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Let $X$ be a complex algebraic variety, possibly singular and/or non-compact. It is well known that if $X$ is smooth then its Euler characteristic is equal to its Euler characteristic with compact support: $\chi(X)=\chi_c(X)$.

Questions. (1) Is the same equality true if $X$ is singular? (I think it is still true for singular curves.)

(2) Is there an elementary proof of this fact at least in the smooth case?

UPDATE: As mentioned by Reladenine Vakalwe below the proof in the smooth case follows immediately from the Poincare duality.

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up vote 4 down vote accepted

Let $e(-)$ denote ordinary Euler characteristic and $\chi_c(-)$ the compactly supported version.

Complex varieties admit Whitney stratifications. In particular, each closed stratum in such a stratification is a strong deformation retract of a tubular neighborhood. It follows (Mayer-Vietoris) that if $Y\subseteq X$ is such a closed strata, then $e(X) = e(Y) + e(X-Y)$.

Added later: Let me try and elucidate the above claim a bit more. For intuition let me make an informal remark: one of the reasons Whitney stratifications are so nice is that if one thinks of strata as spaces glued (or perhaps "linked" is a better word here) together, then in a Whitney stratification the gluing happens using locally trivial fibre bundles.

Now back to the claim above. Let $T$ be the tubular neighborhood of $Y$. This is a fibre bundle over $Y$, with fibre homeomeorphic to the mapping cone $cone(L)$ of a compact space $L$ (the link). Removing $Y$ from this neighborhood amounts to removing the vertex of $cone(L)$ (the "zero section"). So one is reduced to showing that the Euler characteristic of $L$ is $0$. Now the Whitney stratification of $X$ yields a Whitney stratification of $L$ by odd (real) dimensional oriented manifolds. By induction one sees that such a space must have zero Euler characteristic (I think this is originally an observation of Sullivan; the intuition is the same: at each step of your stratification you are attaching a compact space with the above properties to an oriented odd dimensional manifold using locally trivial maps whose fibre satisfies the same properties).

Now, if $X = \bigsqcup_i X_i$ is a Whitney stratification with each $X_i$ smooth, then by Poincare duality, $e(X_i) = \chi_c(X_i)$ for each $i$. Let $Y$ be a stratum of minimal dimension. So $Y$ is closed. By induction on the number of strata we may assume $e(X-Y) = \chi_c(X-Y)$. Now use the observation of the previous paragraph and you are done.

Added even later: I might be getting a bit carried away here (so do point out if you have a counterexample), but I think the following souped up version is true. Let $f\colon X\to Y$ be a morphism between complex varieties. Let $K_0(X)$ and $K_0(Y)$ denote the Grothendieck groups of the bounded derived categories of algebraically constructible sheaves on $X$ and $Y$. Then we have two maps $[f_*], [f_!]\colon K_0(X) \to K_0(Y)$. I want to say that $[f_*] = [f_!]$. I think I have a sketch of a proof of this using resolution of singularities (although maybe using a variant of the stratification method here might do it too). Anyway, food for thought.

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To deduce additivity of Euler characteristic you need to use also that all strata are even dimensional. For instance additivity fails for a point on the real line. The point being that the intersection of the two opens in the M-V sequence is a sphere bundle, and only odd spheres have vanishing Euler characteristic... – Dan Petersen Jun 10 '14 at 14:00
@DanPetersen: The strata are complex varieties, so even real dimension (am I being silly with something?). Maybe I should have said that complex varieties admit algebraic stratifications that satisfy the Whitney conditions. – Reladenine Vakalwe Jun 10 '14 at 14:10
@DanPetersen: I do not see that one gets sphere bundle. My impression is that it should be a bundle with odd dimensional fibers, perhaps even stratified by odd dimensional strata. If the fiber is smooth then its Euler characteristic vanishes anyway. But if it is not smooth then I am less sure. – sva Jun 10 '14 at 16:43
You are of course right (I was too hasty with my comment picturing only smooth points in my head, sorry). However, the Whitney conditions on the strata yield more control on the local picture at each point of the stratum. Instead of trying to explain myself, let me point to notes of MacPherson:, see Theorem 7.2 and 7.3 on page 138. I hope that clears things up? – Reladenine Vakalwe Jun 11 '14 at 17:33
Perhaps, the relevant point here is that the Euler characteristic of the link of the closed stratum is $0$. In fact, I think a variant of this argument yields that the Euler characteristic and compactly supported Euler characteristic with coefficients in any algebraic constructible complex of sheaves on $X$ coincide. This latter statement might actually even be an easier statement to prove (I am thinking along the lines of the proof of Artin vanishing that uses induction on dimension, but it's too late at night for me to think through this properly right now). – Reladenine Vakalwe Jun 12 '14 at 2:26

For any finite type separated scheme over an algebraically closed field and any prime $l$ different from the characteristic, the equality of the $l$-adic Euler characteristic and the compactly supported $l$-adic Euler characteristic has been proved by Laumon, see his paper "Comparaison de caracteristiques d'Euler-Poincare en cohomologie l-adique". Section 1 of Illusie, Zheng "Odds and ends on finite group actions and traces" discusses further generalizations. In case you are using Betti cohomology, you may need to supplement the above by the relevant comparison theorem.

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