# Vector bundle and its definition

Take this definition of vector bundle: its a triple $(V,\pi,M)$ where $V$ is a set, $\pi:V \rightarrow M$ a sutjective map such that for every $m \in M$ the set $\pi^{-1}(m)$ has the structure of real vector space o dimension $r$ and $m$ a real differential variety of finite dimension with a structure of maximal atlas on $V$. Pay attention that i don't assume that $V$ has a topological structure.
Using this assumptions, can i make $V$ a topological space? For example saing that a subset of $V$ is open if its image with $\pi$ is open in $M$.

• If $r$ is finite, then you can canonically topologize $V$ so that $\pi$ is continuous and for each $m$ the inclusion $\pi^{-1}(m)\hookrightarrow V$ is continuous (assuming the usual topology on a finite dimensional vector space). More precisely, consider the largest topology that contains all $\pi$-preimages of open sets and such that all $\pi^{-1}(m)\hookrightarrow V$ are continuous. May 18, 2015 at 15:39
• @Peter Franek The topology that you define is the following: $A$ is open in $V$ iff $A=\pi^{-1}(B)$ where $B$ is opened in $M$. Right? But every elements of $V$ is a linear combination of $r$ vector $v_1,...,v_r$. So if an element $w=a_1 v_1+...+a_r v_r$ i can think at $w$ as the n-tuple $a_1,...,a_r$. With thiss identification can i put on $V$ the euclidian topology saing for example that $A \in V$ is opened iff $\alpha(A)$ is opened in $\mathbb{R}^r$? May 19, 2015 at 13:04
• No, if you define open sets to be just $\pi$-preimages of open sets in $M$, that's not enough. You want more open sets. For example, you want the complement of any closed set in any fiber to be open. Therefore, you need to add as many open sets so that the inclusion of the fibers into $V$ are continuous. But I'm not sure if this is the right track at all. (In most books I have seen there is a local triviality condition that immediately generates a topology) May 19, 2015 at 13:10
• @ Peter Franek because in your comment i cannot understand what do you mean when you say the largest topology of all preimages of open sets ..... How can i define it? As intersection of biggest topology? May 19, 2015 at 13:16
• I thought about a topology generated by all $\pi$-preimages of open sets in $M$ and all sets $U$ such that the intersection of $U$ with each fiber is open (in the vector space). But now I'm not sure if it makes much sense.. will think about it. May 19, 2015 at 13:26

This kind of approach (have the topologisation of manifold descend from a differential atlas) is followed in quite a rich-in-detail way by the Italian book Abate, Tovena, Geometria Differenziale (which I presume you can read). There you can also check the correctness of your requirements. To quote some lemmas and definitions:

• page 62 (on the topologisation of manifolds):

A $n$-dimensional atlas $\mathcal{A}=\{(U_\alpha,\varphi_\alpha)\}$ on a set $M$ induces a topology in which a subset $A\subseteq M$ is open if and only if $\ \forall\alpha\ \ \varphi_\alpha\left(A\cap U_\alpha\right)$ is open. Moreover, it is the only topology allowing all the $U_\alpha$ to be open in $M$ and all the $\varphi_\alpha$ to be topological embeddings.

• page 134 (definition of vector bundle):

A vector bundle of rank $r$ on a manifold $M$ is a manifold $E$ provided with a surjective differential map $\pi:E\to M$ such that:

1. $\forall p \in M,\ \pi^{-1}(p)$ is endowed with a structure of $r$-dimensional vector space.

2. $$\forall p \in M\ \exists\ U\ni p\text{ open set },\ \exists\ \chi:\pi^{-1}(U)\to U\times \mathbb{R}^r\ \text{ diffeomorphism s.t.}\\ \forall p\in U\ \pi_2\circ\chi|_{\pi^{-1}(p)}\text{ is a linear isomorphism } \pi^{-1}(p)\to \pi_2\left(\{p\}\times \mathbb{R}^r\right)$$ $\pi_1$ and $\pi_2$ being the projections on the factors of $U\times \mathbb{R}^r$

• page 136 (on the conditions determining a vector bundle)

Let $M$ a manifold, $E$ a set, $\pi:E\to M$ a surjective function, $\mathcal{A}=\{(U_\alpha,\varphi_\alpha)\}$ an atlas of $M$ and $\chi_\alpha$ maps $\pi^{-1}(U_\alpha)\to U_\alpha\times \mathbb{R}^r$ such that:

1. $\chi_\alpha(\pi^{-1}(p))\subseteq \{p\}\times\mathbb{R}^r$

2. $$\forall U_\alpha\cap U_\beta\neq\emptyset\ \exists g_{\alpha\beta}:U_\alpha\cap U_\beta\to \text{GL}(r,\mathbb{R})\text{ differentiable map s.t. }\\\chi_\alpha\circ\chi_\beta^{-1}:\left(\alpha\cap U_\beta\right)\times\mathbb{R}^r\to\left(\alpha\cap U_\beta\right)\times\mathbb{R}^r\text{ satisfies }\\\chi_\alpha\circ\chi_\beta^{-1}(p,x)=(p,g_{\alpha\beta}(x))$$

Then, $E$ admits exactly one stucture of rank-$r$ vector bundle on $M$ such that $\{(U_\alpha,\phi_\alpha,\chi_\alpha)\}$ satisfies the condition of the point (2) in the definition.

So, as you see, your idea is right in some sense, but the requirements to induce a vector bundle seem quite sharper. The book I'm quoting takes its time (quite the time, actually) to thoroughly deal with these technical lemmas, which seems to me to be useful to your purpose. Of course, I assume it's not the only one in the world.

Added: I don't think I did, but I might have made some mistakes in copying and translating the text.

Added 2: Specifically: let $E, M, U_\alpha,\varphi_\alpha,\chi_\alpha$ as in page 136. Let $\rho_\alpha:\pi^{-1}(U_\alpha)\to \varphi_\alpha(U_\alpha)\times\mathbb{R}^r$ defined by $\rho_\alpha(x)=(\varphi_\alpha\pi_1\chi_\alpha(x),\,\pi_2\chi_\alpha(x))=(\varphi_\alpha\pi(x),\,\pi_2\chi_\alpha(x))$.

$\mathcal{B}=\{(\pi^{-1}(U_\alpha),\rho_\alpha,\varphi_\alpha(U_\alpha)\times\mathbb{R}^r)\}$ is an atlas on the manifold $E$, therefore it induces one and only one topology as of the result in page 62.

• I'm italian so i can read the book in the original language. The traslation is right thanks. May 19, 2015 at 13:27

What is missing from your definition is not only the topological structure of $V$, but the much more important locally-trivial character: there is a covering of $M$ with open sets $U$ such that $\pi^{-1} (U) \simeq U \times \mathbb R ^r$.

• Alex, can you prove that the above definition is not equivalent to the usual one (for some class of real varieties)? May 18, 2015 at 15:40
• so my definition is wrong. I have to add thoose two characters May 18, 2015 at 15:47
• I'm confused because this definition comes from the notes about the theory of relativity. May 18, 2015 at 15:49
• @Dario: Of course you may construct an alternative concept of vector bundle the way you do. But when constructing mathematical objects, ask yourself: "why am I doing it?". Well, we usually want our vector bundles to be differentiable manifolds too, and in order to achieve this we need local similarity to Euclidean spaces. Without the assumption on local triviality as suggested above I do not see how you would prove that your bundles are manifolds themselves. Now, if you don't need them being manifolds, then maybe your definition suffices. It all is about what you want to do with them. May 18, 2015 at 16:00