Recent MathOverflow Questions

On the connection between Faulhaber's formula and identity $n^{2m+1}=\sum_{k=0}^{n-1}\sum_{j=0}^m A_{m,j}k^j(n-k)^j$

Math Overflow Recent Questions - Mon, 10/15/2018 - 15:48

This question is part of series of the questions, as follows:

Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}$,

Coefficients $U_m(n,k)$ in the identity $n^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k$.

Consider the Faulhaber's identities (see D. Knuth, "Johann Faulhaber and Sums of Powers", Knuth, p. 9) for odd powers $n^{2m+1}, \ m\in\mathbb{N}_0$ \begin{equation*} (\diamond)\quad\quad\begin{cases} n^1 &= \binom{n}{1}\\ n^3 &= 6\binom{n+1}{3}+\binom{n}{1}\\ n^5 &= 120\binom{n+2}{5}+30\binom{n+1}{3}+\binom{n}{1}\\ &\vdots\\ n^{2m-1} &= \sum\limits_{1\leq k\leq m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1} \end{cases} \end{equation*} Introduce the $r$ to the lower index of binomial coefficient of $(\diamond)$, then \begin{equation*} (\Box)\quad\begin{cases} r=1: \ \Delta n^1 &= \binom{n}{1-r}=\binom{n}{0}\\ r=1: \ \Delta n^3 &= 6\binom{n+1}{3-r}+\binom{n}{1-r}=6\binom{n+1}{2}+\binom{n}{0}\\ r=1: \ \Delta n^5 &= 120\binom{n+2}{5-r}+30\binom{n+1}{3-r}+\binom{n}{1-r}=120\binom{n+2}{4}+30\binom{n+1}{2}+\binom{n}{0}\\ &\vdots\\ r=1: \ \Delta n^{2m-1} &= \sum\limits_{1\leq k\leq m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1-r} \end{cases} \end{equation*} Continuing similarly, for every $r>0$ we get $\Delta^r n^{2m-1}$.

Therefore, by the Faulhaber's identity $\Delta n^{3}=6\binom{n+1}{2}+\binom{n}{0}$, the perfect cube in $T$ is: \begin{equation*} (1.1)\quad\quad T^3=\sum\limits_{k=0}^{T-1}\Delta(T^3)(k)=\sum\limits_{k=0}^{T-1}6\binom{k+1}{2}+\binom{k}{0}. \end{equation*} Let's rewrite the expression (1.1) in extended view, taking to attention the identity $\tbinom{k+1}{1}=1+2+\cdots+(k+1)$, thus \begin{equation*} (1.2)\quad\quad T^3=(1+6\cdot0)+(1+6\cdot0+6\cdot1)+\cdots+(1+6\cdot0+\cdots+6\cdot(T-1)) \end{equation*} Factorising expression (1.2), we get \begin{equation*} (1.3)\quad\quad T^3=T+(T-0)\cdot6\cdot0+(T-1)\cdot6\cdot1+\cdots+(T-(T-1))\cdot6\cdot(T-1) \end{equation*} Applying the sigma notation on the expression (1.3), we have \begin{equation*} (1.4)\quad\quad T^3=T+\sum\limits_{k=0}^{T-1}6k(T-k) = \sum\limits_{k=0}^{T-1}6k(T-k)+1 \end{equation*} Therefore, we have reached the generating function $6k(T-k)+1$ the case for $m=1$ in the question

Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}$.

The problem: Can we find similar to (1.3), (1.4) expressions for powers $2m+1, \ m>1$, only by means of binomial identities, extended form of the sum of finite difference (in sense of Faulhaber, (1.0), $(\Box)$), of corresponding power and its extended form ?

And, If we can, are they equal to $\sum_{k=0}^{n-1}\sum_{j=0}^m A_{m,j}k^j(n-k)^j$ ?

We assume that pattern $n^{2m+1}=\sum_{k=0}^{n-1}\sum_{j=0}^m A_{m,j}k^j(n-k)^j$ is unknown.

