Can U(N) be endowed with a symplectic structure?

Question

Can we define a symplectic form on the unitary group U(N) so it becomes a symplectic manifold?

Answer 1

No. As Ted Shifrin observes, this is already impossible for $n$ odd because the dimension $\dim U(n) = n^2$ is odd, but here's a uniform proof for all $n$ anyway. It's a classical result that the cohomology $H^{\bullet}(U(n), \mathbb{Z})$ is an exterior algebra on odd generators $\alpha_1, \alpha_3, \dots \alpha_{2n-1}$; in particular, $H^2(U(n), \mathbb{Z})$ vanishes.

But a closed symplectic $2n$-manifold $X$ necessarily has the property that the class of the symplectic form $[\omega] \in H^2(X, \mathbb{R})$ has the property that its $n^{th}$ power $[\omega]^n \in H^{2n}(X, \mathbb{R})$ is nonzero; in particular, $[\omega]$ itself, and hence $H^2(X, \mathbb{R})$, must be nonzero.