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https://mathoverflow.net/questions/427066 | 2 | In their book Topological Vector Spaces (2nd ed.) Lawrence Narici and Edward Beckenstein generalise convex sets for TVS over ultravalued field $K$ as $K$-convex sets. The definition goes as following:
Ultravalue $|\bullet|:K \to \mathbb{R}\_{+}$ can be viewed as a valuation on the field which defines an ultrametric $\rho(x,y) = |x-y|$. On the other hand there are can be ultravalues taking values in arbitrary ordered group with a minimal element $0$, but I think this approach is too general for the book of Narici. I think he assumes all valuations to be real-valued; The key properties are that $|\alpha \beta| = |\alpha||\beta|,|\alpha+\beta| \le \max( |\alpha|, |\beta|)$ and $|\alpha| = 0 \iff \alpha = 0$.
Let $V$ be TVS over ultravalued field $K$. Then the set $A \subset V$ is called absolutely $K$-convex or $K$-disc iff $\mathbb{D}\_K(0,1)A + \mathbb{D}\_K(0,1)A \subset A$, there by $\mathbb{D}\_K(0,1)$ I denote a closed unit ball centered at $0$ with respect to the ultravalue. A set $C \subset V$ is $K$-convex iff it is a translate of an absolutely K-convex set.
Then, there is an exercise 4.202.c : If $\mathrm{char}\; K \neq 2$ then a set $C \subset V$ is $K$-convex iff $\alpha x + (1 - \alpha) y \in C$ for any $x,y \in C$ and any $\alpha \in \mathbb{D}\_K(0,1)$, call this property an A-convexity for further discussion.
Personally, I treated characteristic of the field as hind and used $1/2 \in K$ in the $(\Leftarrow)$ direction of the proof. So, the $K$-convex set is always A-convex even in char 2. But now I'm wondering if the condition on characteristic of the field is just to make the exercise doable by the book's audience, as having $1/2 \in K$ felt very handy. Or is there actually an example of the A-convex set which is not $K$-convex?
I don't have much experience with ultravalued fields of char 2. So, can you suggest an example? I'm not sure where to start.
P.S.
I found a preprint ["Combinatorial properties of non-archimedean convex sets"](https://arxiv.org/abs/2109.04591) by A. Chernikov and A. Mennin. It uses slightly different system of definitions and covers group-valued ultravalues. But the definitions of convex sets are equivalent: see the triple convex combination condition. the correct statement of the theorem is indeed:
>
> If $\mathrm{res} \; K = \frac{\{\alpha \in K : |\alpha|\le 1\}}{\{\alpha \in K : |\alpha|< 1\}} \not \cong \mathbb{F}\_2$ then a set $C \subset V$ is $K$-convex iff $\alpha x + (1 - \alpha) y \in C$ for any $x,y \in C$ and any $\alpha \in \mathbb{D}\_K(0,1)$.
>
>
>
This article also provides nice counter example (see proposition 2.7), a set $\Big \{ (\alpha\_1,\alpha\_2,\alpha\_3) \in \mathbb{D}^3(0,1) : \exists i \in \{1,2,3\} \; . \; |\alpha\_i| < 1 \Big\} \subset K^3$. What is important here is the fact that the equation $1=x+y$ has no nontrivial solutions in a residue field. Actually Will Sawin also uses this in his answer: "In this field, since we must have $α=0$ or $α=1\ldots$". This actually implies that residue field is $\mathbb{F}\_2$.
1. *Narici, Lawrence; Beckenstein, Edward*, Topological vector spaces, Pure and Applied Mathematics (Boca Raton) 296. Boca Raton, FL: CRC Press (ISBN 978-1-58488-866-6/hbk). xvii, 610 p. (2011). [ZBL1219.46001](https://zbmath.org/?q=an:1219.46001).
2. *Chernikov, Artem; Mennin, Alex*, Combinatorial properties of non-archimedean convex sets (<https://arxiv.org/abs/2109.04591>).
| https://mathoverflow.net/users/91850 | Are there "pathological convex sets" over ultravalued fields of char 2? | You can see this already with the ultravalued field $\mathbb F\_2$, with $|1|= 1$, $|0|=0$.
In this field, since we must have $\alpha=0$ or $\alpha=1$, the $\alpha x + (1-\alpha) y$ condition is trivial — every set satisfies it.
But not every set is convex. Since every element of the field is at most $1$, a set is absolutely $K$-convex if and only if it is a vector space, so $K$-convex if and only if it is an affine space.
So there are many sets which satisfy the weaker condition but are not convex, the simplest consisting of three points in the plane.
For an example over a meatier field like $\mathbb F\_2((t))$, simply take a pathological example over $\mathbb F\_2$ and look at its inverse image under the reduction mod $t$ map $\mathbb F\_2[[t]]^n \to \mathbb F\_2^n$.
| 3 | https://mathoverflow.net/users/18060 | 427070 | 173,269 |
https://mathoverflow.net/questions/426996 | 3 | Given a positive integer $n$. For any symmetric $n\times n$ matrix $M$, write $M\_{ik}$ for the $(i,k)$ entry, $M\_k$ for the unordered list (multiset) of entries in row $k$, and $S(M)$ for the unordered list (multiset) of all pairs $(M\_k,M\_{kk})$.
(This is the corrected definition from July 22, 2022, after having read the discussion below.)
(i) For which $n$ does $S(M)$ determine $M$ up to a symmetric permutation (i.e., $P^TMP$ where $P$ is a permutation matrix)?
(ii) For the other $n$, what is the structure of the matrices $M$ that cannot be reconstructed up to a symmetric permutation?
| https://mathoverflow.net/users/56920 | A combinatorial matrix reconstruction problem | The meaning of the question is unclear due to the word "set". As Carlo proposed in a comment, I'll take it that we have an unordered list of $n$ unordered lists of $n$ elements, and we want to identify the symmetric matrices for which those $n$ lists correspond to the rows up to permutation.
Clearly this is possible for $n=1$. For $n\ge 2$, it is impossible in general, as indicated by this example (where $M$ is any symmetric matrix).
$$
\pmatrix{0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & M}
\qquad
\pmatrix{1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & M}
$$
The two matrices are not equivalent under simultaneous row and column permutation since they have different diagonal elements.
Note that this counterexample works even if the rows are given as ordered lists.
Now let's consider the case that all the matrix entries are different, up to the constraints of symmetry. Now the equivalence class of the matrix can always be reconstructed. First, the diagonal entries are those appearing in only one row; place them in arbitrary order down the diagonal, say $d\_1,\ldots,d\_n$. Now the other entries can be identified: for $j\ne k$ the $(j,k)$ entry is the unique value that is both in the same row as $d\_j$ and in the same row as $d\_k$.
In the new formulation, diagonal elements are identified. This does not help for $n\ge 5$. Take two non-isomorphic simple graphs with the same degree sequence. The simplest example is a path of 4 edges, versus a triangle plus an isolated edge. These both give 2 of $\{\!\{ \mathbf{0},1,0,0,0\}\!\}$ and 3 of $\{\!\{ \mathbf{0},1,1,0,0\}\!\}$, where I wrote the diagonal element bold.
For larger sizes, use two non-isomorphic 2-regular graphs; for example , for $n=6$, a 6-cycle and two triangles both give 6 of $\{\!\{ \mathbf{0},1,1,0,0,0\}\!\}$.
| 2 | https://mathoverflow.net/users/9025 | 427082 | 173,272 |
https://mathoverflow.net/questions/426459 | 2 | Let $p\in (0,1)$ and $X\_1, X\_2, ...X\_n \sim \text{Bern}(p)$ be $n$ i.i.d. Bernoulli random variables, where the probability that $X\_i$ is $1$ equals $p$.
Fix $a,b>0$ different from $1$ that satisfy $a^p b^{1-p} = 1$, and define $C\_i = X\_i(a-b)+b$. In other words, $C\_i$ is $a$ when $X\_i$ is $1$, and $b$ when $X\_i$ is $0$.
I am interested in the behavior of the random variables
$$Z\_n = \frac{\sum\_{i = 1}^n\left(\prod\_{j = 1}^iC\_j\right)X\_i}{\sum\_{i = 1}^n\prod\_{j = 1}^i C\_j} $$
as $n\to \infty$. Does $Z\_n$ converge a.s.? Does $Z\_n$ converge in distribution?
Note that the product of the $C\_j$'s is sometimes very small and sometimes very large, as the central limit theorem says that $\frac{1}{\sqrt n} \sum \log C\_i$ converges in distribution to a normal $\mathcal N (0, \sigma^2)$. This follows from the condition that $a^pb^{1-p} = 1$.
I ran some numerical experiments in Mathematica that suggest that $Z\_n$ does converge to a constant, but this constant (presumably the limit of the means of $Z\_n$) is a non-trivial function of $a$, $b$, and $p$. Indeed, the mean of $Z\_n$ is difficult to compute, and does not seem to simplify nicely.
I am not a probabilist, so any resource that deals with this kind of random variable would be helpful.
| https://mathoverflow.net/users/124761 | Self normalized sum of products of i.i.d. random variables | James, using the fact that $X\_i$ only takes on 2 values, write $C\_iX\_i = xC\_i + y$ . Then your numerator is $$\sum^n x \prod^j C\_i + \sum^{n-1} y \prod^j C\_i = \sum^n (x+y) \prod^j C\_i - yC\_n$$. It suffices to show that $$ ~\frac {\prod^n C\_i} {\sum^n \prod^j C\_i } \rightarrow 0$$, where I am going to show the convergence in probability. $$$$
Lemma: if $S\_n$ is a mean 0 random walk, $\frac {e^{S\_n}} {\sum^n e^{S\_i}} \rightarrow 0$ in probability. $$$$Proof: $$\frac {e^{S\_n}} {\sum^n e^{S\_i}} = \frac 1 {1 + e^{-X\_n} + e^{-X\_n - X\_{n-1}} + .... + e^{-X\_n - X\_{n-1} - ... - X\_1}}$$. In the exponents in the denominator is the time reversed random walk, which is also a mean 0 random walk, and it pretty obviously goes to 0 in probability, e.g., break it up into returns to 0.$$$$
So the conclusion is, your expression is converging in probability to x+y. To do this explicitly, $X = (C - b)/(a-b)$, $XC = aX = a(C-b)/(a-b)$ and I seem to be claiming that the limit is a(1-b)/(a-b).
| 1 | https://mathoverflow.net/users/143907 | 427084 | 173,273 |
https://mathoverflow.net/questions/427043 | 2 | Consider the problem
$$(\star) \quad \begin{cases} \frac{d}{dt} X(t,x) = \begin{cases} - 1 & \text{ if } X(t,x) >0, \\
1 & \text{ if } X(t,x) < 0 \end{cases}, &t \in [0,T],\\
X(0,x) = x, &x \in \mathbb R
\end{cases}
$$
This is a typical example of ODE that does not have a solution in the classical sense. The right-hand side is not continuous. However, it is BV, so there exists a regular [Lagrangian flow in the sense of Ambrosio](http://cvgmt.sns.it/media/doc/paper/1612/crippa_bari.pdf). How can it be computed explicitly?
| https://mathoverflow.net/users/122620 | Regular Lagrangian flow for explicit ODE with discontinuous right-hand side | $\newcommand{\Om}{\Omega}\newcommand{\om}{\omega}\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}$As stated in my previous comment, in Theorem 3.1 of the paper linked by the OP about the existence (and uniqueness) of a Lagrangian flow, there is the condition that $D\cdot b$ be absolutely continuous with respect to the Lebesgue measure, where $b$ is the right-hand side of the ODE. Here this condition obviously fails to hold.
Let us show that in your case there is in fact no regular Lagrangian flow. Indeed, suppose the contrary, that there is a regular Lagrangian flow $X\colon[0,T]\times\R\to\R$. Let $\la$ denote the Lebesgue measure over $\R$. Then, according to Definition 4 in the paper linked by the OP, there is a set $\Om\subseteq\R$ such that $\la(\R\setminus\Om)=0$ and the following two conditions hold:
(i) For each $\om\in\Om$ and all $t\in[0,T]$
\begin{equation\*}
X(t,\om)=\om+\int\_0^t b(s,X(s,\om))\,ds, \tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
b(s,x):=1(X(s,x)<0)-1(X(s,x)>0).
\end{equation\*}
(ii) There is some real $L$ such that for all $t\in[0,T]$
\begin{equation\*}
\mu\_t:=X(t,\cdot)\_{\#}\la\le L\la,
\end{equation\*}
so that $\mu\_t$ is the push-forward of the Lebesgue measure $\la$ via the map $X(t,\cdot)$.
Take any $\om\in\Om\cap(0,T)$. Let
\begin{equation\*}
E\_\om:=\{t\in[0,T]\colon X(s,\om)>0\ \forall s\in[0,t)\}
\end{equation\*}
and
\begin{equation\*}
t\_\om:=\sup E\_\om=\max E\_\om. \tag{2}\label{2}
\end{equation\*}
By \eqref{1},
\begin{equation\*}
X(t\_\om,\om)=\om+\int\_0^{t\_\om} (-1)\,ds=\om-t\_\om. \tag{3}\label{3}
\end{equation\*}
Again by \eqref{1}, $X(t,\om)$ is continuous in $t$. So, by \eqref{3} and \eqref{2}, $\om-t\_\om=X(t\_\om,\om)\ge0$.
To obtain a contradiction, suppose that $t\_\om<\om$, that is, $X(t\_\om,\om)>0$. Then $t\_\om<\om<T$ and, again by the continuity of $X(t,\om)$ in $t$, we have $X(s,\om)>0$ for some $u\in(t\_\om,T)$ and all $s\in(t\_\om,u)$. So, $u\in E\_\om$ and $u>t\_\om=\max E\_\om$, which gives the desired contradiction. So,
\begin{equation}
\om-t\_\om=X(t\_\om,\om)=0. \tag{4}\label{4}
\end{equation}
To obtain another contradiction, suppose that $X(t\_1,\om)>0$ for some $t\_1\in(t\_\om,T]$. Again, because $X(\cdot,\om)$ is continuous, there is some $t\_2\in[t\_\om,t\_1]$ such that $X(s,\om)\le X(t\_2,\om)$ for all $s\in[t\_\om,t\_1]$. So, $X(t\_2,\om)\ge X(t\_1,\om)>0$. So, in view of \eqref{4}, $t\_2\ne t\_\om$ and hence $t\_2\in(t\_\om,t\_1]$. Again by the continuity of $X(\cdot,\om)$, there is some $t\_3\in(t\_\om,t\_2)$ such that $X(s,\om)>0$ for all $s\in[t\_3,t\_2]$. It follows by \eqref{1} that
\begin{equation}
X(t\_2,\om)=X(t\_3,\om)+\int\_{t\_3}^{t\_2}(-1)\,ds<X(t\_3,\om),
\end{equation}
which contradicts the condition that $X(s,\om)\le X(t\_2,\om)$ for all $s\in[t\_\om,t\_1]$. So, $X(t,\om)\le0$ for all $t\in(t\_\om,T]$. Similarly, $X(t,\om)\ge0$ for all $t\in(t\_\om,T]$.
So, $X(t,\om)=0$ for all $t\in(t\_\om,T]=(\om,T]$ and all $\om\in\Om\cap(0,T)$. So, for each $t\in(0,T)$,
\begin{equation}
\mu\_t(\{0\})=\la(\{x\in\R\colon X(t,x)=0\})
\ge\la(\{\om\in\Om\cap(0,T)\colon \om<t\})=t>0.
\end{equation}
So, condition $\mu\_t\le L\la$ in (ii) fails to hold. This final contradiction concludes the proof. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 427097 | 173,279 |
https://mathoverflow.net/questions/427092 | 3 | Let $Q$ be a $d$-dimensional Riemannian manifold. A submanifold $M$ of $Q$ is said to be *extrinsically flat* if $R\_{M}(X,Y,Z,W) = R\_{Q}(X,Y,Z,W)$ for all $X,Y,Z,W \in \mathfrak{X}(M)$, where $R\_{M}$ and $R\_{Q}$ are the curvature tensors of $M$ and $Q$, respectively.
It is clear that if the rank of the second fundamental form of $M$ is at most one, then by the Gauss equation $M$ is extrinsically flat.
I am wondering about the existence of such type of submanifolds. For instance, the classical *axiom of planes* of Cartan gives obstructions to the existence of “many” totally geodesic submanifolds, as explained below.
Fix $2 \leq r < d-1$. We say that $Q$ satisfies the *axiom of $r$-planes* if for every $p \in Q$ and each $r$-dimensional subspace $\varSigma$ of $T\_{p}Q$ there exists a totally geodesic submanifold $S$ such that $T\_{p}S = \varSigma$.
**Theorem.** If $Q$ satisfies the axiom of $r$-planes for some $r$, then $Q$ has constant curvature.
**Question**. In general, given a Riemannian manifold $Q$, are extrinsically flat submanifolds of $Q$ plentiful or rare? Is it possible for a Riemannian manifold $Q$ to satisfy the “axiom of extrinsically flat submanifolds” without necessarily having constant curvature?
| https://mathoverflow.net/users/74033 | Extrinsically flat submanifolds of a Riemannian manifold | If you take the simplest case, in which $Q$ is a $3$-dimensional Riemannian manifold, then there are plenty of extrinsically flat surfaces $M\subset Q$. In fact, if one chooses a 'generic' curve $\gamma$ in $Q$ and a 'generic' normal vector field $\nu$ along $\gamma$, then there will be a unique 'extinsically flat' surface $M$ containing $\gamma$ that has $\nu$ as its normal along $\gamma$.
In Cartan's language, the extrinsically flat surfaces in $Q$ depend on two functions of one variable. This does not depend on any knowledge of the curvature of $Q$.
**Addendum 1:** (7/23/22) I thought a little more about this and realized that there is a natural conjecture about the existence of 'extrinsically flat' submanifolds. Here is the statement.
*Conjecture:* For each $r\ge2$, let $D\_r = r^2(r^2{-}1)/12$ be the rank of the Riemann curvature tensor in dimension $r$. For a Riemannian manifold $Q$ of dimension $d\ge r + D\_r$, the PDE for 'extrinsically flat' submanifolds of dimension $r$ is involutive. In particular, such local submanifolds are just as 'plentiful' in $Q$ as they are in flat $\mathbb{R}^d$.
The intuition for the Conjecture is this: 'Extrinsic flatness of $M\subset Q$ is the requirement that, at each point $x\in M$, the Riemann curvature tensor of $M$ at $x$ should be equal to the restriction of the Riemann curvature tensor of $Q$ at $x$ to $T\_xM$. This is $D\_r$ second order PDE on $M$ as a submanifold of $Q$. Since the $r$-dimensional submanifolds of $Q$ can be represented locally as graphs of $d{-}r$ functions of $r$ variables, this PDE system will be formally 'determined' if $d{-}r = D\_r$ and 'overdetermined' if $d{-}r<D\_r$. If the symbol of the PDE system is 'non-degenerate', in the appropriate sense, when $d{-}r = D\_r$, then, at least in the real-analytic category, there will be 'plenty' of local solutions. In fact one would expect the local generality of the solutions in this case to be $2D\_r$ functions of $r{-}1$ variables. Meanwhile, for a 'generic' $Q$ of dimension $d<r + D\_r$, one would expect that it would not have any 'extrinsically flat' submanifolds of dimension $r$.
The above conjecture is easily established for $r=2$. (The case $(r,d)=(2,3)$ is just the statement in the first paragraph.) The case $(r,d)=(3,9)$ (note that $9 = 3 + D\_3$) seems to work, but I haven't checked all of the details, as the algebra is a little tricky.
Also, note that, for the 'overdetermined' non-existence result mentioned above, it is essential that one assume that the metric on $Q$ be 'generic'. As Cartan showed, if $d=2r$ and $Q$ has constant sectional curvature $c$, then the PDE for $r$-dimensional submanifolds of constant sectional curvature $c$ is involutive, and $2r < r + D\_r$ when $r>2$.
| 3 | https://mathoverflow.net/users/13972 | 427102 | 173,281 |
https://mathoverflow.net/questions/427103 | 7 | Let $\mathcal{A}$ be the algebra (in the sense of universal algebra) whose underlying set is the four-element set $\{a,b,c,d\}$ and whose structure consists just of the ternary operation $F$ defined as follows:
* $F(x,x,x)=x$.
* If $x\not=y$ then $F(x,x,y)=F(x,y,x)=F(y,x,x)=y$.
* If $x,y,z$ are pairwise distinct, then $F(x,y,z)$ is the unique element of $\{a,b,c,d\}\setminus \{x,y,z\}$.
This $\mathcal{A}$ is a kind of "toy (and total) model" of the orthocenter, from an equational perspective. In [another question](https://mathoverflow.net/questions/426245/is-the-orthocenter-roughly-equationally-finitely-based) I asked whether the equational theory of the orthocenter (appropriately construed) is finitely axiomatizable; in retrospect that seems more difficult than I'd expected, so I'd like to look at this toy model first:
>
> Is the equational theory of $\mathcal{A}$ finitely based?
>
>
>
| https://mathoverflow.net/users/8133 | Is the equational theory of this "orthocentrish" algebra finitely based? | This algebra is finitely based.
In fact, if you choose any bijection from $\{a,b,c,d\}$ to $\mathbb Z\_2\times \mathbb Z\_2$, then you can transport the operation $F(x,y,z)$ to $\mathbb Z\_2\times \mathbb Z\_2$ it find that it is $x-y+z$. The resulting algebra $\langle \mathbb Z\_2\times \mathbb Z\_2; x-y+z\rangle$ is the square of the algebra $\langle \mathbb Z\_2; x-y+z\rangle$. Since an algebra and its square have the same equational theory, the question reduces to: Is $\langle \mathbb Z\_2; x-y+z\rangle$ finitely based?
Roger Lyndon proved in 1951 that all $2$-element algebras are finitely based. In his paper, the algebra $\langle \mathbb Z\_2; x-y+z\rangle$ is referred to as 'system $L\_4$'. This is one of fifteen systems he deals with in Section I of his paper. Lyndon does not provide an equational basis for this algebra, but writes *``a complete set of axioms [for each of the systems in Section I] can be chosen by inspection from the various familiar sets of axioms for Boolean algebras and Boolean rings''*.
I claim that the following identities suffice:
* $F(x,x,y) = F(x,y,x) = F(y,x,x) = y$.
* $F(F(x\_1,x\_2,x\_3),F(y\_1,y\_2,y\_3),F(z\_1,z\_2,z\_3))=F(F(x\_1,y\_1,z\_1),F(x\_2,y\_2,z\_2),F(x\_3,y\_3,z\_3)).$
The reason these are enough is that $F(x,x,y)=F(y,x,x)=y$ (part of the first bullet point) states that $F$ is a Maltsev operation. The law from the second bullet point states that $F$ commutes with itself. H. P. Gumm showed that any Maltsev operation that commutes with itself on a set $A$ must be $x-y+z$ for some abelian group structure on $A$. Now the law $F(x,y,x) = y$ implies that $-y=y$ for this abelian group. Hence any model of the bulleted identities must be $\langle A; F(x,y,z)\rangle$ where $F(x,y,z) = x-y+z$ for some abelian group of exponent $2$ on $A$. Any two such things have the same equational theory. In fact, if $\langle A; F(x,y,z)\rangle$ and $\langle B; F(x,y,z)\rangle$ are nontrivial and satisfy these identities, then each is embeddable in a power of the other. This is enough to prove that the set of bulleted identities axiomatizes an equationally complete theory, hence forms a basis of identities for any of its nontrivial models.
| 14 | https://mathoverflow.net/users/75735 | 427105 | 173,282 |
https://mathoverflow.net/questions/427109 | 1 | Sorry if this is too easy for MO, but I found it in a research paper, so I thought that it was worth posting here.
I was reading a paper by Rabinowitz([this one](https://apps.dtic.mil/sti/pdfs/ADA093444.pdf) to be more precise) and I came across the following theorem:
>
> **Theorem 2.10:** Suppose $H\in C^1(\mathbb{R}^{2n}, \mathbb{R})$ and satisfies
>
> ($H\_1$) $H(z)\ge 0$
>
> ($H\_2$) $H(z)=o(|z|^2)$ at $z=0$
>
> ($H\_3$) There are constants $r>0$ and $\mu>2$ such that for all $|z|>r$, $$0<\mu H(z)\le (z, H\_z(z))\_{\mathbb{R}^{2n}}.$$
> Then, for any $T>0$, (HS) has a nonconstant $T$ periodic solution.
>
>
>
What I don't understand is one of the remarks after this statement. It says that by integrating the inequality in ($H\_3$) we get that there are constants $a\_1, a\_2>0$ such that $$H(z)\ge a\_1 |z|^{\mu}-a\_2$$ for all $z\in \mathbb{R}^{2n}$.
I don't understand how we actually integrate that inequality. What I am sure is that for $|z|\le r$ I just have to use the fact that $H$ is continuous on this compact set and I have a lower bound. So, the integration part is, as expected, for $|z|>r$.
I thought that maybe I should just write $z=(z\_1, z\_2, ..., z\_{2n})$ and express the scalar product using its definition, i.e. $\displaystyle (z, H\_z(z))\_{\mathbb{R}^{2n}}=\sum\_{j=1}^{2n} z\_j \frac{\partial H}{\partial z\_j}(z)$. From what I understand from the paper, $z$ is a function of time ($t$). So, I think that I should just integrate with respect to $t$ on $[r, |z|]$ and apply integration by parts, but I still don't see how this leads to the inequality in the remark.
So, I would really appreciate your help, I am sure that this is just a routine computation (I saw similar inequalities in other papers of Rabinowitz, but he always just says that they are obtained by integrating some inequality), but I haven't done this before.
| https://mathoverflow.net/users/486258 | How do I integrate this inequality that appears in a paper of Rabinowitz? | For any unit vector $u$ and real $t>0$, let
\begin{equation}
h(t):=H(tu).
\end{equation}
Then
\begin{equation}
h'(t)=H'(tu)\cdot u=\frac{H'(tu)\cdot(tu)}t\ge\frac{\mu H(tu)}t=\frac{h(t)}t.
\end{equation}
Recall that $H>0$ and hence $h(t)>0$ for all real $t$. So, for all real $t\ge1$
\begin{equation}
(\ln h)'(t)\ge\frac1t
\end{equation}
and hence
\begin{equation}
\ln h(t)\ge \ln h(1)+\mu\ln t
\end{equation}
and
\begin{equation}
H(tu)\ge H(u) t^\mu. \tag{1}\label{1}
\end{equation}
The function $H$ is strictly positive and continuous. So, $H(u)\ge c\_1$ for some real $c\_1>0$ and all unit vectors $u$. So, by \eqref{1}, for any vector $z$ with $|z|\ge1$,
\begin{equation}
H(z)\ge c\_1|z|^\mu\ge c\_1|z|^\mu-c\_1.
\end{equation}
Also, the inequality $H(z)\ge c\_1|z|^\mu-c\_1$ trivially holds when $|z|<1$, since $H>0$.
Thus, for all vectors $z$,
\begin{equation}
H(z)\ge c\_1|z|^\mu-c\_1,
\end{equation}
as desired.
| 2 | https://mathoverflow.net/users/36721 | 427113 | 173,283 |
https://mathoverflow.net/questions/427106 | 1 | In "[Sharp threshold phenomena in statistical physics](https://link.springer.com/article/10.1007/s11537-018-1726-x)", *H. Duminil-Copin, Japanese J. of Math. 14, 2019*, a sharp transition of a boolean function is defined as follows:
>
> A sequence of increasing boolean functions $(\mathbf{f\_n})$ undergoes a sharp threshold
> at $(p\_n)$ if there exists a $(\delta\_n)$ tending to 0 such that
> $f\_n(p\_n - \delta\_n) \to 0$ and $f\_n(p\_n + \delta\_n) \to 1$.
>
>
>
An example is $p\_n = 1/n$, for the indicator $f=\mathbb{1}\_{\text{giant component}}$ of the random graph.
What if I write $p\_n = n^{-c}$, for $c<1$ and $c>1$, and consider this the threshold point for $\mathbb{1}\_{\text{giant component}}$? Have I followed the correct definition of a sharp threshold above? That is, if I show that the limit of the indicator $\mathbb{1}\_{\text{giant component}}$ is either 0 or 1 for $c<1$ or $c>1$, have completed an equivalent task to the above definition?
How are these two formulations (the adding $\delta\_n$ method and the use of a power $c$) of the threshold related? Perhaps, I can write $f(n^{-c}) = f(n^{-1} ± \Delta\_n(c))$, and find the corresponding sequence $(\Delta\_n(c))$, and therefore show the two limits above are 0 or 1?
| https://mathoverflow.net/users/90619 | What is the exact definition of a sharp transition? | The parameters $p\_n$ are not arbitrary numerical parameters. They represent the expectation of one binary variable in a product space. Changing them additively is very different from changing the power in the expression $n^{-c}$.
I believe definition 1.1 in the cited paper has a typo, and for $p\_n$ tending to 0, we should require $\delta\_n/p\_n \to 0$ for a sharp threshold, and not just $\delta\_n \to 0$. That is the requirement in the key papers that consider $p\_n \to 0$, e.g. [1].
For the Erdos-Renyi graph $G(n,p)$ and a fixed vertex $v$, consider the property $$A\_n:=\{\, \text{deg} (v) \ge 1\,\}.$$
Then $A\_n$ is an increasing property, and $$f\_n(p)=P\_p(A\_n)=1-(1-p)^{n-1}$$ satisfies
$$\lim\_n f\_n(c/n) =1-e^{-c}$$ so it definitely does not satisfy the definition of sharp threshold that requires $\delta\_n/p\_n \to 0$ (i.e. given that definition of "sharp", since small changes in $c$ don't have the required effect). However,
$\lim\_n f\_n(n^{-c})=0$ for $c>1$ and $\lim\_n f\_n(n^{-c})=1$ for $c<1$.
[1] Friedgut, Ehud, and Jean Bourgain. "Sharp thresholds of graph properties, and the -sat problem." Journal of the American mathematical Society 12, no. 4 (1999): 1017-1054.
| 5 | https://mathoverflow.net/users/7691 | 427116 | 173,284 |
https://mathoverflow.net/questions/427122 | 2 | Let $V\_i$ be a sequence of $k$ dimensional analytic subsets in $\mathbb C^N$. Suppose that the volume of $V\_i$ is uniformly bounded, then Bishop's compactness theorem says that $V\_i$ will convergence by sequence to an analytic subsets $V$.
Q1: What is the precise meaning of "converge" here.
Q2: Is it possible that a sequence of singular point $q\_i\in V\_i$ converge to a smooth point $q\in V$. It seems impossible.
| https://mathoverflow.net/users/78863 | Bishop's compactness theorem and convergence of analytic subset | Convergence is taken [in Hausdorff sense](https://en.wikipedia.org/wiki/Hausdorff_distance),
though you can define the structure of a complex variety (the [Barlet space](http://library.msri.org/books/Book37/files/barlet.pdf))
on the set of cycles, taking every irreducible component with positive integer multiplicity. Barlet convergence is slightly more fine than the Hausdorff convergence, but not by much.
For the second, it is easy to construct an example of singular spaces converging to smooth. Take a sequence of curves in ${\Bbb C}P^2$ with each curve obtained as a union of two projective lines. Assume that this sequence converges to a union of a line with itself. Then the limit is smooth.
| 3 | https://mathoverflow.net/users/3377 | 427126 | 173,286 |
https://mathoverflow.net/questions/427078 | 7 | **Question:**
is there a class of optimization problems for whose solution no efficent algorithm is known, but for which the claimed optimality of a solution can efficiently be verified?
**Edits:**
* There is the publication [Optimality conditions and complete description of polytopes in combinatorial optimization](https://arxiv.org/abs/1809.04363) with optimality criteria for combinatorial optimization problems; the question would be if any of those criteria allow for an efficient evaluation.
-The requested clarification of the is simple: if e.g. the underlying toplogical graph contains multiple \*yes" instances, the question is as to whether it can efficiently be decided whether a specific such instance also is the one with optimal sum of edge weights.
* Ground states of 3D spin glasses are apparently a combinatorial optimization problem for which the optimal *value* can be calculated without actually determining the solution, so it can efficiently be checked whether a "yes" instance of the underlying decision problem also has optimal edge weight sum. cf [Checking for optimal solutions in some NP-complete problems](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjBlO2Iio35AhWSiP0HHTKlBScQFnoECAIQAQ&url=https%3A%2F%2Fwww.researchgate.net%2Fpublication%2F7568085_Checking_for_Optimal_Solutions_in_Some_NP-Complete_Problems&usg=AOvVaw2UlWygGHGH1qmjfgQKvX19)
* I just found the publication[Verifying Integer Programming Results](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwj52J7Tgqv5AhVRgv0HHYHsCtIQFnoECAMQAQ&url=https%3A%2F%2Fopus4.kobv.de%2Fopus4-zib%2Ffiles%2F6104%2FZR-16-58.pdf&usg=AOvVaw2ytzHfdswej5kwigNtPfY9) in which certificates for optimal integer solutions of linear programs are discussed.
| https://mathoverflow.net/users/31310 | Is there an optimization variant of NP completeness | It is unlikely that there is an interesting class of such optimization problems, for the following reason.
Following Chapter 17 of Papadimitriou's book *Computational Complexity*, let EXACT TSP denote the following problem: Given a distance matrix and an integer $B$, is the length of the shortest tour *equal to* $B$? Then Theorem 17.2 of Papadimitriou states that EXACT TSP is $\mathsf{DP}$-complete, where a language $L$ is in the class $\mathsf{DP}$ ([difference polynomial time](https://complexityzoo.net/Complexity_Zoo:D#dp)) if there exists a language $L\_1 \in \mathsf{NP}$ and a language $L\_2 \in \mathsf{coNP}$ such that $L = L\_1 \cap L\_2$. (Note that $\mathsf{DP}$ is *not* the same as $\mathsf{NP}\cap \mathsf{coNP}$.) Papadimitriou goes on to say that most other naturally occurring exact-cost optimization problems are also $\mathsf{DP}$-complete.
If it were easy to verify the optimality of an allegedly optimal solution to TSP, then EXACT TSP would be in $\mathsf{NP}$ (the optimal solution would be a short certificate). In particular, this would imply $\mathsf{DP} \subseteq \mathsf{NP}$. However, $\mathsf{DP} \subseteq \mathsf{NP}$ is certainly not known, and is probably not even true. More generally, no $\mathsf{DP}$-complete problem is going to be an example of what you're looking for, and that rules out a large number of candidates.
| 4 | https://mathoverflow.net/users/3106 | 427127 | 173,287 |
https://mathoverflow.net/questions/427014 | 1 | While I understand that doing the above is not possible *in general*, I would like to know more about how to proceed when it *is* possible. That is, what are the common methods people use to analytically characterize basin of attraction boundaries (i.e. via a closed-form expression) or their (Lebesgue) measures? (Regarding the latter: I do not *necessarily* need an exact expression for their Lebesgue measure; anything that lets me say something *analytically* about the "size" of a basin of attraction would be great!)
Any references or guidance would be immensely appreciated. Thank you very much in advance to anyone who took the time to read this.
**More details:** I have a nonlinear, first-order, autonomous system of $n\in\{2,3,...\}$ ODE defined on $[a,b]^n$, where $-\infty<a<b<\infty$.$^1$ I currently cannot solve for the exact solution, but I have identified the set of locally asymptotically stable (LAS) steady states $\{\vec{s}\_1,...,\vec{s}\_k\}$, where $k<\infty$. Each LAS steady state's basin of attraction seems to be very "well-behaved."$^2$.
P.S. I had originally asked this on [Mathermatics StackExchange](https://math.stackexchange.com/q/4497042/798206), but a kind user recommended that it would be better for me to ask this here!
---
**Footnotes:**
1. That is, this system is only defined at points in $[a,b]^n$, and any trajectory originating in $[a,b]^n$ (and obeying this system) always remains in $[a,b]^n$.
2. They are all simply connected. $k-1$ basins are convex sets; the remaining basin -- while non-convex -- just consists of the remaining space. The boundaries are "simple looking" and "smooth" (both in the colloquial sense). Note that these boundaries are not "flat" in the sense that they cannot be expressed as a hyperplane (or a finite union thereof).
| https://mathoverflow.net/users/486157 | Analytically characterizing basins of attraction boundaries and sizes | Despite having searched for such a reference for quite a while before asking my question, I ironically found a reference that seems to fit the bill! [Alberto and Chiang (2015, p. 198)](https://www.cambridge.org/core/books/stability-regions-of-nonlinear-dynamical-systems/522A4FACB373591173A78C710BF004D8) looks like a great resource for characterizing/analyzing basins of attraction and their boundaries, sizes, etc. The two authors seem to also have other related papers focusing on special cases (e.g. their [2011 paper](https://ieeexplore.ieee.org/document/6070954?reload=true) focuses on general autonomous nonlinear dynamical systems)
These resources seem great, but please do let me know if you know of others! In the meanwhile, I hope this helps any passers by :)
| 0 | https://mathoverflow.net/users/486157 | 427134 | 173,289 |
https://mathoverflow.net/questions/427129 | 0 | Let $\{X\_i, i \in \mathbb{N}\}$ be a sequence of non-lattice i.i.d. centered random variables, $\mathbb{E} |X\_1| ^3 < 0$. Let $S\_n = \sum\limits \_{i=1} ^n X\_i$ be the corresponding random walk and $W^{(n)} \_t = \frac{S\_{\lfloor nt \rfloor}}{\sqrt{n}}$, $t \in [0,1]$.
I am looking for a reference that conditioned on the end point $S\_n$, the normalized trajectory $W^{(n)}$ converges to a Brownian bridge: for a sequence $a\_n$ such that $\frac{a\_n}{\sqrt{n}} \to a \in \mathbb{R}$,
$$
W^{(n)} \big| S\_n \in [a\_n, a\_n+1] \Rightarrow B^a,
$$
where $(B\_t, t\in [0,1])$ is a Brownian bridge and $B^a\_t = B\_t + at$.
In other words, for every continuous functional $f : C[0,1] \to \mathbb{R}$,
$$
\mathbb{E} \Big \{ f (W^{(n)}) \Big| S\_n \in [a\_n, a\_n+1] \Big\} \Rightarrow f(B^a).
$$
| https://mathoverflow.net/users/41071 | Invariance principle: Brownian bridge and random walk conditioned on end point | A more general theorem is proved in [1] for the limits of random walks in the domain of attraction of a stable law. In the case described in the problem, one can also use the strong approximation approach in [2], because the probabilities of deviations of order $\sqrt{n}$ are smaller than the probabilities of the events that are being conditioned on.
[1] Liggett, Thomas M. "An invariance principle for conditioned sums of independent random variables." Journal of Mathematics and Mechanics 18, no. 6 (1968): 559-570.
<https://www.jstor.org/stable/pdf/24901780.pdf>
[2] Csörgo, Miklos, and Pál Révész. Strong approximations in probability and statistics. Academic press, 2014.
| 2 | https://mathoverflow.net/users/7691 | 427136 | 173,290 |
https://mathoverflow.net/questions/427132 | 2 | Let $\Omega \subseteq \mathbb{R}^n$ be a bounded domain. Define the set
$$H = \left\{\nabla f : f \in \mathcal{C}^1(\Omega)\right\}.$$
I suspect that $H$ is a Hilbert space (though I am unsure about completeness) with respect to the inner product
$(F,G) \in H^2 \mapsto \int\_\Omega \langle F(x),G(x)\rangle dx$, where $\langle\cdot,\cdot\rangle$ is the standard inner product on $\mathbb{R}^d$. This inner product is well-defined because functions in $H$ are continuous and $\Omega$ is bounded. Does $H$, or variants thereof, have a name? Are there any references where its properties are studied?
| https://mathoverflow.net/users/111925 | Is there a name for the space of gradients? | The space in question is not complete: take a function $f$ whose derivative is in $L^2$ but is discontinuous, and consider a sequence of smooth functions converging to it in the Sobolev space $W^{1, 2}$. Then the sequence of gradients is clearly Cauchy in your space $H$ and its limit is $\nabla f \notin H$.
This tells you that the space to actually consider is
$$\{\nabla f \in L^2: f \in L^1\_{loc}\}$$
which is a Hilbert space.
The space of $f$ with $\nabla f \in L^2$ modulo constants is denoted $\dot W^{1, 2}$ and is naturally isomorphic to the space of $L^2$ gradients. You could also identify this space of gradients with the space of exact $L^2$, $1$-forms.
I think most of these facts are in Chapter 5 -- Sobolev spaces -- of Evans' book on PDE.
| 2 | https://mathoverflow.net/users/109533 | 427137 | 173,291 |
https://mathoverflow.net/questions/427147 | 4 | My faculty imposes some numerical "recommendations" for promotions. Instead of arguing this sort of recommendation is ridiculous, I think it is wiser to provide some evidence the recommended numbers are far too high for (pure) mathematics.
Is there data available for the average or mean number of PhD completions when applying for Associate Professor and Professor in the at e.g. R1 universities in the USA, or for any other country?
| https://mathoverflow.net/users/130882 | Supervision numbers in pure mathematics | To establish a base line, you could look into [Some Patterns of PhDs in Mathematics Awarded Annually by Institutions of Higher Education in the United States over the Last Two Decades](https://www.ams.org/journals/notices/202201/rnoti-p96.pdf). This lists for each subfield of mathematics how many Ph.D.'s were awarded over a 10 or 20 year period by each major university in the US. You could then normalize that with the faculty of each department to obtain an upper bound on what you might reasonably expect as Ph.D. output per professor.
For example, in number theory a typical math department will award on average three Ph.D's every two years (table 11). These numbers are about the same in other pure math fields. I would expect the outcome of this exercise to be that tenured faculty in pure math delivers on average somewhat less than 1 Ph.D. per year.
| 6 | https://mathoverflow.net/users/11260 | 427149 | 173,294 |
https://mathoverflow.net/questions/427158 | 7 | In provability logic, $\square X \rightarrow X$ is not a theorem.
