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This is a problem i am thinking for a while but did not find an answer. Maybe one of you knows it.

Let $G:= SO(n,\mathbb{R})$, $G\times \mathbb{R}^n\to \mathbb{R}^n$ the standard representation of $G$, and $G\times \mathbb{R}^{n\cdot k}\to \mathbb{R}^{n\cdot k}$ the direct sum of $k$ standard representations of $G$. This action is orthonormal and hence induces a smooth group action $$G\times S^{k\cdot n-1}\to S^{k\cdot n-1}$$ In general this action is not free and the quotient $S^{k\cdot n-1}/G$ is not a manifold anymore but a Whitney stratified space. So my question is:

Does anybody know how to compute the following cohomology groups. $$H^*(S^{k\cdot n-1}/G,\mathbb{Q})=?$$ If one knows a result for the corresponding intersections cohomology groups i would also be pleased to hear about it.

Of course if $n=2$ then $S^{k\cdot n-1}/G\cong \mathbb{C}P^{k-1}$.

The case that $k < n$ is also comparatively simple, in this case the orbits space $S^{k\cdot n-1}/G$ is contractible.

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@O. Straser: I think you mean $\mathbb{C}P^{k-1}$ in the penultimate sentence. Further, I do not understand your last sentence. In any case, the quotient is certainly not contractible when $n=k=2$. – Ricardo Andrade Apr 11 '13 at 9:57
You are absolutely right, i meant $k<n$. Thank you very much for your both corrections! – Oliver Straser Apr 11 '13 at 10:09

The following are some ideas of how to describe the orbit space.

Consider $x=(x_1,x_2,\dots,x_k)\in(\mathbb R^n)^k$. An element $g\in SO(n)$ maps this to $(g.x_1,\dots,g.x_k)$. A basis for the algebra of invariant polynomials on $\mathbb R^{n.k}$ consists of $\langle x_i,x_j\rangle$ for $1\le i\le j\le k$, using the inner product on $\mathbb R^n$. They separate points on the orbit space, since $SO(n)$ is compact. The also form a quadratic invariant mapping $\rho$ from $\mathbb R^{n.k}$ into the space of symmetric $(k\times k)$-matrices whose image is the orbit space. The orbit space is also described by the basis of the algebra of relations between the invariants (the syzygy's) as a real algebraic set. The rank of $\rho(x)$ as a $(k\times k)$-matrix equals the rank of $x$ as a $(n\times k)$-matrix (Wronski). The maximal rank is $k$ if $k< n$ and is $n$ if $k\ge n$. I guess the orbit type stratification of the orbit space is by rank.

You consider the sphere $\lbrace x: \sum \|x_i\|^2=1\rbrace$. This means $Trace(\rho(x))=1$.

If $k\ge n$, then $SO(n)$ acts freely on the orbit through $x$ if and only if $x_1,\dots,x_k$ span $\mathbb R^n$; i.e. the $(n\times k)$-matrix $x$ has maximal rank $n$. These matrices form the open dense regular stratum which consists of principal orbits.

I use facts described in section 29 of here.

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