graded") and $A\_g\cap A^\times\neq\emptyset$ (that's "crossed"). An ordinary group algebra $K[G]$ is an example, a twisted group algebra $K^\xi[G]$ is also an example. Your example is a little bit different than those two, because the degree-1-piece is 2-dimensional over the field, not the field itself. One particularly easy example is the $C\_2$-graded crossed $\mathbb{R}$-algebra $\mathbb{H}=\mathbb{C}\oplus\mathbb{C}j$ (which is also a special case of your assumptions for $G=C\_2$)
Crossed, $G$-graded algebras are nice to have, because you can do Clifford theory with them: The degree-1-piece behaves very much like $K[N]$ behaves inside $K[G]$ for any normal subgroup $N\unlhd G$. In fact: That's one example - Take any $G$-graded algebra and define a $Q:=G/N$-grading by setting $A\_{gN} := \sum\_{h\in N} A\_{gh}$. If you do that with $A=K[G]$, then the degree-1-piece in the $Q$-grading is exactly $K[N]$.
In your example, let's have a look at the normal subgroup $N:=\ker(\sigma)$ of index 2. The degree-1-piece of the $Q$-grading on $A$ is the twisted algebra $A\_N:=\mathbb{C}^\xi[N]$ (just considered as an $\mathbb{R}$-algebra). This is one complication fewer and assuming you understand such algebras well enough, one can count the representations of $A$ by using Clifford theory.
The grading group acts by conjugation on modules of the degree-1-piece. The appropriate generalisation of Clifford's theorem now tells you that for any simple $A$-module $V$, the restriction $V\_{|N}$ of $V$ to $A\_N$ is semisimple and the occurring simple constituents are a single $Q$-orbit that all occur with the same multiplicity $e=e\_V$.
This leaves us with only three cases
1. $U=V\_{|N}$ is itself simple.
2. $V\_{|N} \cong U\oplus U$ for some simple $U$ and conjugation switches the two copies of $U$.
3. $V\_{|N} \cong U\_1\oplus U\_2$ for two non-isomorphic, but conjugated simple modules $U\_1, U\_2$.
In that case $I\_{U\_i} = 1$ and $V$ is obtained by induction from either constituent, i.e. $V=\operatorname{Ind}\_1^Q(U\_i) = A \otimes\_{A\_N} U\_i$. Conversely: The induction of every simple $A\_N$-module $U$ with $I\_U=1$ is a simple $A$-module whose restriction back to $A\_N$ splits into $U$ and its conjugate.
To count real and quaternionic representation, note that $Q$ also acts via conjugation on $\operatorname{End}\_{A\_N}(X)$ for all $A\_N$-modules $X$ and that $\operatorname{End}\_A(X)$ is precisely the space of $Q$-fixed points of this action.
Consider in particular $X=V\_{|N}$ and note that $\operatorname{End}\_{A\_N}(U) = \mathbb{C}$, because $\mathbb{C}\subseteq Z(A\_N)$ so that every module is automatically a $\mathbb{C}$-vector space and the $A\_N$-action is by $\mathbb{C}$-linear maps, so that we can apply Schur's lemma for finite-dimensional algebras over the complex numbers.
Let's look at $X:=V\_{|N}$ in the three cases:
1.) Then $\operatorname{End}\_{A\_N}(X) = \mathbb{C}$. The only possible actions of $Q$ are the trivial action and complex conjugation. $V$ cannot be quaternionic and $V$ is real iff it is the latter.
2.) Then $\operatorname{End}\_{A\_N}(X) = \mathbb{C}^{2\times 2}$. If $Q$ acts $\mathbb{C}$-linearly, then it must act by conjugation by Skolem-Noether. If it acts semilinearly, then it acts by conjugation followed by complex conjugation. Because it switches the two copies of $U$, the conjugating matrix must be of the form $X=\begin{pmatrix}0&\ast\\\ast&0\end{pmatrix}$. The linear case gives $\mathbb{C}$, the semilinear case $\mathbb{H}$ as fixed point space.
3.) Then $\operatorname{End}\_{A\_N}(X) =\operatorname{End}\_{A\_N}(U\_1)\times\operatorname{End}\_{A\_N}(U\_2)=\mathbb{C} \times \mathbb{C}$. The conjugation permutes $U\_1$ and $U\_2$ and so must flip the two copies of $\mathbb{C}$. Again, because there are only two $\mathbb{R}$-automorphisms of $\mathbb{C}$, we only have a limited number of ways, this can happen:
$$(z,w) \mapsto \begin{cases} (w,z) \\ (\overline{w},z) \\ (w,\overline{z}) \\ (\overline{w},\overline{z}) \end{cases}$$
with fixed point space
$$\begin{cases}
\{(z,z) \mid z\in\mathbb{C}\} \\
\{(r,r) \mid r\in\mathbb{R}\} \\
\{(r,r) \mid r\in\mathbb{R}\} \\
\{(z,\overline{z}) \mid z\in\mathbb{C}\}
\end{cases}
$$
In the first and fourth case, $V$ is neither real nor quaternionic. In the second and third case, $V$ is real.
To conclude: The number of real representations of $A$ is the number of simple $A\_N$-modules that are $G/N$-invariant, extend to $A$-modules with the above $Q$-action, plus the number of simple $A\_N$-modules that are not $G/N$-invariant and have the right $Q$-action, divided by two
| 4 | https://mathoverflow.net/users/3041 | 415931 | 169,539 |
https://mathoverflow.net/questions/415935 | 4 | Consider all Pythagorean triangles $a^2 + b^2 = p^2$ in which the hypotenuse $p$ is a prime number. Let $h(x) = \sum\_{p \le x}p^2$, $a(x) = \sum\_{p \le x}ab$ and $r(x) = \sum\_{p \le x}(a+b)^2$. Is it true that:
$$
\lim\_{x \to \infty}\frac{h(x)}{r(x)} = \frac{\pi}{2+\pi}
$$
$$
\lim\_{x \to \infty}\frac{a(x)}{r(x)} = \frac{1}{2+\pi}
$$
| https://mathoverflow.net/users/23388 | Relation between $\pi$, area and the sides of Pythagorean triangles whose hypotenuse is a prime number | **Q:** Is $\lim\_{x \to \infty}\frac{h(x)}{r(x)} = \frac{\pi}{2+\pi} $ ?
**A:** use that $r(x)=h(x)+2a(x)$, hence
$$\frac{h(x)}{r(x)} = \frac{h(x)/a(x)}{2+h(x)/a(x)}$$
and
$$\lim\_{x\rightarrow\infty}\frac{h(x)}{a(x)}=\pi$$
in view of <https://math.stackexchange.com/a/3481801/87355>
| 6 | https://mathoverflow.net/users/11260 | 415948 | 169,542 |
https://mathoverflow.net/questions/415895 | 2 | **Definition.** A semigroup $X$ is called *$E$-separated* if for any distinct idempotents $x,y\in X$ there exists a homomorphism $h:X\to Y$ to a semilattice $Y$ such that $h(x)\ne h(y)$.
Observe that $X$ is $E$-separated if and only if the smallest semilattice congruence on $X$ is idempotent-separating. This seems to be an important notion, so I suggest that it could (and should) be studied in Theory of Semigroups, maybe under some different name. *Do you know any suitable references?*
| https://mathoverflow.net/users/61536 | $E$-separated semigroups | I finally found an answer to my own question: by an old (nontrivial) result of [Putcha and Weissglass](https://msp.org/pjm/1971/39-1/p21.xhtml), a semigroup $X$ is $E$-separated if and only if it is viable.
A semigroup $X$ is *viable* if for any elements $x,y\in X$ with $\{xy,yx\}\subseteq E(X)$ we have $xy=yx$.
However, up to my taste, the notion of viality is a bit less intuitive comparing to the equivalent notion of $E$-separatedness.
Another difference between those (equivalent) notions is that the viality is an internal property whereas the $E$-separatedness is external (defined with the help of external objects). But the $E$-separatedness also can be equivalently defined in internal terms: a semigroup is $E$-separated if the smallest semilattice congruence on $X$ separates idempotents.
Neither $E$-separatedness not the viablity are present in [this Wikipedia page](https://en.wikipedia.org/wiki/Special_classes_of_semigroups).
However they form an important class of semigroups.
| 1 | https://mathoverflow.net/users/61536 | 415962 | 169,545 |
https://mathoverflow.net/questions/415970 | 17 | For $n\in \mathbb{N}$ let $S\_n$ denote the set of permutations (bijections) $\pi: \{0,\ldots,n-1\}\to \{0,\ldots,n-1\}$. A *transposition* swaps exactly $2$ elements and is often denoted by $(i \; k)$ if $i\neq k\in\{0,\ldots,n-1\}$ are the elements being swapped.
For $n\in\mathbb{N}$ let $E\_n$ be the number of elements of $S\_n$ that can be obtained by a composition of all the transposition, such that every transposition is used *exactly once.*
What is the value of $\lim\sup\frac{E\_n}{n!}$?
| https://mathoverflow.net/users/8628 | Fraction of $S_n$ reachable by using every transposition once as $n\to\infty$? | The value is $1/2$.
The problem can be reformulated as follows: How to sort a non-sorted list $(a\_1,...,a\_n)$ that is a permutation of $\{1 .. .n\}$ with the [parity](https://en.wikipedia.org/wiki/Parity_of_a_permutation) the same as $n \choose 2$ by exactly one transposition between every pair of indices?
In this answer the phrase "unsorted element" means an element $a\_p$ that is not equal to $p$.
This is solved by a recursive manner: Suppose $n\geq 4$ and $a\_p\neq p$. Then there's a sequence of transpositions of $a\_p$ with all the other elements such that $a\_p=p$ after the transpositions and there exists an unsorted element in the list. Thus, we can proceed to the subproblem of finding such a sequence on the list with $a\_p$ removed. This recursion stops at $n=3$, but by the parity assumption and the unsorted element assumption, there must be exactly two unsorted elements left, and it's easy to sort them by three transpositions.
Now we only need to prove the existence of such transpositions. We may assume that $a\_1 \neq 1$, $a\_n=1$ and we need to achieve $a\_1=1$ after the transpositions.
Let $(1,2), (1,3), ..., (1,n)$ be the sequence of transpositions. Then $a\_1$ will be equal to $1$ after applying the transpositions in order. If there are unsorted elements, we are done; otherwise the sequence $(1,2), (1,3), ..., (1,n-1), (1,n-2), (1,n)$ will give an unsorted element.
| 25 | https://mathoverflow.net/users/125498 | 415978 | 169,551 |
https://mathoverflow.net/questions/415982 | 4 | [Szőkefalvi-Nagy's theorem](https://en.wikipedia.org/wiki/Sz.-Nagy%27s_dilation_theorem) says the following: if $A$ is a contraction on a Hilbert space $H$, then there exists a unitary $U$ on a Hilbert space $H'\supset H$ for which $A^n=P\_HU^nP\_H$ for all $0\le n$. ($P\_H$ denotes projection onto $H$.)
If $H$ is finite-dimensional and we only need this to be true for $0\le n\le N$ then this can be modified such that $H'$ is also finite-dimensional (cf. [Levy and Shalit - Dilation theory in finite dimensions: the possible, the impossible and the unknown](https://arxiv.org/abs/1012.4514)).
Ando's theorem (see e.g. Levy–Shalit) says the following: if $A$ and $B$ are commuting contractions on $H$, then there exist commuting unitaries $U$ and $V$ on $H'\supset H$ for which $A^nB^m=P\_HU^nV^mP\_H$ for all $0\le n,m$.
The construction in the proof is naturally infinite-dimensional even if $H$ is finite-dimensional so a natural question (Problem C in Levy–Shalit) is: if we only need this to be true for $0\le n+m\le N$ then does an analogous modification work to make $H'$ finite-dimensional?
Levy–Shalit make this out to be wide-open, as of 2011. **My question here is if there are any recent developments in the years since then on this or related problems.**
| https://mathoverflow.net/users/159965 | Status of finite-dimensional Ando's theorem | In their 2013 paper "[Unitary $N$-dilations for tuples of commuting matrices](https://doi.org/10.1090/S0002-9939-2012-11714-9)" McCarthy and Shalit showed that there does indeed exist an $N$-dilation to a finite-dimensional space (see Theorem 1.2 in the paper).
By the way, I found this paper by taking a look at the recent survey article "[Dilation theory: a guided tour](https://zbmath.org/?q=an%3A7393048)" (2021) by Shalit ([link to arXiv](https://arxiv.org/abs/2002.05596)).
Since you're apparently interested in dilation theory, you might also be interested to know that Orr Shalit has a blog [Noncommutative analysis](https://noncommutativeanalysis.wordpress.com/) where he sometimes writes about dilation theory. For instance, the survey article that I mentioned above is announced in the blog post [A survey (another one!) on dilation theory](https://noncommutativeanalysis.wordpress.com/2020/02/14/a-survey-another-one-on-dilation-theory/).
| 4 | https://mathoverflow.net/users/102946 | 415985 | 169,552 |
https://mathoverflow.net/questions/392467 | 4 | I am currently looking at a few simple properties of the Witt ring of a field $K$ (by which I mean the ring of Witt classes of quadratic forms, not the ring of Witt vectors), which are clearly true when the Pythagoras number of $K$ is $1$, and clearly false when it is $3$, but I am not sure what can happen when it is $2$. The two properties I am looking at are:
1. The torsion subgroup of $W(K)$ has exponent (at most) $2$.
2. For any $a,b\in K^\times$ which are sums of squares, the Pfister form $\langle\langle a,b\rangle\rangle$ is hyperbolic.
When $K$ is Pythagorean, 1. is true because $W(K)$ is either $\mathbb{Z}/2\mathbb{Z}$ or is torsion-free, and 2. is obvious since $a$ and $b$ are actually squares by hypothesis.
When the Pythagoras number is at least $3$, we can choose some $a\in K^\times$ which is a sum of squares, but not of $2$ squares. Then $\langle \langle a,a\rangle\rangle = \langle \langle -1,a\rangle\rangle$ is not hyperbolic, which already shows that 2. is false. But $a$ is positive in any real closure of $K$, so $\langle \langle a\rangle\rangle$ becomes hyperbolic in any real closure, and thus by Pfister's local-global theorem $\langle \langle a\rangle\rangle$ is a torsion element in $W(K)$. Since $2\langle \langle a\rangle\rangle = \langle \langle -1,a\rangle\rangle\neq 0$ in $W(K)$, this shows that 1. is also false.
Are those properties [true/false/it depends] when the Pythagoras number is $2$?
| https://mathoverflow.net/users/68479 | Witt ring of a field with Pythagoras number $2$ | Let $p(K)$ be the Pythagoras number of $K$. Suppose that $p(K) = 2$.
If $K$ is a formally real field, then the torsion subgroup of $W(K)$ has exponent 2. If $K$ is a nonreal field, then this is false.
For example, let $K$ be the finite field with $q$ elements where $q \equiv 3 \bmod 4$. Then $p(K) = 2$, but $W(K)$ has exponent 4 because $<1>$ has order 4 in $W(K)$. These results can be found in Lam's textbook on quadratic forms.
Statement (2) is false in general when $p(K) = 2$, but this is more difficult to show. Here are two examples. $K = \mathbb{R}((x,y))$ and
$K = \mathbb{R}((x))(y)$. Proofs of these statements can be found in
K.J. Becher, D.B. Leep, The length and other invariants of a real field, Math. Zeit., Volume 269 (2011), 235-252. See Example 6.6.
Some additional explanation is needed to understand this reference. Statement (2) of the question is equivalent to the condition $(I\_t(K))^2 = 0$ where $I\_t(K)$ is the set of torsion elements in the ideal of even dimensional forms in $W(K)$.
(Note that $(I\_t(K))^n$ is not the same as the more commonly studied object $I\_t^n(K)$, which is the torsion part of $(IK)^n$.) It is shown in Proposition 4.6 of this paper that $\ell(K) \le 2$ if and only if $(I\_t(K))^2 = 0$ where $\ell(K)$ denotes the length of the field, which is defined in section 4.
This explanation of some of the basic notations and terminology of this paper will make it easier to understand Example 6.6 mentioned above.
| 1 | https://mathoverflow.net/users/476997 | 415999 | 169,557 |
https://mathoverflow.net/questions/415934 | 1 | It is well-known that if $O$ is an orthogonal map, then $\Delta u(Ox) = \Delta u$ where $\Delta$ is the Laplacian.
Now, let $A$ be a constant invertible matrix, then we define the weighted Laplacian
$$\Delta\_A = \langle \nabla, A \nabla \rangle.$$
My question is: Does there exist a function $\varphi\_O$ such that $Gf(x):=e^{\varphi\_O(x)}f(Ox)$ satisfies for all $f$ but fixed orthogonal $O$
$$\Delta\_A Gf(x) = G(\Delta\_A f)(x)?$$
| https://mathoverflow.net/users/150564 | Orthogonal invariance of (weighted) Laplacian | $\newcommand{\De}{\Delta}$The answer is: not in general.
Indeed, suppose the contrary. Let
\begin{equation}
A:=\begin{pmatrix}
0&1/2\\
1/2&0
\end{pmatrix},
\end{equation}
so that
\begin{equation}
(\De\_A f)(s,t)=\frac{\partial^2 f(s,t)}{\partial s\,\partial t}=f^{(1,1)}(s,t).
\end{equation}
Let
\begin{equation}
O:=\begin{pmatrix}
0&-1\\
1&0
\end{pmatrix}
\end{equation}
and
\begin{equation}
g:=\varphi\_O,
\end{equation}
so that
\begin{equation}
(Gf)(s,t):=e^{g(s,t)}f(-t,s),
\end{equation}
and the difference between the right-hand side and left-hand side of the equality in question
divided by $e^{g(s,t)}$ is
\begin{equation}
\begin{aligned}
d\_f(s,t)&:=\frac{(G(\De\_A f))(s,t)-(\De\_A(Gf))(s,t)}{e^{g(s,t)}} \\
&=\frac{e^{g(s,t)}f^{(1,1)}(-t,s)
-\dfrac{\partial^2 f(s,t)}{\partial s\,\partial t}
\big(e^{g(s,t)}f(-t,s)\big)}{e^{g(s,t)}} \\
&=-f^{(0,1)}(-t,s) g^{(0,1)}(s,t)+f^{(1,0)}(-t,s) g^{(1,0)}(s,t)
+2 f^{(1,1)}(-t,s) \\
&-f(-t,s)
\left(g^{(0,1)}(s,t) g^{(1,0)}(s,t)+g^{(1,1)}(s,t)\right)=0
\end{aligned}
\end{equation}
for all $f\in C^1$ and all real $s,t$.
Letting now $f\_1(s,t):=s$, $f\_2(s,t):=t$, and $f\_4(s,t):=t^2$, we get
\begin{equation}
0=s d\_{f\_2}(s,t)-d\_{f\_4}(s,t)=s g^{(0,1)}(s,t),
\end{equation}
so that $g^{(0,1)}=0$ and hence $g^{(1,1)}=0$, which implies $0=d\_{f\_1}=g^{(1,0)}$.
So, $g$ is a constant, and hence
$0=d\_f(s,t)=2f^{(1,1)}(-t,s)$ for all $f\in C^1$ and all real $s,t$, which is absurd. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 416005 | 169,559 |
https://mathoverflow.net/questions/416010 | 3 | Let say I have two $n$ x $n$ matrices $A$ and $B$ where all elements are real positive values. I want to find some $n$ x $n$ permutation matrix $P$ such that $\operatorname{tr}(P A P ^T B)$ is minimized. Does there exist such an algorithm or technique?
| https://mathoverflow.net/users/477004 | Algorithm to minimize $\operatorname{tr}(PAP^TB)$? | This problem is NP-hard.
Let $A$ be the adjacency matrix of an $n$-cycle plus the all-ones matrix and $B$ the adjacency matrix of a graph $G$ plus the all-ones matrix.
Then, if $G$ has a Hamiltonian cycle, the maximum trace is achieved when the elements of $PAP^T$ with value $2$ corresponds to a Hamiltonian cycle in $G$. Since finding Hamiltonian cycles is NP-hard, your problem is NP-hard.
| 5 | https://mathoverflow.net/users/125498 | 416020 | 169,562 |
https://mathoverflow.net/questions/415987 | 4 | Usually, I like working with determinants related to the Vandermonde matrix, i.e.
$$\det(x\_j^{i-1})=\prod\_{i<j}(x\_j-x\_i).$$
However, I run into some unusual matrix and its determinant. Define the $(2n)\times (2n)$ matrix $\mathbb{M}\_{2n}$ with entries
$$\mathbb{M}\_{2n}(i,j)=
\begin{cases}
(x\_j-x\_i)(x\_i-x\_{2n}) \qquad \text{if $i<j$}, \\
(x\_j-x\_i)(x\_j-x\_{2n}) \qquad \text{if $i\geq j$}.
\end{cases}$$
>
> **QUESTION.** Is this true?
> $$\det\mathbb{M}\_{2n}=(x\_1-x\_2)^2(x\_2-x\_3)^2\cdots(x\_{2n-1}-x\_{2n})^2(x\_{2n}-x\_1)^2.$$
>
>
>
| https://mathoverflow.net/users/66131 | A determinant of perfect square polynomials | The answer is **yes** for any $n \geq 2$. (The $n=1$ case should be treated separately.)
The determinant, as a polynomial, contains $(x\_m−x\_{m+1})$ $(m=1,2,...,2n-1)$ and $(x\_{2n}-x\_1)$ as factors, because setting $x\_m=x\_{m+1}$ will make the $m$th and $(m+1)$th column of $\mathbb{M}\_{2n}$ equal, and setting $x\_{2n}=x\_1$ will make the $(2n)$th and first column of $\mathbb{M}\_{2n}$ equal, making the determinant zero.
Since [the determinant of a skew-symmetric matrix can always be written as the square of a polynomial in the matrix entries that only depend on the size of the matrix](https://en.wikipedia.org/wiki/Pfaffian), the determinant contains the factor $(x\_1−x\_2)^2(x\_2−x\_3)^2⋯(x\_{2n−1}−x\_{2n})^2(x\_{2n}−x\_1)^2$. As the determinant cannot have degree higher than that of this factor, the only thing left is to plug in values of $x\_k$ such that LHS=RHS≠0 in the last equation stated in the question.
If we let $x\_k=1$ if $k$ is odd and $0$ otherwise, then
$\mathbb{M}\_{2n}(i,j)=
\begin{cases}
\phantom{-}1 \qquad \text{if $i<j$},i \text{ odd}, j \text{ even} \\
-1 \qquad \text{if $i > j$}, i \text{ even}, j \text{ odd}\\
\phantom{-}0 \qquad \text{otherwise}
\end{cases}$.
The $(2n-1)$th row has only one nonzero element, namely $\mathbb M\_{2n-1,2n}=1$, and the $(2n-1)$th column has only one nonzero element, namely $\mathbb M\_{2n,2n-1}=-1$. By applying the [Leibniz formula for determinants](https://en.wikipedia.org/wiki/Leibniz_formula_for_determinants), we see that $\det \mathbb M\_{2n}=\det \mathbb M\_{2n-2}$. By mathematical induction, we have $\det \mathbb{M}\_{2n}=1$ for any $n$. As the RHS also evaluates to $1$, we claim that LHS=RHS for any $x\_k$s.
| 6 | https://mathoverflow.net/users/125498 | 416028 | 169,565 |
https://mathoverflow.net/questions/416032 | 5 | This question is inspired by [Maximal compact subgroup of $\mathrm{SL}(2,\mathbb{H})$](https://mathoverflow.net/questions/195060/maximal-compact-subgroup-of-mathrmsl2-mathbbh). Consider the embedding $\operatorname{U}(n,\mathbb{H})\subset \operatorname{GL}(n,\mathbb{H}) $. Since $\operatorname{U}(n,\mathbb{H})\cong \operatorname{Sp}(n) $ is almost simple, we know that it is actually contained in $\operatorname{SL}(n,\mathbb{H}) $, the kernel of the Dieudonné determinant $\ \det: \operatorname{GL}(n,\mathbb{H}) \rightarrow \mathbb{R}^\*\_{+}\,$. Is there a direct way to prove $\det A=1$ for $A\in \operatorname{U}(n,\mathbb{H}) $?
(Recall that $\ \det\, {}^{t}\overline{A}\neq \det A\ $ for $A$ general in $\operatorname{GL}(n,\mathbb{H}) $.)
| https://mathoverflow.net/users/40297 | Why is $\operatorname{U}(n,\mathbb{H})\subset \operatorname{SL}(n,\mathbb{H}) $? | Yes, relying on the fact that elements of the symplectic group (over $\mathbf{C}$, and hence over $\mathbf{R}$) have determinant 1.
Indeed, an element $g$ of $\mathrm{U}(n,\mathbf{H})$ preserves the canonical Hermitian form $b$. Let $b'$ be any imaginary component of $b$. Then $b'$ is a $g$-invariant real-valued symplectic form on the $4n$-dimensional real space $\mathbf{H}^n$. So the determinant of $g$ as $4n$-dimensional matrix (which is the Dieudonné determinant) is $1$.
**Edit** (abx): actually the determinant $\det\_{\mathbb{R}}(g)$ of $g$ as a real $4n$-dimensional matrix is $(\det\_{D}( g))^4$, where $\det\_D$ is the Dieudonné determinant — since $\det\_D(g)\in\mathbb{R}^\*\_+$, this implies $\det\_D(g)=1$ as required.
Indeed $\det\_{\mathbb{R}}$ is a homomorphism from $\operatorname{GL}(n,\mathbb{H}) $ to $\mathbb{R}^\*$; such a homomorphism is trivial on the derived subgroup $\operatorname{SL}(n,\mathbb{H}) $. Since $\det\_D$ induces an isomorphism of
$\operatorname{GL}(n,\mathbb{H}) /\operatorname{SL}(n,\mathbb{H}) $ onto $\mathbb{R}^\*\_+$, we have $\det\_{\mathbb{R}}=\varphi \circ \det\_{D}$, where $\varphi :\mathbb{R}^\*\_+\rightarrow \mathbb{R}^\*$ is a continuous homomorphism. Such a homomorphism is of the form $x\mapsto x^{a}$ for some $a\in\mathbb{R}$; thus $\det\_{\mathbb{R}}= (\det\_{D})^{a}$. Taking $g= tI\_n$ gives $a=4$.
| 4 | https://mathoverflow.net/users/14094 | 416036 | 169,569 |
https://mathoverflow.net/questions/416007 | 4 | In his book *Introduction to Infinity-Categories,* Land in his Theorem 3.3.16 asserts an equivalence of $\infty$-categories where one of the categories $\mathrm{LFib}(\mathcal C)$ is the full subcategory of $(\mathrm{Cat}\_\infty)\_{/\mathcal C}$ on vertices that are left fibrations $?\to\mathcal C$ ($\mathcal C$ some $\infty$-category). My understanding of this material is that he is following Riehl & Verity’s 2018 article *The comprehension construction.* There, in Notation 6.1.12, is defined a category $\mathrm{coCart}(\mathrm{qCat})\_{/\mathcal C}$, and my understanding is that Land’s $\mathrm{LFib}(\mathcal C)$ should be the full subcategory of $\mathrm{coCart}(\mathrm{qCat})\_{/\mathcal C}$ on vertices which are left fibrations with target $\mathcal C$. The definition of $\mathrm{coCart}(\mathrm{qCat})\_{/\mathcal C}$ makes use of the slice $\infty$-cosmos defined in Riehl-Verity’s *Elements*, Proposition 1.2.22 (here our $\infty$-cosmos is $\mathrm{qCat}$). Per that definition, it seems to me that morphisms between left fibrations $h:\mathcal E\to\mathcal C$ and $f:\mathcal F\to\mathcal C$ in $\mathrm{coCart}(\mathrm{qCat})\_{/\mathcal C}$ are simply functors $\mathcal E\to\mathcal F$ over $\mathcal C$. But 1-simplices in $(\mathrm{Cat}\_\infty)\_{/\mathcal C}$ are the data of a morphism $g:\mathcal E\to\mathcal F$ and a 1-simplex in $\operatorname{Fun}(\mathcal E,\mathcal C)$ between $fg$ and $h$ which is an equivalence in $\operatorname{Fun}(\mathcal E,\mathcal C)$. So these don’t seem like the same categories to me. Can someone familiar with these sources help me reconcile these two definitions?
| https://mathoverflow.net/users/37110 | How to define the $\infty$-category of left fibrations? | First let me point out a small typo : it should be functors over $\mathcal C$ *which send cocartesian edges to cocartesian edges* (this doesn't matter for left fibrations, but for cocartesian fibrations it does).
For your main point, you are right that they are different quasi-categories: the simplicial sets are not isomorphic; but they're the same $\infty$-category, that is, the obvious forgetful functor from one to the other (so the one that uses the trivial $1$-simplex) is a categorical equivalence.
The fact that it is essentially surjective is obvious (the objects are the same), so the point is really about mapping spaces. But this is really a statement about pullbacks, namely the following:
>
> Consider three quasi-categories $C,D,E$ with functors $p: C\to E, q:D \to E$. Suppose $q: D\to E$ is a cocartesian fibration. In this situation, the quasi-category $Fun\_E(C,D)$ defined as a simplicial set by the (ordinary) pullback $Fun(C,D)\times\_{Fun(C,E)}\{p\}$ is equivalent to the homotopy pullback, $Fun(C,D)\times\_{Fun(C,E)}(Fun(C,E)^\simeq)\_{/p}$
>
>
>
Note that the first one is the category of those functors $f :C\to D$ that are strictly over $E$, that is, with $q\circ f = p$, whereas $0$-simplices in the latter are a pair $(f, \sigma)$ where $f : C\to D$ is a functor and $\sigma: q\circ f \to p$ is an equivalence, and you can check that in fact it is what you're thinking of.
