Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2015-01-30T10:59:08Zhttp://mathoverflow.net/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://mathoverflow.net/q/1952630Helmholtz boundary value problem in 2Dusumdelphinihttp://mathoverflow.net/users/664482015-01-30T10:43:36Z2015-01-30T10:48:43Z
<p>I want to solve the Helmholtz equation in 2D with constant nonhomogeneities: $$\nabla^2w+\lambda w=C$$ and with Dirichlet boundary conditions such that $$w(0,0)=0$$ $$w(x,y)\underset{|(x,y)|\rightarrow\infty}{\longrightarrow} w_\infty$$</p>
<p>All the fundamental solutions that I find in the literature do not seem to allow me to impose the first of these two boundary conditions. How do I solve this problem?</p>
http://mathoverflow.net/q/1952620Examples of Bordy hyperbolic affine varieties which are not Kobayashi hyperbolicdiveriettihttp://mathoverflow.net/users/98712015-01-30T10:41:31Z2015-01-30T10:41:31Z
<p>Let $X$ be a complex space. </p>
<p>We say that $X$ is Brody hyperbolic if there is no non-constant holomorphic map $f\colon\mathbb C\to X$. </p>
<p>We say that $X$ is Kobayashi hyperbolic if the Kobayashi pseudo-distance $d_X$ on $X$ is a genuine distance.</p>
<p>Since holomorphic maps are distance-decreasing withe respect to the Kobayashi pseudo-distance, and since $d_\mathbb C\equiv 0$, we immediately see that any Kobayshi hyperbolic complex space is also Brody hyperbolic: if $f\colon\mathbb C\to X$ is any holomorphic map then
$$
d_X(f(z),f(w))\le d_\mathbb C(z,w)=0,
$$
so that $f$ must be constant, $d_X$ being a true distance.</p>
<p>Now, by Bordy's reparametrization lemma, if $X$ is moreover compact, then we also have that Brody hyperbolicity implies Kobayashi hyperbolicity. But in the non-compact case it is well known that Kobayashi hyperbolicity is actually stronger. A standard example is given by the domain $X$ in $\mathbb C^2$ described as follows:
$$
X=\{(z,w)\in\mathbb C^2\mid|z|<1,|zw|<1\}\setminus\{(0,w)\mid|w|\ge 1\}.
$$
This domain $X$ can be quite straightforwardly seen to be not Kobayashi hyperbolic but nevertheless there is no non-constant holomorphic map $f\colon\mathbb C\to X$.</p>
<p>Here is my question:</p>
<p><strong>Question.</strong> Do you know any example of a complex affine variety $X$ which is Brody hyperbolic but not Kobayashi hyperbolic?</p>
<p>If such an example exists, then any compactification $\overline X$ of $X$ is not Kobayashi hyperbolic and so must contain non-constant holomorphic images of the complex plane. But then, all these images must cut the boundary $\overline X\setminus X$ of $\overline X$ in at least two points...</p>
<p>Thanks in advance!</p>
http://mathoverflow.net/q/195260-3Prove that a Graph is connected using eigen values $\lambda$user1590205http://mathoverflow.net/users/665152015-01-30T10:24:14Z2015-01-30T10:24:14Z
<p>Prove that for a graph is connected if and only if $\lambda_{max}$ > $\lambda_{1}$
Prove that for a $d$-regular graph $\lambda_{\max} = \lambda_1 =
\cdots = \lambda_{k-1}$ if and only if the graph has $k$ connected
components. </p>
http://mathoverflow.net/q/1952580Asymmetry of functions defined on the $n$-th roots of the unityPeva Blanchardhttp://mathoverflow.net/users/592392015-01-30T10:07:46Z2015-01-30T10:17:11Z
<p>Let $\mathcal{A} = \{V : \mathbb{U}_n \rightarrow \mathbb{C}\}$ where $\mathbb{U}_n$ is the group of the complex $n$-th roots of the unity. This group naturally acts on $\mathcal{A}$: for any $a \in \mathbb{U}_n$, and any $V \in \mathcal{A}$
$$
\begin{equation*}
(a \star V) : b \mapsto V(a^{-1}b)
\end{equation*}
$$
I define $Fix(V) = \{a \in \mathbb{U}_n,~ a \star V = V\}$ the symmetry group of $V$.</p>
<p>My question is: can we define an equivariant function $\lambda : \mathcal{A} \rightarrow \mathbb{C}$ which measures the "asymmetry" of $V$, i.e., a function such that for all $a,V$
$$
\begin{align*}
\lambda(a \star V) &= a \lambda(V) \\
\lambda(V) = 0 &\Leftrightarrow Fix(V) = \{1\}
\end{align*}
$$</p>
<p>Is it related to some known problem ? Satisfing the first condition is ok, for instance
$$
\begin{equation*}
\lambda_0(V) = \sum_{z \in \mathbb{U}_n} z V(z)
\end{equation*}
$$
but the second condition is not ok. Indeed, it suffices to take $V$ as the indicator function of, say, the union of a triangle and a square (in a suitable $\mathbb{U}_n$). The sum above is zero but this $V$ has a trivial symmetry group.</p>
<p>Thank you.</p>
http://mathoverflow.net/q/1952550A Lie algebra assiciated with a one dimensional foliationAli Taghavihttp://mathoverflow.net/users/366882015-01-30T09:32:11Z2015-01-30T09:32:11Z
<p>A non vanishing vector field $X$ on a manifold is called "well behaved" if for every non vanishing smooth function $f$ we have $$C(X)\simeq C(fX)$$ This means that the centralizer Lie algebras $C(X)$ and $C(fX)$ are isomorphic Lie algebra.</p>
<p>Ex: The vector field $\partial_{x}$ is a well behaved vector field on $\mathbb{R}^{2}$.</p>
