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https://mathoverflow.net/questions/403617
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1
|
Given a complex vector space $V$ of dimension $n>2$, the universal conic $\mathcal C$ of $\mathbb P(V^\*)$ is a divisor in $\mathbb P(t^\*\mathcal E\_3)\overset{\pi}{\rightarrow}\mathbb P({\rm Sym}^2\mathcal E\_3^\*)\overset{t}{\rightarrow} Gr(3,V)$ where $\mathcal E\_3$ is the natural rank $3$ quotient bundle on $Gr(3,V)$.
The Hilbert scheme of conics in $\mathbb P(V^\*)$ can be identified with $\mathbb P({\rm Sym}^2\mathcal E\_3^\*)$.
The class in ${\rm Pic}(\mathbb P(t^\*\mathcal E\_3))$ of $\mathcal C$ is of the form $\pi^\*\mathcal L\otimes \mathcal O\_{\mathbb P(t^\*\mathcal E\_3)}(2)$ for some line bundle $\mathcal L$ on $\mathbb P({\rm Sym}^2\mathcal E\_3^\*)$.
Is there a simple way to identify $\mathcal L$?
|
https://mathoverflow.net/users/85595
|
Class of the universal conic
|
The projective bundle $t \colon \mathbb{P}(\mathrm{Sym}^2\mathcal{E}\_3^\*) \to \mathrm{Gr}(3,V)$ comes with the tautological subbundle
$$
\mathcal{O}\_t(-1) \hookrightarrow t^\*\mathrm{Sym}^2\mathcal{E}\_3^\*.
$$
This embedding gives a global section in
$$
H^0(\mathbb{P}(\mathrm{Sym}^2\mathcal{E}\_3^\*),
\mathcal{O}\_t(1) \otimes p^\*\mathrm{Sym}^2\mathcal{E}\_3^\*)
\cong
H^0(\mathbb{P}(t^\*\mathcal{E}\_3),
\pi^\*\mathcal{O}\_t(1) \otimes \mathcal{O}\_\pi(2)),
$$
which is precisely the equation of the universal conic. So, $\mathcal{L} = \mathcal{O}\_t(1)$.
|
1
|
https://mathoverflow.net/users/4428
|
403622
| 165,575 |
https://mathoverflow.net/questions/403620
|
4
|
It can be proven without any form of infinite choice that the product of two compact spaces (and thus any finite product) is compact, while on the other hand, it is well known that the general form of Tychonoffs theorem implies the axiom of choice.
So a general formulation of this question would be: Is there a proof of Tychonoffs theorem where the use of AC/Zorn's lemma can be reduced to using the axiom of dependent choice up to the ordinal that indexes the product? With the case of countable products being of particular importance.
|
https://mathoverflow.net/users/174368
|
Can Tychonoffs theorem for a countable number of spaces be proven with ZF plus the axiom of (countable) dependent choice?
|
From Herrlich's "The Axiom of Choice", Proposition 4.72 reads as follows:
>
> Each of the following conditions implies the subsequent ones:
>
>
> 1. $\sf DC$.
> 2. Countable products of compact spaces are compact.
> 3. $\sf CC$.
>
>
>
He goes on to remark that (2) is provable from countable choice + the Boolean Ideal theorem, and therefore the implication from (1) to (2) is not reversible in $\sf ZF$. To my understanding, the question of whether (3) implies (2) is open.
(In [**Compactness in countable Tychonoff products and choice**](https://doi.org/10.1002/(SICI)1521-3870(200001)46:1%3C3::AID-MALQ3%3E3.0.CO;2-E) the authors say the theorem is due to Pincus, who did not publish it, and it appears in a paper by Wright.)
|
6
|
https://mathoverflow.net/users/7206
|
403623
| 165,576 |
https://mathoverflow.net/questions/403614
|
4
|
Most often than not, the sheaves appearing in algebraic geometry (with the Zariski topology) are $\mathcal{O}\_X$-modules, instead of simple abelian sheaves.
Now, when dealing with topological spaces (for example, in Verdier duality) and in étale cohomology, it seems that abelian sheaves have the main role.
I wonder why. For example, Verdier duality works just fine as a statement about $\mathcal{O}\_X$-modules (as in Spaltenstein's *Resolution of unbounded complexes*) and we recover the standard topological statements setting $\mathcal{O}\_X=\underline{\mathbb{Z}}$. Even more, since ringed spaces always have fibered products (and the underlying topological spaces are the usual fibered products), base change also works well.
(I'm not sure about the étale case. I would love to know more about it.)
Is there some fundamental reason *not* to consider $\mathcal{O}\_X$-modules in these cases?
|
https://mathoverflow.net/users/131975
|
Why abelian sheaves instead of $\mathcal{O}_X$-modules in topology and étale stuff?
|
We come to the question of what $\mathcal{O}\_X$ should mean, when $X$ is a manifold or topological space. If $\mathcal{O}\_X$ is smooth (or continuous) real valued functions, then we will always wind up studying $H^{\ast}(X, \mathbb{R})$. There is nothing wrong with this and, in fact, de Rham cohomology makes lots of use of sheaves of $\mathcal{O}\_X$-modules, such as the sheaf of differential $k$-forms. But you might want to study $H^{\ast}(X, \mathbb{Z})$.
Alternatively, one could decide that $\mathcal{O}\_X$ is the sheaf of locally constant $\mathbb{Z}$-valued functions. In this case, abelian sheaves are equivalent to $\mathcal{O}\_X$-modules, so it is just a matter of terminology which one you say you are studying.
In algebraic geometry, the sheaf of locally constant $\mathbb{Z}$-valued functions is flasque in the Zarsiki topology (if your underlying space is irreducible), so you always wind up either working in the etale topology, or else studying some variant of de Rham cohomology (D-modules, crystals, etc). It is only in the topological world that you get to work with locally constant sheaves of abelian groups in the ordinary topology and learn something interesting.
|
4
|
https://mathoverflow.net/users/297
|
403631
| 165,581 |
https://mathoverflow.net/questions/403643
|
3
|
**Context.**
*Space of Lipschitz functions.* Denote by $Lip\_0(D)$ the space of all Lipschitz functions on a metric space $D$ vanishing at some base point $e \in D$. The norm in $Lip\_0$ is defined as follows
$$
\|f\|\_{Lip\_0} := Lip(f),
$$
where $Lip(f)$ denotes the Lipschitz constant of $f$.
*Radon-Nikodym property (RNP).* There are many equivalent definitions of the RNP, I will give two of them.
Definition 1. Let $\Sigma$ be the $\sigma$-algebra of subsets of a set $\Omega$. A Banach space $X$ is said to have the RNP if for any measure $\mu \colon \Sigma \to X$ of bounded variation with values in $X$, and any finite positive scalar measure $\lambda \colon \Sigma \to \mathbb R$ such that $\mu$ is absolutely continuous w.r.t. $\lambda$, there exists a $\lambda$-Bochner integrable function $f$ such that $\mu(E) = \int\_E f \,d\lambda$ for all $E \in \Sigma$.
Theorem 1. A Banach space $X$ has the RNP if and only if every Lipschitz function $\mathbb R \to X$ is differentiable almost everywhere.
**Question.** Does the $Lip\_0$ space have the Radon-Nikodym property?
I have tried the following sources, but wasn't able to find an answer: Weaver, Lipschitz Algebras; Ryan, Introduction to Tensor Products of Banach Spaces; Diestel&Uhl, Vector Measures.
Any help will be much appreciated.
|
https://mathoverflow.net/users/160379
|
Does the space of Lipschitz functions have the Radon-Nikodym property?
|
Let $X$ be a metric space consisting of a countable set of points, the distance between any two of which is $2$, together with one additional point $e$ whose distance to any of the other points is $1$. Then ${\rm Lip}\_0(X)$ is isometrically isomorphic to $l^\infty$, which fails the RNP.
Another example: let $X = [0,1]$ with $e = 0$. Then ${\rm Lip}\_0(X)$ is isometrically isomorphic to $L^\infty[0,1]$, which fails the RNP.
|
8
|
https://mathoverflow.net/users/23141
|
403648
| 165,584 |
https://mathoverflow.net/questions/403629
|
20
|
What sort of bounds (explicit of preference) can one give for
$$\int\_T^{2 T} \frac{dt}{|\zeta(1+i t)|^2} \;\;\;\;\;?$$
Some obvious points:
* One can give a pointwise bound $\frac{1}{|\zeta(1+ it)|} \leq C \log t$ (with $C\leq 42.9$ for $t\geq 2$) and deduce a bound of the form $\leq K T (\log T)^2$ on the integral above. If possible, we would like to do better (asymptotically and also as far as the constants involved are concerned).
* To know that the integral is always finite, one needs to know that $\zeta(1+it)\ne 0$ for all $t$. Thus, it is not enought to just apply a mean-value theorem.
|
https://mathoverflow.net/users/398
|
Bound on $L^2$ norm of $1/\zeta(1+i t)$?
|
Problems like this are classical (as noted in Terry's answer), and have been considered more recently with attention to uniformity in the moments. To give a quick indication, one can show that
$$
\zeta(1+it) \approx \prod\_{p\le y} \Big( 1-\frac{1}{p^{1+it}}\Big)^{-1}
$$
for all but a set of values $T\le t\le 2T$ of small measure, provided $y$ is bigger than some power of $\log T$. Such ideas go back to Littlewood (on RH, which can then be removed via zero-density results). See Lemmas 2.1 and 2.2 in [Extreme values of $|\zeta(1+it)|$](https://arxiv.org/pdf/math/0501232.pdf) for example. From this one can work out uniform moments of $\zeta(1+it)$ (positive, negative, complex ...) and understand its distribution. Look in the cited paper, and also work of Lamzouri.
The result is
$$
\frac 1T \int\_T^{2T} |\zeta(1+it)|^{-2} dt \sim \sum\_{n=1}^{\infty} \frac{\mu(n)^2}{n^2} = \frac{\zeta(2)}{\zeta(4)},
$$
as one would expect.
An alternative (which may be useful if one wishes to get explicit results, and avoid zero-density estimates) is to use a (very) long approximation to $1/\zeta(1+it)$. For example, first show that
$$
\frac{1}{\zeta(1+it)} \approx \sum\_{n\le \exp((\log T)^3)} \frac{\mu(n)}{n^{1+it}}.
$$
This would need an explicit zero free region -- I would guess something off-the-shelf from the work of Kadiri et al should be good, and you can make the sum even longer. Usually such a long approximation would be useless, but on the $1$-line you can do whatever you want. Now one can show (easiest to smooth and unsmooth) that
$$
\int\_T^{2T} \frac{dt}{|\zeta(1+it)|^2} =
\sum\_{n\le \exp((\log T)^3) } (T+O(n)) \frac{\mu(n)^2}{n^2}
= \frac{\zeta(2)}{\zeta(4)} T + O((\log T)^3).
$$
It shouldn't be too bad to make this explicit, but not something I'd care to do!
|
13
|
https://mathoverflow.net/users/38624
|
403652
| 165,586 |
https://mathoverflow.net/questions/349409
|
6
|
Let $\mathscr{S}$ be a [limit sketch](https://ncatlab.org/nlab/show/sketch) in a small category $\mathcal{E}$, i.e. just a collection of cones in $\mathcal{E}$. Then its category $\mathbf{Mod}(\mathscr{S})$ of models (i.e. functors $\mathcal{E} \to \mathbf{Set}$ which send the cones in $\mathscr{S}$ to limit cones) is cocomplete, in fact locally presentable. It enjoys the following universal property in the $2$-category of cocomplete categories: If $\mathcal{C}$ is any cocomplete category, then
$$\mathrm{Hom}\_c(\mathbf{Mod}(\mathscr{S}),\mathcal{C}) \simeq \mathbf{Mod}\_{\mathcal{C}}(\mathscr{S}^{\mathrm{op}}).$$
Here, $\mathrm{Hom}\_c$ denotes the category of cocontinuous functors, $\mathscr{S}^{\mathrm{op}}$ denotes the dual *colimit* sketch, and $\mathbf{Mod}\_{\mathcal{C}}$ refers to $\mathcal{C}$-valued models.
A typical example of this universal property is $\mathrm{Hom}\_c(\mathbf{Grp},\mathcal{C}) \simeq \mathbf{CoGrp}(\mathcal{C})$, that is, $\mathbf{Grp}$ is the universal example of a cocomplete category with an internal cogroup object.
The proof is a combination of two well-known results, so I assume that this universal property is also well-known. **Can someone confirm this and point me to literature which I can cite for this result?**
Maybe it follows from some of the theorems in [Kelly's book](http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html) on enriched categories, chapter 6. But I am not sure.
---
Here is a sketch of the two-step proof I was thinking of. Let $G : \mathbf{Mod}(\mathscr{S}) \to [\mathcal{E},\mathbf{Set}] $ be the inclusion and $F : [\mathcal{E},\mathbf{Set}] \to \mathbf{Mod}(\mathscr{S})$ its left adjoint. Then by a "tensor-less" variant of Prop. 2.3.6. in [Tensor functors between categories of quasi-coherent sheaves](https://arxiv.org/pdf/1202.5147.pdf) (as you see, this is not quite the reference I need!) the category $\mathrm{Hom}\_c(\mathbf{Mod}(\mathscr{S}),\mathcal{C})$ is equivalent to the category of those cocontinuous functors $Q^\* : [\mathcal{E},\mathbf{Set}] \to \mathcal{C}$ such that $Q^\* \to Q^\* \circ G \circ F$ is an isomorphism; this is true iff the right adjoint $Q\_\* : \mathcal{C} \to [\mathcal{E},\mathbf{Set}]$ factors over $\mathbf{Mod}(\mathscr{S})$. By the well-known universal property of $[\mathcal{E},\mathbf{Set}]$ as the free cocompletion of $\mathcal{E}^{\mathrm{op}}$ (for instance, Section 4.4 in Kelly's book), $Q^\*$ corresponds to a functor $P : \mathcal{E}^{\mathrm{op}} \to \mathcal{C}$ via $Q^\* = P \otimes\_{\mathcal{E}} -$, and hence $Q\_\* = \mathrm{Hom}(P(-),-)$. By the Yoneda Lemma, the requirement that $Q\_\*$ factors over $\mathbf{Mod}(\mathscr{S})$ exactly means that $P$ is a model of $ \mathscr{S}^{\mathrm{op}}$.
---
I have found a reference: Theorem 2.2.4 in [this paper](https://arxiv.org/abs/1105.3104) (actually, I knew this paper, but I somehow forgot that the theorem is in there). But the authors also explicitly state that this theorem is well-known. I would be happy with a more "classical" or "original" reference.
|
https://mathoverflow.net/users/2841
|
Universal property of the cocomplete category of models of a limit sketch
|
Probably the earliest reference is Theorem 2.5 in
>
> A. Pultr, **The right adjoints into the categories of relational systems**, Reports of the Midwest Category
> Seminar IV. Springer, Berlin, Heidelberg, 1970
>
>
>
Pultr's "relational theories" are exactly small realized limit sketches.
Remark: I have recently generalized the result to large limit sketches ([arXiv:2106.11115](https://arxiv.org/abs/2106.11115)).
|
4
|
https://mathoverflow.net/users/2841
|
403653
| 165,587 |
https://mathoverflow.net/questions/403635
|
5
|
$\newcommand{\w}{\omega}\newcommand{\F}{\mathcal F}\newcommand{\I}{\mathcal I}\newcommand{\J}{\mathcal J}\newcommand{\M}{\mathcal M}\newcommand{\N}{\mathcal N}\newcommand{\x}{\mathfrak x}\newcommand{\cov}{\mathrm{cov}}\newcommand{\lac}{\mathrm{lac}}$[Taras Banakh](https://mathoverflow.net/users/61536/taras-banakh) and I proceed a long quest answering a [question](https://math.stackexchange.com/questions/1203722/approximate-vanishing-in-pontryagin-dual) of ougao at Mathematics.SE.
Recently we encountered a small cardinal $\x\_{\lac}$, which is the smallest cardinality of a family $\F$ of infinite subsets of $\w$ such that for any lacunary set $R$ there exists $F\in\F$ such that $F\cap R$ is finite. Recall that a set $R\subset\w\setminus\{0\}$ is called *lacunary*, if $\inf\{b/a:a,b\in R,\;a<b\}>1$.
It would be ideally for us to find a known small cardinal equal to $\x\_{\lac}$. While $\x\_{\lac}$ remains unknown, we are interested in bounds (especially lower) for it by known small cardinals.
*Our try*. For any family $\I$ of sets let $\cov(\I)=\min\{|\J|:\J\subseteq\I\;\wedge\;\bigcup\J=\bigcup\I\}$. Let $\M$ and $\N$ be the ideals of meager and Lebesgue null subsets of the real line, respectively. We can prove that $\cov(\M)\le \x\_{\lac}$ and are interested whether this bound can be improved and whether $\cov(\N)\le \x\_{\lac}$.
[Lyubomyr Zdomskyy](http://www.logic.univie.ac.at/%7Elzdomsky) suggested that it is consistent that $\mathfrak d<\x\_{\lac}$, where $\mathfrak d$ is the cofinality of $\w^\w$ endowed with the natural partial order: $(x\_n)\_{n\in\w}\le (y\_n)\_{n\in\w}$ iff
$x\_n\le y\_n$ for all $i$. We are interested whether $\x\_{\lac}\le \mathfrak a$, where $\mathfrak a$ is the minimum size of a maximal (with respect to inclusion) pairwise almost disjoint family of infinite subsets of $\omega$.
Thanks.
|
https://mathoverflow.net/users/43954
|
Bounds for a small cardinal
|
**EDIT:** In my original post, I showed that $\mathrm{cov}(\mathcal N) > \mathfrak{x}\_{lac}$ in the random model. Upon further reflection, I think we can prove a stronger result, with an arguably easier (but completely different) proof:
*Theorem:* $\mathfrak{x}\_{lac} \leq \mathrm{non}(\mathcal N)$.
Note that this implies $\mathfrak{x}\_{lac} < \mathrm{cov}(\mathcal N)$ in the random model, and it also implies the consistency of $\mathfrak{x}\_{lac} < \mathfrak{d}$.
In addition to this theorem, let me also point out that it is consistent to have $\mathfrak{a} < \mathrm{cov}(\mathcal M)$. (See Corollary 2.6 in [this](https://arxiv.org/pdf/math/9309205.pdf) paper of Brendle.) Therefore the lower bound $\mathrm{cov}(\mathcal M) \leq \mathfrak{x}\_{lac}$ mentioned in the post already implies $\mathfrak{a}$ is not an upper bound for $\mathfrak{x}\_{lac}$.
*Proof of the theorem:* Suppose we form an infinite set $B$ by choosing from each interval of the form $[2^k,2^{k+1})$ exactly one integer $b\_k$ at random, and then taking $B = \{b\_k :\, k \in \omega \}$. (By "at random" I mean that we choose with the uniform distribution, so each integer in $[2^k,2^{k+1})$ has probability $1/2^k$ of being selected.) I claim that if $A$ is lacunary, then it is almost surely true that $A \cap B$ is finite.
To see this, fix some $c > 1$ and some $n\_0 \in \omega$ such that if $a$ and $b$ are consecutive members of $A$ above $n\_0$, then $b/a > c$. If $k$ is large enough that $n\_0 < 2^k$, then this implies there are at most $log\_c(2)$ members of $[2^k,2^{k+1})$ in $A$. This implies that the probability of choosing $b\_k \in A$ is $\log\_c(2)/2^k$ when $k$ is large enough. It follows that the probability of there being $>K$ members of $A$ in $B$ (when $K$ is large) is $\sum\_{k > K} \log\_c(2)/2^k = \log\_c(2)/2^K$. Since this goes to $0$ as $K$ goes to infinity, the probability of $A \cap B$ being infinite is $0$.
The idea of choosing $B$ randomly, one point at a time, like this can be formalized by defining a probability measure on a Polish space, where points of the space correspond to possible choices of the sequence of $b\_k$'s. What the previous paragraph shows is that in this probability space, the set of all $B$'s with $A \cap B$ infinite form a null set, for any given lacunary set $A$. Hence any non-null subset of this probability space will contain a $B$ with $A \cap B$ finite. Since this holds for any $A$, we see that any non-null subset $X$ of this probability space contains, for *any* lacunary set $A$, some infinite $B$ with $A \cap B$ finite. The smallest possible size of such a set $X$ is $\mathrm{non}(\mathcal N)$.
*QED*
One more observation: The strict inequality $\mathfrak{x}\_{lac} < \mathrm{non}(\mathcal N)$ is also consistent, so this upper bound cannot be improved to an equality. To see this, begin with a model of Martin's Axiom $+ \, \neg \mathsf{CH}$, and then do a legnth-$\omega\_1$, finite support iteration of the eventually different reals forcing. It is not difficult to see that this forcing will make $\mathfrak{x}\_{lac} = \aleph\_1$ in the extension. But the iteration is $\sigma$-centered, and forcing with a $\sigma$-centered poset over a model of $\mathsf{MA}$ does not change the value of $\mathrm{non}(\mathcal N)$. Thus we get $\mathfrak{x}\_{lac} < \mathrm{non}(\mathcal N)$ in the extension.
**Original post:**
It is not provable that $\mathrm{cov}(\mathcal N) \leq \mathfrak{x}\_{lac}$, because $\mathrm{cov}(\mathcal N) > \mathfrak{x}\_{lac}$ in the random model.
To see this, let me sketch an argument that after forcing to add any number of random reals (in the usual way, via a measure algebra), the collection $[\omega]^\omega \cap V$ of infinite subsets of $\omega$ in the ground model has the property described in the definition of $\mathfrak{x}\_{lac}$. That is: for every lacunary set $A \subseteq \omega$ in the extension, there is an infinite $B \subseteq \omega$ in the ground model such that $A \cap B$ is finite.
We work in the ground model. Suppose $\dot A$ is a name for an infinite lacunary set in the extension. There is some fixed $c > 1$ and $n\_0 \in \omega$, and some forcing condition $p$, such that $p \Vdash$ if $a$ and $b$ are consecutive elements of $\dot A$ above $n\_0$, then $b/a > c$.
I claim that for every $\varepsilon > 0$, there are arbitrarily large $n \in \omega$ such that $m(p \wedge [n \in \dot A]) < \varepsilon$.
(Note: Here I'm using the standard notation for forcing with measure algebras. If $\varphi$ is any well-formed formula in the forcing language, then we write $[\varphi]$ to mean the supremum of all the conditions forcing $\varphi$, and $m([\varphi])$ denotes the measure of this supremum. Roughly, you may think of $m([\varphi])$ as the probability that $\varphi$ ends up being true in the forcing extension.)
To prove my claim, suppose, aiming for a contradiction, that it is false. Then there is some $\varepsilon > 0$ and some $N \in \omega$ such that $m([n \in \dot A]) \geq \varepsilon$ for all $n \geq N$. But this is just another way of saying that the "expected size" of $A \cap \{n\}$ is $\geq\varepsilon$ for all $n \geq N$. By the linearity of expectation, this means the expected size of $A \cap \{k,k+1,\dots,\ell-2,\ell-1\}$ is $\geq (\ell-k)\varepsilon$ whenever $\ell > k > N$. But by our choice of $p$, if $k \geq N$ and $\ell < ck$, then $p \Vdash |A \cap \{k,k+1,\dots,\ell-2,\ell-1\}| \leq 1$. Since $c > 1$, this yields a contradiction for sufficiently large $k$, namely $k > N,2/c\varepsilon$.
Using this claim, we can now find an infinite ground model set almost disjoint from the set named by $\dot A$ in the extension. Using the claim, we may find for each $k \in \omega$ some $n\_k > n\_{k-1}$ such that $m([n\_k \in \dot A]) < m(p)/2^{k+2}$. Now let $p' = p - \bigvee\_{k \in \omega}[n\_k \in \dot A]$. Then $m(p') \geq m(p) - \sum\_{k \in \omega}m([n\_k \in \dot A]) > m(p)/2 > 0$, so $p'$ is a condition in our measure algebra, and $p'$ forces $\dot A$ to be disjoint from $\{n\_k :\, k \in \omega \}$ (because it extends the complement of each $[n\_k \in \dot A]$).
This shows that it is impossible to have a name $\dot A$ for a lacunary set such that $\dot A$ is forced to have infinite intersection with every infinite subset of $\omega$ from the ground model. Therefore there is no such set.
|
4
|
https://mathoverflow.net/users/70618
|
403658
| 165,589 |
https://mathoverflow.net/questions/403465
|
1
|
I am looking for ways to evaluate *exactly* (i.e. *analytically* or *semi-analytically*) integrals of the type:
$$
\int\_{-\infty}^{+\infty}B\_{i}^k(u)e^{-\frac{(u-\mu)^2}{2\sigma^2}}du,
$$
where $B\_i^k$ is a spline of order $k$, an element of the B-Spline basis for the linear space of splines of order $k$ on knots $\{t\_i\}$, defined as usual recursively by:
$$
B\_i^k(x)=\frac{x-t\_i}{t\_{i+k}-t\_i}B\_i^{k-1}(x)+\frac{t\_{i+k+1}-x}{t\_{i+k+1}-t\_{i+1}}B\_{i+1}^{k-1}(x),
$$
with
$$
B\_i^0(x)=\begin{cases}
1 & x\in [t\_i;t\_{i+1}) \\
0 & \text{otherwise }
\end{cases}
$$
Of particular interest would be the case of $\mu=0, \sigma=1$.
I am aware of the [Gauss-Hermite quadrature](https://en.wikipedia.org/wiki/Gauss%E2%80%93Hermite_quadrature) :
$$
\int\_{-\infty}^{+\infty}f(x)e^{-\frac{x^2}{2}}\approx \sum\_{i=1}^n w\_i f(x\_i),
$$
where $x\_i$ are the roots of a Hermite polynomial of order $n$ and $w\_i$ are the associated weights. ***Importantly, the approximation sign can be replaced by an exact equality when $f$ is a polynomial of degree $\leq 2n-1$***. (There are versions where the integral is with respect to $e^{-x^2}$ instead of $e^{-\frac{x^2}{2}}$, by changing the type of Hermite polynomial employed).
My question is : is there such an **exact** equality formula for B-spline basis functions?
I am looking to express the integral at the beginning of this question as a sum analogously to the Gauss-Hermite quadrature.
The problem seems to be that even though $B\_i^k$ is known to have finite support, it is not itself a polynomial: each of the restrictions $B\_i^k|\_{(t\_j;t\_{j+1})}$ is a polynomial, without the full function being a polynomial. Otherwise, the answer would have been a trivial application of the Gauss-Hermite quadrature. Is is possible that there is a Gauss-Hermite-type quadrature for integration domains that are compact intervals (as opposed to integration domains that are $\mathbb{R}$) ?
|
https://mathoverflow.net/users/156188
|
Integrating a B-Spline basis function with respect to the standard normal PDF
|
Since a B-spline is a piecewise polynomial function, the question is whether there exists an exact equality formula for the integral $\int\_{-a}^{b}u^pe^{-u^2/2}du$. This integral equals an elementary function of $a$ and $b$ for $p$ an odd integer, while for $p$ an even integer it contains error functions. In general the B-spline will contain both even and odd powers, so no "exact equality formula" in terms of elementary functions will be forthcoming.
For example, the [uniform B-spline](https://en.wikipedia.org/wiki/B-spline#Properties) of order 3 with knots at 0,1,2,3 is given by
$$B(u)=\begin{cases}
0&\text{if}\;\; u<0,\\
u^2/2&\text{if}\;\; 0 \le u < 1,\\
(-2u^2+6u-3)/2&\text{if}\;\; 1 \le u < 2,\\
(3-u)^2/2&\text{if}\;\; 2 \le u < 3,\\
0&\text{if}\;\; u\ge 3,
\end{cases}$$
and the integral $\int\_{-\infty}^\infty B(u)e^{-u^2/2}\,du$ equals
$$\frac{1}{2} \sqrt{\frac{\pi }{2}} \left(6 \,\text{erf}\left(\frac{1}{\sqrt{2}}\right)+10 \,\text{erf}\left(\frac{3}{\sqrt{2}}\right)-15\, \text{erf}\left(\sqrt{2}\right)\right)+\frac{3 \left(1-2 e^{5/2}+e^4\right)}{2e^{9/2}}.$$
No further simplification in terms of elementary functions is possible.
|
0
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https://mathoverflow.net/users/11260
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403661
| 165,591 |
https://mathoverflow.net/questions/403153
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10
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For a partition $\lambda$, let $P\_\lambda$ be the Schur $P$-functions (case $t=-1$ of Hall-Littlewood symmetric functions) and let $p\_\lambda=p\_{\lambda\_1}p\_{\lambda\_1}\cdots p\_{\lambda\_k}$ be the power-sum symmetric functions.
It is known that the space $\Gamma$ spanned by the $P\_\lambda$ for $\lambda$ with distinct parts is the same as the one spanned by the $p\_\rho$ for $\rho$ with odd parts (see e.g. Macdonald's *Symmetric Functions and Hall Polynomials*). I noticed that the coefficients of the transition matrix between the two bases for $\Gamma$ are all in $\mathbb{Z}\_{(2)}$; that is, if $\lambda$ is a partition with distinct parts and
$$P\_\lambda = \sum\_\rho a^\lambda\_\rho p\_\rho$$
then $v\_2(a^\lambda\_\rho) \ge 0$. Is this fact known?
Note that we have $a^\lambda\_\rho=2^{\ell(\rho)-\ell(\lambda)}z\_\rho^{-1}X^\lambda\_\rho(-1)$, where $\ell(\cdot)$ is the number of parts of a partition and $X^\lambda\_\rho \in \mathbb{Z}[t]$ are the Green polynomials (see Macdonald chapter III.7). In particular, the question comes down to showing that
$$v\_2(X^\lambda\_\rho(-1))\ge v\_2(z\_\rho)+\ell(\lambda)-\ell(\rho).$$
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https://mathoverflow.net/users/160416
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$2$-adic valuation of Schur $P$-functions in the power-sum basis
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Here is a proof of the generalization suggested by Richard Stanley in the
comments and even of a more general result (with "odd" replaced by "not
divisible by a given prime $q$"). It is completely different from the argument
I sketched in the comments, and is completely elementary (using no Macdonald
polynomials). Unfortunately, it is also somewhat awkward and way too long
(much of it devoted to a fight with notations).
For any commutative ring $R$, we let $\Lambda\_{R}$ be the ring of symmetric
functions over $R$; this is a commutative $R$-algebra. Let
$\operatorname{Par}$ be the set of all partitions. We shall use the standard
notation $p\_{\lambda}$ for the power-sum symmetric function indexed by a
partition $\lambda$.
Fix a prime $q$. Let $\mathbb{Z}\_{\left( q\right) }$ denote the ring of all
rational numbers that can be written in the form $\dfrac{a}{b}$ for two
integers $a$ and $b$ such that $b$ is coprime to $q$. These numbers are known
as *$q$-integers*. Obviously, $\mathbb{Z}\_{\left( q\right) }$ is a subring
of $\mathbb{Q}$, so that $\Lambda\_{\mathbb{Z}\_{\left( q\right) }}$ is a
subring of $\Lambda\_{\mathbb{Q}}$.
Now, Richard Stanley's generalization (generalized a bit further) claims:
>
> **Theorem 1.** We have
> \begin{align\*}
> \Lambda\_{\mathbb{Z}\_{\left( q\right) }}\cap\mathbb{Q}\left[ p\_{i}
> \ \mid\ i\not \equiv 0\mod q\right] =\mathbb{Z}\_{\left(
> q\right) }\left[ p\_{i}\ \mid\ i\not \equiv 0\mod q\right] .
> \end{align\*}
>
>
>
The proof requires some preparations, which include showing some results of
independent interest.
First, we introduce a few more classical notations from symmetric function
theory: For any partition $\lambda$ and any $i\geq1$, we let $m\_{i}\left(
\lambda\right) $ denote the multiplicity of $i$ in $\lambda$ (that is, the
number of parts of $\lambda$ that equal $i$). For any partition $\lambda$, we
define the positive integer
\begin{align\*}
z\_{\lambda}:=\prod\_{i=1}^{\infty}\left( \left( m\_{i}\left( \lambda\right)
\right) !\cdot i^{m\_{i}\left( \lambda\right) }\right) .
\end{align\*}
Let $V$ be the $\Lambda\_{\mathbb{Z}}$-subalgebra of $\Lambda\_{\mathbb{Q}}$
generated by the fractions $\dfrac{p\_{i}}{i^{k}}$ for all positive integers
$i$ and all nonnegative integers $k$. In other words, let
\begin{align\*}
V=\Lambda\_{\mathbb{Z}}\left[ \dfrac{p\_{i}}{i^{k}}\ \mid\ i>0\text{ and }
k\geq0\right] \subseteq\Lambda\_{\mathbb{Q}}.
\end{align\*}
Now, we claim the following:
>
> **Theorem 2.** We have
> \begin{align\*}
> z\_{\lambda}^{-1}p\_{\lambda}\in V
> \end{align\*}
> for each partition $\lambda$.
>
>
>
To prove this, we need a simple arithmetic lemma:
>
> **Lemma 3.** Let $c$ and $d$ be two integers with $d\neq0$. Then, there exist
> some integers $a$ and $b$ and some nonnegative integer $i$ such that
> $c^i =ad+bc^{i+1}$.
>
>
>
*Proof of Lemma 3.* The ring $\mathbb{Z}/d\mathbb{Z}$ is finite (since
$d\neq0$). For any integer $m$, we let $\overline{m}\in\mathbb{Z}/d\mathbb{Z}$
denote the residue class of $m$ in this ring. The infinitely many residue
classes $\overline{c^{0}},\overline{c^{1}},\overline{c^{2}},\ldots$ all belong
to the finite ring $\mathbb{Z}/d\mathbb{Z}$, and thus cannot all be distinct
(by the pigeonhole principle). In other words, there exist two nonnegative
integers $i$ and $j$ satisfying $i<j$ and $\overline{c^i }=\overline{c^j }$.
Consider these $i$ and $j$. We have $\overline{c^i }=\overline{c^j }$; in
other words, $c^i \equiv c^j \mod d$. Hence, $d\mid c^i
-c^j $. In other words, $c^i -c^j =ad$ for some integer $a$. Consider this
$a$. However, recall that $i<j$. Thus, $j=i+w$ for some positive integer $w$.
Consider this $w$. The integer $w-1$ is nonnegative (since $w$ is positive);
thus, $c^{w-1}$ is an integer. Hence, we can define an integer $b$ by
$b=c^{w-1}$. From $j = i+w = \left(w-1\right) + \left(i+1\right)$, we obtain $c^j = c^{\left(w-1\right) + \left(i+1\right)} = \underbrace{c^{w-1}}\_{= b} c^{i+1} = bc^{i+1}$.
Now, from $c^i -c^j =ad$, we obtain
\begin{align\*}
c^i & =ad+\underbrace{c^j }\_{= bc^{i+1}} =ad+bc^{i+1}.
\end{align\*}
This proves Lemma 3. $\blacksquare$
*Proof of Theorem 2.* We shall use the notation $\ell\left( \lambda\right) $
for the *length* of a partition $\lambda$ (that is, the number of all parts of
$\lambda$). For instance, $\ell\left( \left( 5,2,2\right) \right) =3$. We
shall also use the notation $\left\vert \lambda\right\vert $ for the *size* of
a partition $\lambda$ (that is, the sum of all parts of $\lambda$). We shall
prove Theorem 2 by strong induction on $\left\vert \lambda\right\vert
+\ell\left( \lambda\right) $. Thus, we fix some $N\in\mathbb{N}$, and we
assume (as the induction hypothesis) that Theorem 2 holds for all $\lambda$
with $\left\vert \lambda\right\vert +\ell\left( \lambda\right) <N$. We now
must prove Theorem 2 for all $\lambda$ with $\left\vert \lambda\right\vert
+\ell\left( \lambda\right) =N$.
So let $\lambda$ be a partition satisfying $\left\vert \lambda\right\vert
+\ell\left( \lambda\right) =N$. We must prove that $z\_{\lambda}
^{-1}p\_{\lambda}\in V$.
If $\lambda=\varnothing$, then this is obvious (since $z\_{\lambda}
^{-1}p\_{\lambda}=1$ in this case). Thus, for the rest of this induction step,
we WLOG assume that $\lambda\neq\varnothing$.
We are in one of the following three cases:
*Case 1:* All parts of $\lambda$ are equal to $1$.
*Case 2:* All parts of $\lambda$ are equal, but not equal to $1$.
*Case 3:* Not all parts of $\lambda$ are equal.
Let us consider Case 1 first. In this case, all parts of $\lambda$ are equal
to $1$. In other words, $\lambda=\underbrace{\left( 1,1,\ldots,1\right)
}\_{v\text{ entries}}$ for some positive integer $v$ (since $\lambda
\neq\varnothing$). Consider this $v$. Thus, $p\_{\lambda}=p\_{1}^v $ and
$\left\vert \lambda\right\vert =v$ and $\ell\left( \lambda\right) =v$.
Let $\operatorname{Par}\_{v}$ denote the set of all
partitions of $v$. Thus, $\lambda\in\operatorname{Par}\_{v}$. Now, let $h\_v \in \Lambda\_{\mathbb{Z}}$ denote the $v$-th complete homogeneous symmetric function. A
well-known formula (e.g., (2.5.17) in [Grinberg/Reiner, arXiv:1409.8356v7](https://arxiv.org/abs/1409.8356v7))
yields
\begin{align\*}
h\_{v}=\sum\_{\mu\in\operatorname{Par}\_{v}}z\_{\mu}^{-1}p\_{\mu
}=z\_{\lambda}^{-1}p\_{\lambda}+\sum\_{\substack{\mu\in\operatorname{Par}\_{v};
\\\mu\neq\lambda}}z\_{\mu}^{-1}p\_{\mu},
\end{align\*}
so that
\begin{equation}
z\_{\lambda}^{-1}p\_{\lambda}=h\_v -\sum\_{\substack{\mu\in
\operatorname{Par}\_{v};\\\mu\neq\lambda}}z\_{\mu}^{-1}p\_{\mu}.
\label{darij1.pf.t2.c1.2}
\tag{1}
\end{equation}
However, the only partition of $v$ that has length $\geq v$ is the partition
$\underbrace{\left( 1,1,\ldots,1\right) }\_{v\text{ entries}}=\lambda$. Thus,
if $\mu$ is a partition of $v$ distinct from $\lambda$, then $\mu$ has length
$<v$. In other words, if $\mu\in\operatorname{Par}\_{v}$ satisfies
$\mu\neq\lambda$, then $\ell\left( \mu\right) <v$. Hence, if $\mu
\in\operatorname{Par}\_{v}$ satisfies $\mu\neq\lambda$, then
\begin{align\*}
\underbrace{\left\vert \mu\right\vert }\_{=v=\left\vert \lambda\right\vert
}+\underbrace{\ell\left( \mu\right) }\_{<v=\ell\left( \lambda\right)
}<\left\vert \lambda\right\vert +\ell\left( \lambda\right) =N
\end{align\*}
and therefore $z\_{\mu}^{-1}p\_{\mu}\in V$ (by our induction hypothesis, applied
to $\mu$ instead of $\lambda$). Thus, \eqref{darij1.pf.t2.c1.2} becomes
\begin{align\*}
z\_{\lambda}^{-1}p\_{\lambda}=\underbrace{h\_v}\_{\in\Lambda\_{\mathbb{Z}
}\subseteq V}-\sum\_{\substack{\mu\in\operatorname{Par}\_{v};\\\mu
\neq\lambda}}\underbrace{z\_{\mu}^{-1}p\_{\mu}}\_{\in V}\in V-\sum\_{\substack{\mu
\in\operatorname{Par}\_{v};\\\mu\neq\lambda}}V\subseteq V.
\end{align\*}
Hence, $z\_{\lambda}^{-1}p\_{\lambda}\in V$ has been proved in Case 1.
Let us next consider Case 2. In this case, all parts of $\lambda$ are equal,
but not equal to $1$. In other words, $\lambda=\underbrace{\left(
n,n,\ldots,n\right) }\_{v\text{ entries}}$ for some positive integers $n\neq1$
and $v$ (since $\lambda\neq\varnothing$). Consider these $n$ and $v$. Thus,
$p\_{\lambda}=p\_n^v $ and $z\_{\lambda}=v!\cdot n^v $ and $\left\vert
\lambda\right\vert =nv$ and $\ell\left( \lambda\right) =v$. From $n\neq1$,
we obtain $n>1$ (since $n$ is a positive integer). Thus, $nv > v$ (since $v > 0$). In other words, $v < nv$.
From $p\_{\lambda}=p\_n^v $ and $z\_{\lambda}=v!\cdot n^v $, we obtain
\begin{equation}
z\_{\lambda}^{-1}p\_{\lambda}=\left( v!\cdot n^v \right) ^{-1}p\_n
^v =\dfrac{p\_n^v }{v!\cdot n^v }.
\label{darij1.pf.t2.c2.1}
\tag{2}
\end{equation}
We must prove that $z\_{\lambda}^{-1}p\_{\lambda}\in V$. In other words, we must
prove that $\dfrac{p\_n^v }{v!\cdot n^v }\in V$ (since $z\_{\lambda}
^{-1}p\_{\lambda}=\dfrac{p\_n^v }{v!\cdot n^v }$).
We shall now use Lemma 3 to decompose $\dfrac{p\_n^v }{v!\cdot n^v }$ into
a sum of partial fractions -- one with a denominator of $v!$ and another with
a power of $n$ in the denominator. We will then prove that both of these
fractions belong to $V$.
Indeed, Lemma 3 (applied to $c=n^v $ and $d=v!$) yields that there exist some
integers $a$ and $b$ and some nonnegative integer $i$ such that $\left(
n^v \right) ^i =a\cdot v!+b\left( n^v \right) ^{i+1}$. Consider these
$a$, $b$ and $i$. Multiplying both sides of the equality $\left(
n^v \right) ^i =a\cdot v!+b\left( n^v \right) ^{i+1}$ by $\dfrac
{p\_n^v }{\left( n^v \right) ^{i+1}\cdot v!}$, we obtain
\begin{align}
\dfrac{p\_n^v }{v!\cdot n^v }
&=\left(a\cdot v!+b\left( n^v \right)
^{i+1}\right) \cdot \dfrac{p\_n^v }{\left( n^v \right) ^{i+1}\cdot v!} \\
&=a\cdot\dfrac{p\_n^v }{\left(
n^v \right) ^{i+1}}+b\cdot\dfrac{p\_n^v }{v!}
.
\label{darij1.pf.t2.c2.parfrac}
\tag{3}
\end{align}
Thus, in order to prove that $\dfrac{p\_n^v }{v!\cdot n^v }\in V$, it
suffices to show that $\dfrac{p\_n^v }{\left( n^v \right) ^{i+1}}\in V$
and $\dfrac{p\_n^v }{v!}\in V$ (because $a$ and $b$ are integers, and $V$ is
a ring).
The first of these two claims is easy: We have $\dfrac{p\_n}{n^{i+1}}\in V$
(since $\dfrac{p\_n}{n^{i+1}}$ is one of the designated generators of the
$\Lambda\_{\mathbb{Z}}$-algebra $V$). Hence, $\left( \dfrac{p\_n}{n^{i+1}
}\right) ^v \in V$ (since $V$ is a ring). In other words, $\dfrac{p\_n^v
}{\left( n^v \right) ^{i+1}}\in V$ (since $\left( \dfrac{p\_n}{n^{i+1}
}\right) ^v =\dfrac{p\_n^v }{\left( n^v \right) ^{i+1}}$).
