Why is $U(n,n)= GL(2n, \mathbb{C}) \cap O(2n,2n)$? I have read the statement above. I tried to prove it but I could not. For me, $O(2n,2n)$ already a subgroup of $GL(2n,\mathbb{C})$. 
 A: Usually, $O(2n,2n)$ is a group of $4n$-by-$4n$ matrices, so both $GL(2n,\mathbb C)$ and $U(n,n)$ would need to be mapped to $GL(4n,\mathbb R)$. This is not a problem, though can cause some cognitive dissonance. Any choice of isomorphism $\mathbb C^{2n}\rightarrow\mathbb R^{4n}$ will do, giving imbeddings to $GL(4n,\mathbb R)$.
A common trick with classical groups is the following: let $(,)$ be the hermitian form of signature $n,n$ left invariant by $U(n,n)$. Let $\langle u,v\rangle=\Re(u,v)$ be the real part. Using our chosen isomorphism $\mathbb C^{2n}\rightarrow \mathbb R^{4n}$ gives a symmetric $\mathbb R$-valued from $\langle,\rangle$ on $\mathbb R^{4n}$. By design, the image of $U(n,n)$ preserves $\langle,\rangle$. Also, the signature of $\langle,\rangle$ is simply "double" that of $(,)$, so is $2n,2n$.
Thus, for simple reasons $U(n,n)$ injects to $O(2n,2n)$. Then a dimension-count (looking at the Lie algebra) shows that the dimensions are equal, so at least the connected components of the identity are the same.
