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https://mathoverflow.net/questions/435812 | 2 | The excentral triangle of a reference triangle $ABC$ is the triangle with vertices corresponding to the excenters of $ABC$. Denote with $D$, $E$, $F$ the $A$−, $B$−, $C$− excenters, respectively. Denote with $U$, $V$, $W$ the midpoints of $BC$, $AC$, $AB$, respectively. Let $D'$, $E'$, $F'$ be the reflections of the points $D$, $E$, $F$ with respect to the midpoints $U$, $V$,$W$, respectively. Then $D'E'F'$ has the *same* area as the reference triangle $ABC$ (see Proposition 4 of [Dalcín and Kiss - Some Properties of the García Reflection Triangles](https://www.heldermann.de/JGG/JGG25/JGG251/jgg25010.htm)).
According to the [WOLFRAM Demonstrations Project](https://demonstrations.wolfram.com/InsphereAndFourExspheresOfATetrahedron/#more), a tetrahedron has four ex-spheres, but it does not show how to construct these spheres. I am interested in knowing if reflecting the centers of these 4 ex-spheres with respect to the four centroids of the four faces of the tetrahedron respectively, we would obtain a tetrahedron with the *same* volume as the reference tetrahedron. If so, how to prove it?
| https://mathoverflow.net/users/94729 | Looking for the 3D-analog/extension of a 2D theorem | It's definitely not of the same volume. If you apply your operation to the right tetrahedron you will get a right tetrahedron with smaller edges.
Although it's still possible that the volume is proportional to the volume of original. I haven't checked it.
If you want to check it you should use barycentric coordinates with respect to vertices of your tetrahedron. Let's call it $x\_1 x\_2 x\_3 x\_4$. Non-normalized coordinates of some point $p$ are the signed volumes of tetrahedrons $V(p,x\_2,x\_3,x\_4), V(x\_1,p,x\_3,x\_4), V(x\_1,x\_2, p,x\_4), V(x\_1,x\_2,x\_2,p)$. The sign of the first coordinate depends on which side of hyperplane spanned by $x\_2,x\_3,x\_4$ point $p$ sits. And the same for other coordinates.
If $p$ is excenter then the distance from it to any hyperplane spanned by a face is the same since it's the radius of ex-sphere. So you have $|V(x\_1,p,x\_3,x\_4)| = rS(x\_1,x\_3,x\_4)|$, where $S$ is a usual non-signed area.
Putting this all together we get that that non-normalized barycentric coordinates of an excenter $I\_1$ are $(-S(x\_2,x\_3,x\_4): S(x\_1,x\_3,x\_4): S(x\_1,x\_2,x\_4): S(x\_1,x\_2,x\_3)).$ And to get normalized you should divide each by the sum of coordinates so the new sum equals to $1$.
If you do it for a right tetrahedron you get $(-1/2, 1/2, 1/2, 1/2)$. I will not write the general case.
The centroid of $(x\_2, x\_3, x\_4)$ has coordinates $(0, 1/3, 1/3, 1/3)$. So if we reflect $I\_1$ with respect to it you get $(1/2, 1/6, 1/6, 1/6)$ lets call this point $I^\*\_1$. $I^\*\_1$ is closer to the center of the original tetrahedron than the vertices of the original polyhedron. So the right polyhedron with vertices $I^\*\_1, I^\*\_2, I^\*\_3, I^\*\_4$ will be smaller than the original one.
If you want to compute volume of $I^\*\_1, I^\*\_2, I^\*\_3, I^\*\_4$ in the general case than you just have to organize normalized coordinates of $I^\*\_1, I^\*\_2, I^\*\_3, I^\*\_4$ in a matrix, compute determinant and I think this will be $V(I^\*\_1, I^\*\_2, I^\*\_3, I^\*\_4)/V(x\_1,x\_2,x\_3,x\_4)$.
| 2 | https://mathoverflow.net/users/32454 | 436449 | 176,383 |
https://mathoverflow.net/questions/436451 | 1 | Let $W$ be a standard one dimensional Brownian motion, and let $\mathcal F\_t$ be its completed natural filtration.
Let $\tau$ be an $\mathcal F\_t$ stopping time with $\tau < T$ almost surely for some $T > 0$. Suppose $\xi$ is an $\mathcal F\_\tau$ measurable $L^2$ random variable.
**Question:** Does there exist some $\mathcal F\_t$ predictable process $H$ such that
$$\xi = \mathbb E[\xi] + \int\_0^\tau H\_s \, dW\_s$$
almost surely?
**Idea:**
I tried to proceed as follows - since $\xi$ is $\mathcal F\_\tau$ measurable and $\tau < T$ a.s., $\xi$ is clearly $\mathcal F\_T$ measurable. Now the standard martingale representation theorem gives some predictable $H$ such that
$$\xi = \mathbb E[\xi] + \int\_0^T H\_s \, dW\_s$$
almost surely. But since $\xi$ is $\mathcal F\_\tau$ measurable, it should follow that $H\_s = 0$, $d\mathbb P \times d\mu$ a.e. whenever $s > \tau$, whence
$$\xi = \mathbb E[\xi] + \int\_0^\tau H\_s \, dW\_s$$
as desired.
However, I am not fully sure how to rigorously show the claim $H\_s = 0$ whenever $s > \tau$.
| https://mathoverflow.net/users/173490 | Martingale representation theorem up to a stopping time | If $\xi\in \mathbb D^{1,2}$ is in the Sobolev-Watanabe space then we can apply Clark-Ocone formula to get that
$$\xi=E[\xi]+\int\_0^T E[D\_s\xi|\mathcal F\_s]dW\_s$$
where $D\_s$ is the Malliavin derivative. For $s\in [0,T]$ we may write $\xi=\xi 1\_{\{\tau > s\}}+\xi 1\_{\{\tau \leq s\}}$. Then
\begin{align\*}
\xi&=E[\xi]+\int\_0^T E[D\_s(\xi 1\_{\{\tau > s\}}+\xi 1\_{\{\tau \leq s\}})|\mathcal F\_s]dW\_s\\
&=E[\xi]+\int\_0^T E[D\_s(\xi 1\_{\{\tau > s\}})|\mathcal F\_s]dW\_s+\int\_0^T E[D\_s(\xi 1\_{\{\tau \leq s\}})|\mathcal F\_s]dW\_s
\end{align\*}
$\xi 1\_{\{\tau \leq s\}}$ is $\mathcal F\_s$-measurable so $D\_s (\xi 1\_{\{\tau \leq s\}})=0$. Also for $s>\tau$ we have that $\xi 1\_{\{\tau > s\}}=0$ so $D\_s (\xi 1\_{\{\tau > s\}})=0$ and for $s<\tau$ we have that $\xi 1\_{\{\tau > s\}}=\xi$. So
$$\xi=E[\xi]+\int\_0^\tau E[D\_s\xi|\mathcal F\_s]dW\_s.$$
| 2 | https://mathoverflow.net/users/479223 | 436453 | 176,384 |
https://mathoverflow.net/questions/435080 | 11 | In the paper "Resolutions of unbounded complexes" (Compositio Math., vol. 65, no. 2, pp. 121-154) N. Spaltenstein generalizes the 6 functor formalism to unbounded complexes of sheaves over ringed spaces. The properties which only involve the direct and inverse image functors, Hom and the tensor product are proved in full generality (Theorem A). For statements that also involve the direct and inverse images with compact support (Theorem B) one needs the spaces to be locally compact and Hausdorff, which is not surprising.
However, in this case Spaltenstein also makes the additional assumption that the kernel of the differential in a (possibly unbounded) complex of c-soft sheaves is c-soft in each degree (Condition ($\ast$) on p. 122, which is used in Theorem B). He remarks that this assumption is satisfied for a space $X$ if it is locally finite dimensional, meaning that for every $x\in X$ there is an $n\in\mathbb{Z}$ such that $x\in X$ has an arbitrarily small neighborhood $U$ with the property $H^{>n}\_c(U,S)=0$ for every sheaf $S$ on $U$. So in particular, all locally finite polyhedra have property ($\ast$).
Condition ($\ast$) is needed to construct a class $\mathfrak{S}$ of complexes of sheaves that has the properties listed on p. 150. Namely, if the condition holds, then one can take $\mathfrak{S}=$ the class of complexes of degree-wise c-soft sheaves.
From reading the paper one gets the impression that the condition is a technical shortcut which may not be absolutely necessary. Also, the paper appeared in 1988, which is quite a while ago, so one wonders whether some of the technical issues have since been settled. Namely,
(1) Can one construct a class $\mathfrak{S}$ of complexes that has properties (a)-(f) on p. 150 without using Condition ($\ast$)?
(2) Are there compact Hausdorff spaces that do not satisfy Condition ($\ast$)? Also, is there a wider class of locally compact Hausdorff spaces than locally finite dimensional for which Condition ($\ast$) holds?
| https://mathoverflow.net/users/2349 | Resolutions of unbounded complexes: Condition ($\ast$) in Spaltenstein's paper | Here is a variant of an example due to Lurie (as far as I can tell) [HTT, Counterexample 6.5.4.2] showing that the proper base change theorem [Spaltenstein, Proposition 6.20] does *not* hold for unbounded complexes, even on compact Hausdorff spaces. In particular, as condition 6.14(2) can be replaced by 6.14(a)–(f) or condition (\*), the example shows that a class $\mathfrak S$ with these conditions does not always exist.
In fact, Lurie uses this as one of the motivations why the natural notion of $\infty$-topos is defined in terms of sheaves instead of hypersheaves; see [HTT, Warning 6.5.4.1]. I'm not entirely sure what this translates to in the language of derived categories, but I suspect the "correct" category will somehow look more like "sheaves of spectra" than like chain complexes in $\operatorname{Sh}(X,\mathbf Z)$.
For self-containedness, let me translate the example to the language of Spaltenstein. It turns out that this is a little more involved, roughly because $\operatorname{Hom}(\mathbf Z^{(S)},\mathbf Z^{(T)})$ is harder to compute than $\operatorname{Map}(S,T)$ for sets $S$ and $T$; see the remark after the example.
**Example.** Let $Q = [0,1]^{\mathbf N}$ be the Hilbert cube (where $\mathbf N = \{0,1,\ldots\}$). For $i \in \mathbf N$, write $Q\_i = [0,1]^{\mathbf N\_{> i}}$, so that $Q = [0,1]^{\{0,\ldots,i\}} \times Q\_i = [0,1]^{i+1} \times Q\_i$. For $i \in \mathbf N$, define opens in $X = Q \times [0,1]$ by
\begin{align\*}
U\_i &= (0,1)^i \times [0,1)\times Q\_i \times [0, 2^{-i}),\\
V\_i &= (0,1)^i \times (0,1] \times Q\_i \times [0, 2^{-i}),\\
\end{align\*}
and write $U\_{i,0} = U\_i \cap (Q \times \{0\})$ and $V\_{i,0} = V\_i \cap (Q \times \{0\})$. Then we have $U\_i \cup V\_i \subseteq U\_{i-1} \cap V\_{i-1}$ for all $i \geq 1$, and this induces an equality
$$U\_{i,0} \cup V\_{i,0} = U\_{i-1,0} \cap V\_{i-1,0}.\tag{1}\label{1}$$
Consider the complex $C^\bullet \in \operatorname{Ch}^{\leq 0}(X,\mathbf Z)$ given by
$$\cdots \to \mathbf Z\_{U\_i} \oplus \mathbf Z\_{V\_i} \to \cdots \to \mathbf Z\_{U\_1} \oplus \mathbf Z\_{V\_1} \to \mathbf Z\_{U\_0} \oplus \mathbf Z\_{V\_0},$$
where the differentials are all given by the matrix $\big(\begin{smallmatrix}1 & 1 \\ -1 & -1\end{smallmatrix}\big)$. Consider the base change square
$$\begin{array}{ccc}Q \times \{0\} & \stackrel{q'}\to & X \\ \!\!\!\!\!{\scriptsize f'}\!\downarrow & & \downarrow\!{\scriptsize f}\!\!\! \\ \{0\} & \stackrel q\to & [0,1].\!\end{array}$$
We claim that the natural map $q^\*Rf\_! C^\bullet \to Rf'\_!q'^\* C^\bullet$ is not an isomorphism in $D(\{0\},\mathbf Z) = D(\mathbf Z)$. In fact, we will show that the left hand side is $\mathbf Z[-1]$ and the right hand side $\mathbf Z[0]$, so the map is zero as $\operatorname{Hom}\_{D(\mathbf Z)}(\mathbf Z[-1],\mathbf Z) = \operatorname{Ext}^1(\mathbf Z,\mathbf Z) = 0$.
For the right hand side, restricting to $Q \times \{0\}$ gives the complex
$$\cdots \to \mathbf Z\_{U\_{i,0}} \oplus \mathbf Z\_{V\_{i,0}} \to \cdots \to \mathbf Z\_{U\_{1,0}} \oplus \mathbf Z\_{V\_{1,0}} \to \mathbf Z\_{U\_{0,0}} \oplus \mathbf Z\_{V\_{0,0}},$$
which is acyclic in negative degree by \eqref{1} and the Mayer–Vietoris sequences
$$0 \to \mathbf Z\_{U \cap V} \stackrel{\big(\begin{smallmatrix}1 \\ -1\end{smallmatrix}\big)}\longrightarrow \mathbf Z\_U \oplus \mathbf Z\_V \stackrel{(1\ 1)}\longrightarrow \mathbf Z\_{U \cup V} \to 0.$$
Since $U\_{0,0} \cup V\_{0,0} = Q \times \{0\}$, the Mayer–Vietoris sequence then shows that the natural quotient
$$q'^\*C^\bullet \to \mathbf Z\_{Q \times \{0\}}[0]$$
is a quasi-isomorphism. Thus, $Rf'\_!q'^\* C^\bullet = Rf'\_!\mathbf Z[0] = \mathbf Z[0]$ since $Q$ is compact and contractible.
For the left hand side, we have to do a little work to show that $C^\bullet$ is Postnikov complete, which is the only way I know how to compute cohomology of complexes that are unbounded below on an infinite-dimensional space. (Please let me know if you know a different method!)
For each finite set $I \subseteq \mathbf N \cup \{\infty\}$, write $\pi\_I \colon X \to [0,1]^I$ for the projection (where $\infty$ is the separate factor $[0,1]$ in $X = Q \times [0,1]$). Consider the set $\mathfrak P$ of sets of the form $\pi\_I^{-1}(A)$, where $I \subseteq \mathbf N$ is finite and $A = \prod\_{i \in I} A\_i$ is a product of intervals (open, half-open, or closed). Note that $U\_i$, $V\_i$, $U\_i \cup V\_i$, and $U\_i \cap V\_i$ are in $\mathfrak P$ for all $i$, as are the sets
$$W\_i = \big(U\_i \cap V\_i\big)\setminus\big(U\_{i+1} \cup V\_{i+1}\big) = (0,1)^{i+1} \times Q\_i \times [2^{-i-1}, 2^{-i}).$$
Since the intersection of two intervals is an interval, the set $\mathfrak P$ is closed under finite intersections. For $x \in X$, write $\mathfrak P\_x = \{A \in \mathfrak P\ |\ x \in A^\circ\}$, and note that this forms a neighbourhood basis of $x$ (it will be convenient *not* to restrict solely to open neighbourhoods). Write $\mathfrak B = \operatorname{Open}(X) \cap \mathfrak P$ and $\mathfrak B\_x = \mathfrak B \cap \mathfrak P\_x$; these form a basis for the topology and for the open neighbourhoods of $x$ respectively. Finally, note that if $A \in \mathfrak P$, then $\bar A \in \mathfrak P$.
**Lemma.** *If $A \in \mathfrak P$ and $x \in X$, then $H^i(V,\mathbf Z\_A) = 0$ for all $i > 0$ and $V \in \mathfrak P\_x$ sufficiently small.*
In particular, the opens $\mathfrak B$ and the sheaves $\mathbf Z\_A$ for $A \in \mathfrak P$ satisfy condition 3.12(1) in [Spaltenstein].
*Proof.* Let $B = \bar A$ and $D = B \setminus A$. If $A = \pi\_I^{-1}\big(\prod\_{i\in I} A\_i\big)$ with $\bar A\_i = [a\_i,b\_i]$, then
$$D \subseteq \partial A = \pi\_I^{-1}\big(\prod\_{i \in I} \{a\_i,b\_i\}\big).$$
Call the *faces* of $D$ the subsets of the form $\pi\_i^{-1}(\{a\_i\})$ or $\pi\_i^{-1}(\{b\_i\})$. Consider those $V \in \mathfrak P\_x$ such that $V \cap \partial A$ only meets the faces containing $x$; concretely this can be accomplished by taking $V$ contained in $\pi\_I^{-1}\big(\prod\_{i \in I}[x\_i-\varepsilon,x\_i+\varepsilon]\big)$ for $\varepsilon$ such that $\varepsilon < \lvert x\_i-a\_i\rvert$ whenever $x\_i \neq a\_i$, and likewise for $b\_i$. This means that
$$V \cap D = V \cap \bigcup\_{i \in F} \pi\_i^{-1}(\{x\_i\}) \qquad \text{where } F = \{i \in I\ |\ x\_i\in\{a\_i,b\_i\}\}.$$
Thus, $V \cap D$ is either empty (if $F = \varnothing$) or contractible. Since the locally closed immersion $A\cap V \to V$ factors via the closed immersion $B \cap V \to V$, we get
$$R\Gamma(V,\mathbf Z\_A) = R\Gamma(V,\mathbf Z\_{A \cap V}) = R\Gamma(B \cap V,\mathbf Z\_{A \cap V}).$$
If $F = \varnothing$, then $A \cap V = B \cap V$, so we conclude that $R\Gamma(V,\mathbf Z\_A) = \mathbf Z[0]$ since $B \cap V$ is contractible. Likewise, if $F \neq \varnothing$, then $V \cap D$ is contractible, so the map $R\Gamma(B \cap V,\mathbf Z\_{B \cap V}) \to R\Gamma(B \cap V,\mathbf Z\_{D \cap V})$ is an isomorphism, so we conclude $R\Gamma(B \cap V,\mathbf Z\_{A \cap V}) = 0$. $\square$
Going back to the complex $C^\bullet$ above, we have $\mathscr H^{-i}(C^\bullet) \cong \mathbf Z\_{W\_i}$ by Mayer–Vietoris, since the kernel of $d^{-i}$ is $\mathbf Z\_{U\_i \cap V\_i}$ and the image of $d^{-i-1}$ is $\mathbf Z\_{U\_{i+1}\cup V\_{i+1}}$. Thus the lemma above and [Spaltenstein, Proposition 3.13] (see also [Stacks, Tag [0D63](https://stacks.math.columbia.edu/tag/0D63)]) imply that
$$C^\bullet \to \underset{\substack{\longleftarrow \\ n}}{\operatorname{holim}} \tau\_{\geq -n}C^\bullet$$
is an isomorphism in $D(X,\mathbf Z)$. Then [Stacks, Tag [0D60](https://stacks.math.columbia.edu/tag/0D60)] shows that
$$R\Gamma\big(Q \times [0,a),C^\bullet\big) = \underset{\substack{\longleftarrow \\ n}}{\operatorname{holim}} R\Gamma\big(Q \times [0,a), \tau\_{\geq -n} C^\bullet\big),$$
so we have to compute $R\Gamma(Q \times [0,a),\tau\_{\geq -n}C^\bullet)$. For $n < 0$ we have $\tau\_{\geq -n} C^\bullet = 0$. By [Stacks, Tag [08J5](https://stacks.math.columbia.edu/tag/08J5)] we have distinguished triangles
$$\mathscr H^{-i}(C^\bullet)[i] \to \tau\_{\geq -i} C^\bullet \to \tau\_{\geq -i+1} C^\bullet.\tag{2}\label{2}$$
Recall that $\mathscr H^{-i}(C^\bullet) = \mathbf Z\_{W\_i}$. Thus the same computation as in the lemma (plus the proper base change theorem and the computation of $R\Gamma\_c\big((0,1)^{i+1},\mathbf Z\big)$) gives
$$R\Gamma\big(Q \times [0,a),\mathbf Z\_{W\_i}\big) = \begin{cases}0, & a > 2^{-i},\\ \mathbf Z[-i-1], & 2^{-i-1} < a \leq 2^{-i}, \\ 0, & a \leq 2^{-i-1}.\end{cases}$$
Indeed, in the first case, the closed interval $[2^{-i-1},2^{-i}]$ is contained in $[0,a)$, so again the inclusion
$$(0,1)^{i+1} \times Q\_i \times \{2^{-i}\} \to (0,1)^{i+1} \times Q\_i \times [2^{-i+1},2^{-i}]$$
induces an isomorphism on compactly supported cohomology. In the second case, $W\_i \cap (Q \times [0,a))$ is closed and contractible in $Q \times [0,a)$, and in the third case it is empty.
Given $a \in (0,1]$, there is a unique $i \in \mathbf N$ with $2^{-i-1} < a \leq 2^{-i}$. From \eqref{2} we thus conclude that
$$R\Gamma\big(Q \times [0,a),\tau\_{\geq -n} C^\bullet\big) = \mathbf Z[-1]$$
as soon as $n \geq i$. Thus, the cohomology of $\tau\_{\geq -n}C^\bullet$ is eventually constant for all $n$ and $a$, and we conclude $R\Gamma(Q \times [0,a),C^\bullet) = \mathbf Z[-1]$. Since $R^if\_!C^\bullet = R^if\_\*C^\bullet$ is the sheafification of $U \mapsto H^i(f^{-1}(U),C^\bullet)$ [Stacks, Tag [0D5X](https://stacks.math.columbia.edu/tag/0D5X)] and sheafification preserves stalks, we conclude that
$$q^\*Rf\_!C^\bullet = \mathbf Z[-1],$$
which has cohomology only in degree $1$, and does not agree with $Rf'\_!q'^\*C^\bullet$. $\square$
**Remark.** The example of [HTT, Counterexample 6.5.4.2] is much easier, because it works with sheaves of spaces instead of chain complexes of sheaves of abelian groups. The point is that the sheaf of sets $\pi\_0(\mathscr F)$ of Lurie's example has empty global sections, so $\mathscr F$ itself cannot have global sections. But in abelian categories, nonzero objects can of course map to the zero object, and the vanishing of $q^\*Rf\_\*C^\bullet$ was substantially more work. There might be shortcuts to the above argument using simplicial or homotopical techniques.
**Remark.** One can immediately see that the Hilbert cube is not locally finite-dimensional: opens around an 'interior' point (no coordinate is $0$ or $1$) have higher and higher cohomological dimension, already for compactly supported cohomology with constant coefficients.
On the other hand, checking (\*) directly in this case is a little unclear, but the failure of proper base change shows that it cannot be true. So Spaltenstein wasn't wrong to impose extra hypotheses!
---
**References.**
[HTT] J. Lurie, [*Higher topos theory*](http://dx.doi.org/10.1515/9781400830558). Annals of Mathematics Studies **170**. Princeton University Press, 2009.
[Spaltenstein] N. Spaltenstein, [*Resolutions of unbounded complexes*](http://www.numdam.org/item?id=CM_1988__65_2_121_0). Compos. Math. **65**.2, p. 121-154 (1988).
[Stacks] [The Stacks project](https://stacks.math.columbia.edu).
| 6 | https://mathoverflow.net/users/82179 | 436459 | 176,386 |
https://mathoverflow.net/questions/435913 | 0 | $\newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\ds}{\displaystyle} \newcommand{\Lpn}[2]{\left\lVert#1\right\rVert\_{L^{#2}}}$
$\newcommand{\Lptxy}[3]{\left\lVert#1\right\rVert\_{L^{#2}\_{#3}}}$
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I am verifying the bilinear estimate (Proposition 3.6) in [this paper](https://arxiv.org/pdf/1302.2933.pdf). The first steps were clear except the last one.
$$\Lptxy{\prod\_{j=1}^2 \PQ{u}{j}}{2}{t x y}
= \Lptxy{\F(\prod\_{j=1}^2 \PQ{u}{j})(\tau,\xi,q)}{2}{\tau \xi q}$$
$$=\Lptxy{\ds\circledast\_{j=1}^3 \PQ{u}{j}(\tau,\xi,q)}{2}{\tau \xi q}$$
$$ = \Lptxy{\ds\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z^2}{\prod\_{j=1}^2 \FPQwV{u}{j}} d\nu}{2}{\tau \xi q}$$
$$ \leq \prc{ \ds\int\_\R \ds\sum\_{q} \abs{\int\_{\R^4} \sum\_{q\_1,q\_2 \in \Z^2}d\nu }^2 d\tau d\xi}^\frac{1}{2} \prc{ \ds\int\_\R \ds\sum\_q \abs{{\int\_{\R^4} \sum\_{(q\_1,q\_2) \in \Z^2} \prod\_{j=1}^2 \FPQwV{u}{j} d\nu }}^2 d\tau d\xi }^\frac{1}{2}$$
$$ \leq \sup\_{\tau,\xi,q}\prc{ \int\_{\R^4} \sum\_{q\_1,q\_2 \in \Z^2}d\nu}^\frac{1}{2} \prc{ \ds\int\_\R \ds\sum\_q \abs{{\int\_{\R^4} \sum\_{(q\_1,q\_2) \in \Z^2} \prod\_{j=1}^2 \FPQwV{u}{j} d\nu }}^2 d\tau d\xi }^\frac{1}{2}$$
I could not justify how the last step can be achieved! Namely,
$$\prc{ \ds\int\_\R \ds\sum\_q \abs{{\int\_{\R^4} \sum\_{(q\_1,q\_2) \in \Z^2} \prod\_{j=1}^2 \FPQwV{u}{j} d\nu }}^2 d\tau d\xi }^\frac{1}{2} \lesssim \prod\_{j=1}^2 \Lptxy{\PQ{u}{j}}{2}{t x y}.$$
Any help is appreciated.
| https://mathoverflow.net/users/471464 | Verifying the proof of a bilinear estimate in $L^2$ | $\newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\ds}{\displaystyle} \newcommand{\Lpn}[2]{\left\lVert#1\right\rVert\_{L^{#2}}}$
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$\newcommand{\prdd}{\prod\_{j=1}^2}$
There is a mistake in to which integration applying the Cauchy-Schwarz inequality.
\begin{align\*}
&\Lptxy{\ds\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z^2}{\prod\_{j=1}^2 \FPQwV{u}{j}} d\nu}{2}{\tau \xi q}\\
& \leq \Lptxy{\prc{\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z} d\nu}^\frac{1}{2} \prc{\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z^2} \abs{\prdd \FPQwV{u}{j}}^2 d\nu}^\frac{1}{2} }{2}{\tau \xi q}\\
& \leq \sup\_{\tau,\xi,q}\prc{ \int\_{\R^4} \sum\_{q\_1,q\_2 \in \Z^2}d\nu}^\frac{1}{2} \Lptxy{ \prc{\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z^2} \abs{\prdd \FPQwV{u}{j}}^2 d \nu}^\frac{1}{2} }{2}{\tau \xi q}\\
& \leq \sup\_{(\tau,\xi,q)}\abs{A\_{\tau,\xi,q}}^\frac{1}{2} \prc{\int\_{\R^2}\sum\_{q} \abs{\prc{\int\_{\R^4} \sum\_{q\_1,q\_2 \in \Z} \abs{\prdd \FPQwV{u}{j}}^2 d \nu}}d\tau d\xi}^\frac{1}{2} \\
& \leq \sup\_{(\tau,\xi,q)}\abs{A\_{\tau,\xi,q}}^\frac{1}{2} \prc{\int\_{\R^4}\sum\_{q\_1,q\_2} \abs{\prod\_{j=1}^2\FPQwV{u}{j}}^2\prc{\int\_{\R^2} \sum\_q \abs{\FPQ{u}{3}}^2 d\tau d\xi} d\nu}^\frac{1}{2}\\
&= \sup\_{(\tau,\xi,q)}\abs{A\_{\tau,\xi,q}}^\frac{1}{2} \Lptxy{\FPQ{u}{3}}{2}{\tau \xi q} \prc{\int\_{\R^4}\sum\_{q\_1,q\_2} \abs{\prod\_{j=1}^2\FPQwV{u}{j}}^2d\nu}^2\\
&= \sup\_{(\tau,\xi,q)}\abs{A\_{(\tau,\xi,q)}}^\frac{1}{2} \prod\_{j=1}^3 \Lptxy{\PQ{u}{j}}{2}{t x y}.
\end{align\*}
| 0 | https://mathoverflow.net/users/471464 | 436460 | 176,387 |
https://mathoverflow.net/questions/435319 | 5 | I am having a little confusion in verifying the two dimensional oscillatory integral in Lemma 2.1 in [This paper](https://epubs.siam.org/doi/epdf/10.1137/080739173), namely
$$I\_t (x,y) = \int\_{\mathbb{R}^2} |\xi|^{\epsilon + i \beta} e^{i t(\xi^3 + \xi \eta^2 + x \xi + y \eta)} d\xi\, d\eta.$$
I verified the integration with respect to $\eta$ using the inverse of Fourier transform. I reached the same result on the paper, namely
$$\int\_{\mathbb{R}^2} |\xi|^{\epsilon} e^{i t(\xi^3 + \xi \eta^2 + x \xi + y \eta)} d\xi \,d\eta = \sqrt{\pi} \lim\_{a \to \infty} \int\_\mathbb{R}\frac{|\xi|^\epsilon}{ \sqrt{| t\xi|}} e^{i t(\xi^3+ \xi \eta^2) + x \xi ) -\frac{y^2}{4 t \xi} + \frac{\pi}{4} \operatorname{sgn}{(t \xi)}} \chi\_a(\xi)\, d \xi,$$
where $\chi\_a(\xi) = \chi\_{\{ \xi: |\xi| \leq a\}}(\xi)$.
The author stated that the right hand side of the above equation is bounded uniformly by $|t|^{- \frac{2 + \epsilon}{3}}$.
I think there is a change of variable has involved which made the Ven der Corput's lemma applicable. Can the lemma be used directly? what about the term on the phase function which has $\xi$ on denominator? Any explanation is appreciated. Thanks in advance.
| https://mathoverflow.net/users/471464 | Two dimensional oscillatory integral | Just perform the change of variable $t^{\frac{1}{3}} \xi\mapsto \xi$, then change the variable $a$ such that the integration becomes as in the paper.
| 1 | https://mathoverflow.net/users/471464 | 436461 | 176,388 |
https://mathoverflow.net/questions/436441 | 11 | Construct the $n$-tuple Cartesian product of the ternary set $X\_n=\{0,1,2\}\times\cdots\times\{0,1,2\}=\{0,1,2\}^n$. Define its subset $W\_n$ according to the rule (here $y=(y\_1,\dots,y\_n)$ is made use of)
$$W\_n=\{y\in X\_n: y\_1\leq1, y\_1+y\_2\leq2,\dots,y\_1+\cdots+y\_{n-1}\leq n-1, y\_1+\cdots+y\_n=n\}.$$
Introduce the one-variable polynomial
$$P\_n(x)=\sum\_{y\in W\_n}x^{\prod\_{j=1}^n\binom{2}{y\_j}}.$$
>
> **QUESTION.** If $C\_n$ stands for the Catalan numbers, is this true?
> $$\frac{dP\_n}{dx}(1)=C\_{n+1}.$$
>
>
>
**Examples.** $P\_1(x)=x^2, P\_2(x)=x^4+x, P\_3(x)=x^8+3x^2, P\_4(x)=x^{16}+6x^4+2x$.
| https://mathoverflow.net/users/66131 | And, yet, another evaluation to Catalan numbers | Yes, this is true.
Write each $y\_j = 1 - x\_j$ with $x\_j \in \{-1, 0, 1\}$,
so the condition is that $\sum\_{j=1}^n x\_n = 0$
and no partial sum is negative.
This can be viewed as an n-move king path from $(0,0)$ to $(n,0)$
that never goes below the horizontal axis.
We want the sum over such $(x\_1,\ldots,x\_n)$ of $2^h$
where $h$ is the number of indices $j$ for which $y\_j = 1$, i.e. $x\_j = 0$
(the number of horizontal king steps).
If we drop the nonnegativity condition, then for each $k$
the sum of $2^h$ over paths from $(0,0)$ to $(n,k)$ is
the $t^k$ cofficient in the generating function $(t^{-1} + 2 + t)^n$;
since that's just $(t+1)^{2n} / t^n$, the sum is $2n \choose n+k$.
By a reflection argument familiar from the Catalan enumeration of Dyck paths,
it follows that for $k=0$ restricted to nonnegative paths
is the difference between the unrestricted $k=0$ and $k=-1$ sums,
which is ${2n \choose n} - {2n \choose n-2} = C\_{n+1}$.
| 12 | https://mathoverflow.net/users/14830 | 436463 | 176,389 |
https://mathoverflow.net/questions/436370 | 0 | Given a circle $C$ in the xz-plane which does not intersect the $z$-axis, we can build a smooth 2-torus with surface area $(2\pi a)(2\pi b)$ where $a$ is the radius of the circle $C$ and $b$ is the distance from the $z$-axis to the center of $C$.
Now, a circle has rotational symmetry and the surface area formula in this case becomes a product of arclengths. I am wondering under which conditions a surface area of a surface of revolution remains a product of arclengths. For instance, suppose that rather than $C$ being a circle, we instead take $C$ to be an ellipse, which we can rotate in the $xz$-plane before revolving about the $z$-axis. Will the surface area of the resulting surface $S$ of revolution be invariant under rotations in the $xz$-plane which fix the center of $C$ before revolving $C$ around the $z$-axis [provided that $C$ does not intersect the $z$-axis in the $xz-$plane]? What if instead of $C$ being a smooth (quadratic) curve, we take $C$ to be a convex polygon in the $xz$-plane which does not intersect the $z$-axis. Is the surface area of the resulting surface $S$ of revolution invariant under rotating $C$ in the $xz$-plane while fixing the center point, before revolving around the $z$-axis?
Essentially, I am wondering under which conditions the surface area of a surface $S$ of revolution is given by a product of arclengths. Here, for instance, when we revolve $C$ [contained in the $xz$-plane] around the $z$-axis, the center point of $C$ traverses a circle around the $z$-axis as we revolve. What if rather than using a circle to create a surface of revolution, we “revolve” $C$ around the $z$-axis using an ellipse [which projects to an ellipse in the $xy$-plane]: will the surface area still be a product of arclengths?
It feels like I'm circling (pardon the pun) around a theorem relating curvature to the surface area of $S$. We have, for instance, Gauss–Bonnet: $$ 2\pi \chi = \int\_S \mathcal{K}dS $$ and I'm wondering whether the vanishing of either side of Gauss–Bonnet can tell us that the surface area is a product of arclengths. In the case that $C$ is a convex polytope (polygon) in the $xz$-plane, the curvature of $C$ is concentrated at the vertices, and I'm wondering whether rotating $C$ in the $xz$-plane before we revolve around the $z$-axis can create non-zero curvature on $S$ which therefore tells us that the surface area is no longer a product of arclengths.
Ultimately, I am asking: does there exist a simple closed curve $C$ in the $xz$ plane which does not intersect the $z$ axis for which we can create a surface of revolution $S$ by revolving $C$ around the $z$-axis such that the surface area of $S$ is not the arclength of $C$ multiplied by $2\pi b$ where $b$ is the distance from the $z$-axis to the “center” of $C$. I am especially interested in the case that $C$ is not smooth [just piecewise linear, for instance].
Edit/Update: Let's fix a definition of $b$ as the distance from the $z$-axis to the centroid of the convex hull of $C$ in the $xz$-plane.
Let's fix a definition of “center” as centroid of the convex hull of $C$ in the $xz$-plane.
Another question here is what happens to surface area of $S$ if we rotate $C$ as we revolve [does surface area only “see” integer number of rotations per revolution, for instance? Does it see anything at all?]. But I suppose that is a separate question.
| https://mathoverflow.net/users/131090 | Conditions for surface area of surface of revolution to be product of arclengths | No, the surface area of the surface of revolution $S$ is in general not given by the arc length of $C$ multiplied by $2\pi b$, with $b$ the distance of the centroid of the convex hull of $C$ from the axis. As a simple counterexample, take $C$ to be the union of a semicircle of radius $R$ and a straight segment connecting the endpoints of the semicircle. Place the straight segment parallel to the axis, at a distance $a$ from the latter, and have the semicircle face outward, away from the axis. Then:
$$
b=a+\frac{4}{3\pi } R
$$
The arc length is
$$
L=(2+\pi )R
$$
The surface area of $S$ is
$$
\Sigma = 2\pi (2+\pi ) aR + 4\pi R^2
$$
and therefore $\Sigma \neq 2\pi b L$.
| 1 | https://mathoverflow.net/users/134299 | 436467 | 176,390 |
https://mathoverflow.net/questions/436445 | 3 | I was recently looking into an old problem of Hardy which studies the distribution of integers of the form $2^a 3^b \leq x$, where $a,b\geq 0$. Letting $N(x)$ denote the number of pairs $(a,b)$ satisfying this inequality, one has
$$
N(x) = \frac{\log(2x)\log(3x)}{2\log2\log3} + o\left(\frac{\log x}{\log\log x} \right).
$$
The error term here depends critically on the diophantine nature of the real number $\frac{\log 2}{\log 3}$, and in particular on the growth of the denominators $q\_m$ of its convergents (in the sense of continued fractions). One could obtain a power savings in $\log x$ above if one knew something like $q\_{m+1} \ll q\_m^A$ for some real number $A$, but I suspect proving such a bound is very hard. My knowledge of diophantine approximation is introductory at best, so I would like to know from any experts the following:
1. Are there "standard" conjectures which predict the growth of $q\_m$, at least in the case when one is approximating a quotient of logarithms of integers?
2. Hardy proves that $q\_{m+1} \leq e^{A q\_m}$ for some explicit constant $A$, which he improves (using an old theorem of Pillai) to $e^{\varepsilon q\_m}$ for any $\varepsilon > 0$ with $m > m\_0(\varepsilon)$ . I suspect not much more is known in terms of a better bound unconditionally. Are there improvements to this estimate?
| https://mathoverflow.net/users/307675 | The growth of certain continued fractions | The keyword you are looking for is "irrationality measure" -- I think some authors (such as [Lang](https://mathscinet.ams.org/mathscinet-getitem?mr=1348400)) call it constant type. If you know the irrationality measure of $\alpha$ is $\mu = \mu(\alpha)$, then the convergents of $\alpha$ satisfy $q\_{k+1} \ll q\_k^{\mu -1 + \epsilon}$ for every $\epsilon>0$.
