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.

  • 1
    $\begingroup$ 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. $\endgroup$ Nov 23, 2022 at 19:42
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 23, 2022 at 19:43
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 23, 2022 at 19:46
  • 3
    $\begingroup$ 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! $\endgroup$ Nov 23, 2022 at 19:54
  • 2
    $\begingroup$ 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 $\endgroup$
    – Didier
    Nov 23, 2022 at 20:07

1 Answer 1


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

  • $\begingroup$ 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? $\endgroup$
    – CBBAM
    Nov 24, 2022 at 18:43
  • 1
    $\begingroup$ @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). $\endgroup$
    – Paul Frost
    Nov 24, 2022 at 21:29
  • $\begingroup$ 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. $\endgroup$
    – CBBAM
    Nov 24, 2022 at 23:32

You must log in to answer this question.

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