Complex structure on the set of prequantization line bundles

Question

For geometric quantization, the set of equivalence classes of prequantization line bundles of a quantizable symplectic manifold $(M, ω)$ is parametrized by $H^1(M, S^1)$ which represents the equivalence classes of flat Hermitian line bundles.

When $M$ is a compact Riemann surface, we can construct a complex structure on $H^1(M, \mathbb{R})$ by Hodge $*$ operator and this structure induces a complex structure on $H^1(M, S^1)=H^1(M, \mathbb{R})/H^1(M, \mathbb{Z})$.

My question is as follows:

Does $H^1(M, S^1)$ admit a complex structure when $M$ is a compact Kähler manifold?

If not, which condition $M$ should satisfy to construct a complex structure on $H^1(M, S^1)$?

This post imported from StackExchange MathOverflow at 2014-10-14 10:50 (UTC), posted by SE-user Bo Liu

Comments

Yes. The complex structure on $H^1(M,\mathbb{R})$ can be defined via the Hodge decomposition $H^1(M,\mathbb{R})\otimes \mathbb{C}=H^{1,0}\oplus H^{0,1} $, and this holds on any compact Kähler manifold.

This post imported from StackExchange MathOverflow at 2014-10-14 10:50 (UTC), posted by SE-user abx

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Thank you! For this case, could the complex structure on $H^1(M,\mathbb{R})$ give a complex structure on $H^1(M,\mathbb(S^1))$? And are there some references about this?

This post imported from StackExchange MathOverflow at 2014-10-14 10:50 (UTC), posted by SE-user Bo Liu

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Yes to the first question. Reference: any introduction to complex geometry, e.g. Griffiths-Harris or Voisin.

This post imported from StackExchange MathOverflow at 2014-10-14 10:50 (UTC), posted by SE-user abx

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I see, topologically, $H^1(M,S^1)$ here is just the Jacobian $\mathrm{Pic}^0(M)$, the kernel of $C_1:\mathrm{Pic}(M)\rightarrow H^2(M,\mathbb{Z})$, a complex torus of dimension b_1(M). Is it right?

This post imported from StackExchange MathOverflow at 2014-10-14 10:50 (UTC), posted by SE-user Bo Liu

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Perfectly right.

This post imported from StackExchange MathOverflow at 2014-10-14 10:50 (UTC), posted by SE-user abx