I am reading the paper "Vector Bundles over Elliptic Curves" by Atiyah. I get a problem in understanding the "uniqueness" of the Atiyah bundles.
Let me fix the notation first: $X$ is an elliptic over an algebraic closed field. $\mathcal{E}(r,0)$ is the set of all indecomposable holomorphic vector bundles over $X$, which have rank $r$ and degree $0$
As is stated in Lemma 16 (I am focusing on $d=0$ case):
"For any $E'\in\mathcal{E}(r',0)$ with $\Gamma(E')\neq 0$, there exists $E\in\mathcal{E}(r,0)$, unique up to isomorphism, given by the extension $$0\rightarrow I_1\rightarrow E\rightarrow E'\rightarrow 0,$$ where $I_1$ is the trivial line bundle of $X$."
However, in the simplest case $r'=1$, any self extension of $I_1$ gives such a $E$ and since the space of isomorphism class of self extensions of $I_1$ is given by $$H^1(X,Hom(I_1,I_1))=H^1(X,I_1)=H^{0,1}(X)=\mathbb{C}.$$ Hence every non-trivial element in $H^1(X,Hom(I_1,I_1))$ would correspond an indecomposable self extension of $I_1$ with degree $0$ and therefore such $E$ can't be unique.
I think I must get something wrong but I don't see that. It would be my pleasure if somebody can point out what I get wrong. Thank you very much!
The solution to your paradox is that two extensions $$0\rightarrow I_1\rightarrow E\rightarrow I_1\rightarrow 0 \;\;(\mathcal E)\;\;\operatorname {and}\quad0 \rightarrow I_1\rightarrow E'\rightarrow I_1\rightarrow 0 \;\;(\mathcal E')$$ can be non isomorphic as extensions ($\mathcal E\not \simeq \mathcal E'$) and yet have middle vector bundles which are isomorphic as vector bundles: $E\simeq E'$.
For example if, for some $\lambda \in \mathbb C^*$, $(\mathcal E')=\lambda\cdot(\mathcal E)\in H^1(X,Hom(I_1,I_1))=\mathbb{C}$ we will certainly have $E\simeq E'$ as vector bundles.
This explains why there is up to isomorphism only one non-trivial rank two vector bundle which is an extension of the trivial line bundle by itself.