Why do we need the local triviality condition when working with vector bundles?

Question

I'm currently reading through Adams' paper on the image of the J homomorphism, and wanted to brush up on vector bundles and K-theory before tackling this paper. The definition of (real) vector bundle that I'm working with is as follows:

We have two topological spaces B and E and a continuous surjection $$\pi : E \to B$$ such that each fiber $\pi^{-1}(b)$ has the structure of a vector space. Furthermore, these data satisfy a "local triviality" or "bundle" condition: For every point $p \in B$, there is an open neighborhood $U \subseteq B$ of $p$, an integer $k \geq 0$ and a homeomorphism $\varphi : U \times \mathbb{R}^k \to \varphi^{-1}(U)$, such that $\pi \circ \varphi(x,v) = x$, and the map $v \mapsto \varphi(x,v)$ is an isomorphism from $\mathbb{R}^k$ to $\pi^{-1}(x)$ for each $x \in U$.

My question is why do we need this local triviality condition? If the spirit of a vector bundle is to continuously parameterize a family of vector spaces by B, then the local triviality condition shouldn't be necessary.

I believe that this condition rules out some nasty examples of things that we might not want to think about, like a vector bundle where all of the vector spaces $\pi^{-1}(x)$ are isomorphic, except that one of them has a different orientation than the rest - the local triviality condition forbids this (I think).

I originally believed that this condition was necessary for some of the constructions that we want to do with vector bundles. The direct sum, tensor product, and exterior powers of vector bundles can all be topologized using local trivializations, but we need not appeal to local trivializations in order to define their topologies - we can perform constructions analogous to those we perform in linear algebra and topologize the vector bundle $E_1 \otimes E_2$ in this way, and the two constructions turn out to be homeomorphic.

Other than this, I'm a little lost. This seems like a certain brand of niceness condition, but it's unclear to me exactly what this niceness buys you. If we drop the local triviality condition from the definition of a vector bundle, what constructions and theorems do we lose, and what pathological examples do we admit?

Answer 1

From the perspective of $K$-theory, the point of vector bundles is to be able to classify them. That is, given a space $X$, we want a nice classification of all vector bundles (up to isomorphism, or up to stable equivalence) which can both provide a useful invariant of $X$ and give us information about specific naturally occurring vector bundles we might care about.

The local triviality condition makes classifying vector bundles tractible and have deep relationships to other natural questions in topology. Very crucially, the local triviality condition makes it possible to prove that vector bundles are homotopy invariant (at least assuming all our spaces are paracompact): that is, given a vector bundle on $X\times[0,1]$, the two vector bundles on $X$ you get by restricting to $X\times\{0\}$ and $X\times\{1\}$ are isomorphic. In particular, then, a vector bundle on a contractible space is trivial, so if we cover our space by contractible open sets, we can classify vector bundles by thinking about the possible ways to glue together trivial bundles over those open sets via transition functions. This also makes it possible to classify vector bundles in terms of homotopy classes of maps into Grassmannians, which opens up all sorts of connections to homotopy theory such as Bott periodicity.

If you don't require local triviality, then classifying vector bundles is immensely more complicated and intricately related to the exact topology of your space, rather than just its homotopy type. In particular, for instance, if you partition your space $X$ into arbitrary subsets, you could take any vector bundles of the same rank over each of those subsets and just take their disjoint union to get a vector bundle on $X$. There are many more complicated examples. Basically, classifying vector bundles becomes an incredibly complicated pointset topology question and completely useless for any sort of computable invariant.

Answer 2

In my view the condition is necessary in order to verify that we are paramaterizing a family of vector spaces continuously. The situation you describe with one fiber is actually not continuous, and you can check that a nonzero (local) section of the vector bundle would need to "jump" values. This escapes the intuition one want for a family of vector spaces, and in fact makes no use of the topology in the base space, and would work over a set.

The idea of locally trivial verifies that everything varies continuously, and the local trivializations tell us how to glue these locally defined families together.

