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https://mathoverflow.net/questions/417668 | 1 | Decks are composed of 1 copy of each of $D$ unique cards. The set of cards is $C$ ($|C|=D$), the set of people is $P$ ($|P|=n\geq k$).
Starting with a simpler case (dropping the $k-1$ restriction)
-------------------------------------------------------------
One answer is to give $n-k+1$ full decks to $n-k+1$ people. Then applying pigeonhole principle, any $k$ people will definitely have a full deck. This takes $n-k+1$ decks and $D\times (n-k+1)$ cards. $\blacksquare$
We can show this is minimum. Suppose there are $\leq n-k$ of some card $c\in C$ and hence $\geq n-(n-k)=k$ people who don't have $c$. So, those $k$ people cannot assemble a full deck. Hence, at least $n-k+1$ of each card is required and hence $n-k+1$ full decks. $\blacksquare$
Main Problem
============
Even adding in the restriction that any $k-1$ people cannot together have a full deck, the previous minimum still holds. i.e. we still need $\geq n-k+1$ full decks.
Possible general solution.
--------------------------
There are $\left(\frac{n}{k-1}\right)$ groups of $k-1$ people. This strategy cannot work with $D<\left(\frac{n}{k-1}\right)$.
Partition the set of cards $C$ into $\left(\frac{n}{k-1}\right)$ sets. For extra "niceness" we can easily ensure these parts will have size $\in \left \{ \left \lceil\dfrac{D}{\left(\frac{n}{k-1}\right)}\right \rceil, \left \lfloor\dfrac{D}{\left(\frac{n}{k-1}\right)}\right \rfloor \right \}$
For every group $G\_i$ of $k-1$ people, assign a unique part $C\_i$ to this group, and take away the cards from $C\_i$ from each member of $G\_i$.
For groups $G\_i$ and $G\_j$, $i\neq j$:
----------------------------------------
By construction $G\_i \setminus G\_j \neq \emptyset \implies \exists g\in G\_i, g\notin G\_j$. So $g$ has all the cards of $C\_j$.
So members of $G\_i$ are together only missing the cards of $C\_i$.
By construction $\forall g\notin G\_i\implies g$ has cards of $C\_i$. $\blacksquare$
Number of cards used
--------------------
Number of cards is optimal if $D>\left(\frac{n}{k-1}\right)$ as it achieves the previously computed lower bound. $\blacksquare$
Questions
---------
Im assuming (perhaps prematurely) that my solution is correct.
**Can we prove that there is no solution for the case $D<\left(\frac{n}{k-1}\right)$?**
| https://mathoverflow.net/users/478354 | How to distribute least number of $D$ card decks amongst $n$ people so that any $k$ people have a full deck and no $k-1$ people have a full deck | For each $G\subseteq P$ define $C\_G\equiv\{$**Cards not held by any member of G**$\}$.
Notice $\forall G,H\subseteq P, C\_{G\cup H}=C\_G\cap C\_H$
Let $K\equiv\{G\subseteq P : |G|=k-1\}$. Notice $|K|=\left (\frac{n}{k-1}\right )$ [By definition of $\left (\frac{n}{r}\right )$]
Claim
=====
$\forall G\in K, C\_G\neq \emptyset$
Proof
-----
$|G|=k-1$. But $C\_G=\emptyset\implies G$ has a full deck. $\blacksquare$
Claim
=====
$\forall G,H\in K, G\neq H \implies C\_G \cap C\_H = \emptyset$.
Proof
-----
Let $\exists G,H\in K, G\neq H, C\_G \cap C\_H \neq \emptyset$
Consider $h\in H$.
By construction, $C\_{\{h\}}\supseteq C\_H \supseteq C\_G \cap C\_H \neq \emptyset$.
$\therefore C\_{G\cup\{h\}}=C\_G\cap C\_{\{h\}} \supseteq C\_G \cap C\_G \cap C\_{\{h\}}\neq \emptyset$.
$\implies C\_{G\cup\{h\}} \supset \emptyset \implies C\_{G\cup\{h\}} \neq \emptyset$
$|G\cup\{h\}|=k$. $G\cup\{h\}$ does not have a full deck.$\blacksquare$
Putting it together
===================
Maximum size partition of $C$ is $|C|=D$.
Notice the set $R\equiv\{C\_G : G\in K\}$ forms a partition of $D$. [See 2 above claims]
$|R|=|K| \implies |K|=\left (\frac{n}{k-1}\right )\leq D$ [Pigeonhole principle rears its head once more!] $\blacksquare$
| 0 | https://mathoverflow.net/users/478354 | 417857 | 170,184 |
https://mathoverflow.net/questions/417234 | 10 | Let $G$ be a real connected Lie group. I am interested in its special homotopy properties, which distinguish it from other smooth manifolds
For example
1. $G$ is homotopy equivalent to a smooth compact orientable manifold. In particular, Poincaré duality holds for $G$.
2. $\pi\_1(G)$ is abelian, [$\pi\_2(G) = 0$](https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups)
The impossibly perfect answer to my question is a list of properties that make up a complete homotopy characterization of Lie groups (that is, in every homotopy type (of smooth manifolds) with such properties, there exists a smooth manifold admitting the structure of Lie groups).
P.S. In this question, I am not interested in the homotopy properties of manifolds that distinguish them from other CW-complexes, for this see [resp. question on MO](https://mathoverflow.net/questions/129/how-can-you-tell-if-a-space-is-homotopy-equivalent-to-a-manifold)
| https://mathoverflow.net/users/148161 | Homotopy properties of Lie groups | The problem you mention has a long history. The best homotopy characterization is probably using the notion of finite loop spaces:
A finite loop space is a space $BG$ such that $\Omega BG$ is homotopy equivalent to a finite CW-complex. There are many of those, but one can give a precise homotopy characterization of which ones come from compact Lie groups: They are the ones admitting a "*maximal torus*", defined to be a map $({\mathbb C}P^\infty)^r \to BG$, such that the homotopy fiber is homotopy equivalent to a finite complex.
This was the so-called *maximal torus conjecture*, solved as a consequence of *the classification of $p$-compact groups*, which states that there is a 1-1-correspondence between connected $p$-compact groups and ${\mathbb Z}\_p$-root data, parallel to the classification of connected compact Lie groups, but with $\mathbb Z$ replaced by the $p$-adic integers ${\mathbb Z}\_p$.
As a consequence one can also give a classification of all finite loop spaces: If you pick a connected $p$-compact group for every prime $p$ agreeing over $
\mathbb Q$, there is an explicit double coset space of finite loop space structures with this $p$-local data.
A reference is my ICM survey [The Classification of p–Compact Groups and Homotopical Group Theory](http://web.math.ku.dk/%7Ejg/papers/icm.pdf).
| 12 | https://mathoverflow.net/users/6574 | 417858 | 170,185 |
https://mathoverflow.net/questions/417856 | 8 | If $F$ is a free group then it has cohomological dimension one, which implies that the augmentation ideal $IF=\operatorname{ker}(\epsilon:\mathbb{Z}G\to \mathbb{Z})$ of its group ring is a projective $\mathbb{Z}F$-module. Hence $IF$ is a direct summand of a free $\mathbb{Z}F$-module $M$.
>
> Question: Is it possible to give an explicit construction of such a free module $M$? Perhaps it has a basis related to a free basis of $F$?
>
>
>
| https://mathoverflow.net/users/8103 | Augmentation ideal of a free group | Let $X$ be a free basis for $F$. The Cayley graph of $F$ is a tree $T$ on which $F$ acts freely. The augmented chain complex gives a resolution
$$0\to \mathbb ZF^{(X)}\to \mathbb ZF\to \mathbb Z\to 0$$ (since $T$ is a tree) where we identify $\mathbb ZF^{(X)}$ as the free abelian group on the edges of $T$ and the image of the boundary map sends the edge $(g,x)$ to $gx-x$ and is hence the augmentation ideal (which in any event is clearly the kernel of the augmentation map). So the augmentation ideal is a free module on $X$ with generators $x-1$ with $x\in X$.
| 10 | https://mathoverflow.net/users/15934 | 417860 | 170,187 |
https://mathoverflow.net/questions/417863 | 10 | The question is in the title.
To make the statement more precise, is is true that for any given monoidal category $(\mathcal C, I, \otimes)$ there exists at least one braiding $\beta$? In other words, does the forgetful functor from braided monoidal categories admit a section (not a left adjoint!)?
I strongly suspect the answer to be 'no', or at least 'yes but not functorially'.
| https://mathoverflow.net/users/50376 | Does every monoidal category admit a braiding? | No, sometimes there are even $x,y$'s with no abstract isomorphism $x\otimes y \cong y\otimes x$.
Here are two families of examples:
* Monoids, viewed as discrete categories. The tensor product is just the multiplication, and the existence of a braiding would simply mean that the monoid is commutative, which it typically isn't.
* Functor categories : if $C$ is a category, then $Fun(C,C)$ is monoidal with respect to $\circ$, but it's quite rare for endofunctors to commute.
| 26 | https://mathoverflow.net/users/102343 | 417865 | 170,189 |
https://mathoverflow.net/questions/417864 | 0 | Let $R\_n$ be a random variable with values in $[0,1]$ and $nR\_n$ converges to $\frac{1}{1+C} \chi\_m^2$ in distribution for some constant $C>0$ and $m\in \mathbb{N}$. Is it possible to show that $(1-R\_n)^{-\frac{n-1}{2}} =O\_P(1)$ holds?
I came across this result in the [proof of Theorem 3 of this paper](https://people.eecs.berkeley.edu/%7Ejordan/courses/260-spring10/readings/liang-etal.pdf) but didn't really get it. Some help would be much appreciated.
| https://mathoverflow.net/users/163533 | Random variable is Big O in probability notation | For each natural $n$, let $R\_n$ be random variable with values in $[0,1]$ such that $nR\_n$ converges in distribution, as $n\to\infty$, to a random variable $X$ with a continuous cdf.
Let $T\_n:=(1-R\_n)^{-(n-1)/2}$. Take any real $t>1$. Then for all large enough natural $n$
$$t^{-2/(n-1)}=e^{-2\ln t/(n-1)}<1-\frac{\ln t}n$$
and hence
$$P(T\_n>t)=P(nR\_n>n(1-t^{-2/(n-1)})) \\
\le P(nR\_n>\ln t)\underset{n\to\infty}\longrightarrow P(X>\ln t)
\underset{t\to\infty}\longrightarrow0.$$
So,
$$\lim\_{t\to\infty}\lim\_{n\to\infty}P(T\_n>t)=0;$$
that is (cf. e.g. [equivalence (i)$\iff$(v) in Lemma 1](http://www2.math.uu.se/%7Esvante/papers/sjN6.pdf)), $(1-R\_n)^{-(n-1)/2}=T\_n=O\_P(1)$, as desired.
---
The condition that $X$ has a continuous cdf (which actually holds in your setting) is not needed here, but with it the proof gets simpler just a little bit.
| 2 | https://mathoverflow.net/users/36721 | 417870 | 170,192 |
https://mathoverflow.net/questions/417848 | 3 | Let $K$ be a number field and let $A\_K$ be the adele ring of $K$. Then $K$ sits in $A\_K$ via the diagonal embedding and the quotient $A\_K/K$ is compact. All this is well known. Many proofs of the above fact first reduce the case to that of $K=\mathbb{Q}$ and solves the problem in this case. The proof also provides a fundamental domain for the quotient in terms of a fundamental domain for $\mathbb{Z}$ in $\mathbb{R}$.
I am trying to find a description of the fundamental domain for $A\_K/K$ involving a fundamental domain for $O\_K$, the ring of integers of $K$, in $K\_\infty := \prod\_{v | \infty} K\_v $ (all the notations are standard and I hope that its ok that I don't explain them).
Apparently such a description is possible only if the class number of $K$ is $1$. See the final remarks in [Conrad's notes](http://math.stanford.edu/%7Econrad/676Page/handouts/adelelattice.pdf) for example. I don't understand where exactly the problem occurs if the class number is bigger than $1$. Any help would be greatly appreciated.
| https://mathoverflow.net/users/148866 | Fundamental domain for $A_K/K$ | One does not need that the class number is $1$, the construction in Brian Conrad's notes works in general. See Propositions 6-7 in Ch.V-4 of Weil: Basic Number Theory. The class group enters for $A\_K^\times/K^\times$, not for $A\_K/K$.
| 4 | https://mathoverflow.net/users/11919 | 417874 | 170,195 |
https://mathoverflow.net/questions/417891 | 1 | I would like to find a upper bound of principal polarization of abelian variety in the following stiution:
Suppose $A$ is an abelian variety over a $char=0$ algebraically closed field. And for any two principal polarization $\lambda\_1$, $\lambda\_2$:$A \to A^t$, we we say they are eqivalent if and only if there exsits an isomorphism $\sigma\in Aut(A)$ such that $\lambda\_1=\sigma^t \circ\lambda\_2\circ\sigma$.
Now, let $T$ be the collection of the principal polarization of $A$, is there any upper bound (which is just relative to the dimension $g$) of the cardnarity of $T$/~?
| https://mathoverflow.net/users/478518 | The upper bound of number of the automorphism of principal polarization of abelian variety over algebraically closed field | No, not without some extra assumptions. Take for example $A = E \times E'$ where $E$ and $E'$ are generic elliptic curves connected by an isogeny $E \to E'$ with kernel $\mathbb{Z}/m\mathbb{Z}$, for some squarefree integer $m$. For each factorization $dk = m$, you can find elliptic curves $E\_k, E\_d$ on $A$ which are quotients of $E$ of degree $k, d$ respectively and such that $A \simeq E\_k \times E\_d$. This gives you something like $2^{a-1}$ non-isomorphic (product) principal polarizations, where $a$ is the number of primes dividing $m$.
On the other hand, for a generic abelian variety, the endomorphism ring of $A$ is $\mathbb{Z}$ and the number of principal polarizations is 1.
In general, you can translate this question into a purely algebraic question about the number of conjugacy classes of elements in the endomorphism ring of A. Here, the conjugacy will be defined in a slightly weird way in terms of the Rosati involution. The answer to this algebraic question should ultimately be connected to class numbers of various commutative rings which embed in the endomorphism ring. The case I describe above roughly corresponds to $\mathbb{Z}[x]/(x^2 - m^2)$ embedding in $\mathrm{Mat}\_2(\mathbb{Z})$.
| 3 | https://mathoverflow.net/users/949 | 417897 | 170,200 |
https://mathoverflow.net/questions/417815 | 5 | In "On $\ell$-adic representations attached to modular forms II", Ribet proved that the $\ell$-adic representation $\rho\_{f,\ell}$ attached to a non-CM newform form $f$ satisfies
$${\rm SL}\_2(\mathbb{F}\_\ell)\subset \overline{\rho}\_{f,\ell}(G\_\mathbb{Q}).\qquad\qquad(\*)$$
I am wondering:
(i) Why does $(\*)$ imply that $\overline{\rho}\_{f,\ell|G\_{\mathbb{Q}(\zeta\_\ell)}}$ is absolutely irreducible?
(ii) Let $n$ be a positive integer. Does $(\*)$ imply that ${\rm Sym}^n\overline{\rho}\_{f,\ell|G\_{\mathbb{Q}(\zeta\_\ell)}}$ is irreducible? If the answer is negative, under what condition does this hold?
| https://mathoverflow.net/users/477704 | Irreducibility of the $n$th symetric power of the reduction of the Galois representation of a non-CM newform | Here's an answer that's more work than Kevin Ventullo's, but works for $\ell=2$ and $\ell=3$, and gives you a method that will work in various more general cases.
There is a surjection $\pi\_1\colon G\_\mathbb Q \to \overline{\rho}\_{f,l} (G\_{\mathbb Q})$ and a surjection $\pi\_2 \colon G\_\mathbb Q \to \operatorname{Gal}(\mathbb Q(\zeta\_\ell) / \mathbb Q)$. The group $\overline{\rho}\_{f,l} (G\_{\mathbb Q(\zeta\_\ell)})$ is the image under $\pi\_1$ of the kernel of $\pi\_2$. In other words, it is the intersection of the image of the product map $(\pi\_1,\pi\_2) \colon G\_\mathbb Q \to \overline{\rho}\_{f,l} (G\_{\mathbb Q}) \times \operatorname{Gal}(\mathbb Q(\zeta\_\ell) / \mathbb Q)$ with $G\_{\mathbb Q}) \times 1$.
By Goursat's lemma, the image of the product map is $ \{ (a,b) \mid f\_1(a) =f\_2 (b) \}$ where $f\_1 \colon \overline{\rho}\_{f,l} (G\_{\mathbb Q}) \to Q$ and $f\_2 \colon \operatorname{Gal}(\mathbb Q(\zeta\_\ell) / \mathbb Q) \to Q$ are surjections for some finite group $Q$. Expressed in this language, $\overline{\rho}\_{f,l} (G\_{\mathbb Q(\zeta\_\ell)})$ is the kernel of $f\_1$.
Now because $f\_2$ is surjective, $Q$ is a quotient of a cyclic group of order $\ell-1$, and thus is a cyclic group of order dividing $\ell-1$.
So $\overline{\rho}\_{f,l} (G\_{\mathbb Q(\zeta\_\ell)})$ is the kernel of a map from $\overline{\rho}\_{f,l} (G\_{\mathbb Q(\zeta\_\ell)})$ to a cyclic group of order dividing $\ell-1$. Thus it contains the kernel of a map from $SL\_2(\mathbb F\_\ell)$ to the kernel of a cyclic group of order dividing $\ell-1$.
Since the abelianization of $SL\_2(\mathbb F\_\ell)$ is trivial (for $\ell>3$) or has order relatively prime to $\ell-1$ (for $\ell=2,3$, since $2$ is relatively prime to $1$ and $3$ is relatively prime to $2$), $\overline{\rho}\_{f,l} (G\_{\mathbb Q(\zeta\_\ell)})$ contains $SL\_2(\mathbb F\_\ell)$.
Since $SL\_2(\mathbb F\_\ell$ acts absolutely irreducibly on the standard representation for all $\ell$, it follows that $\overline{\rho}\_{f,l} (G\_{\mathbb Q(\zeta\_\ell)})$ acts absolutely irreducibly on the standard representation, as desired.
It immediately follows that the image of $\operatorname{Sym}^n \overline{\rho}\_{f,\ell} \mid\_{ G\_{\mathbb Q(\zeta\_\ell)}}$ containsthe image of $SL\_2(\mathbb F\_\ell)$ under the $n$th symmetric power representation. Since the $n$th symmetric power representation of $SL\_2(\mathbb F\_\ell)$ is absolutely irreducible for $n \leq \ell-1$, the representation $\operatorname{Sym}^n \overline{\rho}\_{f,\ell} \mid\_{ G\_{\mathbb Q(\zeta\_\ell)}}$ is absolutely irreducible for $n\leq \ell-1$.
| 6 | https://mathoverflow.net/users/18060 | 417900 | 170,201 |
https://mathoverflow.net/questions/417380 | 27 | For positive integers $m$ and $n$, what is the integral of the function $(-1)^{\lfloor x \rfloor + \lfloor y \rfloor}$ on the triangle with vertices $(0,0)$, $(m,0)$, and $(0,n)$?
Pictorially, we are putting a red/black checkerboard coloring on the plane and finding the signed difference between the red region enclosed by the triangle and the black region enclosed by the triangle.
Call this integral $I(m,n)$. Here are some values of $I(m,n)$ (if my calculations of signed sums of areas of little triangles and trapezoids are correct):
$\begin{array}{c|ccccc} & 1 & 2 & 3 & 4 & 5 \\
\hline
1 & 1/2 & 1/2 & 1/2 & 1/2 & 1/2 \\
2 & 1/2 & 0 & -1/6 & 0 & 1/10 \\
3 & 1/2 & -1/6 & 1/2 & 5/6 & 1/2 \\
4 & 1/2 & 0 & 5/6 & 0 & -1/2 \\
5 & 1/2 & 1/10 & 1/2 & -1/2 & 1/2
\end{array}$
It would be good to tabulate more values. Can any computer algebra systems handle such computations for specific $m,n$?
Obviously $I(m,n) = I(n,m)$, and an easy telescoping-sum argument shows that $I(1,n)=1/2$. A symmetry argument shows that $I(m,n)=1/2$ when $m$ and $n$ are both odd, and from this it can be proved that $I(m,n)=0$ when $m$ and $n$ are both even. The values of $I(m,n)$ where $m$ and $n$ have opposite parity seem more subtle.
| https://mathoverflow.net/users/3621 | Area-differences for lattice triangles in a checkerboard | Define
$$ h(x)=(-1)^{\lfloor x\rfloor}\, (x-\lfloor x\rfloor)(x-\lfloor x\rfloor-1) $$
Then we have
$$I(n,m)= \frac{{\rm Mod}(n,2)}{2}+\frac{n}{m} \sum\_{j=1}^{n-1} (-1)^{n-j} \, h\left(\frac{jm}{n}\right)$$
Proof: We use the two primitives
\begin{eqnarray\*}
f(y)=\int\_0^y dx\, (-1)^{\lfloor x\rfloor} &=& {\rm Mod}(y,2)+(-1)^{\lfloor y\rfloor}\,(y-\lfloor y\rfloor)\\
g(z)=\int\_0^z dy \int\_0^y dx\, (-1)^{\lfloor x\rfloor}&=& \frac{z}{2}+(-1)^{\lfloor z\rfloor}\, \frac{(z-\lfloor z\rfloor)(z-\lfloor z\rfloor-1)}{2}
\end{eqnarray\*}
to rewrite
\begin{eqnarray\*}
I(n,m)&=& \int\_0^n dx (-1)^{\lfloor x\rfloor}\, f(m-x \frac{m}{n}) =\frac{n}{m} \int\_0^m dt (-1)^{\lfloor n-t \frac{n}{m}\rfloor}\, f(t)\\
&=&\frac{n}{m} \sum\_{j=1}^{n-1} (-1)^{n-j-1}\, \left\{g\left(\frac{(j+1)m}{n}\right)-g\left(\frac{j m}{n}\right)\right\}\\
&=&\frac{{\rm Mod}(n,2)}{2}+\frac{n}{m} \sum\_{j=1}^{n-1} (-1)^{n-j}\, h\left(\frac{jm}{n}\right)
\end{eqnarray\*}
where we first performed a change of variables $t=m-x m/n$, and then decomposed the integral over intervals $(j\frac{m}{n},(j+1)\frac{m}{n}]$,
over which $\lfloor n-t \frac{n}{m}\rfloor=n-j-1$.
Note that using the 2-periodicity of $h$ we get immediately
$$m \,I(n,m)=(m+2n)\,I(n,m+2n)-n \,{\rm Mod}(n,2).$$
Note also that $h$ is an odd function, so that:
$$m\, I(n,m)+(2n-m)I(n,2n-m)=n\, {\rm Mod}(n,2) .$$
Another consequence of $h(2-x)=-h(x)$ is:
$$I(k n,k(n-s))={\rm Mod}(k,2)\, I(n,n-s)$$
for odd $s$,
and the explicit formula:
$$I(k n,k(n-1))={\rm Mod}(k,2)\, I(n,n-1)=\frac{{\rm Mod}(k,2)}{2}\left( {\rm Mod}(n,2)+(-1)^n \frac{n+1}{3}\right) .$$
Proof: Define $c(n,m,k)=\sum\_{j=k+1}^{k+n-1} (-1)^{n-j}\,h(jm/n)$,
then by the above property of $h$, we have
$$c(n,m,n)+c(n,m,2n)=\sum\_{j=n+1}^{3n-1} (-1)^{n-j}\, h(jm/n)=0$$
by using the $j\to 4n-j$ symmetry, and $h(2m)=0$. We deduce that for $k$ odd:
$$ c(kn,k(n-s),0)=\sum\_{j=1}^{kn-1} (-1)^{n-j}h(j(n-s)/n)=\sum\_{j=1}^{n-1}(-1)^{n-j}h(j(n-s)/n)=c(n,n-s,0)$$
while it vanishes for $k$ even. The above formula follows from $I(n,m)={\rm Mod}(n,2)/2+(n/m)c(n,m,0)$.
Finally, for $s=1$, noting that $\lfloor j(n-1)/n\rfloor]=j-1$ for $1\leq j\leq n-1$, we have:
$$c(n,n-1,0)=\sum\_{j=1}^{n-1}(-1)^{n-j+j-1}\frac{-j(n-j)}{n^2}=(-1)^n\,\frac{n^2-1}{6 n}$$
and the above formula follows.
| 12 | https://mathoverflow.net/users/478524 | 417907 | 170,204 |
https://mathoverflow.net/questions/417894 | 1 | Say that you place an asymptotically Euclidean metric on $\mathbb{R}^3,$ e.g. $\mathbb{R}^3$ is endowed Riemannian metric $g$ such that $\text{supp}(g^{ij}-\delta^{ij})\subseteq\{|x|\leq R\}$ for some large $R>0.$ Does it follow that $(\mathbb{R}^3,g)$ is geodesically complete? This seems to be applied in PDE literature quite often when working with the flow generated by the principle symbol of the Laplace-Beltrami operator, but I am unaware of a reference. It seems intuitively true: $\mathbb{R}^3=\{|x|\leq R\}\cup \{|x|>R\}.$ The first set in the union is compact, and the other is where our metric is just the standard metric.
If this is a simple question, please let me know in the comments and I will delete.
| https://mathoverflow.net/users/371650 | Completeness of asymptotically Euclidean manifolds | Yes, and it follows from [Hopf-Rinow](https://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem) (let's assume $g$ is at least $C^2$ so that there's no ambiguity about the geodesic flow).
Since $g$ and $\delta$ differs only on a compact set, you have the the lengths defined by $g$ and $\delta$ are globally comparable, that is, there exists $\lambda, \Lambda > 0$ such that for every $x\in \mathbb{R}^3$ and $\xi\in T\_x\mathbb{R}^3$ you have
$$ \lambda \delta(\xi,\xi) \leq g(\xi,\xi) \leq \Lambda \delta(\xi,\xi) $$
and so the metric topologies are equivalent. Thus
A set $K$ is closed and bounded w.r.t. $g$
IFF
$K$ is closed and bounded w.r.t $\delta$
IFF
$K$ is compact.
And by Hopf-Rinow you have $g$ is complete.
| 1 | https://mathoverflow.net/users/3948 | 417911 | 170,205 |
https://mathoverflow.net/questions/417723 | 4 | A map $f: X \to X$ preserves an ergodic probability $\mu$, i.e., $\mu \circ f^{-1}=\mu$ and for any $\phi: X \to \mathbb{R}$ with $\int \phi d\mu=0$,
$$\frac{1}{n} \sum\_{i \le n} \phi \circ f^i \to 0 \text{ almost surely and in } L^1(\mu).$$
Therefore, $\max\_{n \ge N} \frac{1}{n} \sum\_{i \le n} \phi \circ f^i \to 0$ almost surely.
Similar to maximal inequality, are there references to quantitatively study
$$\max\_{n \ge N} \frac{1}{n} \sum\_{i \le n} \phi \circ f^i \text{ ?}$$
Similar to martingale inequality, are there references to study
$$\lVert\max\_{n \ge N} \frac{1}{n} \sum\_{i \le n} \phi \circ f^i\rVert\_{L^1} \precsim \lVert\frac{1}{N} \sum\_{i \le N} \phi \circ f^i\rVert\_{L^1} \text{ ?}$$
| https://mathoverflow.net/users/139946 | Maximal ergodic inequality | For a nice introductory discussion to the maximal ergodic inequality, see
[1]. In particular, inequality (5) there is Wiener's maximal ergodic theorem. See also Lemma 15.3 in [2]. A more advanced account, that in particular includes the $L^p$ maximal inequalities, is in the book [3], see Cor 2.2 page 8 for the Wiener inequality and the proof of Theorem 6.3 page 52 for the $L^p$ inequality
$$\lVert\max\_{n \le N} \frac{1}{n} \sum\_{i \le n} \phi \circ f^i\rVert\_{L^p} \le \frac{p}{p-1} \lVert \phi \rVert\_{L^p} \quad (\*) $$
valid for $p>1$. Note that Theorem 6.3 in [3] only states the limiting inequality when $n$ tends to infinity on both sides of $(\*)$, but the proof gives $(\*)$ for every finite $n$ as well, since it is a direct application of Cor. 2.2 and Lemma 6.2.
For $p=1$ the inequality (\*) does not hold, one needs an extra log term on the RHS, see (6.6)on page 52 in [3].
Finally, note that a general inequality of the form proposed,
$$\lVert\max\_{n \ge N} \frac{1}{n} \sum\_{i \le n} \phi \circ f^i\rVert\_{L^p} \precsim \lVert\frac{1}{N} \sum\_{i \le N} \phi \circ f^i\rVert\_{L^p} \quad (\*\*) $$
cannot be valid for any $p$. Consider, for example, $f$ to be a rotation of angle $2\pi(\frac{1}{N}-\epsilon)$ on the unit circle, where $\frac{1}{N}-\epsilon $ is irrational. Let $\phi$ be a function that equals 1 on an arc of length $\pi$ on that circle, and equals $-1$ on the complementary arc of the same length. Then for fixed $N>1$, the RHS of $(\*\*)$ tends to zero as $\epsilon \to 0$, while the LHS is at least $1/(N+1)$.
[1] [http://www-stat.wharton.upenn.edu/~steele/Courses/530/Resources/HopfSLLN.pdf](http://www-stat.wharton.upenn.edu/%7Esteele/Courses/530/Resources/HopfSLLN.pdf)
[2] [https://people.math.wisc.edu/~roch/teaching\_files/275b.1.12w/lect15-web.pdf](https://people.math.wisc.edu/%7Eroch/teaching_files/275b.1.12w/lect15-web.pdf)
[3] Krengel, Ulrich. Ergodic theorems. Vol. 6. Walter de Gruyter, 2011.
<https://books.google.ca/books?hl=en&lr=&id=t4_BDh8gt2kC&oi=fnd&pg=PA1&ots=5R-dupukqq&sig=yx6aMMlq93oOvD4eLotHFdk6q7U&redir_esc=y#v=onepage&q&f=false>
| 5 | https://mathoverflow.net/users/7691 | 417914 | 170,207 |
https://mathoverflow.net/questions/417855 | 1 | Let $a>1$ and define $G\_a(x)=\sum\limits\_{n=0}^{+\infty} \frac{\Gamma(\frac{2n+1}{a})}{\Gamma(2n+1)\Gamma(\frac{1}{a})}x^n$ where $\Gamma$ is the Gamma function. This series is convergent on $\mathbb{R}$ thanks to a ratio test and Stirling's formula.
For $a=2$, using Legendre duplication formula $\Gamma(z)\Gamma(z+\frac{1}{2}) = 2^{1-2z} \Gamma(\frac{1}{2})\Gamma(2z)$ with $z=n+\frac{1}{2}$ shows that $G\_2(x) = e^{\frac{x}{4}}$.
Can we get a somewhat explicit formula for other integer values of $a$, or even for arbitrary real values $a>1$? Has this series been studied somewhere?
For context, I am computing the Laplace transform of generalised Gaussian distributions [1], i.e their log-densities are $\propto -\lvert x\rvert^{a}$ (hence the case $a=2$ corresponds to the usual Gaussian measures). The above power series is related to the moment-generating function of such distributions.
I tried my luck a bit with hypergeometric functions and the multiplication theorem for the Gamma function (generalisation of the duplication formula) for small integer values of $a$ but did not quite make it.
References:
[1] <https://jsdajournal.springeropen.com/track/pdf/10.1186/s40488-018-0088-5.pdf>
| https://mathoverflow.net/users/294161 | Power series of ratio of Gamma functions | For arbitrary real $a > 0$ this is a special case of the generalized $\_p\Psi\_q(A;B;ζ)$ Fox-Wright function, where $A=[(a\_1,\alpha\_1),(a\_2,\alpha\_2),...,(a\_p,\alpha\_p)]$ and $B=[(b\_1,\beta\_1),(b\_2,\beta\_2),...,(b\_q,\beta\_q)]$ being $a\_j, j=1,..,p$ and $b\_k, k=1,..,q$ complex parameters and $\alpha\_j, \beta\_k$ are positive. $$\_p\Psi\_q(A;B;ζ)=\sum\_{n=0}^\infty \frac{ζ^n}{n!}\frac{\prod\_{j=1}^{p}\Gamma(a\_j+\alpha\_jn)}{\prod\_{k=1}^{q}\Gamma(b\_k+\beta\_kn)}$$ None gamma function in the numerator is singular. This means $$a\_{j}+α\_{j}m≠-ℓ,$$ with $j=1,2,..,p ∧ ℓ,m∈ℕ₀$. Series convergence depends on $\kappa, \rho, ϑ$ $$κ=∑\_{j=1}^{q}β\_{j}-∑\_{j=1}^{p}α\_{j}+1$$ $$ρ=∏\_{j=1}^{p}α\_{j}^{-α\_{j}}⋅∏\_{j=1}^{q}β\_{j}^{β\_{j}}$$ $$ϑ=½(q-p)+∑\_{j=1}^{p}a\_{j}-∑\_{j=1}^{q}b\_{j}$$ If $κ>0$ the series has an infinite radius of convergence and $\_p\Psi\_q(ζ)$ is an entire function. Series is uniformly and absolutely convergent for all finite $ζ$ . If $κ<0$ the sum is divergent for all nonzero values of $ζ$ whereas for $κ=0$ the function series has a finite radius of convergence $ρ$. Convergence on the boundary $|ζ|=ρ$ depends on parameter $ϑ$ converging absolutely if $ℜ(ϑ)<-½$.
For |arg$(-ζ)$$|<π-ε$, the Mellin-Barnes Integral $$\_{p}Ψ\_{q}(ζ)=\frac{1}{2πi}\int\_{L}\Gamma(s)⋅\frac{\prod\_{j=1}^{p}\Gamma(a\_{j}-α\_{j}s)}{\prod\_{k=1}^{q}\Gamma(b\_{k}-\beta\_{k}s)}(-ζ)^{-s}ds$$ defines a wider representation of Wright function. $L$ is a contour separating the poles of $\Gamma(s)$ to the left from those of $\Gamma(a\_{j}-α\_{j}s)$ to the right. For contour $L$ going from $-i\infty$ up to $+i\infty$ (possibly non-parallel to the vertical axis) this integral provides an analytical continuation of $\_{p}Ψ\_{q}(ζ)$ in $ζ∈ℂ\backslash [ρ,∞)$ when $ κ=0$.
This function is a special case of FoxH function, (See Wiki's or Wolfram's sites)
$$\_p\Psi\_q(A;B;ζ)=H\_{1+q,p}^{p,1}((1,1),B;A;-ζ^{-1})$$ For this particular case $A=[(1,1),(1/a,2/a)]$ and $B=[(1,2)]$. Thus $G\_a$ function is $$G\_a(x)=\frac{\_2\Psi\_1([(1,1),(1/a,2/a)];[(1,2)];x)}{\Gamma(1/a)}$$ $$G\_a(x)=\frac{H\_{2,2}^{2,1}([[(1,1)],[(1,2)]];[[(1,1),(1/a,2/a)],[\cdot]];-x^{-1})}{\Gamma(1/a)}$$ Note that $κ=2(1-1/a)$ and series converges for $a>1$ to an entire function. You can set this expression using FoxH function in Wolfram's Mathematica v13.0 in symbolic mode to see if there are some explicit formulae for other values of parameter $a$. I suggest try with $a\in \mathbb{Q}$ where $a>1$
| 1 | https://mathoverflow.net/users/141375 | 417935 | 170,213 |
https://mathoverflow.net/questions/417905 | 3 | I am new to the Albanese map so am not sure about its properties.
>
> **Question** Let $k=\mathbb{C}$. Suppose $X$ smooth projective variety, let $\alpha$ be the Albanese map of $X$. Is there a
> descripition of $X$ such that $\dim(X)=\dim(α(X))$?
>
>
>
Any remarks or references would be appreciated.
Cross-Posted on Stack-Exchange <https://math.stackexchange.com/questions/4400133/equidimensional-albenese-map>.
| https://mathoverflow.net/users/99732 | Maximality of Albanese dimension | I think that the varieties you are interested in are called *of maximal Albanese dimension* or *of Albanese general type*.
As remarked in the comments, their Kodaira dimension must be non-negative; however, since the behavior of the Albanese map is related to the $1$-forms and not to the top forms (canonical divisor), there is no clear relation between maximality of the Albanese dimension and maximality of the Kodaira dimension.
For instance, Abelian varieties are of maximal Albanese dimension but not of maximal Kodaira dimension. On the other hand, a surface of general type with irregularity $0 \leq q\leq 1$ is of maximal Kodaira dimension, but not of maximal Albanese dimension.
| 6 | https://mathoverflow.net/users/7460 | 417938 | 170,214 |
https://mathoverflow.net/questions/417936 | 5 | We know
$$
\sum\_{m=0}^\infty \frac{x^m}{(a-m)!m!} = \frac{1}{a!}(1+x)^m
$$
where we understand the factorial as Gamma function $\Gamma(x)$ such that it is divergent if the argument is negative integer.
We also know $$
\sum\_{m=0}^\infty \frac{x^m}{(b+m)!m!} \sim \,\_0F\_1(b,x)
$$
as hypergeometric function while this can be generalized to any $p,q$ for $\_pF\_q(\cdots,x)$.
Now I want to study
$$
f\_{abc}(x)=\sum\_{m=0}^\infty \frac{x^m}{m!(a-m)!(b+m)!(c-m)!}, \qquad a,b,c \in \mathbb{Z}^+
$$
Although the function is defined to sum over infinite numbers of integers, but it is effectively truncated at min$(a,c)$.
I want to ask whether I can find any reference studying the polynomial $f\_{abc}(x)$? Is there any special name for it?
I can imagine one possible manipulation is to use the reflection formula of Gamma function
$$
\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin \pi z}
$$
to change the sum over $(c-m)!$ to $(m-c)! \sin(\pi(m-c))$. And you might then call this function some kind of $\_2F\_1$ type hypergeometric function.
However, it is not that good.
First of all, the relection formula is better used if arguments $z$ are not integer, which is different from my purpose. Second, even after doing that, the function is not fully hypergeometric as there are extra sine functions as coefficients.
I was expecting whether any literature has ever studied these kinds of functions and name them?
| https://mathoverflow.net/users/477870 | Any name for this special function? | This is a standard hypergeometric function. Note that
$$ \frac{1}{(a-m)!} = (-1)^m \frac{(-a)\_m}{a!}\quad\text{and}\quad \frac{1}{(b+m)!} = \frac{1}{b!\,(b+1)\_m}$$
in terms of the rising Pochhammer symbol $(q)\_m = q(q+1)\cdots(q+m-1)$. Hence,
$$f\_{abc} = \frac{1}{a!b!c!} \sum\_{m=0}^\infty \frac{(-a)\_m(-c)\_m}{(b+1)\_m}\frac{x^m}{m!} = \frac{1}{a!b!c!}\,{\_2F\_1}(-a,-c;b+1;x).$$
| 17 | https://mathoverflow.net/users/47484 | 417940 | 170,215 |
https://mathoverflow.net/questions/417913 | 2 |
>
> Is it true that
> $$\sum\_{r=0}^p \sum\_{i=0}^r a\_{n,p,r,i}=0$$
> for all natural $n$ and all natural $p\ge2n$, where
> $$a\_{n,p,r,i}:=\frac{(-1)^r (n+p-r-1)! (n p-i (r-i))}{i!(r-i)! (n-i)!
> (p-r+i)! (n-r+i)! (p-i)!}?
> $$
>
>
>
This **is** true if $n\in\{1,\dots,10\}$ and $p\in\{2n,\dots,2n+10\}$.
(Here it is assumed that $\dfrac1{j!}=0$ for $j\in\{-1,-2,\dots\}$.)
| https://mathoverflow.net/users/36721 | Another combinatorial identity | Subst $k = p - 2n \ge 0$ and $s = r - i$ to get the symmetric
$$\sum\_{s \ge 0,i \ge 0} [s + i \le 2n + k] \frac{(-1)^{s+i} (3n+k-s-i-1)! (2n^2 + nk - is)}{i!(n-i)!(2n + k-i)! s!(n-s)!(2n + k-s)!}$$
But then we see from the $(n-i)!(n-s)!$ in the denominator that the bounds should actually be $0 \le i, s \le n$, and then we only get $x!$ for negative $x$ in the case $n = k = 0$, in the numerator:
$$\sum\_{s=0}^n \sum\_{i=0}^n \frac{(-1)^{s+i} (3n+k-s-i-1)! (2n^2 + nk - is)}{i!(n-i)!(2n + k-i)! s!(n-s)!(2n + k-s)!}$$
---
[Sage gives](https://sagecell.sagemath.org/?z=eJxtjsEKwjAMhu-DvUNuS9JWcZ539RUED5MyNgiVKE3x-S0KSsHj__35yJ88qAfzIDDBM2Yc0pcM1HcnNC9UO8RwoBnNCfEWl3LPEm94ZHUpWJBaMo6s8wgOlBMEEDYi2AP-zhtXQxOrXNXUQmsF-yfUL333yKIFP2t358t1WXORTZZY1jejF7EJRMQ=&lang=sage&interacts=eJyLjgUAARUAuQ==) Wilf-Zeilberger certificate for $s, i$ $$\frac{(2n^2 + kn - is + 3n + k - s - i)i}{(2n^2 + kn - is)(s + 1)}$$ so the inner sum is independent of $s$ and we have
$$n \sum\_{i=0}^n \frac{(-1)^{s+i} (3n+k-s-i-1)! (2n^2 + nk - is)}{i!(n-i)!(2n + k-i)! s!(n-s)!(2n + k-s)!}$$
for an arbitrarily chosen $s$. We could just argue that by taking $s$ outside the support we get zero; alternatively, take $s=n$ and rearrange to
$$\frac{(-1)^n n}{n!(n-1)!(n+k)!} \sum\_{i=0}^n (-1)^i \binom{n}{i}$$
The alternating binomial sum is well known to be zero.
| 4 | https://mathoverflow.net/users/46140 | 417948 | 170,217 |
https://mathoverflow.net/questions/417896 | 22 | Given a connected smooth manifold $M$ of dimension $m>1$, points $p\_1,\dots,p\_n\in M$ and positive values $\{d\_{i,j};1\leq i<j\leq n\}$ satisfying the strict triangle inequalities $d\_{i,j}<d\_{i,k}+d\_{k,j}$,
Can we give $M$ a complete riemannian metric $g$ so that $d\_g(p\_i,p\_j)=d\_{i,j}$, where $d$ is the geodesic distance?
This can fail in dimension $2$, as shown in the [answer](https://mathoverflow.net/a/417951) by André Henriques. I'm pretty sure it has to be true for $m\geq3$, but I have not been able to prove it.
Some comments:
* This occurred to me while answering [Equidistant points on a compact Riemannian manifold](https://mathoverflow.net/questions/417712/equidistant-points-on-a-compact-riemannian-manifold), [my answer](https://mathoverflow.net/questions/417712/equidistant-points-on-a-compact-riemannian-manifold/417722#417722) to that question contains the ideas I tried for $m\geq3$.
* By homogeneity of manifolds you can suppose the points $P\_1,\dotsc,P\_n$ are any set of $n$ points of $M$, and using that it is not hard to reduce the problem to the case of $M$ being diffeomorphic to $\mathbb{R}^m$. In particular if you prove it for $\mathbb{R}^3$ you will have proved it for any manifold of dimension $\geq 3$.
* One of the first ideas which come to mind is trying to somehow imbed $M$ in $\mathbb{R}^N$ for some big $N$, but [triangle inequalities are not sufficient](https://en.wikipedia.org/wiki/Distance_geometry#Characterization_via_Cayley%E2%80%93Menger_determinants) for a finite set to be isometrically imbedded in some $\mathbb{R}^N$.
* What if we change the strict triangle inequalities for the usual ones?
| https://mathoverflow.net/users/172802 | Can we make distances in a finite subset of a manifold whatever we want? | It is not possible to find $5$ points $x\_1,\ldots,x\_5$ on a genus zero Riemannian 2-manifold (a sphere) such that $d(x\_i,x\_j)=1$ for all $i,j$.
The reason is that the complete graph $K\_5$ is not planar.
Assume by contradiction that we have $5$ points $x\_1,\ldots,x\_5$ with $d(x\_i,x\_j)=1$.
Up to permuting the points, we may assume that the minimal geodesic connecting $x\_1$ and $x\_3$ crosses the minimal geodesic connecting $x\_2$ and $x\_4$.
Let $y$ be the point at which these two geodesics intersect.
Then
\begin{align\*}
2&=d(x\_1,x\_3)+d(x\_2,x\_4)\\
&=d(x\_1,y)+d(y,x\_3)+d(x\_2,y)+d(y,x\_4)\\
&=\tfrac12\big((d(x\_1,y)+d(y,x\_2))\\
&\qquad(d(x\_2,y)+d(y,x\_3))+\\
&\qquad(d(x\_3,y)+d(y,x\_4))+\\
&\qquad(d(x\_4,y)+d(y,x\_1))
\big)\\
&\ge\tfrac12\big(d(x\_1,x\_2)+d(x\_2,x\_3)+d(x\_3,x\_4)+d(x\_4,x\_1)\big)=2
\end{align\*}
with equality iff $y$ lies on all six geodesics (between $x\_i$ and $x\_j$ $\forall i,j\in\{1,2,3,4\})$. But if $y$ lies on all six geodesics, then these six geodesics are all part of a single geodesic line (i.e. the points are "aligned"), which is clearly impossible.
The crucial thing that I'm using here is the fact that geodesics admit unique extensions. If the ambient space was a graph, then my argument for deriving a contradiction wouldn't work as I wouldn't be able to conclude that the points are "aligned".
---
In higher dimensions, the answer is yes.
Take the complete graph $K\_n$ on your set of points. Embed it in $M$. Then put a metric on $M$ that agrees with your desired metric in a neighbourhood of the graph, and which is extremely huge away from the graph. Then minimal geodesics will essentially follow the graph.
This solves the problem "up to $\varepsilon$", as the geodesics don't exactly follow the graph, but do so only approximately.
To finish the argument, do the same thing in families, and invoke some version of the intermediate value theorem.
Here's how the argument goes. Let $D$ be the space of metrics on your fixed finite set.
Instead of doing the above construction for a single choice $d\in D$ of distances between the points $x\_i$, imagine that we adapt it to instead construct a *family* of Riemannian metrics on $M$ parametrised by the space $D$. Starting from $d\in D$, the geodesic distance between the $x\_i$ produces another element $d'\in D$. So we get a self-map $D\to D$ which is $\varepsilon$-away from the identity map on $D$. Now, $D$ is itself a manifold, and any self-map that's $\varepsilon$-away from the identity is surjective.
**[added later: the answer is no]**
Error in the above argument: $D$ is in fact a manifold with boundary. My argument works for metrics $d\in D\setminus \partial D$. I.e., metrics where the triangle inequality holds strictly.
A counterexample is provided by the metric on $\{x\_1,x\_2,x\_3,x\_4\}$ given by
$d(x\_1,x\_i)=1$, $d(x\_i,x\_j)=2$ (where $i,j\in\{2,3,4\}$)
**[added even later: all is good]**
Ha ha! I hadn't noticed that you had assumed the strict triangle inequality to hold. So all is good, and this is a valid argument.
| 28 | https://mathoverflow.net/users/5690 | 417951 | 170,218 |
https://mathoverflow.net/questions/417944 | 1 | Given a [pushdown automaton](https://en.wikipedia.org/wiki/Pushdown_automaton) (PDA), we seek a shortest word accepted by it. A standard approach is to map the problem in the corresponding context-free grammar. Can we analyze and solve this problem directly in the PDA?
| https://mathoverflow.net/users/52871 | Shortest word accepted by a PDA | I think you can just solve the "obvious inequalities" to get a polynomial time algorithm. I.e. assume acceptance by empty stack, and for each pair of states $p$, $q$ and a stack symbol $t$, let $T(p, q, t)$ be the minimal word length you need to move from state $p$ to $q$ while erasing $t$ from the stack, without dipping deeper into the stack than the symbol $t$.
There are some obvious inequalities:
$T(p, q, t) \leq c$ if there is a direct transition erasing $t$,
$T(p, q, t) \leq T(r, q', t') + T(q', q, t) + c$ if you have a transition of cost $c$ from state $p$ to state $r$ that writes $t'$ on stack, and
$T(p, q, t) \leq T(r, q, t) + c$ if you have a transition of cost $c$ from state $p$ to state $r$ that writes nothing on stack.
A maximal solution to these inequalities is easy to compute, start with infinite values for all $T(p, q, t)$ and propagate the information.
The number of steps is polynomial, and I think it's easy to see that the maximal solution $T'$ is the "actual truth" about $T(p, q, t)$, by induction on the length of the shortest number of transitions (possibly reading empty words) "realizing" the particular $T(p, q, t)$.
So you can now just take the minimum of $T(q\_0, q, t\_0)$ where $q\_0$ is the initial state and $t\_0$ the initial stack symbol.
| 1 | https://mathoverflow.net/users/123634 | 417960 | 170,219 |
https://mathoverflow.net/questions/417736 | 1 | Can the [Schreier coset](https://en.wikipedia.org/wiki/Schreier_coset_graph) graphs can be seen as a subgraph of Cayley graph on the same groups(neglecting the loop edges) and, hence, have their chromatic numbers bounded by the chromatic numbers of the Cayley graphs on those groups with the same generating set?
Also, how much does the cardinality of the subgroup determine the gap between the chromatic number of schreier coset graphs and the Cayley graphs on those same groups with the same generating set. Like, if the subgroup with respect to which the cosets are taken is large, then is the gap between the chromatic numbers also proportionally large? Thanks beforehand.
| https://mathoverflow.net/users/100231 | Difference in chromatic number between Schreier coset graphs and Cayley graphs | Here's a good example.
Let $G=S\_n$, $S=\{(i,j)\mid 1\leq i<j\leq n \}$. Then $\operatorname{Cay}(G,S)$ is a bipartite graph and so its chromatic number is $2$. Let $H=\{g\in S\_n\mid g(n)=n\}$. It is easy to check that $\operatorname{Sch}(G/H,S)$ is a complete graph (if we forget about loops) and hence its chromatic number is $n$.
| 2 | https://mathoverflow.net/users/173068 | 417963 | 170,220 |
https://mathoverflow.net/questions/417977 | 1 | $\DeclareMathOperator\Norm{Norm}$Suppose $E/\mathbb{Q}(j(E))$ is a CM elliptic curve and $d$ is a non-square. Let $E\_d$ denote the twist of $E$ by $\mathbb{Q}(j(E))(\sqrt{d})$. I know if $d$ is relatively prime to the conductor of $E$, then we have $$N\_{E\_d} = d^2N\_{E}$$
However, by computational investigations, there appear to be certain primes that always divide the norm of the conductor regardless of twist.
As a computational example: I will take a curve with CM by $K=\mathbb{Q}(\sqrt{-5})$. I take the curve with $j$-invariant
$$j(E)= 5887918080 (31261995198\sqrt{5} - 69903946375)$$
(i just picked the one with smallest absolute norm). Taking the absolute norm of the conductor of $E$ gives the factorization
$$\Norm(N\_E)=5^2 \* 7^4 \* 11^4 \* 19^2 \* 41^2 \* 47^4 \* 199^2 $$
Taking $d=77$, I get the twisted curve $E\_d$ which has conductor
$$\Norm(N\_{E\_d}) =5^2 \* 19^2 \* 41^2 \* 47^4 \* 199^2 $$
No matter what integer I twist by, I cannot get rid of these factors. Do I need to twist by something in $\mathbb{Q}(j(E)) =\mathbb{Q}(\sqrt{5})$?
Other examples seem to indicate that there are primes of bad reduction common to all twists. For example for CM elliptic curves over $\mathbb{Q}$ with CM by $\mathbb{Q}(\sqrt{-q})$, where $q$ is an odd prime. $q$ is a prime of bad reduction among all curves with the same $j$-invariant.
My question:
What is known about the primes of bad reduction common to all CM curves with the same $j$-invariant?
| https://mathoverflow.net/users/103423 | Primes of bad reduction for CM elliptic curves | These are exactly the primes where the Neron model of $E$ has fiber type, in [Kodaira's table](https://en.wikipedia.org/wiki/Elliptic_surface), anything other than $I\_0$ and $I\_0^\*$. Here $I\_0$ represents good reduction and $I\_0^\*$ represents a quadratic twist of good reduction by a ramified extension (quadratic twisting by an unramified extension preserves the reduction type).
(At least, this is true away from primes above 2, which are surely more complicated).
The fiber type can be computed by Tate's algorithm, but to tell whether the type is $I\_0$ or $I\_0^\*$, you can simply perform a ramified quadratic twist and check whether either the curve or its quadratic twist has good reduction. The number you quadratic twist by doesn't matter, as long as it is $\pi$-adic valuation, where $\pi$ is the prime of bad reduction, is odd.
For $\pi$ in $\mathbb Q(\sqrt{-5})$ lying over a prime $p$ of $\mathbb Q$, unless $\pi$ is ramified (i.e. $\pi$ lies above $5$), then any number which has $p$-adic valuation odd will do the trick.
| 4 | https://mathoverflow.net/users/18060 | 417978 | 170,225 |
https://mathoverflow.net/questions/417964 | 4 | Let $\mathcal{O}(n)$ and $\mathcal{D}(n)$ denote the set of all integer partitions of $n$ into odd parts and distinct parts, respectively. Let $o(n)=\#\mathcal{O}(n)$ and $d(n)=\#\mathcal{D}(n)$. Euler established that $o(n)=d(n)$.
Introduce the enumerations: $a(n)=$ number of parts in $\mathcal{O}(n)$ and $b(n)=$ number of parts in $\mathcal{D}(n)$. Then, George Beck conjectured that
$a(n)-b(n)=\alpha(n)$
where $\alpha(n)$ counts all partitions of $n$ with odd parts, except one even part (which may be repeated as many times). Soon enough, George Andrews proved this claim and added that $a(n)-b(n)=\beta(n)$ where $\beta(n)$ counts all partitions of $n$ with distinct parts, except one part repeated at least twice. I learned of these from Cristina Ballantine, who gave a seminar this week offering combinatorial proofs as well as further generalizations.
This prompted me to consider a further refinement. So, consider $\alpha\_u(n)$ and $\beta\_u(n)$ as enumerating all partitions of $n$ having odd parts with **one even part appearing $u$-times** and all partitions of $n$ having distinct parts with **the number $u$ being the repeating part**, respectively.
I would now ask:
>
> **QUESTION 1.** Is it true that $\alpha\_u(n)=\beta\_u(n)$ for $u=1,2,3\dots$? **Note.** $u$ is bounded by $n$.
>
>
>
>
> **QUESTION 2.** Is this true too? We have the bivariate generating function
> $$\sum\_{n,u=0}^{\infty}\alpha\_u(n)z^uq^n=\sum\_{j=1}^{\infty}\frac{zq^{2j}}{1-zq^{2j}}\cdot\prod\_{k=1}^{\infty}\frac1{1-q^{2k-1}}
> =\sum\_{n,u=0}^{\infty}\beta\_u(n)z^uq^n.$$
>
>
>
| https://mathoverflow.net/users/66131 | A refinment of Beck's conjecture | Given a partition $\lambda$ with all parts odd except for one even
part $2k$ appearing $u$ times, let $\mu$ be $\lambda$ with the parts
equal to $u$ removed. Apply your favorite bijection to transform $\mu$
into a partition $\nu$ with distinct parts. Then adjoin to $\nu$ $2k$
parts equal to $u$, obtaining a partition $\rho$ with distinct parts
except for the repeating part $u$. (Note that we may have $2k$ or $2k+1$ parts equal to $u$ in $\rho$, depending on whether $\nu$ already had a part equal to $u$.) The map $\lambda\mapsto \rho$ is a bijection, proving Question 1. It is straightforward to verify Question 2.
| 3 | https://mathoverflow.net/users/2807 | 417996 | 170,231 |
https://mathoverflow.net/questions/417888 | 4 | I have the following expression:
$$
\sum\_{k=0}^{\infty}\frac{n!}{(k+n)!}x^k(L\_n^k(x))^2,
$$
where
$$
L\_n^k(x)=\sum\_{j=0}^n(-1)^j\binom{n+k}{n-j}\frac{x^j}{j!}
$$
is the usual associated Laguerre polynomial and $k\in\mathbb N$. In particular $\int\_0^{\infty}e^{-x}x^k(L^k\_n(x))^2=\frac{(k+n)!}{n!}$.
I am trying to figure out a way to simplify this sum. I am not an expert on special functions, and I would appreciate some references or hints. I have been trying quite hard using a few recurrence relations and other formulas found in the literature.
What I would like to prove is the following:
$$
\sum\_{k=0}^{\infty}\frac{n!}{(k+n)!}x^k(L\_n^k(x))^2=e^x+P\_{2n-1}(x),
$$
where $P\_{2n-1}(x)$ is a polynomial of degree $2n-1$ (if $n=0$, we set $P\_{-1}=0)$. This is clearly true when $n=0$, and one can easily prove for small values of $n$ ($n=1,2,...$). An alternative way is to differentiate $2n$ times and prove that the resulting sum gives $e^x$, but this approach remains complicated (at least, for me).
I don't know if this is something known, I would appreciate a reference in that case.
| https://mathoverflow.net/users/205771 | Infinite sum of Laguerre polynomials: is $\sum_{k=0}^{\infty}\frac{n!}{(k+n)!}x^kL_n^k(x)^2=e^x+P(x)$, with $P$ a polynomial of degree $2n-1$? | Everything becomes simpler if add some parameters and start the sum at $k=-n$ instead of $k=0$. Note that if $k$ is a negative integer with $-n\le k \le -1$ then $L\_n^k(x)$ is a polynomial in $x$ divisible by $x^{-k}$.
Then we have the formula
$$\sum\_{k=-n}^\infty \frac{n!}{(k+n)!}z^k L\_n^k(x)L\_n^k(y)=e^z\sum\_{i=0}^n\frac{1}{i!}\binom{n}{i}\left(\frac{(x-z)(y-z)}{z}\right)^i,$$
which can be proved by expanding the right side and simplifying, using Vandermonde's theorem.
Thus
\begin{multline\*}
\sum\_{k=0}^\infty \frac{n!}{(k+n)!}z^k L\_n^k(x)L\_n^k(y)\\
= e^z\sum\_{i=0}^n\frac{1}{i!}\binom{n}{i}\left(\frac{(x-z)(y-z)}{z}\right)^i - \sum\_{k=-n}^{-1} \frac{n!}{(k+n)!}z^k L\_n^k(x)L\_n^k(y).
\end{multline\*}
If we set $z=x$ the terms in the first sum on the right with $i>0$ vanish and we get
$$
\sum\_{k=0}^\infty \frac{n!}{(k+n)!}x^k L\_n^k(x)L\_n^k(y)
=e^x - \sum\_{k=-n}^{-1} \frac{n!}{(k+n)!}x^k L\_n^k(x)L\_n^k(y),
$$
so this gives an explicit formula for the OP's polynomials.
| 8 | https://mathoverflow.net/users/10744 | 417997 | 170,232 |
https://mathoverflow.net/questions/417994 | 2 | Edge coloring of a graph is an assignment of “colors” to the edges of the graph so that no two adjacent edges have the same color with an optimal number of colors. Two edges are said to be adjacent if they are connected to the same vertex.
There is no known polynomial time algorithm for edge-coloring every graph with an optimal number of colors. Nevertheless, is there any efficient algorithm for edge-coloring every complete graph with an optimal number of colors? I bielieve this exits but I cannot find any source for this.
| https://mathoverflow.net/users/148974 | Efficient algorithm for edge-coloring complete graphs | Yes, for all $n$, the edge-chromatic number of $K\_{2n}$ is $2n-1$ and the edge-chromatic number of $K\_{2n+1}$ is $2n+1$. Moreover, it is easy to construct such edge-colourings in polynomial time. For example, see this [Wikipedia](https://en.wikipedia.org/wiki/Graph_factorization) page for an easy method to construct a "$1$-factorization" of $K\_{2n}$. An optimal edge-colouring for $K\_{2n+1}$ can be obtained by applying this method to $K\_{2n+2}$, and then deleting the edges incident to the extra vertex.
| 3 | https://mathoverflow.net/users/2233 | 417998 | 170,233 |
https://mathoverflow.net/questions/417954 | 6 | $\DeclareMathOperator\cd{cd}$Are there any known examples of non-free groups with a property that $\cd(G)+1 = \cd(G \times G)$, or, less restrictive, $G, H$ with $\cd \neq 1, \infty$ such that $\cd(H)+1 = \cd(G \times H)$?
| https://mathoverflow.net/users/81055 | Groups with unusual cohomological dimension of direct product | Let $G=(\mathbb{Q},+)$. Then ${\rm cd}(G)=2$ and ${\rm cd}(G\times G)=3$.
| 6 | https://mathoverflow.net/users/124004 | 418015 | 170,236 |
https://mathoverflow.net/questions/418010 | 0 | Let $P = (V, \sqsubseteq)$ be a partial order and $\mathfrak{D}(P)$ denote the class of downward-closed subsets of the partial order $P$ (i.e, the class of $A \subseteq V$ such that $y\in A \;\&\; x \sqsubseteq y$ implies $x \in A$).
Let a partial order be called *downward complete*, if every non-empty subset has an infimum, *upward complete* if every non-empty subset has an supremum, and *complete* if both downward and upward complete.
In [La Matematica della Verita'](https://books.google.co.uk/books/about/La_matematica_della_verit%C3%A0_Strumenti_ma.html?id=2t2rAAAACAAJ&redir_esc=y) (2006): page 177 , theorem 6.7.7 states that the following three conditions are equivalent:
1. $(V, \sqsubseteq)$ is complete.
2. $(V, \sqsubseteq)$ is downward complete and $\exists z \forall x(x \sqsubseteq z)$
3. $(V, \sqsubseteq)$ is upward complete and $\exists z \forall x(z \sqsubseteq x)$
It is then stated that, conditions 4. and 5. (below), together with the equivalence of 1.-3. just stated, imply that $(\mathfrak{D}(P), \subseteq)$ is a complete partial order:
4. $V \in \mathfrak{D}(P)$ and
5. $\bigcap \mathcal{B} \in \mathfrak{D}(P)$, for any non-empty subfamily $\mathcal{B}$ of $\mathfrak{D}(P)$
I'm wondering if there is a mistake. If not, it is not clear to me how 3. and 4. plus the equivalence of 1.-3. ensure $(\mathfrak{D}(P), \subseteq)$ is a complete partial order.
| https://mathoverflow.net/users/122435 | Partial orders on downward closed sets | Conditions 4 and 5 show that $(\mathfrak{D}(P),{\subseteq})$ satisfies condition 2 in the list: because $V\in\mathfrak{D}(P)$ we have the second part of 2; and 5 says that $(\mathfrak{D}(P),{\subseteq})$ is downward complete, because $\bigcap\mathcal{B}=\inf\mathcal{B}$.
| 1 | https://mathoverflow.net/users/5903 | 418016 | 170,237 |
https://mathoverflow.net/questions/417967 | 7 | Assume that $\omega<\kappa\_1<\dotsb< \kappa\_n$ are infinite cardinals such that for each $1\le i\le n$ there is a $\kappa\_i$-complete, $\kappa\_i^+$-saturated ideal $\mathcal I\_i\subset \mathcal P(\kappa\_i)$. Can you obtain a ZFC model which contains $n$-many measurable cardinals? The natural candidate is $L[\mathcal I\_1,\dotsc, \mathcal I\_n]$.
It is well known that the answer is yes for $n=1$ (see Kunen: [Some applications of iterated ultrapowers in set theory](https://doi.org/10.1016/0003-4843(70)90013-6). Ann. Math. Logic 1 (1970), 179–227.)
| https://mathoverflow.net/users/71011 | Measurable cardinals from saturated ideals | Yes. And if the $\mathscr{I}\_k$s are normal, then the suggested candidate works (I haven't really thought about whether the candidate still works without normality). Here is the argument, which is a typical one:
First, we may assume that each $\mathscr{I}\_k$ is normal, by Jech
Lemma 22.28.
Case 1. There is an proper class transitive inner model of ZFC with $n+1$ measurable cardinals (not just $n$).
Let $M$ be the minimal proper class transitive inner model satisfying ZFC with $n$ (not $n+1$) measurable cardinals $\mu\_1<\ldots<\mu\_n$,
as witnessed by (unique) normal measures $U\_1,\ldots,U\_n$.
Then (this uses the case hypothesis) $\mu\_1,\ldots,\mu\_n$ are countable. So we can iterate $M$ out at its measurables (using the $U\_k$s and their images), eventually sending each $\mu\_k$ to $\kappa\_k$. Let $\mathcal{T}$ be this iteration (consisting of the sequence of iterates $M^{\mathcal{T}}\_\alpha$, iteration maps $i\_{\alpha\beta}^{\mathcal{T}}$, etc), and $M'=M^{\mathcal{T}}\_{\kappa\_n}$ be the final iterate and $i:M\to M'$ the iteration map. Then I claim that $i(U\_k)\subseteq\mathscr{F}\_k$, the filter dual to $\mathscr{I}\_k$. For let $j:V\to M$ be a generic embedding given by $\mathscr{I}\_k$.
So $j(\mathcal{T})$ is an iteration with last model $j(M')$, and $j(\mathcal{T})\upharpoonright(\kappa\_k+1)=\mathcal{T}\upharpoonright(\kappa\_k+1)$, but the $\kappa\_k$th
measure used in $j(\mathcal{T})$ is $i(U\_k)$, and $j(\mathcal{T})$ eventually sends $\kappa\_k$ further out to $j(\kappa\_k)$. Now standard calculations show that the iteration map $i^{\mathcal{j(\mathcal{T})}}\_{\kappa\_kj(\kappa\_k)}$ agrees with $j$ over $\mathcal{P}(\kappa\_k)\cap M^{\mathcal{T}}\_{\kappa\_k}$,
which implies that $j(U\_k)\subseteq G$, the generic filter, and since this is independent of $G$, therefore $j(U\_k)\subseteq\mathscr{F}\_k$, as desired.
Since $i(U\_k)$ is an ultrafilter in $M'$ and $i(U\_k)\subseteq\mathscr{F}\_k$,
it follows that $$L[\mathscr{F}\_1,\ldots,\mathscr{F}\_n]=L[i(U\_1),\ldots,i(U\_n)]\subseteq M',$$
and that $i(U\_k)=\mathscr{F}\_k\cap L[\mathscr{F}\_1,\ldots,\mathscr{F}\_n]$;
let $\bar{\mathscr{F}}\_k$ denote this measure.
Therefore $L[\mathscr{F}\_1,\ldots,\mathscr{F}\_k]\models$"$\bar{\mathscr{F}\_k}$ is a $\kappa\_k$-complete normal measure on $\kappa\_k$". Of course $L[\mathscr{I}\_1,\ldots,\mathscr{I}\_n]=L[\mathscr{F}\_1,\ldots,\mathscr{F}\_n]$, so we are done. (Using the minimality of $M$, can also show that $M'=L[\mathscr{F}\_1,\ldots,\mathscr{F}\_n]$.)
Case 2: Otherwise.
Then the core model $K$ exists. Let $j:V\to M$ be a generic embedding given by $\mathscr{I}\_k$. Then (using core model theory for this level) $j(K)$ is an iterate of $K$ and $j\upharpoonright K$ is the iteration map. But $\mathrm{crit}(j)=\kappa\_k$. Therefore $\kappa\_k$ is measurable in $K$ (as witnessed by its extender sequence), and because $j\upharpoonright K$ is the iteration map, letting $D\_k$ be the unique normal measure on $\kappa\_k$ in $K$ (uniqueness because otherwise we get a measure of Mitchell order 1), we have $D\_k\subseteq\mathscr{F}\_k$. It follows that $L[\mathscr{F}\_1,\ldots,\mathscr{F}\_n]=L[D\_1,\ldots,D\_n]\subseteq K$, and that letting $\bar{\mathscr{F}}\_k=\mathscr{F}\_k\cap L[\mathscr{F}\_1,\ldots,\mathscr{F}\_n]$, then $\bar{\mathscr{F}}\_k$ is a normal, $\kappa\_k$-complete measure on $\kappa\_k$ in $L[\mathscr{F}\_1,\ldots,\mathscr{F}\_n]$,
as desired. (With the case hypotheses as they are, it might be that $L[\mathscr{F}\_1,\ldots,\mathscr{F}\_n]\subsetneq K$.
I could have made the case hypothesis to case 1 be that the sharp for an inner model with $n$ measurables exists, and case 2 its negation. Then
we would get that $K$ and $L[\mathscr{F}\_1,\ldots,\mathscr{F}\_n]$ have the same universe.)
| 4 | https://mathoverflow.net/users/160347 | 418025 | 170,240 |
https://mathoverflow.net/questions/416781 | 1 | Consider an $M/M/1$ queue with the arrival rate $\lambda>0$ and the service rate $\mu>\lambda$ (so that it is stable), in the stationary regime. Let $A\_t$ be the number of arrivals in the time interval $[0,t]$ and $D\_t$ be the number of departures in that time interval; then both $A\_t$ and $D\_t$ have Poisson distribution with mean $\lambda t$ (the second one due to Burke's theorem), but they are of course not independent.
Question: what is $\mathop{\mathrm{Cov}}(A\_t,D\_t)$? (Probably, this is known, but I couldn't find the reference$\dots$)
| https://mathoverflow.net/users/81488 | The input and output processes in a single-server queue | Let $\eta\_t$ be the number of customers in the system at time $t$ and $\rho=\lambda/\mu<1$ be the load. It holds that $\eta\_0+A\_t-D\_t = \eta\_t$, so $A\_t-D\_t = \eta\_t-\eta\_0$. Write
$$
A\_t D\_t = \frac{1}{2}(A\_t^2 + D\_t^2 - (A\_t-D\_t)^2),
$$
from which we obtain (recall that both $A\_t$ and $D\_t$ are Poisson($\lambda t$))
$$
\mathop{\mathrm{Cov}}(A\_t,D\_t) = \lambda t - \frac{1}{2}\mathbb{E}(\eta\_t-\eta\_0)^2,
$$
thus expressing $\mathop{\mathrm{Cov}}(A\_t,D\_t)$ in terms of $\mathop{\mathrm{Cov}}(\eta\_0,\eta\_t)$.
Note that the second term in the above expression is bounded, and moreover converges to $\mathop{\mathrm{Var}} \eta\_0 = \frac{\rho}{(1-\rho)^2}$ as $t\to\infty$ due to asymptotic independence of $\eta\_0$ and $\eta\_t$.
For a fixed $t$, the explicit expression for $\mathop{\mathrm{Cov}}(\eta\_0,\eta\_t)$ can be found in e.g. Section 2 of the following paper:
Reynolds, J. F. (1975). The covariance structure of queues and related processes – a survey of recent work. Advances in Applied Probability, 7(02), 383–415, DOI:10.2307/1426082
| 0 | https://mathoverflow.net/users/81488 | 418026 | 170,241 |
https://mathoverflow.net/questions/417048 | 1 | This is, in a way, a follow up question to [Unipotent orbits and intersection with Levi and pseudo-Levi subgroups](https://mathoverflow.net/questions/415890/unipotent-orbits-and-intersection-with-levi-and-pseudo-levi-subgroups#415890).
I was reading "[A generalisation of the Bala–Carter theorem for nilpotent orbits](https://doi.org/10.1155/S107379289800035X)" by Eric Sommers and was considering the orbit $D\_4(a\_1)$ in $E\_6^\text{ad}$ and there is an issue that I wasn't able to resolve regarding the semi-simple part of the centralizer of this orbit.
According to Sommers, this orbit intersects the principal orbit of the pseudo-Levi subgroup $A\_2^3$ and the pair $(A\_2^3,3A\_2)$ (of pseudo-Levi subgroup and a nilpotent elment of $\mathfrak{e}\_6$ that is distinguished in $\mathfrak{l}=\operatorname{Lie}(A\_2^3$) corresponds (in the generalized Bala–Carter correspondence described there) to $(N,C)$, where $C$ is the conjugacy class in the component group $A(N)=S\_3$ consisting of elements of order $3$: $C=\{(123), (132)\}$.
Now, if I am not mistaken, this pseudo-Levi $L$ of type $A\_2^3$ can be realized in $E\_6^\text{ad}$ as $\left( \operatorname{SL}\_3(\mathbb{C}) \right)^3 / \mu\_3^2$ and its center is $\mu\_3$.
Being the principal orbit of $L$, the centralizer of the orbit in $L$ is the center of $L$ (generated by the elements of $C$). The subgroup $L$ has rank $6$ and so its centralizer should have rank $0$.
On the other side, in the table in Carter's book, it says that the semi-simple part of the centralizer of this orbit has type $T\_2$ with component group $S\_3$.
My question is (assuming my analysis of Sommers' map is correct), how do these two descriptions of the centralizer work together? Where am I missing the rank $2$ torus in the pseudo-Levi description of the orbit?
| https://mathoverflow.net/users/64702 | The identity connected component of centralizers of unipotent orbits | If you decompose the $E\_6$ Lie algebra over the $3A\_2$ subgroup $L$ then, in addition to the adjoint representation, there are two irreducible summands: with the right identifications these are isomorphic to $V^{\otimes 3}$ and its dual ($V$ the natural representation for ${\rm SL} \_3$). So, to describe the centraliser of $N$ this way you have to also consider the (${\rm ad} \, N$) -fixed points in these two subspaces. By inspecting weights, we see that $V^{\otimes 3}$, decomposed over the (diagonally embedded) principal ${\mathfrak sl}\_2$, equals
$$V(6)\oplus V(4)^2\oplus V(2)^3\oplus V(0).$$
In particular, the centraliser of $N$ in each subspace is 7-dimensional, and the centraliser for the principal ${\mathfrak{sl}}\_2$ is one-dimensional. This gives that the reductive part of the centraliser in the whole Lie algebra is 2-dimensional, so must be the Lie algebra of a torus.
| 1 | https://mathoverflow.net/users/26635 | 418030 | 170,243 |
https://mathoverflow.net/questions/418019 | 8 | For each coherent category $C$, let $J\_C$ be the topology on $C$ in which a sieve $\{f\_i\colon U\_i\to X\}\_{i\in I}$ is covering if and only if there exists a *finite* set $I\_0\subseteq I$ such that $\bigcup\_{i\in I\_0} \operatorname{im}(f\_i)=X$ as subobjects of $X$. (This is a Grothendieck topology by Proposition 12 of [Lecture 8 (Grothendieck topologies)](https://www.math.ias.edu/~lurie/278xnotes/Lecture8-Topologies.pdf) of [Lurie's "Categorical logic" notes](https://www.math.ias.edu/%7Elurie/278x.html).)
