1
$\begingroup$

The following definition is given for a vector bundle in Milnor and Stasheff's Characteristic Classes.

A real vector bundle $\xi$ over $B$ consists of the following:

  1. a topological space $E=E(\xi)$ called the total space,
  2. a map $\pi:E \rightarrow B$ called the projection map
  3. for each $b\in B$ the structure of a vector space over the real numbers in the set $\pi^{-1}(b)$/

These must satisfy the following restriction: For each point $b$ of $B$ there should exist a neighborhood $U\subset B$, an integer $n\geq 0$, and a homeomoprhism $$h:U\times \mathbb{R}^n \rightarrow \pi^{-1}(U)$$ so that, for each $b\in U$, the correspondence $x\mapsto h(b,x)$ defines an isomorphism between the vector space $\mathbb{R}^n$ and the vector space $\pi^{-1}(b)$.

I am confused. It seems that there is no compatibility condition imposed on overlaps. I would have expected the definition to include something like the following:

and if $$h_1:U_1\times \mathbb{R}^n \rightarrow \pi^{-1}(U_1)$$ and $$h_2:U_2\times \mathbb{R}^n \rightarrow \pi^{-1}(U_2)$$ are two such homeomorphisms then we have that $$h_1 \mid_{U_1\cap U_2} = h_2 \mid_{U_1\cap U_2}$$

I do not feel comfortable with the definition. I should also try make this post into more of a question.

  1. Is the definition given in Milnor and Stasheff complete?
  2. If the definition is complete, have I simply overlooked where the compatibility is stated?
  3. How does compatibility and transition maps arise from the definition given?
$\endgroup$

2 Answers 2

5
$\begingroup$

The definition you quote is correct. There is no compatibility requirement of the sort you are asking for. Indeed, that is kind of the whole point of the notion of a vector bundle: it is a bundle over $B$ that locally looks like just the product with $\mathbb{R}^n$, but these local fiberwise linear homeomorphisms to $U\times\mathbb{R}^n$ are not necessarily compatible and so cannot necessarily be glued together to give a global fiberwise linear homeomorphism $E\cong B\times\mathbb{R}^n$.

It may be helpful to see where transition maps arise from this definition. The transition map between two of these local trivializations $h_1:U_1\times\mathbb{R}^n\to\pi^{-1}(U_1)$ and $h_2:U_2\times\mathbb{R}^n\to\pi^{-1}(U_2)$ arises simply from the composition $h_1^{-1}\circ h_2$, which maps $(U_1\cap U_2)\times\mathbb{R}^n$ to itself. Since $h_1$ and $h_2$ are linear isomorphisms on each fiber, so is $h_1^{-1}\circ h_2$, so it can be written in the form $(x,v)\mapsto (x,f_{12}(x)v)$ where $f_{12}$ is some function $U_1\cap U_2\to GL_n(\mathbb{R})$ (and this function is what is usually called the "transition map"). Now these transition maps will satisfy the compatibility condition that $f_{12}(x)f_{23}(x)=f_{13}(x)$ for all $x\in U_1\cap U_2\cap U_3$, simply because $(h_1^{-1}\circ h_2)\circ(h_2^{-1}\circ h_3)=h_1^{-1}\circ h_3$. And conversely, it turns out that if you start with a family of functions $f_{ij}:U_i\cap U_j\to GL_n(\mathbb{R})$ satisfying this compatibility condition (for some open cover $(U_i)$ of $B$) then you can use them to build a vector bundle on $B$ for which they are transition maps. But this is very different from asking for a compatibility between the local trivializations $h_i$ themselves.

$\endgroup$
5
$\begingroup$

It would counterproductive to require compatibility conditions as in your question. The requirement

If $h_1:U_1\times \mathbb{R}^n \rightarrow \pi^{-1}(U_1)$ and $h_2:U_2\times \mathbb{R}^n \rightarrow \pi^{-1}(U_2)$ are any two trivializing homeomorphisms, then $h_1 \mid_{U_1\cap U_2} = h_2\mid_{U_1\cap U_2}$

would imply that

  1. For a given open $U \subset B$ we are allowed to choose at most one trivializing homeomorphism $h_U$. For some $U$ there may be no such $h_U$, for other $U$ exactly one.
    This would be unproblematic, we would get a very special "bundle atlas".

  2. The open $U$ which carry a trivializing homeomorphism $h_U$ form an open cover of $B$. By your compatibility condition they can be pasted to a global trivializing homeomorphism $h_B : B \times \mathbb R^n \to \pi^{-1}(B) = E$.
    This means that we would consider only trivial vector bundles.

Milnor and Stasheff's definition is the standard one. Its benefit is that it allows non-trivial bundles. These are needed to get a non-trivial theory. In fact the set of isomorphism classes of vector bundles over $B$ contains essential information about the homotopy type of $B$.

Update:

Eric Wofsey rightly points out that there plenty of naturally occurring vector bundles (like the tangent bundle of a smooth manifold) which are locally trivial, but globally non-trivial. A general theory of vector bundles should of course cover all these objects.

$\endgroup$
1
  • 1
    $\begingroup$ I would add that they are needed not just to get a nontrivial theory (e.g., to make K-theory useful for something) but because many very important naturally occurring bundles (e.g., tangent bundles) are nontrivial and we want to study them within a general framework. $\endgroup$ Commented Mar 8, 2022 at 0:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .