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https://mathoverflow.net/questions/448552 | 4 | The result of LMFDB claims (<https://www.lmfdb.org/EllipticCurve/Q/1640/c/1> )
that (2-part of) Tate-Shafarevich group $\mathrm{Sha}(E/\Bbb{Q})$ of elliptic curve $y^2=x^3-8747x-314874$ has order $16$. In particular, the order of $2$-Selmer group is larger than $16$.
But Magma calculates as following,
```
A:=EllipticCurve([0,0,0,-8747,-314874]);
Sel2:=TwoSelmerGroup(A); Sel2;
```
The output of this is,
```
Abelian Group isomorphic to Z/2 + Z/2 + Z/2 Defined on 3 generators Relations:
2*Sel2.1 = 0
2*Sel2.2 = 0
2*Sel2.3 = 0
```
It seems contradiction because $2$-Selmer group is larger than $2$-part of Tate-Shafarevich group.
Where did I go wrong and how can I correctly calculate order of $2$-Selmer group using Magma ?
| https://mathoverflow.net/users/144623 | Discrepancy in the calculation of $2$-Selmer group by Magma and LMFDB | I don't see any contradiction: the Selmer group also has a contribution of rational points. Indeed, the group of 2-torsion rational points on this elliptic curve is isomorphic to $\mathbb Z/2\mathbb Z$ and generated by $(-54,0)$, as Magma will readily confirm (probably, I used GP/Pari).
UPDATE: As David suspected, the Tate-Shafarevic group of this elliptic curve is isomorphic to $(\mathbb Z/4\mathbb Z)^2$ so 2-descent misses the elements of order 4 in Sha.
| 5 | https://mathoverflow.net/users/2284 | 448554 | 180,589 |
https://mathoverflow.net/questions/448324 | 1 | I am searching a simple example of a finite group $G$, so that the Jacobson radical $J(FG)$, of group algebra $FG$ is commutative, where $F$ is a finite field. I know example for that if $G$ is any finite abelian group, or dihedral group when characteristic of field is $2$ and $p$-groups, or when group ring is semi simple. Please provide me any other example of such finite group $G$ for which corresponding Jacobson radical is commutative. Thank you.
| https://mathoverflow.net/users/81355 | Example of a group algebra with commutative Jacobson radical | Here is a family of examples. Let $F$ have characteristic two, and let $G$ be a finite group of order $2n$ with $n$ odd. Then $J(FG)$ squares to zero, and is therefore commutative.
Another (similar) family of examples is as follows. Let $F$ have characteristic $p$, let $H$ be a $p'$-group, and let $t$ be an automorphism of $H$ of order $p$ which fixes only the identity element. Let $G$ be the semidirect product $H \rtimes \langle t\rangle$. Then every non-principal block of $FG$ has defect zero, and the principal block is commutative, so $J(FG)$ is commutative.
| 3 | https://mathoverflow.net/users/460592 | 448579 | 180,595 |
https://mathoverflow.net/questions/448524 | 1 | Given a topological space $X$ and a Banach space $V$, I wonder for which open sets $U$ it is possible to construct a continuous function $f: X \to V$ such that $f^{-1}[B(0, 1)] = U$ - or maybe there is some nonempty open $V \subseteq U$ such that $f^{-1}[B(0, 1)] = V$. Perhaps some separation properties can ensure this, but I do not see an obvious way to construct this (maybe some Urysohn's lemma argument?).
The main goal is to construct a function that is continuous, but not continuous with respect to an algebra. The definition for this is given an algebra $\mathcal A$ on $X$, we say that $f: X \to V$ is $\mathcal A$-continuous at $x \in X$ if for every $\varepsilon > 0$ there are $A\_1, \dots, A\_n \in \mathcal A$ such that $\bigcup A\_i$ is a neighborhood for $x$ and $\mathrm{diam}(f[A\_i]) < \varepsilon$ for each $i$.
| https://mathoverflow.net/users/161460 | Constructing a continuous function with a prescribed preimage | Open sets with the required property are exactly functionally open sets.
Let us recall that a subset $U$ of a topological space $X$ is *functionally open in* $X$ if $U=f^{-1}[V]$ for some continuous function $f:X\to \mathbb R$ and some open set $V\subseteq \mathbb R$. It is well-known that an open subset $U$ of a normal space $X$ is functionally open if and only if $U$ is of type $F\_\sigma$.
**Theorem.** For a subset $U$ of a topological space $X$, the following conditions are equivalent:
1. $U$ is functionally open in $X$;
2. For any Banach space $Y\ne\{0\}$ there exists a continuous function $f:X\to Y$ such that $f^{-1}[B]=Y$ where $B=\{y\in Y:\|y\|<1\}$ is an open unit ball in $Y$.
*Proof* The implication (2)$\Rightarrow$(1) is trivial becase the real line is a Banach space. To prove the implication (1)$\Rightarrow$(2), assume that the set $U$ is functionally open in $X$. Then $U=g^{-1}[V]$ for some continuous function $g:X\to\mathbb R$ and some open set $V\subseteq \mathbb R$. If $V=\mathbb R$, then $U=X$ and then $U=X=f^{-1}[B]$ for the constant function $f:X\to \{0\}\subset Y$.
So, we assume that $V\ne \mathbb R$. In this case, consider the continuous function $h:\mathbb R\to[0,1]$, $h:t\mapsto \min\{1,\min\_{s\in \mathbb R\setminus V}|t-s|\}$. Observe that $h^{-1}(0)=\mathbb R\setminus V$. Choose any point $y\_1\in Y$ with $\|y\_1\|=1$ and consider the function $f:X\to Y$, $f:x\mapsto (1-h\circ g(x))y\_1$. It is easy to check $f^{-1}[B]=U$. $\square$
| 2 | https://mathoverflow.net/users/61536 | 448581 | 180,596 |
https://mathoverflow.net/questions/448560 | 3 | The question is as in the title.
Let $H$ be a separable Hilbert space and $f : [0,1] \to H$ be a continuous mapping such that
\begin{equation}
f'(t):=\lim\limits\_{\alpha \to 0} \frac{f(t+\alpha)-f(t)}{\alpha} \in H
\end{equation}
exists for almost every $t \in [0,1]$ and $f' : [0,1] \to H$ is Borel-measurable. In the above formula, convergence of the limit is with respect to the norm of $H$.
However, let us further suppose that $H=L^2\bigl([0,1],\mathbb{R}\bigr)$ itself, so that $f(t) : [0,1] \to H$ can be in fact written as $f(t,x) : [0,1] \times [0,1] \to \mathbb{R}$
Then, I am curious about the following:
1. Is it possible for the above derivative $f'(t) : [0,1] \to H$ to be defined almost everywhere on $[0,1] \times [0,1]$ as a real-valued Borel-measurable function?
2. If so, does $f'(t)$ coincide almost everywhere on $[0,1]^2$ with the usual definition of partial derivative of $f(t,x)$ with respect to $t$?
This seems more subtle than I expected and I am quite confused about dealing weith almost everywhere equalities... I posted this question on ME but haven't received any answer yet. all my attempts to figure out myself have failed.. Could anyone please help me?
| https://mathoverflow.net/users/56524 | If $f : [0,1] \to H$ has $t$-derivative with respect to the norm of $H$, and $H=L^2[0,1]$ itself, does the $t$-derivative exist in ordinary sense? | The convergence in $L^2$ of the variation ratios does not yield necessarily a pointwise convergence.
For example, consider the case where $f(t,x) = \mathbb{1}\_{x = 1/t - \lfloor 1/t \rfloor}$ for all $t>0$ and $x \in [0,1]$, and $f(0,x) = 0$. Each function $f(t,\cdot)$ is null almost everywhere, so $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ in $L^2([0,1])$. Yet, for every $x \in [0,1[$ and $n \in \mathbb{N}$, $f((n+x)^{-1},x)=1$, so the convergences $f(t,x) \to 0$ and $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ do not hold almost surely.
Now, if you have a derivative with regard to $t$ in $L^2$ and a derivative with regard to $t$ at almost every $x \in [0,1]$, those derivatives coincide almost everywhere. This is true because both $L^2$ convergence and almost everywhere convergence imply convergence in measure, for which the limit is almost everywhere unique.
Other possible argument: call $g$ the limit in $L^2[0,1]$ of $[f(t,\cdot)-f(0,\cdot)]/t$ as $t \to 0$. If you choose a sequence $(t\_n)$ tending to $0$ sufficiently fast so that the series
$\sum \Vert [f(t\_n,\cdot)-f(0,\cdot)]/t\_n - g \Vert\_2$ converges, then the series $\sum [f(t\_n,\cdot)-f(0,\cdot)]/t\_n$ converges almost surely (and in $L^2$), so $[f(t\_n,\cdot)-f(0,\cdot)]/t\_n - g\to 0$ almost surely.
| 2 | https://mathoverflow.net/users/169474 | 448587 | 180,598 |
https://mathoverflow.net/questions/448594 | 4 | Let $\lambda$ be a partition with at most $p$ parts, let $\mu$ be a partition with at most $q$ parts, and let $\nu$ be a partition with at most $p+q$ parts. Let $m\geq \nu\_1$ be an integer. We denote by $\lambda^\ast$, $\mu^\ast$, $\nu^\ast$ the conjugate partitions, and pad them with $0$'s so the conjugates have $m$ parts. Consider the Littlewood-Richardson coefficients
$$s\_\lambda s\_\mu=\sum c\_{\lambda\mu}^\nu s\_\nu$$
Then I won't bet my life on it, but I think I found the following symmetry:
$$c\_{(p-\lambda^\ast\_{m+1-i})\_{i=1}^m,(q-\mu^\ast\_{m+1-i})\_{i=1}^m}^{(p+q-\nu^\ast\_{m+1-i})\_{i=1}^m}=c\_{\lambda\mu}^\nu$$
Provided this is true, is it already known? I have a proof and I've checked it in several cases with coefficients with unusual values, and it checks out (no sort of comprehensive test yet), but errors are always possible, and it seems like something simple enough that if it's true it would be known.
| https://mathoverflow.net/users/62135 | Is this simple symmetry of Littlewood-Richardson coefficients known? | I think you can understand this if you think of them as characters of $GL\_m$-representations, i.e., $s\_\lambda$ is the character of the irreducible representation which I'll denote by $V\_\lambda$. The convention is that the highest weight is $\lambda$.
First, conjugating (I assume this just means the usual transpose operation of swapping rows and columns of the Young diagram) is a bit of a red herring here since $c\_{\lambda,\mu}^\nu = c\_{\lambda^\*,\mu^\*}^{\nu^\*}$.
Second, the character of $V\_\lambda^\vee$ (dual representation) has highest weight $(-\lambda\_m, -\lambda\_{m-1},\dots, -\lambda\_1)$, which is your reversing operation. To make that into a partition, we can add $p$ to all components (for large enough $p$) and this corresponds to taking the representation $(\det V)^{\otimes p} \otimes V\_{\lambda^\vee}$.
So putting it all together, we start with
$$V\_{\lambda^\*} \otimes V\_{\mu^\*} \cong \bigoplus\_\nu V\_{\nu^\*}^{\oplus c^\nu\_{\lambda, \mu}}$$
Apply duals:
$$V\_{\lambda^\*}^\vee \otimes V\_{\mu^\*}^\vee \cong \bigoplus\_\nu (V\_{\nu^\*}^\vee)^{\oplus c^\nu\_{\lambda, \mu}}$$
Now tensor with $(\det V)^{\otimes (p+q)}$:
$$(V^{\otimes p} \otimes V\_{\lambda^\*}^\vee) \otimes (V^{\otimes q} \otimes V\_{\mu^\*}^\vee) \cong \bigoplus\_\nu (V^{\otimes (p+q)} \otimes V\_{\nu^\*}^\vee)^{\oplus c^\nu\_{\lambda, \mu}}$$
and take characters to get the identity you mentioned.
| 5 | https://mathoverflow.net/users/321 | 448598 | 180,601 |
https://mathoverflow.net/questions/447645 | 3 | *Note: This is a strengthening of the following [result](https://mathoverflow.net/questions/438492/blow-up-limits-for-sde), motivated by the need for strong convergence in applications.*
Let $W$ be a one dimensional standard Brownian motion, and let $X$ be the solution to the SDE
$$dX\_t = \sigma(X\_t) \, dW\_t \, , \, X\_0 = 0$$
with $\sigma: \mathbb R \to \mathbb R$ Lipschitz continuous.
For each $c > 0$, define the process $Y^c$ on $[0, 1]$ by
$$X^c\_t := c^{-1/2} X\_{ct}.$$
Similarly define
$$W^c\_t := c^{-1/2} \, W\_{ct}.$$
**Question:** Is it true that as $c \to 0^+$, we have
$$\mathbb E[\sup\_{0 \leq t \leq 1} |X^c\_t - \sigma(0) W^c\_t|] \to 0?$$
| https://mathoverflow.net/users/173490 | Strong blow up limits for SDE | Yes, this is true. Since $(X\_{t}^{c})\_{t\geq{0}}$ and $(W\_{t}^{c})\_{t\geq{0}}$ are both martingales for fixed $c$, applying Cauchy- Schwarz and then Doob's inequality tells us that:
$$
\mathbb{E}(\sup\_{0\leq{t}\leq{1}}|X\_{t}^{c}-\sigma(0)W\_{t}^{c}|)\leq{\big(\mathbb{E}(\sup\_{0\leq{t}\leq{1}}|X\_{t}^{c}-\sigma(0)W\_{t}^{c}|^{2})\big)^{1/2}}\leq{\sqrt{2}\big(\mathbb{E}|X\_{1}^{c}-\sigma(0)W\_{1}^{c}|^{2}\big)^{1/2}}
$$
By our SDE for $(X\_{t})\_{t\geq{0}}$ we have that:
$$
X\_{1}^{c}=c^{1/2}X\_{c}=c^{1/2}\int\_{0}^{c}\sigma(X\_{u})dW\_{u}=c^{1/2}\int\_{0}^{1}\sigma(X\_{cr})dW\_{cr}=\int\_{0}^{1}\sigma(X\_{cr})dW^{c}\_{r}
$$
Since $(W^{c}\_{t})\_{t\geq{0}}$ is a standard Brownian motion for any $c>0$, by the Ito isometry:
$$
\mathbb{E}(X\_{1}^{c}-\sigma(0)W\_{1}^{c})^{2}=\mathbb{E}\Big(\int\_{0}^{1}(\sigma(X\_{cr})-\sigma(0))dW^{c}\_{r}\Big)^{2}=\int\_{0}^{1}\mathbb{E}(\sigma(X\_{cr})-\sigma(0))^{2}dr
$$
If $L>0$ is the Lipschitz constant of $\sigma$, we have that:
$$
\int\_{0}^{1}\mathbb{E}(\sigma(X\_{cr})-\sigma(0))^{2}dr\leq{L^{2}\int\_{0}^{1}\mathbb{E}X\_{cr}^{2}dr}\leq{L^{2}\sup\_{0\leq{r}\leq{1}}\mathbb{E}X\_{cr}^{2}}\leq{2L^{2}\mathbb{E}X\_{c}^{2}}
$$
The last inequality follows by Doob's inequality, since $(X\_{t})\_{t\geq{0}}$ is a martingale. Finally, recall that by the Gronwall- type argument in [1](https://mathoverflow.net/a/438499/80052) we have that:
$$
\mathbb{E}X\_{c}^{2}\leq{\frac{\sigma^{2}(0)}{L^{2}}\big(e^{2L^{2}c}-1\big)}
$$
Putting all this together, we have that:
$$
\mathbb{E}(\sup\_{0\leq{t}\leq{1}}|X\_{t}^{c}-\sigma(0)W\_{t}^{c}|)\leq{2|\sigma(0)|\sqrt{e^{2L^{2}c}-1}}\rightarrow{0} \hspace{10pt}\text{as}\hspace{10pt}c\rightarrow{0}
$$
| 2 | https://mathoverflow.net/users/80052 | 448602 | 180,603 |
https://mathoverflow.net/questions/448604 | -5 | imagine you have a list of ingredients. Each ingredient has five traits, let's call them A, B, C, D and E. You want a mix of ingredients for which the sum of each trait is known.
for example:
* ingredient 1 has 5 A, 2 B, 0 C, 0 D and 3 E.
* ingredient 2 has 0 A, 3 B, 2 C, 2 D and 0 E
* ingredient 3 has 0 A, 0 B, 6 C, 6 D and 4 E.
* the goal is 10 A, 10 B, 10 C, 10 D and 10 E.
* the algorithm finds that you need 2 x ingredient 1, 2 x ingredient 2 and 1 x ingredient 3.
the algorithm should work
* for any amount of ingredients (my real problem has 206)
* for ingredients which could contain any combination of the five traits
* for any finite sum goal
* if there is no possible combination, the algorithm should return it is impossible
thanks for the help!
Aron
| https://mathoverflow.net/users/506664 | optimization problem: find ingredients that add up to given totals | Let $b\_i$ be the demand for trait $i$. Let $a\_{ij}$ be the number of trait $i$ in ingredient $j$. Let nonnegative integer decision variable $x\_j$ be the number of times ingredient $j$ is used. The problem is to find a feasible solution to the linear equations
$$\sum\_j a\_{ij} x\_j = b\_i \quad \text{for all $i$}$$
You can use an integer linear programming solver to either find a solution or detect infeasibility.
| 0 | https://mathoverflow.net/users/141766 | 448606 | 180,604 |
https://mathoverflow.net/questions/379369 | 3 | The following question appears, more or less, [here](https://math.stackexchange.com/questions/3955627/flatness-of-a-subring-generated-by-two-subrings):
Let $k$ be an algebraically closed field of characteristic zero and let $S$ be a commutative $k$-algebra
(I do not mind to further assume that $S$ is an integral domain).
Let $R\_1,R\_2 \subseteq S$ be two (probably different) $k$-subalgebras of $S$ such that
$R\_i \subseteq S$, $1 \leq i \leq 2$ is:
finitely generated, flat and separable (in other words, $R\_i \subseteq S$ is etale).
Denote the $k$-algebra generated by $R\_1$ and $R\_2$ by $R$.
>
> Is it true that $R \subseteq S$ is flat?
>
>
>
Clearly, $R \subseteq S$ is finitely generated and separable (hence unramified). But what about flatness?
I guess (?) that there exists a counterexample in dimension two.
A non-counterexample: $R\_1=k[x^2], R\_2=k[x^3], S=k[x]$; here $R=k[x^2,x^3] \subseteq k[x]$ is not flat,
but also $R\_i \subseteq S$ are not separable (althouh they are finitely generated and free). See [this](https://mathoverflow.net/questions/371948/separability-of-mathbbcx-over-its-mathbbc-subalgebras) question, which explains that $k[x^2,x^3] \subseteq k[x]$ is not separable, hence $k[x^2] \subseteq k[x]$ and $k[x^3] \subseteq k[x]$ are not separable, since if at least one of them were, then $k[x^2,x^3] \subseteq k[x]$ was separable, by a result concerning separability).
Perhaps [this](https://math.stackexchange.com/questions/1328079/separability-implies-flatness-in-a-special-case) question is relevant.
Any hints and comments are welcome; thank you very much!
| https://mathoverflow.net/users/72288 | Flatness of certain subrings | This is again false. The geometric interpretation is as follows: write $Y = \operatorname{Spec} S$ and $X\_i = \operatorname{Spec} R\_i$. Given étale morphisms $f\_1 \colon Y \to X\_1$ and $f\_2 \colon Y \to X\_2$ of affine schemes, the image factorisation
$$R\_1 \underset k\otimes R\_2 \twoheadrightarrow R \hookrightarrow S$$
corresponds geometrically to taking the (scheme-theoretic) image $Y \to X \hookrightarrow X\_1 \times X\_2$ of the product map $f \colon Y \to X\_1 \times X\_2$. If $Y$ is smooth, then so are the $X\_i$, but $X$ need not be smooth. If $f$ is surjective, it cannot be flat when $X$ is singular [Tags [07NG](https://stacks.math.columbia.edu/tag/07NG) and [00HQ](https://stacks.math.columbia.edu/tag/00HQ)].
We can turn this around to make a counterexample: start with a map $f \colon Y \to X\_1 \times X\_2$ of smooth schemes such that the image is singular but both projections $f\_i \colon Y \to X\_i$ are étale.
**Example.** This is the typical picture of a plane nodal curve: let $Y\_0 = X\_1 = X\_2 = \mathbf A^1$, and consider the morphism $f \colon Y\_0 \to X\_1 \times X\_2$ given by $t \mapsto (t^2-1,t^3-t)$. The scheme-theoretic image is the nodal curve $X\_0 = V(y^2 - x^2(x+1))$, and the map $Y\_0 \to X\_0$ is surjective.
The projections $f\_i \colon Y\_0 \to X\_i$ are not étale, but they become so after removing the points $0$ and $\pm\tfrac{\sqrt 3}{3}$ from $Y\_0$. Let $Y \subseteq Y\_0$ (resp. $X \subseteq X\_0$) be the complement of these points (resp. their images in $X$). This gives the desired counterexample, since we did not remove the singular point $(0,0) = f(-1) = f(1)$ from $X$.
(This parametrisation is explained on [this page](https://www.math.purdue.edu/%7Earapura/graph/nodal.html) of Donu Arapura, which is literally the first result that my favourite search engine produces when I type 'nodal curve'.)
Translated back into algebra, this means $S = k\bigl[t,\tfrac{1}{t},\tfrac{1}{3t^2-1}\bigr]$, with $R\_1 = k[t^2-1] \subseteq S$ and $R\_2 = k[t^3-t] \subseteq S$.
| 2 | https://mathoverflow.net/users/82179 | 448611 | 180,606 |
https://mathoverflow.net/questions/448612 | 0 | As the title suggests, I've been trying to create a lattice (Poset) generator from a positive integer parameter n which represents the number of nodes.
The reason why I'm trying to do this is because I want to create hasse diagrams of lattices for fun, but doing so by hand is a very tiring process and i want an algorithm to help me draw my lattices (I'm using Motion Canvas). Having a pseudocode would be wonderful however I really appreciate any pointer that can help me in my struggle!
So far, I've only been able to create join semilattice generator and I honestly haven't found any useful resources online that could help me.
Here's the code for the curious (not sure how much this will help, though):
```
from random import randint
from itertools import combinations
# join
def build_semilattice(n):
lattice = [[] for i in range(n)]
batches = [[0]] # add the least
step = 1
while step < n-1: # ou step != n + 1
k = randint(step,n-2)
batches.append([i for i in range(step, k+1)])
step = k + 1
batches.append([n-1]) # add the greatest
print(batches)
for batch in range(len(batches) - 1):
for a,b in combinations(batches[batch], 2):
k = randint(batches[batch + 1][0], batches[batch + 1][-1])
lattice[a].append(k), lattice[b].append(k)
return lattice
print(build_semilattice(10))
```
Thank you very much for your answer!
| https://mathoverflow.net/users/506667 | Algorithm to generate a lattice (Poset) from a positive integer parameter | Sage (<https://www.sagemath.org/>) has code to produce a random lattice on $n$ elements: see <https://doc.sagemath.org/html/en/reference/combinat/sage/combinat/posets/poset_examples.html#sage.combinat.posets.poset_examples.Posets.RandomLattice>
I have no idea what algorithm it uses or what distribution it aims to sample from, but it's a start.
| 1 | https://mathoverflow.net/users/25028 | 448613 | 180,607 |
https://mathoverflow.net/questions/448573 | 2 | Consider a homeomorphic embedding $h:S^{n-1}\times [0,1]\rightarrow S^n$ and denote
$$S^{n-1}\_t=h(S^{n-1}\times \{t\}).$$
The generalized Schoenflies theorem states the closure of each connected component of $S^n-S^{n-1}\_{1/2}$ is homeomorphic to the closed $n$-disk. In the last step of Morton Brown's proof (note that I am modifying the notation from the paper slightly), he
defines two sets that are key to the argument:
* $A$ is the connected component of $S^n-S^{n-1}\_1$ that does not contain $S^{n-1}\_0$;
* $B$ is the connected component of $S^n-S^{n-1}\_0$ that does not contain $S^{n-1}\_1$.
By collapsing $\overline A$ and $\overline B$ to two points $a$ and $b$, he gets a quotient map $S^n\rightarrow S^n$ with properties that allow him to conclude the desired result. However, it seems that the definition of this map is implicitly using that $$S^n=A\cup B\cup h(S^{n-1}\times [0,1]),$$
which I struggled to show. The task becomes much easier if we assume that $h$ extends to an embedding $S^{n-1}\times (-\epsilon,1+\epsilon)\rightarrow S^n$ for some $\epsilon >0$, because this facilitates a proof that the above union is clopen in $S^n.$ When it comes to proving the ultimate theorem, this assumption presents no material barrier. But it would still be nice to understand everything as stated. As such, I wonder if someone more versed in these sorts of arguments could explain to me how to prove that the aforementioned union is indeed all of $S^n$?
| https://mathoverflow.net/users/147463 | A detail in Brown's proof of the generalized Schoenflies theorem | **Update:** The argument I previously provided was incorrect. I will now outline a correct argument, followed by the incorrect argument and an explanation of what went wrong.
**Correct Explanation:**
By Alexander duality, we can write decompositions into exactly two connected components, for each of the following two sets:
$$S^n-h\big(S^{n-1}\times (0,1)\big)=C\sqcup D$$
$$S^n-h\big(S^{n-1}\times [0,1]\big)=C'\sqcup D'$$
We set our naming convention so that $C=C'\sqcup S^{n-1}\_1$ and $D=D'\sqcup S^{n-1}\_0\!.$ To see that this makes sense, we must argue that the four connected sets $C'\!,\, D'\!,\, S^{n-1}\_0$ and $S^{n-1}\_1$ (whose disjoint union is precisely $C\sqcup D$) are reasonably distributed between $C$ and $D$ :
* Suppose $S^{n-1}\_0\sqcup S^{n-1}\_1\subseteq C.$ Then we can write $$S^n=C\sqcup h\big(S^{n-1}\times (0,1)\big)\sqcup D=\Big(C\cup h\big(S^{n-1}\times [0,1]\big)\Big)\sqcup D.$$
Because $C$ and $D$ are closed, this contradicts the connectedness of $S^n\!.$ Analogously, we cannot have $S^{n-1}\_0\sqcup S^{n-1}\_1\subseteq D,$ so we can assume that $S^{n-1}\_1\subseteq C$ and $S^{n-1}\_0\subseteq D$.
* Suppose $C'\!\sqcup D'\subseteq C$. Since $C'$ and $D'$ are open, we can see that $$C-D'=C'\sqcup S^{n-1}\_1$$ is a closed set containing $C'\!,$ which is not itself closed. Thus $\overline{C'}$ intersects $S^{n-1}\_1$ nontrivially. We similarly conclude that $\overline{D'}$ intersects $S^{n-1}\_1$ nontrivially. Also note that $$\overline{C'}\cap S^{n-1}\_0\subseteq C\cap D=\emptyset$$ and similarly $\overline{D'}\cap S^{n-1}\_0=\emptyset.$ Thus $$S^n-S^{n-1}\_0=\overline{C'}\cup \overline{D'}\cup h\big(S^{n-1}\times (0,1]\big)$$ is connected, which contradicts the Jordan-Brouwer separation theorem. Analogously, we cannot have $C'\!\sqcup D'\subseteq D,$ so we can assume that $C'\subseteq C$ and $D'\subseteq D$.
Since $C'$ is open and $C=C'\sqcup S^{n-1}\_1$ is closed, we can see that $\overline{C'}$ intersects $S^{n-1}\_1$ nontrivially, so $\overline{C'}\cup h\big(S^{n-1}\times(0,1]\big)$ is connected. Hence, we have a decomposition $$S^n-S^{n-1}\_0=\Big(\overline{C'}\cup h\big(S^{n-1}\times (0,1]\big)\Big)\sqcup D'$$
into two connected sets, which therefore must be the connected components of $S^n-S^{n-1}\_0\!.\,$ This implies that $D'=B$, since the other component contains $S^{n-1}\_1\!.\,$ Analogously, we can conclude that $C'=A.$ Then it is clear that
$$S^n=C'\sqcup D'\sqcup h\big(S^{n-1}\times [0,1]\big)=A\sqcup B\sqcup h\big(S^{n-1}\times [0,1]\big)$$
**Incorrect Explanation:**
Let $q:S^n\rightarrow X$ be the quotient map formed by collapsing $\overline A$ and $\overline B$ to points $a$ and $b$.
It is not too hard to show that $X$ is Hausdorff, so any injective map from a compact space into $X$ is an embedding. Observe that $X$ is locally Euclidean, because:
* $X-\{a,b\}\cong S^n-(\overline A\cup \overline B)$ is an open submanifold of $S^n$;
* The restriction $h:S^{n-1}\times [1/2,1]\rightarrow S^n$ descends to an embedding $D^n\rightarrow X$ of a closed $n$-cell containing $a$ in its interior (we view $D^n$ as coming from $S^{n-1}\times [1/2,1]$ by collapsing $S^{n-1}\times 1$);
* The restriction $h:S^{n-1}\times [0,1/2]\rightarrow S^n$ descends to an embedding $D^n\rightarrow X$ of a closed $n$-cell containing $b$ in its interior (we view $D^n$ as coming from $S^{n-1}\times [0,1/2]$ by collapsing $S^{n-1}\times 0$).
The map $h:S^{n-1}\times [0,1]\rightarrow S^n$ similarly descends to an embedding $p:S^n\rightarrow X$ (where we view $S^n$ as coming from $S^{n-1}\times [0,1]$ by collapsing $S^{n-1}\times 0$ and $S^{n-1}\times 1$ to points). This embedding $p$ must be a homeomorphism, since $S^n$ is compact and $X$ is connected (and both are topological $n$-manifolds). The fact that $p$ is surjective is equivalent to the fact that I had asked about, namely that $$S^n=A\cup B\cup h(S^{n-1}\times [0,1]).$$ The quotient map that I mentioned in my question is then simply $p^{-1}\circ q$.
**What went wrong?**
When I tried to define a chart about $a$ (and similarly for $b$, but I'll stick with one point for clarity), I described an embedding $D^n\rightarrow X$ containing $a$ in the image of the disk's interior, but I actually needed to argue that this image contained a whole *neighborhood* of $a$. My conclusion that $p$ was surjective came from a corollary to "invariance of domain," the proof of which relies crucially on the openness of charts in a topological manifold (although the proof does not require the Hausdorff or second-countable assumptions). For the "chart" that I defined to map onto an open set, we need $A\cup h\big(S^{n-1}\times (1/2,1]\big)$ to be open in $S^n\!.\,$ But if we could show this, then we could argue directly that $A\cup h\big(S^{n-1}\times (0,1]\big)$ is clopen and connected in $S^n-S^{n-1}\_0\!,\,$ which means that it must be the *other* connected component (besides $B$). Then we get
$$S^n-S^{n-1}\_0=A\cup h\big(S^{n-1}\times (0,1]\big)\cup B,$$ which clearly implies the desired result, without having to worry about $X$ being locally Euclidean.
| 2 | https://mathoverflow.net/users/147463 | 448618 | 180,608 |
https://mathoverflow.net/questions/446652 | 12 | **Setting:**
Suppose $\{u\_i\}\_{i=1}^n \subset R^d$ is a collection of unit vectors such that $u\_i^Tu\_j < 0$ for all $i\neq j$, and $w$ is a unit vector such that $u\_i^T w> 0$ for all $i=1,\dotsc,n$. Let $a \in R^d$ be a unit vector.
**Goal:**
We want to show that the following hold or to find a counter-example.
$$\sum\_{i=1}^n \lvert u\_i^T a\rvert u\_i^T w < 1$$
We can prove this statement for the case of $n=2$, but we were not able to prove it for $n>2$.
Any suggestions?
**Update:**
We used **Echo**'s approach from the comment. It worked for the first iteration (when the local maximum is an interior point) from the stationarity condition of the gradient. However, we got stuck at the second iteration (when one of the constraints was active). So now we have a stationarity condition on the projected gradient. We couldn't solve this. Any help would be appreciated.
| https://mathoverflow.net/users/128477 | Prove/disprove a linear algebra inequality | Notation:
$[n] = \{1,\dotsc,n\}$.
For symmetric $n \times n$ matrices $A,B$, $A \succ B$ and $A \succeq B$ means that $A-B$ is positive definite / positive semidefinite.
For a symmetric matrix $A$, $\lambda\_{\min}(A)$ and $\lambda\_{\max}(A)$ denote its minimal and maximal eigenvalues.
Claim:
If $n \ge 2$, and $u\_1, \dotsc, u\_n, w, a \in \mathbb{R}^d$ are unit vectors such that $u\_j^T u\_k < 0$ for every $j \neq k$, and $u\_j^T w > 0$ for every $j$, then $\sum\_j |u\_j^T a| (u\_j^T w) < 1$.
Proof:
If $u\_j^T a = 0$ for every $j$, or $u\_j^T w = 0$ for every $j$, then this is trivial.
Otherwise we can replace $a$ by a scalar multiple of its orthogonal projection onto $\sum\_j \mathbb{R} u\_j$, and the same for $w$.
So we assume that $a, w \in \sum\_j \mathbb{R} u\_j$, and we can then also assume that $\mathbb{R}^d = \sum\_j \mathbb{R} u\_j$.
We prove by induction on $n$.
Let $G = (u\_j^T u\_k)\_{j,k \in [n]} \succeq 0$ be the Gram matrix of $(u\_j)\_{j=1}^n$, and let $M = I-G$.
Then $M\_{j,j} = 0$ for every $j$, and $M\_{j,k} > 0$ for $j \neq k$.
Using the Perron-Frobenius theorem (<https://en.wikipedia.org/wiki/Perron-Frobenius_theorem>) for $\epsilon I + M$ with $\epsilon > 0$, we get that its maximal eigenvalue $\epsilon + \lambda\_{\max}(M)$ has multiplicity $1$, and it has a corresponding eigenvector in $\mathbb{R}\_{>0}^n$, and $\epsilon + \lambda\_{\min}(M) \ge -(\epsilon + \lambda\_{\max}(M))$.
Taking $\epsilon \to 0$, we get $\lambda\_{\min}(M) \ge -\lambda\_{\max}(M)$.
So $\lambda\_{\max}(G) + \lambda\_{\min}(G) \le 2$, and we also see that the eigenvalue $\lambda\_{\min}(G)$ of $G$ has multiplicity $1$.
We could also use the same argument for $(u\_1,\dotsc,u\_n, -w)$ instead of $(u\_1,\dotsc,u\_n)$, and get an $(n+1) \times (n+1)$ Gram matrix $G' \succeq 0$, where $\lambda\_{\min}(G') = 0$ has multiplicity $1$, so $\operatorname{rank}(u\_1,\dotsc, u\_n) = \operatorname{rank}(G') = n$.
So $u\_1, \dotsc, u\_n$ are linearly independent (so $d = n$), and thus $G \succ 0$ and $0 < \lambda\_{\min}(G)$ and $\lambda\_{\max}(G) < 2$, so $-I \prec M \prec I$.
We cannot have $\lambda\_{\min}(G) = \lambda\_{\max}(G) = 1$, because then $M = 0$.
So $\lambda\_{\min}(G) \in (0,1)$.
Note that $G^{-1} = (I-M)^{-1} = I+M+M^2+\dotsm$ (this converges, because $-I \prec M \prec I$), so $(G^{-1})\_{j,k} > 0$ for every $j,k$.
So by the Perron-Frobenius theorem, $G^{-1}$ has a unique unit length eigenvector in $\mathbb{R}\_{>0}^n$, and the corresponding eigenvalue is $\lambda\_{\max}(G^{-1})$ (this will be used later).
We fix linearly independent unit vectors $(u\_j)\_{j \in [n]}$, and choose unit vectors $a, w$ in $\mathbb{R}^d$ so that $S = \sum\_j |u\_j^T a| (u\_j^T w)$ is maximal, only requiring $u\_j^T w \ge 0$ for every $j$.
If $u\_j^T w = 0$ or $u\_j^T a = 0$ for some $j$, then we are done by induction.
So let $u\_j^T w > 0$ and $u\_j^T a \neq 0$ for every $j$.
Then $S > 0$.
Let $a = \sum\_j \alpha\_j u\_j$ and $w = \sum\_j \beta\_j u\_j$.
Then $1 = \|a\|^2 = \alpha^T G \alpha$, $1 = \|w\|^2 = \beta^T G \beta$, and $u\_j^T a = (G \alpha)\_j \neq 0$ and $u\_j^T w = (G \beta)\_j > 0$.
Let $D$ be the diagonal matrix such that $D\_{j,j} = \operatorname{sgn}(u\_j^T a) = \operatorname{sgn}((G \alpha)\_j)$ for every $j$.
Then $S = \sum\_j (G \alpha)\_j D\_{j,j} (G \beta)\_j = \alpha^T G D G \beta$.
Because $(\alpha, \beta)$ give a local maximum for $S$, the differential of the map $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^3$, $(\alpha, \beta) \mapsto (\alpha^T G \alpha, \beta^T G \beta, \alpha^T G D G \beta)$ does not have full rank at our $(\alpha, \beta)$.
So there is an $(A, B, C) \in \mathbb{R}^3 \setminus \{0\}$ such that
$$
2 A G \alpha + C G D G \beta = 2 B G \beta + C G D G \alpha = 0.
$$
Then $0 = \alpha^T (2 A G \alpha + C G D G \beta) = 2 A + C S$ and $0 = \beta^T (2 B G \beta + C G D G \alpha) = 2 B + C S$, so $A = B = -\frac{C}{2} S$, where $C \neq 0$.
So $GDG \beta = S G \alpha$ and $GDG \alpha = S G \beta$.
So if $x\_0 = G \alpha$ and $y\_0 = G \beta$, then $(x\_0,y\_0)$ satisfies $1 = x\_0^T G^{-1} x\_0 = y\_0^T G^{-1} y\_0$, $GD x\_0 = S y\_0$, $GD y\_0 = S x\_0$.
The idea is to study the pairs $(x,y)$ that satisfy these equations, because these could maybe also give the same maximal $S$, so in some cases we could vary $(\alpha, \beta)$ even if $S$ is already maximal.
Note that $(x\_0, y\_0)$, $(-x\_0, -y\_0)$, $(y\_0, x\_0)$, $(-y\_0, -x\_0)$ all satisfy these equations
Idea: If $(x,y)$ is on the ellipse centered at $0$ going through these four points, then it should also satisfy these equations.
In detail:
Let $p = x\_0+y\_0$, $q = x\_0-y\_0$, $P = p^T G^{-1} p$, $Q = q^T G^{-1} q$.
Then $GD p = S p$, $GD q = -Sq$ (so $p,q$ are eigenvectors of $GD$), $p^T G^{-1} q = 0$, $P, Q \ge 0$ and $P + Q = 4$.
Let us take $x = sp + tq$ and $y = sp - tq$ for $s, t \in \mathbb{R}$.
Then $GD x = S y$ and $GD y = S x$ hold, and $x^T G^{-1} x^T = y^T G^{-1} y^T = P s^2 + Q t^2$, so we need this to be $1$.
