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

In this question we consider only cohomology with rational coefficients. All groups will be connected Lie groups. All group actions will be smooth.

Let $M$ be a manifold. Let $G$ be a group acting on $M$. Let $H$ be a normal Lie subgroup of $G$. What can be said about the relation between $H^\ast_{G/H}(M)$ and $H^\ast_G(M)$?

When is $H^\ast_{G/H}(M)$ ring-isomorphic to $H^\ast_{G/H}\otimes_{\mathbb{Q}}H^\ast_G(M)$?

Consider the following particular case: $M$ is the complex projective space $\mathbb{P}^{n-1}$, $G=(\mathbb{C}^*)^n$

acting in the standard way, and $H$ is the sub-torus

$\{(t,\ldots,t):t\in\mathbb{C}^*\}$ of $G$.

Then, $$H^*_G=\mathbb{Q}[\alpha_1,\ldots,\alpha_n],$$ while

$$H^*_{G/H}=\mathfrak{\mathbb{Q}[\alpha_1,\ldots,\alpha_n]}{\alpha_1+\ldots+\alpha_n=0},$$ and

the above ring isomorphism holds.

share|cite|improve this question
What kind of relationship do you want? (I am not being flippant; it sometimes - often? - pays in mathematics to have some idea of what one hopes to be true, before one tries to invent or look up a proof.) – Yemon Choi Jul 9 '12 at 21:25
Assuming that the cohomology is with rational coefficients, I am hoping for some relation that would determine the ring $H^*_{G/H}(M)$ in terms of $H^*_G (M)$ and $H^*_{G/H}$. For instance, when is $H^*_{G/H}(M)$ isomorphic as rings to $H^*_{G}(M)\otimes H^*_{G/H}$ ? – user15512 Jul 9 '12 at 21:38
Could you explain the notation $H^*_{G/H}(M)$? – Ben Webster Jul 9 '12 at 21:59
What does this mean when $G/H$ is not a group ? – DamienC Jul 9 '12 at 22:15
@Ben, this is standard notation for equivariant cohomology, $H_G^\ast(M)=H^\ast(M_G)$ where $M_G$ is the Borel construction. – Chris Gerig Jul 9 '12 at 22:38

If $G$ is compact and $H$ is a closed normal subgroup of $G$ that acts trivially on $M$, then there is a spectral sequence $$E_2^{i,j}=H_{G/H}^i(M;H^j(H;A)) \Rightarrow H^{i+j}_G(M;A)$$ where $A$ is a $G$-module.

If $H$ is central in $G$ and $A$ is a trivial $G$-module then all coefficients are trivial. Otherwise they are understood to be local coefficients.

This is a result of Duflot and can be found in section 3 of her celebrated paper "Depth and equivariant cohomology".

Added: Concerning your "hope" expressed in the comment above: If $A=k$ is a field (with trivial $G$-action) and $H$ is central, then the $E_2$-term becomes $$E_2^{\ast,\ast}=H^\ast_{G/H}(M;k)\otimes_k H^\ast(H;k).$$

By taking $G$ finite and $M$ a point it's obvious that one can't hope for more in general.

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.