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Define a diffeomorphism for $U(m)/U(m-1)\cong S_{2m-1}$.

Looking at the Differentiable Manifolds text by Shahshahani $U(m)/U(m-1)$ looks like a homeomorphism, but I'm skimming Wikipedia's text for diffeomorophisms, orbits, and quotient spaces and not getting any direction.

Would the solution be useful with SU(m) too and instead of $S_{2m-1}$, $\mathbb C \mathbb P(m-1)$.

Michael T Chase
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2 Answers2

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An element of $U(n)$ is an $n$-tuple of orthogonal vectors in $\mathbb{C}^n.$ So, pick a vector (this is the $\mathbb{S}^{2n-1}$ and then do what you will to its orthogonal complement (that is the $U(n-1).$)

Igor Rivin
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  • Thanks for the intuitive approach. – Michael T Chase Apr 29 '21 at 18:08
  • This actually helps a lot. When you said "do what you will" I take it to mean use any topology that will bring about a diffeomorphism between the vector and its orthogonal complement. But FYI without Didier's example I personally would've been lost as I used the example first to connect the dots and then, thank you, used the intuition to fill in the blanks. – Michael T Chase Apr 29 '21 at 18:19
  • Also, I didn't realize that $U(n)$ and $SU(n)$ were just the $\mathbb C$ complement of $O(n)$. Thanks again. – Michael T Chase Apr 29 '21 at 18:31
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A -not so- alternative proof than @IgorRivin' answer.

Consider $S^{2m+1}$ as embedded in $\mathbb{C}^{m+1}$. It is a smooth manifold. Then $U(m+1)$ acts on $S^{2m+1}$ with a Lie-group action. One can show that this action is transitive. Consider $N$ the north pole. Its stabilizer is the set: $$ H=\left\{\begin{pmatrix}M & 0 \\ 0 & 1 \end{pmatrix}\in U(m+1) \right\} \simeq U(m). $$ Hence, $S^{2m+1} \simeq U(m+1) / U(m)$ is a diffeomorphism. This is the map you are looking for. Its explicit writing is: \begin{align} U(m+1)/U(m) & \longrightarrow S^{2m+1} \\ [g]=g H & \longmapsto g\cdot N \end{align}

Didier
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