It remains to prove that $\dfrac{p\_n^v }{v!}\in V$. To do this, we will
need the *Frobenius endomorphism* $\mathbf{f}\_n$. It is defined as follows:
For any commutative ring $R$, we let
\begin{align\*}
\mathbf{f}\_n:\Lambda\_{R}\rightarrow\Lambda\_{R}
\end{align\*}
be the $R$-algebra homomorphism that sends each symmetric function $f$ to
$f\left( x\_{1}^{n},x\_{2}^{n},x\_{3}^{n},\ldots\right) $ (where we regard $f$
as a symmetric formal power series in countably many indeterminates $x\_{1},x\_{2}
,x\_{3},\ldots$). This homomorphism $\mathbf{f}\_n$ is called the $n$-th
*Frobenius endomorphism* and is functorial in $R$ (that is, it commutes with
the morphisms $\Lambda\_{R}\rightarrow\Lambda\_{S}$ induced by ring
homomorphisms $R\rightarrow S$). If you like to think in terms of plethysm,
$\mathbf{f}\_n$ can be described as sending each $f\in\Lambda\_{R}$ to the
plethysm $f\left[ p\_n\right] $.
The functoriality of $\mathbf{f}\_n$ in $R$ entails that the $\mathbf{f}\_n$
defined for $R=\mathbb{Z}$ is a restriction of the $\mathbf{f}\_n$ defined
for $R=\mathbb{Q}$. Thus, we can safely denote both of these maps by
$\mathbf{f}\_n$ without risking confusion. They both are ring homomorphisms
(since they are $R$-algebra homomorphisms for appropriate $R$). Of course,
$\mathbf{f}\_n\left( \Lambda\_{\mathbb{Z}}\right) \subseteq\Lambda
\_{\mathbb{Z}}$ (since the $\mathbf{f}\_n$ defined for $R=\mathbb{Z}$ is a
restriction of the $\mathbf{f}\_n$ defined for $R=\mathbb{Q}$).
It is easy to see that $\mathbf{f}\_n\left( p\_{i}\right) =p\_{in}$ for each
$i>0$. Hence, for each positive integer $i$ and each nonnegative integer $k$,
we have
\begin{align\*}
\mathbf{f}\_n\left( \dfrac{p\_{i}}{i^{k}}\right) =\dfrac{p\_{in}}{i^{k}
}=n^{k}\cdot\dfrac{p\_{in}}{\left( in\right) ^{k}}\in V
\end{align\*}
(since $\dfrac{p\_{in}}{\left( in\right) ^{k}}$ is one of the designated
generators of the $\Lambda\_{\mathbb{Z}}$-algebra $V$). Therefore,
\begin{align\*}
\Lambda\_{\mathbb{Z}}\left[ \mathbf{f}\_n\left( \dfrac{p\_{i}}{i^{k}}\right)
\ \mid\ i>0\text{ and }k\geq0\right] \subseteq V
\end{align\*}
(since $V$ is a $\Lambda\_{\mathbb{Z}}$-algebra).
Now, from $V=\Lambda\_{\mathbb{Z}}\left[ \dfrac{p\_{i}}{i^{k}}\ \mid\ i>0\text{
and }k\geq0\right] $, we obtain
\begin{align\*}
\mathbf{f}\_n\left( V\right) & =\mathbf{f}\_n\left( \Lambda
\_{\mathbb{Z}}\left[ \dfrac{p\_{i}}{i^{k}}\ \mid\ i>0\text{ and }k\geq0\right]
\right) \\
& =\underbrace{\left( \mathbf{f}\_n\left( \Lambda\_{\mathbb{Z}}\right)
\right) }\_{\subseteq\Lambda\_{\mathbb{Z}}}\left[ \mathbf{f}\_n\left(
\dfrac{p\_{i}}{i^{k}}\right) \ \mid\ i>0\text{ and }k\geq0\right] \\
& \qquad\left( \text{since }\mathbf{f}\_n\text{ is a ring homomorphism}
\right) \\
& \subseteq\Lambda\_{\mathbb{Z}}\left[ \mathbf{f}\_n\left( \dfrac{p\_{i}
}{i^{k}}\right) \ \mid\ i>0\text{ and }k\geq0\right] \subseteq V.
\end{align\*}
Let $\mu$ be the partition $\underbrace{\left( 1,1,\ldots,1\right)
}\_{v\text{ entries}}$. Then, $p\_{\mu}=p\_{1}^v $ and $z\_{\mu}=v!\cdot
\underbrace{1^v }\_{=1}=v!$ and $\left\vert \mu\right\vert =v$ and
$\ell\left( \mu\right) =v$. Hence,
\begin{align\*}
\underbrace{\left\vert \mu\right\vert }\_{=v}+\underbrace{\ell\left(
\mu\right) }\_{=v} & =\underbrace{v}\_{< nv} + v\\
& <\underbrace{nv}\_{=\left\vert \lambda\right\vert }+\underbrace{v}
\_{=\ell\left( \lambda\right) }=\left\vert \lambda\right\vert +\ell\left(
\lambda\right) =N.
\end{align\*}
Hence, $z\_{\mu}^{-1}p\_{\mu}\in V$ (by our induction hypothesis, applied to
$\mu$ instead of $\lambda$). In view of $z\_{\mu}=v!$ and $p\_{\mu}=p\_{1}^v $,
this rewrites as $v!^{-1}\cdot p\_{1}^v \in V$. Hence,
\begin{equation}
\mathbf{f}\_n\left( v!^{-1}\cdot p\_{1}^v \right) \in \mathbf{f}
\_n\left( V\right) \subseteq V.
\label{darij1.pf.t2.c2.5}
\tag{4}
\end{equation}
However, since $\mathbf{f}\_n$ is a $\mathbb{Q}$-algebra homomorphism, we
have
\begin{align\*}
\mathbf{f}\_n\left( v!^{-1}\cdot p\_{1}^v \right) =v!^{-1}\cdot\left(
\mathbf{f}\_n\left( p\_{1}\right) \right) ^v =\dfrac{\left(
\mathbf{f}\_n\left( p\_{1}\right) \right) ^v }{v!}=\dfrac{p\_n^v }{v!}
\end{align\*}
(since $\mathbf{f}\_n\left( p\_{1}\right) =p\_n$). Thus,
\eqref{darij1.pf.t2.c2.5} rewrites as $\dfrac{p\_n^v }{v!}\in V$. Hence,
\eqref{darij1.pf.t2.c2.parfrac} becomes
\begin{align\*}
\dfrac{p\_n^v }{v!\cdot n^v }=a\cdot\underbrace{\dfrac{p\_n^v }{\left(
n^v \right) ^{i+1}}}\_{\in V}+b\cdot\underbrace{\dfrac{p\_n^v }{v!}}\_{\in
V}\in V
\end{align\*}
(since $V$ is a ring and since $a$ and $b$ are integers). In view of
\eqref{darij1.pf.t2.c2.1}, this rewrites as $z\_{\lambda}^{-1}p\_{\lambda}\in
V$. Hence, $z\_{\lambda}^{-1}p\_{\lambda}\in V$ has been proved in Case 2.
Let us finally consider Case 3. In this case, not all parts of $\lambda$ are equal.
We need another notation: If $\alpha=\left( \alpha\_{1},\alpha\_{2}
,\ldots,\alpha\_{m}\right) $ and $\beta=\left( \beta\_{1},\beta\_{2}
,\ldots,\beta\_n\right) $ are two partitions, then $\alpha\sqcup\beta$ shall
denote the partition obtained by sorting the tuple $\left( \alpha\_{1}
,\alpha\_{2},\ldots,\alpha\_{m},\beta\_{1},\beta\_{2},\ldots,\beta\_n\right) $
in weakly decreasing order. For instance, $\left( 3,2,2\right) \sqcup\left(
5,3,2\right) =\left( 5,3,3,2,2,2\right) $.
It is easy to see that if $\alpha$ and $\beta$ are two partitions that have no
part in common, then
\begin{equation}
z\_{\alpha\sqcup\beta}=z\_{\alpha}z\_{\beta}.
\label{darij1.pf.t2.c3.zaub}
\tag{5}
\end{equation}
Moreover, if $\alpha$ and $\beta$ are any two partitions, then
\begin{equation}
p\_{\alpha\sqcup\beta}=p\_{\alpha}p\_{\beta}.
\label{darij1.pf.t2.c3.paub}
\tag{6}
\end{equation}
Now, recall that not all parts of $\lambda$ are equal. Hence, we can write
$\lambda$ in the form $\lambda=\alpha\sqcup\beta$ where $\alpha$ and $\beta$
are two nonempty partitions that have no part in common. (Indeed, we can
define $\alpha$ and $\beta$ by choosing an arbitrary part $i$ of $\lambda$,
then letting $\alpha$ be the partition consisting of all parts of $\lambda$
equal to $i$, while $\beta$ is the partition consisting of all remaining parts
of $\lambda$.) Consider these $\alpha$ and $\beta$. It is easy to see that
$\left\vert \alpha\right\vert <\left\vert \alpha\sqcup\beta\right\vert $
(since $\beta$ is nonempty) and $\ell\left( \alpha\right) <\ell\left(
\alpha\sqcup\beta\right) $ (for the same reason). Since $\alpha\sqcup
\beta=\lambda$, these two inequalities rewrite as $\left\vert \alpha
\right\vert <\left\vert \lambda\right\vert $ and $\ell\left( \alpha\right)
<\ell\left( \lambda\right) $. Adding these two inequalities together, we
obtain
\begin{align\*}
\left\vert \alpha\right\vert + \ell\left( \alpha\right)
<\left\vert \lambda\right\vert +\ell\left( \lambda\right) =N.
\end{align\*}
Hence, $z\_{\alpha}^{-1}p\_{\alpha}\in V$ (by our induction hypothesis, applied
to $\alpha$ instead of $\lambda$). Similarly, $z\_{\beta}^{-1}p\_{\beta}\in V$.
However, from \eqref{darij1.pf.t2.c3.zaub} and \eqref{darij1.pf.t2.c3.paub},
we obtain
\begin{align\*}
z\_{\alpha\sqcup\beta}^{-1}p\_{\alpha\sqcup\beta}=\left( z\_{\alpha}z\_{\beta
}\right) ^{-1}p\_{\alpha}p\_{\beta}=\underbrace{z\_{\alpha}^{-1}p\_{\alpha}}\_{\in
V}\cdot\underbrace{z\_{\beta}^{-1}p\_{\beta}}\_{\in V}\in V
\end{align\*}
(since $V$ is a ring). In view of $\alpha\sqcup\beta=\lambda$, this rewrites
as $z\_{\lambda}^{-1}p\_{\lambda}\in V$. Hence, $z\_{\lambda}^{-1}p\_{\lambda}\in
V$ has been proved in Case 3.
We have now proved $z\_{\lambda}^{-1}p\_{\lambda}\in V$ in all three cases 1, 2
and 3. Thus, the induction step is complete.
Thus, Theorem 2 is proved by induction. $\blacksquare$
We are not quite ready to prove Theorem 1 yet. We first need some more notations.
We let $\operatorname{QPar}$ be the set of all partitions that have no part
divisible by $q$. (If $q=2$, these are precisely the partitions into odd parts.)
We let $J$ be the ideal of the ring $\Lambda\_{\mathbb{Q}}$ generated by the
$p\_{i}$ with $i\equiv0\mod q$. In other words, $J=\sum\limits\_{i=1}^{\infty}p\_{iq}\Lambda\_{\mathbb{Q}}$. Recall that the family $\left(
p\_{\lambda}\right) \_{\lambda\in\operatorname{Par}}$ is a basis of the
$\mathbb{Q}$-vector space $\Lambda\_{\mathbb{Q}}$. Thus, the $\mathbb{Q}
$-vector subspace $J$ of $\Lambda\_{\mathbb{Q}}$ has basis $\left( p\_{\lambda
}\right) \_{\lambda\in\operatorname{Par}\setminus\operatorname{QPar}}$
(because multiplying any $p\_{\mu}$ by a $p\_{iq}$ yields a $p\_{\lambda}$ with
$\lambda\in\operatorname{Par}\setminus\operatorname{QPar}$, and conversely,
any $p\_{\lambda}$ with $\lambda \in \operatorname{Par}\setminus\operatorname{QPar}$ can
be obtained in such a way).
A well-known fact (or easy exercise) in abstract algebra says the following:
>
> **Lemma 4.** Let $B$ be a subring of a ring $A$. Let $I$ be a (two-sided)
> ideal of $A$. Then, $B+I$ is a subring of $A$.
>
>
>
Applying Lemma 4 to $A=\Lambda\_{\mathbb{Q}}$, $B=\Lambda\_{\mathbb{Z}\_{\left(
q\right) }}$ and $I=J$, we conclude that $\Lambda\_{\mathbb{Z}\_{\left(
q\right) }}+J$ is a subring of $\Lambda\_{\mathbb{Q}}$. We denote this subring
$\Lambda\_{\mathbb{Z}\_{\left( q\right) }}+J$ by $W$. We note that $W$ is
furthermore a $\mathbb{Z}\_{\left( q\right) }$-subalgebra of $\Lambda
\_{\mathbb{Q}}$ (since $W$ is a subring of $\Lambda\_{\mathbb{Q}}$ and is
preserved under scaling by $\mathbb{Z}\_{\left( q\right) }$).
Next, we observe:
>
> **Proposition 5.** We have $V\subseteq W$.
>
>
>
*Proof of Proposition 5.* We have
\begin{align}
\Lambda
\_{\mathbb{Z}}\subseteq \Lambda
\_{\mathbb{Z}\_{\left( q\right) }}\subseteq \Lambda\_{\mathbb{Z}\_{\left(
q\right) }}+J=W .
\end{align}
Now, $W$ is a commutative ring (since it is a subring of $\Lambda\_{\mathbb{Q}}$) and contains
$\Lambda\_{\mathbb{Z}}$ as a subring (since $\Lambda
\_{\mathbb{Z}}\subseteq W$). Hence, $W$ is a $\Lambda\_{\mathbb{Z}}$-algebra. Thus, in order to prove that $V\subseteq W$, it suffices to show
that $\dfrac{p\_{i}}{i^{k}}\in W$ for each positive integer $i$ and each
nonnegative integer $k$ (by the definition of $V$).
So let us show this. Fix a positive integer $i$ and a nonnegative integer $k$. We must prove that $\dfrac{p\_{i}}{i^{k}}\in W$. If $i\equiv0\mod q$,
then this follows from the obvious fact that $\dfrac{p\_{i}}{i^{k}}\in
J\subseteq\Lambda\_{\mathbb{Z}\_{\left( q\right) }}+J=W$. Thus, we WLOG assume that $i\not \equiv 0\mod q$. Hence, $i$ is coprime to $q$. Hence, $\dfrac{1}{i}\in \mathbb{Z}\_{\left( q\right) }$, so that $\dfrac{1}{i^{k}}\in\mathbb{Z}\_{\left( q\right) }\subseteq W$. Now, $\dfrac{p\_{i}}{i^{k}} =\underbrace{\left( \dfrac{1}{i}\right) ^{k}}\_{\in W} \underbrace{p\_{i}}\_{\in \Lambda\_{\mathbb{Z}} \subseteq W}\in W$ (since $W$ is a ring). This completes our proof of Proposition 5.
$\blacksquare$
At last, we can prove Theorem 1.
*Proof of Theorem 1.* It is clear that $\mathbb{Z}\_{\left( q\right) }\left[ p\_{i}\ \mid\ i\not \equiv 0\mod q\right] \subseteq
\Lambda\_{\mathbb{Z}\_{\left( q\right) }}\cap\mathbb{Q}\left[ p\_{i}
\ \mid\ i\not \equiv 0\mod q\right] $. Hence, it suffices to
prove the reverse inclusion, i.e., to prove that
\begin{align\*}
\Lambda\_{\mathbb{Z}\_{\left( q\right) }}\cap\mathbb{Q}\left[ p\_{i}
\ \mid\ i\not \equiv 0\mod q\right] \subseteq \mathbb{Z}\_{\left( q\right) }\left[ p\_{i}\ \mid\ i\not \equiv 0\mod q\right] .
\end{align\*}
Thus, we fix an arbitrary $f\in\Lambda\_{\mathbb{Z}\_{\left( q\right) }}
\cap\mathbb{Q}\left[ p\_{i}\ \mid\ i\not \equiv 0\mod q\right]
$. We must prove that $f\in \mathbb{Z}\_{\left( q\right) }\left[ p\_{i}\ \mid\ i\not \equiv 0\mod q\right]$.
We have $f\in\Lambda\_{\mathbb{Z}\_{\left( q\right) }}\cap\mathbb{Q}\left[
p\_{i}\ \mid\ i\not \equiv 0\mod q\right] \subseteq
\mathbb{Q}\left[ p\_{i}\ \mid\ i\not \equiv 0\mod q\right] $.
Hence, $f$ is a $\mathbb{Q}$-linear combination of the family $\left(
p\_{\lambda}\right) \_{\lambda\in\operatorname{QPar}}$ (since this family
$\left( p\_{\lambda}\right) \_{\lambda\in\operatorname{QPar}}$ is a basis of
the $\mathbb{Q}$-vector space $\mathbb{Q}\left[ p\_{i}\ \mid\ i\not \equiv
0\mod q\right] $). In other words, we can write $f$ in the form
\begin{equation}
f=\sum\_{\lambda\in\operatorname{QPar}}c\_{\lambda}p\_{\lambda}
\label{darij1.pf.t1.f=}
\tag{7}
\end{equation}
for some $c\_{\lambda}\in\mathbb{Q}$. Consider these $c\_{\lambda}$. We shall prove that they all belong to $\mathbb{Z}\_{\left(q\right)}$.
We let $\left\langle \cdot,\cdot\right\rangle $ denote the Hall inner product
on $\Lambda\_{\mathbb{Q}}$. This is a $\mathbb{Q}$-bilinear form sending
$\Lambda\_{\mathbb{Q}}\times\Lambda\_{\mathbb{Q}}$ to $\mathbb{Q}$ and sending
$\Lambda\_{\mathbb{Z}\_{\left( q\right) }}\times\Lambda\_{\mathbb{Z}\_{\left(
q\right) }}$ to $\mathbb{Z}\_{\left( q\right) }$ (since its restriction to
$\Lambda\_{\mathbb{Z}\_{\left( q\right) }}\times\Lambda\_{\mathbb{Z}\_{\left(
q\right) }}$ is the Hall inner product on $\Lambda\_{\mathbb{Z}\_{\left(
q\right) }}$).
It is well-known (see, e.g., Corollary 2.5.17(b) in [Grinberg/Reiner,
arXiv:1409.8356v7](https://arxiv.org/abs/1409.8356v7)) that the families $\left( p\_{\lambda}\right)
\_{\lambda\in\operatorname{Par}}$ and $\left( z\_{\lambda}^{-1}p\_{\lambda
}\right) \_{\lambda\in\operatorname{Par}}$ are dual bases of $\Lambda
\_{\mathbb{Q}}$ with respect to the Hall inner product. Hence,
\begin{equation}
\left\langle p\_{\lambda},z\_{\mu}^{-1}p\_{\mu}\right\rangle =\delta\_{\lambda
,\mu}
\label{darij1.pf.t1.dualbases}
\tag{8}
\end{equation}
for any $\lambda\in\operatorname{Par}$ and $\mu\in\operatorname{Par}$ (where
the $\delta\_{\lambda,\mu}$ is a Kronecker delta). In other words,
\begin{equation}
\left\langle p\_{\lambda},p\_{\mu}\right\rangle =z\_{\mu}\delta\_{\lambda,\mu
}
\label{darij1.pf.t1.dualbases2}
\tag{9}
\end{equation}
for any $\lambda\in\operatorname{Par}$ and $\mu\in\operatorname{Par}$.
Now, fix a partition $\mu\in\operatorname{QPar}$. Then, Theorem 2 (applied to
$\lambda=\mu$) yields $z\_{\mu}^{-1}p\_{\mu}\in V\subseteq W$ (by Proposition
5). Hence, $z\_{\mu}^{-1}p\_{\mu}\in W=\Lambda\_{\mathbb{Z}\_{\left( q\right) }
}+J$. In other words, we can write $z\_{\mu}^{-1}p\_{\mu}$ in the form $z\_{\mu
}^{-1}p\_{\mu}=w\_{1}+w\_{2}$ for some $w\_{1}\in\Lambda\_{\mathbb{Z}\_{\left(
q\right) }}$ and some $w\_{2}\in J$. Consider these $w\_{1}$ and $w\_{2}$.
Now, it is easy to see that $\left\langle f,w\_{1}\right\rangle \in
\mathbb{Z}\_{\left( q\right) }$. [*Proof:* We have $f\in\Lambda
\_{\mathbb{Z}\_{\left( q\right) }}\cap\mathbb{Q}\left[ p\_{i}\ \mid
\ i\not \equiv 0\mod q\right] \subseteq\Lambda\_{\mathbb{Z}
\_{\left( q\right) }}$ and $w\_{1}\in\Lambda\_{\mathbb{Z}\_{\left( q\right) }}$. Thus, $\left(f, w\_1\right) \in \Lambda\_{\mathbb{Z}\_{\left( q\right) }}\times\Lambda\_{\mathbb{Z}\_{\left( q\right) }}$.
Hence, $\left\langle f,w\_{1}\right\rangle \in\mathbb{Z}\_{\left( q\right)
}$, because the Hall inner product $\left\langle \cdot,\cdot\right\rangle $
sends $\Lambda\_{\mathbb{Z}\_{\left( q\right) }}\times\Lambda\_{\mathbb{Z}
\_{\left( q\right) }}$ to $\mathbb{Z}\_{\left( q\right) }$.]
Furthermore, it is easy to see that $\left\langle f,w\_{2}\right\rangle =0$.
[*Proof:* We have $w\_{2}\in J$; thus, we can write $w\_{2}$ as a $\mathbb{Q}
$-linear combination of the family $\left( p\_{\lambda}\right) \_{\lambda
\in\operatorname{Par}\setminus\operatorname{QPar}}$ (since this family is a
basis of the $\mathbb{Q}$-vector space $J$). In other words, we can write
$w\_{2}$ in the form
\begin{equation}
w\_{2}=\sum\_{\beta\in\operatorname{Par}\setminus\operatorname{QPar}}d\_{\beta
}p\_{\beta}
\label{darij1.pf.t1.w2=}
\tag{10}
\end{equation}
for some coefficients $d\_{\beta}\in\mathbb{Q}$. Consider these $d\_{\beta}$.
From \eqref{darij1.pf.t1.f=} and \eqref{darij1.pf.t1.w2=}, we obtain
\begin{align\*}
\left\langle f,w\_{2}\right\rangle & =\left\langle \sum\_{\lambda
\in\operatorname{QPar}}c\_{\lambda}p\_{\lambda},\sum\_{\beta\in
\operatorname{Par}\setminus\operatorname{QPar}}d\_{\beta}p\_{\beta
}\right\rangle \\
& =\sum\_{\lambda\in\operatorname{QPar}}\ \ \sum\_{\beta\in\operatorname{Par}
\setminus\operatorname{QPar}}c\_{\lambda}d\_{\beta}\underbrace{\left\langle
p\_{\lambda},p\_{\beta}\right\rangle }\_{\substack{=z\_{\beta}\delta
\_{\lambda,\beta}\\\text{(by \eqref{darij1.pf.t1.dualbases2})}}}\\
& =\sum\_{\lambda\in\operatorname{QPar}}\ \ \sum\_{\beta\in\operatorname{Par}
\setminus\operatorname{QPar}}c\_{\lambda}d\_{\beta}z\_{\beta}\underbrace{\delta
\_{\lambda,\beta}}\_{\substack{=0\\\text{(since }\lambda\neq\beta
\\\text{(because }\lambda\in\operatorname{QPar}\\\text{whereas }\beta
\in\operatorname{Par}\setminus\operatorname{QPar}\text{))}}}\\
& =\sum\_{\lambda\in\operatorname{QPar}}\ \ \sum\_{\beta\in\operatorname{Par}
\setminus\operatorname{QPar}}c\_{\lambda}d\_{\beta}z\_{\beta}0=0,
\end{align\*}
qed.]
From $z\_{\mu}^{-1}p\_{\mu}=w\_{1}+w\_{2}$, we obtain
\begin{align\*}
\left\langle f,z\_{\mu}^{-1}p\_{\mu}\right\rangle =\left\langle f,w\_{1}
+w\_{2}\right\rangle =\underbrace{\left\langle f,w\_{1}\right\rangle }
\_{\in\mathbb{Z}\_{\left( q\right) }}+\underbrace{\left\langle f,w\_{2}
\right\rangle }\_{=0}\in\mathbb{Z}\_{\left( q\right) }.
\end{align\*}
On the other hand, from \eqref{darij1.pf.t1.f=}, we obtain
\begin{align\*}
\left\langle f,z\_{\mu}^{-1}p\_{\mu}\right\rangle & =\left\langle \sum
\_{\lambda\in\operatorname{QPar}}c\_{\lambda}p\_{\lambda},z\_{\mu}^{-1}p\_{\mu
}\right\rangle =\sum\_{\lambda\in\operatorname{QPar}}c\_{\lambda}
\underbrace{\left\langle p\_{\lambda},z\_{\mu}^{-1}p\_{\mu}\right\rangle
}\_{\substack{=\delta\_{\lambda,\mu}\\\text{(by \eqref{darij1.pf.t1.dualbases})}
}}=\sum\_{\lambda\in\operatorname{QPar}}c\_{\lambda}\delta\_{\lambda,\mu}\\
& =c\_{\mu}.
\end{align\*}
Hence,
\begin{align\*}
c\_{\mu}=\left\langle f,z\_{\mu}^{-1}p\_{\mu}\right\rangle \in\mathbb{Z}\_{\left(
q\right) }.
\end{align\*}
Forget that we fixed $\mu$. We thus have shown that $c\_{\mu}\in\mathbb{Z}
\_{\left( q\right) }$ for each $\mu\in\operatorname{QPar}$. In other words,
$c\_{\lambda}\in\mathbb{Z}\_{\left( q\right) }$ for each $\lambda
\in\operatorname{QPar}$. Hence, \eqref{darij1.pf.t1.f=} becomes
\begin{align\*}
f=\sum\_{\lambda\in\operatorname{QPar}}\underbrace{c\_{\lambda}}\_{\in
\mathbb{Z}\_{\left( q\right) }}p\_{\lambda}\in\sum\_{\lambda\in
\operatorname{QPar}}\mathbb{Z}\_{\left( q\right) }p\_{\lambda}=\mathbb{Z}
\_{\left( q\right) }\left[ p\_{i}\ \mid\ i\not \equiv 0\mod
q\right]
\end{align\*}
(by the definition of $\operatorname{QPar}$).
Forget that we fixed $f$. We thus have shown that $f\in\mathbb{Z}\_{\left(
q\right) }\left[ p\_{i}\ \mid\ i\not \equiv 0\mod q\right] $
for each $f\in\Lambda\_{\mathbb{Z}\_{\left( q\right) }}\cap\mathbb{Q}\left[
p\_{i}\ \mid\ i\not \equiv 0\mod q\right] $. In other words,
\begin{align\*}
\Lambda\_{\mathbb{Z}\_{\left( q\right) }}\cap\mathbb{Q}\left[ p\_{i}
\ \mid\ i\not \equiv 0\mod q\right] \subseteq \mathbb{Z}\_{\left( q\right) }\left[ p\_{i}\ \mid\ i\not \equiv 0\mod q\right] .
\end{align\*}
As explained, this completes the proof of Theorem 1. $\blacksquare$
|
3
|
https://mathoverflow.net/users/2530
|
403672
| 165,593 |
https://mathoverflow.net/questions/403670
|
2
|
Let $\mathcal{A}=\{A\_1, A\_2, \ldots, A\_m\}$ be a uniformly random set partition of $[n]$.
What can we say about $||\mathcal{A}||\_2 = \sqrt{\sum\_{i=1}^m |A\_i|^2}$? It is clearly upper bounded by $n$, but since the ``typical'' size of these $A\_i$ is more like $\log(n) - \log\log(n)$, it seems reasonable to expect that $||\mathcal{A}||\_2$ is actually more like $O\left(\sqrt{n\log(n)}\right)$, or something like that. What I really need is two things: a) whether $||\mathcal{A}||\_2$ concentrates around its expectation, and b) upper bounds on $\mathbb{E}\left[||\mathcal{A}||\_2\right]$ better than the trivial $n$ bound.
In my search for papers on this topic, I've found the following result: If $Z\_i = |\{j: |A\_j| = i\}|$ is the number of parts of size $i$, then the vector $(Z\_1, Z\_2, \ldots, Z\_n)$ is distributed as $Z\_i \sim \mathrm{Po}\left(\frac{r^i}{i}\right)$ (where $r=r(n)$ is defined by $re^r = n$), independent other than for conditioning on $\sum\_i iZ\_i = n$. So my question could equivalently be phrased as asking about
$$\mathbb{E}\left[\sqrt{\sum\_i i^2 Z\_i}\, \middle|\, \sum\_i i Z\_i = n \right]$$
but it's not clear to me if that is any easier than the original question.
|
https://mathoverflow.net/users/58551
|
"Shape"/"norm" of a uniformly random set partition
|
Since $E[\|\mathcal{A}\|\_{2}^{2}]$ has no square roots, a good strategy for this problem will be to first compute $E[\|\mathcal{A}\|\_{2}^{2}]$ and then deal with $E[\|\mathcal{A}\|\_{2}]$ only after we are comfortable with $E[\|\mathcal{A}\|\_{2}^{2}]$.
We compute $$E[\|\mathcal{A}\|\_{2}^{2}]=B\_{n}^{-1}\sum\_{P\in\mathbb{P}([n])}\sum\_{R\in P}|R|^{2}=B\_{n}^{-1}\sum\_{R\subseteq[n]}\sum\_{P\in\mathbb{P}([n]),R\in P}|R|^{2}$$
$$=B\_{n}^{-1}\sum\_{R\subseteq[n],R\neq\emptyset}|R|^{2}\cdot B\_{n-|R|}=B\_{n}^{-1}\cdot\sum\_{k=1}^{n}\binom{n}{k}B\_{n-k}k^{2}.$$
Here $\mathbb{P}([n])$ denotes the lattice of partitions of $[n]$ while $B\_{r}$ denotes the $r$-th Bell number.
We now look up $B\_{n}\cdot E[\|\mathcal{A}\|\_{2}^{2}]$ from the OEIS, and this is sequence [A175716](https://oeis.org/A175716), and we obtain a simpler formula. From the OEIS, we obtain another formula for the expected value of the norm squared
$$B\_{n}\cdot E[\|\mathcal{A}\|\_{2}^{2}]=n\big{[}(n-1)B\_{n-1}+B\_{n}\big{]}.$$
By Jensen's inequality (or the formula of the variance), we have $E[\|\mathcal{A}\|\_{2}]^{2}\leq E[\|\mathcal{A}\|\_{2}^{2}],$ so
$$E[\|\mathcal{A}\|\_{2}]\leq\sqrt{n[1+(n-1)B\_{n-1}B\_{n}^{-1}]}.$$
For a lower bound of $E[\|\mathcal{A}\|\_{2}]$, we have
$\|\mathcal{A}\|\_{2}\geq\frac{n}{\sqrt{m}}$ where $|\mathcal{A}|=m$. Therefore, by Jensen's inequality, we have
$$E(\|\mathcal{A}\|\_{2})\geq E(\frac{n}{\sqrt{m}})\geq\frac{n}{\sqrt{E(m)}}.$$
However, we know that $E(m)=(B\_{n+1}/B\_{n})-1$.
Therefore, $$\frac{n}{\sqrt{(B\_{n+1}/B\_{n})-1}}\leq E(\|\mathcal{A}\|\_{2})\leq\sqrt{n[1+(n-1)B\_{n-1}B\_{n}^{-1}]}.$$
We therefore conclude that
$$\text{Var}(\|\mathcal{A}\|\_{2})=E(\|\mathcal{A}\|\_{2}^{2})-E(\|\mathcal{A}\|)^{2}\leq n[1+(n-1)B\_{n-1}B\_{n}^{-1}]-\frac{n^{2}}{(B\_{n+1}/B\_{n})-1}.$$
We can therefore conclude that $\|\mathcal{A}\|\_{2}$ does concentrate around its expected value.
|
2
|
https://mathoverflow.net/users/22277
|
403677
| 165,597 |
https://mathoverflow.net/questions/403502
|
1
|
Write $A=(x\_{ij})$ for the generic matrix (comprised of indeterminates) defined over $\mathbb Z[x\_{11},\dots,x\_{nn}]$. In [their constructive commutative algebra book](https://arxiv.org/abs/1605.04832v3), Lombardi and Quitte write that the determinant of the family $(e\_1,Ae\_1,\dots,A^{n-1}e\_1)$ is nonzero. Hence this is a cyclic basis whence the generic matrix is similar to the companion matrix of its characteristic polynomial over the fraction field of $\mathbb Z[x\_{ij}]$. If I understand correctly, being non-zero in the polynomial *domain* ensures the localization at the determinant is injective and there we already have the similarity. On the other hand, the determinant is nonzero because it can be specialized to one which is nonzero.
Now, Chapter III proposition 5.3 "the generic matrix is diagonalizable" has me a little stuck. The proof is as follows.
The authors first claim the coefficients of the characteristic polynomial of $A$ are algebraically independent over $\mathbb Z$. They write "to realize this, it suffices to speicalize $A$ as the companion matrix of a generic monic polynomial". Does this mean passing to a basis in which the representation is the companion matrix of the characteristic polynomial? How does this prove the assertion?
They then claim the discriminant of the characteristic polynomial is nonzero. This makes sense because otherwise every split characteristic polynomial would have a repeated root, but it seems the remainder of the proof makes no mention of the algebraic independence of the roots. What am I missing?
|
https://mathoverflow.net/users/69037
|
Properties of the generic matrix - struggles with constructive proofs
|
Let me answer both of your explicitly asked questions in detail; if you have any further questions on the proof (which is indeed fast-going and slightly handwavy), please add them to your post.
>
> **Question 1.** Why are the $s\_1, s\_2, \ldots, s\_n$ algebraically independent over $\mathbb{Z}$ ? (I am using the notations from the book.)
>
>
>
*Answer:* Let $\mathbf{R}$ be the polynomial ring $\mathbb{Z}\left[c\_0, c\_1, \ldots, c\_{n-1}\right]$ in $n$ new indeterminates $c\_0, c\_1, \ldots, c\_{n-1}$ over $\mathbb{Z}$. Let $C$ be the $n\times n$-matrix
\begin{align}
\begin{pmatrix}
0 & 0 & \dots & 0 & -c\_0 \\
1 & 0 & \dots & 0 & -c\_1 \\
0 & 1 & \dots & 0 & -c\_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & 1 & -c\_{n-1}
\end{pmatrix}
\end{align}
over $\mathbf{R}$. This matrix $C$ is the [companion matrix](https://en.wikipedia.org/wiki/Companion_matrix) of the polynomial $c\_0 + c\_1 T + c\_2 T^2 + \cdots + c\_{n-1} T^{n-1} + T^n \in \mathbf{R}\left[T\right]$ (in fact, I even stole the LaTeX code from the Wikipedia article). Now, let $\phi$ be the $\mathbb{Z}$-algebra homomorphism from $\mathbf{A}$ to $\mathbf{R}$ that sends each entry of the matrix $\left(a\_{i,j}\right)$ to the corresponding entry of $C$ (that is, $a\_{i,j}$ goes to $-c\_{i-1}$ if $j = n$, goes to $1$ if $i = j+1$, and goes to $0$ otherwise). This $\phi$ is uniquely determined, due to the universal property of the polynomial ring $\mathbf{A}$. This homomorphism $\phi$ canonically induces a $\mathbb{Z}\left[T\right]$-algebra homomorphism $\phi\left[T\right] : \mathbf{A}\left[T\right] \to \mathbf{R}\left[T\right]$ (which simply applies $\phi$ to each coefficient independently). This latter homomorphism $\phi\left[T\right]$ sends the characteristic polynomial of the matrix $\left(a\_{i,j}\right)$ to the characteristic polynomial of the matrix $C$ (since $\phi$ sends the matrix $\left(a\_{i,j}\right)$ to the matrix $C$, and since each coefficient of the characteristic polynomial of a matrix is a universal polynomial in the entries of the matrix). In other words, $\phi\left[T\right]$ sends the polynomial $f\left(T\right)$ to the polynomial $c\_0 + c\_1 T + c\_2 T^2 + \cdots + c\_{n-1} T^{n-1} + T^n$. Thus, $\phi$ sends each coefficient $\left(-1\right)^{n-i}s\_{n-i}$ of the former polynomial to the corresponding coefficient $c\_i$ of the latter. Since the $c\_0, c\_1, \ldots, c\_{n-1}$ are algebraically independent over $\mathbb{Z}$ (by their definition as distinct indeterminates!), we thus conclude that the $\left(-1\right)^n s\_n, \left(-1\right)^{n-1} s\_{n-1}, \ldots, -s\_1$ are algebraically independent over $\mathbb{Z}$ as well (because any polynomial relation between them would be mapped by $\phi$ to a polynomial relation between $c\_0, c\_1, \ldots, c\_{n-1}$, which would contradict the algebraic independence of the latter). Hence, the $s\_1, s\_2, \ldots, s\_n$ are algebraically independent over $\mathbb{Z}$ (because changing the order of a bunch of elements and flipping some of their signs clearly cannot damage their algebraic independence).
>
> **Question 2.** Why is the discriminant of $f$ nonzero?
>
>
>
*Answer:* We know that the coefficients $\pm s\_1, \pm s\_2, \ldots, \pm s\_n$ of $f$ are algebraically independent. Thus, $f$ is "as good as" a generic monic polynomial of degree $n$ (in the sense that there is a ring isomorphism from the above-mentioned polynomial ring $\mathbf{R} = \mathbb{Z}\left[c\_0, c\_1, \ldots, c\_{n-1}\right]$ to a subring of $\mathbf{A}$ that sends the generic monic polynomial $c\_0 + c\_1 T + c\_2 T^2 + \cdots + c\_{n-1} T^{n-1} + T^n$ to $f$). Thus, proving that the discriminant of $f$ is nonzero is equivalent to proving that the discriminant of the generic monic polynomial $c\_0 + c\_1 T + c\_2 T^2 + \cdots + c\_{n-1} T^{n-1} + T^n$ is nonzero. But the latter is easy: If the discriminant of the generic monic polynomial $c\_0 + c\_1 T + c\_2 T^2 + \cdots + c\_{n-1} T^{n-1} + T^n$ was zero, then the discriminant of **every** monic polynomial of degree $n$ would be $0$ (since it could be obtained by specializing the $c\_0, c\_1, \ldots, c\_{n-1}$ in the discriminant of the generic monic polynomial); but this would contradict the fact that (for example) the monic polynomial $T^n - 1$ has nonzero discriminant.
|
1
|
https://mathoverflow.net/users/2530
|
403678
| 165,598 |
https://mathoverflow.net/questions/403507
|
1
|
**I asked this question in <https://math.stackexchange.com/q/4236870/528430>, but did not get any help.**
I got stuck with the following while going through the proof of [Lemma 3.21](https://drive.google.com/file/d/1FqoY7rZhEV6miICQVGX-5a9M5b3SoTIB/view?usp=sharing) from the book 'Ergodic Theory: Independence and Dichotomies' by Kerr and Li.
**Problem:** If $(X,\mu,T)$ is a weakly mixing extension of $(Y,\nu,S)$, then $$\{f\in L^2(X|Y):f\circ T=f\}\subseteq L^{\infty}(Y).$$
Here I recall the following definitions which are used for the above problem.
**Definition 1:** Let $(X,\mu)$ and $(Y,\nu)$ be two probability measure space. Let $T:X\rightarrow X$ and $S:Y\rightarrow Y$ be two invertible measure preserving transformations. We say that $(X,\mu,T)$ is an extension of $(Y,\nu,S)$ if there is a $T$ invariant conull set $X'\subseteq X$ and an equivariant measurable map $\pi : X'\rightarrow Y$ such that $\mu (\pi^{-1}(A))=\nu (A)$ for all measurable $A\subseteq Y$.
**Definition 2:** Given an extension $(X,\mu,T)\rightarrow (Y,\nu,S)$, let $\mathbb{E}\_Y:L^2(X)\rightarrow L^2(Y)$ be the conditional expectation. Then $L^2(X|Y)$ is the completion of $L^{\infty}(X)$ with respect to the norm $\|f\|:=\|\mathbb{E}\_Y(f\cdot\overline{f})\|^{1/2}$.
**Definition 3:** An element $f\in L^2(X|Y)$ is said to be conditionally weakly mixing if the mean $$\lim\_{n\rightarrow\infty}\frac{1}{n}\sum\_{s=-n}^{s=n}\|\mathbb{E}\_Y((f\circ T^s)\cdot\overline{f})\|=0.$$
**Definition 4:** The extension $(X,\mu,T)\rightarrow (Y,\nu,S)$ is said to be weakly mixing if every element in $L^2(X|Y)$ orthogonal to $L^{\infty}(Y)$ is conditionally weakly mixing.
Thanks in advance for any help or suggestion.
|
https://mathoverflow.net/users/83956
|
The mean ergodic theorem for weakly mixing extension
|
I suspect that your confusion stems from misinterpreting the norm on $L^2(X|Y)$ in your definition 2. You should note that $\mathbb{E}\_Y$ takes $L^\infty(X)$ to $L^\infty(Y)$ and the new norm taken on $L^\infty(X)$ is with respect to the $L^\infty$-norm on $L^\infty(Y)$, not the $L^2$-norm. That is $\|f\|:=\|\mathbb{E}\_Y(f\cdot\overline{f})\|\_\infty^{1/2}$ for $f\in L^\infty(X)$.
It is now easy to see that $L^2(X|Y)$ decomposes equivariantly as an $L^\infty(Y)$-Hilbert module to $L^\infty(Y)\oplus L^\infty(Y)^\perp$.
Accordingly, you get a corresponding decomposition of $L^2(X|Y)^T$ and you are done by observing that a $T$-invariant weakly mixing element must be 0.
|
1
|
https://mathoverflow.net/users/89334
|
403695
| 165,605 |
https://mathoverflow.net/questions/403702
|
1
|
Let $X$ be a smooth projective variety, let $E$ be a vector bundle of rank $4$ on $X$ and let $L$ be a line budnle on $X$. Consider the projectivization $\mathbb{P}\_X(E):=\mathrm{Proj}Sym(E^\*)$ of the vector bundle $E$. Denote the projection $\mathbb{P}\_X(E)\to X$ by $\pi$ and the Serre sheaf of the projectivization by $\mathcal{O}\_{\pi}(1)$. A section of the vector bundle $S^3E^\*\otimes L$ defines a family of cubic surfaces $p:\mathcal{S}\to X$ as the zero locus of a section the line bundle $\mathcal{O}\_{\pi}(3)\otimes \pi^\*L$ on $\mathbb{P}\_X(E)$. Let $\Delta\subset X$ be the set of points $x \in X$ such that the cubic surface $\mathcal{S}\_x:=p^{-1}(x)$ is singular.
