Hello everyone,

it seems to be "well-known" that $H^0(X;End(V))$ only contains isomorphisms where $X$ is a Riemann surface and $V$ a stable (algebraic) vector bundle over $X$. The usual proof considers (roughly) the (coherent) image sheaf of a non-zero vector bundle morphism $\varphi:V\to V$ and one obtains a contradiction under the hypothesis that $im(\varphi)\neq V$ : $\mu(V)<\mu(im(\varphi))<\mu(V)$ (where $\mu(V)=deg(V)/rk(V)$ denotes the slope of $V$).

Now my question is if this assertion is still true in the complex (differential) geometric context, i.e. when one defines holomorphicity (and hence stability) via del-bar-operators.

The previous proof doesn't seem to work since the image of a (smooth) vector bundle homomorphism $\varphi:V\to V$ is in general NOT a vector (sub)bundle of $V$ (unless the rank of $\varphi$ is (locally) constant). So $\mu(im(\varphi))$ doesn't make sense since the image of $\varphi$ is in general not a subbundle of $V$. Or am I missing something in the holomorphic setup?

Since I'm not an expert in algebraic geometry I have some difficulties in "translating" results concering stable (algebraic) vector bundles over a Riemann surface into the complex geometric (del-bar) approach to stable (holomorphic) vector bundles.

Thanks in advance for any answers, comments and remarks!

Edit: I changed my terminology from "differential geometric (dg)" to "complex geometric" which seems more appropriate (thanks @David Roberts!) and added some comments. Hopefully my questions/confusion is more understandable now.

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    $\begingroup$ You can use a bump function to scale the identity map down to the zero map. So there are tons of self-maps which are not isomorphisms. $\endgroup$ – Jason Starr Aug 5 '11 at 12:44
  • $\begingroup$ @xian - be careful with terminology: dg-categories (ncatlab.org/nlab/show/dg-category) and dg-vector bundles (www.math.umn.edu/~ciocan/rquot.ps) are very different from what you are talking about! Note that dg generally stands for 'differential graded'. $\endgroup$ – David Roberts Aug 6 '11 at 0:22
  • $\begingroup$ @David Roberts: Thanks for pointing this out to me! I've just changed my terminology and added some comments to make things clearer. @Jason Starr: I'm aware that there are tons of endomorphisms of a (smooth) vector bundle. But multiplying by a bump function doesn't work (in general) for holomorphic endomorphisms. Thanks anyway! $\endgroup$ – xian Aug 6 '11 at 6:41
  • $\begingroup$ Doesn't GAGA imply there is no difference between the two situations? $\endgroup$ – Jack Huizenga Aug 6 '11 at 7:33
  • $\begingroup$ I recommend reading the book of S.Kobayashi "Differential geometry of complex vector bundles", if you haven't done so already. For example it is proved there (using the differential geometric approach) that the only holomorphic endomorphisms of a stable bundle over a compact Kahler manifold are multiples of the identity. $\endgroup$ – YangMills Aug 6 '11 at 8:34

Consider the two following potential definitions of stability of a locally free coherent sheaf $E$.

(A) Every subbundle (i.e. locally free subsheaf) has strictly smaller slope.

(B) Every subsheaf has strictly smaller slope.

I claim that over a curve $X$ these two definitions are equivalent. Indeed, suppose (A) holds, and let $F\subset E$ be a subsheaf. Consider the sequence

$$0\to F\to E \to Q\to 0.$$

Here $Q$ may fail to be locally free; Put $G = Q/Q_{tors}$, which is then locally free since $X$ is a curve. Define $K$ by the sequence

$$0\to K \to E \to G\to 0,$$

and observe that $K$ is locally free of the same rank as $F$. Furthermore, $c_1(K) \geq c_1(F)$ since $c_1(Q_{tors})\geq 0$, so $\mu(F) \leq \mu(E)$.

Thus it is equivalent to only consider subbundles for stability, so long as we are working on a curve. Together with GAGA, this shows the two approaches (and any possible combination of definitions) are essentially the same.