In my head[1] this reads as "if X is provable you don't necessarily have a proof of X".
This has lead to the question, what does provable even mean, if not there exists a proof of X? Is there an example of some proposition that is provable but does not have a proof?
Please help
edit: Maybe I didn't make myself clear, $X \rightarrow \square X$ is a theorem in provability logic. This means that $X \not\simeq \square X$. Which means that a proof of X is not equivalent to that proposition being provable, which is very strange to me.
^[1] I come from type theory so I tend to think constructively.
| https://mathoverflow.net/users/476956 | Difference between provability and the existence of a proof? | First note this isn't a constructive logic, so it's wrong to think of "$X$" as "there is a proof of $X$". (Even in constructive logic I find that dubious.)
Second note that if $X$ is provable then $Y \rightarrow X$ is provable for any $Y$. So it's not correct to say that $\square X \rightarrow X$ is not a theorem. The correct statement is that it's a theorem if and only if $X$ is a theorem (assuming your base logic is consistent). So any time you might want to use this implication, because the hypothesis holds, you can in fact use it.
A more interesting statement is that the universal $\forall X : \square X \rightarrow X$ is not a theorem. In other words, we can't prove that any time we prove any statement, that statement is true. This makes it more clear that the issue in question is *trust*. The formal system does not trust that necessarily every statement it can prove is true.
Another perspective on the statement consists of looking at nonstandard models. Every formal system will have nonstandard models that contain "natural numbers" that, from our perspective, are larger than all actual natural numbers. One of these numbers can encode a "proof" of $X$, from the perspective of the model that, in our world, is infinitely long, and therefore is not a real proof. This can happen even if $X$ is not true.
So in a nonstandard models, there can be statements that are provable in the sense that they have a proof encoded by a number understood by that model but don't have a proof in the sense of an actual finite string of symbols.
| 12 | https://mathoverflow.net/users/18060 | 427159 | 173,296 |
https://mathoverflow.net/questions/427157 | 3 | The following problem seems easy at a first glance but I can't see the way to prove it. Actually I don't even know if it's true but it is assumed implicitly in a research paper.
Help highly appreciated.
Given two spd matrices $A$, $B$ with
$x^\top Ax \ge x^\top Bx$
then it follows that
$x^\top A^{-1}x \le x^\top B^{-1}x$.
Any ideas?
| https://mathoverflow.net/users/486297 | Inverse quadratic norms | The proposed result holds **true**.
I am assuming throughout that $spd$ means [*symmetric positive definite*](https://en.wikipedia.org/wiki/Definite_matrix) and that the matrices $A$ and $B$ are $n$-by-$n$ matrices over $\mathbb{R}$ for some $n > 0$.
Indeed, since $B = P^{\top}P$ for some $P \in GL\_n(\mathbb{R})$ by hypothesis (see e.g., [Sylvester's law of inertia](https://en.wikipedia.org/wiki/Symmetric_bilinear_form)), we can assume, without loss of generality, that $B = I\_n$, the $n$-by-$n$ identity matrix.
Let us assume that $A - B = A - I\_n$ is [positive semi-definite](https://en.wikipedia.org/wiki/Definite_matrix) and let $O$ be an $n$-by-$n$ orthogonal matrix over $\mathbb{R}$ such that $O^{\top}AO$ is diagonal. Clearly, the matrices $O^{\top}(A - I\_n)O$ and $O^{\top}(I\_n - A ^{-1})O$ are also diagonal. Conjugating then both sides of the identity $A - I\_n = A(I\_n - A^{-1})$ by $O$, yields immediately that $I\_n - A^{-1}$ is positive semi-definite (check the signs of diagonal coefficients on both sides), hence the result.
| 4 | https://mathoverflow.net/users/84349 | 427166 | 173,297 |
https://mathoverflow.net/questions/401601 | 38 | Let $\mathbb R ^\omega$ be the set of all sequences of real numbers in the product topology.
Let $X$ be the set of all sequences in $\mathbb R ^\omega$ which have at least one 0.
Let $Y$ be the set of all sequences in $\mathbb R ^\omega$ which have at least two 0's.
Are $X$ and $Y$ homeomorphic, and if so, is there a simple proof of this?
| https://mathoverflow.net/users/95718 | Sequences with 0's in $\mathbb R ^\omega$ | For every $n\in\mathbb N$ consider the space
$$X\_n=\{x\in\mathbb R^\omega:\lvert x^{-1}(0)\rvert\ge n\}.$$
**Theorem.** For any positive integer numbers $n<m$, the spaces $X\_n$ and $X\_m$ are not homeomorphic.
*Proof.* The spaces $X\_n$ and $X\_m$ are not homeomorphic because the space $X\_m$ is a $\sigma Z\_{m-1}$-space whereas $X\_n$ is not.
Let us recall that a *closed* subset $A$ of a topological space $X$ is a *$Z\_k$-set* in $X$ if the set $C([0,1]^k,X\setminus A)$ is dense in the function space $C([0,1]^k,X)$ endowed with the compact-open topology. A topological space $X$ is called a *$\sigma Z\_k$-space* if $X$ is the countable union of $Z\_k$-sets.
$Z\_k$-sets are higher-dimensional counterparts of nowhere dense sets and $\sigma Z\_k$ are counterparts of meager spaces.
Since any singleton in $\mathbb R^m$ is a $Z\_{m-1}$-set, the space
$X\_m$ is a $\sigma Z\_{m-1}$-space. Assuming that $X\_m$ is homeomorphic to $X\_n$, we would conclude that $X\_n$ is a $\sigma Z\_{m-1}$-space. Consider the closed subset $A=\{0\}^n\times\mathbb R^{\omega\setminus n}\subseteq X\_n$ of $\mathbb R^\omega$ and observe that $A$ is not a $Z\_n$-set in $\mathbb R^\omega$. Since $X\_n$ is a $\sigma Z\_{m-1}$-space, $A=\bigcup\_{i\in\omega}A\_i$ is a countable union of $Z\_{m-1}$-sets $A\_i$. Since $n\le m-1$, each $Z\_{m-1}$-set $A\_i$ is a $Z\_n$-set and hence the set $C([0,1]^n,X\_n\setminus A\_i)$ is dense in $C([0,1]^n,X\_n)$.
It can be shown that the set $C([0,1]^n,X\_n)$ is dense in $C([0,1]^n,\mathbb R^\omega)$ and hence $C([0,1]^n,X\_n\setminus A\_i)$ is dense in $C([0,1]^n,\mathbb R^\omega)$, which means that $A\_i$ is a $Z\_n$-set in $\mathbb R^\omega$ (being a closed subset of the closed subset $A$ of $\mathbb R^\omega$). By the proof of Lemma 2.7 in the paper [Concerning locally homotopy negligible sets and characterization of $l\_2$-manifolds](https://bibliotekanauki.pl/articles/1364141) of Toruńczyk, the completeness of $\mathbb R^\omega$ implies that the closed $\sigma Z\_n$-set $A=\bigcup\_{i\in\omega}A\_i$ is a $Z\_n$-set in $\mathbb R^\omega$, which is not true. $\quad\square$
| 8 | https://mathoverflow.net/users/61536 | 427167 | 173,298 |
https://mathoverflow.net/questions/427074 | 2 | Let $(X, \Delta)$ be a (projective) klt pair (say over $\mathbb{C}$, but I am also interested in fields of positive characteristic) and $f: X \to Z$ the contraction associated to a $(K\_X + \Delta)$-negative extremal face $F$ of $\overline{NE}(X)$.
>
> Assuming $f$ is a birational morphism, does $Z$ have rational singularities?
>
>
>
If $f$ can be factored as a sequence of divisorial contractions then this is well-known, but I do not know whether this is true if $f$ is a small contraction.
I do not even know if $Z$ is always Cohen-Macaulay (which would be implied by $Z$ having rational singularities).
(As for motivation, the question probably has little or no direct relevance to the MMP; I am interested in rational singularities for their own sake.)
| https://mathoverflow.net/users/519 | Singularities of contractions of extremal faces | In characteristic 0, the answer is well known. By assumption there is an ample divisor $A$ such that $K\_X+\Delta+A$ cuts out $F$ and hence by the BPF theorem $K\_X+\Delta+A\sim \_{\mathbb Q,f}0$ and in fact $K\_X+\Delta +A\sim \_{\mathbb Q}f^\*(K\_Z+B\_Z)$ where $(Z,B\_Z)$ is klt; in the birational case $B\_Z=f\_\*(\Delta+A)$ and otherwise you need to use the canonical bundle formula see Theorem 0.2 of <https://arxiv.org/pdf/math/0308143.pdf>. But then $Z$ has rational singularities by Thm 5.22 in Koll'ar-Mori.
In characteristic $p>0$ see <https://arxiv.org/abs/2006.03571> for some recent results, but note that klt singularities are not always rational or CM (counterexamples in dimension 3 and char 5 are discussed in thm 1.6 of this paper). You may also be interested in <https://arxiv.org/pdf/2206.02674.pdf> and <https://arxiv.org/abs/1703.02269>.
| 5 | https://mathoverflow.net/users/19369 | 427184 | 173,306 |
https://mathoverflow.net/questions/427188 | 2 | Let $J$ denote [Jensen's modification](https://en.wikipedia.org/wiki/Jensen_hierarchy) of the constructible hierarchy. For an ordinal $\alpha$ and an $n\in\mathbb N^+$, let $\rho\_n^{J\_\alpha}$ denote the $\Sigma\_n$-projectum of $J\_\alpha$, the least $\delta\leq\alpha$ such that there is a $\Sigma\_n(J\_\alpha)$-definable subset of $\omega\rho$ not in $J\_\alpha$. (Note that $x<y$ implies $\rho\_x^{J\_\alpha}\geq\rho\_y^{J\_\alpha}$.) For a finite increasing sequence of positive naturals $s=(s\_1,\ldots,s\_m)$, call $s$ a *pattern of projecta* if there is some $\alpha$ where the points of decrease in the sequence $(\rho\_i^{J\_\alpha})\_{1\leq i\leq m}$ are just the integers in $s$, i.e. for any $i\in\mathbb N^+$, $\rho\_i^{J\_\alpha}>\rho\_{i+1}^{J\_\alpha}$ iff $i$ is an entry of $s$. In this way a pattern of projecta encodes when the sequence of $\Sigma\_i$-projecta will "drop" (if $i$ is in the sequence), or will "stay" (if $i$ is not).
What ordering results if we order the patterns of projecta by the size of the least $\alpha$ where $J\_\alpha$ has that pattern? We will call this ordering $<\_\rho$: set $s<\_\rho t$ if the least $\alpha$ where $J\_\alpha$ has pattern of projecta $s$ is less than the least $\beta$ where $J\_\beta$ has pattern of projecta $t$.
>
> What are some of the basic properties of $<\_\rho$, e.g. its order type? If $<\_\rho$ is a well-ordering, what ordinal is it isomorphic to?
>
>
>
What I do know is that all finite sequences of naturals are patterns of projecta, a relevant question is [MO #67933](https://mathoverflow.net/questions/67933), and a relevant paper is "[Patterns of Projecta](https://www.jstor.org/stable/2273621)" by A. Krawczyk (1981). In the MO question's [accepted answer](https://mathoverflow.net/a/81168), Philip Welch gives an explicit construction, for arbitrary sequence $s$, of a $J\_\alpha$ whose pattern of projecta is $s$. However, for this construction to be used in computing $<\_\rho$, we need information about the *minimal* $\alpha$ in particular. Sufficient for this would be if at each step of the Skolem hulling for $1\leq j\leq m-1$, $\pi(H\_m)$ were the minimal $J$-rank with the necessary segment of the pattern so far.
| https://mathoverflow.net/users/479330 | Ordering patterns of projecta by least witness | $<\_\rho$ is a wellorder essentially by definition. The ordertype is $\omega^{\omega}$ (ordinal exponentiation of course).
In fact $s<\_\rho t$ iff either $\mathrm{lh}(s)<\mathrm{lh}(t)$,
or $\mathrm{lh}(s)=\mathrm{lh}(t)$ and $s<\_{\mathrm{lex}}t$,
i.e. letting $i$ be least such that $s(i)\neq t(i)$, we have $s(i)<t(i)$. This easily yields the ordertype $\omega^{\omega}$.
This is just by a slight variant of Philip Welch's construction.
Suppose first $\mathrm{lh}(s)=\mathrm{lh}(t)$.
Let $\beta$ be an ordinal which instantiates the pattern of $t$. Let $i$ be least such that $s(i)<t(i)$. So $\beta$ also instantiates the pattern of $s\upharpoonright i$. But $$\rho\_{t(k)+1}^{L\_\beta}<\rho\_{t(k-1)+1}^{L\_\beta}<\ldots<\rho\_{t(i)+1}^{L\_\beta}<\rho\_{t(i)}^{L\_\beta}=\rho\_{s(i)+1}^{L\_\beta}=\rho\_{s(i)}^{L(\beta)},$$
and the $\rho\_{t(k)+1}^{L\_\beta},\ldots,\rho\_{t(i)+1}^{L\_\beta}$ are each cardinals in $L\_\beta$. Let $\kappa\_j=\rho\_{t(j)+1}^{L\_\beta}$ for $j\in[i,k]$. Now form the hull $$\mathrm{Hull}\_{\Sigma\_{s(i)+1}}^{L\_\beta}(\kappa\_i\cup\{x\}),$$
where $x$ is an appropriate finite set that this is $\Sigma\_{s(i)+1}$-elementary and reflects the relevant information
(including the relevant standard parameters etc). Let $\beta\_i$ be such that $L\_{\beta\_i}$ is the transitive collapse of the hull. Note that $\beta\_i<\beta$, and since $\kappa\_i$ is a cardinal in $L\_\beta$, therefore so
$$\rho\_\omega^{L\_{\beta\_i}}=\kappa\_i=\rho\_{s(i)+1}^{L\_{\beta\_i}}<\rho\_{s(i)}^{L\_{\beta\_i}}$$
and $L\_{\beta\_i}$ instantiates $s\upharpoonright(i+1)$. Since $\kappa\_k<\kappa\_{k-1}<\ldots<\kappa\_i$ and these are cardinals of this model, we can continue forming hulls in this manner at elementarities corresponding to the entries in $s$, and since $\mathrm{lh}(s)=\mathrm{lh}(t)$, there are enough cardinals to do that. This results in an ordinal $\beta\_k$ instantiating $s$, and $\beta\_k<\beta$, as desired.
Now suppose $\mathrm{lh}(s)<\mathrm{lh}(t)$ and let $\beta$ instantiate $t$. Then we get $\mathrm{lh}(t)$-many $L\_\beta$-cardinals $\kappa\_k<\ldots<\kappa\_0<\beta$ as the resulting projecta. Now $\rho\_\omega^{L\_{\kappa\_0}}=\kappa\_0$, so we can proceed like before, taking a $\Sigma\_{s(0)+1}$-elementary hull of $L\_{\kappa\_0}$ in parameters in $\kappa\_1\cup\{x\}$ for an appropriate $x$, etc. This shows that $s<\_\rho t$.
Note also (by calculation as above) that $\sup\_{s}\alpha\_s$,
where $\alpha\_s$ is the least instantiation of $s$, is just the stack of the minimal models of $n$th order arithmetic, for $n<\omega$.
| 3 | https://mathoverflow.net/users/160347 | 427192 | 173,311 |
https://mathoverflow.net/questions/427205 | 2 | $\newcommand{\ur}{\mathrm{ur}}\newcommand{\cris}{\mathrm{cris}}$Let $K$ be a finite extension of $\mathbb{Q}\_p$, $G\_K=\operatorname{Gal}(\overline{K}/K)$ and $I\_K \subset G\_K$ its inertial subgroup. Let $V$ be a finite-dimensional representation of $G\_K$. Assume that $V$ is crystalline as $G\_K$-representation. Is it true that it is crystalline as $I\_K$-representation?
By definition, we need to prove that $$\dim\_{\widehat{K^{\ur}}}(V \otimes B\_{\cris})^{I\_K}=\dim\_{K\_0}(V \otimes B\_{\cris})^{G\_K}=\dim\_{\mathbb{Q}\_p} V$$ where $\widehat{K^{\ur}}$ and $K\_0$ are the maximal complete unramified extension of $\mathbb{Q}\_p$ in $\overline{K}$ and $K$ respectively.
Edit: As remarked by David, $K^{\ur}$ need to be replaced by its completion.
| https://mathoverflow.net/users/146212 | Crystalline when restricted to inertial subgroup | This is purely formal. If $V$ is crystalline, then $V \otimes \mathbf{B}\_{\mathrm{cris}}$ has a basis as a $\mathbf{B}\_{\mathrm{cris}}$-module in which the action of $G\_K$ is trivial. Hence *a fortiori* it has a basis in which the action of $I\_K$ is trivial.
What is much less obvious, but also true, is that the converse holds: if $V$ is crystalline as an $I\_K$-representation, then it's actually crystalline as a $G\_K$-representation. This is because $(B\_{\mathrm{cris}})^{I\_K} = \widehat{K^{\mathrm{nr}}}$ contains the periods of all unramified representations. (Incidentally, the assertion in your question that $(B\_{\mathrm{cris}})^{I\_K} = K^{\mathrm{nr}}$ is incorrect; it is genuinely the completion that you get here.)
| 8 | https://mathoverflow.net/users/2481 | 427207 | 173,314 |
https://mathoverflow.net/questions/427119 | 1 | The theory $\mathsf{TNT}$, introduced by Hao Wang in 1952, adds negative types to simple [Type Set Theory $\mathsf{TST}$](https://en.wikipedia.org/wiki/New_Foundations#The_Type_Theory_TST), so it's written exactly as $\mathsf{TST}$ but with the type indices ranging over $\mathbb Z$ instead of just $\mathbb N$.
Let $\mathsf{ZF}\text{-Reg.}$ be the milieu for models, let $M$ be a transitive non-well-founded model of $\mathsf{ZF}$, by that I mean $(M, \in\_M)$ where $\in\_M$ is not well founded. So as seen from the outside of $M$, there must exist a non-standard infinite ordinal $\zeta$ such that there exists an infinite descending sequence $V\_\zeta, V\_{\zeta-1}, V\_{\zeta-2},\dotsc$ of stages of $\mathsf{ZF}$. Now, take $\mathcal M = \displaystyle\bigcup\_{n \in \mathbb N} V\_{\zeta \ \pm \ n}$ :
>
> Can this this provide a model of $\mathsf{TNT}$, where each sort $i$ range over $V\_{\zeta + i}$, and the membership relation from sort $i$ to sort $i+1$ is the membership relation restricted to $V\_{\zeta+i} \times V\_{\zeta + i + 1}$?
>
>
>
>
> Can we have an omega model this way? I mean the set of naturals in $\mathcal M$ is standard, i.e. externally well-founded finite von Neumann ordinals. Which (if it models $\mathsf{TNT}$) is known to violate $\mathsf{AC}$.
>
>
>
| https://mathoverflow.net/users/95347 | Can $\mathsf{TNT}$ be modeled in non-well-founded models of $\mathsf{ZF}$? | Per comments, the above conditions might not be enough to ensure the result of interpreting $\sf TNT$, however, the following line would work to answer the first quetion:
Suppose we work in $\sf ZF−Reg.$ on a transitive non-$\omega$-non-well-founded model $M$ of $\sf Finite \ \sf ZF$ (i.e. $\sf ZF -\text{ inf.+ every set is finite}$), so the rank of every nonempty set must be a successor rank, now since its non-well-founded then there must exist a non-standard ordinal, i.e. internally looks like a finite von Neumann ordinal but externally it has an infinite predecessor chain subset of it, now working externally [in $\sf ZF-Reg.$] since $M$ is just a set, then pick any such ordinal $\zeta$ and send $V\_\zeta$ to $0$, then send its predecessor stage to $\mathbb N \setminus 1 $ (which captures the integer $−1$), and the predecessor stage of that to $\mathbb N \setminus 2 $, etc...; send each $V\_{\zeta +n}$ to $n$ for each $n \in \mathbb N$. Notice that $\mathbb N$ is the set of all standard naturals in $M$. It's easy to define the restrictions on membership relations, the types are the stages of $M$ that are the preimages of the intergers under the above assignment, and the rest goes through easily.
| 1 | https://mathoverflow.net/users/95347 | 427213 | 173,315 |
https://mathoverflow.net/questions/427124 | 5 | Consider the problem
$$(\star) \quad \begin{cases} \frac{d}{dt} X(t,x) = \sqrt{X(t,x)}, &t \in [0,T],\\
X(0,x) = x, &x \in \mathbb R
\end{cases}
$$
This is the prototype of non-uniqueness for ODEs in the classical sense. I'm tempted to say, however, that the [Lagrangian flow in the sense of Ambrosio](http://cvgmt.sns.it/media/doc/paper/1612/crippa_bari.pdf) exists and should be made of trajectories that don't "stay" at zero. Is this intuition correct? How can the regular Lagrangian flow be computed explicitly in this example?
For the existence and uniqueness of RLF for this example (which is not covered by the classical DiPerna-Lions theory), see [this paper](https://cvgmt.sns.it/media/doc/paper/1272/onedim20.pdf).
| https://mathoverflow.net/users/122620 | Regular Lagrangian flow for "square root example": $\frac{d}{dt} X(t,x) = \sqrt{X(t,x)}$ | Your intuition is right. The key is in the paper you cite, in that they consider uniqueness in the class $L^1\_{\text{loc}}$, which does not allow for concentrations. If you add to this, that the Lagrangian flow conserves mass and that the the classical ODE has uniqueness away from zero, you thus get that the mass transported upwards to 0 from the negative real axis has to immediately passed on to the positive real axis.
So denote by $\Phi(t,x)$ the flow of the ODE with $\Phi(0,x)=x$ and $\partial\_t \Phi(t,x) = \sqrt{|\Phi(t,x)|}$ (I assume the missing absolute values were a typo) as well as the added condition that it is strictly monotone, i.e. at 0 you always pick the solution that immediately continues onwards. This you can write down explicitly, with some tedious case distinctions, depending on if you switch sign or not. If I didn't miscalculate, it should be something like
$$ \Phi(t,x) := \begin{cases} \frac{1}{4}(t+2\sqrt{x})^2 & \text{for } x\ge 0 \\ \frac{1}{4}(t-2\sqrt{|x|})^2 &\text{for } x<0, 2\sqrt{|x|} \le t \\ -\frac{1}{4}(2\sqrt{|x|}-t)^2 &\text{for }x<0,2\sqrt{|x|} > t. \end{cases} $$
Then the solution to the corresponding continuity equation should be given by
$$u(t,\cdot) := u\_0 \circ \Phi(t,\cdot)^{-1} \partial\_x\Phi(t,\cdot)^{-1},$$
which again you can construct explicitly.
Let me however stress again that this is not necessarily "the solution" to the problem. In many contexts, it is equally possible to say that the natural class for solutions is the class of measures. In that case you are allowed to accumulate mass at zero and as a result can construct infinitely many solutions by only passing on some of the mass. So in some sense the reformulation into a continuity equation still retains the non-uniqueness of the ODE and only using some selection principle can you expect uniqueness. It just so happens that in this class of ODEs, disallowing concentrations is precisely enough to obtain uniqueness while still guaranteeing existence.
| 4 | https://mathoverflow.net/users/51695 | 427214 | 173,316 |
https://mathoverflow.net/questions/426737 | 3 | Following Anton Petrunin’s suggestion, I revise the question to make it less vague.
Let $M^{m}$ be an $m$-dimensional Riemannian manifold, and let $\gamma$ be a unit-speed curve $I \to M^{m}$. We say that $\gamma$ is *planar* if there are a unit vector field $N$ along $\gamma$ and a function $\kappa$ such that
$$\begin{aligned}
D\_{t}\gamma' &= \kappa N,\\
D\_{t} N &= -\kappa \gamma',
\end{aligned}$$
where $D\_{t}$ is the covariant derivative along $\gamma$.
I know that, if $M^{m}$ has constant curvature, then studying planar curves in $M^{m}$ is really the same as studying planar curves in $M^{2}$; indeed, any planar curve in $M^{m}$ is contained in a two-dimensional totally geodesic submanifold of $M^{m}$.
I am interested in the case where the curvature is not necessarily constant. Is there any good reason to study planar curves in $M^{m}$ instead of just $M^{2}$?
More precisely, suppose that $\gamma$ is a planar curve in $M^{m}$. Does there exists a two-dimensional Riemannian submanifold $S$ of $M^{m}$ such that $\gamma(I) \subset S$ and such that
$$R\_{S}(X,Y,Z,W) = R\_{M}(X,Y,Z,W)$$
for all $X,Y,Z,W \in \mathfrak{X}(S)$? Here $R\_{S}$ and $R\_{M}$ denote the Riemann curvature tensors of $S$ and $M^{m}$, respectively.
| https://mathoverflow.net/users/74033 | Planar curves in $M^{m}$ vs curves in $M^{2}$ | Here are some comments about the OP's question that don't give a definitive answer to the final question (although the answer may well be 'no', see below), but do provide more information, at least in the simplest possible case, in which $M$ has dimension $3$.
First, by the Gauss equation in dimension $3$, a surface $S\subset M$ has the property $R\_S(W,X,Y,Z) = R\_M(W,X,Y,Z)$ for all tangent vector fields $W,X,Y,Z$ on $S$ if and only if the second fundamental form of $S$ in $M$ has rank at most $1$. [Elsewhere](https://mathoverflow.net/questions/427092/extrinsically-flat-submanifolds-of-a-riemannian-manifold), the OP has called such submanifolds 'essentially flat' (to be abbreviated EF below). To simplify the discussion to follow, I want a name for surfaces $S\subset M$ whose second fundamental form has rank *equal* to $1$ everywhere on $S$. I propose 'nondegenerate essentially flat', to be abbreviated NDEF.
Now, in the real-analytic setting (i.e., all given data are real-analytic, including the metric on $M$), the Cartan-Kähler theorem implies the following result:
**Proposition:** Let $\gamma:I\to M$ be an embedded unit speed curve defined on some interval $I\subset\mathbb{R}$ and suppose that $\nu:I\to TM$ is a unit vector field along $\gamma$ that satisfies $\nu(t)\cdot\gamma'(t) = 0$ while $\nu(t)\cdot \nabla\_{\gamma'(t)}\gamma'(t)\not=0$ for all $t\in I$. Then there exists an embedded NDEF surface $S\subset M$ that contains $\gamma(I)$ and has $\nu(t)\perp T\_{\gamma(t)}S$ for all $t\in I$. Moreover, any other NDEF surface $\tilde S\subset M$ with these properties agrees with $S$ on some neighborhood of $\gamma(I)$ in $M$.
Note that this answers the OP's final question in the affirmative in the real-analytic setting as long as the function $\kappa$ is non-vanishing. However, I suspect that the OP actually wants to ask an even more specific question in the case of a planar curve: *Does there exist a EF surface $S\subset M$ containing $\gamma(I)$ such that both $\gamma'$ and $N$ are tangent to $S$ along $\gamma(I)$?* Even in the real-analytic setting, the Cartan-Kähler theorem cannot be applied to construct such a surface because choosing $\nu = \gamma'\times N$ would have $\nu(t)\cdot \nabla\_{\gamma'(t)}\gamma'(t) = 0$, so that the above Proposition cannot be applied.
Now, the Proposition almost certainly will not hold in the smooth setting without further hypotheses. The reason is that the PDE that the Cartan-Kähler theorem is solving is an initial value problem for a second order PDE that is of degenerate type, i.e., neither elliptic nor hyperbolic and the characteristics of the PDE are of multiplicity $2$. (I hesitate to call this type of system 'parabolic' because, while everyone agrees that $u\_{xx}-u\_t=0$ is a parabolic equation, not everyone agrees that $u\_{xx} = 0$ is a parabolic equation.) While I have not carefully written out the details, I'm pretty sure that I could construct a (real-analytic) Riemannian manifold $M^3$ and a (real-analytic) unit speed curve $\gamma:I\to M$ and a (smooth but not real-analytic) normal vector field $\nu:I\to TM$ along $\gamma$ that satisfy the conditions of the Proposition but for which the desired NDEF surface $S$ does not exist.
Now, in the smooth setting, given an NDEF surface $S\subset M^3$, $S$ will be foliated by the null curves of its second fundamental form, and *all of these null curves will be 'planar' by the OP's definition*. In fact, these null curves are exactly the characteristics of the PDE that defines NDEF surfaces to begin with. Thus, the more `refined' question mentioned above is actually the 'characteristic initial value problem' that is typical of parabolic PDE. (For example, the usual IVP for the heat equation $u\_t = u\_{xx}$ is to specify $u$ along $t=0$, which is clearly a characteristic IVP. It is instructive to note that, even in this very simple setting, one generally gets a solution to the initial value problem only for $t\ge 0$, i.e., there will be no solution matching the initial conditions for $t<0$, even when the specified initial value is real-analytic.)
What I suspect is that the best one can do (in either the real-analytic or smooth setting) for the 'generic' Riemannian $3$-manifold and planar curve $\gamma:I\to M$ would be to prove that, under certain conditions of non-degeneracy that would be tedious to spell out here, there will exist a (real-analytic or smooth) NDEF surface $S\subset M$ with boundary $\gamma(I)$ such that $N$ is tangent to $S$ along $\gamma(I)$, and, 'generically', it will not be possible to extend $S$ as a (real-analytic or smooth) NDEF surface that contains $\gamma(I)$ in its interior.
| 3 | https://mathoverflow.net/users/13972 | 427222 | 173,321 |
https://mathoverflow.net/questions/427223 | 2 | If I am looking at a collection $\mathcal{C}$ of circles $\{C\_1,...,C\_n\}$ all of which have some radii $\{r\_1,...,r\_n\}$ where $r\_i\in\mathbb{R}^{+}$ for each $i \in[n]$. In $\mathcal{C}$, all the circles are pairwise intersecting i.e. each circle intersects another circle. We do not allow touchings (i.e. the circles only touch at one single point). Note that we also do not allow faces in these arrangements that are bounded by two arcs and two intersection points.
We define a $3$-cell in this collection of circles as a cell bounded by three arcs and three intersection points of the circles. Moreover, an empty $3$-cell is a $3$-cell such that no circle passes through it.
I am trying to prove or disprove the following: each circle $C\_i\in\mathcal{C}$ bounds at least two empty $3$-cells so that one of them is inside $C\_i$ and one of them on the outside of $C\_i$.
I have the some of these following ideas: fix $C\_i$ and give it a certain direction and split the intersection points of $C\_i$ with other circles into two sets: $\{p\_1,...,p\_{n-1}\}$ being the first intersections of $C\_i$ with $C\_j$ for $i\neq j$ and $\{q\_1,...,q\_{n-1}\}$ being the set of the second such intersections.
Now, it seems like any set of three circles creates exactly $7$ $3$-cells, however, in the entire $\mathcal{C}$ these are not necessarily empty, and this is the problem. Therefore as a whole, there are the total of $7\cdot \binom{n}{3}$ $3$-cells.
Another thing worth noting looking at is how the circles behave on the inner and the outer side of the fixed $C\_i$ based on the order of their intersection with $C\_i$, but I am not entirely sure how to go from here.
Perhaps
| https://mathoverflow.net/users/485561 | Pairwise intersecting circles in the plane | First, it is sufficient to show that there is always an empty 3-cell inside each circle. This is because we can always do a Möbius transformation to move the outside of a circle inside of it, which will preserve intersections between circles. The same argument can then be used.
Now pick some circle $C$. First, there must exist $C\_i$ and $C\_j$ that form an empty 3-cell in $C$, call it $T$. If this was not the case, no pair of the other circles could intersect each other inside of $C$. But then there would be a face bounded by two arcs in $C$.
Next consider what happens as you draw another circle, say $C\_k$. If it intersects $T$ at 0 or 1 points, $T$ remains a 3-cell. If it intersects two different edges of $T$, you get a smaller 3-cell inside of $T$. If it intersects one edge of $T$ twice, then by the assumption that no faces are bounded by two arcs, there must be another circle $C\_\ell$ so that $\{C\_i,C\_j,C\_k,C\_\ell\}$ contain a 3-cell inside $T$. By repeating this argument until we are out of circles, we find that there there is some empty 3-cell.
| 2 | https://mathoverflow.net/users/141277 | 427237 | 173,327 |
https://mathoverflow.net/questions/427196 | 1 | I have a curious question I stumbled upon that may be interesting to some.
Consider **real**-valued continuous functions on the circle $f\_1(x),f\_2(x),f\_3(x)$ (so they are periodic in $x \mapsto x+2\pi$).
They have the following properties:
1. $\int\_0^{2\pi} \frac{dx}{2\pi} f\_i(x) = 0$
2. $ \int\_0^{2\pi} \frac{dx}{2\pi} f\_i(x) f\_j(x) = \frac{1}{3} \delta\_{ij}$
3. $f\_1(x)^2 +f\_2(x)^2 + f\_3(x)^2 = 1$ for all $x$.
Considered as square-summable vectors on the circle, the first two conditions mean that $f\_i(x)$ are orthogonal to the constant function $1$, and they are mutually orthogonal, each with norm $\frac{1}{\sqrt{3}}$. There are infinitely many functions that satisfy properties 1 and 2.
But is there a solution now also requiring property 3 to hold?
If yes, what is an example of such a set of functions? (Or more generically, how to construct them in general?)
If no, how do we rule them out?
Thanks.
\*\*Edit: if continuity is replaced by smoothness (at least 1-differentiable), does anything change?
| https://mathoverflow.net/users/317106 | Orthogonal functions on circle with constraints | I think we can do it smoothly. Think of $f=(f\_1,f\_2,f\_3)$ as a map from the circle into $\mathbb R^3$. I need it to take values in the unit sphere. Mark the three obvious great circles on the sphere, and note the twelve quarter-circles in this picture.
Here is a continuous solution. As $x$ goes from $0$ to $\pi/6$, $f(x)$ goes from $(1,0,0)$ to $(0,1,0)$ following a great circle path at constant speed. As $x$ goes from $\pi/6$ to $\pi/3$, $f(x)$ follows another such path, from $(0,1,0)$ to, say, $(0,0,1)$. You can keep going like this, in a twelve-part path that eventually traverses each of those quarter-circles once. The average of $f\_i$ is $0$ because four of the twelve terms are $0$ and the remaining eight cancel in pairs. For $i\neq j$ the average of $f\_if\_j$ is $0$ because eight of the twelve terms are $0$ and the remaining four cancel in pairs. The average of $f\_i^2$ is independent of $i$.
This can be modified to a smooth solution by arranging for each of the twelve paths to be not a constant-speed path but rather one that travels along a quarter-circle beginning and ending with a brief interval of staying still, in a symmetrical way.
| 3 | https://mathoverflow.net/users/6666 | 427240 | 173,328 |
https://mathoverflow.net/questions/424141 | 11 | Let $G$ be a (connected) reductive group over a perfect field $k$, and let $\xi\in H^1(k,G)$ be a cohomology class.
By a theorem of Steinberg (Serre, Galois cohomology, Appendix 1 to Chapter III, Theorem 11.1),
*if $G$ is quasi-split,* then there exists a $k$-torus $i\colon T\hookrightarrow G$ such that
$$\xi\in i\_\* \big(H^1(k,T)\big).$$
>
> **Question.** What is an example of $k$, a non-quasi-split reductive $k$-group $G$, and $\xi\in H^1(k,G)$, such that $\xi$ *does not come* from a $k$-torus?
>
>
>
**Remark.** Such $k$ cannot be $\Bbb R$, a $p$-adic field or a number field.
Indeed, let $G$ be a reductive $k$-group, *not necessarily quasi-split*.
If $k$ is $\Bbb R$ or a $p$-adic field, then there exists a $k$-torus $i\colon T\hookrightarrow G$
such that the map
$$i\_\*\colon H^1(k,T)\to H^1(k,G) $$
is surjective.
If $k$ is a number field, then for any finite set $\Xi=\{\xi\_1,\dots,\xi\_n\}\subset H^1(k,G)$ there exists a $k$-torus
$i\colon T\hookrightarrow G$ such that $\Xi\subset i\_\*\big( H^1(k,T)\big)$.
For the last assertion, see M. Borovoi, Abelian Galois cohomology of reductive groups, Mem. Amer. Math. Soc. 132 (1998), no. 626,
Theorem 5.10.
| https://mathoverflow.net/users/4149 | Galois cohomology class of a reductive group not coming from a torus | $\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}$The following example is due to Vladimir Chernousov (private communication).
Let $K={\Bbb Q}(x,y,x',y')$, where $x,y,x',y'$ are variables.
Consider the quadratic forms over $K$
$$ f= \la x,y,-xy\ra\qquad\text{and}\qquad f'=\la x',y',-x'y'\ra.$$
Here $ \la x,y,-xy\ra$ denotes the quadratic form $$f(t\_1,t\_2,t\_3)=xt\_1^2+yt\_2^2-xyt\_3^2.$$
>
> **Lemma 1.** The quadratic form over $K$
> $$ f-f'=\la x,y,-xy, -x',-y', x'y'\ra$$
> is anisotropic (does not represent 0 nontrivially).
>
>
>
*Proof.* The quadratic form $f-f'$ is monomial and multiplicity free
in the sense of Section 4 of Vladimir Chernousov and Jean-Pierre Serre,
*Lower bounds for essential dimensions via orthogonal representations,*
J. Algebra 305 (2006), no. 2, 1055–1070.
By their Proposition 5 on page 1061, the quadratic form $f-f'$ is anisotropic,
as required.
Consider the quaternion $K$-algebras
$$ D=(x,y)\qquad \text{and} \qquad D'=(x',y').$$
Here $(x,y)$ denotes the 4-dimensional associative $K$-algebra with generators $i,j$ and relations
$$i^2=x, \quad j^2=y,\quad ij=-ji.$$
The reduced norm form for $D$ is isomorphic to $\ \la 1\ra-f=\la 1,-x,-y,xy\ra$.
>
> **Lemma 2.** The algebraic $K$-groups $G={\rm PGL}(1,D)$ and $G'={\rm PGL}(1,D')$ have no isomorphic maximal tori.
>
>
>
*Proof.*
Assume for the sake of contradiction that $G$ and $G'$ have isomorphic maximal tori.
Then $D$ and $D'$ have isomorphic maximal subfields, say, $L\subset D$ and $L'\subset D'$.
Write $L=K(\sqrt{a})\subset D$ for $a\in K$.
This implies that there exists a pure quaternion $t\_1 i+t\_2 j+t\_3ij$ such that
$$xt\_1^2+yt\_2^2-xyt
\_3^2=a.$$
Thus the quadratic form $f$ represents $a$.
Similarly, from $L'=K(\sqrt{a})\subset D'$ (with the same $a$) we obtain that $f'$ represents $a$.
Therefore, the quadratic form $f-f'$ represents 0 nontrivially, which contradicts Lemma 1.
>
> **Theorem.** For $G$ and $G'$ as above, let $c\in Z^1(K,G)$ is a 1-cocycle such that $\_c G\simeq G'$.
> Then the cohomology class $[c]\in H^1(K,G)$ does not come from a maximal torus of $G$.
>
>
>
*Proof.*
Assume for the sake of contradiction that there exists a maximal torus
$$ j\colon T\hookrightarrow G\quad\text{such that }\quad [c]\in j\_\* H^1(K,T).$$
Then there exists an embedding
$$j'\colon T\hookrightarrow {}\_c G\simeq G'.$$
Thus $G$ and $G'$ have isomorphic maximal tori, which contradicts Lemma 2.
| 10 | https://mathoverflow.net/users/4149 | 427289 | 173,339 |
https://mathoverflow.net/questions/427272 | 3 | I'm trying to understand variations of Hodge structure. I understand that this is a very broad field, and that many of the concepts have been extended to algebraic geometry over fields other than $\mathbb{C}$, and so forth. In particular, there is a version of the monodromy theorem by Grothendieck, which is rather incomprehensible to me. I am asking this question in the context of variations of Hodge structure for Calabi-Yau manifolds and mirror symmetry, so I hope that the answer can be given with that in mind.
Given a family of projective Kähler manifolds $\pi:\mathcal{X}\to\Delta^\*$, where $\Delta^\*$ denotes the punctured disk, we get a monodromy operator $T\in\text{GL}(H^k(X\_t,\mathbb{C}))$, for $t\in\Delta^\*$. This operator is obtained from the local system associated to the family, which gives a representation $\rho:\pi\_1(\Delta^\*)\to\text{GL}(H^k(X\_t,\mathbb{C}))$, and $T$ is the image of a generator of the fundamental group under this representation. One can then show that $T$ is quasi-unipotent, meaning that there are integers $m,N$ such that $(T^m-I)^N=0$. After pulling back along the map $z\mapsto z^m$, we may as well assume that $m=1$, so that $T$ is in fact unipotent. One then defines the nilpotent operator
$$N=\log T=(T-I)+\dots+(-1)^{N+1}(T-I)^N/N!$$
Associated to this operator is a weight filtration of $H^k(X\_t,\mathbb{Q})$, called the monodromy weight filtration. It is uniquely defined by
$$N(W\_i)\subseteq W\_{i-2}\quad\quad N^k:W\_{N+k}/W\_{N+k-1}\xrightarrow{\cong}W\_{N-k}/W\_{N-k-1}$$
I have two questions about this.
1. Why do we impose this filtration on the cohomology with rational coefficients, rather than with complex coefficients? Does this make a difference, if we restrict our attention to complex manifolds and their variations of Hodge structure?