Then $GDy=Sx$, so $x^T D y = S x^T G^{-1} x = S$.
So if $P s^2 + Q t^2 = 1$, then $(x,y) = (sp + tq, sp - tq)$ gives the same maximal $S$.
If $p = 0$ or $q = 0$ (so $a = \pm w$):
Then $D = (\operatorname{sgn}(u\_j^T a))\_{j=1}^n = \pm I$ and $G y\_0 = S y\_0$, so $G^{-1} y\_0 = S^{-1} y\_0$.
Here $(y\_0)\_j = u\_j^T w > 0$ for every $j$, so $y\_0$ is a Perron-Frobenius eigenvector for $G^{-1}$, thus $S^{-1} = \lambda\_{\max}(G^{-1})$, so $S = \lambda\_{\min}(G) < 1$.
Now let $p,q \neq 0$.
Then $P, Q > 0$, so we get an ellipse.
For $s = t = \frac{1}{2}$ we get $(x,y) = (x\_0, y\_0)$, for $s = t = -\frac{1}{2}$ we get $(x,y) = (-x\_0, -y\_0)$.
So moving along the ellipse, we can get from $(a,w)$ to $(-a, -w)$, without changing $S$.
So there will be a first point where some $u\_j^T a$ or $u\_j^T w$ becomes $0$, and then we are done by induction.
| 8 | https://mathoverflow.net/users/42355 | 448631 | 180,611 |
https://mathoverflow.net/questions/448632 | 1 | I need a good approximations for $H\_p$, for $p \in (0,1) \cap \mathbb{Q}$, the generalization of $H\_n=\sum\_{i=1}^n \frac{1}{i}$ to the real numbers.
I tried $H\_p = p \sum\_{k=1}^\infty \frac{1}{k (k + p)}$ and $H\_p = \sum\_{k=0}^\infty \sum\_{j=0}^\infty (-1)^j (1 + k)^{-2 - j} p^{1 + j}$ for $-\frac{1}{2}<p<\frac{1}{2}$, both converge very slow, just five digit of $H\_\frac{1}{2}$ take ~1,000,000 iterations.
| https://mathoverflow.net/users/506692 | Approximation for interpolation of harmonic numbers | We can compute $H\_p = p \sum\_{k=1}^\infty \frac{1}{k (k + p)}$ very fast and very accurately as follows. Let
\begin{equation\*}
f(x):=\frac p{(x+1)(x+1 + p)}=\frac{1}{x+1}-\frac{1}{x+1+p},
\end{equation\*}
so that
\begin{equation\*}
H\_p=\sum\_{k=0}^c f(k)+\sum\_{k=1}^\infty f\_c(k),
\end{equation\*}
where $c$ is a natural number and $f\_c(x):=f(c+x)$.
If $c$ is large enough, then $\sum\_{k=1}^\infty f\_c(k)$ can be evaluated fast with high precision by the Euler--Maclaurin (EM) formula:
\begin{equation\*}
\sum\_{k=1}^\infty f\_c(k)=-G\_{m,c}+R\_{m,c},
\end{equation\*}
where
\begin{equation\*}
G\_{m,c}:= F(c)
+ \frac{f(c)}2
+
\sum\_{j=1}^{m-1}\frac{B\_{2j}}{(2j)!}f^{(2j-1)}(c),
\end{equation\*}
\begin{equation\*}
F(x):=\ln\frac{x+1}{x+1+p}
\end{equation\*}
(so that $F$ is the antiderivative of $f$ with $F(\infty-)=0$), $m$ is a natural number, the $B\_k$'s are the Bernoulli numbers, and
\begin{equation\*}
|R\_{m,c}|<\frac{2.02}{(2\pi)^{2m-1}}f^{(2m-2)}(c) \\
=\frac{2.02(2m-2)!}{(2\pi)^{2m-1}}((c+1)^{1-2m}-((c+1+p)^{1-2m})
\end{equation\*}
if $m\ge4$.
If we want to get $H(p)$ with $d$ correct decimal digits after the decimal point, then good choices for $m$ and $c$ are as follows:
\begin{equation\*}
m=\Big\lceil\frac d{2\log\_{10}d}\Big\rceil,\quad
c=\Big\lceil\frac1{\pi e}\,m\,10^{d/(2m)}\Big\rceil;
\end{equation\*}
see [Section 6.1](https://link.springer.com/article/10.1007/s00211-018-0978-y) for details or [Section 6.1](https://link.springer.com/article/10.1007/s00211-018-0978-y) for more details.
E.g., if we want to get $H(2/3)$ with $1000$ correct decimal digits after the decimal point, then we can choose $m=167$ and $c=19288$, so that the error $|R\_{m,c}|$ of the corresponding approximation
\begin{equation\*}
H\_{2/3;m,c}:=\sum\_{k=0}^c f(k)-G\_{m,c}
\end{equation\*}
of $H(2/3)$ is $<3\times10^{-1002}$. This approximation, $H\_{2/3;m,c}$, of $H(2/3)$ was computed by Mathematica in just about $0.11$ sec. Since $H\_{2/3}=\frac{3}{2}+\frac{\pi }{2 \sqrt{3}}-\frac{3 \ln3}{2}$, we find that the actual difference $H\_{2/3;m,c}-H\_{2/3}$ between the approximate and true values is $\approx-3.08\times10^{-1003}$. The corresponding Mathematica notebook can be viewed [here](https://www.wolframcloud.com/obj/ipinelis/Published/1.nb).
---
Using the [summation formula](https://link.springer.com/article/10.1007/s00211-018-0978-y) alternative to the EM formula, we can parallelize calculations and thus further reduce the execution time if the desired number $d$ of correct digits is large enough (such as $2000$ or more).
| 5 | https://mathoverflow.net/users/36721 | 448649 | 180,614 |
https://mathoverflow.net/questions/448213 | 3 | I am interested in this non-convex mixed-integer program:
\begin{array}{cl}
\displaystyle\min\_{(x\_{i},y\_{i})\in\mathbb{Z}^{+}\times\mathbb{Z}^{+}}&\displaystyle\sum\_{i=1}^{K}y\_{i}\left(\displaystyle\frac{x\_{i}}{y\_{i}}-\frac{X}{Y}\right)^{2} \\[0.2cm]
\mathrm{s.t.}&\sum\_{i}x\_{i}=X\;(\mathrm{fixed}) \\
&\sum\_{i}y\_{i}=Y\;(\mathrm{fixed}) \\
&x\_{i}\leqslant y\_{i},\quad i=1,2,\ldots,K
\end{array}
subject to the linear constraints shown above. Attempting to simulate this using optimization softwares like $\textsf{cvxpy}$ becomes unfeasible and incompatible, due to the ratio of quadratic over linear term, i.e. $\frac{x\_{i}^{2}}{y\_{i}}$. Most softwares have no platform for such terms, and if they do it becomes unfeasible for large terms $(X,Y)\sim(10^{6},10^{6})$. Thus, my only approach for a proper numerical modeling of this program is to reformulate it either by convexifying the objective function, or by adding envelopes to the objective function so that it becomes easier to model.
>
> **Question**: How can the objective function be reformulated or convexified for the purpose of better numerical testing?
>
>
>
| https://mathoverflow.net/users/393675 | How to convexify or reformulate this non-convex MIP? | As @ManfredWeis noted, the objective is equivalent to minimizing $\sum\_{i=1}^K \frac{x\_i^2}{y\_i}$. You can reformulate this as a (convex) mixed integer second-order cone programming (MISOCP) problem by introducing a new variable $z\_i$ and minimizing $2\sum\_i z\_i$ subject to
\begin{align}
\sum\_i x\_i &= X \\
\sum\_i y\_i &= Y \\
x\_i &\le y\_i &&\text{for all $i$} \\
2 z\_i y\_i &\ge x\_i^2 &&\text{for all $i$}
\end{align}
The new constraint is a rotated second-order cone.
Note that, even in the original problem, relaxing integrality yields unique optimal solution $(x\_i^\*,y\_i^\*)=(X/K,Y/K)$ for all $i$, with objective value $0$. Because the relaxation is convex, every integer optimal solution satisfies $x\_i \in \{\lfloor x\_i^\* \rfloor, \lceil x\_i^\* \rceil\}$ and $y\_i \in \{\lfloor y\_i^\* \rfloor, \lceil y\_i^\* \rceil\}$ for all $i$.
---
An alternative reformulation introduces binary decision variables $u\_{ij}$ for $i\in\{1,\dots,K\}$ and $j\in \{0,\dots, X\}$ and $v\_{ik}$ for $i\in\{1,\dots,K\}$ and $k\in \{1,\dots, Y\}$. The problem is to minimize the quadratic function
$$\sum\_{i=1}^K \sum\_{j=0}^X \sum\_{k=1}^Y \frac{j^2}{k} u\_{ij} v\_{ik} \tag1\label1$$
subject to linear constraints
\begin{align}
\sum\_i x\_i &= X \\
\sum\_i y\_i &= Y \\
x\_i &\le y\_i &&\text{for all $i$} \\
\sum\_j u\_{ij} &= 1 &&\text{for all $i$} \\
\sum\_j j u\_{ij} &= x\_i &&\text{for all $i$} \\
\sum\_k v\_{ik} &= 1 &&\text{for all $i$} \\
\sum\_k k v\_{ik} &= y\_i &&\text{for all $i$}
\end{align}
You can call an MIQP (or BQP) solver.
Alternatively, you can linearize \eqref{1} by introducing nonnegative variables $w\_{ijk}$ to represent the product $u\_{ij} v\_{ik}$.
You can then minimize $$\sum\_{i=1}^K \sum\_{j=0}^X \sum\_{k=1}^Y \frac{j^2}{k} w\_{ijk} \tag{1p}\label{1'}$$
The usual linearization imposes linear constraints
\begin{align}
w\_{ijk} &\ge u\_{ij} + v\_{ik} - 1 \tag2\label2 \\
w\_{ijk} &\le u\_{ij} \tag3\label3 \\
w\_{ijk} &\le v\_{ik} \tag4\label4
\end{align}
But you can omit \eqref{3} and \eqref{4} because the objective will drive these constraints to be satisfied naturally.
A more compact linearization instead replaces \eqref{2} with
\begin{align}
\sum\_j w\_{ijk} &= v\_{ik} &&\text{for all $i$ and $k$} \\
\sum\_k w\_{ijk} &= u\_{ij} &&\text{for all $i$ and $j$}
\end{align}
It is also worth noting that without the linking constraints $\sum\_i x\_i = X$ and $\sum\_i y\_i = Y$ the problem decomposes into independent problems, one for each $i$. So Dantzig-Wolfe decomposition or Lagrangian relaxation might perform well.
| 2 | https://mathoverflow.net/users/141766 | 448651 | 180,615 |
https://mathoverflow.net/questions/448658 | 14 | Suppose Suppose $A$ and $B$ are two entire, non-surjective, functions. This means
$$
A(z)=e^{f(z)}+c\_1
$$
and
$$
B(z)=e^{g(z)}+c\_2
$$
for some entire functions $f$ and $g$, and some complex constants $c\_1$ and $c\_2$.
When is the sum $A(z)+B(z)$ an surjective entire function? More precisely, is there any condition on $f$ and $g$? Clearly the sum may fail to be surjective, for example when $f(z)=g(z)+c$. But also can be surjective, for example when $f(z)=2g(z)$. A naive conjecture would be that the sum is surjective as long as $f-g$ is not constant. Is that actually true?
| https://mathoverflow.net/users/69275 | Sums of non-surjective entire functions | This sum is surjective, with some trivial exceptions. Here is
a general Theorem of Emile Borel: Let $g\_j$ be entire, and
$$e^{g\_1}+\ldots+e^{g\_n}=0.$$
Then the set of indices $\{1,\ldots,n\}$ can be partitioned
into disjoint subsets $I\_k$, such that for $i,j\in I\_k$ we have $g\_i-g\_j$ is constant, and
$$\sum\_{j\in I\_k}e^{g\_j}=0,$$
for every $k$.
In your case, suppose that the sum of two $e^{g\_j}+c\_j$ is not surjective, then we have a relation of the form
$$e^{g\_1}+e^{g\_2}+e^{g\_3}=0,$$
or of the form
$$e^{g\_1}+e^{g\_2}+e^{g\_3}+e^c=0.$$
Applying Borel's theorem, we conclude in the first case that
$g\_i=g+c\_i,$ $1\leq i\leq 3$, $\sum\_{i=1}^3 e^{c\_i}=0$.
And in the second case, one of the $g\_i$ must be constant,
for example, $g\_3=-c$ and if $g\_1$ and $g\_2$ are not constants then their difference must be constant, namely $g\_1=g\_2=\pi i(2n+1)$.
A good reference for Borel's theorem is the book
Serge Lang, Introduction to complex hyperbolic spaces,
Springer, 1987.
| 15 | https://mathoverflow.net/users/25510 | 448678 | 180,621 |
https://mathoverflow.net/questions/448675 | 2 | Let $\beta$ be an element of $\overline{\mathbb F\_q(T)}\setminus\overline{\mathbb F\_q}$.
Is it true that the sequence $(\beta^{q^n}-T)\_n$ admits infinitely many zeros, that is there exist infinitely many distinct places $\mathcal P\_1,\cdots,\mathcal P\_n\cdots,$ of $\mathbb F\_q(T)(\beta)$ such that for every $n\in\mathbb N$, there exists a $k\in\mathbb N$ such that $v\_{\mathcal P\_n}(\beta^{q^k}-T)>0$, where $v\_{\mathcal P\_n}$ is the valuation of $K$ associated to the place $\mathcal P\_n$. That is an analogue of the following number theory problem: Let $\alpha,\beta\ge2$ be an integer. Does the sequence $(\alpha^n-\beta)\_n$ admit infinitely many divisors?
| https://mathoverflow.net/users/33128 | Zeros of a sequence in $\overline{\mathbb F_q(T)}$ | The answer to the first question is yes:
An alternate way to phrase your question is: consider $X = \mathbb P^1\times \mathbb P^1$ with a given curve $C \subset X$ corresponding to $\beta$. We are interested in the geometric points of intersection of $I\_N = C\cap (\phi^n)^\*\Delta$ where $\Delta$ is the diagonal and $\phi: X \to X$ is the partial Frobenius $(x,y) \to (x,y^q)$. This goes back to a paper of Chai and Oort ("Hypersymmetric abelian varieties") where they were thinking of the question in the modular curve context. The arguement is pretty easy so I'll sketch it:
The basic idea is to compute intersection numbers in two different ways and show that, for these to match up, there have to be more and more points of intersection as $n \to \infty$. First, we can easily compute the global intersection number. Since $\beta$ is not constant, it corresponds to a class $(A,B) \in Pic(X) \cong \mathbb Z^2$ with $A \neq 0$ while the class of $(\phi^n)^\*\Delta$ is $(1,q^n)$. Therefore the global intersection number is $q^nA + B \to \infty$ as $n \to \infty$.
On the other hand, for any point $c \in C$, we can show that the local intersection number is bounded. If $C$ is locally cut out by an equation $f(x,y) = 0$, then the local intersection number is computed by the vanishing order of $f(x,x^{q^n}) = 0$ but we know that $f$ has a term of the form $x^k$ for some $k = k(c) \geq 1$ since it is not divisible by $y$. Therefore, the local intersection number is bounded by $k$ for $n \gg 0$.
As for the second problem, I think one can prove it along exactly the same lines but I'll leave it for you to work out the details.
| 3 | https://mathoverflow.net/users/58001 | 448679 | 180,622 |
https://mathoverflow.net/questions/448697 | 2 | I have an integral equation which is not exactly an eigenvalue type equation, but similar:
$$\int d^3 y\, K(x,y,\lambda) f(y) = f(x)$$
Here $\lambda$ can be thought of as an eigenvalue, so it is expected that only for a discrete set of $\lambda$ we have a solution. $K(x,y,\lambda)$ is some fairly simple expression involving ratios of polynomials and also $\lambda$ appears in a rational fashion.
What would be the typical numerical methods to find the lowest eigenvalue $\lambda$ and eigenfunction $f(x)$?
Actually, we can formulate the question generally: how does one find the eigenfunction corresponding to zero eigenvalue of a linear integral operator that depends on some $\lambda$? The task involves finding the particular $\lambda$ for which the operator has a zero eigenvalue to begin with.
$${\cal O}(\lambda) v = 0$$
Generally ${\cal O}(\lambda)$ doesn't have zero eigenvalues, but for a discrete set of $\lambda$ it does. How can I find it numerically? Probably an iterative procedure would be most cost effective. Is this true?
| https://mathoverflow.net/users/41312 | Numerical methods for integral eigenvalue equation | To solve $\int d^3 y\, K(x,y,\lambda) f(y) = f(x)$ I would discretize the coordinates $x\mapsto x\_n$, $y\mapsto y\_m$, $K(x,y,\lambda)\mapsto K(x\_n,y\_m,\lambda)\equiv K\_{nm}(\lambda)$ and solve the determinant equation
$$\text{det}\,[I-K\_{nm}(\lambda)]=0$$
for $\lambda$, when the indices $n,m$ vary over a finite range.
Things would simplify greatly if $K(x,y,\lambda)$ depends only on the difference $x-y$, because then you could Fourier transform and find a $\lambda$ such that $\hat{K}(\xi,\lambda)=1$ for some $\xi$.
| 3 | https://mathoverflow.net/users/11260 | 448698 | 180,630 |
https://mathoverflow.net/questions/448703 | 1 | I have a matrix $A$ $(n\times n)$ with eigenvalues $\lambda\_i$, then I add another matrix to it as: $A+uv^\top$ where $u$ $(n\times 1)$ and $v$ $(n\times 1)$ are column vector. Also, $A=yy^\top$ with $y$ a $(n-1)$ rank matrix, so $A$ is symmetric.
Is there any way to calculate the new eigenvalues and eigenvectors of $(uv^\top+yy^\top)$ using the information from the vector $u$,$v$ and the eigenvalues and eigenvectors of $yy^\top$？
(The eigenvalues could be calculated by the algorithm given in Golub's paper,but it seems that the Bunch-Nielsen-Sorensen formula is unable to calculate the eigenvectors)
| https://mathoverflow.net/users/502666 | Eigenvectors of a non-symmetric rank-one update of a symmetric matrix | You can still use the [Bunch–Nielsen–Sorensen formula](https://en.wikipedia.org/wiki/Bunch%E2%80%93Nielsen%E2%80%93Sorensen_formula), but you will need to distinguish between left and right eigenvectors. These are different, because the rank-one-update is not symmetric. So if you work in the basis where $A$ is diagonal, with elements $a\_k$ on the diagonal, and the vectors $u$ and $v$ in this basis are $\tilde{u}$ and $\tilde{v}$, the $i$-th left and right eigenvectors at eigenvalue $\lambda\_i$ will have elements $\tilde{u}\_k/(a\_k-\lambda\_i)$ and $\tilde{v}\_k/(a\_k-\lambda\_i)$.
The nonsymmetric BNS formula follows from the nonsymmetric
[Sherman–Morrison formula](https://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula), applied to $(A-\lambda I+uv^\top)^{-1}$.
| 2 | https://mathoverflow.net/users/11260 | 448706 | 180,631 |
https://mathoverflow.net/questions/448704 | 9 | We say that a model $M$ of $\mathsf{ZF}$ satisfies *Small Violations of Choice* ($\mathsf{SVC}$) if all (any) of the following apply:
1. There is a model $V\subseteq M$ such that $V\vDash\mathsf{ZFC}$, and $M$ is a symmetric extension of $V$.
2. There is a forcing $\mathbb{P}\in M$ such that $\mathbb{P}\mathrel{\Vdash}\mathsf{AC}$.
3. There is $A\in M$ such that for all $X\in M$, there is an ordinal $\eta$ and a surjection $f\colon A\times\eta\to X$ in $M$.
4. There is $A\in M$ such that for all $X\in M$, there is an ordinal $\eta$ and an injection $f\colon X\to A\times\eta$ in $M$.
The equivalence is true, standard, and very difficult. [1] introduces $\mathsf{SVC}$ and has some work towards proving their equivalence. For the purposes of this question, (1), (3), and (4) are not strictly required, but may be a helpful tool. What I am really interested in is "If (2) then can we force $\mathsf{AC}$ without identifying the cardinalities of distinct ordinals?".
I am particularly interested in condition (2), and in forcing $\mathsf{AC}$ in such a way that if $\lambda\neq\kappa\in M$ are well-ordered cardinals, $\mathbb{P}\mathrel{\Vdash}\check{\lambda}\neq\check{\kappa}$.
**Question 1.** Is this always possible? No! As Asaf Karagila has pointed out, we certainly require that all successors are regular. Any model of $\text{cf}(\omega\_1)=\omega$ would not be able to be extended to a model of $\mathsf{AC}$ without collapsing $\omega\_1$ to $\omega$.
**Question 2.** How much can we save? If $M\vDash\mathsf{DC}\_{{<}\lambda}$, can we always keep cardinals at least the size of $\lambda$ distinct?
[1] *Blass, Andreas*, [**Injectivity, projectivity, and the axiom of choice**](https://doi.org/10.2307/1998165), Trans. Am. Math. Soc. 255, 31-59 (1979). [ZBL0426.03053](https://zbmath.org/?q=an:0426.03053).
| https://mathoverflow.net/users/478588 | Small Violations of Choice: Can we force AC without collapsing the cardinalities of ordinals? | The answer is no.
Firstly, let's get the obvious out of the way. The Feferman–Levy model is a symmetric extension in which $\omega\_1$ is singular. If we force the axiom of choice, we must collapse the ordinal to be countable.
Right. So perhaps we want to require that all successor cardinals are regular. This may be enough? But it is not going to be enough.
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> *Gitik, Moti; Koepke, Peter*, [**Violating the singular cardinals hypothesis without large cardinals**](https://doi.org/10.1007/s11856-012-0028-x), Isr. J. Math. 191, Part B, 901-922 (2012). [ZBL1291.03094](https://zbmath.org/?q=an:1291.03094).
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In that paper, the authors start with a model of $\sf GCH$, for any $\lambda>\aleph\_{\omega+1}$, they build a symmetric extension which does not change cardinals and cofinalities of ordinals, where $\aleph^\*(\mathcal P(\omega\_\omega))>\lambda$. This is a violation of the Singular Cardinals Hypothesis which will require the existence of large cardinals, had it taken place in $\sf ZFC$.
Therefore, constructing the Gitik–Koepke model, where there is no inner model with the necessary large cardinal hypotheses, will result in a model where any extension by forcing to restore choice must collapse cardinals.
To your second question, Fernengel and Koepke had [extended the original Gitik–Koepke model](https://arxiv.org/abs/1812.00960), first to a proper class situation, then for any set-many cardinals with $\sf DC\_{<\lambda}$, under some additional constraints, which we can even show must violate Silver's theorem, and therefore collapse cardinals.
But, if $\sf DC\_{<\lambda}$ holds, then at the very least we can force $\sf AC$ with a $\lambda$-closed forcing and therefore not add any bounded subsets to $\lambda$, thus preserving all cardinals up to $\lambda$ itself.
To see that, note that if $A$ is the seed for $\sf SVC$, and $\sf DC\_{<\lambda}$ holds but $\sf DC\_\lambda$ fails, then $\aleph(A)\geq\lambda$. Simply force with $A^{<\lambda}$, collapsing $A$ to have cardinality $\lambda$.
| 9 | https://mathoverflow.net/users/7206 | 448711 | 180,635 |
https://mathoverflow.net/questions/397001 | 4 | The following surely is kind of a trivial question, but it keeps me confused. It concerns a detail in Lust and Stevens' paper "On depth zero L-packets for classical groups" London Math. Soc. (3) 121 (2020), more precisely about the discussion in the first paragraphs of the 3rd part on depth zero supercuspidal representations. Let me give some more context.
Here, $F$ is an extension of degree at most $2$ of a $p$-adic field $F\_0$ with $p$ odd, and $\sigma$ is the generator of the Galois group $\mathrm{Gal}(F/F\_0)$. We use the letters $\mathfrak o$ and $k$ respectively for the rings of integers and residue fields. For $n\geq 1$, a smooth (complex) representation $\rho$ of $G = \mathrm{GL}\_n(F)$ is said to be *self-dual* if we have an equivalence $\rho^{\sigma} \simeq \rho$ where $\rho^{\sigma}$ is defined by the formula
$$\rho^{\sigma}(g) = \rho\left(\sigma(g^{-1})^T\right)$$
So far so good. Now, assume that $\rho$ is irreducible supercuspidal of depth zero. Thus, there is a well-determined (equivalence class of) representation $\tau$ of the maximal compact subgroup $K = \mathrm{GL}\_n(\mathfrak o\_F)$ such that, on the one hand $\tau$ is the inflation of a cuspidal irreducible representation of the finite group of Lie type $K/K^+ \simeq \mathrm{GL}\_n(k\_F)$, and on the other hand one may write $\rho \simeq \mathrm{c-Ind}\_{F^{\times}K}^{G}\tilde{\tau}$ where $\tilde{\tau}$ is some extension of $\tau$ to a representation of the normalizer $N\_{G}(K) = F^{\times}K$. Here, $F^{\times}$ is identified with the center of $G$. Such an extension $\tilde{\tau}$ is determined by the choice of a character $\chi$ of the center $F^{\times}$ which coincide with the central character of $\tau$ on $F^{\times}\cap K = \mathfrak{o}\_F^{\times}$.
Then, it is claimed that $\rho$ is self-dual if and only if $\tau$, as a representation of $\mathrm{GL}\_n(k\_F)$, is self-dual (that is, it satisfies the same equation as above but with $\sigma$ denoting the generator of $\mathrm{Gal}(k\_F/k\_{F\_0})$).
Alright, at this stage I wanted to check this in the special case $n=1$, and both $\rho$ and $\tau$ are identified with linear characters. I take $\tau$ as the trivial character of $K = \mathfrak o\_F^{\times}$. Since the trivial representation of $\mathrm{GL}\_1(k\_F) = k\_F^{\times}$ is cuspidal and because $\tau$ clearly is self-dual, I am in a special case of the situation described above. Thus, if I choose any character $\chi$ of the center $F^{\times} = G$ (the group $G$ is abelian) which is trivial on $\mathfrak o\_F^{\times}$, then $\rho = \chi$ should be a self-dual (supercuspidal) character of $G$ - the compact induction being trivial in this case.
But now here's the thing. Any such $\chi$ is called an unramified character of $F^{\times}$, and it doesn't look like they are self-dual at all. Indeed, any such $\chi$ may be written $\chi = |\,\cdot\,|\_F^{\alpha}$ for some complex number $\alpha$, where $|\,\cdot\,|$ is the normalized norm on $F$. Any two unramified characters are equivalent if and only if they share they have the same $\alpha$ (they are equal).
Then, since $\sigma$ acts on $F$ as an isometry, for $x\in F^{\times}$ we have
$$\chi^{\sigma}(x) = \chi\left(\sigma(x^{-1})\right) = |\, \sigma(x^{-1})\,|\_F^{\alpha} = |\, x \,|\_F^{-\alpha}$$
So, $\chi$ seems to be self-dual if and only if $\alpha = - \alpha$, that is $\chi$ is trivial... Surely, something is wrong there.
I assume that I have misunderstood something somewhere, but I can't find what exactly. Could somebody point it out to me ?
| https://mathoverflow.net/users/125617 | Confusion with self-dual representations of $\mathrm{GL}_n$ over a $p$-adic field | You are right, there is an error in Lust-Stevens, and any unramified character provides a counter-example. Here is a way to fix their statement: $\rho=\mathrm{cInd}\_{KF^\times}^G\widetilde\tau$ is self-dual if and only if $\tau=\widetilde\tau|\_K$ is self-dual and the central character $\omega\_{\widetilde\tau}\colon F^\times\to\mathbb C^\times$ of $\widetilde\tau$ is such that $\omega\_{\widetilde\tau}(z\sigma(z))=1$ for all $z\in F^\times$ (when $F=F\_0$, this means $\omega\_{\widetilde\tau}^2=1$).
Indeed, $\rho$ being self-dual means $\mathrm{cInd}\_{KF^\times}^F\widetilde\tau\cong\mathrm{cInd}\_{KF^\times}^F\widetilde\tau^\sigma$, which is equivalent to $\widetilde\tau\cong\widetilde\tau^\sigma$ as representations of $KF^\times$.
Now the claim above follows from the observation that an irreducible representation $\widetilde\tau$ is uniquely determined by its central character $\omega\_{\widetilde\tau}$, together with its restriction to $K$.
| 2 | https://mathoverflow.net/users/123673 | 448714 | 180,636 |
https://mathoverflow.net/questions/448712 | 1 | We add a little bit to [On 'fair bisectors' of planar convex regions](https://mathoverflow.net/questions/356279/on-fair-bisectors-of-planar-convex-regions) and [Bisectors and partitioning lines for convex regions defined with respect to the moment of inertia](https://mathoverflow.net/questions/437053/bisectors-and-partitioning-lines-for-convex-regions-defined-with-respect-to-the)
**Definitions:** Given a planar convex region C (could be smooth or polygonal), an area bisector of C
is any line that partitions C into 2 pieces of equal area. A perimeter bisector is a line that partitions C into 2 pieces of equal perimeter. Obviously, thru every point on the boundary of C we can draw an area bisector and a perimeter bisector.
**Question:** Are the following claims easy to prove/counter?
* A planar convex region is centrally symmetric if and only if its area bisectors are all concurrent.
* A planar convex region is centrally symmetric if and only if its perimeter bisectors are all concurrent.
Note: Higher dimensional analogs of these claims are easy to state.
| https://mathoverflow.net/users/142600 | A claim on the concurrency of area bisectors of planar convex regions | * A planar convex region is centrally symmetric if and only if its area bisectors are all concurrent.
Yes. Let all area bisectors of our region $K$ pass through a point $O$. Note that any line $\ell$ through $O$ is an area bisector of $K$: otherwise there exists a parallel area bisector (by continuity), and it does not pass through $O$. Assume that $K$ is not $O$-symmetric. It yields that some chord $AB$ through $O$ is not bisected by $O$. Move it slightly to get a chord $A\_1B\_1$ ($A\_1$ close to $A$). The curvilinear triangles $AOA\_1$ and $BOB\_1$ have unequal area, thus at least one of chords $AB$, $A\_1B\_1$ is not an area bisector.
| 2 | https://mathoverflow.net/users/4312 | 448715 | 180,637 |
https://mathoverflow.net/questions/448717 | 1 | Lets extend $\mathcal L\_{\omega\_1, \omega\_1}$ with axioms of equality and of:
$\sf ZF + Definability+Ture$-$\sf Foundation+True$-$\sf Finiteness $
Where $\sf ZF$ is written, as usual, in $\mathcal L\_{\omega, \omega}$
$\sf Definability$ axiom is written in $\mathcal L\_{\omega\_1, \omega}$ as:
$\textbf{Define: } Dx \iff \bigvee x= \{ y \mid \Phi \}$
where $\Phi$ range over all formulas in $\mathcal L\_{\omega, \omega}$ in which only the symbol "$y$" occurs free, and the symbol "$y$" never occurs bound.
${\sf Definability\!: } \ \forall x Dx$
$\sf True$-$\sf Foundation$ axiom is written in $\mathcal L\_{\omega\_1, \omega\_1}$ as:
$ (\forall v\_n)\_{n \in \omega} \, \exists x: \bigvee\_{n \in \omega} (x=v\_n) \land \bigwedge\_{n \in \omega} (v\_n \not \in x)$
and $\sf True$-$\sf Finiteness$ axiom written in $\mathcal L\_{\omega\_1, \omega}$ as:
$\operatorname {finite}(x) \iff \\ \bigvee\_{n \in w} [\exists v\_0,.., \exists v\_n : x=\{y \mid y \neq y \lor \bigvee\_{i \in n } ( y = v\_i)\}]$
Where $\operatorname {finite}(x)$ is defined in $\mathcal L\_{\omega,\omega}$, as usual, by $x$ having a bijection with an initial set of naturals, i.e. naturals closed under predecessor relation.
Now, this should be arithmetically complete, since it captures the standard model of arithmetic, also it captures the set $\sf HF$ of all standard hereditarily finite sets.
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| https://mathoverflow.net/users/95347 | Can we have ZF + Definability + True Foundation + True Finiteness? Can it be Categorical? | The models of your theory are exactly the pointwise-definable (with respect to $\mathcal{L}\_{\omega,\omega}$ as usual) well-founded models of $\mathsf{ZF}$; incidentally, your true finiteness axiom is unnecessary. Assuming such models exist of course, this means that your theory is consistent and **incomplete** (hence not categorical), since we can always force over a pointwise-definable well-founded model to get a new pointwise-definable well-founded model with e.g. a different belief about the truth value of $\mathsf{CH}$. (This isn't actually trivial since we need to carefully preserve pointwise-definability, but it is true; see [Hamkins/Linetsky/Reitz](https://arxiv.org/pdf/1105.4597.pdf) for more on this. Really, given your questions I strongly recommend reading that paper carefully, I think it will clarify many things!)
| 7 | https://mathoverflow.net/users/8133 | 448718 | 180,638 |
https://mathoverflow.net/questions/448259 | 7 | It is known that for a $\partial\bar\partial$-manifold $X$ (a compact complex manifold satisfies the $\partial\bar\partial$-lemma), the Bott-Chern cohomology $H\_{BC}^{\bullet,\bullet}:=\frac{\ker \partial\cap\ker\bar\partial}{\text{im }\partial\bar\partial}$ is isomorphic to the Dolbeault cohomology $H\_{\bar\partial}^{\bullet,\bullet}(X)=\frac{\ker\bar\partial}{\text{im }\bar\partial}$, and the Frölicher spectral sequence of $X$ degenerates at $E\_1$, then
$$
H^k(X,\mathbb C)=\bigoplus\_{p+q=k}H^{p,q}\_{\bar\partial}(X)=\bigoplus\_{p+q=k}H^{p,q}\_{BC}(X),
$$
so $H^{p,q}\_{\bar\partial}(X)$ can be treated as a subspace of $H^k(X,\mathbb C)$ through the isomorphism between Bott-Chern and Dolbeault cohomology.
If we only assume that the Frölicher spectral sequence of $X$ degenerates at $E\_1$, is there a natural way to treat $H^{p,q}\_{\bar\partial}(X)$ as a subspace of $H^k(X,\mathbb C)$? or is there a natural map $f:H^{p,q}\_{\bar\partial}(X)\to H^k(X,\mathbb C)$ which is injective?
| https://mathoverflow.net/users/99826 | When $H^{p,q}_{\bar\partial}(X)$ can be seen as a subspace of $H^k(X,\mathbb C)$? | I'm not sure that the following answers the question, but at least it shows that it is not always possible to construct a map $H\_{\bar\partial}\to H\_{dR}$ by assigning a $d$-closed representative to each class in $H\_{\bar\partial}$.
Suppose $\alpha$ is a $\bar\partial$-closed form, and we want to find $\beta$ such that $d(\alpha-\bar\partial\beta)=0$. Each summand inside the parentheses is $\bar\partial$-closed, so the equality is equivalent to $\partial\alpha=\partial\bar\partial\beta$. We learn that each class in $H\_{\bar\partial}$ admits a $d$-closed representative if and only if $\partial\ker\bar\partial\subset\operatorname{im}\partial\bar\partial$ — in a more fancy language, the map $H\_{\bar\partial}\to H\_{BC}$ induced by $\partial$ is zero.
As explained in [Ornea, Verbitsky, Vuletescu, Classification of non-Kähler surfaces and locally conformally Kähler geometry](https://arxiv.org/abs/1810.05768), § 2.2, for any non-Kähler compact complex surface, the image of $H^{0,1}\_{\bar\partial}$ by this map is 1-dimensional. So there is a class in $H^{0,1}\_{\bar\partial}$ with no $d$-closed representative, notwithstanding that the Frölicher spectral sequence of any compact complex surface degenerates at the first page.
| 1 | https://mathoverflow.net/users/485324 | 448719 | 180,639 |
https://mathoverflow.net/questions/448727 | 4 |
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> There are meaningful questions we can ask about Euclidean geometry which could not have been posed in the time of Riemann or even of Hilbert, and which would have made no sense at all to Euclid. For example, **does two-dimensional Euclidean geometry emerge as the large-scale limit of a quantum geometry**? The fact that we are able to ask this question today demonstrates that the relevant constellation of absolute presuppositions, scene of inquiry, disciplinary matrix, or however you wish to phrase it, has simply changed.
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I'm trying to understand the highlighted line. What exactly is Quantum Geometry supposed in this context? Better, if the whole bolded question could be explained.
What I found so far:
--------------------
1. This [wiki confuses me more](https://en.wikipedia.org/wiki/Quantum_geometry) since explanations assumes physics context I think.
2. [This MSE thread gives conflicting opinions](https://math.stackexchange.com/questions/2632188/path-to-quantum-geometry-for-mathematicians), in particular, of one user saying this maybe an ambiguous question to ask.
| https://mathoverflow.net/users/159957 | What is Quantum Geometry supposed to be about? | I think it basically goes like this:
1. Take an algebra with a product and a skew product, for example a suitable algebra of functions with a Poisson bracket $\{f,g\}=-\{g,f\}$. You can start with a complex manifold, or mechanical system, or a Lie group, and get a naturally skew-symmetric bracket.
2. Replace multiplication that used to be commutative $fg-gf=0$ with a non-commutative one, with the deviation from commutativity being defined by said Poisson bracket. Define multiplication as $fg-gf=ih\{f,g\}$, where $h$ may be real or rational or imaginary, depending on what you want to study. See [Moyal product](https://en.wikipedia.org/wiki/Moyal_product). The "quantum" here alludes to the $h$ sometimes being interpreted as the Planck's constant.
3. Study that non-commutative algebra and see what cool mathematics would fall out.
| 8 | https://mathoverflow.net/users/38448 | 448728 | 180,640 |
https://mathoverflow.net/questions/448696 | 2 | While reading the well known book *Minimax Methods in Critical Point Theory with Applications to Differential Equations* by Paul Rabinowitz, in the proof of a generalisation of the Mountain Pass Theorem (Theorem 5.29 in the book), I encountered the following abstract result:
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> Let $E$ be a real Hilbert space and consider $b\in C^1(E, \mathbb{R}$) such that $b'$ is compact. Then, $b$ is weakly continuous, i.e. if $(u\_n)\_n\subseteq E$ converges weakly to $u\in E$, then $b(u\_n)\to b(u)$ as $n\to \infty$.
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The reference for this result given in the book is
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> M. A. Krasnoselski, *Topological methods in the theory of nonlinear integral
> equations*, Macmillan, New York, 1964.
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However, even if this is a well renowned book in the field of nonlinear analysis, I do not have access to it. Does anyone know a modern reference for this result? The proof looks nontrivial to me, at least I do not know how to approach it.
