TITLE: Cohomology groups of homogeneous spaces
QUESTION [13 upvotes]: Is there a general method to calculate the cohomology groups of homogeneous spaces ($G/H$), such as $\frac{U(4)}{U(2)\times U(2)}$, $\frac{U(5)}{U(2)\times U(3)}$, $U(4)/U(2)$, etc. If yes, could you give the references for the method? Can this be achieved by computer algebra? Thanks.
In general textbooks, there are only examples for special cases, such as $S^{n-1}=SO(n)/SO(n-1)$, and the method cannot be generalized for all homogeneous spaces.

REPLY [5 votes]: I think that the best way to compute this cohomology is the following. The homogeneous spaces
you are looking at are all of the form $X=G/M$ where $G$ is compact and $M$ is a connected subgroup of $G$ Let $T_M$ be a maximal torus of $M$, embedded into a maximal
torus $T_G$ of $G$ and let $\mathfrak t_M$ $\mathfrak t_G$ be the corresponding (complexified) Lie algebras. Let us first look at the equivariant cohomology $H^*_G(X)$ (say, with $\mathbb C$-coefficients). It is obvious that 
it is the same as $H^*_M(pt)$ (here $pt$ denote "the point") which is known
to be $Sym(\mathfrak t_M^*)^{W_M}$; here $Sym$ means "symmetric algebra",  and $W_M$ means the
Weyl group of $M$. By abstract nonsense it is clear that 
$H^*(X)$ 
is just $H^*_G(X)\underset{H^*_G(pt)}\otimes {\mathbb C}$, 
where $\otimes$ in principle means
"derived tensor product". If $M$ and $G$ have the same rank, then
 you can show that 
$H^*_G(X)$ is always free over $H^*_G(pt)=Sym(\mathfrak t^*)^{W_G}$ (here I denote
$\mathfrak t=\mathfrak t_M=\mathfrak t_G$) hence
you finally get
that $H^*(X)$ 
is equal to  $Sym(\mathfrak t^*)^{W_M}\underset{Sym(\mathfrak t^*)^{W_G}}\otimes{\mathbb C}$.
In the general case, you need to compute the above derived tensor product, which 
in every specific case is usually easy to do.