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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?

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  • $\begingroup$ the key is how two intersecting sets $U,W$ open in $B$ behave related to the maps involved. $\endgroup$
    – janmarqz
    Dec 14, 2022 at 15:22
  • $\begingroup$ @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 $\endgroup$ Dec 14, 2022 at 15:41
  • $\begingroup$ the condition is to be given on $$(U\cap W)\times F\to p^{-1}(U\cap W)\to(U\cap W)\times F$$ $\endgroup$
    – janmarqz
    Dec 14, 2022 at 15:46
  • $\begingroup$ @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 $\endgroup$ Dec 14, 2022 at 15:56
  • $\begingroup$ 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. $\endgroup$ Dec 14, 2022 at 22:57

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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.

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  • $\begingroup$ . . n i c e . . . . . $\endgroup$
    – janmarqz
    Dec 14, 2022 at 21:23

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