parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/437645 | 9 | I am trying to gain a better understanding of Schur-Weyl duality specifically applied to symmetric functions. My motivating example is trying to understand the Frobenius character of the multilinear component of the free Lie algebra (Theorem 8.1 in Reutenauer's book on the subject), but my general confusion is more than just that example. I feel comfortable with representation theory of finite groups and less so with representation theory of Lie groups.
[In](https://qchu.wordpress.com/2012/11/13/four-flavors-of-schur-weyl-duality/) [several](https://math.uchicago.edu/%7Emay/REU2016/REUPapers/Stevens.pdf) [sources](https://en.wikipedia.org/wiki/Schur%E2%80%93Weyl_duality), the statement of Schur-Weyl duality is framed as the $\operatorname{GL}(V)\times \mathfrak{S}\_k$ isomorphism of the tensor algebra $V^{\otimes k}$ with the direct sum of tensor products $\mathbb{S}^\lambda(V) \otimes M^\lambda$ of Specht modules with Schur functors indexed by partitions of $k$ into at most $\dim(V)$-many parts. As I understand it, one can then determine the multiplicity of either a Schur functor (by looking at the dimension of the corresponding Specht module) or the multiplicity of a Specht module (by looking at the dimension of the corresponding Schur functor).
I also seem to understand that, as explained in the second appendix of Stanley's Enumerative Combinatorics Vol. 2, the character value of $A \in \operatorname{GL}(V)$ for an irreducible rational $\operatorname{GL}(V)$ representation $\mathbb{S}^\lambda$ is given by evaluating the Schur polynomial on the eigenvalues of $A$.
Reutenauer's proof of theorem 8.1 (that two particular symmetric functions are equal) is as follows:
>
> This is the Schur-Weyl duality between the representations of the symmetric group and the general linear group (Weyl 1946, Theorem 7.6.F; Macdonald 1979, A7
> in Chapter 1).
>
>
>
My first confusion is that the references I find refer to Schur-Weyl duality as a statement about the tensor algebra, not any other algebra. The universal enveloping algebra of the free Lie algebra can be constructed as a quotient of the tensor algebra, so I'm willing to suspend some disbelief that the statement is still applicable here.
Even with that, I am having trouble connecting the dots here. Reutenauer sets up a $\operatorname{End}(V) \times \mathfrak{S}\_k$-representation, and then looks at the action of a copy of $\mathfrak{S}\_k$ living inside of $\operatorname{End}(V)$ and considers the Frobenius characteristic of the $\mathfrak{S}\_k$-representation found by restricting the action of $\operatorname{End}(V)$. He then compares this to the multivariate generating function of dimensions of homogeneous polynomials of multidegree $\alpha$ which are stable under the $\mathfrak{S}\_k \subseteq \operatorname{End}(V)$ action and asserts they are the same via the proof quoted above.
The character value of the restriction of a representation is just the character value of the original representation. The eigenvalues of any $A \in \mathfrak{S} \leq \operatorname{End}(V)$ should be all $1$. Then I think the character value of $A$ should be the number of semistandard Young tableaux of shape $\lambda$ with entries at most $\dim(V)$ (i.e. $s\_\lambda(1,1,\ldots,1)$). But I am struggling to see why this proves Reutenauer's claim.
Any help or general information about the relationship between Schur-Weyl duality and symmetric functions you could provide would be greatly appreciated.
**Edit**: I made some mistakes in my question including:
* The universal enveloping algebra is the tensor algebra, not a (nontrivial) quotient.
* Eigenvalues of permutation matrices are roots of unity, not 1.
| https://mathoverflow.net/users/497171 | Using Schur-Weyl duality | This answer is a response the the prompt
>
> Any help or general information about the relationship between Schur-Weyl duality and symmetric functions you could provide would be greatly appreciated.
>
>
>
If you have more questions (e.g. about the free Lie algebra), feel free to ask.
For convenience I will let $n = \dim(V)$.
Step 1: A bitrace formula.
As discussed, we have an action of $GL(V) \times S\_k$ on $V^{\otimes k}$. We will compute the trace of an element $(M, g) \in GL(V) \otimes S\_k$ on $V^{\otimes k}$. Conceptually we will think about $GL(V)$ and $S\_k$ as separate (rather than combined into $GL(V) \otimes S\_k$) which is why I use the term "bitrace" (it is the synthesis of two traces).
We compute the trace directly. The matrix $M$ has a Jordan decomposition $M = S + N$ ($S$ semisimple, $N$ nilpotent), and the action of $(M,g)$ on $V^{\otimes k}$ is the sum of the actions of $(S,g)$ and $(N,g)$; since $N$ is nilpotent, $(N,g)$ is not actually an element of $GL(V)\times S\_k$, nevertheless the trace is well defined and equal to zero because it is nilpotent. I say all this only to justify restricting to diagonalisable matrices.
Now, let $v\_1, \ldots, v\_n$ be an eigenbasis of $M$, so that $Mv\_i = x\_i v\_i$ for some complex numbers $x\_i$ (i.e. $M = diag(x\_1,\ldots,x\_n)$).
This induces a basis $V\_I = v\_{i\_1} \otimes \cdots \otimes v\_{i\_k}$ of $V^{\otimes k}$ indexed by words $I = (i\_1, \ldots, i\_k)$ (where $1 \leq i\_j \leq n$). Conveniently, the action of $(M,g)$ on $V\_I$ is easy to compute:
$$
(M,g) \cdot (v\_{i\_1} \otimes \cdots \otimes v\_{i\_k}) = g \cdot (Mv\_{i\_1} \otimes \cdots \otimes Mv\_{i\_k}) = g \cdot (x\_{i\_1}v\_{i\_1} \otimes \cdots \otimes x\_{i\_k}v\_{i\_k}) \\ = x\_{i\_1}x\_{i\_2}\cdots x\_{i\_k} (v\_{g^{-1}(1)} \otimes \cdots \otimes v\_{g^{-1}(n)})
$$
Side note: whether you apply $g^{-1}$ or $g$ to the indices depends on whether you view the symmetric group as having a left or right action, it's not really important.
This value of $(M,g) \cdot v\_I$ is a scalar multiple of another basis element, which we might write $v\_{g(I)}$ by using the induced action of $g \in S\_k$ on tuples of length $k$. To compute the trace of $(M,g)$ we need to sum the "diagonal entries", i.e. the scalars corresponding to those $I$ with $v\_I = v\_{g(I)}$. This computation becomes
$$
\sum\_{g(I) = I} x\_{i\_1} \cdots x\_{i\_k}.
$$
Now, suppose for example that $(2,5,6)$ is a cycle of $g \in S\_k$. Then the condition $g(I) = I$ implies that the equality of indices $i\_2 = i\_5 = i\_6$, but the actual value could be anything in $1, \ldots, n$. This condition also implies that $x\_{i\_2} = x\_{i\_5} = x\_{i\_6}$. This same reasoning extends to cycles of all sizes. The only nonzero "diagonal terms" to be summed are those where all indices acted on by a cycle of $g$ are the same. The actual index associated to each cycle is arbitrary, we we need to sum over those. If the cycles of $g$ have sizes $\mu\_1, \ldots, \mu\_l$, the trace becomes
$$
\sum\_{j\_1=1}^{n} \cdots \sum\_{j\_l=1}^{n} x\_{j\_1}^{\mu\_1} \cdots x\_{j\_l}^{\mu\_l} = (\sum\_{j\_1=1}^{n} x\_{j\_1}^{\mu\_1}) \cdots (\sum\_{j\_l=1}^{n} x\_{j\_l}^{\mu\_l}) = p\_{\mu\_1}(x) \cdots p\_{\mu\_l}(x) = p\_\mu(x)
$$
where I am now using the standard notation for power-sum symmetric functions.
Conclusion: if the eigenvalues of $M$ are $x\_i$ and $g$ has cycle type $\mu$, then the bitrace of $(M,g)$ acting on $V^{\otimes k}$ is $p\_\mu(x)$.
Step 2: Frobenius characteristic and Cauchy identity.
The Frobenius characteristic, $ch$, is an isomorphism between the Grothendieck group of class functions on $S\_k$ and symmetric functions of degree $k$ (here we work over $\mathbb{C}$). It is convenient to define $ch(f)$ for all functions on $S\_k$ (not just class functions) by saying that if $g^\*$ is the indicator function of $g \in S\_k$, then $ch(g^\*) = \frac{1}{k!} p\_\mu(y)$, where $\mu$ is the cycle type of $g$, and I write $y$ for the symmetric function variables in order to distinguish them from the discussion in the previous step. So for example, if $C\_{\mu}^\*$ is the indicator function of the conjugacy class $C\_\mu$ of elements of cycle type $\mu$, then $ch(C\_{\mu}^\*) = \frac{|C\_\mu|}{k!} p\_\mu(y) = \frac{1}{z\_\mu}p\_{\mu}(y)$, where $z\_\mu$ has its usual meaning.
Now if we fix $M \in GL(V)$, then the bitrace of $(M,g)$ may be viewed as a (class) function on $S\_k$, and so we may apply the Frobenius characteristic. If we write $cyc(g)$ for the cycle type of $g$, the result of this calculation is
$$
ch(tr(M,g)) = \sum\_{g \in S\_k} p\_{cyc(g)}(x) \frac{1}{k!} p\_{cyc(g)}(y) = \sum\_{\mu \vdash k} \frac{1}{z\_\mu} p\_\mu (x) p\_\mu(y).
$$
Now, the famous Cauchy identity implies that we have
$$
ch(tr(M,g)) = \sum\_{\mu \vdash k} \frac{1}{z\_\mu} p\_\mu (x) p\_\mu(y) = \sum\_{\lambda \vdash k} s\_\lambda(x) s\_\lambda(y),
$$
where $s\_\lambda$ is the Schur function indexed by $\lambda$.
This may be viewed as a symmetric-function-theoretic formulation of Schur-Weyl duality for the following reasons. Suppose we know that the Frobenius characteristic of the Specht module $S^\lambda$ is the Schur function $s\_\lambda(y)$, and the character of the Schur functor $\mathbb{S}^\lambda(V)$ is the Schur function $s\_\lambda(x)$ (the meaning of which is discussed in the comments to the original post). Then we have found that the bitrace (i.e. $GL(V) \times S\_k$-character) of $V^{\otimes k}$ agrees with that of $\bigoplus\_{\lambda \vdash k} \mathbb{S}^\lambda(V) \otimes S^\lambda$. By semisimplicity, these must be isomorphic.
For example, we can recover the multiplicity of the Specht module $S^\lambda$ in $V^{\otimes l}$ by computing $\dim(\mathbb{S}^\lambda(V))$, which is nothing but the trace of the identity of $GL(V)$. But the identity element of $GL(V)$ (viewed as a matrix) has $1$ as an eigenvalue repeated $n$ times, so the dimension is the evaluation $s\_\lambda(1,\ldots,1)$ (where there are $n$ $1$-s), as you mention in your post.
| 6 | https://mathoverflow.net/users/159272 | 437878 | 176,896 |
https://mathoverflow.net/questions/427315 | 0 | Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Descartes (1638), Frenicle (1657), and subsequently [[Sorli (2003) - Conjecture 3, Chapter 5 on page 89]](https://opus.lib.uts.edu.au/bitstream/10453/20034/2021/02Whole.pdf) conjectured that $k=1$ always holds for an odd perfect number $N = p^k m^2$.
It is fairly easy to [show](https://math.stackexchange.com/questions/4262319) that if $\sigma(p^k)/2$ is prime, then $k = 1$.
Moreover, [Broughan, Delbourgo, and Zhou (2013)](https://www.emis.de/journals/INTEGERS/papers/n39/n39.pdf) show that if $\sigma(p^k)/2$ is a square, then $k = 1$.
An interesting scenario holds when $\sigma(p^k)/2$ is squarefree. Indeed, this assumption implies that
$$H = \gcd(m^2,\sigma(m^2)) = \frac{m^2}{\sigma(p^k)/2} = G \times J^2$$
is **not squarefree**, where $G$ and $J$ are defined as
$$G = \gcd(\sigma(p^k),\sigma(m^2)) = \dfrac{\bigg(\gcd(\sigma(p^k)/2,m)\bigg)^2}{\sigma(p^k)/2}$$
and
$$J = \dfrac{m}{\gcd(\sigma(p^k)/2,m)}.$$
(For the case under consideration, that $\sigma(p^k)/2$ is squarefree, we do in fact have
$$G = \sigma(p^k)/2$$
and
$$J = \dfrac{m}{\sigma(p^k)/2}$$
since $\sigma(p^k)/2 \mid m^2$, and therefore $\sigma(p^k)/2 \mid m$, holds in general.)
Here is my initial question:
**FIRST INQUIRY**
>
> Can you show that $\sigma(p^k)/2$ is squarefree likewise implies that $k=1$?
>
>
>
The reason for this inquiry is because I currently [know](https://math.stackexchange.com/questions/4335854) that $k=1$ likewise implies that $H$ is **not squarefree**.
**LAST INQUIRY**
>
> If $p^k m^2$ is an odd perfect number with special prime $p$, then under what other conditions on $\sigma(p^k)/2$ does $k=1$ follow?
>
>
>
| https://mathoverflow.net/users/10365 | If $p^k m^2$ is an odd perfect number with special prime $p$, then under what other conditions on $\sigma(p^k)/2$ does $k=1$ follow? | This is a partial response, as it does not directly answer the original question that was asked. Additionally, what follows are actually some remarks that would be too long to fit in the *Comments* section.
---
This answer builds on the results in [this MSE question from January 03, 2023](https://math.stackexchange.com/questions/4609954), and shows that
$$H = \frac{\sigma(m^2)}{p^k}$$
is **not a square**, if $\sigma(p^k)/2$ is squarefree.
---
Throughout [this paper](https://arxiv.org/abs/2202.08116), we implicitly rely on the simple equality
$$\sigma(m^2) = \frac{2p^k m^2}{\sigma(p^k)}. \tag{1}$$
Unfortunately, this seems to introduce fractions. To avoid that, we can use prime factorizations, as follows. Write the prime factorization of $m$ as
$$m = {q\_1}^{a\_1} \cdots {q\_n}^{a\_n},$$
for some unique odd primes $3 \leq q\_1 < \ldots < q\_n$, and for some positive integer exponents $a\_1, \ldots, a\_n$. Since both sides of $(1)$ are integers, and since $p \equiv k \equiv 1 \pmod 4$ with $p$ prime, we know that
$$\sigma(p^k) = 2 {q\_1}^{b\_1} \cdots {q\_n}^{b\_n}$$
for some nonnegative integers $0 \leq b\_i \leq 2a\_i$. Thus, we have
$$\sigma(m^2) = p^k {q\_1}^{2a\_1 - b\_1} \cdots {q\_n}^{2a\_n - b\_n}.$$
Suppose that $\sigma(p^k)/2$ is squarefree. This assumption implies that
$$\sigma(p^k) = 2 {q\_1}^{b\_1} \cdots {q\_n}^{b\_n}$$
for $0 \leq b\_i \leq 1$.
Since
$$\frac{\sigma(p^k)}{2} \geq \frac{p^k + 1}{2} \geq 3,$$
then $b\_i = 1$ for at least one $i$.
This means that
$$H = \frac{\sigma(m^2)}{p^k} = {q\_1}^{2a\_1 - b\_1} \cdots {q\_n}^{2a\_n - b\_n}$$
where $2a\_i - b\_i = 2a\_i - 1 \equiv 1 \pmod 2$ for at least one $i$.
Consequently, $H$ is **not a square**, if $\sigma(p^k)/2$ is squarefree.
| 0 | https://mathoverflow.net/users/10365 | 437890 | 176,901 |
https://mathoverflow.net/questions/437911 | 1 | The eigenvalues of a circulant matrix $C$ can be extracted as $$
\Lambda=F^{-1} C F
$$
where the $F$ matrix is a discrete Fourier transform matrix and $\Lambda$ is a diagonal matrix of eigenvalues.
Since $F^{-1}F$=$FF^{-1}=I$, is it possible to write $$
\Lambda=FC F^{-1}
?$$
For example, if our circulant matrix were:
$$C=\begin{pmatrix} 2& -1& 0& 0& 0& 0& -1\\ -1& 2& -1& 0& 0& 0& 0&\\ 0& -1& 2& -1& 0& 0& 0\\ 0& 0& -1& 2& -1& 0& 0\\
0& 0& 0& -1& 2& -1& 0\\ 0& 0& 0& 0& -1& 2& -1\\ -1& 0& 0& 0& 0& -1& 2
\end{pmatrix}$$
There is no practical difference in the numerical values of the eigenvalues from those two approaches, i.e., whether the position of the inverse Fourier matrix is first or the last, it does not seem to matter here.
Is there a mathematical basis for this observation and is this property true for all circulants? $C$ happens to be a symmetric matrix. I am looking for a reference to explain this.
I am not a mathematician, my interest is in signal deconvolution problems for chemistry.
| https://mathoverflow.net/users/142414 | Extracting eigenvalues of a circulant matrix using discrete Fourier matrix | The discrete Fourier transform matrix $F$ is special because it squares to a permutation matrix,$^\ast$ $F^2=P$ with $P^2=I$. I insert $I=F^2 P=P F^{-2}$,
$$F^{-1}\Lambda F= F^{-1}(F^2 P) \Lambda (P F^{-2})F=F(P\Lambda P)F^{-1}.$$
So you see that you can either write $\Lambda=FCF^{-1}$ or $\Lambda'=F^{-1}CF$, the difference is simply that $\Lambda'=P\Lambda P$ is a reordering of the eigenvalues on the diagonal.
---
$^{\ast}$ See, for example, the [DFT wiki page](https://en.wikipedia.org/wiki/Discrete_Fourier_transform#Eigenvalues_and_eigenvectors)
---
In special case of the matrix in the OP, the permutation matrix is
$$P=\left(
\begin{array}{ccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right),$$
which commutes with the circulant matrix $C$. Hence in this case $\Lambda$ and $\Lambda'$ are the same.
| 2 | https://mathoverflow.net/users/11260 | 437919 | 176,907 |
https://mathoverflow.net/questions/437913 | 0 | A *nice* density (on $\mathbb R$) is function $\phi:\mathbb R \to \mathbb R$ such that
* (1) $\phi(x) \ge 0$ for all $x \in \mathbb R$,
* (2) $\int\_{-\infty}^\infty \phi(x) \mathrm{d}x = 1$,
* (3) $\phi$ is even, i.e $\phi(-x) = \phi(x)$ for all $x \in \mathbb R$, and
* (4) $\phi$ is differentiable and $\|\phi'\|\_\infty := \sup\_{x \in \mathbb R}|\phi'(x)|$ is finite.
The 3rd requirement may be dropped in case it courses trouble.
**Question 1.** Given positive $\epsilon$ and $c$, is it possible to find a nice density $\phi$ (which may depend on $\epsilon$ and $c$) such that
* $t\phi(\epsilon/t) \ge c \|\phi'\|\_\infty$ for all positive $t$ ?
**Question 2.** In case **Question 1** has an affirmative answer, can $\phi$ be chosen in the family of (centered) Gaussian densities given by $\phi\_\sigma(x) \propto e^{-x^2/(2\sigma^2)}$, for some positive $\sigma$ ?
| https://mathoverflow.net/users/78539 | Given positive $\epsilon$ and $c$, find a density $\phi$ such that $t\phi(\epsilon/t) \ge c \|\phi'\|_\infty$ for all positive $t$ | Of course, not.
Indeed, suppose that $t\phi(\epsilon/t) \ge c \|\phi'\|\_\infty$ for all positive $t$.
Then, letting $s:=\epsilon/t$, we get $\phi(s)\ge Cs$ for $C:=c \|\phi'\|\_\infty/\epsilon>0$ and all real $s>0$. So, $\int\_{\mathbb R}\phi=\infty$ and hence $\phi$ cannot be a pdf.
| 4 | https://mathoverflow.net/users/36721 | 437920 | 176,908 |
https://mathoverflow.net/questions/437916 | 5 | We denote by $\mathbb{H}^n$ the hyperbolic plane of dimension $n$.
I don't know much about the Ricci flow, so my first question is probably naïve : can one define a normalized Ricci flow such that the hyperbolic metric on $\mathbb{H}^n$ is a stationary solution (constant in time) of the flow ?
My second question,
Let $g$ be a compact perturbation of the hyperbolic metric. By compact I mean a smooth change of the metric on a compact set.
If the previously defined flow exists, does the solution of this Ricci flow with initial data $g$ converge as $t \to \infty$ toward the hyperbolic metric ?
| https://mathoverflow.net/users/497380 | Ricci flow negative curvature | The answer to your first question is Yes. The equation
$$
\tag{\*} \partial\_t g = -2\textrm{Ric} - 2(n-1)g
$$
is a Ricci flow type equation that admits hyperbolic space as a static solution. In fact, if you have solution to (\*) you can rescale time/space to obtain a Ricci flow and vice versa. This is proven in Lemma A.3 [here](https://arxiv.org/pdf/1003.2107.pdf).
For your second question, you need to assume that the change of metric on a compact set is "small" in some sense. Otherwise you could have singularities, etc on the compact set (although in $n=3$ you might be able to construct a flow with surgery that eventually converges back to hyperbolic space plus some other pieces, I don't think this has been done although I think its doable with known technology). However, if you assume $C^0$ closeness, then what you ask is proven by [Schnurer--Schulze--Simon](https://arxiv.org/pdf/1003.2107.pdf) (see also [Bamler](https://arxiv.org/pdf/1011.4267.pdf) for a generalization and further references)
| 8 | https://mathoverflow.net/users/1540 | 437922 | 176,909 |
https://mathoverflow.net/questions/437874 | 6 | Call a set $X$ *hesive* if for every infinite computable set $C$, both $C \cap X$ and $C \setminus X$ are infinite.
It's not hard to see that every hyperimmune degree computes a hesive set, but this isn't a characterization, since also any random set is hesive (in fact, Church stochasticity suffices).
Does every noncomputable degree compute a hesive set?
| https://mathoverflow.net/users/32178 | Sets meeting and avoiding computable sets | $X$ is hesive iff $X$ is bi-immune.
Jockusch showed that a Sacks generic has bi-immune-free degree.
*Jockusch, C. G. Jr.*, [**The degrees of bi-immune sets**](http://dx.doi.org/10.1002/malq.19690150707), Z. Math. Logik Grundlagen Math. 15, 135-140 (1969). [ZBL0184.02002](https://zbmath.org/?q=an:0184.02002).
So no, not every noncomputable degree contains a hesive set.
| 2 | https://mathoverflow.net/users/4600 | 437941 | 176,914 |
https://mathoverflow.net/questions/437914 | 0 | I'm needing references - preferably published papers and books - about this subject. I'm relatively new to the state of the art of fractal geometry and am way too inexperienced to seek for myself at the good and legal sources.
If anyone knows a textbook that contemplates the Bowen formula, that'd be great. I found Iommi's notes, but since I already studied it, now I'm seeking something maybe on the direction of a next step.
| https://mathoverflow.net/users/88795 | Seeking for references - Bowen Formula and a link between dimension theory and thermodynamic formalism | There's Dimension Theory in Dynamical Systems by Yakov Pesin.
| 1 | https://mathoverflow.net/users/152429 | 437943 | 176,916 |
https://mathoverflow.net/questions/437945 | 5 | On 8:38 of [Session 9: Masterclass in Condensed Mathematics](https://www.youtube.com/watch?v=gZ4ES3vjAw4&t=3107s) an $\infty$-category is defined as a simplicial set $\mathcal{S}$ (i.e a functor $\Delta^{op}\rightarrow Sets$) such that for every horn $\wedge\_{i}^{n}\rightarrow \mathcal{S}$ with $1<i<n$ there is at least a lift $\Delta^{n}\rightarrow \mathcal{S}$. As far as I understand, this definition means that there are "multiple ways" of composing morphisms, as opposed to regular categories, in which the lift is unique. My question regards a somehow dual notion of this. That is, let
$\mathcal{S}$ be a simplicial set such that for every horn $\wedge\_{i}^{n}\rightarrow \mathcal{S}$ with $1<i<n$ there is at most a lift $\Delta^{n}\rightarrow \mathcal{S}$. So morphisms are not always composable, but when they are the composition is unique. My questions are the following: Does this concept have a name? Is it of interest? If so, what are its applications?
| https://mathoverflow.net/users/476832 | On the definition of infinity-category | *Partial monoids* (see Definition 2.2) play a useful role in
```
Segal, Graeme
Configuration-spaces and iterated loop-spaces.
Invent. Math. 21 (1973), 213–221.
```
| 2 | https://mathoverflow.net/users/9684 | 437946 | 176,918 |
https://mathoverflow.net/questions/436117 | 4 | I'm interested in a classification of the real forms of the general linear Lie superalgebra $\mathfrak{gl}\_{m|m}(\mathbb{C})$.
The real forms of the *simple* complex Lie superalgebras were classified by Serganova in the paper *Classification of real simple Lie superalgebras and symmetric spaces*. For $\mathfrak{gl}\_{m|n}(\mathbb{C})$, with $m \ne n$, we have a decomposition of Lie superalgebras $\mathfrak{gl}\_{m|n}(\mathbb{C}) = \mathfrak{sl}\_{m|n}(\mathbb{C}) \oplus \mathbb{C}$. Since $\mathfrak{sl}\_{m|n}(\mathbb{C})$ is simple, and the only real form of the abelian complex Lie algebra $\mathbb{C}$ is the abelian real Lie algebra $\mathbb{R}$, we can use Serganova's classification to get all the real forms of $\mathfrak{gl}\_{m|n}(\mathbb{C})$.
The situation for $\mathfrak{gl}\_{m|m}(\mathbb{C})$ is more subtle, since it does not decompose as a direct sum of the corresponding simple Lie superalgebra $\mathfrak{psl}\_{m|m}(\mathbb{C})$ and a center. There are natural enlargements of the real forms of $\mathfrak{psl}\_{m|m}(\mathbb{C})$ to real forms of $\mathfrak{gl}\_{m|m}(\mathbb{C})$. But it's not clear to me if one obtains *all* real forms of $\mathfrak{gl}\_{m|m}(\mathbb{C})$ in this way.
This seems a natural enough question that I expected to find an answer in the literature that somehow bootstraps off the known classification of real forms for the simple Lie superalgebras. However, I haven't been able to locate it. I'm also interested in the analogous question for the Lie superalgebras of types $P$ and $Q$.
| https://mathoverflow.net/users/29738 | Real forms of the general linear Lie superalgebra | It turns out that one *does* get all real forms of $\mathfrak{gl}\_{m|m}(\mathbb{C})$ in the way described in the question. To see this, one can show that a real form of $\mathfrak{gl}\_{m|m}(\mathbb{C})$ is uniquely determined by the induced real form of $\mathfrak{psl}\_{m|m}(\mathbb{C})$. This same method also works for the Lie superalgebras of type $P$ and $Q$. I've written up the details in Appendix B of [this paper](https://arxiv.org/abs/2301.01414).
| 0 | https://mathoverflow.net/users/29738 | 437953 | 176,919 |
https://mathoverflow.net/questions/437948 | 2 | $\DeclareMathOperator\Cov{Cov}$**Backround of my Question**
Let $Y$ be the response variable, $\mathbb{X}$ be the explanatory variables. The ultimate goal of prediction is finding a function $f^{\*}$ that minimize $\mathbb{E}[(Y - f^{\*}(\mathbb{X})^2)]$, we know that the solution is $f^{\*}(\mathbb{X}) = \mathbb{E}[Y | \mathbb{X}]$. Let $\epsilon = Y - \mathbb{E}[Y | \mathbb{X}]$, then we have
$$
Y = f^{\*}(\mathbb{X}) + \epsilon
$$
where $\mathbb{E}[\mathbb{\epsilon}] = 0$, and $\Cov[\mathbb{X}, \epsilon] = 0$.
So, I think the goal of regression somehow become finding a function $g \approx f^\*$ from some hypothesis space (e.g. $g(\mathbb{X}) = \mathbb{X}\beta$ for linear regression).
One way of defining $g \approx f^\*$ (which can persuade myself that it is a good approximation) is
$$
\mathbb{P}[|f^{\*}(\mathbb{X}) - g(\mathbb{X})| > \eta] < \delta
$$
for small $\eta$ and $\delta$.
**Question**
Given that $g \approx f^{\*}$ (in the sense defined above), if I can show that $\Cov[\widehat{Y}, \widehat{\epsilon}] \approx 0$ ($\widehat{Y} = g(\mathbb{X})$, and $\widehat{\epsilon} = Y - g(\mathbb{X})$), then when I see a residual plot like this
[Residual Plot](https://i.stack.imgur.com/7cAYM.png)
I can persuade myself this might be a signal suggesting that $g(\mathbb{X})$ probably a good approximation.
So, my question is how to show that $g \approx f^\* \Longrightarrow \Cov[\widehat{Y}, \widehat{\epsilon}] \approx 0$?
**My Attemp**
If further assume that $\mathbb{X}$ and $\epsilon$ are independent (is $\Cov[\mathbb{X}, \epsilon] = 0$ sufficient?), then $\Cov[f^{\*}\mathbb(X), \epsilon] = 0$, hence
\begin{equation}
\begin{aligned}
|\Cov[\widehat{Y}, \widehat{\epsilon}]| &= |\Cov[f^\*(\mathbb{X}) + (g(\mathbb{X}) - f^\*\mathbb(X)), (f^\*(\mathbb{X}) - g(\mathbb{X})) + \epsilon]| \\
&\leq |\Cov[f^{\*}(\mathbb{X}), f^{\*}(\mathbb{X}) - g(\mathbb{X})]| + |\Cov[g(\mathbb{X}) - f^{\*}(\mathbb{X}), f^{\*}(\mathbb{X}) - g(\mathbb{X})]| + |\Cov[g(\mathbb{X}) - f^{\*}(\mathbb{X}), \epsilon]| \\
& \leq |\Cov[f^{\*}(\mathbb{X}), f^{\*}(\mathbb{X}) - g(\mathbb{X})]| + |\Cov[g(\mathbb{X}) - f^{\*}(\mathbb{X}), \epsilon]| + |Var[f^\*(\mathbb{X}) - g(\mathbb{X})]|
\end{aligned}
\end{equation}
Intuitively, since $f^{\*}(\mathbb{X}) - g(\mathbb{X}) \approx 0$ with high probability, the last three terms may be small (depend on $\epsilon$ and $\delta$). But I don't know how to do it formally.
**Edit**
By @losifpinelis 's construction, $\exists (Y, \mathbb{X})$, for any given $\eta, \delta, M > 0$, there exists a function $g\_N$ such that
$$
\mathbb{P}[|f^{\*}(\mathbb{X}) - g\_N(\mathbb{X})| > \eta] < \delta
$$
but $|\Cov[g\_N(\mathbb{X}), \epsilon\_N]| > M$. Therefore, this is not a proper definition for this problem.
**My next Question** is "Is $\mathbb{E}[(f^{\*}(\mathbb{X}) - g(\mathbb{X}))^2] < \eta$ a proper definition of goodness of approximation?". That is, does
$$
\mathbb{E}[(f^{\*}(\mathbb{X}) - g(\mathbb{X}))^2] < \eta \Longrightarrow \Cov[\widehat{Y}, \widehat{\epsilon}] < SomeFunction(\eta)
$$
hold?
| https://mathoverflow.net/users/497398 | Justification of the use of residual plot | $\newcommand\ep\epsilon\newcommand{\de}{\delta}$Getting rid of the instances of $\approx$, one can state the question as follows:
>
> Let $f(X):=E(Y|X)$, where $X$ and $Y$ are random variables (r.v.'s) ($X$ possibly a multivariate one) such that $EY^2<\infty$. Note that $E\ep=0$ and $Cov(f(X),\ep)=0$, where $\ep:=Y-f(X)$.
>
>
> Suppose that for a sequence $(g\_n)$ of Borel-measurable functions one has $g\_n(X)\to f(X)$ in probability (as $n\to\infty$). Does it then follow that
> \begin{equation\*}
> Cov(g\_n(X),\ep\_n)\to0, \tag{1}\label{1}
> \end{equation\*}
> where $\ep\_n:=Y-g\_n(X)$?
>
>
>
The answer to this question is: Of course, not.
Indeed, let e.g. $X$ be a r.v. uniformly distributed on the interval $[-1,1]$, and let $Y:=X$, so that $f(X)=X$ and $\ep=0$.
Let
\begin{equation\*}
g\_n(X):=X\,1(|X|\ge1/n)+n^2 X\,1(|X|<1/n).
\end{equation\*}
Then $g\_n(X)\to X=f(X)$ in probability.
However, $Eg\_n(X)=0$, $\ep\_n=Y-g\_n(X)=X-g\_n(X)=(1-n^2)X\,1(|X|<1/n)$, and hence
\begin{equation\*}
Cov(g\_n(X),\ep\_n)=Eg\_n(X)\ep\_n=n^2(1-n^2)\,EX^2\,1(|X|<1/n) \\
=(1-n^2)/(3n)\to-\infty\ne0.
\end{equation\*}
So, \eqref{1} fails to hold. $\quad\Box$
---
On the other hand, if $g\_n(X)\to f(X)$ in $L^2$, then it is easy to see that \eqref{1} will hold.
Details on this: We have $\|g\_n(X)-f(X)\|\_2\to0$, where $\|Z\|\_2:=\sqrt{EZ^2}$.
We also have the identity
\begin{equation\*}
g\_n(X)\ep\_n=g\_n(X)(Y-g\_n(X))=f(X)(Y-f(X))+Y(g\_n(X)-f(X))+2f(X)(f(X)-g\_n(X))-(g\_n(X)-f(X))^2.
\end{equation\*}
Taking here the expectations and recalling that $Ef(X)(Y-f(X))=Ef(X)\ep=Cov(f(X),\ep)=0$, we get
\begin{equation\*}
Eg\_n(X)\ep\_n=EY(g\_n(X)-f(X))+2Ef(X)(f(X)-g\_n(X))-E(g\_n(X)-f(X))^2. \tag{2}\label{2}
\end{equation\*}
By the Cauchy--Schwarz inequality and the condition $\|g\_n(X)-f(X)\|\_2\to0$,
\begin{equation\*}
|EY(g\_n(X)-f(X))|\le\|Y\|\_2\,\|g\_n(X)-f(X)\|\_2\to0. \tag{3}\label{3}
\end{equation\*}
Note all that
\begin{equation}
\|f(X)\|\_2\le\|Y\|\_2<\infty \tag{3.5}\label{3.5}
\end{equation}
since $f(X)$ is an orthogonal projection of $Y$ in $L^2$. So,
\begin{equation\*}
|Ef(X)(f(X)-g\_n(X))|\le\|f(X)\|\_2\,\|g\_n(X)-f(X)\|\_2\to0. \tag{4}\label{4}
\end{equation\*}
Also,
\begin{equation\*}
E(g\_n(X)-f(X))^2=\|g\_n(X)-f(X)\|\_2^2\to0. \tag{5}\label{5}
\end{equation\*}
Collecting \eqref{2}--\eqref{5}, we get
\begin{equation\*}
Eg\_n(X)\ep\_n\to0. \tag{6}\label{6}
\end{equation\*}
Next, $\ep\_n=Y-g\_n(X)=Y-f(X)+f(X)-g\_n(X)=\ep+f(X)-g\_n(X)$. Taking here the expectations and recalling that $E\ep=0$, we get
\begin{equation}
|E\ep\_n|=|E(f(X)-g\_n(X))|\le\|g\_n(X)-f(X)\|\_2\to0. \tag{7}\label{7}
\end{equation}
Further, $|Eg\_n(X)-Ef(X)|\le E|g\_n(X)-Ef(X)|\le\|g\_n(X)-Ef(X)\|\_2\to0$, again by the Cauchy--Schwarz inequality and the condition $\|g\_n(X)-f(X)\|\_2\to0$. So,
\begin{equation}
Eg\_n(X)\to Ef(X), \tag{8}\label{8}
\end{equation}
and $|Ef(X)|\le\|f(X)\|\_2<\infty$ by the Cauchy--Schwarz inequality and \eqref{3.5}.
By \eqref{6}--\eqref{8},
\begin{equation}
Cov(g\_n(X),\ep\_n)=Eg\_n(X)\ep\_n-Eg\_n(X)\,E\ep\_n\to0-Ef(X)\times 0=0,
\end{equation}
---
In view of \eqref{3}, \eqref{3.5}, \eqref{4}, \eqref{5}, and \eqref{7}, one can also get an explicit bound on $|Cov(g\_n(X),\ep\_n)|$:
\begin{equation}
|Cov(g\_n(X),\ep\_n)|\le3y\de\_n+2\de\_n^2,
\end{equation}
where $y:=\|Y\|\_2$ and $\de\_n:=\|g\_n(X)-f(X)\|\_2$.
| 2 | https://mathoverflow.net/users/36721 | 437954 | 176,920 |
https://mathoverflow.net/questions/436932 | 6 | Which contractible spaces appear as retracts of the Hilbert cube or of $\Bbb R^\omega$ ?
One wants to think that a sufficiently “nice” contractible space is necessarily
a retract of the Hilbert cube or of $\Bbb R^\omega$ (if non-compact).
Is this view supported by theorems ?
Such a space would have to be both contractible and an absolute retract
(as a retract of a contractible space and of an absolute retract).
>
> What are sufficient “niceness” conditions which guarantee that a contractible space
> is homeomorphic to a retract of the Hilbert cube or of $\Bbb R^\omega$ ?
>
>
>
I would rather not assume compactness. If I understand correctly, the Hilbert cube is a retract of $\Bbb R^\omega$ (contract $(-\infty,0)$ to $0$ and $(1,\infty)$ to $1$) but not vice versa (as retracts preserves being compact), so really the question is about $\Bbb R^\omega$.
I understand that a contractible finite CW complex is necessarily a retract of the Hilbert cube but it is already false that a countable CW complex is necessarily a retract of $\Bbb R^\omega$, as follows from
>
> Let $C\mathbb N$ be the cone over a countably infinite discrete complex (this is a contractible 1-dimensional polyhedron). van Douwen and Pol [van Douwen, Eric K.; Pol, Roman Countable spaces without extension properties. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 25 (1977), no. 10, 987–991 ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=474182))] have constructed a countable regular $T\_2$ space $X$ (which is thus perfectly normal) and a function $A \to C\mathbb N$, defined on a certain closed $A\subseteq X$, which does not extend over any neighboourhood in $X$. In particular, the map of countable complexes $C\mathbb N\to{\*}$ is both a Hurewicz fibration and a homotopy equivalence, but is not soft wrt all perfectly normal pairs. – [Tyrone Nov 20 at 17:55](https://mathoverflow.net/questions/408590/are-acyclic-fibrations-of-nice-spaces-absolute-extensors-for-perfectly-normal-sp#comment1049180_408590)
>
>
>
| https://mathoverflow.net/users/494312 | When is a contractible space a retract of the Hilbert cube or $\Bbb R^\omega$? | Recall that a space $X$ is called an *absolute (neighbourhood) extensor for a property(=class) $\mathcal{P}$ of spaces*, abbreviated $AE(\mathcal{P})$ (respectively, $ANE(\mathcal{P})$), if for every space $Y$ with property $\mathcal{P}$ and every closed subspace $A\subset Y$, every continuous function $f:A\to X$ can be extended over $Y$ (respectively, over a neighbourhood of $A$ in $Y$). An *absolute (neighbourhood) retract for a property(=class) $\mathcal{P}$ of spaces*, abbreviated $AR(\mathcal{P})$ (respectively, $ANR(\mathcal{P})$) is a space $X$ which has $\mathcal{P}$ and moreover has the property that whenever $X$ is embedded as a closed subset of a space $Y$ with property $\mathcal{P}$, then $X$ is a retract of $Y$ (respectively, of a neighbourhood of $X$ in $Y$).
Now, necessary conditions for a space $X$ to be a retract of $\mathbb{R}^\omega$ is that $X$ be a separable, completely metrisable absolute retract. In fact these condition are sufficent.
>
> **Theorem**: Every separable, completely metrisable space embeds in $\mathbb{R}^\omega\cong\ell^2$ as a closed subspace. $\blacksquare$
>
>
>
Recall that a space $Y$ is **normal** if and only if for any closed subspace $B\subseteq Y$, any map $f:B\rightarrow\mathbb{R}$ has a continuous extension over $X$ (this is the Tietze extension theorem). In other words, $Y$ is normal if and only if $\mathbb{R}\in AE(Y)$.
It follows that $\mathbb{R}\in AE(normal)$, and in particular that $\mathbb{R}^\omega\in AE(normal)$, as is any retract of $\mathbb{R}^\omega$.
>
> **Lemma**: A Hausdorff space is an $AR(normal)$ ($ANR(normal)$) if and only if it is a normal $AE(normal)$ ($ANE(normal)$). $\quad\blacksquare$
>
>
>
Thus any retract of $\mathbb{R}^\omega$ is a metrisable $AR(normal)$. That the converse is true is a theorem due to Hanner [1].
>
> **Proposition**: The following statements about a space $X$ are equivalent.
>
>
> 1. $X$ is a separable, completely metrisable, $AR(metric)$.
> 2. $X$ is a metrisable $AR(normal)$.
> 3. $X$ is a retract of $\mathbb{R}^\omega$. $\blacksquare$
>
>
>
Note that an $ANR(metric)$ is an $AR(metric)$ if and only if it is contractible. While every contractible $ANR(normal)$ is an $AR(normal)$, I do not know if the converse holds.
Replacing $\mathbb{R}$ by $I=[0,1]$ we derive similar statements.
>
> **Corollary**: The following statements about a space $X$ are equivalent.
>
>
> 1. $X$ is a compact $AR(metric)$.
> 2. $X$ is a compact, metrisable $AR(normal)$.
> 3. $X$ is a retract of $[0,1]^\omega$. $\blacksquare$
>
>
>
Turning to the additional questions regarding CW complexes:
1. Every finite CW complex is a finite-dimensional, compact $ANR(metric)$. Thus every contractible, finite CW complex is a a retract of a finite-dimensional euclidean disc $D^n$.
2. Every metrisable CW complex belongs to $ANR(metric)$. Since a CW complex is metrisable iff it is locally compact, each metrisable CW complex is completely metrisable. A CW complex is separable if and only if it is countable. Every metrisable CW complex is a topological sum of countable subcomplexes.
3. It follows that a metrisable CW complex $X$ is contractible if and only if it is a retract of $\mathbb{R}^\omega$. If $X$ is finite-dimensional (combinatorial and topological dimensions agreeing in the case), then $\mathbb{R}^\omega$ may be replaced by a finite-dimensional euclidean space $\mathbb{R}^n$ in the last statement.
