Homogeneous space question: a quotient $U(n)/U(n-1)$

Question

One can block-diagonally embed a copy $H$ of the unitary group $U(n-1)$ into $U(n)$ by $$A \mapsto \begin{bmatrix}\det(A)^{-1}&0\\0& A\end{bmatrix}.$$

According to a remark in the example section of Paul Baum's dissertation, the coset space $$U(n)/H \approx S^1 \times \mathbb{C}\mathrm{P}^{n-1}.$$ How can I see this? It probably has some simple orbit–stabilizer–type proof, but I'm just not seeing it right now.

Answer 1

Consider $SU(n)$ acting on $\mathbb C P^{n-1}$ (as a subgroup of $\mathrm{GL}_n(\mathbb C)$. Check that it acts transitively, and that the stabilizer of a point is $U(n-1)$ (embedded as in the OP).

This shows that $SU(n)/U(n-1) = \mathbb C P^{n-1}$. Now we may embed $U(1) = S^1$ into $U(n)$ as $z \mapsto \mathrm{diag}(z,1,\ldots,1)$, and we find an isomorphism (of topological space, not groups) $S^1 \times SU(n) \cong U(n)$. Putting this together with the previous paragraph, we find $S^1 \times \mathbb C P^{n-1} \cong U(n)/U(n-1)$.

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If you don't want to go through the (mild) pain of treating $SU(n)$ first and then breaking up $U(n)$ as $S^1 \times SU(n)$, you could also look at the action of $U(n)$ on $S^1 \times \mathbb C P^{n-1}$ where the action on the first factor is as multiplication by $\det$, and the action on the second factor is via the embedding into $\mathrm{GL}_n(\mathbb C)$.

Then one again finds that the action is transitive and that the stabilizer of a point is $U(n-1)$, and so the required isomorphism follows at once.

Answer 2

I think I can get a related result locally but I'm not sure about the general case:

The generalized Hopf fibration yields a principal bundle $S^1 \to S^{2n-1} \to \mathbb{CP}^{n-1}$ so that locally we have $S^{2n-1} \cong S^1 \times \mathbb{CP}^{n-1}$. We also have the principal bundle $U(n-1) \to U(n) \to S^{2n-1}$, so viewing $U(n-1)$ as a closed Lie subgroup of $U(n)$ (in the way you prescribe) we have that $U(n-1)/U(n) \cong S^{2n-1}$. The result then follows (but again, only locally).