Most irrational numbers you can write down will have irrationality measure $2$. This can be made precise in the sense that a standard probabilistic argument shows that the set of $\alpha$ with $\mu(\alpha)>2$ has null Lebesgue measure. Showing this for explicit numbers, however, can be very hard -- for example, Roth's theorem (that famously probably won him a Fields medal) states that $\mu(\alpha) = 2$ if $\alpha$ is an algebraic irrational.
I'm not sure what is best known for the irrationality measures of quotients of logarithms, but I found this [previous MO question](https://mathoverflow.net/questions/166088/irrationality-measure-of-log2-log6) when looking it up. Lucia seems to think that's the state of the art (at least circa 2014), and that gives something like $q\_{k+1} \ll q\_k^{7.617}$.
| 2 | https://mathoverflow.net/users/37327 | 436470 | 176,391 |
https://mathoverflow.net/questions/436477 | 0 | For $A\subseteq\omega$ we define the *upper density* by $$d\_u(A) = \lim\sup\_{n\to\infty}\frac{|A\cap n|}{n+1}.$$ For $y\in \omega$ we set $A - y:= \{|a\setminus y|:a\in A\}.$ Note that $|a\setminus y|$ equals the difference of $a$ and $y$ if $a\geq y$, and $0$ otherwise. The *upper Banach density* is defined by $$d\_{uB}(A) = \lim\_{n\to\infty}\big(\sup
\{\frac{|(A-y)\cap n|}{n+1}: y\in\omega\}\big).$$
It is easy to see that $d\_{uB}(A) \geq d\_{u}(A)$ for all $A\subseteq \omega$.
**Question.** What is an example of $A\subseteq \omega$ with $d\_{uB}(A) > 0$ and $d\_u(A) = 0$?
| https://mathoverflow.net/users/8628 | Upper density versus upper Banach density on $\omega$ | I think that it is fairly straightforward to get such a set. You can simply get the set as a union of intervals:$\newcommand{\intrvl}[2]{\langle{#1},#2)}\newcommand{\intrvr}[2]{({#1},#2\rangle}\newcommand{\limti}[1]{\lim\limits\_{{#1}\to\infty}}$
$$A=\bigcup\limits\_{n\in\omega} \intrvr{a\_n}{b\_n}.$$
And you choose the intervals in a suitable way. You want them long enough to get $d\_{uB}(A)=1$. And, at the same time, you want start each interval far enough to keep $d\_u(A)=0$. So it suffices to choose them in such way that
\begin{align\*}
\limti n (b\_n-a\_n)=\infty\\
\limti n \frac{\sum\_{k=1}^n (b\_k-a\_k)}{b\_n}=0
\end{align\*}
The first conditions will help you to get upper Banach density (a.k.a. upper uniform density) equal to one. And the second condition will keep the [asymptotic density](https://en.wikipedia.org/wiki/Natural_density) (natural density) zero.
For example, you can take $a\_n=2^n$ and $b\_n=2^n+n$.
Several posts on this site are a bit related - and the answers include some references which might be useful:
* [On the independence of lower and upper asymptotic and Banach densities](https://mathoverflow.net/q/206801)
* [Prescribed values for the uniform density](https://mathoverflow.net/q/103111)
* [Density of a set of natural numbers whose differences are not bounded.](https://mathoverflow.net/q/66191)
* [References on density of subsets of $\mathbb{N}$](https://math.stackexchange.com/q/126745) (on Mathematics Stack Exchange)
| 3 | https://mathoverflow.net/users/8250 | 436480 | 176,396 |
https://mathoverflow.net/questions/436485 | 1 | Let $B$ be a category with products and let $F:A\to B$ be a discrete opfibration.
Let $F^\*:B\to \bf Set$ be the functor corresponding to $F$ under the Grothendieck correspondence.
The following proposition should be true and the proof is rather straightforward:
>
> $F^\*$ preserves products if and only if $A$ has products and $F$ preserves them.
>
>
>
Questions (supposing this is indeed true):
1. Is this stated somewhere in the literature?
2. Is it true for all limits?
3. Is it a consequence of a more generale statement regarding (not necessarily discrete) opfibrations?
| https://mathoverflow.net/users/166165 | Products in discrete fibrations | I don't know about 1., but this is certainly true for all limits. One can easily generalize it to opfibrations with groupoid fibers and get the same result.
For general opfibrations, one direction (the "easy one", namely "$F^\*$ preserves $I$-shaped limits implies $A$ has them and they are preserved by $F$") is still true (and relatively easy), but the converse is typically not. If $A$ has $I$-shaped limits and they are preserved by $F$, the canonical map $F^\*(\lim\_I X)\to \lim\_I F^\*(X)$ (where the second "$\lim\_I$" is to be understood as a pseudolimit) is always a left adjoint, but it is generally neither fully faithful nor essentially surjective.
This is so, even if one adds the natural requirement that cocartesian edges be preserved by products (/$I$-shaped limits) in $A$. I don't know a very reasonable condition. Something like "in limit diagrams in $A$, the projection maps are $F$-cocartesian" (this says something about the co-unit of the afore-mentioned adjunction) seems to be on the right track but not quite enough. If we try to add more conditions, we get closer and closer to a very tautological statement.
A proof could go as follows. Globally, suppose $F$ is an opfibration.
1- Suppose $A$ has $I$-shaped limits, that are preserved by $F$, and let $X: I\to B$ be a diagram. We want to prove that the canonical map $f:F^\*(\lim\_I X)\to \lim\_I F^\*(X)$ is an equivalence of categories. We start by proving that it is fully faithful. For simplicity of notation, let $b:= \lim\_I X$, and $p\_i: b\to X\_i$ the canonical projection maps.
For this, we note that if $a\in A\_b$, then we have a diagram $a\to (p\_i)\_!a, i\in I$ in $A$. I claim that this is a limit diagram in $A$. Let's assume this for a moment.
Then, given $a\_0,a\_1\in A\_b$, we find $$\hom\_{A\_b}(a\_0,a\_1)\cong \lim\_I \hom\_A(a\_0,(p\_i)\_!a\_1)\times\_{\hom\_B(b,X\_i)}\{p\_i\}\cong \lim\_I\hom\_A((p\_i)\_!a\_0,(p\_i)\_!a\_1)\cong \hom\_{\lim\_I F^\*(X)}(f(a\_0),f(a\_1))$$ and it is easy to convince oneself that the composite is the map induced by $f$ (note that $A\_b = F^\*(b)= F^\*(\lim\_IX)$).
This proves fully faithfulness, given the statement that I mentioned. Ok, now, to prove that $a\to (p\_i)\_!a, i\in I$ is a limit diagram.
In a discrete fibration, or more generally, in an opfibration with groupoid fibers, this is easy : we note that because $F$ preserves $I$-shaped limits, the canonical map $a\to \lim\_I (p\_i)\_!a$ is in $A\_b$, and so, because $A\_b$ is a groupoid, it must be an isomorphism.
In fact, this also proves essential surjectivity ! Let $(a\_i)\_{i\in I}$ be an object of $\lim\_I F^\*(X)$ (where I abuse notation and do not mention the isomorphisms $f\_{ij}(a\_j)\cong a\_i$), and take its limit as a diagram in $A$, $a = \lim\_I a\_i$, living over $b=\lim\_I X$, with canonical projection maps $a\to a\_i$ living over $p\_i: b\to X\_i$. But these are all cocartesian, as $F$ has discrete fibers, so that $((p\_i)\_!a)\_{i\in I}\cong (a\_i)\_{i\in I}$.
Now, if $F$ has non-groupoid fibers, both steps can fail (fully faithfulness and essential surjectivity). What you get instead is that the canonical map $F^\*(\lim\_I X)\to \lim\_I F^\*(X)$ is a left adjoint, with right adjoint given by "take limits". The above proof can easily be adapted to show this.
I believe the fibration $LFib\to Cat$ is a counterexample to both fully faithfulness and essential surjectivity: here, $LFib\subset Fun([1],Cat)$ is the full subcategory spanned by discrete opfibrations. It is clearly closed under products, and those are preserved by evaluation at $1$. Furthermore, this is clearly a fibration, and the pullback maps have left adjoints, so it's an opfibration too, with fiber $LFib\_C\simeq Fun(C,Set)$ over $C$. Then the adjunction in question is $Fun(C \times D, Set)\rightleftarrows Fun(C,Set)\times Fun(D,Set)$ which is rarely fully faithful, and is not essentially surjective if $D= \emptyset$ for instance.
2- Conversely, suppose $F^\*$ preserves $I$-shaped limits, and let $X: I\to A$ be a diagram. Let $b := \lim\_I FX$ and consider the canonical map $F^\*(b)\to \lim\_I F^\*(FX)$. In the right hand side, there is an object corresponding to $(FX\_i)\_i$, so if this canonical map is essentially surjective, we can lift it to some $Y\in F^\*(b)$, coming with (because of the nature of the canonical map above) cocartesian projection maps $Y\to X\_i$ lying over the projection maps $p\_i:b\to FX\_i$. The claim is that these $Y\to X\_i$ form a limit diagram, which will prove that $A$ admits limits, and they are preserved by $F$.
So let $Z\in A$ be arbitrary. To prove that $\hom\_A(Z,Y) \to \lim\_I\hom\_A(Z, X\_i)$ is an equivalence, one takes the fibers over some $f\in \hom\_B(FZ, b)\cong \lim\_I\hom\_B(FZ, FX\_i)$, to get $\hom\_{A\_b}(f\_!Z,Y)\to \lim\_I \hom\_{A\_{FX\_i}}((p\_i)\_!f\_!Z,X\_i)$. That this morphism is an isomorphism follows from the fact that the canonical map $F^\*(b)\to \lim\_I F^\*(FX)$ is fully faithful.
This argument is elementary enough, but it gets a bit more complicated for $\infty$-categories, even if it's still true there. Probably the best is to phrase things in terms of cocartesian sections (in the groupoid-fiber case, just sections) of the pullback $I\times\_B A\to I$, and its variant $I^\triangleleft \times\_B A\to I^\triangleleft$.
Given a limit diagram $f:I^\triangleleft\to B$, there is a restriction map $A\_{\lim\_I f}\simeq \Gamma\_{cocart}(I^\triangleleft,I^\triangleleft \times\_B A)\to \Gamma\_{cocart}(I,I \times\_B A)\simeq \lim\_I A\_{f(i)}$, and the claim is then that the map $\Gamma\_{cocart}(I,I\times\_B A)\to Fun(I, A)\times\_{Fun(I,B)}\{f\} \xrightarrow{\lim\_I} A\times\_B \{\lim\_I f\}= A\_{\lim\_I f}$ is a right adjoint to the restriction map.
Once we have that, in the case of $\infty$-groupoid fibers, we are done as both terms are just $\infty$-groupoids, and any adjunction between such is an equivalence, and in the more general case we cannot say much more.
| 1 | https://mathoverflow.net/users/102343 | 436497 | 176,399 |
https://mathoverflow.net/questions/436505 | 3 | I want to show if it's true that $60m^2+6m-1$ is a quadratic residue modulo $6gm+1$ for all $m \in \mathbb{N}$ and $6gm+1$ is prime, for infinitely many positive integers $g$. (I'm not 100% certain this is true, so a proof that it's wrong would be equally helpful).
I'm more looking for a solid method of attacking this sort of problem in general. How can this be shown?
For instance, $(m=1):$ is $65$ a qr mod a prime of the form $6g+1$ infinitely often?
$(m=2):$ is $251$ a qr mod a prime of the form $12g+1$ infinitely often?
$(m=3):$ is $557$ a rq mod a prime of the form $18g+1$ infinitely often?
Is it true for any positive integer $m$?
| https://mathoverflow.net/users/265714 | A polynomial as a quadratic residue mod a prime | Here is what I think you are asking: for each natural number $m$, are there infinitely many primes $p \equiv 1 \bmod 6m$ such that $60m^2 + 6m - 1 \bmod p$ is a quadratic residue?
To avoid being distracted by the algebraic expressions, set $a = 60m^2 + 6m-1$ and $b = 6m$. I think you are asking if there are infinitely many primes $p \equiv 1 \bmod b$ such that $a \bmod p$ is a quadratic residue. I'll show this can be done for arbitrary nonzero integers $a$ and $b$.
For each nonzero integer $n$, quadratic reciprocity implies $n \bmod p$ is a quadratic residue if (not only if) $p \equiv 1 \bmod 4|n|$. So it suffices to find infinitely many primes $p$ such that
$$
p \equiv 1 \bmod b, \ \ \ p \equiv 1 \bmod 4|a|.
$$
These congruences both hold for a prime $p$ such that
$$
p \equiv 1 \bmod 4|a|b.
$$
Dirichlet's theorem tells us there are infinitely many primes $p$ satisfying that last congruence condition, and for all them you'll have
$p \equiv 1 \bmod b$ and $a \bmod p$ is a quadratic residue.
Actually, it is overkill to appeal to Dirichlet's theorem here, because we are seeking primes satisfying a congruence condition of the form $p \equiv 1 \bmod N$, and in that case the infinitude of such primes follows purely algebraically using values of the $N$th cyclotomic polynomial by modifying Euclid's proof of the infinitude of the primes.
| 3 | https://mathoverflow.net/users/3272 | 436510 | 176,403 |
https://mathoverflow.net/questions/436294 | 1 | A lot of texts derive the variational form of a PDE as follows.
First, life begins with a conservation law for the field $q$:
$$\partial\_t \int\_\omega G(q)\;dx + \int\_{\partial\omega} F(q, \nabla q, \ldots)\cdot\nu\;ds = \int\_\omega f\;dx$$
for all control volumes $\omega$, where $\nu$ is the unit outward normal vector to $\partial\omega$, $F$ is the flux function, $G$ takes $q$ to the density of some extensive quantity, and $f$ is a source term.
We then apply the divergence theorem (assuming things are nice enough) to arrive at a strong form
$$\partial\_t G(q) + \nabla\cdot F(q, \nabla q, \ldots) = f.$$
Then we multiply everything by a test function $v$ and push the divergence over onto $v$ using the usual tricks.
So the argument goes conservation law $\rightarrow$ strong form $\rightarrow$ weak or variational form.
I find the indirectness kind of annoying.
**Can we go directly from a conservation law to a variational form instead?**
You could imagine that, for any $u$,
$$\int\_\omega u\;dx = \int\_\Omega u\cdot \mathbf 1\_\omega\;dx$$
where $\mathbf 1\_\omega$ is the indicator function of the set $\omega$.
Likewise, assuming that the boundary of $\omega$ is nice enough, for some reasonable class of vector fields $F$, we can also say that
$$\int\_{\partial\omega} F\cdot\nu\;ds = -\int\_\Omega F\cdot\nabla\mathbf 1\_\omega\;dx$$
in a distributional sense.
Substituting these two relations into our original conservation law, we can then take arbitrary scalar multiples and sums to arrive at a variational form
$$\int\_\Omega\left\{\partial\_tG(q)\cdot v - F(q, \nabla q, \ldots)\cdot\nabla v - f\cdot v\right\}\,dx = \ldots$$
in a distributional sense for any *simple* function $v$, i.e. $v = \sum\_i\alpha\_i\mathbf{1}\_{\omega\_i}$ for some finite collection of control volumes $\{\omega\_i\}$.
(I've left off some boundary terms for brevity.)
We can then take a limit as $v$ approaches a function in, say, $H^1(\Omega)$ or whatever space of test functions is most appropriate.
It's been a few years since I had measure theory or distribution theory.
**Can this argument be made rigorous, without an intermediate assumption that the PDE has a strong solution?**
I'd be perfectly happy if it worked for some common problem or class of problems; for example, the generalized Poisson equation where the conductivity coefficient has a jump discontinuity across some smooth hypersurface.
I don't know that doing so has much practical use, I just find it more satisfying.
Virtually every book I have on Galerkin methods or PDE theory either goes the conservation / strong / variational route or just starts with variational forms and ignores their derivation entirely.
| https://mathoverflow.net/users/49417 | derivation of variational forms of PDE directly from conservation form | Alright guess I'll have to try and do it myself then.
Assumptions: $\Omega$ is a nice enough compact domain that $C^\infty(\Omega)$ is dense in $H^1(\Omega)$, and $u$ is a solution of the conservation form of the diffusion equation -- for all smooth control volumes $\omega$,
$$-\int\_{\partial\omega} k\nabla u\cdot\nu\;ds = \int\_\omega f\; dx$$
where $\nu$ is the unit outward normal vector, $k$ is the (strictly positive) diffusion coefficient, and $f$ is in $L^2(\Omega)$.
We'll take the boundary conditions to be $u|\_{\partial\Omega} = 0$ for simplicity; adding an extra term takes care of the Neumann or Robin cases.
Assume additionally that $u$ is in $H\_0^1(\Omega)$.
Let $v$ be an arbitrary smooth function that vanishes on $\partial\Omega$.
Using the smooth coarea formula, we can write
$$\int\_\Omega k\nabla u\cdot\nabla v\;dx = \int\_\Omega k\nabla u\cdot \frac{\nabla v}{|\nabla v|}|\nabla v|\; dx = \int\_{-\infty}^\infty\int\_{\partial\{x: v(x) \ge t\}}k\nabla u\cdot\frac{\nabla v}{|\nabla v|}\;ds\;dt\ldots $$
but we also know that, if $t$ is a regular value of $v$, the unit outward normal to the hypersurface $v^{-t}(t)$ is equal to $-\nabla v/|\nabla v|$, hence
$$\ldots = -\int\_{-\infty}^\infty\int\_{\partial\{x: v(x) \ge t\}}k\nabla u\cdot\nu\;ds\;dt = \int\_{-\infty}^\infty\int\_{\{x: v(x) \ge t\}}f\;dx\;dt\ldots $$
The latter equality is because of our assumption that $u$ is a solution of the conservation form of the equations and that $v$ is smooth + Sard's theorem.
Finally the latter integral is just a rearrangement of the integral
$$\ldots = \int\_\Omega f\,v\;dx$$
We can then pass to the limit for $v$ an arbitrary element of $H\_0^1(\Omega)$ to say that $u$ is a solution of the variational problem
$$\int\_\Omega k\nabla u\cdot\nabla v\;dx = \int\_\Omega fv\; dx$$
for all $v$, QED.
---
I thought this would directly involve some distribution theory in order to be able to make statements like the gradient of an indicator function being a surface measure.
But the coarea formula lets you shove all that under the hood.
The same idea (+ some extra terms for initial conditions) seems like it should work for more problems as long as they can be expressed as a divergence in spacetime.
Please let me know if there are any mistakes or holes in the argument.
| 0 | https://mathoverflow.net/users/49417 | 436525 | 176,410 |
https://mathoverflow.net/questions/436524 | 7 | There is an old result due to Mycielski and Sierpiński, and popularized in a Monthly article by Taylor and Wagon ([A Paradox Arising from the Elimination of a Paradox](https://doi.org/10.1080/00029890.2019.1559416); see also [this MO answer](https://mathoverflow.net/a/22935)), that can be stated as follows: in [Solovay's model](https://en.wikipedia.org/wiki/Solovay_model), one can partition $2^\omega$ into more than $2^\omega$ nonempty disjoint sets. Or if we let LM denote the axiom that “all subsets of $\mathbb{R}$ are Lebesgue-measurable,” then we can state this: ZF + LM proves $|\mathbb{R}| < |\mathbb{R}/\mathbb{Q}|$. Taylor and Wagon refer to this phenomenon as the **division paradox**.
This result is sometimes used as an argument in favor of the axiom of choice and against LM, but something bothers me about drawing this conclusion. If $A$ and $B$ are sets, then standardly we have the following definition.
>
> We say that $|A|<|B|$ if there is an injection from $A$ to $B$ but no bijection from $A$ to $B$.
>
>
>
But suppose we make the following definition.
>
> W say that $B$ **outnumbers** $A$ if there is an injection from $A$ to $B$ but no surjection from $A$ to $B$.
>
>
>
In the presence of the axiom of choice, $B$ outnumbers $A$ if and only if $|A|<|B|$, but without the axiom of choice they are not equivalent. As a sanity check on whether outnumbering is a reasonable concept, note that [it can be shown](https://cs.nyu.edu/pipermail/fom/2019-June/021567.html) that if $C$ outnumbers $B$ and $B$ outnumbers $A$, then $C$ outnumbers $A$. The question I have is this:
>
> In Solovay's model, let $B$ be a partition of $A$ into nonempty disjoint sets. Can $B$ outnumber $A$?
>
>
>
If the answer is no, then I would be inclined to interpret the division paradox as telling us that in the absence of the axiom of choice, our intuitions about cardinalities are unreliable, and we should be using the concept of outnumbering to compare sizes of sets. In particular, the division paradox gives us no compelling reason to reject Solovay's model.
| https://mathoverflow.net/users/3106 | Partitioning a set of cardinality $\kappa$ into more than $\kappa$ disjoint subsets | Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.
The Division Paradox tells us that our intuition, which usually tells us that injections and surjections tell "the same story" with regards to cardinality, is really reliant on the axiom of choice, in a very significant way.
Note that you don't need to go as far as Mycielski and Sierpinski and the Solovay model, by the way, to get "big partitions" of the real numbers. Given any model of $\sf ZFC$ there is a symmetric extension given by adding Cohen real in which there is a "paradoxical partition" of the reals. (See <http://karagila.org/2020/countable-sets-of-reals/> for details.)
| 11 | https://mathoverflow.net/users/7206 | 436528 | 176,412 |
https://mathoverflow.net/questions/436516 | 5 | Given topological spaces $X$ and $Y$, we define an **open map from $X$ to $Y$** to be a map of sets $f\colon X\to Y$ satisfying the following condition:
* For each $U\in\mathcal{P}(X)$, if $U$ is open in $X$, then $f\_\*(U)$ is open in $Y$.
Here $f\_\*(U):=\{f(x)\in Y\ |\ x\in X\}$ is the direct image of $U$ by $f$, which sits in a triple adjunction
$$f\_\*\dashv f^{-1}\dashv f\_!\colon\mathcal{P}(X)\underset{\leftrightarrows}{\rightarrow}\mathcal{P}(Y),$$
where $f^{-1}$ is the inverse image, and where $f\_!$, the “**direct image with compact support**”, is given by
$$f\_!(U):=\{y\in Y\ |\ f^{-1}(y)\subset U\}.$$
It can be sometimes useful to break down $f\_!(U)$ into two sets:
$$f\_!(U):=f\_{!,\mathrm{im}}(U)\cup f\_{!,\mathrm{cp}}(U),$$
where
\begin{align\*}
f\_{!,\mathrm{im}}(U) &= f\_!(U)\cap\mathrm{Im}(f),\\
f\_{!,\mathrm{cp}}(U) &= f\_!(U)\cap(Y\setminus\mathrm{Im}(f))\\
&= Y\setminus\mathrm{Im}(f).
\end{align\*}
For example, if $f\colon\mathbb{N}\to\mathbb{N}$ is given by $f(n)=2n$, then $f\_{!,\mathrm{im}}(U)=f\_\*(U)$ and $f\_{!,\mathrm{cp}}(U)=\{\text{odd natural numbers}\}$ for $U\subset\mathbb{N}$.
Now, define a “**weird-open map from $X$ to $Y$**” as a map of sets $f\colon X\to Y$ satisfying the following condition:
* For each $U\in\mathcal{P}(X)$, if $U$ is open in $X$, then $f\_!(U)$ is open in $Y$.
Since $f\_!(U)$ contains $Y\setminus\mathrm{Im}(f)$, it's useful to also make the following relative definition: a “**relatively weird-open map from $X$ to $Y$**” as a map of sets $f\colon X\to Y$ such that:
* For each $U\in\mathcal{P}(X)$, if $U$ is open in $X$, then $f\_!(U)\cap\mathrm{Im}(f)$ is open in $\mathrm{Im}(f)$.
**Questions.**
1. Is this an already studied notion (with a proper name)?
2. Are there any useful applications of it? (Be them in topology, analysis, algebraic geometry, etc.)
3. Regarding 2), an immediate property of $f\_!$ that comes to mind is that it induces a further adjoint when passing to presheaves: a continuous map $f\colon X\to Y$ induces adjoint functors
$$f^\*\dashv f\_\*\dashv f\_!\colon\mathsf{PSh}(Y)\underset{\leftrightarrows}{\rightarrow}\mathsf{PSh}(X),$$
and if $f$ is weird-open, there's an extra right adjoint $f\_\dagger$ of $f\_!$. (Here the relative notion is useful, for which we have a "locally defined adjoint" $f\_\dagger\colon\mathsf{PSh}(X)\to\mathsf{PSh}(\mathrm{Im}(f))$.)
Is $f\_\dagger$ useful in practice?
| https://mathoverflow.net/users/130058 | "Weird-open" maps in topology | As suggested in comments, I turn my comment into an answer here.
First of all let me note that in the overwhelming majority of texts I've seen notation is the opposite: $f\_\*$ from the OP is denoted by $f\_!$ and $f\_!$ by $f\_\*$. Still, to avoid further confusion I will stick to the notation of the question.
It is easy to check that for any subset $U$ of $X$ one has $f\_!(U)=Y-f\_\*(X-U)$. It follows that $f$ is weird-open iff it is closed.
Accordingly, $f$ is relatively weird-open iff $f:X\twoheadrightarrow\operatorname{Im}(f)$ is closed.
As for presheaves, if $f$ is open but not necessarily continuous, the map $f\_\*:\operatorname{Opens}(X)\to\operatorname{Opens}(Y)$ has a right adjoint $f^\*$ given by $f^\*(V)=\operatorname{Interior}(f^{-1}(V))$ but $f^\*$ might fail to have further right adjoint. Whereas if $f$ is continuous so that $f^\*(V)=f^{-1}(V)$, but not necessarily open, then $f^{-1}:\operatorname{Opens}(Y)\to\operatorname{Opens}(X)$ may fail to have a left adjoint but it does have a right adjoint called direct image, sending $U$ to $\operatorname{Interior}(f\_!(U))$. And if $f$ is closed then this is the same as $f\_!(U)$.
Note also that for sheaf toposes (more generally, for geometric morphisms between Grothendieck toposes), preservation of some portion of colimits by the direct image is related to propriety [(see the nLab entry for proper geometric morphisms)](https://ncatlab.org/nlab/show/proper+geometric+morphism). In particular, preservation of filtered colimits is called tidiness and preservation of joins of subterminals flatness. Existence of the right adjoint to the direct image is still stronger, such geometric morphisms are called [local](https://ncatlab.org/nlab/show/local+geometric+morphism).
| 6 | https://mathoverflow.net/users/41291 | 436539 | 176,417 |
https://mathoverflow.net/questions/436537 | 2 | Is there a way of endowing the unit ball $B\_X$ of a Banach space $X$ (we may assume that $X$ is an AL space, if that helps) with a topology $\tau$, so that $\tau=\sigma(Y^\*,Y)$ (the weak\* topology) if $X=Y^\*$, for some Banach space $Y$? In other words, is it possible to equip the unit ball of a Banach space $X$ with a topology that corresponds with the weak\* topology if $X$ is a dual space, but is well-defined if $X$ is not a dual space?
As Nik Weaver observes in [this post](https://mathoverflow.net/questions/427447/compactness-of-the-unit-ball-of-a-banach-space-for-topologies-finer-than-the-wea), "... on any dual Banach space there is no locally convex vector space topology strictly stronger than the weak\* topology that makes the unit ball compact." So, given an arbitrary Banach space (or AL space) $X$, could one endow it with something like the *"strongest locally convex vector space topology making the unit ball compact"*?
| https://mathoverflow.net/users/42887 | An abstract characterisation of weak* topologies | This is not possible in general. The obstruction does not come from spaces that are not dual spaces, but from the spaces that appear in several different ways as dual spaces. Indeed, the restriction of the $\sigma(Y^\*,Y)$-topology to the unit ball of $Y^\*$ determines $Y$ uniquely : by the [Krein-Smulian theorem](https://en.wikipedia.org/wiki/Krein%E2%80%93Smulian_theorem), $Y$ coincides with the subspace of elements of $(Y^\*)^\*$ whose restriction to the unit ball is $\sigma(Y^\*,Y)$-continuous. In particular, if $X$ admits several preduals that are not isometrically isomorphic, then there are several non-comparable *maximal locally convex vector space topologies making the unit ball compact*, and there is no strongest such topology.
Having a unique predual is a somewhat exceptional situation (this is the case for von Neumann algebras). The standard example of Banach space with many preduals is $\ell\_1$. It has lots of very wild preduals, including the not-so-wild spaces $C(K)$ for $K$ countable and compact. Very concretely, two non-(isometrically isomorphic) preduals of $\ell\_1$ are given by the space $c$ of converging sequences of complex numbers, and its subspace $c\_0$ of sequences converging to $0$. See [this question](https://math.stackexchange.com/questions/80727/are-these-two-banach-spaces-isometrically-isomorphic/80811#80811). See also the introduction to [this article](https://arxiv.org/abs/1101.5696) for more examples and references.
| 8 | https://mathoverflow.net/users/10265 | 436541 | 176,418 |
https://mathoverflow.net/questions/436534 | 0 | For GOE matrix $A$, we have the following limiting distribution for eigenvalues of $A$ by $\lambda\_N\ge \lambda\_{N-1}\ge \dots \ge \lambda\_1$:
>
> In this [paper][1], if we denote the $k$ largest eigenvalues by $\lambda\_N,\lambda\_{n-1},··· ,\lambda\_{N-k+1}, $ then for Gaussian ensembles the joint distribution function of rescaled eigenvalues has the limit:
> $$
> \lim\_{N\to\infty}P(N^{2/3}(\lambda\_N-2)\le s\_1,\dots, N^{2/3}(\lambda\_{N-k+1}-2)\le s\_k)=F\_{\beta, k}(s\_1,\dots, s\_k)
> $$
>
>
>
Now, if we can order the eigenvalues of $A$ by $|\sigma\_N|\ge |\sigma\_{N-1}|\ge \dots \ge|\sigma\_{1}|$, can we still have the similar results as in [The ratio of spectral edge of the GOE matrix](https://mathoverflow.net/questions/434446/the-ratio-of-spectral-edge-of-the-goe-matrix)?
>
> *[For my notation $\sigma$: For example, the eigenvalues could be the case $1.98>1.1>...>-1.99>-2$ (where I take $\lambda\_N=1.98$, $\lambda\_{N-1}=1.1$ and $\lambda\_2=-1.99$, $\lambda\_1=-2$). Also, we order the eigenvalues by the absolute value: it becomes $|\lambda\_1|>|\lambda\_2|>|\lambda\_N|>....$. So the order changed. I denoted the new notation for $\sigma\_N=\lambda\_1, \sigma\_{N-1}=\lambda\_2, \sigma\_{N-3}=\lambda\_N$.]*
>
>
>
That is
$$
\frac{|\sigma\_{2}|}{|\sigma\_{1}|}=1+O(N^{-2/3})
$$
---
To prove the result in the above question, we need the joint limiting distribution also hold for $|\sigma\_{1}|$ and $|\sigma\_{2}|$. Do we still have $$|\sigma\_{2}|-|\sigma\_{1}|=O(N^{-2/3})?$$
| https://mathoverflow.net/users/168083 | Can we still have the order of ratio result of the two smallest eigenvalues? | The $\sigma\_i$'s are a permutation of the $\lambda\_i\in(-2,2)$, ordered by absolute value in the order $|\sigma\_N|\geq|\sigma\_{N-1}|\geq \cdots\geq|\sigma\_1|$. So $\sigma\_1$ and $\sigma\_2$ are the eigenvalues closest to zero (in the bulk of the spectrum), while $\sigma\_N$ and $\sigma\_{N-1}$ are the eigenvalues closest to $\pm 2$ (near the spectral edge). One has $|\sigma\_1|,|\sigma\_2|={\cal O}(N^{-1})$ in view of the Wigner semicircle, while $|\sigma\_N|,|\sigma\_{N-1}|=2+{\cal O}(N^{-2/3})$, in view of the Tracy-Widom law, hence
$$|\sigma\_2|-|\sigma\_1|={\cal O}(N^{-1}),$$
$$|\sigma\_N|-|\sigma\_{N-1}|={\cal O}(N^{-2/3}).$$
(The OP has this the other way around, perhaps a typo?)
| 1 | https://mathoverflow.net/users/11260 | 436544 | 176,419 |
https://mathoverflow.net/questions/436543 | 4 | We say that a set of natural numbers $A\subseteq \omega$ has *positive upper density* if $$\lim\sup\_{n\to\infty}\frac{|A\cap n|}{n+1} > 0.$$
[Szeméredi's theorem](https://en.wikipedia.org/wiki/Szemer%C3%A9di%27s_theorem) states that every $A\subseteq \omega$ having positive upper density contains arithmetic sequences of arbitrary (finite) length.
For $y\in \omega$ we set $A - y:= \{|a\setminus y|:a\in A\}.$ Note that $|a\setminus y|$ equals the difference of $a$ and $y$ if $a\geq y$, and $0$ otherwise. The *upper Banach density* is defined by $$d\_{uB}(A) = \lim\_{n\to\infty}\big(\sup
\{\frac{|(A-y)\cap n|}{n+1}: y\in\omega\}\big).$$
It is easy to see that $d\_{uB}(A) \geq d\_{u}(A)$ for all $A\subseteq \omega$.
**Question.** If $A\subseteq \omega$ has the property that $d\_{uB}(A)>0$, does $A$ contain arithmetic sequences of arbitrary finite length?
| https://mathoverflow.net/users/8628 | Does Szemerédi's theorem hold for sets with positive upper Banach density? | Yes. As Martin says, this is often how the theorem is stated. It also follows immediately from the also common finitary form:
For all $\delta>0$ and $k\geq 1$, if $N$ is large enough depending on $\delta$ and $k$, $P$ is an arithmetic progression of length $N$, and $A\subseteq P$ has size $\lvert A\rvert\geq \delta N$, then $A$ contains a non-trivial arithmetic progression of length $k$.
(This is e.g. equivalent to the finitary form stated in the Wikipedia article where $P=\{1,\ldots,N\}$, since the property of containing an arithmetic progression is invariant under dilations and translations.)
| 8 | https://mathoverflow.net/users/385 | 436547 | 176,420 |
https://mathoverflow.net/questions/436549 | 4 | Let $a(n)$ be [A227559](http://oeis.org/A227559), i.e., number of partitions of $n$ into distinct parts with boundary size $2$. Be careful here: offset is $3$.
I conjecture that $a(4n+2)=2n+1$ for $n>0$ if and only if $2n+1$ is a prime number.
I guess that my conjecture has no interest, in the event that the generation of the sequence is associated with prime numbers. I visited [A227345](http://oeis.org/A227345), but I never figured out exactly how the sequence is generated.
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Prime numbers and number of partitions of $n$ into distinct parts with boundary size $2$ | There always exist exactly $2n$ partitions of $4n+2$ onto 2 distinct parts. Also, there exists a partition with parts $n-1,n,n+1,n+2$.
Any other partition of $4n+2$ onto distinct parts with boundary size 2, say, with $k>2$ parts, corresponds to a representations $4n+2=x+(x+1)+\ldots+(x+k-1)=k(2x+k-1)/2$, so $k(2x+k-1)=4(2n+1)$.
If $2n+1$ is prime, the number
$4(2n+1)$ has only two factorizations $4(2n+1)=ab$ with factors $a, b$ of distinct parity: $1 \times (8n+4)$ and $4\times (2n+1)$. The first case would mean $k=1$ which is absurd, in the second case $k=4$, which is already counted.
On the other hand, if $2n+1=pq$ is a factorization with $1<p\leqslant q$, we may take $k=p$, $2 x+k-1=4q$ and get an extra partition.
| 6 | https://mathoverflow.net/users/4312 | 436550 | 176,421 |
https://mathoverflow.net/questions/118833 | 12 | $\newcommand\met{\mathrm{met}}$It is a basic topological fact that CW-complexes aren't typically metrizable (they must satisfy a certain local finiteness condition) and the quotient topology is to blame.
**Question:** Suppose $X$ is a CW-complex (possibly with countably many cells and maybe even of finite dimension). Is it possible weaken the topology of $X$ to construct another space $X\_{\met}$ (with the same underlying set), so that the continuous identity function $X\to X\_{\met}$ is a homotopy equivalence?
Update: I will clarify (now much later) that this question has an affirmative answer for simplicial complexes. Given an arbitrary simplicial complex $K$, we have $|K|$, which has the weak topology and is not always metrizable. However, you can give the underlying set of $|K|$ a metrizable topology to form the "metric simplicial complex" $|K|\_m$. The identity $|K|\to |K|\_m$ is continuous and is a homotopy equivalence. A nice proof can be found in Segal and Mardesic's book Shape Theory in Appendix, $\S 1.3$, Theorem 10. As Sergey Melikhov nicely points out in his answer, the same is true for regular CW-complexes, which include simplicial complexes. Using this result, it follows that every CW-complex is homotopy equivalent to *some* metric space. However, my question is a bit more specific.
| https://mathoverflow.net/users/5801 | Making CW-complexes metrizable | The required metric topology on $X$ does exist. This is a consequence of Theorem 2.1 in the paper
>
> Robert Cauty, *Rétractions dans les espaces stratifiables*, Bulletin de la Société Mathématique de France, **102**, (1974), 129-149.
>
>
>
Actually, Cauty's statement is far more general than what is being required. I'll record it in the present context and postpone discussion for afterwards.
>
> **Theorem (Cauty)**: Let $X$ be a CW complex. Then there exists a continuous metric $d$ on $X$ such that;
>
>
> * $X\_{Met}=(X,d)$ is an ANR.
> * The identity $i:X\rightarrow X\_{Met}$ is a homotopy equivalence.
> * For any given compact $K\subseteq X$, a homotopy inverse $j:X\_{Met}\rightarrow X$ to $i$ can be found so that $i\circ j\simeq id\_{X\_{met}}$ and $j\circ i\simeq id\_X$ by homotopies fixing $K$ pointwise at all times. $\quad\blacksquare$
>
>
>
There are no restrictions on the dimension of $X$ or its number of cells. A metric is continuous if and only if it comes from a weaker metric topology.