<p>We conclude from the above definition that this concept is a "foliation" concept. That is to every one dimensional foliation tangent to a well behaved vector field, we associate a Lie algebra.</p>
<p><strong>Two questions:</strong></p>
<blockquote>
<p>1.Is there a well behaved vector field on $\mathbb{R}^{2}$ which foliation is not topological equivalent to the foliation of the above example.</p>
<p>2.What is an example of a well behaved vector field on a <strong>compact</strong> manifold?In particular is a Kronecker foliation on tori, a well behaved foliation?</p>
</blockquote>
http://mathoverflow.net/q/1952540A cohomology associated with a codimension one foliationAli Taghavihttp://mathoverflow.net/users/366882015-01-30T09:19:16Z2015-01-30T09:43:26Z
<p>Let $\alpha$ be a non vanishing one form on a manifold which which defines a codimension one foliation. With this $\alpha$ we define the following complex:</p>
<p>$$\phi:\Omega^{i}(M)\to \Omega^{i+2}(M)\;\;\;\phi(\beta)=d(\alpha\wedge \beta)$$ Obviously $\phi$ satisfies $\phi \circ \phi=0$, so we have cohomologies associated with this complex. The total cohomology is denoted by $H^{*}(\alpha)$</p>
<blockquote>
<p>Are these cohomologies finite dimensional vector space?</p>
<p>Are there some dynamical information in this cohomology?</p>
<p>Is this cohomology independent of choosing the one form $\alpha$ which kernel is tangent to the foliation? This means that: Is it true to say $H^{*}(\alpha) \simeq H^{*}(f\alpha)$ for a non vanishing smooth function $f$?</p>
</blockquote>
http://mathoverflow.net/q/1952503Primes and ParityGil Kalaihttp://mathoverflow.net/users/15322015-01-30T08:16:39Z2015-01-30T08:16:39Z
<p>This problem is motivated by the <a href="http://michaelnielsen.org/polymath1/index.php?title=Finding_primes" rel="nofollow">polymath4</a> project. There, the aim was to find an efficient deterministic algorithm for finding a prime larger than $N$. The hope was to find a polynomial algorithm in $n=\log N$ which can be done assuming the truth of either plausible but intractable number theory conjectures or plausible but intractable problems in computational complexity. What was achieved was an algorithm that runs in time $N^{1/2-c}$ for some small $c>0$ to find the parity of the number of primes in an arbitrary interval of integers smaller than $N$. In view of this, identifying a single interval with an odd number of primes could be useful. Here are some questions regarding primes and parity.</p>
<h3>1)</h3>
<p>Let $N$ be an integer. consider the intervals $[N,2N]$, $[2N,3N], \dots$ $[kN,(k+1)N]$
What is the smallest $k$ that we can guarantee that one of these intervals contain an odd number of primes? </p>
<p>Based on Cramer's probabilistic modeling we can expect $k=a \log N$ to work for every $N$ and some constant $a$. Results about gaps between primes asserts that when $k$ is exponential in $N$ we can find such an interval with one, hence an odd number of, primes. </p>
<p>Is there some hope to prove it for $k=N$? $k=N^{1/2}$? A proof for $k=N^{1/2-c}$ will allow us by divide and conquer to find a prime $p$ larger than $N$ is time $p^{1/2-c}$.</p>
<h3>2)</h3>
<p>For which of the following sequences of intervals $[a(n),2a(n)]$
would it be possible to prove that that (i) there are infinitely many cases of odd number of primes; (ii) this occurs in half the cases? </p>
<p><strong>2.1</strong> $a(n)=n$ or an a.p. (I think (i) is ok);</p>
<p><strong>2.2</strong> $a(n)=p_n$; </p>
<p><strong>2.3</strong> $a_n=n^2$; </p>
<p><strong>2.4</strong> $a_n=2^n$</p>
<p>Is showing that there are infinitely many $n$s for which there are odd number of $n$-digits primes entirely hopeless (like Cramer's conjecture)?</p>
<h3>3)</h3>
<p>Let $p_n$ be the $n$th prime. What can be said/proved about the zeta-like function $$ \prod_{k=1}^\infty {{1}\over{1-(-1)^k(p_k)^s}}$$ </p>
<h3>4)</h3>
<p>Beside polymath4, were such questions about primes and parity considered before?</p>
http://mathoverflow.net/q/1952491Homogeneous polynomial vector fields tangent to the unit sphereDenis Serrehttp://mathoverflow.net/users/87992015-01-30T08:13:21Z2015-01-30T09:54:42Z
<p>This question has something to do with <a href="http://mathoverflow.net/q/194169">that one</a>.</p>
<p>Let $n\ge1$ and $d\ge1$ be two given integers. Consider the polynomial vector fields $v=(v_1,\ldots,v_n)$ whose components $v_j$ are homogeneous of degree $d$ in the indeterminates $X_1,\ldots,X_n$.</p>
<blockquote>
<p>What is the dimension $D(n;d)$of the subspace defined by the equation $$X_1v_1(X)+\cdots+X_nv_n(X)=0\quad ?$$</p>
</blockquote>
<p>I computed this dimension for small dimensions: $D(1;d)=0$, $D(2;d)=d$ and
$D(3;d)=d(d+2)$. I suspect that the problem has been solved a while ago and the formula is simple in terms of binomials. A solution might come by considering a the sequence of morphisms
