A holomorphic vector bundle $E\to M$ over a compact Kähler manifold $M$ with Kähler form $\omega$ is called ***stable*** if for any coherent analytic subsheaf $\mathcal F$ of lower rank of $E$ there holds $\mu(\mathcal F)<\mu(E)$ where $\mu(\mathcal F)=deg\_\omega(\mathcal F)/rank(\mathcal F)$ and $deg\_\omega(\mathcal F)=\int\_MC\_1(\mathcal F)\wedge \*\omega$ where $C\_1$ is the form representing the first Chern class.


My question is where can I find a proof (in case there is one) of the fact that such a stable holomorphic vector bundle is ***simple*** (which means that any holomorphic endomorphism is a constant times the identity). In the famous Uhlenbeck-Yau paper of 1986 it is said that "it is very likely" that it is true but they assume simplicity rather than giving a proof.



are stable holomorphic bundles over compact Kähler manifolds simple?
As far as I can tell the following argument given in the case of $\mathbb{P}^n$ from Okonek-Schneider-Spindler "Vector bundles on complex projective spaces" works here as well. If someone could check this over I'd be grateful, as I'd be surprised if Uhlenbeck-Yau missed it.


First show that if $f:E\_1\to E\_2$ is a homomorphism of stable bundles of the same slope then $f$ has the "expected rank," i.e. it is either injective or generically surjective. For if $I$ is the image sheaf and its rank is smaller than expected, we have by stability


$$\mu(I) > \mu(E\_1) = \mu(E\_2) > \mu(I),$$


a contradiction.


Next suppose $f:E\_1\to E\_2$ is a nonzero homomorphism of stable bundles of the same rank and $c\_1$. By the previous paragraph, $f$ is injective. Then the induced map $\det E\_1\to \det E\_2$ is also injective; this is a map of line bundles of the same $c\_1$, so it is in fact an isomorphism. But then $f$ originally must have been an isomorphism.


In particular, any nonzero endomorphism $f:E\to E$ of a stable bundle is an isomorphism. Now choose a point $x\in X$, and look at the action $f\_x : E\_x\to E\_x$ on the fiber. Choose an eigenvalue $\lambda\in \mathbb{C}$ of $f\_x$. Then $f-\lambda I$ is either zero or an isomorphism, but it isn't an isomorphism at $x$ so we conclude $f = \lambda I$.