1. Is $J\_C$ equivalent to what other people call "[the coherent topology](https://ncatlab.org/nlab/show/coherent+coverage) on $C$"?
2. If $C$ is a coherent category which is Boolean, is the topos $\operatorname{Sh}(C, J\_C)$ Boolean too?
3. Can one find for each Boolean coherent topos $\mathcal E$ a Boolean coherent category $C$ such that both
* $\mathcal E\simeq \operatorname{Sh}(C, J\_C)$ and
* there exists an object $X\in C$ such that every object of $C$ is a subobject of $X^n$?
4. Is the étale topos of the spectrum of any field coherent?
| https://mathoverflow.net/users/478652 | Questions about coherent topology | **Edit :** I should clarify that I've interpreted "Etale topos" to mean the petit/small étale topos everywhere. What I've said about Grothendieck-Galois duality only apply to the petit étale topos. If you are talking about the Gros topos, then these part no longer holds. I actually don't know if the Gros étale topos of a fields has a boolean category of coherent object or not.
1. Yes. I can't give you a proof because as far as I'm concerned this is the definition of the coherent topology. If you see a different definition, maybe edit your question!
2. Essentially no. Take for example a Boolean algebra $B$. It can be seen as a coherent category (I see $B$ as a poset, and every poset as a category in the usual way). Then the associated topos is the topos of sheaves over the Stone spectrum of $B$, and unless $B$ is finite it has plenty of open that are not also closed (in fact the open that are complemented corresponds exactly to the element of $B$). The general case looks like this though: a Boolean coherent category will gives a topos that "looks like" a Stone spectrum.
3. The answer is yes for the first half, no for the second, but only because there are very few coherent boolean topos. I would say, a coherent topos is essentially never Boolean (the only exception being the framework of Galois theory): A coherent topos has always enough points, and it can be proved that a boolean topos with enough points is "atomic", that is a disjoint union of topos of the form $BG\_i$ where the $G\_i$ are localic group. (Here $BG$ is the topos of sets endowed with a continuous action of the localic group $G$.) Adding back the fact that we want this topos to be coherent, we get that the Boolean coherent toposes are exactly the toposes that are finite coproducts of $BG\_i$ where the $G\_i$ are profinite groups.
The second condition you ask for doesn't hold if some of the $G\_i$ are non-discrete though. If $G$ is a profinite group (take $G = \mathbb{Z}\_p$ for example), then a coherent $G$-set $X$ is a finite $G$-set, so there is going to be an open normal subgroup of $G$ that stabilise all the points of $X$, and the things you get as subobjects of $X^n$ will all be stabilised by the same subgroup.
4. Yes. The étale topos of any affine scheme is coherent. (For a general scheme it is locally coherent. I'll let an algebraic geometer give you the precise condition under which we get coherence.) In fact, thanks to Grothendieck Galois duality, the étale topos of a field is one of the rare examples of Boolean coherent toposes: it is $BG$ where $G$ is the absolute Galois group of the field, with its profinite topology.
Answering some of the follow up question in the comments.
1. I would recomand to double checking what I'm going to say here if you plan on using it - I haven't looked at it in enough to details, but I think toposes associated to boolean coherent categories can be characterized as the coherent toposes in which coherent subobject of coherent object have complement. A topos satisfying these condition is clearly the topos of coherent sheaf on a booleancoherent category (by taking all coherent objects) but the converse also seems true. The other condition you have (every object of $C$ is a subobject a power of some fixed object $X$) should corresponds to have a coherent "pre-bound".
2. The class descriebd you in (1) (including both conditions) are the topos that classyfies **single** sorted "boolean" first order theory. This additional condition of having a coherent prebound is not automatic at all as the exemple of the topos $BG$ for $G$ a non discrete profinite group shows.
3. Yes. étale topos of fields are Boolean topos (from their explicit descrition given by Galois duality) so in particular coherent subobject have complements.
| 10 | https://mathoverflow.net/users/22131 | 418044 | 170,247 |
https://mathoverflow.net/questions/417928 | 7 | This is a question that a classmate asked me three years ago.
Let $P(x)=\sum\_{i=0}^n a\_ix^i$ be a polynomial such that each $a\_i>0$. Prove or disprove that there exists a positive integer $r$ such that $P(x)^r=\sum\_{i=0}^{nr} b\_ix^i$ and there exists $0\le j\le nr$ such that $b\_0\le b\_1\le \dots\le b\_{j}$ and $b\_{j+1}\ge\dots\ge b\_{nr}$.
This problem may have a probability problem as its original problem, since the classmate asked this while taking a probability class. What I could do is to try to apply CLT to claim that the "central terms" has only one peak and try to select some $r$ to make the first few terms and last few terms increasing/decreasing (that is, I proved that I can choose an $r$ to make $b\_0\le b\_1\le\dots \le b\_k$ for any $k$, and same at the other side). But I failed to prove the problem... I also tried to factorize it into smaller quadratic polynomials but also failed... Also, I can't come up with a counterexample, too...
Note: if polynomials $p,q$ are single peak, this DOES NOT imply that $pq$ is single peak. For example: $p(x)=1+x+100x^2$ and $q(x)=10000+10000x+10100x^2+9000x^3+9000x^4$. But $p(x)q(x)=900000x^6+909000x^5+1028000x^4+1019100x^3+1020100x^2+20000x+10000$
P.S. This is exactly same as [the problem in MSE](https://math.stackexchange.com/questions/4399939/prove-or-disprove-that-the-power-of-positive-term-polynomial-will-be-eventually). As @ChrisSanders in the comment pointed out, I just ask the question here...
| https://mathoverflow.net/users/170895 | Prove or disprove that the power of positive term polynomial will be eventually single peak | This is answered affirmatively by Odlyzko and Richmond, *On the unimodality of high convolutions of discrete distributions*, Annals of probability (1985) 299--306: all sufficiently large powers of the polynomial (with positive coefficients and no gaps) are strongly unimodal, that is, the coefficients form a log concave sequence. The proof uses estimates of contour integrals over circles of just the right radius.
| 8 | https://mathoverflow.net/users/42278 | 418049 | 170,250 |
https://mathoverflow.net/questions/418028 | 1 | A theorem by Lebesgue, Hausdorff and Banach says the following (Kechris' *Classical Descriptive Set Theory*, p. 192):
>
> Let $X$ be a separable metrizable space and $f: X \rightarrow \mathbb{R}$ be a $\boldsymbol{\Sigma}\_2^0$-measurable function, then $f$ is the pointwise limit of a sequence of continuous functions.
>
>
>
The theorem can be extended to functions with codomain $\mathbb{R}^n$ for each $n\in\mathbb{N}$. My questions are:
1. Does this theorem hold for functions with codomain $\mathbb{R}^\omega$?
2. Is there a counterexample of a $\boldsymbol{\Sigma}\_2^0$-measurable function $f:\mathbb{R}^\omega\rightarrow\mathbb{R}^\omega$ which is *not* pointwise limit of a sequence of continuous functions?
Thanks!
| https://mathoverflow.net/users/141146 | Lebesgue Hausdorff Banach theorem for Baire class $1$ functions on $\mathbb{R}^\omega$ | Is it indeed the case that the statement can be extended to functions with codomain $\mathbb{R}^\omega$.
Suppose we have $f:\mathbb{R}^\omega\rightarrow\mathbb{R}^\omega$ which is $\boldsymbol{\Sigma}\_2^0$-measurable, then in particular the functions $f^n:\mathbb{R}^\omega\rightarrow\mathbb{R}$ with $f^n(x)=f(x)(n)$ are $\boldsymbol{\Sigma}\_2^0$-measurable.
Applying Lebesgue, Hausdorff and Banach theorem we have that for each $n$ there exists a sequence $(g\_m^n)\_{m\in\omega}$ of functions from $\mathbb{R^\omega}$ into $\mathbb{R}$ such that $g\_m^n \longrightarrow g^n$ pointwise as $m\rightarrow \infty$.
Now if we consider the sequence $(g\_m)\_{m\in\omega}$ of functions functions from $\mathbb{R^\omega}$ into $\mathbb{R}^\omega$ with $g\_m(x) = (g\_m^n(x))\_{n\in\omega}$, then we have that they are all continuous and they converge pointwise to $f$.
| 0 | https://mathoverflow.net/users/141146 | 418076 | 170,256 |
https://mathoverflow.net/questions/259663 | 6 | Is there a positive 128-bit integer whose square has all middle bits equal to 1?
(The "middle bits" are naturally the 65th bit through the 192nd bit, defining
the 1st bit as the least significant bit of the full integer.)
| https://mathoverflow.net/users/20757 | Mid-Square with all bits set |
>
> Is there a positive 128-bit integer whose square has all middle bits equal to 1?
>
>
>
***YES***. One is AAAAAAAAAAAAAAAB555555555555555516, which square is 71C71C71C71C71C7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF8E38E38E38E38E3916.
---
It's asked if there are integers $n$ such that $0<n<2^{2k}$ and $(n^2\bmod2^{3k})\ge2^{3k}-2^k$, in the case $k=128/2$.
As noted in [comment by Max Alekseyev](https://mathoverflow.net/posts/comments/1073379?noredirect=1), that holds for $k\equiv 4\pmod6$ (thus including the question's $k=128/2$), because the integer
$$n=\frac{2^{2k+1}+2^{k+1}-1}3$$
does the trick. It's representation in base $4$ consists of $k/2-1$ digits $2$, one digit $3$, and $k/2$ digits $1$. It holds
$$n^2=\frac{8^k-1}{9}2^{k+2}+\frac{8^{k+1}+1}9$$
which representation in base $4$ consists of the digit $1$; the digits $301$ repeated $(k-4)/6$ times; the digit $3$ repeated $1+k$ times; the digit $2$; the digits $032$ repeated $(k-4)/6$ times; the digit $1$. It follows the center $2k$ bits are set, with three more on the left and one more on the right.
We can't use this to conclude exactly how many $n$ there are, or/and for other values of $k$, or/and for an arbitrary value in the middle of the square.
Here is a brute force search for $k\le21$:
| $k$ with a solution | #solution(s) | possible $n$ |
| --- | --- | --- |
| 4 | 1 | 101101012 |
| 5 | 2 | 0010110101201011010102 |
| 10 | 2 | 101010101101010101012111000100100011000112 |
| 11 | 4 | 00101010101101010101012001110001001000110001120101010101101010101010201110001001000110001102 |
| 12 | 3 | 000111000100100011000110201001000111000100010101121100011011100000011110012 |
| 13 | 3 | 000011100010010001100011002001100011011100000011110012011000110111000000111100102 |
| 14 | 1 | 10100010001100101000100110002 |
| 16 | 1 | 101010101010101101010101010101012 |
| 17 | 2 | 0010101010101010110101010101010101201010101010101011010101010101010102 |
| 19 | 1 | 111100100110010001011001001101100000112 |
| 21 | 4 | 0101000001010110111010100110001010110001112011101100111011110011000000000010111000101210100000101011011101010011000101011000111021111011100101110011000010011011101101000012 |
We see and could prove that if $n$ is a solution for $k=k\_a+1$ with $n<2^{2k\_a}$, then $n$ is a solution for $k=k\_a$. And if $n$ is a solution for $k=k\_a+1$ with $n<2^{2k\_a+1}$ and $n$ even, then $n/2$ is a solution for $k=k\_a$. That explains why more often than not, if $n=n\_a$ is a solution for $k=k\_a$, then $n=n\_a$ or/and $n=2n\_a$ is/are a solution for $k=k\_a+1$.
| 7 | https://mathoverflow.net/users/122065 | 418082 | 170,257 |
https://mathoverflow.net/questions/417588 | 6 | Let $V$ be the set of $k$ by $n$ matrices ($k<n$) with entries in $\mathbb{C}$, and let $\mathbb{C}[V]$ denote the set of polynomial functions on $V$. For any subset $I \subseteq [n] = \{1,2,\dotsc, n\}$ with size $k$ let $e\_I$ be the function which evaluates the determinant of the $k$ by $k$ submatrix found by taking only columns indexed by elements of $I$.
Let $A$ be a $k$ by $n$ matrix. It's well known that
$$e\_I(A) = 0 \text{ for all } I \subseteq [n], |I|=k \Rightarrow \operatorname{rank}(A)<k.$$
Moreoever, one can find an example of a matrix $A\_I$ for which $e\_J(A\_I) \neq 0$ if and only if $I=J$, so no proper subset of $S:=\{e\_I: I \subseteq [n], |I|=k\}$ suffices to check the rank of $A$.
If we allow linear combinations of the elements of $S$ we can sometimes get away with fewer functions. The smallest nontrivial example arises when $k=2$, $n=4$. There we have the Plücker relation
$$e\_{1,2}e\_{3,4}-e\_{1,3}e\_{2,4}+e\_{1,4}e\_{2,3} = 0.$$
This implies that the vanishing of
$$\{e\_{1,2}-e\_{3,4},e\_{1,3},e\_{2,4},e\_{1,4},e\_{2,3}\}$$ is sufficient to show a $2$ by $4$ matrix has rank $<2$.
Let's call a set of linear combinations of $e\_I$ with the property above a "rank-detecting set", and write $\beta(k,n)$ for the size of the smallest rank detecting set for $k$ by $n$ matrices. Clearly $\beta(k,n) \leq \binom n k$. It's quite easy to show that $\beta(n-1,n) = n$, and the example above shows that $\beta(2,4) \leq 5$.
Have you come across this notion before? Is anything known in general about the numbers $\beta(k,n)$?
| https://mathoverflow.net/users/38359 | Vanishing linear combinations of minors | This notion is indeed well-known and connected to the arithmetical rank of the ideal $I\_k$ of $C[V]$ generated by the maximal minors.
**Short answer** : that $\beta(k,n) = mk - k^2 +1$ and there is always a set of $mk-k^2+1$ linear combinations of maximal minors which is a *rank detecting set* (with your words).
**Longer answer** : Let $R$ be a commutative ring (say Noetherian) and $I$ a non trivial ideal of $R$. We definie the *arithmetical rank* of $I$ as:
$$ \mathrm{ara}(I) : \mathrm{min} \{m \geq 1, \ \exists f\_1, \ldots, f\_n \in R, \ \textrm{such that} \ \sqrt{I} = \sqrt{(f\_1, \ldots, f\_m)} \}$$
In other words, the arithmetical rank of $I$ is the minimum number of equations sufficient to define, set-theoretically, the same variety as $I$ in $\mathrm{Spec}(R)$.
Let $V$ be the set of $n \times k$ matrices, $\mathbb{C}[V]$ the algebra of polynomial functions over $V$ and $t \leq \mathrm{min}(m,n)$. We denote by $I\_t$ the ideal of $\mathbb{C}[V]$ generated by $t \times t$ minors of $\mathbb{C}[V]$. In [The number of equations defining a determinantal variety](https://www.home.uni-osnabrueck.de/wbruns/brunsw/pdf-article/NumbEqDet.published.pdf), by Brüns and Schwanzl, it was proved that $\mathrm{ara}(I\_t) \geq mk-t^2 +1$. It was furthermore proved before by Bruns, using the theory of algebras with straightening laws (see the references in the paper I gave a link to) that there is a set of linear combinations of minors of size $nk - t^2+1$ which generates an ideal having same ridical as $I\_t$. These sets are studied in details in the chapter 5 of [Determinantal Rings](https://www.home.uni-osnabrueck.de/wbruns/brunsw/detrings.pdf) by Bruns and Vetter.
Note that the cas $t = \mathrm{min}(k,m)$, which seems to be the case of interest for you, has been settled before by Hochster (the precise reference is given in the fist chapter of the book by Bruns and Vetter, but I can't find it on top of my head).
| 8 | https://mathoverflow.net/users/37214 | 418086 | 170,260 |
https://mathoverflow.net/questions/418069 | 4 |
>
> Let $ 0<a<b $, $ f\in C^1\left([0,b]\right)$. Assume that $ f $ is concave on $ [0,a] $ and convex on $ [a,b] $ with $ f'(0)>f'(b) $. Please prove that there exist $ n\_0\in\mathbb{N} $ which is sufficiently large, such that for any $ n\geq n\_0 $ and $ a\_1+a\_2+...+a\_n=b $ with $ a\_i\geq 0 $ ($ i=1,2,...,n $), we have $ \sum\_{i=1}^nf(a\_i)\leq nf\left(\frac{b}{n}\right) $.
>
>
>
Simple observation shows that if $ a\_i\in [0,a] $ for all $ i $, the result is easy to prove. However, I cannot deal with the situation that there exists some $ i\_0 $ such that $ a\_{i\_0}\in(0,b] $. I guess that such case is related with the statement that $ f'(0)>f'(b) $, but I do not know how to continue the proof. Could you give me some references or hints?
| https://mathoverflow.net/users/241460 | How to prove $ \sum_{k=1}^{n}f(a_k)\leq nf\left(\frac{b}{n}\right) $ for sufficiently large $ n $ here? | Denote $b/n=s$. We want a pointwise bound $$f(x)\leqslant f(s)+(x-s)f'(s),\label{1}\tag{$\heartsuit$}$$ then summing \eqref{1} up for $x=a\_1,\ldots,a\_n$ we get the desired inequality. Note that if $s<a$ (that holds for $n>b/a$) we get \eqref{1} on $[0,a]$ by concavity. For proving \eqref{1} on $[a,b]$, by convexity it suffices to verify \eqref{1} for $x=a$ and $x=b$. For $x=a$ this is already done, for $x=b$ it reads as
$$
f(b)\leqslant f(s)+(b-s)f'(s).
$$ When $n$ is large, RHS converges to $f(0)+bf'(0)$. Thus it suffices to check that
$$
f(b)<f(0)+bf'(0).
$$ Assume the contrary:
$$
f(b)\geqslant f(0)+bf'(0).
$$ We have $f(x)\leqslant f(0)+f'(0)x$ for all $x\in [0,a]$ by concavity. Denote by $c$ the endpoint of the maximal segment $[0,c]$ on which we have $f(x)\leqslant f(0)+f'(0)x$. Then $c\in [a,b]$ and we have $f(c)=f(0)+f'(0)c$ (otherwise $c$ is not maximal). This yields
$$
f'(c)=\lim\_{x\to c-0}\frac{f(c)-f(x)}{c-x}\geqslant f'(0).
$$ Since $f'$ increases on $[a,b]$ by convexity we get $f'(b)\geqslant f'(c)\geqslant f'(0)$, a contradiction.
| 3 | https://mathoverflow.net/users/4312 | 418087 | 170,261 |
https://mathoverflow.net/questions/415727 | 1 | Consider a CT Markov Process $X=(X\_t)\_{t\geq0}$ with state space $E\in\mathbb{R}^N$. Are there any general conditions under which a stationary distribution $\pi$ for $X$ is also a limiting distribution and viceversa?
Any useful reference will be much appreciated.
| https://mathoverflow.net/users/86048 | Stationary and limiting distributions | For Markov chains, a very useful condition is Harris recurrence,
see <https://en.wikipedia.org/wiki/Harris_chain>.
This has been generalized to continuous time, see
<https://www.jstor.org/stable/3690386?seq=1#metadata_info_tab_contents>
| 2 | https://mathoverflow.net/users/7691 | 418099 | 170,264 |
https://mathoverflow.net/questions/418092 | 8 | Let $P$ be a partial order on a finite set $S$ (assume that every element is related to at least one other element besides itself…this raises a few quick questions: is this implied by the definition of partial order and if not, what are the "isolated" points called and what is a partial order with no such points called?) and $R$ be the smallest subset of $(S\times S)\setminus P$ such that for all partial orders $Q\supseteq P$, $$R\cap Q=\varnothing\implies P=Q.$$
Is the relation $R$ unique? Does it have a name? Is it equivalent to something else that does?
I asked a similar (perhaps the same) question in an unwieldy way on MSE at [Minimal generating sub-relation of complement of partial order](https://math.stackexchange.com/questions/3480391/minimal-generating-sub-relation-of-complement-of-partial-order) and got no response.
======= edit added by mathematrucker 16 Mar 2022 =======
While searching the literature unsuccessfully for @JosephVanName's nice "one pair extension property of finite partially ordered sets" below it occurred to me that it follows immediately from the fact that the poset under set containment of all unlabeled posets on $n$ points is graded by cardinality. This appears as Lemma 2.1 in [New results from an algorithm for counting posets](https://link.springer.com/article/10.1007/BF00383201) by Culberson and Rawlins (1990), which cites note (3.1) in Aigner, [Producing posets](https://www.sciencedirect.com/science/article/pii/0012365X81901977) (1981). Neither reference supplies a proof, so if anyone knows of one that does, please add it in the comments.
======= edit added by mathematrucker 23 Jul 2022 =======
The following 1988 precursor to Culberson and Rawlins's 1990 paper gives a proof of Lemma 2.1: [On counting posets and the structure of the poset of posets](https://legacy.cs.indiana.edu/ftp/techreports/TR266.pdf). The authors call "isolated" points *singletons* and posets that have none are said to be in *reduced form.*
| https://mathoverflow.net/users/5090 | Smallest relation in complement of partial order that prohibits its extension | I claim that the relation $R$ is in fact unique. The uniqueness follows from the one pair extension property of finite partially ordered sets.
Proposition: Suppose that $X$ is a finite set with partial ordering $P$. Then whenever $Q$ is a partial ordering on $X$ with $P\subseteq Q,P\neq Q$, there exists an ordered pair $(x,y)\in Q\setminus P$ such that $P\cup\{(x,y)\}$ is a partial ordering.
Proof: Suppose that $(x,y)\in Q\setminus P$. Then $x\not\leq\_{P}y$ and
$y\not\leq\_{P}x$. Let $x',y'$ be a pair such that
$x'\leq\_{P}x,y\leq\_{P}y'$, $x'\not\leq\_{P}y'$, and
$$(x''\leq\_{P}x',y'\leq\_{P}y'')\implies(x''\leq\_{P}y''\mbox{ or }x''=x',y''=y').$$
Let $T=P\cup\{(x',y')\}$. Note that since $y\leq y'$, $x'\leq x$, and $y\not\leq\_{P}x$ we have $y'\not\leq\_{P}x'$ and thus $(y',x')\not\in T$.
I claim that $T$ is a partial ordering. The ordering $T$ is clearly reflexive.
Suppose that $r\leq\_{T}s,s\leq\_{T}t$. If $(r,s),(s,t)\in P$, then $r\leq\_{P}t$ by the transitivity of $P$. In the case that $(r,s)=(x',y')=(s,t)$, we have $r=s=t$, so $(r,t)\in T$ by reflexivity. Now, assume that $(r,s)=(x',y')$ and $s\leq\_{P}t$.
Then either $r\leq\_{P}t$ or $s=t$. In either case, $r\leq\_{T}t$. If we assume $(s,t)=(x',y'),r\leq\_{P}s$, then either $r=s$ which would imply that $r\leq\_{T}t$ or we would have $r\leq\_{P}t$ which would imply that $r\leq\_{T}t$ as well. Therefore, $T$ is transitive.
Now, assume that $r\leq\_{T}s\leq\_{T}r$. If $(r,s)\neq(x',y')$ and $(s,r)\neq(x',y')$, then we know that $r\leq\_{P}s\leq\_{P}r$ which implies that $r=s$. On the other hand, we can assume without loss of generality that $(r,s)=(x',y')$. Since it was shown above that $y'\not\leq\_{T}x'$ we conclude that $T$ is antisymmetric.
Hence $T$ is a partial ordering. Observe that $x'\leq\_{Q}x\leq\_{Q}y\leq\_{Q}y'$, so $(x',y')\in Q$. Q.E.D.
Therefore, if we let $R$ be the collection of all ordered pairs $(x,y)\in X^{2}\setminus P$ where $P\cup\{(x,y)\}$ is a partial ordering, then if $Q$ is a partial ordering on $X$ with $P\subseteq Q$ and $Q\cap R=\varnothing$, then $Q=P$ (otherwise a new pair could be added to $R$, contradicting its definition).
| 9 | https://mathoverflow.net/users/22277 | 418100 | 170,265 |
https://mathoverflow.net/questions/418093 | 1 | I am wondering about the following question: A strictly convex (concave) differentiable function $f:\mathcal{R}\to\mathcal{R}$ has the geometrical property that its graph lies completely above (below) the tangential line at any point (except at the point of contact). Now, this is a special property of such functions but is a relaxed version of this also valid for all (not necessarily continuously) differentiable functions? More specifically, let's call $x\_0$ a *convex* point if there is a neighborhood of $x\_0$ such that
$$f(x) > f(x\_0) + f'(x\_0)(x- x\_0)$$
for all $x$ in that neighborhood. Similarly, let's call $x\_0$ concave if the reverse inequality holds for some neighborhood:
$$f(x) < f(x\_0) + f'(x\_0)(x- x\_0)$$
Then, is it true that any non-linear differentiable function has at least one concave or convex point?
Full disclosure: this question was first stated as true on a different forum without any additional information or explanations. It is possible this is well-known and I have never come across this result, or that this is completely wrong. Regardless, I am hoping to get some additional information as I think this is quite a neat result either way.
| https://mathoverflow.net/users/19673 | Convex/concave points of a differentiable function | Let $f$ be non linear in the interval $[-1,1]$. We can suppose $f(-1)=f(1)=0$ and $f(x)>0$ for some $x\in(-1,1)$. Now let $N$ be so big that the closed ball of center $(0,-N)$ and with radius $\sqrt{N^2+1}$ doesn't contain the whole graph of $f$. Let $(x\_0,f(x\_0))$ be one of the points of the graph furthest from $(0,-N)$. Then $f$ is concave at $x\_0$.
Indeed, the whole graph of $f$ is contained in the ball of center $(0,-N)$ and boundary passing through $(x\_0,f(x\_0))$. This implies that the derivative of $f$ at $x\_0$ has to be the slope of the tangent to that ball at $(x\_0,f(x\_0))$, and that tangent only intersects the graph of $f$ at $(x\_0,f(x\_0))$ (in fact it only intersects the ball at that one point).
| 2 | https://mathoverflow.net/users/172802 | 418101 | 170,266 |
https://mathoverflow.net/questions/418051 | 3 | Let $X$ be a complete nonsingular curve and $S$ a scheme over $k$ algebraically closed, and $\cal{F}$ a coherent sheaf on $X \times S$, generated by finitely many global sections and flat over $S$ (via the projection). So my question is the following: is the locus of points $s\in S$ such that $\cal{F}\_s$ is locally free on $X$ open? And if so, is there an easy way to see this?
| https://mathoverflow.net/users/478669 | Is local freeness open for curves? | To elaborate on Jason's comment: consider a morphism of schemes $f:\mathrm{X}\rightarrow\mathrm{S}$, and an $f$-flat coherent sheaf $\mathscr{F}$ on $\mathrm{X}$. Then
(1) the singular locus $\mathrm{Sing}(\mathscr{F})$ of $\mathscr{F}$ (given by the points $x\in\mathrm{X}$ such that $\mathscr{F}$ is not locally free at $x$) is closed - if $r$ is the rank of $\mathscr{F}$, it is cut out by the $r$-th Fitting ideal sheaf of $\mathscr{F}$;
(2) $\mathscr{F}$ is locally free at $x\in\mathrm{X}$ if and only if $\mathscr{F}\_{f(x)}$ is locally free at $x$, where $\mathscr{F}\_{f(x)}$ denotes the restriction of $\mathscr{F}$ to $f^{-1}(f(x))$ - this follows from flatness of $\mathscr{F}$ and the Nakayama lemma.
By (1) and (2), $f(\mathrm{Sing}(\mathscr{F}))$ consists of all $s\in\mathrm{S}$ such that $\mathscr{F}\_{s}$ is not locally free. If $f$ is proper, then of course the latter image is closed in $\mathrm{S}$.
Lemma 2.1.8 in *The Geometry of Moduli Spaces of Sheaves* by Huybrechts & Lehn is a good reference.
| 7 | https://mathoverflow.net/users/104669 | 418106 | 170,267 |
https://mathoverflow.net/questions/418006 | 5 | A category $\mathcal{C}$ is called $\textbf{discrete}$ if the only morphisms are identity morphisms.
Consider the following weaker notion: a category $\mathcal{C}$ is called $\textbf{totally disconnected}$ if $\text{Hom}\_\mathcal{C}(C,D)=\varnothing$ for all $C\neq D$.
Vopenka's principle ($\textbf{VP}$) states that a full large subcategory $\mathcal{D}$ of a locally presentable category $\mathcal{C}$ cannot be discrete. I want to consider an anaologus statement:
($\textbf{VP2}$) Full large subcategory $\mathcal{D}$ of a locally presentable category $\mathcal{C}$ cannot be totally disconnected.
Since discrete categories are totally disconnected, $\textbf{VP2}$ implies $\textbf{VP}$. Hence $\textbf{VP2}$ can't be proven in $\textbf{ZFC}$, because $\textbf{VP}$ can't. Now I would like to know:
(1) Is there a counter-example to $\textbf{VP2}$ in $\textbf{ZFC}$? As far as I know, no counter-example to $\textbf{VP}$ has been found but since $\textbf{VP2}$ is in principle a stronger statement, perhaps we can construct one here?
(2) If not, what is the relation between $\textbf{VP}$ and $\textbf{VP2}$ inside $\textbf{ZFC}$? Are they equivalent? Or does at least consistency of $\textbf{VP}$ implies that of $\textbf{VP2}$?
(I wrote $\textbf{ZFC}$ as my set theory of choice but if there are any problems in using such a weak theory, feel free to consider some other).
Thanks!
| https://mathoverflow.net/users/474727 | Stronger (?) form of Vopenka's principle | VP2 is equivalent to VP because every set carries a rigid binary relation. This is similar as Lemma 6.3 in my book with Adámek. In fact, VP2 is an original formulation of VP (see Jech, Set Theory).
| 6 | https://mathoverflow.net/users/73388 | 418113 | 170,270 |
https://mathoverflow.net/questions/417950 | 2 | By Reflection I mean the following schema:
**Reflection:** $$\forall X \, (\varphi \implies \exists \alpha : \varphi^{V\_\alpha})$$
where $\varphi$ is a first order formula (defined predicates and functions allowed) in which only symbol "$X$" occurs free, $\varphi^{V\_\alpha}$ is the "$\in {V\_\alpha}$" bounded form of $\varphi$.
$V\_\alpha$ is defined in the customary manner as the set of all subsets of elements of the range of a function from an ordinal $\alpha$ where the image of a successor is the power set of the image of its predecessor, and the image of a limit is the union of all images of prior ordinals. And ordinal is defined after von Neumann as a transitive set well ordered by $\in$.
The point here is that we are allowing the use of defined predicate and function symbols, as long as the formula remain first order, i.e. no quantification allowed over them.
Is reflection here a theorem scheme of $\sf ZF$?
| https://mathoverflow.net/users/95347 | Can we allow defined predicates and functions in reflection? | The first answer took care of the case where "Reflection" was a scheme of formulas in the language of ZF. Now we consider what we now think was the intent of the questioner.
We expand the language of ZF by introducing a predicate symbol P and a function symbol F
for each formula in the language of ZF. We also add additional axioms:
∀x1,...,xn(P(x1,...,xn)<-->(x1,...,xn)) where x1,...,xn are the free variables of .
∀x1,...,xn,y(∀z(((x1,...,xn,z)<-->y=z)-->F(x1,....xn)=y))
where x1,...,xn,y are the free variables of
∀x1,...,xn,y(((x1,...,xn,y)∧∃z(z≠y∧(x1,...,xn,z)))-->F(x1,...,xn)=0)
where x1,...,xn,y are the free variables of
Call this theory ZZF.
"Reflection" holds in ZZF.
Proof:Let be a formula in the language of ZZF whose only free variable is X. Let A be the set of formulas such that P is a symbol of . Let B be the set of formulas such that F is a symbol of . To each term t ocurring in we shall associate a variable vt where if s and t are not variables then
vs=vt iff s=t, and vs is not a variable of .
To each term we will also associate a formula ft. We will do this inductively as follows:
if w is a variable then vw=w and fw is w=w.
if s=F(t1,...,tn) then
fs is f(t1)∧...∧f(tn)∧(v(t1),...,v(tn),v(s))
For each subformula θ of we define a formula θ\* in the language of ZF as follows:
(t1=t2)\* is f(t1)∧f(t2)∧v(t1)=v(t2)
(t1∈t2)\* is f(t1)∧f(t2)∧v(t1)∈v(t2)
(P(t1,...,tn))\* is f(t1)∧...∧f(tn)∧(v(t1),...,v(tn))
(θ1∧θ2)\* is (θ1)*∧(θ2)*
(¬θ)\* is ¬(θ\*)
(∃xθ)\* is ∃x(θ\*)
We note that θ\*<-->θ for all subformulas θ of .
Now suppose (X).
By the usual notion of reflection there is an ordinal such that X∈V, V reflects \*, and V reflects all formulas of A and B.
Then \*(X) holds with quantifiers relativized to V.
Therefore (X) holds relativized to V, since relativized to V, (X)<-->\*(X).
| 2 | https://mathoverflow.net/users/133981 | 418114 | 170,271 |
https://mathoverflow.net/questions/418117 | 1 | Let $a(n)$ be [A007306](https://oeis.org/A007306), denominators of Farey tree fractions (i.e., the Stern-Brocot subtree in the range $[0,1]$).
Let $b(n)$ be [A002487](https://oeis.org/A002487), Stern's diatomic series (or Stern-Brocot sequence): $b(0) = 0, b(1) = 1$; for $n > 0$: $b(2n) = b(n), b(2n+1) = b(n) + b(n+1)$.
Also
$$a(n)=b(2n-1)$$
I conjecture that to generate these sequences we can start with $a(1)=b(1)=1$ and then apply
$$a(n)=a(n-1)+b(n-1)-2(a(n-1)\operatorname{mod} b(n-1))$$
$$b(n)=a(n)-b(n-1)$$
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Stern-Brocot tree and subtree | The second half is already given in the question, so really what you're asking is whether $$b(2n-1)=b(2n-3)+b(n-1)-2(b(2n-3)\bmod b(n-1))$$
But as noted in OEIS (quoted with relabelling),
>
> Moshe Newman proved that the fraction b(n+1)/b(n+2) can be generated from the previous fraction b(n)/b(n+1) = x by 1/(2\*floor(x) + 1 - x).
>
>
>
Or, in other words, $$\begin{eqnarray\*}\frac{b(n+2)}{b(n+1)} &=& 2\left\lfloor \frac{b(n)}{b(n+1)} \right\rfloor + 1 - \frac{b(n)}{b(n+1)} \\
&=& 2\frac{b(n) - b(n) \bmod b(n+1)}{b(n+1)} + 1 - \frac{b(n)}{b(n+1)}
\end{eqnarray\*}$$
So $$b(n+2) = b(n) - 2b(n) \bmod b(n+1) + b(n+1)$$
and substituting $n = 2m-3$ and applying $b(2n) = b(n)$ gets us there.
| 3 | https://mathoverflow.net/users/46140 | 418130 | 170,274 |
https://mathoverflow.net/questions/380539 | 19 | [This answer](https://mathoverflow.net/a/69248/150063) says,
>
> IIRC, the calculus of inductive constructions is equi-interpretable with ZFC plus countably many inaccessibles — see Benjamin Werner's *"Sets in types, types in sets"*. (This is because of the presence of a universe hierarchy in the CIC.)
>
>
>
But, I read *"Sets in types, types in sets"* and discovered that the book does not prove this statement. It only conjectures the strength of CIC.
Has "CIC and ZFC + countably many inaccessible cardinals are equiconsistent" been proven or disproven?
| https://mathoverflow.net/users/150063 | Are we sure the calculus of inductive constructions and ZFC plus countably many inaccessible cardinals are equiconsistent? | The situation is a bit subtle. One can interpret CIC in any model of ZFC with infinitely many inacessibles. However, interpreting ZFC in CIC is more subtle. First one needs to assume the law of excluded middle and choice in CIC (and perhaps quotient types depending on how smooth we want things to work). These are very strong assumptions and they increase the consistency strength over plain CIC, which appears to be much weaker.
Once excluded middle and choice are assumed, within each universe level $\mathcal{U}\_1,\mathcal{U}\_2,\ldots$ of CIC we can construct a model of ZFC. This constructs a chain $V\_1 \subseteq V\_2 \subseteq \cdots$ of models of ZFC where each is an end extension of the previous, up to canonical isomorphism. Thus, $V\_2$ has at least one inaccessible, $V\_3$ has at least two inaccessibles, and so on. Thus CIC with choice proves the consistency of ZFC + there are $k$ inaccessibles for any *standard* $k$. Note that I've been careful not to associate universe levels with natural numbers. Indeed, models of CIC and ZFC can have nonstandard natural numbers. However, universe levels are syntactic objects and therefore always standard.
