$U(n)$ and $SU(n)$ are connected smooth submanifold of $M_{n,n}(\mathbb{C})$

Question

How can I prove that $U_n(\mathbb{C})$ and $SU_n(\mathbb{C})$ are smooth submanifolds of $M_{n,n}(\mathbb{C})$ ?

I know that given the manifold $X$, $Y$ is a smooth submanifold of $X$ if $\forall k$ there exist $f=(f_1,...,f_d) \in C^ \infty(A_k) $ such that:

1. $A_k \cap Y = V(f_1,...,f_d)$
2. $J_f$ has rank= $d $

where $\{A_k\}_k$ is a cover of X and $V(f_1,...,f_d)=\{x \in A_k:f_1(x)=f_2(x)=...=f_d(x)=0\}$

but I don't know how to apply this definition.

Am I allowed to say that $U_n(\mathbb{C})$ and $SU_n(\mathbb{C})$ are subgroups of $M_{n,n}(\mathbb{C})$, which is a manifold and then they are submanifolds of $M_{n,n}(\mathbb{C})$?

Thank you for your help

Answer 1

According to Cartan's theorem, every closed subgroup of a Lie group is again a Lie group. And it is easy to check that both $U_n(\mathbb{C})$ and $SU_n(\mathbb{C})$ are closed subgroups of $GL_n(\mathbb{C})$.