EDIT: The fact that $b'$ is compact means that if $A\subseteq E$ is bounded, then the closure of $b'(A)$ is compact. Of course, $b'$ denotes the mapping $x\mapsto b'(x)$, i.e. $b'$ associates with each $x\in E$ the Frechet derivative of $b$ at $x$, which we denote by $b'(x)\in E^{\*}$.
| https://mathoverflow.net/users/124648 | If $b\in C^1(E, \mathbb{R})$ and $b'$ is compact, then $b$ is weakly continuous — a reference request | To obtain a contradiction, suppose that $b(u\_n)\not\to b(u)$. Passing to a subsequence, without loss of generality (wlog) we have $|b(u\_n)-b(u)|\ge c$ for some real $c>0$ and all $n$.
By the mean value theorem, $b(u\_n)-b(u)=b'(v\_n)(u\_n-u)$ for all $n$ and some $v\_n$ on the straight line segment from $u$ to $u\_n$.
Since the sequence $(u\_n)$ is weakly convergent, it is [bounded](https://en.wikipedia.org/wiki/Weak_convergence_(Hilbert_space)#Properties). So, recalling that $b'$ is compact and passing to a subsequence, wlog we have $b'(v\_n)\to B$ for some $B\in E^\*$. Also, $B(u\_n-u)\to0$, since $u\_n\to u$ weakly. Also, $\|b'(v\_n)-B\|\,\|u\_n-u\|\to0$, since $b'(v\_n)\to B$ and the sequence $(u\_n)$ is bounded.
So,
$$|b(u\_n)-b(u)|=|b'(v\_n)(u\_n-u)|\le\|b'(v\_n)-B\|\,\|u\_n-u\|+|B(u\_n-u)|\to0+0=0,$$
which contradicts the condition that $|b(u\_n)-b(u)|\ge c$ for some real $c>0$ and all $n$. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 448730 | 180,641 |
https://mathoverflow.net/questions/448639 | 2 | The *strong multiplicity one theorem* for newforms says the following. Suppose that $f\_1 \in S\_k(\Gamma\_0(N\_1))$ and $f\_2 \in S\_k(\Gamma\_0(N\_2))$ are newforms, where $N\_1, N\_2 \geq 1$ are arbitrary integers. If $a\_{\ell}(f\_1) = a\_{\ell}(f\_2)$ for all primes $\ell$ not dividing a fixed integer $L$, then $f\_1 = f\_2$.
I'm looking for a "mod $p$ version" of this theorem:
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> Let $p$ be a prime. Suppose that $f\_1 \in S\_k(\Gamma\_0(N\_1))$ and $f\_2 \in S\_k(\Gamma\_0(N\_2))$ are newforms with Fourier coefficients in $\mathbf{Z}$, where $N\_1, N\_2 \geq 1$ are arbitrary integers. If $a\_{\ell}(f\_1) \equiv a\_{\ell}(f\_2)$ modulo $p$ for all primes $\ell$ not dividing a fixed integer $L$, then $f\_1 \equiv f\_2$ modulo $p$.
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Is this theorem true, and if so, does anyone know of a reference?
| https://mathoverflow.net/users/394740 | Multiplicity one for newforms modulo $p$ | If by $f\_1\equiv f\_2$ modulo $p$, you mean that $a\_n(f\_1)\equiv a\_{n}(f\_2)$ modulo $p$ for all $n\in\mathbb N$ or maybe for all except finitely many, then this theorem cannot be true.
Let's start with a counterexample. Consider the two newforms
\begin{equation}
f(z)=\underset{n=1}{\overset{\infty}{\Sigma}}a\_{n}q^{n}\in S\_{2}(\Gamma\_0(11)),\ g(z)=\underset{n=1}{\overset{\infty}{\Sigma}}b\_{n}q^{n}\in S\_{2}(\Gamma\_0(77))
\end{equation}
attached respectively to the elliptic curves $X\_0(11)$ and $E:y^2+xy=x^3+x^2-51x+110$. Then $a\_{\ell}(f)\equiv a\_{\ell}(g)$ modulo 3 for all primes $\ell\neq 7$ yet $a\_7(f)=-2$ while $a\_7(g)=-1$ so $a\_7(f)\not\equiv a\_7(g)$ modulo 3, and in fact typically $a\_n(f)\not\equiv a\_n(g)$ modulo 3 if $n$ is a strictly positive integer which is divisible by 7.
What's going on? The problem is that $a\_\ell$ is the trace of the image of the Frobenius morphism at $\ell$ through the Galois representation attached to the newform provided $\ell$ does not divide the conductor of the newform. The fact that $a\_{\ell}(f)\equiv a\_\ell(g)$ for almost all primes implies that the residual representations $\bar{\rho}\_f$ and $\bar{\rho}\_g$ are isomorphic (by Cebotarev's density theorem). If $\ell$ divides neither the conductor of $f$ nor the conductor of $g$, then $a\_\ell(f)$ modulo $p$ and $a\_\ell(g)$ modulo $p$ are both the trace of the image of the Frobenius morphism, so they are equal. However, 7 divides the conductor of $g$ but not the Artin conductor of the residual representation $\bar{\rho}\_g$ nor the conductor of $f$ so the relation between $a\_7(g)$ and $a\_7(f)\equiv\mathrm{tr}\bar{\rho}\_g(\mathrm{Fr}(7))$ breaks down.
This will happen generally speaking whenever you have a level-raising (or equivalently level-lowering) phenomenon, that is to say congruences between forms which are new of different levels.
| 2 | https://mathoverflow.net/users/2284 | 448753 | 180,646 |
https://mathoverflow.net/questions/448751 | 4 | Consider based maps $f : X \to Z$ and $g : Y \to Z$, which induces the following map at the based loop space level : $$\theta := \mu\_Z \circ \big(\Omega f \times \Omega g) : \Omega(X\times Y) = \Omega X \times \Omega Y \to \Omega Z \times \Omega Z \to \Omega Z,$$
where $\mu\_Z$ is the standard choice of loop concatenation. Suppose we are given that $\theta$ is an $A\_2$ map in the sense that there is a homotopy $$\mu\_Z \circ (\theta \times \theta) \simeq \theta \circ \mu\_{X\times Y},$$
where $\mu\_{X\times Y}$ is the standard concatenation in $\Omega(X\times Y)$.
**Question:** Can we claim that $\theta$ is an $A\_\infty$-map? In particular, assuming suitable connectivity of $X,Y,Z$, can we deloop $\theta$ to get a map $X\times Y \to Z$?
I understand it is probably too much to ask for! But here is my approach. I will consider the interval operad $\mathcal{E}\_1$ as the $A\_\infty$-operad acting on the loop spaces. From this, I am able to prove that the following diagrams are *homotopy* commutative for each $n$ :
$$\require{AMScd}
\begin{CD}
\mathcal{E}\_1(n) \times \big(\Omega(X\times Y)\big)^n @>{}>> \Omega(X\times Y)\\
@V{1\times \theta^n}VV @VV{\theta}V\\
\mathcal{E}\_1(n) \times \big(\Omega Z\big)^n @>>> \Omega Z
\end{CD}$$
This is weaker than the notion of $\mathcal{E}\_1$-map as in [The Geometry of Iterated Loop Spaces](https://www.math.uchicago.edu/%7Emay/BOOKS/geom_iter.pdf) by J. P. May, and so I am unable to apply the recognition principle directly. On the other hand, I have come across articles in the algebraic context (e.g., [this one](https://arxiv.org/abs/2004.06652)), where the author considers the above as the definition of an $E\_\infty$-coalgebra map. Also, at the very end of the same monograph by May, the author notes that the theory can possibly be weaken to include *homotopy* morphisms. In the article Strong homotopy algebras over monads by T. Lada, the author introduced a notion of strong homotopy morphism, which asks for higher homotopy relations, and proved the recognition principle. But the $\theta$ map is not a strong homotopy morphism (in the sense of T. Lada) either. Can we still expect a recognition principle without higher homotopy relations?
Any comment or reference regarding this will be highly appreciated.
| https://mathoverflow.net/users/38824 | Delooping a weak $E_1$ map by bar construction | No, take $Z$ a connected space for which $\Omega Z$ is homotopy commutative but $Z$ has no $A\_\infty$ multiplication, e.g. $Z$ could be an $H$-space which doesn't have the homotopy type of a loop space. If $f=g=\mathrm{Id}\_Z$, then the multiplication $\Omega Z \times \Omega Z \rightarrow \Omega Z$ is an $A\_2$-map by the hypothesis of homotopy commutativity and and Eckmann-Hilton argument. If it were an $A\_\infty$ map, we could apply the bar construction to achieve a map $B(\Omega Z \times \Omega Z)=B\Omega Z \times B\Omega Z \rightarrow B \Omega Z$ which is an $A\_\infty$ multiplication. Since $Z$ is connected $B\Omega Z \simeq Z$, so we have produced an $A\_\infty$ multiplication on $Z$ which contradicts our assumption.
| 4 | https://mathoverflow.net/users/134512 | 448759 | 180,649 |
https://mathoverflow.net/questions/448512 | 0 | This posting is a continuation to an earlier one titled "[Can Foundation be captured in $\mathcal L\_{\omega\_1, \omega}$ ?](https://mathoverflow.net/questions/448486/can-foundation-be-captured-in-mathcal-l-omega-1-omega)"
It appears that capturing foundation is problematic at every $\mathcal L\_{\alpha, \omega}$ and even at $\mathcal L\_{ \infty, \omega}$. [see [this](https://mathoverflow.net/a/448488/95347) and comments below it]
However, can there be particular cases where this fails? I mean where we can capture Foundation for some particular $\mathcal L\_{\omega\_1, \omega}$ theory.
I'm specifically referring to the theory [$\sf ZF + Def$](https://mathoverflow.net/questions/448369/does-adding-definability-axiom-expressed-in-infinitary-language-to-zf-let-all-m). This theory has all of its models being exactly the pointwise definable models of $\sf ZF$.[[Hamkins](https://mathoverflow.net/a/448434/95347)]
So, [it proves $\sf V=HOD$](https://mathoverflow.net/questions/448130/are-all-constructible-from-below-sets-parameter-free-definable/448132#448132), and accordingly there is a finitary formula $\phi$ in two free variables that defines a binary relation $\leqslant$ that well orders the universe.
Now can we use this feature to capture Foundation?
$\textbf{Define: } x \in^{\leqslant} y \iff x \in y \land \forall m \in y \, ( x \leqslant m )$
That is, $x$ is the element of $y$ of the least order according to $\leqslant$.
Now, we define:
$ \textbf{Define: } y \in^\leqslant\_K x \iff y \in^\leqslant x \cap K $
Where, as usual: $ x \cap K =\{y \mid y \in x \land y \in K\}$
$\textbf{Foundation: } \forall K \forall x:\\ \neg [ \bigwedge\_{n \in \omega} (\exists v\_0,..,\exists v\_n: \bigwedge\_{i \in n} (v\_{i+1} \in^\leqslant\_K v\_i) \land v\_0 \in^\leqslant\_K x)] $
Now, the idea is that if there exists a nonempty set $K$ such that every element of $K$ has an element of it that is an element of $K$, then there would exist $x \in K$ that violates Foundation.
>
> Would that succeed in capturing Foundation, thereby rendering all models of the resulting theory well founded?
>
>
>
| https://mathoverflow.net/users/95347 | Can Foundation be captured for some $\mathcal L_{\omega_1, \omega}$ theories? |
>
> **Lemma.** A model of ZF satisfies the “Foundation” axiom if and only if it is an $\omega$-model, i.e., iff it satisfies the simpler $\mathcal L\_{\omega\_1,\omega}$-sentence $\forall x\in\omega\,\bigvee\_{n\in\omega}x=n$.
>
>
>
**Proof:**
Right-to-left: Working inside the model, given $x$ and $K$, define by recursion a partial function $f\colon\omega\to K$ such that $f(0)=x$ and if $f(n)$ is defined and intersects $K$, then $f(n+1)$ is the $\le$-least element of $f(n)\cap K$. If $f$ is total, then $\{f(n):n\in\omega\}$ contradicts the usual axiom of foundation of ZF. Thus, let $n\in\omega$ be minimal such that $f(n)$ is undefined. Assuming $n$ is standard, we have $\neg\exists v\_0,\dots,v\_n \cdots$ as in the axiom of “Foundation”.
Left-to-right: Define by recursion a function $f\colon\omega\to V$ such that $f(0)=0$ (say) and $f(n+1)=\{f(n)\}$. Let $K=\{f(n):n\in\omega\}$, and $x=f(n)$ for a fixed nonstandard $n\in\omega$. Then “Foundation” fails for $K$ and $x$.
>
> **Corollary.** The following are equivalent (provably in ZF):
>
>
> 1. Your theory is consistent.
> 2. ZF (or ZFC) has an $\omega$-model.
> 3. The closure of ZF (or ZFC) under the $\omega$-rule is consistent.
>
>
>
**Proof:** The equivalence of 2 and 3 holds for arbitrary countable FO theories by a well-known consequence of the omitting types theorem. Clearly, 1 implies 2 by the Lemma. On the other hand, if $M$ is an $\omega$-model of ZF, then $L^M$ is an $\omega$-model of ZF + V=L, and the set of parameter-free definable elements of $L^M$ is its elementary submodel, thus an $\omega$-model of ZF + Def.
>
> **Corollary:** If the theory is consistent, then it has a non-well-founded model.
>
>
>
**Proof:** Assume the theory has a model. If it is not well founded, we are done, thus we may assume there exists a well-founded model of ZF. It follows that there exists a least ordinal $\alpha$ such that $L\_\alpha$ is a model of ZF. Now, $L\_\alpha$ satisfies “the closure of ZF under the $\omega$-rule is consistent”, as the closure as computed in $L\_\alpha$ is included in the true closure. Thus, there exists $M\in L\_\alpha$ such that $L\_\alpha$ satisfies “$M$ is a point-wise definable $\omega$-model of ZF”. Since $L\_\alpha$ is itself a transitive model of ZF, $M$ really is a point-wise definable $\omega$-model of ZF, i.e., a model of your theory. But since $M\in L\_\alpha$, $M$ cannot be well founded by the minimality of $\alpha$.
| 1 | https://mathoverflow.net/users/12705 | 448767 | 180,651 |
https://mathoverflow.net/questions/448780 | 2 | **Background**: Let $X$ be a Fano variety over number field $K$, where its anticanonical bundle $K\_X^{-1}$ is ample. Let $i: X \to \mathbb{P}^n$ be the anticanonical embedding, where $K\_X^{-m} \cong i\_\*O(1)$. The standard (exponential) height $H$ on $\mathbb{P}^n$ is induced by standard metric on $O(1)$, and pull back to a height $i^\*H$ on $X$.
Batyrev-Manin and Peyre conjectured a formula to count number of rational points of height up to $B$:
$$N(X, K\_X^{-1}, B) := \#\{x \in X(K): i^\* H(x) \leq B\} \sim c(X, K\_X^{-1}) B^{a(X, K\_X^{-1}} (\log B)^{b(X, K\_X^{-1})}$$
(Strictly speaking, one needs to focus on sufficiently small Zariski open subset of $X$, after removing accumulating subvarieties)
[Batyrev-Tschinkel](https://math.nyu.edu/%7Etschinke/papers/yuri/97dec/dec97.pdf) extended the conjecture to a more general setup, where the datum of $(X, L)$ is considered: here $X$ is a projective variety over $K$, and $L$ some suitable ample line bundle. For my question here, I only care about strongly saturated, $L$-primitive varieties $X$ in their language; this effectively means I do not need to remove any accumulating subvarieties. In this case, one expects an asymptotic formula
$$N(X, L, B) := \#\{x \in X(K): H\_L(x) \leq B\} \sim c(X, L) B^{a(X, L)} (\log B)^{b(X, L)}$$
as $B \to \infty$, where
* $a(X, L)$ is (effectively)
$$a(X, L) := \inf\{t \in \mathbb{Q}: tL + K\_X \in \Lambda\_{eff}(X)\}$$
(Definition 2.2.4)
* (Unimportant, including for completeness) $b(X, L)$ is (effectively)
$$b(X, L) := rank(Pic(X, L))$$
(Definition 2.3.11) Here if $L^{a(X, L)} \otimes K\_X$ is thought of as an effective divisor with support on irreducible components $D\_1, \cdots, D\_l$, then $Pic(X, L) := Pic(X - \bigcup\_{i=1}^l D\_i)$.
+ As a sidenote, there is a natural restriction map $\rho: Pic(X) \otimes \mathbb{R} \to Pic(X, L) \otimes \mathbb{R}$. We will use $\Lambda\_{eff}(X, L)$ to denote the image of $\Lambda\_{eff}(X)$ under $\rho$. (Definition 2.3.11)
* $c(X, L)$ is of the shape
$$c(X, L) = \frac{\gamma(X, L)}{a(X, L) (b(X, L) - 1)!} \delta(X, L) \tau(X, L)$$
(page 32, middle of step 4). Here the extra constants are:
+ $\gamma(X, L)$ is some invariant for the triple $(Pic(X, L), Pic(X, L) \otimes \mathbb{R}, \Lambda\_{eff}(X, L))$ (Definition 2.3.14)
+ $\delta(X, L)$ is the cardinality of $H^1(Gal(\overline{K}/K), Pic(X, L))$ (Definition 3.4.3)
+ $\tau(X, L)$ is some Tamagawa measure of $\overline{X(K)}$ in $X(\mathbb{A}\_K)$. (Definition 3.3.10)
**Question**
I am interested in compatibility of the conjecture for $(X, L)$ and $(X, L^{\otimes k})$ for some positive integer $k$, particularly on the leading constant term $c(X, L)$. By checking definition,
* I am pretty sure that the constants/invariants $b, \gamma, \delta, \tau, Pic$ are the same for $(X, L)$ and $(X, L^{\otimes k})$.
* I think that $a(X, L^{\otimes k}) = \frac{1}{k} a(X, L)$. This should then imply
$$c(X, L^{\otimes k}) = \frac{1}{k} c(X, L)$$
* At the same time, the induced heights should satisfy $H\_{X, L^{\otimes k}} = H\_{X, L}^k$, which should then mean $$N(X, L^{\otimes k}, B) \sim N(X, L, B^{1/k})$$
But then there seems to be a compatibility issue in the leading constant term: the conjecture for $(X, L)$ would predict
$$N(X, L, B^{1/k}) \sim c(X, L) B^{a(X, L)/k} (\log B)^{b(X, L)}$$
However, the conjecture for $(X, L^{\otimes k})$ would predict
$$N(X, L, B^{1/k}) \sim N(X, L^{\otimes k}, B) \sim c(X, L^{\otimes k}) B^{a(X, L^{\otimes k})} (\log B)^{b(X, L^{\otimes k})} = \frac{1}{k} c(X, L) B^{a(X, L)/k} (\log B)^{b(X, L)}$$
which is off by a factor of $\frac{1}{k}$.
Where in my calculation was wrong? Thank you!
| https://mathoverflow.net/users/479911 | Leading constant in Batyrev-Tschinkel's refinement of Manin conjecture | Your calculation is off in a couple places but neither removes the discrepancy.
First, the exponent of $(\log B)$ is $b(X,L)-1$.
Second, substituting $B^{1/k}$ for $B$ in the term $(\log B)^{ b(X,L)-1}$ produces $$(\log B^{1/k} )^{ b(X,L)-1}= (\log B)^{ b(X,L)-1} / k ^{ b(X,L)-1}$$ instead of $(\log B)^{ b(X,L)-1} $.
Plugging this into your formulas, we get a discrepancy of $\frac{1}{k^{b(X,L)}}$ rather than $\frac{1}{k}$. This discrepancy would be exactly cancelled if $[\mathcal L]$ were used instead of $[K\_X]$ in the definition of $\gamma\_{\mathcal L}(V)$. So I wonder if this is what's meant.
| 4 | https://mathoverflow.net/users/18060 | 448789 | 180,655 |
https://mathoverflow.net/questions/448778 | 2 | Let $G=(V, E)$ be the graph on vertices $V = \{0, \cdots, k\}^n$, where vertices $(v\_1, \cdots, v\_n)$ and $(w\_1, \cdots, w\_n)$ share an edge iff $\lvert v\_i - w\_i\rvert \leq 1$ for all $i$.
A *walk of length t* is a sequence $X\_0, \cdots, X\_t\in V$ where $(X\_i, X\_{i+1})\in E$. Among these walks of length $t$, choose $X\_0, \cdots, X\_t$ uniformly at random. As far as I see, this is *not* simply a Markov process, as we choose the $(X\_0, \cdots, X\_k)$ globally at random, rather than performing a random walk.
Is there a way to prove some of the following intuitive statements?
1. The distribution of $X\_t$ given $X\_0$ converges to something independent of $X\_0$ for $t\to\infty$?
2. Say, the weight of a vertex is given by $$w(v\_1, \cdots, v\_n) := v\_1 + \cdots + v\_n$$ and the weight of the entire walk by $$w(X\_0, \cdots, X\_t) := w(X\_0) + \cdots + w(X\_t).$$ Then, $\mathbb{E}[w(X\_0, \cdots, X\_t)\vert X\_0 = x, X\_t = y]$ is asymptotically independent of $x$ and $y$ when $t\to\infty$.
| https://mathoverflow.net/users/97160 | Randomly chosen walk of fixed length | The right setup here is that of **topological Markov chains**. This is essentially the same as a directed graph, i.e., a (finite, for simplicity) set of states (vertices) $A$ endowed with a $\{0,1\}$-valued $A\times A$ admissibility matrix $\Sigma$. Then one looks at the sequences (finite or infinite) of states $x\_0,x\_1,\dots$ subject to the condition $\Sigma\_{x\_i x\_{i+1}}=1$ for any pair of adjacent states.
A natural **mixing condition** on the matrix $\Sigma$ (satisfied in your situation) is that there exists a power $n$ with the property that all entries of $\Sigma^n$ are positive (i.e., any two states can be joined by a length $n$ admissible path). Under this condition the behaviour of various uniform distributions assocaited with the topological Markov chain is described by the "ordinary" Markov chain on $X$ whose transition probabilities are determined by the Perron - Frobenius eigenvectors of the admissibility matrix. In particular, the answers to both your questions are "yes" (and the "something" in question 1 is precisely the stationary distribution of the aforementioned Markov chain). For instance, see Chapter 3 of the [Brin - Stuck textbook](https://mathscinet.ams.org/mathscinet-getitem?mr=3558919).
| 1 | https://mathoverflow.net/users/8588 | 448791 | 180,656 |
https://mathoverflow.net/questions/283483 | 12 | As the title says, my question is on a specific argument in [Kirillov - Skew divided difference operators and Schubert polynomials](https://arxiv.org/abs/0705.4546) ([journal](http://www.emis.de/journals/SIGMA/2007/072), [MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=2322799)) on positivity of divided difference operators. I recall the relevant passages of the [paper](https://arxiv.org/abs/0705.4546) in the following.
Let $m\in\mathbb{Z}\_{>0}$ and $n=m-1$. Denote by $\mathcal{S}\_m$ the symmetric group on $m$ letters. For $1\leq\mu\leq n$, let
$$
\mathcal{S}\_m^{(\mu)}=\{w\in\mathcal{S}\_m\mid w(\mu+1)<w(\mu+2)<\cdots<w(m)\}\,.
$$
For $w\in\mathcal{S}\_m$, denote by $\mathfrak{S}\_w\in\mathbb{Z}[x\_1,\ldots,x\_n]$ the Schubert polynomial associated to $w$. For $1\leq a<b\leq m$, denote by $\partial\_{ab}$ the divided difference operator with respect to the variables $x\_a$ and $x\_b$.
The author wants to show that $\partial\_{ab}\mathfrak{S}\_w\in\mathbb{Z}\_{\geq 0}[x\_1,\ldots,x\_m]$. We clearly can assume that $b\leq n$ and $b-a>1$ (since $\mathfrak{S}\_w$ depends only on $x\_1,\ldots,x\_n$ and since $\partial\_{i,i+1}\mathfrak{S}\_w$ is either zero or itself a Schubert polynomial with positive coefficients where $1\leq i\leq n$).
Let $\bar x=(x\_1,\ldots,x\_n),\bar x\_1=(x\_1,\ldots,x\_{a-1}),\bar x\_2=(x\_a,\ldots,x\_b),\bar x\_3=(x\_{b+1},\ldots,x\_n)$. By a theorem in [Macdonald's notes on Schubert polynomials](http://www.ms.unimelb.edu.au/~ram/Resources/NOSPContents.html) (namley Theorem (4.19) in Chapter IV which is restated as Lemma 1 in [Kirillov's paper](https://arxiv.org/abs/0705.4546)), we can write
$$
\mathfrak{S}\_w(\bar x)=\sum\_{u\_1,u\_2,u\_3}d\_{u\_1u\_2u\_3}^w\mathfrak{S}\_{u\_1}(\bar x\_1)\mathfrak{S}\_{u\_2}(\bar x\_2)\mathfrak{S}\_{u\_3}(\bar x\_3)
$$
where the sums runs over $u\_1\in\mathcal{S}\_m^{(a-1)},u\_2\in\mathcal{S}\_m^{(b-a+1)},u\_3\in\mathcal{S}\_m^{(n-b)}$ and where $d\_{u\_1u\_2u\_3}^w\in\mathbb{Z}\_{\geq 0}$.
**Note.** Neither Kirillov nor Macdonald actually show that the elements $u\_1,u\_2,u\_3$ are contained in $\mathcal{S}\_m$ if $w\in\mathcal{S}\_m$. But I think this part is "obvious" because of Theorem (4.11) in [Macdonald's notes](http://www.ms.unimelb.edu.au/~ram/Resources/NOSPContents.html).
If we now apply $\partial\_{ab}$ to this expression, we find that
$$
\partial\_{ab}\mathfrak{S}\_w(\bar x)=\sum\_{u\_1,u\_2,u\_3}d\_{u\_1u\_2u\_3}^w\mathfrak{S}\_{u\_1}(\bar x\_1)(\partial\_{ab}\mathfrak{S}\_{u\_2}(\bar x\_2))\mathfrak{S}\_{u\_3}(\bar x\_3)\,.
$$
Hence, we can assume that $a=1$ and that $w\in\mathcal{S}\_m^{(b)}$.
I come now to the question which concerns page 10 of [Kirillov's paper](https://arxiv.org/abs/0705.4546): He says that we can even assume $b=n$. My question is why we can assume this.
**Thoughts.** (1) If $w\in\mathcal{S}\_m^{(b)}$, it does not mean that $w\in\mathcal{S}\_b$. It is very easy to give examples for this.
(2) We can also assume that $w$ has last descent at $b$. Otherwise the positivity is obvious.
(3) In terms of the Schubert element $\mathfrak{S}^{(m)}(\bar x)$ in the nilCoxeter algebra (cf. Fomin and Stanley's paper), it suffices to prove that
$$
\partial\_{1b}\mathfrak{S}^{(m)}(x\_1,\ldots,x\_b,0,\ldots,0)e\_{b+1}\cdots e\_n e\_{b+1}\cdots e\_{n-1}\cdots e\_{b+1}e\_{b+2}e\_{b+1}
$$
has coefficients in $\mathbb{Z}\_{\geq 0}[\bar x]$. Here, $e\_1,\ldots,e\_n$ denote the generators of the nilCoxeter algebra.
(4) For me, the case $b<n$ looks essentially different from the case $b=n$. Is it possible that there is a mistake in [Kirillov's paper](https://arxiv.org/abs/0705.4546)?
I tried very much to understand the reduction or to give a new proof which clearly distinguishes the case $b<n$ and $b=n$. I had not much success. Every idea or comment is welcome!
| https://mathoverflow.net/users/66288 | Question on a reduction in Kirillov's paper on positivity of divided difference operators | I had the same concern about Kirillov's paper. For $b=n$, $x\_b$ can have power at most $1$, whereas for $b<n$ it could have a larger power. I have an alternative proof that uses different methods than Kirillov. It remains combinatorial, but it uses combinatorial results that appear not to be published (I'm trying to find out; see [Pulling out a variable from a Schubert polynomial](https://mathoverflow.net/questions/448607/pulling-out-a-variable-from-a-schubert-polynomial)). For $\partial\_{ab}$, there is a positive formula for an expression as a product of a Schubert polynomial in $x\_1,\ldots,x\_a,x\_b,\ldots,x\_n$ and some polynomial in $x\_{a+1},\ldots,x\_{b-1}$, with nonnegative coefficients, in which case you can apply a divided difference to the $a$ index, which affects $x\_a$ and $x\_b$, yielding a positive result.
The positive formula actually fits in a MathOverflow answer. It involves double Schubert polynomials $\mathfrak{S}\_{w\_0}(x;y)$ and the Cauchy formula, as well as Sottile's Pieri formula. Define the factorial elementary symmetric polynomial
$$E\_p(x;y\_i)=\prod\_{j=1}^p(x\_j-y\_i)$$
This has an alternative expansion as
$$E\_p(x;y\_i)=\sum\_{j=0}^p (-y\_i)^{p-j}e\_{j,p}(x)$$
Where $e\_{j,p}(x)=e\_j(x\_1,\ldots,x\_p)$ is the elementary symmetric polynomial of degree $j$ in $p$ variables. Then
$$\mathfrak{S}\_{w\_0}(x;y)=\prod\_{i=1}^n E\_{n+1-i}(x;y\_i)$$
Let's say we want to pull out the variable at index $j$. Then there is a permutation $\mu$ such that
$$\mathfrak{S}\_{w\_0}(x;y)=\mathfrak{S}\_\mu(x;y\_1,\ldots,y\_{j-1},y\_{j+1},\ldots,y\_n)E\_{n+1-j}(x;y\_j)$$
Apply the Cauchy formula, and expand the factorial elementary symmetric polynomial, obtaining
$$\mathfrak{S}\_{w\_0}(x;y)=\sum\_{u,q}\mathfrak{S}\_{u\mu}(x)\mathfrak{S}\_u(-y\_1,\ldots,-y\_{j-1},-y\_{j+1},\ldots,-y\_n)(-y\_j)^{n+1-j-q}e\_{q,n+1-j}(x)$$
Use Sottile's Pieri formula to multiply by the elementary symmetric polynomial to get
$$\mathfrak{S}\_{w\_0}(x;y)=\sum\_{u,q}\sum\_{u\mu\xrightarrow[q]{n+1-j} ww\_0}\mathfrak{S}\_{ww\_0}(x)\mathfrak{S}\_u(-y\_1,\ldots,-y\_{j-1},-y\_{j+1},\ldots,-y\_n)(-y\_j)^{n+1-j-q}$$
Then apply the Cauchy formula
$$\mathfrak{S}\_{w\_0}(x;y)=\sum\_w \mathfrak{S}\_{ww\_0}(x)\mathfrak{S}\_w(-y)$$
to get
$$\mathfrak{S}\_w(x)=\sum\_{u\mu\xrightarrow[q]{n+1-j} ww\_0}x\_j^{n+1-j-q}\mathfrak{S}\_u(x\_1,\ldots,x\_{j-1},x\_{j+1},\ldots,x\_n)$$
Apply this repeatedly to get the full result.
| 3 | https://mathoverflow.net/users/62135 | 448798 | 180,658 |
https://mathoverflow.net/questions/448529 | 1 | I'm reading "Topological entropy bounds measure-theorettic entropy", by L.W. Goodwyn.
[enter link description here](https://www.ams.org/journals/proc/1969-023-03/S0002-9939-1969-0247030-3/S0002-9939-1969-0247030-3.pdf)
After Proposition 2, he mentions that "finite closed cover can yield entropy strictly greater than topological entropy". I was looking for the reference to this (Topological entropy and expansive cascades - Dissertation) but i couldn't find it.
I would appreciate if can help me with an example or the reference.
| https://mathoverflow.net/users/506604 | Example of finite closed cover with entropy strictly greater than topological entropy | You can find an example of a cascade which has topological entropy equal to zero, and which has a finite closed cover having entropy equal to $\log 2$ in the [article](https://www.jstor.org/stable/1995916) "The Product Theorem for Topological Entropy" by L.W. Goodwyn.
Let $\hslash (T) = \sup h(\alpha, T)$, where the supremum is taken over all finite closed covers. We have that $\hslash (T)$ may not be equal to $h(T)$. In fact, that's what usually happens.
I don't know if the example presented in the dissertation is the same as in the article. But the idea of the example should be the same.
For a positive integer $m$, let $C\_m$ denote the set of all bisequences $(\ldots, x(- 1), x(0), x(1),\ldots)$ of points in $I^m$, i.e., $C\_m = (I^m)^{\mathbb{Z}}$. We let $C\_m$ have the product topology and we define $\sigma\_m: C\_m \to C\_m$ by
$$\sigma(x)(n)=x(n+1),$$
and let $\pi\_m :C\_m \to I^m$ be the projection $\pi\_m(x) = x(0)$ for $x \in C\_m$.
The example you are looking for will be a subcascade of the sequence cascade $(C\_1, \sigma\_1)$. For each positive integer $m$, we define $X\_m$ to be the set of all bisequences $x= (\ldots, -x(-1), x(0), x(1),\ldots) \in C\_1$ satisfying the requirement that there are no more than $m$ integers $i$ such that $|x(i) -\frac{1}{2}|>1/m$. Every $X\_m$ is closed and $\sigma\_1$-invariant. Define
$$X= \bigcap\_{m=1}^{\infty} X\_m,\qquad T=\sigma\_1|\_{X}\quad\text{and}\quad \pi=\pi\_1|\_X.$$
So you have $h(\alpha,T)=0$ (*check*), the topological entropy of $T$ is zero. To see that $\hslash (T) > h(T)$, define $F\_0 = \{x \in X: x(0) \in [0,\frac{1}{2}]\}$ and $F\_1 = \{x\in X : x(0) \in [\frac{1}{2},1]\}$, and let $\beta=\{F\_0,F\_1\}$. We have $h(\beta,T)=\log 2$.
| 1 | https://mathoverflow.net/users/482407 | 448806 | 180,660 |
https://mathoverflow.net/questions/448805 | 4 | I have a very soft question which might be very standard in textbooks or literature but I haven't seen it.
To a fixed group $G$ we may attach different topologies to make it different topological groups with the same underlying groups.
**My question 1 is**: Are those classifying spaces associated to different topological groups (with the same underlying groups) the same? (For this question my guess is no)
For example, $G=\mathbb{Z}/2$ is a group. $X$ is the topological group $G$ with discrete topology, $Y$ is $G$ equipped with trivial topology (only open sets are empty set and the whole set). Then $X$ is disjoint union of singletons, $Y$ is contractible. What are classifying spaces $BX,BY$? Are they the same? I know $BX$ may be chosen as $\mathbb{RP}^\infty$, what about $BY$?
It is well-known that if $G$ is discrete, then $BG$ is the classifying space of a small category or EM space $K(G,1)$. What if $G$ is not discrete, and if the answer to my question 1 is no, then is $BG$ still the classifying space of some small category?
**My question 2 (if the answer to question 1 is no) is**: When are these different classifying spaces comparable or identical? To be more specific, if $G,H$ are different topological groups with the same underlying groups and a continuous group isomorphism $G\to H$ (e.g. $G$ is discrete), do we have a principal bundle morphism? That is maps $EG\to EH, BG\to BH$ that commute with bundle maps. Is there any criterion for these maps to be homotopy equivalences?
A corollary of question 2 is to understand what can be said about the relationships between group homology of a Lie group and homology of its classifying space.
**My question 3 is**: Given a (connected) Lie group $L$, we have three interesting homologies (with coefficients to be determined)
1. the algebraic homology/group homology $H\_{g,\*}(L)$ which is also topological homology of classifying space of discrete underlying group of $L$, $H\_\*(BL^\delta)$;
2. ordinary topological homology of the Lie group as a topological space $H\_\*(L)$;
3. topological homology of the classifying space of $L$, $H\_\*(BL)$.
I'm wondering what is the relation between them.
For 2 and 3 we have Eilenberg-Moore spectral sequence $E^2\_{p,q}=\text{Tor}\_{p,q}^{H\_\*(L;k)}(k,k)\Rightarrow H\_{p+q}(BL;k)$ where $k$ is a field and also Serre spectral sequence. If the answer to question 2 is comparable then it would relate 3 to 1.
**Background:** Homology 1 in question 3 could provide important invariants for scissors congruence theory in non-Euclidean spaces; Homology 2 is important in its own interest; Homology 3 is nothing but characteristic classes.
| https://mathoverflow.net/users/471160 | Classifying space of a non-discrete group and relationship between group homology and topological homology of Lie groups | You may want to look at the classical paper of Jack Milnor, "On the homology of Lie groups made discrete." The Friedlander-Milnor conjecture states that the map $BG^\delta \to BG$ (where $G$ is a Lie group and $G^\delta$ is $G$ made discrete) is a mod $p$ cohomology equivalence. Also chase papers referring to this one, but I think the conjecture is still open. With integer coefficients, this is very false, as the paper explains, by the work of Sah and Wagoner.
| 8 | https://mathoverflow.net/users/460592 | 448807 | 180,661 |
https://mathoverflow.net/questions/448804 | 2 | I need help proving that the set $B \cap L^2\big( (0,T) \times (0,1)\big)$ is a closed subset of $L^2\big( (0,T) \times (0,1)\big)$, where $B$ is defined as:
$$B=\Big\{x \in L^{\infty}\big(0,T;L^1(0,1)\big) \ : \|x(t)\|\_{L^1(0,1)} \le 1 , \ \ \text{a.a in } \ (0,T) \Big\} $$
Any suggestions or proofs would be greatly appreciated!
| https://mathoverflow.net/users/473534 | Is $B \cap L^2\big( (0,T) \times (0,1)\big)$ closed in $L^2\big( (0,T) \times (0,1)\big)$? | $\newcommand{\R}{\mathbb R}\newcommand{\C}{\mathbb C} $To attach a meaning to the intersection $C:=B\cap L^2\big((0,T)\times(0,1)\big)$ of the subset $B$ of the space $L^{\infty}\big(0,T;L^1(0,1)\big)$ with the space $L^2\big((0,T)\times(0,1)\big)$, we need to identify functions $(0,T)\times(0,1)\ni(t,s)\mapsto x(t,s)\in\C$ in $L^2\big((0,T)\times(0,1)\big)$ with functions $(0,T)\ni t\mapsto x\_t\in L^{\infty}\big(0,T;L^1(0,1)\big)$, which we will do by the standard formula $x\_t(s):=x(t,s)$.
Let $(x^{(n)})$ be a sequence in $C$ converging to some $x\in L^2\big((0,T)\times(0,1)\big)$ in $L^2\big((0,T)\times(0,1)\big)$:
\begin{equation}
d\_n:=\int\_0^T dt\,\int\_0^1 ds\, |x^{(n)}(t,s)-x(t,s)|^2\to0
\end{equation}
(as $n\to\infty$). We have to show that then $x\in B$, so that $x$ is actually in $C$.
Using the Cauchy--Schwarz inequality, we have
\begin{equation}
\begin{aligned}
&\int\_0^T dt\,\int\_0^1 ds\, |x^{(n)}(t,s)-x(t,s)| \\
&\le\sqrt{T\times1}\Big(\int\_0^T dt\,\int\_0^1 ds\, |x^{(n)}(t,s)-x(t,s)|^2\Big)^{1/2}
=\sqrt T\sqrt{d\_n}\to0.