**The question**: Assume a general cubic surface $\mathcal{S}\_x:=p^{-1}(x)$ is smooth. Is it true that $\Delta$ is either a divisor in $X$ or empty? Is there a formula for the divisor $\Delta\subset X$ in terms of $E$ and $L$?
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https://mathoverflow.net/users/98256
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Singular locus of a family of cubic surfaces
|
The discriminant of a degree $d$ polynomial in $n$ variables has degree $n (d-1)^{n-1}$, so the discriminant of a cubic in four variables is $4 \cdot 2^3 = 32$.
The discriminant is, by construction, invariant under $SL\_4$. If we look at scalars in $GL\_4$, they act on cubic polynomials by multiplication by the inverse cube of the scalar, so they act on polynomials of degree $32$ in the coefficients of a cubic polynomial by multiplication by the inverse $3 \times 32$ power.
Since the discriminant is invariant under $SL\_4$ and scalars act by the power $-96$, it must be equivariant under $GL\_4$ for the character $\det^{-24}$.
Thus, for a vector bundle $E$, and line bundle $L$, the discriminant of a section of $L \otimes S^3 E^\*$ is a section of $L^{32}\otimes \det E^{ - 24}$.
For a consistency check, note that if we tensor $E$ with a line bundle and $L$ with the third power of that line bundle, then both $L \otimes S^3 E^\*$ and $L^{32}\otimes \det E^{ - 24}$ are preserved.
|
3
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https://mathoverflow.net/users/18060
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403705
| 165,606 |
https://mathoverflow.net/questions/403696
|
0
|
**EDIT**: Let $\Omega\subset \mathbb{R}^n$ be a bounded domain with smooth boundary. Let $f\colon \mathbb{R}^n\backslash \Omega \to \mathbb{R}$ be a continuous function which is harmonic in $\mathbb{R}^n\backslash \bar\Omega$ and vanishes on $\partial \Omega$. Let us also assume that $f$ vanishes at infinity.
**What assumptions on the decay rate at infinity imply that $f\equiv 0$?**
The case $n=3$ is of special interest to me.
**ADD:** The motivation of my question comes from the very classical problem from electrostatics ($n=3$) which is probably solved. Assume the domain $\Omega$ is filled with a conductor and electrified with a charge. All the charged is necessarily accumulated on the surface of the domain. The potential of the created electric field in the space is a harmonic function outside if the domain and is constant on the boundary. It decays at infinity like $1/r^2$. **Is such potential unique? Equivalently, is distribution of charge on the surface is unique?**
Probably this is a very well studied question, but I am not a specialist.
|
https://mathoverflow.net/users/16183
|
Harmonic functions in infinite domain in Euclidean space
|
Assuming "vanish at infinity" means that
>
> for every $\epsilon > 0$ there exists a sufficiently large ball $B\_\epsilon$ such that $\big|f|\_{\mathbb{R}^n \setminus B\_\epsilon}\big| < \epsilon$
>
>
>
then you can just apply the maximum principle to $B\_\epsilon \setminus \Omega$ and conclude that $f$ is bounded by $\epsilon$ on that set.
Take $\epsilon \to 0$ you get that $f$ must vanish identically.
No decay rate needed.
|
2
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https://mathoverflow.net/users/3948
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403709
| 165,608 |
https://mathoverflow.net/questions/403289
|
4
|
One of the generalizations of algebraic geometry is provided by the theory of semiring schemes, cf. [Lorscheid 2012](https://arxiv.org/abs/1212.3261). The theory follows the same set up of scheme theory, but we use semirings instead of rings.
Given a semiring $R$, we have a semiringed space $\mathrm{Spec}(R)$, defined by mimicking the usual definition for rings. This gives a functor $\mathrm{Spec}$ from the opposite of the category semirings to that of affine semiring schemes.
Conversely, there's also a global sections functor $\Gamma:\mathrm{AffSemiSch}^\circ\to\mathrm{Semiring}$ sending a semiringed space $(X,\mathcal{O}\_X)$ to $\Gamma(X,\mathcal{O}\_X)$.
Does the pair $(\mathrm{Spec},\Gamma)$ give as in ordinary algebraic geometry a contravariant equivalence of categories $\mathrm{Semiring}\cong\mathrm{AffSemiSch}^\circ$?
|
https://mathoverflow.net/users/130058
|
Are affine semiring schemes equivalent to semirings?
|
The thesis Algebraic geometry over semi-structures and hyper-structures of characteristic one by Jaiung Jun [accessible here](https://jscholarship.library.jhu.edu/bitstream/handle/1774.2/37850/JUN-DISSERTATION-2015.pdf?sequence=1&isAllowed=y) gives half of the proof in proposition 2.2.6, the remaining half being easy.
It states that $\mathrm{Spec}:\mathrm{Semiring}\to\mathrm{AffSemiSch}^\circ$ is fully faithful. Since it is essentially surjective by construction, it is an equivalence of categories.
Also $\Gamma(\mathrm{Spec}(R),\mathscr{O}\_{\mathrm{Spec}(R)})\cong R$ by construction, and $\mathrm{Spec}$ is left adjoint to $\Gamma$ by the usual proof ([here is one](https://math.stackexchange.com/a/1526426)). So since $\Gamma$ is right adjoint to an equivalence of categories, it follows that $\Gamma$ itself must be an equivalence, and therefore $\mathrm{Spec}$ and $\Gamma$ are mutually inverse.
---
Edit: a better reference is [Čech cohomology of semiring schemes](https://www.sciencedirect.com/science/article/pii/S0021869317302132), item (3) of proposition 2.1, which explicitly states "The opposite category of affine semiring schemes is equivalent to the category of semirings."!
|
4
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https://mathoverflow.net/users/130058
|
403710
| 165,609 |
https://mathoverflow.net/questions/403659
|
5
|
I'm interested in Laplace Beltrami operators $$-\Delta\_g:\ \ D(-\Delta\_g) \longrightarrow L^2\left(M,\sqrt{|g|}dx\right)$$
on a smooth compact Riemannian Manifold (M,g). Let us fix a unique metric $g$ on $M$.
For any other smooth metric $\widetilde g$ on $M$, we can identify the square integrable functions with respect to its associated volume form with our original $L^2$ above, via the unitary map
$U:f\longmapsto \sqrt{|g|/|\widetilde g|}f.$
Under this unitary identification the Laplace-Beltrami operators corresponding to the various smooth metrics on $M$ yield a family of operators on a common domain of definition.
It has been [proved](https://projecteuclid.org/journals/tohoku-mathematical-journal/volume-35/issue-2/Generic-properties-of-the-eigenvalue-of-the-Laplacian-for-compact/10.2748/tmj/1178229047.full) that the eigenvalues of this family depend continuously on the metrics.
Is this true for the eigenfunctions or eigenprojections as well?
I feel like the answer should be yes, arguing in coordinates that if the coefficients of the operators are close, their resolvents and hence their spectral projections are, but I am not sure how to prove it rigorously.
I know that if I consider an analytic one parameter family of metrics the result holds essentially by Kato's perturbation theory.
But I am ultimately interested in proving that a certain composition of maps from functions on the manifold to $\mathbb R$ is robust to arbitrary, small (in the $C^\infty$-topology) changes in the metric and one of the maps in the composition is an operator of the form $f(-\Delta\_g)$.
This is my first question here and also a duplicate of [this](https://math.stackexchange.com/questions/4246996/do-laplace-beltrami-eigenfunctions-vary-continuously-with-the-metric) question on stackexchange, so if I'm transgressing against any etiquette rules, please do let me know. :)
|
https://mathoverflow.net/users/366530
|
Do Laplace-Beltrami eigenfunctions vary continuously with the metric?
|
EDITED: Added clarification, as pointed out by @TerryTao.
Let $g\_1$, $g\_2$ be Riemannian metrics and $\Delta\_1$, $\Delta\_2$ their respective Laplacians. Let $\lambda\_1$ be an eigenvalue of $\Delta\_1$ and $\lambda\_2$ an eigenvalue of $\Delta\_2$.
I think what can be proved is the following: Given an eigenfunction $u\_2$ of $\Delta\_2$ with eigenvalue $\lambda\_2$, there exists an eigenfunction $u\_1$ of $\Delta\_1$ with eigenvalue $\lambda\_1$ such that
$$ \|u\_2-u\_1\|\_2 \le C(|\lambda\_2-\lambda\_1| + \|g\_2-g\_1\|\_{2,2})\|u\_2\|\_2, $$
where $C$ depends on both $g\_1$ and $\lambda\_1$ (specifically, the spectral gap between $\lambda\_1$ and the other eigenvalues of $\Delta\_1$), $\|\cdot\|\_{2,2}$ is the $W^{2,2}$ Sobolev norm with respect to a background metric $g\_0$, and $\|\cdot\|\_2$ is the $L^2$ norm with respect to $dV\_1$.
Here's a sketch of my proof:
Let $\lambda\_2$ be an eigenvalue of $\Delta\_2$ and $u\_2$ a nontrival eigenfunction. Let $\lambda\_1$ be an eigenvalue of $\Delta\_1$ and $u\_1$ a nontrival eigenfunction.
A straighforward calculation shows that
$$
(-\Delta\_1+\lambda\_1)(u\_2-u\_1)
= (\Delta\_2-\Delta\_1)u\_2 - (\lambda\_2-\lambda\_1)u\_2.
$$
Choose $u\_1$ so that, for any eigenfunction $u$ of $\Delta\_1$ with eigenvalue $\lambda\_1$,
$$ \int u(u\_2-u\_1)\,dV\_1 = 0. $$
The claim now follows by elliptic estimates.
|
3
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https://mathoverflow.net/users/613
|
403715
| 165,612 |
https://mathoverflow.net/questions/370814
|
2
|
For $u \in L^\infty(\mathbb R)$ and $\eta\_\epsilon$ [mollifier](https://math.stackexchange.com/questions/1366268/standard-mollifier-comparing-the-definition-in-evans-and-wiki), it is well-known that for the (distributional) derivative it holds that $(u \ast \eta\_\epsilon)' = u'\ast \eta\_\epsilon$.
Is it also true for the fractional Laplacian that
$$(-\Delta)^\alpha (u \ast \eta\_\epsilon) = (-\Delta)^\alpha u \ast \eta\_\epsilon$$
holds? Where can I find a proof of this?
|
https://mathoverflow.net/users/122620
|
Fractional Laplacian and convolution $(-\Delta)^\alpha (u \ast \eta_\epsilon) = (-\Delta)^\alpha u \ast \eta_\epsilon$?
|
Answering the question asked in a comment recently: you can read more about distributional definition of the fractional Laplacian in the paper by Luis Silvestre [1] (or in his PhD thesis, if I remember correctly), in an excellent book by Stefan Samko [2], or in Section 5 of my survey [3]. I do not think any of these references has the statement that you are looking for written explicitly, though.
---
As for the proof, it is in fact a relatively simple consequence of the definition: the distribution $f = (-\Delta)^s u$ is given by
$$ \langle f, w \rangle = \langle u, (-\Delta)^s w \rangle $$
for all infinitely smooth $w$ such that all derivatives of $w$ are absolutely integrable (and $(-\Delta)^s w$ is defined by any of the usual definitions). Now simply apply this to $w(y) = \eta\_\epsilon(x - y)$ for a fixed $x$ to get
$$ ((-\Delta)^s f) \* \eta\_\epsilon(x) = \langle f, w \rangle = \langle u, (-\Delta)^s w \rangle = u \* (-\Delta)^s \eta\_\epsilon(x) = \ldots $$
Now the right-hand side can be written as
$$ \ldots = \int\_{\mathbb R} \int\_{\mathbb R} u(x - y) (\eta\_\epsilon(y + z) - \eta\_\epsilon(y) - z \cdot \nabla \eta\_\epsilon(y) \mathbb 1\_B(z)) dz dy = \ldots $$
(with $B$ denoting the unit ball $(-1, 1)$), and it is not very difficult to see that we may apply Fubini's theorem to get
$$ \begin{aligned} \ldots & = \int\_{\mathbb R} \int\_{\mathbb R} u(x - y) (\eta\_\epsilon(y + z) - \eta\_\epsilon(y) - z \cdot \nabla \eta\_\epsilon(y) \mathbb 1\_B(z)) dy dz \\
& = \int\_{\mathbb R} (u \* \eta\_\epsilon(x + z) - u \* \eta\_\epsilon(x) - z \cdot (u \* \nabla \eta\_\epsilon)(y) \mathbb 1\_B(z)) dz = \ldots \end{aligned} $$
We already know that $u \* \nabla \eta\_\epsilon = \nabla (u \* \eta\_\epsilon)$, so we eventually get
$$ \ldots = (-\Delta)^s (u \* \eta\_\epsilon)(x) ,$$
as desired.
---
If you are familiar with distributional convolution, you can alternatively write $(-\Delta)^s u$ as $L \* u$ for an appropriate distribution $L$, and use associativity of convolution to immediately get
$$ (-\Delta)^s u \* \eta\_\epsilon = (L \* u) \* \eta\_\epsilon = L \* (u \* \eta\_\epsilon) = (-\Delta)^s (u \* \eta\_\epsilon) .$$
Convolution is indeed associative, because $L$ and $\eta\_\epsilon$ are integrable distributions, and $u$ is a bounded distribution.
---
References:
* [1] L. Silvestre, *Regularity of the obstacle problem for a fractional power of the Laplace operator*, Commun. Pure Appl. Math., 60 (2007), 67–112, [DOI:10.1002/cpa.20153](https://doi.org/10.1002/cpa.20153)
* [2] S. Samko, *Hypersingular Integrals and Their Applications*, CRC Press, London–New York, 2001, [DOI:10.1201/9781482264968](https://doi.org/10.1201/9781482264968)
* [3] M. Kwaśnicki, *Fractional Laplace Operator and its Properties*, in: A. Kochubei, Y. Luchko, *Handbook of Fractional Calculus with Applications. Volume 1: Basic Theory*, De Gruyter Reference, De Gruyter, Berlin, 2019, [DOI:10.1515/9783110571622-007](https://doi.org/10.1515/9783110571622-007)
|
1
|
https://mathoverflow.net/users/108637
|
403718
| 165,613 |
https://mathoverflow.net/questions/403674
|
2
|
I seem to recall that the prime number theorem (PNT) is **equivalent** to the fact that the Riemann zeta function $\zeta(s)$ is non-zero on all of $\text{Re}(s) = 1$ (see <https://math.stackexchange.com/questions/1379583/why-is-zeta1it-neq-0-equivalent-to-the-prime-number-theorem> or <https://math.stackexchange.com/questions/706934/riemann-zeta-function-non-vanishing-on-the-line-mathrmre-z-1>), where I suppose "**equivalent**" means that PNT and the non-vanishing of $\zeta(s)$ along $\text{Re}(s)=1$ can be both be proven in a relatively short manner from the other.
Is it the case that Dirichlet's theorem on primes in arithmetic progressions is also **equivalent** to the fact that all the Dirichlet $L$-functions $L(s,\chi)$ are non-zero at $s=1$, in any sense similar to the equivalence described above? Perhaps another way of phrasing the question is: is it "possible" that $L(s,\chi)$ could equal 0 (for some non-trivial character $\chi$) but still Dirichlet's theorem could hold?
It has come to my attention that this has already been discussed here [Analytic equivalents for primes in arithmetic progressions](https://mathoverflow.net/questions/356251/analytic-equivalents-for-primes-in-arithmetic-progressions?rq=1). Apologies all, and thanks for your comments.
|
https://mathoverflow.net/users/112504
|
How essential is the vanishing of the Dirichlet $L$-functions to Dirichlet's theorem on primes in arithmetic progressions?
|
**Theorem**: *Fix a positive integer $m$. The following two conditions are equivalent*:
(1) $L(1,\chi) \not= 0$ *for all nontrivial Dirichlet characters* $\chi \bmod m$.
(2) *For all $a \in (\mathbf Z/m\mathbf Z)^\times$, the set of primes $\{p \equiv a \bmod m\}$ has Dirichlet density $1/\varphi(m)$.*
*Proof*. When $m$ is $1$ or $2$, the condition (1) is vacuously true (there are no nontrivial Dirichlet characters mod $m$) and condition (2) is obvious, so from now on we can let $m \geq 3$.
The usual proof of Dirichlet's theorem shows (1) implies (2). It remains to show (2) implies (1).
For each Dirichlet character $\chi \bmod m$, $L(s,\chi)$ is analytic for ${\rm Re}(s) > 0$ except that $L(s,\mathbf 1\_m)$ has a simple pole at $s = 1$, where $\mathbf 1\_m$ denotes the trivial character mod $m$ (we have $L(s,\mathbf 1\_m) = \zeta(s)\prod\_{p \mid m} (1-1/p^s)$ and $\zeta(s)$ is analytic on ${\rm Re}(s) > 0$ except for a simple pole at $s = 1$).
Set
$$
r\_\chi := {\rm ord}\_{s=1}(L(s,\chi))
$$
so $r\_{\mathbf 1\_m} = -1$ (the simple pole at 1) and $r\_\chi \geq 0$ for all nontrivial $\chi$. Condition (1) is equivalent to $r\_\chi = 0$ for all nontrivial $\chi$, so we want to show condition (2) implies the numbers $r\_\chi$ vanish for all nontrivial $\chi$.
For a Dirichlet character $\chi \bmod m$ and ${\rm Re}(s) > 1$, define
$$
\log L(s,\chi) := \sum\_{p,k} \frac{\chi(p^k)}{kp^{ks}} =
\sum\_p \frac{\chi(p)}{p^s} + \sum\_{p,k\geq 2} \frac{\chi(p^k)}{kp^{ks}}.
$$
This is a logarithm of $L(s,\chi)$ (meaning the exponential of that series is $L(s,\chi)$. The second series on the right is absolutely convergent for ${\rm Re}(s) > 1/2$, with absolute value bounded by $\sum\_{p,k \geq 2} 1/(kp^{k\sigma})$, so for $s > 1$ we can say
$$
\log L(s,\chi) = \sum\_p \frac{\chi(p)}{p^s} + O(1),
$$
where the $O$-constant is $\sum\_{p,k \geq 2} 1/(kp^k)$. In the usual proof of Dirichlet's theorem, for each $a \in (\mathbf Z/m\mathbf Z)^\times$ and $s > 1$ we write
$$
\sum\_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)}\sum\_{\chi \bmod m} \overline{\chi}(a)\left(\sum\_p \frac{\chi(p)}{p^s}\right),
$$
so
$$
\sum\_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)}\sum\_{\chi \bmod m} \overline{\chi}(a)\log L(s,\chi) + O(1),
$$
where the $O$-constant on the right is an overall term (outside the sum).
Now let's bring in the order of vanishing $r\_\chi$. For all $s$ near $1$, $L(s,\chi) = (s-1)^{r\_\chi}f\_\chi(s)$ where $f\_\chi(s)$ is a holomorphic function in a neighborhood of $s = 1$ (in fact it is holomorphic on ${\rm Re}(s) > 0$, or even $\mathbf C$) and $f\_\chi(1) \not= 0$. Therefore $f\_\chi(s)$ has a logarithm around $s = 1$ (well-defined up to adding an integer multiple of $2\pi$), so for $s > 1$
$$
\log L(s,\chi) = r\_\chi\log(s-1) + \ell\_{f\_\chi}(s)
$$
where $\ell\_{f\_\chi}(s)$ is a suitable logarithm of $f\_\chi(s)$. Thus
$\log L(s,\chi) = r\_\chi\log(s-1) + O(1)$ for $s$ near $1$ to the right, and plugging this into the formula for
$\sum\_{p \equiv a \bmod m} 1/p^s$ we can say
$$
\sum\_{p \equiv a \bmod m} \frac{1}{p^s} =
\frac{1}{\varphi(m)}\sum\_{\chi \bmod m} \overline{\chi}(a)(r\_\chi\log(s-1)) + O(1).
$$
Let's extract the term for the trivial character mod $m$: since
$r\_{\mathbf 1\_m} = -1$,
$$
\sum\_{p \equiv a \bmod m} \frac{1}{p^s} = -\frac{1}{\varphi(m)}\log(s-1) +
\frac{1}{\varphi(m)}\left(\sum\_{\chi \not= \mathbf 1\_m} \overline{\chi}(a)r\_\chi\right)\log(s-1) + O(1).
$$
In order to bring in a Dirichlet density,
we want to divide both sides by $\sum\_p 1/p^s$ for
$s$ near $1$ to the right. For such $s$,
$$
\log \zeta(s) = -\log(s-1) + O(1)
$$
from the simple pole of $\zeta(s)$ at $s = 1$ and
$$
\log \zeta(s) = \sum\_p \frac{1}{p^s} + O(1)
$$
from the Euler product for $\zeta(s)$ when $s > 1$. Therefore $\sum\_p 1/p^s = -\log(s-1) + O(1)$ as $s \to 1^+$, so $-\log(s-1) \sim \sum\_p 1/p^s$ as $s \to 1^+$.
Dividing through the last (big) formula above for $\sum\_{p \equiv a \bmod m} 1/p^s$ by $\sum\_p 1/p^s$ and letting $s \to 1^+$, we get
$$
\frac{\sum\_{p \equiv a \bmod m} 1/p^s}{\sum\_p 1/p^s} \to \frac{1}{\varphi(m)} -
\frac{1}{\varphi(m)}\left(\sum\_{\chi \not= \mathbf 1\_m} \overline{\chi}(a)r\_\chi\right)
$$
as $s \to 1^+$. So we have shown, without assuming condition (1) in the theorem, that for all $a \in (\mathbf Z/m\mathbf Z)^\times$ the set of primes $\{p \equiv a \bmod m\}$ has Dirichlet density
$$
\frac{1}{\varphi(m)}\left(1 - \sum\_{\chi \not= \mathbf 1\_m} \overline{\chi}(a)r\_\chi\right).
$$
Finally it is time to assume condition (2) in theorem, which implies
$$
\sum\_{\chi \not= \mathbf 1\_m} \overline{\chi}(a)r\_\chi = 0
$$
for all $a \in (\mathbf Z/m\mathbf Z)^\times$.
When $\chi = \mathbf 1\_m$, $\overline{\chi}(a)r\_\chi = 1(-1) = -1$, so condition (2) implies
$$
\sum\_{\chi} \overline{\chi}(a)r\_\chi = -1
$$
for all $a \in (\mathbf Z/m\mathbf Z)^\times$,
where the sum runs over all Dirichlet characters mod $m$ (including the trivial character). We want to show the above equation, for all $a$, implies $r\_\chi = 0$ for all nontrivial $\chi \bmod m$.
Using vectors indexed by the Dirichlet characters mod $m$, let
$\mathbf r\_m = (r\_\chi)\_\chi$ and
$\mathbf v\_a = (\chi(a))\_\chi$ for each $a \in (\mathbf Z/m\mathbf Z)^\times$. The space of all complex vectors $\mathbf z = (z\_\chi)\_\chi$ has a Hermitian inner product
$\langle \mathbf z, \mathbf w\rangle = \frac{1}{\varphi(m)}\sum\_{\chi} z\_\chi\overline{w\_\chi}$ for which the vectors $\mathbf v\_a$ are an orthonormal basis by the usual orthogonality of Dirichlet characters mod $m$. The equation
$\sum\_\chi \overline{\chi}(a)r\_\chi = -1$ above says
$\langle \mathbf r\_m,\mathbf v\_a\rangle = -1/\varphi(m)$ for
all $a$ in $(\mathbf Z/m\mathbf Z)^\times$, so
$$
\mathbf r\_m = \sum\_{a} \langle \mathbf r\_m,\mathbf v\_a\rangle\mathbf v\_a =
-\frac{1}{\varphi(m)}\sum\_{a}\mathbf v\_a.
$$
For nontrivial Dirichlet characters $\chi \bmod m$, the $\chi$-component of $\sum\_{a} \mathbf v\_a$ is
$\sum\_a \chi(a)$, which is $0$ (the $\mathbf 1\_m$-component is $\varphi(m)$, but that's irrelevant). Since
$\mathbf r\_m$ has $\chi$-component $r\_\chi := {\rm ord}\_{s=1}L(s,\chi)$, we have
$r\_\chi = 0$ for all nontrivial $\chi$, so
$L(1,\chi) \not= 0$ for all nontrivial $\chi$. QED
|
9
|
https://mathoverflow.net/users/3272
|
403725
| 165,615 |
https://mathoverflow.net/questions/403665
|
4
|
Given two positive constants $c\_1,c\_2$ and two independent standard normal random variables $a,b$, how to calculate the following expected value
$$
\mathbb{E}\left[\frac{a^2}{c\_1a^2+c\_2b^2}\right]
$$
Thank you.
|
https://mathoverflow.net/users/366173
|
Expected value of a ratio of squared normal and linear combination of squared normal
|
$\newcommand\E{\mathscr E}$The expectation in question is
$I/c\_1$, where
$$I:=E\frac{a^2}{a^2+cb^2},\quad c:=c\_2/c\_1>0.$$
In turn, using polar coordinates, we get
$$I=\frac1{2\pi}\int\_0^{2\pi}\frac{\cos^2 t\, dt}{\cos^2 t+c\sin^2 t},\quad c:=c\_2/c\_1>0.$$
Further, writing
$$I=\frac4{2\pi}J$$
for
$$J:=\int\_0^{\pi/2}\frac{\cos^2 t\, dt}{\cos^2 t+c\sin^2 t}$$
and using the standard substitution $t=\arctan u$, so that $\cos^2 t=1/(1+u^2)$ and $\sin^2 t=u^2/(1+u^2)$, we have
$$J=\int\_0^\infty\frac{du}{(1+cu^2)(1+u^2)}.$$
Using partial fraction decomposition to compute $J$ and collecting the pieces, we finally get that the expectation in question is
$$\E(c\_1,c\_2):=\frac1{c\_1+\sqrt{c\_1c\_2}}.$$
---
"Sanity" checks: (i) $\E(tc\_1,tc\_2)=\E(c\_1,c\_2)/t$ for all real $t>0$; (ii) $\E(1,0)=1$; (iii) $\E(1,1)=1/2$; (iv) $\E(c\_1,c\_2)\to\infty$ as $c\_1\downarrow0$.
|
4
|
https://mathoverflow.net/users/36721
|
403735
| 165,620 |
https://mathoverflow.net/questions/403720
|
12
|
The irreducible decomposition of the tensor product of two irreducible representations of GL(n) is described by the Littlewood-Richardson rule. This same rule also governs the decomposition of the product of two Schur polynomials into a linear combination of Schur polynomials. In both cases, we label the components of the product and the decomposition with Young diagrams, or integer partitions.
We noticed that the Young diagrams in the decomposition always seem to form a lattice w.r.t. dominance order. That is, for each pair of diagrams in the decomposition, the least upper bound ("join") and the greatest lower bound ("meet") of the pair is also part of the decomposition.
For example, looking at the following decomposition (where the Young diagrams are denoted by the corresponding integer partitions):
$$(2,1) \otimes (2,1)=(4,2) \oplus (4,1,1) \oplus (3,3) \oplus 2 (3,2,1) \oplus (3,1,1,1) \oplus (2,2,2) \oplus (2,2,1,1)$$
The two unordered pairs in this decomposition are:
(2,2,2), (3,1,1,1); and (3,3), (4,1,1).
The join and meet of the first pair are (3,1,1,1) and (2,2,1,1) respectively.
For the second pair, these are (4,2) and (3,2,1). In both cases, the join and meet diagrams are parts of the decomposition. We checked it for many larger examples with extensive symbolic calculations, and the product had this lattice property in all cases.
Thus my question is: Applying the Littlewood-Richardson rule to a pair of Young diagrams, does the set of Young diagrams appearing in the result (with nonzero multiplicity) always form a lattice wrt dominance order?
|
https://mathoverflow.net/users/12897
|
Lattice structure (wrt dominance order) on the set of Young diagrams appearing in the decompositions given by the Littlewood-Richardson rule
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Please excuse that I answer with a link, I only have a phone right now.
<http://www.findstat.org/MapsDatabase/Mp00192/>
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4
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https://mathoverflow.net/users/3032
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403740
| 165,622 |
https://mathoverflow.net/questions/403736
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4
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Let $k$ be a field and $M$ the subgroup of $\operatorname{GL}(n,k)$ consisting of permutation matrices. We say a subgroup $G$ of $M$ linearly independent if $G$ is a linearly independent subset of $M\_n(k)$. It is well-known that there is an isomorphism between $M$ and the full symmetric group $S\_n$ with respect to the canonical basis of the affine $n$-dimensional space $k^n$, I identify them in this way. It is clear that any subgroup of a regular subgroup of $S\_n$ is linearly independent. Is there a full description of all linearly independent subgroups of $S\_n$?
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https://mathoverflow.net/users/134942
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Linearly independent subgroups of permutation matrices
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This just deals with the complex case: Here is an initial remark: Since we know that the character afforded by the natural permutation representation of $S\_{n}$ is the sum of the trivial character an a degree $n-1$ irreducible, we know that the $n \times n$ permutation matrices span a space of dimension $1+(n-1)^{2}.$ Hence this is an upper bound for the order of a subgroup $H$ consisting of permutation matrices which are irreducible. But this upper bound can never be attained if $n >2$. If it were, then the $n-1$-dimensional representation of $S\_{n}$ would have to remain irreducible on restriction to $H$, so that $|H|$ would have order divisible by $n-1$, so making $|H| = 1 +(n-1)^{2}$ impossible.
But now $|H| = (n-1)^{2}$ is also impossible for $n >2$, since if $|H| = (n-1)^{2} >1$, $H$ can have no complex irreducible character of degree $n-1$, as such a character would be non-trivial. Thus $H| < (n-1)^{2}$ when $n >2$.
Now it is easy to check via character theory that $|H| \leq 2 + (n-2)^{2}$ when $n >2$, and this can only occur if the restriction of the degree $n-1$ irreducible character to $H$ decomposes as the sum of an irreducible of degree $n-2$ and an irreducible of degree $1$.
This upper bound can be attained for $n = 3$ with $H \cong A\_{3}.$ But the bound can only be attained for $n \leq 4$ since equality forces both $(n-2) | |H|$ and $(n-2) | 2.$ However, even the case $n = 4$, it is not difficult to check that the upper bound of $6$ is not attained.
Thus we have $|H| \leq 1 +(n-2)^{2}$ if $n > 3.$ I am not sure if there is a systematic way to improve this bound for larger $n$.
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1
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https://mathoverflow.net/users/14450
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403751
| 165,624 |
https://mathoverflow.net/questions/403685
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3
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Let $\mathcal P(\mathbb R)$ be the set of probability measures. Set for $\mu,\nu\in\mathcal P(\mathbb R)$
$$d(\mu,\nu) := \inf\left\{\varepsilon>0:~ F\_{\mu}(x-\varepsilon)-\varepsilon \le F\_{\nu}(x)\le F\_{\mu}(x+\varepsilon)+\varepsilon,~ \forall x\in\mathbb R\right\}$$
and
$$\rho(\mu,\nu) := \sup\left\{\int fd\mu- \int fd\nu:~ f \mbox{ is } 1-\mbox{Lipschitz and uniformly bounded by } 1 \right\},$$
where $F\_{\mu}$ (resp. $F\_{\nu}$) denotes the cumulative distribution function of $\mu$ (resp. $\nu$). It is known that the weak convergence on $\mathcal P(\mathbb R)$ is equivalent to the convergence under $d$ and the convergence under $\rho$. Does there exist $C>0$ s.t.
$$\frac{d(\mu,\nu)}{C}~\le~ \rho(\mu,\nu)~\le~ Cd(\mu,\nu),\quad \forall \mu,\nu\in\mathcal P(\mathbb R)?$$
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https://mathoverflow.net/users/261243
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On the weak convergence of probability measures on $\mathbb R$
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For every $\varepsilon\in(0,1)$ there are $\mu$, $\nu$ such that $d(\mu,\nu)=\varepsilon$ and $\rho(\mu,\nu)=\varepsilon^2$.
For example, $\mu=\varepsilon\delta\_0 + (1-\varepsilon)\delta\_2$ and
$\nu=\varepsilon\delta\_\varepsilon + (1-\varepsilon)\delta\_2$.
Hence there is no $C>0$ such that $d(\mu,\nu)\leq C \rho(\mu,\nu)$ for all $\mu,\nu$.
(Addition to address GJC20's comment:) However, $\frac{2d^2}{2+d}\leq\rho\leq 3d$; see Theorem 8.10.43 in Bogachev, *Measure Theory* (Springer 2007).
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3
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https://mathoverflow.net/users/95282
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403756
| 165,627 |
https://mathoverflow.net/questions/403758
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0
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I think that Minimum weight vertex cover problem is NP-easy. However I don't know how to prove that. Does anyone know how to prove it?
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https://mathoverflow.net/users/178444
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Is minimum weight vertex cover problem NP-easy?
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Consider the following decision problem:
>
> Given a vertex-weighted graph $G$, a subset $S$ of its vertices, and a number $k$, is there a vertex cover of $G$ extending $S$ with total weight at most $k$?
>
>
>
This problem is clearly in NP. Let us see how to solve the minimum weight vertex cover problem using it as an oracle.
First, we find the least possible total weight of a vertex cover. This can be done using a bisection method, by repeatedly querying with $S=\varnothing$ and using varying $k$, starting with the sum of weights of all vertices (say). Call this minimal possible value $k\_0$.
Now querying the oracle on $G,k\_0$ and some set $S$ tells us whether there is a minimal weight cover containing the subset $S$, which lets us build a minimum weight cover inductively: first let $S$ range over all $1$-element subsets of vertices. After we find one which works, say $S=\{u\_1\}$, run the oracle for all subsets of the form $\{u\_1,u\_2\}$, and so on. After a number of queries at most quadratic in the number of vertices, we find a minimum weight vertex cover of $G$.
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1
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https://mathoverflow.net/users/30186
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403761
| 165,630 |
https://mathoverflow.net/questions/403750
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2
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I recently calculated the number (possible multidegrees) of Fano complete intersections of dimension $n$ , because I wanted to make the remark that it grows "very rapidly" as $n \rightarrow \infty$
The calculation itself is an "exercise" and in particular it is surely well-known. I would like to know how number theorists/combinatorists would think about the growth of this function.
By the adjunction formula a smooth complete intersection in $\mathbb{P}^k$ of multidegree $(d\_{1}, \ldots d\_{m})$ is Fano $\iff$ $d\_1 + \ldots + d\_m \leq k$.
This implies that the number of possible multidegrees of complete intersection Fano $n$-folds is $$\sum\_{i=0}^{n-1} P(i) $$, where $P$ is that "partition function" studied by Euler and others.
I am aware that many authors don't consider $\mathbb{P}^n$ as a complete intersection. In which case we can take:
$$\sum\_{i=1}^{n-1} P(i) .$$
**Question:** This function of course grows "very rapidly" as $n \rightarrow \infty$. How would a number theorist express that?
Q1. for example is there a "nicest" function $f(n)$ which approximates this?
Q2. Is there some technical term which describes precisely "how fast" this grows?
Also if there is a proof from first principles which is reasonably accessible for someone from another area then that would be great!
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https://mathoverflow.net/users/99732
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The growth of the number of Fano complete intersection families
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Hardy and Ramanujan obtained the asymptotic $$P(n) \approx \frac{1}{4n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}}}$$ which can be summarized as saying that $P(n)$ grows roughly as the exponential of the square root of $n$.
Summing from $1$ to $n$ clearly increases the asymptotic by a factor of at most $n$, so we can still say $P(n)$ grows roughly as the exponential of the square root of $n$, or we can be more precise as follows
$$P (n-c) = \frac{1}{4 (n-c) \sqrt{3}} e^{ \pi \sqrt{\frac{2(n-c)}{3}} } = \frac{1}{4 (n-c) \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}} - \frac{ \pi c}{ \sqrt{6 n }} + O\left( \frac{c^2}{ n^{3/2}}\right) } $$
which for $c = o(n^{3/4})$ is $$\approx \frac{1}{4 n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}} - \frac{ \pi c}{ \sqrt{6 n }} }$$ and for $c> o (n^{3/4})$ is exponentially smaller than $P(n)$ and can be ignored.
$$ \sum\_{i=1}^{n-1} P(i) = \sum\_{c=1}^{n-1} p(n-c) \approx \sum\_{c=1}^{\infty} \frac{1}{4 n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}} - \frac{ \pi c}{ \sqrt{6 n }} }= \frac{1}{4 n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}}} \sum\_{c=1}^{\infty} e^{ - \frac{ \pi c}{ \sqrt{6 n }} } \approx \frac{1}{4 n \sqrt{3}} e^{ \pi \sqrt{\frac{2n}{3}}} \frac{ \sqrt{6n}}{\pi} = \frac{1}{ \sqrt{8n } \pi} e^{ \pi \sqrt{\frac{2n}{3}}} $$
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2
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https://mathoverflow.net/users/18060
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403764
| 165,631 |
https://mathoverflow.net/questions/403745
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1
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Let $k$ be a field of characteristic zero, for example $k=\mathbb{R}$ or $k=\mathbb{C}$.
Of course, $k(x^2,x^3)=k(x)$, since $x=\frac{x^3}{x^2}$.
Let $f\_1,\ldots,f\_n,g\_1,\ldots,g\_m \in k[x]$, $n,m \geq 1$.
Denote: $F(T)=f\_nT^n+\cdots+f\_1T+x^2$
and $G(T)=g\_mT^m+\cdots+g\_1T+x^3$.
Let $h \in k[x]$.
>
> Is it possible to characterize
> $f\_1,\ldots,f\_n,g\_1,\ldots,g\_m \in k[x]$, $n,m \geq 1$,
> such that for *every* $h \in k[x]$ we have: $k(F(h),G(h))=k(x)$?
>
>
>
I could not find such $f\_i,g\_j,n,m$.
**For example:**
$F(T)=T+x^2$, $G(T)=T+x^3$.
For $h=x^3$ we get $F(h)=x^3+x^2$, $G(h)=x^3+x^3=2x^3$,
so clearly, $k(F(h),G(h))=k(x)$.
For $h=x^4-x^3$ we get $F(h)=x^4-x^3+x^2$, $G(h)=x^4-x^3+x^3=x^4$,
abd then, $k(F(h),G(h)) \subsetneq k(x)$.
I think that for arbitrary fixed $f\_i,g\_j,n,m$, there exists $h \in k[x]$ such that
$k(F(h),G(h)) \subsetneq k(x)$, but not sure how to prove this.
I am familiar with D-resultant, but it seems difficult to apply it.
The theorem relevant here is that
$D-\operatorname{Res\_s}(\frac{(F(h))(s)-F(h)}{s-x},\frac{(G(h))(s)-G(h)}{s-x})$
is a non-zero element of $k[x]$ if and only if $k(F(h),G(h))=k(x)$.
See [this paper](https://www.academia.edu/29672226/Jacobian_pairs_D_resultants_and_automorphisms_of_the_plane).
If we apply this to $x^2,x^3$ we obtain that their D-resultant is $x^2$, and indeed
$k(x^2,x^3)=k(x)$.
Also, I wonder if algebraic geometry could help solve my question.
**In addition:**
**(1)** I do not mind to further assume that for every $h \in k$,
$k(F(h),G(h)) = k(x)$.
In the above example, this is satisfied.
**(2)** $f\_1,\ldots,f\_n,g\_1,\ldots,g\_m \in k[x]$,
not necessarily in $k$.
Any comments are welcome; thank you!
Also asked in [MSE](https://math.stackexchange.com/questions/4245950/a-variation-on-kx2-x3-kx) with no comments.
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https://mathoverflow.net/users/72288
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A variation on $k(x^2,x^3)=k(x)$
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Such $F, G$ do exist, unless I missed a criterion.
One possibility is to take $F(T)$ a polynomial such that $F(h)$ is never zero, such as $T^3 + x^2$, and then $G(T) = x F(T)$. Then you can use the same $x = G(T)/F(T)$ trick.
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1
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https://mathoverflow.net/users/18060
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403767
| 165,632 |
https://mathoverflow.net/questions/403699
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0
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For any simple, undirected graph $G=(V,E)$, we denote by $\chi(G)$ the smallest cardinal $\kappa$ such that there is a coloring $c:V \to \kappa$.
We say that $v\neq w\in V$ are *incompatible* if $\{v,w\}\notin E$, and for any coloring $c: V\to \chi(G)$ we have $c(v) \neq c(w)$.
It is easy to see that if $v\neq w\in V$ are incompatible, they must lie in the same connected component of $G$.
What is an example of a graph containing an incompatible pair of vertices?
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https://mathoverflow.net/users/8628
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"Incompatible" pairs with respect to graph coloring
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If $R$ is a poset, then a function $C:R\rightarrow R$ is said to be a closure operator if $r\leq C(r)=C(C(r))$ and $(r\leq s)\rightarrow(C(r)\leq C(s))$ whenever $r,s\in R$. We say that a subset $C\subseteq R$ is a closure system if for all $r\in R$, there is a least $s\in C$ where $r\leq s$. If $R$ is a complete lattice, then $C\subseteq R$ is a closure system if and only if $C$ is closed under arbitrary greatest lower bounds including the empty greatest lower bound. The closure operators and closure systems on a poset are in a one-to-one correspondence. If $C$ is a closure system, then let $C^{\*}:R\rightarrow R$ be the mapping where if $r\in C$, then $C^{\*}(r)$ is the least element $s$ in $C$ with $r\leq s$. Then $C^{\*}$ is a closure operator. Similarly, if $C$ is a closure operator, then define $C^{\*}=\{r\in R\mid r=C(r)\}$. Then $C^{\*}$ is a closure system. Furthermore, if $C$ is a closure operator or closure system, then $C=(C^{\*})^{\*}$.
Suppose that $R,S$ are posets. Then a pair of mappings $f:R\rightarrow S,g:S\rightarrow R$ is said to be a Galois connection if
$s\leq f(r)\leftrightarrow r\leq g(s)$. If $f:R\rightarrow S,g:S\rightarrow R$ is a Galois connection, then the functions $f\circ g,g\circ f$ are closure operators. Furthermore, the maps $f,g$ restrict to order reversing inverse bijections between the closure systems $(f\circ g)^{\*},(g\circ f)^{\*}$.
Suppose that $X,Y$ are sets. Let $R\subseteq X\times Y$ be a relation. Then
define functions $F:P(X)\rightarrow P(Y),G:P(Y)\rightarrow P(X)$ by letting
$F(A)=\{y\in Y\mid\forall x\in A,(x,y)\in R\}$ and
$G(B)=\{x\in X\mid\forall y\in B,(x,y)\in R\}$. Then the functions $F,G$ form a Galois connection.
Let $\lambda$ be a cardinal, and let $V$ be a vertex set. Then define a relation
$R\_{\lambda,V}\subseteq[V]^{2}\times^{V}\lambda$ where if $u,v\in V,u\neq v$ and
$f:V\rightarrow\lambda$ is a function, then $\{u,v\},f)\in R\_{\lambda,V}$ if and only if $f(u)\neq f(v)$. Then from this relation, we define a Galois connection
$F\_{\lambda,V}:P([V]^{2})\rightarrow P(^{V}\lambda),G\_{\lambda,V}:P(^{V}\lambda)\rightarrow P([V]^{2})$ the standard way by setting $F\_{\lambda,V}(R)=\{s\in^{V}\lambda\mid\forall r\in R,(r,s)\in R\_{\lambda,V}\}$ and $G\_{\lambda,V}(S)=\{r\in[V]^{2}\mid\forall s\in S,(r,s)\in R\_{\lambda,V}\}$.