Now, as mentioned above, Cauty's actual statement is far more general.
>
> *The theorem holds verbatim when $X$ is any ANR for stratifiable spaces*.
>
>
>
The term *stratifiable* is meant in the sense of [Borges](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-17/issue-1/On-stratifiable-spaces/pjm/1102994724.full), who established already that a stratifiable space belongs to ANR(stratifiable) if and only if it belongs to ANE(stratifiable). Ceder had previously established the result that all CW complexes are stratifiable, and Cauty subsequently showed that all CW complexes belong to ANR(stratifiable).
Ignoring all references to stratifiability, we can at the very least use the extra generality to extend the theorem stated above to include the cases $(i)$ $X$ is the product of finitely many CW complexes, $(ii)$ $X$ is an open subset of a CW complex, $(iii)$ $X$ is a closed subset of a CW complex $Y$ whose inclusion $X\subseteq Y$ is a cofibration, $(iv)$ $X$ is the loop space of a CW complex.
Finally, let us comment on the last listed property regarding the compact $K$. Any compact subset $K\subseteq X$ has the same topology as its image in $X\_{Met}$. The statement says that $i:X\rightarrow X\_{Met}$ is a homotopy equivalence under $K$. This is clearly true when $K$ is a finite subcomplex of $X$, since then its inclusions $K\subseteq X$ and $K\subseteq X\_{Met}$ are cofibrations (the latter following since $K$ is an ANR for metric spaces). That it should remain true for any $K$ (say a Cantor set or compact fractal) is quite surprising!
| 4 | https://mathoverflow.net/users/54788 | 436560 | 176,424 |
https://mathoverflow.net/questions/436551 | 4 | Given an integer $a$, I would like to build a table of entries $(p, \text{ord}\_p(a))$, where $p$ runs over the prime numbers not dividing $a$ and not exceeding a fixed parameter $P$, and $\text{ord}\_p(a)$ is the multiplicative order of $a$ modulo $p$.
I known that computing the multiplicative order is a difficult problem, since it is related to the discrete logarithm problem, so I'm not expecting a very efficient algorithm to build such a table. However, I would like to know if there is some trick to build the table more efficiently than just computing $\text{ord}\_p(a)$ for each $p$. For instance, can some of the computation done to compute $\text{ord}\_p(a)$ be used to speed up the computation of $\text{ord}\_q(a)$ for $q \neq p$? Can $\text{ord}\_p(a)$ be computed in parallel for more primes $p$? The only thing I could come up is that working modulo $p\_1 \cdots p\_k$, one can in fact compute the multiplicative order of $a$ modulo $p\_i$ in parallel, but $p\_1 \cdots p\_k$ gets very large so it does not seem a great advantage.
**Notes:**
* I'm interested in $a$ and $p$ in the range of a 32-bits / 64-bits integers.
* I would like to use only operations for 32-bits / 64-bits integers, no fancy arbitrary-size-integer arithmetic.
* I'm assuming that the list of primes $p$ is precomputed.
| https://mathoverflow.net/users/488969 | "Efficient" way to build a table of multiplicative orders modulo $p$ of a fixed integer $a$ | Order computations are generally easier than discrete logarithms, and they are *much* easier if you know the factorization of the group order.
If you're dealing with a precomputed list of 32- or 64-bit primes, then you can precompute the factorization of $p-1$ for each $p$. Given this factorization, order computations mod $p$ can be very efficient: Chapter 7 ("Fast order algorithms") of [Sutherland's thesis](https://math.mit.edu/%7Edrew/sutherland-phd.pdf) is a good reference for this.
| 11 | https://mathoverflow.net/users/156215 | 436561 | 176,425 |
https://mathoverflow.net/questions/436529 | 5 | An entire function $F: \mathbb C \to \mathbb C$ belongs to the Fock space $\mathcal F^2$ if
$$
\int\_{\mathbb C} |F(z)|^2e^{-|z|^2} \, dA(z) < \infty.
$$
It is well-known that every $F \in \mathcal F^2$ has order $\rho$ which satisfies $\rho \leq 2$, that is,
$$
\rho = \limsup\_{r \to \infty} \frac{\log \log M(r)}{\log r} \leq 2, \quad M(r) = \sup\_{\theta \in \mathbb R} |F(re^{i\theta})|.
$$
Moreover, if $\rho = 2$ then its type $\lambda$ satisfies $\lambda \leq 1/2$ (see e.g. [here](https://link.springer.com/content/pdf/10.1007/BF03321762.pdf?pdf=button)) where
$$
\lambda = \limsup\_{r \to \infty} \frac{\log M(r)}{r^\rho}.
$$
Since $F$ is of finite order $\rho \leq 2$, we can look at its Hadamard factorization
$$
F(z)=z^m e^{az^2+bz+c} P(z), \quad a,b,c \in \mathbb C,
$$
with $P$ the canonical product corresponding of the zeros of $F$.
Questions: Is it true that the constant $a$ must satisfy $|a| < 1/2$? Or can $a$ be equal to $1/2$ and $F$ is still in the Fock space, i.e. it satisfies the integrability condition above?
| https://mathoverflow.net/users/223636 | Hadamard factorization of a function in the Fock space | Yes, one can have $a=1/2$.
Let $P$ be an entire function of order $3/2$, normal type, whose indicator
$h$ has the properties $h(0)<0$, $h(\pi)<0$. Such a function exists, see for example
B. Ya. Levin, Distribution of zeros of entire functions, AMS, 1980, Chap II, section 4.
Then by continuity, the indicator is negative for $|\theta|<\delta$ and for $|\theta-\pi|<\delta$, with some positive $\delta$, which implies exponential decrease in these sectors. Since the exponential growth on any ray is at most $\exp|z|^{3/2}$,
we conclude that $\exp(z^2/2)P(z)$ is in the Fock class.
Remark. Actually $a$ can be any complex number.
I will assume wlog that $a$ is positive.
Then, $f(z)=e^{az^2}P(z)$ will be in the Fock space if the indicator $h$ of $P$
satisfies
$$h(\theta)<1-a\cos2\theta.$$
For example, one can take $h(\theta)=-a\cos2\theta$.
To construct a canonical product $P$ with such indicator, one chooses zeros $a\_n$ satisfying
the following conditions:
$$\lim\_{r\to\infty}\sum\_{n: |a\_n|\leq r}a\_n^{-2}=-2a,$$
$$\sum |a\_n|^{-2}=\infty,$$
and $\sqrt{n}=o(|a\_n|)$. Such a sequence is easy to construct. In view of the second condition, the canonical product with zeros at $a\_n$ will be of genus $2$, and the third condition assures that zeros have zero density. Now the first condition
implies that $P$ has the required indicator
by Theorem 2 in section 1, Chap. II of Levin's book.
| 3 | https://mathoverflow.net/users/25510 | 436566 | 176,428 |
https://mathoverflow.net/questions/436577 | 5 | Suppose $f: \mathbb C \times (-1,1) \to \mathbb C$ is a smooth function that satisfies $f(0,t)=1$ for all $t\in (-1,1)$. Assume that for any $k\in \mathbb N$, any $z \in \mathbb C$ and any $t \in (-1,1)$ there holds
$$ \frac{\partial^k f}{\partial t^k}(z,0)=z^k.$$
Does there exist constants $c,\delta>0$ such that
$$ e^{-c|z|} \leq |f(z,t)| \leq e^{c|z|},$$
for all $|t|\leq \delta$ and all $z\in \mathbb C$?
| https://mathoverflow.net/users/50438 | Family of functions with prescribed derivatives | A counterexample:
$$f(z,t):=e^{tz}[1+(e^{|z|^2}-1)h(t)],$$
where $h(t):=e^{-1/|t|}$ for $t\ne0$, with $h(0):=0$.
Then all the assumptions on $f$ hold, but the conclusion
$$|f(z,t)|\le e^{c|z|}\ \;\forall z\in\mathbb C \tag{1}\label{1} $$
fails to hold for any real $t\ne0$ and any real $c$.
---
If now
$$f(z,t):=e^{tz}[1+e^{|z|^2}\sin(|z|^2)\,h(t)],$$
with the same $h$ as before, then all the assumptions on $f$ hold, but both the conclusion \eqref{1} and the conclusion
$$e^{-c|z|}\le|f(z,t)|\ \;\forall z\in\mathbb C \tag{2}\label{2} $$
will fail to hold for any real $t\ne0$ and any real $c$.
---
In the counterexamples above, we can replace $e^{|z|^2}$ and $\sin(|z|^2)$ by $e^{z^2}$ and $\sin z$, respectively (and then consider only the real values of $z$ to see that the conclusions \eqref{1} and \eqref{2} will still fail to hold for any real $t\ne0$ and any real $c$). The bonus provided by this replacement is that then $f(z,t)$ will be analytic in $z$, which will address the [comment by G. Fougeron](https://mathoverflow.net/questions/436577/family-of-functions-with-prescribed-derivatives/436578#comment1125207_436577).
| 9 | https://mathoverflow.net/users/36721 | 436578 | 176,434 |
https://mathoverflow.net/questions/90046 | 15 | Is it true that every finitely generated (topologically) torsion-free nilpotent pro-$p$ group is isomorphic to a subgroup of $U\_d(\mathbb{Z}\_p)$, the group of $d\times d$-upper triangular matrices with 1's in the diagonal, for some $d$?.
This question is the analogous of this well known result: every finitely generated torsion-free nilpotent group is isomorphic to a subgroup of $U\_d(\mathbb{Z})$ for some $d$.
| https://mathoverflow.net/users/15235 | Linear embeddings of nilpotent pro-$p$ groups | Yes, this is true. In the argument below, all references are to the book [*Analytic Pro-$p$ Groups*](https://doi.org/10.1017/CBO9780511470882) by Dixon–Du Sautoy–Mann–Segal.
Let $G$ be a finitely generated torsion-free nilpotent pro-$p$ group. We'll deal first with the special case that $G$ is *uniform*, meaning here that the quotient of $G$ by the closed subgroup $G^p$ generated by $p$th powers is abelian [Theorem 4.5]. In this case, one can endow $G$ with the structure of a Lie algebra over $\mathbb Z\_p$, as explained in [Sections 4.3 & 4.5]. That is, for any elements $g,h\in G$, the element $g^{p^n}h^{p^n}$ is the $p^n$th power of a unique element of $G$, and the element $g^{p^n}h^{p^n}g^{-p^n}h^{-p^n}$ is the $p^{2n}$th power of a unique element of $G$. So we can define an addition and Lie bracket on $G$ by
$$
g + h = \lim\_{n\to\infty} (g^{p^n}h^{p^n})^{1/p^n} \quad\text{and}\quad [g,h] = \lim\_{n\to\infty} (g^{p^n}h^{p^n}g^{-p^n}h^{-p^n})^{1/p^{2n}} \,.
$$
These make $G$ into a Lie algebra $L(G)$ over $\mathbb Z\_p$, which is finitely generated free as a $\mathbb Z\_p$-module and for which the pro$-$p topology on $G$ coincides with the natural $\mathbb Z\_p$-module topology [Proposition 4.16, Theorems 4.17 & 4.30]. Moreover, the Baker–Campbell–Hausdorff power series converges on $L(G)$ and recovers the group law on $G$ (follows from [Lemma 7.12]).
Now the fact that $G$ is nilpotent implies that $L(G)$ is a nilpotent Lie algebra. (This can be proved inductively by writing $G$ as a central extension of a finitely generated torsion-free nilpotent pro-$p$ group $G'$ — which is also uniform — by $\mathbb Z\_p$ and noting that the induced sequence on Lie algebras is also a central extension.) So $\mathbb Q\_p\otimes\_{\mathbb Z\_p}L(G)$ is a finite-dimensional nilpotent Lie algebra over $\mathbb Q\_p$, and the Baker–Campbell–Hausdorff power series makes it into the $\mathbb Q\_p$-points of a unipotent group $U(G)$ over $\mathbb Q\_p$.
The upshot of all this is that $G$ admits a continuous embedding in the $\mathbb Q\_p$-points of a unipotent group $U(G)$, namely via the composite
$$
G \cong L(G) \subset \mathbb Q\_p\otimes\_{\mathbb Z\_p}L(G) = U(G)(\mathbb Q\_p) \,.
$$
But every unipotent group embeds as a closed algebraic subgroup of the group $U\_n$ of upper triangular matrices, and hence we get a continuous embedding of $G$ as a subgroup of $U\_n(\mathbb Q\_p)$, i.e. $G$ has a faithful unipotent continuous representation defined over $\mathbb Q\_p$. It is now not hard to show that we can rescale coordinates on this representation so as to make the representation take values in $U\_n(\mathbb Z\_p)$, and we are done.
---
We've now dealt with the case that $G$ is uniform. In the general case, we know that $G$ has a uniform open normal subgroup $H$ [Corollary 4.3], and this subgroup will again be finitely generated, torsion-free and nilpotent. So we know from the above that $H$ admits a continuous embedding $f\colon H\hookrightarrow U(H)(\mathbb Q\_p)$ into the $\mathbb Q\_p$-points of a unipotent group. We want to show that $f$ extends uniquely to a continuous group homomorphism $f'\colon G\to U(H)(\mathbb Q\_p)$ — such an $f'$ would automatically be an embedding since $G$ is torsion-free and we'd be done.
Since $U(H)(\mathbb Q\_p)$ is a uniquely divisible group, there is only one possibility for $f'$: we must take $f'(x) = f(x^n)^{1/n}$ where $n$ is chosen so that $x^n\in H$. We want to show that this $f'$ is a group homomorphism, for which we use the Hall–Petresco identity [Appendix A]. This states that for any two elements $x,y$ in a group $G$, there is a sequence of elements $c\_1=xy,c\_2,c\_3,\dotsc$ of $G$, with each $c\_i$ lying in the $i$th step of the descending central series of $G$, such that
$$
x^ny^n = (xy)^nc\_2^{n\choose2}c\_3^{n\choose3}\dots c\_{n-1}^nc\_n = (xy)^n\prod\_{i=2}^{\infty}c\_i^{n\choose i}
$$
for all $n\geq0$ (the product on the right-hand side has only finitely many terms different from $1$). In our case, since $G$ is nilpotent, we only need to take the product on the right up to some fixed $N$. So if we choose $n$ divisible by a sufficiently large power of $p$, then all of the terms $x^n$, $y^n$, $(xy)^n$ and all $c\_i^{n\choose i}$ in the Hall–Petrescu identity lie in the subgroup $H$. So we obtain
$$
f'(x)^nf'(y)^n = f'(xy)^n \prod\_{i=2}^Nf'(c\_i)^{n\choose i}
$$
for all such $n$. Now since $U(H)$ is a unipotent group, we know that the multiplication and exponentiation in $U(H)$ are given coordinatewise by polynomials with coefficients in $\mathbb Q\_p$. So both sides of the above equality are given coordinatewise by polynomials in $n$. Since the equality holds for infinitely many $n$, it in fact must hold for *all* $n$. Setting $n=1$ we recover
$$
f'(x)f'(y)=f'(xy)
$$
so that $f'$ is a group homomorphism and we are done.
| 4 | https://mathoverflow.net/users/126183 | 436581 | 176,435 |
https://mathoverflow.net/questions/436594 | 1 | Let $(X, \mathcal X, \mu)$ and $(Y, \mathcal Y, \nu)$ be $\sigma$-finite measure spaces. Let $\overline{\mathbb R} := \mathbb R \cup \{\pm \infty\}$.
* $f:X \to \overline{\mathbb R}$ is called **$\mu$-simple** if it has the form $f = \sum\_{i=1}^n a\_i 1\_{A\_i}$ where $(a\_i) \subset \mathbb R \setminus \{0\}$ and $(A\_i) \subset \mathcal X$ is pairwise disjoint such that $\mu(A\_i) < \infty$ for all $i = 1, \ldots,n$.
* $f:X \to \overline{\mathbb R}$ is called $\mu$-**measurable** if it is a $\mu$-a.e. pointwise limit of a sequence of $\mu$-simple functions.
Let $\mathcal X \otimes \mathcal Y$ the product $\sigma$-algebra on $X \times Y$. Let $\pi^X:X \times Y \to X$ and $\pi^Y:X \times Y \to Y$ be the projection maps. Let $\gamma$ be a measure on $\mathcal X \otimes \mathcal Y$ with marginals $\mu$ on $X$ and $\nu$ on $Y$, i.e., $(\pi^X)\_\sharp \gamma = \mu$ and $(\pi^Y)\_\sharp \gamma = \nu$. s
>
> Is $c\_x$ is $\nu$-measurable for $\mu$-a.e. $x \in X$? If not, is there $x \in X$ such that $c\_x$ is $\nu$-measurable?
>
>
>
The answer is affirmative for a very particular case below.
---
>
> **Theorem** If $\mu(X)=\nu(Y)=1$ and $\gamma = \mu \otimes \nu$, then $c\_x$ is $\nu$-measurable for $\mu$-a.e. $x \in X$.
>
>
>
*Proof* Let $(c\_n)$ be a sequence of $\gamma$-simple functions such that $c\_n (x, y) \to c(x, y)$ as $n \to \infty$ for $\gamma$-a.e. $(x, y) \in X \times Y$. This implies there is $N \in \mathcal X \otimes \mathcal Y$ such that $\gamma(N)=0$ and
$$
c\_n (x, y) \to c(x, y) \quad \forall (x,y) \in S := (X \times Y) \setminus N.
$$
For $x \in X$, let $S\_x := \{y \in Y : (x, y) \in S\}$ be the $x$-section of $S$. Then $S\_x \in \mathcal Y$. By Fubini's theorem, $\nu (S\_x) =1$ for $\mu$-a.e. $x \in X$. WLOG, we assume $c\_n (x, y) = \sum\_{i=1}^{\varphi\_n} a\_{ni} 1\_{A\_{ni}} (x, y)$. Let $A\_{ni,x}$ be the $x$-section of $A\_{ni}$. Then $A\_{ni, x} \in \mathcal Y$. Let
$$
c\_{n, x} : Y \to \overline{\mathbb R}, y \mapsto \sum\_{i=1}^{\varphi\_n} a\_{ni} 1\_{A\_{ni, x}} (y).
$$
Then $c\_{n, x}$ is $\nu$-simple for every $(n,x) \in \mathcal N \times X$. Clearly, $c\_{n, x} \to c\_x$ on $S\_x$ for $\mu$-a.e. $x\in X$. This completes the proof.
| https://mathoverflow.net/users/99469 | Let $c: X \times Y \to \overline{\mathbb R}$ be $\gamma$-measurable. Is $c_x:Y \to \overline{\mathbb R}, y \mapsto c(x, y)$ $\nu$-measurable? | No to both. Let $\gamma$ be the uniform distribution on $\Delta=\{(x,x)\mid x\in [0,1]\}$, the diagonal of $[0,1]^2$. The marginals are simply the uniform distribution on $[0,1]$. Fix some function $g:[0,1]\to\mathbb{R}$ that is not Lebesgue measurable. Define $c$ by $c(x,y)=g(y)$ if $x\neq y$ and $c(x,y)=0$ for $x=y$. Then $c$ is $\gamma$-measurable, but each section agrees almost everwhere with $g$ and is, therefore, not Lebesgue-measurable. Here, this is the same as not being $\nu$-measurable.
| 2 | https://mathoverflow.net/users/35357 | 436596 | 176,438 |
https://mathoverflow.net/questions/436554 | 3 | Let $n,a,b$ be integers such that $n$ and $a$ are coprime, and $n$ and $b$ are also coprime. According to the Prime number theorem for arithmetic progressions, the primes which are $a\mod n$ have the same asymptotic density as the primes which are $b\mod n$. Is the same true for semiprimes?
| https://mathoverflow.net/users/394725 | Density of semiprimes in arithmetic progression | Yes "Riemann", your hypothesis is true.
Landau showed the count of all semiprimes grows as follows:
$$
|\{pq \leq x\}| \sim \frac{x}{\log x}(\log \log x).
$$
If $\gcd(a,n) = 1$ then a special case of the answer by Lucia [here](https://mathoverflow.net/questions/156982/chebotarev-density-theorem-for-k-almost-primes) says
$$
|\{pq \leq x : pq \equiv a \bmod n\}| \sim
\frac{1}{\varphi(n)}\frac{x}{\log x}(\log \log x)
$$
because Lucia's result implies that, for each $b, c \in (\mathbf Z/n\mathbf Z)^\times$, the number of semiprimes $pq \leq x$ where $p \equiv b \bmod n$ and $q \equiv c \bmod n$ is asymptotic to
$(1/\varphi(n)^2)(x/\log x)(\log\log x)$. You have to be careful about avoiding duplicate counting of the products $pq$ depending on whether or not $b \equiv c \bmod n$. Therefore
\begin{align\*}
|\{pq \leq x : pq \equiv a \bmod n\}| & =
\sum\_{\substack{(b,c) \bmod n \\ bc \equiv a \bmod n}} |\{pq \leq x : p \equiv b \bmod n, q \equiv c \bmod n\}| \\
& =
\sum\_{\substack{b \bmod n}} |\{pq \leq x : p \equiv b \bmod n, q \equiv b^{-1}a \bmod n\}| \\
& \sim
\sum\_{\substack{b \bmod n}} \frac{1}{\varphi(n)^2}\frac{x}{\log x}(\log \log x) \\
& = \frac{1}{\varphi(n)}\frac{x}{\log x}(\log \log x).
\end{align\*}
There is a more general result. For fixed $k \geq 1$, Landau showed
$$
|\{p\_1\cdots p\_k \leq x\}| \sim \frac{x}{\log x}\frac{(\log \log x)^{k-1}}{(k-1)!}
$$
(that is counting $k$-almost primes $p\_1\cdots p\_k$, not the $k$-tuples $(p\_1,\ldots,p\_k)$, so order and multiplicity matter when passing between these two methods of counting), and one then expects when $\gcd(a,n) = 1$ that
$$
|\{p\_1\cdots p\_k \leq x : p\_1\cdots p\_k \equiv a \bmod n\}| \sim
\frac{1}{\varphi(n)}\frac{x}{\log x}\frac{(\log \log x)^{k-1}}{(k-1)!}.
$$
Such an estimate follows from Lucia's result by estimating
$$
|\{p\_1\cdots p\_k \leq x : p\_1 \equiv b\_1 \bmod n, \ldots, p\_k \equiv b\_k \bmod n\}|
$$
and summing those estimates over the $\varphi(n)^{k-1}$ different $k$-tuples
$(b\_1,\ldots,b\_k)$ from $(\mathbf Z/n\mathbf Z)^\times$ where
$b\_1\cdots b\_k \equiv a \bmod n$ (let $b\_1, \ldots, b\_{k-1}$ be arbitrary units mod $n$ and then $b\_k \bmod n$ is determined by the congruence $b\_1\cdots b\_k \equiv a \bmod n$).
| 4 | https://mathoverflow.net/users/3272 | 436607 | 176,443 |
https://mathoverflow.net/questions/436601 | 3 | All representable functors are continuous. This makes it possible to associate additional natural operations with them, which are absent for arbitrary presheaves.
>
> 1. What are the sufficient and what are the necessary conditions on the subcanonical site $I$ for all sheaves to be continuous?
> 2. Is there a canonical way to introduce the structure of such a site on an arbitrary small category? (by canonical I mean functorial with respect to category equivalences)
>
>
>
| https://mathoverflow.net/users/148161 | Necessary and sufficient conditions for all sheaves on a site to be continuous functors? | Given a category $C$, and a familly of co-cone in $C$ (you can take all colimit cocone in $C$ if you want - the family doesn't even have to be small) there is a smallest topology on $C$ so that sheaves for this topology sends these cocone to limit cones.
The detail of the construction below should also give some answer to your question 1.
For each of the special cocone $F:I^\triangleright \to C$, we consider the map $colim\_I F \to F(\*)$ in the category $P(C)$ of presheaves over $C$. A presheaf sends this cocone to a colimit if and only if it is right orthogonal to this map. A Grothendieck topology is the same as a left exact localization of $P(C)$, so we are looking for the smallest left exact localization that inverse these maps.
A Grothendieck topology invert a map $f:X \to Y$ if and only if it invert the two monomorphism $Im(f) \to Y$ and $X \to X \times\_Y X$ and it inverts a monomorphism $A \hookrightarrow X$ if and only if for each representable $c$ and each map $c \to X$ (that each element of $X(c)$) the Sieve on $c$ obtain as the pullback of $A \to X$ is a covering sieve.
So, putting everything together, saying that some class of maps $colim\_I F \to F(\*)$ is inverted by the topology (that is that sheaves sends the corresponds cocone to a limit cone) can be written as a fact that a bunch of sieve are covering sieve. and hence there is a smallest topology which satisfies these conditions. (the Grothendieck topology generated by those sieve).
Of course this topology has no reason to be subcanonical in general. For example, assuming that the category $C$ has pullback, this topology being subcanonical implies in particular that colimits in $C$ are universal (and this is probably not sufficient).
| 4 | https://mathoverflow.net/users/22131 | 436614 | 176,446 |
https://mathoverflow.net/questions/436613 | 4 | There is a $q$-binomial identity that I encountered in one paper I am reading (<https://arxiv.org/abs/1910.06193>) which probably admits a very simple proof that I do not see: for two nonnegative integers $a,b$, we have
$$
q^{ab}=\sum\_{k\ge 0}(-1)^kq^{\binom{k}2}\binom{a}{k}\_q\binom{b}{k}\_q(q;q)\_k,
$$
where $(q;q)\_k=(1-q)(1-q^2)\cdots(1-q^k)$, $\binom{n}{k}\_q=\frac{(q;q)\_n}{(q;q)\_k(q;q)\_{n-k}}$ for $0\le k\le n$ and $\binom{n}{k}\_q=0$ for $k>n$.
I am sure that this can be proved by a version of the A=B method, but for connecting this to my work not the identity as such but rather a conceptual proof or including it into some context is desirable. Since it is not an identity "with positive coefficients", it is not obvious if I should ask for a combinatorial interpretation, or some sort of Euler characteristic result, so I will just leave it slightly open-ended: have you seen any version of this identity?
| https://mathoverflow.net/users/1306 | (Conceptual) proof and/or interpretation of a $q$-binomial identity | Note that it will be enough to check the identity when $q$ is a prime power, in which case we can choose a field $F$ with $|F|=q$, and vector spaces $A$ and $B$ of dimension $a$ and $b$ over $F$. In this context it is more natural to consider the function $\pi\_q(k)=\prod\_{i=1}^k(q^i-1)=(-1)^k(q;q)\_k$ rather than $(q;q)\_k$, because $\pi\_q(k)$ is a positive integer and is more closely related to counting problems. We then have
$$ {\binom{n}{k}\_q}=\frac{\pi\_q(n)}{\pi\_q(k)\pi\_q(n-k)} $$
and the stated identity becomes
$$ q^{ab} = \sum\_{k\geq 0} q^{\binom{k}{2}}\binom{a}{k}\_q\binom{b}{k}\_q\pi\_q(k). $$
The left hand side is $|\text{Hom}(A,B)|$. To give a homomorphism $\alpha\colon A\to B$ we have to choose a rank $k$, a kernel $U\leq A$ of dimension $a-k$, an image $V\leq B$ of dimension $k$, and an isomorphism $\alpha\_1\colon A/U\to V$. The number of choices for $U$ is $\binom{a}{a-k}\_q=\binom{a}{k}\_q$. The number of choices for $V$ is $\binom{b}{k}\_q$. The number of choices for $\alpha\_1$ is $|GL\_k(F)|=q^{\binom{k}{2}}\pi\_q(k)$.
| 6 | https://mathoverflow.net/users/10366 | 436616 | 176,447 |
https://mathoverflow.net/questions/427389 | 6 | Does there exist an upper bound of the analytic rank of the modular Jacobian varieties $J\_1(N)$?
(Or more generally of $J\_\Gamma$ for a congruence subgroup $\Gamma\_0 \subseteq \Gamma \subseteq \Gamma\_1$.)
I want one like $ rank J\_1(N) < C \dim J\_1(N)$, for some nice small constant $C$.
($N$ is an arbitrary positive integer, or it’s ok to assume that it is a prime number)
I know there’s such upper bound for $J\_0(p)$ and for $C$ smaller than $1.2$.
(See Kowalski, E., Michel, P., The analytic rank of J 0 ( q ) and zeros of automorphic L -functions, theorem 1,
and Kowalski, E., Michel, P., VanderKam, J. M., Nonvanishing of higher derivatives of
automorphic L-function.)
Can we generalize them?
I am not familiar with the analytic number theory, the symmetric square of modular forms, and every analytic things mentioned in the papers at all.
| https://mathoverflow.net/users/128235 | Upper bound of the analytic rank of the modular Jacobian varieties $J_1(N)$ | I remember discussing this with Emmanuel Kowalski not long ago. The short answer is that generalising the result to $J\_1(N)$ is an open problem, and seems to be very difficult.
| 2 | https://mathoverflow.net/users/2481 | 436628 | 176,451 |
https://mathoverflow.net/questions/435037 | 1 | $\newcommand{\diff}{ \, \mathrm d}$
Let
* $X,Y$ be Polish spaces,
* $\mathcal C\_b(X)$ the space of all real-valued bounded continuous functions on $X$,
* $\mathcal P(X)$ the space of Borel probability measures on $X$,
* $\mu \in \mathcal P(X)$ and $\nu \in \mathcal P(Y)$.
* $L\_1 (\mu)$ the space of all $\mu$-integrable functions $\varphi:X \to \mathbb R \cup \{-\infty\}$,
* $\Pi(\mu, \nu)$ a subset of $\mathcal P(X \times Y)$ that contains all measures whose marginal on $X$ is $\mu$ and that on $Y$ is $\nu$, and
* $c:X \times Y \to [0, +\infty]$ measurable.
Let $\Phi\_c$ (resp. $\Phi'\_c$) be the collection of all $(\varphi, \psi) \in \mathcal C\_b(X) \times \mathcal C\_b(Y)$ (resp. $(\varphi, \psi) \in L\_1 (\mu) \times L\_1 (\nu)$) such that $\varphi (x)+\psi(y) \le c(x, y)$ for all $(x,y) \in X \times Y$. Let
$$
\begin{align}
\mathbb J (\varphi, \psi) &:= \int \varphi \diff \mu + \int \psi \diff \nu &&\forall (\varphi, \psi) \in L\_1(\mu) \times L\_1(\nu),\\
\mathbb K (\gamma) &:= \int c \diff \gamma &&\forall \gamma \in \Pi(\mu, \nu).
\end{align}
$$
The Kantorovich and its dual problems are
$$
\begin{align}
(\mathrm{KP}) &: \quad \inf \left \{ \mathbb K (\gamma) : \gamma \in \Pi(\mu, \nu) \right \}, \\
(\mathrm{DP}) &: \quad \sup \left \{ \mathbb J (\varphi, \psi) : (\varphi, \psi) \in \Phi\_c \right \}, \\
(\mathrm{DP'}) &: \quad \sup \left \{ \mathbb J (\varphi, \psi) : (\varphi, \psi) \in \Phi'\_c \right \}.
\end{align}
$$
Clearly,
$$
\Phi\_c \subset \Phi\_c'
\quad \text{and} \quad
\sup \mathrm{DP} \le \sup \mathrm{DP'} \le \inf \mathrm{KP}.
$$
The central definition leading to the existence of solutions of above problems is $c$-concavity, i.e.,
>
> **Definition 2.33** A function $\varphi: X \rightarrow \mathbb{R} \cup\{-\infty\}$ is said to be **$c$-concave** if there exists $\psi: Y \to \mathbb{R} \cup\{-\infty\}$ such that $\psi \not \equiv-\infty$ and that
> $$
> \varphi(x) = \psi^c (x) := \inf \_{y \in Y}[c(x, y)-\psi(y)] \quad \forall x \in X.
> $$
> Here $\varphi$ is called the **$c$-conjugate** of $\varphi$.
>
>
>
I'm able to prove that
>
> **Theorem** Let $c$ be real-valued and lower semi-continuous. Assume there exists $(c\_X, c\_Y) \in L\_1 (\mu) \times L\_1(\nu)$ such that they are real-valued and that $c(x,y) \le c\_X(x) + c\_Y (y)$ for all $(x,y) \in X \times Y$. Then $\mathrm{DP'}$ admits a solution.
>
>
>
At the bottom of page 87 of Villani's **Topics in Optimal Transport**, there is **Exercise 2.36** to prove that $\mathrm{DP'}$ admits a maximizer, i.e.,
>
> **Exercise 2.36** Let $c$ be lower semi-continuous. Assume there exists $(c\_X, c\_Y) \in L\_1 (\mu) \times L\_1 (\nu)$ that are non-negative such that
> $$c(x,y) \le c\_X (x) + c\_Y(y) \quad \forall (x, y) \in X \times Y.$$
> Then $\mathrm{DP'}$ admits a solution of the form $(\varphi, \varphi^c) \in \Phi\_c'$.
>
>
>
**My attempt:** To make it easier, for **Exercise 2.36** I assumer further that $c, c\_X, c\_Y$ are real-valued. By **Theorem**, $\mathrm{DP'}$ admits a solution $(\varphi, \psi) \in \Phi'\_c$. Now we **assume** that $\varphi^c \in L\_1 (\nu)$. By definition,
$$
\varphi^c (y) \le c(x, y)-\varphi(x) \quad \forall (x, y) \in X \times Y.
$$
So $(\varphi, \varphi^c) \in \Phi'\_c$. Also,
$$
\varphi^c (y) = \inf\_{x \in X} (c(x, y)-\varphi(x)) \ge \inf\_{x \in X} (\psi (y)) = \psi (y).
$$
So $\mathbb J (\varphi, \varphi^c) \ge \mathbb J (\varphi, \psi)$. Then $(\varphi, \varphi^c)$ satisfies the requirement. This is called by the author as *double convexification trick*.
---
**My question:** The first issue is to prove that $\varphi^c$ is measurable, and the second one is to prove that $\varphi^c$ is $\nu$-integrable. However, the measurability of $\varphi^c$ is [subtle and non-trivial](https://mathoverflow.net/questions/434969/optimal-transport-how-is-the-use-of-disintegration-theorem-valid-in-this-constr). Above exercise is exactly **Theorem 4.10** in [this lecture note](https://drive.google.com/file/d/1-6FpiqPG4GVDEBk1qgsGU141oVKyxVG-/view) in which the measurability of $\varphi^c$ is **not** proved.
In optimal transport, this is a fundamental result in both theory and practice. Could you elaborate on how to finish **Exercise 2.36**?
Thank you so much for your elaboration!
| https://mathoverflow.net/users/99469 | Optimal transport: the existence of an optimal pair of $c$-conjugate functions | I have found a related paper [Existence and stability results in the $L^1$ theory of optimal transportation](https://www.math.ucdavis.edu/%7Esaito/data/emd/ambrosio-pratelli.pdf) by Luigi Ambrosio and Aldo Pratelli. Step 2 of the proof of Theorem 3.2 is
>
> Step 2. Now we show that $\psi:=\varphi^c$ is $\nu$-measurable, real-valued $\nu$-a.e. and that
> $$
> \varphi+\psi=c \text { on } \Gamma \text {. }
> $$
> It suffices to study $\psi$ on $\pi\_Y(\Gamma)$ : indeed, as $\gamma$ is concentrated on $\Gamma$, the Borel set $\pi\_Y(\Gamma)$ has full measure with respect to $\nu=\pi\_{Y \#} \gamma$. For $y \in \pi\_Y(\Gamma)$ we notice that $(10)$ gives
> $$
> \psi(y)=c(x, y)-\varphi(x) \in \mathbb{R} \quad \forall x \in \Gamma\_y:=\{x:(x, y) \in \Gamma\}
> $$
> In order to show that $\psi$ is $\nu$-measurable we use the disintegration $\gamma=\gamma\_y \otimes \nu$ of $\gamma$ with respect to $y$ (see the appendix) and notice that the probability measure $\gamma\_y$ is concentrated on $\Gamma\_y$ for $\nu$-a.e. $y$, therefore
> $$
> \psi(y)=\int\_X c(x, y)-\varphi(x) d \gamma\_y(x) \quad \text { for } \nu \text {-a.e. } y
> $$
> Since $y \mapsto \gamma\_y$ is a Borel measure-valued map we obtain that $\psi$ is $\nu$-measurable.
>
>
>
In the proof, $\psi$ is proved to be $\nu$-measurable. It seems to me being $\nu$-measurable is equivalent to being [Bochner measurable](https://en.wikipedia.org/wiki/Bochner_measurable_function), i.e., $\psi$ is a $\nu$-a.e. pointwise limit of a sequence of simple functions. So $\psi$ is $not$ necessarily Borel measurable.
| 1 | https://mathoverflow.net/users/99469 | 436635 | 176,453 |
https://mathoverflow.net/questions/436153 | 4 | $\DeclareMathOperator\GL{GL}$I'm reading the proof of Serre's open image theorem from his book *"Abelian $\ell$-adic representations and elliptic curves"*. This is chapter IV Section 2.2 of the book. Let's assume that $E$ is an elliptic curve without CM over a number field $K$ and $\rho:G\_K \rightarrow \GL(V\_\ell)$ the associated Galois representation on $ V\_\ell = T\_\ell \otimes \mathbb{Q} $ where $ T\_\ell $ is the Tate module. Then he proves that the image of the Galois representation is open in $\GL(V\_\ell)$ w.r.t the $\ell$-adic topology. The only part that I don't understand from the proof is in the very beginning when he deduces that the ($\ell$-adic) Lie algebra of the image contains $\mathfrak{sl}\_2$ from the fact that it's centralizer is $\mathbb{Q}\_\ell$. Is he using some sort of classification for the Lie subalgebras here?
My main question is this: To me this seems like a Lie algebra analogue of the double centralizer theorem for simple subalgebras of central simple algebras. I was wondering if such a result exists, namely if from the centralizer of the Lie subalgebra being small we can deduce that the subalgebra itself is large in some sense. I'm not sure what should be the conditions and what's the right formulation but I feel like $\GL\_2$ being reductive and $\operatorname{SL}\_2$ being semi-simple might play a role here.
| https://mathoverflow.net/users/496065 | Double centralizer theorem for ($\ell$-adic) Lie algebras | $\newcommand{\g}{\mathfrak{g}}\newcommand{\sl}{\mathfrak{sl}}$The crucial point in the proof is that the absolute Galois group $G\_K$ acts irreducibly on $V\_{\ell}$, which is based on a nontrivial Shafarevich finiteness theorem (Sect. 1.4). Applying this result to all finite algebraic extensions of $K$, one gets that the $\ell$-adic Lie algebra $\g\_{\ell}$ of the image acts irreducibly (and faithfully) on $V\_{\ell}$.
