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Let $X$ be a smooth projective, complex variety. Denote by $\bar X$ its analytification, i.e. $X$ with the "classical" topology of a complex manifold. Now do these spaces have the same Euler characteristic $\chi(X)=\chi(\bar X)$? Why (not)?

Edit: To clarify, by $\chi(X)$ I mean the topological Euler characteristic of the underlying topological space of the scheme $X$, not the Euler characteristic $\chi(X,\mathcal{O}_X)$ of its structure sheaf.

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up vote 7 down vote accepted

On the topological side, we have the Fröhlicher relation (where $\Omega^p_X$ is the sheaf of regular algebraic $p$-forms on $X)$

$$\chi(X^{top })=\sum_{p,q} (-1)^{p+q}dim_{\mathbb C} H^q(X,\Omega^p_X)$$ It is quite remarkable in that the left-hand side can be computed if you just know the underlying topological space of $X$, whereas the right hand side, involving coherent sheaf cohomology, requires for its computation the algebraic structure of $X$.

On the algebraic side, given a smooth algebraic variety $X$ of dimension $n$, its Euler characteristic $\chi(X)$ may be taken to mean the integer $\chi(X)=\sum_{i} (-1)^{i}dim_{\mathbb C} H^i(X,\mathcal O_X)$, an invariant closely related to the arithmetic genus $p_a(X)$ by the formula $p_a(X)=(-1)^n(-1+\chi (\mathcal O_X) )$ .
For a curve Riemann-Roch tells you that $\chi (\mathcal O_X)=1-g$, where $g$ is the good old genus of that curve, and so we have $p_a(X)=(-1)^1(-1+(1-g))=g$, which is reassuring!
For higher dimensional varieties we have the formula ( a special case of Hirzebruch's Riemann-Roch formula) $$\chi (\mathcal O_X)=t_{2n}[Todd(X)]$$

The right-hand side, the Todd genus, takes some time to unravel, but the amazing thing is that it only depends on the differentiable structure $X^{diff}$ of $X$, whereas the individual terms of the alternating sum on the left-hand side depend on the algebraic structure of $X$.
For $n=2$, the case of a surface, this unpacks to Max Noether's formula

$$ \chi (\mathcal O_X)=\frac{c_1^2(X)+c_2(X)}{12} $$
where the $c_i$'s are the Chern classes of the differentiable tangent bundle.
Note that it is a priori not in thre least evident that the right-hand side is an integer, i.e. that its numerator is divisible by $12$ !

(Emmy Noether, of noetherian fame, was Max's daughter by the way)

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By $\chi(X)$, I do not mean $\chi(\mathcal{O}_X)$, but the actual topological Euler characteristic of the topological space $X$ (with its 'locally Zariski' topology). Those should be very different, if I am not mistaken? – Jesko Hüttenhain Aug 19 '11 at 15:18
If $X$ is irreducible, the constant sheaf $\mathbb Z$ will be flabby, hence acyclic and so in this sense $\mathbb\chi(X,\mathbb Z)=1$. If you insist on singular homology for the Zariski toplogy, I doubt that you will find anything interesting: otherwise Grothendieck wouldn't have invented étale cohomology... – Georges Elencwajg Aug 19 '11 at 15:37
I'm glad about that, rattle. – Georges Elencwajg Aug 19 '11 at 15:43
Is $\chi(X,\mathbb{Z})$ the same as the euler characteristic of $X$, defined via singular homology with coefficients in $\mathbb{Z}$? If yes, then all my questions have been answered. – Jesko Hüttenhain Aug 19 '11 at 15:44
I suppose I wasn't clear: I think that $\chi_{sing} X^{Zar}\neq \chi_{sing} X^{classical}$, where on both sides you have singular homology, but on the left-hand side you take $X$ with its non-Hausdorff Zariski topology (=all open sets huge) and on the right-hand side you take $X$ with its classical, Hausdorff, metric topology (=some open sets very small). – Georges Elencwajg Aug 19 '11 at 15:57

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