Recent MathOverflow Questions

Matrix eigenvalues inequality (2)

Math Overflow Recent Questions - Fri, 02/01/2019 - 19:39

Suppose that $A$ is a $n\times n$ positive matrix, whose eigenvalues are $a_1\ge a_2\ldots \ge a_n>0;$ $B$ is a $m \times m$ positive matrix, whose eigenvalues are $b_1\ge b_2\ldots \ge b_m>0;$ $C$ is a $p \times p$ positive matrix, whose eigenvalues are $c_1\ge c_2\ldots \ge c_p>0.$ For $n\times p$ full rank matrix $X$ with $n\ge p,$ a $n\times m$ full rank matrix $Y$ with $n\ge m,$, and $m\ge p,$

Question: How to prove it? $$det\Big(X'(A+YBY')^{-1}X+C\Big)\ge l(X,Y)\prod_{i=1}^p\Big(\frac{1}{a_i+b_{i}}+c_{p-i+1}\Big),$$ where $l(X,Y)$ is a positive constant that only depends on $X,Y.$

Can a bijection between function spaces be continuous if the space's domains are different?

Math Overflow Recent Questions - Fri, 02/01/2019 - 18:46

It is well-known that any bijection $\mathbb{R} \rightarrow \mathbb{R}^2$ cannot be continuous. But suppose we have the two spaces $A = \{f(x):\mathbb{R^2}\rightarrow \mathbb{R} \}$ and $B = \{f(x):\mathbb{R}\rightarrow \mathbb{R} \}$ and and a map between them $A \xrightarrow{\phi} B$. (perhaps adding regularity conditions on the functions in $A,B$ so they're Hilbert or Banach spaces and continuity can be defined for $\phi$). The question is, can $\phi$ be both continuous and bijective?

This question is inspired by trying to invert the Radon transform for tensor fields, as in https://arxiv.org/pdf/1311.6167.pdf, but can be formulated outside of this context.

EDIT: The conditions I want on $A,B$ should be function spaces of continuously differentiable functions, or even stronger like analytic functions, equipped with some $L^p$ norm. These regularity conditions are to model something like the Radon or tensor transform in the sense ensuring that small perturbations to the data lead to small perturbations in the reconstructions.

Why is $\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $?

Math Overflow Recent Questions - Fri, 02/01/2019 - 11:07

This question is an old question from mathstackexchange.

Let $f_- (n) = \Pi_{i=0}^n ( \sin(i) - \frac{5}{4}) $

And let

$ f_+(m) = \Pi_{i=0}^m ( \sin(i) + \frac{5}{4} ) $

It appears that

$$\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $$

Why is that so ?

Notice

$$\int_0^{2 \pi} \ln(\sin(x) + \frac{5}{4}) dx = Re \int_0^{2 \pi} \ln (\sin(x) - \frac{5}{4}) dx = \int_0^{2 \pi} \ln (\cos(x) + \frac{5}{4}) dx = Re \int_0^{2 \pi} \ln(\cos(x) - \frac{5}{4}) dx = 0 $$

$$ \int_0^{2 \pi} \ln (\sin(x) - \frac{5}{4}) dx = \int_0^{2 \pi} \ln (\cos(x) - \frac{5}{4}) dx = 2 \pi^2 i $$

That explains the finite values of $\sup $ and $ \inf $.. well almost. It can be proven that both are finite. But that does not explain the value of their product.

Update

This is probably not helpful at all , but it can be shown ( not easy ) that there exist a unique pair of functions $g_-(x) , g_+(x) $ , both entire and with period $2 \pi $ such that

$$ g_-(n) = f_-(n) , g_+(m) = f_+(m) $$

However i have no closed form for any of those ...

As for the numerical test i got about $ln(u) (2 \pi)^{-1}$ correct digits , where $u = m + n$ and the ratio $m/n$ is close to $1$.

Assuming no round-off errors i ended Up with $1.2499999999(?) $. That was enough to convince me.

I often get accused of " no context " or " no effort " but i have NOO idea how to even start here. I considered telescoping but failed and assumed it is not related. Since I also have no closed form for the product I AM STUCK.

I get upset when people assume this is homework. It clearly is not imho ! What kind of teacher or book contains this ?

——-

Example : Taking $m = n = 8000 $ we get

$$ max(f_-(1),f_-(2),...,f_-(8000)) = 1,308587092.. $$ $$ min(f_+(1),f_+(2),...,f_+(8000)) = 0,955226916.. $$

$$ 1.308587092.. X 0.955226916.. = 1.249997612208568.. $$

Supporting the claim.

