In the paper "Sato-Tate Distributions and Galois Endomorphism Modules in Genus 2" (arxiv: <http://arxiv.org/abs/1110.6638>), the authors use the singular homology $H\_1(A\_\mathbb{C}^{top},\mathbb{Q})$ ($A$ is an abelian variety). I have seen this construction other places, but most recently here. I am not that well versed in homolgy but I looked up what the singular homology is and I think I understand it. However I don't see what it represents in the abelian variety case. 


In the paper they state that if $V\_\ell(A)$ is the rational Tate-module then 


$$V\_\ell(A) \cong H\_1(A\_\mathbb{C}^{top},\mathbb{Q}) \otimes \mathbb{Q}\_\ell$$


I have studied a bit on abelian varieties over finite fields and in that case we get


$$V\_\ell(A) \cong T\_\ell(A) \otimes \mathbb{Q}\_\ell$$


where $T\_\ell(A)$ is the (non-rational) Tate-module. From there we get that 


$$A(\mathbb{F}\_q)\_\ell \cong T\_\ell(A) / (1-F)T\_\ell(A)$$


where $F$ is the Frobenius. Morever $V\_\ell(A) \cong V\_\ell(B)$ if and only if $A \sim B$. That is, the relationship between $T\_\ell(A)$ and $V\_\ell(A)$ gives a way to look at the group that can appear inside a isogeny classes.


Is the relationship between $V\_\ell(A)$ and $H\_1(A\_\mathbb{C}^{top},\mathbb{Q})$ similar for abelian varieties not over finite fields as the relationship between $V\_\ell(A)$ and $T\_\ell(A)$? Obviously it won't be exactly the same as we do not have a Frobenius endomorphism. If not, what is the relevance of $H\_1(A\_\mathbb{C}^{top},\mathbb{Q})$ and what can we interpret it as?



Interpretation of $H_1(A_\mathbb{C}^{top},\mathbb{Q})$
$A\_\mathbb C$ is not defined in the finite field case because you can't base change from a finite field to $\mathbb C$.


$H^1( A\_\mathbb C^{top})$ is a vector space whose dimension is $2g$. One can define $g$ in terms of the Tate module if you'd like, (or even more simply using the fact that the number of $\ell$-torsion points is $\ell^{2g}$), so this gets you $H^1(A^{top}\_\mathbb C)$ up to isomorphism.


To not just define it up to isomorphism but construct it explicitly you must go beyond algebraic tools like $T\_\ell$ and $V\_\ell$. The reason for this is that those constructions, because they are algebraic, carry an action of $\operatorname{Gal}(\mathbb C|\mathbb Q)$. (assuming $A$ is defined over $\mathbb Q$.) But as $\operatorname{Gal}(\mathbb C|\mathbb Q)$ is a profinite group, any action of it on a finite-dimensional $\mathbb Q$-vector space must factor through a finite group. But there is no reasonable action of this type on $H^1$ (You can see this because the isomorphism you wrote down should be Galois-equivariant, but the action of Galois on the Tate module does not factor through a finite group.)


So you must use a non algebraic construction - in particular, complex analysis and topology. At this point, you notice that $A\_{\mathbb C}^{top}$ is topologically equal to a torus, and you can define it in terms of the torus, as in abx's answer.


But there is no purely algebraic way to achieve the same goal.


