$\operatorname{SU}(n)$ as manifold

Question

I am trying to do this has a while, but I cannot use correctly the regular value theorem to do so!

I appreciate any help. The problem is that I cannot choose the function to take $SU_n$ as a regular value...

I was wondering about the domain as $M_n(R)$ and the codomain as $M_n(R)\times \mathbb{R}.$ The function would be something like $f(A) = (AA^*,\det A)$, then $SU_n = \det^{-1}(I,1).$

Answer 1

Suppose you wanted to prove first that $\operatorname{U}(n)$ was a manifold. One can show something is a manifold by showing it is an embedded submanifold of a bigger manifold. I claim that $\operatorname{U}(n)$ is an embedded submanifold of $\operatorname{GL}(n,\Bbb C)$. Indeed, there is a map $\Phi: \operatorname{GL}(n,\Bbb C)\longrightarrow \operatorname{GL}(n,\Bbb C)$ that sends $A$ to $AA^\ast$. This map is in fact equivariant with respect to the smooth action $B\cdot A = BAB^\ast$, so it has constant rank $n^2$ by the following:

The image of its differential at the identity is the set of matrices of the form $A+A^*$ which has dimension $n^2$, so $\operatorname U(n)=\Phi^{-1}(\mathrm {id})$ is an embedded submanifold of dimension $2n^2-n^2=n^2$.

To compute $D\Phi(B)$ note that $\Phi(A+h) =AA^\ast+hA^\ast+Ah^\ast+hh^\ast$ and thus $\Phi(A+h)-\Phi(A)-T(A)(h)=hh^\ast$ is of quadratic order where $T(A)(h) = hA^\ast+Ah^\ast$. This is a linear transformation, and by uniqueness of the differential, it must be the case $D\Phi(A)(h) =hA^\ast+Ah^\ast$, in paritcular $D\Phi({\rm id})(h) = h+h^\ast$, as claimed.

Now show that $\operatorname{SU}(n)$ is an embedded submanifold of $\operatorname{U}(n)$.

Answer 2

Hint Note that any matrix of the form $B = AA^*$ satisfies $B^* = (AA^*)^* = (A^*)^* A^* = AA^* = B$, so the first component of the candidate function, $$f(A) := (AA^*, \det A),$$ for which $SU_n = f^{-1}(I_n, 1)$, actually maps into the proper ($n^2$-dimensional) subspace of Hermitian matrices.