# Euler character of a numerically trivial divisor

Let $X$ be a complex projective variety (might be singular) and $D$ be a Cartier divisor on $X$. Suppose $D$ is ${numerically}$ trivial, then is the Euler character $\chi(X,D)= \chi(X, \mathcal{O}_X)$?

Here numerically trivial means the intersection number of the divisor $D$ with any curve is zero.

I saw somebody mentioned this result followed from Riemann-Roch, but did not see the reason (even for the smooth case).

• What does $\chi(X,D)$ mean? – Georges Elencwajg Aug 24 '13 at 17:27
• @GeorgesElencwajg $\chi(X,D)= \sum_{i=0}^{\infty}(-1)^{i}h^{i}(X,D)$. – Li Yutong Aug 24 '13 at 19:54
• Of course, but what is $H^i(X,D)$ ? – Georges Elencwajg Aug 24 '13 at 20:47
• Well... I really got confused. $H^{i}(X,D)$ is the i-th cohomology of $D$. Is there anything wrong with that? – Li Yutong Aug 25 '13 at 0:45
• I suppose this means cohomology of $\mathcal O(D)$ ? – Georges Elencwajg Aug 25 '13 at 6:26

By $H^i(X, D)$ I presume you mean $H^i(X, O_X(D))$, where $O_X(D)$ is the invertible sheaf associated to $D$.
Hirzebruch-Riemann-Roch says that the Euler characteristic of any vector bundle $E$ on a smooth projective variety $X$ is $\int_X ch(E)td(X)$, i.e. $ch(E)td(X)\cap [X]$ where $ch(E)$ is the chern character of $E$ and $td(X)$ is the todd class of $X$. Anyways, it doesn't really matter for the question at hand what the exact formula is, the point is the Euler characteristic can be expressed in terms of the topological data of $E$ and $X$, i.e. the chern classes of $E$ and of $X$. In our case $E=O_X(D)$ so the only possible non-zero chern class is $c_1(E)=D$. Let $n=dim X$.When we expand out $\int_X ch(E)td(X)$ the terms that contribute are of the form $D^i\alpha \cap[X]$ where $\alpha \in A^{n-i}(X)$ (Chow theory, or whatever cohomology theory you are working in) and by the numerically trivial hypothesis, any such term for $i>0$ is zero. And what we are left with is the exact same things we would have when we take $E=O_X$, since $c_1(O_X)=0$.
For $X$ singular, I don't have the answer. I heard that there are versions of Riemann Roch for singular $X$, but I don't know the precise statement and perhaps there is an alternative way, e.g. using resolution of singularities, but I don't know.
• Thank you! Could you elaborate on why " by the numerically trivial hypothesis" $D^i \alpha^{n-i}, i>0$ is zero? – Li Yutong Aug 25 '13 at 10:50
• @LiYutong I used poor notation, I fixed it to read $D^i\alpha \cap [X]=0$ where $\alpha \in A^{n-i}(X)$ (or $H^{2n-2i}(X, \mathbb{Z})$), for $i>0$. Let's take $i=1$. Then $\alpha \in A^{n-1}(X)$ , which means $\alpha \cap [X] \in A_1(X)$ is a $1$-cycle. And so $D \cap(\alpha \cap [X])=0$ since numerically trivial means the intersection of $D$ with a curve (or any 1-cycle as 1-cycles are linear combinations of curves) is zero. – ykm Aug 25 '13 at 23:55
• Thank you so much! I understand it now. So basically $\chi(X,D)$ only depends on the numerical class of $D$. – Li Yutong Aug 26 '13 at 22:11