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Let $X$ be a compact complex surface.

This definition is from Donaldson, Kronheimer: The Geometry of 4-Manifolds, p. 209:

Definition: A holomorphic $SL(2,\mathbb{C})$ bundle $E$ over $X$ is called stable if the following holds: For each line bundle $L$ over $X$ we have $$h^0(Hom(L,E)) \neq 0 \Rightarrow deg (L)<0.$$

Denote by $K_X$ the canonical bundle of $X$, let $F$ be the ideal sheaf of a finite set of points in $X$. Let $E$ be a bundle over $X$ fitting into the following short exact sequence:

$$0 \rightarrow \mathcal{O}_X \rightarrow E \rightarrow K_X \otimes F \rightarrow 0.$$

The following lemma is a consequence of Lemma 10.3.7 in Donaldson, Kronheimer: The Geometry of 4-Manifolds:

Lemma: Let $X$ be such that $Pic(X)=\langle K_X \rangle$, $deg(K_X)>0$. Then $E$ is stable.

In the book, the proof goes by showing that $h^0(E \otimes K_X^{-1})=0$.

Question 1: How can I see that the definition of stability given above is equivalent to the usual definition of slope stability, i.e.: every subbundle $E'$ shall have strictly smaller than $E$?

Question 2: $h^0(Hom(L,E)) \neq 0 \Rightarrow deg (L)<0$ is equivalent to $deg(L)\geq 0 \Rightarrow h^0(Hom(L,E)) = 0$. All line bundles with positive degree on $X$ are powers of $K_X$, so we must check that $h^0(Hom(K_X,E))=h^0(E \otimes K_X^{-1})=0$ and $h^0(Hom(\mathcal{O}_X,E))=0$. Why is the second condition not checked in the proof in the book? (Also it doesn't seem true, which can be seen if I take global sections in the short exact sequence above) How does stability follow from $h^0(E \otimes K_X^{-1})=0$?

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1 Answer 1

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Q1: First, in algebraic geometry (and other fields), subbundles have a different meaning than what you assert. A subbundle $E'\subset E$, where $E',E$ are bundles mean the quotient $E/E'$ is a bundle. This works well for non-singular curves, but not in higher dimension. So, the slope stability for higher dimensional situation is, given any proper subsheaf $E'\subset E$, $\deg E'<\deg E$. In the situation above (your quotation of definition), the bundle in question has degree zero (being an $SL(2,\mathbb{C})$ bundle) and rank two. Staring with your definition, let $L\subset E$ a proper subsheaf. I will let you convince yourself that the only case of interest is when rank of $L=1$. The fact that $L\subset E$ implies, $h^0(L^{-1}\otimes E)\neq 0$ and thus $\deg L<0$. The converse is equally straight forward.

Q2: Why do you think one should check $h^0(\operatorname{Hom}(\mathcal{O}_X,E)\neq 0$, since it is clearly false from your exact sequence?

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  • $\begingroup$ Thank you, I understand the answer to question 1. For question 2: In order to establish the stability of $E$, I wanted to check if it satisfies the definition given above, which would necessitate that $h^0(Hom(\mathcal{O}_X,E)=0$. The mistake was probably that $E$ is in general not an $SL(2,\mathbb{C})$ bundle, so the above definition does not apply. I still don't understand how $h^0(E \otimes K_X^{-1})=0$ implies that $E$ is stable, though. $\endgroup$
    – user505117
    May 21, 2020 at 12:20
  • $\begingroup$ I added the sentence "How does stability follow from $h^0(E \otimes K_X^{-1})=0$?" to the second question. $\endgroup$
    – user505117
    May 21, 2020 at 12:22

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