8
$\begingroup$

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.

$\endgroup$
3
+50
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.