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$\require{AMScd}$ As I understand it, a rank $k$ vector bundle is a pair of topological spaces with a map between them $$ E\xrightarrow{p}B $$ such that there exists an open cover $(U_\alpha)$ of $B$ over which $E$ locally looks like $U_\alpha\times\mathbb R^k$. Precisely, we have homeomorphisms $$ \phi_\alpha\colon p^{-1}(U_\alpha)\to U_\alpha\times\mathbb R^k $$ such that the diagram \begin{CD} p^{-1}(U_\alpha) @>\phi_\alpha>> U_\alpha\times\mathbb R^k\\ @VVbV @VV\text{pr}_1V\\ U_\alpha @>\text{id}>> U_\alpha \end{CD} commutes (I can't do diagonal arrows).

A section of the vector bundle is a continuous map $s\colon B\to E$ such that $p\circ s=\text{id}_B$. A basic fact about vector bundles is that every vector bundle admits at least one section, namely the zero section, obtained by identifying each $p^{-1}(U_\alpha)$ with $U_\alpha\times\mathbb R^k$ and taking $s(b)=0$. Explicitly, we are taking: $$ s(b)=\phi_\alpha^{-1}((b,0)) $$ for $b\in U_\alpha$.

But in order to get a well defined continuous map, we need to ensure that $\phi_\alpha^{-1}(b,0)=\phi_\beta^{-1}(b,0)$ if $b\in U_\alpha\cap U_\beta$. Now one thing we certainly can't do is require that $\phi_\alpha$ and $\phi_\beta$ actually agree on the intersection - then we could glue them all together to get a global trivialization of the bundle, which we can't do in general. Why then are we able to insist that they agree at a certain point? To me, it seems equivalent to the existence of a global section, and hence completely circular.


There must be something obvious I'm missing. In the meantime, here's an equivalent formulation of the problem: we define an affine bundle $E\xrightarrow{p}B$ to be the same as a vector bundle, but we identify the fibres with $k$-dimensional affine linear subspaces of some fixed real vector spaces. So they needn't contain a special zero point.

Since affine spaces are homeomorphic to vector spaces of the same dimension, any rank $k$ affine bundle is isomorphic to some rank $k$ vector bundle (though I'm not at all sure about whether this is true, for similar reasons). So an affine bundle should admit a global section, but there is no easy way to see how to get one.

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  • $\begingroup$ I recommend changing the title to "$\mathbb{R}^n$-fiber bundle" or "affine bundle". $\endgroup$
    – Kyle
    Commented Jun 16, 2015 at 16:42
  • $\begingroup$ @squirrel I strongly disagree. The reason for my question was confusion as to the definition of a vector bundle. The title 'Why can we always take the zero section of an $\mathbb R^n$-fibre bundle?' is a worse title because 1) We can't always take the zero section of an $\mathbb R^n$-bundle, and more importantly there is no reason to suppose that we can, and because 2) most of the answers below, which specifically address my confusion with the definition of a vector bundle, would stop making sense. $\endgroup$ Commented Jun 17, 2015 at 14:24
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    $\begingroup$ Ah, I agree with you now. I thought your question was "Why do bundles of the described form admit zero-sections?", as opposed to "Why do vector bundles, which I believe are defined as below, have zero-sections?" I didn't notice your acceptance of Dan's answer, so I thought you were still looking for more answers like PabloDC's or something. $\endgroup$
    – Kyle
    Commented Jun 17, 2015 at 15:13

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In your definition you are missing the condition that the transition maps for a vector bundle are supposed to be linear fiberwise. Once you assume that, the problem you're facing in gluing will go away.

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If you had a canonical zero element in each fiber, you would see the section right? So but that is already true, since you usually require in the definition that your fibers admit a vector space structure and you also require that the local trivializations are isomorphisms i.e. also preserve zero. That gives you a continous choice of zeros everywhere.

