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Let $X$ be a (EDIT) variety with a group structure. For $a\in A$, let $t_a$ be the translation on $X$: $t_a(x) = a+x$.

Why is the function $f:X\to \mathbf{C}$ given by $f(a) = \sum_{i} (-1)^i \mathrm{Tr}(t_a^\ast, H^i(X,\mathbf{C}))$ continuous?

Here I consider the usual singular cohomology with $\mathbf{C}$-coefficients. (The coefficients don't really matter. You can even take $\mathbf{Q}_{\ell}$-coefficients and work with $\ell$-adic cohomology.)

I call the function $f$ on $X$ the trace function. Note that one can use the Lefschetz trace formula to see that the image of $f$ lies in $\mathbf{Z}$.

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    Yes, my variety is proper and smooth by assumption. (I should have included that!) The point is that I want to show that the Euler characteristic zero and from there conclude that my variety is$a$complex torus. This should be possible without using the theory in Mumford...(For example, suppose that $X$ is of dimension 1. Then it is well-known that the Euler char. is $2-2g$ where $g$ is the genus. You see that $g=1$ once you prove that the above trace function is continuous. Of course, you can also use more direct methods to show that $g=1$...This is what you're doing.)2012-02-04

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