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:
- 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.
- 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.
Best Answer
The linked question Redundancy in the definition of vector bundles? and your statement
plus your comment including the question
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
Why do we need the local triviality condition when working with vector bundles?
Alternative characterization of local triviality of vector bundles?
Where is the compatibility condition in the definition of a vector bundle?