# Why is the trace map on an abelian variety continuous

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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I may be missing something, but isn't the pullback morphism $t^*_a$ the identity map on each $H^i(X,\mathbb C)$? This should follow from considering either the canonical isomorphism $H^i(X,\mathbb C) = \bigwedge^i T_X^*$ or from the De Rham isomorphism via forms. In any case the function $f$ is then constant, equal to the topological Euler characteristic of $X$. – Gunnar Þór Magnússon Feb 4 '12 at 10:29
Actually, this is exactly what I'm trying to prove. Namely, I have a variety $X$ with a (compatible) group structure. I want to prove that its Euler characteristic is zero. I know that the trace of $t_a^\ast$ is zero (by the Lefschetz trace formula) since $t_a$ has no fixed points when $a$ is not the identity element in $X$. Then, by the continuity I want to conclude that the trace of $t_0^\ast = \textrm{id}_X^\ast$ is zero. In particular, the Euler char. is zero. I'm not sure if I can use.... – Ali Feb 4 '12 at 10:57
.. that $H^i = \Lambda^i H^1$ assuming that $X$ is a variety with a group structure. (In the question I wrote that $X$ was an abelian variety. I don't think I should have done that.) – Ali Feb 4 '12 at 10:58
Hmm... is your variety an algebraic group? Is it compact? If so then it is a complex torus (see Chapter 1 of Mumford's "Abeliean varieties") and everything works as before. If the variety is not compact then I don't really know what happens, but perhaps you could argue by saying that translations correspond to the flow of holomorphic vector fields, which are (almost by definition) homotopic to the identity and thus act trivially on cohomology. This is modulo any trouble you may have with possibly infinite cohomology groups in this case. – Gunnar Þór Magnússon Feb 4 '12 at 12:50
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.) – Ali Feb 4 '12 at 13:24