2. What does this filtration represent, geometrically? Up until this point, there has been a nice geometric interpretation for the concepts which were introduced (e.g. the Hodge bundles, Gauss-Manin connection, the period map, etc.), but I have no idea how to think about this filtration in geometric terms.
| https://mathoverflow.net/users/168781 | What does does the monodromy weight filtration represent? | Given a nilpotent endomorphism $N$ of a finite dimension vector space $V$, Jordan canonical form implies that we can decomponse $V$ into a sum of "blocks" on which we can find bases satisfying $Ne\_1 = e\_{2}, Ne\_2=e\_3,\ldots Ne\_k=0$. To make it basis independent, pass to the filtration
$$ \langle e\_k\rangle\subset \langle e\_k, e\_{k-1}\rangle \subset \ldots$$
After suitable indexing, this is the monodromy weight filtraton $W\_\bullet$.
The point I wanted to make is that $W$ may seem complicated, but it isn't that bad. Furthermore the construction works over $\mathbb{Q}$ (because $N$ has zero eigenvalues, and so has a Jordan form over this field).
This is also useful further along in the story, when one gets to limit mixed Hodge structures: part of the definition of a mixed Hodge structure is that $W$ is rational.
Regarding your second question, I don't really have a good answer. It is clear that $W$ is quite natural from the point of linear algebra, and Schmid gave an analytic interpretation in terms of certain growth conditions. However, the geometric meaning seems much more elusive.
| 4 | https://mathoverflow.net/users/4144 | 427294 | 173,342 |
https://mathoverflow.net/questions/427301 | 3 |
>
> In $\mathbb{F\_p}$, $p$ prime what is the larget subset $S$ of quadratic residues with no pair of elements differing by 1?
>
>
>
In [this](https://mathoverflow.net/q/427035/7113) related question Seva gives an example:
"...assuming $p\equiv\pm3\pmod 8$, consider the set of all those quadratic residues $r$ such that if $r'$ is the smallest quadratic non-residue exceeding $r$, then $r'-r$ is odd."
This gives a density $|A|/p = 1/3 + o(1)$ for any $p$ (the condition $p\equiv\pm3\pmod 8$ is not necessary for this problem). Can one do better?
| https://mathoverflow.net/users/7113 | Largest subset of quadratic residues with no pair of elements differing by 1 | No.
Let $T$ be any subset of $\mathbb F\_p$. We can consider the general problem of finding a set $S \subseteq T$ with no pair of elements differing by $1$ that maximizes $|S|$.
The same construction works, i.e. $\{x \in T \mid x-y$ is odd for $y$ the largest nonmember of $T$ less than $x \}$ attains the maximal cardinality.
Proof: Consider another solution $S'$. For each $x\in S$, only one of $x$ and $x+1$ can lie in $S'$. Every $x \in T$ lies in one of these pairs since if $x-y$ is not odd then it's even and $x-1-y$ is odd (and $x-1 \in T$ since $x-y >0$ and is even and thus is $\geq 2$).
So the maximum size of $S'$ is at most the number of such pairs. Since these pairs don't overlap, and every pair contains one element of $S$, this is also the size of $S$, QED.
There's also an intuitive/visual proof by example. If $T$ is the set of stars
```
_***_**_*_*****_
```
then S is the set of x's
```
_x*x_x*_x_x*x*x_
```
and this is clearly optimal
| 2 | https://mathoverflow.net/users/18060 | 427305 | 173,346 |
https://mathoverflow.net/questions/427274 | 5 | Consider a smooth 2d-manifold $M$ and let $g$ be a smooth $(0,2)$-tensor satisfying $rk(g)\geq1$ everywhere. Obviously if $rk(g)=2$ at a point $p\in M$ then $g$ is locally diagonalisable (i.e. there exists a local coordinate system in which $g$ is diagonal). The same conclusion holds if $rk(g)=1$ in a neighbouhood of a point. But:
**Question**: If $rk(g)=1$ at a point $p\in M$ but $rk(g)=2$ elsewhere, is it possible to find a system a local coordinate system such that $g$ is diagonal, i.e.
$$
g=\begin{pmatrix}\lambda\_1&0\\0&\lambda\_2\end{pmatrix}
$$
with $\lambda\_1(p)=0$ and $\lambda\_2(p)\neq0$.
| https://mathoverflow.net/users/153070 | Local diagonalisation of a degenerated 2d metric tensor | The answer is 'yes'. Here is how one can see this: Suppose that $g$ is a $(0,2)$ form on a neighborhood of the origin in the $xy$-plane such that the rank of $g$ is $1$ at the origin and $2$ everywhere else. Let $h = \mathrm{d}x^2 + \mathrm{d}y^2$ and let
$$
g = E(x,y)\,\mathrm{d}x^2 + 2 F(x,y)\,\mathrm{d}x\,\mathrm{d}y + G(x,y)\,\mathrm{d}y^2.
$$
The symmetric matrix
$$
S(x,y) = \begin{pmatrix} E(x,y) & F(x,y)\\ F(x,y) & G(x,y)\end{pmatrix}
$$
then has the property that it has two distinct eigenvalues at the origin $(x,y) = 0$ and hence has two distinct eigenvalues on an open disk $U$ containing the origin. Consequently, the eigenvalues are smooth functions of $(x,y)$ on $U$, as are their corresponding eigenvectors, which can be thought of as orthogonal unit vector fields with respect to $h$. Thus, we can write
$$
g = \mu\_1(x,y)\,\alpha^2 + \mu\_2(x,y)\,\beta^2
$$
where the functions $\mu\_i$ on $U$ are the eigenvalues of $S$, with the property that $\mu\_1(0,0) = 0$, but $\mu\_2(0,0)\not=0$ and $\alpha$ and $\beta$ are nonvanishing $1$-forms on $U$ such that $\alpha^2 + \beta^2 = h$.
Now, by shrinking $U$ if necessary, we can find nonvanishing functions $u$ and $v$ on $U$ such that $\alpha = p\,\mathrm{d}u$ and $\beta = q\,\mathrm{d}v$. (This is the classical fact that every nonvanishing $1$-form on the plane is locally the multiple of an exact form, i.e., the existence of 'integrating factors'.) Then we have
$$
g = \lambda\_1\,\mathrm{d}u^2 + \lambda\_2\,\mathrm{d}v^2,
$$
where $\lambda\_1 = p^2\mu\_1$ and $\lambda\_2 = q^2\mu\_2$, as desired.
| 11 | https://mathoverflow.net/users/13972 | 427307 | 173,347 |
https://mathoverflow.net/questions/427302 | 4 | In category theory, a dagger category is a precategory $\mathcal{C}$ such that for every pair of objects $A:\mathcal{C}$ and $B:\mathcal{C}$ there is a function $(-)^{\dagger\_{A,B}}:\mathrm{Mor}(A,B) \to \mathrm{Mor}(B,A)$ such that
* for all objects $A:\mathcal{C}$, $\mathrm{id}\_A^{\dagger\_{A,A}} = \mathrm{id}\_A$,
* for all objects $A:\mathcal{C}$, $B:\mathcal{C}$, and $C:\mathcal{C}$, and morphisms $f:\mathrm{Mor}(A,B)$ and $g:\mathrm{Mor}(B,C)$, $(g \circ f)^{\dagger\_{A,C}} = f^{\dagger\_{A,B}} \circ g^{\dagger\_{B,C}}$
* ${(f^{\dagger\_{A,B}})}^{\dagger\_{B,A}} = f$
and which satisfies an equivalent Segal condition using unitary isomorphisms as categories do using isomorphisms.
In the context of higher category theory, does the concept of $(n,1)$-dagger categories make sense, and if so, how does one go about defining them?
| https://mathoverflow.net/users/483446 | $(n,1)$-dagger categories | Well, it is easy to give definitions, the problem is finding the "right" one.
Here "right" can mean that gives the correct notion up to homotopy (many definition will be equivalent) but also it can mean the one that let you the most easily talk about the examples and constructions you care about ( this much more subjective and it depends on why you want this definition)
I'll propose two (edit, three) definitions - there is a good chance that they are equivalent in some appropriate homotopical sense given they are both fairly reasonable, but I don't know if they are.
If you don't mind I'll work with $(\infty,1)$ instead of $(n,1)$ because it is simpler. If you are interested in $n=2$ or $n=3$ more specifically, then that's definitely a different story - and here we might come up with a long but explicit definition.
Before we start, a little bit of introspection on the definition is needed. I think in the case of Dagger category that has been nicely done [here](https://mathoverflow.net/questions/220032/are-dagger-categories-truly-evil). The problem with the definition of Dagger categories is that it break the "equivalence principle" because it impose an equation $x^\*=x$ on objects, or more fundamentally because it is not invariant under equivalence of categories (a category equivalent to a Dagger category is not automatically a dagger category), so even if you phrase the definition in a way that avoid writting $x^\* = x$ something else will break the equivalence principle.
The MO discussion [linked](https://mathoverflow.net/questions/220032/are-dagger-categories-truly-evil) above discuss this point (it's interesting, read it !), and arrive at the conclusion ([this answer](https://mathoverflow.net/a/220111/22131)) that the good way to conceptualize a dagger category is a pair of a category $C$ together with a groupoid $C\_u \to C$ where $C\_u$ is the groupoids of object and dagger-isomorphism between them. The structure of Dagger category can be then added in a "non evil" way on top of this.
This idea of a category $C$ together with a groupoid $C\_u$ and an essentially surjective functor $C\_u \to C$ is conveniently represented in $\infty$-category theory by a [Segal space](https://ncatlab.org/nlab/show/Segal+space) not satisfying the completeness condition.
So it is natural to define Dagger $(\infty,1)$-categories as Segal spaces with structure.
This leads to:
**First definition :**
For $n \in \mathbb{N}$ we denote as usual by $[n]$ the poset (seen as a category)
$$ 0 \to 1 \to \dots \to n $$
and we denote by $[n]\_\dagger$ the free dagger category on $[n]$. Note that $[n]\_\dagger$ is relatively easy to describe explicitely : $[n]$ is free on the graph with only arrow $i \to i+1$, so $[n]\_\dagger$ is free on the graph
$$ 0 \leftrightarrows 1 \leftrightarrows \dots \leftrightarrows n $$
and so arrows in $[n]\_\dagger$ are justs path in this fairly simple graph.
I denote by $\Delta\_\dagger$ the full subcategory of Cat on the objects $[n]\_\dagger$, it admit $\Delta$ as a non-full subcategory.
**Definition :** A Dagger $(\infty,1)$-precategory is a presheaf of spaces on $\Delta\_\dagger$ whose restriction to $\Delta$ is a Segal space.
A Dagger $(\infty,1)$-category should then be a Dagger $(\infty,1)$-precategory that satisfies an appropriate Segal condition, but a version of the Segal condition that uses the Dagger structure: given a Dagger $(\infty,1)$-precategory $X$ one should construct a space of "Dagger isomorphisms" which factor the diagonal $X([0]) \to X\_I \to X([0])^2$ and define a Dagger $(\infty,1)$-category as being a precategory such that the map $X([0]) \to X\_I$ is an equivalence.
Some additional thought is needed to determine what exactly should be the correct definition of $X\_I$, but I would guess something like the space of $f$ such that $f^\* f \sim Id$ and $f$ is an equivalence. Or maybe more conceptually, take $I$ to be the Dagger version of the walking isomorphism and define $X\_I$ as the space of maps $I \to X$ in the natural model structure for Dagger $(\infty,1)$-precategories (the Bousfiled localization of the projective/injective model structure).
*Note :* the intuition behind that definition is that Dagger categories are monadic over categories (in the 1-categorical sense) and that monads is "Nervous" in the sense of [Bourke and Garner](https://arxiv.org/abs/1805.04346) for the category of arities $\Delta$, so following the idea suggested by Meadow myself [here](https://arxiv.org/abs/2106.02706) this provides a natural way to extend the definition to an $\infty$-categorical one. That's exactly what the above definition of Dagger $(\infty,1)$-precategory above is. To say this more naively, $X([n])$ represents the space of string of n composable arrows in $X$, and the category $\Delta\_\dagger$ encodes all the way you can transform such string in a dagger categories.
**Second definition:** One can also follow a much more naive route. One takes $\infty$-categories to mean quasi-categories, and then it actually make sense to ask for a quasi-category with an identity on object involution.
**Definition 2 :** A Dagger simplicial set is a simplicial set $X$ with for each $n$ an involutive map $\dagger: X(n) \to X(n)$ such $\dagger$ is the identity on $X(0)$ and collectively they form a morphism of simplicial sets $X \to X^{op}$ (where $X^{op}$ is defined by taking $X^{op}(n) = X([n]^{op})$ )
Then I can for exemple define a Dagger Quasi-category as a Dagger simplicial set which is a quasi-category.
**Are they equivalent ?**
Well, the question don't quite make sense yet, because the second definition is missing a key ingredient: *what are the equivalences of Dagger quasi-categories ?* It is not clear it is reasonable to define the equivalence as the morphism of dagger-simplicial sets that are equivalence of quasi-categories, nor to use an explicit definition with morphism having a "dagger inverse".
For the first definition, the notion of equivalence is easy as we have a nice Quillen Model structures on the category of presheaves of spaces on $\Delta\_\dagger$ by starting with either the projective or the injective model structure and then localising so that the fibrant objects are the Dagger $(\infty,1)$-categories.
So I'm proposing:
**Conjecture :** There exists a model structure on the category of Dagger simplicial sets whose fibrant objects are (certain) Dagger quasi-categories and which is Quillen equivalent to the model structure for Dagger $(\infty,1)$-categories.
I'm phrasing it as a conjecture, because I tend to think this is true - and probably within reach with a bit of work, but there are also definitely a few things that could go wrong making it false.
---
**Edit: A third definition**
So the suggestion by Achim Krause doesn't quite work, but one can adjust it so that it does work.
Taking the "opposite category" produce an action of the two elements group $C\_2$ on $Cat\_\infty$. Now, as a generalization of the fact that every groupoid is equivalent to its opposite through the functor which is the identity on object and send every arrow to its inverse, the restriction of this action to $Gpd\_\infty$ should admit a trivialization (I don't know if this has been explicitely constructed anywhere though). In particular, we have functor $Gpd\_\infty \to Cat\_\infty^{C\_2}$.
**Definition:** The $(\infty,1)$-category of Dagger $(\infty,1)$-precategories is defined as the pullback $Gpd\_{\infty} \times\_{Cat\_\infty^{C\_2}} E^{C\_2}$ where $E$ is the full subcategory of $(Cat\_\infty)^{[1]}$ of essentially surjective functor and its map to $Cat\_\infty$ is the source functor.
(the $C\_2$ exponent means the space of homotopy fix point)
Informally, we are saying that a Dagger precategory is an $\infty$-precategory (in the sense of an essentially surjective functor $G \to C$ where $G$ is an $\infty$-groupoid) endowed with an involutive equivalence with $G^{op} \to C^{op}$ (that is what an element of $E^{C\_2}$ is ) such that the component of this equivalence on $G$ identifies (as an involution) the "obvious" equivalence $G \simeq G^{op}$ coming fron the fact that $G$ is an $\infty$-groupoid.
This last condition implies both that the dagger is the identity on object and that arrows in $G$ get sent to dagger-isomorphism.
Now this is only a Dagger precategory, one further needs to impose a Segal type condition that will enforce that $G$ is the $\infty$-groupoid of dagger-isomorphism. So, one needs to give a definition of dagger isomorphism in this structure, build a natural map from the Hom space of $G$ to the spaces of dagger isomorphisms, and ask this map to be an equivalence.
It also seems reasonable to conjecture that this $\infty$-category is equivalent to the other two definitions.
| 12 | https://mathoverflow.net/users/22131 | 427322 | 173,355 |
https://mathoverflow.net/questions/427330 | 1 | Let
* $sd : sSet \to sSet$ denote barycentric subdivsion;
* $cosk\_1 : sSet \to sSet$ denote 1-coskeletalization.
**Question:** Let $X$ be a graph or simplicial set. If the homotopy type of $cosk\_1(X)$ is known, then what can be said about the homotopy type of $cosk\_1(sd(cosk\_1(X)))$?
I'm interested in this question primarily in the case where my simplicial sets are in fact abstract simplicial complexes. This special case of the question can be read as follows. Let
* $sd X$ denote the barycentric subdivision of an abstract simplicial complex $X$;
* $cosk\_1(X)$, for an abstract simplicial complex $X$, denote the largest abstract simplicial complex with the same vertices and edges as $X$.
**Question (bis):** Let $X$ be a graph or abstract simplicial complex. If the homotopy type of $cosk\_1(X)$ is known, then what can be said about the homotopy type of $cosk\_1(sd(cosk\_1(X)))$?
| https://mathoverflow.net/users/2362 | Barycentric subdivision and 1-coskeletalization | I don't know anything about simplicial sets, but I think the question for simplicial complexes is easy.
A complex is said to be ["flag"](https://en.wikipedia.org/wiki/Clique_complex) (also known as a clique complex) if every subset $S$ for which every pair $u, v\in S$ form an edge is in fact a face. Evidently a flag complex is determined by a graph. So when you are asking about "the largest abstract simplicial complex with the same vertices and edges as $X$," you really just mean the flag complex determined by the graph (i.e., 1-skeleton) of $X$.
To answer your question, we may as well replace $X$ by the flag complex determined by its graph. Then you are asking about how $X$ compares to the flag complex determined by the graph of $\mathrm{sd}(X)$, where $\mathrm{sd}(X)$ is the [barycentric subdivision](https://en.wikipedia.org/wiki/Barycentric_subdivision) of $X$. But it is well known that a barycentric subdivision is already flag (indeed, I think this is where the terminology "flag" comes from: the faces of the barycentric subdivision are precisely flags of faces of the original complex). So, in your language, we have $\mathrm{sd}(X) = \mathrm{cosk}\_1(\mathrm{sd}(X))$ as simplicial complexes, and because $\mathrm{sd}(X)$ is homotopy equivalent (in fact, homeomorphic) to $X$, certainly $X$ has the same homotopy type as $\mathrm{cosk}\_1(\mathrm{sd}(X))$.
| 2 | https://mathoverflow.net/users/25028 | 427338 | 173,356 |
https://mathoverflow.net/questions/427313 | 7 | $A$ and $B$ are two linked (geometric) circles in $\Bbb{R}^3$. (Let, for definiteness, both have radius = 1, the first lies in the $z=0$ plane and its center is the origin of coordinates $(0,0,0)$, the second lies in the $y=0$ plane and its center is the point $(1,0,0)$.)
**Is it true that the circles $A$ and $B$ cannot be separated by a set that is homeomorphic to the $2$-sphere?** (The homeomorphism can be arbitrarily “bad”, as in the case of the horned Alexander sphere.)
(A set $S$ *separates* $A$ and $B$ iff $A$ and $B$ are subsets of different сonnected components of $\mathbb{R}^3\setminus S$.)
I know how to solve this problem in a "smooth" case using knot theory. But this solution doesn't work when the embedding of the $2$-sphere is arbitrarily "bad".
| https://mathoverflow.net/users/486410 | Is it possible to separate two linked (geometric) circles in $\Bbb R^3$ by a set homeomorphic to the 2-sphere (with arbitrarily “bad” homeomorphism)? | By Alexander duality, $\mathbb{S}^3\setminus S$ has two connected components with trivial homologies.
On the other hand, $A$ is nontrivial in $H\_1(\mathbb{S}^3\setminus B)$ — a contradiction.
| 9 | https://mathoverflow.net/users/1441 | 427339 | 173,357 |
https://mathoverflow.net/questions/427156 | 2 | I'm reading a proof of **Theorem 1.37** from Santambrogio's *Optimal transport for applied mathematicians: calculus of variations, PDEs, and modeling*. First, I quote related definitions. Let $X,Y$ be Polish spaces and $\overline{\mathbb{R}} := \mathbb R \cup \{\pm \infty\}$.
---
* **Definition 1.10.** Given a function $\chi: X \rightarrow \overline{\mathbb{R}}$ we define its $c$-transform $\chi^{c}: Y \rightarrow \overline{\mathbb{R}}$ by $\chi^{c}(y)=\inf \_{x \in X} c(x, y)-\chi(x)$. We also define the $\bar{c}$-transform of $\zeta: Y \rightarrow \overline{\mathbb{R}}$ by $\zeta^{\bar{c}}(x)=\inf \_{y \in Y} c(x, y)-\zeta(y)$. Moreover, a function $\psi$ defined on $Y$ is $\bar{c}$-concave if there exists $\chi$ such that $\psi=\chi^{c}$ (and, analogously, a function $\varphi$ on $X$ is said to be $c$-concave if there is $\zeta: Y \rightarrow \overline{\mathbb{R}}$ such that $\left.\varphi=\zeta^{\bar{c}}\right)$.
* **Definition 1.36.** Once a function $c: \Omega \times \Omega \rightarrow \mathbb{R} \cup\{+\infty\}$ is given, we say that a set $\Gamma \subset$ $\Omega \times \Omega$ is $c$-cyclically monotone (briefly $c$-CM) if, for every $k \in \mathbb{N}$, every permutation $\sigma$ and every finite family of points $\left(x\_{1}, y\_{1}\right), \ldots,\left(x\_{k}, y\_{k}\right) \in \Gamma$ we have
$$
\sum\_{i=1}^{k} c\left(x\_{i}, y\_{i}\right) \leq \sum\_{i=1}^{k} c\left(x\_{i}, y\_{\sigma(i)}\right).
$$
Below is the theorem of my interest.
>
> **Theorem 1.37.** If $\Gamma \neq \emptyset$ is a $c$-CM set in $X \times Y$ and $c: X \times Y \rightarrow \mathbb{R}$ (note that $c$ is required not to take the value $+\infty)$, then there exists a c-concave function $\varphi: X \rightarrow$ $\mathbb{R} \cup\{-\infty\}$ (different from the constant $-\infty$ function) such that
> $$
> \Gamma \subset\left\{(x, y) \in X \times Y: \varphi(x)+\varphi^{c}(y)=c(x, y)\right\}.
> $$
>
>
>
**Proof.** We will give an explicit formula for the function $\varphi$, prove that it is well-defined and that it satisfies the properties that we want to impose. Let us fix a point $\left(x\_{0}, y\_{0}\right) \in \Gamma:$ for $x \in X$ set
$$
\begin{aligned}
\varphi(x)=\inf \left\{c\left(x, y\_{n}\right)\right.&-c\left(x\_{n}, y\_{n}\right)+c\left(x\_{n}, y\_{n-1}\right)-c\left(x\_{n-1}, y\_{n-1}\right)+\cdots+\\
&\left.+c\left(x\_{1}, y\_{0}\right)-c\left(x\_{0}, y\_{0}\right): n \in \mathbb{N},\left(x\_{i}, y\_{i}\right) \in \Gamma \text { for all } i=1, \ldots, n\right\}
\end{aligned}
$$
Since $c$ is real-valued and $\Gamma$ is non-empty, $\varphi$ never takes the value $+\infty$. If we set, for $y \in Y$,
$$
\begin{aligned}
-\psi(y)=\inf \left\{-c\left(x\_{n}, y\right)\right.&+c\left(x\_{n}, y\_{n-1}\right)-c\left(x\_{n-1}, y\_{n-1}\right)+\cdots+c\left(x\_{1}, y\_{0}\right)+ \\
&\left.-c\left(x\_{0}, y\_{0}\right): n \in \mathbb{N},\left(x\_{i}, y\_{i}\right) \in \Gamma \text { for all } i=1, \ldots, n, y\_{n}=y\right\} .
\end{aligned}
$$
Note that from the definition we have $\psi(y)>-\infty$ if and only if $y \in\left(\pi\_{y}\right)(\Gamma)$. Moreover, by construction we have $\varphi=\psi^{\bar{c}}$. This proves that $\varphi$ is $c$-concave ${ }^{8}$. The fact that $\varphi$ is not constantly $-\infty$ can be seen from $\varphi\left(x\_{0}\right) \geq 0$ : indeed, if we take $x=x\_{0}$, then for any chain of points $\left(x\_{i}, y\_{i}\right) \in \Gamma$ we have
$$
\sum\_{i=0}^{n} c\left(x\_{i+1}, y\_{i}\right) \geq \sum\_{i=0}^{n} c\left(x\_{i}, y\_{i}\right)
$$
where we consider $x\_{n+1}=x\_{0}$. This shows that the infimum in the definition of $\varphi\left(x\_{0}\right)$ is non-negative.
To prove $\varphi(x)+\varphi^{c}(y)=c(x, y)$ on $\Gamma$ it is enough to prove the inequality $\varphi(x)+$ $\varphi^{c}(y) \geq c(x, y)$ on the same set, since by definition of $c$-transform the opposite inequality is always true. Moreover, since $\varphi^{c}=\psi^{\bar{c} c}$ and $\psi^{\bar{c} c} \geq \psi$, it is enough to check $\varphi(x)+\psi(y) \geq c(x, y)$
Suppose $(x, y) \in \Gamma$ and fix $\varepsilon>0$. From $\varphi=\psi^{\bar{c}}$ one can find a point $\bar{y} \in \pi\_{y}(\Gamma)$ such that $\color{blue}{c(x, \bar{y})-\psi(\bar{y})<\varphi(x)+\varepsilon}$. In particular, $\psi(\bar{y}) \neq \pm \infty$. From the definition of $\psi$ one has the inequality $-\psi(y) \leq-c(x, y)+c(x, \bar{y})-\psi(\bar{y})$ (since every chain starting from $\bar{y}$ may be completed adding the point $(x, y) \in \Gamma)$.
Putting together these two informations one gets
$$
-\psi(y) \leq-c(x, y)+c(x, \bar{y})-\psi(\bar{y})<-c(x, y)+\varphi(x)+\varepsilon,
$$
which implies the inequality $c(x, y) \leq \varphi(x)+\psi(y)$ since $\varepsilon$ is arbitrary.
---
**My question:** In previous paragraphs, the author said that
* $\varphi: X \to \mathbb R \cup \{-\infty\}$ is proper, i.e., $\varphi$ is not identical to $-\infty$. In particular, $\varphi (x\_0) \ge 0$.
* $\psi(y) \neq -\infty$ if and only if $y \in \pi\_{y} (\Gamma)$ where $\pi\_{y}:X \times Y \to Y$ is the projection map.
My concern lies within the sentence
>
> Suppose $(x, y) \in \Gamma$ and fix $\varepsilon>0$. From $\varphi=\psi^{\bar{c}}$ one can find a point $\bar{y} \in \pi\_{y}(\Gamma)$ such that $\color{blue}{c(x, \bar{y})-\psi(\bar{y})<\varphi(x)+\varepsilon}$.
>
>
>
With $x\_{n+1} := x\_0$, we have
\begin{align}
\varphi (x) &:= \inf \left \{ c(x, y\_n) - c (x\_n, y\_n) + \sum\_{i=0}^{n-1} [ c(x\_{i+1}, y\_i) - c(x\_{i}, y\_i) ] \,\middle\vert\, n \in \mathbb N^\*,(x\_i, y\_i)\_{i=1}^n \subset \Gamma \right \} \\
&= \inf \left \{ c(x, y\_n) -c(x\_0, y\_n) + \sum\_{i=0}^{n} [ c(x\_{i+1}, y\_i) - c(x\_{i}, y\_i) ] \,\middle\vert\, n \in \mathbb N^\*,(x\_i, y\_i)\_{i=1}^n \subset \Gamma \right \}.
\end{align}
Notice that $(x\_i, y\_i)\_{i=0}^n \subset \Gamma$, so $\sum\_{i=0}^{n} [ c(x\_{i+1}, y\_i) - c(x\_{i}, y\_i) ] \ge 0$. Then we have an estimate
$$
\varphi (x) \ge \inf \{ c(x, y\_n) -c(x\_0, y\_n) \mid y\_n \in \pi\_{y} (\Gamma)\}.
$$
As the author said
$$
\varphi (x) = \psi^{\bar{c}} (x) = \inf\_{y \in Y} [c(x, y) - \psi (y)].
$$
>
> Then I think the inequality $\color{blue}{c(x, \bar{y})-\psi(\bar{y})<\varphi(x)+\varepsilon}$ is only valid if $\varphi (x) \neq -\infty$. But I could not get how $\varphi (x) \neq -\infty$ in case $(x, y) \in \Gamma$. Could you elaborate on my confusion?
>
>
>
| https://mathoverflow.net/users/477203 | Optimal transport for applied mathematicians: how does $\varphi (x) = \inf_{y \in Y} [c(x, y) - \psi (y)] \neq -\infty$ follow in Theorem 1.37? | I hope I did not misunderstand the question, but it seems $\varphi(x) > - \infty$ holds as follows if $(x, y) \in \Gamma$:
For any $(x\_i, y\_i) \in \Gamma$, $i=1, \dots, n$, we see that
\begin{align}
&c(x, y\_{n}) - c(x\_n, y\_n) + \sum\_{i=0}^{n-1} c(x\_{i+1}, y\_i) - c(x\_i, y\_i) \\
&= c(x, y\_{n}) - c(x\_n, y\_n) + \sum\_{i=0}^{n-1} c(x\_{i+1}, y\_i) - c(x\_i, y\_i) \\&+ c(x\_0, y) - c(x, y) -c(x\_0, y) + c(x, y) \\
&= - c(x, y) + c(x, y\_n) - c(x\_n, y\_n) + c(x\_n, y\_{n-1}) - ... + c(x\_1, y\_0) - c(x\_0, y\_0) + c(x\_0, y) \\
&+c(x, y) - c(x\_0, y) \\
&\geq c(x, y) - c(x\_0, y) > -\infty,
\end{align}
where the last inequality holds since $(x, y) \in \Gamma$ and by $c$-cyclical monotonicity of $\Gamma$.
Thus, $\varphi(x)$ is bounded from below by $c(x, y) - c(x\_0, y)$.
| 2 | https://mathoverflow.net/users/106046 | 427350 | 173,359 |
https://mathoverflow.net/questions/427351 | 3 | I just want to know if Tangled Type Theory $\mathsf{TTT}$ of Randall Holmes ([see: [Holmes - NF is consistent, p:11](https://arxiv.org/abs/1503.01406), [Holmes - The equivalence of NF-style set theories with “tangled” type theories; the construction of $\omega$-models of predicative NF (and more), p:4-5](https://randall-holmes.github.io/Papers/tangled.pdf)]) can be written completelly without any reference to sequences, i.e. written only in terms of types.
>
> In particular does the following theory have exactly the same axioms of $\mathsf{TTT}$?
>
>
>
Language multi-sorted FOL, with sorts (types) indexed by the naturals, equality symbol restricted to same type, while membership symbol restricted from lower to higher types, i.e. $x\_i^n = x\_j^m$ is well formed formula only when $ n=m$, and $x\_i^n \in x\_j^m$ is well formed only when $n < m$.
**Extensionality:**
$i=1,2,3,\dotsc; j=0,1,2,\dotsc, j<i \\ \forall x^i \forall y^i: \forall z^j (z^j \in x^i \iff z^j \in y^i) \to x^i=y^i$
**Comprehension:**
$i=1,2,3,\dotsc; j=0,1,2,\dotsc, j< i \\ \exists x^i \forall y^j ( y^j \in x^i \iff \phi ) $
where $\phi $ is a well formed formula in which $ x^i $ is not free.
| https://mathoverflow.net/users/95347 | Can we write Tangled Type Theory without reference to type sequences? | No it is not the same. The type sequences are absolutely needed, in some form.
--Randall Holmes (author of the theory in question)
The theory you describe is provably inconsistent.
The formula has to be the result of replacing the variables in an axiom of ordinary TST (indexed by type in the usual order) with variables of types determined using a strictly increasing sequence of TTT types. This is unavoidable. Finite sequences could be used in place of the infinite ones I use, but this is a complication, not a simplification.
| 9 | https://mathoverflow.net/users/485780 | 427355 | 173,360 |
https://mathoverflow.net/questions/403896 | 1 | Let $m = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
It is [known](https://mathoverflow.net/questions/280897) that
$$\gcd(\sigma(q^k),\sigma(n^2)) = \frac{(\gcd(n,\sigma(n^2)))^2}{\gcd(n^2,\sigma(n^2))}$$
and therefore that
$$\gcd(\sigma(q^k),\sigma(n^2)) = \gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} \geq 3$$
if and only if $\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$.
Lastly, it is [known](https://math.stackexchange.com/a/4249583/28816) that
$$\gcd(\sigma(q^k),\sigma(n^2))=1$$
implies $k=1$.
Here is my:
>
> **QUESTION:** Under what conditions is it true that
> $$\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))?$$
>
>
>
I know that the **GCD** equation is true, for example, when $\sigma(n^2) = q^k n$ (and therefore, $\sigma(q^k) = 2n$). Are there other conditions under which the **GCD** equation is true?
| https://mathoverflow.net/users/10365 | On odd perfect numbers and a GCD - Part III | Let $p^s Q^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv s \equiv 1 \pmod 4$ and $\gcd(p,Q)=1$.
I did some more digging on when the equations
$$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$
$$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$
$$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$
simultaneously hold. Note that we have the identity
$$\gcd(\sigma(Q^2), \sigma(p^s)) \gcd(Q^2, \sigma(Q^2)) = \left(\gcd(Q, \sigma(Q^2))\right)^2.$$
**Hence, when exactly one of the three equations above holds, then the other two equations follow.**
---
In particular, note that
$$\gcd(Q^2, \sigma(Q^2)) = \gcd(\sigma(Q^2), \sigma(p^s))$$
is equivalent to
$$\frac{Q^2}{\sigma(p^s)/2} = \frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2}$$
which, in turn, is equivalent to
$$Q = \gcd(\sigma(p^s)/2, Q).$$
This last GCD equation holds if and only if $Q \mid \sigma(p^s)/2$.
---
Furthermore, in particular, note that
$$\gcd(Q, \sigma(Q^2)) = \gcd(Q^2, \sigma(Q^2))$$
is equivalent to
$$\left(\frac{Q}{\sigma(p^s)/2)}\right)\cdot\gcd(\sigma(p^s)/2, Q) = \frac{Q^2}{\sigma(p^s)/2}$$
which, in turn, is equivalent to
$$\gcd(\sigma(p^s)/2, Q) = Q.$$
This last GCD equation holds if and only if $Q \mid \sigma(p^s)/2$.
---
Lastly, in particular, note that
$$\gcd(\sigma(Q^2), \sigma(p^s)) = \gcd(Q, \sigma(Q^2))$$
is equivalent to
$$\frac{\left(\gcd(\sigma(p^s)/2, Q)\right)^2}{\sigma(p^s)/2} = \left(\frac{Q}{\sigma(p^s)/2}\right)\cdot\gcd(\sigma(p^s)/2, Q)$$
which, in turn, is equivalent to
$$\gcd(\sigma(p^s)/2, Q) = Q.$$
This last GCD equation holds if and only if $Q \mid \sigma(p^s)/2$.
---
Thus, if we set
$$G = \gcd(\sigma(Q^2), \sigma(p^s))$$
$$H = \gcd(Q^2, \sigma(Q^2))$$
$$I = \gcd(Q, \sigma(Q^2))$$
then we get the biconditional
>
> $$G = H = I \iff Q \mid \sigma(p^s)/2.$$
>
>
>
Of course, as a sanity check, when $\sigma(p^s) = 2Q$, then we obtain the conjunction
$$Q \mid \sigma(p^s)/2$$
and
$$\sigma(p^s)/2 \mid Q,$$
which by [Conjunction Elimination](https://en.wikipedia.org/wiki/Conjunction_elimination) yields
$$Q \mid \sigma(p^s)/2$$
and hence, that
$$G = H = I.$$
| 0 | https://mathoverflow.net/users/10365 | 427357 | 173,361 |
https://mathoverflow.net/questions/427346 | 3 | Let $X$ be a smooth projective variety (say over $\mathbb{C}$). An object $F \in D^b(X)$ is said to be rigid if $\mathrm{Ext}^1(F,F)=0$. I was wondering if we can always find a rigid object on a projective variety of dimension bigger or equal to $2$ (see the edit below for comments on the dimensional hypothesis). Ideally, I would also like the Chern character of this object to be non-zero.
In case $H^1(\mathcal{O}\_X) =0$, any line bundle will do the job. On the other hand, if $H^1(\mathcal{O}\_X) \neq 0$, the existence of the trace maps shows that the rank of such an object must be zero. I have some specific examples in mind (mostly structure sheaves of rigid subvarieties of some special varieties), but I would like to know if such objects exist in general on any smooth projective variety.
**Edit:** as Johan elliptically points out in the comments, the Grothendieck-Riemann-Roch Theorem shows that $\chi(F,F) =0$ for $F \in D^b(X)$, when $X$ is an elliptic curve. In particular, if $F$ is a coherent sheaf, the non vanishing of $\mathrm{Hom}(F,F)$ implies necessarily that $\mathrm{Ext}^1(F,F) \neq 0$, as there are no higher Ext's. On the other hand, we know that that an object in the derived category of an elliptic curve is quasi-isomorphic to the direct sum of its shifted cohomology sheaves. From this, we can deduce that all objects have non vanishing $\mathrm{Ext}^1$.
This seems however a very specific phenomenon related to curve (as the category $Coh(X)$ is then hereditary and any object in the derived category is quasi-isomorphic to a sum of shifted coherent sheaves). This is why I will make an assumption on $\dim X$.
| https://mathoverflow.net/users/37214 | Existence of rigid objects in the derived category of a smooth projective variety | I am writing up as *one* answer the comments by @Johan, by @Libli, and by myself. If either of them prefers to write an answer, I am happy to delete this answer.
Let $A$ be an Abelian variety. For every scheme $S$ and every $S$-valued point $x\in A(S)$, denote by $\mu\_x$ the associated translation automorphism of the $S$-scheme $S\times A$, i.e., $\mu\_x(y) = x+y$.
Denote by $\widehat{A}$ together with the invertible sheaf $\mathcal{P}$ on $\widehat{A}\times A$ the relative $\text{Pic}^0$ of $A$, normalized so that $\mathcal{P}|\_{\widehat{A}\times\{0\}}$ is the structure sheaf on $\widehat{A}$. Of course $\widehat{A}\times A$ is a commutative group scheme with its structure as the product of two commutative group schemes. Denote by $G$ the noncommutative group scheme structure on $\widehat{A}\times A$ defined by $$ ([\mathcal{L}],x)\bullet([\mathcal{M}],y) = ([\mathcal{L}\otimes \mu\_x^\*\mathcal{M}],x+y).$$ There is an "action" of $G$ on the bounded derived category of coherent sheaves on $A$ that associates to each $([\mathcal{L}],x)$ in $G$ and each bounded complex $C^\bullet$ of coherent sheaves on $A$ the associated bounded complex of coherent sheaves, $\mathcal{L}\otimes \mu\_x^\*(C^\bullet).$ (According to an article of Orlov, this action identifies $G$ with the identity component of the group of autoequivalences of the bounded derived category of coherent sheaves on $A$.)
If $C^\bullet$ is not quasi-isomorphic to the zero complex, i.e., if it is not an exact complex, there there exists an integer $p$ such that the cohomology sheaf $h^p(C^\bullet)$ is nonzero. In general, a "flat deformation" of the complex $C^\bullet$ does not necessarily give rise to a flat deformation of the coherent sheaf $h^p(C^\bullet)$, since base change is not left exact. However, for a connected, smooth group scheme $G$, for a $G$-equivariant family of deformations over $G$, the coherent sheaves $h^p$ are compatible with base change: this holds over a dense open of $G$ (since $G$ is reduced), and this dense open is $G$-invariant, thus it is all of $G$. Therefore, the deformations of $C^\bullet$ arising from the action of $G$ give rise to a deformation of $h^p(C^\bullet)$.
By hypothesis, the coherent sheaf $h^p(C^\bullet)$ on $A$ is nonzero. If the rank is positive, then the action of the normal subgroup $\widehat{A}\times\{0\}$ of $G$ on this sheaf is nontrivial by considering "det" of the coherent sheaf. If the rank of the sheaf is zero, i.e., if the support of the sheaf is a proper closed subscheme of $A$, then the action of the subgroup $\{[\mathcal{O}\_A]\}\times A$ of $G$ on the sheaf is nontrivial since it "moves" this proper closed subscheme. Either way, the action of the group scheme $G$ on the sheaf is nontrivial.
Since the action of $G$ already produces nontrivial deformations of the sheaf $h^p(C^\bullet)$, it also produces nontrivial deformations of the complex $C^\bullet$. Thus, the only rigid complexes in the bounded derived category of $A$ are quasi-isomorphic to zero.
| 1 | https://mathoverflow.net/users/13265 | 427395 | 173,370 |
https://mathoverflow.net/questions/427372 | 2 | Given $L$ variables $k\_i$ where each $k\_{i} \in \{1, 2, 3, \ldots, N\}$ I want to obtain how many different sums $k\_{1}+k\_{2}+\cdots+k\_{L}$ are generated and the value of these sums.
There are $L^N$ possible sums but many give the same result, e.g. for $L=2$ and $N=3$ we have
* 1 + 1 = 2 (one solution that gives 2)
* 1 + 2 = 2 + 1 = 3 (two solutions that give 3)
* 1 + 3 = 3 + 1 = 2 + 2 = 4 (three solutions that give 4)
* 2 + 3 = 3 + 2 = 5 (two solutions that give 5)
* 3 + 3 = 6 (one solution that gives 6)
Is there a general formula that given a result $s$ outputs the number of combinations of the $k\_i$ for $i=1,\ldots,L$ variables with possible values $\{1,\ldots,N\}$ which sum gives $s$?
