Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Question: why are there no non-trivial extensions of $\mathbb{G}_m$ by abelian varieties?

Specifically, let $A$ be an abelian variety over a field $k$. Then it seems to be well known that any extension $$ 0 \to A \to G \to \mathbb{G}_m \to 0$$ in the category of group varieties over $k$ is split. For instance, this is mentioned in Deligne's Theorie de Hodge I page 426 just before Section 3. Chevalley's structure theorem of algebraic groups certainly implies this.

Question: Is there a simple/direct proof?

Note that there are non-trivial extensions of abelian varieties by tori (semiabelian varieties).

Thanks.

share|improve this question
2  
I would have thought that Chevalley's theorem on the structure of connected algebraic groups would be an ideal answer to the question. To me, it seems easier than trying to compute an Ext group... –  Pete L. Clark May 10 '11 at 4:07
2  
How, does Chevalley's theorem imply this? I can see how it implies that the extension splits after pulling back along some $n$'th power map $\mathbb{G}_m \to \mathbb{G}_m$ but not that it is actually trivial. –  Torsten Ekedahl May 10 '11 at 4:18
    
@Pete: Thanks for the comment. @Ekedahl: You are right, I misspoke! Thanks for clarifying! –  SGP May 10 '11 at 13:06
add comment

1 Answer

Using the Kummer exact sequence $0\rightarrow\mu_n\rightarrow\mathbb{G}_m\rightarrow\mathbb{G}_m\rightarrow0$ we get a long exact sequence $$ 0\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)\rightarrow\mathrm{Hom}(\mathbb{G}_m,A) \rightarrow\mathrm{Hom}(\mu_n,A)\rightarrow\mathrm{Ext}^1(\mathbb{G}_m,A). $$ As $\mathrm{Hom}(\mathbb{G}_m,A)=0$ this gives an embedding $\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and at least over an algebraically closed field in which $n$ is invertible we always have that $\mathrm{Hom}(\mu_n,A)$ is non-trivial and consequently so is $\mathrm{Ext}^1(\mathbb{G}_m,A)$.

This does not contradict Deligne's claim as he seems to be (somewhat implicitly it must be admitted) speaking of extensions up to isogeny (in Principle 2.1 this is made more clear). In fact Chevalley's theorem implies that this is true: Consider a connected affine subgroup $G'\hookrightarrow G$ for which the quotient is an abelian variety. Then the kernel of the composite $G'\rightarrow\mathbb{G}_m$ is affine and embeds in $A$ and thus is finite. On the other hand $G'\rightarrow\mathbb{G}_m$ must be surjective as otherwise $G'$ would be finite and hence trivial so $G$ would be an abelian variety which is not possible. The kernel of $G'=\mathbb{G}_m\rightarrow\mathbb{G}_m$ is then some $\mu_n$ which shows that the extension is in the image of $\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and in particular is trivial up to isogeny.

The final conclusion is that $\mathrm{Ext}^1(\mathbb{G}_m,A)=\mathrm{Hom}(\hat{\mathbb Z}(1),A)$.

Addendum: I just realised that I didn't answer the actual question about whether there is a simpler proof (to the isogeny statement would be my interpretation). I doubt it (depending of course somewhat on your definition of simpler). A proof (arguably less simple) avoiding the use of Chevalley's theorem (but assuming to be on the safe side that the base field is algebraically closed of characteristic $0$) would be to prove that for some $n$ the pull back of the extension along the $n$'th power on $\mathbb{G}_m$ is trivial as an $A$-torsor because then the extension would be described by a $2$-cocycly $\mathbb{G}_m\times\mathbb{G}_m\rightarrow A$ and any map of that form is constant. That in turn will follow from the fact that for some $m$ the class of this extension as an $A$-torsor would be in the image of $H^1(\mathbb{G}_m,A[m])\rightarrow H^1(\mathbb{G}_m,A)$. This follows from the fact that $H^1(\mathbb{G}_m,A)$ is torsion and then the conclusion follows as any class of $H^1(\mathbb{G}_m,C)$ for any finite locally constant sheaf $C$ is killed by pulling back along some $n$'th power map.

share|improve this answer
    
You are being to kind: it does contradict Deligne's statement. He should have said it is trivial modulo torsion (which is all he needs). –  mephisto May 10 '11 at 13:01
    
Many thanks for your clarifications and very informative answer. I did not realise that the principle of Deligne was only up to isogeny. Thanks! –  SGP May 10 '11 at 13:07
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.