Define closure operators $C\_{\lambda,V}:P([V]^{2})\rightarrow P([V]^{2}),D\_{\lambda,V}:P(^{V}\lambda)\rightarrow P(^{V}\lambda)$ by letting
$C\_{\lambda,V}=G\_{\lambda,V}\circ F\_{\lambda,V},D\_{\lambda,V}=F\_{\lambda,V}\circ G\_{\lambda,V}$.
If $G=(V,E)$, then the following are equivalent:
1. $\chi\_{G}\leq\lambda$.
2. $F\_{\lambda,V}(E)\neq\emptyset$.
3. $C\_{\lambda,V}(E)\neq[V]^{2}$.
We observe that if $\lambda,\mu$ are cardinals with
$\lambda\leq\mu$, then $C\_{\mu,V}(E)\subseteq C\_{\lambda,V}(E)$.
In particular, if $\lambda$ is a cardinal and $F\_{\lambda,V}(E)\neq\emptyset$, then every edge in $C\_{\lambda,V}(E)\setminus E$ is an incompatible edge. In particular, the graph $(V,E)$ has an incompatible edge if and only if $E$ is non-closed and non-dense.
**Finite chromatic number examples**
If $(V,E)$ is a connected bipartite graph with bipartition $A,B$, then
$C\_{2,V}(E)=\{\{a,b\}\mid a\in A,b\in B\}$. Connected incomplete bipartite graphs therefore have incompatible edges.
More generally, as was mentioned by Tony Huynh in the comments, we say that a graph $G=(V,E)$ is uniquely colorable with $k$ colors if it is not colorable with $k-1$ colors and where there is only one coloring $f:V\rightarrow k$ up to permutations (i.e. if $f,g:V\rightarrow k$ are colorings, then there is a permutation $h\in S\_{k}$ with $g=h\circ f$). Every complete graph is uniquely colorable. Furthermore, one can easily construct new uniquely colorable graphs from old uniquely colorable graphs. Suppose that $(V,E)$ is a uniquely colorable graph with $k$-colors. Let $f:V\rightarrow k$ be the unique (up-to-permutation) coloring of $(V,E)$. Then let $v$ be a vertex with $v\not\in V$. Then let $v\_{1},\dots,v\_{k}$ be vertices where $f(v\_{1}),\dots,f(v\_{k})$ are distinct. Then $(V\cup\{v\},E\cup\{\{v,v\_{i}\}\mid 1\leq i\leq k\})$ is also uniquely colorable. In particular, every [k-tree](https://en.wikipedia.org/wiki/K-tree) is uniquely colorable with k colors.
If $(V,E)$ is a uniquely colorable graph with $k$ colors, and $f:V\rightarrow k$ is the unique coloring, then whenever $\{u,v\}\not\in E$ and $f(u)\neq f(v)$, the edge $\{u,v\}$ is incompatible with $(V,E)$. In particular, if $$|E|<|V|\cdot(|V|-1)/2-\sum\_{i=0}^{k-1}|f^{-1}[\{i\}]|\cdot(|f^{-1}[\{i\}]|-1)/2,$$
then there is some edge $\{u,v\}$ incompatible with $(V,E)$.
**Deletion-contraction**
There is a deletion-contraction formula for the number of $k$-colorings of a graph, and this deletion-contraction formula applies for infinite graphs as well with infinitely many colorings.
Suppose that $G=(V,E)$ is a graph. If $\simeq$ is an equivalence relation on $V$, then let $G/\simeq$ denote the graph with vertex set $V/\simeq$ and with edge set $E/\simeq=\{\{[u],[v]\}\mid\{u,v\}\in E,u\not\simeq v\}$.
If $\{u,v\}$ is a pair, then let $G/\{u,v\}$ denote the graph $G/\simeq$ where $x\simeq y$ if and only if $x=y$ or $\{x,y\}=\{u,v\}$.
If $(V,E)$ is a graph where $u,v,\in V,u\neq v,\{u,v\}\not\in E$, then by using the same notation as before, we define a function
$$L:F\_{k,V}(E\cup\{\{u,v\}\})\cup F\_{k,V/\{u,v\}}(E/\{u,v\})\rightarrow F\_{\lambda,V}(E)$$
by letting $L(f)=f$ whenever $f\in F\_{k,V}(E\cup\{\{u,v\}\})$ and
$L(f)=f\circ\pi$ where $\pi:V\rightarrow V/\{u,v\}$ is the quotient mapping.
Therefore, $F\_{k,V}(E\cup\{\{u,v\}\})=F\_{k,V}(E)$ if and only if
$F\_{k,V/\{u,v\}}(E/\{u,v\})=\emptyset$. In particular, if $\{u,v\}\in[V]^{2}\setminus E$, then $\{u,v\}$ is an incompatible edge for $(V,E)$ if and only if $$\chi(G/\{u,v\})>\chi(G).$$
**infinite chromatic topology**
Suppose that $\lambda$ is an infinite cardinal. Let $\text{pc}(\lambda)$ be the least cardinal with $\lambda^{\text{pc}(\lambda)}>\lambda$.
Proposition: $C\_{\lambda,V}^{\*}$ is a topological closure system. In the corresponding topology, the union of less than $\text{pc}(\lambda)$ many closed sets is closed, and the intersection of less than $\text{pc}(\lambda)$ many open sets is open.
If $P$ is a partition of a set $V$, then let $U\_{P}=\{\{u,v\}\mid u\neq v,u=v(P)\}$.
In the $C\_{\lambda,V}$-topology, the basic open sets are precisely the sets of the form $U\_{P}$ where $|P|\leq\lambda$.
If $\lambda=|V|$, then the $C\_{\lambda,V}$-topology is discrete.
Proposition: A non-empty subset $A\subseteq[V]^{2}$ is open in the $C\_{\lambda,V}$-topology if and only if $U\_{P}\subseteq A$ for some partition $P$ with
$|P|\leq\lambda$.
Proof: The direction $\rightarrow$ has already been established. For the direction, $\leftarrow$, suppose that $P$ is a partition of $V$ into at most $\lambda$ many sets. Then we shall show that $U\_{P}\cup\{\{p,q\}\}$ is open for each pair $p,q$ with $p\neq q$. The case where $p=q(P)$ is trivial, so assume that $p\neq q(P).$
Let $$Q=(\{\{p,q\}\}\cup\{R\setminus\{p,q\}|R\in P\})\setminus\{\emptyset\}.$$ Then
$$U\_{P}\cup\{\{p,q\}\}=U\_{P}\cup U\_{Q}.$$
Q.E.D.
Therefore, if $A\subseteq[V]^{2}$, then either $A$ is dense in the $C\_{\lambda,V}$ topology or $A$ is closed in the $C\_{\lambda,V}$ topology. Therefore, for each graph $(V,E)$, we know that $C\_{\lambda,V}(E)=[V]^{2}$ or $V\_{\lambda,V}(E)=E$. Therefore, the graph $(V,E)$ can never have an incompatible edge whenever $(V,E)$ has infinite chromatic number.
You can also argue that if $G$ has infinite chromatic number, then $G$ cannot have an incompatible edge by using the facts that that
$\chi(G/\{u,v\})\leq\chi(G)+2$ but in order for $\{u,v\}$ to be incompatible, we must have $\chi(G/\{u,v\})>\chi(G)$.
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2
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https://mathoverflow.net/users/22277
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403771
| 165,633 |
https://mathoverflow.net/questions/403763
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2
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A coalgebra is a triple $(A,\Delta,\epsilon)$ consisting of a vector space, a coproduct, and a counit. Now as we all know, just like the unit in an algebra, the counit of a coalgebra is unique, i.e. if there exists another $\epsilon'$ satisfying the counit axiom, then since
$$
\epsilon(a) = \epsilon(a\_{(1)}\epsilon'(a\_{(2)})) = \epsilon'(\epsilon(a\_{(1)})a\_{(2)})=\epsilon'(a).
$$
So if there is no "choice" of counit, would it not be better define a coalgebra to be a coassociative pair $(A,\Delta)$ for which there exists a counit. This is how we define a unital algebra and a group, so why the different approach for coalgebras?
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https://mathoverflow.net/users/352001
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What is a coalgebra?
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Is it how we define a unital algebra? I’d define a unital algebra as coming endowed with a unit, rather than just asserting one exists. For one reason, then the natural notion of map is a unital map. But at any rate there’s just no distinction here between the definitions algebras and coalgebras, people who define algebras using a unit property will do so for coalgebras too, and people who define algebras using a unit structure will do so for coalgebras too.
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8
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https://mathoverflow.net/users/22
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403776
| 165,636 |
https://mathoverflow.net/questions/403598
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1
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By the central limit theorem (CLT), I mean the Lindeberg-Lévy CLT that says if $X\_1,X\_2,\ldots$ are i.i.d. random variables with $\mathbf{E}[X\_1] = 0$ and $\mathbf{E}[X\_1^2] = 1$, then
$$ \frac{X\_1+\cdots+X\_n}{\sqrt{n}}$$
converges in distribution to the standard normal random variable.
The de Moivre-Laplace theorem, which is historically the first instance of the CLT, is essentially the special case of CLT where $X\_1$ has only two possible values. In other words, it states that in certain cases, binomial distributions can be approximated by normal distributions. The de Moivre-Laplace theorem can be proved by direct computation, where one uses Stirling's formula.
I found a (probably) new proof of the CLT that derives it from de Moivre-Laplace theorem. The proof is fairly elementary and avoids using the characteristic function. I wrote a 5-page note on it and submitted to the American Mathematical Monthly. However, they rejected it, saying that "the proof in this article is new to the Board, and "clever". BUT the argument gives neither conceptual insight into why the result is true, nor potential for generalization (the remarkable feature of the CLT is that in fact it is true in much greater generality). It's "just a proof" and we do not think our Monthly readers will find it interesting."
I mostly agree to what they said, but still think this proof is worth putting out somewhere. Could anyone suggest journals where it would be appropriate to submit this kind of notes? I should say that this note is not elementary to the degree that undergraduate students would easily read it.
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https://mathoverflow.net/users/365252
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Where should I submit a derivation of the CLT from the de Moivre-Laplace theorem?
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I would be personally interested to see your proof. You should compare it to the Lindeberg proof that also does not use characteristic functions but replaces the variables one by one by Gaussians. You can find an exposition of this proof in the book [1] or in [Chin - A Short and Elementary Proof of the Central Limit Theorem by Individual Swapping](https://arxiv.org/abs/2106.00871).
Regarding your query: [Expositiones Mathematicae](https://www.journals.elsevier.com/expositiones-mathematicae) might be a relevant journal. See also the long list and discussion in [Which journals publish expository work?](https://mathoverflow.net/questions/15366/which-journals-publish-expository-work?rq=1)
[1] Breiman, Leo. "[Probability](https://doi.org/10.1137/1.9781611971286), Classics in Applied Mathematics, vol. 7, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 1992." Corrected reprint of the 1968 original. MR1 163370.
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5
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https://mathoverflow.net/users/7691
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403777
| 165,637 |
https://mathoverflow.net/questions/403697
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3
|
This question comes from P13 and P17 of the book [Andrei N.Borodin and Paavo Salminen](https://books.google.de/books?hl=en&lr=&id=8cZrllyXfOUC&oi=fnd&pg=PR9&dq=Andrei%20N.Borodin%20and%20Paavo%20Salminen&ots=VDvtxpIWm1&sig=_C4_7rZfM7zYWLroOkzG4NVQjyM&redir_esc=y#v=onepage&q=Andrei%20N.Borodin%20and%20Paavo%20Salminen&f=false).
Page [P13](https://i.stack.imgur.com/AHq9u.png) defines the speed measure $m(dx)$, the scale function $s$, and the killing measure $k(dx)$.
Case 9 on [P17](https://i.stack.imgur.com/VaGmK.png):
"We consider here the special case in which the basic characteristics are absolutely continuous with respect to the Lebesgue measure and have smooth derivatives. In other words,
$m(dx)=m(x)dy,k(dx)=k(x)dx,s(x)=\int^x s'(y)dy$,
where $m,s'$ are continuous and positive, and $k$ is continuous and non-negative. Morever, if $s''$ is continous, then the second order infinitesimal generator $$\mathcal{G}f(x)=\frac{1}{2}a(x)^2f''(x)+b(x)f'(x)-c(x)f(x),$$
The functions $a, b, c$ are the infinitesimal parameters of $X$.
"
My question is, based on the statement above, how to get the following results:
$m(x)=2a^{-2}(x)e^{B(x)}, s'(x)=e^{-B(x)},k(x)=2a^{-2}(x)c(x)e^{B(x)}, B(x)=\int^x 2a^{-2}(y) b(y)dy$?
|
https://mathoverflow.net/users/147009
|
How to get speed measure $m(dx)$, scale function $s$, and killing measure $k(dx)$ of a diffusion from the infinitesimal generator?
|
Intuitively, you are equating coefficients in the two different representations of the generator:
$$
{1\over m(x)}\left[\left({f'(x)\over s(x)}\right)'-k(x)f(x)\right] = {1\over 2}a(x)^2f''(x)+b(x)f'(x)-c(x)f(x).
$$
The left side expands out to
$$
{1\over m(x)}\left[{f''(x)\over s(x)}-{s'(x)f'(x)\over s(x)^2}-k(x)f(x)\right].
$$
Therefore
$$
{a^2\over 2} = {1\over ms},\quad b=-{s'\over ms^2},\quad c={k\over m}.
$$
Dividing the first two of these:
$$
(\log s)'={s'\over s} =-{2b\over a^2},
$$
so
$$
s(x) = C\_1\exp\left(-2\int^xb(t)/a(t)^2 dt\right)=C\_1e^{-B(x)}.
$$
Choosing the constant of integration $C\_1$ to be $1$, the rest follows easily.
|
2
|
https://mathoverflow.net/users/42851
|
403781
| 165,639 |
https://mathoverflow.net/questions/403762
|
1
|
>
> Let $ u\in H^1(2B) $ be a weak solution of $ \Delta u=0 $ in $ 2B $, where $ B=B(0,1) $ is a ball with center $ 0 $ and radius $ 1 $. Then there exists some $ p>2 $ such that
>
> \begin{eqnarray}
> \left(\frac{1}{|B|}\int\_{B}|\triangledown u|^p dx\right)^{1/p}\leq C\left(\frac{1}{|2B|}\int\_{2B}|\triangledown u|^2 dx\right)^{1/2}.
> \end{eqnarray}
> where $ C $ is an absolute constant.
>
>
>
I recently saw this problem and I want to get the solution of this problem. However, I meet with some troubles in it. Here is my try. First as $ u-\frac{1}{|2B|}\int\_{2B}u $ is also a weak solution for the Laplace equation, then by using integration by parts on the function, I can obtain that
\begin{eqnarray}
\frac{1}{|B|}\int\_{B}|\triangledown u|^2 dx\leq C\left\{\int\_{2B}\left|u-\frac{1}{|2B|}\int\_{2B}u\right|^2dx\right\}.
\end{eqnarray}
where $ C $ is an absolute constant. Then, by using the Sobolev-Poincaré inequality I have
\begin{eqnarray}
\left(\frac{1}{|B|}\int\_{B}|\triangledown u|^2 dx\right)^{1/2}\leq C\left(\frac{1}{|2B|}\int\_{2B}|\triangledown u|^q dx\right)^{1/q}
\end{eqnarray}
where $ q=\frac{2d}{d+2} $. I think it is quite similar to the final result. But I cannot go further. Can you give me some hints or references?
|
https://mathoverflow.net/users/241460
|
How to prove the reverse Hölder inequality for Laplace equations?
|
The inequality is scale invariant and holds for a ball of any radius. It follows by a standard argument that is the inductive step in what's known as Moser iteration.
The constant $C$ below can change from line to line but always depends only on the dimension. Let $B = B(0,r)$ and $2B = B(0,2r)$. Let $\chi$ be a smooth compactly supported function on $2B$ that is identically $1$ on $B$ and satisfies
$$ \|\nabla\chi\|\_\infty \le \frac{C}{r} $$
The Sobolev inequality states that there exists a constant $C$, for any smooth compactly supported function $f$ on $2B$,
$$
\left(\int\_{2B} |f|^{\frac{2d}{d-2}}\right)^{\frac{d-2}{d}} \le C\int\_{2B} |\nabla f|^2
$$
Suppose $\Delta u = 0$. First, observe that, if you integrate by parts, then for any constant $a > 0$,
\begin{align\*}
\int\_{2B} \chi^2|\nabla u|^2 &= \int\_{2B} - 2(a^{-1} u\nabla\chi)\cdot(a\chi\nabla u) \\
&\le a^{-2}\int\_{2B} \chi^2|\nabla u|^2 + a^{2}\int\_{2B} u^2|\nabla\chi|^2.
\end{align\*}
In particular, if we set, say, $a = 2$, then
$$
\int\_{2B} \chi^2|\nabla u|^2
\le C\int\_{2B} u^2|\nabla\chi|^2.
$$
It now follows that
\begin{align\*}
\left(\int\_{B} |u|^{\frac{2d}{d-2}}\right)^{\frac{d-2}{d}} &\le \left(\int\_{2B} |\chi u|^{\frac{2d}{d-2}}\right)^{\frac{d-2}{d}}\\
& \le C\int\_{2B} |\nabla (\chi u)|^2\\
&\le C\int\_{2B} \chi^2|\nabla u|^2 + |\nabla\chi|^2u^2\\
&\le C\int\_{2B} |\nabla\chi|^2u^2\\
& \le C\|\nabla\chi\|\_\infty^2\int\_{2B} u^2\\
&= \frac{C}{r^2}\int\_{2B} u^2.
\end{align\*}
Since $\Delta(\partial\_ku) = 0$, the estimate above holds for $\partial\_ku$. The desired estimate can be derived from this.
|
3
|
https://mathoverflow.net/users/613
|
403785
| 165,640 |
https://mathoverflow.net/questions/403726
|
1
|
Let $u,\phi:\mathbb R \to \mathbb R$ be smooth functions and $\Omega\_\epsilon$ be a bounded domain in $\mathbb R$ with diameter $\epsilon>0$ (consider for exaple the ball $B\_{\epsilon/2}(0)$). Is it true that
$$\frac{1}{|\Omega\_\epsilon|}\int\_{\Omega\_\epsilon} \phi (-\Delta)^s u dx - \left( \frac{1}{|\Omega\_\epsilon|}\int\_{\Omega\_\epsilon} (-\Delta)^s u dx\right)\left( \frac{1}{|\Omega\_\epsilon|}\int\_{\Omega\_\epsilon}
\phi dx\right) \to 0 $$ as $\epsilon \to 0$? Here $(-\Delta)^s$ denotes the fractional Laplacian operator
|
https://mathoverflow.net/users/122620
|
Averaging and fractional Laplacian
|
(I believe this may be too basic for this site, but too long for a comment.)
All that we need to assume is that $\phi$ and $(-\Delta)^s u$ are uniformly continuous and bounded. (Continuity suffices if we additionally know that $\Omega\_\epsilon$ are all contained in a bounded region.)
If $|f(x)-f(y)|\leqslant\delta$ whenever $|x-y|\leqslant\epsilon$ and the diameter of $\Omega$ is no greater than $\epsilon$, then — for an arbitrary $x\_0 \in \Omega$ — we have $$\biggl| \frac{1}{|\Omega|} \int\_{\Omega} f(x) dx - f(x\_0)\biggr|\leqslant \delta.$$ Applying this three times, with $f = \phi$, $f = (-\Delta)^s u$ and $f = \phi (-\Delta)^s u$, we find that
$$\begin{aligned} & \biggl| \frac{1}{|\Omega\_\epsilon|} \int\_{\Omega\_\epsilon} \phi (-\Delta)^s u - \frac{1}{|\Omega\_\epsilon|} \int\_{\Omega\_\epsilon} \phi \times \frac{1}{|\Omega\_\epsilon|} \int\_{\Omega\_\epsilon} (-\Delta)^s u \biggr| \\ & \qquad \leqslant \biggl| \frac{1}{|\Omega\_\epsilon|} \int\_{\Omega\_\epsilon} \phi (-\Delta)^s u - \phi(x\_\epsilon) (-\Delta)^s u(x\_\epsilon) \biggr| \\ & \qquad \qquad + \biggl| \frac{1}{|\Omega\_\epsilon|} \int\_{\Omega\_\epsilon} \phi \times \frac{1}{|\Omega\_\epsilon|} \int\_{\Omega\_\epsilon} (-\Delta)^s u - \phi(x\_\epsilon) (-\Delta)^s u(x\_\epsilon) \biggr| \\ & \qquad \leqslant 2 M \delta + 2 M \delta = 4 M \delta,\end{aligned}$$
Here $M$ is an upper bound for $|\phi|$ and $|(-\Delta)^s u|$, $x\_\epsilon$ is an arbitrary point of $\Omega\_\epsilon$, and, for a given $\delta > 0$, $\epsilon$ is small enough as in the definition of uniform continuity of $\phi$ and $(-\Delta)^s u$. In the last inequality we used twise the standard estimate
$$ |a A - b B| \leqslant |a| |A - B| + |B| |a - b| . $$
|
0
|
https://mathoverflow.net/users/108637
|
403790
| 165,642 |
https://mathoverflow.net/questions/403774
|
8
|
Is it true that a finite group with squarefree order has periodic group cohomology (with trivial coefficients)?
I cannot see why this would be the case, but I'm looking at a paper which seems to implicitly say it's true [Vogel, ON STEENROD'S PROBLEM FOR NON ABELIAN FINITE GROUPS, p.1].
|
https://mathoverflow.net/users/125639
|
Finite group with squarefree order has periodic cohomology?
|
All subgroups are then squarefree, and by FTAG all such abelian subgroups are cyclic. Now, a group has periodic cohomology iff all its abelian subgroups are cyclic (Theorem VI.9.5 of Brown's book) -- this is proved by first reducing to Sylow p-subgroups.
|
10
|
https://mathoverflow.net/users/12310
|
403793
| 165,643 |
https://mathoverflow.net/questions/403791
|
6
|
Let $\chi\_{\mu}^{\lambda}$ denote a value of an irreducible character of the symmetric group $\frak{S}\_n$, where $\mu, \lambda\vdash n$. When $\mu=(n)$, then it's known that
$$\sum\_{\lambda\vdash n}\chi\_{\mu}^{\lambda}=\delta\_\text{odd}(n).$$
I like to ask:
>
> **QUESTION 1.** Is there a formula for the enumeration (cardinality) of $a\_n=\#\{\lambda\vdash n: \chi\_{(n)}^{\lambda}=0\}$? How about $b\_n=\#\{\lambda\vdash n: \chi\_{(n)}^{\lambda}=1\}$ or $c\_n=\#\{\lambda\vdash n: \chi\_{(n)}^{\lambda}=-1\}$?
>
>
>
Encouraged by Mark Wildom's prompt/neat [reply](https://mathoverflow.net/questions/403791/counting-pm-1-and-0s-in-the-character-tables#comment1034199_403791), I like to update the question to counting all $0$'s.
>
> **QUESTION 2.** How about the cardinality (or generating function) of the below?
> $$d\_n=\#\{(\lambda,\mu): \lambda\vdash n, \mu\vdash n, \chi\_{\mu}^{\lambda}=0\}.$$
>
>
>
|
https://mathoverflow.net/users/66131
|
Counting $\pm 1$ and $0$'s in the character tables of $\frak{S}_n$
|
It is an open problem to determine how many of the entries of the character table are zero — or, indeed whether the proportion of zeros tends to a positive constant, or to zero. One should be careful about the exact problem considered: if a character is picked at random, and a group element is picked at random (so $p(n)$ choices for the character and $n!$ choices for the group element) then almost all entries are zero. This is the partial result mentioned in Stanley's [comment](https://mathoverflow.net/questions/403791/counting-pm-1-and-0s-in-the-character-tables#comment1034207_403791) to the question. However this is a very different statistic from picking the character at random and a conjugacy class at random (the problem in the question). Miller ([On parity and characters of symmetric groups](https://www.armiller.org/Parity.pdf)) has done some numerics on this (see Table 3), and conjectured that almost all values were multiples of any given number $d$. This last conjecture has been established for all prime numbers $d$, in the work of Peluse ([On even entries in the character table of the symmetric group](https://arxiv.org/abs/2007.06652)) and Peluse and Soundararajan ([Almost all entries in the character table of the symmetric group are multiples of any given prime](https://arxiv.org/abs/2010.12410)) (consequently, $\pm 1$ — or indeed any fixed non-zero integer — appears with density zero in the character table). As for zero entries, the best lower bound that I know (which follows from these ideas of using Murnaghan—Nakayama and the structure of random partitions) is that at least a proportion $c/\log n$ of the entries are zero.
|
9
|
https://mathoverflow.net/users/38624
|
403796
| 165,644 |
https://mathoverflow.net/questions/351751
|
8
|
The standard simplices $\Delta^n \subset \{\mathbf{x}\in\mathbb{R}^{n+1}\mid x\_0 + \ldots + x\_n =1 \} =: \mathbb{A}^n$ carry two natural sorts of smooth differential forms:
1. Those differential forms on the interior of $\Delta^n$ that extend smoothly to a neighbourhood of $\Delta^n$ in $\mathbb{A}^n$ (this definition is used for instance by Dupont in his book *Curvature and Characteristic Classes*). This is implicitly using Whitney's extension theorem, I think.
2. Consider $\Delta^n$ as a diffeological subspace of $\mathbb{A}^n$, where a function $\phi\colon \mathbb{R}^k \to \Delta^n$ is a plot if and only if the composite $\mathbb{R}^k \to \Delta^n \hookrightarrow \mathbb{A}^n$ is smooth. A differential form $\omega$ on $\Delta^n$ then consists of the data of a differential form $\phi^\*\omega$ for each plot $\phi$ with a compatibility condition when a plot factors through another via a smooth map between Euclidean spaces.
The first of these is more of a "maps out" viewpoint, and probably corresponds to a natural smooth space structure defined via smooth real-valued functions. The second is a "maps in" viewpoint. Note that the $D$-topology arising from the diffeology in 2. above is the standard topology on the simplex. We then get a cochain complex of differential forms of each type, as exterior differentiation can be defined in the more-or-less obvious way in each case.
>
> My question is: how do these relate? Is one a subcomplex of the other? Or are they quasi-isomorphic, via a third cochain complex?
>
>
>
The motivation is that Dupont's simplicial differential forms on semisimplicial manifolds $X\_\bullet$ look like they should be differential forms on the fat geometric realisation considered as a diffeological space, since a simplicial differential form is more or less descent data for the sheaf of differential forms and the 'cover' $\coprod\_{n\geq 0} \Delta^n\times X\_n \to ||X\_\bullet||$, assuming *the first definition* above. However, if his differential forms on $\Delta^n$ (or more precisely on $\Delta^n\times X\_n$) aren't diffeological differential forms, they don't give a form on $||X\_\bullet||$ as a diffeological space. I guess all we really need is a map of cochain complexes from the first to the second given above.
|
https://mathoverflow.net/users/4177
|
Differential forms on standard simplices via Whitney extension vs diffeological structure
|
The two chain complexes are isomorphic for any $n≥0$.
Fix some $n≥0$, $k≥0$ and consider $k$-forms on the $n$-simplex.
I will use the notations $Ω\_e^k$ and $Ω\_d^k$ for forms of type 1 and 2 respectively.
I also use the notation $Δ\_d^n$ for the diffeological $n$-simplex
so that $\def\Hom{\mathop{\rm Hom}} \def\R{{\bf R}} \def\d{\,{\rm d}} Ω\_d^k=\Hom(Δ\_d^n,Ω^k)$.
First, we have a canonical map $ι\colon Ω\_e^k→Ω\_d^k$ that
sends $ω∈Ω\_e^n$ to the map $Δ\_d^n→Ω^k$
that sends a smooth map $φ\colon S→Δ\_d^n$ to the $k$-form $φ^\*ω∈Ω^k(S)$.
Secondly, the map $ι$ is injective,
which follows from the following two observations.
By the Yoneda lemma the restriction of $ι(ω)\colon Δ\_d^n→Ω^k$ to the sheaf represented by the open interior of $Δ\_d^n$ equals the restriction of $ω$ to the open interior of $Δ^n$.
Furthermore, any two forms in $Ω\_e^k$ whose restrictions to the open interior of $Δ^n$ coincide must be equal (by continuity).
Thirdly, the map $ι$ is surjective.
This can be shown using a two-step construction:
in the first step we construct a certain element $ω$ of $Ω\_e^k$
starting from some given element $ψ$ of $Ω\_d^k$
and in the second step we show that $ι(ω)=ψ$.
First, let's briefly examine the simplest nontrivial case $d=n=1$.
We have a form $ψ∈Ω\_d^1(Δ\_d^1)=Ω\_d^1([0,1])$,
and we already know its restriction $f(x) \d x$ to the open interval $(0,1)$.
In particular, $f$ is a smooth function on $(0,1)$.
Pulling back $ψ$ along the plot $\R→\R$ ($x↦x^2$)
yields some 1-form $g(x) \d x$, where for any $x≠0$ we have $g(x)=2xf(x^2)$
and $g$ is a smooth function on $(-1,1)$.
Since $g$ is odd, we have $g(0)=0$,
so the even function $h$ given by $h(x)=g(x)/(2x)$ is smooth on $(-1,1)$
and we have $h(x)=f(x^2)$ for all $x≠0$.
Set $f(0)=h(0)$.
We claim $f$ is smooth on $(-1,1)$.
Indeed, $h'(x)=2xf'(x^2)$ for $x≠0$,
and since $h'$ is odd, we have $h'(0)=0$
and $h'(x)/(2x)$ is a smooth function on $(-1,1)$
whose value at $x=0$ is $f'(0)$.
We repeat this step by induction, proving $f$ has derivatives of all orders at 0.
The general case is nothing else than a multivariable paremetrized
version of the above argument.
More precisely, we argue as follows.
For the first step, suppose we are given some $ψ∈Ω\_d^k$.
Given some point $x∈Δ^n$, we would like to define $ω(x)$.
Denote by $d≥0$ the codimension of the stratum of $Δ^n$ that contains $x$.
Parametrize the simplex $Δ^n$ using $n$ coordinates
in such a way that $x\_1=⋯=x\_d=0$
and the other $n-d$ coordinates of $x$ are strictly positive,
so that the points $y$ of the stratum containing $x$ are defined by the relations $y\_i≥0$ for $1≤i≤d$.
Decompose $$ω=∑\_I g\_I ∏\_{i∈I} \d x\_i$$ into its individual components with respect to this coordinate system.
Pick one such component $g\_I ∏\_{i∈I} \d x\_i$; we would like to define $g\_I$ as a smooth function on some open neighborhood $U$ of the given stratum of $Δ^n$, which is parametrized by the remaining $n-d$ coordinates $x\_{d+1}$, …, $x\_n$.
Consider the smooth map $U→Δ^n$ that (in the coordinates introduced above)
sends $x\_i↦x\_i^2$ for $1≤i≤d$ and $x\_i↦x\_i$ for $i>d$.
This defines a morphism $τ\colon U→Δ\_d^n$, so we have a form $τ^\*ψ∈Ω^k(U)$.
Now take the coefficient $h$ of $τ^\*ψ$ before $∏\_{i∈I} dx\_i$.
Take the partial derivative of $h$ with respect to all coordinates $x\_i$
such that $i∈I$ and $i≤d$.
Divide the resulting function by $2^{\#(I∩\{1,…,d\})}$.
This is the function $g\_I$.
This argument also proves that the resulting $k$-form $ω$ is smooth.
For the second step, we have to show that $ι(ω)=ψ$.
Reusing the notation of the previous paragraph,
consider some arbitrary plot $φ\colon S→Δ\_d^n$ such that $φ(s)=x$ for some $s∈S$.
The first $d$ coordinates of $φ$ must be nonnegative in a neighborhood $U$ of $s$.
Since $φ$ is differentiable, the first derivatives of these $d$ coordinates must vanish at the point $s$.
Thus, taking the square root of each of the first $d$ coordinates produces a smooth map $ε\colon U→\R^n$ such that $φ=τε$, where the map $τ$ was defined in the previous paragraph.
Now $φ^\*ι(ω) = ε^\* τ^\* ι(ω) = ε^\* τ^\*ψ = φ^\*ψ$,
where $τ^\*ι(ω)=τ^\*ψ$ by definition of $ω$.
|
3
|
https://mathoverflow.net/users/402
|
403801
| 165,647 |
https://mathoverflow.net/questions/403802
|
2
|
Suppose that $f : X \rightarrow Y$ is a morphism of schemes.
Let $Z \hookrightarrow Y$ the scheme-thereotic image of $f$.
Under what conditions is the morphism $X \rightarrow Z$ an epimorphism?
If both $X = \mathrm{Spec} \, B$ and $Y = \mathrm{Spec} \, A$ are affine, then $Z = \mathrm{Spec}\, A/I$, where $I$ is the kernel of $A \rightarrow B$. Then $A/I \rightarrow B$ is injective, hence a monomorphism. Therefore $X \rightarrow Z$ is an epimorphism.
I would think that if $f$ is quasi-compact or $X$ is integral, the scheme theoretic image has nice properties ($f(X)$ is dense in $Z$) and hopefully the result is true.
|
https://mathoverflow.net/users/122284
|
Epi-mono factorisations in schemes via scheme-theoretic image
|
This already fails in the affine case. For example, if $Y = \mathbf A^1$ and $X = \mathbf A^1 \setminus \{0\}$, then $Y$ is the scheme-theoretic image of $X \hookrightarrow Y$. But if $Z$ is the line with two origins, then the two inclusions $Y \to Z$ agree on $X$ but are not identical.
There is a positive result when $f$ has closed image, for then $X \to \operatorname{im}(f)$ is surjective (see also the introduction to [this question](https://mathoverflow.net/q/56564/82179)). For schemes of finite type over a Jacobson ring (e.g. $\mathbf Z$ or a field $k$), this is the only way $X \to \operatorname{im}(f)$ can be an epimorphism:
**Lemma.** *Let $f \colon X \to Y$ be an epimorphism of schemes, and let $y \in Y$ be a closed point. Then there exists $x \in X$ with $f(x) = y$. In particular, if $Y$ is Jacobson and $f$ of finite type, then $f$ is surjective.*
*Proof.* For the first statement, consider the open $U = Y\setminus\{y\}$ and let $Z$ be two copies of $Y$ glued along $U$. The two natural inclusions $Y \to Z$ differ, so their restrictions to $X$ differ, showing that $y$ is in the (set-theoretic) image of $f$. The second statement follows from Chevalley's theorem. $\square$
However, there exist epimorphisms between non-Jacobson schemes that are not surjective:
**Example.** Let $A$ be a local domain of dimension $\geq 2$; for example $A = k[x,y]\_{(x,y)}$. Let $A \subseteq B$ be a local homomorphism to a discrete valuation ring $B$; for example by blowing up the closed point of $A$ and localising at a generic point of the exceptional fibre. Let $X = \operatorname{Spec} B$ and $Y = \operatorname{Spec} A$. Then $f \colon X \to Y$ is not surjective (it only hits the generic point $\eta$ and the closed point $s$), but it is an epimorphism. Indeed, if $g,h \colon Y \rightrightarrows Z$ agree on $X$, then $g(s) = h(s)$. If $U \subseteq Z$ is an affine open neighbourhood of $g(s)$, then $g$ and $h$ both land in $U$, since the only open subset of $Y$ containing $s$ is $Y$ itself. The result then follows since $A \subseteq B$ is a monomorphism of rings. $\square$
|
4
|
https://mathoverflow.net/users/82179
|
403804
| 165,648 |
https://mathoverflow.net/questions/403803
|
4
|
Intuitively, a constructively irrational number is one for which we can effectively separate it from any rational number in terms of the latter's denominator. More formally, a constructively irrational number is a number $x$ such that there is a known primitive recursive function $f : \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$ such that $|x - \frac{p}{q}| > \frac{1}{f(q)}$ for all $p, q \in \mathbb{Z}$ such that $q \neq 0$. This corresponds closely to what is meant by "irrational" in constructive mathematics, where "irrational" is interpreted as stronger than "not rational" (indeed, strictly stronger).
For example, $\sqrt{2}$ is constructively irrational, given that $|\sqrt{2} - \frac{p}{q}| > \frac{1}{3q^2}$.
Which theorems of irrational number theory are known to also hold for constructively irrational numbers? In particular, are all algebraic irrational numbers constructively irrational? Are $\pi$ and $e$ constructively irrational? What about $e^n$ for $n \in \mathbb{Z}$ and $n \neq 0$? Or $\ln{n}$ for $n \in \mathbb{Z}^+$ and $n \geq 2$? And finally, is $\log\_p{n}$ constructively irrational for $p$ a prime, $n \in \mathbb{Z}^+$, and $n$ not an integer power of $p$?
Answers to any of these questions would be greatly appreciated.
|
https://mathoverflow.net/users/163672
|
What is known about constructively irrational numbers?
|
All algebraic numbers are by this definition constructively irrational. You can adopt Liouville's proof that Liouville numbers are transcendental and turn it in the other direction to get a function of the sort you want given an algebraic number and its corresponding polynomial. Explicit versions of [Baker's theorem](https://en.wikipedia.org/wiki/Baker%27s_theorem) also can be thought of as a similar statement. Baker's sort of methods also give you your desired result for a lot of logarithms.
What you are interested in is also closely connected to the idea of irrationality measure. In particular, Mahler's theorem on the irrationality measure of $\pi$ may be enough to show that $\pi$ is constructively irrational. For the best bounds currently on that, [see this paper by Doron Zeilberger and Wadim Zudilin](https://arxiv.org/abs/1912.06345). I say "may" here because there are epsilons floating around there and I haven't checked to see if they can be made explicit.
A quick aside about a number you didn't ask about but I'm now wondering about: I don't know if Apery's proof that $\zeta(3)$ is irrational can be turned into a proof of constructive irrationality. My guess is that things there are much too delicate.
|
5
|
https://mathoverflow.net/users/127690
|
403807
| 165,650 |
https://mathoverflow.net/questions/403797
|
12
|
Is there a fixed-point theorem that implies the following result?
>
> Let $F$ be a nonempty convex set of functions on a discrete group with values in $[0,1]$. Suppose $F$ is invariant with respect to left shifts and closed with respect to the pointwise convergence. Then $F$ contains a constant function.
>
>
>
This statement looks like [Ryll-Nardzewski fixed point theorem](https://en.wikipedia.org/wiki/Ryll-Nardzewski_fixed-point_theorem), but it does not seem to follow from the theorem.
|
https://mathoverflow.net/users/1441
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A variation of the Ryll-Nardzewski fixed point theorem
|
The claim does not hold. Let $F$ be the free group on $\{a,b\}$ and $X⊂F$ be the words whose last letter is $a$ or $b$. Let $\xi=\delta\_a+\delta\_b−\delta\_{a^{-1}}−\delta\_{b^{-1}}\in\ell\_1(F)$. Then for any $s\in F$, one has $\langle 1\_X,s\xi\rangle\geq1$. (To see this, view $F$ as the $4$-regular tree and $s\xi$ as a signed characteristic function of the $1$-neighborhood of $s\in F$.) Hence the pointwise-topology closed convex hull of $F1\_X$ in $\ell\_\infty(F)$ cannot contain constant functions.
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12
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https://mathoverflow.net/users/7591
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403815
| 165,654 |
https://mathoverflow.net/questions/403799
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3
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My general question is how to construct an isotropic random vector field $\vec f: \mathbb{R}^3 \to \mathbb{R}^3$ with a given mean magnitude $\mathbb{E}[\|\vec f(\vec x)\|]=\mu$ and with vector magnitudes and direction correlated up to some length scales $l$ (beyond which the correlation goes to zero). I prefer a construction satisfying $\nabla \cdot \vec f=0$, but a construction without that condition is already interesting.
More precisely, we are given a vector $\vec{\mu}$ in $\mathbb{R^3}$ and a matrix-valued correlation function $C: \mathbb{R^3} \times \mathbb{R^3} \to M\_3(\mathbb{R})$ which is isotropic, i.e $C(\vec v,\vec w)=C(\|\vec v-\vec w\|)$. One may define a Gaussian Process $f(\vec x) \sim GP(\vec{\mu},C)$ such that $\mathbb{E}[\vec f(\vec x)]=\vec\mu$ and $Cov(\vec f(v),\vec f(w))=C(\vec v,\vec w)$. I believe that constructions of such a random field $f$ are known. The equivalent problem for a scalar field $f$ seems well-studied: one can first draw the field in Fourier space by normalizing a white noise field with the appropriate power spectrum $P(k)$, and then transform back to real space. Here I am looking for a simple procedure for $\mathbb{R}^3$; I think this is known but I haven't found a clear description anywhere.
**Question**: How can one generate such a random field $f$ with a mean vector of zero ($\vec{\mu}=\vec 0$), and a mean *magnitude* of $\mu$ ($\mathbb{E}[\|\vec f(\vec x)\|]=\mu$)? And what if we also impose $\nabla \cdot \vec f=0$?
|
https://mathoverflow.net/users/115409
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Divergence-free Gaussian vector field with given mean magnitude and correlation function
|
One construction uses that divergence free fields are precisely the rotation fields:
Choose any isotropic matrix valued covariance function $C':\mathbb{R}^3\times\mathbb{R}^3\to \mathbb{R}^{3\times3}\_{\ge0}$. Then, the push forward $\nabla\times g=GP(0,\nabla\times C'\times\nabla)$ of the Gaussian process $g=GP(0,C')$ is such an isotropic covariance function with divergence free realizations.
The details of the construction of the covariance $C:=\nabla\times C'\times\nabla$ are that the rotation operator from the left is applied to the first argument and the rotation operator from the left is applied to the second argument.
If you include a factor in $C'$, you can fit it afterwards to get the intended mean magnitude. If you choose $C'$ "general enough", than $C$ is "general enough".
(Shameless self-promotion:) See also <https://arxiv.org/abs/1801.09197> (or <https://arxiv.org/abs/2002.00818>).
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1
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https://mathoverflow.net/users/25523
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403818
| 165,655 |
https://mathoverflow.net/questions/403792
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3
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In the paper "*Silting mutation in triangulated categories*" by *Aihara and Iyama*, I stumbled upon this nice definition( **Definition 2.1**) of a tilting/silting subcategory of a triangulated category $\mathcal{T}$. Let $\mathcal{M}$ be a subcategory of a triangulated category $\mathcal{T}$:
(1) We say that $\mathcal{M}$ is a silting subcategory of $\mathcal{T}$ if $Hom\_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[>0]) = 0$ and **thick**$(\mathcal{M}) = \mathcal{T}$.
(2) We say that $\mathcal{M}$ is a tilting subcategory of $\mathcal{T}$ if it is silting and $Hom\_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[< 0]) = 0$.
It is clear to me as it is stated on the paper that **thick**$(\mathcal{M})$ is the smallest thick subcategory of $\mathcal{T}$ containing $\mathcal{M}$. Where a thick subcategory is triangulated subcategory of $\mathcal{T}$ closed under direct summands. What is not clear to me is the notion of $Hom\_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[>0]) = 0$ (respectively $Hom\_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[\neq 0]) = 0$).
