## New answers tagged abelian-varieties

4

Well, this is surely false for elliptic curves, i.e. when $g=1$.
What is true is that, for any $t \in \mathbb{N}$, the number of elliptic curves in $A$ with $\deg _{\mathcal{L}} C \leq t$ is finite. However, it may happen that there exist elliptic curves on $A$ whose $\mathcal{L}$-degree is arbitrarily large.
This already happens in the case $A=E \times ...

2

$A_\mathbb C$ 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 ...

1

I am not sure I fully understand your question, but : $A_{\mathbb{C}}$ is the quotient of a complex vector space $V$ by a lattice $\Gamma $, which is canonically isomorphic to $H_1(A_{\mathbb{C}},\mathbb{Z})$. From this you get canonical isomorphisms $A[\ell^n]\cong H_1(A_{\mathbb{C}},\mathbb{Z})\otimes _{\mathbb{Z}}\mathbb{Z}/\ell^n\ $, $\ T_{\ell}(A)\cong ...

3

The fiber of the projection to $E'_i$, say $F_i$, is isomorphic to $E_j$ ($j\neq i$), not to $E'_j$. We have indeed $(F_1.F_2)=2$, and $F_1,F_2$ generate the group of divisors on $A$ up to numerical equivalence, because $E_1$ and $E_2$ generate the analogous group on $E_1\times E_2$.

1

Kleiman's paper "Algebraic Cycles and the Weil Conjectures", in the book "Dix Exposes Sur La Cohomologie des Schemas" is, I think, what you're looking for.

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