(taking the previous example, for $L=2$ and $N=3$ there are 3 possible ways to obtain $s=4$, 2 possible ways to obtain $s=5$, etc)
***Edit (addendum):*** What if instead of a sum of $L$ terms now I want to obtain the difference between two of them? i.e. given $k\_1=1,...,N$ and $k\_2=1,...,N$, how many times I obtain each possible difference $k\_{1}-k\_2$? I am not sure how can I generalize the problem...
| https://mathoverflow.net/users/486480 | Different sum combinations of $L$ identical lists of consecutive natural numbers | You are asking for the number of compositions of $s$ into $L$ parts,
with largest part at most $N$. This is a classical problem,
equivalent to finding the coefficient of $x^s$ in the polynomial
$(x+x^2+\cdots+x^N)^L$. Write this polynomial as
$x^L(1-x^N)^L/(1-x)^L$. Expand the numerator and denominator by the
binomial theorem, multiply the two series, and extract the coefficient
of $x^s$ to get a formula for what you want expressed as a finite
sum. It is also possible to obtain this formula using the Principle of
Inclusion-Exclusion. A closely related result (where 0 is allowed as a
summand) appears in *Enumerative Combinatorics*, vol. 1, second ed.,
Exercise 1.28.
| 8 | https://mathoverflow.net/users/2807 | 427396 | 173,371 |
https://mathoverflow.net/questions/427394 | 1 | Let $k$ be a field of characteristic not 2 or 3. Then the set of elliptic curves over $k$ can be parametrized by the affine variety $S=D(4a^3+27b^2)\subset\mathbb{A}^2\_k$ via the family $E\to S$ where $E\subset\mathbb{P}^2\_k\times S$ given is by the equation $y^2=x^3+ax+b$.
1. Is this in some mild sense universal? For example is every family of elliptic curves $E'\to S'$ over some smooth $k$-variety $S'$ the pull-back of $E\to S$ via a (necessarily non-unique) morphism $S'\to S$? If not, at least locally?
2. Are there analogous constructions for abelian varieties of higher dimension? Is there a family of abelian varieties of dimension $g$ over some irreducible (affine? smooth?) variety $S$ over $k$ such that every abelian variety of dimension $g$ over $k$ is a fiber over a rational point of $S$?
In the literature I have only found (stronger) results over algebraically closed fields or when we additionally require something like a $\Gamma\_1(N)$ structure or something.
| https://mathoverflow.net/users/36563 | What's a right parameter space of abelian varieties over a non algebraically closed fields? | 1. This is true in a Zariski-local sense. Any elliptic curve together with a nowhere vanishing differential can be put in this form. For an elliptic curve over $S$, the differentials form a line bundle over $\operatorname{Spec} S$, so we can get a nowhere vanishing differential over a Zariski open cover where this line bundle is trivial.
2. For principally polarized abelian varieties, you can choose $n$ such that $GL\_n$ admits $Sp\_{2g}( \mathbb Z/3)$ as a subgroup and consider the moduli space parameterizing pairs of a principally polarized abelian variety of genus $g$ together with a function from the set of full level 3-structures on the abelian variety compatible with the polarization to $GL\_n$ which is $Sp\_{2g}(\mathbb Z/3)$-equivariant. This moduli space will be a scheme because it is a bundle over $GL\_n/Sp\_{2g}(\mathbb Z/3)$ with fiber the moduli space of abelian varieties with full level $3$ structure, and both are schemes, and every family of abelian varieties factors through this Zariski-locally since the obstruction to doing that is a $GL\_n$-torsor which splits locally.
But there is almost no reason to do this in practice, and you are better off working with the moduli stack of abelian varieties for most purposes.
| 8 | https://mathoverflow.net/users/18060 | 427398 | 173,373 |
https://mathoverflow.net/questions/418931 | 5 | Let $A$ be real a positive semidefinite matrix of dimension $n$ and with $1$s on the diagonal. Those matrices are sometimes referred to as *correlation matrices*. From the positivity of the minors, we know that each matrix element $a\_{ij}$ satisfies $-1 \leq a\_{ij} \leq 1$. The problem is the following: considering that some fixed off-diagonal elements are zero, what is the maximal possible largest eigenvalue $\lambda\_1$ of $A$?
Remark: since $\lambda\_1(A) \leq \lambda\_1(|A|)$, where $|A|$ is the entry-wise absolute value of $A$, we can restrict ourselves to $0\leq a\_{ij} \leq 1$.
As an example, let’s take
$$
A= \begin{pmatrix}
1 & a\_{12} & 0 & 0 \\
a\_{12} & 1 & a\_{23} & 0 \\
0 & a\_{23} & 1 & a\_{34} \\
0 & 0 & a\_{34} & 1
\end{pmatrix}.
$$
From numerical optimisation, I know that the maximal largest eigenvalue reachable with the constraint that $A$ is positive semidefinite is $2$. This is obtained e.g. by taking $a\_{12}=a\_{34}=1$ and $a\_{23}=0$. Furthermore, if we did not have the PSD constraint, the maximum would always be reached at an extreme point (i.e., at some attribution of $0$ or $1$ to each entry and in this case, $1$ everywhere). My intuition is that this is still true for PSD matrices, i.e., it would be attained by a PSD $\{0,1\}$-matrix. Note that the only PSD $\{0,1\}$-matrices are block-diagonal matrices with blocks consisting of only ones (up to some permutation).
Is there a way to prove this, without using the explicit from of the eigenvalues, which can become very cumbersome for general matrices?
| https://mathoverflow.net/users/479369 | Maximal eigenvalue of a correlation matrix with some entries fixed as zeros | The question can be solved by considering the Lovász number of a graph whose adjacency matrix has entries $0$ if $a\_{ij} \neq 0$ and $1$ everywhere else. This is proven (up to a typo) in Section 33 [here](https://arxiv.org/pdf/math/9312214.pdf).
| 0 | https://mathoverflow.net/users/479369 | 427402 | 173,375 |
https://mathoverflow.net/questions/427403 | 8 | **Background.** Suppose that $M$ is an oriented, connected, closed manifold of dimension $d$ with fundamental class $\mu \in H\_d(M;\Bbb Z)$. Let $\Delta : M \to M \times M$ be the diagonal map. Then the pushforward $\Delta\_\ast\mu \in H\_d(M\times M)$ is defined.
If we apply the intersection pairing to $\Delta\_\ast\mu$ with itself, we obtain the self-intersection class
$$\Delta\_\ast \mu \cdot \Delta\_\ast \mu \in H\_0(M\times M) \cong \Bbb Z .$$
If I'm not mistaken, then $\Delta\_\ast \mu \cdot \Delta\_\ast \mu$ is the Euler characteristic
of $M$ multiplied by a suitable generator.
I know of two proofs, but both of them use a certain amount of geometry. The first proof
interprets the intersection of the cycles representing $\Delta\_\ast \mu$ (i.e., the image of the diagonal map) as the transversal self-intersection of the
diagonal with itself. The latter is supported in a tubular neighborhood of the diagonal.
Then one makes use of the tubular neighborhood theorem to interpret the that self-intersection as the transversal intersection of the zero section of tangent bundle with itself. Lastly, one uses the Poincaré–Hopf index theorem to interpret the latter as the Euler characteristic.
The other proof I know of is along the lines of Milnor and Stasheff Chapter 11. However, the latter would also seem to require the tubular neighborhood theorem to interpret the pullback $\Delta^\ast(u'')$ of the diagonal cohomology class $u'' \in H^d(M\times M)$
as the Euler class of the tangent bundle of $M$.
In summary, as far as I know, there is no proof that $\Delta\_\ast \mu \cdot \Delta\_\ast \mu$ is the Euler characteristic of $M$ which doesn't make use of the tubular neighborhood theorem. Note too that the intersection $\Delta\_\ast \mu \cdot \Delta\_\ast \mu$ does not require transversality to define—one may take the Poincaré dual of $\Delta\_\ast \mu$, then apply the cup square and then apply Poincaré duality again.
**My Question.** Suppose now that $M$ is merely an oriented connected Poincaré duality space
of dimension $d$, with fundamental class $\mu$.
*Does $\Delta\_\ast \mu \cdot \Delta\_\ast \mu$ coincide with the Euler characteristic of $M$?*
Note that the analogue of the tubular neighborhood theorem is not generally known to hold in the Poincaré case (i.e., there may not be a Poincaré embedding of the diagonal map).
| https://mathoverflow.net/users/8032 | On the Euler characteristic of a Poincaré duality space | It follows from Poincaré duality in $M\times M$. I'll suppress gradings and be vague about signs here.
Let $x\_i$ be a rational homology basis for $M$. Let $a\_i$ be the cohomology basis that is dual to this basis in the linear algebra sense: $\langle a\_i,x\_j\rangle=\delta\_{ij}$. Let $y\_i$ be the homology basis that corresponds to $a\_i$ by Poincaré duality. In other words, the intersection product of $x\_i$ and $y\_j$ is $\pm \delta\_{ij}$.
Then $\Delta\_\ast\mu$ is the sum of $\pm x\_i\times y\_i$, and it's also the sum of $\pm y\_j\times x\_j$, so the intersection product of $\Delta\_\ast\mu$ with itself is the sum, over $i$ and $j$, of $\pm\delta\_{ij}\delta\_{ji}$, i.e. the sum of $\pm\delta\_{ii}$. This is the Euler characteristic, if you get the signs right.
| 14 | https://mathoverflow.net/users/6666 | 427404 | 173,376 |
https://mathoverflow.net/questions/419755 | 5 | Let $R$ be a commutative unital ring and let $M$ be a unital $R$-module. A non-Archimedean ring seminorm on $R$ is a map $|\cdot| \colon R \rightarrow \mathbb{R}\_{\geq 0}$ which satisfies
$$ | 0\_R| = 0, \quad |1\_R| \leq 1, \quad |r - s| \leq \max \{ |r|, |s|\}, \quad |r \cdot s|\leq | r|\cdot |s| $$
for all $r,s \in R$. A non-Archimedean module seminorm on $M$ over $\left( R, |\cdot| \right)$ is a map $\| \cdot \| \colon M \rightarrow \mathbb{R}\_{\geq 0}$ which satisfies
$$ \| 0\_M \| = 0, \quad \| m - n \| \leq \max \{ \|m\|, \|n\| \}, \quad \|r \cdot m\| \leq |r| \cdot \| m \|.$$
Given two non-Archimedean seminormed $\left( R, |\cdot| \right)$-modules $\left( M, \|\cdot\|\_M \right)$ and $\left( N, \|\cdot\|\_N \right)$ one can define a non-Archimedean seminorm on the
tensor product $M \otimes\_R N$ by setting
$$ \| z \|\_{M \otimes N} := \inf \, \left \{ \max\_{i \in I} \| m\_i \|\_M \cdot \| n\_i \|\_N \, \bigg| \, z = \sum\_{i \in I} m\_i \otimes n\_i, \quad I \textrm{ is finite } \right \}. $$
It is written or hinted in various places (for example, in [Introduction To Berkovich Analytic Spaces](http://berkovich-2010.institut.math.jussieu.fr/lecture-notes/Temkin-lecture-notes.pdf) or in [Rigid Analytic Geometry and Its Applications](https://link.springer.com/book/10.1007/978-1-4612-0041-3)) that even if $R$, $M$ and $N$ are normed (or even complete), it is possible for $\| \cdot \|\_{M \otimes N}$ to be only seminorm and not a norm but I couldn't find an example of this.
Can someone describe an example or provide a reference which contains an explicit example? Does the situation change if we assume in addition that the norms are multiplicative (i.e $|x \cdot y| = |x| \cdot |y|$ and $\| r \cdot m \| = |r| \cdot \| m \|$)?
| https://mathoverflow.net/users/471381 | An example where the non-Archimedean tensor product of normed modules is only seminormed? | Let $k$ be a field. Pick some $r \in (0,1)$ and let $R$ be the ring $k[[t]]$ endowed with the absolute value that is trivial on $k$ and such that $\vert t\vert =r$. Let $M$ be $k((t))$ endowed with the absolute value that extends that on $R$. Pick $s \in (0,r)$ and let $N$ be the ring $k[[t]]$ endowed with the absolute value that is trivial on $k$ and such that $\vert t\vert =s$. In this situation, the tensor seminorm on $M\otimes\_R N$ is only a seminorm. Indeed, we can write $1 = t^{-N} \otimes t^N$ for any integer $N$, hence $\| 1\|\_{M\otimes\_R N} \le r^{-N} s^N$, which tends to 0 when $N$ goes to $\infty$.
| 3 | https://mathoverflow.net/users/4069 | 427406 | 173,377 |
https://mathoverflow.net/questions/427447 | 7 | Let $(\mathcal{X} , \|\cdot \|\_\mathcal{X})$ be a Banach space and $\mathcal{X}'$ its topological dual. We denote by $\| \cdot \|\_{\mathcal{X}'}$ the dual norm and define also the topological dual $\mathcal{X}''$ of the Banach space $(\mathcal{X}',\|\cdot\|\_{\mathcal{X}'})$. The unit ball of $\mathcal{X}'$ is denoted by
$$\mathcal{B} = \{ y \in \mathcal{X}', \ \| y\|\_{\mathcal{X}'} \leq 1\}.$$
We consider three topologies on $\mathcal{X}'$, on which we recap basic facts:
* The norm topology, for which $\mathcal{B}$ is not compact as soon as $\mathcal{X}$ is infinite dimensional (Riesz' theorem).
* The weak\* topology, which is the coarsest topology such that the linear functionals $y \mapsto y(x)$ are continuous for any $x \in \mathcal{X}$. The Banach-Alaoglu theorem states that $\mathcal{B}$ is compact for the weak\*-topology.
* The weak topology, which is the coarsest topology such that the linear functionals $y \mapsto z(y)$ are continuous for $z \in \mathcal{X}''$.
The weak\* topology is weaker than the weak topology, which is weaker than the norm topology. Moreover, the unit ball $\mathcal{B}$ is not compact for the weak topology as soon as the space is not reflexive (otherwise, the weak and weak\* topologies coincide).
My questions are the following: Are there intermediate topologies between the weak\* and the weak topology for which the unit ball $\mathcal{B}$ is compact? Or can we say in some sense that the weak\* topology is the finest for which the unit ball is compact?
I am not expecting a unique answer for every non-reflexive infinite dimensional Banach spaces, but possibly characterisations of the spaces for which the weak\* is indeed the only topology between the weak\* and the weak topology.
If it helps, the same questions can be considered for the specific cases:
* $(\mathcal{X},\mathcal{X}',\mathcal{X}'') = (c\_0(\mathbb{Z}), \ell\_1(\mathbb{Z}), \ell\_\infty(\mathbb{Z}))$ where $c\_0(\mathbb{Z})$ is the space of vanishing sequences endowed with the norm $\|\cdot\|\_\infty$.
* $(\mathcal{X},\mathcal{X}',\mathcal{X}'') = (\mathcal{C}(\mathbb{T}), \mathcal{M}(\mathbb{T}), \mathcal{M}'(\mathbb{T}))$ where $\mathbb{T}$ is the torus, $\mathcal{C}(\mathbb{T})$ the space of continuous periodic functional endowed with the supremum norm, and $\mathcal{M}(\mathbb{T}) the space of finite Radon measure.
(Motivation: I try to understand what is the largest topology for which $\mathcal{B}$ is compact beyond the weak\* topology in order to use the Krein-Millmann theorem ensuring the existence of extreme points for convex compact sets.)
| https://mathoverflow.net/users/39261 | Compactness of the unit ball of a Banach space for topologies finer than the weak* topology | As pointed out by Goulifet, my previous answer was wrong. In fact almost the exact opposite is true: 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. That's because any stronger topology for which the unit ball is compact would have to agree with the weak${}^\*$ topology on the unit ball (if two compact Hausdorff topologies are comparable, they are equal), so by Krein-Smulian they would have the same continuous linear functionals (anything continuous for the new, stronger topology would be continuous for the weak\* topology on the unit ball and therefore weak\* continuous).
We conclude that the new topology would have to equal the weak\* topology using an argument kindly supplied by Jochen Wengenroth in the comments.
| 10 | https://mathoverflow.net/users/23141 | 427450 | 173,387 |
https://mathoverflow.net/questions/427474 | 3 | Suppose that $D\_{KL}(p\_1\parallel q)<1$ and $D\_{KL}(p\_2\parallel q)<1$. I'm trying to show that either $D\_{KL}(p\_1\parallel p\_2)$ or $D\_{KL}(p\_2\parallel p\_1)$ will have an upper bound close to 1 provided ${q}$ is fixed. It seems intuitive that if $p\_1$ and $q$ are similar enough and if $p\_2$ and $q$ are also similar enough then $p\_2$ can reasonably approximate $p\_1$ or vice versa. Is this actually true?
| https://mathoverflow.net/users/486746 | An inequality relating the Kullback-Leibler divergence of two discrete distributions with constant reference distribution | The answer is no. E.g., let $p\_1,p\_2,q$ be pdf's on $[0,1]$ such that $q=1$, $p\_1=\frac1{1-h}\,1\_{[0,1-h]}$, and $p\_2=\frac1{1-h}\,1\_{[h,1]}$, for some $h\in(0,1)$. Then
$$D\_{KL}(p\_1\parallel q)=D\_{KL}(p\_2\parallel q)=\ln\frac1{1-h}\to0$$
as $h\downarrow0$, whereas
$$D\_{KL}(p\_1\parallel p\_2)=D\_{KL}(p\_2\parallel p\_1)=\infty$$
for all $h\in(0,1)$.
| 4 | https://mathoverflow.net/users/36721 | 427476 | 173,390 |
https://mathoverflow.net/questions/427481 | 0 | Given some variables $x\_1, x\_2, \dots, x\_n$, the Vandermonde determinant is given by
$$V\_n(x\_1,\dots,x\_n):=\det(x\_j^{i-1})\_{i,j=1}^n=\prod\_{i<j}(x\_j-x\_i).$$
One can take as special cases: $x\_j=j$ or $x\_j=q^j$.
I was playing around with the following variation $A\_n=((i+j-1)^{i-1})\_{i,j=1}^n$. It turns out that (numerically) $\det(A\_n)=\det(j^{i-1})\_{i,j=1}^n$.
In addition, if $B\_n=(q^{(i+j-1)(i-1)})\_{i,j=1}^n$ then $\det(B\_n)$ and $\det(q^{j(i-1)})\_{i,j=1}^n$ differ only by a factor of a power of $q$, that is, $q^c$ for some $c$.
On the other hand, if $C\_n=(x\_{i+j-1}^{i-1})\_{i,j=1}^n$ then (unfortunately) $\det(C\_n)$ and $V\_n$ are vastly different; in fact, $\det(C\_n)$ does not factor as $V\_n$.
>
> **QUESTION 1.** Is this true? $\det(A\_n)=\det(j^{i-1})\_{i,j=1}^n$.
>
>
>
>
> **QUESTION 2.** What worked right for $A\_n$ andd $B\_n$ and failed for $C\_n$?
>
>
>
| https://mathoverflow.net/users/66131 | A variant of numeric Vandermonde which failed symbolically? | Matrix $A\_n$ can be generalized to $\big[(j+f(i))^{i-1}\big]\_{i,j=1}^n$ for any function $f(i)$ not just $f(i)=i-1$.
Since
$$(j+f(i))^{i-1} = \sum\_{k=0}^{i-1} c\_{i,k}\cdot j^k,$$
where $c\_{i,k} := \binom{i-1}k f(i)^{i-1-k}$, we have
$$\big[(j+f(i))^{i-1}\big]\_{i,j=1}^n = \begin{bmatrix}
c\_{1,0} & 0 & 0 & \dots & 0\\
c\_{2,0} & c\_{2,1} & 0 & \dots & 0\\
c\_{3,0} & c\_{3,1} & c\_{3,2} & \dots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
c\_{n,0} & c\_{n,1} & c\_{n,2} & \dots & c\_{n,n-1}
\end{bmatrix}\cdot \big[j^{i-1}\big]\_{i,j=1}^n.$$
Taking determinants, we get
$$\det\big[(j+f(i))^{i-1}\big]\_{i,j=1}^n = c\_{1,0}c\_{2,1}\cdots c\_{n,n-1}\cdot \det\big[j^{i-1}\big]\_{i,j=1}^n=\det\big[j^{i-1}\big]\_{i,j=1}^n.$$
---
For $B\_n$ things work even for a simpler reason:
$$q^{(i+j-1)(i-1)} = q^{(i-1)^2} q^{j(i-1)}$$
and we can take out a factor of $q^{(i-1)^2}$ from the $i$-th row of matrix $B\_n$.
However for matrix $C\_n$ neither binomial expansion work, nor we can take out any common factors from its rows. So, despite visual similarity $C\_n$ does not possess the properties of $A\_n$ or $B\_n$.
| 7 | https://mathoverflow.net/users/7076 | 427482 | 173,392 |
https://mathoverflow.net/questions/427465 | 12 | *This is related to a couple recent MO/MSE questions of mine, namely [1](https://mathoverflow.net/questions/427135/what-algebraic-properties-are-preserved-by-mathbbn-leadsto-beta-mathbbn),[2](https://math.stackexchange.com/questions/4498457/which-ultrafilters-are-sort-of-commutators). Belatedly, I've tweaked this post to remove an overly-ambitious secondary question; see the edit history if interested.*
Let $\beta\mathbb{Z}$ be the set of all ultrafilters on $\mathbb{Z}$, and as usual conflate $n$ and $\{A\subseteq\mathbb{Z}:n\in A\}$ for $n\in\mathbb{Z}$. We can extend any binary operation on $\mathbb{Z}$ to a semicontinuous analogue on $\beta\mathbb{Z}$, at the cost of many (most?) algebraic properties. Ultrafilter *addition* is quite well studied (see e.g. [Hindman/Strauss](https://people.dm.unipi.it/dinasso/ULTRA2015/BUMN/Hindman_Strauss%20-%20Algebra%20in%20the%20Stone-Cech%20compactification%20(2012%20-%202nd%20ed).pdf)), but I've been able to find much less about ultrafilter *subtraction*: $$\mathcal{U}\widehat{-}\mathcal{W}=\{A\subseteq\mathbb{Z}: \{k:\{a: a-k\in A\}\in\mathcal{U}\}\in\mathcal{W}\}.$$
Say that an ultrafilter $\mathcal{W}$ is **zeroid** iff $\mathcal{W}=\mathcal{U}\widehat{-}\mathcal{U}$ for some $\mathcal{U}$. My question is:
>
> Which ultrafilters are zeroid?
>
>
>
To make this actually answerable, I tentatively guess that $(i)$ $0$ is the only zeroid principle ultrafilter but not the only zeroid ultrafilter and $(ii)$ $p$-points are not zeroid; are these guesses true?
| https://mathoverflow.net/users/8133 | Ultrafilter subtraction and "zero" | Both your guesses are correct. To see this, it's helpful to reformulate the way you're thinking about the subtraction operator on $\beta \mathbb Z$. Beginning with subtraction on $\mathbb Z$, you can first extend this to an operator $\beta \mathbb Z \times \mathbb Z \rightarrow \beta \mathbb Z$ by setting $\mathcal U - n = \{B-n :\, B \in \mathcal U\} = \{A \subseteq \mathbb Z :\, A+n \in \mathcal U\}$. Notice that this agrees with your definition of subtraction on $\beta \mathbb Z$ when we identify $n$ with the principle ultrafilter at $n$. So this is the "right" way to think of subtracting an integer from an ultrafilter. But then there is only one way to extend this to a semi-continuous operation $\beta \mathbb Z \times \beta \mathbb Z$: we must define $$\mathcal U - \mathcal W = \textstyle {\mathcal W}\text{-}\!\lim\_n (\mathcal U-n).$$ This again agrees with your definition (it must!), but personally I find it much more intuitive to think of $\mathcal U - \mathcal W$ as a topological limit of the sequence $\mathcal U, \mathcal U - 1, \mathcal U - 2, \dots$.
This description of what's going on makes your two guesses plainly true. For $(i)$, note that if $\mathcal U$ is nonprincipal then so is $\mathcal U - n$ for all $n$, which means (because $\mathbb Z$ is open in $\beta \mathbb Z$) that the limit ${\mathcal W}\text{-}\!\lim\_n (\mathcal U-n)$ is in $\beta Z \setminus \mathbb Z$ for any $\mathcal W$ (including $\mathcal W = \mathcal U$). For $(ii)$, note that ${\mathcal W}\text{-}\!\lim\_n (\mathcal U-n)$ is a limit of a countable sequence of points in $\beta \mathbb Z \setminus \mathbb Z$, hence not a P-point. In fact, let me observe that $\mathcal U - \mathcal W$ is never a weak P-point (for the same reason), which shows that not every ultrafilter is representable as the difference of two others.
| 16 | https://mathoverflow.net/users/70618 | 427486 | 173,393 |
https://mathoverflow.net/questions/427479 | 7 | Let $X$ be a normed vector space with its dual $X^\*$. Let $S^\*$ be the unit sphere of $X^\*$. We have known that if $X$ (or $X^\*$ ) is reflexive then the weak-star and weak topology of $X^\*$ coincide and thus the weak-star closure (or weak closure) of $S^\*$ is the unit ball. I have the following questions:
1. If is reflexive, what is the weak-star sequential closure of $S^\*$?
2. If $X$ is non-reflexive, what are the weak-star and weak-star sequential closure of $S^\*$
| https://mathoverflow.net/users/487196 | Weak star closure of unit sphere in dual space | Suppose the weak$^\*$ sequential closure of $S^\*$ contains $0$. So there is a sequence $(f\_n) \subseteq X^\*$ with $\|f\_n\|=1$ for each $n$, and with $f\_n\rightarrow 0$ weak$^\*$. For $f\in B^\*$ the unit ball, we seek a bounded sequence $(t\_n)$ of scalars such that $f+t\_nf\_n \in S^\*$ for each $n$, as then $f+t\_nf\_n \rightarrow f$ weak$^\*$. We can find such $t\_n$ from the triangle inequality,
$$ \big| \|f\| - |t\_n| \big| \leq \|f + t\_nf\_n\| \leq \|f\| + |t\_n|, $$
and using that $t\mapsto \|f+tf\_n\|$ is continuous. Conclude: the weak$^\*$-sequential closure of $S^\*$ is all of $B^\*$.
So, when is $0$ in the weak$^\*$ sequential closure of $S^\*$?
Here is an easy argument if $X$ is separable. Then $X$ hash a countably dense subset $\{x\_k\}$, then for each $n$ by Hahn-Banach we can find $f\_n\in S^\*$ with $f\_n(x\_k)=0$ for $k\leq n$. Given $x\in X$ and $\epsilon>0$ there is $k$ with $\|x-x\_k\|<\epsilon$, and so for $n\geq k$ we have $|f\_n(x)| = |f\_n(x-x\_k)| < \epsilon$. Conclude that $f\_n\rightarrow 0$ weak$^\*$.
As [Dirk Werner](https://mathoverflow.net/users/127871/dirk-werner) points out, in the general case, we can use the Josefson-Nissenzweig Theorem which exactly says that any (infinite-dimensional) Banach space has the property we need: $X^\*$ has a weak$^\*$-null sequence of norm one vectors. Here is one source for this theorem: [Extract from Diestel, Sequences and Series in Banach Spaces](https://link.springer.com/chapter/10.1007/978-1-4612-5200-9_13)
| 3 | https://mathoverflow.net/users/406 | 427491 | 173,394 |
https://mathoverflow.net/questions/427399 | 11 | Let us say that three consecutive positive integers $(m-1,m,m+1)$ have a big prime factor if the largest prime factor $p$ of $N:=(m-1)m(m+1)$ satisfies $e^p>N$.
I ckecked that it is true for all $m<10^8$, except for $m=3,9,15,49,55,99,351,441,2431$ where $p=3,5,7,7,11,11,13,17,19$, respectively.
**Question**: Is it true for all $m>2431$?
*Application*: A positive answer would solve a problem in Section D25 of Richard Guy's [*Unsolved Problems in Number Theory*](https://doi.org/10.1007/978-0-387-26677-0).
>
> Simmons notes that $n! = (m - 1)m(m + 1)$ for $(m, n) = (2, 3)$, $(3, 4)$, $(5, 5)$, and $(9, 6)$, and asks if there are other solutions. More generally he asks if there are any other solutions of $n! + x = x^k$. This is a variation on the question of asking for $n!$ to be the product of $k$ consecutive integers in a nontrivial way ($k \ne n + 1 - j!$). Compare **B23**.
>
>
>
Note that $n! > e^n$ iff $n \ge 6$. But I checked for $m \le 2431$ that $n! = (m-1)m(m+1)$ iff $$(m,n) = (2,3), (3,4), (5,5), (9,6).$$ So for $m>2431$, if $e^p>N$ and $N = n!$, then $p! \ge n!$, but $p<n$ because $p$ is a prime factor of $n!$ (and $n>2$), contradiction.
---
Our question for three consecutive positive integers can be generalized to $k \ge 3$ consecutive positive integers, where the bound $e^p$ could be optimized. Now for $k=2$ it is not clear that $e^p$ (or even $p!$) works because I found large exceptions.
```
sage: m=123200
sage: factor(m*(m+1))
2^6 * 3^6 * 5^2 * 7 * 11 * 13^2
sage: factorial(13)>m*(m+1)
False
sage: m=2697695
sage: factor(m*(m+1))
2^5 * 3^2 * 5 * 7^3 * 11^2 * 13 * 17 * 19 * 29
sage: ceil(exp(29))>m*(m+1)
False
```
On the contrary, for $k=2$ we can wonder whether, for a given prime $p>2$, there are infinitely many $m$ such that the biggest prime factor of $m(m+1)$ is $\le p$. The case $p=3$ reduces to the existence of infinitely many couple of non-negative integers $(a,b)$ with $\lvert 2^a-3^b\rvert = 1$, but it is false (see why at [distance between powers of 2 and powers of 3](https://mathoverflow.net/q/116840/34538)). Now what if $p$ is large *enough*? In other words:
**Bonus question**: Is there an integer $n$ such that the set of integers $m(m+1)$ whose prime factors are less than $n$, is infinite?
| https://mathoverflow.net/users/34538 | Do consecutive integers have a big prime factor? | Thanks to Szpiro’s/$abc$ conjecture, the answer to your question is true for sufficiently large $N$, up to a factor of $3$ times the primes $p$ (which factor, I believe, can be removed if one is a bit careful in the estimation of the prime factors of $N$ below $3\log m$ in the proof below). Note that the problem you wish to solve with the answer to your question can still be solved (for sufficiently large numbers) if you weaken your hypothesis to $e^{kp}>N$ for some $k\ge1$ (indeed, $p\gg\log N$ suffices for your problem as $\log n!\gg n$). Here, I’d simply resort to $k=3$ (actually, any $k>9/4=2.25$ suffices) to allow for a simpler proof.
For a rational elliptic curve $E$ with minimal discriminant $\Delta\_E$ and conductor $f\_E$, let
$$\sigma\_E:=\frac{\log|\Delta\_E|}{\log f\_E}\,.$$ Szpiro’s conjecture states that for every $\varepsilon>0$, there are only finitely many elliptic curves with $\sigma\_E\ge 6+\varepsilon$.
Now, consider the rational elliptic curve
$$E\colon y^2=x(x-1)(x-m^2)$$
for a given natural number $m$. By well-known/standard results, the above is a minimal model and so the minimal discriminant is $$\Delta\_E=(4m^2(m^2-1))^2\,.$$
Note that, in your notation, $N=m^3-m$, and so $\Delta\_E=(4mN)^2$. In particular, $E$ has multiplicative reduction at all odd prime factors of $\Delta\_E$ but additive reduction at $2$; thus, the conductor $f\_E$ is $$f\_E=2^{n}\prod\_{2<p|N}p\,,$$ for some $n\in\{2,3,4,5,6\}.$ Now, working asymptotically (we write $\sim$ for *asymptotically equal to* and $\gtrsim$ for *asymptotically greater than or equal to*), then assuming to the contrary that $e^{kp}\le N$ for $k=3$, or better still some $k>9/4$, we obtain
\begin{align}
\log f\_E&= n\log 2 +\sum\_{2<p|N}\log p\\
&\le n\log 2+ \sum\_{p\le\frac{1}{k}\log N}\log p\\
&\sim n\log 2+ \sum\_{p\le\frac{3}{k}\log m}\log p\\
&\sim n\log 2+\frac{3}{k}\log m\,,
\end{align}
where we have used the prime number theorem $\sum\_{p\le x}\log p\sim x$ in the final step. This implies that
\begin{align}
\sigma\_E=\frac{\log|\Delta\_E|}{\log f\_E}&=\frac{\log 16m^4(m^2-1)^2}{\log f\_E}\\
&\sim\frac{\log 16m^8}{\log f\_E}\\
&\gtrsim\frac{4\log2+8\log m}{n\log 2+ \frac{3}{k}\log m}\\
&\sim \frac{8}{3}k>6,
\end{align}
which contradicts Szpiro’s conjecture for sufficiently large $N$.
| 6 | https://mathoverflow.net/users/166628 | 427500 | 173,396 |
https://mathoverflow.net/questions/427484 | 2 | I am reading [a paper by Szekeres and Peters](https://www.cambridge.org/core/journals/anziam-journal/article/computer-solution-to-the-17point-erdosszekeres-problem/0EC7876789232266D60439A4C00D86D9) on computing the 17-point case of [the Erdős–Szekeres conjecture](https://en.wikipedia.org/wiki/Happy_ending_problem). The conjecture states that the minimum number of points in the plane (in general position, no three points collinear) such that any arrangement will contain a convex subset of $n$ points is
$$
2^{n-2}+1
$$
The paper describes an algorithm which tests if a subset of points is convex by checking the orientation of each ordered triple of points. The signature function $\sigma$ maps an ordered triple of points $(a,b,c)$ to $\{+,-\}$, if the three points are oriented clockwise or counterclockwise respectively.
On page 6, the paper gives a convexity condition from 4 relations on the signatures of each ordered triple of points,
>
> Let abcde be any (ordered) set of five points of $S\_9$. It forms a convex 5-subset if and
> only if its Q = 10 triples satisfy one of the four relations (termed convex relations):
> $$
> R\_1: \sigma(abc)=\sigma(bcd)=\sigma(cde)\\
> \quad R\_2: \sigma(abc) = \sigma(bce) = -\sigma(ade)\\
> \quad R\_3: \sigma(abd) = \sigma(bde) = -\sigma(ace)\\
> \quad R\_4: \sigma(acd) = \sigma(cde) = -\sigma(abe)
> $$
>
>
>
And on page 9, there is a similar set of 8 relations for a 6-subset,
>
> Let abcdef be any (ordered) set of six points of $S\_{17}$. It forms a convex 6-subset if and
> only if its ${6 \choose 3} = 20$ triples satisfy one of the eight convex relations:
> $$
> R\_1: \sigma(abc)=\sigma(bcd)=\sigma(cde)=\sigma(def)\\
> \quad R\_2: \sigma(abc) = \sigma(bcd) = \sigma(cdf) = -\sigma(aef) \\
> \quad R\_3: \sigma(abc) = \sigma(bce) = \sigma(cef) = -\sigma(adf) \\
> \quad R\_4: \sigma(abd) = \sigma(bce) = \sigma(def) = -\sigma(acf) \\
> \quad R\_5: \sigma(acd) = \sigma(cde) = \sigma(def) = -\sigma(abf) \\
> \quad\quad R\_6: \sigma(abc) = \sigma(bcf) = -\sigma(ade) = -\sigma(def) \\
> \quad\quad R\_7: \sigma(abd) = \sigma(bdf) = -\sigma(ace) = -\sigma(cef) \\
> \quad\quad R\_8: \sigma(acd) = \sigma(cdf) = -\sigma(abe) = -\sigma(bef)
> $$
>
>
>
From these two quotes, it seems that there are a set of $2^{n-3}$ relations for $n$ point convexity. How are these convex relations derived? And how can the set of convex relations for an arbitrary number of points be found?
| https://mathoverflow.net/users/487612 | Determining if a polygon is convex using relations on orientation of each ordered triple of points | It's important to note that when it talks about *ordered points*, this is ordered by $x$-coordinate and not (as one might otherwise suppose) by traversing the edges of the polygon.
Suppose we have $n$ points, $p\_1$ to $p\_n$, ordered by $x$-coordinate. Observe that $p\_1$ and $p\_n$ are on the convex hull. Now draw a line $p\_1 - p\_n$ and partition the points $p\_2, \ldots, p\_{n-1}$ by which side of the line they fall. Call the subset of points which are below or on the line $q$ and the subset of points which are above or on the line $r$.
For the polygon to be convex, $q$ must be a chain of anti-clockwise turns, and $r$ must be a chain of clockwise turns. Therefore $$\sigma(q\_1 q\_2 q\_3) = \sigma(q\_2 q\_3 q\_4) = \cdots = \sigma(q\_{|q|-2} q\_{|q|-1} q\_{|q|}) = -\sigma(r\_1 r\_2 r\_3) = \cdots = -\sigma(r\_{|r|-2} r\_{|r|-1} r\_{|r|})$$
We get $2^{n-3}$ cases by considering which subset of $p\_3, \ldots, p\_{n-1}$ is in the same chain as $p\_2$.
| 3 | https://mathoverflow.net/users/46140 | 427502 | 173,397 |
https://mathoverflow.net/questions/427499 | 3 | It seems to me that a proof of $\alpha\_{n}=o(n)$ where the quantity $\alpha\_{n}$ is defined in [About Goldbach's conjecture](https://mathoverflow.net/questions/61842/about-goldbachs-conjecture) together with the main result of <https://kyushu-u.pure.elsevier.com/en/publications/exceptional-zeros-and-the-goldbach-problem> and Terry Tao's answer to this former question of mine: [Does the existence of a Landau-Siegel zero imply the existence of a complex zero off the critical line?](https://mathoverflow.net/questions/404340/does-the-existence-of-a-landau-siegel-zero-imply-the-existence-of-a-complex-zero) would entail that if every large enough even integer is the sum of two primes, then GRH (the analogue of the Riemann Hypothesis for Dirichlet L-functions) holds true.
Can this be established rigorously?
| https://mathoverflow.net/users/13625 | Does asymptotic Goldbach imply GRH? | Granville (2008) proved that a sufficiently strong average error term for the Goldbach-Hardy-Littlewood conjecture is equivalent to RH, and a refinement of it is equivalent to GRH. See MR2357316 and MR2492859. A closely related result was proved by Bhowmik-Ruzsa (2018) and Bhowmik-Halupczok-Matsumoto-Suzuki (2018). See MR3788238 and MR3867327.
Friedlander-Goldston-Iwaniec-Suriajaya (2022) proved that a weak form of the Goldbach-Hardy-Littlewood conjecture rules out exceptional zeros of Dirichel $L$-functions. See MR4356845.
| 12 | https://mathoverflow.net/users/11919 | 427505 | 173,399 |
https://mathoverflow.net/questions/427469 | 3 | Let $X$ be a complex manifold endowed with a holomorphic closed 2-form $\omega$ whose associated map $\omega : TX \to T^\*X$ is invertible. Can we always embed $X$ as an open subset of a compact complex manifold $Y$ endowed with a holomorphic Poisson structure $\pi : T^\*Y \to TY$ such that $\pi|\_X = \omega^{-1}$? We may assume that $X$ is quasi-projective, or other reasonable assumptions, if that helps.
Note that this holds, for example, if $X$ is the cotangent bundle of a compact manifold.
| https://mathoverflow.net/users/479013 | When does a holomorphic symplectic manifold compactify to a Poisson manifold? | No, not even under the nicest possible algebraicity assumptions like quasiprojective.
Let $Z$ be the product of two curves of genus $\geq 2$. Choose a nonzero $2$-form $\omega$ on $Z$, the wedge of a nonzero one-form on each of the two curves. Let $X$ be obtained from $Z$ by removing the locus where $\omega$ vanishes. Then $X$ clearly has a nowhere vanishing (and thus invertible as $\dim X=2$) two-form.
Now any compactification $Y$ of $X$ is a compact complex manifold birational to the algebraic variety $Z$, thus is Moishezon, and because it's a surface, [must be projective algebraic](https://mathoverflow.net/a/51752/18060). Since $Z$ is a minimal surface, $Y$ is the blow-up of $Z$ at finitely many points. So if the Poisson structure on $X$ extends to $Y$, then it extends to $Z$ less finitely many points, hence to $Z$.
But $Z$ does not have a nontrivial Poisson structure since $T Z \otimes TZ $ is a sum of line bundles of negative degree and thus has no global sections.
| 8 | https://mathoverflow.net/users/18060 | 427509 | 173,402 |
https://mathoverflow.net/questions/427506 | 1 | Suppose we have i.i.d. samples $x\_i\sim N(0,\Sigma)$ and $y\_i\sim x\_i^T\omega^\*+\xi\_i,\xi\_i\sim N(0,1)$ where $\omega^\*$ is the fixed point of:
$$\omega\_{i+1} = \omega\_i − \eta\nabla\_\omega f(\omega\_i, x\_i
, y\_i), \quad \omega\_0 = 0$$
where $f(\omega, x\_i, y\_i) = \frac{1}{2}|y\_i − \omega^T x\_i|^2 $and $\eta$ is the step size. Formulate an appropriate range
of $\eta$ so that $\omega\_i$ will become an ergodic process. Also find mean and covariance under the equilibrium distribution.
So, I tried to calculate it and got another form of the expression:
$$\omega\_{i+1}-\omega^\*=(I-\eta x\_i x\_i^T)(\omega\_i-\omega^\*)+\eta\xi\_ix\_i$$
which could help to directly express $\omega\_i$ by a successive multiplication of $(I-\eta x\_i x\_i^T)$ and $\omega^\*$. Then I wanted to calculate $\frac{\sum \omega\_i}{n}$ to get the ergodicity, but I didn't make it.
| https://mathoverflow.net/users/348579 | Using gradient descent in probability case | Lets first assume $\zeta\_i=0$ and ask the following:
* under which conditions $w^\*$ is a stable fixed point?
