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Everywhere i see definition of vector bundle as triple $(E, p, B)$, $B$ and $E$ are manifold and local trivialization condition holds. For example see the definition here. .

Local trivialization gives manifold structure on $E$, then why initial assumption on $E$ that $E$ is manifold is necessary.

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You can avoid assuming that $E$ is a smooth manifold, but you must then ensure that the trivializations you've put forth induce a smooth structure on $E$. In other words, if $\varphi\colon p^{-1}(U) \to U \times \mathbb R^k$ and $\psi\colon p^{-1}(V) \to V \times \mathbb R^k$ are to be charts for $E$ then $\psi \circ \varphi^{-1}$ must be a diffeomorphism from $(U \cap V) \times \mathbb R^k$ to itself. This leads to the idea of specifying vector bundles via transition functions.

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  • $\begingroup$ I assumed that you wanted something smooth because of the tags; let me know if that's wrong. If you just want topological manifolds then I don't think there's any issue. [I would like to cook up an example of this actually failing in the smooth category; I admit that I can't think of something offhand, but it does not seem to me that smoothness is automatic.] $\endgroup$ Commented Aug 30, 2012 at 19:08
  • $\begingroup$ this is okk... i got the point... thanks for the answer... $\endgroup$
    – zapkm
    Commented Aug 30, 2012 at 19:10
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All you need is that $E$ and $B$ are topological spaces Perhaps a consequence of the local triviality condition is that $E$ is a topological manifold, but I've never seen it presented as an axiom.

Take a look at this definition.

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  • $\begingroup$ can i not take $E$ as a set and then $\pi$ as surjective map and topology of $E$ is manifold topology... So minimum condition is needed: $E$ as set which satisfies all other vector bundle condition, which in turn give $E$ as manifold. $\endgroup$
    – zapkm
    Commented Aug 30, 2012 at 18:43
  • $\begingroup$ You need $E$ to be a topological space so that $\pi : E \twoheadrightarrow B$ can be continuous. If $E$ is just a set then what does it mean to say $\pi : E \twoheadrightarrow B$ is continuous? $\endgroup$ Commented Aug 30, 2012 at 18:46
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In the link you provide, $E,B$ are only assumed to be topological spaces, and $p:E\to B$ is a continuous surjective map.

If you are interested in smooth vector bundles then assuming that $E$ only is a topological space and $B$ is a smooth manifold is enough, since local trivialization of $p$ will give $E$ the structure of a smooth manifold. But you still need $p$ to be assumed continuous, otherwise you cannot form a smooth atlas.

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