most recent 30 from http://mathoverflow.net 2013-06-19T00:02:14Z http://mathoverflow.net/feeds/user/21399 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130195/alternating-sum-with-barnes-g-functions JM Landsberg 2013-05-09T19:49:47Z 2013-05-09T19:54:55Z

<p>Let $B(n)=(n-2)!(n-3)!\cdots 1!$ denote the Barnes G-function. I am pretty sure that $$ \sum_{m=0}^{k^2-1} (-1)^m\binom{k^2-1}m \frac{G(k+n-m+1)}{G(n-m+1)G(k+1)(k^2)!} = n-2k^2-2k $$ when $k$ is odd and is $n-\frac 12(k^2-1)$ when $k$ is even, but I lack a proof. I'd already be happy for some references for identities regarding the Barnes function, as this is all new to me. </p>

http://mathoverflow.net/questions/128108/a-hypersurface-in-the-grassmannian-of-endomorphisms JM Landsberg 2013-04-19T17:06:14Z 2013-04-19T17:06:14Z

<p>Let $V$ be complex vector space of dimension $n$, let ${\frak s\frak l }(V)$ denote the traceless endomorphisms, and let $G(k,W)$ denote the Grassmannian. Let $E\in G(2q,{\frak s\frak l} (V))$ and take a basis $X_1,...,X_{2q}$ of $E$. The polynomial $E\mapsto det_{2qn}([X_i,X_j])$, where $[X_i,X_j]$ is the commutator and the size $2qn$ matrix consists of size $n$ blocks indexed by pairs $(i,j)$, is well defined up to scale. I would like a geometric interpretation of the hypersuface it determines in $G(2q,{\frak s\frak l} (V))$, ideally in terms of representation theory. (I only know it is well defined indirectly, because it comes from a generalization of Strassen's equations on tensors.)</p>

http://mathoverflow.net/questions/114854/equations-of-the-secant-variety/115640#115640 JM Landsberg 2012-12-06T20:34:20Z 2012-12-06T20:34:20Z

<p>If $I(X)$, the ideal of $X$ is empty in degrees less than $d$, then there can be no equations of the secant variety until degree $d+1$, and the ideal in degree $d+1$ consists of all polynomials $P$ such that all partials of $P$ are in $I_d(X)$. There is a similar description for the ideal of the secant variety in higher degrees which I call "multi-prolongation", but one does not know when one has generators for the ideal by this method, and it becomes very difficult to compute. In your case, since you have a hypersurface the termination problem does not arise - your cubic is the unique cubic all of whose partial derivatives are in the ideal of $X$.</p>

http://mathoverflow.net/questions/115639/fibers-of-the-secant-map JM Landsberg 2012-12-06T20:26:05Z 2012-12-06T20:26:05Z

<p>Let $X\subset CP^N$ be a homogeneous (or perhaps just smooth) complex subvariety and let $S^r(X)$ denote its abstract $r$-th secant variety (the incidence variety in $X\times \cdots \times X\times CP^N$), and assume the map to the actual secant variety in $CP^N$ (the projection onto the last factor) is generically finite to one. Is it possible that there is some point in the actual secant variety whose fiber has a component that is an isolated point and another component that is positive dimensional? (I just care about the case where $X$ is a triple Segre product if that makes any difference.) </p>

http://mathoverflow.net/questions/88573/alternating-sum-of-binomial-coefficients JM Landsberg 2012-02-15T22:52:37Z 2012-03-21T02:00:11Z

<p>I would like to know a closed formula for $\sum_{j=0}^{p-n } (-1)^j\binom{n^2}{p-n-j} \binom{n+j-1}j\binom{2n+j}{n+j+1}$, especially in the case $p$ is near $n^2/2$. Similarly, I would like a closed formula for: setting $q=2\cdot\lceil\frac{n(n+1)}{4}\rceil -1$, and setting $p=\lceil\frac{q}{2}\rceil-1$, what is the sum $ \sum_{j=0}^{p-n } (-1)^j\binom{q}{p-n-j} \binom{n+j-1}j\binom{2n+j}{n+j+1} $? </p> <p>In either case I would be happy for an estimate of the growth of the sum (divided by $\binom {n^2-1}p$ in the first case, and divided by $\binom{q-1}p$ in the second).</p>

http://mathoverflow.net/questions/88435/distribution-of-young-diagrams JM Landsberg 2012-02-14T15:16:42Z 2012-02-14T15:16:42Z

<p>Consider $\Lambda^p(C^n\otimes C^n)=\oplus_{\pi}S_{\pi}C^n\otimes S_{\pi'}C^n$ as a $GL_n\times GL_n$-module. This space has dimension $\binom {n^2}p$. I would like any information on the shapes of pairs of diagrams $(\pi,\pi')$ that give the largest contribution to the dimension asymptotically. I am most interested in the case where $p$ is near $n^2/2$. Is there a slowly growing function $f(n)$ such that partitions with fewer than $f(n)$ steps contribute negligibly? If so, can the fastest growing such $f$ be determined?</p>

http://mathoverflow.net/questions/130195/alternating-sum-with-barnes-g-functions JM Landsberg 2013-05-10T22:47:07Z 2013-05-10T22:47:07Z

Sorry, B(n)=G(n) - typo.

http://mathoverflow.net/questions/129857/relating-two-characterizations-of-mathfrak-sl-n-2-among-simple-lie-algeb JM Landsberg 2013-05-06T21:48:59Z 2013-05-06T21:48:59Z

I'm not sure this will help, but 1. may be rephrased geometrically as: among all complex simple Lie algebras, sl_n, n&gt;2, is the unique series such that the orbit of a highest weight line X in P(g) is such that the tangent space T_{Id}X is reducible (as a module for the stabilizing parabolic), and 2. may be phrased geometrically as: sl_n is the unique case where there is a G-invariant cubic hypersurface in P(g), as the form in 2. is completely symmetric.

http://mathoverflow.net/questions/128108/a-hypersurface-in-the-grassmannian-of-endomorphisms JM Landsberg 2013-04-22T20:29:53Z 2013-04-22T20:29:53Z

Not a satisfactory one - when q=1 even the rank of the matrix is well defined and beyond the obvious &quot;rank 0=abelian&quot; I lack a good interpretation.

http://mathoverflow.net/questions/88573/alternating-sum-of-binomial-coefficients/91526#91526 JM Landsberg 2012-03-20T19:33:13Z 2012-03-20T19:33:13Z

I checked on a better computer, this diverges from the sum when n&gt;70 unfortunately.

http://mathoverflow.net/questions/88573/alternating-sum-of-binomial-coefficients/91526#91526 JM Landsberg 2012-03-20T16:13:38Z 2012-03-20T16:13:38Z

I can't prove this, but it works for as large as my computer can handle. A proof, or estimates of the asymptotic behavior (again, as p~ n^2/2) and n goes to infinity) would be most welcome. Thanks!