I'm not sure if $sup f_+ = 7,93.. $ or the average of $f_+ $ ( $ 3,57..$ ) relate to the above $1,308.. $ and $0,955..$ or the truth of the claimed value $5/4$.

In principle we could write the values $1,308..$ and $0,955..$ as complicated integrals. By using the continuum product functions $f_-(v),f_+(w)$ where $v,w$ are positive reals.

This is by noticing $ \sum^t \sum_i a_i \exp(t \space i) = \sum_i a_i ( \exp((t+1)i) - 1)(\exp(i) - 1)^{-1} $ and noticing the functions $f_+,f_-$ are periodic with $2 \pi$.

Next with contour integration you can find min and max over that period $2 \pi$ for the continuum product functions.

Then the product of those 2 integrals should give you $\frac{5}{4}$.

—-

Maybe all of this is unnecessarily complicated and some simple theorems from trigonometry or calculus could easily explain the conjectured value $\frac{5}{4}$ .. but I do not see it.

——

—— Update This conjecture is part of a more general phenomenon.

For example the second conjecture :

Let $g(n) = \prod_{i=0}^n (\sin^2(i) + \frac{9}{16} ) $

$$ \sup g(n) \space \inf g(n) = \frac{9}{16} $$

It feels like this second conjecture could somehow follow from the first conjecture since

$$-(\cos(n) + \frac{5}{4})(\cos(n) - \frac{5}{4}) = - \cos^2(n) + \frac{25}{16} = \sin^2(n) + \frac{9}{16} $$

And perhaps the first conjecture could also follow from this second one ?

Since these are additional questions and I can only accept one answer , I started a new thread with these additional questions :

https://math.stackexchange.com/questions/3000441/why-is-inf-g-sup-g-frac916

random Young Tower

Math Overflow Recent Questions - Fri, 02/01/2019 - 05:55

the following setting is the same as Baladi's paper "ALMOST SURE RATES OF MIXING FOR I.I.D. UNIMODAL MAPS" (2002), define random Young tower ($\Delta_{\omega})_{\omega\in \Omega}$ sharing the same base $(\Lambda, Leb)$:

For each sample $\omega \in \Omega$ with $\sigma: \Omega \to \Omega $, we have dynamic $F_{\omega}: \Delta_{\omega} \to \Delta_{\sigma \omega}$, return time $R_{\omega}$ on $\Lambda$, separation time $s_{\omega}$ on $\Lambda \times \Lambda$, partition of $\Lambda=\bigcup_{j} \Lambda_j(\omega)$, s.t. $F_{\omega}^{R_{\omega}}(\Lambda_j(\omega))=\Lambda$ with uniform distortion:

$\log \frac{JF_{\omega}^{R_{\omega}}(x)}{JF_{\omega}^{R_{\omega}}(y)} \le C \cdot \beta^{s_{\omega}(F_{\omega}^{R_{\omega}}(x),F_{\omega}^{R_{\omega}}(y))}$ where $C, \beta<1$ are constant, $x,y \in \Lambda_j(\omega) $.

Moreover, for convenience, we assume $Leb(R_{\omega}=1)>0$(mixing condition), and assume existence of quasi-invariant measure on $\Delta_{\omega}$: $(\mu_{\omega})_{\omega \in \Omega}$ s.t. $(F_{\omega})_{*} \mu_{\omega}= \mu_{\sigma \omega} $.

My question is

if we have uniform tail estimate: $Leb(R_{\omega}>n) \le C \cdot \frac{1}{n^{\alpha+1}}$ where C is constant does not depend on $\omega$,

do we have uniform decay of correlation:

$|\int \phi \circ F^n \psi d \mu_{\omega}-\int \phi d \mu_{\sigma^n \omega} \int \psi d\mu_{\omega} | \le C \cdot \frac{1}{n^{\alpha}} $ where $\phi, \psi $ have usual regularity, $C$ independent of $\omega$?

Rademacher theorem

Math Overflow Recent Questions - Thu, 01/31/2019 - 17:04

If $f:\mathbb{R}^n\to\mathbb{R}^m$ is of class $C^1$ and $\operatorname{rank} Df(x_o)=k$, then clearly $\operatorname{rank} Df\geq k$ in a neighborhood of $x_o$. It is not particularly difficult to prove the following counterpart of this result for Lipschitz mappings (very nice exercise). Recall that by the Rademacher theorem Lipschitz mappings are differentiable a.e.

