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https://mathoverflow.net/questions/402178
|
5
|
Euclid proof of the infinitude of primes can be extended into this.
Assuming there is a finite number of primes, $k$, sort them in increasing order and split the series after any prime at $t$. Create the difference between the products of each group.
$$ \left | \prod\_{t< m \leq k} p\_m - \prod\_{1 \leq n \leq t} p\_n \right | $$
The result cannot be divisible by any used prime, therefore it has to be $1$.
There are $k-1$ differences in the form:
$$ P\_k(t) = \prod\_{t< m \leq k} p\_m - \prod\_{1 \leq n \leq t} p\_n $$
and obviously:
$$ P\_{k}(1) > P\_{k}(2) > \cdots > P\_k(k-1) $$
However, in this strictly monotonic series, we can have only once $-1$ and $1$, which means that as long as there are at least four primes there cannot be a finite number of primes. And we know four initial primes.
The question which comes naturally is:
Can you write every prime as a difference between split products of the full set of first $k$ primes?
$$ p\_N = \left | \prod\_{n \in J} p\_n - \prod\_{m \notin J} p\_m \right |, J \subset \{1,2,...,k\}$$
It seems as we are adding more primes we have more combinations of the differences, closing to $2^m$ differences. However primorial function, the product of primes, grows as $m^m$ so it seems that it will quickly start skipping as we are going towards larger primes.
(I think there is only some hope if we loose the restriction and say that we are allowed to use two products of different set of primes, not necessarily the complete set, since then the number of combinations is somewhat larger, but it still does not look sufficient.)
Anyway, does the proof sound correct and what extension or restriction of this might still give us each prime?
(The proof here explicitly elaborates Euclid proof in the sense of what if we take that $p\_1 p\_2 \cdots p\_n + 1=1$ which is actually the only option, but without using $+1$, which is, albeit trivial actually, *constructed* for the purpose of the proof, and then it discusses the actual number of initial primes we need in order to claim that there is an infinite number of them.)
(Euclid's proof has a small gap that is rarely mentioned. Notice that when you create $p\_1p\_2p\_3...p\_n$ you implicitly assume that there is *at least* one prime number and that all primes are positive. $p\_1p\_2p\_3...p\_n+1=1$ is perfectly valid in case you have no prime numbers, and if primes can be negative everything falls apart. This is all elementary, but it is still a small gap.)
|
https://mathoverflow.net/users/nan
|
Extended Euclid proof and primes in form $|\prod\limits_{n \neq m} p_n -\prod\limits_{m \neq n} p_m|$
|
See Guy, Lacampagne, and Selfridge, [Primes at a Glance](https://www.ams.org/journals/mcom/1987-48-177/S0025-5718-1987-0866108-3/S0025-5718-1987-0866108-3.pdf), Mathematics of Computation, volume 48, number 177, January 1987, pages 183-202.
Abstract. Let $N = B - L$, $B \ge L$, $\gcd(B,L) = 1$, $p \mid BL$ for all primes $p\le\sqrt N$. Then $N$ is $0,1$ or a prime. Writing $N$ in this form suggests a primality and a squarefreeness test. If we also require that when the prime $q \mid BL$ and $p < q$ then $p \mid BL$, we say that $B - L$ is a *presentation* of $N$. We list all presentations found for any $N$. We believe our list is complete.
There is also Takashi Agoh, Paul Erdős and Andrew Granville, Primes at a (somewhat lengthy) glance, The American Mathematical Monthly Vol. 104, No. 10 (Dec., 1997), pp. 943-945.
|
1
|
https://mathoverflow.net/users/3684
|
402201
| 165,061 |
https://mathoverflow.net/questions/402207
|
3
|
Suppose that $\Omega \subset \Bbb R^d$ is a sufficiently nice domain. From the examples of orthogonal bases in Hilbert space cases (or looking at a wavelets basis), it seems natural to me that one may expect the elements of a (normalized) Schauder basis $\{u\_n\}\_{n=1}^\infty$ of $W^{1,p}\_0(\Omega)$ (for $p>1$) to be more and more "oscillatory" as $n\to\infty$. I wonder if that is true in the sense that
$$
\lim\_{n\to\infty} \mathscr L^d(\{ x\in\Omega : |\nabla u\_n(x)| \le \varepsilon \}) = 0
$$
for any fixed $\varepsilon>0 $?
If the above is not true, is it possible to construct such a basis that satisfies the above property?
**Edit:** Thanks to a comment, I realized that the proposed notion of oscillation is not true (almost trivially). I want to modify it a little bit to capture what I originally had in mind. For any fixed $\varepsilon>0$, is it true that
$$
\lim\_{n\to\infty} \int\_{\{|\nabla u\_n| \le \varepsilon \}} |\nabla u\_n|^p \,dx = 0 \ ?
$$
|
https://mathoverflow.net/users/80191
|
Must a Schauder basis for $W^{1,p}_0(\Omega)$ be oscillatory?
|
Even the modified question does not hold.
Let $u\_n$ be a basis such that $\mathcal{L}^d(\operatorname{spt} u\_n) \to 0$, e.g. a wavelet basis and let $\phi \in C\_0^\infty(\Omega)$ a function such that $$\int\_{\{|\nabla \phi| \leq \varepsilon\}} |\nabla \phi|^p dx > 0$$ for all $\varepsilon > 0$, e.g. a smooth bump.
Now construct a modified basis
$$\tilde{u}\_n = a\_n( \phi + u\_n), $$
where the $a\_n >0$ are chosen in such a way that this is again normalized. By the triangle-inequality
$$ 1 = \| \tilde{u\_n} \|\_{W^{1,p}} \leq a\_n( \| u\_n \|\_{W^{1,p}} + \| \phi \|\_{W^{1,p}}) = a\_n(1 + \| \phi \|\_{W^{1,p}}), $$
so the $a\_n$ are uniformly bounded from below by $a\_-:= \frac{1}{1+\| \phi \|\_{W^{1,p}}}$. On the other hand, since $\int\_{\operatorname{spt}u\_n} |\nabla \phi|^p dx \to 0$, it is easy to show that $a\_n \to a\_-$ and that in particular $a\_n$ is bounded from above by some $a\_+$.
But then
$$ \int\_{\{|\nabla \tilde{u}\_n|\} \leq \varepsilon} |\nabla\tilde{u}\_n|^p dx \geq \int\_{\{|\nabla \tilde{u}\_n|\leq \varepsilon\} \setminus \operatorname{spt} u\_n } |\nabla\tilde{u}\_n|^p dx = a\_n^p \int\_{\{|\nabla (a\_n \phi)|\leq \varepsilon\} \setminus \operatorname{spt} u\_n } |\nabla\phi|^p dx \\\geq a\_-^p \int\_{\{|\nabla (a\_+\phi)|\leq \varepsilon\} \setminus \operatorname{spt} u\_n } |\nabla\phi|^p dx \to a\_-^p \int\_{\{|\nabla \phi|\leq \varepsilon/a\_+\}} |\nabla\phi|^p dx > 0.$$
|
3
|
https://mathoverflow.net/users/51695
|
402213
| 165,064 |
https://mathoverflow.net/questions/402140
|
2
|
I asked the following [question](https://math.stackexchange.com/questions/4223194/a-uniform-continuity-type-condition-on-an-integral) on MathStackExchange, but I have not received the answer that I'm looking for. Although it may not be a research-level question, I thought I could ask it here.
---
I'm currently reading this [paper](https://www.ams.org/journals/proc/1996-124-02/S0002-9939-96-03154-1/S0002-9939-96-03154-1.pdf) (and working on a similar one). Specifically, I'm trying to improve the two hypotheses (2.5) and (2.6) of [O'Regan's paper](https://link.springer.com/article/10.1007/s000130050243).
The main goal is to study the Hammerstein integral equation (in $\mathcal{C}(I,E))$:
$$x(t) = \int\_{0}^{t} K(t,s)f\big(s,x(s)\big)ds,\quad t\in I;$$
where $I=[0,1]$, $K $ is a scalar kernel, $E$ a Banach space and $f:I \times E \rightarrow E$ is a given function.
**Suppose that $s \mapsto K(t, s)$ is integrable and $t \mapsto K(t, s)$ is continuous.**
**My goal is to see if the following is true:**
\begin{array}{l}
\text { For each } \epsilon>0 \text { , there exists } \delta>0 \text { such that, for any } t\_{1}, t\_{2} \in I, \text { if }\left|t\_{1}-t\_{2}\right|<\delta \text { then }\\
\int\_{t\_1}^{t\_2} K(t\_2,s) d s< \epsilon
\end{array}
---
|
https://mathoverflow.net/users/102228
|
A "uniform continuity" type condition on a Hammerstein integral equation
|
This does certainly not follow from your other hypotheses, as what you want to conclude is not much weaker than the equi-integrability of $\{K(t,\cdot):t\in I\}$ (sometimes also called absolute continuity in the $L\_1$-norm), but what you assume is only a continuity in one variable which implies nothing about the $L\_1$-norm, not even the boundedness.
A counterexample might look as follows: Around the line $t=s/2$ with $t\in(0,1)$, say, for $s\in[t/3,t/2]$, the function $K(t,s)$ assumes huge values (for instance, $1/t^2$). Then you extend the function to a nonnegative function preserving your condition such that $K(t,s)=0$ for $t\in\{0,1\}$ or $s\ge t$.
Then you have even $\sup\_{\lvert t\_1-t\_2\rvert\le\delta}\int\_{t\_1}^{t\_2}K(t\_2,s)ds=\infty$ for every $\delta>0$.
|
1
|
https://mathoverflow.net/users/165275
|
402214
| 165,065 |
https://mathoverflow.net/questions/402218
|
6
|
It is known that the logarithm of the modulus of an *analytic* function $f: D \subset \mathbb C \rightarrow \mathbb C$ ($D$ is a domain) is subharmonic. I have two questions:
(1) Are there some weaker conditions than analyticity that ensure the same result?
(2) Is there any characterization of functions $f$ such that $\log |f|$ is subharmonic?
|
https://mathoverflow.net/users/151918
|
$\log |f|$ is subharmonic
|
Conditions for a log-subharmonic function $f$ on $D\in\mathbb{R}^n$ are described by Mochizuki in [A Class of Subharmonic Functions and Integral Inequalities](https://www.jstage.jst.go.jp/article/iis/10/2/10_2_153/_pdf) (2004).
The conditions are phrased in terms of an inequality for the volume average $A\_p$ of $|f|^p$ and the surface average $M$ of $|f|$, in the form $A\_p\leq M^p$ for any closed ball in $D$.
If $n\geq 2$ the condition $A\_{1+2/n}\leq M^{1+2/n}$ is sufficient for $\log|f|$ to be subharmonic. If $n=2$ this condition is also necessary, if $n>2$ it is not.
|
9
|
https://mathoverflow.net/users/11260
|
402219
| 165,066 |
https://mathoverflow.net/questions/402211
|
4
|
Let $\mathbb{R}\_\*=\mathbb{R}^\omega/\mathcal U$ for some ultrafilter $\cal U$. In the definitions of [this question](https://mathoverflow.net/q/72612/118366) and assuming ZFC + CH there are only three types of cuts in $\mathbb{R}\_\*$: $(\omega,\omega\_1),~(\omega\_1,\omega),~(\omega\_1,\omega\_1)$. And only $(\omega\_1,\omega\_1)$ cut could be filled. For an ordered field $\mathbb F$ lets say that a cut $\mathbb{F}=A\coprod B$ is *good* iff for any $c>0$ there are $a\in A$ and $b\in B$ such that $b-a<c$. Any good cut is of type $(\lambda,\lambda)$ where $\lambda$ is cofinality of $\mathbb F$. If any good cut is filled lets call the field *quasi-complete*. Any ordered field can be embedded into quasi-complete field with appropriate universal properties ([see here](https://mathoverflow.net/a/140962/118366)).
Can we proof that (may be for special ultrafilter $\cal U$):
1. $\mathbb{R}\_\*$ is quasi-complete ?
2. If $\mathbb{F}$ is quasi-complete then $\mathbb{F}\_\*=\mathbb{F}^\omega/\mathcal{U}$ is quasi-complete ?
3. Any $(\omega\_1,\omega\_1)$ cut in $\mathbb{R}\_\*$ is good ?
|
https://mathoverflow.net/users/118366
|
On a completeness property of hyperreals
|
This is also called Cauchy-completeness, and it coincides for non-Archimedean ordered fields with the natural valuation to the valuation-theoretic notion of completeness. Also, this is the same as having no proper dense ordered field extension.
I will say that an ordered pair $(A,B)$ of subsets of an ordered field $F$ is a *cut generator* if $A<B$ and there is no $x \in F$ with $A<x<B$. If $(A,B)$ is a cut generator, then the pair $(A',B')$ where $A'$ is the set of lower bounds of elements of $A$ and $B'$ is the set of upper bounds of elements of $B$ is a cut in $F$ whose type is $(\operatorname{cf}(A,<),\operatorname{cf}(B,>))$.
The answer to your first question is negative. In ZFC+CH, the field $\mathbb{R}\_\*$ is unique up to isomorphism to be real-closed, of cardinality continuum / $\aleph\_1$ and without $(\omega,\omega)$ type cuts (in particular, the choice of ultrafilter doesn't matter). So it is isomorphic to the field $\mathbf{No}(\omega\_1)$ of surreal numbers with countable birth day. I use the latter because its elements can be represented in a more explicit way, and this helps find good cuts in $\mathbf{No}(\omega\_1)$. For instance, using the sign-sequence presentation of surreal numbers, you have the cut generator $(A,B)$ where $A$ is the set of numbers whose sign-sequence is a concatenation of $(+-)$, and $B$ is the set of numbers obtained by adding $+$ at the end of the sign-sequence of elements of $A$. So $A=\{(),(+-),(+-+-),...\}$ and $B=\{(+),(+-+),...\}$ (the dots hide uncountably many numbers). One can also define $a\_{\gamma}:=\{a\_{\rho} :\rho<\gamma\ | \ b\_{\rho}:\rho<\gamma \}$ and $b\_{\gamma}:= \{a\_{\gamma} \ | \ b\_{\rho}:\rho<\gamma \}$ by induction on $\gamma<\omega\_1$ and obtain $A=\{a\_{\gamma} \ : \ \gamma<\omega\_1\}$ and $B=\{b\_{\gamma} \ : \ \gamma<\omega\_1\}$.
It should be possible to prove the result directly in $\mathbb{R}\_\*$ using $\mathbb{N}\_\*$-indexed sums of fastly growing sequences, but by transfer it's actually easy to obtain a convergent sum, and thus not a cut!
Also, I don't know if ZFC alone proves that $\mathbb{R}\_\*$ is not quasi-complete. In fact I may have asked this exact question on MSE or MO.
---
Using the same isomorphism $\mathbb{R}\_\* \cong \mathbf{No}(\omega\_1)$, one can obtain cuts of type $(\omega\_1,\omega\_1)$ that are not good. This again implies some familiarity with surreal numbers, so you can admit the existence of a map $x\mapsto \omega^x: \mathbf{No}(\omega\_1)\rightarrow \mathbf{No}(\omega\_1)^{>0}$, sometimes called the $\omega$-map, which is strictly increasing, with
>
> $\forall x\in \mathbf{No}(\omega\_1)^{>0},\exists ! d\_x \in
> > \mathbf{No}(\omega\_1), \exists r \in \mathbb{R}^{>0}, r^{-1} \omega^{d\_x}<x<r\omega^{d\_x}$
>
>
>
Then $(\omega^A,\omega^B)$ (where $\omega^X=\{\omega^x \ : \ x \in X\}$ is a cut generator, whose corresponding cut has type $(\omega\_1,\omega\_1)$. Indeed if there were a number $c \in \mathbf{No}(\omega\_1)$ between $\omega^A$ and $\omega^B$, then $d\_c$ would have to lie between $A$ and $B$. The corresponding cut is not good, because we have $x+1<y$ for all $(x,y) \in \omega^A\times \omega^B$.
Everything I wrote requires some justification so feel free to ask if you want me to elaborate.
|
7
|
https://mathoverflow.net/users/45005
|
402223
| 165,068 |
https://mathoverflow.net/questions/402227
|
25
|
A famous result of Galois, in his letter to Auguste Chevalier, is that for $p$ prime $>11$ the group $\operatorname{PSL}(2,\mathbb{F}\_p) $ does not embed in the symmetric group $\mathfrak{S}\_p$. The standard proof nowadays goes through the classification of subgroups of $\operatorname{PSL}(2,\mathbb{F}\_p) $ (Dickson's theorem), which is far from trivial. Does anyone know a simpler argument?
|
https://mathoverflow.net/users/40297
|
$\operatorname{PSL}(2,\mathbb{F}_p) $ does not embed in $\mathfrak{S}_p$ for $p>11$
|
A few months ago Péter Pál Pálfy has given a talk about this exact topic. The abstract of the talk was the following:
>
> In his "testamentary letter" Galois claims (without proof) that PSL(2,p) does not have a subgroup of index p whenever p>11, and gives examples that for p = 5, 7, 11 such subgroups exist. The attempt by Betti in 1853 to give a proof does not seem to be complete. Jordan's proof in his 1870 book uses methods certainly not known to Galois. Nowadays we deduce Galois's result from the complete list of subgroups of PSL(2,p) obtained by Gierster in 1881.
>
>
> In the talk I will give a proof that might be close to Galois's own thoughts. Last October I exchanged a few e-mails on this topic with Peter M. Neumann. So the talk is in some way a commemoration of him.
>
>
>
The recording of the talk is available [here](https://drive.google.com/file/d/10T_AuJb-J12movAkKJNXbIz87sw1-wM9/view?usp=sharing). The presentation of the proof begins around 32 minutes in.
The proof is elementary, but is itself nontrivial. We easily reduce to showing $G=\operatorname{PSL}(2,\mathbb{F}\_p)$ has no index $p$ subgroup for large $p$. The idea is to study the natural doubly transitive action of $G$ on the projective line and its interrelation between an index $p$ subgroup and the subgroup of affine transformations inside $G$. The bounds on $p$ come out of realizing that all elements of $\mathbb F\_p^\times$ must satisfy quadratic relations coming from that action.
|
31
|
https://mathoverflow.net/users/30186
|
402234
| 165,072 |
https://mathoverflow.net/questions/401083
|
7
|
This is related to [my earlier (unanswered) MO post](https://mathoverflow.net/questions/400879/fibonacci-embedded-in-catalan). Preserve notations from there.
We take advantage of the one-to-one correspondence between the $(s,s+1)$-core partitions and $(s,s+1)$-Dyck paths. Let $\mathcal{F}\_s$ denote the set of $(s,s+1)$-Dyck paths whose associated core partitions have **distinct parts**.
Also, we recall the two notions [**area** and **bounce**](https://www.sciencedirect.com/science/article/pii/S0195669804000320) statistics defined on Dyck paths. We write $[n]\_t=\frac{1-t^n}{1-t}$ and the $t$-binomials $\binom{n}k\_t=\frac{[n]\_t}{[k]\_t[n-k]\_t}$.
Experimental observations lead to the following
>
> **QUESTION 1.** Is this true?
> $$\sum\_{\lambda\in\mathcal{F}\_s}q^{\text{area}(\lambda)}\,t^{\text{bounce}(\lambda)}
> =\sum\_{j\geq0}\binom{s-j}j\_tq^j\,t^{\binom{s}2-j(s-j)}.$$
>
>
>
>
> **QUESTION 2.** Alternatively, if $H\_s(q,t):=\sum\_{\lambda\in\mathcal{F}\_s}q^{\text{area}(\lambda)}\,t^{\text{bounce}(\lambda)}$ then does this hold?
> $$H\_{s+1}(q,t)=t^{s+1}\,H\_s(q,t)+q\,t^s\,H\_{s-1}(q,t)$$
> with initial conditions $H\_1(q,t)=1$ and $H\_2(q,t)=q+t$.
>
>
>
|
https://mathoverflow.net/users/66131
|
$(q,t)$-Fibonacci polynomials: area & bounce statistics
|
I mentioned in [my previous answer](https://mathoverflow.net/questions/400879/fibonacci-embedded-in-catalan/402188#402188) that the order ideals corresponding to $(s,s+1)$-cores with distinct parts are precisely the subsets of $\{1,2,\dots,s-1\}$ that contain no consecutive elements. This means that the Dyck paths under investigation are all of height 2, with peaks at each element of the order ideal. Since there are no consecutive elements, the paths will always be "bouncing", in the sense that the associated bounce path is the same as the Dyck path itself.
Let's restrict our attention to the Dyck paths with exactly $j$ peaks. There are exactly $\binom{s-j}{j}$ of these. They will all have area $j$, so it remains to prove that the generating function for their bounce statistic is given by
$$t^{\binom{s}{2}-j(s-j)}\binom{s-j}{j}\_t=t^{\binom{j}{2}+\binom{s-j}{2}}\binom{s-j}{j}\_t.$$
To each subset $\{a\_1,a\_2,\dots,a\_j\}$ of $\{1,2,\dots,s-1\}$ we can associate the subset $\{b\_1,b\_2,\dots,b\_j\}$ of $\{0,1,\dots,s-j-1\}$ given by $b\_i=a\_i-i$. We can check that the bounce statistic of this subset is equal to $\binom{s-j}{2}+b\_1+b\_2+\cdots+b\_j$, therefore the generating function we want is the coefficient of $x^j$ in the expression
$$t^{\binom{s-j}{2}}\prod\_{i=0}^{s-j-1}(1+t^ix).$$
From here the desired result follows from the q-binomial theorem.
|
3
|
https://mathoverflow.net/users/2384
|
402250
| 165,076 |
https://mathoverflow.net/questions/402255
|
3
|
I'm interested in the following operator $T$, close relative of the standard logarithmic derivative:
$$f(x)\to Tf(x)=\frac{\text{d}(\log {f})}{\text{d}(\log {x})}=\frac{xf'}{f},$$
where $f$ is an increasing, positive $C^{\infty}(\mathbb R^+)$ function.
Does anyone know whether this pops up, say, in functional analysis or probability? If so, in which context?
|
https://mathoverflow.net/users/167834
|
On the operator $f\to xf'/f$
|
You see this in the discussion of modular forms and related topics. When complex variable $\tau$ is in the upper half-plane $\operatorname{Im} \tau > 0$, the related complex variable $q = e^{2\pi i \tau}$ is in the (punctured) unit disk $0 < |q| < 1$.
An important derivation in this setting [call it say $\vartheta$] is defined as follows. If $f$ is a function of $q$, equivalently a function of $\tau$ with period $1$,
$$
\vartheta f = \frac{1}{2\pi i}\frac{d}{d\tau} f \qquad\text{in terms of }\tau
$$
or
$$
\vartheta f = q\frac{d}{dq} f \qquad\text{in terms of } q
$$
Thus, the logarithmic derivative in terms of $\tau$ is essentially your operator $T$ in terms of $q$:
$$
\frac{\vartheta f}{f} = \frac{1}{2\pi i}\frac{df/d\tau}{f} = q \frac{df/dq}{f} = T [f] .
$$
---
Some random examples (i)
$$
E\_2 = 24\frac{\vartheta \eta}{\eta} = 24 \;T [\eta]
$$
where $\eta$ is the Dedekind eta function and $E\_2$ is an Eisenstein series:
$$
\eta(q) = q^{1/24}\prod\_{n=1}^\infty(1-q^n)
\\
E\_2(q) = 1 - 24\sum\_{k=1}^\infty \sigma(k) q^k
$$
And (ii)
$$
T[j\_{3B}](\tau) = \frac{1}{2}E\_2(\tau) - \frac{3}{2}E\_2(3\tau)
$$
where
$$
j\_{3B}(\tau) = \frac{\eta(\tau)^{12}}{\eta(3\tau)^{12}}
$$
is a Hauptmodul for modular curve $X\_0(3)$. See [A030182](http://oeis.org/A030182).
---
Plug. These two examples copied from the appendix of [arXiv:2005.10733](https://arxiv.org/abs/2005.10733)
|
6
|
https://mathoverflow.net/users/454
|
402259
| 165,081 |
https://mathoverflow.net/questions/401905
|
0
|
Consider a $6\times 1$ continuous random vector
$$
\eta\equiv (\eta\_1,\eta\_2,..., \eta\_6)
$$
satisfying the following property:
$$
\underbrace{\begin{pmatrix}
\eta\_1\\
\eta\_2\\
\eta\_3
\end{pmatrix}}\_{\equiv x\_1} \sim \underbrace{\begin{pmatrix}
\eta\_1\\
\eta\_4\\
\eta\_5
\end{pmatrix}}\_{\equiv x\_2} \sim \underbrace{\begin{pmatrix}
\eta\_2\\
\eta\_4\\
\eta\_6
\end{pmatrix}}\_{\equiv x\_3} \sim \underbrace{\begin{pmatrix}
\eta\_3\\
\eta\_5\\
\eta\_6
\end{pmatrix}}\_{\equiv x\_4} \sim G
$$
where ``$\sim$'' denotes "distributed as" and $G$ is some distribution with support $\subseteq \mathbb{R}^3$.
**Question:** I am looking for necessary and sufficient conditions on the support of $G$ such that there exists a $4\times 1$ random vector $\epsilon \equiv (\epsilon\_0, \epsilon\_1,\epsilon\_2,\epsilon\_3)$ having an absolutely continuous distribution with full support on $\mathbb{R}^4$ and such that
$$ (\*) \quad \quad
\begin{aligned}
\eta\_1= \epsilon\_1-\epsilon\_0\\
\eta\_2= \epsilon\_2-\epsilon\_0\\
\eta\_3= \epsilon\_3-\epsilon\_0\\
\eta\_4= \epsilon\_1-\epsilon\_2\\
\eta\_5= \epsilon\_1-\epsilon\_3\\
\eta\_6= \epsilon\_2-\epsilon\_3\\
\end{aligned}
$$
In particular, I would like to understand the following:
A) if the distribution of $\epsilon$ has full support on $\mathbb{R}^4$, then $G$ CANNOT have full support on $\mathbb{R}^3$. Is this correct? Why?
B) if A is correct, then there should be certain boxes in
$\mathbb{R}^3$ where $G$ is zero. Can we characterise those boxes?
|
https://mathoverflow.net/users/42412
|
Conditions for existence of a distribution with full support
|
$\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$There is no necessary and sufficient condition **in terms of the support of $G$** for the following: there exists a $4\times 1$ random vector $\ep:=(\epsilon\_0, \ep\_1,\ep\_2,\ep\_3)$ having an absolutely continuous distribution with full support on $\R^4$ and such that for
\begin{equation}
\begin{aligned}
\eta\_1&:= \ep\_1-\ep\_0, \\
\eta\_2&:= \ep\_2-\ep\_0, \\
\eta\_3&:= \ep\_3-\ep\_0, \\
\eta\_4&:= \ep\_1-\ep\_2, \\
\eta\_5&:= \ep\_1-\ep\_3, \\
\eta\_6&:= \ep\_2-\ep\_3
\end{aligned}
\tag{1}
\end{equation}
we have
\begin{equation}
x\_1:=\begin{pmatrix}
\eta\_1\\
\eta\_2\\
\eta\_3
\end{pmatrix} \sim \begin{pmatrix}
\eta\_1\\
\eta\_4\\
\eta\_5
\end{pmatrix} \sim \begin{pmatrix}
\eta\_2\\
\eta\_4\\
\eta\_6
\end{pmatrix} \sim \begin{pmatrix}
\eta\_3\\
\eta\_5\\
\eta\_6
\end{pmatrix} \sim G.
\tag{2}
\end{equation}
Indeed, as was noted in Martin Hairer's comment, the map given by the first three definitions in (1) that maps the random vector $\ep$ to the random vector $x\_1$ is surjective and continuous. So, it is necessary that (the distribution $G$ of) $x\_1$ have full support on $\R^3$, because $\ep$ has full support on $\R^4$. (Indeed, take any nonempty open ball $B\_3\subset\R^3$. The preimage of $B\_3$ under the mentioned continuous map contains a nonempty open ball $B\_4\subset\R^4$. So, $P(x\_1\in B\_3)\ge P(\ep\in B\_4)>0$.)
On the other hand, (2) implies that $\eta\_1\sim\eta\_2$. Letting now $G$ be any probability distribution on $\R^3$ with full support whose first two one-dimensional marginals are not the same, we see that the condition $x\_1\sim G$ in (2) cannot be satisfied.
So, there is no necessary condition on the support of $G$ that would also be sufficient.
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1
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https://mathoverflow.net/users/36721
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402261
| 165,082 |
https://mathoverflow.net/questions/402049
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2
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Let $C$ be a smooth projective curve of genus $g$ with an involution $\iota: C \to C$. We have the quotient map $\pi: C \to C/\iota$, with $C/\iota$ a smooth curve of genus $h$.
The pullback map $\pi^{\*}: \text{Jac}(C/\iota) \to \text{Jac}(C)$ in this case is injective, and we define the Prym variety as the cokernel:
$$0 \to \text{Jac}(C/\iota) \to \text{Jac}(C) \to \text{Prym}(\pi) \to 0$$
Note that $\text{Prym}(\pi)$ is an Abelian variety of genus $g-h$. Now consider the action by $\iota^{\*}$ on $\text{Jac}(C)$. The induced action on $\text{Jac}(C/\iota)$ is trivial, i.e. line bundles pulled back from the quotient are $\iota$-invariant. I am nearly certain that the induced action on $\text{Prym}(\pi)$ is by $\pm 1$ and therefore has $2^{2g-2h}$ fixed points. But I am struggling to prove this. Anyone have any tips?
I'm unsure how to prove that a particular involution on an Abelian variety is simply the standard $\pm 1$ action.
(I think an equivalent statement is that the only deformations of invariant degree $0$ line bundles on $C$ come from deformations on $C/\iota$. But there must be a more direct way to argue.)
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https://mathoverflow.net/users/105661
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Induced action on Prym variety
|
Kapil's suggestion to express the tangent space as $H^1(C,\mathcal O\_C)$ is great in characteristic $0$. In characteristic $p$, and in particular characteristic $2$, the statement is still true, but you can't detecet it from the tangent space.
Instead, note that the statement is equivalent to the claim that for $L$ a line bundle of degree $0$ on $C$, $L \otimes \iota^\* L$ is isomorphic to the pullback of a line bundle from $C/\iota$.
One can check explicitly that $L \otimes \iota^\* L$ is isomorphic to the pullback of $\det (\pi^\* L )\otimes \det (\pi^\* \mathcal O)^{-1}$. To do this, note that $\pi^\* \pi\_\* L$ is an extension of $\iota^\* L (-D)$ by $L$, for $D$ the branch divisor, so its determinant is $L \otimes \iota^\* L(-D)$, and the $ \det (\pi^\* \mathcal O)^{-1}$ term cancels the $(-D)$.
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2
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https://mathoverflow.net/users/18060
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402280
| 165,086 |
https://mathoverflow.net/questions/402268
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1
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I am reading Louigi's lecture note on random trees and graphs [here](http://problab.ca/louigi/notes/ssprob2021.pdf). I get stuck on part (b), Exercise 1.2.3 on page 19, which says the following:
>
> Let $T\_n$ be uniformly drawn from $\mathcal{T}\_n$, the set of $n$-vertices labeled rooted trees, and $c(v;t)$ be the number of children of vertex $v$ in a fixed tree $t$. Then $\Pi\_n(c):=\frac{1}{n}\sharp\{u\in[n]:c(u;T\_n)=c\}$ converges in probability to $\mathbb{P}(Poisson(1)=c)$.
>
>
>
In part (a) he proved that $c(1;T\_n)$ converges in distribution to Poisson(1) distribution, which is clear to me using the so-called "line-breaking" construction in the lecture note. Then the above, as far as I'm concerned, is just a WLLN, so I want to check the Markov's condition:
$$\frac{1}{n^2}Var(\sum\limits\_{u\in[n]}1(c(u;T\_n)=c))\rightarrow 0.$$
However I am having problem controlling the "crossing-terms":
$$\frac{1}{n^2}\sum\limits\_{u\not=v}[\mathbb{P}(c(u;T\_n)=c(v;T\_n)=c)-\mathbb{P}^2(c(u;T\_n)=c)].$$
By line-breaking construction I can write out the probability of the first, by counting the number of labeled rooted trees with vertices 1 and 2 each has exactly c children:
\begin{aligned}
\mathbb{P}(c(1;T\_n)=c(2;T\_n)=c) &= \frac{\sharp\{t\in\mathcal{T}\_n:c(1;t)=c(2;t)=c\}}{n^{n-1}}\\
&= \frac{\sharp\{v\in[n]^{n-1}:\text{ there are exactly }c\text{ copies of 1,2 in }v=(v\_1,\cdots,v\_{n-1})\}}{n^{n-1}}\\
&= \frac{1}{n^{n-1}}{n-1\choose c}{n-c-1\choose c}(n-2)^{n-2c-1}\\
&= \frac{1}{(c!)^2}(1-\frac{2}{n})^{n-2c-1}\mathop{\Pi}\limits\_{k=1}^{2c}(1-\frac{k}{n}),
\end{aligned}
and the second probability, without the square, is(as proved in earlier context in the lecture note)
$$\mathbb{P}(c(1;T\_n)=c)=\mathbb{P}(Bin(n-1,\frac{1}{n})=c)={n-1\choose c}\frac{1}{n^c}(1-\frac{1}{n})^{n-1-c}.$$
Then I have no idea how to bound the difference. Any help is appreciated.
|
https://mathoverflow.net/users/174600
|
Empirical degree distribution of random $n$ vertices labeled rooted tree converges to Poisson distribution
|
Essentially, you want to show that
$$p\_{12}-p\_1^2\to0,\tag{1}$$
where
$$p\_{12}:=P(c(1;T\_n)=c(2;T\_n)=c),\quad p\_1:=P(c(1;T\_n)=c).$$
We have
$$p\_1=\binom{n-1}c\frac1{n^c}\Big(1-\frac1n\Big)^{n-1-c}
=\Big(1-\frac1n\Big)^{n-1-c}\frac1{c!}\prod\_{j=1}^c\Big(1-\frac jn\Big)\to\frac{e^{-1}}{c!}.$$
If your calculation of $p\_{12}$ were correct, then we would have
$$p\_{12}
=\Big(1-\frac2n\Big)^{n-2c-1}\prod\_{k=1}^{2c}\Big(1-\frac kn\Big)\to e^{-2},$$
so that (1) would fail to hold.
So, apparently the factor $1/(c!)^2$ is missing in your ultimate expression for $p\_{12}$.
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1
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https://mathoverflow.net/users/36721
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402283
| 165,087 |
https://mathoverflow.net/questions/402066
|
3
|
I have an intriguing and probably simple question: reading the articles and books of Wolfgang Arendt on Semigroups of Linear operators I found on many places properties of the Neumann Laplacean.
In W. Arendt, *Semigroups and evolution equations: functional calculus, regularity and kernel estimates*, Handbook of Differential Equations: Evolutionary Equations. Vol. 1. North-Holland, 2002, pages 1-85 (it can be seen [here](https://pdf.sciencedirectassets.com/275750/1-s2.0-S1874571704X80014/1-s2.0-S1874571704800033/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEHoaCXVzLWVhc3QtMSJGMEQCIGr7B1Ys0MtGCGlRjYjdliXo%2B%2BYZNaLUT5GZgfDgNeloAiArjt6HRoR5puTLQreuJufjgP%2BHghYmGilHIbK0f8VLGyqDBAij%2F%2F%2F%2F%2F%2F%2F%2F%2F%2F8BEAQaDDA1OTAwMzU0Njg2NSIMxFoN563xJJfTAHFyKtcDjZaYJo13RtY%2Bic1%2FU6AGKL1G%2BOgsnCAw1kIuVo3upp6LTB%2F1O5rKT35vmHUHlVhPRrITyX2%2BSAs0iCllHHUXaXgTfwPBCSCjVgyGeH9Z6PVsAyZe28kBWYl8P%2B8Rnlmj6GfEhv1zHTvdSxVs4390eqY%2FT67Mxh3xb2QYpgSbe3%2Fxe0U9SA8Fnkmr74kI8%2FKzDlwdxso0owdrkiF8HOtZf4ln3emQbYLQ%2FKnQrAbamDksJgJe6jTvx3VkGpDoNFgcjyW8c1KN%2B5YgfzbJlRYl9J2WxACfUx0lNMz0zh%2BFHye%2F0Mp8sUsxenO7AN%2Ff1FFvUPU%2F0lNkZlfZJ5UEJGn0%2FIpVVsC7OQZnX%2FSo8V64HVEbwA6TcD6Ol6bNx6dqTsRLeXlHPaXXezYYqdiYvBxMHp35YsCgNEQLj%2Bx2JD2xrlJZAvXMDATlTsrUhG6IQ8kbu1qs3oXomiCDAF2CBeqe%2FLucHnmd9gtTOWY5ShPEgG5%2F5x2009zC1JHZGgewzzzZ2%2FGitRu3gxtUDXF2CMN3vL%2FTM81w10Mc6IJlcM%2F%2FDKQB%2BFNMHdz2Z%2F7t5aQzgItoLD0oOmRlj9iA7P0hATqVJaDVo%2FWNnG68BXfLMBgw5io257IIDJM9MJ7Y%2BIgGOqYBwWzFrP5bKt64cy1cV%2FOfwY4XAze79x0RKsYwxFk2sjXy%2B2DDVdFNW2DiGdvyE1Mt6b6AUDcKJQgiu3Z1a377%2BpUQw71xbxF9T6UyG%2F8YX828RHudQdXka8FfobSrJ6R%2FL3%2BgiudgDRb6vtf7M3TKx%2B63Eh8vaRC2tmEBtqZRQq1WqSl8U8FOZpmnW38mJTTZ5lodAZYUGYTjB4tlgEeBZAt2%2BlHc9A%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20210819T111640Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYXTP6BI6O%2F20210819%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=2e2f5db38ce340b9ccb54fdc9707dfa7c41e08ce4db583d70228765e4c54d162&hash=04113dd051fb45f73f1cda5e3e3cf9adf04ab704daeb12de710164836b83e713&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S1874571704800033&tid=spdf-9d3b6b70-5436-4746-b235-e7030149c934&sid=415ad98c96b3e44254187a148604ba4e4751gxrqb&type=client)),
on page 69 we found an assertion that the Neumann Laplacean generates an ultracontractive semigroup. This means that the below property is satisfied only for any $1\geq t>0$ (see the definition from page 65).
**Here is my question**: How can we prove that the property is true for any $t\in (0,\infty)$? And why the definition of ultracontractivity is only for $t\in (0,1]$?
>
> For an open, bounded, connected and with an uniform Lipschitz boundary $\Omega\subseteq\mathbb{R}^2$ consider the semigroup of linear operators $S(t)\_{t\geq 0}$ generated by the **Neumann Laplacean**:
>
>
>
\begin{equation}
\Delta\_N:D(\Delta\_N)\to L^2(\Omega),\ D(\Delta\_N)=\left\{u\in H^1(\Omega)\ \big |\ \Delta u\in L^2(\Omega),\ \dfrac{\partial u}{\partial\nu}=0, \ \mathcal{H}^{1}\text{- a.e. on}\ \partial\Omega\right\}.
\end{equation}
>
> Then for any $1\leq p\leq q\leq +\infty$ there is a constant $c=c(\Omega,p,q)$ that possess the following property (called *ultracontractivity*):
>
>
>
\begin{equation}
\Vert S(t)\phi\Vert\_{L^q(\Omega)}\leq c t^{-\frac{N}{2}\left (\frac{1}{p}-\frac{1}{q}\right )}\Vert\phi\Vert\_{L^p(\Omega)},\ \forall\ \phi\in L^p(\Omega),\ \forall\ t\geq 0
\end{equation}
**P.S. It's a natural question, since in many other books like *Cazenave & Haraux - An introduction to semilinear evolution equations* (page 44) or *Barbu Viorel - Analysis and Control of Nonlinear Infinite Dimensional Systems* (page 31) the above property is proved for any $t>0$ in the case of Dirichlet Laplacean.**
|
https://mathoverflow.net/users/61629
|
Contractivity of Neumann Laplacean
|
It might be helpful to point out the following conceptual reasons why ultracontractivity estimates are mainly interesting for times close to $0$.
Let us consider the following general setting: We have a finite measure space $(\Omega,\mu)$, two integrability indices $1 \le p < q \le \infty$ and a $C\_0$-semigroup $(S(t))\_{t \ge 0}$ on $L^p(\Omega,\mu)$.
Generally speaking, we are interested in situations where one of the operators $S(t)$ (or all of them for $t > 0$) map $L^p$ into $L^q$ (which can be interpreted as somekind of "smooting" property). Ultracontractivity means that we have specific estimates of the operator norm from $L^p$ to $L^q$ for all $t > 0$.
Now here are some useful observations:
**The long-time behaviour does not depend one the choice of spaces.**
If, for some $t\_0 > 0$, we have $S(t\_0)L^p \subseteq L^q$, then there are constants $c\_1,c\_2,c\_3,c\_4 > 0$ such that
\begin{align\*}
\tag{1}
\|S(t)\|\_{L^p \to L^p} & \le c\_1 \|S(t)\|\_{L^p \to L^q} \le c\_2 \|S(t-t\_0)\|\_{L^q \to L^q} \\
& \le c\_3 \|S(t-2t\_0)\|\_{L^q \to L^p} \le c\_4 \|S(t-2t\_0)\|\_{L^p \to L^p}
\end{align\*}
for all $t > 2t\_0$. The first and the fourth inequality follow from the fact that $L^q$ embeds continuously into $L^p$ (since the measure space is finite), and the second and the third inequality follow from the semigroup law and the fact that $S(t\_0)$ is a bounded operator from $L^p$ to $L^q$ due to the closed graph theorem.
