# Definition of a coordinate vector bundle

Consider the following definition of a coordinate vector bundle.

Let $M$ be a smooth manifold of dimension $m$, and $\{(f, U_f)\}$ an atlas of compatible charts for $M$. A smooth coordinate vector bundle of rank $n$ over $M$ relative to this atlas consists of a smooth manifold $B$, and a smooth surjective mapping $\pi \colon B\to M$, and diffeomorphisms $\phi_f\colon U_f\times \mathbb{R}^n\to \pi^{-1}(U_f)$ such that

1. $\pi\phi_f(p, v)=p$ for all $(p, v)\in U_f\times \mathbb{R}^n$.
2. the smooth maps $\phi_{f, p}\colon \mathbb{R}^n\to \pi^{-1}(U_f)$ defined for $p\in U_f$ by $\phi_{f, p}(v)=\phi_f(p, v)$ are such that $\phi^{-1}_{f', p}\circ \phi_{f, p}\colon \mathbb{R}^n\to \mathbb{R}^n$ is in $\operatorname{GL}(n, \mathbb{R})$ for each $f$ and $f'$ and all $p\in U_f\cap U_{f'}$.
3. the map $g_{f'f}\colon U_f\cap U_{f'}\to \operatorname{GL}(n, \mathbb{R})$ defined by $\phi^{-1}_{f', p}\circ \phi_{f, p}$ is smooth.

Although it is built into the usual definition of vector bundles that each fiber $\pi^{-1}(p)$ is a vector space and the coordinate maps $\phi_{f, p}$ are linear, this definition doesn't make these two properties explicit. My question is whether these two properties can be extracted from the above definition readily. (I can assume that they still have to hold in this case, because the transition maps satisfy the Cech cocycle relations.)

• By 1., certainly each fibre of $B$ can be set-theoretically identified with $\mathbb{R}^n$ over $U_f$, using $\phi_f$. By 2., the induced vector space structure doesn't depend on your choice of $f$. – uncookedfalcon Aug 24 '14 at 3:15
• @uncookedfalcon I agree that by 1, $\phi_{f,p}$ is indeed a bijection $\mathbb{R}^n\to \pi^{-1}(p)$, but I can't follow the 2nd part of your comment. Does your argument also imply that $\phi_{f, p}$ is also linear? – EPS Aug 24 '14 at 3:50

this is in response to your comment; since it's a bit long I write it as an answer. The idea is the following: given two points $v,w$ in the fiber $\pi^{-1}(p)$, we want to define $v+w$. Using a chart $U_f$, we borrow the vector space structure from $\mathbb{R}^n$, i.e. we take $$v +w := \phi_{f, p} ( \phi_{f, p}^{-1}(v) + \phi_{f, p}^{-1}(w))$$To check well definition, if we used another chart $U_{f'}$, we need that $$\phi_{f', p}(\phi_{f', p}^{-1}(v) + \phi_{f', p}^{-1}(w)) = \phi_{f, p} ( \phi_{f, p}^{-1}(v) + \phi_{f, p}^{-1}(w))$$Equivalently: $$\phi_{f, p}^{-1} \phi_{f', p}(\phi_{f', p}^{-1}(v) + \phi_{f', p}^{-1}(w)) = \phi_{f, p}^{-1}(v) + \phi_{f, p}^{-1}(w)$$Apply linearity of $\phi^{-1}_{f, p} \phi_{f', p}$ (point 2.) on the LHS and cancel some adjacent inverses, and we get addition is well defined. Scalar multiplication is similar.
These things are linear by definition, because the vector space structure on $\pi^{-1}(p)$ just comes from carrying the structure over via $\phi_f$. The only thing to check is that this is well-defined, which is what the condition about $GL_n$ does.
In other words, if you have an arbitrary set $S$ and a vector space $V$ and a bijection $M : V \to S$ then you can make $S$ into a vector space by saying that $s + t := M(M^{-1}(s) + M^{-1}(t))$ and similarly for scalar multiplication.