# Understanding the definition of vector bundle

I'm currently studying from Husemaller Vector Bundles and I'm having some problems understanding the definition given and the conventions used by the author. I think that the book gives a definition which is very general and then the reader should specify it to the needed context. I don't understand when I have to disambiguate the latter.

For example proposition $$1.5$$ p.$$25$$ states:

Proposition 1.5: Let $$\xi = (E,p,B)$$ be a $$k-$$dimensional vector bundle. Then $$p$$ is an open map. The fibre preserving functions $$a : E \oplus E \longmapsto E$$ and $$s: F \times E \longmapsto E$$ defined by the algebraic operations $$a(x,x') = x+x', s(k,x) = kx, k \in F$$, are continuos.

What I don't understand are the following: why should I ask continuity of cross sections in the first place? the definition of vector bundle given is the following :

Definition: A $$k$$-dimensonal vector bundle $$\xi$$ over $$F$$ is a bundle $$\xi = (E,p,B)$$ together with the structure of a $$k-$$dimensional vector space over $$F$$ on each fibre $$p^{-1}(b)$$ such that the following local triviality condition is satisfied. Each point of $$B$$ has an open neighborhood $$U$$ and a $$U-$$isomorphism $$h:U \times F^k \longmapsto p^{-1}(U)$$ such that the restriction $$b \times F^k \longmapsto p^{-1}(b)$$ is a vector space isomorphism for each $$b \in U$$.

Here the word $$U-$$isomorphism seems undefined. Does the author mean homeomoprhism? the word bundle came with an other definition which doesn't seem to involve continuity, in other words a bundle is described as a triple $$(E,p,B)$$ where $$p: E \longmapsto B$$ is a map.

I don't understand what's peculiar with this definition, it seems that random maps are bundles. Could someone with more experience with me help me clarify?

In the end I don't understand what kind of structure $$E$$ has in order to well define sum and "scalar" multiplication, i.e $$a,s$$.

Any help would be appreciated.

• there is a very strong condition on the map $p$ in the definition, a random map usually don't satisfy these condition, usually even the fibers are not vector space, let alone the part about local triviality of the map
– ali
Oct 11, 2021 at 16:19

first of all if you have maps $$f:X\to S$$,$$g:Y\to S$$,by a $$S$$ morphism between $$h:X\to Y$$ we mean that $$f=h\circ g$$(so the two way you can go from $$X$$ to $$S$$ should be the Same). In your example you have maps $$E_{|U}\to U$$,$$U\times V\to U$$ so you can talk about a $$U$$ morphism between $$E_{|U}$$ and $$U\times V$$(the point of a $$S$$ morphism $$h$$ is basically that you want $$h$$ to sends fiber over $$s$$ in $$X$$ to the fiber over $$s$$ in $$Y$$ ).

When you define a vector space, you don't mention continuity because you don't even have a topology, but it is an easy exercise that if you have a topological field(like $$\mathbb{R},\mathbb{C}$$) if you put the product topology on $$V$$, the sum and scalar product become continuous.

about the relationship between definition and proposition: when you have $$p: E\to B$$ by definition you have a sum and scalar product on each fiber so you get maps $$E\times E\to E$$, and $$F\times E\to E$$ and these are continuous because $$E$$ locally looks like $$U\times V$$ and for $$U\times V$$ you know that these maps are continuous(by previous paragraph).

the idea of the bundle is that it is a space $$E$$ over $$B$$(meaning that there is a map $$E\to B$$) that locally looks like the product of $$B$$ with a fixed space. if this fixed space is a vector space, we call $$E$$ a vector bundle. but you can also talk about a circle-bundle or a $$F$$-bundle for any fixed space you pick.

• Thanks, this answer clarifies a lot, I'm surprise by the definition of $U-$isomorphism, on the notes given by Hatcher [here](pi.math.cornell.edu/~hatcher/VBKT/VB.pdf) it seems to require homeomorphism. Are the two notions equivalent? Added: on $U \times V$ you put the product topology to exploit second paragraph of your answer? Oct 11, 2021 at 16:25
• for your first question: this definition says $U$-isomorphism I assume that by an isomorphism he means a homomorphism that respect the addition and scalar product on fibers. being a $U$-isomorphism is as I said means that the fiber over every point of $U$ in $E$ goes to the fiber over that point in $U\times V$. hatcher notes also has this condition. an $S$ morphism is a standard notation in math(it comes from category theory)
– ali
Oct 11, 2021 at 17:12
• @jacopoburelli for your second question, yes. as a topological space $U\times V$ has product topology
– ali
Oct 11, 2021 at 17:13
• Okay, thanks for the comment, I wasn't familiar with these aspects of category theory. Oct 11, 2021 at 17:14
• What's the map $E_{|_{U}}$ you'reffering to? $U \subseteq B$. Oct 11, 2021 at 20:07