Inequality (1) shows that estimates for the norm $\|S(t)\|\_{L^p \to L^q}$ are not particularly interesting for large $t$, since they are not different from the estimates that one can prove for $\|S(t)\|\_{L^p \to L^p}$. For instance, if the semigroup is generated by the Dirichlet Laplacian, then $\|S(t)\|\_{L^p \to L^p}$ converges exponentially to $0$, and hence so does $\|S(t)\|\_{L^p \to L^q}$ (with the same rate).
It is also worthwhile to note that the long-term behaviour of $\|S(t)\|\_{L^p \to L^p}$ (and hence of $\|S(t)\|\_{L^p \to L^q}$) can be changed by rescaling the semigroup: if one substracts $a\operatorname{id}$ (for a real number $a$) from the semigroup generator, then the semigroup gets multiplied by $e^{-ta}$, so we are always able to enforce exponential decay of the $\|\cdot\|\_{L^p \to L^q}$-norm as $t \to \infty$, simply by a scalar shift of the generator.
**The short-time behaviour depends heavily on the choice of spaces.**
This is probably easiest to explain by considering an explicit example, say the Neumann-Laplace operator $\Delta\_N$ on a bounded domain, which generates a semigroup $(S(t))\_{t \ge 0}$.
The semigroup satisfies $\|S(t)\|\_{L^\infty \to L^\infty} \le 1$ for all $t \ge 0$ (this follows, for instance, from the positivity of the semigroup and the fact that the constant one-function is a fixed point of the semigroup) as well as $\|S(t)\|\_{L^1 \to L^1} \le 1$ for all $t \ge 0$ (this follows, for instance, by duality from the aforementioned $L^\infty$-estimate). Hence, by interpolation, $\|S(t)\|\_{L^p \to L^p} \le 1$ for all $p \in [1,\infty]$ and all $t \ge 0$. In other words, we have boundedness of the semigroup for all times when we consider the operator norm on a fixed $L^p$-space.
However, the situation changes significantly if we consider the norm $\|\cdot\|\_{L^p \to L^q}$ for $p > q$. To see this, fix a function $f$ which is in $L^p$ but not in $L^q$. By the strong continuity of the semigroup on $L^p$ we have $S(t)f \to f$ in $L^p$ as $t \downarrow 0$, but we must have $\|S(t)f\|\_{L^q} \to \infty$ since $f \not \in L^q$. This proves that $\|S(t)\|\_{L^p \to L^q} \to \infty$ as $t \downarrow 0$.
In other words, while the "smoothing effect" caused by the operator $S(t)$ is still present for small $t$ (i.e., $S(t)L^p \subseteq L^q$ for all $t > 0$), the effect becomes quantitively weaker in the sense that the operator norm from $L^p$ to $L^q$ explodes as $t \downarrow 0$.
So, *the interesting thing about an ultracontractivity estimate is that it gives us an explicit quantititve bound for the blow-up of $\|S(t)\|\_{L^p \to L^q}$ as $t \downarrow 0$.*
**Relation to Sobolev embeding theorems.**
I think the following relation to Sobolev embedding theorems is quite illuminating. As pointed out in the question, for the semigroup $(S(t))\_{t \ge 0}$ generated by the Neumann Laplace operator, one has the ultracontractivity estimate
$$
\tag{2}
\|S(t)\|\_{L^p \to L^q} \le c t^{-\frac{N}{2}(\frac{1}{p} - \frac{1}{q})} \quad \text{for } t \in (0,1],
$$
where $N$ is the dimension of the the underlying domain.
Now, if $\lambda > 0$ is a real number, then the resolvent of $\Delta\_N$ on $L^p$ is given by the integral
$$
R(\lambda,\Delta\_N)f = \int\_0^\infty e^{-\lambda t} S(t) f \, dt,
$$
for each $f \in L^p$; the integral converges as a Bochner integral in $L^p$. Now let us analyse whether the integral converges also converges in $L^q$, even if $f$ is only in $L^p$. To this end, we split the integral into the integrals $\int\_0^1$ and $\int\_1^\infty$:
* It follows from $(1)$ that $\int\_1^\infty \|e^{-\lambda t} S(t)f\|\_{L^q} \, dt < \infty$ for each $f \in L^p$.
* It follows from the ultracontractivity estimate $(2)$ that $\int\_0^1 \|e^{-\lambda t} S(t)f\|\_{L^q} \, dt < \infty$ for each $f \in L^p$ if the additional condition $\frac{N}{2}(\frac{1}{p}-\frac{1}{q}) < 1$ (equivalently, $\frac{2}{N} > \frac{1}{p} - \frac{1}{q}$) is satisfied.
Hence, if $\frac{2}{N} > \frac{1}{p} - \frac{1}{q}$, then the resolvent $R(\lambda,\Delta\_N)$ maps $L^p$ into $L^q$; since the range of the resolvent equals the domain of $\Delta\_N$, this means that the domain of the Neumann Laplace operator on $L^p$ is contained in $L^q$ if $\frac{2}{N} > \frac{1}{p} - \frac{1}{q}$.
If the underlying domain $\Omega$ in $\mathbb{R}^N$ has sufficiently smooth boundary, the domain of $\Delta\_N$ in $L^p$ is the Sobolev space $W^{2,p}$; so in this case, what we have shown before amounts to saying that the Sobolev embedding theorem $W^{2,p} \subseteq L^q$ holds if $\frac{2}{N} > \frac{1}{p} - \frac{1}{q}$. Since all we needed for the proof was the ultracontractivity property $(2)$, we conclude that ultracontractivity gives rise to an abstract version of the Sobolev embedding theorem.
More information in this direction can, for instance, be found in Chapter 2 of Davies' book "*Heat kernels and spectral theory*" (1989).
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5
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https://mathoverflow.net/users/102946
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402285
| 165,089 |
https://mathoverflow.net/questions/402274
|
5
|
Let $h(n,t) = \sum\limits\_{j = 0}^n {\binom
{\lfloor {\frac{n}{2}} \rfloor }{j}\binom
{\lfloor {\frac{n+1}{2}}\rfloor }{j}t^j \\ }.$
I am interested in the Hankel determinants $${D\_k}(n,t) = \det \left( {h(k + i + j,t)} \right)\_{i,j = 0}^{n - 1}.$$
These can easily be computed for $0 \leq k \leq 3.$
It seems that $${D\_4}(n,t) = {t^{\lfloor {\frac{{{n^2}}}{4}} \rfloor }}b(n,t)$$ with $b(n,t) = \sum\limits\_{j = 0}^{2n} \min({\binom{3+j}{3},\binom{2n+3-j}{3}})t^j.$
In order to prove this, I need the identity
$$a{(n,t)^2} = b(n - 1,t)\sum\limits\_{j = 0}^{n - 1} {{t^j}} - tb(n - 2,t)\sum\limits\_{j = 0}^n {{t^j}} $$
with $a(n,t) = \sum\limits\_{j = 0}^{n - 1} {(j + 1){t^j}} .$
Any idea how to prove this identity?
|
https://mathoverflow.net/users/5585
|
An interesting Hankel determinant
|
Denote $a\_n=a(n,t)$ and $b\_n=b(n,t)$. To help avoiding the **min** function, write
$$b\_n=\binom{n+3}3t^n+\sum\_{j=0}^{n-1}\binom{3+j}3\left[t^{2n-j}+t^j\right].$$
Notice that $a\_n=\frac{nt^{n+1}-(n+1)t^n+1}{(1-t)^2}$ and $\sum\_{j=0}^nt^j=\frac{1-t^{n+1}}{1-t}$. Your identity takes the form
$$(nt^{n+1}-(n+1)t^n+1)^2=(1-t)^3[(1-t^n)b\_{n-1}-t(1-t^{n+1})b\_{n-2}].$$
Now, as Mark Widon mentioned, try to read-off the coefficients of $t^k$.
**UPDATE.** Resorting back the original formulation of the claim
$$a\_n^2=b\_{n-1}\sum\_{j=0}^{n-1}t^j-t\,b\_{n-2}\sum\_{j=0}^nt^j,$$
I was able (after lots of routine algebraic simplification and reorganization) to rewrite the right-hand side as
\begin{align\*}
\left(\sum\_{j=0}^{n-1}(j+1)t^j\right)^2
&=\sum\_{j=0}^{n-1}\binom{3+j}3t^j+\sum\_{j=0}^{n-2}\left[\beta\_n-\beta\_{j+1}-\binom{n-j}3\right]\,t^{n+j} \\
&=\sum\_{j=0}^{n-1}\binom{3+j}3t^j+\sum\_{j=0}^{n-2}
\frac{(n - j - 1)(j^2 + 4jn + n^2 + 5j + 7n + 6)}6\, t^{n+j}
\end{align\*}
where $\beta\_k=\frac{k(k+1)(2k+1)}6$ (the sum of squares function).
Once we got this far, the next step is to compare the coefficients of $t^k$.
|
5
|
https://mathoverflow.net/users/66131
|
402289
| 165,090 |
https://mathoverflow.net/questions/402277
|
6
|
Does simple theory of types + ambiguity prove axiom of infinity?
The simple theory of types known as $\sf TST$ is a multi-sorted first order theory, syntactical restrictions include $\in$ being a dyadic symbol where the symbol on the right of it is one sort (type) higher than the one on the left, while the two symbols linked by $=$ must be of the same type.
*Axioms* (on top of multi-sorted axioms of identity):
**Extensionality:** $\forall x^{i+1} \, \forall y^{i+1}:\\ \forall z^i (z^i \in x^{i+1} \leftrightarrow z^i \in y^{i+1}) \to x^{i+1}=y^{i+1}$
**Comprehension:** if $\phi$ is a formula in which
$x^{i+1}$ doesn't occur, then: $$\exists x^{i+1} \, \forall y^i (y^i \in x^{i+1} \leftrightarrow \phi)$$
Now the schema of ambiguity is:
**Ambiguity:** If $\phi^+$ is the formula obtained from formula $\phi$ by raising all type indices in $\phi$ by one, then: $$\phi \iff \phi^+$$ It's well known that $\sf TST +Ambg$ is equi-interpretable with Quine's $\sf NF$ [[Specker](https://arxiv.org/pdf/1503.01406.pdf)]. The latter is known to prove *Infinity* [[Specker](https://arxiv.org/pdf/1503.01406.pdf)].
>
> Would that entail that the fomer must also prove *Infinity*?
>
>
>
That was the first question.
The second question is if the answer is to the positive, then clearly the above theory is purely motivated by a logic of types, and types are only needed in a relative manner, and clearly this doesn't need the particular values of types to matter other than their relative positions. There is a remote resemblance between ambiguity and axiom of reducibility of Russell's, although of course they are not the same principle. If Infinity is provable from the above purely logically motivated axioms for set theory, then in some sense this could be seen as a motivation for *logicism*, since infinity, a clearly mathematical axiom, is not axiomatized here! So my second question is:
>
> Had $\sf TST + Ambg$ been considered as motivating the program of logicism?
>
>
>
|
https://mathoverflow.net/users/95347
|
Does simple theory of types + ambiguity prove axiom of infinity?
|
The development of TST + Amb is arguably motivated by the logicist program, ultimately.
But Amb is not a purely logical principle, it is a conjecture past the logically provable facts, with unexpected consequences.
I dont think that TST + Amb can be taken to motivate logicism. It is a by-product of this program.
TST + Ambiguity proves every stratified theorem of NF, so it proves Infinity, disproves Choice, and so forth.
|
3
|
https://mathoverflow.net/users/345616
|
402292
| 165,092 |
https://mathoverflow.net/questions/402262
|
12
|
This is the second in a pair of questions. For the other see [Are representations in computable analysis the equivalent to countably-generated condensed sets?](https://mathoverflow.net/questions/402260/are-representations-in-computable-analysis-the-equivalent-to-countably-generated).
Dustin Clausen and Peter Scholze have a theory of *condensed sets* (see [Lectures in condensed mathematics](https://www.math.uni-bonn.de/people/scholze/Condensed.pdf)), which is a slightly different take on topology. For most cases, the behavior of the usual topology and the “condensed” one, align. However, for some quotient spaces, like $\mathbb{R} / \mathbb{Q}$, the usual quotient topology is indiscrete, so every map $f : \mathbb{R} / \mathbb{Q} \to \mathbb{R} / \mathbb{Q}$ is continuous. However, as a condensed set, more structure is preserved. Indeed, the “continuous” maps are exactly those coming from continuous functions $\mathbb{R} \to \mathbb{R}$ which commute with the equivalence relation. (See [this answer](https://math.stackexchange.com/questions/4044728/examples-of-the-difference-between-topological-spaces-and-condensed-sets/4199337#4199337) to [Examples of the difference between Topological Spaces and Condensed Sets](https://math.stackexchange.com/questions/4044728/examples-of-the-difference-between-topological-spaces-and-condensed-sets) for more motivation.)
I’m trying to understand condensed sets better and how they relate to topology as it comes up in constructive mathematics.
In constructive theories, for example dependent type theory, there is an interesting phenomenon that I’ve always found fascinating. Just from the axioms of type theory, we automatically get topological structure on types. For example, every constructively definable function $f : 2^\mathbb{N} \to 2^\mathbb{N}$ is continuous for the usual product topology applied to $2^\mathbb{N}$.
Now, many constructive theories also allow quotients. For example, in Coq you can add quotients, and similarly in Homotopy Type Theory quotients are implied by univalence and higher inductive types. While quotients may break computability in some theories, if I understand correctly, they don’t break continuity. (For example, you still can’t construct a term for a non-continuous function $f : 2^\mathbb{N} \to 2^\mathbb{N}$ in a constructive type theory with quotients, right?)
Now my question is as follows:
* **Does it seem that the induced topological structure on quotient spaces in constructive type theory is actually the condensed set structure of Clausen and Scholze?**
Here are some ways to make my question more formal:
1. Consider the quotient space $2^\mathbb{N} / \text{fin}$ of all binary sequences, mod two sequences being equivalent if they agree on all but finitely many sequences. Can we (in say HoTT with univalence, or MLTT with quotients) construct a term for a function $f : 2^\mathbb{N} / \text{fin} \to 2^\mathbb{N} / \text{fin}$ which is not a condensed set endomorphism? I assume not.
2. Is it possible to make this “induced topology” theory formal? For example, I’ve heard folks say that it is consistent with constructive mathematics that every function is continuous, but I don’t know what that means formally. For example, can we add an axiom to MLTT or some other type theory that says that every $f : A \to B$ is continuous for the *induced topologies* on $A$ and $B$? Or better, is there a topos model of MLTT where the types are topological spaces and the functions are continuous? (Sorry, if I’m not using the right terms here. I’m still getting used to categorical logic.)
3. If there is a way to make this formal, can this formalism be extended to type theories with quotients, and then can one replace topological spaces with condensed sets?
4. (Bonus) In HoTT, if indeed, one can endow all sets (homotopy 0-types) with a condensed set structure, what about higher homotopy types? If the sets can be though of as condensed sets, can say the n-groupoids be interpreted as condensed n-groupoids and so forth?
|
https://mathoverflow.net/users/12978
|
Are the “topologies” arising from constructive type theories with quotients actually condensed sets?
|
Not quite a complete answer, but:
There are models of constructive mathematics where sets are $T\_0$ second-countable (optionally zero-dimensional) topological spaces equipped with equivalence relations, morphisms are continuous maps which respect the equivalence relation, and pointwise-equivalent morphisms are equal. In these two models, quotients are taken simply by inflating the equivalence relation, and the result satisfies all of the usual rules of MLTT + Quotient Types. (We need to add an equivalence relation structure so that the quotients are well-behaved enough for exponents to exist.)
In particular, each functions $f : 2^{\mathbb{N}}/\operatorname{fin} \to 2^{\mathbb{N}}/\operatorname{fin}$ arises from some continuous function $g:2^{\mathbb{N}} \to 2^{\mathbb{N}}$
These models don't have a set of truth values (hence no power-set operation), but they can be upgraded to contain one. (Roughly speaking - the above models arise as categories of PERs over two different PCAs, and there's a general technique for upgrading these, called realisability topoi.) This is the smallest change I know to the category of topological spaces that results in a model of MLTT.
---
As for what "every function is continuous" means - there are three ways that I know of. The first is, to every set, assign a topology (we could say that the opens are the unions of preimages of arbitrary functions into $\mathbb{R}$, for instance), and then every function is continuous with respect to this topology. This could also be done in classical mathematics, but there all sets would be given the discrete topology. Constructively, it may be the case that we get nontrivial topologies, and may even happen to get the 'correct' topologies for spaces such as $2^{\mathbb{N}}$ or $\mathbb R$.
The second way gives a number of examples where this holds. It is consistent that all functions $2^{\mathbb{N}} \to \mathbb{R}$ are continuous. In the presence of countable choice (which is consistent with this; iirc both (and MP, mentioned below) hold in Johnstone's topological topos, which has FirstCountableTop as a full subcategory and which I *suspect* is a full subcategory of that of condensed sets), this implies the stronger theorem that *If $M$, $N$ are metric spaces with $M$ complete, then all functions $f:M \to N$ are continuous*. (If MP ($\not \geq\_{\mathbb{R}}$ implies $<\_{\mathbb{R}}$) holds, we can weaken "complete" to "complement of a subset of a complete metric space".)
The third way is to generalise to grothendieck topologies, which is the main way to make (grothendieck) topoi, which model MLTT + quotients. Here, we specify a small category of "test spaces" and specify some families of morphisms to be "coverings" (for example, we could take all compact second-countable spaces for test spaces, and finite joint surjections as coverings) satisfying some technical conditions. The resulting grothendieck topos includes the spaces from the test category, and doesn't add any new morphisms between them - so if all of your morphisms between test spaces were continuous, they will continue to all be in the larger topos. The condensed sets are formed very similarly to this, except that the category of test spaces is no longer small, so we need to deal with size issues to get a grothendieck topos, like Simon mentioned.
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4
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https://mathoverflow.net/users/100508
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402293
| 165,093 |
https://mathoverflow.net/questions/402295
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3
|
Let $d\_{KS}(F,G)= \sup\_{x} |F(x) -G(x)|$ be the Kolmogorov-Smirnov distance between two cdfs $F$ and $G$.
Question: Let $F\_m$ be a cdf of distribution with $m$ atoms and let $\Phi$ bet the distribution of standard normal. What can we say about
\begin{align}
g(m)=\inf\_{F\_m} d\_{KS}(F\_m,\phi)
\end{align}
In other words, we are looking for the best approximation of Gaussian by $m$ atoms.
I am especially interested if we know the lower bound on $g(m)$. I highly suspect that this problem is either solved or there are good bounds, but I couldn't find a solution in the literature.
|
https://mathoverflow.net/users/69661
|
Best approximation of normal with $m$ atoms in Kolmogorov-Smirnov distance
|
The best difference is $\frac{1}{2m}$, attained by the distribution with $m$ atoms, each with mass $\frac{1}{m}$, at the points where the cdf of the normal distribution takes the values $\frac{2i-1}{m}$ for $i$ from $1$ to $m$.
Proof that this is optimal: One atom must have mass at least $\frac{1}{m}$. Call this $x$. Then
\begin{align}
\frac{1}{m} & \leq F\_m(x+\epsilon) - F\_m(x-\epsilon) \\
& \leq |F\_m(x+\epsilon) -\Phi(x+\epsilon) | + | \Phi(x+\epsilon) -\Phi(x-\epsilon)| + |\Phi(x-\epsilon) - F\_m (x-\epsilon)|\\
& \leq d\_{KS} (F\_m, \Phi)+ | \Phi(x+\epsilon) -\Phi(x-\epsilon)| + d\_{KS}(F\_m,\Phi)
\end{align}
and as $\epsilon$ goes to $0$, $| \Phi(x+\epsilon) -\Phi(x-\epsilon)|$ goes to $0$ because $\Phi$ is continuous, so
$$\frac{1}{m} \leq 2 d\_{KS}(F\_m,\Phi)$$ showing the desired optimality.
Of course, this has nothing to do with the Gaussian, and works for any continuous distribution.
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4
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https://mathoverflow.net/users/18060
|
402298
| 165,095 |
https://mathoverflow.net/questions/402245
|
3
|
Here's a fairly easy fact from point-set topology that I'm having trouble finding a reference for. Say $X$ is a total order satisfying the least-upper bound property, and $S$ is a closed subset of it. Then the subspace topology on $S$ and the order topology on $S$ coincide.
Does anyone know what would be a reference I can cite for this fact so I don't have to take up space in my paper to include a proof?
Actually, I'm pretty certain I've seen a proof of this statement given either here or else on math.stackexchange, but I haven't been able to find it. Those aren't ideal references, obviously, but still seem preferable to taking up space including a proof myself...
Thank you all!
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https://mathoverflow.net/users/5583
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A closed subset of a Dedekind-complete order has subspace topology equal to order topology
|
While I agree that it's pretty direct to show, I was unable to find a reference for a proof of this fact myself (I thought it was in Willard, but I thumbed through my copy and failed to find it). So here's a proof in any case.
Let $S$ be a closed subset of a linear order $X$ with the least upper bound property, that is, every subset of $X$ with an upper bound has a least upper bound, and consequently, every subset of $X$ with a lower bound has a greatest lower bound. Note that any least upper bound or greatest lower bound for a subset of $S$ must belong to $S$ due to it being closed. Subbasic open sets for the subspace $S$ are given by $U\_b=(\leftarrow,b)\cap S$ and $V\_a=(a,\rightarrow)\cap S$ for $a,b\in X$; we will only show $U\_b$ is open in $S$'s restricted order topology, since the proof for $V\_a$ is identical (with orders reversed).
If $U\_b=S$ or $U\_b=\emptyset$ we are done; otherwise note that $U\_b$ is non-empty and bounded above, so it has a least upper bound $b'\in S$, and likewise $S\setminus U\_b=[b,\rightarrow)\cap S$ is nonempty and bounded below, so it has a greatest lower bound $b''\in S$.
If $b'\not\in U\_b$, then $U\_b=(\leftarrow,b')\cap S$ and we're done. If $b'\in U\_b$, then $b'<b\leq b''$, and $(b',b'')\cap S=\emptyset$. Thus $U\_b=(\leftarrow,b']\cap S=(\leftarrow,b'')\cap S$, finishing the proof.
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3
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https://mathoverflow.net/users/73785
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402304
| 165,098 |
https://mathoverflow.net/questions/402287
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6
|
**Question 1.** Let $\epsilon > 0$ and $V > 0$. Is there always a complete connected Riemannian manifold $M$ with
$$
\operatorname{diam} M < \epsilon\quad\text{ (small diameter)} \quad \text{and} \quad \operatorname{vol}M > V\quad\text{(large volume)}?
$$
In other words, can we construct worlds with room for arbitrarily many planets, but where any two planets are arbitrarily close to each other?
Note that $M$ has to be compact by Hopf-Rinow. I'm almost certain that the answer is 'yes', but I'm having a hard time finding an explicit sequence $M\_1, M\_2, M\_3,\ldots$ such that $\operatorname{diam}(M\_n)$ goes to zero while $\operatorname{vol}(M\_n)$ goes to infinity.
**Question 2.** (Fixed dimension) Is there a sequence of complete connected Riemannian manifolds $M\_1, M\_2, M\_3, \ldots$ with
$$ \dim M\_n = m, \quad \lim\_{n \to \infty}\operatorname{diam} M\_n = 0, \quad \text{and} \quad \lim\_{n \to \infty}\operatorname{vol}M\_n = \infty
$$
for some fixed $m \in \mathbb{N}$.
For this one, I have less intuition, but I'm more inclined towards 'no'. Note that for $m = 0, 1$, this is clearly impossible. For $m = 2$ my intuition is very strongly in favour of 'no'.
|
https://mathoverflow.net/users/123207
|
Superconnected spaces
|
The answer to both questions is positive, even in dimension 2.
Take a round sphere of diameter $\epsilon$, and make many, say $N$ little holes in it.
Then take $N$ spheres of diameter $\epsilon$ and make one little hole in each.
Then glue these $N$ spheres to the first sphere along the boundaries of the holes. The diameter of the resulting surface is about $3\epsilon$ while the volume (area) is approximately $(N+1)\epsilon$. Now make $\epsilon$ as small as you wish, while $N>V/\epsilon$.
Here is another construction. Take a round sphere $S$ of diameter $\epsilon$, think of it as the Riemann sphere, and consider the
$N$-fold ramified covering $S\_1\to S,\;z\mapsto z^n$. Here $S\_2$ is
also a sphere, and equip it with the pullback of the metric from $S$.
Then the solume of $S\_1$ will be $N\epsilon$ while diameter is at most
$2\epsilon$. If a smooth metric is desirable, approximate it with a smooth one.
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6
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https://mathoverflow.net/users/25510
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402307
| 165,100 |
https://mathoverflow.net/questions/402315
|
4
|
Does there exist an analog of Lagrange inversion formula in positive characteristic? Obviously, the formula is still valid for coefficient with index not divisible by the characteristic, but for the other ones I did not manage to find one.
|
https://mathoverflow.net/users/33128
|
Lagrange inversion formula in positive characterisic
|
There are many forms of Lagrange inversion. The ones that don't involve division by integers are valid in positive characteristic. For example:
Given a power series $R(t)$, there is a unique power series $f=f(x)$ such that
$f(x) = x R(f(x))$, and for any Laurent series $\phi(t)$ and $\psi(t)$ and any integer $n$ we have
$$[x^n]\phi(f)=[t^n]\bigl(1-tR'(t)/R(t)\bigr)\phi(t)R(t)^n$$
and
$$[x^n]\frac{\psi(f)}{ 1-xR'(f)}=[t^n]\psi(t)R(t)^n.$$
|
7
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https://mathoverflow.net/users/10744
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402318
| 165,103 |
https://mathoverflow.net/questions/402333
|
4
|
This is a repost of the [same question on math.SE](https://math.stackexchange.com/questions/4227470/unbounded-set-in-vg-has-an-unbounded-subset-in-v), which received several comments but no answers/comments on the first question.
---
Suppose $\kappa$ is a cardinal preserved in the generic extension $V[G]$. Let $Y \subseteq \kappa$ be an unbounded set in $V[G]$. Does there always exist an $X \in V$ such that $X \subseteq Y$ and $X$ is unbounded?
This question comes from Section 18 of James Cummings' [*Singular Cardinal Arithmetic*](https://projecteuclid.org/journalArticle/Download?urlId=10.1305%2Fndjfl%2F1125409326). In this context, he forced with the Levy collapses $\operatorname{Col}(\omega,\delta^{+\omega}) \times \operatorname{Col}(\delta^{+\omega+2},<\kappa)$, and fixed "$X \subseteq \gamma$ unbounded of order type $\delta\_V^{+\omega+1}$...". In the next line, he then says that "However, it is easy to see that there is $Y \in V$ with $Y \subseteq X$ unbounded".
This observation is certainly not easy for me, so my questions are:
1. Why do such a $Y$ exist?
2. Is the existence of such a $Y$ limited to forcing with Levy collapse?
Any help is appreciated.
**Note:** Andreas Blass provided a counterexample to the second question, which is a Mathias real.
|
https://mathoverflow.net/users/146831
|
Unbounded set in $V[G]$ has an unbounded subset in $V$?
|
Let $\mathbb{P} = \textsf{Col}(\delta^{+ \omega + 2}, < \kappa)$ and $\mathbb{Q} = \textsf{Col}(\omega, \delta^{+ \omega})$. Then $\mathbb{P}$ is $< \delta^{+ \omega + 2}$-closed so it doesn't add any new set of ordinals of order type $\leq \delta^{+ \omega + 1}$. Put $V\_1 = V^{\mathbb{P}}$ and note that all cardinals $\leq \delta^{+ \omega + 2}$ are preserved in $V\_1$. Suppose $V\_1^{\mathbb{Q}} \models \mathring{Y}$ has order type $\delta^{+ \omega + 1}$. Since $|\mathbb{Q}| < \delta^{+ \omega + 1}$, we can find a condition $p \in \mathbb{Q}$ such that $p$ forces $ \delta^{+ \omega + 1}$ many ordinals into $\mathring{Y}$. Put $X = \{\alpha: p \Vdash\_{\mathbb{Q}} \alpha \in \mathring{Y}\}$. Then $X \in V\_1$ and therefore by the previous observation, $X \in V$.
|
3
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https://mathoverflow.net/users/345950
|
402337
| 165,108 |
https://mathoverflow.net/questions/402339
|
8
|
It is well known that all even perfect numbers are of the form $N=(2^{q}-1).2^{q-1}$ with $M\_{q}:=2^{q}-1$ a Mersenne prime.
As the very defining property of such a perfect number is to fulfill the equality $\sigma(N)=2N$, one can see that this value is almost the sum of a geometric series. But another conceptual framework can emerge from this: namely that $\sigma(N)$ is (close to) a $2$-adic series.
As Euler showed unrigorously, one has $\sum\_{k=0}^{\infty}2^{k}=-1$, the latter being interpreted as convergence in $\mathbb{Z}\_{2}$, the ring of $2$-adic integers. So a proof of the existence of infinitely many even perfect numbers should be morally equivalent to the convergence of the sequence $(2N)\_{N}$ where $N$ runs over the even perfect numbers towards $0$ (edited after Wojowu's comment, I wrote $-1$ at first) in $\mathbb{Z}\_{2}$.
My question is: can this be generalized to hypothetical odd perfect numbers? Namely, should $\sigma(N)$ for $N$ an OPN be "close" to some $p$-adic series for some $p>2$? Is there some hint that this should indeed be the case?
Edited after JoshuaZ' supposedly incomplete but insightful answer:
The number $2$ in the perfect number defining equality $\sigma(N)=2N$ may be viewed as the Euler factor at $p=2$ of $\zeta(s)$ as $s$ tends to $1$, that is, $N=\sigma(N)\lim\_{s\to 1}\frac{\zeta\_{2}(s)}{\zeta(s)}$ where $\zeta\_{2}$ is the $2$-adic zeta function. Could we be able to prove that any perfect number $N$ whose smallest prime factor is $p$ fulfills $N=\sigma(N)\lim\_{s\to 1}\frac{\zeta\_{p}(s)}{\zeta(s)}$, this would imply $p=2$ and that no OPN exists.
Second edit: one can define the "numerical $p$-adic transform" of an L-function $F$ whose sequence of Dirichlet coefficients is $(a\_{n}(F))\_{n>0}$, so that $F(s)=\sum\_{n>0}a\_{n}(F).n^{-s}$ whenever $\Re(s)>1$, by $\mathcal{L}\_{p}(F):=\sum\_{n>0}a\_{n}(F)p^{n}$ in $\mathbb{Z}\_{p}$. That way perhaps a link between $\mathcal{L}\_{2}(\zeta)$ and RH could be made.
|
https://mathoverflow.net/users/13625
|
Can perfect numbers be seen $p$-adically?
|
(Not a complete answer but a bit too long for a comment.)
There's a fundamental difficulty here in proving the sort of result you envision. If there were some prime $p$ which had to divide every odd perfect number and larger odd perfect numbers had to be divisible by higher powers, that would work. But we can't right now rule out now even that for any given odd prime $p$, there are infinitely many odd perfect numbers with smallest prime factor $p$. As far as I'm aware, the closest thing we have to a restriction on that [are the results in this recent paper of mine, especially Theorem 8 on page 47](http://math.colgate.edu/%7Eintegers/v76/v76.pdf).
The closest thing in the literature might be some of what Tomohiro Yamada has done. One goal that Tomohiro Yamada has been working on has been to rule out certain families of exponents. One of [his strongest results of this sort is the following](https://arxiv.org/abs/1706.09341): Let $N$ be an odd perfect number and write $$N= q^ap\_1^{2e\_1}p\_2^{2e\_2}\cdots p\_{k-1}^{2e\_{k-1}}$$ where $q \equiv a \equiv 1$ (mod 4) and $p\_1$, $p\_2 \cdots $ $p\_{k-1}$ are distinct primes none equal to $q$. (Note that by Euler's theorem for odd perfect numbers we can always put one in this form.) Then if all the $e\_i$ are the same. That is, $e\_1=e\_2...=e\_{k-1}=e$, then we have $k \leq 2e^2 + 8e+3$. Note that since there are bounds on the size of an odd perfect number in terms of its number of distinct prime factors, this automatically says that there are only finitely many odd perfect numbers which all share the same exponent for all but one prime.
The obvious stronger goal to aim for is a theorem of the same flavor as above, but where instead of assuming that all the exponents are the same, construct a function $f(x)$ such that if all the $e\_i$ are bounded above by some constant $E$, then $k \leq f(E)$. The results from Tomohiro use high powered sieve theory, but it seems like not many people have thought that hard about extending them, so I can't speak about how plausible this goal is given the current power of machinery. It might be the sort of thing that a group of very dedicated people might prove or it might be completely out of reach with current technology.
That said, if one did have a theorem of the sort envisioned in the last paragraph, one would then have as a corollary that for any $m$ there are only finitely many odd perfect numbers not divisible by a perfect $m$th power of a prime. So this would move sort of in the direction you want. But even this would be a very weak move in that direction because there's no guarantee that the primes are the same in each case.
Finally, let me add one piece of almost completely unfounded speculation, of how one might be able to get a result of the sort you want. It isn't hard to write down a Dirichlet series in terms of $\zeta(s)$ where the series has a non-zero $n$th term if and only if $n$ is an odd perfect number. Sketch of argument: write down the Dirichlet series for $(\sigma(n)-2n)^2$ using that
$$\sum\_{n=1}^{\infty} \frac{\sigma(n)^2}{n^s} = \frac{\zeta(s)\zeta(s-1)^2\zeta(s-2)}{\zeta(2s-2)}. $$
(The above is a special case of a general theorem for a Dirichlet series for $\sigma\_a(n)\sigma\_b(n)$. See e.g. Theorem 305 in Hardy and Wright.)
Then remove all the even terms from the Dirichlet series, and you now have a series which has a term which is zero exactly where the odd perfect numbers are. One would hope that one could then maybe turn analytic methods to estimate the coefficients of this Dirichlet series. I've never been successful in getting anything non-trivial out of this, but it is possible that someone who is better at complex analysis than I am could make this work. But your question brings up a related idea had not occurred to me until now and which might be worth someone thinking about: how does the resulting series behave $p$-adically? Vague hand-wave to Kubota-Leopoldt goes here. If that could work it might be possible to go in the other direction, do some sort of $p$-adic approach to the Dirichlet series in question and see if that gives us new information about odd perfect numbers.
|
8
|
https://mathoverflow.net/users/127690
|
402351
| 165,114 |
https://mathoverflow.net/questions/402341
|
1
|
This is related to discrete dynamical systems, with the initial condition $X\_1$ being a random variable with a non singular distribution. The system is driven by the iteration $X\_{n+1} = g(X\_n)$ for some rather smooth mapping $g$. The purpose here is to find a mapping $g$ so that the invariant distribution (also called attractor or invariant measure or fixed point distribution solution of an integral equation) is pre-specified. This is an inverse problem in the sense that typically $g$ is known and we are looking for the invariant distribution. Here the opposite is true: we assume the invariant distribution is known, and we want to find $g$ that will guarantee that no matter what the non-singular distribution of $X\_1$ is, we end up with the the target invariant distribution. There are typically many possible solutions for $g$. The reason I am asking this question is because I believe I got the wrong solution in a simple case, and would like to know how to get it right.
The most basic example is when the invariant distribution is uniform on $[0, 1]$, that is, $F\_X(x) = x$. In that case, $g(x) = \{ bx \}$ (the fractional part of $bx$), regardless of $b$ ( an integer $>1$) will work. Here I am trying to solve the case where the invariant distribution (the CDF) is $F\_X(x) = x^2$, also with the same support domain being $[0, 1]$. Of course, this is a fist step towards dealing with much more complicated $F\_X(x)$.
So, here is my investigation about $F\_X(x)=x^2$, and my question is about what is wrong with my analysis. I know something is wrong. First, $g$ must be a many-to-one mapping. The easiest case is when $g$ is a two-to-one mapping, as in the logistic map where $g(x)=4x(1-x)$, with $x\in [0, 1]$. Besides that, if the density $f\_X$ is symmetric around $x\_0$, then $g(\cdot)$ can not be symmetric around $x\_0$. The reason I say this is because I am focusing on mappings $g$ of $[0, 1]$ that are symmetric around some point $x\_0$, and in this case, $x\_0=\frac{1}{2}$. That is, $g(x)=g(1-x)$.
**My wrong solution that needs a fix**
Let $g$ be a two-to-one mapping, with $g(x)=g(1-x)$. If the (known) invariant distribution is $F\_X(x)$, then $g$ must satisfy the functional equation $f\_X(x)= f\_X(h\_1(x))-f\_X(h\_2(x))$ with $h\_1(x)=g^{-1}(x)$ being one of the two inverses of $g(x)$ when $x<1/2$, and $h\_2(x)=1-h\_1(x)$ being the other inverse. Based on this, we must have, using the notation $y=g^{-1}(x)$,
$$x^2 = y^2 - (1-y^2)$$
which results in $y = (x^2+1)/2$. Inverting this again, we get $x=\sqrt{2y-1}$, valid if $y\geq \frac{1}{2}$. It follows that $g(x)=\sqrt{2x-1}$ if $\frac{1}{2}\leq x \leq 1$, and $g(x) = \sqrt{1-2x}$ if $0\leq x \leq \frac{1}{2}$. Yet the true mean of the invariant distribution ($F\_X(x) = x^2$ on $[0, 1]$) does not seem to match that resulting from $g(\cdot)$ at equilibrium, when $n=\infty$. The system in question is ergodic.
|
https://mathoverflow.net/users/140356
|
Invariant distributions for iterated random variables (stochastic dynamical systems)
|
To have here the invariant distribution with cdf $F$ given by $F(x)=x^2$ for $x\in[0,1]$, all that is needed is a change of variables.
More generally, let $F$ be the cdf of any non-atomic distribution supported on an interval $I$ in $\mathbb R$. Let $F^{-1}$ denote the inverse of the restriction of $F$ to $I$. Let $U$ be a random variable uniformly distributed on $[0,1]$. Then the cdf of the random variable
$$X:=F^{-1}(U)$$
is $F$, and
$$U=F(X).$$
Suppose that a function $g\_\*$ is such that the uniform distribution on $[0,1]$ is invariant for the dynamical system $U\_{n+1}=g\_\*(U\_n)$ -- that is, $U\overset D=g\_\*(U)$, where $\overset D=$ means the equality in distribution. Then $F(X)\overset D=g\_\*(F(X))$, that is,
$$X\overset D=g(X),$$
where
$$g:=F^{-1}\circ g\_\*\circ F.$$
So, the distribution with the prescribed cdf $F$ is invariant for the dynamical system $X\_{n+1}=g(X\_n)$, as desired.
In particular, if $F(x)=x^2$ for $x\in[0,1]$ and
$g\_\*(u)=\{bu\}$ for $u\in[0,1]$, then
$$g(x)=\sqrt{\{bx^2\}}$$
for $x\in[0,1]$.
|
3
|
https://mathoverflow.net/users/36721
|
402354
| 165,116 |
https://mathoverflow.net/questions/402353
|
1
|
I am looking for a classification of irreducible real representations of $\mathrm{SL}(2,\mathbb{R})$ of finite dimension (in the following by "representation" I mean a representation of finite dimension). There is a complete classification of complex representations of $\mathrm{SL}(2,\mathbb{C}).$ More precisely, if $V$ is an irreducible complex representation of $\mathrm{SL}(2,\mathbb{C})$, then $V$ is isomorphic to $\mathrm{Sym}^a(\mathbb{C}^2)$ for some $a \geq 0.$
Is there a similar classification in the case of real representations of $\mathrm{SL}(2,\mathbb{R})$? Is any irreducible real representation of $\mathrm{SL}(2,\mathbb{R})$ isomorphic to $\mathrm{Sym}^a(\mathbb{R}^2)$ for some $a \geq 0$?
|
https://mathoverflow.net/users/346215
|
Irreducible real representations of $\mathrm{SL}(2,\mathbb{R})$
|
First, the classification of *complex* representations of $\mathrm{SL}\_2(\mathbb{R})$ is the same as that of $\mathrm{SL}\_2(\mathbb{C})$. This is because any representation of the Lie group $\mathrm{SL}\_2(\mathbb{R})$ gives a representation of the Lie algebra $\mathfrak{sl}\_2(\mathbb{R})$. But by the universal property of complexification any such representation factors through $\mathfrak{sl}\_2(\mathbb{C})$. A priori these representations may lift only to the universal cover, but since all these representations factor through the map $\widetilde{\mathrm{SL}\_2(\mathbb{R})} \rightarrow \mathrm{SL}\_2(\mathbb{C})$ they factor through $\mathrm{SL}\_2(\mathbb{R})$.
Now, we still need to sort out the real forms of these representations, that is we need to sort out whether the irreps support an invariant symmetric bilinear form, an antisymmetric bilnear form, or are not self-dual. In particular, it's enough to just observe that every representation has a real form. But $\mathrm{Sym}^n(\mathbb{R}^2)$ is a direct construction of a real form of each irrep.
So the classification is exactly the same as in the complex case.
|
3
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https://mathoverflow.net/users/22
|
402362
| 165,119 |
https://mathoverflow.net/questions/402336
|
3
|
From what I understand, the [Borel construction](https://ncatlab.org/nlab/show/Borel+construction) takes a $G$-space $X$ and produces a topological space $X\times\_{G}\mathbf{E}G$―the homotopy quotient $X/\!\!/G$ of $X$ by $G$ in the $\infty$-category of spaces $\mathcal{S}$―satisfying
$$\mathrm{H}^\*(X\times\_G\mathbf{E}G;R)\cong\mathrm{H}^\*\_G(X;R).$$
One partiularly important example is given by $\mathbf{B}G$, the Borel construction / homotopy quotient of the point: $\mathbf{B}G\simeq\mathbf{E}G\times\_G\*$.
