Non-Abelian Duistermaat-Heckman Measure (not just a reference request) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:54:34Z http://mathoverflow.net/feeds/question/74398 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74398/non-abelian-duistermaat-heckman-measure-not-just-a-reference-request Non-Abelian Duistermaat-Heckman Measure (not just a reference request) Michael 2011-09-02T21:23:45Z 2011-09-04T13:19:08Z <p>Let $V$ be a representation of some torus $T$. It is then well-known that the Duistermaat-Heckman measure for $P(V)$ is the weak limit of the properly rescaled distribution of multiplicities of weights in $\mathrm{Sym}^n(V)$.</p> <p>I've seen many allusions to the fact that the analogous statement is true for general compact Lie groups $K$ (i.e., one pushes further forward to a positive Weyl chamber and compares with the irrep distribution), e.g., in Allen Knutson's reply to <a href="http://mathoverflow.net/questions/70615/example-in-guillemin-sternbergs-convexity-paper" rel="nofollow">my last question</a> and in the appendix of Guillemin-Prato's 1990 paper, but could not find an explicit statement of this in the literature.</p> <p>Do you know whether this statement is true at all, and do you maybe even have a reference?</p> http://mathoverflow.net/questions/74398/non-abelian-duistermaat-heckman-measure-not-just-a-reference-request/74514#74514 Answer by Allen Knutson for Non-Abelian Duistermaat-Heckman Measure (not just a reference request) Allen Knutson 2011-09-04T13:19:08Z 2011-09-04T13:19:08Z <p>Hello again. Yes, it's true.</p> <p>The more general statement you want is, let $X$ be projective with a $K$-equivariant ample line bundle ${\mathcal O}(1)$. For each $n$, let $\mu_n$ be a measure on ${\mathfrak t}^*_+$, $$ \mu_n := \sum_{\lambda \in {\mathfrak t}^*_+} \frac{\dim Hom_K(V_{n\lambda}, \Gamma(X;{\mathcal O}(n))}{n^{\dim X}} \delta_{\lambda}. $$ (Note that $V_{n\lambda}$ only means anything if $n\lambda$ is integral.) Then $\lim_{n\to \infty} \mu_n$ is the nonabelian DH measure for $K$ acting on $X$. </p> <p>One approach to proving this to degenerate $X$ to $X' := (X//N \times G//N) // T$, where $G = K^{\mathbb C}$ and $N$ is a maximal unipotent group. Then the nonabelian DH measure of $X$ is the nonabelian DH measure of $X'$ is the {\em abelian} DH measure of $X//N$. (Here $G//N$ goes by the name "Gel$'$fand variety", $X//N$ by "imploded cross-section", and $X'$ by "Vinberg asymptotic cone".) At that point the theorem has to be checked once and for all for $G//N$.</p> <p>Really, the point is that on both the symplectic and quantum sides one can obtain the nonabelian measure from the abelian, which sounds surprising but is just another way of saying that characters characterize representations. On the quantum (resp. symplectic) side, take the weight multiplicity function (resp. abelian DH measure) and apply differencing operators (resp. directional derivatives) in the directions of positive roots.</p>