If it's not a stable fixed point for noise-free case, then you won't end up with a fixed mean or stationary distribution after adding additive noise. This means $w^\*$ being a fixed point is a necessary condition for ergodicity. In the case of isotropic Gaussian noise, it might also be sufficient (needs checking)
Example
-------
Consider 2-dimensional problem with the following covariance:
\begin{equation}
\Sigma=\left(
\begin{matrix}
1 & 0 \\
0 & 2
\end{matrix}
\right)
\end{equation}
One condition on step size $\eta=\alpha$ which guarantees convergence
$$\frac{2}{\alpha}<\text{Tr}(\Sigma)+2\|\Sigma\|$$
This equation is derived in many places, ie 3.30 in Diniz "Adaptive Filtering". This produces the following condition on step size in our example:
$$\alpha<\frac{2}{7}$$
However, this condition is too strict and larger values of $\alpha$ work. The necessary and sufficient condition turns out to be the following:
$$\alpha<\frac {4} {9 + \sqrt {17}}$$
I extend the expression above for a general Gaussian in this [note](http://yaroslavvb.com/convergence.pdf). I haven't seen anyone else derive this, please correct me if this already occurs in literature.
General solution
----------------
For $x\_i$ sampled from a zero-centered Gaussian with covariance $\Sigma$, the necessary and sufficient condition for $w^\*$ to be a fixed point in a noiseless case is that
$$\alpha<\frac{2}{\rho(A)}$$
Where $\rho$ denotes spectral radius and $A$ is defined below. Let $s=(s\_1, s\_2, s\_3, \ldots)$ be the eigenvalues of $\Sigma$, then
\begin{equation}\label{defa}
A=
2 \left(
\begin{array}{cccc}
s\_1 & 0 & 0 & \ldots \\
0 & s\_2 & 0 & \ldots \\
0 & 0 & s\_3 & \ldots \\
\ldots & \ldots & \ldots & \ldots
\end{array}
\right)
+
\left(
\begin{array}{cccc}
s\_1 & s\_1 & s\_1 & \ldots \\
s\_2 & s\_2 & s\_2 & \ldots \\
s\_3 & s\_3 & s\_3 & \ldots \\
\ldots & \ldots & \ldots & \ldots
\end{array}
\right)
\end{equation}
I haven't found a simpler closed form expression for this, but a sequence of bounds can be derived by bounding spectral radius of $A$ with norm of $A^k$ for various values of $k$, [discussion](https://math.stackexchange.com/q/3846464/998).
I suspect that convergence to fixed point in noiseless case also implies convergence to stationary distribution in the case of isotropic additive noise. In the case of non-isotropic noise, may need to consider the ratio of $\Sigma$ and covariance of noise matrix, like is done in NQM [paper](https://proceedings.neurips.cc/paper/2019/hash/e0eacd983971634327ae1819ea8b6214-Abstract.html)
Existing results for non-Gaussian case:
---------------------------------------
It can be shown that in the case of 1 dimension and deterministic $x$, the following condition on $\alpha$ is necessary and sufficient for convergence
\begin{equation}\label{supersimple}
\alpha x^4 < 2 x^2
\end{equation}
Since we have $h=x^2$ for Hessian $h$, this reduces to the well known bound on convergent learning rate: $\alpha < 2/h$
In the case of stochastic x, the following is necessary and sufficient
\begin{equation}\label{eq:0}
\alpha E[x^4] < 2 E[x^2]
\end{equation}
For the case of $x$ being distributed as standard normal, this gives $2/(3h)$ for the largest learning rate, three times smaller than what's allowed in deterministic case
For the case of $d$ dimensions, the following is a sufficient condition, with $\prec$ indicating Loewner order\footnote{assumption A.6 in Bach [paper](https://www.di.ens.fr/%7Efbach/aistats_defossez_bach_with_supp.pdf)
\begin{equation}\label{eq:1}
\alpha E[xx'xx'] \prec E[xx']
\end{equation}
The right-hand side can be tightened to
\begin{equation}\label{eq:1x}
\alpha E[xx'xx'] \prec 2 E[xx']
\end{equation}
Bach, Deffosez2015 [showed](https://www.di.ens.fr/%7Efbach/aistats_defossez_bach_with_supp.pdf) that the following optimization over symmetric matrices gives sufficient condition for convergence, and conjectured it to also be necessary (Lemma 1 of Defossez2015)
\begin{equation}\label{eq:2}
\frac{1}{\alpha} < \sup\_{A\in \mathcal{S}(R^d)} \frac{E[(x'Ax)^2]}{2 E[x'A^2 x] }
\end{equation}
We can show this to be equivalent to the following positive semi-definite constraint
\begin{equation}\label{eq:3}
\alpha E[xx' \otimes xx'] \prec E[xx'\otimes I] + E[I\otimes xx']
\end{equation}
Most recently, [Jain](https://arxiv.org/abs/1610.03774) generalized last Eq to batch sizes beyond 1 and formally showed it to be a necessary condition for monotonic convergence. When applied to Gaussian case, this is equivalent to $\text{Tr}(\Sigma)+2\|H\|$ condition derived earlier.
| 1 | https://mathoverflow.net/users/7655 | 427511 | 173,403 |
https://mathoverflow.net/questions/427510 | 6 | Let $V$ be a finite dimensional vector space. Let $\Lambda$ be a collection of subspaces of $V$ such that, if $X$ and $Y$ are in $\Lambda$, then $X\cap Y$ and $X+Y$ are in $\Lambda$. This makes $\Lambda$ into a lattice; we impose that $\Lambda$ is a distributive lattice, meaning that
$$X \cap (Y+Z) = (X \cap Y) + (X \cap Z) \qquad \forall\ X,\ Y,\ Z \in \Lambda$$
I want a reference for the following statement:
>
> There is a basis $B$ for $V$ such that, for every $X \in \Lambda$, the intersection $B \cap X$ is a basis of $X$.
>
>
>
The only reference I have been able to find are these [notes on Koszul algebras](https://www.math.uni-bielefeld.de/birep/activities/topics/files/ws05-hille-koszul-algebras-and-distributive-lattices.pdf), where it is described as "a standard result in lattice theory". Those notes cite Gratzer, "Lattice theory", 1971, Section 2.7, Theorem 19 but, in fact, Theorem 19 of Section 2.7 in Gratzer does not mention vector spaces at all.
| https://mathoverflow.net/users/297 | Distributive lattice of subspaces | Just to not leave this open, a proof can be found in Proposition 7.1 of Chapter 1 of Quadratic Algebras by Alexander Polishchuk and Leonid Positselski as alluded to by Mariano on MSE <https://math.stackexchange.com/questions/63493/non-distributivity-of-subspaces>.
| 7 | https://mathoverflow.net/users/15934 | 427513 | 173,404 |
https://mathoverflow.net/questions/425082 | 12 | Some of the earliest writings on the [Joyal model structure on simplicial sets](https://ncatlab.org/nlab/show/model+structure+for+quasi-categories) include Jacob Lurie's account in [Higher Topos Theory](https://arxiv.org/abs/math/0608040v1) from 2006,
as well as Joyal's own account in [The Theory of Quasi-Categories and its Applications](https://mat.uab.cat/%7Ekock/crm/hocat/advanced-course/Quadern45-2.pdf) from 2008.
I have seen unsourced claims that this model structure may originate in 1980s, possibly in the correspondence between Joyal and Grothendieck. However, an examination of the only available [letter from Joyal to Grothendieck](http://webusers.imj-prg.fr/%7Egeorges.maltsiniotis/ps/lettreJoyal.pdf) does not reveal any material on quasicategories. I was unable to find any other correspondence between Joyal and Grothendieck.
In his 2008 notes cited above, Joyal says “The results presented here are the fruits of a long term research project which began around thirty years ago.”, which appears to indirectly corroborate the above unsourced claims.
**When did the Joyal model structure on simplicial sets originate? Is there any written source that confirms its existence in 1980s?**
| https://mathoverflow.net/users/402 | When did the Joyal model structure on simplicial sets originate? | Here is what André Joyal wrote in an email to me:
>
> No, I have not discovered the model structure for quasi-categories in the 1980's.
> I became interested in quasi-categories (without the name) around 1980
> after attending a talk by Jon Beck on the work of Boardman and Vogt.
> I wondered if category theory could be extended to quasi-categories.
> In my mind, a crucial test was to show that a quasi-category is a Kan complex
> if its homotopy category is a groupoid.
> All my attempts at showing this have failed for about 15 years, until I stopped trying hard!
> I found a proof after extending to quasi-categories a few basic notions of category theory.
> This was around 1995. The model structure for quasi-categories was discovered soon after.
> I did not publish it immediately because I wanted to show that it could
> be used for proving something new in homotopy theory.
> I am a bit of a perfectionist (and overly ambitious?).
> I was hoping to develop a synthesis between category theory and homotopy theory
> (hence the name quasi-categories).
> I met Lurie at a conference organised by Carlos Simpson in Nice (in 2001?).
> I gave a talk on the model structure and Lurie asked for a copy of my notes afterward.
> I intuitively understood that he could develop the
> theory of quasi-categories more and better than I could.
> He was young and a better mathematician than I was.
> I do not regret it.
>
>
>
| 15 | https://mathoverflow.net/users/402 | 427514 | 173,405 |
https://mathoverflow.net/questions/384709 | 6 | The following identity is not hard to prove:
$$
\sum\_{1\leq i\_1<i\_2<\ldots <i\_{2n}\leq N} (-1)^{i\_1+\ldots+i\_{2n}}\frac{(1-x\_{i\_1})(1-x\_{i\_3})\ldots(1-x\_{i\_{2n-1}})}{(1-x\_{i\_2})(1-x\_{i\_4}) \ldots (1-x\_{i\_{2n}})\ \ }=
\frac{(x\_{1}-x\_2)(x\_{2}-x\_{3}) \ldots (x\_{N-1}-x\_{N})}{(1-x\_{2})\ \ldots\ \ \ \ (1-x\_{N-1})(1-x\_{N})}.
$$
I am curious if it appeared anywhere before and if it is a part of a certain family of similar identities. As a corollary, one immediately sees that LHS is a power series starting with degree $N-1.$ This fact leads to a family of functional equations for multiple polylogarithms.
| https://mathoverflow.net/users/21620 | An identity for rational functions leading to equations for multiple polylogarithms | If we denote $1-x\_i=y\_i$, this reads as $$\sum\_{1\leq i\_1<i\_2<\ldots <i\_{2n}\leq N} (-1)^{i\_1+\ldots+i\_{2n}}\frac{y\_{i\_1}y\_{i\_3}\ldots y\_{i\_{2n-1}}}{y\_{i\_2}y\_{i\_4} \ldots y\_{i\_{2n}}}=
\frac{(y\_{2}-y\_1)(y\_{3}-y\_{2}) \ldots (y\_{N}-y\_{N-1})}{y\_{2}\ldots y\_{N-1}y\_{N}}\\
=(1-y\_1/y\_2)(1-y\_2/y\_3)\ldots (1-y\_{N-1}/y\_N),$$
which may be proved by expanding the brackets. And such expanding was certainly done before. I am less sure about $1-y\_i=x\_i$ change of variables.
The coefficients of products of differences often appear, perphaps the most known is Vandermonde determinant $$\prod\_{0\leqslant i<j\leqslant n-1}(y\_j-y\_i)=\sum\_{\pi}{\rm sign}\,(\pi)\prod y\_i^{\pi\_i},$$
where $\pi$ runs over all $n!$ permutations of the set $\{0,\ldots,n-1\}$. There are many interesting cases, when *all* coefficients can not be found, but *some* of them can, like [Dyson's conjecture](https://en.wikipedia.org/wiki/Dyson_conjecture) (I have no idea whether it can lead to anything interesting about polylogarithms.) If you need complete expansions only, there are not so many cases when they are known, apart of above I can suggest the cycle graph instead of the path. If you allow not only differences, there are complete expansions of, for example, $(\sum x\_i)^N \prod\_{i<j}(x\_j-x\_i)$.
| 3 | https://mathoverflow.net/users/4312 | 427523 | 173,406 |
https://mathoverflow.net/questions/427462 | 3 | Work in ZFC with no large cardinal assumptions. Say that a (parameter-definable) class $X \subseteq ORD$ is *club* if it is closed and unbounded in the sense that:
1. For each $\beta \in ORD$, there exists $\gamma \geq \beta$ such that $\gamma \in X$, and
2. For each $\delta, \epsilon \in ORD$, and each increasing function $\beta : \delta \to \epsilon$, if $\beta(\delta') \in X$ for all $\delta' < \delta$, then $\sup\_{\delta' < \delta} \beta(\delta') \in X$.
Now let $X \subseteq ORD \times ORD$ be a (parameter-definable) class. Assume that for each $\alpha \in ORD$, the (parameter-definable) class $X(\alpha) := \{\beta \in ORD \mid (\alpha,\beta) \in X\}$ is club in the above sense.
Consider the diagonal intersection $\Delta\_{\alpha \in ORD} X(\alpha) := \{\alpha \in ORD \mid \alpha \in \cap\_{\beta < \alpha} X(\beta)\}$. This is again a parameter-definable class.
**Question:** Does ZFC prove (as a schema in $X$) that the above diagonal intersection $\Delta\_{\alpha \in ORD} X(\alpha)$ is club in the above sense?
| https://mathoverflow.net/users/2362 | Is the definable club filter normal? | Work in ZF. As Asaf was probably mentioning in his comment, you have asked two distinct questions, one in the title (referring to a filter), and one in the main body of the question (just referring to clubs). Moreover, the notion asked about in the title hasn't been defined clearly. Given what you wrote in the body of the question, I will presume the following definition:
Let's define "the definable club filter is normal" to mean that for each meta-integer $n$ and $X\subseteq\mathrm{Ord}\times\mathrm{Ord}$ which is $\Sigma\_n$-definable in parameters, if for each $\alpha\in\mathrm{Ord}$ there is a club proper class $C\subseteq X\_\alpha$ such that $C$ is also $\Sigma\_n$-definable in parameters, then there is a club $C$ which is definable in parameters with $C\subseteq$ the diagonal intersection of $\left<X\_\alpha\right>\_{\alpha\in\mathrm{Ord}}$.
Theorem: The definable club filter is normal (under ZF).
(We will in fact end up getting $C$ to be $\Sigma\_{n+3}$-or-so-definable in the same parameter used to define $X$, and the conversion from the formula defining $X$ to the formula defining $C$ will be recursive.)
Lemma: the answer to the question in the main body of the question is "yes", in fact just assuming ZF.
Proof: This is what I was referring to in my comment above; the proof here is essentially the usual one: the fact that the diagonal intersection is closed, is (as usual) immediate. Let's show it's unbounded. Let $f:\mathrm{Ord}\to\mathrm{Ord}$ be the function where $f(\alpha)$ is the least $\beta>\alpha$ such that $X\_\gamma\cap[\alpha,\beta)\neq\emptyset$ for each $\gamma<\alpha$.
(Note indeed $f(\alpha)\in\mathrm{Ord}$.) Then $f$ is definable in parameters,
and $\alpha<\alpha'\implies f(\alpha)\leq f(\alpha')$. For each $\alpha\in\mathrm{Ord}$ and each $n<\omega$, $f^n(\alpha)$ exists and is in $\mathrm{Ord}$. (By induction on $n$.) Note that $(\alpha,n)\mapsto f^n(\alpha)$ (with domain $\mathrm{Ord}\times\omega$) is definable from parameters. Thus, by Collection, for each $\alpha\in\mathrm{Ord}$ we can find $\beta\in\mathrm{Ord}$ such that $f^n(\alpha)<\beta$ for each $n<\omega$. Now fix $\alpha\in\mathrm{Ord}$ and let $\beta=\sup\_{n<\omega}f^n(\alpha)\in\mathrm{Ord}$. Note that $\beta\in\Delta\_{\gamma\in\mathrm{Ord}}X\_\gamma$, which proves the diagonal intersection is unbounded, as desired.
Proof of theorem:
Fix $n,X$. Given a pair $(\varphi,p)$ where $\varphi$ is a $\Sigma\_n$ formula with two free variables and $p$ some set, let $C\_{\varphi,p}=\{x\bigm|\varphi(x,p)\}$.
Say $(\varphi,p)$ is *good* if $C\_{\varphi,p}$ is a club proper class of ordinals. Let $\mathscr{G}$ be the class of good pairs.
Let $f:\mathrm{Ord}\to V$ be the class function where $f(\alpha)$ is the least $\beta\in\mathrm{Ord}$ such that for all $\gamma<\alpha$, there is a good $(\varphi,p)\in V\_\beta$ such that $C\_{\varphi,p}\subseteq X\_\gamma$. For $\alpha\in\mathrm{Ord}$ let $D\_\alpha=\bigcap\_{(\varphi,p)\in\mathscr{G}\cap V\_{f(\alpha)}}C\_{\varphi,p}$. Note that $D\_\alpha$ is club proper class, much like in the proof of the lemma.
Note that $\mathrm{Lim}\cap\Delta\_{\alpha\in\mathrm{Ord}}D\_\alpha\subseteq\Delta\_{\alpha\in\mathrm{Ord}}X\_\alpha$. But $\Delta\_{\alpha\in\mathrm{Ord}}D\_\alpha$ is club proper class by the lemma, and so so is $\mathrm{Lim}\cap\Delta\_{\alpha\in\mathrm{Ord}}D\_\alpha$.
| 6 | https://mathoverflow.net/users/160347 | 427524 | 173,407 |
https://mathoverflow.net/questions/427153 | 1 | Let $W$ be a standard one dimensional Brownian motion, and consider the SDE
$$dX\_t = \sigma(X\_t) \, dW\_t, \, \, \, X\_0 = 1 \, \text {a.s.}$$
Assume $\sigma$ is regular enough that the above SDE admits a globally defined solution.
Suppose $|\sigma(x)| \to 0$ as $x \to 0$.
**Question:** Is it true that almost surely, $X\_t > 0$ for all $t$?
It seems like the Dambis-Dubins-Schwarz theorem may help, but I’m not sure how to turn it into a proof.
| https://mathoverflow.net/users/173490 | Is the solution to this SDE always positive? | To complement Nawaf's answer ([1](https://mathoverflow.net/questions/427153/is-the-solution-to-this-sde-always-positive#comment1098413_427153) [2](https://mathoverflow.net/questions/427153/is-the-solution-to-this-sde-always-positive#comment1098415_427153)), I thought I'd present a short argument how with some additional regularity assumptions, the answer is in fact yes.
Suppose that $\sigma\in{C^{2}(\mathbb{R})}$ with $\sigma(0)=0$. Then $\sigma$ is locally Lipschitz, which is sufficient to give us existence and pathwise uniqueness for the solution of the SDE:
$$
dX\_{t}=\sigma(X\_{t})dB\_{t}, \hspace{10pt}X\_{0}=x\_{0}
$$
for any choice of $x\_{0}\in{\mathbb{R}}$ (see theorem 6.9 of [Miller - Stochastic calculus](http://www.statslab.cam.ac.uk//%7Ejpm205/teaching/lent2016/lecture_notes.pdf)). Since $X\_t=0$ is a perfectly valid solution to the above SDE when $X\_0=0$, it is actually *THE* solution. By the strong Markov property, if $X\_t$ ever hits $0$, it is stuck there. Thus, to show that $X\_{t}>0$ for all $t\geq{0}$ when $X\_{0}=1$, it suffices to show that $X\_{t}$ cannot hit $0$ in finite time.
Consider the process $U\_t=\log(\sigma(X\_t))$. Notice that $X\_t$ hits a zero of $\sigma$ in finite time iff $U\_t$ diverges to $-\infty$ in finite time. By Itô's formula,
$$
dU\_t = \sigma'(X\_t)dB\_t + \frac{1}{2}\big(\sigma''(X\_t)\sigma(X\_t) - (\sigma'(X\_t))^2\big)dt.
$$
This is a diffusion with bounded coefficients when $X\_t$ lies in a neighborhood of $0$ (as per our regularity assumptions on $\sigma$). Thus, while $X\_t$ lies in a neighborhood of $0$, $U\_t$ cannot run away to $-\infty$ in finite time and so $X\_{t}$ cannot hit $0$ in finite time.
| 3 | https://mathoverflow.net/users/80052 | 427532 | 173,410 |
https://mathoverflow.net/questions/427544 | 6 | Let $X$ be an object in one of the well-known symmetric monoidal model categories of spectra. E.g., an $\mathtt S$-module in the sense of EKMM, or an orthogonal spectrum, or a $\Gamma$-space, etc.
One can form the spectrum $X\wedge X$, which is a "naive" spectrum with an action of $\Sigma\_2$, and one can take the categorical fixed points to obtain the spectrum $(X\wedge X)^{\Sigma\_2}$. What can be said about the homotopy type of this spectrum as a functor of $X$?
More specifically I would like to know the following:
* Does this construction have good homotopical properties. For example, does it preserve weak equivalences between cofibrant spectra?
* Does the homotopy type of $(X\wedge X)^{\Sigma\_2}$ (say for cofibrant $X$) depend on the specific model of smash product?
* Is it true that if $X$ is a suspension spectrum then $(X\wedge X)^{\Sigma\_2}\simeq X$? If it is not always true, is it true in some cases, i.e., for some models of smash product?
* Can the homotopy type of $(X\wedge X)^{\Sigma\_2}$ be described in other cases?
| https://mathoverflow.net/users/6668 | What are the naive fixed points of a non-naive smash product of a spectrum with itself? | In both the orthogonal world and the EKMM world one can set things up so that $G$-spectra are just spectra with an action of $G$, and the naive and genuine equivariant stable categories are the homotopy categories for two different model structures on the same underlying geometric category. I will take that point of view.
I will write $\kappa^G(X)$ for the categorical fixed points of a spectrum $X$ with $G$-action, i.e. the largest subspectrum on which the action is trivial. We also have the Lewis-May fixed point functor which I will denote by $\lambda^G(\cdot)$, and the geometric fixed point functor $\phi^G(\cdot)$. There are natural maps $\kappa^G\to\lambda^G\to\phi^G$.
Here are some facts about the EKMM category:
* The functor $\Sigma^\infty$ preserves smash products on the nose, but $\Sigma^\infty X$ is not cofibrant unless $X$ is a point.
* The functor $\Sigma^\infty$ also commutes with $\kappa^G$, so $\kappa^{C\_2}(X\wedge X)=X$ whenever $X$ is a suspension spectrum.
* Now let $T$ be the cofibrant replacement of $S^0$. This is an $\mathbb{L}$-spectrum in the sense of EKMM, so it has a kind of action of the monoid $\mathcal{L}(1)=\mathcal{L}(\mathcal{U},\mathcal{U})$ of linear isometric embeddings of the universe $\mathcal{U}=\mathbb{R}^\infty$ in itself. If we just consider the underlying Lewis-May spectrum, it can be identified with $\Sigma^\infty\mathcal{L}(\mathcal{U},\mathcal{U})\_+$. Similarly, $T\wedge T$ is just $\Sigma^\infty\mathcal{L}(\mathcal{U}^2,\mathcal{U})\_+$. From this it is easy to see that $\kappa^{C\_2}(T\wedge T)=0$.
* Similarly, I am pretty sure that $\kappa^{C\_2}(X\wedge X)=0$ whenever $X$ is cofibrant.
* We can also define $T\_G$ to be $\Sigma^\infty\mathcal{L}(\mathcal{U}\otimes\mathbb{R}[G],\mathcal{U})\_+$, equipped with appropriate action of $\mathcal{L}(1)$. This is the cofibrant replacement of $S^0$ in the genuine model structure, and in the case $G=C\_2$ we just have $T\_G=T\wedge T$. The Lewis-May fixed point functor can be described as $\lambda^G(X)=\kappa^G(F(T\_G,X))$. John Rognes's answer describes $\lambda^{C\_2}(X\wedge X)$ in various cases.
I think that the story in the orthogonal world is similar, but the analysis is a bit more complicated and I do not have the details to hand.
| 7 | https://mathoverflow.net/users/10366 | 427546 | 173,413 |
https://mathoverflow.net/questions/427517 | 2 | Let $G$ be a complex semisimple group and $\mathcal{O} \subset G$ a conjugacy class, i.e. $\mathcal{O} = \{gag^{-1} : g \in G\}$ for some $a \in G$. Let $\Omega$ be the Cartan 3-form on $G$ defined by
$$
\Omega(x^L, y^L, z^L) = \langle x, [y, z] \rangle,\quad (x,y,z \in \mathrm{Lie}(G))
$$
where $\langle,\rangle$ is the Killing form, and $x^L, y^L, z^L$ are the left-invariant vector fields generated by $x,y,z$.
**Question:** Is the pullback of $\Omega$ to $\mathcal{O}$ trivial in cohomology?
| https://mathoverflow.net/users/385475 | Is the restriction of the Cartan 3-form on conjugacy classes exact? | Yes, it is exact, and there is in fact a canonical 2-form on each conjugacy class whose derivative is your $\Omega$. This was an important observation when studying D-branes in WZW models, see, e.g. <https://arxiv.org/abs/hep-th/0008038> or <https://arxiv.org/abs/hep-th/0205233>.
| 4 | https://mathoverflow.net/users/3473 | 427570 | 173,417 |
https://mathoverflow.net/questions/427549 | 7 | In the algebraic group $G = \operatorname{PGL}\_4(\mathbb{C})$, let $E$ denote the subgroup of elements of order dividing 2 in the diagonal maximal torus; it is generated by the images of the three matrices
$$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1 \\
\end{bmatrix},\quad
\begin{bmatrix}
1 & 0 & 0\\
0 & -1 & 0 \\
0 & 0 & I\_2
\end{bmatrix},\quad
\begin{bmatrix}
-1 & 0 \\
0 & I\_3
\end{bmatrix}.
$$
Direct computation shows that $N\_{G}(E)/C\_{G}(E) \cong S\_4\,$, which is the Weyl group of $G$. Analogous result can be obtained for $n = 3, 5, 6, 7, 8, 9, 10$ in $\operatorname{PGL}\_{n}(\mathbb{C})$ by direct calculations. Is this observation true for any $n$? Is there any reference relevant?
| https://mathoverflow.net/users/488802 | $N_{G}(E)/C_{G}(E)$ is the Weyl group of $G$? | $\newcommand{\ZZ}{{\mathcal Z}\_G}
\newcommand{\NN}{{\mathcal N}\_G}
\newcommand{\zz}{{\mathfrak z}\_G}
\newcommand{\Lie}{{\rm Lie\,}}
\renewcommand{\tt}{{\mathfrak t}}
\renewcommand{\gg}{{\mathfrak g}}
\newcommand{\X}{{\sf X}}
\newcommand{\Z}{{\Bbb Z}}$
**Yes,** this is true for any $n\ge 3$.
Let $G$ be a semisimple group *of adjoint type*
over an algebraically closed field $k$ of characteristic 0.
Let $T\subset G$ be a maximal torus.
Write
$$E=T^{(2)}:= \{t\in T(k)\ |\ t^2=1\}.$$
We wish to compute the centralizer $\ZZ(E)$ and the normalizer $\NN(E)$.
Observe that $\NN(E)\supseteq \ZZ(E)\supseteq T$.
We compute $\zz(E):=\Lie \ZZ(E)$.
>
> **Lemma 1.** $\zz(E)=\tt:=\Lie T$.
>
>
>
*Proof.*
Since $\ZZ(E)\supseteq T$, we have $\zz(E)\supseteq\tt$.
Write the root decomposition
$$ \Lie G=\tt\oplus\bigoplus\_{\beta\in R}\gg\_\beta\,,$$
where $R=R(G,T)\subset \X^\*(T)$ is the root system.
Since $\zz(E)\supseteq\tt$, we have
$$ \zz(E)=\tt\oplus\bigoplus\_{\beta\in M}\gg\_\beta$$
for some subset $M\subseteq R$.
Here
$$M=\{\beta\in R\ |\ \beta(t)=1\ \forall t\in E\}.$$
Let $S\subset R$ be a system of simple roots (a basis of $R$).
Since $G$ is of adjoint type, the set $S\subset R\subset \X^\*(T)$
is a basis of the character group $\X^\*(T)$ of $T$.
It follows that for any simple root $\alpha\in S$,
there exists $t\in E=T^{(2)}$ such that $\alpha(t)=-1$.
Let $W=W(G,T)=\NN(T)/T$ be the Weyl group.
The group $W$ acts on $E$ and on $R$, and $W\cdot S=R$.
Therefore, for any root $\beta\in R$, there exists $t\in E$
such that $\beta(t)=-1$, and therefore
$M=\varnothing$ and $\zz(E)=\tt$, as required.
We compute $\NN(E)$. Since ${\rm char}(k)=0$,
it follows from Lemma 1 that the identity component $\ZZ(E)^0$ of $\ZZ(E)$ is $T$.
We see that $\NN(E)$ normalizes $\ZZ(E)$, and hence it normalizes $\ZZ(E)^0=T$.
It follows that $\NN(E)\subseteq \NN(T)$.
On the other hand, $\NN(T)$ normalizes $T$, and therefore, it normalizes $E=T^{(2)}$,
whence $\NN(T)\subseteq \NN(E)$.
Thus $\NN(E)=\NN(T)$.
We wish to compute $\ZZ(E)$.
Consider the Weyl group $W=\NN(T)/T$.
Set $$W\_E=\ZZ(E)/T\subseteq \NN(T)/T=W.$$
Then $W\_E$ is a finite group, the kernel of the natural homomorphism
$W\to{\rm Aut\,} E$,
and $\ZZ(E)$ is the preimage of $W\_E$ in $\NN(T)$.
>
> **Lemma 2.** Let $W'=S\_n$ (the permutation group on $n$ symbols)
> naturally acting on the set
> $$E'=\ker\, \Sigma\colon (\Z/2\Z)^n\to \Z/2\Z,$$
> where $\Sigma(x\_1,x\_2,\dots,x\_n)=x\_1+x\_2+\cdots+x\_n\,.$
> If $n\ge 3$, then this action is effective (has trivial kernel).
>
>
>
*Proof.* Left to OP and the reader.
Observe that Lemma 2 is false for $n=2$, when ${\rm Aut\,} E'=\{1\}$
whereas $S\_2\neq \{1\}$.
>
> **Theorem.** Let $G={\rm PGL}\_n$ with $n\ge 3$.
> Then $\NN(E)/\ZZ(E)=\NN(T)/T=W$.
>
>
>
*Proof.*
We have seen that $\NN(E)=\NN(T)$. By Lemma 2,
if $n\ge 3$, then $W\_E=\{1\}$, whence $\ZZ(E)=T$, and the theorem follows.
**Edit.** Observe that the analogues of Lemma 2 and the theorem do not hold for the adjoint group ${\rm SO}(2n+1)$ of type ${\sf B}\_n$ for $n\ge 1$.
Indeed, then
$$W\simeq(\Z/2\Z)^n\rtimes S\_n\,,$$
and the normal subgroup $(\Z/2\Z)^n\subseteq W$ acts on $E=T^{(2)}\cong (\pm1)^n$ trivially.
Therefore, $\ZZ(E)\neq T$.
| 9 | https://mathoverflow.net/users/4149 | 427574 | 173,419 |
https://mathoverflow.net/questions/427575 | 15 | **Motivation.** As I was travelling in the UK, I used a physical copy of the "A-Z Road Atlas BRITAIN" for getting around. I was impressed that whenever I wanted to go from the map segment shown on page 23, say (showing a part of the West Midlands), to the segment east of the one I was looking at, I only had to turn unexpectedly few pages. (For every page, the page showing the next part of the country South, East, West, or North, was indicated.) Which motivated the following concept.
**Formalization.** We regard any non-negative integer $n$ as an ordinal, so $n$ is the set of all its predecessors, i.e. $0 = \emptyset$, and $n = \{0,\ldots,n-1\}$ for any positive integer $n$.
Let $n$ be a positive integer, and let $G = (V,E)$ be a finite, simple, undirected graph with $|V| = n$. We define the "page-turning number" of $G$ by $$\pi(G) = \min\big\{\max\{|\varphi(v) - \varphi(w)|: \{v,w\}\in E\} \; : \; \varphi: V \to n \text{ is bijective}\}.$$ The intuition behind this is the following: we
regard $\varphi\in S\_n$ as a "page-number assignment" to the vertices of the graph and the $\max$ part denotes the largest number of pages we have to turn to get from one vertex to any adjacent one. We want to minimize on this $\max$ part.
Note that for the complete graph $K\_n$ we have $\pi(K\_n) = n-1$. It seems that the page-turning number might be loosely related to coloring, but it is certainly not the same. I would be grateful for hints to an official name for $\pi(G)$.
**Question.** For any positive integer $n$, let the *$n\times n$-grid graph* be given by $G\_n = (n\times n, E)$ where $$E = \big\{\{(a,b), (c,d)\}: a,b,c,d \in n \text{ and } |a-c| + |b-d| = 1\big\}.$$ In terms of $n$, what is the value of $\pi(G\_n)$?
| https://mathoverflow.net/users/8628 | Page-turning number of a graph | The page-turning number of a graph $G$ is also known as the *bandwidth* of $G$ (<https://en.wikipedia.org/wiki/Graph_bandwidth>).
The Wikipedia page also contains values of the bandwidth for some special graphs, including the $m\times n$ grid graph, in which case it is equal to $\min\{m,n\}$. This was proved by Jarmila Chvátalová: <https://doi.org/10.1016/0012-365X(75)90039-4>.
| 17 | https://mathoverflow.net/users/24076 | 427582 | 173,422 |
https://mathoverflow.net/questions/427558 | 1 |
>
> What concepts in the real world can be described by adjunctions?
>
>
>
For example, parents and children are adjoint to one another. Specifically, work in $ZFC$ plus a finite class of atoms $\mathscr{X}$ (one for each person) and let $${\sf Par}:\mathcal{P}(\mathscr{X})\to\mathcal{P}(\mathscr{X})$$ $${\sf Chi}:\mathcal{P}(\mathscr{X})\to\mathcal{P}(\mathscr{X})$$ the functions sending a collection of atoms to the collection of all the atoms corresponding to their parents or children, respectively. Then $${\sf Par\dashv Chi}$$ since we have a (covariant) Galois connection between the poset $\big(\mathcal{P}(\mathscr{X}),\subseteq\big)$ and itself given by the above functions (your parent's children includes you and your children's parents also includes you\*), and a covariant Galois connection [is](https://ncatlab.org/nlab/show/Galois+connection) an adjunction. What are some other examples?
---
\*Technically, those of us without children (myself currently included) form a hole in the argument as it stands -- we get sent to the empty set by ${\sf Chi}$, then back to the empty set by ${\sf Par}$ again instead of a set of people containing ourselves. To fix this, add an additional atom ${\sf f}$ to the theory called *free time*, and let $\mathcal{L}$ be a relation called *having a life* which is the reflexive transitive symmetric closure of the relation on $\mathcal{P}(\mathscr{X})\cup\{{\sf f}\}$ associating all atoms corresponding to people without children to ${\sf f}$. Now define $$\widehat{{\sf Par}}:\mathcal{P}(\mathscr{X})\to\mathcal{P}(\mathscr{X})\cup\{{\sf f}\}\big/\mathcal{Life}$$ $$\widehat{{\sf Chi}}:\mathcal{P}(\mathscr{X})\cup\{{\sf f}\}\big/\mathcal{Life}\to\mathcal{P}(\mathscr{X})$$ as before except have $\widehat{\sf Par}$ send the empty set to $[{\sf f}]$ and $\widehat{\sf Chi}$ send $[{\sf f}]$ to the empty set. Now, we genuinely have that $$\widehat{\sf Par}\dashv\widehat{\sf Chi}.$$ (All of this is well intended humor; my wife and I look forward to joining the ranks of parents with no free time.)
| https://mathoverflow.net/users/92164 | Adjunctions in the real world | If you're willing to admit your example as "describing a concept by an adjunction", then I would argue that *any* binary relation can be "described by an adjunction", namely [the Galois connection that it induces](https://ncatlab.org/nlab/show/Galois+connection#InducedFromARelation).
To be sure, your example isn't quite the Galois connection induced by the relation of "is a parent of". For one thing, the latter is contravariant; it consists of
* $\rm Par'$ that sends a set $C$ of people to the set of people who are parents of *everyone* in $C$ (thus, $\rm Par'(C)$ is empty if $C$ doesn't consist of a group of siblings), and
* $\rm Chi'$ that sends a set $P$ of people to the set of people who are children of *everyone* in $C$ (thus, $\rm Chi'(P)$ is empty if $P$ has cardinality greater than 2, or is a pair of people who have never had children together).
However, suppose we consider instead the relation "is *not* a parent of". Then we get another contravariant Galois connection consisting of
* $\rm Par''$ that sends a set $C$ of people to the set of people $x$ such that no one in $C$ is a child of $x$.
* $\rm Chi''$ that sends a set $P$ of people to the set of people $y$ such that no one in $P$ is a parent of $x$.
Now compose this Galois connection on one side with the contravariant "complementation" automorphism of the set of sets of people, and we get a *covariant* adjunction consisting of
* $\rm Par$ that sends a set $C$ of people to the set of people who are a parent of at least one person in $C$.
* $\rm Chi$ that sends a set $P$ of people to the set of people *both* of whose parents are in $P$.
This is the same $\rm Par$ as yours, but a different $\rm Chi$. As appropriate for a covariant adjunction, we have $C \subseteq {\rm Chi}({\rm Par}(C))$ (if you are in the set $C$, then both of your parents are in the set of all parents of people in $C$) and ${\rm Par}({\rm Chi}(P)) \subseteq P$ (any parent of someone both of whose parents are in $P$ must be in $P$). The childless are no exception: if no one in $P$ has any children, then ${\rm Chi}(P) = \emptyset$, hence ${\rm Par}({\rm Chi}(P)) = \emptyset \subseteq P$.
I could be misunderstanding, but I don't think your $\rm Chi$ is adjoint to $\rm Par$ in any sense, even after correcting for childlessness. You seem to be using the relations $a\le g(f(a))$ and $b \le f(g(b))$ that hold for a contravariant Galois connection but applying them to covariant functors, but I don't think that's any kind of adjunction: if you dualize one of the categories to make the functors covariant, you also need to flip the direction of one of the inequalities.
Anyway, since binary relations abound in the real world, so do the Galois connections they generate.
| 5 | https://mathoverflow.net/users/49 | 427587 | 173,424 |
https://mathoverflow.net/questions/427551 | 1 | Consider $N \times N$ matrices
$$A = \begin{bmatrix}
0 & 0 & \cdots & 0 & 1 \\
1 & 0 & 0 & & 0 \\
\vdots & 1 & 0 & \ddots & \vdots \\
0 & & \ddots & \ddots & 0 \\
0 & 0 & \cdots &1 & 0 \\
\end{bmatrix}$$
and
$$B=\operatorname{diag}( \cos(2\pi\cdot 0/N),...,\cos(2\pi\cdot (N-1)/N)).$$
Does anybody know why the eigenvalues of $i(A+A^T)+2B$ are invariant under 90° rotations?- Numerics seem to imply this. What I mean by this is that if $\lambda$ is an eigenvalue, then also $e^{i \frac{\pi}{2}} \lambda.$
| https://mathoverflow.net/users/119875 | Eigenvalues invariant under 90° rotation | Assume $N$ is even (this is false when N is odd).
Let $X=2B, Y=A+A^T$.
Let
$$P = \begin{bmatrix}
1 & 1 & 1 & \cdots & 1 \\
1 & \zeta & \zeta^2 & \cdots & \zeta^{N-1} \\
1 & \zeta^2 & \zeta^4 & \cdots & \zeta^{2(N-1)} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \zeta^{N-1} & \zeta^{2(N-1)} & \cdots & \zeta^{(N-1)^2} \\
\end{bmatrix}$$
$$Q = \text{diag}(1, -1, 1, -1, \ldots, 1, -1)$$
where $\zeta=e^{\frac{2i\pi}{N}}$.
One can easily check that
$$PXP^{-1}=Y$$
$$PYP^{-1}=X$$
$$QXQ^{-1}=X$$
$$QYQ^{-1}=-Y$$
from which it follows that
$$P(X+iY)P^{-1}=Y+iX=i(X-iY)$$
$$Q(X+iY)Q^{-1}=X-iY$$
This shows that $X+iY$ is conjugate to $i(X+iY)$, so its eigenvalues are invariant under multiplication by $i$.
| 5 | https://mathoverflow.net/users/160416 | 427608 | 173,431 |
https://mathoverflow.net/questions/425687 | 4 | We call a ring extension (where $R$ and $S$ are commutative) $R \subset S$ essential if for every ideal $I$ of $S$ we have
that $I \cap S \neq 0 \implies I \cap R \neq 0$.
Suppose now that $R \subset S$ is an extension that is flat. Can we find a ring homomorphism $S \to T$ such that the composite $R \to S \to T$ is an essential extension that is also flat?
| https://mathoverflow.net/users/145417 | Flat essential ring extensions | **Edit**: I have added some additional remarks to the answer below.
As already [mentioned](https://mathoverflow.net/questions/425687/flat-essential-ring-extensions#comment1099107_425687), when $R$ is a domain and $S$ is a flat overring, we can find $S \to T$ such that $T$ is an overring of $R$ which is both flat and essential over $R$. Indeed, if $S$ itself is not essential over $R$, by Zorn's Lemma it has an ideal $I$ that is $\subseteq$-maximal with respect to $I \cap R$ $=$ $0$. Because $R$ is an integral domain, $I$ is a prime ideal of $S$. Then $T$ $=$ $Q(S/I)$, the quotient field of $S/I$, is clearly essential over $R$. It is easy to verify that $T$ is flat over $R$ by means of the following criterion, which appeared as exercise 22 to Ch.I, §2 of Bourbaki's Commutative Algebra and is often useful when it comes to flatness. For, let $\mathfrak{a}$ $\le$ $R$ $\ni$ $a$ and $t$ $\in$ $T$ such that $at$ $\in$ $\mathfrak{a}T$. If $a$ $\in$ $I$ then $a$ $=$ $0$, so that $(\mathfrak{a}:a)$ $=$ $R$. And if $a$ $\notin$ $I$ then $a$ is a unit in $T$, and hence $t$ $\in$ $\mathfrak{a}T$ $\subseteq$ $(\mathfrak{a}:a)T$.