Moving to the pointed setting, one has the [pointed Borel construction](https://mathoverflow.net/questions/105161), which takes a pointed $G$-space $X$ and returns $\mathbf{E}G\_+\wedge\_{G}X$, the homotopy quotient of $X$ by $G$ in the $\infty$-category of pointed spaces $\mathcal{S}\_\*$. Concretely, it is given by
\begin{align\*}
\mathbf{E}G\_+\wedge\_{G}X &\overset{\mathrm{def}}{=} \frac{\mathbf{E}G\times\_G X}{\mathbf{E}G\times\_G\*},\\
&\cong \frac{\mathbf{E}G\times\_G X}{\mathbf{B}G}.
\end{align\*}
Now, $\mathbf{E}G\_+\wedge\_G\*\cong\*$, rather than $\mathbf{B}G$. But while $\*$ is the monoidal unit of $\mathcal{S}$, it is not that of $\mathcal{S}\_\*$, which is $S^{0}$. Hence it would be interesting to know:
**Question.** What is the pointed Borel construction $\mathbf{E}G\_+\wedge\_G S^0$ of the $0$-sphere? Is it related to $\mathbf{B}G\_+$?
|
https://mathoverflow.net/users/130058
|
What is the pointed Borel construction of the $0$-sphere?
|
Let's apply your definition (which I think has typos on the RHS - the two "+" subscripts on the EG should not be there I think). Let's model everything as topological spaces and do the calculation there. Let's model $S^0$ as the discrete set $\{p,q\}$ with basepoint $p$.
$$EG\_+\wedge\_GS^0:=\frac{EG\times\_G\{p,q\}}{EG\times\_G\{p\}}$$
$$\simeq \frac{EG\times\_G\{p\}\coprod EG\times\_G\{q\}}{EG\times\_G\{p\}}$$
$$\simeq \frac{BG\coprod BG}{BG}$$
$$\simeq \text{pt} \coprod BG $$
$$\simeq BG\_+$$
|
2
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https://mathoverflow.net/users/163893
|
402378
| 165,125 |
https://mathoverflow.net/questions/402375
|
4
|
Let $A$ be a ring. Is the sequence \begin{align} \textstyle A \to \prod\_{\mathfrak{p}} A\_{\mathfrak{p}} \rightrightarrows \prod\_{\mathfrak{p}\_{1},\mathfrak{p}\_{2}} A\_{\mathfrak{p}\_{1}} \otimes\_{A} A\_{\mathfrak{p}\_{2}} \end{align} exact? Here the products are over all prime ideals of $A$.
*Thoughts:*
1. Since $A \to \prod\_{\mathfrak{p}} A\_{\mathfrak{p}}$ is faithfully flat, the sequence \begin{align} \textstyle A \to \prod\_{\mathfrak{p}} A\_{\mathfrak{p}} \rightrightarrows (\prod\_{\mathfrak{p}} A\_{\mathfrak{p}}) \otimes\_{A} (\prod\_{\mathfrak{p}} A\_{\mathfrak{p}}) \end{align} is exact, so it is enough to show that \begin{align} \textstyle (\prod\_{\mathfrak{p}} A\_{\mathfrak{p}}) \otimes\_{A} (\prod\_{\mathfrak{p}} A\_{\mathfrak{p}}) \to \prod\_{\mathfrak{p}\_{1},\mathfrak{p}\_{2}} A\_{\mathfrak{p}\_{1}} \otimes\_{A} A\_{\mathfrak{p}\_{2}} \end{align} is injective.
2. If $A$ is an integral domain, the question is equivalent to asking that $A = \bigcap\_{\mathfrak{p}} A\_{\mathfrak{p}}$ inside its fraction field. See e.g. [this](https://math.stackexchange.com/questions/630752/an-integral-domain-a-is-exactly-the-intersection-of-the-localisations-of-a-a) and a related question [here](https://mathoverflow.net/questions/293224/localization-and-intersection).
3. Here I'm asking more-or-less whether sections of $\mathcal{O}\_{X}$ satisfies descent with respect to the morphism $\coprod\_{x \in X} \operatorname{Spec} \mathcal{O}\_{X,x} \to X$ (which is not an fpqc cover). It's at least true that morphisms of quasi-coherent sheaves do not satisfy descent for such maps, see [this](https://mathoverflow.net/questions/127362/counter-example-to-faithfully-flat-descent).
|
https://mathoverflow.net/users/15505
|
Descent for the "localizations at all primes" ring map
|
Let $S$ be a compact, totally disconnected topological space whose topology is not discrete.
Let $k$ be a field and let $A$ be the ring of locally constant $k$-valued functions on $S$.
Then I claim the kernel of $\prod\_{\mathfrak p} A\_{\mathfrak p} \to \prod\_{\mathfrak p\_1,\mathfrak p\_2} A\_{\mathfrak p\_1} \otimes A\_{\mathfrak p\_2}$ is the ring of $k$-valued functions on $S$ with no local constancy condition and thus is isomorphic to $A$.
To see this, note that every point of $S$ defines a prime ideal of $A$, and all prime ideals arise this way. For these prime ideals, $A\_{\mathfrak p}$ is just $k$, so $\prod\_p A\_{\mathfrak p}$ is the ring of $k$-valued functions on $S$ with no local-constancy condition.
Because these localizations are just fields, $A\_{\mathfrak p\_1} \otimes A\_{\mathfrak p\_2}$ vanishes unless $\mathfrak p\_1=\mathfrak p\_2$, in which case it is the field $k$ again, but the map to such a $k$ is subtracting the two pullbacks, i.e. subtracting the two identity maps $k \to k$, and thus vanishes.
Thus, the arrow is zero and so the kernel is again the ring of functions with no local-constancy condition, as desired.
Maybe with a Noetherian hypothesis it's true?
|
7
|
https://mathoverflow.net/users/18060
|
402379
| 165,126 |
https://mathoverflow.net/questions/402372
|
6
|
Consider this ODE on $[1, \infty)$
$(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - ({4a} + m(m+1))f(r) = -4af(1) $
with initial conditions
$\frac{a}{1-2a} f(1) + f'(1) = C, \qquad \lim\_{r\to \infty} f(r) = 0$
where $0\leq a < \frac{1}{2}$, $m$ is a positive integer, and $C \in \mathbb{R}$.
I want to ask if there exists a unique solution (at least for $a$ small enough).
If $a=0$, then this becomes the Euler equation:
$r^2f''(r) + 2r f'(r) - (m(m+1))f(r) = 0 $
$ f'(1) = C, \qquad \lim\_{r\to \infty} f(r) = 0$
which we know the unique solution is:
$f(r) = \frac{-C}{\alpha r^{\alpha}}$
where $\alpha = \frac{1}{2} + \frac{\sqrt{1+4m(m+1)}}{2}$
Can I prove existence and/or uniqueness for the $a>0$ case using some kind of continuity method?
I know for instance that injectivity is a continuous property for elliptic operators, and one has the method of continuity to prove surjectivity of a 1-parameter family of elliptic operators.
Is there something similar in this context?
Any help or references is appreciated.
$\textbf{EDIT} $: I wrote the equations incorrectly above. I apologize for that. I allowed the right side to decay to $0$ and so I believe it's possible to prove existence now. Here are the correct equations:
$(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - \left(\frac{4a^2}{r(r-2a)} + m(m+1)\right)f(r) = -\frac{4a^2}{r(r-2a)}f(1)+ \frac{4a(1-2a)}{(1-a)r(r-2a)} C $
with initial conditions
$f'(1) = C, \qquad \lim\_{r\to \infty} f(r) = 0$
There is an ODE that is somehow related to the above non-local differential equation.
$(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - \left(\frac{4a^2}{r(r-2a)} + m(m+1)\right)f(r) = -\frac{2a}{r(r-2a)} D $
with initial conditions
$\frac{2a}{1-2a}f(1) + \frac{2}{1-a}f'(1) = D, \qquad \lim\_{r\to \infty} f(r) = 0$
where $D$ is any real number.
|
https://mathoverflow.net/users/138705
|
Existence and uniqueness of an Euler-type ODE with varying parameters
|
As Iosif said, in general the system you specified does not admit a solution. Here we will give a more pedestrian argument using only comparisons.
Monotonicity
------------
**Claim**: if a solution exists, and $f(1) > 0$, then the function is monotonically decreasing; if $f(1) < 0$, then the function is monotonically increasing.
*Proof*: we will focus on the positive case. The negative case is similar.
Let $\zeta = \frac{4a}{4a + m(m+1)} \in (0,1)$ (if $a\in (0,\frac12)$). The second derivative test shows that $f$ cannot have a local maximum with $f(r) > \zeta f(1)$ or a local minimum with $f(r) < \zeta f(1)$. This immediately implies monotonicity in light of $f(1) > 0 = \lim f(r)$.
>
> *More details on monotonicity proof*: Given $f(1) > f(\infty) = 0$. Suppose $f$ were not monotonic. Then there exists $r\_m, r\_M$ with $1 < r\_m < r\_M < \infty$ such that $f(r\_m) < f(r\_M)$.
>
>
> I claim that $f(r\_M) > 0$ and $f(r\_m) < f(1)$. Suppose not: if $f(r\_M) \leq 0$ then $f(r\_m) < 0$ and $f$ would have a negative local minimum, in contradiction to the second derivative test. If $f(r\_m) \geq f(1)$ then $f$ would have a local maximum with value $> f(1)$, against in contradiction to the second derivative test.
>
>
> Therefore there must exist a local minimum $s\_m\in (1,r\_M)$ with $f(s\_m) \leq f(r\_m)$ and a local maximum $s\_M\in (r\_m,\infty)$ with $f(s\_M) \geq f(r\_M)$.
>
>
> We have therefore established
> $$ \zeta f(1) \leq f(s\_m) < f(s\_M) \leq \zeta f(1) $$
> which is a contradiction.
>
>
>
In particular, we must have $f' \leq 0$ on $[1,\infty)$.
Comparison
----------
We have then
$$ f'(1) - f'(r) = -\int\_1^r f''(s) ~ds = \int\_1^r \frac{4a}{s^2 - 2as} f(1) - \frac{4a + m(m+1)}{s^2 - 2a s} f(s) + \frac{2(s-a)}{s^2 - 2as} f'(s) ~ds $$
Since $f'$ is signed, we know that it is absolutely integrable on $[1,\infty)$. Since $f$ is monotonic (and hence bounded) the second integrand is also absolutely integrable. We conclude then that $\lim\_{r\to\infty} f'(r)$ exists.
Since $\lim\_{r\to\infty} f(r) = 0$, we must have also $\lim\_{r\to\infty} f'(r) = 0$.
But now writing $f'(r) = - \int\_r^\infty f''(s) ~ds$ using the above formula, we see that asymptotically $|f'(r)| \sim \frac1r$ (coming from the $4a f(1)$ term if it is non-zero; the other two terms can both be bounded by $O(1/r) f(r) = o(1/r)$). But this contradicts the integrability of $f'(r)$.
And hence we have proved:
**Claim**: no solution can exist with $f(1) \neq 0$.
Uniqueness
----------
When $f(1) = 0$, the same maximum principle argument shows that $f$ must be identically zero. This shows that
**Theorem** The only solution to your system is $f \equiv 0$, with $f(1) = f'(1) = 0$ and $C = 0$.
Final remark
------------
Heuristically, if you want to look for asymptotically constant solutions to your equation, you probably want it to converge to $\zeta f(1)$ in the limit.
|
6
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https://mathoverflow.net/users/3948
|
402384
| 165,129 |
https://mathoverflow.net/questions/402385
|
1
|
Let us define $M\_0=2^n$ for $n\in\mathbb{N}$. Let $\ell\in\mathbb{N}$ be the number of random variables we are working with. For $1\leqslant i\leqslant\ell$, we define $M\_i$ to be a random variable following a binomial distribution with parameters $M\_{i-1}$ and $2^{-n}$.
I'm interested in computing $\mathbb{P}\left[M\_\ell=1\middle|M\_\ell\geqslant1\right]$. To be fair, I only needed to lower-bound this probability, which is easy to do by computing $\mathbb{P}\left[M\_1=1\middle|M\_1\geqslant1\right]\approx\frac{\mathrm{e}^{-1}}{1-\mathrm{e}^{-1}}>\frac12$, but, out of curiosity, I was wondering if it was possible to get a closed-form for $\mathbb{P}\left[M\_\ell=1\middle|M\_\ell\geqslant1\right]$, and I do not know of any theorem that would help me with this.
Intuitively, this probability should decrease exponentially fast to $1$, and it might be possible to get the closed-form expression of the probability for $\ell\leqslant3$ by hand, but it will quickly become tedious to operate that way. Is there any clever way I'm missing for this?
|
https://mathoverflow.net/users/178595
|
Distribution of a random variable obtained by chaining distribution
|
Abbreviate $N=M\_0=2^n$ and $b=1/N$. As described, the conditional distribution of $M\_i$, given that $M\_{i-1}=n$, is binomial with parameters $n$ and $b$. The conditional generating function of $M\_i$ is therefore
$$
E[s^{M\_i}\mid M\_{i-1}]=(1-b+bs)^{M\_{i-1}}.
$$
Using $g\_i$ to denote the generating function of $M\_i$, this leads to the recursion
$$
g\_{i}(s) = g\_{i-1}(1-b+bs),
$$
and, because $g\_0(s) = s^N$, by recursion,
$$
g\_i(s) = (1-b^i+b^is)^N, \qquad0\le s\le 1.
$$
In other words, $M\_i$ has the binomial distribution with parameters $N$ and $b^i$.
In particular,
$$
P[M\_i=1\mid M\_i\ge 1] = {Nb^i(1-b^i)^{N-1}\over 1-(1-b^i)^N}.
$$
|
1
|
https://mathoverflow.net/users/42851
|
402386
| 165,130 |
https://mathoverflow.net/questions/402374
|
2
|
Assume $\varepsilon \in [0,1/2]$. Consider the discrete-time random walk $X\_0 = 0$, $X\_{t+1} - X\_t \sim f(X\_t) \delta\_0 + (1-f(X\_t))\operatorname{Rademacher}$, where $\delta\_0$ is the Dirac delta on zero and $\operatorname{Rademacher}$ is the Rademacher distribution, and
$$
f(x) = \begin{cases} 1-\varepsilon & x > 0 \\ \frac{1}{2} & x=0 \\ \varepsilon & x<0. \end{cases}
$$
I am interested in the probability $O\_T = \mathbb{P}[X\_T > 0]$, and would like to show that $O\_T > 1/2$ for sufficiently large $T$.
I tried to prove this via a continuum approximation to SDE and generalized arcsine laws and it seemed very hard for me.
I am wondering whether there is more general theory, as it seems to me that higher diffusion in some part of the space and appropriate boundary behavior should lead to the random walk spending more time in the "less diffusive" part.
I'm grateful for any pointers or proof ideas.
|
https://mathoverflow.net/users/346420
|
Occupation time of non-stationary random walk
|
Here is a back of envelope computation that you should be able to make rigorous without too much trouble.
Let's condition upon the last hit of $0$. The passing time from it to $T$ is large with probability close to $1$ (all you need to show for this is that the probability to be in any particular position tends to $0$). Let that time be $S$.
Let $p$ be the probability of moving when you are to the right of $0$ and let $q$ be the probability of moving when you are to the left of $0$. Then the probability to end up on the right divided by the probability to end up on the left under such conditioning is the same as the probability to stay above $0$ for $S$ steps in a lazy random walk with the probability $p$ of movement divided by the same probability with $q$ instead of $p$.
The first random walk is effectively the standard random walk with about $pS$ steps and the second one is the standard random walk with about $qS$ steps. The corresponding probabilities for the standard random walk are well-known to be approximately proportional to the inverse square root of the number of steps, so we get $\frac{P(X\_T>0)}{P(X\_T<0)}\to\sqrt{\frac qp}=\sqrt{\frac{1-\varepsilon}{\varepsilon}}$ as $T\to\infty$ in your setting.
It all should be well-known with quantitative estimates of the convergence speed but I leave it to better experts than I to provide a reference.
|
3
|
https://mathoverflow.net/users/1131
|
402393
| 165,132 |
https://mathoverflow.net/questions/402397
|
9
|
I am looking for a proof of the following claim:
>
> Let $H\_n$ be the nth [harmonic number](https://en.wikipedia.org/wiki/Harmonic_number). Then,
> $$\frac{\pi^2}{12}=\ln^22+\displaystyle\sum\_{n=1}^{\infty}\frac{H\_n}{n(n+1) \cdot 2^n}$$
>
>
>
The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJw9yj0KgDAMQOG9J0la0SZr6Sg4eoIOBZFCjWLN_f0Z3B6PT3Mk730wEwjGoxUQR-hGrcsZTItNN5BIneYOXoIDcBL7lP0kYjDHWeQCqPsKjIldQ0v9_-eSeCDGcAMuIR0B&lang=gp&interacts=eJyLjgUAARUAuQ==).
|
https://mathoverflow.net/users/88804
|
An infinite series involving harmonic numbers
|
Denoting $H\_0=0$, we have $$\sum\_{n=1}^\infty \frac{H\_n}{n(n+1)2^n}=\sum\_{n=1}^\infty \left(\frac1n-\frac1{n+1}\right)\frac{H\_n}{2^n}=\sum\_{n=1}^\infty \frac{1}n\left(\frac{H\_n}{2^n}-\frac{H\_{n-1}}{2^{n-1}}\right)\\=\sum\_{n=1}^\infty\frac1{n^22^n}-\sum\_{n=1}^\infty\frac{H\_n}{(n+1)2^{n+1}}.$$
It is well known that $\sum\_{n=1}^\infty\frac1{n^22^n}=\frac{\pi^2}{12}-\frac{\log^2 2}2$ (see the value of ${\rm Li}\_2(1/2)$ [here](https://en.wikipedia.org/wiki/Polylogarithm)). Thus, it remains to show that $$\sum\_{n=1}^\infty\frac{H\_n}{(n+1)2^n}=\log^2 2.$$
For this, we take the square of the series $$
\log 2=-\log\left(1-\frac12\right)=\frac12+\frac1{2\cdot 2^2}+\frac1{3\cdot 2^3}+\ldots$$
to get $$\log^2 2=\sum\_{a,b=1}^\infty \frac1{ab2^{a+b}}=\sum
\_{a,b=1}^\infty \frac1{(a+b)2^{a+b}}\left(\frac1a+\frac1b\right)=\sum\_{n=2}^\infty\frac1{n2^n}2H\_{n-1}=\sum\_{n=1}^\infty\frac{H\_n}{(n+1)2^{n}}.$$
|
17
|
https://mathoverflow.net/users/4312
|
402405
| 165,134 |
https://mathoverflow.net/questions/401735
|
5
|
Let $W^{2,2}(\mathbb R)\newcommand{\R}{\mathbb R}\newcommand{\C}{\mathbb C}\newcommand{\N}{\mathbb N}$ denote the Sobolev space as defined in chapter 5 of [Evans' PDE book](https://bookstore.ams.org/gsm-19-r) and consider the linear operator
\begin{equation\*}\begin{split}T: D(T)&\to L^2(\mathbb R), \\ \phi&\mapsto \phi''.\end{split}\end{equation\*}
Here, $D(T):=W^{2,2}(\mathbb R)$ is a dense subset of $L^2(\mathbb R)$ and $T$ will be considered as a densily defined operator on $L^2(\mathbb R)$.
---
**My question.** Is $T$ self-adjoint? By self-adjoint I mean that $\langle T\psi, \tilde\psi\rangle = \langle \psi, T\tilde\psi\rangle$ for all $\psi,\tilde\psi\in D(T)$, and that $D(T)=D(T^\*)$, where $D(T^\*)$ is defined as the set of all $\tilde\psi\in L^2(\mathbb R)$ such that the linear operator $$T^\*\_{\tilde\psi}: D(T)\to\mathbb R, \psi\mapsto\langle{T\psi, \tilde\psi}\rangle$$ is bounded.
---
**My attempt.** A proof of the first property: Let $\psi, \tilde\psi\in W^{2,2}(\R)$. We thus need to show that
\begin{equation\*}
\int\_{\R} \psi''\tilde\psi=\int\_\R \psi\tilde\psi''.
\end{equation\*}
Let $(\phi\_n)\_{n\in\N}$ be a sequence of functions in $C\_{\text c}^\infty(\R)$ converging in $W^{2,2}$ to $\tilde\psi$. (Such a sequence exists, see for instance Lemma 23 of <https://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/.>)
Now, by Definition of the weak derivative,
\begin{equation}\label{eq:phi n prime prime}\tag{1}
\int\_{\R}\psi''\phi\_n = \int\_\R \psi\phi\_n''
\end{equation}
for all $n\in\N$. Now, from the Cauchy-Schwarz inequality, we get
\begin{equation\*}
\left\lvert\int\_\R \psi'' \phi\_n - \int\_\R \psi'' \tilde\psi\right\rvert \le \int\_\R \lvert\psi''\rvert\lvert\phi\_n-\tilde\psi\rvert\le \lVert\psi''\rVert\_{L^2} \lVert\phi\_n-\tilde\psi\rVert\_{L^2}.
\end{equation\*}
Analogously,
\begin{equation\*}
\left\lvert\int\_\R \psi \phi\_n''-\int\_\R \psi \tilde\psi''\right\rvert\le \int\_\R \lvert\psi\rvert \lvert\phi\_n''-\tilde\psi''\rvert\le \lVert\psi\rVert\_{L^2} \lVert\phi\_n''-\tilde\psi''\rVert\_{L^2}.
\end{equation\*}
But $\phi\_n\to \tilde\psi$ in $W^{2,2}$ implies that $\lVert\phi\_n-\tilde\psi\rVert\_{L^2}\to 0$ and $ \lVert\phi\_n''-\tilde\psi''\rVert\_{L^2}\to 0$ as $n\to\infty$. Therefore, we do indeed have
\begin{equation\*}
\int\_{\R} \psi''\tilde\psi=\int\_\R\psi\tilde\psi''.
\end{equation\*}
---
However, how can one prove that $D(T)=D(T^\*)$ ?
|
https://mathoverflow.net/users/129831
|
Is the second weak derivative a self-adjoint operator?
|
In Theorem 4.7b) of [these lecture notes](https://www.math.kit.edu/iana3/%7Eschnaubelt/media/st-skript15.pdf), the following general result is proved:
>
> Let $X$ be a Hilbert space and let $A$ be densely defined, closed and symmetric. Then $A$ is self-adjoint if and only if the spectrum of $A$ is contained in the real axis.
>
>
>
In Example 4.8d) of the same lecture notes, this is applied to prove the self-adjointness of the Laplacian $A = \Delta$ on $L^2(\mathbb{R}^d)$ with domain $D(A) = W^{2, 2}(\mathbb{R}^d)$, which specializes to the second derivative for $d=1$.
|
1
|
https://mathoverflow.net/users/136913
|
402406
| 165,135 |
https://mathoverflow.net/questions/402399
|
4
|
I have seen the claim that Beilinson Lichtenbaum implies that higher algebraic $K$ groups coincides with etale ones integrally in high enough degrees. Is this statement accurate? What conditions are required and how to derive it?
|
https://mathoverflow.net/users/127776
|
Etale $K$ theory coincides with algebraic one in high enough degrees
|
To my knowledge the most general known statement has been proven by Clausen and Mathew in their paper [Hyperdescent and étale K-theory](https://arxiv.org/abs/1905.06611) as Theorem 1.2. The precise conditions on your commutative ring (or more generally algebraic space) are a bit technical to summarize, but are very general and give explicit bounds. Away from the residue chacteristics one can apply the Voevodsky-Rost norm residue theorem and at the residue characteristics they manage to apply a reduction to topological cyclic homology ($TC$). I recommend looking at their paper for a more detailed explanation.
|
5
|
https://mathoverflow.net/users/2039
|
402408
| 165,136 |
https://mathoverflow.net/questions/402308
|
11
|
In ZF(C), one can easily get a class partition of $V$, we can even get an $\mathrm{Ord}$-partition using the Cumulative hierarchy: $P=\{V\_{α+1}\setminus V\_α\mid α∈\mathrm{Ord}\}$, such a partition let us do stuff like Scott's trick: given a class $A$, we can look at $A∩V\_β$ where $β$ is the minimal $β$ so that intersection is not empty.
But the fact that $\bigcup P=V$ is equivalent to the axiom of regularity.
I remember somewhere reading that in $ZF\text{-regularity}$ we can't have Scott's trick like trick.
We can formulate Scott's trick as:
>
> There exists an $\mathrm{Ord}$-partition of $V$ (or equivalently - there exists cumulative hierarchy that sums up to the universe).
>
>
>
In a sense this version of Scott's trick is a "global trick".
---
While the intuition tells me that $\text{regularity}$ does not follow from this version of Scott's trick, is this true?
|
https://mathoverflow.net/users/113405
|
Scott's trick without regularity
|
Yes! Extend $\sf ZF - Reg.$ with the existence of a unique Quine atom $\sf Q=\{Q\}$, take $V$ to be the hierarchy over $\sf Q$, that is: $$\begin{align} & V\_\emptyset = \sf Q \\ & V\_{\alpha+1}= \mathcal P (V\_{\alpha}) \\ & V\_\lambda= \bigcup\_{\alpha < \lambda} V\_\alpha, \text {for limit } \lambda \\ & V= \bigcup \_{\alpha \in \mathrm{Ord}} V\_\alpha \end {align}$$
clearly this $V$ has an $\mathrm{Ord}$-partition.
|
5
|
https://mathoverflow.net/users/95347
|
402412
| 165,138 |
https://mathoverflow.net/questions/402366
|
0
|
Given two freely independent random hermitian matrices $A$ and $B$ following laws $\mu, \nu$, one can compute the empirical spectral distribution of $AB$ by their free multiplicative convolution $\mu\boxtimes\nu$ using the $S$-transform. Is there a way to compute the empirical spectral distribution of other products of $A$ and $B$, such as $AB^{-1}$?
(I am new to random matrix theory so the question might sound naive.)
|
https://mathoverflow.net/users/346351
|
Free multiplicative convolution of two random matrices
|
There is the possibility of dealing with any non-commutative rational function in A and B, by using more general (operator-valued) versions of free probability. Usually the corresponding equations cannot be solved analytically, but they are nice fixed-point equations which can be addressed numerically. See for example, chapter 10 in the book
[Mingo, Speicher: Free Probability and Random Matrices](https://rolandspeicher.com/literature/mingo-speicher/)
or the article
[Helton, Mai, Speicher: Applications of Realizations (aka Linearizations) to Free Probability](https://arxiv.org/pdf/1511.05330.pdf)
|
2
|
https://mathoverflow.net/users/112626
|
402413
| 165,139 |
https://mathoverflow.net/questions/402401
|
0
|
In the case of linear PDE, say $$Lu=0$$ if we have its green function say $G(x,y)$ then using that one can give solution of non homogenous PDE i.e. $Lu\_f=f$ where $u\_f=G\*f$.
Is the same thing hold for non-linear PDE? Even if not, I wanted to know if we have quasilinear PDE is that holds? If this is not true at all, then what is the use of the green function for nonlinear PDE?
Any help or reference will be appreciated.
|
https://mathoverflow.net/users/119011
|
Application of Green function for non linear PDE
|
The equality $u\_f= G\ast f$ uses linearity in an essential way since
$$
L\_y G(x,y)= \delta(y-x).
$$
The function $f$ is a superposition of $\delta$'s
$$
f(\bullet)=\int f(x)\delta(\bullet-x) dx.
$$
On the other hand the Green function has indirect uses in nonlinear equations. The solution on the [Yamabe problem](https://www.ams.org/journals/bull/1987-17-01/S0273-0979-1987-15514-5/S0273-0979-1987-15514-5.pdf) is one such situation.
|
1
|
https://mathoverflow.net/users/20302
|
402421
| 165,143 |
https://mathoverflow.net/questions/402388
|
1
|
I asked this question on MSE a while ago but didn't receive any useful answers.
Suppose I have a $1$-parameter family continuous maps $f\_t: \mathbb{S}^2\rightarrow \mathbb{C}P^1$ from a topological $2$-sphere to the Riemann sphere which is a local homeomorphism away from isolated points (for example, imagine a continuous family of rational functions $\mathbb{C}P^1 \longrightarrow \mathbb{C}P^1$ and then "forget" the complex structure on the domain). Suppose that $f\_t$ gives us a complex structure on $\mathbb{S}^2$ pulls back a complex structure on $\mathbb{S}^2$ for each $t$. With this new complex structure on $\mathbb{S}^2$, there exists a uniformising parameter $z\_t: \mathbb{C}P^1 \longrightarrow \mathbb{S}^2$ (up to Mobius transformations) such that $f\_t\circ z\_t$ is holomorphic for each $t$.
>
> **Question:** Does there exist a continuous choice of $z\_t$ so that $f\_t\circ z\_t: \mathbb{C}P^1 \longrightarrow \mathbb{C}P^1$ is a continuous family of holomorphic maps?
>
>
>
**Some comments**
If $f\_t$ is a homeomorphism for all $t$, then clearly this is true because $f\_t$ itself gives the parameter $z\_t$.
I am not familiar with the proof of the uniformisation theorem well enough to answer this. An equivalent question would be: If you have a continuous family of atlases $\mathcal{A}\_t$ on $\mathbb{S}^2$, then can you choose a continuous family of biholomorphisms $z\_t : \mathbb{C}P^1 \longrightarrow (\mathbb{S}^2,\mathcal{A}\_t)$? (although, I am not exactly sure how to define a "continuous family of atlases" since the domains and codomains of the coordinate charts may change.)
|
https://mathoverflow.net/users/125534
|
Existence of continuous family of uniformising parameters
|
That your $f\_t$ are local homeomorphisms away from isolated points is not sufficient for the conclusion you want. Your $f\_t$ must be at least topologically holomorphic. (A continuous map is called topologically holomorphic if it is open and discrete).
Now one needs some stronger restrictions on $z\mapsto f(z)$ (how do you pull back the conformal structure via just local homeomorphism?)
If $z\mapsto f\_t(z)$ safisfy a mild regularity condition, namely that they are
quasiregular (with respect to some fixed conformal structure on $S^2$), then the proof can proceed as follows: Let $\mu\_t=(f\_t)\_{\overline{z}}/(f\_t)\_z$
be the Beltrami coefficient and suppose that $t\mapsto \mu\_t$ is continuous
(as a function with values in $L^\infty$). Then to find homeomorphisms $\phi\_t$ such that $f\_t\circ\phi^{-1}\_t$ are holomorphic, one has to solve the Beltrami equation
$$\phi\_{\overline z}=\mu\_t\phi\_z,$$
and there is a theorem which guarantees the existence of a unique normalized
homeomorphic solution of Beltrami equation which depends continuously on $t$,
see
>
> Ahlfors, Lectures on quasiconformal mappings, Ch. V, C.
>
>
>
This involves stronger restrictions on both $z\mapsto f\_t(z)$ and $t\mapsto f\_t(z)$. Perhaps they can be somewhat relaxed but I do not think that the conclusion of continuity can be obtained under your restrictions.
|
1
|
https://mathoverflow.net/users/25510
|
402426
| 165,144 |
https://mathoverflow.net/questions/402294
|
2
|
Let $D\subset\mathbb{R}^2$ be a planar domain (maybe simply connected) and consider all the mappings $f:D\to\mathbb{R}^2$ with constant, fixed, positive singular values. Let $E=(E\_1,E\_2)$ be the orthonormal frame on $D$, such that at each point $p$ the image vectors $\mathrm{d}f\bigl(E\_i(p)\bigr)$ are orthogonal and have constant norm.
Calling $\kappa\_1$ and $\kappa\_2$ the curvatures of the $E\_1$ and $E\_2$ integral curves, then
>
> Can $\kappa\_1$ and $\kappa\_2$ change sign on $D$ ?
>
>
>
(When $D$ is simply connected) the nets of such integrals lines (called HP-nets and/or slip-line nets) have important applications in elasticity and plasticity theory, and there's a great deal of physical evidence towards a negative answer ($\kappa\_1$ and $\kappa\_2$ cannot change sign), however I have no knowledge of a proof.
|
https://mathoverflow.net/users/171439
|
Signs of curvatures of integrals lines of frames with constant principal values
|
Here's how one can construct a specific example to illustrate what can happen:
First, recall from my answer to [this question](https://mathoverflow.net/questions/351546/are-all-maps-mathbbr2-to-mathbbr2-with-fixed-singular-values-affine) that, if you have a smooth map $f:D\to\mathbb{R}^2$ with constant positive singular values, then, letting $(e\_1,e\_2)$ be the orthonormal frame field on $D$ such that $\bigl(f'(e\_1),f'(e\_2)\bigr)$ are orthogonal with $|f'(e\_i)|=\sigma\_i$ with $0<\sigma\_1<\sigma\_2$, then the dual $1$-forms $(\omega\_1,\omega\_2)$ satisfy
$$
\mathrm{d}\omega\_1 = \kappa\_1\,\omega\_1\wedge\omega\_2\qquad
\mathrm{d}\omega\_2 = \kappa\_2\,\omega\_2\wedge\omega\_1\,\tag1
$$
where $\kappa\_i:D\to\mathbb{R}$ is the curvature function of the flow lines of $e\_i$. Moreover, we know that
$$
\mathrm{d}(\kappa\_1\,\omega\_1) = \mathrm{d}(\kappa\_2\,\omega\_2) = 0.\tag2
$$
Conversely, given a surface $S$ endowed with a coframing $\omega = (\omega\_1,\omega\_2)$ and functions $\kappa\_1$ and $\kappa\_2$ that satisfy equations (1) and (2), one can construct a corresponding $f:D\to\mathbb{R}^2$ with constant singular values $\sigma\_i$.
Now, it's not hard to construct such data $(\omega\_1,\omega\_2,\kappa\_1,\kappa\_2)$ for which $\kappa\_1$ and $\kappa\_2$ vanish along some curves. For example, on a domain $S$ in the $uv$-plane that contains the origin, assume that $\lambda\_1$ and $\lambda\_2$ are nonzero functions that satisfy the linear hyperbolic system
$$
\frac{\partial\lambda\_1}{\partial v} + u\lambda\_2 =
\frac{\partial\lambda\_2}{\partial u} + v\lambda\_1 = 0\,.\tag3
$$
For example, $\lambda\_1 = \lambda\_2 = \mathrm{e}^{-uv}$ satisfies (3) on the entire $uv$-plane. Then the data
$$
\omega\_1 = \lambda\_1\,\mathrm{d}u,\quad
\omega\_2 = \lambda\_2\,\mathrm{d}v,\quad
\kappa\_1 = \frac{u}{\lambda\_1}\,,\quad
\kappa\_2 = \frac{v}{\lambda\_2}
$$
satisfy (1) and (2) while $\kappa\_1$ vanishes along $u=0$ and $\kappa\_2$ vanishes along $v=0$. Applying the integration method from the above-mentioned answer, one can now use this data to construct the desired $f$.
Taking the solution $\lambda\_1=\lambda\_2 = \mathrm{e}^{-uv}$ of (3), one finds that, letting $F:\mathbb{C}\to\mathbb{C}$ be the entire holomorphic function that satisfies $F(0)=0$ and $F'(z) = \mathrm{e}^{iz^2/2}$, and, for each real $\sigma>1$, letting $S\_\sigma:\mathbb{C}\to\mathbb{C}$ satisfy $S\_\sigma(x+iy)= x + i\sigma y$, and letting $D\subset\mathbb{C}$ be a domain containing $0\in\mathbb{C}$ on which $F$ has an inverse $F^{-1}:D\to\mathbb{C}$ satisfying $F^{-1}(0)=0$, then the mapping $f\_\sigma:D\to\mathbb{C}$ defined by $f\_\sigma(z) = F\bigl(S\_\sigma\bigl(F^{-1}(z)\bigr)\bigr)$ has constant singular values $1$ and $\sigma$ while the corresponding functions $\kappa\_1$ and $\kappa\_2$ change sign along curves that meet transversely at the origin in $D$.
More generally, it is useful to switch points of view and let $F$ be the set of triples $(x;e\_1,e\_2)$ where $x\in\mathbb{R}^2$ and $(e\_1,e\_2)$ are an oriented orthonormal basis of $\mathbb{R}^2$. Then $F$ is a smooth $3$-manifold embedded naturally in $(\mathbb{R}^2)^3 = \mathbb{R}^6$.
Define the canonical $1$-forms $\omega\_i = e\_i\cdot \mathrm{d}x$ and $\omega\_{12} = e\_1\cdot\mathrm{d}e\_2$, and note that they satisfy
$$
\mathrm{d}\omega\_1 = -\omega\_{12}\wedge\omega\_2\,,\quad
\mathrm{d}\omega\_2 = \omega\_{12}\wedge\omega\_1\,,\quad
\mathrm{d}\omega\_{12} = 0.\quad
$$
Now, on $F\times\mathbb{R}^2$ with projection $(\kappa\_1,\kappa\_2):F\times\mathbb{R}^2\to\mathbb{R}^2$ onto the second factor, consider the exterior differential ideal $\mathcal{I}$ generated by the $1$-form $\theta = \omega\_{12}+\kappa\_1\,\omega\_1 -\kappa\_2\,\omega\_2$ (which is a contact form on the $5$-manifold $F\times\mathbb{R}^2$) and the pair of $2$-forms
$$
\Upsilon\_1 = (\mathrm{d}\kappa\_1-{\kappa\_1}^2\,\omega\_2)\wedge\omega\_1
\qquad
\Upsilon\_2 = (\mathrm{d}\kappa\_2-{\kappa\_2}^2\,\omega\_1)\wedge\omega\_2
$$
Then $\mathrm{d}\theta \equiv \Upsilon\_1 - \Upsilon\_2 \mod \theta$. Then $\mathcal{I}$ is involutive for the independence condition $\omega\_1\wedge\omega\_2\not=0$.
In particular, each of the integral curves of the rank 4 system
$$
\theta = \mathrm{d}\kappa\_1-\omega\_1-{\kappa\_1}^2\,\omega\_2
= \mathrm{d}\kappa\_2-\omega\_2-{\kappa\_2}^2\,\omega\_1
= \omega\_1-\omega\_2 = 0
$$
on $F\times\mathbb{R}^2$ can be thickened to an integral surface of $\mathcal{I}$ on which $\omega\_1\wedge\omega\_2$ is nonvanishing. For example, we have
the integral curve
$$
(x;e\_1,e\_2,\kappa\_1,\kappa\_2) = \bigl((t,t);\,\partial/\partial x,\,\partial/\partial y, \,\arctan(t), \,\arctan(t)\bigr),
$$
which extends to an integral manifold of $\mathcal{I}$ on an open domain $D$ containing the origin $(0,0)$. The corresponding solution $f$ will have each of $\kappa\_1$ and $\kappa\_2$ changing sign along a curve that meets the 'diagonal' transversely. (One could probably solve this initial value problem explicitly, but I don't have time to attempt that right now.)
|
2
|
https://mathoverflow.net/users/13972
|
402432
| 165,146 |
https://mathoverflow.net/questions/402442
|
2
|
Given a finite number of algebraic curves over $\mathbb{Q}$ is there a curve that covers all of them?
|
https://mathoverflow.net/users/343850
|
Any finite number of curves over $\mathbb{Q}$ have a common cover
|
**If we are talking about branched covers, yes**: Let the curves be $C\_1$, $C\_2$, ..., $C\_r$. Take the product $\prod C\_i$, embed it into projective space, and intersect with a general codimension $r-1$ plane $H$. Then (by Bertini) $H \cap \prod C\_i$ will be a smooth curve and, for $G$ chosen generically, the projection onto each of the $C\_i$ will be nonzero.
**If we are talking about unbranched covers, then the literal answer is no**: Curves of genus zero have no non-trivial unbranched covers, so we'd need to exclude those. Also, for curves of genus one, all unbranched covers are also of genus one, so two curves of genus one have a common branched cover if and only if they are isogneous.
A harder question is if, given curves over $\mathbb{Q}$ of genuses $\geq 2$, they have a common unbranched cover.
I strongly expect the answer to be "no", and I can prove it for curves over $\mathbb{C}$. Fix integers $(d\_1, d\_2, g\_1, g\_2)$ with $d\_1 (g\_1-1) = d\_2 (g\_2-1)$ and let $X(d\_1, d\_2, g\_1, g\_2) \subset M\_{g\_1} \times M\_{g\_2}$ be the moduli space of pairs of curves $(C\_1, C\_2)$ with a common cover $C$, such that $C \to C\_i$ is degree $d\_i$. Given a curve $C\_i$, it has only finitely many covers of given degree $d\_i$, so the projection of $X(d\_1, d\_2, g\_1, g\_2)$ to either factor is finite to one, and thus the $\dim X(d\_1, d\_2, g\_1, g\_2) \leq \max \left( \dim M\_{g\_1}, \dim M\_{g\_2} \right) < \dim \left( M\_{g\_1} \times M\_{g\_2} \right)$.
So each $X(d\_1, d\_2, g\_1, g\_2)$ has positive codimension in $M\_{g\_1} \times M\_{g\_2}$, and a complex variety can't be the union of countably many positive dimensional subvarieties.
Over $\mathbb{Q}$, or even $\overline{\mathbb{Q}}$, this argument fails, but I would bet the conclusion is still right.
|
3
|
https://mathoverflow.net/users/297
|
402445
| 165,148 |
https://mathoverflow.net/questions/402435
|
10
|
Why are [W-types](https://ncatlab.org/nlab/show/W-type) called "W"?
Probably "W" means either "wellordered" or "wellfounded". ([Martin-Löf](https://www.csie.ntu.edu.tw/~b94087/ITT.pdf) uses the term "wellorder".) But these are notions associated to order theory, whereas W-types don't directly have to do with order relations (if at all).
|
https://mathoverflow.net/users/347155
|
Why are W-types called "W"?
|
You write:
>
> Probably "W" means either "wellordered" or "wellfounded". […] But these are notions associated to order theory, whereas W-types don't directly have to do with order relations (if at all).
>
>
>
I don’t know an official source for this, but I’ve always assumed W stands for “well-founded” as you suggest. *(Edit: I asked Per Martin-Löf, and he confirmed that this was what he meant by it.)* The justification for this is several-fold.
On the one hand, W-types do naturally carry well-founded partial orders (the *structurally smaller* relation), and the recursion/induction principles of W-types can be seen as well-founded induction over these orders.