Finally, I will address a question from the comments. One direction in the following statement is clear, while the other follows from the fact that any Tychonoff space embeds into a product of intervals of potency no greater than its weight.
>
> **Proposition** The following statements about a Hausdorff space $X$ are equivalent.
>
>
> 1. $X$ is a compact $AR(normal)$ of weight $\leq\kappa$.
> 2. $X$ is a retract of $[0,1]^\kappa$. $\blacksquare$
>
>
>
Note that $\mathbb{R}^\kappa$ is not normal when $\kappa$ is uncountable.
[1] O. Hanner, *Solid spaces and absolute retracts*, Arkiv för Matematik **1**, (1951), 375-382.
| 3 | https://mathoverflow.net/users/54788 | 437956 | 176,921 |
https://mathoverflow.net/questions/437957 | 5 | Within ZFC, the [von Neumann hierarchy](https://en.wikipedia.org/wiki/Von_Neumann_universe) consists of sets $V\_\alpha$ indexed by ordinals, subject to the following conditions:
* $V\_0=\varnothing$.
* $V\_{\alpha+1}=\mathcal P(V\_\alpha)$.
* $V\_\lambda=\bigcup\_{\beta<\lambda}V\_\beta$ for limit $\lambda$.
My question is: **what is the formal justification for the last step?**
The [axiom of union](https://en.wikipedia.org/wiki/Axiom_of_union) would allow us to construct $V\_\lambda$ if we already had a set $\{V\_\beta:\beta<\lambda \}$. However, the existence of $V\_\beta$ for $\beta<\lambda$ does not obviously guarantee the existence of this set: $V\_{\omega\cdot 2}$ in ZF famously models itself minus the [axiom of replacement](https://en.wikipedia.org/wiki/Axiom_schema_of_replacement). This of course suggests that with replacement, this set could be constructed as the image of $\lambda$ (which under the von Neumann ordinal construction is the set of all lower ordinals) under the class function $V$.
But then $V$ would have to be a class function somehow defined through transfinite induction. Since predicates can't refer to themselves, and since we can't just assert the existence and uniqueness of a class function as a theorem the way we can with normal functions, the way this works eludes me.
| https://mathoverflow.net/users/147705 | Replacement axiom and the von Neumann hierarchy | This is really just a long comment, but the phrasing on the wikipedia page for replacement is verbose and perhaps obscuring how to use it here. Consider this version of replacement:
>
> For any set $X$ and binary predicate $\phi(-,-)$ such that for each element $x\in X$ there exists a unique set $y\_x$ such that $\phi(x,y\_x)$ is true, there exists a set $Y$ whose members are precisely the sets $y\_x$ such that there exists some $x\in X$ with $\phi(x,y\_x)$ true. We denote the set $Y$ guaranteed by this axiom together with a set $X$ and binary predicate $\phi(-,-)$ by $$\{y\_x:x\in X\}.$$
>
>
>
This is equivalent to all other standard phrasings of replacement over the rest of the $ZFC$ axioms, and is easier to use in the situation you outiline above. Specifically, take $X=\lambda$ and let $$\phi(-,-)=\text{All sets up to rank $-$ are members of $-$, and nothing else.}$$ which accepts ordinals in the first argument and arbitrary sets in the second. For each $\beta<\lambda$ we have that $\phi(\beta,V\_\beta)$ is true, and if $\phi(\beta,Z)$ is true for some other set $Z$ then $Z=V\_\beta$ by extensionality, so $V\_\beta$ is unique satisfying $\phi(\beta,V\_\beta)$ for all $\beta<\lambda$. Consequently $$\{V\_\beta:\beta\in\lambda\}$$ is a set by replacement, and the union of this set is $V\_\lambda$.
| 6 | https://mathoverflow.net/users/92164 | 437962 | 176,923 |
https://mathoverflow.net/questions/436938 | 2 | Since there are only finitely many residues mod $n$, there is some function $H(n) \le n$ such that for all integers $n>1$, $r$, and $e>H(n)$, if $r$ is an $e$-th power mod $n$ then there is some $p \le H(n)$ such that $r$ is a $p$-th power mod $n$. (As usual, $p$ denotes a prime number.)
Is $H\_0$ such a function, where $H\_0(n)=p$ is defined by
$$
(p-1)\# \le n \le p\#
$$
with the primorial $x\#=\prod\_{p\le x}p=\exp(\vartheta(x))$?
For example, 10 is not a square, cube, or fifth or seventh power mod 36, so this would say that it is not an $e$-th power for any $e>7$.
Are any such residues coprime to their modulus? (I assume not.)
| https://mathoverflow.net/users/6043 | Mod n, are all higher powers also lower powers? | So your question is as follows: let $n$ be a positive integer, $q$ the smallest prime such that $q\sharp > n$, $e>q$ be an integer and $r$ be an $e$-th power mod $n$. Then is $r$ a $p$-th power for some $p\leq q$?
In general, *no*. Take a large prime $q$ and $n$ be the greatest power of $2$ less than $q\sharp$: then $n \geq e^{(1+o(1))q}$ by the PNT, so we can find (PNT again) some prime $s>q$ such that $2^s < n$. Take $r=2^s$ and $e=s$, assume that $r$ is a $p$-th power for some $p \leq q$. But this implies that the $2$-adic valuation of $r$ is divisible by $p$, which is impossible.
---
Now, let’s state a simple condition under which the statement *holds* (albeit for mostly trivial reasons).
Note that $q\sharp= e^{\theta(q)} \in [c^{-1},c]e^q$ where $c=e^{0.007q/\ln{q}}$ (by eg [this](https://en.m.wikipedia.org/wiki/Chebyshev_function)): so if $n \geq 6$, we see that $n \leq q\sharp < 3^q$.
Our assumption is that we also have $v\_2(n) \leq q$.
If $e$ has a prime divisor $p \leq q$, then $p$ works. Otherwise, replacing $e$ with one of its prime divisors, we can assume that $e$ is prime.
As $\varphi(n) < n \leq q\sharp$, there is some $p\leq q$ not dividing $\varphi(n)$, so that taking the $p$-th power in $(\mathbb{Z}/d\mathbb{Z})^{\times}$ is an isomorphism for all $d |n$.
Moreover, if $s$ is a prime power dividing $n$ and coprime to $n/s$, such that $r$ and $s$ are not coprime, then the valuation of $s$ is at most $q$ (by the assumption), so the only noninvertible $e$-th power mod $s$ is zero (every noninvertible element to the $e$-th power is zero). Thus $r$ is zero mod $s$ (hence a $p$-th power).
It follows by CRT that $s$ is a $p$-th power mod $n$, QED.
| 1 | https://mathoverflow.net/users/165657 | 437966 | 176,925 |
https://mathoverflow.net/questions/437324 | 1 | Let $R$ be a commutative ring with unit. Let $M\_A$ denote the submodule of $R^m$ generated by columns of a matrix $A$ with entries in $R$. Suppose we are given two matrices $A,B \in R^{m \times k}$. I want to know if the following statement is true:
1. $M\_A = M\_B$ if and only if there exists an invertible matrix $P \in R^{k \times k}$ such that $A = BP$.
Note that one direction is super-easy: if $A = BP$ for invertible $P$, then the columns of both $A$ and $B$ generate the same submodule of $R^m$. So really the question is about the other direction. This is what I have tried for the "only if" direction: since $M\_A = M\_B$, we know that one can find matrices $P, Q \in R^{k \times k}$ such that $A = BP$ and $B = AQ$. From this we deduce that $A(I - QP)= B(I - PQ) = 0$. From here on, I don't really know what to do, but most likely this approach is just too naive.
However, I think the answer is true for a ring $R$ when $R$ is a principal ideal ring. Does anybody have a reference to this result (the particular case I want is when $R = \mathbb{Z}\_d$, so a reference to this will also suffice)? More generally, I'd like to know for which rings does this property hold, and where can I find a proof of this result.
EDIT: Actually now I'm beginning to think that the result may not be true in general for PIRs. However, some special cases are true. I know the following facts:
(i) If you replace PIR by PID (principal integral domain), then the result is true.
(ii) If you have a PIR, and both $A,B$ have one row, then the result is true. Proof goes like this: Since a PIR is a Hermite ring, there exists an invertible matrices $K,K'$ such that $AK = (s,0,\dots,0)$ and $BK' = (s',0,\dots,0)$. Thus $s,s'$ generate the same ideal, and thus they must be associates (which is a property of PIRs by a result of I.Kaplansky). The result now follows.
(iii) If you are allowed to add zero columns to both $A,B$, then also the result is true, because of the existence of the Howell normal form. (see [this](https://en.wikipedia.org/wiki/Howell_normal_form) and the references therein).
I'd like to stress that I'm interested in the case when the ring is $\mathbb{Z}\_d$.
| https://mathoverflow.net/users/151406 | Condition for equality of modules generated by columns of matrices | Interpreting the various matrices as maps between free modules in the usual way, the question becomes:
If $M$ is a submodule of $R^m$, and $\alpha,\beta: R^k\to M$ are epimorphisms, then must $\alpha$ and $\beta$ be equivalent, in the sense that there is an automorphism $f:R^k\to R^k$ with $\alpha = \beta f$?
For $R=\mathbb{Z}/d\mathbb{Z}$, or indeed any artinian $R$, the answer is yes, since then for finitely generated modules we have projective covers and a Krull-Schmidt theorem, so both $\alpha$ and $\beta$ are equivalent to the same epimorphism $\gamma:R^k=P(M)\oplus Q\to M$, where $P(M)$ is the projective cover of $M$, and $\gamma$ restricts to the usual map $P(M)\to M$ on $P(M)$ and the zero map on $Q$.
However, the answer is no if $R$ has a stably free module $P$ that is not free. For then, suppose $P\oplus R^n\cong R^k$. Then we have two epimorphisms $R^k\to R^n$, one with kernel $P$, and one with kernel $R^{k-n}$. Since they have non-isomorphic kernels, these epimorphisms can't be equivalent.
| 2 | https://mathoverflow.net/users/22989 | 437972 | 176,927 |
https://mathoverflow.net/questions/437992 | 2 | The Hecke group of level two, $\Gamma\_{0}(2)$, is an index-$2$ subgroup of the Fricke group of level two, $\Gamma\_{0}^{+}(2)$, i.e. $\left[\Gamma\_{0}^{+}(2):\Gamma\_{0}(2)\right] = 2$. The index of $\Gamma\_{0}(2)$ in the modular group, $\text{SL}(2,\mathbb{Z})$ is $\left[\text{SL}(2,\mathbb{Z}):\Gamma\_{0}(2)\right] = 3$. Using the multiplicative property of indices, we find that the index of $\Gamma\_{0}^{+}(2)$ in the modular group turns out to be fractional, i.e.
$$\mu = \left[\text{SL}(2,\mathbb{Z}):\Gamma\_{0}^{+}(2)\right] = \frac{\left[\text{SL}(2,\mathbb{Z}):\Gamma\_{0}(2)\right]}{\left[\Gamma\_{0}^{+}(2):\Gamma\_{0}(2)\right]} = \frac{3}{2}.$$
How can this be given that the indices of all other Fricke groups (of prime levels) in the modular group are integers?
| https://mathoverflow.net/users/99716 | Fractional group index? | $\Gamma\_0^+(2)$ is not a subgroup of ${\rm SL}\_2(\mathbb{Z})$: the elements of $\Gamma\_0^+(2)$ not in $\Gamma\_0(2)$ are fractional linear transformations such as $z \mapsto -1/(2z)$ that are represented by integer matrices (such as $\left(0\;-1\atop2\;\phantom-0\right)$) of determinant $2$. [Likewise for $\Gamma\_0^+(p)$ with $p$ an odd prime, even though the "index" $(p+1)/2$ turns out to be an integer in that case.]
| 3 | https://mathoverflow.net/users/14830 | 437996 | 176,937 |
https://mathoverflow.net/questions/437987 | 2 | I'm looking for a reference for the following:
Suppose that $f\_1,f\_2\colon S^n\rightarrow S^n$ are smooth maps. Let $i\colon S^n\rightarrow \mathbb{R}^{n+1}$ be the inclusion, and suppose that $F\colon S^n\times I \rightarrow \mathbb{R}^{n+1}$ is a smooth homotopy of $i\circ f\_1$ and $i\circ f\_2$. Moreover, suppose that 0 is a regular value for $F$. Then $$\deg(f\_2)-deg(f\_1)=d(F),$$ where $d(F)$ is the Brouwer degree of $F$ (i.e., $d(F)$ is $F^{-1}(0)$ counted with signs given by the Jacobian).
When $F^{-1}(0)=\emptyset$ this follows from the classical Hopf theorem, but I'm not sure how to prove this in general (though it should be equally classical).
| https://mathoverflow.net/users/497433 | Calculating degree via homotopy | I believe this follows from the fact that 'bordant' maps induce the same map on homology. So, for example: suppose $W$ is an oriented $(n+1)$-manifold with boundary and $W \to \mathbb{R}^{n+1}-\{0\}$ is a map. Then the composite $\partial W \to W \to \mathbb{R}^{n+1}-\{0\}$ is trivial on $H\_n$.
In your case you'd like to apply this in the following way:
1. Remover a tubular neighborhood of $F^{-1}(0)$ to get a manifold $W$ with boundary given (with orientations) by $\overline{S^n} \amalg S^n \amalg K$ where $K$ is the sphere bundle associated to the normal bundle of $F^{-1}(0)$.
2. Observe that $F^{-1}(0)$ was a finite collection of points, so $K$ is a union of spheres. The resulting maps have degree as you indicated (since you may shrink the original tubular neighborhood until the map $F$ is well approximated by its derivative).
To prove the fact, observe that the boundary map $H\_{n+1}(W,\partial W) \to H\_n(\partial W)$ takes the fundamental class to the fundamental class, hence is surjective, so that the next map in the sequence $H\_n(\partial W) \to H\_n(W)$ is zero.
| 6 | https://mathoverflow.net/users/6936 | 437999 | 176,938 |
https://mathoverflow.net/questions/437993 | 1 | Suppose $I(X;Y)$ denotes mutual information and on the other hand there is a relationship as follows.
\begin{align}
|p(y)-p(y|x)|<\delta p(y),\qquad\forall x,y.
\end{align}
Then we can say about mutual information:
\begin{align}
I(X;Y)=\sum\_{x,y}p(x,y)\log\frac{p(y|x)}{p(y)}\leq\sum\_{x,y}p(x,y)\log\frac{(1+\delta)p(y)}{p(y)}=\log(1+\delta)\leq\delta.
\end{align}
Is the opposite true? That is, if mutual information is less than $\epsilon$, can it be said that there is a $\delta$ in which first relationship is established?
| https://mathoverflow.net/users/68835 | Does bounding mutual information restrict the defined meter? | $\newcommand\ep\varepsilon$Of course not.
E.g., suppose that $P(X=0,Y=0)=t\ep$, $P(X=0,Y=1)=1/2-t\ep$, $P(X=1,Y=0)=(1-t)\ep$, and $P(X=1,Y=1)=1/2-(1-t)\ep$, where $t$ and $\ep$ are in the interval $(0,1)$. Then
$$I(X;Y)\sim c\_t\ep\to0$$
as $\ep\downarrow0$, where $c\_t:=\ln2+t\ln t+(1-t)\ln(1-t)$.
However, here $P(Y=0|X=0)=2t\ep$, which is not relatively close to $P(Y=0)=\ep$ if (say) $t=1/4$. (Or you can let $t\downarrow0$; then $c\_t\to\ln2$.)
---
The idea of this counterexample is quite simple: it is to let the conditional probability mass function (pmf) of $Y$ given $X$ differ substantially from its unconditional pmf only on a set of values rather unlikely to be taken by $Y$. This will affect the mutual information only little.
| 1 | https://mathoverflow.net/users/36721 | 438002 | 176,939 |
https://mathoverflow.net/questions/429008 | 11 | Given a well-ordering of $\mathbb{R}$, there is a natural way to define an associated partition of pairs of real numbers into two pieces: one assigns the value $0$ to a pair $r<s$ if the well-ordering agrees with the standard ordering on the pair, and gives it the value $1$ if not.
This construction is due originally to Sierpiński, and it is an important example/counterexample in Ramsey Theory because this coloring cannot have an uncountable homogeneous set: the separability of $\mathbb{R}$ tells us there must be a rational number between any two consecutive elements of a well-ordered (or reverse well-ordered) subset of $\mathbb{R}$.
In the square-bracket notation (discussed in my [last question](https://mathoverflow.net/questions/427961/does-2-aleph-0-rightarrow-aleph-12-3-require-that-the-continuum-is-weak)), Sierpiński's result says
$$2^{\aleph\_0}\nrightarrow [\aleph\_1]^2\_2.$$
Now Galvin and Shelah state the following in their paper [1]
>
> We remark that an easy generalization of Sierpiński's
> proof shows that $2^{\aleph\_0}\nrightarrow [\aleph\_1]^r\_{r!(r-1)!}$ for every positive integer r.
>
>
>
Their conclusion says that one can color the increasing $r$-tuples of reals with $r!(r-1)!$ colors in such a way that for any uncountable $X\subseteq\mathbb{R}$ each of these colors is realized by some $r$-tuple drawn from $X$.
I was unsuccessful at verifying their claim, as the most straightforward (to my mind) generalization of Sierpiński's proof provides a weaker coloring of $r$-tuples using $r!$-colors: one takes a well-ordering $<^\*$ of $\mathbb{R}$ and the coloring assigns to an $r$-tuple $a\_0<\dots<a\_{r-1}$ (in the usual ordering) the permutation of the index set $r$ that arises in the natural way once you rewrite your $r$-tuple in $<^\*$-increasing order. Once again, the separability of finite powers of $\mathbb{R}$ is what allows you to conclude that every color appears on every uncountable set. So, this argument establishes only
$$2^{\aleph\_0}\nrightarrow [\aleph\_1]^r\_{r!}$$
and my question is how does one improve to obtain the stronger result mentioned by Galvin and Shelah?
It is not out of the question that there is a typo in their paper, but I am also aware that their definition of ''easy generalization'' is probably not the same as my own.
[1] *Galvin, Fred; Shelah, Saharon*, [**Some counterexamples in the partition calculus**](https://doi.org/10.1016/S0097-3165(73)80004-4), J. Comb. Theory, Ser. A 15, 167-174 (1973). [ZBL0267.04006](https://zbmath.org/?q=an:0267.04006).
| https://mathoverflow.net/users/18128 | Higher-dimensional Sierpiński partitions | We may assume $r\ge3$. Let $\prec$ be a well-ordering of $\mathbb R$. For $n\in\mathbb N$ let $S\_n$ denote the set of all permutations of the set $[n]=\{1,2,\dots,n\}$.
Consider a set $X=\{x\_1,\dots,x\_r\}\in\binom{\mathbb R}r$ with $x\_1\lt x\_2\lt\cdots\lt x\_r$ and let $d\_i=|x\_i-x\_{i+1}|$ for $1\le i\le r-1$. Put $X$ in the class $C\_\sigma$ ($\sigma\in S\_r$) if $x\_{\sigma(1)}\prec x\_{\sigma(2)}\prec\cdots\prec x\_{\sigma(r)}$, and put $X$ in the class $D\_\tau$ ($\tau\in S\_{r-1}$) if $d\_{\tau(1)}\gt d\_{\tau(2)}\gt\cdots\gt d\_{\tau(r-1)}$. We have defined $r!(r-1)!$ disjoint classes $C\_\sigma\cap D\_\tau$ ($\sigma\in S\_r$, $\tau\in S\_{r-1}$). I claim that every uncountable subset of $\mathbb R$ contains members of each of these classes.
Let $A$ be an uncountable subset of $\mathbb R$. We may assume that $A$ has order type $\omega\_1$ in the well-ordering $\prec$, and that $U\cap A$ is uncountable whenever $U$ is open and $U\cap A\ne\varnothing$.
We recursively choose $r$ distinct points in $A$. The first two points are chosen arbitrarily. Now suppose $n$ points have been chosen, $2\le n\le r-1$. We designate one of the previously chosen points as a target, and the next point we choose is unequal to but very close to the target; say, at a distance less that $1/4$ of the minimum distance between any two previously chosen points. Let $X=\{x\_1,\dots,x\_r\}$ be the set of points chosen in this way, $x\_1\lt x\_2\lt\cdots\lt x\_r$; the indices do not, of course, represent the order in which the points were chosen.
It is easy to see that, by picking our targets appropriately, the set $X$ constructed in this way can be made to belong to any given class $D\_\tau$. E.g., if the first two points are $a\lt b$, the third point will be chosen close to $a$ if $\tau(2)\lt\tau(1)$, close to $b$ if $\tau(1)\lt\tau(2)$.
Next, for each $i\in[r]$ choose a very small neighborhood $U\_i$ of $x\_i$. Since $U\_i\cap A$ is uncountable, we can successively choose $y\_{\sigma(1)}\in U\_{\sigma(1)}\cap A$, $y\_{\sigma(2)}\in U\_{\sigma(2)}\cap A$, etc., so that $y\_{\sigma(1)}\prec y\_{\sigma(2)}\prec\cdots\prec y\_{\sigma(r)}$. Thus the set $Y=\{y\_1,\dots,y\_r\}\in\binom{\mathbb R}r$ belongs to $C\_\sigma$; moreover, it still belongs to $D\_\tau$ if the neighborhoods $U\_i$ were chosen small enough.
| 3 | https://mathoverflow.net/users/43266 | 438013 | 176,940 |
https://mathoverflow.net/questions/438001 | 14 | Let $C\_p$ b the cyclic group of order $p$, with $p$ a prime.
Is is possible for $C\_p$ to act (cellularly) on a rationally acyclic finite dimensional CW complex $X$ with no fixed points?
Standard Smith Theory implies that for this to be possible, $H\_\*(X;\mathbb Z/p)$ would have to be infinite dimensional. (So, e.g., with $p=2$, $X$ might be homotopy equivalent to an infinite wedge of $\mathbb RP^2$s.)
I am rather hoping that an example like this exists. But a proof that this can't happen would be fine too!
| https://mathoverflow.net/users/102519 | Can a cyclic group of prime order act on a rationally acyclic finite dimensional complex and have no fixed points? | Yes, I think you can make an example like this (for $p=2$, but it generalizes).
Let $R$ be the group ring $\mathbb Z[C\_2]=\mathbb Z[x]/(x^2-1)$. Make a chain complex of free $R$-modules
$$
M\_0 \leftarrow M\_1 \leftarrow M\_2
$$
as follows.
$M\_0=R$.
$M\_1$ has a basis $(a\_n)$ indexed by $n\ge 0$.
$M\_2$ has a basis $(b\_n)$ also indexed by $n\ge 0$.
$\partial a\_0=x-1$, $\partial a\_n=0$ when $n>0$, $\partial b\_n=(1+x)a\_n+(1-x)a\_{n+1}$.
The homology is such that $H\_0\cong\mathbb Z$ (trivial action), $H\_2=0$, and $H\_1$ has exponent $4$.
Now build a cell complex with free $C\_2$-action, having this as its complex of cellular chains. There is one orbit of $0$-cells, say $e^0$ and $xe^0$. There are $1$-cells $e^1\_n$ and $xe^1\_n$. The cell $e^1\_0$ is attached to $e^0$ and $xe^0$, while for $n>0$ both ends of $e^1\_n$ are attached to $e^0$. There are $2$-cells $e^2\_n$ and $xe^2\_n$, with the attaching map for $e^2\_n$ representing the appropriate $1$-dimensional homology class of this $1$-skeleton.
| 17 | https://mathoverflow.net/users/6666 | 438014 | 176,941 |
https://mathoverflow.net/questions/437607 | 1 | Let $X$ be completely regular space, $\beta X$ be Stone-Čech
compactification of $X$, and $\upsilon X$ be Hewitt realcompactification of $X$.
Then $X\subset \upsilon X\subset \beta X$.
If the remainder $\beta X\setminus X$ is countably compact space, then can $%
\beta X\setminus \upsilon X$ be countably compact?
Under what conditions is it countably compact?
Does anyone have an article or book recommendation on this?
| https://mathoverflow.net/users/86099 | A question about realcompact spaces | There exists a Tychonoff space $X$ such that $\beta X\setminus X$ is countably compact but $\beta X\setminus \upsilon X$ is locally compact, $\sigma$-compact and not countably compact. Such a space can be constructed as follows.
In the Stone-$\check{\mathrm C}$ech remainder $\omega^\*=\beta\omega\setminus \omega$ choose a sequence $(U\_n)\_{n\in\omega}$ of pairwise disjoint nonempty open sets. In each $U\_n$ choose a point $x\_n\in U\_n$ and using the Tychonoff property of $\omega^\*$, find a continuous function $f\_n:\omega^\*\to [0,1]$ such that $f\_n(x\_n)=1$ and $f\_n[\omega^\*\setminus U\_n]=\{0\}$. Then $K\_n=f\_n^{-1}(1)$ is a nonempty compact $G\_\delta$-set in $\omega^\*$ and $(K\_n)\_{n\in\omega}$ is a discrete sequence of compact $G\_\delta$ sets in $\omega^\*$ whose union $K=\bigcup\_{n\in\omega}K\_n$ is a $\sigma$-compact $G\_\delta$-set in $\omega^\*$ and $K$ is open in its closure $\overline K$ in $\omega^\*$. Consider the set $Y=\omega^\*\setminus K$ and observe that $Y$ is of type $F\_\sigma$ and $G\_\delta$ in $\omega^\*$.
It is well-known that the space $\omega^\*$ has exactly $2^{\mathfrak c}$ closed sets and every infinite closed subset of $\omega^\*$ has cardinality $2^{\mathfrak c}$. Using these two facts, it is easy to construct (by transfinite induction of length $2^{\mathfrak c}$) a Bernstein-like set $B\subseteq Y$ such that for every infinite closed subset $F\subseteq Y$ the sets $F\cap B$ and $F\setminus B$ are nonempty. Then the subspace $X=B\cup\omega$ of the space $\beta\omega$ has the desired property.
Indeed, its Stone-$\check{\mathrm C}$ech compactification coincides with $\beta \omega$ (because every bounded continuous function $f:X\to\mathbb R$ uniquely extends to a continuous function on $\beta\omega$). The space $\beta X\setminus X=\beta\omega\setminus X=K\cup(Y\setminus B)$ is countably compact because for every countable infinite discrete subset $C\subseteq K\cup(Y\setminus B)$ either for some $n\in\omega$ the intersection $C\cap K\_n$ is infinite and hence has an accumulation point $x\in K\_n\subset \beta X\setminus X$ or else for every $n\in\omega$ the intersection $K\_n\cap C$ is finite and the set $\overline C\setminus C$ of accumulation points of $C$ is contained in $Y$. Since the closed set $\overline C$ is infinite, it has cardinality $2^{\mathfrak c}$ and hence $(\overline C\setminus C)\setminus B$ is non-empty, which means that the set $C$ has an accumulation point in $Y\setminus B\subseteq \beta X\subseteq X$ and $\beta X\setminus X$ is countably compact.
Next, we show that the Hewitt completion $\upsilon X$ coincides with $\beta\omega\setminus K$. Indeed, for every $x\in K$ we can find $n\in\omega$ such that $x\in K\_n$ and hence $K\_n$ is a $G\_\delta$-subset of $\beta\omega$ that contains $x$ and is disjoint with $X$, witnessing that $x\notin\upsilon X$.
On the other hand, for every $x\in Y$, every $G\_\delta$-set $G\subseteq\beta\omega$ that contains $x$ has nonempty intersection with the Bernstein-like set $B\subseteq X$ (because $G$ contains a non-empty and hence infinite closed $G\_\delta$-subset, which intersects $B$ by the choice of $B$). This means that $Y\subseteq\upsilon X$ and hence $\upsilon X=\beta\omega\setminus K$.
It remains to observe that the complement $\beta X\setminus\upsilon X=K$ is locally compact, $\sigma$-compact and not countably compact, even not pseudocompact.
| 4 | https://mathoverflow.net/users/61536 | 438023 | 176,943 |
https://mathoverflow.net/questions/438011 | 11 | Nirenberg's paper [*On elliptic PDEs*](http://www.numdam.org/item/?id=ASNSP_1959_3_13_2_115_0) claims that if a function $f$ on $\mathbb{R}^n$ tends to zero at infinity or is in $L^q$ for any $q < \infty$ then the "interpolation" inequality
$$
\lVert∇ f \rVert\_{L^{2p}} ≤ C \left(\lVert f\rVert\_{L^\infty} \lVert ∇^2 f \rVert\_{L^p}\right)^{1/2}
$$
holds (theorem on p. 125).
This is proved by establishing the inequality
$$
\lVert∇ f \rVert\_{L^{q}} ≤ C \left(\lVert f\rVert\_{L^r} \lVert ∇^2 f \rVert\_{L^p}\right)^{1/2} \label{1}\tag{1}
$$
(equation (2.5) in the proof of that theorem)
for any $1 < r,p < \infty$ and $\frac{2}{q} = \frac{1}{r} + \frac{1}{p}$ with a constant $C$ **independent of $r, p$**, and the taking the limit $r \to \infty$.
I haven't really looked at Nirenberg's proof, but as far as I can see it is "calculus-based".
On the other hand inequality \eqref{1} looks like the interpolation inequality
$$
\lVert f \rVert\_{W^{1,q}} ≤ C\_{p, r} \left(\lVert f\rVert\_{L^r} \lVert f \rVert\_{W^{p, 2}}\right)^{1/2} \label{2}\tag{2}
$$
for $1 < r, p < \infty$ and $q$ as above,
that you get by using harmonic analysis (the Mihlin multiplier theorem and the Littlewood--Paley characterization of Sobolev spaces, to be precise) to establish that $W^{1,q}$ is the $\frac{1}{2}$-interpolation space between $W^{2,p}$ and $W^{0,r} = L^r$ (e.g. Bergh--Löfström, *An introduction to interpolation spaces*, Thm 6.4.5).
Is there some way to get \eqref{1} using the same methods as for \eqref{2} (maybe only with $\lVert f \rVert\_{W^{p,2}}$ on the RHS instead of $\lVert ∇^2 f \rVert\_{L^p}$), but with the constant independent of $p, r$, so that you can conclude the case $r = \infty$?
Or even better, is there some way to modify the interpolation arguments to give the $r = \infty$ case directly?
| https://mathoverflow.net/users/123448 | Is it possible to obtain the inequality $\|\nabla f\|_{L^{2p}} \leq C (\|f\|_{L^\infty} \|f\|_{W^{2, p}})^{1/2}$ from interpolation/harmonic analysis? | Here is a Littlewood-Paley + interpolation style proof. If $P\_k$ denotes a Littlewood-Paley projection to frequencies $|\xi| \sim 2^k$, then the standard Littlewood-Paley characterisations of Sobolev spaces give
$$ \| \nabla f \|\_{L^{2p}} \sim \| (\sum\_k 2^{2k} |P\_k f|^2)^{1/2} \|\_{L^{2p}}$$
and
$$ \| \nabla^2 f \|\_{L^{p}} \sim \| (\sum\_k 2^{4k} |P\_k f|^2)^{1/2} \|\_{L^{p}}$$
while
$$ \| \sup\_k |P\_k f| \|\_{L^\infty} \lesssim \|f\|\_{L^\infty}$$
so by Holder's inequality it suffices to establish the pointwise bound
$$ (\sum\_k 2^{2k} |P\_k f(x)|^2)^{1/2}
\lesssim A(x)^{1/2} B(x)^{1/2}$$
where
$$ A(x) := (\sum\_k 2^{4k} |P\_k f(x)|^2)^{1/2}$$
and
$$ B(x) := \sup\_k |P\_k f(x)|.$$
In fact we have the stronger estimate
$$
{\sum\_k 2^{k} |P\_k f(x)| \lesssim A(x)^{1/2} B(x)^{1/2}} \label{3} \tag{$\ast$}
$$
since from the triangle inequality we have
$$ \sum\_{k \leq k\_0} 2^{k} |P\_k f(x)| \lesssim 2^{k\_0} B(x)$$
and from Cauchy-Schwarz we have
$$ \sum\_{k > k\_0} 2^{k} |P\_k f(x)| \lesssim 2^{-k\_0} A(x)$$
and the claim follows by optimising in $k\_0$. (The estimate \eqref{3} also likely follows from some interpolation lemma on weighted $\ell^p$ spaces in Bergh-Lofstrom, but I don't have the reference handy.)
This argument in fact shows that we can replace the $W^{1,p}$ norm by the slightly stronger Triebel-Lizorkin norm $F^{1,p}\_1$ if desired.
| 12 | https://mathoverflow.net/users/766 | 438037 | 176,948 |
https://mathoverflow.net/questions/438038 | 1 | Let $\mu$ be *a finite complex valued measure* on $\mathbb{R}$ and let $\hat{\mu}$ be it's Fourier–Stieltjes transform
$$
\hat{\mu}(\omega)= \int\_{\mathbb{R}} e^{it\omega} d \mu(t)
$$
**Question:** Does $\hat{\mu}$ uniquely determine $\mu$? I am fairly sure that it does. However, I was not able to locate my standard references. Is there a good place where I can find proof of this fact?
| https://mathoverflow.net/users/69661 | Uniqueness of Fourier–Stieltjes transform for finite complex valued measures | I assume that $\mu$ is a regular complex Borel measure. Assume that $\widehat{\mu}=0$. Let $f \in \mathcal{S}(\mathbb{R})$ be a Schwartz class function. Then, writing $f=\widehat{g}$ for $g\in \mathcal{S}(\mathbb{R})$, we have from Fubini's theorem that
$$
\int\_{\mathbb{R}}f(t)d\mu(t) = \int\_{\mathbb{R}}\int\_{\mathbb{R}}g(s)e^{-its}dsd\mu(t) = \int\_{\mathbb{R}}g(s)\widehat{\mu}(-s)ds = 0.
$$
This implies $\mu$ is orthogonal to the Schwartz class, and by density, it's orthogonal to the space $C\_0(\mathbb{R})$ of continuous functions vanishing at infinity, which implies by Riesz's theorem that $\mu=0$.
| 4 | https://mathoverflow.net/users/78745 | 438039 | 176,949 |
https://mathoverflow.net/questions/438010 | 1 | suppose we have a $p$-stabilized newform $f$ of classical level $\Gamma\_0(p)$; then there is a unique ordinary deformation into a Hida family $\mathbf{f}$ defined over a finite flat extension $R/\Lambda$, and a corresponding representation $\rho: G\_{\mathbb{Q}}\to \text{GL}\_2(\text{Frac}(R))$ on a big vector space $\mathbb{V}$. Under some mild conditions, there exists a free rank-two $R$-lattice $\mathbb{T}\subset \mathbb{V}$.
I believe there should be some kind of autoduality for $\mathbb{V}$, and maybe $\mathbb{T}$ as well, in the form of a perfect pairing
$$
\mathbb{T} \times \mathbb{T} \to R(1)
$$
or similar, which is equivariant. Does anyone have a reference where this kind of thing is spelled out?
| https://mathoverflow.net/users/120548 | Reference for auto-duality of nearly ordinary deformations associated to Hida families | [Section 1.6 in Fukaya-Kato](https://www.math.ucla.edu/%7Esharifi/sharificonj.pdf) spells it out very nicely; the same material appeared earlier in section 4.1 of Ohta's *On the $p$-adic Eichler-Shimura isomorphism for $\Lambda$-adic cusp forms*.
| 1 | https://mathoverflow.net/users/120548 | 438043 | 176,950 |
https://mathoverflow.net/questions/437909 | 5 | $\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$I'm reading [this article by Boyd and Chua [1]](https://web.stanford.edu/%7Eboyd/papers/pdf/fading_volterra.pdf), in which they prove the approximability of arbitrary time-invariant (TI) operators (or filters) with polynomials of some basis filters with separation and *fading memory* (FM) properties. This shows that those operators, indeed have fintie Volterra expansion approximates.
Consider the following:
* $C(\Bbb R)$ as the space of bounded continuous functions $u: \Bbb R \to \Bbb R$,
* $\norm u = \sup\_{t \in \Bbb R} |u(t)|$ being their norm,
* $\Bbb R\_-$ as $\{t : t \le 0 \}$, with analogous definitions for $C(\Bbb R\_-)$ and $\norm\cdot$ as above,
* and time-invariant functionals $F: C(\Bbb R\_-) \to \Bbb R$ and time-invariant operators $N: C(\Bbb R) \to C(\Bbb R)$.
* $K=\bigl\{u\in C(\Bbb R): \abs{u(t)}\le M\_1\land\abs{u(s)-u(t)}\le M\_2(s-t)\;\forall t, s\in\Bbb R, t<s\bigr\}$: explicitly, $K$ is the space of bounded uniformly Lipschitz continuous functions. Functions belonging to $K$ are called *signals*.
* $K\_-=\{u\in K : u(t)=0\text{ if }t>0\}$: Boyd & Chua prefer to define $K\_-$ by using a "projection" operator $P$ such that
$$\DeclareMathOperator{\dmu}{d\!}
Pu(t) =
\begin{cases}
u(t) & t\le 0\\
0 & t>0
\end{cases}
$$ and then noting that $K\_- =P K$. Be it noted that $K\_-$ is compact in $C(\Bbb R)$ but only with respect to the weighted $\sup$ norm
defined as
$$
\norm u\_w=\sup\_{t\le 1} \abs{u(t)w(-t)}
$$
where $w:\Bbb R\_+\to(0,1]$ is a weight function such that $\lim\_{t\to\infty} w(t)=0$ (see below).
* For every TI operator $N$, an associated $F$ is defined by $Fu=Nu\_e(0)$, in which $u\_e$ is some extension of $u$ from $C(\Bbb R\_-)$ to $C(\Bbb R)$.
* Furthermore, the fading memory (FM) property is defined for TI operators on $N \in K \subset C(\Bbb R)$ as the existence of some weight function $w: \Bbb R\_+ \to (0,1]$ with $\lim\_{t \to \infty} w(t) = 0$ that makes $K\_-$ closed w.r.t. the weighted norm $w$. i.e. for any $v,u \in K$, one can find a $\delta$ for any $\epsilon$ such that
$$\norm{u-v}\_w:= \sup\_{t\le 0} \abs{u(t)-v(t)}w(-t) < \delta \implies \abs{Nu(0) - Nv(0)} < \epsilon$$
*Lemma 2* states that there are some functionals on $K\_-$ that separate points. To prove this, a class of functionals $G \in \mathbf G \subset K\_-$ is defined by
$$\mathbf{G}=\left\{G(u)=\int\limits\_0^{+\infty}g(\tau)u(-\tau)\dmu\tau :u\in K\_-\land\int\limits\_0^{+\infty}\abs{g(\tau)}w(\tau)^{-1}\dmu\tau\right\}$$
which are (shown to be) continuous w.r.t. the weighted norm $w$ (thanks to the condition introduced above). The authors further construct functions $g\_0$ defined by:
$$g\_0(t):= [u(-t)-v(-t)] w(t)\exp (-t)$$
and their associated $G\_0$ **assuming** that they belong to $\mathbf G$. Then, they show that $g\_0$ indeed separates points on $K\_-$. However, **I don't understand why $G\_0 \in \mathbf G$ in the first place.** $g\_0$, by construction, dependends on $u$ and $v$. Thus, it does not look to me that $G\_0$ is even time-invariant (simply shifting $u$ and $v$ in time will yield a different $g\_0$). So my question is why is the proof of this lemma correct?
**Reference**
[1] Stephen Boyd, Leon O. Chua, "[Fading memory and the problem of approximating nonlinear operators with Volterra series](https://web.stanford.edu/%7Eboyd/papers/pdf/fading_volterra.pdf)" (English), IEEE Transactions on Circuits and Systems 32, 1150-1161 (1985), [MR0809696](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0809696), [Zbl 0587.93028](https://zbmath.org/0587.93028), doi:[10.1109/TCS.1985.1085649](https://doi.org/10.1109/TCS.1985.1085649).
| https://mathoverflow.net/users/123104 | Boyd & Chua 1985: Is the proof of Lemma 2 correct? | Don't let the bad wording of the paper fool you. The statement of lemma 2 in [1] above does not mean that the functional $G$ constructed by the kernel $g\_0$ separates all the points $u, v\in K\_-$ such that $u\neq v$: *it means only that, for any such two functions, you can construct a kernel such that the associated linear functional constructed as above is such that $Gu\neq Gv$*. At first I was fooled by the same thought but then I read *Remark 4* in the same page ([1], §IV p. 1153), which contains the statement quoted below
>
> Suppose $E$ is a compact metric space and $\mathbf G$ a set of continuous functionals on $E$ which separate points, that is, for any distinct $u,v\in E$ there is a $G\in\mathbf G$ such that $Gu \neq Gv$.
>
>
>
Then I simply realized that the authors want only to show that the functional
$$\DeclareMathOperator{\dmu}{d\!}G\_0(q)=\int\limits\_0^{+\infty}g\_0(\tau)q(-\tau)\dmu\tau \quad \forall q\in K\_-
$$
with the kernel $g\_0$ constructed as described above is such that $G\_0 u\neq G\_0v$ for **any fixed couple of different functions chosen for its construction**. If $u\_1, v\_1\in K\_-$ are a different couple such that $u\_1\neq v\_1$, you do not know if $G\_0u\_1\neq G\_0 v\_1$ (an in general this is not true): however you can construct another functional, say $G\_1$, whose kernel is defined as exactly as $g\_0$ i.e.
$$
g\_1(t):= [u\_1(-t)-v\_1(-t)] w(t)\exp (-t)
$$
and separates the new points but obviously possibly not the points $u$ and $v$.
| 6 | https://mathoverflow.net/users/113756 | 438052 | 176,954 |
https://mathoverflow.net/questions/435828 | 0 | I am interested in the low-rank approximation of a binary(01) third-order tensor. Does anyone know how to define its tensor nuclear norm based on whatever tensor decomposition methods? Could anyone provide some useful references?
| https://mathoverflow.net/users/65795 | Tensor nuclear norm for a binary 3rd-order tensor | I apologize that I'm not more knowledgeable about tensor calculus, but I did find this on arXiv: [Friedland and Lim - Nuclear Norm of Higher-Order Tensors](http://arxiv.org/abs/1410.6072). I hope you find an answer to your question.
| 0 | https://mathoverflow.net/users/497473 | 438056 | 176,955 |
https://mathoverflow.net/questions/438053 | 0 | Let $\kappa>0$ and $d,k$ be a positive integers with $k\ge d$. For $k\in \mathbb{Z}^+$ large enough, can one find a geodesically complete and simply connected $d$-dimensional Riemannian submanifold $(M\_{\kappa},g)$ of the $k$-fold product of the hyperbolic plane $\prod\_{i=1}^k\, \mathbb{H}^2$ with sectional curvature bounded in $[-\kappa,0)$?
