As is well known, every Kaehler manifold can canonically be given the structure of a symplectic manifold. Is it naive to assume that holomorphic vector bundles over a Kaehler manifold can be given the structure of a symplectic vector bundle?
1 Answer
$\begingroup$
$\endgroup$
2
This seems to have nothing to do with Kahler manifolds, at least not how you have stated it. Any $C^{\infty}$-complex vector bundle over a paracompact smooth manifold admits a Hermitian metric, by employing a partition of unity. The imaginary part of this Hermitian metric is a skew-symmetric, non-degenerate bilinear form on each fiber which makes that complex vector bundle into a symplectic vector bundle.
-
1$\begingroup$ A reformulation using the structure groups: the complex vector bundle has structure group $GL(n,\mathbb{C})$. The partition of unity yields a reduction of structure group to $U(n)$. But $U(n)=O(2n)\cap Sp(2n,\mathbb{R})\cap GL(n,\mathbb{C})$ so that the bundle acquires, in particular, a symplectic structure. $\endgroup$ Nov 4, 2014 at 17:02
-
2$\begingroup$ Yet another way to say it is that $\text{U}(n)$ is the maximal compact subgroup of both $\text{GL}_n(\mathbb{C})$ and of $\text{Sp}_{2n}(\mathbb{R})$, so all three classifying spaces are homotopy equivalent. $\endgroup$ Nov 4, 2014 at 18:09