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Let $F$ be a stable vector bundle of degree $d$ and rank $r$, with $(r,d)$ coprime and $X$ an elliptic curve. I know that I can construct an extension $$0 \to H^0(F) \otimes O_X \to G \to F \to 0 $$ such that the boundary map of the associated long sequence in cohomology is given by $Id: H^0(F) \to H^0(F)$. Now know that G is a vector bundle of rank $r+d$ and degree $d$. What I'd like to do is proving that $G$ is stable (here I have problems), because it'll let me proceed by induction with the computation of $M(r,d)$.

dolce
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ArthurStuart
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1 Answers1

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edit: I'm guessing, since you say there is an induced LES, that you mean the sequence is exact. Note there is an isomorphism between U(r,d) and U(r+d,d) given by 4.9 which is giving you the extension (the proof there does the isomorphism starting from the other direction however). In any case, assuming this is how you get the extension, $G$ is indecomposable. Now an indecomposable vector bundle on an elliptic curve is stable iff $(r,d)=1$ by 4.19. Now since $(r,d) = 1$, then $k | (r + d) \implies (k, r) = (k, d) = 1$ so any divisors of $\ r + d$ cannot divide $\ d$ so $(r + d, d) = 1$

aegbert
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  • First of all, thank you for the answer. Now, I know that it is sufficient to prove that $G$ is indecomposable because $r$ and $r+d$ are coprime. But what I can't understand is the reason why $G$ is indecomposable. Why does this extension force $G$ to be indecomposable. And, is it true that $H^0(F) \otimes O_X$ is semistable? – ArthurStuart Apr 05 '14 at 14:12
  • (1) As far as I can tell, $G$ does not have to be indecomposable, if you compare with the last paragraph or two of $4.9$ proof mentioned above. For the second question, I think you can just use the slope criterion and look at subbundles of $O_X^d$ which should be $O_X^{e<d}$, since they all have degree zero, then it should be semistable. Finally, I think you would usually do the classification by first doing indecomposables, and then saying that decomposables are direct sums of these. – aegbert Apr 05 '14 at 14:43