On the other hand, well-foundedness was originally defined and developed to analyse and explain induction principles — it ended up becoming part of “order theory”, but it was motivated by Cantor’s analysis of induction. W-types give an alternative analysis of induction principles, not starting with an order relation — so as such, they can be seen as an alternative exploration of the original intention of well-foundedness.
And on the third hand, an important way of viewing W-types is as *types of well-founded labelled trees*, in exactly the traditional order-theoretic sense of *well-founded trees*. This can be made precise in two directions. If you are modelling type theory with W-types inside set theory (or indeed any other foundation without native inductive types), one way to model W-types is to define them as sets of isomorphism classes of suitably-labelled well-founded trees. On the other hand, inside type theory, you can associate to each element of a W-type a well-founded labelled tree, and show that elements of the W-type correspond precisely to isomorphism classes of such trees.
|
17
|
https://mathoverflow.net/users/2273
|
402453
| 165,151 |
https://mathoverflow.net/questions/402443
|
3
|
I am working on some non-local differential equations that appear in geometric analysis.
One of which I posted [here](https://mathoverflow.net/questions/402372/existence-and-uniqueness-of-an-euler-type-ode-with-varying-parameters) and was answered by @WillieWong and @losifPinelis.
Consider this non-local differential equation on $[1,\infty)$.
$(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - \left(\frac{4a^2}{r(r-2a)} + m(m+1)\right)f(r) = -\frac{4a^2}{r(r-2a)}f(1)+ \frac{4a(1-2a)}{(1-a)r(r-2a)} C $
with initial conditions
$f'(1) = C, \qquad \lim\_{r\to \infty} f(r) = 0$
where $C$ is any real number, $m$ is a non-negative integer, and $a\in [0,\frac{1}{2})$
The follwoing is an ODE that is somehow related to the above non-local differential equation:
$(r^2 - 2ar)f''(r) + 2(r-a) f'(r) - \left(\frac{4a^2}{r(r-2a)} + m(m+1)\right)f(r) = -\frac{2a}{r(r-2a)} D $
with initial conditions
$\frac{2a}{1-2a}f(1) + \frac{2}{1-a}f'(1) = D, \qquad \lim\_{r\to \infty} f(r) = 0$
where $D$ is any real number.
If $a=0$, then both equations become the known Euler equation:
$r^2f''(r) + 2r f'(r) - (m(m+1))f(r) = 0 $
$ f'(1) = C, \qquad \lim\_{r\to \infty} f(r) = 0$
which we know the unique solution is:
$f(r) = \frac{-C}{\alpha r^{\alpha}}$
where $\alpha = \frac{1}{2} + \frac{\sqrt{1+4m(m+1)}}{2}$
Is there an explicit solution to these equations?
Can I prove existence and/or uniqueness for the $a>0$ case using some kind of continuity method?
I know for instance that injectivity is a continuous property for elliptic operators, and one has the method of continuity to prove surjectivity of a 1-parameter family of elliptic operators.
Is there something similar in this context?
Any help or references is appreciated.
|
https://mathoverflow.net/users/138705
|
Existence and uniqueness of an Euler-type ODE with varying parameters part 2
|
Reformulate
-----------
Supposing $f$ is a solution to your first formulation, with $f(1) = \lambda$. Let $\tilde{f}(r) = \lambda^{-1} f(r)$. Then $\tilde{f}$ solves the differential equation
$$ \tag{1} r(r-2a) \tilde{f}'' + 2(r-a) \tilde{f}' - m(m+1) \tilde{f} = \frac{4a^2}{r(r-2a)} (\tilde{f} - 1) + \frac{4a(1-2a)}{(1-a)r(r-2a)} \tilde{C} $$
with
$$\tilde{f}'(1) = \tilde{C} \text{ and } \tilde{f}(1) = 1 \tag{2}$$
(Here $\tilde{C} = \lambda^{-1} C$.)
Hence your question is equivalent to asking
1. whether there exists some parameter $\tilde{C}$ such that the initial value problem (1), (2) admits a solution with a finite limit
2. whether said $\tilde{C}$ is unique
3. and what is the finite limit.
---
For convenience of typing, I will drop all tilde signs below. Note that the $f$ I will refer to is therefore different from the $f$ in your original question.
---
Third question
--------------
Let's answer the third question first. Suppose for a moment that a solution exists with a non-zero limit $f\_\infty = \lim f(r)$. Suppose first that $f\_\infty > 0$ (the argument will be similar if $f\_\infty < 0$.) Then there exists $R > 0$ such that for all $r > R$, we have
$$ r(r-2a) f'' + 2(r-a) f' > \frac{m^2}{2} f\_\infty$$
and $f(r) > \frac12 f\_\infty$. This shows that $f$ has no local maxima and hence must eventually be monotonic.
The same argument from my previous answer then shows that $\lim f'(r)$ exists (and hence must be zero) and that $|f'(r)|$ is at least size $1/r$ asymptotically. This contradicts the integrability of $f'(r)$.
**This shows that if a solution were to exist with a finite $f\_\infty$, then $f\_\infty = 0$ necessarily.**
---
Second question
---------------
Decompose $f = g + h\_C$, where $g$ solves
$$ r(r-2a) g'' + 2(r-a) g' - m(m+1) g - \frac{4a^2}{r(r-2a)} g = - \frac{4a^2}{r(r-2a)} $$
with $g'(1) = 0$ and $g(1) = 1$, and $h\_C$ solves
$$ r(r-2a) h'' + 2(r-a) h' - m(m+1) h - \frac{4a^2}{r(r-2a)} h = \frac{4a(1-2a)}{(1-a)r(r-2a)} C$$
with $h'(1) = C$ and $h(1) = 0$.
We see that the mapping $C \mapsto h\_C$ is linear. So let's concentrate on the case $C = 1$.
Then $m(m+1) h + \frac{4a^2}{r(r-2a)} h + \frac{4a(1-2a)}{(1-a)r(r-2a)} > 0$ whenever $h \geq 0$. So $h$ cannot have a positive local maximum. Since $h$ is initially increasing and $h(1) = 0$, this means that $h$ is strictly increasing. And hence $\lim h(r)$ exists in $(0,\infty]$ by the monotone convergence theorem; in particular, $\lim h(r) \neq 0$.
**This answers question 2**. Suppose with the value $C\_0$, we find $f$ a solution to the original problem. If $D \neq 0$, then the solution $\hat{f}$ using the value $C\_0 + D$ will have limit $\lim \hat{f}(r) = \lim f(r) + \lim h\_D(r) \neq 0$. And hence $C\_0$ is unique. Furthermore, since we established that there can be no finite limit besides $0$, this means that $\lim h\_D$ is necessarily $\pm \infty$.
---
First Question (Partial answer)
-------------------------------
I will use the functions $g$ and $h\_C$ as the previous part.
The equation satisfied by $g$ can be written as
$$ [(r^2 - 2ra)g']' = m(m+1) g + (g-1) \frac{4a^2}{r(r-2a)} $$
Note that if $g(r) \geq 1$ then the RHS is positive. And once $g'$ turns positive, then $g$ is increasing. So it is not too hard to argue in fact that $(r^2-2ra)g'$ is strictly increasing on $[1,\infty)$, and hence $g$ is also monotonically increasing.
As we noted before, $h\_1$ (and hence $h\_C$ for any $C > 0$) is also monotonically increasing.
1. First, I claim that for any $C > 0$ we have that $h\_C$ and $g\_C$ intersects at most once. Suppose the two curves intersect the first time at $r\_0$. Then since $g(1) > h\_C(1)$ we have that $h'\_C(r\_0) \geq g'(r\_0)$. But their difference satisfies the equation
$$ [ r(r-2a) (h'\_C - g')]' = [ m(m+1) + \frac{4a^2}{r(r-2a)} ] (h\_C - g) + \frac{4a(1-2a)}{(1-a)r(r-2a)} C + \frac{4a^2}{r(r-2a)} $$
which shows that after $r\_0$ the value $h'\_C - g'$ will be always positive.
2. For $r > 1$, let $C(r)$ be the unique value such that $h\_{C(r)}(r) = g(r)$; in other words let $C(r) = \frac{g(r)}{h\_1(r)}$. The analysis in step 1 indicates that $C(r)$ is a decreasing function of $r$.
3. By the monotone convergence theorem $\lim C(r)$ converges to a finite, non-negative limit $C\_\infty$.
I believe that $f = g - h\_{C\_\infty}$ is the desired solution.
* It is clear from construction that $g - h\_{C\_\infty}$ is everywhere strictly positive. And hence for any $C < C\_\infty$, we have that $\lim\_{r\to\infty} g - h\_{C} = +\infty$, using that $\lim h\_1 = +\infty$.
* It is also clear from the construction that $\lim\_{r\to\infty} g - h\_C = -\infty$ for any $C > C\_\infty$.
* So if there exists a solution, it must correspond to $C\_\infty$.
I don't have the time to work out all the details, so here's a sketch for what needs to be done to prove that $C\_\infty$ works.
1. Suppose that at some point $r$ you have that $f(r) > \frac{4a}{m(m+1)r(r-2a)}[ a + \frac{1-2a}{1-a} C\_\infty]$. and $f'(r) > 0$. Then for some sufficiently small $\epsilon$ we have $f - h\_\epsilon$ is still everywhere positive. Here we use that we can choose $\epsilon$ small enough and take advantage of the strict positivity of $f$ on $[1,r]$, and that on the interval $[r,\infty)$ the initial conditions $f(r) - h\_\epsilon(r) > \frac{4a}{m(m+1)r(r-2a)}[ a + \frac{1-2a}{1-a} (C\_\infty + \epsilon) ]$ and $f'(r) - h\_\epsilon'(r) > 0$ implies that $f - h\_\epsilon$ is strictly increasing beyond that point. But this would contradict the definition of $C\_\infty$.
2. So for every $r$ we must have at least one of $f'(r) \leq 0$ or $f(r) < \frac{4a}{m(m+1)r(r-2a)}[ a + \frac{1-2a}{1-a} C\_\infty]$ holds. This can be seen to imply that the larger of $f(r)$ and the decaying envelope (RHS of the inequality for $f(r)$) is a decreasing function. By the monotone convergence theorem this has a limit.
3. If the limit is non-zero, this means $f$ is eventually larger than the envelope, and hence the limit is the limit of $f$. But this contradicts the answer to Question 3. And hence the limit is 0. By comparison this shows that this is also the limit of the positive function $f$.
|
2
|
https://mathoverflow.net/users/3948
|
402461
| 165,155 |
https://mathoverflow.net/questions/402383
|
4
|
Consider a collection of $C^\*$-algebras $\{A\_i\}\_{i \in I}$. We can form the direct sum $$\bigoplus\_{i \in I}^{c\_0} A\_i:= \left\{(a\_i)\_{i \in I} \in \prod\_{i\in I} A\_i: \lim\_{i \in I} \|a\_i\| = 0\right\}$$
which is an ideal in the $C^\*$-algebra
$$\bigoplus\_{i \in I}^{\ell^\infty} A\_i:= \left\{(a\_i)\_{i \in I} \in \prod\_{i\in I} A\_i: \sup\_{i \in I} \|a\_i\| <\infty\right\}.$$
Given a $C^\*$-algebra $A$, denote its multiplier algebra by $M(A)$. One possible realisation of the multiplier algebra is by setting $M(A):= \mathcal{L}\_A(A)$, the adjointable operators when we view $A$ as a (right) Hilbert $C^\*$-module over itself.
I believe I have proven that
$$M\left(\bigoplus\_{i \in I}^{c\_0}A\_i\right) \cong \bigoplus\_{i \in I}^{\ell^\infty} M(A\_i)\cong M\left(\bigoplus\_{i \in I}^{\ell^\infty}A\_i\right)$$
but I can't find a reference for this statement. So, is my assertion true?
---
Here is a proof sketch:
We use the implementation of the multiplier $C^\*$-algebra as adjointable operators. We then have natural maps
$$M\left(\bigoplus\_{i \in I}^{c\_0}A\_i\right) \to \bigoplus\_{i \in I}^{\ell^\infty} M(A\_i): t \mapsto (\iota\_i ^\* t\iota\_i)\_{i \in I}$$ where
$\iota\_i: A\_i \hookrightarrow \bigoplus\_{i \in I}^{c\_0} A\_i$ is the inclusion map and
$$\bigoplus\_{i \in I}^{\ell^\infty} M(A\_i) \to M\left(\bigoplus\_{i \in I}^{c\_0}A\_i\right) : (t\_i)\_{i \in I} \mapsto [(a\_i)\_i \mapsto (t\_i(a\_i))]$$
These are easily checked to be $\*$-isomorphisms that are inverse to each other, and this establishes the isomorphism $M\left(\bigoplus\_{i \in I}^{c\_0}A\_i\right) \cong \bigoplus\_{i \in I}^{\ell^\infty} M(A\_i)$. The other isomorphism is shown similarly.
|
https://mathoverflow.net/users/216007
|
Direct sum of multiplier algebras
|
Let's try to flesh out your "sketch". Set $A=c\_0-\oplus\_i A\_i$ and consider this as a Hilbert $C^\*$-module over itself. Let $\iota\_i:A\_i\rightarrow A$ be the inclusion, and $\jmath\_i:A\rightarrow A\_i$ the left inverse to $\iota\_i$. These are both non-degenerate $\*$-homomorphisms, and so extend to unital $\*$-homomorphisms $\overline\iota\_i:M(A\_i)\rightarrow M(A)$ and $\overline\jmath\_i:M(A)\rightarrow M(A\_i)$. Also, notice that by definition of the $A$-valued inner-product on $A$,
$$ (b|\iota\_i(a)) = \iota\_i\big( (\jmath\_i(b)|a) \big)
\qquad (b\in A, a\in A\_i). $$
Given $T\in M(A)$ set $T\_i = \overline\jmath\_i(T)\in M(A\_i)$. By definition, $T\_i(\jmath\_i(a)) = \jmath\_i(T(a))$ for $a\in A$, and as $\jmath\_i$ is a $\*$-homomorphism, also $T\_i^\*(\jmath\_i(a)) = \jmath\_i(T^\*(a))$. For $a\in A\_i$ and $b=(b\_j)\in A$,
$$ (T\iota\_i(a)|b) = (\iota\_i(a)|T^\*(b))
= \iota\_i\big( (a|\jmath\_i(T^\*(b))) \big)
= \iota\_i\big( (a|T\_i^\*(\jmath\_i(b))) \big) $$
while
$$ (\iota\_i T\_i(a)|b) = \iota\_i\big( (T\_i(a)|\jmath\_i(b)) \big)
= \iota\_i\big( (a|T\_i^\*(\jmath\_i(b))) \big). $$
Thus $T\iota\_i = \iota\_iT\_i$ for each $i$.
As the linear span of the images of the $\iota\_i$ are dense in $A$, it now follows that
$$ T(a) = \sum\_i T\iota\_i\jmath\_i(a) = \sum\_i \iota\_i\big( T\_i\jmath\_i(a) \big), $$
and the isomorphism $M(A) \cong \ell^\infty-\oplus\_i M(A\_i)$ now follows.
---
I actually find arguing using Hilbert $^\*$-modules a bit cumbersome. Instead, let $A\_i\subseteq\mathcal B(H\_i)$ non-degenerately, for each $i$, for some suitable Hilbert space $H\_i$. Set $H = \oplus\_i H\_i$, so naturally $A$ acts non-degenerately on $H$. Set $B=\ell^\infty-\oplus\_i A\_i$, so also $B$ acts non-degenerately on $H$. I'll now consider $B$, but much the same argument works for $A$.
We know that $M(B)\subseteq B''\subseteq \mathcal B(H)$, and indeed
$$ M(B) = \big\{ T\in B'' : Tb, bT\in B \ (b\in B) \big\}. $$
Given $b=(b\_i)\in B$ and $\xi=(\xi\_i)\in H$, by definition, $b(\xi) = (b\_i(\xi\_i))$. As such, with $p\_i\in\mathcal B(H)$ the projection onto the factor $H\_i$, we see that $b p\_i = p\_i b$ so $p\_i\in B'$.
Thus, any $T\in M(B)\subseteq B''$ commutes with each $p\_i$. By linearity and continuity, there is $(T\_i) \in \ell^\infty-\oplus\_i \mathcal B(H\_i)$ with $T(\xi) = (T\_i(\xi\_i))$ for each $\xi\in H$. Using the inclusions $A\_i\rightarrow B$, we can now show that each $T\_i\in M(A\_i)$.
We have hence showed that $M(B) \cong \ell^\infty-\oplus\_i M(A\_i)$. [This seems surprising to me, but I believe this 2nd argument.]
|
4
|
https://mathoverflow.net/users/406
|
402462
| 165,156 |
https://mathoverflow.net/questions/402346
|
4
|
Let $V:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be a $C^{\infty}$ vector field. Fix a *(single)* real number $d$ such that
$$
1\leq d\leq n
.
$$
Under what conditions is the flow map $\Phi\_V$ defined as sending any $x\_0\in \mathbb{R}^n$ to the time $1$-solution of autonomous system of ODEs:
$$
\dot{x(t)} = V(x(t)) \qquad x(0)=x\_0,
$$
satisfies:
$$
\mathcal{H}^d(K)= \mathcal{H}^d(\Phi(K)) \qquad \mbox{for all compact }K\subseteq \mathbb{R}^n
;
$$
where $\mathcal{H}^d$ is the $d$-dimensional Hausdorff (outer) measure on $\mathbb{R}^n$?
---
*I'm not looking for a general characterisation (though I'd be happy to have one) but just a simple ''non-trivial'' necessary condition.*
|
https://mathoverflow.net/users/346124
|
When is a smooth field's flow map volume preserving diffeomorphism
|
I would like to know why you are interested in the specific $d$-dimensional measures given by the Hausdorff ones. For $d=n$ such a choice is understandable since the Lebesgue measure in $\mathbb{R}^{n}$ is proportional to $\mathcal{H}^{n}$. But (to the best of my knowledge) there is no nice description in the case of $d$-dimensional Hausdorff measures. A difference even appears in the definition of Hausdorff measures if we change coverings by arbitrary subsets to, say, balls (we obtain the so-called spherical Hausdorff measures). I'd like to mention that
1. Dynamics usually have nothing to do with the original geometry of the phase space. Thus, for most purposes in dynamics it is better to let ourselves vary measures (for example, by considering different metrics in tangent spaces or adding a weight (density) function to a given measure).
2. To study evolution of measures via smooth dynamics (i. e. when we are armed with linearization)
it is better to consider measures (or pseudo measures) or just quantities based on coverings by sets which behave well under linear transformations. But these (pseudo) measures usually provide only a two-sided majorant for the Hausdorff measures $\mathcal{H}^{d}$.
There are two standard types of nice sets: (A) parallelepipeds (which are mapped by linear transformations also to parallelepipeds) and (B) balls (which are mapped by linear transformations to ellipsoids).
(A) For the evolution of $d$-dimensional volumes of a parallelepiped with edges $\xi\_{1,0},\ldots,\xi\_{d,0}$ under linearization near a point $x\_{0}$ over a time $t$ there is the Liouville trace formula:
$$ |\xi\_{1}(t) \wedge \ldots \wedge \xi\_{d}(t)|\_{\bigwedge^{d}\mathbb{R}^{n}} = |\xi\_{1}(0) \wedge \ldots \wedge \xi\_{d}(0)|\_{\bigwedge^{d}\mathbb{R}^{n}} \cdot \operatorname{exp}\left( \int\_{0}^{t} \operatorname{Tr} \left[ A(s) \circ \Pi(s) \right]ds \right),$$
where $A(t)=DV(x(t;x\_{0}))$; $\dot{\xi}\_{i}(t)=A(t)\xi\_{i}(t)$ with $\xi\_{i}(0)=\xi\_{i,0}$; $\Pi(t)$ is the orthogonal projector onto the space spanned by $\xi\_{1}(t),\ldots,\xi\_{d}(t)$ and $\operatorname{Tr}$ is the trace. So, this formula shows linear evolution of natural volumes of the simplest $d$-dimensional objects. So, in some sense the condition
$$\int\_{0}^{t} \operatorname{Tr} \left[ A(s) \circ \Pi(s) \right]ds = 0$$
is necessary for natural $d$-dimensional volumes to be preserved under the time-$t$ map. Note that for $d=n$ there is nothing more than $\operatorname{Tr}A(s)=\operatorname{div}V(x(s,x\_{0}))$ under the integral. To obtain a proper pseudo measure (for which we may have volume preserving), I may suggest to consider a $d$-dimensional pseudo measure in $\mathbb{R}^{n}$ given by coverings by $n$-dimensional parallelepipeds for which we sum their $d$-dimensional volumes given by the maximum over all $d$-dimensional volumes of its $d$-dimensional edges-parallelopipeds. The main problem to solve is to include a parallelepiped, which is perturbed by a small ball, to a larger parallelepiped, which have $d$-dimensional volume (in the introduced sense) close to the unperturbed one. I do not know if this may work and need some time to resolve it (maybe someone can do it before me).
(B) The image of the unit ball $\mathcal{B}$ under a linear transformation $L$ is an ellipsoid $\mathcal{E}$ with semi-axes $\alpha\_{1} \geq \ldots \geq \alpha\_{n}$, which are given by the singular values of $L$, i. e. the eigenvalues of $L^{\*}L$ arranged in non-decreasing order. We can introduce a $d$-dimensional volume of the ellipsoid in the case of a non-integer $d=k+s$, where $k$ is a non-negative integer and $s \in (0,1]$, as
$$\omega\_{d}(\mathcal{E}):=\alpha\_{1} \cdot \ldots \cdot \alpha\_{k}. \alpha^{s}\_{k+1}$$
Also $\omega\_{d}(L):=\alpha\_{1} \cdot \ldots \cdot \alpha\_{k} \alpha^{s}\_{k+1}$ is called the singular value function of $L$. In fact, for integer $d$ we have
$$\omega\_{d}(L) = \sup \frac{|L\xi\_{1,0} \wedge \ldots \wedge L\xi\_{d,0}|\_{\bigwedge^{d}\mathbb{R}^{n}}}{|\xi\_{1,0} \wedge \ldots \wedge \xi\_{d,0}|\_{\bigwedge^{d}\mathbb{R}^{n}}} = \sup\_{|\xi\_{i,0}| \leq 1, i=1,\ldots,d} |L\xi\_{1,0} \wedge \ldots \wedge L\xi\_{d,0}|\_{\bigwedge^{d}\mathbb{R}^{n}}.$$
So, if the approach from (A) works, $\omega\_{d}(D\Phi\_{V})=1$ (for integer $d$) should be a natural necessary condition.
|
3
|
https://mathoverflow.net/users/85336
|
402488
| 165,164 |
https://mathoverflow.net/questions/402471
|
3
|
This [paper](https://homepages.inf.ed.ac.uk/gdp/publications/trace.pdf) says that $FinVect\_k$ is collectively complete for traced symmetric monoidal categories, in the sense that given distinct arrows in the free traced SMC (over some generating monoidal signature) there exists a strong functor (from the free traced SMC) into $FinVect\_k$ distinguishing them.
Is there an analogous result for Cartesian categories? It would seem intuitive to me that $Set$ might be collectively complete for Cartesian categories in the sense above -- but I can't find a reference anywhere. Is this true, and is there somewhere I could find a proof?
|
https://mathoverflow.net/users/156811
|
Is Set (collectively) complete for Cartesian categories?
|
If I understand what you are asking, the answer is yes.
Indeed:
**Proposition.**
Let $\mathcal{C}$ be a locally small cartesian monoidal category and let $f\_0, f\_1 : X \to Y$ be a parallel pair of morphisms in $\mathcal{C}$.
Then there is a cartesian monoidal functor $F : \mathcal{C} \to \textbf{Set}$ such that $F f\_0 = F f\_1$ implies $f\_0 = f\_1$.
*Proof.* Take $F = \mathcal{C} (X, -)$ and consider $\textrm{id}\_X \in F (X)$. ◼
(This is basically a small part of the Yoneda lemma.)
|
7
|
https://mathoverflow.net/users/11640
|
402492
| 165,166 |
https://mathoverflow.net/questions/402501
|
1
|
Let $f:X\rightarrow Y$ be a morphism of projective varieties. We may assume that $X$ and $Y$ are smooth, and $f$ is flat of relative dimension one. Fix an ample divisor $A$ on $X$.
I would like to ask if there exists a compact moduli space $\overline{M}\_{g,d}(X)$ such that all points of $\overline{M}\_{g,d}(X)$ represent curves $C\subset X$ of arithmetic genus $g$ and degree $d$ (with respect to $A$) with the following additional property:
$(\star)$ none of the irreducible components of $C$ is contracted by $f$.
Thank you.
|
https://mathoverflow.net/users/14514
|
Moduli spaces of horizontal curves
|
It is not possible to have such a moduli space that contains all the smooth curves of genus $g$ and degree $d$ and over which the universal family of curves is proper.
Let $X = \mathbb P^1 \times \mathbb P^1$ with coordinates $x,y$, $Y= \mathbb P^1$, $f$ the projection onto the $x$ coordinate. Let $C\_t$ be given by $y=tx$. In the limit as $t \to \infty$, this converges to the curve with the vertical component $x=0$ (in addition to the horizontal component $y=\infty$). So if the moduli space is compact and the universal family is proper, the limit as $t \to \infty$ of $C\_t$ will contain that component.
You could try to do something non-proper but if you want the universal family to be flat you will still have problems, as then $C\_{\infty}$ will have to be contained in the union of $x=0$ and $y = \infty$, so to have the same degree as $C\_t$ according to an ample line bundle will have to contain both $x=0$ and $y=\infty$.
|
3
|
https://mathoverflow.net/users/18060
|
402503
| 165,169 |
https://mathoverflow.net/questions/402446
|
6
|
Let $\mathcal{C}$ and $\mathcal{D}$ be two tensor categories (if necessary, assume they are fusion categories). Is the canonical braided monoidal functor $$\mathcal{Z}(\mathcal{C})\boxtimes\mathcal{Z}(\mathcal{D})\rightarrow\mathcal{Z}(\mathcal{C}\boxtimes\mathcal{D})$$ an equivalence?
NB: The two monoidal categories $\mathcal{Z}(\mathcal{C})\boxtimes\mathcal{Z}(\mathcal{D})$ and $\mathcal{Z}(\mathcal{C}\boxtimes\mathcal{D})$ have the same Frobenius-Perron dimension so it would be enough to show that the above functor is either injective or surjective in the sense of [EGNO](http://www-math.mit.edu/%7Eetingof/egnobookfinal.pdf).
|
https://mathoverflow.net/users/105094
|
Drinfeld center of a Deligne tensor product
|
By Cororllary 3.26 of arxiv:1009.2117, any braided tensor functor out of a non-degenerately braided fusion category is automatically fully faithful. Since $Z(\mathcal{C})\boxtimes Z(\mathcal{D})$ is non-degenerate when $\mathcal{C}, \mathcal{D}$ are fusion, the result follows immediately from your FP dimension observation.
|
7
|
https://mathoverflow.net/users/351
|
402509
| 165,170 |
https://mathoverflow.net/questions/402495
|
-1
|
From *Surface subgroups of Coxeter and Artin groups* (Gordon, Long and Reid, 2003) [DOI link](https://doi.org/10.1016/j.jpaa.2003.10.011), we can read that (Theorem 1.1) a Coxeter group is either virtually free or contains a surface group ($\pi\_1$ of a closed orientable surface of genus $\ge 1$).
My question is:
Are all Coxeter groups virtually free or virtually surface groups?
|
https://mathoverflow.net/users/197544
|
Are all Coxeter groups virtually free or virtually surface groups?
|
When asking a question about Coxeter groups, it might be useful to focus on right-angled Coxeter groups first: usually, they are easier to handle. Below are a few criteria you can play with in order to create various examples of (right-angled) Coxeter groups.
Let $\Gamma$ be a finite simplicial graph. The right-angled Coxeter group $C(\Gamma)$ is:
* finite if and only if $\Gamma$ is complete;
* virtually free if and only if $\Gamma$ is chordal (i.e. it does not contain an induced cycle of length $\geq 4$);
* virtually abelian of rank $n \geq 1$ if and only if $\Gamma$ decomposes as a join of a complete graph and $n$ copies of $K\_2^{\mathrm{opp}}$ (= two isolated vertices);
* virtually a hyperbolic surface group if and only if $\Gamma$ decomposes as a join of a complete graph and a cycle of length $\geq 5$;
* hyperbolic if and only if $\Gamma$ is square-free (i.e. it does not contain an induced cycle of length $4$);
* a free product if and only if $\Gamma$ is disconnected;
* multi-ended if and only if $\Gamma$ contains a separating complete subgraph (possibly empty).
|
4
|
https://mathoverflow.net/users/122026
|
402510
| 165,171 |
https://mathoverflow.net/questions/402499
|
2
|
Given a field $k$ with characteristic $p$ and a finite cyclic $p$-group $G$ of order $p^a$, it is well-known that all the indecomposable representations of $kG$ are given by mapping a generator $x$ of $G$ to the Jordan matrix $J\_s\in M\_s(k)$ with all eigenvalues one for $1\leq s\leq p^a$. If we replace $k$ by a commutative ring $R$ with characteristic $p$, then what are the indecomposable representations of $RG$? Is it the same as in the situation of $kG$?
|
https://mathoverflow.net/users/134942
|
Indecomposable representations for group ring $RG$ over commutative ring $R$ with characteristic $p$
|
One might ask whether one can classify all indecomposable $RG$-modules when one knows all indecomposable $R$-modules but the example $R=K[x,y]/(x^2,y^2)$ shows that this is not possible.
The answer will in general be that one can not classify the indecomposable representations as those algebras are most often of "wild" representation type (see for example <https://www.tandfonline.com/doi/abs/10.1080/00927879108824178> ).
If $G$ is a non-trivial cyclic group and $R$ is a representation-infinite finite dimensional $K$-algebra (for example $R=K[x,y]/(x^2,y^2)$) then $RG \cong R \otimes\_K K[x]/(x^n)$ for some $n$ and this will have wild representation type since the quiver of $RG$ will have at least three loops.
In the example of $R=K[x,y]/(x^2,y^2)$ one can classify all indecomposable $R$-modules but for $RG$ this is a wild problem already.
|
2
|
https://mathoverflow.net/users/61949
|
402511
| 165,172 |
https://mathoverflow.net/questions/402497
|
27
|
In so-called 'natural unit', it is said that physical quantities are measured in the dimension of 'mass'. For example, $\text{[length]=[mass]}^{-1}$ and so on.
In quantum field theory, the dimension of coupling constant is very important because it determines renormalizability of the theory.
However, I do not see what exactly the mathematical meaning of 'physical dimension' is. For example, suppose we have self-interaction terms $g\_1\cdot \phi\partial^\mu \phi \partial\_\mu \phi$ and $g\_2 \cdot \phi^4$, where $\phi$ is a real scalar field, $g\_i$ are coupling constants and we assume $4$ dimensional spacetime.
Then, it is stated in standard physics books that the scalar field is of mass dimension $1$ and so $g\_1$ must be of mass dimension $-1$ and $g\_2$ is dimensionless. But, these numbers do not seem to play any 'mathematical' role.
To clarify my questions,
1. What forbids me from proclaiming that $\phi$ is dimensionless instead of mass dimension $1$?
2. What is the exact difference between a dimensionless coupling constant and a coupling constant of mass dimension $-1$?
These issues seem very fundamental but always confuse me. Could anyone please provide a precise answer?
|
https://mathoverflow.net/users/56524
|
How do we give mathematical meaning to 'physical dimensions'?
|
Mathematically, the concept of a physical dimension is expressed using one-dimensional vector spaces and their tensor products.
For example, consider mass.
You can add masses together and you know how to multiply a mass by a real number.
Thus, masses should form a one-dimensional real vector space $M$.
The same reasoning applies to other physical quantities, like length, time, temperature, etc.
Denote the corresponding one-dimensional vector spaces by $L$, $T$, etc.
When you multiply (say) some mass $m∈M$ and some length $l∈L$,
the result is $m⊗l∈M⊗L$.
Here $M⊗L$ is another one-dimensional real vector space,
which is capable of “storing” physical quantities of dimension mass times length.
Multiplicative inverses live in the dual space:
if $m∈M$, then $m^{-1}∈M^\*$, where $\def\Hom{\mathop{\rm Hom}} \def\R{{\bf R}} M^\*=\Hom(M,\R)$.
The element $m^{-1}$ is defined as the unique element in $M^\*$
such that $m^{-1}(m)=1$, where $-(-)$ denotes the evaluation
of a linear functional on $M$ on an element of $M$.
Observe that $m ⊗ m^{-1} ∈ M⊗M^\* ≅ \R$, where the latter canonical isomorphism
sends $(f,m)$ to $f(m)$, so $m^{-1}$ is indeed the inverse of $m$.
Next, you can also define powers of physical quantities,
i.e., $m^t$, where $m∈M$ is a mass and $t∈\R$ is a real number.
This is done using the notion of a *density* from differential geometry.
(The case $\def\C{{\bf C}} t\in\C$ works similarly, but with
complex one-dimensional vector spaces.)
In order to do this, we must make $M$ into an *oriented* vector space.
For a one-dimensional vector space, this simply means that
we declare one out of the two half-rays in $M∖\{0\}$ to be positive,
and denote it by $M\_{>0}$.
This makes perfect sense for physical quantities like mass, length, temperature.
Once you have an orientation on $M$,
you can define $\def\Dens{\mathop{\rm Dens}} \Dens\_d(M)$
for $d∈\R$ as the one-dimensional (oriented) real vector space
whose elements are equivalence classes of pairs $(a,m)$,
where $a∈\R$, $m∈M\_{>0}$.
The equivalence relation is defined as follows:
$(a,b⋅m)∼(a b^d,m)$ for any $b∈\R\_{>0}$.
The vector space operations are defined as follows:
$0=(0,m)$ for some $m∈M\_{>0}$,
$-(a,m)=(-a,m)$,
$(a,m)+(a',m)=(a+a',m)$,
and $s(a,m)=(sa,m)$.
It suffices to add pairs with the same
second component $m$ because the equivalence relation allows you to change the second component arbitrarily.
Once we have defined $\Dens\_d(M)$, given $m∈M\_{>0}$ and $d∈\R$,
we define $m^d∈\Dens\_d(M)$ as the equivalence class of the pair $(1,m)$.
It is easy to verify that all the usual laws of arithmetic,
like $m^d m^e = m^{d+e}$, $m^d n^d = (mn)^d$, etc.,
are satisfied, provided that multiplication and reciprocals are interpreted as explained above.
Using the power operation operations we just defined,
we can now see that the equivalence class of $(a,m)$
is equal to $a⋅m^d$, where $m∈M\_{>0}$, $m^d∈\Dens\_d(M)\_{>0}$,
and $a⋅m^d∈\Dens\_d(M)$.
This makes the meaning of the equivalence relation clear.
In particular, for $d=-1$ we have a canonical isomorphism $\Dens\_{-1}(M)→M^\*$
that sends the equivalence class of $(1,m)$ to the element $m^{-1}∈M^\*$ defined above,
so the two notions of a reciprocal element coincide.
If you are dealing with temperature without knowing about the absolute zero,
it can be modeled as a one-dimensional real *affine* space.
That is, you can make sense of a linear combination
$$a\_1 t\_1 + a\_2 t\_2 + a\_3 t\_3$$
of temperatures $t\_1$, $t\_2$, $t\_3$
as long as $a\_1+a\_2+a\_3=1$,
and you don't need to know about the absolute zero to do this.
The calculus of physical quantities can be extended
to one-dimensional real affine spaces without much difficulty.
None of the above constructions make any noncanonical choices of
physical units (such as a unit of mass, for example).
Of course, if you do fix such a unit $μ∈M\_{>0}$, you can construct
an isomorphism $\R→\Dens\_d(M)$ that sends $a∈\R$ to $aμ^d$,
and the above calculus (including the power operations)
is identified with the usual operations on real numbers.
In general relativity, we no longer have a single one-dimensional
vector space for length.
Instead, we have the *tangent bundle*,
whose elements model (infinitesimal) displacements.
Thus, physical quantities no longer live in a fixed one-dimensional
vector space, but rather are sections of a one-dimensional
vector bundle constructed from the tangent bundle.
For example, the volume is an element of the total space
of the line bundle of 1-densities $\Dens\_1(T M)$,
and the length is now given by the line-bundle of $λ$-densities $\Dens\_λ(T M)$, where $λ=1/\dim M$.
|
55
|
https://mathoverflow.net/users/402
|
402515
| 165,174 |
https://mathoverflow.net/questions/400747
|
9
|
I was reading a [physics paper](https://www.sciencedirect.com/science/article/abs/pii/0550321386901550) where it was mentioned that the basic framework of Connes' differential non-commutative geometry (or actually, a slight modification of Connes in that paper) would need some extensions in order to arrive at a theory of non-commutative complex geometry, where one now works with complex manifolds.
Are there specific technical obstacles which would make such an extension of non-commutative geometry particularly difficult or involved? What would be the main issues with trying to do this?
|
https://mathoverflow.net/users/119114
|
Non-commutative complex geometry
|
I don't think there are any obvious ``technical obstacles'' to extending noncommutative geometry to the a theory of noncommutative complex geometry. Instead I would say that there are a number of differing points of view on what form noncommutative complex geometry should take, and it's not completely clear how the different approaches relate to each other.
Let's start with Connes' theory of noncommutative geometry, in particular spectral triples, which are usually thought of as ``noncommutative Riemannian manifolds''. A basic spectral triple is an unbounded representative of a $K$-homology class of a unital $C^\*$-algebra. Such unbounded representatives exist for all classes of any unital $C^\*$-algebra, so in particular, they exist for the $K$-homology classes of a compact Hausdorff space. Now in general a compact Hausdorff space will may not admit a differential structure, so in the commutative case a spectral triple is a more general structure than a smooth structure. To address this Connes introduced a number of higher axioms, and then proved his *reconstruction theorem*, showing that spectral triples satisfying these extra axioms are equivalent to compact Riemannian manifolds. The question of how well-suited these higher axioms to the noncommutative setting is the subject of debate . . . but that is a discussion for another day.
So following this point of view, a "complex spectral triple" would be spectral triple (satisfying the higher axioms) plus "something extra". Connes' proposal was to look at positive Hochschild cocycles since in the classical case, for surfaces, such cocycles are equivalent to complex structures. This is explained in Section VI.2 of Connes' book.
The motivating noncommutative example is, as usual, the noncommutative torus $\mathbb{T}\_{\theta}^2$. This is a $\theta$-deformation of a classical complex manifold (in fact a Calabi--Yau manifold) and since its structure is quite close to the classical situation, a lot of the classical complex and Kähler geometry carries over.
In another direction, there is the work of [Fröhlich, Grandjean, and Recknagel:](https://arxiv.org/abs/math-ph/9807006). They start with a spectral triple, and then try to build a noncommutative version of complex geometry on the associated differential graded algebra. Their approach takes globally defined classical identities in complex geometry and makes them into noncommutative axioms. For example, the Lefschetz identities and the Kähler identities play a major role in their work. Their main example is again the noncommutative torus.
In a very different approach one should mention noncommutative projective algebraic geoemtry. This is an entirely algebraic approach to noncommutative geometry based on Serre's characterisation of the quasi-coherent sheaves over a projective variety. This approach has been hugely successful in recent years, see [here](https://books.google.ie/books?id=HsTkPOj0iusC&pg=PA364&lpg=PA364&dq=q-gr%20category&source=bl&ots=zDF5YBFXun&sig=ACfU3U2XWYe0ycbG9cl7PzYY3Nk8Ms7rOw&hl=fr&sa=X&ved=2ahUKEwiQppmBz8zyAhUEgFwKHcRSDY8Q6AF6BAgQEAM#v=onepage&q=q-gr%20category&f=false) for a nice introduction. A connection with the differential geometric approach to noncommutative complex geometry (analogous to the classical GAGA correspondence) has been postulated, see [here](https://arxiv.org/abs/1209.3595) for example. However, we are still very far from any kind of a noncommutative GAGA.
A large and very important family of motivativing examples comes from Drinfeld--Jimbo quantum groups: the quantum flag manifolds. As shown in the seminal [papers](https://arxiv.org/abs/math/0307402) of Heckenberger and Kolb, the irreducible quantum flag manifolds admit a direct $q$-deformation of their classical de Rham complex. These complexes have a remarkable structure, $q$-deforming the classical Kähler geometry of the flag manifolds. These examples are crucial for understanding both the noncommutative geoemtry of the quantum groups, and as a forum for reconciling the various approaches to noncommutative complex geometry.