If I'm right $Hom\_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[> 0]) = 0$ means that for every $Hom\_{T}(X,Y) =0$ for every $X \in \mathcal{M}$ and $Y \in \mathcal{M}[>0]$ where $\mathcal{M}[>0]$ stands for $\Sigma^{i}(\mathcal{M})$ for $i \in \mathbb{Z}^{+}, i \neq 0$ and $\Sigma$ is the translation functor associated to the triangulated category $T$. Is my understanding of this definition right? Also as a first example of tilting subcategory the stalk of complexes of a ring $A$ in the triangulated category $K^{b}(A)$ is considered. Can someone recall me the proper definition of a stalk complex of a ring in a bounded homotopy category of complexes? Thanks for the support!
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https://mathoverflow.net/users/369159
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On the definition and an example of silting/tilting subcategories in a triangulated categories according to a paper by Aihara and Iyama
|
$\operatorname{Hom}\_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[>0]) = 0$ means that $\operatorname{Hom}\_{\mathcal{T}}\left(X, \Sigma^i(Y)\right) = 0$ for all objects $X,Y$ of $\mathcal{M}$ and all integers $i>0$.
A stalk complex is a complex with only one nonzero term. So "$A$ considered as a stalk complex" means the complex with $A$ in degree zero and with $0$ in all other degrees.
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2
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https://mathoverflow.net/users/22989
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403823
| 165,658 |
https://mathoverflow.net/questions/403752
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10
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Something I've been thinking about for a while that I'm not sure I understand is why $\mathcal{Z}$ stability, as opposed to say $\mathcal{O}\_\infty$-stability or even $\mathcal{K}$-stability is so important to representation theory. I know that the Jiang-Su algebra has a lot of interesting properties such as being strongly self-absorbing, or projectionless, simple, KK-equivalent to $\mathbb{C}$. I can certainly see that it is an interesting object, but I think I struggle to understand the relevance for classification.
I was wondering if someone a bit closer to classification might be able to explain a bit is why $\mathcal{Z}$-stability is the type of stability we are interested in for Elliot classification. One thing I remember hearing in the YMC\*A minicourse Chris Schafhauser gave this year is that it's the analogue to the hyperfinite $II\_1$ factor in von Neumann algebras. If anyone could expand on this I would be really interested.
Bit of a soft question I suppose, is more out of interest than anything. Probably there is an abstract in a paper/introduction which explains this and pointing me in the right direction would be good.
|
https://mathoverflow.net/users/208800
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What is the significance of the Jiang Su algebra in classification of C$^*$ -algebras?
|
Diego's answer in the comments above is related to why we would expect any classifiable $C^\ast$-algebra to satisfy $A\cong A\otimes \mathcal Z$:
Since $\mathcal Z$ is separable, nuclear, unital, simple and UCT (the properties of $C^\ast$-algebras we wish to classify by $K$-theory and traces),
tensoring with any other $C^\ast$-algebras with these properties will have all the same properties.
And since $\mathcal Z$ has the same $K$-theory and traces as $\mathbb C$, one doesn't change $K$-theory and traces by tensoring with $\mathcal Z$.
Hence if we expect to have classification by $K$-theory and traces, we would expect $A \cong A \otimes \mathcal Z$ for all $C^\ast$-algebras classified by $K$-theory and traces.
Now, I really want to address a misconception which I think is quite common amongst people not working in the
classification or structure programme of nuclear C\*-algebras:
*$\mathcal Z$-stability (essentially) has nothing to do with the Jiang-Su algebra $\mathcal Z$!*
For instance, the UHF algebras, the irrational rotation algebras, and the Cuntz algebras are all $\mathcal Z$-stable, but they have nothing to do with $\mathcal Z$.
This is similar to how McDuff factors\* don't really have anything to do with $\mathcal R$,
or how $\mathcal O\_\infty$-stable $C^\ast$-algebras have nothing to do with $\mathcal O\_\infty$\*\*
One should consider the very natural and frequently occuring regularity property "$\mathcal Z$-stability" as being equivalent
(by very deep, surprising, and non-constructive theorems!) to "$A \cong A\otimes \mathcal Z$".
Similar to the McDuff property, $\mathcal Z$-stability can be characterised by the (norm-)central sequence algebra $\frac{\prod\_{\mathbb N} A}{\bigoplus\_{\mathbb N} A} \cap A'$ (for separable unital $A$)
being suitably non-trivial, e.g. by requiring that it contains a unital copy of $Z(2,3)$;
the $C^\ast$-algebra of continuous functions $f\colon [0,1] \to M\_2(\mathbb C) \otimes M\_3(\mathbb C)$ with $f(0) \in M\_2(\mathbb C) \otimes 1$ and $f(1) \in 1 \otimes M\_3(\mathbb C)$.
Unfortunately, it would be much too technical to explain exactly how this property is used in the classification thereom.
A key technique is that these "almost central homotopies" coming from $Z(2,3)$ can for instance be used to show that $\mathcal Z$-stable $C^\ast$-algebras have cancellation of full projections by $K\_0$, that $K\_1(A) = U(A)/U\_0(A)$
(this was proved in an unpublished paper by Jiang in the late 90's; the paper can be found on arXiv), and to prove that the Cuntz semigroup is almost unperforated (Rørdam).
$\mathcal Z$-stability turns out to be a very mild property that can be verified in an abundance of examples, such as through the Toms-Winter conjecture in the cases where this is known to hold, such as separable, simple, non-type I $C^\ast$-algebras with finite nuclear dimension.
In fact, it is very hard to construct separable simple nuclear non-type I $C^\ast$-algebras which are not $\mathcal Z$-stable (cf. Villadsen, Rørdam, Toms).
So if you have any natural construction that gives you a separable, nuclear, simple $C^\ast$-algebra, you are almost certain to obtain a $\mathcal Z$-stable $C^\ast$-algebra
unless you tried very very hard to construct one without this property.
---
(\*) McDuff factor: separable $II\_1$-factors such that $M \cong M\overline{\otimes} \mathcal R$. Equivalently, their ($W^\ast$-)central cequence algebras are non-abelian.
(\*\*) For separable, nuclear, simple $C^\ast$-algebras being $\mathcal O\_\infty$-stable is equivalent to being purely infinite by a theorem of Kirchberg.
In particular, the Cuntz algebras $\mathcal O\_n$ all satisfy $\mathcal O\_n \cong \mathcal O\_n \otimes \mathcal O\_\infty$ (this is quite spectacular; you can't express this isomorphism explicitly!).
Being $\mathcal O\_\infty$-stable is equivalent to the central sequence algebra $\frac{\prod\_{\mathbb N} A}{\bigoplus\_{\mathbb N} A} \cap A'$ (for separable unital $A$)
containing two isometries $s\_1,s\_2$ such that $s\_1^\ast s\_2 = 0$.
And this is the property which is used for classifying purely infinite $C^\ast$-algebras by "moving things around" in a paradoxical way.
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15
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https://mathoverflow.net/users/126109
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403830
| 165,661 |
https://mathoverflow.net/questions/403828
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6
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In Kechris' book "Classical Descriptive Set Theory" there is the following theorem (12.16):
>
> Let $X$ be a Polish space and $E$ an equivalence relation such that every equivalence class is closed and the saturation of any open set is Borel. Then $E$ admits a Borel selector.
>
>
>
So under the hypotheses of the theorem, one gets a set of representatives that is Borel.
There are also stronger forms of this theorem that weaken the assumptions for instance to the classes being $G\_\delta$ instead of closed (see Miller (1980)).
I have the feeling that a theorem with an even weaker assumption as follows should be true.
**Question:** Let $X$ be a Polish space and $E$ an equivalence relation such that every equivalence class is *Borel* and the saturation of any open set is Borel. Does $E$ always admit a Borel selector?
I am not an expert in this field and therefore also happy if you can just give a reference.
Thank you in advance!
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https://mathoverflow.net/users/120027
|
A strong Borel selection theorem for equivalence relations
|
Let $X$ be the Cantor space $2^\omega$, and let $E$ be the relation of "equivalence mod $\mathrm{Fin}$" -- i.e., $xEy$ if and only if $\{n \in \omega :\, x(n) \neq y(n) \}$ is finite. The equivalence classes for this relation are countable (hence Borel, and even $F\_\sigma$). If $U \subseteq 2^\omega$ is open, then $U$ contains a basic clopen subset of $2^\omega$, which (by how these are defined) contains a representative of every equivalence class of $E$. So the saturation of any nonempty open set with respect to $E$ is $2^\omega$ itself. However, $E$ does not admit a Borel selector: any selector for $E$ is non-measurable with respect to the usual Haar measure on $2^\omega$, by Vitali's classical argument.
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12
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https://mathoverflow.net/users/70618
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403840
| 165,666 |
https://mathoverflow.net/questions/403825
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4
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In the classical setting, we can define automorphic forms on $\text{SL}\_n(\mathbb{R})$ with respect to any lattice $\Gamma$. In fact, for $n \geq 3$, all lattices are arithmetic subgroups.
I have encountered the lifting of automorphic forms to the adeles (so to automorphic representations) for $\Gamma$ being a congruence subgroup or, more generally, the unit group of an order in a quaternion algebra. I am wondering what are the precise conditions that make this lift possible in general.
Can we "lift" any classical arithmetic subgroup to a compact open subgroup over the adeles? I understand how this might happen when the subgroup is associated to an order in an algebra (the matrix algebra or a division algebra). But there are other constructions of lattices $\Gamma$, especially in higher rank. What I am maybe asking is whether simply arithmeticity of the subgroup is enough to make full use of adelic lifts (if they even exist).
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https://mathoverflow.net/users/168129
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Adelization for any classical arithmetic subgroup
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For a subgroup to have a meaningful lift to the adeles, it is necessary and sufficient for the subgroup to be a congruence subgroup in the sense that for some $N$, the subgroup contains all elements congruent to the identity mod $N$.
Given an element of $SL\_n(\mathbb A\_{\mathbb Q})$ (or the same thing for the norm 1 unit group of a central simple algebra)that is integral at every place, it makes sense to consider its congruence class mod $N$ (by ignoring all places not dividing $N$ and modding out the local ring by $n$ at each place dividing $n$) and by the Chinese remainder theorem this defines a congruence class in $SL\_n(\mathbb Z/N)$ or the appropriate analogue. Thus we can define the subgroup of adelic elements that are congruent mod $N$ to an element of the group, and this gives an adelic interpretation for automorphic forms.
The property of every finite index subgroup being a congruence group is called the *congruence subgroup property*, and it is known for non-anisotropic arithmetic groups of real rank $\geq 2$, i.e. every arithmetic lattice in $SL\_n(\mathbb R)$ except for unit groups of division algebras or $n=2$. For $n=2$ it is known to fail badly and for unit groups of division algebras it is (according to Wikipedia) open.
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6
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https://mathoverflow.net/users/18060
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403842
| 165,668 |
https://mathoverflow.net/questions/403410
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8
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Let $T\_R$ be the first-order theory of [real closed fields](https://en.wikipedia.org/wiki/Real_closed_field). This is precisely the theory over the language $\{0,1,+,\times\}$ such that the theorems are the formulas that hold in $\Bbb R$. It can be effectively axiomatized by saying that it's a field in which every odd-degree polynomial has a root, and that for all $r$ either $r$ or $-r$ has a square root, but that $-1$ is not a square.
Define $T\_C$ to be the analogous theory for $\Bbb C$, but where we include complex conjugation in the language. So it's precisely the first-order theory over the language $\{0,1,+,\times,\bar{}\}$ such that a formula is a theorem if and only if it holds in $\Bbb C$. This can be axiomatized by saying that it's an algebraically closed field in which $z\mapsto\bar z$ is a self-inverse automorphism and $z\bar z=-1$ has no solutions.
Then these two theories can interpret each other. The theory $T\_R$ is interpretable in $T\_C$ by considering the fixed points of $z\mapsto\bar{z}$. In the other direction, and $T\_C$ is interpretable in $T\_R$ by equipping pairs $(r\_0,r\_1)$ with operations to make them act like $r\_0 + r\_1i$.
I'm interested in whether this pair of interpretations rises to the level of biinterpretation, as defined by Joel David Hamkins [here](https://arxiv.org/pdf/2001.05262.pdf). This definition requires that when we compose the two interpretations above (to get interpretations of $T\_R$ and $T\_C$ in themselves) these composite interpretations are isomorphic to the identity representations, in a way that is definable and provable within the theories themselves.
At first it seemed to me that this would be true, since applying the two interpretations to a model does indeed yield the original model. And Hamkins himself seemingly calls this a biinterpretation [here](https://mathoverflow.net/questions/345410/model-theory-of-the-complex-numbers-with-conjugation).
What has me doubting is that the nLab says [here](https://ncatlab.org/nlab/show/interpretation#biinterpretations_of_theories) that biinterpretations of theories induce equivalences between the categories of models. That can't be true here, since the above interpretations send $\Bbb R$ and $\Bbb C$ to each other, and yet $\Bbb R$ has only the trivial automorphism, whereas $\Bbb C$ also has complex conjugation. An equivalence of categories should preserve automorphism groups.
So is this a biinterpretation or not?
---
I also asked this question over on the [category theory Zulip](https://categorytheory.zulipchat.com/#narrow/stream/229199-learning.3A-questions/topic/biinterpretations.20and.20automorphisms). The answers weren't conclusive, but they lead me to believe the subtlety lies in the interpretation of $T\_C$ in itself. If we adjoin a square root of $-1$ to the fixed points of $z\mapsto\bar{z}$ then we would want to define an isomorphism back to the original field by sending this fixed point to $i$. But how do we define $i$ separately from $-i$?
Is this indeed the problem, and is it unavoidable? Does it also prevent $\Bbb R$ and $\Bbb C$ from being biinterpretable as models?
|
https://mathoverflow.net/users/4613
|
Are these theories of real and complex number biinterpretable?
|
$\DeclareMathOperator\Th{Th}\def\R{\mathbb R}\def\C{\mathbb C}\DeclareMathOperator\Aut{Aut}$Basically, all the statements you mention are correct, we just have to be careful about the details to avoid the apparent contradiction. So let us review how these interpretations work:
* There is a 1-dimensional interpretation $\rho$ of $\Th(\R,0,1,+,\cdot)$ in $\Th(\C,0,1,+,\cdot,\bar{})$ with domain $\{z:z=\bar z\}$ and all operations absolute.
* There is a 2-dimensional interpretation $\kappa$ of $\Th(\C,0,1,+,\cdot,\bar{})$ in $\Th(\R,0,1,+,\cdot)$ defined in the usual way so that a pair $(x,y)$ represents $x+iy$.
* The interpretation $\rho\circ\kappa$ of $\Th(\R,0,1,+,\cdot)$ in itself is definably isomorphic to the identity interpretation, the isomorphism being $(x,0)\mapsto x$.
* The interpretation $\kappa\circ\rho$ of $\Th(\C,0,1,+,\cdot,\bar{})$ in itself is isomorphic to the identity interpretation, with the isomorphism being $(x,y)\mapsto x+iy$. Now, here comes the trouble: this isomorphism is only definable *with a parameter $i$* (which is not definable without parameters, as it is indistinguishable from $-i$). You can read this in two ways:
+ $\Th(\R,0,1,+,\cdot)$ is bi-interpretable with $\Th(\C,0,1,i,+,\cdot,\bar{})$.
+ $\Th(\R,0,1,+,\cdot)$ is bi-interpretable with $\Th(\C,0,1,+,\cdot,\bar{})$ via a bi-interpretation with parameters. (Actually, the bi-interpretation as such is still parameter-free, only the connecting isomorphism needs a parameter, as explained above.)Note that different people have different conventions about what exactly is the default definition of an interpretation, and in particular, whether it allows parameters. I’d guess that for syntactically minded people, it is more natural to consider paarameter-free interpretations, whereas for model-theorists, it is more natural to allow parameters.
* $\Th(\R,0,1,+,\cdot)$ is *not* bi-interpretable with $\Th(\C,0,1,+,\cdot,\bar{})$ without parameters. An interpretation $\iota$ of $T\_1$ in $T\_0$ induces a construction of a model $i(M)\models T\_1$ from any model $M\models T\_0$, and it is easy to check that if $\iota$ is a parameter-free bi-interpretation, then $\Aut(\iota(M))\simeq\Aut(M)$. Now, $\Th(\R,0,1,+,\cdot)$ has rigid models (i.e., with $\Aut(M)=1$) such as $\R$ itself, whereas every model of $\Th(\C,0,1,+,\cdot,\bar{})$ has at least one nontrivial automorphism: $\bar{}$.
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10
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https://mathoverflow.net/users/12705
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403859
| 165,674 |
https://mathoverflow.net/questions/403858
|
3
|
Cross-posted from Math Stackexchange.
In an older question to which I provided an [answer](https://math.stackexchange.com/questions/4119317/let-fx-fracx-ln-x1x-evaluate-lim-m-to-f-leftx-0-right-fracx-1/4135364#4135364) it was asked how to compute a particular limit involving the roots of a transcedental function around its extremum. This limit required the evaluation of several terms of a power series for each one of the two roots the function possesses around the minimum, which is reminiscent of a procedure akin to the Lagrange inversion theorem. However, as explicitly stated in the assumptions of the theorem, the derivative of the function has to be non-zero for the theorem to apply (the function has to be locally invertible). That requirement however, did not stop me from deriving a term by term expansion for both functional inverses around the maximum.
The reason why I was surprised is because I couldn't find any references on the subject, and this calculation seems to be a low-hanging fruit of a simple generalization to a well-known theorem.
*My question* is twofold, but an answer to either components will suffice:
1. Assuming that around a point $x\_0$, where $f'(x\_0)=0, f''(x\_0)\neq 0$ and given that $f(x)=\sum\_n a\_n(x-x\_0)^n$, what are the coefficients $b^{\pm}\_{n}$ of the series expansions of the two functional inverses $r^\pm(x)=\sum\_{n=0}^{\infty}b\_n^{\pm}(x-f(x\_0))^{n/2}$, that satisfy $f(r^\pm (x))=x$? They also can be seen as the two solutions to the equation $f(x)=c$ in a neighborhood around $c=f(x\_0)$. Is there a general formula for them (however formal?) Here I define $r^+(x)$ to be the root satisfying $r^+(x)>x\_0$ and the other one such that $r^-(x)<x\_0$.
2. Has this structure been studied in the literature before? Does it come under a certain name? If not, what is it that makes this theorem difficult to establish or uninteresting?
*My work:* Experimenting with a couple of simple functions and their behavior around double zeroes indicates that these coefficients are related by $b\_n^-=(-1)^n b\_n^+$. Also, these series expansions are obviously one-sided: they are defined for $x\in [f(x\_0),R)$ if $f''(x\_0)>0$ and for $x\in (R, f(x\_0)]$ if $f''(x\_0)<0$, for some value of $R$ representing a radius of convergence. I also computed the first few coefficients for arbitrary $f$ with a minimum at $x\_0$
$$b\_0=x\_0,~~ b\_1=\sqrt{\frac{2}{f''(x\_0)}}, ~~ b\_2=-\frac{f'''(x\_0)}{3(f''(x\_0))^2}$$
I also noticed that in a further generalization of the problem, when I demand that $f'(x\_0)=f''(x\_0)= \dots =f^{(n-1)}(x\_0)=0, f^{(n)}(x\_0)\neq 0$, there are now $n$ functional inverses, but most of them represent complex roots around the extremum ($n-1$ if $n$ is odd and $n-2$ if $n$ is even). This hint may be useful if one actually tries to formalize the theorem, since in the complex plane all branches will be included.
|
https://mathoverflow.net/users/126536
|
"Lagrange inversion" around an extremum
|
This is easily reduced to the ordinary Lagrange inversion.
Indeed, without loss of generality, $x\_0=0=f(x\_0)=f'(x\_0)<f''(x\_0)$, so that for all real $z$ close enough to $0$ we have $f(z)=z^2 h(z)$, where $h$ is a function analytic near $0$ such that $h(0)>0$. So, the equation $f(z)=w$ (for real $z$ near $0$ and small enough $w\ge0$) can be rewritten as
$$g(z):=z\sqrt{h(z)}=\sqrt w\,\text{sgn}\,z,$$
so that $g$ is analytic near $0$: $g(z)=\sum\_{n=0}^\infty a\_n z^n$ for real $z$ near $0$. The $a\_n$'s can be obtained by Faà di Bruno's formula.
Using now the ordinary Lagrange inversion, we get
$$z=\sum\_{n=0}^\infty b\_n w^{n/2}\text{sgn}^n z$$
for some explicitly written $b\_n$'s and all real $w$ close enough to $0$.
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3
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https://mathoverflow.net/users/36721
|
403864
| 165,676 |
https://mathoverflow.net/questions/403870
|
8
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Can there be an uncountable set $S\subseteq\mathbb R$ such that for each subset $D\subseteq S$, there is an open set $U$ with $D=S\cap U$?
I'm asking merely out of curiosity, but I'll mention that this would imply $2^{\aleph\_1}=2^{\aleph\_0}$.
|
https://mathoverflow.net/users/4600
|
VC dimension of standard topology on the reals
|
An uncountable $S\subseteq \mathbb{R}$ has an accumulation point $x\in S$. Then for $D=\{x\}$ there is no such open set $U$.
|
12
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https://mathoverflow.net/users/24076
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403878
| 165,680 |
https://mathoverflow.net/questions/403868
|
2
|
$\DeclareMathOperator\Ent{Ent}\newcommand{\prior}{\mathrm{prior}}\newcommand\Data{\mathrm{Data}}$I came across this paper on the optimality of Bayes' theorem
<https://sinews.siam.org/Portals/Sinews2/Issue%20Pdfs/sn_July-August2021.pdf>
and could not figure out where one inequality comes from.
Denote by $m$ the parameter to estimate and by $D\_{KL}$ the KL distance.
One statement at the end is that from "a duality formulation of variational inference" in a given reference, the following inequality holds
$$
- \log E\_{\pi\_{\prior}}[e^{f(m)}] \leq E\_\rho[f(m)] + D\_{KL}(\rho | \pi\_{\prior}) \;.
$$
with mild conditions on $\pi\_{\prior}(m)$ and $\rho(m)$ and a large class of $f(m)$.
(When $f(m) = \log\_{\pi\_{like}}(\Data|m)$, the equality holds only when $\rho=$ Bayes posterior distribution.)
Q: Where does the inequality comes from? In the reference [http://www.cmap.polytechnique.fr/~merlet/articles/probas\_massart\_stf03.pdf](http://www.cmap.polytechnique.fr/%7Emerlet/articles/probas_massart_stf03.pdf) ,
Section 2.3.1 Duality and variational formulas gives
$$
\Ent\_P[Y] = \sup \{ E\_P[UY] , \, U:\Omega\rightarrow \bar{R},\, E[e^U] = 1 \} \;.
$$
I still don't see how this helps with deriving the inequality.
|
https://mathoverflow.net/users/157812
|
An inequality in the optimality of Bayes' theorem
|
First, let us do some cleaning here.
1. Let $\pi:=\pi\_{\mathrm{prior}}$, $Y:=\rho/\pi$, and $F:=f(m)$.
2. You copied the inequality in question incorrectly. In your first-linked paper, the inequality is
$-\ln E\_\pi e^F\le E\_\rho(-F) + D\_{KL}(\rho|\pi)$, that is,
$$E\_\rho F-\ln E\_\pi e^F\le D\_{KL}(\rho|\pi). \tag{0}$$
3. Your first-linked paper uses some facts from your second-linked book. However, the definition of "the Kullback-Leibler information" in that book is non-standard: it is
$$K(\rho,\pi):=E\_\pi Y\ln Y-E\_\pi Y\,E\_\pi\ln Y=E\_\pi(Y\ln Y-\ln Y)
=E\_\rho\ln\frac\rho\pi+E\_\pi\ln\frac\pi\rho,$$
which is the symmetrized version of the standard Kullback--Leibler (KL) information $D\_{KL}(\rho||\pi)=E\_\rho\ln\frac\rho\pi=E\_\pi Y\ln Y$ (your first-linked paper uses the standard definition of the KL information). To avoid further confusion, we are not going to use the book in the rest of this answer.
Thus, the corrected version of your inequality in question is this:
$$E\_\rho F-\ln E\_\pi e^F\le E\_\pi Y\ln Y$$
or, equivalently,
$$E\_\pi YF-\ln E\_\pi e^F\le E\_\pi Y\ln Y. \tag{1}$$
Since $E\_\pi Y=1$, the left-hand side of (1) will not change if $F$ is replaced by $F-c$, for any real constant $c$. So, without loss of generality $E\_\pi e^F=1$, and then (1) becomes
$$E\_\pi YF\le E\_\pi Y\ln Y. \tag{2}$$
By convexity,
$$e^F\ge e^u+(F-u)e^u$$
for all real $u$. Substituting here $u=\ln Y$ and taking the expectations, we get
$$E\_\pi Y=1=E\_\pi e^F\ge E\_\pi Y+E\_\pi (F-\ln Y)Y,$$
so that (2) follows.
So, (0) holds, as desired.
Of course, (0) will continue to hold with $K(\rho,\pi)$ in place of $D\_{KL}(\rho|\pi)$, because the symmetrized KL information $K(\rho,\pi)$ is no less than the standard KL information $D\_{KL}(\rho|\pi)$.
|
2
|
https://mathoverflow.net/users/36721
|
403880
| 165,681 |
https://mathoverflow.net/questions/403885
|
3
|
Many examples of Calabi-Yau manifolds are constructed as algebraic varieties in weighted projective space, or more generally as complete intersection Calabi-Yau (CICY) manifolds. Are there such realizations of compact hyperkahler manifolds besides K3? If no, is there a fundamental obstruction?
|
https://mathoverflow.net/users/138208
|
Compact hyperkahler manifold as algebraic variety in weighted projective space?
|
If $X$ is a hypersurface and $\dim(X) > 2$ then by Lefschetz hyperplane theorem $H^{2,0}(X) = 0$, hence $X$ can't by hyperkahler.
|
8
|
https://mathoverflow.net/users/4428
|
403892
| 165,686 |
https://mathoverflow.net/questions/403638
|
6
|
Let $X$ be a preperfectoid space over $\mathrm{Spa}(\mathbb{Q}\_p,\mathbb{Z}\_p)$. It has several associated sites, with successively finer topologies: $$X\_{an} \subset X\_{et} \subset X\_{proet} \subset X\_v.$$
I was wondering: **what is the relationship between vector bundles on these different sites**? Here, by a vector bundle I mean a locally finite free $\mathcal{O}\_X$-sheaf for $X\_{an}, X\_{et}$ and $X\_v$, and a locally finite free $\widehat{\mathcal{O}}\_X$-sheaf for $X\_{proet}$.
If $X$ is perfectoid, it is known that the categories of vector bundles on all of these sites are equivalent, by Theorem 3.5.8 of Kedlaya-Liu's paper "Relative p-adic Hodge Theory II".
Is this also true for preperfectoid spaces? What about sousperfectoid spaces? If this statement is not true, is it still true that some of these categories are equivalent? I expect that at the very least the pullback functor from $X\_{an}$ to $X\_v$ is fully faithful. My apologies if there is some silly counterexample.
Thanks!
|
https://mathoverflow.net/users/143589
|
Vector bundles on the various sites of a preperfectoid
|
There is some relevant [work of Ben Heuer](https://arxiv.org/abs/2012.07918v2) on this.
In short, analytic and etale vector bundles agree on all sousperfectoid adic spaces (and much more generally, for all "etale sheafy" adic spaces), while proetale and v-vector bundles also agree (because proetale locally, the space is perfectoid, and so this reduces to the assertion for perfectoids). The pullback functor from etale vector bundles to proetale vector bundles is fully faithful. So in summary, there are two distinct categories (etale vector bundles and proetale vector bundles), one being a full subcategory of the other.
However, etale and proetale vector bundles are very different. This is already a well-known phenomenon for $p$-adic fields $K$ (complete discretely valued with perfectly residue field), where letting $C=\widehat{\overline{K}}$ it is the difference between $K$-vector spaces and $G\_K$-equivariant $C$-vector spaces, which forms the subject of Sen--Tate theory.
By the way, there is some recent line of work (by Heuer, Mann, Werner, ...) that takes a fresh look at the $p$-adic Simpson correspondence from the perspective of proetale vector bundles. Roughly speaking, proetale vector bundles are what Faltings called "generalized representations" in his first paper on the $p$-adic Simpson correspondence, and should be more-or-less equivalent to Higgs bundles in case $X$ is smooth over an algebraically closed nonarchimedean $C/\mathbb Q\_p$. Noth that Higgs bundles also contain usual vector bundles on $X$ fully faithfully, as those with vanishing Higgs field.
(On the other hand, if $X$ is proper, then proetale vector bundles contain $C$-local systems fully faithfully, by the primitive comparison isomorphisms.)
|
7
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https://mathoverflow.net/users/6074
|
403895
| 165,687 |
https://mathoverflow.net/questions/177232
|
10
|
If a finite projective plane $\pi\_1$ of order $m$ contains, as a sub plane, a
finite projective plane $\pi\_2$ of order $n$, then $m \geq n^2$ with equality holding only in the case of a Baer sub plane. Otherwise $m \geq n^2 + n$. (This is a theorem of Bruck that can be found in Hall's Group Theory book, I believe.) My
question is, how nearly can this bound be achieved?
For example, we have cases for $m=9$ and $n=2$, where $\pi\_1$ is one of the non-Desaurgian planes of order $9$, and $\pi\_2$ is the Fano plane. Is anything known about how close we can come to the bound in the general case? Has the bound been improved?
|
https://mathoverflow.net/users/39684
|
Subplanes of Finite Projective Planes
|
No, the bound has not been improved. Even a subplane of order 3 of a projective plane of order 12 has not yet been ruled out. (You can see this
in the arguments in the work on possible collneations of a projective plane of order 12, such as Janko-Van Trung (1980) [On projective planes of order 12 which have a subplane of order 3. I. J. Combin. Theory Ser. A 29, 254–256].) There's a result of Roth from 1964 that if the subplane is fixed by a non-trivial collineation then $n > m^2+m+1$ [Roth, Richard (1964). Collineation groups of finite projective planes. Mathematische Zeitschrift, 83(5), 409-421, Corollary 2 on page 418]. Of course, if you found a subplane with equality in the Bruck bound, you would have a projective plane of order not a prime power, and have solved a much bigger problem, by disproving the prime power conjecture for finite projective planes.
|
1
|
https://mathoverflow.net/users/369536
|
403902
| 165,689 |
https://mathoverflow.net/questions/403888
|
8
|
Can there be an uncountable set $S\subseteq\mathbb R$ such that for each subset $D\subseteq S$, there is a Borel set $U$ with $D=S\cap U$?
I'm asking merely out of curiosity, but I'll mention that this would imply $2^{\aleph\_1}=2^{\aleph\_0}$. This is a hopefully more interesting adaption of a recent [too easy question](https://mathoverflow.net/q/403870/4600).
|
https://mathoverflow.net/users/4600
|
VC dimension of Borel sets
|
Yes! Martin's Axiom implies that if $S \subseteq \mathbb R$ and $|S| < \mathfrak{c}$, then every subset $D$ of $S$ is a relative $G\_\delta$ in $S$: i.e., there is a $G\_\delta$ set $X \subseteq \mathbb R$ with $X \cap S = D$. (And let me note that $2^{\aleph\_0} = 2^{\aleph\_1}$ is another consequence of Martin's Axiom.)
|
9
|
https://mathoverflow.net/users/70618
|
403903
| 165,690 |
https://mathoverflow.net/questions/403893
|
1
|
>
> Let $K$ be a number field and $v$ be it's one of $K$'s non-archimedian valuation.
> Then, I would like to prove $K\_v(a^{1/m}) /K\_v$ is unramified if
> only if $v(a)≡0 \pmod m$.
>
>
>
This is from Silverman's 'the arithmetic of elliptic curves', p213.
I know unramified extension of local field is in bijection with extensions of the residue field. Thus, the unramified extension is generated by roots of unity of order prime to the character of residue field of local field.
But I don't have tactics to judge given extension is unramified or not.
Thank you in advance.
|
https://mathoverflow.net/users/144623
|
$K_v(a^{1/m}) /K_v$ is unramified if only if $v(a)≡0 \pmod m$
|
If $v(a)\not\equiv 0\pmod m$, ramification is easy: just consider the valuation of the element $a^{1/m}$.
The converse is a little subtler than you make it seem, and depending on how exactly you phrase it, it need not be true: for instance, if $m=p$ coincides with the residue characteristic of $v$, $a=1$ and we interpret $a^{1/m}$ as a primitive $p$-th root of unity, the extension *will* be ramified.
Silverman includes a number of assumptions which exclude this case: he takes $v\not\in S$, and a couple paragraphs above it is stated that we assume this imples $v(m)=0$. Assuming this is the case and that $v(a)\equiv 0\pmod m$, note that, by multiplying $a$ by some $m$-th power, we may assume $v(a)=0$. Now, the polynomial $X^m-a$ has discriminant $\pm m^ma^{m-1}$, which under all these assumptions has valuation zero. This implies the discriminant of the extension $K\_v(\sqrt[m]{a})/K\_v$ is a $v$-adic unit, which implies the extension is unramified.
|
3
|
https://mathoverflow.net/users/30186
|
403905
| 165,692 |
https://mathoverflow.net/questions/403729
|
2
|
Let $u,v:\mathbb R \to \mathbb R$ and $\phi: \mathbb R \to \mathbb R\_+$ be smooth bounded functions. Assume also $\phi' \ge 0$. Assume that $u(0) - v(0) = 0$ and that $0$ is a strict global minimum of $u-v$. Let us assume $D\_\epsilon = \{x: u(x)-v(x) < \epsilon\} \subset B\_1(0)$. Under these assumptions, is it possible to determine the sign of $$\int\_{D\_\epsilon} \phi \, \Big([(-\Delta)^s](u-v)\Big) $$
that is $$\int\_{D\_\epsilon} \phi(x) \left(\int\_{\mathbb R} \frac{u(x+z) -v(x+z) -v(x)-u(x)}{|z|^{1+2s}} dz\right) dx $$
positive or negative?
Note that, if we had $$\int\_{D\_\epsilon} \phi(x)\partial\_x(u-v)dx$$ instead of the fractional Laplacian, I would compute
$$\int\_{D\_\epsilon} \phi(x)\partial\_x(u-v)dx = \int\_{D\_\epsilon} \phi(x)\partial\_x(\min\{u-v-\epsilon,0\}) dx = \int\_{B\_1} \phi(x)\partial\_x(\min\{u-v-\epsilon,0\}) dx\ge 0$$
(becasue the function is continuous and identically zero on a neighborhood of $\partial B\_1$).
|
https://mathoverflow.net/users/122620
|
Determine the sign (positive or negative) of an integral with the fractional Laplacian
|
The sign can be arbitrary already for $s = 1$. In this case we can take $u(x) - v(x) = 1 - \cos (\pi x)$ for $|x| \leqslant 1$ and $u(x) - v(x) = 2$ when $|x| > 1$, and $\epsilon = 2$. Then the integral becomes
$$ I := \int\_{-1}^1 \phi(x) (-2 \pi^2 \cos(\pi x)) dx = -2 \pi^2 \int\_{-1}^1 \phi(x) \cos(\pi x) dx .$$
Now it is easy to cook up $\phi$ so that the above expression is either positive or negative. To be specific:
* If $\phi(x) = 0$ for $x < \tfrac12$ and $\phi(x) > 0$ for $x > \tfrac12$, then clearly $I > 0$.
* On the other hand, if $\phi(x) = 0$ for $x < \tfrac12$ and $\phi(x) = 1$ for $x > 0$, then it is easy to see that $I < 0$.
---
The same construction will work for $s \in (0, 1)$ sufficiently close to $1$. A similar argument (but with a less explicit $u - v$) should also work for a general $s \in (0, 1)$, but I did not attempt to work out the details.
|
1
|
https://mathoverflow.net/users/108637
|
403906
| 165,693 |
https://mathoverflow.net/questions/403898
|
2
|
Let $X$ be a paracompact topological space or a manifold (which is not a particular case since the structure sheaves are different). It is well-known that vector bundles (more generally, $\mathcal{O}\_X$-modules) are soft, hence acyclic for $\Gamma$.
**I wonder if the same holds for $\Gamma\_c$.**
|
https://mathoverflow.net/users/131975
|
Are vector bundles acyclic for $\Gamma_c$?
|
Yes, at least if $X$ is locally compact.
A useful reference is Bredon's book. In brief, soft sheaves are in particular $c$-soft (they satisfy the softness condition for *compact* subsets of $X$ as opposed to all closed subsets of $X$). On a locally compact space, $c$-soft sheaves are $\Gamma\_c$-acyclic. Bredon states a more general version of this for an arbitrary "family of supports" $\Phi$, where "locally compact" becomes "the family of supports $\Phi$ is paracompactifying".
|
4
|
https://mathoverflow.net/users/1310
|
403910
| 165,695 |
https://mathoverflow.net/questions/403926
|
3
|
Let $R$ be a ring and let $M$ be a right module over $R$. We say that $M$ is faithfully flat as a right module if the functor $M \otimes\_R -$ from left $R$-modules to abelian groups that preserves and reflects exact sequences. Faithful flatness for a left $R$-module is defined analogously.
What is an example of an $R$-bimodule that is faithfully flat as a right module, but not faithfully flat as a left module? Or what is an example of an $R$-bimodule that is faithfully flat as a left module, but not faithfully flat as a right module?
|
https://mathoverflow.net/users/371382
|
Faithful flatness for rings
|
Let $G$ be a non-trivial finite group and let $R=\mathbb ZG$. We can view $M=\mathbb ZG$ as an $R$-bimodule via the right regular module structure on the right and the trivial module structure on the left (so left multiplication by $g\in G$ fixes $M$). Then $M$ is faithfully flat as a right module because $M\otimes\_R ()$ is the underlying abelian group functor. But $M$ is not flat as a left $R$-module. Indeed $M$ is finitely presented (cf. the bar resolution) and so if it were flat it would be projective. But then the trivial module $\mathbb Z$ would be projective which is never the case for a non-trivial finite group as no idempotent $e$ satisfies $ge=e$ for all $g\in G$ in a group algebra unless the order of $G$ is a unit in the coefficient ring.
Of course you can switch the roles of right and left.
|
3
|
https://mathoverflow.net/users/15934
|
403948
| 165,707 |
https://mathoverflow.net/questions/403927
|
6
|
There are a variety of characterizations of spin structures on the tangent bundle of a manifold. Two facts about them:
1. Spin structures on $TM$ are an affine space over $H^1(M; \mathbb{Z}/2\mathbb{Z})$, but in general there's no canonical way to identify them with $H^1$.
2. Spin structures on $TM$ are the same as trivializations of $TM$ along the $1$-skeleton that extend to trivializations on the $2$-skeleton.
I'm interested in the case where $M = S^3 \setminus L$ is a knot or link complement. (I'm not completely clear whether I want to think about $M$ as compact with torus boundary or non-compact, but I guess my proof below uses ideal triangulations.) I am suspicious that the following is true:
**Theorem:** Spin structures on $S^3 \setminus L$ are in one-to-one correspondence with orientations of $L$, hence with $H^1(S^3 \setminus L; \mathbb{Z}/2\mathbb{Z})$.
Proof idea: From an oriented diagram of $L$ we can use the octahedral decomposition to get an ideal triangulation of $S^3 \setminus L$. For such a triangulation it's possible to orient everything so that the obvious trivialization of the $1$-skeleton (coming from its orientation) extends over the $2$-skeleton.
This is pretty vague, but I think the "compatible orientations" are a branching, in the sense of [1]. I believe [1] works out the details, since they have a "spin calculus" for triangulations of spin $3$-manifolds.
**My question**: Is the theorem true? If so, is there a more direct proof than what I've given? It seems like there should be a more direct path than choosing some special triangulation.
[1] R. Benedetti, C. Petronio, *Branched Standard Spines of 3-manifolds*, Lect. Notes Math. 1653, Springer (1997)
EDIT: Some motivation for the question is the "well known" fact that spin structures on a hyperbolic knot complement are in natural 1-1 correspondence with lifts of the holonomy representation $\rho : \pi\_1(S^3 \setminus K) \to \operatorname{PSL}\_2(\mathbb C)$ to $\operatorname{SL}\_2(\mathbb C)$. (The argument has to do with identifying $\mathbb H^3$ as $\operatorname{PSL}\_2(\mathbb C)/\operatorname{SO}\_3$.) Then for a hyperbolic knot it's straightforward to pick a canonical lift/spin structure, because in one lift the trace of the meridian is $2$ and in the other it's $-2$. It doesn't matter which meridian you pick, because $\mathfrak m$ and $\mathfrak m^{-1}$ have the same trace.
However, upon further thought we *can't* identify this choice of lift with a choice of meridian ($\mathfrak m$ versus $\mathfrak m^{-1}$), which is the same as an orientation of the knot. This agrees with the answers that say that orientations are not naturally the same as spin structures.
|
https://mathoverflow.net/users/113402
|
Is a spin structure on a knot complement the same thing as an orientation of the knot?
|
Is the theorem true? There is an non-natural bijection. There is no natural bijection.
A link exterior is homotopy-equivalent to a $2$-complex, so a trivialization of the tangent bundle over the $2$-skeleton is simply a map:
$$S^3 \setminus L \to SO\_3.$$
This is because there is a canonical trivialization of $TS^3$ coming from, say, a left-invariant framing.
Similarly, if you had a trivialization over the $1$-skeleton that admits an extension to the $2$-skeleton, any two extensions $S^3 \setminus L \to SO\_3$ have to be homotopic, due to $\pi\_2 SO\_3$ being trivial.
So spin structures on link exteriors are canonically in bijection with
$$\pi\_0 Maps(S^3 \setminus L, SO\_3).$$
$SO\_3$ is diffeomorphic to $\mathbb RP^3$, which is the $3$-skeleton of $\mathbb RP^\infty$, which is a $K(\mathbb Z\_2,1)$-space.
So you have a natural map between the spin structures on the link exteriors and
$$\pi\_0 Maps(S^3 \setminus L, K(\mathbb Z\_2,1)) \equiv H^1(S^3 \setminus L, \mathbb Z\_2).$$
You can argue this map is injective. The argument that comes to mind is that $\pi\_1 \mathbb RP^3 \to \pi\_1 \mathbb RP^\infty$ is an isomorphism.
I think the relation to Kevin Walker's observation is that diffeomorphisms of the link exterior act trivially on some of the spin structures, but not on all. A diffeomorphism that reverses a link component does not affect the spin structure. The only way a diffeomorphism can act non-trivially on spin structures is if they permute components of the link. Similarly the diffeomorphism group of a knot exterior acts trivially on the spin structures, even for invertible knots (where diffeomorphisms can reverse the knot). The action of the diffeomorphism group on the spin structures factors through the permutation representation, i.e. the map $\Sigma\_n \to GL\_n \mathbb Z\_2$ given by the $n \times n$ matrices that permute the rows and columns, and that permutation representation is the permutation of the link components.