So, the consistency of CIC with choice and excluded middle implies the $\Pi\_1$ statement $$\forall k\,\operatorname{Con}(ZFC + k\text{-many inaccessibles})\tag{\*}.$$
This is strictly weaker than the consistency of ZFC with infinitely many inaccessibles. However, $(\*)$ is actually enough to construct a full model of CIC! By compactness, we can construct a model $V$ of ZFC with $k$ inaccessibles, where $k$ is nonstandard. In this model, we can list the first standardly many inaccessibles as $\kappa\_1 < \kappa\_2 < \cdots$. This hierarchy allows us to construct a corresponding sequence of universes for a model of CIC. Note that this is *not* an interpretation per se since we can only witness externally that $k$ is nonstandard.
Nevertheless, from the above it follows that the consistency strength of CIC with excluded middle and choice is exactly $(\*)$ and therefore strictly weaker than ZFC with infinitely many inaccessibles.
| 9 | https://mathoverflow.net/users/2000 | 418133 | 170,276 |
https://mathoverflow.net/questions/418111 | 7 | Let $\{X\_i\}\_{i=1}^{\infty}$ be i.i.d. random variables such that: i) $X\_i > 0$; and ii) $\textrm{Pr}[X\_i > x] \sim x^{-\alpha}$ at large $x$ for some $\alpha \in (0, 1)$. Define the quantities
\begin{equation}
S\_n \equiv \frac{X\_1 + \cdots + X\_n}{n^{1/\alpha}}.
\end{equation}
By Kolmogorov's zero-one law (right?), there is some $\mu \in [0, \infty]$ such that $\liminf\_{n \rightarrow \infty} S\_n = \mu$ almost surely. **I just want to know whether $\mu$ is zero, finite, or infinite (and see a proof).**
I haven't been able to find an answer to this question anywhere (note that I'm specifically interested in $\alpha < 1$ and that $S\_n$ is normalized differently than the usual mean). For a bit of context, I can easily show/find in textbooks that $\limsup\_{n \rightarrow \infty} S\_n = \infty$. But those textbooks never seem to mention the liminf. The closest result I could find is that when $\alpha = 1$, $\liminf\_{n \rightarrow \infty} S\_n = \infty$. But this is shown via truncation + SLLN, and I don't see a way to generalize it to $\alpha < 1$.
More generally, one could define
\begin{equation}
S\_n^{(\beta)} \equiv \frac{X\_1 + \cdots + X\_n}{n^{\beta}}.
\end{equation}
Then all the following are not too difficult to prove: i) for any $\beta < 1/\alpha$, $\lim\_{n \rightarrow \infty} S\_n^{(\beta)} = \infty$; ii) for any $\beta > 1/\alpha$, $\lim\_{n \rightarrow \infty} S\_n^{(\beta)} = 0$; and iii) for $\beta = 1/\alpha$, $\limsup\_{n \rightarrow \infty} S\_n^{(\beta)} = \infty$ (all of these holding almost surely). Thus the question of the liminf when $\beta = 1/\alpha$ is the one missing (to me) piece to understanding the scaling of the partial sums, and I'm really surprised that I haven't been able to find it anywhere.
| https://mathoverflow.net/users/478720 | What is the liminf of a sum of i.i.d. random variables with heavy tails? | $\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$Here we implement the idea proposed in mike's answer. The proof below works for all $\alpha\in(0,2]$.
Let $a:=1/\alpha$. Let
\begin{equation\*}
Z\_n:=X\_1+\cdots+X\_n,
\end{equation\*}
so that
\begin{equation\*}
S\_n=Z\_n/n^a.
\end{equation\*}
Take any real $\ep>0$.
We have
\begin{equation\*}
S\_n\to S \tag{1}\label{1}
\end{equation\*}
in distribution as $n\to\infty$ for some nonnegative random variable (r.v.) $S$
such that $P(S\ge\ep)<1$, so that
\begin{equation\*}
q:=\frac{1+P(S\ge\ep)}2<1; \tag{2}\label{2}
\end{equation\*}
see e.g. [Attraction domain of a stable distribution](https://encyclopediaofmath.org/wiki/Attraction_domain_of_a_stable_distribution) and [Stable distribution](https://encyclopediaofmath.org/wiki/Stable_distribution).
For each natural $k$, let $C\_k\in(0,\infty)$ be such that $P(S\ge C\_k)\le1/k^2$.
It follows from \eqref{1} that for some strictly increasing sequence $(n\_k)$ of natural numbers and all natural $k$ we have
\begin{equation\*}
n\_{k+1}-n\_k\to\infty \tag{3}\label{3}
\end{equation\*}
as $k\to\infty$,
\begin{equation\*}
P(S\_{n\_k}>C\_k)\le P(S\ge C\_k)+1/k^2\le2/k^2,
\end{equation\*}
and
\begin{equation\*}
\ep\_k:=\frac{2\ep n\_{k+1}^a-C\_k n\_k^a}{(n\_{k+1}-n\_k)^a}>\ep. \tag{4}\label{4}
\end{equation\*}
Introducing now the event
\begin{equation\*}
A\_m:=\{\forall k\ge m\ S\_{n\_k}\le C\_k\},
\end{equation\*}
we have
\begin{equation\*}
1-P(A\_m)\le\sum\_{k\ge m}P(S\_{n\_k}>C\_k)\le\sum\_{k\ge m}2/k^2\to0 \tag{5}\label{5}
\end{equation\*}
as $m\to\infty$.
Moreover, letting
\begin{equation\*}
T\_k:=\frac{Z\_{n\_{k+1}}-Z\_{n\_k}}{(n\_{k+1}-n\_k)^a},
\end{equation\*}
we have
\begin{equation\*}
\begin{aligned}
& P(A\_m,\forall k\ge m\ S\_{n\_k}>2\ep) \\
&=P(A\_m,\forall k\ge m\ Z\_{n\_k}>2\ep n\_k^a) \\
&\le P(\forall k\ge m\ Z\_{n\_{k+1}}-Z\_{n\_k}>2\ep n\_{k+1}^a-C\_k n\_k^a) \\
&=P(\forall k\ge m\ T\_k>\ep\_k) \\
&\le P(\forall k\ge m\ T\_k>\ep) \\
&=\prod\_{k=m}^\infty P(T\_k>\ep),
\end{aligned}
\tag{6}\label{6}
\end{equation\*}
by \eqref{4} and the independence of the $X\_i$'s.
By \eqref{3}, $T\_k\to S$ as $k\to\infty$ (cf. \eqref{1}), so that, by \eqref{2}, $P(T\_k>\ep)\le q<1$ for all large enough $k$.
So, by \eqref{6}, for all natural $m$ we have $P(A\_m,\forall k\ge m\ S\_{n\_k}>2\ep)=0$ and hence for
\begin{equation}
B\_m:=\{\forall k\ge m\ S\_{n\_k}>2\ep\}
\end{equation}
we have
\begin{equation\*}
P(B\_m)\le1-P(A\_m)\to0
\end{equation\*}
as $m\to\infty$, by \eqref{5}. But $B\_1\subseteq B\_2\subseteq\cdots$. So, $P(B\_m)=0$ for all natural $m$ and hence
\begin{equation\*}
P(\exists m\ \forall k\ge m\ S\_{n\_k}>2\ep)=P(B\_1\cup B\_2\cup\cdots)=0.
\end{equation\*}
So,
\begin{equation\*}
P(\liminf\_{n\to\infty}S\_n\le2\ep)\ge P(\forall m\ \exists k\ge m\ S\_{n\_k}\le2\ep)=1,
\end{equation\*}
for any real $\ep>0$.
Thus,
\begin{equation\*}
P(\liminf\_{n\to\infty}S\_n=0)=1.
\end{equation\*}
| 3 | https://mathoverflow.net/users/36721 | 418140 | 170,279 |
https://mathoverflow.net/questions/311958 | 5 | I found a very interesting problem in the [Open Problem Garden](http://www.openproblemgarden.org/op/a_nowhere_zero_point_in_a_linear_mapping), which I am surprised is not as well-known as I would think it would be:
Prove that If $p>3$ is prime and $A$ is an invertible $n \times n$ matrix with entries in ${\mathbb Z\_p}$, then there are column vectors $x,y \in {\mathbb Z\_p}^n$ which have no coordinates equal to zero such that $Ax=y$.
It is easy to show that this conjecture is false when $p=2,3$. Also, when $p$ is a little larger than $n$, this conjecture has been [proven true](https://doi.org/10.1007/BF01215347). However, the case when $n$ is large relative to $p$ seems not only difficult to prove, but also it seems to require nontrivial computational resources to even get a sense of what is going on.
Has any work been done to try to computationally understand this problem? In particular, is this conjecture known to be true for say $p=5$, $n=15$?
| https://mathoverflow.net/users/7089 | A nowhere-zero point in a linear mapping conjecture | This is the Alon-Jaeger-Tarsi conjecture first stated in 1981 and [resolved](https://arxiv.org/pdf/2107.03956.pdf) very recently (for $p\ge 83$) by János Nagy and Péter Pál Pach.
| 2 | https://mathoverflow.net/users/9924 | 418150 | 170,280 |
https://mathoverflow.net/questions/418145 | 1 | Consider $n$ labeled balls, $k$ of which are red and $(n-k)$ blue. Given a permutation of these balls, we tick $n-1$ times. For the $i$-th tick, if the $i$-th ball in the permutation is red, then it paints the $(i+1)$-th ball in the permutation blue (if the latter is already blue, then it remains blue).
We secretly mark one of the red balls at the beginning. How many permutations are there in which our marked ball becomes blue by the end of the process?
I want to prove that the answer is
$$
\sum\_{j=0}^{k-2}(-1)^j\frac{n!2^{k-2-j}{{k-1}\choose{j}}(k-1-j)}{n-j}.
$$
To show this, I was trying to use the inclusion-exclusion principle without success. How could we derive this formula?
ps.: There are other (maybe nicer) formulas that would work, but I am particularly interested in the one given above.
**Edit:** Thank you for all the great answers so far! My main question, however, remains open: how to prove that the formula I proposed is correct?
| https://mathoverflow.net/users/478035 | Counting permutations defined by a simple process | Let us put an additional blue ball in position $0$, to the left of the $n$ balls.
The condition on the permutations of the $n$ balls is then that the marked red ball be preceded by an odd number (say $2r-1$) of red balls, which in turn must be preceded by a blue ball. Let us refer to such permutations as good.
Let $p\_{n,k}$ denote the number of good permutations of the $k$ red balls and $n-k$ blue ones.
Let $j$ be the position of the marked red ball in a good permutation. Then $j\ge2r$.
If $j=2r$, then $j$ is even and the only blue ball to the left of the marked red ball is the additional blue ball in position $0$. So, for any given even $j\in[n]:=\{1,\dots,n\}$, the number of good permutations with $2r=j$ is
\begin{equation\*}
\Big(\prod\_{i=0}^{j-2}(k-1-i)\Big)(n-j)!=\frac{(k-1)!(n-j)!}{(k-j)!}.
\end{equation\*}
(If $j>k$, then the latter fraction is understood as $0$.)
Similarly counted are the good permutations with $j>2r$, where we must use one of the $n-k$ blue balls to place it immediately to the left of the $2r-1$ red balls preceding the marked red ball.
Thus,
\begin{equation\*}
\begin{aligned}
p\_{n,k}&=\sum\_{j\in[n]}\Big(
1(j\text{ is even})\frac{(k-1)!(n-j)!}{(k-j)!} \\
&+\sum\_{1\le r<j/2}(n-k)\frac{(k-1)!(n-2r-1)!}{(k-2r)!}
\Big) \\
&=(k-1)!\sum\_{j\in[k]}
1(j\text{ is even})\frac{(n-j)!}{(k-j)!} \\
&+(k-1)!(n-k)
\sum \_{r=1}^{\lfloor k/2\rfloor } \frac{(n-2 r)!}{(k-2 r)!} \\
&=(k-1)!(n-k+1)
\sum\_{r=1}^{\lfloor k/2\rfloor } \frac{(n-2 r)!}{(k-2 r)!}.
\end{aligned}
\end{equation\*}
---
This very simple expression is easy to analyze. Indeed, consider what is, according to the OP's comment, the case of interest: $n=2k-1$. Then
\begin{equation\*}
p\_{n,k}=q\_k:=p\_{2k-1,k}=k!\sum\_{r=1}^{\lfloor k/2\rfloor } \frac{(2k-1-2 r)!}{(k-2 r)!}.
\end{equation\*}
The OP wanted to show that
\begin{equation\*}
P\_k:=\frac{q\_k}{(2k-1)!}
\end{equation\*}
is $\le1/3$ and $P\_k\to1/3$ as $k\to\infty$.
To prove this, write
\begin{equation\*}
P\_k=\sum\_{r=1}^\infty a\_{k,r}, \tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
a\_{k,r}:=\frac{k!}{(2k-1)!} \frac{(2k-1-2 r)!}{(k-2 r)!};
\end{equation\*}
the latter fraction is understood as $0$ if $2r>k$. We can also write
\begin{equation\*}
a\_{k,r}=\prod\_{i=0}^{2r-1}\frac{k-i}{2k-1-i}
=\frac{k}{2k-1}\prod\_{i=1}^{2r-1}\frac{k-i}{2k-1-i}
\le \frac{k}{2k-1}\frac1{2^{2r-1}}.
\end{equation\*}
It also follows that $a\_{k,r}\to\frac1{2^{2r}}$ as $k\to\infty$, for each natural $r$. So, by \eqref{1} and dominated convergence,
\begin{equation\*}
P\_k\to\sum\_{r=1}^\infty \frac1{2^{2r}}=\frac13, \tag{2}\label{2}
\end{equation\*}
as was desired.
Next, it is easy to see that, for each natural $r\ge2$, $a\_{k,r}$ is increasing in natural $k\ge2r$. A little complication here is that $a\_{k,1}$ is decreasing in $k$. However, it is rather easy to see that $\sum\_{r=1}^3 a\_{k,r}$ is increasing in natural $k\ge5$. So, by \eqref{1}, $P\_k$ is increasing in natural $k\ge5$.
So, by \eqref{2}, $P\_k<1/3$ for $k\ge5$. It also easy to see that $P\_k<1/3$ for $k\in\{1,3,4\}$ and $P\_2=1/3$.
Thus, $P\_2=1/3$ and $P\_k<1/3$ for $k\in\{1,3,4,5,6,\dots\}$, as was also desired.
| 2 | https://mathoverflow.net/users/36721 | 418151 | 170,281 |
https://mathoverflow.net/questions/418149 | 3 | Let $\phi, a \in C^{\infty}([0,1])$ and assume $a(0)=1$. Suppose that
$$
\int\_0^1 e^{\tau \,\phi(t)}\,a(t)\,dt =0 \qquad \text{for all $\tau \in \mathbb R$}.
$$
Does it follow that $\phi$ is a constant in a neighborhood of $t=0$?
If the answer to the above question is affirmative I have a follow up question as follows: Suppose additionally that $b \in C^{\infty}([0,1]$ with $b(0)=1$ and instead of the latter equation there holds
$$\Bigg |\int\_0^1 e^{\tau \,\phi(t)}\,a(t)\,dt\Bigg| \leq \frac{1}{|\tau|} \Bigg|\int\_0^1 e^{\tau \,\phi(t)}\,b(t)\,dt\Bigg|$$
Does it again follow that $\phi(t)$ is constant in a neighborhood of $t=0$?
| https://mathoverflow.net/users/50438 | On an asymptotic integral | No. E.g., let $\phi(t)=(t-1/2)^2$ (so that $\phi$ is even about $1/2$) and $a(t)=2(1/2-t)$ (so that $a$ is odd about $1/2$).
| 2 | https://mathoverflow.net/users/36721 | 418154 | 170,283 |
https://mathoverflow.net/questions/417518 | 5 | In C-H Sah's book *Hilbert's third problem: scissors congruence*, the author defines the data for *abstract scissors congruence* in order to prove Zylev's theorem by combinatorial means in great abstraction. I expect that most of the book is over my head as an undergrad but I'm looking for clarification as to wether the definition Sah gives is actually correct. My advisor and I could not figure out how to resolve the apparent discrepancies.
The definition is given as follows on page 5:
>
> The abstract scissors congruence data consists of a distinguished family of nonempty subsets (to be called $n$-simplices where $n$ is to be interpreted as the dimension) of a nonempty set $X$ and a specified equivalence relation (called congruence) among the $n$-simplices. We need a few definitions before stating the addition condition to be satisfied by the $n$-simplices.
>
>
> Two $n$-simplices $A$ and $B$ are said to be **interior disjoint** if the following conditions hold:
>
>
> * (D1) $A \cap B$ contains no $n$-simplices, and
> * (D2) if $C$ is an $n$-simplex contained in $A \cup B$, then $C \subset A$ if and only if $C \cap B$ contains no $n$-simplices.
>
>
> A **polyhedron** $P$ is understood to be a **finite** pairwise interior disjoint union of $n$-simplices. The concept of interior disjoint union can then be extended to polyhedra with $n$-simplices replaced by nonempty polyhedra. We will use $\coprod$ to denote pairwise interior disjoint unions (as well as direct sum when there is no chance of confusion). We omit the proof of the following elementary result:
>
>
> **Lemma 2.1**. If $P$, $Q$ and $R$ are polyhedra with $P \coprod R = Q \coprod R$, then $P = Q$.
>
>
> If $A$, $B$, $C$ are $n$-simplices with $A = B \coprod C$, we say $A$ is **simply subdivided** into $B$ and $C$. If $P = \coprod P\_i$ is a polyhedron where each $P\_i$ is an $n$-simplex, then a **subdivision** of $P$ is understood to be a **finite** succession of simple subdivisions such that each simple subdivision is performed on one of the $n$-simplices exhibited in the preceeding step. For example, the first step may be a simple subdivision of $P\_1$ into $Q$ and $R$; the second step may be a simple subdivision of $Q$ or $R$ or any $P\_j$ with $j > 1$; and so on.
>
>
> To complete the abstract scissors congruence data, we impose the following condition:
>
>
> * (S) Let $A$ and $B$ be $n$-simplices. Then there is at least one subdivision of $A$, say $A = \coprod\_{1 \le i \le t} A\_i$, such that $A \cap B = \coprod\_{j \in J} A\_j$ for some $J \subset \{ 1,\dots,t \}$.
>
>
>
My issue is with the axiom (S). It seems that the expression $A \cap B = \coprod\_{j \in J} A\_j$ is either not well-defined or implicitly states the idea that $A \cap B$ must be a polyhedron. Taking the second interpretation, if we consider the usual case of Euclidean scissors congruence then this axiom does not hold, as the intersection of two triangles joined side to side is not a polygon.
Another example is given on the next page: for $n \ge 0$, let the $n$-simplices be arbitrary $(n+1)$-element subsets of an infinite set $X$, and congruence is equinumerosity. If $n = 0$ then the polyhedra are just finite sets and interior disjointness is set-theoretic disjointness. Sah then states that if $n>0$ then no $n$-simplices are interior disjoint (because D2 never holds) and moreover he states that '(S) is clear and cardinality is a complete invariant'. As far as I can tell (S) does not actually hold in this case either, as the intersection of two $(n+1)$-element sets may have fewer elements but not be empty.
Keeping the Euclidean simplices in mind, the most obvious way to fix the statement of axiom (S) would be to require $A$ and $B$ to either be interior disjoint or else satisfy the current definition of (S). This still does not fix the finite subset case, though, as no two simplices are interior disjoint.
Presumably the entire book is not built on nonsense, so what is the proper definition used by Sah and other authors that refer to this book? Another definition of abstract scissors congruence in terms of category theory (specifically the idea of *double categories*) is given by Inna Zakharevich in her thesis *Scissors Congruence and K-Theory*, but it strikes me as a bit more complicated, and I was hoping for a more elementary definition requiring less heavy machinery.
| https://mathoverflow.net/users/478205 | Looking for clarification of C-H Sah's definition of abstract scissors congruence | I agree that Sah's definition is not quite correct, but it's not difficult to fix. The key point is this: when you have an intersection that contains $0$ simplices, you declare it to be empty. If you look at the definition of an assembler (which was partially designed to fix Sah's definition) it uses the categorical structure of the assembler to keep track of this. The point of axiom (S) here is that "you can always refine a subdivision." If you have two different subdivisions of a polytope, you can find a common refinement; if you have a subpolytope of a polytope and a subdivision of the large one, then you can refine the subdivision so that it restricts to a subdivision of the smaller one.
To fix the definition you need to interpret intersection as "take the union of all simplices contained in the intersection." So in both cases that you're worried about, the intersection is actually empty and everything holds automatically. This is actually the point of the definition of an assembler: since you take the "intersection" inside the assembler (the categorical definition of pullback) you automatically disregard any non-valid sets which are contained in the intersection.
Also: quick explanation of what an assembler is. It's the partial order on valid simplices. The "covering families" inside the Grothendieck topology are exactly Sah's families of covers. Axiom (R) says any two covers have a common subcover. Axiom (E) says that the empty set can be covered by nothing. Disjointness says that there are no valid simplices contained in the intersection of two simplices. That's pretty much it. The goal is to fix Sah's definition and to also allow it to be analyzed by $K$-theoretic means, but the idea is strongly inspired by Sah.
| 1 | https://mathoverflow.net/users/1378 | 418162 | 170,286 |
https://mathoverflow.net/questions/418178 | 7 | Here we choose the definition of "is a cardinal" as there is no surjective map from a smaller ordinal to it.
It's easy to prove that, if $L\_{\alpha+1} \vDash\ \alpha\text{ is inaccessible}$, then $L\_\alpha \vDash ZFC$. Also it's easy to prove that if $L\_\alpha \vDash ZFC$, then α is a cardinal in $L\_{\alpha+1}$. However, I don't know how to prove confinally many cardinals won't collapse at the next step of L.
If the answer is false, then is it true at the least such α?
| https://mathoverflow.net/users/170286 | If $L_\alpha \vDash ZFC$, then do we have $L_{\alpha+1} \vDash \alpha\text{ is inaccessible}$? | Yes. The elements of $L\_{\alpha+1}$ are exactly those subsets of $L\_\alpha$ which are definable from parameters over $L\_\alpha$. But $L\_\alpha\models\mathrm{ZFC}$, so from here we can just use the usual proof that second order ZFC implies inaccessibility. That is:
(i) every bounded subset of $L\_\alpha$ which is in $L\_{\alpha+1}$, is already in $L\_\alpha$, and so $L\_{\alpha+1}$ agrees with $L\_\alpha$ regarding power sets and cardinalities of elements of $L\_\alpha$; that is, for each $x\in L\_\alpha$, we have $\mathcal{P}(x)^{L\_{\alpha+1}}=\mathcal{P}(x)^{L\_\alpha}$, and $\mathrm{card}(x)^{L\_{\alpha+1}}=\mathrm{card}(x)^{L\_\alpha}$; in particular, $L\_{\alpha+1}\models$"$\alpha$ is a strong limit cardinal", and
(ii) for every $\beta<\alpha$ and every $f:\beta\to\alpha$ which is in $L\_{\alpha+1}$, $f$ is bounded in $\alpha$; that is, $L\_{\alpha+1}\models$"$\alpha$ is regular".
| 14 | https://mathoverflow.net/users/160347 | 418186 | 170,294 |
https://mathoverflow.net/questions/418194 | 0 | Here $B\_r$ presents the open ball of radius r in $R^3$. So I hope to know how to prove the following inequality.
$$
\int\_{B\_r} |u|\le Cr^{\frac{9}{5}}\left(\int\_{B\_r} |u|^2\right)^{\frac{1}{5}}\left(\int\_{B\_r} |u|^3\right)^{\frac{1}{5}}
$$
I did not figure out where these exponents come from. Could anyone show me?
| https://mathoverflow.net/users/295572 | Using Hölder's Inequality to prove the following equation | I think that I have got the answer… it is quite a simple result from the Hölder inequality,
$u^{\frac{2}{5}}u^{\frac{3}{5}}1$ with exponents $\frac{1}{5}$, $\frac{1}{5}$ and $\frac{3}{5}$….
| 0 | https://mathoverflow.net/users/295572 | 418195 | 170,297 |
https://mathoverflow.net/questions/418185 | 4 | Let $M$ be a Ricci-flat Riemannian manifold and $N \subset M$ a totally geodesic submanifold. Is $N$ also Ricci-flat?
A partial result in that direction is that the Ricci curvature of $N$ is given by
$$\operatorname{Ric}^N(Y, Z) = \operatorname{tr}(TN \ni X \mapsto R(X, Y)Z \in TN),$$
where $R$ is the Riemmanian curvature tensor of $M$. So $\operatorname{Ric}^N(Y, Z)$ is the trace of the restriction of $X \mapsto R(X, Y)Z$ to $TN$. But the restriction of a trace-free map is not necessarily trace-free.
| https://mathoverflow.net/users/409915 | Ricci curvature of totally geodesic submanifold | You can find an explicit counterexample in the Riemannian Schwarzschild solution.
Let $(z,r,\omega) \in \mathbb{R} \times (1,\infty) \times \mathbb{S}^2$, denote by $h$ the standard sphere metric. Consider the following Riemannian metric
$$ ds^2 = (1- r^{-1})~dz^2 + (1-r^{-1})^{-1} ~dr^2 + r^2 h $$
One can explicitly compute that this metric is Ricci-flat using:
* The metric is a warped product of $\mathbb{R}\times (1,\infty)$ against the sphere,
* Exercise 5.5 from O'Neill's *Semi-Riemannian Geometry*
* Standard formulae for the Gauss curvature of a diagonal metric in two dimensions.
As a warped product, fixing any $\omega\in \mathbb{S}^2$, the submanifold $\mathbb{R}\times (1,\infty) \times \{\omega\}$ is totally geodesic, but it has non-vanishing Gauss curvature $K = \frac{1}{r^3}$ and so is not Ricci flat.
(I think you also have a co-dimension one example if you look at a $z$-level set, which is totally geodesic due to the reflection symmetry in $z$. I am pretty sure that this slice also has non-vanishing Ricci curvature.)
Because any $z$-level set is asymptotically flat, if such a set were not Ricci flat then Bishop-Gromov comparison would imply that the level set is isometric to Euclidean space. This is clearly false, as can be seen by, say, calculating the Riemann curvature, or by computing the ADM mass of the level set.
| 8 | https://mathoverflow.net/users/3948 | 418198 | 170,299 |
https://mathoverflow.net/questions/418193 | 7 | Indirect method (associated with a certain problem of electrostatics) indicates that $$\sum\limits\_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!}=\frac{2}{3\pi}.$$ Is this result known?
| https://mathoverflow.net/users/32389 | Infinite series for $1/\pi$. Is it known? | Using the [standard power series for the complete elliptic integral of the second kind](https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_second_kind) $$E(k) = \frac{\pi}{2} \sum\_{j=0}^\infty \left(\frac{(2j)!}{2^{2j}(j!)^2}\right)^2 \frac{k^{2j}}{1-2j},$$
we find
\begin{align\*}
\sum\limits\_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!} k^{2j}&=\sum\_{j=1}^\infty\frac{-j}{j+1} \left(\frac{(2j)!}{2^{2j}(j!)^2}\right)^2 \frac{k^{2j}}{1-2j} \\
&= -\frac{1}{k^2} \int\_0^k\mathrm{d}k\,k^2 \frac{\mathrm{d}}{\mathrm{d}k}\left(\frac{2}{\pi}E(k)\right)\\
&= \frac{2}{3}\frac{k^2-1}{k^2}\frac{2}{\pi}K(k) - \frac{k^2-2}{3k^2}\frac{2}{\pi}E(k).
\end{align\*}
In the limit $k\to 1$ only the second term survives with $E(1)=1$ and therefore
\begin{align\*}
\sum\limits\_{j=1}^\infty \frac{(2j-3)!!\,(2j-1)!!}{(2j-2)!!\,(2j+2)!!} &=\frac{2}{3\pi}.
\end{align\*}
| 19 | https://mathoverflow.net/users/47484 | 418199 | 170,300 |
https://mathoverflow.net/questions/418180 | 3 | For $\varepsilon > 0$, we say a function $f: \mathbb R^n \to \mathbb R^m$ is (pointwise) $(1 - \varepsilon)$-Hölder continuous at $x \in \mathbb R^n$ if
$$ \lim\_{ y \to 0} \frac{f(x + y) - f(x)}{\lVert y \rVert^{1 - \varepsilon}} = 0.$$
**Question:** Suppose $f$ is $(1 - \varepsilon)$-Hölder continuous at all $x \in \mathbb R^n$ for every $0 < \varepsilon < 1$. Is it true that $f$ is differentiable everywhere?
| https://mathoverflow.net/users/173490 | Pointwise Hölder continuity of order $1-\varepsilon$ | One counter example is the Takagi function $\tau$, see Theorems 8.1 and 8.2 in the survey [2] (As noted yesterday at [4]).
More detail: Takagi [1] showed that his function $\tau$ is nowhere differentiable. On the other hand, It is easy to verify that for $0 \le x <x+h\le 1$, we have
$$|\tau(x+h)-\tau(x)| \le 2 h\log\_2(h)\,,$$
so $\tau$ is Holder continuous of order $\alpha$ for all $\alpha \in (0,1)$.
See Theorem 8.1 page 15 in [2] (For refinements, see Kono [3].)
[1] T. Takagi, A simple example of the continuous function without derivative, from Proceedings of the Physico-Mathematical Society of Japan, ser II, Vol 1. 1903, pp 176-177.
[Collected Papers of Teiji Takagi (S. Iyanaga, Ed), Springer Verlag, New York 1990].
[2] <https://arxiv.org/pdf/1112.4205.pdf>
[3] N. Kono, On generalized Takagi functions, Acta Math. Hung. 49 (1987), No 3-4, 315–324.
[4] <https://www.facebook.com/groups/1923323131245618/posts/3414086925502557/?comment_id=3414118705499379&reply_comment_id=3414119175499332¬if_id=1647334661196681¬if_t=group_comment&ref=notif> )
| 7 | https://mathoverflow.net/users/7691 | 418212 | 170,302 |
https://mathoverflow.net/questions/418211 | 7 | Is it true that every group quasi-isometric to the Heisenberg group admits a proper cocompact action by isometries on Nil?
| https://mathoverflow.net/users/159356 | Discrete cocompact group of isometries of Nil | This is true. The first step is that the group $\Gamma$ then has a subgroup of finite index that embeds as a lattice in the Heisenberg group $H$. This step is sketched [in this answer](https://mathoverflow.net/a/228717/14094) (I repeated the argument here below "original answer"). To deduce the general case, let me relax assumptions.$\DeclareMathOperator\Aut{Aut}$
>
> **Proposition.** Let $\Gamma$ be a discrete group with a finite index subgroup $\Lambda$ that embeds as a lattice into a simply connected nilpotent Lie group $G$. Let $K$ be a maximal compact subgroup of $\Aut(G)$. Let $W$ be the finite radical of $\Gamma$ (= the maximal finite normal subgroup, which exists here). Then $\Gamma/W$ embeds as a lattice into $K\ltimes G$.
>
>
>
Proof: first step: there exists a Lie group $H$ of the form $H^0\rtimes L$ with $L$ finite, $H^0$ simply connected nilpotent, in which $\Gamma$ maps as a lattice with finite kernel.
Fix an embedding of $G$ into the upper unipotent subgroup of $\mathrm{GL}\_n(\mathbf{R}$ for some $n$. This defines a representation of $\Lambda$. Induce it to $\Gamma$. We obtain a representation $f$ of $\Gamma$, for which $\Lambda$ acts unipotently. Then the Zariski closure of $f(\Lambda)$ is a simply connected nilpotent Lie group in which $f(\Lambda)$ is a lattice. Let $H$ be the Zariski closure of $f(\Gamma)$. Then $H^0$ is the Zariski closure of $f(\Lambda)$, and has finite index in $H$. Let $L$ be a maximal compact subgroup of $H$. Then $LH^0=H$ (this is true for every virtually connected Lie group, result of Mostow). Since $L\cap H^0=1$, we deduce $H=H^0\rtimes L$ and $L$ is finite. So the first step is proved.
Now conclude. Consider $H=H^0\rtimes L$ as in the first step. Without loss of generality, we can suppose that $L$ acts faithfully on $H$ (otherwise, mod out by the kernel). So $L\subset\Aut(H)$. Hence $L$ is contained in a maximal compact subgroup of $\Aut(H)$. We are now done (the assumption then now force the kernel to be exactly the finite radical of $\Gamma$).
>
> **Corollary.** There exists a left-invariant Riemannian metric on $G$ and a geometric (=proper isometric cocompact) action of $\Gamma$ on $G$.
>
>
>
Proof: choose a left-$G$-invariant right-$K$-invariant Riemannian metric on $G$. Then the action of $G\ltimes K$ on $G$ is geometric (=continuous proper isometric cocompact), and hence so is that of $\Gamma$.
---
Original answer (which passes to a finite index subgroup)
It (essentially) directly follows from Gromov's theorem that a f.g. group quasi-isometric to $\mathsf{Nil}$ is virtually isomorphic to the integral Heisenberg group. (No need to assume the existence of an action.)
Indeed, such a group has polynomial growth, so is virtually nilpotent and hence virtually isomorphic to a lattice in some simply connected nilpotent Lie group $G$.
Once we know that $G$ is the 3-dimensional Heisenberg group, one is done since every lattice therein contains with finite index a copy of the integral Heisenberg group.
One can directly invoke Pansu's 1988 theorem to deduce that the associated Carnot Lie algebra of $G$ is isomorphic to the Heisenberg group, and hence $G$ itself as well (since $G$ is 2-step-nilpotent it is Carnot).
Alternatively, one can avoid Pansu's theorem, saying that $G$ has growth $n^4$ which leaves only 2 candidates: $\mathbf{R}^4$ and the Heisenberg group. So one just has to know that they are not quasi-isometric and this follows form various arguments, e.g., using asymptotic cones (which is homeomorphic to $\mathbf{R}^4$ for $\mathbf{R}^4$ and to $\mathbf{R}^3$ for the other).
| 7 | https://mathoverflow.net/users/14094 | 418214 | 170,303 |
https://mathoverflow.net/questions/418208 | 4 | Some statements that are true for ordinary groupoids fail for topological groupoids (by which I mean groupoids internal to the category of topological spaces): for instance, every ordinary groupoid is equivalent to the disjoint union of one-object groupoids, but [not every topological groupoid is equivalent to a disjoint union of topological groups](https://math.stackexchange.com/questions/2954664/is-every-topological-groupoid-equivalent-to-a-disjoint-union-of-topological-grou).
Thinking about a different statement that is true for ordinary groupoids (each ordinary groupoid is equivalent to a skeletal groupoid) and [Butz-Moerdijk's paper](http://citeseer.ist.psu.edu/viewdoc/download;jsessionid=756C6C16BD72D7117B5F1AAA6D1679D1?doi=10.1.1.12.6700&rep=rep1&type=pdf "Carsten Butz and Ieke Moerdijk: Representing topoi by topological groupoids") has led me to the following questions:
1. Is every topological groupoid equivalent to a skeletal topological groupoid?
For the record, a *functor* between two topological groupoids $\mathcal C$ and $\mathcal D$ is an ordinary functor $$(\mathcal F\_0\colon \rm Ob(\mathcal C)\to \rm Ob(\mathcal D), \mathcal F\_1\colon \rm Mor(\mathcal C)\to\rm Mor(\mathcal D))$$ between the underlying ordinary groupoids such that $\mathcal F\_1$ and $\mathcal F\_2$ are continuous. A *natural transformation* between two functors $\mathcal F, \mathcal G\colon \mathcal C\to\mathcal D$ of topological groupoids is a natural transformation $$\eta\colon \rm Ob(\mathcal C)\to\rm Mor(\mathcal D)$$ between the underlying ordinary functors such that $\eta$ is continuous. We say $\eta$ is a *natural isomorphism* if there is a natural transformation $\epsilon\colon \mathcal G\to\mathcal F$ such that for all $X\in\rm Ob(\mathcal C)$, $\eta\_X\circ \epsilon\_X=\rm id$ and $\epsilon\_X\circ \eta\_X=\rm id$. Two topological groupoids $\mathcal C$ and $\mathcal D$ are said to *equivalent* if there are functors $\mathcal F\colon\mathcal C\to\mathcal D$ and $\mathcal G\colon\mathcal D\to\mathcal C$ such that $\mathcal F\circ \mathcal G$ and $\mathcal G\circ\mathcal F$ are naturally isomorphic to the identity functors.
2. It feels to me the following condition could be relevant in (1): say that two points $x$ and $y$ of a topological space $X$ are *topologically equivalent* if for each open set $U\subseteq X$, $x\in U$ if and only if $y\in U$.
If a topological groupoid is equivalent to a skeletal topological groupoid, are isomorphic objects topologically equivalent?
(Side questions: Is there a name in the literature for "topologically equivalent" points? Has the condition "all topologically equivalent points are isomorphic" any relevance in the study of topological groupoids?)
