Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2015-07-05T19:51:45Zhttp://mathoverflow.net/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://mathoverflow.net/q/2108722Distribution of trivial subset sumsIgor Rivinhttp://mathoverflow.net/users/111422015-07-05T18:34:10Z2015-07-05T18:34:10Z
<p>Suppose I have a set $S$ of $n$ integers picked uniformly at random from $[-L, L].$ Let $z(S)$ be the number of subsets of $S$ which sum to zero. The question is: what is the distributon of the variable $z(S)$ (I am interested in $n$ fixed and $L$ growing)?</p>
http://mathoverflow.net/q/210871-6Must a proof of the asymptotic Goldbach conjecture be effective to imply GRH?Sylvain JULIENhttp://mathoverflow.net/users/136252015-07-05T18:14:02Z2015-07-05T18:14:02Z
<p>It was shown by Hardy and Littlewood that GRH (i.e. the Generalized Riemann Hypothesis for Dirichlet L-functions) implies that every large enough odd number is the sum of three primes. Later on (circa 1937), Vinogradov managed to remove the assumption of GRH. Much more reccently (1997), Deshouillers et al. showed that GRH implies that every odd number greater than 6 is the sum of three primes, and, once again, the assumption of GRH was removed (by Helfgott in 2013). One can therefore wonder whether the asymptotic binary Goldbach conjecture itself implies GRH (since the binary Goldbach conjecture implies Helfgott's result) or not, and whether the proof has to be effective (therefore give an explicit value of the number $x_0$ such that every even integer greater than $x_0$ is the sum of two primes) to do so. </p>
<p>Hence my question: must a proof of the asymptotic Goldbach conjecture be effective to imply GRH?</p>
<p>Thanks in advance.</p>
http://mathoverflow.net/q/210870-4Proof of probabilistic model of "normal" [on hold]snarkhttp://mathoverflow.net/users/757322015-07-05T18:12:49Z2015-07-05T18:12:49Z
<p>In <em>A New Look at Anomaly Detection</em> there is a claim for the proof of probabilistic definition of normal is as follows,</p>
<p>a guess of the probability for event i is $\pi_i$, the true probability is $p_i$ then "if we make the average value of $-\log \pi_i$ as small as possible then we can prove the estimated probabilities are as close as possible to the underlying $p_i$" (direct quote)</p>
<p>in particular that
$\max_{\pi} \sum_i{p_i\log{\pi_i}}=\sum_i{p_i\log{p_i}}$</p>
<p>I don't follow the logic of why this is.</p>
http://mathoverflow.net/q/2108674Which journals publish short communications?Sergei Akbarovhttp://mathoverflow.net/users/189432015-07-05T17:07:13Z2015-07-05T18:04:23Z
<p>Perhaps, somebody asked this already, excuse me in this case.</p>
<p>Can anybody advise mathematical journals that publish short communications? (I mean little papers without proofs.)
It sometimes happens that a proof is so long that it takes years to review and few journals are willing to accept it. I think it would be useful to be able to announce the result in a short communication (and to post it in arXiv).</p>
<p>I am working in Functional analysis and in Geometry.</p>
http://mathoverflow.net/q/2108661Semigroup solution via Lumer-Phillipsuser75730http://mathoverflow.net/users/02015-07-05T16:33:56Z2015-07-05T16:33:56Z
<p>Let $a$ be a coercive, bounded bilinear form on $H^1(\Omega)$, where $\Omega$ is some sufficiently "nice" region. I defined an operator $A:H^1(\Omega)\mapsto H^1(\Omega)^*$ by:
$$
(Au)[v]=a(u,v)\,\,\,\,\,\,\forall u,v\in H^1(\Omega)
$$</p>
<p>I would like a mild solution for the equation
\begin{cases}
u'(t)=Au(t)+f(t)\\
u(0)=u_0
\end{cases}
where $f\in L^2([0,T],H^1(\Omega)^*)$. My idea was to embed $H^1(\Omega)$ in $H^1(\Omega)^*$ via the $H^1(\Omega)$ scalar product and then to define a scalar product on $H^1(\Omega)^*$ via the Riesz isomorphism $J:H^1(\Omega)\mapsto H^1(\Omega)^*$:
$$
(f,g)_{H^1(\Omega)^*}:=(J^{-1}(f),J^{-1}(g))_{H^1(\Omega)}.
$$
I was then able to verify the Lumer-Phillips conditions for the operator $A$, where the underlying Hilbert space is $H^1(\Omega)^*$. This gives me the mild solution:
$$
u(t)=e^{At}u_0+\int_0^t e^{A(t-s)}f(s)\,ds.
