Geometric structures which always exist

79 Views Asked by At

It is well known fact that Riemannian metric always exist on any manifold. Another way to say that is that any manifold always admit $O(n)$ structure i.e. $O(n)$ reduction of structure group of frame bundle $L(M)$.

One way to see that (next to the classical argument with partition of unity) is that $Gl(n)/O(n)$ is contactible, existance of a $G$ structure is equivalent to the existance of section of a fibre bundle $L(M)/G$ and fibre bundles with contractible fibre always possess one.

My questions are:

$\bullet$ What are another closed subgroups $G \subset Gl(n)$ with $Gl(n)/G$ contractible ?

$\bullet$ What are another closed subgroups $G \subset Gl(n)$ such that any manifold admits $G$ structure?

1

There are 1 best solutions below

2
On

For compact case, there is essentially one such subgroup up to conjugation, which is $O(n)$. Indeed, one can apply Corollary 1.4 in this paper, which says that for a compact subgroup $H$ in an almost connected locally compact group $G$, $G/H$ is contractible if and only if $H$ is maximal compact. In the case $G = GL(n,\mathbb R)$ they are conjugate of $O(n)$.