**Lemma**: a module $M$ over a commutative ring $A$ is flat iff for every $a\in A$ and every ideal $\mathfrak{a}$ of $A$ one has $(\mathfrak{a}M:a)$ $=$ $(\mathfrak{a}:a)M$. Here, $(\mathfrak{a}M:a)$ $=$ $\{x\in M$ $\mid$ $ax\in \mathfrak{a}M\}$ and $(\mathfrak{a}:a)$ $=$ $\{b\in A$ $\mid$ $ab\in \mathfrak{a}\}$. So this comes down to: if $ax\in \mathfrak{a}M$ then $x$ $=$ $by$ for some $y\in M$ and $b\in A$ with $ab$ $\in$ $\mathfrak{a}$.
**Proof**: if $M$ is flat, then $0 \to A/(\mathfrak{a}:a)
\xrightarrow{a\cdot} A/\mathfrak{a}$ remains exact after tensoring with $M$. That is, $0 \to M/(\mathfrak{a}:a)M \xrightarrow{a\cdot} M/\mathfrak{a}M$ is exact. For the converse, it suffices to show that for every ideal $\mathfrak{a}$ of $A$, the natural map $\mathfrak{a} \otimes\_A M \to \mathfrak{a}M$ is injective. Let $\vartheta$ $=$ $\sum\_{i=1}^n a\_i\otimes x\_i$ with $\sum\_{i=1}^n a\_ix\_i$ $=$ 0 (where $a\_i$ $\in$ $\mathfrak{a}$ and $x\_i$ $\in$ $M$). We will rewrite $\vartheta$ as the sum of $n-1$ tensors. By induction, $\vartheta$ will then be equal to the empty sum of tensors, that is: to zero. Putting $\mathfrak{b}$ $=$ $\sum\_{i=1}^{n-1} a\_iA$, we have $x\_n$ $\in$ $(\mathfrak{b}M:a\_n)$ $=$ $(\mathfrak{b}:a\_n)M$, so that $x\_n$ $=$ $by$ for suitable $y$ $\in$ $M$ and $b$ $\in$ $A$ for which we can write $a\_nb$ $=$ $\sum\_{i=1}^{n-1} a\_ic\_i$ with the $c\_i$ $\in$ $A$. Then $a\_n \otimes x\_n$ $=$ $a\_n \otimes by$ $=$ $a\_nb \otimes y$ $=$ $(\sum\_{i=1}^{n-1} a\_ic\_i) \otimes y$ $=$ $\sum\_{i=1}^{n-1} a\_i \otimes c\_iy$, and hence $\vartheta$ $=$ $\sum\_{i=1}^{n-1} a\_i \otimes (x\_i+c\_iy)$, $\square$.
(Note that one can even limit oneself to considering only finitely generated ideals $\mathfrak{a}$.)
Here is a **counterexample** in the case of non-domains. Let $R=\mathbb{F}\_2[x]/(x^2)$. Using the Lemma, one sees that an $R$-module $M$ is flat over $R$ iff $\ker(x\cdot:M\to M)$ $=$ $xM$. So the overring $S$ $=$ $(R[y]/(y^2-x))\_{(x,y)}$ is flat over $R$. If $I$ denotes the non-zero ideal $xyS$ of $S$, one has $I\cap R$ $=$ $0$, and the ideal $I$ is maximal with respect to this property. (The residue classes of $S$ modulo $I$ have representatives 1, $x$, $y$ and $x+y$. Adding $x$ $+$ $y$ to the ideal yields $xy$ $+$ $y^2$ $=$ $xy$ $+$ $x$, so that 0 $\ne$ $x$ will be in the ideal. The same goes when $1$, $x$ or $y$ is added to $I$, of course.)
Assume $S\to T$ is an $S$-algebra such that $R\rightarrowtail T$ and $T$ is flat and essential over $R$. Then $IT$ is an ideal of $T$. If $IT$ $=$ $0$, we have $xy=0$ in $T$, so $y=xt$ for some $t\in T$, because $T$ is $R$-flat. But then $x=y^2=x^2t^2=0$ in $T$, contradicting $R\rightarrowtail T$. And if $IT$ $\ne$ $0$ then, by essentiality, $0$ $\ne$ $IT\cap R$ $=$ $IT\cap S\cap R$. Since $I\cap R$ $=$ $0$, this means that $IT\cap S\supsetneq I$. So one of the elements $1,x,y$ and $x+y$ of $S$ must be in $IT$. But one easily checks that if $s$ is any of these four, having $s=xyt$ for a $t\in T$ implies $x=0$ in $T$. (E.g., if $x+y=xyt$, then $0$ $=$ $x^2t$ $=$ $xy^2t$ $=$ $xy+y^2$ $=$ $xy+x$, so $x$ $=$ $xy$, and therefore $x$ $=$ $xy^2$ $=$ $x^2$ $=$ $0$.)
Notice that $R$ is local and noetherian, and even artinian.
**Edit2**: In the case that $R$ is a domain, the proof did not even use the fact that $S$ is flat over $R$. The statement is therefore likely to be capable of some generalization, though not to a great extent. For when $S$ is not essential over $R$, you will need to pass to a quotient of $S$ first, and this does not sit well with preserving flatness. The statement obviously also holds when $R$ is VNR (Von Neumann regular, i.e. every $R$-module is flat). A common generalization of domains and VNR rings is formed by the class of **pp rings** (principal ideals of $R$ are projective as $R$-modules). If $R$ is pp and $a$ $\in$ $R$, then $\mathrm{ann}\_R(a)$ $=$ $eR$ for an idempotent $e$ of $R$ (this characterizes pp rings). One can write $R$ $=$ $R\_e \times R\_{1-e}$, and $a$ becomes zero in $R\_e$ and regular (not a zero divisor) in $R\_{1-e}$. Every $R$-module $M$ $=$ $M\_e \times M\_{1-e}$. In particular, this holds for ideals of $R$ and of $R$-algebras. The category of $R$-modules is simply the product of the category of $R\_e$-modules and that of the $R\_{1-e}$-modules, and properties merely have to be verified on both "sides" independently. Again take $I$ $\leq$ $S$ maximal with $I \cap R$ $=$ $0$, and let $T$ $=$ $Q(S/I)$, the total ring of quotients of $S/I$. To see that $T$ is flat over $R$, let $\mathfrak{a}$ $\leq$ $R$, $a$ $\in$ $R$, and $t$ $\in$ $T$ such that $at$ $\in$ $\mathfrak{a}T$. We may assume that $a$ is either zero or regular in $R$. If $a$ $=$ $0$, then $(\mathfrak{a}:a)$ $=$ $R$, so that $t$ $\in$ $(\mathfrak{a}:a)T$. And if $a$ is regular in $R$, it remains regular in $S/I$. For assume that $as$ $\in$ $I$ for an $s$ $\in$ $S$ with $s$ $\notin$ $I$. Then $(sS+I) \cap R$ $\ne$ $0$, so we have $0$ $\ne$ $b$ $=$ $ss\_1+i$ $\in$ $R$ for an $s\_1$ $\in$ $S$ and an $i$ $\in$ $I$. But then $0$ $\ne$ $ab$ $=$ $ass\_1+ai$ $\in$ $I \cap R$, contradiction. Therefore, $a$ is a unit in $T$, and hence $t$ $\in$ $\mathfrak{a}T$ $\subseteq$ $(\mathfrak{a}:a)T$. Again, flatness of $S$ never enters into it, which makes one wonder whether that assumption is ever helpful at all for the problem at hand.
| 2 | https://mathoverflow.net/users/31923 | 427609 | 173,432 |
https://mathoverflow.net/questions/427585 | 6 | The Chebyshev polynomial $U\_n(x)$ of the second kind is characterized by
$$
U\_n(\cos\theta)=\frac{\sin(n+1)\theta}{\sin(\theta)}.
$$
It seems that
$$\operatorname\*{Res}\_x \left( U\_n(x)+tU\_{n-1}(x),\sum\_{k=0}^{n-1}U\_k(x) \right) =(-1)^{\frac{n(n-1)}{2}} t^{\left\lfloor\frac{k}{2} \right\rfloor}2^{n(n-1)},$$
where Res denotes the resultant of two polynomials.
I do not know how to prove this "equality". Is it a known result?
Similar results appeared in Dilcher and Stolarsky [1] theorem 2 and in Jacobs, Rayes and Trevisan [2].
**References**
[1] Karl Dilcher and Kenneth Stolarsky, "[Resultants and discriminants of Chebyshev and related polynomials](https://doi.org/10.1090/S0002-9947-04-03687-6)", Transactions of the American Mathematical Society, 357, pp. 965-981 (2004), [MR2110427](https://www.ams.org/mathscinet-getitem?mr=2110427), [Zbl 1067.12001](https://zbmath.org/?q=an%3A1067.12001).
[2] David P. Jacobs and Mohamed O. Rayes and Vilmar Trevisan, "
[The resultant of Chebyshev polynomials](https://www.cambridge.org/core/services/aop-cambridge-core/content/view/D3FA1C0EBDAB83534C86375A0D711BF7/S0008439500017677a.pdf/the-resultant-of-chebyshev-polynomials.pdf)",
Canadian Mathematical Bulletin 54, No. 2, 288-296 (2011), [MR2884245](https://mathscinet.ams.org/mathscinet-getitem?mr=MR2884245), [Zbl 1272.12006](https://zbmath.org/?q=an%3A1272.12006).
| https://mathoverflow.net/users/120597 | Resultant of linear combinations of Chebyshev polynomials of the second kind | Since $U\_n + t U\_{n-1}$ is of degree $n$ and $\sum\_{k=0}^{n-1} U\_k$ is of degree $n-1$ with leading coefficient $2^{n-1}$, the resultant factors as
$$ 2^{n(n-1)} (-1)^{n(n-1)} \prod\_{j=1}^{n-1} (U\_n(x\_j) + t U\_{n-1}(x\_j))$$
where $x\_1,\dots,x\_{n-1}$ are the zeroes of $\sum\_{k=0}^{n-1} U\_k$.
Fortunately, these zeroes can be located explicitly using the usual trigonometric addition and subtraction identities. Telescoping the trig identity $\sin k \theta = -\frac{\cos\left(k+\frac{1}{2}\right) \theta - \cos\left(k-\frac{1}{2}\right) \theta}{2 \sin \frac{\theta}{2} }$ we conclude that
$$ \sum\_{k=0}^{n-1} U\_k(\cos \theta) = -\frac{\cos\left(\left(n+\frac{1}{2}\right) \theta\right) - \cos\left(\frac{\theta}{2}\right)}{2 \sin \theta \sin \frac{\theta}{2}} = \frac{\sin\left(\frac{n}{2} \theta\right) \sin\left(\frac{n+1}{2} \theta\right)}{2 \cos \frac{\theta}{2} \sin^2 \frac{\theta}{2}}$$
and so the $n-1 = \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n-1}{2} \rfloor$ zeroes of $\sum\_{k=0}^{n-1} U\_k$ take the form $\cos( \frac{2\pi j}{n+1} )$ for $1 \leq j < (n+1)/2$ and $\cos( \frac{2\pi j}{n} )$ for $1 \leq j < n/2$.
Since the first class $\cos( \frac{2\pi j}{n+1} )$ of zeroes are also zeroes of $U\_n$, and the second class $\cos( \frac{2\pi j}{n} )$ are zeroes of $U\_{n-1}$, the resultant therefore simplifies to
$$ 2^{n(n-1)} (-1)^{n(n-1)} t^{\lfloor \frac{n}{2} \rfloor}
\prod\_{1 \leq j < \frac{n+1}{2}} U\_{n-1}\left( \cos\left( \frac{2\pi j}{n+1} \right) \right)
\prod\_{1 \leq j < \frac{n}{2}} U\_{n}\left( \cos\left( \frac{2\pi j}{n} \right) \right).$$
But
$$ U\_{n-1}\left( \cos\left( \frac{2\pi j}{n+1} \right) \right) = \left. \sin\left(n\frac{2\pi j}{n+1}\right) \right/ \sin\left(\frac{2\pi j}{n+1}\right) = -1$$
and similarly
$$ U\_{n}\left( \cos\left( \frac{2\pi j}{n} \right) \right) = \left. \sin\left((n+1)\frac{2\pi j}{n}\right) \right/ \sin\left(\frac{2\pi j}{n}\right) = +1$$
and the claim then follows after counting up the signs.
| 12 | https://mathoverflow.net/users/766 | 427618 | 173,433 |
https://mathoverflow.net/questions/427273 | 10 | It is known from Serre's classical result that every p-torsion occurs in the homotopy groups of every sphere. Is it known: do elements of every order occur in homotopy groups of spheres?
| https://mathoverflow.net/users/148161 | Do elements of every order occur in homotopy groups of spheres? | As others suggested, I am posting my earlier [comment](https://mathoverflow.net/questions/427273/do-elements-of-every-order-occur-in-homotopy-groups-of-spheres#comment1098754_427273) as an answer:
---
$\DeclareMathOperator\denom{denom}$Sure. Let $n$ be a positive integer. Let $N$ be the product of $2n$ and Euler $\phi(2n)$. Then for each prime $p$ that divides $n$, the number $p-1$ divides $N$. So for each prime divisor $p$ of $n$, the denominator of the Bernoulli number $B\_N$ is divisible by $p$. So none of the prime factors of $n$ cancel with factors in the numerator of $B\_N/N$. So $\denom(B\_N/N)$ is divisible by $n$. Now $\denom(B\_N/N)$ or $2\denom(B\_N/N)$ is the order of a cyclic summand in the image of the J-homomorphism in the $(2N-1)$st stable homotopy group of $S^0$. So by Freudenthal, there is an unstable homotopy group of a sphere with an element of order $n$.
| 17 | https://mathoverflow.net/users/nan | 427625 | 173,435 |
https://mathoverflow.net/questions/427520 | 3 | Let $0<\theta\_1,\theta\_2<\pi/2$. Suppose $\psi$ is a smooth real-valued function with compact support.
Consider the oscillatory integral
$$I(t):=\int\_{0}^{1}\frac{1}{(y-e^{\dot{\imath}\theta\_1})
(y-e^{\dot{\imath}\theta\_2})}\int\_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx dy.$$
I am trying to find the asymptotic behavior of $I(t)$ as $t\rightarrow \infty$.
Since the phase $x\mapsto t x^2 r^2$ has a non-degenerate stationary point at $x=0$, one is tempted to try the stationary phase method that gives
$$\int\_{\mathbb{R}}\psi(x)e^{\dot{\imath} t x^2y^2}dx= \frac{C \psi(0)}{\sqrt{t}y}+O(\frac{1}{t y^2}),$$
with some constant $C$. But then we are faced with the singularities $y^{-1}$ and $y^{-2}$.
If we use Fubini's theorem then apply the stationary phase method we end up with
$$\int\_{0}^{1}\frac{e^{\dot{\imath} t x^2y^2}}{(y-e^{\dot{\imath}\theta\_1})
(y-e^{\dot{\imath}\theta\_2})}dr= \frac{\widetilde{C} }{\sqrt{t}x}+O(\frac{1}{t x^2}),$$
which is again not integrable against $\psi$.
Another approach is to change variables $s=\sqrt{t}r$ to write
$$I(t)=\frac{1}{\sqrt{t}}
J(t)$$
where
$$J(t):=\int\_{0}^{\sqrt{t}}\frac{1}{(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta\_1})
(\frac{y}{\sqrt{t}}-e^{\dot{\imath}\theta\_2})}\int\_{\mathbb{R}}\psi(x)e^{\dot{\imath} x^2y^2}dx dy.$$
The idea now is to check whether $\lim\_{t\rightarrow \infty}J(t)$
is a constant independent of $t$. But, by Fubini's theorem, we have
$$\lim\_{t\rightarrow \infty}J(t)=c\int\_{\mathbb{R}}
\frac{\psi(x)}{x}dx$$ which may diverge.
| https://mathoverflow.net/users/116555 | Asymptotic behavior of a double oscillatory integral | $\newcommand{\R}{\mathbb R}\newcommand\sgn{\operatorname{sgn}}\newcommand{\vpi}{\varphi}$Obviously, for $a:=\sqrt{2\pi}\,\psi(0)$ we have
\begin{equation\*}
\psi(0)=f(0),
\end{equation\*}
where
\begin{equation\*}
f:=a\vpi
\end{equation\*}
and $\vpi$ is the standard normal density. Letting now $g(x):=\frac{\psi(x)-f(x)}x$ for $x\ne0$, with $g(0):=\psi'(0)$, we see that $g$ is a smooth integrable function and
\begin{equation\*}
\psi(x)=f(x)+xg(x) \tag{1}\label{1}
\end{equation\*}
for all real $x$.
So,
\begin{equation\*}
I(t)=\frac{J\_f(t)+J\_h(t)}{\sqrt t}, \tag{2}\label{2}
\end{equation\*}
where $h(x):=xg(x)$,
\begin{equation\*}
J\_f(t):=
\int\_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)}
\int\_{\R}dx\,f(x)e^{ix^2y^2},
\end{equation\*}
\begin{equation\*}
z\_1:=e^{i\theta\_1},\quad z\_2:=e^{i\theta\_2},
\end{equation\*}
with $J\_h(t)$ defined similarly. Here and in what follows, $t$ is any real number $>0$, unless specified otherwise.
Note that
\begin{equation\*}
\Im z\_1\ne0\quad\text{and}\quad\Im z\_2\ne0. \tag{3}\label{3}
\end{equation\*}
Note also that
\begin{equation\*}
\int\_{\R}dx\,f(x)e^{ix^2y^2}=\frac a{\sqrt{1-2 i y^2}}
\end{equation\*}
and hence
\begin{equation\*}
\begin{aligned}
J\_f(t)&=
a\int\_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)\sqrt{1-2 i y^2}}.
\end{aligned}
\tag{4}\label{4}
\end{equation\*}
For any fixed real $A>0$, by dominated convergence (which holds in view of \eqref{3}),
\begin{equation\*}
\int\_0^A\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)\sqrt{1-2 i y^2}}
\to\frac1{z\_1z\_2}\int\_0^A\frac{dy}{\sqrt{1-2 i y^2}}\ll1
\end{equation\*}(as $t\to\infty$).
We write $E\ll F$ to mean $|E|=O(F)$.
So, letting $A$ go to $\infty$ slowly enough, we will have $A=o(\sqrt t)$ and
\begin{equation\*}
\int\_0^A\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)\sqrt{1-2 i y^2}}
=o(\ln t).
\end{equation\*}
Also, we will have
\begin{equation\*}
\int\_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)\sqrt{1-2 i y^2}} \\
\sim\int\_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)y\sqrt{-2 i}}
\sim\frac{1+i}{4z\_1z\_2}\,\ln t. \tag{5}\label{5}
\end{equation\*}
The latter asymptotic expression in \eqref{5} can be obtained by taking the latter integral in \eqref{5} in closed form, by partial fraction decomposition. It can also be obtained by writing $\int\_A^{\sqrt t}=\int\_A^{t^b}+\int\_{t^b}^{t^{1/2}}$ for $b\in(0,1/2)$ such that $b$ is close to $1/2$.
So, assuming $\psi(0)\ne0$, by \eqref{4},
\begin{equation\*}
J\_f(t)\sim
a\frac{1+i}{4z\_1z\_2}\,\ln t
=\frac{1+i}{4z\_1z\_2}\,\sqrt{2\pi}\,\psi(0)\ln t.
\tag{6}\label{6}
\end{equation\*}
Next,
\begin{equation\*}
J\_h(t)= \int\_{\R}dx\,\sgn(x)g(x)K(t,x), \tag{7}\label{7}
\end{equation\*}
where
\begin{equation\*}
K(t,x):=\int\_0^{\sqrt t}\frac{dy\,e^{ix^2y^2}|x|}{(\frac y{\sqrt t}-z\_1)(\frac y{\sqrt t}-z\_2)}
=\frac12\int\_0^{r^2}dv\, e^{iv}H\_r(v),
\end{equation\*}
\begin{equation\*}
r:=|x|\sqrt t,
\end{equation\*}
\begin{equation\*}
H\_r(v):=\frac1{\sqrt v\,(\frac{\sqrt v}r-z\_1)(\frac{\sqrt v}r-z\_2)}.
\end{equation\*}
Further,
\begin{equation\*}
2K(t,x)=K\_1+K\_2, \tag{8}\label{8}
\end{equation\*}
where
\begin{equation\*}
K\_1:=\int\_0^{1\wedge r^2}dv\, e^{iv}H\_r(v),\quad K\_2:=\int\_{1\wedge r^2}^{r^2}dv\, e^{iv}H\_r(v),
\end{equation\*}
$u\wedge w:=\min(u,w)$.
For $v\in(0,1\wedge r^2)$, we have $H\_r(v)\ll\frac1{\sqrt v}$ and hence
\begin{equation\*}
K\_1\ll1. \tag{9}\label{9}
\end{equation\*}
Note that $K\_2=0$ if $r\le1$. If now $r>1$, then
\begin{equation\*}
H'\_r(v)=-\frac{H\_r(v)}{2v}\,\Big(1+\frac1{\sqrt v-rz\_1}+\frac1{\sqrt v-rz\_2}\Big)
\ll \frac{|H\_r(v)|}v\ll\frac1{v^{3/2}}
\end{equation\*}
for $v>0$; so, integrating by parts, we see that
\begin{equation\*}
K\_2\ll1. \tag{10}\label{10}
\end{equation\*}
Collecting \eqref{7}--\eqref{10} and recalling that $g$ is an integrable function, we see that
\begin{equation\*}
J\_h(t)\ll1. \tag{11}\label{11}
\end{equation\*}
Collecting \eqref{2}, \eqref{6}, and \eqref{10}, we conclude that
\begin{equation\*}
I(t)\sim
\frac{1+i}{4z\_1z\_2}\,\sqrt{2\pi}\,\psi(0)\frac{\ln t}{\sqrt t}, \tag{12}\label{12}
\end{equation\*}
as $t\to\infty$.
---
As seen from the above proof, for \eqref{12} to hold it is enough that the function $\R\setminus\{0\}\ni x\mapsto\frac{\psi(x)-\psi(0)}x$ be integrable, with $\psi(0)\ne0$. (Of course, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)
---
Concerning the case $\psi(0)=0$: Then the bounds on $H\_r(v)$ and $H'\_r(v)$ developed above provide for dominated convergence. So, taking into account that for each real $x\ne0$
\begin{equation}
H\_r(v)\to\frac1{z\_1z\_2\sqrt v},
\end{equation}
we have
\begin{equation}
K(t,x)\to\frac1{2z\_1z\_2}\int\_0^\infty\frac{dv\,e^{iv}}{\sqrt v}
=\frac{1+i}{2z\_1z\_2}\,\sqrt{\frac\pi2}
\end{equation}
and
\begin{equation\*}
J\_h(t)\to C\_\psi:= c\_\psi \frac{1+i}{2z\_1z\_2}\,\sqrt{\frac\pi2},
\end{equation\*}
where
\begin{equation}
c\_\psi:=\int\_{\R}dx\,\sgn(x)g(x)=\int\_{\R}dx\,\frac{\psi(x)}{|x|}.
\end{equation}
Therefore and because here $f=0$ and hence $J\_f(t)=0$, we conclude that
\begin{equation\*}
I(t)\sim \frac{C\_\psi}{\sqrt t},
\end{equation\*}
as $t\to\infty$, provided that $\psi(0)=0$, $c\_\psi\ne0$, and the function $g$ is integrable. (In the case when $\psi(0)=0$, we have $g(x)=\psi(x)/x$ for $x\ne0$. As was already noted, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)
If $\psi(0)=0$ and $c\_\psi=0$, then one has to dig deeper yet, possibly ad infinitum.
However, if $\psi$ is a "smooth bump", as you said in a comment, then apparently $\psi\ge0$ and hence $c\_\psi>0$ (unless $\psi=0$ and hence $I(t)=0$).
| 6 | https://mathoverflow.net/users/36721 | 427631 | 173,438 |
https://mathoverflow.net/questions/427620 | 13 | Let $X$ be a compact Hausdorff topological space such that for every continuous $f,g:X\to\mathbb{R}$ with $0\le f\le g$ there is a continuous $h:X\to\mathbb{R}$ such that $f=gh$.
>
> What can be said about $X$?
>
>
>
Is there a special term for such a property? Any characterizations in terms of the inner structure of $X$ or perhaps some properties of $C(X)$?
It is not hard to show that if $X$ is extremally disconnected this property is satisfied.
In my specific situation I have a compact space, but of course this property makes sense for an arbitrary topological space. Perhaps there is some information in this general case as well.
| https://mathoverflow.net/users/53155 | When can we divide continuous functions? | The kind of completely regular space you are looking for is an $F$-space.
Suppose that $X$ is a completely regular space. Then we say that a subset $A\subseteq X$ is $C^\*$-embedded if whenever $f:A\rightarrow\mathbb{R}$ is bounded and continuous, there is a continuous $g:X\rightarrow\mathbb{R}$ with $g|\_A=f$. We say that a subset of $X$ of the form $f^{-1}[\{0\}]^c$ for some continuous $f:X\rightarrow\mathbb{R}$ is a cozero set. An $F$-space is a space where every cozero set is $C^\*$-embedded.
Proposition: Let $X$ be a completely regular space. Then the following are equivalent:
1. $X$ is an $F$-space
2. whenever $f,g:X\rightarrow\mathbb{R}$ are continuous with $0\leq f\leq g$ there is some continuous $h:X\rightarrow\mathbb{R}$ with $f=gh$.
3. whenever $f,g:X\rightarrow\mathbb{R}$ are bounded and continuous with $0\leq f\leq g$ there is some continuous $h:X\rightarrow\mathbb{R}$ with $f=gh$.
Proof: $2\rightarrow 3$ follows since in (3) we strengthen the premises.
$1\rightarrow 2$ Suppose that $X$ is an $F$-space, and $0\leq f\leq g$. Then let $\ell:g^{-1}[\{0\}]^c\rightarrow\mathbb{R}$ be the function defined by letting
$\ell(x)=f(x)/g(x)$. Then $g^{-1}[\{0\}]^c$ is a cozero set, and $\ell$ is bounded, so $\ell$ extends to a bounded continuous function $h:X\rightarrow\mathbb{R}$. Observe that $f=gh$.
$3\rightarrow 1$. Let $U$ be a cozero set, and let $\ell:U\rightarrow[0,1]$ be a continuous function. Let $g:X\rightarrow[0,1]$ be a continuous function where $U=g^{-1}[\{0\}]^c$. Then let $f:X\rightarrow[0,1]$ be the continuous function such that $f(x)=g(x)\ell(x)$ whenever $x\in U$ and $f(x)=0$ elsewhere. We observe that $0\leq f\leq g\leq 1$, so there must be a continuous function $h:X\rightarrow[0,1]$ with $f=gh$. However, this means that $h(x)=\ell(x)$ whenever $x\in U$, so $h$ is a continuous extension of the function $\ell$.
Q.E.D.
There are plenty of other characterizations of the $F$-spaces. If $X$ is a completely regular space, then let $C(X)$ denote the ring of all continuous functions $f:X\rightarrow\mathbb{R}$. If $A,B\subseteq X$, then we say that $A,B$ are completely separated if there is a continuous function $f:X\rightarrow[0,1]$ with $A\subseteq f^{-1}[\{0\}],B\subseteq f^{-1}[\{1\}]$.
The following fact can be found in the standard text Rings of Continuous Functions by Gillman and Jerison.
Let X be a completely regular space. The following are equivalent:
1. $X$ is an $F$-space.
2. The Stone-Cech compactification $\beta X$ is an $F$-space.
3. Every finitely generated ideal in $C(X)$ is principal.
4. Every ideal in $C(X)$ is convex.
5. For all $f\in C(X)$, there is a $g\in C(X)$ with $f=g\cdot |f|$.
6. If $f\in C(X)$, then the sets $\{x\in X\mid f(x)>0\}$ and $\{x\in X\mid f(x)<0\}$ are completely separated.
7. If $M\subseteq C(X)$ is a maximal ideal, then the set of all prime ideals $P$ in $C(X)$ with $P\subseteq M$ is linearly ordered.
8. If $f,g\in C(X),$ then $\langle f,g\rangle=\langle|f|+|g|\rangle.$
**Comparison with other properties**
If $X$ is an $F$-space, then every countable subset of $X$ is $C^\*$-embedded. Therefore, every compact $F$-space has a homeomorphic copy of $\beta\omega$. A basically disconnected space is a completely regular space where every cozero set has open closure. Every basically disconnected space is an $F$-space, but $(\beta\omega)\setminus\omega$ is an $F$-space that is not basically disconnected. There are also some connected $F$-spaces.
| 16 | https://mathoverflow.net/users/22277 | 427638 | 173,440 |
https://mathoverflow.net/questions/426630 | 8 | Let $\Omega$ be a bounded domain with a Lipschitz boundary. Consider the Dirichlet-to-Neumann map $\Lambda:H^{\frac{1}{2}}(\partial \Omega)\to H^{-\frac{1}{2}}(\partial \Omega)$ defined via
$$ \langle \Lambda f, h\rangle = \int\_{\Omega} \nabla u\cdot \nabla v \, dx,$$
for any $f,h \in H^{\frac{1}{2}}(\partial \Omega)$ where $u\in H^{1}(\Omega)$ is the unique solution to $\Delta u=0$ with Dirichlet data $f$ on $\partial \Omega$ while $v\in H^{1}(\Omega)$ is any function with trace $h$ on $\partial \Omega$.
My question is whether it is true that $\Lambda$ is continuous from $H^1(\partial \Omega)$ to $L^2(\partial \Omega)$. Moreover, is it also continuous from $H^s \to H^{s-1}$ for all $s$ in the closed interval $[0,1]$?
| https://mathoverflow.net/users/50438 | Dirichlet-to-Neumann map on Lipschitz domains | Let $u$ be the solution of the Dirichlet problem for Laplacian in a Lipschitz domain with boundary data $g$. Then, for every $s\in [1/2,3/2]$,
$$
\| u \|\_{H^{s}\,(U)} \leq C \| g \|\_{H^{s-1/2}\,\,\,\,(\partial U)} .
$$
This is a classical result of [Jerison & Kenig](https://www.sciencedirect.com/science/article/pii/S0022123685710671). See the remarks below Theorem 0.5 of that paper.
I believe this estimate and a duality argument implies the result you are looking for.
---
Edit: Since this answer received a downvote, let me add the details of the little duality argument. If I am making a mistake, I would welcome specific comments pointing out my error.
Let $f \in H^1(\partial \Omega)$ and $h\in L^2 (\partial \Omega)$. Let $u$ and $v$ be their respective harmonic extensions to $\Omega$. Then
\begin{align\*}
\bigl| \bigl\langle \Lambda f, h \bigr\rangle \bigr|
=
\biggl| \int\_{\Omega} \nabla u \cdot \nabla v \biggr|
& \leq
\| \nabla u \|\_{H^{1/2}\,\,(\Omega)} \| \nabla v \|\_{H^{-1/2}\,\,\,(\Omega)}
\\ &
\leq
C\| u \|\_{H^{3/2}\,\,(\Omega)} \| v \|\_{H^{1/2}\,\,(\Omega)}
\\ &
\leq
C\| f \|\_{H^1\,(\partial\Omega)} \| h \|\_{L^2\,(\partial\Omega)} \,\,,
\end{align\*}
where in the last line we used the Jerison-Kenig estimate at both endpoints: $s=1/2$ for $v$ and $s=3/2$ for $u$.
By duality, we obtain the estimate
$$
\| \Lambda f \|\_{L^2\,(\partial \Omega)}
\leq
C \| f \|\_{H^1\,(\Omega)},
$$
which concludes the argument that $\Lambda$ is continuous from $H^1(\Omega)$ to $L^2(\partial \Omega)$.
I have made this argument for $s=1$, but it is easy to check it works for $s$ precisely in the range $[0,1]$.
You would apply the Jerison-Kenig estimate with $(s+1/2)$ to $u$ and $(-s+3/2)$ to $v$.
| 4 | https://mathoverflow.net/users/5678 | 427643 | 173,442 |
https://mathoverflow.net/questions/427661 | 1 | It is stated in [Caruso - An introduction to $p$-adic period rings](https://arxiv.org/abs/1908.08424) (the remarks following equation *(2)*) that the $p$-adic étale cohomology groups of an algebraic variety $X$ over a finite extension $K$ of $\mathbb{Q}\_p$ are not $\mathbb{C}\_p$-admissible. In fact, $\mathbb{C}\_p\otimes\_{\mathbb{Q}\_p}H\_\text{ét}^r(X\_{\overline{K}},\mathbb{Q}\_p)$ is isomorphic to the graded module of $\mathbb{C}\_p\otimes\_K H\_{\text{dR}}^r(X)$ for the de Rham filtration, but not to $\mathbb{C}\_p\otimes\_K H\_{\text{dR}}^r(X)$ itself. Is there a reference for this result or a quick proof using a characterization of $\mathbb{C}\_p$-admissibility (at least to prove that they are not $\mathbb{C}\_p$-admissible) ?
| https://mathoverflow.net/users/170999 | $p$-adic étale cohomology groups are not $\mathbb{C}_p$-admissible | The characterization of $\mathbb C\_p$-admissibility in the notes is that the action of the inertia group factors through a finite quotient.
A standard calculation in étale cohomology shows that for $\mathbb P^1$, the action of the Galois group on $H^2( \mathbb P^1\_{\overline{K}}, \mathbb Q\_p)$ is by the inverse of the $p$-adic cyclotomic character.
Since the $p$-adic cyclotomic character does not have the finite inertia property, this shows that it fails for at least one variety.
| 4 | https://mathoverflow.net/users/18060 | 427665 | 173,445 |
https://mathoverflow.net/questions/427343 | 3 | We work over an algebraically closed field $k$ of characteristic $2$. Let $X$ be a cubic threefold realized as a conic bundle via $f: X \to \mathbb{P}^2$ (after blowing up some line). Let $C \to \mathbb{P}^2$ be a plane quintic curve describing the locus of degenerate conics of $f$, and let $\pi: \widetilde{C} \to C$ be the etale double cover parametrizing the lines on these degenerate conics. Let $H$ be the hyperplane section on $C$, and consider $M = \pi^\*H$. I need to work out the dimension $h^0(M)$. If $k$ had characteristic not $2$ then we know $h^0(M) = 4$, since $h^0(M) = h^0(\pi\_\*M) = h^0(H) + h^0(H + \eta)$ where $\eta$ is some line bundle satisfying $\eta^2 \cong \mathcal{O}\_C$, and we know $h^0(H) = 3$, and $h^0(H + \eta) = 1$ as in Beauville's paper on singularities of the Theta divisor.
However this approach does not work in characteristic $2$. Here we simply get a short exact sequence
$$0 \to \mathcal{O}\_C \to \pi\_\*\mathcal{O}\_{\widetilde{C}} \to \mathcal{O}\_C \to 0$$
because in characteristic $2$ we deal with Artin-Schreier extensions. After twisting by $H$ and taking long exact sequence, this bounds $3 \leq h^0(M) \leq 6$ for us. It would suffice in my case to know that this quantity is even, so essentially I want to ensure that even in characteristic $2$, the theta divisor is $\textbf{even}$.
I know that $det(\pi\_\*M) = 2H$ has $h^0(2H) = 11$ since $C$ is genus $6$ and $2H$ is the canonical divisor, but I am not sure if this is of any use. I also tried to manually compute the global sections from cocycle data but that seems like a nightmare.
| https://mathoverflow.net/users/160814 | Is the theta characteristic attached to an etale double cover of a plane quintic arising from a cubic threefold even in characteristic two? | I'll show the rank is at most 4 in characteristic 2, modulo some claims that I think can be justified.
Your exact sequence
$$0 \to \mathcal{O}\_C \to \pi\_\*\mathcal{O}\_{\widetilde{C}} \to \mathcal{O}\_C
\to 0$$
induces a long exact sequence on cohomology
$$0 \to H^0(C, \mathcal{O}\_C(1) ) \to H^0(C, \pi\_\*\mathcal{O}\_{\widetilde{C}}(1) )\to H^0(C, \mathcal{O}\_C(1) ) \to H^1(C ,\mathcal O\_C(1))$$
and it suffices to show that the rank of $H^0(C, \mathcal{O}\_C(1) ) \to H^1(C ,\mathcal O\_C(1))$ is at least $2$.
It's not hard to check that this map is given by cup product with the extension class of the extension $\pi\_\* \mathcal O\_{\tilde{C}} $ of $\mathcal O\_C$ with $\mathcal O\_C$ in $\operatorname{Ext}^1(C,\mathcal O\_C) = H^1 ( C, \mathcal O\_C) = H^0 (C, K\_C)^\vee$.
Now by adjunction, $K\_C = \mathcal O\_C(2)$ and $H^0(C, K\_C) = H^0(C,\mathcal O\_C(2) ) = H^0( \mathbb P^2, \mathcal O\_{\mathbb P^2}(2))$ is the space of quadratic polynomials in three variables. So the extension class is a linear form on quadratic polynomials in three variables, i.e. a three-by-three symmetric matrix.
The cup product $H^0(C, \mathcal O\_C(1)) \times H^0 (C, K\_C)^\vee \to H^1(C,\mathcal O\_C(1)) = H^0(C, K\_C(-1))$ is dual to the product map $H^0(C,\mathcal O\_C(1)) \times H^0(C, K\_C(-1)) \to H^0(C,K\_C)$ so the rank of the cup product map is equal to the rank of this symmetric matrix. So it suffices to prove the symmetric matrix can't have rank $\leq 1$.
The symmetric matrix is certainly nonzero as the extension is nontrivial, so it suffices to prove it can't have rank exactly $1$. By $GL\_3$-symmetry, we can rule out a specific $3\times 3$ matrix, i.e. $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$, that corresponds to the linear form on quadratic polynomials that extracts the coefficient of $x^2$.
What do we know about this extension class? By the Artin-Schreier exact sequence $H^1(C, \mathbb F\_2) \to H^1(C,\mathcal O\_C) \to H^1(C, \mathcal O\_C)$, it must be stable under the Frobenius map $H^1(C,\mathcal O\_C) \to H^1(C,\mathcal O\_C)$, which is dual to the Verschiebung map $H^0(C, K\_C) \to H^0(C,K\_C)$.
How does Verschiebung act on quadratic polynomials in $x,y,z$? I claim it acts by multiplying by the degree $5$ polynomial $f$ defining your curve, then ignoring all monomials where the exponent of $x,y$, or $z$ is even, then dividing by $xyz$, so all the exponents are even, and then taking the square root. This sends a degree $2$ polynomial to a degree $\frac{ 2+5 - 3}{2} =2$ polynomial.
If the linear form that extracts the coefficient of $x^2$ is stable under Frobenius, then the coefficient of $x^4 \cdot xyz$ in $fq$ must match the coefficient of $x^2$ in $q$ for an arbitrary quadratic polynomial $q$. This can only happen if the coefficients of $x^5, x^4y , x^4z$ in $f$ all vanish, since otherwise we could multiply by $yz, xz, xy$ respectively and get a contradiction. But this forces the vanishing locus of $f$ to have a singularity at $(1:0:0)$, contradicting the assumption that it is smooth.
So indeed the rank of the matrix is $\geq 2$, making the rank of global sections $\leq 4$.
| 2 | https://mathoverflow.net/users/18060 | 427668 | 173,446 |
https://mathoverflow.net/questions/426076 | 7 | Let $\mathbb P$ be a forcing that does not collapse $\omega\_1$, $\theta$ sufficiently large and regular and $X\prec H\_\theta$ a countable elementary substructure with $\mathbb P\in X$ as well as $p\in \mathbb P\cap X$. This is a typical situation if one deals with (semi)proper forcings. Let $\delta=X\cap\omega\_1$. Assume further that $\dot S\in X$ is a $\mathbb P$-name for a subset of $\omega\_1$. I would like to find a $(X, \mathbb P)$-semigeneric condition $q\leq p$ and simultaneously control whether $\delta\in\dot S^G$ or not. Lets say we ask for such $q$ so that
$$q\Vdash \check\delta\notin\dot S$$
There are obvious obstructions to this: If there is $T\in \mathcal P(\omega\_1)\cap X$ with $\delta\in T$ and
$$p\Vdash\check T\subseteq \dot S \mod \mathrm{NS}\_{\omega\_1}$$
then there cannot be such $q$. I can prove that if the Strong Reflection Principle ($\mathrm{SRP}$) holds, this is the "only reason" for the nonexistence of such $q$:
>
> Assume $\mathrm{SRP}$ holds. Let $\mathbb P, X, p, \dot S$ as above. Exactly one of the following holds:
>
>
> 1. There is $q\leq p$ that is $(\mathbb P, X)$-semigeneric and $q\Vdash\check\delta\notin \dot S$
> 2. There is $T\in \mathcal P(\omega\_1)\cap X$ with $\delta\in T$ and $$p\Vdash\check T\subseteq \dot S \mod \mathrm{NS}\_{\omega\_1}$$
>
>
>
The proof of $\neg 1.\Rightarrow 2.$ is similar to the the proof of "stationary set preserving forcings are semiproper" (under $\mathrm{SRP}$). Let us say the name dichotomy holds for $\mathbb P$ if exactly one of 1. and 2. holds for any $X, p, \dot S$ as above. One can hope that if a property is provable for the class of $\omega\_1$-preserving forcings under $\mathrm{SRP}$ then maybe this property holds in general for more "well-behaved" classes of forcings. So my question is:
>
> Is it provable (in $\mathrm{ZFC}$) that the name dichotomy holds for
>
>
> * $\sigma$-closed forcings?
