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Let $K$ be an algebraically closed field. Assume for simplicity that $K$ has characteristic zero. Let $A/K$ be a semiabelian variety. Let $n$ be an integer coprime to $char(K)$. Denote by $\pi_1(A)$ the etale fundamental group of $A$. Is it true that $Hom(\pi_1(A), {\mathbb{Z}}/n)$ is isomorphic to $A[n]^\vee$ (edit)? What happens, if I replace $A$ by an arbitrary connected algebraic group over $K$?

Remark: The answer is ``yes'' in the case where $A/K$ is proper. (The key ingredient to the proof of this fact is the theorem of Serre-Lang, that every finite etale cover $X/A$ carries the structure of an abelian variety, provided $A/K$ itself is an abelian variety.)

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What is $A^\vee[n]$ for a semiabelian variety $A$? –  Mikhail Borovoi Feb 21 '11 at 17:40
    
Sorry, I probably meant $A[n]^\vee(K)$, where $-^\vee$ in the Cartier dual. I will edit the question tomorrow accordingly. (I only have the mobile phone at the moment) –  Sebastian Petersen Feb 21 '11 at 20:50
    
OK, what is $A[n]$? Is it the group of points of order dividing $n$? –  Mikhail Borovoi Feb 21 '11 at 21:35
    
I know you restricted to char. 0, but it's false in char. $p>0$ already for $\mathbb{G}_m$, which has a ton of $p$-covers. –  Felipe Voloch Feb 21 '11 at 21:54
    
In my notation $A[n]$ stands for the kernel of multiplication with $n$ on $A$. It is a finite etale group scheme over $K$. I sometimes identify a finite etale group scheme over an alg closed field with its points. So I ask whether $Hom(\pi_1(A), Z/n)$ is $Hom(A[n], Z/n)$, where $A[n]$ are the $n$-torsion points. –  Sebastian Petersen Feb 21 '11 at 22:21
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1 Answer

I can give a partial answer, or at least a strategy towards my question, myself. I hope that this will also put my question into the right context, and increase its chances of being answered completely. (I admit that it was not so well-formulated at the beginning.)

Let $K$ be algebraically closed of characteristic zero (for simplicity). Let $X/K$ be a separated connected algebraic $K$-scheme. Then we have $H^1(X, \mathbb{Z}/n)=Hom(\pi_1(X), \mathbb{Z}/n)$. On the other hand we have an exact sequence $$1\to \Gamma(X, {\cal{O}}_X)^\times/n\to H^1(X, \mathbb{Z}/n)\to H^1(X, {\mathbb{G}}_m)[n]\to 1\ (*),$$ associated to the short exact sequence of etale sheaves $$1\to \mu_n\to \mathbb{G}_m\to \mathbb{G}_m\to 1$$ (and noting that $\mu_n=\mathbb{Z}/n$ because $K$ is algebraically closed).

Furthermore $H^1(X, \mathbb{G}_m)=Pic(X)$, the group of invertible sheaves on $X$. If $X$ is proper in addition, then $\Gamma(X, {\cal{O}}_X)^\times=K^\times$. Hence there is an isomorphism $$Hom(\pi_1(X), \mathbb{Z}/n)=H^1(X, \mathbb{Z}/n)\cong Pic(X)[n]$$ for every connected proper $K$-scheme.

Now consider the special case where $A/K$ is an abelian variety. Then $$Pic(X)[n]=A^\vee[n]=A[n]^\vee=Hom(A[n], \mathbb{Z}/n);$$ hence we obtain in fact a canonical isomorphism $$Hom(\pi_1(A), \mathbb{Z}/n)=Hom(A[n], \mathbb{Z}/n).$$ This is where the "well-known" isomorphism $\pi_1(A)\cong \prod_\ell T_\ell(A)$ comes from. (The fact that $\pi_1(A)$ is abelian has to be shown in addition.)

One also sees that $H^1(A, Z_{\ell})$ is dual to the Tate module $T_\ell(A)$ ($\ell$ a prime number).

Now let $B$ be a semiabelian variety. I asked the above question, because I wanted to know, whether the situation is similar in the case of a semiabelian variety. For example, I wanted to know:

i) Is $Hom(\pi_1(B), \mathbb{Z}/n)$ (canonically) isomorphic to $Hom(B[n], \mathbb{Z}/n)$?

ii) Is there a useful relation between $\pi_1(B)$ and $\prod_\ell T_\ell(B)$, where $T_\ell(B)=lim_i B[\ell^i]$ is the Tate module (defined in a naive way analoguos to the proper case). Are these groups canonically isomorphic?

iii) Is there a useful relation between $H^1(B, Z_\ell)$ and $T_\ell(B)$? Is the first $\mathbb{Z}_\ell$-module canonically isomorphic to the dual of the second?

I think, this case of semiabelian varieties is somewhat different, because $\Gamma(B, {\cal{O}}_B)^\times/n$ does not vanish any more, unless $B$ is proper.

But nevertheless these questions still make sense to me. Further comments / answers are appreciated very much.

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