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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.

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    $\begingroup$ 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 $\endgroup$
    – ali
    Oct 11, 2021 at 16:19

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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.

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  • $\begingroup$ 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? $\endgroup$ Oct 11, 2021 at 16:25
  • $\begingroup$ 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) $\endgroup$
    – ali
    Oct 11, 2021 at 17:12
  • $\begingroup$ @jacopoburelli for your second question, yes. as a topological space $U\times V$ has product topology $\endgroup$
    – ali
    Oct 11, 2021 at 17:13
  • $\begingroup$ Okay, thanks for the comment, I wasn't familiar with these aspects of category theory. $\endgroup$ Oct 11, 2021 at 17:14
  • $\begingroup$ What's the map $E_{|_{U}}$ you'reffering to? $U \subseteq B$. $\endgroup$ Oct 11, 2021 at 20:07

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