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    $\begingroup$ If my answer is not helping you, then I would want you to please write down your precise definition of vector bundles. A wonderful read on this topic is Milnor's Characeristic Classes, by the way. $\endgroup$ Commented Apr 1, 2015 at 18:09
  • $\begingroup$ I gave a definition at the beginning of my post. A more concise statement would be - a rank $k$ vector bundle is a fibre bundle with fibre $\mathbb R^k$. This definition is different, but perhaps it's the correct one. $\endgroup$ Commented Apr 1, 2015 at 19:23
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    $\begingroup$ Okay then you are definitely having a very non-conventional definition of vector bundles. All authors I know usually require extra structure on vector bundles! An equivalent definition for this usual vector bundle in terms of fiber bundles would be: a fiber bundle with fiber $\mathbb R^k$ and structure group $GL_k(\mathbb R)$. $\endgroup$ Commented Apr 1, 2015 at 19:43
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I think it's more straightforward than you're currently thinking. Every fiber $F_b$ has the structure of a vector space, and therefore has a zero element $0_b$. Sending an element $b$ to the zero vector $0_b$ defines the (continuous) zero section you're after.

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    $\begingroup$ Each fibre has the structure of a vector space, but not in a unique way. So there's no canonical way to choose a zero element. And since we want the zero section to be continuous, we need to be careful that we choose the zero elements in such a way that they all match up in a continuous way. It's not obvious to me how to do that. $\endgroup$ Commented Apr 1, 2015 at 17:52
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    $\begingroup$ @Donkey_2009 What on earth could it mean that the fiber has the structure of a vector space "not in a unique way"? Each fiber is a vector space, period. $\endgroup$ Commented Apr 1, 2015 at 18:01
  • $\begingroup$ Perhaps @Donkey_2009 means that a given abstract vector has no canonical isomorphism with $\mathbb{R}^n$? $\endgroup$
    – PeterJL
    Commented Apr 1, 2015 at 18:16
  • $\begingroup$ @SpamIAm That is indeed what I mean. You've seen my definition of a vector bundle, and (at least according to that definition) there is no canonical isomorphism of each fibre with $\mathbb R^n$. My definition is that of a fibre bundle with fibre $\mathbb R^n$, but I now see that isn't correct. $\endgroup$ Commented Apr 1, 2015 at 19:24
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As was explained in the other answers, for vector bundles one requires that the fibers are equipped with a vector space structure, and this immediately gives a (continuous) section. For affine bundles (in your sense), as you point out, there is no preferred origin so one cannot just define the zero section using the local charts. Nonetheless, assuming the existence of partitions of the unity for the base one can construct a section as follows.

Fix a (locally finite) affine bundle atlas $\{\phi_i:U_i\times A\rightarrow E\}$ and let $\{\rho_i\}$ be a partition of the unity associated to the covering $\{U_i\}$ of $B$. In an affine space convex combinations of vectors do not depend of the base point, so you simply define

$ s(x)=\sum \rho_i(x)\phi_i(x,0). $

In this way you get a continuous section.

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  • $\begingroup$ $\phi_i(x,0)$ is an element in $E$, right? So what's the meaning of multiplying it by $\rho_i(x)$? $\endgroup$
    – rmdmc89
    Commented Feb 5, 2019 at 21:29
  • $\begingroup$ They all live in the same fiber $E_x$ which has a vector space structure $\endgroup$
    – User203940
    Commented May 26, 2021 at 15:46
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Viktor Kleen is right that your definition is missing something, but I think the usual formulation of what you're missing is a bit more than what Viktor wrote. Part of the structure of a vector bundle is an assignment, to each fiber, of a vector-space structure on that fiber; furthermore the local coordinate maps $\phi_\alpha$ in your definition are required to be linear on every fiber.

The vector-space structure on the fibers includes, in particular, a choice of zero vector in every fiber, so you have a function assigning to each point $b\in B$ the zero vector of the fiber over $b$. This function is continuous on each $U_\alpha$ because it corresponds, under the homeomorphism $\phi_\alpha$, to the continuous function $U_\alpha\to U_\alpha\times \mathbb R^k$ sending each $b$ to $(b,0)$ (where $0$ means the zero vector in $\mathbb R^k$). Finally, a function continuous on each set $U_\alpha$ of an open cover is continuous on the whole space, so you have a continuous zero section.

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