Answer 3

You may of course consider bundles of a very general type. See for example Chapter 2 of

Husemoller, Dale. Fibre bundles. Vol. 5. New York: McGraw-Hill, 1966.

Husemoller also performs some general contructions with bundles, e.g. products. In that sense you are right.

If you come to vector bundles, you will of course require that all fibres $p^{-1}(x)$ are topological vector spaces. Now consider the projection $p : E = B \times \mathbb R^n \to B$. For each $x \in B$ choose a subspace $V_x \subset \mathbb R^n$. We do not make any assumption concerning the dimension of $V_x$. Then $E' = \bigcup_{x \in B} \{x\} \times V_x$ is a subspace of $E$ and $p$ restricts to $p' : E' \to B$. This would be a vector bundle in your sense. It may look completely erratic. A classification of these bundles up to isomorphism is practically impossible, and there is certainly no connection with the space $B$. In fact, the topology of $B$ does not play any role for the choice function $x \mapsto V_x$.

I would even say that your general bundles are in effect bundles of sets $E, B$. The situation is similar for functions between spaces: You may consider arbitrary functions or continuous functions, but obviously you cannot say much about general functions. Only restriction to smaller classes of functions will produce something interesting. Another example are groups: You cannot say much about general groups, but if you restrict to special classes like finitely generated abelian groups, you get very interesting results.

Thus we should "choose wisely" which type of objects we want to consider. In my opinion general vector bundles would not be a wise choice.

Moreover, many "naturally occurring" vector bundles are locally trivial. Examples are tangent bundles of smooth manifolds.

Two final remarks:

1. Your example using different orientations does not work. An orientation is an additional structure on a vector space.

2. "If the spirit of a vector bundle is to continuously parametrize a family of vector spaces by $B$, then the local triviality condition shouldn't be necessary." Try to define what a continuous parametrization should be. See Andres Mejia's answer.

Edited:

The present question motivated me to think about the meaning of "the spirit of a vector bundle is to continuously parametrize a family of vector spaces by $B$". I posted a new question Alternative characterization of local triviality of vector bundles?

This contains a suggestion what a continuous parametrization could be. If you accept the corresponding definition, you will see that local triviality is a consequence of being continuously parametrized.

Answer 4

"If the spirit of a vector bundle is to continuously parameterize a family of vector spaces by B, then the local triviality condition shouldn't be necessary."

I think the spirit of a vector bundle is more than that.

A manifold $M$ of dimension $n$ is a topological space that "locally resembles $\mathbb{R}^n$." By "locally resembles $\mathbb{R}^n$," we mean that for each $x \in M$, there exists an open set $U \subset M$ and a homeomorphism $\varphi \colon U \to \mathbb{R}^n$.

A vector bundle $\pi \colon E \to B$ of rank $k$ is a map "whose fibers are vector spaces" and that "locally resembles the projection $p \colon B \times \mathbb{R}^k \to B$." By this we mean that for each $x \in B$, the fiber $\pi^{-1}(x) \subset E$ is a $k$-dimensional $\mathbb{R}$-vector space, and there is an open set $U \subset B$ and a homeomorphism $\varphi \colon \pi^{-1}(U) \to U \times \mathbb{R}^k$ such that:

• The map $\pi|_{\pi^{-1}(U)} \colon \pi^{-1}(U) \to U$ is just the map $p|_{U \times \mathbb{R}^k} \colon U \times \mathbb{R}^k \to U$ after applying $\varphi$. That is: $$\pi|_{\pi^{-1}(U)} = p|_{U \times \mathbb{R}^k} \circ \varphi$$
• For each $x \in U$, the map $\varphi$ identifies $\pi^{-1}(x) \subset E$ with the $\mathbb{R}^k$-factor of $\{x\} \times \mathbb{R}^k$. That is: The map \begin{align*} \varphi|_{\pi^{-1}(x)} \colon \pi^{-1}(x) & \to \{x\} \times \mathbb{R}^k \end{align*} is a vector space isomorphism.