The reason for this equivalence is that $Fun(C,D)\to Fun(C,E)$ is also a cocartesian fibration, so in particular a fibration in the Joyal model structure, and so pull-back along it preserves weak equivalences. It then suffices to observe that the inclusion $\{p\}\to (Fun(C,E)^\simeq)\_{/p}$ is an equivalence, because $Fun(C,E)^\simeq$ is a Kan complex.
The idea is that because $D\to E$ is a cocartesian fibration, you have "enough room" to "strictify" any equivalence $q\circ f\simeq p$ to an actual equality $q\circ f = p$ up to changing $f$.
(note that the latter, the one with $q\circ f\simeq p$, is the one that you would want to have as an $\infty$-category, because $=$ is too strict a notion. It just happens that in this case you can actually strictify)
| 3 | https://mathoverflow.net/users/102343 | 416040 | 169,573 |
https://mathoverflow.net/questions/415793 | 6 | A famous conjecture in topology asserts:
*The Euler characteristic of a closed aspherical $2n$-manifold $M$ satisfies $(-1)^n\chi(M) \geq 0$*.
This was conjectured by Hopf for manifolds with non-positive sectional curvature and (much later) by Thurston for all aspherical manifolds.
By the classification of surfaces, it holds in dimension $2$.
I am curious about the status of this conjecture in dimension 4.
It seems as if the special case for manifolds with negative sectional curvature was settled by Milnor, as explained in this paper by [Chern](https://link.springer.com/content/pdf/10.1007/BF02960745.pdf). Undoubtedly, the more general conjecture by Thurston is much younger. I wonder if the general version has some kind of reformulation in less manifoldy and more algebraic terms. More precisely, I am curious about the following.
* For which fundamental groups of aspherical $4$-manifolds is the $4$-dimensional Thurston conjecture settled?
One approach to the Thurston conjecture that I am aware of is the employment of $\ell^2$ invariants: the Singer conjecture [implies](https://people.math.osu.edu/davis.12/papers/singer-hopf.pdf) the Hopf conjecture in general. However, the Singer conjecture is open.
| https://mathoverflow.net/users/14233 | Status of the Hopf-Thurston sign conjecture in dimension 4 | There has been a lot of work on cases of this conjecture connected to Coxeter groups. M. Davis and R. Charney made a conjecture that comes from these cases in 1995 in The Euler characteristic of a non-positively curved piecewise Euclidean manifold. Also
have a look at the article Vanishing theorems and conjectures for the $\ell^2$-homology of right-angled Coxeter groups by M Davis and B Okun. They prove the case when $M^4$ has a non-positively curved piecewise Euclidean cubical structure. This is a far larger class than it sounds: every flag triangulation of the 3-sphere gives rise to such a manifold via the Davis construction.
Davis-Okun also give a conjecture for the $\ell^2$-homology of groups that act properly cocompactly on contractible manifolds that extends the Singer conjecture, which is equivalent to the case when the group acts freely properly cocompactly.
| 8 | https://mathoverflow.net/users/124004 | 416043 | 169,575 |
https://mathoverflow.net/questions/416049 | 4 | In my recent [MO question](https://mathoverflow.net/questions/415987/a-determinant-of-perfect-square-polynomials), Darij Grinberg mentioned a closely related (structure-wise) determinant, that is,
$$\det\left(x\_{\min\{i,j\}}\right)\_{i,j}^{1,m}=x\_1(x\_2-x\_1)(x\_3-x\_2)\cdots(x\_m-x\_{m-1}).$$
In particular, $\det(\min\{i,j\})=1$. This prompted me to ask:
>
> **QUESTION.** Is this true? The characteristic polynomial of the matrix $\mathbf{N}\_m=(\min\{i,j\})\_1^m$ equals
> $$P\_m(\lambda)=\sum\_{k=0}^m(-1)^{m-k}\binom{2m-k}k\,\lambda^k.$$
>
>
>
**Remark 1.** It's worth pointing out that
$$P\_m(\lambda^2)=(-1)^m\lambda^{2m}\,U\_{2m}\left(\frac1{2\lambda}\right),$$
where $U$ is [*Chebyshev polynomial of the second kind.*](https://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html)
**Remark 2.** Needless to say, $P\_m(\lambda)$ implies a formula for all elementary polynomials $\mathbf{e}\_j$ of the roots of characteristic polynomial for the given matrix $\mathbf{N}\_m$.
**Remark 3.** As a "fun" aside, the entries of the matrix $\mathbf{N}\_m$ offer a systematic way of calculating the total number of squares (of all sizes) in an $m\times m$ grid; i.e.
$$\sum\_{i,j=1}^m\min\{i,j\}=1^2+2^2+\cdots+m^2.$$
| https://mathoverflow.net/users/66131 | Characteristic polynomial of a simple matrix: Chebyshev? | Yes, the characteristic polynomial is given by $(-1)^m U\_{2m}(1/2\lambda ) \lambda^{2m} $.
The inverse matrix is given by $$\begin{pmatrix} 2 & -1 & 0 & 0 & \dots \\ -1 & 2 & -1 & 0 & \dots \\ 0 & -1 & 2 & -1 & \dots \\ 0 & 0 & -1 & 2 & \dots % \\ 0 & 0 & 0 & -1 & \dots
\\ \dots & \dots & \dots & \dots & 1\end{pmatrix}$$
where the $1$ in the bottom
-right corner denotes that the bottom-right entry (and no other diagonal entry) is a $1$. (This can be seen by writing your matrix as $A^T A$, where $A$ has $1$s on the diagonal and upper triangle and $0$s in the lower triangle, and taking the inverse of $A$.)
The characteristic polynomial is $Q\_m(\lambda) + Q\_{m-1}(\lambda)$, where $Q\_m$ is the characteristic polynomial of the $m \times m$ matrix with $2$s on the diagonal, $-1$s adjacent to the diagonal, and $0$s elsewere.
Laplace expansion gives $$Q\_m(\lambda)= (\lambda-2) Q\_{m-1}(\lambda) - Q\_{m-2} (\lambda)$$ which gives the generating function $$\sum\_{m=0}^{\infty} Q\_m(\lambda) t^m = \frac{1}{ 1 - (\lambda -2) t + t^2} $$ so the characteristic polynomial of this matrix has the generating function $$\frac{1+t }{ 1 - (\lambda -2) t + t^2}.$$
Since the determinant is $1$, we can obtain the characteristic polynomial of the inverse matrix by substituting $\lambda^{-1}$ for $\lambda$ and multiplying by $(-\lambda)^m$, i.e. substituting $(-\lambda t)$ for $t$, getting
$$\frac{1 - \lambda t }{ 1+ (1 -2\lambda ) t + \lambda^2 t^2}$$
as the generating function for the characteristic polynomial of your matrix.
The Chebyshev polynomial has the generating function $$ \sum\_{n=0}^{\infty} U\_n(x) t^n = \frac{1}{ 1- 2x t+ t^2}$$ so
$$ \sum\_{m=0}^{\infty} U\_{2m}(x) t^{2m}=\frac{1}{2} \left( \frac{1}{ 1- 2x t+ t^2} + \frac{1}{ 1+ 2xt + t^2}\right) = \frac{ 1 + t^2 } { 1 + 2t^2 + t^4 - 4 x^2 t^2 } $$
and thus
$$ \sum\_{m=0}^{\infty} (-1)^m U\_{2m}(1/2\lambda ) \lambda^{2m} t^{2m} = \frac{ 1 - \lambda^2 t^2 } { 1 - 2\lambda^2 t^2 + \lambda^4 t^4 + t^2 } $$
which is the same after substituting $\lambda^2$ for $\lambda$ and $t^2$ for $t$.
| 8 | https://mathoverflow.net/users/18060 | 416053 | 169,578 |
https://mathoverflow.net/questions/416022 | 1 | **Background:**
I am reading the paper: *Best constant in Sobolev inequality* by Talenti (see [here](https://math.jhu.edu/%7Ejs/Math646/talenti.sobolev.pdf)) and I am trying to understand the following step.
On p. 365, the author is arguing that the solutions to the following equation
$$\left(r^{m-1}\left|u^{\prime}\right|^{\dot{p}-1} \operatorname{sgn} u^{\prime}\right)^{\prime}+C r^{m-1}|u|^{q-1} \operatorname{sgn} u=0 \quad(C=\text { a positive constant })$$
take the form $(a+br^p)^{1-m/p}.$ Here $m$ is the dimension of the space, $q=\frac{2m}{m-2}$ and $1<p<m.$ The above ODE corresponds to the radial solutions for the Euler Lagrange equation associated to the Sobolev inequality.
He begins by considering the case $p=2$. Then we have,
$$\left(r^{m-1}\left|u^{\prime}\right| \operatorname{sgn} u^{\prime}\right)^{\prime}+C r^{m-1}|u|^{q-1} \operatorname{sgn} u=0.\label{1}\tag{28}$$ By considering $u(r)=r^{1-m/2}v(r)$ we obtain the following ODE for $v$,
$$r\left(r v^{\prime}\right)^{\prime}=(1-m / 2)^{2} v-C v^{q-1}.$$
By multiplying the above ODE by $v$ and integrating this is equivalent to solving,
$$\left(r v^{\prime}\right)^{2}=(1-m / 2)^{2} v^{2}-(2 C / q) v^{q}+\text { constant }.\label{2}\tag{$\ast$}$$
From this the author concludes that all the solutions of \eqref{1} that are positive decreasing and satisfy the following decay conditions
$$u(r)=o(r^{1-m/2}),u'(r)=o(r^{-m/2})$$
must be of the form $u(r)=(a+br^2)^{1-m/2}$ for constants $a$ and $b$ related to $C$.
**Question:** I am not sure how looking at the ODE \eqref{2} allows the author to deduce that the solutions must take the form $u(r)=(a+br^2)^{1-m/2}$?
| https://mathoverflow.net/users/68232 | Extremizers of the Sobolev inequality | Take any extremizer $u$ as in the [paper you linked](https://math.jhu.edu/%7Ejs/Math646/talenti.sobolev.pdf).
From the conditions $u(r)=r^{1-m/2}v(r)$, $u(r)=o(r^{1-m/2})$, and $u'(r)=o(r^{-m/2})$ (as $r\to\infty$), we deduce $v(r)=r^{m/2-1}u(r)$, $v(r)=o(1)$, and $v'(r)=(m/2-1)r^{m/2-2}u(r)+r^{m/2-1}u'(r)=o(1/r)$.
So, it follows from your ODE $(\*)$ that the unnamed constant there is $0$, so that
\begin{equation\*}
(r v'(r))^2=(1-m/2)^2 v(r)^2-(2C/q) v(r)^q. \tag{$\*\*$}
\end{equation\*}
Since $v$ is decreasing, we have $v'\le0$ and hence
\begin{equation\*}
r v'(r)=-\sqrt{(1-m/2)^2 v(r)^2-(2C/q) v(r)^q}. \tag{$\*\*\*$}
\end{equation\*}
Note that $2=p<m$, and hence $m>2$.
Take any real $r\_\*>0$ and then any real $U\_\*>0$ such that $U\_\*^2-4c\_1 r\_\*^2>0$, where
\begin{equation\*}
c\_1:=\frac C{m(m-2)}.
\end{equation\*}
Let then
\begin{equation\*}
a:=a(r\_\*):=\frac{U\_\*-\sqrt{U\_\*^2-4c\_1 r\_\*^2}}2\quad\text{and}\quad
b:=b(r\_\*):=\frac{U\_\*+\sqrt{U\_\*^2-4c\_1 r\_\*^2}}{2r\_\*^2},
\end{equation\*}
so that $a$ and $b$ are positive real numbers, solving the system of equations $a+br\_\*^2=U\_\*$ and $abm(m-2)=C$.
Moreover, it is now straightforward to check that, for such $a$ and $b$, the formulas
\begin{equation\*}
v\_\*(r)=r^{m/2-1}u\_\*(r)\quad\text{and}\quad u\_\*(r)=(a+br^2)^{1-m/2} \tag{1}
\end{equation\*}
define a solution $v\_\*$ of ODE $(\*\*\*)$ on $(0,\infty)$ with the "initial" condition $u\_\*(r\_\*)=U\_\*^{1-m/2}$.
Recall that the extremizer $u$ is a solution of $(\*\*\*)$. Moreover, condition (25b) in that paper implies that $u(r)=o(r^{1-m/2})$ as $r\downarrow0$, and hence
$U(r)/r\to\infty$ as $r\downarrow0$, where
\begin{equation\*}
U(r):=u(r)^{1/(1-m/2)}.
\end{equation\*}
So, for any small enough real $r\_\*>0$, choosing at this point $U\_\*:=U(r\_\*)$, we see that the condition $U\_\*^2-4c\_1 r\_\*^2>0$ holds.
So, the functions $u$ and $u\_\*$ are both solutions of $(\*\*\*)$ on $(0,\infty)$ satisfying the same "initial" condition: $u\_\*(r\_\*)=U\_\*^{1-m/2}=u(r\_\*)$. So, by the uniqueness, $u=u\_\*$, as desired.
| 3 | https://mathoverflow.net/users/36721 | 416058 | 169,580 |
https://mathoverflow.net/questions/416054 | 5 | Let $E$ be an elliptic curve over $\mathbb Q$. Let $F$ be the rational function field of $E$.
The $K\_2$ group of $F$ [may be described](https://en.wikipedia.org/wiki/Algebraic_K-theory#Matsumoto%27s_theorem) by elements in $F^\times ⊗\_\mathbb{Z} F^\times$ quotiented by the relations $\langle f ⊗ (1 − f)\rangle (f ∈ F^\times, f\neq 0,1)$.
For $P ∈ C(\bar {\mathbb Q})$ and $\{g, h\}\in K\_2(F)$, the tame symbol is defined as
$$T\_P (\{g, h\}) = (−1)^{\text{ord}\_P(g)\text{ord}\_P(h)}\frac{g^{\text{ord}\_P (h)}}{h^{\text{ord}\_P (g)}}(P)$$
The tame kernel $K\_2^T(E)$ is the subgroup of $K\_2(F)$ whose elements $\{g,h\}$ satisfies $T\_p(\{g,h\})=1$ for every $P ∈ E(\bar {\mathbb Q})$.
**Question:** How to compute\* the torsion of
* $K\_2(E)$, or
* $K\_2^T(E)/K\_2(\mathbb Q)$?
\*It's not known whether the torsion of the groups above are finitely generated. So it could be possible that there is no algorithm to compute the torsion because the torsion is infinite. Having this in mind, the word "compute" should be interpreted as follows:
* Is there an algorithm to determine whether a symbol $\{g,h\}$ is trivial in one of the K-groups?
* What is the most efficient way to find as many as possible torsion elements in one of the K-groups?
| https://mathoverflow.net/users/125498 | Computation of the torsion of K-groups related to elliptic curves | The most efficient way I know to detect whether an element of $K\_2(E)$ is torsion is to use the elliptic dilogarithm. This relies however on a conjecture, and it is not an exact method, in the sense that it uses floating-point arithmetic.
Consider the map
\begin{align\*}
\beta : F^\times \otimes F^\times & \to \mathbb{Z}[E(\overline{\mathbb{Q}})] \\
g \otimes h & \mapsto \sum\_{P,Q \in E(\overline{\mathbb{Q}})} \mathrm{ord}\_P(f) \mathrm{ord}\_Q(g) [P-Q].
\end{align\*}
Bloch has defined a regulator map
\begin{equation\*}
\mathrm{reg}\_\infty \colon K\_2(E) \to \mathbb{R},
\end{equation\*}
which can be computed using the elliptic dilogarithm $D\_E : E(\mathbb{C}) \to \mathbb{R}$. Namely, given $x \in K\_2(E)$, whose image in $K\_2^T(E)$ is written as $\sum\_i \{g\_i,h\_i\}$, we have (up to a constant factor)
\begin{equation\*}
\mathrm{reg}\_\infty(x) = \sum\_i D\_E(\beta(g\_i \otimes h\_i))
\end{equation\*}
where $D\_E$ is extended by linearity by defining $D\_E(\sum n\_P [P]) = \sum n\_P D\_E(P)$. The elliptic dilogarithm can be computed very rapidly: writing $E(\mathbb{C}) = \mathbb{C}^\times/q^{\mathbb{Z}}$, one has $D\_E([x]) = \sum\_{n \in \mathbb{Z}} D(xq^n)$. Here $D$ is the Bloch-Wigner dilogarithm, implemented e.g. in PARI/GP as $\texttt{polylog(2,x,2)}$. Since $|q|<1$, the series for $D\_E$ converges exponentially fast.
One should also take into account the bad places of $E$. For each prime $p$, there is a residue map $K\_2(E) \to K'\_1(\mathcal{E}\_p)$, where $\mathcal{E}\_p$ is the fiber at $p$ of the minimal regular model of $E$ over $\mathbb{Z}$. It turns out that $V\_p := K'\_1(\mathcal{E}\_p) \otimes \mathbb{Q}$ is nonzero precisely when $E$ has split multiplicative reduction at $p$, in which case $\operatorname{dim}(V\_p)=1$. It is also possible to compute the residue map, see Bloch, Grayson, [$K\_2$ and $L$-functions of elliptic curves - Computer calculations](http://math.stanford.edu/%7Econrad/BSDseminar/refs/BlochK2.pdf) and Rolshausen, Schappacher, [On the second $K$-group of an elliptic curve](http://irma.math.unistra.fr/%7Eschappa/NSch/Publications_files/1998b_KRNSch.pdf).
Now, it is conjectured that the extended regulator map
\begin{equation\*}
\mathrm{reg} \colon K\_2(E) \to \mathbb{R} \oplus \bigoplus\_p (V\_p \otimes \mathbb{R})
\end{equation\*}
is an isomorphism after tensoring $K\_2(E)$ with $\mathbb{R}$. In particular, and conjecturally, an element $x$ of $K\_2(E)$ is torsion precisely when it is in the kernel of the extended regulator map.
If the image of $x$ appears numerically to be 0, then one may try to ascertain that $x$ is torsion in $K\_2^T(E)$ by finding Steinberg relations (this becomes a linear algebra problem in the group $F^\times \otimes F^\times$). On the other hand, if the image appears to be nonzero, this can in principle be proved by computing with enough accuracy.
Regarding the torsion in $K\_2(E)$ and $K\_2^T(E)/ K\_2(\mathbb{Q})$, it is known that $K\_2(\mathbb{Q})$ embeds in $K\_2(E)$ by means of the structural morphism $E \to \operatorname{Spec} \mathbb{Q}$, since this morphism has a section. The group $K\_2(\mathbb{Q})$ is infinitely generated (for a description, see Milnor's book *Introduction to algebraic $K$-theory*). So the torsion of $K\_2(E)$ is infinite. I don't know about the torsion of $K\_2^T(E)/ K\_2(\mathbb{Q})$ in general, but Goncharov and Levin have given a complex computing $K\_2^T(E)/K\_2(k)$ for an elliptic curve $E/k$, at least when $k$ is algebraically closed, see Theorem 1.5 in [Zagier's conjecture on $L(E,2)$](https://arxiv.org/abs/alg-geom/9508008). For $k=\overline{\mathbb{Q}}$, this shows (if my computation is correct) that $K\_2^T(E)/K\_2(\overline{\mathbb{Q}})$ contains a copy of the group $E(\overline{\mathbb{Q}})\_{\mathrm{tors}}$.
| 5 | https://mathoverflow.net/users/6506 | 416069 | 169,582 |
https://mathoverflow.net/questions/415784 | 4 | Let $\mathbf{F}$ be a discrete Fourier transform (DFT) matrix such that
\begin{align}
F\_{m,n}=e^{-j2\pi(m-1)(n-1)/N},\quad m,n=1,\ldots,N.
\end{align}
What we can say about the singular value decomposition (SVD) of truncated DFT matrix (the following matrix)?
\begin{align}
\tilde{\mathbf{F}} =
\begin{bmatrix}
\mathbf{I}\_k,\mathbf{0}
\end{bmatrix}
\mathbf{F}
\begin{bmatrix}
\mathbf{I}\_k\\\mathbf{0}
\end{bmatrix},
\end{align}
where $\mathbf{I}\_k$ is the identity matrix of size $k<N$.
| https://mathoverflow.net/users/68835 | Singular value decomposition of truncated discrete Fourier transform matrix | Let me insert a factor $N^{-1/2}$, so that the Fourier transform is unitary:
$$U\_{mn}=N^{-1/2}e^{-2\pi i(m-1)(n-1)/N},\quad m,n=1,\ldots,N.$$
We truncate the $N\times N$ matrix $U$ to the $k\times k$ upper left corner,
$$U^{(k)}\_{mn}= N^{-1/2}e^{-2\pi i(m-1)(n-1)/N},\quad m,n=1,\ldots,k\leq N.$$
A characterisation of the singular values of $U^{(k)}$ is given in [The Eigenvalue Distribution of Discrete Periodic Time-Frequency Limiting Operators](https://arxiv.org/abs/1707.05344) and in [The Future Fast Fourier Transform?](https://epubs.siam.org/doi/pdf/10.1137/S1064827597316266).
>
> **Of order $k^2/n$ of the singular values are close to unity and
> $k-k^2/n$ are close to 0.**
>
>
>
singular values squared of $U^{(k)}$ for $N=1024$, $k=256$.
**Comment:** Actually a total of $\max(0,2k-N)$ of the singular values of $U^{(k)}$ are *precisely* equal to 1, as Noam Elkies was kind enough to explain to me [here.](https://mathoverflow.net/q/416100/11260)
| 3 | https://mathoverflow.net/users/11260 | 416071 | 169,583 |
https://mathoverflow.net/questions/416047 | 3 | I encountered the following problem in one of my research projects which can be encapsulated as follows. Let's say we have a set $\mathcal{C}$ of functions $f$ defined from $\mathbb R\_+$ to $\mathbb R$, and we have two functionals $A = A(f) \,\colon\, \mathcal{C} \to \mathbb{R}\_+$ and $B = B(f) \,\colon\, \mathcal{C} \to \mathbb{R}\_+$. Assume that we have the following inequality
$$ A + (\lambda + \gamma)^2 + \lambda^2 \leq 2\,\sqrt{B + 2f(0)(\lambda - \gamma) + \lambda^2 + \gamma^2}\,\sqrt{A + (\lambda + \gamma)^2 + \lambda^2} - \left(\gamma - f(0)\right)^2 \tag{1},$$ which holds for every $\lambda,\gamma \in \mathbb R$. Additional, we also know that $f^2(0) \leq \min\{\frac{3}{4}A, \frac{1}{3}B \}$ (so $f(0)$ is not really "free"). The problem is to find the smallest fixed (or universal) constant $C > 0$ for which $$ A \leq C\,B \tag{2}$$ holds independently of the choice of $f \in \mathcal{C}$. For instance, one baby special case is when $\lambda = \gamma = 0$, and we can easily see that $(2)$ holds with $C = 4$. So my question boils down to what is the best possible constant in (2) that one can deduce from (1) by tuning $\lambda$ and $\gamma$ while keeping in mind that we have upper bounds on $f^2(0)$ in terms of $A = A(f)$ and $B = B(f)$?
---
Remark: I think I have obtained a solution myself and maybe I can post my own solution later.
| https://mathoverflow.net/users/163454 | Obtaining the "best possible" inequality by tuning hyper-parameters | $\newcommand{\ga}{\gamma}$Letting $\ga\to\infty$ (with $A,B,\lambda,f(0)$ fixed), we see that the left-hand side of your inequality
\begin{equation}
\begin{aligned}
&A + (\lambda + \gamma)^2 + \gamma^2 \\
&\leq 2\,\sqrt{B + 2f(0)(\lambda - \gamma) + \lambda^2 + \gamma^2}\,\sqrt{A + (\lambda + \gamma)^2 + \gamma^2} \\
&- \left(\gamma - f(0)\right)^2
\end{aligned}
\tag{1}
\end{equation}
is $\sim2\ga^2$, whereas the right-hand side of your inequality is $\sim(2\sqrt2-1)\ga^2<2\ga^2$.
So, there are no values of $A,B,\lambda,f(0)$ such that your inequality (1) holds for all real $\ga$.
Therefore, the inequality $A\le CB$ -- as well as, e.g., the inequality $A>CB$ -- hold for all real $C$ and all (actually nonexistent) $A,B$ such that your inequality (1) holds for some $\lambda,f(0)$ and all real $\ga$.
| 3 | https://mathoverflow.net/users/36721 | 416073 | 169,584 |
https://mathoverflow.net/questions/416056 | 4 | This question is motivated by a real-world application related to an art project that involves displaying images, but my search hit a dead end after finding the wikipage about [**Kirkman systems**](https://en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem) (other related terms include [**Steiner systems**](https://en.wikipedia.org/wiki/Steiner_system) and the [**Social golfer problem**](https://en.wikipedia.org/wiki/Social_golfer_problem)) and looking over references linked there. A few people have written programs for this specific question that established lower bounds (so: more than $2100$) but none has found an exact answer.
**The question is:**
>
> Given a set of $40$ elements, what is the maximum number of subsets, each with $4$ elements, that can be created such that no triple appears more than once?
>
>
>
(For example, if the set includes $A,B,C,D,E$ as elements, then one cannot include in the collection of subsets *both* $\{A,B,C,D\}$ *and* $\{A,B,C,E\}$ since, in this scenario, we would have the triple $A,B,C$ appearing more than once.)
As an excerpt from the [**History**](https://en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem#History) section of the aforelinked wikipage, there is under the first bullet point a question attributed to Wesley Woolhouse (1844):
>
> "Determine the number of combinations that can be made out of $n$ symbols, $p$ symbols in each; with this limitation, that no combination of $q$ symbols, which may appear in any one of them shall be repeated in any other."
>
>
>
followed by the formula:
$$\frac{n!}{q!(n-q)!)} \div \frac{p!}{q!(p-q)!}$$
Unfortunately, one finds that this formula is *false* already for small examples (e.g. $n=5, p=4, q=3$) and, indeed, reading further on that page indicates that the formula only holds in certain scenarios.
Rephrased, I am looking for an answer to Woolhouse's question for the case of $n=40, p=4, q=3$ either by a counting argument, an effective program, or a reference. Please tag/retag as appropriate; thanks!
| https://mathoverflow.net/users/22971 | Enumerating subsets with no triple appearing together more than once | In fact, Wikipedia article on [Steiner systems](https://en.wikipedia.org/wiki/Steiner_system) that you linked already provides an answer to your question:
>
> An $S(3,4,n)$ is called a **Steiner quadruple system**. A necessary and sufficient condition for the existence of an $S(3,4,n)$ is that $n \equiv 2\ \text{or}\ 4 \pmod6$.
>
>
>
Notice that $40\equiv 4\pmod{6}$, and so Steiner system $S(3,4,40)$ does exist. It is formed by $\frac{\tbinom{40}{3}}{\tbinom43} = 2470$ blocks (quadruples) that contain every triple *exactly* once.
Actual blocks of such a system can be seen in the La Jolla Covering Repository at [this link](https://ljcr.dmgordon.org/show_cover.php?v=40&k=4&t=3).
| 9 | https://mathoverflow.net/users/7076 | 416074 | 169,585 |
https://mathoverflow.net/questions/370332 | 4 | Murthy numbers, in a given base, are positive integers, such as 2009 in base 10, which are not relatively prime to their reversal, that is, the number written backwards (in base 10 such numbers are [AO71249](http://oeis.org/A071249) in the OEIS).
In base 10, numbers from 8432 to 8440 are all Murthy numbers. Are there arbitrarly long runs of consecutive numbers in base 10 all of which are Murthy numbers? In other bases?
| https://mathoverflow.net/users/60732 | Runs of consecutive numbers all of which are Murthy numbers | Let $m+1, \dots, m+n$ be a sequence of $n$ consecutive Murthy numbers such that each $m+i$ shares with its reversal $\overline{m+i}$ a prime factor $p\_i\equiv 3\pmod4$ such that 10 is a quadratic nonresidue modulo $p\_i$. We will show how to construct such a sequence of $n+1$ consecutive Murthy numbers.
Define $t := p\_1\cdots p\_n(10^{\frac12\mathrm{lcm}(p\_1-1,\dots,p\_n-1)}+1)$ and notice that the product $p\_1\cdots p\_n$ divides both $t$ and its reversal. (Any smaller $t$ with this property will also do the job.)
The new sequence will have the form:
$$t\cdot (10^k + 1) 10^l + m, \dots, t\cdot (10^k + 1) 10^l + m+n,$$
where integers $k,l$ (larger than the length of $t$ and $m+n$) are to be determined.
First we notice that the last $n$ numbers in this sequence are Murthy since $t\cdot (10^k + 1) 10^l + m+i$ shares with its reversal the same prime factor $p\_i$.
So it remains to enforce Murthyness on $t\cdot 10^k + m$. Let $q\equiv 3\pmod{4}$ be a prime having $10$ as a primitive root. We require that both $t\cdot (10^k+1)10^l + m$ and its reversal are divisible by $q$, that is
$$\begin{cases}
t\cdot 10^{k+l} + t\cdot 10^l + m \equiv 0\pmod{q},\\
\overline{m}\cdot 10^{d+k+l} + \overline{t}\cdot (10^k+1)\equiv 0\pmod{q},
\end{cases}
$$
where $d$ is the difference in decimal lengths between $t$ and $m$. This system can be solved by first eliminating the terms $10^{k+l}$ and expressing $10^l$ in terms of $10^k$, and then obtaining a quadratic equation w.r.t. $10^k$. If it's not solvable, we can change the value of $q$ to make it solvable. Then the values of $k,l$ are obtained by taking discrete logarithms (thanks to $10$ being a primitive root modulo $q$).
---
**Example.** For $m=8434$ and $n=3$, we have $p\_1=7$, $p\_2=3$, $p\_3=11$, and we can take $t=1617$. Then the system is solvable for $q=29$ with solutions $(10^l,10^k)\equiv (12,20)\pmod{29}$ or $(10^l,10^k)\equiv (8,16)\pmod{29}$. Correspondingly, $(l,k)\equiv (21,12)\pmod{28}$ or $(l,k)\equiv (5,16) \pmod{28}$. The latter produces the following sequence of consecutive Murthy numbers:
$$1617\cdot (10^{16}+1)\cdot 10^5 + 8434 + i,\qquad i=0,1,2,3.$$
P.S. We should have infinite supply of primes having 10 as a primitive root by [Artin's conjecture](https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots).
| 3 | https://mathoverflow.net/users/7076 | 416079 | 169,586 |
https://mathoverflow.net/questions/416090 | 8 | In [this recent](https://arxiv.org/abs/2004.04550) preprint, the authors construct a certain uncountable family of non-finitely presented FP groups. Recall that group is an FP group if the trivial $\mathbb Z[G]$-module $\mathbb Z$ has a finite projective resolution by finitely generated $\mathbb Z[G]$-projectives.
In fact, if I understand correctly, the family that they construct is a family of groups of cohomological dimension $2$; but this specific point does not seem to be the focus of their paper. Furthermore, theirs is not the first example of a non-finitely presented FP group, and so it leads me to my question:
>
> Were there earlier examples of non finitely-presented FP groups of cohomological dimension $2$ ?
>
>
>
I want to be liberal with the meaning of the word "example": my question is specifically whether there was an earlier proof that such things existed, whether or not some explicit examples were given.
(NB : this is not quite my field of research, so maybe this is (very?) classical : any pointers to classical literature on the topic where this question is discussed would be helpful)
| https://mathoverflow.net/users/102343 | Non-finitely presented FP groups with cohomological dimension $2$ | The Bestvina-Brady construction of non-finitely presented groups of type FP produces groups of cohomological dimension two. Bestvina-Brady groups are parametrized by finite flag simplicial complexes. The Bestvina-Brady group is of type FP iff the flag complex is acyclic (=has the same ordinary homology as a point), is finitely generated iff the flag complex is connected, is finitely presented iff the flag complex is simply connected. For an acyclic flag complex the cohomological dimension of the Bestvina-Brady group is equal to the dimension of the flag complex.
So if you apply the Bestvina-Brady construction to any non-simply connected 2-dimensional acyclic complex you get a group having the properties that you asked for.
My `generalized Bestvina-Brady groups' allow you to construct uncountable families of groups of type FP, each having cohomological dimension two. The main novelty in my article with Tom Brown that you referenced in your question is that we use very different methods, whereas previous constructions all used CAT(0) cubical techniques.
As for some references: for a discussion of the question before it was answered, look at K S Brown's book Cohomology of Groups, Chapter VIII, especially VIII.5-VIII.8. The article by Bestvina and Brady is the best place to look for the construction of the examples.
| 11 | https://mathoverflow.net/users/124004 | 416092 | 169,587 |
https://mathoverflow.net/questions/416104 | 0 | Let us consider the measure algebra $M(\mathbb{R})$ consisting of all Radon measures on reals.
Let $\mu$ be a Radon measure in $M(\mathbb{R})$ and $\delta\_0$ be the point mass measure concentrated on 0, which is also the multiplicative identity of $M(\mathbb{R})$.
>
> Q. Does there exist any strictly positive real number $r>0$ and Radon measures $\mu\_1,\cdots \mu\_n$ satisfying the following idenity? $$\mu^\*\*\mu=r\delta\_0+\sum\_1^n\mu\_i^\*\*\mu\_i.$$
>
>
>
Remark. $M(\mathbb{R})=C\_0(\mathbb{R})^\*$ is a unital dual Banach $\*$-algebra. The involution on $M(\mathbb{R})$ is defined as follow
$$\langle \mu^\*,f\rangle=\int f(t)\overline{d\mu(-t)}.$$
| https://mathoverflow.net/users/84390 | An equation in the convolution measure algebra on reals | The answer is no in general. Indeed, let $\hat\mu$ denote the Fourier transform of $\mu$, so that $\hat\mu(t)=\int e^{itx}\mu(dx)$ for real $t$. Then the equality
$$\mu^\*\*\mu=r\delta\_0+\sum\_1^n\mu\_i^\*\*\mu\_i$$
would imply
$$|\hat\mu|^2=r+\sum\_1^n|\hat\mu\_i|^2\ge r>0.$$
Taking now any $\mu$ with $\hat\mu(t)\to0$ as $t\to\infty$ gives a contradiction.
| 4 | https://mathoverflow.net/users/36721 | 416112 | 169,592 |
https://mathoverflow.net/questions/416077 | 8 | Suppose that $M$ is a *compact, real*
algebraic subset of $\mathbb R^n$ and $f:\mathbb R^n \to \mathbb R^m$ is the projection to the first $m$ coordinates. If $f$ maps $M$ *bijectively* unto its image $f(M)$, is it true that $f(M)$ is algebraic?
| https://mathoverflow.net/users/13842 | Projections of compact real algebraic sets | The answer is no. (Although the previous example I gave was bad.)
Let $C$ be the curve $y^2 = x^2 (x-1)(2-x)$, so $C$ has a smooth component with $1 \leq x \leq 2$, and also a node at $(0,0)$. Let $M$ be the normalization of $C$; explicitly,
$$M = \{ (x,y,z) : z^2 = (x-1)(2-x),\ y=xz \}.$$
Then the projection $M \to C$ is $1$-to-$1$ over the smooth component but misses the node.
---
However, such examples can only miss a set of lower dimension than $M$, and cannot occur if all connected components of the Zariski closure of $f(M)$ have the same dimension! This follows from
*Bialynicki-Birula, A.; Rosenlicht, M.*, [**Injective morphisms of real algebraic varieties**](http://dx.doi.org/10.2307/2034464), Proc. Am. Math. Soc. 13, 200-203 (1962). [ZBL0107.14602](https://zbmath.org/?q=an:0107.14602).
At the start of section 2, they prove the following result:
>
> Let $V$ and $W$ be real algebraic sets and let $f: V \to W$ be an injective morphism such that $f(V)$ is Zarsiki-dense in $W$. Then $f(V)$ contains a Zariski-open Zariski-dense subset of $W$.
>
>
>
Note that this is very false for $f$ non-injective; consider the map $x \mapsto x^2$ from $\mathbb{R}$ to itself.
In our setting, take $V = M$ and let $W$ be the Zariski-closure of $f(M)$. For simplicity, let $M$ be irreducible, so $W$ will be as well.
Bialynicki-Birula and Rosenlicht's result shows that $f(M)$ must contain a Zariski-open Zariski-dense subset $U$ of $W$, and thus $W \setminus U$ must be a proper Zariski closed subset of $W$. In particular, $W \setminus U$ must have dimension lower than $\dim W = \dim M$. If we now assume that $W$ is irreducible and all its connected components have the same dimension, then $U$ must be dense in $M$ for the analytic topology.
But, also, $f(M)$ is closed in the analytic topology since $M$ is compact. We have shown that $f(M)$ is closed for the analytic topology and contains a dense set (namely $U$) for the analytic topology, so $f(M) = W$.
| 9 | https://mathoverflow.net/users/297 | 416123 | 169,594 |
https://mathoverflow.net/questions/416089 | 10 | Let $P(n)$ of a sequence $s(1),s(2),s(3),...$ be obtained by leaving $s(1),...,s(n)$ fixed and reverse-cyclically permuting every $n$ consecutive terms thereafter; apply $P(2)$ to $1,2,3,...$ to get $PS(2)$, then apply $P(3)$ to $PS(2)$ to get $PS(3)$, then apply $P(4)$ to $PS(3)$, etc. The limit of $PS(n)$ is $a(n)$ ([A057063](https://oeis.org/A057063)).
The sequence begins
$$1, 2, 4, 6, 3, 10, 12, 7, 16, 18, 11, 22, 13, 5, 28$$
Some examples:
$$1,2,(4,3),(6,5),(8,7),(10,9),(12,11),(14,13),(16,15),(18,17)$$
$$1,2,4,(6,5,3),(7,10,8),(12,11,9),(13,16,14),(18,17,15)$$
$$1,2,4,6,(3,7,10,5),(12,11,9,8),(16,14,18,13)$$
$$1,2,4,6,3,(10,5,12,11,7),(8,16,14,18,9)$$
I conjecture that $a(n)+1$ is prime if and only if $a(n)=2(n-1)$.
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Prime numbers from permutation | **1. Answer to the question.**
**Claim 1.** $a(n)=2(n-1)$ iff the number $d=a(n)$ moves only leftwards (while it moves).
*Proof.* At each move, every moving number moves either to the right or $1$ left.
The number $d=a(n)$ came to the $n$th position during $P(n-1)$, moving $1$ left. If $d$ moved leftwards through all $P(2),\dots, P(n-1)$, then it started at position $n+(n-2)=2(n-1)$ (so $d=2(n-1)$). Otherwise it shifted leftwards by a smaller distance, so it was smaller than $2(n-1)$. $\quad\square$
**Claim 2.** A number $d$ moves only to the left (while it moves) iff $d+1$ is prime.
*Proof.* While $d$ moves to the left, it appears at position $d-i+2$ before $P(i)$. Then, at $P(i)$, it moves to the right iff $d-i+2>i$ and $d-i+2\equiv 1\pmod i$, that is, if $d\equiv -1\pmod i$ and $d+1\geq 2i$, or in other wirds, $i\mid d+1$ and $(d+1)/i>1$.
So, if $p\leq d/2$ is the least prime divisor of $d+1$, then at $P(p)$ the number $d-p+2=(d+1)-(p-1)>p$ shifts to the right (if it did not shift earier --- in fact, it did not). Otherwise, $d$ is prime, and it cannot shift right. $\quad\square$
The two claims yield the result.
**2. Proof that the resulting arrangement is a permutation.**
We fix a number $d$ and describe how it moves. We aim at proving that it shifts rightwards only finitely many times; this clearly yields that $d$ will eventually stop.
Let $a\_n$ denote the position of $d$ before $P(n)$, and put $b\_n=a\_n+n-1$. Informally speaking, $b\_n-1$ denotes a position from which $d$ would come to its current position moving only to the left.
If $d$ moves at $P(n)$, we have $a\_{n+1}=a\_n-1$ if $a\_n\not\equiv 1\pmod n$ and $a\_{n+1}=a\_n+n-1$ otherwise. Therefore, $b\_{n+1}=b\_n$ if $n\nmid b\_n$, and $b\_{n+1}=b\_n+n$ otherwise (in the latter case we call $n$ a *crucial* index); we want to show that there are finitely many crucial indices.
Let $n<m$ be two consecutive crucial indices. Put $c\_n=b\_n/n$; if $c\_n=1$, then $b\_n=n$, and $d$ has just stopped (and $m$ does not exist). Otherwise, if $c\_n>1$, we have $b\_{n+1}=n(c\_n+1)$, and hence $c\_n+1>b\_{n+1}/(n+1)\geq b\_{n+1}/m=b\_m/m=c\_m$. Since $c\_m$ is an integer, we conclude that $c\_m\leq c\_n$.
Number all crucial indices as $n\_1<n\_2<\dots$; put $B\_i=b\_{n\_i}$ and $C\_i=c\_{n\_i}$. Those sequences act as follows:
$$
C\_i=B\_i/n\_i\leq C\_{i-1}; \quad B\_{i+1}=B\_i+n\_i=n\_i(C\_i+1).
$$
So, while $C\_i$ preserves the value $k>1$, we have $B\_{i+1}=B\_i\cdot \frac{k+1}k$. This may happen only finitely many times if $k>1$, since all the $B\_i$ are integers.
Thus, the (non-increasing) sequence $C\_i$ cannot preserve any value $k>1$ indefinitely, so it decreases from time to time, and eventually it reaches $1$, as desired.
| 9 | https://mathoverflow.net/users/17581 | 416125 | 169,595 |
https://mathoverflow.net/questions/415108 | 3 | The Schur multipliers of finite simple groups are known and easily accessible:
<https://en.wikipedia.org/wiki/List_of_finite_simple_groups>
Moreover, as a consequence of the second Whitehead's Lemma, if $L$ is a finite-dimensional simple Lie algebra over a field $\mathbb{F}$ of characteristic zero, then its Schur multiplier $H\_2(L, \mathbb{F})$ is trivial. This is no longer true in positive characteristic. Now, according to the Block-Wilson-Strade-Premet classification, every simple finite-dimensional Lie algebra over an algebraically closed field of characteristic p > 3 is of classical, Cartan, or Melikian
type. For my purposes, I would appreciate if I could have need any information or possible references about the following:
>
> **QUESTION** Is there any explicite description of $H\_2(L,\mathbb{F})$, where $L$ is a finite-dimensional simple Lie algebra over an algebraically closed field $\mathbb{F}$ of characteristic $p>3$?
>
>
>
I am mainly interested in the case in which $L$ is a restricted simple Lie algebra of Cartan type.
| https://mathoverflow.net/users/17582 | Schur multiplier of finite-dimensional simple Lie algebras in positive characteristic | As $L$ is simple, it has trivial abelianization, so one has $H\_1(L,\mathbb{F})=0$. Therefore, it follows from the Universal Coefficient Theorem for Lie algebras that $H\_2(L, \mathbb{F})\cong H^2(L, \mathbb{F})$. For a finite-dimensional graded Lie algebra of Cartan type over an algebraically closed field of characteristic $p>3$, the second cohomology space of $L$ with coefficients in the trivial module $\mathbb{F}$ is determined in the paper
<https://www.sciencedirect.com/science/article/pii/0021869392900046>
| 2 | https://mathoverflow.net/users/14653 | 416133 | 169,599 |
https://mathoverflow.net/questions/416106 | 9 | I'm really interested in Topological Quantum Field Theory (TQFT) and am currently planning to focus on it in my undergraduate thesis. My university, unfortunately, does not allow double majors in mathematics and physics or even a minor in physics, hence I fear that I do not have enough background in physics. I have taken the standard physics series though and have also read that a high level of understanding of quantum field theory is not a hard prerequisite to doing mathematical research in TQFTs. My questions then are primarily:
>
> How then should I start learning and doing undergraduate-level research on TQFTs? Can I also ask for recommended readings on TQFTs for undergraduates and/or directions for undergraduates hoping to do research in the field?
>
>
>
We were told that we are not really expected to come up with a unique result or proof of our own given our lack of mathematical knowledge, and I'm really hopeful that this is doable. Although I will do my best to independently learn what I need to, I will understand if this is too advanced of a topic for just an undergraduate. Thank you!
| https://mathoverflow.net/users/476977 | Undergraduate research in Topological Quantum Field Theory | I am not sure if this is an answer or a request for clarification.
There are a lot of different topics in math (and in physics) that go by the name 'topological quantum field theory'. Beyond the initial hand-waving about '(some) observables not depending on the spacetime manifold's geometric structure', they aren't always that closely related. You can't reasonably cover all of them in a thesis.
So your first task would be narrow your list of topics down. Where do you want to focus on? Are you interested in higher category theory? Operads? Knots? Quantum groups? Vertex algebras? Group cohomology? Geometric representation theory? The existence of smooth structures on manifolds? The topology of moduli spaces? The algebraic geometry of moduli spaces? This isn't anything like an exhaustive list. (And it's probably 10 years out of date!)
I don't think you need to *know* physics, but it's helpful to at least try try reading physics lectures notes. I've found Greg Moore's various reviews have a mathematician-friendly perspective, FWIW.
| 10 | https://mathoverflow.net/users/35508 | 416136 | 169,601 |
https://mathoverflow.net/questions/416147 | 6 | Let $C$ be an irreducible curve of arithmetic genus $1$ over a field $k$ and with a double $k$-point $p\in C$.
Is $C$ rational over $k$?
If $C$ is a plane cubic the answer is positive since we can parametrize $C$ with the lines through $p$.
In general I would embed $C$ in some projective space $\mathbb{P}^n$ and I would project $C$ until I get a plane cubic with a node. I think this should be ok if $k$ is infinite but if $k$ is finite I do not know how to make sure that $C\subset \mathbb{P}^n$ can be birationally projected onto a plane nodal cubic.
Thank you.
| https://mathoverflow.net/users/14514 | Singular curves of genus 1 | There's no problem over finite fields, but there is a problem over fields that have a nontrivial Brauer class. If you take a genus $0$ curve that's not rational (say a plane quadric), it will always have points over a degree $2$ extension (intersect with a line), and then you can glue two of them together to get a nodal curve of arithmetic genus $1$.
This curve will not be a nodal cubic, but you can embed it as a degree $4$ curve in $\mathbb P^3$.
For example, let $a$ be a quadratic nonresidue mod $p$, consider the curve in $\mathbb P^3$ with coordinates $x,y,z,w$ given by the equations $x^2 -a y^2 - p z^2$ and $xy = zw$ over $\mathbb Q\_p$. The only rational point of this curve is $(x:y:z:w)=(0:0:0:1)$, which is a node, so the curve has arithmetic genus $1$, has a node, and isn't rational.
| 10 | https://mathoverflow.net/users/18060 | 416150 | 169,603 |
https://mathoverflow.net/questions/416145 | 6 | We can consider the generalized Harmonic numbers $$H\_{n,m} := \sum\_{k=1}^{n} \frac{1}{k^{m}} $$ as a partial version of the Riemann zeta function, because $$\lim\_{n \to \infty} H\_{n,m} = \zeta(m). $$
We could also define the "partial Sophomore's Dream function" $$S\_{r} := \sum\_{q=1}^{r}\frac{1}{q^{q}} ,$$
as we have $$\lim\_{r \to \infty} S\_{r} := S= \int\_{0}^{1}x^{-x}dx ,$$ where the integral on the right is equal to the first Sophomore's Dream constant.
>
> **Question**: while studying the series $$A := \sum\_{r=1}^{\infty}(S-S\_{r}) \approx 0.3371877158, $$ I was wondering whether the function $S\_{r}$ has already been studied, and if someone has coined a name for it that is hopefully less awkward than mine. I am looking for alternative representations of $S\_{r}$ that might help finding a closed form of $A$.
>
>
>
**Note**: this question was previously [asked](https://math.stackexchange.com/questions/4369181/has-the-partial-sophomores-dream-function-been-studied-before) on MSE.
| https://mathoverflow.net/users/93724 | Has the "partial Sophomore's Dream function" been studied before? | We have that $S=\sum\_{n=1}^\infty \frac 1 {n^n}$ and $S\_r=\sum\_{n=1}^r \frac 1 {n^n}$. Thus,
$A=\sum\_{r=1}^\infty \sum\_{n=r+1}^\infty \frac 1 {n^n} = \sum\_{n=2}^\infty \sum\_{r=1}^{n-1} \frac 1 {n^n}$
But then, $A=\sum\_{n=2}^\infty \frac {n-1} {n^n}$. To evaluate this, consider the integral $\int\_0^1 t^{-tx} dt = \sum\_{n=1}^\infty \frac {x^{n-1}} {n^n}$. Differentiating with respect to $x$, we get $\sum\_{n=2}^\infty \frac {(n-1)x^{n-2}} {n^n} = \int\_0^1 -t\ln t \cdot t^{-xt} dt$. And substituting $x=1$ finally gives
$A=\int\_0^1 -\ln t \cdot t^{1-t} dt$.
Note that a WolframAlpha calculation gives 0.3371877 as the value of the integral, in excellent agreement with the value you obtained.
| 6 | https://mathoverflow.net/users/114143 | 416154 | 169,605 |
https://mathoverflow.net/questions/416131 | 2 | Let $\mathsf{T}$ be a rigid abelian tensor category and suppose that we're given fiber functors $\omega\_M:\langle M\rangle \to \mathsf{vect}\_k$ for every object $M$ of $\mathsf{T}$. Is there a canonical way to obtain a fiber functor on $\mathsf{T}$?
| https://mathoverflow.net/users/131975 | Given fiber functors on all the subcategories of the form $\langle M\rangle$, can we obtain a fiber functor on the whole category? | I think the answer is no in general.
Consider, for example, a tannakian category $M$ over a field $k$, and
suppose for simplicity that it is a
union $M=\bigcup M\_{n}$, $M\_{n}\subset M\_{n+1}$, of neutral tannakian
categories. Is $M$ neutral? In other words, if we assume that each $M\_{n}$ has
a $k$-valued fibre functor, does this imply that $M$ does?
Suppose first that $k$ is algebraically closed, and choose a $k$-valued fibre
functor $\omega\_{n}$ on $M\_{n}$. Because $k$ is algebraically closed,
$\omega\_{n+1}|M\_{n}$ is isomorphic to $\omega\_{n}$. In fact, given $\omega
\_{n}$, we can modify $\omega\_{n+1}$ so that $\omega\_{n+1}|M\_{n}=\omega\_{n}$.
Thus there exists a fibre functor $\omega$ on $M$ such that $\omega
|M\_{n}=\omega\_{n}$ (axiom of dependent choice!).
When we try to do this with $k$ not algebraically closed, we obtain a sequence
of torsors $Hom(\omega\_{n},\omega\_{n+1}|M\_{n})$. Of course, by making a
different choice of fibre functors, we get a different sequence of torsors,
but if, for example, the fundamental groups of $P\_{n}$ of the $M\_{n}$ are
commutative, then we get in this way a well-defined element of
$\varprojlim^{1}H^{1}(k,P\_{n})$, which is an obstruction to $M$ being neutral.
Consider the category $M$ of motives of weight 0 over the field with $p$
elements ($p$ prime) and assume the Tate conjecture. We can write
$M=\bigcup M\_{n}$ as above. It is not known whether the $M\_{n}$ are neutral,
even though the obstructions coming from the Brauer group of $k$ vanish.
Assuming they are neutral, Kontsevich has given a heuristic argument that, $M$ will not be neutral.
This answer has been abstracted from arXiv:math/0607569, which see for more details.
| 2 | https://mathoverflow.net/users/nan | 416161 | 169,606 |
https://mathoverflow.net/questions/416121 | 0 | Let $h$ be a random variable and $g(h)$ be a real-valued function of $h$.
We know that if h is a real-random variable then:
$E\_h[g(h)] = \int\_{-\infty}^{\infty} f(h) g(h) dh$ where f(h) is the PDF of h.
I want to learn if there is an integral to express the expectation over $h$ when $h$ is a complex-random variable ($g(h)$ is still a real-valued function). The integral cannot be the same as the real-valued function as the limits of integral ($-\infty$ to $\infty$) do not cover the complex plain.
| https://mathoverflow.net/users/477072 | Integral form of expectation with respect to complex random variables | There is an unfortunate mismatch between the notations and concepts in probability used by mathematicians and by practitioners of many engineering fields. In the latter fields, it is conventional to refer to an expectation "with respect to" a particular random variable; in the former one may refer to $\operatorname E(g(H))$ where $H$ is a random variable, and hence $g(H)$ is a random variable. The expectation is defined as
$$
\operatorname E(g(H)) = \int\_\Omega g(H(\omega)) P(d\omega)
$$
where $\Omega$ is the set of all outcomes $\omega$ of a random experiment, each of which outcomes yields a particular value of $g(H),$ and $P$ is a measure that assigns probabilities $P(E)$ to sets $E\subseteq\Omega$ of outcomes.
Now first consider a simple case where, rather than spreading $g(H)$ about the whole complex plane $\mathbb C,$ we have $\Pr(H=1\text{ or } H=2 \text{ or } H=3)=1.$ Now the expected value is
\begin{align}
\operatorname E(g(H)) & = g(1)\Pr(H=1)+g(2)\Pr(H=2) + g(3)\Pr(H=3) \\[8pt]
& = \sum\_{h\,=\,1}^3 g(h)\Pr(H=h).
\end{align}
**Important:** Notice the way in which I distinguish between capital $H,$ the random variable, and lower-case $h,$ the possible value of $H.$ Without that distinction, even the simple expression $\Pr(H=h)$ cannot be understood. Mathematicians may perceive this present remark as a clarification to set a confused undergraduate straight, but I see books written by engineering professors that neglect this distinction and write probability mass functions and probability density functions as $f\_h(h)$ instead of $f\_H(h),$ and in fact that question posted above uses the lower-case $h$ for both of these two different things, where it ought instead to say
$$
\operatorname E(g(H)) = \int\_{-\infty}^{+\infty} g(h) f(h)\, dh
$$
with capital $H$ for the random variable and lower-case $h$ for the bound variable running through the set of all possible values of the random variable (capital) $H.$ The reason I'm writing this answer instead of telling you to post in some other place such as math(dot)stackexchange(dot)com or stats(dot)stackexchange(dot)com, as seems to be suggested by some votes to close the question, is that it is so often professors in engineering fields rather than confused undergraduates who make this mistake, and also mathematicians so seldom seem aware of the need to set this record straight.
If the value of $H$ is in $\mathbb C$ rather than $\mathbb R,$ then it is perfectly correct to say
$$
\operatorname E(g(H)) = \int\_\Omega g(H(\omega)) P(d\omega) = \int\limits\_\text{plane} g(h) f\_H(h) \, dh
$$
where $dh$ is the "element of area in the plane" and $f\_H(h)$ is the probability density of $H$ with respect to area in the plane. And here again, in the first integral I have capital $H$ and in the second lower-case $h$ except in the subscript, for a reason. Inattention to this distinction can actually render one unable to follow standard undergraduate textbook exercises.
| 4 | https://mathoverflow.net/users/6316 | 416166 | 169,607 |
https://mathoverflow.net/questions/416183 | 1 | Given an undirected graph $G = (V, E)$, a *vertex k-cut* of $G$ is a vertex subset of $V$ the removing of which disconnects the graph in at least $k$ connected components
(from <https://cris.unibo.it/handle/11585/713744>).
Given a biconnected graph, the minimal cardinality of a vertex 2-cut is 2, by definition.
My conjecture is that the above property can be generalised to:
>
> The minimal cardinality of a vertex $k$-cut of a biconnected graph
> is $k$
>
>
>
In other words, can one get $k$ connected components by removing less than $k$ vertices from a biconnected graph?
| https://mathoverflow.net/users/68336 | Minimal cardinality of a vertex $k$-cut of a biconnected graph | Question (if I got it right):
>
> can one get $k$ connected components by removing less than $k$ vertices from a biconnected graph?
>
>
>
Answer:
Yes. Consider the graph $G=(V,E)$ with $V=\{a,b,v\_1,...,v\_n\}$ and $E=\{av\_i,bv\_i : i=1,...,n\}$. It is 2-connected. Removing nodes $a$ and $b$ you get $n$ components, where $n$ can be arbitrarily large.
| 3 | https://mathoverflow.net/users/477125 | 416189 | 169,612 |
https://mathoverflow.net/questions/416187 | 3 | Let $(M^2,g)$ be a noncompact orientable Riemannian surface without boundary. Let $A \in \Gamma(\operatorname{Sym}(TM))$ be a section of the bundle of symmetric endomorphisms of $TM$, that is, for each $p \in M$, the linear map $A\_p : T\_p M \to T\_p M$ is symmetric with respect to the inner product $g\_p$. Assume that the eigenvalues $\lambda\_1(p), \lambda\_2(p) \in \mathbb{R}$ of $A\_p$ are distinct for all $p \in M$.
>
> Do there exist two smooth globally defined unit vector fields $E\_1, E\_2$ on $M$ such that $E\_i(p)$ is an eigenvector of $A\_p$ associated with $\lambda\_i(p)$ for $i=1,2$ and all $p \in M$?
>
>
>
>
> Does the topology of $M$ impose any restrictions on the existence of such vector fields?
>
>
>
| https://mathoverflow.net/users/85934 | Global choice of eigenvectors on an open surface | Not necessarily. To construct a counter-example, start from the other direction. Suppose that the tangent bundle of $M$ can be split as the direct sum $TM = L\_1\oplus L\_2$ where $L\_1$ and $L\_2$ are non-trivial smooth line bundles. Then you can easily construct a metric $g$ on $M$ such that $L\_1$ and $L\_2$ are $g$-orthogonal. Let $A:TM\to TM$ be the linear transformation that is the identity on $L\_1$ and minus the identity on $L\_2$. Then $A$ is symmetric with respect to $g$, but the two eigenbundles of $A$ have no non-trivial sections.
A simple example of such a splitting is to let the (orientable) cylinder $M$ be the quotient of the plane $\mathbb{R}^2$ by the translation $(x,y)\mapsto (x{+}\pi,y)$, and let $L\_1$ be the bundle spanned by the vector field $Z = \cos x\,\partial\_x + \sin x\,\partial\_y$ (which is only defined up to a sign on $M$), while $L\_2$ is spanned by the vector field $W = -\sin x\,\partial\_x + \cos x\,\partial\_y$ (again defined only up to a sign on $M$).
| 7 | https://mathoverflow.net/users/13972 | 416192 | 169,614 |
https://mathoverflow.net/questions/416201 | 8 | The question is in the title:
>
> **Question:** Which inscribed $n$-dimensional polytope (inscribed in the unit sphere) with $2^n$ vertices has the largest possible volume?
>
>
>
Is it the $n$-dimensional cube? If not, how much larger can its volume be?
| https://mathoverflow.net/users/156219 | Inscribed $n$-polytope with $2^n$ vertices of maximal volume | For $n=3$, the maximal volume polytope with 8 vertices is described in that paper.
