2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ @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 $n=k=2$. $\endgroup$ Apr 11, 2013 at 9:57
  • $\begingroup$ You are absolutely right, i meant $k<n$. Thank you very much for your both corrections! $\endgroup$ Apr 11, 2013 at 10:09

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.