Theorem. If $f:\mathbb{R}^n\to\mathbb{R}^m$ is Lipschitz, differentiable at $x_o$ and $\operatorname{rank} Df(x_o)=k$, then in any neighborhood of $x_o$ the set of points satisfying $\operatorname{rank} Df\geq k$ has positive measure.

This result is so natural that I am sure it has been observed before.

Question. Do you know a reference for this result?

I might use this result in my research, and I would prefer to quote it rather than prove it.

A question on the Faulhaber's formula

Math Overflow Recent Questions - Thu, 01/31/2019 - 02:53

Consider the Faulhaber's formula Knuth, page 9: \begin{equation}\label{knuth1} (2.1) \quad \quad n^{2m-1}=\begin{cases} n^1 = \binom{n}{1}, \ &\mathrm{if} \ m=1;\\ n^3 = 6\binom{n+1}{3}+\binom{n}{1}, \ &\mathrm{if} \ m=2;\\ n^5 = 120\binom{n+2}{5}+30\binom{n+1}{3}+\binom{n}{1}, \ &\mathrm{if} \ m=3;\\ \vdots\\ n^{2m-1} = \sum\limits_{1\leq k\leq m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1}, \ &\mathrm{if} \ m\in\mathbb{N}; \end{cases} \end{equation} The coefficients $(2k-1)!T(2m,2k)$ in the Faulhaber's identities (2.1) can be calculated using following formula \begin{equation} (2.2) \quad \quad (2k-1)!T(2n,2k)=\frac{1}{r}\sum_{j=0}^{r}(-1)^j\binom{2r}{j}(r-j)^{2n}, \end{equation} where $r=n-k+1$ and $T(2n,2k)$ is central factorial number, see A303675. Consider the partial case of (2.1) for $m=2$, for instance \begin{equation} (2.3) \quad \quad n^3=\sum_{k=1}^{n} 6\binom{k}{2}+1=\sum_{k=1}^{n}6k(n-k)+1 \end{equation} The l.h.s of identity (2.3) is reached by means of Hockey Stick pattern in terms of Binomial coefficients \begin{equation} \sum_{k=s}^{n}\binom{k}{s}=\binom{n+1}{s+1} \end{equation} It follows from the identity (2.3) that \begin{equation} \sum_{k=1}^{n}\binom{k}{2}=\sum_{k=1}^{n}k(n-k) \end{equation} Now, lets keep our attention to the identity (2.3) again. As we said, this is the partial case of Faulhaber's formula for $m=2$, and we can notice that the following structure of (2.3) can be assumed \begin{equation*} n^{3}=\sum_{k=1}^{n}\sum_{j=0}^{m=1} A_{m,j}k\strut^j(n-k)\strut^j \end{equation*} and, consequently, for each non-negative integer $m$ \begin{equation*} (2.4)\quad \quad n^{2m+1}=\sum_{k=1}^{n}\sum_{j=0}^{m} A_{m,j}k\strut^j(n-k)\strut^j \end{equation*} The coefficients $A_{m,j}$ in (2.4) are terms of sequences: A302971 - numerators; A304042 - denominators. More info concerning the identity $n^{2m+1}=\sum_{k=1}^{n}\sum_{j=0}^{m} A_{m,j}k\strut^j(n-k)\strut^j$ can be found at this link.