\end{aligned}
\end{equation}
That is, $\int\_0^T dt\, g^{(n)}(t)\to0$, where $g^{(n)}(t):=\|x^{(n)}\_t-x\_t\|\_{L^1(0,1)}$. That is, the sequence $(g^{(n)})$ of nonnegative functions converges to $0$ in $L^1(0,T)$ and hence in measure over the interval $(0,T)$. So, passing to a subsequence, without loss of generality we may assume that the sequence $(g^{(n)})$ converges to $0$ almost everywhere on $(0,T)$. That is, $\|x^{(n)}\_t-x\_t\|\_{L^1(0,1)}\to0$ for a.a. $t\in(0,T)$. Recall that for each $n$ we have $x^{(n)}\in C\subseteq B$, whence $\|x^{(n)}\_t\|\_{L^1(0,1)}\le1$ for a.a. $t\in(0,T)$. So,
\begin{equation}
\|x\_t\|\_{L^1(0,1)}=\lim\_n\|x^{(n)}\_t\|\_{L^1(0,1)}\le1
\end{equation}
for a.a. $t\in(0,T)$. Thus, $x\in B$. Thus, $x\in B$. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 448812 | 180,663 |
https://mathoverflow.net/questions/448735 | 6 | [On nlab](https://ncatlab.org/nlab/show/doctrine#as_2monads) I read
>
> For instance, there is a morphism of theories from the theory of commutative rings to the theory of abelian groups which sends a ring to its multiplicative group of units, but this is not induced by any morphism of monads because it does not preserve the underlying set.
>
>
>
Can anyone tell me: what theories and what morphisms are we talking about? From the point of view of Lawvere single-sorted theories, there is no such morphism either, because the functors induced by them also preserve the underlying set (actually, the category of Lawvere theories is equivalent to the category of finitary monads on $\mathrm{Set}$).
Probably, we are talking about theories with some more powerful logic? (like geometric?)
| https://mathoverflow.net/users/148161 | Multiplicative group of a ring as a morphism of theories | The functor sending a (not necessarily commutative) ring to its group of units is induced by a morphism of *cartesian* (= finite limit) theories.
More generally, suppose given (small!) cartesian theories $\mathcal{T}$ and $\mathcal{T}'$.
Let $\mathcal{M}$ and $\mathcal{M}'$ be the respective categories of models.
These are locally finitely presentable categories, so given a functor $G : \mathcal{M} \to \mathcal{M}'$; the following are equivalent:
* $G$ preserves (small) limits and filtered colimits.
* $G$ has a left adjoint that sends finitely presentable objects in $\mathcal{M}'$ to finitely presentable objects in $\mathcal{M}$.
But $\mathcal{T}^\textrm{op}$ (resp. $\mathcal{T}'{}^\textrm{op}$) is equivalent to the full subcategory of finitely presentable objects in $\mathcal{M}$ (resp. $\mathcal{M}'$), so we deduce the following is also equivalent:
* $G$ is induced by a morphism $\Phi : \mathcal{T}' \to \mathcal{T}$ of cartesian theories, in the sense that $G \cong \Phi^\*$, where $\Phi^\* (M) = M \Phi$.
(Note the contravariance!)
| 8 | https://mathoverflow.net/users/11640 | 448815 | 180,666 |
https://mathoverflow.net/questions/448817 | 5 | This question is about a technical imprecision which is easily avoidable but whose details I'd like to understand better. When we refer to "the consistency strength of $\mathsf{PA}$" (say) we aren't properly referring to $\mathsf{PA}$ as a set; rather, we care about the specific way in which $\mathsf{PA}$ is enumerated. "The set of sentences which $\mathsf{ZFC}$ proves are $\mathsf{PA}$-axioms" is another way to define the same (hopefully!) set, but properly speaking it is equiconsistent with $\mathsf{ZFC}$.
It occurs to me that I don't know whether the usual description of the $\mathsf{PA}$ axioms is optimal with respect to consistency strength. Below, let $(T\_e)\_{e\in\omega}$ be some appropriate enumeration of the c.e. theories. Suppose WLOG that $T\_0$ is the usual c.e. presentation of $\mathsf{PA}$.
>
> **Question**: Is there an index $e$ such that ("extensionally") $T\_e=\mathsf{PA}$ but $\mathsf{PA}+\mathsf{Con}(T\_e)\not\vdash \mathsf{Con}(T\_0)$?
>
>
>
One natural way to try to get a positive answer to this question is to enumerate the axioms of $\mathsf{PA}$ very slowly, waiting for contradictions, along the following lines: given a $\mathsf{PA}$-provably-total computable function $f$, let $T\_f$ be the c.e. theory which enumerates the $n$th axiom of $\mathsf{PA}$ in the usual ($T\_0$) sense iff there is no contradiction in $\mathsf{PA}$ using fewer than $f(n)$-many symbols. As long as $f$ is reasonably-fast-growing (say, dominating $n\mapsto 2^n$) it seems "$\mathsf{PA}$-believable" that $T\_f$ detects a contradiction before falling prey to it. On the other hand, I don't see how to show the consistency of $\mathsf{PA}+\mathsf{Con}(T\_f)+\neg\mathsf{Con}(T\_0)$ for any specific $f$.
([This older question](https://mathoverflow.net/questions/417484/proving-short-consistency-can-we-do-better-than-brute-force-search) seems thematically similar, but I don't actually see a concrete connection.)
| https://mathoverflow.net/users/8133 | Is the usual enumeration of $\mathsf{PA}$ "minimal for consistency strength"? | No (to the title), yes (to the question), see
Sy-David Friedman, Michael Rathjen, and Andreas Weiermann: *Slow consistency*, Annals of Pure and Applied Logic 164 (2013), no. 3, pp. 382–393, doi [10.1016/j.apal.2012.11.009](https://doi.org/10.1016/j.apal.2012.11.009).
The idea is that an axiom of usual PA of Gödel number $n$ is padded to length $f(n)$, where $f$ is a recursive function growing faster than all provably total recursive functions of PA (namely, $F\_{\epsilon\_0}$).
| 13 | https://mathoverflow.net/users/12705 | 448822 | 180,668 |
https://mathoverflow.net/questions/448835 | 0 | Let $X = \text{Spec}A$ be an affine scheme. It is well known that if $U$ is an open subset which contains $\text{SpecMax}A$, then $U\supseteq X$. The previous statement generalizes to arbitrary schemes, due to the nature of the Zariski topology of a scheme, being locally affine.
When talking about stacks, one typically replaces the notion of closed points with points of finite type. Again, say $\mathcal{X}$ is a stack, and assume $U\subseteq \mathcal{X}$ is an open substack which contains all points of finite type in $\mathcal{X}$. Is it true that $U\supseteq \mathcal{X}$?
I am not really sure how to approach such a question, as my intuition from the Zariski topology, having a basis of basic open subsets of the topology doesn't generalize at all.
Thanks in advance!
| https://mathoverflow.net/users/174655 | Covering a stack by an open substack that contains all points of finite type | Assuming you mean that $\mathscr X$ is an *algebraic* stack (e.g. in the sense of [Tag [026N](https://stacks.math.columbia.edu/tag/026N)]), then this is true, and follows relatively straightforwardly from the case of schemes.
Although you did not make a precise statement in the scheme case, it is indeed true that if $U \subseteq X$ is an open subscheme containing all *locally closed points*, then $U = X$. Moreover, locally closed points are the same thing as points that are (locally) of finite type [Tag [01TA](https://stacks.math.columbia.edu/tag/01TA)].
Now let $\mathscr U \subseteq \mathscr X$ be an open substack, i.e. $j\colon\mathscr U \to \mathscr X$ is representable by open immersions. In particular, if $x \colon \operatorname{Spec} k \to \mathscr X$ is a point of finite type, the pullback along $\mathscr U \to \mathscr X$ gives an open immersion $U \hookrightarrow \operatorname{Spec} k$, and we say $\mathscr U$ *contains $x$* if $U \neq \varnothing$. If this holds for all finite type points, the question is whether $\mathscr U \to \mathscr X$ is an isomorphism.
By assumption, there exists a smooth surjection $f \colon X \to \mathscr X$ where $X$ is a scheme. Checking that a morphism is an isomorphism is smooth-local, so it suffices to check that the base change $U \to X$ of $j$ along $f$ is an isomorphism. Note that $U \to X$ is again an open immersion, so by the above it suffices to check that $U$ contains all points of finite type of $X$. If $x \colon \operatorname{Spec} k \to X$ is a point that is (locally) of finite type, then so is the composition $\operatorname{Spec} k \to X \to \mathscr X$, since $f$ is locally of finite type [Tags [06MH](https://stacks.math.columbia.edu/tag/06MH) and [03YJ](https://stacks.math.columbia.edu/tag/03YJ)]. By assumption, the pullback of $U \to X$ along $x$ is an isomorphism, i.e. $x \in U$. $\square$
| 4 | https://mathoverflow.net/users/82179 | 448847 | 180,673 |
https://mathoverflow.net/questions/448848 | 4 | I am aware that a Frechet space $V$ is a topological vector space whose topology is induced by countably many seminorms $\{ \lVert \cdot \rVert\_k \}$ such that
1. it must be Hausdorff
2. it must be complete
In the [relevant Wikipedia entry](https://en.wikipedia.org/wiki/Fr%C3%A9chet_space)
it is stated that when constructing a Frechet space, one needs to check that for any sequence $\{ x\_n \}$ Cauchy with respect to each seminorm $\lVert \cdot \rVert\_k$, there exists some $x \in V$ such that $x\_n \to x$ w.r.t. each seminorm.
However, I am quite confused here in that, what if one only finds out that there exists $x^k \in V$ for "each" $\lVert \cdot \rVert\_k$ such that $\lVert x\_n-x^k \rVert\_k \to 0$ as $n \to \infty$? That is, there is no guarantee that $x^k = x^{k'}$ for $k \neq k'$.
I tried to resolve this issue by the Hausdroff condition, but cannot find a way out. Could anyone please help me?
| https://mathoverflow.net/users/56524 | Confusion with the definition of a Frechet space regarding completeness and uniqueness of a limit | Consider the space $X$ of sequences $x=(x\_k)\_k \in \mathbf{C}^{\mathbf N}$ with finite support, and let $\|x\|\_k = |x\_k|$. $X$ is clearly not a Fréchet space (one not so elementary reason is that a Fréchet space cannot have countably infinite dimension by Baire's theorem). However, it is easy to see that your condition holds, simply because $\mathbf{C}$ is complete.
| 6 | https://mathoverflow.net/users/10265 | 448850 | 180,674 |
https://mathoverflow.net/questions/448845 | 1 | Consider a multivariate normal distribution. What are necessary and sufficient conditions (esp. on the covariance matrix) under which this distribution is exchangeable?
| https://mathoverflow.net/users/506883 | Under what conditions is a multivariate normal distribution exchangeable? | $\newcommand\Si\Sigma\newcommand\si\sigma\newcommand\1{\mathbf1}$Let $D:=N(\mu,\Si)$ be the $n$-variate normal distribution with mean vector $\mu=[\mu\_1,\dots,\mu\_n]^\top$ and covariance matrix $\Si=[\si\_{ij}\colon i,j=1,\dots,n]$.
Suppose that $D$ is the distribution of an exchangeable random vector $X=[X\_1,\dots,X\_n]^\top$. Then (i) $\mu\_i=EX\_i=EX\_j=\mu\_j=:a$ and $\si\_{ii}=Var\,X\_i=Var\,X\_j=\si\_{jj}=:b^2$ for all $i$ and $j$ and (ii) $\si\_{ij}=Cov(X\_i,X\_j)=Cov(X\_k,X\_l)=\si\_{kl}=:c$ for all $i,j,k,l$ such that $i\ne j$ and $k\ne l$. So,
$$\mu=a\1\quad\text{and}\quad\Si=c\1\1^\top+(b^2-c)I,\tag{1}\label{1}$$
where $\1:=[1,\dots,1]^\top$ and $I$ is the identity matrix.
For such a covariance matrix $\Si$, $\1$ is an eigenvector belonging to eigenvalue $b^2+(n-1)c$, and any vector orthogonal to $\1$ is an eigenvector belonging to eigenvalue $b^2-c$. Since $\Si$ is positive definite, we conclude that
$$b^2+(n-1)c>0\quad\text{and}\quad b^2-c>0. \tag{2}\label{2}$$
Vice versa, suppose that conditions \eqref{1} and \eqref{2} hold. Then
$$\Si^{1/2}=u\1\1^\top+vI,$$
where
$$v:=\sqrt{b^2-c}\quad\text{and}\quad
u:=\frac{\sqrt{b^2+(n-1)c}-\sqrt{b^2-c}}n.$$
So, letting
$$X:=a\1+\Si^{1/2}Z=a\1+u(\1^\top Z)\1+vZ, \tag{3}\label{3}$$
where $Z$ is a standard normal random vector in $\mathbb R^n$, we see that $X$ is an exchangeable normal random vector in $\mathbb R^n$ with the mean vector $\mu$ and the covariance matrix $\Si$ as in \eqref{1}.
Thus, the conjunction of conditions \eqref{1} and \eqref{2} is necessary and sufficient for the $n$-variate normal distribution $N(\mu,\Si)$ to be exchangeable. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 448858 | 180,675 |
https://mathoverflow.net/questions/448859 | 0 | Suppose $K$ is an algebraic number field. We want to determine whether, for all even $n$, there always exists a cyclic Galois extension $L$ of degree $n$ over $K$ such that the intermediate field $K\_2$ does not contain any primitive $2^i$-th root of unity for $i \geq 3$ where $ K \_2 $ is the intermediate field.
If $ K = \mathbb{Q} $ then it is true. Since for all $ n $ there exist infinitely many primes $ p $ such that $ n $ divides $ n -1 $. As $ p$-th cyclotomic polynomial is irreducible over $ \mathbb{Q} $ then using Galois corresponding theorem we get a cyclic galois extension $ L $ of degree $ n $. Now since $ \phi(2^{i}) > 2 $ where $ i \geq 3 $ so any primitive $ 2^i $-th root of unity can not lie in $ K\_2 $.
| https://mathoverflow.net/users/215016 | Not containing $ 2^{i}$-th primitive roots of unity in cyclic galois extension of number field | It's not quite clear from the post what $K\_2$ is, but I'll just pretend $K\_2=L$, which gives the strongest result.
Those infinitely many extensions constructed over $\mathbb{Q}$ are all linearly disjoint (e.g. since in each one, the respective prime $p$ is the only ramified prime), and thus there are in particular infinitely many $L/\mathbb{Q}$ among them which are linearly disjoint to $K(\zeta\_8)$. Then firstly $LK/K$ is still cyclic of degree $n$, and also $LK$ has no $2^i$-th roots of unity which weren't already in $K$. So yes, it's possible unless in the trivially impossible case where $K$ itself contains such roots of unity.
| 2 | https://mathoverflow.net/users/127660 | 448860 | 180,676 |
https://mathoverflow.net/questions/448856 | 0 | Let $W=(W\_t)\_{t\ge 0}$ be a standard Brownian motion starting from zero and $Z>0$ be an independent random variable. Fix $T>0$ and $C>0$. Denote by $\mathcal A$ the set of progressively measurable processes $\alpha=(\alpha\_t)\_{t\ge 0}$ such that
$$\alpha\_t\ge 0 \quad \mbox{and}\quad \int\_0^T\alpha\_s ds \le C.$$
Define further for every $\alpha\in \mathcal A$
$$X^\alpha\_t:=Z + W\_t + \int\_0^t \alpha\_s ds, \quad \forall t\ge 0.$$
What is the optimiser for the optimisation problem
$$\sup\_{\alpha\in\mathcal A} \mathbb P\big[\inf\_{0\le t\le T}X^\alpha\_t>0\big]?$$
My guess is to allocate the whole drift $C$ at time zero, i.e. $X^\*\_t:=Z+C+W\_t$ (of course this strategy can be approximated by a suitable sequence $(\alpha^n)\_{n\ge 1}\subset \mathcal A$), while I don't know how to prove it.
Any answer or comments are highly appreciated.
| https://mathoverflow.net/users/493556 | Maximise the probability that a drifted Brownian motion doesn't hit zero prior to $T$ | The process you are describing is simply bigger than all the other: by the condition on $\alpha$
$$Z + W\_t + \int^t \alpha\_s ds \le Z + W\_t + C $$
| 3 | https://mathoverflow.net/users/143907 | 448868 | 180,679 |
https://mathoverflow.net/questions/448865 | 2 | Suppose $A \cong M\_n(\mathbb{D})$ where $A$ is a simple algebra over division ring $D$. We want to find an explicit isomorphism between $A$ and $M\_n(\mathbb{D})$. I read from Ivanyos et al. (2012) that if we have a rank 1 element $e \in A$ then the left action of $A$ on $Ae$ provides an isomorphism.
When I take a look at other literature, this seems to be a trivial issue, but I am not sure how the construction works. Could anyone explain how this works or refer me to some sources that details this construction.
Thank you!
| https://mathoverflow.net/users/506897 | Solving the explicit isomorphism problem | The element $e\in A$ being of rank one is equivalent to asking $Ae$ to have dimension $n$ over $D$. Now, there is a homomorphism $A\cong End\_D(Ae)$ taking $a\in A$ to the endomorphism $be\mapsto abe$, which is readily seen to be $D$-linear. The map is injective, since otherwise the kernel $\{a\in A:abe=0,\forall b\in A\}$ would be a nontrivial two-sided ideal of $A$, contradicting the assumption that $A$ is simple.
Choosing a $D$-basis of $Ae$ identifies $End\_D(Ae)$ with $M\_n(D)$.
| 2 | https://mathoverflow.net/users/123673 | 448874 | 180,682 |
https://mathoverflow.net/questions/448863 | 1 | I know that it must be a simple consequence of the Kővári–Sós–Turán (and Erdős–Stone) theorem, but I am struggling to formulate a proof: Let $H$ be a fixed-size $r$-chromatic graph. Then there exists $\varepsilon = \varepsilon(H)$ s.t. if $G$ is an $n$-vertex graph that contains $\geq n^{r-\varepsilon}$ copies of $K\_r$, it must also contain a (not necessarily induced) copy of $H$.
Thanks in advance.
| https://mathoverflow.net/users/506896 | Extremal graph theory - many copies of $K_r$ imply a copy of $r$-chromatic $H$ | This follows from Proposition 2.1 of the paper [Many $T$ copies in $H$-free graphs](https://arxiv.org/abs/1409.4192) by Alon and Shikhelman.
**Theorem** (Alon and Shikelman)
Let $T$ be a fixed graph with $t$ vertices. Then $ex(n,T,H)=\Omega(n^t)$ if and only if $H$ is not a subgraph of a blow-up of $T$. Otherwise, $ex(n,T,H) \le n^{t-\varepsilon(T,H)}$ for some $\varepsilon(T,H) > 0$.
Here, $ex(n,T,H)$ is the maximum number of copies of $T$ in an $n$-vertex $H$-free graph.
To derive your result, we apply Alon-Shikelman with $T=K\_r$ and $H$ a fixed $r$-chromatic graph. Thus, $H$ is a subgraph of a blow-up of $K\_r$. So, we may choose $\varepsilon(H)$ to be any positive constant less than $\varepsilon(K\_r, H)$.
| 3 | https://mathoverflow.net/users/2233 | 448876 | 180,683 |
https://mathoverflow.net/questions/448867 | 3 | $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ has two interesting properties: on one hand it is non-compact, but on the other hand it admits a unique $\text{SL}(d,\mathbb R)$-invariant *finite* measure on $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$.
I wonder if there exists a bounded measurable subset $F$ of $\text{SL}(d,\mathbb R)$ (with subspace topology, of course) that is Borel isomorphic to $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$?
By "Borel isomorphic" I mean: There exists a bijection $f:F\subset \text{SL}(d,\mathbb R) \to \text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ such that both $f$ and $f^{-1}$ are Borel measurable, with $F$ inheriting the subspace topology from $\text{SL}(d,\mathbb R)$. Of course, "up to a subset of measure zero" is always allowed.
If this is true, then it is good to know a constructive proof but the proof for the existence of such an $F$ will also be greatly appreciated!
---
Here I am not assuming $F$ is a "fundamental domain" for $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$, whose definition itself has many interpretations and I asked $F$ to be bounded and thus *may not be homeomorphic to* $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ (but as a possible approach one may want to construct such an $F$ whose closure is homeomorphic to the one-point compactification of $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ modulo a set of measure zero). I only want this $F$ to be *measure theoretically isomorphic to* $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ but not topologically. Disproving this claim seems also very hard.
| https://mathoverflow.net/users/506835 | Existence of a bounded measurable subset of $\text{SL}(d,\mathbb R)$ that is Borel isomorphic to $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$? | By the Iwasawa decomposition, $\text{SL}(d,\mathbb R)$ is homeomorphic to $\mathbb{R}^n\times\text{SO}(d,\mathbb R)$, where $n=(d^2+d-2)/2$. As $\mathbb{R}^n$ is homeomorphic to any open ball in $\mathbb{R}^n$, it follows that $\text{SL}(d,\mathbb R)$ is homeomorphic to a bounded Borel subset of $\text{SL}(d,\mathbb R)$; this homeomorphism can be constructed explicitly. Now take a nice fundamental domain of $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$; e.g. take a suitable Borel subset. We see that this fundamental domain is also homeomorphic to a bounded Borel subset of $\text{SL}(d,\mathbb R)$, and we are done.
| 3 | https://mathoverflow.net/users/11919 | 448877 | 180,684 |
https://mathoverflow.net/questions/448853 | 1 | Given $M$ a continuous local martingale, and $M^\text{\*} = \sup\_{0 \leq s \leq t} M\_s$ its running maximum, we consider the finite variation integral
$$
I\_T:= \int\_0^T (M^\text{\*}\_s - M\_s) \, \text{d}M^\text{\*}\_s
$$
I'm seeking to show **this integral is 0** a.s.. Intuitively, this is since $(M^\text{\*}\_s - M\_s) \, \text{d}M^\text{\*}\_s$ is zero: if we're currently at our maximum then the bracket term zero, and otherwise $M^\text{\*}\_s$ is constant near $s$.
A little bit more rigorously, we have that is
$$
I\_T = \lim\_{n \to \infty} \sum\_{[s,t]\in \pi\_n} (M^\text{\*}\_{t,s} - M\_{t,s})M^\text{\*}\_{t,s}
$$
(where the limit is in probability, and for a sequence of partitions $\pi\_n$ with mesh tending to 0, and where we write $M\_{t,s} = M\_t - M\_s$ and similarly for $M^\text{\*}$) and that eventually we'll have "sufficiently small" (this is the part which I think needs more details) $t-s$. Then either we're reaching a new maximum, so that $M^\text{\*}\_t = M\_t$ and $M^\text{\*}\_s = M\_s$ and the difference in brackets is $0$. Or we're not reaching a new maximum, and so $M^\text{\*}\_{t,s} = 0$.
I think this isn't sufficient at the moment, since even if we could suppose $M$ is monotone on sufficiently small intervals we still need to choose our partitions $\pi\_n$ uniformly in $\omega \in \Omega$ (for the convergence in probability).
I've also tried applying Itô's on $MM^\text{\*}$, but it reduces this to a similar problem of reasoning for sufficiently small $t-s$ for a partition dependent on the outcome for the $\int M^\text{\*} \, \text{d}M$ term.
(I've asked this question on MSE [here](https://math.stackexchange.com/questions/4716163/integral-of-m-text-m-the-difference-of-running-maximum-and-current-value-w), but gotten little traction. I thought Overflow might be more appropriate since I'm seeking to formalize and fill the gaps of the intuitive ideas I have - let me know if this isn't appropriate, for instance if the question particularly belongs on either site)
| https://mathoverflow.net/users/506885 | Integral of $M^\text{*} - M$ with respect to $M^\text{*}$ is zero for $M^\text{*}$ the running maximum of $M$ a continuous local martingale | The integral with regard do $\mathrm{d}M^\*$ is a pathwise Stieltjès integral, so the question is an analysis problem.
Let $f : \mathbb{R}\_+ \to \mathbb{R}$ be any continuous function, $F$ its current maximum, and $\mu$ the Stieltjès measure associated to $F$
One checks that $F$ is also continuous, so $O := \{s \in \mathbb{R}\_+ : F(s)-f(s)>0\}$ is open subset in $\mathbb{R\_+}$ and contained in $\mathbb{R\_+}^\*$ since $F(0)=f(0)$. Hence it is an at most countable union of disjoint open intervals. On each one of these open intervals, $F$ remains constant, so the $\mu$-measure of this interval is $0$. As a result $\mu(O)=0$, so $F-f$ is null $\mu$-almost everywhere and the integral
$\int(F-f) \mathrm{d}\mu$ is $0$.
| 4 | https://mathoverflow.net/users/169474 | 448882 | 180,686 |
https://mathoverflow.net/questions/448879 | 1 | We consider the heat kernel
$$
g :\mathbb R\_{>0} \times \mathbb R^d \to \mathbb R,\quad (t, x) \mapsto \frac{1}{(4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ).
$$
Then
$$
\partial\_t g(t, x) = \Delta g(t, x) = \left(\frac{|x|^2-2 d t}{4 t^2}\right) g(t, x).
$$
In a previous [thread](https://mathoverflow.net/questions/448491/is-there-f-such-that-int-0t-fs-mathrm-d-s-infty-and-partial-t-g), I got that
>
> There do *not* exist a constant $C \ge 1$ and a measurable function $f:(0, \infty) \to \mathbb R\_+$ such that
>
>
> * $\int\_0^t f(s) \, \mathrm d s < +\infty$ for all $t>0$.
> * $|\partial\_t g| (t, x) \le f(t)g(Ct, x)$ for all $t>0$ and $x \in \mathbb R^d$.
>
>
>
I would like to ask if the following relaxation holds, i.e.,
>
> There exist measurable functions $f, h:(0, \infty) \to \mathbb R\_+$ such that
>
>
> * $h$ is Lipschitz.
> * $\int\_0^t f(s) \, \mathrm d s < +\infty$ for all $t>0$.
> * $|\partial\_t g| (t, x) \le f(t)g(h(t), x)$ for all $t>0$ and $x \in \mathbb R^d$.
>
>
>
Thank you so much for your help!
| https://mathoverflow.net/users/477203 | Are there $f,h$ such that $h$ is Lipschitz, $\int_0^t f(s)\,\mathrm d s<\infty$ and $|\partial_t g| (t, x) \le f(t)g(h(t), x)$? | This is impossible for **any** positive function $h$. Indeed, for $|x|=\sqrt t$, the inequality $|\partial\_t g| (t, x) \le f(t)g(h(t), x)$ implies
$$f(t)\ge\frac Ct\,u^{-d/2}e^{cu}\ge\frac Bt$$
for all real $t>0$, where $C,c,B$ are certain positive real numbers not depending on $t$, and $u:=t/h(t)$. So, $\int\_0^t f=\infty$ for all real $t>0$.
| 3 | https://mathoverflow.net/users/36721 | 448885 | 180,688 |
https://mathoverflow.net/questions/448881 | 9 | Let $G$ be a group acting doubly-transitively on a set $X$. Then the vector space $V\_X$ of functions $f\colon X\to\mathbb C$ with finite support such that $\sum\_{x\in X}f(x)=0$ carries an action of $G$.
>
> Is $V\_X$ necessarily irreducible? Is it indecomposable?
>
>
>
When $G$ is finite, then there is an easy [character-theoretic argument](https://math.stackexchange.com/a/3471289). When $\mathbb F$ is any field and $G=\mathrm{GL}\_2(\mathbb F)$ acts on $X=\mathbb P\_{\mathbb F}^1$, this is answered [here](https://mathoverflow.net/q/228359). What can be said in general?
---
**EDIT (6/15):** Although Noam gave a great example of a reducible $V\_X$, the question still remains of whether $V\_X$ can be decomposable. I suspect this can be proven by proving the only $G$-homomorphisms $V\_X\to V\_X$ are scalars, and such homomorphisms are uniquely determined by where $[y]-[x]$ is mapped to, where $x\ne y\in X$ and $[x]$ denotes the characteristic function of $\{x\}$. Then, the condition for such a map to be a homomorphism reduces to a complicated cocycle condition.
**Further update:** Per YCor's suggestion, I have posted a follow-up [here](https://mathoverflow.net/questions/448953/do-doubly-transitive-actions-give-rise-to-indecomposable-representations-for-inf) regarding indecomposability.
| https://mathoverflow.net/users/123673 | Do doubly-transitive actions give rise to irreducible representations for infinite groups? | $V\_X$ can fail to be irreducible. If $X \subseteq {\bf C}$ is a field and $G$ is the group of affine linear transformations $x \mapsto ax+b$ (any $a,b \in X$ with $a \neq 0$) then $V\_X$ has the proper subrepresentation consisting of functions $f : X \to {\bf C}$ of finite support such that $\sum\_x f(x) = \sum\_x x f(x) = 0$.
| 17 | https://mathoverflow.net/users/14830 | 448887 | 180,689 |
https://mathoverflow.net/questions/448864 | 3 | I'm trying to understand module classes that are defined as the kernels of higher Ext functors (e.g., arising [here](https://www.sciencedirect.com/science/article/pii/S0168007207000632?via%3Dihub); as this paper suggests, I'm coming at this problem outside of module theory). One way to do this, of course, is to understand the higher Ext functors and then specialize to the case when they become zero. However, I'm wondering if there's a more direct route?
That is, in the $n=1$ case, $Ext^1(M, N)$ correspond to the extensions of $M$ by $N$ under a natural notion of equivalence, and the equivalence class of $0$ is represented the trivial extension, i.e., $M + N$.
Is there a similar construction for $n\geq 2$? Is there a canonical/trivial longer exact sequence that we can always build and that will represent the equivalence class of 0 in $Ext^n(M,N)$?
| https://mathoverflow.net/users/178292 | Zeros of higher Ext functors | Elements of $\operatorname{Ext}^i(M,N)$ for $i \geq 1$ can be represented by *Yoneda extensions*: exact sequences $$E = \big(0 \to N \to Z\_{i-1} \to \ldots \to Z\_0 \to M \to 0\big)$$
modulo the equivalence relation $E \sim E'$ if there exists another Yoneda extension $E''$ with morphisms of chain complexes $E \leftarrow E'' \to E'$ that are the identity on $M$ and $N$. See for instance [Tag [06XP](https://stacks.math.columbia.edu/tag/06XP)] for details.
The idea is that $E$ gives a quasi-isomorphism $\sigma\_{\leq 0} E \stackrel\sim\to M$ (where $\sigma$ is the brutal truncation [Tag [0118](https://stacks.math.columbia.edu/tag/0118)]), so the composition
$$M \cong \sigma\_{\leq 0} E \twoheadrightarrow \sigma\_{\leq -i} E = N[i]$$
gives the required morphism $M \to N[i]$ in the derived category. If you want to produce the zero morphism, you want $\sigma\_{\leq 0} E$ to be the direct sum complex
$$\big(0 \to N \stackrel{\text{id}}\to N \to 0 \to \ldots \to 0\big) \oplus \big(0 \to 0 \ldots \to 0 \to M\big),$$
which is for instance the case when
$$E = \big( 0 \to N \stackrel{\text{id}}\to N \to 0 \to \ldots \to 0 \to M \stackrel{\text{id}}\to M \to 0 \big).$$
(If $i = 1$, then we see that the middle $M$ and $N$ sit in the same degree, so we should really consider the split extension $0 \to N \to N \oplus M \to M \to 0$.)
| 5 | https://mathoverflow.net/users/82179 | 448889 | 180,691 |
https://mathoverflow.net/questions/448841 | 4 | I have a sequential, hereditarily Lindelöf topological space $\mathcal{X}$, and some subset $A \subseteq \mathcal{X}$. I am interested in the following properties:
1. There is some compact set $B$ with $A \subseteq B \subseteq \mathcal{X}$.
2. Every sequence $(a\_n)\_{n \in \mathbb{N}}$ in $A$ has a subsequence which converges in $\mathcal{X}$.
3. For every open cover $\bigcup\_{n \in \mathbb{N}} U\_n \supseteq \mathcal{X}$ there is some $N \in \mathbb{N}$ with $A \subseteq \bigcup\_{n \leq N} U\_n$.
I know that each of these implies the next, but do the implications reverse?
(Trying to Google sources is made difficult by mentions of relatively compact dominating; and since I don't wish to assume Hausdorffness, that is not the notion I am looking for.)
| https://mathoverflow.net/users/15002 | Being contained in a compact set | It seems that [Gutik's hedgehog](https://mathoverflow.net/questions/296395/what-is-the-genuine-name-for-the-gutik-hedgehog) is a required counterexample.
I recall that Gutik's hedgehog is the set $$X=\{(0,0)\}\cup\{(\tfrac1n,0):n\in\mathbb N\}\cup\{(\tfrac1n,\tfrac1{nm}):n,m\in\mathbb N\}$$
endowed with the topology $\tau$ consisting of subsets $U\subseteq X$ satisfying the following two conditions:
(1) if $(\frac1n,0)\in U$ for some $n\in\mathbb N$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $k\ge m$;
(2) if $(0,0)\in U$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $n\ge m$ and all $k\in\mathbb N$
It is easy to see that Gutik's hedgehog is countable, second-countable, and Hausdorff but not regular.
In Gutik's hedgehog consider the subset $A=\{(0,0)\}\cup\{(\frac 1n,\frac1{nm}):n,m\in\mathbb N\}$.
It is easy to see that every nontrivial sequence in $A$ has a subsequence that converges in $X$ (to a point in the set $\{(0,0)\}\cup\{(\frac1n,0):n\in\mathbb N\}$). On the other hand, no subspace of $X$ containing $A$ is compact. So, the property (2) is satisfied but (1) does not.
The conditions(2) and (3) are equivalent for sequential spaces. This equivalence follows from two lemmas.
**Lemma 1.** Let $A$ be a subset of a topological space $X$ such that every open cover of $X$ contains a finite subfamily covering $A$. Then every sequence in $A$ has an accumulation point in $X$.
*Proof.* Assuming that $A$ contains a sequence $(a\_n)\_{n\in\omega}$ without accumulation points in $X$, for every $x\in X$, we can find a neighborhood $U\_x\subseteq X$ of $X$ such that the set $\{n\in\omega:a\_n\in U\_x\}$ is finite. Then for every finite subfamily $\mathcal F$ of the open cover $\{U\_x:x\in X\}$ of $X$ the set $\{n\in\omega:a\_n\in \bigcup\mathcal F\}$ is finite and hence $\{a\_n:n\in\omega\}\not\subseteq\bigcup\mathcal F$ and $A\not\subseteq \bigcup\mathcal F$. $\quad \square$
**Lemma 2.** If a sequence $(x\_n)\_{n\in\omega}$ in a sequential space $X$ has an accumulation point in $X$, then $(x\_n)\_{n\in\omega}$ has a subsequence that converges to some point of $X$.
*Proof.* Let $a$ be an accumulation point of the sequence $(x\_n)\_{n\in\omega}$ in $X$.
For every point $x\in X$ let $I\_x$ be the intersection of all neighborhoods of $x$ in $X$. It is easy to see that for every $x\in X$ and $y\in I\_x$ we have $I\_y\subseteq I\_x$.
For a subset $A\subseteq X$, consider the set $\dot A:=\{x\in X:I\_x\cap A\ne\emptyset\}$ and observe that $A\subseteq \dot A$.
If for some $x\in X$ the set $\Omega\_x:=\{n\in\omega:x\_n\in I\_x\}$ is infinite, then $(x\_n)\_{n\in\Omega\_x}$ is a required convergent (to $x$) subsequence of the sequence $(x\_n)\_{n\in\omega}$ and we are done. So, assume that for every $x\in X$ the set $\Omega\_x$ is finite. In this case, the set $A=\{x\_n:n\in\omega\setminus\Omega\_a\}$ is not closed in $X$ as $a\in\overline{A}\setminus A$.
Moreover, the set $\dot A$ is not closed as $A\subseteq \dot A\subseteq\bar{A}$ and $a\notin \dot A$. By the sequentiality of $X$, the set $\dot A$ contains a sequence $(a\_k)\_{k\in\omega}$ that converges to some point $b\in X\setminus \dot A$. If $a\_k\in I\_b$ for some $k\in\omega$, then $I\_{a\_k}\subseteq I\_b$ and $a\_k\in \dot A$ imply $I\_b\cap A\supseteq I\_{a\_k}\cap A\ne\emptyset$ and hence $b\in \dot A$, which contradicts the choice of $b$. This contradiction implies that $I\_b\cap\{a\_k:k\in\omega\}=\emptyset$ and hence the set $\{a\_k:k\in\omega\}$ is infinite. Since $\{a\_k:k\in\omega\}\subseteq \dot A$, for every $k\in\omega$ there exists a number $i\_k\in\omega$ such that $x\_{i\_k}\in I\_{a\_k}$. We claim that $b$ is the limit of the sequence $\{x\_{i\_k}:k\in\omega\}$. Indeed, for every open neighborhood $O\_b$ of $b$ in $X$, there exists $m\in\omega$ such that $\{a\_k:k\ge m\}\subseteq O\_b$. Then for every $k\ge m$ we have $x\_{i\_k}\in I\_{a\_k}\subseteq O\_b$, which means that the sequence $(x\_{i\_k})\_{k\in\omega}$ converges to $b$.
Assuming that the set $F=\{x\_{i\_k}:k\in\omega\}$ is finite and taking into account that $b\notin \dot A$, we can find a neighborhood $O\_b$ such that $O\_b\cap F=\emptyset$, which contradicts the convergence of the seqeunce $(x\_{i\_k})\_{k\in\omega}$ to $b$. This contradiction shows that the set $F$ is infinite. Then we can choose an increasing number sequence $(n\_k)\_{k\in\omega}$ such that $F=\{x\_{n\_k}\}\_{k\in\omega}$ and $x\_{n\_i}\ne x\_{n\_j}$ for $i\ne j$. For every $k\in\omega$ find $p\_k\in\omega$ such that $n\_k=i\_{p\_k}$. The convergence of the sequence $(x\_{i\_p})\_{p\in\omega}$ to $b$ implies the convergence of the sequence $(x\_{n\_k})\_{k\in\omega}=(x\_{i\_{p\_k}})\_{k\in\omega}$ to $b$. Therefore, $(x\_{n\_k})\_{k\in\omega}$ is a required convergent subsequence of the sequence $(x\_n)\_{n\in\omega}$. $\quad\square$
| 2 | https://mathoverflow.net/users/61536 | 448893 | 180,693 |
https://mathoverflow.net/questions/448895 | 30 | For two variables, their maximum
$\max\{x\_1,x\_2\}$ can be expressed using one $|\cdot|$ operation:
$$
\max\{x\_1,x\_2\} = \frac12(x\_1+x\_2+|x\_1-x\_2|).
$$
For $3$ variables, it seems fairly clear that three $|\cdot|$ operations are necessary, though I can only demonstrate sufficiency:
$$
\max\{x\_1,x\_2,x\_3\}
=
\frac14\left[
2x\_1+x\_2+x\_3+|x\_2-x\_3|+2\left|x\_1-\frac12(x\_2+x\_3+|x\_2-x\_3|)\right|
\right]
.
$$
For $5$ variables, it has been claimed here
<https://twitter.com/ereliuer_eteer/status/1669081421007454209>
that the minimum number of
$|\cdot|$ operations is 54, though I have verified neither sufficiency nor necessity.