3. Suppose the anwer to the first question in (2) is "yes". Let $\mathcal C$ and $\mathcal D$ be topological groupoids in which isomorphic objects are topologically equivalent. Are $\mathcal C$ and $\mathcal D$ equivalent if and only if their skeletons are *isomorphic* as topological groupoids?
4. For a topological groupoid $\mathcal C$ is there a canonical site $(C\_\mathcal C, J\_\mathcal C)$ such that the category of equivariant sheaves on $\mathcal C$ is equivalent to the category of sheaves on $(C\_\mathcal C, J\_\mathcal C)$?
If two topological groupoids $\mathcal C$ and $\mathcal D$ are equivalent, does it follow that the sites $(C\_\mathcal C, J\_\mathcal C)$ and $(C\_\mathcal D, J\_\mathcal D)$ are equivalent? Does the converse hold?
5. If $\mathcal E$ and $\mathcal E'$ are equivalent topoi, is the Butz-Moerdijk groupoid associated to $\mathcal E$ *equivalent* or *isomorphic* to the Butz-Moerdijk groupoid associated to $\mathcal E'$, as topological groupoids?
**Remark on how to answer.** I feel a bit guilty for writing such an overwhelming amount of questions. (1)-(3) have mainly the purpose to make sure that there aren't any pitfalls or phenomena that occur in the topological groupoid world that don't occur in the ordinary groupoid world (it seems this is a topic one could easily make mistakes). So for these questions a simple yes/no would totally suffice - I can work out the proofs as an exercise.
| https://mathoverflow.net/users/478652 | Topological groupoids and equivariant sheaves | First note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use. It is really too much to ask for a continuous quasi-inverse functor. Moerdijk has papers with the "right" notion of Morita equivalence of topological groupoids.
Here is an answer to question 1. Let $X$ be a topological space and let $R$ be an equivalence relation. Then you get a topological groupoid $\mathcal G$ with object space $X$ and arrow space $R$ with multiplication $(x,y)(y,z) = (x,z)$ for $(x,y), (y,z)\in R$. Here $y$ is the domain of $(x,y)$ and $x$ the range. I claim this groupoid is equivalent to a skeletal groupoid in your sense if and only if the quotient map $\pi\colon X\to X/R$ admits a continuous section. Such a section need not exist, e.g., the quotient map from the cantor space to the interval $[0,1]$.
Indeed, if $s\colon X/R\to X$ is a continuous section, then the subcategory with objects $s(X/R)$ and arrows the identities at these objects is a skeletal subcategory equivalent to $\mathcal G$. The functor sends $x$ to $s(\pi(x))$ on objects and $(x,y)\mapsto (s\pi(x),s\pi(y))$ on arrows. The natural isomorphism is given by $\eta\_x = (s\pi(x),x)$.
Conversely, any skeletal groupoid equivalent to $\mathcal G$ in your sense would have to be a groupoid with only identities, so essentially a space $Y$. Moreover, the functor $F\colon \mathcal G\to Y$ would have to send objects in the same $R$-class to the same element of $Y$ so would induce a continuous bijection $f\colon X/R\to Y$. Moreover, the "inverse" functor $F'\colon Y\to \mathcal G$ would have image a collection of objects making a cross section to $R$ and $F'\circ f\circ \pi$ would be a continuous section of $\pi$.
| 6 | https://mathoverflow.net/users/15934 | 418216 | 170,304 |
https://mathoverflow.net/questions/418217 | 3 | Let $C \subset \mathbb P^3$ be a smooth complete intersection curve given by $2$ hypersurfaces of degree at least $4$ in $\mathbb P^3$. Then can it happen that:
1. $C$ is a $2$-sheeted covering of a curve of low genus (say $\leq 3$)?
2. $C$ is a $3$-sheeted covering of an elliptic curve?
Note that, in both the cases by Riemann–Hurwitz theorem we get a value of the degree of the ramification divisor. So the question boils down to whether there really exist ramification divisors having degrees those values.
Are there any other ways to give a bound on the degree of the ramification divisor in these two situations which contradicts the previously obtained value?
Any indication on whether this can happen (or can't happen) is appreciated.
| https://mathoverflow.net/users/474635 | On $k$-sheeted covering of curves | **Edit:** As pointed out by Sasha, my original argument was not complete for case (2). One can argue as follows: suppose $C$ is a $(p,q)$ complete intersection, with $p\geq q$. The maximum number $\ell$ of points of $C$ lying on a line is $\leq p\ $ — otherwise the line is contained in $C$.
By B. Basili, *Indice de Clifford des intersections complètes de l’espace*, Bull. S. M. F. **124**, no. 1 (1996), p. 61-95, the gonality of $C$ is $pq-\ell$, hence $\geq p(q-1)$. But if $C$ is a 2-sheeted cover of a curve $D$ of genus $\leq 3$, a $g^{1}\_3$ on $D$ lifts to a $g^1\_6$ on $C$; similarly if $C$ is a 3-sheeted cover of
an elliptic (or more generally hyperelliptic) curve $E$, a $g^1\_2$ on $E$ lifts to a $g^1\_6$ on $C$. Hence the gonality of $C$ is $\leq 6$, which is less than $p(q-1)$ if $q\geq 4$.
| 6 | https://mathoverflow.net/users/40297 | 418219 | 170,305 |
https://mathoverflow.net/questions/323708 | 6 | Let $(V,\*)$ be an algebra and denote $A\_\*\in \text{Hom}(V^{\otimes 3},V)$ the associator of the binary product $\*\in \text{Hom}(V^{\otimes 2},V)$ defined as $A\_\*(a,b,c):=(a\*b)\*c-a\*(b\*c)$.
The associator $A\_\*$ is assumed to enjoy the following property:
$A\_\*(a,b,c)+A\_\*(b,c,a)-A\_\*(b,a,c)=0$.
**Question**: Does this "weak associativity" condition have a name and are there some references discussing it?
| https://mathoverflow.net/users/104743 | Weak associativity | Let me assume that the characteristic of the ground field is different from two. Let me start by replacing your identity by something where the existing symmetries are a bit more apparent. I claim that your identity for the associator (which I will denote simply $(a,b,c)$, dropping $A\_\*$) is equivalent to the following two identities:
$$
\begin{gather}
(a,b,c)+(b,c,a)+(c,a,b)=0,\\
(a,b,c)+(c,b,a)=0.
\end{gather}
$$
First, they clearly imply your identity: $$0=(a,b,c)+(b,c,a)+(c,a,b)=(a,b,c)+(b,c,a)-(b,a,c).$$ Second, your identity, if we set $a=b=c$, becomes third power associativity $(a,a,a)=0$, and multilinearizing that, we get
$$
(a,b,c)+(a,c,b)+(b,c,a)+(b,a,c)+(c,a,b)+(c,b,a)=0.
$$
Using your identity $(a,b,c)+(b,c,a)=(b,a,c)$ and the identity $(a,c,b)+(c,b,a)=(c,a,b)$ obtained by acting by the transposition $b\leftrightarrow c$, we obtain the identity $(b,a,c)+(c,a,b)=0$, and we use the same calculation as before:
$$0=(a,b,c)+(b,c,a)-(b,a,c)=(a,b,c)+(b,c,a)+(c,a,b).$$
These two identities, in turn, are manifestly equivalent to the system of identities
$$
\begin{gather}
(a,b,c)+(a,c,b)+(b,c,a)+(b,a,c)+(c,a,b)+(c,b,a)=0,\\
(a,b,c)+(c,b,a)=0.
\end{gather}
$$
The first of them defines Lie admissible algebras, and the second defines flexible algebras. Search for papers that include "flexible Lie-admissible algebras" in the title brings 15 matches on MathSciNet, including, for instance, [Benkart, Georgia M.; Osborn, J. Marshall. Flexible Lie-admissible algebras.
J. Algebra 71 (1981), no. 1, 11–31.](https://www.sciencedirect.com/science/article/pii/0021869381901034) I understand that this class of algebras was studied because it contains the so called "Okubo algebra" arising in one of the constructions of octonions.
| 2 | https://mathoverflow.net/users/1306 | 418227 | 170,308 |
https://mathoverflow.net/questions/418159 | 1 | I am a PhD student. During my researches, I often have to deal with inequalities involving sums of binomial coefficients, where the sums are indexed by some set of integer compositions. For example, let $l,k$ be positive integers such that $l \leq k$ and either $l$ is odd or both $l$ and $k$ are even. One of my results can be proven if I show that for any $i \leq k \lfloor \frac{l}{2}\rfloor - 1$:
$$
\displaystyle \sum\_{\substack{\alpha \models i+k} \\ \ell(\alpha) = k \\ \alpha\_a \leq \lfloor \frac{l}{2} \rfloor + 1} \prod\_{a=1}^k \binom{l}{2(\alpha\_a - 1)}
\leq
\displaystyle \sum\_{\substack{\alpha \models i+l} \\ \ell(\alpha) = l \\ \alpha\_a \leq \lfloor \frac{k}{2} \rfloor + 1} \prod\_{a=1}^l \binom{k}{2(\alpha\_a - 1)},
$$
where the notations under the left sum means that we take the sum over all $\alpha = (\alpha\_1,...,\alpha\_k) \subseteq \{1,...,\lfloor \frac{l}{2} \rfloor + 1\}^k$ such that $\alpha\_1 + ... + \alpha\_k = k+i$, and analogously for the other.
My question here is not about solving this problem in particular (although if the solution is easy feel free to give it).
My question is rather: how do you work with such sums? I am not used to these kinds of formulas, and while it is manageable when $l$ and $k$ are small, they become hard very rapidly. So, is there a trick to understand these kind of sums? Is there some already known results? Is there formulas? Combinatorial arguments?
| https://mathoverflow.net/users/73667 | Inequalities between sums of products of certain binomial coefficients | Take the LHS:
$$\sum\_{\substack{\alpha \models i+k} \\ \ell(\alpha) = k \\ \alpha\_a \leq \lfloor \frac{l}{2} \rfloor} \prod\_{a=1}^k \binom{l}{2(\alpha\_a - 1)}$$
Firstly, we can simplifying by rolling the subtraction of one into the definition of $\alpha$:
$$\sum\_{\substack{\sum \alpha\_a = i \\ \ell(\alpha) = k \\ 0 \leq \alpha\_a < \lfloor \frac{l}{2} \rfloor}} \prod\_{a=1}^k \binom{l}{2\alpha\_a}$$
The constraint on the sum of the $\alpha\_a$ can be done with coefficient extraction:
$$\begin{eqnarray\*}
& [x^{2i}] \sum\_{\substack{\ell(\alpha) = k \\ 0 \leq \alpha\_a < \lfloor \frac{l}{2} \rfloor}} \prod\_{a=1}^k \binom{l}{2\alpha\_a} x^{2\alpha\_a} \\
=& [x^{2i}] \left( \sum\_{j=0}^{\lfloor \frac{l}{2} \rfloor - 1} \binom{l}{2\alpha\_a} x^{2\alpha\_a} \right)^k \\
=& [x^{2i}] \left( \frac{ (1+x)^l + (1-x)^l }2 - \binom{l}{l - l \bmod 2}x^{l - l \bmod 2} \right)^k
\end{eqnarray\*}$$
One (untested) idea to show that the coefficients of the polynomial corresponding to the LHS don't exceed those of the RHS would be to take the $2i^{\textrm{th}}$ derivative and evaluate it at zero.
| 2 | https://mathoverflow.net/users/46140 | 418230 | 170,309 |
https://mathoverflow.net/questions/418228 | 8 | In page-479 of Visual Complex Analysis, Tristan Needham derives the flux of a vector field in Geometric form:
$$ \nabla \cdot X = \partial\_s |X| + \kappa\_p |X|$$
The $\partial\_s$ is a derivative along streamlines of the vector field $X$ an $\kappa\_p$ is the curvature of the orthogonal streamline at the point we are taking divergence at.
Where did this formula originate from, and, what other sources discuss it? There is nothing of it on wikipedia. The book itself claims it couldn't find any reference to the formula in Modern literature, but is it still true in 2022 ?
| https://mathoverflow.net/users/159957 | Geometric definition of divergence using curvature mentioned in Tristan Needham | Given any hypersurface of a Riemannian manifold with a unit normal vector field $\nu$, extend $\nu$ to be unit length. Then
$$\operatorname{div}(u\nu)=u\operatorname{div}\nu+\text{d}u(\nu).$$
This proves that if $X$ is any nonvanishing vector field on a neighborhood of the hypersurface, which is orthogonal to the hypersurface, then
$$\operatorname{div}X=|X|H+\text{d}|X|(\nu)$$
where $H$ is the mean curvature. And so, when only given a Riemannian manifold, this holds also when $X$ is a hypersurface-orthogonal vector field (also called "complex lamellar vector field"); $H$ is the mean curvature of the orthogonal hypersurface and $\nu$ is $\frac{X}{|X|}$. Every vector field on a two-dimensional space is hypersurface-orthogonal, so this fully contains Needham's formula as a special case.
| 10 | https://mathoverflow.net/users/156492 | 418232 | 170,310 |
https://mathoverflow.net/questions/418144 | 3 | The Hom scheme of two projective varieties over some field is constructed as an open subfunctor of the Hilbert scheme of the product of the two schemes by Grothendieck. So it is a countable union of quasi-projective varieties. For example even in the simplest case $\mathcal{Hom}(\mathbb{P}\_k^1, \mathbb{P}\_k^1)$ there seems to be infinitely many components. So my question is threefold:
1. Is there some general condition that ensures it is a finite union of quasi-projective varieties?
2. Is there some general condition that ensures each component in the countable union is a projective variety?
3. For what type of morphisms $f: X\rightarrow Y$ can one say the map $\mathcal{Hom}(Z,X)\rightarrow \mathcal{Hom}(Z,Y)$ is a proper morphism on the connected components? I know that if $f$ is a closed immersion, the induced map is also closed. I was speculating that if $f$ is finite then the induced map is going to be proper, is something like this true?
| https://mathoverflow.net/users/127776 | When Hom scheme has projective components? | Let $k$ be an algebraically closed field of characteristic zero and let $X$ be a projective variety over $k$. To try and answer your first two questions, let us try formalize the property that Hom-schemes $\mathrm{Hom}(Y,X)$ are finite unions of quasi-projective schemes.
**Definition.** We say that $X$ is *bounded over $k$* if, for every normal projective variety $Y$ over $k$, the Hom-scheme $\mathrm{Hom}(Y,X)$ is of finite type over $k$.
We can then show the following result; see [1] and [2].
(I think that your question is only concerned with 1 and 6 in the following result, but the rest might also be useful to you.)
**Theorem.** The following are equivalent.
1. $X$ is bounded over $k$.
2. For every algebraically closed field $L$ containing $k$, the projective variety $X\_L$ is bounded over $L$.
3. For every smooth projective curve $C$ over $k$, the Hom-scheme $Hom(C,X)$ is of finite type.
4. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ has only finitely many connected components.
5. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ is quasi-projective.
6. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ is a projective scheme over $k$.
7. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ is a projective scheme over $k$. The subscheme of non-connected morphisms $Hom^{nc}(Y,X)$ is a **bounded** projective scheme of dimension $<\dim X$ which maps finitely to $X$.
Boundedness is most likely equivalent to "hyperbolicity". In fact, it is implied by hyperbolicity (Kobayashi, Brody, ...). More precisely:
>
>
> >
> > Theorem. Assume $k=\mathbb{C}$. If $X$ is hyperbolic (i.e., every holomorphic map $\mathbb{C}\to X^{an}$ is constant), then $X$ is bounded.
> >
> >
> >
>
>
>
>
>
> >
> > Conjecture. Assume $k=\mathbb{C}$. If $X$ is bounded over $\mathbb{C}$, then $X$ is hyperbolic.
> >
> >
> >
>
>
>
This should answer your first question. For your second question: if $X$ has no rational curves, then every Hom-scheme $Hom(Y,X)$ has projective components (but it can have infinitely many components). Conversely, if all Hom-schemes $Hom(Y,X)$ have (only) projective components, then $X$ has no rational curves. (Use that the components of $Hom(\mathbb{P}^1,\mathbb{P}^1)$ are affine varieties of increasing dimension.)
Abelian varieties have no rational curves and give examples of non-finite type Hom-schemes with each component projective. Note that non-trivial abelian varieties are far from being hyperbolic, so there's no contradiction to the above conjecture.
References.
[1] R. van Bommel, A. Javanpeykar, L. Kamenova.
*Boundedness in families with applications to arithmetic hyperbolicity*
<https://arxiv.org/abs/1907.11225>
[2] A. Javanpeykar and L. Kamenova
*Demailly's notion of algebraic hyperbolicity: geometricity, boundedness, moduli of maps*
<https://arxiv.org/abs/1807.03665>
| 3 | https://mathoverflow.net/users/4333 | 418236 | 170,311 |
https://mathoverflow.net/questions/418222 | 4 | Let $x$ be a $n$-dimensional Gaussian random vector, i.e., $x \sim \mathcal{N}(0,\sigma^2 I\_n)$. What is its probability of falling in a cone? Say a cone $C = \{ x \in \mathbb R^n: \frac{\langle x - v, -v \rangle}{\|x-v\|\_2 \|v\|\_2} \leq \cos \theta \}$ is parameterized by $v \in \mathbb R^n$ and $\theta \in (0,\pi / 2)$. What is $\mathbb P(x \in C)$? Is there a closed form?
| https://mathoverflow.net/users/478836 | Probability of a Gaussian random vector in a cone | Let
\begin{equation\*}
Z=(Z\_1,\dots,Z\_n):=x/\sigma\sim N(0,I\_n),
\end{equation\*}
\begin{equation\*}
Y\_n:=Z\_2^2+\dots+Z\_n^2\sim\chi^2\_{n-1},
\end{equation\*}
\begin{equation\*}
c:=\|v\|\_2/\sigma>0,\quad t:=\cos\theta\in(0,1),\quad u:=\frac t{\sqrt{1-t^2}}=\cot\theta\in(0,\infty).
\end{equation\*}
By the spherical symmetry, without loss of generality $v=c\sigma(1,0,\dots,0)$. So,
\begin{equation\*}
\begin{aligned}
P(x\in C)&=1-P(c-Z\_1>t\sqrt{(Z\_1-c)^2+Y\_n}) \\
&=1-P(Z\_1<c-u\sqrt Y\_n).
\end{aligned}
\tag{1}\label{1}
\end{equation\*}
Note that the random variables (r.v.'s) $Z\_1$ and $Y\_n$ are independent. So,
\begin{equation\*}
P(x\in C)=1
-\frac{2^{(1-n)/2}} {\Gamma ((n-1)/2)}
\int\_0^\infty \Phi(c-u\sqrt y)
e^{-y/2} y^{(n-3)/2}\,dy, \tag{2}\label{2}
\end{equation\*}
where $\Phi$ is the standard normal cdf.
Mathematica cannot do anything with the latter integral. So, it is unlikely that it can be expressed in closed form.
---
However, using \eqref{1} or \eqref{2}, one can easily find various approximations to $P(x\in C)$, depending on how $n,c,u$ vary.
For instance, suppose that $u$ is fixed and $n\to\infty$. Then, by the central limit theorem and the [delta method](https://en.wikipedia.org/wiki/Delta_method#Univariate_delta_method),
\begin{equation}
V\_n:=\sqrt2\,(\sqrt Y\_n-\sqrt{n-2})\to V
\end{equation}
in distribution, where $V$ is a standard normal r.v., which let us choose to be independent of $Z\_1$. Then, by \eqref{1},
\begin{equation\*}
\begin{aligned}
P(x\in C)&=P(Z\_1\ge c-u\sqrt Y\_n) \\
&=P(Z\_1+\tfrac u{\sqrt2}\,V\_n\ge c-u\sqrt{n-2}) \\
&\to P(Z\_1+\tfrac u{\sqrt2}\,V\ge c\_0) \\
&=1-\Phi\Big(\frac{c\_0}{\sqrt{1+u^2/2}}\Big)
\end{aligned}
\tag{3}\label{3}
\end{equation\*}
if $c$ varies with $n$ so that $c-u\sqrt{n-2}$ converges to some real $c\_0$. Similarly, if $u$ and $c$ are fixed whereas $n\to\infty$, then $P(x\in C)\to1$.
Of course, one can also use various asymptotic expansions to obtain more detailed asymptotics.
---
In fact, we have a lower bound on $P(x\in C)$ that is
essentially the same as the limit in \eqref{3}:
\begin{equation\*}
\begin{aligned}
P(x\in C)
&=P(Z\_1+\tfrac u{\sqrt2}\,V\_n\ge c-u\sqrt{n-2}) \\
&>P(Z\_1+\tfrac u{\sqrt2}\,V\ge c-u\sqrt{n-2}) \\
&=1-\Phi\Big(\frac{c-u\sqrt{n-2}}{\sqrt{1+u^2/2}}\Big).
\end{aligned}
\tag{4}\label{4}
\end{equation\*}
The inequality in \eqref{4} follows by [formula (2.6)](https://projecteuclid.org/journals/annals-of-statistics/volume-22/issue-1/Extremal-Probabilistic-Problems-and-Hotellings-T2-Test-Under-a-Symmetry/10.1214/aos/1176325373.full), which means that $V\_n$ is strictly stochastically greater than $V$.
| 4 | https://mathoverflow.net/users/36721 | 418237 | 170,312 |
https://mathoverflow.net/questions/417637 | 1 | Consider the sparse linear regression model $y=X\theta^\*+w$, where $w\sim N(0, \sigma^2 I\_{n\times n})$ and $\theta^\*\in R^d$ is supported on a subset $S$. Suppose that the sample covariance matrix $\hat{\Sigma}=X^TX/n$ has its diagonal entries uniformly upper bounded by 1, and that for some parameter $\gamma>0$, it also satisfies an $\ell\_{\infty}$-curvature condition of the from
$$\|\hat{\Sigma}\Delta\|\_\infty\ge \gamma\|\Delta\|\_\infty, \, \Delta\in C\_3(S)
$$
where $C\_3(S):=\{\Delta\in R^d: \|\Delta\_{S^c}\|\_1\le 3 \|\Delta\_{S}\|\_1\}$
| https://mathoverflow.net/users/168083 | Lasso of sparse linear regression model | The proof you are looking for should be in
>
> Theorem 1 of *Sup-norm convergence rate and sign concentration property of Lasso and Dantzig estimators* by Karim Lounici (<https://arxiv.org/abs/0801.4610>).
>
>
>
Check the relationship of your curvature condition with Assumption 2 of that paper. The proof there should give you the main ideas, in particular the standard argument to prove that $\hat \theta-\theta^\*\in C\_3(S)$.
This proof fails for random design matrices when $|S|>>\sqrt n$ because the curvature condition in $\ell\_\infty$ norm cannot hold. It is possible to overcome this difficult for Gaussian random design matrices, see
>
> Theorem 5.1 in *De-Biasing The Lasso With Degrees-of-Freedom Adjustment*
> by Bellec and Zhang (<https://arxiv.org/abs/1902.08885>)
>
>
>
| 1 | https://mathoverflow.net/users/141760 | 418243 | 170,314 |
https://mathoverflow.net/questions/418229 | 2 | Consider the following: $r\_1,...,r\_t$ are iid symmetric signs taking value $\pm1$, independent of $B\sim Binomial(p, q)$ with integer $p$ and $q=p^{-1.01}$.
**Question**: Consider $t$ as a non-decreasing function of $p$. For which such non-decreasing function of $p$ does there exist a constant $\lambda>0$ independent of $p$ such that the following holds:
$$E[\exp(\lambda B \sum\_{u=1}^t \frac{r\_u}{\sqrt t})] \le 1.05,$$
for $p$ large enough.
**Remark:**
Note that $\tilde Z = \sum\_u\frac{r\_u}{\sqrt t}$ has variance 1 and is very close to $N(0,1)$, for instance by Tusnady's inequality. If $\tilde Z$ were replaced by 1 inside the exponential, we could proceed as follows: using the exact formula for the MGF of the binomial, and using $(1+x/p)^p\le e^x$,
$$
E[\exp(\lambda B)=
(1-q + qe^\lambda)^p
\le (1 + qe^\lambda)^p \le \exp(pq e^\lambda) \le \exp(p^{-0.01} e^{\lambda}),
$$
so that for any constant $\lambda$ (say, $\lambda=1$), the RHS is bounded from above by 1.05 or any other constant strictly greater than 1. If $t$ is constant, not growing to $\infty$ simultaneously as $p$, then we can use this remark to answer the question.
Edit: **Another remark**: the previous argument shows
$$E[\exp(\lambda \tilde Z B)=
(1-q + qe^{\lambda\tilde Z})^p
\le (1 + qe^{\lambda\tilde Z})^p \le \exp(pq e^{\lambda\tilde Z}) \le \exp(p^{-0.01} e^{\lambda\tilde Z})
$$
and since $\tilde Z\le \sqrt t$ always holds,
$\sqrt t = \log(p^{0.009})\asymp \log p$ with $\lambda=0$ works. The harder question is with $t$ varying more rapidly than this.
| https://mathoverflow.net/users/141760 | Bound on the MGF of the product of two independent binomial, one being centered | $\newcommand{\la}{\lambda}$The estimate of $t$ that you got cannot be improved. Indeed, let
\begin{equation}
S\_t:=\sum\_{u=1}^t \frac{r\_u}{\sqrt t}.
\end{equation}
Suppose, as you did, that
\begin{equation}
E\exp(\la BS\_t)\le1.05
\end{equation}
for some real $\la>0$. Then
\begin{equation}
1.05\ge\frac{\la^{2p}}{(2p)!}\,EB^{2p}\,ES\_t^{2p}
\gg \frac{\la^{2p}}{p^{2p}}\,EB^{2p}\,ES\_t^{2p},
\end{equation}
where $a\gg b$ means that $a\ge c^p b$ for some universal real constant $c>0$.
Using Corollaries 1 and 2 by [Latała](https://projecteuclid.org/journals/annals-of-probability/volume-25/issue-3/Estimation-of-moments-of-sums-of-independent-real-randomvariables/10.1214/aop/1024404522.full), we get
\begin{equation}
EB^{2p}\gg\Big(\frac p{\ln p}\Big)^{2p}
\end{equation}
and
\begin{equation}
ES\_t^{2p}\gg \min[t^p,p^p].
\end{equation}
Thus,
\begin{equation}
1.05\gg \frac{\la^{2p}}{p^{2p}}\,\Big(\frac p{\ln p}\Big)^{2p}\,\min[t^p,p^p]
=\la^{2p}\Big(\frac{\min[t,p]}{\ln^2 p}\Big)^p,
\end{equation}
which implies $\sqrt t=O(\ln p)$, as claimed.
| 1 | https://mathoverflow.net/users/36721 | 418245 | 170,315 |
https://mathoverflow.net/questions/418256 | 4 | Let $\hat{D} := D \backslash \{0\}$ be a ball in $R^n$ with the origin $\{0\}$ removed. Assume that $\hat{D}$ has a structure as an orbifold (may be distinct from its standard manifold structure). Is it possible to extend the orbifold structure from $\hat{D}$ to $D$ ?
| https://mathoverflow.net/users/35716 | Extension of an orbifold structure from punctured balls to balls | No. This is not true in dimension three (nor in any higher dimension).
The three-dimensional orbifold structures allowed at a point are controlled by the list of finite subgroups of $\mathrm{SO}(3)$. In particular, there are at most three one-dimensional loci that converge at any point.
Let $S^2(2^m)$ be the two-orbifold which is topologically a two-sphere, decorated with $m$ orbifold points of order two. Suppose that $m > 3$. We form $\hat{D} = S^2(2^m) \times (0, 1]$. This is the desired example.
---
Another way to think about this: the link of a point $x$ in an $n$-dimensional orbifold must be a spherical orbifold in dimension $n-1$. In the example above, (for $m > 4$) we found a particular hyperbolic orbifold structure on $S^2$. However *any* euclidean or hyperbolic orbifold structure on $S^2$ would have served as well.
| 2 | https://mathoverflow.net/users/1650 | 418258 | 170,318 |
https://mathoverflow.net/questions/418250 | 4 | Let $\mathrm{G}$ be any infinite discrete group, and $\mathrm{H}$ be any finite index subgroup of $\mathrm{G}.$ I do not know if the Frobenius reciprocity theorem is true for the infinite groups. I want to say that, given any irreducible finite-dimensional complex representation $\rho$ of $\mathrm{G}$ there exists an irreducible finite-dimensional complex representation $\psi$ of $\mathrm{H}$ such that $\text{Ind}\_{\mathrm{H}}^{\mathrm{G}}(\psi)$ contains $\rho.$ How to prove this then?
| https://mathoverflow.net/users/100578 | Frobenius reciprocity theorem for infinite groups | A semi-answer, too long for a comment.
"the Frobenius reciprocity theorem" for finite groups is just a special case of the Hom-Tensor adjunction if you phrase it as
$$\operatorname{Hom}\_{\mathbb{C}[H]}(X,\operatorname{Res}\_H^G(Y)) = \operatorname{Hom}\_{\mathbb{G}[G]}(\operatorname{Ind}\_H^G(X),Y)$$
because $\operatorname{Ind}\_H^G(X) = \mathbb{C}[G] \otimes\_{\mathbb{C}[H]} X$. So in this sense it is also true for infinite groups (and any number of other algebraic objects) because it is "just abstract nonsense".
But the answer to your question depends on what you mean by "contains". If you mean "is a subquotient" then the usual proof using that adjunction goes through unchanged; in fact you get an epimorphism $\operatorname{Ind}\_H^G(\psi)\twoheadrightarrow \rho$ if you choose $\psi$ to be any (irreducible) $H$-submodule of the restriction of $\rho$.
If "contains" means "is a submodule" however, then things get tricky, because representations of infinite group need not be semisimple. Even uncomplicated groups like $G=\mathbb{Z}$ have indecomposable, but not irreducible modules: Jordan blocks.
And because of that, in some areas of mathematics, people restrict the meaning of "representation" to mean only "unitary representation" or something like that so that more representations are semisimple. But again, not all infinite groups have this nice property. Non-amenable groups have already been mentioned as a source of problems here.
| 4 | https://mathoverflow.net/users/3041 | 418262 | 170,319 |
https://mathoverflow.net/questions/418045 | 12 | It's a [standard exercise](https://math.stackexchange.com/a/3289290/28111) to show that every *countable* first-order theory has an irredundant axiomatization. For *uncountable* first-order theories, the result is much more difficult and was [proved by Reznikoff](https://arxiv.org/abs/1108.5171). For $\mathcal{L}\_{\omega\_1,\omega}$, I believe the state-of-the-art for the analogous question is due to [Hjorth/Souldatos](https://arxiv.org/abs/1012.3422) following [X. Caicedo](https://math.uniandes.edu.co/archivos/publicaciones/IndependentsetsofaxiomsinL-kappa,alpha.pdf), *Independent sets of axioms in $L\_{\kappa,\alpha}$*, Canadian Mathematical Bulletin 22 (1981), 219–223.
I don't recall seeing anything, however, about *second-order* logic:
>
> Suppose $T$ is a second-order theory - of arbitrary cardinality, in an arbitrary language. Must there be a second-order theory $S$ in the same language such that $(i)$ $S$ and $T$ are semantically equivalent (= have the same classes of models) but $(ii)$ no proper subset of $S$ is semantically equivalent to $S$?
>
>
>
Again the countable case is easy, but the uncountable case is unclear to me (at a glance I don't think Reznikoff's argument generalizes). More generally, I'd be interested in any sources treating the "irredundance property" in a broader class of logics than just the $\mathcal{L}\_{\kappa\lambda}$s.
| https://mathoverflow.net/users/8133 | Do second-order theories always have irredundant axiomatizations? | Here is a proof of Reznikoff’s theorem I found among my notes, not quite following Reznikoff’s proof. I stared at it for a while, and it seems to apply to second-order logic just the same; in fact, the only property of first-order logic it uses is that any sentence contains only finitely many non-logical symbols, and there are only countably many sentences in any given finite language. Let me know if I missed something.
We say that a theory $T$ in a language $L$ *depends* on a predicate or function symbol $s\in L$ if there are models $A\models T$, $B\let\nmodels\nvDash\nmodels T$ such that $A\let\res\restriction\res(L\let\bez\smallsetminus\bez\{s\})=B\res(L\bez\{s\})$. Let $L\_T$ denote the set of symbols $s\in L$ on which $T$ depends. If $L'\subseteq L$, let $T\res L'$ be the set of $L'$-sentences valid in $T$.
>
> **Lemma 1:** If $A\models T$ and $A\res L\_T=B\res L\_T$, then $B\models T$.
>
>
>
**Proof:** Fix $\phi\in T$, we will show $B\models\phi$. Let $\{s\_i:i<n\}$ enumerate the symbols of $L\bez L\_T$ occurring in $\phi$, and for $i\le n$, let $A\_i$ denote $A$ with the interpretation of $s\_j$ changed to $s\_j^B$ for each $j<i$. Applying $n$ times the definition of $L\_T$, we see that $A\_i\models T$ for each $i\le n$, thus $A\_n\models\phi$. But $A\_n$ and $B$ agree on all symbols that occur in $\phi$, hence $B\models\phi$. **QED**
>
> **Lemma 2:** $T\res L\_T$ is equivalent to $T$.
>
>
>
**Proof:**
For every $L$-sentence $\phi$, we define an $L\_T$-sentence $\phi'$ by replacing all predicates $P\in L\bez L\_T$ with $\bot$, and all functions $F\in L\bez L\_T$ by a fixed fresh variable $x$, and putting a universal quantifier $\forall x$ in front of the whole formula. Similarly, if $A$ is an $L\_T$-model and $c\in A$, we define an $L$-model $A\_c$ expanding $A$ by $P^{A\_c}=\varnothing$ and $F^{A\_c}(\vec a)=c$. Clearly,
$$A\models\phi'\iff\forall c\in A\:A\_c\models\phi.$$
If $T\models\phi$ and $A\models T$, then $(A\res L\_T)\_c\models T$ for each $c\in A$ by Lemma 1, hence $(A\res L\_T)\_c\models\phi$, and $A\models\phi'$. That is, if $T\models\phi$, then $\phi'\in T\res L\_T$.
Consequently, if $A\res L\_T\models T\res L\_T$, then $A\res L\_T\models\{\phi':T\models\phi\}$, i.e., $(A\res L\_T)\_c\models T$ for any $c\in A$, and therefore $A\models T$ by Lemma 1. Thus, every model of $T\res L\_T$ is a model of $T$. **QED**
**NB:** Alternatively, Lemma 2 can be proved using the Craig interpolation lemma.
>
> **Lemma 3:** If $T$ depends on $\kappa\ge\omega$ symbols, there exist models $\{A\_\alpha:\alpha<\kappa\}$ and sentences $\{\phi\_\alpha:\alpha<\kappa\}$ valid in $T$ such that
>
>
> $$A\_\alpha\models\phi\_\beta\iff\alpha\ne\beta\tag1$$
>
>
> for all $\alpha,\beta<\kappa$, and
>
>
> $$T\models\phi\implies\{\alpha<\kappa:A\_\alpha\nmodels\phi\}\text{ is finite}\tag2$$
> for all sentences $\phi$.
>
>
>
**Proof:**
Assume that $T$ depends on all symbols in $\{s\_\alpha:\alpha<\kappa\}$, and pick models $A\_\alpha\nmodels T$, $A'\_\alpha\models T$ such that $A\_\alpha$ and $A'\_\alpha$ differ only in the interpretation of $s\_\alpha$. For each $\alpha<\kappa$, fix a sentence $\xi\_\alpha$ such that $T\models\xi\_\alpha$ and $A\_\alpha\nmodels\xi\_\alpha$.
Since $A'\_\alpha\models T$, we see that $A\_\alpha\nmodels\phi$ for $T\models\phi$ can only happen when $s\_\alpha$ appears in $\phi$. Since only finitely many symbols appear in $\phi$, this implies (2). Moreover, considering $\phi=\xi\_\alpha$, it implies that for each $\alpha<\kappa$, only finitely many $A\_\beta$ can satisfy the same sentences as $A\_\alpha$, thus there are $\kappa$ inequivalent models on the list; w.l.o.g., we may assume that $A\_\alpha\not\equiv A\_\beta$ whenever $\beta\ne\alpha$.
Fix $\alpha<\kappa$. We already know that $\{\beta\ne\alpha:A\_\beta\nmodels\xi\_\alpha\}$ is finite; let us enumerate it as $\{\beta\_{\alpha,i}:i<n\_\alpha\}$. For each $i<n\_\alpha$, we fix a sentence $\zeta\_{\alpha,i}$ such that $A\_{\beta\_{\alpha,i}}\models\zeta\_{\alpha,i}$ and $A\_\alpha\nmodels\zeta\_{\alpha,i}$. Then we put
$$\phi\_\alpha=\xi\_\alpha\lor\bigvee\_{i<n\_\alpha}\zeta\_{\alpha,i},$$
and observe that it makes (1) hold. **QED**
>
> **Theorem:** Every theory $T$ has an independent axiomatization.
