Definition of vector bundle

Let $$E$$, $$B$$, $$F$$ be topological spaces, and $$p:E\to B$$ a continuous surjection. These data define a fiber bundle with fiber $$F$$ if, for every $$b\in B$$, exists an open set $$b\in U\subset B$$ and a homeomorphism $$h:p^{-1}(U)\to U\times F$$ such that (letting $$\pi:U\times F\to U$$ be the canonical projection and writing $$p_\cdot$$ for the map $$p^{-1}(U)\to U$$ induced by $$p$$) $$p_\cdot=\pi\circ h$$.

From what I understand, the fiber bundle above is a $$n$$-dimensional vector bundle iff $$F$$ has also the structure of $$n$$-dimensional $$k$$-vector space for some field $$k$$, and ($$*$$) the map $$p^{-1}(b)\to F$$ induced by $$h$$ is an isomorphism of vector spaces for any $$b\in B$$.

Is this the actual definition? I'm quite sure that this is what is written in my notes, but how do I get a structure of vector space on $$p^{-1}(b)$$, if not using the structure on $$F$$ and the fact that $$p^{-1}(b)\to F$$ is bijective? And using such structure on $$p^{-1}(b)$$, the requirement ($$*$$) becomes trivial, so that any fiber bundle with fiber a vector space would be a vector bundle. Would you please enlighten me?

• the key is how two intersecting sets $U,W$ open in $B$ behave related to the maps involved. Dec 14, 2022 at 15:22
• @janmarqz if $b$ is contained in two open sets $U,W\subset B$, we get two homeomorphisms $p^{-1}(b)\to F$, so two structures of vector space on $p^{-1}(b)$. However I don't see any way to require a linear homomorphism between these structures, if this is what you meant (probably not, but I couldn't interpret your hint otherwise). Thanks Dec 14, 2022 at 15:41
• the condition is to be given on $$(U\cap W)\times F\to p^{-1}(U\cap W)\to(U\cap W)\times F$$ Dec 14, 2022 at 15:46
• @janmarqz in the sense that I should require the homeomorphism $(U\cap W)\times F\to(U\cap W)\times F$ also to be linear? But $(U\cap W)\times F$ has not vector space structure, from what I understand Dec 14, 2022 at 15:56
• Let me also give a useful example to keep in mind. If you consider the tangent bundle of a smooth manifold, you have the diffeomorphisms $\psi_U$ between an open subset $U$ of your manifold and $R^n$. In this case, the $t_{UV}(x)$ I mentioned below turns out to be $D(\psi_U^{-1}\circ\psi_V)$ which we know is linear. Dec 14, 2022 at 22:57

It is not true that any fiber bundle with fiber a vector space is a vector bundle. Let me first define a vector bundle in a precise manner. Let $$p:E\xrightarrow{}B$$ be a fiber bundle with fibers homeomorphic to $$\mathbb{R}^n$$. Let us denote $$p^{-1}(x)$$ by $$F_x$$ for $$x\in B$$. Consider trivializing neighbourhoods $$U,V\subset B$$ such that $$U\cap V$$ is nonempty. That is, we have homeomorphisms $$\varphi_U:p^{-1}(U)\xrightarrow{}U\times \mathbb{R}^n$$ and $$\varphi_V:p^{-1}(V)\xrightarrow{}V\times \mathbb{R}^n$$. As mentioned in the comments, these homeomorphisms restrict to a homeomorphism $$(U\cap V)\times \mathbb{R}^n \xrightarrow{\varphi_U^{-1}} p^{-1}(U\cap V)\xrightarrow{\varphi_V}(U\cap V)\times \mathbb{R}^n$$ which homeomorphically maps $$(x,F_x)$$ to itself. In particular, we can write this assignment as $$(x,v)\longmapsto \big(x,t_{UV}(x)(v)\big)$$ where $$t_{UV}(x):\mathbb{R}^n\xrightarrow{} \mathbb{R}^n$$ is a homeomorphism for each $$x\in U\cap V$$. Therefore, we can also consider this $$t_{UV}$$ as a map $$t_{UV}:U\cap V \xrightarrow{} \text{Homeo}(\mathbb{R}^n)\hspace{2cm} x\longmapsto t_{UV}(x)$$ where $$\text{Homeo}(\mathbb{R}^n)$$ is the set of homeomorphisms of $$\mathbb{R}^n$$ to itself. The idea is that every intersection of trivializing neighbourhoods induce a homeomorphism from $$F_x$$ to itself. We say that this fiber bundle is a vector bundle if the image of every $$t_{UV}$$ is contained in $$GL_n(\mathbb{R}) \subset \text{Homeo}(\mathbb{R}^n)$$. So we want the local trivializations to match in a linear fashion.