Prove that $U(n)$ is a manifold

Solution 1:

Note that for your function $A\mapsto A^*A$, the derivative can never be sujective as $A^*A$ are hermitian for all $A \in M_n(\mathbb C)$. So instead we consider the mapping $F: M_n(\mathbb C) \to H_n(\mathbb C)$, where $F(A) = A^*A$ and $H_n(\mathbb C)$ is the space of Hermitian matrices. Then

$$DF_A(V) = \frac{d}{dt} \bigg|_{t=0} (A+tV)^* (A+ tV) = V^*A + A^*V.$$

Now let $A\in U(n)$. Then we want to know if $DF_A : M_n(\mathbb C) \to H_n(\mathbb C)$ is surjective. Indeed it is: Let $B \in H_n(\mathbb C)$, then $W = \frac 12 AB \in M_n(\mathbb C)$ satisfies

$$DF_A(W) = \frac 12 B^*A^*A + \frac 12 A^* AB = \frac 12 B^*+ \frac 12 B = B$$

as $A^*A = I$ and $B^* = B$. Thus $I\in H_n(\mathbb C)$ is a regular value of $F$, and so $U(n)= F^{-1}(I)$ is a smooth manifold.

Solution 2:

Hint Denote the function $A \mapsto A^* A$ by $\Phi$. Then, since $$\Phi(A)^* = (A^* A)^* = A^* A^{**} = A^* A = \Phi(A),$$ we can regard $\Phi$ as a map $$M(n, \Bbb C) \to H(n, \Bbb C),$$ where here $$H(n, \Bbb C) := \{A \in M(n, \Bbb C) : A^* = A\}$$ denotes the (real) subspace of all Hermitian matrices. So, one can prove that $I$ is a regular value of $\Phi$ (so regarded) by showing that that the tangent map $T_A \Phi$ of $\Phi$ at every point $A \in U(n)$ is surjective. (As usual, we can again regard as a map $M(n, \Bbb C) \to H(n, \Bbb C)$ via the usual canonical identification of the tangent space to a vector space at a point with the vector space itself.)