Equivariant cohomology of nilpotent orbits - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:11:29Z http://mathoverflow.net/feeds/question/79624 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79624/equivariant-cohomology-of-nilpotent-orbits Equivariant cohomology of nilpotent orbits Yuji Tachikawa 2011-10-31T13:16:58Z 2011-10-31T14:11:52Z <p>Let $\mathfrak{g}$ be a simple Lie algebra over $\mathbb{C}$, and let $N$ be a nilpotent orbit of $\mathfrak{g}$. What is the equivariant cohomology of its closure, $H^*_G(\overline{N})$, with respect to the group $G$ for $\mathfrak{g}$?</p> <p>Also, <code>$H^*_G(\overline{N})$</code> has a natural "inner product" which takes value in the quotient field $S$ of <code>$H^*_G(pt)$</code>, defined via the equivariant integration. It would be nice to know this structure too.</p> <p>I would be happy if I know the answer for the minimal nilpotent orbit for the simply-laced $\mathfrak{g}$.</p> <hr> <p>Let me give the background to my question. Let $H^*_G(pt)=\mathbb{C}[t_1,\ldots,t_r]$ so that $t_1$ has degree 2, ..., $t_r$ has degree $h^\vee$ ($r$ is the rank of $\mathfrak{g}$. As $\mathrm{Spec} H^*_G(pt)= \mathfrak{h}/W$, the standard flat metric on $\mathfrak{h}$ determines a metric on $\mathfrak{h}/W$, which we denote by $\langle .,. \rangle$. Note that the vector field $\partial/\partial t_r$ is unique up to a scalar multiplication. So, $\langle \partial/\partial t_r,\partial/\partial t_r\rangle$ determines a rational function on $\mathfrak{h}/W$, i.e. an element of $S$ (unique up to a scalar multiplication).</p> <p>Let $N$ be the minimal nilpotent orbit of a simply-laced $\mathfrak{g}$. My collaborators and I calculated $\int_{\overline{N}} 1$. And it equaled $\langle \partial/\partial t_r,\partial/\partial t_r\rangle$.</p> <p>This suggests that $H^*_G(\overline{N})$ has a natural basis corresponding to $\partial/\partial t_i$, and the inner product given by the equivariant integral equals $\langle.,.\rangle$.</p> <p>Is this something known in the literature?</p> http://mathoverflow.net/questions/79624/equivariant-cohomology-of-nilpotent-orbits/79632#79632 Answer by Alexander Braverman for Equivariant cohomology of nilpotent orbits Alexander Braverman 2011-10-31T14:06:51Z 2011-10-31T14:11:52Z <p>First, since $\overline N$ is contractible its equivariant cohomology is the same as for $pt$. The Poincare pairing is uniquely determined by $\int_{\overline N} 1$ (since it is linear with respect to $H^*_G(pt)$). </p> <p>More precisely, any cohomology class of $\overline N$ has the form $\alpha\cdot 1$ where $\alpha$ is an equivariant cohomology class of $pt$ and $1$ denotes the unit cohomology class in $\overline N$ and we have $$\langle \alpha\cdot 1,\beta\cdot 1\rangle =\alpha\beta\int_{\overline N} 1.$$</p> <p>I don't know a good way to compute $\int_{\overline N} 1$ for arbitrary $N$ - other than replacing $\overline N$ by a resolution and using fixed point localization.</p> <p>By the way, if $\overline N$ is the minimal orbit, then ${\mathbb C}^2\times {\overline N}$ is the same as the Uhlenbeck space of $\mathbb C^2$ of second Chern class 1 - that should give you another way to compute that integral (is it obvious that you get the same answer?)</p>