a question on Euler characteristic of normal crossing divisors

Question

Let $X$ be a smooth, projective complex algebraic variety. Let $D$ be a simple normal crossings divisor on $X$, with irreducible components $D_i$, for $i \in I$. For each non-empty subset $J \subset I$, put $$ D_J=\bigcap_{i \in J} D_i $$ Define also $D(0)=X$ and, for $m \geq 1$, $$ D(m)=\bigsqcup_{|J|=m} D_J $$ Then $D(m)$ is a smooth closed subvariety of $X$ of codimension $m$. Put also $U=X-D$. Finally, denote for any smooth, complex algebraic variety $Z$ $$ \chi(Z)=\sum (-1)^k \dim H^k(Z, \mathbb{Q}) $$

One knows from mixed Hodge theory that the Euler characteristic of $U$ is determined by the alternating sum of the Euler characteristics of $D(m)$, that is $$ \chi(U)=\sum_{m \geq 0} (-1)^m \chi(D(m)) $$

My question is: is there a formula computing

$$ \sum (-1)^m m \chi(D(m))=-\chi(D(1))+2\chi(D(2))-3\chi(D(3))+\cdots $$

somehow in terms of $U$ or just the first $D(m)$?

I tried to derivate the Hodge polynomial but I didn't get anything. Thanks for your help!

Answer 1

Your formula is a little bit wrong. You want $D(0)= X$, not $D$, and your summation should start at $m=0$.

This shouldn't work. Let $X$ be $\mathbb P^3$ with $2$ lines on a quadric surface blown up, becoming copies of $\mathbb P^1 \times \mathbb P^1$. Let $D$ be the union of the quadric surface and the exceptional divisors. Then $X- D$ is the same in both instances.

Then if the two lines intersect: $\chi(D (0) ) = 6$, $\chi(D (1)) = 8$, $\chi(D(2))=2$, $\chi(D(3))=0$.

If the two lines do not intersect: $\chi(D(0))=8$, $\chi(D(1)) = 8$, $\chi(D(2))=0$, $\chi(D(3))=0$.

So you get $-4$ in the first case but $-8$ in the second. And there should be no way to deal with this in general. Unless you know enough about $U$ that you can reconstruct the whole arrangement, or the whole arrangement up to some operation that, like blowing up points on a surface, just happens to leave this quantity unchanged, you will not be able to compute this number.

Edit: Let $T$ be the total Chern class of $\Omega_X$. Then by the adjunction formula, the total Chern class of $\Omega_{D_J}$ is $T$ divided by the Chern class of the ideal of functions vanishing on $D_J$. Since $D_J$ is a complete intersection, this is isomorphic to $\prod_{i \in J} (1- c_i)$. To get $\chi(D_J)$ , we extract the top Chern class by hitting it with the fundamental class of $D_J$, which is the fundamental class of $X$ times the cycle class of $D_J$. Since $D_J$ is a transverse intersection, its cycle class is $\prod_{i \in J} c_i$. So the generating function:

$$ \sum_m \chi(D(m)) (-t)^{m} = \sum_{J \subset I} (-t)^m P\left( T \cup \prod_{i \in J} \frac{c_i}{1-c_i}\right)=P\left(T \cup \sum_{J \subset I} \prod_{i \in J} \frac{-t c_i}{1-c_i} \right) = P \left (T \cup \prod_{i \in I} \left ( 1 - \frac{t c_i}{1-c_i} \right) \right) $$

To get the Euler characteristic of $U$, we just take the generating function and evaluate it at $t=1$. To get your sum, we just differentiate the generating function at $t=1$.

I think there might also be an elegant way to express this using Grothendieck-Hirzebruch-Riemann-Roch.