Finally, we note [the approach of Pirkovskii](https://arxiv.org/pdf/1311.0309.pdf) based on topological algebras. The motivating examples here are the quantum $n$-polydisk and the quantum $n$-ball.
|
7
|
https://mathoverflow.net/users/3072
|
402517
| 165,175 |
https://mathoverflow.net/questions/402523
|
6
|
For any fixed $\frac{1}{2} < \sigma < 1$, let
$$\int\_0^T \frac{|\zeta(\sigma+it)|^2}{\sqrt{1+t^2}} \ dt = O(T^\theta), \qquad T \uparrow \infty. $$
It is clear that $\theta > 0$, since we have the classical asymtotic
$$\int\_0^T \frac{|\zeta(\sigma+it)|^2}{T} \ dt \sim \zeta(2\sigma), \qquad T \uparrow \infty. $$
Is there more precise information about the value of $\theta$?
|
https://mathoverflow.net/users/345624
|
Asymptotic estimate for an integral involving the squared modulus of the Riemann zeta function
|
Let us introduce the notation
$$M(T):=\int\_0^T|\zeta(\sigma+it)|^2\,dt.$$
Then
$$\int\_0^T \frac{|\zeta(\sigma+it)|^2}{\sqrt{1+t^2}} \,dt=\int\_0^T\frac{dM(t)}{\sqrt{1+t^2}}=\frac{M(T)}{\sqrt{1+T^2}}+\int\_0^T\frac{tM(t)}{(1+t^2)^{3/2}}\,dt$$
by writing this as a Riemann-Stieltjes integral and then integrating by parts. Using that $M(t)$ is asymptotically $\zeta(2\sigma)t$, we conclude that the left-hand side is asymptotically $\zeta(2\sigma)\log T$.
|
9
|
https://mathoverflow.net/users/11919
|
402526
| 165,179 |
https://mathoverflow.net/questions/402534
|
9
|
Suppose that $\mathcal{V}$ is a symmetric monoidal model category, and that $\mathcal{C}$ is a $\mathcal{V}$-enriched model category. Write $\Bbb{R}\!\operatorname{Hom}(-,-)$ for the derived Hom functor
$$
\Bbb{R}\!\operatorname{Hom}(-,-) : \operatorname{Ho}(\mathcal{C})^{\textrm{op}}\times\operatorname{Ho}(\mathcal{C})\to\mathcal{V}.
$$
Suppose also that $\mathcal{C}$ (respectively $\mathcal{V}$) has functorial cofibrant and fibrant replacement functors, denoted $Q\_\mathcal{C}$ and $R\_\mathcal{C}$ (respectively $Q\_\mathcal{V}$ and $R\_\mathcal{V}$), so that $\Bbb{R}\!\operatorname{Hom}(X,Y)$ may be computed as $\mathcal{C}(Q\_\mathcal{C}(X),R\_\mathcal{C}(Y)).$
Let $X\in\mathcal{C},$ let $F : D\to\mathcal{C}$ be a diagram in $\mathcal{C},$ and suppose that $\operatorname{holim}F$ and $\operatorname{holim}\Bbb{R}\!\operatorname{Hom}(X,-)\circ F$ exist.
**Question:** Is it true that
$$
\operatorname{holim}\left(\Bbb{R}\!\operatorname{Hom}(X,-)\circ F\right)\simeq\Bbb{R}\!\operatorname{Hom}(X,\operatorname{holim}F)?
$$
If this is true, what is the proof (preferably a proof using the language of model categories, as opposed to a proof using $\infty$-categories). If this is not true in general, I would be interested in knowing what general hypotheses could be placed on the objects/functors/categories involved which would guarantee that there is such an equivalence.
I've heard that in the language of $\infty$-categories, we do have commutativity of limits and homs, which makes me suspect that this statement at the level of model categories should hold. However, I'm not sure if the translation is so direct, and it makes me a bit suspicious that I have not been able to find a statement like the above anywhere in my searches of the literature, and my attempts to prove the statement have not been fruitful.
Thanks!
|
https://mathoverflow.net/users/29322
|
Does derived hom commute with homotopy limits?
|
Yes, this is always true.
Replacing $X$ by its cofibrant replacement if necessary, we can assume $X$ to be cofibrant.
In this case, $\def\Hom{\mathop{\rm Hom}} \Hom(X,-)\colon C→V$ is a right Quillen functor.
The right derived functor of this right Quillen functor computes the derived hom $\def\RHom{\mathop{\rm RHom}} \RHom(X,-)$.
We can now conclude by invoking the fact that right derived functors of right Quillen functors preserve homotopy limits.
The latter fact can be proved as follows.
For simplicity of exposition, suppose that the injective model structure on D-indexed diagrams valued in $C$ and $V$ exists.
(This is true in almost all practical cases, and the assumption
can be eliminated anyway by working with more complicated models for homotopy limits.)
Replacing $F$ by its fibrant replacement if necessary, we can assume $F$ to be injectively fibrant.
Then $\RHom(X,-)∘F$ is also injectively fibrant.
Homotopy limits of injectively fibrant diagrams can be computed as ordinary limits.
Thus, the statement becomes
$$\lim(\Hom(X,-)∘F) ≅ \Hom(X,\lim F),$$
which is a true 1-categorical fact.
|
10
|
https://mathoverflow.net/users/402
|
402539
| 165,182 |
https://mathoverflow.net/questions/402540
|
0
|
I have a set of equations with some inequality constraints that I expect generally does not have a unique solution.
The equations take the form below:
$$\alpha/N+(1-\alpha)x\_1=a\_1$$
$$\alpha/N+(1-\alpha)x\_2=a\_2$$
$$\vdots$$
$$\alpha/N+(1-\alpha)x\_N=a\_N$$
$$x\_1+x\_2+\dots+x\_N=1$$
$$0<x\_i<1$$
$$1>\alpha>0$$
where $N$ is fixed and $a\_i$ are known and satisfy $a\_i>0$ and $a\_1+\dots a\_N=1$.
If I toy around in Mathematica I get that if I take $N=3$ and $(a\_1,a\_2,a\_3)=(1/2,3/10,1/5)$ the feasible solutions are given as a family of solutions for $x\_1$, $x\_2$, and $x\_3$ all depending on $\alpha<3/5$. I expect in general that one can get some estimated range for $\alpha$ that depends on the values of $a\_i$, computationally for instance with enough time.
I am wondering if there is any structure I can utilize for this set of equations to simplify things, or if there is a simple reduction of this problem to a simpler one.
It seems from the examples that $\alpha<1-2\min\{a\_i\}$ is the best bound, but I don't see why this is true immediately.
P.S. The motivation for looking at these equations is related to probability, and estimation of a distribution from its marginals.
|
https://mathoverflow.net/users/69486
|
Bounding parameter satisfying a collection of inequalities
|
Let $n:=N$ and $t:=\alpha$. We have
$$0<x\_i=\frac{a\_i-t/n}{1-t}<1$$
for all $i\in[n]:=\{1,\dots.n\}$ --
or, equivalently,
$$t<t\_{n,a}:=\min\_{i\in[n]}\min\Big(\frac{1-a\_i}{1-1/n},na\_i\Big)
=\min\Big(\frac{1-a\_{\max}}{1-1/n},na\_{\min}\Big),$$
where $a\_{\max}:=\max\_{i\in[n]}a\_i$ and $a\_{\min}:=\min\_{i\in[n]}a\_i$.
So, $t\_{n,a}$ is the best bound on $t$.
If, as in your example, $n=3$ and $(a\_1,a\_2,a\_3)=(1/2,3/10,1/5)$, then $t\_{n,a}=3/5$, as you found. If, as in your comment, $n=4$ and $(a\_1,a\_2,a\_3,a\_4)=(1/2,1/6,1/6,1/6)$, then $t\_{n,a}=2/3$, as you also found.
|
3
|
https://mathoverflow.net/users/36721
|
402541
| 165,183 |
https://mathoverflow.net/questions/402536
|
8
|
Given matrices $A, B \in \mathbb{R}^{n\times n}$, I would like to solve the following optimization problem,
$$\begin{array}{ll} \underset{v \in \mathbb{R}^n}{\text{maximize}} & \|Av\|\_2+\|Bv\|\_2\\ \text{subject to} & \|v\|\_2 = 1\end{array}$$
I'm hoping to solve this with some sort of convex optimization approach, such as SDP or SOCP. When $B=0$, the problem reduces to simply asking for the largest singular value of $A$, which can be solved by an SDP (see [here](https://math.stackexchange.com/questions/2136401/spectral-norm-minimization-via-semidefinite-programming) for instance although it's very well known). It can also be computed with SVD of course.
I've tried lots of different approaches but can't seem to write it in a convex way. I would be okay with an SVD-like solution, that is, one that is iterative not a convex program, but I would much prefer a convex program because ultimately I would like to use this as an inner program to something else, ideally.
---
As a note on one attempt that got a decent ways, I did manage to establish that it was equivalent to the problem,
$$\begin{array}{ll} \underset{u, v \in \mathbb{R}^n}{\text{maximize}} & \|A^T u + B^T v\|\_2\\ \text{subject to} & \|u\|\_2 = \|v\|\_2 = 1\end{array}$$
|
https://mathoverflow.net/users/97603
|
Maximizing sum of vector norms
|
I have no doubt that someone will come with some brighter idea but here are my 2 cents anyway.
If you don't aim at something very fast, I would just use the inequality $(a+b)^2\le (ta^2+(1-t)b^2)(t^{-1}+(1-t)^{-1})=F\_t(a,b)$ and try to find $\min\_{t\in[0,1]}\max\_v F\_t(\|Av\|,\|Bv\|)$. The maximum inside is just $(t^{-1}+(1-t)^{-1})$ times the maximal eigenvalue of $tA^\*A+(1-t)B^\*B$ and the function $F\_t$ is convex in $t$, so the maximum in $v$ is also convex, which makes simple one-dimensional tools for finding the minimum like bisection quite efficient.
The reason it will work is simple: the maximal unit eigenvector $v$ will move continuously (when you have a multiple eigenvalue, you can slowly move it from one limiting position to the other keeping the parameter $t$ constant), so there will be a moment in that process where $\|Av\|/\|Bv\|=(1-t)/t$. At that moment the inequality in the Cauchy-Schwarz will become equality, so the sum of the norms will achieve the upper bound coming from the quadratic relaxation (I assume that $A,B$ are both non-degenerate in this argument). Thus, the minimax we created really equals the maximin.
Note that it will all work nicely if you want *just the value of the maximum, not the maximizing vector itself*. The difficulty with the latter is that you may achieve the minimum in $t$ where the maximal eigenvalue is multiple, in which case not *every* eigenvector will be good but only the one for which the ratio of the norms is just right. Fortunately, generically it should not happen. However, you should be ready for this nuisance.
|
5
|
https://mathoverflow.net/users/1131
|
402544
| 165,185 |
https://mathoverflow.net/questions/402508
|
6
|
I am interested in the complexity of multiplying two matrices $A$ and $B$, i.e. to compute $AB$.
From [Le Gall and Urrotia], I know that:
* if $A$ and $B$ are square-matrices of size $n$, then this can be done in $O(n^{\omega})$ where $\omega\approx 2.372$.
* if $A$ has size $n\times n^{k}$ and $B$ has size $n^k \times n$ this can be done in $O(n^{\omega(k)})$ with $\omega(k)<\omega$ for $k<1$ (typically, $\omega(k)=2$ for $k\le0.3$).
I am wondering if there an efficient algorithm when $A$ has size $n^k\times n$ and $B$ has size $n\times n$ (or when $B$ has size $n\times n^{k}$), for $k\in(0,1)$.
Remarks:
* when I say "efficient" algorithm, I mean an algorithm that a complexity smaller than the naïve $O(n^{2+k})$ algorithm. I am not refering to an actual implementation, just the existence of an algorithm.
* I suppose that additing or multiplying two entries of a matrix is $O(1)$.
[Le Gall and Urrotia] [Improved Rectangular Matrix Multiplication using Powers of the
Coppersmith-Winograd Tensor by François Le Gall† Florent Urrutia](https://epubs.siam.org/doi/pdf/10.1137/1.9781611975031.67)
|
https://mathoverflow.net/users/90045
|
Complexity of rectangular matrix multiplication
|
Assuming that *efficient* means better than the naive $O(n^{2+k})$ multiplication, let us review some possibilities.
**Padding.** For $k > \omega-2$, just pad $A$ with $n-n^k$ zero or garbage rows, perform square matrix multiplication in $O(n^\omega)$ time, and discard the extra rows from the output. This is already efficient because $\omega < 2+k$. (For $k < \omega-2$ the naive method would be better.)
**Naive splitting.** Following Knight (1995), we note that $\langle m,n,p \rangle$ matrix multiplication, with $m$ the smallest of the three dimensions, can be done in $O(m^{\omega-2}np)$ time. This is by splitting $A\_{m \times n}$ into $1 \times (n/m)$ blocks of size $m \times m$, and splitting $B\_{n \times p}$ into $(n/m) \times (p/m)$ blocks of same size. Then we naively perform the block matrix multiplication of size $\langle 1, \, n/m, \, p/m \rangle$ by performing $np/m^2$ square multiplications of size $m$, each in time $O(m^\omega)$, for a total of $O(m^{\omega-2}np)$. Applying to the current question with input size $\langle n^k, n, n \rangle$ we achieve $O(n^{k(\omega-2)}n^2) = O(n^{2+k(\omega-2)})$. This is efficient and better than padding for all $0<k<1$.
These are just some straightforward transformations that show that "efficient" is possible for the task given. I believe newer developments (Le Gall and others) may be provide more efficient methods.
**Tensor powering.** One thing to note is that the splitting method above used fast MM inside the square blocks, but naive MM to combine them. In fact one can use fast methods on both levels, and the blocks need not be square. Coppersmith and Winograd (1982) note that if we have two rectangular matrix multiplication methods,
* $\langle m,n,p \rangle$ in $K$ multiplications and
* $\langle m',n',p' \rangle$ in $K'$ multiplications,
then we can do
* $\langle mm', nn', pp' \rangle$ in $KK'$ multiplications.
This is a powerful tool for combining MM methods of different dimensions. I am not sure, but possibly some of Le Gall & Urrutia's "symmetric" $\langle n, n^k, n \rangle$ rectangular methods that the OP cites could be "tensored up" in this way to solve the "asymmetric" $\langle n^k, n, n\rangle$ case for the current question.
**Almost quadratic.** Coppersmith (1982) reports that if $k = 0.17227$, then we can perform the $\langle n^k, n, n\rangle$ multiplication in $O(n^2 \, (\log n)^2)$ time, that is, almost in quadratic time! Note that the naive splitting would give an exponent of $2 + k(\omega-2) \le 2.0643$, using $\omega \le 2.373$.
Bibliography
------------
*Knight, Philip A.*, [**Fast rectangular matrix multiplication and QR decomposition**](http://dx.doi.org/10.1016/0024-3795(93)00230-W), Linear Algebra Appl. 221, 69-81 (1995). [ZBL0827.65044](https://zbmath.org/?q=an:0827.65044).
*Coppersmith, D.*, [**Rapid multiplication of rectangular matrices**](http://dx.doi.org/10.1137/0211037), SIAM J. Comput. 11, 467-471 (1982). [ZBL0486.68031](https://zbmath.org/?q=an:0486.68031).
*Coppersmith, D.; Winograd, S.*, [**On the asymptotic complexity of matrix multiplication**](http://dx.doi.org/10.1137/0211038), SIAM J. Comput. 11, 472-492 (1982). [ZBL0486.68030](https://zbmath.org/?q=an:0486.68030).
|
8
|
https://mathoverflow.net/users/171662
|
402552
| 165,187 |
https://mathoverflow.net/questions/100650
|
7
|
Quantum motivation
------------------
[Noncontextuality inequalities](http://arxiv.org/abs/1102.0264) (and in particular Bell inequalities) can be [mapped into graphs](http://arxiv.org/abs/1010.2163), in such a way that its relevant properties can be calculated via some simple graph-theoretical functions. In particular, the maximal quantum violation of a noncontextuality inequality is upperbounded by the [Lovász function](http://en.wikipedia.org/wiki/Lovasz_number) of its graph.
A family of inequalities that interests myself particularly is represented by the [Möbius ladder](http://en.wikipedia.org/wiki/M%C3%B6bius_ladder); I was unable to find its Lovász function in the literature, but from [physical and mathematical grounds](http://arxiv.org/abs/1206.3212) we managed to prove that
$$\frac{n}{2}\bigg(1+\cos\frac{\pi}{n}\bigg) \le\vartheta(M\_{2n}) \le \frac{n\big(2\cos(\pi/n)+1\big)}{2+\cos(\pi/n)}.$$
Note that both bounds coincide in the asymptotic limit. Furthermore, my physical intuition claims that the function is always equal to this lower bound. I cannot prove it, however.
Quantum-free question
---------------------
Let $M\_{2n}$ be the [Möbius ladder](http://en.wikipedia.org/wiki/M%C3%B6bius_ladder) graph. I conjecture ts [Lovász function](http://en.wikipedia.org/wiki/Lovasz_number) to be
$$\vartheta(M\_{2n}) = \frac{n}{2}\bigg(1+\cos\frac{\pi}{n}\bigg).$$
But I can not prove this. I hope that this problem would be easy to a graph-theoretician, or even buried in the literature somewhere. Anybody knows?
|
https://mathoverflow.net/users/9211
|
Lovász function of the Möbius ladder
|
Now, 9 years later, this conjecture has been proven by [Bharti et al.](https://arxiv.org/abs/2104.13035) (one of the authors is Adán Cabello, who was in the paper that originally posed the conjecture).
The technique used was semidefinite programming (SDP) duality: computing the Lovász function is an SDP, and as all SDPs it has a dual SDP that upperbounds the solution of the primal SDP. They managed to find a feasible solution of the dual for all even $n$ that coincides with the already known lower bound that came from a feasible solution of the primal, proving that both solutions are optimal and give the Lovász function.
|
1
|
https://mathoverflow.net/users/9211
|
402555
| 165,188 |
https://mathoverflow.net/questions/402554
|
1
|
Is there an example of a non-affine scheme $X$ such that every short exact sequence of vector bundles over $X$ splits? If there are such examples then what if we ask it to be true of all (not necessarily finite rank) locally free $\mathcal{O}\_X$-modules
|
https://mathoverflow.net/users/99988
|
Are there nonaffine schemes over which every exact sequence of vector bundles is split?
|
Affine plane with double origin works.
|
5
|
https://mathoverflow.net/users/348986
|
402559
| 165,189 |
https://mathoverflow.net/questions/298248
|
1
|
Let $m \in \mathbb{N}\setminus \{0,1\}$, $\alpha \in ]0,1[$.
Let $\Omega$ be a bounded open subset of $\mathbb{R}^n$ of class $C^{m,\alpha}$.
It is known that if $f \in C^{\frac{m-2+\alpha}{2},m-2+\alpha}([0,T]\times \mathrm{cl}\,\Omega)$, $g \in C^{\frac{m+\alpha}{2};m+\alpha}([0,T]\times\partial\Omega)$, $u\_0 \in C^{m,\alpha}(\mathrm{cl}\, \Omega)$ (satisfying some compatibility conditions at $t=0$), then there exists a unique solution $u$ in $C^{\frac{m+\alpha}{2};m+\alpha}([0,T]\times \mathrm{cl}\,\Omega)$ of
\begin{cases}
\partial\_t u -\Delta u = f &\mbox{ in }[0,T]\times \mathrm{cl}\,\Omega,\\
u=g & \mbox{ on } [0,T] \times \partial\Omega,\\
u(0,\cdot) = u\_0 & \mbox{ in }\mathrm{cl} \, \Omega.
\end{cases}
For the elliptic case holds a very similar result, but in this case we allow $m$ to be also $1$, that is $m \in \mathbb{N}\setminus \{0\}$. If $f \in C^{m-2,\alpha}(\mathrm{cl}\,\Omega)$, $g \in C^{m,\alpha}(\partial\Omega)$, then there exists a unique solution $u$ in $C^{m,\alpha}(\mathrm{cl}\,\Omega)$ of
\begin{cases}
\Delta u = f &\mbox{ in }\mathrm{cl}\,\Omega,\\
u=g & \mbox{ on } \partial\Omega.
\end{cases}
In this case, for $m=1$, the space $C^{-1,\alpha}(\mathrm{cl}\,\Omega)$ is the space of distributions which euqals the divergence of an element in $C^{0,\alpha}(\mathrm{cl}\,\Omega,\mathbb{C}^n)$, and the Laplacian is to be intended in the weak sense.
Then my question is the following:
There exists an analog of the case $m=1$ for the heat (parabolic) equation?
|
https://mathoverflow.net/users/58541
|
Schauder regularity heat equation
|
For a partial answer, see Theorem 6.48 in *Second order parabolic differential equations*, 1996, by Gary Lieberman.
There is an additional assumption that $f$ belongs to the Morrey space $M^{1,n+1+\alpha}$ defined page 130 of the book. In particular, $L^{\infty}\subset M^{1,n+1+\alpha}\subset L^1$.
I do not know if you can remove this additional assumption or not.
|
1
|
https://mathoverflow.net/users/349090
|
402570
| 165,193 |
https://mathoverflow.net/questions/402549
|
6
|
Let $f:\mathbb{R}\times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ be a smooth function and $G\subset \operatorname{SO}(n)$ be a $1$-dimensional compact Lie group (diffeomorphic to the circle). Moreover let $G$ act on $\mathbb{R}^{n}$ by standard left multiplication. We assume that $f$ is equivariant with respect to $G$, i.e. for all $g\in G$ and all $(t,x)\in \mathbb{R}\times \mathbb{R}^{n}$ we have $f(t,g\cdot x) = g\cdot f(t,x)$. Let now $(t\_{0},x\_{0}) \in \mathbb{R}\times \mathbb{R}^{n}$ such that $x\_{0} \not= 0$, $f(t\_{0},x\_{0}) = 0$ and
$$\ker \left ( \frac{\partial f}{\partial x}(t\_{0},x\_{0}) \right ) = T\_{x\_{0}}(G\cdot x\_{0}),$$
i.e. the kernel of the Jacobian of $f$ is only in the direction of the action.
**QUESTION:** Is there any version of the implicit function theorem in this setting? Does one need more additional conditions to be able to construct near-by solutions? If yes, which conditions are these?
Thanks in advance.
|
https://mathoverflow.net/users/348912
|
Equivariant implicit function theorem
|
The equivariant version of the implicit function theorem is the following.
>
> Let $f: \mathbb{R}^p \times \mathbb{R}^n \to \mathbb{R}^m$ be a smooth function (possibly only defined on open neighborhoods) which is equivariant with respect to the action of a compact Lie group $G$ on $\mathbb{R}^n$ and $\mathbb{R}^m$. Let $(t\_0, x\_0)$ be such that $f(t\_0, x\_0) = 0$. Assume that the stabilizer of $x\_0$ is trivial and that $0$ is a fix point for the action on $\mathbb{R}^m$ (these assumptions are not essential but simplify the argument, see below). Consider the so-called deformation complex $$\mathfrak{g} \to \mathbb{R}^n \to \mathbb{R}^m,$$
> where the first arrow is the action of the Lie algebra at the point $x\_0$, i.e. $\xi \mapsto \xi \,. x\_0$ and the second arrow is the differential of $f$ at $(t\_0, x\_0)$ with respect to the second slot (i.e. the Jacobian).
> If this complex is exact, then there exist a smooth function $x: \mathbb{R}^p \to \mathbb R^n$ such that
> $$
> \{ (t, g \cdot x(t)) | t \in \mathbb R^p, g \in G \} = f^{-1}(0).
> $$
>
>
>
(I'm a bit sloppy here and in the proof below: everything needs to be restricted to open neighborhoods of $t\_0$, $e \in G$ and $0 \in \mathbb{R}^m$ etc).
Remark: Your assumption about the derivative is equivalent to the exactness of the complex in the first arrow. However, since you assume that $n = m$, the complex is never exact in the second arrow. What you could do is apply this result to the function $pr \circ f$, where $pr$ is the projection onto the image of the Jacobian.
Proof: Since $G$ is compact, and the action is free at $x\_0$, there exist slice coordinates around $x\_0$, i.e. there is a map $\iota: \mathbb{R}^{n-d} \to \mathbb{R}^n$ such that $\iota(0) = x\_0$ and such that $\chi: G \times \mathbb{R}^{n-d} \to \mathbb{R}^n, (g, y) \mapsto g \cdot \iota(y)$ is a local diffeo (here $d$ is the dimension of $G$). Define the map $F: \mathbb{R}^p \times \mathbb{R}^{n-d} \to \mathbb{R}^{n}$ by $F(t, y) = f(t, \iota(y))$. The assumption about the exactness of the deformation complex is equivalent to the invertibility of the Jacobian of $F$. Thus, using the ordinary implicit function theorem, there exist $y(t)$ such that $F(t, y(t)) = 0$ and every such point in the zero level set is of this form. Set $x(t) = \iota(y(t))$ and the claim follows as $\chi$ is a local diffeo.
Final remark: The statement generalizes directly to actions on manifolds, and properness of the action is enough (instead of compactness of $G$). In fact, they generalize even to the infinite-dimensional setting. Moreover, the assumptions about the stabilizers of $x\_0$ and $f(t\_0, x\_0)$ can be relaxed. You can find the details in my PhD thesis <https://arxiv.org/abs/1909.00744> and in the paper <https://arxiv.org/abs/2010.10165>.
|
7
|
https://mathoverflow.net/users/17047
|
402581
| 165,197 |
https://mathoverflow.net/questions/402422
|
15
|
There is a general result which holds for the rational numbers $ \mathbb Q $ (as well as number fields in general):
>
> For any completion $ K $ of $ \mathbb Q $ and any finite extension $ L/K $ of degree $ n $, the function $ L \to \mathbb R $ defined by $ x \to \sqrt[n]{|N\_{L/K}(x)|} $ gives a norm on $ L $.
>
>
>
The nontrivial part is to prove that the norm thus defined satisfies the triangle inequality. $ K $ is either $ \mathbb R $ or $ \mathbb Q\_p $ for a prime $ p $, and for the latter case one can argue using Hensel's lemma that there is an equivalence between having norm $ \leq 1 $ and being integral over the $ p $-adic integers, which combined with the ultrametric property of the $ p $-adic norm is sufficient to prove the claim. I'm satisfied with this argument in the sense that it seems to give a "moral" explanation for why we should expect the claim to be true.
However, the argument for $ K = \mathbb R $ is simply to note that the claim is obviously true for the only nontrivial finite extension of $ \mathbb R $, which is $ \mathbb C/\mathbb R $. While this clearly is sufficient for a proof, I don't understand why one should expect this to be true in advance. Is there a better motivation for why we should expect this result to hold, perhaps an argument which treats all completions symmetrically instead of having one argument for the finite primes and another for the prime at infinity?
|
https://mathoverflow.net/users/131052
|
Why does the field norm on the field extension $ \mathbb C/\mathbb R $ induce a vector space norm?
|
The map $|N(\cdot)|^{1/n}$ is a continuous multiplicative extension of $|\cdot|$.
By a multiplicative function I mean a function $\chi:L\to [0,\infty)$
such that $\chi(0)=0$, $\chi(1)=1$ and for every $x,y\in L$, $\chi(xy)=\chi(x)\chi(y)$.
A multiplicative function which satisfies for every $x,y\in L$, $\chi(x+y)\leq \chi(x)+\chi(y)$ is called an absolute value.
**Theorem:** Let $K$ be a field and $|\cdot|$ an absolute value on $K$.
Assume $K$ is complete with respect to the induced metric.
Then for every finite field extension of $K$, every continuous multiplicative extension of $|\cdot|$ is an absolute value.
In fact, there exists a unique such continuous multiplicative extension of $|\cdot|$, which is $|N(\cdot)|^{1/n}$.
The uniqueness follows at once from the fact that all norms are equivalent on finite dimensional vector spaces over complete fields: for multiplicative functions $\chi\_1,\chi\_2$, the function $\chi\_1/\chi\_2$ (defined on the multiplicative group) is multiplicative too, hence must be unbounded or trivial.
The less trivial part of the theorem is its first part.
In my other answer I gave a proof of this fact which indeed holds in the generality of complete fields. This post is to give an easy proof I found, under the assumption that the fields are locally compact.
We consider a local field $K$, endowed with an absolute value $|\cdot|$,
a finite field extension $L$ of $K$ and a continuous multiplicative extension $\chi$ of $|\cdot|$. We argue to show that $\chi$ is an absolute value on $L$.
Consider $L$ as a $K$-vector space and consider the corresponding space $\Omega$ consisting of $K$-norms on $L$.
Consider the multiplicative group $L^\*$ as a locally compact group (see below for justification) and let it act on $\Omega$ by $\|\cdot\|\mapsto \chi(x)^{-1} \|x\cdot\|$
for $x\in L^\*$. Note that for $x\in K^\*$,
$$ \chi(x)^{-1} \|x\cdot\|=\chi(x)^{-1} |x|\|\cdot\|=\|\cdot\|,$$
thus the $L^\*$-action on $\Omega$ factors via $L^\*/K^\*$,
which is a compact group, as it is homeomorphic to a projective space.
This action admits a fixed point. Indeed, for every norm $\|\cdot\|\in \Omega$,
the map
$$ L \ni v \mapsto \int\_{L^\*/K^\*} \chi(x)^{-1}\|xv\|~ \text{dHaar}\_{L^\*/K^\*}(x) \in [0,\infty)$$
is easily seen to be an $L^\*$-fixed norm on $L$.
We let $\|\cdot\|$ be such a fixed point which is normalized to satisfy $\|1\|=1$. Then for every $x\in L^\*$,
$$ \|x\|=\|x\cdot 1\|=\chi(x)\cdot \chi(x)^{-1}\|x\cdot 1\|=
\chi(x)\cdot \|1\|=\chi(x). $$
Thus $\|\cdot\|=\chi$ and in particular, we conclude that $\chi$ is indeed a norm.
This finishes the proof.
---
To see that $L^\*$ is a topological group we need to verify the continuity of the inversion map $x\mapsto x^{-1}$ on $L^\*$.
Its continuity on $K^\*$ follows by a standard argument from the existence of the absolute value $|\cdot|$.
From the continuity of the inversion on $K^\*$ we get its continuity on $\text{GL}\_n(K)$, as inversion is polynomial in the matrix entries and $\det(\cdot)^{-1}$, thus also its continuity on $L^\*$.
|
14
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https://mathoverflow.net/users/89334
|
402589
| 165,198 |
https://mathoverflow.net/questions/402598
|
3
|
Is there an open connected orientable 3-manifold $M$ with the following properties:
1. $M$ admits a complete hyperbolic metric with finite hyperbolic volume.
2. $H\_{i}(M,\mathbb{Z})=0$ for any $i>0$.
|
https://mathoverflow.net/users/17895
|
Special kind of 3-manifolds
|
No. Suppose that $M$ is a finite volume oriented hyperbolic three-manifold. In the closed case, as $M$ is oriented, we have $H\_3(M, \mathbb{Z}) \cong \mathbb{Z}$ generated by the fundamental class. In the open case, $M$ has torus cusps. Appealing to "one-half lives, one-half dies" we find that $M$ has non-trivial (in fact infinite) $H\_1$.
[See Lemma 3.5 of Hatcher's [notes on three-manifolds](https://pi.math.cornell.edu/%7Ehatcher/3M/3Mdownloads.html) for the statement and proof of "one-half lives, one-half dies".]
|
5
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https://mathoverflow.net/users/1650
|
402601
| 165,202 |
https://mathoverflow.net/questions/402623
|
11
|
By Serre's theorem, we know the only nontorsion parts of the homotopy groups of spheres occur as $\pi\_n(S^n)$ and $\pi\_{4n-1}(S^{2n})$. The first of these are trivial to describe, but the second have very interesting, symmetric incarnations, they are the generalised hopf fibrations, at least for $n=1,2,4$, associated to the real normed division algebras.
Are there a similar explicit descriptions for representatives of these higher nontorsion elements too?
Even if we don't have explicit descriptions, do we know anything about the values of the hopf invariants associated to these maps?
|
https://mathoverflow.net/users/128502
|
Is there a concrete description of the nontorsion elements in the homotopy groups of spheres?
|
For $n \neq 1,2,4$, the minimal positive Hopf invariant of an element of $\pi\_{4n-1}(S^{2n})$ is $2$.
An explicit element of Hopf invariant $2$ can be constructed as follows: consider the attaching map $S^{4n-1} \to S^{2n} \vee S^{2n}$ of the $4n$-cell of the CW-complex $S^{2n} \times S^{2n}$ and compose it with the codiagonal $S^{2n} \vee S^{2n} \to S^{2n}$.
That no elements of Hopf invariant $1$ exist outside the "Adams dimensions" $n = 1,2,4$ is known as the "Hopf invariant 1 problem". It was studied and solved by Adams. Later Adams and Atiyah found a shorter proof. For a very accessible write-up, see this [REU paper](http://math.uchicago.edu/%7Emay/REU2016/REUPapers/Banerjee.pdf).
|
21
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https://mathoverflow.net/users/14233
|
402624
| 165,205 |
https://mathoverflow.net/questions/402618
|
4
|
[**Semiring categories**](https://ncatlab.org/nlab/show/rig+category), also called **rig categories** or **bimonoidal categories**, are pseudomonoids in the symmetric monoidal bicategory $(\mathsf{SymMonCats},\otimes\_{\mathbb{F}},\mathbb{F})$¹. These are a categorification of semirings, the monoids in $(\mathsf{CMon},\otimes\_{\mathbb{N}},\mathbb{N})$, and carry two monoidal structures, one additive and one multiplicative, with the multiplicative one being coherently bilinear over the additive monoidal structure. A great introduction for these is [Johnson–Yau's *Bimonoidal Categories, $E\_n$-Monoidal Categories, and Algebraic $K$-Theory*](https://nilesjohnson.net/drafts/Johnson_Yau_ring_categories.pdf).
Similarly, **ring categories** are pseudomonoids in $(\mathsf{2Ab},\otimes\_{\mathbb{S}},\mathbb{S})$. They are a categorification of rings, the monoids in $(\mathsf{Ab},\otimes\_{\mathbb{Z}},\mathbb{Z})$, and are more simply those semiring categories having (weak) additive inverses (i.e. for each object $A$ in a ring category, there exists an object $-A$ such that $A\oplus(-A)\cong\mathbf{0}\_{\mathcal{C}}$ via a coherent isomorphism).
It is well-known that given a monoidal category $(\mathcal{C},\otimes,\mathbf{1}\_{\mathcal{C}})$, we can use [Day convolution](https://ncatlab.org/nlab/show/Day+convolution) to get a monoidal structure on presheaves on $\mathcal{C}$, obtaining a monoidal category $(\mathsf{PSh}(\mathcal{C}),\circledast,\mathsf{h}\_{\mathbf{1}\_{\mathcal{C}}})$.
**Question 1.** Given a semi/ring category $(\mathcal{C},\oplus,\otimes,\mathbf{1}\_{\mathcal{C}},\mathbf{0}\_{\mathcal{C}},\ldots)$, is the tuple $(\mathcal{C},\circledast^{\oplus},\circledast^{\otimes},\mathsf{h}\_{\mathbf{1}\_{C}},\mathsf{h}\_{\mathbf{0}\_{C}},\ldots)$ obtained by applying Day convolution to both monoidal structures also a semi/ring category?
**Question 2.** Day convolution gives a bijection
$$
\{\text{promonoidal structures on $\mathcal{C}$}\}
\cong
\{\text{biclosed monoidal structures on $\mathsf{PSh}(\mathcal{C}$})\}.
$$
Assuming the statement in question 1 holds, is there an analogue of this bijection for semi/ring categories?
---
¹More or less―for nonsymmetric bimonoidal categories, 19 of the 22 nonsymmetric bimonoidal category axioms of [Johnson–Yau, Definition 2.1.2](https://nilesjohnson.net/drafts/Johnson_Yau_ring_categories.pdf) hold. The exceptions are 2.1.13, 2.1.15, and 2.1.16.
|
https://mathoverflow.net/users/130058
|
Day convolution for bimonoidal categories
|
Regarding Q2: probably there is a way to avoid going deep into coherence conditions: instead of proving by hand the equivalence between promonoidal structures on $C$ and biclosed monoidal structures on $\hat C$, one can resort to a more conceptual pov.
What happens for pro/monoidal categories is that there is a pseudomonad $S$ on $\sf Cat$ with the property that $S$ lifts to a pseudomonad $\hat S$ on $\sf Prof$ (the Kleisli bicategory of $P=\hat{(-)} = [(-)^{op},{\sf Set}]$), and pseudo-$S$-algebra structures correspond to pseudo-$\hat S$-algebra structures (this is an equivalence of categories, in the appropriate sense; see [here](https://mathoverflow.net/questions/352843/promonoidal-categories-as-s-algebras)).
I believe a similar argument holds for every (almost every?) monad $S$ equipped with a distributive law over $P$ (the presheaf construction); this does not fall short from an equivalence
$$
\{S\text{-algebra structures on } PX\} \cong \{\hat S\text{-algebra structures on } X\}
$$ where $PX$ is regarded as an object of $\sf Cat$, and $X$ as an object of ${\sf Kl}(P)$.
Regarding Q1: have you tried to find the distributive and annullator morphisms for the putative 2-rig structure on $\widehat{C}$?
I was trying to find at least one distributive morphism, and I have no idea how to reduce $F\hat{\otimes}(H\hat{\oplus} K)$ to/from $F\hat{\otimes} H \,\hat{\oplus}\, F\hat\otimes K$, if $F,H,K : \widehat{C}$. If I'm not wrong (this is very back-of-the-envelope coend calculus),
$$\begin{align\*}
F\hat\otimes H &= \int^{UA}FU\times HA\times [\\_, U\otimes A]\\
F\hat\otimes K &= \int^{U'B}FU'\times KB\times [\\_, U'\otimes B]
\end{align\*}$$
whereas
$$\begin{align\*}
F \hat\otimes \,(H\hat\oplus K) &= \int^{UV} FU \times (H\hat\oplus K)V \times [\\_, U\otimes V] \\
&=\int^{UVAB} FU \times HA \times KB \times [V, A\oplus B] \times [\\_, U\otimes V] \\
&=\int^{UAB} FU \times HA \times KB \times [\\_, U\otimes (A\oplus B)] \\
&=\int^{UAB} FU \times HA \times KB \times [\\_, U\otimes A \oplus U\otimes B] \\
\end{align\*}$$
...and now we're stuck, unless we have either
* R1. a compatibility between $\oplus$ and $\times$, perhaps *another* distributive morphism;
* R2. a siftedness condition ensuring that
$$ \int^{UA}FU\times HA\times [\\_, U\otimes A] \oplus \int^{U'B}FU'\times KB\times [\\_, U'\otimes B]$$
can be reduced to an integral on just $U$.
Actually, you need both in order for the computation to proceed; but the conjunction of R1 and R2 is quite strong, as you can see.
---
Edit: the situation with annullators (for Laplaza, morphisms ${\bf 0}\otimes X \to {\bf 0}$ and $X\otimes {\bf 0} \to \bf 0$) is even worse!
Let's open $F \hat\otimes {\bf 0}$ recalling that in this case $\bf 0$ is the representable $y{\bf 0}$ on the additive unit of $C$:
$$\begin{align\*}
\int^{UV} FU \times [V,{\bf 0}] \times [\\_,U\otimes V]
&=\int^U FU \times [\\_, U\otimes {\bf 0}] \\
&\overset{\rho\_U}\to\int^U FU \times [\\_, {\bf 0}]\\
&=\varinjlim F \times [\\_, {\bf 0}]
\end{align\*}$$
the cartesian structure on $\sf Set$ now entails that this is $\bf 0$ if and only if either factor is empty, but I see no way in which this can be or even map into $y{\bf 0}$ again, as it should.
|
4
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https://mathoverflow.net/users/7952
|
402628
| 165,207 |
https://mathoverflow.net/questions/402191
|
5
|
Perhaps the characteristic feature of the theory of ends is that they are extremely useful for computing sets of transformations between two functors. For example, one has the formulas
\begin{align\*}
\mathrm{Nat}(F,G) &\cong \int\_{A\in\mathcal{C}}\mathrm{Hom}\_{\mathcal{C}}\left(F\_{A},G\_{A}\right),\\
\mathrm{DiNat}(F,G) &\cong \int\_{A\in\mathcal{C}}\mathrm{Hom}\_{\mathcal{C}}\left(F^{A}\_{A},G^{A}\_{A}\right),
\end{align\*}
see [Coend Calculus](https://arxiv.org/abs/1501.02503), Theorem 1.4.1 and Example 1.4.4.
Is there a similar end formula for the set $\mathrm{Nat}^\otimes(F,G)$ of *monoidal* natural transformations between two strong monoidal functors $F,G\colon\mathcal{C}\rightrightarrows\mathcal{D}$?
|
https://mathoverflow.net/users/130058
|
End formulas for sets of monoidal natural transformations
|
### Monoidal ends
Let $(\mathcal{C},\otimes)$ be a monoidal category and let $(\mathcal{D},\times)$ be a *cartesian* monoidal category. Let $(X,\eta,\mu) : (\mathcal{C}^{\mathrm{op}},\otimes) \times (\mathcal{C},\otimes) \to (\mathcal{D},\times)$ be a lax monoidal functor. A wedge to $(X,\eta,\mu)$ is a wedge to the underlying functor $X : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathcal{D}$, thus consisting of an object $T \in \mathcal{D}$ and a family of morphisms $(w\_A : T \to X(A,A))\_{A \in \mathcal{C}}$, such that the following properties hold:
1. $w\_1 : T \to X(1,1)$ is equal to
$$T \xrightarrow{\exists!} 1 \xrightarrow{\eta} X(1,1).$$
2. $w\_{A \otimes B} : T \to X(A \otimes B, A \otimes B)$ is equal to
$$T \xrightarrow{~(w\_A,w\_B)~} X(A,A) \times X(B,B) \xrightarrow{~~\mu~~} X(A \otimes B, A \otimes B).$$
A universal wedge, i.e. end, is defined as usual. It is easy to see that if $\mathcal{C}$ is small and $\mathcal{D}$ is complete, then any lax monoidal functor has an end. We can denote it by $\int (X,\eta,\mu)$.
Since it is a common practice (*sigh*) to ignore forgetful functors and just write $X$ both for the functor and the monoidal functor, some people will prefer to call this a "monoidal wedge" and a "monoidal end", the latter then being denoted by something like $\int^{\otimes} X$. I do not know if this concept has appeared elsewhere, I just made it up to answer the question below.
