Question: 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 $dim( H^0(J,\cal O(\Theta)))$?

Answer: 
Since my comment was too cryptic, I will spell it out as an answer. 

A polarization on an abelian variety $A$ is an ample divisor $D$ (modulo linear equivalence). A polarization is principal if the self-intersection $D^g$ is equal to $g!$, where $g = \dim A$. It is well-known that the theta divisor is a principal polarization of the Jacobian. I am not sure what is a good reference, maybe Griffiths-Harris.

The Riemann-Roch theorem for abelian varieties (proved e.g. in Mumford) says that the Euler characteristic of the line bundle corresponding to a divisor $D$ is $D^g/g!$, so it is $1$ in the case of a principal polarization. Now, it is also shown in Mumford that only one $\dim H^i$ is non-zero, so if the divisor is ample it has to be $H^0$. Putting it all together $\dim H^0 =1$.

Edit: I was informed by email that my definition of polarization is too restrictive, so not quite right. The theta divisor defines a principal polarization anyway and this is proved in some of the other answers.
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 $X$).