$$
My question is, is there a better way of doing this? I am not so sure that it is a good idea to identify $H^1(\Omega)$ with $H^1(\Omega)^*$.</p>
<p>Thanks</p>
http://mathoverflow.net/q/2108650void probability for Poisson point processHosseinhttp://mathoverflow.net/users/741562015-07-05T16:31:25Z2015-07-05T16:31:25Z
<p>Assume we have a Poisson point process (PPP) on 2D space with density $\lambda$. Let $d_i$ be the distance of each node respect to the origin. Assume that we mark each point $i$, independent of other points, with black color with probability $p(d)$, where $p(d)$ is a positive strictly decreasing function of $d$. Consider a disk with radius $r$ centered at the origin. What is the probability of having no black points inside the disk? </p>
http://mathoverflow.net/q/210863-1Distance between point and plane [on hold]Pilpelhttp://mathoverflow.net/users/757292015-07-05T16:19:31Z2015-07-05T16:19:31Z
<p>So according to <a href="http://stackoverflow.com/questions/3860206/signed-distance-between-plane-and-point">this</a>, the signed distance between a point will be the dot product of the plane's normal vector (does it have to be a unit vector?) and the point-in-plane minus the point vector. </p>
<p>I searched everywhere and I can't find a good explanation on <em>why</em> does the dot product give the correct answer. I even studied a little bit more about the dot product itself and I came to know that the dot product of <code>a * b</code> is like multiplying the magnitudes of the vectors that go <a href="https://www.youtube.com/watch?v=KDHuWxy53uM" rel="nofollow">on the same direction</a>. This still doesn't help me understand my problem.</p>
<p>If it matters, I encountered this problem as a programmer.</p>
http://mathoverflow.net/q/2108600Construction of a path of quadratic variationkennethhttp://mathoverflow.net/users/56562015-07-05T14:52:44Z2015-07-05T14:52:44Z
<p>This question has been posted to Stack Exchange earlier, and no answer is available yet.</p>
<p>Consider a path $x: [0,1] \to \mathbb R$. it's $p$-variation on an interval
is defined by
$$V_{p}(x, [a, b]) = \lim_{|\Pi| \to 0} \sum_{i=1}^{n}|x(t_{i}) - x(t_{i-1})|^{p}$$
where $\Pi = \{a= t_{0}< t_{1}, \ldots, < t_{n} =b\}$ is a partition and $|\Pi| =
\max_{i} |t_{i} - t_{i-1}|$ is its mesh size. It is well known that
a Brownian motion is almost surely quadratic variation path in any subinterval of $[0,1]$, i.e. $V_{2}(B, [a, b]) \in (0,1)$ almost surely for arbitrary $0\le a < b \le 1$.</p>
<p>(Q). Is there any simple and explicit construction of a single path, which has
finite positive quadratic variation in any subinterval of $[0,1]$? </p>
http://mathoverflow.net/q/2108590schatten 1-norm of rank $k$ matrixuser58955http://mathoverflow.net/users/486092015-07-05T14:52:30Z2015-07-05T14:59:57Z
<p>I am looking for a high-probability lower bound for the following rank-$k$ matrix
$$
X = u_1 v_1^T + u_2 v_2^T + \cdots + u_k v_k^T,
$$
where $u_1,\dots,u_k,v_1,\dots,v_k$ are independent $N(0,I_n)$ vectors (standard multidimensional Gaussian).</p>
<p>I think the Schatten 1-norm of $X$ should be concentrated around $kn$ because the form is already close to singular value decomposition (if normalised by a factor of $1/n$) but I am not sure how to argue that formally. Perhaps one can use min-max theorem and use the fact that the inner product of two random gaussian vectors have magnitude $1/\sqrt{n}$. Is there a better way?</p>
http://mathoverflow.net/q/2108584Geometric generic fibrepotentially densehttp://mathoverflow.net/users/756162015-07-05T14:34:39Z2015-07-05T19:19:32Z
<p>This is a pretty elementary question about schemes, but it came up in the course of research, so let's try it here rather than MSE.</p>
<blockquote>
<p><strong>Question:</strong> Are the fibres of a family of complex varieties isomorphic <em>as schemes</em> to the geometric generic fibre, outside of a union of countably many subfamilies?</p>
</blockquote>
<p>That seems outlandish, so let me explain below the reasoning that led me to ask. If my argument is flawed I would be glad to know why; if, by contrast, this is known, I would be glad of a reference. </p>
<hr>
<p>I fix the following simple setup for concreteness. </p>
<p>Let $f: X \rightarrow \mathbf A^1$ be a family of varieties over $\mathbf C$. </p>
<p>Let $k = \mathbf C(t)$, the function field of the base, and $K=\overline{k}$. Let $G$ be the geometric generic fibre of $f$, that is, the $K$-variety$ X \times_{\mathbf A^1} \operatorname{Spec K}$. </p>
<p>Now $K$ is algebraically closed, has characteristic zero, and has the same cardinality as $\mathbf C$. So there is a field isomorphism $\alpha: \mathbf C \simeq K$. (As I understand it, this depends on the axiom of choice, but that's alright.) So base change by $\alpha$ turns $G$ into a variety $G_\alpha$ over $\mathbf C$, isomorphic to $G$ as a scheme. (OK so far?)</p>
<p>Now suppose for concreteness that $f$ is a family of hypersurfaces in projective space $\mathbf P^n$, so it is given by a form $F(x_0,\ldots,x_n;t)$ where the $x_i$ are coordinates coordinates on $\mathbf P^n$ and $t$ is the coordinate on $\mathbf A^1$. </p>
<p>Now pick any number $z \in \mathbf C$ which is algebraically independent from all the coefficients of $F$. Then we can choose our field isomorphism so that $\alpha^{-1}$ fixes all the coefficients of $F$, and $\alpha^{-1}(t)=z$. Then base change by $\alpha$ just has the effect of substituting $z$ in place of $t$ in the form $F$: in other words, $G \simeq G_\alpha \simeq G_z$, the fibre of $f$ over $z \in \mathbf A^1$.</p>
