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I am self studying smooth manifolds and I have encountered the notion of a vector bundle. My current intuition for these objects is that they somehow generalize what we encountered when studying tangent bundles, which is each point in the base space is in some way associated with a vector space. However, looking at the formal definition of a vector bundle I do not understand the role of a local trivialization and why we need it. For reference, Lee's Introduction to Smooth Manifolds defines vector bundles as follows:

Let $M$ be a topological space. A (real) vector bundle of rank $k$ over $M$ is a topological space $E$ together with a surjective continuous map $\pi: E \rightarrow M$ satisfying the following conditions:

  1. For each $p \in M$, the fiber $E_p = \pi^{-1}(p)$ over $p$ is endowed with the structure of a $k$-dimensional real vector space.
  2. For each $p \in M$, there exist a neighorhood $U$ of $p$ in $M$ and a homeomorphism $\phi: \pi^{-1}(U) \rightarrow U \times \mathbb{R}^k$ (called a local trivialization of $E$ over $U$) satisfying the following conditions:
  • $\pi_U \circ \phi = \pi$ (where $\pi_U: U \times \mathbb{R}^k \rightarrow U$ is the projection);
  • for each $q \in U$, the restriction of $\phi$ to $E_q$ is a vector space isomorphism from $E_q$ to $\{q\} \times \mathbb{R}^k \cong \mathbb{R}^k$.

I do not have a good feel for what a local trivialization is. More specifically, my confusion is regarding the second item in the definition. What is this conditions saying exactly and why do we need it? I must be missing something, but what extra information does it give us that first item in the defintion does not?

I took a look at the following question:

Redundancy in the definition of vector bundles?

but I still do not understand what the second item in the definition (and specifically local trivializations) is saying on an intuitive level.

CBBAM
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    You are looking at this backwards. Vector bundles pre exist the definition of vector bundles, and that definition is designed to capture the salient features of a whole class of objects. We define vector bundles to be locally trivial because vector bundles are locally trivial. – Mariano Suárez-Álvarez Nov 23 '22 at 19:42
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    The definition incorporates the features of vector bundles that are needed to do what people down with them — what they already did with them even before anyone gave a definition. And that is why you question is best resolved by learning more about vector bundles and seeing what people do with them. – Mariano Suárez-Álvarez Nov 23 '22 at 19:43
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    Just as you learned to walk by walking, you'll get a feel of why local triviality is useful by seeing it being used to prove things. I am not going to go o tempora o mores but really, this expectation of having an intuitive feeling for why things are in the way they are upon first encounter is only a source of frustration. – Mariano Suárez-Álvarez Nov 23 '22 at 19:46
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    @MarianoSuárez-Álvarez Thank you for your comment, when you say that vector bundles are locally trivial do you mean locally they resemble trivial bundles? I suppose I need to take the definition for what it is for the time being and move on. – CBBAM Nov 23 '22 at 19:50
  • @MarianoSuárez-Álvarez I am still not 100% clear on the definition, but I will take your advice and move on in the book. Hopefully things will become clear with more exposure to them. – CBBAM Nov 23 '22 at 19:53
  • Locally trivial essentially means: its section locally look like vector-valued functions, so we can develop differential calculus for these objects. – Quaere Verum Nov 23 '22 at 19:54
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    Lee wrote an extraordinarily great book which is, among other things, a way to convince you of the sense of those definitions. You will have noticed that one thing the book is not is short. You did not learn to walk in your first day, either! – Mariano Suárez-Álvarez Nov 23 '22 at 19:54
  • @QuaereVerum I have not gotten to sections yet, so perhaps this is what I am missing. – CBBAM Nov 23 '22 at 19:57
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    Let me add that differential geometry is the part of mathematics where you only have one recipe, which is gluing trivial things together, and you are looking at how many dishes you can make with this recipe by just switching the ingredients (what you consider as trivial). Sometimes you end up with a very good meal – Didier Nov 23 '22 at 20:07
  • Roughly speaking, finding a local trivialization means being able to "flatten your vector bundle out locally". The point is that at a local level, everything is trivial; you'll only get nontrivial behavior globally. A good example of this to keep in mind is the mobius band vs cylinder as bundles over the circle. – Mr. Brown Nov 24 '22 at 00:16
  • @ZackFox By local do you mean homeomorphic to $\mathbb{R}^n$? I suppose what is confusing me is if the vector bundle already requires the surjective map $pi$ to be such that $\pi_{-1}({p})$ is a vector space, isn't a trivialization $\phi$ saying the same thing? What does part 2 of the definition say that 1 already doesn't? – CBBAM Nov 24 '22 at 03:15
  • I mean that your vector bundle is locally homeomorphic to $U\times \mathbb{R}^n$ where $U$ is some open set in your base space. The first condition tells you that the fiber of every point is a vector space. The second condition describes the local picture (i.e. in an open neighborhood of a point). – Mr. Brown Nov 24 '22 at 18:08
  • I encourage you to, as others in this thread are, to not think too much about the technical definition yet in a vacuum. Think of the mobius band vs the cylinder; you want to say both these are vector bundles over the circle. Draw some pictures and identify what the axioms are saying in these two specific instances, and you'll start to get a feel for these technical definitions. – Mr. Brown Nov 24 '22 at 18:11
  • @ZackFox That example is very helpful and may have given me the intuition I am looking for. In the case of a cylinder, it trivially holds that every fiber looks like $\mathbb{R}^n$ and things vary in a continuous fashion. For a mobius band the same cannot be said globally as once we reach the end points there is a "jump" in vector spaces as we vary the point around the circle. The only question I have about this situation is what would a trivializing open set around the end point $0 \sim 1$ look like, as one side the vector spaces are "flipped" whereas the other side they are not. – CBBAM Nov 24 '22 at 18:51
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    At this gluing point, you just choose another local cover. The point of local trivializations is exactly that you can “locally unflip” these vectors but not globally. – Mr. Brown Nov 24 '22 at 19:24
  • @ZackFox I think I have a good intuition as to why we need local trivialization now, thanks a lot for your help! – CBBAM Nov 24 '22 at 19:28