*Berman, J. D.; Hanes, K.*, [**Volumes of polyhedra inscribed in the unit sphere in (E^3)**](http://dx.doi.org/10.1007/BF01435416), Math. Ann. 188, 78-84 (1970). [ZBL0187.19604](https://zbmath.org/?q=an:0187.19604).The link is paywalled, but a summary with the optimal shape drawn is available in [this answer](https://math.stackexchange.com/a/988708/248217).
It is combinatorially very different from a cube: it is simplicial, and each vertex has degree 4 or 5. The correct optimal polyhedron also has only $D\_2$ symmetry.
**Edit.** An inscribed polytope which maximizes the volume given the number of vertices is always simplicial (see Lemma 1 in
*Horváth, Ákos G.; Lángi, Zsolt*, [**Maximum volume polytopes inscribed in the unit sphere**](http://dx.doi.org/10.1007/s00605-016-0949-2), Monatsh. Math. 181, No. 2, 341-354 (2016). [ZBL1354.52016](https://zbmath.org/?q=an:1354.52016).), so the answer is "no" as well in any dimension $>2$.
The gap is even quantitative. Let $\mu\_n$ be the largest volume of an $n$-dimensional polytope with $2^n$ vertices inscribed in the ball of radius $\sqrt{n}$ (this normalization is better since it counterbalances the fact that the ball of unit radius has a tiny volume for large $n$). The cube gives the lower bound $\mu\_n \geq 2^n$. By taking direct products, we have $\mu\_{m+n} \geq \mu\_m\mu\_n$, and therefore $\mu\_{3n} \geq \mu\_3^n \geq (2+\epsilon)^{3n}$, exponentially better that the cube. We cannot do much better since the upper bound $\mu\_n \leq C^n$ for some constant $C$ holds trivially by comparing with the ball itself.
| 17 | https://mathoverflow.net/users/908 | 416203 | 169,618 |
https://mathoverflow.net/questions/416168 | 6 | The equation $z^2=\overline{z}$ has four zeros and this example motivates us to generalize the problem to this form;
How many zeros does the equation $P(z)=\overline{z}$ have if $P(z)$ is a polynomial of degree $n>1?$ Can we find the bound for the number of zeros of this problem? The example motivate us to conjecture that it may be at most $2n,$ if not at most $2n+n-2=3n-2.$ I am suggesting mere by intuition! May I request you to share your thoughts on this?
| https://mathoverflow.net/users/128472 | Fixed points of a function $z\mapsto\overline{P(z)}$ of a complex variable | Function $z\mapsto\overline{P(z)}$ has at most $3d-2$ fixed points, where $d\geq 2$
is the degree of $P$, and this is best possible. This remarkable result is due to Khavinson and Świa̧tek,
[MR1933331](https://mathscinet.ams.org/mathscinet-getitem?mr=1933331)
Khavinson, Dmitry, Świa̧tek, Grzegorz ,
[On the number of zeros of certain harmonic polynomials](https://doi.org/10.1090/S0002-9939-02-06476-6),
Proc. Amer. Math. Soc. 131 (2003), no. 2, 409–414.
and it was later generalized to rational functions, and to some transcendental
functions. There is a survey of related results:
D. Khavinson and G. Neumann, [From the fundamental theorem of algebra to astrophysics:
a “harmonious” path](https://www.ams.org/journals/notices/200806/tx080600666p.pdf), Notices Amer. Math. Soc. 55 (2008), no. 6, 666–675.
Let me mention a major unsolved problem: let $p$, $q$ be polynomials
of degrees $m>n$. How many solutions can the equation
$$\overline{p(z)}=q(z)$$
have? Can one do better than the Bézout estimate $mn$?
| 8 | https://mathoverflow.net/users/25510 | 416204 | 169,619 |
https://mathoverflow.net/questions/416023 | 1 | For which properties (P) [of groups] does the following hold:
given a group $G$ which has a finite presentation with at most $n$ relations of length at most $\ell$, there is a $R(n,\ell)$ so that, if the ball of radius $R$ in the [labelled/unlabelled] Cayley graph of $G$ [w.r.t. to the generating set of said presentation] is given, then one can conclude whether (P) hold or not.
There are quite a few caveats/comments to the question:
* by "decided" I mean one can always deduce whether (P) holds or not.
* the ball of radius $R$ cannot be computed if the word problem is not solvable for $G$. So the ball of radius $R$ needs to be given.
* there might be two questions: one for the labelled Cayley graph and one for the unlabelled Cayley graph.
* in the sense of the above question, the presentation is not an input of the problem, only $n$ and $\ell$ (but if you have the labelled Cayley graphs, I think [but maybe I'm wrong] you can read off the presentation from a large enough ball)
* one might like to consider the presentation as an input. Then if it has solvable word problem, the question is still interesting (since it puts a limit on how large the balls need to be computed)
* the only property which I believe[!] I know this is true is hyperbolicity.
* there are some results which enable to conclude that the group has property (T), but they do not always conclude (at least, not the ones I am aware of).
* you only get one ball, not all balls of radius $R$. So for example it's not clear to me that properties related to the growth of the group answer this question.
| https://mathoverflow.net/users/18974 | Which properties can be read off the balls of a Cayley graph? | This question gestures in a few different directions. Since I think they may all be of some interest, I'll attempt to summarise what's known in a few of these directions here. Hopefully some of this helps!
**Finitely generated groups**
For a finitely generated group $G$ with a given generating set $S$, we can ask whether there is an $R$ such that $P$ is determined by the labelled isomorphism type of the ball $B(R)$ in the Cayley graph. These are exactly the properties that are open in the *Gromov--Grigorchuk space of marked groups* -- see, for instance, [this](https://arxiv.org/abs/math/0511714) paper of Cornulier--Guyot--Pitsch, and many others.
To give a flavour, for finitely presented groups, these properties $P$ are closed under passing to quotients. So examples include finiteness, Property (T), and the *isolated* groups studied in the above-mentioned paper.
**Finitely presented groups**
The original version of the question gives us a finitely presented group $G=\langle a\_i\mid r\_j\rangle$, and a constant $R$ that depends on the presentation. However, as noted in comments, if we take $R$ greater than the lengths of all the relators, then $G$ is determined up to isomorphism by the (again labelled) $R$-ball in the Cayley graph, since the loops in this ball generate all the relators. So $G$ itself, and hence every property of $G$, is trivially determined by this ball. However...
**Recursively enumerable properties**
Just because the ball determines $P$ doesn't mean that "one can always deduce whether (P) holds or not" (in the words of the question), in the sense that we may not know which $R$-balls correspond to $P$ and which $R$-balls correspond to not $P$.
This brings us into the world of algorithmic properties of groups. Properties $P$ that can be read off a finite ball are very close to the recursively enumerable properties: $P$ is *recursively enumerable* if there's a Turing machine that takes as input a finite presentation and terminates if and only if the resulting group has $P$.
If there's a Turing machine $T(P)$ that takes as input a finite labelled graph $B$ and terminates only if a group with $R$-ball $B$ has $P$ then $P$ is recursively enumerable: indeed, given your group presentation $G$, one can in parallel try to compute all balls $B\_R$ and input them into $T(P)$. If any of these eventually terminate then we know that $G$ has $P$.
So although this isn't a perfect match for the condition stated in the question (which asks for the algorithm to decide whether **or not** $G$ has $P$ based on the $R$-ball), I think recursively enumerable properties are a good approximation to what the question asks for.
Here's a brief list of some properties that we know are recursively enumerable:
* Hyperbolicity is recursively enumerable by a theorem of Papasoglu [Papasoglu, 'An algorithm detecting hyperbolicity'. *Geometric and computational perspectives on infinite groups (Minneapolis, MN and New Brunswick, NJ, 1994)*.].
* Automaticity is recursively enumerable by a theorem of Epstein et al [Epstein et al. *Word processing in groups*. Jones and Bartlett Publishers, Boston, MA, 1992. xii+330 pp].
* Property (T) is recursively enumerable by the [theorem of Ozawa](https://arxiv.org/abs/1312.5431) mentioned above.
* Limit groups in the sense of Sela are recursively enumerable by a theorem of [Groves and myself](https://arxiv.org/abs/0704.0989).
...
In practice, almost all of these results involve checking some condition on some suitably large ball in the Cayley graph, so fall into at least the spirit of the question.
**Recursive properties modulo the word problem**
A property $P$ is *recursive* if both $P$ and not $P$ are recursively enumerable. The [Adyan--Rabin theorem](https://en.wikipedia.org/wiki/Adian%E2%80%93Rabin_theorem) implies that this fails for every Markov property, meaning that very few non-trivial properties are recursive. There one or two exceptions, the most notable being anything that depends on the abelianisation.
The failure of most properties to be recursive can be traced back to the existence of finitely presented groups with unsolvable word problem (the [Novikov--Boone theorem](https://en.wikipedia.org/wiki/Presentation_of_a_group#Novikov%E2%80%93Boone_theorem)). So, as in the original question, it makes sense to ask whether assuming a solution to the word problem makes any properties $P$ recursive.
Groves, Manning and I called a property $P$ *recursive modulo word problem* if $P$ can be determined inside any class of group presentations in which the word problem is uniformly solvable. (There are other definitions that might encapsulate this idea better, as proposed for instance by [Rauzy](https://arxiv.org/abs/2111.01190).) Remarkably, quite a few properties turn out to be recursive modulo the word problem, including:
* abelian-ness, and more generally nilpotence of class $k$ for any $k$;
* freeness, being the fundamenal group of a closed surfaces, and limit groups in the sense of Sela ([Groves--W.](https://arxiv.org/abs/0704.0989)),
* admitting a non-trivial free splitting ([Touikan](https://arxiv.org/abs/0906.3902))
* being a geometric 3-manifold group ([Groves--Manning--W.](https://arxiv.org/abs/1210.2101))
| 3 | https://mathoverflow.net/users/1463 | 416206 | 169,621 |
https://mathoverflow.net/questions/416198 | 4 | Let $\varphi(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$ be the Gaussian density and
$f:\mathbb{R}\to\mathbb{R}$ another measurable function.
Under what conditions can $f$ be recovered from its convolution with $\varphi$? In other words, under what conditions does $f\ast\varphi=0$ imply that $f$ is zero a.e?
If $f\in L^1(\mathbb{R})$, then it has a Fourier transform and the statement follows since $\varphi$ has a Fourier inverse. What about other conditions on $f$? For example, what if it is bounded by a polynomial? Or a subexponential function?
| https://mathoverflow.net/users/477138 | Recovering a function from its Gaussian convolution | $\newcommand{\vpi}{\varphi}\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}\newcommand{\ep}{\varepsilon} $The minimal condition
\begin{equation\*}
|f|\*\vpi<\infty \tag{1}\label{1}
\end{equation\*}
is already enough for the recovery of $f$.
Indeed, since $\vpi(x-u)=\vpi(u)e^{xu}e^{-x^2/2}$, condition \eqref{1} can be rewritten as
$\int\_\R|f(u)|\vpi(u)e^{xu}\,du<\infty$ for all real $x$ or, equivalently, as
$\int\_\R|f(u)\vpi(u)e^{zu}|\,du<\infty$ for all complex $z$. Letting now
\begin{equation\*}
g(z):=\int\_\R f(u) \vpi(u) e^{zu}\,du,
\end{equation\*}
we have an entire function $g\colon\C\to\C$ such that
\begin{equation\*}
g(x)=e^{x^2/2}\int\_\R f(u) \vpi(x-u)\,du=e^{x^2/2}(f\*\vpi)(x)=0
\end{equation\*}
for all real $x$.
So, $g=0$. In particular,
\begin{equation\*}
0=g(it)=\int\_\R f(u) \vpi(u) e^{itu}\,du
\end{equation\*}
for all real $t$ -- that is, the Fourier transform of the integrable function $f\vpi$ is $0$. It follows that $f\vpi=0$ almost everywhere (a.e.) and thus $f=0$ a.e., as desired.
| 4 | https://mathoverflow.net/users/36721 | 416212 | 169,622 |
https://mathoverflow.net/questions/412037 | 7 | Let $(A, \Delta)$ be a compact quantum group in the sense of Woronowicz. Is it true that the comultiplication $\Delta : A \to A \otimes A$ always injective?
This is true for both the universal (because one has a counit) and the reduced (because the Haar state is faithful) version, but the general case seems more delicate.
In particular, is it true for the dual of a discrete group $\Gamma$ realized as a compact $C^\*$-algebraic quantum group using a $C^\*$-norm that lies strictly between the universal and the reduced $C^\*$-norms? What about the case where $\Gamma$ is the free group on two generators?
| https://mathoverflow.net/users/128540 | Is the comultiplication of a compact quantum group always injective? | No, the comultiplication need not be injective. When $\Gamma$ is a countable group and $\pi : \Gamma \to \mathcal{U}(H)$ is a faithful unitary representation with the property that $\pi \otimes \pi$ is weakly contained in $\pi$, we write $A = C^\*\_\pi(\Gamma)$ and there is a unique unital $\*$-homomorphism $\Delta : A \to A \otimes A$ satisfying $\Delta(\pi(g)) = \pi(g) \otimes \pi(g)$ for all $g \in \Gamma$. Then, $(A,\Delta)$ is a compact quantum group in the sense of Woronowicz.
Now $\Delta$ is faithful if and only if $\pi$ is weakly contained in $\pi \otimes \pi$. That need not be the case, as the following example with $\Gamma = \mathbb{F}\_2$ shows.
For every $0 < \rho < 1$, we consider the function $\varphi\_\rho : \mathbb{F}\_2 \to \mathbb{R}$ given by $\varphi\_\rho(g) = \rho^{|g|}$, where we use the word length $|g|$. By [Haa, Lemma 1.2], $\varphi\_\rho$ is a positive definite function and we define $\pi\_\rho$ as the associated cyclic representation. Denote by $\lambda$ the regular representation. By [Haa, Theorem 3.1], we have that $\pi\_\rho$ is weakly contained in $\lambda$ if and only if $\rho \leq 1/\sqrt{3}$.
We now claim that with $\rho = 2/3$, the representation $\pi = \pi\_\rho \oplus \lambda$ provides a counterexample for the faithfulness of $\Delta$. By construction, $\pi \otimes \pi$ is weakly equivalent with $(\pi\_\rho \otimes \pi\_\rho) \oplus \lambda$. Since $\rho^2 < 1/\sqrt{3}$, it follows from [Haa, Theorem 3.1] that $(\pi \otimes \pi) \sim \lambda \prec \pi$. But since $\rho > 1/\sqrt{3}$, we have $\pi\_\rho \not\prec \lambda$, so that $\pi \not\prec \pi \otimes \pi$.
[Haa] U. Haagerup, An example of a nonnuclear C$^\*$-algebra, which has the metric approximation property. Invent. Math. 50 (1978/79), 279-293.
| 6 | https://mathoverflow.net/users/159170 | 416221 | 169,623 |
https://mathoverflow.net/questions/416119 | 4 | $\DeclareMathOperator\SL{SL}$I am currently looking at the paper titled "$\SL(2,\mathbb{C})$ quotients de $(\mathbb{P^1})^n$" by Marzia Polito. The author has considered diagonal action of $\SL(2,\mathbb{C})$ over $(\mathbb{P^1})^n$ given by $(g,(x\_1, x\_2,\cdots,x\_n))=(gx\_1, gx\_2,\cdots,gx\_n)$ for $g\in \SL(2,\mathbb{C})$ and $(x\_1, x\_2,\cdots,x\_n)\in (\mathbb{P^1})^n $. The author has considered linearization with respect to the ample line bundle $L(m)=O(m\_1)\otimes O(m\_2)\cdots O(m\_n)$ where $m=(m\_1,m\_2,\cdots m\_n)$ and $|m|=\sum\_1^nm\_i$ and also defined for $x\in (\mathbb{P^1})^n$ the set $I\_k(x)=\{i\in(1,2\cdots ,n)|x\_i=x\_k\}$. Then in Proposition 1 the author claims
$x\in (\mathbb{P^1})^n$ is semistable iff for all $k$, $\sum\_{i\in I\_k(x)}m\_i\leq\frac{|m|}{2}$. The author has not given any proof and I cannot do it myself. Any proof or comment will be highly appreciated.
| https://mathoverflow.net/users/211682 | Question regarding semistability of a point of GIT quotient | The details of this proof can be found in introductory texts on geometric invariant theory (GIT). I recommend Dolgachev's textbook Lectures on Invariant Theory, Chapter 11, where one can find a proof via the Hilbert-Mumford numerical criterion. The technical overhead to learning the Hilbert-Mumford numerical criterion is a bit high, but the benefit is that the same technique applies to many other GIT problems.
The only drawback of Dolgachev's treatment is that it steps through a chain of inequalities that, to a student, will probably appear unmotivated. I present a geometrically motivated sketch of the proof below. The full proof along these lines is given in Theorem 3.6 of a [preprint](https://arxiv.org/abs/2111.06351) of mine. (Many thanks to user [afh](https://mathoverflow.net/users/339730/afh) for linking to it in the comments!)
Informally, one expects an object in projective space to be GIT unstable if that object has a high order of contact with a complete flag in projective space. In low-dimensional situations, instead of talking about flags, it may be enough to talk about order of contact with points and lines. For instance, a plane cubic curve is non-stable if it is singular, and having a singularity is the same as having a point of higher-than-expected multiplicity. In the case of configurations of points on $\mathbb{P}^1$, this philosophy says that if the configuration has a point of high multiplicity, the configuration should be unstable.
The way we compute this ``order of contact'' between a complete flag and our object, in practice, is with the Hilbert-Mumford criterion, and the closely related concept of the weight polytope (Chapter 9 in Dolgachev's textbook). In your setting, the case of an ${\text{SL}}\_2$-action acting linearly on some variety $V$ embedded in projective space over $\mathbb{C}$ via some ample ${\text{SL}}\_2$-linearized line bundle $\mathcal{L}$, the data of a complete flag is nothing more than a choice of 1-dimensional subspace of $\mathbb{C}^2$. To any complete flag $F$ in $\mathbb{C}^2$ and any point $v \in V$, we can associate a weight polytope $\Pi(v, F, \mathcal{L})$. The point $v \in V$ is semistable for $\mathcal{L}$ if and only if, for every flag $F$, the weight polytope $\Pi(v, F, \mathcal{L})$ contains the origin.
If ${\text{SL}}\_2$ acts on two varieties $V$ and $W$, then the weight polytopes for the product action on $V \times W$, as embedded in projective space with the Segre embedding, are computed by
$$\Pi((v,w),F, \mathcal{L}) = \Pi(v,F, \mathcal{L}) + \Pi(w,F, \mathcal{L}),$$
where $+$ denotes Minkowski sum.
Also, the linear ${\text{SL}}\_2$-action on $V$ induces ${\text{SL}}\_2$-actions on the Veronese embeddings of $V$ in larger projective spaces, and this scales the weight polytope: for any $m \in \mathbb{N}$, we have
$$\Pi(v,F, \mathcal{L}^{\otimes m}) = m \Pi(v, F, \mathcal{L}).$$
This is good news for your problem, because it means that you just need to understand the possible weight polytopes for the ${\text{SL}}\_2$-action on $\mathbb{P}^1$ and sheaf $\mathcal{O}(1)$. The weight polytope in this case is either a point or a closed interval in $\mathbb{R}$, and it can be computed as follows. Given a flag in $\mathbb{C}^2$ associated to a one-dimensional subspace $H$ of $\mathbb{C}^2$, and a point $v \in \mathbb{P}^1,$ there are three possibilities for the weight polytope $\Pi = \Pi(v, H, \mathcal{O}(1))$. If $v \in H$, then $\Pi = \{1\}$; if $v \in H^\perp$, then $\Pi = \{-1\}$; otherwise $\Pi = [-1,1]$. Returning to the case of your problem, when we take Minkowski sums of such intervals with equal weights, the only way that the flag will detect instability is if more than $n/2$ points lie on $H$ or $H^\perp$. With possibly distinct weights, one may still compute the endpoints of the weight polytope to get the result.
I suggest carefully tracing through the definition of weight polytope in Dolgachev's book to understand the calculation above. The three possibilities for $\Pi$ correspond to the possible forms of the coordinates of $v$ when written in a basis with first vector taken from $H$.
I'll also mention that the problem of finding quotients of spaces of $n$ points on $\mathbb{P}^1$ was also addressed in the foundational text Geometric Invariant Theory (Chapter 3), as well as in the more approachable Introduction to the Theory of Moduli by Mumford and Suominen (Section 2).
| 2 | https://mathoverflow.net/users/477081 | 416224 | 169,624 |
https://mathoverflow.net/questions/415991 | 6 | The Computational Diffie Hellman (CDH) problem is to compute $g^{XY}$ given $g^X$ and $g^Y$ where $g$ generates the group. The Discrete Logarithm (DLOG) problem is to compute $X$ given $g^X$. The latter in $P$ implies the former in $P$.
>
> Are there groups where CDH problem is easy and in $P$ while DLOG is difficult and not known to be in $P$?
>
>
>
| https://mathoverflow.net/users/10035 | Groups in which Computational Diffie Hellman is in $P$ but Discrete Logarithm is not known to be in $P$ | Some background for this answer:
1. A computation is *non-uniform* if, in addition to an input $x$, one gets an "advice string" that depends solely on the *length* of $x$. One can view this as precomputation, although this advice string need not be computable (this should not be relevant here though).
In this answer, non-uniformity is used to ensure one has access to parameters of certain elliptic curves, and potentially a factorization of a certain number.
2. The *generic group model* is an abstract model in cryptography where, rather working directly with group elements, one is given oracles for the relevant group operations. Algorithms in the generic group model therefore work for *all* groups with efficiently-computable operations, and do not exploit the particular structure of any single group. There are lower bounds in the generic group model --- solving DLOG in $\Theta(\exp(\sqrt{\log |G|}))$ operations is provably optimal iirc.
The only "mainstream" groups used within cryptography that are plausibly generic are Elliptic Curve groups ($\mathbb{F}\_p^\times$ and related groups admit *non-generic* algorithms that go broadly by the name of "index calculus" algorithms).
>
> The paper you quote there requires $|G|$ to be $\mathsf{polylog}|G|$ smooth according to abstract. Are you sure it applies to cyclic multiplicative group mod p a prime? Please post full answer for this group.
>
>
>
It does not, as $\mathbb{F}\_p^\times$ is not thought to be a "generic" group --- one can speed up DLog via appealing to index calculus methods.
The paper works in the generic group model, so is mainly applicable to groups we think are generic, e.g. elliptic curve groups.
Still, I will include two of the papers results, as your understanding of the paper from the abstract is wrong, and it can handle the case of $|G| = p$ a large prime.
Here is corollary 5 of the paper
>
> **Corollary 5**: If the smoothness assumption is true, then there exists a polynomial-time generic algorithm computing discrete logarithms in cyclic groups of order $n$, making calls to a DH oracle for the same group, if and only if all the multiple prime factors of $n$ are of order $(\log n)^{O(1)}$.
>
>
>
This result is non-uniform, and requires a certain number theoretic assumption.
It clearly handles the case of $|G| = p$ though.
Removing the smoothness assumption (and I believe the non-uniformity assumption, but I have not carefully read), corollary 7 states that
>
> **Corollary 7**. Let $P$ be a fixed polynomial, let $G$ be a cyclic group with generator $g$, and let $B := P (\log |G|)$. Then there exists a list of expressions $A(p)$ in $p$ with the following property: if every prime factor $p$ of $|G|$ greater than $B$ is single and if for
> every such prime factor at least one of the expressions $A(p)$ is $B$-smooth, then breaking the DH protocol in $G$ with respect to $g$ is polynomial-time equivalent to computing discrete logarithms in $G$ to the base $g$. The list contains the following expressions:
>
>
> 1. $p − 1 , p + 1 , p + 1 \pm 2a$,
> if $p \equiv 1 (\mod 4)$, where $p = a^2 + b^2$,
> 2. $p + 1 \pm 2a, p + 1 \mp a \pm 2b , p + 1 \pm(a + b)$, if $p \equiv 1 \bmod 3$, where $p = a^2 − ab + b^2, a \equiv 2 \bmod 3$, and $b \equiv 0 \bmod 3$,
> 3. $\frac{(p^k)^l − 1}{p^k − 1} = (p^k)^{l−1} + · · · + p^k + 1$,
> where $k, l = (\log p)^{O(1)}$, and $f(p)$, where $f(x) \in \mathbb{Z}[x]$ is a nonconstant polynomial
> dividing $x^n − 1$ for some $n = O(1)$.
>
>
>
these technical conditions are related to constructions of what the authors call "auxillary groups", which are the non-uniform information the authors need for their reduction.
This is to say that the result can definitely be improved, for example
1. removing the smoothness assumption for cor. 4, or
2. relaxing the technical conditions with respect to the construction of auxillary groups,
Of main interest to you though, the work is in the generic group model, and the example you have mentioned in the comments is not a generic group.
| 2 | https://mathoverflow.net/users/101207 | 416238 | 169,625 |
https://mathoverflow.net/questions/416099 | 0 | Mathematica gives me the following solution to $\int{x^{-a} (b -cx^{-d})^e }$:
$$\int{x^{-a} (b -cx^{-d})^e dx} = -\frac{b^{e}x^{1-a} \, \_2F\_1\left(\frac{a-1}{d},-e;\frac{a+d-1}{d};\frac{c x^{-d}}{b}\right)}{a-1},$$
where $a,b,c,d,e > 0$.
I'd like to figure out what are the steps involved in this solution.
I started by finding the Taylor's expansion of the integrand, but it doesn't seem to be the right way to follow.
| https://mathoverflow.net/users/103291 | What are the steps involved in the solution to $\int{x^{-a} (b -cx^{-d})^e }dx$? | The proof starts by applying the change of variable $z=(c/b)x^{-d}$ and then using the following identity
$$\int (1-z)^{u} z^{v} dz = \frac{z^{v+1}}{v+1} \, \_2F\_1(-u, v+1; v+2; z), |x|<1,$$
the proof is concluded.
| 1 | https://mathoverflow.net/users/103291 | 416243 | 169,627 |
https://mathoverflow.net/questions/416241 | 9 | Let $B$ be the closed unit ball in $\mathbb R^3$ centered at the origin and let $U= \{x\in \mathbb R^3\,:\, \frac{1}{2}\leq |x| \leq 1\}.$ Let
$$ S\_U= \{u \in C^{\infty}(U)\,:\, \Delta u =0 \quad\text{on $U^{\textrm{int}}$}\},$$
and
$$ S\_B= \{u \in C^{\infty}(B)\,:\, \Delta u =0 \quad\text{on $B^{\textrm{int}}$}\}.$$
Is the following statement true? Given any $\epsilon>0$ and any $u \in S\_U$, there exists an element $v \in S\_B$ such that $\|v-u\|\_{L^2(U)} \leq \epsilon$.
| https://mathoverflow.net/users/50438 | Density of restrictions of harmonic functions inside a ball | No. If $\varphi \in C^\infty\_c(B)$ is a bump function equal to $1$ in $\lvert x\rvert \leq 1/2$ then from Green's theorem we have
$$ \int\_B u \Delta \varphi = 0$$
for all $u \in S\_B$, but the same is not true in general for typical $u \in S\_U$, which by the Cauchy–Schwarz inequality implies that $u$ is a positive distance away from $S\_B$ in the $L^2(U)$ norm. For instance, if we take $u = K\rvert\_U \in S\_U$ where $K(x) = \frac{-1}{4\pi \lvert x\rvert}$ is the Newton potential (the fundamental solution to $\Delta K = \delta$) then
$$ \int\_B u \Delta \varphi = \int (\Delta K) \varphi = 1 \neq 0$$
and hence $u$ is a positive distance from $S\_B$.
One can create similar obstructions using functions $\varphi$ which behave like a specified spherical harmonic in the angular variable (instead of being constant in the angular variable, which is basically what is being done here).
| 20 | https://mathoverflow.net/users/766 | 416245 | 169,628 |
https://mathoverflow.net/questions/416230 | 2 | The Fourier transform of the (indicator function of the) unit ball is well known to be given by the Bessel functions (see [Fourier transform of the unit sphere](https://mathoverflow.net/questions/149692/fourier-transform-of-the-unit-sphere)).
What can be said about the $\ell\_1$ ball, that is the set of points in $\mathbb{R}^n$ with $\ell\_1$ norm bounded by radius $r$? More generally one can ask about $\ell\_p$ ball for other values of $p$. The interesting fact about $\ell\_2$ is that if the function is radially symmetric, the same can be said about its Fourier transform and effectively the transform becomes a uni-variate one. It's not clear if the same be said about the $\ell\_p$ ball for $p \neq 2$. On a related note, is there an orthogonal/useful transform which is as "friendly" to the $\ell\_1$ metric as the Fourier transform is to the $\ell\_2$ metric? (probably not, by the fact that "orthogonal" is an $\ell\_2$ notion.)
| https://mathoverflow.net/users/7581 | Fourier transform of the unit ball in L1 metric | The Fourier transform $F(\mathbf{k})$ of the unit ball in L1 metric can be evaluated as follows$^\ast$.