So,
$\g\_{\ell}$ is an irreducible Lie subalgebra of $\mathrm{End}\_{\mathbb{Q}\_{\ell}}(V\_{\ell})$, whose centralizer consists of scalars $\mathbb{Q}\_{\ell}\mathrm{Id}$, i.e., the natural faithful 2-dimensional representation of $\g\_{\ell}$ in $V\_{\ell}$ is absolutely irreducible. Hence, $\g\_{\ell}$ is reductive, i.e., splits into a direct sum
$$\g\_{\ell}=\g\_{\ell}^{0}\oplus c\_{\ell}$$
of a semisimple Lie algebra $\g\_{\ell}^{0}$ and the center $c\_{\ell}$. The absolute irreducibility implies that $c\_{\ell}$ is either $0$ or $\mathbb{Q}\_{\ell}\mathrm{Id}$. In both cases $V\_{\ell}$ is an absolutely irreducible representation of $\g\_{\ell}^{0}$; in particular, $\g\_{\ell}^{0}\ne \{0\}$. The semisimplicity of $\g\_{\ell}^{0}$ implies that
$$\g\_{\ell}^{0}\subset \sl(V\_{\ell})\cong \sl\_2(\mathbb{Q}\_{\ell}).$$
Now it follows easily that
$\g\_{\ell}^{0}= \sl(V\_{\ell})$, because no proper Lie subalgebras of $\sl\_2$ are semisimple.
| 4 | https://mathoverflow.net/users/9658 | 436645 | 176,457 |
https://mathoverflow.net/questions/436638 | 3 | Let $D$ be the unit disk in $\mathbb R^2$ centered at the origin. Given any $\lambda \in \mathbb R$, let $u\_\lambda$ be the unique solution to the semilinear elliptic equation
$$ -\Delta u + u^3=0 \quad \text{on $D$},$$
subject to the constant Dirichlet data $u|\_{\partial D} =\lambda$.
Prove that $u(0)$ is uniformly bounded in $|\lambda|$ and that more generally given any $h\in (0,1)$, the solution $u(x)$, with $|x|<h$ is uniformly bounded in $|\lambda|$.
| https://mathoverflow.net/users/50438 | Boundedness of solutions to a semilinear PDE | Let me give a positive answer perhaps omitting some details.
Fact 1. Let $u'' \geq ku^\alpha$ in $[c,\ell[$ with $k>0, \alpha>1$ and $u,u' \geq 0$. Let $A=u(c)$, then $ \ell \to c$ as $A \to \infty$.
This follows by multiplying by $u'$ and integrating. One obtains $$u' \geq k \sqrt{u^{\alpha+1}-A^{\alpha+1}} \geq k \sqrt{(u-A)u^\alpha}$$ and integrating again
$$
\ell-c \leq k^{-1}\int\_A^\infty \frac{ds}{\sqrt{(s-A)s^\alpha}} =k^{-1} \int\_0^\infty \frac{dt}{(t+A)^{\frac \alpha 2}\sqrt t} \to 0
$$
as $A \to \infty$.
Assume now that $u''+\frac{N-1}{r}u'=u^3$ with $u'(0)=0$ and $u(1)=\lambda >0$. The maximum principle yields $u \geq 0$ and $r^{1-N} (r^{N-1} u')'=u^3$ with $u'(0)=0$ also $u' \geq 0$. Multiplying the equation by $u'$ and integrating
$$
\frac 12 u'^2(r)+(N-1)\int\_0^r \frac{u'^2}{t} dt= \frac 14 (u^4(r)-u(0)^4)
$$
and then $u'^2 \leq \frac 12 u^4$. If $A=u(0)$ is sufficiently large, then $u(r) \geq A$ gives $u'' \geq \frac 12 u^3$ for $r \geq \frac 12$ and one concludes by Fact 1 with initial point $\frac 12$ that the solution explodes before reaching $r=1$.
| 3 | https://mathoverflow.net/users/150653 | 436648 | 176,458 |
https://mathoverflow.net/questions/436651 | 4 | Given an entire function $f : \mathbb{C} \to \mathbb{C}$, $\log |f|$ is subharmonic. Globally, this means that for any disk $D\_r(c)$ we have the submean property
$$\log |f(c)| \le \frac{1}{\mu(D\_r(c))} \int\_{D\_r(c)} \log |f(z)|~dz$$
If there are no zeros in a disk, this follows from Cauchy's theorem applied to the analytic function $\log f$, as $\log |f| = \Re \log f$, and indeed we have equality. Exactly at a zero, the inequality is trivial as $\log |f| = \log 0 = -\infty$. In the general case, if there are zeros, the inequality follows immediately from [Jensen's formula](https://en.wikipedia.org/wiki/Jensen%27s_formula).
Alternatively, one can use the first two facts to prove subharmonicity locally (away from zeros, and exactly at zeros), and then prove that local subharmonicity implies global subharmonicity.
**Question:** Is there is a more direct proof of the global submean property, without using Jensen's formula or that local subharmonicity implies global?
**Motivation:** I am playing with complex analysis in Lean and Mathlib, and want the global submean property (including around zeros) without having to prove Jensen's theorem or local-to-global. It feels like there is a chance that a direct trick involving Jensen's inequality (as opposed to Jensen's formula) or other properties of subharmonic functions would suffice, but I haven't found one yet.
| https://mathoverflow.net/users/22930 | Direct proof of the global submean property for $\log |f|$ | One way to derive the global inequality is the Principle of Harmonic Majorant: If $D$ is a bounded region, $u$ is harmonic
in $\overline{D}$, $f$ is analytic in $\overline{D}$, then the inequality $\log|f(z)|\leq u(z),\, z\in\partial D$ implies the same in $D$. To prove this, just apply the Maximum principle to the harmonic function
$\log|f|-u$ in the region $D$ minus small neighborhoods of zeros of $f$; that the inequality holds on the boundaries of these small neighborhoods is clear since $\log|z|\to-\infty$
when $z\to$ a zero of $f$.
Now to prove your global inequality, use as $u$ the solution of the Dirichlet problem with the boundary values $\log|f|$.
The solution of the Dirichlet problem for a disk is completely elementary, and the value at the center is just the average of the values on the circumference.
So we only used the Maximum Principle for harmonic functions and solvability of the Dirichlet problem for a disk.
| 3 | https://mathoverflow.net/users/25510 | 436654 | 176,459 |
https://mathoverflow.net/questions/436624 | 6 | Let $f : \mathbb{N} \rightarrow \mathbb{N}$ be a non-decreasing function and let $X\_f$ be the class of graphs where every $n$-vertex graph $G$ is $(C\_3, C\_4, \ldots, C\_{f(n)})$-free, i.e. $G$ contains no cycles of length at most $f(n)$.
It is known that if $f$ is constant, then graphs in $X\_f$ can have a superlinear number $n^{1+\epsilon(f)}$ of edges. On the other hand, if $f : n \mapsto n$, then $X\_f$ is the class of forests, which have a linear number of edges.
I'm wondering if there is a transition from the superlinear to the linear regime.
For example, is there a known function $f^\*$ such that
1. for any $g \in o(f^\*)$, the class $X\_g$ has $n$-vertex graphs with a superlinear number ($\omega(n)$) of edges;
2. for any $g \in \omega(f^\*)$, the graphs in $X\_g$ have a linear number of edges?
| https://mathoverflow.net/users/83519 | Graphs without short cycles and with linear number of edges | Threshold is $\log n$.
1. If the graph has at least, say, $2n$ edges, it has a cycle of length at most, say, $2\log\_2 n$ (proof: remove vertices of degree at most 2 while it is possible. After each step, $|E|\geqslant 2|V|$ property is preserved. So, you get a graph with all degrees at least 3,and considering DFS from any vertex you find a cycle of above mentioned length.)
2. For any $C>0$, there exists $\varepsilon>0$ such that there exists a graph with $Cn$ edges and all cycles not shorter than $\varepsilon \log n$. (proof: take a random graph $G(n, 5C/n)$ and destroy all short cycles.)
| 5 | https://mathoverflow.net/users/4312 | 436671 | 176,462 |
https://mathoverflow.net/questions/436678 | 2 | Let $W$ be a standard Brownian motion. Define the upper left Dini derivative $D^-W$ by
$$D^-W\_t := \limsup\_{h \to 0^-} \frac{W\_{t+h} - W\_t}{h}.$$
Fix $a > 0$, and define the stopping time $\tau$ by
$$\tau := \inf \{t > 0 \, | \, W\_t \geq a\}$$
**Question:** Is it true that $D^- W\_\tau = +\infty$, almost surely?
| https://mathoverflow.net/users/173490 | Upper left Dini derivative of Brownian motion at a hitting time | The derivative is indeed infinite, basically because Brownian motion does not have points that are too regular in a sense.
A *slow point* for a realisation of Brownian motion is a time $t$ such that
$$ \sup\_{h\to0^+}\frac{|W\_{t+h}-W\_t|}{h^{1/2}}<\infty. $$
For the purposes of this answer, let me call a *very slow point* a time $t$ such that the above supremum is (strictly) less than 1.
>
> **Theorem** (Davis) (Greenwood & Perkins)**.**
>
>
> Almost surely, the set of very slow points is empty.
>
>
>
It follows by time reflection that at all points, either the upper or lower left Dini derivative of Brownian motion is infinite. Since the lower left Dini derivative is zero at $\tau$, then the upper one must be infinite.
---
*A conditioned limit theorem for random walk and Brownian local time on square root boundaries,* P. Greenwood and E. A. Perkins (1983). [Link.](https://doi.org/10.1214/aop/1176993594)
*On Brownian slow points,* B. Davis (1983). [Link.](https://doi.org/10.1007/BF00532967)
| 3 | https://mathoverflow.net/users/129074 | 436681 | 176,463 |
https://mathoverflow.net/questions/436682 | 6 | Is the set of all solutions $x > 0$ to the equation $\pi(x) = \operatorname{li}(x)$ unbounded? Is $\liminf\_{x \to \infty} |\pi(x)-\operatorname{li}(x)|$ equal to $0$?
Here, $\pi(x)$ denotes the prime counting function and $\operatorname{li}(x) = \int\_0^x \frac{dt}{\log t}$ denotes the logarithmic integral function.
| https://mathoverflow.net/users/17218 | Is the set of all solutions $x > 0$ to $ \pi(x) = \operatorname{li}(x)$ unbounded? | $\newcommand{\li}{\operatorname{li}}$Yes, those values of $x$ are unbounded. As JoshuaZ indicates in a comment, the key here is Littlewood's result on sign changes, and because of the way $\pi(x)$ grows, it is easy to deal with jump discontinuities
without any difficult transcendence results. Here are the details.
Littlewood has proven that the difference $\pi(x)-\li(x)$ changes sign infinitely often, so there will be arbitrarily large $x\_0$ such that $\pi(x\_0)-\li(x\_0)>0$. If we then consider the set of values on which this difference becomes nonpositive, this set will also be nonempty. Let $x\_1$ be the infimum of this set. $\pi(x)-\li(x)$ is a right continuous function, so $\pi(x\_1)-\li(x\_1)\leq 0$. Moreover, if $x\_1$ happened to be a jump discontinuity, then the left limit of $\pi(x)-\li(x)$ at $x\_1$ would be $1$ smaller than its right limit, which would contradict $x\_1$ being the infimum of the above set. So $x\_1$ satisfies $\pi(x\_1)=\li(x\_1)$.
| 7 | https://mathoverflow.net/users/30186 | 436693 | 176,467 |
https://mathoverflow.net/questions/436695 | 0 | Let $A$ be a noncommutative Koszul algebra (see [here](https://en.wikipedia.org/wiki/Koszul_algebra) for a definition of Koszul) and let $c \in A$ be a central element. Will the quotient of $A$ by the ideal generated by $c$ again be Koszul. If not what is a counter example and what else could I require to ensure Koszulity?
| https://mathoverflow.net/users/491434 | Quotients of Koszul algebras | With no assumptions, obviously the answer is no. You didn't even require that c is homogeneous.
If $c$ has any components of degree $>2$, then I think the answer is that the quotient is never Koszul: Koszul algebras are quadratic.
If $c$ has degree 0,1 or 2, then certainly there are cases where $A/(c)$ will be Koszul:
* $A=K\langle x,y\rangle$ the free algebra on 2 variables over any field $K$ is Koszul, and $c= xy-yx$ gives $A/c=K[x,y]$ which is again Koszul.
* If $A$ is free or polynomial, the quotient by any degree 1 element will be free or polynomial on one fewer generators.
* If you have a Koszul quotient of the path algebra of a quiver, you can sometimes kill a vertex and get something Koszul. For example, if $x,y$ are the edges of an oriented 2-cycle, the quotient $xy=0$ is 5-dimensional and Koszul, and killing either vertex gives the base field.
I suspect that this is not "typical" behavior (for example, this paper shows that there are quadratic quotients of polynomial rings which are not Koszul: <https://arxiv.org/pdf/1605.09145.pdf>), but don't have the time/energy to come up with a bunch of counter-examples right now.
| 4 | https://mathoverflow.net/users/66 | 436698 | 176,469 |
https://mathoverflow.net/questions/435307 | 2 | Let $\mathfrak{g}$ be a simple complex Lie algebra. Let $V\_1,\cdots, V\_n$ be the fundamental representations (the irreducible ones with fundamental weights $\omega\_1,\cdots,\omega\_n$). Take a $k$-tensor product of these representations: $V\_{\lambda\_1}\otimes\cdots\otimes V\_{\lambda\_k}$ (with each $\lambda\_i\in\{\omega\_1,\cdots,\omega\_n\}$).
Decompose this product into irreducible representations. Let $\lambda=\sum n\_i\omega\_i$ be the highest weight of such a simple summand. Can we conclude $\sum n\_i\leq k$?
I can show it holds for type $A\_n$ and $C\_n$.
| https://mathoverflow.net/users/169800 | Tensor product of fundamental representations | Actually one can show that if $\sum n\_i \lambda\_i$ is a highest weight in a $k$-tensor product of fundamental representations, we have $\sum n\_i\leq \beta \cdot k$ for some $\beta$ uniquely determined by the simple type of $\mathfrak{g}$.
| 0 | https://mathoverflow.net/users/169800 | 436699 | 176,470 |
https://mathoverflow.net/questions/436716 | 1 | Let $n\_1, n\_2, ...$ be a sequence of natural numbers such that $\{n\_i: i \in \mathbb{N}\}$ as a set is all of natural numbers. Let $k$ be a positive integer. Is is possible to obtain a lower bound of the form
$$
\# (\{ n\_i + i^k: i \in \mathbb{N} \} \cap [1,T]) \gg T^c
$$
for some $c>0$? I'm not sure where to start so any reference and comments are appreciated!
| https://mathoverflow.net/users/84272 | Cardinality of $\{ n_i + i^k: i \in \mathbb{N} \} \cap [1,T]$ where $\{n_i \}$ is all natural numbers in some order | Such a lower bound is not possible, even when $k=1$. Indeed, there exists a sequence $(n\_1,n\_2,\dotsc)$ containing every natural number such that for infinitely many positive integers $N$ we have $n\_i=2^i$ for $i\in\{N+1,N+2,\dotsc, 2^N\}$. Then, for such an $N$ we have that
$$\{ n\_i + i: i \in \mathbb{N} \} \cap [1,2^N]\subset [1,N],$$
so that no $c>0$ satisfies the desired bound.
| 7 | https://mathoverflow.net/users/11919 | 436718 | 176,476 |
https://mathoverflow.net/questions/435932 | 4 | Let $G$ be a finite group, $G',H$ be its subgroups and $H'=G'\cap H$. For each $g\in G$, we create a map $f\_g:G'/H'\rightarrow G/H: aH'\rightarrow gaH$. It's easy to see that the map is well defined and injective. Let $S$ be a subset of $G/H$, assume that there is no $g\in G$ such that $f\_g(G'/H')\subset S$.
***Question***: I want to estimate and find some properties of $M(G,G',H)=\max|S|$ and $\alpha(G,G',H)=\max\frac{|G/H|}{|G/H|-|S|}$, for all or some particular cases of groups $G,G',H$. Note that we have $M(G,G',H)=\left(1-\frac{1}{\alpha(G,G',H)}\right)|G/H|$.
***Some results that I have found:***
a) $1\leq\alpha(G,G',H)\leq |G'/H'|$. The first inequality is trivial. Assume that $|S|> \left(1-\frac{1}{|G'/H'|}\right)|G/H|$ then:
$\mathbb{E}\_{g\in G}[|f\_g(G'/H')\cap S|]=|G'/H'|\mathbb{P}\_{a\in G'/H',g\in G}[f\_g(a)\in S]=\frac{|G'/H'|}{|G/H|}|S|>|G'/H'|-1 $
So there exists $g\in G$ such that $|f\_g(G'/H')\cap S|=|G'/H'|\Rightarrow f\_g(G'/H')\subset S$.
b) If $G'\subset H\Rightarrow H'=G'$, then by a), $\alpha(G,G',H)=1,M(G,G',H)=0$.
c) If $H,H'$ is a trivial group then $\alpha(G,G',H)=|G'|$ because each coset of $G/G'$ contains at most $|G'|-1$ elements of $S$, so $|S|\leq (|G'|-1)|G/G'|$
d) If $N$ is a normal subgroup of $G,H$ and $N'=N\cap G'$ then $\alpha(G/N,G'/N',H/N)=\alpha(G,G',H),M(G/N,G'/N',H/N)=M(G,G',H)$, because the two set of left cosets and the set of all function $f\_g$ are still the same under isomorphism after taking quotent by $N$ and $N'$.
e) By c), d), if $G'$ is a commutative group then $\alpha(G,G',H)=\alpha(G/H,G'/H',{e})=|G'/H'|$.
f) If $N$ is a subgroup of $G,G'$ such that $N\cap H$ be the trivial group and $nh=hn,\forall n\in N, h\in H$ then $\alpha(G,G',H)=|N|\alpha(G,G',NH)$ (I proof by dividing $G/NH,G'/NH'$ into $|N|$ parts but the proof is quite complicative so let's omit it).
g) If $H'$ is a trivial group and $hg=gh, \forall h\in H, g\in G'$ then $\alpha(G,G',H)=|G'|\alpha(G,G',G'H)=|G'|$ by f), b).
h) $M(G,G',H)=|G/G'H|M(G'H,G',H), \alpha(G,G',H)=\alpha(G'H,G',H)$
so we can assume $G=G'H$.
***Motivation***: Let $P\_S$ be a group of permutation of the set $S$. Let $\{1,2,...,n\},n\geq 3$ be the sets of vertices of the complete graph $K\_n$, we have a bijection from the set of left cosets $P\_{\{1,2,...,n\}}/(P\_{\{3,4,...,n\}}\times P\_{\{1,2\}})$ to the set of edges of $K\_n$ by map the coset $\sigma(P\_{\{3,4,...,n\}}\times P\_{\{1,2\}})$ to the edge $(\sigma(1),\sigma(2))$.
For each $\sigma\in P\_{\{1,2,...,n\}}$, we see that the map $f\_{\sigma}: P\_{\{1,2,...,r\}}/(P\_{\{3,4,...,r\}}\times P\_{\{1,2\}})\rightarrow P\_{\{1,2,...,n\}}/(P\_{\{3,4,...,n\}}\times P\_{\{1,2\}})$ corresponds to the complete subgraph $K\_r$ of $K\_n$ with vertices $\sigma(1),\sigma(2),...\sigma(r)$.
The subset $S$ of $P\_{\{1,2,...,n\}}/(P\_{\{3,4,...,n\}}\times P\_{\{1,2\}})$ such that there is no $\sigma\in P\_{\{1,2,...,n\}}$ such that $f\_{\sigma}(P\_{\{1,2,...,r\}}/(P\_{\{3,4,...,r\}}\times P\_{\{1,2\}}))\subset S$ creates a $K\_r$-free graph of $n$ vertices, so $M(P\_{\{1,2,...,n\}},P\_{\{1,2,...,r\}},P\_{\{3,4,...,n\}}\times P\_{\{1,2\}})= \left(1-\frac{1}{r}+o(1)\right)\frac{n^2}{2}\Rightarrow \alpha(P\_{\{1,2,...,n\}},P\_{\{1,2,...,r\}},P\_{\{3,4,...,n\}}\times P\_{\{1,2\}})=r+o(1)$ with $r$ fixed and $n$ increase by Turán's theorem.
***More questions***:
1. When $\alpha(G,G',H)=|G'/H'|$?
2. Improve the lower bound of $\alpha(G,G',H)$ which is only depend on $|G'/H'|$ or show such bound doesn't exist.
We can view the bipartite graph $K\_{\{1,2,...,r\},\{r+1,r+2,...,2r\}}$ as $Aut(K\_{\{1,2,...,r\},\{r+1,r+2,...,2r\}})/Aut(K\_{\{1,2,...,r-1\},\{r,r+1,...,2r-2\}})$, then by Erdős–Stone-Simonovits theorem we have
$M(P\_{\{1,2,...,n\}},Aut(K\_{\{1,2,...,r\},\{r+1,r+2,...,2r\}}),P\_{\{3,4,...,n\}}\times P\_{\{1,2\}})=o(n^2)\Rightarrow \alpha(P\_{\{1,2,...,n\}},Aut(K\_{\{1,2,...,r\},\{r+1,r+2,...,2r\}}),P\_{\{3,4,...,n\}}\times P\_{\{1,2\}})=1+o(1)$
as $r$ fixed and $n$ increase, so such bound in question 2 doesn't exist and we have a new question:
2'. Is it true that for all group $G'$, subgroup $H'$ of $G'$ and $\varepsilon>0$, there exists group $G$ and subgroup $H$ of $G$ such that $M(G,G',H)$ is defined and $M(G,G',H)<\varepsilon|G/H|$?
| https://mathoverflow.net/users/432274 | Turán's theorem for cosets of groups | ***Question 2'***: We choose $G=P\_{G'/H'\cup A},H=P\_{\{(G'/H')-\{eH'\})\cup A}$, use natural acting of $G'$ on $G'/H'$, we can view $G'$ as subgroup of $G$ and $G/H=G'/H'\cup A$. We have the stabilizer subgroup with respect to the left coset $eH'$ of $G$ and $G'$ are $H$ and $H'$ respectively so $H'=G'\cap H$. Let $S\in G/H$, if $|S|\geq |G'/H'|$, because $G$ is the permutation group of $G/H$ so there exists $g\in G$ such that $f\_g(G'/H')\subset S$. So $M(G,G',H)=|G'/H'|-1$, now we just need to choose $A$ such that $\varepsilon|G/H|>|G'/H'|-1$.
***Question 1***:
Assume $\alpha(G,G',H)=|G'/H'|$, take $S$ statisfies the condition and $|S|=(1-\frac{1}{|G'/H'|})|G/H|$. We have:
$\mathbb{E}\_{g\in G}[|f\_g(G'/H')\cap S|]=\frac{|G'/H'|}{|G/H|}|S|=|G'/H'|-1$
but $|f\_g(G'/H')\cap S|\leq|G'/H'|-1$ so $|f\_g(G'/H')\cap S|=|G'/H'|-1,\forall g\in G$. Because $gS$ also has that properties of $S$ for $g\in G$, so we can assume $\{eH\}\notin S$.
Let $R=G/H-S$, then $eH\in R$,$|R|=\frac{|G/H|}{|G'/H'|}$ and $|f\_g(G'/H')\cap R|=1, \forall g\in G$. If $\{r\}=f\_g(G'/H')\cap R$ then we take $h(r)=f\_{hg}(G'/H')\cap R$. We see that this is a transitive action of $G$ on $R$. We have $g(eH)=eH$ if and only if $gH\in G'/H'\Rightarrow g\in \{xy|x\in G',y\in H\}$ so $\{xy|x\in G',y\in H\}$ must be a subgroup of $G$ (stable group of $eH\in R$).
Now if $\{xy|x\in G',y\in H\}$ is a subgroup of $G$, we have $G'H=\{xy|x\in G',y\in H\}$ and $|G'H|=\frac{|G'||H|}{|H'|}$ ([Example 3.25 Page 15](https://kconrad.math.uconn.edu/blurbs/grouptheory/gpaction.pdf)). Take $L$ be the left transversal for $G'H$ in $G$ and take $R=\{lH|l\in L\}$, it can be check that $|R|=\frac{|G/H|}{|G'/H'|}$ and $|f\_g(G'/H')\cap R|=1, \forall g\in G$, we take $S=G/H-R$ then we have $\alpha(G,G',H)=|G'/H'|$.
So $\alpha(G,G',H)=|G'/H'|$ if and only if $\{xy|x\in G',y\in H\}$ is a subgroup of $G$. This result might suggest duality property of this problem.
| 0 | https://mathoverflow.net/users/432274 | 436728 | 176,479 |
https://mathoverflow.net/questions/436735 | 7 | Define a compact Kähler surface $X$ to be a K3 surface if $X$ is simply connected, $K\_X \simeq \mathcal{O}\_X$, and $h^{0,1}=0$. If $X$ is projective, then a theorem typically attributed to Bogomolov and Mumford, asserts that $X$ admits a rational curve. It has recently been shown that projective K3 surfaces admit infinitely many rational curves.
>
> 1. Do non-projective K3 surfaces admit rational curves?
> 2. Do non-projective K3 surfaces admit infinitely many rational curves?
>
>
>
| https://mathoverflow.net/users/105103 | Do non-projective K3 surfaces have rational curves? | Some of them do, and some don't.
Indeed, by global Torelli theorem, there is a K3 surface $X$ with $\mathrm{Pic}(X) = 0$. Such $X$ has no curves, in particular no rational curves.
On the other hand, there is a K3 surface $X$ such that $\mathrm{Pic}(X)$ is generated by a single class with square $-2$; such a class (up to sign) is represented by a smooth rational curve.
| 12 | https://mathoverflow.net/users/4428 | 436736 | 176,481 |
https://mathoverflow.net/questions/436730 | 9 | Recall the homotopy excision theorem, as stated in Hatcher (Theorem 4.23): Let $X$ be a CW complex decomposed as the union of subcomplexes $A$ and $B$ with nonempty connected intersection $C = A \cap B$. If $(A,C)$ is $m$-connected and $(B,C)$ is $n$-connected for some $n,m \geq 0$, then the map $\pi\_i(A,C) \rightarrow \pi\_i(X,B)$ is an isomorphism for $i<m+n$ and a surjection for $i=m+n$.
The proof of this theorem in Hatcher is fairly complicated (but elementary). The other sources I've looked at (e.g. May's book) give similar proofs. Are there other proofs? As an example, the first application Hatcher gives is to prove the Freudenthal suspension theorem, and my favorite proof of this uses the Serre spectral sequence. More generally, I often find proofs of basic homotopy theoretic results clearer if they use the Serre spectral sequence or other such things rather than being overly "elementary". But I'd also be interested in alternate elementary proofs.
The only other proof I know of is Rezk's proof using homotopy type theory, but I can't make heads or tails of it.
There is an earlier MO question [here](https://mathoverflow.net/questions/262207/are-there-intuitively-clear-and-not-technical-proofs-of-homotopy-excision-theore) that is sort of along the same lines, but it includes desiderata like the proof being "ideologically profound" that I certainly am not looking for (in fact, I don't even know what this means, to be honest).
| https://mathoverflow.net/users/496513 | Alternate proofs of homotopy excision theorem | The proof of Theorem 9.3.5 (especially the part on page 486) in Spanier's "Algebraic Topology" may be more to your liking. It presumes you have already established the relative Hurewicz theorem, e.g. by Serre spectral sequence methods.
| 10 | https://mathoverflow.net/users/9684 | 436740 | 176,482 |
https://mathoverflow.net/questions/436726 | 10 | I am currently writing an expository paper on gauge theory and gravity and throughout the gauge theory part I have been able to mostly stay away from coordinates unless I wished to
provide specific examples. I am trying to continue this practice through the section on general relativity but am struggling at writing a coordinate free proof of the variation of the Einstein-Hilbert action.
Ideally what I am looking for is a way to take the Einstein-Hilbert action :
$$
S[g]=\int\_M\langle g,Rc\rangle\_g\text{dvol}\_g
$$
(where $\langle\cdot,\cdot\rangle$ is the induced inner product on the bundle of symmetric tensors, and $\text{dvol}\_g$ is the volume form) and then vary it with a section of the symmetric tensor bundle $h$:
$$
\frac{d}{dt}\Big|\_{t=0}S[g+th]=0
$$
A similar question was asked [here](https://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-act) many years ago, and I have tried to follow up on their recommended sources, namely Besse's Einstein Manifolds and I am honestly struggling to figure out how they are obtaining the results they are obtaining as they do not provide many details.
For concrete examples of where I am struggling, take the variation of the volume form. Besse states that:
$$
\begin{align}
\frac{d}{dt}\Big|\_{t=0}\text{dvol}\_{g+th}=\frac{1}{2}\text{tr}\_g(h)\text{dvol}\_g
\end{align}
$$
and I agree, but I can only get there in coordinates by writing:
$$
\begin{align}
\frac{d}{dt}\Big|\_{t=0}\text{dvol}\_{g+th}=&\frac{d}{dt}\Big|\_{t=0} \sqrt{\det(g+th)}dx^1\wedge \cdots
\wedge dx^n\\
=& \frac{1}{2}\frac{1}{\sqrt{\det(g)}}\frac{d}{dt}\Big|\_{t=0}\det(g+th)dx^1\wedge \cdots dx^n\\
=&\frac{1}{2}\frac{1}{\sqrt{\det(g)}}\det(g)\text{tr}(g^{-1}h)dx^1\wedge \cdots \wedge dx^n\\
=&\frac{1}{2}\sqrt{\det(g)}g^{ij}h\_{ij}dx^1\wedge \cdots \wedge dx^n\\
=&\frac{1}{2}\text{tr}\_g(h)\text{dvol}\_g
\end{align}
$$
Furthermore, when varying the scalar curvature, $s=\langle g,Rc\rangle\_g$ we writes that this most easily found by noting that $s=\text{tr}\_g(Rc)$ and then applying the product rule:
$$
\begin{align}
\frac{d}{dt}\Big|\_{t=0}\text{tr}\_{g+th}(Rc)=\langle h , Rc\rangle \_g+\text{tr}\_g\left(\frac{d}{dt}\Big|\_{t=0}Rc\_{g+th}\right)
\end{align}
$$
I can see how the second term comes about as the trace of $Rc$ with respect to $g$ is just $g^{-1}\lrcorner Rc$, but the first term is harder for me to come by without coordinates. With coordinates I can write that:
$$
\begin{align}
\frac{d}{dt}\Big|\_{t=0}\text{tr}\_{g+th}(Rc)=&\frac{d}{dt}\Big|\_{t=0}(g+th)^{ij}Rc\_ij\\
=&-g^{ik}h\_{kl}g^{jl}Rc\_{ij}+g^{ij}\frac{d}{dt}\Big|\_{t=0}Rc\_ij\\
=&\langle h , Rc\rangle \_g+\text{tr}\_g\left(\frac{d}{dt}\Big|\_{t=0}Rc\_{g+th}\right)
\end{align}
$$
But is there any way to see this without coordinates? Perhaps some notation that allows one to write the inner product in terms of $g$ without reference to coordinates? I feel like if I can past my initial uneasiness here I'll be able to follow the rest of the argument in the book, albeit with some work, so any help would be appreciated.
| https://mathoverflow.net/users/496509 | Variation of the Einstein Hilbert action in a coordinate-free way | This is really just a long commentary about your question. First, it is always possible to write everything without using coordinates, because the indices can refer to a (moving) frame of tangent vectors. If I understand correctly, your main goal is to not use indices.
I've always preferred index-free formulas over ones with indices. But usually only for the final formula. It is unusual in Riemannian geometry to be able to write full calculations without indices.
If you want to compute the variation of something, call it $\mathrm{Blah}$, with respect to something else, say $\mathrm{Else}$, then you need a precise definition of $\mathrm{Blah}$ with respect to $\mathrm{Else}$. Gauge theory is in a sense less nonlinear than Riemannian geometry. So it is often possible to write rigorous formulas without indices and that can be differentiated with respect to a variation in the connection relatively easily.
However, this simply isn't true in Riemannian geometry, and your two examples demonstrate this well. The concepts of determinant and trace (with respect to a Riemannian metric) are awkward to define rigorously using index-free notation, i.e., without using a basis of tangent vectors.
So before you can even differentiate the volume form without indices, you need a formula for it that does not use indices.
Another awkward issue is when you want to contract a tensor of higher order with a lower order tensor. Compare
$$ g^{jk}\nabla\_jR\_{kl}\,dx^l $$
to
$$ g^{jk}\nabla\_lR\_{jk}\,dx^l. $$
Each is a contraction of the tensors $g^{-1}$ and $\nabla R$. So what notation can you use do distinguish between these two possibilities without using indices?
It might be possible to invent notation that allows you to do calculation without using indices, but I don't know of any successful effort to do this. The closest I know of is [Penrose's abstract index notation](https://en.wikipedia.org/wiki/Abstract_index_notation).
$\newcommand\Hom{\operatorname{Hom}}$ ADDED: You can of course define everything functorially.
In particular, if $T = T\_xM$, then $g \in S^2T^\*$ defines functorially a map $g: T \rightarrow T^\*$, which induces functorially a map $$\det g: \Lambda^nT \rightarrow \Lambda^nT^\*.$$ This means that $\det g \in \Lambda^nT^\*\otimes\Lambda^nT^\*$. If $g$ is positive definite (i.e., $g(v,v) > 0$ for any $v \ne 0$), then for any nonzero $\omega \in \Lambda^nT$, $$(\det g)(\omega,\omega) > 0.$$ Since $\dim \Lambda^nT^\* = 1$, this implies that there exists, unique up to sign, $\omega \in \Lambda^nT^\*$ such that $$\omega\otimes \omega = \det g.$$ Then $$dV\_g = \omega.$$
I'm pretty sure you could do all the variation calculations using this definition of $dV\_g$, but, as far as I can tell, it isn't worth the trouble.
| 12 | https://mathoverflow.net/users/613 | 436762 | 176,487 |
https://mathoverflow.net/questions/436760 | 3 | $N$-player discounted stochastic games with finite state and action spaces possess a Nash equilibrium in stationary strategies. This has been proved by Fink (1964) and a closely related result by Takahashi (1964). Various generalizations have appeared since then.
I have read Fink's proof and understood it (at least I think so). To prove equilibrium, he must compare his proposed strategy to all other possible strategies. However it seems to me that he only considers the set of stationary strategies.
I can see one reason that looking only at stationary strategies is sufficient. Namely, if every other player uses a stationary strategy (this is the candidate Nash equilibrium), then the $n$-th player must solve a Markov Decision problem and it has been proved (e.g., look at Puterman's or Derman's books) that for the MDP nothing is lost by using a stationary strategy.
However, I have looked and have not been able to find some work in which the above or some other argument is used to justify considering only stationary strategies for the $N$-player case.
So my **question** is this: *do you know of any work that addresses the above concerns*? Any help will be greatly appreciated.
| https://mathoverflow.net/users/496530 | Existence of stationary Nash equilibrium of discounted stochastic game | The point that stationary best responses to stationary strategies are best responses without any further restrictions was already made in the very first paper on stochastic games in the context of zero-sum games in
>
> Shapley, Lloyd S. "[Stochastic games.](https://doi.org/10.1073/pnas.39.10.1095)" *Proceedings of the
> National Academy of Sciences* 39.10 (1953): 1095-1100.
>
>
>
For a more explicit statement of the argument, which is exactly the one you gave, you can look at
>
> Sobel, Matthew J. "[Noncooperative stochastic games.](https://doi.org/10.1214/aoms/1177693059)" *The Annals
> of Mathematical Statistics* 42.6 (1971): 1930-1935.
>
>
>
| 2 | https://mathoverflow.net/users/35357 | 436763 | 176,488 |
https://mathoverflow.net/questions/436759 | 4 | I am trying to generalize an algorithm for the construction of a certain linear combinations of functions $\boldsymbol{f}:x\in\mathbb{R}\mapsto \boldsymbol{f}(x)\in\mathbb{R}^n$ that utilizes Wronskian matrices, to the multidimensional version $\boldsymbol{f}:\boldsymbol{x}\in\mathbb{R}^m\mapsto\boldsymbol{f}(\boldsymbol{x})\in\mathbb{R}^n$ utilizing "some" multidimensional generalization of Wronskian matrices.
Some online research turned up [A. I. Petrov, A multidimensional generalization of the
Wronskian, Uspekhi Mat. Nauk, 1964, Volume 19, Issue 5, 194–
196](https://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=rm&paperid=9022&option_lang=eng), which is written in Russian language.
>
> **Question:**
>
>
> is there an English (or German) translation of the above linked paper freely available online?
