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If I have two real/complex vector bundles $E$ and $F$ over a smooth manifold $M$ and I can exhibit an isomorphism for each fiber $E_p \to F_p$ does this imply that there exists a bundle isomorphism $E \to F$?

Danlo
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    Briefly, any $n$-dimensional real vector space is isomorphic to $\Bbb R^n$, so your reasoning would imply that every vector bundle is trivial. – Ted Shifrin Jan 25 '24 at 18:30

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No it does not, here is a counter example. Let $S^2$ be the two sphere, and $TS^2$ it's tangent bundle. Let $E=S\times \mathbb R^2$, the trivial vector bundle. Then for each fibre $T_pS^2$, any vector bundle chart $\phi:TS^2|_U\rightarrow U\times \mathbb R^2$ containg $p$ gives me an isomorphism $T_pS^2\rightarrow (S\times \mathbb R^2)_p$. However, these are clearly not isomorphic vector bundles, as $TS^2$ cannot be trivial by the hairy ball theorem.

Chris
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To elaborate on the answer given by Chris: it's true that a pointwise isomorphism $E_p\cong F_p$ does not given an isomorphism $E\cong F$. This is the whole point of bundles; otherwise everything would be of the form $M\times\mathbb{R}^r$.

It is, however, true that given a vector bundle morphism $\varphi:E\to F$ such that $\varphi_p:E_p\to F_p$ is an isomorphism for all $p\in M$, then $\varphi$ is an isomorphism of vector bundles. And this fact is often very useful.

Example: let $J:TM\to TM$ be the complex structure of a complex manifold $M$. Let $T^{1,0}M\subset TM\otimes\mathbb{C}$ be the $+i$-eigenspace, also called the holomorphic tangent bundle. Then $TM\cong T^{1,0}M$. Why? There is a natural embedding $TM\hookrightarrow TM\otimes\mathbb{C}$. At the level of the fibres, this map sends a basis vector to itself, now viewed as a basis vector for a complex vector space. And these basis vectors are a (real) basis for the $+i$-eigenspace of $J$. Hence the embedding gives a fibrewise isomorphism $T_pM\cong T^{1,0}_pM$, and we conclude $TM\cong T^{1,0}M$ (feel free to work out the details).

Quaere Verum
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Take the tangent bundle of a manifolds $M$ and $M\times \mathbb{R}^{\text{dim }M}$, and you'll see why this doesn't hold.

Example: $T\mathbb{S}^2$ and $\mathbb{S}^2\times\mathbb{R}^2$. The tangent bundle of $T\mathbb{S}^2$ is not trivial.

A Name
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