$$\cdots\rightarrow\Lambda_{n-2}({\mathbb R}^n)\otimes {\rm Hom}_n^{d-1}\rightarrow\Lambda_{n-1}({\mathbb R}^n)\otimes {\rm Hom}_n^{d}\rightarrow\Lambda_{n}({\mathbb R}^n)\otimes {\rm Hom}_n^{d+1},$$
where ${\rm Hom}_n^d$ denotes the space of homogeneous polynomials of degree $d$, and each arrow is of the form $V\mapsto X\wedge V$.</p>
http://mathoverflow.net/q/1952483Co-HausdorffificationDominic van der Zypenhttp://mathoverflow.net/users/86282015-01-30T08:12:48Z2015-01-30T10:16:56Z
<p>Given a topological space $(X,\tau)$ we can define the "$T_2$-ification" of $X$ by setting $T_2(X,\tau) = X/\simeq$ where $x\simeq y$ in $X$ if and only if for every open neighborhood of $x$ has nonempty intersection with every open neighborhood of $y$. Then $T_2(X,\tau)$ has the following universal property:</p>
<p>For every $T_2$-space $Z$ and continuous function $f: X\to Z$, there is $\bar{f}: T_2(X,\tau) \to Z$ such that $f = \bar{f} \circ pr$ where $pr: X\to X/\simeq$ is the canonical projection.</p>
<p>A <em>co-$T_2$-ification</em> would be a space with a similar property as above, but all arrows reversed. (I hope this description is clear enough.)</p>
<p>Does every space have a co-$T_2$-ification?</p>
<p>(There are similar constructions for $T_i$-ifications at least for $i\in \{0,1\}$, so the same question could be asked for those.)</p>
http://mathoverflow.net/q/1952451Connected sum of chiral manifoldsJens Reinholdhttp://mathoverflow.net/users/142332015-01-30T06:38:28Z2015-01-30T06:38:28Z
<p>Let $M,N$ be two closed, smooth, orientable manifolds of the same dimension and assume that these manifolds are chiral, i.e. they do not admit an orientation reversing automorphism. Then there are two essentially different ways to form the conncected sum $M\#N$. In concrete examples, what methods can be used to show that the two resulting manifolds are not diffeomorphic? Is there an example where the two manifolds are (accidently) diffeomorphic?</p>
http://mathoverflow.net/q/1952410Parabolic subgrouplaylahttp://mathoverflow.net/users/665012015-01-30T03:03:32Z2015-01-30T04:59:28Z
<p>I have a question about root set corresponding to $P_θ ∩ M_Ω$
where $θ$ is a subset of simple roots, $Ω=θ∪{α}$ where $α$ is a simple root and not in $θ$, $P_θ$ is a parabolic subgroup corresponding to $\theta$ and $M_Ω$ is a Levi factor of the parabolic subgroup $P_Ω$.</p>
<p>What is $P_θ ∩ M_Ω$?</p>
<p>Thanks!</p>
http://mathoverflow.net/q/1952151An expectation of the product of random unitariesAtnaphttp://mathoverflow.net/users/664712015-01-29T19:55:13Z2015-01-30T09:23:03Z
<p>I want to find the answer of
$$\int dU \ U^m X \ U^{\dagger m}$$</p>
<p>Where $m\in\mathbb{N}$ and $U$'s are unitary matrices in $U(n)$ and $dU$ is a normalized Haar measure. $X$ is a given self-adjoint matrix. </p>
<p>I used Schur lemma and found that the answer is of the form</p>
<p><em>(Edit: the correct form is )</em>
$$pX+(1-p)tr(X)\frac{I}{n}$$</p>
<p>with $p\in[0,1]$, but if this is true then by comparing above expressions and using the results obtained in <a href="http://mathoverflow.net/questions/195186/expectation-of-trace-of-nth-power-of-unitary-matrices">my last question</a>, $p$ can be found as
$$p=\frac{m-1}{n^2-1}$$
which is greater than 1 for $m>n^2$.</p>
<p>I don't know what is going wrong here. Does anyone have an idea about my solution or another way for solving this integral?</p>
http://mathoverflow.net/q/19520710A possibly surprising appearance of Lucas numbersClark Kimberlinghttp://mathoverflow.net/users/614262015-01-29T17:46:39Z2015-01-30T10:55:03Z
<p>Let $S$ be the set of polynomials defined as follows: $0$ is in $S$, and if $p$ is in $S$, then $p + 1$ is in $S$ and $x \cdot p$ is in $S$, so that $S$ "grows" in generations: $g(0)=\{0\}$, $g(1)=\{1\}$, $g(2)=\{2,x\}$, $g(3)=\{3,2x,x+1,x^2\}$, and so on, with $|g(n)|=2^{n-1}$. Let $S^*$ be the set obtained from $S$ by substituting $r=\sqrt{2}$ for $x$ and keeping only the first appearance of each duplicate. Successive generations $G(n)$ now begin with $\{0\}$, $\{1\}$, $\{2,r\}$, $\{3,2r,r+1\}$, with $|G(n)|$ starting with $1,1,2,3,4,7,11,18,29,...$; i.e., Lucas numbers beginning at the 4th term, and checked for 30 generations. Can someone prove that $|G(n)|=L(n-1)$ for $n \geq 4$? </p>
http://mathoverflow.net/q/1951690Uniform $L^p-L^{p'}$ bound of a Fourier multipliershanlinhttp://mathoverflow.net/users/357022015-01-29T12:06:06Z2015-01-30T08:32:16Z
<p>Let $(\tau,\xi)\in\mathbb{R}\times \mathbb{R}^n$, and consider the function
$$
m_{\epsilon}(\tau,\xi)=\frac{1}{\tau+|\xi|^4+\epsilon|\xi|^2+i}
$$
in $\mathbb{R}^{n+1}$. My first question is that does the function $m_{\epsilon}(\tau,\xi)$ define a $L^p(\mathbb{R}^{n+1})-L^{p'}(\mathbb{R}^{n+1})$ multiplier with $p=\frac{2(n+4)}{n+8}$, $p'=\frac{2(n+4)}{n}$ for all $\epsilon\in\mathbb{R}$? Or equivalently, do we have the following estimates
$$
\|\mathcal{F^{-1}}(\frac{\hat{f}(\tau,\xi)}{\tau+|\xi|^4+\epsilon|\xi|^2+i})\|_{L^{p'}(\mathbb{R}^{n+1})}\leq C\|f\|_{L^p(\mathbb{R}^{n+1})}~~~~ ?