In other words, the problem asks to pass through the case $r=1$ in $(\Box)$: $$(\star)\quad \quad \Delta n^{2m+1}=\Sigma_{k=1}^{m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1-r}$$ to (in sense of $(\star)$): $$n^{2m+1}=\sum_{k=0}^{n-1}\Delta n^{2m+1}(k)=\sum_{k=0}^{n-1}\sum_{j=0}^m A_{m,j}k^j(n-k)^j$$.

Please, note that formulae notation in present question may be different in some places from notation in question Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}$. More details on this question, as well as detailed derivations are available at this link.

Classification of smooth algebraic surfaces with a smooth morphism to $\Bbb P^1$

Math Overflow Recent Questions - Mon, 10/15/2018 - 11:29

Let $k$ be an algebraically closed field, it is well known that $\mathbb P^1$ is simply connected, but how about smooth projective surfaces $X$ with a smooth morphism to $\Bbb P^1$?

Except the case $X$ is a product of curves or a projective bundle like Hirzebruch surfaces, could we classify all such $X \rightarrow \Bbb P^1$ ? What about higher dimensional cases?

Motivation: Shafarevich conjecture over function field.

Are there enough meromorphic functions on a compact analytic manifold?

Math Overflow Recent Questions - Mon, 10/15/2018 - 10:34

Let $X$ be a compact complex analytic manifold, $D\subset X$ an irreducible smooth divisor, given as zeroes of a global meromorphic function $f\in {\mathfrak M} (X)$. Are there enough other meromorphic functions defining $D$?

Here is a precise question: can one find, for each $x\in X$, a meromorphic function $g\in {\mathfrak M} (X)$ such that $g$ is defined (i.e., not $\infty$) at $x$ and $D={\mathrm{zeroes}}(g)$?

Explicit proof that $c_0$-module $\ell_\infty$ is not projective

Math Overflow Recent Questions - Mon, 10/15/2018 - 07:42

It is well known in narrow circles that the homological dimension (in the sense of relative Banach homology) of $c_0$-module $\ell_\infty$ is 2. As the corollary, this module is not projective. This proof is rather involved, its main ingridient is a lack of a right inverse for the mapping: $$ \Delta:c_0\;\hat{\otimes}\;c_0\to(c\;\hat{\otimes}\;c_0)\oplus(c_0\;\hat{\otimes}\;\ell_\infty): x\;\hat{\otimes}\;y\mapsto (x\;\hat{\otimes}\;y)\oplus(x\;\hat{\otimes}\;y) $$ in the category of left Banach $c_0$-modules.

I would like to see a more direct proof of non-projectivtity. The standard route would be to show that there is no right inverse $c_0$-morphism for the mapping $\pi:c \;\hat{\otimes}\; \ell_\infty\to \ell_\infty \colon a\; \hat{\otimes}\; x\mapsto a\cdot x$, where $c$ is the Banach space of convergent sequences.

Does anyone have an idea how to prove non-projectivity more or less directly?

Product of arithmetic progressions

Math Overflow Recent Questions - Mon, 10/15/2018 - 02:17

Let $(a_1,a_2\ldots,a_n)$ and $(b_1,b_2,\ldots,b_n)$ be two permutations of arithmetic progressions of natural numbers. For which $n$ is it possible that $(a_1b_1,a_2b_2,\dots,a_nb_n)$ is an arithmetic progression?

The sequence is (trivially) an arithmetic progression when $n=1$ or $2$, and there are examples for $n=3,4,5,6$:

$n=3$: $(11,5,8), (2,3,1)$

$n=4$: $(1,11,6,16), (10,4,13,7)$

$n=5$: $(8,6,4,7,5), (4,9,19,14,24)$

$n=6$: $(7,31,19,13,37,25), (35,11,23,41,17,29)$

Sum of Gaussian matched by Brownian Motion?