Now we get a structure, which seems to be direct consequence of the Faulhaber's formula, with only one difference, the corresponding binomial coefficients are replaced by the polynomials $k^j(n-k)^j$. So, for the moment, the task is to represent the binomial coefficients in terms of polynomial $k^j(n-k)^j$. Let's introduce the function \begin{equation} (2.5)\quad \quad F_s(n,k)=k\binom{n-k}{s-1}=k\sum_{j=0}^{n-k-1}\binom{j}{s-2} \end{equation} For instance, \begin{equation} \begin{split} s=1: \ F_1(n,k) &=0\\ s=2: \ F_2(n,k) &=k(n-k) \\ s=3: \ F_3(n,k) &=k(n-k)(n-k-1)/2\\ s=4: \ F_4(n,k) &=k(n-k)(n-k-1)(n-k-2)/6 \end{split} \end{equation} And most importantly \begin{equation} (\star)\quad \quad \sum_{k=0}^{n}\binom{k}{s}=\sum_{k=0}^{n}F_s(n,k) \end{equation} As we already have the run-algorithm for $m=2$ in Faulhaber's formula that is $n^3$, lets take another example, for $m=3$ and corresponding result $n^5$, we will apply recently received results (1.6) and (1.7). By Faulhaber's formula, the fifth power is \begin{equation} (2.6)\quad \quad n^5 = 120\tbinom{n+2}{5}+30\tbinom{n+1}{3}+\tbinom{n}{1} \end{equation} Our main aim is to compile expression (2.6) to the form $n^{5}=\sum_{k=1}^{n}\sum_{j=0}^{2} A_{2,j}k\strut^j(n-k)\strut^j$. By the identity $(\star)$ and definition (2.5) we have \begin{equation} \begin{split} &\sum_{k=1}^{n}\tbinom{k+1}{4}=\sum_{r=1}^{n}F_4(n,k+1)=\sum_{r=1}^{n}-(1/6) k (-1 + k - n) (k - n) (1 + k - n)\\ &=1/6(kn^3-3k^2n^2+3k^3n-k^4-kn+k^2) \\ &\sum_{k=1}^{n}\tbinom{k}{2}=\sum_{k=1}^{n}F_2(n,k)=\sum_{k=1}^{n}k(n-k)\\ &\sum_{k=1}^{n}\tbinom{k-1}{0}=\tbinom{n}{1}=n \end{split} \end{equation} Let be $a=n-k$ in $F_4(n,k)=1/6k(n-k)(n-k+1)(n-k-1)$, thus \begin{equation} \begin{split} F_4(n,k) &=1/6k(a^3-a)\\ &=1/6k((n-k)^3-(n-k))\\ &=1/6k[n^3-3kn^2+3k^2n-k^3-n+k]\\ &=1/6(kn^3-3k^2n^2+3k^3n-k^4-kn+k^2) \end{split} \end{equation} Let's substitute $F_2(n,k)$, and $F_4(n,k)$ to the equation (2.6), respectively \begin{equation} \begin{split} n^5 &= 120\sum_{k=1}^{n}1/6(kn^3-3k^2n^2+3k^3n-k^4-kn+k^2)+30\sum_{k=1}^{n}kn-k^2+\sum_{k=1}^{n}1\\ &= 20\sum_{k=1}^{n}(kn^3-3k^2n^2+3k^3n-k^4-kn+k^2)+30\sum_{k=1}^{n}kn-k^2+\sum_{k=1}^{n}1\\ &= 20\sum_{k=1}^{n}(kn^3-3k^2n^2+3k^3n-k^4)-20\sum_{k=1}^{n}kn+k^2+30\sum_{k=1}^{n}kn-k^2+\sum_{k=1}^{n}1\\ &= 20\sum_{k=1}^{n}k(n-k)^3+10\sum_{k=1}^{n}k(n-k)+\sum_{k=1}^{n}1\\ &= \sum_{k=1}^{n}20k(n-k)^3+10k(n-k)+1 \end{split} \end{equation} As we can see, the form of the our transformation is different from $n^{5}=\sum_{k=1}^{n}\sum_{j=0}^{2} A_{2,j}k\strut^j(n-k)\strut^j$, but still is near to the result $n^{5}=\sum_{k=1}^{n}30k^2(n-k)^2+1$. We can see that if the variable of polynomial at $n^5=\sum_{k=1}^{n}20k(n-k)^3+10k(n-k)+1$ would be $k^2(n-k)^2$ then we get desired result $n^{5}=\sum_{k=1}^{n}30k^2(n-k)^2+1$.

The problem: Revise the function $F_s(n,k)$ such way, that the construction $n^{5}=\sum_{k=1}^{n}\sum_{j=0}^{m} A_{m,j}k\strut^j(n-k)\strut^j$ directly follows when corresponding binomial coefficient sums in Faulhaber's formula (2.1) is replaced by sums of $F_s(n,k)$.

