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I'm having trouble understanding the failure of splitting of holomorphic vector bundle. Following are my thoughts on this issue: In general, for a s.e.s. within the same category (by which I mean $E,F,G$ are objects of same category and $f,g$ are permitted morphisms) $$0 \rightarrow E \xrightarrow{f} F\xrightarrow{g} G \rightarrow 0$$ I have an "intuitive proof" for the splitting:

1.We have a direct sum decomposition $F\cong f(E)\bigoplus F/f(E)$.

2.Since $f$ is injective by exactness, $F \cong E \bigoplus F/f(E)$.

3.We have $F/Kerg \cong G$.

4.By exactness, $Kerg\cong Imf$, thus $F \cong E \bigoplus G$.

Once I came up with this argument, I started to think what could go wrong. (I'm not pretty familiar with category theory, so please correct me if I say something wrong)

  1. The quotient object is not always defined. (i.e. sit inside the same category)
  2. The direct sum decomposition might not hold true. (this decomposition is valid as sets, but probably not valid as objects in the category, i.e. the projections might not be morphism)
  3. $F/Kerg \cong G$ minght not hold true. (again, such map might not be morphism)

Except for the above three points, I did not see where could go wrong with this argument

Now I try to apply this "intuitive" argument to the following cases:

  1. vector bundles (in smooth category). It seems to me that in this case all the criterion are valid. But all of the proofs I found for the splitting of s.e.s. of smooth vector bundles make use of a fiberwise metric on $F$ to produce an orthogonal complement to prove. I wonder why we need this extra structure to prove.

  2. holomorphic vector bundles. I thik in this case the problem happens for reason 1, i.e. the quotient object is not always defined as a holomorphic vector bundle, but I'm not pretty sure.

So my questions overall are:

  1. Does the intuitve argument I presented serve as the usual argument of splitting of s.e.s? If not, what is the problem with the argument?

  2. Why do we need a fiberwise metric to have a decomposition that is "the same" as the decomposition in intuitve proof 1 rather than just simply using the "canonical decomposition" ?

  3. I understand that the splitting in holomorphic vecotr bundle case fails to happen since we cannot find a fiberwise metric by piece together local ones using bump functions. But this seems a techique-wise reason, not the essential reason. Any different way to see this? (probably from category view?)

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    $\begingroup$ Your direct sum decomposition 1. is equivalent to splitting, so this seems very circular. $\endgroup$
    – Thorgott
    Commented May 9, 2022 at 17:21
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    $\begingroup$ The quotient $F/f(E)$ need not sit inside $F$ as a holomorphic subbundle (Condition 2 fails), so the first step of the intuitive argument is not true. One simple example is the tautological line bundle $E$ over the projective line, which is a subbundle of the trivial rank-two bundle $F$, but the quotient (isomorphis to $E^*$) is not a holomorphic subbundle of $F$. (To see why $E^*$ is not a holomorphic subbundle is another matter. One standard justification is the principle Griffiths and Harris call "curvature decreases in subbundles and increases in quotient bundles.") $\endgroup$ Commented May 12, 2022 at 2:53
  • $\begingroup$ @AndrewD.Hwang I think you should expand on this slightly and post it as an answer. I addressed the issue in this post and probably you have, somewhere, too. $\endgroup$ Commented May 14, 2022 at 22:50
  • $\begingroup$ Thanks @Ted, especially for your linked answer: Commenting was easier than typing up those details. In fact, the other supporting answer that turned up in a search is yours, which certainly predates whatever comparable answers I've given. :) $\endgroup$ Commented May 15, 2022 at 13:26

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$\newcommand{\Proj}{\mathbf{CP}}$tl; dr: Condition 2 fails; the first step of the intuitive argument is not true.


In the "real" situation, we can take orthogonal complements using a metric and remain in the smooth category. With complex vector bundles, by contrast, a metric needs to be conjugate-linear in one variable in order to "behave decently" from a linear-algebraic perspective, e.g., to be positive-definite. (See also the exercise below.) The orthogonal complement of a holomorphic subbundle $E \subset F$ is therefore "usually not" a holomorphic subbundle of $F$.

In the same sense, the quotient $F/f(E)$ is "usually not" isomorphic to a holomorphic subbundle of $F$. One standard justification is the principle Griffiths and Harris call curvature decreases in subbundles and increases in quotient bundles.

A basic example is the tautological line bundle $E$ over a complex projective space $\Proj^{n}$, which is a subbundle of the trivial rank $(n + 1)$ bundle $F$, but the quotient (isomorphic to $E^{*} \simeq T\Proj^{n} \otimes \mathcal{O}_{\Proj^{n}}(-1)$) is not a holomorphic subbundle of $F$.


Exercise: A complex-bilinear form $h$ on a finite-dimensional complex vector space is never positive-definite.

Proof: For every vector $v$ we have $h(iv, iv) = -h(v, v)$.

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