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If $L$ is a complex line bundle on a topological space $X$, let $c_1(L)$ denote the image of its Chern class in $H^2(X;C)$. A complex manifold structure on $X$ [ok which is also compact and say algebraic] splits $H^2(X,C)$ into $H^{2,0}$, $H^{1,1}$, and $H^{0,2}$. If $L$ can be given a compatible complex structure, then $c_1(L)$ belongs to $H^{1,1}$. It's rewarding to study the moduli space of isomorphism classes of compatible complex structures on a given $L$--for instance, this is another finite-dimensional complex manifold in a natural way.

One not-very-specific way to explain what a "compatible complex structure" is, is to say that it's a tensor on the total space of L, satisfying a differential equation. Is there a natural kind of differential-geometric structure like this that one can put on a line bundle that guarantees that c_1(L) belongs to $H^{2,0}$ or $H^{0,2}$? Is there an interesting moduli space of such structures?

Tim points out that the question is silly, because $H^{2,0}$ and $H^{0,2}$ do not intersect H^2(X;Z) (which classifies line bundles per Donu), but that there might be some integral classes in $H^{2,0} + H^{0,2}$, or the part of it that's stable by complex conjugation. Is there a structure on a line bundle that lands its Chern class there?

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Um... Assume $X$ is Kaehler so that I have a Hodge decomposition. Doesn't complex conjugation then map $H^{(2,0)}$ into $H^{(0,2)}$? So isn't it the case that the image of $H^2(X;\mathbb{Z})$ in $H^2(X;\mathbb{C})$ intersects the $(2,0)$ part trivially? (One could ask about the real part of $H^{(2,0)}+H^{(0,2)}$, though.) – Tim Perutz Jun 8 '10 at 19:18

If you want a Hodge decomposition, you need assume something about $X$, such as being compact and Kaehler. Let's say it is. Then I infer from your question that "complex line bundle" means complex $C^\infty$ with $\mathbb{C}$ as fibre. In this case, you can get anything in $H^2(X,\mathbb{Z})$ as a first Chern class by using the exponential sequence

$$1 \to \mathbb{Z} \to O_X\to O_X^*\to 1$$

where I'm using $O_X$ etc for the sheaf of complex valued $C^\infty$ functions. The sheaf is fine, so it implies what I said above.

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When $(X,g,\omega)$ is a compact Kaehler surface, the integral classes representable as the real part of a (2,0)-form can be represented as the curvatures of anti-instantons.

One has two splittings of the $g$-harmonic 2-forms $\mathcal{H}^2_g(X)$: the Hodge-theoretic one, and the one determined by the Hodge star, $$ \mathcal{H}^2_g(X) = \mathcal{H}^+_g(X) + \mathcal{H}^-_g(X), $$ into self-dual and anti-self-dual harmonic forms.

The relation with the Hodge decomposition is that $$ \mathcal{H}^+ = \mathbb{R}\omega \oplus (\mathcal{H}^{2,0}\oplus \mathcal{H}^{0,2})_{\mathbb{R}} $$ and $$ \mathcal{H}^- = (\mathcal{H}^{1,1}_0)_{\mathbb{R}}$$ (trace-free part of the real (1,1)-forms). We can check this pointwise, where it's linear algebra. The wonderful thing about Hodge theory is that it then implies the corresponding cohomology-level statement. (Note that $\mathcal{H}^\pm$ are maximal positive-definite (+) and negative-definite (-) subspaces of the wedge-product form on $H^2(X;\mathbb{R})$. So as a by-product we get the Hodge index theorem.)

Suppose that there's an integral class $c$ in $ (\mathcal{H}^{2,0}\oplus \mathcal{H}^{0,2})_{\mathbb{R}}$. It's represented by a hermitian line bundle $L_c$, and (by Hodge theory again) we can choose a unitary connection in $L_c$ whose curvature is harmonic and hence self-dual.

Corrected: This connection, which is called an abelian anti-instanton, is not quite unique, because we could add to it $i$ times any closed 1-form; but if this closed form is exact, or represents $2\pi$ times an integral cohomology class, then the new connection is gauge-equivalent to the original one. Hence, the space of anti-instantons mod gauge is an torsor for the torus $H^1(X;S^1)$. The catch, I think, is that for generic Kaehler metrics in a fixed Kaehler class, there will be no such integral classes unless $\mathcal{H}^{1,1}_0=0$.

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I don't think I've understood it yet, but this is a cool interpretation. Tell me more about your catch: presumably to talk about whether or not a harmonic differential form is integral you are using the identification $\mathcal{H}^{0,2} + \mathcal{H}^{2,0} = H^{0,2} + H^{2,0}$. Isn't the right hand side, as a subspace of $H^2(X;C)$, independent of the Kahler form and even the Kahler class chosen? – David Treumann Jun 9 '10 at 3:11
Ah yes, you're right, these subspaces depend only on the complex structure. Apologies for the rash afterthought. What I had in mind was that, as one varies the (conformal class of the) Riemannian metric on a 4-manifold, one can realize arbitrary variations in the splitting of the harmonic 2-forms into (anti-)self-dual parts. So, if both summands of the splitting are non-zero, they generically miss all the lattice points. But my attempt to adapt this to Kaehler metrics was misguided... – Tim Perutz Jun 9 '10 at 3:45

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