I understand the notion of a nontrivial fiber bundle with fiber $F$ over a base manifold $B$, as defined in terms of the projection map $\pi$: for any sufficiently small region $U \subset B$, the preimage $\pi^{-1}(U)$ is homeomorphic to the product space $U \times F$, but the preimage $\pi^{-1}(B)$ itself (the total space) is not homeomorphic to $B \times F$. The Mobius strip is a standard example for visual intuition.

However, physicists like myself often think of a fiber bundle in terms of its sections rather than its projection map. Is there an equivalent definition of a nontrivial bundle formulated in terms of its sections $\sigma$ (the right-inverses of $\pi$)? I.e. a statement of the form "a fiber bundle is nontrivial iff (some section $\sigma$ has)/(all sections $\sigma$ have) property $X$"? If not, is there any intuition for what the sections of a nontrivial bundle "look like"? I know that a principle bundle is nontrivial iff it does not admit any global section, but I'm curious how things work for general fiber bundles.


Is there an equivalent definition of a nontrivial bundle formulated in terms of its sections σσ (the right-inverses of ππ)?

Yes. See below.

I.e. a statement of the form "a fiber bundle is nontrivial iff (some section σσ has)/(all sections σσ have) property XX"?

No. See below for why I technically said no.

If not, is there any intuition for what the sections of a nontrivial bundle "look like"?

Individual sections can "look like" anything.

Answer: Each $\pi^{-1}(U)$ comes with a (not unique) homeomorphism: $t_i: \pi^{-1}(U_i) \to U_i \times F$. Thus over each intersection, we obtain transition functions $f_{ij}: U_{ij} \to Aut(F)$, which satisfy the cocycle condition: $$f_{ij}f_{jk} = f_{ik}.$$

So your bundle is trivial iff your choices of transition functions $t_i$ could be altered so that $f_{ij} =1$ for all ${ij}$. In fancy language, your bundle's triviality is measured by the Cech Cohomology generated by these $f_{ij}$.

What does this have to do with sections? Well sections could be used to create the trivializations. So start with sections, $s_i$, build associated trivializations, $t_i$, then build associated transition functions $f_{ij}$, and then your bundle is non trivial if you could have used different sections $s'_i$ so that all of the transition functions were the identity.

So the reason why I said no to the second question I answered is that it is not a matter of talking about some section having a property or all sections having a property. But rather it is about the collection of sections having a property.

  • $\begingroup$ Isn't "a set of sections $s_i'$ such that all the transition functions [between the associated trivializations] are the identity" morally speaking the same thing as a global section? Yet a nontrivial non-principal vector bundle can still admit a global section. (I apologize for the naive question, I don't know very much about fiber bundles.) $\endgroup$ – tparker Oct 17 '17 at 3:39
  • $\begingroup$ A vector bundle always has a global section: the zero section. So whenever you say "global section" it can come with a caveat in the vector bundle case, or without a caveat in the principal bundle case. So there is no general notion of "non-trivial global section". But yes, morally speaking what I said is the same as a "non-trivial global section". I just don't think there's another way to say it. $\endgroup$ – cheyne Oct 17 '17 at 16:02
  • $\begingroup$ Okay, so you're saying that roughly speaking, a fiber bundle is nontrivial iff it admits a nontrivial global section - where coming up with an exact general definition of "nontrivial global section" would probably be rather involved, but basically it means "any global section" in the case of a principal bundle, and "a global section which is never zero" in the case of a vector bundle. Is that the general picture? $\endgroup$ – tparker Oct 17 '17 at 16:56
  • $\begingroup$ Exactly, which, as you pointed out, is a locally well-defined section which transitions properly. Note that if you have a zero-valued section, you can not define the transition function t_i / 0! $\endgroup$ – cheyne Oct 17 '17 at 19:12

This answer discusses smooth fiber bundles.

In a way, every fiber bundle with structure group $G$ can be represented by a principal $G$-bundle, where $G$ can be any Lie group. So, if you understand principal bundles, all I need to do is explain the above claim.