>
>
> What else can be recommended as freely accessible resources for the definition of multidimensional Wronskian matrices
>
>
>
**Addendum:**
from the statement "**In the case of functions of several variables, there is no one determinant which may properly be taken as the generalization of the wronskian**" in M. Green's 1916 paper [The Linear Dependence of Functions of Several Variables, and Completely Integrable Systems of Homogeneous Linear Partial Differential Equations](https://www.jstor.org/stable/1988834#metadata_info_tab_contents) it became clear that questions about *the* multidimensional Wronskian are ill posed.
| https://mathoverflow.net/users/31310 | English translation of “A multidimensional generalization of the Wronskian” | The solution to the mystery of how A. I. Petrov wrote a paper in 1964, when he was a student writing what appears to be his first paper in 1971, is that mathnet.ru has mistakenly taken his name to be A. I. Petrov, when in fact he is A. I. Perov, as you can see in the Wronskian paper. Moreover, Perov is still working at Voronezh State University, and I can send you his email address if you contact me at [email protected]. I don't know if he will help you, but the paper was part of a special collection of abstracts, so was not translated in the usual Uspekhi translations into English. The paper is pretty clearly translated using DeepL, so I don't think you really need a German or English reference.
| 12 | https://mathoverflow.net/users/13268 | 436770 | 176,489 |
https://mathoverflow.net/questions/436777 | 9 | Is there a generalization of the notion of braided monoidal category that does not force the braiding $\gamma$ to be an isomorphism? I mean, it is of course possible to define such a kind of category, but is this a common notion with an established name?
| https://mathoverflow.net/users/25527 | Is there a generalization of braided monoidal category without the isomorphism requirement? | Day, Panchadcharam, and Street have a [paper](http://science.mq.edu.au/%7Estreet/laxcentre.pdf) on lax braidings, though I don't think it could be called a *common* notion. Anyway, "lax" seems to be the obvious terminology to try here.
| 14 | https://mathoverflow.net/users/43000 | 436778 | 176,493 |
https://mathoverflow.net/questions/436756 | 1 | Let $F$ be a nonarchimedean local field of characteristic zero, and $E$ an extension of $F$ with $[E:F]=2^n$ for some $n$. Is it always possible to find a quadratic extension $M$ of $F$ such that $F\subseteq M\subseteq E$ ?
| https://mathoverflow.net/users/32746 | Quadratic extension of local field | * Try $f(x)=x^4+2x+2\in \Bbb{Q}\_2[x]$ and $K=\Bbb{Q}\_2[a]/(f(a))$.
Then $f(x) = (x-a)g(x)\in K[x]$ where $g$ is irreducible ($a^{-3}g(ax-a/3)$ is Eisenstein). So $f$ has only one root in $K$, $Aut(K/\Bbb{Q}\_2)$ is trivial and hence there is no subfield $L$ such that $[K:L]=2$ ie. no subfield such that $[L:\Bbb{Q}\_2]=2$.
* If $F$ is a finite extension of $\Bbb{Q}\_p$ with $p$ odd and $[E:F]=2^n,n\ge 1$ then yes there is always a quadratic subfield. Assume the opposite, take an uniformizer $\pi\_E$, the residue field of $F$ is $\Bbb{F}\_q$, that of $E$ is $\Bbb{F}\_{q^m}$, it must be that $m | 2^n$, and it must be that $m=1$ as otherwise $F(\zeta\_{q^2-1})/F$ is a quadratic subfield.
Therefore, $E = F(\pi\_E)$ where $\pi\_E^{2^n} = u \pi\_F$ for some $u\in O\_E^\times$.
So $u=\zeta\_{q-1}^r (1+c\pi\_E)$ with $c\in O\_E$,
$\varpi\_E =(1+c\pi\_E)^{-1/2^n} \pi\_E\in E$ is again an uniformizer and $\varpi\_E^{2^n}=\zeta\_{q-1}^r \pi\_F\in F$.
ie. $F(\varpi\_E^{2^{n-1}})/F$ is a quadratic subextension.
| 3 | https://mathoverflow.net/users/84768 | 436786 | 176,495 |
https://mathoverflow.net/questions/436782 | 0 | Define for real valued random variable $X\in L^p$, the $p$-statistic
$$X\_p:=\arg\min\_{c\in \mathbb R}E[|X-c|^p].$$
For example $X\_1$ is the median of $X$, $X\_2$ is the mean of $X$ and also $X\_\infty$ is the midpoint of the range of $X$.
Let $X,Y\in L^\infty$ be two real valued random variables so that $X\_p=Y\_p$ for all $p>0$. Then is it true that $X=Y$ in distribution? What if we assume some regularity on $X,Y$?
This seems like an analogue to moment characterization.
| https://mathoverflow.net/users/479223 | Do these $L^p$ type statistics characterize distributions? | No, it looks like many different sufficiently symmetric distributions with enough concentration at $0$ will have $\arg\min\_c E[|X-c|^p] = 0$ for all $p > 0$.
Concrete family of examples: if
$$ X = \begin{cases} 0 & \text{w.prob $2/3$} \\
t & \text{w.prob $1/6$} \\
-t & \text{w.prob $1/6$} , \end{cases} $$
then, regardless of our choice of $t > 0$, for all $p > 0$ we have $0 = \arg\min\_c E[|X-c|^p]$. We can also replace $2/3$ with anything greater than $1/2$.
Proof sketch: by symmetry and monotonicity, we can suppose $0 \leq c \leq t$ in the minimization.
\begin{align}
E[|X-c|^p] &= \frac{1}{6} (t + c)^p + \frac{2}{3} c^p + \frac{1}{6} (t-c)^p .
\end{align}
For all $p \geq 1$, this expression is strictly increasing in $c$, implying it is minimized at $c=0$.
For all $0 < p < 1$, the expression is concave in $c$ (the second derivative is negative), so it is minimized at one of the endpoints, and $c=0$ results in a value of $\frac{1}{3}t^p$, while $c=t$ results in a larger value of at least $\frac{2}{3}t^p$.
| 4 | https://mathoverflow.net/users/29697 | 436787 | 176,496 |
https://mathoverflow.net/questions/436789 | 12 | According to Hilton-Milnor theorem for $n\geq 2$
$$
\pi\_k(\mathbb{S}^n\vee\mathbb{S}^n)=
\pi\_k(\mathbb{S}^n)\oplus
\pi\_k(\mathbb{S}^n)\oplus
\bigoplus\_{i=1}^\infty
\pi\_k(\mathbb{S}^{m\_i}),
$$
where $m\_i$ is a sequence of integers that tend to $\infty$. (Correct me if this statement is incorrect.)
>
> Is there an explicit formula for the sequence $m\_i$?
>
>
>
>
> For what values of $k$ is $\pi\_k(\mathbb{S}^n\vee\mathbb{S}^n)$ infinite?
>
>
>
If it is difficult to find an explicit formula for $m\_i$ for an arbitrary $n$, I would like to see some examples where the answers to both of the above questions can be explicit.
| https://mathoverflow.net/users/121665 | $\pi_k(\mathbb{S}^n\vee\mathbb{S}^n)$ | Yes, there is an explicit formula. It even describes the torsion elements in the homotopy groups of $S^n \vee S^n$. One statement is
$$\Omega \Sigma(A \vee B) \simeq \Omega \Sigma A \times \Omega \Sigma B \times \Omega \Sigma(\bigvee\_{i,j \geq 1} A^{\wedge i} \wedge B^{\wedge j})$$
where $\wedge$ is the smash product, and $\wedge i$ means the $i$-fold smash, e.g. $A^{\wedge 3} = A \wedge A \wedge A$.
So if you want to know the homotopy groups of $\Sigma(A \vee B)$ through a range, you need to apply the above formula inductively until the wedge product on the right side is sufficiently highly connected.
For example, if you apply this to $\pi\_4 (S^2 \vee S^2)$ you get a group of the form
$$\Bbb Z\_2 \oplus \Bbb Z\_2 \oplus \Bbb Z\_2 \oplus \Bbb Z \oplus \Bbb Z.$$
The copies of $\Bbb Z\_2$ are instances of $\pi\_4 S^2$ and $\pi\_4 S^3$ respectively, while the copies of $\Bbb Z$ are instances of $\pi\_4 S^4$ via the above formula.
You get $\pi\_k(S^n \vee S^n)$ finite when the iterated Whitehead brackets of the non-trivial elements in rational homotopy of $S^n$ do not appear in dimension $k$. So those would be all dimensions not in the list $n, 2n-1, 3n-2, 4n-3, \dotsc$.
edit: The level of explicitness you can get will depend on some constraints. If you let $k$ be relatively small, this can be done. But if you want to know the sequence $m\_i$ that's valid for all $k$, given that the sequence is a solution to a type of partition problem, there likely isn't a cute closed-form expression available. I suppose it goes without saying that the sequence $m\_i$ does not depend on $k$ -- the terms when $m\_i > k$ just don't matter to the answer.
| 15 | https://mathoverflow.net/users/1465 | 436790 | 176,497 |
https://mathoverflow.net/questions/436665 | 8 | I am looking for some literature with some (counter) examples of the following fact (though I don't know if the fact is true or not):
>
> Let $M, M'$ be two non-compact connected $3$-manifolds with the same
> proper homotopy type. Then $M$ is homeomorphic to $M'$.
>
>
>
Recall that two spaces $X$ and $Y$ are said to have the same proper homotopy type if there are proper maps $f\colon X\to Y$ and $g\colon Y\to X$ such that both $f\circ g$ and $g\circ f$ are properly homotopic to the identity maps. A proper homotopy from $X$ to $Y$ is a homotopy, that is, a map $H\colon X\times [0,1]\to Y$, which is a proper map.
| https://mathoverflow.net/users/363264 | Non-compact three-manifolds with the same proper homotopy type are homeomorphic? | Take a look at the example shown at Remark 5.9 b) in the paper "A topological equivalence relation for finitely presented groups."by M. Cárdenas, F.F. Lasheras, A.Quintero and R.Roy. Journal of Pure and Applied Algebra, DOI 10.1016/j.jpaa.2019.106300 . The numerably punctured spaces $\mathbb R^3$ and the "semispace" $\mathbb R^3\_+$ are properly homotopic but not homeomorphic. (It works for $n\geq 2$).
| 9 | https://mathoverflow.net/users/42904 | 436801 | 176,500 |
https://mathoverflow.net/questions/436806 | 0 | I was researching upon the Collatz conjecture, and I was reading all the research work done by mathematicians including Terry Tao's. I had read that before Terry Tao's research it was proven that almost all Collatz sequences eventually end up below the number you start from. Is it like other sequences which are exceptions are the ones that may grow infinitely?
If proven that all Collatz orbits attain bounded values (hypothetical example), can the statement above be proven for all orbits?
| https://mathoverflow.net/users/496145 | If proven that all Collatz sequences attain bounded values, is it also proven that all sequences end up below the number you start from? | This does not follow. Nothing in that work rules out that there might even be infinitely many distinct cycles from the Collatz function, all spread out a lot. So even if you had that sort of boundedness claim it would not follow.
| 1 | https://mathoverflow.net/users/127690 | 436807 | 176,503 |
https://mathoverflow.net/questions/436685 | 10 | Let $G$ be a discrete amenable, residually finite, ICC(i.e. each non-trivial conjugacy class is infinite) group. Let $C^\*\_r(G)$ be the reduced group $C^\*$-algebra of $G$. Since $G$ is ICC the (faithful) canonical trace $\tau$ that maps every non-trivial group element to 0 is an extreme trace.
Is there a group $G$, satisfying the above properties and an extreme trace $\tau’\neq\tau$ on $G$ that is faithful on $C^\*\_r(G)$, i.e. $\tau’(x^\*x)>0$ for all non-zero $x\in C^\*\_r(G)$?
I would be interested in an example where any of the adjectives amenable,RF or ICC (in this case we just want two distinct extreme faithful traces) are dropped but my main interest is when all three adjectives are present.
| https://mathoverflow.net/users/34640 | Faithful extreme traces on group C*-algebras | The lamplighter group $G = (\mathbb{Z}/2\mathbb{Z}) \wr \mathbb{Z}$ is such an example. The group is amenable, ICC and residually finite. The C$^\*$-algebra $C^\*\_r(G)$ can be identified with the crossed product of $\mathbb{Z}$ acting by the shift on the Cantor space $X = \{0,1\}^{\mathbb{Z}}$. For every $t \in (0,1)$, we have the $\mathbb{Z}$-invariant probability measure $\mu\_t$ on $X$ given by the infinite product of the probability measure on $\{0,1\}$ that assigns measure $t$ to $\{0\}$. The measure $\mu\_t$ uniquely extends to a trace on $C^\*\_r(G)$ that is extremal and faithful.
| 9 | https://mathoverflow.net/users/159170 | 436810 | 176,505 |
https://mathoverflow.net/questions/436811 | 4 | Assuming $f(x)=e^{-x^2}$ for $x$ in $[-10,10]$, I have tried the following:
1. Fourier transform $\mathcal{F}$: $\frac{d}{dx}$ can be diagonalized as $\mathcal{F}^{-1} i\omega \mathcal{F}$. Therefore, $\sin(\frac{d}{dx}) f(x) = \sin(\mathcal{F}^{-1} i\omega \mathcal{F})f(x)=\mathcal{F}^{-1}\sin(i\omega)\*\mathcal{F}f(x)=\mathcal{F}^{-1}\sinh(\omega)\*\mathcal{F}f(x)$. I computed the Fourier transform of $f(x)$, multiplied it with $\sinh(\omega)$ and then computed the inverse Fourier transform.
2. Cauchy integral formula: Generated a matrix $D$ by approximating $\frac{d}{dx}$ using the forward differences and computed the following contour integral: $\sin(D)f=\frac{1}{2\pi i} \oint\limits\_\Gamma \sin(z)(zI-D)^{-1}f \, dz$. Since $D$ is an upper triangular matrix with similar values on the main diagonal, it has one eigenvalue repeated $N$ (size of matrix D) times. Therefore, I chose $\Gamma$ to be a small circle around that eigenvalue.
3. Taylor series: Expanded $\sin(\frac{d}{dx})$ as $\frac{d}{dx}-\frac{d^3}{3!\,dx^3}+\cdots$ and performed discrete differentiations.
4. Scaling and squaring of matrix: Rewrote $\sin(\frac{d}{dx})=\frac{e^{id/dx}-e^{-id/dx}}{2i}$ and used matrix $D$ as defined above.
The calculations in all cases were very unstable. Is there any other (stable) approach that I can try?
Edit: In general, what are the conditions of $g(x)$ and $f(x)$ for $g(\frac{d}{dx})f(x)$ to exist?
| https://mathoverflow.net/users/478084 | How to compute $\sin(\frac{d}{dx})f(x)$? | As noted in the OP, $\sin (d/dx) = (\exp (id/dx) - \exp (-id/dx))/(2i)$, which casts the operator as a combination of two shift operators,
$$
\sin (d/dx) f(x) = \frac{1}{2i} (f(x+i) - f(x-i))
$$
The convergence radius of the Taylor expansion of $f$ around $x$ will have to include $x+i$ and $x-i$.
| 15 | https://mathoverflow.net/users/134299 | 436812 | 176,506 |
https://mathoverflow.net/questions/436723 | 4 | Let
\begin{equation\*}
\begin{split}
M\_m
&=\begin{pmatrix}
-\binom{1}{0} & \binom{2}{0} &-\binom{3}{0} &\dotsm & (-1)^{m-1}\binom{m-1}{0} & (-1)^m\binom{m}{0}\\
0 & \binom{2}{1} &-\binom{3}{1} &\dotsm & (-1)^{m-1}\binom{m-1}{1} & (-1)^m\binom{m}{1}\\
0 & 0 &-\binom{3}{2} &\dotsm & (-1)^{m-1}\binom{m-1}{2} & (-1)^m\binom{m}{2}\\
\vdots & \vdots &\vdots &\ddots & \vdots & \vdots\\
0 & 0 & 0 &\dotsm & (-1)^{m-1}\binom{m-1}{m-2} & (-1)^m\binom{m}{m-2}\\
0 & 0 & 0 &\dotsm & 0 & (-1)^m\binom{m}{m-1}
\end{pmatrix}\_{m\times m}\\
&=(M\_{i,j})\_{m\times m},
\end{split}
\end{equation\*}
where
\begin{equation\*}
M\_{i,j}=
\begin{cases}
(-1)^{j}\dbinom{j}{i-1}, & 1\le i\le j\le m;\\
0, & 1\le j<i\le m.
\end{cases}
\end{equation\*}
For $m=5$, by the famous software Mathematica, we obtain
$$
\begin{pmatrix}
-1 & 1 & -1 & 1 & -1 \\
0 & 2 & -3 & 4 & -5 \\
0 & 0 & -3 & 6 & -10 \\
0 & 0 & 0 & 4 & -10 \\
0 & 0 & 0 & 0 & -5 \\
\end{pmatrix}^{-1}
=
\begin{pmatrix}
-1 & \frac{1}{2} & -\frac{1}{6} & 0 & \frac{1}{30} \\
0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{4} & 0 \\
0 & 0 & -\frac{1}{3} & \frac{1}{2} & -\frac{1}{3} \\
0 & 0 & 0 & \frac{1}{4} & -\frac{1}{2} \\
0 & 0 & 0 & 0 & -\frac{1}{5} \\
\end{pmatrix}.
$$
What is the inverse of the triangular matrix $M\_m$ for $m\in\mathbb{N}=\{1,2,\dotsc\}$?
The matrix $M\_m$ comes from the recursive relation
\begin{equation}\label{beta(m+1minus1)}
\sum\_{k=j+1}^{m}(-1)^{k}\binom{k}{j}\beta\_{m+1,k}
=(-1)^{j+1} \beta\_{m,j}, \quad 0\le j\le m-1,
\end{equation}
where the first few $\beta\_{m,j}$ are
\begin{align\*}
\beta\_{1,0}&=1, & & & & & &\\
\beta\_{2,0}&=\frac{5}{3}, & \beta\_{2,1}&=1, & & & &\\
\beta\_{3,0}&=\frac{11}{5}, & \beta\_{3,1}&=\frac{13}{6}, & \beta\_{3,2}&=\frac{1}{2}, & &\\
\beta\_{4,0}&=\frac{93}{35}, & \beta\_{4,1}&=\frac{101}{30}, & \beta\_{4,2}&=\frac{4}{3}, & \beta\_{4,3}&=\frac{1}{6}.
\end{align\*}
We can also derive
\begin{align\*}%\label{beta(m+1)m-form}
\beta\_{m,m-1}&=\frac{1}{(m-1)!}, \quad m\ge1,\\
\beta\_{m,m-2}&=\frac{3m+4}{6(m-2)!}, \quad m\ge2,\\
\beta\_{m,m-3}&=\frac{15 m^2+35 m+24}{120(m-3)!}, \quad m\ge3,\\
\beta\_{m,m-4}&=\frac{105 m^3+315 m^2+364 m+176}{5040(m-4)!}, \quad m\ge4.
\end{align\*}
We guess that
\begin{equation\*}
\beta\_{m,m-k}=\frac{1}{(2k-1)!(m-k)!}\sum\_{\ell=0}^{k-1}\theta\_{k,\ell} m^\ell, \quad m\ge k,
\end{equation\*}
where $\theta\_{k,\ell}$ is a sequence of positive integers.
What is the explicit or closed-form expression of the sequence $\theta\_{k,\ell}$ for $0\le\ell\le n-1$?
What is the explicit or closed-form expression of the sequence $\beta\_{m,j}$ for $0\le j\le m-1$?
| https://mathoverflow.net/users/147732 | What is the inverse of a triangular matrix whose nonzero elements are binomial coefficients? What is the closed-form solution to a recursive relation? | This may answer the question, the sequence is the only implicit thing. I consider $Q=\begin{pmatrix}Q\_{i,j}\end{pmatrix}\_{n\times n}$ for $n\ge 3$, where
$$Q\_{i,j}=
\begin{cases}
\dbinom{j}{i-1}, & 1\le i\le j\le n;\\
0, & 1\le j<i\le n.
\end{cases}
$$
This is the same as the one defined in the question up to multiplying it by a $\pm 1$ diagonal matrix.
Define sequence $u\_k$ by $u\_1=1$ and
$$
u\_k=-\sum\_{i=0}^{k-2}\dfrac{u\_{i+1}}{(k-i)!}
$$
for every $k>1$. The $n\times n $ matrix $Q^{-1}=\begin{pmatrix}q\_{i,j}\end{pmatrix}\_{n\times n}$ is given by
$$
\begin{cases}
q\_{i,j}=0,&1\le j<i\le n;\\
q\_{i,i}=\dfrac{1}{i},&1\le j=i\le n;\\
q\_{i,i+1}=-\dfrac{1}{2},&2\le j=i+1\le n;\\
q\_{i,i+k}=u\_{k+1}(i+1)\cdots(i+k-1),&k+1\le i+k\le n, k\ge 2.
\end{cases}
$$
The proof is formal and should be direct, also the sequence $u\_k$ has some zero entries.
| 4 | https://mathoverflow.net/users/121643 | 436814 | 176,507 |
https://mathoverflow.net/questions/436817 | 4 | This question has been posted on History of Science and Mathematics stack exchange, but there was no answer or comments there.
In Weierstrass notation, the principal elliptic function $\wp$ is a solution of the equation
$$(\wp')^2=4\wp^3-g\_2\wp-g\_3.$$
The case when $g\_3=0$ is called [lemniscatic](https://en.wikipedia.org/wiki/Lemniscate_elliptic_functions) (it corresponds to a square lattice), and the case $g\_2=0$ is called [equianharmonic](https://en.wikipedia.org/wiki/Equianharmonic) (it corresponds to a hexagonal lattice). The origin of the first term is clear: it comes from the problem on the length of the [Bernoulli's lemniscate](https://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli).
What is the origin of the term "equianharmonic"?
The term "equianharmonic" seems out of date: checking Google ngram shows a strange pattern: its usage experienced a peak around 1860 and was much more frequent than "lemniscatic" until 1980s, and nowadays it is used about 10 times less frequently than "lemniscatic". Also the sizes and details of Wikipedia articles confirm this.
| https://mathoverflow.net/users/25510 | The origin and use of the term "equianharmonic" (elliptic function) | An answer to remove this question from the "unanswered list": The term "equianharmonic" refers to "equal anharmonic ratio", as explained by Wiener in 1901, see this earlier [MO post](https://mathoverflow.net/a/385121/11260).
| 3 | https://mathoverflow.net/users/11260 | 436822 | 176,510 |
https://mathoverflow.net/questions/436833 | 8 | It is well known that if $f(z)$ and $g(z)$ are both holomorphic on a (path-)connected open set $C$ and $\lvert f(z)\rvert=\lvert g(z)\rvert$ on $C$ then $f(z)=cg(z)$ on $C$ for some constant $c$. Do we have a robust version of this theorem like the following?
>
> Suppose for simplicity $C$ is the unit disk and $\operatorname{dist}$ is some distance measure such as L2-distance. If $d=\operatorname{dist}\_C(\lvert f\rvert,\lvert g\rvert)$ is small then $d'=\operatorname{dist}\_{C}(f,cg)$ is also small for some $c$.
>
>
>
**Update.** Furthermore, are we able to upper bound $d'$ by $d^{O(1)}$?
Any relevant information will be greatly appreciated!
| https://mathoverflow.net/users/22954 | A robust version of "a holomorphic function is determined by its modulus" | You can get a robust version for free by a precompactness argument: e.g., if $\|f\|,\|g\|\leq 1$, then for any $\epsilon>0$, there exists $\delta>0$ such that $\||f|-|g|\|\_{\mathbb{D}}<\delta$ implies $\|f-cg\|\_{(1-\epsilon)\mathbb{D}}<\epsilon$ for some $c$. Here the norm may be $L^\infty$ or $L^2$-norm or $H\_2$-norm.
Assume the contrary; then there are sequences $f\_i,g\_i$ and an $\epsilon>0$ such that
$$
\||f\_i|-|g\_i|\|\to 0;\quad
\|f\_i-cg\_i\|\_{(1-\epsilon)\mathbb{D}}>\epsilon\quad\forall c.
$$
Norm-boundedness for analytic function implies uniform boundedness on compact subsets of $\mathbb{D}$, so by passing to a subsequence we may assume that $f\_i\to f$ and $g\_i\to g$ uniformly on compact subsets of $\mathbb{D}$. But then $|f|\equiv|g|$ but $\|f-cg\|\_{(1-\epsilon)\mathbb{D}}\geq\epsilon$ for all $c$, a contradiction.
**Update** there is a lazy way to obtain the following more concrete bound:
>
> for each $f$ analytic in the unit disc, and each $\epsilon>0$, there exist $C>0$ such that $$\inf\_{c\in\mathbb{C}}\|f-cg\|\_{(1-\epsilon)\mathbb{D}}\leq C\||f|-|g|\|\_\mathbb{D}.$$ Let's work with supremum norms for concreteness.
>
>
>
To illustrate the idea, assume first that $f$ has no zeros. Then, $g$ also has no zeros in $(1-\epsilon)\mathbb{D}$ provided $d=\||f|-|g|\|\_\mathbb{D}<\inf\_{(1-\epsilon)\mathbb{D}}|f|$. This implies that $\log f$ and $\log g$ are analytic functions, and moreover for any $r\in(1-\epsilon/2,1)$, we have $$\sup\_{|z|=r}|\Re (\log f(z) - \log g(z))|\leq C\_1 d$$
for some constant $C\_1$ depending only on $\inf\_{|z|=r}|f(x)|$. By Harnack's estimates for the derivatives of harmonic functions, this means
$$\sup\_{|z|\leq 1-\epsilon}|(\log f(z)-\log g(z))'|\leq C\_2 d,$$
with $C\_2$ depending only on $C\_1$ and $\epsilon$. Finally, we integrate this inequality and note that the exponential is a locally Lipschitz function. This gives
$$
\sup\_{|z|\leq 1-\epsilon}\left|f(z) - \frac{f(0)}{g(0)}g(z)\right|\leq C\_3 d,
$$
where $C\_3$ now may depend on $\epsilon$ and on $f$, via $\sup\_{|z|\leq 1-\epsilon}|f(z)|$ and $\inf\_{|z|\leq r}|f(z)|$).
If $f$ has zeros, it does not really matter since near them both $|f|$ and $|g|$ are small. We choose $r$ in the above proof so that $f$ has no zeros on the circle $|z|=r$. For $d$ small enough, the level line of $|f|=2d$ has a small loop around each zero. Removing the interior of each loop, we run the same argument over the resulting domain. The only difference is that we don't know a priori that $\log f-\log g$ is single-valued, but that doesn't matter, as it is enough to estimate the difference over some fixed sheet. If $d$ is not "small enough", then the bound is trivial by taking $C$ large enough.
| 8 | https://mathoverflow.net/users/56624 | 436834 | 176,512 |
https://mathoverflow.net/questions/436828 | 4 | Let $\zeta\_p$ be a $p$-th root of unity for a prime $p$, let $L:=\mathbb{Q}(\zeta\_p)$ and $K$ the maximal totally real subfield of $L$, i.e. $K:=\mathbb{Q}(\zeta\_p+\zeta\_p^{-1})$. I am trying to prove that the narrow class number of $K$ divides the class number of $L$ i.e. $h\_{K}^+\mid h\_L$.
I was trying to show that if $F$ is an extension of $K$ which is unramified at all finite primes, then $F(\zeta\_p)$ will be an extension of $L$ which is unramified at all primes. However, I am not sure how to prove this. Any help would be much appreciated!
| https://mathoverflow.net/users/478525 | Class numbers of cyclotomic fields and their maximal totally real subfields | To prove that $F(\zeta\_p)$ is an extension of $L$ which is unramified at all primes, it is enough to show that $F(\zeta\_p)$ is an extension of $L$ which is unramified at all finite primes, because all finite primes are the only primes that can ramify.
To show that $F(\zeta\_p)$ is an extension of $L$ which is unramified at all finite primes, we can use the following criterion: If $F$ is a Galois extension of $K$ which is unramified at all finite primes, then $F(\zeta\_p)$ is an extension of $L$ which is unramified at all finite primes. This criterion can be proved as follows:
Let $\mathfrak{p}$ be a prime of $L$ which lies above a prime $\mathfrak{P}$ of $F$. We want to show that $\mathfrak{p}$ is unramified in $F(\zeta\_p)$.
Since $F$ is a Galois extension of $K$, the decomposition group of $\mathfrak{P}$ in $F$ is equal to the Galois group of $F/K$. Since $F$ is unramified at all finite primes, the decomposition group of $\mathfrak{P}$ in $F$ is a subgroup of the inertia group of $\mathfrak{P}$ in $F$.
On the other hand, the Galois group of $F(\zeta\_p)/L$ is equal to the Galois group of $F/K$ by the Galois correspondence, so the decomposition group of $\mathfrak{p}$ in $F(\zeta\_p)$ is equal to the Galois group of $F/K$.
Since the decomposition group of $\mathfrak{p}$ in $F(\zeta\_p)$ is equal to the decomposition group of $\mathfrak{P}$ in $F$, which is a subgroup of the inertia group of $\mathfrak{P}$ in $F$, it follows that the decomposition group of $\mathfrak{p}$ in $F(\zeta\_p)$ is a subgroup of the inertia group of $\mathfrak{p}$ in $F(\zeta\_p)$. This means that $\mathfrak{p}$ is unramified in $F(\zeta\_p)$.
| 1 | https://mathoverflow.net/users/496584 | 436837 | 176,514 |
https://mathoverflow.net/questions/436849 | 11 | Let us call a nonempty topological space a *topological tree* if it is Hausdorff and for two distinct points there is a continuous injective path connecting the points, which is unique up to reparametrisation.
This should be equivalent to the following definition:
The space $X$ is a *topological tree* if it is Hausdorff, pathwise connected and whenever $\gamma\_1,\gamma\_2:[0,1]\to X$ are topological embeddings with $\gamma\_1(0)=\gamma\_2(0)$ and $\gamma\_1(1)=\gamma\_2(1)$, it is true that $\gamma\_1([0,1])=\gamma\_2([0,1])$.
I am not reqiring the space to be paracompact nor to have the homotopy type of a CW-complex or anything like that, just a Hausdorff space.
Examples of such *topological trees* are $\mathbb R$, a hedgehog space (<https://en.wikipedia.org/wiki/Hedgehog_space>), or the Long Line (or the Long Ray).
What one could expect that such a space is always contractible as it is called a *tree* and trees should be contractible but the example of the Long Line shows that this is not the case.
However, it **should** be true that such a space is *weakly contractible*, i.e. every homotopy group is trivial. As $\pi\_0$ is trivially trivial, I tried to show that for the fundamental group $\pi\_1$ but was not able to do so.
Maybe this is a hard question like the one why two different points in a pathwise connected Hausdorff space can always be connected by an injective arc. I know where to find a proof for that fact, but is it frustrating that there seems to be not *simple* proof for that fact that seems obvious but is not.
By the way: I do not know if that defintion of a topological tree is already in use under a different name. I just took the notion of an $\mathbb R$-tree from metric geometry and removed the metric geometry out of the definition.
Another question would be if a *topological tree* is always 1-dimensional, using different notions of topological dimensions, but that should be a different question for a different day.
| https://mathoverflow.net/users/153400 | A topological tree is weakly contractible | Let $X$ be a a "topological tree" by your definition. Then $X$ is uniquely arcwise connected and Hausdorff. Let $f:S^n\to X$ be a map from the $n$-sphere where $n\geq 1$. It follows from the Hahn-Mazurkiewicz Theorem that the image $f(S^n)$ is a uniquely arcwise connected Peano continuum. This is equivalent to being a dendrite and [dendrites are contractible](https://wildtopology.com/2019/08/06/shape-injectivity-of-the-hawaiian-earring-part-ii/). Thus $f$ contracts in $f(S^n)$ and it follows that $\pi\_n(X)$ is trivial for all $n\geq 0$.
The results I'm using here are part of "Continuum Theory." Nadler's book has a great chapter on dendrites.
| 8 | https://mathoverflow.net/users/5801 | 436864 | 176,521 |
https://mathoverflow.net/questions/436873 | 0 | We know that a Poisson distribution can be approximated by a binomial distribution. More exactly, let $(X\_{jn})\_{1\leq j \leq n}$ be a i.i.d. triangular array such that
$$P[X\_{jn}= 1 ] = p\_n = 1- P[X\_{jn}=0]$$
and:
1. $p\_n \to 0$ as $n \to \infty$;
2. $np\_n \to \lambda$ as $n \to \infty$
So we have the following convergence in distribution:
$$S\_n = \sum\_{j=1}^n X\_{jn} \overset{d}{\to} N= \sum\_{j=1}^N 1, \quad N \sim \hbox{Poisson}(\lambda)$$
Thus, if we want approximate $N\sim \hbox{Poisson}(\lambda)$ by a binomial distribution, we can set $X\_{jn} \sim \hbox{Bernoulli}(\lambda/n)$ or $S\_n \sim \hbox{Binomial}(\lambda/n , n)$
Now, given $(\xi\_j)\_{j=1}^\infty$ a i.i.d. sequence of random variables independent of $N$. Consider $Y = \sum\_{j=1}^N \xi\_j$. Something tells me that I can find a sum $S\_n$ that converges in distribution to $Y$:
$$S\_n \overset{d}{\to} Y = \sum\_{j=1}^N \xi\_j, \quad N \sim \hbox{Poisson}(\lambda)$$
Since $Y$ is infinitely divisible, so it there is a triangular array $(X\_{ij})$ such that $S\_n$ converges in distribution to $Y$, But I think I should adjust some weights in the summations: $S\_n= \sum\_{j=1}^n w\_j X\_{jn}$, where $X\_{jn} \sim \hbox{Bernoulli}(\lambda/n)$.
Is there any constructive way to express this sum $S\_n$?
| https://mathoverflow.net/users/478920 | Approximation of a random sum of random variables (infinitely divisible distribution) by a triangular array | $\newcommand\la\lambda$Let
$$S\_n:=\sum\_{j=1}^n\xi\_j X\_{j,n},$$
where the $\xi\_j$'s are iid random variables (r.v.'s) and, for each $n$, the $X\_{j,n}$'s are iid r.v.'s independent of $\xi\_j$'s and such that each $X\_{j,n}$ has the Bernoulli distribution with parameter $p\_n$.
Suppose that $n\to\infty$ and $np\_n\to\la$ for some real $\la>0$.
Then
\begin{equation\*}
S\_n\to Y:=\sum\_{j=1}^N\xi\_j \tag{1}\label{1}
\end{equation\*}
in distribution, where
$N$ is a Poisson r.v. with parameter $\la$ independent of the $\xi\_j$'s.
This follows easily by the method of characteristic functions: If $f(t):=Ee^{it\xi\_1}$ for real $t$, then
\begin{equation\*}
Ee^{itS\_n}=(Ee^{it\xi\_1 X\_{1,n}})^n
=(1-p\_n+p\_n f(t))^n\to e^{\la(f(t)-1)} \tag{2}\label{2}
\end{equation\*}
and
\begin{equation\*}
Ee^{itY}=\sum\_{n=0}^\infty P(N=n)f(t)^n
=\sum\_{n=0}^\infty \frac{\la^n}{n!}\,e^{-\la}f(t)^n
=e^{\la(f(t)-1)}, \tag{3}\label{3}
\end{equation\*}
so that $Ee^{itS\_n}\to Ee^{itY}$.
The result you quoted in the beginning of your post is the special case of \eqref{1} with $\xi\_j=1$ for all $j$.
---
**Details on \eqref{2}:**
\begin{equation\*}
\begin{aligned}
Ee^{it\xi\_1 X\_{1,n}}&=Ee^{it\xi\_1 X\_{1,n}}\,1(X\_{1,n}=0)+Ee^{it\xi\_1 X\_{1,n}}\,1(X\_{1,n}=1) \\
&=E1(X\_{1,n}=0)+Ee^{it\xi\_1}\,1(X\_{1,n}=1) \\
&=1-p\_n+Ee^{it\xi\_1}\,E1(X\_{1,n}=1) \\
&=1-p\_n+f(t) p\_n.
\end{aligned}
\end{equation\*}
Here we used the equality $Ee^{it\xi\_1}\,1(X\_{1,n}=1)=Ee^{it\xi\_1}\,E1(X\_{1,n}=1)$, which holds because $\xi\_1$ and $X\_{1,n}$ are independent.
**Details on \eqref{3}:**
\begin{equation\*}
\begin{aligned}
Ee^{itY}&=\sum\_{n=0}^\infty E1(N=n)e^{itY} \\
&=\sum\_{n=0}^\infty E1(N=n)\exp\Big(it\sum\_{j=1}^n\xi\_j\Big) \\
&=\sum\_{n=0}^\infty E1(N=n)\,E\exp\Big(it\sum\_{j=1}^n\xi\_j\Big) \\
&=\sum\_{n=0}^\infty P(N=n)f(t)^n.
\end{aligned}
\end{equation\*}
Here we used the equality $E1(N=n)\,\exp\big(it\sum\_{j=1}^n\xi\_j\big)=E1(N=n)\,E\exp\big(it\sum\_{j=1}^n\xi\_j\big)$, which holds because the $\xi\_j$'s and $N$ are independent.
| 1 | https://mathoverflow.net/users/36721 | 436882 | 176,527 |
https://mathoverflow.net/questions/436832 | 1 | Let $f:\mathbb{P}^3\dashrightarrow\mathbb{P}^2$ be a dominant rational map defined over a field $k$ (not necessarily algebraically closed) of characteristic zero.
Consider a resolution $\widetilde{f}:X\rightarrow\mathbb{P}^2$ of $f$ and assume that a general fiber of $\widetilde{f}$ is the strict transform of a conic in $\mathbb{P}^3$ while $S = \widetilde{f}^{-1}([1:0:0])$ is a surface.
Under these hypotheses can we say something on the birational type of $S$? For instance assuming that $S$ has a point defined over the base field $k$ how far can $S$ be from being rational over $k$?
| https://mathoverflow.net/users/14514 | Geometry of contracted divisors | The surface $S$ is uniruled. Indeed, considering $X$ as a family of conics in $\mathbb{P}^3$ parameterized by an open subset of $\mathbb{P}^2$, we obtain a rational map
$$
\phi \colon \mathbb{P}^2 \dashrightarrow
\mathrm{Hilb}(\mathbb{P}^3)
$$
to the Hilbert scheme of conics in $\mathbb{P}^3$. Resolving this map by an appropriate blowup $\pi \colon Y \to \mathbb{P}^2$, we obtain a morphism
$$
\tilde\phi \colon Y \to \mathrm{Hilb}(\mathbb{P}^3).
$$
Moreover, if $Z \subset Y \times \mathbb{P}^3$ is the corresponding family of conics, the projection $\pi$ induces a surjective morphism $Z \to X$, such that
$$
Z\_0 := Z \times\_Y \pi^{-1}(0)
$$
dominates $S = X \times\_{\mathbb{P}^2} \{0\}$. So, it remains to note that $Z\_0 \to \pi^{-1}(0)$ is a family of conics.
| 2 | https://mathoverflow.net/users/4428 | 436889 | 176,529 |
https://mathoverflow.net/questions/436879 | 1 | Let $f:\mathbb R^n \to \mathbb R$ and $E := \{x \in X : f \text{ not Fréchet differentiable at }x\}$. Then $E$ [is Borel measurable](https://math.stackexchange.com/questions/3307418/let-f-mathbbrn-rightarrow-mathbbr-is-the-set-of-points-at-which-f). It is well-known that
>
> [Theorem](https://www.pmf.ni.ac.rs/filomat-content/2017/31-18/31-18-27-4617.pdf) If $f$ is convex, then the Hausdorff dimension of $E$ is at most $n-1$.
