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

AndresB
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    First of all, when $N$ is odd, this is an odd-dimensional manifold. In general, do you know something about $H^2(U(N))$? – Ted Shifrin May 22 '19 at 16:31
  • Having a look at this could help... https://math.stackexchange.com/questions/722079/no-symplectic-structure-on-s2n-n1?rq=1 – N. Ciccoli May 27 '19 at 07:38
  • I'm sure I replied to @TedShifrin comment saying that I don't know what $H^{2}$ means. – AndresB May 27 '19 at 19:58
  • @N.Ciccoli So I guess it can't be done except,maybe, for SU(2)? – AndresB May 27 '19 at 19:59
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    $SU(2)$ is $S^3$, so certainly that won't work. $H^2$ stands for second (deRham) cohomology. You need to know about the basics of differential forms here. A symplectic form gives a generator for $H^2$, whose $n$th wedge product with itself then generates $H^{2n}$, the top cohomology. – Ted Shifrin May 27 '19 at 20:41

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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.

Qiaochu Yuan
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