Stiefel manifold as a fibre bundle over another Stiefel manifold.

Question

I want to show $V_n(\mathbb R^k)$ is a fibre bundle over $V_m(\mathbb R^k)$ were $n>m$

Here $V_n(\mathbb R^k)$ is the Stiefel manifold, that is the set of all $n$ orthogonal frames in $\mathbb R^k$. We have the projection map $\pi:V_n(\mathbb R^k)\rightarrow V_m(\mathbb R^k)$ given by $\pi(x_1,...,x_m,...,x_n)=(x_1,...,x_m)$.

I'm not entirely sure what the fibres are. It seems $\pi^{-1}((x_1,...,x_m))$ is all $(x_{m+1},...,x_n)$ which are form an orthogonal frame with $(x_1,...,x_m)$. I would guess that this should be $O(n-m)$ but I'm not sure why this is true.

If this is true also don't see how to show that this is locally trivial.

Answer 1

Hint: To show that $V_n(\mathbb{R}^k)\rightarrow V_m(\mathbb{R}^k)$ is a fibration, used the fact it is a submersion and use Ehresmann lemma since $V_n(\mathbb{R}^k.)$ is compact

To find the fibre of $(x_1,..,x_m)$, consider the orthogonal $V$ of $(x_1,..,x_m)$, it is a vector space isomorphic to $\mathbb{R}^{k-m}$ and $(x_1,..,x_m,x_{m+1},..,x_n)$ is an element of this fibre if $(x_{m+1},..,x_n)$ is in $V$, so the fibre is $V_n(\mathbb{R}^{k-m})$.

Answer 2

Tou can think of these objects as homogeneous paces. $V_n(R^k)= O(k)/O(k-n)$. Note that if $m<n$ $O(k-n) < O(k-m)$. The natural map $V_n(R^k)\to V_m(R^k)$ is equivariant for the natural action of $O(k)$. Therefore its derivative is of constant rank, and is therefore surjective. It is a fibration , and the fiber is $O(k-m)/O(k-n)= V_{k-n}(R^{k-m})$.

This is a general fact about Lie groups and homogenous spaces. Let $K<H<G$ three Lie groups such that $H,K$ are closed in $G$. Then the natural map $G/K\to G/H$ is a submersion with fiber $H/K$.(same argument).