>
>
>
I would like to ask for a reference of the following statement, i.e.,
>
> If $f$ is locally Lipschitz, then the Hausdorff dimension of $E$ is at most $n-1$.
>
>
>
My closest search is [the following](https://pagine.dm.unipi.it/alberti/ricerca/2010-12/acp-icm2010.pdf)
>
> **Definition 1.2.** A set $E \subset \mathbb{R}^n$ is porous at a point $x \in E$ if there is a $c>0$ and there is a sequence $y\_n \rightarrow 0$ such that the balls $B\left(x+y\_n, c\left|y\_n\right|\right)$ are disjoint from $E$. The set $E$ is porous if it is porous at each of its points, and it is called $\sigma$-porous if it is a countable union of porous sets.
>
>
> **Theorem 1.3.** Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a Lipschitz function. Then the set of those points at which $f$ is not differentiable but it is **(directionally) differentiable** in $n$ linearly independent directions is $\sigma$-porous.
>
>
>
We have the Hausdorff dimension of a $\sigma$-porous subset of $\mathbb R^n$ [is at most $n-1$](https://www.pmf.ni.ac.rs/filomat-content/2017/31-18/31-18-27-4617.pdf). However, the Lipschitz function in **Theorem 1.3.** has one more restriction, i.e., it must be directionally differentiable in $n$ linearly independent directions.
Could you elaborate on such a reference?
| https://mathoverflow.net/users/99469 | Hausdorff dimension of the non-differentiability set of a locally Lipschitz function | You can't find a reference because it's false. Rademacher's theorem (Lebesgue-almost everywhere differentiability) is the best one can do.
In fact, *for every Lebesgue-null set $E \subset \mathbf{R}$, you can construct a Lipschitz function $f: \mathbf{R} \to \mathbf{R}$ that is not differentiable at any point of $E$.* [ACP10]
The reference for this is Theorem 1.1 in... the same paper you linked in your question. The result is on page 1. I don't think they give a proof there, but elsewhere, Preiss gives the argument in some [notes](https://homepages.warwick.ac.uk/%7Emasfay/talks/helsinki_07_08_slides.pdf) from a talk given in Helsinki. The proof basically goes as follows.
*Proof.*
Construct some nested sequence of open sets $\mathbf{R} = G\_0 \supset G\_1 \supset \cdots \supset E$ that are rapidly shrinking: for every connected component $C$ of $G\_k$, the next open set has
\begin{equation}
\lvert G\_{k+1} \cap C \rvert \leq 2^{-k-1} \lvert C \rvert.
\end{equation}
(Let us say additionally that $\lvert G\_1 \rvert \leq 1$.) You can find such a sequence because the Lebesgue measure is outer regular.
The sets $(G\_k \setminus G\_{k+1} \mid k \in \mathbf{N})$, together with $E$, partition the real line, and we define the function
$\psi: x \mapsto (-1)^k$ if $x \in G\_k \setminus G\_{k+1}$. This is bounded and measurable. (We need not define $\psi$ on $E$ as it is a null set.)
Then the desired function is $f: x \in \mathbf{R} \to \int\_0^x \psi$. This is Lipschitz, but not differentiable at any point of $E$.
Essentially, the reason is the following:
take an arbitrary point $x \in E$, and let, for each $k \in \mathbf{N}$, $(a\_k,b\_k)$ be the connected component of $G\_k$ containing $x$. Then the derivative of $f$ on $(a\_k,b\_k)$ is equal to $(-1)^k$ on a large portion of the interval. Certainly this is the case on $(a\_k,b\_k) \setminus G\_{k+1}$, and therefore
\begin{equation}
\lvert f(b\_k) - f(a\_k) - (-1)^k (b\_k - a\_k) \rvert
\leq 2 \lvert (a\_k,b\_k) \setminus G\_{k+1} \rvert
\leq 2^{-k} (b\_k - a\_k).
\end{equation}
If you now repeat the same calculation for the next term, you find that
\begin{equation}
\lvert f(b\_{k+1}) - f(a\_{k+1}) - (-1)^{k+1} (b\_{k+1} - a\_{k+1}) \rvert \leq 2^{-k-1} (b\_{k+1} - a\_{k+1}).
\end{equation}
If you divide this equation through by $b\_{k+1} - a\_{k+1}$, respectively the previous equation through by $b\_k - a\_k$, you will see that the two combined are incompatible with differentiability of $f$ at $x$. Q.E.D.
[ACP10] G. Alberti, M. Csörnyei, and D. Preiss. *Differentiability of lipschitz functions, structure of null sets, and other problems.* In Proceedings of the ICM 2010, pages 1379-1394.
| 4 | https://mathoverflow.net/users/103792 | 436892 | 176,531 |
https://mathoverflow.net/questions/348751 | 6 | Let $(M,g)$ be a Riemannian manifold such that for each $C>0$ there is $p\in M$ and $X,Y\in T\_pM$ unitary such that $K(X,Y) > C.$ Does this imply that the diameter of $(M,g)$ is infinite?
I just have an intuition about it, for example, by the neck singularity on Ricci flow, or by looking to the Gabriel's Horn: [Gabriel](https://www.daviddarling.info/encyclopedia/G/Gabriels_Horn.html).
I searched a lot for a counter-example and possible known theorem's on Petersen's book and other references, but I could conclude nothing, does anyone has a clue?
| https://mathoverflow.net/users/94097 | Unbounded sectional curvature implies infinite diameter? | If $(M,g)$ is complete, then yes. Since sectional curvature is continuous, if it is unbounded then $(M,g)$ is not compact, and by the contrapositive of the Hopf-Rinow theorem, a complete non-compact Riemannian manifold has infinite diameter.
If $(M,g)$ is not complete, then no. You can take something like $(0,1)^n$ and modify it with little dimples that become smaller but sharper as you approach the boundary, so that their sectional curvature becomes arbitrarily large but the manifold's diameter does not increase significantly.
| 7 | https://mathoverflow.net/users/4832 | 436894 | 176,532 |
https://mathoverflow.net/questions/436785 | 6 | I am interested in *Lagrangian correspondences* in the context of symplectic manifolds, namely Lagrangian submanifolds $L\_{12}$ of $M\_1\times \bar M\_2$ where $M\_1$ and $M\_2$ are symplectic manifolds with symplectic forms $\omega\_1$ and $\omega\_2$, and $\bar M\_2$ has its symplectic form reversed (so the symplectic form on the product is written $\omega\_1-\omega\_2$, if we omit tedious pullbacks).
As experts know better than I, these correspondences are to be thought of as morphisms in a putative symplectic category. Unfortunately, producing a correspondence $L\_{13}$ given correspondences $L\_{12}$ and $L\_{23}$ is not obvious, unless the latter two intersect transversally.
As of ~10 years ago, there were a few candidate approaches to resolve the issue, which were summarised by Weinstein here:
*Weinstein, Alan*, [**Symplectic categories**](https://doi.org/10.4171/PM/1866), Port. Math. (N.S.) 67, No. 2, 261-278 (2010). [ZBL1193.53173](https://zbmath.org/?q=an:1193.53173).
**What is the state of the art today?** Has there been any progress in finding a useful and versatile definition that does not suffer from (as Weinstein calls it) this nontransversality problem?
| https://mathoverflow.net/users/22757 | Progress on composition of Lagrangian correspondences/definition of symplectic categories? | Though not about solving the nontransversality problem, Fukaya's paper [Unobstructed immersed Lagrangian correspondence and filtered A infinity functor](https://arxiv.org/abs/1706.02131) is the state of the art in why nontransversality isn't a problem for most interesting purposes. It's been known for a long time that generic compositions are immersed, so Fukaya just works with immersed Lagrangians equipped with bounding cochains and shows that this data can be carried through a correspondence to give an $A\_\infty$-functor on Fukaya categories.
You should also check out the work of Nate Bottman, much of which is motivated by trying to construct higher categorical versions of Weinstein's category. The survey article by Abouzaid and Bottman should give you a very up to date picture of the field: [Functoriality in categorical symplectic geometry](https://arxiv.org/abs/2210.11159).
| 5 | https://mathoverflow.net/users/10839 | 436905 | 176,536 |
https://mathoverflow.net/questions/436874 | 4 | In short: Baire 2 functions are often assumed to be given by a double sequence of continuous functions, thought this is not the exact definition. Does one need the Axiom of Choice (or related) to connect these two definitions?
Longer version: we are working over the real numbers. As is well-known, Baire 0 functions are the continuous ones, while Baire $n+1$ functions are the pointwise limits of Baire $n$ functions.
I call a function $f$ from reals to reals *effectively Baire 2* in case it is the double limit of a double sequence $(f\_{n,m})\_{n, m \in \mathbb{N}}$, i.e. for all reals $x$, we have
$$
(\forall \epsilon>0)(\exists n'\in \mathbb{N})(\forall n> n')(\exists m'\in \mathbb{N})(\forall m>m')( |f\_{m,n}(x)-f(x)|<\epsilon). \qquad (\*)
$$
As pointed out by Baire himself already, Baire 2 functions can be *represented* by effectively Baire 2 functions, though Baire did not use the latter terminology.
The notion (\*) is essentially the definition of Baire 2 used in reverse mathematics/second-order arithmetic.
My question is then whether one needs the Axiom of Choice (or related) to show that for a general Baire 2 function, there is a double sequence satisfying (\*).
| https://mathoverflow.net/users/33505 | The difference between Baire 2 and 'effectively Baire 2' | It's provable in ZF that every Baire-2 function is effectively Baire-2. It suffices to prove the following:
(ZF) There is an explicit function which maps each Baire-1 function $f: \mathbb{R} \rightarrow \mathbb{R}$ to a sequence of rational polynomials that pointwise converge to it.
The first step is to construct a sequence of reals $\langle r\_n \rangle$ such that each $r\_n$ codes a rational-valued function $f\_n$ such that $|f\_n(x)-f(x)| \le \frac{1}{n}$ for all $x \in \mathbb{R}.$ We'll construct $r\_1,$ which immediately generalizes to other $n.$
Let $U\_k$ enumerate the basic open sets. We use transfinite recursion to define a descending sequence of closed sets $\langle C\_{\alpha}: \alpha<\omega\_1 \rangle$:
1. $C\_0=\mathbb{R},$
2. $C\_{\alpha+1}=C\_{\alpha} \setminus \bigcup \{U\_k: \forall x, y \in C\_{\alpha} \cap U\_k (|f(x)-f(y)|\le 1)\},$
3. For limit $\alpha,$ $C\_{\alpha}=\cap\_{\xi<\alpha} C\_{\xi}.$
If $C\_{\alpha} \neq \emptyset,$ then $f \restriction\_{C\_{\alpha}}$ is continuous at some $x \in C\_{\alpha}$ by Baire's Characterization Theorem. Then $x \not \in C\_{\alpha+1},$ so $C\_{\alpha+1} \subsetneq C\_{\alpha}.$ Define a surjective partial map $g: \omega \rightharpoonup\beta = \{\alpha: C\_{\alpha} \neq \emptyset\}$ by sending $k$ to the greatest $\alpha$ such that $U\_k \cap C\_{\alpha} \neq \emptyset,$ and a well-founded partial ordering $(\omega, \prec)$ by $i \prec j$ if $g(i) < g(j).$ Define $h: \text{dom}(g) \rightarrow \mathbb{Z}$ by $h(k)=\lfloor \sup\_{U\_k \cap C\_{g(k)}} f \rfloor.$
Define $f\_1: \mathbb{R} \rightarrow \mathbb{Z}$ as follows: for given $x,$ let $j=\min\{k<\omega: x \in U\_k \wedge \forall \alpha (x \in C\_{\alpha} \leftrightarrow U\_k \cap C\_{\alpha} \neq \emptyset)\}.$ Set $f\_1(x) = h(j).$
Let $r\_1$ be a real which encodes $(\omega, \prec)$ and $h.$ From $(\omega, \prec),$ one can determine the sequence $\langle C\_{\alpha}\rangle$ and hence $g.$ Thus, $f\_1$ can be determined from $r\_1.$ This completes the construction of $r\_1$ (and hence all $r\_n$).
Let $r$ encode $\langle r\_n \rangle$ (and thus $f$). By Shoenfield absoluteness relativized to $r,$ we have in $L[r]$ that $r$ encodes a Baire-1 function $\hat{f}.$ Enumerate the rational polynomials by $\langle q\_n \rangle.$ A Stone-Weierstrass argument shows that there is $\langle n\_k \rangle$ such that $\langle q\_{n\_k} \rangle$ converges pointwise to $\hat{f}$ in $L[r].$ Let $\langle n\_k \rangle$ be the $L[r]$-least such sequence. Applying Shoenfield absoluteness upwards relative to a real encoding $(r, \langle n\_k \rangle),$ we see in $V$ that $\langle q\_{n\_k} \rangle$ converges pointwise to the Baire-1 function coded by $r,$ namely $f.$ This completes the construction.
Some additional remarks: The above construction can be carried out in $Z\_2.$ I don't know enough about subsystems of analysis to say more with confidence. By analyzing how much of $L[r]$ is really needed in the above, you can probably get this down to $\Pi^1\_1-CA\_0.$
As I showed in the comments, it is not provable in ZF that every Baire-3 function is effectively Baire-3, since if $\mathbb{R}$ is a countable union of countable sets, then every indicator function is Baire-3.
| 5 | https://mathoverflow.net/users/109573 | 436916 | 176,539 |
https://mathoverflow.net/questions/436798 | 14 | Good morning,
I hope this question is not too far out of the scope of the forum. I am posting it here because this doesn't seem to be a very standard problem.
Yesterday we were calculating the equivalent resistances of various polyhedra between adjacent vertices and we noticed this:
Consider this bijection $f$ between edges of a polyhedron and its dual: if edge $AB$ on polyhedron $X$ separates faces $\alpha$ and $\beta$, $f(AB)$ on polyhedron $X\star$ connects the dual vertices of faces $\alpha$ and $\beta$.
Now construct a circuit of resistances all equal to $1$, each occupying an edge of polyhedron $X$ and do the same for $X\star$. We noticed that the equivalent resistance across $AB$ on $X$ and $f(AB)$ on $X\star$ sum to $1$ in all cases we calculated.
We could prove this works for platonic solids but the proof relies on too many symmetries to be easily generalised. However, we calculated this to work on a few other solids (EDIT: all pyramids and prisms and a few Archimedean and Catalan solids) which do not possess the same symmetries.
Is this true in general? If not, for what classes of solids is it true?
| https://mathoverflow.net/users/472669 | Dual polyhedra and electric circuits | Here is the [proof](https://www.ejgta.org/index.php/ejgta/article/view/657) for any planar graph.
| 3 | https://mathoverflow.net/users/472669 | 436921 | 176,541 |
https://mathoverflow.net/questions/436920 | 0 | I would like to extend (cubic or higher degrees) [spline wavelets](https://en.wikipedia.org/wiki/Spline_wavelet) to complex domain. First, does this continuation exist? Second, I appreciate it if anyone could point me to some references.
| https://mathoverflow.net/users/478084 | Analytic continuation of spline wavelets (reference request) | An overview with pointers to the literature is [Complex B-splines](https://www.ee.cuhk.edu.hk/~tblu/monsite/pdfs/forster0601.pdf), by Forster, Blu, and Unser; see page 262.
One recent application is
[Parameter characterization of complex wavelets and its use in 3D reconstruction,](https://link.springer.com/chapter/10.1007/978-3-319-62434-1_38) by Lopez et al.
| 0 | https://mathoverflow.net/users/11260 | 436926 | 176,542 |
https://mathoverflow.net/questions/436876 | 4 | Let $f\in \mathbb{R}(x\_1,\ldots,x\_n)$ be a rational function. Suppose that $f$ is continuous on $\mathbb{R} ^n$. Must it be Lipschitz on the unit ball?
This question might be related to [Are continuous rational functions arc-analytic?](https://mathoverflow.net/questions/278008/are-continuous-rational-functions-arc-analytic)
| https://mathoverflow.net/users/4690 | Is a continuous rational function Lipschitz? | Consider the polynomial
$p(x,y)=(y^3-x^5)^2+(y-x^2)^8$ in the neighborhood of $(0,0)$. Apart from the strip $y^3/x^5\in(1/2,2)$, it is bounded from below by $C(x^2+y^2)^5$; within the strip it is bounded by $C|x|^{40/3}$, and this estimate is sharp. So the function
$$
\frac{(x^2+y^2)^7}{p(x,y)}
$$
has a continuous extension, but on the curve $y^3=x^5$ it behaves like $x^{2/3}(1+o(1))$, hence it is non-Lipschitz in any neighborhood of $(0,0)$.
| 4 | https://mathoverflow.net/users/17581 | 436927 | 176,543 |
https://mathoverflow.net/questions/436925 | 18 | When writing a paper, it's possible that some auxiliary results hold in more generality or in a stronger version than what's actually needed to prove the main results of the article. And so here comes the question:
**Should one state and prove the exact auxiliary result that is used, or should one sharpen it to its best possible version?**
I can think of pros and cons of both approaches: Proving better results cannot be a bad thing in itself, but spending time proving a too strong and not-so-interesting Lemma might be distracting and not worth the effort. Even if it's not so hard to improve the Lemma it might be confusing to the reader to use a weaker version of what's stated.
### Example:
Suppose I need to use a Lemma of the form:
>
> *For every $\varepsilon>0$ there exists a sequence $(x\_n)\_n$ with property $(P)$ such that $|x\_n|<\varepsilon$ for all $n\in\mathbb{N}$.*
>
>
>
However, looking at the proof of this Lemma I (and most likely the referee and the reader) noticed that slightly changing the proof a stronger version holds:
>
> *For every sequence of positive numbers $(\varepsilon\_n)\_n$ there exists a sequence $(x\_n)\_n$ with property $(P)$ such that $|x\_n|<\varepsilon\_n$ for all $n\in\mathbb{N}$.*
>
>
>
Which version should I include if I only need the first (and weaker) statement?
| https://mathoverflow.net/users/123450 | Should one state the sharpest version of a Lemma even if only a weaker version is needed? | It depends on context. Here are some relevant considerations: How much more difficult is the stronger Lemma to prove? If the proof is nearly identical, then stating the strongest one may be a good idea. But it may not be helpful to spend a lot of time on making a minor Lemma slightly stronger if it takes a lot of effort and distracts from the main exposition. One thing I've occasionally done is included two versions of a Lemma, and explicitly called a stronger one a "Proposition" and make clear that it is a separate question of just how far we can push the Lemma. (Which may be an interesting question in its own right.)
One thing to also keep in mind is that at least stating the strongest version may help others in two other ways. First, if they extend, generalize, or improve your result, having a stronger version of the Lemma may also be helpful. Second, if someone is trying to understand what the limiting steps are in your main result (e.g. why does this only apply to p-groups but not nilpotent groups, or why can some general inequality not be strengthened, etc.) then having a Lemma which is stronger than you need it can help understand that that Lemma is not the where whatever obstruction there is to making the result stronger.
Another consideration is that a stronger Lemma may benefit you later. If you come back to a problem years later, you might not remember the stronger version, or might remember it but might not remember the proof. So if you decide not to include the proof in the final paper, it may be a good idea to keep a written up version in your own copy, or possibly just commented out in the LaTeX.
| 23 | https://mathoverflow.net/users/127690 | 436928 | 176,544 |
https://mathoverflow.net/questions/436918 | 1 | I'm reading about *subdifferentiable function* at page 232 of Villani's *Optimal Transport: Old and New*.
---
**Definition 10.5** (Subdifferentiability, superdifferentiability). Let $U$ be an open set of $\mathbb{R}^n$, and $f: U \rightarrow \mathbb{R}$ a function. Then:
* (i) $f$ is said to be *subdifferentiable* at $x$, with subgradient $p$, if
$$
f(z) \geq f(x)+\langle p, z-x\rangle+o(|z-x|) .
$$
The convex set of all subgradients $p$ at $x$ will be denoted by $\nabla^{-} f(x)$.
* (ii) $f$ is said to be *uniformly subdifferentiable* in $U$ if there is a continuous function $\omega: \mathbb{R}\_{+} \rightarrow \mathbb{R}\_{+}$, such that $\omega(r)=o(r)$ as $r \rightarrow 0$, and
$$
\forall x \in U \quad \exists p \in \mathbb{R}^n ; \quad f(z) \geq f(x)+\langle p, z-x\rangle-\omega(|z-x|).
$$
* Corresponding notions of *superdifferentiability* and *supergradients*
are obtained in an obvious way by just reversing the signs of the inequalities.
---
**My understanding** I think $o(|z-x|)$ is the Landau symbol, i.e., $g (x) = o(|z-x|)$ means that $\lim\_{z \to x} \frac{g(x)}{|z-x|} = 0$.
>
> Could you explain what $f(z) \geq f(x)+\langle p, z-x\rangle+o(|z-x|)$ means?
>
>
>
I'm not sure if it means that the limit
$$
\lim\_{z\to x} \frac{f(z) - f(x)-\langle p, z-x\rangle}{|z-x|}
$$
exists and is non-negative.
| https://mathoverflow.net/users/99469 | What does Landau symbol mean in an inequality? | Landau $o(\cdot)$ notations should be interpreted in inequalities as inferior/superior limits.
In this case in particular, it $f(z)\ge f(x)+⟨p,z−x⟩+o(|z−x|)$ is equivalent to
$$
\liminf\_{z \to x} \frac{f (z) - f (x) - \langle p, x - z\rangle}{\vert z - x\vert} \ge 0;
$$
the corresponding limit does not need to exist.
| 2 | https://mathoverflow.net/users/42047 | 436930 | 176,545 |
https://mathoverflow.net/questions/436880 | 6 | **Background:**
A tantalizing conjecture of Lovasz is the following:
>
> Let $G$ be a (finite) connected vertex-transitive graph. Then $G$ contains a Hamiltonian cycle or is one of $5$ counter-examples.
>
>
>
(technically, Lovasz conjectured something about finding Hamiltonian paths in such $G$, but this strengthening has been raised in later literature)
On the [Wikipedia article](https://en.wikipedia.org/wiki/Lov%C3%A1sz_conjecture#Special_cases) for this conjecture, it is said that it is still open whether all connected Cayley graphs of dihedral groups contain Hamiltonian cycles, but it is known for special cases of generating sets. No reference is given for what "special cases" are known, so I'm curious if the following is open.
**Question:**
Let $D$ be a (finite) dihedral group, and let $H\le D$ be its subgroup of rotations. Suppose $S\subset D\setminus H$ generates $D$. Does the Cayley graph $\Gamma(D,S)$ have a Hamiltonian cycle?
I think this special case is nice, because with a bit of clever relabelling, one can show that our Cayley graph $\Gamma$ is closely related to a weakly-connected directed Cayley graph $\Gamma'$ defined over $H$, which allows us to conclude that $\Gamma$ has a Hamiltonian *path* due to the fact that $\Gamma'$ has a directed Hamiltonian path.
| https://mathoverflow.net/users/130484 | Lovasz's conjecture for dihedral Cayley graphs | It turned out to be such a long commentary.
Here is what is known about Hamiltonian cycles of dihedral groups:
0. **Conjecture.**
Every connected Cayley graph on a dihedral group has a Hamilton
cycle (W. Holszty´nski and R. F. E. Strube, 1978).
1. If $p$ is a prime, then every cayley graph dihedral group $D\_{p}$ is hamiltonian (W. Holszty´nski and R. F. E. Strube, 1978).
2. All cubic cayley graphs over dihedral groups are hamiltonian (Brian Alspach and Cun-quan Zhang, 1989).
3. If $X$ is a connected Cayley graph on the dihedral group $D\_n$, $n$ even, then
$X$ has a Hamilton cycle (Brian Alspach, C. C. Chen, Matthew Dean, 2010).
An interesting story about this problem happened in 2018.
H. Zhou and B.Xia
posted a short article in the arXiv (arXiv:1810.13311) in which they claimed to prove that
every connected Cayley graph on a generalized dihedral group has a
Hamilton cycle.
For a nontrivial abelian group A the generalized dihedral group $\operatorname{Dih}(A)$
of $A$ is the semidirect product of $A$ by $Z\_2$ with $Z\_2$ acting on $A$ by inverting elements.
But as of today, that article has been removed from the arXiv.
| 5 | https://mathoverflow.net/users/173068 | 436931 | 176,546 |
https://mathoverflow.net/questions/436899 | 1 | I was trying to understand the behaviour of the primitive equality (=) in the axiomatization of category, which takes morphisms as primitives and objects as derivatives in bijection to identity morphisms, and based on the definitions I have found (which are all in the same vein), it seems that trying to set up two different automorphisms over one derived object A is impossible because they necessarily collapse to the identity morphism determining A via the idempotence axioms 1) and 2) below. This would mean that to introduce multiple automorphisms, which we need for example for groups characterized as one object categories, structure is essential. And if we want structure in purely categorical terms, we must move into a higher-order category, in order to enable this construction. So we might take a category of categories, where functors encode the structure of source into target in the obvious way.
Is there any way to set up the definition of category with derived objects so that there can exist multiple different automorphisms, and how would this be achieved? It might not be necessary for practical purposes, but as a consequence of this setup, the elementary equality check x ?= y seems to behave oddly due to axiom 5) - interacts with different levels of morphisms as though they were the same.
According to nLab definition <https://ncatlab.org/nlab/show/single-sorted+definition+of+a+category#references>
A category (single-sorted version) is a collection C, whose elements are called morphisms, together with two functions s,t:C→C and a partial function ∘:C×C→C, such that:
1. s(s(x))=s(x)=t(s(x))
2. t(t(x))=t(x)=s(t(x))
3. x∘y is defined if and only if s(x)=t(y).
4. If x∘y is defined, then s(x∘y)=s(y) and t(x∘y)=t(x).
5. x∘s(x)=x and t(x)∘x=x (both composites are always defined, because of the first two axioms)
6. (x∘y)∘z=x∘(y∘z), if either is defined (in which case the other is defined by the axiom 3).
| https://mathoverflow.net/users/496639 | Is it possible to set up multiple automorphisms over a structureless object inside single-sort defined category? | You might want to think about a concrete example: let's for a moment take $C$ to be a monoid. How would the 1-sorted definition work?
Let's be more precise: let $(M, e, \cdot)$ be a monoid (with carrier set $M$, identity element $e$ and binary operation $\cdot$). We want to create a category (and present it as a 1-sorted theory, as in the nlab page you linked) such that *composition* $\circ$ corresponds to the monoid operation $\cdot$.
First of all, notice how $x \cdot x$ is defined for all $x \in M$, so $s(x) = t(x)$ and hence (since our choice of $x$ was arbitrary) $s = t$: this is because of axiom $3$. But we also know that $x \cdot y$ is defined for all $x, y \in M$: this, together with the previous fact, forces $s(x) = s(y)$ (and likewise for $t$): the two function are also *constant*. To reiterate, all of this follows from axiom $3$ *alone*.
But at this point, axioms $1$ and $2$ are trivially satisfied, as well as axiom $4$. Axiom $6$ also holds, but this time because monoids are associative by definition.
The only thing left to figure out is what the (only) element $a$ in the image of $s$ (and hence, of $t$) is: to figure this out, we need to look at axiom $5$. After applying everything we already know about $s$ and $t$, axiom $5$ becomes
$$\forall x (x \cdot a = x = a \cdot x)$$
As you probably know, this is the defining property of the identity element $e$: we have now proven that $s(x) = t(x) = e$ for every $x \in M$ (up to proving that every monoid has a unique identity element, a well-known and easy-to-prove fact).
You might notice that never, in the discussion above, have we ever implied that all monoids are the trivial monoid $\{\*\}$: in fact, the example we've been looking at is the prototypical 1-object category with (possibly) multiple automorphisms.
I can't tell what got you to think that if $s(f) = t(f)$ then $f$ is an identity; this is of course *not* the case. I can only guess: maybe you thought that if $s(f) = t(f)$, the fact that axiom $1$ and $2$ become the same statement somehow implies that $s(f) = t(f) = f$; if that's the case, it clearly isn't true.
Hopefully this clears things up a bit.
| 3 | https://mathoverflow.net/users/111265 | 436936 | 176,548 |
https://mathoverflow.net/questions/436941 | 5 | Let $\mathbb{F}$ be a non-Archimedean local field. Let $\{T\_a\}\_{a=1}^\infty$ be a sequence of linear operators $\mathbb{F}^n\to\mathbb{F}^n$ of rank $n$. After a choice of subsequence, is it possible to construct sequences of vectors $\{e^i\_a\}\subset \mathbb{F}^n$, $i=1,\dots, n,$ such that the following properties are satisfied:
(1) for any $a$ the vectors $e^1\_a,\dots, e^n\_a$ form a basis of $\mathbb{F}^n$ and when $a\to \infty$ they converge to another basis of $\mathbb{F}^n$;
(2) the vectors $T\_a(e^1\_a),\dots, T\_a(e^n\_a)$ also form a basis of $\mathbb{F}^n$ and when $a\to \infty$ appropriate multiples of these vectors converge to another basis of $\mathbb{F}^n$?
Remark. In the Archimedean case the answer is positive. Indeed fix a Euclidean (resp. Hermitian) metric. $T\_a$ admits an orthonormal basis $e^1\_a,\dots, e^n\_a$ such that $T\_a(e^1\_a),\dots, T\_a(e^n\_a)$ are pairwise orthogonal vectors. After a choice of subsequence all the required properties will be satisfied.
| https://mathoverflow.net/users/16183 | A question on linear algebra over non-Archimedean local field | $\def\FF{\mathbb{F}}\def\GL{\text{GL}}$This is basically the same thing YCor sketches in his comment: Let $R$ be the ring of integers of $\FF$. Let $K$ be the group of invertible matrices with entries in $R$ whose inverses are also in $R$, and let $T$ be the group of diagonal matrices.
The ring $R$ is a dvr so, by the [Smith normal form theorem](https://en.wikipedia.org/wiki/Smith_normal_form), each of your matrices $T\_a$ can be factored as $U\_a D\_a V\_a$ with $D\_a \in T$ and $U\_a$, $V\_a \in K$.
The group $K$ is compact in the non-archimedean topology. Proof: $K$ can be written as $\{ (g,h) : gh = \text{Id}\_n,\ g,h \in \text{Mat}\_{n \times n}(R) \}$. This is a closed subspace of $\text{Mat}\_{n \times n}(R)^2$, and $\text{Mat}\_{n \times n}(R)^2 \cong R^{2n^2}$ is obviously compact. So we can extract a subsequence where the $U\_a$ approach a limit $U$ and the $V\_a$ approach a limit $V$.
Then we take $(e\_a^1, e\_a^2, \ldots, e\_a^n)$ to be the columns of $V\_a^{-1}$, just as in the archimedean case.
I learned the slogan "Smith normal form is non-archimedean singular value decomposition" from Kiran Kedlaya.
| 6 | https://mathoverflow.net/users/297 | 436944 | 176,550 |
https://mathoverflow.net/questions/436952 | 0 | Let $G$ be a reductive group over $\mathbb{C}$ and $H\subseteq G$ a reductive subgroup. Let $\rho$ be a faithful irreducible finite dimensional representation of $G$ over $\mathbb{C}$. Assume that $\rho|\_H$ is reducible. Then is it always the case that the centralizer $Z\_G(H)$ is strictly larger than the center $Z(G)$?
| https://mathoverflow.net/users/32746 | Centralizer of a reductive subgroup | $\DeclareMathOperator\GL{GL}$No. Take $H=\GL\_2$ embedded diagonally into $G=\GL\_2\times \GL\_2$ and take
$\rho$ equal to $\mathbb C^2 \otimes (\mathbb C^2)^\*$ with the natural action of $\GL\_2$ on $\mathbb C^2$.
| 1 | https://mathoverflow.net/users/496679 | 436953 | 176,552 |
https://mathoverflow.net/questions/436655 | 1 | Suppose that $(M,X)$ is a simply connected complete Riemannian manifold with pinched sectional curvature between $[a,0]$. Let $r>0$ and fix any point $p\in M$. Is there a bound on the local Lipschitz constant of the Riemannian exponential map $\exp\_p$ restricted to the Euclidean ball at the origin of radius $r$, written in terms of these sectional curvature bounds?
| https://mathoverflow.net/users/36886 | Local Lipschitz constant of exponential map for Hadamard manifolds | Here is an argument that gives a sharp estimate.
Given $x \in M$, $v \in T\_xM$, and $(e\_1, \dots, e\_n)$ an orthonormal basis of $T\_xM$,
$$
(d\exp\_x(v))e\_k = J\_k(1),
$$
where $J\_k$ is the Jacobi field along the constant speed geodesic
$$(t) = \exp\_x(tv),\ 0 \le t \le 1,
$$
that satisfies $J\_k(0) = 0$ and $\nabla\_{c'}J\_k(0) = e\_k$. Therefore, it satisfies the Jacobi equation
$$
\nabla^2{c'c'}J\_k = R(c',J\_k)c'.
$$
Parallel transport $(e\_1, \cdots, e\_n)$ along the curve $c$ and let $J(t)$ be the matrix such that along $c$,
$$
J\_k = e\_jJ^j\_k
$$
From the Jacobi equation, it follows that
[
J'' + KJ = 0,
]
where $K$ is the symmetric matrix given by
$$
K\_k^j = e\_j\cdot R(e\_k,c')c'.
$$
If the sectional curvature is bounded from below by $-\kappa^2$, then
$$
K \ge -\kappa^2|v|^2 I,
$$
because $|c'| = |v|$.
Let $A = J'J^{-1}$, which satisfies
\begin{align\*}
A' + A^2 + K &= 0,\ 0 \le t \le 1\\
A(t) &= t^{-1}I + O(t)\\
A^T &= A.
\end{align\*}
As an aside, $A$ is the second fundamental form at $\exp\_xv$ of the geodesic sphere of radius $|v|$ centered at $x$ written with respect to the frame $(e\_1, \dots, e\_n)$.
If
$$
a(t) = \frac{\kappa|v|\cosh \kappa |v|t}{\sinh \kappa |v|t},
$$
then
\begin{align\*}
a' + a^2 - \kappa^2|v|^2 = 0.
\end{align\*}
If $B = A-aI$,
then a straightforward calculation shows that
\begin{align\*}
B' &\le -2aB.
\end{align\*}
It is easy to check that
\begin{align\*}
\lim\_{t\rightarrow 0} B(t) &= 0.
\end{align\*}
It follows that $B \le 0$, i.e., $A \le aI$., Therefore,
\begin{align\*}
|J|' &= \frac{J\cdot J'}{|J|} = \frac{JAJ}{|J|}
\le a|J|.
\end{align\*}
This implies that
$$
|J|(t)| \le \frac{\sinh \kappa|v|t}{\kappa |v|}.
$$
This gives the sharp estimate that if the sectional curvature is bounded from below by $-\kappa^2$, where $\kappa > 0$, then
$$
|d\exp\_x(v)| \le \frac{\sinh \kappa|v|}{\kappa|v|}.
$$
| 3 | https://mathoverflow.net/users/613 | 436959 | 176,555 |
https://mathoverflow.net/questions/436958 | 1 |
>
> **Question:**
>
>
> which $n$ and $k$ satisfy $\frac{k^n-1}{2^n-1}\in\mathbb{N}$?
>
>
>
The motivation for the question is a constraint on the cardinality of interpolation-constraints for the $2^n$ corners of a hypercube $[0,1]^n$:
* for one of the corners, e.g. $(1\_1,\,\dots,\,1\_n)$ there shall be exactly one constraint
* for all other corners the number of constraints shall be equal
$n=2; k=2m, k=6m\pm1$ are examples of valid solutions
| https://mathoverflow.net/users/31310 | Solutions to diophantine equation related to an interpolation problem on hypercubes | So we are trying to solve
$$k^n - 1 \equiv 0 \mod 2^n - 1 $$
I.E
$$ k^n \equiv 1 \mod 2^n - 1 $$
A class of solutions can be found by looking at the carmichael function. Namely
$$ k^{\lambda(2^n - 1)} \equiv 1 \mod 2^n - 1 $$
So we want that $\lambda(2^n - 1) = n$
Just going through this table: <https://en.wikipedia.org/wiki/Carmichael_function>
We see here that $\lambda(2^4 - 1) = 4$ so for any choice of $k>2$ coprime to 15 and $n=4$ this identity holds. So you can find some cool tricks like $$ \frac{11^4 - 1}{2^4 - 1} = 976$$
The next instance is $\lambda(2^6 - 1) = \lambda(63) = LCM(\lambda(3), \lambda(7)) = LCM(2,6) = 6 $
I wrote some code to find more instances, up to $n=50$ the only values are: 4, 6, 12. Found using the python 3.8 code below:
```
import math
prime_list = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,57,67,71,73,79,83,89,97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]
def lcm(a, b):
return a*b // math.gcd(a, b)
def carm_prime_power(prime, power):
if prime == 2 and power > 2:
return 2**(power - 2)
return (prime-1)*prime**(power - 1)
def carmichael(n):
factor_list = {}
for primes in prime_list:
if n%primes == 0:
factor_list[primes] = 1
n = n//primes
while n%primes == 0:
factor_list[primes] += 1
n = n//primes
current_lcm = 1
for primes in factor_list:
current_lcm = lcm(current_lcm, carm_prime_power(primes, factor_list[primes]))
return current_lcm
if __name__ == '__main__':
n = 4
while n < 50:
if carmichael(2**n - 1) == n:
print(n)
n+=1
```
When $n \ne \lambda(2^n -1)$ it probably gets a lot more complex to characterize solutions. If you're interested I can write some code to search for such solutions.
| 2 | https://mathoverflow.net/users/46536 | 436961 | 176,557 |
https://mathoverflow.net/questions/436945 | 3 | In the paper 'On the Holder continuity of solutions of second order elliptic equations in two variables' by Piccinini and Spagnolo, they prove the following estimate:
$$
\begin{array}{ll}
\left(\int\_S p\_{11} |u\_T|^2 \right)^{\frac12}\left(\int\_S \frac{\langle P(x) (u\_N,u\_T), (1,0) \rangle^2}{p\_{11}} \right)^{\frac12} &\leq & \sqrt{\frac{\Lambda}{\lambda}} \left(\int\_S \lambda |u\_T|^2 \right)^{\frac12}\left(\int\_S \frac{\langle P(x) (u\_N,u\_T), (1,0) \rangle^2}{p\_{11}} \right)^{\frac12} \\
& \leq & \sqrt{\frac{\Lambda}{\lambda}}\frac12 \left(\int\_S \lambda |u\_T|^2 + \frac{(p\_{11} u\_N + p\_{12}u\_T)^2}{p\_{11}} \right)\\
& \leq & \sqrt{\frac{\Lambda}{\lambda}}\frac12 \left(\int\_S \left( p\_{22} - \frac{p\_{12}^2}{p\_{11}}\right) |u\_T|^2 + \frac{(p\_{11} u\_N + p\_{12}u\_T)^2}{p\_{11}} \right)\\
& = & \sqrt{\frac{\Lambda}{\lambda}}\frac12 \left(\int\_S \langle P(x) (u\_N,u\_T),(u\_N,u\_T)\rangle \right)
\end{array}
$$
Here $P(x)$ is a symmetric positive definite $2\times2$ matrix with entries $(p\_{ij})$ and $\lambda |\xi|^2 \leq \langle P(x)\xi,\xi\rangle \leq \Lambda |\xi|^2$ and $(u\_N,u\_T)$ are the normal and tangential derivative with $S$ being the unit circle in $\mathbb{R}^2$.