Monoidal natural transformations
--------------------------------
Now let $(\mathcal{C},\otimes)$, $(\mathcal{C}',\otimes)$ be two monoidal categories and $(F,\eta\_F,\mu\_F),(G,\eta\_G,\mu\_G) : (\mathcal{C},\otimes) \to (\mathcal{C}',\otimes)$ be two strong monoidal functors. Consider the functor
$X : \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathbf{Set}$ defined on objects by
$$X(A,B) := \mathrm{Hom}(F(A),G(B)).$$
We equip it with the following lax monoidal structure:
$$\eta\_X : 1 \to X(1,1)$$
corresponds to the isomorphism $\eta\_G \circ \eta\_F^{-1} : F(1) \to 1 \to G(1)$, and
$$\mu\_X : X(A,A') \times X(B,B') \to X(A \otimes B,A' \otimes B')$$
maps a pair of morphisms $f : F(A) \to G(A')$, $g : F(B) \to G(B')$ to the morphism
$$F(A \otimes B) \xrightarrow{\mu\_F^{-1}} F(A) \otimes F(B) \xrightarrow{f \otimes g} G(A') \otimes G(B') \xrightarrow{\mu\_G} G(A' \otimes B').$$
One needs to check the coherence conditions in the definition of a lax monoidal functor, I will not do this here.
It is straight forward to check that $\int (X,\eta,\mu)$ is the set of morphisms $(F,\eta\_F,\mu\_F) \to (G,\eta\_G,\mu\_G)$ (aka monoidal natural transformations).
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2
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https://mathoverflow.net/users/2841
|
402630
| 165,208 |
https://mathoverflow.net/questions/402637
|
19
|
The one-point compactification $\mathbb{N}\_\infty$ of $\mathbb{N}$ is obtained from the discrete space $\mathbb{N}$ by adjoining a limit point $\infty$. It may be identified with the subspace of Cantor space
$$
\mathbb{N}\_\infty = \{ \alpha \in \{0,1\}^\mathbb{N} \mid
\forall n \,.\, \alpha\_n \geq \alpha\_{n+1} \}.
$$
Indeed, we may embed $\mathbb{N} \to \mathbb{N}\_\infty$ by mapping $n$ to the sequence
$$\overline{n} = \underbrace{1 \cdots 1}\_n 0 0 \cdots,$$
and taking $\infty = 1 1 1 \cdots$.
Classically of course adjoining a single point to a countable set has no effect on countability. How about the computable version? If we adjoin the new point as an isolated one then of course we again obtain a countable set. This question is about adjoining $\infty$ as a limit point in the sense of metric spaces.
Let $\varphi$ be a standard enumeration of partial computable maps.
>
> **Question:** Do there exist a total computable map $q$ and a partial computable map $s$ such that:
>
>
> 1. $\varphi\_{q(n)} \in \mathbb{N}\_\infty$ for all $n \in \mathbb{N}$
> 2. For all $k \in \mathbb{N}$, if $\varphi\_k \in \mathbb{N}\_\infty$ then $s(k)$ is defined and $\varphi\_{q(s(k))} = \varphi\_k$.
>
>
>
The map $q$ realizes an enumeration $\mathbb{N} \to \mathbb{N}\_\infty$, and $s$ the fact that $q$ is surjective.
**Clarification:** The following map $q : \mathbb{N} \to \mathbb{N}\_\infty$ comes to mind:
$$q(n)(k) =
\begin{cases}
1 & \text{if $T\_n$ has not terminated within $k$ steps of execution}\\
0 & \text{if $T\_n$ has terminated within $k$ steps of execution}
\end{cases}
$$
However, it seems hard to get the corresponding map $s$ witnessing surjectivity of $q$.
(I should say that $q$ works as a computable enumeration for yet a third way of adjoining a point to $\mathbb{N}$, namely $$\mathbb{N}\_\bot =
\{ S \subseteq \mathbb{N} \mid \forall i, j \in S \,.\, i = j \}.$$
We embed $n \in \mathbb{N}$ into $\mathbb{N}\_\bot$ as a singleton $\{n\}$, while the extra point is $\emptyset$. Think of $\mathbb{N}\_\bot$ as the set of enumerable subsets of $\mathbb{N}$ with at most one element.
|
https://mathoverflow.net/users/1176
|
Is the one-point compactification of $\mathbb{N}$ computably countable?
|
The answer is no. Suppose that there are computable functions $q$ and $s$ as you describe.
Let $k$ be a program that performs the following task. It starts enumerating $1$s at the start of the sequence until it discovers that $s(k)$ is defined. (We use the Kleene recursion theorem to know that there is such a self-referential program $k$.) Note that this must eventually happen, since otherwise $\varphi\_k$ would be $\infty$, in which case $s(k)$ should be defined.
When it finds that $s(k)$ is defined, then the program pauses the enumeration of its output and starts computing $\varphi\_{q(s(k))}$. This will definitely produce an element of $\mathbb{N}\_\infty$. And so the program waits until either it produces more $1$s than we have put on $\varphi\_k$, in which case program $k$ switches to $0$s immediately, causing $\varphi\_{q(s(k))}\neq\varphi\_k$; or else $\varphi\_{q(s(k))}$ produces a $0$, in which case we can let $\varphi\_k$ produce all $1$s, again causing $\varphi\_{q(s(k))}\neq\varphi\_k$.
|
17
|
https://mathoverflow.net/users/1946
|
402641
| 165,210 |
https://mathoverflow.net/questions/402577
|
4
|
Let $M$ be a closed $6$-dimensional Riemannian manifold with a spin$^{\mathbb{C}}$ structure. It is known that real $4$-forms on $M$ act on the positive-spinors as trace-free hermitian endomorphisms by Clifford multiplication (say, denoted by $\gamma$). Now for a real $3$-form $\beta$ on $M$, one can see that $\gamma(\beta)^2-|\beta|^2\mathrm{Id}$ is trace-free and hence must represent a $4$-form on $M$. Can we identify this $4$-form?
|
https://mathoverflow.net/users/131004
|
Identifying a $4$-form on a $6$-dimensional manifold
|
$\newcommand{\R}{\mathbb{R}}$As you state, there is a $SO(6)$-equivariant map $\delta:\operatorname{Sym}^2(\Lambda^3\R^6)\to \Lambda^4 \R^6$ such that $\gamma(\delta(\beta^{\otimes 2})) = \gamma(\beta)^2 - |\beta|^2\operatorname{id}$ on the positive spinors (the central $U(1)\subset\operatorname{Spin}^c(6)$ cancels, so that it suffices to work with the quotient $SO(6)$). By adjunction, this corresponds to an invariant tensor in $\operatorname{Sym}^2(\Lambda^3 \R^6)\otimes\Lambda^4\R^6\subset (\R^6)^{\otimes 10}$. It is a classical fact, known as the first fundamental theorem for the orthogonal group (compare [this answer](https://mathoverflow.net/questions/238788/invariant-polynomials-under-diagonal-action-of-the-orthogonal-group)), that these invariants have a basis spanned by perfect matchings of the ten indices into five pairs, which are then contracted with the metric tensor. The (anti-)symmetry constraints then exhibit the representation $\operatorname{Sym}^2(\Lambda^3 \R^6)\otimes\Lambda^4\R^6$ as a quotient, in which many of these pairings vanish by antisymmetry; with a little work, one finds that the space of invariants is $1$-dimensional, corresponding to the map $\operatorname{Sym}^2(\Lambda^3\R^6)\to \Lambda^4 \R^6$ which contracts two tensor indices of the two inputs with the metric and then antisymmetrizes the remaining 4 indices. So the map $\delta$ must be a constant multiple of this map, and the constant can be easily determined by considering $\beta = e\_1\wedge e\_2\wedge e\_3 + e\_1\wedge e\_4\wedge e\_5$.
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2
|
https://mathoverflow.net/users/35687
|
402642
| 165,211 |
https://mathoverflow.net/questions/402631
|
0
|
Let $\mathbf{v}\_1, \mathbf{v}\_2$ be two vectors in $\mathbb{R}^n$. I would like to compute the following singular integral:
$$\int\_{-\infty}^{ \infty} \int\_{-\infty}^{\infty}
\int\_{[-1,1]^n}
e(\theta\_1 \mathbf{v}\_1.\mathbf{x} +\theta\_2 \mathbf{v}\_2.\mathbf{x} ) d \mathbf{x} d \theta\_1 d\theta\_2$$
I vaguely understand that this is a volume of the subset of $[-1, 1]^n$ where
$$\mathbf{v}\_1.\mathbf{x} = \mathbf{v}\_2.\mathbf{x} = 0$$ and $.$ denotes the dot product. But I'm struggling to show this rigorously... Any reference/suggestions is appreciated!
We use the notation $e(z) = e^{2 \pi i z}$
|
https://mathoverflow.net/users/84272
|
How to compute $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{[-1,1]^n}\exp[2\pi i(\theta_1 v_1.x+\theta_2v_2.x)]d^nx d\theta_1d\theta_2$
|
First integrate over $\theta\_1,\theta\_2$. Use the delta function representation (for $k\in\mathbb{R}$)
$$\int\_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$
to evaluate
$$\int\_{-\infty}^\infty \int\_{-\infty}^\infty e^{2\pi i (v\_1\cdot x)\theta\_1+2\pi i(v\_2\cdot x)\theta\_2}\,d\theta\_1 d\theta\_2=\delta(v\_1\cdot x)\delta(v\_2\cdot x).$$
Next for the integral over the vector $x\in\mathbb{R}^n$ use the [coarea formula](https://en.wikipedia.org/wiki/Dirac_delta_function#Properties_in_n_dimensions)
$$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int\_{g(x)=0}\frac{f(x)}{|\nabla g(x)|}\,d^{n-1}\sigma(x),$$
once with $g(x)=v\_1\cdot x$, $f(x)=\delta(v\_2\cdot x)$, then once more with $g(x)=v\_2\cdot x$, $f(x)=1$, to arrive at
$$\int\_{[-1,1]^n}\delta(v\_1\cdot x)\delta(v\_2\cdot x)\,d^nx=\frac{1}{|v\_1||v\_2|}\int\_{[-1,1]^n\cap v\_1\cdot x=0\cap v\_2\cdot x=0}d^{n-2}\sigma(x).$$
So the integral in the OP equals the area of the $n-2$-dimensional region in $[-1,1]^n$ where $v\_1\cdot x=0=v\_2\cdot x$, divided by the product $|v\_1||v\_2|$ of the lengths of the two vectors $v\_1,v\_2$.
Dimensional check: if $x$ has dimension of length $L$ and $\theta\_1,\theta\_2$ are dimensionless, then the integral should have dimension $L^n$. The area of the $n-2$ dimensional region has dimension $L^{n-2}$, and since the product $|v\_1||v\_2|$ has dimension of $L^{-2}$ the dimensions match.
|
2
|
https://mathoverflow.net/users/11260
|
402646
| 165,212 |
https://mathoverflow.net/questions/398496
|
8
|
Corollary 12.14 of Digne-Michel's book *Representations of finite groups of Lie type* gives various decompositions of the regular representation $\operatorname{reg}\_G$ in terms of the Deligne-Lusztig characters. The two I am primarily interested in are $$\operatorname{reg}\_G = \frac{1}{|G^F|\_p} \sum\_{T \in \mathcal{T}} \epsilon\_G \epsilon\_T R^G\_T(\operatorname{reg}\_T) = \frac{1}{|G^F|\_p} \sum\_{\substack{T \in \mathcal{T}\\ \theta \in \operatorname{Irr}(T^F)}} \epsilon\_G \epsilon\_T R^G\_T(\theta).$$
My question is: can someone give an intuitive reason for why these decompositions exist? I am not asking for a proof, but for a moral reason that this *should be true*.
For context: the Deligne-Lusztig characters form a core part of my thesis on the representation theory of finite groups of Lie type, but I do not have the space to completely develop all of the results. I have therefore chosen a few proofs which I feel to be illustrative of the techniques in the field, and for the rest I am trying to offer an intuitive explanation. If I had to guess a decomposition of the regular representation then this would certainly be it (possibly modulo the signs), but I am having difficulty coming up with a way to justify this to readers.
An example of the type of argument I am looking for: the Mackey decomposition for Deligne-Lusztig induction/restriction can be interpreted as `pushing forward' the Bruhat decomposition of a finite group of Lie type onto its representations. Indeed, you can make this argument rigorous (at least on the level of characters, I'm not focusing on the actual representations) by looking at the action of $G^F$ on a certain finite affine variety related to the double cosets, then using properties of the Lefschetz number to make combinatorial simplifications, so to me this gives a good intuition for the Mackey decomposition.
|
https://mathoverflow.net/users/175051
|
Intuitive reason that the regular representation is a uniform function
|
This is an interesting question, but the kind of geometric or structural intuition your are looking for may not exist. To put it another way, the reason behind the fact in the OP is a non-trivial combination of several other central facts, but can't conceptually be reduced to either of them.
If we want to guess that some virtual character such as
$$ \frac{1}{\lvert G^F\rvert\_p} \sum\_{T, \theta} \epsilon\_G \epsilon\_T R^G\_T(\theta)$$
equals the regular character $\operatorname{reg}\_G$, then, as a first step, we had better make sure that they have the same degree. That this is true is non-trivial and the proof actually uses some of the same steps as the proof that the two characters are equal. More precisely, to compute the degree of the above character, one has to, as far as we know, at some point use the fact that *the Steinberg character $\mathrm{St}$ is zero on all non-semisimple elements*. This is based on properties of Bruhat decomposition/BN-pairs and Curtis's alternating sum formula for $\mathrm{St}$.
The above fact about the values of $\mathrm{St}$ together with the fact that values of $\sum\_{\theta} R^G\_T(\theta)$ are Lefschetz numbers and hence zero on non-trivial semisimple elements, gives an easy expression for the inner product of $\mathrm{St}$ and $\sum\_{\theta} R^G\_T(\theta)$. The alternating sum formula for $\mathrm{St}$ together with the fact that $\mathrm{St}(1)=|G^F|\_p$, can then be used to show that
$$\sum\_{\theta} \epsilon\_G \epsilon\_T R^G\_T(\theta)(1) = \frac{|G^F|}{\lvert G^F\rvert\_p}.$$
Finally, a non-trivial result of Steinberg says that there are $|G^F|\_p^2$ $F$-stable maximal tori, so $\sum\_{T,\theta} \epsilon\_G \epsilon\_T R^G\_T(\theta)$ indeed has degree $|G^F|$.
One might ask whether lifting the question to isomorphism of representations or some other categorified objects would make the fact in the OP more 'geometric'. It is not clear to me that this can be done and the fact about the values of the character $\mathrm{St}$ highlighted above is used in all of the proofs I know of the statement in the OP. Note that the proof of 12.14 in Digne--Michel uses this fact (when it refers to 9.4), so in particular I am not aware of any character-free proof (as in, does not use anything about character values).
In summary, I think the result in the OP can be made plausible by combining the observation in LSpice's answer with a summary of why the two characters have the same degree (e.g., as given above), but this doesn't reduce the result to any straightforward conceptual or geometric principle. Instead, like many non-trivial proofs, it is a combination of several main ingredients. In this case the main ingredients are the existence and properties of the Steinberg character (the alternating sum formula, character values), Bruhat decomposition, the character formula for Deligne--Lusztig characters, etc.
|
5
|
https://mathoverflow.net/users/2381
|
402653
| 165,213 |
https://mathoverflow.net/questions/402632
|
1
|
I want to formulate something using the language of (possibly higher) category theory, but my knowledge in category theory is what most graduate students have learned in a first course in algebraic topology. So hopefully someone can help me out.
Let me provide some background to my question. A paper I am reading says "Let Cat be an $(\infty, 1)$-category of (small) categories, obtained by localizing at categorical equivalences. Define the $\infty$-functor
$$\textrm{Vect}\_\nabla:\textrm{Man}^{\textrm{op}}\to\textrm{Cat}$$
where $\textrm{Man}$ is the category of smooth manifolds, **so that for each manifold $M$, $\textrm{Vect}\_\nabla(M)$ is the category of vector bundles over $M$ with connections.**" Okay, I understand only the bold part.
My question is, let say I want to define a "functor" $T$ so that for each manifold $M$, $T(M):\textrm{Vect}\_\nabla(M)\to\Omega(M)$ is a functor from the category of vector bundles over $M$ with connections to the functor $\Omega^\*:\textrm{Man}^{\textrm{op}}\to\textrm{Set}$ from the opposite category of $\textrm{Man}$ to the category of sets, which assigns to each manifold $M$ the set $\Omega(M)$ of differential forms on $M$ (probably with more algebraic structures, but anyway) such that for each $(E, \nabla)$, $T(M)(E, \nabla)\in\Omega(M)$. What the object should $T$ be? I expect it has something to do with $\infty$-functor?
Of course one can think of $T$ as Chern character form, where the Chern character is a natural transformation between the $K$-theory to ordinary cohomology as functors. But I cannot use K-theory here for some reason. Moreover, the "functor" $T$ must take a manifold $M$ first, and then take a vector bundle over $M$ with a connection.
Google shows that there are several books, notes and papers about higher category theory and related areas, but I am not certain where to start and it seems that most of them are too abstract for me, while I just need to learn all the necessary mathematics to formulate my question. You are very welcome to recommend any good source with concrete examples.
Thank you.
Edit: Thanks to AT0 for correcting a mistake.
|
https://mathoverflow.net/users/41686
|
A question about possibly $\infty$-category or functors
|
$T$ can be formalized as a natural transformation $\def\Vect{{\rm Vect}} \def\Vectc{\Vect\_\nabla} \Vectc→Ω^n$ of functors $\def\Man{{\sf Man}} \def\op{{\sf op}} \def\Grpd{{\sf Grpd}} \Man^\op → \Grpd$.
The functor $\Vectc$ sends a smooth manifold $M$ to the groupoid $\Vectc(M)$
of vector bundles with connection over $M$ and connection-preserving isomorphisms. It also sends a smooth map $f\colon M→M'$ of smooth manifolds to the corresponding pullback functor $$\Vectc(f)\colon \Vectc(M')→\Vectc(M).$$
There are many ways to make $\Vectc$ preserve composition (as required by the definition of a functor).
Some of the easiest approaches are (1) strictify $\Vectc(M)$ by adding formal pullbacks; (2) make minor adjustments to the definition of vector bundles and pullbacks, ensuring the pullback preserves compositions on the nose; or (3) use the site of cartesian manifolds (diffeomorphic to $\def\R{{\bf R}} \R^n$), which yields an equivalent category of ∞-sheaves.
The last approach is the one used most often in practice.
Observe that on cartesian manifolds, $\Vectc$ can be defined as the groupoid of connection 1-forms (every bundle on $\R^n$ is trivial), and differential forms pull back strictly.
The functor $Ω^n$ sends a smooth manifold $M$ to the set of differential $n$-forms on $M$, which is turned into a discrete groupoid by adding identity morphisms.
It sends a smooth map $f\colon M→M'$ to the pullback map $$Ω(f)=f^\*\colon Ω(M')→Ω(M).$$
Now $T\colon \Vectc→Ω^n$ is a natural transformation, whose components are given by functors $$T(M)\colon \Vectc(M)→Ω^n$$ that send a vector bundle with connection over $M$ to the corresponding differential $n$-form (given by the Chern–Weil homomorphism, for example) and a connection-preserving isomorphism gets mapped to an equality of differential forms.
|
2
|
https://mathoverflow.net/users/402
|
402658
| 165,214 |
https://mathoverflow.net/questions/402661
|
0
|
Is there a definition Df(g) of uniform continuity of g, without using the notion of metric?
Let $(E,d\_E)$ and $(F, d\_F)$ metrics spaces, $f$ continuous fonction of $E$ to $F$
We must have :
Df$(f)$ iff $f$ is uniform continue.
|
https://mathoverflow.net/users/110301
|
About uniform continuity
|
The notion of uniform continuity does not require all of the structure behind metric spaces. Every metric space is automatically a uniform space, and the uniformly continuous functions are the morphisms in the category of uniform spaces.
The two most common ways of defining uniform spaces are in terms of entourages or also in terms of uniform covers. The entourage definition of a uniform space is simpler while the uniform cover definition of a uniform space generalizes better to point-free topology and gives some very good motivation to the notion of paracompactness.
We shall now give the entourage definition of a uniformity. Let $X$ be a set. Then a uniformity is a filter $\mathcal{U}$ on the set $X\times X$ that satisfies the following properties:
1. If $E\in\mathcal{U}$, then $\{(x,x)\mid x\in X\}\subseteq E$.
2. If $E\in\mathcal{U}$, then $E^{-1}\in\mathcal{U}$.
3. If $E\in\mathcal{U}$, then there exists some $F\in\mathcal{U}$ where
$F\circ F\subseteq E$.
If $\bigcap\mathcal{F}=\{(x,x)\mid x\in X\}$, then we say that the uniformity $\mathcal{F}$ is separated.
Every metric space $(X,d)$ is endowed with a uniformity $\mathcal{F}\_{d}$ such that
$E\in\mathcal{F}\_{d}$ if and only if there exists an $\epsilon>0$ where
$\{(x,y)\mid d(x,y)<\epsilon\}\subseteq E$.
The $\epsilon-\delta$-definition of uniform continuity generalizes to uniform spaces quite easily. If $(X,\mathcal{F}),(Y,\mathcal{G})$ are uniform spaces, then we say that a mapping $f:X\rightarrow Y$ is uniformly continuous if for all $E\in\mathcal{G}$, there exists some $D\in\mathcal{F}$ such that if $(x,y)\in D$, then $(f(x),f(y))\in E$.
**More facts about uniform spaces**
Each uniform space $(X,\mathcal{F})$ is automatically endowed with a topology where a set $U$ is open if and only if for each $x\in U$, there exists an entourage $E$ where $E[x]\subseteq U$. Furthermore, every separated uniform space is automatically completely regular in this topology.
Every completely regular space can be endowed with a compatible uniformity.
Every compact Hausdorff space has a unique compatible uniformity, and if $C$ is a compact space, and $X$ is a uniform space, then every continuous mapping
$f:C\rightarrow X$ is uniformly continuous.
Topological groups can also be endowed with a three compatible uniformities, namely the left uniformity, the right uniformity, and the intersection of these two uniformities.
**References**
General topology textbooks (such as Willard Ch 9, and Kelley as was observed by Willie Wong or by Bourbaki Chapter II as noted by abx) often have a chapter on uniform spaces. There are a couple good general topology texts on uniform spaces including the textbook by John Isbell (which uses the uniform cover definition of uniformity) as well as the book by I.M. James (which uses the entourage definition of a uniformity).
|
4
|
https://mathoverflow.net/users/22277
|
402664
| 165,217 |
https://mathoverflow.net/questions/402643
|
2
|
Let $X$ be a closed complex manifold. Let $L$ be the trivial holomorphic line bundle. Can there be a short exact sequence of holomorphic line bundles $0\to L\to L\oplus L\to L\to 0$ that does not split?
|
https://mathoverflow.net/users/349872
|
Short exact sequence of trivial holomorphic line bundles not splitting
|
For any line bundle $L$, such exact sequence splits. Indeed the map $L\rightarrow L$ induced by the surjection $p:L\oplus L\rightarrow L$ must be nonzero on one of the summands, say the first one; hence it is the multiplication by a nonzero scalar $\alpha $. Then the map $L\xrightarrow{\ (\alpha ^{-1},0)\ } L\oplus L$ is a section of $p$.
|
4
|
https://mathoverflow.net/users/40297
|
402667
| 165,218 |
https://mathoverflow.net/questions/402663
|
3
|
In a 2017 article [More on supersymmetric and 2d analogs of the SYK model](https://link.springer.com/article/10.1007%2FJHEP08%282017%29146) by Murugan, Stanford and Witten, the authors take a model called the SYK model (named after Sachdev, Ye and Kitaev) and study supersymmetric versions in dimensions one and two.
The physics motivation for the original SYK model in condensed matter physics can be found [here](https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.70.3339) and in related works (it originally models the spin-$S$ quantum Heisenberg magnet with Gaussian-random, infinite-range exchange interactions). At the end of the article, it is mentioned that constructing a supersymmetric SYK model in dimension three is still an open problem.
Such a model would be more realistic and would not suffer from some of the problems which one encounters with global symmetries in dimensions one and two, so I was curious if this had been looked into in the meantime. Would there be some peculiar difficulties involved in doing this, or has no-one studied it because it would necessitate very involved calculations and not many new ideas which are not already in the paper I cite above?
|
https://mathoverflow.net/users/119114
|
Supersymmetric SYK Model in 3D?
|
The disordered SYK model in three dimensions with supersymmetry was studied by Fedor Popov in [Supersymmetric tensor model at large $N$ and small $\varepsilon$.](https://arxiv.org/abs/1907.02440)
The complications are discussed in [A 3d disordered superconformal fixed point](https://arxiv.org/abs/2108.00027):
>
> There are two problems with the generalization of SYK-like models with
> disorder to dimensions higher than two. Firstly, pure fermionic models
> do not generically have relevant operators (a four-Fermi interaction
> being marginally irrelevant in two dimensions). Secondly, including
> bosonic degrees of freedom is problematic, since disordered
> Hamiltonians fail to be generically positive definite.
>
>
>
|
2
|
https://mathoverflow.net/users/11260
|
402668
| 165,219 |
https://mathoverflow.net/questions/402607
|
0
|
Let $X$ be a compact and convex space and let $T=[0,1]$ be some parameter space. Let $F:X\times T\rightrightarrows X$ be a correspondence that is compact-valued, convex, and upper-hemicontinous. By Kakutani's fixed point theorem, there is a fixed point $x(t)=F(x(t),t)$ for each parameter $t\in T$.
Suppose we also know that
1. the set $F(x, t')$ converges $F(x, t)$ as $t'\to t$ for any $x\in X$ (in the sense
that the indicator functions for the two sets converge
pointwise), and
2. there is a unique fixed point $x(t)$ for each $t\in T$.
Question: Is this enough to determine that $x(t)$ is a continuous function?
Here is my very informal attempt at a proof:
Consider sequences $(x\_n, t\_n)\to (x, t)$ such that $x\_n\in F(x\_n, t\_n)$. Suppose $x\notin F(x, t)$. Since, $F(\cdot, t)$ and $F(\cdot, t\_n)$ are close for large enough $n$, then $x\notin F(x,t\_n)$. By UHC of $F$, for any open set $V$ containing $F(x, t\_n)$, I can find an open set $U$ of $x$ such that for all $x'\in U$, $F(x', t\_n)\in V$. I can pick $V$ such that it excludes $x$, and by consequence, any $x\_n$ for $n$ large enough. However, this would imply that there is an $x\_n\in U$ and $x\_n\notin V$, which implies $x\_n\notin F(x\_n, t\_n)$. Hence, $x\in F(x, t)$, and the graph
$$
\{(x,t): x\in F(x, t) \}
$$
is closed. This along with the uniqueness should be sufficient for showing that $x(t)$ is a continuous function.
|
https://mathoverflow.net/users/121674
|
Continuity of Kakutani fixed points
|
I assume that you mean that $F$ is upper semicontinuous on the product space. Then in particular (since $X$ is compact, Hausdorff and $F$ has closed values), $F$ has a closed graph. This implies that also the set $\{(x,t):x\in F(x,t)\}$ is closed. This means that the multimap $$t\mapsto\{x:x\in F(x,t)\}$$ has a closed graph and assumes closed values. Since $X$ is a compact space, it follows that this multimap is upper semicontinuous.
Of course, in the single-valued case upper semicontinuity and continuity are equivalent.
In particular, your hypothesis 1 superfluous. (Well, actually it follows from the upper semicontinuity of $F$ with respect to both variables - it is the special case that you fix one variable.) Note that you really have to require the upper semicontinuity with respect to both variables, even in the metric case: Your first part of the proof becomes wrong unless you assume some sort of locally uniform convergence (which implies of course again the upper semicontinuity).
|
1
|
https://mathoverflow.net/users/165275
|
402673
| 165,222 |
https://mathoverflow.net/questions/402596
|
2
|
This question is related to the matrices described in [Deyi Chen's recent MO post](https://mathoverflow.net/questions/402572/euler-numbers-and-permanent-of-matrices) (look at some examples there). The main difference: we are asking for a determinant evaluation instead of a permanent, plus we have added variables. Define the sequence
$$f\_n=\mathrm{det}\left[x\_{i-j}\cdot\operatorname{sgn} \left(\tan\frac{(i+j)\pi}{2n+1} \right)\right]\_{1\le i, \,j\le 2n},$$
where $n\geq1$ and $\operatorname{sgn}$ is the sign-function; i.e. $\operatorname{sgn}(y)=1$ if $y>0$; $\operatorname{sgn}(y)=-1$ if $y<0$; $\operatorname{sgn}(0)=0$.
If we set all $x\_k=1$, user44191 offers an alternative description of the matrix: $n−1$ antidiagonals of $1$'s, $n$ antidiagonals of $-1$'s, an antidiagonal of $0$'s, $n$ antidiagonals of $1$'s, and $n-1$ antidiagonals of $-1$'s, in the given order.
>
> **QUESTION 1.** Is this true? The determinant equals $f\_n=(-1)^nP(\dots,x\_{-2},x\_{-1},x\_0,x\_1,x\_2,\dots)^2$ for some **squared** polynomial $P$ of several variables (coefficients in $\mathbb{Z}$) so that each of its monomials is of the form $x\_{i\_1}x\_{i\_2}\cdots x\_{i\_m}$ with $i\_1+i\_2+\cdots+i\_m=0$.
>
>
>
>
> **QUESTION 2 (specialization).** Is this true? If we set $x\_k=1$, for all $k$, then $f\_n=(-1)^n$.
>
>
>
**Example.** For instance,
$$f\_2=
\det\left[ \begin {array}{cccc}
x\_0&-x\_{-1}&-x\_{-2}&0\\
-x\_1&-x\_0&0&x\_{-2} \\
-x\_2&0&x\_0&x\_{-1}\\
0&x\_2&x\_1&-x\_0
\end {array} \right]
=(x\_{-2}x\_2-x\_{-1}x\_1-x\_0^2)^2.$$
**Example.** The matrix for $n=3$ and its determinant:
$$
\left[ \begin {array}{cccccc} x\_0&x\_{-1}&-x\_{-2}&-x\_{-3}&-x\_{-4}&0\\
x\_1&-x\_0&-x\_{-1}&-x\_{-2}&0&x\_{-4}\\
-x\_2&-x\_1&-x\_0&0&x\_{-2}&x\_{-3}\\
-x\_3&-x\_2&0&x\_0&x\_{-1}&x\_{-2}\\
-x\_4&0&x\_2&x\_1&x\_0&-x\_{-1}\\
0&x\_4&x\_3&x\_2&-x\_1&-x\_0
\end {array} \right],
$$
\begin{align\*}
f\_3&=-(x\_{-4}x\_0x\_4 - x\_{-4}x\_{1}x\_3 + x\_{-4}x\_2^2 - x\_{-3}x\_{-1}x\_4 + x\_{-3}x\_0x\_3 \\
& \qquad + x\_{-3}x\_1x\_2 + x\_{-2}^2x\_4 + x\_{-2}x\_{-1}x\_3 - 2x\_{-2}x\_0x\_2 \\
& \qquad - x\_{-2}x\_1^2 - x\_{-1}^2x\_2 - x\_0^3)^2.
\end{align\*}
|
https://mathoverflow.net/users/66131
|
Determinants of striped Hankel matrices
|
**to Question 1:** Yes.
To prove this, let me fix a positive integer $n$ and denote your matrix (whose
determinant $f\_{n}$ is) by $A$. The notation $\left[ k\right] $ shall be
used for the set $\left\{ 1,2,\ldots,k\right\} $ whenever $k$ is an integer.
The notation $M\_{i,j}$ will be used for the $\left( i,j\right) $-th entry of
any matrix $M$. Thus,
\begin{equation}
A\_{i,j}=x\_{i-j}\cdot\operatorname\*{sgn}\left( \tan\dfrac{\left( i+j\right)
\pi}{2n+1}\right)
\label{eq.darij1.1}
\tag{1}
\end{equation}
for any $i,j\in\left[ 2n\right] $.
Let $B$ be the $2n\times2n$-matrix obtained by "turning $A$ upside down",
i.e., reversing the order of the rows of $A$. Explicitly, this means that
\begin{equation}
B\_{i,j}=A\_{2n+1-i,j}\qquad\text{for all }i,j\in\left[ 2n\right]
.
\label{eq.darij1.2}
\tag{2}
\end{equation}
We note that $B$ can be obtained from $A$ by $n$ row-swaps (i.e., by $n$
steps, where each step swaps a pair of rows). Indeed, all we need to do is to
swap the $1$-st and the last row, then to swap the $2$-nd and the
$2$-nd-to-last row, etc., until we reach the middle of the matrix. Since each
of these swaps multiplies the determinant by $-1$, this entails that
\begin{equation}
\det B=\left( -1\right) ^{n}\det A.
\label{eq.darij1.3}
\tag{3}
\end{equation}
Now, I claim that the matrix $B$ is alternating -- i.e., that
\begin{equation}
B\_{i,i}=0\qquad\text{for all }i\in\left[ 2n\right]
\label{eq.darij1.4}
\tag{4}
\end{equation}
and
\begin{equation}
B\_{i,j}=-B\_{j,i}\qquad\text{for all }i,j\in\left[ 2n\right]
.
\label{eq.darij1.5}
\tag{5}
\end{equation}
Indeed, in order to prove \eqref{eq.darij1.4}, it suffices to observe that
\begin{align\*}
B\_{i,i} & =A\_{2n+1-i,i}=x\_{\left( 2n+1-i\right) -i}\cdot\operatorname\*{sgn}
\left( \tan\dfrac{\left( \left( 2n+1-i\right) +i\right) \pi}
{2n+1}\right) \\
& =x\_{2n+1-2i}\cdot\underbrace{\operatorname\*{sgn}\left( \tan\dfrac{\left(
2n+1\right) \pi}{2n+1}\right) }\_{=\operatorname\*{sgn}\left( \tan\pi\right)
=\operatorname\*{sgn}0=0}=0.
\end{align\*}
The proof of \eqref{eq.darij1.5} is not much harder (using the fact that
$\dfrac{\left( 2n+1-i+j\right) \pi}{2n+1}=\pi-\dfrac{\left( i-j\right)
\pi}{2n+1}$ and therefore
\begin{align}
\tan\dfrac{\left( 2n+1-i+j\right) \pi}{2n+1}=\tan\left( \pi-\dfrac{\left(
i-j\right) \pi}{2n+1}\right) =-\tan\dfrac{\left( i-j\right) \pi}{2n+1},
\end{align}
and furthermore $\tan$ is an odd function).
Thus, we know that the matrix $B$ is alternating. Hence, as for any
alternating $2n\times2n$-matrix, its determinant is the square of its
[Pfaffian](https://en.wikipedia.org/wiki/Pfaffian). In other words,
\begin{equation}
\det B=\left( \operatorname\*{Pf}B\right) ^{2},
\label{eq.darij1.6}
\tag{6}
\end{equation}
where $\operatorname\*{Pf}B$ denotes the Pfaffian of $B$. The latter Pfaffian
is a polynomial in the entries of the matrix with coefficients in $\mathbb{Z}
$. Since the entries of the matrix belong to $\mathbb{Z}\left[ \ldots
,x\_{-2},x\_{-1},x\_{0},x\_{1},x\_{2},\ldots\right] $, we thus conclude that the
Pfaffian belongs to $\mathbb{Z}\left[ \ldots,x\_{-2},x\_{-1},x\_{0},x\_{1}
,x\_{2},\ldots\right] $ as well. In other words,
\begin{equation}
\operatorname\*{Pf}B\in\mathbb{Z}\left[ \ldots,x\_{-2},x\_{-1},x\_{0},x\_{1}
,x\_{2},\ldots\right] .
\label{eq.darij1.7}
\tag{7}
\end{equation}
Now, \eqref{eq.darij1.3} yields
\begin{align}
\det A=\left( -1\right) ^{n}\det B=\left( -1\right) ^{n}\left(
\operatorname\*{Pf}B\right) ^{2}
\end{align}
(by \eqref{eq.darij1.6}). Because of \eqref{eq.darij1.7}, this shows that
$\det A$ equals $\left( -1\right) ^{n}\cdot P^{2}$ for some polynomial
$P\in\mathbb{Z}\left[ \ldots,x\_{-2},x\_{-1},x\_{0},x\_{1},x\_{2},\ldots\right] $
(namely, for $P=\operatorname\*{Pf}B$), exactly as claimed in Question 1.
In order to complete the answer to Question 1, we now need to show that each
monomial in $P=\operatorname\*{Pf}B$ is of the form $x\_{i\_{1}}x\_{i\_{2}}\cdots
x\_{i\_{n}}$ with $i\_{1}+i\_{2}+\cdots+i\_{n}=0$. This can be done in various
ways, but the easiest is probably the following: Let us equip the polynomial
ring $\mathbb{Z}\left[ \ldots,x\_{-2},x\_{-1},x\_{0},x\_{1},x\_{2},\ldots\right]
$ with a $\mathbb{Z}$-grading in which each indeterminate $x\_{i}$ is
homogeneous of degree $i$. Now, recall the explicit formula for the Pfaffian
as a sum over all perfect matchings on the set $\left[ 2n\right] $ (see Definition 3 in [Michel Goemans, *18.438 in Spring 2014, Lectures 4 and 6*](http://www-math.mit.edu/%7Egoemans/18438S14/lec4-algmat.pdf), or any
good textbook on Pfaffians). If
\begin{equation}
M=\left\{ \left\{ a\_{1},b\_{1}\right\} ,\left\{ a\_{2},b\_{2}\right\}
,\ldots,\left\{ a\_{n},b\_{n}\right\} \right\}
\label{eq.darij1.9o}
\tag{9}
\end{equation}
is such a perfect matching, then the corresponding addend in
$\operatorname\*{Pf}B$ is
\begin{equation}
\pm B\_{a\_{1},b\_{1}}B\_{a\_{2},b\_{2}}\cdots B\_{a\_{n},b\_{n}}.
\label{eq.darij1.9}
\tag{10}
\end{equation}
Each of the $n$ factors $B\_{a\_{i},b\_{i}}$ in this product can be rewritten as
\begin{align}
B\_{a\_{i},b\_{i}}=A\_{2n+1-a\_{i},b\_{i}}=x\_{\left( 2n+1-a\_{i}\right) -b\_{i}
}\cdot\left( 1\text{ or }-1\text{ or }0\right) ,
\end{align}
and thus (using our weird grading) is homogeneous of degree $\left(
2n+1-a\_{i}\right) -b\_{i}=2n+1-a\_{i}-b\_{i}$. Hence, the entire product
\eqref{eq.darij1.9} is homogeneous of degree
\begin{align\*}
\sum\_{i=1}^{n}\left( 2n+1-a\_{i}-b\_{i}\right) & =n\left( 2n+1\right)
-\underbrace{\sum\_{i=1}^{n}\left( a\_{i}+b\_{i}\right) }
\_{\substack{=1+2+\cdots+2n\\\text{(since \eqref{eq.darij1.9o} is
a}\\\text{perfect matching of }\left[ 2n\right] \text{)}}}\\
& =n\left( 2n+1\right) -\left( 1+2+\cdots+2n\right) =0.
\end{align\*}
This means that this product is a $\mathbb{Z}$-linear combination of monomials
of the form $x\_{i\_{1}}x\_{i\_{2}}\cdots x\_{i\_{n}}$ with $i\_{1}+i\_{2}
+\cdots+i\_{n}=0$. Clearly, the same must therefore holds for the polynomial
$\operatorname\*{Pf}B$ (since this polynomial is a sum of such products). This
concludes the answer to Question 1.
Answering Question 2 requires proving that $\det B=1$ when all $x\_{i}$ are set
to $1$. This should be easy given that $\operatorname\*{sgn}\left( \tan
\dfrac{\left( i+j\right) \pi}{2n+1}\right) $ can be explicitly computed
(and the matrix $B$ becomes a circulant when all $x\_{i}$ are $1$); but it's
late here and I have too many things on my list until the quarter begins. Sorry!
|
4
|
https://mathoverflow.net/users/2530
|
402683
| 165,225 |
https://mathoverflow.net/questions/402682
|
2
|
In Silverman's book AEC, question 7.6 asks to prove $E\_0(K)$ has finite index in $E(K)$ for $K$ a local field. For part (a), I know the topology on $P^{n}(K)$ is the quotient topology on $K^{n+1}$, and the topology on $K^{n+1}$ is induced by the absolute value. However, I do not know how to prove the compactness of $P^{n}(K)$ in this topology.
|
https://mathoverflow.net/users/350297
|
Why is $P^n(K)$ compact, when $K$ is a local field?
|
Because the quotient mapping $K^{n+1} - \{\mathbf 0\} \to {\mathbf P}^n(K)$ (note $\mathbf 0$ is not in the domain) is continuous by the definition of the quotient topology, it suffices to show there is a compact subset $C$ of $K^{n+1} - \{\mathbf 0\}$ such that every element of ${\mathbf P}^n(K)$ is hit by some element of $C$, as that makes the natural mapping $C \to {\mathbf P}^n(K)$ continuous and surjective, so the target is compact . We want to show every line through the origin in $K^{n+1}$ contains an element of $C$. What can $C$ be?
When $K = \mathbf R$ (not a local field, but a good warm-up example), for each $\mathbf v \in \mathbf R^{n+1} - \{\mathbf 0\}$ the vector $\mathbf v/||\mathbf v||$ is equal to $\mathbf v$ in ${\mathbf P}^n(\mathbf R)$, so as $C$ we can use the ordinary unit sphere
$$
S^n = \{\mathbf v \in \mathbf R^{n+1} : ||\mathbf v|| = 1\},
$$
which is compact.
When $K$ is a local field, let $||\cdot||\_\infty$ be the sup-norm on $K^{n+1}$, so its values are in $|K|$: for each $\mathbf v \in \mathbf K^{n+1}$ there is $c \in K$ such that $||\mathbf v||\_\infty = |c|$.
When $\mathbf v \not= \mathbf 0$, $\mathbf v/c$ is on the same line through the origin as $\mathbf v$ and $||\mathbf v/c||\_\infty = 1$, so as $C$ we want to take the "sup-norm unit sphere"
$$
\{\mathbf w \in K^{n+1} : ||\mathbf w||\_\infty = 1\}.
$$
Note this is a subset of $K^{n+1}- \{\mathbf 0\}$, and superficially its definition resembles the unit sphere in $\mathbf R^{n+1}$, but it is different in one sense: it is an open subset. (Its subspace topology in $K^{n+1}$ or $K^{n+1} - \{\mathbf 0\}$ is the same.) Why is it compact?
