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most recent 30 from mathoverflow.net 2019-05-17T22:14:05Z

Representing simplicial homotopy classes cubically?

Sat, 02/02/2019 - 10:08

Let $(X,x_0)$ be a pointed simplicial set. Assume if you like that $X$ is the nerve of a category but do not assume that $X$ is a Kan complex.

Because $Ex^\infty X$ is a Kan complex, every homotopy class $\alpha \in \pi_n(X,x_0)$ may be represented by a map $sd^k \Delta[n] \to X$ such that the restriction $sd^k \partial \Delta[n] \to X$ is constant at $x_0$. I'm wondering about different "normal forms" for homotopy classes.

For instance, consider subidivided cubes. In dimension 2, I think they should look like this:

$\require{AMScd} D^2_0 = \begin{CD} \bullet \end{CD} \\ D^2_1 = \begin{CD} \bullet @>>> \bullet @<<< \bullet \\ @VVV @VVV @VVV \\ \bullet @>>> \bullet @<<< \bullet \\ @AAA @AAA @AAA \\ \bullet @>>> \bullet @<<< \bullet \\ \end{CD} \\ D^2_2 = \begin{CD} \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet\\ @VVV @VVV @VVV @VVV @VVV \\ \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet\\ @AAA @AAA @AAA @AAA @AAA \\ \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet\\ @VVV @VVV @VVV @VVV @VVV \\ \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet\\ @AAA @AAA @AAA @AAA @AAA \\ \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet \end{CD} \\ D^2_3 = \dots $

Questions:

  1. Can every $\alpha \in \pi_n(X, x_0)$ be represented by a map $D^n_k \to X$ sending the boundary to the constant at $x_0$?
  2. If not, is there a better definition of subdivided cubes for which the answer to (1) becomes "yes"?

It's nice that with the above definition, $D^n_{k+1}$ can be obtained by gluing together a bunch of copies of $D^n_k$ in an easy way. But perhaps this is too good to be true.

A trivial observation is that for $n=1$ the above cubical subdivision is basically the same as barycentric subdivision and the answer to (1) comes out as "yes".

Upper and lower bounds on the number of certain subsets of the power set

Sat, 02/02/2019 - 09:55

Let $A$ be a set with $n$ elements. Call a subset $C$ of the power set of $A$ "good" if

  • Each element of $C$ has at least three elements.

  • If $P, Q\in C$ and $P\cap Q$ has more than one element, then $P=Q$.

I've been interested in finding good upper and lower bounds of the number of good collections, but I haven't made any headway. Does anyone know of any?

Proving two inequalities involving the gamma and digamma functions

Sat, 02/02/2019 - 09:02

I'm having trouble proving the following inequality:

$$\forall p>1 \quad \forall m\geq 0 \quad \dfrac{m^2\Gamma(\dfrac{2m}{p})\Gamma(\dfrac{2m}{q})}{\Gamma(\dfrac{2m+2}{p})\Gamma(\dfrac{2m+2}{q})}\geq\dfrac{1}{4}p^2(p-1)^{\frac{2}{p}-2},$$ where as usual $q=\dfrac{p}{p-1}$. In fact, it seems clear from Mathematica that for a fixed $p$, the LHS is a decreasing function of $m$ (strictly unless $p=2$, in which case it's constant). The RHS can be seen to be the limit as $m\to \infty$. I actually only care about integer $m\geq 0$, but I don't find that helpful.

I have tried both a direct approach (three known inequalities that are nice enough to apply here, but lead to wrong inequalities) and working with the derivative, which naturally involves instances of the digamma function. Proving that the LHS is decreasing is equivalent to the following inequality: $$\forall p>1 \quad \forall m\geq 0 \quad \dfrac{1}{m}+\dfrac{1}{p}(\psi(\dfrac{2m}{p})-\psi(\dfrac{2m+2}{p}))+\dfrac{1}{q}(\psi(\dfrac{2m}{q})-\psi(\dfrac{2m+2}{q}))\leq0,$$ which again seems to be correct (if you're wondering, the limit as $m\to 0$ is negative for $p\neq2$). Much like before, I tried using two inequalities (for the digamma function), as well as the series representation. They seemed promising at first, but the inequalities gave me positive upper bounds, while the series converges too slowly to be useful (I suspect that any partial sum is positive for large enough $m$).

Any advice would be much appreciated. I'll be glad to explain more about the inequalities I've tried if requested.

Is there a natural transformation from Lists to Domains of Lists

Sat, 02/02/2019 - 08:47

The list domain, $(L, \mu_L, \eta_L)$, on $Set$, takes a set to its set of lists, with $\mu_L : L \cdot L \rightarrow L$ being concatenation of lists. Given a set of lists, there is a natural way to define a domain (dcpo) of lists, by partial initial fragment $l_1 \le l_2 $ if $ l_2 = Concat(l_1, l^{′}_{2})$. I am wondering if this means there is a functor $D$ that takes a set of lists on a set $X$ and gives back the set of domains on that set of lists. Further, is there a natural transformation from $L$ to $D$. Further, is this a monad map, making a monad $(D, \mu_D, \eta_D)$.