Technically to get the above factorization you need to know that the link has no split Hopf link components. When you have a split Hopf link you get a $GL\_2(\mathbb Z)$ in the mapping class group and this action would not belong to the above representation.
|
7
|
https://mathoverflow.net/users/1465
|
403950
| 165,708 |
https://mathoverflow.net/questions/403912
|
1
|
This is a cross-post from Stackexchange Mathematics (<https://math.stackexchange.com/questions/3893961/h%c3%b6lder-continuous-dependence-on-parameters-for-solutions-of-ode>).
We have the following standard result for continuous dependence of the initial value for ODEs with a continuous right-hand side (Satz 8.18 in [https://www.mathematik.hu-berlin.de/~baum/Skript/DGL-2012.pdf](https://www.mathematik.hu-berlin.de/%7Ebaum/Skript/DGL-2012.pdf)):
>
> Let $F:U \subset \mathbb{R}^n \times \mathbb{R} \rightarrow \mathbb{R}^n$ on $U$ be continuous and Lipschitz continuous with respect to the $\mathbb{R}^n$-variable with Lipschitz constant $L$.
> Let $(x\_0,t\_0),(x\_0^\*,t\_0) \in U$ and
> $\varphi\_{x\_0},\varphi\_{x\_0^\*} : [t\_0 - \epsilon, t\_0 + \epsilon] \rightarrow \mathbb{R}^n$
> be solutions of the ODE $x'=F(x,t)$ with initial values $\varphi\_{x\_0}(t\_0)=x\_0$ and $\varphi\_{x\_0^\*}(t\_0)=x\_0^\*$.
> Then:
> $$| \varphi\_{x\_0}(t)-\varphi\_{x\_0^\*}(t) |
> \leq |x\_0-x\_0^\*| \cdot e^{L|t-t\_0|}
> \forall t \in [t\_0-\epsilon,t\_0+\epsilon].$$
>
>
>
Question:
is there a similar result for Holder continuity?
My dream result would be
>
> Let $F:U \subset \mathbb{R}^n \times \mathbb{R} \rightarrow \mathbb{R}^n$ on $U$ be smooth and $\alpha$-Holder continuous with respect to the $\mathbb{R}^n$-variable, with $\alpha$-Holder norm bounded by $L$.
> Let $(x\_0,t\_0),(x\_0^\*,t\_0) \in U$ and
> $\varphi\_{x\_0},\varphi\_{x\_0^\*} : [t\_0 - \epsilon, t\_0 + \epsilon] \rightarrow \mathbb{R}^n$
> be solutions of the ODE $x'=F(x,t)$ with initial values $\varphi\_{x\_0}(t\_0)=x\_0$ and $\varphi\_{x\_0^\*}(t\_0)=x\_0^\*$.
> Then there exist a universal constant $c$ independent of $F$, and $\beta \in (0,1)$ such that
> $$| \varphi\_{x\_0}(t)-\varphi\_{x\_0^\*}(t) |
> \leq c|x\_0-x\_0^\*|^{\beta} \cdot e^{L|t-t\_0|}
> \forall t \in [t\_0-\epsilon,t\_0+\epsilon].$$
>
>
>
Note that I am happy to assume my function is smooth in order to have a unique solution.
I also note that I can use a $C^k$-bound for $F$ to get a $C^k$ bound for the solution depending on the initial value.
But what I need is a $C^{0,\beta}$-bound for the solution that only depends on the $C^{0,\alpha}$-bound for $F$.
I can imagine something like this exists for $\beta=\alpha/2$, but maybe even $\beta=\alpha$ is possible.
I know that the proof of cited theorem cannot be adapted to prove my dream result.
I also found "Agarwal, Lakshmikantham: Uniqueness and nonuniqueness criteria for ordinary differential equations" to make some statements about uniqueness of the solution for a Holder-continuous right-hand side $F$.
But I did not find anything resembling the estimate I need.
Context:
I have two metrics on a compact manifold, $g\_1, g\_2$, satisfying the estimate $||g\_1-g\_2||\_{C^{1,\alpha},g\_1}<c\_1$.
I also have a vector $\eta$ satisfying $|| \eta ||\_{C^{1,\alpha},g\_1} < c\_2$.
I would like to have an estimate $|| \eta ||\_{C^{1,\beta},g\_2} < F(c\_1,c\_2)$, where $\beta$ can depend on $\alpha$, but should not depend on $\eta$, and $F(c\_1,c\_2)$ is some universal expression in $c\_1$ and $c\_2$.
I believe that my dream ODE result from above would give me such an estimate.
|
https://mathoverflow.net/users/164084
|
Hölder continuous dependence on parameters for solutions of ODE
|
No, what you want is not possible. (Basically, you can approximate the nonunique example with smooth examples which have slightly moved initial data. This gives you two nearby initial data points that move apart arbitrarily fast.)
Consider the case $n = 1$, let $F\_0(x) = \begin{cases} 0 & x \leq 0 \newline \sqrt{x} & x> 0\end{cases}$. Let $F\_\delta$ be smooth functions that equal $F\_0$ outside $(-\delta,\delta)$; you can choose $F\_\delta$ to have uniform $C^{\alpha}$ norm for $\alpha < \frac12$. (Possibly also $\alpha = \frac12$, have to check.)
If you take $\delta \ll \epsilon^2$, and let $x\_0 = -\delta$ and $x\_0^\* = +\delta$, you have that $\varphi\_{x\_0} \equiv -\delta$ but $\varphi\_{x\_0^\*}(\epsilon) > \frac14 \epsilon^2$. Taking $\delta\to 0$ you see an arbitrarily large factor of inflation even compared against $(2\delta)^\beta$ for any $\beta > 0$.
|
1
|
https://mathoverflow.net/users/3948
|
403951
| 165,709 |
https://mathoverflow.net/questions/403919
|
2
|
LLPO is the statement $\forall x \in \mathbb R. x \leq 0 \vee x \geq 0.$ The statement should be understood as a fragment of the Law of Excluded Middle, rather than a statement about the ordering of the real line. The fragment is usually considered large enough to not be constructive in any reasonable sense.
I have a conjecture on the relation of LLPO to universal statements about dense subspaces, and am looking for a proof or refutation of that conjecture.
As background, consider the following non-constructive statements, which are all provable using LLPO:
1. For every pair of unit vectors $u, v$ in $\mathbb R^3$, there is a unit vector $w \in \mathbb R^3$ such that $u \perp w \perp v$.
2. The Intermediate Value Theorem: For every continuous function $f \in C([0,1])$ with $f(0) \leq 0 \leq f(1)$, there is an $x \in [0,1]$ such that $f(x) = 0$.
3. Every square matrix with entries in $\mathbb C$ has an eigenvector.
Each of these has a universal quantifier over some topological space $T$, which can be replaced with a dense subspace $\hat T$ for which a constructive proof is possible:
1. $T$ = pairs of unit vectors, $\hat T$ = pairs of non-parallel unit vectors. Given non-parallel vectors $u$ and $v$, we use the cross product to set $w := \frac{u \times v}{|u \times v|}$. However, perturbing nearly-parallel vectors slightly can radically alter the set of vectors perpendicular to them.
2. $T$ = continuous functions, $\hat T$ = piecewise-linear functions with rational slopes. However, perturbing a function slightly can radically alter its set of roots.
3. $T$ = matrices, $\hat T$ = matrices whose characteristic polynomials have only simple roots. However, perturbing such matrices slightly can radically alter their eigenspaces.
Has this phenomenon been studied in any generality?
Full statement of conjecture
----------------------------
Given some theory of constructive topology,
* Let $T$ be overt.
* Let $K$ be compact.
* Let $H$ be Hausdorff.
* Let $f, g : T \times K \to H$ be continuous functions.
Then the following two claims entail each other:
$$\text{LLPO} \implies \forall x \in T. \exists y \in K. f(x,y) = g(x,y)$$
$$\exists \hat T \text{ dense in }T. \forall x \in \hat T. \exists y \in K. f(x,y) = g(x,y)
$$
Does anyone have a proof or refutation for this conjecture?
|
https://mathoverflow.net/users/75761
|
LLPO as constructivity/computability for dense subsets
|
A counterexample to "Stuff provable by $\mathrm{LLPO}$ is constructively true on some dense set" can be built as follows:
We start with two recursively inseparable disjoint c.e. set $A, B \subseteq \mathbb{N}$. Using $\mathrm{LLPO}$ countably many times, we can actually built a "decidable" set $C \subseteq \mathbb{N}$ with $A \subseteq C$ and $B \cap C = \emptyset$. We just use $\mathrm{LLPO}$ to pick a true case between $n \notin A$ and $n \notin B$. Our domain is $\mathbb{N}$, which is as nice as it could be, and has no non-trivial dense subsets.
To put this example into the specified format, we use $T = \mathbb{N}$, $K = \{0,1\}$ and $H = \mathbb{R}$. The function $g$ is constant $0$. To define $f$, we use effective enumerations $(a\_n)\_{n \in \mathbb{N}}$ of $A$ and $(b\_n)\_{n \in \mathbb{N}}$ (which exist since $A$ and $B$ are c.e.). We now let $f(n,0) = 2^{-\min \{i \mid a\_i = n\}}$, $f(n,0) = 0$ if $n \notin A$, $f(n,1) = 2^{-\min \{i \mid b\_i = n\}}$, $f(n,1) = 0$ if $n \notin B$.
The difference between how I've used $\mathrm{LLPO}$ here versus how it is used in your examples is that in your examples the union of the two open sets defined by "Only first answer is correct" and "Only second answer is correct" was dense. But this is not necessary, as the set of instances leading to "Both answers are correct" can have non-empty interior. And getting this interior as an open set is not constructive, but actually needs countably many $\mathrm{LPO}$'s / $\mathrm{ACA}\_0$ / $\operatorname{lim}$.
If we use $\mathrm{LLPO}$ just once and use a sufficiently classical metatheory, we can get a dense open set on which the construction is continuous though. Working in a Baire space, we can then even use $\mathrm{LLPO}$ countably many times and get a dense set on which stuff is still continuous.
|
2
|
https://mathoverflow.net/users/15002
|
403962
| 165,711 |
https://mathoverflow.net/questions/403939
|
30
|
Some background first.
I recently graduated with a master's degree in applied mathematics. During graduate school I began working on a paper, which I continued to work on post-graduation. A complete working copy of the paper is done and I have posted it on the arXiv [here](https://arxiv.org/abs/2106.14958). The work contained in the paper is completely original and solves an open problem.
It was of my opinion that the paper contained publishable material. To verify, I emailed a professor at my alma mater with a copy of the current draft (current as of approximately six months ago). The professor did reply stating that the work was publishable and even suggested some Q1/Q2 journals that might accept this type of work. While this was useful feedback, I received the reply in just a few days, so I doubt the professor in question had the ability to read my paper in depth.
The problem:
I have published a couple of papers before and thus have some experience in the world of academic publishing. That said, the scale and complexity of this paper is something I have never dealt with before so I am not comfortable with proceeding to publish it without help/guidance, i.e. on my own. In particular, I suspect I am going to have to divide the work into small portions and publish a few separate papers, but I don't know how best to do this and do not want to sink a lot of additional time into this without any direction. Also, the paper is very dense and I am concerned that its readability is not optimal. Given that I do not have a ton of experience with large papers like this and am essentially working in a vacuum, I also really desire to get feedback on the quality of my proofs, which I suspect are not as concise as they could be. My situation seems a little unusual and I suspect that the feedback I am looking for would typically be provided by an adviser in a Ph.D. program.
**Given that I do not have an adviser that can provide detailed feedback, what should I do?**
I have tried reaching out to researchers/experts with relevant backgrounds and offering authorship in exchange for the help I need. Since what I'm seeking entails a significant amount of work, this proposition seemed reasonable; however, my efforts have not lead to much fruit.
**As mathematicians in academia, how are these types of requests viewed and how might I go about reaching out for help? I do not have funding and so I considered the offering of authorship as a a reasonable incentive to get the help I need. Is this practice frowned upon and is there anything else I can do to increase my chances of getting a researchers attention?**
|
https://mathoverflow.net/users/125801
|
How can I seek help in preparing a very long research article for publication?
|
First of all, I would consider it against the ethics of scientific publishing to accept an offer as a co-author when you were not involved in the research. So I don't think that is viable route.
What you have achieved is quite unusual, you have on your own identified and developed a research direction and produced a set of results that advance the state of the art. Isn't that what a Ph.D. is all about? Rather than seeking a co-author, I would seek a Ph.D. advisor. Contacting an expert in the field, asking for a chance to present your results – with the objective to expand this into a Ph.D. thesis – might very well succeed.
Preparing a seminar in which you present your work would also help you to focus on the essential innovation, which is difficult to extract from the arXiv paper. You might even find that this seminar can be converted into a paper that would be more suitable for publication. In the mean time, by posting your work on arXiv you have established your priority, so a journal publication is not at all urgent.
|
37
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https://mathoverflow.net/users/11260
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403968
| 165,712 |
https://mathoverflow.net/questions/403958
|
4
|
Following Takahashi (*"On the concrete construction of hyperbolic structure of 3-manifolds"*), I was able to construct the Euclidean cusp cross-section for the 5\_2 knot complement (please see <http://kias.dyndns.org/topogeoimages/5_2.cusp.png>), and determine the vertex invariants
```
x = 0.12256116687665364 + 0.7448617666197441 i
y = 0.6623589786223728 + 0.5622795120623012 i and
z = 0.7849201454990264 + 1.3071412786820453 i.
```
I would like to be able to compute the cusp shape from x, y and z, but as a retired physicist and learning this on my own, I have not been able to come up with an algorithm to do this. I have been able to compute the cusp shape following Yokota ("On the cusp shape of hyperbolic knots"), but this is from the knot, not from the cusp cross-section. I feel this should be easy, but the SnapPy documentation (and code, for that matter) is not entirely clear to me, nor is Coulson et. al. ("Computing Arithmetic Invariants of 3-Manifolds").
If I could just see the cusp shape in terms of x1, x2, x3, y1, y2, y3, z1, z2 and z3, I am reasonably sure I can work backwards to an algorithm I can use. Thanks in advance.
**Edit:**
In reference to Sam Nead's answer:
I believe that the image in the link ("5.2.cusp.png") is topologically the correct tiling. Using Mathematica to compute the angles around each edge in the cusp tiling, and then to draw the associated rays along each angle, I pieced these together to make an approximate geometrical sketch of the tiling (please see <http://kias.dyndns.org/topogeoimages/5_2.geometric.png>). This matches closely with Takahashi's figure 25.
Then I did what any physicist would do: I took measurements of the sides of the bounding parallelogram. The ratio of those lengths (arbitrarily in pixels, but since similar parallelograms have equal side ratios, that should not be a problem) is approximately 452/153 = 2.9425. The acute angle of the bounding parallelogram (corresponding to x1, for instance) is 80.6562 degrees or 1.40772 radians. As per Adams, Hildebrand and Weeks ("Hyperbolic invariants of knots and links"), I should obtain the homological cusp shape as
(longitude/meridian) . Exp(Pi.I.angle\_radians)
For this I get 0.479646 + 2.91505 I.
In checking this with SnapPy (which gives a cusp shape of -2.4902446675 + 2.9794470665 I), I noticed that, while I got the same tetrahedra shapes as Takahashi, SnapPy disagrees with those (my y is the difference between SnapPy's first two shapes).
Further, <https://knotinfo.math.indiana.edu> gives translations yielding a cusp shape of 2.49024 - 2.97945 I. I am assuming this is simply a choice of orientation (the real part of the meridian differs by a sign). When I followed Yokota, I got SnapPy's result.
I know that there can exist inequivalent cusp tilings, but it seems to me that the tetrahedra shapes should be invariant, and certainly the cusp shape is an invariant. I am now pretty much thoroughly confused.
**Further edit:**
Thank you!
Now, with some judicious use of the law of sines, I find for the magnitude of the cusp shape
Csc[Arg[y2]] Csc[Arg[z3]] Sin[Arg[y3]] Sin[Arg[z1]] +
Csc[Arg[z1]] Sin[Arg[z2]] +
Csc[Arg[x2]] (Sin[Arg[x3]] +
Csc[Arg[z3]] Sin[Arg[x1]] Sin[Arg[z2]]) = 3.01951
which with the complement of the acute angle (Arg[z1]+Arg[x3]) = 99.3438 degrees, gives precisely the result of the shape of m015 from SnapPy.
One more question: when I ask SnapPy for the cusp shape with and without specifying shortest peripheral curve, the answers seem to differ by an integer. For 5\_2, the integer was 2; for 6\_1 it is also 2; for 6\_2, it is 7. And when I compute the cusp shape for m004, my answer is exactly 2 greater than SnapPy's. Is there a simple reason for this (i.e., is the shape defined mod Z)?
PS:
Out of SnapPy (2.8) CensusKnots, 1267 were single cusp and for only 81 were the difference between cusp shapes not integers.
|
https://mathoverflow.net/users/371704
|
Computation of cusp shape from vertex invariants
|
It looks like you have computed the "tetrahedra shapes" for the three ideal hyperbolic tetrahedra making up the knot complement. Each of these gives a (euclidian!) shape to the four "cusp triangles" that "cut off" the four ideal vertices of the tetrahedron. To compute the shape of the cusp, you need to understand how it is tiled by the (in total 12) cusp triangles. To do this by hand takes some work! To get started, draw one of the tetrahedra $t$ in the upper-half-space model (with a vertex at infinity), and consider how the other three tetrahedra, that also meet infinity, and that also meet $t$, lie in the model. Each of these has just one cusp triangle that separates the body of the tetrahedron from infinity. You want to use the shapes $x\_i$, $y\_i$, and $z\_i$ to tile these outward... (say on the horoplane at height one).
**EDIT**: Very nice pictures! Here is a bit of a snappy session which may be helpful to you.
>
> In[1]: M = Manifold("5\_2")
>
>
> In[2]: M.cusp\_info()
>
>
> Out[2]: [Cusp 0 : complete torus cusp of shape -2.49024466751 +
> 2.97944706648\*I]
>
>
> In[3]: M.set\_peripheral\_curves("shortest")
>
>
> In[4]: M.cusp\_info()
>
>
> Out[4]: [Cusp 0 : complete torus cusp of shape -0.49024466751 +
> 2.97944706648\*I]
>
>
>
The thing to notice here is that the cusp shape (that is the "shape of the fundamental parallelogram") depends on the choice of a basis for the cusp group $P$. The cusp group $P$ is a copy of $\mathbb{Z}^2$. Each element of $P$ gives a parabolic Mobius transformation. Depending on the generating set (for $P$) that we use, we will get different cusp shapes!
Since $5\_2$ is a knot, the cusp group has one natural basis coming from the topological meridian (bounds a disk in $S^3$) and the topological longitude (bounds a Seifert surface in $S^3$). Since the complement of $5\_2$ is a hyperbolic manifold (called m015 in the snappy census), the cusp group has another natural basis coming from the two elements of $P$ which have smallest (parabolic) translation.
If you ask snappy for the cusp shape of the hyperbolic manifold, it agrees with you, up to a sign.
>
> In[5]: M.identify()
>
>
> Out[5]: [m015(0,0), 5\_2(0,0), K3\_2(0,0), K5a1(0,0)]
>
>
> In[6]: N = Manifold("m015")
>
>
> In[7]: N.cusp\_info()
>
>
> Out[7]: [Cusp 0 : complete torus cusp of shape -0.4902446675 +
> 2.9794470665\*I]
>
>
>
|
3
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https://mathoverflow.net/users/1650
|
403971
| 165,714 |
https://mathoverflow.net/questions/403957
|
4
|
Let $\chi\_{\mu}^{\lambda}$ denote a value of an irreducible character of the symmetric group $\frak{S}\_n$, where $\mu, \lambda\vdash n$. When $\mu=(n)$, then it's known that
$$\sum\_{\lambda\vdash n}\chi\_{\mu}^{\lambda}=\delta\_\text{odd}(n).$$
This time, I'm interested to know:
>
> **QUESTION 1.** Is there anything known about this total sum:
> $$\frak{s}\_n:=\sum\_{\mu\vdash n}\sum\_{\lambda\vdash n}\chi\_{\mu}^{\lambda}=?$$
>
>
>
LeechLattice supplied the OEIS link [here](http://oeis.org/A082733) with values $\frak{s}\_n=$$1, 2, 5, 13, 31, 89, 259, 842, 2810, \dots$
>
> **QUESTION 2.** The following appears true. Is it? $\frak{s}\_n$ is even iff $p(n)$ is even. Here, $p(n)$ stands for the [number of partitions of](https://oeis.org/A000041) $n$.
> Incidentally, one observes that it is not known "how often" $p(n)$ is even or odd.
>
>
>
|
https://mathoverflow.net/users/66131
|
Total sum of characters of the symmetric group $\frak{S}_n$
|
The sum $\sum\_\mu \chi^\lambda\_\mu$ over partitions $\mu$ of $n$ is the multiplicity of the irreducible $\chi^\lambda$ in the character afforded by $\mathfrak S\_n$ acting on itself by conjugation. If $\psi$ is the character for conjugation, then $\psi(g)$ is the size of the centralizer of $g$, so
$$\langle \chi^\lambda,\psi\rangle =\sum\_{\mu\vdash n} \frac{\chi^\lambda\_\mu z\_\mu}{z\_\mu},$$ where $z\_\mu$ is the size of the centralizer of any permutation with cycle type $\mu$. So your sum is counting how many irreducibles the conjugation representation breaks up into.
I should mention that the individual row sums $\sum\_{\mu\vdash n} \chi^\lambda\_\mu$ are not completely understood. It is already non-trivial to prove that they are greater than $0$ (this should be in "An explicit model for the complex representations of $S\_n$" by Inglis, Richardson and Saxl.) This is unlike the situation for the group acting on itself by left multiplication. However, when you take your double sum, then there is a good combinatorial description. In fact, summing down any column we know:
$$\sum\_\lambda \chi^\lambda\_\mu = \# \{g\in \mathfrak S\_n : g^2=h\}$$
for any fixed $h\in\mathfrak S\_n$ with cycle type $\mu$. This works because each irreducible character of $\mathfrak S\_n$ is the character of a real representation. In general, there is the work of Frobenius--Schur. (This is what Lucia was already quickly and correctly observing. Indeed, the sum of degrees will dominate.)
For your second question, it comes from the paper linked to by Lucia in your previous post. Miller proved the statement you are observing (see Theorem 2 and its proof, particularly the last few lines). The number of even entries plus the number of odd entries (denoted $E\_n$ and $O\_n$) is congruent to the number of partitions $p(n)$, and $E\_n$ is always even. So $O\_n\equiv p(n)$ mod 2.
|
11
|
https://mathoverflow.net/users/371843
|
403974
| 165,715 |
https://mathoverflow.net/questions/403937
|
15
|
It is clear that the term [*Moore graph*](https://en.wikipedia.org/wiki/Moore_graph) was coined by Hoffman and Singleton in their paper [On Moore graphs with diameters $2$ and $3$](https://doi.org/10.1147/rd.45.0497), where they write
>
> E. F. Moore has posed the problem of describing graphs for which equality holds in (2). We call such graphs "Moore graphs of type $(d,k)$".
>
>
>
My question is: does anyone know where Moore posed this problem?
|
https://mathoverflow.net/users/2663
|
Reference request: Moore graphs
|
Moore posed this problem to Hoffman at a conference, so it is not in print. Hoffman makes the following remark (from "Selected Papers of Alan Hoffman with Commentary", pp. 367):
>
> After I discussed the preceding paper at an IBM summer workshop, E.F. Moore
> raised the graph theory problem described in the paper, and my GE colleague Bob
> Singleton and I pondered it. Moore told me the problem because he thought the
> eigenvalue methods I was using might find another "Moore graph" of diameter
> 2 besides the pentagon and the Petersen graph. Indeed, we found the Hoffman-Singleton graph with 50 nodes (and showed it was unique) and that any other Moore
> graph of diameter 2 had to have 3,250 nodes (and to this day, no one knows if such
> a graph exists). Moore declined joint authorship, so we thanked him by giving his
> name to the class of graphs. When it was later proved by Damerell, and also by
> Bannai and Ito, that there were no other Moore graphs other than the trivial odd
> cycles, I felt a twinge of guilt in giving Moore's name to such a small set. But I
> was wrong: Moore graphs, Moore geometries, etc. continue to be discussed in the
> profession.
>
>
>
|
18
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https://mathoverflow.net/users/2384
|
403975
| 165,716 |
https://mathoverflow.net/questions/380983
|
1
|
I am doing some research in combinatorics, and I found that I have to consider the following binomial coefficient :
$$ \binom{\binom{i}{j}}{k} $$
(In fact, I have to take the product for fixed $i,k$ and odd $j$’s, but to make this product, I have to manipulate those coefficients, and I don’t know how.)
Is there a way to write it in terms of other binomial coefficients, power series, or other combinatorial tools?
|
https://mathoverflow.net/users/73667
|
Binomial coefficient in a binomial coefficient
|
*(In order that this question appears as answered I'm converting Darij Grinberg's comment into an answer.)*
For $j$ and $k$ (that will remain fixed) in $\mathbb N$, and a set $X$, denote $\mathcal F(X):=\mathcal{P}\_k\mathcal{P}\_j(X)$, the set of all $k$-element sets $\mathcal U:=\{U\_1,\dots,U\_k\}$ of $j$-element subsets of $X$. Denote $\mathcal C(X)\subset \mathcal F(X)$ the set of those elements $\mathcal U$ of $\mathcal F(X)$ which are coverings of $X$, that is $\displaystyle \bigcup\_{U\in \mathcal U}U =X$. Every $\mathcal U\in\mathcal F(X)$ is a covering of exactly one subset $Y$ of $X$, namely $Y:=\displaystyle \bigcup\_{U\in \mathcal U}U$. Therefore
$$\mathcal F(X)=\bigsqcup\_{Y\subset X } \mathcal C(Y)$$
$$\mathcal C(X)=\Big(\bigcup\_{x\in X } \mathcal F\big(X\setminus\{x\}\big)\Big)^c,$$
and note also that
$$ \bigcap\_{x\in Y} \mathcal F\big(X\setminus\{x\}\big)= \mathcal F\big(X\setminus Y\big).$$
As to cardinalities, if $|X|=n$, we have of course $\big|\mathcal F(X)\big|=\begin{pmatrix} n\\j \\ k \end{pmatrix} := \bigg( {{n\choose j}\atop k}\bigg) $, and, if we denote $\begin{bmatrix} n\\j \\ k \end{bmatrix}:=\big|\mathcal C(X)\big| $ we have from the above disjoint union
$$\begin{pmatrix} n\\j \\ k \end{pmatrix}=\sum\_{m=0}^n{n\choose m}\begin{bmatrix} m\\j \\ k \end{bmatrix},$$
which can be inverted giving
$$\begin{bmatrix} m\\j \\ k \end{bmatrix}=\sum\_{n=0}^m(-1)^{m-n}{m\choose n}\begin{pmatrix} n\\j \\ k \end{pmatrix}.$$
The latter, of course, can be seen as an instance of the inclusion-exclusion formula: $$\big|\mathcal C(X)\big| =\bigg|\Big(\bigcup\_{x\in X } \mathcal F\big(X\setminus\{x\}\big)\Big)^c\bigg|=\sum\_{Y\subset X}(-1)^{|Y|}\big|\mathcal F\big(X\setminus Y\big)\big|=\sum\_{Y\subset X}(-1)^{|X\setminus Y|}\big|\mathcal F(Y)\big|$$
|
2
|
https://mathoverflow.net/users/6101
|
403981
| 165,718 |
https://mathoverflow.net/questions/310920
|
5
|
Let $\lambda$ be the partition of integer $d$. The Frobenius coordinate of $\lambda$ is given
$$ (a\_1 ,\ldots,a\_{d(\lambda)}|b\_1,\ldots,b\_{d(\lambda)}),$$
where $d(\lambda)$ denote the diagonal of $\lambda$.
Let $a\_i'= a\_i +\frac12$ and $b\_i'= b\_i +\frac12$ are modified Frobenius coordinates.
By a classical theorem of Frobenius, it states that
$f(\lambda)=\frac{|C\_{2,1,\ldots,1}|}{dim \lambda}\chi\_{2,1,\ldots,1}^{\lambda}=\frac12\sum\_{i=1}^{d(\lambda)}(a\_i')^2-(b\_i')^2$
Notice that the above example we treat the L.H.S as a function of $\lambda$ as we fix the partition of the form $2,1^{d-2}$
If there is any generalisation of the above theorem? That is if we replace
$(2,1,\ldots,1)$ with any other partition of $d$ then say $\mu$ then what will be
$$\frac{|C\_{2^{d/2}}|}{dim \lambda}\chi\_{2^{d/2}}^{\lambda}$$
For example say $\mu$ is of the form $2^{d/2}$
in Frobenius coordinate? If it is already studied can anyone please site the reference.
|
https://mathoverflow.net/users/45170
|
Frobenius coordinate expansion of character
|
There is a generalisation of the character formula, although it is usually stated in terms of contents rather than Frobenius coordinates. Also, it applies to conjugacy classes labelled by partitions $(1^{m\_1} 2^{m\_2} \cdots)$, where $m\_2, m\_3, \ldots$ are fixed and $m\_1$ varies with $n$ (as in the example you gave, where there is one part of size $2$, and $n-2$ parts of size 1). So in particular, it does not apply directly to the partitions $(2^{n/2})$.
For the benefit of the uninitiated, let me state right away that $\frac{|C\_\mu| \chi\_{\mu}^\lambda}{dim(\lambda)}$ is a *central character*. The sum of all elements of $S\_n$ of cycle type $\mu$ is a central element of the group algebra of $S\_n$, and the central character is the scalar by which the central element acts on the irreducible representation labelled by $\lambda$. Central characters are important in modular representation theory, which might provide a little motivation for why the quantity in question is a natural thing to consider.
Suppose that we have a partition $\lambda = (\lambda\_1, \ldots, \lambda\_l)$. Suppose that $d$ is the largest integer such that $\lambda\_d \geq d$ (the *Durfee size* of $\lambda$). Then the Frobenius coordinates of $\lambda$ are $(\alpha\_1, \ldots, \alpha\_d | \beta\_1, \ldots, \beta\_d)$, where $\alpha\_i = \lambda\_i - i$ and $\beta\_i = \lambda\_i^\prime - i$ (here $\lambda^\prime$ is the transpose partition of $\lambda$). So the $\alpha\_i$ and $\beta\_i$ count the number of boxes in past the diagonal in the $i$-th row and $i$-th column respectively (and the number of each is $d$). On the other hand, the contents of a box in row $j$ and column $i$ is $i-j$. It is not difficult to see that the multiset of contents of $\lambda$ consists of $0$ with multiplicity $d$, and $\{1,2,\ldots, \alpha\_i\}$ for each $i$ together with $\{-1, -2, \ldots, -\beta\_i\}$ for each $i$. So in particular, the sum of $r$-th powers of the contents of $\lambda$ can be expressed as polynomials in the $\alpha\_i$ and $\beta\_i$ by [Faulhaber's Formula](https://en.wikipedia.org/wiki/Faulhaber%27s_formula). For example, the sum of the contents ($r=1$) is
$$\sum\_i (1 + 2 + \cdots + \alpha\_i) - (1 + 2 + \cdots + \beta\_i) = \sum\_i {\alpha\_i + 1 \choose 2} - {\beta\_i + 1 \choose 2},$$
which is precisely the quantity appearing in the example you gave. We write $cont(\lambda)$ for the multiset of contents of $\lambda$.
Now, for a partition $\mu = (\mu\_1, \mu\_2, \cdots, \mu\_k)$, let $\bar{\mu} = (\mu\_1 - 1, \mu\_2 - 1, \ldots, \mu\_k - 1)$ be the partition obtained from $\mu$ by subtracting 1 from each nonzero part of $\mu$ (and ignoring trailing zeros). For example, if $\mu = (2,1, \ldots, 1)$ as in your example, then $\bar{\mu} = (1)$.
**Theorem** There exists a family $f\_{\rho}$ of elements of $\mathbb{Q}[n] \otimes \Lambda$ (where $n$ is a polynomial variable) and $\Lambda$ is the ring of symmetric functions with the property that
$$\frac{|C\_\mu| \chi\_{\mu}^\lambda}{dim(\lambda)} = f\_{\bar{\mu}}(|\lambda|, cont(\lambda))$$
By this notation I mean that we evaluate $f\_{\bar{\mu}}$ at $n=|\lambda|$ and its symmetric function variables at $cont(\lambda)$.
It turns out that $f\_{(1)} = p\_1$ (the first power-sum symmetric function) which doesn't actually depend on $n$. Then putting this together, we see
$$\frac{|C\_{(2,1,\ldots,1)}| \chi\_{(2,1,\ldots,1)}^\lambda}{dim(\lambda)} = p\_1(cont(\lambda)) = \sum\_i {\alpha\_i + 1 \choose 2} - {\beta\_i + 1 \choose 2},$$
which recovers the result you stated. In general, you can take $f\_{\bar{\mu}}$, express it as a polynomial in $n$ and the power-sums $p\_1, p\_2, \ldots \in \Lambda$, and upon evaluating at the contents of $\lambda$ you will obtain a polynomial in $n$ and the Frobenius coordinates of $\lambda$. For example,
$$f\_{(2)} = p\_2 - {n \choose 2},$$
which tells us that the central character corresponding to the conjugacy class $(3,1,\ldots, 1)$ may be computed using the formula
$$\left( \sum\_{i} \frac{\alpha\_i(\alpha\_i + 1)(2\alpha\_i + 1)}{6} + \frac{\beta\_i (\beta\_i + 1)(2\beta\_i + 1)}{6} \right) - {|\lambda| \choose 2}.$$
There is a lot more to be said about the $f\_\rho$. For example, they form a $\mathcal{R}$-basis of $\mathcal{R} \otimes \Lambda$, wher $\mathcal{R}$ is the subring of $\mathbb{Q}[n]$ consisting of integer-valued polynomials. But in this post I worked over $\mathbb{Q}$ because I wanted to use power-sum symmetric functions to turn contents-evaluaion into polynomials in Frobenius coordinates.
References you might find helpful (each of which contains an explanation of the above theorem):
---
*Corteel, Sylvie; Goupil, Alain; Schaeffer, Gilles*, [**Content evaluation and class symmetric functions**](http://dx.doi.org/10.1016/j.aim.2003.09.010), Adv. Math. 188, No. 2, 315-336 (2004). [ZBL1059.05104](https://zbmath.org/?q=an:1059.05104).
---
Section 5.4 of
*Ceccherini-Silberstein, Tullio; Scarabotti, Fabio; Tolli, Filippo*, Representation theory of the symmetric groups. The Okounkov-Vershik approach, character formulas, and partition algebras., Cambridge Studies in Advanced Mathematics 121. Cambridge: Cambridge University Press (ISBN 978-0-521-11817-0/hbk). xv, 412 p. (2010). [ZBL1230.20002](https://zbmath.org/?q=an:1230.20002).
(although this is based on the material of the above paper)
---
Section 3 of my own paper (which doesn't appear in the citation engine, so I'm providing an arXiv link instead) [Stable Centres I: Wreath Products](https://arxiv.org/abs/2107.03752)
|
3
|
https://mathoverflow.net/users/159272
|
403984
| 165,719 |
https://mathoverflow.net/questions/403953
|
2
|
In the book **Neverending fractions** from Borwein, van der Poorten, Shallit and Zudilin, there is the so called **distance formula** (Theorem 2.45, p. 43) stated:
$$\alpha\_1\alpha\_2\cdot...\cdot\alpha\_n=\frac{(-1)^n}{p\_{n-1}-q\_{n-1}\alpha}$$
for $n\geq 0$ with $p\_{-1}=1$ and $q\_{-1}=0$, where $$\alpha=[a\_0;a\_1,...,a\_{j-1},\alpha\_j]$$ ($\alpha\_j$ being the tail of the continued fraction development of $\alpha$) and, as usual, $$\frac{p\_n}{q\_n}=[a\_0;a\_1,...,a\_{n}]$$
In the book, the authors explain the name with
>
> It turns out that one may usefully think of $\left|\log(\left|p\_{n-1}-q\_{n-1}\alpha\right|)\right|$ as measuring the weighted distance that the continued fraction has traversed in moving from $\alpha$ to $\alpha\_n$
>
>
>
My questions are:
1. Does this distance formula only hold for *quadratic* irrationals?
2. Where can I find additional literature / information on this formula? (I couldn't find any other references for it so far)
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https://mathoverflow.net/users/108459
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Distance formula for continued fractions
|
1. This formula is valid for any continued fraction. But for quadratic irrationals it is especially useful because it allows to express fundamental unit of corresponding field in terms of continued fraction expansion of $\sqrt{n}$. For a reduced quadratic irrational $\omega=[0;\overline{a\_1,\ldots,a\_n}]$ with period $n=\mathrm{per}(\omega)$ we write
$$
{\mathrm{per}}\_e(\omega)=\begin{cases}
n, &\text{if $n={\mathrm{per}}(\omega)$ is even;} \\
2n,&\text{if $n={\mathrm{per}}(\omega)$ is odd.}
\end{cases}
$$
In this case, the fundamental unit can be found using Smith’s formula
$$
\varepsilon\_0^{-1}(\omega)=\omega T(\omega)T^2(\omega)\ldots
T^{\mathrm{per}\_e(\omega)-1}(\omega),
$$
where $T(\alpha)$ stands for the Gauss map:
$T(\alpha)=\left\{{1}/{\alpha}\right\}$.
2. The Smiths article (*Smith H. J. S. Note on the Theory of the Pellian
Equation and of Binary Quadratic Forms of a Positive
Determinant. Proc. London Math. Soc., 1875, s1-7,
196-208.*) is [here.](http://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/plms/s1-7.1.196)
A good book about this topic is *B. A. Venkov, Elementary number theory, ONTI, Moscow 1931; English transl.,
Wolters-Noordhoff Publ., Groningen 1970.* (see [zbmath](https://zbmath.org/?q=an:0204.37101) for review, and chapter 2 for continued fractions).
Applications to quadratic irrationals (Gauss-Kuz'min statistica and distribution of lengths) can be found in [Spin chains and Arnold's problem on the Gauss-Kuz'min statistics for quadratic irrationals.](http://iam.khv.ru/articles/Ustinov/nth34_eng.pdf)
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4
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https://mathoverflow.net/users/5712
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403990
| 165,721 |
https://mathoverflow.net/questions/403995
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3
|
If positive integer $n$ such $n\mid2^n-2$,where $n>1$, we called $n$ is Poulet number, see: <https://en.wikipedia.org/wiki/Super-Poulet_number>
I found if $n>2$ is Poulet number, then $\dfrac{2^n-2}{n}$ always is composite number
Is this a known result? if $n$ is odd number, so $\frac{2^n-2}{n}$ is even number, and $\frac{2^n-2}{n}>2$,so $\frac{2^n-2}{n}$ is composite number.
But for $n$ is even number. I can't prove it
|
https://mathoverflow.net/users/38620
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Prove that $\frac{2^n-2}{n}$ is composite number
|
Let $n=2k>2$ be a Poulet number, then $k$ divides $2^{2k-1}-1$ (in particular, $k$ is odd) and we must prove that $q:=\frac{2^{2k-1}-1}k$ is composite.
Let $s$ be minimal positive integer for which $k$ divides $2^s-1$ (i.e., $s$ is a multiplicative order of 2 modulo $k$). Then all numbers $t$ for which $k$ divides $2^t-1$ are divisible by $s$. In particular, both $2k-1$ and $\varphi(k)$ are divisible by $s$, thus $s\leqslant \varphi(k)<k$, and $2k-1=sw$ for an integer $w\geqslant 3$. Therefore $$q=\frac{2^{sw}-1}k=\frac{2^s-1}k\cdot (1+2^s+\ldots+2^{(w-1)s}).$$
If $q$ is prime, the first multiple must be equal to 1. Thus $k=2^s-1$, and $$2^{sw}-1=qk=q(2^s-1).$$
But we may factorize $2^{sw}-1$ as $2^{sw}-1=\prod\_{d|sw} \Phi\_d(2)$, where $\Phi\_n$ is $n$-th cyclotomic polynomial. For $d>1$ we get $\Phi\_d(2)=\prod\_{\xi}|2-\xi|>1$, where $\xi$ runs over primitive roots of unity of degree $d$. Thus
$$
q=\frac{2^{sw}-1}{2^s-1}=\prod\_{d\, \text{divides}\,sw\, \text{but not}\, s } \Phi\_d(2)
$$
is composite unless $sw$ has unique divisor not dividing $s$ (this divisor is of course $sw$). If $s$ has a prime divisor $p\ne w$, then $sw/p$ would be another divisor of $sw$ not dividing $s$.
Therefore $w$ is prime and $s=w^j$
for certain integer $j\geqslant 1$, $k=2^s-1$ and $2^{w^j+1}-2=2^{s+1}-2=2k-1=sw=w^{j+1}$. But modulo $w$ we get $2^{w^j+1}\equiv 4$ that yields a contradiction.
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17
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https://mathoverflow.net/users/4312
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403996
| 165,724 |
https://mathoverflow.net/questions/403834
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3
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Let $M$ be a closed $4$-d Riemannian manifold and $Z$ be its twistor space of $M$, i.e., the bundle of almost complex structures on $M$. Let $V$ be a Spin$^{\mathbb{C}}$ bundle, $V\_+$ denote the positive spin bundle. We know $Z$ admits more than one almost complex structure. So it can have canonical Spin$^{\mathbb{C}}$ bundle once we fix an almost complex structure and we can have other Spin$^{\mathbb{C}}$ bundles by twisting it with a complex line bundle. Now let $\pi:Z\rightarrow M$ be the projection. Can we realize $\pi^\*(V\_+)$ as a sub-bundle of some Spin$^{\mathbb{C}}$ bundle of positive spinors on $Z$ and if yes how?
|
https://mathoverflow.net/users/131004
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Pull back of Spin$^{\mathbb{C}}$ bundle
|
$\newcommand{\spinors}{\mathbb{S}}$The tangent bundle $TZ$ fits into an exact sequence
$$
0\to T\_{/M}Z\to TZ\xrightarrow{\pi\_\*}\pi^\*TM\to 0
$$
where the fiberwise tangent bundle $T\_{/M}Z$ is two-dimensional and equipped with a canonical complex structure. Explicitly, the fiber of $Z\to M$ over a point $m$ are the (oriented) complex structures on $T\_mM$, which are isomorphic to the projective line of the two-dimensional complex bundle $\spinors\_+(M)\to M$ of positive spinors (which you denote by $V\_+$). Note that this projective line is defined without the choice of a spin structure, as the action of $\operatorname{Spin}(4)\cong SU(2)\times SU(2)$ on $\mathbb{CP}^1$ factors through projection of the first factor to $PSU(2)$ and therefore descends to $SO(4)$. The isomorphism sends a complex structure to the line of spinors which are annihilated by Clifford multiplication with vectors $v\in T\_mM\otimes\mathbb C$ which are antiholomorphic with respect to the chosen complex structure.
The exact sequence splits using the Euclidean metric on $TZ$, whose existence you take for granted when you talk about the spinor bundle of $Z$; to be precise, the splitting can be obtained via the Levi-Civita connection, with the metric on $TZ$ defined as the orthogonal sum of the induced metric on $T\_{/M}Z$ and the pulled back metric on $M$. The fiberwise tangent bundle has a complex structure from the description of the fiber as the projective line of a complex vector bundle, and the pullback $\pi^\*TM$ has a canonical complex structure by definition of $Z$. It follows that the sum $TZ$ has a (almost) complex structure as well.
