Show that the intersection of $U(n)$ with symmetric matrices is a submanifold of $U(n)$.

Question

$U(n)$ is the group of unitary matrices, and $S(n)=\{A\in M_n(\mathbb{C})|A^T=A\}$ are the symmetric matrices, both $U(n)$ and $S(n)$ are viewed as submanifolds of $\mathbb{C}^{n^2}$. My question is how to show that their intersection $U(n)\cap S(n)$ is a submanifold of $U(n)$.

Answer 1

Use the metric induced from $\mathbb{C}^{n^2}$, then $A\rightarrow A^t$ is an isometry on $U(n)$. To see that $U(n)\cap S(n)$ is connected, write $U\in U(n)\cap S(n)$ as $U=A+iB$, $U^∗U=I,U=U^t$ implies $A,B$ are real and symmetric, and $AB=BA$, so $A,B$ can be diagonalized simutaneously by a real orthogonal matrix $P$, and $P^tUP=D, D=diag\{e^{i\theta_1},\cdots,e^{i\theta_n}\}$ . So we can show $U=PDP^t=PD^{\frac{1}{2}}P^tPD^{\frac{1}{2}}P^t=VV^t,V=PD^{\frac{1}{2}}P^t$,$V$ is unitary, a path from $V$ to I in $U(n)$ gives a path from $VV^t$ to $I$ in $U(n)\cap S(n)$, this show that that $U(n)\cap S(n)$ is connected. By theorem in Riemann geometry that the fixed points of an isometry is a totally geodesic submanifold, I get what I try to prove.