Math Overflow Recent Questions - Sun, 10/14/2018 - 09:02

Given independent Gaussian $d$ dimensional vectors $G_i$,

If $\mathbb{E}(\sum_{i \le n} G_i) \cdot (\sum_{i \le n} G_i)^T=n \cdot I_{d \times d} + o(n^{1-\epsilon})$.

there exists Brownian motion $B_t$ s.t. $\sum_{i \le n} G_i=B_n+o(n^{\frac{1}{2}-\epsilon})$?

one dimension is simple, true, I guess it is possible in high dimension, BM should be constructed from Gaussian vectors.

On two congruences modulo peime p

Math Overflow Recent Questions - Sat, 10/13/2018 - 20:16

How to prove the following two congruences?

Question1: Let $p\equiv 1 \pmod 3$ be a prime, then $$\sum_{k=0\atop k\neq(p-1)/3}^{(p-1)/2}\frac{\binom{2k}k}{3k+1}\equiv 0 \pmod p.$$

Question2: For any odd prime $p>3$, we have
$$\sum_{k=1}^{(p-1)/2}\frac{1}{k\binom{2k}k}\sum_{j=1}^k\frac{\binom{2j}j}{j}\equiv \frac{1}3B_{p-2}\left(\frac{1}3\right) \pmod p,$$ where $B_n(x)$ are the Bernoulli polynomials defined by $$\sum_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}=\frac{te^{xt}}{e^t-1}.$$

Is every stable matrix orthogonally similar to a $D$-skew-symmetric matrix?

Math Overflow Recent Questions - Sat, 10/13/2018 - 19:27

Let $\mathrm{diag}(A)$ denote the diagonal matrix with diagonal entries of $A\in\mathbb{R}^{n\times n}$ and let $\succeq$ denote the standard partial order in the cone of (symmetric) positive definite matrices. Let me start by recalling a definition from [1].

Definition ($D$-skew-symmetric matrix). A matrix $A\in\mathbb{R}^{n\times n}$ is said to be $D$-skew-symmetric is there exists a diagonal matrix $D\succ 0$ such that $(A-\mathrm{diag}(A))D$ is skew-symmetric.

Now, let $A\in\mathbb{R}^{n\times n}$ be a matrix with eigenvalues $\{\lambda_i\}_{i=1}^n$ such that $\mathrm{Re}(\lambda_i)<0$.

My Question: Is $A$ orthogonally similar to a $D$-skew symmetric matrix with non-positive diagonal?

My question is motivated by the fact that in the $2\times 2$ case it is possible to prove that the answer is in the affirmative.

Proof for $n=2$. After an orthogonal diagonalization of the symmetric part of $A$, we can assume that $A$ is in the form $$ A=\begin{bmatrix}a_{11} & a_{12} \\ -a_{12} & a_{22} \end{bmatrix}, $$ where $a_{ij}\in\mathbb{R}$, $a_{11}<0$, $a_{11}+a_{22}<0$. Further, since $A$ has eigenvalues with negative real part, $\det(A)=a_{11}a_{22}+a_{12}^{2}>0$. We distinguish two cases:

  1. If $a_{22}\le 0$, then $A+A^{\top}\preceq 0$. Hence, $A$ is $D$-skew-symmetric with negative diagonal for $D=I$.
  2. If $a_{22}> 0$, then $A+A^{\top}\not \preceq 0$. Define $$ T:=\frac{1}{\sqrt{a_{22}-a_{11}}}\begin{bmatrix} \sqrt{-a_{11}} & -\sqrt{a_{22}}\\ \sqrt{a_{22}}& \sqrt{-a_{11}} \end{bmatrix} $$ and note that $T$ is an orthogonal matrix. It holds $$ T^{\top} A T = \begin{bmatrix} a_{11}+a_{22} & a_{12}+\sqrt{-a_{11} a_{22}}\\ -a_{12}+\sqrt{-a_{11} a_{22}}& 0 \end{bmatrix}. $$ Define $$ D:=\begin{bmatrix} 1 & 0\\ 0 & \frac{a_{12}-\sqrt{-a_{11} a_{22}}}{a_{12}+\sqrt{-a_{11} a_{22}}} \end{bmatrix}, $$ where $D\succ 0$ since $a_{11}a_{22}+a_{12}^{2}>0$. Thus, $T^\top AT$ is $D$-skew-symmetric with non-positive diagonal.

