I'm reading through Atiyah's paper that classifies vector bundles over an elliptic curve, and I'm a little confused about one of his proofs.

Lemma 15(i) states that if $E \in \mathcal{E}(r,d)$ is a vector bundle of rank $r$ and degree $d\geq0$ over $X,$ then $s:=h^0(X,E)=d$ if $d>0$ and $s=0$ or $1$ if $d=0.$

For the proof, if $d\geq r,$ choose a maximal splitting $E=(L_1 ,...,L_r)$ with each $L_i>1$ (where 1 denotes the trivial line bundle). But in Lemma 11 he has the equation

$d=deg(E)=\sum^{r}_{1} deg(L_i).$ If each $L_i>1$ doesn't that mean $deg(L_i )\geq2$ so $deg(E)\geq2r?$ But we are only assuming $d\geq r.$

Anyways, If $deg(L_i)\geq 2 $ then it's clear that $H^1 (X,L_i)=0$. Is that how he claimed that $H^1 (X,L_i)=0$ in the next line or is he using something else?

EDIT: Actually, we don't necessarily have $deg(L_i)\geq 2$ since $1$ denotes the trivial line bundle which has degree $0$. So if $L_i >1$ it just means that $deg(L_i)>0.$ Still, how do we know that $L_i >1$? And how does this imply $H^1 (X,L_i)=0$?

We say that $L_1 \geq L_2$ if $\Gamma Hom(L_2 ,L_1)\neq 0.$ It implies that $deg(L_1)\geq deg(L_2).$

$\Gamma Hom(L_2,L_1) $ is the space of global sections of the bundle $Hom(L_2,L_1)≅L^*_2 \otimes L_1$

  • $\begingroup$ Even if deg(L_i) = 1,h^1 = 0. I don't understand your notation, $Li>1$. Can you explain. $\endgroup$ – aginensky Apr 27 '15 at 23:03
  • $\begingroup$ I added a line to my question. What do you mean by your first statement? $\endgroup$ – 010110111 Apr 27 '15 at 23:09
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    $\begingroup$ @010110111 In the last line above, I think you want $L_1\geq L_2$. $\endgroup$ – Ben Lim Apr 28 '15 at 10:17
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    $\begingroup$ @010110111 Also, you might wanna add that $\Gamma Hom(L_2,L_1)$ is the space of global sections of the bundle $Hom(L_2,L_1) \cong L_2^\ast \otimes L_1$. $\endgroup$ – Ben Lim Apr 28 '15 at 10:19

I'm pretty sure that by $L_i>1$ he means that the line bundle has sections. Now this either means that $L_i=1$ or $\deg L_i\geq 1$. On an elliptic curve, any line bundle (divisor) of positive degree is non-special so for any of these $L_i$ with $\deg L_i\geq 1$, we must have $H^1(L_i)=0$. Also in general on a curve, any line bundle $L$ of degree 0 with a section is precisely 1. So this answers your other questions. About why $L_i>1$, if you look above in the paper at Lemma 10, then you see that if $L_1=1$, where $L_1$ is by definition the sub line bundle of maximal degree, then any section $\phi\in\Gamma(E)$ has $div(\phi)=0$, and therefore $\Gamma(E)$ generates a trivial sub bundle $\mathcal O^s$, where $s=\dim\Gamma(E)$. But as $s\geq d$ by Riemann-Roch, this is a contradiction to $d\geq r$. From the definition of maximal splitting, the line bundles $L_i$ are constructed inductively, so $L_i\geq L_{i-1}$, and thus we see that all of the $L_i$ have $h^1=0$. Hope this helps. Also, maybe you should have explained in the question what the maximal splitting is since the vector bundles you're talking about are indecomposable and thus it's not a real splitting.


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