Why is $SU(n)/SU(n-1)$ the $2n-1$-sphere?

Question

I am looking at Fomenko, Fuchs' book on "Homotopical Topology" and they claim that we have the isomorphism $$ SU(n)/SU(n-1) \cong S^{2n-1} $$ Why is this true? Here is what I have so far:

If I have a matrix $A \in SU(n-1)$, then we can embed $SU(n-1)$ into $SU(n)$ by constructing the block-diagonal matrix $$ \begin{bmatrix} A & 0 \\ 0 & 1 \end{bmatrix} $$ From here it is clear that the resulting dimension of the homogenenous space is $2n-1$ since there are $2n-1$ free entries in the matrix from the embedding. I'm not sure though how to use the unitary property of these matrices.

score 11 · Accepted Answer · edited Feb 12 '25 at 15:28

11

The group $SU(n)$ has a natural action on $\Bbb C^n$ and also on the unit sphere within, the set $\{z\in\Bbb C^n:\|z\|=1\}$ which we can identify with $S^{2n-1}$. Then $SU(n)$ acts transitively on $S^{2n-1}$. For a suitable point $z_0$ of $S^{2n-1}$ the stabiliser of $z_0$ is $SU(n-1)$. Therefore we may identify $S^{2n-1}$ with the coset space $SU(n)/SU(n-1)$.

edited Feb 12 '25 at 15:28

Dan Asimov

1,177

answered Jun 30 '17 at 04:31

Angina Seng

161,540

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Why do we need to only pick one stabilizer? – 54321user Jun 30 '17 at 04:52
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@MoarCake559 The stabilizer of any point in $S^{2n-1}$ will be isomorphic to $\text{SU}(n-1)$. – Andreas Blass Jun 30 '17 at 05:15

Why is $SU(n)/SU(n-1)$ the $2n-1$-sphere?

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