>
>
>
**Proof:**
Put $\kappa=|L\_T|$, and assume first that $\kappa$ is infinite. Let $\{A\_\alpha:\alpha<\kappa\}$ and $\{\phi\_\alpha:\alpha<\kappa\}$ be as in Lemma 3. Enumerate $T\res L\_T$ as $\{\chi\_\alpha:\alpha<\kappa\}$, and define
$$\begin{align\*}
\psi\_\alpha&=\Bigl(\let\ET\bigwedge\ET\_{\beta\colon A\_\beta\nmodels\chi\_\alpha}\phi\_\beta\Bigr)\to\chi\_\alpha,\\
S&=\{\phi\_\alpha\land\psi\_\alpha:\alpha<\kappa\}.
\end{align\*}$$
Since $T\equiv T\res L\_T$ by Lemma 2, it is clear that $S\equiv T$. Moreover, $A\_\alpha\models\phi\_\beta\land\psi\_\beta$ for $\beta\ne\alpha$, but $A\_\alpha\nmodels\phi\_\alpha$, hence $S$ is independent.
If $\kappa$ is finite, $T$ has a countable axiomatization $\{\phi\_n:n<\omega\}$ by Lemma 2. Put
$$\begin{align\*}
\psi\_n&=\Bigl(\ET\_{i<n}\phi\_i\Bigr)\to\phi\_n,\\
S&=\{\psi\_n:n\in\omega,\text{ $\psi\_n$ is not logically valid}\}.
\end{align\*}$$
Since $\ET\_{i\le n}\psi\_i$ is equivalent to $\ET\_{i\le n}\phi\_n$, $S\equiv T$. Moreover, since $\psi\_n\in S$ is not valid, there is $A\_n\models\neg\psi\_n$. This means $A\_n\models\phi\_i$ for $i<n$ and $A\_n\models\neg\phi\_n$, which implies $A\_n\models\{\psi\_i:i\ne n\}$. Thus, $S$ is independent. **QED**
| 9 | https://mathoverflow.net/users/12705 | 418270 | 170,323 |
https://mathoverflow.net/questions/418247 | 3 | [This question is looking at the paper
* Yau, S.-T., *On The Ricci Curvature of a Compact Kähler Manifold and the Complex Monge-Ampére Equation, I*, Comm. Pure Appl. Math., **31** (1978) 339-411, doi:[10.1002/cpa.3160310304](https://doi.org/10.1002/cpa.3160310304), ([pdf](https://jasonpayne.webs.com/Math5339/On%20the%20Ricci%20Curvature%20of%20a%20Compact%20Kahler%20Manifold%20and%20the%20Complex%20Monge-Ampere%20Equation%20I,%20S.T.%20Yau.pdf))]
My problem arises from (2.43)
$$
\Delta \varphi=f
$$
where
$$
-m \leqq f \leqq C\_{1} \exp \{C \sup \varphi\} \exp \left\{-\inf \_{M} \varphi\right\}.
$$
The paper already has a estimation of $\sup\varphi$ then the paper gets:
$$
\sup \_{M}|\nabla \varphi| \leqq C\_{6}\left(\exp \{-C \inf \varphi\}+\int\_{M}|\varphi|\right)
$$
How does this step use the Schauder estimate? Why is there a $\int\_{M}|\varphi|$ on the right handside, should it be $\sup|\varphi|$?
I'm so confused about this step, thanks for helping.
| https://mathoverflow.net/users/469129 | A problem of using Schauder estimate in the paper of Yau's proof of calabi conjecture | There are many non-equivalent formulations of the Schauder estimates; the version you are suggesting is the most common. For the version he needs, Yau gives a precise page & theorem reference (p.156, formula 5.5.23) to Morrey's book. In the case of the Euclidean Laplacian you can also see p.69-70 of Gilbarg & Trudinger.
| 1 | https://mathoverflow.net/users/156492 | 418274 | 170,325 |
https://mathoverflow.net/questions/417810 | 0 | Define a set $C\_3(S):=\{\Delta\in R^d: \|\Delta\_{S^c}\|\_1\le 3\|\Delta\_S\|\_1\}$. Suppose we form a random design matrix $X\in R^{n\times d}$ with rows drawn iid from a $N(0,\Sigma)$ distribution and that for all $\Delta \in C\_3(S)$,
$$
\|\Sigma\Delta\|\_{\infty}\ge \gamma\|\Delta\|\_{\infty}
$$
| https://mathoverflow.net/users/168083 | Sparse linear regression model | The result is completely independent of the regression model ($y, w$ have no role and need not be involved). $\alpha$ in the definition of $C\_3(S)$ is not defined, I will assume it is $\alpha=3$.
For each $\Delta\in C\_3(S)$, there exists $k=1,...,p$ such that
$$|e\_k^TX^TX/n \Delta| - |e\_k^T\Sigma\Delta| + |e\_k^T\Sigma\Delta|
\ge \gamma |\Delta|\_\infty - |e\_k^T(X^TX/n-\Sigma)\Delta|.$$
Now, using $|u^Tv|\le \|u\|\_\infty\|v\|\_1$,
$$
|e\_k^T(X^TX/n-\Sigma)\Delta|
\le \max\_{j=1,...,p} |e\_k^T(X^TX/n -\Sigma)e\_j| \|\Delta\|\_1
\le \max\_{j=1,...,p} |e\_k^T(X^TX/n -\Sigma)e\_j| ~ 6 \|\Delta\_S\|\_1
$$
and $\|\Delta\_S\|\_1 \le s\|\Delta\|\_\infty$. By the Hanson-Wright inequality, for each $j,k$, assuming $\max\_{i}\Sigma\_{ii}\le 1$,
$$
P(|e\_k^T(X^TX/n -\Sigma)e\_j| > c(\sqrt{x/n} + {x/n})) ) < e^{-x}.
$$
By the union bound over all pairs $i,k$ and taking $x=\log(p^3)$, we obtain that
$$
\max\_{j=1,...,p} |e\_k^T(X^TX/n -\Sigma)e\_j| \le C \sqrt{\log (p)/n}
$$
with probability at least $1/p$. This provides the desired bound when
$s\sqrt{\log(p)/n}$ is smaller than a constant.
| 1 | https://mathoverflow.net/users/141760 | 418275 | 170,326 |
https://mathoverflow.net/questions/418210 | 4 | I'm fixing some type of structure $\Sigma$ (possibly multi-sorted, with functions symbols and relation symbols, though assuming it single sorted with only relation symbols wouldn't change anything). Let $A$ and $B$ be two $\Sigma$-structure. Is there a connection between the following to notion :
1. The locale $[A,B]\_\Sigma$ classyfing morphisms of $\Sigma$-structure form $A$ to $B$.
2. Morphism of $\Sigma$-structure from $A$ to an ultrapower of $B$.
Intuitively both are way to make formal the idea that "their ought to be a morphism from $A$ to $B$, up to potential (infinite) cardinality obstruction"
As a first concrete step, the question would be:
**Is it true that $[A,B]\_\Sigma \neq \varnothing$ if and only if there exists a morphism of $\Sigma$-structure from $A$ to an ultrapower of $B$ ?**
But Ideally (and if the above is indeed true) I would like something more concrete that explain how to go back and forth between a morphism to an ultrapower and some sort of witness that the locale is non-trivial (like maybe points of a compactification or something like this).
If you are a model theorist, I guess you can replace $[A,B]\_\Sigma \neq \varnothing$ by the existence of a forcing extention of the base set theory in which there is a map from $A$ to $B$ and that should give something that can be translated back to my question.
I'm giving an explicit example in order to fix the idea :
Let $\Sigma$ be single sorted with a single binary relation $R$, and take $A$ and $B$ two $\Sigma$ structure where $R$ is interpreted as the relation $\neq$. Then $[A,B]\_{\Sigma}$ is the classifying locale of injection from $A$ to $B$ wich is non trivial as soon as "$B$ is infinite or $|A| \leqslant |B| < \infty$ ". And this is also the condition under which (I'm assuming choice here) you will be able to get an injection from $A$ to some ultrapower of $B$.
| https://mathoverflow.net/users/22131 | The locale of morphisms vs a morphism to an ultrapower? | I believe there are situations where the relevant locale is trivial, but there are many homomorphisms post-ultrapower.
For example, take $\mathcal{B}=(\mathbb{N};<)$ and let $\mathcal{A}$ be a nontrivial ultrapower of $\mathcal{B}$. There are no genuine homomorphisms from $\mathcal{A}$ to $\mathcal{B}$, and moreover this remains true in any forcing extension (since no homomorphism of strict linear orders can "move" infinitely-far-apart points to within finite distance of each other); per the comments, I *think* this means that the relevant locale is trivial too, but I'm not very familiar with locales. On the other hand, we trivially have lots of homomorphisms from $\mathcal{A}$ into ultrapowers of $\mathcal{B}$.
---
EDIT: Well after the fact, I've run into a result which while not directly related to this question may still be of similar interest, so I've decided to mention it here. John Bell's paper [*Isomorphism of structures in $S$-toposes*](https://www.jstor.org/stable/2273748). Bell shows that for structures $\mathcal{A},\mathcal{B}$ in the same language, the following are equivalent:
* $\mathcal{A}\cong\mathcal{B}$ in some forcing extension (that is, $\mathcal{A}\cong\_{\infty\omega}\mathcal{B}$ by earlier results).
* $\mathcal{A}$ and $\mathcal{B}$ "become isomorphic" in some topos over $\bf{Set}$ (which Bell calls "$S$").
I think this result adds some nuance to the observations above, although I'm a bit out of my element here.
| 2 | https://mathoverflow.net/users/8133 | 418287 | 170,330 |
https://mathoverflow.net/questions/418269 | 4 | I have a monad on an ind-category (specifically, my ind-category has a monoidal structure and I have an algebra object, so the monad is tensoring with it). It would be very useful in my work if the Eilenberg-Moore category of the monad it itself an ind-category. Is there a known criterion for the monad for this to be true? Further, is there a handy description for a generating sub-category of modules? I'm also happy to use the dual case of a comonad on a pro-category if there's a break in symmetry that makes that an easier question.
| https://mathoverflow.net/users/112314 | When is the Eilenberg-Moore category of a monad on an ind-category itself an ind-category? | Let $T$ be a monad on an [accessible category](https://ncatlab.org/nlab/show/accessible+category) (i.e. an $\mathbf{Ind}$-category). If the underlying endofunctor of $T$ is [finitary](https://ncatlab.org/nlab/show/finitary+monad) (i.e. preserves filtered colimits), then the Eilenberg–Moore category of $T$ is also accessible. This is proven (in slightly more generality) in Theorem 2.78 of Adámek–Rosický's [Locally presentable and accessible categories](https://www.cambridge.org/core/books/locally-presentable-and-accessible-categories/94CB48295B6AF097FC232313A57BDE17), for instance.
| 8 | https://mathoverflow.net/users/152679 | 418289 | 170,332 |
https://mathoverflow.net/questions/418265 | 6 | We consider block matrices
$$\mathcal A = \begin{pmatrix} 0 & A\\A^\* & 0 \end{pmatrix}$$ and
$$\mathcal B = \begin{pmatrix} 0 & B\\C & 0 \end{pmatrix}.$$
Then we define the new matrix
$$T(t) = \begin{pmatrix} \mathcal A+t & \mathcal B \\ \mathcal B^\* & \mathcal A-t\end{pmatrix}.$$
Numerical experiments seem to show that the eigenvalues of $[0,\infty) \ni t\mapsto T(t)$ have the property that their absolute values are monotonically increasing in $t \ge 0.$ However, I do not have a proof of this, does anybody know how this follows? (The eigenvalues of $T(t)$ seem to come in pairs $\pm \lambda$ with $\lambda = \lambda(t) \ge 0$, i.e. $+\lambda(t)$ is increasing, while $-\lambda(t)$ is decreasing.
To illustrate the effect, consider
$$T(t)=\begin{pmatrix}
t & 1& 0& 2\\
1 & t & 0& 0\\
0 & 0& -t & 1\\
2& 0 & 1 & -t
\end{pmatrix},$$ then the eigenvalues of $T(t)$ are
$$ \pm 1 \mp \sqrt{2+t^2}.$$
Please let me know if anything is unclear.
| https://mathoverflow.net/users/119875 | Monotonicity of eigenvalues | The idea is to apply a unitary congruence $$U=\dfrac{1}{\sqrt{2}}\begin{pmatrix}I&-I\\I&I\end{pmatrix}.$$ I consider here $\mathcal{B}$ to be hermitian, $\mathcal{B}=\mathcal{B}^\*$, whereas the general case may be 'different'.
So $$R=UT(t)U^\*=\begin{pmatrix}\mathcal{A}-\mathcal{B}&tI\\tI&\mathcal{A}+\mathcal{B}\end{pmatrix}.$$
Similarly $$
R-\lambda I=\begin{pmatrix}-\lambda I&(A-B)&tI&0\\(A^\*-C)&-\lambda I&0&tI\\tI&0&-\lambda I &(A+B)\\0&tI&(A^\*+C)&-\lambda I\end{pmatrix}.
$$
Using the well known determinant formula for block matrices with a commuting off-diagonal block, you obtain that the eigenvalues $\lambda$ satisfy $$\det\left(\begin{pmatrix}-\lambda I&F\\G&-\lambda I\end{pmatrix}\begin{pmatrix}-\lambda I&2A-F\\2A^\*-G&-\lambda I\end{pmatrix}-t^2I\right)=0.$$
Equivalently, $$\det\begin{pmatrix}(\lambda^2-t^2)I +F(2A^\*-G)&-2\lambda A\\-2\lambda A^\*&(\lambda^2-t^2)I+G(2A-F)\end{pmatrix}=0.$$
The monotonicity of the eigenvalues follows and $-\lambda$ is also an eigenvalue since $$\begin{pmatrix}Q&X\\Y&Z\end{pmatrix}$$ is unitarily congruent to $$\begin{pmatrix}Q&-X\\-Y&Z\end{pmatrix}.$$
If $\mathcal{B}$ is not hermitian the block matrix $$T(t)=\begin{pmatrix}tI&A&0&B\\A^\*&tI&C&0\\0&C^\*&-tI&A\\B^\*&0&A^\*&-tI\end{pmatrix}$$ may not be similar to $-T(t)$ for $2\times 2$ blocks, $t=2$, $A=\begin{pmatrix}1&0\\0&1\end{pmatrix}, B=\begin{pmatrix}0&0\\0&5\end{pmatrix}$ and $C= \begin{pmatrix}0&0\\3&0\end{pmatrix}.$
The property seems to hold for $T(t)\in \mathbb{M}\_4(\mathbb{C})$, consider the characteristic polynomial of $T(t)$ and $-T(t)$.
| 6 | https://mathoverflow.net/users/121643 | 418290 | 170,333 |
https://mathoverflow.net/questions/418244 | 2 | A ray is a continuous one-to-one image of the half-line $[0,\infty)$.
If $f:[0,\infty)\to \mathbb R ^2$ is continuous and one-to-one, then we say that the ray $X=f[0,\infty)$ limits onto itself if for each $x\in X$ there exists a sequence $(a\_n)\in [0,\infty)^\omega$ such that $a\_n\to\infty$ and $f(a\_n)\to x$. This is equivalent to saying that each initial segment of the ray is nowhere dense in the ray.
A connected space is said to be indecomposable if it cannot be written as the union of two proper closed connected subsets.
>
> **Question.** Is every plane ray which limits onto itself indecomposable?
>
>
>
F.B. Jones has shown that every locally connected plane ray is locally compact. Note that a ray limiting onto itself must be first category and therefore not locally compact. So a ray limiting onto itself is not locally connected.
S. Curry has shown that if the closure of the ray is one-dimensional and non-separating, then the ray (as well as its closure) is indecomposable.
If it helps, assume that the ray is nowhere dense in the plane. Then we at least know that the closure is one-dimensional.
*Jones, F. Burton*, [**One-to-one continuous images of a line**](http://dx.doi.org/10.4064/fm-67-3-285-292), Fundam. Math. 67, 285-292 (1970). [ZBL0192.60102](https://zbmath.org/?q=an:0192.60102).
*Curry, Stephen B.*, [**One-dimensional nonseparating plane continua with disjoint (\epsilon)- dense subcontinua**](http://dx.doi.org/10.1016/0166-8641(91)90014-D), Topology Appl. 39, No. 2, 145-151 (1991). [ZBL0718.54042](https://zbmath.org/?q=an:0718.54042).
| https://mathoverflow.net/users/95718 | A plane ray which limits onto itself | It seems like there are decomposable rays which limit onto themselves.
Let $E=\bigcup\_{n\geq0}E\_n$ and $O=\bigcup\_{n\geq0}O\_n$, with $E\_n=[2n,2n+1]$ and $O\_n=[2n+1,2n+2]$. We want to construct a ray $f:[0,\infty)$ which limits onto itself and such that $f(E),f(O)$ are closed in the ray $X$ and connected.
For $f(E)$ to be connected it will be enough to ensure that $\forall x\in f(E)\;\forall\varepsilon>0\;\exists N$ such that $\forall n>N$, $d(x,f(E\_n))<\varepsilon$. Indeed, if that happens any clopen of $f(E)$ which contains $x$ will have to contain $f(E\_n)$ for $n$ big enough, so it will contain all $f(E)$.
For $f(E)$ to be closed in $X$ it will be enough to assign a small tubular neighborhood $B\_n$ to the set $f(O\_n^o)$ (where $O\_n^o$ is the interior of $O\_n$) such that $f(E)$ doesn't intersect $B\_n$.
The construction will be inductive. Set $f([0,2])$ as some arbitrary smooth curve. Now given $n\geq1$, suppose that we have defined $f$ in $[0,2n]$ injectively and we have nhoods $A\_i$ of $f(E\_i^o)$ and $B\_i$ of $f(O\_i^o)$, for $i=1,\dots, n-1$, such the sets $A\_i$ don't intersect $f(O\cap[0,2n])$ and the $B\_i$ don't intersect $f(E\cap[0,2n])$. Suppose also that the $A\_i$ are pairwise disjoint and bounded by simple closed curves (and the same for the $B\_i$).
This means that the set $R\_n=\mathbb{R}^2\setminus (f([0,2n])\cup \cup\_{i=1}^{n-1}\overline{B\_i})$ is homeomorphic to the annulus $(0,1)\times\mathbb{S}^1$, and its boundary is a non injective curve formed by the $f(E\_i)$ and the boundaries of the $B\_i$ (similarly for $S\_n=\mathbb{R}^2\setminus (f([0,2n])\cup \cup\_{i=1}^{n-1}\overline{A\_i})$).
We will define $f(E\_n)$ as a smooth injective curve in $R\_n$ (except for the initial point $f(2n)$). We can define this curve so that it passes at distance $<2^{-n}$ of every point of the set $f(E\cap[0,2n])$, which is contained in the boundary of $R\_n$. Similarly, we define $f(O\_n)$ as a smooth curve in $S\_n\setminus f(E\_n)$ (which is homeomorphic to an annulus) which passes at distance $<2^{-n}$ of every point of $f(O\cap[0,2n])$. Then define $A\_n,B\_n$ to be small tubular neighborhoods of $f(E\_n^o)$ and $f(O\_n^o)$ such that the conditions above are satisfied.
This could be modified to make the curve nowhere dense: at each step we can enlarge $A\_n$ and $B\_n$ so that they contain some point $p\_n$ which is not in $f([0,2n+2])$. If we take a sequence $p\_n$ dense in the plane, the ray won't intersect a nhood of each $p\_n$, so it will be nowhere dense. If for some $n$, $p\_n$ is already in $\overline{A\_i}$ for some $i<n$, we wouldn't need to enlarge $A\_n$, and the same for $B\_n$ (this is to keep the $A\_i$ pairwise disjoint).
| 1 | https://mathoverflow.net/users/172802 | 418291 | 170,334 |
https://mathoverflow.net/questions/418307 | 3 | Apologies if this question is too basic for MO.
I think it should be the case that
>
> for any decreasing $f \colon [A,\infty) \to [0,\infty)$ and $k \geq 0$, if $\int\_A^\infty f(x) e^{kx} \, dx < \infty$ then $f(x)e^{kx} \to 0$ as $x \to \infty$.
>
>
>
**Proof** [I think]**.** Write $g(x)=f(x)e^{kx}$. If $f$ is differentiable, argue by contrapositive that if $g$ is integrable but $\{x:g(x) \geq \varepsilon\}$ is unbounded for some $\varepsilon>0$, then $f$ is not decreasing: Take sequences $x\_n,y\_n \to \infty\,$ with $\,x\_n - y\_n \downarrow 0$ such that $g(x\_n) = \varepsilon$, $g \leq \varepsilon$ on $[y\_n,x\_n]$ and $g(y\_n) \leq \frac{\varepsilon}{2}$; then we can find $z\_n \in (y\_n,x\_n)$ such that $g'(z\_n) \geq \frac{\varepsilon}{2(x\_n-y\_n)}$ and hence $f'(z\_n)=e^{-kz\_n}(g'(z\_n)-kg(z\_n)) \geq \varepsilon e^{-kz\_n}\!\left(\frac{1}{2(x\_n-y\_n)}-k\right)$, which is clearly positive for sufficiently large $n$, and so $f$ is not decreasing.
If $f$ is non-differentiable, let $\varphi$ be a nonnegative unit-integral $C^1$ bump function supported on an interval $[0,\delta]$, and define $\tilde{f} \colon [A+\delta,\infty) \to [0,\infty)$ by $\tilde{f}= f \ast \varphi$; then writing $\tilde{g}(x)=\tilde{f}(x)e^{kx}$, we have
$$ g(x) \leq \tilde{g}(x) \leq e^{k\delta}g(x-\delta)\text{,} $$
and so the differentiable case applied to $\tilde{f}$ gives the result. $\ \square$
>
> If the result is correct, it seems like it should be a standard result: either a result with a name, or an immediate special case of a result with a name. Is it so?
>
>
>
| https://mathoverflow.net/users/15570 | Is it a named result (or consequence thereof) that decreasing functions integrable against $e^{kx}$ decay faster than $e^{-kx}$? | Here is a simple proof. Since $f$ is nonnegative and decreasing, there is a limit $c:=\lim\_{x\to\infty}f(x)\ge0$, and $f\ge c$. So, $\infty>\int\_A^\infty f(x)e^{kx}\,dx
\ge\int\_A^\infty ce^{kx}\,dx=\infty$ if $c>0$. So, $c=0$ and hence
\begin{equation}
f(x+)=\int\_{(x,\infty)} \mu\_f(dt)
\end{equation}
for all $x\in[A,\infty)$, where $\mu\_f$ is the Lebesgue--Stieltjes measure based on the decreasing function $f$, so that $\mu\_f((a,b])=f(a+)-f(b+)$ for all $a$ and $b$ in $[A,\infty)$ such that $a\le b$.
Hence, by the Tonelli theorem, for any $y\in[A,\infty)$,
\begin{equation}
\begin{aligned}
J(y)&:=\int\_y^\infty f(x)e^{kx}\,dx \\
&=\int\_y^\infty f(x+)e^{kx}\,dx \\
&=\int\_y^\infty e^{kx}\,dx\, \int\_{(x,\infty)} \mu\_f(dt) \\
&=\int\_{(y,\infty)}\, \mu\_f(dt) \int\_y^t e^{kx}\,dx \\
&\ge\int\_{[y+1,\infty)}\, \mu\_f(dt) \int\_y^t e^{kx}\,dx \\
&\ge\int\_{[y+1,\infty)}\, \mu\_f(dt) \int\_y^{y+1} e^{kx}\,dx \\
&=f((y+1)-) e^{k(y+1)}h(k) \\
&\ge f(y+1) e^{k(y+1)}h(k),
\end{aligned}
\tag{1}\label{1}
\end{equation}
where $h(k):=(1-e^{-k})/k$ if $k>0$, with $h(0):=1$, so that $h(k)>0$ for each real $k\ge0$.
The condition $\int\_A^\infty f(x)e^{kx}\,dx<\infty$ implies $J(y)\to0$ and hence, by \eqref{1}, $f(y+1) e^{k(y+1)}\to0$ as $y\to\infty$. Thus, the desired conclusion holds.
---
Since this proof is very simple and transparent, it is unlikely that it is published anywhere as a theorem. It is quite possible that it appears in some, perhaps quite unlikely, papers as a lemma. Such statements are much easier to prove than to find in literature.
| 2 | https://mathoverflow.net/users/36721 | 418311 | 170,342 |
https://mathoverflow.net/questions/418312 | 0 | So, I am in need of an indication of literature or where to start.
I am having a problem consisting of reading a vector of characters (for example, there are n=4 possible characters {A,B,C,D}) using a moving window in the format [1 0 1].
For example:
The vector is:
>
> ABBCDE
>
>
>
and results in a reading of
>
> (A,B)(B,C)(B,D)(C,E)
>
>
>
I am having trouble calculating what length of character vector is needed to contain all possible combinations of characters. Considering a random vector.
PS: a difficulty I have is in the overlapping of the patterns.
| https://mathoverflow.net/users/478901 | Probability of getting all pattern combinations in moving window over a vector of characters | The expected length needed to see all patterns of length $k$ in an alphabet of size $a$ is $k \ln(a) a^k (1+o(1))$.
See Propositions 11.9 and 11.10 in [1] for the case $a=2$.
[1] Levin, David A., and Yuval Peres. Markov chains and mixing times. Vol. 107. American Mathematical Soc. second ed., with contributions by E. Wilmer, 2017.
<https://pages.uoregon.edu/dlevin/MARKOV/markovmixing.pdf>
| 0 | https://mathoverflow.net/users/7691 | 418335 | 170,348 |
https://mathoverflow.net/questions/418341 | 0 | Consider the following finite version Hindman theorem:
For every sufficiently large $N\in\omega$ and 2-partition of $N=N\_0\cup N\_1$, there are $i<2,a,b,c\in N\_i$ such that $a+b=c$.
The only proof I know for this is by iteratively using Hales-Jewett theorem. What are the alternative proofs?
| https://mathoverflow.net/users/74918 | Finite Hindman theorem | This is Schur's theorem, it follows from Ramsey theorem: consider the complete graph with vertices $1,\ldots,N+1$ and color the edge between vertices $i$ and $j$ with color $s\in \{0,1\}$ iff $|i-j|\in N\_i$. A monochromatic triangles provides a monochromatic solution of $a+b=c$.
| 9 | https://mathoverflow.net/users/4312 | 418342 | 170,349 |
https://mathoverflow.net/questions/418351 | 5 | Let $f:\mathbb{CP}^n\to\mathbb{CP}^n$ be a holomorphic map; I am interested in what the subvariety of critical points could be.
More specifically, let $J=\{p\in \mathbb{CP}^n\ :\ \det\mathrm{Jac}(f)=0\}$:
1. can $J$ be a smooth submanifold of $\mathbb{CP}^n$?
2. can $J$ be an irreducible subvariety of $\mathbb{CP}^n$?
3. is there a way to describe the subvarieties $J$ that can be obtained this way?
Obviously, for $n=1$, the answer to 1,2 is easy, because $J$ is a union of isolated points.
Regarding 3, if we want a constructive answer (i.e., given $J$, we want to produce $f$) it is not immediate to answer to 3, already for $n=1$.
On a further level, I am also wondering if it can happen that $J$ is, e.g., smooth but $(\det\mathrm{Jac}(f))$ is not reduced (i.e., locally $\det\mathrm{Jac}(f)=(g)^k$ for some $g$ holomorphic and $k>1$).
I tried to write down the equations explicitly fixing the degree of $f$ (as a homogeneous polynomial on $\mathbb{C}^{n+1}$), but the result is not easily manageable, at least for me; I also tried to look at it from and algebraic viewpoint, noticing a link with the concept of *socle* of a Gorenstein algebra, but, as much as it was interesting, it did not give me any hint on how to exlpicitly describe $J$ (or $f$ given $J$, or given the equation I want for $J$).
I believe this kind of question could already appear somewhere in the literature, but it evaded my googleing efforts, up to now.
| https://mathoverflow.net/users/17111 | Geometry of critical points of holomorphic maps in projective space | For $n=2$, the locus $J$ is smooth and irreducible for a general $f$; i.e., these $f$ form a Zariski dense subset of the parameter space of such $f$. For $n\ge3$ and for general $f$, the locus $J$ will be (mildly) singular, irreducible, and of general type. [See Theorems 14 and 15 in the linked paper.](https://arxiv.org/abs/1910.11290) (The proof of Theorem 14, which proves general type, was shown to the authors by Jason Starr.)
| 9 | https://mathoverflow.net/users/11926 | 418352 | 170,350 |
https://mathoverflow.net/questions/418367 | 4 | **Definition** A fine measure on $P\_\kappa(\lambda)$ is a non-principal ultrafilter on $P\_\kappa(\lambda)$ which contains all upper cones $\uparrow{x}=\{y\in P\_\kappa(\lambda)|x\subset y\}$, for all $x\in P\_\kappa(\lambda)$.
Bagaria and Magidor defined a cardinal $\kappa$ to be $\aleph\_1$-strongly compact if there exists a non-principal $\aleph\_1$-complete fine measure on $P\_\kappa(\lambda)$, for all $\lambda\ge\kappa$. This is a refinement of the usual definition of strongly compacts where $\kappa$-completeness has been relaxed to $\aleph\_1$-completeness.
I want to further relax the assumption that all upper cones belong to $U$.
**Definition** Let $F$ be a subset of $\lambda$ (which maybe empty).
An ultrafilter on $P\_\kappa(\lambda)$ is $F$-fine if for every $a\in \lambda$, $\uparrow{\{a\}}\in U$ if and only if $a\in F$.
If $F=\lambda$ then we recover the definition of a fine measure.
**Question**
Assume $\kappa$ is an $\aleph\_1$-strongly compact cardinal. Let $F\subset \lambda$ be of size $<\lambda$. Can we prove there is a non-principal $F$-fine $\aleph\_1$-complete ultrafilter on $P\_\kappa(\lambda)$?
| https://mathoverflow.net/users/13694 | $\aleph_1$-complete fine measures on $P_\kappa(\lambda)$ | Yes, if $F$ is large enough. Let $j : V \to M$ be the embedding derived from an $\aleph\_1$-strongly compact ultrafilter $U$ on $P\_\kappa(\lambda)$. Let $[\mathrm{id}]$ be the set in $M$ represented by the identity function. Define $W \subseteq P(P\_\kappa(\lambda))$ by:
$$A \in W \Leftrightarrow [\mathrm{id}] \cap j(F) \in j(A).$$
Then $W$ is a countably complete ultrafilter. If $\alpha \in F$, then $j(\alpha) \in [\mathrm{id}] \cap j(F)$, so $[\mathrm{id}] \cap j(F) \in j({\uparrow}\{\alpha\})$. If $\alpha\notin F$, then $j(\alpha) \notin j(F)$, so $[\mathrm{id}] \cap j(F) \notin j({\uparrow}\{\alpha\})$. Thus $W$ is $F$-fine.
It remains to check that $W$ is nonprincipal. This holds when $|F| \geq \kappa$. To see this, suppose on the contrary that there is $z\_0$ such that $A \in W$ iff $z\_0 \in A$. Then $[\mathrm{id}] \cap j(F) = j(z\_0)$. But $|[\mathrm{id}] \cap j(F)| \geq |j[\lambda] \cap j(F)| \geq j(\kappa)$, which means $|z\_0| \geq \kappa$, which is false.
On the other hand, if $|F|<\mathrm{crit}(j)$, then $j(F) = j[F] = [\mathrm{id}] \cap j[F]$, so $A \in W$ iff $j(F) \in j(A)$, thus $A \in W$ iff $F \in A$. In this case, $W$ is principal.
I'm not sure what happens when $\mathrm{crit}(j) \leq |F| < \kappa$.
| 5 | https://mathoverflow.net/users/11145 | 418368 | 170,354 |
https://mathoverflow.net/questions/418357 | 4 | The CHSH game is the standard example of a game where two cooperating players Alice and Bob who cannot communicate, but who nevertheless can get an advantage by measuring an entangled quantum state in some basis which depends on its input.
Detailed information is available here: <https://en.wikipedia.org/wiki/CHSH_inequality#CHSH_game>
The optimal strategy in the game needs measurement in quite complicated bases, which are not just $\{ \vert 0 \rangle , \vert 1 \rangle \} $ combined with $\{ \vert + \rangle , \vert - \rangle \} $. I can verify the computation and it makes perfect sense to me, which this is optimal for the particular example of the CHSH game.
However, by measuring the quantum state $ \frac{ \vert 00 \rangle + \vert 11 \rangle}{\sqrt{2}}$ in the two bases mentioned I can also get some correlations that I cannot reproduce classically without communication. Then why doesn't there exist a standard example of a quantum game using these simpler bases?
| https://mathoverflow.net/users/143779 | Why does the CHSH game need complicated bases to show advantage? | It's not that hard to understand in terms of the [Bloch sphere](https://en.wikipedia.org/wiki/Bloch_sphere). If Alice measures along vector $\alpha$ on the Bloch sphere, and Bob measures along $\beta$, then the correlation between their answers is the dot product $\alpha \cdot \beta$. In other words, the probability that they give the same answer is $\tfrac{1+(\alpha\cdot \beta)}{2} =\tfrac{1+\cos \theta}{2} = \cos^2 \tfrac{\theta}{2}$ where $\theta$ is the angle between $\alpha$ and $\beta$.
Let Alice measure along $\alpha\_0$ if she receives $0$ and along $\alpha\_1$ if she receives $1$, and likewise for Bob along $\beta\_0$ and $\beta\_1$. Then we want the angles $\angle(\alpha\_0, \beta\_0)$, $\angle(\alpha\_0, \beta\_1)$ and $\angle(\alpha\_1, \beta\_0)$ to be as acute as possible, and we want $\angle(\alpha\_1, \beta\_1)$ to be as obtuse as possible. Fiddle around and you'll convince yourself that you want $(\alpha\_1, \beta\_0, \alpha\_0, \beta\_1)$ to lie on a great circle at $\pi/4$ apart (so the angle between $\beta\_1$ and $-\alpha\_1$ is also $\pi/4$).
---
That said, I originally learned Bell's inequality from a book that used the following game: The referee sends Alice and Bob signals $a$ and $b$ from $\{ 1,2,3 \}$, and their goal is to always agree if $a=b$ and disagree as often as possible if $a \neq b$. Then the quantum strategy is to take $\alpha\_1 = \beta\_1$, $\alpha\_2 = \beta\_2$ and $\alpha\_3=\beta\_3$ at angles of $\tfrac{2 \pi}{3}$, which gives the probability of agreement $\cos^2 \tfrac{\pi}{3} = \tfrac{1}{4}$ when $a \neq b$.
The best classical strategy is for them both to choose the same surjection $f: \{ 1,2,3 \} \longrightarrow \{ 0,1 \}$ and return $f(a)$ and $f(b)$; this achieves $\tfrac{1}{3}$. (Or, more realistically, they pre-agree on the same sequence of surjections $f\_1$, $f\_2$, $f\_3$, ... from $\{ 1,2,3 \}$ to $\{ 0,1 \}$, and both play function $f\_k$ on turn $k$. The probability is still $\tfrac{1}{3}$, but the referee is less likely to detect the strategy.)
The book explained Bell's inequality as "if you have three socks, each black or white, and a referee demands you reveal two of them at random, the odds they will match are at least $\tfrac{1}{3}$, but quantumly it can be as low as $\tfrac{1}{4}$." I always found that variant more mnemonic than the CHSH version. (I wish I could remember which book this was!)
| 5 | https://mathoverflow.net/users/297 | 418370 | 170,355 |
https://mathoverflow.net/questions/418326 | 1 | In a previous question [here](https://mathoverflow.net/questions/418265/monotonicity-of-eigenvalues), I asked the question below for block matrices and received an answer showing the question is true if $\mathcal B$ is hermitian and false, in general if $\mathcal B$ is non-hermitian. However, numerical experiments suggest it is still true if we are talking only about matrices rather than block matrices and this is the content of this question.
We consider matrices
$$\mathcal A = \begin{pmatrix} 0 & a\\\bar a& 0 \end{pmatrix}$$ and
$$\mathcal B = \begin{pmatrix} 0 & b\\c & 0 \end{pmatrix}$$
with $a,b,c \in \mathbb C.$
Then we define the new matrix
$$T(t) = \begin{pmatrix} \mathcal A+t & \mathcal B \\ \mathcal B^\* & \mathcal A-t\end{pmatrix}.$$
Numerical experiments seem to show that the eigenvalues of $[0,\infty) \ni t\mapsto T(t)$ have the property that their absolute values are monotonically increasing in $t \ge 0.$ However, I do not have a proof of this, does anybody know how this follows? (The eigenvalues of $T(t)$ seem to come in pairs $\pm \lambda$ with $\lambda = \lambda(t) \ge 0$, i.e. $+\lambda(t)$ is increasing, while $-\lambda(t)$ is decreasing.
