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If we quotient $U(N)$ by $U(N-1)$ we get the odd dimensional sphere $S^{2N-1}$. (Here the quotient is in the sense of embedding $U(N-1)$ in the bottom right hand corner (with 1 as the (1,1) entry and zero everywhere else) and taking its orbits as the set of new objects.) If we quotient now by $U(1)$ (embedded on the diagonal) we get ${\mathbb CP}^{N-1}$.

More generally, if we quotient $U(N)$ by $U(N-k)$, for some $k < N$ (with an analagous embedding), and then quotient by $U(k)$ (embedded again on the diagonal) we get the $k$-Grassmannian $G_k({\mathbb C}^N)$.

My question is: What is the object we obtain when we quotient by $U(N-k)$? As we saw, it is the sphere for $k=1$. However, I cannot identify it with a familar object for higher $k$.

Also, more generally, if $F$ is a generalised flag manifold of signature $(d_1, \ldots ,d_k)$, then quotienting $U(N)$ by $$ U(N-d_1) \times \cdots \times U(N-d_k), $$ and then by $$ U(d_1) \times \cdots \times U(d_k), $$ gives $F$. What is the object we get from the first quotienting?

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    $\begingroup$ Putting a name to the space, it's a complex Stiefel manifold. See en.wikipedia.org/wiki/Stiefel_manifold#As_a_homogeneous_space . $\endgroup$ Commented Sep 22, 2010 at 14:12
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    $\begingroup$ The homogeneous spaces you describe are called Stiefel manifolds. They're well known in algebraic topology, but not always isomorphic to more elementary manifolds. $\endgroup$
    – user1504
    Commented Sep 22, 2010 at 14:13
  • $\begingroup$ Great, thanks for that. Charles, seeing as you got there first, enter your response as an answer and I'll mark as accepted. $\endgroup$ Commented Sep 22, 2010 at 14:18
  • $\begingroup$ Or maybe the first to answer the generalised version... $\endgroup$ Commented Sep 22, 2010 at 14:39
  • $\begingroup$ It is a different Grassmanian. $\endgroup$ Commented Sep 22, 2010 at 15:29

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Putting a name to the space, it's a complex Stiefel manifold. See http://en.wikipedia.org/wiki/Stiefel_manifold. (But I wasn't the first.)

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