> * proper forcings?
> * semiproper forcings?
>
>
>
| https://mathoverflow.net/users/125703 | For which class of forcings does the "name dichotomy" hold? | It turns out that it is not provable in $\mathrm{ZFC}$ that even $\sigma$-closed forcings satisfy the name dichotomy. The answer to all three questions is thus no (unfortunately).
I will show that under $V=L$, $\mathbb P:=\mathrm{Add}(\omega\_1, 1)$ does not satisfy the name dichotomy:
Consider the function $f:\omega\_1\rightarrow \omega\_1$ defined via $f(\alpha)$ is the least $\beta$ so that $\alpha$ is countable in $L\_{\beta+1}$ (this is the wellknown example of a function in $L$ not bounded by any canonical function).
If $G$ is $\mathbb P$-generic, then in $V$ we can define the following set:
$$S=\{\alpha<\omega\_1\mid G\upharpoonright\alpha\text{ is generic over }L\_{f(\alpha)}\}$$
Here, "$G\upharpoonright \alpha$ is generic over $L\_{f(\alpha)}$" means that $G\upharpoonright \alpha=\{p\in G\mid \mathrm{dom}(p)\subseteq\alpha\}$ is generic for $\mathrm{Add}(\alpha, 1)^{L\_{f(\alpha)}}$ over $L\_{f(\alpha)}$ (in particular $G\upharpoonright\alpha\subseteq L\_{f(\alpha)}$). Let $\dot S$ be a name in $V$ for this set $S$.
>
> Claim: $S$ is stationary costationary in $V[G]$.
>
>
>
Proof: It is not difficult to see that $S$ is stationary. To see that $S$ is costationary, let $\dot C$ be a name for a club in $V$. Let $\theta$ be large enough, regular and $X\prec H\_\theta$ countable with $\dot C\in X$. There is a descending sequence $\vec p:=\langle p\_\alpha\mid \alpha<\omega\_1\rangle\in H\_\theta$ of conditions in $\mathbb P$ so that for all $\alpha$ there is $\alpha\leq\beta$ so that $p\_\alpha\Vdash\check\beta\in \dot C$. By elementarity, we may assume $\vec p\in X$. Let $q=\bigcup\_{\alpha<\delta^X} p\_\alpha$. It is clear that $q$ is *not* $(X,\mathbb P)$-generic, but we have $q\Vdash\check \delta^X\in \dot C$. It follows that $q\Vdash\check\delta^X\in\dot C-\dot S$. $\square$
The argument above shows even a little more: If $T\in V$ is stationary in $\omega\_1$, then $T\cap S$, $T-S$ are stationary in $V[G]$ (just choose $X$ with $\delta^X\in T$). It follows that 2. of the name dichotomy fails for $\dot S$ (for any appropriate $\theta$, $X$, $p$).
To get a failure of 1. as well, it is enough to find a countable $X\prec H\_\theta$ so that if $X\cong L\_\gamma$ then $\mathcal P(\delta)\cap L\_\gamma=\mathcal P(\delta)\cap L\_{f(\delta)}$ for $\delta=\delta^X$: Any $q$ that is $(X, \mathbb P)$-semigenric is actually $(X, \mathbb P)$-generic (as $\mathbb P$ has size $\omega\_1$) and hence as $L\_\gamma$ and $L\_{f(\delta)}$ have the same dense subsets of $\mathrm{Add}(\delta, 1)$, such $q$ forces $G\upharpoonright \delta$ to be generic over $L\_{f(\delta)}$.
The following example of such an $X$ is due to Ralf Schindler:
Let $X\_0=\mathrm{Hull}^{H\_\theta}(\emptyset)$ and for $n<\omega$ let $X\_{n+1}=\mathrm{Hull}^{H\_\theta}(\{X\_n\})$. Put $X=\bigcup\_{n<\omega} X\_n$. I claim that this $X$ works. Let $\delta=\delta^X$. Let $X\cong L\_\gamma$.
>
> Claim: $f(\delta)=\gamma+1$.
>
>
>
Proof: $L\_\gamma$ is a model of $\mathrm{ZF}^-$, $\delta$ is uncountable in $L\_\gamma$ and so $f(\delta)>\gamma$. For $n<\omega$ let $X\_n\cong L\_{\gamma\_n}$. Then $\langle \gamma\_n\mid n<\omega\rangle$ is definable over $L\_{\gamma+1}$: $\gamma\_0$ is the least $\xi$ with $L\_\xi\prec L\_{\gamma}$ and $\gamma\_{n+1}$ is the least $\gamma\_n<\xi$ with $L\_\xi\prec L\_{\gamma}$. Thus $\langle \delta^{X\_n}\mid n<\omega\rangle$ is definable over $L\_{\gamma+1}$ and as $\delta=\mathrm{sup}\_{n<\omega} \delta^{X\_n}$, $\delta$ is countable in $L\_{\gamma+2}$. $\square$
As $L\_\gamma$ is a model of $\mathrm{ZF}^{-}$ we have
$$\mathcal P(\delta)\cap L\_\gamma=\mathcal P(\delta)\cap L\_{\gamma+1}$$
so that $X$ has the desired property.
| 0 | https://mathoverflow.net/users/125703 | 427669 | 173,447 |
https://mathoverflow.net/questions/427673 | 4 | I am reading the survey paper: "The de-Rham Witt complex and Crystalline cohomology" by Luc Illusie.
In math line (2.1.12), Illusie considers the pairing $\langle-,-\rangle:\Omega\_{X/S}^1\times T\_{X/S}\longrightarrow \mathcal{O}\_X$ of the tangent and cotangent bundles on a scheme $X$ of characteristic $p$, relative to a morphism $X\longrightarrow S$. ($S$ also has characteristic $p$).
Earlier we defined a bunch of maps, which would be necessary to describe in order for me to ask my question. The first one is the Cartier operation, $C$, which sends a closed form $w\in \Omega\_{X/S}^1$ to a 1-form $Cw \in \Omega\_{X^{(p)}/S}^1$, where $X^{(p)}$ is the base change of $X\longrightarrow S$ with respect to the absolute Frobenius on $S$. Denote by $W$ the canonical projection $W:X^{(p)}\longrightarrow X$.
Illusie then wants to say something about the pairing
$$\langle Cw,W^\*D\rangle$$
which happens on $\Omega\_{X^{(p)}/S}^1\times T\_{X^{(p)}/S}\longrightarrow \mathcal{O}\_{X^{(p)}}$. Namely, he gives the identity
$$\langle Cw,W^\*D\rangle^p = \langle w,D^p\rangle - D^{p-1}\langle w,D\rangle.$$
My questions are:
1. What does the $p$-th power of a tangent vector, $D^p$, mean? Does the tangent space have a ring structure?
2. The pairing apriori should have values in $\mathcal{O}\_{X^{(p)}}$, once it is raised to the $p$'th power we regard it as having values in $\mathcal{O}\_X$ as this is. For this reason it seems that Illusie is regarding $D^{p-1}$ as an element of $\mathcal{O}\_X$, which is consistent with math line (2.1.13), why is that? I guess this goes back to the first question.
3. Illusie also writes $D\_i^p = 0$, where $D\_i = \partial/\partial x\_i$, for some etale basis $(x\_i)$, why is that true? I guess that this should all be clear once I understand this notation.
Thanks in advance!
| https://mathoverflow.net/users/174655 | Pairing of cotangent and tangent bundles | For (1), recall that if $R$ is a ring, then a derivation $D: R \to R$ satisfies the Leibniz rule, which by induction on $n$ implies that if $D^n$ denotes the $n$-fold iterate of $D$, then
$$D^n(fg) = \sum\_{i=0}^n \binom{n}{i} D^i(f) D^{n-i}(g).$$
Since $\binom{p}{i} \equiv 0$ for $0<i<p$, this implies that if $R$ is an $\mathbf{F}\_p$-algebra, then $D^p(fg) = D^p(f) g + f D^p(g)$. In other words, $D^p$ is a derivation.
Let me answer (3) before (2). If you work locally, i.e., consider the derivation $\partial\_x$ on $\mathbf{F}\_p[x]$, then $(\partial\_x)^p x^n$ is zero for $n<p$, and is $p! \binom{n}{p} x^{n-p}$ for $n\geq p$, which is zero. So the derivation $(\partial\_x)^p$ is identically zero.
Let's now discuss (2); since everything is local on $X$ and $X$ is smooth, we can assume that $X$ is etale over $\mathbf{A}^n\_S$ with basis $x\_1, \cdots, x\_n$. Let $\omega$ be a closed $1$-form on $X$. Because of the Cartier isomorphism $\mathfrak{C}: \mathcal{H}^i(F\_\ast \Omega^\bullet\_{X/S}) \xrightarrow{\sim} \Omega^i\_{X^{(p)}/S}$, we can write $\omega = df + \sum\_{i=1}^n F^\ast(g\_i) x\_i^{p-1} dx\_i$ for some functions $g\_i$ and $f$ on $X$. Both $\langle \mathfrak{C} \omega, D\rangle^p$ and $\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle$ kill $df$, so by Frobenius semilinearity, we can assume that $\omega = x\_i^{p-1} dx\_i$. We'll also just assume $n=1$ and write $x$ instead of $x\_1$.
Then $\langle \mathfrak{C} \omega, D\rangle = \langle dx^{(p)}, D\rangle = D(x)$, and $\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx)$. So we need to prove that
$$(Dx)^p = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx),$$
which is an identity due to Hochschild. See <https://joshuamundinger.github.io/assets/notes/hochschild-identity.pdf> for a cute argument; it reduces to using the multinomial analogue of the Leibniz rule for $D^p(x^p)$ and a multinomial analogue of the binomial coefficient vanishing.
---
Clarification from comments: let's again just consider a single variable $x$ and assume $X = \mathbf{A}^1\_{\mathbf{F}\_p}$. The most general closed $1$-form is $\omega = df + g(x^p) x^{p-1} dx$. Then $\mathfrak{C}(df) = 0$, so we can take $a\_i$ to be $g(x^p) x^{p-1}$. Now, $\partial\_x$ is $\mathbf{F}\_p[x^p]$-linear, so
$$\partial\_x^{p-1} (g(x^p) x^{p-1}) = g(x^p) \cdot (\partial\_x^{p-1} x^{p-1}) = (p-1)! g(x^p) = -g(x^p).$$
This means that $\mathfrak{C}(g(x^p) x^{p-1} dx) = g(x^{(p)}) dx^{(p)}$ can be written as $-\partial\_x^{p-1} (g(x^p) x^{p-1}) dx^{(p)}$, as desired.
| 8 | https://mathoverflow.net/users/102390 | 427677 | 173,448 |
https://mathoverflow.net/questions/427453 | 7 | Convex polyhedra are rigid by Cauchy’s theorem. Steffen’s polyhedron is an example of a non-convex polyhedron that is flexible (i.e., non-rigid). However, it appears to have edges of different lengths. My question: are there flexible polyhedra with equilateral triangular faces? I am interested in both finite and (non-trivial) infinitesimal flexibility.
Motivation: I have a bars-and-balls magnetic construction set and I would like to build a flexible polyhedron. But all the bars I have are equal in length.
Cross-posted on [math.stackexchange](https://math.stackexchange.com/questions/4474324/are-equilateral-polyhedra-with-triangular-faces-rigid)
| https://mathoverflow.net/users/486571 | Are polyhedra with equilateral triangular faces rigid? | This depends on how do you define a "polyhedron". If you accept a doubly covered lozenge (two copies of two adjacent equilateral triangles), then no. But under reasonable nondegeneracy conditions the answer is mostly likely yes. We worked on this problem in our recent [Domes over curves](https://arxiv.org/abs/2005.02555) paper, joint with Alexey Glazyrin (see $\S$5.2). If there was a flexible polyhedron of this type this would disprove some conjectures in the area that we believe are true. Anyway, your question is formalized as Conjecture 5.5 where we also have plenty of context on the subject.
P.S. Both Alexey and I gave several Zoom talks on this paper. You can find the links [on my website](https://www.math.ucla.edu/%7Epak/papers/research.htm#gc).
| 8 | https://mathoverflow.net/users/4040 | 427679 | 173,450 |
https://mathoverflow.net/questions/427657 | 2 | Let $M$ be a compact Riemannian manifold. I'll call a set of non-empty subsets $C\_1,\dots,C\_N$ a *geodesic tiling* of $M$ if:
* Each $C\_n$ is closed (geodesically) convex hull of a finite number of $\{p\_{i,n}\}\_{i=1}^{k\_n}$ in $C\_n$; i.e. the smallest geodesically convex subset of $M$ containing the points $\{p\_{i,n}\}\_{i=1}^{k\_n}$,
* $\cup\_{n=1}^N\, C\_n=M$
* If $n\neq n'$ then $C\_n$ and $C\_{n'}$ have disjoint interiors.
It seems to me intuitive that such a collection of sets $\{C\_n\}\_{n=1}^N$ must exist. But does it? And where can I find a reference?
| https://mathoverflow.net/users/36886 | Do all compact manifolds admit geodesic tiling | If dimension $=2$, then "yes".
If dimension $\ge 3$, then "no".
If such a tiling would exist, then the common boundary of a pair of $C$'s would contain a geodesic hypersurface, but generic Riemannian manifold does not have such hypersurfaces.
For more on convex hulls in Riemannian world, see [our paper](https://arxiv.org/abs/2103.15189).
| 3 | https://mathoverflow.net/users/1441 | 427680 | 173,451 |
https://mathoverflow.net/questions/427594 | 7 | Let $X$ be a Polish space and $T\colon X\to X$ be a continuous map. We say that a point $x\in X$ is *quasi-regular* if for every bounded continous function $\varphi\colon X\to\mathbb{R}$ the sequence $A\_n\varphi(x)$ of Birkhoff averages converges as $n\to\infty$, where $$A\_n\varphi(x)=\frac1n\sum\_{j=0}^{n-1} \varphi(T^j(x)).$$
Setting $$L\_x(\varphi)=\lim\_{n\to\infty}A\_n\varphi(x)$$ we obtain a bounded linear functional on the space $\mathcal{C}\_b(X)$ of bounded continous functions $\varphi\colon X\to\mathbb{R}$. Clearly, $L\_x(\varphi)=L\_x(\varphi\circ T)$, but does $L\_x$ have to be given by $$L\_x(\varphi)=\int\_X\varphi\,\text{d}\mu$$ for some $T$-invariant Borel probability measure $\mu$ on $X$?
Comments
--------
Clearly, the answer is yes if we assume that $X$ is compact (by the Riesz theorem). Furthermore, if $\hat{X}$ is any compactification of $X$, the $L\_x$ must be given by some Borel probability measure $\hat{\mu}$ on $\hat{X}$, but since it may be impossible to extend $T$ to $\hat{X}$ we cannot claim that $\hat{\mu}$ is $T$-invariant.
| https://mathoverflow.net/users/24676 | Are all quasi-regular points on Polish spaces generic points? | $A\_n \varphi (x) = \int\_X \varphi \mathrm{d} \mu\_n$ for a Borel probability measure $\mu\_n$, in fact a measure with finite support
$\{ T^j(x) \mid j=0,1,\ldots,n-1\}$.
By the assumption about the point $x$, the sequence of $\mu\_n$ is weakly Cauchy, where "weakly" refers to the duality between the space $M\_\sigma(X)$ of finite Borel measures on $X$ and the space $C\_b(X)$.
The space $M\_\sigma(X)$ is weakly sequentially complete (Bogachev, ``Measure Theory'', Theorem 8.7.1).
Therefore there is a measure $\mu\in M\_\sigma(X)$ such that
$\int\_X\varphi\mathrm{d}\mu=\lim\_n\int\_X\varphi\mathrm{d}\mu\_n$
for every $\varphi\in C\_b(X)$.
Then $L\_x (\varphi) = \int\_X \varphi \mathrm{d} \mu$.
The measure $\mu$ is $T$-invariant because
$L\_x(\varphi)=L\_x(\varphi\circ T)$ for every $\varphi\in C\_b(X)$.
| 4 | https://mathoverflow.net/users/95282 | 427690 | 173,453 |
https://mathoverflow.net/questions/427686 | 6 | This starts with a vaguely-recalled result (which may be false!): that if $\mathcal{U}$ is a measure on the least measurable cardinal $\kappa$, then every elementary $j: L[\mathcal{U}]\rightarrow M$ such that $(j,M)$ is first-order definable over $L[\mathcal{U}]$ with parameters from $L[\mathcal{U}]$ "comes from" iterating the elementary embedding associated to $\mathcal{U}$ itself. That's a bit vague of course, but it just occurred to me that I don't know how to whip up a situation where the opposite extreme holds:
>
> Is it consistent with $\mathsf{MK}$ (relative to large cardinals) that for each $n$ there is a definable pair $(j,M)$ such that $j:V\rightarrow M$ is elementary but $(j,M)$ is not $\Sigma\_n$-definable?
>
>
>
Here "definable" means "first-order definable over $V$ with set parameters." Both $j$ and $M$ are proper-class-sized objects ($M$ literally is a proper class, $j$ is a class function) so this is nontrivial; correspondingly, I'm using $\mathsf{MK}$ as my base theory here so that I can directly phrase the principle in question.
I strongly suspect that the answer to the question is yes but I don't know how to produce a model in which this holds.
| https://mathoverflow.net/users/8133 | Can there be no complexity bound on the definable elementary $V\rightarrow M$? | Yes, this is possible. If there is a proper class of measurable cardinals and $V = \text{HOD}$, then any class of ordinals $A$ is definably encoded by the iterated ultrapower $j\_A : V\to M$ that hits the $\text{HOD}$-least measure on the $\alpha$-th measurable iff $\alpha\in A$. You can recover $A$ by looking at the generators of $j\_A$, the ordinals $\alpha$ such that $\alpha\notin H^M(j[V]\cup \alpha)$. So if $A$ is definable but not $\Sigma\_{n}$-definable, roughly the same will hold for $j\_A$.
Conversely, if there is an embedding $j :V \to M$ that is not $\Delta\_2$-definable from a parameter, then there is an inner model with a proper class of measurable cardinals. In fact, assuming there is no inner model with a proper class of measurable cardinals, then every elementary embedding $j:V\to M$ is the ultrapower of $V$ by a set-sized extender. (Such an ultrapower embedding is always $\Delta\_2$-definable.) To see this, build the core model $K$, and note that $j$ restricts to an iterated ultrapower $i :K\to N$. If $\lambda$ is the supremum of the measurable cardinals of $K$, then all generators of $i$ must be below $i(\lambda)$, or in other words, $i$ is equal the ultrapower of $K$ by the extender of length $i(\lambda)$ derived from $i$. It follows that every ordinal is in $H^N(i[K]\cup i(\lambda)) \subseteq H^M(j[V]\cup j(\lambda))$, which means that $j$ is equal to the ultrapower of the extender of length $j(\lambda)$ derived from $j$.
| 7 | https://mathoverflow.net/users/102684 | 427700 | 173,456 |
https://mathoverflow.net/questions/427708 | 3 | **Starting point.** The struggle for a magic square consisting of distinct square numbers is still ongoing, but it has produced an amusing landmark result called the [Parker square](https://www.bradyharanblog.com/the-parker-square). One of the issues is that square numbers are very scarce - which leads to the following question.
**Problem.** A *$3\times 3$ magic square* is a $3\times 3$ integer matrix with all entries being distinct, such that all row sums, column sums, and diagonal sums are equal.
We call a set $D\subseteq \mathbb{N}$ *dense* if $$\lim\inf\_{n\to\infty}\frac{|D\cap \{1,\ldots, n+1\}|}{n+1} > 0.$$
If $D\subseteq\mathbb{N}$ is dense, is there always a $3\times 3$ magic square with all the entries lying in $D$?
| https://mathoverflow.net/users/8628 | $3\times 3$ magic squares consisting of entries of a dense set $D\subseteq \mathbb{N}$ | Yes.
By Szemerédi's theorem, your set contains an arithmetic progression of arbitrary length. In particular, it contains a progression of length 9, say it's $d\_1,\ldots,d\_9$. Then
$$
\begin{pmatrix}
d\_2 & d\_7 & d\_6\\
d\_9 & d\_5 & d\_1\\
d\_4 & d\_3 & d\_8
\end{pmatrix}
$$
is a magic square.
| 10 | https://mathoverflow.net/users/101078 | 427709 | 173,458 |
https://mathoverflow.net/questions/280429 | 9 | Given $n\in\mathbb{N}$, consider the numbers $\{1,\ldots,n^2\}$ and a permutation $\pi\in S\_{n^2}$. It induces pairs $(1,\pi(1))$, $\ldots$, $(n^2,\pi(n^2))$.
Consider an $n\times n$ grid. How many possibilities are there to place the pairs in the grid such that each row grows from left to right in the first value of the pair, while each column grows from top to bottom in the second value?
e.g. $n=3$, some $\pi\in S\_9$, a valid placement would be:
$\begin{align\*}
(1,2) && (2,1) && (3,3)\\
(4,4) && (5,5) && (9,6)\\
(6,7) && (7,8) && (8,9)
\end{align\*}$
For the the $\pi$, the following placement is not valid as the lower left $(8,9)$ has a larger first-coordinate (8) than the lower middle $(6,7)$ has (6).
$\begin{align\*}
(1,2) && (2,1) && (3,3)\\
(4,4) && (5,5) && (9,6)\\
(8,9) && (6,7) && (7,8)
\end{align\*}$
It seeems to me that it is somewhat hard to find an exact value for the number of valid placements for a given permutation $\pi$. Therefore, a hopefully easier, less accurate question: Is the above number always smaller or equal than the number for the specific permutation $\pi=id$?
In case of $\pi=id$, the number of valid placements is given by [this sequence](https://oeis.org/A039622), which denotes the number of Young Tableaus of shape $(n,\ldots,n)$ or the number of linear extensions of the $n\times n$ lattice (see [this question](https://mathoverflow.net/questions/280316/linear-extension-of-the-n-times-n-lattice)).
| https://mathoverflow.net/users/41187 | Bound on number of nxn grids with lexicographical ordering / poset structure | We have indeed found a proof based on an injection of the set of valid placements into the set of valid placements of the identity permutation. Find details in Theorem 9 of Version 4 of this ArXiv paper: <https://arxiv.org/pdf/1710.03435.pdf>.
| 0 | https://mathoverflow.net/users/41187 | 427712 | 173,461 |
https://mathoverflow.net/questions/427728 | 2 | There are facts in Mathematics that are so "obvious" and "well-known" that no-one includes a proper proof. An example is:
Theorem: If polynomial $P(x,y)$ with rational coefficients is irreducible over ${\mathbb Q}$ but not absolutely irreducible, then the equation $P(x,y)=0$ has at most finitely many rational solutions, and there is an algorithm for listing them all.
This theorem is very important and used (explicitly or implicitly) in hundreds of papers and books devoted to study rational points on curves $P(x,y)=0$, because it allows to assume without loss of generality that $P(x,y)$ is absolutely irreducible, define genus, and proceed in standard way. However, I am not able to find the full proof with all details. At best, there are short sentences like
"In this case all rational points are singular and the statement follows from Bézout's Theorem"
or
"If it not absolutely irreducible, then the absolute factors are conjugate, so any rational solution to one of them satisfies them all, so one can solve the system to find the finitely many rational points and check whether any of them are integer points."
or
"if $P(x,y)$ is irreducible over ${\mathbb Q}$ but not absolutely irreducible then the action of Galois acts transitively on the Q-irreducible components, but rational points are fixed by Galois, so X(Q) is contained in the intersection of the Q-irreducible components; thus in this case we
reduce to the 0-dimensional problem."
However, I cannot find anywhere a proper proof with all details and with definitions of all concepts involved. Why exactly all rational points are singular? Bézout's Theorem is the statement about finiteness but not about algorithm for computing all points. What exactly is meant by "factors are conjugate" or "action of Galois acts transitively", and why this is true?
So, the question is to present a more detailed proof of this theorem or point me to a reference with a proper detailed proof.
| https://mathoverflow.net/users/89064 | Rational solutions to $P(x,y)=0$ for $P$ reducible over ${\mathbb C}$ | The main idea of the proof already appears in what you've written, but here are some more details.
Factor $P(x,y) = Q\_1(x,y) \cdots Q\_n(x,y)$ into irreducibles, where the factorization takes place over a number field $K \supsetneq \mathbb Q$. Let $(a,b)$ be a rational solution to $P(x,y) = 0$. Then $Q\_i(a,b) = 0$ for some $i$. Since $Q\_i$ is not defined over $\mathbb Q$, then there is some element $\sigma \in {\rm Gal}(K/\mathbb Q)$ such that $Q\_i^\sigma \ne Q\_i$. (Note that $\sigma$ is just acting on the coefficients of $Q\_i$.) But $P^\sigma = P$ since $P$ is defined over $\mathbb Q$, and $Q\_i^\sigma$ is still an irreducible factor of $P = P^\sigma$, so $Q\_i^\sigma = Q\_j$ for some $j \ne i$.
Using the fact that $(a,b)$ is rational, we have
$$ Q\_j(a,b) = Q\_i^\sigma(a,b) = \left(Q\_i(a,b)\right)^\sigma = 0^\sigma = 0, $$
so $(a,b)$ lies on the intersection $\{Q\_i = 0\} \cap \{Q\_j = 0\}$. The fact that $(a,b)$ is a singular point on the curve now essentially follows from the product rule from calculus.
| 4 | https://mathoverflow.net/users/66491 | 427741 | 173,470 |
https://mathoverflow.net/questions/427704 | 33 | I submitted a short paper and received a positive review and a negative review. The editor (he) briefly wrote the following things:
1. He thinks my original result could be mistaken because of XYZ
2. He presents an alternative theorem (not entirely in mathematical language but with a combination of math and English), which gives a finer result than mine
3. He presents a sketch of proof
4. He presents an example
5. He adds that, if I think he is wrong, and I was right, please revise my manuscript addressing concerns of reviewers and submit again.
I thought I was right, but after working on this topic for a few more months, I find that he is actually correct. So I wrote down his theorem, proofs, and example in details. I am about to submit, but an idea jumps into my head: he should own the copyright, not me. What should I do here?
*Background*: he is a very smart and cutting-edge researcher in my field, but he does not work on the problems in my direction, so I won't be plagiarizing his papers.
| https://mathoverflow.net/users/122649 | The editor wrote the paper for me | [Comments combined into a community wiki answer.]
*Copyright* is the wrong word in this context; the correct word is *authorship*. A reasonable course of action is to propose to the editor that you and he be coauthors of the paper. If the editor agrees, then the paper would need to be re-submitted to a different journal; it would be a conflict of interest for the editor to accept his own paper for publication. If the editor declines, then the editor's contribution should be formally acknowledged in the paper, and you can proceed with the publication process.
| 56 | https://mathoverflow.net/users/3106 | 427744 | 173,472 |
https://mathoverflow.net/questions/427380 | 6 | Let $\Gamma=(V,E)$ be a finite connected graph.
*Pretty standard notation.* Given a set $S\subset V$, write $\Gamma|\_S$ for the restriction of $\Gamma$ to $S$, i.e., the subgraph $(S,\{\{v,w\}\in E: v,w\in S\})$. Write $\partial\_{\textrm{edge}} S$ for the set of edges $\{v,w\}\in E$ with $v\in S$, $w\notin S$. By $|A|$ we mean the number of elements of a set $A$. We denote by $V\setminus S$ the complement $\{v\in V: v\notin S\}$. A *neighbor* of $v\in V$ is a vertex $w$ such that $\{v,w\}\in E$.
**Question.** Is there an absolute constant $c>0$ such that the following statement (Statement A) is true?
*Statement **A**.* Let $\Gamma=(V,E)$ be a graph whose every vertex has degree $\geq 3$. Then there is an $S\subset V$ such that $\Gamma|\_S$ is connected and
$$|\partial\_{\textrm{edge}} S|\ \geq\ c\cdot|E|.$$
---
It is not hard to reduce statement A to the following:
*Statement **B**.* Let $\Gamma=(V,E)$ be a graph whose every vertex has degree $\geq 3$. Then there is an $S\subset V$ such that:
* $\Gamma|\_S$ is connected,
* every $v\in V\setminus S$ has a neighbor in $S$,
* the sum of the degrees of the elements of $\ V\setminus S\ $ is $\, \geq\ c\cdot|E|$.
---
I will give below (as an answer) a proof of Statement B for regular graphs, and a proof of the reduction of Statement A to Statement B.
| https://mathoverflow.net/users/398 | Existence of connected set with large edge boundary | If I am not mistaken, then the following recursive construction disproves statement A:
* Let $G\_1$ consist of a $K\_8$ and an additional vertex $r\_1$ connected to half of the vertices of this $K\_8$
* For $k > 1$ take $2^{2^k}$ copies of $G\_{k-1}$, connect each pair of copies of $r\_{k-1}$ by an edge, and add a vertex $r\_k$ which is incident to half of the copies of $r\_{k-1}$
(connecting $r\_k$ only to half of the copies has the sole purpose of making the numbers in the induction below a bit prettier)
**Claim 1:** $G\_k$ has $\frac{k+3}{2} 2^{2^{k+1}}$ edges and $2\cdot 2^{2^{k+1}}$ of these edges are contained in copies of $G\_1$.
We use induction on $k$. For $k=1$ note that $G\_1$ has ${8 \choose 2} + 4 = 32 = \frac{1+3}{2} 2^{2^{1+1}}$ edges and all of them are contained in a copy of $G\_1$.
For the induction step, note that the $2^{2^k}$ copies of $r\_{k-1}$ form a complete graph. Together with the $\frac{2^{2^k}}{2}$ edges incident to $r\_k$ the number of edges of $G\_k$ which are not contained in any copy of $G\_{k-1}$ is thus $${2^{2^k}\choose 2} + \frac{2^{2^k}}{2} = \frac 12 (2^{2^k})^2 = \frac{2^{2^{k+1}}}{2}.$$
Since $G\_k$ contains $2^{2^k}$ copies of $G\_{k-1}$ each of which by induction hypothesis contains $\frac{k+2}{2} 2^{2^{k}}$ edges, the total number of edges in $G\_k$ is
$$
\frac{2^{2^{k+1}}}{2} + 2^{2^k} \frac{k+2}{2} 2^{2^{k}} = \frac{k+3}{2} 2^{2^{k+1}}
$$
as claimed. For the number of edges contained in a copy of $G\_1$ simply note that all of them are contained in one of the $2^{2^k}$ copies of $G\_{k-1}$ each of which contains $2\cdot 2^{2^k}$ such edges, giving
$$
2^{2^k} \cdot 2 \cdot 2^{2^k} = 2\cdot 2^{2^{k+1}}
$$
such edges in total. This finishes the proof of Claim 1.
**Claim 2:** Let $S$ be a connected set of vertices of $G\_k$ with maximal edge boundary.
Then $V(G\_k) \setminus S$ only consists of vertices in copies of $G\_1$ and $r\_k$, and $r\_1$ is only in $V(G\_k) \setminus S$ if $k=1$. This immediately implies that
$$
|\partial\_{\text{edge}} S| \leq 2\cdot 2^{2^{k+1}} + \frac{2^{2^k}}{2}.
$$
To show this, we first note that $G\_k$ contains a set $S'$ with $|\partial\_{\text{edge}} S| > 2^{2^{k+1}}$ because in every copy of $G\_1$ we can pick a connected set containing the respective copy of $r\_1$ whose edge boundary contains $19$ of the $32$ edges and then add all vertices not contained in any copy of $G\_1$. In particular, the set $S$ whose edge boundary is maximal must contain at least one of the copies of $r\_{k-1}$, otherwise $S$ would be contained in some copy of $G\_k$ and thus by induction its edge boundary would contain at most $2\cdot 2^{2^{k}} + 2^{2^{k-1}} < 2^{2^{k+1}}$ edges.
Now assume that some copy of $r\_{k-1}$ is not contained in $S$. This copy of $r\_{k-1}$ has at most $2^{2^k}$ neighbours in $S$. Thus adding this copy and a connected subset of the corresponding copy of $G\_{k-1}$ whose edge boundary contains strictly more than $2^{2^{k}}$ edges to $S$ increases the edge boundary while the set stays connected, thereby contradicting maximality of $S$. We have thus shown that $S$ contains all copies of $r\_{k-1}$, and by iterating this argument we see that $S$ must contain all copies of $r\_{j}$ for every $j<k$.
| 4 | https://mathoverflow.net/users/97426 | 427750 | 173,473 |
https://mathoverflow.net/questions/427739 | 1 | Suppose we are given a reductive group $G$, its closed subgroup $H$ (not necessarily reductive), an affine $G$-variety $X$ and its closed subvariety $Y$ such that
(1) The $G$ action on $X$ is free and each orbit is closed.
(2) For $g\in G$, $gY=Y$ if and only if $g\in H$ if and only if $gY\cap Y\neq \emptyset$.
(3) The intersection of each $G$-orbit and $Y$ is nonempty.
Is it true that $X//G$ is a geometric quotient of $Y$ with respect to the $H$ action?
| https://mathoverflow.net/users/488976 | Free closed group action on varieties | It is true if $X$ is normal and $Y$ is irreducible: Let $S:=X//G$. Since $X$ is affine and $G$ is reductive, the morphism $X\to S$ is a principal $G$-bundle (consequence of Luna's slice theorem, see Luna's original paper). This implies that also the geometric quotient $T:=X/H$ exists and $X\to T$ is a principal $H$-bundle. Since $Y\subseteq X$ is closed and $H$-stable, the geometric quotient $U:=Y/H$ exists and is a closed subvariety of $T$. Your assumptions imply that $Y$ intersects each $G$-orbit in an $H$-orbit. Thus $f:U\to S$ is bijective. Because $X$ is normal, also $S$ is normal. The irreducibility of $Y$ now implies that $f$ is an isomorphism, as claimed. This is a consequence of Zariski's main theorem: $f$ is birational because of $char=0$. Let $U\hookrightarrow \tilde S\to S$ be the factorization ${\rm finite}\circ{\rm open\ embedding}$. Then $\tilde S\to S$ is an isomorphism because $S$ is normal and $\tilde S$ is irreducible. Thus $f$ is a bijective open embedding, hence an isomorphism.
Remark: The irreducibility of $Y$ is essential: Take $G={\bf G}\_m$, $X:={\bf A}^1\times{\bf G}\_m$ with $G$ acting trivially on the first factor, and $H:=1$. Now put $Y:=\{xt=1\}\cup\{(0,1)\}$. Then $Y\to X/G={\bf A}^1$ is bijective but not an isomorphism.
| 2 | https://mathoverflow.net/users/89948 | 427754 | 173,475 |
https://mathoverflow.net/questions/427707 | 2 | Let $\{X\_k\}$ be a sequence of random variables, with $X\_k\in\{+1, -1\}$ for $k>0$, generated as follows.
First, define $S\_n=X\_1+\dots +X\_n$, with $X\_0=S\_0=0$, and let $0<\beta<\frac{1}{2}$.
Then, if $|S\_n|<n^\beta$, let
$P[X\_{n+1}=1]= P[X\_{n+1}=-1]=\frac{1}{2}$. Otherwise:
* If $S\_n> n^\beta$, then $P[X\_{n+1}=1]= \frac{1}{2}-\epsilon,
P[X\_{n+1}=-1]=\frac{1}{2}+\epsilon$
* If $S\_n< -n^\beta$, then $P[X\_{n+1}=1]= \frac{1}{2}+\epsilon,
P[X\_{n+1}=-1]=\frac{1}{2}-\epsilon$
Here $0\leq\epsilon<\frac{1}{2}$. If $\epsilon=0$, then regardless of $\beta$, this is your typical iid Bernouilli trials. I am interested in the asymptotics related to this sequence, when $\epsilon>0$ and $\beta<\frac{1}{2}$. In particular do we have a result, a modified version of the central limit theorem, that would state something like this: $S/n^\gamma$ is normally distributed (asymptotically) with finite variance $>0$. Or something else? I have no idea what $\gamma$ might be, I am curious to know.
The goal, in the end, is to produce random sequences that are random enough to fool all known tests of randomness, yet not truly random. Or to put it in other words, to benchmark tests of randomness for pseudo-random number generators. Clearly, my sequence violates the law of the iterated logarithm, and was designed for that very purpose.
Note that $S\_n$, when properly rescaled, gives rise to a sequence that behaves like a Brownian motion, but is obviously non-Brownian since its variance is an order of magnitude smaller than that of a Brownian motion (my guess).
**Update**
A related type of sequence is investigated by Drezner & Farnum (1993). See more recent version [here](https://www.d.umn.edu/%7Eyqi/mydownload/12letters.pdf). I was unable to have my sequence fit in that framework, I don't think it does.
That said, let $p\_n(m)=P(S\_n=m)$, with $-n\leq m \leq n$. Also, let $S\_0=0$ and $p\_0(0)=1$. Then $p\_n(m)$ can be recursively computed using some modified version of the Pascal triangle recursion:
$$
p\_{n+1}(m)=\Big[\frac{1}{2}+\epsilon \cdot A\_n(m-1)\Big]p\_n(m-1)+\Big[\frac{1}{2}-\epsilon\cdot A\_n(m+1)\Big]p\_n(m+1),
$$
with
$$
A\_n(m)=\chi(m<-n^\beta)-\chi(m>n^\beta).
$$
Here $\chi$ is the indicator function. It should be easy to prove that
$\text{E}[S\_n]=0$ due to symmetry. Also, when $\epsilon=0.05$ and $\beta=0.45$ I get the almost perfect fit
$\sqrt{\text{Var}[S\_n]}\approx 1.39 \cdot n^{0.35}$. Below is the Python code, just in case there's a typo in my formula (the Python code is correct).
```
import random
import math
random.seed(1)
epsilon=0.05
beta=0.45
Prob={}
Prob[(0,0)]=1
nMax=2001
def psi(n,m):
p=0.0
if m>n**beta:
p=-1
if m<-n**beta:
p=1
return(p)
OUT=open("rndproba.txt","w")
for n in range(1,nMax):
Exp=0
Var=0
for m in range(-n,n+1,2):
if m+1>n:
Prob[(n-1,m+1)]=0
if m-1<-n:
Prob[(n-1,m-1)]=0
Prob[(n,m)]=(0.5+epsilon*psi(n-1,m-1))*Prob[(n-1,m-1)]+(0.5-epsilon*psi(n-1,m+1))*Prob[(n-1,m+1)]
Exp=Exp+m*Prob[(n,m)]
Var=Var+m*m*Prob[(n,m)]
Var=Var**0.5
line=str(n)+"\t"+str(beta)+"\t"+str(epsilon)+"\t"+str(Exp)+"\t"+str(Var)+"\n"
print("**",n,Exp,Var)
OUT.write(line)
OUT.close()
```
**Formula for the variance**
Taking advantages of symmetries, we have for $n\geq 0$:
$$
\text{Var}[S\_{n+1}]=\text{Var}[S\_n]+1-\delta\_n, \quad \text{with }\delta\_n=8\epsilon\cdot \sum\_{m>n^\beta} m \cdot p\_n(m).
$$
The sum for $\delta\_n$ is finite since $p\_n(m)=0$ if $m>n$. This is an exact formula. Of course,
$\text{Var}[S\_0]=0$. Also, if $\epsilon=0$, the sequence is perfectly random and $\delta\_n=0$
| https://mathoverflow.net/users/140356 | Bernoulli trials with small dependencies: asymptotics (central limit theorem, law of the iterated logarithm) | For $0<\beta \le 1/2$, any limiting distribution of the rescaled process $S\_n/n^\beta$ will be fully supported in $[-1,1]$, so it will not be normal.
The downward drift will imply (using Hoeffding's inequality and a union bound over the largest $m<n$ such that $S\_m<m^\beta$) that $P(S\_n>n^\beta+k) \le Ce^{-c\_\epsilon k}$.
I expect that for $ 0<\beta < 1/2$, the limiting distribution of process $S\_n/n^\beta$ will be uniform in $[-1,1]$, while for $\beta=1/2$, it will be a truncated normal distribution, i.e., it will have the normal density restricted to $[-1,1]$ and normalized to have integral 1.
| 1 | https://mathoverflow.net/users/7691 | 427771 | 173,480 |
https://mathoverflow.net/questions/427769 | 3 | Let $X$ be a compact manifold with boundary $\partial X$ with $
\dim X\setminus \partial X=n$. Moreover, $X$ and $\partial X$ are both aspherical. Then what's the $H^n(X\cup\_{\Sigma\subset \partial X} C(\Sigma))$, where each $\Sigma $ is a boundary component? By Lefschetz duality, $H^n(X,\partial X)=\mathbb Z$. I am trying to apply the (relative) Mayer–Vietoris sequence to $H^n(X,\partial X)$ and $H^n(X\cup\_{\Sigma\subset \partial X} C(\Sigma))$ to deduce the $H^n(X\cup\_{\Sigma\subset \partial X} C(\Sigma))=\mathbb Z$, but it seems it's not very successful so far. Is there an easy way/reference to show $H^n(X\cup\_{\Sigma\subset \partial X} C(\Sigma))=\mathbb Z$?
| https://mathoverflow.net/users/104837 | Cohomology of the coned off space | You're going to need some more assumptions about your manifold if you want $H^n(X,\partial X)=\mathbb{Z}$ (think about connectedness and orientability).
In general (even without those assumptions), $H^n(X\cup\_{\Sigma\subset \partial X} C(\Sigma))\cong H^n(X,\partial X)$ for $n>1$:
Using the long exact sequence of the pair, $H^n(X\cup\_{\Sigma\subset \partial X} C(\Sigma))\cong H^n(X\cup\_{\Sigma\subset \partial X} C(\Sigma), \cup v\_i)$, where $v\_i$ are the cone points. Then by homotopy equivalence, this is isomorphic to $H^n(X\cup\_{\Sigma\subset \partial X} C(\Sigma), \cup C(\Sigma))$. If they weren't already, let $C(\Sigma)$ be the closed cones, and now apply excision.
