TITLE: A question on deformations of Theta divisor in the Jacobian of a complex curve
QUESTION [12 upvotes]: Suppose $C_g$ is a smooth compact complex curve (of genus $g$), and let $J$ be its Jacobian.  Recall that the Jacobian $J$ of a curve $C_g$ is a complex torus that can by obtained by contractions of all rational curves on the $g$-th symmetric power of $C_g$, e.i.,  $Sym^g(C_g)$. Recall also that there is a theta divisor $\Theta$ in $J$, 
depending on a point $p\in C_g$.
The divisor $\Theta$ is the image in $J$ of the set of points $(p,p_1,...,p_{g-1})$ with $p$ fixed.
Question. How to calculate the dimension of the set of divisors on $J$ linearly equivalent to $\Theta$? In other words, what is $dim( H^0(J,\cal O(\Theta)))$?

REPLY [8 votes]: This can also be done in a purely algebraic fashion. Take the given description of $\Theta$ as the locus of effective classes on $X=Jac^{g-1}C$. Put $Y=Jac^gC$. Pick a point $D$ on $Y$ with $h^0(C,D)=1$; Abel's theorem (that $C^{(g)}\to Y$ is birational) ensures that this holds for all $D$ outside some locus of codimension at least $2$. There is an embedding $i_D:C\to X$ given by $i_D(x)=K_C-D+x$. Then $i_D(C)\cap\Theta$ is the set of $x$ in $C$ such that $K-D+x$ is effective; by Serre duality, this is equivalent to $h^0(C,D-x)>0$. But $h^0(C,D)=1$, so the only such points $x$ are the points in the unique effective divisor $D_1$ in the linear system $|D|$, so $i_D(C)\cap\Theta$ is exactly the divisor $D_1$, regarded as a subscheme of $C$. So if $h^0(X,\Theta)=r+1$, then there is an $r$-dimensional subspace of sections in $H^0(X,\Theta)$ that vanish along $i_D(C)$.
We aim to prove that $r=0$. Consider the incidence scheme $W\subset Y\times |\Theta|$ consisting of pairs $(D,\Phi)$ with $i_D(C)\subset\Phi$. The fibers of $pr_1:W\to Y$ have dimension at least $r-1$, so $\dim W\ge g+r-1$, and so the fiber $pr_2^{-1}(\Theta)$ has dimension at least $g-1$. By Abel's theorem, as before, there is a point $E$ in $pr_2^{-1}(\Theta)$ with $h^0(C,E)=1$. Then $i_E(C)$ does not lie in $\Theta$, by the previous argument, contradiction.
You can then prove that $\Theta$ is ample, by showing that it is non-degenerate: the set of points $a$ on $A=Jac^0C$ such that $t_a^*\Theta$ is linearly equivalent to $\Theta$ is trivial. On any torsor under an abelian variety a non-degenerate line bundle with non-vanishing $H^0$ is ample (I'm going to give a blanket reference to Mumford at this point).
Corollary: $\Theta^g=g!$ (from Riemann-Roch on $X$).