How to compute the (co)homology of orbit spaces (when the action is not free)? Suppose a compact Lie group G acts on a compact manifold Q in a not necessarily free manner. Is there any general method to gain information about the quotient Q/G (a stratified space)? For example, I would be interested in its (co)homology groups.
To be a little more concrete - I am interested in G=SU(2) acting by diagonal conjugation on the space Q=G^N (N a positive integer).
 A: Your example ($G^N/G, G=SU(2)$) is the space of conjugacy classes of $SU(2)$ representations of the free group of rank $N$. I seem to recall there are some articles out there that calculate its cohomology for small values of $N$, so you might look for those (of course, when $N=1$, you get a 1-simplex). 
Eg. "The topology of moduli spaces of free group representations"
http://www.ams.org/mathscinet-getitem?mr=2529483
But I think I recall some explicit results for $G=SU(2), N=2,3$ somewhere. Since the free group is a punctured surface group, you might look at articles about moduli space of flat connections on surfaces, and perhaps the Narasimhan-Sheshadri-type results which relate $U(n)$ and $SU(n)$ flat moduli spaces to moduli of holomorphic bundles over Riemann surfaces. I dont know if the punctures can be dealt with in that context.  
You might look also at Akbulut-McCarthy's book, and I'm sure Andy P. and Jeffrey G. can direct you to other articles in this direction.
A: For the concrete example you have in mind, I believe there is an explicit computation of the Poincare polynomial in "A perfect Morse function on the moduli space of flat connections" by Michael Thaddeus, Topology 39, 2000, pp. 773--787 (MR1760428). 
A: Computing the cohomology of the quotient by a non-free action $X/G$ is generally quite difficult.  As a first step, one can switch to a problem that is somewhat easier, namely that of computing the cohomology of the homotopy quotient, $X \times_G EG$.  Here $EG$ is a contractible space with a free action of $G$. The idea here is that $X \times EG$ is $G$-equivariantly homotopy equivalent to $X$ and now the action is free. When $X$ is a point then the homotopy quotient is just $EG/G = BG$.
The cohomology of $X \times_G EG$ is what is known as the (Borel) equivariant cohomology of $X$, written $H_G^*(X)$. This is now a nice cohomology theory satisfying appropriate axioms. An important feature of it is that it is a module over the ring $H_G^*(pt) = H^*(BG)$.
If the group $G$ is a torus then there are some quite powerful tools for computing equivariant cohomology in terms of things like the set of nonfree orbits.  A beautiful place to start is the paper of Aityah and Bott, The moment map and equivariant cohomology. Topology 23 (1984), no. 1, 1-28.
One of the theorems you will find explained in there is that the inclusion of the subspace of non-free orbits $S$ into $X$ becomes an isomorphism on equivariant cohomology once you localize (in the ring-theoretic sense) by inverting the equivariant Euler classes of the normal bundles of the components of $S$.   Fortunately, the map from equivariant cohomology to the localized equivariant cohomology is usually injective so doing this doesn't lose any information.
Finally, there is a map from $X \times_G EG \to X/G$ given by collapsing $EG$ down to a point.  This map gives rise to a spectral sequence that allows one to compute the cohomology of the quotient from the equivariant cohomology and additional geometric data about the action and its fixed points and stabilizers.
A: In general, this is a difficult question. Here are a couple of related facts that I know. Consider a discrete group $G$ acting on a space $X$, which we will assume is a simplicial complex (and that the action is simplicial). Moreover, to simplify things assume that the stabilizer of a simplex stabilizes that simplex pointwise (this can be arranged by subdividing).
1) If $X$ is simply-connected, then $X/G$ is simply connected if and only if $G$ is generated by elements that stabilize vertices. More generally, let $H$ be the subgroup of $G$ generated by vertex stabilizers (observe that this is normal!).  There is then an exact sequence
$$1 \to H \to G \to \pi_1(X/G) \to 1.$$
This is a theorem of M.A. Armstrong; see his paper
MR0187244 (32 #4697)
Armstrong, M. A.
On the fundamental group of an orbit space.