Ideally, can we take it to have constant sectional curvature $-\kappa$?
| https://mathoverflow.net/users/491352 | Hadamard submanifolds of $k$-fold product of hyperbolic plane | For the $i$th factor ${\mathbb H}^2$ in the product of hyperbolic planes, pick a complete geodesic $c\_i$, $i=1,...,k$. The product
$$
F=c\_1\times ... \times c\_k\subset X=\prod\_{i=1}^k {\mathbb H}^2
$$
is a $k$-flat, i.e. a totally-geodesic (although we do not need this) isometrically embedded Euclidean subspace of
dimension $k$. Let $(M,g)$ be a $d$-dimensional Riemannian manifold (no restrictions whatsoever). By the Nash isometric embedding theorem, there exists an isometric embedding
$$
f: (M,g)\to F,
$$
provided that $k$ is much larger than $d$. (Specifically, you can take any $k\ge (3d+11)/2$.) Then the composition of $f$ with the identity embedding $F\to X$, gives an isometric embedding $(M,g)\to X$. The image of this embedding (with the Riemannian metric induced from $X$) is a Riemannian submanifold of $X$ isometric to $(M,g)$. Now, if you wish, take $(M,g)$ to be say, the $d$-dimensional hyperbolic space (complete, simply-connected, constant curvature $-1$).
Edit. It is possible that you are using a nonstandard notion of a Riemannian submanifold and what you really mean is a **totally geodesic** submanifold. Then the only complete negatively curved totally geodesic submanifolds of $X$ (of dimension $\ge 2$) are hyperbolic planes.
| 4 | https://mathoverflow.net/users/39654 | 438058 | 176,956 |
https://mathoverflow.net/questions/438067 | 3 | A topological space $X$ is called a **$\sigma$-space** if every $F\_{\sigma}$-subset of $X$ is $G\_{\delta}$.
A topological space $X$ is called a **$Q$-space** if any subset of $X$ is $F\_{\sigma}$.
**Definition.** A topological space $X$ is called a **hereditary $\sigma$-space** if every subset of $X$ is $\sigma$-space.
**Question.** Is there a hereditary $\sigma$-space $X$ such that it is not $Q$-space ?
| https://mathoverflow.net/users/112417 | Is there a hereditary $\sigma$-space $X$ such that it is not $Q$-space? | Every $S\_1(B\_\Gamma,B\_\Gamma)$ space is a $\sigma$-space, and the property $S\_1(B\_\Gamma,B\_\Gamma)$ is hereditary for subsets (B. Tsaban and M. Scheepers, [The combinatorics of Borel covers](https://doi.org/10.1016/S0166-8641(01)00078-5), Topology and its Applications 121 (2002), 357-382.)
For example, a Sierpiński set satisfies $S\_1(B\_\Gamma,B\_\Gamma)$.
On the other hand, it is very difficult to be a Q-set. For example, Q-sets have Lebesgue measure zero. In particular, they cannot be Sierpiński.
| 4 | https://mathoverflow.net/users/2415 | 438073 | 176,960 |
https://mathoverflow.net/questions/438070 | -2 | For given $N$ and $K$ I need to compute:
$\displaystyle\sum\_{}\prod\_{K\_1}^{K\_N}\binom{N}{k\_i}$ such that $\displaystyle\sum\_{1}^Nk\_i=K$
The outer $\displaystyle\sum\_{}$ is to indicate that I need sum of all such products,
for example if $N=3$ and $K=5$
the following are sets of $k\_i$ = {0,0,5}, {0,5,0}, {5,0,0}, {0,1,4},{0,4,1}, {1,0,4},{4,0,1},{4,1,0},{1,4,0}, {0,2,3}, {0,3,2}, {2,0,3}, {3,0,2}, {2,0,3}, {3,0,2}, {1,2,2}, {2,2,1}, {2,1,2}, {3,1,1} {1,3,1}, {1,1,3}
Then I need,
$\binom{3}{0}\binom{3}{0}\binom{3}{5}
+\binom{3}{0}\binom{3}{5}\binom{3}{0}
+\binom{3}{5}\binom{3}{0}\binom{3}{0}
+\binom{3}{0}\binom{3}{1}\binom{3}{4}
+\binom{3}{0}\binom{3}{4}\binom{3}{1}
+\binom{3}{4}\binom{3}{1}\binom{3}{0}
+\binom{3}{1}\binom{3}{0}\binom{3}{4}
+\binom{3}{4}\binom{3}{0}\binom{3}{1}
+\binom{3}{3}\binom{3}{0}\binom{3}{2}
+\binom{3}{0}\binom{3}{3}\binom{3}{2}
+\binom{3}{3}\binom{3}{2}\binom{3}{0}
+\binom{3}{2}\binom{3}{2}\binom{3}{1}
+\binom{3}{2}\binom{3}{1}\binom{3}{2}
+\binom{3}{3}\binom{3}{1}\binom{3}{1}
+\binom{3}{1}\binom{3}{3}\binom{3}{1}
+\binom{3}{1}\binom{3}{1}\binom{3}{3}$
| https://mathoverflow.net/users/497485 | How can I efficiently compute the following series | It is a coefficient of $x^K$ in $(1+x)^N\cdot (1+x)^N\cdots (1+x)^N=(1+x)^{N^2}$, thus ${N^2\choose K} $
| 4 | https://mathoverflow.net/users/4312 | 438074 | 176,961 |
https://mathoverflow.net/questions/438059 | -1 | Let $N = q^k n^2$ be an odd perfect number with *special prime* $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the *classical sum of divisors* of the positive integer $x$ by $\sigma(x)=\sigma\_1(x)$.
By the definition of a *perfect number* $N$, we have $\sigma(N)=2N$. Since $\gcd(q,n)=1$ and because the divisor sum $\sigma$ is a multiplicative function, it follows that
$$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2 q^k n^2.$$
Now, just like when solving equations in terms of one variable, one should solve the equality
$$\sigma(q^k)\sigma(n^2)=2 q^k n^2$$
in terms of one of the expressions.
If one solves for $n$, that introduces a square-root, and takes one out of the integers. If one solves for $q$, it introduces a $k$th root, and also takes one out of the integers. One should not solve for $\sigma(q^k)$, since that quantity can already be re-expressed in terms of $q$ and $k$ as
$$\sigma(q^k)=\frac{q^{k+1} - 1}{q - 1},$$
if needed. So, the more natural quantity to solve for is $\sigma(n^2)$.
Indeed, throughout [this paper](https://arxiv.org/abs/2202.08116), we implicitly rely on the simple equality
$$\sigma(n^2) = \frac{2q^k n^2}{\sigma(q^k)}. \tag{1}$$
Unfortunately, this seems to introduce fractions. To avoid that, we can use prime factorizations, as follows. Write the prime factorization of $n$ as
$$n = {p\_1}^{a\_1} \cdots {p\_m}^{a\_m},$$
for some unique odd primes $3 \leq p\_1 < \ldots < p\_m$, and for some positive integer exponents $a\_1, \ldots, a\_m$. Since both sides of $(1)$ are integers, and since $q \equiv k \equiv 1 \pmod 4$ with $q$ prime, we know that
$$\sigma(q^k) = 2 {p\_1}^{b\_1} \cdots {p\_m}^{b\_m}$$
for some nonnegative integers $0 \leq b\_i \leq 2a\_i$. Thus, we have
$$\sigma(n^2) = q^k {p\_1}^{2a\_1 - b\_1} \cdots {p\_m}^{2a\_m - b\_m}.$$
Note that the only other facts about odd perfect numbers that we use in [this paper](https://arxiv.org/abs/2202.08116) are that $b\_i < 2a\_i$ for at least one index $i$, and that $q$ is different from the $p\_i$'s.
---
With this information, we immediately see that
$$G := \gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\left(2 {p\_1}^{b\_1} \cdots {p\_m}^{b\_m},q^k {p\_1}^{2a\_1 - b\_1} \cdots {p\_m}^{2a\_m - b\_m}\right)$$
$$= {p\_1}^{\min(b\_1,2a\_1 - b\_1)} \cdots {p\_m}^{\min(b\_m,2a\_m - b\_m)},$$
$$H := \gcd\left(n^2,\sigma(n^2)\right) = \gcd\left({p\_1}^{2a\_1} \cdots {p\_m}^{2a\_m}, q^k {p\_1}^{2a\_1 - b\_1} \cdots {p\_m}^{2a\_m - b\_m}\right)$$
$$= {p\_1}^{2a\_1 - b\_1} \cdots {p\_m}^{2a\_m - b\_m},$$
and
$$I := \gcd\left(n,\sigma(n^2)\right) = \gcd\left({p\_1}^{a\_1} \cdots {p\_m}^{a\_m}, q^k {p\_1}^{2a\_1 - b\_1} \cdots {p\_m}^{2a\_m - b\_m}\right)$$
$$= {p\_1}^{\min(a\_1,2a\_1 - b\_1)} \cdots {p\_m}^{\min(a\_m,2a\_m - b\_m)}.$$
---
Here is our:
>
> **QUESTION:** What is the (simplified) prime factorization for $\gcd(G,J)$, where
> $$G = \gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\left(2 {p\_1}^{b\_1} \cdots {p\_m}^{b\_m},q^k {p\_1}^{2a\_1 - b\_1} \cdots {p\_m}^{2a\_m - b\_m}\right)$$
> $$= {p\_1}^{\min(b\_1,2a\_1 - b\_1)} \cdots {p\_m}^{\min(b\_m,2a\_m - b\_m)},$$
> and
> $$J = \frac{H}{I} = \frac{I}{G} = \frac{n}{\gcd\left(\sigma(q^k)/2,n\right)}?$$
>
>
>
(Note that we have the identity $H = G \times J^2$.)
---
**OUR ATTEMPT**
We realized that, since the prime factorizations for $H$ and $I$ are given as follows:
$$H = {p\_1}^{2a\_1 - b\_1} \cdots {p\_m}^{2a\_m - b\_m}$$
and
$$I = {p\_1}^{\min(a\_1,2a\_1 - b\_1)} \cdots {p\_m}^{\min(a\_m,2a\_m - b\_m)},$$
then we have
$$J = \frac{H}{I} = \frac{n}{\gcd\left(\sigma(q^k)/2,n\right)} = \frac{{p\_1}^{2a\_1 - b\_1} \cdots {p\_m}^{2a\_m - b\_m}}{{p\_1}^{\min(a\_1,2a\_1 - b\_1)} \cdots {p\_m}^{\min(a\_m,2a\_m - b\_m)}}$$
$$= {p\_1}^{2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1)} \cdots {p\_m}^{2a\_m - b\_m - \min(a\_m,2a\_m - b\_m)}.$$
Therefore, we have
$$\gcd\left(G,J\right)=\gcd\left({p\_1}^{\min(b\_1,2a\_1 - b\_1)} \cdots {p\_m}^{\min(b\_m,2a\_m - b\_m)},{p\_1}^{2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1)} \cdots {p\_m}^{2a\_m - b\_m - \min(a\_m,2a\_m - b\_m)}\right)$$
$$={p\_1}^{\min\left(\min(b\_1,2a\_1 - b\_1),2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1)\right)} \cdots {p\_m}^{\min\left(\min(b\_m,2a\_m - b\_m),2a\_m - b\_m - \min(a\_m,2a\_m - b\_m)\right)}.$$
We guess it all boils down to evaluating
$$\min\left(\min(b,2a - b),2a - b - \min(a,2a - b)\right)$$
for *arbitrary integers* $a$ and $b$. **Alas, this is where we get stuck!**
We do know that
$$\min(b,2a-b) + 2a-b = 2\min(a,2a-b)$$
holds; however, this does not seem to help in this scenario.
| https://mathoverflow.net/users/10365 | Given that $H = \frac{n^2}{\sigma(q^k)/2} = G \times J^2$, where $q^k n^2$ is an odd perfect number, then what is the value of $\gcd(G, J)$? | **This is a partial answer, which uses the ideas in my earlier comments.**
---
We compute
$$\gcd(G,J)={p\_1}^{\min\left(\min(b\_1,2a\_1 - b\_1),2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1)\right)} \cdots {p\_m}^{\min\left(\min(b\_m,2a\_m - b\_m),2a\_m - b\_m - \min(a\_m,2a\_m - b\_m)\right)}$$
$$={p\_1}^{\min\left(2a\_1 - b\_1,b\_1,2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1)\right)} \cdots {p\_m}^{\min\left(2a\_m - b\_m,b\_m,2a\_m - b\_m - \min(a\_m,2a\_m - b\_m)\right)}$$
$$={p\_1}^{\min\left(b\_1,\min(2a\_1 - b\_1,2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1))\right)} \cdots {p\_m}^{\min\left(b\_m,\min(2a\_m - b\_m,2a\_m - b\_m - \min(a\_m,2a\_m - b\_m))\right)}$$
$$={p\_1}^{\min\left(b\_1,2a\_1 - b\_1 - \min(a\_1,2a\_1 - b\_1)\right)} \cdots {p\_m}^{\min\left(b\_m,2a\_m - b\_m - \min(a\_m,2a\_m - b\_m)\right)}.$$
**Alas! This is the furthest I could go with this technique.**
| 0 | https://mathoverflow.net/users/10365 | 438080 | 176,965 |
https://mathoverflow.net/questions/437979 | 3 | Two years ago I evaluated some integrals related to $\Gamma(1/4)$.
First example:
$$(1)\hspace{.2cm}\int\_{0}^{1}\frac{\sqrt{x}\log{(1+\sqrt{1+x})}}{\sqrt{1-x^2}} dx=\pi-\frac{\sqrt {2}\pi^{5/2}+4\sqrt{2}\pi^{3/2}}{2\Gamma{(1/4)^{2}}}.$$
The proof I have is based on the following formula concerning the elliptic integral of first kind (integrating both sides with carefully).
$$i \cdot K(\sqrt{\frac{2k}{1+k}})=K(\sqrt{\frac{1-k}{1+k}})-K(\sqrt{\frac{1+k}{1-k}})\cdot\sqrt{\frac{1+k}{1-k}}$$
for $0<k<1$.
\begin{align}
(2)\hspace{.2cm}\int\_{0}^{1}\frac{\sqrt{2x-1}-2x \arctan{(\sqrt{2x-1})}}{\sqrt{x(1-x)}(2x-1)^{3/2}}dx=\frac{\sqrt{2}\pi^{5/2}}{\Gamma{(1/4)}^2}-\frac{\sqrt{2\pi}\Gamma{(1/4)}^2}{8}.
\end{align}
\begin{align}
(3)\hspace{.2cm}\int\_{0}^{\pi/2}\frac{\sin{x}\log{(\tan{(x/2))}+x}}{\sqrt{\sin{x}}(\sin{x}+1)}dx=\pi-\frac{\sqrt{2\pi}\Gamma{(1/4)}^{2}}{16}-\frac{\sqrt{2}\pi^{5/2}}{2\Gamma{(1/4)}^{2}}.
\end{align}
Could you find a solution to (2) and (3) employing only Beta function or other method?
I've tried with Mathematica, Mapple, etc and seems that this evaluations are not so well known.
Question is an improvement of (1) that has been proved in an elementary approach.
| https://mathoverflow.net/users/nan | Some Log integrals related to Gamma value | I also played around with this integral. My solution is a bit shorter than the OPs: First use the trick by @Claude and define
$$\tag{1}
I(a)=\int\_0^1 \mathrm dx \sqrt\frac{x}{1-x^2}\log(a+\sqrt{1+x}),
$$
such that
$$\tag{2}
I(1) = I(0) + \int\_0^1 \mathrm da \, I'(a).
$$
Partial fraction decomposition of $I'(a)$ gives
\begin{align}
I'(a) &= \int\_0^1 \mathrm dx \frac{\sqrt\frac{x}{1-x^2}}{a+\sqrt{1+x}}
\,\frac{a-\sqrt{1+x}}{a-\sqrt{1+x}}\tag{3a}\\
&=\int\_0^1 \mathrm dx \frac{\sqrt\frac{x}{1-x}}{1+x-a^2}
+ \int\_0^1 \mathrm dx \frac{a\sqrt\frac{x}{1-x^2}}{a^2-x-1}\tag{3b}.
\end{align}
Now we exchange the integration order in the second term only and also move $I(0)$ into the second term, to get the result
\begin{align}
I(1) &=
\int\_0^1 \mathrm da \int\_0^1 \mathrm dx \frac{\sqrt\frac{x}{1-x}}{1+x-a^2}
&+&
\int\_0^1 \mathrm dx \int\_0^1 \mathrm da \frac{a\sqrt\frac{x}{1-x^2}}{a^2-x-1}+I(0) \tag{4a}\\
&= \int\_0^1 \mathrm da \, \pi\left(1-\sqrt{\tfrac{1-a^2}{2-a^2}}\right)
&+&
\int\_0^1 \mathrm dx \sqrt{\tfrac{x}{1-x^2}} \log\sqrt x\tag{4b}\\
&= \pi-\pi^{3/2}\frac{\Gamma\left(\tfrac{3}{4}\right)}{\Gamma\left(\tfrac{1}{4}\right)}
&+&\,\,
\frac{(\pi-4)\sqrt\pi \,\Gamma\left(\tfrac{3}{4}\right) }{2\Gamma\left(\tfrac{1}{4}\right)}\tag{4c}\\
&=\pi-\frac{(\pi+4)\sqrt\pi \,\Gamma\left(\tfrac{3}{4}\right) }{2\Gamma\left(\tfrac{1}{4}\right)}.\tag{4d}
\end{align}
Note that in (4c) the Beta function integral as motivated by @Carlo was used.
| 4 | https://mathoverflow.net/users/90413 | 438084 | 176,966 |
https://mathoverflow.net/questions/438068 | 1 | Let $\mathcal{L}(E)$ be the algebra of all bounded linear operators on a complex Hilbert space $E$.
On $\mathcal{L}(E)^2$, we have two equivalent norms:
\begin{eqnarray\*}
N\_1(A,B)
&=&\sup\left\{\sqrt{\|Ax\|^2+\|Bx\|^2},\;x\in E,\;\|x\|=1\;\right\},
\end{eqnarray\*}
and
$$N\_2(A,B)=\sqrt{\|A\|^2+\|B\|^2}.$$
Clearly, $N\_1(A,B)\leq N\_2(A,B)$.
>
> In general $N\_1\neq N\_2$. I want to find sufficient conditions for equality.
>
>
>
| https://mathoverflow.net/users/113054 | Sufficient condition for two norms to be equal | $\newcommand\la\lambda\newcommand\ep\varepsilon\newcommand\ip[2]{\langle #1,#2\rangle}\newcommand\Span{\operatorname{span}}$As noted in a comment by [Yemon Choi](https://mathoverflow.net/questions/438068/sufficient-condition-for-two-norms-to-be-equal/438089#comment1129532_438068), the question is not well posed.
Apparently, the OP wanted to have a *good*, nontrivial sufficient condition, preferably close to necessity. Such a sufficient condition will be provided here, and the sufficient condition will also be necessary when $E$ is finite dimensional.
Let
\begin{equation}
U:=A^\*A=\int\_{[0,\|A\|^2]}\la\,dP^A\_\la\quad\text{and}\quad
V:=B^\*B=\int\_{[0,\|B\|^2]}\la\,dP^B\_\la \tag{0}\label{0}
\end{equation}
be the [spectral decompositions](https://en.wikipedia.org/wiki/Spectral_theorem#Spectral_subspaces_and_projection-valued_measures) of the self-adjoint bounded linear operators $A^\*A$ and $B^\*B$. (Note that the families $(P^A\_\la)$ and $(P^B\_\la)$ of orthoprojectors are uniquely determined by condition \eqref{0}.)
Let
$$Q^A\_\ep:=\int\_{[\|A\|^2-\ep,\|A\|^2]}dP^A\_\la, \quad Q^B\_\ep:=\int\_{[\|B\|^2-\ep,\|B\|^2]}dP^B\_\la,$$
$$E^A\_\ep:=Q^A\_\ep E,\quad
E^B\_\ep:=Q^B\_\ep E. $$
Then the condition
\begin{equation}
E^A\_\ep\cap E^B\_\ep\ne\{0\}\quad\forall\ep>0 \tag{1}\label{1}
\end{equation}
is sufficient for $N\_1(A,B)=N\_2(A,B)$.
---
Indeed, if \eqref{1} holds, then for each real $\ep>0$ there is a unit vector
$$u\_\ep\in E^A\_\ep\cap E^B\_\ep,$$
so that $Q^A\_\ep u\_\ep=u\_\ep=Q^B\_\ep u\_\ep$.
So,
$$\begin{aligned}
\|Au\_\ep\|^2&=\ip{A^\*Au\_\ep}{u\_\ep} \\
&=\int\_{[0,\|A\|^2]}\la\,d\ip{P^A\_\la u\_\ep}{u\_\ep} \\
&=\int\_{[0,\|A\|^2]}\la\,d\ip{P^A\_\la Q^A\_\ep u\_\ep}{Q^A\_\ep u\_\ep} \\
&=\int\_{[\|A\|^2-\ep,\|A\|^2]}\la\,d\ip{P^A\_\la Q^A\_\ep u\_\ep}{Q^A\_\ep u\_\ep} \\
&\ge(\|A\|^2-\ep)\int\_{[\|A\|^2-\ep,\|A\|^2]}d\ip{P^A\_\la Q^A\_\ep u\_\ep}{Q^A\_\ep u\_\ep} \\
&=(\|A\|^2-\ep)\ip{Q^A\_\ep u\_\ep}{Q^A\_\ep u\_\ep} \\
&=(\|A\|^2-\ep)\ip{u\_\ep}{u\_\ep}
=\|A\|^2-\ep.
\end{aligned}$$
Similarly, $\|Bu\_\ep\|^2\ge\|B\|^2-\ep$. So,
$$N\_1(A,B)^2\ge\|Au\_\ep\|^2+\|Bu\_\ep\|^2
\ge\|A\|^2+\|B\|^2-2\ep=N\_2(A,B)^2-2\ep$$
for any real $\ep>0$.
Thus, $N\_1(A,B)\ge N\_2(A,B)\ge N\_1(A,B)$. $\quad\Box$
---
*Remark 1:* Condition \eqref{1} is necessary for $N\_1(A,B)=N\_2(A,B)$ when $E$ is finite dimensional. Indeed, suppose that $N\_1(A,B)=N\_2(A,B)$. Then for some sequence $(x\_n)$ on the unit sphere $S$ in $E$ we have
\begin{equation}
\|A\|^2+\|B\|^2\ge\|Ax\_n\|^2+\|Bx\_n\|^2\to\|A\|^2+\|B\|^2.
\end{equation}
It follows that $\|Ax\_n\|^2\to\|A\|^2$ and $\|Bx\_n\|^2\to\|B\|^2$. The unit sphere $S$ in the finite-dimensional Hilbert space $E$ is compact. So, passing to a subsequence, without loss of generality we have $x\_n\to x$ for some $x\in S$. So, $\|Ax\|^2=\|A\|^2$ and $\|Bx\|^2=\|B\|^2$, which implies \eqref{1}. $\quad\Box$
*Remark 2:* Condition \eqref{1} is not in general necessary for $N\_1(A,B)=N\_2(A,B)$ when $E$ is infinite dimensional. E.g., let $E:=\ell^2$, with the standard basis $(e\_1,e\_2,\dots)$.
Let $U$ be the positive-semidefinite self-adjoint linear operator on $E$ whose matrix (in the standard basis) is the block-diagonal matrix with the $2\times2$ diagonal blocks
\begin{equation}
D\_j^U:=\frac j{j+1}\,P\_j^U,\quad\text{where}\quad
P\_j^U:=\begin{bmatrix}1&0\\0&0 \end{bmatrix}
\end{equation}
for integers $j\ge1$.
The factor $\frac j{j+1}$, strictly increasing to $1$ in $j$, was introduced to force gliding to $\infty$ in search of a (nonexistent) maximizer $x\in S$ of $\ip{Ux}x$.
Similarly, let $V$ be the positive-semidefinite self-adjoint linear operator on $E$ whose matrix (in the standard basis) is the block-diagonal matrix with the $2\times2$ diagonal blocks
\begin{equation}
D\_j^V:=\frac j{j+1}\,P\_j^V,\quad\text{where}\quad
P\_j^V:=\frac1{j^2+1}\,\begin{bmatrix}j^2&j\\j&1 \end{bmatrix}
\end{equation}
for integers $j\ge1$.
Note that $P\_j^U$ and $P\_j^V$ are orthoprojector matrices of rank $1$ each, with the respective column spaces $\Span\Big\{\begin{bmatrix}1\\0 \end{bmatrix}\Big\}$ and $\Span\Big\{\begin{bmatrix}j\\1 \end{bmatrix}\Big\}$.
In accordance with \eqref{0}, let $A:=\sqrt{U}$ and $B:=\sqrt{V}$. Then $\|A\|^2=\|U\|=1$ and $\|B\|^2=\|V\|=1$. Moreover, letting an integer $i$ go to $\infty$, we get $$N\_1(A,B)\ge\|Ae\_{2i}\|^2+\|Be\_{2i}\|^2=\ip{Ue\_{2i}}{e\_{2i}}+\ip{Ve\_{2i}}{e\_{2i}} \\ \to1+1=\|A\|^2+\|B\|^2=N\_2(A,B),$$
which implies that $N\_1(A,B)=N\_2(A,B)$.
However, for any integer $j\ge1$ and any $\ep\in(\frac1{j+2},\frac1{j+1}]$,
\begin{equation}
E\_\ep^A=\Span\{e\_{2i}\colon i\ge j\}\quad\text{and}\quad
E\_\ep^B=\Span\{e\_{2i}+\tfrac1{2i+1}\,e\_{2i+1}\colon i\ge j\},
\end{equation}
so that \eqref{1} fails to hold. $\quad\Box$
---
It appears that the following condition is necessary and sufficient for $N\_1(A,B)=N\_2(A,B)$:
>
> There exist sequences $(a\_n)$ and $(b\_n)$ such that $a\_n-b\_n\to0$,
> $a\_n\in S\cap E\_{1/n}^A$ for all $n$, and $b\_n\in S\cap E\_{1/n}^B$ for all $n$.
>
>
>
This conjecture is probably easy to prove, but ....
| 3 | https://mathoverflow.net/users/36721 | 438089 | 176,968 |
https://mathoverflow.net/questions/438020 | 4 | Do we have an example of a smooth action $S^1 \curvearrowright T^n$ which is faithful, locally free but not free?
I know such an action must induce an injection $\rho:\pi\_1(S^1)\to\pi\_1(T^n)$.
Another related question is: Is the image of $\rho$ saturated?
Thanks in advance!
| https://mathoverflow.net/users/110093 | Faithful locally free circle actions on a torus must be free? | This follows from Theorem 9.3 (page 216) of the book "Compact tranformation groups" by Bredon (note that it is freely available online).
More is true: Any effective group action of a torus on a torus is free. The proof starts along the lines you mentioned via fundamental group considerations.
The result is originally due to Connor and Montgomery:
Conner, P. E., and Montgomery, DC. Transformation groups on a K(n, l), I. Michigan Math. J. 6 (1959), 405-412.
| 10 | https://mathoverflow.net/users/99732 | 438101 | 176,976 |
https://mathoverflow.net/questions/438064 | 1 | I encountered in my research on dynamical systems a problem, which considers for some $L>0$ on the $C\_n=[0,L]^n$ the set $\mathcal{C}\_n=\{(x\_1,\ldots,x\_n)\mid\exists j,k:\,x\_{i\_j}=x\_{i\_k}\}$. I am looking for an interpretation of $\mathcal{C}\_n$ on the $n$-torus, when imposing periodic boundary conditions on $C\_n$. For $n=2$ this is a simple closed curve which winds once around the torus. For $n=3$, I am already struggeling what $\mathcal{C}\_3$ "looks like". In particular, I am wondering how curves of the form $\gamma:t\mapsto x+t\, e\_j$, with $(e\_j)\_i=\mathrm{1}\_{i=j}$, intersect with $\mathcal{C}\_n$.
In general, literature recommendations on the $n$ torus are highly appreciated!
| https://mathoverflow.net/users/333230 | Curves on $n$-torus analogous to curve implied by diagonal in square for torus | I suppose I should add a caveat before turning my comment into an answer. In your definition, you have the $\exists j,k$ qualifier, which I believe means that $\mathcal C\_n$ is the union of $n \choose 2$ tori, i.e. one for every choice of $j \neq k$. If you allow $j=k$ then I suppose you would have $\mathcal C\_n = C\_n$, but I doubt you meant that. Finally, if you take the statement $\exists j,k$ out of your set definition, you would have a single torus. With that caveat, I'll expand my comment.
The observation follows from your equation $x\_{i\_j} = x\_{i\_k}$, this is basically stating your set is the graph of a function.
A slightly different perspective on a related idea. Any $n \times n$ matrix $A$ with integer coefficients and $\det(A) = \pm 1$ induces an affine-linear automorphism of $\mathcal C\_n$ (thought of as a torus).
Provided you are okay with that, notice that the matrix
$$\pmatrix{1 & 1 \\ 0 & 1}$$
sends $\{0\} \times [0,L]$ to $C\_2$.
i.e. we have turned the equation $x\_1=x\_2$ into the functional expression $(x\_1, x\_2) = (x\_2, x\_2)$, turning $\mathcal C\_2$ into the image of the function $x\_2 \longmapsto (x\_2,x\_2)$, or the graph of $x\_1(x\_2) = x\_2$, i.e. $x\_1$ as a function of $x\_2$ is equal to $x\_2$.
You can do the same with $C\_3$, writing it as the union of the graphs of the functions
$$\pmatrix{x\_1\\x\_2\\0} \longmapsto \pmatrix{x\_1 \\x\_2\\x\_2}$$
$$\pmatrix{x\_1\\0\\x\_3} \longmapsto \pmatrix{x\_1 \\x\_1\\x\_3}$$
$$\pmatrix{0\\x\_2\\x\_3} \longmapsto \pmatrix{x\_3 \\x\_2\\x\_3}$$
${{}}$
| 1 | https://mathoverflow.net/users/1465 | 438103 | 176,978 |
https://mathoverflow.net/questions/378539 | 5 | Let $(M,g)$ be a Riemannian manifold. Let $S\_g$ be the corresponding Sasaki metric on $TM$. For every $p\in M$, $V\_p\in T\_pM$, is it true and obvious that $0\_p$ is the closest point of the zero section to $V\_p$?
With some abuse of terminology a rephrase of the question would be: Is the height of a right triangle shorter than its hypotenuse?
| https://mathoverflow.net/users/36688 | The distance to the zero section of $TM$ | *Let me expand my comment to remove the question from unanswered.*
Any point in $w\in\mathrm{T}M$ is a pair $w=(V,p)$ where $p\in M$ and $V\in\mathrm{T}\_pM$.
Let $t\mapsto w(t)=(V(t),\gamma(t))$ be a curve in $\mathrm{T}M$.
Let $V=V(0)$ and $p=\gamma(0)$.
Note that
$$w'(0)=\nabla\_{\gamma'(0)}V\oplus \gamma'(0)\in \mathrm{T}\_{p}M\oplus\mathrm{T}\_{p}M=\mathrm{T}\_{(V,p)}\mathrm{T}M.$$
Therefore
$$
\begin{aligned}
\langle V\oplus 0,w'(0)\rangle
&=\langle V,\nabla\_{\gamma'(0)}V\rangle=
\\
&=\tfrac12\cdot\langle V,V\rangle'(0).
\end{aligned}
$$
It follows that $(V,p)\mapsto V\oplus 0$ is the gradient of the function $f\colon(V,p)\mapsto \tfrac12\cdot\langle V,V\rangle$.
Whence the statement follows.
| 2 | https://mathoverflow.net/users/1441 | 438108 | 176,980 |
https://mathoverflow.net/questions/438111 | 5 | Suppose one is given an odd prime $p$, a generator $g$ of $(\mathbb Z/p \mathbb Z)^\*$ and two integers $a$ and $b$. Is there an efficient method to determine whether $\log\_g a < \log\_g b$? (Here we are defining $\log\_g x$ as between one and $p-1$ inclusive.)
Bonus question added later: What if we restrict the problem to $b=-a$?
| https://mathoverflow.net/users/7089 | Discrete log problem modified | If there were a way of doing this in time polynomial in $\log(p)$, you could solve discrete logarithm in time polynomial in $\log(p)$ by doing a binary search. That is, to find $\log\_g(a)$, first see if $\log\_g a < \log\_g (g^{(p-1)/2})$. If yes, next compare $\log\_g(a)$ to $\log\_g (g^{\lfloor (p-1)/4\rfloor})$, if no to $\log\_g(g^{\lfloor 3 (p-1)/4 \rfloor})$, etc.
| 7 | https://mathoverflow.net/users/13650 | 438116 | 176,981 |
https://mathoverflow.net/questions/422392 | 2 | I found myself with the following integral
$$ \int\_{b\_1}^{b\_2} \sqrt{\frac{(b-b\_1)(b\_2-b)(b\_3-b)}{(b\_4-b)}} \ db $$
with $ b\_1 < b\_2 < b\_3 < b\_4 $. I know that
$$ \int\_{b\_1}^{b\_2} \frac{db}{\sqrt{(b-b\_1)(b\_2-b)(b\_3-b)(b\_4-b)}} $$
is equal to
$$ \frac{2}{(b\_4-b\_2)(b\_3-b\_1)} K(k) $$
where $K(k)$ is the complete elliptic integral of first kind, so I suspect that this integral is somehow reducible to a linear combination of elliptic integrals, but I can't find the right way.
| https://mathoverflow.net/users/482168 | Definite integral of the square root of a polynomial ratio | The expression obtained by @Robert can be simplified with the Imaginary-Argument Transformation from [DLMF 19.7.7](https://dlmf.nist.gov/19.7). Then, the limit $t\to 1^-$ can be performed. The result can be further simplified and written in many ways due to the large number of elliptic functions relations. The simplest expression I found reads
\begin{align}
&\int\_0^1 \mathrm db \,\sqrt{\frac{b(1-b)(b\_3-b)}{b\_4-b}}\\
&=\frac{1}{4 \sqrt{b\_4}}\left[
b\_4 (3 b\_4-b\_3-1) E(k)
- b\_3 (b\_4+b\_3-1) K(k)\\
\quad+ \frac{b\_3}{\sqrt{b\_3-1}}
\left[(b\_3-1)^2 + 2(b\_3+1)b\_4 - 3 b\_4^2\right] \Pi
\left(\tfrac{1}{1-b\_3},k\right)
\right]\tag{1}
\end{align}
with parameter and elliptic modulus
$$
m=k^2=\frac{b\_4-b\_3}{\left(1-b\_3\right) b\_4}.\tag{2}
$$
One can rewrite this expression with the Imaginary-Modulus Transformation [DLMF 19.7.5](https://dlmf.nist.gov/19.7) to get $0<\tilde m<1$ and real $0<\tilde k<1$, i.e., ${\tilde k}{}^2=\frac{b\_4-b\_3}{b\_3(b\_4-1)}$, however the expressions become more complicated. For $b\_3=2, b\_4=3$ this last formulation gives the representation obtained in the comment by @Robert.
| 3 | https://mathoverflow.net/users/90413 | 438127 | 176,986 |
https://mathoverflow.net/questions/438105 | 4 | Working in $\sf ZF$
Define: $W\_0 = \emptyset \\ W\_{\alpha+1} = H\_{\leq |W\_\alpha|} =\{x \mid \forall y: y \in \operatorname {trcl} (\{x\}) \ |y| \leq |W\_\alpha| \} \\ W\_\lambda= \bigcup W\_{\alpha < \lambda}, \text { for limit ordinal } \lambda$
Where cardinality "| |" is defined after Scott.
This cumulative size hierarchy is also indexed by ordinals.
Now if we define ordinal definable sets in terms of those stages istead of the usual $V\_\alpha$ stages of the cumulative hierarchy. That is, we use exactly the same definition of ordinal definability but only replace the symbol $V$ by $W$. Designate that as $\operatorname {OD}^\*$, then:
>
> Is $\sf HOD=HOD^\*$?
>
>
>
| https://mathoverflow.net/users/95347 | Is ordinal definability in terms of stages of cumulative size hierarchy equivalent to the usual one? | The answer is yes, in a very general way.
What I claim, first, is that the [Lévy-Montague reflection theorem](https://en.wikipedia.org/wiki/Reflection_principle) holds in ZF for any definable continuous cumulative hierarchical representation of the set-theoretic universe $V$. That is, if you have defined sets $U\_\alpha$ for every ordinal $\alpha$, such that
* the sequence $\alpha\mapsto U\_\alpha$ is definable, without parameters
* the sequence is monotone $\alpha\leq\beta\implies U\_\alpha\subseteq U\_\beta$
* the sequence is continuous, $U\_\lambda=\bigcup\_{\alpha<\lambda}U\_\alpha$ for limit ordinals $\lambda$
* the sequence accumulates to the entire universe, $V=\bigcup\_\alpha U\_\alpha$
Then for every formula $\varphi(x)$ in the first-order language of set theory, there is some ordinal $\alpha$ such that $\varphi$ is absolute between $U\_\alpha$ and $V$. Indeed, there is a closed unbounded class of such ordinals.
The proof is nearly identical to the usual proof of the reflection theorem — these hypotheses are what is used about the $V\_\alpha$ hierarchy.
Once one has reflection, then one can define HOD using the hierarchy, as you suggest, by saying a set $a$ is *ordinal definable*, if there is some $\alpha$ and some ordinal parameters below $\alpha$, such that $a$ is definable in $U\_\alpha$ from those parameters. A set is \*hereditarily ordinal definable, if $a$ and every hereditary member of $a$ is ordinal definable.
In any model $V$ of ZF, if a set is ordinal definable, then by reflection it is ordinal definable in that sense. And conversely, if a set is ordinal definable in that sense, in some $U\_\alpha$ using some formula $\varphi$ with ordinal parameters $\vec \beta$, then in $V$ we may define the set using parameters $\langle\alpha, \vec\beta,\varphi\rangle$ as parameters. (And note the very subtle point that we must sometimes use the formula $\varphi$ itself as a parameter, meaning its Gödel code, since in an $\omega$-nonstandard model the formula $\varphi$ used to define $a$ in $U\_\alpha$ might be nonstandad. I view it as a kind of lucky miracle that ordinal-definability doesn't stumble on this problem.)
So this version aligns with the usual version of HOD.
**Conclusion.** It doesn't matter which definable continuous hierarchy you use when defining HOD. They all give the same class.
Another way to see this is to observe the following:
**Theorem.** For any such hierarchy $U\_\alpha$ as above, there is a closed unbounded class of ordinals $\theta$ for which $$U\_\theta=V\_\theta.$$ In particular, any two such hierarchies agree on a class club.
**Proof.** This is the typical club argument. First, by continuity, the class of such ordinals $\theta$ is closed. And it is unbounded, since we can start in any $U\_{\alpha\_0}$, find a $V\_{\alpha\_1}$ containing all those elements, and then a $U\_{\alpha\_2}$ containing all those elements, and so on, building an alternating chain
$$U\_{\alpha\_0}\subseteq V\_{\alpha\_1}\subseteq U\_{\alpha\_2}\subseteq\cdots$$
If $\theta=\sup\_n\alpha\_n$ is the supremum, then $U\_\theta=V\_\theta$ by continuity. So this is a closed unbounded class. $\Box$
Since reflection also works on a class club, we see in this way that every formula $\varphi$ reflects to a ordinal on which the two hierarchies agree.
| 7 | https://mathoverflow.net/users/1946 | 438131 | 176,988 |
https://mathoverflow.net/questions/438133 | 2 | Let $\mathbb{N}$ denote the set of positive integers. We define a relation $R \subseteq \mathbb{N}^4$ in the following way:
>
> $(p,q,n,s)\in R$ if and only if there is $S\subseteq [0,1]^n$ with $|S| = s$ such that for all $x\in [0,1]^n$ there is $y\in S$ such that $\| x-y \|< \frac{p}{q}$.
>
>
>
**Question.** Is $R\subseteq \mathbb{N}^4$ computable?
| https://mathoverflow.net/users/8628 | Computability of fillability of unit cube in $\mathbb{R}^n$ by $k$ $\varepsilon$-balls | The question whether $(p,q,n,s)\in R$ in any instance can be expressed as a sentence in the language of the structure $\langle\mathbb{R},+,\cdot,0,1,<\rangle$, a real-closed field, and by Tarski's theorem on real closed fields, there is a computable uniform decision procedure to decide the truth of all such sentences.
See further description at [my answer to a related question](https://mathoverflow.net/a/134260/1946).
| 5 | https://mathoverflow.net/users/1946 | 438135 | 176,989 |
https://mathoverflow.net/questions/438129 | 6 | Can choice be proved with ZF+[Tarski axiom](https://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory)?
| https://mathoverflow.net/users/51212 | Can the axiom of choice be proved with ZF+Tarski axiom? | Following the link found in the Wikipedia article about the [Tarski–Grothendieck set theory](https://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory), the required proof (by Tarski himself!) can be found beginning on p.181 of his article ["On the well-ordered subsets of any set"](http://matwbn.icm.edu.pl/ksiazki/fm/fm32/fm32115.pdf) published in 1939 in "Fundamenta Mathematicae" (in fact, Tarski shows that his axiom implies the well-ordering theorem/axiom, which is known to be equivalent to the axiom of choice).
| 12 | https://mathoverflow.net/users/54780 | 438137 | 176,990 |
https://mathoverflow.net/questions/438124 | 5 | Let $G\_{\mathbb{Q}\_p}$ denote the absolute Galois group of the $p$-adic field $\mathbb{Q}\_{p}$. Also, their structure as abstract groups is completely known.
It is well known that this group embeds into the absolute Galois group of the rationals, $G\_{\mathbb{Q}}$.
My question is: are there any known relations between the two absolute Galois groups of two $p$-adic fields, say $\mathbb{Q}\_{p\_1}$ and $\mathbb{Q}\_{p\_2}$, where $p\_1$ and $p\_2$ are different primes, regarded as subgroups of the absolute Galois group of the rationals?
| https://mathoverflow.net/users/174655 | Relation between $G_{\mathbb{Q}_p}$ for different primes | As noted in the remarks above, there are different (although conjugate) embeddings $G\_{\mathbb{Q}\_p}\rightarrow G\_\mathbb{Q}$. Let us fix one of them and denote its image by $G\_p$.