Question: How many $|\cdot|$ operations are necessary for $n$ variables?
To be explicit about the rules, in addition to $|\cdot|$, the 4 arithmetic operations are allowed.
Update. That twitter post I linked is actually computing the median, not the max of 5 variables. So the analogous question for abs-operation complexity, but now for the median, is quite natural.
| https://mathoverflow.net/users/12518 | Minimum number of $|\cdot|$ operations necessary to express $\max$ | Let $f(k)$ be the minimum number of $|\cdot|$ operations to express the maximum of $2^k$ variables. Using the identities
$$\max\{x\_1,x\_2\} = (x\_1+x\_2+|x\_1-x\_2|)/2,$$
$$\max\{x\_1,\ldots,x\_{2^{k+1}}\} =
\max\{\max\{x\_1,\ldots,x\_{2^k}\},\max\{x\_{2^k+1},\ldots,x\_{2^{k+1}}\}\},$$
it is clear that $f(1)=1$ and $f(k+1)\leq 4f(k)+1$. Hence it follows by induction that
$$f(k)\leq (4^k-1)/3.$$
Now let us consider the case of $n$ variables. If $2^{k-1}<n\leq 2^k$, then the minimal number of corresponding $|\cdot|$ operations is less than $4^k/3$, which is less than $(4/3)n^2$.
**Added.** In fact that minimal number of $|\cdot|$ operations is less than $(3/8)n^2$. For details, see AspiringMat's response and the comments below it.
| 27 | https://mathoverflow.net/users/11919 | 448896 | 180,694 |
https://mathoverflow.net/questions/448911 | 3 | Let $p\_i \in [0,1]$ and $\sum\_{i} p\_i = 1$, and furthermore let $a\_i$ and $b\_i$ be positive real numbers. Is the inequality
$$
\sum\_{i} p\_i \frac{a\_i}{b\_i} \leq \frac{\sum\_{i} p\_i a\_i}{\sum\_{i} p\_i b\_i}$$
true?
| https://mathoverflow.net/users/506925 | Is the inequality $\sum_{i} p_i \frac{a_i}{b_i} \leq \frac{\sum_{i} p_i a_i}{\sum_{i} p_i b_i}$ true? | **No**. This reads
$${\mathbb E}[f(a,b)]\le f({\mathbb E}[a,b])$$
for an arbitrary discrete probability. This is true if and only if $f$ is a concave function (Jenssen Inequality). But $f(x,y)=\frac xy$ is not concave (nor convex).
You might prefer the convex function $g(x,y)=\frac{x^2}y$, for which we thus have
$${\mathbb E}\left[\frac{a^2}b\right]\ge \frac{{\mathbb E}[a]^2}{{\mathbb E}[b]}.$$
| 7 | https://mathoverflow.net/users/8799 | 448913 | 180,701 |
https://mathoverflow.net/questions/448910 | 2 | Let $f:\mathbb{Z}\_{p}\rightarrow\mathbb{C}$ be a function such that $\delta\_{a}f:\mathbb{Z}\_{p}\rightarrow\mathbb{C}$ is a locally constant function for any $a\in\mathbb{Z}\_{p}$, where $\delta\_{a}f(x):=f(x+a)-f(x)$. Does it necessarily imply that $f$ is locally constant?
| https://mathoverflow.net/users/140298 | Smooth function on $\mathbb{Z}_{p}$ | No, $f$ does not have to be locally constant. Here is an example. Take a nonzero $\mathbb{Q}$-linear map $L \colon \mathbb{Q}\_p \to \mathbb{Q}$, and let $f = L|\_{\mathbb{Z}\_p}$. Then $\delta\_a f(x) = L(x+a)-L(x) = L(a)$, so $\delta\_a f$ is a constant function for every $a$. However $f$ is not locally constant: take a $z \in \mathbb{Z}\_p$ with $f(z) \neq 0$, then $|f(p^n z)| = p^n |f(z)| \to \infty$ in $\mathbb{C}$, while $p^n z \to 0$ in $\mathbb{Z}\_p$.
If we also assume that $f$ is continuous, then indeed $f$ has to be locally constant. Proof:
We can replace $f$ with $f-f(0)$, so let $f(0) = 0$.
For $n \in \mathbb{Z}\_{\ge 0}$ let us call a function $g \colon \mathbb{Z}\_p \to \mathbb{C}$ $n$-locally constant, if $g(x)=g(y)$ for every $x,y \in \mathbb{Z}\_p$ with $|x-y|\_p \le p^{-n}$. So $g$ is locally constant if and only if it is $n$-locally constant for some $n \ge 0$. Let $A\_n$ be the set of $a \in \mathbb{Z}\_p$ such that $\delta\_a f$ is $n$-locally constant.
Then $\bigcup\_{n \ge 0} A\_n = \mathbb{Z}\_p$, and each $A\_n$ is closed in $\mathbb{Z}\_p$, because
$$A\_n = \bigcap\_{x,y \in \mathbb{Z}\_p : \, |x-y| \le p^{-n}} \{a \in \mathbb{Z}\_p: \, f(x+a)-f(x) = f(y+a)-f(y) \}.$$
So by the Baire category theorem, there is an $n$ such that $A\_n$ contains an open ball $c + p^m \mathbb{Z}\_p$ for some $m \ge n$.
Then $f(x+c) = f(x)+f(c)$ for every $x \in p^m \mathbb{Z}\_p$, and $f(x+c+y) - f(x) = f(c+y) = f(y) + f(c)$ for every $x,y \in p^m \mathbb{Z}\_p$.
Using $f(x+y+c) = f(x+y)+f(c)$, we get that $f(x+y) = f(x) + f(y)$ for every $x,y \in p^m \mathbb{Z}\_p$.
So if $x \in p^m \mathbb{Z}\_p$, then $p^k f(x) = f(p^k x) \to f(0) = 0$ as $k \to \infty$, hence $f(x) = 0$.
It easily follows that $f$ is locally constant.
| 5 | https://mathoverflow.net/users/42355 | 448916 | 180,702 |
https://mathoverflow.net/questions/448430 | 4 | For a proof for an article I would need the following result:
If $A\_\Gamma$ is an Artin group such that the $K(\pi,1)$-conjecture holds for it and $\Gamma'\subset\Gamma$ is an induced subgraph, then the $K(\pi,1)$-conjecture holds for $A\_{\Gamma'}$.
I've searched in all the references I know about this conjecture, but none of them seems to prove this fact. My question is if you know whether this has been proved or not, and in case it its where I can find it.
Thanks in advance.
| https://mathoverflow.net/users/482329 | $K(\pi,1)$-conjecture ofr Artin groups behave well with respect to special subgroups. Reference-Request | Yes, it has been proved by Godelle and Paris, it is cited as Theorem 5.5 in the article <https://arxiv.org/pdf/1211.7339> by Paris (and Theorem 3.1 for the equivalence with the K(pi,1) conjecture).
| 4 | https://mathoverflow.net/users/48519 | 448923 | 180,704 |
https://mathoverflow.net/questions/448849 | 1 | The following question on polynomials arose as a potentially helpful intermediate step on a proof of a Theorem that I want to demonstrate. Its statement is quite elementary, and I can think of a number of applications it might have to the study of Hilbert polynomials in commutative algebra and certain combinatorial sequences arising in discrete geometry.
>
> **Conjecture 1**: Let $(c\_0,\ldots,c\_n)$ be non-negative integers and consider the polynomial $P(x) = \sum\_{i=0}^n c\_i \binom{x+n-i}{n}$. Assume that $P(x)$ has at least one negative real-root and denote by $-r$ the largest (in absolute value) negative real-root of $P(x)$. Then either $\lfloor r \rfloor = 0$ or $-\lfloor r\rfloor$ is a root of $P(x)$ too.
>
>
>
A weaker statement that could also be useful:
>
> **Conjecture 1'**: Let $(c\_0,\ldots,c\_n)$ be non-negative integers and consider the polynomial $P(x) = \sum\_{i=0}^n c\_i \binom{x+n-i}{n}$. If $P(-1)\neq 0$, then the negative real roots of $P(x)$ (if any) lie in the interval $(-1,0)$.
>
>
>
| https://mathoverflow.net/users/147861 | Location of the negative real roots of certain integer-valued polynomials | Conjecture 1' looked a bit easier to test. Unless I've misinterpreted the question (not impossible), I believe the conjecture is false. If we look at the case $n=2$, the polynomial is
$$\begin{align}
P(x)&=c\_0\binom{x+2}{2}+c\_1\binom{x+1}{2}+c\_2\binom{x}{2} \\
&=\frac12\left(c\_0(x+2)(x+1)+c\_1 (x+1)x + c\_2 x(x-1)\right) \\
&=\frac12\left(\left(c\_0+c\_1+c\_2 \right)x^2+\left(3c\_0+c\_1-c\_2 \right)x+2c\_0 \right)
\end{align}$$
Taking $\left(c\_0,c\_1,c\_2\right)=\left(18,1,1\right)$ gives
$$P(x)=10x^2+27x+18=(2x+3)(5x+6)$$
with two negative roots, each less than $-1$.
Is there a reformulation that would be worth checking?
| 2 | https://mathoverflow.net/users/506194 | 448937 | 180,712 |
https://mathoverflow.net/questions/448920 | 1 | Let $(M,g)$ be a compact Riemannian manifold (e.g. $M=S^3$ the 3-sphere) and let $\Delta$ be the metric Laplacian on $M$. Then $\Delta$ has an $L^2(M)$ basis of eigenfunctions $\pi\_m$, $$ \Delta \pi\_m = - \lambda^2\_m \pi\_m. $$ Define the square root of the Laplacian, $(-\Delta)^{1/2}$, by $$ (-\Delta)^{1/2} \pi\_m = \lambda\_m \pi\_m. $$ Then I understand that $(-\Delta)^{1/2}$ is a pseudodifferential operator of order 1 on $M$. My question is, does $(-\Delta)^{1/2}$ satisfy the Leibniz rule $$ (-\Delta)^{1/2}(fg) = f(-\Delta)^{1/2} g + g (-\Delta)^{1/2} f, $$ perhaps up to a smoothing operator of some kind? Additionally, intuitively it feels like $(-\Delta)^{1/2}$ should essentially be the gradient (i.e. exterior derivative $d$ on functions), but of course it is a scalar operator unlike the exterior derivative. Is there a sense in which $$ (-\Delta)^{1/2} f = d f + \text{stuff} ?$$ Any references would be appreciated.
| https://mathoverflow.net/users/104213 | Leibniz rule for square root of Laplacian | For the Leibniz rule, it is not true that
\begin{equation} (-\Delta)^{\frac12}(fg) = f (-\Delta)^{\frac12}g + g (-\Delta)^{\frac12}f \end{equation}
even in the simplest possible case, i.e. in $\mathbb{R}$. I think the easiest way to see it is to consider $f,g$ sufficiently "good" functions, i.e. Schwarz functions in this case, then applying the Fourier transform, we have
\begin{align} \widehat{(-\Delta)^{\frac12}(fg)}(\xi) - \widehat{ f (-\Delta)^{\frac12}g}(\xi) - \widehat{ g (-\Delta)^{\frac12}f}(\xi) & = |\xi| \int\_\mathbb{R} \hat{f}(\omega)\hat{g}(\xi-\omega)d\omega \\
& - \int\_\mathbb{R} \hat{f}(\omega)\hat{g}(\xi-\omega)|\xi-\omega|d\omega \\ &- \int\_\mathbb{R} \hat{f}(\omega)|\omega|\hat{g}(\xi-\omega)d\omega \\
& = \int\_\mathbb{R} \hat{f}(\omega)\hat{g}(\xi-\omega)[|\xi|-|\omega|-|\xi-\omega|]d\omega. \end{align}
Then I let someone else verify that you in any reasonable sense this is not a smoothing operator. For example one cannot control the $L^1(\mathbb{R})$ norm of the above expression in terms of the $L^2(\mathbb{R})$ norms of $f,g$.
What is reasonable to expect and in fact it is true in $\mathbb{R}^n$ and it could be true also in manifolds (where you have good boundedness properties of the Riesz transforms) is that you can control some kind of norm of $(-\Delta)^{\frac12}(fg)$ in terms of norms of $f, g, (-\Delta)^\frac12 f, (-\Delta)^\frac12 g$. See for example, Chapter 7.6, of the book of Grafakos, Modern Fourier Analysis, (3rd Edition)
| 2 | https://mathoverflow.net/users/153260 | 448938 | 180,713 |
https://mathoverflow.net/questions/448919 | 8 | Let $G$ be a topological (or simplicial) group, let $X$ and $Y$ be $G$-spaces, and let $f,f':X\to Y$ be $G$-maps which are homotopic as maps of spaces. In general, $f$ and $f'$ may (of course) fail to be equivariantly homotopic. For instance, we can just take $X$ to be a point and $Y$ to be any path-connected $G$-space having two distinct fixed points of the $G$-action, with $f$ and $f'$ given by the inclusion of two fixed points. But **is there any example of such $f$ and $f'$ if we assume that $X$ and $Y$ are total spaces of principal $G$-bundles?**
| https://mathoverflow.net/users/144250 | Homotopic but not equivariantly homotopic maps | For any $G$-space the $G$-equivariant maps $[EG,X]\_G$, also known as the homotopy fixed points $X^{hG}$, are a Borel homotopy invariant of $X$ meaning that it is an invariant of $G$-equivariant maps for which the underlying map is a weak equivalence. If $Z$ has the trivial action, this means we can compute $[EG,Z \times EG]\_G$ as $[EG,Z]\_G=[EG/G,Z]=[BG,Z].$ On the other hand, the nonequivariant maps $[EG,Z] \cong \pi\_0(Z)$ because $EG$ is contractible. So it suffices to find a space, $Z$ such that $[BG,Z] \not= \pi\_0(Z)$. The universal example (if $G$ is not contractible) is $Z=BG$.
Explicitly, if $q:EG \rightarrow BG$ is the quotient, this produces homotopic maps $EG \xrightarrow{q \times \mathrm{Id}} BG \times EG$ and $EG \xrightarrow{\* \times \mathrm{Id}} BG \times EG$ which are not equivariantly homotopic.
| 11 | https://mathoverflow.net/users/134512 | 448946 | 180,715 |
https://mathoverflow.net/questions/448795 | 2 | *This is a follow-up question to [Can I wrap a suitcase with hair ties](https://mathoverflow.net/questions/424215/can-i-wrap-a-suitcase-with-hair-ties).*
Now we know that it is possible to wrap a suitcase with hair ties without tying them together,
>
> but can you do it with large rotational symmetry?
>
>
>
In other words, I am looking for a nontrivial link $L$ in a solid torus made by short circles that is trivial in the ambient euclidean space and such that $L$ is invariant with respect to a large cyclic group of rotations of the solid torus.
| https://mathoverflow.net/users/1441 | Wrapping a suitcase with large rotational symmetry | **Edit: my previous answer was incorrect**
No, this one you cannot do. If one had such an arrangement of circles in the solid torus inside $R^3$, and quotiented by a rotation, then by the [equivariant Dehn's lemma](https://link.springer.com/article/10.1007/BF02566211) there would be a collection of essential disks in the complement of the split link that are invariant under the symmetry.
Since the link components get permuted, no disk can have a fixed point of the cyclic action. So they are disjoint from the axis of symmetry of the rotation.
Hence the components are unknotted in the solid torus, a contradiction
| 5 | https://mathoverflow.net/users/1345 | 448948 | 180,717 |
https://mathoverflow.net/questions/448949 | 4 | Let $A$ be an algebra over a field $K$. Loosely speaking, an algebra is said to be tame if for each $d \in \mathbb{Z}\_{>0}$ all but finitely-many of the indecomposable $A$-modules of $K$-dimension $d$ occur within a finite number of one-parameter families (indexed over $K$). (At least, this is often considered the definition when $K$ is algebraically closed... if there is a different definition for more general fields, please let me know.)
A standard (and extremely important) example of a tame algebra is $K[x]$, where $K$ is algebraically closed. As a principal ideal domain, its finite-dimensional indecomposable modules are isomorphic to quotients $K[x] / (p)$, where each $(p)$ is a primary ideal. In particular, the primary ideals of $K[x]$ are powers of irreducible elements of $K[x]$. Since $K$ is algebraically closed, these are just linear polynomials $x - \lambda$ with $\lambda \in K$. Hence, the indecomposable modules of dimension $d$ are given by the quotients $K[x] / (x - \lambda)^d$. So $K[x]$ is tame whenever $K$ is algebraically closed.
What happens if we remove the assumption that $K$ is algebraically closed? The situation seems to deteriorate somewhat. For example consider $\mathbb{Q}[x]$. This is still a principal ideal domain, so most of the above reasoning applies. However since $\mathbb{Q}$ is not algebraically closed, we have (for example) infinitely many irreducible polynomials $x^2 + p x +q$, where $p,q \in \mathbb{Q}$. Unless I am mistaken, this appears to produce a two-parameter family of indecomposable $\mathbb{Q}[x]$-modules of dimension 2: one can think of this from the perspective of representations of $\mathbb{Q}[x]$ by considering $2 \times 2$ matrices (up to commuting $\mathbb{Q}$-isomorphism/similarity over $\mathbb{Q}$) whose eigenvalues are roots of $x^2 + p x +q$ for each pair of rationals $(p,q)$ that yields an irreducible polynomial.
This appears to indicate that $\mathbb{Q}[x]$ is not tame, and thus (by a theorem of Drozd), is semi-wild (i.e. badly behaved from a representation-theoretic perspective). I find this extremely surprising to the point where I feel like there must be something wrong with my reasoning. This is especially considering that we have a 'classification' of indecomposables in terms of powers of irreducible polynomials of $\mathbb{Q}[x]$... though perhaps the problem here is a lack of classification of the irreducible polynomials themselves? The problem seems to disappear if we consider finite fields, because then there are only finitely many choices of coefficients for each degree of polynomial.
Can anyone with more experience in this area confirm my reasoning here and/or offer further illumination on this matter?
| https://mathoverflow.net/users/95742 | Are polynomial algebras over fields (that are not algebraically closed) tame? | The definition of tame representation type usually assumes that you're dealing with a finite dimensional algebra $A$ over a field $k$. In this case, the definition of a one parameter family of modules is that there is an $A$-$k[T]$-bimodule $X$ which is finitely generated and free over $k[T]$, and then the modules in the family are all required to be of the form $X \otimes\_{k[T]} M$ for some $k[T]$-module $M$. The requirement is then that in any particular (finite) dimension over $k$, all but a finite number of indecomposable $A$-modules belong to a finite set of one parameter families. In other words, $k[T]$ is the model for tame representation type.
For an infinite dimensional algebra, I guess one can use the same definition, and then $k[x]$ would be tame. But you should beware that the finite/tame/wild trichotomy has only been proved for finite dimensional algebras, and it's not clear that one should try to apply the same definitions to infinite dimensional algebras. For example, there are algebras with no finite dimensional representations at all, and then do you want to say that those have finite representation type? It's all a bit unsatisfactory.
| 7 | https://mathoverflow.net/users/460592 | 448951 | 180,719 |
https://mathoverflow.net/questions/448871 | 5 | Are there modern good lecture notes/book about Brill-Noether theory of curves.
Most interesting theorems here are proved via limit linear series, which I found no lecture notes on (instead there is the original paper from the 80's which references several older papers and while well written is hard to read due to this).
The other class of interesting theorems are proved via topicalization methods on which I also haven't found good sources
Thanks!
| https://mathoverflow.net/users/135743 | Modern references to Brill-Noether theory on curves? | I recommend also Geometry of Algebraic Curves, Volume I, (1984), chs. V and VII, by Arbarello, Cornalba, Griffiths and Harris, as well as Volume II, (2011), ch. XXI, of the same title, by Arbarello, Cornalba, and Griffiths. (The authors present in Volume II, a "simplified" version, due to Pareschi, of Lazarsfeld's proof.) The history of the problem and proofs are exposited very thoroughly in the bibliographical notes to the cited chapters.
I would add to this history that the "Brill Noether matrix" appears even earlier in Roch's paper proving his half of the Riemann-Roch theorem, and that the "Brill-Noether number" is a generalization, for r > 1, of the formula d ≥ (g/2) + 1, in the section numbered 5. of the Erste Abtheilung, of Riemann's paper on Abelian Functions, which Riemann asserts to hold whenever a general curve of genus g has a non constant meromorphic function with only d poles.
Here is another modern source, in which a masters student from Holland generalizes the results of ACGH to arbitrary alg closed fields. <http://abarbon.com/assets/Andrea%20Barbon%20-%20Algebraic%20Brill-Noether%20Theory.pdf>
but the webpage is "not secure".
| 6 | https://mathoverflow.net/users/9449 | 448958 | 180,721 |
https://mathoverflow.net/questions/448953 | 6 | *This is a follow-up to [this question](https://mathoverflow.net/questions/448881/do-doubly-transitive-actions-give-rise-to-irreducible-representations-for-infini).*
Let $G$ be a group acting doubly-transitively on a set $X$. Then the vector space $V\_X$ of functions $f\colon X\to\mathbb C$ with finite support such that $\sum\_{x\in X}f(x)=0$ carries an action of $G$.
>
> There are examples where $V\_X$ can be reducible. Is $V\_X$ necessarily indecomposable?
>
>
>
One possible approach is to prove the only $G$-homomorphisms $\varphi\colon V\_X\to V\_X$ are scalars. Such a homomorphism is determined by the image $f\in V\_X$ of $[x]-[y]$, where $x\ne y\in X$ and $[x]$ is the characteristic function of $\{x\}$. Now, $G$-equivariance in particular implies a strange cocyle condition:
>
> Let $g\in G\_x$. Then $f-\pi(g)f=\pi(h)f$ for any $h\in G\_y$ such that $hx=gy$ (which exists by double-transitivity).
>
>
>
I wonder if this is enough to prove $\mathrm{supp}(f)=\{x,y\}$.
| https://mathoverflow.net/users/123673 | Do doubly-transitive actions give rise to indecomposable representations for infinite groups? | Yes.
Let $\varphi$ be a $G$-homomorphism $V\_X \to V\_X$. As mentioned in the question, it suffices to show $\varphi$ is a scalar.
For $x, y\in X$ with $x\neq y$, let $c\_{x,y}$ be obtained by evaluating the function $\varphi([x]-[y])$ at $x$. Then $c\_{g(x),g(y)}$ is the evaluation of $$\varphi([g(x)]-[g(y)]) = \varphi( g \cdot [x] - g\cdot[y])=\varphi( g\cdot ([x]-[y]) = g\cdot \varphi([x]-[y])$$ at $g(x)$ and hence equals $c\_{x,y}$.
Since the action of $G$ on $X$ is doubly-transitive, this implies $c\_{x,y}$ is a constant function of $x,y\in X$ with $x \neq y$, say $c\_{x,y}=c$ for all $x,y$.
Then the value at $x$ of $\varphi([x]-[y])$ is $c$, the value at $y$ is $-c$ since $[x]-[y]= - ([y]-[x])$, and the value at $z$ with $z\neq x,z\neq y$ is $0$ since $[x]-[y]=([z]-[y])-([z]-[x])$.
So $\varphi( [x]-[y]) = c ([x]-[y])$ and since these elements span $V\_X$, $\varphi$ is scalar multiplication by $c$, as desired.
| 8 | https://mathoverflow.net/users/18060 | 448960 | 180,723 |
https://mathoverflow.net/questions/448942 | 5 | For a Polish space $(X,d)$ its hyperspace $(K(X),d\_H)$ is also a Polish space. (Here $K(X)$ denotes the set of all nonempty compact subsets of $X$, and the Hausdorff metric $d\_H$ is defined by $d\_H(K,L)=\inf \{\varepsilon>0:\ K\subseteq L\_\varepsilon,\ L\subseteq K\_\varepsilon\}$, where $A\_\varepsilon$ is the open $\varepsilon$-neighborhood of $A$).
**Question:** Is there any nontrivial Polish space $(X,d)$ that is isometric to its hyperspace $(K(X),d\_H)$?
What I know so far:
* There is [this MO question](https://mathoverflow.net/questions/117893/a-question-about-a-hierarchy-of-metric-spaces-arising-from-an-operation-defined) with a couple of comments but no real answer.
* Trivial examples: the empty space, a singleton space, a countably infinite discrete space.
* $X$ cannot be totally bounded. In particular, it cannot be compact. (Proof: if $X$ has at least 2 points and it is totally bounded, then for some $\varepsilon>0$ we have that $N\_X(\varepsilon)=\max\{|F|:\ F\subseteq X \text{ is finite and } d(x,y)>\varepsilon \text{ for any two distinct points }x,y\in F\}$
is finite but at least 2, thus $N\_{K(X)}(\varepsilon)>N\_X(\varepsilon)$, which distinguishes the two spaces.
* A natural candidate would be [the Urysohn universal space](https://en.wikipedia.org/wiki/Urysohn_universal_space) $U$. Unfortunately, the above property fails for $U$ because $K(U)$ is not homogeneous. In fact, there exists a four-point metric space $A$, a point $a\in A$, and an isometric embedding $f: A\setminus{a}\to K(U)$ that does not extend to $A$.
An example: $A=\{a\_1,a\_2,a\_3,a\_4\}$, the pairwise distances are $1,1,1,1,1,2$ with only $d\_A(a\_1,a\_4)=2$. Pick two points $x,y\in U$ with $d\_U(x,y)=1$ and let $K\_1=\{x\}$, $K\_2=\{y\}$, $K\_3=\{x,y\}$. Let $f(a\_i)=K\_i$ for every $i\leq 3$, which is an isometric embedding of $A\setminus \{a\_4\}$ into $K(U)$. It is easy to check that no suitable image exists for $a\_4$.
* If we replace *isometry* with *homeomorphism*, then there are many examples including the Cantor space, the Baire space, and the Hilbert cube. In this case, the interesting problem seems to be classifying such Polish spaces. This is probably too much to ask. A more reasonable goal is classifying some subclass (e.g. compact 0-dimensional Polish spaces).
| https://mathoverflow.net/users/479121 | Polish space isometric to its hyperspace | It is much easier to deal with zero dimensional spaces and ultrametrics than connected metrics. $\omega^\omega$ with the standard ultrametric is isometrically isomorphic to its own hyperspace.
In particular, we give $\omega^\omega$ the ultrametric where if $f,g:\omega\rightarrow\omega$, then let $d(f,g)=2^{-n}$ where $n$ is the least natural number with $f(n)\neq g(n)$ and where $d(f,f)=0$.
We observe that a metric space $(X,d)$ is isometrically isomorphic to $\omega^\omega$ if and only if $(X,d)$ is a Polish ultrametric space where
$d[X\times X]=\{0\}\cup\{2^{-n}\mid n\in\omega\}$ and $x\in X,n\in\omega$ implies that there is an infinite family $(x\_r)\_{r\in\omega}$ where
$d(x\_r,x\_s)=2^{-n}$ for $r<s$ and where $x\_0=x$.
It is easy to see that $d\_H[K(\omega^\omega)\times K(\omega^\omega)]=\{0\}\cup\{2^{-n}\mid n\in\omega\}$.
Now suppose that $C\in K(\omega^\omega)$ is non-empty compact and $n\in\omega$. Let $f\in C$. Then let $C\_0=C$. By compactness, the set $\pi\_n[C]$ is finite, so let $v$ be an natural number with $v>\pi\_n(c)=c(n)$ for each $c\in C$. If $r>0$, then let $f\_r\in\omega^\omega$ be a function with $f\_r(n)=v+r$ and where $f\_r(m)=f(m)$ for each $m<n$. Then let $C\_r=C\cup\{f\_r\}$. Then we have $d\_H(C\_r,C\_s)=2^{-n}$ whenever $r,s\in\omega,r\neq s$. Therefore, $\omega^\omega$ is isometrically isomorphic to $K(\omega^\omega)$.
| 2 | https://mathoverflow.net/users/22277 | 448961 | 180,724 |
https://mathoverflow.net/questions/448971 | 6 | Let $a\_n$ be a nonnegative sequence that Cesaro converges to $K > 0$. We recall this means
$$\frac{1}{N} \sum\_{n = 1}^N a\_n \to K$$
as $N \to \infty$.
Suppose $a\_{\phi\_n}$ with $\phi: \mathbb N \to \mathbb N$ is a rearrangement of the $a\_n$ such that for some $\varepsilon > 0$, we have
$$|\phi(n) - n| \leq \varepsilon n$$
for all $n \in \mathbb N.$
Is it true that there exists some $C\_\varepsilon$ depending only on $\varepsilon$ (and not $a\_n$ or $K$) with $C\_\varepsilon \to 0$ as $\varepsilon \to 0$ such that the partial averages of $a\_{\phi\_n}$ satisfy
$$(1-C\_\varepsilon) K \leq \liminf\_{N \to \infty} \frac{1}{N} \sum\_{n = 0}^{N-1} a\_{\phi\_n} \ \leq \limsup\_{N \to \infty} \frac{1}{N} \sum\_{n = 0}^{N-1} a\_{\phi\_n} \ \leq (1 + C\_\varepsilon)K?$$
| https://mathoverflow.net/users/173490 | Does control on the “magnitude” of the rearrangement give control of the rearranged Cesaro sums? | $\newcommand\ep\varepsilon$We have $\{\phi\_0,\dots,\phi\_{N-1}\}\subseteq\{0,\dots,\lfloor(N-1)(1+\ep)\rfloor\}$. Therefore and because $a\_n\ge0$ for all $n$,
$$\sum\_{n=0}^{N-1} a\_{\phi\_n}\le \sum\_{0\le n\le(N-1)(1+\ep)} a\_n.$$
So,
$$\limsup\_{N\to\infty}\frac{1}{N}\sum\_{n = 0}^{N-1} a\_{\phi\_n} \le (1+\ep)K.$$
Similarly,
$$\{\phi\_0,\dots,\phi\_{N-1}\}\supseteq\{0,\dots,\lfloor(N-1)(1-\ep)\rfloor\}; \tag{1}\label{1}$$
see details on this at the end of this answer.
Therefore and because $a\_n\ge0$ for all $n$,
$$\sum\_{n=0}^{N-1} a\_{\phi\_n}\ge \sum\_{0\le n\le(N-1)(1-\ep)} a\_n.$$
So,
$$\liminf\_{N\to\infty}\frac{1}{N}\sum\_{n = 0}^{N-1} a\_{\phi\_n} \ge (1-\ep)K.$$
**Details on \eqref{1}:** We have $|\phi\_n-n|\le\ep n$ for all $n$, which can be rewritten as $|k-\psi\_k|\le\ep\psi\_k$ for all $k$, where $\psi$ is the inverse of $\phi$, so that $\phi\_{\psi\_k}=k$ for all $k$. So, $\psi\_k\le\dfrac k{1-\ep}$ for all $k$. So, if $k\le (1-\ep)(N-1)$, then $\psi\_k\le N-1$. That is, $\{\psi\_0,\dots,\psi\_{\lfloor(N-1)(1-\ep)\rfloor}\}\subseteq\{0,\dots,N-1\}$. That is, $\{0,\dots,\lfloor(N-1)(1-\ep)\rfloor\}\subseteq\{\phi\_0,\dots,\phi\_{N-1}\}$, so that \eqref{1} holds.
| 5 | https://mathoverflow.net/users/36721 | 448973 | 180,726 |
https://mathoverflow.net/questions/448983 | 2 | In [this](https://en.wikipedia.org/wiki/Hele-Shaw_flow) Wikipedia article, Hele-Shaw flow is discussed in some detail. I'd like to find a textbook that discusses Hele-Shaw flow in greater detail. Thanks
| https://mathoverflow.net/users/506997 | Fluid dynamics textbook discussing Hele-Shaw flow | A mathematics-oriented text book is [Conformal and Potential Analysis in Hele-Shaw Cells](https://www.google.com/books/edition/Conformal_and_Potential_Analysis_in_Hele/y68nodht9RwC?hl=en&gbpv=0), by Gustafsson and Vasil'ev (2006).
>
> This monograph aims at giving a presentation of recent and new ideas
> that arise from the problems of planar fluid dynamics and which are
> interesting from the point of view of geometric function theory and
> potential theory. In particular, this book is concerned with geometric
> problems for Hele-Shaw flows. Also Hele-Shaw flows on parameter spaces
> (e.g., the Teichmüller space) are treated and connections with string
> theory are revealed. Ultimately, the interaction between several
> branches of complex and potential analysis, and planar fluid mechanics
> is discussed. For most parts of this book the background provided by
> graduate courses in real and complex analysis, in particular, the
> theory of conformal mappings and in fluid mechanics is assumed. There
> are some historical remarks concerning the people that have
> contributed to the topic. The book is as self-contained as possible.
>
>
>
| 3 | https://mathoverflow.net/users/11260 | 448987 | 180,730 |
https://mathoverflow.net/questions/448992 | 1 | Let $\mathcal{D}^b(X)$ be the bounded derived category of constructible $\mathbb{C}$-sheaves on $X$.
(1) For any closed embedding $i:Z\rightarrow X$ and $A\in \mathcal{D}^b(X)$, there is a natural morphism $i^!(A)\rightarrow i^\*(A)$ which induces a distinguished triangle in $\mathcal{D}^b(Z)$,$$i^!(A)\rightarrow i^\*(A)\rightarrow B\xrightarrow{+1}.$$
(2) For any vector bundle $f:X\rightarrow Y$ and $A\in \mathcal{D}^b(X)$, there is a natural morphism $f\_!(A)\rightarrow f\_\*(A)$ which induces a distinguished triangle in $\mathcal{D}^b(Y)$,$$f\_!(A)\rightarrow f\_\*(A)\rightarrow C\xrightarrow{+1}.$$
Is it possible to describe the cones $B$ and $C$ explicitly? Even in the case that $A=\underline{\mathbb{C}}$ is the constant sheaf on $X$? Any comments will help! Thanks!
| https://mathoverflow.net/users/507011 | Cones of natural morphisms $i^!\rightarrow i^*$ and $f_!\rightarrow f_*$ | For (2), every vector bundle can be compactified to a projective space bundle (the Proj of the graded ring of symmetric powers of the dual bundle plus a rank one trivial bundle). Let $j$ be the open immersion into the projective space bundle and $\pi$ the projection from the projective space bundle to $Y$. Then $f= \pi \circ j$ so $f\_! A = \pi\_! j\_! A = \pi\_\* j\_! A$ since $\pi$ is proper while $f\_\* A = \pi\_\* j\_\* A$ so $C$ is $\pi\_\*$ of the mapping cone of $j\_! A \to j\_\* A$, which is just the restriction of $j\_\* A$ to the divisor at $\infty$.
Depending on $A$, this might be easy or difficult to compute.
However, for $A$ the constant sheaf, this is not the best way to describe the mapping cone. This is because, in this case, $f\_\* A$ is simply a constant sheaf and $f\_!A $ is the constant sheaf shifted by twice the dimension $d$ of the vector bundle, so the mapping cone is simply an extension of two constant sheaves, associated to a cohomology class in $H^{2d}(Y,\mathbb Q)$.
The relevant cohomology class is the Euler class or top Chern class of the bundle $X$. The generator for the cohomology of $f\_! \mathbb Q$ can be represented by a section of the bundle $X \to Y$, and the isomorphism between $f\_\* \mathbb Q$ and $\mathbb Q$ can be calculated by pulling back to the zero section, so the cohomology class obtained by sending a generator on the map $f\_\* \mathbb Q \to f\_\* \mathbb Q$ is the pullback of a section to the zero section, i.e. the intersection of an arbitrary section with the zero section.
| 4 | https://mathoverflow.net/users/18060 | 448995 | 180,733 |
https://mathoverflow.net/questions/448969 | 2 | This seems like something one might find in a book so I would be grateful for any references you think may be helpful.
I am interested in the rate at which of a function integrated against the $N$th Fejer Kernel approaches it's value at zero. In other words, I am looking to understand the big oh term below
$$\int\_{-\pi}^\pi f(t)F\_N(t) dt - f(0) = O(1/N^?)$$
where
$$F\_N(t) = \frac{1}{2\pi}\cdot\frac{\sin^2(Nt/2)}{N\sin^2(t/2)} = \sum\_{n = -N+1}^{N-1}\left(1-\frac{|n|}{N}\right)\cos(nt)$$
is the $N$th Fejer Kernel and a delta-sequence. I can't find this result anywhere. Thank you in advance.
| https://mathoverflow.net/users/506972 | Rate of convergence of Fejer kernel to the Dirac delta function | For any $\delta,x$ such that $0<\delta\leq |x|<\pi$ one has $|F\_N(x)|\leq[2\pi(N+1)\sin^2(\delta/2)]^{-1}$, so the error in the delta-function approximation is of order $1/N$.
Two explicit examples, if $f(t)=\cos t$ one has
$$\int\_{-\pi}^\pi f(t)F\_N(t) dt =1-\frac{1}{N}+{\cal O}(N^{-2}).$$
and if $f(t)=\sin^2(t/2)$ one has
$$\int\_{-\pi}^\pi f(t)F\_N(t) dt =\frac{1}{2N}.$$
| 4 | https://mathoverflow.net/users/11260 | 448996 | 180,734 |
https://mathoverflow.net/questions/448998 | 12 |
>
> If we omit parameters in the definition of $L$ would the result still be $L$?
>
>
>
That is, we define a successor stage $L\_{\alpha+1}$ in the constructible universe $L$, without including parameters; as:
$L\_{\alpha+1} =\{\{ y \mid y \in L\_\alpha \land (L\_\alpha, \in) \models \phi(y) \}\mid \phi \text { is a first order formula}\} $
| https://mathoverflow.net/users/95347 | Can $L$ be defined without parameters? | Yes, the parameter-free version of $L$ gives rise to the same constructible universe $L$. You will still get all of $L$ this way, but it will come more slowly.
The reason is that at stage $\alpha+1$, you in effect have $\alpha$ as a parameter, since this is definable as the largest ordinal. So you can use $\alpha$ as a parameter. Every finite sequence of ordinals is coded by a single ordinal, and so in this way you can have any finitely many ordinal parameters, and this is enough to pick out any object in the resulting hierarchy. So eventually you will get every set in $L$.
It seems that the two hierarchies will catch up to each other at every cardinal stage, $L\_\kappa$ for any cardinal $\kappa$. At the finite stages this is clear, since the finite sets in $L\_n$ are all definable. For any ordinal $\xi$, if the set $L\_\xi$ appears at some stage $\lambda$, then it will be definable in the later stages from ordinal parameter $\xi$, and any definable subset of it will be definable in $L\_\xi$ from some ordinal parameter $\beta$, and hence will be definable at stage $\alpha+1$ in your hierarchy, if $\alpha$ codes the pair $\langle\xi,\beta\rangle$. Inductively, this will all happen before the next cardinal stage, and therefore $L\_{\xi+1}$ will arrive also before the next cardinal stage.
| 14 | https://mathoverflow.net/users/1946 | 449001 | 180,737 |
https://mathoverflow.net/questions/448972 | 0 | I am trying to understand why I am getting an almost singular matrix in a problem I have.