<hr>
<p>This argument has the disturbing (to me) consequence that all but countably many fibres of $f$ are isomorphic, albeit in a weird way, as schemes. (Of course, it doesn't claim that they are isomorphic as varieties over $\mathbf C$, which would be absurd.) But maybe this just shows that my intuition about scheme isomorphism is lacking. Either way, I would be glad to know!</p>
http://mathoverflow.net/q/2108559Counter examples for strengthening Whitehead's theorem?KotelKanimhttp://mathoverflow.net/users/504092015-07-05T13:23:09Z2015-07-05T19:43:26Z
<p>Let $f:X\to Y$ be a pointed map of pointed connected $n$-dimensional CW complexes. Whitehead's theorem says that if $f_*:\pi_qX\to \pi_qY$ is an isomorphism for $q\le n$ and a surjection for $q=n+1$, then $f$ is a homotopy equivalence (e.g. Theorem (Whitehead) on p.75 of May's "Concise Course in Algebraic Topology"). </p>
<p>I am interested in counter examples to this when you drop the surjectivity condition for $q=n+1$. That is,</p>
<blockquote>
<p><strong>Question:</strong> What examples are there for a map $f:X\to Y$ of pointed connected $n$-dimensional CW complexes that induces isomorphism on $\pi_q$ for $q\le n$, but is not a homotopy equivalence?</p>
</blockquote>
<p>I would also like to know what are the "minimal" examples to this. For example, it seems impossible for $n\le 2$ (the induced map on universal covers is a homology isomorphism by Hurewicz theorem, hence a weak equivalence and thus induces isomorphsim on all higher homotopy groups). I also wonder if there is an example with <em>finite</em> complexes.</p>
http://mathoverflow.net/q/2108548A question of Erdos on entire functionsAshutoshhttp://mathoverflow.net/users/26892015-07-05T12:56:40Z2015-07-05T15:51:51Z
<p>At the end of the following <a href="http://www.renyi.hu/~p_erdos/1964-04.pdf">paper</a>, Erdos asked if there is a family $F$ of entire functions of size continuum such that for every $z \in \mathbb{C}$, $\{f(z) : f \in F\}$ has size less than continuum. He also showed how to construct such a family under CH.</p>
<p>Did someone solve it?</p>
http://mathoverflow.net/q/2108531The representation theory for the fake Heisenberg groups over non-perfect local fieldm07klhttp://mathoverflow.net/users/94012015-07-05T12:55:27Z2015-07-05T12:55:27Z
<p>Let $K$ be a local field of characteristic $p$, where $p$ is a prime number greater than 2. In particular, $(x+y)^p=x^p+y^p$ for $x,y\in K$.</p>
<p>The fake Heisenberg group is defined to be
$$
G=\{\begin{pmatrix}
1&a&b\\
0&1&a^p\\
0&0&1\\
\end{pmatrix}: a,b\in K\}$$
It is a non-abelian, $K$-split connected algebraic group, which is also unipotent with nilpotent length two. Consider its Lie ring
$$
\mathfrak{g}=\text{Log}(G)=\{\begin{pmatrix}
0&x&y\\
0&0&x^p\\
0&0&0\\
\end{pmatrix}: x,y\in K\}$$
Since $K$ is not perfect, $\mathfrak{g}$ is not stable under scalar multiplication. In particular, $\mathfrak{g}$ is not a vector space.</p>
<p>Q.1: Do we know the unitary dual $\hat{G}$ of $G$?
Notice that $G$ is a central non-split extension of the additive group $K$: $$0\rightarrow K\rightarrow G \rightarrow K \rightarrow 0.$$
I can image to apply the Machey Machine to this group extension similar to the real Heisenberg groups.</p>
<p>Q.2: Is $G$ CCR? or type I?</p>
<p>Q.3: Let $\hat{\mathfrak{g}}$ be the Pontrjagin dual of the additive group $\mathfrak{g}$ and let $G$ act on $\hat{\mathfrak{g}}$ by the coadjoint action. Are all orbits closed or locally closed in $\hat{\mathfrak{g}}$?</p>
<p>Dear all, my motivation is about the Kirillov orbit method for unipotent Groups over a local field of positive characteristic (see Corollary 7.3 and Exapmple 8.3 in <a href="http://arxiv.org/pdf/1107.5486v1.pdf" rel="nofollow">A general Kirillov Theory for locally compact nilpotent groups</a>). It describes the unitary dual of $G$ by its coadjoint orbits. In order to do this, we need to know Q.3 has an affirmative answer according to Corollary 7.3 and Exapmple 8.3 in op. cit. </p>
<p>So if the associated Lie ring $\mathfrak{g}$ is a vector space (this is always the case if char(K)=0), then we can identify its Pontrygagin dual with its linear dual, on which the unipotent group $G$ acts algebraiclly by the coadjoint action. Since unipotent Groups act algebraiclly on an affine variety, all its orbits are closed. Then the orbit method (Corollary 7.3 in op. cit.) implies that the group $G$ is CCR and the Kirillov-orbit map is a homeomorphism. Hence, we have affirmative answers for all my questions.</p>
<p>For unipotent Groups over a local field $K$ with char(K)=p. I find the above obstruction. However, if we start with a Lie algebra $\mathfrak{g}$ (in particular, a vector space) with nilpotent length less that $p$ and consider the group $G=\exp(\mathfrak{g})$ obtained by exponentiating it, then we have affirmative answers for all my questions by exactly same argyments.</p>
http://mathoverflow.net/q/2108510What are constructions for induced $C_5$-free graphs?domotorphttp://mathoverflow.net/users/9552015-07-05T09:27:54Z2015-07-05T17:18:57Z
<p>During a recent workshop, the question came up whether there are some constructions for graphs that are induced $C_5$-free, but they contain "everything else," so we don't want to forbid $C_5$'s, other induced cycles and so on.
What are some non-trivial constructions for such graphs?