[1] Kaszkurewicz, E., and L. Hsu. "On two classes of matrices with positive diagonal solutions to the Lyapunov equation." Linear algebra and its applications 59 (1984): 19-27.

What are the monomorphisms of ($\infty$-)toposes?

Math Overflow Recent Questions - Sat, 10/13/2018 - 19:18

There are standard notions of "surjections" and "embeddings" of toposes. However, not every surjection is an epimorphism, and not every regular monomorphism is an embedding.

Let $f: \mathcal Y \to \mathcal X$ be a geometric morphism. Recall that $f$ is said to be

  • surjective if $f^\ast$ is conservative (or equivalently, $f^\ast$ is comonadic.)

  • an embedding if $f_\ast$ is fully faithful (equivalently, $f^\ast$ is a localization).

I'd say the correct notion of monomorphism / epimorphism is the $(\infty,1)$-categorical one: $f$ is a monomorphism iff the canonical square $f \circ 1 = f \circ 1$ is a pullback, and dually for epimorphisms. Since $(\infty,1)$-colimits of topoi are computed by taking $(\infty,1)$-limits of the inverse image functors between the underlying categories, $f$ is an epimorphism iff $f^\ast$ is a monomorphism of $(\infty,1)$-categories. That is,

  • $f$ is an epimorphism iff $f^\ast$ is a replete subcategory inclusion, i.e. $f^\ast$ reflects the property of being isomorphic and is an inclusion of path components on hom-spaces.

(Note that the coalgebras for any accessible left exact comonad on $\mathcal Y$ is an $\infty$-topos $\mathcal X$ which admits a canonical surjection from $\mathcal Y$ which will typically not be an epimorphism.)

As for monomorphisms, clearly if $f$ is an embedding, then it is a monomorphism. But not even every regular monomorphism is an embedding. For example, if $F,G: C \to D$ are functors, then the $(\infty,1)$-equalizer of the induced geometric morphisms $Psh(C) \to Psh(D)$ is presheaves on the iso-inserter of $F$ and $G$. The canonical map $Psh(IsoIns(F,G)) \to Psh(C)$ typically fails to be an embedding. Anyway, this leaves me with the question:

Question: What are the monomorphisms of topoi, or of $\infty$-topoi?

I expect this may be complicated, given how complicated monomorphisms of affine schemes are (a category which behaves in some ways similarly to the category of toposes).

Note that because every embedding is a monomorphism, by the surjection / embedding factorization system it suffices to determine which surjections are monomorphisms.

fiber of a map into Grassmanian

Math Overflow Recent Questions - Sat, 10/13/2018 - 18:10

Suppose $R\subset K=K_0\supset K_1\supset K_2\supset...\supset K_{n-1}\supset K_n=\{0\}$ are all vector spaces with $dim R\cap K_i=r_i$ where $r_i$ are some fixed numbers. Suppose $O\subset Gr(r_0,\sum l_i)$is the set of all $R$ satisfies these conditions, where $Gr(r_0,\sum l_i)$ is the set of all $r$ dimensional subspaces in a vector space of dimension $\sum l_i$. Then we have a map$$O\longrightarrow Gr(r_{n-1},l_n)\times Gr(r_{n-2}-r_{n-1},l_{n-1})\times...\times Gr(r_0-r_1,l_1)\\ R\longmapsto R\cap K_{n-1},(R\cap K_{n-2})/(R\cap K_{n-1}),...,, (R\cap K_2)/ (R\cap K_1),R/(R\cap K_1)$$ Obviously this is a projection and the author claimed it is an affine fibration of rank $$t=(r_{n-2}-r_{n-1})(l_n-r_{n-1})+(r_{n-3}-r_{n-2})(l_{n-1}+l_n-r_{n-2})+...+(r_0-r_1)(l_2+...+l_n-r_1)$$ But I have no idea how to get this, hope somebody can help me.