PS For those who interested, we collected a few questions concenring, everybody are invited to participate, visit https://kolosovpetro.github.io/math_project/

Reference Request: simple graph vertex labelings with balanced induced edge weights

Math Overflow Recent Questions - Wed, 01/30/2019 - 20:43

Say that the edge weight induced by a vertex labeling is the sum of the weights of the two vertices comprising it. Here is the problem of interest: given a simple $d$-regular graph $G = (V,E)$, find a (bijective) vertex labeling $\ell: V \to \{1, \dots, |V|\}$ which minimizes the difference-sum; that is, the difference between the weight of the highest-weighted edge and the weight of the lowest-weighted edge. That is:

difference-sum$(\ell)$ = max-sum$(\ell)$ $-$ min-sum$(\ell)$, where

max-sum$(\ell) = \max_{\{v_i,v_j\} \in E} \{\ell(v_i) + \ell(v_j) \}$

min-sum$(\ell) = \min_{\{v_i,v_j\} \in E} \{\ell(v_i) + \ell(v_j) \}$

I have been unable to find any literature which addresses this problem, so I am hoping that someone can point me to some papers that address this matter.

Is there a notion of a connection for which the horizontal lift of a curve depends on its orientation?

Math Overflow Recent Questions - Wed, 01/30/2019 - 17:04

Given a fiber bundle $\pi:E\to M$, a curve $\gamma:[0,1]\to M$, and a point $p \in \pi^{-1}(\gamma(0))$, a connection on the bundle allows us to uniquely lift $\gamma$ to a horizontal curve in E through $p$. In almost all situations I have encountered, the horizontal lift does not depend on the orientation of $\gamma$. To be precise, the two curves $t\to \gamma(t)$ and $t\to\gamma(1-t)$ have the same horizontal lift through $p$.

I have a fiber bundle for which I would like to have a type of parallel transport which depends on which direction one is moving in the base. So my question is: what is the best way to formulate a connection which is orientation dependent, and so enables this type of parallel transport?

Extending to non-monotonic functions

Math Overflow Recent Questions - Wed, 01/30/2019 - 16:44

I wish to enquire about examples of situations similar to what happened to me some time ago.

As I was trying to characterise all real-valued functions which preserve a certain geometric inequality (in an attempt to generalise a fun result I came across in a paper), I realised that it was simple to characterise all monotonic functions that satisfy the requirement, and that there existed interesting (i.e. non-pathological) non-monotonic examples.

However, generalising the characterisation to non-monotonic functions has eluded me.

Are you aware of interesting examples of situations where generalising characterisation to non-monotonic functions appears to be several orders of magnitude more elusive than 'monotonic characterisation'?

The complex topological $K$-theory spectrum is not an $H\mathbb{Z}$-module

Math Overflow Recent Questions - Wed, 01/30/2019 - 15:42

Why the nontriviality of the first $k$-invariant of $ku$ implies that $H\mathbb{Z}$ is not a $H\mathbb{Z}$-module.?

Hartog's Number of the Reals and $\Theta$ without choice

Math Overflow Recent Questions - Wed, 01/30/2019 - 14:17

There are two important numbers that in some meaningful sense describe "how well-orderable" the reals are:

  1. Hartog's Number $H(\Bbb R)$, the least ordinal/well-ordered cardinal that doesn't inject into $\Bbb R$
  2. The ordinal $\Theta$, the least ordinal/well-ordered cardinal that $\Bbb R$ doesn't surject onto

The first number $H(\Bbb R)$ can be thought of as describing the supremum of the cardinalities of all well-orderable subsets of $\Bbb R$, whereas the second number $\Theta$ can be thought of as describing the supremum of the cardinalities of all well-orderable equivalence classes of $\Bbb R$.

With choice, these are all equal to $\mathfrak c^+$, the cardinal after $\mathfrak c$.

My question: without choice, do we have any results regarding:

  1. How large each of these numbers can be?
  2. How small each of these numbers can be?
  3. Which number is larger or if they can be equal?

I know that $AD$ determines some of these strongly enough to relate them to large cardinals (I believe Woodin cardinals), but I'm also interested in what the possibilities are without something that strong.

Enriched cartesian closed categories

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:56

Let $V$ be a complete and cocomplete cartesian closed category. Feel free to assume more about $V$ if necessary; in my application $V$ is simplicial sets, so it is a presheaf topos and hence has all sorts of nice properties you might want (except that its underlying-set functor $V(1,-) : V \to \rm Set$ is not faithful or conservative).

