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I have a question relating to an answer on MathOverflow.net. The cited answer says:

Let $X$ be a topological space for which [the Euler characteristic] $\chi(X)$ is defined and behaves in the expected way for unions, Cartesian products, and quotients by a finite free action. ... [Then] $$\chi(X^{(2)}) = \frac{\chi(X \times X) - \chi(\operatorname{Diag}(X))}{2} + \chi(X) = \frac{\chi(X)^2 + \chi(X)}{2}$$ [where $X^{(2)}$ denotes the symmetric square of $X$].

Question: Does anyone know a reference for this result, or, failing that, a short proof? For the application that I have in mind I need the result for algebraic varieties over an algebraically closed field (whose characteristic may be positive), but a more general result would be nice to see.

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up vote 4 down vote accepted

The Cartesian product is the union of the diagonal and its complement, and a finite group $C_2$ acts freely on that complement with quotient the symmetric square minus a copy of $X$, so $$\chi(X \times X) = \chi(\text{Diag}(X)) + 2 (\chi(X^{(2)}) - \chi(X))$$

and the conclusion follows. (This is essentially combinatorics; run through the argument for $X$ a finite discrete space if this part is unclear.)

In practice, though, it seems to me that all of the work goes into verifying the "behaves in the expected way" hypothesis, and if that's what you were asking about then I have nothing useful to say (except that if the hypothesis is difficult to verify for the definition of Euler characteristic that you're using then consider using a different one).

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Thanks! I believe a slight generalisation of the statement should also be true, namely $\chi(\mathscr{O}_{X\times X}) = \chi(\mathscr{O}_{\operatorname{Diag}(X)}) + 2(\chi(\mathscr{O}_{X^{(2)}}) - \chi(\mathscr{O}_X))$ where $X$ is an algebraic variety and $\chi$ is the Euler characteristic of the sheaf cohomology. Does this generalisation follow easily (I don't immediately see how), or is it much more complicated? –  Hamish Aug 13 '12 at 17:07
    
@Hamish: I don't know enough about sheaf cohomology to say. Have you tried writing down a suitable exact sequence? –  Qiaochu Yuan Aug 13 '12 at 17:17
    
That was my first thought too. If $X = \operatorname{Spec}(A)$ is affine for example, then the only candidate sequence I can think of is $0 \rightarrow (A \otimes A)^{C_2} \rightarrow A \otimes A \stackrel{\phi}{\rightarrow} B$ where $B = \bigoplus_{g\in C_2} (A \otimes A) = (A \otimes A) \oplus (A\otimes A)$ and $\phi(a\otimes b) = (0, b\otimes a - a \otimes b)$. But this is not surjective. I think I'll ask it as a new question. –  Hamish Aug 13 '12 at 19:20
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