I suppose the question also makes sense for other graphs.</p>
<p>Just to clarify, by construction I mean something very explicit, like the vertices correspond to length $n$ zero-one sequences and we connect two vertices if their inner product is a prime number.</p>
http://mathoverflow.net/q/2108470Worst case difference in rank by column-row swappingTurbohttp://mathoverflow.net/users/100352015-07-05T03:18:21Z2015-07-05T17:13:01Z
<p>Given a matrix $m\in\{-1,+1\}^{n\times n}$. Consider $m^\sigma$ to be collection of all matrices obtained from $m$ by permuting rows and columns.</p>
<p>Consider $\mathscr{M}[m^\sigma]$ to be collection of all $n\times n$ matrices obtained from matrices in $m^\sigma$ by swapping an equal number of rows for an equal number of columns of same indices.</p>
<p>As an example, say you pick row/column indices $i$ and $j$. Then you include matrix where you first replace $i$th row with transpose of $i$th column and vice versa followed by similar operation on $j$th row and column.</p>
<p>What is worst case difference between least rank and largest rank of any matrix in $\mathscr{M}[m^\sigma]$?</p>
<p>Can we say anything about their ratios (such as bound based on some intrinsic property of the matrix)?</p>
http://mathoverflow.net/q/2108164Is the $\infty$-category of presentable $\infty$-categories presentable?Zippyhttp://mathoverflow.net/users/757072015-07-04T15:11:50Z2015-07-05T19:22:31Z
<p>Let $\mathit{Pr}^L$ be the $\infty$-category of presentable $\infty$-categories and continuous functors in some universe. Is it presentable itself a larger universe?</p>
http://mathoverflow.net/q/2107713Non-Forking and Related Conceptskav11http://mathoverflow.net/users/756852015-07-03T19:34:28Z2015-07-05T19:10:25Z
<p>Is the importance of developing forking machinery in the way we set it up, or is it in the fact that it allows us to come up with a notion of independence via the properties of non-forking? I'm currently reading Baldwin's Fundamentals of Stability Theory and would like to know if I need to have a very deep understanding of how to set up the machinery or whether it is enough to have a rudimentary understanding so that I can come back to the details in a second reading.</p>
<p>Also, are there any expository papers detailing the use of non-forking from an at least somewhat historical viewpoint? I'm thinking of something in the spirit of Baldwin's first chapter.</p>
<p>Thank you.</p>
http://mathoverflow.net/q/2107685Upper bound for number of prime numbers in a rangeAlexeyhttp://mathoverflow.net/users/313562015-07-03T18:46:47Z2015-07-05T19:04:44Z
<p>Theorem 3.2 in <a href="http://arxiv.org/pdf/1405.2593.pdf">http://arxiv.org/pdf/1405.2593.pdf</a> shows that for any $x$ there are $\gg x\exp(-\sqrt{\log x})$ integers $x_0 \in [x; 2x]$ such that $\pi(x_0 + \log x) - \pi(x_o) \gg \log\log x$. </p>
<p>Is there an upper bound for number of such $x_0$? I think it must be $<x(\log x)^{-c}$ for any $C$.</p>
http://mathoverflow.net/q/2106582Volume of arithmetic quotients of symmetric spacesdulal naru gopalhttp://mathoverflow.net/users/692892015-07-02T09:16:05Z2015-07-05T16:28:22Z
<p>Now let $\textbf{G}$ be some connected semisimple linear algebraic group over a number field $F$. Let $G_{\infty}$ be $\textbf{G}(\mathbb{R}\otimes_{\mathbb{Q}} F)$. Let $K_{\infty}$ be a maximal compact subgroup of $G_{\infty}$. We fix an embedding $G\hookrightarrow GL_N$ for some $N$. Let $\mathfrak{P}$ denote a prime of $\mathcal{O}_F$ lying over $p$. Let $G(\mathfrak{P})$ be the intersection of $G_{\infty}$ with the congruence subgroup of $GL_N(\mathcal{O}_F)$ at level $\mathfrak{P}$. Let $\Gamma$ be an arithmetic lattice of $G_{\infty}$.<br>
Let us denote<br>
$\Gamma (\mathfrak{P}):=\Gamma \cap G(\mathfrak{P})$.
Let us denote by $e$ and $f$ the ramification and inertia degrees of $\mathfrak{P}$ in $F$. Hence we get that $[F_{\mathfrak{P}}:\mathbb{Q}_p]=ef$ where $F_{\mathfrak{P}}$ is the completion of $F$ at $\mathfrak{P}$. We define $G_k:=G \cap (1+p^kM_{efN}(\mathbb{Z}_p))$ where $G:=\varprojlim_k\Gamma /\Gamma({\mathfrak{P}}^k)$
For $k\geqslant 0$, we define $Y_k:=\Gamma (\mathfrak{P}^{ek})\backslash G_{\infty} /K_{\infty}$. </p>
<p>Then let $vol(\mathfrak{P}^{ek})$ be the volume of $\Gamma(\mathfrak{P}^{ek})\backslash G_{\infty}$. </p>
<p>Let $\Gamma(\mathfrak{P}^{ek})\backslash G_{\infty}$ be compact. </p>
<p>Then I want to show that $vol(\mathfrak{P}^{ek})\sim [G:G_k]$. Here $\sim$ denotes that both of them are similar order as $k \rightarrow \infty$. It will be helpful if I can get some hints or some reference on this fact. Thank you in advance.</p>
http://mathoverflow.net/q/2106552Subsets of $\mathbb{N}$ whose lower density respects complementsDominic van der Zypenhttp://mathoverflow.net/users/86282015-07-02T08:45:20Z2015-07-05T17:52:53Z
<p>The lower density of $A\subseteq\mathbb{N}$ is defined to be $\lambda(A)=\lim\text{inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$. We set $${\cal C} = \{A\subseteq \mathbb{N}: \lambda(\mathbb{N}\setminus A) = 1 - \lambda(A)\}.$$</p>
<p>Do both ${\cal C}$ and ${\cal P}(\mathbb{N})\setminus {\cal C}$ have cardinality $2^{\aleph_0}$?</p>
http://mathoverflow.net/q/21065310Algorithms for calculating R(5,5) and R(6,6)Emile Okadahttp://mathoverflow.net/users/202002015-07-02T07:57:35Z2015-07-05T12:29:56Z
<p>Calculating the <a href="https://en.wikipedia.org/wiki/Ramsey%27s_theorem#Ramsey_numbers">Ramsey numbers</a> R(5,5) and R(6,6) is a notoriously difficult problem. Indeed ErdÅ‘s once said:</p>
<blockquote>
<p>Suppose aliens invade the earth and threaten to obliterate it in a year's time unless human beings can find the Ramsey number for red five and blue five. We could marshal the world's best minds and fastest computers, and within a year we could probably calculate the value. If the aliens demanded the Ramsey number for red six and blue six, however, we would have no choice but to launch a preemptive attack. </p>
</blockquote>
<p>I am curious what algorithm we would employ if such a situation were to occur. I know analytic results have been used to put bounds on R(5,5) and R(6,6), but I am mostly interested in the problem from a computational perspective. If we were to set a computer to the task and let it run for however long it might take, what algorithm would we use? How many operations might we expect it to take/what would it's time complexity be?</p>
<p><strong>Edit:</strong> I should clarify that I am seeking the best classical algorithm. It was after reading the paper that Carlo Beenakker cites using quantum annealing that I became interested in finding the best classical alternative.</p>
http://mathoverflow.net/q/2105575What is Jantzen's formula for the determinant of the Shapovalov form in the case of generalized Verma modules?João Penedoneshttp://mathoverflow.net/users/755732015-07-01T08:16:16Z2015-07-05T15:37:30Z
<p>The best reference I found is
<a href="http://www.sciencedirect.com/science/article/pii/0001870879900665" rel="nofollow">[Kac, Kazhdan '79]</a>
which extends the results of Shapovalov and Jantzen to the case of infinite dimensional Lie algebras.
