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I'm reading Thomas Friedrich's "Dirac Operators in Riemannian Geometry," where the following is stated (in the Remark on page 42 before section 2.2 begins, if anyone is following along with the reference at home):

"[L]et $P\rightarrow X=\mathbb{RP}^2$ be an $Spin(n)$-principal bundle. Since $$\dim(\mathbb{RP}^2)=2, \: \pi_1(Spin(n))=0, $$ this bundle has a section and is thus trivial."

He also remarks that, "The same effect occurs if we replace $\mathbb{RP}^2$ by an arbitrary 2-dimensional manifold."

Now, I am aware of the thread Every principal $G$-bundle over a surface is trivial if $G$ is compact and simply connected: reference? in which it is shown using quite a bit of machinery that

If $G$ is a compact and simply connected Lie group and $\Sigma$ is a compact orientable surface, then every principal $G$-bundle over $\Sigma$ is trivial.

(Aside: It seems from the answer given that one does not need the orientability hypothesis on $\Sigma$ which is relevant to the remark in Friedrich, as $\mathbb{RP}^2$ is not orientable.)

This is shown using classifying spaces and shows directly that the bundle is trivial. But Friedrich's phrasing seems to indicate that we could show the bundle to be trivial by constructing a section of the bundle, i.e. by lifting the identity map $X\rightarrow X$ to $X\rightarrow P\rightarrow X$. Is there some (restricted) lifting criterion for principal bundles that I am unaware of in this case that produces this section? Any insight or references are appreciated.

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Brian Klatt
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  • There's an obstruction theory that describes how to do this, and in this case it says that there are no obstructions. This is essentially the point of the "quite a bit of machinery" in the answer you linked to. – Qiaochu Yuan Dec 10 '15 at 19:50
  • So in your estimation producing a section requires machinery which is essentially equivalent to the machinery which shows that the bundle is trivial? I was hoping there was a slightly different perspective which doesn't go into classifying spaces and the like, which I feel uncomfortable with. But, I'm aware of your faculties in this area so I'm inclined to accept your judgement if you say so. – Brian Klatt Dec 10 '15 at 19:55
  • I think in low dimensions more geometric options are available, but I'm not familiar with them. The obstruction theory is fully general in the same way that covering space theory is fully general. You can see an example of it at work here: http://math.stackexchange.com/questions/1107682/elementary-proof-of-the-fact-that-any-orientable-3-manifold-is-parallelizable – Qiaochu Yuan Dec 10 '15 at 19:58
  • I was attempting to understand your response to that question just this week, and was mortified by how little I could cling to... thanks for the reference though, I have much to learn. – Brian Klatt Dec 10 '15 at 20:01
  • Can you quickly recommend a few references to learn about this obstruction theory and maybe the stuff about classifying spaces? I'm never sure where to turn when I run up against this material. – Brian Klatt Dec 10 '15 at 20:09
  • Me neither, unfortunately. – Qiaochu Yuan Dec 10 '15 at 20:38
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    One good source is a chapter in Davis and Kirk's algebraic topology book. –  Dec 11 '15 at 00:56

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You can phrase this in the language of obstruction theory if you want, but you could also avoid the language in this specific case.

You have a 2-dimensional CW complex $X$ and a principal $G$-bundle over it. You need to demonstrate the existence of a global section. For convenience of notation, I assume there's one 0-cell and one 2-cell, but this makes no difference.

First, pick an element in the fiber above each 0-cell. Now, pick some 1-cell. The $G$-bell pulls back to a trivial bundle over the 1-cell $[0,1]$ (more precisely, consider the characteristic map $[0,1] \to X$ and pull back the $G$-bundle along this map). The bundle is trivial over $[0,1]$. Then your choice of section over the 0-cells gives you an element above 0 and 1. Now, the assumption that the group is path-connected implies you can find a section above $[0,1]$ that extends the given section. Path-connectedness is essential even if you only have one 0-cell: there is a nontrivial $\Bbb Z/2$-bundle over $S^1$.

Now do this for every 1-cell. We now want to do the same thing for each 2-cell: for each, we have a map $D^2 \to X$ that we pull back the bundle (and section) along and trivialize the bundle over $D^2$. So now we have a map from $S^1$ to $G$ and want to extend across the 2-cell. This is where we use that $G$ is simply connected! Now extend across every 2-cell as before. We have constructed a section over the whole manifold.

Last note: because a Lie group has $\pi_2(G) = 0$, you can automatically push the above argument one dimension up; every simply connected $G$ has all $G$-bundles over a 3-dimensional complex trivial.

  • Exactly the kind of argument I was hoping for. I also appreciate you pointing out the subtlety of extending over the 1-skeleton even if there's only one 0-cell. It's a point I hadn't thought through clearly in the past. – Brian Klatt Dec 12 '15 at 19:32
  • @BrianKlatt: Sure, glad to. I should not that the non-simply connected case is not significantly harder to deal with; one needs to do some obstruction theory now, but I think you get that $G$-bundles over surfaces or 3-manifolds, where $G$ is path-connected, are classified by an invariant in $H^2(X;\pi_1(G))$. For $SO(3)$ this is $w_2$; for $U(2)$ this is $c_1$. (I fought through this computation last night, so figured it was probably worth stating explicitly in case someone else finds it useful...) –  Dec 12 '15 at 19:41
  • Do you have a go-to source for this material about obstruction theory? – Brian Klatt Dec 12 '15 at 20:00
  • @BrianKlatt: Just the one I mentioned in the comments above: Davis and Kirk's algebraic topology book. I think it's pretty basic, but it's all I know or have needed. –  Dec 12 '15 at 20:01