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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!

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One does not need 'mixed Hodge theory' for the alternating sum formula, just the long exact sequence in compactly supported cohomology and the fact that the compactly supported cohomology has the same Euler characteristic as the usual cohomology for algebraic varieties. Over $\mathbb{C}$ this can be seen from the fact that links of points on algebraic varieties have Euler number zero, which in turn one can see by using resolution of singularities to reduce to the same statement for odd dimensional spheres. –  Vivek Shende May 11 '13 at 4:17

1 Answer 1

up vote 1 down vote accepted

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.

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Sorry for the typo Will: it is corrected now. And thanks for your example! Do you think there is some kind of Riemann-Roch computation of this number, using that $\chi(Z)$ is up to sign the degree of the top Chern class of $\Omega_Z$? –  hodgpol May 10 '13 at 6:09
    
You might be able to, in terms of the Chern classes of the $D_i$. You can use the adjunction formula to compute the total Chern class of $\Omega_{D_J}$, and use the fact that the divisors are normal crossings to compute the fundamental class of $D_J$ as the cup product of the classes of $D_i, i\in J$. Then you can sum over all intersections. I don't think you will see the kind of cancellation that makes the formula for the simple alternating sum so nice, though. –  Will Sawin May 10 '13 at 6:41
    
Do you have a reference where this kind of computation is done? –  hodgpol May 10 '13 at 7:03
    
I don't have a reference. Does this help? –  Will Sawin May 11 '13 at 2:25
    
Hey Will, thanks a million for your help! Could you just explain a bit more your notation? For instance, what is P? –  hodgpol May 11 '13 at 11:38

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