The crucial part of the above calculation is the inequality $\lambda \leq \left( p\_{22} - \frac{p\_{12}^2}{p\_{11}}\right)$.
My question is the following: Is there a different way to prove this estimate without explicitly writing down the expressions and instead proceed using only matrix inequalities?
| https://mathoverflow.net/users/100801 | Matrix inequality in a paper by Piccinini-Spagnolo | Let $0<\lambda\_1\le\lambda\_2$ be the eigenvalues of the symmetric positive $2\times2$ matrix $P$. Then $$\lambda\le\lambda\_1=\min\_{|\xi|=1} \langle P\xi,\xi\rangle \le\langle Pe\_1,e\_1\rangle=p\_{11}\le \lambda\_2 =\max\_{|\xi|=1} \langle P\xi,\xi\rangle,$$
so $\lambda p\_{11}\le \lambda\_1\lambda\_2=\det P=p\_{11}p\_{22}-p\_{12}^2$, whence the inequality, since $\langle Pe\_1,e\_1\rangle=p\_{11}>0$.
| 4 | https://mathoverflow.net/users/6101 | 436964 | 176,558 |
https://mathoverflow.net/questions/436805 | 5 | Let $\rho:G\_{\mathbb{Q}}\rightarrow \mathbf{Gl}\_{n}(\mathbb{Q}\_{p})$. I would like to understand in depth why the local Langlands correspondence for $\rho\_{|\mathbb{Q}\_{p}}$ must consider $p$-adic representations instead of a complex representation. As far as I know, the $p$-adic representations
1. retrieve information that loses complex representations (like $\mathcal{L}$ invariant),
2. also allows for local-global correspondence.
Can someone give examples of 1. and 2. in $Gl\_{1}$ and $Gl\_{2}$ case?
| https://mathoverflow.net/users/169282 | Understand the $p$-adic local Langlands correspondence with examples | Let's look at the case of representations associated to modular forms. I'm going to switch the roles of $\ell$ and $p$, because I find $\ell$-adic Hodge theory disturbing; so I'm going to look at $\rho\_{f, \ell} |\_{G\_{\mathbb{Q}\_p}}$.
For $\ell \ne p$, we can attach a Weil--Deligne representation to $\rho\_{f, \ell} |\_{G\_{\mathbb{Q}\_p}}$ via Grothendieck abstract monodromy. This WD representation has coefficients in $\overline{\mathbb{Q}}\_\ell$, but the definition of a WD representation doesn't depend on the topology on the coefficient field, so we can transport it along a field isomorphism $\overline{\mathbb{Q}}\_\ell \cong \mathbb{C}$ to get a complex-valued Weil--Deligne representation. The local-global compatibility theorem (due to Carayol in this case) tells us that this WD representation is the one associated by local Langlands to the smooth complex $GL\_2(\mathbb{Q}\_p)$-representation $\pi\_{f, p}$ (the local factor of $f$ at $p$).
Now, there are multiple ways of extending this to cover $\ell = p$. One approach is the following: via Fontaine's functor $D\_{\mathrm{pst}}$, we can attach a Weil--Deligne representation to $\rho\_{f, p}|\_{G\_{\mathbb{Q}\_p}}$; and it is known (by a theorem of Saito, IIRC) that this Weil--Deligne representation (again, transported via a field isomorphism $\overline{\mathbb{Q}}\_p \cong \mathbb{C}$) is the one associated to $\pi\_p$ by local Langlands. So that's a meaningful statement of "local-global compatibility for $\ell = p$" which doesn't involve p-adic Banach spaces.
However, this formulation isn't quite the whole story, because:
* Fontaine's $D\_{\mathrm{pst}}$ functor only applies to a subclass of $p$-adic representations of $G\_{\mathbb{Q}\_p}$ (the de Rham ones); this includes all the ones from modular forms, but it misses lots of other interesting objects (e.g. representations associated to non-classical overconvergent eigenforms).
* Even when $V$ is de Rham, the Weil--Deligne representation associated to $V$ doesn't uniquely determine $V$ up to isomorphism, because it forgets the Hodge filtration, and in some cases there are multiple non-isomorphic choices of filtrations for a given $V$. (This is precisely what the $\mathcal{L}$-invariant parametrises.)
This motivates the formulation of $p$-adic Langlands in terms of Banach-space representations, which are "rich enough" to match up with the whole category of p-adic representations of $G\_{\mathbb{Q}\_p}$. Some Banach-space representations are completions of smooth (or locally-algebraic) representations with respect to an invariant norm, which gives the link with the classical formulation of Langlands; but many Banach-space representations aren't of this type, and more subtly, in some cases the same smooth representation can admit multiple invariant norms (giving different Banach representations as the completions), which is, again, parametrised by the $\mathcal{L}$-invariant.
| 1 | https://mathoverflow.net/users/2481 | 436968 | 176,560 |
https://mathoverflow.net/questions/436967 | 4 | The question has been motivated by the fact that the $1+1$ massless bosonic free field suffers the infrared problem as a "tempered distribution".
The reason is essentially that $\int\_{\mathbb{R}} \frac{dp}{\lvert p \rvert}$ is logarithmically divergent.
Since this is a infrared problem, I am curious whether the issue will be resolved by introducing a infrared cutoff, which is mathematically interpreted as compact supports in the spacetime variable.
In other words, the $1+1$ massless bosonic free field can be defined as a "just distribution" instead of being tempered?
Or more concretely, does the following integral converges for an arbitrary compactly supported smooth function $f(x,y)$ on $\mathbb{R}^2$?:
\begin{equation}
\int\_{\mathbb{R}}\frac{dp}{\lvert p \rvert} \int\_{\mathbb{R}^2} dxdy
f(x,y)e^{i(-\lvert p \rvert x+ py)}
\end{equation}
It seems nontrivial to evaluate the above integral for me. Could anyone please help?
| https://mathoverflow.net/users/56524 | Making sense of $1+1$ massless bosonic free field as a "distribution" rather than tempered | The massless GFF is well-defined as a random tempered distribution modulo constants, i.e. an element of the dual of the space of Schwartz test functions with vanishing integral. If you want it to be defined as a "normal" random tempered distribution, then you have to arbitrarily fix the zero mode somehow. For example, you could enforce that testing against the indicator function of the centred unit ball gives zero.
| 7 | https://mathoverflow.net/users/38566 | 436972 | 176,561 |
https://mathoverflow.net/questions/436895 | 4 | I'm reading *Theorem 1.17.* and its proof at page 14 of Santambrogio's [Optimal transport for applied mathematicians](https://drive.google.com/file/d/1udEnhpR3Y_LCMtsk-hMyHbxDgIIAf7GK/view?usp=share_link). The content is not hard but a little bit long (because of related detail). Please save your time by scrolling down to *Theorem 1.17.* if you are familiar with optimal transport.
---
Let $X=Y=\Omega$ be a compact subset of $\mathbb{R}^d$. Let the cost function $c:X \times Y \to [0, \infty)$ be of the form $c(x, y)=h(x-y)$ for a strictly convex function $h: \mathbb R^d \to [0, \infty)$. By [Kantorovich duality](https://mathoverflow.net/questions/422491/what-is-the-role-of-of-continuity-in-this-proof-of-kantorovich-duality), there exist an optimal transport plan $\gamma$ and a Kantorovich potential $\varphi:X \to \mathbb R \cup \{-\infty\}$ such that
$$
\varphi(x)+\varphi^c(y) \leq c(x, y) \text { on } \Omega \times \Omega \text { and } \varphi(x)+\varphi^c(y)=c(x, y) \text { on } \operatorname{spt}(\gamma) .
$$
Here $\varphi^c:Y \to \mathbb R \cup \{-\infty\}$ is the [$c$-conjugate](https://mathoverflow.net/questions/435037/optimal-transport-the-existence-of-an-optimal-pair-of-c-conjugate-functions) of $\varphi$ and $\operatorname{spt}(\gamma)$ the support of $\gamma$. Because $c$ is Lipschitz, $\varphi, \varphi^c$ are real-valued and Lipschitz. Let us fix a point $\left(x\_0, y\_0\right) \in \operatorname{spt}(\gamma)$. One may deduce from the previous computations that
$$
x \mapsto \varphi(x)-c\left(x, y\_0\right) \quad \text { is minimal at } x=x\_0.
$$
If $\varphi$ and $h$ are differentiable at $x\_0$ and $x\_0-y\_0$, respectively, and $x\_0 \notin \partial \Omega$, one gets $\nabla \varphi\left(x\_0\right)=\nabla h\left(x\_0-y\_0\right)$. This works if the function $h$ is differentiable, if it is not we shall write $\nabla \varphi\left(x\_0\right) \in \partial h\left(x\_0-y\_0\right)$ (using the subdifferential of $h$, see **Box 1.12**). For a strictly convex function $h$ one may inverse the relation passing to $(\nabla h)^{-1}$ thus getting
$$
x\_0-y\_0=(\nabla h)^{-1}\left(\nabla \varphi\left(x\_0\right)\right) .
$$
Notice that the expression $(\nabla h)^{-1}$ makes sense for strictly convex functions $h$, thanks to the considerations on the invertibility of $\partial h$ in **Box 1.12**.
This formula gives the solution to the transport problem with this cost, provided $\varphi$ is differentiable a.e. with respect to $\mu$. This is usually guaranteed by requiring $\mu$ to be absolutely continuous with respect to the Lebesgue measure, and using the fact that $\varphi$ may be proven to be Lipschitz. Then, one may use the previous computation to deduce that, for every $x\_0$, the point $y\_0$ (whenever it exists) such that $\left(x\_0, y\_0\right) \in \operatorname{spt}(\gamma)$ is **unique** (i.e. **$\gamma$ is of the form $\gamma\_{\mathrm{T}}:=(\mathrm{id}, \mathrm{T})\_{\#} \mu$ where $\mathrm{T}\left(x\_0\right)=y\_0$**). Moreover, this also gives uniqueness of the optimal transport plan and of the gradient of the Kantorovich potential.
>
> **Box 1.12.**
>
>
> * For every convex function $f: \mathbb{R}^d \rightarrow \mathbb{R} \cup\{+\infty\}$ we define its subdifferential at $x$ as the set
> $$
> \partial f(x)=\left\{p \in \mathbb{R}^d: f(y) \geq f(x)+p \cdot(y-x) \forall y \in \mathbb{R}^d\right\}.
> $$
> It is possible to prove that $\partial f(x)$ is never empty if $x$ lies in the interior of the set $\{f<+\infty\}$. At every point where the function $f$ is differentiable, then $\partial f$ reduces to the singleton $\{\nabla f\}$.
> * If $h$ is strictly convex then $\partial h$, which is a multi-valued map, can be inverted and is uni-valued, thus getting a map $(\partial h)^{-1}$, that should use in the statement of Theorem $1.17$ instead of $(\nabla h)^{-1}$.
>
>
>
We may summarize everything in the following theorem:
>
> **Theorem 1.17.** Given $\mu$ and $v$ probability measures on a compact domain $\Omega \subset \mathbb{R}^d$ there exists an optimal transport plan $\gamma$ for the cost $c(x, y)=h(x-y)$ with h strictly convex. It is unique and of the form (id, $T)\_{\#} \mu$, provided $\mu$ is absolutely continuous and $\partial \Omega$ is negligible. Moreover, there exists a Kantorovich potential $\varphi$, and $\mathrm{T}$ and the potentials $\varphi$ are linked by
> $$
> \mathrm{T}(x) = x-(\nabla h)^{-1}(\nabla \varphi(x)) .
> $$
>
>
> **Proof.**
>
>
> * The previous theorems give the existence of an optimal $\gamma$ and an optimal $\varphi$. The previous considerations show that if we take a point $\left(x\_0, y\_0\right) \in \operatorname{spt}(\gamma)$ where $x\_0 \notin \partial \Omega$ and $\nabla \varphi\left(x\_0\right)$ exists, then necessarily we have $y\_0=x\_0-(\nabla h)^{-1}\left(\nabla \varphi\left(x\_0\right)\right)$.
> * The points $x\_0$ on the boundary are negligible by assumption. The points where the differentiability fails are Lebesgue-negligible by Rademacher's theorem. Indeed, $\varphi$ shares the same modulus of continuity of $c$, which is a Lipschitz function on $\Omega \times \Omega$ since $h$ is locally Lipschitz continuous and $\Omega$ is bounded. Hence, $\varphi$ is also Lipschitz.
> * From the absolute continuity assumption on $\mu$, these two sets of "bad" points (the boundary and the nondifferentiability points of $\varphi$ ) are $\mu$-negligible as well. This shows at the same time that every optimal transport plan is induced by a transport map and that this **transport map** is $x \mapsto x-(\nabla h)^{-1}(\nabla \varphi(x))$. Hence, it is uniquely determined (since the potential $\varphi$ does not depend on $\gamma$ ). As a consequence, we also have uniqueness of the optimal $\gamma$.
>
>
>
---
**My question** I'm fine with the fact that there is a $\mu$-null Borel subset $N$ of $\Omega$ such that the gradient $\nabla \varphi: N^c \to \mathbb R^d$ is Borel measurable. Of course, the transport map $T$ must be Borel measurable. However, it's possible that [a non-measurable map has a measurable graph](https://mathoverflow.net/questions/376493/is-there-an-example-of-a-non-measurable-function-with-a-measurable-graph).
>
> Could you elaborate on how the inverse $(\nabla h)^{-1}$ is Borel measurable?
>
>
>
| https://mathoverflow.net/users/99469 | Optimal Transport: how is this transport map Borel measurable? | $\newcommand\p\partial\newcommand{\R}{\mathbb R}\newcommand\ep\varepsilon$Let $h\colon\R^d\to\R$ be a strictly convex function.
Consider the set $S:=2^{\R^d}$ of all subsets of $\R^d$ endowed with the [Hausdorff distance](https://en.wikipedia.org/wiki/Hausdorff_distance#Definition) $d\_H$. As usual, let $\p h(x)$ denote the [subdifferential](https://encyclopediaofmath.org/wiki/Subdifferential) of the function $h$ at a point $x\in\R^d$; we will identify $(\R^d)^\*$ with $\R^d$.
Thus, we have the map
\begin{equation\*}
\R^d\ni x\mapsto\p h(x)\in R, \tag{1}\label{1}
\end{equation\*}
where $R:=\{\p h(x)\colon x\in\R^d\}\subseteq S$.
It suffices to prove
>
> **Proposition 1:** The map $\p h$ in \eqref{1} is invertible and its inverse is continuous.
>
>
>
*Proof:* Using horizontal shifts, without loss of generality (wlog) it is enough to show that, for any sequence $(x\_k)$ in $\R^k$,
\begin{equation\*}
\text{if $d\_H(\p h(x\_k),\p h(0))\to0$ then $x\_k\to0$.} \tag{2}\label{2}
\end{equation\*}
By subtracting an appropriate affine function from $h$, wlog we may and will assume that
\begin{equation\*}
h(0)=0\quad\text{and}\quad 0\in\p h(0),\quad\text{so that}\quad h\ge0.
\end{equation\*}
To obtain a contradiction, suppose that \eqref{2} is false. Then there exist a real $\ep>0$, a sequence $(x\_k)$ in $\R^d$, and a sequence $(a\_k)$ in $\R^d$ such that for all $k$
\begin{equation\*}
a\_k\in\p h(x\_k)\quad\text{and}\quad |x\_k|\ge\ep, \quad\text{whereas}\quad a\_k\to0.
\end{equation\*}
Since $h$ is convex and $a\_k\in\p h(x\_k)$, for all $k$ we have $0=h(0)\ge h(x\_k)+a\_k\cdot(0-x\_k)$ and hence
\begin{equation\*}
0\le h(x\_k)\le a\_k\cdot x\_k, \tag{3}\label{3}
\end{equation\*}
with $\cdot$ standing for the dot product. Let also $|\cdot|$ denote the Euclidean norm.
Letting now
\begin{equation\*}
y\_k:=\frac\ep{|x\_k|}\,x\_k=\frac\ep{|x\_k|}\,x\_k+\Big(1-\frac\ep{|x\_k|}\Big)0
\end{equation\*}
we have $|y\_k|=\ep$ and,
using the convexity of $h$ again and looking back at \eqref{3}, we get
\begin{equation\*}
0\le h(y\_k)\le\frac\ep{|x\_k|}\,h(x\_k)
\le\frac\ep{|x\_k|}\,a\_k\cdot x\_k
=a\_k\cdot y\_k\to0,
\end{equation\*}
so that $h(y\_k)\to0$. Passing to subsequences, wlog we have $y\_k\to y$ for some $y\in\R^d$ with $|y|=\ep$, so that $y\ne0$.
On the other hand, the convex function $h\colon\R^d\to\R$ is [necessarily continuous](https://math.stackexchange.com/a/2961784/96609). So,
$h(y)=\lim\_k h(y\_k)=0$. So, for all $t\in[0,1]$, once again by the convexity of $h$, we have
$0\le h(ty)\le(1-t)h(0)+th(y)=0$, so that $h(ty)=0$ for all $t\in[0,1]$, which contradicts the strict convexity of $h$. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 436980 | 176,564 |
https://mathoverflow.net/questions/436969 | 5 | For $X\subseteq\mathfrak{M}\models \mathsf{TA}$, say that *$X$ is $\mathfrak{M}$-disruptive* iff there is some formula $\varphi$ in the language of arithmetic + a new unary predicate symbol $U$ such that
* the expansion of $\mathfrak{M}$ by interpreting $U$ as $X$ satisfies $\varphi$, but
* there is no expansion of the standard model $\mathbb{N}$ which satisfies $\varphi$.
Basically, disruptive sets witness $\mathbb{N}\not\preccurlyeq\_{\Pi^1\_1}\mathfrak{M}$ (meanwhile we're assuming $\mathbb{N}\preccurlyeq\mathfrak{M}$). I'm curious whether proper cuts are (essentially) the *only* way for a set to be disruptive:
>
> Suppose $X$ is $\mathfrak{M}$-disruptive. Must there be a proper definable cut in the expansion $(\mathfrak{M};X)$?
>
>
>
| https://mathoverflow.net/users/8133 | Does visible nonstandardness imply visible ill-foundedness? | Assume for contradiction that the answer is positive. Let $\def\ind{\mathrm{IND}\_{L(U)}}\ind\def\N{\mathbb N}\DeclareMathOperator\th{Th}\def\fM{\mathfrak M}$ denote full induction in the language of arithmetic with a new predicate $U$. I claim that for any $L(U)$ formula $\phi(U)$, we have
$$\N\models\forall X\,\phi(X)\iff\th(\N)+\ind\vdash\phi(U).\tag1$$
The right-to-left implication is obvious, as the theory $\th(\N)+\ind$ is sound. Conversely, if $\th(\N)+\ind\nvdash\phi(U)$, there exists a model $(\fM,X)\models\th(\N)+\ind+\neg\phi(U)$. Since $(\fM,X)$, being a model of $\ind$, has no definable proper cut, the assumption ensures $X$ is not $\fM$-disruptive. Thus, $\N$ expands to a model of $\neg\phi(U)$, i.e., $\forall X\,\phi(X)$ is false.
But $(1)$ is impossible, as it reduces $\Pi^1\_1$ truth to $\varnothing^{(\omega+1)}$.
| 4 | https://mathoverflow.net/users/12705 | 436994 | 176,569 |
https://mathoverflow.net/questions/436992 | 0 | I just want to understand the embedding behind Reinhardt's cardinals. We have an elementary embedding $j: V \to V$. Let the background theory be $\sf MK - Choice$. We know that $V$ itself is a class stage of the cumulative hierarchy, i.e. there is a class ordinal $\kappa$ such that $V=V\_\kappa$, but this means that the range of $j$, being structure preserving, would also be a stage of the cumulative hierarchy, this would be $V\_{j(\kappa)}$. So, either $j(\kappa)= \kappa$, by then $j$ would be an automorphism, but $V$ is a well-founded transitive class model of $\sf ZF$, and I think it can't admit automorphisms over it. So $j(\kappa) \neq \kappa$, but this leads to $j(\kappa) < \kappa$, and so $V\_{j(\kappa)} \in V$, but this would be inconsistent since $j$ is freely used in Replacement and Separation.
>
> Where is the flaw in the above argument?
>
>
>
| https://mathoverflow.net/users/95347 | Where do the universe embedds to in Reinhardt's cardinals setting? | An elementary embedding is not an automorphism, although an automorphism *would be* an elementary embedding. But an elementary embedding is not requiring to be surjective.
An elementary embedding just means that the images satisfy the same properties of their origin, in the codomain space that is. In the case of an elementary embedding $V\to V$ it just means that $x$ and $j(x)$ have the same first order properties. And indeed, if $\kappa<j(\kappa)$, then there is no reason to expect that some $\alpha$ is mapped to $\kappa$ at all. We do not expect that in the case of measurable cardinal, or even weaker theories, like the transfer principle from the reals to the hyperreals.
However, much like any other function that the universe is closed under, it will have fixed points. Many of them. So having $j(x)=x$ also tells you very little about the general behaviour of $j$. So, again, it is not enough to deduce that it is surjective just because we found some $\lambda$ such that $j(\lambda)=\lambda$.
Your main problem is overusing $\kappa$. $V$ is just $V$, if you *really must insist* on thinking of it as a stage in the von Neumann hierarchy, then it is $V\_{\rm Ord}$. The elementary embedding is defined on sets, so $j(\mathrm{Ord})$ is not a set, and while we can define $j$ on class as $j(X)=\bigcup\_{\alpha\in\rm Ord}j(X\cap V\_\alpha)$, there is no reason to expect $j$ to behave exactly as it does with sets.
| 5 | https://mathoverflow.net/users/7206 | 436997 | 176,570 |
https://mathoverflow.net/questions/437011 | 17 | Does the set of squares $S = \{n^2: n\in\omega\}$ adhere to [Benford's law](https://en.wikipedia.org/wiki/Benford%27s_law) for the first digit in every base $b\geq 2$?
**Precise formulation of what it means for a set $T\subseteq \omega$ to "adhere to Benford's law".** Let $b \geq 2$ be an integer. For $x\in\omega$, let $f^1\_b(x)$ be the first digit of $x$ in $b$-ary representation. We say that a set $T\subseteq \omega$ *adheres to Benford's law in base $b$* if for every $d\in\{0,\ldots , b-1\} = b$ we have $$\lim \sup\_{n\to\infty} \frac{|\{t\in (T\cap n): f^1\_b(t) = d\}|}{|T\cap n| + 1} \; = \; \log\_b\Big(1 + \frac{1}{d}\Big).$$
| https://mathoverflow.net/users/8628 | Does the set of square numbers adhere to Benford's law in every base? | No. Benford's law works well for sequences that grow exponentially, and the squares grow too slowly.
In particular, fix a base $b \geq 3$, consider the case of $d = 1$, and choose $n = 2 \cdot b^{2k}$. For this $n$, we have
$$
|\{ t \in (S \cap n) : f\_{b}^{1}(t) = 1 \}| = \sum\_{r=1}^{2k} \lfloor b^{r/2} \sqrt{2} \rfloor - \lfloor b^{r/2} \rfloor + 1,
$$
while the size of $|T \cap n| + 1 = \lfloor b^{k} \sqrt{2} \rfloor + 1$.
This leads to a value for the ratio which tends to
$$
\frac{(\sqrt{2} - 1) \left(1 + \frac{1}{\sqrt{b}} + \frac{1}{b} + \cdots\right)}{\sqrt{2}} = \frac{\frac{\sqrt{2} - 1}{\sqrt{2}}}{\frac{\sqrt{b} - 1}{\sqrt{b}}}
$$
as $k \to \infty$. We have that
$$ \frac{\frac{\sqrt{2} - 1}{\sqrt{2}}}{\frac{\sqrt{b} - 1}{\sqrt{b}}} > \log\_{b}(2) $$
for all $b \geq 3$.
| 26 | https://mathoverflow.net/users/48142 | 437013 | 176,575 |
https://mathoverflow.net/questions/437010 | 2 | This question is about finding the number of samples in a sequence required for the convergence of a series as a function of an error tolerance $\epsilon$. I want to show what I have tried so far.
The function is
$$ \sum\_{q=1}^{N-1} \exp(-q^2 \sigma^2/2)(1 - q/N) $$ for $\sigma > 0$ and $N \geq 1$
It is confirmed that this function converges with a large N. I want to find a function for $N$ at which the error is $\epsilon$.
So, the sum of this expression from $N$ to $\infty$ is less than $\epsilon$. The sum of this from $N$ to $\infty$ can be represented as,
$$ \sum\_{q=1}^{\infty} \exp(-q^2\sigma^2/2) - \sum\_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) < \epsilon $$
For the first term, I tried something like this. If $q \geq 1$, $q(q-1) \geq 0$.
So,
$$ \sum\_{q=1}^{\infty} \exp(-q(q-1)\sigma^2/2) \exp(-q \sigma^2/2) \leq \sum\_{q=1}^{\infty} \exp(-q \sigma^2/2) = \frac{\exp(-\sigma^2/2)}{(1 - \exp(-\sigma^2/2))} $$
For the second term,
$$ \sum\_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) $$
I used the same trick as above, but now I say that as $q \leq N-1, \quad q(q-1) \leq (N-1)(N-2) $
For the second term, as there is a negative in equation (2), we will test for greater than equal to conditional properties of that function.
$$ \sum\_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) =\sum\_{q=1}^{N-1} \exp(-q(q-1)\sigma^2/2) \exp(-q \sigma^2/2) (1-q/N) $$
$$ \sum\_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) \geq \exp(-(N-1)(N-2)\sigma^2/2) \sum\_{q=1}^{N-1} (1-q/N) \exp(-q \sigma^2/2) $$
I typed this sum on wolfram, and the final expression I have is this,
$$ \sum\_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) \geq \exp(-(N-1)(N-2)\sigma^2/2) \frac{\exp(-\sigma^2/2) (N-1 + \exp(-N \sigma^2/2) - N \exp(-\sigma^2/2))}{((1 - \exp(-\sigma^2/2))^2 N}$$
Rearranging a bit,
$$ \sum\_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) \geq \exp(-(N^2-3N+3)\sigma^2/2) \frac{(N-1 + \exp(-N \sigma^2/2) - N \exp(-\sigma^2/2))}{((1 - \exp(-\sigma^2/2))^2 N}$$
If I plug this into the original inequality, things become complicated.
$$ \sum\_{q=1}^{\infty} \exp(-q^2\sigma^2/2) - \sum\_{q=1}^{N-1} \exp(-q^2\sigma^2/2) (1-q/N) \leq \frac{\exp(-\sigma^2/2)}{(1 - \exp(-\sigma^2/2))} - \exp(-(N^2-3N+3)\sigma^2/2) \frac{(N-1 + \exp(-N \sigma^2/2) - N \exp(-\sigma^2/2))}{((1 - \exp(-\sigma^2/2))^2 N} < \epsilon $$
I don't know how to proceed from here.
| https://mathoverflow.net/users/489481 | Convergence as a function of error for the following function | $\newcommand\ep\epsilon\newcommand{\si}{\sigma} $"I want to find a function for $N$ at which the error is $\epsilon$."
This question is stated very poorly.
Indeed, let $n:=N$ (there is no reason to use $N$ where $n$ will do.) The $n$th error is
\begin{equation\*}
\ep\_n:=s-s\_n=\ep\_{1n}+\ep\_{2n},
\end{equation\*}
where
\begin{equation\*}
s:=\sum\_{q=1}^\infty a\_q,\quad s\_n:=\sum\_{q=1}^{n-1} a\_q(1-q/n),
\quad a\_q:=e^{-q^2\si^2/2},
\end{equation\*}
\begin{equation\*}
\ep\_{1n}:=\sum\_{q=n}^\infty a\_q,\quad \ep\_{2n}:=\frac1n\,\sum\_{q=1}^{n-1} a\_q q.
\end{equation\*}
Clearly, $\ep\_n$ takes only countably many values; so, the equality $\ep\_n=\ep$ can hold only for countably many values of $\ep$. Also, a closed-form expression for $\ep\_n$ is not available. So, solutions of the equation $\ep\_n=\ep$ for $n$ are not available in closed form, even when such solutions exist.
>
> However, for any real $\ep>0$, we can provide an explicit lower bound $n\_{\si,\ep}$ on $n$ such that $\ep\_n\le\ep$ for $n\ge n\_{\si,\ep}$ -- and this is what appears to have actually been tried to do in most of the OP.
>
>
>
Indeed, note that
\begin{equation\*}
r\_q:=\frac{a\_{q+1}}{a\_q}=e^{-(q+1/2)\si^2}
\end{equation\*}
is decreasing in $q$. So,
\begin{equation\*}
\ep\_{1n}\le\sum\_{q=n}^\infty a\_n r\_n^{q-n}=\frac{a\_n}{1-r\_n}=\frac{e^{-n^2\si^2/2}}{1-e^{-(n+1/2)\si^2}}
\le2e^{-n^2\si^2/2}
\end{equation\*}
if
\begin{equation\*}
n\ge\frac{\ln2}{\si^2}-\frac12.
\end{equation\*}
Next,
\begin{equation\*}
\ep\_{2n}\le\frac1n\,\sum\_{q=1}^\infty a\_q q
\le\frac{h(\si)}n,
\end{equation\*}
where
\begin{equation}
h(\si):=\sum\_{q=1}^\infty a\_1 r\_1^{q-1} q
=\frac{e^{5 \si ^2/2}}{(e^{3 \si^2/2}-1)^2}.
\end{equation}
Thus, if
\begin{equation}
n\ge n\_{\si,\ep}:=\max\Big(\frac{\ln2}{\si^2}-\frac12,\sqrt{\frac2{\si^2}\,
\max\Big(0,\ln\frac3\ep\Big)},
\frac{h(\si)}{\ep/3}\Big),
\end{equation}
then $\ep\_n\le\ep$. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 437018 | 176,576 |
https://mathoverflow.net/questions/436982 | 7 | The following is written in section 1.6 (p.7) of this paper: <https://arxiv.org/pdf/1010.6257.pdf>.
($\cdots$) Which lens spaces bound a smooth, simply-connected 4-manifold $W$ with $b\_2(W)=1$? ($\cdots$) The answers to these questions are unknown. By contrast, the situation in the topological category is much simpler: a lens space $L(p,q)$ bounds a topological, simply-connected 4-manifold with $b\_2=b\_2^+=1$ iff $-q$ is a square mod $p$.
I am curious about the proof of the last statement, but there is no proof or reference about the last statement. How can this be proved? Or is there a reference for this?
| https://mathoverflow.net/users/164671 | Lens space bounding a topological, simply-connected 4-manifold with $b_2=1$ | This follows from the relationship between the $\mathbb{Q}/\mathbb{Z}$ linking form of a $3$-manifold and the intersection form of a $4$-manifold that it bounds. Suppose that $W$ is 1-connected ($H\_1=0$ would suffice) with $\partial W = Y$ where $Y$ is a rational homology sphere; everything is oriented here. Let $q\_W$ be the intersection form on $H\_2(W)$ with respect to a basis and let $\lambda\_Y$ be the linking form.
From the long exact sequence $ 0 \to H\_2(W) \to H\_2(W,Y) \to H\_1(Y) \to 0$ and Poincaré duality, we get that $H\_1(Y)$ is the cokernel of $q\_W$, and with some additional work you learn that $\lambda\_Y = -(q\_W)^{-1}$ where one is using a basis for $H\_1(Y)$ coming from the chosen basis for $H\_2(W).
Now if $W$ has $b\_2=b\_2^+ = 1$, the map $q\_W$ must be multiplication by some integer $p > 0$. So the recipe above says that $H\_1(Y) = \mathbb{Z}/p$, and the linking form is isomorphic to $(-1/p)$. The forms on $\mathbb{Z}/p$ isomorphic to $-1/p$ are exactly those of the form $-n^2/p$ for some $n$ relatively prime to $p$. Now the linking form of $L(p,q)$ is isomorphic to $q/p$ (with respect to some orientation convention that I'm certainly not going to work out here!). Hence we see one direction.
To go the other way, suppose that $Y$ is a rational homology sphere with $H\_1(Y) = \mathbb{Z}/p$ and linking form $\lambda\_Y(g,g) = -1/p$ for some generator. Then add a handle along a loop representing $g$ to get a $b\_2=b\_2^+ = 1$ cobordism between $Y$ and some $Y'$. If you do the framing correctly, then $Y'$ will be a homology sphere. So far everything has been smooth, and in general that's as far as you can go. But in the topological category, $Y'$ bounds a contractible manifold; gluing that to the cobordism gives the desired $W$.
Notes:
1. I haven't been careful about orientations but it should all work out as stated.
2. It takes a little work to determine what framing (and how to describe it) you need in the second argument. The original source for this is Kervaire-Milnor, Groups of Homotopy Spheres.
| 7 | https://mathoverflow.net/users/3460 | 437024 | 176,581 |
https://mathoverflow.net/questions/437025 | 8 | Let $A$ be an abelian category (you can assume additional conditions for its goodness). Let $\mathrm{Seq}(A) = \mathrm{Func}(\mathbb{Z}, A)$, where $\mathbb{Z}$ is the standard order category on integers. Let $\mathrm{Chain}(A)$ be a subcategory of complexes in it.
>
> Is $\mathrm{Chain}(A)$ a reflexive or coreflexive subcategory of $\mathrm{Seq}(A)$?
>
>
>
As far as I can see, the inclusion functor $\mathrm{Chain}(A) \to \mathrm{Seq}(A)$ preserves all limits and colimits. I'm also interested in answers for variations like complexes bounded on one side.
| https://mathoverflow.net/users/148161 | Is the category of chain complexes a reflexive or coreflexive subcategory of the category of functors? | Yes, it's reflective and coreflective, under mild assumptions on the codomain category $\mathcal A.$ The adjoints are given, by definition, by Kan extension along the quotient from the abelian group-enriched category freely generated by $\mathbb Z$ to the abelian group-enriched category $\mathbb Z\_\partial$, Ab-functors out of which define chain complexes. (This map imposes the relations $d\_{i-1}d\_i=0$ for every $i.$) So it's sufficient to assume $\mathcal A$ is complete and cocomplete.
Explicitly, I believe the reflection $B\_\bullet$ of a sequence $\cdots A\_i \stackrel{f\_i}{\to} A\_{i-1}\cdots$ is given by $B\_{i-1}=A\_{i-1}/f\_if\_{i+1}(A\_{i+1})$ and the coreflection $C\_\bullet$, by $C\_{i+1}=f\_{i+1}^{-1}(\mathrm{ker}f\_i),$ both with the differential induced by the $f\_i.$ So any abelian $\mathcal A$ should be fine.
| 14 | https://mathoverflow.net/users/43000 | 437026 | 176,582 |
https://mathoverflow.net/questions/436611 | 1 | I want to perform a Morlet Wavelet transform analysis (WTA) on a sequence of binary data (0, 1), length about 19000 observations. The result seems reasonable, but I have my doubts whether WTA can be performed on a binary dataset, not technically, but mathematically.
I used the most simple code in R with the `WaveletComp` package after filtering out the high frequency signal with the `astrochron` package
`res=bandpass(dat, flow=1/100000000, fhigh=1/10, win=0)`
`res_every10=analyze.wavelet(my.data= res, my.series =2, loess.span = 0, dt = 1, dj = 1/12, method="AR", params = list(AR=list(p=1)))`
I also attached a sample dataset [`here`](https://docs.google.com/spreadsheets/d/1SxdH5QJE6pJqP3KcavtvFkjJAQA8OWK3/edit?rtpof=true&sd=true#gid=1021614223). Thank you for the suggestions.
| https://mathoverflow.net/users/496411 | Morlet wavelet transform of binary dataset in R | Yes, this is a perfectly valid operation.
Here is one reference where a Morlet wavelet analysis has been applied to a binary data set: [Morlet wavelet transforms of heart rate variability for autonomic nervous system activity](https://www.researchgate.net/publication/280319349_Monet_wavelet_transforms_of_heart_rate_variability_for_autonomic_nervous_system_activity), see section 2.1.
| 1 | https://mathoverflow.net/users/11260 | 437034 | 176,585 |
https://mathoverflow.net/questions/377616 | 8 | Recall that the Kronecker product
$s\_\lambda \* s\_\mu$ of two Schur functions $s\_\lambda$ and $s\_\mu$ is the symmetric function
whose expansion (in terms of Schur functions) is given by
\begin{equation}
\sum\_{\nu \, \vdash \, n} g\_{\lambda \mu}^\nu \, s\_\nu
\end{equation}
where $\lambda$, $\mu$, and $\nu$ are partitions of $n$ and
$g\_{\lambda \mu}^\nu$ is the Kronecker coefficient, which famously
counts the multiplicity of $V\_\nu$ in the tensor product $V\_\lambda \otimes V\_\mu$ of the irreducible representations of the symmetric group $S\_n$.