If we fix prime numbers $p\_1,\dots,p\_n$ (not necessarily distinct), then
the set of $\sigma=(\sigma\_1,\dots,\sigma\_n)\in G\_\mathbb{Q}^n$ for which the closed subgroup
$\left< G\_{p\_1}^{\sigma\_1},\dots,G\_{p\_n}^{\sigma\_n}\right>$ is the profinite free product of $G\_{p\_1},\dots,G\_{p\_n}$ has Haar measure $1$.
This is (a special case of) Theorem 4.1 in W.-D. Geyer, Galois groups of intersection of local fields, Israel Journal of Mathematics 30, 1978.
Geyer also notes in section 4.4 that one cannot expect this to hold for *every* $\sigma$, but his example is for archimedean places. A purely group theoretic result of Haran occurring as Proposition 4.2.3 in Neukirch-Schmidt-Wingberg's Cohomology of Number Fields however shows that also for non-archimedean places it does not hold for *every* $\sigma$.
| 3 | https://mathoverflow.net/users/50351 | 438138 | 176,991 |
https://mathoverflow.net/questions/437668 | 1 | Consider the following setting: suppose that $X$ is a smooth variety and let $f:X\rightarrow \Delta$ be a smooth morphism outside the origin $0$. Let the central fiber $X\_0$ be a reduced (Cartier) divisor with simple normal crossings. Here a reduced divisor is a Weil divisor $D=\sum\_{i}n\_i D\_i$ with all $n\_i=1$. Take some successive blowing ups along the singular locus of $X\_0=\sum \_{i\in I}X\_i$ to get a proper modification $\mu:V\rightarrow X$ from a smooth variety $V$ to $X$. We then write the total transform$\mu^\*{X\_0}=\sum \_{i\in I}\mu\_\*^{-1}X\_i+\sum\_{j\in J}k\_j V\_j^\prime$ with $k\_j\geq 1$. Here, we can ensure the strict transform $\sum \_{i\in I}\mu\_\*^{-1}X\_i$ to be smooth.
Here is my question, can we determine the ramification divisor of the morphism $\mu$? Namely, if we write $$K\_V\equiv\mu^\*K\_X+\text{Ram}(\mu),$$
then what is $\text{Ram}(\mu)$, should it be written as $t\_j V^\prime\_j$? If so, what is the relationship between $t\_j$ and $k\_j$?
| https://mathoverflow.net/users/141609 | Determine the coefficient of the exceptional divisor | Ok, here's an expanded version of what I said in the comment.
SNC case
--------
First, I'm going to create a special log resolution of the pair $(X, X\_0)$. We assume $d = \dim X$. We are assuming that $X$ is nonsingular and $X\_0$ is simple normal crossings.
We begin by letting $S\_{n}$ denote the stratum of this pair of dimension $n$. In other words, $S\_0$ is the points that are intersections of components of $X\_0$, $S\_1$ is the curves that are components of $X\_0$, etc. Let $Y^0 \to X$ be th blowup of all the points in $S\_0$. This separates the strict transforms of the components of $S\_1$. We let $S\_{1}^0$ denote the set of those strict transforms (likewise with $S\_{2}^0$, etc). Let $E\_{0,i}^0 \subseteq Y\_0$ denote the set of exceptional divisors.
Now, we can write
$$
K\_{Y^0} = \pi\_0^\* K\_X + (d-1) \sum E\_{0,i}^0
$$
since we are blowing up nonsingular centers of codimension $d$ in a nonsingular variety. We also ahve
$$
\pi\_0^\* X\_0 = (\pi\_0)^{-1}\_\* X\_0 + d \sum E\_{0,i}^0.
$$
since each $E\_{0,i}^0$ maps to a point in $S\_0$ which is the intersection of $d$ components of $X\_0$.
Next, we blowup the curves in $S\_1^0$ (these are curves in $Y^0$). Note, each curve may intersect some $E\_{0,i}^0$, but it cannot contain any, and likewise no $E\_{0,i}^1$ contains any curve in $S\_1^0$.. Let $E\_{1, i}^1$ denote the exceptional divisors of this new blowup and $E\_{0,i}^1$ the strict transforms of the old exceptional divisors. In this case we have
$$
K\_{Y^1} = \pi\_{1,0}^\* K\_{Y^0} + (d-2) \sum {E\_{1,i}^1} = \pi\_1^\* K\_X + (d-1) \sum E\_{0,i}^1 + (d-2) \sum E\_{1,i}^1.
$$
Here we crucially used the fact that no element of the the $S\_1^0$ were contained in $E\_{0,i}^0$.
By a similar consideration, we also have
$$
\pi\_1^\* X\_0 = (\pi\_1)^{-1}\_\* X\_0 + d \sum E\_{0,i}^1 + (d-1) E\_{1,i}^1.
$$
Continuing in this way, we eventually obtain a log resolution that separates the components of $X\_0$, $\pi : Y \to X$ and
$$
K\_{Y} = \pi^\* K\_X + (d-1) \sum E\_{0,i} + \dots + 1 \sum E\_{d-2,i}
$$
where the $E\_{0,i}$ are the strict transforms of the exceptional divisors from $Y^0 \to X$, $E\_{1,i}$ are the strict transforms of the exceptional divisors from $Y^1 \to Y^0$, etc.
and also
$$
\pi^\* X\_0 = (\pi)^{-1}\_\* X\_0 + d \sum E\_{0,i} + (d-1) \sum E\_{1,i} + \dots + 2 \sum E\_{d-2, i}
$$
In other words for each $E$ an excpetional divisor on $Y$ (or an element of the strict transform), the coefficient of $\pi^\* X\_0$ is exactly one bigger than the coefficient in the relative canonical divisor (what you called Ram).
SLC case
--------
Let's now assume that $X\_0$ is semi-log canonical (SLC) but not necessarily SNC. In this case, since $X\_0$ is Gorenstein, this is equivalent to it being Du Bois (see work of Kovács and Doherty). It is also equivalent to the pair $(X, X\_0)$ being log canonical by inversion of adjunction.
Anyways, since $(X, X\_0)$ is log canonical, for any resolution of singularities, $\pi : V \to X$, if we write
$$
K\_V + D = \pi^\* (K\_X + X\_0)
$$
we know that all the coefficients of $D$ are $\leq 1$ (this is the definition of log canonical). In other words, if we write
$$
K\_V = \pi^\* K\_X + \mathrm{Ram}\;\;\;\; \text{ and }\;\;\;\; \pi^\* X\_0 = (\pi)^{-1}\_\* X\_0 + \sum b\_i E\_i
$$
where $E\_i$ are the exceptional divisors (note $\mathrm{Ram} = \sum c\_i E\_i$ for some $c\_i$)
then we have that
$$
D = \pi^\* K\_X + \pi^\* X\_0 - K\_V = \pi^\* K\_X + \pi^\* X\_0 - \pi^\* K\_X - \mathrm{Ram} = (\pi)^{-1}\_\* X\_0 + \sum b\_i E\_i - sum c\_i E\_i.
$$
In other words, for each exceptional divisor $E\_i$, we have that the coefficient of
$$
D = \pi^\* X\_0 - \mathrm{Ram}
$$
is $b\_i - c\_i$ and
$$
b\_i - c\_i \leq 1
$$
or in other words $b\_i \leq c\_i + 1$.
Note, we had equality in the SNC case when I picked a very special resolution of singularities.
| 1 | https://mathoverflow.net/users/3521 | 438145 | 176,994 |
https://mathoverflow.net/questions/437885 | 4 | Consider a torus $T$ over $\mathbb{C}$. Let $\rho: T\rightarrow \operatorname{GL}\_{n}(\mathbb C)$ be a finite dimensional complex representation.
Is there an elementary way (undergrad level) to see that $\rho(T)$ is diagonalizable?
The standard way uses algebra of functions on $k[T]$, tensor products and symmetric algebras, but it is too involved, can we see it directly?
We already know by Lie–Kolchin that the image is trigonalizable.
| https://mathoverflow.net/users/27398 | Action of complex torus on a vector space | You need to put some niceness hypothesis on the function $\rho$, such as "algebraic" or "analytic". Otherwise, the map $\mathbb{C}^{\ast} \to \text{GL}\_2(\mathbb{C})$ by $z \mapsto \begin{bmatrix} 1 & \log |z| \\ 0 & 1 \end{bmatrix}$ is a representation and not diagonalizable. Since you didn't tell me which hypothesis to use, I will take $\rho$ to be algebraic, meaning that each of the $n^2$ entries in the matrix $\rho(z\_1, z\_2, \dots, z\_d)$ is a Laurent polynomial in $z$. (You used $n$ for both the dimension of $T$ and the dimension of the representation, but I assume you didn't want that; I'll put $d = \dim T$.)
So we can write
$$\rho(z\_1, z\_2, \dots, z\_d) = \sum\_{(k\_1, k\_2, \ldots, k\_d)} P\_{(k\_1, k\_2, \ldots, k\_d)} z\_1^{k\_1} z\_2^{k\_2} \cdots z\_d^{k\_d}$$
where each $P\_k$ is an $n \times n$ matrix.
From the equation $\rho(xy) = \rho(x) \rho(y)$, we deduce that
$$\sum\_{(k\_1, \ldots, k\_d)} P\_{(k\_1, \ldots, k\_d)} x\_1^{k\_1} \cdots x\_d^{k\_d} y\_1^{k\_1} \cdots y\_d^{k\_d} = \sum\_{(i\_1, \ldots, i\_d),\ (j\_1, \ldots, j\_d)} P\_{(i\_1, \ldots, i\_d)} P\_{(j\_1, \ldots, j\_d)} x\_1^{i\_1} \cdots x\_d^{i\_d} y\_1^{j\_1} \cdots y\_d^{j\_d}.$$
Comparing the coefficient of $x\_1^{i\_1} \cdots x\_d^{i\_d} y\_1^{j\_1} \cdots y\_d^{j\_d}$ on each side, we deduce that
$$P\_{(i\_1, \ldots, i\_d)} P\_{(j\_1, \ldots, j\_d)} = \begin{cases} P\_{(i\_1, \ldots, i\_d)} & (i\_1, \ldots, i\_d) = (j\_1, \ldots, j\_d) \\ 0 & (i\_1, \ldots, i\_d) \neq (j\_1, \ldots, j\_d)\\ \end{cases}.$$
In other words, the $P$'s are mutually orthogonal idempotents.
Moreover, from the equation $\rho(1) = \text{Id}\_n$, we deduce that $\sum P\_{(k\_1, \ldots, k\_d)} = \text{Id}\_n$. So the $P$'s are a complete set of mutually orthogonal idempotents.
Now use your favorite proof that mutually orthogonal idempotents are simultaneously diagonalizable.
| 6 | https://mathoverflow.net/users/297 | 438146 | 176,995 |
https://mathoverflow.net/questions/437039 | 1 | Let $X=Y = \mathbb R^d$ and $c:X \times Y \to [0, \infty)$ be Borel measurable. Let $\mu, \nu$ be Borel probability measures on $X,Y$ respectively. Let $\mathcal T$ be the set of all Borel measurable maps $T:X \to Y$ such that $T\_\sharp \mu = \nu$. Let
$$
\mathbb M (T) := \int\_X c(x, T(x)) \mathrm d \mu(x) \quad \forall T \in \mathcal T.
$$
Then we are interested in the Monge's transportation problem
$$
\mathrm{MP} : \quad\inf\_{T \in \mathcal T} \mathbb M (T).
$$
The existence and uniqueness of the solution of $\mathrm{MP}$ is guaranteed if $c$ is strictly convex and the supports of $\mu, \nu$ are compact [[1](http://www.math.toronto.edu/mccann/assignments/477/Caffarelli96.pdf)]. We can remove the assumption of compact supports by stronger conditions on $c$ [[2](https://projecteuclid.org/journals/acta-mathematica/volume-177/issue-2/The-geometry-of-optimal-transportation/10.1007/BF02392620.full)].
Let $\Pi (\mu, \nu)$ be the set of Borel probability measures on $X \times Y$ with marginals $\mu$ on $X$ and $\nu$ on $Y$. Recently, I have seen a strong theorem from this lecture [note](https://www.math.leidenuniv.nl/%7Evangaans/TA2011-gradient-mei11.pdf), i.e.,
>
> **Theorem 3.14.** Assume
>
>
> 1. $h:\mathbb R^d \to [0, \infty)$ is strictly convex and $c(x, y) := h(x-y)$ for all $(x, y) \in X \times Y$.
> 2. $\int\_{X \times Y} c \mathrm{d} \gamma <\infty$ for some $\gamma \in \Pi(\mu, \nu)$,
> 3. $\mu\left(\left\{x \in X: \int\_Y c(x, y) \mathrm{d} \nu(y)<\infty\right\}\right)>0$,
> 4. $\nu\left(\left\{y \in Y: \int\_X c(x, y) \mathrm{d} \mu(x)<\infty\right\}\right)>0$,
> 5. $\mu$ is absolutely continuous with respect to Lebesgue measure.
>
>
> Then $\mathrm{MP}$ has a unique (up to $\mu$-a.e. solution).
>
>
>
The conditions 2, 3, 4 are very mild and just to ensure the dual of the dual of the corresponding Kantorovich's problem has a solution in a form of a pair of [$c$-conjugates](https://mathoverflow.net/questions/435037/optimal-transport-the-existence-of-an-optimal-pair-of-c-conjugate-functions). **Theorem 3.14.** is striking because it does not require the supports of $\mu, \nu$ to be compact nor any condition on $c$ besides strict convexity.
Could you elaborate if there are some references of **Theorem 3.14.**?
---
[[1](http://www.math.toronto.edu/mccann/assignments/477/Caffarelli96.pdf)] Caffarelli, Luis A. "Allocation maps with general cost functions." Partial differential equations and applications. Routledge, 2017. 29-35.
[[2](https://projecteuclid.org/journals/acta-mathematica/volume-177/issue-2/The-geometry-of-optimal-transportation/10.1007/BF02392620.full)] Gangbo, Wilfrid, and Robert J. McCann. "The geometry of optimal transportation." Acta Mathematica 177.2 (1996): 113-161.
| https://mathoverflow.net/users/99469 | Strict convexity of the cost function is enough to ensure the existence and uniqueness of the optimal transport map | First a comment: you write (before stating your Theorem 3.14) that "The existence and uniqueness of the solution of the Monge Problem is guaranteed if $c$ is strictly convex and the supports of $\mu,\nu$ are compact", but this is absolutely not true: you really need some conditions on the starting point, i-e that $\mu$ does not charge "small sets" in some sense (this is exactly assumption 5 in your theorem 3.14, but this can be relaxed). If $\mu=\delta\_x$ is a Dirac delta there exists no transport map $T$ from $\mu$ to $\nu$, unless $\nu=\delta\_{T(x)}$, so clearly the Monge problem is ill-posed in general.
Next, my real answer: this precise statement is often called the Brenier-McCann theorem. You can find an extremely general version in [1], Theorem 10.38.
Note in particular that this is stated without any compactness assumptions or behaviour at infinity, and that the strict convexity $c(x,y)=h(|x-y|)$ is not needed (only the so-called and weaker twist condition). This result is usually credited to Brenier, Rachev and Rüschendorf for the quadratic cost in Euclidean spaces, and then R. McCann extended to Riemannian manifolds.
[1] Villani, C. (2009). Optimal transport: old and new (Vol. 338, p. 23). Berlin: springer.
| 2 | https://mathoverflow.net/users/33741 | 438148 | 176,996 |
https://mathoverflow.net/questions/427565 | 8 | In W.S.Massey's singular homology (Graduate Texts in Mathematics 70 Springer (1980)) there is a formula for the boundary of a double slant product on page 176
$$\partial(\phi\backslash\backslash a\otimes b\otimes g) = (\delta\phi)\backslash\backslash a\otimes b\otimes g + (-1)^{|\phi|}\phi\backslash\backslash\partial(a\otimes b\otimes g)$$
Am I right in calculating that this is missing a minus sign and should be
$$\partial(\phi\backslash\backslash a\otimes b\otimes g) = -(\delta\phi)\backslash\backslash a\otimes b\otimes g + (-1)^{|\phi|}\phi\backslash\backslash\partial(a\otimes b\otimes g)?$$
If the minus sign is required here, then it is surely also required subsequently, for example in the formula
$$\partial(u\backslash v)=(\delta u)v+(-1)^{|u|}u\backslash\partial(v)$$
which should then be
$$\partial(u\backslash v)=-(\delta u)v+(-1)^{|u|}u\backslash\partial(v).$$
I have been for some years puzzled by this as the formulae in the book give the appearance of having signs in the expected places. I would be very happy to be put right on this issue.
| https://mathoverflow.net/users/124943 | Calculation of boundary of slant product in W. S. Massey's Singular Homology textbook | You and Tyrone are correct that this is a sign mismatch.
Just to be clear: Massey writes this sign for the double slant product because it is the desirable one. It would imply that we have a map of chain complexes
$$C^\*(Y, G\_1) \otimes C\_\*(X) \otimes C\_\*(Y) \otimes G\_2 \to C\_\*(X) \otimes G\_1 \otimes G\_2$$
where this is the tensor product of chain complexes, using the Koszul sign convention.
This is not compatible with the other things that Massey writes. In particular, a couple of lines later he writes that we should take the convention
$$(\delta \phi)(b) = (-1)^{|\phi|} \phi(\partial b)$$
(which does not alter the cohomology groups) and this is the source of the difficulty. This definition does not make the pairing
$$C^\*(Y, G\_1) \otimes C\_\*(Y) \to G\_1$$
into a chain map.
Many people (myself included) would prefer to use the sign convention
$$(\delta \phi)(b) = (-1)^{|\phi|+1} \phi(\partial b)$$
for the coboundary operator $\delta$ precisely to fix this type of problem. This *looks* like it violates the Koszul sign convention, but duality has a tendency to do that. If you make this change, I believe Massey's sign for the double slant product works out.
(One possible explanation is this. In the definition of $\partial$ you take an alternating sum of faces, starting at the 0'th and ending at the n'th. The sign convention I've written above for $\delta$ on cochains corresponds to taking an alternating sum over faces, starting at the n'th and ending at the 0'th. Duality tends to interact better with this order reversal of the faces.)
| 6 | https://mathoverflow.net/users/360 | 438150 | 176,997 |
https://mathoverflow.net/questions/430546 | 3 | I am reading this paper
>
> "A JKO splitting scheme for Kantorovich-Fisher-Rao gradient flows"
> (<https://arxiv.org/abs/1602.04457>)
>
>
>
and in the proof of Proposition 2.2, basically, if the measure $\rho$ is smooth such that $\rho = \rho(x)dx$, i.e., we think about the measure as its density then one can define
$$\mathbf{v}^\varepsilon = \mathbf{v} + \varepsilon \frac{\mathbf{w}}{\rho}$$
where $\mathbf{v}$ is a given vector field and $\mathbf{w}$ is any divergence-free vector field, and the proof carries through.
**My question is:** if $\rho$ is not that nice, how to define such a thing? I guess some kind of approximations but not really know what to do?
| https://mathoverflow.net/users/124759 | Equivalent definition of the Kantorovich-Fisher-Rao distance | Well, when we wrote the paper we were not really concerned with full rigor at this stage, all we wanted to emphasize was that the "KFR" distance (by now rather the WFR or HK distance, as in Wasserstein-Fisher-Rao or Hellinger-Kantorovich) constructed in 3 independent papers was really the same, at least formally. To the best of my knowledge this still hasn't been proved fully rigorously, but it is clear to everybody in the field (yes, I know how bad this may sound...)
Regarding your precise question: if you wish, our proposition 2.2. is a purely formal result, so the proof is formal too. It is not one of these proofs where the statement is fully rigorous, you first prove it under stronger (regularity/positivity) assumptions, and then relax the assumptions to get full generality. Our statement is formal to begin with (and I would agree that we should have mentioned this very explicitly when writing the paper), hence so is the proof.
For a partial attempt at answering more mathematically: even in classical optimal transport, the statement that the velocity field $v\_t=\nabla r\_t$ can be taken as a gradient is to be understood in a very delicate and particular sense. (Our whoe point in the paper was that the $r\_t$ in question is the reaction term, so horizontal displacement and vertical creation/anihilation of mass are related.) One of the reasons for this is that, if one thinks of an (optimal, kinetic-energy-minimizing) velocity field $v\_t$ as representing a "tangent vector" $T\_{\mu\_t}\mathcal P$ at a point $\mu\_t$ in the Wasserstein space, then the natural functional space to have it live in is the weighted $L^2(\mu)t)$ space. As a consequence, that "$v\_t$ must be a gradient" should rather be replaced by the condition that
$$
v\_t\in \overline{\{\nabla\phi,\,\phi\in C^1\}}^{L^2(\mu\_t)},
$$
where the completion of smooth gradients is taken in the above and natural functional setting. This is exactly how the "tangent space" is constructed for optimal transport, I can suggest for example to look into [1], sections 8.4 and 8.5.
Taking the completion precisely takes care of the "vacuum issue" (that $\rho$ may not be smooth and positive).
[1] Ambrosio, L., Gigli, N., & Savaré, G. (2005). Gradient flows: in metric spaces and in the space of probability measures. Springer Science & Business Media.
| 1 | https://mathoverflow.net/users/33741 | 438151 | 176,998 |
https://mathoverflow.net/questions/438157 | 2 | I am looking at the convergence of the series
$$ \cos(t\theta) = \frac{\sin(\pi t)}{\pi} \cdot \Bigg[\frac{1}{t} + 2t \sum\_{k=1}^\infty (-1)^k \frac{\cos(k\theta)}{t^2 - k^2}\Bigg].$$
Here $t\in\mathbb{R}$. The above equality is rather trivial, but the convergence of the right side towards the left side is not straightforward to me. Based on empirical observations, we have convergence to the target function if $|\theta| <\pi$, regardless of $t$ (even if $t\in\mathbb{Z}$, by taking the limit). If $|\theta|>\pi$, we still have convergence, but towards a different function rather than $\cos(t\theta)$.
Is that correct? How to prove it or find the conditions on $\theta,t$ for the convergence to $\cos(t \theta)$?
**Purpose**
This was part of a bigger project to find an analytic continuation of $\zeta^\*(s)$ where $\zeta^\*(s) = \zeta(s)$ if $\Re(s)$ is an integer. The hope was that the analytic continuation would coincide with $\zeta(s)$. You may replace $\zeta,\zeta^\*$ by the Dirichlet eta function or related functions. You could do the same using $\Im(s)$ instead of $\Re(s)$.
The first step is establish the following and check when it is valid, for an arbitrary even function $f(t)$:
$$ f(t) = \frac{\sin(\pi t)}{\pi} \cdot \Bigg[\frac{f(0)}{t} + \phi'(t)\sum\_{k=1}^\infty (-1)^k \frac{f(k)}{\phi(t) - \phi(k)}\Bigg].$$
Here I used $\phi(t)=t^2$. Then have a similar formula for odd functions, and by combining both, a formula for any function regardless of parity.
I did get some pretty decent approximation (analytic extension) when working with $\Re(s)$ fixed and $t$ is the variable. But not an exact continuation. However, it does not lead to anything interesting, even if the approximation was exact. It's less accurate anyway as $t$ increases.
What I really wanted is the same thing but with $\Im(s)$ fixed instead, and $\sigma$ playing the role of $t$. There I failed; it would have been an exciting development otherwise. I am not finished yet, as there are plenty of options for $\phi(t)$, not just $\phi(t)=t^2$. I tried many low-hanging fruits; all failed. There has to be some $\phi$ that will work, I guess, though it won't be a simple function. You could choose $\phi$ by solving an integral equation in $\phi$ so that the RHS matches the LHS. Anyway, that's where I am now. Not sure if I will pursue the idea.
| https://mathoverflow.net/users/140356 | Convergence of series related to partial fraction expansion of cotangent function | Here's a complex analysis proof. For $|\theta|\leq\pi,$ we have that
$$
F(t)=\frac{\cos(\theta t)\pi}{\sin (\pi t)}
$$
is an odd meromorphic function for $t\in\mathbb{C}$ with simple poles at $k\in\mathbb{Z}$, which moreover is bounded as $|\Im t|\to\infty$.
We claim that the series in brackets
$$
G(t)=\frac{1}{t} + 2t \sum\_{k=1}^\infty (-1)^k \frac{\cos(k\theta)}{t^2 - k^2}
$$
has all the same properties. Indeed, we have
$$
\left|(-1)^k\frac{\cos (k\theta)}{t^2-k^2}\right|\leq
\frac{1}{|(\Re t)^2-(\Im t)^2-k^2|}\leq \frac{2}{(\Im t)^2+k^2},
$$
if either $(\Im t)^2\geq 2(\Re t)^2$, or $k^2\geq 2(\Re t)^2$. From the latter case, the series converges absolutely and uniformly on compact subsets of $\mathbb{C}\setminus \mathbb{Z}$, and the former case can be used to show, e.g., by comparing with the integral, that $|G(t)|$ is bounded over $|\Im t|\geq 10$, say. The poles at $\pm k$ only come from $k$-th term.
Now we simply note that $F$ and $G$ have the same residues at the poles, so $F-G$ is a holomophic function in the whole plane $\mathbb{C}.$ It is enough to show that $F-G$ is bounded, then it is constant by Liouville's theorem, and since it is odd, the constant is zero.
To check boundedness, by maximum principle, it suffices to upper-bound $|F|$ and $|G|$ separately on the boundary of each box $$R\_m=\left\{m-\frac12<\Re t<m+\frac12,|\Im t|\leq 10\right\},$$
by a constant independent of $m$. For $F$, we just upper-bound $|\cos t\theta|\leq e^{10|\theta|}$ and lower-bound $\sin(\pi t)$ by its minimum (which is independent of $m$ by periodicity). For $G$, we notice that we can actually write, for any $m$,
\begin{multline}
G(t)=\lim\_{N\to\infty}\sum\_{k=-N}^N(-1)^k\cos(k\theta)\frac{1}{t-k}=\lim\_{N\to\infty}\sum\_{k=-N+m}^{N+m}(-1)^k\cos(k\theta)\frac{1}{t-k}\\=\frac{(-1)^m}{t-m}+2(t-m)\sum\_{k=1}^\infty(-1)^{k-m}\frac{\cos((k-m)\theta)}{(t-m)^2-k^2},
\end{multline}
and bounding the absolute value term by term, as above, on the boundary of $R\_m$, yields an estimate independent of $m$.
If $\theta>\pi$, then $F(t)$ is no longer bounded as $t\to\infty$, while the RHS still is, so the equality cannot hold. As noted by Conrad in the comments, the RHS does not change under replacing $\theta\mapsto \theta+2\pi m$, so by choosing an appropriate $m$ we can reduce this to the previous case.
| 3 | https://mathoverflow.net/users/56624 | 438165 | 177,002 |
https://mathoverflow.net/questions/438081 | 17 | Is there a c.e. theory $T⊢\text{ZFC}$ in the language of set theory such that the minimum transitive model of $T$ exists but does not satisfy $V=L$?
You may assume that ZFC has transitive models. Note that $M$ is minimal iff $∀M' \, M'⊈M$ and minimum iff $∀M' \, M'⊇M$.
It may be tempting to consider ZFC + $0^\#$ (assuming large cardinal axioms), but while this theory has a minimum inner model (i.e. $L[0^\#]$), it has incomparable minimal transitive models. Model comparability uses iterability, but transitiveness does not suffice for iterability. Moreover, for every c.e. $T⊢\text{ZFC\P}+0^\#$ having a model $M$ with $On^M = α < ω\_1$, the intersection of all such $M$ equals $L\_α$, and furthermore a subset of $L\_α$ is definable (with parameters) in all such $M$ iff it is in $L\_{α^{+,\mathrm{CK}}}$. To see this (briefly), $0^\#$ allows $M$ to 'continue' $L$ beyond $α$, and $L\_{α^{+,\mathrm{CK}}}⊆(L\_{α^{+,\mathrm{CK}}})^M$ (the well-founded part of any model of KP being admissible), and so $L\_{α^{+,\mathrm{CK}}}∩V\_α=L\_α$. Also, existence of $M$ is $Σ^1\_1(α)$, so the intersection of all $M$ is at most $L\_{α^{+,\mathrm{CK}}}$.
A minimum transitive model of ZFC + $A$ for a *statement* $A$ cannot be produced through set forcing either. Also, I think that minimal models need not satisfy $V=HOD$; not sure about minimum models.
Thus, minimum transitive models usually either satisfy $V=L$ or do not exist. However, I suspect that exceptions exist, including in the form $L\_α[r]$ with $r∈ℝ$, but require an interesting coding argument. Perhaps there is such an $r$ computable from the theory of the least transitive model of ZFC $L\_α$, with $r$ self-verifying (in $L\_α[r]$) and yet weak enough for $L\_α[r]⊨\text{ZFC}$.
*Update:* I accepted Farmer Schlutzenberg's positive answer (see also the partial answers by Hamkins and Enayat). A remaining open problem is whether such a $T$ can be obtained by extending ZFC with a single statement.
| https://mathoverflow.net/users/113213 | Minimum transitive models and V=L | Yes, I claim you can in fact get one whose minimum model is a set forcing extension of a segment of $L$. Let $L\_\alpha$ be least modelling ZFC. Let $\mathbb{P}=\mathbb{P}^{L\_\alpha}$ be Jensen's forcing for adding a $\Pi^1\_2$-singleton, as defined over $L\_\alpha$; ZFC proves that forcing with $\mathbb{P}^L$ over $L$ adds a unique $L$-generic filter. Working in $V$, define the sequence $\left<p\_n\right>\_{n<\omega}$ through $\mathbb{P}$, which will be $L\_\alpha$-generic, as follows. Fix a recursive enumeration $\left<\varphi\_k\right>\_{k<\omega}$ of all formulas in the language of set theory in one free variable. Now let $p\_0=\emptyset$. Suppose we have defined $p\_n$. Let $\alpha\_n$ be the least $\beta$ such that $L\_\beta\preccurlyeq\_{\Sigma\_n}L\_\alpha$. Let $p\_{n+1}$ be the least $p\in\mathbb{P}$ such that $p\leq p\_n$ and $p$ is in all open dense subsets $D$ of $\mathbb{P}$ which are defined over some $L\_{\alpha\_i}$ by $\varphi\_j$, for some $(i,j)$ with $i,j\leq n$. (That is, for some such $(i,j)$, $D$ is the unique $D'\in L\_{\alpha\_i}$ such that $L\_{\alpha\_i}\models \varphi\_j(D')$.) This $p$ exists since there are only finitely many dense sets we consider here. This defines the sequence.
Let $G$ be the filter generated by $\left<p\_n\right>\_{n<\omega}$.
Since $L\_\alpha$ is pointwise definable, $G$ is $L\_\alpha$-generic. So $L\_\alpha[G]\models$ ZFC.
Now let $T$ be the following recursive (not just r.e.) theory extending ZFC, which will correspond to the construction above: for $n<\omega$ let $T\_n$ be ZFC + "$V=L[g]$ for some $(L,\mathbb{Q})$-generic filter $g$, where $\mathbb{Q}$ is Jensen's forcing in $L$, and defining $q\_n$ with respect to $L$ as in the construction above, we have $q\_n\in g$". (That is, note that there is a recursive sequence $\left<\psi\_n\right>\_{n<\omega}$ of formulas such that for each $n$, we have that $p\_n$ is the unique $p\in L\_\alpha$ such that $L\_\alpha\models\psi\_n(p)$. Have $T\_n$ assert that there is $q\in g$ (where $g$ is as above) such that $L\models\psi\_n(q)$. Thus, the sequence $\left<T\_n\right>\_{n<\omega}$ is recursive. Of course, the complexity of $\psi\_n$ increases with $n$. Note that the "$g$" is a bound variable, not some new constant, so I am working with only the language of set theory.) Now set $T=\bigcup\_{n<\omega}T\_n$. So $T$ is recursive.
Claim: $L\_\alpha[G]$ is the minimum transitive model of $T$.
Proof: Let $P$ be any transitive model of $T$. Then certainly $L\_\alpha\subseteq P$, and easily if $\alpha<\mathrm{OR}^P$ then $L\_\alpha[G]\subseteq P$. So we may assume $\alpha=\mathrm{OR}^P$. So $L^P=L\_\alpha$. But then letting $G'$ be the (unique) $(L\_\alpha,\mathbb{P})$-generic filter such that $P=L\_\alpha[G']$, we get $p\_n\in G'$ for each $n<\omega$, since $P\models T\_n$ for each $n$. But therefore $G'=G$, so $P=L\_\alpha[G]$, which suffices.
Remark: This construction is somewhat related to that for Proposition 34 of "On a Conjecture Regarding the Mouse Order for Weasels", arXiv:2207.06136, joint with Jan Kruschewski; (that proposition is stated rather generally, but in its simplest instantiation it gives an example of $G$ which is (only just) Cohen generic over $L\_{\omega\_1^{\mathrm{ck}}}$ but with KP failing in $L\_{\omega\_1^{\mathrm{ck}}}[G]$).
Remark 2: The question is rather related to a question of Harvey Friedman's, on which Woodin and Koellner made recent (boldface) progress. The question was (if I recall it precisely) whether there can be an ordinal $\alpha$ and a *single sentence* $\varphi$ such that there is a unique transitive model $M$ such that $\mathrm{OR}^M=\alpha$ and $M\models$ ZFC + "$V\neq L$" + $\varphi$. It was already known that any such model must satisfy "$0^\sharp$ does not exist", and I think also that it must satisfy "$V=\mathrm{HOD}$".
| 9 | https://mathoverflow.net/users/160347 | 438178 | 177,005 |
https://mathoverflow.net/questions/437901 | 3 | If we add to the wholeness axiom, the axiom of stratified\acyclic replacement, what would be the consistency state and strength of the resulting theory?
The wholeness axiom $\sf WA$, [introduced](http://pcorazza.lisco.com/papers/MVS/WholenessAxiom.pdf) by Paul Corazza, found to be [consistent with $\sf V=HOD$](https://arxiv.org/pdf/math/9902079.pdf), is axiomatized in the first order language with signature $\{ \in , j\}$, where $j$ is a primitive total one place function symbol. An "$\in$-formula" is a formula in which $j$ doesn't occur.
$\sf WA$ is the statement that "$j$ is a non-trivial elementary embedding from $V$ to $V$ over signature $\{ \in \}$".
More explicity:
* $\exists x: j(x) \neq x$
* if $\varphi(x\_1,..,x\_n)$ is an $\in$-formula, then: $$ \forall x\_1,.., \forall x\_n \\(\varphi(x\_1,..,x\_n) \iff \varphi(j(x\_1),..,j(x\_n)))$$
Now, the theory is:
$\sf ZC + \sf Rep^\in + WA$
where $\sf Rep^\in$ is replacement scheme restricted to $\in$-formulas.
To be especially noticed that there is no restriction on $\sf Z$, so $j$ can be used in instances of Separation.
Now, stratified replacement "$\sf Rep^\equiv $" is Replacement schema restricted to stratified formulas.
[Stratification](https://en.wikipedia.org/wiki/New_Foundations#Axioms_and_stratification) criterion is defined after Quine as in Stratified Comprehension, plus the requirement that $j(x)$ is one type higher than $x$.
[Equivalently](https://arxiv.org/pdf/2009.13274.pdf) $\sf Rep^\equiv$ can be formalized by the restriction of replacement schema to acyclic formulas, with an edge stipulated to occur between $x$ and $j(x)$ in the [definition of acyclic formula](https://arxiv.org/pdf/2010.14949.pdf).
The rationale beyond this is that the stratification\acyclicity criterion precludes Kunen's known proof of the critical sequence $\langle \kappa\_n | n \in \omega \rangle $, defined as usual by $\kappa\_0 = \kappa = \operatorname{cp}(j)$ and $\kappa\_{n+1} = j(\kappa\_n)$, being a set.
So, formally the question is:
>
> Would $\sf ZC + \sf Rep^\in + Rep^\equiv + WA$, be stronger than, $\sf ZC + \sf Rep^\in + WA$ ?
>
>
>
**[After note]:** Stratified replacement proved to be inconsistent, since it does capture the critical sequence as shown in the answer by Greg Kiramyer. But, still the Acyclic variant remains viable. I initially thought it to be equivalent to the stratified one; I'm realizing now that the proof of equivalence of acyclicity with stratification is not carried for the language using $j$ with this particular specifications of stratification, so they might not be equivalent!? Therefore ***Acyclic Replacement*** might stand a chance?!
| https://mathoverflow.net/users/95347 | If we add stratified\acyclic replacement to the wholeness axiom, would that increase its consistency strength? | The statement beginning "The rationale beyond" is not quite correct. The critical sequence can be constucted by exploiting the facts that $\bigcup j(x)$ will be assigned the same number as $x$ in a stratification assignment, $\bigcup x=x$ for any limit ordinal, and $\omega$ is well ordered by proper subset relation $\subsetneq$.
Let $F(W,K,x,y)$ be the "formula":
$x∈W \land \exists f: f \text { is a function } \land \\ dom(f)=\{t \in W \mid t \subseteq x\} \land \\ f(0)=K \land f(x)=y\land \\ \forall t((t≠0 \land t∈ dom(f)) \to \\f(t) = \bigcup j(f(\bigcup \{s∈W \mid s\subsetneq t\}))))$
Then for all $n \in \omega $, there is a unique $y$ such that $F(\omega,\kappa,n,y)$, the $n$-th element of the critical sequence, where $\kappa$ is the critical point of $j$.
| 5 | https://mathoverflow.net/users/133981 | 438182 | 177,006 |
https://mathoverflow.net/questions/438164 | 2 | Let $\Omega\_1$ and $\Omega\_2$ be (simply connected) domains on $\mathbb{R}^2$, with coordinates $(x,y)$ and $(X,Y)$ respectively. Given a (smooth) function $Z(X,Y)$ such that $Z\left(\partial \Omega\_2 \right)=0$, consider the system of PDEs
\begin{align}
\left(\frac{\partial X}{\partial x} \right)^2+\left(\frac{\partial Y}{\partial x} \right)^2+\left(\frac{\partial Z\left(X,Y \right)}{\partial x} \right)^2=1
\end{align}
\begin{align}
\left(\frac{\partial X}{\partial y} \right)^2+\left(\frac{\partial Y}{\partial y} \right)^2+\left(\frac{\partial Z\left(X,Y \right)}{\partial y} \right)^2=&1,
\end{align}
subjected to
\begin{align}
\left(X,Y\right)\left(\partial \Omega\_1 \right)=\partial \Omega\_2.
\end{align}
Does a solution exist ?
This problem comes as a particular case of Chebyshev mappings, where a surface is covered by a net of lines whose tangents are unit vectors at all its points. In this case the boundary of the surface is constrained to lie on plane. Generically, the (local) existence of solutions is shown by obtaining an equivalent system that is explicitly hyperbolic (by cross-differentiation), so the problem is hyperbolic in nature. In this case the condition is not hyperbolic but elliptic, so the existence of solutions is not secured
| https://mathoverflow.net/users/171439 | Hyperbolic system of PDEs with elliptic-like boundary contions | Here is an example for which there is no solution: Let $\Omega\_1$ be defined by $x^2+y^2\le 1$ and $\Omega\_1$ be defined by $X^2+Y^2\le R^2$, where $R>0$ is large. Take $Z(X,Y) = 0$. Then one is asking for a map $f:\Omega\_1\to\Omega\_2$ that preserves lengths of $x$-lines and $y$-lines and carries the circle $x^2+y^2=1$ onto the circle $X^2+Y^2=R^2$. However, because any two points in $\Omega\_1$ can be joined by a piecewise curve composed of $x$-segments and $y$-segments of total length at most $3$ (say), it follows that the image of $f$ must be contained in a disk of radius at most $3$. When $R$ is sufficiently large, there will be points on the boundary of $\Omega\_2$ that are farther apart than that.
(Note that, if one allows $Z(X,Y)$ to be nonzero, this only makes the problem worse.)
*Remark:* To get a sense of what these restrictions look like. Consider the case where $Z(X,Y)=0$ and $\Omega\_1$ is defined by $x^2+y^2\le 1$. Then each solution to the above equations for $f:\Omega\_1\to\mathbb{R}^2$ is of the form
$$
f(x,y) = \bigl(X(x,y),Y(x,y)\bigr) = \bigl(p\_1(x)+q\_1(y),p\_2(x)+q\_2(y)\bigr),
$$
where $p=(p\_1,p\_2):[-1,1]\to\mathbb{R}^2$ and $q=(q\_1,q\_2):[-1,1]\to\mathbb{R}^2$ are unit speed curves. To describe the domains $\Omega\_2$ such that $f(\partial\Omega\_1) = \partial\Omega\_2$, you need to be able to determine which closed curves $\partial\Omega\_2$ are of the form $\bigl\{f(\cos t,\sin t)\ |\ 0\le t\le 2\pi\ \bigr\}$.
| 3 | https://mathoverflow.net/users/13972 | 438183 | 177,007 |
https://mathoverflow.net/questions/438184 | 10 | Given two sets $A$ and $B$, the function set $B^A$ is characterized by the universal property that the functor $(-)^A:\mathrm{Set} \to \mathrm{Set}$ is the right adjoint of the functor $(-)\times A:\mathrm{Set} \to \mathrm{Set}$. What is the universal property of the set of injections between two sets $A$ and $B$ in the category of sets, if any such universal property exists?
| https://mathoverflow.net/users/483446 | Universal property of the set of injections in the category of sets | $\newcommand{\Inj}{\operatorname{Inj}}\newcommand{\Set}{\mathrm{Set}}$**Maps $X \to \Inj(A,B)$ correspond to monomorphisms $A \times X \to B \times X$ in the slice $\Set/X$, which can be thought of as “$X$-indexed monomorphisms $A \to B$”.**
This isn’t hard to check: the main tool is that for any objects over $X$, say $A' \to X$ and $B' \to X$, maps $f : A' \to B'$ over $X$ correspond to $X$-indexed families of maps between their fibers $f\_x : A'\_x \to B'\_x$, and such $f$ is a mono in $\Set/X$ precisely if each $f\_x$ is a mono (equivalently, injection) in $\Set$.
