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Let $X$ be a fixed smooth, connected, compact Riemann surface of genus $g$. The Jacobian variety $\mbox{Jac}(X)$, which parametrises isomorphism classes of holomorphic degree $0$ line bundles on $X$, is the quotient $H^1(X;\mathcal O)/H^1(X;\mathbb Z)$ of cohomology groups, where $\mathcal O$ is the sheaf of holomorphic functions on $X$ and $\mathbb Z$ is the sheaf of locally-constant integer-valued functions on $X$. This construction of the Jacobian as a $g$-dimensional complex torus comes from the cohomology of the exponential sequence, as explained in most standard references.

The surface $X$ may admit more than one complex structure, if $g$ is positive. Exactly how does the choice of complex structure on $X$ influence the complex structure on $\mbox{Jac}(X)$? Can this be seen somehow in the quotient $H^1(X;\mathcal O)/H^1(X;\mathbb Z)$? (Does the embedding of the lattice $H^1(X;\mathbb Z)\cong\mathbb Z^{2g}$ into $H^1(X;\mathcal O)\cong\mathbb C^g$ determine the complex structure on $\mbox{Jac}(X)$, and is this embedding determined by the complex structure on $X$?)

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    $\begingroup$ Very nice question and very mature formulation. Bravo! $\endgroup$ – Georges Elencwajg Jan 2 '15 at 10:51
  • $\begingroup$ Thanks --- I do try. I appreciate the encouragement. Happy new year! :-) $\endgroup$ – MathsByTheSea Jan 2 '15 at 16:48
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All complex tori of dimension $g$ are diffeomorphic to $(S^1)^{2g}$, but as you say, there can be many complex structures. The embedding $j:\mathbb{Z}^{2g}\hookrightarrow\mathbb{C}^g$ does determine the complex structure, assuming we want the projection $\mathbb{C}^g\to\mathbb{C}^g/j(\mathbb{Z}^{2g})$ to be holomorphic.

The complex structure on $X$ determines the complex structure on its Jacobian in a simple way: the complex structure is already encoded in the sheaf $\mathcal{O}_X$, and therefore in $H^1(X,\mathcal{O}_X)$. So basically, by defining the Jacobian you are already assuming that $X$ has a complex structure attached to it.

The Torelli Theorem says precisely that the Jacobian of a Riemann surface completely determines the Riemann surface. One might naively think this means that there is a correspondence between complex structures on a surface of genus $g$ and complex structures on $(S^1)^{2g}$. However, there are two subtleties involved here:

  1. The Jacobian variety comes with a natural choice of an ample line bundle (its first Chern class can be seen as a positive definite Hermitian form on $H^1(X,\mathcal{O}_X)$ whose imaginary part is just the intersection pairing), and when talking about Jacobians, one usually assumes that you're talking about the torus along with (the first Chern class of) this line bundle. So really, it's not the complex structure of the Jacobian that determines the Riemann surface, but it's actually the complex structure along with this natural ample line bundle (called a polarization). There are examples of isomorphic Jacobian varieties that come from non-isomorphic curves. Of course this isomorphism will not respect polarizations.

  2. For $g>2$ there are complex tori which are not Jacobians (this is the famous Schottky Problem).

So basically, the complex structure of the curve determines the complex structure on the Jacobian, but not the other way around.

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  • $\begingroup$ This is a really nice answer! I appreciate the level of detail. If I can press just a little bit further, how does the complex structure on $X$ determine the embedding $j$? $\endgroup$ – MathsByTheSea Jan 2 '15 at 4:29
  • $\begingroup$ I will add this to my answer. $\endgroup$ – rfauffar Jan 2 '15 at 4:31
  • $\begingroup$ Hope that clears it up a bit! $\endgroup$ – rfauffar Jan 2 '15 at 4:35
  • $\begingroup$ Indeed, it does --- thanks!! $\endgroup$ – MathsByTheSea Jan 2 '15 at 4:37
  • $\begingroup$ Just a small clarification: $(S^1)^{2g}$, right? $\endgroup$ – MathsByTheSea Jan 3 '15 at 18:45

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