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A Kähler–Hodge manifold $M$ can be defined as a Kähler manifold whose Kähler form $\omega$ is integral, namely $\omega\in H^{2}(M,\mathbb{Z})$. It is known then that there always exists a Hermitian line bundle $L\to M$ whose Chern class satisfies $c_{1}(L) = [\omega]$. I have the following questions:

  • Does anyone know a reference where the statements above are proven?

  • How can this line bundle $L$, given a particular integral Kähler form $\omega$, be explicitly constructed?

  • What happens if $\omega$ is not integral? Can one still construct some kind of $\mathbb{R}$-bundle whose curvature is $\omega$?

Thanks.

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2 Answers 2

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At least when $M$ is compact, which I assume here, the standard reference for this subject is probably the book

  • Algebraic Geometry, Griffiths & Harris

but the following books are excellent references as well :

A proof could run as follows:

  • Hodge's theorem implies that there exists a complex line bundle $L$ on $M$ with first Chern class $c_1(L)=\omega$;
  • Endow $L$ with an hermitian metric $(L,\|\cdot\|_0)$; its curvature form $c_1(L,\|\cdot\|_0)$ is a $(1,1)$-form cohomologous to $\omega$. By the $dd^c$-lemma, there exists a function $\phi$ such that $\omega= c_1(L,\|\cdot\|_0)+dd^c\phi$; then $\|\cdot\|=e^{-\phi}\|\cdot\|_0$ is a hermitian metric on $L$ with curvature form equal to $\omega$.

For an explicit construction of $L$: the standard proof of Hodge's theorem is cohomological, and uses a long exact sequence in cohomology. Tracing down the arguments gives an actual construction if one is given an open covering of $M$ by contractible open sets.

The arguments really require that $\omega$ be integral. A generic complex torus of dimension $>1$ carries no complex line bundle, but has many Kähler forms.

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  • $\begingroup$ It seems that you are assuming that $M$ is compact, a restriction which was not in the question. $\endgroup$
    – abx
    Commented Feb 18, 2015 at 10:52
  • $\begingroup$ @abx: Indeed... I'll edit the answer. $\endgroup$
    – ACL
    Commented Feb 19, 2015 at 10:02
  • $\begingroup$ Thanks for the explanation. How crucial is that $M$ is a compact manifold? Does a similar result hold in for a non-compact base $M$? $\endgroup$
    – Bilateral
    Commented Mar 14, 2015 at 23:00
  • $\begingroup$ Dear @ACL, I have difficulty to understand why a generic complex torus of dimension $>1$ has no complex line bundle. Let $T^n$ be a complex torus, then $Pic^0(T)$ is the dual torus of $T$ of complex dimension $n$. In particular, there exist many non-isomorphic topologically trivial holomorphic line bundles. $\endgroup$
    – Doug Liu
    Commented Jan 16, 2023 at 17:38
  • $\begingroup$ You're perfectly right. What I had in mind is that the Neron-Severi group is in general trivial, so that all complex line bundles are topologically trivial. $\endgroup$
    – ACL
    Commented Jan 17, 2023 at 20:44
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If we consider a usual sympleetic manifold with $[\omega]\in H^2(M,\mathbb Z)$ integral then the Weil's theorem guarantees that there exists a Hermitian line bundle $L$ over $X$ with a conneetion $V$ such that $V$ is compatible with the Hermitan strueture on $L$ and induces the symplectic form by $\omega = curv(V)$.

For the second part of your question , The proof is in section "12.2 Integral closed 2-forms and line bundles" http://www.math.toronto.edu/~jeffrey/mat1312/lec14.gq.pdf

In third part of your question: Manifolds which are not quantizable then there is no such line bundle $L$ which first chern class be proportional to kahler class

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