The problem is a simple as
$$
\min\_{X \in \mathbb{R}^{m,n}} \left\lVert AX - B \right\rVert\_F^2
$$
Obvioulsy in constructing $A$ and $B$ there're few steps. But I can overly simplify with a toy example the step I believe is causing the issue.
Suppose you have $f(x) = e^{-\frac{x^2}{2\sigma^2}}$ and define $T\_y$ as $ (T\_yf)(x) = f(x - y)$.
Suppose you have now $Y = \left\{ y\_1, \ldots, y\_n \right\}$.
Essentially I think my choice of $\sigma$ and $Y$ is such that I get something which is (numerically) linearly dependent. In theory I don't think this would happen but in practice that's my problem (numerically).
Is there maybe a theoretical/numerical method that tells me *how much*
$
\left\{ T\_y f\right\}\_{y \in Y}
$
are linearly indepedent?
(You can imagine that you later fix $\left\{ x\_1,\ldots,x\_m \right\}$ and then
$$
A =
\begin{pmatrix}
T\_{y\_1}(x\_1) & T\_{y\_2}(x\_1) & \ldots & T\_{y\_n}(x\_1) \\
T\_{y\_1}(x\_2) & T\_{y\_2}(x\_2) & \ldots & T\_{y\_n}(x\_2) \\
\vdots & \vdots & \ddots & \vdots \\
T\_{y\_1}(x\_m) & T\_{y\_2}(x\_m) & \ldots & T\_{y\_n}(x\_m)
\end{pmatrix}
$$
This in my calculation (numerically) is never full rank.
**Note**: $m >> n$.
Is there some study or measure I can use to work out better choices of $Y$ and $\sigma$?
| https://mathoverflow.net/users/152487 | Least square error problem ill conditioning | At least when $c:=x\_i-y\_i$ does not depend on $i$, any $n\times n$ submatrix of $A$ is of the form $M=(f\_c(u\_i-u\_j)\colon i,j=1,\dots,n\}$, where $f\_c(x):=f(x+c)$. So, $\det M>0$, since the function $f\_c$ (being the characteristic function of a normal distribution) is positive definite.
However, $f\_c=f\_{c;2}$ is only "borderline" positive definite, in the sense that for any real $p>0$, the function $f\_{c;p}$ defined by the formula $f\_{c;p}(x):=\exp(-|(x+c)/\sigma|^p/2)$ is positive definite if and only if $p\le2$. So then, $\det M$ may be very close to $0$ and hence $A$ may be very close to a non-full rank matrix.
The same will hold if for some permutation $\pi$ of the set $[n]:=\{1,\dots,n\}$ we have that $x\_i-y\_{\pi(i)}$ does not depend on $i$. A similar effect can be expected if $x\_i-y\_{\pi(i)}$ does not much depend on $i$ for for some permutation $\pi$ of $[n]$. Because there a great number ($n!$) of such permutations, it is likely that there is one with this "does not much depend on $i$" property.
---
There also is the general tendency for large $n\times n$ matrices to be close to singular. Indeed, the determinant is the product of the eigenvalues (repeated according to their multiplicities). So, if even one eigenvalue is very close to $0$, then the determinant may be close to $0$. This effect is exponentially strong in the number of eigenvalues close to $0$.
That is why different regularization methods, such as [Tikhonov regularization](https://en.wikipedia.org/wiki/Least_squares#Tikhonov_regularization) and [Lasso method](https://en.wikipedia.org/wiki/Least_squares#Lasso_method), are used with the method of least squares.
| 2 | https://mathoverflow.net/users/36721 | 449004 | 180,740 |
https://mathoverflow.net/questions/448977 | 3 | The Riemann mapping theorem in $\mathbb{R}^2$ is known not to generalize well in higher dimensions and is basically trivial in lower dimensions.
I’m interested in how it generalizes for fractional dimensional sets particular with dimensions between 2 and 3.
We can start by considering dimensions between 1 and 2. If you take a simply connected fractal such as the vicsek fractal with dimension $\log\_3{5}$ then it’s easy to see (though perhaps hard to prove) that any simply connected open subset of this can be invertibly continuously mapped to any other simply connected open subset. Moreover any such map trivially must be conformal since the only angles of intersection possible are right angles and a continuous map would send intersections to intersections.
Based on the idea that conformal maps exist between simply connected open subsets for the prototypical spaces of dimension 1, $\log\_3(5)$ and 2 I’m willing to wager \$10 that such maps exist for all dimensions between 1 and 2. I’d be shocked if a simply connected set with Hausdorff dimension < 2 failed to have a Riemann mapping theorem analogue.
Where it starts to get interesting for me is dimensions greater than 2 but less than 3.
Does anyone know of any results in this space for sets with dimensions between 2 and 3
| https://mathoverflow.net/users/46536 | Do we have uniformization theorems for fractional dimensional spaces? | So I believe the situation is a bit complex. I think Vicsek-like fractals made from squares, cubes, hypercubes etc... all support a riemann mapping theorem and respectively have dimensions $\log\_3(5), \log\_3(7), \log\_3(9) ... \log\_3(2n+1) ... $ which grows without bound.
It seems like dimension isn't really the important thing here. It's rather the "shape of the set" which decides whether conformal mapping between open sets is possible or not. Its weird to write it but "euclidean space is quite pathological compared to vicsek like spaces and possibly other fractals killing off Riemann mapping theorems"
| 2 | https://mathoverflow.net/users/46536 | 449016 | 180,744 |
https://mathoverflow.net/questions/448982 | 2 | Fix a positive integer $r$. Describe the solutions to the system of equations given by:
$$\begin{equation}\sum\_{1\leq i\leq r}X\_i^2\equiv0\pmod{X\_k}(1\leq k\leq r)\end{equation}$$
**Example:** In the case that $r=3$, $(X\_1,X\_2,X\_3)=(1,5,13)$ is a solution, because $1^2+5^2+13^2=195=3\cdot 5\cdot 13$ is divisible by $1$, $5$ and $13$.
**Motivation:** The degrees of the irreducible characters $\text{Irr}(G)$ of a finite group $G$ satisfy the following relation:
$$\begin{equation}\sum\_{\chi\in\text{Irr}(G)}\chi(1)^2=|G|\end{equation}$$
Furthermore, for every $\chi\in\text{Irr}(G)$, $\chi(1)$ divides $|G|$. So the broader question is: is there some way to arithmetically characterize sequences of positive integers which are the irreducible degrees of some finite group?
The example I gave above does *not* qualify. No group has irreducible degrees $(1,5,13)$. Such a group would have $195$ elements, but since $195$ is a squarefree number, our hypothetical group with $195$ elements would have to be solvable. But nontrivial solvable groups must have nonprincipal linear characters, so $(1,5,13)$ cannot be its set of irreducible degrees.
(27th Aug)
**Example:**
In the case that $r=4$, $(X\_1,X\_2,X\_3,X\_4)=(1,1,2,4)$ is a solution, because $1^2+1^2+2^2+4^2=22=2\cdot 11$. But this does not correspond to the irreducible degrees of some finite group $G$. Suppose it did, so $|G|=22$. Then $G$ must be non-abelian, so $G\cong C\_{11}\rtimes C\_2$. However, every irreducible character of $C\_{11}\rtimes C\_2$ has either degree $1$ or $2$, which contradicts the assumption that $G$ has an irreducible character of degree $4$.
| https://mathoverflow.net/users/502468 | An arithmetic problem involving a system of equations | Even for $r=3$ the question isn't easy. For instance a special case is
[Markov's diophantine equation](https://en.wikipedia.org/wiki/Markov_number) $a^2+b^2+c^2=3abc$, for which one infinite family of solutions is $(1, F\_{2n-1}, F\_{2n+1})$, where $F\_i$ is the $i$th Fibonacci number.
| 2 | https://mathoverflow.net/users/18739 | 449020 | 180,745 |
https://mathoverflow.net/questions/449022 | 5 | Do complex-normed spaces exist? Is there an extension of $p$-norms to $p\in\Bbb C\setminus\Bbb R$?
A while ago I thought of extending $L^p$-spaces to the complex-normed setting. After some discussions, we came up with variations involving $$\int\_S|f|^{|p|}\arg f^p\,d\mu$$ and exponentiated versions. The argument would convey a sense of "spin", however the choice of terms seems a bit arbitrary and there wasn't any natural interpretation of it. Note that, for this specific variation, $p=i$ would get us very close to the Shannon entropy.
The only information I could find on this extension was [an unresolved MSE question](https://math.stackexchange.com/questions/3576808/are-complex-valued-p-norms-possible).
| https://mathoverflow.net/users/113397 | Is there a meaningful interpretation of an $L^i$-space? | Not only do $\def\L{{\sf L}}\L^p$-spaces make sense for all complex $p$, but their noncommutative generalizations play a crucial role in the Tomita–Takesaki modular theory.
One construction is described in the answer to the question [Is there an introduction to probability theory from a structuralist/categorical perspective?](https://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical/20820#20820).
The input data is a measurable space $(X,M)$ together with a σ-ideal $N$ of negligible (alias measure 0) sets. This is less data than a traditional measure space, since $\L^p$-spaces do not depend on the choice of a measure, only on the σ-ideal of negligible sets.
Below, it is convenient to replace $p$ by its reciprocal $1/p$. Thus, what used to be $\L^2$ is now denoted by $\L^{1/2}$ and $\L^∞$ is denoted by $\L^0$.
Now the space $\L^p(X,M,N)$ can be constructed as the quotient of the set of pairs $(f,μ)$ by an equivalence relation $\sim$.
Here $f$ is an equivalence class of bounded measurable functions on $(X,M,N)$ (modulo equality almost everywhere), $μ$ is a measure on $(X,M,N)$ (required to vanish on elements of $N$), and $μ$ is required to be finite if $\Re p>0$.
The equivalence relation is generated by $$(fg^p,μ)\sim (f,gμ),$$
where $g$ is an equivalence class of a strictly positive real measurable function on $(X,M,N)$ and $gμ$ denotes the multiplication of a measure by a function, as in the Radon–Nikodym theorem.
Equivalently, if $\Re p>0$, we can drop the boundedness condition on $f$ and finiteness condition on $μ$ and instead require that the measure $|f|^{1/\Re p}μ$ is finite. This construction produces the same $\L^p$-space because
$$(f,μ)=(\arg f\cdot (|f|^{1/\Re p})^{-\Im p}(|f|^{1/\Re p})^p,μ)\sim(\arg f\cdot (|f|^{1/\Re p})^{-\Im p},|f|^{1/\Re p}μ),$$
and in the last pair, the first component has absolute value at most 1 (hence is bounded), whereas $|f|^{1/\Re p}μ$ is a finite measure.
Here and below we write $p=\Re p+\Im p$ for the real and purely imaginary parts of $p$.
The $\L^p$-norm in this description is simply the $\Re p$-th power of the integral of $|f|^{1/\Re p}μ$.
To see that the resulting $\L^p$-space is isomorphic to the traditional $\def\LL{{\rm L}}\LL^p$-space $$\LL^p(X,M,μ)=\left\{f\biggm| \int|f|^{1/p} dμ<∞\right\},$$
observe that for any faithful measure $μ$ the map
$$f↦(f,μ)$$
yields an isomorphism $$\LL^p(X,M,μ)→\L^p(X,M,N),$$
where the right side does not depend on the choice of $μ$.
As a special case,
$\L^0(X,M,N)$ is the commutative von Neumann
algebra of equivalence classes of bounded measurable functions on $(X,M,N)$,
whereas $\L^1(X,M,N)$ is the Banach space of finite complex-valued measures on $(X,M,N)$.
The condition $f≥0$ singles out the positive subspace $\L^p\_{≥0}⊂\L^p$, and the map $$μ↦(1,μ),\qquad \L^1\_{≥0}→\L^p\_{≥0}$$
is a bijection.
Here $\L^1\_{≥0}$ can be identified with finite (positive real) measures on $(X,M,N)$.
Furthermore, given a complex number $q$, we can define the power operation $$\L^p\_{≥0}(X,M,N)→\L^{pq}(X,M,N),\qquad (1,μ)↦(1,μ),$$
using which one can show that $(f,μ)=fμ^p$, justifying the above construction and the equivalence relation $\sim$.
The Hölder inequality yields a canonical multiplication map
$$\L^p(X,M,N)⊗\L^q(X,M,N)→\L^{p+q}(X,M,N)$$
given by
$$((f,μ),(g,μ))↦(fg,μ),$$
where we used the equivalence relation $\sim$ to pick representatives for elements of $\L^p$ and $\L^q$ with the same measure $μ$ as the second component.
The induced map
$$\L^p(X,M,N)⊗\_{\L^0(X,M,N)}\L^q(X,M,N)→\L^{p+q}(X,M,N)$$
is an isomorphism, where $⊗$ denotes the usual algebraic tensor product, which happens to be automatically complete.
Thus,
$$\L^p(X,M,N)≅\L^{\Re p}(X,M,N)⊗\_{\L^0(X,M,N)}\L^{\Im p}(X,M,N),$$
so the study of $\L^p$-spaces for complex $p$ reduces to the case of real $p$, which is well-known, and the case of purely imaginary $p$.
For a purely imaginary $p$, given a choice of a faithful measure $μ$, the map $$\L^0(X,M,N)→\L^p(X,M,N),\qquad f↦fμ^p$$
is an isomorphism, with its inverse being the map $f↦fμ^{-p}$.
Thus, for traditional measure spaces, we do not see something radically new, although one can find some satisfaction in the fact that the above constructions of $\L^p$-spaces are independent of the choice of a measure, whereas the isomorphism $\L^0→\L^p$ requires such a noncanonical choice, so from a conceptual point of view, there is some value in $\L^p$-spaces for purely imaginary $p$.
All of the above generalizes to the case of noncommutative von Neumann algebras.
For a purely imaginary $p$,
the isomorphism $\L^0→\L^p$ is now only an isomorphism of *left* modules over a (noncommutative) von Neumann algebra.
Indeed, the *bimodules* $\L^0$ and $\L^p$ are *not* isomorphic for a type III factor.
This essentially the Tomita–Takesaki modular theory of von Neumann algebra, see the answer to [Making sense of "every non-commutative algebra has its own internal time evolution (aka a one-parameter group)"?](https://mathoverflow.net/questions/386464/making-sense-of-every-non-commutative-algebra-has-its-own-internal-time-evoluti/386481) for more information.
| 8 | https://mathoverflow.net/users/402 | 449026 | 180,747 |
https://mathoverflow.net/questions/449024 | 28 | Is the space of Jordan curves in $\textbf{R}^2$ contractible? In other words, is there a canonical or continuous way to deform each Jordan curve to the unit circle $\textbf{S}^1$.
If the curves are smooth one can do this via curve shortening flow/rescaling. I think this approach even works for rectifiable curves, by a [paper](https://link.springer.com/article/10.1007/s00039-013-0248-1) of Lauer. But I do not know a reference for the general topological case. In the smooth case, is there a way to do this without using flows?
By the space of Jordan curves here I mean one-to-one continuous maps from $\textbf{S}^1$ to $\textbf{R}^2$, modulo homeomorphisms of $\textbf{S}^1$. So two Jordan curves are close provided that they admit parameterizations which are pointwise close.
| https://mathoverflow.net/users/68969 | Contractibility of the space of Jordan curves | It seems that the answer is "yes".
Let us identify $\mathbb{R}^2$ with $\mathbb{S}^2\setminus\{n\}$.
Each Jordan curve $\gamma$ bounds a disc containing $n$.
This disc admits a conformal parametrization by the unit disc $\mathbb{D}$ such that its center goes to $n$.
This parametrization is unique up to rotation of $\mathbb{D}$.
In particular, the image $\gamma\_r$ of the circle of radius $r$ in $\mathbb{D}$ is completely determined by $\gamma$.
Note that there is a homotopy from that sends $\gamma$ to $\gamma\_{1/2}$.
The continuity at $t=1$ follows from Thm 15 (VIII, §81) in "Automorphic Functions" by Lester R Ford.
So the question is reduced to the smooth case, and you know it already.
| 25 | https://mathoverflow.net/users/1441 | 449029 | 180,749 |
https://mathoverflow.net/questions/449030 | 4 | *Note: All sets and functions defined below are assumed measurable. $\mu$ denotes the Lebesgue measure.*
Let $D$ be a dense subset of $[0, 1]$, and $f: D \to \mathbb R$ a function. Given $\varepsilon > 0$, say that $g: [0, 1] \to \mathbb R$ is an $\varepsilon$-almost extension of $f$ if
$$\mu(\overline{\{x \in D \, | \, g(x) \neq f(x)\}}) \leq \varepsilon.$$
**Question:** Suppose $f: D \to \mathbb R$ is continuous. Does there exist for every $\varepsilon > 0$, a continuous $\varepsilon$-almost extension of $f$?
| https://mathoverflow.net/users/173490 | Can a function that is continuous on a dense set be almost extended to a continuous function? | Let $A,B$ be disjoint countable dense subsets of $[0,1]$. Let $p:B\rightarrow(0,\infty)$ be a function where $\sum\_{b\in B}p(b)\leq 1$. Define a function $f:A\rightarrow[0,1]$ by letting $f(a)=\sum\_{b\in B,b<a}p(b)$. Then $f$ is a continuous function. Suppose now that $g:[0,1]\rightarrow\mathbb{R}$ is a continuous function. I claim that
$\{a\in A:f(a)\neq g(a)\}$ is dense in $A$ (and hence dense in [0,1]). Suppose to the contrary that $\{a\in A:f(a)\neq g(a)\}$ is not dense in $A$. Then there is an open subset $U\subseteq(0,1)$ where if $a\in A\cap U$, then $f(a)=g(a)$. Now, suppose that
$b\in B\cap U$. Then $g(b)=\lim\_{a\in A,a\rightarrow b}g(a)=\lim\_{a\in A,a\rightarrow b}f(a)$. But this limit does not exist since $\lim\_{a\in A,a\rightarrow b^-}f(a)<\lim\_{a\in A,a\rightarrow b^+}f(a)$ which is our contradiction. Therefore, $\{a\in A:f(a)\neq g(a)\}$ is dense in $A$.
We can get everything to work using Lusin's theorem if we slightly changed the problem by weakening the amount that $f,g$ have to agree with each other by just requiring that $\mu(\{x\in D:f(x)\neq g(x)\})\leq\epsilon$. If $D$ is a subset of $[0,1]$, then whenever $f:D\rightarrow[0,1]$ is continuous, we can extend $f$ to an upper semicontinuous $f\_1:[0,1]\rightarrow[0,1]$, and by Lusin's theorem, for each $\epsilon>0$ there will be a continuous $g:[0,1]\rightarrow[0,1]$ and a compact $E\subseteq[0,1]$ with $\mu([0,1]\setminus E)<\epsilon$ and $g(x)=f\_1(x)$ for $x\in E$; in particular, $g(x)=f(x)$ for $x\in E\cap D$.
| 10 | https://mathoverflow.net/users/22277 | 449032 | 180,751 |
https://mathoverflow.net/questions/449018 | 0 | Recall that a right module $M\_R$ is called **semiartinian** if every nonzero homomorphic image has nonzero socle. It's well known that the following two statements are equivalent:
1. $M$ is semiartinian.
2. Every nonzero homomorphic image of $M$ has essential socle.
It's clear that $(2)$ implies (1). But how can I prove the converse?
| https://mathoverflow.net/users/498775 | A characterization of semiartinian modules | Argue by contradiction.
Assume that $M$ is semiartinian and that $N$ is a nonzero homomorphic image of $M$ whose socle $A\_0$ is not essential. There will exist a nonzero submodule $B\leq N$ that is disjoint from $A\_0$. Enlarge the submodule $A\_0$ to a submodule $A\_1$ that is maximal for the condition $A\_1\cap B = 0$. The module $N/A\_1$ is nonzero, since
$$0\neq B\cong B/(A\_1\cap B)\cong (A\_1+B)/A\_1\leq N/A\_1.$$
$N/A\_1$ is a quotient of $N$, which is isomorphic to a quotient of $M$, so $N/A\_1$ is isomorphic to a quotient of $M$. Since $M$ is semiartinian, the nonzero quotient $N/A\_1$ must have a nonzero simple submodule. This means that $N$ has a submodule $A\_2$ covering $A\_1$ (i.e., $A\_1\lneq A\_2$, and there is no intermediate submodule $A\_1\lneq C\lneq A\_2$). By the maximality condition on $A\_1$ we have $A\_2\cap B\neq 0$ and by the modularity $\textrm{Sub}(N)$ (= the submodule lattice of $N$) we even get that $S:=A\_2\cap B$ covers $0$. Hence $S$ is a simple submodule of $N$ contained in $B$. All simple submodules of $N$ are contained in the socle $A\_0$ which is contained in $A\_1$. We get that $S\subseteq A\_1\cap B=0$. This is a contradiction, since $S\neq 0$.
| 1 | https://mathoverflow.net/users/75735 | 449036 | 180,752 |
https://mathoverflow.net/questions/448770 | 19 | $\def\RR{\mathbb{R}}$Let $\omega$ be a $2$-form on $\RR^{2n}$, where $\RR^{2n}$ has the usual Euclidean norm. The **comass** of $\omega$ is defined to be $\max\_{|u|, |v| \leq 1} \omega(u,v)$. Here is the main question:
>
> If $\omega\_1$, $\omega\_2,\ldots,\omega\_n$ are $2$-forms of comass $\leq 1$ on $\RR^{2n}$, what is the maximum possible value of
> $$|\omega\_1 \wedge \omega\_2 \wedge \cdots \wedge \omega\_n|?$$
> In particular, if $\omega\_1 = \omega\_2 = \cdots = \omega\_n = e\_1 \wedge e\_2 + e\_3 \wedge e\_4 + \cdots + e\_{2n-1} \wedge e\_{2n}$, then
> $\omega\_1 \wedge \omega\_2 \wedge \cdots \wedge \omega\_n = n! (e\_1 \wedge \cdots \wedge e\_{2n})$, is this $n!$ optimal?
>
>
>
This is a followup to previous posts on [Mathoverflow](https://mathoverflow.net/questions/448599/how-big-can-a-wedge-of-2-forms-be) and [math.SE](https://math.stackexchange.com/questions/4713029/how-big-can-a-wedge-of-2-forms-be); here are observations from those posts (plus a few more new ones):
1. Define the skew-symmetric matrix $A^k$ by $A^k\_{ij} = \omega\_k(e\_i, e\_j)$. Then the comass of $\omega\_k$ is the spectral radius (largest norm of an eigenvalue) of $A^k$. Recall that the eigenvalues of a skew symmetric matrix are of the form $\pm \lambda\_1 \sqrt{-1}$, $\pm \lambda\_2 \sqrt{-1}, \ldots, \pm \lambda\_n \sqrt{-1}$, so the condition that the comass is $\leq 1$ is equivalent to $-1 \leq \lambda\_1, \lambda\_2, \ldots, \lambda\_n \leq 1$.
2. In fact, we may assume that, for each $k$, the eigenvalues of $A^k$ are $\pm \sqrt{-1}$, see [David Speyer's comment on the math.SE post](https://math.stackexchange.com/posts/comments/9985908). In this case, the condition on $A^k$ is equivalent to saying that $A^k$ is an orthogonal matrix with $(A^k)^2 = - \text{Id}$.
3. $\omega\_1 \wedge \omega\_2 \cdots \wedge \omega\_n$ is multilinear in the coefficients $A^k\_{ij}$ of the skew-symmetric matrices. Specifically, it is the unique symmetric multilinear polynomial $P(A^1, A^2, \ldots, A^n)$ in $n$ skew-symmetrix matrices such that $P(A, A, \ldots, A)$ is $n! \operatorname{Pfaffian}(A)$. If $A$ has eigenvalues $\pm \lambda\_1 \sqrt{-1}, \ldots, \pm \lambda\_n \sqrt{-1}$, then $\operatorname{Pfaffian}(A) = \pm \lambda\_1 \lambda\_2 \cdots \lambda\_n$, so, in the case $A^1 = A^2 = \cdots = A^n$, equality holds exactly when all the eigenvalues are $\pm \sqrt{-1}$.
4. Previous posts have proved the $n!$ bound in the cases $n=2$ and $n=3$. There is also an easy $n^n$ bound; see [David Speyer's comment on the previous MO post](https://mathoverflow.net/posts/comments/1159176).
5. A speculation from David Speyer: If $A$ is a skew symmetric matrix with eigenvalues $\pm \lambda\_1 \sqrt{-1}, \ldots, \pm \lambda\_n \sqrt{-1}$, then $\sum\_{i<j} A\_{ij}^2 = \sum \lambda\_i^2$. I wonder whether the $n!$ bound might continue to be correct if all that we assume about $A^1, A^2, \ldots, A^n$ is that $\sum\_{i<j} (A^k\_{ij})^2 \leq n$. (Note that $\sum\_{i=1}^n \lambda\_i^2 \leq n$ implies $\prod \lambda\_i \leq 1$ by AM-GM, so the case $A^1 = A^2 = \cdots = A^n$ still works with this weaker hypothesis.) The case of $n=2$ and two distinct two-forms also works.
A related article appeared [here](https://doi.org/10.1007/s00022-023-00685-3).
| https://mathoverflow.net/users/28128 | How big can a wedge of $n$ 2-forms in $\mathbb{R}^{2n}$ be? | Yes! Here is a proof.
Define the norm $|\omega|$, for a $p$-form on $\mathbb R^m$, to be the least upper bound of $|\omega(v\_1,\dots ,v\_p)|$ for vectors $v\_i$ of length one. For any $p$ and $q$ we can ask for the smallest number $C$ such that for every $p$-form $\alpha$ and every $q$-form $\beta$ on $\mathbb R^m$ we have $|\alpha\wedge \beta|\le C|\alpha||\beta|$. Call it $C\_{p,q}$. It does not depend on $m$, because if $C$ is valid for forms in $\mathbb R^{p+q}$ and if $v\_1,\dots,v\_{p+q}$ span $V\cong \mathbb R^{p+q}$ in $\mathbb R^m$ then
$$
|(\alpha\wedge \beta)(v\_1,\dots ,v\_{p+q})|\le C|\alpha\_V||\beta\_V||v\_1|\dots |v\_{p+q}|\le C|\alpha||\beta||v\_1|\dots |v\_{p+q}|,$$
where $\alpha\_V$ and $\beta\_V$ are the restrictions to $V$.
I claim that $C\_{2,2n-2}\le n$. This implies that for $2$-forms $\omega\_1,\dots ,\omega\_n$ we have $|\omega\_1\wedge \dots \wedge \omega\_n|\le n|\omega\_1\wedge \dots \wedge \omega\_{n-1}||\omega\_n|$, and therefore by induction on $n$ we have $|\omega\_1\wedge \dots \wedge \omega\_n|\le n!|\omega\_1|\dots |\omega\_n|$.
To prove the claim, first observe that $C\_{p,q}$ has this other interpretation: it is the smallest $C$ such that the inner product of two $p$-forms on $\mathbb R^{p+q}$ always satisfies $\alpha\cdot \beta\le C|\alpha||\beta|$. This is so because the norm satisfies $|\beta|=|\ast\beta|$ (Hodge star) and $|\alpha\wedge \ast\beta|=|\alpha\cdot\beta|$.
So a restatement of the claim to be proved is that for $2$-forms on $\mathbb R^{2n}$ we have $\alpha\cdot\beta\le n|\alpha||\beta|$. This can be seen by making a change of orthogonal basis so that
$$\alpha = a\_1e\_1\wedge e\_2+\dots a\_ne\_{2n-1}\wedge e\_{2n}$$
for some $a\_1,\dots ,a\_n$. Write
$$\beta=b\_1e\_1\wedge e\_2+\dots b\_ne\_{2n-1}\wedge e\_{2n}+\dots ,$$
where the other terms will not matter. Then
$$|\alpha|=max|a\_i|$$
$$|\beta|\ge max |b\_i|$$
$$\alpha\cdot\beta=\sum\_i a\_ib\_i\le n\ max|a\_ib\_i|\le n|\alpha||\beta|$$.
I did not say why this norm is invariant under Hodge star. Suppose that $\beta$ is a $p$-form in $\mathbb R^{p+q}$. The claim is that $|(\ast \beta)(v\_1,\dots,v\_q)|\le |\beta||v\_1|\dots |v\_q|$. Without loss of generality the vectors $v\_i$ are orthogonal and of length one. Complete them to an orthonormal basis by vectors $w\_1,\dots ,w\_p$. Then
$$
|(\ast \beta)(v\_1,\dots,v\_q)|= |\beta(w\_1,\dots ,w\_p)|=|\beta|.$$
| 14 | https://mathoverflow.net/users/6666 | 449042 | 180,753 |
https://mathoverflow.net/questions/449049 | 3 | $\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$I am trying to find the commensurator of $\SL\_{2}(\mathbb{Z})$ on $\GL\_{2}^+(\mathbb{R})$.
So far I have been able to prove that $\GL\_{2}^+(\mathbb{Q})$ is included in the commensurator by looking at the congruence subgroups, and my intuition tells me that this inclusion could in fact be an equality, but I am not able to prove it.
Is there a way to prove the other inclusion, or to find (in case it is not $\GL\_2^+(\mathbb{Q})$) the commensurator in another way?
By $\GL\_2^+$ I mean invertible and positive determinant matrices.
| https://mathoverflow.net/users/507084 | Commensurator of $\mathrm{SL}_2(\mathbb{Z})$ on $\mathrm{GL}_2^+(\mathbb{Q})$ | It's not harder to compute the commensurator in $\mathrm{GL}\_2(\mathbf{C})$. Since it contains the scalars, it is enough to compute the commensurator in $\mathrm{SL}\_2(\mathbf{C})$.
If a matrix $A=\begin{pmatrix}a & b\\c & d\end{pmatrix}$ of determinant 1 commensurates $\mathrm{SL}\_2(\mathbf{Z})$, then for some $n>0$, the matrices $A(I+nE\_{12})A^{-1}$, $A(I+nE\_{21})A^{-1}$, $A^{-1}(I+nE\_{12})$, $A^{-1}(I+nE\_{21})$ are all integral. Doing the computation, this implies that $a^2$, $b^2$, $c^2$, $d^2$, $ab$, $ac$, $cd$, $bd$ are all rational. One easily deduces that $A$ is a scalar multiple of a rational matrix.
Conversely, every scalar multiple of a rational matrix commensurates $\mathrm{SL}\_2(\mathbf{Z})$.
For your precise question, the answer is then $\{\lambda A:\lambda\in\mathbf{R}\_{>0},A\in\mathrm{SL}\_2(\mathbf{Q})\}$
| 5 | https://mathoverflow.net/users/14094 | 449055 | 180,755 |
https://mathoverflow.net/questions/449054 | 1 | What are classical/simple examples of smooth projective surfaces $S$ (over $\mathbb C$) for which all the $1$-forms have a common zero (and $h^{1,0}(S)>0$)?
| https://mathoverflow.net/users/85595 | reference quest: surface whose $1$-forms have a common zero | If $S$ admits a global holomorphic 1-form without zeros, $S$ would admit a $C^\infty$ real closed 1-form which has no zeros, equivalently, $S$ would admit a $C^\infty$ fibre bundle structure over the circle.
In this way, the Euler number $\chi(S)$ and the signature $\sigma (S)$ vanish: see [Holomorphic one-forms, fibrations over the circle, and characteristic numbers of Kähler manifolds](https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/holomorphic-oneforms-fibrations-over-the-circle-and-characteristic-numbers-of-kahler-manifolds/DAC4529082DBF6F70474970ACFA70426), Lemma 2.
So a simple choice may be a surface with either $\chi(S)$ or $\sigma (S)$ non-vanishing, and $h^{1,0}(S)=1$. A concrete example may be the [Cartwright-Steger surface](https://www.maths.usyd.edu.au/u/donaldc/cs-surface/cs-surface.pdf). See also [Enumeration of the 50 fake projective planes](https://www.sciencedirect.com/science/article/pii/S1631073X09003987).
| 2 | https://mathoverflow.net/users/125498 | 449057 | 180,756 |
https://mathoverflow.net/questions/449053 | 5 | Let $X$ be a connected compact metric space.
**Question:** Is there a sequence of compact hyperbolic surfaces (the curvature may differ between surfaces) that converges to $X$ in the Gromov-Hausdorff metric?
| https://mathoverflow.net/users/125498 | Can hyperbolic surfaces approximate every connected compact metric space? | The answer is "yes".
Moreover, given an integer $n\geqslant 2$, there a sequence of compact hyperbolic $n$-manifolds that converges to $X$ in the sense of Gromov--Hausdorff.
See V. Šahović. “Approximations of Riemannian manifolds with linear curvature constraints.” Thesis. Univ. Münster, 2009.
| 7 | https://mathoverflow.net/users/1441 | 449058 | 180,757 |
https://mathoverflow.net/questions/448956 | 5 | Crossposted from [MSE](https://math.stackexchange.com/questions/4690683/expositions-of-the-carlitz-scoville-vaughan-theorem-in-combinatorics):
I recently came across Ira Gessel's [slides](https://people.brandeis.edu/%7Egessel/homepage/slides/csv.pdf) on a theorem he says should "be considered one of the fundamental theorems of enumerative combinatorics."
**The Carlitz-Scoville-Vaughan Theorem:** Let $A$ be an alphabet, and let $R$ be a relation on $A$, that is, a
subset of $A × A = A^2 $. Let $A^{(R)}$ be the set of words $a\_1 · · · a\_n$ in
$A^∗$ such that $a\_1 R a\_2 R · · · R a\_n$ . Note that the empty word 1
and all words of length one are in $A^{(R)}$ . Let $\bar{R} = A^2 − R.$
Then $$\sum\_{w\in A^{(\bar{R})}}w = \left(\sum\_{w\in A^{({R})}}(-1)^{l(w)}w \right)^{-1}$$
Here $l(w)$ is the length of $w,$ and we are working in the ring of
formal power series in noncommuting variables.
This does look like a pretty powerful theorem; it seems to yield combinatorial interpretations to several important sequences and yet I can't find anything like this in the standard enumerative combinatorics books, which suggests I'm not looking in the right places.
So I have the following questions:
1. What is this theorem saying, intuitively? How could one have been led naturally to this? How would I recognize when it makes sense to apply it?
2. Where can I find other reasonably elementary expositions of this or similar results? Other than these slides and some references in a couple of PhD theses I can't seem to find it. Does it usually go by another name? What other techniques are used to prove the results that can be proved by this theorem?
| https://mathoverflow.net/users/506963 | Intuitive explanations of the Carlitz-Scoville-Vaughan theorem | This is more of an extended comment than an answer, but maybe you will find it helpful. I agree with [Sam Hopkins](https://mathoverflow.net/questions/448956/intuitive-explanations-of-the-carlitz-scoville-vaughan-theorem#comment1160085_448956) that it is fruitful to think of the Carlitz–Scoville–Vaughan (CSV) theorem as a combinatorial reciprocity theorem; it is called that in Jair Taylor's 2014 *Electronic Journal of Combinatorics* paper, [Counting Words with Laguerre Series](https://doi.org/10.37236/3500), for example. Curiously, though, it does not seem to be mentioned in Matthias Beck and Raman Sanyal's book, *[Combinatorial Reciprocity Theorems](https://bookstore.ams.org/view?ProductCode=GSM/195)*, so I think there is room for someone to explain more thoroughly the connection between CSV and other combinatorial reciprocity theorems.
I first encountered CSV when working on my Ph.D. thesis. One of my main results was a combinatorial reciprocity theorem relating the directed paths in a directed graph $D$ with the directed paths in its complement $\overline D$. A corollary was a simple formula (which I called the [rook reciprocity theorem](https://doi.org/10.37236/1234)) relating the rook numbers of complementary boards; decades ago, Riordan had written down an equivalent formula, but the reciprocity point of view led to a much nicer formulation. Ira Gessel was on my thesis committee, and quickly recognized that directed graphs are equivalent to relations, and he came up with a different proof—using CSV—of (a special case of) my reciprocity result. You can find Gessel's proof starting on page 35 of my Ph.D. thesis, [Symmetric function generalizations of graph polynomials](https://arxiv.org/abs/math/0103229).
Marcelo Aguiar and his collaborators have suggested approaching combinatorial reciprocity theorems by way of Hopf algebras. The reciprocity theorem in my thesis, like many other reciprocity theorems, lives in the world of symmetric functions. It takes a combinatorially significant symmetric function, applies the involution $\omega$, and interprets the result combinatorially. The involution $\omega$ is essentially the antipode of the Hopf algebra of symmetric functions. In their paper [Hopf monoids and generalized permutahedra](https://arxiv.org/abs/1709.07504), Marcelo Aguiar and Federico Ardila show how to derive various combinatorial reciprocity theorems using the Hopf algebra point of view. If you are not familiar with Hopf algebras, they may seem very abstract, but a lowbrow way of phrasing the main idea is that if you have some combinatorial structures that compose (the product) and decompose (the coproduct) in a "compatible" way, then chances are there will be a combinatorial reciprocity theorem (the antipode) available. But again, Aguiar and Ardila don't mention CSV. I'd be interested in seeing whether CSV can be interpreted in terms of the antipode of some Hopf algebra.
| 3 | https://mathoverflow.net/users/3106 | 449063 | 180,760 |
https://mathoverflow.net/questions/449060 | 6 | Let $G$ be a Galois group and $H$ be its normal subgroup. Let $M$ be a $G$-module.
Consider the restriction map $res: H^1(G,M) \to H^1(H,M)$. Its kernel is given by $H^1(G/H,M^H)$.
On the other hand, there exists the corestriction map $cores: H^1(H,M) \to H^1(G,M)$.
What is known about the kernel of $cores$? Are there any known results that explicitly describe the kernel or embed it into certain cohomology?
**I am particularly interested in the following case**: Let $K$ be a quadratic number field. Consider $G=Gal(\overline{K}/\Bbb{Q})$ and $H= Gal(\overline{K}/K)$. Let $E$ be an elliptic curve defined over $\Bbb{Q}$. Let $M = E(\overline{K})$. For the restriction map, we can express $ker(res)$ as $H^1(Gal(K/\Bbb{Q}), E(K))$ as mentioned above. However, I am facing difficulty in controlling $ker(cores)$.
| https://mathoverflow.net/users/144623 | Ker of corestriction of Galois cohomology | (Not sure why this question comes up naturally. The more interesting question, and the one analogue to the kernel of restriction, is to ask what is the cokernel of corestriction. That turns up a lot. Like in class field theory where the cokernel of the norm map is a central question, while the kernel of the norm map is large and not of the same level of interest.)
