Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm studying manifold theory and I've got to the point of discussing the definition of a vector bundle. The definition is quite long and a bit confusing and I was wondering if someone with a bit more intiution in this topic could explain it prehaps using a nice example.

For completeness this is the definition I have:

A smooth real vector bundle $(E,M,\pi)$ of rank $k\in\mathbb{N}_0$ is a smooth manifold $E$ of dimension $m+k$ ($E$ is the total space), another smooth manifold $M$ of dimension $m$ ($M$ is the base manifold) and a smooth surjective map $\pi:E\rightarrow M$ (projection map) such that

  1. $\exists$ an open cover $\{U_{\alpha}\}$ of $M$ and diffeomorphisms $\psi_\alpha:\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times\mathbb{R}^k$,

  2. $\forall p \in M$, $\psi_\alpha(\pi^{-1}(p)) = \{p\}\times\mathbb{R}^k,$

  3. Whenever $U_\alpha\cap U_\beta \not= \emptyset$, the map $$\psi_\alpha\circ\psi_\beta^{-1}:(U_\alpha\cap U_\beta) \times\mathbb{R}^k\rightarrow(U_\alpha\cap U_\beta) \times\mathbb{R}^k$$ has to take the form $(x,v)\rightarrow(x,A_{\alpha\beta}v)$, where $A_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow GL_{\mathbb{R}}(k)$ is smooth.

Thanks in advance.

share|cite|improve this question
Let us start with the most trivial example of a vector bundle one can think of, a vector bundle over a point. This is (essentially) just a vector space. Then what about vector bundles over more interesting spaces. I like to think of it as a "parameterized vector space" where the space of parameters is the base space. – Baby Dragon Mar 27 '13 at 1:07

2 Answers 2

up vote 6 down vote accepted

A rank $k$ vector bundle over a manifold $M$ can be viewed as a twisted version of the Cartesian product $M \times \Bbb R^k$. Given an appropriate cover $U_\alpha$ of $M$, the transition functions $A_{\alpha\beta}$ tell us how to glue together the local products $U_\alpha \times R^k$ to get the total space $E$. If all transition functions map all overlap points to the identity $I \in \mathrm{GL}(k, \Bbb R)$, then there is no twisting and we just get the Cartesian product $M \times \Bbb R^k$.

One example is as follows. Consider the circle $S^1$, thought of as the set $\Bbb R/2\pi \Bbb Z$. We will construct two rank $1$ vector bundles on $S^1$. Cover $S^1$ by the open sets $U_1 = (0, 2\pi)$ and $U_2 = (-\pi, \pi)$. We have that $$U_1 \cap U_2 = (0, \pi) \cup (\pi, 2\pi).$$ Assume that on the component $(0, \pi)$, the transition function is given by $$A_{12}(x) = 1 \text{ for all } x \in (0, \pi).$$ Since a transition function must be continuous, we see that the only possibilities for $A_{12}|_{(\pi, 2\pi)}$ are \begin{align*} \text{(a)} & \quad A_{12}(x) = 1 \text{ for all } x \in (\pi, 2\pi), \\ \text{(b)} & \quad A_{12}(x) = -1 \text{ for all } x \in (\pi, 2\pi). \end{align*} Case (a) gives the trivial rank $1$ vector bundle over $S^1$, $E = S^1 \times \Bbb R$.. In case (b), we have that the total space $E$ is the (infinite) Möbius band.

share|cite|improve this answer
@Stephen: Yes, I'm using the fact that $U_1 \cap U_2$ is disconnected, so we can choose two different values for $A_{12}$ on each component. In general, $A_{12}$ does not need to be constant on each component, I just made it that way for this simple example. I've also used the fact that transition functions determine the bundle. – Henry T. Horton Mar 27 '13 at 1:31
@HenryTHorton I see this makes a lot more sense now. Thanks. – user67881 Mar 27 '13 at 1:36

Theres another way to see that a bundle is a kind of 'twisted product' in a very general kind of way; and this is to use a little category theory.

First, a bundle in its most general form is simply an arrow $E \rightarrow M$. Now consider the categorical definition of the product $F \leftarrow E \rightarrow M$; this defines $E$ as the product of $F$ and $M$ in the category we are using (suppose for definiteness it is the category of manifolds, then the product$F \times M$ is the usual product in manifolds).

Now, when we remove the left hand arrow we think of the remainder $E \rightarrow M$ as a generalised product - a twisted product.

In this generality this is difficut to work with; so other axioms are demanded to make it amenable.

For example:

a. that it has fibres: the category has a terminal object and the pullback $E_p$ over it exists; This object is the fibre where $p:1 \rightarrow M$ lands. Note that fibres may differ over $M$.

b. That the fibres are all isomorphic; then we say the fibre is typical.

c. local triviality: this requires that we work in a topological context, so our bundle is at least equipped with a topology. Then we demand each pullback $E_U$ exists (for $U$ sufficiently small); it is locally trivial if it is a product of $U$ and another object $F$; note that over different places in $M$ we do not get a typical fibre. This is simply a modification of the first of the above.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.