In John Lee's classic Introduction to Smooth Manifolds, the following definition of vector bundle is given.

Definition. 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 $\pi:E\to M$ satisfying the following conditions:

(i) 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.

(ii) For each $p\in M$, there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $\Phi:\pi^{-1}(U)\to U\times\Bbb{R}^k$ (called a local trivialization of $E$ over $U$*), satisfying the following conditions:

  1. $\pi_U\circ\Phi=\pi$ (where $\pi_U:U\times\Bbb{R}^k\to U$ is the projection);

  2. for each $q\in U$, the restriction of $\Phi$ to $E_q$ is a vector space isomorphism from $E_q$ to $\{q\}\times\Bbb{R}^k\cong\Bbb{R}^k$.

But if we skip conditions (i) and 2, can't we just define the vector space structure on $E_p$ by using its set-theoric bijection with $\{p\}\times\Bbb{R}^k$?

In other words:

Question: Let $E$ and $M$ be topological spaces and $\pi:E\to M$ a continuous map such that for each $p\in M$ there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $\Phi:\pi^{-1}(U)\to U\times\Bbb{R}^k$ such that $\pi_U\circ\Phi=\pi$.

Is $E$ is vector bundle?

  • 3
    $\begingroup$ The problem is you don't know whether all trivializations will induce the same vector space structure on the fiber. For this to work, you'd have to use trivializations whose change of coordinates is linear on fibers. $\endgroup$
    – Pedro
    Jan 3, 2016 at 13:59

2 Answers 2


No Look at $[0,1]\times \mathbb{R}$, now identify $\{0\} \times \mathbb{R}$ and $\{1\} \times \mathbb{R}$ via the map $$f:\{0\} \times \mathbb{R}\rightarrow \{1\} \times \mathbb{R}$$ $$f(0,x)=(1,x^3)$$ a non linear map. The base space is then $S^1$. Basically in a vector bundle the maps between the fibres must be linear.


As already pointed out, your definition is incomplete in that it lacks compatibility conditions among the different trivializations. The original definition guarantees this compatibility using a god-given vector space structure on the fibers, but it's true that instead you could take your new definition and add the following condition:

Given $U$ and $V$ admitting a trivialization, the map $$\Phi_U \circ \Phi_V^{-1}: (U \cap V) \times \mathbb{R}^k \to (U \cap V) \times \mathbb{R}^k$$ is given by $$(x,w) \mapsto (x,g_{UV}(x)(w))$$ where $g_{UV}: U \cap V \to GL_n(\mathbb{R})$ is smooth.

P.S.: I am not completely sure on how to prove that the original definition implies the smoothness of the $g_{UV}$... I hope that somebody will point it out. It should be the only point missing to show the equivalence of the two definitions.

Edit: As pointed out by Karl Kronenfeld in the comment, if we define smooth functions

$$ f_i: U\cap V \to (U \cap V) \times \mathbb{R}^k, \ x \mapsto (x, e_i),$$ $$ g_j: (U\cap V)\times \mathbb{R}^k \to \mathbb{R}, \ (x,v) \mapsto v_j,$$

then we obtain $g_{UV}^{ij}$ as the composition $g_j \circ \Phi_U \circ \Phi_V^{-1}\circ f_i$.

  • $\begingroup$ As for smoothness of $g_{UV}$: the $i$th coordinate projection $(U\cap V)\times \Bbb R^k\to\Bbb R$ and the map $(U\cap V)\to(U\cap V)\times\Bbb R^k$ sending $x\mapsto (x,e_j)$ are smooth, so the $ij$th component of the matrix $g_{UV}(x)$ is smooth by composition. $\endgroup$ Jan 28, 2019 at 15:30
  • $\begingroup$ Thanks, I've edited my answer to include this! :) $\endgroup$
    – 57Jimmy
    Jan 28, 2019 at 17:33

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