$$
Where $\mathcal{F^{-1}}$ denotes the Fourier inversion in $\mathbb{R}^{n+1}$.. </p>
<p>If $\epsilon=0$, this is true, which is essentially due to the generalized Strichartz estimates (see e.g. Stein's Harmonic Analysis, p. 369). I'm interested in the case where $\epsilon\ne 0$, which can be thought of some kind of perturbations of the case $\epsilon=0$. In particular, if the above question is valid, then can one show that the bound $C$ is actually uniform with $\epsilon$? </p>
<p>Thanks for any comments or references.</p>
http://mathoverflow.net/q/1951687How to make the Capelli's identity less mysterious?semyon aleskerhttp://mathoverflow.net/users/161832015-01-29T12:01:56Z2015-01-30T06:29:16Z
<p>The formulation of the Capelli's identity is very elementary; it has important applications in invariant theory and representation theory, see <a href="http://en.wikipedia.org/wiki/Capelli%27s_identity" rel="nofollow">http://en.wikipedia.org/wiki/Capelli%27s_identity</a></p>
<p>To remind it, it requires some notation. Let $x_{ij}$, $1\leq i,j\leq n$, be commuting variables. Define differential operators on the space of functions on $n\times n$ matrices:
$$E_{ij}=\sum_{a=1}^n x_{ia}\frac{\partial}{\partial x_{aj}},\, 1\leq i,j\leq n.$$ </p>
<p>The Capelli identity states that
$$\det\left[\begin{array}{cccc}
E_{11}+n-1&\dots&E_{1,n-1}&E_{1n}\\
\vdots&\vdots&\dots&\vdots\\
E_{n-1,1}&\dots&E_{n-1,n-1}+1&E_{n-1,n}\\
E_{n,1}&\dots&E_{n,n-1}&E_{n,n}+0
\end{array}\right]=\\\det\left[\begin{array}{ccc}
x_{11}&\dots&x_{1n}\\
\dots&\dots&\dots\\
x_{n1}&\dots&x_{nn}
\end{array}
\right]\cdot
\det\left[\begin{array}{ccc}
\frac{\partial}{\partial x_{11}}&\dots&\frac{\partial}{\partial x_{1n}}\\
\dots&\dots&\dots&\\
\frac{\partial}{\partial x_{n1}}&\dots&\frac{\partial}{\partial x_{nn}}
\end{array}\right].
$$</p>
<p>Note that in the right hand side of the equality the two matrices have commuting entries, while in the left hand side the entries do not commute. Hence one has to be careful to define the determinant. <strong>The convention for the determinant of such matrices is
$$\det(a_{ij})=\sum_{\sigma\in S_n}sgn(\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)},$$
where the order of terms is important.</strong></p>
<p><strong>This order of terms in the determinant is the most mysterious point for me. Is there any reason for it? What happens if one chooses some other ordering of terms: will the Capelli identity be modified somehow or it will not work at all?</strong></p>
<p>There are more recent generalizations of the Capelli identity, see e.g. the above link to Wikipedia. However unfortunately they do not clarify to me the original Capelli idenity, but rather use it as a basic inspiration for further extensions (may be I am missing something).</p>
http://mathoverflow.net/q/1950702How many points does 'the-most-point-contained-circle' contain at least?mathlovehttp://mathoverflow.net/users/344902015-01-28T11:07:23Z2015-01-30T10:10:14Z
<blockquote>
<p><strong>Question</strong> : Given any $n$ distinct points $S$, consider the $\binom n2$ discs $D_{pq}$ formed by picking a pair of points $p,q$ from $S$ and using them as a diameter. For each disc $D_{pq}$, let $N_{pq}$ be the number of points of $S$ on or inside $D_{pq}$. Let $N_S(n)$ be the largest $N_{pq}$, i.e.