Bockstein homomorphism and Square Operations: Their consistency formulas

Math Overflow Recent Questions - Sat, 10/13/2018 - 17:59

Here are various ways to define "Bockstein homomorphism:"

  1. Let $\beta_p:H^*(-,\mathbb{Z}_p) \to H^{*+1}(-,\mathbb{Z}_p)$ be the Bockstein homomorphism associated to the extension $$\mathbb{Z}_p\to\mathbb{Z}_{p^2}\to\mathbb{Z}_p,$$ it is an element of the mod $p$ Steenrod algebra $A_p$ where $p$ is a prime. If $p=2$, then $\beta_2=Sq^1$.

  2. Let $\beta_p'$:$H^*(-,\mathbb Z_p)\to H^{*+1}(-,\mathbb Z)$ be the Bockstein homomorphism associated to the extension $$\mathbb Z\stackrel{\cdot p}{\to}\mathbb Z\to\mathbb Z_p.$$

  3. Let $\beta_{2^n}: H^*(-,\mathbb Z_{2^n})\to H^{*+1}(-,\mathbb Z_2)$ be the Bockstein homomorphism associated to the extension $$\mathbb Z_2\to\mathbb Z_{2^{n+1}}\to\mathbb Z_{2^n}.$$ By the naturality of connecting homomorphism, $\beta_{2^{n+k}}\cdot2^k=\beta_{2^n}$ where $\cdot2^k: H^*(-,\mathbb Z_{2^n})\to H^*(-,\mathbb Z_{2^{n+k}})$ is induced from $\mathbb Z_{2^n}\stackrel{\cdot2^k}{\to}\mathbb Z_{2^{n+k}}$.

  • question (i) Since $\beta_2=Sq^1$ coincides with the Steenrod square, are there other additional coincidences of other "Generalized Square" (Pontryagin Square, Postnikov Square, etc) coincide with the above "Bockstein homomorphism" $\beta_p$, $\beta_p'$, $\beta_{2^n}$?

  • question (ii) Are there useful consistency formulas for these above "Bockstein homomorphism"?

For example, for Steenrod square, the total Stiefel-Whitney class $w=1+w_1+w_2+\cdots$ is related to the total Wu class $u=1+u_1+u_2+\cdots$ through the total Steenrod square $$ w=Sq(u),\ \ \ Sq=1+Sq^1+Sq^2+ \cdots . $$ Therefore, $w_n=\sum_{i=0}^n Sq^i (u_{n-i})$. The Steenrod squares have the following properties: $$ Sq^i(x_j) =0, \ i>j, \ \ Sq^j(x_j) =x_jx_j, \ \ Sq^0=1, $$

Do we have something similar for thse "Bockstein homomorphism?" $\beta_p$, $\beta_p'$, $\beta_{2^n}$?

Constructing a feature map

Math Overflow Recent Questions - Sat, 10/13/2018 - 17:47

Constructing a feature map $f:R\rightarrow R^{\inf}$ such that $$f(x)^Tf(z) = exp(-|x-z|)$$

Anyone knows the answer? Thanks in advance!

Pontryagin square on spin and non-spin manifold

Math Overflow Recent Questions - Sat, 10/13/2018 - 17:11

The Pontryagin square, maps $x \in H^2({B}^2\mathbb{Z}_2,\mathbb{Z}_2)$ to $ \mathcal{P}(x) \in H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4)$. Precisely, $$ \mathcal{P}(x)= x \cup x+ x \cup_1 2 Sq^1 x. $$ The $\cup_1$ is a higher cup product. The $Sq^1 x= x \cup_1 x$.

