MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M = GL(n,\mathbb{C})$ be the set of non-degenerate bilinear forms on $\mathbb{C}^n$ (not necessarily symmetric). The general linear group $G = GL(n,\mathbb{C})$ acts on $M$ in the usual way $$G\times M \rightarrow M,~~~~~~(A,X) \mapsto AXA^T.$$ What is the equivariant cohomology $H^*_G(M)$?

share|cite|improve this question
I mildly object to denoting the set of non-degenerate bilinear forms $GL(m,\mathbb C)$. :-/ – Mariano Suárez-Alvarez Jul 5 '13 at 22:03

I'll assume that you mean the Borel cohomology $H^*(EG\times_GM)$. If so, the answer is $$ \mathbb{Z}[\![c_1,\dotsc,c_n]\!]\otimes\Lambda^*(a_1,\dotsc,a_n), $$ where $|c_k|=2k$ and $|a_k|=2k-1$. To see this, first note that you can replace everything by maximal compact subgroups without changing the homotopy type. This means that the relevant space is $EU(n)\times_{U(n)}U(n)^{\text{ad}}$, where $U(n)^{\text{ad}}$ refers to the space $U(n)$ with $U(n)$ acting on it by conjugation. This can be described in a different way as follows. For any complex vector bundle $V$ (with Hermitian inner product) over a space $X$, we can consider the fibre bundle $U(V)$ whose points are pairs $(x,g)$, where $x\in X$ and $g$ is a unitary automorphism of $V_x$. If we take $V$ to be the tautological bundle over $BU(n)$ then $U(V)$ is easily identified with $EU(n)\times_{U(n)}U(n)^{\text{ad}}$. For any $V$ and $X$ it is known that $$ H^*(U(V)) = H^*(X) \otimes\Lambda^*(a_1,\dotsc,a_n), $$
and my claim above is a special case of this. The key ingredient in the proof is Miller's stable splitiing theorem for $U(n)$, in the equivariant form proved by Nitu Kitchloo; some additional details are in my paper "Common subbundles and intersections of divisors".

share|cite|improve this answer
Hi Neil, thanks for your answer. Unfortunately it is not quite what I am looking for. My action is $(A,X) \mapsto AXA^T$ where $A^T$ is the regular transpose, not the conjugate transpose. Restricting to the unitary group, the action is $(A,X) \mapsto A X A^T = A X \bar{A}^{-1}$, where the bar means entry-wise complex conjugation. To illustrate in the case $n=1$, the adjoint action of $U(1)$ on $U(1)$ is trivial, but the action I'm thinking about is by $(a,x) \mapsto axa = a^2x$ which is transitive with stabilizer $\pm 1$. The homotopy quotient is then $BZ_2 = \mathbb{R} P^{\infty}$. – Tom Baird Jul 7 '13 at 19:22

Update: I managed to partially answer this question after posing it. For coefficient fields relatively prime to $n$!, there is an isomorphism $ H_G^*(M) \cong H^*(BLO(n))$ where $BLO(n)$ is the classifying space of the continuous loop group $LO(n)$, while in characteristic 2 there is an isomorphism $ H_G^*(M) \cong H^*(BLU(n))$.

Details are found in Corollary 6.4 of

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.