*Notation:* $\theta(x)$ is the unit step function, $\delta(x)$ is the delta function, and ${\cal P}$ denotes the principal value of the integral.
$$F(k\_1,\ldots k\_n)=\int \cdots\int e^{i\mathbf{k\cdot x}}\theta\left(1-\sum\_{p=1}^n|x\_p|\right)dx\_1\cdots dx\_n$$
$$\qquad=\frac{1}{2\pi i}{\cal P}\int\_{-\infty}^\infty \frac{ds}{s}\left(1-e^{-is}\right)\prod\_{p=1}^n\left(\pi\delta(k\_p-s)+\pi\delta(k\_p+s)+\frac{2is}{s^2-k\_p^2}\right)\qquad \mathbf{(1)}$$
$$\qquad =\frac{1}{2\pi i}{\cal P}\int\_{-\infty}^\infty \frac{ds}{s}\left(1-e^{-is}\right)\prod\_{p=1}^n\left(\frac{2is}{s^2-k\_p^2}\right)$$
$$\qquad\qquad\qquad\qquad+\,\text{Re}\,\sum\_{q=1}^n\left[\frac{1-e^{-ik\_q}}{ik\_q}\prod\_{p=1,p\neq q }^n\left(\frac{2ik\_q}{k\_q^2-k\_p^2}\right)\right].\qquad \mathbf{(2)}$$
Then finally I arrive at the result
>
> $$F(k\_1,k\_2,\ldots k\_n)=2\,\text{Re}\,\sum\_{q=1}^n\left[\frac{1-e^{-ik\_q}}{ik\_q}\prod\_{p=1,p\neq q }^n\left(\frac{2ik\_q}{k\_q^2-k\_p^2}\right)\right].\qquad \mathbf{(3)}$$
>
>
>
I checked that for $n=2$ and $n=3$ this result agrees with the direct evaluation of the integrals (expressions given in the comment).
---
$^\ast$ Details of the calculation:
I first replace $\theta(1-\sum\_p|x\_p|)$ by $\theta(\xi-\sum\_p|x\_p|)$, denote the $\xi$-dependent integral by $F\_\xi$, and take the derivative with respect to $\xi$, $F\_\xi\mapsto F'\_\xi$, so that the step function becomes the delta function $\delta(\xi-\sum\_p|x\_p|)$. Then I Fourier transform with respect to $\xi$, $F'\_\xi\mapsto G\_s$. The $n$-fold integral of $G\_s$ over $x\_1$, $x\_2, \ldots x\_n$ factorizes, so I can carry it out explicitly, using the identity $\int\_0^\infty e^{ikx}\,dx=\pi\delta(k)+i{\cal P}k^{-1}$; finally I Fourier transform back $G\_s$ to the $\xi$ variable, to obtain $F'\_\xi$. One more integration of the $\xi$ variable, using $F\_{\xi=0}=0$, and then setting $\xi=1$ gives the first integral expression (1).
For the second integral expression I may safely assume that all $k\_p$'s are distinct, so products of delta functions do not contribute. I can then integrate out the single delta functions, obtaining the second integral expression (2).
The integral can then be performed by closing the contour in the lower half of the complex plane, the poles are shifted to the lower half of the complex plane, and the residue is reduced by a factor of two to account for the principal value. It then turns out that the integral equals exactly the contribution from the delta functions, so I end up with the final result (3).
| 4 | https://mathoverflow.net/users/11260 | 416254 | 169,630 |
https://mathoverflow.net/questions/171703 | 14 | My question is about a variant of the usual notion of relative constructibility, $\le\_c$ (which an earlier version of this question confusingly denoted "$\le\_L$"), in set theory.
Fix a countable transitive model $W\models\mathsf{ZFC+V\not=L}$. By a theorem of Barwise, given any set $A\in W$ there is a (possibly-ill-founded) end extension $W'\supseteq\_{end}W$ such that $W'\models A\in L$. In light of this, we can consider the following preorder on elements of $W$: $$A\le\_{L,end}B\quad:=\quad\forall W'\supseteq\_{end}W[W'\models \mathsf{ZFC}+B\in L\implies W'\models A\in L].$$
Broadly speaking, I'm interested in whether there is a nice description of $\le\_{L, end}$ (or an interesting subrelation, such as $\le\_{L, end}$ restricted to $\mathbb{R}^W$) without referring to end extensions - and in general, anything we can say about $\le\_{L, end}$, or the induced degree structure.
To keep things reasonably concrete, the following specific question seems natural:
>
> Is $\le\_{L, end}$ the same as $\le\_c$?
>
>
>
I'm especially interested in the situation where $V$ satisfies strong large cardinal axioms (say, "There is a proper class of Woodins) and $W$ is "far from $L$" (say, $\mathbb{R}^W$ is closed under sharps).
| https://mathoverflow.net/users/8133 | When is $A$ "$L$-ish" whenever $B$ is "$L$-ish"? | Remarks:
(i) I'm interpreting the definition of $\leq\_{L,\mathrm{end}}$ as quantifying over set models $W'$, not proper classes.
(ii) I'm considering the main question (comparing the two orders), particularly in the case that $V$ has large cardinals, but mainly not in the case of "particular interest", i.e. where $W$ is far from $L$; some remarks on the latter case are made in the "Edit" at the bottom.)
Claim: Assume ZF + there are ordinals $\kappa<\lambda$ such that $L\_\kappa\models$ZFC and $L\_\lambda\models$ZFC and $\kappa$ is a cardinal in $L\_\lambda$. Let $\psi$ be the statement
"there is a countable transitive $W\models$ZFC such that defining $\leq\_{L,\mathrm{end}}$ w.r.t. $W$, then $\leq\_{L,\mathrm{end}}$ is different to $\leq\_{\mathrm{c}}\upharpoonright W$". Then:
(i) There is a forcing extension $V[G\_1]$ of $V$ such that $V[G\_1]\models\psi$, and
(ii) If every real has a sharp (in particular, if there is a measurable cardinal), then $\psi$ holds in $V$.
In fact, we will get the two orders to disagree over $\mathbb{R}^W$.
Proof: The proofs of (i) and (ii) are almost the same, so I'll deal with them simulatenously.
Let $\lambda$ be least as hypothesized, and $\kappa$ the corresponding ordinal. Then $L\_\lambda$ is pointwise definable by condensation etc, so $\lambda<\omega\_1$,
and there is a bijection $\pi:\omega\to\lambda$ with $\pi\in L\_{\lambda+2}$.
If every real has a sharp,
then we can find a sequence $\vec{y}=\left<y\_\alpha\right>\_{\alpha<\lambda}$ of reals such that $y\_\alpha<\_{\mathrm{c}}y\_\beta$ for $\alpha<\beta<\lambda$, and in fact with $\vec{y}\upharpoonright \beta\in L[y\_\beta]$ for each $\beta<\lambda$. In any case, we can easily force the existence of such a sequence of reals. So from now on we assume that there is such a sequence.
We will find a sequence $G=\left<x\_n\right>\_{n<\omega}$ which
is generic over $L\_\lambda$ for the $\omega$-fold finite support product $\mathbb{C}^{<\omega}$ of Cohen forcing, such that setting $W=L\_\kappa[G]$ (not $W=L\_\lambda[G]$), we have $\leq\_{\mathrm{c}}\upharpoonright X$ is a wellorder of ordertype $\lambda$, where $X=\{x\_n\}\_{n<\omega}$, and hence this order is not in $W$
(in fact not in $L\_\lambda[G]$).
However, we do have $\leq\_{L,\mathrm{end}}\upharpoonright X\in W$. For this, it suffices to see it is in $L\_\lambda[G]$. And for this, it suffices to see it is in $L\_\lambda[G,H]$ whenever $H$ is $L\_\lambda[G]$-generic for $\mathrm{Coll}(\omega,\kappa)$ (by Solovay's theorem on this business; alternatively use homogeneity of the forcing and the uniformity of the next sentence). For the latter, we observe that $\leq\_{L,\mathrm{end}}\upharpoonright X$ is $\Pi^1\_1(\{z\})$ whenever $z$ is a real coding $L\_\kappa[G]$, and since $L\_\lambda[G,H]\models$ZFC and is transitive, it is therefore in $L\_\lambda[G,H]$. For the $\Pi^1\_1(\{z\})$-definability, it is easy assuming ZF+DC, as if there is an uncountable counterexample $W'$ to the $\forall W'$ quantifier, we can get a countable one by taking a countable hull (cf. Remark (i) at the start). Without DC we can still do a variant of this. Suppose $W'$ is a counterexample. Let $W''=L^{W'}$, and note that $W''$ is still a counterexample. Now let $W'''$ be the definable hull in $W''$ of parameters in $W\cup\{W\}$ (recalling $A,B\in W$); since $W''$ models "$V=L$", this gives an elementary substructure, and it is countable, so $W'''$ is a countable counterexample, as desired.
So we need to construct $G$.
Let $T\subseteq{^{<\omega}}2$ be a perfect tree.
Recall that $t\in T$ is a \emph{splitting node} of $T$
iff $t\frown(0)\in T$ and $t\frown(1)\in T$ (this is defined in the same manner for finite trees below). Given a real $x$,
let $b\_x$ be the infinite branch
through $T$ determined by using $x(n)$ as the bit of $b\_x$ following the $n$th splitting node along $b\_x$. So the map ${^\omega}2\to[T]$ (the codomain is the set of branches of $T$) sending $x\mapsto b\_x$ is a bijection.
Note that if $T\in L$,
then $x\equiv\_{\mathrm{c}}b\_x$.
Lemma: There is a perfect tree $T\subseteq {^{<\omega}}2$, with $T\in L$,
such that for every finite
sequence $\vec{b}=(b\_0,\ldots,b\_{m-1})$ of pairwise distinct branches $b\_i$ through $T$, $\vec{b}$ is $L\_\lambda$-generic
for $\mathbb{C}^m$ (the $m$-fold product of Cohen forcing).
Proof: For $m\in[1,\omega)$, let
$\mathscr{D}\_m$ be the set of all
open dense $D\subseteq\mathbb{C}^m$
such that $D\in L\_\lambda$.
Let $\mathscr{D}=\bigcup\_{m\in[1,\omega)}\mathscr{D}\_n$.
Fix an enumeration $\vec{D}=\left<D\_n\right>\_{n<\omega}$
of $\mathscr{D}$, such that each $D\in\mathscr{D}$ gets repeated infinitely often, and such that $\vec{D}\in L$. Let $m\_n$
be the arity of $D\_n$ (that is,
$D\_n\subseteq\mathbb{C}^{m\_n}$).
We construct a sequence $\left<T\_n\right>\_{n<\omega}$ of finite trees $T\_n\subseteq{^{k\_n}2}$
where $k\_n<\omega$, such that:
(i) $T\_0=\{\emptyset\}$ and $k\_0=0$,
(ii) for all maximal nodes $t$ of $T\_n$,
we have $\mathrm{lh}(t\_n)=k\_n$,
(iii) for all maximal nodes $t$ of $T\_n$,
there are exactly $n$
splitting nodes $s$ of $T\_n$
such that $s\subseteq t$ (so $s\subsetneq t$, since $t$ is maximal),
(iv) for each $i<n$, we have $T\_i=\{t\upharpoonright k\_i\bigm|t\in T\_n\}$, and
each maximal node of $T\_i$ is a splitting node of $T\_n$ (so $k\_i<k\_n$), and
(v)
for each $i<n$,
letting $m=m\_i$,
we have $\vec{t}\in D\_i$ for each
$m$-tuple $\vec{t}=(t\_0,\ldots,t\_{m-1})$
of pairwise distinct maximal nodes $t\_i$ of $T$.
The construction is straightforward by density: given $T\_n$,
construct $T\_{n+1}$ by first adding $t\frown(0)$ and $t\frown(1)$ to $T\_{n+1}$ for each maximal $t$ of $T\_n$, and then letting $m=m\_n$,
successively extend the $m$-tuples
of distinct so-far-maximal nodes so as to get them into $D\_n$. Since there are only finitely many such $m$-tuples, this is achieved after finitely many extensions, and then we just extend every node further up to some common length $k\_{n+1}$ (note there the longest splitting nodes of $T\_{n+1}$ are the maximal nodes of $T\_n$).
Setting $T=\bigcup\_{n<\omega}T\_n$, we claim this works. For let $\vec{b}=(b\_0,\ldots,b\_{m-1})$ be an $m$-tuple of distinct branches through $T$. Let $n\_0$ be large enough that $b\_i\upharpoonright k\_{n\_0}\neq b\_j\upharpoonright k\_{n\_0}$ whenever $0\leq i<j<m$. Then at every stage $n\in[n\_0,\omega)$
such that $m\_n=m$,
we ensured that $(b\_0\upharpoonright k\_{n+1},\ldots,b\_{m-1}\upharpoonright k\_{n+1})\in D\_n$.
Since every open dense $D\subseteq\mathbb{C}^m$
in $L\_\lambda$ gets repeated infinitely often in the enumeration,
it follows that $(b\_0,\ldots,b\_{m-1})$ is $L\_\lambda$-generic, as desired.
This completes the proof of the lemma.
Now let $z\_n=y\_{\pi(n)}$ for $n<\omega$ (recall
$\left<y\_\alpha\right>\_{\alpha<\lambda}$ and $\pi:\omega\to\lambda$ from earlier). Let $b\_n=b\_{z\_n}$ be the branch induced by $z\_n$ (as described earlier). So ($\*$) for all finite $m$-tuples of distinct integers $(n\_0,\ldots,n\_{m-1})$, $(b\_{n\_0},\ldots,b\_{n\_{m-1}})$ is generic over $L\_\lambda$ for $\mathbb{C}^m$.
However, we don't know that the
full sequence $\left<b\_n\right>\_{n<\omega}$ is generic over $L\_\lambda$ for the $\omega$-fold finite support product $\mathbb{C}^{<\omega}$ of $\mathbb{C}$. But by a standard trick,
using ($\*$),
we can modify each $b\_n$ on at most finitely many digits, producing a sequence $\left<b'\_n\right>\_{n<\omega}$
which is $L\_\lambda$-generic for $\mathbb{C}^{<\omega}$.
(Enumerate the dense subsets of $\mathbb{C}^{<\omega}$ in $L\_\lambda$
as $\left<E\_n\right>\_{n<\omega}$,
and progressively extend conditions
$p\_n$ getting into $E\_n$,
such that, letting $p\_{ni}$
be the projection of $p\_n$
to the $i$th component,
and letting $A\_i=\mathrm{dom}(p\_{ni})$,
we ensure that for $n'>n$,
we have that $p\_{n'i}$ agrees
with $b\_n$ outside of $A\_i$.
This can be achieved using (\*),
since $\mathbb{C}^{<\omega}$ factors
nicely.)
Now let $G=\left<b'\_n\right>\_{n<\omega}$. So $G$ is generic over $L\_\lambda$ for
$\mathbb{C}^{<\omega}$, and note
that since $b'\_n$ eventually agrees with $b\_n$,
we have $b'\_n\equiv\_{\mathrm{c}} b\_n\equiv\_{\mathrm{c}}z\_n=y\_{\pi(n)}$. Therefore
letting $X=\{b'\_n\}\_{n<\omega}$,
the restriction $\leq\_{\mathrm{c}}\upharpoonright X$ is just a wellorder of length $\lambda$,
as desired.
---
Edit: Consider now the case that $W$ is closed under sharps, assuming that $V$ is closed under sharps for reals. Then the preceding argument
does not work, and in fact we have the following (related to some things Hamkins wrote in his answer above):
Theorem 2: Assume ZF + for every real $x$, $x^\#$ exists.
Let $W$ be a countable transitive model of ZFC which is closed under (true) sharps.
Define $\leq\_{L,\mathrm{end}}$ w.r.t. $W$.
Let $A,B\in W$.
Then $A\leq\_{\mathrm{c}}B$ implies $A\leq\_{L,\mathrm{end}}B$.
Proof: Suppose $A\leq\_{\mathrm{c}}B$, i.e. $A\in L(B)$. Then since $B^\#\in W$,
it follows that $A\in L^W(B)=L\_{\mathrm{OR}^W}(B)$. Let $W'$ be any end-extension of $W$ which models ZFC with $B\in L^{W'}$. If $\mathrm{OR}^{W'}=\mathrm{OR}^W$
then $L^{W'}=L^W$,
so $B\in L^W$, so $L^W(B)=L^W$,
so $A\in L^W=L^{W'}$. Suppose instead
that $\mathrm{OR}^W\neq\mathrm{OR}^{W'}$. Let $\gamma$ be the least $W'$-ordinal such that $B\in L^{W'}$ (note $\gamma$ might be illfounded). Let $\alpha\in\mathrm{OR}^W$ be such that $A\in L\_\alpha(B)$. Then
$\alpha\in\mathrm{OR}^{W'}$,
so $\gamma+\alpha$ makes sense in $W'$ and $\gamma+\alpha\in\mathrm{OR}^{W'}$, which easily implies that $L\_\alpha(B)\subseteq L^{W'}$,
so $A\in L^{W'}$, as desired.
| 3 | https://mathoverflow.net/users/160347 | 416259 | 169,632 |
https://mathoverflow.net/questions/416249 | 1 | Let $f(x)$ be a strictly increasing function such that $\lim\limits\_{x\to\pm\infty}f(x)=\pm\infty$ and $\lim\limits\_{x\to+\infty}f'(x)e^{f(x)-x}=+\infty$. If $f(a)=0$ for some prescribed $a\in\Bbb R$, does a minimum exist for $$\int\_{\Bbb R}xe^{f(x)-x^2}\,dx$$ and if so, what would be the minimiser $f^\*$? The presence of the exponential terms comes from a log-normal distribution integral.
The given conditions mean that $f$ has asymptotic order $x^\alpha$ for some $\alpha\in[1,2)$ as $x\to+\infty$. The goal is to maximise the negative area on the negative reals (whilst minimising the positive area) so perhaps we may need to define $f\in C^1$ piecewise, as there is no constraint on the rate of increase on the negative side.
| https://mathoverflow.net/users/113397 | Does a minimiser exist for this Gaussian-like functional? | $\newcommand\R{\mathbb R}$For any $a\in\R$, there is no minimizer of
\begin{equation\*}
I(f):=\int\_\R xe^{f(x)-x^2}\,dx \tag{-1}\label{-1}
\end{equation\*}
over all $f\in F\_a$, where $F\_a$ is the set of all strictly increasing functions $f\colon\R\to\R$ such that $\lim\_{x\to\pm\infty}f(x)=\pm\infty$, $\lim\_{x\to\infty}f'(x)e^{f(x)-x}=\infty$, and $f(a)=0$. Also,
\begin{equation\*}
\inf\_{f\in F\_a}I(f)=0 \tag{0}\label{0}
\end{equation\*}
for any $a\in\R$.
Indeed, take any $a\in\R$ and any $f\in F\_a$.
Since $f$ is strictly increasing, we have
\begin{equation\*}
xe^{f(x)-x^2}>xe^{f(0)-x^2} \tag{1}\label{1}
\end{equation\*}
for all nonzero real $x$.
Consider now the following two possible cases:
*Case 1: $a\le0$.* Then $f(x)<f(a)=0$ and $x<0$ for real $x<a$ and hence
\begin{equation\*}
\begin{aligned}
\int\_{-\infty}^a xe^{f(x)-x^2}\,dx&>\int\_{-\infty}^a xe^{-x^2}\,dx.
\end{aligned}
\tag{2}\label{2}
\end{equation\*}
Also, by \eqref{1},
\begin{equation\*}
\begin{aligned}
\int\_a^\infty xe^{f(x)-x^2}\,dx&>\int\_a^\infty xe^{f(0)-x^2}\,dx \\
&=e^{f(0)}\int\_a^\infty xe^{-x^2}\,dx \\
&\ge e^{f(a)}\int\_a^\infty xe^{-x^2}\,dx \\
&=\int\_a^\infty xe^{-x^2}\,dx,
\end{aligned}
\tag{3}\label{3}
\end{equation\*}
because $f$ is increasing, $a\le0$, $\int\_a^\infty xe^{-x^2}\,dx\ge0$,
and $f(a)=0$.
In view of \eqref{-1}, \eqref{2}, \eqref{3}, and the equality $\int\_{-\infty}^\infty xe^{-x^2}\,dx=0$,
\begin{equation\*}
I(f)>0 \tag{4}\label{4}
\end{equation\*}
in Case 1.
*Case 2: $a>0$.* Then, again by \eqref{1},
\begin{equation\*}
\begin{aligned}
\int\_{-\infty}^a xe^{f(x)-x^2}\,dx&>\int\_{-\infty}^a xe^{f(0)-x^2}\,dx \\
&=e^{f(0)}\int\_{-\infty}^a xe^{-x^2}\,dx \\
&>e^{f(a)}\int\_{-\infty}^a xe^{-x^2}\,dx \\
&=\int\_{-\infty}^a xe^{-x^2}\,dx,
\end{aligned}
\tag{5}\label{5}
\end{equation\*}
because $f$ is increasing, $a>0$, $\int\_{-\infty}^a xe^{-x^2}\,dx<0$,
and $f(a)=0$.
Also, here $f(x)>f(a)=0$ and $x>0$ for real $x>a$ and hence
\begin{equation\*}
\begin{aligned}
\int\_a^\infty xe^{f(x)-x^2}\,dx&>\int\_a^\infty xe^{-x^2}\,dx.
\end{aligned}
\tag{6}\label{6}
\end{equation\*}
In view of \eqref{-1}, \eqref{5}, \eqref{6}, inequality \eqref{4} holds
in Case 2 as well.
Take now any $k$ and $A$ in $(0,\infty)$, and let
\begin{equation\*}
f\_{a,k,A}(x):=k(x-a)\,1(x\le a+A)+(kA+2(x-A))\,1(x>a+A)
\end{equation\*}
for real $x$. Then $f\_{a,k,A}\in F\_a$ and, by dominated convergence,
\begin{equation\*}
\lim\_{k\downarrow0}\lim\_{A\to\infty}I(f\_{a,k,A})
=\lim\_{k\downarrow0}\int\_\R xe^{k(x-a)-x^2}\,dx
=\int\_\R xe^{-x^2}\,dx=0.
\end{equation\*}
So, $\inf\_{f\in F\_a}I(f)\le0$. Now \eqref{0} follows by \eqref{4}, which also shows that there is no minimizer of $I(f)$ over $f\in F\_a$. $\quad\Box$.
| 1 | https://mathoverflow.net/users/36721 | 416260 | 169,633 |
https://mathoverflow.net/questions/416271 | 4 | I am interested in extensions $A\leq B$ of commutative rings with the property that for all ideals $I\leq A$ we have $IB\cap A=I$. Is there a standard name for this property, or a standard reference for results about it?
| https://mathoverflow.net/users/10366 | Terminology/literature for $\forall I\leq A,\; IB\cap A=I$ | Such extension is called "cyclically pure". An extension is called pure if the induced map $A\otimes\_A M\to B\otimes\_A M$ is injective for any $A$ module $M$. If the map $A\to B$ splits as map of $A$-modules, then it is pure, and the converse holds if $B$ is finitely presented as an $A$-module.
Also clearly, purity implies cyclic purity (taking $M=A/I$), and a classic paper by Hochster addressed the converse:
[Cyclic purity versus purity in excellent Noetherian rings](https://www.ams.org/journals/tran/1977-231-02/S0002-9947-1977-0463152-5/S0002-9947-1977-0463152-5.pdf).
| 3 | https://mathoverflow.net/users/2083 | 416283 | 169,643 |
https://mathoverflow.net/questions/416269 | 8 | Let $d$ be a positive number.
There is a two dimensional recurrence relation as follow:
$$R(n,m) = R(n-1,m-1) + R(n,m-d)$$
where
$R(0,m) = 1$ and $R(n,0) = R(n,1) = \cdots = R(n, d-1) = 1$ for all $n,m>0$.
How to analyze the asymptotics of $R(n, kn)$ for fixed $k$?
It is easy to see that
$$R(n, kn) = O\left( c\_{k,d}^{n} \cdot (n+k+d)^{O(1)} \right)$$
Is there a way (or an algorithm) to find $c\_{k,d}$ given $k$ and $d$?
PS: I have calcuated the bivariate generating function of $R(\cdot, \cdot)$:
\begin{align}
f(x,y)
&= \frac{1 - xy - y^{d} + xy^{d}}{(1 - x)(1 - y)(1 - xy - y^{d})} \\
&= \frac{1}{(1 - x)(1 - y)} + \frac{xy^{d}}{(1 - x)(1 - y)(1 - xy - y^{d})} \\
\end{align}
| https://mathoverflow.net/users/115910 | How to find the asymptotics of a linear two-dimensional recurrence relation | Your generating function $f(x,y)$ is convergent on a polydisk $|x|<\epsilon$, $|y|< \delta$, for some $\epsilon, \delta < 1$. We can reduce $\epsilon$ so that $\epsilon \ll \delta^k$ and the domain of $f(x,y)$ to the product of the polydisk $|x|<\epsilon$ and the annulus $\delta' < |y| < \delta$, with $\delta' \sim \delta$. Then $g(z,y) = f(z/y^k, y)$ is well-defined and analytic on the product domain of $|z|\lesssim \epsilon/\delta^k$ and $\delta' < |y| < \delta$. The generating function $g(z,y)$ has a unique expansion that is a power series in $z$ and a Laurent series in $y$. Since $x^n y^m = (z/y^k)^n y^m = z^n y^{m-kn}$, the coefficients $R(n,kn)$ are the coefficients of $z^n y^0$ in this double expansion. Hence, using the Cauchy integral formula, the generating function for $R(n,kn)$ is now given by the contour integral
$$
\sum\_{n=0}^\infty R(n,kn) z^n = \frac{1}{2\pi i} \oint\_{|y|\sim \delta} g(z,y) \frac{dy}{y} .
$$
Of course, the integration contour is now allowed to deform, as long as it doesn't hit any singularities of the integrand.
The above integrand has a convenient partial fraction expansion expansion with respect to $z$,
$$
\frac{g(z,y)}{y} = \frac{y^k}{(1-y-y^d)(y^k-z)}
- \frac{y^{d+k-1}(1-y^d)}{(1-y)(1-y-y^d) (y^{k-1} (1-y^d)-z)} .
$$
Expanding in $z$ under the integral gives the formula
$$
R(n,kn) = \frac{1}{2\pi i} \oint \frac{dy}{y^{kn} (1-y-y^d)}
- \frac{1}{2\pi i} \oint \frac{y^d dy}{(1-y)(1-y-y^d) y^{n(k-1)} (1-y^d)^n} ,
$$
whose leading asymptotics can be estimated by residues or steepest descent.
Using residues on the first term, all the contributions come from the roots of $1-y-y^d=0$. The root with the smallest magnitude $y\_\*$ will give the leading contribution. Some experimentation shows that $y\_\*$ is the unique positive real root. The leading asymptotic term then looks like
$$
C\_\* y\_\*^{-kn} .
$$
In the special case $d=1$, $y\_\* = 1/2$. In the special case $k=1$, the second term has no poles inside its contour and hence evaluates to zero. So in the sequel we can assume that $k>1$.
Applying [steepest descent](https://en.wikipedia.org/wiki/Method_of_steepest_descent) to the second integral, we find the stationary phase points at the roots $y\_\star$ of $y^d = (k-1)/(k+d-1) + O(1/n)$. Let $y\_\star$ be that root (actually we just need its $n\to \infty$ limit) which minimizes the magnitude of $y\_\star^{k-1} (1-y\_\star^d)$. It is easy too see that $y\_\star = \sqrt[d]{(k-1)/(d+k-1)}$. Then the leading asymptotic term is
$$
C\_\star n^{-1/2} y\_\star^{-n(k-1)} (1-y\_\star^d)^{-n}
= C\_\star n^{-1/2} w\_\star^{-kn} ,
$$
where $w\_\star = \left(\frac{k-1}{d+k-1}\right)^{\frac{1-1/k}{d}} \left(\frac{d}{d+k-1}\right)^{1/k} = \frac{s^{s/k}}{(s+1)^{(s+1)/k}}$ with $s=(k-1)/d$, as pointed out in the comments.
The conclusion is that
$$
R(n,kn) = O\left((\min(y\_\*,w\_\star)^{-k})^n\right) .
$$
Experimentation suggests that $y\_\* < w\_\star$ in all cases, except for $(k,d)=(2,1)$, when $y\_\* = w\_\star = 1/2$. Otherwise the $y\_\*$ contribution always seems to dominate over the $w\_\star$ contribution. At least that's how it seemed for not too large values of $d$.
Doing some rudimentary asymptotic calculations for large $d$, it seems that $y\_\* \sim 1- \frac{\log d}{d} + o(\log(d)/d)$. A better approximation is
$$
y\_\* \sim 1 - \frac{W(d)}{d} + \frac{W(d)^3}{2(W(d)+1) d^2} + O(W(d)^3/d^3) ,
$$
where $W(d)$ is the [Lambert W function](https://en.wikipedia.org/wiki/Lambert_W_function). On the other hand, $w\_\star$ has a minimum as a function of $k$ (treating it as a continuous variable) around $k\_\star \sim W(d) + 1 + O(W(d)^2/d)$, the minimum reaches roughly
$$
w\_\star \sim 1 - \frac{W(d)}{d-1} \lesssim y\_\* .
$$
One should push the asymptotics of $k\_\star$ one more order to get a better estimate of the difference, but experiments do show that the minimum of $w\_\star$ does dip below $y\_\*$, for instance this happens near $(k,d) = (5, 205)$ or $(6,700)$. So it looks like the $w\_\star$ asymptotic contribution will dominate in small ranges of $k \sim k\_\star(d)$ if it happens to fall near an integer. To get the size of that window, one could go to a quadratic Taylor approximation of $w\_\star$.
With a bit of extra work, one could also extract the coefficients $C\_\*$ and $C\_\star$ from the integral formula.
| 3 | https://mathoverflow.net/users/2622 | 416295 | 169,647 |
https://mathoverflow.net/questions/416289 | 0 | This question is an "outgrowth" of <https://math.stackexchange.com/questions/4380919/> which led to a numerically-generated two-parameter function $f\_b(n)$, where $b$ is the number base $2,3,4,\ldots$, and $n$ is as follows, paraphrasing the original question...
What is the number of $n$-character words consisting of $0$'s and $1$'s (and *etc* for bases greater than $2$) such that the number of $1$'s, counting from the left, is at all times greater than the number of $0$'s? When $b\gt2$, the number of $1$'s is greater than the combined number of all other digits. (Note: the original question only asked about the $b=2$ case.)
