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

Hi,

Given two (smooth, projective) curves $X$ and $Y$ over a field $k$, define a correspondence to be a line bundle $L$ on $X\times Y$. A trivial correspondence is a correspondence of the form $p_1^*L_1\otimes p_2^*L_2$. Define a function $N$ on the set of correspondences by $$N(L) = -\chi(L)+\chi_1(L)\chi_2(L),$$ where $\chi_1(L)$ (resp. $\chi_2(L)$) is the Euler characteristic of $L\mid X\times y$ (resp. $L\mid x\times Y$) for any $y\in Y$ (resp. $x\in X$) (the Euler characteristic is constant). Then it turns out that $N$ is well defined on the set of correspondences modulo trivial corresopndences, and it is a non-degenerate quadratic form. By this I mean that $N(L)\geq 0$ and $N(L) = 0$ if and only if $L$ is a trivial correspondence ('$N$ is non-degenerate') and the form $$ [L_1,L_2] = N(L_1\otimes L_2) - N(L_1) - N(L_2) - N(\mathcal O_{X\times Y}) $$ is bilinear ('$N$ is quadratic').

  • Question: what is going on here? It just seems magic that this works. Does it work more generally? In higher dimensions? Is there a complex analytic analog of this?

Thanks!

share|improve this question
1  
Is this just intersection theory on $X \times Y\phantom.$? By the index theorem, the intersection pairing on ${\rm NS}(X \times Y)$ is negative definite on the subgroup orthogonal to $X \times y$ and $x \times Y$, and I guess you're reconstructing this pairing multiplied by $-1$. –  Noam D. Elkies Nov 19 '11 at 4:38

1 Answer 1

up vote 2 down vote accepted

The Riemann--Roch says that there is a cohomology class $a \in H^{4}(X\times Y,Q)$ such that $$ \chi(L) = a - c_1(L)((g_X-1)[Y] + (g_Y - 1)[X]) + c_1(L)^2/2. $$ Let $c_1(L) = d_X[Y] + d_Y[X] + c_{10}(L)$ be the Kunneth decomposition. Note that $d_X = c_1(L)\cdot [X]$ and $d_Y = c_1(L)\cdot [Y]$. Then $$ \chi_1(L) = d_X - g_X + 1, \qquad \chi_2(L) = d_Y - g_Y + 1. $$ Substituting all this we see that $$ \chi(L) - \chi_1(L)\chi_2(L) = a - d_Y(g_X - 1) - d_X(g_Y - 1) + (2d_Xd_Y + c_{10}(L)^2)/2 - d_Xd_Y -d_X(g_Y - 1) - d_Y(g_X + 1) + (g_X - 1)(g_Y - 1). $$ One can see in fact that everything cancels with the only exception of the term $c_{10}(L)^2$ which gives your quadratic form. Now you can see that it is important that we work on a product (we use the Kuneth decomposition) and that the dimension of the product is 2 (for higher dimension there would be other terms of higher degree in Riemann--Roch). So, basically this fact is true only for product of curves.

On the other hand, you kight have something similar for two varieties of odd dimension $X$ and $Y$ such that $H^{odd}(X,Q)$ and $H^{odd}(Y,Q)$ are concentrated in degrees $\dim X = \dim Y$.

share|improve this answer
    
Thanks for your useful answer, Sasha. –  unknown Nov 19 '11 at 14:52

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.