# non degenerate quadratic form on the group of correspondences on an algebraic curve?

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!

• 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

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 $\dim X = \dim Y$.