Let $E$ be a vector bundle of rank $3$ on a smooth, projective surface and let $\varphi:E\to E$ be a nilpotent endomorphism. If $E$ is indecomposable, is it true that $\varphi$ has rank $2$? My claim is suggested by the fact that, in the case of a vector space $V$ of dimension $3$, if $\varphi:V\to V$ is an automorphism with a unique eigenvalue $\lambda$ and $\varphi-\lambda Id$ has rank $1$, then there are two Jordan blocks. I am interested in this problem since I am studying endomorphisms of non simple Mukai-Lazarsfeld bundles on a $K3$ surface.



Nilpotent endomorphisms of indecomposable vector bundles on a surface.
The answer is no, in fact you can already construct a counter-example on an elliptic curve $E$:


A non-zero element in $\mathrm{Ext}^1(O\_E, O\_E)$ produces th unique indecomposable 2-dimensional vector bundle $F$ (the Atiyah bundle) fitting into the short exact sequence
$O\_E \into F \onto O\_E$. A short exact sequence computation will show you that $\mathrm{Ext}^1(O\_E, F)$ is still one-dimensional, and thus you get three-dimensional vector bundle $G$. It is indecomposable and has a surjection $G \onto O\_E$ and an injection $O\_E \into F \into G$, and thus it has a rank one endomorphism.


(If you pull this vector bundle to the product $C \times E$ for any curve $C$, you get a counter-example on a surface.)


Why is your Jordan block argument wrong? The reason is that the Jordan decomposition is not canonical, i.e. it depends on choices.