The spinor bundle of the sum of two even-dimensional vector spaces is the tensor product of the respective spinor bundles. This shows that $\spinors\_+(TZ)\cong \spinors\_+(T\_{/M}Z)\otimes \pi^\*\spinors\_+(TM)\oplus\spinors\_-(T\_{/M}Z)\otimes \pi^\*\spinors\_-(TM)$. Thus we can realize $\pi^\*\spinors\_+(TM)$ as a subbundle of the positive spinors on $Z$ if $\spinors\_+(T\_{/M}Z)$ is trivial, which we can achieve by a judicious choice of Spinᶜ-structure on the vertical tangent bundle, coming from a $PU(2)$-equivariant Spinᶜ-structure on $\mathbb{CP}^1$; it is an easy exercise that such Spinᶜ-structures are in bijection with the integers, with the positive spinor bundle given by the line bundle $\mathcal O(k)$ and the negative spinor bundle given by the line bundle $\mathcal O(k+2)$, which can be made equivariant iff $k$ is even, so that we may take $k = 0$.
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1
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https://mathoverflow.net/users/35687
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403997
| 165,725 |
https://mathoverflow.net/questions/404011
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1
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Consider the initial-value problem associated to the PDE $u^\epsilon\_t + u^\epsilon\_x - \epsilon u^\epsilon\_{xx} + \epsilon u^\epsilon\_{xxx} = 0$.
To prove that, as $\epsilon \to 0$, the weak solution $u^{\epsilon}$ converges in $L^2$ to the weak solution $u$ of the IVP for $u\_t + u\_x=0$ (with the same initial data), does it suffice to observe that $$\sup\_{t\in (0,T)}\Vert u^\epsilon(t,\cdot)\Vert\_{L^2(\mathbb R)} \le \Vert u(0,\cdot) \Vert\_{L^2}$$
and use weak convergence? Or do we need something more?
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https://mathoverflow.net/users/nan
|
Limit of $u^\epsilon_t + u^\epsilon_x - \epsilon u^\epsilon_{xx} + \epsilon u^\epsilon_{xxx} = 0$ as $\epsilon \to 0$
|
Since your equation is linear, this is sufficient. The convergence holds in the sense of distributions.
Notice the following: the limit, as a solution of the hyperbolic equation $u\_t+u\_x=0$ is unique, being actually $u(t,x)=u(0,x-t)$. In particular
$$\|u(t)\|\_{L^2({\mathbb R})}\equiv\|u(0)\|\_{L^2({\mathbb R})}.$$
Since on the other hand
$$\|u\_\epsilon(t)\|\_{L^2({\mathbb R})}\le\|u(0)\|\_{L^2({\mathbb R})}$$
and $u\_\epsilon\rightharpoonup u$, we conclude that the weak convergence is actually a strong one. More precisely $\|u\_\epsilon(t)-u(t)\|\_{L^2({\mathbb R})}\rightarrow0$ for almost every $t$.
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4
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https://mathoverflow.net/users/8799
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404017
| 165,731 |
https://mathoverflow.net/questions/403954
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1
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Consider the notion of a split coequalizer (see the [nLab](https://ncatlab.org/nlab/show/split+coequalizer) for the definition). Note that the definition seems to be non-symmetric. Are there any conditions on the ambient category such that it becomes symmetric?
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https://mathoverflow.net/users/145805
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Conditions such that split coequalizers are a symmetric notion
|
The definition of split coequaliser is essentially a transcriptions of that of an algebra for a monad, ie an object in the Eilengberg--Moore construction of the maximal category with an adjunction that yields the given monad.
Yes, it's asymmetric. The construction is awkward for other reasons. For example, the composite of two monadic adjunctions (ie of the Eilenberg-Moore kind) need not be monadic.
Sorry.
Any of the textbook accounts of monads will tell you about this, for example "Toposes, Triples and Theories" by Michael Barr and Charles Wells.
|
0
|
https://mathoverflow.net/users/2733
|
404021
| 165,733 |
https://mathoverflow.net/questions/403914
|
1
|
According to numerical simulation, the relationship
$$\sum^{\infty}\_{N=1}\frac{N^{-\alpha} \, x^{N-1}}{\Gamma(N)}\sim e^{x}x^{-\alpha}$$
where $\Gamma$ is the Gamma function seems to be true.
Do you have any idea how to show this relationship using asymptotic or exact methods?
|
https://mathoverflow.net/users/nan
|
Is this relationship, $\sum^{\infty}_{N=1}\frac{N^{-\alpha} \, x^{N-1}}{\Gamma(N)}\sim e^{x}x^{-\alpha}$, true?
|
Put $\mu:=\alpha-1$; then $\sum^{\infty}\_{n=1}\frac{ x^{n-1}}{(n-1)!n^\alpha} = x^{-1}\sum^{\infty}\_{n=1}\frac{ x^n}{n!n^\mu}=x^{-3/2}I\_\mu(x)$ for the function $I\_\mu(x)$ given in Johannes Trost’s answer to [Asymptotic expansion of $\sum\limits\_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}} }$](https://mathoverflow.net/questions/264595/asymptotic-expansion-of-sum-limits-n-1-infty-fracx2n1n-sqrtn/266265#266265), with a complete asymptotic series.
|
6
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https://mathoverflow.net/users/6101
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404022
| 165,734 |
https://mathoverflow.net/questions/404009
|
5
|
Forgive me if this question turns out to be too elementary-then feel free to move it to stack exchange. I believe that this should be very basic fact from topos theory nevertheless being not familiar with topos theory let me ask it.
One of the highlights of topos theory is the possibility to consider an analogue of the power set in the form of $Sub(A)=\{all \ subobjects \ of \ A \}$ (which are defined as equiavlence classes of monic arrows). In any topos there is a distinguished object $\Omega$ called the subobjects classifier. Moreover in any topos there is an analogue of the operation of taking all functions from $A \to B$-this is called exponentation. Given those facts we have a one to one correspondence $Sub(A) \cong \Omega^{A}$. Am I right in saying that:
>
> The left hand side of this correspondence is a priori not an object of our topos $\mathcal{T}$ but the right hand side is: therefore this natural correspondence is the way of making $Sub(A)$ again an object in our category $\mathcal{T}$
>
>
>
Please corect me if I'm wrong
|
https://mathoverflow.net/users/24078
|
Subobjects as an object in a topos
|
The general notion you're looking for is a **representable functor**. For example:
$${\mathcal E}(X\times{-},Y) \sim {\mathcal E}({-},Y^X)$$
$${\textsf{Sub}}({-}) \sim {\mathcal{E}}({-},\Omega)$$
The thing on the left is a general contravariant functor from the category to $\mathbf{Set}$.
The thing on the right is of the form ${\mathcal E}({-},R)$,
where $R$ is the "representing object" for the functor.
The definition of a **préfaisceau représentable** is given in SGA4 Exposé I, remark 1.4.2. Presumably the idea is then used, maybe extensively, in SGA4 and beyond, but I will leave others to search for it.
This is also one of the many equivalent ways of expressing an adjunction.
Of course, whether or not the (topos or other) category *has* a representing object for a particular functor is very much up for debate in the topic in question.
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9
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https://mathoverflow.net/users/2733
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404023
| 165,735 |
https://mathoverflow.net/questions/403978
|
2
|
Let $G\_k$ be the graph obtained by applying the following procedure k-times:
1. Start with a graph with single vertex $v$ (Call this graph $H$)
2. Add a vertex $u$ such that $u$ is not adjacent to any vertex of $H$ (i.e., $K:= H \cup \{u\}$) union of two graphs
3. Add a vertex $w$ such that $w$ is adjacent to all the vertices of $K$ (i.e., $J := K \vee \{w\}$) join of two graphs
4. Set $H = J$
5. Goto step 2.
My question is, is there a name for the class of graphs $\{G\_k\}\_{k\ge1}$? Please provide some references. Thank you.
|
https://mathoverflow.net/users/33047
|
Name of an inductively defined sequence of graphs
|
It's not quite the same question, but the graphs that can be obtained by repeating either of the two operations (add a disjoint vertex or a dominating vertex), not necessarily in strict alternation, are called [threshold graphs](https://en.wikipedia.org/wiki/Threshold_graph).
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4
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https://mathoverflow.net/users/440
|
404031
| 165,738 |
https://mathoverflow.net/questions/404027
|
1
|
I think the following inequality might be true and was hoping somebody might spot it or know a proof:
Suppose $f:\mathbb R\to \mathbb R$ is convex and suitably nice so that
$$\int\_{\mathbb R} e^{-f(x)} dx = 1$$
Then is it true that
$$\int\_{\mathbb R} f(x) e^{-f(x)} dx\ge 0$$
?
I also wonder what area of mathematics this might fit into or be a baby case of (perhaps the theory of logarithmically concave distributions)?
Thanks in advance.
|
https://mathoverflow.net/users/9202
|
Elementary inequality about integrals of exponentials of concave functions (possibly connected to log concave distributions)
|
The answer is no.
E.g., take any real $b>1$ and let $a:=2e^b$. Let $f(x):=a|x|-b$ for all real $x$.
Then $\int\_{\mathbb R}e^{-f(x)} dx=1$ but $$\int\_{\mathbb R} f(x) e^{-f(x)} dx
=1-b<0.$$
|
1
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https://mathoverflow.net/users/36721
|
404033
| 165,739 |
https://mathoverflow.net/questions/404029
|
1
|
What is the Lp norm of the $N$-dimensional Hadamard matrix $H = ((-1)^{i \cdot j})\_{i,j}$ for $p > 2$? I know that $\|H\|\_1 = N$, $\|H\|\_2 = \sqrt{N}$, $\|H\|\_\infty = N$ but I can't figure out what it is for other values of $p$. Can we at least give a good upper-bound on it?
Here I consider the induced norm: $\|H\|\_p = \max\_{x : \|x\|\_p = 1} \|Hx\|\_p$
(a previous version of this question incorrectly said that $\|H\|\_\infty = 1$)
|
https://mathoverflow.net/users/92442
|
Lp norm of Hadamard matrix
|
**Important Edit**: As J.J Green pointed out, the OP contains an incorrectly stated value for $\|H\|\_{\infty}$, which I copied without checking below. Interpolating between $(1,\infty)$ using the corrected version would give the trivial bound $\|H\|\_p \leq N$. You regain the sharp bound by interpolating instead between $(1,2)$ and $(2,\infty)$.
---
I assume $L^p$ norm means the operator norm on $\mathbb{R}^N$ with the $\ell\_p$ norm.
Then by Riesz-Thorin-Stein interpolation, you have
$$ \|H\|\_p \leq \|H\|\_1^{1/p} \|H\|\_\infty^{1-1/p} = \sqrt[p]{N} $$
By testing on the vector $(1,0,0,\ldots,0) \mapsto (-1,1,-1,1,\ldots)$ you have
$$ \|H\|\_p \geq \sqrt[p]{N} $$
and hence $\sqrt[p]{N}$ is the value.
|
6
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https://mathoverflow.net/users/3948
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404034
| 165,740 |
https://mathoverflow.net/questions/404032
|
7
|
In my [earlier MO post](https://mathoverflow.net/questions/403957/total-sum-of-characters-of-the-symmetric-group-fraks-n), I proposed the double sum $\sum\_{\mu\vdash n}\sum\_{\lambda\vdash n}\chi\_{\mu}^{\lambda}$ regarding characters of the symmetric group $\mathfrak{S}\_n$. Soon after, I started considering the sum of squares $\sum\_{\mu\vdash n}\sum\_{\lambda\vdash n}(\chi\_{\mu}^{\lambda})^2$ hoping to gain a better formula. A further look into older MO posts [here](https://mathoverflow.net/questions/162428/the-simultaneous-conjugacy-problem-in-the-symmetric-group-s-n#comment414650_162428) and also [here](https://mathoverflow.net/questions/41337/a-general-formula-for-the-number-of-conjugacy-classes-of-mathbbs-n-times-m?rq=1) shows a Burnside-type Lemma
$$\frac{1}{n!} \sum\_{\alpha \in \frak{S}\_n} \left( \sum\_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$
After comparing the last two sums, I got prompted to ask:
>
> **QUESTION.** The numerics suggest the below equality. Why is this true?
> $$\sum\_{\mu\vdash n}\sum\_{\lambda\vdash n}(\chi\_{\mu}^{\lambda})^2
> =\frac{1}{n!} \sum\_{\alpha \in \frak{S}\_n} \left( \sum\_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$
>
>
>
|
https://mathoverflow.net/users/66131
|
Total sum of squares of characters of the symmetric group $\mathfrak{S}_n$
|
The sum $\sum\_\chi \chi(\alpha)^2$ is the size of the centralizer $z\_\mu=\frac{n!}{|K\_\mu|}=1^{m\_1}m\_1! 2^{m\_2}m\_2!\cdots n^{m\_n} m\_n!$ if $\alpha$ has cycle type $\mu=1^{m\_1}2^{m\_2}\ldots$, so
$$\frac{1}{n!} \sum\_{\alpha\in \mathfrak S\_n} \left(\sum\_\chi \chi(\alpha)^2\right)^2=
\frac{1}{n!} \sum\_{\mu\vdash n} |K\_\mu| z\_\mu^2=\sum\_{\mu\vdash n} z\_\mu
=\sum\_{\mu\vdash n}\sum\_{\lambda\vdash n} (\chi^{\lambda}\_\mu)^2,$$
as desired.
Although this identity is clear, there is something interesting here when you stand back. When you first come to representations, you'd maybe first think of $\mathfrak S\_n$ acting on itself by left multiplication, which you quickly check affords the character $\rho$ which gives $|\mathfrak S\_n|$ at the identity element and $0$ elsewhere. So, using the inner product, you immediately decompose this representation into irreducibles and find
$$\rho=\sum\_\chi \chi(1)\chi,$$
i.e. each irreducible shows up with multiplicity its dimension.
Now you might try $\mathfrak S\_n$ acting on itself by conjugation, which affords the character $\psi$ with value $z\_\mu$ at any element of cycle type $\mu$. How does this one decompose? As we saw before,
$$\langle \psi,\chi^\lambda\rangle=\sum\_{\mu\vdash n}\chi^\lambda\_\mu.$$
But using that
$$\sum\_{\lambda\vdash n} (\chi^\lambda\_\mu)^2=z\_\mu,$$
we also have the expression
$$\psi=\sum\_{\chi} \chi^2,$$
since evaluating the RHS at an element with cycle type $\mu$ indeed gives $z\_\mu$.
I should mention that the individual products $\chi^2$ are not generally well understood. For example, there is a conjecture that if $\lambda$ is a staircase shape $(n,n-1,n-2,\ldots,1)$, then the square of $\chi^\lambda$ contains at least one copy of each irreducible, i.e.
$$\langle (\chi^\lambda)^2,\chi^\nu\rangle\geq 1$$
for all $\nu\vdash \binom{n+1}{2}$.
This is a folklore conjecture due to Jan Saxl. (Of course, if one knew how to decompose arbitrary products $\chi\phi$, then one would know how to decompose the squares $\chi^2$, and if one knew how to decompose squares, then one would know how to decompose the conjugation representation.)
|
11
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https://mathoverflow.net/users/371843
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404035
| 165,741 |
https://mathoverflow.net/questions/404039
|
3
|
Suppose we have a sequence of entire functions $f\_n$ such that $$\text{$f\_n(z)\to0$ for each natural $z$}\tag{1}$$
(as $n\to\infty$).
Is it possible to give general additional conditions on the sequence $(f\_n)$ ensuring that (1) implies
$$\text{$f\_n(z)\to0$ for each complex $z$?}\tag{2}$$
As a minimum, I would like such general additional conditions to hold and be easily verifiable when
$$f\_n(z)=-1+\frac1{n^z}\sum\_{k=0}^\infty k^z\,\frac{n^k}{k!}\,e^{-n};$$
cf. [this answer](https://mathoverflow.net/a/404024/36721).
|
https://mathoverflow.net/users/36721
|
On convergence of entire functions
|
Edited.
1. For your general question,
one sufficient condition is that your functions are of exponential type $<\pi$.
This is best possible since anything convergent to $\sin\pi z$ would be a counterexample, and $\sin\pi z$ has exponential type exactly $\pi$.
One can slightly improve this. For example when all $L^2(R)$ norms of your functions are bounded, independently of $n$. Then one can apply the Sampling theorem (which is usually credited to Nyquist in the West and Kotelnikov in the former Soviet Union, but in fact goes back to Cauchy), which estimates the
$L^2(R)$ norm in terms of $\ell^2(Z)$ norm, provided that the exponential
type is $\leq\pi$.
2. However your specific function does not have finite exponential type,
so these general results will not work, and you need to use some specific properties of your function. You say that yout functions satisfy $f\_n(z)\to 0$ for every integer $z$. Why is this so, and why your proof does not work for non-integer $z$?
|
3
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https://mathoverflow.net/users/25510
|
404043
| 165,744 |
https://mathoverflow.net/questions/373336
|
5
|
I'm sure it is well-known how many edges you must delete in a (highly linked) graph to destroy all cycles. Is it also known how many edges you must delete to destroy only all triangles? And even, how many you need if additionally, you may not use an edge to destroy more than one triangle? (Note the latter doesn't mean you can't touch a common edge, but after deleting this edge only one triangle "counts" as destroyed. Thus you need 4 edges for the graph $K\_4$, and $K\_6$ has no solution.)
A reference would be welcome, especially if the problem would be in NP (but a non-NP lower bound around $O(n^2)$ would also be fine).
|
https://mathoverflow.net/users/11504
|
Deleting triangles in a graph
|
Given a graph $G$, the problem of determining the minimum size $\tau(G)$ of a set of edges $X$ such that $G-X$ is triangle-free is indeed NP-complete. This was proved by [Yannakakis](https://epubs.siam.org/doi/abs/10.1137/0210021?journalCode=smjcat). Regarding bounds for $\tau(G)$, we can consider the following dual problem. Let $\nu(G)$ be the maximum number of edge-disjoint triangles of $G$. Clearly $\nu(G) \leq \tau(G) \leq 3\nu(G)$. A famous conjecture of Tuza asserts:
>
> $\tau(G) \leq 2\nu(G)$, for all graphs $G$.
>
>
>
The complete graph $K\_4$ shows that this bound is best possible. Tuza's conjecture is still open, but it has been shown to hold for many restricted graph classes. For example it holds for threshold graphs by this recent [paper](https://arxiv.org/abs/2105.09871) of Bonamy, Bożyk, Grzesik, Hatzel, Masařík, Novotná, and Okrasa. The best general bound is by [Haxell](https://www.sciencedirect.com/science/article/pii/S0012365X98001836), who proved that $\tau(G) \leq \frac{66}{23} \nu(G)$ for all graphs $G$.
|
2
|
https://mathoverflow.net/users/2233
|
404056
| 165,747 |
https://mathoverflow.net/questions/382050
|
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$.
Define the *abundancy index*
$$I(x)=\frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the classical *sum of divisors* of $x$.
Since $q$ is prime, we have the bounds
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1},$$
which implies, since $N$ is perfect, that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
We now prove the following claim:
>
> **CLAIM:** $$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg)$$
>
>
>
**PROOF:** We know that
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k}=\gcd(n^2,\sigma(n^2)),$$
(since $\gcd(q^k,\sigma(q^k))=1$).
However, we have
$$\sigma(q^k) - q^k = 1 + q + \ldots + q^{k-1} = \frac{q^k - 1}{q - 1},$$
so that we obtain
$$\frac{\sigma(n^2)}{q^k}=\frac{\bigg(q - 1\bigg)\bigg(2n^2 - \sigma(n^2)\bigg)}{q^k - 1}=\sigma(n^2) - \bigg(q - 1\bigg)\bigg(2n^2 - \sigma(n^2)\bigg)$$
$$=q\sigma(n^2) - 2(q - 1)n^2.$$
Dividing both sides by $qn^2$, we get
$$I(n^2) - \frac{2(q-1)}{q} = \frac{I(n^2)}{q^{k+1}} > \frac{1}{q^{k+1}}\cdot\frac{2(q-1)}{q},$$
which implies that
$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg).$$
**QED.**
To illustrate the improved bound:
**(1)** Unconditionally, we have
$$I(n^2) > \frac{2(q-1)}{q} \geq \frac{8}{5} = 1.6.$$
**(2)** Under the assumption that $k=1$:
$$I(n^2) > 2\bigg(1 - \frac{1}{q}\bigg)\bigg(1 + \left(\frac{1}{q}\right)^2\bigg) \geq \frac{208}{125} = 1.664.$$
**(3)** However, it is known that under the assumption $k=1$, we actually have
$$I(q^k) = 1 + \frac{1}{q} \leq \frac{6}{5} \implies I(n^2) = \frac{2}{I(q^k)} \geq \frac{5}{3} = 1.\overline{666}.$$
Here are my questions:
>
>
> >
> > **(A)** Is it possible to improve further on the unconditional lower bound for $I(n^2)$?
> >
> >
> >
>
>
>
>
>
> >
> > **(B)** If the answer to Question **(A)** is **YES**, my next question is "*How*"?
> >
> >
> >
>
>
>
I did notice that
$$\frac{2(q-1)}{q}+\frac{2}{q(q+1)}=I(n^2)=\frac{2q}{q+1}$$
when $k=1$.
|
https://mathoverflow.net/users/10365
|
Improving the lower bound $I(n^2) > \frac{2(q-1)}{q}$ when $q^k n^2$ is an odd perfect number
|
Here is a way to come up with an improved lower bound for $I(n^2)$, albeit in terms of $q$ and $n$:
We write
$$I(n^2) - \frac{2(q - 1)}{q} = \frac{I(n^2)}{q^{k+1}} = \frac{\sigma(n^2)}{q^k}\cdot\frac{1}{qn^2} > \frac{1}{qn^2},$$
from which it follows that
$$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{qn^2}.$$
This improved lower bound for $I(n^2)$, which does not contain $k$, can then be used to [produce an upper bound](https://arxiv.org/abs/2108.07697) for $k$.
|
0
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https://mathoverflow.net/users/10365
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404066
| 165,748 |
https://mathoverflow.net/questions/404058
|
1
|
According to Wikipedia, a function $f: \mathbb{R}^n \to \mathbb{R} \cup \{-\infty, +\infty\}$ is called coercive if,
$$f(x) \to +\infty \text{ as } \|x\| \to +\infty$$
and it is super-coercive if
$$\lim\_{\|x\| \to \infty} f(x)/\|x\| \to +\infty$$
My question is, does the Fenchel dual $f^\star$ share the same property?
Can assume $f$ is convex, lower semicontinuous, etc. on top.
|
https://mathoverflow.net/users/114734
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Does coercivity/supercoercivity conjugates?
|
No. The conjugate of the constant zero is the indicator of zero (and vice versa). More generally, the conjugate of the indicator of a closed convex set is positively one-homgeneous (and hence, not super coercive).
|
0
|
https://mathoverflow.net/users/9652
|
404067
| 165,749 |
https://mathoverflow.net/questions/404076
|
4
|
I am looking for a proofs of the following two claims:
>
> **Claim 1.**
> $$\frac{2\pi}{\sqrt{3}}=\displaystyle\sum\_{n=1}^{\infty}\frac{(-1)^{\Omega\_1(n)}}{n}$$ where $\Omega\_1(n)$ is the number of prime factors of the form $p \equiv 1 \pmod{6}$ of $n$ .
>
>
>
The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJw9TcsKwjAQvOc_hN0abVPBS9hv0LtUaCWpS23UvC6i32566VyGeTCTBlLNAi1Osxl7BQ7pw9ToTLa_xacvhraFJlLyYdwY75B_KNlCvkxSdZsjlYSJt4tuO0TtTUzeAaP-ikAhzeBKJQ0SYKfwuh5hXbbFy7OLECq1X0VbnbkObx_hUOb-A88xLQ==&lang=gp&interacts=eJyLjgUAARUAuQ==).
>
> **Claim 2.**
> $$\frac{\sqrt{3}\pi}{2}=\displaystyle\sum\_{n=1}^{\infty}\frac{(-1)^{\Omega\_5(n)}}{n}$$ where $\Omega\_5(n)$ is the number of prime factors of the form $p \equiv 5 \pmod{6}$ of $n$ .
>
>
>
The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJxFjc0KwjAQhO99D2G3RttU6iXsM-hdKrSS1KU2av4uos9uetG5DN_AzMSBZL1IFYdZj30LFunFVKtEpr-Eu8uBMtkmkuKm7RiukD4o2EA6TUJ2qz1RK5h4vXDTISqnQ3QWGNW78OTjDDaX4yAANhLPvyOs8nbxcGwD-FJu__B0AXZYHrlqUH0BCzcxOQ==&lang=gp&interacts=eJyLjgUAARUAuQ==).
|
https://mathoverflow.net/users/88804
|
Two conjectural infinite series for $\pi$
|
You recently asked a similar [question](https://math.stackexchange.com/questions/4246647/a-conjectural-infinite-series-for-frac-pi2) for modulus $4$ on math.stackexchange. I just used the exact same technique.
---
For $i\in\{1,5\}$, define $f\_i:\mathbb{Z}\_{\ge 1}\to \mathbb{C}^\times$ by $f\_i(n)=(-1)^{\Omega\_i(n)}$, then $f\_i$ is completely multiplicative and
$$L(s,f\_i)=\left(1-2^{-s}\right)^{-1}\left(1-3^{-s}\right)^{-1}\prod\_{p\equiv i\pmod 6}\left(1+p^{-s}\right)^{-1}\prod\_{p\equiv -i\pmod 6}\left(1-p^{-s}\right)^{-1}.$$
Hence,
$$L(s,f\_1)L(s,f\_5)=\left(1-2^{-s}\right)^{-2}\left(1-3^{-s}\right)^{-2}\prod\_{p\ge 5}\left(1-p^{-2s}\right)^{-1}=\frac{2^{s}+1}{2^s-1}\cdot\frac{3^s+1}{3^s-1}\cdot \zeta(2s).$$
Therefore, if both $L(1,f\_1)$ and $L(1,f\_5)$ converge, their product is $6\zeta(2)=\pi^2$. Now, let $\chi$ be the non-trivial Dirichlet character modulo $6$, then
$$L(1,\chi)=\prod\_{ p\equiv 1\pmod 5}(1-p^{-1})^{-1}\prod\_{p\equiv 5\pmod{6}}\left(1+p^{-1}\right)^{-1}=\frac13L(1,f\_5).$$
Therefore, it suffices to show that $L(1,\chi)=\frac{\pi}{2\sqrt{3}}.$
Let $\chi\_2$ be the non-trivial Dirichlet character modulo $3$, then $\chi(p)=\chi\_2(p)$ for all primes $p\neq 2$ and
$$L(1,\chi)=\frac{L(1,\chi\_2)}{(1+2^{-1})^{-1}}=\frac32 \cdot \frac{\pi}{3\sqrt 3}=\frac{\pi}{2\sqrt{3}}$$
Where we use [this](https://math.stackexchange.com/questions/906424/analogue-of-zeta2-frac-pi26-for-dirichlet-l-series-of-mathbbz-3/906443#906443) answer.
|
10
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https://mathoverflow.net/users/165478
|
404082
| 165,751 |
https://mathoverflow.net/questions/403980
|
5
|
It is stated by Douglas Bridges in [Constructive mathematics: a foundation for computable analysis](https://doi.org/10.1016/S0304-3975(98)00285-0) that the following property, which I will call the zero product property:
If $x,y \in \mathbb{R}$ and $xy = 0$, then $x = 0$ or $y = 0$.
is equivalent to the Lesser Limited Principle of Omniscience (LLPO):
For each binary sequence $(a\_n)$ with at most one term equal to 1, either $a\_{2n} = 0$ for all $n$ or else $a\_{2n+1} = 0$ for all $n$.
The LLOP is non-algorithmic and a trivial consequence of the Law of Excluded Middle (LEM); as such it is often rejected by constructive mathematicians. In light of this revelation how do construcivists operate, in general situations, without use of the zero product property?
Edit: The following example is referenced in some of the answers below but has been removed from the original post because it is receiving more attention than the fundamental question.
$$x^2-4 = 0 \implies (x-2)(x+2) = 0 \implies x-2 = 0 \ \text{or} \ x+2 = 0 \implies x = 2 \ \text{or} \ x = -2.$$
I realize the example given is fairly easy to reconcile without the zero product property. I'm looking for answers that give more general techniques employed to operate with out the zero product property. I've gathered that typically one must add extra assumptions (e.g. apartness of x and y.)
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https://mathoverflow.net/users/312621
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How do working constructivists get by with out the zero product property?
|
There are a couple of very similar statements that all work out:
1. if $x$ and $y$ are both apart from $0$, then $xy$ is apart from $0$.
2. If $x$ is apart from $y$ and $xy = 0$, then $x = 0$ or $y = 0$ (as mentioned by Andreas Blass in the comments).
3. $xy = 0$ iff $\inf \{|x|,|y|\} = 0$.
A more round-about way which works at least in computable analysis even allows to make case distinctions here. (Classical metatheory from here on). We can use the three-valued space $\mathbb{T}$ with underlying set $\{\bot,0,1\}$ and topology generated by $\{\{0\},\{1\}\}$. Given $x,y \in \mathbb{R}$ with $xy = 0$, we can compute $c(x,y) \in \mathbb{T}$ where $c(x,y) = 0$ if $y \neq 0$, $c(x,y) = 1$ if $x \neq 0$ and $c(0,0) = \bot$.
For any reasonable[1] space $\mathbf{X}$, the map $\operatorname{Merge}\_\mathbf{X} : \subseteq \mathbb{T} \times \mathbf{X} \times \mathbf{X} \to \mathbf{X}$ defined as $\operatorname{Merge}\_\mathbf{X}(0,x,y) = x$, $\operatorname{Merge}\_\mathbf{X}(1,x,y) = y$ and $\operatorname{Merge}\_\mathbf{X}(\bot,x,x) = x$ will be computable. Note that $\operatorname{Merge}\_\mathbf{X}(\bot,x,y)$ is undefined if $x \neq y$.
Thus, we are allowed to make a case distinction according to the zero product property in computable analysis, as long as we end up with the same result no matter which case we pick if they are both true (and as long as the result lives in a reasonable[1] space). This is not really specific to the zero product property, but more a general case of being able to get away with something that may at first glance look like using $\mathrm{LLPO}$.
[1] Reasonable spaces include the computably admissible ones, as well as any other example I'm aware of anyone would want to be working with. It doesn't work for everything though.
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3
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https://mathoverflow.net/users/15002
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404116
| 165,758 |
https://mathoverflow.net/questions/404119
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0
|
For whatever reason, I have always defined matrices as being $n \times m$, and that is how I have been defining matrices throughout my dissertation. Recently however, I have noticed that nearly every other source primarily defines matrices as being $m \times n$. Is the later more formal notation? Should I go through my whole dissertation to change the notation? How important is it?
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https://mathoverflow.net/users/152336
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Is it improper to define matrices as being $n \times m$ rather than $m \times n$?
|
It is generally a good idea to keep things alphabetical, unless there is a good reason to do otherwise.
With matrices, it can be okay because an $n\times m$ matrix is really a linear transformation from $\mathbb{F}^m\to \mathbb{F}^n$ (where $\mathbb{F}$ is your field of definition), and so viewed this way, the $m$-dimensional space comes before the $n$-dimensional space.
So I think you are okay sticking with what you have.
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5
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https://mathoverflow.net/users/3199
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404120
| 165,760 |
https://mathoverflow.net/questions/401746
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16
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Many papers refer to an untitled manuscript of Jon Beck (Cornell, 1966) for the origin of the monadicity theorem (originally called a "tripleability theorem"). An early proof is in Manes's 1967 thesis *A Triple Miscellany: Some Aspects of the Theory of Algebras over a Triple* (Theorem 1.2.9). Manes cites Beck's 1967 thesis *Triples, Algebras, and Cohomology* as a reference, but the monadicity theorem does not actually appear there.
Where can one find a copy (preferably digitised) of the untitled manuscript of Beck containing the monadicity theorem? (Considering that the manuscript is cited, presumably a copy exists and was circulated, rather than passed on by word of mouth.)
Evidence for the existence of the manuscript is given by an email of Marta Bunge on the categories mailing list (dated 4th November 2007):
>
> There is an unpublished (untitled and undated) four-pages manuscript which
> John Beck gave to me (and I supposed also to many ohers) when he was at
> McGill. In it, he states and proves two theorems, the CTT (crude
> tripleableness theorem), and the PTT (precise tripleableness theorem). There
> is a connection between triples and descent implicit in the PTT. But this is
> not the same connection with descent as the Benabou-Roubaud theorem.
>
>
>
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https://mathoverflow.net/users/152679
|
Jon Beck's untitled manuscript containing the "tripleability theorem" (i.e. the monadicity theorem)
|
After reaching out to every researcher who cited the manuscript, John Kennison was kind enough to find and scan his copy of the untitled manuscript containing the crude and precise monadicity theorems. I have uploaded it to the nLab for posterity: [Jon Beck's untitled manuscript](https://ncatlab.org/nlab/files/Untitled+manuscript.pdf). This copy was distributed at the *Conference Held at the Seattle Research Center of the Battelle Memorial Institute* in June – July 1968, though evidence from citations suggests it was first distributed as early as 1966.
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23
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https://mathoverflow.net/users/152679
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404123
| 165,763 |
https://mathoverflow.net/questions/404094
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12
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Are the fibers of a surjective holomorphic submersion $\mathbb{C}^n\to\mathbb{C}$ all homeomorphic?
For $n=1$ this means that a surjective entire function $\mathbb{C}\to\mathbb{C}$ without critical points assumes each value infinitely often. Is this obvious?
|
https://mathoverflow.net/users/374739
|
Are the fibers of a surjective holomorphic submersion $\mathbb{C}^n\to\mathbb{C}$ all homeomorphic?
|
The answer is no for $n=1$.
**Lemma**: Suppose $f$ is an entire function with $f^{-1}(z\_0)$ finite non-empty for some $z\_0 \in \mathbb{C}$. Then $f$ is surjective.
Proof: By Picard, $f$ misses at most one value. Up to translating $f$ by a scalar (which obviously preserves the hypothesis), we may assume $f$ misses $0$. Then $f = e^g$ for some entire $g$. By assumption, $z\_0 \neq 0$, so $\exp^{-1}(z\_0)$ is infinite, so by Picard, $g^{-1}(\exp^{-1}(z\_0)) = f^{-1}(z\_0)$ is infinite, which is a contradiction.
Now take $q(z) := \sum\_{n \geq 1} \frac{z^n}{n\cdot n!}$ and $f(z) := ze^{q(z)}$. I claim this $f$ yields a contradiction.
Observe that $f(z) = 0$ if and only if $z = 0$. Therefore, the lemma applies, and we see $f$ is surjective. Moreover, we see that not all fibers of $f$ are in bijection: $f^{-1}(0)$ is a singleton, but by (great) Picard (applied to $f(\frac{1}{z})$), $f^{-1}(z)$ is infinite for any other value of $z$.
Finally, let's check that $f$ is a submersion. We have:
$$df = (1+zq')e^q dz.$$
We have $1+zq' = e^z$ by choice of $q$, so this is $e^{z+q}dz$, which is clearly a nowhere vanishing differential.
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14
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https://mathoverflow.net/users/144469
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404127
| 165,764 |
https://mathoverflow.net/questions/403088
|
8
|
I think that I might have spotted I small mistake (a missing $5$-defective Lehmer pair) in the classification of terms of Lehmer sequences without primitive divisors given in:
*[1](https://www.google.com/search?q=Bilu%2C%20Hanrot%2C%20and%20Voutier%2C%20Existence%20of%20primitive%20divisors%20of%20Lucas%20and%20Lehmer%20numbers) Bilu, Hanrot, and Voutier, Existence of primitive divisors of Lucas and Lehmer numbers, Journal für die reine und angewandte Mathematik (Crelles Journal), 2001.* ([available on the web](https://www.google.com/search?q=Bilu%2C%20Hanrot%2C%20and%20Voutier%2C%20Existence%20of%20primitive%20divisors%20of%20Lucas%20and%20Lehmer%20numbers))
Since I might very much be wrong myself, I would like to know if my reasoning is correct or not. I recall below the main definitions of [1](https://www.google.com/search?q=Bilu%2C%20Hanrot%2C%20and%20Voutier%2C%20Existence%20of%20primitive%20divisors%20of%20Lucas%20and%20Lehmer%20numbers).
A *Lehmer pair* is a pair of complex numbers $(\alpha, \beta)$ such that $(\alpha + \beta)^2$ and $\alpha\beta$ are non-zero coprime integers and $\alpha / \beta$ is not a root of unity. Two Lehmer pairs $(\alpha\_1, \beta\_1)$ and $(\alpha\_2, \beta\_2)$ are said to be *equivalent* if $\alpha\_1 / \alpha\_2 = \beta\_1 / \beta\_2 \in \{-1,+1,\sqrt{-1},-\sqrt{-1}\}$. Given a Lehmer pair $(\alpha, \beta)$, the associated Lehmer sequence is
$$\widetilde{u}\_n(\alpha, \beta) := \begin{cases} (\alpha^n - \beta^n) / (\alpha - \beta) & \text{ if $n$ is odd} \\
(\alpha^n - \beta^n) / (\alpha^2 - \beta^2) & \text{ if $n$ is even}\end{cases}$$
for every positive integer $n$ (it is an integer sequence).
A prime number $p$ is a *primitive divisor* of $\widetilde{u}\_n(\alpha, \beta)$ if $p$ divides $\widetilde{u}\_n(\alpha, \beta)$ but does not divide $(\alpha^2 - \beta^2)^2 \widetilde{u}\_1(\alpha, \beta) \cdots \widetilde{u}\_{n-1}(\alpha, \beta)$. If $\widetilde{u}\_n(\alpha, \beta)$ has no primitive divisor then the Lehmer pair $(\alpha, \beta)$ is *$n$-defective.*
One of the claims of Theorem 1.3 of [1](https://www.google.com/search?q=Bilu%2C%20Hanrot%2C%20and%20Voutier%2C%20Existence%20of%20primitive%20divisors%20of%20Lucas%20and%20Lehmer%20numbers) is that, up to equivalence, all $5$-defective Lehmer pairs are of the form $((\sqrt{a} - \sqrt{b})/2, (\sqrt{a} + \sqrt{b})/2)$ with
$$(1) \qquad (a, b) = (\phi\_{k-2\varepsilon}, \phi\_{k-2\varepsilon} - 4\phi\_k) \quad (k \geq 3)$$
or
$$(2) \qquad (a, b) = (\psi\_{k-2\varepsilon}, \psi\_{k-2\varepsilon} - 4\psi\_k) \quad (k \neq 1) ,$$
where $k$ is a nonnegative integer, $\varepsilon \in \{-1, +1\}$, $(\phi\_n)$ is the sequence of Fibonacci numbers, and $(\psi\_n)$ is the sequence of Lucas numbers.
**Claim 1:** *The Lehmer pair $(\alpha\_0, \beta\_0) := ((1 - \sqrt{5}) / 2, (1 + \sqrt{5}) / 2)$ is $5$-defective.*
First, note that $(\alpha\_0 + \beta\_0)^2 = 1$ and $\alpha\_0\beta\_0 = -1$ are non-zero coprime integers and $\alpha\_0 / \beta\_0 = (\sqrt{5}-3) / 2$ is not a root of unity, so that $(\alpha\_0, \beta\_0)$ is indeed a Lehmer pair. Second, for the associated Lehmer sequence we have $\widetilde{u}\_5 = 5$ and $(\alpha\_0^2 - \beta\_0^2)^2 = 5$, thus $\widetilde{u}\_5$ has no primitive divisor and $(\alpha\_0, \beta\_0)$ is $5$-defective.
**Claim 2:** *The Lehmer pair $(\alpha\_0, \beta\_0)$ is not equivalent to a pair of the form $(\alpha, \beta) = ((\sqrt{a} - \sqrt{b})/2, (\sqrt{a} + \sqrt{b})/2)$ with $(a, b)$ as in (1) or (2).*
For the sake of contradiction suppose $(\alpha\_0, \beta\_0)$ is equivalent to a pair of the form $(\alpha, \beta) = ((\sqrt{a} - \sqrt{b})/2, (\sqrt{a} + \sqrt{b})/2)$ with $(a, b)$ as in (1) or (2). Then $(a - b) / 4 = \alpha\beta = \pm \alpha\_0 \beta\_0 = \pm 1$ so that $a - b = \pm 4$. In case (1), we have $a - b = 4\phi\_k \geq 8$, because $k \geq 3$. In case (2), we have $a - b = 4\phi\_k \geq 8$, because $k \neq 1$. Absurd.
**Possible source of the error:** I think that the missing $5$-defective pair is lost in the last paragraph of case $n = 5$ in section "Small $n$" of [1](https://www.google.com/search?q=Bilu%2C%20Hanrot%2C%20and%20Voutier%2C%20Existence%20of%20primitive%20divisors%20of%20Lucas%20and%20Lehmer%20numbers). It is said that:
*"By (28), we have $k \geq 3$ in the case (34), and $k \neq 1$ in the case (35)."*
But, in case (35), $k = 1$ (and $\varepsilon = 1$) are not in contradiction with (28).
In other words, (2) should allow $k = 1$ (and $\varepsilon = 1$).
This would lead to the $5$-defective pair $((\sqrt{-1} + \sqrt{-5}) / 2, (\sqrt{-1} - \sqrt{-5}) / 2)$, which is equivalent to $(\alpha\_0, \beta\_0)$.
Thank in advance to anyone who takes the time to check.
|
https://mathoverflow.net/users/357523
|
Possible small mistake in Bilu-Hanrot-Voutier paper on primitive divisors of Lehmer sequences (?)
|
Yes, there are some omissions in the lists in the original BHV article. I think all of them were fixed by Mourad Abouzaid
Mourad Abouzaid, Les nombres de Lucas et Lehmer sans diviseur primitif, J. Théor. Nombres Bordeaux 18 (2006), no. 2, 299–313.
|
9
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https://mathoverflow.net/users/138069
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404133
| 165,765 |
https://mathoverflow.net/questions/404131
|
7
|
$\DeclareMathOperator\lcm{lcm}$Let $p\_k$ be the $k$th prime number. Set $$L(n) = \lcm(p\_1-1, p\_2-1, \dotsc, p\_n-1). $$
What can we say about the growth of $L(n)$? Trivially, one has that $L(n) < p\_1p\_2 \dotsb p\_n$.
One can do better than that since after $2$, every prime is odd. Thus, if $n \geq 3$, one has $$L(n) \leq \frac{p\_1p\_2 \cdots p\_n}{2^{n-1}}.$$ And one can continue with larger primes, using explicit versions of Dirichlet's theorem on arithmetic progressions, to put in powers of 3, and then 5, and so on in the denominator in the same way.
In the other direction, one can use Linnik's theorem to get lower bounds on $L(n)$ but these are weak.
$L(n)$ is [A058254](https://oeis.org/A058254) in the OEIS, but that doesn't give any non-trivial bounds, even heuristically.
Question: Can we get an asymptotic to $\log L(n)$, ideally with explicit bounds? Closely related: is it true that $\log L(n) = o(p\_n)$? This is a natural bound to ask about because $\log \lcm (p\_1, p\_2 \cdots p\_n ) = \sum\_{p \leq p\_n} \log p \sim p\_n$ by the Prime Number Theorem.