I couldn't see how to rigorously set up the problem and derive a solution. So instead, I just programmed the problem, generating the following sequences for different bases
```
n: 1 2 3 4 5 6 7 8 9 10 11 12
base 2: 1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 462
base 3: 1, 1, 3, 5, 15, 29, 87, 181, 543, 1181, 3543, 7941
base 4: 1, 1, 4, 7, 28, 58, 232, 523, 2092, 4966, 19864, 48838
base 5: 1, 1, 5, 9, 45, 97, 485, 1145, 5725, 14289, 71445, 185193
base 6: 1, 1, 6, 11, 66, 146, 876, 2131, 12786, 32966, 197796
```
Googling those sequences immediately finds
```
base 2: https://oeis.org/A001405
base 3: https://oeis.org/A126087
base 4: https://oeis.org/A128386
base 5: https://oeis.org/A121724
base 6: ????? (couldn't google anything for the b=6 sequence)
```
For base 2, that straightforward *oeis* answer is simply the binomial coefficient $f\_2(n)=\left(n-1 \atop \lfloor\frac{n-1}2\rfloor\right)$ (which immediately answered the original question). But then we get entirely different mathematical functions for each different base $b$. At least I'm not seeing any sensible relationship between them.
So, the overall result is that we have our $f\_b(n)$ generated in exactly one numerical way, but identified with apparently entirely unrelated mathematical functions for $b=2,3,4,5$ (and for $b=6$ I couldn't google anything). So perhaps we're looking at some new more general function, for which the already-known $b=2,3,4,5$ functions are just special cases. Anyway, that's the conjecture I'm asking about, and suggesting might be worth investigating further.
And in case it's of any interest the small **C** program is below. To generate results for base $b$ between $n\_1\le n\le n\_2$ run it with the three command-line arguments **n1 n2 b** (note that it just uses $32$-bit signed ints, so run it with $b^{n\_2}\lt2^{31}$)
```
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
int main ( int argc, char *argv[] ) {
int n1 = ( argc>1? atoi(argv[1]) : 1 ),
n2 = ( argc>2? atoi(argv[2]) : 10 ),
base = ( argc>3? atoi(argv[3]) : 2 );
int n = n1;
int ipow(), binky(), nwords=0, ichar=0, iword=0,
isprefmax(char*,char,int), imax[99],nmax[99],nprintmax=1;
char *itoa(), basechars[99] =
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz#$*";
for ( n=n1; n<=n2; n++ ) {
nwords = ipow(base,n); nprintmax = 1;
for ( ichar=0; ichar<base; ichar++ ) imax[ichar] = 0;
for ( iword=0; iword<nwords; iword++ ) {
char *aword = itoa(iword,base,n);
for ( ichar=0; ichar<base; ichar++ )
if ( isprefmax(aword,basechars[ichar],base) ) imax[ichar] += 1; }
printf("n=%2d, nwords=%8d, #max=", n,nwords);
for ( ichar=1; ichar<base; ichar++ )
if ( imax[ichar] != imax[ichar-1] ) nprintmax=base;
nmax[n] = (nprintmax==1?imax[0]:-999);
for ( ichar=0; ichar<nprintmax; ichar++ ) printf("%6d",imax[ichar]);
if (base==2) printf(", (%2d,%2d)=%6d", n-1,(n-1)/2,binky(n-1,(n-1)/2));
printf("\n");
} /* --- end-of-for(n) --- */
for ( n=n1; n<=n2; n++ ) printf("%d%s",nmax[n],(n<n2?", ":"\n"));
exit(0); }
int isprefmax ( char *a, char c, int base ) {
int i=0, nc=0, nother=0, ichar=0; int ismax=0;
char basechars[99] =
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz#$*";
if ( a == NULL ) goto end_of_job;
for ( i=0; a[i]!='\000'; i++ ) {
if ( a[i] == c ) nc++; else nother++;
if ( nother >= nc ) goto end_of_job; }
ismax = 1;
end_of_job: return ( ismax ); }
int ipow ( int base, int exp ) {
int basetoexp = 1;
while ( 1 ) {
if ( exp&1 ) basetoexp *= base;
exp >>= 1;
if ( !exp ) break;
base *= base; }
return ( basetoexp ); }
int binky ( int n, int k ) {
/* --- allocations and declarations --- */
int this_binky= (-1), /* init for error */
binky_max = INT_MAX/2; /* assert binky <= binky_max */
/* --- check args for "convergence" --- */
if ( n<k || k<0 ) return ( -1 ); /* argument error */
if ( n==k /* default=1 if n == k */
|| n<1 || k<1 ) return ( 1 ); /* default=1 if n or k == 0 */
/* --- recurse (in case one curse isn't enough) --- */
this_binky = binky(n-1,k-1) + binky(n-1,k);
if ( this_binky > binky_max ) this_binky = (-1); /* overflow check??? */
return ( this_binky );
} /* --- end-of-function binky() --- */
char *itoa ( int i, int base, int len ) {
static char a[99], digits[99] = /* up to base 65 */
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz#$*";
int n=97;
memset(a,'0',99); a[98] = '\000';
while ( 1 ) {
a[n--] = digits[i%base];
if ( (i/=base) < 1 ) break; }
if ( len > 97-n ) n = 97-len;
return ( a+n+1 ); }
/* --- end-of-file --- */
```
| https://mathoverflow.net/users/476477 | What is this numerically-generated function? | First, we notice that the first character must be 1. Let's take it off and focus on the remaining $n-1$ characters, where the number of 1's in each prefix must be at least as many as the number of the other $b-1$ digits.
Replacing each 1 with `[` and each other digit with `]`, we get a truncated [Dyck word](https://en.wikipedia.org/wiki/Dyck_language), where a few `]]...]` at the end are truncated. The number of such words with $n-1-k$ `[` and $k$ `]` equals
$\frac{n-2k}{n}\binom{n}{k} = \binom{n}{k} - 2\binom{n-1}{k-1}$. It remains to distribute $b-1$ digits $\ne 1$ among the $k$ positions, resulting in the formula:
\begin{split}
f\_b(n) &= \sum\_{k=0}^{\lfloor (n-1)/2\rfloor} \frac{n-2k}{n}\binom{n}{k} (b-1)^k \\
&= \sum\_{k=0}^{\lfloor (n-1)/2\rfloor} \left( \binom{n}{k} - 2\binom{n-1}{k-1}\right) (b-1)^k.
\end{split}
I doubt there is a simple expression for $b>2$, but the generating function can be derived from here.
| 0 | https://mathoverflow.net/users/7076 | 416299 | 169,649 |
https://mathoverflow.net/questions/320814 | 4 | A functor $N\colon\mathrm{Cat}\_{A\_\infty}\longrightarrow\mathrm{Cat}\_\infty$ is constructed in a paper [1] by Faonte. This gives a way to get an $\infty$-category by starting with an $A\_\infty$-category.
Going the other way, is it possible to *define* linear $A\_\infty$-categories as special $\infty$-categories?
---
References
----------
[1] Simplicial nerve of an A-infinity category (Giovanni Faonte, [arXiv:1312.2127)](https://arxiv.org/abs/1312.2127), suggested by DamienC in an answer to [MO152370](https://mathoverflow.net/questions/152370/are-infty-1-categories-a-infty-categories).
| https://mathoverflow.net/users/130058 | Is it possible to define linear $A_\infty$-categories as special $\infty$-categories? | An affirmative to the "conjecture" above is fully recorded in this work:
<https://arxiv.org/abs/2003.05806>
In Remark 1.2, for example, we comment how Gepner-Haugseng's results imply that k-linear A-infinity categories are precisely k-chain-complex-enriched infinity-categories.
This passes through the infinity-categorical equivalence between k-linear dg-categories and k-linear A-infinity categories. The proof is completely in line with Rune's comments--in fact, it was based on a discussion I had with Rune back before COVID.
By the way, if you define a k-linear infinity-category to be an infinity-category enriched over k-chain-complexes, you're fine. (Re: Denis's comment.) But some people define k-linear infinity-categories as those with an action of the infinity-category of k-linear chain complexes; then you're not fine. Not all A-infinity-categories have (even finite) colimits, for example. I think this is what Yonatan points out in the comments.
So, to address OP's original question: No, k-linear A-infinity-categories are not a full subcategory of infinity-categories.
| 4 | https://mathoverflow.net/users/3593 | 416305 | 169,651 |
https://mathoverflow.net/questions/416302 | 15 | First of all, I am interested in the general case of a non-orientable manifold but let's for now consider the projective plane $\mathbb{R}P^2.$ In short, I am curious if there is any relation between the diffeomorphism group $\text{Diff}(\mathbb{R}P^2)$ of the projective plane and the diffeomorphism group $\text{Diff}(S^2)$ of its orientation double cover.
As I understand, any diffeomorphism $\mathbb{R}P^2\to \mathbb{R}P^2$ can be lifted to a diffeomorphism $S^2\to S^2$ of the orientation bundle (same in the general case). That means that we can consider $\text{Diff}(\mathbb{R}P^2)$ as a subgroup in $\text{Diff}(S^2).$ I know that there are a lot of results on geometry and topology of this group and I wonder if some results remain true for $\text{Diff}(\mathbb{R}P^2).$ I tried to search for some literature on this topic but didn't find anything useful for me (partially because I am not sure what I am looking for). My final goal for now is to compute (or find results on) the curvature for a right-invariant metric on the group of volume-preserving diffeomorphisms $\text{SDiff}(\mathbb{R}P^2).$ I would be very much interested to learn anything on this topic.
Thanks.
| https://mathoverflow.net/users/131858 | Diffeomorphism group of the projective plane | Two different answers using almost identical techniques! Allen's response got me to think through my response more carefully. Let me edit in a comment to point out my sloppiness, as it points out a useful detail in the machinery we are using.
$\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\Emb{Emb}\DeclareMathOperator\SO{SO}$One approach to studying $\Diff(\Bbb RP^2)$ would be to look at the bundle
$$\Diff(D^2) \to \Diff(\Bbb RP^2) \to \Emb(S^1, \Bbb RP^2).$$
**Edit:** technically this is not the fiber. Since the $D^2$ is not embedded, the fiber is the diffeomorphisms of $\Bbb RP^2$ that restrict to the identity on the embedded curve. If we blow up the curve, you can think of this as the group of diffeomorphisms of $D^2$ that are either the identity on the boundary, or the antipodal map on the boundary. Sticking with the convention that $Diff(D^2)$ is diffeomorphisms of the $2$-disc that are the identity on the boundary, the fiber would be $Diff(D^2) \times \Bbb Z\_2$. So Smale's theorem tells us $Diff(\Bbb RP^2)$ is (up to a homotopy-equivalence) a $2$-sheeted covering space of the component of $\Bbb RP^1$ in $Emb(S^1, \Bbb RP^2)$. A more systematic way to see this would be to consider the space of embeddings of a tubular neighbourhood of $\Bbb RP^1$ in $\Bbb RP^2$. There is the embedded circle it contributes, but there are also the automorphisms of the tubular neighbourhood, switching the directions of the fibers of the M"obius band. The fiber of the restriction map $Diff(\Bbb RP^2) \to Emb(M, \Bbb RP^2)$ where $M$ is the M"obius band, now is literally a copy of $Diff(D^2)$.
The bundle is not onto the base space, it is onto the subspace of embeddings whose complement is a disc, i.e. embedded curves whose normal bundles are Moebius bands.
I believe this embedding space has the homotopy-type of the subspace of linear embeddings. The linear embeddings has the homotopy-type of the unit tangent bundle of $\Bbb RP^2$, which could be thought of as
$$ UTS^2 / \Bbb Z\_2 $$
where in the action one negates the base-point and the tangent vector simultaneously.
$UTS^2$ can be thought of as a copy of $\SO\_3$, which I believe would make the quotient a copy of the lens space $L\_{4,1}$ — the action of $\Bbb Z\_2$ on $\SO\_3$ performs a rotation by $\pi$ on the first two column vectors. **Edit:** Passing to the covering space, we recover that $Diff(\Bbb RP^2)$ is homotopy-equivalent to $\SO\_3$.
Let me know if that makes sense or not. Sometimes I am a little rusty when writing answers in the morning.
| 15 | https://mathoverflow.net/users/1465 | 416310 | 169,653 |
https://mathoverflow.net/questions/416220 | 2 | I originally asked this on MSE, but did not get an answer there.
Let $M$ be a von Neumann algebra. Let $\varphi: M\_+ \to [0, \infty]$ be a weight on $M$. Consider
\begin{align\*}&\mathfrak{p}\_\varphi:= \{x\in M\_+: \varphi(x) < \infty\}\\
&\mathfrak{n}\_\varphi:= \{x \in M: \varphi(x^\*x) < \infty\}\\
& \mathfrak{m}\_\varphi:=\mathfrak{n}\_\varphi^\* \mathfrak{n}\_\varphi=
\left\{\sum\_{j=1}^ny\_j^\*x\_j: x\_j ,y\_j \in \mathfrak{n}\_\varphi\right\}\end{align\*}
In Takesaki's second volume, chapter VII, (the proof of) lemma 1.9, the following is claimed:
If $x =x^\* \in \mathfrak{m}\_\varphi$ and $x= u|x|$ is its polar decomposition, then $|x| \in \mathfrak{m}\_\varphi$.
**Question**: Why is this the case?
**Attempt**: We have $|x|= x^+ + x^-$, so I tried to show that $x^+, x^- \in \mathfrak{m}\_\varphi$. Now, we can write $x=p-q$ where $p,q \in \mathfrak{p}\_\varphi= \mathfrak{m}\_\varphi^+$ so if we would have $p \ge x^+$ and $q \ge x^-$, we would be done by the hereditary property. I think these inequalities are true when $p$ and $q$ commute, but I don't know if we can choose this decomposition with $pq = qp$.
Thanks in advance for any help or suggestions!
| https://mathoverflow.net/users/216007 | Decomposition of an element as a difference of positive elements in the definition subalgebra of a weight (Takesaki) | I don't know what Takesaki had in mind for the proof, but what you're asking is incorrect. Here is a counterexample where $\phi$ in the counterexample is a normal faithful semifinite (n.f.s.) weight.
Let $H$ be a separable infinite dimensional Hilbert space, and $M = B(H\oplus H)$. Fix a faithful normal state $\psi$ on $B(H)$ and let $\mathrm{Tr}$ be the standard trace on $B(H)$. Define $\phi ((x\_{ij})\_{i,j=1,2}) = \mathrm{Tr}(x\_{11}) + \psi(x\_{22})$. This is clearly a n.f.s. weight.
Fix a positive contraction $a\in B(H)$ which is trace-class but so that $a^{1/2}$ is not trace-class. Let $p = \left( \begin{array}{cc} a & (a-a^2)^{1/2} \\ (a-a^2)^{1/2} & 1-a \end{array} \right)$ and $q= \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)$ which are projections that are both in $\mathfrak m\_{\phi}$. Hence $x=p-q \in \mathfrak m\_\phi$ is self-adjoint. A straightforward computation gives $x^2 = \left( \begin{array}{cc} a & 0 \\ 0 & a \end{array} \right)$ and thus $|x| = \left( \begin{array}{cc} a^{1/2} & 0 \\ 0 & a^{1/2} \end{array} \right)$. As $a^{1/2}$ is not trace-class we have $\phi(|x|) = \infty$ so $|x| \notin \mathfrak m\_\phi$.
| 4 | https://mathoverflow.net/users/126109 | 416318 | 169,656 |
https://mathoverflow.net/questions/416246 | 1 | I am interested in the following situation: Given any $n>1$ suppose I have a codimension-1 foliation of $R^n\_{++}$ (i.e. the subset of strictly positive $n$-vectors) arising from an $(n-1)$-dimensional, involutive, plane field $F$. Further, suppose that the normal to the subspace spanned by the vectors of each $F\_p$ is either strictly positive for all $p \in R^n\_{++}$ or strictly negative for all $p \in R^n\_{++}$ (so I take it to be the former).
My understanding is that this situation can arise from considering the inverse images of $f(p),p \in R$, where f is a smooth real-valued function on $R^n\_{++}$ (with surjective differential everywhere). My question is: Does such a foliation *have* to arise in this manner? Does there always exist such an $f$? The reason I ask is because I would like to be able to construct a complete ordering of the leaves of this type of foliation; if my question has an answer in the affirmative then I can do it by means of $f$. Otherwise, I would have to find a different way (say by considering "positive" line segments going from one leaf to another). I've looked at a paper by Novikov dealing with partial orderings of codimension-1 foliations but the foliation itself is too general for my needs.
Any references to material dealing with such a question would be welcome (as would any corrections to my understanding of the subject matter).
| https://mathoverflow.net/users/92328 | Codimension-1 foliations of Euclidean space with strictly positive normal bundle | I believe the answer is yes. It's enough to assume that the last coordinate of each normal is positive. Then each leaf is a graph $x\_n=x\_n(x\_1,\ldots, x\_{n-1})$ on some domain (depending on the leaf) in $\mathbb R^{n-1}$. This function must go to infinity at the boundary of the domain (if the boundary is nonempty) which implies that all leaves are proper. If they are proper then each leaf separates $\mathbb R^n$. This seems clear but there is also a reference (see section 2 in [this paper](https://link.springer.com/article/10.1007%2FBF02621903)). This allows you to get a linear ordering on the space of leaves given your condition on the normals.
Edit: sorry my example claiming that the leaf space might be non Hausdorff was wrong so I removed it. You always get that in this situation the leaf space is Hausdorff. Given any two leaves $F\_1<F\_2$ the space of leaves between them $\{F\_1<F<F\_2\}$ is open and such sets can separate any two leaves. Then the space of leaves in $\mathbb R$ and the projection from $\mathbb R^n$ to that $\mathbb R$ is the desired function $f$. I am not sure what can be said about the regularity of this function however.
Edit: The question was changed to a foliation on the first quadrant $\mathbb R^n\_{++}$. This case easily reduces to the case of a foliation on $\mathbb R^n$ considered above. Take a increasing diffeomorphism $\phi: (0,\infty)\to \mathbb R$. Then use $\Phi:\mathbb R^n\_{++}\to\mathbb R^n$ given by $\Phi(x\_1,\ldots, x\_n)=(\phi(x\_1),\ldots, \phi(x\_n))$. Push forward the hyperplane field to $\mathbb R^n$ using this map. The assumption on the normals is preserved.
| 1 | https://mathoverflow.net/users/18050 | 416327 | 169,660 |
https://mathoverflow.net/questions/416258 | 14 | Erdős and Selfridge (1971) state that the following is "implied by an unpublished result of Rosser" which they claim appears in a forthcoming book on sieve methods by Halberstam and Richert.
I guess I could try to find something implying it in that 400-page book (called "Sieve Methods", available online in unsearchable format) but I thought I'd ask if maybe somebody knows a precise reference [or a proof!] for this result?
>
> For every $\epsilon>0$, there is a constant $c=c(\epsilon)$ such that at least $c\frac{k}{\log k}$ integers between $n$ and $n+k$ have all their prime factors greater than $k^{1/2-\epsilon}$.
>
>
>
| https://mathoverflow.net/users/4600 | Unpublished result of Rosser in Sieve Methods book | This is part of the theory of the "linear sieve." Chapter 8 of the book of Halberstam and Richert deals with this topic. Alternatively you could look at Iwaniec's paper [On the error term in the linear sieve](http://matwbn.icm.edu.pl/ksiazki/aa/aa19/aa1911.pdf). The result you want can be extracted from the lower bound (1.4) on page 2 of Iwaniec's paper, together with the formula for $f(s)$ at the top of that page.
| 10 | https://mathoverflow.net/users/38624 | 416328 | 169,661 |
https://mathoverflow.net/questions/416333 | 2 | Let $A$ be real matrix with $M > 1$ rows and $K > 2$ columns, and each entry $a\_{m,k} \in (0,1)$, with each row summing to $1$. For all $m$
$$
\sum\_{k=1}^{K} a\_{m,k} = 1
$$
I want to find out if for any row $m$
$$
\sum\_{k=1}^{K} \frac{(a\_{m,k})^2}{\sum\_{i=1}^{M} a\_{i,k}} \geq \frac{1}{M}
$$
I believe that I have proof that it holds for $K = 2$ columns. For $K > 2$ I have tried to used gradient descent to find counterexamples but I have not found any.
* I notice $\sum\_{k=1}^{K} \sum\_{m=1}^{M} a\_{m,k} = M$.
* It seems to be a quite tight bound without further assumptions, when looking at SGD cases.
The next thing I want to look at is whether the left hand side is convex in $A$.
* Maybe I can reduce problems with $M > 2$ rows to a simpler problem with $M = 2$ rows, where some rows are summed. However, I think it is difficult to maintain the constraints on the values for the second row in that case.
I will be grateful for any pointers. Thanks.
(Context: I am thinking of the rows as agents having beliefs about a discrete rv with $K$ outcomes. Can they make a bet where they split a pot of $1$ dollar according to the fraction of probability mass they have assigned to the outcome, and if so what is the expected fraction for an agent $m$ and when if ever would it be less than $1/M$ dollar?)
| https://mathoverflow.net/users/477213 | Inequality for matrix with rows summing to 1 | If I am not missing something, this seems a direct application of [Titu's lemma](https://brilliant.org/wiki/titus-lemma/)
$$
\sum\_{k=1}^K \frac{x\_k^2}{y\_k} \geq \frac{\left(\sum\_{k=1}^K x\_k \right)^2}{\sum\_{k=1}^K y\_k}, \quad x\_k \geq 0, y\_k > 0,
$$
which is quickly proved using Cauchy-Schwarz's inequality (as shown in the link above). Just set
$$
x\_k = a\_{m,k}, \quad y\_k = \sum\_{i=1}^M a\_{i,k}
$$
and simplify the RHS using the two relations you already wrote in your question.
| 1 | https://mathoverflow.net/users/1898 | 416337 | 169,665 |
https://mathoverflow.net/questions/407857 | 4 | It is known that [fully faithful functors are closed under pushouts in Cat](https://ncatlab.org/nlab/show/full+and+faithful+functor) (e.g. Lemma 4.9 of [this paper](https://lmcs.episciences.org/742/pdf)). Are locally fully faithful 2-functors closed under (strict) 2-pushouts in the 2-category 2-Cat of 2-categories, (strict) 2-functors, and 2-natural transformations? I expect this to be true, but giving an explicit description of a 2-pushout is daunting. Is there a simpler way to reason to prove this by reasoning entirely locally (i.e. in the hom-categories)?
I expect the fully weak setting to be more difficult, but if it is known that locally fully faithful pseudofunctors are closed under pseudopushouts in a bicategory of bicategories, this would also answer my question.
| https://mathoverflow.net/users/152679 | Are locally fully faithful 2-functors closed under 2-pushout in 2-Cat? | No. I'm not sure whether you're asking about pushout along an arbitrary 2-functor, or just about pushing out a locally fully faithful functor along *another* locally fully faithful functor. Either way the answer is no.
Consider the inclusion of 1-categories into 2-categories as the locally discrete ones. This functor is fully faithful. It has a right adjoint, so it preserves pushouts. And it identifies the faithful functors between 1-categories with the locally fully faithful functors between essentially discrete 2-categories. So it suffices to show that faithful functors of 1-categories are not closed under pushout in $Cat$.
So it suffices to show that injective homomorphisms of monoids are not closed under pushout in $Mon$.
For this, if you're asking about pushout along an arbitrary map, then it suffices to consider the pushout of the injection $\mathbb N \to \mathbb Z$ along the map $\mathbb N \to \mathbb N / (2=1)$, which is the non-injection $\mathbb N / (2=1) \to \ast$.
Otherwise, we're pushing out an injective homomorphism along another injective homomorphism. The result need not be an injective homomorphism -- see [here](https://mathoverflow.net/questions/249589/pushouts-of-injective-monoid-homomorphisms?) (the first-linked paper at Benjamin Steinberg's answer there gives an example with $\leq 4$-element semigroups; adding disjoint unit elements these are $\leq 5$-element monoids).
| 1 | https://mathoverflow.net/users/2362 | 416341 | 169,667 |
https://mathoverflow.net/questions/416339 | 6 | Suppose that $\boldsymbol{t}\sim \mathcal{N}(\boldsymbol{u};\boldsymbol{0},\boldsymbol{M})=f\_{\boldsymbol{t}}(\boldsymbol{u})$, where $\boldsymbol{t}$ is a $N$-dimensional gaussian random vector, and
\begin{equation}
\boldsymbol{M}=
\left[
\begin{matrix}
1&m\_{12}&\cdots& m\_{1N}\\
m\_{21}&1&\cdots& m\_{2N}\\
\vdots&\vdots&&\vdots\\
m\_{N1}&m\_{N2}&\cdots& 1
\end{matrix}
\right]
\end{equation}
where $m\_{i,j}=m\_{j,i}\leq1$, $i=1,2,\cdots,N$, and $j=1,2,\cdots,N$;
Let
\begin{equation}
F=\int\limits\_{\boldsymbol{u}\in\mathcal{Y}(\boldsymbol{\gamma})} f\_{\boldsymbol{t}}(\boldsymbol{u}) \,d\boldsymbol{u}
\end{equation}
where $\mathcal{Y}(\boldsymbol{\gamma})=\left(-\infty,\gamma\_1\right)\times\cdots\times \left(-\infty,\gamma\_N\right)$
Now, if let the $(i,j)\text{th}$ element of $\boldsymbol{M}$ be an independent variable $a,$ $i\neq j$, i.e., we have $m\_{i,j}=m\_{j,i}=a<1$.
Then we let $F(a)=\int\limits\_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f\_{\boldsymbol{t}}(\boldsymbol{u}) \, d\boldsymbol{u}$.
**My question is:**
Is $F(a)$ a monotonically increasing function of $a$?
| https://mathoverflow.net/users/163557 | Is this function monotonically increasing? | Yes, this is a special case of [Slepian's inequality](https://en.wikipedia.org/wiki/Slepian%27s_lemma).
| 5 | https://mathoverflow.net/users/36721 | 416343 | 169,668 |
https://mathoverflow.net/questions/416242 | 1 | How to prove that
$$\int (1-z)^{u} z^{v} dz = \frac{z^{v+1}}{v+1} \, \_2F\_1(-u, v+1; v+2; z)?$$
| https://mathoverflow.net/users/103291 | How to prove that $\int (1-z)^{u} z^{v} dz$ is equal to $\frac{z^{v+1}}{v+1}_2F_1(-u, v+1; v+2; z)$? | The generalized binomial coefficients are
$$
\binom{u}{n} := \frac{(-1)^n(-u)\_n}{n!} ,
$$
here we used the notation $(-u)\_n$ for the rising factorial: $(-u)\_n := \prod\_{k=0}^{n-1}(-u+k)$.
Taylor's expansion of $(1-z)^u$ around $z=0$ is
$$
(1-z)^u = \sum\_{n=0}^\infty \binom{u}{v}(-z)^n = \sum\_{n=0}^\infty \frac{(-u)\_n}{n!} z^n ,\quad |z| < 1.
$$
Thus we have
$$\int (1-z)^{u} z^{v} dz=
\sum\_{n=0}^{\infty}\frac{(-u)\_{n}}{n!}\int z^{n+v}dz = z^{v+1}\sum\_{n=0}^{\infty}\frac{(-u)\_{n}}{n! (n+v+1)}z^{n}.$$
Next, $\frac{1}{n+v+1}$ can be expressed as
$$\frac{1}{n+v+1}=\frac{\Gamma(n+v+1)}{\Gamma(n+v+2)} = \frac{\Gamma(v+1)(v+1)\_n}{\Gamma(v+2)(v+2)\_n} = \frac{(v+1)\_n}{(v+1)(v+2)\_n}.$$
Then
$$\int (1-z)^{u} z^{v} dz=
\frac{z^{v+1}}{v+1}\sum\_{n=0}^{\infty}\frac{(-u)\_{n}(v+1)\_n}{n!(v+2)\_n}z^{n}=\frac{z^{v+1}}{v+1}\,\_2F\_1(-u,v+1;v+2;z).$$
| 0 | https://mathoverflow.net/users/103291 | 416364 | 169,670 |
https://mathoverflow.net/questions/416346 | 1 | Define
$$
F(\lambda,x):=g(x)+\lambda\int\limits\_{\partial\_e Y}f(y)d\mu\_x(y)
$$
where
* $Y$ is a convex space and $w^\*$-compact, moreover $Y$ forms a Bauer simplex (in particular $Y$ corresponds to some subset of some algebraic state space),
* its extreme boundary $\partial\_e Y$ is $w^\*$-compact,
* the points $x$ correspond to elements in $Y$,
* the measures $\mu\_x$ are probability measures (depending on $x$) and have support in $\partial\_e Y$,
* $g$ is a functional ($w^\*$-continuous and affine) on $Y$,
* $f$ is a continuous function on $\partial\_e Y$,
* $\lambda$ is a real parameter,
* $F(0,\cdot)$ is assumed to be a $w^\*$-continuous, affine functional on $Y$.
Then I consider
$$
R(\lambda):=\inf\_{x\in Y} F(\lambda,x),
$$ which can be shown to exist.
I am interested in computing the derivative of $R$ with respect to $\lambda$. For this, one clearly needs some smoothness of $\inf$.
**My question is**: under what precise conditions this derivative can be taken, and moreover which result will it produce? For example, given that $R$ is differentiable in $\lambda$ and that the set of minimizers corresponding to $R(0)$ (i.e. the points $x\in Y$ for which $R(0)=F(0,x)$) is nonempty, what can we say?
