A question on deformations of Theta divisor in the Jacobian of a complex curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:15:02Z http://mathoverflow.net/feeds/question/95288 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95288/a-question-on-deformations-of-theta-divisor-in-the-jacobian-of-a-complex-curve A question on deformations of Theta divisor in the Jacobian of a complex curve aglearner 2012-04-26T19:14:49Z 2012-05-02T02:19:02Z <p>Suppose $C_g$ is a smooth compact complex curve (of genus $g$), and let $J$ be its <em>Jacobian</em>. 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 <em>theta divisor</em> $\Theta$ 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.</p> <p><strong>Question.</strong> How to calculate the dimension of the set of divisors on $J$ linearly equivalent to $\Theta$? In other words, what is $dim( H^0(J,\cal O(\Theta)))$?</p> http://mathoverflow.net/questions/95288/a-question-on-deformations-of-theta-divisor-in-the-jacobian-of-a-complex-curve/95318#95318 Answer by Felipe Voloch for A question on deformations of Theta divisor in the Jacobian of a complex curve Felipe Voloch 2012-04-27T02:14:41Z 2012-05-02T02:19:02Z <p>Since my comment was too cryptic, I will spell it out as an answer. </p> <p>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.</p> <p>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 $\dim H^0 =1$.</p> <p>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.</p> http://mathoverflow.net/questions/95288/a-question-on-deformations-of-theta-divisor-in-the-jacobian-of-a-complex-curve/95326#95326 Answer by Sebastian for A question on deformations of Theta divisor in the Jacobian of a complex curve Sebastian 2012-04-27T07:12:41Z 2012-04-27T07:12:41Z <p>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 $\Gamma=$ 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 $J.$</p> http://mathoverflow.net/questions/95288/a-question-on-deformations-of-theta-divisor-in-the-jacobian-of-a-complex-curve/95333#95333 Answer by Francesco Polizzi for A question on deformations of Theta divisor in the Jacobian of a complex curve Francesco Polizzi 2012-04-27T08:34:09Z 2012-04-27T12:34:13Z <p>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 $\Lambda$. </p> <p>By elementary linear algebra there is a basis of $\Lambda$ with respect to which $E$ is given by the matrix $$\left(\begin{matrix}0 &amp; D \cr - D &amp; 0 \end{matrix}\right),$$ 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 $i=1, \ldots ,g-1$.</p> <p>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.</p> <p>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 <em>principal polarization</em>, i.e. $D$ is the identity matrix. This can be seen by taking a standard homology basis for $H_1(C, \mathbb{Z})$. </p> <p>It follows $h^0(X, \Theta)=1$.</p> <p>See [Birkenhake-Lange, Complex Abelian Varieties, Chapters 3 and 11] for further details. </p> http://mathoverflow.net/questions/95288/a-question-on-deformations-of-theta-divisor-in-the-jacobian-of-a-complex-curve/95342#95342 Answer by inkspot for A question on deformations of Theta divisor in the Jacobian of a complex curve inkspot 2012-04-27T10:17:24Z 2012-04-27T10:17:24Z <p>This can also be done in a purely algebraic fashion. Take the given description of $\Theta$ as the locus of effective classes on <code>$X=Jac^{g-1}C$</code>. Put $Y=Jac^gC$. Pick a point $D$ on $Y$ with $h^0(C,D)=1$; Abel's theorem (that <code>$C^{(g)}\to Y$</code> is birational) ensures that this holds for all <code>$D$</code> outside some locus of codimension at least $2$. There is an embedding <code>$i_D:C\to X$</code> given by <code>$i_D(x)=K_C-D+x$</code>. 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 <code>$D_1$</code> in the linear system $|D|$, so $i_D(C)\cap\Theta$ is exactly the divisor <code>$D_1$</code>, 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 <code>$i_D(C)$</code>.</p> <p>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.</p> <p>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).</p> <p>Corollary: $\Theta^g=g!$ (from Riemann-Roch on $X$).</p>