There’s a general philosophical point here. Interesting universal properties are often most naturally described in terms of *slices* of the ambient category; an “$X$-indexed family of whatsits” will usually be a whatsit in $\mathcal{E}/X$. Objects of $\mathcal{E}/X$ are viewed as $X$-indexed families of objects of $\mathcal{E}$ — this “relative viewpoint” was famously exploited and popularised by Grothendieck, in the category of schemes, i.e. the “spaces” of algebraic geometry — and if you started with plain objects of $\mathcal{E}$ (like your original $A$ and $B$), then to mention them in $\mathcal{E}/X$, you use the *constant* families of objects $A \times X \to X$, as above.
| 16 | https://mathoverflow.net/users/2273 | 438186 | 177,008 |
https://mathoverflow.net/questions/438189 | 1 | **Disclaimer.** Not sure this is MO-level but would really appreciate some help with this. Thanks in advance. Moved from SE.
Let $a,b,c \ge 0$, and define a function $g:\mathbb R \to \mathbb R$ by $g(t) := \sqrt{(t-1)^2 + a^2} + b|t|$. It is clear that $g$ is convex.
**Question.** What is an analytic formula for the value $M(a,b,c)$ of $\min\_{t \in \mathbb R} g(t)$ subject to $|t-1| \le c $ ?
**N.B.:** In the special case where $a=0$, one can easily compute $$ M(0,b,c) = \min\_{|t-1| \le c}|t-1| + b|t| = m(a,b):= \begin{cases}b,&\mbox{ if }b \le 1,\\ 1+(1-c)\_+ (b-1),&\mbox{ else.}\end{cases}.$$
**An approximation.** Also note that for any $a$, one has $g(t) \le a + |t-1| + b|t|$ for all $t \in \mathbb R$, and so $\max(a,M(0, b,c)) \le M(0,b,c) \le a+M(0,b,c)$. Thus, we deduce that
$$
M(a,b,c) \asymp \max(a,m(b,c)) \asymp a + m(b,c),
$$
where $m (b,c)$ is as given above, and $u \asymp v $, means that $u$ and $v$ are within absolute constant multiples of one another.
| https://mathoverflow.net/users/78539 | Analytic expression for the min value of $g(t):= \sqrt{(t-1)^2 + a^2}+ b|t|$ subject to $|t-1| \le c$ | First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and
$$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$
Since $g(t)$ is increasing in $t\ge1$, wlog $t\le1$.
Next, change the variables and the constants according to the formulas
$$u=(t-1)^2,\quad A:=a^2,\quad c\_2:=\min(1,c^2),$$
so that $u\in[0,1]$ and $t=1-\sqrt u$. So, the problem reduces to minimizing
$$h(u):=\sqrt{u + A} + b(1-\sqrt u)$$
in $u\in[0,c\_2]$ given $A\ge0$, $b\ge0$, and $c\_2\in[0,1]$.
Note that $h'(u)$ has the same sign for real $u>0$ as $\frac u{A+u}-b^2$, which increases from $-b^2$ to $1-b^2$ as $u$ increases from $0$ to $\infty$. So, the only critical point of $h$ on $(0,\infty)$ is
$$u\_\*:=A\frac{b^2}{1-b^2}$$
if $0\le b<1$, and no critical points of $h$ on $(0,\infty)$ if $b\ge1$.
Letting now $h\_{\min}$ denote the minimum of $h$ on $[0,c\_2]$, we see that the desired minimum of $g$ is
$$
h\_{\min}=
\begin{cases}
h(u\_\*)=b+\sqrt{(1-b^2)A}&\text{ if }0\le b<1\ \&\ u\_\*\le c\_2, \\
h(c\_2)=\sqrt{c\_2 + A} + b(1-\sqrt{c\_2})&\text{ if }b\ge1\ \text{or}\ (0\le b<1\ \&\ u\_\*>c\_2).
\end{cases}
$$
This can be simplified:
$$
h\_{\min}=
\begin{cases}
b+a\sqrt{1-b^2}&\text{ if }b^2\le\dfrac{c\_2}{a^2+c\_2}, \\
\sqrt{c\_2 + a^2} + b(1-\sqrt{c\_2})&\text{ otherwise}.
\end{cases}
$$
| 4 | https://mathoverflow.net/users/36721 | 438213 | 177,015 |
https://mathoverflow.net/questions/436898 | 8 | Let $X$ be an algebraic stack of finite type over a field.
Is there an intrinsic way to calculate the minimum of the dimensions of all atlases of $X$?
By intrinsic here I mean using constructions such as the inertia stack, the stabilizer group construction, etc.
A natural conjecture is that this minimum should be something like the dimension of $X$ plus the dimension of the largest stabilizer of any point of $X$, as one can see using classifying stacks, for example.
| https://mathoverflow.net/users/496640 | Smallest atlas for algebraic stack | Professor Jarod Alper provided me an answer to this question in separate correspondence. When the stabilizer at a point of the stack is smooth, my conjecture above is true by the theory of miniversal deformations. This is Theorem 3.6.1 in Professor Alper's notes here: [https://sites.math.washington.edu/~jarod/moduli.pdf](https://sites.math.washington.edu/%7Ejarod/moduli.pdf).
| 1 | https://mathoverflow.net/users/496640 | 438223 | 177,018 |
https://mathoverflow.net/questions/438218 | 2 | The *Euclidean distortion* of a metric space $X$, denoted $c\_2(X)$, is the infimum of $c$ for which there exists a map $f\colon X\to\ell^2$ such that
$$d\_X(x,y) \leq \|f(x)-f(y)\|\_{\ell^2} \leq c\cdot d\_X(x,y).$$
In the introduction of their paper "[Nonembeddability theorems via Fourier analysis](https://arxiv.org/abs/math/0510547)", Khot and Naor mention the following in passing:
>
> The Euclidean distortion of a metric space $X$ is relatively well
> understood: it is enough to understand the distortion of finite
> subsets of $X$, and for finite metrics there is a simple semidefinite
> program which computes their Euclidean distortion [39].
>
>
>
Here, reference [39] is Linial, London, and Rabinovich's "[The geometry of graphs and some of its algorithmic applications](https://users.math.msu.edu/users/iwenmark/Teaching/CMSE890/EMBEDDING_The_Geometry_of_Graphs_and_Some_of_Its_Algorithmic.pdf)", which introduced the semidefinite program mentioned above. I'm interested in the first part of Khot and Naor's claim, which appears to suggest that
$$c\_2(X) = \sup\_{\substack{T\subseteq X \\ |T|<\infty}}c\_2(T)$$
for every metric space $X$. I.e., Euclidean distortion is *finitely determined*.
**Question:** Under what conditions on the metric space $X$ is $c\_2(X)$ finitely determined?
Here's what I know:
* The inequality $\geq$ holds since restricting a bilipschitz function $X\to\ell^2$ to $T$ can only improve the bilipschitz constants.
* The equality is known to hold if $X$ is locally finite. This follows from the main result of Ostrovskii's "[Embeddability of locally finite metric spaces into Banach
spaces is finitely determined](https://arxiv.org/abs/1103.0748)".
* The equality happens to hold for $X=\mathbb{R}/\mathbb{Z}$ with the quotient metric. This is a consequence of (the proof of) Claim 2.1 in Linial and Magen's "[Least-Distortion Euclidean Embeddings of Graphs: Products of Cycles and Expanders](https://core.ac.uk/download/pdf/82714902.pdf)", which establishes that $n$ equally spaced points in $X$ have distortion $\frac{n}{2}\sin\frac{\pi}{n}$. (This limits to the distortion of the obvious embedding of $X$ in $\mathbb{R}^2$.)
* The equality does not hold in pathological cases. For example, if $X$ is the power set of $\ell^2$ with trivial metric, then every finite subset of $X$ isometrically embeds into $\ell^2$ as the vertices of a regular simplex, but Cantor's theorem implies $c\_2(X)=\infty$.
| https://mathoverflow.net/users/29873 | When is Euclidean distortion finitely determined? | The problem is that you misdefined "Euclidean distortion". The infimum is over all maps into arbitrary Hilbert spaces, not just $\ell\_2$. Of course, you can use only $\ell\_2$ if the metric space is separable, but not when it is non separable. Now if every finite subset $F $of an infinite metric space $X$ embeds into $\ell\_2$ with constant $C+ |F|^{-1}$, then $X$ C-embeds into an ultra power of $\ell\_2$, and ultra powers of a Hilbert space are Hilbert spaces.
| 5 | https://mathoverflow.net/users/2554 | 438226 | 177,019 |
https://mathoverflow.net/questions/438199 | 3 | I am looking at trying to show that a complex symmetric matrix always has a complex symmetric square root. Showing a square root exists is fairly easy if the matrix is also invertible by using the Jordan Canonical Form.
I have seen on here that showing that the square root of a matrix A is a (Hermite) polynomial in A proves that if A is symmetric then so is its square root. My question is, why is this true?
The reference for this would be Function of Matrices, Defn 1.2 (Matrix Function using Jordan Canonical Form) and Defn 1.4 (Matrix Function using Hermite Interpolation) and Theorem 1.12 (which shows that the two definitions given are equivalent).
| https://mathoverflow.net/users/497608 | Why is the square root of a complex symmetric matrix also complex symmetric | If your complex symmetric matrix $A$ is not invertible, it might not have a square root at all, e.g. $$ \pmatrix{i & 1\cr 1 & -i\cr}$$
If $A$ is invertible, let $\lambda\_j$ be the eigenvalues of $A$ and $m\_j$ their multiplicities. Let $P(z)$ be a polynomial such that $P(z)$ and the first $m\_j-1$ of its derivatives agree with some branch of $\sqrt{z}$ and the first $m\_j-1$ of its derivatives
at each $\lambda\_j$. Then the polynomial $P(z)^2 - z$ and the first $m\_j - 1$ of its derivatives are $0$ at $\lambda\_j$, implying that $P(z)^2 - z$ is divisible by the characteristic polynomial of $A$, and so the Cayley-Hamilton theorem implies $P(A)^2 = A$, i.e.
$P(A)$ is a square root of $A$.
Of course if $A$ is symmetric, $P(A)$ is symmetric.
| 6 | https://mathoverflow.net/users/13650 | 438232 | 177,020 |
https://mathoverflow.net/questions/438222 | 8 | It's well known that sentences about the real closed field can be decided by algorithm and the complexity of this is about $d^{2^{O(n)}}$ where $d$ is the product of the degrees of polynomials in the sentence. See [here](https://en.wikipedia.org/wiki/Real_closed_field#Complexity_of_deciding_%F0%9D%98%9Brcf)
Now adding an operation such as the $\sin$ function to the field structure leads to the ability to define natural numbers (and so now its undecidable).
Other operations such as adding an $\exp$ appear to be open problems (we can't prove if it is decidable or not, let alone having an algorithm which provably runs in finite time).
So I was curious, do we know ANY other *relations* which can be added to the underlying structure such that we still end up with a decidable theory AND we are able to prove new sentences which were not expressible before.
I would expect it should be possible to construct an ordinal-indexed hierarchy of such decidable theories, each of whose runtime for deciding expressions is naturally related to the ordinal it is indexed by and whose limit approaches the theory of the natural numbers.
But that is just wishful thinking for now ^
I guess I'm just curious if anyone knows how to make a "bigger" theory than the real closed field which is still decidable.
| https://mathoverflow.net/users/46536 | What theories are larger than the real closed field but still decidable? | Thanks to a 1958 paper by Abraham Robinson (whose impetus was a question of Alfred Tarski), an example of such a theory that properly extends RCF is the theory of the structure $(\mathbb{R},~+,~\cdot,~A)$, where $A$ is the collection of algebraic real numbers.
More generally, Robinson's proof shows that the theory of structures of the form $(\mathbb{R},~+,~\cdot,~F)$ is decidable for any *real closed* subfield $F$ of the field or reals.
A free copy of Robinson's paper can be found [here](http://matwbn.icm.edu.pl/ksiazki/fm/fm47/fm47111.pdf).
| 12 | https://mathoverflow.net/users/9269 | 438235 | 177,021 |
https://mathoverflow.net/questions/438035 | 9 | Given the [*Ramanujan theta function*](https://en.wikipedia.org/wiki/Ramanujan_theta_function),
$$f(a,b) = \sum\_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$
Let $q = e^{2\pi i \tau}$ and assume $\tau = \sqrt{-d}$.
---
**I. Degree 5**
\begin{align}
a &= q^{11/60}\;\frac{f(-q,-q^4)}{f(-q)} \,=\, q^{11/60}\, \prod\_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}\\
b &= q^{-1/60}\,\frac{f(-q^2,-q^3)}{f(-q)} = q^{-1/60}\prod\_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}\\
\end{align}
where $a,b$ surprisingly are ***radicals***. They obey the nice relation,
$$a\,b^{11}-a^{11}\,b = 11a^6\,b^6 +1$$
Equivalently,
$$\Big(\frac{b}{a}\Big)^5-\Big(\frac{a}{b}\Big)^5 = \frac1{(ab)^6}+11$$
The integer $60$ reflects the order of the icosahedron.
---
**II. Degree 7**
\begin{align}
a &= -q^{61/168}\;\frac{f(-q,-q^6)}{f(-q^2)} = -q^{61/168}\prod\_{n=1}^\infty \frac{(1-q^{7n-1})(1-q^{7n-6})(1-q^{7n})}{(1-q^{2n})}\\
b &= \,q^{13/168}\;\frac{f(-q^2,-q^5)}{f(-q^2)} \,=\; q^{13/168}\;\prod\_{n=1}^\infty\frac{(1-q^{7n-2})(1-q^{7n-5})(1-q^{7n})}{(1-q^{2n})}\\
c &= q^{-11/168}\;\frac{f(-q^3,-q^4)}{f(-q^2)} = q^{-11/168}\,\prod\_{n=1}^\infty \frac{(1-q^{7n-3})(1-q^{7n-4})(1-q^{7n})}{(1-q^{2n})}\\
\end{align}
where $a,b,c$ again are radicals. They obey,
$$a^3b+b^3c+c^3a = 0$$
and the integer $168$ reflects the order of the Klein quartic. (*See also [this post](https://mathoverflow.net/questions/437994/using-the-rogers-selberg-identities-to-solve-certain-septics).*)
---
**III. Degree 9**
\begin{align}
a &= \;q^{5/9}\;\frac{f(-q,-q^8)}{f(-q^3)} = q^{5/9}\prod\_{n=1}^\infty \frac{(1-q^{9n-1})(1-q^{9n-8})(1-q^{9n})}{(1-q^{3n})}\\
b &= \;q^{2/9}\;\frac{f(-q^2,-q^7)}{f(-q^3)} \,=\, q^{2/9}\prod\_{n=1}^\infty \frac{(1-q^{9n-2})(1-q^{9n-7})(1-q^{9n})}{(1-q^{3n})}\\
c &= -q^{-1/9}\;\frac{f(-q^4,-q^5)}{f(-q^3)} = -q^{-1/9}\prod\_{n=1}^\infty \frac{(1-q^{9n-4})(1-q^{9n-5})(1-q^{9n})}{(1-q^{3n})}
\end{align}
where $a,b,c$ still are radicals. They now obey,
$$\color{blue}{a^2b+b^2c+c^2a = 0}$$
$$a^2c+b^2a+c^2b = 1$$
(*Added later.*) The first is equivalent to,
$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c} = 0$$
so the ratio of the proper functions sum to zero which does not happen with the lower degrees. More on "[*Nonic Analogues of the Rogers-Ramanujan functions*](https://core.ac.uk/download/pdf/82473664.pdf)" (though they didn't affix the factor $q^{m/9}$ to make them into radicals).
---
**IV. Degree 11**
Studied by G. Andrews in 1975, the [first of five](https://mathworld.wolfram.com/Rogers-RamanujanIdentities.html) has the form,
$$a\_1=\frac{f(-q,-q^{10})}{f(-q^4)}$$
To make it a radical, I affixed a factor $q^{m/n}.\,$ I hoped it would involve $n = 660$ thus manifesting [*V. Arnold's trinities*](https://en.wikipedia.org/wiki/ADE_classification#Trinities) regarding PSL(2,5), PSL(2,7), and PSL(2,11) with orders $60, 168, 660$, but it was only $n = 264 = 11\times24$. (*Sigh*.)
---
**V. Questions**
1. Is there anything else special about $a^2b+b^2c+c^2a = 0$?
2. In what other contexts does it appear?
| https://mathoverflow.net/users/12905 | On the Klein quartic and the similar $a^2b+b^2c+c^2a$? | The questions are:
>
> 1. Is there anything else special about $a^2b+b^2c+c^2a = 0$?
> 2. In what other contexts does it appear?
>
>
>
For question 1, Given the $q$-series of degree $9$,
$$ \frac ac=a^3-b^3,\quad\frac ba=b^3-c^3,\quad\frac cb=c^3-a^3 $$
which leads to the telescoping sum
$$ \frac ac+\frac ba+\frac cb = 0 $$
as noticed already. Another feature is multi-section. Let
$$ A := -q^{-1/9}\frac{f(-q^{1/3})}{f(-q^3)} = a + b + c $$
where $a,b,c$ form a 3-section of the $q$-series $A$.
For question 2, refer to
L. E. Dickson, *History of the Theory of Numbers*, Volume II, Chapter XXI, section "Impossibility of $x^3+y^3=z^3$",
pp. $545$-$550$. On page $546$ it states
>
> $\quad$ J. A. Euler$^9$ noted that, if $\,p^3+q^3+r^3=0\,$ is
> possible, $\,x=p^2q,$ $y=q^2r,$ $z=r^2p\,$ satisfy
> $\,x/y+y/z+z/x=0\,$ or $\,x^2z+y^2x+z^2y=0.$
>
>
>
Something similar happens with the Klein quartic
curve and the Fermat septic curve.
| 3 | https://mathoverflow.net/users/113409 | 438238 | 177,023 |
https://mathoverflow.net/questions/438225 | 1 | So on nLab the definition of a trivial (Grothendieck) topology is the following: "The Grothendieck topology on any category for which only the identity morphisms are covering is the trivial topology. Its sheaves are all the presheaves."
I am having trouble understanding what is meant exactly by only the "identity morphisms are covering." After reading a bit, I was under the impression that those are of the form $\hom(-,X)$, but then I don't understand what the link is with what I traditionally understand as "identity morphisms", which to me are of the form $1\_X$ for objects $X\in\operatorname{Obj}(C)$. Am I missing something obvious or is the definition of identity morphisms different in this context?
| https://mathoverflow.net/users/497629 | Trivial Grothendieck topology and identity morphisms | Technically, a Grothendieck topology is specified by its covering sieves, not its covers, so it would have been more accurate to say the covering sieves of the trivial topology are those generated by identity morphisms. The cover of $X$ by the sieve $hom(-,X)$ is indeed the sieve generated by the identity morphism of $X$. Taken literally, most categories do not admit a Grothendieck topology in which the *only* covers are identity maps, since whenever an identity morphism is a cover so is every isomorphism with the same target.
| 2 | https://mathoverflow.net/users/32 | 438239 | 177,024 |
https://mathoverflow.net/questions/438234 | 3 | By some numerical tests, we can see that the following function is negative on $(0,1)$:
$$\small f(x)=\int\_0^\infty\frac{s^{x-1} e^{-2 s} (\pi \cos(\pi x) (s^{2 x}+(0.1)^2)-\sin(\pi x) \ln(s) (s^{2 x}-(0.1)^2)+2 \pi (0.1) s^x )}{(2 (0.1) s^x \cos(\pi x)+s^{2 x}+(0.1)^2)^2} ds.$$
Note that the integrated function change sign for specific values of $x$.
Is there any technique to show that such an integral is negative?
| https://mathoverflow.net/users/149793 | Integral of a function changing sign | In fact, this conjecture fails to hold for $x$ in a right neighborhood of $0$.
Indeed, let $g(x,s)$ denote the integrand. Note that $g(x,s)$ is continuous in $x\in[0,1)$ for each real $s>0$, and
$$g(0+,s)=g(0,s)=\frac{100 \pi e^{-2 s}}{121 s}$$
for all real $s>0$. So, by the Fatou lemma,
$$f(0+)\ge\int\_0^\infty g(0+,s)\,ds=\infty>0,$$
which contradicts the conjecture that $f<0$ on $(0,1)$.
---
The above reasoning is not quite correct -- because the functions $g(x,\cdot)$ with $x\in(0,1)$ do not seem to be all bounded from below by an integrable function, and therefore the Fatou lemma cannot be applied.
The answer still remains negative, though, but we have to work harder to get it.
In what follows, $x\in(0,1/2)$ and $s\in(0,\infty)$, so that $\cos\pi x>0$ and $\sin\pi x>0$.
Removing manifestly positive terms in the numerator of the ratio expressing $g(x,s)$, we get
\begin{equation}
\frac{g(x,s)}{\sin\pi x}\ge g\_1(x,s):=g\_2(x,s)e^{-2s}, \tag{1}\label{1}
\end{equation}
where
\begin{equation}
g\_2(x,s) :=
\frac{s^{x-1} (s^{2 x}-1/100) \ln s}{(c s^x/5+s^{2 x}+1/100)^2},\quad c:=\cos\pi x>0. \tag{2}\label{2}
\end{equation}
So,
\begin{equation}
\frac{f(x)}{\sin\pi x}\ge\int\_0^\infty g\_1(x,s)\,ds \tag{3}\label{3}
\ge f\_1(x):=\int\_0^1 g\_1(x,s)\,ds,
\end{equation}
since $g\_1(x,s)\ge0$ if $s\ge1$. In what follows, $x\in(0,1/2)$ and $s\in(0,1)$.
Next,
\begin{equation}
\int\_0^1 g\_2(x,s)\,ds=
10\frac{ \tan ^{-1}\frac{c+10}{\sqrt{1-c^2}}-
\tan^{-1}\frac{c}{\sqrt{1-c^2}}}{x^2\sqrt{1-c^2}}
\sim\frac{100}{11x^2} \tag{4}\label{4}
\end{equation}
as $x\downarrow0$.
By \eqref{1} and \eqref{2}, $|g\_1(x,s)-g\_2(x,s)|\le2s|g\_2(x,s)|\le2\times10^4|\ln s|$. Since $\int\_0^1 |\ln s|\,ds<\infty$, it follows from \eqref{3} and \eqref{4} that
\begin{equation}
\frac{f(x)}{\sin\pi x}\ge\frac{100-o(1)}{11x^2}
\end{equation}
as $x\downarrow0$, which implies that $f>0$ in a right neighborhood of $0$. $\quad\Box$
| 5 | https://mathoverflow.net/users/36721 | 438245 | 177,027 |
https://mathoverflow.net/questions/352806 | 5 | The answer to another question ([Upper bound of the fraction of Gamma functions](https://mathoverflow.net/q/313103/125166)) gave an asymptotic upper bound for an expression with Gamma functions:
$$\left(\frac{\Gamma(a+b)}{a\Gamma(a)\Gamma(b)}\right)^{1/a}\!\leq \,C\,\frac{a+b}a, \forall a,b\geq\frac12$$
What is the best possible value for the constant $C$ in that statement?
| https://mathoverflow.net/users/122182 | The exact constant in a bound on ratios of Gamma functions | The optimal $C$ is $\mathrm{e}$.
**Proof**:
We have
$$\ln C \ge \ln a - \ln(a + b)
+ \frac{\ln \Gamma(a + b)
-\ln a - \ln\Gamma(a) - \ln\Gamma(b)}{a}.$$
Let
$$F(a, b) := \ln a - \ln(a + b)
+ \frac{\ln \Gamma(a + b)
-\ln a - \ln\Gamma(a) - \ln\Gamma(b)}{a}.$$
We have
$$\frac{\partial F}{\partial b}
= - \frac{1}{a+b} + \frac{\psi(a + b) - \psi(b)}{a} \ge 0 \tag{1}$$
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$.
The proof of (1) is given at the end.
Fixed $a\ge 1/2$, we have
$$G(a) := \lim\_{b\to \infty} F(a, b) =
\ln a
+ \frac{
-\ln a - \ln\Gamma(a)}{a}
$$
where we have used
$$\lim\_{b\to \infty} -\ln(a + b) + \frac{\ln \Gamma(a + b) - \ln\Gamma(b)}{a} = 0.$$
(*Note*: Use $\sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x} \le \Gamma(x) \le \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}\mathrm{e}^{\frac{1}{12x}}$ for all $x > 0$.)
We have
$$
a^2 G'(a) = a - 1 - a\psi(a) + \ln a + \ln \Gamma(a) \ge 0.\tag{2}
$$
The proof of (2) is given at the end.
We have
$$\lim\_{a\to \infty} G(a) = 1.$$
(*Note*: Use $\sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x} \le \Gamma(x) \le \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}\mathrm{e}^{\frac{1}{12x}}$ for all $x > 0$.)
Thus, the optimal $C$ is $\mathrm{e}$.
---
**Proof of (1)**:
Using Theorem 5 in [1]: for all $u > 0$,
$$\ln u - \frac{1}{2u} - \frac{1}{12u^2} < \psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u+1/14)^2},$$
we have
\begin{align\*}
&- \frac{1}{a+b} + \frac{\psi(a + b) - \psi(b)}{a}\\
\ge{}&- \frac{1}{a+b} + \frac{1}{a} \left(\ln (a+b) - \frac{1}{2(a+b)} - \frac{1}{12(a+b)^2}\right)\\
&\qquad
- \frac{1}{a}\left(\ln b - \frac{1}{2b} - \frac{1}{12(b+1/14)^2}\right)\\
={}& \frac{1}{a}\ln(1 + a/b) - \frac{1}{a+b} - \frac{1}{2a(a+b)} - \frac{1}{12a(a+b)^2} + \frac{1}{2ab} + \frac{1}{12a(b+1/14)^2}\\
\ge{}& \frac{1}{a}\left(\ln(1 + a/b) - \frac{a/b}{1 + a/b}\right)
+ \frac{1}{2ab} - \frac{1}{2a(a+b)} + \frac{1}{12a(b+1/14)^2} - \frac{1}{12a(a+b)^2}\\
\ge{}&0
\end{align\*}
where we use $\ln(1+x) \ge \frac{x}{1+x}$ for all $x \ge 0$.
We are done.
$\phantom{2}$
**Proof of (2)**:
Using
$\Gamma(x) \ge \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}$ and
$\psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u+1/14)^2}$ for all $u > 0$ (Theorem 5 in [1]),
we have
\begin{align\*}
&a - 1 - a\psi(a) + \ln a + \ln \Gamma(a)\\
\ge{}& a - 1 - a \left(\ln a - \frac{1}{2a} - \frac{1}{12(a+1/14)^2}\right) + \ln a + \frac12\ln(2\pi) + (a-1/2)\ln a - a\\
={}& \frac12\ln(2\pi a) - \frac12 + \frac{a}{12(a+1/14)^2}\\
\ge{}& 0.
\end{align\*}
We are done.
---
*Reference*
[1] L. Gordon, “A stochastic approach to the gamma function”, Amer. Math. Monthly, 9(101), 1994, 858-865.
| 2 | https://mathoverflow.net/users/141801 | 438254 | 177,031 |
https://mathoverflow.net/questions/438258 | 35 | (This *must* have been asked before and exist somewhere in Community Wiki, but I can't find it...)
Where can you post open (math) problems? And what are the advantages and disadvantages?
* Example: This place (and Math StackExchange), duh.
* Example: The journal AMM had a corner "Unsolved Problems", but no longer. (Editor Stan Wagon told me that he doesn't know any math journal accepting unsolved problems.)
* Example: But then, probably any journal will accept a small "Open Questions" section after a research article.
* Example: USENET once had an useful group. The group still exists. (Yes, I'm *that* old...)
* Example: Other social media, e.g. Reddit has a math sub.
| https://mathoverflow.net/users/11504 | Places where one can post open problems | If you can motivate the problem and make some partial progress on it, you can try and publish it as a paper in a specialized journal, or at the very least upload it to the arXiv.
If you only have empirical evidence, there are journals that are receptive to this kind of this ("Mathematics of Computation" and "Experimental Mathematics" spring to mind).
If it concerns elementary mathematics and is relevant to a wide enough audience you can try a popular journal such as "American Mathematical Monthly".
Other than that, you can
* put in on your website, if you have one.
* share it with experts, as they have the highest chance of solving it, and at the very least of assessing its importance and difficulty and possibly guiding you towards relevant literature or a proof.
* share it in the problem session of a relevant conference. Problems from such sessions tend to be published in the form of conference proceedings.
| 22 | https://mathoverflow.net/users/31469 | 438262 | 177,034 |
https://mathoverflow.net/questions/430766 | 3 | Is there any result like the [Bramble-Hilbert lemma](https://en.wikipedia.org/wiki/Bramble%E2%80%93Hilbert_lemma) for Bochner spaces?
More specifically: let $H$ be a (e.g.) Hilbert space, $I\subset \mathbb R$ a bounded interval, and $L \in \mathcal L(H^k(I;H), Y)$ for $Y$ a normed space. If $L$ vanishes on $k-1$ polynomials, do we necessarily have $||L v ||\_Y\leq C\_k ||\partial\_t^{k} v||\_{L^2(I,H)}$?
| https://mathoverflow.net/users/155442 | Approximation in Bochner spaces | Writing $$v(t)=p(t)+\frac{1}{(k-1)!}\int\_0^t (t-s)^{k-1}v^{(k)}(s)\, ds$$ with $p$ the Taylor polynomial at $0$ of degree $k-1$, then $\|v-p\|\_{H^k} \leq C\|v^{(k)}\|\_{L^2}$. It follows that
$$\|Lv\|=\|L(v-p)\| \leq \|L\|\|v-p\|\_{H^k} \leq C\|L\|\|v^{(k)}\|\_{L^2}.
$$
| 2 | https://mathoverflow.net/users/150653 | 438275 | 177,037 |
https://mathoverflow.net/questions/438269 | 6 | Let $f: \mathbb R^n \to \mathbb R$ be a measurable function.
We say $f$ is *precise* if for every $x \in \mathbb R^n$ and every compact subset $K$ of $\mathbb R^n$ such that for $|K \cap B\_\delta (x)|>0$ for every $\delta>0$, we have :
* $\lim\_{\delta \to 0^+} \text{essinf}\_{K \cap B\_\delta (x)} f = \lim\_{\delta \to 0^+} \inf\_{K \cap B\_\delta (x)} f$
* $\lim\_{\delta \to 0^+} \text{esssup}\_{K \cap B\_\delta (x)} f = \lim\_{\delta \to 0^+} \sup\_{K \cap B\_\delta (x)} f$.
**Question:** Is it true that $f$ is precise if and only if $f$ is continuous?
| https://mathoverflow.net/users/173490 | A characterisation of continuous real functions | Edit: the proof can be made a little simpler.
Yes, this condition is equivalent to $f$ being continuous. The reverse direction is easy because if $f$ is continuous at $x$ then all of the limits in question equal $f(x)$. For the forward direction, suppose $f$ is not continuous at some point $x$. Wlog $f(x) > \lim\_{\delta \to 0}{\rm inf}\_{B\_\delta(x)} f$. Choose $\epsilon$ so that $f(x) > f(x) - \epsilon > \lim\_{\delta \to 0} {\rm inf}\_{B\_\delta(x)} f$ and set $A = \{y: f(y) \leq f(x) - \epsilon\}$. If for some $\delta > 0$ the set $A \cap B\_\delta(x)$ has measure zero, then ${\rm essinf}\_{B\_{\delta'}} f \geq f(x) - \epsilon$ for all $\delta' \leq \delta$, which means that the limit of the essential infs is $\geq f(x) - \epsilon$, which is strictly greater than the limit of the infs, showing that $f$ is not precise. Otherwise $A \cap B\_\delta(x)$ has positive measure for all $\delta$. For each $n \in \mathbb{N}$ let $K\_n$ be a positive measure compact subset of $A \cap B\_{1/n}(x)$; then $K = \{x\} \cup \bigcup K\_n$ is a compact set for which $\lim\_{\delta \to 0}{\rm esssup}\_{K \cap B\_\delta(x)} f \leq f(x) - \epsilon$ (since the sup on $K \cap B\_\delta(x) \setminus\{x\}$ is at most $f(x) - \epsilon$), whereas $\lim\_{\delta \to 0}\sup\_{K \cap B\_\delta(x)} f \geq f(x)$. So the condition is violated again.
| 6 | https://mathoverflow.net/users/23141 | 438276 | 177,038 |
https://mathoverflow.net/questions/438280 | 14 | A continuum is a compact connected metrizable topological space.
Given a cardinal $\kappa$, a topological space $X$ is called $\kappa$-connected if it is not possible to write $X$ as the disjoint union of more than one and at most $\kappa$ many closed subsets. In particular a space is connected iff it is $2$-connected iff it is $n$-connected for all $2\leq n<\aleph\_0$.
It is a standard result that continua are $\aleph\_0$-connected and, by writing a continuum as a union of singletons, it is clear that continua are $\mathfrak c$-disconnected.
Given a continuum $X$ let $\mathrm{disc}(X)$ denote the smallest $\kappa$ such that $X$ is $\kappa$-disconnected.
Is it consistent with $\mathsf{ZFC}$ to have a continuum $X$ with $\aleph\_0<\mathrm{disc}(X)<\mathfrak c$?
Is it a theorem of $\mathsf{ZFC}$ that if $X$ and $Y$ are nontrivial continua, then $\mathrm{disc}(X)=\mathrm{disc}(Y)$?
Assuming a positive answer to the previous two questions, is $\mathrm{disc}(X)$ equal to some standard cardinal invariant of the continuum?
| https://mathoverflow.net/users/49381 | How “disconnected” can a continuum be? | The answer to all three of your questions is yes.
The cardinal $\mathrm{disc}([0,1])$ is discussed in [this](https://mathoverflow.net/questions/285780/are-the-sierpi%C5%84ski-cardinal-acute-mathfrak-n-and-its-measure-modification?rq=1) MO question of Taras Banakh. He calls this cardinal the *Sierpiński cardinal* and denotes it $\acute{\mathfrak n}$. The comments and answers there give a good bit of information about this cardinal, including that it is consistent to have $\acute{\mathfrak n} < \mathfrak{c}$. This consistency result was first proved by Stern, who showed that it is true in the random real model.
The Sierpiński cardinal also goes by another name in the literature: $\mathfrak{a}\_T$. This is because it can be characterized as the smallest size of an infinite maximal family of almost disjoint subtrees of the infinite binary tree $2^{<\omega}$.
Arnie Miller shows in Theorem 3 of [this](https://people.math.wisc.edu/%7Eawmille1/res/cov.pdf) paper that $\mathfrak{a}\_T = \aleph\_1$ if and only if every uncountable Polish space can be partitioned into $\aleph\_1$ closed sets. As every continuum is Polish, this shows that $\mathrm{disc}(X) = \mathfrak{a}\_T = \aleph\_1$ for every continuum $X$, provided that $\mathfrak{a}\_T = \aleph\_1$. This almost answers your second question.
Miller's argument does not readily generalize to cardinals $>\!\aleph\_1$. Recently I wrote a paper about partitions of Polish spaces into closed sets (and Borel sets, more generally), and I found a a way to extend Miller's result to cardinals $>\!\aleph\_1$ (via a different approach): see Theorem 2.4 [here](https://arxiv.org/pdf/2112.00535.pdf). So it is true that $\mathrm{disc}(X) = \mathfrak{a}\_T$ for any continuum $X$; and more generally, $\mathfrak{a}\_T$ is equal to the smallest size of any partition of a Polish space $X$ into uncountably many closed sets (or compact sets, or $F\_\sigma$'s).
| 15 | https://mathoverflow.net/users/70618 | 438283 | 177,043 |
https://mathoverflow.net/questions/438287 | 1 | * Suppose with have a topological manifold $X$ and a group $G$, is there a way to compute the fundamental group of $X/G$ in function of $\pi(X)$ and $\pi(G)$?
* are there any settings on X that can simplify the computing of $\pi(X/G)$ in function of $\pi(X)$ and $\pi(G)$?
could you please suggest to me a reference where can I find related topics?
| https://mathoverflow.net/users/497616 | fundamental group of $X/\mathbb{R}^n$ | See Proposition 8.10 in Chapter I of [Transformation Groups and Algebraic K-Theory](https://link.springer.com/book/10.1007/BFb0083681) by Lück. When a Lie group $G$ acts properly on a connected manifold $X$, the proposition provides an exact sequence
$$
\pi\_1(X,x\_0)\rightarrow\pi\_1(X/G,x\_0G)\rightarrow\pi\_0(G)/M\rightarrow 0
$$
where $M$ is the normal subgroup of $\pi\_0(G)$ generated by $\pi\_0$ of the stabilizers. A description of the kernel of the first morphism $\pi\_1(X,x\_0)\rightarrow\pi\_1(X/G,x\_0G)$ can also be found there.
| 3 | https://mathoverflow.net/users/128556 | 438289 | 177,045 |
https://mathoverflow.net/questions/438290 | 8 | By the downward Lowenheim-Skölem theorem we can find two countable ordinals $\alpha < \beta$ such that $L\_\alpha \prec L\_{\omega\_1}$ and $L\_\beta \prec L\_{\omega\_1}$. That is, $L\_\alpha$ and $L\_\beta$ are elementary submodels of $L\_{\omega\_1}$. Consequently, $L\_\alpha \prec L\_\beta$.
Hence my question, the other way around:
>
> Given countable ordinals $\alpha$ and $\beta$ such that $L\_\alpha \prec L\_\beta$, do we always have $L\_\alpha \prec L\_{\omega\_1}$ ? If not, how do construct/prove the existence of such a $\alpha$ and $\beta$?
>
>
>
| https://mathoverflow.net/users/138089 | Elementary countable submodels in Gödel's universe | Very clearly not. Take some countable elementary submodel $M\_0$ of $L\_{\omega\_2}$, and take $M\_1$ to be another one, but with $M\_1$ a end extension of $M\_0$. We can find such models by first finding two uncountable $\gamma<\delta$ such that $L\_\gamma\prec L\_\delta\prec L\_{\omega\_2}$, and then taking $M\_1$ be a countable elementary submodel of $L\_\delta$ with $L\_\gamma$ added as a predicate to the language, and then letting $M\_0$ be $M\_1\cap L\_\gamma$.
Now, since both are well-founded models of enough set theory and $V=L$, their transitive collapses are some $L\_\alpha$ and $L\_\beta$ which are countable, and the collapses agree on $M\_0$.
Consequently we have that $L\_\alpha\prec L\_\beta$. But it is also clear that $L\_\alpha$ (and $L\_\beta$) think that they have an uncountable ordinal, whereas $L\_{\omega\_1}$ thinks that all the ordinals are countable.
| 7 | https://mathoverflow.net/users/7206 | 438292 | 177,046 |
https://mathoverflow.net/questions/438306 | 3 | Call a family of sets $\mathcal{F} \subseteq [\omega]^\omega$ *maximal* if there does not exist some $X \in [\omega]^\omega \setminus \mathcal{F}$ such that $X$ is almost disjoint with all elements of $\mathcal{F}$. Let $\mathcal{A} \subseteq \mathcal{F}$ be an almost disjoint family. Does there always exist some $\mathcal{A} \subseteq \mathcal{A}' \subseteq \mathcal{F}$ such that $\mathcal{A}'$ is a mad (maximal almost disjoint) family?
I am also interested in the weaker question: Does every maximal family contain a mad family?
| https://mathoverflow.net/users/146831 | Extending almost disjoint family in a maximal set | Every family $\mathcal{F}$ containing $\omega$ itself as a member is trivially maximal, since no infinite set is almost disjoint from $\omega$. But the family could otherwise consist of an almost disjoint non-maximal family $\mathcal{A}$. That is, $\mathcal{F}=\mathcal{A}\cup\{\omega\}$ is a counterexample.
| 3 | https://mathoverflow.net/users/1946 | 438318 | 177,055 |
https://mathoverflow.net/questions/438312 | 2 | Let $W\_t$ be a Wiener process with $W\_0=0$, and let $L=\{at+by=c\}$ be a line with $c/b<0$ (i.e. the line crosses the $Y$-axis below $0$).
Assume that $W\_t$ stayed above $L$ up to time $T$. What is the PDF of $W\_T$ under this assumption? Does it have a closed form?
| https://mathoverflow.net/users/64302 | Density of $W_t$ assuming it stayed above a line $L$ | The problem can be restated as follows:
>
> For a real $a>0$ and a real $b$, let
> $$X\_t:=a+bt+W\_t$$
> for real $t\ge0$, where $W$ is a standard Wiener process. Let
> $$\tau:=\inf\{t>0\colon X\_t=0\}.$$
> For a real $t>0$, find the joint distribution of $X\_t$ and $\tau$.
>
>
>
The answer is well known:
$$P(X\_t\in dx,\tau>t)=f\_t(x-a-bt)(1-e^{-2ax/t})\,1(x>0)\,dx, \tag{1}\label{1}$$
where $f\_t$ is the pdf of $W\_t$. A derivation of this formula can be found [here](https://works.bepress.com/iosif-pinelis/24/).
| 3 | https://mathoverflow.net/users/36721 | 438322 | 177,059 |
https://mathoverflow.net/questions/438227 | 2 | One of the performance metrics calculated in the analysis of telecommunications systems is the ergodic channel capacity, $C\_{\rm erg}$. During one of my studies, I found the expression below for such a metric considering a Rayleigh channel and AWGN.
\begin{align}\label{eq:CAPACIDADEERGODICA}
C\_{\rm erg} & =
\frac{ z^2 }{2 \bar{\gamma} h\_{\rm l}^{2}A\_{0}^{2}\log(2)} {\rm H}\_{3,4}^{4,1}\left[ \frac{1}{\bar{\gamma}h\_{\rm l}^{2}A\_{0}^2} \ \middle\vert \begin{array}{c}
(-1, 1), (\frac{z^2}{2}, 1), (0, 1) \\
(0, 1), (\frac{z^2}{2}-1, 1), (-1, 1), (-1, 1)
\end{array}\right]
\end{align}
To provide more insights into high SNR regimes ($\bar{\gamma} \to \infty$), the asymptotic behavior of this metric is often required.
Therefore, I'd like to know the asymptotic expression of the ergodic capacity, given by the above equation when $\bar{\gamma} \to \infty$.
OBS.: All the other variables can be considered constant.
**UPDATE 10/01/2023**
Additional information on how the previous equation is derived.