  1. If I only consider the $\mathbb{Z}_2$ normal subclass as $$H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_2) \subset H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4),$$ then the $\mathbb{Z}_2$ generator I believe can be written as $$ x \cup x =x \cup (w_2(M)+w_1(M)^2) $$

  2. If I consider the full $\mathbb{Z}_4$ class as $H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4),$ then the $\mathbb{Z}_4$ generator, can it be written as

    $$ \mathcal{P}(x)= x \cup (w_2(M)+w_1(M)^2) + x \cup_1 2 Sq^1 x? $$ in general?

  3. Therefore, on the spin-manifold $w_2(M)=w_1(M)=0$, or the Pin$^-$-manifold, the $ (w_2(M)+w_1(M)^2)=0$, so the first term disappear completely?

So on the spin-manifold or the Pin$^-$-manifold, the $\mathbb{Z}_4$ class of $ \mathcal{P}(x)$ is reduced to a $\mathbb{Z}_2$ subclass generated by

$$ x \cup_1 Sq^1 x= \frac{1}{2}(\mathcal{P}(x) -x^2) ? $$ Is this precise?

Cohomology of complex manifold vs cohomology of its complex submanifold.

Math Overflow Recent Questions - Sat, 10/13/2018 - 16:26

Let $X$ be a smooth complex analytic manifold. Let $Z\subset X$ be a smooth compact analytic submanifold. Let $A$ be a holomorphic vector bundle over $X$. Assume that $$H^i(Z, A|_Z)=0 \mbox{ for any } i>0. $$

Is it true that for any open neighborhood $U$ of $Z$ there exists an open neighborhood $V\subset U$ of $Z$ such that $$H^i(V, A)=0 \mbox{ for any } i>0.$$

A reference would be helpful.

An orientable compact even dimensional manifolds whose all even cohomologies do not vanish but it does not admit any symplectic structure

Math Overflow Recent Questions - Sat, 10/13/2018 - 15:30

What is an example of an orientable compact $2n$ dimensional manifold $M$ whose all even dimensional De Rham cohomology groups $H_{\mathrm{DeR}}^{2i}(M)$ are nonzero, but $M$ does not admit any symplectic structure?

Lengths of proofs and quasilinear time

Math Overflow Recent Questions - Sat, 10/13/2018 - 15:19

Length of proofs depends not only on the theory but also on its axiomatization. Once an axiomatization is fixed, typical proof systems are equivalent up to a polynomial factor. But what if we care about polynomial factors and want the proof system to be quasilinear time efficient?

Thus for example, for $∀x_1 ∀x_2 ... ∀x_n \, 0=0$ we want a size $\tilde{O}(n)$ proof rather than only $\tilde{O}(n^2)$ that we get using sequent calculus (without optimizations).

For all I know, there might not be a single preferred proof system for the predicate calculus up to quasilinear size equivalence. However, ZFC and many other natural axiomatic systems have a certain closure that allows defining quasilinear time completeness as follows.

Let $P$ be a proof system for the predicate calculus whose soundness is provable in ZFC, and such that (constructibly provably in ZFC) $P$-proofs can be verified in quasilinear time. (A verification checks whether a given string is a $P$ proof of a given statement.) Let a ZFC $P$ proof of $B$ be a $P$ proof of $A⇒B$ together with the string $A⇒B$, where $A$ is a conjunction of ZFC axioms and $B$ is in the language of set theory.

Let us say that $P$ is quasilinear time complete for ZFC iff for every proof system $P'$ satisfying the above properties, there is a quasilinear time algorithm that given $B$ and a ZFC $P'$ proof of $B$ returns a ZFC $P$ proof of $B$.

Question: What are some natural examples of proof systems that are quasilinear time complete for ZFC?

To see that proof systems that are quasilinear time complete for ZFC exist, note that a ZFC $P$ proof of $B$ can be obtained as follows:
* Start with a $P'$ proof of $A⇒B$ ($A$ is a conjunction of ZFC axioms) and use soundness of $P'$ and a particular quasilinear verification of $P'$ proofs to get a ZFC $P$ proof that $A⇒B$ is provable in the predicate calculus.
* Use reflection over the predicate calculus to get a ZFC $P$ proof of $A⇒B$, and hence a ZFC $P$ proof of $B$.

