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 contradictible.

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.

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 $S^{k\cdot n-1}/G\cong \mathbb{C}P^{n-1}$.

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

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 contradictible.