Proc. Cambridge Philos. Soc. 61 1965 639--646. 
A related theorem can be found in my paper "Obtaining presentations from group actions without making choices".
2) As far as homology goes, there is a whole theory of equivariant homology here.  A good first place to read about it is Brown's book "Cohomology of Groups", Chapter VII, and a more comprehensive introduction is tom Dieck's book "Transformation Groups"
As you will see if you read the above sources, the answer comes down to "It's complicated!". In concrete settings, you are probably better off trying to get a good topological/geometric understanding of the orbit space with your "bare hands".
A: Here's a procedure to compute the homology of $X/G$ for finite $G$ for specific examples “with your bare hands”.
I'm interested in the homology of the space of unlabeled networks with non-negative edge weights with coefficients in $\mathbb{Z}$. In this case, $X=\mathbb{R}_{\ge 0}^{N}$, $G=\Sigma_n \hookrightarrow \Sigma_N$, $N=\binom{n}{2}$ is the number of edge weights, and $G$ is the group of edge permutations induced by relabeling vertices. I compactify $X$ by intersecting with the plane $x_1+\dotsb+x_N=1$.
A while back I wrote some Sage code to compute Dirichlet fundamental domains. This gives a list of half-plane inequalities yielding a convex polyhedron $F(x_0)$, and $G$ is represented as a group of permutation matrices. I recently realized this code is fairly general, and works for any convex polyhedral space $X$ and finite matrix group $G$.
https://github.com/jacksonwalters/orbit-space-homology
I then attempted to compute the cellular homology using the polyhedron faces as cells, keeping track of orientations, and whether a map gluing the faces together preserves or reverses orientation.
An unresolved issue that came up was what to do when a face $f \subset F(x_0)$ is glued to itself non-trivially, so $gf = \pm f$. Do you count as usual, then divide by the size of the stabilizer? Doing so yields a complex, but the homology seems random and varies with $x_0$, which is no good. For example, if there is a square face, and there is a 4-rotation generated by $g$ which preserves orientation, you'd get $(f+f+f+f)/4=f$. Working over $\mathbb{Z}/2\mathbb{Z}$ and foregoing the division (also yielding a complex with bizarre homology) you get 0. For a flip ($gf=-f$), you get 0 in either case.
One fix is to sub-divide so that there are no faces with non-trivial self-gluings (so $G$ permutes cells and yields an appropriate complex in the sense of Bredon homology). Sage does allow barycentric subdivision, but 1) there are too many faces 2) there's an odd parameter subdivision_frac which cannot be set to zero and seems to puff out the faces, and causes the old vertices to not be contained in the new vertices.
An alternative is to label the vertices and compute simplicial homology instead, which I recently implemented. To avoid the non-trivial self-gluing, I add some new vertices which are fixed points of the $G$ action. Fixed point subspaces are just the $\lambda=1$ eigenspaces computed as $\ker([g]-\operatorname{Id}_N)$. Intersecting these with the polyhedron gives the new vertices.
In the above example, this has the effect of adding a new vertex in the middle of the square, cutting it into four triangles. We use these faces and make sure to throw out the big square.
In my example, this adds 4 new vertices to the original 7. The computation is slow, but I find that $H_k(X/G,\mathbb{Z}) = \mathbb{Z}$ in degree 0, and 0 otherwise.
This is consistent with my expectations, as I'd be very curious to see an example where $X$ is contractible, $G$ is finite, and $H_k(X/G,\mathbb{Z})$ has torsion. I tried a few by hand, and they all yield contractible spaces, e.g. a triangle with $\mathbb{Z}/3\mathbb{Z}$ acting by rotation is a dunce cap, a triangular pyramid with the same rotation acting around an axis going through the top vertex is a solid cone, and my n=3 example $\left(\mathbb{R}_{\ge 0}^3 \cap \{x_1+x_2+x_3=1\}\right)/\Sigma_3$ is a triangular fundamental domain with no gluings, which is contractible.
According to this answer to cohomology of the orbit space of a group action, over a field the cohomology should be trivial when $X$ is a contractible space, but that doesn't answer the question over $\mathbb{Z}$.
