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If there is a short exact sequence of holomorphic vector bundles, $$0 \overset{a_1}{\to} W \overset{a_2}{\to} V \overset{a_3}{\to} F \overset{a_4}{\to} 0,$$ then one can expect a $C^{\infty}$ splitting $$V \cong W \oplus F$$ rather than a holomorphic splitting.


I know that a s.e.s. needs consecutive maps to equal $1$, and that for exactness that $im(a_i) = ker(a_{i+1})$. I also know that a vector bundle is just a manifold with the fiber as a vector space (complex here). For a shorthand of notation of a vector bundle, I use $\pi: E \to B$ where $B \times V$ is the product space and $\pi$ is the fiber bundle. Written like a s.e.s., this is $$V \to E \overset{\pi}{\to} B.$$ Also $a_2$ is injective and $a_3$ is surjective.

So is the reason why the splitting is only $C^{\infty}$, and not holomorphic, because the maps, either $a_2^{-1}$ or $a_3^{-1}$ are not injective?

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2 Answers 2

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On a (paracompact) complex manifold all short exact sequences of $C^{\infty}$ vector bundles are $C^{\infty}$ split, so it is enough to exhibit an exact sequence of holomorphic vector bundles that doesn't holomorphically split.

The simplest example is the exact sequence on $\mathbb P_\mathbb C^1$:
$$0\to \mathcal O(-2) \to \mathcal O(-1) \oplus \mathcal O(-1)\to \mathcal O\to 0$$ It does not split because the bundles $\mathcal O(-1) \oplus \mathcal O(-1)$ and $\mathcal O(-2) \oplus \mathcal O$ are not isomorphic: the second has nonzero holomorphic sections but the first doesn't.

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  • $\begingroup$ Would you mind being more specific on how you defined this short exact sequence? $\endgroup$
    – Ben
    Commented Oct 27, 2015 at 19:59
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    $\begingroup$ Oh that's just Euler exact sequence. I see. $\endgroup$
    – Ben
    Commented Oct 27, 2015 at 20:10
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Note that unless $W$ or $F$ is zero, neither $a_2$ nor $a_3$ will be invertible and so there are no such maps $a_2^{-1}$ or $a_3^{-1}$.

Given such a short exact sequence, there exists a $C^\infty$ splitting essentially because there exist $C^\infty$ partitions of unity: take a cover of $B$ by open sets over each of which you can trivialize $V$, and put a smooth hermitian metric on each of those trivialized bundles. You can then use a partition of unity to patch those metrics together into a metric on the entire bundle $V$. Then you have a hermitian metric on each fibre $V_x$, and so taking the orthogonal complement of $W_x$ you get a splitting $V_x = W_x \oplus W^\perp_x$, which gives a splitting $V = W\oplus W^\perp$ globally since the metric varies smoothly. Finally, we have $W^\perp\cong F$ since both are isomorphic to the quotient bundle $V/W$.

This whole story fails to work in the holomorphic world because in general there are no holomorphic partitions of unity - there just aren't enough holomorphic functions, in contrast to the smooth case.

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  • $\begingroup$ What's an easy example of a SES of holomorphic vector bundles that fails to split (holomorphically)? $\endgroup$ Commented Oct 22, 2012 at 18:13
  • $\begingroup$ @Eric, thanks. So the prime indicates the complex conjugate of the vector bundle? The fact that there are not that many holomorphic functions may seem obvious but I'm not appreciative of it yet (even though I know a complex derivative is a very strong criteria).... $\endgroup$
    – nate
    Commented Oct 22, 2012 at 22:05
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    $\begingroup$ @nate the point with there "not being enough holomorphic functions" is that to construct $C^\infty$ partitions of unity you use bump functions, which are functions constantly equal to 1 on some open set and constantly equal to 0 outside a larger set. There can be no holomorphic bump functions since a holomorphic that is 0 on an open set must be 0 everywhere (by analytic continuation). $\endgroup$ Commented Oct 23, 2012 at 0:06
  • $\begingroup$ @nate: I've changed the notation from $F'$ to $W^\perp$. $\endgroup$
    – bradhd
    Commented Oct 23, 2012 at 1:06

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