The Tate spectral squence (Cohomology of number fields, printed version, Theorem 2.1.11) involves the corestriction map (loc cit Theorem 2.1.12). The resulting five term exact sequence answers the question. Let $K/k$ be a **finite** extension with Galois group $G$ and let $M$ be a Galois module for $\operatorname{Gal}(\bar k/k)$ whose order is a unit in $k$. Then this is the dual of the said 5-term exact sequence:
$$
0 \leftarrow H\_1\bigl(G, H^2(K,M)\bigr) \leftarrow
H^1(k,M)\overset{\operatorname{cor}}{\leftarrow} H\_0\bigl(G,H^1(K,M)\bigr)
\leftarrow H\_2\bigl(G,H^2(K,M)\bigr) \leftarrow H^0(k,M)
$$
This holds if $H^i(k,M)=0$ for all $i>2$.
If $k$ is local, then this condition holds for all $M$ as $\operatorname{scd}(k)=2$ by Corollary 7.2.5.
In the case of a global field, I would switch to the restricted cohomology with respect to a finite set $S$ of places such that $K/k$ and $M$ are unramified outside $S$ and $S$ contains all places above primes dividing the order of $M$ and all infinite places.
Write $G\_S(k)$ for the Galois group of the maximal extension of $k$ which is unramified outside $k$.
One obtains the above 5-terms sequence with cohomology with respect for $G\_S(K)$ and $G\_S(k)$ as long as $H^i\bigl(G\_S(k), M)=0$ for all $i>2$.
This condition holds for all finite $M$ as $\operatorname{cd}(G\_S(k))=2$, neglecting issues for $p=2$ by adding the assumption that $k$ has no real places.
One expects that $\operatorname{scd}(G\_S(k))=2$ and this is known if the Leopoldt conjecture holds for $k$ as explained in Corollary 10.3.8.
For an elliptic curve $E$ defined over $k$, there is a surjective map $H^3\bigl( G\_S(k), E[m]\bigr) \to H^3\bigl( G\_S(k), E\bigr)[m]$ for all $m$. Hence for odd $m$ we also have the required vanishing of $M=E(\bar k)$ and for even $m$ one needs to add the hypothesis that $k$ has no real place, to impose local conditions at these infinite places.
For infinite $G$, one has to replace the inner cohomology groups by limits taken over corestriction maps for finite subextensions as in loc.cit.
(All references in Cohomology of number fields are to the printed version on my desk, the numbering of theorems in the online version may be different.)
| 7 | https://mathoverflow.net/users/5015 | 449064 | 180,761 |
https://mathoverflow.net/questions/449048 | 7 | Let $A$ be an associative algebra over $\mathbb{C}$ with irreducible finite-dimensional representations on $V$ and $W$. Then is the tensor product of representations on $V \otimes W$ semi-simple?
The below question is concerning representations of a group and in this case the answer is positive due to a result of Chevalley. I tried looking in the literature but I could not find any equivalent results for representations of an algebra.
[Tensor product of simple representations](https://mathoverflow.net/questions/57997/tensor-product-of-simple-representations)
| https://mathoverflow.net/users/165204 | Tensor product of irreducible representations of an algebra | By a famous theorem of R. Steinberg, STEINBERG, R. Complete sets of representations of algebras. PROC. Am. Math. Soc. 13 (1962), 746-747, if $V$ is a faithful representation for a monoid $M$, then $T(V)=\bigoplus\_{n\geq 0}V^{\otimes n}$ is a faithful representation for the monoid algebra $KM$ for any field $K$. If $M$ is finite, then $KM$ is finite dimensional and so it follows easily that $\bigoplus\_{n=0}^rV^{\otimes n}$ is a faithful $KM$-module for some $r$.
So you can get a counterexample by taking a finite monoid $M$ with a faithful complex semisimple representation which does not have a semisimple complex algebra. There are many such examples. Here is an easy one.
Consider $M$ the monoid of $3\times 3$ complex matrices consisting of zero, the identity, all matrix units $E\_{ij}$ and the nilpotent matrix $\begin{bmatrix}0 & 1 &0\\
0 &0 &1 \\ 0 & 0 &0 \end{bmatrix}$. Then $\mathbb CM$ has a faithful irreducible representation $V$, namely the one I used to define it. Irreducibility follows since all the matrix units are in the image of the representation, and so the map $\mathbb CM\to M\_3(\mathbb C)$ is sujective.
But $\mathbb CM$ does not have a semisimple algebra. If you factor out the ideal of $\mathbb CM$ spanned by the matrices of rank at most 1, you get an algebra with a nilpotent ideal (spanned by the coset of the nilpotent matrix). So $\mathbb CM$ has a nonsemisimple quotient and hence is not semisimple. It follows that not all tensor powers of $V$ are semisimple by R. Steinberg's theorem.
**Added.** I think already $V\otimes V$ is not semisimple. It contains as a submodule the exterior power $\Lambda^2(V)$. This is annihilated by all the rank $\leq 1$ matrices but not by the nilpotent matrix. So the image of $\mathbb CM$ under this exterior power representation is not semisimple (it has a nilpotent ideal spanned by the image of the nilpotent matrix) and so this subrepresentation is not semisimple and hence $V\otimes V$ is not semisimple.
**Added.** Here is a two-dmensional example. Take the monoid consting of the 2x2 identity matrix and the matrices $\pm \begin{bmatrix} 1 & 1\cr 0 & 0\end{bmatrix}, \pm \begin{bmatrix} 1 & -1\cr 0 & 0\end{bmatrix}, \pm \begin{bmatrix} 0 & 0\cr 1 & 1\end{bmatrix}, \pm \begin{bmatrix} 0 & 0\cr 1 & -1\end{bmatrix}$. It generated as a monoid by $\begin{bmatrix} -1 & -1\cr 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0\cr 1 & -1\end{bmatrix}$. It is easy too see there is no invariant subspace since these rank $1$ matrices span all $2\times 2$ matrices. So this is an irreducible represention $V$.
But $V\otimes V$ has a nonsemisimple submodule. The vectors $e\_1\otimes e\_1$ and $e\_2\otimes e\_2$ span an invariant subspace with $e\_1\otimes e\_1-e\_2\otimes e\_2$ spanning an invariant subspace $W$ with no complement. A complement would have to be fixed by all of the matrices (since $W/\mathbb C(e\_1\otimes e\_1-e\_2\otimes e\_2)$ is the trivial module) but no such vector exists.
| 10 | https://mathoverflow.net/users/15934 | 449067 | 180,762 |
https://mathoverflow.net/questions/449052 | 1 | Take the language $\mathcal L(=,\in)\_{\omega\_1,\omega\_1}$, if we restrict infinite quantification strings, i.e. of the forms $``(\forall v\_i)\_{i \in \omega}"; ``(\exists v\_i)\_{i \in \omega}"$, to bounded forms, that is $``\forall v\_0 \in x, \forall v\_1 \in x,..."$ and $``\exists v\_0 \in x , \exists v\_1 \in x,..."$ abbreviated as $``(\forall v\_i \in x)\_{i \in \omega}"; ``(\exists v\_i \in x)\_{i \in \omega}"$ respectively, or even if we further demand the bound to be a definable constant or the value of a function. Call the resulting language as "*bounded*-$\mathcal L(=,\in)\_{\omega\_1, \omega\_1}$", and denote it by $\mathcal L(=,\in)\_{\omega\_1, [\omega\_1]}$
>
> How would $\mathcal L(=,\in)\_{\omega\_1, [\omega\_1]}$ differ from $\mathcal L(=,\in)\_{\omega\_1, \omega\_1}$ as regards basic properties of logic like Completeness, Compactness, etc..
>
>
>
As a clear example, if to $\mathcal L(=,\in)\_{\omega\_1, [\omega\_1]}$ we add all axioms of $\sf ZFC$ written, as usual, in the fragment $\mathcal L(=,\in)\_{\omega,\omega}$ and suppose we add an axiom of Foundation but in a bounded form, that is:
$\textbf{Foundation: } \\\forall \alpha \, (\forall v\_i \in V\_\alpha)\_{i \in \omega} \, \exists x: \bigvee\_{i \in \omega} (x=v\_i) \land \bigwedge\_{i \in \omega} (v\_i \not \in x)$
Would that theory be different from theory written in $\mathcal L(=,\in)\_{\omega\_1,\omega\_1}$, axiomatized by $\sf ZFC$ axioms in $\mathcal L(=,\in)\_{\omega,\omega}$ , and Foundation written as above but unbounded.
| https://mathoverflow.net/users/95347 | Is bounded-$\mathcal L_{\omega_1,\omega_1}$ significantly different from $\mathcal L_{\omega_1,\omega_1}$? | Your new formulation of foundation is equivalent to the corresponding unbounded form. The reason is that every failing instance of the unbounded formulation will lead to a failing instance of the bounded formulation, since if there is a countable collection of $v\_i$ with no $\in$-minimal element, then by the usual ZFC axioms it follows that $v\_0$ is in some $V\_\alpha$, and so the $v\_i$ that appear within this $V\_\alpha$ will fall under the bounded form, and an $\in$-minimal for that part of the family will be $\in$-minimal for the whole family since $V\_\alpha$ is transitive.
| 5 | https://mathoverflow.net/users/1946 | 449073 | 180,764 |
https://mathoverflow.net/questions/449068 | 5 | In almost every introductory notes on [Tits buildings](https://en.wikipedia.org/wiki/Building_(mathematics)) these are motivated
as structures capturing/ sharing several features of [symmetric spaces](https://en.wikipedia.org/wiki/Symmetric_space). Could somebody elaborate what are precisely the main structural similarities between buildings and
symmetric spaces motivating the vievpoint that buildings can
be regarded in some sense as generalizations of symmetric spaces?
(so is there something more 'substantial' beyond the sloppy slogan
that both objects carry 'high degree
of symmetry'; so what are the 'characteristic' features buildings and
symmetric spaces have in common in deeper sense?)
The central construction in the theory of buildings is certain simplicial complex $\Sigma= \Sigma(W,S)$ for appropriate Coxeter group $W$ ( the Weyl group of the buildings) with generators S gouverning the symmetry of the complex. Is there a kind of 'Weyl group' which naturally reflects the symmetry of symmetric spaces?
| https://mathoverflow.net/users/108274 | Buildings as generalizations of symmetric spaces | (Wanted to post as a comment but didn't have enough reputation)
I have a partial answer only for your first question.
Let $F$ be a non-Archimedean field. For p-adic(or $\pi-$adic) symmetric space $\Omega^r \subset \mathbb{P}^{r-1}\_F$(now also called Drinfeld period domain), Drinfeld found a very clear relation between the symmetric space and the Bruhat-Tits building for $GL\_r(F)$, $\mathcal{BT}^\bullet$(Tits building being its boundary) via a certain *reduction map* $$\rho:\Omega^r \rightarrow |\mathcal{BT}^\bullet|$$ (|.| means the geometric realization of the simplicial complex) and he used this map to construct an admissible covering of the symmetric space s.t. the nerve of the covering is the barycentric subdivision of the $\mathcal{BT}^\bullet$. You can see this in Drinfeld's seminal paper on "Elliptic modules"(Section 6) or also in Chapter 3 of
*Deligne, Pierre; Husemoller, Dale H.*, Survey of Drinfel’d modules, Current trends in arithmetical algebraic geometry, Proc. Summer Res. Conf., Arcata/Calif. 1985, Contemp. Math. 67, 25-91 (1987). [ZBL0627.14026](https://zbmath.org/?q=an:0627.14026).
If you want to know a specific relationship of the symmetric space with the Tits building, then one such is contained in the work of P. Schneider and U. Stuehler(in section 3)
*Schneider, P.; Stuhler, U.*, [**The cohomology of (p)-adic symmetric spaces**](https://doi.org/10.1007/BF01232257), Invent. Math. 105, No. 1, 47-122 (1991). [ZBL0751.14016](https://zbmath.org/?q=an:0751.14016).
contains some results relating the cohomology of symmetric space and certain generalized Tits building .
| 6 | https://mathoverflow.net/users/495875 | 449078 | 180,766 |
https://mathoverflow.net/questions/445569 | 2 | Is it easy to construct examples of smooth complex projective surfaces $S$ of general type such that $h^0(\Omega\_S)>1$, $alb\_S:S\rightarrow Alb(S)$ is generically finite (unto its image) and
$${\rm Bs}(\Omega\_S):=\{x\in S,\ \omega\_x(T\_xS)=0 \ \forall \omega\in H^0(\Omega\_S)\}$$
is non-empty?
| https://mathoverflow.net/users/85595 | Base locus of cotangent bundle of a surface | I am just posting my comments as one answer.
Let $A$ be an Abelian variety of dimension three. Denote by $\Theta$ an ample divisor class on $A$. Let $S’$ be a smooth effective Cartier divisor on $A$ in the algebraic equivalence class of $n\Theta$ for very positive $n$. Let $f:S\to S’$ be the blowing up of $S’$ at a point $p$, precomposed with a blowing up of a point on the exceptional divisor over $p$. Denote the exceptional locus of $f$ by $E$. It is a union of two rational curves that intersect at a point $q$.
By direct computation, the pullback map is an isomorphism from $\Omega\_{S’}|\_p$ to $H^0(E,\Omega\_S|\_E)$. Thus every $1$-form on $S$ agrees on $E$ with the pullback of a unique $1$-form on $S’$. The difference is a $1$-form that is zero along $E$. This is the same as a global section of the coherent sheaf $\Omega\_{S’}$ that vanishes at $p$.
By Serre vanishing and duality, both $h^1(A,\mathcal{O}(-S’))$ and $h^2(A,\mathcal{O}(-2S’))$ equal zero. Thus, the conormal sheaf of $S’$ also has vanishing $h^1$. Thus, by the fundamental short exact sequence, every global section of $\Omega\_{S’}$ lifts uniquely to a global section on $S’$ of $\Omega\_A|\_{S’}$. Because $h^1(A,\Omega\_A(-S’))$ is zero, this lifts further to a unique global section on $A$ of $\Omega\_A$. Of course the only global section of this free $\mathcal{O}\_A$-module that vanishes at $p$ is the zero section. Altogether, every global $1$-form on $S$ is the pullback of a unique global $1$-form on $A$. Of course all of these vanish at $q$.
| 0 | https://mathoverflow.net/users/13265 | 449087 | 180,769 |
https://mathoverflow.net/questions/449059 | 3 | Consider a discrete time Markov chain $(X\_t)$ on a finite state space $\mathcal{S}$, with transition matrix $P$. Assume that the chain admits a stationary distribution $\pi$, which I will identify with its probability mass function $(\pi(i))\_{i\in \mathcal{S}}$. In this case I know that, given any function $f:\mathcal{S}\to\mathbb{R}$, the following ergodic theorem holds
$$\mathbb{P}\left(\lim\_{n\to\infty}\frac{1}{n}\sum\_{t=0}^n f(X\_t)=\sum\_{i\in\mathcal{S}}f(i)\pi(i)\right)=1$$
I would be interested to know whether, and under which hypotheses, the result generalize to the case when:
* $\mathcal{S}$ is a compact metric space (in particular $\mathcal{S}=[0,1]$). Of course, here the chain is described by a transition kernel $P:\mathcal{S}\to\Delta(\mathcal{S})$ (with the notation $\Delta(\mathcal{S})$ I mean the space of probability distributions on $\mathcal{S}$) and its stationary distribution, if it exists, is $\pi\in\Delta(\mathcal{S})$ such that $\pi=\int\_{\mathcal{S}}P(s)d\pi(s)$
* Instead of considering the time average of a function $f:\mathcal{S}\to\mathbb{R}$, we consider the time average of the (independent) realizations of a family of random variables whose law depends on the state of the chain, namely a Markov kernel $\Phi:\mathcal{S}\to\Delta(\mathbb{R})$.
It makes sense to ask that, for all $s\in\mathcal{S}$ it holds $\int\_{\mathbb{R}}xd\Phi(s)(x)<\infty$, so that each variable in the family has finite expectation. Denote by $\phi(s)$ a realization of $\Phi(s)$.
My **question** is: (under what hypotheses) does it still hold that
$$\mathbb{P}\left(\lim\_{n\to\infty}\frac{1}{n}\sum\_{t=0}^n \phi(X\_t)=\int\_\mathcal{S}\left(\int\_{\mathbb{R}}xd\Phi(s)(x)\right)d\pi(s)\right)=1$$
**Observation**: what I am able to observe if a version of the ergodic theorem holds for continuous state spaces, then, trivially, defining $F:\mathcal{S}\to\mathbb{R}$ such that $F(s)=\int\_{\mathbb{R}}xd\Phi(s)(x)$, my statement of interest is true if
$$\lim\_{n\to\infty}\frac{1}{n}\sum\_{t=0}^n \phi(X\_t)=\lim\_{n\to\infty}\frac{1}{n}\sum\_{t=0}^n F(X\_t)$$
This seems to point to a relation between the kernels $P$ and $\Phi$. But I don't really know which.
Any help or reference would be greatly appreciated. I beg your pardon in advance in case my notation is not precise, and I am willing to clarify it.
| https://mathoverflow.net/users/143913 | "Ergodic theorem" for Markov kernels | I understand your setup in the following way (I will somewhat modify your notation). One is given a
measurable state space $S$ and two "kernels" $\{\pi\_s\}, \{\varphi\_s\}$ (i.e., families of probability measures indexed by $S$ and subject to the usual measurability conditions). The measures $\pi\_s$ are on $S$, and the measures $\{\varphi\_s\}$ are on $\mathbb R$. One runs the Markov chain $(S\_t)$ with the transition probabilities $\{\pi\_s\}$ and at each moment $t$ independently samples a random variable $X\_t$ from the distribution $\varphi\_{S\_t}$. The question is about the ergodic averages of the sequence $(X\_t)$ under the assumption that the chain $(S\_t)$ has a stationary probability measure $\pi$.
First of all, since you are asking about convergence of ergodic time averages to the spatial average, one has to assume that the measure $\pi$ is ergodic (otherwise one has to pass to its ergodic decomposition). [Of course, this assumption has to be imposed in the case of a finite state space as well.]
An answer to your question follows from the observation that the sequence $(S\_t,X\_t)$ is also Markov with the transition probabilities
$$
\widetilde \pi\_{s,x} = \int \delta\_\sigma\otimes\varphi\_\sigma\,d\pi\_s(\sigma) \;.
$$
It has a stationary measure $\widetilde\pi$ obtained by the same lifting from the stationary measure $\pi$
$$
\widetilde\pi = \int \delta\_{\sigma}\otimes\varphi\_\sigma\,d\pi(\sigma) \;.
$$
The key point is that ergodicity of $\pi$ implies that of $\widetilde\pi$ (it follows, for instance,
from the characterization of ergodicity in terms of absence of non-constant bounded harmonic functions in combination with the that the harmonic functions of the lifted chain depend on the first coordinate only, and therefore are just the lifts of the harmonic functions of the original chain). Thus, one just has to apply the usual ergodic theorem for the time shift on the path space of the lifted chain to the function $F(s,x)=x$ of the time 0 state.
The final comment is that this question is purely measure theoretical and does not require any continuity or compactness assumptions.
| 2 | https://mathoverflow.net/users/8588 | 449091 | 180,770 |
https://mathoverflow.net/questions/449094 | 0 | It is known since Bochner that for M compact with negative Ricci curvature, the group of isometries is discreet and hence finite. Are there any generalizations to compact non-positive sectional curvature manifolds? Of course, $T^n$ is a counterexample for strictly 0 curvature case. But what about sectional curvature <0 on a non-empty open set, and vanishing elsewhere?
As I understand Bochner's argument does break down in this case.
Known counterexamples?
| https://mathoverflow.net/users/16877 | Isometries of manifolds with non-positive sectional curvature | Bochner's theorem extends to nonpositive Ricci to give:
>
> If $(M,g)$ is compact and has $\textrm{Ric}\leq 0$ then any Killing vector $X$ is parallel and $\textrm{Ric}(X,X) = 0$.
>
>
>
See Petersen (3rd ed) Theorem 8.2.2.
Thus if $(M,g)$ is compact and has $\textrm{Ric}\leq 0$ with $<0$ then $X = 0$.
| 3 | https://mathoverflow.net/users/1540 | 449097 | 180,772 |
https://mathoverflow.net/questions/449075 | 4 | I am investigating a function defined in terms of the singular values of matrices. Initially, I simplified the problem by focusing on the eigenvalues of $2 \times 2$ Hermitian, positive-definite matrices.
For given $p, q \geq 1$, the function $N\_{(p,q)}$ is defined as follows:
$$ N\_{(p,q)} (A) := \inf \left\lbrace \epsilon > 0 :\left(\frac{a}{\epsilon}\right)^p + \left(\frac{b}{\epsilon}\right)^q \leq 1 \right\rbrace.$$
Here, $a$ and $b$ denote the singular values of $A$, with $a$ being the largest. How can one show that this function satisfies the triangle inequality?
I am particularly interested in connections to Horn's inequalities, as they concern the singular values of sums of matrices and seem highly relevant to the problem at hand. Furthermore, the underlying function $$\Theta\_{(p,q)} (a,b) = a^p + b^q$$ is a convex modular functional, which is pertinent.
I am seeking guidance or references that could provide some insight into this problem. Has anyone encountered a similar problem or does anyone have suggestions on potential approaches?
| https://mathoverflow.net/users/105743 | A potential new norm for matrices and Horn's inequalities | By the standard classification of unitarily invariant norms (see e.g., [this blog post](https://nhigham.com/2021/02/02/what-is-a-unitarily-invariant-norm/)), the expression $N\_{(p,q)}$ is a norm if and only if the function
$$ \| (x,y) \| := \inf \left\{ \varepsilon > 0: \left(\frac{\max(|x|, |y|)}{\varepsilon}\right)^p + \left(\frac{\min(|x|, |y|)}{\varepsilon}\right)^q \leq 1 \right\}$$
is a norm, or equivalently if the set
$$ S := \{ (x,y): \max(|x|,|y|)^p + \max(|x|,|y|)^q \leq 1 \}$$
is convex (certainly it is symmetric and has non-empty interior). In the upward sector $\{ (x,y): |x| \leq y \}$, this set is convex from the convexity of $y^p+|x|^q$ already observed by the OP, and at the endpoint $(a,a)$ of the set on the upper right edge of the sector, where $a>0$ is the unique positive solution to the equation $a^p+a^q=1$, implicit differentiation of the condition $y^p + |x|^q = 1$ reveals that the boundary of this set has a slope of $-\frac{q a^{q-1}}{p a^{p-1}}$ at this point, which is less than or equal to $-1$ iff we have the condition
$$ p a^p \geq q a^q. \quad (1)$$
By symmetry along the y-axis, at the opposite endpoint $(-a,a)$, the slope is at most $+1$ iff the same condition (1) holds. As $S$ is symmetric across the diagonals $x=y$ and $x=-y$, we conclude that the entire set $S$ is convex iff (1) holds, and so $N\_{(p,q)}$ is a matrix norm iff (1) holds also.
To obtain an explicit counterexample to the triangle inequality when (1) fails: observe that $\mathrm{diag}(a,a)$ is in the convex hull of the matrices $\mathrm{diag}(a+\delta,a-\delta)$, $\mathrm{diag}(a-\delta,a+\delta)$ for any $0 < \delta < a$, so by symmetry and the triangle inequality one sees that if one wishes $N\_{(p,q)}$ to be a norm, one needs
$$ N\_{(p,q)}(\mathrm{diag}(a,a)) \leq N\_{(p,q)}(\mathrm{diag}(a+\delta,a-\delta))$$
which is equivalent to
$$ (a+\delta)^p + (a-\delta)^q \geq a^p + a^q = 1.$$
Taking $\delta$ to be small and performing a Taylor expansion, we see that this fails when (1) fails for $\delta$ small enough.
To see when (1) holds, we make the change of variables $a^p = \theta$, $a^q = 1-\theta$ for some $0 < \theta < 1$, then
$$ \frac{p}{q} = \frac{\log \theta}{\log(1-\theta)}$$
and the condition (1) rearranges to
$$ \theta \log \theta \leq (1-\theta) \log (1-\theta).$$
This turns out to hold precisely when $\theta \geq 1/2$, or equivalently $p \geq q$ (the difference $\theta \log \theta - (1-\theta) \log (1-\theta)$ is odd, vanishes at $\theta=1/2,1$, and concave for $1/2 \leq \theta \leq 1$).
| 8 | https://mathoverflow.net/users/766 | 449103 | 180,774 |
https://mathoverflow.net/questions/449089 | 3 | I have a Banach manifold $\mathcal{M}$ and I have a Lie group $G$, that is finite dimensional, such that $G$ acts freely on $\mathcal{M}$. I would like to know if $\mathcal{M} / G$ is a Banach manifold as well or not.
I know that this is true if $\mathcal{M}$ is a finite dimensional smooth manifold, and we have the quotient manifold theorem (for example in Lee's book). But I'm struggling to come up with a reference that discussess the infinite dimensional case. Can someone point me to a source that discusses these issues? I looked at Lang's book "Fundamentals of Differential Geometry", but did not find this discussed (assuming I did not miss something).
| https://mathoverflow.net/users/151406 | Quotient by freely acting group on Banach manifold | In finite dimensions, properness of the action is all you need for a slice theorem (and thus for the manifold structure of the quotient). However, in the Banach realm, properness is not enough. For example, the translation action of $c\_0 \subseteq l\_\infty$ of sequences converging to zero on the Banach space $l\_\infty$ of
bounded sequences is proper but $c\_0$ doesn't have a topological complement (which would be the tangent space of the slice).
If I may reference my own work, in [Slice theorem and orbit type stratification in infinite
dimensions](https://arxiv.org/pdf/1812.04698.pdf) we discuss these and other problems in infinite dimensions. Theorem 3.5 gives a general slice theorem (which even holds for manifolds modeled on locally convex spaces, not necessarily Banach ones). It requires four conditions, which in the Banach setting you can verify as follows:
* the stabilizer is a principal Lie subgroup of the acting group: this follows from Theorem IV.3.16. of Neeb's [Towards a Lie theory of locally convex groups](https://link.springer.com/article/10.1007/s11537-006-0606-y). This is stated without proof there and references the still unpublished book by Neeb and Glöckner; I'm sure they will send you their existing manuscript.
* The orbit is a locally closed submanifold: use the inverse function theorem (cf Proposition 3.17 which discusses the way harder case of tame Fréchet manifolds)
* Existence of a local addition and invariant topological metric: if you have a strong invariant Riemann metric, then this should be automatic. If not, you need some other additional structure that provides you a "local linearization" (e.g. equivariantly embed your manifold in a Banach space with a linear action).
I think Bourbaki "Lie groups and Lie algebras" also discusses the Banach case, but I don't have it hand right now to double check.
| 5 | https://mathoverflow.net/users/17047 | 449112 | 180,775 |
https://mathoverflow.net/questions/448944 | 2 | The dihedral group $D\_8$ can be presented as $(\mathbb{Z}\_2\times \mathbb{Z}\_2)\rtimes \_{\rho}\mathbb{Z}\_2$, where the last factor acts on $\mathbb{Z}\_2\times \mathbb{Z}\_2$ as
$$
\rho\_1(a,b)=(b,a) \ .
$$
I am interested in explicit 3-cocycle of the group cohomology $H^3(D\_8,\mathbb{R}/\mathbb{Z})$. Let me be more explicit about the presentation I would like to see, and which part of the group-cohomology I want to represent. Given a manifold $X$ a class in $H^1(X,\mathbb{D}\_8)$ (the $H^1$ is the only one existing with non-abelian coefficients) can be decomposed into classes $A\in H^1(X,\mathbb{Z}\_2)$ and $(B\_1,B\_2)\in H\_{\rho}^1(X,\mathbb{Z}\_2\times \mathbb{Z}\_2)$ (this denotes the cohomology with local coefficients, understanding $\mathbb{Z}\_2\times\mathbb{Z\_2}$ as a $\mathbb{Z}\_2$ module for $\rho$). I know that there exist a (unique) order-two class in $H^3(D\_8,\mathbb{R}/\mathbb{Z})$ which when pulled back using $A, (B\_1,B\_2)$ produces
$$
A\cup B\_1\cup B\_2
$$
where the cup product is the standard one for $\mathbb{Z}\_2$. I think this is essentially the class corresponding to the subgroup $H^1\_{\rho}(\mathbb{Z}\_2,H^2(\mathbb{Z}\_2\times \mathbb{Z}\_2,\mathbb{R}/\mathbb{Z}))\cong \mathbb{Z}\_2$. Does this make sense?
I look for a generalization replacing $\mathbb{Z}\_n\times \mathbb{Z}\_n$ with $n$ even. $\rho$ still acts as $\rho\_1(a,b)=(b,a)$ and there is a group $(\mathbb{Z}\_n\times \mathbb{Z}\_n)\rtimes \_{\rho}\mathbb{Z}\_2$. A natural guess is to take the $\mathbb{Z}\_2\times \mathbb{Z}\_2$ subgroup and thus write something like
$$
A\cup(n/2) B\_1\cup (n/2)B\_2
$$
but I think this is not correct, since this is actually the pull back of a class of the direct product $\mathbb{Z}\_n\times \mathbb{Z}\_n\times \mathbb{Z}\_2$.
If it would be easier it is also fine to provide the class algebraically in group-cohomology.
| https://mathoverflow.net/users/495347 | Explicit 3-cocycle of group cohomology of dihedral group and generalization to other semidirect products | I think I found a solution to my question for any semidirect product $G\_1\rtimes \_{\rho}G\_2$, $G\_1,G\_2$ abelian. Essentially I was looking for the class of $H^3(B G\_1\rtimes \_{\rho}G\_2,\mathbb{R}/\mathbb{Z})$ corresponding to the subgroup $H^1\_{\rho}(BG\_2,H^2(BG\_1,\mathbb{R}/\mathbb{Z}))$. A class in this group is a twisted homomorphsim $\mu: G\_1\rightarrow H^2(BG\_1,\mathbb{R}/\mathbb{Z})$, where the target can be understood as the group of alternating bicharacters on $G\_1$. Given $A\in H^1(X,G\_2)$ one can check that $\mu(A) \in H^1\_{\rho}(X,H^2(BG\_1,\mathbb{R}/\mathbb{Z}))$, and it can be used to construct a sensible cup product which gives a class $\mu(A)\cup B\cup B\in H^3(X,\mathbb{R}/\mathbb{Z})$ for any $B\in H^1\_{\rho}(X,G\_1)$
| -1 | https://mathoverflow.net/users/495347 | 449116 | 180,776 |
https://mathoverflow.net/questions/449106 | 2 | The von Mises distribution has the highest entropy for a given first circular moment. It seems that the converse is true: for a fixed entropy, the von Mises distribution has the highest first circular moment (in magnitude). This seems like something that should already be proven somewhere, but I haven't been able to find it, nor have my efforts at proving it met with success so far. Does anyone know where a proof of this is (or failing that, how to prove it)?
| https://mathoverflow.net/users/507129 | von Mises Distribution property | Yes, this is true.
Indeed, the first [circular moment](https://en.wikipedia.org/wiki/Von_Mises_distribution#Moments) for a probability density $f$ on the interval $[0,2\pi]$ is
$$m\_1(f):=\int\_0^{2\pi}e^{ix}f(x)\,dx.$$
So,
$$\begin{aligned}&|m\_1(f)| \\
&=\max\_{u\in[0,2\pi]}\Big(\cos u\,\int\_0^{2\pi}\cos x\,f(x)\,dx
+\sin u\,\int\_0^{2\pi}\sin x\,f(x)\,dx\Big) \\
&=\max\_{u\in[0,2\pi]}\int\_0^{2\pi}\cos(x-u)\,f(x)\,dx.
\end{aligned}$$
So, a density $f$ maximizing $|m\_1(f)|$ given the entropy $H:=-\int\_0^{2\pi}f\ln f$ is maximizing
$$I\_u(f):=\int\_0^{2\pi}\cos(x-u)\,f(x)\,dx$$
for some $u\in[0,2\pi]$, again given the entropy. Using Lagrange multipliers, we see that for such a maximizer $f$ we have
$$\rho\cos(x-u)=\lambda\ln f(x)+\mu$$
for some real $\rho,\lambda,\mu$ with $\rho^2+\lambda^2+\mu^2>0$ and almost all (a.a.) $x\in[0,2\pi]$. So, for some real $a$ and $b$ and a.a. $x\in[0,2\pi]$ we have $\ln f(x)=a\cos(x-u)+b$ and hence $f(x)=ce^{a\cos(x-u)}$ for $c:=e^b$. Thus, $f$ is a [von Mises density](https://en.wikipedia.org/wiki/Von_Mises_distribution#Definition). $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 449124 | 180,778 |
https://mathoverflow.net/questions/449114 | 1 | Consider two [multilinear](https://en.wikipedia.org/wiki/Multilinear_polynomial) Polynomials $A(x\_1,x\_2,x\_3,\dotsc,x\_n)$ and $B(x\_1,x\_2,x\_3,\dotsc,x\_n)$ of $n > 2$ variables $x\_i \in \mathbb{R}$ and their ratio
\begin{equation\*}
F(x\_1,x\_2,x\_3,\dotsc,x\_n) = \frac{A(x\_1,x\_2,x\_3,\dotsc,x\_n)}{B(x\_1,x\_2,x\_3,\dotsc,x\_n)}
\end{equation\*}
Define $G$ to be the function obtained from $F$ by substituting $(x\_1,x\_2)=(t,-t)$ and maximizing over $t$:
\begin{equation\*}
G(x\_3,\dotsc,x\_n) := \max\_t F(t,-t,x\_3,\dotsc,x\_n)
\end{equation\*}
Here the maximum is assumed to exist in a local sense, i.e. assume there is a function
$t = f(x\_3,\dotsc,x\_n)$ that locally parametrizes the solution set of
\begin{equation}
\frac{\partial}{\partial t} F(t,-t,x\_3,\dotsc, x\_n)=0.
\end{equation}
such that we have
\begin{equation}
G(x\_3,\dotsc,x\_n) := F(f(x\_3,\dotsc,x\_n),-f(x\_3,\dotsc,x\_n),x\_3,\dotsc, x\_n)
\end{equation}
**Question**: Can $G$ be written as
\begin{equation\*}
G(x\_3,\dotsc,x\_n) = \frac{\widetilde{A}(x\_3,\dotsc,x\_n)}{\widetilde{B}(x\_3,\dotsc,x\_n)}
\end{equation\*}
with multilinear polynomials $\widetilde{A},\widetilde{B}$ of the remaining $n-2$ variables?
| https://mathoverflow.net/users/507148 | Maximizing the ratio of multilinear polynomials | Not in general. E.g., let $n=3$, $A(x\_1,x\_2,x\_3):=1-x\_1 x\_2+2 x\_3 x\_2+x\_1 x\_3$, and $B(x\_1,x\_2,x\_3):=2-x\_1 x\_2+x\_3 x\_2+2 x\_1 x\_3$. Then for $x\_3\in(0,2\sqrt2)$
$$G(x\_3)=\max\_{t\in\mathbb R}\frac{A(t,-t,x\_3)}{B(t,-t,x\_3)}
=\frac{x\_3^2+2 \sqrt{6 x\_3^2+1}+6}{8-x\_3^2},$$
which is not the ratio of multilinear polynomials in $x\_3$.
| 1 | https://mathoverflow.net/users/36721 | 449126 | 180,779 |
https://mathoverflow.net/questions/449128 | 3 | It is well-known that Chi-squared distribution $X\_n$ with degree-$n$ freedom has an approximate formula for its median as $n\left(1-\frac{2}{9n}\right)^3$. Or $(X\_n/n)^{\frac{1}{3}}$ is approximately normally distributed with mean $1-\frac{2}{9n}$ and variance $\frac{2}{9n}$. But unfortunately I cannot find any rigorous proof for this approximation formula.
For example I have read the original 1931 article <https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1076144/pdf/pnas01728-0064.pdf> here but the approach is not very mathematically rigorous.
I wonder if there are any solid work estimating the *difference* between the median of $X\_n$ and $n\left(1-\frac{2}{9n}\right)^3$ in terms of $n$ *explicitly*. Or if there are any quantitative result for measuring *how close* between $(X\_n/n)^{\frac{1}{3}}$ and normal distribution with mean $1-\frac{2}{9n}$ and variance $\frac{2}{9n}$.
| https://mathoverflow.net/users/489992 | Quantitative results (with formal proof) on the median approximation of Chi-squared distribution | Such an approximation follows by the [Cornish--Fisher asymptotic expansion](https://en.wikipedia.org/wiki/Cornish%E2%80%93Fisher_expansion) for the quantiles of a probability distribution.
The error estimate can be obtained by noting that here the Cornish--Fisher asymptotic expansion can be obtained by inverting the [Edgeworth asymptotic expansion](https://en.wikipedia.org/wiki/Edgeworth_series#The_Edgeworth_series) for the distribution function of the sum of i.i.d. random variables.
---
I have finally got around to supplying the straightforward but tedious details.
Recall that $\chi^2\_n$ is the distribution of the sum of $n$ i.i.d. copies of $Z^2$, where $Z\sim N(0,1)$. So, if $S\_n\sim\chi^2\_n$,
we can use the Edgeworth asymptotic expansion for the distribution function, say $F\_n$, of $S\_n$ as e.g. presented (with an explicit expression for the remainder) in Theorem 1 of Section VI.3 of [Petrov's book](https://link.springer.com/book/10.1007/978-3-642-65809-9), so that
\begin{equation\*}
F\_n(x)=\Phi(z)+\phi(z)\sum\_{j=1}^{k-2}P\_j(z)n^{-j/2}+O(n^{-(k-1)/2}) \tag{10}\label{10}
\end{equation\*}
uniformly in real $x$, where $z:=\dfrac{x-n}{\sqrt{2n}}$;
$\Phi$ and $\phi$ are the c.d.f. and p.d.f. of $N(0,1)$; $k$ is any integer $\ge3$; and the $P\_j$'s are certain polynomials with coefficients depending on the moments of the underlying distribution (in our case, the moments of the distribution of $Z^2$).
In our case, after long calculations, we have \eqref{10} with $k=7$ and
\begin{equation\*}
\begin{aligned}
P\_1(z)&=\frac{\sqrt 2(1-z^2)}3, \\
P\_2(z)&=-\frac{ z (3 - 11 z^2 + 2 z^4)}{18}, \\
P\_3(z)&=\frac{\sqrt{2} \left(3+69 z^2-399 z^4+145 z^6-10 z^8\right)}{810}, \\
P\_4(z)&=\frac{z \left(135-1035 z^2+7947 z^4-4167 z^6+560 z^8-20 z^{10}\right)}{9720}, \\
P\_5(z)&=-\frac{\sqrt{2}}{204120}\,(3375+3105 z^2-16470 z^4+142218 z^6 \\
&\qquad\qquad\quad-94941 z^8+18501 z^{10}-1288 z^{12}+28
z^{14}),
\end{aligned}
\tag{20}\label{20}
\end{equation\*}
with a remainder $O(n^{-(7-1)/2})=O(n^{-3})$.