Theorem 1 of this paper gives the Shapovalov formula for the determinant of the bilinear Shapovalov form on the modules of a contragredient Lie algebra.
In the last equation of this paper, the authors give a generalization of the determinant formula for generalized Verma modules based on parabolic subalgebras. This is exactly the result I am looking for. However, the formula given does not seem to be correct because it does not reduce to the standard Shapovalov determinant when we choose the minimal (Borel) parabolic subalgebra. I tried reading the original paper of [Jantzen '77]
but it is written in German. I also looked into the book <a href="http://www.ams.org/bookstore-getitem/item=GSM-94" rel="nofollow">[Representations of Semisimple Lie Algebras in the BGG Category O, Humphreys, 2008]</a>. Jantzen's determinant formula is mentioned in the beginning of section 9.13 but not given explicitly.
It would be great if you could point me to an english reference that contains the correct version of this determinant formula.</p>
<p>Alternatively, I would also be satisfied with a refinement of Jantzen's simplicity criterion, as given in section 9.13 of the book mentioned above. I would like to know what are the submodules of $M_I(\lambda)$ when $M_{I}(\lambda)$ is not simple. I expect $M_I(s_\beta \cdot \lambda)$, at least for one $\beta \in \Psi_\lambda^+$, to be a submodule of $M_I(\lambda)$ when $M_{I}(\lambda)$ is not simple. However, I don't know what is the precise statement.</p>
<p>I am interested in the case of the conformal group in $d>2$ dimensions which is isomorphic to $SO(d+2)$.
The interesting representations for conformal field theory are generalized Verma modules, where the highest weight belongs to an irreducible finite dimensional representation of $SO(d)$ and is an eigenstate of the generator of $SO(2)$ (here I am considering the subgroups $SO(2)\times SO(d)\subset SO(d+2)$) . Then, the other $2d$ generators of $SO(d+2)$ are associated with $d$ positive roots and $d$ negative roots that are used to create the generalized Verma module.</p>
http://mathoverflow.net/q/2104601Subgroup of Projective general linear group on complete discrete valuation ringuser75536http://mathoverflow.net/users/755362015-06-30T10:02:42Z2015-07-05T15:48:50Z
<p>Let $R$ be a complete dvr and $k$ its residue field of positive characteristic.
Let $H$ be a finite subgroup of $PGL_2(k)$ such that the order of $H$ is prime with $char(k)$.</p>
<p>Is there some elementary way to show that $H$ is a subgroup of $PGL_2(R)$?</p>
http://mathoverflow.net/q/2102436Does the proof of Picard's theorem become simpler by increasing the number of points that are not attained?Erfan Salavatihttp://mathoverflow.net/users/512032015-06-26T20:59:27Z2015-07-05T19:22:10Z
<p>Let $f$ be an entire analytic function which attains all but $k$ complex numbers $z_1,\ldots,z_k$. Is there any elementary proof, for some $k$, that $f$ is constant?</p>
http://mathoverflow.net/q/2102183Braid relations $n_\alpha n_\beta n_\alpha \ldots = n_\beta n_\alpha n_\beta \ldots $ in arbitrary reductive groupsNicolas Schmidthttp://mathoverflow.net/users/38242015-06-26T15:02:41Z2015-07-05T16:34:06Z
<p>I'm currently trying to prove or disprove the following claim. First let me set up some notation.</p>
<p>Let $G$ be a connected reductive group over a field $K$, let $S \leq Z \leq N \leq G$ be respectively a maximal split torus, its centralizer and its normalizer. Let $\Phi$ be the root system of the pair $(G,S)$, and for $\alpha \in \Phi$ let $U_\alpha \leq G$ be the associated root subgroup (possibly noncommutative, when $G$ is not split). Given $u \in U_\alpha(K)-\{1\}$, there exist unique elements $m(u) \in N(K)$ and $v,v' \in U_{-\alpha}(K)$ such that $vuv' = m(u)$. The image of the element $m(u)$ in the Weyl group $W_0 = N(K)/Z(K)$ is the reflection $s_\alpha$ associated to the root $\alpha$, where $W_0$ is naturally identified with the Weyl group of $\Phi$ via the action of $N(K)$ on $X^\ast(S)$.</p>
<p><strong>Now for the claim</strong>: For any two roots $\alpha, \beta \in \Phi$ with $\alpha,\beta$ not parallel (an obvious necessary condition), and <strong>any</strong> $u \in U_\alpha(K)-\{1\}$ and $v \in U_\beta(K)-\{1\}$ the braid relation</p>
<p>$n_\alpha n_\beta n_\alpha \ldots = n_\beta n_\alpha n_\beta \ldots$</p>
<p>in $N(K)$ holds true, where the number of factors on both sides equal the order of $s_\alpha s_\beta$, and $n_\alpha = m(u)$ and $n_\beta = m(v)$.</p>