I leave what happens in the cases $n=0,1$ as an exercise.
| 4 | https://mathoverflow.net/users/6646 | 427778 | 173,482 |
https://mathoverflow.net/questions/427777 | 0 | In a research paper, when one quotes a numbered equation, is it necessary to put the definite article *the* in front of *equation* like "the equation (3.2)", or is it ok to omit *the* and simply write "equation (3.2)"?
| https://mathoverflow.net/users/32746 | How to quote an numbered equation in a paper | In ordinary English it is correct to omit the definite article 'the' because 'equation (3.2)' serves as the proper name for the equation.
| 3 | https://mathoverflow.net/users/124943 | 427779 | 173,483 |
https://mathoverflow.net/questions/421450 | 2 | $\newcommand{\suc}{\operatorname{succ}}\newcommand{\IsPrime}{\operatorname{IsPrime}}$I'm self learning Homotopy type theory reading the HoTT book. I understand that if $A+B$ and $\neg A :\equiv A\rightarrow 0$ are inhabited, then there is an inhabitant of $B$.
**Proof** : Compose $A\rightarrow 0$ with $0\rightarrow B$ to obtain an inhabitant of $A\rightarrow B$; one can then use the identity of $B$ to obtain an inhabitant of $B\rightarrow B$.
From $A\rightarrow B$ and $B\rightarrow B$, we obtain the type $A+B \rightarrow B$ which combined with $A+B$ gives $B$.
As an exercise, I try to prove that there is an infinite number of prime numbers.
The type I try to find an inhabitant of is
$$\prod\_{n:\mathbb{N}}\sum\_{p:\mathbb{N}}\IsPrime(p)\times(p>n)$$
I managed to find an inhabitant of
$$\prod\_{n:\mathbb{N}}\prod\_{p:\mathbb{N}} (p\leq n) + (p > n)$$
and one of
$$\prod\_{n:\mathbb{N}}\sum\_{p:\mathbb{N}}\IsPrime(p) \times (p\mid n! + 1)$$
From these two last types, I think I should be able to prove that $p>n$.
My strategy to do this is to deduce the type $0$ from
$$\IsPrime(p) \times (p\mid n!+1) \times (p \leq n)$$
and deduce that $p>n$ from it using the lemma at the beginning of my post.
My problem is that I only managed to find an inhabitant of $0=\suc(0)$ (here $0:\mathbb{N}$) but I cannot go further and find in inhabitant of the type $0$ from it. Is there a way to construct an inhabitant of
$$(0=\suc(0)) \rightarrow 0$$
I see no rule about the type $\mathbb{N}$ which could help me. Is it implicit in the definition of $\mathbb{N}$ or do we have to add a rule to deduce it ? In fact, as a consequence of $0=\suc(0)$, I find that every terms are equal but cannot deduce an inhabitant of $0$ from it. I think I am missing something obvious. Can someone help me ?
Thanks a lot.
| https://mathoverflow.net/users/33937 | Homotopy type theory : how to disprove that $0=\operatorname{succ}(0)$ in the type $\mathbb{N}$ | $\def\opn{\operatorname}$
$\def\mb{\mathbb}$
To prove this result, we shall define the observational equality principle for $\mb{N}$.
---
The observational equality $\opn{Eq}\_{\mb{N}} : \mb{N} \rightarrow \mb{N} \rightarrow \mathcal{U}$ satisfies the following equations :
\begin{align}
\opn{Eq}\_{\mb{N}}(0\_\mb{N},0\_\mb{N}) &:\equiv \mathbf{1} & \opn{Eq}\_{\mb{N}}(\opn{succ}\_{\mb{N}}(n), 0\_{\mb{N}}) &:\equiv \emptyset \\
\opn{Eq}\_{\mb{N}}(0\_{\mb{N}}, \opn{succ}\_{\mb{N}}(n)) &:\equiv \emptyset & \opn{Eq}\_{\mb{N}}(\opn{succ}\_{\mb{N}}(n),\opn{succ}\_{\mb{N}}(m)) &:\equiv \opn{Eq}\_{\mb{N}}(n, m)
\end{align}
Please note that it is also an inductive type :
\begin{align}
\opn{Eq}\_{\mb{N}}(0\_{\mb{N}}, m) &:\equiv \opn{E}\_0(m) \\
\opn{Eq}\_{\mb{N}}(\opn{succ}\_{\mb{N}}(n), m) &:\equiv \opn{E}\_S(n, \opn{Eq}\_{\mb{N}}(n), m)
\end{align}
with $\opn{E}\_0 : \mb{N} \rightarrow \mathcal{U}$ being defined as
\begin{align}
\opn{E}\_0(0\_{\mb{N}}) &:\equiv \mathbf{1} \\
\opn{E}\_0(\opn{succ}\_{\mb{N}}(n)) &:\equiv \emptyset
\end{align}
and $\opn{E}\_s$ defined as
\begin{align}
\opn{Eq}\_S(n, X, 0\_{\mb{N}}) &:\equiv \emptyset \\
\opn{Eq}\_S(n, X, \opn{succ}\_{\mb{N}}(m)) &:\equiv X(m)
\end{align}
---
This observational equality of the natural numbers is important because it can be used to prove equalities and inequalities.
The reflexive property of this relation, enables us to do so
$$\opn{refl-Eq}\_{\mb{N}} : \prod\_{n:\mb{N}} \opn{Eq}\_{\mb{N}}(n, n)$$
which is again defined by induction on $\mb{N}$
\begin{align}
\opn{refl-Eq}\_{\mb{N}}(0) &:\equiv \star \\
\opn{refl-Eq}\_{\mb{N}}(\opn{succ}\_{\mb{N}}(n)) &:\equiv \opn{refl-Eq}\_{\mb{N}}(n)
\end{align}
Giving this definition, we can easily prove by induction that for $m, n : \mb{N}$
$$(m = n) \iff \opn{Eq}\_{\mb{N}}(m, n)$$
Hence, it follows there is a type family $$(0\_{\mb{N}} = n) \rightarrow \opn{Eq}\_{\mb{N}}(0\_{\mb{N}}, n)$$ indexed by $n : \mb{N}$. Thus, since $\opn{Eq}\_{\mb{N}}(0\_{\mb{N}}, \opn{succ}\_{\mb{N}}(n)) :\equiv \emptyset$ we have $$(0\_\mb{N} = \opn{succ}\_{\mb{N}}(n)) \rightarrow \emptyset$$
| 2 | https://mathoverflow.net/users/368143 | 427786 | 173,484 |
https://mathoverflow.net/questions/427675 | 3 | $\DeclareMathOperator\Sp{Sp}$Let $X=\Sp(A)$ be a connected smooth affinoid rigid space over a discretely valued non-archimedean field $K$. Let $\mathcal{R}$ be a valuation ring of $K$, and fix a uniformizer $\pi$. If $Y=\Sp(B)\subset X$ is a connected affinoid subdomain, the conditions above imply that the rings of power-bounded elements $A^{\circ}, B^{\circ}$ are noetherian affine formal models of $A$ and $B$ respectively. Furthermore, we have an inclusion $A^{\circ}\rightarrow B^{\circ}$, and we can denote its quotient by $C=B^{\circ}/A^{\circ}$. This quotient will, in general, not be flat as a $\mathcal{R}$-module as it will have $\pi$-torsion. For example, if $X=\mathbb{B}^{1}\_{\vert \frac{1}{\pi}\vert}$ is the disk of radius $\vert \frac{1}{\pi}\vert$ and $Y=Sp(K\langle t\rangle)$ is the unit disc, then $t$ is power bounded on $Y$ and not on $X$. However, $\pi t$ is power-bounded in the ring of rigid functions of both spaces. My question is whether $C$ has bounded $\pi$-torsion. That is if there is a natural $n$ such that every element killed by a power of $\pi$ is already killed by $\pi^{n}$. I am not sure if this is the case, as a counterexample to this would be finding a family of functions $f\_{n}$ on $\mathbb{B}^{1}\_{\vert \frac{1}{\pi}\vert}$
such that they are power-bounded on $\mathbb{B}^{1}\_{1}$ and such that there is a point $x\in \mathbb{B}^{1}\_{\vert \frac{1}{\pi}\vert}$ such that $\vert f\_{n}(x)\vert \geq \vert \frac{1}{\pi^{n}}\vert$, which seems like a natural thing to happen. I would also be interested in knowing if there are any conditions on $Y$ that would imply this kind of behaviour.
| https://mathoverflow.net/users/476832 | Bounded torsion of quotients of affine formal models | As it turns out, the quotient $C$ does not necessarily have bounded $\pi$-torsion. For example, let $X=Sp(K\langle t\rangle)$ and $Y=Sp(K\langle \frac{t}{\pi}\rangle)$ be the disc of radius $\pi$. Then all the elements of the form $\frac{t^{n}}{\pi^{n}}$ are power bounded in $K\langle \frac{t}{\pi}\rangle$. Furthermore, their equivalence classes in $C$ are $\pi^{n}$-torsion but not $\pi^{n-1}$ torsion for each $n$. Thus, in general, the answer seems to be no. I would, however, still be interested in knowing any conditions on an admissible affinoid subdomain that would imply this kind of behavior.
| 1 | https://mathoverflow.net/users/476832 | 427789 | 173,486 |
https://mathoverflow.net/questions/427804 | 29 | One has the nice identities
$${xy\choose 1}={x\choose 1}{y\choose 1},$$
$${xy+1\choose 2}={x+1\choose 2}{y+1\choose 2}+{x\choose 2}{y\choose 2}$$
and
$${xy+2\choose 3}={x+2\choose 3}{y+2\choose 3}+4{x+1\choose 3}{y+1\choose 3}+{x\choose 3}{y\choose 3}.$$
(The proof is essentially trivial by interpreting ${z\choose k}$ as a polynomial of degree $k$.)
This sequence of identities stops: There seems to be no nice expression of ${xy+k-1\choose k}$ as a linear combination of
${x+k-i\choose k}{y+k-i\choose k},i=1,\ldots,k$ for $k\geq 4$.
*Is there a good reason for this breakdown?* (Probably a better question is: Is there a reason for these identities to hold for $k=2$ and, especially, for $k=3$?)
| https://mathoverflow.net/users/4556 | Reason for breakdown of a nice binomial identity | $\def\des{\operatorname{des}}$Let $\des(\pi)$ be the number of descents of the permutation $\pi$. Then for any permutation $\pi$ in $S\_k$, we have
\begin{equation\*}\binom{xy+k-\des(\pi)-1}{k} =\sum\_{\sigma\tau=\pi}\binom{x+k-\des(\tau)-1}{k}
\binom{y+k-\des(\sigma)-1}{k}.\tag{$\*$}\label{star}
\end{equation\*}
If $\pi$ is the identity then $\des(\pi)=0$, so the left side of \eqref{star} is $\binom{xy+k-1}{k}$ and on the right, $\tau=\sigma^{-1}$. The OP's identities correspond to the fact that if $k\le 3$ then the number of descents of $\pi$ is the same as the number of descents of $\pi^{-1}$, but this is not true for $k>3$. This also explains the occurrence of the Eulerian numbers as coefficients for $k\le3$.
As far as I know, the earliest proof of \eqref{star} is in
Bogdan Mielnik and Jerzy Plebański,
*[Combinatorial approach to Baker–Campbell–Hausdorff exponents](http://www.numdam.org/item/?id=AIHPA_1970__12_3_215_0)*,
Annales de l’I. H. P., section A, tome 12, no 3 (1970), pp. 215–254 (see equation (11.10)); further references can be found in Jason Fulman and T. Kyle Petersen,
*[Card shuffling and P-partitions](https://doi.org/10.48550/arXiv.2004.01659)*, arXiv:2004.01659 [math.CO].
| 65 | https://mathoverflow.net/users/10744 | 427813 | 173,492 |
https://mathoverflow.net/questions/427815 | 13 | Let $n,m\in\mathbb{N}$. Is there a formula for the number of subgroups of index $n$ in $\mathbb{Z}^m$? Perhaps in terms of the divisors of $n$?
| https://mathoverflow.net/users/41644 | Number of finite index subgroups in a free abelian group | Yes. This is given by OEIS sequence [A160870](https://oeis.org/A160870). The number of subgroups of index $n$ in $\mathbf{Z}^m$ is there denoted $T(n,m)$. There is a recursive formula in terms of the divisors of $n$ given at this page. The initial conditions are
$$ T(n,1) = 1 \quad \textrm{ for all } n\in \mathbb{N}$$
and recursively for $m > 1$, we have either
$$
\quad T(n,m) = \sum\_{d \mid n} \left(\frac{n}{d}\right)^{m-1} \cdot T(d, m-1)
$$
or equivalently,
$$
\quad T(n,m) = \sum\_{d \mid n} d \cdot T(d, m-1)
$$
Note that we can solve this recurrence to get the "explicit" formula
$$ T(n,m) = \sum\_{\substack{(d\_0,d\_1,\ldots,d\_m)}} d\_1 \cdots d\_{m-1}$$
where the sum is over all sequences of integers $(d\_0,d\_1,\ldots,d\_m)$ with $d\_0=1$, $d\_m=n$, and $d\_i \mid d\_{i+1}$ for all $i=0,\ldots,m-1$.
For example, there are $T(4,5) = 651$ subgroups of index $4$ in $\mathbf{Z}^5$.
| 19 | https://mathoverflow.net/users/120914 | 427822 | 173,495 |
https://mathoverflow.net/questions/427819 | 4 | In Section 4.2.4 of [1], the authors write
>
> In this section we consider a causal linear process
> $$
> X\_t = \sum\_{j = 0}^\infty a\_j \varepsilon\_{t - j}, \quad t \in \mathbb{N},
> $$
> where, without loss of generality, $\sum\_{j = 0}^\infty a\_j^2 = 1$ and $\varepsilon\_t$ $(t \in \mathbb{Z})$ are i.i.d. zero mean random variables with Var$(\varepsilon\_1) = \sigma^2\_\varepsilon < \infty$. Thus, Var$(X\_1) = \sigma^2\_X = \sigma^2\_\varepsilon$. Note that
> Gaussian processes are included in this definition, but the class is much more general.
>
>
>
I read the last sentence as "all Gaussian processes are causal linear processes". And [2] seems to support that. It argues that *"...the Mallows MA closure is exhausted by three types of processes. The first type is the set of stationary Gaussian processes with mean zero..."*.
However, a counter-example can be constructed by a Gaussian process which is defined as $X(t) = Y$ for all $t \in \mathbb{N}$ where $Y$ is a zero-mean Gaussian random variable. This is an (admittedly degenerate) Gaussian process and I fail to see how it can be expressed as a linear process.
So, is my counter-example wrong? Or do we need additional assumptions for a Gaussian process to be a (causal) linear process? [1] often works with regularly varying auto-covariance functions resp. spectral densities. Has this maybe been assumed implicitly? If so, how do these assumptions help to prove the claim?
References
==========
[1] J. Beran, Y. Feng, S. Ghosh, and R. Kulik. Long-memory processes. Springer, 2016.
[2] <https://www.pnas.org/doi/pdf/10.1073/pnas.93.22.12128>
| https://mathoverflow.net/users/302666 | Why is every Gaussian process a linear process? | $\newcommand\ep\varepsilon\newcommand\si\sigma\newcommand\N{\mathbb N}$Your counterexample is correct. Indeed, if
$$X\_t=\sum\_{j=0}^\infty a\_j \ep\_{t-j} \tag{1}\label{1}$$
for $t\in\N$, $\sum\_{j=0}^\infty a\_j^2=1$, and the $\ep\_t$'s are iid zero-mean random variables with variance $\si^2\in(0,\infty)$, then for $u\in\N$
$$EX\_tX\_{t+u}=\si^2\sum\_{j=0}^\infty a\_ja\_{j+u}\to0$$
as $u\to\infty$, by the Cauchy--Schwarz inequality.
This contradicts the condition that $X\_t=Y$ for all $t$, because then we would have $EX\_tX\_{t+u}=EX\_1^2=\si^2>0$ for all $u\in\N$.
---
Generally (see e.g. [Theorem 2, p. 263](https://link.springer.com/content/pdf/bfm%3A978-3-642-61943-4%2F1.pdf)), a Gaussian process $(X\_t)$ is of the form \eqref{1} if and only if $(X\_t)$ is a zero-mean stationary process with $EX\_t^2<\infty$ which has an absolutely continuous spectral measure and its spectral
density $f$ satisfies the condition
$$\int\_{-\pi}^\pi\ln f(t)\,dt>-\infty$$
(so that $f$ does not get too close to $0$ on any set of significant enough Lebesgue measure).
(In your counterexample, the spectral measure is degenerate, a multiple of the Dirac measure supported on the singleton set $\{0\}$, and hence not all absolutely continuous.)
| 6 | https://mathoverflow.net/users/36721 | 427824 | 173,497 |
https://mathoverflow.net/questions/427818 | 2 | In ${\mathbb R}^n$, a vector $a=(a\_1,\ldots,a\_n)$ is said to *majorize* another vector $b=(b\_1,\ldots,b\_n)$ if for any convex function $f\colon\mathbb R\to\mathbb R$, we have
$$\sum\_{i=1}^nf(a\_i)\ge \sum\_{i=1}^nf(b\_i).$$
Note that this does not depend on the order of the coordinates. This natural pre-order has been massively studied, and has found many applications.
Now, it is natural to extend this notion to $n\times n$ symmetric matrices as follows: say that $A$ *spectrally majorizes* $B$ if the spectrum of $A$ majorizes that of $B$.
My question is: has this spectral notion been investigated? In particular, is there any nice characterization? Or, at least, some simple (non-spectral) sufficient conditions? Or some known contexts in which such ordered pairs (A,B) arise naturally?
| https://mathoverflow.net/users/477827 | Spectral majorization for symmetric matrices | Yes. This notion of majorization of Hermitian matrices has been investigated before in the context of quantum information theory, and there are several good characterizations of this notion of majorization. This notion of majorization is quite natural since it gives a notion of whether one density operator is 'more mixed' than another density operator.
If $A$ is a matrix, then write $\sigma(A)$ for the spectrum of $A$ (with multiplicity included). If $C,D$ are tuples of real numbers, then write $C\preceq D$ if $D$ majorizes $C$.
If $V$ is a finite dimensional complex inner product space, then let $L(V)$ denote the collection of all linear mappings from $V$ to $V$. We say that a linear mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is trace preserving if $\text{Tr}(\mathcal{E}(A))=\mathcal{E}(A)$ for each $A\in L(V)$. A linear mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is positive if whenever $P\in L(V)$ is positive semidefinite, then $\mathcal{E}(P)$ is also positive semidefinite. A linear mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be completely positive if $\mathcal{E}\otimes 1\_V:L(V\otimes V)\rightarrow L(W\otimes V)$ is positive. A channel is a completely positive trace preserving mapping from some $L(V)$ to some $L(W)$. A linear mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be unital if $\mathcal{E}(1\_V)=1\_W.$ Observe that if $\mathcal{E}$ is both unital and trace preserving, then $\dim(V)=\dim(W)$.
A proof of (1-4) from the following fact can be found in the text The Theory of Quantum Information by John Watrous.
Theorem (Uhlmann): Let $A,B$ be Hermitian mappings. Then the following are equivalent.
1. $\sigma(A)\preceq\sigma(B)$.
2. There is a mixed unitary channel $\mathcal{E}$ with $A=\mathcal{E}(B)$.
3. There is a unital channel $\mathcal{E}$ with $A=\mathcal{E}(B)$.
4. There is a positive, trace preserving, and unital map $\mathcal{E}$ with $A=\mathcal{E}(B)$.
5. For every convex function $f:\mathbb{R}\rightarrow\mathbb{R}$, we have
$\text{Tr}(f(A))\leq\text{Tr}(f(B))$.
| 1 | https://mathoverflow.net/users/22277 | 427827 | 173,500 |
https://mathoverflow.net/questions/427436 | 8 | There is a well known Morita equivalence between the group C\*-algebra $C^\*(H)$ and $C\_0(G/H) \rtimes G$, where $H$ is a subgroup of $G$. The corresponding equivalence of representations is an incarnation of Mackey's imprimitivity theorem, see Rieffel's `Morita Equivalence for Operator Algebras', Example 1.
In topology, one can essentially treat a $H$-space $X$ as equivalent to the $G$ space $X \times\_H G$. In particular, the $H$-equivariant K-theory of $X$ is isomorphic to the $G$-equivariant K-theory of $X \times\_H G$. When $X$ is a point, this corresponds to the Morita equivalence of the previous paragraph.
My question is whether this Morita equivalence is a special case of an equivalence between $C\_0(X) \rtimes H$ and $C\_0(X \times\_H G) \rtimes G$. When $X$ is a point, one uses $C^\*(G)$ as the equivalence bimodule, but I don't know what to replace this with in the more general setting.
In case this is true: is there a general construction in $C^\*$-algebras that is analogous to the construction of $X \times\_H G$ from $X$. Explicitly, if $A$ is a $C^\*$-algebra with $H$-action, is there a canonical $C^\*$-algebra $\tilde A$ with $G$-action such that $A \rtimes H$ is Morita equivalent to $\tilde A \rtimes G$?
| https://mathoverflow.net/users/151844 | Generalisation of the equivalence between $C^*(H)$ and $C_0(G/H) \rtimes G$; induction of group actions on C*-algebras | To answer the question in the final paragraph: yes, there is such a construction. If $H$ is a closed subgroup of $G$, and if $H$ acts on a $C^\*$-algebra $A$, then one defines the induced $C^\*$-algebra $\operatorname{Ind}\_H^G A$ to be the collection of all continuous, bounded functions $f:G\to A$ satisfying:
1. $f(gh)=h^{-1}f(g)$ for all $h\in H$ and $g\in G$; and
2. the function $gH\mapsto \lVert f(g)\rVert$ vanishes at infinity on $G/H$.
$\operatorname{Ind}\_H^G A$ is a $C^\*$-algebra under pointwise operations and the supremum norm, and it carries an action of $G$ by $\*$-automorphisms (coming from the action of $G$ on itself by left translation). This construction really is a generalisation of the situation considered in the earlier part of the question: if $X$ is a locally compact $H$-space then we have $\operatorname{Ind}\_H^G C\_0(X) \cong C\_0(G\times\_H X)$, $G$-equivariantly.
Green ([The local structure of twisted covariance algebras, Zbl 0407.46053](https://zbmath.org/?q=an%3A0407.46053)) proved, essentially, that there is a canonical Morita equivalence between the crossed products $(\operatorname{Ind}\_H^G A)\rtimes G$ and $A\rtimes H$. An equivalence bimodule can be constructed from a suitable completion of the space of compactly supported continuous functions from $G$ to $A$, similarly to what is done for $A=\mathbb{C}$.
(Incidentally, a small comment on the second-last paragraph of the question: when $X$ is a point I believe that the imprimitivity bimodule is the one that implements unitary induction of representations from $H$ to $G$, as constructed by Rieffel. When $H=G$ this is indeed $C^\*(G)$ but I'm not sure this holds in general. For instance, when $H$ is the trivial subgroup the Morita equivalence is between $\mathbb{C}$ and $C\_0(G)\rtimes G$, with the equivalence bimodule being $L^2(G)$ (on which $C\_0(G)\rtimes G$ acts faithfully as the full $C^\*$-algebra of compact operators, per Mackey's generalisation of the Stone–von Neumann theorem).
A good place to learn about all of this, including the history and many related results, is Echterhoff's survey now published as Chapter 2 in [Cuntz, Echterhoff, Li, and Yu - $K$-theory for group $C^\*$-algebras and semigroup $C^\*$-algebras, Zbl 1390.46001](https://zbmath.org/?q=an%3A1390.46001) (also available on the [arXiv](https://arxiv.org/pdf/1006.4975.pdf)). See Theorem 2.6.4 in the published version, and Theorem 6.4 in the arXiv version.
| 3 | https://mathoverflow.net/users/85913 | 427832 | 173,501 |
https://mathoverflow.net/questions/427842 | 86 | Consider the following Turing machine $M$: it searches over valid ZFC proofs, in lexicographic order, and if it finds a proof that $M$ halts, then it halts.
If we fix a particular model of Turing machine (say single-tape Turing machine), and if we fix an algorithm to verify that a given string is a valid ZFC proof of the fact that $M$ halts, this should constitute an unambiguous description of a Turing machine $M$.
(Standard arguments in computability theory, i.e., Kleene's recursion theorem, allows $M$ to compute functions of its own description).
Does $M$ halt?
I find this question puzzling because there's no apparent logical contradiction either way. There could be a proof, in which case it will halt. If there is no proof, then it doesn't halt. What would the answer "depend" on? $M$ either halts or doesn't halt, but could its behavior be independent of ZFC?
I should note that a closely related Turing machine $M'$ can be used to give a simple proof of Godel's incompleteness theorem. It's much more "rebellious" in its behavior, where if it finds a proof that it halts, it doesn't halt, and if it finds a proof that it doesn't halt, it halts. It follows that there cannot be a proof of its halting or non-halting in ZFC (unless ZFC is inconsistent).
However $M$ is just earnestly trying to figure out its fate. Which is it?
| https://mathoverflow.net/users/5534 | Will this Turing machine find a proof of its halting? | It is a very nice question. The answer is yes, the machine will find a proof of its own halting nature, and it will halt when it does so.
I claim this is a consequence of [Löb's theorem](https://en.wikipedia.org/wiki/L%C3%B6b%27s_theorem). Let $M$ be a Turing machine such as you describe. Note that it is not quite correct to say "the" Turing machine that does what you say, since there will be infinitely many different machines $M$ that search for proofs that they themselves halt. It may not be clear initially that they all have the same behavior, but let me show that indeed they do all halt.
Let $\psi$ be the assertion "$M$ halts." Thus, we can prove in ZFC that if $\psi$ is provable, then it is true, since $M$ would discover the proof. Thus, ZFC proves $\text{Pr}\_{ZFC}(\ulcorner\psi\urcorner)\to\psi$. But this is exactly the situation that Löb's theorem is about, and it tells us that we can prove $\psi$ directly in ZFC. So we can prove in ZFC that $M$ halts, as I claimed. It follows that we can prove in PA and much less that $M$ halts, since once we have the actual ZFC proof that it halts, then we can prove in a very weak theory that the actual Turing machine computation halts in whatever specific number of steps it would take to verify the finding of it.
That argument uses the ZFC version of Löb's theorem, but we can get by with the standard PA version, even though M is searching for proofs in ZFC. The reason is that in PA we can prove that $\text{Pr}\_{PA}(\ulcorner\psi\urcorner)\to\psi$, since if PA proves that $M$ halts, then we can prove that ZFC will prove it as well, and so $M$ will halt. Thus, we need only the standard PA version of Löb's theorem to see that PA proves that $M$ halts.
Incidentally, regarding the negated version and the proof of the incompleteness theorem you mention at the end of the post, these ideas are also the basis of the universal algorithm. See my paper [The modal logic of arithmetic potentialism and the universal algorithm](https://arxiv.org/abs/1801.04599).
| 75 | https://mathoverflow.net/users/1946 | 427846 | 173,506 |
https://mathoverflow.net/questions/427857 | 9 | For any simple, undirected graphs $G, H$, let $G\times H$ denote their [category-theoretical product](https://en.wikipedia.org/wiki/Tensor_product_of_graphs).
What is an example of an infinite connected graph $G$ with $G \cong G \times G$?
(Note that the totally disconnected graph $G = (\omega, \emptyset)$ has $G \cong G \times G$.)
| https://mathoverflow.net/users/8628 | Example of a connected graph $G$ with $G \cong G \times G$ | The category of graphs and homomorphisms has countable powers. Given a graph $H = (V(H), E(H))$, its countable power is the graph $H^\omega = (V(H^\omega), E(V^\omega))$ where $V(H^\omega) = V(H)^\omega$ is the countable power of vertices, and $E(H^\omega)$ is defined for $a, b \in V(H)^\omega$ by
$$\{a, b\} \in E(H^\omega) \iff
\forall i \in \omega \,.\, \{a\_i, b\_i\} \in E(H).
$$
For each $i \in \omega$ there is a projection $\pi\_i : H^\omega \to H$, defined by $\pi\_i(a) = a\_i$.
With these projections we really get a power:
$\mathrm{Hom}(L, H^\omega) \cong \mathrm{Hom}(L, H)^\omega$
holds in virtue of the isomorphism taking $f : L \to H^\omega$ to $(\pi\_i \circ f)\_{i \in \omega}$.
For any graph $H$ we have $H^\omega \cong H^\omega \times H^\omega$, with isomorphism taking $a \in V(H^\omega)$ to $((a\_{2 i})\_{i \in \omega}, (a\_{2 j + 1})\_{j \in \omega})$.
Now consider the graph $G = K\_3^\omega$ where $K\_3$ is the complete graph on three vertices $V(K\_3) = \{0, 1, 2\}$. Given $a, b \in \{0,1,2\}^\omega$, we have
$$\{a, b\} \in E(K\_3^\omega) \iff \forall i \in \omega \,.\, a\_i \neq b\_i.$$
The only question remaining is whether $G$ is connected. Take arbitrary vertices $a, b \in \{0,1,2\}^\omega$ and let $c \in \{0,1,2\}^\omega$
be defined by $c\_i = \min (\{0, 1, 2\} \setminus \{a\_i, b\_i\})$. Then $c\_i \neq a\_i$ and $c\_i \neq b\_i$, for all $i \in \omega$, therefore $a$ and $b$ are connected via $c$. The graph $G$ is connected.
| 18 | https://mathoverflow.net/users/1176 | 427861 | 173,509 |
https://mathoverflow.net/questions/427200 | 2 | Let $W$ be a standard one dimensional Brownian motion, and let $X$ be the solution to the SDE
$$dX\_t = \sigma(X\_t) \, dW\_t \;, \quad X\_0 = 1 \;.$$
where $\sigma:\mathbb R \to \mathbb R$ is a Lipschitz continuous function.
For every $M > 0$, let $A\_M$ denote the event
$$\{\underset{0 \leq t \leq 1}{\text{max}} W\_t \geq M\} \;, $$
and let $\mathbb P^M$ be the probability measure given by
$$\mathbb P^M (E) = \frac{\mathbb P(E \cap A\_M)}{\mathbb P(A\_M)} \;, $$
for all events $E$.
We denote by $\mathbb E\_{\mathbb P^M}$ the expectation under $\mathbb P^M$.
Consider the solution to the deterministic ODE
$$dY\_t = \sigma(Y\_t) \, dt \; , \quad Y\_0 = 1.$$
**Question:** Is it true that
$$\lim\_{M \to \infty} \, \mathbb E\_{\mathbb P^M} \big [\underset{0 \leq t \leq 1}{\sup} |X\_t - Y\_{Mt}| \, \big] = 0?$$
| https://mathoverflow.net/users/173490 | Large noise limit for SDE with general volatility coefficients | The answer is no, as can be seen in the case $\sigma(u) = u$, so that $X\_t = \exp(W\_t - t/2)$. For the result to be true, Markov's inequality implies that the law of $W$, conditional on $A\_M$, would need to give probability $1/2$ to the event $\sup\_{t \le 1}|W\_t - Mt + t/2| < K\exp(-M/2)$ for some fixed $K>0$. By large deviations, this event has probability smaller than $c\exp(-c\exp(M))$ for some $c>0$, while $A\_M$ has probability larger than $c \exp(-cM^2)$ for some $c>0$, yielding a contradiction.
| 5 | https://mathoverflow.net/users/38566 | 427865 | 173,511 |
https://mathoverflow.net/questions/317712 | 6 | Let $A,B \in \mathrm{SL}\_3(\mathbb{Z})$. Set
$$S = \langle A \rangle \cdot \langle B \rangle = \{A^mB^n : m,n \in \mathbb{Z}\}.$$
>
> Is $S$ closed in the profinite topology on
> $\mathrm{SL}\_3(\mathbb{Z})$ ?
>
>
>
Equivalently (using the congruence subgroup property), I am asking whether for every $C \in \mathrm{SL}\_3(\mathbb{Z})$ for which $$C \equiv A^{m\_k}B^{n\_k} \pmod k$$ holds for any $k$, we necessarily have $C = A^mB^n$ for some $m,n \in \mathbb{Z}$.
| https://mathoverflow.net/users/38889 | Are double cosets of cyclic subgroups separable in a special linear group? | Although this is an older question, I think that the answer may still be of interest.
I realised that a positive answer follows from the work of Grunewald and Segal [*Grunewald, Fritz; Segal, Daniel*, [**Conjugacy in polycyclic groups**](http://dx.doi.org/10.1080/00927877808822268), Commun. Algebra 6, 775-798 (1978).] In fact, the following is true:
**If H and K are virtually polycyclic subgroups of $SL\_n(\mathbb{Z})$ then the double coset $HK$ is closed in the congruence topology.**
The proof is essentially contained in Exercise 13 on p. 63 of the book of Dan Segal [**Polycyclic groups**](https://doi.org/10.1017/CBO9780511565953), Cambridge Tracts in Mathematics, 82. Cambridge University Press, Cambridge, 1983.
The idea is as follows: consider $SL\_n(\mathbb{Z})$ as a subset of the $n \times n$ integer matrices $M\_n(\mathbb{Z})$. Here $M\_n(\mathbb{Z})$ is a group under matrix addition, so it is isomorphic to $\mathbb{Z}^{n^2}$. A key observation is that *the profinite topology on $M\_n(\mathbb{Z})$ induces precisely the congruence topology on its subset $SL\_n(\mathbb{Z})$* (see Exercise 12 on p. 62 in Segal's book).
Now, consider the action of the direct product $H \times K$ on $M\_n(\mathbb{Z})$ defined as follows: $$(h,k) \circ m=h m k^{-1}, \text{ for all } (h,k) \in H \times K,~m \in M\_n(\mathbb{Z}).$$ The resulting semidirect product $G=M\_n(\mathbb{Z}) \rtimes (H \times K)$ is a virtually polycyclic group, so it is *conjugacy separable* by a theorem of Remeslennikov/Formanek. Note that the conjugacy class of the identity matrix $i \in M\_n(\mathbb{Z})$ in $G$ is precisely the subset $HK \subseteq SL\_n(\mathbb{Z}) \subseteq M\_n(\mathbb{Z})$. It follows that $HK$ is closed in the profinite topology on $M\_n(\mathbb{Z})$, hence it is also closed in the congruence topology on $SL\_n(\mathbb{Z})$.
Finally , let me note that this does not extend to triple cosets: indeed, in
[*Lennox, John C.; Wilson, John S.*, [**On products of subgroups in polycyclic groups**](http://dx.doi.org/10.1007/BF01222760), Arch. Math. 33, 305-309 (1980). ] the authors construct an example of 3 cyclic subgroups of $SL\_3(\mathbb{Z})$ whose product is not profinitely closed.
| 3 | https://mathoverflow.net/users/7644 | 427882 | 173,518 |
https://mathoverflow.net/questions/427889 | 0 | Since the Johnson graph/triangular graph $J(n,2)$ is the complement of the Kneser graph $K(n,2)$, which is also incidentally the line graph of the complete graph $K\_n$, I thought whether the same can be said about the line graphs of the complete hypergraphs $H\_n^k$. That is, can we say that the line graphs of the complete hypergraphs $H\_n^k$ are isomorphic to the complements of Kneser graphs $K(n,k)$? The Baranyai's theorem and the Erdos-Ko-Rado theorems seem to be pointing out in a similar direction. Any hints? Thanks beforehand.
| https://mathoverflow.net/users/100231 | Line graphs of complete hypergraphs as complement of Kneser graphs | Yes, the correspondence goes the following way: The complete k-uniform Hypergraph has n vertices and the edges are given by all k-element subsets of {1,...,n}. Thus, the line graph has those k-element subsets as vertices and they are adjacent if and only if they intersect non-trivially. The complement graph, thus, has the same vertex set but the vertices are adjacent if and only if they intersect trivially. This is the definition of the Kneser graph.
| 2 | https://mathoverflow.net/users/486126 | 427892 | 173,519 |
https://mathoverflow.net/questions/427874 | 0 | **Note**: This is a simplified version of the following [question](https://math.stackexchange.com/questions/4504786/determine-if-an-integral-expression-is-in-l2-mathbbr). I did not get a full response and realized can make it simpler to have my main interrogation answered. I decided to write it as a separate question because, if I just edit my previous question, it will not get any attention.
**Question**: Consider
$$
v(x)\equiv e^x\int\_x^\infty \frac{f(x')}{e^{x'}}dx'.
$$
where $f\in H^1(\mathbb{R}^+)\cap C^1(\mathbb{R}^+)$. I want to prove that
$v\in L^2(\mathbb{R}^+)$.
**Context**: I am computing the resolvent set of a differential operator and I computed the solution of the resolvent equation. Using variation of parameters, I end up with expressions similar to what is above. Among other things, I need to determine if those expressions give me functions that are in $L^2(\mathbb{R}^+)$.
**Reasoning**: Considering the behavior at $x=\infty$, it "seems" fine to me also since the integral will go to zero exponentially fast and fast enough to cancel the behavior of the exponential outside and make $v$ be in $L^2(\mathbb{R})$. Now, obviously, this last sentence is a heuristic argument, which needs to be made formal, if true at all. I need help with that.
| https://mathoverflow.net/users/51290 | Determine if an integral expression is in $L^2(\mathbb{R})$ | From Theorem 0.3.1 of [1], we have that if a kernel operator
$$
T(f)\equiv \int\_0^\infty K(x,x') f(x')dx',\;\;K(x,y)=e^{x-x'}\chi\_{y>x>0}
$$
is such that
$$
\sup\_{x'}\left(\int\_0^\infty K(x,x') dx\right),\;\;
\sup\_{x}\left(\int\_0^\infty K(x,x') dx'\right)\leq C
$$
then
$$
\|T(f)\|\_{L^2(\mathbb{R}^+)}\leq C \|f\|\_{L^2(\mathbb{R}^+)}.
$$
Thus $T$ defines a continuous operator on $L^2(\mathbb{R}^+)$. In my specific case, $K(x,y)=e^{x-x'}\chi\_{y>x>0}$, $C=1$ and thus that answers my question!
A big thank you to Christian Remling to point me toward that theorem.
[1] *Sogge, Christopher D.*, Fourier integrals in classical analysis, Cambridge Tracts in Mathematics. 105. Cambridge: Cambridge University Press. x, 237 p. (1993). [ZBL0783.35001](https://zbmath.org/?q=an:0783.35001).
| 2 | https://mathoverflow.net/users/51290 | 427904 | 173,525 |
https://mathoverflow.net/questions/427900 | 3 | How to prove the following identity?
Let $r = (r\_1, r\_2, \ldots, r\_d)$ and $c = (c\_1, c\_2, \ldots, c\_d)$ be sequences of natural numbers such that $s = r\_1 + r\_2 + \cdots + r\_d = c\_1 + c\_2 + \ldots + c\_d$.
Denote by $\mathcal{M}(r,c)$ the set of matrices whose rows sums and column sums are $r$ and $c$ respectively.
Then
$$
\sum\_{M = (m\_{i,j}) \in \mathcal{M}(r,c)} \prod\_{i,j} \frac{1}{m\_{i,j}!} = \frac{s!}{(\prod\_{i=1}^d r\_i!) (\prod\_{j=1}^d c\_j!)}.
$$
I remember that once I saw this identity, but I could not find the source now.
| https://mathoverflow.net/users/44193 | Combinatorial identity concerning integral matrices with prescribed row sums and column sums | Here is a quick way to see this with generating functions. By using the notation $[\mathbb x^r]f(\mathbb x)$ for the coefficient of a monomial $\mathbb x^r$ in a formal power series $f(\mathbb x)$, we can write:
$$LHS=\left[\prod\_{i,j}x\_i^{r\_i}y\_j^{c\_j}\right]\prod\_{i,j}e^{x\_iy\_j}=\left[\prod\_{i,j}x\_i^{r\_i}y\_j^{c\_j}\right]e^{(x\_1+\cdots+x\_d)(y\_1+\cdots+y\_d)}$$
$$=\frac{1}{s!}\binom{s}{r\_1,r\_2,\dots,r\_d}\binom{s}{c\_1,c\_2,\dots,c\_d}=RHS$$
| 9 | https://mathoverflow.net/users/2384 | 427907 | 173,526 |
https://mathoverflow.net/questions/427879 | 0 | Let $A=\mathbb{R}[x\_1,\dots,x\_n]$ be the algebra of real polynomials in $n$ variables. Fix polynomials $p\_1,\dots,p\_k\in A$.
Consider the subset
$$M:=\{(q\_1,\dots,q\_k)\in A^k|\, p\_1q\_1+\dots+p\_kq\_k=0\}.$$
Clearly $M$ is an $A$-submodule of $A^k$. Necessarily $M$ is finitely generated.
**I am wondering if there exists a software which allows to compute explicit generators of $M$ as an $A$-module.**
In my case $n=16,k=10$, and all $p\_i$'s are explicit homogeneous polynomials of second degree.
| https://mathoverflow.net/users/16183 | Software to compute generators of a module over polynomial ring | What you compute is the "*syzygy module*" of $p\_1,\ldots,p\_k$.
You can try the following M2 script to check the computation times in Macaulay 2 for different $n$'s and $k$'s.
For me $n=16, k=6$ was a matter of minutes, but $k > 6$ seems to take much longer.
If you decrease the sparsity of your $p\_i$ by setting Density to 0.1, 0.5 or higher, you will probably be out of luck with your computation.
```
n = 16
k = 10
R=QQ
--- or R=ZZ/23 or any other prime
A=R[x_1..x_n]
mat = random(A^1, A^{k:-2}, Density=>0.05)
erz = gens ker mat
```
| 2 | https://mathoverflow.net/users/21940 | 427908 | 173,527 |
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