For the application I'm interested in, I'd like to then use this to get explicit bounds on $\lcm (p\_{n}-1, p\_{n+1}-1, p\_{n+2}-1, \dotsc, p\_{n+m}-1)$ where $m$ is at least about $n^2$, and I would like that $\log \lcm (p\_{n}-1, p\_{n+1}-1, p\_{n+2}-1, \dotsc, p\_{n+m}-1)$ grows slower than $\sum\_{p\_n \leq p \leq p\_{n+m}} \log p \sim p\_{n+m}$. So it may be possible to get a useful bound even without an asymptotic for $L(n)$.
|
https://mathoverflow.net/users/127690
|
Asymptotics of $\operatorname{lcm} ((2-1), (3-1), (5-1), (7-1), (11-1), \dotsc, p_n-1 )$
|
It is true that $\ln L(n)=o(p\_n)$ for $n\to+\infty$. To prove it, let us get a bound for contribution of large primes into $\ln L(n)$ and then estimate the contribution of the rest trivially. Choose a parameter $R<\sqrt p\_n$. Let $M(n)=\mathrm{lcm}[1,2,\ldots,p\_n]$. Then we have $\ln M(n)=\psi(p\_n)\sim p\_n$, where $\psi$ is Chebyshev's function. Now,
$$
\sum\_{p^k\mid L(n), p<p\_n/R}\ln p\leq \sum\_{p^k\mid M(n),p<p\_n/R}\ln p=O(p\_n/R).
$$
Next, if $p\geq p\_n/R$ contributes to $\ln L(n)$, then there is a natural number $k\leq R$ and a prime $q\leq p\_n$ such that $q-1=kp$. We will evaluate the terms that come from different $k$. As $q\leq p\_n$, for a given $k$ we should have $p<p\_n/k$. The number of primes $p<x$ such that $kp+1$ is prime is known to be bounded by
$$
C\frac{k}{\varphi(k)}\frac{x}{\ln^2 x}
$$
(see, for example, H. Iwaniec, "Sieve methods", Section 2.6). Applying this to our case, we get the bound of the form
$$
O\left(\frac{1}{\varphi(k)}\frac{p\_n}{\ln^2 p\_n}\right)
$$
for the number of primes we are interested in, and their contribution is, therefore,
$$
O\left(\frac{1}{\varphi(k)}\frac{p\_n}{\ln p\_n}\right)
$$
for any $k\leq R$. Summing over all $k$ and using the fact that
$$
\sum\_{k\leq R}\frac{1}{\varphi(k)}=C\_1\ln R+O(1),
$$
we obtain
$$
\sum\_{p^k\mid L(n), p\geq p\_n/R}\ln p=O\left(\frac{p\_n\ln R}{\ln p\_n}\right).
$$
Combining the first and the second estimate and choosing $R=\ln p\_n$, we get
$$
\ln L(n)=O\left(\frac{p\_n\ln\ln p\_n}{\ln p\_n}\right).
$$
The results used in the proof do not rely on anything particularly non-explicit, so the constants can be made completely explicit. Also, the bound for the number of primes of the form $kp+1$ is conjectured to be sharp, so the bound should probably also be asymptotically sharp, unless I am missing something.
|
11
|
https://mathoverflow.net/users/101078
|
404135
| 165,766 |
https://mathoverflow.net/questions/404130
|
7
|
Giuga's conjecture (1950), which is still open and has strong numerical support, reads :
>
> Let $n$ be a positive integer. If $1+\sum\_{k=1}^{n-1}k^{n-1} \equiv 0\pmod{n}$ then $n$ is prime.
>
>
>
What would an analog for function fields be?
|
https://mathoverflow.net/users/469
|
What is a function field analog of Giuga's conjecture?
|
The integers 1 to $n-1$ are the non-zero elements of $\mathbb Z/n\mathbb Z$. So one could take an ideal $I$ in a Dedekind domain $R$ and ask whether
$$
(\*)\qquad
1+\sum\_{\substack{a\in R/I\\ a\ne0\\}} a^{\#R/I - 1} \equiv 0 \pmod{I}
$$
is equivalent to $I$ being a prime ideal. I have no idea whether this is a reasonable question, since I have not done any experiements (I'll let you try), but this or something similar seems like a natural generalization.
**Addendum:** I did some experiments with $R=\mathbb F\_p[x]$ and $I=(f(x))$. For the cases
$$
p=2~\text{and}~\deg(f)\le8,\qquad
p=3~\text{and}~\deg(f)\le4,\qquad
p=5~\text{and}~\deg(f)\le3,
$$
it is true that
$$
\text{$(\*)$ is true}\quad\Longleftrightarrow\quad
\text{$f(x)$ is irreducible in $\mathbb F\_p[x]$.}
$$
|
10
|
https://mathoverflow.net/users/11926
|
404141
| 165,768 |
https://mathoverflow.net/questions/404125
|
2
|
Suppose we have a random walk $S\_n$ with i.i.d. steps $X\_i$. We assume that
$$\mathbb{E}[X\_i] = -\mu, \text{Var}[X\_i] = 1,$$
where $\mu$ is close (or going) to zero. We also assume that the moment generating function and its derivatives $M\_{X\_i}, M\_{X\_i}', M\_{X\_i}'', M\_{X\_i}'''$ are all bounded in $(-\epsilon, \epsilon)$ by a constant $C$, where $\epsilon, C$ are independent of $\mu$ (which is going to zero).
Fix a constant $a\geq 1$, are there any estimates in the literature for the probability of the random walk always stays below $a$, i.e.
$$\mathbb{P}\big\{{\max\_{n\geq 0} S\_n \leq a}\big\}?$$
(I believe the upper bound should be $Ca\mu$.)
For the special case where we replace $a$ by $0$, then
$$\mathbb{P}\big\{{\max\_{n\geq 0} S\_n \leq 0}\big\} \leq C\mu$$
which essentially follows from Sparre-Andersen theorem together with Berry-Esseen bound.
|
https://mathoverflow.net/users/49551
|
Random walk always stays below a level $a$
|
Let
$$p(a):=P\big(\max\_{n\ge0} S\_n\le a\big).$$
Assume that $c\_3:=E|X\_1-EX\_1|^3<\infty$.
By the improvement by [Sakhanenko](https://epubs.siam.org/doi/pdf/10.1137/1119047) of Lemma 8 by [S. Nagaev](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=tvp&paperid=1854&option_lang=eng) (the improvement consisting in removing an extra factor $c\_3$) and trivial time-rescaling,
$$p(a)-p(0)\le C\mu(a+c\_3)$$
for $\mu\ge0$ and $a\ge0$; everywhere here, $C$ denotes various universal positive real constants. Also, you know that $p(0)\le C\mu$. So, for all $a\ge c\_3$
$$p(a)\le C\mu a.\tag{1}$$
---
Concerning a lower bound on $p(a)$, Corollary 1 to Theorem 16 of $\S$23 of the book by [Borovkov](https://link.springer.com/book/10.1007/978-1-4612-9866-3#:%7E:text=The%20object%20of%20queueing%20theory,on%20the%20direction%20of%20investigation.) implies
$$p(a)\ge1-e^{-Qa} \tag{2}$$
for $a\ge0$, where $Q:=\sup\{t\colon M(t)\le1\}$, $M(t):=Ee^{tX}$, and $X:=X\_1$.
If e.g. $|X|\le b$ almost surely for some real $b>0$, then for all $t\in[0,Q]$ we have $M''(t)=EX^2e^{tX}\le b^2 M(t)\le b^2$, by the convexity of $M$. So, $1=M(Q)\le M(0)+M'(0)Q+b^2Q^2/2=1-\mu Q+b^2Q^2/2$, whence $Q\ge c\mu$, where $c:=\frac2{b^2}$. If now $Qa$ is large, then (2) implies $p(a)\approx1$, so that the lower bound in (2) is as good as it can be. If, finally, $Qa$ is not large, then the lower bound in (2) is $\asymp Qa\ge c\mu a$, which matches (1).
|
3
|
https://mathoverflow.net/users/36721
|
404145
| 165,769 |
https://mathoverflow.net/questions/404097
|
1
|
For a family of probability measures sharing the same form of distribution function $F(x; p)$ with different parameters (i.e., $p$'s), if the parameter falls in a compact subset of real line, can we say these probability measures constitute a compact subset in Wasserstein space?
For example, considering exponential distributions $\text{Exp}(\lambda)$ and $\lambda \in [a, b] \subset \mathbb{R}$, intuitively, it seems these exponential distributions should be in a compact subset in the sense of Wasserstein metric. If so, I am just wondering whether there is any rigorous proof for it.
|
https://mathoverflow.net/users/374718
|
Does the compactness of parameter of distribution function imply the compactness of the distribution (or probability measure) in Wasserstein space?
|
Since the continuous image of a compact set is compact, it suffices to determine whether the mapping $p \to F(x,p)$ is continuous. This is the case for most natural parametrized families of distributions, but needs to be verified in each case. For instance, for the exponential family, the transformation $x \mapsto(1+\epsilon) x$ takes the Exp$(\lambda)$ distribution $\mu$ (which has mean $1/\lambda$) to the Exp$(\frac{\lambda}{1+\epsilon})$ distribution $\nu$. The Wasserstein distance can be bounded using this coupling:
$$W\_1(\mu,\nu) \le \int |\epsilon x| \, d\mu(x)=|\epsilon/\lambda| \,.$$
This suffices to establish continuity of the mapping $\lambda \mapsto$ Exp$(\lambda)$.
|
1
|
https://mathoverflow.net/users/7691
|
404153
| 165,770 |
https://mathoverflow.net/questions/402482
|
4
|
Let $k$ be a field of characteristic zero and $A$ be a graded commutative dg-algebra over $k$ with differential of degree $+1$ satisfying $H^0(A)=k, H^i(A)=0$ for $i<0$. Denote by $\mathcal J$ a dg ideal of generated by $A^k, k<0$ ($A^k$ denotes elements of degree $k$). Is it true that $\mathcal J$ is acyclic? In other words, is the natural projection $f\_\mathcal J\colon A\to A/\mathcal J$ quasi-isomorphism?
If there is a quasi-isomorphism $A\to B$ with $B$ concentrated in non-negative degrees, then the map $f\_\mathcal J$ should be quasi-isomorphism.
If you know some general facts about the cohomology of $\mathcal J$ or when this statement is true it also would be very interesting!
It is easy to construct a counterexample when we do not pose condition $H^0(A)=k$. Let $A$ be generated by $x,y,h$ with $\deg x=-1, \deg y=1$ and $\deg h=0$ (so $x^2=y^2=0$). The differential is defined by the formula $d(x)=xyh, d(h)=d(y)=0.$ Then the element $xy\in\mathcal J$ closed but not exact. But in this case $H^0(A)$ is very big.
|
https://mathoverflow.net/users/nan
|
$A$ is a commutative connective dg-algebra satisfying $H^0(A)=k$. Is it true that a dg ideal generated by elements of negative degree acyclic?
|
It is not necessarily true. I can provide a counterexample but don't know any general statement about when this is true.
Consider $$A=k[x,dx,y]/\langle x^2y, ydx\rangle$$ with $x$ in degree $-2$ and $y$ in degree $3$.
The obvious map from $k[x,dx]$ to $A$ is a dg-algebra map which induces an isomorphism below degree $1$, which implies that $H^0(A)\cong k$ and $H^i(A)=0$ for $i<0$.
But the dg ideal $\mathcal{J}$ consists of $A^{<0} \oplus k[xy]$; since $[xy]$ is in degree $1$ and $\mathcal{J}$ contains nothing in degrees $0$ or $2$, we conclude that $\mathcal{J}$ cannot be acyclic.
|
2
|
https://mathoverflow.net/users/3075
|
404160
| 165,772 |
https://mathoverflow.net/questions/404108
|
3
|
Let $p$ and $q$ be integers.
Let $f(n)$ be [A007814](https://oeis.org/A007814), the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Then we have an integer sequence given by
\begin{align}
a(0)=a(1)&=1\\
a(2n)& = pa(n)+qa(2n-2^{f(n)})\\
a(2n+1) &= a(n-2^{f(n)})
\end{align}
I conjecture that $a(\frac{4^n-1}{3})$ is always a cube of an integer
for any $p,q \in \mathbb{Z}$, $n\in \left\lbrace0, \mathbb{N}\right\rbrace$.
Is there a way to prove it?
|
https://mathoverflow.net/users/231922
|
Subsequence of the cubes
|
Experimenting with a CAS suggests an induction. In order to handle the induction, we need to consider the forms of the numbers involved. $\frac{4^m-1}{3} = 1 + 2^2 + 2^4 + \cdots + 2^{2m-2}$ alternates $1$ and $0$ bits. The map $2n + 1 \to n - 2^{f(n)}$ drops the rightmost bit (which is $1$) and clears the next least significant bit. The map $2n \to n$ drops the rightmost bit (which is $0$). And the map $2n \to 2n - 2^{f(n)}$ moves the least significant bit right one place. So the numbers which occur in the evaluation tree of $\frac{4^m-1}{3}$ are of the form $2^i + 2^j + 2^{j+2} + \cdots + 2^{j+2k}$ where $i \le j - 2$ and $k \ge 0$; and (when we get down to one bit) the form $2^i$.
Starting with the simplest case, when $i > 0$, $a(2^i) = pa(2^{i-1}) + qa(2^{i-1})$ so by induction $a(2^i) = (p+q)^i$.
For the more general case, let $A(i,j,k) = a(2^i + 2^j + 2^{j+2} + \cdots + 2^{j+2k})$ subject to the aforementioned constraints on $i,j,k$.
For $i > 0$, we have an even argument and use the second case of the recursion:
\begin{eqnarray\*}
A(i,j,k) &=& a(2^i + 2^j + 2^{j+2} + \cdots + 2^{j+2k}) \\
&=& pa(2^{i-1} + 2^{j-1} + 2^{j+1} + \cdots + 2^{j+2k-1}) + qa(2^{i-1} + 2^j + 2^{j+2} + \cdots + 2^{j+2k}) \\
&=& pA(i-1, j-1, k) + qA(i-1, j, k)
\end{eqnarray\*}
Then it's a trivial proof by induction that $$A(i,j,k) = \sum\_{u=0}^i \binom{i}{u} p^u q^{i-u} A(0, j-u, k)$$
For $i = 0$, we have an odd argument and use the third case of the recursion:
\begin{eqnarray\*}
A(0,j,k) &=& a(1 + 2^j + 2^{j+2} + \cdots + 2^{j+2k}) \\
&=& a(2^{j+1} + \cdots + 2^{j+1+2(k-1)}) \\
&=& \begin{cases}
a(0) & \textrm{if } k=0 \\
a(2^{j+1}) & \textrm{if } k=1 \\
A(j+1, j+3, k-2) & \textrm{otherwise}
\end{cases} \\
&=& \begin{cases}
1 & \textrm{if } k=0 \\
(p+q)^{j+1} & \textrm{if } k=1 \\
\sum\_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} A(0, j+3-u, k-2) & \textrm{otherwise}
\end{cases}
\end{eqnarray\*}
Further CAS experimentation suggests that the theorem we need to prove is $$A(0, j, 2v-1) = A(0, j, 2v) = \left(p\frac{q^v-1}{q-1} + q^v\right)^{j+1} A(0, 2, 2v-2)$$
The first of those equalities is easy: $$A(0,j,2) = \sum\_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} A(0, j+3-u, 0) = \sum\_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} = (p+q)^{j+1}$$ so $A(0,j,1) = A(0,j,2)$ and since the only occurrence of $k$ in the third case is in the parameter $k-2$ the rest follows by induction on $v$.
The second equality is the interesting one. Again, by induction on $v$:
\begin{eqnarray\*}
A(0,j,2v) &=& \sum\_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} A(0, j+3-u, 2v-2) \\
&=& \sum\_{u=0}^{j+1} \binom{j+1}{u} p^u q^{j+1-u} \left(p\frac{q^{v-1}-1}{q-1} + q^{v-1}\right)^{j+4-u} A(0, 2, 2v-4) \\
&=& \left[ \sum\_{u=0}^{j+1} \binom{j+1}{u} p^u \left(pq\frac{q^{v-1}-1}{q-1} + q^v\right)^{j+1-u} \right] \color{Blue}{\left(p\frac{q^{v-1}-1}{q-1} + q^{v-1}\right)^3 A(0, 2, 2v-4)} \\
&=& \left( p + pq\frac{q^{v-1}-1}{q-1} + q^v \right)^{j+1} \color{Blue}{A(0, 2, 2v-2)} \\
&=& \left(p\frac{q^v-1}{q-1} + q^v\right)^{j+1} A(0, 2, 2v-2)
\end{eqnarray\*}
as desired.
The answer to the original question now drops out: $a\left(\tfrac{4^m-1}{3}\right) = A(0, 2, m-2)$, but $A(0, 2, k)$ has the form of a cube times $A(0, 2, k-2)$ so by induction it's always a cube.
|
8
|
https://mathoverflow.net/users/46140
|
404182
| 165,776 |
https://mathoverflow.net/questions/404169
|
4
|
Are the fibers of a surjective polynomial submersion $\mathbb{C}^n\to\mathbb{C}$ all homeomorphic?
|
https://mathoverflow.net/users/374739
|
Are the fibers of a surjective polynomial submersion $\mathbb{C}^n\to\mathbb{C}$ all homeomorphic?
|
The following example is taken from Section 7 of Nguyen, "[A Remark on Polynomial Mappings from $\mathbb{C}^n$ to $\mathbb{C}^{n-1}$ and an Application of the Software Maple in Research](http://dx.doi.org/10.4236/am.2016.715154)": Map $\mathbb{C}^2$ to $\mathbb{C}$ by $f(z,w) = z+z^2w$.
We compute $\tfrac{\partial f}{\partial z} = 1+2zw$ and $\tfrac{\partial f}{\partial w} = z^2$; these two polynomials have no common zero, so $f$ is a submersion.
We have $f^{-1}(0) = \{ z(1+zw) = 0 \}$, which is the disjoint union of the once-punctured genus zero curve $z=0$ and the twice-punctured genus zero curve $zw=-1$.
For $\lambda \neq 0$, the equation $z^2 w + z = \lambda$ is a twice punctured genus zero curve. It can be parameterized as $z \mapsto (z, (\lambda-z)/z^2)$ for $z \neq 0$.
It looks like Nguyen's paper has other interesting examples as well.
---
Other interesting references:
Artal-Bartolo, Cassou-Nogues and Luengo-Velasco, [On polynomials whose fibers are irreducible with no critical points](https://mathscinet.ams.org/mathscinet-getitem?mr=1282228), construct submersions $\mathbb{C}^2 \to \mathbb{C}$ whose generic fibers have arbitrarily large genus. I suspect that not all the fibers are homeomorphic, but don't understand their construction, so I don't know.
If I understand the [Mathscinet review](https://mathscinet.ams.org/mathscinet-getitem?mr=1895930) of "Polynomials with general $C^2$-fibers are variables" correctly, then Abyhankar and Moh "[Embeddings of the line in the plane](https://mathscinet.ams.org/mathscinet-getitem?mr=379502)" and Suzuki "[Propriétés topologiques des polynômes de deux variables complexes, et automorphismes algébriques de l'espace $C^2$](https://mathscinet.ams.org/mathscinet-getitem?mr=338423)" show that, if $f(x,y)$ is an irreducible polynomial such that $\{ f(x,y) = 0 \}$ is isomorphic to $\mathbb{C}$, then there is an automorphism of $\mathbb{C}^2$ taking $f$ to one of the coordinate functions; this would then imply that $\{ f(x,y) = \lambda \}$ is isomorphic to $\mathbb{C}$ for all $\lambda$. I have not attempted to read the papers in question.
|
11
|
https://mathoverflow.net/users/297
|
404187
| 165,777 |
https://mathoverflow.net/questions/404183
|
2
|
Consider the following process on $\mathbb{C}$:
* Start at the point 1.
* At each step, move by adding $e^{i\theta}$, where $\theta$ is uniformly drawn from $\mathbb{S}^1$.
* Stop at the first positive time $T$ where you move inside the *open* unit disc.
What is the distribution of the stopping times?
*Some notes*:
* $T = 1$ happens 1/3 of the time, by simple geometry.
* $T = 2$ happens 1/9 of the time, by evaluating an explicit integral. (Let $g(T,x)$ be the probability that starting at $x$ you will hit the unit disc with exactly stopping time $T$, you can write an integral relating $g(T+1,x)$ to $g(T,x+z)$ with $z\in \mathbb{S}^1$. When $T = 1$ the function is explicitly known by simple geometry. When $T = 2$ and $|x| = 1$ the value can be obtained by direct integration, but for $|x| > 1$ I don't know how to evaluate the integral involved. And hence I also cannot get the exact value for $T = 3$ at $x = 1$. [Later this weekend I may try to evaluate these chain of integrals numerically and provide some more numeric data.])
* Numerical simulations of the random walk seems to suggest that $P(T)$ follows a power rule $T^\alpha$ where $\alpha \approx \log\_2(0.4)$. But I only checked up to around $T = 1000$, as the numerical simulation is not very efficient.
|
https://mathoverflow.net/users/3948
|
Distribution of stopping time for a 2D random walk
|
For the continuous counterpart, if a 2-D Brownian particle is started at $x$ with $|x| > 1$, the density function of the hitting time of the unit disk decays as
$$ \frac{1}{t (\log t)^2} $$
as $t \to \infty$. More precisely, it is comparable to
$$ \frac{|x|-1}{|x|} e^{-(|x|-1)^2/(2t)} \frac{(|x|+t)^{1/2}}{t^{3/2}} \frac{1 + \log |x|}{(1 + \log(1 + \tfrac{t}{|x|})) (|x| + \log t)} \, . $$
with absolute constants in both bounds. This incredibly precise estimate, as well as similar results in higher dimensions, were proved in:
* T. Byczkowski, J. Małecki, M. Ryznar, *Hitting Times of Bessel Processes*. Potential Anal 38, 753–786 (2013). <https://doi.org/10.1007/s11118-012-9296-7>
I would expect the large-time decay of $P(T)$ is thus similar to $T^{-1} (\log T)^{-2}$.
|
2
|
https://mathoverflow.net/users/108637
|
404189
| 165,778 |
https://mathoverflow.net/questions/404191
|
1
|
>
> Dirichlet problem for Laplace equation as follows
> \begin{eqnarray}
> \Delta{u}&=&0\text{ in }B\_r(0)\\
> u&=&g\text{ on }\partial B\_{r}(0),
> \end{eqnarray}
> where $ g $ is continuous.
>
>
>
It is already known that $ u(x)=C\_n\int\_{\partial B\_{r}(0)}\frac{r^2-\lvert x\rvert^2}{\lvert x-y\rvert^n}g(y)dS(y) $ by the construction of Green function in balls, where $ C\_n $ is a constant depending only on $ n $. From this formula, we can see that $ u(x)\in C^{\infty}(B\_r(0)) $ and $ u(x)\rightarrow g(\xi) $ if $ x\rightarrow \xi $ with $ \xi\in\partial B\_r(0) $. I am thinking about the problem when $ g $ is not continuous but $ g\in L^2(\partial B\_{r}(0)) $. In that case the integration $ u(x)=C\_n\int\_{\partial B\_{r}(0)}\frac{r^2-\lvert x\rvert^2}{\lvert x-y\rvert^n}g(y)dS(y) $ still makes sense and $ u(x) $ is still smooth. Can I say that $ u(x) $ actually solves the Dirichlet problem? In other words, can I show that for a.e. $ \xi\in\partial B\_{r}(0) $, $ u(x)\rightarrow g(\xi) $ if $ x\rightarrow \xi $?
|
https://mathoverflow.net/users/241460
|
Is the Poisson formula valid when the boundary condition is $ L^2 $?
|
This is clearly false as stated, since a necessary condition is that $g(x)\to g(\xi)$ as $x\to\xi$ a. e. in the sphere, but if $g$ is merely $L^2$, this may well fail for every point.
The convergence holds in a weaker sense. For example, it is true that if $g\_\rho(x)=g(\rho x)$ for $x\in\partial B\_r(0)$ and $\rho<1$, then $g\_\rho\to g$ a. e. and in $L^2$ as $\rho\to 1.$ Moreover, non-tangential limits (i. e., limits when $x\to\xi$ staying in the cone $\mathrm{dist}(x;[0,\xi])<\alpha |x-\xi|$ for some fixed $\alpha<1$) exist almost everywhere. See, for example, Axler, Boudron, Ramey, "Harmonic function theory."
|
4
|
https://mathoverflow.net/users/56624
|
404195
| 165,779 |
https://mathoverflow.net/questions/404193
|
7
|
Given a coalgebra $C$, can there exist more than one algebra structure on $C$ giving it the structure of a bialgebra? I will also ask the same question for Hopf algebras.
|
https://mathoverflow.net/users/351872
|
Different Bialgebra/Hopf algebra structures on coalgebras
|
Yes: if $k$ is a field of characteristic 2, let $C$ be the coalgebra over $k$ spanned by 1, $x$, $y$, and $z$ with $x$ and $y$ primitive, $\Delta z = z \otimes 1 + 1 \otimes z + x \otimes y + y \otimes x$ — in the algebra structure, $z$ is going to equal $xy = yx$, so $\Delta z$ has to equal $(\Delta x)(\Delta y)$. Then put algebra structures on this as follows:
* $xy=yx$
* $y^2 = 0$
* either $x^2 = 0$ or $x^2 = y$
This gives two graded (with $\deg x = 1$, $\deg y = 2$) connected cocommutative bialgebras, and so by a theorem of Milnor and Moore, there is a unique antipode making such a bialgebra into a Hopf algebra.
|
9
|
https://mathoverflow.net/users/4194
|
404198
| 165,781 |
https://mathoverflow.net/questions/404194
|
1
|
Let $t\_1 < t\_2 < \cdots <t\_m$ be real, and $X = \cup\_{i=1}^{m-1} (t\_i, t\_{i+1})$ be a union of real open intervals. Let $f:X \rightarrow \{-1, 1\}$ be any piecewise constant function of form
$$
f(x) =
\begin{cases}
a\_1 & \text{ if } t\_1 < x < t\_2 \\
a\_2 & \text{ if }t\_2 < x < t\_3 \\
\vdots \\
a\_{m-2} & \text{ if } t\_{m-2} < x < t\_{m-1} \\
a\_{m-1} & \text{ if } t\_{m-1} < x < t\_m
\end{cases}
$$
Where $a\_i \in \{-1, 1\}$, and $a\_{i} = -a\_{i+1}$ for $i = 1, ..., m-1$.
I have a number of questions regarding polynomial approximations of such a function $f$:
1. Can we always find a sequence of polynomials $(p\_n)\_{n=1}^\infty$ so that $(p\_n)\_{n=1}^\infty$ converge pointwise to $f$, and $|p\_n(x) - f(x)| \leq 1$ for all $x \in X$ and $n \in \mathbb{N}$?
2. If so, are such polynomials easy to find and construct (i.e. do we have closed form solutions)?
3. How quickly do we get convergence?
I am aware that, upon picking a suitable inner product, we can use any collection of orthonormal polynomials to make approximations of functions. For example I know the Chebyshev, Bernstein, Jacobi etc. polynomials can be used to approximate continuous functions on bounded intervals, but I have found no theorem that says we can use these to construct approximations for arbitrary piecewise constant functions like the one given above.
Indeed, it is easy to find a polynomial approximation for the Heaviside Step function for example, however it is unclear how, or if this an be done for more complicated step functions.
|
https://mathoverflow.net/users/129192
|
Determining polynomial approximations of piecewise constant functions
|
The polynomial $Q(y):=\frac43y-\frac13 {y^3}$ is increasing on the interval $[-1,1]$; it has fixed points $0$ and $\pm1$, and $\text{sgn }( Q(y)-y )=\text{sgn}y$. Thus the iterates of $Q$ starting from any $y\in [-1,1]\setminus\{0\}$ converge monotonically to $\text{sgn } y$ (in fact with exponential rate given by $Q'(\pm1)=\frac13$, and uniformly away from $0$
).
Assuming w.l.o.g $a\_i=(-1)^i$, your step function can be written $f(x):= \text{sgn} P(x)$ with $P(x):= \prod\_{i=1}^m(t\_i-x)$, for any $x\in X$. If we choose any $M\ge \|P(x)\|\_{\infty,X}$ the polynomial sequence of iterates $Q^{n}\big( {P(x)}/ M\big)$ converges to $f$ on $X$; in fact increasing/decreasing between consecutive nodes, and uniformly on compacts set of $X$.
|
3
|
https://mathoverflow.net/users/6101
|
404199
| 165,782 |
https://mathoverflow.net/questions/404213
|
11
|
Clearly this is impossible for $p$ of even degree, and I imagine that Cardano’s formula quickly reveals it to be impossible in the cubic case, although I have not checked in detail. My guess is that no such $p$ exists. Does one exist? If so, is there an explicit example?
Failing a general yes or no answer, are there sufficient conditions to identify a non-surjective polynomial function?
|
https://mathoverflow.net/users/351164
|
Does there exist some $p(x) \in \mathbb{Q}[x]$, deg$(p) > 1$, which maps $\mathbb{Q}$ onto itself surjectively?
|
No, this can't happen. One way to prove this is via Hilbert irreducibility: The polynomial $p(x) - t$ is irreducible over $\mathbb Q[x,t]$, so there are infinitely many specializations $t = c$ with $c \in \mathbb Q$ such that $p(x) - c$ is irreducible in $\mathbb Q[x]$. Since the degree of $p(x)$ is greater than 1, it follows that for each such $c$ the polynomial $p(x) - c$ has no rational roots.
|
32
|
https://mathoverflow.net/users/66491
|
404217
| 165,787 |
https://mathoverflow.net/questions/404216
|
2
|
I would like to construct a sequence of discrete random variable $X\_2, X\_3,...,X\_n,...$, where $X\_n \in\{0,1,2,...,n-1\}$. Given any $\epsilon \in (0,1)$, its Shannon entropy and min-entropy should satisfy the following relationships
\begin{cases}
H(X\_n)\geq(1-\epsilon)\log\_2(n)\\
H\_{min}(X\_n)=const
\end{cases}
for all $n\geq\mathbb{N}\_{\epsilon}$ and some $const > 0$.
My understanding is that the Shannon entropy indicates the underlying distribution should be approximately uniform. And the min-entropy suggests that the largest possibility of $X\_n$ should be $2^{-const}$. But I am stuck with coming up with such a distribution. Is there anyone who could provide some hints?
|
https://mathoverflow.net/users/376483
|
Difference between Shannon entropy and min-entropy
|
Since the OP may be interested in what $\epsilon$ are achievable I am providing this alternative to the other answer, where it is correctly stated:
>
> If you fix the maximal atom (say, $p$) of a distribution $\mu$ supported by $n$ points, then its entropy is maximal when all the remaining atoms have the same weight $(1-p)/(n-1)$.
>
>
>
Also note that
$$
p\geq \frac{1-p}{n-1} \iff np\geq 1\iff p\geq \frac{1}{n}
$$
so that $p$ is indeed the maximal atom of $\mu.$
This means that one actually obtains the *equality* below for the Shannon entropy:
$$
H(\mu) = -p\log p - (1-p)\log(1-p) + (1-p)\log(n-1),
$$
when $p$ is fixed which gives
$$
H(\mu) = H\_2(p) + \left(1-p\right)\log(n-1)
$$
or
$$
H(\mu) \geq \left(1-p\right)\log(n-1)\sim \left(1-p\right)\log n \quad (1)
$$
where $f(n)\sim g(n)$ denotes that $\lim\_{n\rightarrow \infty} \frac{f(n)}{g(n)}=1.$ This means that for $n$ large enough you can pick any $\epsilon \geq p$ for which
$$
H(\mu)\geq (1-\epsilon) \log n
$$
is indeed satisfied.
Depending on what application you have in mind, this may suffice.
|
3
|
https://mathoverflow.net/users/17773
|
404223
| 165,789 |
https://mathoverflow.net/questions/404210
|
3
|
**The classical setting.**
Given a monoid $A$, there's a category $\mathbf{B}A$, called the **[delooping](https://ncatlab.org/nlab/show/delooping#delooping_of_a_group_to_a_groupoid)** of $A$, having a single object $\star$ and satisfying $\mathrm{Hom}\_{\mathbf{B}A}(\star,\star)\overset{\mathrm{def}}{=}A$, with composition given by multiplication and the sole identity $\mathrm{id}\_{\star}$ given by $1\_A$.
This construction is characterised by the following property: for any other category $\mathcal{C}$, we have a bijection of sets
$$
\left\{
\begin{gathered}
\text{functors}\\
\mathbf{B}A\to\mathcal{C}
\end{gathered}
\right\}
\cong
\left\{
\begin{aligned}
&\text{pairs $(X,\phi)$ with}\\
&\,\,\,\,\,\,\,\text{- $X$ an object of $\mathcal{C}$;}\\
&\,\,\,\,\,\,\,\text{- $\phi$ a morphism of monoids}\\
&\text{from $A$ to $\left(\mathrm{Hom}\_{\mathcal{C}}(X,X),\circ,\mathrm{id}\_{X}\right)$.}
\end{aligned}
\right\}.
$$
For example:
* A functor $\mathbf{B}\mathbb{N}\to\mathcal{C}$ is the same as an *endomorphism* $A\to A$ of $\mathcal{C}$;
* A functor $\mathbf{B}\mathbb{Z}\to\mathcal{C}$ is the same as an *automorphism* $A\to A$ of $\mathcal{C}$;
* A functor $\mathbf{B}\mathbb{Z}/2\to\mathcal{C}$ is the same as an *involution* $A\to A$ of $\mathcal{C}$;
* A functor $\mathbf{B}\mathbb{B}\to\mathcal{C}$ is the same as an *idempotent* $A\to A$ of $\mathcal{C}$, where $\mathbb{B}=(\{0,1\},\text{OR},1)$.
**The $\infty$-categorical setting.**
**Preliminary Question.** Given an $\infty$-category $\mathcal{C}$, is there a natural monoidal $\infty$-groupoid structure on $\mathrm{Hom}\_{\mathcal{C}}(X,X)$?
**Question.** Is there an analogue of deloopings for $(\infty,1)$-categories, where we start with a monoidal $\infty$-groupoid $\mathcal{C}$ and construct an $(\infty,1)$-category $\mathbf{B}\mathcal{C}$ such that
* A functor $\mathbf{B}\mathcal{C}\to\mathcal{D}$ from $\mathbf{B}\mathcal{C}$ to another $(\infty,1)$-category $\mathcal{D}$;
is the same thing as
* An object of $\mathcal{D}$ together with a functor of monoidal (?) $\infty$-groupoids $\mathcal{C}\to\mathrm{Hom}\_{\mathcal{D}}(X,X)$,
with $\mathrm{Hom}\_{\mathcal{D}}(X,X)$ the [morphism space](https://kerodon.net/tag/01J3) of $\mathcal{D}$ from $X$ to itself, and where this bijection can be made into a full-fledged isomorphism of (appropriate) $\infty$-groupoids?
|
https://mathoverflow.net/users/130058
|
Delooping monoidal $\infty$-groupoids into $\infty$-categories
|
I assume that with ``monoidal ∞-groupoid'' you mean an $E\_1$-space. In this case the answer is yes. It is well known that $E\_1$-spaces can be modeled by functors
$$X:\Delta^{\mathrm{op}}\to \operatorname{Space}$$
satisfying the Segal conditions. Now if you are given an $\infty$-category $\mathcal{C}$ you can define a simplicial space
$$s(\mathcal{C}): [n]\mapsto\operatorname{Map}\_{\operatorname{Cat}\_∞}(\Delta^n,\mathcal{C})\,.$$
In fact this functor is fully faithful and identifies $\operatorname{Cat}\_∞$ with the category of complete Segal spaces. For the following we won't need all this though - we will use only that it takes values in Segal spaces (which follows immediately from $\Delta^n\amalg\_{\Delta^0} \Delta^m\simeq\Delta^{n+m-1}$ in $\operatorname{Cat}\_∞$).
Now let $x\in\mathcal{C}$ be an object of $\mathcal{C}$. Then we can define the simplicial space
$$ \operatorname{End}\_{\mathcal{C}}(x):\Delta^{\mathrm{op}}\to \operatorname{Space}\qquad [n]\mapsto \{x\}\times\_{\operatorname{Map}(\{0,\dots,n\},\mathcal{C})} \operatorname{Map}\_{\operatorname{Cat}\_∞}(\Delta^n,\mathcal{C})\,.$$
That is it sends $[n]$ to the (∞-)groupoid of functors $F:\Delta^n\to \mathcal{C}$ that sends all objects to $x$. It is easy now to see that $\operatorname{End}\_{\mathcal{C}}(x)$ satisfies the Segal conditions and so it is an $E\_1$-space.
---
This takes care of your preliminary question. To go back to your main question, the functor $(\mathcal{C},x)\mapsto \operatorname{End}\_{\mathcal{C}}(x)$ obviously preserves all limits and filtered colimits, and so it has a left adjoint $B$ exactly as you wanted. To get a more ``concrete'' description $B$ sends an $E\_1$-space $X$ to the ∞-category corresponding to the completion of $X$ seen as a Segal space. That is
$$BX:=\int^{[n]\in\Delta^{\mathrm{op}}} X([n])\times \Delta^n$$
where the coend is computed in $\operatorname{Cat}\_∞$.
With more care one can show that $B:E\_1-\operatorname{Space}\to(\operatorname{Cat}\_∞)\_{\Delta^0/}$ is fully faithful with essential images those arrows $\Delta^0\to\mathcal{C}$ that are essentially surjective (that is such that $\mathcal{C}$ has only one equivalence class of objects). Indeed this is a special case of the equivalence between the ∞-category of Segal spaces and the ∞-category of flagged ∞-categories.
|
2
|
https://mathoverflow.net/users/43054
|
404229
| 165,792 |
https://mathoverflow.net/questions/403965
|
4
|
Let $A$ be a $C^\*$-algebra, $E$ be a (right) Hilbert $A$-module and $t \in \mathcal{L}\_A(E)$ be an adjointable operator satisfying $t=t^\*$. Is it true that
$$\|t\| = \sup\_{z \in E, \|z\| = 1} \|\langle tz,z\rangle\|\_A?$$
Obviously, if we denote the supremum on the right by $M$, we have $M \le \|t\|$ by the Cauchy-Schwarz inequality, so the main interest lies in the other inequality.
When $A= \mathbb{C}$ (and thus $E$ is a Hilbert space), the result is well-known, but neither of the proofs I know seem to generalise to the framework of Hilbert $C^\*$-modules. Perhaps a slick application of the spectral theorem does the job (as seems to be suggested in a comment?).
|
https://mathoverflow.net/users/216007
|
$\|t\| = \sup_{\|z\| \le 1} \|\langle tz,z\rangle\|$ when $t=t^*$
|
I'll post an answer based on the comments above.
Since $t$ is self-adjoint, we can write $t= t\_+-t\_-$ where $t\_+$ and $t\_-$ are positive elements with $t\_+ t\_- = 0$. The latter condition ensures that $\|t\| = \max\{\|t\_+\|, \|t\_-\|\}$.
Assume, without loss of generality, that $\|t\| = \|t\_+\|$.
Let $\epsilon > 0$. By definition of the operator norm, there is an element $z$ in the unit ball of $E$ such that $\|t\_+^{1/2}\| \le \epsilon + \|t\_+^{1/2}z\|.$
Consider the isometric unital $\*$-morphism
$$C(\sigma(t)) \to C^\*(1,t): f \mapsto f(t)$$
given by continuous functional calculus. Define $f \in C(\sigma(t))$ to be $f=0$ on $\sigma(t)\cap [-\infty,0]$ and to be $1$ on $\sigma(t)\cap [\delta,\infty]$ and $\operatorname{Im}f\subseteq [0,1]$, where $\delta>0$ is chosen such that $2s^{1/2}< \epsilon$ when $0 \le s \le\delta$, and consider the element $f(t)\in C^\*(1,t)\subseteq \mathcal{L}\_A(E)$. Then we define $x:= f(t)z \in E$, and we note that
$\|x\| \le \|f(t)\|\|z\| \le \|f\|\_\infty =1.$ By definition of $t\_-$, we see that $t\_-f(t) = 0$. Hence,
$$\|\langle tx,x\rangle\| = \|\langle t\_+ x,x\rangle\| = \|t\_+^{1/2}x\|^2.$$
Let $g(s) = \max\{s,0\}$. Then $$\|g^{1/2}f-g^{1/2}\|\_\infty = \sup\_{s \in [0,\delta]} |s^{1/2}f(s)-s^{1/2}| \le \sup\_{s \in [0,\delta]} 2s^{1/2}\le \epsilon.$$
It follows that
$$\|t\_+^{1/2}x-t\_+^{1/2}z\|=\|t\_+^{1/2}f(t)z-t\_+^{1/2}z\| \le \epsilon.$$
Hence, $$\|t\_+^{1/2}\| \le 2\epsilon + \|t\_+^{1/2}x\|=2\epsilon + \|\langle tx,x\rangle\|^{1/2} \le 2 \epsilon + M^{1/2}.$$
Letting $\epsilon \to 0$, we obtain $\|t\_+^{1/2}\| \le M^{1/2}$. Squaring both sides and invoking the $C^\*$-identity, we obtain
$$\|t\| = \|t\_+\| \le M$$
and the other inequality is proven.
|
2
|
https://mathoverflow.net/users/216007
|
404233
| 165,793 |
https://mathoverflow.net/questions/404226
|
8
|
It was known that the James space $J$ has separable second conjugate, is non-reflexive and isometric to its second conjugate. I want to know whether there are Banach spaces $X$ with separable second conjugates $X^{\*\*}$, but $X$ is not a dual space (the James space $J$ is a dual space). Furthermore, are there any references about Banach spaces with separable second conjuates ?
Thank you !
|
https://mathoverflow.net/users/41619
|
Banach spaces whose second conjugates are separable
|
Yes, there are such spaces. To see this, first note that Joram Lindenstrauss [showed](https://link.springer.com/article/10.1007/BF02771677) that for every separable Banach space $Y$ there exists a Banach space $X$ such that $X^{\ast\ast}$ is separable and $X^{\ast\ast}/X$ is isomorphic to $Y$. Apply this result with $Y$ a separable Banach space that is not isomorphic to a subspace of a separable dual space (e.g., $c\_0$ or $L\_1$; see for example Theorem 6.3.7 from Albiac and Kalton's book *Topics in Banach Space Theory* for details). Then, for such $Y$, the space $X$ prescribed by Lindenstrauss' construction cannot be isomorphic to a dual space. Indeed, if it were, then $X^{\ast\ast}/X$ would be isomorphic to a subspace of $X^{\ast\ast}$ by virtue of the fact that that every dual space is complemented in its bidual (in particular, $X^{\ast\ast}$ would be isomorphic to $(X^{\ast\ast}/X)\oplus X$), and thus $Y$ would also be isomorphic to a (complemented) subspace of the separable dual space $X^{\ast\ast}$ - a contradiction.
(As it happens, the aforementioned book of Albiac and Kalton includes Lindenstrauss' result in the chapter with the title 'Important Examples of Banach Spaces'.)
|
11
|
https://mathoverflow.net/users/848
|
404243
| 165,798 |
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