$$N_S(n)=\max\{N_{pq} :p,q\in S,p\not=q\}.$$
Let $f(n)$ be the minimum value of $N_S(n)$ when $S$ varies over the set of $n$ points. Then, how can we represent $f(n)$ by $n$ ?</p>
</blockquote>
<p><strong>Remark</strong> : This question has been <a href="http://math.stackexchange.com/questions/829017/how-many-points-does-the-most-point-contained-circle-contain-at-least">asked previously on math.SE</a> with receiving only a partial answer.</p>
<p>On math.SE, a user achille hui got the following bounds for $n\gt 3$ :
$$\left\lceil \frac{n}{3}\right\rceil + 1 \le f(n) \le \left\lceil\frac{2n}{3}\right\rceil.$$</p>
<p>This fixes $f(4)$ to $3$. (We have $f(3)=2$ trivially.)</p>
<p>However, no more answers have been received.</p>
<p>The question comes from that I changed 'rectangle' to 'circle' in the following question.</p>
<p>"When we place $n$ distinct points on the $xy$-plane, prove that there exists a rectangle, whose diagonal is closed by two of the $n$ points and whose edges are parallel to either $x$-axis or $y$-axis, has at least $\lfloor(n+1)/5\rfloor+1$ points inside or on the edges."</p>
<p>I would like to know how to find $f(n)$ (or better bounds), and any relevant references.</p>
http://mathoverflow.net/q/1948991Probabilistic statement on matrix ranksTurbohttp://mathoverflow.net/users/100352015-01-26T09:06:30Z2015-01-30T05:13:53Z
<p>Given $A\in\{0,1\}^{n\times n}$ with $\operatorname{rank}(A)=r=2^{O((\log_2n)^{\frac{1}{c+1}})}$.</p>
<p>Denote $\mathsf{1_n}\in\{0,1\}^n$ as vector with $1$s.</p>
<p>Does
$$\lim_{n\rightarrow\infty}\mathsf{P_{A\in\{0,1\}^{n\times n}}}\Bigg(\mathsf{1_n}'A\mathsf{1_n}>\frac{(r-1)r^{(\log_2r)^c}}{2\log_2r}\Bigg)=1\quad?$$</p>
<p>Infact is there a matrix (family) such that $$\mathsf{1_n}'A\mathsf{1_n}<\frac{(r-1)r^{(\log_2r)^c}}{2\log_2r}\quad?$$</p>
<p>$c\geq1$ fixed.</p>
http://mathoverflow.net/q/1948405Some calculus in the orthogonal group $O(n)$Ali Taghavihttp://mathoverflow.net/users/366882015-01-25T16:02:56Z2015-01-30T10:07:45Z
<p>How can one compute each of the following matrices, explicitly:</p>
<p>$$\int_{O(n)} e^{g}dg$$ or
$$\int_{O(n)} g^{n}dg \;\;\;\;n\in \mathbb{N} \;\;n>1$$
What is the explicite entries of the resulting matrices, for $n=2$?</p>
<p>Moreover, for $n,m\in \mathbb{Z}$ define the linear operator $T_{n,m}$ on $M_{n}(\mathbb{R})$ as follow:
$$T_{n,m}(A)=\int_{O(n)}g^{n}Ag^{m}$$
Under what suficcient and necessary condition, $T_{n,m}$ is conjugate to $T_{n',m'}$? Is there an explicit formulation for $T_{n,m}$?</p>
<p>The integration is based on the Haar measear defined on orthogonal group $O(n)$.</p>
http://mathoverflow.net/q/1948070Exactness of total complex [on hold]Userhttp://mathoverflow.net/users/663292015-01-25T05:47:29Z2015-01-30T04:45:16Z
<p>The acyclic assembly lemma (pages 59/60 of An introduction to homological algebra, Weibel) establish that if I have a bounded double complex with exact rows (or columns) then the total complex is also exact.</p>
<p>By bounded I mean that the non-zero terms of the double complex can be enclosed by determinate square.</p>
<p>The questions is: if the rows are exact excep in the last non-zero level of each row (whiches coincides), the total complex is also exact except in the last non-zero level?</p>
<p>I would appreciate a proof without spectral sequence's tools.</p>
http://mathoverflow.net/q/19478810On the boundary of the twindragonNikita Sidorovhttp://mathoverflow.net/users/81312015-01-24T21:48:36Z2015-01-30T03:43:13Z
<p>Let $\mathcal T$ be the famous <em>twindragon</em>, i.e.,
$$
\mathcal T=\left\{\sum_{n=0}^\infty a_n\left(\frac{1+i}2\right)^n : a_n\in\{0,1\}\right\}.
$$</p>
<p><img src="http://i.stack.imgur.com/1Gy4C.gif" alt="enter image description here"></p>
<p>Then, as is well known, $\mathcal T$ has a non-empty interior, whereas $\partial\mathcal T$ is indeed a fractal whose Hausdorff dimension is known as well - see, e.g., <a href="http://ecademy.agnesscott.edu/~lriddle/ifs/heighway/heighwayBoundary.htm">this survey</a> (it's on the Heighway dragon, but the twindragon is just two of those placed back to back). </p>
<p>Now let $X\subset\{0,1\}^{\mathbb N}$ be defined as follows:
$$
\partial \mathcal T=\left\{\sum_{n=0}^\infty b_n\left(\frac{1+i}2\right)^n : b_n\in X\right\}.