I submitted this over at math stack, but got no response.

symplectic sum of two copies of $Bl_{p}(\mathbb{CP}^{2})$

Sat, 02/02/2019 - 08:35

Let $M^{4}= \mathbb{P}(\mathcal{O} \oplus \mathcal{O}(1))$ and $\omega$ the symplectic form on $M^{4}$ given by the anticanonical polarisation.

Suppose we form a symplectic (Gompf) sum of two copies of $(M^{4},\omega)$ along a fibre of the $\mathbb{P}^{1}$-bundle (where the orientation reversing identification of the normal bundles is the obvious one i.e. we just pick an orientation reversing isomorphism of $\mathbb{R}^{2}$ and let the isomorphism of normal bundles be equal to this pointwise).

Question: Does the resulting symplectic $4$-manifold have a compatible Kahler structure?

Real rank 0 implies stable rank 1 on $C^\ast$-algebras?

Sat, 02/02/2019 - 05:27

A $C^\ast$ algebra has defined stable rank (https://www.univie.ac.at/nuhag-php/bibtex/open_files/2079_Rieffel-StableRank.pdf) and real rank (https://core.ac.uk/download/pdf/82123484.pdf), which are two different non-commutative versions of Lebesgue covering dimension.

In particular an algebra $A$ has

  • Stable rank 1 if and only if every element of $A$ can be approximated by invertibles.
  • Real rank 0 if and only if every self-adjoint element of $A$ can be approximated by self-adjoint invertibles.

It's known that $rr(A)\leq 2sr(A)-1$ (see second paper). Is there any example of a $C^\ast$-algebra with real rank $0$ and stable rank $>1$? If so, is there a simple one?

$P$ is projective if and only if $P\otimes N\cong Hom(Hom(P,R),N)$

Sat, 02/02/2019 - 00:50

Let $R$ be a commutative Noetherian ring and $P$ be finitely generated $R$-module.

How to prove the following.

$P$ is projective if and only if $P\otimes N\cong Hom(Hom(P,R),N)$ for all finitely generated $R$-modules $N$.

Poset dimension and width (Dilworth's theorem)

Fri, 02/01/2019 - 22:57

For a given poset $P$, let $\mathrm{dim}(P)$ denote the least cardinal $\kappa$ such that there exists a $\kappa$-sized collection of linear extensions of $P$, say $\mathcal{L}$, such that $\leq_P = \bigcap_{L\in \mathcal{L}}\leq_L$. Let $\mathrm{width}(P)$ denote the least cardinal $\lambda$ such that every antichain $A\subset P$ has size $<\lambda$.

Note that the definition of width here is slightly different from the usual definition seen in combinatorics textbooks but we do this in the set-theoretic convention to deal with the possibility that there is no antichain with maximum cardinality.

Dilworth (https://www.jstor.org/stable/1969503?seq=1#metadata_info_tab_contents) proved that if $\mathrm{width}(P)$ is finite, then $\mathrm{dim}(P)<\mathrm{width}(P)$. The proof goes through the fact that if the width of $P$ is $k$, then $P$ can be decomposed into a union of $<k$ many chains. This fact is not true for $k\geq \aleph_0$. However, the counter-example (Perles' example https://link.springer.com/article/10.1007%2FBF02759806) has dimension 2 so it satisfies $\mathrm{dim}(P)<\mathrm{width}(P)$. Are there examples violating $\mathrm{dim}(P)<\mathrm{width}(P)$?

EDIT: Suggested by bof, the width is better defined as the supremum of the sizes of antichains. So with the new definition, Dilworth's theorem states $\mathrm{dim}(P)\leq \mathrm{width}(P)$ for $P$ of finite width and my question will be modified to whether this is true in general.

Matrix eigenvalues inequality (2)

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?

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} $?

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

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 $\omega$?

Continuous functions of three variables as superpositions of two variable functions

Thu, 01/31/2019 - 18:14

Could we always locally represent a continuous function $F(x,y,z)$ in the form of $g\left(f(x,y),z\right)$ for suitable continuous functions $f$, $g$ of two variables? I am aware of Vladimir Arnold's work on this problem, but it seems that in that context $F(x,y,z)$ is written as a sum of several expressions of this form. Can one reduce it to just a single superposition $g\left(f(x,y),z\right)$; or does anyone know a counter example?

For a related posts see: Kolmogorov superposition for smooth functions and Kolmogorov-Arnold theorem for (just-)functions

Rademacher theorem

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

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

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?

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

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

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

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.

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