Method 1: the sup-norm on $K^{n+1}$ is continuous, so the above set is closed (inverse image of a point) and bounded in $K^{n+1}$, and thus is compact because $K$ is locally compact.
Method 2: the above set is the set-theoretic difference
$$
\{\mathbf w \in K^{n+1} : ||\mathbf w||\_\infty \leq 1\} -
\{\mathbf w \in K^{n+1} : ||\mathbf w||\_\infty < 1\}
$$
where the bigger set is compact in $K^{n+1}$ (here we use that $K$ is a *local* field) and the set being removed is open in $K^{n+1}$, so its complement in $K^{n+1}$ is closed. Thus this set-theoretic difference is closed inside a compact set and thus is compact.
|
7
|
https://mathoverflow.net/users/3272
|
402687
| 165,226 |
https://mathoverflow.net/questions/402689
|
6
|
Can you prove or disprove the following claim:
First, define the function $\xi(n)$ as follows: $$\xi(n)=\begin{cases}-1, & \text{if }\varphi(n) \equiv 0 \pmod{4} \\
1, & \text{if }\varphi(n) \equiv 2 \pmod{4} \\ 0, & \text{if otherwise }
\end{cases}$$
where $\varphi(n)$ denotes [Euler's totient function](https://en.wikipedia.org/wiki/Euler%27s_totient_function). Then,
$$\frac{\pi^2}{72}=\displaystyle\sum\_{n=1}^{\infty}\frac{\xi(n)}{n^2}$$
The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJwrTbI1NQACa66KTI08TdvqzDSN1NKc1KKCDBBf1cTW1kgn2dZQB1PcACiuawgkDDQ1rYtSS0qL8jSSNa1ruYpti0tzNfKAmkqTdMDG6mvkxRkBVXEVFGXmlWgUaxnqwTkBmXFG-uZGmtYAT1go5g==&lang=gp&interacts=eJyLjgUAARUAuQ==).
|
https://mathoverflow.net/users/88804
|
An infinite series involving the mod-parity of Euler's totient function
|
The only odd values of $\phi(n)$ are $\phi(1)=\phi(2)=1$.
$\phi(n)$ is even but not divisible by $4$ when:
1. $n=4$
2. $n=2^{\left\{0,1\right\}}p^m$, where $p=4k+3$ is prime, $m=1,2,3,...$
We have
$$
\frac{\pi^2}{6}=1+\frac14+\sum\_{\substack{n=1\\\phi(n)\equiv 0}}^\infty\frac{1}{n^2}+\sum\_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}.
$$
(congruences are modulo $4$)
The claim in the question reads
$$
\frac{\pi^2}{72}=-\sum\_{\substack{n=1\\\phi(n)\equiv 0}}^\infty\frac{1}{n^2}+\sum\_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}.
$$
Combining these two we get a hypothetical identity
$$
\sum\_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}=\frac{13\pi^2}{144}-\frac{5}{8}.
$$
However
$$
\sum\_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}=\frac{1}{16}+\left(1+\frac14\right)\sum\_{m=1}^\infty\sum\_{p\equiv 3}\frac{1}{p^{2m}}=\frac{1}{16}+\frac54\sum\_{p\equiv 3}\frac{1}{p^2-1}.
$$
Thus the conjecture is equivalent to
$$
\sum\_{p\equiv 3}\frac{1}{p^2-1}=\frac{1}{20} \left(\frac{13 \pi ^2}{9}-11\right).\tag{\*}
$$
The article on sums over primes on mathworld <https://mathworld.wolfram.com/PrimeSums.html> does not list any sums of this kind. Numerical checks show that the claim is false. For example summing over the first $N=500000$ primes gives for the ratio of the LHS of (\*) to the RHS:
$$
1.000153116
$$
with the truncation error term of the order
$$
\int\_{N}^\infty \frac{dx}{x^2\ln^2(x)}\sim\frac{1}{N\ln^2(N)}\sim 10^{-10}.
$$
|
8
|
https://mathoverflow.net/users/82588
|
402696
| 165,229 |
https://mathoverflow.net/questions/402303
|
7
|
Let $\mathcal C$ be a category whose skeleton has $\lambda$-many objects and $\kappa$-many morphisms. Then the skeleton of the endofunctor category $\mathcal C^{\mathcal C}$ has at most $\kappa^{3 \times \kappa}$ many morphisms.
My guess is that in most cases, this upper bound is achieved.
**Question:** What is an example of a category $\mathcal C$ (other than the terminal category or equivalents) whose skeleton has $\lambda$ many objects and $\kappa$ many morphisms, but the skeleton of whose endofunctor category $\mathcal C^{\mathcal C}$ has
1. *Strictly fewer* than $\kappa^\kappa$ many morphisms?
2. *Strictly fewer* than $\kappa^\kappa$ many objects?
3. As few as $\kappa$-many morphisms?
4. As few as $\lambda$-many objects?
**EDIT:** Neil Strickland's example of $B\mathbb N$ in the comments below affirmatively answers (1), (2), and (3). So it remains to see about (4).
Note that $\mathcal C^{\mathcal C}$ always has (skeletally) at least $\lambda+1$ many objects and $\kappa+1$-many morphisms, given by constant morphisms between constant endofunctors, along with the identity functor.
Note also that if $\mathcal C$ is accessible, and has as many objects and morphisms as the size of the universe, then the number of *accessible* endofunctors and morphisms between them is also the size of the universe. So if my guess is correct, then *most* endofunctors are usually non accessible.
See also [here](https://mathoverflow.net/questions/334266/if-mathcal-c-mathcal-c-is-equivalent-to-mathcal-c-is-mathcal-c-nece/334962#334962) for an argument which shows that if $\mathcal C$ has small products and coproducts and is not a preorder, then $\mathcal C^{\mathcal C}$ has (skeletally) at least $2^\kappa$-many objects, where $\kappa$ is the size of the universe.
|
https://mathoverflow.net/users/2362
|
Category with few endofunctors?
|
An example of (1),(2),(3),(4) with $\kappa=\lambda=|\mathbb R|$ is to take $\mathcal C$ to be the posetal category $\mathbb R.$ This is already skeletal. Its category of endofunctors is the poset of non-decreasing functions $f:\mathbb R\to\mathbb R,$ with the pointwise order. There are at most $|\mathbb R^{\mathbb Q}|=|\mathbb R|$ such functions: $f$ is determined by the preimages $f^{-1}((-\infty,q))$ for $q\in\mathbb Q,$ each of these intervals is convex, and there are at most $|\mathbb R|$ convex subsets of $\mathbb R.$
(I don't know if there are, unconditionally, examples with $\lambda$ and $\kappa$ regular. I think the $\mathbb R$ example generalizes to $2^\mu$ with the lex order, where $\mu$ is an ordinal satisfying $|2^{2^{<\mu}}|=|2^\mu|,$ for example $\mu=\beth\_\alpha$ with $\alpha$ a limit ordinal. Here $2^{<\mu}$ means the functions $\mu\to 2$ with bounded support.)
|
5
|
https://mathoverflow.net/users/164965
|
402698
| 165,231 |
https://mathoverflow.net/questions/402621
|
4
|
I am looking to find real quadratic fields whose Hilbert class field is abelian over $\Bbb Q$. Then I learned about genus numbers and genus field of the number field. It is enough to find a number field whose class number is equal to the genus number. In [YOSHIOMI FURUTA article](https://projecteuclid.org/download/pdf_1/euclid.nmj/1118802021)
I found the formula, but I have a problem with computing the order of the quotient.
So I was wondering, is there a way to find genus number using Sage just like the class number.
Any help is very appreciated
|
https://mathoverflow.net/users/131448
|
How to calculate genus number of number field using sage?
|
I may as well promote my comment to an answer, so this is closed. In a short paper, H. Hasse [1] computed the genus field and genus number of every real quadratic field. In particular, he shows that, given a real quadratic field $\Omega = \mathbb{Q}(\sqrt{d})$, the genus number of $\Omega$ is $2^{r-1}$, where $r$ is the number of distinct prime divisors of (the square-free) $d$. Thus, for example, the genus number of $\mathbb{Q}(\sqrt{210})$ is $8$, as $210 = 2 \cdot 3 \cdot 5 \cdot 7$.
Thus the genus number is (rather trivially) computable in Sage, for example by the command `2^(len(set(d.factor()))-1)`, where $d$ is as above.
[1] *Hasse, Helmut*, [**Zur Geschlechtertheorie in quadratischen Zahlkörpern**](http://dx.doi.org/10.2969/jmsj/00310045), J. Math. Soc. Japan 3, 45-51 (1951). [ZBL0043.04002](https://zbmath.org/?q=an:0043.04002).
|
2
|
https://mathoverflow.net/users/120914
|
402704
| 165,234 |
https://mathoverflow.net/questions/402505
|
7
|
Let $\Gamma$ be a discrete group and $A$ be a $C^\*$-algebra. Consider an action $\alpha: \Gamma \to \operatorname{Aut}(A)$. There is a notion of amenability for such an action (see e.g. Brown and Ozawa's book "C\*-algebras and finite-dimensional approximations", section 4.3), but how should one think intuitively about an amenable action? How was the definition originally motivated?
|
https://mathoverflow.net/users/216007
|
Amenable action intuition
|
It is impossible to understand the motivation behind the definition of an amenable action without first understanding the definition of amenable groups, so let me first talk about groups (for simplicity, just countable ones).
Finite groups are precisely the ones for which there is an invariant probability measure (the uniform one). Infinite groups also have invariant measures (the counting ones), but they can not be normalized and made finite (for topological groups one should talk about the Haar measures instead of the counting ones).
There are two natural ways of relaxing the above condition of the existence of a finite invariant measure, and therefore of extending the class of finite groups. One consists in extending the space of measures and looking for invariant objects in this extended space. In our context it amounts to passing from the "usual" sigma-additive probability measures to finitely additive ones (called means). This is precisely the original 1929 definition of von Neumann: a group is amenable if it has an invariant mean.
The other way consists in keeping measures, but replacing, as a trade off, exact invariance with an approximate one. In spite of being more constructive, this approach was only developed in the 50s, and turned out to be equivalent to the original one (I skip the historical details and just mention that the most important contributor was Day who also coined the modern term "amenability"). More precisely, a group $G$ is amenable if and only if it carries an *approximately invariant* sequence of probability measures $m\_n$, i.e., such that
$$
\| g m\_n - m\_n \| \to 0 \qquad\forall\,g\in G \;.
$$
This is the *Day condition* (sometimes one also adds the name of Reiter, but this is not really correct from the historical point of view).
In order to pass to group actions, it is actually more convenient to go a bit further and first talk about *groupoids* instead. They are like groups with the only difference that the multiplication is defined only subject to a certain condition. Very succinctly, a groupoid is a "small category with invertible morphisms". In plain language it means that there is a set of points ("objects") and a set of arrows ("morphisms") between objects endowed with a composition operation with the same properties as for groups and with the only additional requirement that for composing two arrows the endpoint of the first must coincide with the starting point of the second. In particular, a part of the formal definition of a groupoid is the presence of a *target map* from morphisms to objects which assigns to any arrow its endpoint. Therefore, over each object $x$ of a groupoid $\mathbf G$ there is the corresponding fiber $\mathbf G^x$ of the target map which consists of all arrows ending at $x$. A group $G$ can be considered as the groupoid $\mathbf G$ for which there is only one object $\bullet$, so that all arrows are composable, and the fiber $\mathbf G^\bullet$ is just the whole group $G$.
One can easily see that the groupoid moves around the fibers of the target map, namely, given any arrow $\mathbf g$ with the starting point $x$ and the endpoint $y$, one can compose $\mathbf g$ with any arrow whose endpoint is $x$, and the result will be an arrow whose endpoint is $y$. Formally,
$$
\mathbf g \mathbf G^x = \mathbf G^y \qquad \forall\,\mathbf g:x\to y \;.
$$
One can now talk about the systems $(m^x)$ of measures on the target fibers of a groupoid invariant in the sense that $\mathbf g m^x=m^y$. They are called *Haar systems*, being a straighforward generalization of the Haar measures in the group case.
In precisely the same way as for groups, there are two ways of relaxing the condition of the existence of a *finite* Haar system: either to talk about systems of means instead of measures, or to replace exact invariance with an approximate one and to require the existence of a sequence of probability measures $m\_n^x$ on the fibers $\mathbf G^x$ such that
$$
\| \mathbf g m\_n^x - m\_n^y \| \to 0 \qquad\,\forall\mathbf g:x\to y \;.
$$
In the same way as for groups, these two definitions happen to be equivalent (Renault 1980), and produce what is called *amenable groupoids* (I skip some technical details here).
Now back to actions. An action of a group $G$ on a space $X$ gives rise to the *action groupoid* whose objects are the points from $X$, and the arrows are the triples $(x, g ,y)$ with $y=gx$, so that for any $y\in X$ the corresponding fiber $\mathbf G^y$ can be identified just with the group $G$, and the aforementioned action of the groupoid on the fibers of the target map amounts just to the left action of the group on itself. Thus, the amenability of the action groupoid amounts precisely to the existence of a system of means on the group indexed by the action space and invariant with respect to the group action, or, in the approximate terms, to the condition that there exists a sequence of systems $(m\_n^x)$ of probability measures on $G$ indexed by the action space, and such that
$$
\| g m\_n^x - m^{gx}\_n \| \to 0 \;.
$$
This is precisely the definition from Brown - Ozawa you are asking about (I skip the details concerning the difference between the definitions in the topological and in the measure "worlds").
One more comment. The way the notion of an amenable action was first introduced by Zimmer in 1977 is actually different. His motivation was the fixed point characterization of amenable groups (the existence of a fixed point for any continuous affine action on a compact) - this is the main *application* of amenability, but is not so convenient for *establishing* amenability - and given in pretty convoluted terms of existence of invariant sections for certain Banach bundles over the action space (this is the reason some of the papers on amenable action from that period are so excessively long).
For more historical and motivational details see the [book](https://mathscinet.ams.org/mathscinet-getitem?mr=1799683) by Anantharaman-Delaroche and Renault or a [recent preprint](https://arxiv.org/pdf/2005.13752.pdf) of Bühler and Kaimanovich.
|
14
|
https://mathoverflow.net/users/8588
|
402712
| 165,237 |
https://mathoverflow.net/questions/402639
|
4
|
Let $M$ be a compact $3$-manifold such that no component of $\partial M$ is $S^2$ and one component $F$ of $\partial M$ is the projective plane.
If $i\_\*:\pi\_1(F) \to \pi\_1(M)$ is an isomorphism, can we prove that $M$ is homeomorphic to $F \times [0,1]$?
|
https://mathoverflow.net/users/280895
|
3-manifold with boundary containing a projective plane
|
Yes, the result follows from a theorem of [Livesay](https://doi.org/10.2307/1970543) and the Poincaré Conjecture that if $M$ is a compact connected non-orientable 3-manifold with $\pi\_1(M)$ finite, then $M$ is homeomorphic to $P^2\times I$ minus a collection of disjoint open 3-balls.
---
The first answer to the original question that I wrote here was flawed. Here is a corrected version, thanks to the comments below.
First note that $M$ is non-orientable since it contains a 2-sided projective plane. Let $N$ be the orientation double cover of $M$, and let $p:N\to M$ be the covering map. Then $N$ is simply-connected since $\pi\_1(M)\cong\mathbb{Z}/2\mathbb{Z}$.
If some component of $\partial M$ had non-positive Euler characteristic, then the same would be true of some component of $\partial N$, but this cannot happen since $N$ is simply-connected. By assumption, no component of $\partial M$ is a 2-sphere. Hence $\partial M$ is a collection of projective planes. Let $k$ be the number of components of $\partial M$.
By the Poincaré Conjecture, $N$ is homeomorphic to a 3-sphere with the interiors of $k$ disjoint 3-balls removed. Let $q:N\to N$ be the covering transformation, an orientation reversing involution. Now $H\_2(N;\mathbb{Q})\cong\mathbb{Q}^{k-1}$ and $H\_2(N;\mathbb{Q})$ is generated by any $k-1$ components of $\partial N$. Since $q$ acts on each component of $\partial N$ as an orientation reversing homeomorphism, it follows that the Lefschetz number $\Lambda\_q$ is given by
$$\Lambda\_q=1-0+(-(k-1))=2-k.$$
But $q$ has no fixed points, so $\Lambda\_q=0$ and $k=2$. See [Epstein](https://doi.org/10.1112/plms/s3-11.1.469) for the original (and more general) proof of this step.
Therefore $N$ is homeomorphic to $S^2\times I$, and it follows by Livesay or [Rubinstein](https://doi.org/10.1090/S0002-9939-1976-0420625-3) that $M$ is homeomorphic to $P^2\times I$. The argument I had in mind here involves doubling $N$ along its boundary to get $S^2\times S^1$. The covering map $p$ also extends to the double. Then the [classification](https://doi.org/10.1112/blms/15.5.401) of Seifert fibered spaces covered by $S^2\times\mathbb{R}$ tells us that the only compact quotient of $S^2\times S^1$ which contains a 2-sided projective plane is $P^2\times S^1$. Hence $M$ is homeomorphic to $P^2\times I$.
|
3
|
https://mathoverflow.net/users/126206
|
402723
| 165,240 |
https://mathoverflow.net/questions/402717
|
13
|
Let $S(\alpha) = \sum\_{n\leq N}f(n) e^{2\pi i \alpha n}$ for some arithmetic function $f$. Suppose $\alpha\_1, \ldots, \alpha\_R$ are real numbers that are $\delta$-spaced modulo $1$, for some $0 < \delta < 1/2$. The large sieve inequality then gives
$$
\sum\_{r=1}^R \left| S\left(\alpha\_r \right)\right|^2 \ll (N + \delta^{-1}) \sum\_{n\leq N}|f(n)|^2.
$$
The $N$ term is satisfactory when the support of $f$ is dense enough in $[1,N]$. However it becomes worse the sparser the support of $f$ is.
Are there any methods or inequalities or examples known in literature to tackle the case when the support of $f$ is sparse, say when $\#(\text{supp}(f) \cap [1,N]) \asymp \sqrt{N}$. Even in this extreme of $\sqrt{N}$ the large sieve can give non-trivial cancellation, but can one do better?
|
https://mathoverflow.net/users/75932
|
Large sieve inequality for sparse trigonometric polynomials
|
If $f$ is $M$-sparse, then from Cauchy-Schwarz one has $|S(\alpha)|^2 \leq M \sum\_{n \leq N} |f(n)|^2$ which gives the bound
$$ \sum\_{r=1}^R |S(\alpha\_r)|^2 \leq R M \sum\_{n \leq N} |f(n)|^2$$
which is superior in the regime $RM \leq N$ (i.e., below the range of the Heisenberg uncertainty principle). For $RM \geq N$ one is now consistent with the uncertainty principle and one cannot hope to do better than the large sieve in general. Indeed, if $M$ divides $N$ and $f$ is the indicator function of the multiples $\{ mN/M: m=1,\dots,M\}$ of $N/M$, then $S(\alpha)=M$ for $\alpha$ a multiple of $M/N$, so if one chooses $\alpha\_1,\dots,\alpha\_R$ to contain the $N/M$ different multiples of $M/N$ mod $1$ (and sets $\delta$ to be anything less than or equal to $M/N$) then equality in the large sieve is essentially attained.
Note that in this example $f$ had a lot of additive structure (in particular the support was basically closed under addition). If $f$ is additively unstructured (e.g., if it is supported on a dissociated set) then one can do better by taking higher moments. For instance by applying the large sieve to $f\*f$ rather than to $f$ one has
$$ \sum\_{r=1}^R |S(\alpha\_r)|^4 \ll (N + \delta^{-1}) \sum\_{n \leq 2N} |f\*f(n)|^2$$
and hence by Cauchy-Schwarz
$$ \sum\_{r=1}^R |S(\alpha\_r)|^2 \ll R^{1/2} (N + \delta^{-1})^{1/2} (\sum\_{n \leq 2N} |f\*f(n)|^2)^{1/2}.$$
If $f$ is not very additively structured, then $f\*f$ will be quite spread out and this can be a superior bound to the previously mentioned bounds. Similarly if one plays with higher convolutions such as $f\*f\*f$. Basically one is convolving one's way out of sparsity in order to make more efficient use of the large sieve. (A similar technique is frequently employed to study large values of Dirichlet polynomials, taking advantage of the fact that the logarithms $\log n$ of the natural numbers only have a very limited amount of additive structure, thanks to the divisor bound.)
As a variant of the above strategy, if $f$ is supported in some set $E$ on which good bounds are known on exponential sums $\sum\_{n \in E} e(\alpha n)$ (or weighted versions of such sums) - which is a measure of lack of additive structure in $E$ - then one can sometimes get good bounds from a duality argument (sometimes known as a "large values" argument, particularly in the context of controlling the large values of Dirichlet polynomials). See for instance
*Green, Ben; Tao, Terence*, [**Restriction theory of the Selberg sieve, with applications**](http://dx.doi.org/10.5802/jtnb.538), J. Théor. Nombres Bordx. 18, No. 1, 147-182 (2006). [ZBL1135.11049](https://zbmath.org/?q=an:1135.11049).
for an example of this (where the function $f$ is supported on something like the set of primes).
|
16
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https://mathoverflow.net/users/766
|
402734
| 165,248 |
https://mathoverflow.net/questions/402722
|
7
|
Let $S$ be a subset of ${1,2,...,n}$ such that for every $a,b$ in $S$ the numbers of form $a^k+b^k$ are distinct ($k$ is positive integer)
What is the maximum cardinality of $S$
|
https://mathoverflow.net/users/174530
|
Sidon sets with k >1
|
The $k=2$ case was considered by Alon and Erdős (*European J. Combin.* 6 (1985) 201-203, MR0818591) and improved by Lefmann and Thiele (*Combinatorica* 15 (1995) 379-408, MR1357284).
They expressed the problem as looking for the largest Sidon set of integers squared. The 1995 result: There exists a Sidon set $S \subset \{1^2, 2^2, \ldots, n^2\}$ with $$|S| \ge c \cdot n^{2/3}$$ where $c>0$ is constant. Alon and Erdős explain that a result of Landau on the density of the sums of two squares implies that, for any Sidon set $S$, $$|S| \le \frac{c'n}{(\log n)^{1/4}}.$$
Scanning the MathSciNet citations from references for the two articles, I don't see subsequent improvements of these bounds or treatments of $\{1^k, 2^k, \ldots, n^k\}$ for $k \ge 3$. (In the related case of starting from an infinite set of integers to a fixed power, there are more results; see a recent [arXiv article](https://arxiv.org/abs/2006.02783) by Kiss and Sándor and the references there.)
|
10
|
https://mathoverflow.net/users/14807
|
402736
| 165,250 |
https://mathoverflow.net/questions/402741
|
6
|
Let $X$ be a topological space and $\mathcal{F}$ be a sheaf of commutative topological groups on $X$. I am interested in the following question:
* Is there a natural way to introduce topology on $H^i(X, \mathcal{F})$?
My guess is that for each open covering $\mathcal{U}$ the space of Čech cochains $\check{C}(\mathcal{U}, \mathcal{F})$ can be endowed with the compact-open topology. This topology can be restricted to the space of closed cochains, descends to $H^i(\mathcal{U}, \mathcal{F})$ and induce direct limit topology on $H^i(X, \mathcal{F})$. But does this construction make sense? Say, is it functorial? Is it true that $H^i(X, \mathcal{F})$ are commutative topological groups (with respect to the natural group operation)? Are the other reasonable choices of topology on Čech complex? Can they lead to other topologies?
And what is the best reference on this topic?
I don't think that this is important, but in the situation I am interested in, the space $X$ is a (finite-dimensional) manifold and $\mathcal{F}$ is a sheaf of Lie groups.
|
https://mathoverflow.net/users/82309
|
Topology on cohomology of a sheaf of topological groups
|
Both cases ($F$ is a sheaf of abelian topological groups or abelian Lie groups) can be treated using the same machinery.
The Yoneda embedding embeds abelian Lie groups as a fully faithful subcategory of the category of sheaves of abelian groups on the site of smooth manifolds, and the embedding functor preserves small limits.
I refer to the latter category as the category of *abelian smooth groups*.
This category is complete and cocomplete.
It is, in fact, better behaved than the category of abelian topological groups, since the latter category is not an abelian category:
a morphism of abelian topological groups can have a trivial kernel and cokernel, without being an isomorphism.
On the other hand, the category of smooth groups is abelian.
From now on, we work either with presheaves of chain complexes of abelian topological groups or presheaves of chain complexes of smooth groups.
The category of such chain complexes can be equipped with the projective
model structure transferred from simplicial topological spaces
respectively simplicial sheaves on the site of smooth manifolds.
The category of presheaves of such chain complexes can itself be equipped
with the projective model structure,
which can then be further localized with respect to Čech nerves of open covers.
Fibrant objects in the resulting model category are presheaves
of chain complexes that satisfy the homotopy descent condition.
The fibrant replacement functor computes the (hyper)cohomology of sheaves.
More precisely, if $F→RF$ is a fibrant replacement of the sheaf $F$,
then $H^i(X,F)=H^i(Γ(X,RF))$.
The resulting cohomology theory is essentially (a reformulation of) the Segal–Mitchison cohomology in the topological case,
or its [smooth version by Brylinski](https://arxiv.org/abs/math/0011069).
In particular, the cohomology group $H^i(X,F)$ is by definition
a smooth (respectively topological) group, since $H^i$ is computed in the category of smooth (respectively topological) groups.
|
5
|
https://mathoverflow.net/users/402
|
402744
| 165,253 |
https://mathoverflow.net/questions/402714
|
7
|
Let $K\_{n,n}$ be a complete bipartite graph with two parts $\{u\_1,u\_2,\ldots,u\_n\}$ and $\{v\_1,v\_2,\ldots,v\_n\}$, and let $K^-\_{n,n}$ be the graph derived from $K\_{n,n}$ by delete a perfect matching $\{u\_1v\_1,u\_2v\_2,\ldots,u\_nv\_n\}$.
Since $K^-\_{n,n}$ is now $(n-1)$-regular, it has $n-1$ disjoint perfect matchings. My question is whether the edges of $K^-\_{n,n}$ with $n\geq 4$ can be decomposed into $n-1$ disjoint perfect matchings in such a way that in each matching $M$, if $u\_iv\_j\in E(M)$ then $v\_iu\_j\not\in E(M)$.
|
https://mathoverflow.net/users/148974
|
Disjoint perfect matchings in complete bipartite graph
|
The answer is that this is possible for all $n>4$.
Your question is equivalent to asking whether there exists a unipotent Latin square $L$ of order $n$ with $L\_{ij}\ne L\_{ji}$ for $i\ne j$. The equivalence is obtained by using $L\_{ij}$ to record the index of the matching that contains the edge $u\_i v\_j$ (and putting $L\_{ii}=n$ for each $i$).
The existence of such a Latin square follows from a stronger property. It is a theorem (collectively due to the work of Kotzig, McLeish, Turgeon and others) that for all $n\notin\{2,4\}$ there exists a Latin square of order $n$ that has no intercalates (an intercalate is a $2\times2$ submatrix that is itself a Latin square). This property of Latin squares is called $N\_2$ in the literature.
If you permute the rows of any $N\_2$ Latin square you can make every symbol on the main diagonal equal to $n$. You then have what you need.
|
8
|
https://mathoverflow.net/users/351290
|
402749
| 165,254 |
https://mathoverflow.net/questions/402688
|
16
|
The Cayley-Menger determinant gives the squared volume of a simplex in $\mathbb{R}^n$ as a function of its $n(n+1)/2$ edge lengths:
$$v\_n^2 = \frac{(-1)^{n+1}}{(n!)^2 2^n}
\begin{vmatrix}
0&d\_{01}^2&d\_{02}^2&\dots&d\_{0n}^2&1\\
d\_{01}^2&0&d\_{12}^2&\dots&d\_{1n}^2&1\\
d\_{02}^2&d\_{12}^2&0&\dots&d\_{2n}^2&1\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
d\_{0n}^2&d\_{1n}^2&d\_{2n}^2&\dots&0&1\\
1&1&1&\dots&1&0
\end{vmatrix}$$
where $d\_{i j}$ is the distance from vertex $i$ to vertex $j$ of the simplex.
For the regular simplex with all edge lengths 1, i.e. $d\_{i j}=1$ for all $i,j$, we have:
$$(n!)^2 2^n v\_n^2 = (-1)^{n+1}
\begin{vmatrix}
0&1&1&\dots&1&1\\
1&0&1&\dots&1&1\\
1&1&0&\dots&1&1\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
1&1&1&\dots&0&1\\
1&1&1&\dots&1&0
\end{vmatrix}
= n+1$$
This follows from the fact that the $(n+2)\times(n+2)$ matrix here has eigenvector $(1,1,1,\dots,1)$ with eigenvalue $n+1$, along with $n+1$ eigenvectors of the form $(1,0,0,\dots,-1,0,0,\dots)$ with eigenvalue $-1$.
Any odd integer $2k+1$ has a square that is equal to 1 modulo 8, since:
$$(2k+1)^2 = 4k(k+1)+1$$
and either $k$ or $k+1$ must be even. It follows that, for any simplex whose edge lengths are all odd integers, the quantity $(n!)^2 2^n v\_n^2$ must equal $n+1$ modulo 8.
When $n+1$ is *not* a multiple of 8, this means a simplex in $\mathbb{R}^n$ whose edge lengths are all odd integers cannot be degenerate, i.e. it cannot have zero volume.
However, this does not settle the same question when $n=8k-1$.
Are there known examples of degenerate simplices in $\mathbb{R}^{8 k-1}$ with odd integer edge lengths? Or is there a proof for their nonexistence that completes the proof that applies in other dimensions?
Acknowledgement: This question arose from [a discussion on Twitter between Thien An and Ian Agol](https://twitter.com/thienan496/status/1430773058218827778).
**Edited to add**: For all cases where $n = 7$ mod 16, it is possible to rule out a degenerate simplex by working modulo 16, where any squared odd integer must equal either 1 or 9. Computing the determinant when $x$ is added to any single squared edge length gives a quadratic in $x$ that has **even** coefficients for $x$ and $x^2$ (given that all the original entries are integers), from which it follows that adding 8 to any squared edge length preserves the determinant modulo 16. Since the determinant when all squared edge lengths are equal to 1 is $n+1$, changing any number of the squared edge lengths from 1 to 9 can never yield a determinant divisible by 16.
|
https://mathoverflow.net/users/23829
|
Is there a degenerate simplex in $\mathbb{R}^{8 k-1}$ with odd integer edge lengths?
|
This is answered in a paper by R. L. Graham, B. L. Rothschild & E. G. Straus ["Are there $n+2$ Points in $E\_n$ with Odd Integral Distances?"](https://sci-hub.ru/https://www.tandfonline.com/doi/abs/10.1080/00029890.1974.11993491). Such simplexes exist iff $n+2 \equiv 0 \pmod {16}$. They also consider the related problem of integral distances relatively prime to 3 and 6.
|
12
|
https://mathoverflow.net/users/795
|
402750
| 165,255 |
https://mathoverflow.net/questions/402691
|
17
|
Recall that a compact Riemann surface/algebraic curve $C$ is **hyperelliptic** if it admits a branched double cover $C \to \mathbb P^1$, where $\mathbb P^1$ is the complex projective line/Riemann sphere. Among those curves of hyperbolic type ($g \ge 2$), the only genus that admits such a double cover in general is $g = 2$. While this fact is essentially trivial from the perspective of Hartshorne's fourth chapter, I have always found this proof to feel like a nuclear flyswatter (at least over $\mathbb C$).
Intuitively, a hyperelliptic Riemann surface of genus $g > 2$ is one that can be "drawn conformally" (whatever exactly that may mean) in such a way that all $g$ of its holes are "lined up." As such, my intuition suggests that the proposition in the title is somehow analogous to the trivial statement from Euclidean plane geometry that every configuration of $n > 1$ points is collinear if and only if $n=2$, where the holes in the surface are treated as analogous to points in the plane. So my question is this: is there a "holomorphic geometric topological" proof of said proposition that proceeds roughly as described in this paragraph rather than using any big theorems like Riemann-Roch?
|
https://mathoverflow.net/users/27219
|
How would a topologist explain "every Riemann surface of genus $g$ is hyperelliptic if and only if $g=2$"?
|
Here is a small variant on Eremenko's answer.
The "Fenchel-Nielsen" coordinates on the space of hyperbolic metrics on a surface $\Sigma\_g$ can be described via a pants decomposition. This is a decomposition of the surface along a collection of curves, that split the surface into a union of disjoint $3$-punctured spheres.
Hyperbolic metrics on 3-punctured spheres making the boundary into totally geodesic curves are specified by the cuff lengths, i.e. three real parameters.
So if you have a surface $\Sigma\_g$, take a pants decomposition. Notice that you can choose pants decomposition equivariant with respect to a hyperelliptic involution. So if you think about the constraints on Fenchel-Nielsen coordinates coming from the surface being hyperelliptic, these only occur for $g > 2$. With $g=2$ all the curves are preserved and pants exchanged via the hyperelliptic involution, i.e. there are no additional constraints to be hyperelliptic.
But when $g>2$, there are curves off the hyperelliptic axis in your pants decomposition, so there are cuff lengths that are constrained by the hyperelliptic involution.
This is a long version of my comment.
|
8
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https://mathoverflow.net/users/1465
|
402752
| 165,256 |
https://mathoverflow.net/questions/402746
|
3
|
We know for morphisms of schemes $f:X \rightarrow S, u:S’ \rightarrow S, X’=X \times\_S S’$, (coherent) sheaf $F$ over X,we have natural morphism
$$u^\* R^q f\_\*F\rightarrow R^q u’\_\* f’^\*F$$
where $f’, u’$ are basechanges. (Or using derived category if you like.)
The base change theorems says under certain restrictions of the objects and morphisms, it’s isomorphism. Now I’m wondering in more general cases, if there’re some ‘obstruction’ (like certain cohomology groups, spectral sequences) to measure how far it is from isomorphism?
|
https://mathoverflow.net/users/170335
|
What‘s the obstruction to base change
|
The composition of functors $Rf\_\* \colon D(X) \to D(S)$ and $Lu^\* \colon D(S) \to D(S')$ is a Fourier--Mukai functor given by an explicit object $K \in D(X \times S')$. It is easy to check that it has no cohomology sheaves in positive degrees, its zero cohomology sheaf is isomorphic to the structure sheaf of the fiber product, but is also may have some cohomology sheaves in negative degrees. The morphism of functors in base change is induced by the natural morphism
$$
K \to \mathcal{H}^0(K).
$$
Consequently, the cone of the morphism in base change is controlled by the Fourier-Mukai functor with kernel $\tau\_{\le -1}(K)$. In particular, base change gives an isomorphism if and only $\tau\_{\le -1}(K) = 0$ if and only if $K = \mathcal{H}^0(K)$.
By the way, the negative cohomology sheaves of $K$ are responsible for the derived structure of the fiber product. So, this answer is essentially equivalent to the comment of @dhy.
|
11
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https://mathoverflow.net/users/4428
|
402755
| 165,257 |
https://mathoverflow.net/questions/402761
|
6
|
Let $A$ a square real matrix such that the largest singular value $\sigma\_\text{max}(A) = \sigma < 1$. I want to find a lower bound on $\langle (I + A)^{-1}x, x\rangle$ where $x$ is a vector of euclidean norm $1$: $\langle x, x\rangle=1$.
I empirically find that a seemingly tight lower bound is
$$
\langle (I + A)^{-1}x, x\rangle \geq \frac{1}{1+\sigma}
$$
which is reached for $A= \sigma I$. I cannot prove the above result.
Note that it is pretty straightforward to prove that $\sigma\_\text{min}((I + A)^{-1}) \geq \frac1{1+\sigma}$ but that does not suffice to conclude, since I do not assume that $A$ is symmetric.
PS: $\sigma\_\text{max}$ and $\sigma\_\text{min}$ are the largest and smallest singular values: $\sigma\_\text{max} = \sqrt{\lambda\_\text{max}(AA^T)}$ is the operator norm of $A$.
|
https://mathoverflow.net/users/173967
|
Lower bound $\langle (I + A)^{-1}x, x \rangle$ given that $\sigma_\text{max}(A) < 1$
|
The map $f(z)=(1+z)^{-1} - (1+\sigma)^{-1}$ maps the disk of radius $\sigma$ into the right half plane as a function of one complex variable.
Therefore, essentially by von Neumann's inequality, we get that $$\frac{f(A)+f(A)^\*}{2}=\mathrm{Re }f(A)\geq 0$$ since $\|A\|\leq \sigma.$ Assuming $A$ has real entries, this implies the claim as
$$\langle (1+A)^{-1}x,x\rangle = \langle \mathrm{Re} (1+A)^{-1}x,x\rangle.$$
To see the calculation with von Neumann's inequality more explicitly, let $\psi(z) = \frac{z-1}{z+1}.$ Note $\psi$ takes the right half plane to the disk. So, $\psi \circ f$ takes the disk of radius $\sigma$ into the unit disk.
Therefore, von Neumann's inequality states that
$\|\psi \circ f(A)\|\leq \sup\_{z\in \sigma\mathbb{D}} |\psi\circ f(z)| \leq 1.$
Note that $$1-(\psi \circ f(A))^\*(\psi \circ f(A)) \geq 0.$$ (Here by $T\geq 0$ we mean that $T$ is positive semi-definite.)
Writing out what that means
$$1-(f(A)^\*+1)^{-1}(f(A)-1)^\*(f(A)-1)(f(A)+1)^{-1} \geq 0.$$
So,
$$(f(A)^\*+1)(f(A)+1)-(f(A)^\*-1)(f(A)-1)=2(f(A)+f(A)^\*)\geq 0.$$
Results of the above form (positivity of noncommutative rational functions) always have to have "algebraic proofs," many of which can be done algorithmically. See, e. g.,
[Helton, Klep, and McCullough - The convex Positivstellensatz in a free algebra](https://www.sciencedirect.com/science/article/pii/S0001870812001892) and
[Pascoe - Positivstellensätze for noncommutative rational expressions](https://www.ams.org/journals/proc/2018-146-03/S0002-9939-2017-13773-3/home.html).
|
6
|
https://mathoverflow.net/users/32470
|
402770
| 165,261 |
https://mathoverflow.net/questions/402713
|
8
|
In [Zhu's seminal paper](https://www.ams.org/journals/jams/1996-9-01/S0894-0347-96-00182-8/), he proves (5.3.2) that if $V$ is a vertex algebra the character of all of its modules are ***modular forms***! (This is not literally true- there are conditions).
I have always found this statement very mysterious. How on earth do modular forms/elliptic curves come into the picture?
1. Is there a proof using the moduli space of elliptic curves $\mathcal{M}\_{1,1}$ more directly? e.g. if $M$ is a vertex algebra module, it can be localised onto any elliptic curve $E$. Is Zhu's proof saying we can vary $E$ and thereby get some structure living over $\mathcal{M}\_{1,1}$, from which we can take a character (which will just be a modular form because it's a section of a line bundle on $\mathcal{M}\_{1,1}$)?
I am also very confused about the conditions ($C\_2$-finiteness, rationality).
2. Are these conditions necessary for the theorem? If they are, is there any "geometric" meaning to them, say in terms of the associated chiral algebras, or the structure in 1. living over $\mathcal{M}\_{1,1}$?
Hopefully, with all the progress on vertex/chiral algebras in the 25 years since the proof was published, there is a clean modern answer to the above.
|
https://mathoverflow.net/users/119012
|
Why are VOA characters modular forms (geometrically)?
|
I'm sure someone here can handwave the intuition behind these statements much better than me. This handwaving version goes like this, one is interested in computing vacuum 1-point functions on a torus, but you can cut a torus along an $S^1$ making it into a cylinder, put a module `M` on the boundaries, now you start with a vector $m \in M$, you *evolve* it with $q^{L\_0}$ and get a vector $m' \in M$. Gluing back the torus corresponds to taking the trace.
Now to actually answer the different parts of the question.
>
> if MM is a vertex algebra module, it can be localised onto any elliptic curve EE. Is Zhu's proof saying we can vary EE and thereby get some structure living over M1,1M1,1, from which we can take a character
>
>
>
There are a number of misinterpretations here. It is true that given a vertex algebra module (with very mild conditions) you can get a sheaf on an elliptic curve corresponding to it (a module for the corresponding chiral algebra). However, this has nothing to do with coinvariants of $V$. Coinvariants of $V$, or conformal blocks, have to do with the bundle/sheaf that you get on an elliptic curve by localizing $V$ itself, and not a module.
The statement from Zhu can be translated in a way very similar to what you say though: For each genus g curve $X$ with $n$ marked points $\{x\_i\} \subset X$, one can construct a vector space $C(X, V, \{x\_i\})$ of coinvariants. In the language of Frenkel-Ben-Zvi cited in the comments these are coinvariants of $V$ itself: you put the vacuum module $V$ at each of the marked points $x\_i$. So far no module $M$ is involved. Now as you move $X$ in the moduli space $\mathcal{M}\_{g,n}$, these spaces arrange into a twisted $\mathcal{D}$-module on this moduli space.
In the particular case of 1-marked elliptic curves, ie. $g=1$ and $n=1$, under some conditions (below we'll deal with that part of your question) these vector spaces are finite dimensional and you get an actual vector bundle with a flat connection. Notice that this is simply the statement that the twisted $D$-modules are coherent sheaves, since the space is one dimensional, this has to be a vector bundle with a flat connection. Zhu's statement is that **characters of irreducible representations of $V$ are flat sections of these bundles (rather their duals), and moreover they form a basis of the space of flat sections**.
So the role of the module $M$ is to produce a section of a geometric object constructed from $V$, not from $M$ directly.
>
> Is there a proof using the moduli space of elliptic curves M1,1M1,1 more directly?