The ergodic capacity is found as the expectation of $\log\_2(1+\gamma)$:
\begin{equation}\label{eq:Capacidade}
C\_{\rm erg} = \int\_{0}^{\infty}\log\_2(1+\gamma)f\_{\Gamma}(\gamma){\rm d}\gamma,
\end{equation}
where
\begin{align}\label{eq:PDFsnr}
f\_{\Gamma}(\gamma) & =
\frac{ z^2}{2 \bar{\gamma} h\_{\rm l}^{2}A\_{0}^{2}} {\rm H}\_{1,2}^{2,0}\left[ \frac{\gamma}{\bar{\gamma} h\_{\rm l}^{2}A\_{0}^2} \ \middle\vert \begin{array}{c}
(\frac{z^2}{2},1) \\
(0, 1), (\frac{z^2}{2}-1,1)
\end{array}\right]
\end{align}
is the PDF of the instantaneous SNR.
| https://mathoverflow.net/users/103291 | Asymptotic analysis of an expression involving a Fox's H function | Your expression can be simplified as follows: As $m=q=4$, the denominator parameter $(0,1)$ cancels with the numerator parameter $(0,1)$. Furthermore, as all linear coefficients are one, the Fox H function reduces to the simpler Meijer G function. I'll denote
$$\tag{1}
\xi = \frac 1 {\bar\gamma h\_1^2 A\_0^2},\quad \zeta = \frac{z^2}{2},
$$
such that
\begin{align}\tag{2a}
C\_{\rm erg} & =
\frac{ \zeta \xi}{\log 2} {\rm G}\_{2,3}^{3,1}\left[ \xi \, \middle\vert \begin{array}{c}
-1, \zeta \\
-1, -1, \zeta-1
\end{array}\right] \\
&= \frac{\zeta}{\log 2} {\rm G}\_{2,3}^{3,1}\left[ \xi \, \middle\vert \begin{array}{c}
0, \zeta+1 \\
0, 0, \zeta
\end{array}\right]\tag{2b}\\
&= \frac{1}{\log 2}\left(
\frac {\pi \, \Gamma(1-\zeta)} {\sin(\pi\zeta)} \xi^\zeta
-\gamma\_\mathrm E - \zeta^{-1} - \log \xi + \mathcal O(\xi)
\right),\tag{2c}
\end{align}
with [Euler's constant](https://en.wikipedia.org/wiki/Euler%27s_constant) $\gamma\_\mathrm E$, the [Gamma function](https://en.wikipedia.org/wiki/Gamma_function) $\Gamma$, and with the usual big $\mathcal O$ [notation](https://en.wikipedia.org/wiki/Big_O_notation). The calculation was done in *Mathematica*.
| 5 | https://mathoverflow.net/users/488821 | 438324 | 177,060 |
https://mathoverflow.net/questions/438263 | 5 | Is there a concrete example of a $4$ tensor $R\_{ijkl}$ with the same symmetries as the Riemannian curvature tensor, i.e.
\begin{gather\*}
R\_{ijkl} = - R\_{ijlk},\quad R\_{ijkl} = R\_{jikl},\quad R\_{ijkl} = R\_{klij}, \\
R\_{ijkl} + R\_{iklj} + R\_{iljk} = 0.
\end{gather\*}
for which there is no metric for which it is the Riemannian curvature tensor?
The existence of such a curvature was already shown by [Robert Bryant](https://mathoverflow.net/questions/202211/equations-satisfied-by-the-riemann-curvature-tensor), however, I'm looking for a concrete example.
| https://mathoverflow.net/users/497575 | Example of a curvature with no associated metric | A simple example (which just uses Deane Yang/Robert Bryant's idea) is to consider any space of dimension at least three and consider the tensor field
$$ R\_{ijkl} = f(x)(\delta\_{ik}\delta\_{jl}-\delta\_{il}\delta\_{jk})$$
where $f(x)$ is your favorite function which changes sign and whose derivative is non-vanishing when $f=0$ (thanks to Robert Bryant for the correction). When $f=0$, this curvature tensor has constant sectional curvature $0$, no matter what metric we choose.
However, we can now apply the proof of [Schur's lemma](https://en.wikipedia.org/wiki/Schur%27s_lemma_(Riemannian_geometry)) (i.e., trace the second Bianchi identity twice) to see that when $f$ vanishes, we have that
$$dR=\frac{n}{2}dR$$
where $R$ is the scalar curvature.
As such, the differential of the scalar curvature must vanish when $f=0$. However, no matter which metric we pick, there is no way to make this happen if the differential of $f$ is nonzero as it goes from being positive to negative.
Edit: My original answer had a mistake which was pointed out in the comments. Here is a revised version which uses the same idea which should (hopefully) work.
| 6 | https://mathoverflow.net/users/125275 | 438332 | 177,062 |
https://mathoverflow.net/questions/438273 | 1 | A variety $ X $ is $ F $-split if there exists an $ \mathcal{O}\_{X} $-linear map $ \phi: F\_{\ast}(\mathcal{O}\_{X}) \to \mathcal{O}\_{X} $ such that $ \phi \circ F^{\sharp} = \operatorname{id}\_{\mathcal{O}\_{X}} $. Such a map $ \phi $ is called a splitting. A closed sub-scheme $ Y $ of $ X $ is compatibly split if there exists a splitting $ \phi $ such that $ \phi(F\_{\ast}(\mathcal{I}\_{Y})) \subseteq \mathcal{I}\_{Y} $. Note that $ \mathcal{I}\_{Y} \subseteq \phi(F\_{\ast}(\mathcal{I}\_{Y})) $ already.
Toric varieties are $ F $-split. Does anyone know of an example of an affine, toric, variety $ U\_{\sigma} $ and a point $ x \in U\_{\sigma} $ such that $ x $ is not compatibly split? Here I am using Hartshorne's definition of a "variety" as an integral, separated, scheme of finite type over an algebraically closed field.
| https://mathoverflow.net/users/470753 | Does there exist a point $ x $ of an affine toric variety $ U_{\sigma} $ such that $ x $ is not compatibly split? | Actually, on an affine variety, if it is $F$-split, then it is compatibly $F$-split with every point (possibly changing the splitting). This is not true in the projective case (an ordinary projective elliptic curve is $F$-split, but not compatibly split with any point). It is also not true for non-closed points (for instance the generic point of a cusp in $\mathbb{A}^2$ cannot be compatibly split, since that would force the cusp itself to be $F$-split).
**Lemma.** *If $R$ is a $F$-split $F$-finite ring, then for any maximal ideal $Q \subseteq R$, there is a Frobenius splitting compatibly split with $m$.*
*Proof.* Suppose $R$ is an $F$-split ring and $Q \in Spec R$ is any closed point. Let $\phi : F\_\* R \to R$ be an $F$-splitting. If $\phi$ is compatible with $Q$ we are done, so suppose its not compatible with $Q$. Thus $\phi(F\_\* Q) \not\subseteq Q$. Since outside of $Q$, $\phi$ is surjective, we actually see that $\phi(F\_\* Q) = R$. Thus there exists $x\_1 \in Q$ with $\phi(F\_\* x\_1) = 1$. Let $\phi\_1 = \phi \circ (\cdot F\_\* x)$. In other words, $\phi\_1(F\_\* y) = \phi(F\_\* xy)$. Note $\phi\_1$ is also a Frobenius splitting (since it also sends $1$ to $1$).
If $\phi\_1$ is compatible with $Q$, we are also done, so we repeat the process. Continuing in in this way, we eventually come to a Frobenius splitting $\phi\_{n}$ where $n > p d + 1$ where $d$ is the number of generators of the ideal $Q = (f\_1, \dots, f\_d)$. Now,
$$1 = \phi\_n(F\_\* 1) = \phi(F\_\* x\_1 x\_2 \dots x\_n).$$
By construction, $x\_1 x\_2 \dots x\_n \in Q^{pd + 1} \subseteq (f\_1^p, \dots, f\_d^p) =: Q^{[p]}$, that containment is basically the pigeonhole principal. On the other hand, an easy computation shows that $\phi(F\_\* Q^{[p]}) \subseteq Q$ and so we see that $1 = \phi\_n(F\_\* 1) \in Q$, a contradiction. Thus at some earlier point we must have had a Frobenius splitting $\phi\_i$ that was compatible with $Q$.
| 2 | https://mathoverflow.net/users/3521 | 438337 | 177,064 |
https://mathoverflow.net/questions/438196 | 4 | **Background:** The equation
$$a^4+b^4+c^4=2d^4$$
has infinitely many positive integral solutions if we take $c=a+b$ and $a^2+ab+b^2=d^2$ further assuming that $GCD(a,b,c)=1$.
**Main problem:** Find some positive integral solutions to the equation
$$a^4+b^4+c^4=2d^4$$
with $a\lt b\lt c\ne a+b$ and $GCD(a,b,c)=1$.
| https://mathoverflow.net/users/nan | On the equation $a^4+b^4+c^4=2d^4$ in coprime positive integers $a\lt b\lt c$ such that $a+b\ne c$ | Let $A=\frac{a}{d}, B=\frac{b}{d}, C=\frac{c}{d},$ then we get
$$A^4+B^4+C^4=2\tag{1}$$
Let $A=x+y, B=x-y, z=C^2$ then
$$2x^4+12y^2x^2+2y^4+z^2 = 2\tag{2}$$
Hence
$$y^2 = -3x^2 \pm \frac{\sqrt{32x^4+4-2z^2}}{2}\tag{3}$$
So we find the rational solutions of $(4)$.
$$v^2=32x^4+4-2z^2\tag{4}$$
$(4)$ can be parameterized to $(5)$ using $(z,v)=(4x^2,2)$ with $w=(k^2+2)C$.
$$w^2 = 4(k^2+2)(-2+k^2)x^2-4(k^2+2)k\tag{5}$$
Thus we must find the rational solutions of simultaneous equations $(3),(5)$.
First we get a parametric solution of $(5)$ using giving some rational number $k$.
Take $k=\frac{-9}{13}$. Then we get:
$$x = \frac{3}{142}\frac{28561m^2-430732-7081100m}{28561m^2+430732}$$
$$w = \frac{-1257}{11999}\frac{-10768300+173732m+714025m^2}{28561m^2+430732}$$
$$C = \frac{-3}{71}\frac{-10768300+173732m+714025m^2}{28561m^2+430732}$$
Hence $(3)$ becomes
$$y^2 = \frac{10083247442281m^4-41255619857608m^2+2788930240200m^3+2293337020040464-42060204482400m}{20164(28561m^2+430732)^2}$$
Hence we must find the rational solutions $(m,V)$ of $(6)$
$$V^2 = 10083247442281m^4+2788930240200m^3-41255619857608m^2-42060204482400m+2293337020040464\tag{6}$$
Quartic equation $(6)$ is birationally equivalent to the elliptic curve using a known solution $(m,V)=(360/169,-47445892)$ as follows.
$$Y^2 = X^3 -X^2 -1097465452X+ 3288951361780\tag{7}$$
Though I couldn't find the generator of elliptic curve , I got one solution
$P(X,Y)=(\cfrac{-97636631990}{5536609}, \cfrac{53963430434179560}{13027640977})$.
Since $(7)$ has rank $1$, we get infinitely many solutions of $(7)$ using group law as follows.
For instance, we get a solution of $(6)$,
$2Q(m,V)=(\cfrac{-762617488871059540}{36474714629699307},\cfrac{-63980049963485293932709632365019549028}{46581170383319934672021751209})$
We get $(x,y)=(\cfrac{4317501713916820330332746705607}{16486256234272723829845043984402},\cfrac{12346241358503326020929860043561}{16486256234272723829845043984402})$
Finally, we get
$a = 4014369822293252845298556668977$
$b = 8028813922964684804294250901215$
$c = 8331871536210073175631303374584$
$d = 8243128117136361914922521992201$
| 13 | https://mathoverflow.net/users/150249 | 438347 | 177,068 |
https://mathoverflow.net/questions/438342 | 7 | In 2-category theory we often encounter notions with noninvertible coherence 2-cells, in which case there is a choice about which way the 2-cell should point. Generally, one of these directions is called "lax" and the other "colax" (or "oplax"). We have to make at least one arbitrary choice for which of the two to call "lax" and which to call "colax", but once we've made that choice in *one* place, the choices everywhere else are determined. One general place to phrase the choice is as defining lax and colax morphisms of algebras for a 2-monad; everything else can be seen as either an instance of that or a generalization of that, and that tells you which direction is lax and which is colax (at least if you want to be consistent, which not everyone does).
At least, that's what I used to think. But today I started thinking about lax [dinatural transformations](https://ncatlab.org/nlab/show/dinatural+transformation), which should be like ordinary dinatural transformations but with a 2-cell inhabiting the dinaturality hexagon. There are two directions that that 2-cell can go in, but I can't figure out a way to relate those directions to any other situation that has a lax and a colax version.
Since dinatural transformations don't even compose in general, there's no clear way to see them as morphisms of algebras for some monad. The obvious thing to do would be to specialize a dinatural transformation to a natural one by making the two functors either both totally covariant or both totally contravariant, and then compare to the usual terminology in that special case. But there are two ways to do this (covariant or contravariant), and the same direction of 2-cell in a dinaturality hexagon specializes to a lax natural transformation in one case and a colax one in the other case. I suppose we could choose to name it according to the covariant direction because "covariant functors are more basic than contravariant ones", but that feels more arbitrary than I'd like.
You can also specialize a dinatural transformation to an [extranatural transformation](https://ncatlab.org/nlab/show/extranatural+transformation). But I've looked at the references cited [here](https://mathoverflow.net/q/67083/49) about lax extranatural transformations, and as far as I can see none of them gives any reason for choosing which direction is lax and which is colax.
It is, of course, true that if you change the domain category to its opposite, permuting the domain to regard a functor $F:C^{\rm op}\times C\to D$ as a functor $F:(C^{\rm op})^{\rm op}\times C^{\rm op}\to D$, then the direction of the 2-cell in a dinaturality hexagon switches. This is not the case for ordinary natural transformations, and suggests that maybe there is more arbitrariness in the dinatural and extranatural case than in the ordinary one.
But I still have to ask, in case anyone else can see something I can't: Is there any principled way to decide which direction of a 2-cell in a dinaturality hexagon (or an extranaturality wedge) is "lax" and which is "colax"?
| https://mathoverflow.net/users/49 | Which direction does a lax dinatural transformation go? | This is a *very* loose answer since it's based on something that, as far as I know, has not been fully formalized. The key idea is that of what I might call a "dependent arrow" between two objects.
The closest analogy is to that of a dependent path in HoTT, where if we have $a, a' : A$, $p : a = a'$, $b : B(a)$ and $b' : B(a')$, then even though $b$ and $b'$ have different types, we can still talk about paths between them using $p$. Specifically, the type of dependent paths is $p \# b = b'$, where $\#$ denotes transportation. As you probably, know, if $f : \Pi\_{a : A} B(a)$, then $f$ maps a path $p : a = a'$ to a dependent path $apD\_f(p) : p \# f(a) = f(a')$.
In a hypothetical directed (homotopy) type theory, we might have the same information, but with $p$ replaced with a directed morphism from $a$ to $a'$ and $B$ somehow functorial. Now transporting $b$ along $p$ means $B(p)(b)$, where $B(p)$ is the functorial action of $B$ on morphisms in $A$. A dependent arrow from $b$ to $b'$ over $p$ is a morphism $B(p)(b) \to b'$ in $B(a')$.
If $B$ is instead contravariant, then we have to transport $b'$ instead. That's because $B(p) : B(a') \to B(a)$ now, so we can't apply it to $b$ anymore. Thus, a dependent arrow from $b$ to $b'$ over $p$ should be a morphism $b \to B(p)(b')$.
In general, as we'll see below, $B$ might depend both covariantly and contravariantly on $a$, so we have to transport both $b$ and $b'$ to make sense of a morphism between them. Again, I've not seen a formalization of this, so forgive me if it's a bit fuzzy.
We'll denote this type of "dependent arrows over $p$" as $b \to\_p b'$.
Let's see how this works with an example. Let $F, G: C \to D$ be functions (functors). If a directed morphism from $F$ to $G$ is to be analogous to a path from $F$ to $G$, one might expect such a morphism to be a pointwise morphism $\alpha\_c : F(c) \to G(c)$ for each $c : C$.
What about the action of $\alpha$ on morphisms? For $f : c \to c'$ in $C$, this should be a dependent arrow $\alpha\_c \to\_f \alpha\_{c'}$.
In this case, the type of $\alpha\_c : F(c) \to G(c)$ depends on $c$ both covariantly and contravariantly. To get a dependent arrow from $\alpha\_c$ to $\alpha\_{c'}$ over $f$, we need to transport both sides. For $\alpha\_c$, we can post-compose with $G(f)$ and for $\alpha\_{c'}$, we can pre-compose with $F(f)$. Thus, a dependent path in this case is a 2-cell $G(f) \circ \alpha\_c \Rightarrow \alpha\_{c'} \circ F(f)$.
If we're thinking of $D$ as a 1-category, the 2-cells are just identities, so this reduces to naturality. If D is instead a 2-category, then what we get here is a *lax* natural transformation. If we want an oplax transformation, then what we should do is instead map $f : c \to c'$ to a dependent map $\alpha\_{c'} \to\_f \alpha\_c$.
---
Let's try this out for dinatural transformations. This is where things get (at least for me) very fuzzy. We want to repeat the above, but allow $F$ and $G$ to be both covariant and contravariant at once. I think of $F$ and $G$ has "normally" only having a single variable, but then having a two-variable "extension".
In our hypothetical directed type theory, this sort of thing shows up all the time. For example, the type of $id\_c : \hom\_C(c, c)$ is not covariant or contravariant, but $\hom\_C$ can be extended to a two-variable functor $C^{op} \times C \to Type$, which makes it both covariant and contravariant.
In any case, we can try our best. A morphism from $F$ to $G$ is again pointwise, but rather than $\alpha\_{c, c'} : F(c, c') \to G(c, c')$, we want to think of $F$ and $G$ as being single-variable here, to get $\alpha\_c : F(c, c) \to G(c, c)$.
The action of $\alpha$ on morphisms takes a morphism $f : c \to c'$ in $C$ to a morphism $\alpha\_f : \alpha\_c \to\_f \alpha\_c'$. As in the case of natural transformations, both sides need to be transported. The covariant transport takes $\alpha\_c$ to $G(1, f) \circ \alpha\_c \circ F(f, 1) : F(c', c) \to G(c, c')$. The contravariant transport takes $\alpha\_{c'}$ to $G(f, 1) \circ \alpha\_{c'} \circ F(1, f) : F(c', c) \to G(c, c')$. Sketching this out, we get the hexagon for dinaturality.
Thus, if we follow the same convention as with natural transformations, a lax dinatural transformation has a 2-cell $\alpha\_c$ to $G(1, f) \circ \alpha\_c \circ F(f, 1) \Rightarrow G(f, 1) \circ \alpha\_{c'} \circ F(1, f)$ and an oplax dinatural transformation has the reverse direction. The key here is that the lax version maps $f$ to $\alpha\_c \to\_f \alpha\_{c'}$.
| 3 | https://mathoverflow.net/users/178774 | 438361 | 177,073 |
https://mathoverflow.net/questions/438352 | 5 | (This is a fairly basic question, not really research level, except that I am a research mathematician working on other things who is trying to understand more topology for use in my own work.)
Let $\Sigma$ be a smooth, connected, compact 2d manifold, henceforth simple a "surface". We don't assume anything more (not necessarily closed or oriented).
$\Sigma$ has a tangent bundle $T\Sigma$, to which we can associate the "unit sphere bundle" $\mathbf{S}=\mathbf{S}(T\Sigma)$.
There is a map $\Sigma \rightarrow BS^{1}$ classifying this circle bundle, and so a class $c\_{1}(\mathbf{S}) \in H^{2}(\Sigma,\mathbf{Z})$, which is the obstruction to trivialising/finding a section of $\mathbf{S}$.
However, if $\Sigma$ is not closed, or not orientable, then $H^{2}(\Sigma,\mathbf{Z})=0$, so the circle bundle $\mathbf{S}$ should admit a section, and hence there should be a nowhere vanishing section of $T\Sigma$, that is a nowhere vanishing vector field on $\Sigma$.
But doesn't that contradict the Poincar'e-Hopf index theorem, saying that the Euler characteristic should be the sum of the indices of zeroes of the vector field? For example, $\mathbf{RP}^{2}$ has Euler characteristic $1$, but the above argument seems to show that there is a nowhere vanishing vector field on $\mathbf{RP}^{2}$ (which I don't believe).
So what gives?
| https://mathoverflow.net/users/497719 | Meaning of the first Chern class of the unit tangent bundle of a surface | The Poincaré-Hopf holds for non-orientable manifolds also. For a closed non-orientable surface $\Sigma$ the structure group of the sphere tangent bundle does not reduce to $SO(2)=S^1$, so it's classified by a map $\Sigma\to BO(2)$ rather than a map $\Sigma\to BS^1$. The obstruction to finding a section of the sphere tangent bundle is an element of $H^2(\Sigma; \mathbb{Z}^{w\_1})$, the cohomology with coefficients in the local system of integers twisted by the first Stiefel-Whitney class $w\_1:\pi\_1(\Sigma)\to \mathbb{Z}/2\cong \operatorname{Aut}(\mathbb{Z})$. By Poincaré duality with local coefficients, this group is isomorphic to $H\_0(\Sigma;\mathbb{Z})\cong\mathbb{Z}$.
In the case of manifolds with boundary, the statement of Poincaré-Hopf has to be modified to include some boundary conditions, such as that the vector field points outwards on the boundary. Think of a disk $D^2$, which has Euler characteristic $1$. It admits a nowhere zero vector field; but it does not admit one that points outward on the boundary.
| 6 | https://mathoverflow.net/users/8103 | 438370 | 177,075 |
https://mathoverflow.net/questions/438369 | 2 | Not counting equivalent braids, are there finite or infinite numbers of 3-braids whose closures are trivial knot or links? If the answer is infinite, are there some patterns in those infinite numbers of braids, e.g. there exists some repeated parts?
| https://mathoverflow.net/users/492606 | Are there infinite number of 3-braids with trivial closure? | Suppose that $\beta$ is a three-braid whose closure is trivial. Suppose that $\gamma$ is any three-braid. Then $\gamma \beta \gamma^{-1}$ is again a three-braid with trivial closure. However, if $\beta$ is non-trivial, then for "generic" $\gamma$, the braids $\beta$ and $\gamma \beta \gamma^{-1}$ will not be equivalent.
To explain the word generic, note that $\beta$ commutes with its powers (and its roots, if any). Thus taking $\gamma$ to be a power (or a root, or a power of a root) of $\beta$ will result in $\gamma \beta \gamma^{-1}$ being equivalent to $\beta$. More generally, $\gamma$ should not lie in the centraliser of $\beta$. However, the centraliser is typically a very small subset of the braid group. (One notable exception to this is when $\beta$ lies in the centre of the braid group.)
---
Here is a simple version of HJRW's suggestion: Let $\beta$ be the braid $\sigma\_1 \sigma\_2^{-1}$. This is a three-strand pseudo-Anosov braid. Thus its centraliser (in the braid group) is virtually $\mathbb{Z}^2$ (as it contains $\beta$ and the central element $\Delta$).
A bit surprisingly, this is also the *only* pseudo-Anosov three-braid (conjugacy class) closing to the unknot, due to the work of Birman-Menasco (cited in the comments).
| 2 | https://mathoverflow.net/users/1650 | 438372 | 177,076 |
https://mathoverflow.net/questions/438368 | 6 | A quick search for "diagonalisable matrix" on Wikipedia immediately gives the result that the set of real-diagonalisable matrices is not dense in the set of real matrices.
I need, however, to know whether the set of complex-diagonalisable real matrices, i.e. matrices with purely real entries that become diagonalisable when treated as complex matrices, is dense in the set of real matrices or not.
This is for my personal research that explicitly involves the reality (or lack thereof) of the trace of a complex analytic, not necessarily entire, function (real when restricted over the reals) of square matrices- should the above result be true, such a trace would be automatically real for any real matrix, by mixing the above result, the trivial fact that the set of mutually distinct N-(complex)number pairs is dense in the set of N-number pairs, and that the Frobenius cofactors appearing in the form of Sylvester's formula for diagonalisable matrices must all have trace 1 for matrices with distinct eigenvalues.
My question is now clear- is the above true?
P.S. I'm a literal mathematics junior(going to take a leap year before becoming a senior), so I know what I am talking about.
| https://mathoverflow.net/users/113020 | Is the set of purely real square matrices, that are complex-diagonalisable, dense in the set of real matrices? | The answer is yes. Recall that the discriminant of a polynomial $x^n + a\_{n-1} x^{n-1} + \cdots + a\_1 x + a\_0$ is a polynomial $\Delta(a\_0, a\_1, \ldots, a\_{n-1})$ which vanishes if and only if the complex roots of $x^n + a\_{n-1} x^{n-1} + \cdots + a\_1 x + a\_0$ are not mutually distinct. Recall also that, if the characteristic polynomial of a matrix has distinct roots, then it is diagonalizable (this isn't if and only if, but it is the direction I need).
Therefore, if the discriminant of the characteristic polynomial of $A$ is nonzero, then $A$ is diagonalizable over the complex numbers. The discriminant of the characteristic polynomial will be a huge multivariate polynomial, but all we need to know is that it is not the zero polynomial, which is true because it is not zero if $A$ is diagonal with distinct diagonal elements.
The locus in $\mathbb{R}^N$ where a polynomial is nonzero (other than the zero polynomial) is always dense.
| 9 | https://mathoverflow.net/users/297 | 438380 | 177,080 |
https://mathoverflow.net/questions/438343 | 5 | I have been thinking about the validity of the following inequality:
if $P(z)=\sum\_{k=0}^na\_kz^k, a\_n\neq 0$ and $P(z)$ is non-zero in $|z|<1, $ then for $\theta \in [0, 2\pi],$ and $p>0$
\begin{align\*}
&\int\_{0}^{2\pi}\left|n\left(1-\frac{|a\_0|-|a\_n|}{n(|a\_0|+|a\_n|)}\right)P(e^{i\theta})+(\eta-e^{i\theta})P'(e^{i\theta})\right|^{p}d\theta\\
&\qquad\qquad\quad \leq \left[n\left(1-\frac{|a\_0|-|a\_n|}{n(|a\_0|+|a\_n|)}\right)\right]^{p}\int\_{0}^{2\pi}|P(e^{i\theta})|^{p}d\theta,
\end{align\*}
for any $\eta$ lying in the closed unit disc.
In fact the above inequity is motivated by the inequality from Melas and Rubinstein [2] who proved that
if $P(z) $ is a polynomial of degree $n$ then for $\theta \in [0, 2\pi],$
\begin{align\*}
&\int\_{0}^{2\pi}\left|nP(e^{i\theta})+(1-e^{i\theta})P'(e^{i\theta})\right|^{p}d\theta\\
&\qquad\qquad\quad \leq n^{p}\int\_{0}^{2\pi}|P(e^{i\theta})|^{p}d\theta.
\end{align\*}
Alternatively the above result [due to Melas] can be established directly through an inequality due to Arestov [1] involving admissible operators on the class of polynomials. Is my above claim right? Your suggestions are of great help.
**References**
[1] Vitaliĭ Vladimirovich Arestov, "[On integral inequalities for trigonometric polynomials and their derivatives](https://doi.org/10.1070/IM1982v018n01ABEH001375)" (Russian original)
(English translation) Mathematics of the USSR, Izvestiya 18, 1-17 (1982), [Zbl 0517.42001](https://zbmath.org/0517.42001); Izvestiya Akademii Nauk SSSR. Seriya Matematicheskaya, 45, 3-22 (1981), [MR607574](http://www.ams.org/mathscinet-getitem?mr=607574), [Zbl 0538.42001](https://zbmath.org/0538.42001).
[2] Antonios D. Melas and Zalman Rubinstein, "[Problem 10255 and solution](https://doi.org/10.1080/00029890.1996.12004751)" American Mathematical Monthly, 103, 177-181 (1996).
| https://mathoverflow.net/users/128472 | An inequality for polynomials | $\newcommand\ep\varepsilon$This inequality does not hold in general.
Indeed, by continuity/denseness, the condition $a\_n\ne0$ can be dropped (in response to a comment by the OP, details on this are now given at the end of this answer).
Now just take $p=2$, $\eta=-1$, $n=2$, $P(z)=1-z+0z^2$ (so that $a\_0=1$ and $a\_n=a\_2=0$). Then the left- and right-hand sides of the conjectured inequality are $8\pi$ and $4\pi$, respectively. $\quad\Box$
---
**Details on the mentioned continuity/denseness:** Let again $p=2$, $\eta=-1$, $n=2$, but now $P(z)=1-\ep-z+\ep z^2$ with $\ep<0$ -- so that $a\_0=1-\ep$ and $a\_n=a\_2=\ep\ne0$). Then the roots $1$ and $1/\ep-1$ of $P(z)$ are outside the open unit disk, so that $P(z)\ne0$ if $|z|<1$. Moreover, with $\ep\uparrow0$, the left- and right-hand sides of the conjectured inequality will converge to $8\pi$ and $4\pi$, respectively. More specifically, the difference between the left- and right-hand sides of the conjectured inequality will then be $4(1-\ep)\pi\to4\pi>0$.
$\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 438382 | 177,081 |
https://mathoverflow.net/questions/438174 | 6 | A map with $(2n-1)$ rows and $(2n-1)$ columns of square cells is generated randomly as follows: the value of each cell is chosen from {1, -1} randomly and independently with probability {$p$, $1-p$}. The starting position of the player is at the center (i.e. at the $n$th row and the $n$th column) of the map. For each step, the player can choose a cell next to him (i.e. choose one of the four neighbour cells) and go there. Once the player goes to a new cell with value 1 (resp. -1), the score of the game will increase (resp. decrease) by 1. If the player goes to a cell that has already been visited before, then the score does not change. The initial score is 0 and the goal of the game is to reach the boundary of the map with a positive score. For the above setting, the probability that there exists a strategy to achieve the goal of the game is denoted as $f(n,p)$.
Q1: What's the infimum of $p$ such that $\lim\_{n\rightarrow\infty}f(n,p)>0$?
A quick answer is $0.089\pm0.002$ because I played the game and achieved the goal of the game for $n=200,p=1/11$ but unable to achieve the goal for $n=200,p=2/23$. I need some references for a more acurate answer.
Q2: If the value of each cell is chosen from {1, $-x$} instead and the other conditions remains the same, then the answer of Q1 is a function of $x$ (denoted as $p(x)$) instead. Have anyone found the function $p(x)$ till now?
A quick answer for $x\rightarrow\infty$ is $p(\infty)=0.592746050792...$ because this case is a percolation problem and have been solved by many researchers. But I haven't got the answer for a finite $x$ yet. Is there any references for this problem?
| https://mathoverflow.net/users/114036 | What's the minimum ratio of positive cells such that the player has a positive probability to reach the boundary of a large random map? | So the following is a bit naive, but perhaps can be a starting point and an actual percolation theorist can do more.
The event you're considering is $E(n,p)$ = there exists a connected component $C$ containing the origin and some site on the boundary of $[-n,n]^2$ whose sum is positive. (This seems equivalent to your formulation since repeated visits don't count further towards score, and so for any connected component your walker can visit every site and the net score is just the sum of the numbers in $C$.
Clearly $E(n,p)$ is contained in the union over all such components $C$ of $E(C,p)$, the event that $C$ has positive sum. We can further simplify by just counting all connected components (often called lattice animals) containing the origin of size at least $k$, even though some of those won't touch the boundary. The number of those is roughly (up to a subexponential factor, which again won't matter) $\alpha^k$. Unfortunately I don't know how well $\alpha$ is bounded; I erroneously first thought it was the so-called connectivity constant on the square lattice, which is known to a couple of decimals, but this constant I'm having more trouble searching for. A (probably bad) bound is that $\alpha < 4.65$.
But for any $C$ of size $k$, this is just the event that a binomial RV with mean $kp$ and variance $kp(1-p)$ is greater than $k/2$ (i.e. that more than half of the trials are successes/positive). The probability of this is asympotically something like
$P(X > k/2) \approx P(Z > \frac{k/2 - kp}{\sqrt{kp(1-p)}})$ for a standard normal $Z$. And this probability is (up to a polynomial, which won't matter)
$exp(-(\frac{k/2 - kp}{\sqrt{kp(1-p)}})^2/2) = exp(-\frac{k(1-2p)^2}{8p(1-p)})$.
So, we see that if $\frac{(1-2p)^2}{8p(1-p)} > \log \alpha$, then $P(E(C,p))$ will decay exponentially more quickly than $\alpha^k$, meaning that a union bound over all $C$ of size at least $k$ will give an exponentially decaying geometric series as an upper bound for $f(n,p) = P(E(n,p))$.
I was too lazy to exactly solve the quadratic, but graphing indicates that this happens when $p < .065$ (using the $4.65$ upper bound for $\alpha$). So, it should be the case that your infimum is at least $.065$.
Upper bounds for critical thresholds are traditionally much harder; again, probably a percolation theorist might see a good way to go here.
| 1 | https://mathoverflow.net/users/116357 | 438383 | 177,082 |
https://mathoverflow.net/questions/438243 | 2 | Let $G$ be a linearly reductive algebraic group, and let $X$ be an irreducible affine variety, over an algebraically closed field $\mathbb{K}$, with a regular action of $G$. By Rosenlicht's result, we know that there's a $G$-invariant open set $U\subseteq X$ such that the geometric quotient $\Phi:U\rightarrow U/\!/G$ exists. So, "generic" orbits are contained in $U$. But is there something more specific about these "generic" orbits known?
**My question:**
>
> This $U$ contains orbits of maximal dimension, but is there more specific information for which $x\in X$, the orbit $\mathcal{O}(x)\subset U$?
>
>
>
I'm interested in the setting of quiver representations:
Let $\mathbb{K}$ be an algebraically closed field of characteristic $0$. Let $A$ be a finite-dimensional (associative and unital) algebra over $\mathbb{K}$. Assume there is a quiver $Q=(Q\_0,Q\_1)$, where $Q\_0$ is the set of vertices and $Q\_1$ is the set of arrows, associated to $A$, i.e., $A\cong\mathbb{K}Q/I$, with $0 \neq I\subset \mathbb{K}Q$ an admissible ideal. Further assume that the algebra $A$ is such that the quiver $Q$ is acyclic.
For a dimension vector $\beta$, the representation space
$$\DeclareMathOperator\rep{rep}\DeclareMathOperator\GL{GL}
\rep\_{\beta}(Q,I):=\biggl\lbrace M\in\prod\_{a\in Q\_1}\operatorname{Mat}\_{\beta(ha)\times\beta(ta)}(\mathbb{K}) \;\bigg|\; \text{$M(r) = 0$ for all $r \in I$}\biggr\rbrace
$$ parametrizes $\beta$-dimensional representations of $(Q,I)$.
The linear algebraic group
$$
\GL\_{\beta}:=\prod\_{x\in Q\_0}\GL\_{\beta(x)}(\mathbb{K})
$$
acts on $\rep\_{\beta}(Q,I)$ by change of basis.
Now take an irreducible component $\mathcal{C}\subseteq\rep\_{\beta}(Q,I)$ and restrict the action of $\GL\_{\beta}$ to $\mathcal{C}$.
So my question now becomes:
>
> Is there any information for which representations $V\in\mathcal{C}$, the orbit $\mathcal{O}(V)\subset U$, other than saying $V$ has to be in "general" position?
>
>
>
| https://mathoverflow.net/users/338456 | Orbits in the open set given by Rosenlicht's result | Let me add more details to my [comment](https://mathoverflow.net/questions/438243/orbits-in-the-open-set-given-by-rosenlichts-result#comment1130011_438243) above. Let $S$ be a scheme. Let $\overline{X}$ be a proper $S$-scheme, and let $X\subset \overline{X}$ be a dense Zariski open subscheme.
A closed subset $R\subset X\times\_S X$ is an *algebraic equivalence relation* if it contains the diagonal (i.e., it is reflexive), if it is invariant under the involution $(\text{pr}\_2,\text{pr}\_1)$ of $X\times\_S X$ (i.e., it is symmetric), and if it is transitive, i.e., $R$ contains the image of the following composition, $$R\times\_{\text{pr}\_2,X,\text{pr}\_1} R \hookrightarrow (X\times\_S X) \times\_{\text{pr}\_2,X,\text{pr}\_1} (X\times\_S X) = X\times\_S X\times\_S X \xrightarrow{\text{pr}\_1,\text{pr}\_3} X\times\_S X.$$ For instance, for a group scheme $G$ over $S$ with an $S$-action on $X$, the closure $Z$ of the image of the associated map is an algebraic equivalence relation,
$$\Psi:G\times\_S X \to X\times\_S X, \ (g,x) \mapsto (g\cdot x,x).$$
Denote by $\overline{R}$ the closure of $R$ in $X\times\_S \overline{X}$.
For the projection, $\text{pr}\_1:\overline{R}\to X$, there is a maximal open subscheme $U$ of $X$ over which the projection is flat. If $X$ is reduced and Noetherian, then $U$ is a dense open subscheme by Grothendieck's generic flatness / generic freeness theorem. Thus, there is an induced $S$-morphism from $U$ to the relative Hilbert scheme, $$f\_{\overline{R}}:U \to \operatorname{Hilb}\_{\overline{X}/S}.$$ This is the "modern" take on the classical construction of a quotient of $R$ as a "rational map".
In fact, in terms of making $U$ as big as possible, it is usually better to work with the Chow scheme (due to Angeniol in characteristic $0$, and due to David Rydh in positive characteristic and mixed characteristic). The maximal open subscheme $V$ of $X$ over which $\overline{R}$ is a "good algebraic cycle", and thus defines a morphism from $V$ to the relative Chow scheme, always contains $U$ by the existence of the Hilbert–Chow morphism.
As I mentioned in my [comment](https://mathoverflow.net/questions/438243/orbits-in-the-open-set-given-by-rosenlichts-result#comment1130130_438243), this construction is in Kollár's book [Rational curves on algebraic varieties](https://doi.org/10.1007/978-3-662-03276-3), specifically in Section 4 of Chapter IV. I do not believe that he explicitly singles out the application to forming group quotients, since that is not his main concern (he wants to construct the maximal quotient by the family of rational curves, not group quotients).
| 2 | https://mathoverflow.net/users/13265 | 438396 | 177,086 |
https://mathoverflow.net/questions/438305 | 19 | Let's denote the Fibonacci numbers by $F\_0=0,F\_1=1,F\_{n+2}=F\_{n+1}+F\_n \; \forall n \ge 0$. According to [Zeckendorf's theorem](https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem), every positive integer can be represented uniquely as the sum of some (at least $1$) distinct Fibonacci numbers where no two consecutive Fibonacci numbers occur.
Suppose positive integers $a=\sum\_{r=1}^{p}F\_{i\_r}$ and $b=\sum\_{s=1}^{q}F\_{j\_s}$ are in their Zeckendorf forms, i.e. $i\_1 \ll \cdots \ll i\_p,j\_1 \ll \cdots \ll j\_q$, where $x \ll y \Leftrightarrow x+2 \le y$. Since $F\_0=0$ and $F\_1=F\_2=1$, we can require that $i\_1,j\_1 \ge 2$ if both $a$ and $b$ are non-zero. Define the **Fibonacci product** of $a$ and $b$ by $a \circ b=\sum\_{r=1}^{p}\sum\_{s=1}^{q}F\_{i\_r+j\_s}$ (this result is not necessarily in its Zeckendorf form), and the Zeckendorf form of $0$ to be $0=F\_0$. It can be verified that $0$ is the identity element of the Fibonacci product. By a theorem of [Knuth](https://www.cs.umb.edu/%7Ervetro/vetroBioComp/compression/FIBO/KnuthFibb.pdf), the Fibonacci product on non-negative integers is both commutative and associative (the introduction of an identity does not break these), which makes the non-negative integers a commutative monoid.
In the [same paper](https://www.cs.umb.edu/%7Ervetro/vetroBioComp/compression/FIBO/KnuthFibb.pdf), Knuth concluded that Fibonacci product is monotonically increasing in each variable, which means it has the cancellation property. Thus the monoid embeds into its Grothendieck group $G\_F$. Consider the submonoid generated by $\{ F\_i \}\_{i \ne 1}$ we get that $\Bbb{Z} \vartriangleleft G\_F$ ($F\_i \circ F\_j=F\_{i+j} \; \forall i,j \ge 2$), but the other elements of $G\_F$ are not clear to me (*e.g. I don't even know if $G\_F$ has any torsion element*). **Has the group $G\_F$ been studied somewhere, or the structure of $G\_F$ is too trivial to be looked into?** Any referrence or direct description is welcome.
**Correction:** There's a typo in the "carry rules" $(8)$ & $(9)$ in Knuth's paper. The correct carry rules should be: $$\overline{0(d+1)(e+1)} \rightarrow \overline{1de} \\ \overline{0(d+2)0e} \rightarrow \overline{1d0(e+1)}$$ for $d,e \ge 0$. Knuth seemed to use these correct rules in his following discussion so it did no harm to his conclusion.
| https://mathoverflow.net/users/166298 | Grothendieck group of the Fibonacci monoid | This operation is not mysterious at all! The monoid $(\mathbf N,\circ)$ is isomorphic to a multiplicative submonoid $T$ of the commutative ring $\mathbf Z[\varphi] = \mathbf Z[t]/(t^2-t-1)$, where $\varphi = \tfrac{1+\sqrt{5}}{2}$ is the golden ratio. In particular, it is isomorphic to a submonoid of $\mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$, where $\mathcal P$ is the set of prime ideals in $\mathbf Z[\varphi]$, and the first factor is the powers of the fundamental unit $\varphi$. In fact, the groupification $T^{\text{gp}} \hookrightarrow \mathbf Q(\varphi)^\times \cap \mathbf R\_{>0} \cong \mathbf Z \oplus \mathbf Z^{\oplus \mathcal P}$ is an *isomorphism*; see the corollary below.
I first wrote this as an answer with full details, but a small literature search (starting from Knuth's paper on MathSciNet and looking for forward references) turned up some original sources that are a little slicker than what I wrote. So in the interest of readability, I have removed some of the proofs.
The main reference is the 2-page paper [Arnoux, 1989], explaining the relation to $\mathbf Z[\varphi]$. In addition, [Zhuravlev, 2007] notes that every nonzero element of $\mathbf Z[\varphi]$ can be written (non-uniquely) as $\pm\varphi^{-n}t$ for $t \in T$ and $n \in \mathbf N$, giving the promised computation of $T^{\text{gp}}$ of the corollary below.
**Notation.** Let $f \colon \mathbf Z[t] \twoheadrightarrow \mathbf Z[\varphi]$ be the quotient map $t \mapsto \varphi$, and let $g \colon \mathbf Z[\varphi] \to \mathbf Z$ be the *group* homomorphism $a+b\varphi \mapsto b$. Note that $g$ is not a ring homomorphism; in fact $g(\varphi^n) = F\_n$ by Binet's formula. Write $h$ for the composition $g \circ f$; this exhibits a polynomial $\sum a\_it^i$ as a radix-F expansion of its image $\sum a\_i F\_i$.