The median, say $m\_n$, of $\chi^2\_n$ is the unique root $x$ of the equation $F\_n(x)=1/2$. Using now \eqref{10} and \eqref{20}, expanding in powers of $n^{-1/2}$, and letting
\begin{equation\*}
m\_{n;4}:=n+\sum\_{j=-1}^4 c\_j n^{-j/2}
\end{equation\*}
for some real constants $c\_j$, we see that
\begin{equation\*}
m\_n=m\_{n;4}+O(n^{-5/2})
\end{equation\*}
iff
\begin{equation\*}
(c\_{-1},\dots,c\_4)=\Big(0,-\frac23,0,\frac{32}{405},0,\frac{1472}{25515}\Big),
\end{equation\*}
so that
\begin{equation\*}
m\_{n;4}=n-\frac{2}{3}+\frac{32}{405 n}+\frac{1472}{25515 n^2}.
\end{equation\*}
The approximation of the median $m\_n$ in question was
\begin{equation\*}
\tilde m\_{n;4}:=n\Big(1-\frac 2{9n}\Big)^3=n-\frac{2}{3}+\frac{4}{27 n}-\frac{8}{729 n^2}.
\end{equation\*}
We see that already the coefficient of $n^{-1}$ in $\tilde m\_{n;4}$ is not the best possible one. So, there can hardly be a solid reason for the estimate $\tilde m\_{n;4}$ of the median $m\_n$.
The approximation $m\_{n;4}$ of the median $m\_n$ is much better than $\tilde m\_{n;4}$ even for $n$ as small as $5$ or $10$. E.g.,
\begin{equation\*}
m\_{5;4}-m\_5\approx-0.0000167\quad\text{versus}\quad \tilde m\_{5;4}-m\_5\approx0.0111,
\end{equation\*}
\begin{equation\*}
m\_{10;4}-m\_{10}\approx-6.28\times10^{-6}\quad\text{versus}\quad \tilde m\_{10;4}-m\_{10}\approx0.00622.
\end{equation\*}
---
For somewhat related results, see [this paper](https://www-sciencedirect-com.services.lib.mtu.edu/science/article/pii/S0167715220303308?via%3Dihub) and references there. (Recall that the chi-squared distribution is a subspecies of the gamma family of distributions).
| 4 | https://mathoverflow.net/users/36721 | 449129 | 180,781 |
https://mathoverflow.net/questions/449096 | 7 | Note 1. Early I posted a related question [Set-theoretic tautologies](https://mathoverflow.net/questions/176732/set-theoretic-tautologies). But the answer did not contain any concrete references to the literature. So I posted this, more precisely formulated question, hoping to receive a concrete reference to the literature.
In any case, I am thankful to François G. Dorais and Emil Jeřábek for their answers.
Note 2. Because the 2-element Boolean algebra $\mathbf2$ is isomorphic to the powerset algebra on a singleton $V$, we can use the set-theoretic
notations $\emptyset, V, \cup, \cap, C, \subseteq$ instead of normally used in $\mathbf2$ the notations $0,1,+,.,',\leq$.
Let $B$ be a power set Boolean algebra.
Let $P$ be an unquantified formula constructed using variables and the symbols $\emptyset, V$ (the universe),
$\cup, \cap, C, \subseteq, =$ (union, intersection, complement, inclusion, equality, respectively) such as, for example, $A \subseteq A \cup X$.
[It is well known](https://mathoverflow.net/questions/176732/set-theoretic-tautologies), that such a formula is valid in $B$ iff it is valid in the 2-element Boolean algebra $\mathbf2$.
So instead of proving $P$ in $B$, we can check that $P$ is true in $\mathbf2$ - by checking $2^n$ possible cases ($n$ is the number of variables in $P$) substituting instead of each variable $\emptyset$ or $V$ and calculating the result, using the obvious rules:
$\emptyset \cup \emptyset = \emptyset$,
$\emptyset \cup V = V$,
$V \cup \emptyset = V$,
$V \cup V = V$ and so on for $\cap, C, \subseteq$.
If the result is $\top$ in all cases then $P$ is true in $\mathbf2$ and, therefore, in $B$.
Note that this method also works for some unquantified formulas containing also some propositional connectives - in addition, after calculating all set terms in $P$ we should check that the resulting formula(not containing variables) is true.
For example, $A \subseteq B \rightarrow A \cup X \subseteq B \cup X$ is true in $\mathbf2$ and $B$.
But there are some formulas(for example, $X \subseteq Y \lor Y \subseteq X$) that are valid in $\mathbf2$ but not in $B$.
So my question is:
What is the widest class of unquantified formulas in $B$ containing variables, the above set-theoretic symbols and
propositional connectives, which are valid simultaneously in $\mathbf2$ and in $B$?
| https://mathoverflow.net/users/5761 | Formulas that are valid simultaneously in a power set Boolean algebra $B$ and the 2-element Boolean algebra $\mathbf2$ | The class of formulas you're looking for contains all quasi-identities:
$$a\_1 = b\_1 \land \cdots \land a\_k = b\_k \to a = b$$
where $k \geq 0$ and $a,a\_1,\ldots,a\_k,b,b\_1,\ldots,b\_k$ are terms formed using $\varnothing,V,{\complement},{\cap},{\cup}$. (We can omit ${\subseteq}$ because $a \subseteq b \iff a \cap b = a$, for example.) I believe these may be exactly the class you're looking for, up to logical equivalence modulo the theory of Boolean algebras.
Some general context: The "well-known fact" from the question is a consequence of [Birkhoff's HSP Theorem](https://en.wikipedia.org/wiki/Variety_(universal_algebra)#Birkhoff%27s_theorem) which says that a nonempty class of algebras is closed under homomorphisms (H), subalgebras (S) and (arbitrary) products (P) if and only if it is axiomatized by identities (i.e., $a = b$ where $a,b$ are terms). There is a generalization of this theorem to quasi-identities which says that a nonempty class of algebras is closed under isomomorphisms (I), subalgebras (S), products (P) and ultraproducts (U) if and only if it is axiomatized by quasi-identities.
A first-order formula $\phi$ (in the language $\varnothing,V,{\complement},{\cap},{\cup},{=}$) with property ISP will have the required property that $\phi$ is valid in $\mathbf{2}$ iff it is valid in every powerset algebra. The forward direction follows from IP since the powerset algebras are isomorphic to powers of $\mathbf{2}$. The reverse direction follows from IS because every nontrivial Boolean algebra contains an isomorphic copy of $\mathbf{2}$.
Note that every first-order formula has property IU by [Łoś's Theorem](https://en.wikipedia.org/wiki/Ultraproduct#%C5%81o%C5%9B%27s_theorem), and every quantifier-free formula has property S. So perhaps the quasi-identities are what you're looking for?
---
*Addendum*: What if the powerset algebras are not meant to include the trivial Boolean algebra? Then the above can be modified to include [Horn formulas](https://en.wikipedia.org/wiki/Horn_clause) of the form
$$a\_1 = b\_1 \land \cdots \land a\_k = b\_k \to \bot$$
as well, provided that property P is restricted to *nonempty* products everywhere.
However, this is not a very substantial extension since
$$a\_1 = b\_1 \land \cdots \land a\_k = b\_k \to V = \varnothing$$
is a quasi-identity which is equivalent to the above for all nontrivial Boolean algebras.
| 7 | https://mathoverflow.net/users/2000 | 449138 | 180,783 |
https://mathoverflow.net/questions/449135 | 1 | While studying the "[coin-flipping degree](https://mathoverflow.net/questions/448538/bounds-on-the-coin-flipping-degree)" problem I have come across the following conjecture. It gives bounds on the power coefficients of a polynomial that maps the unit interval to itself. If true, this could contribute to solving the "coin-flipping degree" problem.
**Conjecture:** *Let $p(\lambda)=a\_0 \lambda^0 + ... + a\_n\lambda^n$ be a polynomial that maps the closed unit interval to itself and satisfies $0\lt p(\lambda)\lt 1$ whenever $0\lt\lambda\lt 1$. Then $|a\_i|\le |b\_i|$, where $b\_i$ is a power coefficient of the following polynomial: $$q(\lambda) = b\_0 \lambda^0 + ... + b\_n\lambda^n = (T\_n(1-2\lambda)+1)/2,$$ and where $T\_n(x)$ is the Chebyshev polynomial of the first kind of degree $n$.*
Has this conjecture been studied or proved anywhere? If not, is the conjecture true?
| https://mathoverflow.net/users/171320 | Coefficient bounds for polynomials that map the unit interval to itself | This is correct. If $0=c\_0<c\_1<\ldots<c\_n=1$ are extrema of the polynomial $q$, that is, $q(c\_j)=(-1)^j$ for $j=0,1,\ldots,n$, you may interpolate $p$ at nodes $c\_0,\ldots,c\_n$ to get
$$
p(x)=\sum\_{j=0}^n p(c\_j)\frac{\prod\_{k\ne j}(x-c\_k)}{\prod\_{k\ne j}(c\_j-c\_k)},
$$
that gives for the coefficient $a\_m$ (for $m=0,1,\ldots,n$) the formula
$$
a\_m=\sum\_{j=0}^n p(c\_j)\frac{[x^m]\prod\_{k\ne j}(x-c\_k)}{\prod\_{k\ne j}(c\_j-c\_k)}=\sum\_{j=0}^n p(c\_j)\cdot \alpha\_{m,j},
$$
where the sign of $\alpha\_{m,j}(-1)^j$ is the same for all $j=0,1,\ldots,n$. Thus $|a\_m|$ under constraints $|p(c\_j)|\leqslant 1$ is maximal for $p=q$.
| 4 | https://mathoverflow.net/users/4312 | 449148 | 180,786 |
https://mathoverflow.net/questions/449139 | -4 | Let $F(g(f))$
be a functional that sends functions of a vector variable (from $n$-dimensional vector space) to $\mathbb R$.
$g$
is some function of scalar valued functions $f$.
I'm interested in a coordinate free calculation of the first and second derivatives of functionals on the spaces of real valued functions of a vector variable, where the vector space dimension can be more than 3.
It can be [shown](https://en.m.wikipedia.org/wiki/Functional_derivative#Formula) using vector calculus that the first (order) functional derivative $\partial F$
is equal to the partial derivative of $g$
with respect to $f$
minus the divergence of the gradient of $g$.
\begin{gather\*}
F=\int {f(x)} ^{4} dx^{4} \\
\partial F= \frac{\partial{g}}{\partial{f}} - \nabla \cdot\nabla g(f).
\end{gather\*}
Divergence of the gradient should be equal to $4$, and if I plug in the above functional into the formula on Wikipedia it should work just as well for 4d and I get the result: $ \partial F=−12f^{3}$.
Can I just plug this into the formula again and get the second derivative of the functional?
| https://mathoverflow.net/users/492578 | Coordinate free computation of the second derivative of a functional | For
$$
F=\int d^4 z \ f^4 (z) \ ,
$$
the first functional derivative is
$$
\frac{\delta F}{\delta f(x)} = \int d^4 z \ 4 f^3 (z)\ \delta^4 (x-z)
=4 f^3 (x)
$$
and the second functional derivatives are
$$
\frac{\delta^{2} F}{\delta f(x) \delta f(y)} = 12 f^2 (x) \ \delta^4 (x-y)
$$
| 0 | https://mathoverflow.net/users/134299 | 449153 | 180,788 |
https://mathoverflow.net/questions/449123 | 5 | Let $k$ be an algebraically closed field of characteristic $p>0.$ How can I construct two projective curves $C\_1,C\_2$ of genus $ g \geq 2$ so that the abelianizations $\pi\_1(C\_i)^{ab},i=1,2$ are isomorphic, but $\pi\_1(C\_1) \not \equiv \pi\_1(C\_2)?$
One could try to construct curves $C\_1$ and $C\_2$ and try to arrange it so that a degree $n$ cover has different abelianization, but I failed to write something concrete.
| https://mathoverflow.net/users/472750 | Two curves of genus $g \geq 2$ in characteristic $p >0 $ with isomorphic abelianizations | See
*Nakajima, Shoichi*, On generalized Hasse-Witt invariants of an algebraic curve, Galois groups and their representations, Proc. Symp., Nagoya/Jap. 1981, Adv. Stud. Pure Math. 2, 69-88 (1983). [ZBL0529.14016](https://zbmath.org/?q=an:0529.14016).
The classical Hasse-Witt invariant, $\gamma$, is the rank of the $p$-torsion in the Jacobian of $C$, it is an integer between $0$ and $g$. We always have
$$\pi\_1(C)^{\text{ab}} \cong \prod\_{\ell \neq p} \mathbb{Z}\_{\ell}^{2g} \times \mathbb{Z}\_p^{\gamma}.$$
So two curves (both over an algebraically closed field of characteristic $p$) have isomorphic $\pi\_1(C)^{\text{ab}}$ if and only if they have the same genus and same Hasse-Witt invariant.
The "generalized Hasse-Witt" invariants of this paper control covers of the curve with Galois group $C\_n \ltimes C\_p^m$ where $n|p^m-1$. (Here $C\_n$ is the cyclic group of order $n$.)
In particular, in Section 6, the author considers the genus $2$ curve
$$y^2+y = Ax + \frac{B}{x} + \frac{C}{x+1} \qquad ABC \neq 0$$
over a field of characteristic $2$. This curve always has $g=\gamma = 2$, but the author shows that the $C\_3 \ltimes C\_2^2$ covers (in other words, $A\_4$ covers) can be either $40$, $39$ or $38$, according to the values of $(A,B,C)$.
---
I discovered this paper while researching how to compute a similar answer I thought of; I'll record how that answer would work. Let $p \neq 2$ and let $X$ be a curve of genus $g$. Then $X$ has a unique $C\_2^{2g}$ cover, call that cover $Y$. Then $X$ will have a cover with Galois group $C\_2^{2g} \ltimes C\_p^r$ iff $r$ is less than or equal to the Hasse invariant of $Y$. (How $C\_2^{2g}$ acts on $C\_p^r$ will be determined by how $C\_2^{2g}$ acts on the $p$-torsion in $J(Y)$, but all I'll need is the rank.) I therefore set out to find two curves $X\_1$ and $X\_2$ with the same genus and Hasse-Witt invariant, but where the corresponding covers $Y\_1$ and $Y\_2$ have different Hasse-Witt invariant.
Suppose that $X$ is hyperelliptic of the form $y^2 = \prod\_{i=1}^{2g+2} (x-\alpha\_i)$. For $S$ any subset of $\{ \alpha\_1, \alpha\_2, \ldots, \alpha\_{2g+2} \}$ of cardinality $2k+2$, let $C\_S$ be the genus $k$ hyperelliptic curve $y^2 = \prod\_{\alpha\_i \in S} (x-\alpha\_i)$. I get that $J(Y)$ is isogenous to $\prod\_S J(C\_S)$.
In particular, I took $p=11$ and the two genus $2$ curves
$$X\_1 := \{y^2 = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)\}$$
and
$$X\_2 := \{ y^2 = (x-1)(x-2)(x-3)(x-4)(x-5)(x-7) \}.$$
I compute that both of these are ordinary (Hasse-Witt invariant $2$). However, of the $\binom{6}{4}$ genus $1$ curves of the form $C\_S$ in the product above, I compute that $11$ of them are ordinary in the $X\_1$ case and only $10$ are for $X\_2$.
Thus, $X\_1$ should have a $C\_2^{4} \ltimes C\_{11}^{11+2}$ cover, but $X\_2$ should not.
| 5 | https://mathoverflow.net/users/297 | 449158 | 180,789 |
https://mathoverflow.net/questions/449147 | 5 | A function $F: \mathbb R\_+ \rightarrow \mathbb R\_+$ is said to be long tailed if $F(\infty)=0$ and for all $y \geq 0$ $$\frac{F(x+y)}{F(x)} \rightarrow 1, \quad x\rightarrow \infty.$$
Let $\mu$ be a measure on $[0, \infty)$ with full support and denote $f(x)=\mathcal L(\mu,x)$ its Laplace transform. I am trying to figure out whether or not it is true that $f$ is long tailed however given $\mu$.
* If $\mu (x) \sim x^\gamma g(x) dx$, $\gamma>-1$ and $g$ analytic in 0, $g(0)\neq 0$ then there should be no problem since we can essentially invoke Watson's lemma, and then $f(x) \sim g(0)\Gamma(\gamma+1)x^{-\gamma-1}$.
* There are also things like $\mu(x)=e^{-1/x}dx$. Then $f(x) =2 K\_1(2 \sqrt{x})/\sqrt{x}$ with $K\_\nu$ the modified Bessel function, and $f(x) \sim \sqrt{\pi}e^{-2 \sqrt{x}}x^{-3/4}$. Even though the decay is faster than every polynomial order the long tail property is verified.
Purely discrete measures are ruled out by definition, I would initially be happy with the absolutely continuous case only (or a counterexample).
| https://mathoverflow.net/users/97651 | Long tail property of Laplace transforms | Yes, this is true, assuming that
\begin{equation\*}
f(s):=\int\_{[0,\infty)} e^{-sx}\mu(dx)<\infty
\end{equation\*}
for some real $s\_0$ and all real $s\ge s\_0$. As in the OP, we are also assuming that $\mu$ is a measure on $[0, \infty)$ with full support. We want to show that then for each real $y\ge0$ we have
\begin{equation\*}
r\_y(s):=\frac{f(s-y)}{f(s)}\to1 \tag{10}\label{10}
\end{equation\*}
(as $s\to\infty$).
To do that, take any real $h>0$ and write
\begin{equation\*}
f(s)=I\_1(s)+I\_2(s),
\end{equation\*}
where
\begin{equation\*}
I\_1(s):=I\_{h,1}(s):=\int\_{[0,h)} e^{-sx}\mu(dx),\quad I\_2(s):=I\_{h,2}(s):=\int\_{[h,\infty)} e^{-sx}\mu(dx).
\end{equation\*}
So, for real $s\ge s\_0$,
\begin{equation\*}
r\_y(s)\le\frac{I\_1(s-y)}{I\_1(s)}\Big(1+\frac{I\_2(s-y)}{I\_1(s-y)}\Big). \tag{20}\label{20}
\end{equation\*}
Next,
\begin{equation\*}
I\_1(s-y)=\int\_{[0,h)} e^{yx}e^{-sx}\mu(dx)\le e^{yh}I\_1(s) \tag{30}\label{30}
\end{equation\*}
and, for real $s\ge s\_0$,
\begin{equation\*}
I\_2(s)\le e^{-(s-s\_0)h}f(s\_0)=Ce^{-sh},
\end{equation\*}
where $C:=e^{s\_0h}f(s\_0)\in[0,\infty)$. On the other hand,
\begin{equation\*}
I\_1(s)\ge \int\_{[0,h/2)} e^{-sx}\mu(dx)\ge ce^{-sh/2},
\end{equation\*}
where $c:=\int\_{[0,h/2)} \mu(dx)>0$, since $\mu$ is a measure on $[0,\infty)$ with full support. So, for real $s\ge s\_0$ we have $I\_2(s)\le \frac Cc\, I\_1(s)\,e^{-sh/2}$.
So, $\dfrac{I\_2(s-y)}{I\_1(s-y)}\to0$ (as $s\to\infty$).
It now follows from \eqref{20} and \eqref{30} that $L\_y:=\limsup\_{s\to\infty}r\_y(s)\le e^{yh}$ for any real $h>0$, and hence $L\_y\le1$ for any real $y\ge0$. On the other hand, the function $f$ is nonincreasing and hence $r\_y(s)\ge1$ for any real $y\ge0$ and $s\ge s\_0$. Thus, \eqref{10} follows. $\quad\Box$
---
It is easily seen from the above proof that the condition that $\mu$ is a measure on $[0, \infty)$ with full support can be relaxed to the condition that $0$ is in the support of $\mu$.
| 4 | https://mathoverflow.net/users/36721 | 449160 | 180,790 |
https://mathoverflow.net/questions/449107 | 4 | Let $\Omega$ be a bounded domain, $f\in L^{\infty}(\Omega),$ and $0\leq u\in H\_0^1(\Omega)$ is a non-negative solution of $\Delta u=f$. My question is as follows:
* Can we conclude that $u\in C^0(\bar{\Omega})$ without any boundary smoothness assumption on $\Omega$? If not, could you please provide a counterexample?
By employing De Giorgi iteration and the Harnack inequality, we can establish that $u$ belongs to $L^{\infty}(\Omega)$ and exhibits Hölder continuity in $\Omega$. Given that $u\in H\_0^1(\Omega)$, my intuition is that $u$ remains continuous throughout the entire closure $\bar{\Omega}$ and equals zero on the boundary. However, since no assumptions have been made regarding the boundary $\partial \Omega$, I have been unable to prove this conjecture.
| https://mathoverflow.net/users/166368 | Continuous up to the boundary without boundary smoothness | Here is an example in the plane which is genuinely discontinuous up to the boundary. I try to keep it as elementary as possible. There is an if and only if criterion for continuity up to the boundary of harmonic functions, due to Wiener: this example is essentially how one shows the only if part.
Take $B\_1$, and remove from it a collection of balls $\bar{B}\_{r\_k}(2^{-k}),$ with $k \geq 1$, centered at points on the real axis, of radii to be chosen but with $r\_k \ll 2^{-k}$. Set $U\_k = B\_1 \setminus \cup\_{j = 1}^k\bar{B}\_{r\_j}(2^{-j})$, and $U = (\cap\_k U\_k) \setminus \{0\}$ ($0$ is also removed to make $U$ open). Solve for $u\_k$ satisfying $\Delta u\_k = - 1$ on $U\_k$ and $u\_k \in H^1\_0(U\_k)$, and similarly $u$ with $\Delta u = -1$ on $U$ with $u \in H^1\_0(U)$ (there isn't any issue with solving this using standard energy methods, even on $U$).
From the maximum principle, $u\_{k+1}\leq u\_k$ on $U\_{k+1}$, and $u \leq u\_k$ on $U$. Let us note here that $u\_0$ (with no balls removed) has $u\_0 \leq \frac{1}{4} < 1$ (just by directly computing $u\_0 = (1 - |x|^2)/4$).
We also have that $u\_k \rightarrow u$ pointwise on $U$: indeed, the $u\_k$ are uniformly bounded in $H^1(U)$ by integrating by parts,
$$
\int\_{U\_k} |\nabla u\_k|^2 = \int\_{U\_k} u\_k \leq C \|\nabla u\_k\|\_{L^2(U\_k)}.
$$
Extract a weakly convergent subsequence $u\_k \rightharpoonup v \in H^1\_0(B\_1)$. It is straightforward to check that $v$ solves $\Delta v = -1$ on $U$, and that $v \in H^1\_0(U\_k)$ for every $k$. I claim (see below) that $v \in H^1\_0(U)$; then from uniqueness/weak maximum principle it follows that $v = u$, that $u\_k \rightarrow u$ in $L^2(U)$, and from regularity the convergence is pointwise, or even locally in $C^j$, etc.
Consider, then, $w = u\_k - u\_{k + 1}$. This function is harmonic on $U\_{k+1}$, and vanishes on the boundary except along $\partial B\_{r\_{k+1}}(2^{-k})$--along which it has $w < 1$. The function $h = 1 - \frac{\log (|x - 2^{-k}|/r\_k)}{\log (2/ r\_k)}$ is also harmonic on $U\_k$, is nonnegative on $B\_1$, and is $1$ along $\partial B\_{r\_{k+1}}(2^{-k})$: by the maximum principle, $w \leq h$. For $x$ in the interval $[-1, 0]$, we have that
$$
w \leq 1 - \frac{\log (2^{-k}/r\_k)}{\log (2/r\_k)} = \frac{k\log 2}{\log (2/r\_k)}.
$$
Now, we get to choose $r\_k$ as small as we like: for example, we can choose it to be $2e^{-t 2^k}$ for a small $t$, so that $w \leq 2k t 2^{-k}$ along this entire line segment [the main point here is to choose it so that $u\_k - u\_{k + 1}$ is summable with small sum].
Then summing this, we see that for any $x \in [-1, 0)$,
$$
|u - u\_0| \leq \sum\_{k = 0}^\infty |u\_k - u\_{k + 1}| \leq C t.
$$
Choosing $t$ small enough and using that $u\_0 > 1/8$ along $[-1/2, 0]$, this gives $ u > 1/16$ on this segment.
On the other hand, on each of the boundaries of balls composing $\partial U \setminus \{0\}$, we have $u = 0$. We conclude that $u$ had to be discontinuous at $0$.
**To verify the claim**: let $h\_r = \frac{[\log|x| - \log r^2)]\_+}{|\log r|}$. This function vanishes for $|x| \leq r^2$, and for $|x| \geq r$ it has $h\_r \geq 1 > v$. In between, we can integrate
$$
\int\_{r^2 < |x| < r} |\nabla h\_r|^2 = \int\_{r^2}^r \frac{1}{s |\log r|^2}ds \leq \frac{C}{|\log r|}.
$$
The point of this is, set $v\_r = \min\{v, h\_r\}$. Then $v\_r \in H^1\_0(U)$ (it vanishes on $B\_{r^2}$, no issues near $0$ here), and $v\_r = v$ outside of $B\_r$. We then have
$$
\int |\nabla (v\_r - v)|^2 = C \int\_{|x|< r} |\nabla v|^2 + |\nabla h\_r|^2,
$$
with the first term going to $0$ because $|\nabla v|^2$ is integrable, while the second term is controlled by $1/|\log r|\rightarrow 0$ from above. So $v\_r \rightarrow v$ in $H^1(B\_1)$, and $v\_r \in H^1\_0(U)$: it follows that $v \in H^1\_0(U)$.
| 4 | https://mathoverflow.net/users/378654 | 449161 | 180,791 |
https://mathoverflow.net/questions/449172 | 1 | We have the following term:
$$ (e^{-a h}+e^{-b h})^n / 2^n$$
Now we take the limit:
$$ h\to 0, n\to \infty $$
What relation of $h$ and $n$ must be satisfied for the following limit to hold?
$$\lim\_{h\to 0, n \to \infty}(\frac{e^{-a h}+e^{-b h}}{2})^n$$
$$=\lim\_{h\to 0, n \to \infty}(1-\frac{1}{2}ah-\frac{1}{2}bh)^n $$
$$=\lim\_{h\to 0, n \to \infty}e^{-\frac{a+b}{2}h n} $$
For example, if we let $\frac{e^{-h n}}{h^2}$ keep fixed, can the above hold?
| https://mathoverflow.net/users/503932 | calculating a double limit | For $h\to0$, we have
$$\frac{e^{-ah}+e^{-bh}}2=1-\frac{a+b}2\,h+O(h^2)
=\exp\Big(-\frac{a+b}2\,h+O(h^2)\Big)$$
and hence
$$\Big(\frac{e^{-ah}+e^{-bh}}2\Big)^n
=\exp\Big(-\frac{a+b}2\,nh+O(nh^2)\Big).$$
So, if $h\to0$ and $nh\to c\in\mathbb R$, then $nh^2\to0$ and hence
$$\lim\Big(\frac{e^{-ah}+e^{-bh}}2\Big)^n
=\lim\exp\Big(-\frac{a+b}2\,nh\Big) \\
=\exp\Big(-\frac{a+b}2\,c\Big).$$
---
Also, if $a>0$, $b>0$, $h\to0$ and $nh\to\infty$, then
$$-\frac{a+b}2\,nh+O(nh^2)=nh\Big(-\frac{a+b}2\,+O(h)\Big)\sim-nh \frac{a+b}2\to-\infty$$ and hence
$$\lim\Big(\frac{e^{-ah}+e^{-bh}}2\Big)^n
=\lim\exp\Big(-\frac{a+b}2\,nh\Big)=0.$$
---
If $a>0$, $b>0$, $h\to0$ and $nh\to-\infty$, then similarly
$$\lim\Big(\frac{e^{-ah}+e^{-bh}}2\Big)^n
=\lim\exp\Big(-\frac{a+b}2\,nh\Big)=\infty.$$
---
If $a>0$, $b>0$, $h\to0$, $n\to\infty$, but $nh$ does not converge to a limit, then
$$\lim\exp\Big(-\frac{a+b}2\,nh\Big)$$
does not exist.
| 1 | https://mathoverflow.net/users/36721 | 449181 | 180,799 |
https://mathoverflow.net/questions/449182 | 8 | I'm mainly following references such as Kelly, Loregian and the nLab, and it seems customary there to generalize (co)ends to the enriched context (over a symmetric monoidal category $\mathcal{V}$) by first defining the $\mathcal{V}$-valued (co)ends, for $\mathcal V$-functors
$$P\colon \mathcal C^\mathsf{op}\boxtimes\mathcal{C}\longrightarrow\mathcal{V},$$
and then define the general case of - let's say end - for a $\mathcal V$-functor $T\colon\mathcal{C}^\mathsf{op}\boxtimes\mathcal{C}\to\mathcal{D}$ by the object $\int\_cT(c,c)$ of $\mathcal D$ together with its extranatural transformation $\{ p\_a\colon\int\_cT(c,c)\to T(a,a) \}$ (which actually is $\{p\_a\colon \mathbb{1}\to\mathcal{D}(\int\_cT(c,c),T(a,a))\}$) exhibiting via $\mathcal{D}(d,p\_a)$, for every $d$ in $\mathcal{D}$, the end of the $\mathcal V$-functor
$$\mathcal D(d,T(-,-))\colon\mathcal{C}^\mathsf{op}\boxtimes\mathcal{C}\to\mathcal{V}.$$
Now, this makes sense to me, and it clearly gives by definition the formula $\int\_c\mathcal{D}(d,T(c,c))\cong\mathcal{D}(d,\int\_cT(c,c))$, but why do we do that? Is this step (passing through the $\mathcal V$-valued case) necessary for some deep reason I'm not seeing or it's just convenience? My feeling would be that we can generalize verbatim the definition of (co)ends to the enriched case (as a initial/terminal enriched extranatural transformation), just remembering what we actually mean when talking about morphisms in an enriched category. But then I struggle to prove the above continuity formula. Could anyone enlighten me about that?
Thank you so much in advance.
| https://mathoverflow.net/users/170683 | Why are enriched (co)ends defined like that? | The $\newcommand{\V}{\mathcal{V}}\newcommand{\D}{\mathcal{D}}\newcommand{\C}{\mathcal{C}}$universal property of being an “initial/terminal cowedge” — even strengthened to “$\V$-initial/-terminal” — is weaker than the standard definition, and too weak for many purposes. In particular, it’s not strong enough to imply any universal property of the form $\D(\int^\C P, X) \cong \ldots$
The difference shows up already in (co)limits — even in (co)products. For a counterexample, take $\V = \mathrm{Set^2}$, and $\C$ to be the $\V$-category with four objects $a,b,p,q$, generated by “global arrows” from $a$ and $b$ to $p$, and “half-present arrows” from $a$ and $b$ to $q$; so explicitly, $\C(x,x) = (1,1)$ for each $x$, $\C(a,p) = \C(b,p) = (1,1)$, $\C(a,q) = \C(b,q) = (1,0)$, and $\C(x,y) = (0,0)$ otherwise. Then the arrows $a,b \to p$ form a $\V$-initial cospan under $a,b$, since it’s the only such cospan and has no non-trivial endomorphisms. But $p$ is *not* a $\V$-coproduct for $a$ and $b$, since we have $\C(p,q) = (0,0)$, not $(1,0)$ as a $\V$-coproduct should have.
The problem is that the $\V$-category of (co)cones/wedges under a diagram only contains the *global* (co)cones/wedges as its objects — in the example, it can’t see the “half-present” cocone to $q$. This could be correcting by considering it as a $\V$-indexed category, but that requires a whole extra layer of technology.
When a (co)limit/end for a diagram exists, it will be an initial/terminal (co)cone/wedge, and so then any initial/terminal (co)cone/wedge will be isomorphic to it, and hence will be a $\V$-(co)limit/end. In particular, in a (co)-complete category, the weaker property of being an initial/terminal (co)cone/wedge does imply the full universal property. But outside the (co)complete setting, they diverge, as the counterexample shows, and the full universal property is what one generally wants.
| 11 | https://mathoverflow.net/users/2273 | 449185 | 180,800 |
https://mathoverflow.net/questions/449156 | 2 | Compute Fourier transforms of homogeneous functions of the form,
$$
\frac{1}{|x|^{n+d}}P\_d(x)
$$
where $P\_d$ is a homogenous harmonic polynomial of degree $d$ in $n+1$ variables.
| https://mathoverflow.net/users/122182 | Fourier transforms of homogeneous functions | Your function is, with $P\_d$ homogeneous harmonic polynomial of degree $d$ in $n$ variables,
$$
u(x)=\frac{P\_d(x)}{\vert x\vert^{n+d}}.
\tag{1}
$$
This is an homogeneous distribution of degree $-n$ on $\mathbb R^n\backslash\{0\}$. The first thing to do is to extend that distribution to a distribution on $\mathbb R^n$. Let us first check for a test function $\phi\in C\_c^\infty(\mathbb R^n\backslash\{0\})$, the absolutely converging integral
\begin{multline}
\langle T, \phi\rangle=
\int\_{\mathbb R^n}\frac{P\_d(x)}{\vert x\vert^{n+d}}\phi(x) dx
=\int\_0^{+\infty} r^{n-1}\int\_{\mathbb S^{n-1}}
\frac{P\_d(r\omega)}{r^{n+d}}\phi(r\omega)d\sigma(\omega) dr
\\=
\int\_{0}^{+\infty}\int\_{\mathbb S^{n-1}}
{P\_d(\omega)}\phi(r\omega)d\sigma(\omega)\frac{dr}r
\end{multline}
$\mathbf {\text{Claim:}}$ Let $\psi$ be a function in the Schwartz class of $\mathbb R$ and let $H=\mathbf 1\_{\mathbb R^+}$.
We have
$$\int\_0^{1}\frac{\psi(r)-\psi(0)}{r} dr
+\int\_1^{+\infty}\frac{\psi(r)}{r} dr
=
\langle \frac{d}{dr}\bigl(
H(r) \ln r
\bigr),\psi(r)\rangle\_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R) }.
$$
Indeed we have,
\begin{align}
\langle \frac{d}{dr}\bigl(
H(r) \ln r
\bigr),\psi(r)\rangle\_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R) }
&=-\int\_0^{+\infty}\psi'(r)
\ln r\ dr
=\lim\_{\epsilon \rightarrow 0\_+}
-\int\_\epsilon^{+\infty} \psi'(r)
\ln r\ dr,
\\
&\text{and}\hskip25pt
\\
-\int\_\epsilon^{+\infty} \psi'(r)
\ln r\ dr&=\bigl[\psi(r)
\ln r\bigr]^{\epsilon}\_{+\infty}
+\int\_{\epsilon}^{+\infty}\psi(r) r^{-1} dr
\\&=\psi(\epsilon)
\ln \epsilon
+\int\_{\epsilon}^{+\infty}
\psi(r) r^{-1} dr
\\
&=
\int\_{1}^{+\infty}\psi(r) r^{-1} dr
+\int\_{\epsilon}^{1}\psi(r) r^{-1} dr
-\int\_{\epsilon}^{1}\psi(\epsilon)r^{-1} dr
\\
&=
\int\_{1}^{+\infty}\psi(r) r^{-1} dr
+\int\_{\epsilon}^{1}
\bigl[\psi(r) -\psi(\epsilon)\bigr]r^{-1}
dr,
\end{align}
proving the claim.$\ \mathbf{\square}$
We may thus extend $T$ as a temperate distribution $\tilde T$
on $\mathbb R^n$ with the definition
for $\phi\in\mathscr S(\mathbb R^n
)$,
$$
\langle \tilde T, \phi\rangle\_{\mathscr S'(\mathbb R^n),\mathscr S(\mathbb R^n) } =
\int\_{\mathbb S^{n-1}}
{P\_d(\omega)}
\langle \frac{d}{dr}\bigl(
H(r) \ln r
\bigr),\phi(r\omega)\rangle\_{\mathscr S'(\mathbb R\_r),\mathscr S(\mathbb R\_r) }
d\sigma(\omega).
\tag{2}$$
However the distribution $\tilde T$ fails to be homogeneous on $\mathbb R^n$, as it is seen when you calculate
$
\langle \tilde T(\lambda x), \phi(x)\rangle.
$
Note that an homogeneous distribution of degree $\mu$
is temperate and its Fourier transform is also homogeneous with degree $-\mu-n$.
That does not mean that you cannot calculate the Fourier transform of $\tilde T$. In particular Theorem 7.1.18 in the ALPDO of Lars Hörmander (Grundlehren 256) shows that
the Fourier transform of ${\tilde T}$ is smooth outside of the origin. It is also possible to calculate directly
$$
\langle \widehat{\tilde T}, \phi\rangle=\langle \tilde T, \hat \phi\rangle,
$$
when $\hat \phi(0)=\int \phi(x) dx =0$,
but in general you will need to use the Fourier transform of
the one-dimensional distribution
$\frac{d}{dr}\bigl(
H(r) \ln r
\bigr)$.
| 2 | https://mathoverflow.net/users/21907 | 449198 | 180,804 |
https://mathoverflow.net/questions/449208 | 4 | I am misunderstanding something in Theorem 2.1.9 in Dimca’s Sheaves in Topology:
Let $X$ be a real smooth manifold. Then the natural morphism from the constant sheaf to the de Rham complex
$$\mathbb{R}\_X \rightarrow \Omega\_X^\bullet$$
is an acyclic resolution of $\mathbb{R}\_X$.
I am getting confused about in which sense $\Omega\_X^\bullet$ is acyclic: if it is acyclic as a complex, then it has no cohomology above $H^0$ by definition. But, of course a smooth real manifold can still have cohomology above $H^0$!
So I would like to confirm that acyclic here means a resolution using $\Gamma$-acyclic sheaves, i.e., something like injective or flasque sheaves.
Edit: Maybe we want hypercohomology of both sides, since Poincar\’e’s lemma says that the de Rham complex $\Omega\_X^\bullet$ really is acyclic as a complex. But then the statement
$$ H^k(X,\mathbb{R}) \cong \frac{\text{ker } d: \Omega^k(X) \rightarrow \Omega^{k+1}(X)}{\text{Im }d: \Omega^{k-1}(X) \rightarrow \Omega^k(X)} $$
makes no sense to me, because the right hand side here should be zero for $k>0$ if $\Omega^\bullet(X)$ is acyclic as a complex
| https://mathoverflow.net/users/131090 | Clarification on smooth de Rham theorem | The following statements are true:
* The complex $0 \to \mathbb{R}\_X \to \Omega^0\_X \to \Omega^1\_X \to \dots$ is an acyclic complex (of sheaves), i.e. it is exact at each step. This is the content of the Poincare lemma, and it is what it means for $\Omega^\bullet$ to be a resolution of $\mathbb{R}\_X$.
* Each individual $\Omega^i$ is flasque, and therefore an acyclic sheaf (i.e. its sheaf cohomology vanishes in positive degree). This is a different meaning of the word "acyclic" from the above; but specifying "acyclic resolution" means this sense of the word acyclic (a resolution by acyclic sheaves), since the other kind of acyclicity is already implicit in the use of the word "resolution". (I don't blame you for finding this confusing!)
However, it is *not* true that either of these statements has anything to do with the exactness of the sequence of global sections $\Omega^0(X) \to \Omega^1(X) \to \dots$, which very frequently has interesting cohomology in all possible degrees (up to $\dim X$). (But why would you expect it to be this? There is much more information in a sheaf than just its global sections.)
| 10 | https://mathoverflow.net/users/2481 | 449210 | 180,807 |
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