I thank you in advance!
| https://mathoverflow.net/users/145631 | Derivatives of infimum in variational problem | Let $t:=\lambda$, so that
$$R(t):=\inf\_{x\in Y}F(t,x).$$
Suppose that $\int\_{\partial\_e Y}f\,d\mu\_x$ is lower-semicontinuous in $x$ (with respect to the appropriate topology, which you appear to assume to be the $w^\*$-topology with respect to some unspecified duality). Then the set
$$Y\_t:=\{x\in Y\colon F(t,x)=R(t)\}$$
is a nonempty compact set, for each real $t$. Further, let
$$L\_t:=\Big\{\int\_{\partial\_e Y}f\,d\mu\_x\colon x\in Y\_t\Big\}.$$
Take now any real $t$.
If $L\_t$ is a singleton set $\{l\_t\}$, then $R$ is differentiable at $t$ and $R'(t)=l\_t$.
If $L\_t$ is not a singleton set, then $R$ is not differentiable at $t$.
These statements follow e.g. from Theorems 0.25 and 1.5 in the book by Levin "Convex analysis in spaces of measurable functions and its applications in mathematics and economics", Moscow: Nauka, 1985. (I have this book only in Russian.)
| 0 | https://mathoverflow.net/users/36721 | 416373 | 169,676 |
https://mathoverflow.net/questions/416311 | 16 | A Schwartz function on $\mathbb R^d$ is a $C^\infty$ function, such that all differentials of order $k \ge 0$ decay faster than any polynomial. They include the class $C^\infty\_c(\mathbb R^d)$ of compactly supported, smooth functions.
I would like two know, if for every Schwartz function $f$, there are Schwartz functions $g,h$ such that $f(x)=g(x)h(x)$ for all $x \in \mathbb R^d$. If $f \in C^\infty\_c$, we can choose $g=f$ and $h$ as a cutoff function, such that $f(x) \neq 0 \implies h(x)=1$ to get such a (trivial) representation. For a general Schwartz function, I was unable to find a construction or a counter example.
| https://mathoverflow.net/users/123409 | Is every Schwartz function the product of two Schwartz functions? | Yes, such a decomposition exists. More general given a compact set $B\subset\mathcal{S}(\mathbb{R}^{n})$ there is a function $\varphi\in\mathcal{S}(\mathbb{R}^{n})$ and a compact set $C\subset\mathcal{S}(\mathbb{R}^{n})$ with $B=\varphi C$. This property is called the *compact strong factorisation property* by J. Voigt. Details can be found in the following paper:
J. Voigt: [Factorization in some Fréchet algebras of differentiable functions](https://eudml.org/doc/218540). Studia Math. 77 (1984): 333-347
| 16 | https://mathoverflow.net/users/83700 | 416374 | 169,677 |
https://mathoverflow.net/questions/416100 | 2 | This question was motivated by [Singular value decomposition of truncated discrete Fourier transform matrix](https://mathoverflow.net/q/415784/11260)
Consider for integers $1\leq k\leq N$, $1\leq n\_0\leq N-k+1$ the $k\times k$ sub-unitary matrix $W(N,k,n\_0)$ with elements
$$[W(N,k,n\_0)]\_{nm}=N^{-1/2}e^{2\pi i(n-1)(m-1)/N},\;\;n\_0\leq n,m\leq n\_0+k-1.$$
This is a principal submatrix of the unitary [discrete Fourier transform matrix](https://en.wikipedia.org/wiki/DFT_matrix) $W(N,N,1)$. It is sub-unitary, so its singular values lie in the interval $[0,1]$.
**Conjecture:** A total of $\max(2k-N,0)$ singular values are precisely equal to 1.
I have not found this statement in the literature (cited [here](https://mathoverflow.net/a/416071/11260)), which focuses on the accumulation of $k^2/N$ singular values *near* unity. Can one provide a proof of the conjecture?
Evidence for the conjecture follows from small-$N$ cases I examined, for example, the singular values squared of
$W(8,6,1)$ equal $\left\{1,1,1,1,\frac{1}{8} \left(\sqrt{\sqrt{2}+2}+2\right),\frac{1}{8} \left(2-\sqrt{\sqrt{2}+2}\right)\right\}$, and those of
$W(8,5,2)$ equal $\left\{1,1,\frac{1}{16} \left(\sqrt{16 \sqrt{2}+25}+7\right),\frac{1}{4},\frac{1}{16} \left(7-\sqrt{16 \sqrt{2}+25}\right)\right\}$, while those of
$W(8,4,3)$ equal $\left\{\frac{1}{4} \left(\sqrt{\sqrt{\sqrt{3}+2}+2}+2\right),\frac{1}{8} \left(\sqrt{2 \left(\sqrt{2}-\sqrt{6}+4\right)}+4\right),\frac{1}{8} \left(4-\sqrt{2 \left(\sqrt{2}-\sqrt{6}+4\right)}\right),\frac{1}{4} \left(2-\sqrt{\sqrt{\sqrt{3}+2}+2}\right)\right\}$.
| https://mathoverflow.net/users/11260 | Unit singular value conjecture for discrete Fourier transform submatrix | The conjecture is true. Moreover we show that
if $W(N,N,1)$ is replaced by any unitary $N \times N$ matrix
then $\max(2k-N,0)$ is still a lower bound on the multiplicity,
and is usually sharp (though for $k<N$ there are easy exceptions
such as the $N \times N$ identity matrix).
First observe that if $T$ is any "sub-unitary" matrix
(i.e. $|Tv| \leq |v|$ for all $v$)
then the vectors $v$ such that $|Tv| = |v|$ form a vector space
whose dimension is the multiplicity of $1$ as a singular value of $T$.
Indeed that multiplicity is the dimension of the $1$-eigenspace of $T^\* T$;
but if $T^\* T v = v$ then $|Tv|^2 = (Tv,Tv) = (v, T^\* T v) = |v|^2$,
and conversely if $|v| = |Tv|$ then
$|v|^2 = (v, T^\* T v) \leq \left|v\right| \left|T^\*T v\right| \leq |v|^2$
(since $\|T^\* T\| \leq 1$), with equality only if $T^\* T v = v$.
Now suppose $T$ is obtained as the principal $k \times k$ minor of
an $N \times N$ unitary matrix $U$. Then for any $v \in {\bf C}^k$
we find $Tv$ by padding $v$ by $N-k$ zeros to a vector $\bar v \in {\bf C}^n$,
applying $U$, and discarding the last $N-k$ entries of the result $U \bar v$.
Therefore $|Tv| \leq |v|$ with equality if and only if
$U\bar v$ is supported on the first $k$ coordinates.
This identifies the space $\{ v : |Tv| = |v| \}$ with
the kernel of the $k \times (N-k)$ submatrix of $U$
formed by intersecting the first $k$ rows with the last $N-k$ columns.
Since this submatrix has rank at most $\min(N-k,k)$,
its kernel has dimension at least $\max(2k-N,0)$.
In the special case that $U$ is the matrix of the discrete Fourier transform,
any square submatrix supported on the first $m$ rows or the first $m$ columns
is a Vandermonde matrix or its transpose, and is thus nonsingular.
Therefore our $k \times (N-k)$ submatrix has a nonsingular square submatrix
of order $\min(N-k,k)$. Hence it attains the upper bound $\min(N-k,k)$
on the rank of a $k \times (N-k)$ matrix, and the singular value $1$
has the lowest possible multiplicity $\max(2k-N,0)$, **QED**.
To show that this holds for "typical" unitary $U$, we need only observe that
$m \times n$ matrices of rank $\min(m,n)$ form a nonempty open subset of
the space of $m \times n$ matrices. Hence unitary matrices $U$ whose
relevant $k \times (N-k)$ submatrix satisfies this condition form
the preimage of an open set under a continuous map, and therefore
constitute an open subset of the group ${\rm U}\_N$ of unitary matrices.
This open subset is nonempty because it contains the discrete
Fourier transform, and is thus dense because ${\rm U}\_N$ is connected.
| 3 | https://mathoverflow.net/users/14830 | 416382 | 169,679 |
https://mathoverflow.net/questions/416146 | 3 | Let $V$ be a $\mathbb{C}$-vector space of dimension $N \geq 2$, let $d$ be a positive integer, let $l < N$ be a positive integer, and let $U \subseteq S^d(V)$ be a linear subspace of codimension $k=\binom{l+d-1}{d}$. Suppose that there exists a basis of $U$ of the form $\{u\_1^{\vee d},\dotsc, u\_r^{\vee d}\}$ such that
$$
\operatorname{span}\{u\_1^{\vee d-1},\dotsc, u\_r^{\vee d-1}\}=S^{d-1}(V),
$$
where $r=\binom{N+d-1}{d}-k$.
Does there exist a linear subspace $W \subseteq V$ of dimension $l$ for which $S^d(V)=U \oplus S^d(W)$?
Here, $\vee$ denotes the symmetric product.
This is a variant of my previous question [Given a subspace $U \subseteq S^d(V)$, does there always exist a complement of the form $S^d(W)$?](https://mathoverflow.net/q/415782/150898), which did not assume any particular form for $U$.
| https://mathoverflow.net/users/150898 | Given a subspace $U \subseteq S^d(V)$ of a particular form, does there always exist a complement of the form $S^d(W)$? | The answer is still no and a variant of [my previous counter-example](https://mathoverflow.net/a/415807/37214) will provide a counter-example for this new question.
Let $V = \mathbb{C}^4$ with basis $\{x,y,z,t\}$ and $\mathbb{C}^3 \subset V$ with basis $\{x,y,z\}$. Put $U = S^2 \mathbb{C}^3 \oplus \mathbb{C}\cdot t^2$, a basis of which is given by $\{x^2, y^2, z^2, (x+y)^2, (x+z)^2, (y+z)^2, t^2\}$. This basis satisfies the hypothesis in the question as:
$$ \operatorname{span} \{x,y,z,x+y,x+z,y+z,t \} = V.$$
Note that $\dim U = 7$, $\dim S^2V = 10$ and that for any $W \subset V$ of dimension $2$ (so that $\dim S^2 W = 3$), we have:
$$S^2{\left(\mathbb C^3 \cap W\right)} \subset \left(U \cap S^2W \right),$$
with $\dim \mathbb{C}^3 \cap W \geq 1$.
| 1 | https://mathoverflow.net/users/37214 | 416401 | 169,683 |
https://mathoverflow.net/questions/416331 | 9 | This might be an easy question, maybe the example I'm looking for is common knowledge. As always, recall that a topological space $X$ is *scattered* if and only if every non-empty subset $Y$ of $X$ contains at least one point which is isolated in $Y$. It is known that any first countable, $T\_3$, Lindelöf and scattered space is countable. Is there an example of an uncountable, first countable, Hausdorff, Lindelöf and scattered space?
| https://mathoverflow.net/users/146942 | Example of an uncountable scattered space with some properties | There exists an example for this question under Continuum Hypothesis, more precisely, under the assumption $\mathfrak b=\omega\_1$. In this case by Theorem 10.2 in the van Douwen's survey paper ``[The integers and Topology](https://www.sciencedirect.com/science/article/pii/B9780444865809500069)'', there exists an uncountable set $C\subseteq\mathbb R\setminus\mathbb Q$, which is *concentrated* at rationals in the sense that for every open set $U\subseteq\mathbb R$ with $\mathbb Q\subseteq U$ we the complement $C\setminus U$ is countable.
Consider the uncountable space $X=\mathbb Q\cup C$ endowed with the topology $\tau$ consisting of the sets $W\subseteq X$ such that for every $q\in W\cap\mathbb Q$ there exists $\varepsilon>0$ such that $\{x\in C:|x-q|<\varepsilon\}\subseteq W$.
It is easy to see that the space $(X,\tau)$ is (functionally) Hausdorff, first-countable, Lindelof and scattered. More precisely, the subspace $C=X\setminus\mathbb Q$ is open and discrete and $\mathbb Q$ is closed and discrete in $X$. So, $X$ is scattered of finite scattered height.
To see that $X$ is Lindelof, observe that for every open cover $\mathcal U$ there exists a countable subfamily $\mathcal V\subseteq\mathcal U$ such that $\mathbb Q\subseteq \bigcup\mathcal V$. Since $C$ is concentrated at $\mathbb Q$, the difference $C\setminus\bigcup\mathcal V$ is countable and hence we can choose a countable subfamily $\mathcal V'\subseteq \mathcal U$ such that $C\setminus \bigcup\mathcal V\subseteq\bigcup\mathcal V'$. Then $\mathcal V\cup\mathcal V'$ is a required countable subcover of $\mathcal U$, witnessing that $X$ is Lindelof.
---
The above example cannot be constructed in ZFC because of the following
**Theorem.** Under $\mathfrak b>\omega\_1$, every scattered Lindelof $T\_1$-space of countable pseudocharacter and finite scattered height is countable.
Prior writing down the proof of this theorem, let us recall the definition of the scattered height.
Given a subset $A$ of a topological space $X$, let $A^{(1)}$ be the set of nonisolated points of $A$. Observe that for a closed subset $A$ of $X$, the set $A^{(1)}$ is closed in $A$. Let $X^{(0)}=X$ and for every nonzero ordinal $\alpha$ let
$$X^{(\alpha)}=\bigcap\_{\beta<\alpha}\big(X^{(\beta)}\big)^{(1)}.$$
By transfinite induction it can be shown that $(X^{(\alpha)})\_\alpha$ is a decreasing sequence of closed subsets of $X$, so it should stabilize at some ordinal $\alpha$. The smallest ordinal $\alpha$ such that $X^{(\alpha+1)}=X^{(\alpha)}$ is called the *scattered height* of $X$ and is denoted by $\hbar(X)$. A topological space $X$ is scattered if and only if $X^{(\hbar(X))}=\emptyset$.
*Proof of Theorem.* Assume that $\mathfrak b>\omega\_1$. By Mathematical Induction we shall prove that *for every natural number $n$, every Lindelof scattered space $X$ of countable pseudocharacter and scattered height $\hbar(X)\le n$ is countable.*
For $n=0$, each scattered space $X$ with $\hbar(X)\le 0$ is empty and hence countable. Assume that for some positive integer number $n$, every Lindelof scattered space $X$ of countable pseudocharacter and scattered height $\hbar(X)<n$ is countable. Take a Lindelof scattered space $X$ of countable pseudocharacter and scattered height $\hbar(X)=n$. Then $X^{(n)}=\emptyset$ and $X^{(n-1)}$ is a nonempty discrete subspace of $X$. Being a closed subspace of the Lindelof space $X$, the discrete space $X^{(n-1)}$ is Lindelof and countable. Write $X^{(n-1)}$ as $\{x\_k\}\_{k\in\omega}$. Since $X$ has countable pseudocharacter, for every $k\in\omega$ there exists a decreasing sequence $(U\_{k,m})\_{m\in\omega}$ of open sets in $X$ such that $\bigcup\_{m\in\omega}U\_{k,m}=\{x\_k\}$. Assuming that the space $X$ is uncountable, choose a transfinite sequence $\{y\_\alpha\}\_{\alpha\in\omega\_1}$ consisting of pairwise disjoint elements of $X\setminus X^{(n-1)}$. For every $\alpha\in\omega\_1$, let $f\_\alpha:\omega\to\omega$ be the function assigning to each $k\in\omega$ the smallest number $f\_\alpha(k)$ such that $y\_\alpha\notin U\_{k,f\_\alpha(k)}$. The number $f\_\alpha(k)$ exists since $y\_\alpha\notin\{x\_k\}=\bigcap\_{m\in\omega}U\_{k,m}$. Since $\mathfrak b>\omega\_1$, there exists a countable subfamily $\mathcal F\subseteq\omega^\omega$ such that for every $\alpha\in\omega\_1$ there exists a function $f\in\mathcal F$ such that $f\_\alpha\le f$. For every $f\in\mathcal F$, consider the open neighborhood $U\_f=\bigcup\_{k\in\omega}U\_{k,f(k)}$ of the set $\{x\_k\}\_{k\in\omega}=X^{(n-1)}$. The choice of $\mathcal F$ guarantees that $\{y\_\alpha\}\_{\alpha\in\omega\_1}\cap\bigcap\_{f\in\mathcal F}U\_f=\emptyset$. By the Pigeonhole Principle, for some $f\in\mathcal F$, the set $\Omega=\{\alpha\in\omega\_1:y\_\alpha\notin U\_f\}$ is uncountable. On the other hand, $Y=X\setminus U\_f$ is a closed (and hence Lindelof) subspace of $X$. It follows from $Y\cap X^{(n-1)}=\emptyset$ that $\hbar(Y)<n$ and hence $Y$ is countable, by the inductive hypothesis. But this contradicts the uncountability of the set $\{y\_\alpha\}\_{\alpha\in\Omega}\subseteq Y$.
---
Can the theorem be generalized to scattered spaces of arbitrary scattered height?
**Problem.** *Assume that $\mathfrak b>\omega\_1$. Is every scattered Lindelof $T\_1$-space of countable pseudocharacter countable?*
---
**Remark.** Thanks to the intervention of Will Brian, I realized that the equality $\mathfrak d=\omega\_1$ in the initial redaction of my answer can be replaced by the weaker equality $\mathfrak b=\omega\_1$. So, now we have the following characterization.
**Theorem.** *The following conditions are equivalent:*
1. $\mathfrak b>\omega\_1$.
2. *Every Lindelof scattered $T\_1$-space of countable pseudocharacter finite scattered height is countable.*
3. *Every first-countable Lindelof scattered Hausdorff space of scattered height 2 is countable.*
The answer of Will Brian shows that this theorem does not extend to scattered Lindelof spaces of infinite scattered height.
| 10 | https://mathoverflow.net/users/61536 | 416404 | 169,685 |
https://mathoverflow.net/questions/416405 | 2 | If $p : X \rightarrow Y$ is the dominant surjective finite morphism of varieties between smooth projective varieties over of same dimension over $\mathbb{C}$, then do we know about the properties of the pull-back maps $p^{\star} : H^{k,k}(Y) \rightarrow H^{k,k}(X)$ is surjective/injective? If someone could provide reference to this?
| https://mathoverflow.net/users/476114 | When does a morphism between varieties induce surjection/injection morphism on cohomology? | For a finite map between smooth varieties, the induced pullback map on cohomology is injective (hence it is somewhat rarely surjective - only when it's an isomorphism).
This is because there a is a trace map $H^{k,k}(X) \to H^{k,k}(Y)$, and the composition of the two $H^{k,k}(Y) \to H^{k,k}(X) \to H^{k,k}(Y)$ is multiplication by the degree of $p$.
This holds in the greater generality of any surjective morphism between smooth projective varieties, but there it requires more powerful tools to prove (it follows from the decomposition theorem, but I'm not sure if that's really needed).
| 8 | https://mathoverflow.net/users/18060 | 416408 | 169,686 |
https://mathoverflow.net/questions/416411 | 6 | Let $X$ be a compact complex manifold. What are obstructions to the existence of a real subbundle $V$ of $TX$ such that $TX = V \otimes \mathbb{C}$?
For example, does $\mathbb{CP}^n$ have such a structure?
| https://mathoverflow.net/users/392184 | Existence of a real structure on the tangent bundle of a complex manifold | For a vector bundle to have real structure, it must be isomorphic to its complex conjugate, hence it's $i$th Chern class must be equal to its own negation (i.e. $2$-torsion) for all odd $i$. So $2c\_1, 2c\_3, 2 c\_5, \dots$ are all obstructions.
Since $2c\_1$ is nonzero for the tangent bundle of projective space, $\mathbb C\mathbb P^n$ has no such structure.
| 7 | https://mathoverflow.net/users/18060 | 416412 | 169,688 |
https://mathoverflow.net/questions/416344 | 2 | It is well known that each $n\in\mathbb N=\{0,1,2,\ldots\}$ can be written as $2w^2+x^2+y^2+z^2$ with $w,x,y,z\in\mathbb N$. Furthermore,
$$\{2w^2+x^2+y^2:\ w,x,y\in\mathbb N\}=\mathbb N\setminus\{4^k(16m+14):\ k,m\in\mathbb N\}.$$
Motivated by this, here I pose the following novel question.
**Question 1.** Can each $n\in\mathbb N$ be written as $2w^2+x^2+y^2+z^2+xyz$ with $x,y,z\in\mathbb N$?
I guess that the answer is positive, which has been verified for $n\le 10^6$.
Similarly, I have the following four questions.
**Question 2.** Are $7$ and $487$ the only natural numbers which cannot be written as $w^2+x^2+y^2+z^2+xyz$ with $w,x,y,z\in\mathbb N$?
**Question 3.** Is it true that each $n\in\mathbb N$ with $n\not\equiv 3\pmod4$ can be written as
$4w^2+x^2+y^2+z^2+xyz$ with $w,x,y,z\in\mathbb N$?
I also believe that the answers to Questions 2 and 3 should be positive, which have been verified for $n\le 10^6$. It is easy to see that $$x^2+y^2+z^2+xyz\not\equiv3\pmod4$$ for any $x,y,z\in\mathbb Z$.
**Question 4.** Is $23$ the only natural number which cannot be written as $w^2+x^2+y^2+z^2+3xyz$ with $w,x,y,z\in\mathbb N$?
I guess that the answer is positive. I have checked this for natural numbers up to $2\times10^6$.
**Question 5.** Is it true that each $n\in\mathbb N$ with $n\not\equiv3\pmod4$ can be written as $4w^3+x^2+y^2+z^2+xyz$ with $w,x,y,z\in\mathbb N$? Are $7,\,87$ and $267$ the only natural numbers which cannot be written as $w^3+x^2+y^2+z^2+xyz$ with $w,x,y,z\in\mathbb N$?
It seems that Question 5 should also have a positive answer; I have checked this for natural numbers up to $10^5$.
Any ideas to the above new questions? Your comments are welcome!
| https://mathoverflow.net/users/124654 | Can each natural number be represented by $2w^2+x^2+y^2+z^2+xyz$ with $x,y,z\in\mathbb N$? | The answer to question 1 is yes - the other questions seem to me to be more difficult.
If $n$ is odd, then there are non-negative integers $w$, $x$ and $y$ so that $n = 2w^{2} + x^{2} + y^{2}$. One way to see this is that the class number of this quadratic form $Q\_{1} = 2w^{2} + x^{2} + y^{2}$ is $1$, and so every locally represented integer is represented. The only local obstructions are at $2$, and this implies that every integer not of the form $14 \cdot 4^{k} \pmod{4^{k+2}}$ for $k \geq 0$ is represented by $Q\_{1}$.
If $n$ is even, then there are non-negative integers $w$, $x$ and $y$ so that $n = 2w^{2} + x^{2} + xy + y^{2} + 1$. The quadratic form $Q\_{2} = 2w^{2} + x^{2} + xy + y^{2}$ also has class number $1$, and it represents all integers not of the form $10 \cdot 4^{k} \pmod{4^{k+2}}$ for some $k \geq 0$. In particular, $Q\_{2}$ represents every odd number and so every even $n$ can be written in the form $2w^{2} + x^{2} + xy + y^{2} + 1$.
Note that if $m \in \mathbb{N}$ and $m = x^{2} + xy + y^{2}$ for some $x, y \in \mathbb{Z}$, then there are $x'$ and $y'$ in $\mathbb{N}$ so that $m = (x')^{2} + (x')(y') + (y')^{2}$. This can be seen by applying one of the six automorphisms of the quadratic form $x^{2}+xy+y^{2}$.
| 5 | https://mathoverflow.net/users/48142 | 416426 | 169,693 |
https://mathoverflow.net/questions/416437 | 0 | If I choose a principal bundle, let us say $G\rightarrow P \rightarrow B$, with $G=U(1)$, $P=S^1 \times S^1$ and $B=S^1$. Can I follow the identity element of the group over a curve at the base. How isn't this equivalent to a preferred Horizontal subspace (and hence a canonical connection) ?
| https://mathoverflow.net/users/477214 | Non existence of preferred Horizontal subspace on a bundle | 1. A trivial principal bundle, $P= G\times B\to B$ has a preferred connection (which is also flat). There is an obvious map $B\to \{1\}\times B\subset P$ and you can take as horizontal bundle the image of its differential.
2. If your principal bundle $P\to B$ is not trivial, but is trivializable, i.e. exists a section $\sigma:B\to P$, then you can use this section to trivialize it, i.e. construct a isomorphism with the trivial bundle, $\Phi\_\sigma: P\overset{\simeq}{\to} G\times B$. Over $G\times B$ you have the preferred connection of item 1) and you can use $\Phi\_\sigma$ to pull it back to $P$.
You now have a connection on $P$, but the latter depends on the choice of the trivialization $\sigma$. How many choices do you have? Well, the space of sections is acted transitively and freely by (hence is in bijection with) the gauge group $\mathcal G(P) \simeq \operatorname{Maps}(B, G)$ as you can alter $\sigma$ by multiplying it with an element of $\operatorname{Maps}(B, G)$. This tells you that the possible choices that you have are $\mathcal{G}(P)$.
Notice that all these connections will be flat.
3. If the principal bundle $P$ is not trivializable, then you cannot apply any of the above and you have a space of connection which is an *affine* space modelled over $\Omega^1(B; \operatorname{ad} (P))$ where $\operatorname{ad} (P) = P\times\_{\operatorname{ad}}\mathfrak{g} $ is the bundle associated to the adjoint action $\operatorname{ad}:G\to \operatorname{Aut}(\mathfrak{g})$ of $G$ on its Lie algebra $\mathfrak{g}$.
As an example of non-trivializable bundle you can take $P$ equal to the bundle of oriented orthonormal frames of the tangent bundle of $\mathbb{S}^2$ (after you choose an orientation for $\mathbb{S}^2$ of course); this is an $\mathbb{S}^1\simeq \operatorname{SO}(2)$ bundle over $\mathbb{S}^2$ which does not admit any section.
As a side note, even in this case the gauge group $\mathcal{G}(P)$ acts on the set of connections, however it is no more in bijection with $\operatorname{Maps}(B,G)$ but instead with $\Gamma(B; \operatorname{Ad}(P))$ i.e. sections of the bundle associated to the action $\operatorname{Ad}: G\to \operatorname{Aut}(G) $, $\operatorname{Ad}(g)\cdot p = gpg^{-1}$.
In the case of a trivial bundle $\operatorname{Ad}(P)\simeq G\times B$ hence in item 2) we were able to identify $\mathcal{G}(P)$ with $\operatorname{Maps}(B, G)$.
| 2 | https://mathoverflow.net/users/99042 | 416442 | 169,697 |
https://mathoverflow.net/questions/416430 | 4 | I can deduce some results about this from the prime number theorem or from results about the [primorial function](https://en.wikipedia.org/wiki/Primorial#Characteristics) $p\#$ but I'm wondering what the state of the art is:
Given integers $0\le a\_1<a\_2<\dots <a\_c$, what bound can we put on the least modulus $m$ such that for all $i\ne j$, we have $a\_i\not\equiv a\_j$ (mod $m$)?
| https://mathoverflow.net/users/4600 | Least modulus distinguishing some integers | I'm not sure about the state of art, but here is a rough estimate in terms of $c$ and $\ell:=a\_c-a\_1$ in the case $c\ll \ell$.
First, we notice that for an integer $M$ satisfying
$$M\# ~>~ \big(\frac{c}{2(c-1)}\ell\big)^{c(c-1)/2} ~\geq~ \prod\_{1\leq i<j\leq c} (a\_j-a\_i),$$
where the second inequality follows from AM-GM, there exists a prime $m\leq M$ that does not divide the r.h.s.
Second, taking $M$ as smallest as possible and estimating [primorial](https://en.wikipedia.org/wiki/Primorial) $M\#$ as $e^M$, we have
$$m\leq M\approx \frac{c}2 + \frac{c^2}2\log\frac{\ell}2.$$
I believe this rough estimate can be made rigorous if needed.
| 3 | https://mathoverflow.net/users/7076 | 416457 | 169,703 |
https://mathoverflow.net/questions/416301 | 3 | **Update:**
Now I know why my method fails. But I still wanna know how to work out the original question, that is to show the exactness of the chain complex $C\_\*(X)$ except for two positions $n=0,N-1$.
---
Here is a little problem about the normalization theorem in the theory of simplicial objects.
Given a finite set $X$ with $|X|=N>2$, we can construct a simplicial free abelian group $C\_\*(X)$ defined as follows: for each $n\geq0$, $C\_n$ is defined to be the free abelian group generated by the set of all $(n+1)$-tuples $(x\_0,x\_1,\dots,x\_n)$ of **distinct** elements in $X$; and set $C\_n(X)=0$ for $n\geq N$. The differential map is defined as $d(x\_0,x\_1,\dots,x\_n)=\sum\_{i=0}^n(-1)^i(x\_0,\dots,\hat{x\_i},\dots,x\_n)$.
An exercise in Weibel's book (see Weibel *The K-Book: An introduction to algebraic K-theory*, VI.5, Ex.5.1., p.550 or [p.35](https://sites.math.rutgers.edu/%7Eweibel/Kbook/Kbook.VI.pdf)) asserts that $H\_n(C\_\*)=0$ for $n\neq 0,N-1$.