$$
Clearly, $X$ is a subshift. </p>
<p><strong>QUESTION.</strong> Is there a closed description of $X$? In particular, is $X$ a sofic subshift (or even a subshift of finite type)? </p>
<p>The closed formula for its dimension - $\log\lambda/\log\sqrt2$ with $\lambda$ being a root of $2x^3-x+1$ - suggests so. The proof from the link uses the Hutchinson formula for some self-similar IFS whose attractor is precisely $\partial\mathcal T$, which is nice, but I'd like it to be in the form $h(X)/\log\sqrt2$, where $h(X)$ is the topological entropy of the subshift $X$. </p>
http://mathoverflow.net/q/19465218What are the products $\prod_{A\subset{\mathbb F}_p\colon |A|=n} \sum_{a\in A} \zeta^a$ equal to?Sevahttp://mathoverflow.net/users/99242015-01-23T10:52:56Z2015-01-30T07:23:25Z
<p>This is a somewhat more explicit version of a <a href="http://mathoverflow.net/questions/194343/prime-factors-of-sum-i-in-i-zeta-pi">question</a> I have recently asked.</p>
<p>Let $p$ be an odd prime, and write $\zeta:=\exp(2\pi i/p)$ (any other primitive $p$th root of unity will do as well). For integer $n\in[0,p]$, the products
$$ {\mathcal P}_p(n) := \prod_{\substack{A\subseteq{\mathbb F}_p \\ |A|=n}}\ \sum_{a\in A} \zeta^a $$
are rational integers; can they be found "explicitly"? </p>
<p>It is immediately seen that ${\mathcal P}_p(0)=0$ and ${\mathcal P}_p(1)=1$, and I can show that ${\mathcal P}_p(2)=\left(\frac2p\right)$ (it is not difficult to see that ${\mathcal P}_p(2)=\pm 1$, but to determine the sign is trickier). Also, we have ${\mathcal P}_p(p-n)=(-1)^{\binom{p}{n}}{\mathcal P}_p(n)$, so that ${\mathcal P}_p(p)=0$, ${\mathcal P}_p(p-1)=-1$, and ${\mathcal P}_p(p-2)=\pm 1$. </p>
<blockquote>
<p>$\quad$ What are the values of ${\mathcal P}_p(n)$ for $n\in[3,p-3]$?</p>
</blockquote>
<p>Some numerical data (thanks to Talmon Silver for the programming):</p>
<p>$\quad {\mathcal P}_5(3)=-1$<br>
$\quad {\mathcal P}_7(3)=-2^7$<br>
$\quad {\mathcal P}_{11}(3)=23^{11}$<br>
$\quad {\mathcal P}_{13}(3)=159^{13}$<br>
$\quad {\mathcal P}_{17}(3)=-24617^{17}$<br>
$\quad {\mathcal P}_{19}(3)=-611009^{19}$<br>
$\quad {\mathcal P}_{23}(3)=1265401351^{23}$ </p>
<blockquote>
<p>$\quad$ If finding the individual values ${\mathcal P}_p(n)$ is difficult, can we at least find explicitly the product
$$ {\mathcal P}_p(1){\mathcal P}_p(2)\dotsb{\mathcal P}_p(p-2){\mathcal P}_p(p-1)
= \prod_{\varnothing\ne A\subsetneq{\mathbb F}_p} \sum_{a\in A} \zeta^a \ ?$$</p>
</blockquote>
<p>Denoting this product by ${\mathcal P}_p$, </p>
<p>$\quad {\mathcal P}_3=-1$<br>
$\quad {\mathcal P}_5=-1$<br>
$\quad {\mathcal P}_7=-2^{14}$<br>
$\quad {\mathcal P}_{11}=-(3\cdot 23^4 \cdot 67\cdot 89)^{22}$<br>
$\quad {\mathcal P}_{13}=-(3^{12}\cdot 5\cdot 53^6 \cdot 79^4\cdot 131^2 \cdot 157^2 \cdot 313\cdot 547\cdot 599\cdot 911)^{26}$ </p>
<hr>
<p>The problem can be restated in a purely combinatorial way, as hinted to in Ofir's comment below. Write $N:=\binom pn$, let $A_1,\dotsc,A_N$ be all the $n$-element subsets of ${\mathbb F}_p$, and for $z\in{\mathbb F}_p$ denote by $r_n(z)$ the number of representations $z=a_1+\dotsb+ a_N$ with $a_1\in A_1,\dotsc, a_N\in A_N$. We have then ${\mathcal P}_p(n)=\sum_{z\in{\mathbb F}_p}r_n(z)\zeta^z$, and from the fact that ${\mathcal P}_p(n)$ is an integer, it follows that $r_n(z)$ are actually equal to each other for all $z\in{\mathbb F}_p\setminus\{0\}$; as a result, we have, say, ${\mathcal P}_p(n)=r_n(0)-r_n(1)$. On the other hand,
$$ r_n(0)+(p-1)r_n(1) = \sum_{z\in{\mathbb F}_p} r_n(z) = |A_1|\dotsb|A_N| = n^{\binom pn}. $$
This yields
$$ {\mathcal P}_p(n) = \frac1{p-1} \left( p\,r_n(0)-n^{\binom pn}\right). $$
Thus, the problem boils down to finding $r_n(0)$, the number of all zero-sum $N$-tuples
$(a_1,\dotsc,a_N)$ with the components $a_i$ representing each of the $n$-element subsets of ${\mathbb F}_p$.</p>
http://mathoverflow.net/q/1945675Induced subgraphs of small strongly regular graphsJernejhttp://mathoverflow.net/users/17372015-01-22T13:43:26Z2015-01-30T09:48:42Z
<p>Consider a strongly regular graph $G$ with parameters $(76,30,8,14).$ Hoffman's bound tells us that $\overline{G}$ has an independent set of size at most $4$ and its not hard to see there are indeed $4$ independent vertices in $\overline{G}.$</p>
<p>So if $x \in V(G)$ then the graph $H$ induced by $N(v)$ is a $8$-regular, $K_4$-free graph on $30$ vertices and known bounds tell us that $\alpha(H) \ge 5.$</p>
<p>Hence $G$ must must contain $K_{1,5}$ as an induced subgraph. I do not see any way to extend this subgraph without introducing cases.</p>
<p>What I am wondering is </p>
<blockquote>
<p><strong><em>Question 1.</em></strong> Can someone construct larger graphs that must be present as induced subgraphs of $G$</p>
</blockquote>
<p>and</p>
<blockquote>
<p><strong><em>Question 2.</em></strong> Can someone find large induced subgraphs for some of the missing SRG's on less than 100 vertices?</p>