Switch now to the quasi-symmetric world: Given a composition $\alpha = (\alpha\_1, \dots, \alpha\_k)$ of $n$ let $L\_\alpha$ be the fundamental quasi-symmetric functions defined by
\begin{equation}
L\_\alpha = \sum x\_{\ell\_1} \cdots \,x\_{\ell\_k}
\end{equation}
where the sum is taken over all sequences
$1 \leq \ell\_1 \leq \cdots \leq \ell\_k$ such that $\ell\_i < \ell\_{i+1}$ whenever
$i = \alpha\_1 + \cdots + \alpha\_j$ for some $1 \leq j \leq k-1$.
The space of symmetric functions within the $\mathrm{QSym}\_n :=\Bbb{Q}$-span of $\{ L\_\alpha \, \big| \, \alpha \, \models \, n \}$ coincides with the
$\mathrm{Sym}\_n:= \Bbb{Q}$-span of $\{ s\_\lambda \, \big| \, \lambda \, \vdash \, n \}$
the latter of which is endowed with the Kronecker $\*$-product.
**Question:**
Can the Kronecker $\*$-product on $\mathrm{Sym}\_n$ be extended to all
of $\mathrm{QSym}\_n$ so that there exist non-negative
integers $\tilde{g}\_{\alpha,\beta}^{\, \gamma}$ for each
triple $\alpha$, $\beta$, $\gamma$ of compositions of $n$
satisfying
\begin{equation}
L\_\alpha \* L\_\beta =
\sum\_{\gamma \, \models \, n} \tilde{g}\_{\alpha,\beta}^{\, \gamma}
\, L\_\gamma \quad \text{?}
\end{equation}
p.s. Covertly, I am asking whether or not there is some kind of tensor product structure (as in a symmetric tensor category) on the projective indecomposable representations of the 0-Hecke algebra $H\_n(0)$. Any thoughts on that would also be appreciated.
thanks, ines.
| https://mathoverflow.net/users/70119 | "Kronecker Product" for quasi-symmetric functions | In the world of symmetric functions, Kronecker coefficients give the structure constants for both the inner multiplication and the inner comultiplication. While the natural introduction of the inner multiplication uses the representations/characters of the symmetric group, the inner comultiplication has a very straightforward description: the structure constants are obtained via the expansion
$$s\_{\lambda}(x\_1y\_1,...,x\_iy\_j,...)=\sum\_{\nu ,\mu\, \vdash \, n} g\_{\nu \mu}^\lambda \, s\_\nu (x)s\_{\mu}(y).$$
Similarly we can ask whether $L\_{\alpha}(xy)$ has such an expansion, and the answer is positive! There exist nonnegative integers $\tilde{g}\_{\beta, \gamma}^{\, \alpha}$ such that
$$L\_{\alpha}(xy)=\sum\_{\beta, \gamma \, \models \, n}\tilde{g}\_{\beta, \gamma}^{\, \alpha}L\_{\beta}(x)L\_{\gamma}(y).$$
This gives a natural inner *comultiplication* for quasisymmetric functions. Moreover, by dualizing, these allow us to define a genuine inner multiplication for the dual $\widehat{\mathrm{QSym}\_n}$, which is isomorphic to the descent algebra of Solomon.
Originally this was shown in Gessel's paper, where you will find the details and combinatorial interpretations
>
> I. Gessel, "Multipartite P-partitions and inner products of skew
> Schur functions", Contemporary Mathematics, 34:289–317, 1984
>
>
>
You can also read about it in
>
> C. Malvenuto, C. Reutenauer
> "Duality between quasi-symmetric functions and the Solomon descent algebra"
> J. Algebra, 177 (1995), pp. 967-982
>
>
>
---
Now, as far as a natural inner *multiplication* for $\mathrm{QSym}$, this is an open problem, at least according to Hazewinkel in
>
> M. Hazewinkel, "Symmetric functions, noncommutative symmetric functions, and quasisymmetric functions", Acta Appl. Math. 75 (2003), 1-3, 55–83
>
>
>
The same paper mentions that there is no known structure on $H\_n(0)$ that induces the inner comultiplication from above, so the answer to the last question in the postscript is also missing at the moment.
| 5 | https://mathoverflow.net/users/2384 | 437042 | 176,588 |
https://mathoverflow.net/questions/437037 | 4 | Suppose $\Omega\subset \mathbb R^2$ is a bounded domain with smooth boundary and suppose that
$$ F: \Omega \to \Omega,$$
is a diffeomorphism that fixes $\partial \Omega$ (i.e $F|\_{\partial \Omega}$ is equal to the identity map) and such that the pull back of the Euclidean metric under $F$, namely $F^\star e$ is again equal to the Euclidean metric $e$. Can we conclude that $F$ is the identity map?
| https://mathoverflow.net/users/50438 | On diffeomorphisms that preserve the metric | The expression "fixes $\partial\Omega$" is ambiguous. Do you mean that $f(\partial\Omega)=\partial\Omega$ or that $f(z)=z$
for all $z\in\partial\Omega$?
For the first, weaker condition, all exceptional exceptional domains $\Omega$ and functions can be easily described.
Preservation of Euclidean metric implies that $f$ is conformal, therefore it is complex analytic, and $|f'(z)|=1$
everywhere. By Complex Analysis, this easily implies that
$f(z)=cz+b,$ where $|c|=1$. Now it is easy to describe all domains $\Omega$ which can mapped homemoprphically onto themselves by such functions.
If $c$ is not a root of unity, they must be disks, including the whole plane, and any disk with center $b$ is mapped onto itself by $f(z)=cz+b-cb$. If $c$ is a root of unity
but not $1$, then $\Omega$ must have a rotational symmetry of some finite order, and again all such domains possess such homeomorphisms. Finally if $c=1$, a domain must be periodic (with period $b$).
These are all exceptional domains.
So for the second condition (fixing the boundary pointwise) we have only one exceptional domain, namely the plane, and $f(z)=cz+b$ with arbitrary $c, |c|=1$.
| 5 | https://mathoverflow.net/users/25510 | 437045 | 176,590 |
https://mathoverflow.net/questions/437029 | 1 | Let $\chi\_X:\{-1,1\}^n\to \{0,1\}$ be the characteristic function of a subset $X\subseteq \{-1,1\}^n$, which is randomly drawn from all subsets with exactly $k$ elements.
Is the support of the Fourier transform (Walsh-Hadamard transform) $\hat{\chi}\_X$ large ($\geq c2^n$ for a constant $c>0$) with high probability?
| https://mathoverflow.net/users/111720 | Support of Fourier transform of random characteristic function | Impose any bijection $\pi:u\mapsto t$ between the sets $\{\pm 1\}^n$ and $\{0,1,\ldots,2^n-1\}$ so that you can write the derived binary function $$f\_X:\{0,\ldots,2^n-1\}\rightarrow \{0,1\},$$
via
$$
f\_X(t)=\chi\_X(\pi^{-1}(t)).$$
Rueppel (R. A. Rueppel. *Analysis and Design of Stream Ciphers* Springer-Verlag, 1986, Chapter 4) has shown that when we consider *all binary sequences* of period $N$ under the uniform probability distribution, the expected linear complexity is $N/2,$ with a small variance, see for example [this paper](https://link.springer.com/chapter/10.1007/3-540-39805-8_21). Here we have $N=2^n.$ Linear complexity of a sequence is the length of the shortest linear feedback shift register that will generate it and it can be computed via the recursive Berlekamp-Massey algorithm.
Blahut's theorem (google it) states that the linear complexity of a binary sequence is equal to the Hamming weight (number of nonzero coefficients) of its Fourier/Hadamard transform.
Certainly if $k$ is near $N/2$ the lower bound of the form you want holds for some $c>0.$
If $k$ is near $N,$ or near zero, we can use the idea of $k-$ error linear complexity to relate the linear complexity of the given sequence to that of the constant (either all zero or all 1) sequence. There are some recent bounds on the $k-$error linear complexity. See for example [this paper](https://arxiv.org/pdf/1108.5793.pdf) which may help you obtain some bounds.
| 2 | https://mathoverflow.net/users/17773 | 437051 | 176,593 |
https://mathoverflow.net/questions/437027 | 1 | Say that a compact convex polytope is *rational* if is the intersection of half-spaces whose bounding hyperplanes are the zero-sets of affine functions of the coordinates with rational coefficients. Say that a continuous function $f$ from a rational compact convex polytope $K$ into itself is *rational continuous piecewise affine* if there exists a dissection of $K$ into finitely many rational polytopes $K\_1,\dots,K\_n$ (whose interiors are disjoint and whose union is $K$) such that for all $1 \leq i \leq n$ the restriction of $f$ to $K\_i$ is an affine map with rational coefficients. (People sometimes write "piecewise linear" where I write "piecewise affine" and abbreviate "continuous piecewise linear" as "cpl".)
The Brouwer fixed point theorem assures us that such a map $f$ must have a fixed point. But must it have a fixed point with rational coordinates?
| https://mathoverflow.net/users/3621 | Fixed points of rational continuous piecewise affine maps | A fixed-point is obtained as the intersection of two affine subspaces (a piece of the graph of $f$ and the identity graph) whose equations are defined over $\mathbb{Q}$. Therefore Cramer's formula (applied to a maximal rank subsystem) ensures that if there exists a solution then there exists a rational one.
| 1 | https://mathoverflow.net/users/24309 | 437052 | 176,594 |
https://mathoverflow.net/questions/437057 | 1 | Let $X$ be an irreducible subvariety of dimension $d$ in $\mathbb P^n$. Can we find a linear projection $\pi:\mathbb P^n \dashrightarrow \mathbb P^{d+1}$ such that $\pi: X \to \pi(X)$ is a finite and regular birational map?
| https://mathoverflow.net/users/16323 | Existence of a linear projection | By induction it is enough to check that if $n \ge d + 2$ there is a linear projection $\mathbb{P}^n \dashrightarrow \mathbb{P}^{n-1}$ that has all required properties. For this consider the subvariety
$$
S = \{(p,\xi) \in \mathbb{P}^n \times X^{[2]} \mid
p \in \langle \xi \rangle \}.
$$
Here $X^{[2]}$ is the Hilbert square of $X$, so that $\xi$ is a subscheme of $X$ of length $2$, and $\langle \xi \rangle$ is the unique line in $\mathbb{P}^n$ containing $\xi$. Note that
$$
\dim(S) \le \max(2d + 1, d + n - 1) = d + n - 1
$$
(meaning that every irreducible component of $S$ enjoys this bound). Indeed, the open subset in $S$ that corresponds to $\xi = \{x\_1,x\_2\}$ with $x\_1 \ne x\_2$ has dimension $2d - 1$, and the closed subset where $\xi$ is a tangent vector has dimension at most $d + n - 1$ ($d$ parameters for a point and at most $n - 1$ for the tangent direction).
Now consider the morphism
$$
S \to \mathbb{P}^n,
\qquad
(p,\xi) \mapsto p.
$$
By the dimension bound there is a point $p \in \mathbb{P}^n \setminus X$ where the fiber has dimension at most
$$
d + n - 1 - n = d - 1.
$$
The linear projection from such a point works, because at most $(d - 1)$-dimensional set of points in the image of $X$ have more than one point in the preimage.
| 3 | https://mathoverflow.net/users/4428 | 437058 | 176,596 |
https://mathoverflow.net/questions/437032 | 2 | Problem Statement
=================
Let $g:\mathbb R^{d}\to \mathbb R,d\in\{2,3\}$ be an integrable function (assumption **I1**). Suppose $\mathcal T$ is a rotation, and $f:\mathbb R^d\to\mathbb C$ (assumption **C**) is an integrable function (assumption **I2**) such that $\mathcal T\mathcal F f=\mathcal F f$ (assumption **S**). Finally, let assumption(s) **U** be some further, unknown restrictions on $\mathcal T$, $f$, and $g$. Then, I would like to prove the following:
\begin{equation}
\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal F\mathcal Tg\right)=\mathcal T\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right). \label{1}\tag{1}
\end{equation}
From the convolution theorem, the following is equivalent:
\begin{equation}
f\*\left(\mathcal Tg\right)=\mathcal T\left(f\*g\right).
\label{2}\tag{2}
\end{equation}
In words, this says that if the convolution kernel $f$ is invariant to rotations in Fourier space, then the convolution is equivariant to rotations, i.e., rotating $g$ and then performing the convolution gives the same result as performing the convolution and then rotating the result.
My question is to define **U** and show how these assumptions can be used to prove equation \eqref{1}.
Assumptions
===========
I will now summarize the motivation for each assumption. I believe each to be necessary to prove this result.
Assumption **I1** and **I2**
----------------------------
From the convolution theorem:
$$
\mathcal F^{-1}\left(\mathcal Ff\cdot\mathcal Fg\right)=f\*g.
$$
If $f$ and $g$ are integrable, then the convolution is defined ([[source]](https://math.stackexchange.com/questions/519034/show-that-convolution-of-two-measurable-functions-is-well-defined "Show that convolution of two measurable functions is well-defined")).
Assumption **C**
----------------
Since the transform of a real-valued function is Hermitian, and Hermitian functions are not invariant to rotations, $f$ must be complex-valued to satisfy assumption **S**.
Assumption **S**
----------------
Empirically, I have observed that this is a necessary assumption for equation \eqref{1} to be achieved. In my experiments for $d=2$ ($d=3$), $\mathcal T$ is replaced with $90^\circ$ rotations, $\mathcal F$ is replaced with the discrete Fourier transform in 2 (3) dimensions, and $f$ and $g$ are matrices ($3$-dimensional tensors).
| https://mathoverflow.net/users/495707 | Proof of covariant convolution for a kernel function that is rotation symmetric in Fourier space | I would think that no extra assumption **U** is needed, assumption **S** suffices. The key thing to observe is that the Fourier transform of a rotated function is equal to a rotated version of the Fourier transform of that function, see for example [Appendix A: Rotation property of Fourier transforms](https://link.springer.com/content/pdf/bbm:978-1-4613-1333-5/1.pdf) of the book [Jakowatz, Wahl, Eichel, Ghiglia, and Thompson - Spotlight-Mode Synthetic Aperture Radar: A Signal Processing Approach](https://doi.org/10.1007/978-1-4613-1333-5) for a proof.
Denote by $R$ the operation that rotates the vector $x$, then assumption **S** that $\mathcal T\mathcal F f=\mathcal F f$ implies that $f(Rx)=f(x)$. Now apply this to the convolution,
$$(f\ast {\cal T}g)(x)=\int dz\, f(x-z)g(Rz)=\int dz\, f(Rx-Rz)g(Rz)$$
$$=\int dRz\, f(Rx-Rz)g(Rz)=(f\ast g)(Rx)={\cal T}(f\ast g)(x),$$
which is the desired identity. (When switching from $z$ to $Rz$ as integration variable I have used that the Jacobian for this transformation is unity.)
| 2 | https://mathoverflow.net/users/11260 | 437061 | 176,597 |
https://mathoverflow.net/questions/402193 | 1 | Lately I've been trying (and have failed) to find an electronic copy of Huber's *Bewertungsspektrum und rigide Geometrie*, which (from what I understand) is the original reference developing the basics of the theory of adic spaces. Is it available online somewhere?
| https://mathoverflow.net/users/130058 | Looking for an electronic copy of Huber's Bewertungsspektrum und rigide Geometrie | An electronic copy of "*Bewertungsspektrum und rigide Geometrie*" is available [from Wuppertal University](http://www2.math.uni-wuppertal.de/~huber/preprints/Bewertungsspektrum%20und%20Rigide%20Geometrie.pdf).
*[year-old post, bumped to the front page by a spammer, answered for the record]*
| 4 | https://mathoverflow.net/users/11260 | 437071 | 176,601 |
https://mathoverflow.net/questions/141993 | 7 | Throughout the question, we only consider primes of the form $3k+1$. A reference for cubic reciprocity is Ireland & Rosen's **A Classical Introduction to Modern Number Theory**.
>
> How can I count the relative density of those $p$ (of the form $3k+1$) such that the equation $2=3x^3$ has no solutions modulo $p$?
>
>
>
Really, even pointers on how to say anything meaningful about these $p$ are welcome. Originally I also asked about how to count the density of *all* $p$ (not just those of the form $3k+1$) such that $2$ (or $3$) is not a cubic residue modulo $p$, but Felipe Voloch's comment quickly addresses how to deal with them, via Chebotarev's density theorem.
The difference between the question and these easier problems is that here I am asking that $k+1$ is not a cube modulo the prime $3k+1$, so the same approach does not seem to apply.
>
> Finally, if it turns out that the density is not zero, how does one go about finding the density of those $p=3k+1$ that satisfy that none of the equations $x^3=2$, $x^3=3$, $x^3=k+1$ have solutions?
>
>
>
(Ideally, the techniques lift to other situations, such as studying fifth powers modulo primes $p=5k+1$, etc, but even methods exclusive to the case of cubes are very welcome.)
| https://mathoverflow.net/users/6085 | Relative densities | A solution can be obtained as suggested by Keith Conrad in the comments, via Chebotarëv's theorem. Details can be found in $\S3.4$ of
>
> Coloring the $n$-Smooth Numbers with $n$ Colors
>
>
> Andrés Eduardo Caicedo, Thomas A. C. Chartier, Péter Pál Pach
>
>
> The Electronic Journal of Combinatorics **28 (1)** (2021), #P1.34, 79 pp. DOI: <https://doi.org/10.37236/8492>
>
>
>
Many thanks to both Felipe Voloch and David E Speyer for their patient explanations.
| 1 | https://mathoverflow.net/users/6085 | 437080 | 176,603 |
https://mathoverflow.net/questions/436906 | 0 | I want to understand an approximation of a compound Poisson distribution in this [paper](https://link.springer.com/content/pdf/10.1023/A:1022616601841.pdf?pdf=button).
First, let's set the environment. Consider $\mathcal{P}$ the class of distributions of real-valued and strictly stationary processes with expectation zero and finite variance. According to this [topic](https://mathoverflow.net/questions/428657/show-that-the-set-of-strictly-stationary-mean-zero-and-finite-variance-stochast), $\mathcal{P}$ is closed with respect to the Mallows metric $d$ (see the topic for a formal definition of $d$). Abusing the notation, we write $X \in \mathcal{P}$ to say that the law of $X$ is in $\mathcal{P}$.
Given $N \sim \hbox{Poisson}(\lambda)$. Let $\xi = (\xi\_t)\_{t \in \mathbb Z}$ be an ergodic process in $\mathcal{P}$ and $\xi\_1,\xi\_2, \xi\_3,... \overset{iid}{\sim} \xi$ sequence of stochastic process independent of $N$. Define the compound Poisson stochastic process $Y = (Y\_t)\_{t \in \mathbb Z}$ in $\mathcal{P}$:
\begin{equation}
Y\_t = \sum\_{j=1}^N \xi\_{t;j}, \quad N \sim \hbox{Poisson}(\lambda)
\end{equation}
(warning: $Y$ is not the classic [Compound Poisson process](https://en.wikipedia.org/wiki/Compound_Poisson_process))
This is a particular case with $(W\_t)\_{t \in \mathbb{Z}}\equiv 0$, $k=1$ of the Step 2 of the proof of item (ii) of Theorem 1 from the same [paper](https://link.springer.com/content/pdf/10.1023/A:1022616601841.pdf?pdf=button) cited above (The statement of the theorem is on page 454 and its proof on page 465). According to the same paper, we can approximate $Y$ by the following sequence of linear processes $(X^{(n)} , n \geq 1)$ (See equation 5.11 on the [paper](https://link.springer.com/content/pdf/10.1023/A:1022616601841.pdf?pdf=button) ):
\begin{equation}
X^{(n)}\_t = \sum\_{j =1 }^n \bar{\xi}\_j U\_{t - j ;n}
\end{equation}
where $(\bar{\xi}\_j)\_{j\in \mathbb Z}$ is a fixed realization of $\xi = (\xi\_t)\_{t \in \mathbb Z}$ and $(U\_{t;n})\_{t \in \mathbb Z} \overset{iid}{\sim} \hbox{Bernoulli}(\lambda/n)$ independent of $\xi$. The approximation or the convergence is with respect to the Mallows metric $d$. The following characterization is useful: $d(X^{(n)},Y) \to 0,\,(n \to \infty)$ is equivalent to:
1. $X\_{t\_1,...,t\_m}^{(n)}\implies Y\_{t\_1,...,t\_m}\, (n \to \infty)$ for all $t\_1,...,t\_m \in \mathbb{Z}$ and all $m \in \mathbb{N}$. This is a convergence in distributions, and the method of characteristic functions can be used.
2. $E[|X^{(n)}\_{t}|^2] \to E[|Y\_{t}|^2], (n \to \infty)$ for any $t$.
For me, the proof is strange and confusing because $(\bar{\xi}\_j)\_{j\in \mathbb Z}$ sometimes is fixed realization, and sometimes it is treated as random. This causes conflict with items 1 and 2 above. For example:
To demonstrate the first item, the convergence of the characteristic functions $\varphi^{(n)}(s) \to \varphi\_Y(s)$ has to be point-wise, and the paper delivers a convergence in probability. Still in this case, I tried to adapt this [answer](https://mathoverflow.net/questions/436873/approximation-of-a-random-sum-of-random-variables-infinitely-divisible-distribu/), but I don't have $(\bar{\xi}\_j)\_{j\in \mathbb Z}$ iid. So it seems that necessarily I have to first assume $(\bar{\xi}\_j)\_{j\in \mathbb Z}$ as a fixed realization and then treat it as random. Very strange!
For the second item, the convergence of expectations would have to be a convergence of real numbers, but the paper once again delivers a convergence in probability, first treating $(\bar{\xi}\_j)\_{j\in \mathbb Z}$ as fixed and then as random. See equation (5.13) on the [paper](https://link.springer.com/content/pdf/10.1023/A:1022616601841.pdf?pdf=button) .
Any clarification?
| https://mathoverflow.net/users/478920 | Understanding the approximation of a random sum of random processes | The paper seems to be written rather carelessly. In particular, it is indeed unhelpful to denote a random object and its realizations by the same symbol, leaving the job of figuring out which is which here or there to the reader.
Further, it is clear that any sequence (say, the zero sequence) can be a realization of a random sequence $\xi$ (if e.g. $\xi$ is the sequence of iid normal random variables).
So, the result should be true, not for all realizations of $\xi$, but for **almost** all of them (excluding a set of realizations of probability $0$).
Indeed, the central point in the proof seems to be the use of the ergodic theorem in the middle of page 466 of the paper, where the convergence must of course be in the almost sure sense (but this is not specified on the paper).
| 1 | https://mathoverflow.net/users/36721 | 437091 | 176,608 |
https://mathoverflow.net/questions/437084 | 1 | Consider this PDE:
$\begin{cases}u\_t+f(u)u\_x=0\\ u(x,0)=\varphi(x)\end{cases}$
**Has this PDE *weak solutions* whatever is $f$ or $\varphi$? I want to find an existence theorem and bibliography about that?**
Can anyone help me?
Thanks in advance!
| https://mathoverflow.net/users/146723 | Existence theorem of weak solutions of $u_t+f(u)u_x=0$ | The existence and uniqueness of a generalized solution $u=u(t,x)$ to the Cauchy problem for such kind of equations and for the more general one
$$
u\_t + \sum\_{i=1}^n \frac{\partial}{\partial x\_i}\varphi\_i(t,x,u)+\psi(t,x,u)=0
$$
have been proved more than fifty years ago by Kruzhkov in his paper [1].
Kruzhkov builds $u$ in the class of bounded measurable function by using the method of vanishing viscosity introduced earlier by H. Hopf: the differentiability conditions required for the functions $\varphi(t,x,\cdot)$, $i=1,\ldots, n$ and $\psi(t,x,\cdot)$ are mild.
Moreover, the first paragraph of the paper gives a nice historical introduction to this field of research, supported by the bibliography at its end.
**Reference**
[1] Stanislav Nikolaevich Kruzhkov, "First order quasilinear equations in several independent variables" (English, Russian original), Mathematics of the USSR, Sbornik, vol. 10 pp. 217-243 (1970), [DOI:SM1970v010n02ABEH002156](https://doi.org/10.1070/SM1970v010n02ABEH002156), [MR0267257](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0267257), [Zbl 0215.16203](https://zbmath.org/0215.16203).
| 3 | https://mathoverflow.net/users/113756 | 437092 | 176,609 |
https://mathoverflow.net/questions/437072 | 2 | Consider two d-uplets $u = (u\_1,...,u\_d)$ and $v = (v\_1, ..., v\_d)$ both living in $\mathbb{N}^d$ with $d$ a positive integer. They both verify $$(\*) \sum\_{i=1}^d u\_i = \sum\_{i=1}^d v\_i = k$$ with $k$ a positive integer, $k\geq d$ supposed to tend to infinity later on.
Given a constant $c \in \mathbb{N}$ such that $c<k$ ($c$ could also depend on $k$, e.g. $ c= \lceil k/d \rceil$), I would like to upper bound the cardinal of the following sets:
$$\{(u,v) | u \neq v, (\*), \sum\_{i=1}^d |u\_i -v\_i| \leq c \}. $$
It could also be written as bounding the cardinal of the set $$\{ u | u \neq v, (\*), \sum\_{i=1}^d |u\_i -v\_i| \leq c \} $$ for any $v$.
My knowledge of combinatorics is very low, I'd be glad to receive any reference or idea that could lead to the solution of this problem or similar ones. I do not even have intuition for the magnitude order of this quantity.
| https://mathoverflow.net/users/335858 | Counting the number of pair of d-uplets with upper bounded distance | This is not the best upper bound, but for the second set, ($\ast$) tells us about the norm of $u$ and $v$ and $\sum\_{i=1}^{d}|u\_i-v\_i|$ is the $L\_1$ distance between $u$ and $v$ (Manhattan metric). Ignoring $\ast$ we get an upper bound by counting the lattice points in the ball of radius $c$. Now considering $\ast$, we intersect this ball with the surface of a radius $k$ sphere (in this metric), this sharpens our bound. Hope this help you having a linear programming perspective.
| 2 | https://mathoverflow.net/users/496618 | 437093 | 176,610 |
https://mathoverflow.net/questions/436556 | 4 | Let $K$ be a non-archimedean field complete with respect to a discrete valuation with ring of integers $\mathcal{R}$, uniformizer $\pi$ and residue field $k$. Consider an affinoid analytic $K$-variety $X=Sp(A)$ with an affine formal model $\mathfrak{X}=Spf(A^{\circ})$ where $A^{\circ}\subset A$ is the set of power-bounded elements. Let $f\in A^{\circ}$ be an element such that its clas in $A^{\circ}/\pi$ is non-zero. As the generic fiber functor sends complete localizations on $\mathfrak{X}$ to Laurent subdomains in $X$, it follows that the affinoid subdomain $X(\frac{1}{f})=\{ x\in X \text{ such that } \vert f(x)\vert \geq 1 \}$ has an affine formal model given by the formal spectrum of $A^{\circ}\langle f^{-1}\rangle$. Let $\mathfrak{X}\_{k}$ denote the special fiber of $\mathfrak{X}$. By definition, the special fiber associated to the formal scheme $\mathfrak{X}(\frac{1}{f})=Spf(A^{\circ}\langle f^{-1}\rangle)$ is a Zariski open of $\mathfrak{X}\_{k}$. Hence, all properties of the special fiber $\mathfrak{X}\_{k}$ which are local in the Zariski topology (irreducibility, smoothness etc) will also hold in $\mathfrak{X}(\frac{1}{f})\_{k}$. My question is as follows: does this generalize to more general Laurent subdomains of $X$? For example, consider $f\in A^{\circ}$ as before, and consider subdomains of the form $Y=X(\frac{\pi^{n}}{f},\frac{f}{\pi^{n}})=\{ x\in X \text{ such that } \vert f(x)\vert = n \}$. As far as I know, these are not necessarily the generic fiber of an affine open subspace of $\mathfrak{X}$, so we cannot argue as above to construct affine formal models of $Y$ with topological properties similar to those of $\mathfrak{X}$. By Raynaud's theory we know there is an admissible formal blow-up $\mathfrak{X}'\rightarrow \mathfrak{X}$ such that there is an open $Z\subset \mathfrak{X}'$ such that its generic fiber is $Y$. However, admissible formal blow-up seems to have a weird behaviour at the topological level, so I don't know which properties of the topological space assocaited to $\mathfrak{X}$ are preserved under these kinds of maps. I would like to know if the fact that $X$ admits an affine formal model such that its special fiber has some topological property (mainly interested in irreducibility) implies that an affinoid subdomain of the form $X(\frac{\pi^{n}}{f},\frac{f}{\pi^{n}})$ also admits such an affine formal model. What about subspaces of the form $X(\frac{\pi^{n}}{f})$? Would something like this hold for Weierstrass subdomains of $X$?
| https://mathoverflow.net/users/476832 | On the local properties of rigid analytic varieties | For an affinoid $X=\mathrm{Sp}A$, the number of Shilov points of $X$ is a lower bound for the number of irreducible components of the special fiber of *any* formal model of $X$. This follows, for instance, from Proposition 2.2 of [this paper](https://arxiv.org/abs/2101.09759).
So then it's easy to make concrete examples: $X=\mathrm{Sp}K \langle T \rangle$ has an obvious formal model with irreducible special fiber, but any formal model of the Laurent domain $U=\mathrm{Sp}K \langle T,\pi/T \rangle \subset X$ will have at least two irreducible components in its special fiber, since $U$ has two Shilov points.
Hope this helps!
| 2 | https://mathoverflow.net/users/496798 | 437099 | 176,612 |
https://mathoverflow.net/questions/437105 | 1 | Let $A$ be an irreducible non-negative matrix. Is it true that the eigenvectors of $A$ can span the $R^n$ ?
Or are all the eigenvalues of $A$ distinct?
| https://mathoverflow.net/users/116579 | The dimension of the eigenvector space of non-negative irreducible matrices | You don't need to search for complicated counterexamples; just consider the matrix with all elements equal to 1.
[EDIT: removed a second counterexample after a comment pointed out it was reducible. If you want an example with all distinct eigenvalues, you can take the cyclic shift matrix.]
| 1 | https://mathoverflow.net/users/1898 | 437112 | 176,615 |
https://mathoverflow.net/questions/437114 | 10 | Mertens' Theorem states that
$$\sum\_{p \leq x}\frac{1}{p} = \log \log x + M + O(1/\log x).$$
This is weaker than the prime number theorem; in fact according to the [Wikipedia page](https://en.wikipedia.org/wiki/Mertens%27_theorems), the prime number theorem is equivalent to
$$\sum\_{p \leq x}\frac{1}{p} = \log \log x + M + o(1/\log x).$$
However no reference or proof is given, it just simply says that "Although this equivalence is not explicitly mentioned there, it can for instance be easily derived from the material in chapter I.3 of: G. Tenenbaum. Introduction to analytic and probabilistic number theory."
So what is a proof/reference for this fact? I naturally tried to prove it myself using partial summation, and it is easy to see where the $\log \log x$ comes from, however the constant $M$ and $o(1/\log x)$ are a bit more mysterious.
Standard proofs in prime number theory often proceed by introducing a logarithmic weight via the Von Mangoldt function, then proving an asymptotic and going back again. I would prefer to avoid such an approach as I have a different problem in mind coming from counting rational points on varieties where I can't introduce a logarithmic weight.
| https://mathoverflow.net/users/5101 | Proving Mertens' theorem using the prime number theorem | Well, one can always say that the PNT is equivalent to
$$\sum\_{p \leq x}\frac{1}{p} = \log \log x + M + o\left(\frac{1}{\log x}\right),\tag{$\ast$}$$
because both results are true (with better error terms). This is of course not what is meant by the Wikipedia page. Instead, the idea is that the equivalence PNT$\,\Leftrightarrow(\ast)$ can be established in a simpler way than either PNT or $(\ast)$. On the other hand, "simpler" is a subjective word, e.g. I usually find Tauberian arguments tricky.
At any rate, the first three exercises in Section 8.1.1 of Montgomery-Vaughan: Multiplicative number theory I address this question. For example, the PNT easily implies the relation $\psi(x)\sim x$ (logarithmic weights!), which then implies rather nontrivially (using Theorem 8.1 = Axer's theorem) that
$$\sum\_{p\leq x}\frac{\log p}{p}=\log x+C+o(1)$$
for some constant $C$. From here, $(\ast)$ follows easily by partial summation.
| 15 | https://mathoverflow.net/users/11919 | 437116 | 176,616 |
https://mathoverflow.net/questions/437038 | 4 | I have a general question about techniques used in [@Emerton's proof](https://mathoverflow.net/q/62405), sketched below, in the answer to [$\mathbb{P}^n$ is simply connected](https://mathoverflow.net/q/62282).
Given a finite étale map $\pi: Y \to \mathbb P^n$ (we regard all involved schemes as $k$-schemes for some fixed base field $k$), a base point $x \in \mathbb P^n$ and a point $y \in Y$ lying over $x$. Then naïvely/on the level of sets we construct a map $ \mathbb P^n \setminus \{x\} \to Y \setminus \{y\}$, $x' \mapsto y'$ by taking a unique line $L$ joining $x'$ with $x$, choosing the unique component $L'$ of the preimage $ \pi^{-1}(L)$ which contains $y$ and letting $y' \in Y'$ be the point lying over $x'$.
Now the funny thing is that $x' \mapsto y'$ is *algebraic*, i.e., a morphism in the category of schemes and that sounds quite surprising to me at first glance.
Emerton's argument was to observe that this map is realized as the composition of three maps, which can presumably be recognized as morphisms on their own:
1. $x' \mapsto \pi^{-1}(L)$, which is a map from $\mathbb P^n \setminus \{x\}$ to the Hilbert scheme of $Y$
2. picking out of $L' \subset \pi^{-1}(L)$ the component containing $y$, which should be a morphism from a locally closed subset of the Hilbert scheme of $Y$ to the Hilbert scheme of $Y$ itself and finally
3. mapping $L'$ to $L' \cap \pi^{-1}(x')$.
My naïve question is why the steps 1–3 are all morphisms in sense of algebraic geometry?
There are two standard ways in algebraic geometry to specify a morphism $f: X \to Y$ between schemes: ‘old school style’ by writing down an explicit polynomial equations with variables living in projective spaces which implicitly contain $X $ and $Y$.
The more modern functorial approach is to constuct a natural transformation
between the functors $F\_X= \operatorname{Hom}(-,X)$ and $F\_Y= \operatorname{Hom}(-,Y)$ which the schemes $X$, $Y$ represent. That amounts to associating a family
a $(f\_S: F\_X(S) \to F\_Y(S)\_{S \in (\mathrm{Sch}/k)}$ of maps indexed by the class $(\mathrm{Sch}/k)$ of $k$-schemes which commute with morphisms $m: S \to T$ in compatible way (= natural transform).
Seemingly morphisms 1–3 above were specified via the second approach. But the aspect which irritates me is that seemingly all three maps were only specified on the level of $k$-points,
for example the first map was seemingly specified just as a map $(\mathbb P^n \setminus \{x\})(k) \to \operatorname{Hilb}\_Y(k)$, but in the spirit of the functorial construction above it is expected to specify this map as a *family of maps* $(\mathbb P^n \setminus \{x\})(S) \to \operatorname{Hilb}\_Y(S)$ for all $k$-schemes $S$.
Or is it in this special situation sufficient to specify everything on level of $k$-points only? E.g., naïvely one could be tempted to interpret $ x'$ not as a $ k $-point, but an $ S $-point etc. But on the other hand it seems to make no sense to talk about a line between a $ k $-point (= $x$) and an $ S $-point, so I'm a bit skeptical if the construction can be ‘prolonged’ naturally to $ S $-points….
Update: as Marsault Chabat [pointed out](https://mathoverflow.net/questions/437038/construct-morphisms-of-schemes-on-level-of-associated-functors#comment1126407_437038) the constructions go through fine as long as $ S $ are spectra of fields, but I am not sure why these should work for arbitrary $ S $. Clearly it's wrong that a morphism is already determined by what it does on finite field extensions of ground field (“geometric points”).
| https://mathoverflow.net/users/108274 | Construct morphisms of schemes on level of associated functors | I can't speak for Matt Emerton specifically, but my understanding is that it is conventional to describe maps in terms of $k$ points in such a way that there is a clear extension of the definition to $S$ points. This is perhaps less rigorous, but if you know how to fill in the details, it removes clutter and leaves the essential geometric idea intact.
A partial justification for this practice is that a map between varieties over an algebraically closed field is determined by its value on $\overline k$ points, so in "nice" cases this description is not too lossy.
In your specific case, note that the $k$ point $x$ yields a constant $S$ point $S \to {\rm Spec} k \to \mathbb P^n$. There is a map that takes a pair of disjoint points of $\mathbb P^n$ to a line in $\mathbb P^n$. One way of describing it on $S$ points is as follows. Let $\mathcal O\_S^{\oplus n+1} \to L\_1$ and $\mathcal O\_S^{\oplus n + 1} \to L\_2$ be the surjections corresponding to two $S$ valued points $x\_1, x\_2$. Then dualizing, we obtain an injection $L\_1^{\vee} \oplus L\_2^{\vee} \to \mathcal O\_S^{\oplus n+1}$ (this follows from Nakayama and the fact that $x\_1, x\_2$ are disjoint). Then take the subscheme of $\mathbb P^n\_S$ defined by the vanishing of the homogeneous ideal of ${\rm Sym}(\mathcal O\_S^{\oplus n+1})$ generated by $L\_1^{\vee} \oplus L\_2^{\vee}$.
This yields the map $(\mathbb P^n-x)(S) \to {\rm Hilb}(\mathbb P^n)(S)$. Then pullback the subscheme along the flat map $Y \to \mathbb P^n$ to get the first map in Emerton's answer.
As you can see, this description more verbose. In my opinion, it obscures the key point: given two disjoint lines you can take their span and get a plane!
(One final nitpick: I don't think your dichotomy "old school" v.s. "modern" is very accurate. Classical algebraic geometers were comfortable with defining maps without always writing down explicit equations.)
| 1 | https://mathoverflow.net/users/494541 | 437129 | 176,621 |
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