For natural proof systems, I expect that the exact reasonable choice of ZFC axioms does not matter (with different axiomatizations being quasilinear time equivalent), and that the proof systems will also be quasilinear time complete for other appropriate axiomatic systems such as PA and Z2. Also, ZFC, PA, Z2 and other typical strong theories that are not finitely axiomatizable prove all instances of reflection over the predicate calculus, and I expect that with reasonable axiomatizations, the proofs of reflection instances have quasilinear size.

A curious process with positive integers

Math Overflow Recent Questions - Sat, 10/13/2018 - 14:46

Let $k > 1$ be an integer, and $A$ be a multiset initially containing all positive integers. We perform the following operation repeatedly: extract the $k$ smallest elements of $A$ and add their sum back to $A$. Let $x_i$ be the element added on $i$-th iteration of the process. The question is: is there a simple formula describing $x_i$, or can they be computed faster than simulating the process? One can easily see that for $k = 2$ we have $x_i = 3i$, but no simple pattern is evident for $k > 2$.

Flatness of direct image sheaf over local artinian ring

Math Overflow Recent Questions - Sat, 10/13/2018 - 13:13

Let $\pi:X \to \mbox{Spec}(\mathbb{C}[t]/(t^2))$ be a smooth, projective morphism and $L$ be an invertible sheaf on $X$. Denote by $L_0$ the restriction of $L$ to the closed fiber, say $X_0$ of $\pi$. Suppose that the natural morphism $H^0(L) \to H^0(L_0)$ is surjective. Can we conclude that $\pi_*L$ is flat over $\mbox{Spec}(\mathbb{C}[t]/(t^2))$?

A link between hooks and contents: Part II

Math Overflow Recent Questions - Sat, 10/13/2018 - 12:52

This is a question in the spirit of an earlier problem.

Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$.

Recall also the notation for the content of a cell $u=(i,j)$ in a partition: $c_u=j−i$. Further, let $h_u$ denote the hook-length of the cell $u$.

The below identity is rather cute for which I don't remember a reference or a proof, so here I ask.

Identity. Given a partition $\lambda\vdash n$, it holds that $$\sum_{u\in\lambda}h_u^2=n^2+\sum_{u\in\lambda}c_u^2.$$

For example, if $\lambda=(4,3,1)\vdash 8$ then the hooks are $\{6,4,3,1,4,2,1,1\}$ and the contents are $\{0,1,2,3,-1,0,1,-2\}$. Hence \begin{align} LHS=6^2+4^2+3^2+1^2+4^2+2^2+1^2+1^2&=84 \\ RHS=\mathbf{8^2}+0^2+1^2+2^2+3^2+(-1)^2+0^2+1^2+(-2)^2&=84. \end{align}

Remainder term in an integral linked to the Riemann zeta function

Math Overflow Recent Questions - Sat, 10/13/2018 - 12:51

Sorry if this is not research level, but the following problem occurs in my own research: it is trivial to show that for $k\ge2$ integral we have $\zeta(k)=(1/(k-1)!)\int_0^\infty t^{k-1}/(e^t-1)\,dt$ (valid in fact also for complex $k$ with $\Re(k)>1$). Now let $m\ge1$ be a fixed integer, and for $N\ge1$ consider the following expression $$\dfrac{1}{(k-1)!N!}\sum_{0\le j\le N}(-1)^{N-j}\binom{N}{j}(j+1)^N\int_0^\infty\dfrac{t^{k-1}}{e^t-1}(1-e^{-m(j+1)t})\,dt\;,$$ which occurs as a remainder term for the computation of $\zeta(k)$. I would like to know the asymptotic behavior of this as $N\to\infty$. It is experimentally of the form $c(m)^{-N}$ for some $c(m)$ depending on $m$ but not on $k$, but what is the constant $c(m)$ ?


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