</blockquote>
<p>The motivating factor for this problem is that getting an induced subgraph of order 19 not having $2$ as an eigenvalue is enough to reconstruct $G.$</p>
http://mathoverflow.net/q/1917170Some quantities which definitions are (somehow) similar to the classical DivergenceAli Taghavihttp://mathoverflow.net/users/366882014-12-29T16:43:08Z2015-01-30T06:12:39Z
<p>Motivated by classical formulas $L_{X}=d\circ i_{X}+i_{X}\circ d$ and $L_{X} \Omega=Div(X) \Omega$ and the essential role of the diff operator $d$ in definition of divergence, we define some quantities, say $\overline{Div}$, based on the adjoint operator $d^{*}=\pm *d*$, where $*$ is the Hodge star operator.</p>
<p>In this way our main question is that:</p>
<blockquote>
<p>What are some geometric or physical interpretations for $\overline{Div}$? What are some calculus identities for this quantity?In particular is it true that for a closed manifold $M$, with volum form $\Omega$, we have $\int_{M} \overline{Div}(X)\Omega=0$? </p>
<p>Moreover what is the dynamical interpretation of $\overline{Div}(X)=0$. This is motivated by classical case: If $Div(X)=0$ then $X$ has no an attractor, since the flow of $X$ generates a one parameter family of volume preserving diffeomorphisms. So we ask: Is there a vector field $X$ which has a compact attractor invariant set but $\overline{Div}(X)$ is identically zero?</p>
</blockquote>
<ol>
<li>For a vector field $X$ on a $2$ dimensional surface with volum form $\Omega$ define:</li>
</ol>
<p>$$\overline{Div}(X)=(i_{X}\circ d^{*}+d^{*}\circ i_{X})(\Omega)$$</p>
<ol start="2">
<li><p>A vector field $X$ on a Riemannian manifold $(M,g)$ defines a one form $\alpha$. Now $\overline{Div}(X)$ is defined as a unique function with $$\alpha \wedge d^{*}(\Omega)=\overline{Div}(X). \Omega $$</p></li>
<li><p>For a symplectic manifold $(M,\omega)$, $\overline{Div}$ is the unique function with $$ (i_{X}\circ d^{*}+d^{*}\circ i_{X})(\Omega)\wedge \omega=\overline{Div}(X).\Omega$$ where $\Omega$ is the corresponding volume form.</p></li>
</ol>
http://mathoverflow.net/q/18552710Riemann's formula for the metric in a normal neighborhoodMartin Gisserhttp://mathoverflow.net/users/91612014-10-27T19:20:11Z2015-01-30T07:02:53Z
<p>I would love to understand the famous formula $g_{ij}(x) = \delta_{ij} + \frac{1}{3}R_{kijl}x^kx^l +O(||x||^3)$, which is valid in Riemannian normal coordinates and possibly more general situations.</p>
<p>I'm aware of 2 proofs: One using Jacobi fields [cf. e.g. S.Sternberg's "Curvature in Mathematics and Physics" from which the question title and formula is stolen :-) or cf. S.Lang's "Differential and Riemannian Manifolds"]. The other proof involves computing that $\partial_k\partial_lg_{ij}(x)$ shares some symmetries of curvature [cf. M.Spivak's "A Comprehensive Introduction to Differential Geometry, Vol. 2" where it is a several page "hairy computation" or cf. H.Weyl's 1923 edition of Riemann's Habilitationsvortrag (reprinted in a recent German book by Jürgen Jost) which I find uncomprehensible.]</p>
<p><strong>Are you aware of any other proof? Are normal coordinates necessary?</strong></p>
<p>While the Jacobi fields proof is short and elegant enough, it irks me that it requires "higher technology" not involved in the endproduct. Somehow the formula should be provable by pure calculus. Indeed, it is stated as an exercise in P.Petersen's "Riemannian Geometry": From the context I guess he thinks it should follow from the expression of $\partial_lg_{ij}$ as a sum of 2 Christoffel symbols and the simplified expression for curvature at $x=0$ where the Christoffel symbols vanish. Alas my attempts at this go in circles...</p>
<p>I find the situation quite amazing: Not many textbooks treat this fundamental and historic formula. (Estimating from the sample on my shelf it is 3/17. E.g. it seems it's not even in Levi-Civita's classic.)</p>
<p><strong>Update/Scholium:</strong></p>
<p>In classical language: The knackpoint seems to be a "differential Bianchi formula" for the Christoffel symbols at $0$. This follows from the geodesic equation. I see no other way yet.</p>
<p>A more modern approach minimizing (but not eliminating) the role of geodesics is in A.Gray's Tubes book. (Noted in comments. I'm waiting for www.amazon.de to deliver this treasure.)</p>
<p>$\bullet$ While geodesics are very geometric and normal coordinates are very practical, methinks the formula is a tad ungeometric. <em>What I'm hoping/asking for is a coordinate-independent formula for the second derivative of $g$ in terms of a suitable "reference connection".</em></p>
http://mathoverflow.net/q/1634555Structure of an intersection of $L^p$-spacesGoulifethttp://mathoverflow.net/users/392612014-04-15T14:45:26Z2015-01-30T08:42:43Z
<p>In what follows, $L^p$ denotes the space of functions from $\mathbb{R}$ to $\mathbb{R}$ such that $\int_{\mathbb{R}} |f(x)|^p\mathrm{d}x < \infty$.