1 Answers1

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The linked question Redundancy in the definition of vector bundles? and your statement

However, looking at the formal definition of a vector bundle I do not understand the role of a local trivialization and why we need it.

plus your comment including the question

What does part 2 of the definition say that 1 already doesn't?

indicate that you do not understand the purpose of requiring the existence of local trivializations or that you even think that this requirement could be redundant.

I think when you begin to learn new concepts and start with reading definitions quite often the purpose of all requirements occurring in the definitions is not really clear. Usually this undergoes a change when you learn the theory based on the definitions and an "aha" experience arises.

Anyway, let me make some remarks.

Remark 1.

There is the concept of a pre-vector bundle (or a family of vector spaces) which is defined via Lee's condition 1.

Unfortunately pre-vector bundles may look completely erratic - both topologically and algebraically.

As an example consider $M = [0,1]$. Then the projection $\pi : M \times \mathbb R^2 \to M$ gives us a nice trivial vector bundle. Now consider a function $f$ assigning to each point $p \in M$ a one-dimensional subspace $f(p)$ of $\mathbb R^2$. Define $E_f = \bigcup_{p \in M} \{p\} \times f(p) \subset M \times \mathbb R^2$. Then $\pi_f = \pi \mid{E_f} : E_f \to M$ is a one-dimensional pre-vector bundle. Now consider the function $f(p) = \mathbb R \times \{0\}$ for rational $p$ and $f(p) = \{0\} \times \mathbb R$ for irrational $p$. Then $E_f$ is not locally trivial; indeed no open $U \subset [0,1]$ admits any homeomorphism $\phi : \pi_f^{-1}(U) \to U \times \mathbb R$ (even if we drop both bullet points in condition 2.).

Another example is this: Take again $M = [0,1]$ and $\pi : E = M \times \mathbb R \to M$. Now consider a function $h$ assigning to each point $p \in M$ a homeomorphism $h(p) : \mathbb R \to \mathbb R$. Let $\mathbb R_{h(p)}$ be the real vector space with addition $x + y = h^{-1}(h(x) + h(y))$ and scalar multiplication $\alpha x = h^{-1}(\alpha h(x))$. Let $E_h$ be the space $M \times \mathbb R$, but give each $\{p\} \times \mathbb R$ the vector space structure $\{p\} \times \mathbb R_{h(p)}$. Now consider the function $h(p)(x) = x$ for rational $p$ and $h(p)(x) = x+1$ for irrational $p$. Then $E_h$ is topologically locally trivial, but not algebraically. Indeed no open $U \subset [0,1]$ admits any homeomorphism $\phi : \pi^{-1}(U) \to U \times \mathbb R$ such that $\pi_U \circ \phi = \phi_U$ and $\phi_q : \{q\} \times \mathbb R_{h(q)} \to \{q\} \times \mathbb R$ being a vector space isomorphism for all $q \in U$.

Remark 2.

The "naturally occurring" vector bundles like the tangent bundle of a smooth manifold are locally trivial. This suggests that local triviality is an important property distingishing "interesting bundles" from general pre-vector bundles.

Remark 3.

There are a number of related questions in this forum, for example

Paul Frost
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  • Thank you for your reply! You are exactly right that the trivialization condition is what is the source of my confusion, I suppose I need to read further ahead to get a better feel for it. In your first example, what makes the vector bundle $\pi_f: E_f \rightarrow M$ erratic and fail the trivialization condition? Is it because as we vary $p$ the corresponding subspace jumps around in a possibly non-continuous way? Since we are working with $k < \infty$, aren't all all non-trivial vector (sub) spaces isomorphic to $\mathbb{R}^k$, and so how does continuity fail here? – CBBAM Nov 24 '22 at 18:43
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    @CBBAM Yes, the "jumping" of $f(p)$ makes $E_f$ a space which is not locally compact. But trivial vector bundles over locally compact spaces are locally compact which proves my assertion of the non-existence of any homeomorphism. There are uncountably many functions $f$ with an even more erratic behavior, and all these give erratic pre-vector bundles. See the second of my links if you want to understand in what sense such functions are non-continuous (this is not elementary and perhaps comes too early for you). – Paul Frost Nov 24 '22 at 21:29
  • Thank you again, I think I have a good feeling for the local trivialization condition now and why we need it. I hope the rigorous details come with time. – CBBAM Nov 24 '22 at 23:32