To illustrate the effect, consider
$$T(t)=\begin{pmatrix}
t & 1& 0& 2\\
1 & t & 0& 0\\
0 & 0& -t & 1\\
2& 0 & 1 & -t
\end{pmatrix},$$ then the eigenvalues of $T(t)$ are
$$ \pm 1 \mp \sqrt{2+t^2}.$$
Please let me know if you have any questions.
| https://mathoverflow.net/users/119875 | Monotonicity of eigenvalues II | The characteristic polynomial is even in both $X$ and $t$ : $P\_t(X)=Q(X^2,t^2)$ where
$$Q(Y,s)=(Y-s)^2-(2|a|^2+|b|^2+|c|^2)(Y-s)-4|a|^2s+|a^2-b\bar c|^2.$$
The variation of $s\mapsto Y(s)$ is given by the derivative
$$2(Y-s)(Y'-1)-(2|a|^2+|b|^2+|c|^2)(Y'-1)-4|a|^2=0.$$
The sign of $Y'$ changes when $s$ crosses the value
$$s\_0=\frac1{4|a|^2}(2|a|^2+|b|^2+|c|^2)(2|a|^2-|b|^2-|c|^2).$$
This value is admissible for $t\_0^2$ if it is positive, that is is $2|a|^2>|b|^2+|c|^2$.
In conclusion, the absolute values of the eigenvalues are monotonous functions of $s$ whenever $2|a|^2\le|b|^2+|c|^2$. If on the contrary $2|a|^2>|b|^2+|c|^2$, then one of the eigenvalues is not monotone, and the change happens when $t$ crosses
$$t\_0=\frac1{2|a|}\sqrt{(2|a|^2+|b|^2+|c|^2)(2|a|^2-|b|^2-|c|^2)}.$$
| 4 | https://mathoverflow.net/users/8799 | 418372 | 170,356 |
https://mathoverflow.net/questions/418359 | 7 | **Background.** (Can be skipped if you already know what is the Eguchi-Hanson metric.) The Eguchi-Hanson metric $g$ is a complete Ricci-flat Riemannian metric on the cotangent bundle of the 2-sphere, $T^\*S^2$. By removing the zero-section, we can identify it with $S^3/\mathbb{Z}\_2 \times (0, \infty) = (\mathbb{R}^4\setminus\{0\}) / \mathbb{Z}\_2$, where $\mathbb{Z}\_2$ acts by antipodal reflections. It then has the explicit description
$$g = \frac{r^2}{\sqrt{1+r^4}}(dr^2 + r^2 \alpha\_1^2) + \sqrt{1 + r^4}(\alpha\_2^2+\alpha\_3^2),$$
where $r^2 = x\_0^2 + x\_1^2 + x\_2^2 + x\_3^2$,
$$\alpha\_1 = \frac{1}{r^2} (x^0 dx^1 - x^1 dx^0 + x^2 dx^3 - x^3 dx^2)$$
and $\alpha\_2, \alpha\_3$ are defined with the same formula by cyclic permutations of $(1, 2, 3)$. It has also been described by Calabi as a Kähler metric on $T^\*\mathbb{CP}^1$, where the Kähler form is $\pi^\*\omega\_{FS} + i\partial\bar{\partial} (u \circ t)$, where $\omega\_{FS}$ is the Fubini-Study Kähler form on $\mathbb{CP}^1$, $t : T^\*\mathbb{CP}^1 \to \mathbb{R}$ is the squared-norm function with respect to the Fubini-Study metric, and $u(t) = 4\sqrt{1+t} - 4\log(1+\sqrt{1+t})$.
**Question.** *Is the bundle map $\pi : T^\*S^2 \to S^2$ a Riemannian submersion?*
In other words, does $\pi$ restrict to isometries $d\pi : (T\_\xi(T^\*S^2))^{\mathrm{horizontal}} \to T\_xS^2$? I tried to work this out explicitly using the above coordinate expressions, but it is very messy, and I couldn't do it. I was wondering if there is another argument, or perhaps a way to see that it's not a submersion using asymptotic properties of the metric.
| https://mathoverflow.net/users/409915 | Is the bundle map of the Eguchi-Hanson metric a Riemannian submersion? | No. The reason is basically the same why if you take flat $\mathbb R^4\setminus 0$ and quotient by the standard Hopf $\mathbb S^1$ action you get a punctured cone over $\mathbb S^2$ but the projection to that $\mathbb S^2$ is not a Riemannian submersion because the horizontal spaces scale with $r$. The Euguchi-Hanson metric is asymptotic to this one and the same exact thing happens.
What IS true for the Euguchi-Hansen metric is that if you fix $r$ then the projection from $\mathbb S^3/\mathbb Z\_2$ to $\mathbb S^2$ is a Riemannian submersion **up to scaling**. But the scaling changes with $r$.
This can be seen as follows.
The forms $\alpha\_1,\alpha\_2,\alpha\_3$ form the standard left invariant orthornormal basis in the round binivariant metric on $\mathbb S^3$ (and its factor $SO(3)=\mathbb S^3/\mathbb Z\_2$). Here $\alpha\_1$ can be thought of as the inner product (with respect to the binivariant metric) with the Hopf vector field $X$ generating a free isometric $\mathbb S^1$ action on $SO(3)$. For any fixed $r$ the difference between the round metric on $SO(3)$ and the Euguchi Hanson one is that this circle is scaled by a function depending on $r$. But this does not affect the quotient metric at all and the projections $SO(3)\to \mathbb S^2$ are Riemannian submersions (up to scaling) with respect to both metrics on $SO(3)$. However, as I said you get a scaling depending on $r$ so the global map $TS^2\to S^2$ is not a Riemannian submersion.
| 9 | https://mathoverflow.net/users/18050 | 418379 | 170,360 |
https://mathoverflow.net/questions/418347 | 3 | Suppose $\mathcal{E} \to \mathcal{C}$ is a cocartesian fibration over (the nerve of) a classical category, and there is a section on zero simplices that sends $C$ to $s(C)$ such that, for every edge $f\colon C \to C'$ in $\mathcal{C}$, the Kan complex
$\mathrm{Hom}\_{\mathcal{E}\_{C'}}(f\_!(s(C)), s(C'))$
is discrete (i.e. all higher $\pi\_i$ vanish).
Is it true that data of an extension $s$ to a full section $\mathcal{C} \to \mathcal{E}$ is equivalent to data of morphisms
$t\_f\colon [f\_!](s(C)) \to s(C')$
in the homotopy category $h\mathcal{E}\_{C'}$ such that, for every two composable morphisms $f\colon C \to C'$ and $g\colon C' \to C''$,
$t\_g \circ [g\_!](t\_f) ~= t\_{g\circ f}$
in the homotopy category $h\mathcal{E}\_{C''}$ (under the canonical identification of $[g\_!]\circ [f\_!]$ with $[(g\circ f)\_!]$).
| https://mathoverflow.net/users/115211 | Constructing sections of a cocartesian fibration | Yes. Consider the full subcategory $\mathcal S$ of $\mathcal E$ spanned by the objects $s(C)$.
It is clear that a section extending $s$ is the same thing as a section $\mathcal C\to \mathcal S$ extending $s$.
But I claim that $\mathcal S$ is a $1$-category: indeed the fiber of $\hom\_\mathcal E(s(C),s(C'))\to \hom\_\mathcal C(C,C')$ over $f$ is $\hom\_{\mathcal E\_C'}(f\_!s(C),s(C'))$, and so is discrete, and the base $\hom\_\mathcal C(C,C')$ is also discrete. Of course if $E\to B$ is a map to a discrete space where each fiber is discrete, $E$ must be discrete too (say by the long exact sequence of homotopy groups)
In particular, a section $\mathcal C\to \mathcal S$ extending $s$ is just an ordinary functor between ordinary categories : you have to pick images $s(f)$'s for every $f$ that compose well. Because $\mathcal S \subset \mathcal E$ and the latter is coCartesian over $\mathcal C$, you can reformulate the data of $s(f)$ in the way you did, and "compose well" accordingly.
| 1 | https://mathoverflow.net/users/102343 | 418382 | 170,361 |
https://mathoverflow.net/questions/418366 | 3 | Let $\mathcal{H}\_A\otimes\mathcal{H}\_B$ be a finite-dimensional bipartite Hilbert space, $P\_A$ a positive semi-definite operator on $\mathcal{H}\_A$, $P\_B$ a positive semi-definite operator on $\mathcal{H}\_B$, and $U\_{AB}$ a unitary operator on $\mathcal{H}\_A\otimes\mathcal{H}\_B$. In some roundabout way involving integrals over the unitary group it seems I can prove the following inequality:
$$\mathrm{tr}\_A\left(\mathrm{tr}\_B((P\_A\otimes P\_B)U\_{AB})\mathrm{tr}\_B((P\_A\otimes P\_B)U\_{AB}^\dagger)\right)\leq\mathrm{tr}(P\_A^2)\mathrm{tr}(P\_B)^2.$$
I have been unable to find a more ``conventional'' proof of this inequality, e.g. using von Neumann's trace inequality or Holder's inequality, so I am curious if this is a known result or if someone can provide a proof which uses only linear algebra.
| https://mathoverflow.net/users/98045 | Proving a bipartite trace inequality? | You can rewrite the left-hand side $\mathrm{tr}\_A\left(P\_AX P\_AX^\*\right)$,
where $X=\mathrm{tr}\_B((1\otimes P\_B)U\_{AB})$. By Hölder's inequality this is less than $\|P\_A\|\_2^2 \|X\|\_\infty^2$, so your inequality is just that $\|X\|\_\infty \leq \mathrm{tr}(P\_B)$, which is certainly "conventional". For a justification use Hölder's inequality to obtain $\mathrm{tr}\_A(CX) = \mathrm{tr}((C\otimes P\_B) U\_{AB}) \leq \|C\otimes P\_B\|\_1 = \|C\|\_1 \mathrm{tr}(P\_B)$, so this conventional inequality follows by taking the supremum over all $C$ with $\|C\|\_1 =1$.
| 4 | https://mathoverflow.net/users/10265 | 418388 | 170,363 |
https://mathoverflow.net/questions/418392 | 6 | Consider the equation
$x^5-2x^2+z=0$
How do you derive the Lagrange inversion theorem series solution for it? I know it exists because the answer is here for any trinomial <https://arxiv.org/pdf/0910.2957.pdf>
I am trying to figure out how derive the series solution.
There is this well known result, which I think is needed:
$$ h(f^{-1}(z))=h(0)+\sum\_{n\geq 1}\frac{z^n}{n}\cdot [z^{n-1}]\left(h'(z)\cdot\left(\frac{z}{f(z)}\right)^n\right).$$
What functions should I use for f(z), h(z)
There is no information about this online. Most sources just give the series solution for $x^5-x+a=0$ which has $f(x)=x^5-x$ and then the bring radical solution is as follows by applying the above formula.
the best I could come up with is:
$z=2y-y^{5/2}$ taking the square root of the solution of this gives the solution to the original equation
so $h(z)$ would be a square root
$$ h(f^{-1}(z))=\sum\_{n\geq 1}\frac{z^n}{n}\cdot [y^{n-1}]\left(1/2y^{-1/2}\cdot\left(\frac{1}{2-y^{3/2}}\right)^n\right).$$
This gets messy and does not generate integer y's , so it does not work. Any ideas.
| https://mathoverflow.net/users/470659 | Series solution for general trinomial | One pair of solutions will
be Puiseux series $ax^{1/2}+bx+cx^{3/2}+\cdots$ (where $x^{1/2}$ can
have one of two signs). Thus set
$u=x^{1/2}$ in the solution, giving a series $F(u)$ satisfying
$F(u)^5-2F(u)^2+u^2=0$. Hence $\sqrt{-F(u)^5+2F(u)^2}=u$. By ordinary
Lagrange inversion, $$ [u^n]F(u) =[u^{n-1}]\frac
1n\left(\frac{1}{\sqrt{2-u^3}}\right)^n. $$
**Addendum.** The series $F(x^{1/2})$ and $F(-x^{1/2})$ give two
solutions to $x^5-2x^2+z=0$. The other three solutions $G(x)$ are
given by
$$ [x^n]G(x) = \frac 1n[x^{n-1}]\left(\frac{x}{2(x+\alpha)^2
-(x+\alpha)^5}\right)^n, $$
for $n\geq 1$, and $G(0)=\alpha$, where $\alpha=2^{1/3}$ (three
different values).
**Addendum 2.** The solution above can easily be generalized. Let
$P(t)\in\mathbb{C}[t]$. The fractional power series (Puiseux series)
$y$ satisfying $P(y)=x$ are given as follows: let $\alpha$ be a zero
of $P(t)$ of multiplicity $m$. Then $y=\alpha + \sum\_{n\geq 1} a\_n x^{n/m}$, where
$$ a\_n = \frac 1n[u^{n-1}]\left(
\frac{u}{P(u+\alpha)^{1/m}}\right)^n. $$
There are $m$ values of $P(u+\alpha)^{1/m}$ (differing by multiplication by an $m$th root of unity), so we get $m$ solutions corresponding to $\alpha$.
| 10 | https://mathoverflow.net/users/2807 | 418396 | 170,366 |
https://mathoverflow.net/questions/418394 | 1 | Suppose you have a number
$$
N = p\_1^{e\_1}p\_2^{e\_2}\cdots p\_k^{e\_k}
$$
and are looking for the largest divisor $d|N$ such that $d^2<N$ (that is, [A060775](https://oeis.org/A060775)$(N)$.) How can I efficiently find this $d$?
If $N$ has a small number of divisors, you can just iterate through them all (keeping the best at each step). If $e\_i$ is large for some $i$, then you can search the divisors $d$ of $N/p\_i^{e\_i}$ less than $\sqrt{N}$ and find
$$
e=\min\left(\left\lfloor\frac{\log(\lfloor\sqrt{N-1}\rfloor/d)}{\log p\_i}\right\rfloor, e\_i\right)
$$
(this would be
```
s=sqrtint(N-1);
f=factor(N);
e=min(logint(s\d, f[i,1]), f[i,2])
```
in [PARI/GP](https://pari.math.u-bordeaux.fr/); presumably lots of languages have efficient implementations)
giving $p\_i^ed$ as the number to test. But for the general case where $k$ is large neither is practical. Is there some method which takes significantly fewer than $\tau(N)$ steps? You can assume that the factorization is given.
| https://mathoverflow.net/users/6043 | Efficiently finding the largest divisor of N less than sqrt(N) | I believe the Schroeppel & Shamir algorithm [1] can be adapted for use here, with time something like $O\left(\sqrt{\tau(n)}\omega(n)\right)$ and space something like $O\left(\sqrt[4]{\tau(n)}\right)$. The basic idea is splitting the prime powers into four sets with roughly the same number of divisors each, listing the divisors, and generating divisors from two pairs of two sets dynamically using min-heaps, and combining as expected.
[1] Richard Schroeppel and Adi Shamir, [A $T = O(2^{n/2})$, $S = O(2^{n/4})$ algorithm for certain NP-complete problems](https://epubs.siam.org/doi/abs/10.1137/0210033), *SIAM Journal on Computing*, Vol. 10, Iss. 310, pp. 456–464.
| 6 | https://mathoverflow.net/users/6043 | 418399 | 170,367 |
https://mathoverflow.net/questions/418371 | 1 | I cross-post a question that has not been answered on MSE, see [here](https://math.stackexchange.com/questions/4391605/question-about-milnor-hypersurface).
Consider the Milnor hypersurface $H\_{ij}$, i.e., the smooth hypersufrace in $\mathbb CP^i \times \mathbb CP^j$ for fix pair of integers $j \ge i\ge 0$ defined by
$$H\_{ij}= \left\{{z\_0: \ldots :z\_i)} \times {(\omega\_0: \ldots :\omega\_j) \in \mathbb CP^i \times \mathbb CP^j \; | \; z\_0\omega\_0+ \ldots +z\_i\omega\_i=0}\right\}. $$
>
> **Question.** How can one prove that $H\_{11}$ is homeomorphic to $\mathbb CP^1$?
>
>
>
| https://mathoverflow.net/users/170441 | Milnor hypersurface | Since the question was not answered on MSE, I will provide a short answer here.
The expression $H\_{ij}$ is a bi-homogeneous polynomial of bi-degree $(1, \, 1)$, hence it defines an effective divisor in the complete linear system $|\mathcal{O}\_{\mathbb{P}^i \times \mathbb{P}^j}(1, \, 1)|$. Moreover, a simple computation with derivatives shows that this divisor is smooth.
In particular, when $i=j=1$, we have a smooth divisor $D$ in the complete linear system $|\mathcal{O}\_{\mathbb{P}^1 \times \mathbb{P}^1}(1, \, 1)|$. The rational map associated with this linear system is the usual embedding of $\mathbb{P}^1 \times \mathbb{P}^1$ as a smooth quadric in $\mathbb{P}^3$. Thus, the curve $D$ corresponds to a smooth hyperplane section of this quadric, namely, a smooth conic, which is isomorphic (not just homeomorphic) to $\mathbb{P}^1$.
| 5 | https://mathoverflow.net/users/7460 | 418414 | 170,371 |
https://mathoverflow.net/questions/418408 | -1 | I have a task:
Lat's take independent variables $X\_{i}$ with Poisson distribution $Poiss(a)$. Distribution of $a$ has density $p(a)=\frac{8}{3}a^3e^{-2a}, a\ge0$.
Calculate:
$E(a|X\_1=3, X\_2=2, X\_3=5, X\_4=5, X\_5=1, X\_6=4)$.
I tried to calculate this in this way:
$E(a|X\_1=3, X\_2=2, X\_3=5, X\_4=5, X\_5=1, X\_6=4)=\int{af(a|X\_1=3, X\_2=2, X\_3=5, X\_4=5, X\_5=1, X\_6=4)da}= \int{a \frac{f(X\_1=3, X\_2=2, X\_3=5, X\_4=5, X\_5=1, X\_6=4|a) \cdot f(a)}{P(X\_1=3, X\_2=2, X\_3=5, X\_4=5, X\_5=1, X\_6=4)}}da$, but i do not know how to calculate $P(X\_1=3, X\_2=2, X\_3=5, X\_4=5, X\_5=1, X\_6=4)$. Could you help me in this way or show other solution?
Thanks in advance.
| https://mathoverflow.net/users/478987 | Poisson distribution and conditional expected value | Your problem seems to be a special case of <https://web.stanford.edu/class/stats200/Lecture20.pdf>
about prior and posterior distributions in Bayesian analysis. By assumption $X\_1,\ldots,X\_6 \sim Poisson(a)$ and $a \sim \Gamma(\alpha,\beta) = \Gamma(4,2)$, the a priori distribution. Then with $s = x\_1+\ldots+x\_n$ where $x\_1 = 3, \ldots , x\_6 = 4$, hence $n = 6$ we get that the a posteriori distribution of $a$ is $\Gamma(s+\alpha,n+\beta) = \Gamma(s+4,n+2) = \Gamma(20+4,6+2) = \Gamma(24,8)$. In particular $\mathbb{E}a = 24/8 = 3$.
| 1 | https://mathoverflow.net/users/100904 | 418422 | 170,372 |
https://mathoverflow.net/questions/418432 | 1 | This problem, which is in the book "Probability Theory"(fourth edition) written by M. Loeve, states as follows,
Rule: In order to compute $PB$, $B = f(A\_{1}, A\_{1}^{c}, \cdots, A\_{m}, A\_{m}^{c})$, take the following steps:
1. Reduce the operations on events to complementations, intersections,
and sums;
2. Replace each event with its indicator, expand, and take
the expectation.
In this way find
\begin{equation\*}
P(\bigcup\_{j=1}^{m} A\_{j}) \text{ and } P(\bigcap\_{j=1}^{k} A\_{j} \bigcup\_{j=k+1}^{m} A\_{j}^{c}) \text { in terms of } P(\bigcap\_{j=1}^{r} A\_{j})\text{'s}.
\end{equation\*}
I was wondering how can $\bigcup\_{j=1}^{m} A\_{j}$ can be represented only by a set of $\bigcap\_{j=1}^{r} A\_{j}$ since the sequence $\{\bigcap\_{j=1}^{r} A\_{j}\}$ is decreasing with the largest value is $A\_{1}$.
Any hint for solving it? Thanks.
| https://mathoverflow.net/users/138147 | The method of indicators for probability | For any event $B$, let $1\_B$ and $B^c$ denote the indicator and the complement of $B$, respectively. Let $A:=\bigcup\_{j=1}^{m} A\_j$.
Let $\binom{[m]}r$ denote the set of all subsets of cardinality $r$ of the set $[m]:=\{1,\dots,m\}$.
Then
$$1\_A=1-1\_{A^c}=1-\prod\_{j=1}^m 1\_{A\_j^c}
=1-\prod\_{j=1}^m (1-1\_{A\_j})
=1-\sum\_{r=0}^m(-1)^r \sum\_{J\in\binom{[m]}r}\prod\_{j\in J} 1\_{A\_j}
=\sum\_{r=1}^m(-1)^{r-1} \sum\_{J\in\binom{[m]}r}\prod\_{j\in J} 1\_{A\_j}.$$
Taking now the expectations, we get
$$P(A)
=\sum\_{r=1}^m(-1)^{r-1} \sum\_{J\in\binom{[m]}r}P\Big(\bigcap\_{j\in J} A\_j\Big).$$
---
In a comment, the OP insisted that $P(\bigcup\_{j=1}^m A\_j)$ be expressed only in terms of the probabilities of the form $P(\bigcap\_{j=1}^r A\_j)$. However, this is impossible to do, even for $m=2$. Indeed, let $x:=P(A\_1)$, $y:=P(A\_1\cap A\_2)$, and $z:=P(A\_2\setminus A\_1)$. Then $P(A\_1\cup A\_2)=x+z$, which is clearly impossible to express in terms of $x$ and $y$.
Indeed, one can let e.g. $x=y=0$ and then take any $z\in[0,1]$. So, given $P(A\_1)=0$ and (hence) $P(A\_1\cap A\_2)=0$, the probability $P(A\_1\cup A\_2)$ can take any value in $[0,1]$.
| 1 | https://mathoverflow.net/users/36721 | 418435 | 170,375 |
https://mathoverflow.net/questions/415755 | 6 | I have two questions that seem to be related.
I wonder if there is a user-friendly algorithm (starting from ribbon/slice presentation of knots/disks) for the construction of double (in general $p$-fold) coverings of $B^4$ along ribbon/slice disks.
Using the handle decomposition of ribbon/slice disks can we control the handle decomposition of double (in general $p$-fold) coverings of $B^4$ along these disks?
| https://mathoverflow.net/users/475366 | Double ($p$-fold) coverings of $B^4$ along ribbon/slice disks | I would like to popularize the lecture notes of Brendan Owens about the double branched cover of knots in $S^3$ and of ribbon/slice disks in $B^4$.
He both discussed constructions and obstructions. Also, it includes the classical references listed by Professor Ruberman.
>
> Owens, Brendan. "Knots and 4-manifolds." Winter Braids Lecture Notes 6 (2019): 1-26.
>
>
>
It is avaliable online [here](https://wbln.centre-mersenne.org/item/?id=WBLN_2019__6__A2_0).
| 4 | https://mathoverflow.net/users/131172 | 418443 | 170,378 |
https://mathoverflow.net/questions/264622 | 5 | Given an $\omega\_1$-tree $T$ in the ground model, can Laver forcing add a cofinal branch to $T$? Assume GCH in the ground model.
Definitions:
An $\omega\_1$-tree is a well-founded tree of height $\omega\_1$ with all levels countable. A cofinal branch of $T$ is a maximal chain of type $\omega\_1$. Laver forcing $\mathbb{L}$ consists of subtrees $T$ of $\omega^{<\omega}$ where, for some $x\in T$, every $y<\_T x$ has only one child and every $y\geq\_T x$ has $\aleph\_0$ children. $q\leq\_{\mathbb{L}} p$ iff $q\subset p$.
Related results:
Silver proved in 1971 that a countably closed forcing cannot add a cofinal branch to $T$: given $p\_\varnothing$ forcing a new cofinal branch $B$, construct conditions $(p\_s:s\in 2^{<\omega+1})$ where $p\_t\leq p\_s$ for $s\subset t$ but for $s\perp t$ nodes $b\_s\perp b\_t$ of $T$ are respectively forced into $B$ by $p\_s$ and $p\_t$. This contradicts countability of any level $T\_\delta$ chosen high enough that $b\_s\in T\_{<\delta}$ for all finite $s$.
I can modify the above argument to show that Sacks forcing (perfect subtrees of $2^{<\omega}$) does not add a cofinal branch to $T$. Given $s\in 2^n$ and $p\_s$, I find, for each $y\in p\_s$ minimal with respect to having $n$ splitting nodes below it, conditions $$q\_{sy0},q\_{sy1}\leq p\_s|y=\{x\in p\_s: x\not\perp y\}$$ that respectively force nodes $c\_{sy0},c\_{sy1}$ into $B$ such that $c\_{sy0}\perp c\_{sz1}$ for all $y,z$. Then $p\_{si}=\bigcup\_y q\_{syi}$ preserves the first $n$ 'levels' of splitting nodes of $p\_s$, ensuring that $\{p\_{f|n}:n<\omega\}$ has a common extension $q\_f$ for each $f\in 2^{\omega}$. Finally, choose $\delta$ sufficiently high and then extend each $q\_f$ to $p\_f$ deciding $B$ at level $\delta$. If $f\not=g$, then $p\_f$ and $p\_g$ disagree about $B$ at level $\delta$.
The above fusion argument also applies to Miller forcing; just replace 'splitting' with 'infinitely splitting.' However, for Laver forcing, $p\_{si}=\bigcup\_y q\_{syi}$ above may fail to be a condition because the 'trunks' of the $q\_{syi}$ may be too long.
| https://mathoverflow.net/users/12106 | Does Laver forcing add cofinal branches to $\omega_1$-trees? | No.
In our paper <https://arxiv.org/abs/1409.4596> with Jindra Zapletal on Y-stuff we prove that Laver forcing is Y-proper. Consequently it has the $\omega\_1$-approximation property (i.e., does not add fresh sets of size $\omega\_1$ in Joel's terminology) and hence adds no new branches to trees.
PS: Thanks to Wolfgang for pointing me to this question.
| 4 | https://mathoverflow.net/users/33363 | 418446 | 170,381 |
https://mathoverflow.net/questions/418240 | 1 | I am trying to find an asymptotic formula for the following sum as $T \to \infty$.
$$ \sum\_{t = 1}^{T} \prod\_{\substack{p \; \textrm{prime} \\ p | t}} \rho(p) \frac{1 - \frac{1}{p^2}}{1 - \frac{\rho(p)}{p^2}}$$
where for a fixed even $c$, $\rho$ is defined on the primes as follows,
$$ \rho(p) = \begin{cases}
1 + \left( \frac{c}{p} \right) & (c, p) = 1\\
0 & \textrm{otherwise}
\end{cases} $$
So $\rho(p) = 2$ if $c$ is a quadratic residue of $p$ and $0$ otherwise. Note that the summand is multiplicative.
I tried breaking it into sub sums based on the number of prime factors and whether $t$ is a quadratic residue of small primes. But did not achieve much success.
I did some experiments on the computer, and it shows that the sum is asymptotic to $AT$ for some constant $A = A(c)$.
Edit: I am considering the sum only for $c$ not a perfect square.
| https://mathoverflow.net/users/167999 | Sum of an arithmetic sequence involving Euler factors | Let $c = 2^k m$ with $m$ odd. The analysis is slightly different depending on the parity of $k$ and on whether $m\equiv 1$ or $3\ \text{mod} 4$. For simplicity, I'll assume $k$ is even, $m\equiv 1\ \text{mod 4}$, and $m$ not a perfect square. In this case, quadratic reciprocity gives
$$
\rho(p) = 1 + \chi(p)
$$
whenever $(p,2m)=1$. Here $\chi$ denotes the quadratic Dirichlet character $(\frac{\cdot}{m})$. Let $f(n)$ denote the summand and consider the associated Dirichlet series
$$
D(s) = \sum\_{n=1}^\infty \frac{f(n)}{n^s}.
$$
By Perron's formula, we expect
$$
\sum\_{t=1}^T f(t) \sim \underset{s=1}{\text{Res}}\ \frac{D(s)T^s}{s}.
$$
Since $f$ is multiplicative, $D(s)$ admits an Euler product, and the goal now is to manipulate this Euler product into an expression involving known $L$-functions, along with some (potentially complicated) arithmetic factors. In this case, it will involve the Dirichlet $L$-function associated to $\chi$, the Riemann zeta function, and some other arithmetic factors. To see this, note that we can write, for $(n,2m)=1$,
$$
f(n) = \prod\_{p\mid n} \left(1+\chi(p) + \frac{\rho(p)}{p^2-\rho(p)}\right).
$$
This can be rewritten as a divisor sum
$$
\tag{1}
f(n) = \sum\_{d\mid n} \mu^2(d) g(d),
$$
where $g$ is the multiplicative function defined by
$$
g(p) = \chi(p) + \frac{\rho(p)}{p^2-\rho(p)}.
$$
Using convolution notation, we have $f = 1\ast \mu^2 g$. The nice thing about this is that if $f = g\ast h$ and $D\_f(s), D\_g(s), D\_h(s)$ are the Dirichlet series associated with $f,g,h$, respectively, then $D\_f(s) = D\_g(s) D\_h(s)$. Thus the representation (1) shows that our Dirichlet series $D(s)$ has the form
$$
D(s) = \zeta(s) \prod\_{p\mid 2m} \left(1-\frac{1}{p^s}\right) \prod\_{(p,2m)=1} \left(1 + \frac{\chi(p)}{p^s} +\frac{\frac{\rho(p)}{p^2-\rho(p)}}{p^s} \right).
$$
The last term in the second product behaves nicely for $\text{Re}(s) > -1$, so we rewrite this in the form
$$
D(s) = \zeta(s) \prod\_{(p,2m)=1} \left(1 + \frac{\chi(p)}{p^s}\right) H(s),
$$
where $H(s)$ is given by
$$
\begin{aligned}
H(s) &= \prod\_{p\mid 2m} \left(1-\frac{1}{p^s}\right) \prod\_{(p,2m)=1} \left(1 + \frac{\chi(p)}{p^s} +\frac{\frac{\rho(p)}{p^2-\rho(p)}}{p^s} \right) \left(1 + \frac{\chi(p)}{p^s}\right)^{-1} \\
&= \prod\_{p\mid 2m} \left(1-\frac{1}{p^s}\right)\prod\_{(p,2m)=1} \left(1 + \frac{\rho(p)}{(p^2-\rho(p)(p^s + \chi(p))} \right).
\end{aligned}
$$
The main feature of $H$ is that is converges absolutely in the half plane $\text{Re}(s) > -1$. We now write
$$
\begin{aligned}
\prod\_{(p,2m)=1} \left(1 + \frac{\chi(p)}{p^s}\right) &= \prod\_{(p,2m)=1} \left(1 - \frac{\chi(p)^2}{p^{2s}}\right)\left(1 - \frac{\chi(p)}{p^{s}}\right)^{-1} \\
&= \frac{L(s,\chi)}{\zeta(2s)} \left(1 - \frac{\chi(2)}{2^{s}}\right)\prod\_{p\mid 2m} \left(1 - \frac{1}{p^{2s}}\right)^{-1}.
\end{aligned}
$$
Thus
$$
D(s) = \frac{L(s,\chi)}{\zeta(2s)} G(s),
$$
where $G(s)$ is given by
$$
G(s) = H(s)\left(1 - \frac{\chi(2)}{2^{s}}\right)\prod\_{p\mid 2m} \left(1 - \frac{1}{p^{2s}}\right)^{-1}.
$$
Since $m$ is not a square, $L(s,\chi)$ has no pole at $s=1$, and so $D(s)$ has only a simple pole at $s=1$. Therefore
$$
\tag{2}
\underset{s=1}{\text{Res}}\ \frac{D(s)T^s}{s} = T \left(\frac{L(1,\chi)}{\zeta(2)}\right) G(1).
$$
The calculation is slightly different in the other cases. This comes from the fact that quadratic reciprocity works out slightly differently in these cases.
I should note that all of this can be made fully rigorous, so in the case dealt with above, your sum is (provably) asymptotic to the expression on the right of (2).
| 2 | https://mathoverflow.net/users/307675 | 418455 | 170,383 |
https://mathoverflow.net/questions/418436 | 0 | I have a question on [Lawler – Notes on the Bessel process](http://www.math.uchicago.edu/%7Elawler/bessel18new.pdf), on page 4. Let $X\_t$ be one-dimensional Brownian motion, and we want to use $N\_t$ as a measure-changing (local) martingale, defined as $$N\_t=\left(\frac{X\_t}{X\_0}\right)^a\exp\left(-\frac{a(a-1)}{2}\int\_0^t\frac{ds}{X\_s^2}\right),\quad t<T\_0,$$
up to the first visit to zero. It satisfies the SDE $$dN\_t=\frac{a}{X\_t}N\_tdX\_t,\quad N\_0=1.$$
Then it is supposed in the note that $0<\varepsilon<X\_0<R$, and $\tau=T\_{\varepsilon}\wedge T\_R$, as up to the first visit to $\varepsilon$ or $R$. Then it claims that $N\_s$ stopped by $\tau$, $$dN\_{t\wedge\tau}=\frac{a}{X\_{t\wedge\tau}}1(t<\tau)N\_{t\wedge\tau}dX\_t,$$
is uniformly bounded. **But why is this true?**
I have a rough thought that $N\_{t\wedge\tau}=N\_0+\int\_0^t\frac{a}{X\_{s\wedge\tau}}N\_{s\wedge\tau}dX\_{s\wedge\tau}\leq N\_0+\frac{a}{\varepsilon}\int\_0^tN\_{s\wedge\tau}dX\_{s\wedge\tau}$, a.s. as $t<\tau$. Then **if we have the stochastic version of Grönwall's inequality,** i.e. if we can read $N\_{t\wedge\tau}\leq N\_0\exp(\frac{a}{\varepsilon}X\_{t\wedge\tau})$, a.s., then it is uniformly bounded for sure. But I am quite pessimistic about whether this holds or not.
| https://mathoverflow.net/users/174600 | Uniform boundedness of this SDE? And possibly a stochastic Grönwall inequality? | On the time interval $[0,\tau]$, the exponential factor in the definition of $N\_t$ is bounded below by $0$ and above by $\exp(Kt)$ for some constant $K=K(\epsilon,R)\ge 0$. Therefore $0<N\_{t\wedge\tau}\le(R/\epsilon)^a\exp(K\cdot (t\wedge \tau))$. This is "uniform" provided you ignore the dependence on $t$.
| 2 | https://mathoverflow.net/users/42851 | 418456 | 170,384 |
https://mathoverflow.net/questions/418386 | 2 | Let $V$ be a nonempty, irreducible, smooth projective variety over $\mathbf{C}$.
>
> Is there a smooth projective variety $X$ over $\mathbf{C}$, a surjective map $X\to V$ of varieties over $\mathbf{C}$, such that $X$ contains as a dense open $\mathbf{C}$-subscheme some $\mathbf{C}$-group scheme $G$ of finite type?
>
>
>
Abelian varieties and toric varieties are of course examples with $X=V$.
| https://mathoverflow.net/users/nan | Compactifications of group varieties | Smooth projective curves of genus $>1$ are counterexamples. To see this, you can use the following lemma.
**Lemma.** Let $X$ be a proper integral variety over $\mathbb{C}$. Then the following are equivalent.
1. For every abelian variety $A$, every morphism $A\to X$ is constant.
2. For every finite type connected group scheme $G$ over $\mathbb{C}$, every morphism $G\to X$ is constant.
A proof of this lemma is given in Lemma 2.5 of arxiv.org/abs/1807.03665
To show that smooth projective curves of genus $g>1$ give counterexamples, we can argue as follows:
>
>
> >
> > Let $X$ be a smooth projective curve of genus $g>1$. Then, every morphism from an abelian variety $A$ to $X$ is constant. This can be seen by using the uniformisation of $A$ by affine $\dim(A)$-space or by pulling-back differentials. Now, by the above Lemma, every morphism $G\to X$ is constant, where $G$ is any finite type connected group scheme over $\mathbb{C}$. But this implies that every morphism $Y\to X$ is constant, where $Y$ is a variety containing a dense open isomorphic to the variety underlying a finite type connected group scheme.
> >
> >
> >
>
>
>
Note: Any hyperbolic variety gives a counterexample. For example, the moduli space of genus $q$ ($q>1$) smooth proper curves with level $N$ ($N>3$) structure.
| 3 | https://mathoverflow.net/users/4333 | 418460 | 170,387 |
https://mathoverflow.net/questions/418375 | 2 | Let $L\_1$ and $L\_2$ be two nonintersecting picewise-linear or smooth knots in $\mathbb R^3$. Suppose they are ambient isotopic. Does there exist an embedded surface $f: S^1\times[0,1]\to \mathbb R^3$ such that $L\_1=f(\cdot,0)$ and $L\_2=f(\cdot,1)$. This is similar to concordance but stronger -- this "concordance" must be realised in $\mathbb R^3$ instead of $\mathbb R^3 \times [0,1]$.
If necessary, assume that $L\_1,L\_2$ are not linked or separated far from each other.
In other words is there a Seifert surface that is homeomorphic to a cylinder for a link consisting of two isotopic knots.
| https://mathoverflow.net/users/13842 | An equivalence relation on knots similar to concordance | Take a knot, and another copy of the same knot far away. It is then a nice instructive exercise to prove that if they are not unknots they do not cobound annuli.
| 2 | https://mathoverflow.net/users/6205 | 418472 | 170,393 |
https://mathoverflow.net/questions/418490 | -1 | I am trying to understand the set
$$\mathcal{A}=\{f:f\text{ non-negative, measurable}, \text{vanishes at infinity and } f(x)=f^\*(x)\}$$
where $f:\mathbb{R}^n\to \mathbb{R}$ is vanishing at infinity if for all $t>0$ such that $\text{Vol}(\{x:f(x)>t\})<+\infty$ and $f^\*$ denotes the symmetric decreasing rearrangement.