>
>
>
Yes indeed, and this was already included in Emile's comments, but I'll repeat them essentially here: We have a vector bundle with a flat connection on the moduli space of elliptic curves given by the dual bundle to coinvariants of $V$. This flat connection is rather explicit (I'll describe it below), so to have a flat section is to solve an explicit differential equation. This in turn leads to prove that flat sections give rise to a solution of an explicit ODE. This ODE has a singularity in the boundary of the moduli space at $q = 0$, but it is a mild singularity in the sense that it is a regular singular point. What you do is prove that the formal character of a module $M$ gives a formal power series that solves this differential equation (there is a subtlety here between taking $\tau \in \mathbb{H}$ as the parameter or $q = e^{2 \pi i \tau}$). Once you have this, the general theory of Frobenius expansions of solutions of these ODEs tell you that these characters converge (the hard part) and therefore they form vector valued modular forms (the easy part: we already knew that they were sections of a bundle on the moduli space of elliptic curves).
>
> I am also very confused about the conditions (C2C2-finiteness, rationality).
> Are these conditions necessary for the theorem?
>
>
>
Not exactly. What $C\_2$ gives you is the finite dimensionality of the space of coinvariants (in fact you need less than this, as we'll see below), what rationality gives you is the fact that you can construct **all** flat sections starting from irreducible modules. The idea here is geometric and goes back to studying the behavior of flat sections near the nodal curve limit $q=0$.
This idea roughly goes as follows: If you give me a flat section, it gives a solution to an ODE as we talked above. As such you can expand it in power series of $q$ (and in theory possibly a polynomial in $\tau = \log q$). The lowest coefficient of this series defines a symmetric function of the Zhu algebra $Z(V)$ of $V$. This is a remarkable associative algebra that has a few properties: The list of its irreducible modules up to isomorphism is in bijection with the irreducible modules V. And in addition under the finitenes conditions listed above $Z(V)$ happens to be finite dimensional and semisimple. So the conformal block being a symmetric function, is a linear combination of traces of irreducible $Z(V)$-modules. This is the direct connection: one can start from an irreducible finite dimensional module $M\_0$ for the Zhu algebra $Z(V)$ of $V$. One induces a module $M$ for $V$ having $M\_0$ as it lowest graded piece. Then by what we have talked above we know that the character of $M$ is a flat section, and its restriction to its lowest graded piece $M\_0$ is the symmetric function we started from. There are some holes here and there in the description I just gave, but it is essentially correct.
Now, if $V$ is not rational, but it is $C\_2$ cofinite, you still have finite dimensional space of coinvariants and finitely many irreducibles. Just that now you may have extensions between these modules. Miyamoto carried out Zhu's program in this scenario: instead of looking at irreducible $V$ modules induced from $Z(V)$ we need to look at modules induced from *projective modules of higher Zhu algebras*.
Finite dimensionality of coinvariants on elliptic curves.
---------------------------------------------------------
The $C\_2$ condition and its role in having finite dimensionality of coinvariants is a bit confusing still (at least to me). It appears in different ways that are somewhat independent. I'll describe in detail the situation for elliptic curves which is what your question is about.
The space of coinvariants of the elliptic curve $X\_q = \mathbb{C}^\* / \mathbb{Z} \simeq C / \mathbb{Z} + \mathbb{Z}\tau$ (where the action of $Z$ is given by multiplication by $q$), with coefficients in the chiral algebra associated to a vertex algebra $V$ with supports in its vacuum module ($V$ itself) in the marked point $0 \in X\_q$ is explicitly given by the cokernel of the map
$$ d: V \otimes V \otimes \Gamma(\mathcal{O}\_{X\_q}, X\_q \setminus 0) \rightarrow V, \qquad a \otimes b \otimes f(z) \mapsto \mathrm{res}\_z f(z) Y(a,z) b.$$
So these functions $f(z)$ are biperiodic functions of $z$ with possible poles at $m + n \tau$ with $m,n \in \mathbb{Z}$. It turns out that $d(V \otimes V \otimes f') \subset d(V\otimes V \otimes f)$ so it is enough to kill a basis of functions modulo total derivatives. This is the top de Rham cohomology of $X\_q \setminus 0$ and this has a basis given by the constant function $1$ and the Weierstrass $\wp(z,q)$ function. So the space of coinvariants is simply
$$ V / \langle \mathrm{res}\_z \wp(z) Y(a,z)b, \, \mathrm{res}\_z Y(a,z)b \rangle $$
I am lying slightly here in that the vertex operation that is needed is denoted by $Y[a,z]$ in Zhu's paper and has to do with the fact that we are using an isomorphism $\mathbb{C}^\*/\mathbb{Z} \simeq \mathbb{C} / \mathbb{Z}^2$, but that doesn't really change the exposition.
The key point is that **every vertex algebra** is filtered, the relevant filtration is known as the Li filtration. And the above two-term complex is compatible with this filtration. So by looking at the associated graded we get an immediate bound for the homology of that two term complex in degree 0. It turns out that the operation
$$ a \otimes b \mapsto \mathrm{res}\_z \wp(z;q) Y(a,z)b$$ in this associated graded is simply the operation $a \otimes b \mapsto a\_{(-2)}b = \mathrm{res}\_z z^{-2} Y(a,z)b$, that is, in this filtration, we only care about the singular part of $f(z)$, not $f(z)$ itself.
There you go, just killing $d(V \otimes V \otimes \wp(z))$ we get, in the associated graded, the $C\_2$ quotient of $V$. So if $V$ is $C\_2$ cofinite, we know that the space of coinvariants is finite dimensional.
But there is more. We didn't kill the constant functions $f(z)= 1$. Zhu didn't really care since he needed rationality and finite dimensionality of $Z(V)$, but in his paper is clear that what actually is enough to prove finite dimensionality of coinvariants is not $R\_V := V/C\_2(V)$ being finite dimensional, but rather the quotient of $R\_V$ by the image of $a \otimes b \mapsto \mathrm{res}\_z Y(a,z)b = a\_{(0)}b$.
Now $R\_V$ is a Poisson algebra, with Poisson bracket given by $\{a,b\} = a\_{(0)}b$ so **the actual sufficient condition to have finite dimensionality of coinvariants is that the zeroth Poisson homology of $R\_V$, $R\_V / \{ R\_V, R\_V\}$ is finite dimensional**.
This condition fits nicely in a much wider context: it was proved that the finite dimensionality of the first Poisson homology of $R\_V$ also plays a role in the finite dimensionality of the first chiral homology of the elliptic curve with coefficients in $V$ (there is a complex whose degree $0$ homology is coinvariants, the general homology is known as *chiral homology*). We expect this to be the case for arbitrary chiral homology in fact, not only degree $0$ and $1$.
Now that I've set up the above notation, I can quickly tell you about the flat connection that I promised above it was explicit. What is a class in the dual to the coinvariants: it's a functional $\varphi: V \rightarrow \mathbb{C}$ that satisfies
$$ \varphi ( \mathrm{res}\_z f(z) Y(a,z)b ) = 0$$
For every biperiodic $f$ with possible poles at $z=0$ and every $a,b \in V$. Now there is the Weierstrass $\zeta$ function, which is not really bi-periodic, but it almost is. It's derivative is the Weierstrass $\wp$ function. The flat sections of coinvariants are $\varphi$ that satisfy the equation above and in addition satisfy:
$$ \frac{d}{d\tau } \varphi(a) = \varphi ( \mathrm{res}\_z \zeta(z) Y(\omega, z)a ) $$
where $\omega \in V$ is the conformal vector. Notice that I never talked about $V$ being conformal up to this point and indeed the conformal vector only appears in the geometric picture just to give us the flat connection.
$C\_2$ appears in a different way as a sufficient condition to prove finite dimensionality of coinvariants in arbitrary genus. I'll point you to recent work of Damiolini, Gibney and Tarasca. They use this condition to check that you indeed have a vector bundle in the moduli space. But I have a strong suspicion that this can be relaxed as well to finite dimensionality of the Poisson homology.
A final remark in an already long answer. When you ask:
>
> say in terms of the associated chiral algebras,
>
>
>
Not as far as I know actually. The key point was the Li filtration. And the Li filtration is a filtration of Vertex algebras, but it is not compatible with the chiral algebra structure. There is no literature that I am aware about this filtration in relation to chiral algebras. And up to not long ago at least one of the leading experts in the field of chiral algebras was not aware of the existence of this filtration. It is picking up pace and there have been many exciting results in the last couple of years, so I expect to see some new idea popping up in the arxiv soon regarding this.
|
11
|
https://mathoverflow.net/users/17980
|
402786
| 165,264 |
https://mathoverflow.net/questions/402718
|
0
|
Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described [here](https://mathoverflow.net/questions/236186/how-to-define-the-injectivity-radius-of-manifolds-with-boundary/236194#236194), is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference for that statement?
**Added question**: Is it essential that we assume that $\overline{E}$ is of bounded geometry and not $E$? What happends if we just assume that $E$ is of bounded geometry?
Thanks in advance!
|
https://mathoverflow.net/users/163368
|
Is this a manifold of bounded geometry?
|
Unless I am terribly mistaken, the answer is yes and the proof strategy is rather simple. The crucial observation is that the definition of bounded geometry depends on quantities that are continuous.
Consider first the injectivity radius function of the boundary, $r\_{b}\colon\delta X\to \mathbb{R}$,
$$r\_b(x)=\sup\{\ t>0\mid \exp\colon B\_{\delta X}(0\_x,t)\to \delta X \quad \text{is a diffeomorphism}\ \}.$$ This is a continuous and positive function on $\delta X$ and it is bounded away from zero on $\delta X\setminus K$, so it must be bounded away from zero over all $\delta X$ because $K$ is compact. Denote the positive lower bound by $r\_b(X)$; this is the injectivity radius of $X$.
Now define the ''normal collar injectivity radius'' $r\_C\colon \delta X\to\mathbb{R}$ as $$r\_c(x)=\sup\{\ t>0\mid \kappa\colon B\_{\delta X}(x,r\_b(X))\times[0,t)\to X\quad \text{is well-defined and a diffeomorphism} \ \},$$
where $\kappa(x,t)=\exp(t\nu\_x)$ and $\nu\_x$ is the unit inward normal vector. Again, this is continuous and bounded away from zero outside a compact set, hence bounded away from zero; we have obtained that $X$ satisfies the normal collar condition in Schick's definition.
A similar argument works for the remaining two conditions.
**Answer to the added question**: If $E$ is a manifold with boundary of bounded geometry with the metric that is the restriction of that of $\overline{E}$, then $E=\overline{E}$.
Assume for the sake of contradiction that $E$ is of bounded geometry and there is a point $x\in \overline{E}\setminus E$. Consider first the case where $x$ is in the closure of $\delta E$, so $x\in \delta \overline{E}$ and there is a sequence $(y\_n)$ in $\delta E$ with $y\_n\to x$. The injectivity radius $r\_b$ of $\delta E$ is positive, so $B\_{\delta \overline{E}}(y\_n,r\_b)\subset \delta E$ and $d\_{\delta \overline{E}}(x,y\_n)>r\_b$ for all $n$, a contradiction.
Similar arguments apply when $x$ is a limit of points in the normal collar or a limit of points not in the normal collar.
|
2
|
https://mathoverflow.net/users/44172
|
402791
| 165,267 |
https://mathoverflow.net/questions/402767
|
4
|
The Hankel determinants of the Catalan numbers are well known and can be written as
$d(k,n)= \det \left( C\_{k + i + j} \right)\_{i,j = 0}^{n - 1}=\prod\_{i=1}^{k-1}\frac{\binom{2n+2j}{j}}{\binom{2j}{j}}$ with $d(k,0)=1.$
Computations suggest that
$$D\_k(x)=\sum\_{n\geq 0}d(k,n)x^n=\frac{A\_{k}(x)}{(1-x)^{\binom{k}{2}+1}}$$
where $A\_{k}(x)$ is a palindromic and unimodal polynomial of degree $\binom{k-1}{2}.$
Moreover it seems that $A\_{k}(x)$ is gamma-nonnegative, i.e. a linear combination of polynomials $x^j(1+x)^{\binom{k-1}{2}-2j}$ with positive coefficients.
Is this known? Any idea how to prove this?
|
https://mathoverflow.net/users/5585
|
Generating functions for Hankel determinants of Catalan numbers
|
I'm upgrading my comments to an answer.
As I've mentioned in comments/answers to some of your previous MO questions (e.g. [Number of bounded Dyck paths with negative length as Hankel determinants](https://mathoverflow.net/questions/372811/number-of-bounded-dyck-paths-with-negative-length-as-hankel-determinants) and [Some nice polynomials related to Hankel determinants](https://mathoverflow.net/questions/389657/some-nice-polynomials-related-to-hankel-determinants)), this Hankel determinant of Catalan numbers counts fans of nested Dyck paths (see, e.g., Section 3.1.6 of <https://arxiv.org/abs/1409.2562>), which are the same as plane partitions of staircase shape with bounded entries, which were first counted by Proctor (see his "Odd symplectic groups" paper <https://doi.org/10.1007/BF01404455>).
At any rate, this interpretation means that your generating function $D\_k(x)$ is the same as the generating function $\sum\_{m \geq 0}\Omega\_P(m)x^m$ where $\Omega\_P(m)$ counts the number of order preserving maps $P\to \{0,1,\ldots,m\}$ for a certain poset $P$ (namely, the staircase partition shape poset; equivalently, the Type A root poset). The general theory of $P$-partitions, as developed by Stanley, thus says that $D\_k(x) = \sum\_{L} x^{\mathrm{des}(L)}/(1-x)^{\binom{k}{2}-1}$, where the sum is over linear extensions of $P$, when it is naturally labeled (see e.g. Theorem 3.15.8 of EC1). Thus your $A\_k(x)$ is the $P$-version of an Eulerian number generating function, and a general result of Brändén (see <https://doi.org/10.37236/1866> or <https://arxiv.org/abs/1410.6601>) says that such poset linear extension descent genearting functions are $\gamma$-nonnegative as long as the poset in question is graded, which it is in this case.
|
6
|
https://mathoverflow.net/users/25028
|
402795
| 165,269 |
https://mathoverflow.net/questions/402797
|
1
|
What is the name of the following combinatorial game:
Two players, moving in turn.
Positions: $0,1,2,\ldots$.
Moves: $n\longmapsto n-1$ or $n\longmapsto \lfloor n/2\rfloor$
if $n>0$.
No move for $0$ which loses.
The determination of the winning strategy is easy:
$n=2^k(2m+1)>0$ wins if and only if $k$ is even.
Determinating the winning strategy for the misere convention ($0$ wins) is also easy:
$2^k$ wins for $k$ odd and loses for $k$ even,
$2^k(2m+1)$ with $m\geq 1$ wins for $k$ even and loses for $k$ odd.
This game has certainly been described somewhere. Does it has a name? Does somebody have a reference?
|
https://mathoverflow.net/users/4556
|
Name for an easy combinatorial game
|
This is "Mark", supposedly due to Mark Krusemeyer; see the first sentence of the introduction to <https://arxiv.org/abs/1509.04199> and section 2 of <https://doi.org/10.37236/2015>.
|
3
|
https://mathoverflow.net/users/3075
|
402805
| 165,272 |
https://mathoverflow.net/questions/402789
|
9
|
$\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}$A number of algebraic structures can be defined as monoids in some appropriate monoidal category:
* A **monoid** is a monoid in $(\mathsf{Sets},\times,\pt)$;
* A **semiring** is a monoid in $(\mathsf{CMon},\otimes\_{\N},\N)$;
* A **ring** is a monoid in $(\mathsf{Ab},\otimes\_\Z,\Z)$;
* An **$R$-algebra** is a monoid in $(\mathsf{Mod}\_R,\otimes\_R,R)$;
* A **graded $R$-algebra** is a monoid in $(\mathsf{Gr}\_\Z\mathsf{Mod}\_R,\otimes\_R,R)$;
* A **differential graded $R$-algebra** is a monoid in $(\mathsf{Ch}\_\bullet(\mathsf{Mod}\_R),\otimes\_R,R)$.
Is this also the case for [differential rings](https://en.wikipedia.org/wiki/Differential_algebra)?
|
https://mathoverflow.net/users/130058
|
Are differential rings monoids in a monoidal category?
|
$\newcommand{\defeq}{\overset{\mathrm{def}}{=}}\newcommand{\id}{\mathrm{id}}\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\d}{\mathrm{d}}\newcommand{\dAb}{\mathsf{End}(\mathsf{Ab})}$DGAs are monoids in chain complexes. To get differential rings as monoids in some monoidal category, it suffices to remove the grading and the $\d^{2}=0$ condition.
In detail, consider the category $\mathsf{End}(\mathsf{Ab})\defeq\mathsf{Fun}(\mathbf{B}\N,\mathsf{Ab})$ whose
* Objects are pairs $(A,\d)$ with $A$ an abelian group and $d\colon A\to A$ a morphism of abelian groups.
* Morphisms $(A,\d\_A)\to(B,\d\_B)$ are morphisms of abelian groups preserving the derivation, i.e. such that the diagram
$$
\require{AMScd}
\begin{CD}
A @>\d\_A>> A\\
@V f V V @VV f V\\
B @>>\d\_B> B
\end{CD}
$$
commutes.
We can then put a monoidal structure $\otimes\_\Z$ on $\dAb$ by defining
$$(A,\d\_A)\otimes(B,\d\_B)=(A\otimes\_\Z B,\d\_A\otimes\_\Z1\_B+1\_A\otimes\_\Z\d\_B),$$
where the unit is given by the pair $(\Z,\d\_\Z)$ with $\d\_\Z\overset{\mathrm{def}}{=} 0$. Note that a morphism in $\dAb$ from $(\Z,\d\_\Z)$ to $(A,\d\_A)$ is just a "constant" element of $A$, i.e. an element with $\d\_A a = 0$.
A monoid in $(\dAb,\otimes\_\Z,(\Z,\d\_\Z))$ will then be a triple $((A,\d),\mu,\eta)$ with
* $(A,\d)$ an object of $\dAb$; this accounts for the underlying additive abelian group of a differential ring and the derivation $\d$, which is $\Z$-linear;
* $\mu\colon(A,\d\_A)\otimes\_\Z(A,\d\_A)\to(A,\d\_A)$ a morphism of $\dAb$; this accounts for the multipication and the Leibniz rule: asking for the diagram
$$
\require{AMScd}
\begin{CD}
A\otimes\_\Z A @>\d\_A\otimes\_\Z1\_A+1\_A\otimes\_\Z\d\_A>> A\otimes\_\Z A\\
@V \mu V V @VV \mu V\\
A @>>\d\_A> A
\end{CD}
$$
to commute is equivalent to asking
$$\d(ab)=\d(a)b+a\d(b)$$
to hold for all $a,b\in A$;
* $\eta\colon(\Z,\d\_\Z)\to(A,\d\_A)$ a morphism of $\dAb$, determining an element $1\_A$ of $A$;
such that the usual associativity and unitality diagrams commute, which makes $(A,\mu,\eta)$ into a ring, and together with $\d$, this makes the quadruple $((A,\d),\mu,\eta)$ into a differential ring.
|
7
|
https://mathoverflow.net/users/130058
|
402816
| 165,274 |
https://mathoverflow.net/questions/402801
|
2
|
Let $\frak{g}$ be a complex simple Lie algebra of rank $l$. For $\frak{h}$ a choice of Cartan subalgebra, let $\alpha\_1, \cdots, \alpha\_r$ be the corresponding choice of simple roots, $X\_{\alpha\_i}, H\_{\alpha\_i}, X\_{-\alpha\_i}$ the Cartan--Weyl basis, and $\pi\_1, \cdots, \pi\_l$ the fundamental weights. For the irreducible $\frak{g}$-module $V(\pi\_k)$ let $v \in V(\pi\_k)$ be a highest weight vector, i.e.
$$
X\_{\alpha\_i}v = 0 ~~ \forall i=1, \dots, r.
$$
How will the elements $X\_{-\alpha\_i}$ act on $v$? Initial experiments suggest that
$$
X\_{-\alpha\_i}v = 0, ~~ \forall i \neq k.
$$
However I cannot seem to see a general proof. Any help is very appreciated.
|
https://mathoverflow.net/users/352001
|
Action of the negative Cartan-Weyl generators on a highest weight element
|
Let α be a simple root that is not αk. Associated to this simple root is a subalgebra isomorphic to $\mathfrak{sl}\_2$. For this subalgebra, an element of weight ωk has weight zero. In the fundamental representation you are considering, such an element is highest weight by assumption and therefore by the classificaition of representations of $\mathfrak{sl}\_2$ is annihilated by Fα.
|
4
|
https://mathoverflow.net/users/425
|
402820
| 165,277 |
https://mathoverflow.net/questions/402847
|
4
|
I'm trying to compute the following Haar integral over the unitary group:
$$
\int\limits\_{\mathbb{U}(d)}\dfrac{1}{\sum\_{k,l=1}^d u\_{ik}\overline{u\_{il}}c\_{kl}}dU.
$$ Is there anything known about the value of such integrals? I haven't been able to find anything that studies the integrals of either non polynomial functions or non class functions of unitaries.
|
https://mathoverflow.net/users/71033
|
Haar integral of rational function of unitaries
|
There exist no closed-form expressions for arbitrary $d$ for the integral over the unitary group $\mathbb{U}(d)$ of a rational function of the matrix elements.
There are asymptotic results for large $d$, see for example [J. Math. Phys. 37, 4904 (1996)](https://arxiv.org/abs/cond-mat/9604059). The leading order term for $\text{tr}\,C$ of order $d$ is
$$\int\limits\_{\mathbb{U}(d)}\dfrac{1}{\sum\_{k,l}u\_{ik}\overline{u\_{il}}c\_{kl}}dU=\frac{d}{\text{tr}\,C}+{\cal O}(1/d),$$
independent of the index $i$.
Alternatively, if $d$ is not large but the matric $C$ is close to the unit matrix, $c\_{kl}=\delta\_{kl}+\epsilon\_{kl}$, one can expand
$$
\begin{split}
\int\limits\_{\mathbb{U}(d)}\dfrac{1}{\sum\_{k,l}u\_{ik}\overline{u\_{il}}c\_{kl}}dU & =1-\int\limits\_{\mathbb{U}(d)}\sum\_{k,l}u\_{ik}\overline{u\_{il}}\epsilon\_{kl}\,dU+{\cal O}(\epsilon^2)\\
& =1-\frac{1}{d}\sum\_{k}\epsilon\_{kk}+{\cal O}(\epsilon^2).
\end{split}$$
|
7
|
https://mathoverflow.net/users/11260
|
402848
| 165,284 |
https://mathoverflow.net/questions/402802
|
3
|
Informally, an $\infty$-category should be the following data:
* A collection of objects
* A space of morphisms between any two objects
* Weak associativity rules: Coherent homotopies between all of the different ways to compose morphisms
As far as I understand, simplicially enriched categories can be used as a somewhat strict version model for that: The mapping simplicial-sets are indeed models for mapping spaces (by that I mean a simplicial category is fibrant if the mapping simplicial-sets are Kan complexes), but the composition rule should still be strict.
Is there a model that spells out all the details required for defining an $\infty$-category using the same ingredients as in a simplicial category but with weak composition rules instead?
|
https://mathoverflow.net/users/125868
|
Weak composition rule for simplicial categories
|
The most obvious approach is to consider simplicial $\def\Ai{{\sf A}\_∞}\Ai$-categories, where $\Ai$ denotes a nonsymmetric operad in simplicial sets that is weakly equivalent to the terminal operad, i.e., the associative operad.
In such a structure, instead of the usual compositions we have maps
$$\Ai(n)⨯C(X\_{n-1},X\_n)⨯⋯⨯C(X\_0,X\_1)→C(X\_0,X\_n)$$
that satisfy the usual identities for algebras over operads.
As it turns out, the resulting notion is in some sense equivalent to the usual simplicial categories: every simplicial $\Ai$-category can be rectified to a simplicial category by Proposition 9.2.4
in [arXiv:1410.5675v3](https://arxiv.org/abs/1410.5675v3).
Other generalizations that extend simplicial categories using less obvious notions of coherence for composition
include [Segal categories](https://ncatlab.org/nlab/show/Segal+category) and [complete Segal spaces](https://ncatlab.org/nlab/show/complete+Segal+space).
All of these are also equivalent to simplicial categories.
The book by Bergner (*The Homotopy Theory of (∞,1)-Categories*) contains an exposition of these equivalences.
|
6
|
https://mathoverflow.net/users/402
|
402849
| 165,285 |
https://mathoverflow.net/questions/402851
|
4
|
$\newcommand{\R}{\mathbb R}$Consider the following [construction](https://www.mathcounterexamples.net/pointwise-convergence-not-uniform-on-any-interval/). For real $u$, let
\begin{equation}
f(u):=\frac{2u^2}{1+u^4},
\end{equation}
so that the function $f\colon\R\to\R$ is continuous, $0\le f\le1=f(\pm1)$, and $f(u)\to0=f(0)$ as $|u|\to\infty$.
Let $Q$ be any countable dense subset of $\R$. Let $(w\_q)\_{q\in Q}$ be any family of (strictly) positive real numbers such that $\sum\_{q\in Q}w\_q<\infty$.
Finally, for each natural $n$ and all real $x$, let
\begin{equation}
g\_n(x):=\sum\_{q\in Q}w\_q\, f(n(x-q)).
\end{equation}
The latter series converges uniformly in $x$, and hence the function $g\_n$ is continuous. Moreover, by dominated convergence, $g\_n\to0$ pointwise (as $n\to\infty$).
Take now any nonempty open interval $I\subset\R$. Then $r\in I$ for some $r\in Q$. Moreover, $\{r+1/n,r-1/n\}\cap I\ne\emptyset$ eventually -- that is, for all large enough $n$ (depending on $I$). So,
\begin{equation}
\liminf\_n\,\sup\_I g\_n\ge w\_r \liminf\_n f(n(r\pm1/n-r))=w\_r>0.
\end{equation}
Thus, we have a sequence of continuous functions $g\_n$ converging to $0$ pointwise on $\R$ while for any nonempty open interval $I\subset\R$ we have $\liminf\limits\_n\,\sup\limits\_I g\_n>0$.
---
This brings us to
>
> **Question:** Does there exist a sequence of continuous functions $g\_n\colon\R\to\R$ converging to $0$ pointwise on $\R$ such that for any nonempty open interval $I\subset\R$ we have $\liminf\limits\_n\,\sup\limits\_I g\_n\ge1$?
>
>
>
(Of course, here we can replace $\ge1$ by $=1$.)
|
https://mathoverflow.net/users/36721
|
How bad can pointwise convergence in $C$ be?
|
This is just a standard application of the Baire category theorem.
Proposition: Suppose that $X$ is a topological space where the Baire category theorem holds and $g\_{n}:X\rightarrow[0,\infty]$ for each natural number $n$. Suppose that $\overline{\lim}\_{n}\sup\_{I}g\_{n}\geq 1$ for each non-empty open set $I$. Then there is a point $x\in X$ such that $\overline{\lim}\_{n}g\_{n}(x)\geq 1$.
Proof: If $0<\delta<1$ and $n$ is a natural number, then let $U\_{n,\delta}=g\_{n}^{-1}(\delta,\infty)$. Then each $U\_{n,\delta}$ is open. Let $O\_{n,\delta}=\bigcup\_{k=n}^{\infty}U\_{k,\delta}$. Then $O\_{n,\delta}$ is open and dense. Therefore, $\bigcap\_{m,n}O\_{n,1-1/m}$ is dense by the Baire category theorem. If $x\in\bigcap\_{m,n}O\_{n,1-1/m}$, then for each $m>0$, we have $x\in O\_{n,1-1/m}$ for all $n$. Therefore, since $x\in O\_{n,1-1/m}$, we know that $x\in U\_{k,1-1/m}$ for infinitely many $k$. Thus, $g\_{k}(x)>1-1/m$ for infinitely many $k$. Therefore, $\overline{\lim}\_{n\rightarrow\infty}g\_{n}(x)\geq 1-1/m$, so
since $m$ is arbitrary, we conclude that $\overline{\lim}\_{n\rightarrow\infty}g\_{n}(x)\geq 1$ for $x\in\bigcap\_{m,n}O\_{n,1-1/m}$. Q.E.D.
|
12
|
https://mathoverflow.net/users/22277
|
402855
| 165,288 |
https://mathoverflow.net/questions/402742
|
3
|
Consider the split monic $f=\prod\_{i=1}^n(x-x\_i)\in \mathbb Z[x\_1 ,\dots ,x\_n,x]$. Its discriminant is usually defined as $$(-1)^{n(n-1)/2}\prod\_{i=1}^nf^\prime(x\_i)=\prod\_{1\leq i<j\leq n}(x\_i-x\_j)^2.$$
What is the reason for taking this definition as opposed to $\prod\_{i=1}^nf^\prime(x\_i)$? The product of the derivatives at the roots "feels" to me more canonical than the product on the RHS.
|
https://mathoverflow.net/users/69037
|
Why the sign in the definition of the discriminant?
|
The reason is that the formula on the *right side* should be considered more fundamental, not the formula on the left, when seeking a symmetric expression in the roots. Don't use a product of anything "at" the roots, but a symmetric expression in the roots that vanishes if any pair of roots are equal. That explains the factors $(x\_i-x\_j)^2$. Do you consider the simplest polynomial with a double root at 0 to be $x^2$ or $-x^2$? The product on the left (without the sign) is also interesting and has a name: it is called the *resultant* of $f(x)$ and $f'(x)$.
|
7
|
https://mathoverflow.net/users/3272
|
402862
| 165,292 |
https://mathoverflow.net/questions/402824
|
2
|
>
> **Question 1.** Let $R$ be a polynomial ring over a field $k$. Assume that $R$ is graded in the usual way (i.e., each variable has degree $1$). Let $I$ and $J$ be two ideals of $R$ such that $I$ is generated in degree $\leq a$ (that is, there is a set of generators of $I$ that have degrees $\leq a$) and $J$ is generated in degree $\leq b$. Is it true that the ideal $I \cap J$ is generated in degree $\leq a+b$ ?
>
>
>
When $R$ is a univariate polynomial ring, this is simply claiming that the lcm of two polynomials of degrees $\leq a$ and $\leq b$ is a polynomial of degree $\leq a+b$; of course this is correct. But this kind of logic does not generalize to more variables. (It is [easy to see](https://web.archive.org/web/20150512110802/https://crazyproject.wordpress.com/2011/02/27/the-intersection-of-two-monomial-ideals-is-a-monomial-ideal/) that Question 1 has a positive answer when $I$ and $J$ are monomial ideals.)
If true, Question 1 would give a new (inductive) proof for the following theorem (part of the Subspace arrangement theorem of Derksen and Sidman -- see Theorem 2.1 in [their *A sharp bound for the Castelnuovo-Mumford regularity of subspace arrangements*, arXiv:math/0109035v1](https://arxiv.org/abs/math/0109035v1)):
>
> **Theorem 2.** Let $R$ be a polynomial ring over a field $k$. Assume that $R$ is graded in the usual way (i.e., each variable has degree $1$). Let $I\_1, I\_2, \ldots, I\_t$ be $t$ homogeneous ideals of $R$ that are all generated in degree $1$ (that is, each $I\_i$ has the form $I\_i = V\_i R$ for some $k$-vector subspace $V\_i$ of the degree-$1$ homogeneous component of $R$). Then, the ideal $I\_1 \cap I\_2 \cap \cdots \cap I\_t$ is generated in degree $\leq t$.
>
>
>
Some more questions suggest themselves:
>
> **Question 3.** (If the answer to Question 1 is negative:) Does the answer to Question 1 become positive if we replace "ideal" by "homogeneous ideal"?
>
>
>
(This would still be enough to reprove Theorem 2.)
Truth be told, I'm less interested in the polynomial case than I am in the exterior algebra case:
>
> **Question 4.** What if we replace the polynomial ring by an exterior algebra?
>
>
>
(You can assume $k$ has characteristic $0$ for simplicity.)
Finally, if **all** these questions have negative answers, here is what I am really looking for:
>
> **Question 5.** Is there an elementary proof of the analogue of Theorem 2 for exterior algebras (which is part of Theorem 9 in [Francesca Gandini, *Degree bounds for invariant skew polynomials*, arXiv:2108.01767v1](https://arxiv.org/abs/2108.01767v1))?
>
>
>
"Elementary" means no use of nontrivial commutative algebra for me (I'm actually fine with Schur functors, although I have a hunch that they too can be avoided).
I'm putting the invariant theory tag on this question because the ultimate use of Question 5 is in Francesca Gandini's proof of the analogue of the Noether bound for the exterior algebra, which I have asked about in [Noether's bound for anticommutative invariant theory (diff. forms instead of polynomials)?](https://mathoverflow.net/questions/38565/) and am now trying to understand...
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https://mathoverflow.net/users/2530
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Intersecting a ideal generated in degree $\leq a$ with one generated in degree $\leq b$ in a polynomial ring
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Take $I=(a^3,b^3)$ and $J=(ac^2-bd^2)$. Then according to Macaulay2, $I\cap J$ has generators in degrees $7,8,9$, for instance $a^3c^6-b^3d^6$. So the answers to Question 3 and 1 are no.
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2
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https://mathoverflow.net/users/2083
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402871
| 165,297 |
https://mathoverflow.net/questions/402872
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1
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Let $X,Y$ be Banach spaces and let $A:X\rightarrow Y$ be a linear operator. Does it suffice to show that there exists a sequence $x\_n\in X$ such that $\lim\_{n\rightarrow\infty}Ax\_n = 0$ with $||x\_n||=1\quad\forall n\in\mathbb{N}$ to proof non-injectivity of $A$?
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https://mathoverflow.net/users/153356
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Showing non-injectivity
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There are examples of Banach spaces $X,Y$ along with bounded linear mappings $L:X\rightarrow Y$ and sequences $(x\_{n})\_{n}$ of elements in $X$ such that
$^{\lim}\_{n\rightarrow\infty}L(x\_{n})=0$ in the metric space induced by the norm on $Y$ but where $\|x\_{n}\|=1$ for each $n$. For instance, if $X=Y$ and $X$ is a separable Hilbert space with orthonormal basis $(e\_{n})\_{n\geq 1}$, and
$A:X\rightarrow Y$ is the bounded linear operator defined by letting
$L(e\_{n})=e\_{n}/n$ for $n\geq 1$, then $\lim\_{n\rightarrow\infty}L(e\_{n})=0$ in the metric topology induced by the norm, but $\|e\_{n}\|=1$ for each $n$.
The open mapping theorem for Banach spaces states that if $X,Y$ are Banach spaces and $L:X\rightarrow Y$ is a surjective bounded linear mapping, then the mapping $L$ is an open mapping. As a consequence, if $X,Y$ are Banach spaces and $L:X\rightarrow Y$ is a bijective linear continuous mapping, then $L$ is a homeomorphism (i.e. $L^{-1}$ is also continuous). As a consequence, if $L:X\rightarrow Y$ is a continuous linear surjection between Banach spaces with where $\lim\_{n\rightarrow\infty}L(x\_{n})=0$ with respect to metric generated by the norm but where $\|x\_{n}\|=1$ for each $n$, then the mapping $L$ cannot be injective.
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4
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https://mathoverflow.net/users/22277
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402875
| 165,299 |
https://mathoverflow.net/questions/397542
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12
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There is a [construction](https://iopscience.iop.org/article/10.1070/IM1971v005n04ABEH001121/meta) of the algebraic K-theory groups $K\_i(R)$ of a ring $R$ by Volodin. He gave an explicit construction of the plus-construction $BGL(R)^+$ as the quotient of the bar construction $BGL(R)$ by the union $\bigcup\_{n,\sigma} BU\_n(R)^\sigma$ of the classifying space of the group $U\_n(R)$ upper-triangular matrices with ones on the diagonal, conjugated by a permutation matrix $\sigma$.
>
> What are the uses of this construction in modern approaches to algebraic K-theory? What are its advantages and disadvantages?
>
>
>
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https://mathoverflow.net/users/157284
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Uses of Volodin's construction of algebraic K-theory
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It has several uses:
1. Volodin $K$-theory was used by Igusa in the late 1970s/Early 1980s to define $K$-theoretic invariants of families of pseudoisotopies.
2. In the 1980s, the Volodin construction (actually a variant of it) was used by Goodwillie to relate rational relative K-theory to rational relative THH. This was further developed by Dundas and McCarthy in their famous result that stable K-theory is isomorphic to THH. This is the genesis of trace methods in algebraic K-theory.
3. In the early 1990s, Suslin and Wodzicki used Volodin's space to obtain excision results for algebraic $K$-theory.
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12
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https://mathoverflow.net/users/8032
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402876
| 165,300 |
https://mathoverflow.net/questions/402821
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7
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Let $M$ be a submanifold of a symmetric space $Q$. The normal bundle $NM$ is called *abelian* if $\exp(N\_{p}M)$ is contained in some totally geodesic and flat submanifold of $Q$ for all $p \in M$; see Terng & Thorbergsson, "Submanifold geometry in symmetric spaces", J. Differential Geom. 42 (1995), 665–718.
It is clear that every codimension-one submanifold (i.e., hypersurface) of $Q$ has abelian normal bundle.
I am interested in the case where $Q$ is a Lie group $G$ equipped with a bi-invariant metric.
**Questions**: Are submanifolds with abelian normal bundle (and codimension greater than one) plentiful or rare in $G$? In particular, do two-dimensional examples exist?
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https://mathoverflow.net/users/74033
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Submanifolds of Lie groups with abelian normal bundle
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One class of examples in a compact, connected Lie group $G$ of rank $r$ are the conjugacy classes of codimension $r$. Taking an element $a\in G$ whose centralizer is a maximal torus (a generic condition), the conjugacy class of $a$ is a submanifold $C\_a\subset G$ of codimension $r$ whose normal plane at $a$ is the tangent at $a$ to $Z\_a$, the centralizer of $a$ in $G$, which is a flat, totally geodesic submanifold. By the $G$-homogeneity of $C\_a = G/Z\_a\subset G$, the fact that this holds at $a$ implies that it holds at all point of $C\_a$.
This gives an $r$-parameter family of mutually noncogruent examples.
There are other examples: When $G=\mathrm{SU}(n)$, and $a = \mathrm{diag}(\lambda\_1,\lambda\_2,\ldots,\lambda\_n)$ is a diagonal element for which the $\lambda\_i^2$ are all distinct, the submanifold
$$
M\_a = \{\ g\_1 a g\_2\ |\ g\_1,g\_2\in\mathrm{SO}(n)\ \}\subset\mathrm{SU}(n)
$$
is homogeneous under the isometry group of $\mathrm{SU}(n)$ (endowed with its biïnvariant metric), and its tangent plane at $a$ is orthogonal to the diagonal maximal torus $T\subset\mathrm{SU}(n)$, so, by homogeneity, its tangent plane at any point is orthogonal to a flat, totally geodesic submanifold of $\mathrm{SU}(n)$. Hence it has an abelian normal bundle.
This gives another $r=n{-}1$ parameter family of mutually noncongruent examples of codimension $r$, distinct from the conjugacy classes.
Using the methods of exterior differential systems, one can show that, when $n=3$, these two families account for all of the codimension $2$ submanifolds of $\mathrm{SU}(3)$ with abelian normal bundle, in the sense that any connected submanifold $M^6\subset\mathrm{SU}(3)$ with abelian normal bundle is, up to ambient isometry, an open subset of one of the examples listed above. The argument that I have written out is a calculation using exterior differential systems and the moving frame, but, when I have time, I can sketch the proof, if there is interest.
**Addendum 1:** I had a flight with some time to look at the other two compact simple rank 2 groups. It turns out that all of the connected codimension two submanifolds with abelian normal bundle in $\mathrm{SO}(5)$ and $\mathrm{G}\_2$ are (open subsets of) homogeneous compact ones as well. In each case, there is one additional $2$-parameter family of examples beyond the principal conjugacy classes.
**Addendum 2:** I also checked the rank $r=3$ case $G = \mathrm{SU}(4)$, and found that every connected codimension $3$ submanifold of $G$ with abelian normal bundle is an open subset of one of the two types of homogeneous examples listed above. Since any codimension $2$ submanifold of $\mathrm{SU}(4)$ that is foliated by codimension $3$ submanifolds of $\mathrm{SU}(4)$ with abelian normal bundle will have abelian normal bundle, it follows that there are many non-homogeneous codimension $2$ submanifolds of $\mathrm{SU}(4)$ with abelian normal bundle.
In fact, it now seems likely that, for any compact simple group $G$ of rank $r>1$, there are two $r$-parameter families of homogeneous codimension $r$ submanifolds with abelian normal bundle and every connected codimension $r$ submanifold with abelian normal bundle is, up to ambient isometry, an open subset of a homogeneous one. The two families are as follows: The first family is the family of principal conjugacy classes in $G$, and the second family is the set of principal orbits of $K\times K$ acting by left and right multiplication in $G$ where $G/K$ is a symmetric space of rank $r$. For example, when $G=\mathrm{SU}(n)$ $(n\ge3)$, $K=\mathrm{SO}(n)$; when $G = \mathrm{SO}(n)$ $(n\ge 5)$, $K = \mathrm{SO}(p)\times\mathrm{SO}(n{-}p)$ where $p = \bigl[\tfrac12 n\bigr]$; when $G=\mathrm{Sp}(n)$ $(n\ge3)$, $K=\mathrm{U}(n)$; when $G=\mathrm{G}\_2$, $K=\mathrm{SO}(4)$; when $G=\mathrm{F}\_4$, $K=\mathrm{Sp}(3)\mathrm{SU}(2)$; when $G=\mathrm{E}\_6$, $K=\mathrm{Sp}(4)$; when $G=\mathrm{E}\_7$, $K=\mathrm{SU}(8)$, and when $G=\mathrm{E}\_8$, $K=\mathrm{SO}'(16)$.
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6
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https://mathoverflow.net/users/13972
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402877
| 165,301 |
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