<p>I think I have a proof of this in the case when $\Phi$ is reduced. So a counterexample should involve non-reduced root systems. I tried to disprove the claim using the example $G = SU(h)$ of a quasi-split unitary group (example 1.15 in Tits Corvallis article), but unfortunately (or fortunately) the claim seems to hold in this case. (<strong>edit</strong>: Even for unitary groups over division algebras the claim seems to hold true.)</p>
<p>So if you either have a counterexample to the above claim or a proof (with or without reference), it would help me a lot.</p>
http://mathoverflow.net/q/2088677Eliminating Gibb's phenomenon, and approximating with jumping functions in Fourier Analysis : An attempt and a question in this regardRajesh Dhttp://mathoverflow.net/users/144142015-06-09T18:44:21Z2015-07-05T12:41:21Z
<blockquote>
<p>Physical Motivation : Hear to these audio files <a href="https://soundcloud.com/rajesh-d-1/s_f-using-spectrum-in-0-4khz/s-VkzHY" rel="nofollow">S_f</a> and <a href="https://soundcloud.com/rajesh-d-1/p_f-constructed-using-spectrum-in-04khz/s-NfyFH" rel="nofollow">P_f</a>. S_f is Fourier partial
sum and P_f is the new reconstruction, both use spectrum only in the
region (0,4KHz) for reconstructing the signal. Both files are at sample rate of 64kHz.</p>
<p>This problem seems like a nightmare to me. I tried to expand
$K_{\omega}^f(t)$, but I am clueless of getting some kind of a closed
form or some kernel like structure. If I try to take derivative, its
even worse. I desperately need some clues on line of attack I should
do. Please help me in this regard. Thanks.</p>
</blockquote>
<h2>Main Question</h2>
<p>(this section is fully self contained and does not need anything that comes further, for answering the question).</p>
<p>Let $f:\mathbb{R}\to\mathbb{R}$, $f(t) = 0, t<0$. Let $f \in L^2(\mathbb{R})$ and is locally BV. Let its jump set be $\{x_i\},i=1,2,3,...,x_i>0$, and the corresponding jump amounts set be $\{D_i\}$.</p>
<p>Let $F(\omega)$ be the Fourier transform of $f$. Let $$x_t(\omega) = \int_0^{\omega}R(\Omega)\cos(\Omega t + \Phi(\Omega))d\Omega$$ and $$y_t(\omega) = \int_0^{\omega}R(\Omega)\sin(\Omega t + \Phi(\Omega))d\Omega,$$ defined only for $\omega
\ge 0$, where $R(\omega) = |F(\omega)|$ and $\Phi(\omega) = \angle F(\omega)$.</p>
<p>Consider the curve $\alpha_t(\omega) \equiv (x_t(\omega),y_t(\omega)) $ and let $(X_t(s),Y_t(s))$ be its arc length parametrization, where the transformation is given as $s(\omega) = \int_0^{\omega}R(\Omega)d\Omega$.</p>
<p>We define momemnt of inertia about center of mass, of the curve segment between $0$ and $\omega$ as $$I^f_{\omega}(t) = \int_{0}^{s(\omega)} ((X_t(\rho)-X^t_{cm}(s(\omega)))^2 + (Y_t(\rho)-Y^t_{cm}(s(\omega)))^2) d\rho$$, where $X^t_{cm}(\eta) = \frac{1}{\eta}\int_{0}^{\eta}X_t(\rho)d\rho$ and $Y^t_{cm}(\eta) = \frac{1}{\eta}\int_{0}^{\eta}Y_t(\rho)d\rho$.</p>
<p>Let $$\require{enclose}
\enclose{horizontalstrike}{K_{\omega}^f(t) = \sqrt{\frac{I^f_{\omega}(t)}{s(\omega)^3}}} $$</p>
<p>$${K_{\omega}^f(t) = \frac{1}{\log\omega}\sqrt{\frac{I^f_{\omega}(t)}{s(\omega)}}} $$</p>
<p>(<strong>edit</strong> : if this converges, to make it converge to right value, I just replace $s(\omega)^3$ with $s(\omega){\log^2(\omega)}$, this does not affect convergence as we know $s(\omega) \thicksim O(\log\omega)$ ).</p>
<p><strong>I'd like some help to prove the following statement</strong></p>
<p>Given any open interval $(a,b)\subset(0,\infty)$ and $a,b\notin\{x_i\}$ $$\lim_{\omega\to\infty} V_a^b(K^f_{\omega}) = k \sum_{i/x_i\in (a,b)}|D_i|$$ </p>
<p>where $k$ is some constant independent of the function $f$. Here $V_a^b(f)$ is the variation of the function $f$ in $(a,b)$.</p>
<h2><strong>Note :</strong> What follows this section is my motivation/goal as to why I am interested in this problem. It is not needed to answer the question and please read further only if you are interested/ generally curious.</h2>
<hr>
<h2>Motivation</h2>
<p>Consider a BV function $f$ with jumps, the Fourier partial integral function (analogous to Fourier partial sum in periodic case) as $\omega\to\infty$ does converge to $f$ pointwise except possibly at jumps. But the total variation of the partial integral function does not converge to the total variation of the function $f$, more over it goes to infinity. (attributed to Gibb's phenomenon). To overcome Gibb's phenomenon, alternate summation methods were suggested, like Cesaro summation but their convergence is very slow especially when $f$ has a jump.</p>