Let $\pi:E\to B$ be a fiber bundle with fiber $F$ and structure group $G$. This means that $G$ is a Lie subgroup of $\mathrm{diff}(F)$, the diffeomorphism group of $F$, and that we have the following data:

$1)$ An open covering of $B$ by the sets $\{U_\alpha\}_{\alpha\in I}$.

$2)$ A trivialization $\psi_\alpha:U_\alpha\times F\to \pi^{-1}(U_\alpha)$ for every $\alpha\in I$, such that all the induced transition maps $\varphi_{\alpha\beta}:U_\alpha\cap U_\beta\to \mathrm{diff}(F)$ admit values in $G$.

The associated principal $G$-bundle, $P\to B$, can be described, as a set, as follows. For a point $b\in B$, the fiber $P_b$ consists of all the diffeomorphisms $F\to E_b$ which agree with the given bundle data. Namely, let $\alpha\in I$ be such that $b\in U_\alpha$. So $\psi_\alpha|_b$ is a diffeomorphism $F\to E_b$, and the fiber of $P$ over $b$ is defined by $$ P_b:=\psi_\alpha|_b\cdot G. $$ This is well-defined, as all transition maps are in $G$.

We now describe a smooth structure for $P$. By construction, the set $P|_{U_\alpha}$ is isomorphic to the set $U_\alpha\times G$ (an isomorphism is given by the trivialization $\psi_\alpha$). Equip $P$ with the smooth structure that turns all those set isomorphisms into diffeomorphisms. By construction, there is a natural $G$ action on $P$, which turns $P$ into a principal $G$-bundle.

Let us think of vector bundles as examples for the above construction. A real vector bundle $E\to B$ of rank $k$ is a fiber bundle with fiber $\mathbb{R}^k$ and structure group $GL_k(\mathbb{R})$. The above procedure leads in this case to the frame bundle of $E$. If $E$ is equipped with metrics on the fibers, then it becomes a fiber bundle with structure group $O(k)$. Then, the procedure leads to the orthonormal frame bundle. Similarly, any additional structure on $E$ which can be described in terms of the structure group leads to a smaller subbundle of the frame bundle.

Edit: This is a continuation of the previous paragraph. Say we have a vector bundle $E\to B$, and we want to determine whether this is a trivial bundle or not. As explained above, the vector bundle $E$ is encoded by its frame bundle $P$, which is a principal $GL_k(\mathbb{R})$-bundle. We know that $P$ is trivial if and only if it admits a global section. This means that $E$ is trivial if and only if it admits a global frame. In a similar way, one can understand any fiber bundle by examining its associated principal bundle.

  • $\begingroup$ I'm afraid I don't see how this answer relates to my question. $\endgroup$ – tparker Oct 15 '17 at 23:16
  • $\begingroup$ @tparker In your post, you say that you understand when a principal bundle is trivial and when it is not. In my answer, I explain how every fiber bundle can be encoded by a principal bundle. Now, the original fiber bundle is trivial if and only if the associated principal bundle is. Furthermore, two fiber bundles with structure group $G$ are isomorphic to one another if and only if their associated principal $G$-bundles are. Altogether, this means that if you understand principal bundles (as you claim you do), you should be able to understand general fiber bundles. $\endgroup$ – Amitai Yuval Oct 16 '17 at 8:02
  • $\begingroup$ @tparker I even added a paragraph to my answer. I think the relation to your question should be clear now. $\endgroup$ – Amitai Yuval Oct 16 '17 at 8:27
  • $\begingroup$ @AmitaiYuval I think the issue is: why can we assume an arbitrary fiber bundle comes with a structure group? $\endgroup$ – cheyne Oct 16 '17 at 16:06
  • $\begingroup$ @cheyne One can always take the Lie subgroup of $\mathrm{diff}(F)$ generated by the transition maps with respect to an arbitrary collection of trivializations. In the "worst" case, this is all $\mathrm{diff}(F)$. $\endgroup$ – Amitai Yuval Oct 17 '17 at 5:31

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