Let $C$ be a complete and cocomplete $V$-enriched category with powers and copowers (a.k.a. cotensors and tensors), hence all $V$-enriched weighted limits. And suppose that the underlying ordinary category $C_0$ is cartesian closed, i.e. we have natural isomorphisms of hom-sets

$$ C_0(X\times Y, Z) \cong C_0(X, Z^Y). $$

Is it necessarily the case that $C$ is also cartesian closed in the $V$-enriched sense, i.e. we have $V$-natural isomorphisms of $V$-valued hom-objects

$$ C(X\times Y, Z) \cong C(X, Z^Y)? $$

I can't decide whether I think this is likely to be true or not. I used to believe that it was true, and I believe I have implicitly used it (or more precisely its generalization to the locally cartesian closed case, which should follow from the simple one applied to enriched slice categories) in a published paper or two. However, right now I can't see how to prove it, and when stated clearly it sounds unlikely, since usually it is an extra condition on an adjunction between enriched categories to be an enriched adjunction (even when one of the functors is known to be an enriched functor, as is the case here for the cartesian product). On the other hand, it is true when $C=V$ itself: a cartesian closed category, defined by an isomorphism of hom-sets, automatically enhances to an isomorphism of hom-objects $Z^{X\times Y} \cong (Z^X)^Y$.

Does there exist a curve which avoids a given countable union of small subsets?

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:53

Let $X$ be a projective variety over $\mathbb{C}$. Let $X_1, X_2, \ldots$ be proper closed subsets. Then $\cup_i X_i \neq X(\mathbb{C})$. However, I am interested in a stronger statement.

Assume that, for all $i$, we have that $\mathrm{codim}(X_i)\geq 2$.

Then, does there exist a smooth projective curve $C\subset X$ such that $C$ and $\cup_i X_i$ are disjoint?

The condition on the codimension is clearly necessary. (Take $X_1$ to be an ample divisor to see this.)

Rewriting integral via higher derivatives

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:34

Let $f $ be an infinitely smooth function on $(0,\infty)$. Suppose further that $f $ and all of its derivatives are bounded. I have the following integral $$b\int_0^1f (a+bx)\; dx, $$ where $ a$ and $ b$ are real numbers. The number $b $ is of interest and I seek bounded, smooth function $g $ such that $$b\int_0^1f (a+bx)\; dx = b^2\int_0^1 f^{(\ell)}(a+bx)g (x) \; dx,$$ where $f^{(\ell)}$ is some derivative of $f$.

For example, I have used integration by parts and a linear $g (x)= x-1$ on the quantity $\int f(a+bx)g(x)$ to get $$b\int_0^1f (a+bx)\; dx =-bf(a)+ b^2\int_0^1 f'(a+bx)g (x) \; dx.$$

But I would like to have the intgral by itself. I haven't made an attempt for higher powers. I thought I'd ask here to see if anyone knows are results of this type, or can give me some pointers. Thanks

Objects with trivial automorphism group

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:31

Suppose I am working with a category of objects such that each object $X$ in the category has as a trivial automorphism group. An example of such an object would be genus zero curves with $n$-punctures, i.e., $(\mathbb{P}_k^1, (s_i)_{i=1}^n \in k )$.

Let $F(S)/{\sim}$ represent families of these objects over an affine scheme $S=\operatorname{Spec} A$ where $A$ is a $k$-algebra and $k$ an algebraically closed field. As an example, consider families of genus zero curves with $n$-punctures over $S$ and fibers $X= X_s \cong (\mathbb{P}_k^1, (s_i: k \to \mathbb{P}_k^1)_{i=1}^n) $. Then $H^1(X, TX)=0$ and $Aut(X)=0$.

Motivation for question: It seems that since both $H^1(X, TX)=0$ and $Aut(X)=0$ we can conclude that every family of genus zero curves with $n$-punctures is trivial, i.e any family over $S$ is isomorphic to the trivial family $(\mathbb{P}_A^1, (s_i: A \to \mathbb{P}_A^1)_{i=1}^n)$.

Question: Now going back to the beginning, suppose my family of objects over $S$ is such that all fibers are isomorphic to some objet $X$, $H^1(X, TX)=0$ and $Aut(X)=0$, does this imply that all families are trivial?

Or even more generally, suppose $H^1(X, TX)=n$ and $Aut(X)=0$, does this imply that there exists $n$ distinct families $F(S)/{\sim}$?

I know there are probably many conditions on $X$ I should include, however, just assume that my object $X$ has any property that makes it "nice" to work with.

When is it okay to intersect infinite families of proper classes?

Math Overflow Recent Questions - Wed, 01/30/2019 - 13:14

For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite collections of proper classes all the time.

