# equivariant cohomology

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.

• 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.
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.