The Fibonacci product $\circ$ has an obvious extension to $\mathbf N \times \mathbf N \to \mathbf N$ by setting $n \circ 0 = n = 0 \circ n$ for all $n \in \mathbf N$ (which does not affect associativity). We view $\mathbf Z[\varphi]$ as a subring of $\mathbf R$ in the obvious way, and it has a conjugation $\overline{(-)} \colon \mathbf Z[\varphi] \to \mathbf Z[\varphi]$ taking $\varphi$ to $1-\varphi = \varphi^{-1} = \tfrac{1-\sqrt{5}}{2}$.
**Definition.** Define the subset $Z \subseteq t^2\mathbf Z[t]$ of elements of the form $P(t)=\sum\_{i=2}^r a\_it^i$ with all $a\_i \in \{0,1\}$ and $a\_ia\_{i+1} = 0$ for all $i$. We remove $0$ from this set and add $1$, since this will be our multiplicative unit. Let $T \subseteq \mathbf Z[\varphi]$ be the image of $Z$ under $f$. Then Zeckendorf's theorem shows that $f$ and $g$ give bijections
$$Z \stackrel\sim\to T \stackrel\sim\to \mathbf N.$$
By definition, the map $h$ has the property $h(xy) = h(x)h(y)$ for all $x,y \in Z$. But $Z$ is not closed under multiplication, so this doesn't produce a monoid isomorphism $h \colon Z \stackrel\sim\to \mathbf N$. The key point is:
**Proposition** [Arnoux, 1989] *The set $T$ is closed under multiplication. In particular, $g \colon T \to \mathbf N$ is a monoid isomorphism for the Fibonacci product on $\mathbf N$, i.e.*
$$g(xy) = g(x) \circ g(y)\qquad \text{for all } x, y \in T.$$
This also gives a conceptual proof of associativity of $(\mathbf N,\circ)$ (which is not used in the proof). So we see that $\mathbf Z[\varphi]$ interpolates nicely between $\mathbf Z[t]$ and $(\mathbf N,\circ)$.
The proposition can easily be shown by hand using the multiplicative structure of $f$ together with Lemma 3 of Knuth; see the revision history of this post for such a proof. Instead, Arnoux deduces it from the following:
**Lemma** [Arnoux, 1989]. *The set $T \setminus \{1\}$ is given by the elements $t=a+n\varphi$ with $a,n \in \mathbf Z\_{>0}$ such that $\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$. The map $g^{-1}$ is given by $n \mapsto a\_n + n\varphi$, where*
$$a\_n = \left\lfloor (n+1)\tfrac{-1+\sqrt{5}}{2} \right\rfloor$$
*is Hofstadter's $G$-sequence (OEIS [A005206](https://oeis.org/A005206)).*
See Lemmas 2 and 3 in Arnoux. The final statement is not there, but can easily be deduced from the first. This shows that $T$ is closed under multiplication as $(\varphi-2,\varphi-1) \subseteq (-1,1)$. In addition, it gives the clean formula
$$n \circ m = nm + na\_m + ma\_n.$$
Note that in the first statement, we don't need the assumption $a > 0$. Indeed, if $a \leq 0$ we get $\overline{\!\ t\ \!} = a+n\overline\varphi \leq \overline\varphi < \varphi-2$ since $\overline\varphi < 0$ and $n \geq 1$.
Finally, we need the following observation that appears to be due to [Zhuravlev, 2007]; see Proposition 4.1.
**Lemma.** *Every element $x \in \mathbf Z[\varphi]$ can be written as $\pm\varphi^{-n} \cdot t$ for some $t \in T$ and $n \in \mathbf N$.*
Since $T$ and $\varphi$ are positive, the sign agrees with the sign of $x$. The proof (and the whole paper) is notationally heavy (and logically hard to follow), so let me include an argument here.
*Proof.* Since $\lvert \overline\varphi \rvert < 1$, we get $\pm\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$ for $n \gg 0$. If $t = a+n\varphi$, we necessarily have $n \neq 0$, for otherwise $t$ and therefore $x$ is $0$. Wihout loss of generality, we may assume $n > 0$. By the previous lemma (and the discussion after), we get $t \in T$, so $x = \pm\varphi^{-n} \cdot t$ as desired. $\square$
(Notational note: what Zhuravlev calls $\delta(n)$ is related to my $g^{-1}(n)$ via $-\delta(n)/\varphi = \overline{g^{-1}(n)}$. Zhuravlev's Fibonacci sequence is off by $1$ compared to Knuth.)
**Corollary.** *There exists a choice of generators of each prime ideal of $\mathbf Z[\varphi]$ inducing an injection $\psi \colon T \hookrightarrow \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$. For any such choice, the map $\psi^{\text{gp}} \colon T^{\text{gp}} \to \mathbf Z \oplus \mathbf Z^{\oplus \mathcal P}$ is an isomorphism, and the quotient map $T \to \mathbf N^{\oplus \mathcal P}$ is surjective.*
*Proof.* We saw that $T$ is a submonoid of $(\mathbf Z[\varphi]\setminus\{0\},\times)$. Since $\mathbf Z[\varphi]$ is a real quadratic principal ideal domain, we can produce an isomorphism $(\mathbf Z[\varphi]\setminus\{0\},\times) \cong \mathbf Z/2 \oplus \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$ by picking generators for each prime ideal. By the lemma, we may pick a generator in $T$ for each prime ideal. Since all elements of $T$ and all chosen representatives are positive, we don't need the sign factor $\mathbf Z/2$, giving an embedding $T \hookrightarrow \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$. The final two statements follow immediately from the lemma since $\varphi^n \in T$ for all $n \geq 2$. $\square$
While subgroups of free abelian groups are free, the same does not hold for commutative monoids, i.e. there is no unique factorisation into irreducible elements. For instance, the elements $\varphi^n$ for $n \geq 2$ and $n=0$ are in $T$, but $\varphi$ is not (as $g(\varphi) = 1 = g(\varphi^2)$ and $g$ is injective on $T$). So $\varphi^6$ factors both as $(\varphi^2)^3$ or $(\varphi^3)^2$. The result above is probably the most precise you are going to get (also because it depends on infinitely many choices).
Note that negative powers of $\varphi$ are not in the image. In fact, $(\mathbf N,\circ)$ is a sharp monoid: if $a,b \neq 0$, then $a \circ b \neq 0$. I don't know if there exists a choice of generators for which no element picks up a negative power of $\varphi$ (i.e. the image is in $\mathbf N \oplus \mathbf N^{\oplus \mathcal P} \subseteq \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$), but this seems unlikely to me.
---
**References.**
[Arnoux, 1989] P. Arnoux, *Some remarks about Fibonacci multiplication*. Appl. Math. Lett. **2**.4, p. 319-320 (1989). DOI:[10.1016/0893-9659(89)90078-5](https://doi.org/10.1016/0893-9659(89)90078-5)
[Zuravlev, 2007] V. G. Zhuravlev, *Sums of squares over the Fibonacci $\circ$-ring*. Zap. Nauchn. Semin. POMI **337**, p. 165-190 (2006). Translation in J. Math. Sci., New York **143**.3, p. 3108-3123 (2007). DOI:[10.1007/s10958-007-0195-1](https://doi.org/10.1007/s10958-007-0195-1)
| 9 | https://mathoverflow.net/users/82179 | 438398 | 177,088 |
https://mathoverflow.net/questions/438410 | 0 | Let
* $X := \mathbb R^n$,
* $C\_b(X)$ the space of all real-valued bounded continuous,
* $C\_c(X)$ the space of all real-valued continuous functions with compact supports, and
* $C\_c^\infty(X)$ the space of all real-valued smooth functions with compact supports.
Let $\mu, \mu\_n$ be Borel probability measures on $X$. We say that $\mu\_n$ converges to $\mu$ *weakly* iff
$$
\mu\_n \rightharpoonup \mu \overset{\text{def}}{\iff} \int\_X f \mathrm d \mu\_n \to \int\_X f \mathrm d \mu \quad \forall f \in C\_b(X).
$$
Because $\mathbb R^n$ is locally compact and separable, we [have](https://math.stackexchange.com/a/4570642/1019043)
$$
\mu\_n \rightharpoonup \mu \iff \int\_X f \mathrm d \mu\_n \to \int\_X f \mathrm d \mu \quad \forall f \in C\_c(X).
$$
>
> Can we further restrict the space of test functions to $C\_c^\infty (X)$?
>
>
>
Thank you so much for your elaboration!
| https://mathoverflow.net/users/477203 | Can we further restrict the space of test functions to $C_c^\infty (X)$ in weak convergence? | Any $f\in C\_c(X)$ can be uniformly approximated by functions $f\_n\in C\_c^\infty(X)$, say by convolving $f$ with appropriate mollifiers $\psi\_n\in C\_c^\infty(X)$.
So, your desired conclusion indeed follows.
| 4 | https://mathoverflow.net/users/36721 | 438412 | 177,092 |
https://mathoverflow.net/questions/438407 | 8 | When it is said that Kunen inconsistency theorem proves that given $\sf ZFC$ no elementary embedding can exist from the universe to itself. Most references quote full choice in stating that result, and contemplate a salvage by forsaking choice altogether like in Reinhardt's cardinals setting.
>
> Is there a known weaker form of choice, like dependent or countable choice, that can evade this theorem?
>
>
>
| https://mathoverflow.net/users/95347 | Is there a form of choice that can elude Kunen's inconsistency theorem? | [Work of Usuba](https://arxiv.org/abs/2004.01515) combined with [work of Woodin](https://www.worldscientific.com/doi/10.1142/S021906131000095X) shows that if there is a Reinhardt cardinal $\kappa$ that is a limit of Lowenheim-Skolem cardinals, then there is a forcing extension in which $\kappa$ remains a Reinhardt cardinal but $\text{DC}\_\kappa$ holds. This means one has $\text{DC}\_\lambda$ where $\lambda = \sup\_{n < \omega} j^n(\kappa)$ where $j : V\to V$ is elementary with critical point $\kappa$. In this context, $\lambda^+$ must be a measurable cardinal, so one really cannot have much more choice than $\text{DC}\_\lambda$; e.g., $\text{DC}\_{\lambda^+}$ and even $\text{AC}\_{\lambda^+}$ fail.
The consistency proof uses Woodin's Easton iteration of collapse forcings as in SEM 1 Theorem 226 but substituting Usuba's Proposition 4.7. One only iterates up to $\lambda$ (so one is not forcing full choice, which of course would kill the Reinhardt). Then one has to lift $j$ to the forcing extension, which uses the master condition argument for rank-to-rank embeddings which can be found in Section 5 of Hamkins's ["Fragile measurability"](https://www.jstor.org/stable/2275264) paper, in the context of $I\_1$.
There are also "global" forms of choice that are consistent with Reinhardts relative to stronger principles. For example, the axiom [WISC](https://ncatlab.org/nlab/show/WISC) follows from a Reinhardt cardinal plus a proper class of Lowenheim-Skolem cardinals, and so in particular, Reinhardt plus WISC follows from a super-Reinhardt cardinal (or just a Reinhardt and a proper class of supercompacts).
In particular, from choiceless large cardinal axioms beyond a Reinhardt (e.g., a Berkeley cardinal) one can prove the consistency of Reinhardts with weak forms of choice. It is an open question whether, for example, Reinhardt plus DC can be proved consistent starting from a single Reinhardt.
| 22 | https://mathoverflow.net/users/102684 | 438413 | 177,093 |
https://mathoverflow.net/questions/438405 | 5 | For any geometric morphism $f:\mathcal{F} \to \mathcal{E}$ of Grothendieck 1-topoi, there exists a functor of small categories $\ell :D\to C$ and left exact localizations $\mathcal{F} \hookrightarrow \mathcal{P}D$ and $\mathcal{E} \hookrightarrow \mathcal{P}C$ such that the inverse image functor $f^\* : \mathcal{E} \to \mathcal{F}$ is the composite
$$ \mathcal{E} \hookrightarrow \mathcal{P} C \xrightarrow{\ell^\*} \mathcal{P}D \to \mathcal{F}.$$
Is this also true for $\infty$-topoi?
The proof of the 1-topos fact that I know doesn't seem to generalize obviously. It's in section C2.5 of *Sketches of an Elephant*: we represent $\mathcal{F}$ as internal sheaves on an internal site in $\mathcal{E}$, then apply a Grothendieck construction to externalize that site relative to some site for $\mathcal{E}$, put a topology on it, and show that the resulting fibration is cover-reflecting and induces the given geometric morphism. The use of Grothendieck topologies is what it's not immediately clear how to generalize, since not every $\infty$-topos is a topos of sheaves in the straightforward sense.
| https://mathoverflow.net/users/49 | Fibrations of sites for $\infty$-topoi | Here is an argument for the 1-categorical version that essentially bypass the use of internal site and should be much easier to generalize to the $\infty$-categorical case. ( I mean you can still see internal site barely hidden in plain sight, but the point is you don't need to see them to follow the proof)
As a first approximation, let's cheat a little, and allow ourselves to work with big sites - and we will deal with size issues at the end.
So how would we do this? A naive answer is to take $C = \mathcal{E}$ and $D = \mathcal{F}$... but that actually doesn't work at all.
The right way to do it is inspired from this idea of internal sites: One takes $C = \mathcal{E}$, but we take $D$ to be the comma category whose objects are triplets $X \in \mathcal{E}$, $Y \in \mathcal{F}$ and an arrow $u: Y \to f^\*X$.
(note that it corresponds to taking the internal site $\mathcal{F}$ seen as an $\mathcal{E}$-indexed category and applying the Grothendieck construction to it, exactly as in the argument you explained)
$l$ is the obvious forgetful functor $D$ to $C$, and the important point is that it has fully faithful right adjoint $i$ sending $X$ in $C = \mathcal{E}$ to the triplet $f^\* X \to f^\* X$.
Note that presheaves on $\mathcal{F}$ can be seen as special presheaves on $D$, the one that only depends on the $Y$ component, and (ignoring the obvious size problems) the left adjoint to this inclusion can be seen to be left exact (**edit:** this is because it can be identified with $f\_!$ for $f:D \to \mathcal{F}$ the forgetful functor which is itself left exact). In particular, "sheaves" over $\mathcal{F}$ are a left exact localization of presheaves over $D$.
Now, There is an obvious commutative square with vertically $i^\*$ and $f\_\*$ and horizontally the inclusion of $\mathcal{E}$ and $\mathcal{F}$ to $\mathcal{P}C$ and $\mathcal{P}D$.
The key point now is because $i$ is right adjoint to $l$, then $i^\*$ is right adjoint to $l^\*$, that is $i^\* = l\_\*$.
Taking the left adjoint of all the functors in the square above (and ignoring the size problem it creates at this stage) you get a commutative square with the two sheafification functors $\mathcal{P} C \to \mathcal{E}$ and $\mathcal{P} D \to \mathcal{F}$ horizontally and $l^\*$ and $f^\*$ vertically.
Restricting this to an object of $\mathcal{P} C$ that is already a sheaf, you get the desired decomposition.
Finally, let's deal with the size problems: as usual in this kind of argument, one can simply take $\kappa$ a regular cardinal such that:
* both $\mathcal{E}$ and $\mathcal{F}$ are locally $\kappa$-presentable.
* The category $\mathcal{E}\_\kappa$ and $\mathcal{F}\_\kappa$ of $\kappa$-presentable objects are closed under finite limits.
* $f^\*$ sends $\kappa$-presentable objects to $\kappa$-presentable objects.
Then taking $C = \mathcal{E}\_\kappa$ and $D$ the comma category for $f^\*: \mathcal{E}\_\kappa \to \mathcal{F}\_\kappa$ and you can make all the arguments above.
Obviously there are some difficulties to make all this works in $\infty$-category theory and I feel it would be too much works to to everything here. But I don't see any major obstacle, let me know if I missed one, I'd happy to look into it!
| 5 | https://mathoverflow.net/users/22131 | 438417 | 177,095 |
https://mathoverflow.net/questions/438418 | -5 | This is more of a curiosity than a research question, but I could not find it answered anywhere. What is the largest $N$ for which the statement in the title is true? I have recently read that the largest known prime is $2^{82589933} − 1$, but I imagine it is not known if, for example, $2^{82589933} − 3$ is prime.
A closely related question: if $\pi(x)$ is the prime counting function, that is, $\pi(x)$ is the number of primes not exceeding $x$, what is the largest value of $n$ for which $\pi(n)$ is known exactly?
| https://mathoverflow.net/users/66323 | It is known if $n$ is prime for all $n\leq N$ | As explained in the comments, there is no well-defined answer to this question, because there is no organized effort to test primality of all numbers up to a certain bound.
However, there is a (more or less) well-defined answer (at any given point in time) to a related question: What is the largest $N$ for which the primality status of $2^n - 1$ is known for all $n\le N$? The answer, as of this writing, is [57885161](https://www.mersenne.org/primes/) (though some "insiders" in the GIMPS project may be able to report a slightly higher number than that—but not higher than 74207281).
| 1 | https://mathoverflow.net/users/3106 | 438429 | 177,098 |
https://mathoverflow.net/questions/438404 | 3 | Let $p$ be a prime. The *minimal ramification problem* is to ask whether or not every finite $p$-group $G$ can be realized as the Galois group of a tamely ramified extension of $\mathbb{Q}$ with exactly $r(G)$ ramified primes where $r(G)$ is the minimal number of generators of $G$. This problem has an affirmative solution for some family of $p$-groups, e.g. all Sylow $p$-subgroups of the symmetric group and of the classical groups over finite fields of characteristic prime to $p$, cf. [On the minimal ramification problem for ℓ-groups by Hershy Kisilevsky and Jack Sonn](https://www.cambridge.org/core/journals/compositio-mathematica/article/on-the-minimal-ramification-problem-for-groups/141781DB0334A033F0AC48F68B22FF43). In general, it's still open.
I'm interested in the following refined quesiton: Let $ p $ be an odd prime. Is there a finite Galois extension $ L $ of $ \mathbb{Q} $ such that
1. the Galois group $ G:=\text{Gal}(L/\mathbb{Q}) $ is a finite $ p $-group with order $ |G|> p^{9} $;
2. the extension $ K/\mathbb{Q} $ is ramified at exactly $ 3 $ primes $ q\_{1},q\_{2},q\_{3} $ where $ q\_{i}\equiv 1~\text{mod}~p $ but $ q\_{i}\not \equiv 1~\text{mod}~p^{2} $ for $ i=1,2,3 $?
Note that $L/\mathbb{Q}$ is tamely ramified at $q\_1,q\_2,q\_3$ if and only if $ q\_{i}\equiv 1~\text{mod}~p $. Moreover, the condition that "$ q\_{i}\not \equiv 1~\text{mod}~p^{2} $" is essential here. Without this condition, the answer to question is Yes from the known case of the minimal ramification problem as above.
| https://mathoverflow.net/users/492970 | On the refined minimal ramification problem for $p$-groups | The answer is affirmative at least for $p\ge 11$ by the following construction (for the smaller primes, the extension constructed is not of the demanded degree $>p^9$, but surely there will be some alternative construction):
Let $q\_1\equiv 1$ mod $p$, $q\_1\ne 1$ mod $p^2$, and let $K/\mathbb{Q}$ be the $C\_p$-subextension of $\mathbb{Q}(\zeta\_{q\_1})$ (unramified outside $q\_1$). Next, let's construct a suitable $C\_p$-extension $F/K$, ramified exactly at two (suitable) prime ideals $\nu\_2, \nu\_3$.
Due to the following, we want $\nu\_2, \nu\_3$ to extend (different) rational primes $q\_2, q\_3$, both split in $K(\zeta\_p)$, but not in $K(\zeta\_{p^2})$ (in particular, they are then $\equiv 1$ mod $p$, but not mod $p^2$.
A (special case of a) theorem by Gras and Munnier (<https://pmb.centre-mersenne.org/item/10.5802/pmb.a-91.pdf>) now gives sufficient conditions (on $q\_2,q\_3$) for such an extension to exist. Namely, there exists a certain finite elementary-abelian $p$-extension $L/K(\zeta\_p)$ such that a $C\_p$-extension $F/K$ as desired exists as soon as the Frobenius elements of (primes extending) $q\_2$ and $q\_3$ in $L/K(\zeta\_p)$ generate the same cyclic subgroup. But we have only imposed a condition on the Frobenius in $K(\zeta\_{p^2})/K(\zeta\_p)$ to be nontrivial, so by Chebotarev's density theorem, we can find plenty such primes $q\_2,q\_3$ (if $\zeta\_{p^2}\notin L$, the Frobenius elements could even both be chosen trivial).
So we have a (non-Galois) extension $F/\mathbb{Q}$ of degree $p^2$, ramified at exactly three primes of the prescribed shape; the same is therefore true for its Galois closure $\Omega/\mathbb{Q}$. The Galois group of this embeds into the wreath product $C\_p\wr C\_p$ (of order $p^{p+1}$) by construction, and due to the splitting of $q\_2, q\_3$ in $K/\mathbb{Q}$, we have constructed the inertia groups at $q\_2, q\_3$ to (lie in the block kernel and) act non-trivially on exactly one block of the imprimitive wreath product. It is then an easy exercise in group theory to show that the group generated by all the inertia groups is indeed the full $C\_p\wr C\_p$.
Note, however, that $r(C\_p\wr C\_p)=2$ (as remarked in the comments), so while this construction does fulfill all the requirements of the question, it is not a ``minimal ramification" realization for this particular group.
| 5 | https://mathoverflow.net/users/127660 | 438437 | 177,101 |
https://mathoverflow.net/questions/438451 | 0 | Axiality has been studied under a definition given here: <https://en.wikipedia.org/wiki/Axiality_(geometry)>
Consider an *alternative definition* of axiality as follows: For a convex region C, consider a chord L and the set of chords of C that are perpendicular to L. For each perpendicular chord, consider the ratio: length of smaller segment to length of larger segment into which L divides it. Now, for chord L, define an 'axiality ratio' as the *minimum* of these ratios over all its perpendicular chords. Now, the best axis of C is that chord for which axiality ratio is a *maximum* and that value of the axiality ratio could be called the axiality of C itself.
**Question:** What shape of C *minimizes* axiality under this new definition?
Note: Some more related thoughts are recorded at <http://nandacumar.blogspot.com/2023/01/axiality-of-planar-convex-regions.html> and <http://nandacumar.blogspot.com/2023/01/centralness-of-convex-planar-regions.html>
| https://mathoverflow.net/users/142600 | On 'axiality' of planar convex regions | Too many bodies have vanishing axiality ratio.
Indeed, if the ratio for $L$ is positive, then the whole body projects to $L$.
So you may start with convex smooth body, locate these exceptional chords and modify the body in a small neighborhood of its ends making the ratio = zero.
In particular, you get a body with zero ratio arbitrarily close to any given body (in the sense of Hausdorff).
| 0 | https://mathoverflow.net/users/1441 | 438455 | 177,105 |
https://mathoverflow.net/questions/438181 | 12 | Roughly speaking, given a set-sized logic $\mathcal{L}$ let $\mathcal{L}'$ be gotten by adding to $\mathcal{L}$ the ability to quantify over $\mathcal{L}$-definable relations. (The details are a bit tedious, so I've put the rigorous definition at the end of this question.) Note that $\mathcal{L}'$ is always strictly stronger than $\mathcal{L}$: let $\kappa$ be some cardinal greater than the set of $\mathcal{L}$-formulas in the language $\{<\}$, and consider the least $\mathcal{L}$-undefinable element of the structure $(\kappa;<)$. We can iterate this jump operation up to any ordinal $\alpha$ by taking unions at limit stages as usual (since we're not worrying about syntax there is no "notation"-flavored issue) to get $\mathcal{L}^{(\alpha)}$. Even iterating through all the ordinals makes sense; call the result $\mathcal{L}^{(\infty)}$.
I'm interested in pinning down $\mathsf{FOL}^{(\infty)}$. Here's what I know so far:
* $\mathsf{FOL}^{(\infty)}$ is (always up to expressive strength) a sublogic of $\mathcal{L}\_{\infty,\omega}^L=\mathcal{L}\_{\infty,\omega}\cap L$. More generally, if $M$ is an inner model and $\mathcal{L}$ is a set-sized sublogic of $\mathcal{L}\_{\infty,\omega}^M$ which is appropriately definable in $M$, then $\mathcal{L}'\subseteq\mathcal{L}^M\_{\infty,\omega}$ as well.
* $\mathsf{FOL}^{(\infty)}$ is strictly weaker than $\mathcal{L}\_{\infty,\omega}^L$: if $(r\_i)\_{i\in\omega}$ are sufficiently mutually Cohen generic (say, generic over a countable elementary submodel of $L\_{\omega\_1}$), then the $\{<,\in\}$-structures $\omega\sqcup\{r\_{2i}:i\in\omega\}$ and $\omega\sqcup\{r\_{2i+1}: i\in\omega\}$ are $\mathsf{FOL}^{(\infty)}$-equivalent but not $\mathcal{L}\_{\infty,\omega}^L$-equivalent.
* The argument of the previous bulletpoint implies that the $\mathsf{FOL}^{(\infty)}$-definable relations on $(\omega;<)$ fall well short of the constructible ones. On the other hand, it's also easy to see that they reach well past the hyperarithmetic: the $\mathsf{FOL}^{(\omega\_1^{CK})}$-definable relations are exactly the hyperarithmetic ones, and $\mathcal{O}$ is then $\mathsf{FOL}^{(\omega\_1^{CK}+1)}$-definable since a computable linear order is a well-order iff it supports a hyperarithmetic jump sequence.
* Finally, despite its clear weaknesses, for every countable structure $\mathfrak{A}$ the automorphism orbit relation on elements (or $n$-tuples for fixed $n$) of $\mathfrak{A}$ is $\mathsf{FOL}^{(\infty)}$-definable (this is just Scott's argument).
My main question is whether there is a better description of this logic as a whole:
>
> **Question 1**: Is there a snappier description of $\mathsf{FOL}^{(\infty)}$?
>
>
>
The definition of $\mathsf{FOL}^{(\infty)}$ is "bottom-up" (a more snappy version: it's the smallest possibly-class-sized logic containing $\mathsf{FOL}$ and containing $\mathcal{L}'$ whenever $\mathcal{L}$ is a set-sized logic it contains). I would be especially be interested in a "top-down" definition, such as "the largest logic containing $\mathsf{FOL}$ with *[tameness properties]*." Lindstrom's theorem is one motivating example of such a definition, but there are others - e.g. [Barwise's analysis of $\mathcal{L}\_{\infty,\omega}$](https://www.sciencedirect.com/science/article/pii/0003484372900022).
My secondary question, for which I at least have a conjecture, is about its arithmetic power specifically:
>
> **Question 2**: What are the $\mathsf{FOL}^{(\infty)}$-definable relations on the naturals?
>
>
>
My guess is that these are exactly the relations in $L\_{\beta\_0}$, where ${\beta\_0}$ is the smallest "gap ordinal" (= such that $L\_{\beta\_0}\cap\mathbb{R}=L\_{\beta\_0+1}\cap\mathbb{R}$). This ${\beta\_0}$ is known to also be the least ordinal such that $L\_{\beta\_0}\models\mathsf{ZFC^-}$, or such that $L\_{\beta\_0}\cap\mathbb{R}\models\mathsf{Z}\_2$. See [Putnam](https://projecteuclid.org/journals/notre-dame-journal-of-formal-logic/volume-4/issue-4/A-note-on-constructible-sets-of-integers/10.1305/ndjfl/1093957652.full), [Marek/Srebrny](https://www.sciencedirect.com/science/article/pii/0003484374900059/), or [Madore (entry 2.17)](http://www.madore.org/%7Edavid/math/ordinal-zoo.pdf).
---
Details
-------
Here's a more precise definition of $\mathcal{L}'$. We take the closure of the $\mathcal{L}$-formulas under the addition of a whole family of new relation quantifiers $$\bigtriangledown^{\mathcal{L},p}\_{\delta,\eta,\rho\_1,...,\rho\_n}.$$ Here the $\mathcal{L}$-superscript indicates the logic with respect to which we are quantifying over definable relations, $p\in\omega$ indicates the arity of the relations being quantified over, and $\delta,\eta,\rho\_1,...,\rho\_n$ are $\mathcal{L}$-formulas with respective arities $d, e, r\_1,...,r\_n$ such that $e=2d$ and each $r\_i$ is divisible by $d$. The semantics of these new quantifiers is given as follows:
* In any structure $\mathfrak{A}$, if $\eta^\mathfrak{A}$ is an equivalence relation on $\delta^\mathfrak{A}$ and each $\rho\_i^\mathfrak{A}$ is well-defined on $\delta^\mathfrak{A}/\eta^\mathfrak{A}$ in the obvious sense, then "$\bigtriangledown^{\mathcal{L},p}\_{\delta,\eta,\rho\_1,...,\rho\_n}X[\mathit{stuff}]$" (with $X$ a $p$-ary relation variable) means
>
> "For every $\eta^\mathfrak{A}$-invariant relation $X\subseteq(\delta^\mathfrak{A})^p$ such that $X/\eta^\mathfrak{A}$ is $\mathcal{L}$-definable in the related structure $(\delta^\mathfrak{A}/\eta^\mathfrak{A}; \rho\_1,...,\rho\_n)$, it is the case that $[\mathit{stuff}]$."
>
>
>
* On the other hand, if $\delta,\eta,\rho\_1,...,\rho\_n$ do not satisfy the above conditions in $\mathfrak{A}$, then we interpret $\bigtriangledown^{\mathcal{L},p}\_{\delta,\eta,\rho\_1,...,\rho\_n}$ in some fixed trivial way (say, always outputting $\perp$).
Note that when we iterate $'$ we get multiple $\bigtriangledown$-quantifiers; so, for example, $\mathsf{FOL}^{(\omega)}$ has a separate quantifier (family) $\bigtriangledown^{\mathsf{FOL}^{(n)}}$ for each $n\in\omega$. *(Incidentally, a stronger version of this construction was suggested in [this MSE post](https://math.stackexchange.com/questions/4605261/non-henkin-non-full-semantics-for-second-order-logic/4605604#4605604), but I've ultimately decided that the more limited approach here is more natural.)*
| https://mathoverflow.net/users/8133 | What is the "iterated definability" limit of first-order logic? | If I understand the question properly (I'm not sure whether I do), then it looks like your conjecture for question 2 is correct, i.e. $L\_{\beta\_0}\cap\mathcal{P}(\omega)$. Here is a hastily written sketch.
For in fact, letting $N\_\alpha$ be the set of $\mathcal{L}^{(\alpha)}$ relations on $\omega$, then $N\_\alpha=L\_{\omega+\alpha}\cap\mathcal{P}(\omega)$ for all $\alpha\leq\beta\_0$, and $\beta\_0$ is a closure point.
The small issue here is to see that over $N\_\alpha$, we can simulate $L\_\alpha$ definably from parameters, assuming that $N\_\alpha=L\_\alpha\cap\mathcal{P}(\omega)$.
Suppose $\alpha=\beta+1$. Since $L\_\beta$ projects to $\omega$, there is a real $x\in N\_\alpha$ which codes $L\_\beta$.
We can refer to $x$, since this is given by some $\mathcal{L}^{(\alpha)}$ relation. Then in fact for all ordinals in $\gamma\in[\alpha,\kappa\_\alpha)$,
where $\kappa\_\alpha$ is the least admissible $\kappa>\alpha$,
we can define simulate $L\_\gamma$ over $N\_\gamma$ by looking at all models $M$
coded by reals which satisfy ``$V=L[x]$''; all of these are wellfounded,
because otherwise the wellfounded part is a model of KP, but it easily enough follows then that $N\notin L\_{\kappa\_\alpha}$, so $N\notin N\_\gamma$. So over $N\_\gamma$ we can simulate $L\_\gamma$, and hence define (from parameters) all reals in $L\_{\gamma+1}$.
So we get that $N\_{\kappa\_\alpha}=L\_{\kappa\_\alpha}\cap\mathcal{P}(\omega)$.
Now suppose that $\alpha$ is a limit of admissibles, and we have $N\_\alpha=L\_\alpha\cap\mathcal{P}(\omega)$. Then over $N\_\alpha$, we can simulate $L\_\alpha$ by looking at all models $M$
coded by reals, which satisfy ``$V=L$'',
and are wellfounded in $N\_\alpha$,
i.e. have no descending sequence through their ordinals which is in $N\_\alpha$.
(Any illfounded such model $M$ gets into some $L\_\kappa$ for some admissible $\kappa<\alpha$, but then we get a descending sequence before $\alpha$.)
So we can simulate $L\_\alpha$ over $N\_\alpha$ as desired.
| 6 | https://mathoverflow.net/users/160347 | 438464 | 177,107 |
https://mathoverflow.net/questions/438442 | 3 | The Kock-Lawvere axiom for a topos $\mathcal{E}$ states that given a specified commutative ring object $R \in \mathcal{E}$, for all local Artinian $R$-algebra objects $A \in \mathcal{E}$, the morphism
$$A \to R^{\mathrm{Spec}\_R(A)}$$
is an isomorphism.
Now, we don't assume that the Kock-Lawvere axiom holds for the topos $\mathcal{E}$. Assuming that in the topos $\mathcal{E}$ one could define the power series ring $R[[\epsilon]]$ of $R$ as the inverse limit of the local Artinian $R$-algebras $R[\epsilon]/(\epsilon^n)$ and the formal spectrum $\mathrm{Spf}(R[[x]])$ of $R[[\epsilon]]$ as the inductive limit of the spectra $\mathrm{Spec}(R[\epsilon]/(\epsilon^n))$, is it consistent to assume that the morphism
$$R[[\epsilon]] \to R^{\mathrm{Spf}(R[[\epsilon]])}$$
is an isomorphism?
| https://mathoverflow.net/users/483446 | Analogue of Kock-Lawvere axiom for power series rings? | Yes, it is consistent, it even follows from the Kock-Lawvere axiom, as follows.
We defined $\mathrm{Spf}(R[[\epsilon]]) := \mathrm{colim}\_n \mathrm{Spec}(R[\epsilon]/(\epsilon^n))$, so we have
$$R^{\mathrm{Spf}(R[[\epsilon]])} = R^{\mathrm{colim}\_n \mathrm{Spec}(R[\epsilon]/(\epsilon^n))} = \mathrm{lim}\_n R^{\mathrm{Spec}(R[\epsilon]/(\epsilon^n))}$$
by the universal property of the colimit. By the Kock-Lawvere axiom we have $R^{\mathrm{Spec}(R[\epsilon]/(\epsilon^n))} = R[\epsilon]/(\epsilon^n)$, so we get exactly $\mathrm{lim}\_n R[\epsilon]/(\epsilon^n)$, which is how we defined $R[[\epsilon]]$.
(However, I might misunderstand you question. Do you ask for a topos where the Kock-Lawvere axiom *does not* hold but the morphism in question is still an isomorphism?)
| 4 | https://mathoverflow.net/users/166281 | 438467 | 177,109 |
https://mathoverflow.net/questions/438462 | 6 | Consider a Euclidean space $X$ of large dimension $N$. For a measurable subset $G\subseteq X$ and $\varepsilon>0$ let
$$G\_\varepsilon:=\{x\in G\mid B\_\varepsilon(x)\subseteq G\}$$
be the set of all points in $G$ with distance at least $\varepsilon$ from the boundary of $G$.
I am looking for results of the form
$$\mathrm{Vol}(G\_\varepsilon)/\mathrm{Vol}(G)\leq \exp(-N\rho(\varepsilon))$$
for bounded measurable sets (or similar results).
This is to make precise the intuition that, as the dimension of the space increases a larger fraction of the volume of a set is concentrated near the boundary.
For example let $G=[0-\varepsilon,1+\varepsilon]^N$, then
$G\_\varepsilon=[0,1]^N$ and the volume fraction above becomes $(1+2\varepsilon)^{-N}$ hence the conjecture for exponential decrease.
Are there any results available already?
| https://mathoverflow.net/users/81206 | Concentration of volume towards the boundary | $\newcommand\ep\varepsilon\newcommand\R{\mathbb R}$Suppose that $G$ is a measurable subset of $\R^N$ with volume $|G|>0$ such that
$$|G|\le C^N|B|,$$
where $C>0$ is a real constant and $B$ is the unit ball in $\R^N$.
Without loss of generality (wlog), $|G\_\ep|>0$. Also, $G\_\ep+\ep B\subseteq G$ for any real $\ep>0$. So, in view of the [Brunn--Minkowski inequality](https://en.wikipedia.org/wiki/Brunn%E2%80%93Minkowski_theorem#Statement),
$$|G|\ge|G\_\ep+\ep B|\ge(|G\_\ep|^{1/N}+\ep|B|^{1/N})^N
=|G\_\ep|\Big(1+\ep\Big(\frac{|B|}{|G\_\ep|}\Big)^{1/N}\Big)^N
\ge|G\_\ep|\Big(1+\ep\Big(\frac{|B|}{|G|}\Big)^{1/N}\Big)^N
\ge|G\_\ep|\Big(1+\frac\ep C\Big)^N.
$$
Thus,
$$\frac{|G\_\ep|}{|G|}\le e^{-N\rho(\ep)},\tag{1}\label{1}$$
where $\rho(\ep):=\ln(1+\frac\ep C)$.
---
Here is another (better?) statement:
Suppose that $|G|>0$ and the values of the (outer) measure of the projections of $G$ onto the $N$ coordinate axes of $\R^N$ are no greater than positive numbers $L\_1,\dots,L\_N$, respectively. Then
$$\frac{|G\_\ep|}{|G|}\le\prod\_{i=1}^N(1-2\ep/L\_i)\_+,$$
where $u\_+:=\max(0,u)$.
For $N=1$, this follows almost immediately from the definition of the outer measure. For $N>1$, this follows by induction, using one-dimensional cross-sections and the Tonelli theorem.
In particular, if $L\_i=L>0$ for all $i$, then
$$\frac{|G\_\ep|}{|G|}\le(1-2\ep/L)\_+^N\le e^{-2N\ep/L}. \tag{2}\label{2}$$
A further improvement can be obtained by replacing $L\_i$, for each $i$, by an upper bound on the measures of all one-dimensional cross-sections of $G$ in the direction of the $i$th coordinate axis.
By considering $G=CB$ and $G=[0,L]^N$, it is clear that the bounds in \eqref{1} and \eqref{2} are optimal, each in its own way.
| 5 | https://mathoverflow.net/users/36721 | 438481 | 177,110 |
https://mathoverflow.net/questions/438426 | 5 | Setup: Let $k$ be an algebraically closed field. Let $C$ be a smooth connected curve over $k$. Let $K(C)$ be the function field of $C$.
Tsen's Theorem implies that every $\mathbb{G}\_m$-gerbe over $K(C)$ splits, i.e., the etale cohomology $H^2\_{et}(K(C),\mathbb{G}\_m)=0.$
Now, let $T$ be a *not-necessarily split* torus over $K(C).$
**Question:** Do we always have that $H^2\_{et}(K(C),T)=0$?
Tsen's theorem implies that we do have the desired vanishing if $T$ is split or quasi-split, i.e., an induced torus.
A paper that I am reading seems to suggest that the desired vanishing is a piece of common sense. However, I cannot find an explicit reference for it, nor can I prove it. Any help is greatly appreciated.
| https://mathoverflow.net/users/497756 | Torus gerbes over curves | I am just posting my comment as one answer. Let $K$ be a field, and let $T$ be a $K$-group scheme such that there exists a field extension $K'/K$ that is finite and separable (i.e., étale), and there exists an isomorphism $i$ of $k'$-group schemes from $T\_{K'}:=\text{Spec}\ K'\times\_{\text{Spec}\ K}T$ to the split torus $\mathbb{G}\_{m,K'}^d$, i.e., $T$ is a torus.
By adjointness, there is a natural morphism of $K$-group schemes to the Weil restriction, $T\to R\_{K'/K}T\_{k'}.$ The composition with $R\_{K'/K}i$ gives a morphism of $K$-schemes, $$T\to R\_{K'/K}\mathbb{G}\_{m,K'}^d.$$ Since it is true after basechange to $K'$, also this morphism of $K$-group schemes is a closed immersion whose quotient, $Q$, is also a torus. Therefore we have a short exact sequence of $K$-group schemes, $$1 \to T \to R\_{K'/K}\mathbb{G}\_{m,K'}^d \to Q \to 1.$$
The associated long exact sequence of étale cohomology gives the following, $$H^1\_{\text{et}}(\text{Spec}\ K',\mathbb{G}\_{m,K'})^d \to H^1\_{\text{et}}(\text{Spec}\ K,Q) \to H^2\_{\text{et}}(\text{Spec}\ K,T)\to H^2\_{\text{et}}(\text{Spec}\ K',\mathbb{G}\_{m,K'})^d.$$
By Hilbert's Theorem 90, the first group in this sequence is trivial. Standard results about Brauer groups prove that the last group is trivial if and only if the Brauer group of $K'$ is trivial if and only if every $\textbf{PGL}\_\ell$-torsor over $K'$ has a $K'$-point (for all positive integers $\ell$).
Thus, the Galois cohomology group $H^2\_{\text{et}}(\text{Spec}\ K, T)$ is trivial if every $\textbf{PGL}\_\ell$-torsor over $K'$ has a $K'$-point and every $Q$-torsor over $K$-has a $K$-point. For $K$ and $K'$ equal to function fields of curves over an algebraically closed field $k$, this can be proved directly, cf., Tsen's original proof.
In fact, since all such torsors are dense Zariski open subschemes of smooth projective varieties that are separably rationally connected, this also follows from the "rationally connected fibration theorem", i.e., the Kollár-Miyaoka-Mori conjecture.
| 3 | https://mathoverflow.net/users/13265 | 438491 | 177,113 |
https://mathoverflow.net/questions/438492 | 3 | 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
$$Y^c\_t := c^{-1/2} X\_{ct}$$
**Question:** Is it true that as $c \to 0$, $Y^c$ converges in law to $\sigma(0) B\_t$ for a Brownian motion $B$?
| https://mathoverflow.net/users/173490 | Blow up limits for SDE | The answer to your question is yes, at least for the one- dimensional marginals. That said, I'm confident that the convergence you want actually holds on the level of stochastic processes (i.e. $c^{-1/2}X\_{ct}\xrightarrow{(d)}B\_t$ as functions on $C[0,T]$ for any $T>0$), but this requires some additional work. That said, here's the argument that proves convergence in distribution of the one dimensional marginals.
