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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.

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2 Answers 2

up vote 7 down vote accepted

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 $f = \lambda I$.

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Could you please explain to a newbie like me why $\mu(I)>\mu(E_1)$? –  Mircea Jul 26 '13 at 5:21
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Because $I$ is a quotient of $E_1$ of smaller rank and $E_1$ is stable. (Using an exact sequence $0\to K \to E_1 \to I\to 0$, we have $\mu(I) > \mu(E_1)$ iff $\mu(K) < \mu(E_1)$.) –  Jack Huizenga Jul 26 '13 at 5:29

It seems that the answer is very simple (and should have been known to Uhlenbeck-Yau). Let $E$ be a stable bundle, equipped with an Hermitian-Einstein metric and connection, and $End(E)$ its authomorphism bundle, which is also Hermitian-Einstein, with slope 0. Then any automorphism is parallel, because a Hermitian-Einstein bundle with slope zero cannot have non-parallel sections. Then the corresponding eigenbundles are parallel with respect to the connection. A parallel sub-bundle of a Hermitian bundle obviously splits.

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How do you equip a stable bundle with a Hermitian-Einstein metric? –  Mircea Jul 27 '13 at 9:02
    
You equip it by Uhlenbeck and Yau, indeed... That's the point of Misha's answer I think! On the other hand, this is certainly not the easy direction in the Kobayashi-Hitchin correspondence! –  diverietti Jul 27 '13 at 10:00
    
but as I said Uhlenbeck and Yau assume simplicity in their definition of stability.. –  Mircea Jul 27 '13 at 14:31
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Understood. Now simplicity is never assumed in the definition of stability (see Lubcke-Teleman). Sorry for a confusion. –  Misha Verbitsky Jul 28 '13 at 11:08

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