I am interested to understand the structure we can put on intersection of $L^p$-spaces. </p>
<p>For $I$ an interval of $[1,+\infty]$, we define $$L^I = \bigcap_{p\in I} L^p.$$
Obviously, because $L^a \cap L^b \subset L^c$ for $1\leq a\leq c\leq b \leq \infty$, we have that $L^{[a,b]} = L^a \cap L^b$. Thus, $L^I$ is a Banach space for the norm $\lVert \cdot \lVert_a + \lVert \cdot \lVert_b$.</p>
<p><strong>Questions:</strong> What can we say of the structure of $L^I$ if $I$ is an open (or semi-open) interval of $[1,\infty]$? What is the more natural topology on it that make it complete? Are these spaces studied somewhere, for instance as a subpart of a more general theory?</p>
<p>Extensions:</p>
<ul>
<li><p>I am also interested by spaces of the form $$L^p_+ = \bigcup_{\epsilon >0} \bigcap_{0<r<\epsilon} L^{r+p},$$ with similar questions as previously. </p></li>
<li><p>What happens if $I \subset (0,\infty]$, knowing that $L^p$-spaces for $p<1$ are quasi-Banach spaces?</p></li>
</ul>
<p>Thank you by advance for attention.</p>
http://mathoverflow.net/q/13118577Philosophy behind Yitang Zhang's work on the Twin Primes Conjecturepagemanhttp://mathoverflow.net/users/71262013-05-20T03:24:56Z2015-01-30T05:06:51Z
<p>Yitang Zhang recently published a <a href="https://simonsfoundation.org/features/science-news/unheralded-mathematician-bridges-the-prime-gap/">new attack on the Twin Primes Conjecture</a>. Quoting Andre Granville :</p>
<blockquote>
<p>“The big experts in the field had
already tried to make this approach
work,” Granville said. “He’s not a
known expert, but he succeeded where
all the experts had failed.”</p>
</blockquote>
<p>What new approach did Yitang Zhang try & what did the experts miss in the first place?</p>
http://mathoverflow.net/q/1276390How would I apply Wick's theorem to the time-ordered product of three fields?AnonymousPersonhttp://mathoverflow.net/users/331292013-04-15T16:21:10Z2015-01-30T10:36:29Z
<p>I think I know how to apply Wick's theorem in order to expand the time-ordered product of quantum fields, but I just want to verify my understanding. Could someone perform it for the arbitrary product:</p>
<p>$$\mathcal{T}[\phi(x_1)\phi(x_2)\phi(x_3)]$$</p>
<p>Thank you for any clarification that you might provide.</p>
http://mathoverflow.net/q/1242125zero-dimensional completely regular space with $\sigma$-complete clopen algebraFred Dashiellhttp://mathoverflow.net/users/203002013-03-11T08:50:28Z2015-01-30T10:32:44Z
<p>Suppose $X$ is a zero-dimensional completely regular space (clopen sets form a base) such that the Boolean algebra of clopen sets is a $\sigma$-complete Boolean algebra. Must $X$ be basically disconnected? That is, must every cozero set in $X$ have open closure?</p>
<p>I thought this was known, but I can't find it. The point is to show that such $X$ is strongly zero-dimensional, i.e., $\beta(X)$ is zero-dimensional, so that every cozero set is a countable union of clopens.</p>
<p>It should be noted that the analogous statement for extremally disconnected spaces is true: A zero-dimensional space is extremally disconnected (open sets have open closures) if and only if the Boolean algebra of all its clopen sets is a complete Boolean algebra. The proof is an easy exercise. </p>
http://mathoverflow.net/q/1162927From Shortest Paths to Manifold StructureAerykhttp://mathoverflow.net/users/34002012-12-13T15:53:18Z2015-01-30T09:30:42Z
<p>I'm relatively green in the differential geometry area, so my apologies if what I'm asking is ill-posed and/or not research-level.</p>
<p>I have a situation where I know the shortest path between any two points in the plane. Is there a way to reconstruct a corresponding 2D-manifold such that the shortest path between points on the manifold with respect to the Euclidean metric projects to the given shortest path on the plane? I'm only interested (for the moment) in the nicest cases (i.e. dense, locally compact, analytic, etc.).</p>
<p>I can only think of the answer in the trivial example: if the shortest path is always a straight line, then the corresponding manifold is just the plane. But what is the calculation that shows this must be true?</p>
<p>And what if the shortest path between two points is given by the unique exponential curve $y=C_1+C_2e^x$ through those points? What if it's the unique monic parabola through the points?</p>
http://mathoverflow.net/q/1145303How to implement linear constraints that include several absolute valuesHugohttp://mathoverflow.net/users/294302012-11-26T14:41:00Z2015-01-30T09:02:17Z
<p>Dear all,</p>
<p>I am trying to implement a linear constraint that includes several absolute values in the form: Abs(A) + Abs(B) + Abs(C) + Abs(D) + ... = 1</p>
<p>Since the minimization problem includes quite a lot of variables (~100) it is not feasible to implement a linear constraint for each potential +/- combination. Currently I am using ALGLIB with the MINBLEIC Function. Hence, I think it is also not possible to use additional 0/1 indicator variables (i) and to estimate sth. like (2*i-1)*A.</p>
<p>Every help is very much appreciated!
Hugo</p>