<p>In my work, I propose a construction of a sequence of functions, just like partial sums, each one denoted as $P^f_{\omega}$ using only Fourier spectrum in the interval $(0,\omega)$. The function sequence is intended to converge to $f$ pointwise except at jump points, and not just that, but also overcome Gibb's phenomenon, there by variation of $P^f_{\omega}$ in any given open interval, converging to that of the function $f$ as $\omega\to\infty$.</p>
<p>More interesting part is that the function $P^f_{\omega}$ can possess jump discontinuties. (even when $f$ does not have jumps). I also predict that the convergence is much faster than Cesaro partial sums.</p>
<p>Another interesting part is that in Fourier analysis we try to approximate even functions that jump, with smooth functions. But here we use jumping functions to approximate jumping functions. (This problem could be formulated in Fourier series but I choose to do it for Fourier transform setup).</p>
<h2>Goal</h2>
<p>Let $f:\mathbb{R}\to\mathbb{R}$, $f(t) = 0, t<0$. Let $f \in L^2(\mathbb{R})$ and is locally BV. Let its jump set be $\{x_i\},i=1,2,3,...,x_i>0$, and the corresponding jump amounts set be $\{D_i\}$. </p>
<p>We construct a two functions for each $\omega$ denoted as $R^f_{\omega}(t)$ and $K_{\omega}^f(t)$ using the knowledge of Fourier transform only in $(0,\omega)$ (procedure described in later section) with following properties.</p>
<ol>
<li><p>As $\omega\to\infty$, $R^f_{\omega}\to J_f$ pointwise. The function $J_f(t)$ is defined as $$J_f(t) = 0, t \notin\{x_i\}$$ and $$J_f(t) = D_k, t = x_k \in \{x_i\}$$.</p></li>
<li><p>Given any open interval $(a,b)\subset(0,\infty)$ and $a,b\notin\{x_i\}$ $$\lim_{\omega\to\infty} V_a^b(K^f_{\omega}) = \sum_{i/x_i\in (a,b)}|D_i|$$ here $V_a^b(f)$ is the variation of the function $f$ in $(a,b)$.</p></li>
</ol>
<p>Let $S_{\omega}^f(t)$ be the Fourier partial integral of the function $f$. Let $M_{\omega}$ be the set of maxima of the function $K^f_{\omega}$. We construct the function $$L^f_{\omega}(t) = \sum_{y\in M_{\omega}} R^f_{\omega}(y)\delta(t-y)$$ and there by the step function $$\Gamma^f_{\omega}(t) = \int_0^t L^f_{\omega}(\tau)d\tau$$</p>
<p>Let $S_{\omega}^{\Gamma^f_{\omega}}(t)$ be the Fourier partial integral of the step function $\Gamma^f_{\omega}(t)$.</p>
<p>Consider the function $$P^f_{\omega}(t) = S^f_{\omega}(t) - S_{\omega}^{\Gamma^f_{\omega}}(t) + \Gamma^f_{\omega}(t)$$</p>
<p><strong>Two statements can be proven</strong></p>
<ol>
<li><p><strong>As $\omega\to\infty$, $P^f_{\omega}\to f$ pointwise</strong> except at $\{x_i\}$.</p></li>
<li><p><strong>Given any open interval $(a,b)\subset(0,\infty)$ and $a,b\notin\{x_i\}$ , $$\lim_{\omega\to\infty}V_a^b(P^f_{\omega}) = V_a^b(f)$$</strong></p></li>
</ol>
<hr>
<h2>Construction of the function $R^f_{\omega}$</h2>
<p>Let $F(\omega)$ be the Fourier transform of $f$. Let $$x_t(\omega) = \int_0^{\omega}R(\Omega)\cos(\Omega t + \Phi(\Omega))d\Omega$$ and $$y_t(\omega) = \int_0^{\omega}R(\Omega)\sin(\Omega t + \Phi(\Omega))d\Omega,$$ defined only for $\omega
\ge 0$, where $R(\omega) = |F(\omega)|$ and $\Phi(\omega) = \angle F(\omega)$.</p>
<p>Consider the curve $\alpha_t(\omega) \equiv (x_t(\omega),y_t(\omega)) $ and let $(X_t(s),Y_t(s))$ be its arc length parametrization, where the transformation is given as $s(\omega) = \int_0^{\omega}R(\Omega)d\Omega$.</p>
<p>We define moemnt of inertia of the curve segment between $0$ and $\omega$ as $$I^f_{\omega}(t) = \int_{0}^{s(\omega)} ((X_t(\rho)-X^t_{cm}(s(\omega)))^2 + (Y_t(\rho)-Y^t_{cm}(s(\omega)))^2) d\rho$$, where $X^t_{cm}(\eta) = \frac{1}{\eta}\int_{0}^{\eta}X_t(\rho)d\rho$ and $Y^t_{cm}(\eta) = \frac{1}{\eta}\int_{0}^{\eta}Y_t(\rho)d\rho$.</p>
<p>Let $$\require{enclose}
\enclose{horizontalstrike}{K_{\omega}^f(t) = k\sqrt{\frac{I^f_{\omega}(t)}{s(\omega)^3}}} $$ $${K_{\omega}^f(t) = \frac{k}{\log\omega}\sqrt{\frac{I^f_{\omega}(t)}{s(\omega)}}} $$where $k$ is a constant independent of the function $f$. Derivation of its value, I push it to later. </p>
<p>It can be seen that (<a href="http://mathoverflow.net/a/175862/14414">with possible help from here</a>), as $\omega\to \infty$, $K^f_{\omega}\to H_f$ pointwise. The function $H_f(t)$ is defined as $$H_f(t) = 0, t \notin\{x_i\}$$ and $$H_f(t) = |D_k|, t = x_k \in \{x_i\}$$</p>
<p>To get the polarity of the jump we do this, define $$b_{\omega}^f(t) = signum(\frac{d}{dt}(S_{\omega}^f(t))$$ and finally $$R_{\omega}^f(t) = b_{\omega}^f(t)K^f_{\omega}(t) $$</p>
<h2><strong>How question is linked to motivation</strong></h2>
<p>As per my motivation I need to prove that $K_{\omega}^f(t)$ satisfies the property 2 (convergence of variation) as described in the early part of description of the goal. Hence I'd like some help to prove the following statement</p>
