Timeline for Example of a manifold which is not a homogeneous space of any Lie group
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5 events
| when toggle format | what | by | license | comment | |
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| Feb 24, 2012 at 17:42 | comment | added | Qiaochu Yuan | @MTS: sorry, I meant a $\text{Stab}(x)$-invariant inner product. | |
| Feb 24, 2012 at 4:45 | comment | added | MTS | Qiaochu, I think what you are saying is the following: if $M = G/H$, then the tangent bundle of $M$ is the homogeneous vector bundle induced by the adjoint representation of $H$ on $\mathfrak{g}/\mathfrak{h}$. If $H$ is compact then we can integrate with respect to the Haar measure of $H$ to get an invariant inner product on $\mathfrak{g}/\mathfrak{h}$. Then translating around with $G$ gives the invariant metric on $M$. Is that right? I don't think you can choose a $G$-invariant inner product on $T_x(M)$ since $G$ doesn't preserve the point $x$ in general. | |
| Feb 24, 2012 at 1:36 | comment | added | Qiaochu Yuan | @MTS: if $M$ is a homogeneous space for a Lie group $G$ and $x \in M$ is a point with compact stabilizer, then you can choose a $G$-invariant inner product on $T_x(M)$ and homogeneity gives you a $G$-invariant Riemannian metric on $M$, doesn't it? I guess it is possible that there are no points with compact stabilizers, but in any case this shows at least that any compact surface of genus at least two is not a homogeneous space of a compact Lie group. | |
| Feb 24, 2012 at 1:05 | comment | added | MTS | This is a nice argument, but I am not asking for a homogeneous Riemannian metric. This doesn't rule out the possibility of having a transitive action of a Lie group that is not an action by isometries for any Riemannian metric. | |
| Feb 24, 2012 at 0:29 | history | answered | Roberto Frigerio | CC BY-SA 3.0 |