Prove that U(n) is an embedded submanifold.

Question

I am having trouble proving that U(n) is a submanifold of $M_n(\Bbb{C})$. I thought about using the regular value theorem. I used the function $F:GL_n(\Bbb{C})\to M_n(\Bbb{C})$ by sending a matrix $A$ to $AA^*-I$ where $A^*$ is the adjugate of $A$. Then I saw for sure that $F^{-1}(Z) = U(n) = \{A\in GL_n(\Bbb C) : AA^*=I\} $ (Z is the zero matrix.) So the next step I did was trying to show that $(dF)_A$ is surjective $\forall A\in U(n)$. $(dF_A)(X) = \lim_{t \to 0} {F(A+tX)-F(A) \over t}$=$\lim_{t \to 0 }\frac{(A+tX)(A+tX)^* - I - AA^* +I}{t}=*$, since $A\in U(n)$ we get that $AA^*=I$ so we get the following: $*=\lim_{t \to 0} \frac{F(A+tX)}{t}$. From here I don't know how to continue computing the derivative. I am wondering if I could prove the same with another function $F$, or how to compute the derivative further. I think I can prove surjectivity myself if I have the derivative.

Answer 1

For the derivative you have $$ F(A+tX) -F(A) = (A+tX)(A+tX)^*-AA^* = t(XA^*+AX^*)+t^2XX^* $$ so that $$ dF_A(X) = XA^*+AX^*. $$

However note that $F:GL_n(\mathbb C)\to Mat_n(\mathbb C)$ is not a submersion (i.e. the differential is not everywhere surjective). Note also that if that were the case, the preimage $U_n$ would be zero-dimensional...

A wiser choice of range for the map $F$ yields the proof that you are after. What special properties does the matrix $AA^*$ have?