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

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Riemann-Roch for abelian varieties. See Mumford's book. The dimension is one since the theta divisor is a principal polarization. – Felipe Voloch Apr 26 '12 at 22:39
Felipe, thanks for your comment! It is a bit cryptic though for me to understand why it is 1. Do I need to read the whole book of Mumford to understand the answer? Also, unfortunately I was never able to understand what is "principle polarization". – aglearner Apr 26 '12 at 23:30

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.

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Is there some simple way to explain that theta divisor is principal ? and ample ? I cannot remember now, it seems to me there was such an argument... Riemann-Roch for abelian varieties - we can use just Hircebruch's version and we are lucky that canonical class is equal to zero for any abelian variety so we get just ch( D) which is exp(D) = \sum_i D^i /i! - so the g-dimensional component is $D^g/g! $. – Alexander Chervov Apr 27 '12 at 6:56
Felipe, thank you for giving more details! Once we know that $D$ is ample we can indeed use Kodaira vanishing to deduce $H^i(A,O(D))=0$ for $i>0$. – aglearner Apr 27 '12 at 15:25

In general, let $L$ be a line bundle on a complex torus $X=V/ \Lambda$ of dimension $g$ and let $H$ be the hermitian form corresponding to the first Chern class $c_1(L)$. The imaginary part $E:= \textrm{Im}(H)$ is an alternating form which is integer-valued on the lattice $\Lambda$.

By elementary linear algebra there is a basis of $\Lambda$ with respect to which $E$ is given by the matrix $$\left(\begin{matrix}0 & D \cr - D & 0 \end{matrix}\right),$$ where $D=\textrm{diag}(d_1, \ldots, d_g)$ and the $d_i$ are strictly positive integers satisfying $d_i|d_{i+1}$ for all $i=1, \ldots ,g-1$.

If $L$ is positive-definite then one shows that $$h^0(X, L)=\textrm{Pf}(E)=\det(D).$$ The proof consists in explicitly writing a basis for $H^0(X, L)$ by using canonical theta functions, as in Sebastian's answer.

If $X=J(C)=H^0(\omega_C)^*/H_1(C, \mathbb{Z})$ is the Jacobian of a smooth curve, then the theta divisor $\Theta$ is a principal polarization, i.e. $D$ is the identity matrix. This can be seen by taking a standard homology basis for $H_1(C, \mathbb{Z})$.

It follows $h^0(X, \Theta)=1$.

See [Birkenhake-Lange, Complex Abelian Varieties, Chapters 3 and 11] for further details.

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Dear Francesco, thank you very much for the answer, it is very helpful! – aglearner Apr 27 '12 at 10:51

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 $X$).

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Since the fact in the corollary is purely topological it also has a topological proof. From the definition of Theta given above, its homology class is Poincare dual to the deRham class dX1^dY1 +...+dXg^dYg, where the one - forms dXi and dYj are Poincare dual to a symplectic homology basis on the curve. Hence the self intersection number of Theta is the coefficient of the g fold wedge product of this two form, i.e. g! – roy smith Apr 27 '12 at 19:30
Dear Roy, I have to say, I spent a lot of time today trying to understand how one could prove topologically that $\Theta^g=g!$, and I failed... Why is the Poincare dual given by the form you wrote down? – aglearner Apr 28 '12 at 0:28
I was afraid you would ask that. Ok, first the curve embeds in its jacobian with homology class A1xB1+...AgxBg, where Ai,Bj is a symplectic homology basis of H^1 of the curve. Then the theta divisor is the image of the symmetric product of the curve g-1 times, which is covered by a degree (g-1)! map by the cartesian product of the curve. I claim it follows that the theta divisor has homology class = to 1/(g-1)! times the (g-1) fold pontyragin product of the class of the curve, i.e. A1xB1x...xAg-1xBg-1 + .... +A2xB2x...AgxBg. Then you have to check this is Poincare dual to what I said. – roy smith Apr 28 '12 at 3:20
A reference should be chapter 11.2, p. 328, of Birkenhake and Lange's Complex abelian varieties. – roy smith Apr 28 '12 at 3:30
I am glad this helped. Note also that a similar argument, taking the g fold pontryagin product of the class of the curve, and dividing by g!, shows that the class of the image of the abel map from C^(g) is the fundamental class of the jacobian. Thus the g fold abel map has degree one. This implies "Jacobi inversion" by topological methods as well. By the easy connectedness theorem for fibers of this map to a smooth target, and the weak direction of abel's theorem, and Riemann Roch, the fibers are finite unions of projective spaces, so the converse of abel follows too. – roy smith Apr 28 '12 at 15:48

There also exist a Riemann surface approach to this question, as explained for example in Narasimhan's bok on compact Riemann surfaces, or also Griffiths-Harris: By Riemann's theorem, 'your' theta divisor is (up to translation) the same as the divisor of a holomorphic section in a bundle given by factor's of automorphy explicitly: Let $J=\mathbb{C}^g/\Gamma$ and $L\to J$ be a holomorphic line bundle. Since $\pi^*L$ is trivial whence pulled back to $\mathbb{C}^g$ there exists holomorphic functions $\varphi_\lambda,$ for $ \lambda\in\Gamma$ without zeros such that the trivilaisations $$ L_{\pi(z)}=(\pi^*L)_ z\cong \mathbb C $$ and $$L_{\pi(z)}= (\pi^*L)_ {z+ \lambda} \cong \mathbb C$$ differ by $\varphi_\lambda.$ The $\varphi_\lambda$ for the theta-bundle are, if we identify $\Gamma=<e_1,..,e_g,B_1,..,B_g>$ as usual,given by $\varphi_{e_l}=1$ and $$\varphi_{B_l}(z)=\exp^{-2\pi i z_l-\pi i B_{l,l}},$$ where $z=(z_1,..,z_g)$ and $B_l=(B_{1,l},..,B_{g,l}).$ All holomorphic sections in the bundle $L\to J$ are therefore given by functions $\theta$ on $\mathbb C^g$ satisfying $\theta(z+e_l)=\theta(z)$ and $\theta(z+B_l)=\varphi_l(z)\theta(z),$ but one can easily show, that there is only one such function up to multiplication by a constant, the famous theta function of $J.$

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The last sentence " but one can easily show, that there is only one such function up to multiplication by a constant, the famous theta function of J", does not make me happy. How to "easily see" ? – Alexander Chervov Apr 27 '12 at 7:18
I am sorry, I should have said: It is not that difficult, especially one does not need much theory. You make a Fourier expansion $f(z)=\Sum_{n\in\mathbb Z^g}a_n \exp^{2\pi i<n,z>}$ and than you use the 'periodicity' in the $B_l$ direction to show that all $a_n$ are determined by $a_0.$ – Sebastian Apr 27 '12 at 7:24

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