My strategy goes as follows: Consider the unnormalized complex $U\_\*$ with $U\_n(X)$ defined to be the free abelian group generated by the unnormalized $(n+1)$-tuple $(x\_0,x\_1,\dots,x\_n)$ of $n$ elements in $X$ which are allowed to have repetitions whose differentials are defined as above. Then an easy computation shows that the chain complex $U\_\*$ is acyclic for $n\geq1$(See for example Mac Lane's book *Homology*, Ex.1 in the end of Section 7, Chapter VIII, p. 238). And the normalization theorem(MacLane *Homology*, VIII.6, p.236 or the Theorem 3.3 in [Normalization](https://ncatlab.org/nlab/show/Moore+complex#NormalizedChainComplexOnGeneralGroup) or [thisQuestion](https://mathoverflow.net/questions/127151/simplicial-chain-complex-with-ordered-simplices)) gives that the two chain complexes $C\_\*, U\_\*$ are quasi-isomorphic. So the acyclicity of $U\_\*$ implies the exactness of $C\_\*$. But $C\_{N}=0, C\_{N-1}$ is a free abelian group of rank $N!$ which injects into $C\_{N-2}$ which also has rank $N!$. This forces the $(N-2)$-th differential $d\_{N-2}:C\_{N-2}\to C\_{N-3}$ to be the zero map, which is impossible. I'm wondering which step goes wrong?
| https://mathoverflow.net/users/471160 | The exactness of the associated chain complex of a simplicial free abelian group over a finite set and the normalization theorem | Leaving an answer so that this question can be marked resolved.
Your chain complex is called the "complex of injective words" and proving that it is highly connected is maybe not so direct. Here is a paper by an author I respect <https://arxiv.org/pdf/1608.04496.pdf> with a "simple" proof that looks pretty involved to me.
| 2 | https://mathoverflow.net/users/9068 | 416459 | 169,704 |
https://mathoverflow.net/questions/416466 | 0 | Let $V$ be a $\mathbb{C}$-vector space, and let $f\_1,\dots,f\_n \in S^d(V^\*)$ be homogeneous polynomials of degree $d$ for which $V(f\_1,\dots, f\_n)=\{0\}$.
Must there exist a positive integer $k\geq d$ such that for all $v \in S^k(V^\*)$ there exists $g\_1,\dots, g\_n \in S^{k-d}(V^\*)$ for which $v=\sum\_{i=1}^n g\_i f\_i$?
I am inclined to guess that this is true, based on the projective Nullstellensatz, which tells us that for any finite set of linear forms $L=\{l\_1,\dots, l\_m\}$ that spans $V^\*$, there exists a positive integer $k\geq d$ such that for all $l\in L$ there exists $g\_1,\dots, g\_n \in S^{k-d}(V^\*)$ for which $l^k=\sum\_{i=1}^n g\_i f\_i$.
| https://mathoverflow.net/users/150898 | A variation on the projective Nullstellensatz | Since $V(f\_1,\dots,f\_n)=0$, by the Nullestellensatz
$$ \sqrt{ (f\_1,\dots, f\_n)} = I ( V( f\_1,\dots, f\_n)) = I(0) = (x\_1,\dots, x\_r)$$
where $x\_1,\dots, x\_r$ are a basis for $V^\*$.
So for each $i$ from $1$ to $r$, $x\_i^{e\_i} \in (f\_1,\dots, f\_n)$ for some $e\_i \in \mathbb N$.
It follows that $$(x\_1,\dots,x\_r)^{\sum\_{i=1}^n e\_i} \subseteq (f\_1,\dots, f\_n) $$ as each monomial in $(x\_1,\dots,x\_r)^{\sum\_{i=1}^r e\_i}$ is a multiple of $x\_i^{e\_i}$ for some $i$. (In fact, we can subtract $r-1$ from the exponent if desired).
So taking $k =\sum\_{i=1}^r e\_i$, for $n \geq k$, and $v$ homogeneous of degree $n$, we have $v\in (x\_1,\dots,x\_r)^n \subseteq (x\_1,\dots,x\_r)^k$, so $v = \sum\_{i=1}^n g\_i v\_i$, and we can take the degree $n-k$ part of each $g\_i$.
| 4 | https://mathoverflow.net/users/18060 | 416469 | 169,708 |
https://mathoverflow.net/questions/416468 | 7 | Theorem 2 in [these notes](https://www.math.ias.edu/%7Elurie/278xnotes/Lecture16-Enumerations.pdf)[1] states that, roughly, that each Grothendieck topos can be built (using limits and colimits) from localic topoi. To what extent is that related to the [theorem](https://mathoverflow.net/questions/412504/an-extension-of-the-galois-theory-of-grothendieck) of Joyal and Tierney which states that each Grothendieck topos is equivalent to the topos of equivariant sheaves on a groupoid in the category of locales?
1. Jacob Lurie, 2018, lecture notes from Math 278X Categorical Logic, [https://www.math.ias.edu/~lurie/278x.html](https://www.math.ias.edu/%7Elurie/278x.html), Lecture 16 *Enumerations*
| https://mathoverflow.net/users/477332 | Every Grothendieck topos can be built from localic topoi | The groupoid representation of Joyal and Tierney may be identified with the truncated simplicial diagram appearing in the statement of Theorem 2 of the Lurie notes mentioned in the question. That is, a groupoid object is precisely a diagram of the shape indicated in Theorem 2 satisfying some axioms, which are automatically satisfied by construction when one takes pseudopullbacks as indicated in the diagram. In Joyal and Tierney's theorem, the groupoid in localic toposes is precisely this groupoid: it has topos of objects $U$, topos of 1-cells $U \times\_X U$, and so forth.
To elaborate a bit, the diagram displayed in Theorem 2, which looks like
$U \times\_X U \times\_X U ^\to\_\to \to U \times\_X U ^\to\_\to U$
(implicitly there are also some maps going back in the reverse direction) is really just the first few stages of the [Cech nerve](https://ncatlab.org/nlab/show/%C4%8Cech+nerve) of the map $U \to X$. The Cech nerve as a whole is indexed by the dual simplex category $\Delta^{op}$, and here we just have the part on the 3 smallest objects of $\Delta^{op}$, namely $[0], [1], [2]$. It may be more familiar that when you truncate all the way to just two objects $[0],[1]$, you get the maps $U \times\_X U ^\to\_\to U$ -- the kernel pair of the map $U \to X$, which is an equivalence relation. The higher analog of the fact that a kernel pair is always an equivalence relation is that the Cech nerve of a map is always a groupoid object -- $[0]$ corresponds to the objects of the groupoid, $[1]$ corresponds to the morphisms, $[2]$ corresponds to composable pairs of morphisms (*not* to 2-cells), and in general $[n]$ corresponds to composable $n$-tuples of morphisms.
You can more generally think of an internal category object (not just an internal groupoid object) in a category as a certain type of simplicial object, as described [here](https://ncatlab.org/nlab/show/internal+category#internal_nerve).
| 10 | https://mathoverflow.net/users/2362 | 416470 | 169,709 |
https://mathoverflow.net/questions/416467 | 2 | I am reading a paper about the curve shortening flow which make use of one inequality but I don't know where does it come from where f(x,t) is a smooth function and C is a constant depending on time t. Since the curve $\gamma$ is closed and bounded, it also compact,$$\underset{\gamma}{\max}|f|^2\le C\int\_\gamma|f'|^2+f^2$$ My friend told me it maybe from Morrey's inequality but I cant see it.Can anyone tell me what it is? Paper:<https://projecteuclid.org/journals/journal-of-differential-geometry/volume-23/issue-1/The-heat-equation-shrinking-convex-plane-curves/10.4310/jdg/1214439902.full> [Corollary 4.4.4]
| https://mathoverflow.net/users/470164 | Question of an inequality from curve-shortening flow | $\newcommand{\thh}{\theta}$The inequality in the proof of Corollary 4.4.4 in the linked paper is stated there (without proof) literally as follows:
\begin{equation}
\max|f|^2\le C\int|f'|^2+f^2. \tag{1}\label{1}
\end{equation}
The paper does not appear to specify the meaning of $C$ or $\int$ or $f$. The proof only says that \eqref{1} is then to apply to $k''$, where $k$ is the curvature of a convex planar curve; $k$ is a function of "the angle $\thh$ of the tangent line" (with values in $[0,2\pi)$) and a parameter $t$. The prime $'$ denotes the partial derivative with respect to $\thh$.
So, apparently $\int|f'|^2+f^2$ means $\int\_0^{2\pi}(|f'|^2+f^2)$, and it is enough to show that
\begin{equation}
\max|f|^2\le C\int\_0^{2\pi}(|f'|^2+f^2), \tag{2}\label{2}
\end{equation}
where $C$ is a universal positive real constant and $f\in C^1[0,2\pi]$.
Without loss of generality, $\max|f|=\max f$ (otherwise, replace $f$ by $-f$). So,
\begin{equation}
\max|f|=\max f\le\min f+\int\_0^{2\pi}|f'|
=\int\_0^{2\pi}\Big(|f'|+\frac{\min f}{2\pi}\Big)
\le \int\_0^{2\pi}\Big(|f'|+\frac{|f|}{2\pi}\Big)
\le\int\_0^{2\pi}(|f'|+|f|)
\end{equation}
and hence, by the Cauchy--Schwarz inequality,
\begin{equation}
\max|f|^2
\le2\pi\int\_0^{2\pi}(|f'|+|f|)^2
\le4\pi\int\_0^{2\pi}(|f'|^2+f^2).
\end{equation}
So, \eqref{2} follows.
| 4 | https://mathoverflow.net/users/36721 | 416478 | 169,712 |
https://mathoverflow.net/questions/416362 | 1 | Consider in 1D the operator given by
$$
\mathcal{L} = \frac{d^2}{dx^2} - V'(x)\frac{d}{dx},
$$
where $V(x)$ is a convex, sufficiently quickly growing potential, so that $\mathcal{L}$ has a complete set of eigenfunctions $\psi\_n(x)$ (orthogonal with respect to a stationary distribution $\rho(x)dx$).
Let $\psi\_1$ be the first nonconstant eigenfunction. Is it true that $\psi\_1$ is monotone? This is the case in the simple case of e.g. $V(x)=x^2$, where the eigenfunctions are the Hermite polynomials and $\psi\_1(x) = x$.
| https://mathoverflow.net/users/2192 | Monotonicity of the top eigenfunction of the generator of a diffusion | This is rather a way to get a (positive) answer than a complete one, some missing details will be clear in a moment.
Let me use $A$ for $\mathcal L$ and $B=A-V''$. The symmetrizing measure (for both) is $e^{-V}\, dx$. Both operators are negative and self-adjoint in $L^2(e^{-V}\, dx)$ and the first eigenvalue of $A$ is zero with eigenfunction $1$.
Assume that $\lambda<0$ is an eigenvalue for $A$ with eigenfunction $u$. Then $u$ is not constant and $u'$ is an eigenfunction of $B$ corresponding to the same $\lambda$, by differentiating. Of course, domain problems arise in this point and require some assumptions on $V$.
Conversely, if $0 \neq u'$ is an eigenfunction of $B$, then $\lambda u-Au=c$ and $(\lambda-A)(u-\frac{c}{\lambda})=0$ and $u-\frac{c}{\lambda} \neq 0$ since $u' \neq 0$. Domain and integrability problems here arise again.
Summing up, and assuming compactness of both resolvents, $\sigma (B)= \sigma (A) \setminus \{0\}$ ($0$ is not an eigenvalue of $B$, since $V'' >0$). But then the second eigenvalue of $A$ is the first of $B$ which, by general arguments, is simple and has a positive eigenfunction.
| 1 | https://mathoverflow.net/users/150653 | 416480 | 169,714 |
https://mathoverflow.net/questions/409095 | 8 | Say that a logic $\mathcal{L}$ satisfies the **weak test property** iff for all $\mathfrak{A}\subseteq\mathfrak{B}$ we have $(1)\implies(2)$ below:
1. For each $\mathcal{L}$-formula $\varphi$ with parameters from $\mathfrak{A}$ we have $$\vert\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}\vert=\vert\varphi^\mathfrak{B}\vert.$$ (In this case write "$\mathfrak{A}\trianglelefteq\_{\mathcal{L}}^{\mathsf{Card}}\mathfrak{B}$.")
2. $\mathfrak{A}\preccurlyeq\_\mathcal{L}\mathfrak{B}$.
This is a massive weakening of the Tarski-Vaught test, which says that we get elementarity merely from $\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}$ being nonempty whenever $\varphi^\mathfrak{B}$ is nonempty. By contrast, $\mathfrak{A}\trianglelefteq\_\mathcal{L}^\mathsf{Card}\mathfrak{B}$ is a highly restrictive hypothesis (and so the corresponding implication is weaker): as long as $\mathcal{L}$ is "reasonable" it immediately implies, for example, that $\vert\mathfrak{A}\vert=\vert\mathfrak{B}\vert$ via the formula $x=x$.
My question is:
>
> Does second-order logic have the weak test property?
>
>
>
Producing interesting instances of $\trianglelefteq\_{\mathsf{SOL}}^\mathsf{Card}$, even before trying to also prevent $\preccurlyeq\_{\mathsf{SOL}}$, seems very difficult; on the other hand, I see absolutely no reason why $\mathsf{SOL}$ *should* have the weak test property.
*In fact there is a whole spectrum of variants of the test property which seem interesting to me. For each class $X$ of cardinals and pair of structures $\mathfrak{A}\subseteq\mathfrak{B}$, say $\mathfrak{A}\trianglelefteq\_\mathcal{L}^X\mathfrak{B}$ iff for each $\mathcal{L}$-formula $\varphi$ with parameters from $\mathfrak{A}$ and each $\kappa\in X$ we have $\vert\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}\vert<\kappa\iff \vert\varphi^\mathfrak{B}\vert<\kappa$; then the weak test property at $X$ is the implication $\trianglelefteq\_\mathcal{L}^X\implies \preccurlyeq\_\mathcal{L}$. The Tarski-Vaught test itself corresponds to $X=\{1\}$, while the weak test property corresponds to $X=\mathsf{Card}$. If the main question above happens to have a positive answer - which would surprise me quite a bit! - I would be further interested in which $X$s are "sufficient" to ensure $\preccurlyeq\_\mathcal{L}$.*
| https://mathoverflow.net/users/8133 | Does "agreement on cardinalities" imply second-order elementary substructurehood? | No, second order logic does not have the weak test property: let $\mathfrak{B}=(\mathbb{R},{<})$ (that is, the real numbers with the only predicate being the usual "less than" order) and let $\mathfrak{A}=(\mathbb{R}\backslash\{0\},{<})$. Then $\mathfrak{B}\models$"I am a complete linear order" (completeness as in "for every $<$-downward closed set $X$ such that $X\neq\mathbb{R}$, there is a least upper bound for $X$, and likewise symmetrically"), whereas $\mathfrak{A}$ does not satisfy this, so $\mathfrak{A}\not\equiv\_{\mathrm{SOL}}\mathfrak{B}$, and hence $\mathfrak{A}\not\preccurlyeq\_{\mathrm{SOL}}\mathfrak{B}$. But property 1 does hold for $(\mathfrak{A},\mathfrak{B})$. For for simplicity let's first consider the case that the arity of $\varphi$ is 1. Let $x\_1<x\_2<\ldots<x\_n$ be elements of $\mathbb{R}\backslash\{0\}$. Then the only subsets $X\subseteq\mathbb{R}$ which are second-order definable over $(\mathbb{R},{<})$ from $(x\_1,\ldots,x\_n)$ are finite unions of intervals with endpoints in $\{-\infty,x\_1,\ldots,x\_n,\infty\}$, and therefore, if $0\in X$, then there is an non-empty open interval $(-\varepsilon,\varepsilon)\subseteq X$, so $\varphi^{\mathfrak{B}}$ and $\varphi^{\mathfrak{B}}\cap\mathfrak{A}$ both have cardinality continuum. (E.g. if $\varepsilon>0$ is small enough then for each $x\in(-\varepsilon,\varepsilon)$ we can produce an automorphism $\pi:\mathfrak{B}\to\mathfrak{B}$ which fixes $x\_1,\ldots,x\_n$ but with $\pi(0)=x$.) For arity $k>0$ it is similar, with $k$-dimensional rectangles.
| 5 | https://mathoverflow.net/users/160347 | 416485 | 169,715 |
https://mathoverflow.net/questions/416348 | 3 | Assuming that $i \geq 0$, let $p\_i$ denote an $i$-th prime: $p\_0 = 2, p\_1 = 3, \ldots$ Then $b\_i$ denotes the second-to-last bit of $p\_i$, i.e. $b\_i = \left\lfloor p\_i/2 \right\rfloor \bmod 2.$
The sequences $B\_0$ and $B\_1$ are constructed by the following algorithm: if $b\_i$ is equal to $i \bmod 2$, append $b\_i$ to $B\_0$; otherwise (i.e. if $b\_i$ is not equal to $i \bmod 2$), append $b\_i$ to $B\_1$. Thus $$\begin{array}{l}
B\_0 =
1010101101010101010101101010101010011001001111001101001011001010\ldots,\\
B\_1 =
1101010101011001011010101100101010101010010101011101010101010101\ldots
\end{array}$$
Basically, both of these sequences are $\ldots1010\ldots$, slightly interspersed with runs of two or more identical bits. For example, the first $10^6$ bits of $B\_1$ do not even contain ten consecutive zeros. What is the explanation of this phenomenon?
| https://mathoverflow.net/users/122796 | How to explain a particular property of the second-to-last bits of primes? | I think the phenomena you observe are explained more by the way you construct $B\_0$ and $B\_1$ than by the properties of primes (though properties of primes seem to play a role). If we suppose instead that $b\_i$ were random, obtained by a coin flip, it's easy to see that for each bit of $B\_0$ or $B\_1$, the probability that the next bit is the same is $\frac{1}{3}$, and the probability that the next bit is different is $2/3$.
Indeed, say $b\_i \equiv i \mod 2$ for some $i$, so at time $i$ we have just added $b\_i$ to $B\_0$.
Then if the next $k-1$ bits are not congruent to their position mod $2$, followed by the $k$th bit congruent to its position, $i+k$, mod $2$, we next append the bit $i+k \mod 2$ to $B\_0$. This occurs with probability $\frac{1}{2^k}$, and in each sequence this occurs for exactly one value of $k$.
So we get a different bit with probability $\frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + \dots = \frac{2}{3}$, and the same bit with probability $\frac{1}{2^2} + \frac{1}{2^4} + \frac{1}{2^6} + \dots = \frac{1}{3}$. The $B\_1$ case is identical.
So the probability of getting $10$ consecutive zeroes starting at any given point is $\frac{1}{2} \cdot \frac{1}{3^9}$. However, the length of time we expect to take before seeing the first run of $10$ consecutive zeroes is a little longer than the inverse of that, $3^{10}-1=59048$, by a classical argument.
So we waited about 17 times longer than one would expect to find a run of ten zeroes. (But you had 4 choices to pick to find such a long run without zeroes). That's not completely surprising, but it's a lot less strange than in the naive random model, where the expected wait time is $2^{11}-1$, so you would have waited about $489$ times longer than expected.
---
More generally, if we replace $b\_i$ with a Markov chain with probability of taking the same value on adjacent steps $p$, then the probability that $B\_0$ takes the same value on adjacent steps should be $$p^2 + p^2(1-p)^2 + p^2 (1-p)^4 + \dots = \frac{ p^2}{ 1- (1-p)^2} = \frac{p^2 }{ 2p -p^2} = \frac{p}{ 2-p}$$ since to get two of the same bit in $B\_0$, $b\_i$ must be same-same or same-different-different-same or same-different-different-different-different-same.
To explain $\frac{ p}{2-p} = \frac{1}{2} - .28 = .22$ as in dvitek's analysis, we need $p= \frac{2}{ 1/.22 +1} =.361$, i.e. there is a discrepancy of about $.14$ to explain in the more natural statistic of how often two adjacent primes have distinct second-to-last bits.
| 5 | https://mathoverflow.net/users/18060 | 416487 | 169,716 |
https://mathoverflow.net/questions/416481 | 6 | It can be found, at Hartshorne exercise 4.2.7 for example, that in the case where $\operatorname{char}(k) \neq 2$ we have a nice correspondence between etale degree 2 covers of a curve $C$ and 2-torsion in $JC$. This correspondence, as constructed in Hartshorne, breaks down in the case where $\operatorname{char}(k) = 2$.
However it has been suggested in [Pries and Stevenson - A survey of Galois theory of curves in characteristic $p$](https://arxiv.org/abs/1004.2267), at the bottom of section 4.7, that "remarks following proposition 4.13 in chapter 3 of Milne's Étale cohomology" show that some correspondence between $JC[2](k)$ and etale degree 2 covers of $C$ still exists. Does anyone know how to construct such a correspondence explicitly? I believe it is erroneous (there is no proposition 4.13, it is a lemma!).
Otherwise, do étale degree 2 covers of $C$ correspond to anything else? Artin–Schreier theory suggests that it should be $H^1(C,\mathcal{O}\_C)^F$. However from such a nontrivial cover $\pi:X \to C$ we also obtain a nontrivial extension of $\mathcal{O}\_C$-algebras
$$0 \to \mathcal{O}\_C \to \pi\_\*\mathcal{O}\_X \to \mathcal{O}\_C \to 0$$
by looking at affines of $C$ and writing explicit Artin–Schreier extensions, and observing that the cocycles on $\pi\_\*\mathcal{O}\_X$ are unipotent $2\times2$ matrices. Is there any way to relate this to $JC$?
| https://mathoverflow.net/users/160814 | Degree-2 étale covers of curves in characteristic 2 vs torsion points on the Jacobian | There is a duality between degree 2 coverings and two-torsion points on the Jacobian — i.e. both form elementary abelian 2-groups, and these groups are naturally dual.
This is the Artin–Milne Poincaré duality pairing in fppf cohomology (Corollary 4.9 of [Duality in the Flat Cohomology of Curves](https://eudml.org/doc/142402))
$$ H^1(C\_{\overline{k}}, \mathbb Z/2) \times H^1 (C\_{\overline{k}},\mu\_2) \to H^2(C\_{\overline{k}}, \mu\_2) \to \mathbb Z/2 $$
with $H^1(C\_{\overline{k}}, \mathbb Z/2)$ classifying degree 2 coverings and $H^1 (C\_{\overline{k}},\mu\_2)$ classifying 2-torsion points of $J$.
Concretely, we can express this by noting that each covering of $C$ comes from a unique covering $A \to J$ of the same degree, and setting the pairing of this covering with a point in $J[2]$ to be zero if and only if that point lies in the image of $A[2]$.
More generally, for $C$ in characteristic $p$ and a $\mathbb Z/p$-covering of $C$ (and thus an isogeny $0 \to \mathbb Z/p \to A \to J\to 0$), we define the pairing with a $p$-torsion point by taking any lift of that point, multiplying by $p$, and obtaining an element of the kernel $\mathbb Z/p$. This doesn't depend on our choice of preimage because two lifts differ by an element of $\mathbb Z/p$ which is killed by multiplication by $p$.
This is linear in the $A[p]$ variable since every step is compatible with multiplication by $p$. This is linear in the covering variable since the addition map on coverings is fiber product followed by adding the kernels, and the product of two lifts is a lift to the fiber product.
To check this pairing has no kernel in the covering variable, the main point is to check that $A[p]$ and $J[p]$ have the same number of elements (since the rank is the multiplicity of the slope 0 of the Newton polygon, which is an isogeny invariant). Since $A \to J$ is a degree $p$ finite étale covering, its kernel contains a single $p$-torsion point, so the image of $A[p] \to J[p]$ has codimension $1$, forcing the pairing to be nontrivial.
To check it has no kernel in the other variable, factor the multiplication by $p$ map $A \to A$ (which kills all the $p$-torsion) into a totally inseparable morphism (which preserves points and thus $p$-torsion and an étale morphism (which therefore must kill all the $p$-torsion). The étale morphism is a fiber product of $\mathbb Z/p$-coverings, so if a point was in the image of $p$-torsion on each $\mathbb Z/p$-covering it would have to be in the image of $p$-torsion on this covering, forcing it to be trivial.
You're correct on your interpretation of Artin–Schreier theory. (I think you meant unipotent instead of unitary.) I don't know how to connect that perspective to the duality with the Jacobian — in particular, I don't think the Poincaré duality pairing above plays nicely with the Artin–Schreier map $\mathbb Z/2 \to \mathcal O\_C$.
| 6 | https://mathoverflow.net/users/18060 | 416504 | 169,721 |
https://mathoverflow.net/questions/414531 | 3 | **Does every three dimensional compact solvmanifold admit either Euclidean, nil, or sol geometry?**
definitions/motivation/background:
A solvmanifold is a manifold $ M $ admitting a transitive action by a solvable Lie group $ G $. In other words
$$
M \cong G/H
$$
where $ G $ is solvable and $ H $ is a closed subgroup of $ G $.
In dimension $ 2 $ there is strange coincidence where $ M $ is a solvmanifold if and only if it admits a transitive action by the Euclidean group $ E\_2 $ if and only if it is a flat manifold. This coincidence arises because $ SO\_2 $ is abelian so special Euclidean group $ SE\_2 $ is solvable.
Plenty of solvmanifolds are flat i.e. Euclidean. Some examples in dimension $ 3 $ include
the three torus $ T^3 $ (since abelian groups are solvable) as well as more exotic flat manifold like the one constructed from
$$
G=SE\_2= \{
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1
\end{bmatrix} : a,b,x,y \in \mathbb{R}, a^2+b^2=1 \}
$$
and
$$
H=
\{
\begin{bmatrix}
-1 & 0 & n \\
0 & -1 & m \\
0 & 0 & 1
\end{bmatrix} : n,m \in \mathbb{Z} \}
$$
the resulting compact flat 3 dimensional solvmanifold $ G/H $ has fundamental group $ \mathbb{Z}^2 \rtimes \mathbb{Z} $ where the semi direct product is with respect to
$$
n \mapsto \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}^n
$$
and the abelianization is $ H\_1 \cong \mathbb{Z} $ (thus this manifold is certainly not the 3 torus). Indeed five of the six possible flat compact orientable 3-manifolds ( the sixth is addressed in the EDIT) can be constructed this way where $ H $ is the semidirect product of a lattice in $ \mathbb{Z}^2 $ with a finite (orders $ 1,2,3,4,6 $ respectively, this corresponds to the monodromy) cyclic subgroup of $ SL\_2(\mathbb{Z}) $ preserving that lattice.
Another geometry solvmanifolds admit is nil geometry (since every niloptent group is solvable), for example
$$
G= \{
\begin{bmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{bmatrix}
: x,y,z \in \mathbb{R} \}
\, , \,
H= \{
\begin{bmatrix}
1 & k & n \\
0 & 1 & m \\
0 & 0 & 1 \\
\end{bmatrix}
: k,n,m \in \mathbb{Z} \}
$$
yielding $ G/H $ the Heisenberg nilmanifold.
And, finally, some solvmanifolds have sol geometry. For example, the quotient $ G/H $ for
$$
G=
\{
\begin{bmatrix}
a & 0 & x \\
0 & b & y \\
0 & 0 & 1
\end{bmatrix} : ab=1
\}
$$
and
$$
H=
\{
\begin{bmatrix}
\beta^k & 0 & n+m \beta \\
0 & \beta^{-k} & n+m \beta^{-1} \\
0 & 0 & 1
\end{bmatrix}
: k,n,m \in \mathbb{Z} \}
$$
where $ \beta $ is the root of $ x^2+3x+1 $ (or $ x^2+dx+1 $ for any integer $ d $ such that the roots of the polynomial are real and not integers ( $ |d| \geq 3 $ )). The fact that this discrete subset of matrices forms a subgroup follows from the fact that the unimodular companion matrix for the polynomial $ x^2+dx+1 $ satisfies the following matrix equation
$$
\begin{bmatrix} \beta & 0 \\ 0 & \beta^{-1} \end{bmatrix}
\begin{bmatrix} 1 & \beta \\ 1 & \beta^{-1} \end{bmatrix}
=\begin{bmatrix} 1 & \beta \\ 1 & \beta^{-1} \end{bmatrix}
\begin{bmatrix} 0 & -1 \\ 1 & -d \end{bmatrix}
$$
Note that in the reference <https://arxiv.org/abs/0903.2926> they construct the lattice $ H $ using $ d=-3 $ and $ \beta= \frac{3+\sqrt{5}}{2} $. This compact three dimensional solvmanifold with sol geometry has fundamental group $ \mathbb{Z}^2 \rtimes \mathbb{Z} $. Here the semidirect product is with respect to the map
$$
n \mapsto \begin{bmatrix} 0 & -1 \\ 1 & -d \end{bmatrix}^n
$$
and the abelianization is $ H\_1 \cong \mathbb{Z} $.
EDIT: there was originally a second part to this question asking if every flat compact 3 manifold is a solvmanifold. The answer is no. The Hantzsche-Wendt Manifold, mentioned in the comment by HJRW, is a flat compact 3-manifold that does not admit a transitive action by any Lie group so in particular there is no transitive action by a solvable Lie group. See
<https://math.stackexchange.com/questions/4374758/transitive-lie-group-action-on-the-hantzsche-wendt-manifold>
| https://mathoverflow.net/users/387190 | 3 dimensional solvmanifolds and Thurston geometries | Yes.
First observe that every compact 3d solvmanifold admits a Thurston geometry. To see why: Let $ M $ be a compact solvmanifold. Then by theorem of Auslander $ M $ is finitely covered by a compact special solvmanifold $ M' $ (a compact special solvmanifold is the quotient of a simply connected solvable Lie group $ G $ by a cocompact lattice). So $ G $ counts as a model geometry (with trivial stabilizer) for $ M' $. Since $ M' $ covers $ M $ then $ M $ also has model geometry $ G $.
A solvmanifold is aspherical and has solvable fundamental group. Since it is aspherical the geometry cannot be $ S^3 $ or $ S^2 \times E^1 $. Since it has solvable fundamental group it certainly has virtually solvable fundamental group and thus the geometry cannot be $ \tilde{SL\_2} $ or $ H^2 \times E^1 $ or $ H^3 $. Thus every three dimensional compact solvmanifold admits either $ E^3 $ Nil or Solv geometry.
| 2 | https://mathoverflow.net/users/387190 | 416509 | 169,724 |
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