  • $\begingroup$ Thank you very much, Jack! Maybe this is trivial for algebraic geometers but coming from a differential geometric background, I didn't see that. $\endgroup$ – xian Aug 9 '11 at 6:12

Although this is rather a comment (or at best a partial answer) than an (honest) answer, I don’t write it as such. But it’ll contain some thoughts about my question and so it seems appropriate to write it as an answer (at least to me…).

@Jason Starr: Of course you’re completely right. Sorry, I was probably too sloppy by only writing “Riemann surface” than “compact (connected) Riemann surface”. Maybe I hoped that it would be clear from the context. In any case, your (and Jack’s) comment sparked my interest in GAGA and how it really applies.

@YangMills: Today I was able to read some passages of Kobayashi’s “Differential geometry of complex vector bundles” and it clarified some of my confusions. Thanks!

So here are my thoughts about my question:

Let $X$ be a compact Riemann surface and denote by $O_X$ the sheaf of holomorphic functions on $X$. Then there is a bijection between holomorphic vector bundles of rank $r$ over $X$ and locally free $O_X$-modules of rank $r$. But what are the morphisms on each side?

The morphisms of locally free $O_X$-modules (of rank $r$) are simply sheaf homormorphisms.

Maybe all authors (at least for example Huybrechts or Kobayashi) define a holomorphic vector bundle morphism $\varphi:V\to W$ (where $V$ and $W$ are holomorphic vector bundles over $X$) as a holomorphic map $\varphi:V\to W$ such that the restriction of $\varphi|x$ to any fiber $V(x)$ maps complex linearly to $W(x)$ AND the rank of $\varphi|x$ is independent of $x\in X$ (call this definition “definition B”).

In my question I used an alternative definition (“definition A”) of a holomorphic vector bundle morphism, i.e. I didn’t require the rank to be constant. If one uses the definition B instead, the image of a holomorphic vector bundle morphism $\varphi: V\to V$ is indeed a holomorphic subbundle of $V$ and one can proof that $H^0(End(V))=H^0(Iso(V))$ just as in the “sheaf case”.

But with definition A one can prove this fact as well by using sheaf theory because we can define stability (resp. slope $\mu(P)$) for every torsion-free coherent sheaf $P$ over $X$. Without going into detail I shall just mention that stability for $P$ means that $$\mu(P')<\mu(P)$$ for every coherent subsheaf $P'$ with $0 < rank(P') < rank(P)$.

Now every holomorphic vector bundle homomorphism $\varphi:V\to V$ (w.r.t definition A) induces a sheaf homorphism $\phi:S\to S$ of the underlying locally free $O_X$-sheaf $S$ and vice versa. (This fails in definition B.) The image of $S$ under $\phi$ is in general no longer locally free (claim: this is precisely the case when $\varphi$ has constant rank). But we can construct the (coherent) image sheaf $im S$ of $S$ so that the slope of $im S$ is defined. Then we can proceed as I indicated in my question (cf. Kobayashi’s “Differential geometry of complex vector bundles”) and we see that $\phi$ and hence $\varphi$ is an isomorphism.

Even if this seems quite plausible I’m still not really satisfied because the above stability condition seems to be much stronger than the one for a holomorphic vector bundle $V$ since one considers not only locally free subsheaves (i.e. holomorphic subbundles of $V$) but also coherent subsheaves of the corresponding locally free sheaf.

So my feeling is that if one uses definition A (i.e. omitting constant rank) then $H^0(End(V))=H^0(Iso(V))$ is in general only true under this stronger stability condition.

(But it’s also possible that this is complete non-sense…)

  • $\begingroup$ For the purpose of GAGA, the morphisms between holomorphic vector bundles are your type (A), the sheafy kind. $\endgroup$ – Jack Huizenga Aug 8 '11 at 19:36
  • $\begingroup$ Actually, Huybrechts emphasises the existence of two definitions on p.72 (the "Warning" after Example 2.2.20). For curves this doesn't affect the notion of stability, as Jack has explained. $\endgroup$ – Peter Dalakov Aug 8 '11 at 20:25

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