<p>Given any open interval $(a,b)\subset(0,\infty)$ and $a,b\notin\{x_i\}$ $$\lim_{\omega\to\infty} V_a^b(K^f_{\omega}) = \sum_{i/x_i\in (a,b)}|D_i|$$ </p>
http://mathoverflow.net/q/2084720Change of grading used in the paper "The diagonal subring and the Cohen-Macaulay property of a multigraded ring" by Eero HyryCusphttp://mathoverflow.net/users/94852015-06-04T21:22:41Z2015-07-05T13:55:01Z
<p>I am currently reading the paper "The diagonal subring and the Cohen-Macaulay property of a multigraded ring" by Eero Hyry. I do not understand the following part in Lemma 1.1. <a href="http://www.ams.org/journals/tran/1999-351-06/S0002-9947-99-02143-1/S0002-9947-99-02143-1.pdf" rel="nofollow">here</a>.</p>
<p>Let $T=\bigoplus_{\underline n\in \mathbb Z^r}T_{\underline n}$ be an $r$-graded ring defined over a local ring. Let $S=\bigoplus_{n_j=0}T_{\underline n}$ and $\mathfrak M$ be maximal homogeneous ideal of $S.$ </p>
<p>I do not understand how $T_{\mathfrak M}$ can be considered as $\mathbb Z$-graded ring defined over the local ring $S_{\mathfrak M}.$</p>
<p>If I consider $S=\bigoplus_{n_i,n_j=0}T_{\underline n}$ and $\mathfrak M$ is maximal homogeneous ideal of $S,$ then also the statement is true? </p>
<p>Please explain the change of grading used in the proof.</p>
http://mathoverflow.net/q/1982264Does the stable category of a nice exact category embed in (the underlying category of) a derivator?Kevin Carlsonhttp://mathoverflow.net/users/430002015-02-23T05:07:38Z2015-07-05T13:41:12Z
<p>In <em>Derivators, Pointed Derivators, and Stable Derivators</em>, Moritz Groth gives as an example of a non-invertible morphism with trivial cone an inclusion $f:X\to I$. Here $X$ is an object of injective dimension $1$ in an exact category $\mathcal{E}$ with enough injectives, $I$ is an injective, and the stable category $\underline{\mathcal{E}}$ has a "suspended structure" given in the same way as the triangulation of the stable category of a Frobenius category but without the invertibility of suspension. </p>
<p>This is fine as far as it goes, but it's not obvious to me that this is an example of a cone <em>in a pointed derivator,</em> what he's just defined, and Groth doesn't give any indication of how to see it as such.</p>
<p><strong>Question:</strong> how can I realize this example as a cone in a derivator?</p>
<p>Having looked around, I don't see any evidence that $\underline{\mathcal{E}}$ is known to be the underlying category of a derivator, even with extra assumptions a la a paper of Stovicek to make $\mathcal{E}$ more like a Grothendieck category. Is it, in fact, such an underlying category, given some extra assumptions? Or can we, perhaps, give it an exact embedding in a homotopy category, and extend to a derivator in that way?</p>
http://mathoverflow.net/q/3247926What are some mathematical sculptures?Gerald Edgarhttp://mathoverflow.net/users/4542010-07-19T12:37:37Z2015-07-05T18:43:55Z
<p>Either intentionally or unintentionally.
Include location and sculptor, if known.</p>
http://mathoverflow.net/q/3040226Parabolic envelope of fireworksJoseph O'Rourkehttp://mathoverflow.net/users/60942010-07-03T12:50:53Z2015-07-05T12:40:15Z
<p>The envelope of parabolic trajectories from a common launch point is itself a parabola.
In the U.S. this weekend many will have a chance to observe this fact directly, as the 4th of July is traditionally celebrated with fireworks.</p>
<p>If the launch point is the origin, and the trajectory starts off at angle $\theta$ and velocity $v$, then under unit gravity it follows the parabola
$$
y = x \tan \theta - [x^2 /(2 v^2)] (1 + \tan^2 \theta)
$$
and the envelope of all such trajectories is another parabola:
$$
y = v^2 /2 - x^2 / (2v^2)
$$
<br>
<img src="http://cs.smith.edu/~orourke/MathOverflow/TrajectoriesEnvelope.jpg" alt="alt text">
<br></p>
<p>These equations are not difficult to derive.
I have two questions.
First, is there a way to see that the envelope of parabolic trajectories is itself a parabola, without computing these equations?
Is there a purely geometric argument?
Perhaps there is a way to nest cones and obtain the above picture through conic sections, but I couldn't see it.</p>
<p>Second, of course the trajectories are actually pieces of ellipses, not parabolas, if we follow the true inverse-square law of gravity.
Is the envelope of these elliptical trajectories also an ellipse?
(I didn't try to work out the equations.)
Perhaps the same geometric viewpoint (if it exists) could apply, e.g. by slightly tilting the sections.</p>
<p>I ask these questions in a weekend recreational spirit.</p>
<p><em>4 July 2015</em>: And touched again on its 5<sup>th</sup> anniversary.</p>