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I am confused on what it takes to be a vector bundle morphism. The definition I am working with is from Lee(Intro to smooth manifolds): let $\pi:E \to M$ and $\pi':E' \to M$ be vector bundles, then a bundle homomorphism is a continuous map $F: E \to E'$ such that $\pi \circ F = \pi$ and whose restriction to each fibre is linear.

Now my question comes from working through Huybrechts complex geometry, I will include the entire proposition but the details aren't that important.

Proposition: Let X be an almost integrable complex manifold. Then there exists a direct sum decomposition $$T_\mathbb{C}X \cong T^{1,0}X \oplus T^{0,1} $$ of complex vector bundles on X.

The proof states that the isomorphism holds from the decomposition on all fibres, I agree with the canonical vector space isomorphism. My struggle however is when do isomorphic fibres gives isomorphic bundles? For example, the tangent bundle of $S^2$ and the trivial bundle $S^2 \times R^2$ both have fibres $R^2$. However, the tangent bundle of $S^2$ is twisted and the trivial bundle is not, so there is no bundle isomorphism here. I think when fibres are canonically isomorphic we are in luck for a bundle isomorphism but unsure how to build up a bundle map $F$ from this information and prove that it in fact commutes.

I admit to being very liberal with whether I am talking about smooth or holomorphic vector bundles but the general idea still holds I think.

jake walsh
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It is all about if you have a bundle map to start with, then you can discuss if this bundle map is an isomorphism.

For your example of $TS^2$ and $S^2\times R^2$, you don't even have a clear nontrivial bundle map between them, let alone talking about if they are isomorphic. (You are very correct in your statement that you don't know how to build up a bundle map in this case, from the fiber isomorphisms.)

For Huybrechts' example, I believe $T^{1, 0}X$ and $T^{0, 1}X$ are defined in $T_{\mathbb C}X$ using an almost complex structure (which by definition is a bundle map), so there IS a concrete bundle map for Huybrechts to talk about. Then he just has to check whether the map is a linear isomorphism on the fiber.

I also want to say that this is related to the concept of natural bundles, see for example https://ncatlab.org/nlab/show/natural+bundle#:~:text=A%20type%20of%20fiber%20bundle,the%20corresponding%20bundles%20covering%20them.

So the tangent bundle of a smooth manifold is natural, and linear algebra constructions give many other natural bundles such as various tensor bundles.

Three aggies
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  • What is this map explicitly though? I am having a hard time writing it down. – jake walsh Dec 23 '23 at 18:17
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    For Huybrechts? I think $T^{1, 0}X$ and $T^{0, 1}X$ are the eigen-subbundle of $J$ for eigenvalue $i$ and $-i$. So using inclusion, we have $T^{1, 0}X\oplus T^{0, 1}X\to T_{\mathbb C}X$. In the other way, you can have projection. For example, $T_{\mathbb C}X\to T^{1, 0}X; v \mapsto \frac{1}{2}(v - i Jv)$. Just think about what $J$ does on each subspace. – Three aggies Dec 23 '23 at 18:44