Suppose $(X\_{t})\_{t\geq{0}}$ solves the stochastic differential equation:
$$
dX\_{t}=\sigma(X\_{t})dB\_{t}, \hspace{5pt} X\_{0}=0
$$
where $\sigma$ is Lipschitz with Lipschitz constant $K>0$. That is $|\sigma(x)-\sigma(y)|\leq{K|x-y|}$ for all $x,y\in{\mathbb{R}}$. First, we will show that $\sup\_{t\in{[0,T]}}\mathbb{E}X\_{t}^{2}<\infty$ for all $T>0$.
To this effect, consider the stopping time $T\_{n}:=\inf\{t\geq{0}:|X\_{t}|=n\}$. For each $n\in{\mathbb{N}}$ we define the function $f\_{n}:[0,\infty)\rightarrow{\mathbb{R}}$ as follows:
$$
f\_{n}(t)=\mathbb{E}X\_{t\wedge{T\_{n}}}^{2}
$$
Since $|X\_{t\wedge{T\_{n}}}|\leq{n}$ by the definition of $T\_{n}$, $|f\_{n}(t)|\leq{n^{2}}$ for all $t>0$. By Ito's formula applied to $X\_{t}$,
$$
X\_{t\wedge{T\_{n}}}^{2}=\int\_{0}^{t\wedge{T\_{n}}}\sigma(X\_{s})X\_{s}dB\_{s}+\int\_{0}^{t\wedge{T\_{n}}}\sigma^{2}(X\_{s})ds
$$
Taking expectations we have that:
$$
\mathbb{E}X\_{t\wedge{T\_{n}}}^{2}=\mathbb{E}\int\_{0}^{t\wedge{T\_{n}}}\sigma^{2}(X\_{s})ds\leq{\int\_{0}^{t}\mathbb{E}\sigma^{2}(X\_{s\wedge{T\_{n}}})ds}
$$
Using the fact that $\sigma$ is Lipschitz we have that for any $x\in{\mathbb{R}}$, $|\sigma(x)|\leq{|\sigma(0)|+ K|x|}$.This in turn implies that $\sigma^{2}(x)\leq{2\sigma^{2}(0)+2K^{2}|x|^{2}}$. Applying this to the inequality above gives us:
$$
\mathbb{E}X\_{t\wedge{T\_{n}}}^{2}\leq{\int\_{0}^{t}\big(2\sigma^{2}(0)+2K\mathbb{E}X\_{s\wedge{T\_{n}}}^{2}\big)ds}=2\sigma^{2}(0)t+2K^{2}\int\_{0}^{t}\mathbb{E}X\_{s\wedge{T\_{n}}}^{2}ds
$$
In other words, for all $n\in{\mathbb{N}}$ and $t\in{[0,\infty)}$, $f\_{n}$ satisfies the following inequality:
$$
f\_{n}(t)\leq{2\sigma^{2}(0)t+2K^{2}\int\_{0}^{t}f\_{n}(s)ds}
$$
By a Gronwall- type argument, it follows that for all $n\in{\mathbb{N}}$ and $t\in{[0,\infty)}$ we have that:
$$
f\_{n}(t)\leq{\frac{\sigma^{2}(0)}{K^{2}}\left(e^{2K^{2}t}-1\right)}
$$
By Fatou's lemma,
$$
\mathbb{E}X\_{t}^{2}\leq{\liminf\_{n\rightarrow{\infty}}\mathbb{E}X\_{t\wedge{T\_{n}}}^{2}}\leq{\frac{\sigma^{2}(0)}{K^{2}}\left(e^{2K^{2}t}-1\right)}
$$
Hence, we have that $\mathbb{E}X\_{t}^{2}\leq{\frac{\sigma^{2}(0)}{K^{2}}\left(e^{2K^{2}t}-1\right)}<\infty$ for all $t>0$. From here, the proof is routine. We're interested in the process $Y^{c}\_{t}=c^{-1/2}X\_{ct}$ where $c>0$ is small.
$$
X\_{ct}=\int\_{0}^{ct}\sigma(X\_{s})dB\_{s}
$$
By Brownian scaling we have that $(c^{-1/2}B\_{ct})\_{t\geq{0}}\overset{d}{=}(\widetilde{B}\_{t})\_{t\geq{0}}$ where $(\widetilde{B}\_{t})\_{t\geq{0}}$ is a Brownian motion. Hence, "changing variables" in the stochastic integral above, we have that:
$$
X\_{ct}=c^{1/2}\int\_{0}^{t}\sigma(X\_{cu})d\widetilde{B}\_{u}
$$
Fix $\varepsilon>0$. Then:
\begin{align\*}
\mathbb{E}(c^{-1/2}X\_{ct}-\sigma(0)\widetilde{B}\_{t})^{2}&=\mathbb{E}\left(\int\_{0}^{t}\left(\sigma(X\_{cu})-\sigma(0)\right)d\widetilde{B}\_{u}\right)^{2}=\int\_{0}^{t}\mathbb{E}(\sigma(X\_{cu})-\sigma(0))^{2}du \\
&\leq{K^{2}\int\_{0}^{t}\mathbb{E}X\_{cu}^{2}du} \leq{K^{2}\varepsilon^{2}t+K^{2}\int\_{0}^{t}\mathbb{E}X\_{cu}^{2}1\_{(|X\_{s}|>\varepsilon \hspace{2pt} \text{for some $s\in{[0,ct]}$})}du}
\end{align\*}
Since the process $(X\_{t})\_{t\geq{0}}$ is continuous and square integrable, the second term on the last line goes to $0$ as $c$ tends to $0$. Since $\varepsilon>0$ is arbitrary, we conclude that for any $t>0$,
$$
c^{-1/2}X\_{ct}\xrightarrow{(d)}\sigma(0)B\_{t} \hspace{5pt} \text{as $c\rightarrow{0}$}
$$
**Bonus:** It turns out "upgrading" our argument to prove that:
$$
c^{-1/2}X\_{ct}\xrightarrow{(d)}\sigma(0)B\_t \hspace{5pt} \text{as functions on $C[0,T]$ for all $T>0$}
$$
is not that hard. Observe that it suffices to prove two things:
1. Convergence of the finite- dimensional marginals of the processes $(c^{-1/2}X\_{ct})\_{t\geq{0}}$ to those of $(\sigma(0)B\_{t})\_{t\geq{0}}$.
2. Precompactness of the measures on $C[0,T]$ induced by the processes $(c^{-1/2}X\_{ct})\_{t\geq{0}}$.
The argument for (1) is analogous to the one- dimensional case that I've written out above, so I won't bother writing it out in detail. For precompactness, we make use of a neat theorem from Billingsley's "Convergence of Probability Measures." Namely, theorem 8.3 of this book gives us a condition for a collection of measures $(\mathbb{P}\_{n})\_{n\geq{1}}$ on $C[0,T]$ to be precompact:
**Theorem:** A sequence of measures $(\mathbb{P}\_{n})\_{n\geq{1}}$ on $C[0,T]$ is tight if:
1. For any $\varepsilon>0$ there exist $M>0$ and $n\_{0}\in{\mathbb{N}}$ such that for all $n\geq{n\_{0}}$,
$$
\mathbb{P}\_{n}(\{x:|x(0)|\geq{M}\})\leq{\varepsilon}
$$
2. For all $\varepsilon>0$ there exists $\delta>0$ and $n\_{0}\in{\mathbb{N}}$ such that for all $n\geq{n\_{0}}$,
$$
\delta^{-1}\mathbb{P}\_{n}(\{x:\max\limits\_{t\leq{s}\leq{t+\delta}}|x(s)-x(t)|\geq{\varepsilon}\})\leq{\varepsilon}
$$
for every $t\in{[0,T]}$.
Observe that the processes $Y^{c}\_{t}=c^{-1/2}X\_{ct}$ are all $0$ at $0$, so the first criteria is satisfied. For the second criteria, repeating our Gronwall- type argument for the 4th moment in place of the 2nd moment tells us that:
$$
C(T):=\max\_{t\in{[0,T]}}\mathbb{E}X\_{t}^{4}<\infty
$$
for any $T>0$. Using a Tchebyshev- type argument we have:
\begin{align\*}
\mathbb{P}(\max\limits\_{t\leq{s}\leq{t+\delta}}|c^{-1/2}X\_{cs}-c^{-1/2}X\_{ct}|\geq{\varepsilon})&=\mathbb{P}\left(\max\limits\_{t\leq{s}\leq{t+\delta}}|\int\_{t}^{t+s}\sigma(X\_{cr})d\widetilde{B}\_{r}|\geq{\varepsilon}\right) \\
&\leq{\frac{\mathbb{E}\Big(\max\limits\_{t\leq{s}\leq{t+\delta}}|\int\_{t}^{t+s}\sigma(X\_{cr})d\widetilde{B}\_{r}|^{4}\Big)}{\varepsilon^{4}}} \\
&\leq{C\hspace{2pt}\mathbb{E}\Big(\int\_{t}^{t+\delta}\sigma(X\_{cr})d\widetilde{B}\_{r}\Big)^{4}\varepsilon^{-4}} \\
&\leq{C'\hspace{2pt}\mathbb{E}\Big(\int\_{t}^{t+\delta}\sigma^{2}(X\_{cr})dr\Big)^{2}\varepsilon^{-4}}
\end{align\*}
where $C,C'>0$ are absolute constants. The inequality on the third line follows by Doob's inequality, using the fact that the process $\Big(\int\_{t}^{t+s}\sigma(X\_{cr})\widetilde{B}\_{r}\Big)\_{s\geq{0}}$ is a martingale. The inequality on the fourth line follows by applying the Burkholder- Davis- Gundy inequality (<https://almostsuremath.com/2010/04/06/the-burkholder-davis-gundy-inequality/>). Using the fact that $\sigma$ is Lipschitz,
$$
\int\_{t}^{t+\delta}\sigma^{2}(X\_{cr})dr\leq{\int\_{t}^{t+\delta}\big(2\sigma^{2}(0)+2K^{2}X\_{cr}^{2}\big)dr}\leq{2\delta\sigma^{2}(0)+2\delta{K^{2}}\max\_{t\leq{r}\leq{t+\delta}}X\_{cr}^{2}dr}
$$
$$
\Big(\int\_{t}^{t+\delta}\sigma^{2}(X\_{cr})dr\Big)^{2}\leq{8\delta^{2}\sigma^{4}(0)+8\delta^{2}{K^{4}}\max\_{t\leq{r}\leq{t+\delta}}X\_{cr}^{4}dr}
$$
Thus:
\begin{align\*}
\mathbb{P}(\max\limits\_{t\leq{s}\leq{t+\delta}}|c^{-1/2}X\_{cs}-c^{-1/2}X\_{ct}|\geq{\varepsilon})\leq{8C'\varepsilon^{-4}\delta^{2}\Big(\sigma^{4}(0)+K^{4}\mathbb{E}\max\_{t\leq{r}\leq{t+\delta}}X\_{cr}^{4}\Big)}
\end{align\*}
Finally, since $(X\_{t})\_{t\geq{0}}$ is a martingale, by Doob's inequality,
$$
\mathbb{E}\big(\max\_{r\in{[t,t+\delta]}}X^{4}\_{cr}\big)\leq{\mathbb{E}\big(\max\_{r\in{[0,T]}}X^{4}\_{r}\big)}\leq{C\hspace{2pt}\mathbb{E}X\_{T}^{4}}\leq{C\cdot{C(T)}}
$$
Thus, we see that taking $\delta<C''\varepsilon^{4}$ for some constant $C''>0$ we have that:
$$
\mathbb{P}(\max\limits\_{t\leq{s}\leq{t+\delta}}|c^{-1/2}X\_{cs}-c^{-1/2}X\_{ct}|\geq{\varepsilon})\leq{\delta\cdot{\varepsilon}}
$$
for any $t\in{[0,T]}$. This completes our proof.
| 3 | https://mathoverflow.net/users/80052 | 438499 | 177,116 |
https://mathoverflow.net/questions/438477 | 3 | Let $F$ be a non-archimedean field and let $\pi$ be a self-dual supercuspidal representation of $\mathrm{GL}\_n(F)$ (which, by a result of Adler exists only when $n=1$ or $n$ is even). Then, under LLC for $\mathrm{GL}\_n$ we obtain a $n$-dimensional irreducible self-dual representation $\varphi\_\pi$ of the Weil group $W\_F$.
>
> Is there an interpretation of the Frobenius-Schur indicator of $\varphi\_\pi$ in terms of the representation $\pi$?
>
>
>
It is *not* naively the Frobenius-Schur indicator of the representation $\pi$, since Corollary 9 of Mishra "A note on sign of a self-dual representation" says the Frobenius-Schur indicator of $\pi$ is always $1$.
| https://mathoverflow.net/users/123673 | Frobenius-Schur indicator of a self-dual L-parameter | Prasad and Ramakrishnan ([arXiv link](https://arxiv.org/abs/0807.0240)) study how the signs of (discrete series) representations behave along the local Langlands correspondence, not just for GL($n$), but for all inner forms. Assume the base field has characteristic 0.
If $n$ is odd, $\pi$ and its Langlands parameter have the same sign.
If $n$ is even, you need to use the Jacquet-Langlands correspondence (at least to use Prasad-Ramakrishnan's result to get the sign of the parameter). Since $\pi$ is supercuspidal, then it corresponds to a representation $\pi\_D$ on a division algebra $D$, which does not always have the same sign as $\pi$. Here the signs of $\pi\_D$ and the Langlands parameter are opposite.
| 1 | https://mathoverflow.net/users/6518 | 438505 | 177,117 |
https://mathoverflow.net/questions/317633 | 4 | Let $Y$ be a simplicial complex and let $\{Y\_i\}\_{i\in I}$ be a set of subcomplexes of $Y$ such that $\bigcup\_{i\in I}Y\_i=Y$. Let $\mathcal N$ be the nerve of this covering, and assume that for each finite $J\subset I$, we have that $\bigcap\_{j\in J}Y\_j$ is either empty or contractible.
One version of the *nerve theorem* says that, in the above situation, $Y$ is homotopy equivalent to $\mathcal N$.
I'm interested in the proof in the specific situation where we *cannot assume* that each simplex of $Y$ is contained in finitely many of the $Y\_i$.
The theorem is stated in the above generality in [1], as Theorem 10.6. In the proof given there, it is assumed "for convenience" that the preceding local finiteness assumption holds, and that in the general case one uses a "slightly different" argument.
I tried a bit to deduce it from other versions of the nerve theorem (see e.g. [2]), say by trying to replace $\{Y\_i\}$ by an open covering with a similar nerve. (I had some difficulty finding a more detailed proof in the literature, and I would like to see one to see if it's possible to modify it slightly.)
What is the "slightly different" argument mentioned in [1], or where can it be found?
[1]: *Björner, Anders*, Topological methods, Graham, R. L. (ed.) et al., Handbook of combinatorics. Vol. 1-2. Amsterdam: Elsevier (North-Holland). 1819-1872 (1995). [ZBL0851.52016](https://zbmath.org/?q=an:0851.52016).
[2]: Proposition 4G.2 and Corollary 4G.3 in:
*Hatcher, Allen*, Algebraic topology, Cambridge: Cambridge University Press (ISBN 0-521-79540-0/pbk). xii, 544 p. (2002). [ZBL1044.55001](https://zbmath.org/?q=an:1044.55001).
| https://mathoverflow.net/users/76590 | Nerve theorem for locally infinite covers by subcomplexes | I asked myself exactly this question the other day (while looking back at Björner's handbook article), and I poked around in Björner's papers looking for an answer. My guess is that Björner was referring to the argument, attributed to Quillen, that is found on p. 92 of his article [Homotopy type of posets and lattice complementation](https://www.sciencedirect.com/science/article/pii/009731658190042X).
Quillen's argument is a great little microcosm of homotopy theoretical ideas. While reminiscent of the one in Björner's Handbook article, instead of using a map that goes directly between the original complex and the nerve, Quillen cooks up a third space that maps to both by homotopy equivalences (a tried and true technique). The third space consists of all pairs $(\sigma, F)$ where $\sigma$ is a simplex in $K$ and $F$ is an element of the nerve containing $\sigma$ in its intersection (that is, $F$ is a finite subset of $I$, and $\sigma\in Y\_i$ for all $i\in I$); this is a poset, via the natural orderings on $K$ and on $\mathcal{N}$ by set-theoretic inclusion. Note that this poset is kind of like the graph of a multifunction version of the map Björner uses in the Handbook article: Björner shows that the map sending $\sigma$ to the largest such subset $F$ is a homotopy equivalence, but when no largest $F$ exists, instead of making a choice, Quillen just accepts the full swath of choices. As is so often the case in homotopy theory, the indeterminacy is in some sense contractible (the poset consisting of all $F$ containing $\sigma$ is the face poset of an infinite simplex), and so everything works out. Of course Quillen needs a version of his Fiber Theorem (aka Theorem A).
| 3 | https://mathoverflow.net/users/4042 | 438506 | 177,118 |
https://mathoverflow.net/questions/438488 | 7 | Let $0<\alpha<1$ and define
$$Tf(x):=\int e^{\dot{\imath} x y} \frac{f(y)}{|x-y|^{\alpha}}dy.$$
The Hardy-Littlewood-Sobolev inequality characterizes $L^p-L^q$ boundedness of
$Hf(x):=\int \frac{f(y)}{|x-y|^{\alpha}}dy$ which majorizes $|Tf(x)|$ by the triangle inequality, when $f>0$. It asserts that
$$ \|Tf\|\_{q}\leq C \|f\|\_{p}\qquad (1)$$
where $C$ is a constant independent of $f$,
if and only if
$$1<p,q<\infty\quad\frac{1}{p}-\frac{1}{q}=1-\alpha\qquad (2)$$
The necessity of (2) for (1) can be easily deduced by scaling.
My question:
Is (2) necessary for the inequality
$$ \|Tf\|\_{q}\leq C\_{1} \|f\|\_{p}\qquad (3)\qquad ?$$
It seems that $T$ does not have any scaling properties. I cannot come up with counterexamples that show that (3)
is false outside the range of exponents described by (2).
Is the boundedness of $T$ known ?
| https://mathoverflow.net/users/116555 | $L^p-L^q$ boundedness of this simple singular oscillatory integral operator | Let me consider instead the operator
$$Sf(x):=\int e^{-i x y} \frac{f(y)}{|x-y|^{\alpha}}dy$$
which has the same properties as $T$ since $Tf(x)=Sf(-x)$. Writing
$$e^{-i x y}=e^{i|x-y|^{2}/2}e^{-i|x|^{2}/2}e^{-i|y|^{2}/2}$$
the operator $S$ can be written
$$Sf(x)=e^{-i|x|^{2}/2}\int \frac{e^{i|x-y|^{2}/2}}{|x-y|^{a}}
e^{-i|y|^{2}/2}f(y)dy.$$
Thus $T:L^{q}\to L^{p}$ is bounded iff convolution with the oscillating kernel $|x|^{-a}e^{i|x|^{2}/2}$ is bounded. This kind of kernel has been studied extensively, for instance in: P.Sjolin, Convolution with Oscillating Kernels, Indiana University Mathematics Journal Vol. 30, No. 1 (January–February, 1981), pp. 47-55.
The range of values of $p,q$ is larger than (2), e.g. Sjolin proves boundedness $L^{p}\to L^{p}$ for $p\in[p\_{0},p\_{0}']$ where $p\_{0}=\frac{2n}{n+a}$, provided $0\le a<n$. Thus the answer to your question seems to be: no, condition (2) is not necessary for boundendess of $T$.
| 8 | https://mathoverflow.net/users/7294 | 438507 | 177,119 |
https://mathoverflow.net/questions/438371 | 5 | Firstly, a small question of nomenclature. If $(S,\bullet)$ is a magma, is there good terminology to relate $a$ to $b$ when
$$a\bullet b=b=b\bullet a?$$
Can we say that $b$ *absorbs* $a$? Can we say that $a$ is $b$-invariant? Or $b$ is $a$-invariant?
---
Now, for my question.
Let $C(\mathbb{G})$ be an algebra of continuous functions on a compact quantum groups with comultiplication $\Delta$. There is no assumption that that $C(\mathbb{G})$ is any particular completion of the algebra of regular functions, so in particular we don't have that the Haar state is faithful (in fact in my application it will be in the case that the Haar state is not faithful).
Let $\mathbb{H}\subseteq \mathbb{G}$ be a quantum subgroup in the sense that there is a quotient map $\pi:C(\mathbb{G})\to C(\mathbb{H})$ such that:
$$\Delta\_{C(\mathbb{H})}\circ \pi=(\pi\otimes \pi)\circ \Delta.$$
Where $h\_{\mathbb{H}}$ is the Haar state on $C(\mathbb{H})$, the state $\phi\_{\mathbb{H}}:=h\_{\mathbb{H}}\circ \pi$ on $C(\mathbb{G})$ is called a *Haar idempotent*, in particular:
$$\phi\_{\mathbb{H}}\star \phi\_{\mathbb{H}}=\phi\_{\mathbb{H}}\text{ where }\varphi\_1\star \varphi\_2=(\varphi\_1\otimes \varphi)\Delta.$$
Consider a state $\varphi$ on $C(\mathbb{G})$ such that:
$$\varphi\star \phi\_{\mathbb{H}}=\phi\_{\mathbb{H}}=\phi\_{\mathbb{H}}\star \varphi.$$
Is it the case that there exists a state $\varphi\_0$ on $C(\mathbb{H})$ such that:
$$\varphi=\varphi\_0\circ \pi?$$
---
I am happy to provide further definitions and facts around this question if required.
| https://mathoverflow.net/users/35482 | States "absorbed" by a Haar idempotent on a compact quantum group | Without making the assumption that $C(\mathbb{H})$ is the universal C$^\*$-algebra of the compact quantum group $\mathbb{H}$, the answer is negative. Whenever $\mathbb{G}$ is not co-amenable, we could take $C(\mathbb{H}) = C\_r(\mathbb{G})$ and $C(\mathbb{G}) = C\_u(\mathbb{G})$. We define $\pi : C\_u(\mathbb{G}) \to C\_r(\mathbb{G})$ as the canonical quotient homomorphism. When $h$ denotes the Haar state on $C\_r(\mathbb{G})$, the state $h\_1 = h \circ \pi$ is the Haar state on $C\_u(\mathbb{G})$. Therefore, the equality $\varphi \ast h\_1 = h\_1 = h\_1 \ast \varphi$ holds for all states $\varphi$ on $C\_u(\mathbb{G})$. Since $\pi$ is not faithful, there are states on $C\_u(\mathbb{G})$ that do not factor through $\pi$.
However, if we assume that $C(\mathbb{H})$ is the universal C$^\*$-algebra of $\mathbb{H}$, the answer is positive. The result and proof go as follows. It combines some analytic and algebraic considerations. As far as I know, such a result is not available in the literature, but other users might know a reference!
**Proposition.** Let $\mathbb{G}$ and $\mathbb{H}$ be compact quantum groups and $\pi : C(\mathbb{G}) \to C(\mathbb{H})$ a surjective unital $\*$-homomorphism satisfying $(\pi \otimes \pi) \circ \Delta = \Delta \circ \pi$. Assume that $C(\mathbb{H})$ is the universal enveloping C$^\*$-algebra of the Hopf algebra $\operatorname{Pol}(\mathbb{H})$. Denote by $h$ the Haar state on $\mathbb{H}$ and write $h\_1 = h \circ \pi$. Let $\varphi$ be a state on $C(\mathbb{G})$. Then the following statements equivalent.
1. There exists a state $\psi$ on $C(\mathbb{H})$ such that $\varphi = \psi \circ \pi$.
2. We have that $\varphi \ast h\_1 = h\_1 = h\_1 \ast \varphi$.
3. We have that $\varphi \ast h\_1 = h\_1$.
**Proof.** The implications $1 \Rightarrow 2 \Rightarrow 3$ are trivial. Assume that 3 holds. Denote by $c\_0(\widehat{\mathbb{G}})$ the $c\_0$ direct sum of the matrix algebras $B(H\_\alpha)$, $\alpha \in \operatorname{Irrep}(\mathbb{G})$. We denote by $\ell^\infty(\widehat{\mathbb{G}})$ the $\ell^\infty$ direct sum. Denote by $W \in M(C(\mathbb{G}) \otimes c\_0(\widehat{\mathbb{G}}))$ the corresponding direct sum of unitary representations of $\mathbb{G}$. We similarly define $V \in M(C(\mathbb{H}) \otimes c\_0(\widehat{\mathbb{H}}))$ for the compact quantum group $\mathbb{H}$. The morphism $\pi$ dualizes to a morphism $\widehat{\pi} : \ell^\infty(\widehat{\mathbb{H}}) \to \ell^\infty(\widehat{\mathbb{G}})$ satisfying $(\pi \otimes \text{id})(W) = (\text{id} \otimes \widehat{\pi})(V)$.
Denote by $p\_\varepsilon \in c\_0(\widehat{\mathbb{H}})$ the minimal central projection that corresponds to the trivial representation of $\mathbb{H}$. Write $q = \widehat{\pi}(p\_\varepsilon)$. Note that $(h\_1 \otimes \text{id})(W) = q$. Define $T \in \ell^\infty(\widehat{\mathbb{G}})$ by $T = (\varphi \otimes \text{id})(W)$. Since $(\Delta \otimes \text{id})(W) = W\_{13} W\_{23}$ and $\varphi \ast h\_1 = h\_1$, we get that $T = T q$. Write $R = W(1 \otimes q) - (1 \otimes q)$. Because
$$R^\* R = 2 (1 \otimes q) - (1 \otimes q) W (1 \otimes q) - (1 \otimes q) W^\* (1 \otimes q) \; ,$$
we conclude that $(\varphi \otimes \text{id})(R^\* R) = 0$. Since $\varphi$ is a state on a C$^\*$-algebra, it follows that
$$(\varphi \otimes \text{id} \otimes \text{id})(W\_{12} R\_{13}) = 0 \; .$$
This means that
$$(\varphi \otimes \text{id} \otimes \text{id})(W\_{12} W\_{13}) \, (1 \otimes q) = T \otimes q \; .$$
Write $T$ as the direct sum of the matrices $T\_\alpha \in B(H\_\alpha)$ for $\alpha \in \operatorname{Irrep}(\mathbb{G})$. So, for all $\alpha,\beta,\gamma \in \operatorname{Irrep}(\mathbb{G})$ and for all $X \in \operatorname{Mor}\_\mathbb{G}(\alpha,\beta \otimes \gamma)$ and $Y \in \operatorname{Mor}\_\mathbb{H}(\varepsilon,\gamma)$, we get that
$$T\_\alpha \, X^\*(1 \otimes Y) = X^\*(1 \otimes Y) \, T\_\beta \; .$$
By taking linear combinations, the same holds if $\gamma$ is any finite dimensional unitary representation of $\mathbb{G}$, not necessarily irreducible.
We now deduce that $T\_\alpha \, Z = Z \, T\_\beta$ for all $\alpha,\beta \in \operatorname{Irrep}(\mathbb{G})$ and $Z \in \operatorname{Mor}\_\mathbb{H}(\beta,\alpha)$. To prove this, choose solutions of the conjugate equations for $\alpha,\beta \in \operatorname{Irrep}(\mathbb{G})$, given by $t \in \operatorname{Mor}\_\mathbb{G}(\varepsilon,\overline{\beta} \otimes \beta)$ and $s \in \operatorname{Mor}\_\mathbb{G}(\varepsilon,\beta \otimes \overline{\beta})$. Writing $\gamma = \overline{\beta} \otimes \alpha$, $X = s \otimes 1$ and $Y = (1 \otimes Z)t$, we get that $X^\*(1 \otimes Y) = Z$ and the claim follows.
This means that $T = \widehat{\pi}(S)$ for some $S \in \ell^\infty(\widehat{\mathbb{H}})$.
Every $a \in \operatorname{Pol}(\mathbb{G})$ is of the form $a = (\text{id} \otimes \omega)(W)$ where $\omega$ is a uniquely determined, ``finitely supported'' functional on $\ell^\infty(\widehat{\mathbb{G}})$. Note that $\pi(a) = 0$ if and only if $\omega \circ \widehat{\pi} = 0$. Since $(\varphi \otimes \text{id})(W) = T = \widehat{\pi}(S)$, we conclude that there is a well defined linear functional $\psi : \operatorname{Pol}(\mathbb{H}) \to \mathbb{C}$ such that $\psi(\pi(a)) = \varphi(a)$ for all $a \in \operatorname{Pol}(\mathbb{G})$.
Then $\psi(1) = \varphi(1) = 1$ and, because $\pi : \operatorname{Pol}(\mathbb{G}) \to \operatorname{Pol}(\mathbb{H})$ is surjective, $\psi(b^\* b) \geq 0$ for all $b \in \operatorname{Pol}(\mathbb{H})$. Since $C(\mathbb{H})$ is the universal C$^\*$-algebra of the compact quantum group $\mathbb{H}$, it follows that $\psi$ uniquely extends to a state on $C(\mathbb{H})$ that we still denote as $\psi$. By construction, $\varphi = \psi \circ \pi$.
| 2 | https://mathoverflow.net/users/159170 | 438517 | 177,125 |
https://mathoverflow.net/questions/438523 | 2 | An $\mathbf{Ab}$-category is a category enriched over the category of abelian groups. What is an example of a category that can be enriched over abelian groups in more than one way?
An abelian category is a very special type of $\mathbf{Ab}$-category. So can a category be given the structure of an abelian category in more than one way, or is being abelian unique and hence a property of the category?
| https://mathoverflow.net/users/491434 | Can a category be enriched over abelian groups in more than one way? | You can easily find examples among categories with one element: a category with one element is a (multiplicative) monoid, and $Ab$-enrichment over it is a choice of an addition which turns it into a ring. And there can be multiple such additive structures: you can for instance consider pullback along a permutation which preserves multiplication but not addition.
More interestingly, if we instead work with additive categories rather than preadditive ones (meaning we require all finite products to exist), then it turns out $Ab$-enrichment is necessarily unique. This in particular applies to abelian categories. Indeed, given two morphisms $f,g:A\to B$, we can characterize $f+g$ as the composition
$$A\to A\oplus A\to B\oplus B\to B,$$
where the first map is the diagonal map, the last map is the codiagonal, and the middle map is the map which you can think of being $f,g$ on the components. See [Wikipedia](https://en.wikipedia.org/wiki/Additive_category#Internal_characterisation_of_the_addition_law) for some more details.
| 15 | https://mathoverflow.net/users/30186 | 438525 | 177,126 |
https://mathoverflow.net/questions/438504 | 2 | $\DeclareMathOperator\SL{SL}$Let $p$ be an odd prime and let $ \SL^1\_2(\mathbb{Z}\_p)$ denote the kernel of the natrual surjective morphism $\SL\_2(\mathbb{Z}\_p)\rightarrow \SL\_2(\mathbb{Z}\_p/p\mathbb{Z}\_p)$. Then the abelianization of $\SL^1\_2(\mathbb{Z}\_p)$ is $(\mathbb{Z}/p\mathbb{Z})^3$. Recall that there are only two nonabelian groups with order $p^3$ up to isomorphism as follows:
1. $G\_1=\langle x,y\mid x^{p^2}=y^p=1,~y^{-1}xy=x^{1+p}\rangle$.
2. $G\_2=\langle x,y,z\mid x^p=y^p=z^p=1,~xz=zx,~zy=yz,~xy=yxz\rangle$.
**Question:** Is there a continuous surjective morphism $\SL^1\_2(\mathbb{Z}\_p)\to G\_i$ for $i=1,2$?
| https://mathoverflow.net/users/149460 | Finite quotients of $p$-adic congruence subgroups of $\operatorname{SL}_2$ | $\DeclareMathOperator\SL{SL}\newcommand{\Z}{\mathbf{Z}}$Indeed, $G\_1$ is a quotient, but not $G\_2$.
First, one checks that the third term in the lower central series is the kernel of $\SL\_2^1(\Z\_p)\to\SL\_2^1(\Z/p^3\Z)$. So this converts this into a question about quotients of a group of order $p^6$, namely $\SL\_2^1(\Z/p^3\Z)$.
For this group, the derived subgroup is central and generated by $p$-powers. This already implies that every quotient of $\SL\_2^1(\Z/p^3\Z)$ of exponent $p$ is abelian, and hence $G\_2$ is not a quotient.
One gets $G\_1$ by modding out by the normal subgroup of elements of the form $\begin{pmatrix}1+p^2a & 0\\ pc & 1-p^2a\end{pmatrix}$ for $a\in\Z/p\Z$, $c\in\Z/p^2\Z$.
| 4 | https://mathoverflow.net/users/14094 | 438526 | 177,127 |
https://mathoverflow.net/questions/436414 | 1 | Suppose we have a sympletic toric manifold $(M,\omega)$ of dimension $4$ and let $\triangle$ be its corresponding Delzant polytope. Suppose that this polytope is "nice" enough so that we are able to defined a map, using action-angle coordinates $\Psi:\triangle \times \mathbb{T}^2\rightarrow \triangle \times \mathbb{T}^2$
\begin{equation}
\Psi(x\_1,x\_2,\theta\_2,\theta\_2)=(-x\_1,-x\_2,-\theta\_1,-\theta\_2).
\end{equation}
Examples of Delzant polytopes where we could define this map would be a square, octagon, etc.… We would just need to have enough symmetries. We are assuming that the polygon is centered at the origin.
Now this map will be a symplectomorphism, and what I have been wondering is if this could be a Hamiltonian symplectomorphism ? I tried some approaches to prove this but I got nowhere. Basically my idea was to use Banyaga's theorem, since we know that $H\_1(M)=\{0\}$, we would just need to prove that $\Psi$ is in the connected component of the identity in $\operatorname{Symp}(M)$, however I was not able to construct such a family of maps.
Then, I remebered the Arnol'd conjecture, and when we can talk about Floer homology, this basically gives us a lower bound on the number of fixed points of an Hamiltonian symplectomorphism in terms of Betti numbers of $M$. So here $\Psi$ has $4$ fixed points, and so if the topology of $M$ is complicated enough we could come into some trouble, assuming that we were in a position to talk about Floer homology.
So right now I'm inclining to the fact that generally this won't be a Hamiltonian symplectomorphism because I'm not sure how $H\_2(M)$ behaves, but I would like to hear some input on this. What do you think ?
| https://mathoverflow.net/users/155363 | Could this be a Hamiltonian symplectomorphism on a symplectic toric manifold | Sometimes it is, sometimes it isn't.
The spheres living over the edges generate the second homology, so you can read off the action on $H\_2$ from that. For $S^2\times S^2$ (square) the action on homology is trivial (because opposite edges are homologous) and the symplectomorphism is indeed Hamiltonian. For a hexagon (3-point blow up of $P^2$) the action on homology is nontrivial.
| 2 | https://mathoverflow.net/users/10839 | 438531 | 177,130 |
https://mathoverflow.net/questions/438203 | 6 | Suppose that $\mathbb{R}^n$ is the maximal group that can act properly on a manifold $N$ and $\mathbb{R}^m$ is the maximal group that can act properly on a manifold $M$ ( i.e, $\mathbb{R}^{m+1}$ can't act properly on $M$ ).
Question: is $\mathbb{R}^{n+m}$ the maximal group that can act properly on the manifold $M\times N$ ? ( i.e $\mathbb{R}^{n+m+1}$ can't acy properly on $M\times N$ )
| https://mathoverflow.net/users/497616 | Proper action on product manifold | First, let's formulate the question properly:
Given a topological space $X$, define be
$$
d(X):=\sup \{ n: X~ \hbox{is homeomorphic to} ~Y\times {\mathbb R}^n\}.
$$
**Lemma.** The following quantities are equal when $X$ is a manifold:
(1) $d(X)$
(2) P(X):=$\max\{n: \exists ~ \hbox{a principal}~ {\mathbb R}^n\hbox{-bundle with the total space}~X\}$
(3) p(X):=$\max\{n: \exists ~ \hbox{a proper}~ {\mathbb R}^n\hbox{-action on }~X\}$, where ${\mathbb R}^n$ is equipped with the standard topology.
Proof. A principal ${\mathbb R^n}$-bundle with the total space $X$ is the same thing as a proper ${\mathbb R^n}$-action on $X$. At the same time, each principal ${\mathbb R^n}$-bundle with the total space $X$ is trivial. qed.
Remark. One can avoid using this lemma in order to justify the example below, just I find the definition $d(X)$ cleaner.
Thus, working in the topological category, you are asking if there are manifolds $X, Y$ such that $d(X\times Y)> d(X)+ d(Y)$.
Now, here is an example: Let $X$ be the Whitehead manifold (or any other contractible 3-manifold not homeomorphic to ${\mathbb R}^3$). Then $d(X)=0$. (This is not entirely trivial, but for the purpose of a counter-example we just need to know that $p(X)<3$ which is obvious since $X$ is not homeomorphic to ${\mathbb R}^3$.) On the other hand, $X\times {\mathbb R}$ is homeomorphic to ${\mathbb R}^4$ (see for instance [here](https://mathoverflow.net/questions/424165/how-to-prove-the-product-of-whitehead-manifold-and-mathbbr-is-homeomorphic)), hence, $d(X\times {\mathbb R})=4$.
The same example works in the smooth category.
---
Update. It appears that you are now interested in proper ${\mathbb Z}^n$-actions. The answer in this setting is the same. You similarly define the invariant $c(X)$, detecting the highest rank of a discrete free abelian group acting properly on $X$. Then, let $X$ again be the Whitehead manifold. It turns out that $c(X)=0$. This is a nontrivial result of Bob Myers:
*Myers, Robert*, [**Contractible open 3-manifolds which are not covering spaces**](http://dx.doi.org/10.1016/0040-9383(88)90005-5), Topology 27, No. 1, 27-35 (1988). [ZBL0658.57007](https://zbmath.org/?q=an:0658.57007).
(One can also appeal to an older theorem of Waldhausen, but it only proves that $c(X)\le 2$, which suffices for our purposes but is suboptimal.)
At the same time, $c(X\times {\mathbb R})=4$, but $c(X)
+ c({\mathbb R})=1$.
| 8 | https://mathoverflow.net/users/39654 | 438532 | 177,131 |
https://mathoverflow.net/questions/438494 | 0 | I'm wondering what the solutions of complex linear difference equations like
\begin{equation}
f(z+\eta\_1)+f(z+\eta\_2)+\ldots+f(z+\eta\_n)=0,\ \ \ \eta\_1 \cdots\eta\_n \in \mathbf{C}
\end{equation}
look like. It seems easy when $\eta\_i$ are all integers, but I don't know the general case.
If we cannot get the exact form of solutions, can we get the order?
**remark.** the order of a meromorphic function is define to be $\rho(f) = \overline{\lim}\frac{\log T(r,f)}{\log r}$, where $T(r,f)$ represents the Nevanlinna characteristic of $f(z)$. If $f(z)$ is an entire function, $\rho(f) = \overline{\lim} \frac{\log\log M(r,f)}{\log r}$ where $M(r,f) = \sup\_{|z| < r} f(z)$.
| https://mathoverflow.net/users/497814 | Solutions of complex linear difference equations | Look for a solution of the form $f(z)=e^{\lambda z}$. Plugging this to your equation, you obtain that
$\lambda$ must be a zero of the entire function
$$F(\lambda)=\sum\_{j=1}^n e^{\lambda\eta\_j}.$$
When there are at least two distinct $\eta\_j$, this entire function $F$ has infinitely many zeros $\lambda\_k$.
Then any finite linear combination
$$f(z)=\sum\_{k} c\_ke^{\lambda\_k z}$$ is
a solution. Also limits of such functions will be solutions.
If $F$ has a multiple root $\lambda\_k$ there will be also solutions
of the form $P\_k(z)e^{\lambda\_k z}$, where $P$ is a polynomial.
There is a theorem of Malgrange (Theorem 16.4.1 in Hormander's Analysis of linear partial differential operators) which says that these are all solutions.
(Linear combinations of exponentials are dense in the set of all solutions).
Since the coefficients $c\_j$ can be arbitrary (subject to the condition that the series converges), you can obtain entire solutions of arbitrarily fast growth.
| 2 | https://mathoverflow.net/users/25510 | 438533 | 177,132 |
https://mathoverflow.net/questions/438511 | 4 | Let $A$ be an $n\times n$ random matrix with i.i.d entries (say standard Gaussian) $A\_{ij}$. I know that there is a CLT type result known for the determinant of $A$. More precisely there is a CLT for $\log(|\det(A)|)$. For any fixed $(i, j)$, If we look at the minor $M\_{ij}$ of $A$, it follows that $\log(|M\_{ij}|)$ also satisfies the same CLT.
I want to know if anything is known about the joint distribution of $(M\_{11}, M\_{22})$? I would be more generally interested in understanding the joint law of $(M\_{ij}: 1\leq i, j\leq k)$ for an arbitrary (but fixed) $k$. In particular, can one expect any kind of asymptotic independence between $( M\_{11}, M\_{22})$ as $n\to \infty$?
| https://mathoverflow.net/users/69849 | Joint distribution of minor of Wigner Hermitian matrices | There is certainly no asymptotic independence between $\det M\_{11}, \det M\_{22}$. From the base times height formula for parallelepipeds we see that
\begin{align\*} \frac{|\det M\_{12}|}{|\det M\_{22}|} &= \frac{|v\_1 \wedge v\_3 \wedge \dots \wedge v\_n|}{|v\_2 \wedge v\_3 \wedge \dots \wedge v\_n|}\\
&= \frac{\mathrm{dist}(v\_1, V)}{\mathrm{dist}(v\_2,V)}\\
&= \frac{|v\_1 \cdot u|}{|v\_2 \cdot u|}
\end{align\*}
where $v\_i = (A\_{i1}, A\_{i3}, \dots, A\_{in})$, $V \subset {\bf R}^{n-1}$ is the span of $v\_3,\dots,v\_n$, and $u$ is a unit normal to $V$. For fixed $V$, $v\_1 \cdot u, v\_2 \cdot u$ are independent gaussians, and so we conclude that $\log |\det M\_{12}| - \log |\det M\_{22}|$ is bounded in probability, and similarly for $\log |\det M\_{12}| - \log |\det M\_{11}|$, hence
$\log |\det M\_{11}| - \log |\det M\_{22}|$ is also bounded in probability. On the other hand, as you observed it is known that $\log |\det M\_{11}|$, $\log |\det M\_{22}|$ are asymptotically gaussian with variance comparable to $\log n$, so there is no asymptotic independence here.
| 9 | https://mathoverflow.net/users/766 | 438535 | 177,133 |
Subsets and Splits