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 $Hom(L\_2,L\_1)≅L^\*\_2 \otimes L\_1$



Atiyah's vector bundles over an elliptic curve
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. 


