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Given an abelian surface $A$ as the product of two non isogenous elliptic curves $E_1 $ and $E_2$ over $\mathbb{C}$. We also have a smooth 2 dimensional moduli space $M$ of sheaves on $A$ with some prescribed Chern classes.

I'm interested in finding some irreducible components of $M$. I know $M$ is not empty, i.e. there is a sheaf $E$ on $A$, representing a point in $M$. The sheaf is given as $E=O_A\oplus L_1 \oplus L_2\oplus (L_1\otimes L_2)$, where $L_i$ is a 2-torsion line bundle coming from a 2-torsion bundle on $E_i$.

My first idea was to embed $A$ into $M$ via its Picard variety, since $Pic^0(A)=A$ and these line bundles have vanishing Chern classes in $NS(A)$ . If $p\in A$ and $K_p$ is the associated line bundle, i looked at $\phi: A \rightarrow M$, $p \mapsto E\otimes K_p$. By Atiyah's version of the Krull-Schmidt theorem we have $E\cong E\otimes K_p$ if and only if $K_p \in$ { $O_A,L_1,L_2,L_1\otimes L_2$ }.

So the map $p \mapsto E\otimes K_p$ is not injective. If two line bundles $P$ and $Q$ satisfy $P\cong L_1\otimes Q$ or $P\cong L_2\otimes Q$ or $P\cong (L_1\otimes L_2)\otimes Q$, then we get $\phi(p)=\phi(q)$ if $P$~$K_p$ and $Q$~$K_q$.

So my idea was to "kill" the action of $L_1, L_2$ and $L_1\otimes L_2$. So define the group $G:=${$O_A,L_1,L_2,L_1\otimes L_2$}$\subset Pic^0(A)=A$.

Questions: Can we look at $A/G$? Is this an irreducible smooth projective surface? So is $A/G$ an irreducible component in $M$? Is it even an abelian surface, maybe the product product of two other elliptic curves? Or is this idea not so good at all?

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If I understand correctly what you are doing, $A/G$ is just $C_1\times C_2$, where $C_i$ is the quotient of $E_i$ b the subgroup of order 2 generated by $L_i$. So it is again a product of 2 elliptic curves. –  rita Dec 2 '10 at 19:06
    
Is it really this simple? I'm not so familiar with taking qoutients. But from what i read it can be quite complicated, so for example why should $E_1/C_1$ be smooth? –  TonyS Dec 2 '10 at 20:52
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You are dividing an abelian variety by a finite subgroup, so the quotient is again an abelian variety (in particular $G$ acts freely on $A$ and $A/G$ is smooth). In addition in your case the group $G$ is equal to $G_1xG_2$, were $G_i$ generated by $L_i$, and the action is the product of the action of $G_i$ on $E_i$, so $A/G=E_1/G_1\times E_2/G_2$. –  rita Dec 2 '10 at 22:21
    
Ah, interesting. Do you know some literature where i can read more about quotients of abelian varieties? PS: You can make this an answer, then i can click to set this as my accepted answer. –  TonyS Dec 3 '10 at 10:25
    
@TonyS: Thanks! –  rita Dec 3 '10 at 12:10
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up vote 6 down vote accepted

You are dividing a complex abelian variety by a finite subgroup, so the quotient is again an abelian variety (in particular $G$ acts freely on $A$ and $A/G$ is smooth). This is explained in any text on abelian varieties, for instance in Birkenhake-Lange.

In addition in your case the group $G$ is equal to $G_1\times G_2$, were $G_i$ is generated by $L_i$i, and the action is the product of the action of $G_i$ on $E_i$, so $A/G=E_1/G_1\times E_2/G_2$.

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So I finally found an irreducible component. Thanks a lot for your help. –  TonyS Dec 3 '10 at 14:35
    
You're welcome! –  rita Dec 3 '10 at 19:09
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