Here is a simple example. Let a ring $R$ be called $n$-Dedekind-finite if $ab=1\implies ba=1$ for $a,b\in \mathbb{M}_n(R)$ (the $n\times n$ matrix ring over $R$). A ring $R$ is said to be stably finite if it is $n$-Dedekind-finite for every integer $n\geq 1$. The class of stably finite rings is the intersection (over positive integers $n$) of the classes of $n$-Dedekind-finite rings.

So my question is a straightforward one. What principle makes it okay for me to intersect classes in this manner?

Phrased a little differently my question is the following. Suppose we have an $\mathbb{N}$-indexed collection of sentences $S_n$ in the first-order language of rings. Why is it valid (in my day-to-day work) to form the class of rings satisfying $\land_{n\in \mathbb{N}}S_n$, even though the corresponding class doesn't necessarily exist if each $S_n$ is a sentence in the language of ZFC?

[Part of my motivation for this question is the idea that much of "normal mathematics" can be done in ZFC. That's how I've viewed my own work. I'm happy to think of my rings living in $V$, subject to all the constraints of ZFC. (In some cases I might go further, by invoking the existence of universes or the continuum hypothesis, but that is not the norm.) But I'm unsure how to justify my use of infinite Boolean operations on proper classes. Especially since, in some cases, my conditions on the rings are conditions on sets, from the language of ZFC!]

Is this consequence of the invariant subspace problem known?

Math Overflow Recent Questions - Wed, 01/30/2019 - 12:59

An interesting fact popped out of a paper I'm writing: if the invariant subspace problem for Hilbert space operators has a positive solution, then every $A \in B(\mathcal{H})$ can be made "upper triangular" in the sense that there is a maximal chain of closed subspaces of $\mathcal{H}$, each of which is invariant for $A$.

The proof is quite easy, almost trivial, yet I had never heard of it before. (It isn't mentioned in the answers to this question, for example.) It was a surprise to notice that the ISP has such a strong consequence for the structure of arbitrary operators because one thinks of is as merely the first, most basic question on that topic.

Surely this is known?

Marginal Distribution of Partition Matrix

Math Overflow Recent Questions - Wed, 01/30/2019 - 12:56

Assume that $X\sim IW_{p}(n,I_p)$ has an inverse Wishart distribution, which probability density function is $$f(X\mid n)\propto |X|^{-\frac{n+p+1}{2}}exp\Big(-\frac{1}{2}tr( X^{-1})\Big),~~\qquad (1)$$ Partition the matrices $X$ with $$X=\pmatrix{X_{11}&X_{12}\\X_{21}&X_{22}},$$ where $X_{ij}$ is $p_i\times p_j$ matrices. From the definition of IW, we know $X_{11}\sim IW_{p_1}(n-p_2,I_{p_1}).$

Qustion: If random $p\times p$ positive matrice $Y$ satisfies $$f(Y\mid n)\propto |Y|^{-\frac{n+p+1}{2}}exp\Big(-\frac{1}{2}tr( Y^{-1})\Big)\Big[\prod_{i=1}^p(\lambda_i-\lambda_j)\Big]^{-1},~~\qquad (2)$$ where $\lambda_1>\lambda_2>\cdots>\lambda_p>0$ are the ordered eigenvalues.

Partition the matrices $Y$ with $$Y=\pmatrix{Y_{11}&Y_{12}\\Y_{21}&Y_{22}}.$$ What is the marginal distribution of $Y_{11}???$

Help with signal analysis question [on hold]

Math Overflow Recent Questions - Wed, 01/30/2019 - 12:44

I am having trouble understanding the question and would like some help please,

Q) determine the number of terms required such that the ratio of the power of the non-zero nth harmonic to the power of the fundamental is < 1%.

Answer is 11

but how do you do it I have no idea.

see attached pics of the question also. square wave form

amplitude spectrum

"Almost planar graphs"

Math Overflow Recent Questions - Wed, 01/30/2019 - 12:39

A graph $G$ is "Almost Planer" if it has a subgraph $H$ and an embedding in the plane such that $H$ is planar in this embedding and all the crossing edges are cords of faces of $H$ (all crossings and crossing edges are contained inside the faces of $H$.)
Is there any literatures on this class of graphs? Are they classified any other name? What kind of graphs are in this class? Are there graphs which are not Almost Planar?

Any information on this will be appreciated.

Pages

Subscribe to curious little things aggregator