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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 $\mathfrak{g}$?

Also, $H^*_G(\overline{N})$ has a natural "inner product" which takes value in the quotient field $S$ of $H^*_G(pt)$, defined via the equivariant integration. It would be nice to know this structure too.

I would be happy if I know the answer for the minimal nilpotent orbit for the simply-laced $\mathfrak{g}$.

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).

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 $\langle \partial/\partial t_r,\partial/\partial t_r\rangle$.

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 $\langle.,.\rangle$.

Is this something known in the literature?

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Perhaps you really mean equivariant Borel-Moore homology? That isn't homotopy invariant, and thus doesn't have to be trivial on an orbit closure... – Ben Webster Oct 31 '11 at 15:18
Hmm, that's the word people (e.g. Nakajima) uses in the papers, indeed. What's the difference? I'm sorry for the ignorance. – Yuji Tachikawa Oct 31 '11 at 15:32
up vote 3 down vote accepted

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 $H^*_G(pt)$).

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 $$ \langle \alpha\cdot 1,\beta\cdot 1\rangle =\alpha\beta\int_{\overline N} 1. $$

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.

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?)

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Ah, thank you. (That shows how little I know about equivariant cohomology... I need to learn the basics.) In principle you can compute $\int_{\overline N}1$ using the result by McGovern, . I haven't done the calculation for non-minimal orbits. I knew the relation to the Uhlenbeck space, but does it give another way to calculate? – Yuji Tachikawa Oct 31 '11 at 15:03
I guess I can re-ask my question by saying "is there natural bundles/sheaves on nilpotent orbits whose characteristic classes or something would naturally correspond to $\partial/\partial t_i$?" As per the rule of MathOverflow, may be I should ask a separate question, or maybe I should rewrite my question. Thank you very much anyway. – Yuji Tachikawa Oct 31 '11 at 15:06
In fact, I don't understand your computation in the minimal case: the integral $\int_{\overline N} 1$ is supposed to be a rational function of degree $-\dim N=-2r$ whereas your function (if I understand the notation correctly) has degree $-2h^{\vee}$. What am I missing? – Alexander Braverman Oct 31 '11 at 15:19
The minimal nilpotent orbit has complex dimension $2h^\vee-2$, so it gies a function of degree $2-2h^\vee$. On the function side, note that the inner product on $\mathfrak{h}$ has degree $-2$, because $\langle v,w\rangle$ for $v,w\in\mathfrak{h}$ is of degree zero. So $\langle \partial/\partial t_r,\partial/\partial t_r\rangle$ is of degree $2-2h^\vee$. – Yuji Tachikawa Oct 31 '11 at 15:35
Apparently $\partial/\partial t_r$ is called the primitive vector, see . It's denoted by $\partial/\partial P_\ell$ there; $\langle \rangle$ is denoted by $I(.,.)$. – Yuji Tachikawa Oct 31 '11 at 15:39

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