Unitary matrices form a manifold

Question

If $f : \mathbb R^m \times \mathbb R^n \to \mathbb R^n$ is a smooth function, such that $a \in \mathbb R^n$ is a regular value of $f$, then $f^{-1}(a) $ is a smooth manifold of dimension $m$.

A common application of this result, is to show that the set of orthogonal matrices $O_n$ form a manifold, because they are the inverse image of $0$, which is a regular value, under the function $X \to X^TX - I_n$, which goes from $\mathbb M_{n \times n}$ to $S_{n \times n}$, the vector space of symmetric matrices.

Suppose we want to extend this to the set of unitary matrices. That is, I want to show that $U_n$, as a subset with topology from $\mathbb C^{n^2}$, forms a manifold.

1 : Is there a variant of the above theorem with $\mathbb C$ in place of $\mathbb R$? Does the statement even change?

2 : Suppose it doesn't, then can we say that as unitary matrices are the inverse image of $0$ under the mapping $X \to \overline{X^T}X - I_n$, and $0$ is a regular value, that $U_n$ is also a manifold? (The dimension will be clear once we use the result).

3 : Suppose I can't do the above. Then what can I do?

score 2 · Accepted Answer · answered Aug 22 '17 at 05:12

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You don't need a complex version. Just identify $\Bbb C$ with $\Bbb R^2$ in the usual way. Then think of the complex $n$-by-$n$ matrices as $\Bbb R^{2n^2}$. As you say $X\mapsto \overline{X^t}X-I$ will have $O$ as a regular value, considered as a map to the Hermitian matrices, and then $U_n$ will be a smooth manifold.

answered Aug 22 '17 at 05:12

Angina Seng

158,341

1

Ok, so the theorem above is true for $\mathbb C$, simply because we treat $\mathbb C = \mathbb R^2$, so everything else falls in position. Thank you, this is really the confirmation that I needed. Hermitian matrices have the dimension $n^2 - n$ over $\mathbb R$, so that $U_n$ will have dimension $n^2 + n$. Is that right? – Sarvesh Ravichandran Iyer Aug 22 '17 at 10:15
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@астонвіллаолофмэллбэрг Make those dimensions both $n^2$. – Angina Seng Aug 22 '17 at 10:18
Wouldn't Hermitian matrices have dimension $n^2 - n$? Similar to how we show that symmetric matrices have such a dimension. – Sarvesh Ravichandran Iyer Aug 22 '17 at 10:22
@астонвіллаолофмэллбэрг Similar to the way we treat symmetric matrices, we find that Hermitian matrices have dimension $n^2$. – Angina Seng Aug 22 '17 at 10:23
But then what does this page say? https://math.stackexchange.com/questions/150643/basis-for-hermitian-matrices – Sarvesh Ravichandran Iyer Aug 22 '17 at 10:24
@астонвіллаолофмэллбэрг It says "what is $n$ plus $n(n-1)/2$ plus $n(n-1)/2$?" – Angina Seng Aug 22 '17 at 10:25
OH. All right, I have found the mistake. Thank you for the answer. – Sarvesh Ravichandran Iyer Aug 22 '17 at 10:27

Unitary matrices form a manifold

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