From elementary properties of $f^\*$ (see [Burchard - A short course on rearrangement inequalities](http://www.math.utoronto.ca/almut/rearrange.pdf)), we know that $f^\*$ satisfies
1. $f^\*$ is a lower semi-continuous, radial, and decreasing function. Thus $\mathcal{A}$ consists of lower semi-continuous functions.
2. I also know that the operator which sends a function to its symmetric rearrangement is not always continuous on the space of $W^{1,p}$ functions.
I am curious to know if there are any results that characterize functions in the set $\mathcal{A}$ or in other words what are the possible functions that are equal to their rearrangement?
I understand that there can be many examples of such functions so perhaps it makes sense to ask if we can classify functions in the set $\mathcal{A}\cap L^p$, $\mathcal{A}\cap W^{k,p}$, etc.
| https://mathoverflow.net/users/68232 | What functions are equal to their symmetric decreasing rearrangement? | It should be straightforward to verify that $\mathcal A$ consists *exactly* of the lower semi-continuous radial and decreasing functions (which are negative and vanish at $\infty$): You already know that every $f\in\mathcal A$ has these properties, so you only have to verify that if $f$ has these properties, then $f=f^\*$. If the definition is of any value, this should be rather straightforward to do (start with the classical case $n=1$).
| 3 | https://mathoverflow.net/users/165275 | 418491 | 170,396 |
https://mathoverflow.net/questions/418489 | 3 | I have a question concerning certain elements having zero trace in a finite field extension and I do have the feeling that additive characters should play a role, but I am not sure how. I am stating the problem slightly more generally than what I am really interested in.
Let $q$ be prime power, let $\mathbb{F}\_{q}$ the finite field of cardinality $q$ and let $n$ be a prime with $n$ dividing $q+1$. I denote with $\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}:\mathbb{F}\_{q^{2n}}\to \mathbb{F}\_{q^2}$ the trace map.
I am interested in the cardinality of the set $$\{x\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(x)=0 \hbox{ and } x=y^n \textrm{ for some }y\in \mathbb{F}\_{q^{2n}}\}.$$
For instance when $n=3$, I have a bit of evidence that that this set has cardinality $(q+1)^2(q^2-1)/3$.
| https://mathoverflow.net/users/45242 | Number of certain elements in a finite field having zero trace | We have
$$ \# \{x\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(x)=0 \hbox{ and } x=y^n \textrm{ for some }y\in \mathbb{F}\_{q^{2n}}\} = \frac{\#\{y\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n)=0 \} +n-1}{ n}$$
since $\mathbb F\_{q^n}$ contains all the $n$th roots of unity, so it suffices to calculate.
\begin{align\*}
\# \{y\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n)=0 \} &= \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2} } \sum\_{ y \in \mathbb F\_{q^{2n}} }\psi\_2 (\lambda \mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n) ) \\ &= \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2} } \sum\_{ y \in \mathbb F\_{q^{2n}} }\psi\_{2n} (\lambda y^n) ) \\& = q^{2n-2} + \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2}^\times } \sum\_{ y \in \mathbb F\_{q^{2n}} }\psi\_{2n} (\lambda y^n) ) \\ &=q^{2n-2} + \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2}^\times } \sum\_{ x \in \mathbb F\_{q^{2n}} } \sum\_{\substack{ \chi \colon \mathbb F\_{q^{2n}}^\times \to \mathbb C^\times \\ \chi^n=1 } } \psi\_{2n} (\lambda x) \chi(x) \\ &=q^{2n-2} + \frac{1}{q^2} \sum\_{\lambda \in \mathbb F\_{q^2}^\times } \sum\_{ x \in \mathbb F\_{q^{2n}} } \sum\_{\substack{ \chi \colon \mathbb F\_{q^{2n}}^\times \to \mathbb C^\times \\ \chi^n=1 } } \psi\_{2n} (x) \chi(\lambda^{-1} x) \end{align\*}
where $\psi\_2 \colon \mathbb F\_{q^2} \to \mathbb C^\times$ is a nondegenerate additive character and $\psi\_{2n} = \psi\_2 \circ \mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}$
Now since $\chi$ has order $n$ and the quotient group $\mathbb F\_{q^{2n}}^\times / \mathbb F\_{q^2}^\times $ has order $q^{2n-2} + q^{2n-4} + \dots + 1$ which is divisible by $n$, $\chi$ must factor through this quotient group so $\chi(\lambda^{-1})=1$, giving
$$\#\{y\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n)=0 \} = q^{2n-2} + \frac{q^2-1}{q^2} \sum\_{\substack{ \chi \colon \mathbb F\_{q^{2n}}^\times \to \mathbb C^\times \\ \chi^n=1 } } \sum\_{ x \in \mathbb F\_{q^{2n}} } \psi\_{2n} (x) \chi( x) .$$
Here the inner sum $\sum\_{ x \in \mathbb F\_{q^{2n}} } \psi\_{2n} (x) \chi( x)$ vanishes for $\chi$ trivial and is a Gauss sum of absolute value exactly $q^{n}$ for $\chi$ trivial. This gives an estimate with main term and error term
$$ \left| \#\{y\in \mathbb{F}\_{q^{2n}}|\mathrm{Tr}\_{\mathbb{F}\_{q^{2n}}/\mathbb{F}\_{q^2}}(y^n)=0 \} -q^{2n-2} \right| \leq (n-1) (q^2-1) q^{n-2} $$
| 4 | https://mathoverflow.net/users/18060 | 418508 | 170,403 |
https://mathoverflow.net/questions/418266 | 17 | I'm interested in pairs $A=(a\_{i,j})\_{i,j=0,1,\ldots}$ and $B=(b\_{i,j})\_{i,j=0,1,\ldots}$ of infinite matrices for which:
* They are uni-lower-triangular, i.e., $a\_{i,i}=1$ for all $i$ and $a\_{i,j}=0$ for all $j>i$.
* They are inverses, i.e. their product $AB$ is the identity.
* The generating function of the $k$th column of $A$ is the reciprocal of the generating function of the $(k+1)$st row of $B$, that is,
$$ \sum\_{j=k}^{\infty}a\_{j,k} \; x^{j-k}= \big(\sum\_{i=0}^{k+1}b\_{k+1,k+1-i} \; x^{i}\big)^{-1}.$$
Notice how for the generating function of columns we start at the $1$ on the diagonal and go down, while for the generating function of rows we start at the $1$ on the diagonal and go left (and all g.f.'s are normalized to start with the constant term).
I have a handful of examples of this occurring with matrices of combinatorial significance:
**Example 1**. We let $a\_{i,j}=\binom{i}{j}$ and $b\_{i,j}=(-1)^{i-j}\binom{i}{j}$. Note the generating function of the $k$th row of $B$ is $(1-x)^{k}$, and the generating function of the $(k-1)$th column of $A$ is $1/(1-x)^{k}$.
**Example 2**. We let $a\_{i,j}=S(i+1,j+1)$ and $b\_{i,j} = s(i+1,j+1)$, where $S(n,k)$ and $s(n,k)$ are the Stirling numbers of the 2nd and 1st kind, respectively (the shift by one is just to match my convention that indexing of rows/columns starts at $0$). This looks like
$$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 1 & 1 & 0 & 0 \\ 1 & 3 & 1 & 0 \\ 1 & 7 & 6 & 1 \\ \vdots & & & & \ddots \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ -1 & 1 & 0 & 0 \\ 2 & -3 & 1 & 0 \\ -6 & 11 & -6 & 1 \\ \vdots & & & & \ddots \end{pmatrix}$$
Note that the generating function of the $k$th row of $B$ (and the reciprocal of the generating function of the $(k-1)$th column of $A$) is $(1-x)(1-2x)\cdots(1-kx)$.
**Example 3**. The previous examples were over $\mathbb{Z}$, this example is over $\mathbb{Z}[q]$; actually it is a $q$-analog of Example 1. We let $a\_{i,j} = \binom{i}{j}\_q$ be the usual $q$-binomial $\binom{i}{j}\_q = \frac{(1-q^i)(1-q^{i-1})\ldots(1-q^{i-j+1})}{(1-q^j)(1-q^{j-1})\ldots(1-q)}$, and we let $b\_{i,j} = (-1)^{i-j} \; q^{\binom{i-j}{2}} \; \binom{i}{j}\_q$. This looks like:
$$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 1 & 1 & 0 & 0 \\ 1 & q+1 & 1 & 0 \\ 1 & q^{2}+q+1 & q^{2}+q+1 & 1 \\ \vdots & & & & \ddots \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ -1 & 1 & 0 & 0 \\ q & -(q+1) & 1 & 0 \\ -q^3 & q^3+q^2+q & -(q^2+q+1) & 1 \\ \vdots & & & & \ddots \end{pmatrix}$$
Note that the generating function of the $k$th row of $B$ (and the reciprocal of the generating function of the $(k-1)$th column of $A$) is $(1-x)(1-qx)\cdots(1-q^{k-1}x)$.
**Question 1**. What is going on here? Why are these pairs of matrices of combinatorial sequences inverse "in two ways"? How is being inverse in one way related to being inverse in the other way?
Note that all of these examples come from sequences of uniform posets: they are the Whitney numbers of the 2nd and 1st kind of these posets. Hence, this question is related to [my previous question](https://mathoverflow.net/questions/417198/interesting-uniform-posets). (But it's easy to come up with sequences of uniform posets whose matrices are not "inverse in the 2nd way.")
Also note that in all the examples, the generating function of the $(k+1)$st column of $A$ is obtained from the generating function of the $k$th column by multiplying by a simple factor; but (except for Example 1), it is not exactly the same factor every time, so these matrices are not quite [Riordan arrays](https://en.wikipedia.org/wiki/Riordan_array).
**Question 2**. Can you find some more examples of matrices of combinatorially significant sequences with these properties?
**EDIT**: One further observation is that you can take any such pair $A=(a\_{i,j})$, $B=(b\_{i,j})$ of matrices and get another one $A'$, $B'$ by choosing some constant $\kappa$ and setting $a'\_{i,j}=\kappa^{i-j} \; a\_{i,j}$, $b'\_{i,j}=\kappa^{i-j} \; b\_{i,j}$. So for example from the Pascal's triangle example you can (basically) get the $f$-vectors of the cross polytope this way.
| https://mathoverflow.net/users/25028 | Matrices of combinatorial sequences that are inverse in two ways | Following up on David's nice answer, there is a different parametrization that makes the pattern much more obvious. Namely, let $s\_1,s\_2,\dots$ be arbitrary, then you can write
$$A=\Big(h\_{i-j}(s\_1,s\_2,\dots, s\_{j+1})\Big)\_{i,j=0}^{\infty}$$
$$B=\Big((-1)^{i-j}e\_{i-j}(s\_1,s\_2,\dots, s\_{i})\Big)\_{i,j=0}^{\infty}$$
And checking that the matrices and the corresponding generating functions are inverses becomes a little more clear. The $h\_i, e\_i$ denote the complete homogeneous and elementary symmetric functions respectively, with the convention $h\_0, e\_0 =1$.
| 15 | https://mathoverflow.net/users/2384 | 418510 | 170,404 |
https://mathoverflow.net/questions/418466 | 2 | Consider the periodic Riccati equation $y'(x)=y(x)^2+q(x)$ on the real line $\mathbb{R}$, where $q\in C^\infty(\mathbb{R})$ is a periodic function with period $T=1$. Suppose $q(x)$ can take both positive and negative values but $\int\_0^1 q(x)dx<0$. Does a periodic solution to this Riccati equation exist? Any references? Thanks a lot.
| https://mathoverflow.net/users/119968 | Existence of periodic solutions to scalar Riccati equations | The answer is negative.
The MSN review says that this is the contents of
MR1466035
Tang, Fenjun
The periodic solutions of Riccati equation with periodic coefficients.
Ann. Differential Equations 13 (1997), no. 2, 165–169.
I looked at this paper and the argument there makes no sense.
Setting $y=-w'/w$ we obtain $w''+qw=0$, and periodicity
of $y$ means $w(x+T)=\lambda w(x)$ for some $\lambda\neq 0$, so that $w$ is an eigenfunction of the shift operator. For generic $q$ there will be two such eigenfunctions (with distinct eigenvalues), and the question is whether at least one of them is free of zeros on $[0,T]$.
Here is a simple counterexample. Let $T=2\pi$, and
$q(x)=1$ for $0\leq x\leq 3\pi/2$, and then extend $q$ to a smooth
$2\pi$ periodic function arbitrarily, but so that
$$\int\_0^{2\pi}q(x)dx<0.$$
Then the general solution $w$ of our linear equation will
be of the form $w(x)=A\cos(x-\alpha),\; 0<x<3\pi/2$, so every solution
has at least one zero. This means that the corresponding Riccati equation has no smooth real solutions $y=-w'/w$ defined on the real line, thus no
periodic solutions.
| 4 | https://mathoverflow.net/users/25510 | 418513 | 170,405 |
https://mathoverflow.net/questions/418142 | 8 | Let $G = \text{GL}\_n(\mathbb{C})$ and let $N\_+$ be the subgroup of upper triangular matrices with $1$'s on the diagonal. Let $w$ be a permutation, let $B\_+ w B\_+$ be the Bruhat cell and let $\overline{B\_+ w B\_+}$ be its closure in $G$. I want to describe the ring of functions on $\overline{B\_+ w B\_+}$ which are invariant for $N\_+ \times N\_+$ acting on the left and right. I believe I know what the answer is, and I would like to know if I am right and get references to previous work.
Additional notation: Let $[n] := \{ 1,2,\ldots, n \}$. Let $T$ be the torus of diagonal matrices in $G$, and let $(t\_1, t\_2, \dots, t\_n)$ be the entries of the diagonal matrix. Recall that the permutation matrix $w$ has $1$'s in position $(w(j), j)$.
Here is what I believe the answer to be. For any $I$, $J \subset [n]$ with $|I| = |J|$, and $g \in \text{GL}\_n$, let $\Delta\_{I,J}(g)$ be the minor with rows $I$ and columns $J$. Define a partial order $\prec$ on $[n]$ by $j\_1 \prec j\_2$ iff $j\_1 < j\_2$ and $w(j\_1) > w(j\_2)$. So, for $w= w\_0$, this is the total order $1 \prec 2 \prec \cdots \prec n$ and, for $w = e$, all the elements of $[n]$ are incomparable.
Let $J$ be any lower order ideal for this partial order. Then one can verify that the minor $\Delta\_{w(J), J}(g)$ is invariant for the $N\_+ \times N\_+$ action on $\overline{B\_+ w B\_+}$. We'll call this $\Delta(J)$. In addition, we have $\Delta([n]) = \Delta\_{[n], [n]}(g)=\det(g)$, so $\Delta([n])$ is a unit and we have to include $\Delta([n])^{-1}$ in our ring of invariants. So, question:
>
> Is the ring of $N\_+ \times N\_+$ invariant functions genearted by the $\Delta(J)$'s, and by $\Delta([n])^{-1}$? What is a reference for this?
>
>
>
I'll close by noting there is a good, explicit description of the ring generated by the $\Delta(J)$'s. If we restrict $\Delta(J)$ to $wT \subset B\_+ w B\_+$, we see that $\Delta(J) = \pm \prod\_{j \in J} t\_j$, so we can think of this as the monomial $\prod\_{j \in J} t\_j$ on $wT \cong N\_+ \backslash B\_+ w B\_+ / N\_+$. Since each orbit of $N\_+ \times N\_+$ on $B\_+ w B\_+$ contains a unique point in $wT$, the relations between the $\Delta(J)$'s are exactly the same as the relations between these monomials on $T$. Any product of the $\Delta(J)$'s gives a monomial of the form $\prod t\_j^{a\_j}$ where $j\_1 \prec j\_2$ implies $a\_{j\_1} \geq a\_{j\_2}$. In other words, the ring generated by the $\Delta(J)$'s is the semigroup ring corresponding to the semigroup of linear extensions of $([n], \prec)$, and is closely related to [Stanley's order polyope](http://dedekind.mit.edu/%7Erstan/pubs/pubfiles/66.pdf) $\mathcal{O}(\prec)$.
| https://mathoverflow.net/users/297 | Coordinates on $N_+ \backslash \overline{B_+ w B_+} / N_+$ | I have proved that the invariant ring is as I expected. As discussed in the question, $N\_+ \backslash B\_+ w B\_+ / N\_+ \cong T$, so the ring of $N\_+ \times N\_+$ invariants on $B\_+ w B\_+$ is the coordinate ring of $T$, which is the Laurent polynomial ring in the $t\_i$.
Note, let $R$ be the ring of $N\_+ \times N\_+$ invariants on $\overline{B\_+ w B\_+}$, so $R$ is a subring of $\mathbb{C}[t\_1^{\pm}, t\_2^{\pm}, \ldots, t\_n^{\pm}]$. Moreover, the $T$ action on $\overline{B\_+ w B\_+}$ normalizes $N\_+ \times N\_+$, so we get a $T$-action on $R$, so $R$ must be a $T$-invariant subspace of $\mathbb{C}[t\_1^{\pm}, t\_2^{\pm}, \ldots, t\_n^{\pm}]$. So $R$ is a monomial subring of $\mathbb{C}[t\_1^{\pm}, t\_2^{\pm}, \ldots, t\_n^{\pm}]$. We need to show that a monomial $t\_1^{a\_1} t\_2^{a\_2} \cdots t\_n^{a\_n}$ extends to a function on $\overline{B\_+ w B\_+}$ if and only if $j\_1 \prec j\_2$ implies $a\_{j\_1} \geq a\_{j\_2}$. The question shows that, if $a$ obeys these inequalities, then $t\_1^{a\_1} t\_2^{a\_2} \cdots t\_n^{a\_n}$ does extend to a function on $\overline{B\_+ w B\_+}$ , so what we need is the reverse implication.
So, suppose that there are $j\_1$ and $j\_2$ with $j\_1 \prec j\_2$ and $a\_{j\_1} < a\_{j\_2}$. We may assume that $(j\_1, j\_2)$ is a cover of $\prec$, meaning that there is no $j$ with $j\_1 \prec j \prec j\_2$. In this case, $w$ covers $w (j\_1 j\_2)$ in the strong Bruhat order, so $B\_+ w (j\_1 j\_2) B\_+$ is a divisor in $\overline{B\_+ w B\_+}$. We will show that the function $\prod t\_j^{a\_j}$ does not extend to $B\_+ w (j\_1 j\_2) B\_+$.
Put $i\_1 = w(j\_1)$, $i\_2 = w(j\_2)$. Consider matrices $X$ whose only nonzero entries are in positions $(w(j), j)$, together with $(i\_1, j\_2)$ and $(i\_2, j\_1)$. As long as $X\_{i\_1 j\_1}$ is nonzero, this is in $B\_+ w B\_+$; when $X\_{i\_1 j\_1}$ becomes $0$, we pass into $B\_+ w (j\_1 j\_2) B\_+$. We evaluate the functions $t\_j$ on such a matrix: We have $t\_j = X\_{w(j) j}$ for $j \neq j\_1$, $j\_2$, we have $t\_{j\_1} = X\_{i\_1 j\_1}$, and we have
$t\_{j\_2} = \frac{X\_{i\_2 j\_1} X\_{i\_1 j\_2} - X\_{i\_1 j\_1} X\_{i\_2 j\_2}}{X\_{i\_1 j\_1}}$.
We assumed that $a\_{j\_1} < a\_{j\_2}$. Plugging the above formulas into the product $\prod t\_j^{a\_j}$, we see that $X\_{i\_1 j\_1}$ appears with exponent $a\_{j\_2} - a\_{j\_1}$ in the denominator. So, as $X\_{i\_1 j\_1} \to 0$, the monomial $\prod t\_j^{a\_j}$ blows up, and thus doesn't extend to $B\_+ w (j\_1 j\_2) w B\_+$.
Still glad to hear references!
| 2 | https://mathoverflow.net/users/297 | 418515 | 170,406 |
https://mathoverflow.net/questions/418499 | 1 | Let $n$ and $d$ be large positive integers with $n \le d^\gamma$, for some absolute constant $\gamma>0$; i.e., $n$ is at most polynomial in $d$. Let $x\_1,\ldots,x\_n,x\_{n+1}$ be drawn iid from the uniform distribution on the sphere $S\_{d-1}(\sqrt{d})$ of radius $\sqrt{d}$ in $\mathbb R^d$, and consider the random variable $\Delta(n,d) := \min\_{1 \le i \le n}\|x\_{n+1}-x\_i\|$. Let $\alpha \in (0,1)$.
>
> **Question.** *What is a good upper-bound for $\Delta(n,d)$, perhaps valid with probability at least $\alpha$ ?*
>
>
>
One would expect $\Delta(n,d)=o(\sqrt{d})$, i.e., $\Delta(n,d)/\sqrt{d} \to 0$ in the limit $d \to \infty$ w.h.p.
A crude (and possibly very bad) estimate
----------------------------------------
Because the sphere $S\_{d-1}(\sqrt{d})$ can be covered with $N\_d(\varepsilon) \le (\sqrt{d}/\varepsilon)^d$ (euclidean) balls of radius $\varepsilon$, it is clear that $\Delta(n,d) \le \sqrt{d}/n^{1/d} \asymp \sqrt{d}$ with probability tending to $1$ with $d$. However, this bound is very far from my target, namely $o(\sqrt{d})$.
| https://mathoverflow.net/users/78539 | For $x_1,...,x_n$ iid random on sphere of radius $\sqrt{d}$ in $R^d$, what is a good upper-bound on min distance of $x_{n}$ from the other $x_i$'s? | $\newcommand{\De}{\Delta}\newcommand{\R}{\mathbb R}\newcommand{\ga}{\gamma}\newcommand{\Ga}{\Gamma}$This is to provide a detalization on Will Sawin's comment. Specifically, let us show that the best upper bound on $\De:=\De(n,d)$ is $\sim\sqrt{2d}$ (as $d\to\infty$ and $d^\ga\ge n\to\infty$).
Indeed, for real $m>0$, let
\begin{equation\*}
p:=P(\De>m)=P(\min\_{1\le i\le n}\|x\_{n+1}-x\_i\|>m). \tag{1}\label{1}
\end{equation\*}
We want to choose $m$ so that $p$ be bounded away from $0$. In other words, writing
\begin{equation\*}
p=e^{-u},
\end{equation\*}
we want to choose $m$ so that
$$u=O(1).$$
Let $e\_1:=(1,0,\dots,0)\in\R^d$. Note that $x\_1/\sqrt d$ equals $G/\|G\|$ in distribution, where $G=(G\_1,\dots,G\_n)$ is a standard Gaussian random vector in $\R^d$.
So, by spherical symmetry,
\begin{equation\*}
\begin{aligned}
e^{-u}=p&=P(\min\_{1\le i\le n}\|\sqrt d\, e\_1-x\_i\|>m) \\
&=P(\|\sqrt d\, e\_1-x\_1\|>m)^n \\
&=P\Big(1-\frac{e\_1\cdot x\_1}{\sqrt d}>\frac{m^2}{2d}\Big)^n \\
&=P\Big(1-\frac{G\_1}{\|G\|}>\frac{m^2}{2d}\Big)^n \\
\end{aligned}
\end{equation\*}
where $e\_1:=(1,0,\dots,0)\in\R^d$ and $\cdot$ is the dot product. So,
\begin{equation\*}
P\Big(1-\frac{G\_1}{\|G\|}>\frac{m^2}{2d}\Big)=e^{-u/n}=1-(1+o(1))\frac un. \tag{2}\label{2}
\end{equation\*}
If $m\ge\sqrt{2d}$, then $P\big(1-\frac{G\_1}{\|G\|}>\frac{m^2}{2d}\big)\le P(G\_1<0)=1/2$, which contradicts \eqref{2} for all large enough $n$. So, without loss of generality,
\begin{equation\*}
0< m<\sqrt{2d} \tag{3}\label{3}
\end{equation\*}
and hence
\begin{equation\*}
\begin{aligned}
&P\Big(1-\frac{G\_1}{\|G\|}>\frac{m^2}{2d}\Big) \\
&=P\Big(G\_1<0,1-\frac{G\_1}{\|G\|}>\frac{m^2}{2d}\Big) \\ &+P\Big(G\_1>0,1-\frac{G\_1}{\|G\|}>\frac{m^2}{2d}\Big) \\
&=P(G\_1<0)+P\Big(G\_1>0,1-\frac{|G\_1|}{\|G\|}>\frac{m^2}{2d}\Big) \\
& =\frac12+\frac12\,P\Big(1-\frac{|G\_1|}{\|G\|}>\frac{m^2}{2d}\Big).
\end{aligned}
\end{equation\*}
So,
\begin{equation\*}
P\Big(1-\frac{G\_1}{\|G\|}>\frac{m^2}{2d}\Big)=\frac12+\frac12\,P(Y>z),
\end{equation\*}
where
\begin{equation\*}
Y:=1-\frac{G\_1^2}{\|G\|^2},\quad z:=1-\Big(1-\frac{m^2}{2d}\Big)^2. \tag{4}\label{4}
\end{equation\*}
So, in view of \eqref{2}, $P(Y>z)=1-(2+o(1))\frac un$, that is,
\begin{equation\*}
P(Y\le z)\sim\frac{2u}n\ge\frac{2u}{d^\ga}. \tag{5}\label{5}
\end{equation\*}
Next, $Y$ has the beta distribution with parameters $\frac{d-1}2,\frac12$. So, for
\begin{equation\*}
c\_d:=\frac{\Ga(d/2)}{\Ga(1/2)\Ga((d-1)/2)}\sim\sqrt{\frac{d}{2\pi}}, \tag{6}\label{6}
\end{equation\*}
we have
\begin{equation\*}
\begin{aligned}
P(Y\le z)&=c\_d \int\_0^z y^{(d-3)/2}(1-y)^{-1/2}\,dy \\
&\le c\_d z^{(d-3)/2}\int\_0^z (1-y)^{-1/2}\,dy \\
&\le 2c\_d z^{(d-1)/2}.
\end{aligned}
\tag{7}\label{7}
\end{equation\*}
By \eqref{4}, \eqref{3}, \eqref{7}, \eqref{5}, and \eqref{6},
\begin{equation\*}
1\ge z^{(d-1)/2}\gtrsim\frac{u\sqrt{2\pi}}{d^{\ga+1/2}},
\end{equation\*}
which implies $z\to1$.
Thus, by \eqref{4}, $m\sim\sqrt{2d}$, as claimed.
| 5 | https://mathoverflow.net/users/36721 | 418520 | 170,408 |
https://mathoverflow.net/questions/418532 | 5 | For each first-order theory $T$ there is an associated [weak syntactic category](https://www.math.ias.edu/~lurie/278xnotes/Lecture2-Syntax.pdf), sometimes also called "the category of definable sets of $T$" and denoted $\mathrm{Def}(T)$.
Also, for each theory $T$ there is an associated theory $T^\mathrm{eq}$ which can be obtained from $T$ by adding quotient sorts for each definable equivalence relation, see e.g. [here 2.3](http://math.univ-lyon1.fr/~jimenez/Topics_in_Stability_Theory.pdf).
The nLab [claims](https://ncatlab.org/nlab/show/elimination+of+imaginaries#relation_to_indrepresentable_power_objects) that $\mathrm{Def}(T^\mathrm{eq})$ has coproducts.
**Question:** Why? Intuitively, coproducts correspond to the existence of *sum types*. But in $T^\mathrm{eq}$ we only added quotient types.
| https://mathoverflow.net/users/478652 | Why does the category of definable sets of $T^\mathrm{eq}$ have coproducts? | In general, a theory is called proper if every sort is nonempty (in every model) and there is a sort which has (in every model) at least two elements. Then we have the following theorem: if $\mathbb{T}$ is any proper theory, the syntactic category of $\mathbb{T}^{eq}$ is a pretopos. This is a consequence of the following (non trivial) theorem, which is a generalization of the result found in [1]: call a coherent category proper if every object is a subobject of a nonempty one and if there is a decidable object $D$ such that $D$ and $\neg D$ are nonempty, where $\neg D$ is the complement of the diagonal $\Delta: D \to D \times D$. Then the theorem says that any proper coherent exact category is a pretopos.
In the case of a classical first-order theory, whose models have at least two elements, the syntactic category of $\mathbb{T}^{eq}$ is an exact coherent category, and it will be proper, since $\mathbb{T}$ is. Hence, it is a pretopos and thus it has disjoint coproducts. In fact, for a proper coherent category, the exact completion and the coproduct completion are equivalent.
[1] Victor Harnik: "Model theory vs. categorical logic: two approaches to pretopos completion". In Bradd Hart et al., editor, Models, logics, and higher-dimensional categories: a tribute to the work of Mihaly Makkai, volume 53 of CRM Proceedings and Lecture Notes. American Mathematical Society, Providence, R.I., 2011.
| 6 | https://mathoverflow.net/users/12976 | 418535 | 170,409 |
https://mathoverflow.net/questions/418528 | 2 | $\newcommand{\loc}{\mathrm{loc}}$Let $\Omega$ be a bounded open set (smooth as we wish if necessary) in $\mathbb{R}^n$, $(\omega\_k)$ a sequence of open subsets whose closure is contained in $\Omega$ and whose union covers $\Omega$. Let $(u\_k)\_{k\in\mathbb{N}}$ a sequence in $H^1(\Omega)$ and assume the uniform bound
$$
\| u\_k \|\_{H^1(\omega\_k)} \leq M
$$
for every $k\in\mathbb{N}$, with the positive constant $M>0$ independent of $k$.
By a diagonal argument, we can infer the existence of a subsequence (not relabeled here) $(u\_k)\_{k\in\mathbb{N}}$ which weakly converge in $H^1\_{\loc}(\Omega)$ to an element $u\in H^1(\Omega)$. Therefore
$$
(\*)\qquad \qquad \int\_\omega \nabla u\_k \cdot \psi \to \int\_\omega \nabla u \cdot \psi
$$
for every open set $\omega$ whose closure is contained in $\Omega$, and every $\psi\in H^1(\Omega,\mathbb{R}^n)$.
**Question.** Given that both $(u\_k)\_{k\in\mathbb{N}}$ and the weak limit $u$ belong to $H^1(\Omega)$, is this sufficient to infer that we have weak convergence in $H^1$ instead of $H^1\_{\loc}$? In other words, is it true that $(\*)$ can be replaced by$$
(\*\*)\qquad \qquad \int\_\Omega \nabla u\_k \cdot \psi \to \int\_\Omega \nabla u \cdot \psi
$$
for every $\psi\in H^1(\Omega,\mathbb{R}^n)$?.
**Update** I realized that the way it is formulated is trivial. I was trying to minimize the number of formulas to write and I came up with this formulation that I was thinking equivalent to my real question but it is not. Nevertheless I am going to accept the answer because it answers to the question I wrongly asked.
| https://mathoverflow.net/users/41568 | Weak convergence in $H^1_{\mathrm{loc}}$ | The sequence $(u\_k)$ need not converge in $H^1(\Omega)$ under the given hypotheses. Indeed, recall that a weakly convergent sequence is bounded. Therefore, a sequence whose terms have $\lVert u\_k \rVert\_{1;\Omega} \to \infty$ but $\lVert u\_k \rVert\_{1;\omega\_k} \leq M$ would not converge weakly in $\Omega$.
| 2 | https://mathoverflow.net/users/103792 | 418536 | 170,410 |
https://mathoverflow.net/questions/418325 | 4 | I am a PhD student and during my research I was presented to the claim that
**For a positive definite function $f:\mathbb{R}\to \mathbb{R}$ continuous in $0$, with $0$ a stable point at $t=0$ for $x$, one has
$${\lim\inf}\_{t\to\infty} f(x(t))=0\Longrightarrow {\lim\inf}\_{t\to\infty} \|x(t)\|=0.$$**
In this context,
>
> We say that a function $f:\mathbb{R}\to \mathbb{R}$ is positive definite if $f(x)\geq 0$ and $f(x)=0\iff x=0.$
>
>
>
>
> We say that $p$ is a stable point at $t\_0$ if, for any neighborhood $H$ of $p$, there's a neighboorhood $V$ of $p$ s.t. if $x(t\_0)\in V$, then $x(t)\in H$ for all $t>t\_0.$
>
>
>
I thought this is not true without more hypothesis. Could anyone have an ideia to (dis)prove this? If this is not true, any ideia about extra hypothesis? Thank you.
| https://mathoverflow.net/users/172600 | $\lim\inf_{t\to \infty} f(x(t))=0\Longrightarrow \lim\inf_{t\to\infty} \|x(t)\|=0$ | Here is my take on your question. Let $f:\mathbb R\to\mathbb R\_+$ be a function such that $\{x:f(x)=0\}=\{0\}$ and $f$ is continuous at zero. Let $\phi:\mathbb R\to\mathbb R$ be locally Lipschitz continuous, so that the differential equation $x'=\phi(x)$ admits a unique local solution for any given initial condition $x\_0$. We denote the solution by $X:(t,x\_0)\mapsto X\_t(x\_0)$. Suppose also that $0$ is stable for this equation, in the sense that there exists a neighbourhood of zero from which all solutions are defined for all times, and $\lim\_{r\to0}\sup\_{|x\_0|\leq r}\sup\_{t\geq0}|X\_t(x\_0)|=0$. It is equivalent to your notion of stability, at least if you consider solutions to ODEs.
Consider the following property.
$$ (P)=(P)\_{f,\phi}:\text{For all }x\_0\in\mathbb R\text{ such that }\liminf\_{t\to\infty}|f(X\_t(x\_0))|=0,\text{ we have }\liminf\_{t\to\infty}|X\_t(x\_0)|=0. $$
>
> **Question 1.**
> Do we always have $(P)$?
>
>
>
Clearly this is not the case. Take any such $f$ that goes to zero at infinity (say $f:x\mapsto x^2/(1+x^4)$) and any such $\phi$ that admits solutions going to infinity (say $\phi:x\mapsto (x^3-x)/(1+x^4)$). Then any solution going to infinity satisfies the first but not the second condition (with the given $f$ and $\phi$, the set of $x\_0$ satisfying the first condition is $\mathbb R\setminus\{-1,1\}$, while for the the second it is $(-1,1)$).
>
> **Question 2.**
> Does $(P)$ hold if we suppose that $f$ is [radially unbounded](https://en.wikipedia.org/wiki/Radially_unbounded_function)?
>
>
>
No. It would be obvious that it fails if $f$ were allowed to have another point $x$ such that $f(x)=0$. But it also works with something like
$$f:x\mapsto \begin{cases}x^2(x-2)^2 & \text{for }x\neq1,\\1&\text{for }x=1.\end{cases}$$
Then just take $\phi:x\mapsto-x(x-1)(x-2)/(1+x^4)$.
>
> **Question 3.**
> Does $(P)$ hold for $f$ lower semicontinuous and eventually bounded below\*?
>
>
>
\* Say that we mean $\lim\_{R\to\infty}\inf\_{|x|\geq R}f(x)>0$, whereas radially bounded meant that this limit was $\infty$.
Yes. If the limit inferior of $t\mapsto f(X\_t(x\_0))$ is zero, then by definition there exists a sequence of times $t\_n\to\infty$ such that $f(X\_{t\_n}(x\_0))$ converges to zero. By the boundedness condition, there exists some $R,\varepsilon>0$ such that $|x|\geq R$ implies $f(x)>\varepsilon$; in particular we eventually have $|X\_{t\_n}(x\_0)|<R$. By compactness, we can extract a subsequence $n\mapsto\sigma(n)$ for which $X\_{t\_{\sigma(n)}}(x\_0)$ converges, say to some limit $x\_\infty\in[-R,R]$. Using the lower semicontinuity, we know that $\lim\_{n\to\infty}f(X\_{t\_{\sigma(n)}}(x\_0))\geq f(x\_\infty)$, so $f(x\_\infty)=0$ and $x\_\infty=0$. This means that $\liminf\_{t\to\infty}|X\_t(x\_0)|\leq\liminf\_{n\to\infty}|X\_{t\_{\sigma(n)}}(x\_0)|=|x\_\infty|=0$ as expected (and in fact we have $\liminf\_{t\to\infty}|X\_t(x\_0)|=0$ by stability).
>
> **Question 4.**
> Does $(P)$ hold if $\phi$ admits no solution going to infinity?
>
>
>
No; see question 2.
>
> **Question 5.**
> Does $(P)$ hold if $\phi$ admits only one fixed point?
>
>
>
Yes. 0 has to be a fixed point to be stable, so we can easily see by the intermediate value theorem that $\phi$ is non zero and has constant sign on $(-\infty,0)$ and $(0,+\infty)$. By stability, we see that its signs there are respectively positive and negative, so in fact we always have $\liminf\_{t\to\infty}|x(t)|=\lim\_{t\to\infty}|x(t)|=0$.
| 2 | https://mathoverflow.net/users/129074 | 418539 | 170,411 |
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