Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
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
add comment

4 Answers 4

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.

share|improve this answer
    
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
add comment

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.

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

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 $J.$

share|improve this answer
    
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
add comment

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 $X$).

share|improve this answer
    
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
2  
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
1  
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
3  
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.