Let $G$ be a connected Lie group, and $P \rightarrow M$ a principal Gbundle. Is $P$ trivial over the 1skeleton of $M$. If yes, what is a reference? More generally, over the $n$skeleton, $G$bundles are classified by $\pi_{n1}(G)$. How does one see this, or is there a good reference?

2$\begingroup$ Did you look in Steenrod's "The topology of fibre bundles"? $\endgroup$– José FigueroaO'FarrillMay 18, 2011 at 14:52

3$\begingroup$ The statement about the classification of $G$ bundles by $\pi_{n1}(G)$ is wrong... $\endgroup$– Thomas NikolausMay 18, 2011 at 17:02

1$\begingroup$ You need two facts: 1skeleton is homotopy equivalent to a wedge of circles, and principal $G$bundles over $S^n$ are classified by $\pi_{n−1}(G)$. Since $G$ is connected, $π_0(G)$ is trivial so any pricipal $G$bundle over a circle is trivial, and hence the same holds for 1skeleton. This argument fails for nskeleton because it is generally not homotopy equivalent to the wedge of nspheres: consider e.g. 2skeleton of the 2torus. $\endgroup$– Igor BelegradekMay 19, 2011 at 2:03
2 Answers
This can be seen using obstruction theory. Therefore consider the Whitehead tower of BG:
$$ EG \to ... \to BG\langle3\rangle\to BG\langle2\rangle \to BG\langle1\rangle \to BG $$
Now a $G$ bundle over $M$ is classified by a map $f: M \to BG$. The bundle is trivial iff we can lift $f$ to a map $M \to EG$. Since the fibres in the Whitehead tower are EilenbergMacLane spaces we can write down the obstruction to lift $f$ successively through the $BG\langle n\rangle$'s:
Obstruction against lift to $BG\langle 1\rangle$ lies in $H^1(M,\pi_0(G))$
Obstruction against lift to $BG\langle2\rangle$ then lies in $H^2(M,\pi_1(G))$
Obstruction against lift to $BG\langle3\rangle$ then lies in $H^3(M,\pi_2(G))$, etc...
In your case $\pi_0(G) = 0$ and therefore the first obstruction vanishes. On the 1skeleton all the higher obstructions vanish for dimensional reasons, thus all bundles are trivial.
If you moreover assume that $G$ is simply connected and finite dimensional (e.g. SU$(n)$, Spin$(n)$,...) then also $\pi_2(G) =0$. Hence there are no nontrivial $G$bundles over manifolds or cell complexes of dimension $\leq 3$.

$\begingroup$ Oh, I was very confused by the second sentence of your last paragraph. What you mean is that all highenough obstructions vanish. Certainly $H^2(M,\pi_1(G))$ does not vanish in general. $\endgroup$ May 18, 2011 at 17:08

$\begingroup$ I hope it is less confusing now... $\endgroup$ May 18, 2011 at 17:23
Hi,
I don't know your background knowledge, so I'll try to post an elementary explanation to your question, which is also the way I understand it. I am not 100% reliable, and there might be more or less serious mistakes in what follows, but I think the main ideas are there.
You know that a principal $G$ bundle is trivial $\mathrm{iff}$ it admits a section. So suppose you have a (locally finite) cell complex $X$, put $X_n$ the $n$squeleton of $X$ and let $P$ be a principal $G$bundle over $X$. We'll also call $P_n$ the restriction of $P$ to $X_n$, which is still a princial $G$ bundle. You want to know why $P_1$ is trivial, and how (non) triviality of the $P_n$ is related to the fundamental groups of $G$.
Well, what is $P_0$? It is nothing more than a discrete collection of (homeomorphic) copies of $G$, one above each vertex of $X_0$. Now, for each vertex $x\in X_0$, chose a point $g_x\in G$. This gives you a trivialisation of $P_0$ (the (continuous) map $X_0\rightarrow P_0,~x\mapsto (x,g_x)$), and $P_0$ is trivial.
Similarly what is $P_1$? It is a graph with, along each line, tightly packed (homeomorphic) copies of $G$. To show it is trivial we need only construct a section of $P_1$. Let's start with the section of $P_0$ we already constructed. This was nothing more than a collection of pretty random points in the fibres over the vertives of $X$. What would it mean to extend this (very partial) section of $P_1$ (only defined on the zero squeleton $X_0$) to the whole of $X_1$? It means constructing, for each edge $e$ with endpoints $x$ and $y$ a continuous map from $e$ (which is basically $[1,+1]$) to $P_1$ that takes on the values $g_x$ at $x$ and $g_y$ at $y$ (that is with values imposed on $\lbrace 1,+1\rbrace=\partial [1,+1]$. This is, a priori, possible $\mathrm{iff}$ $g_x$ and $g_y$ lie in the same path component of $G$, but such will always be the case [EDIT: actually, it's not, consider the $\lbrace 1,+1\rbrace$ principal fibre bundle over the circle which is trivial on the circle minus $1$ and on the circle minus $+1$, and has a cocycle that is $1$ on the left and $+1$ on the right (this is also $\mathbb{S}^1\rightarrow\mathbb{S}^1,~z\mapsto z^2$): it is not trivial].
You can, for instance cut the edge $e$ into many little closed intervals that each trivialize the (induced) bundle [Edit: and only overlap at their extremities], and glue sections together, imposing the first and final value of the section, and working your way from $1$ to $+1$. [EDIT: What may happen now is that the final two values you have to link within $G$ may not lie in the same path component, and so there is a dependence on $\pi_0 (G)$] Now we have a map from the $1$squeleton $X_1$ to $P_1$ that is a section of the bundle, and is continuous along each edge, and prolongs the first section we constructed earlier on. But because $X$ is locally finite, this map is continuous. (Actually I'm pretty sure that condition is unnecessary, because you can impose your sections to be 'locally constant around the vertices')
So $P_1$ is trivial,always, regardless of any restrction. This technique can be carried on. Suppose you have shown $P_n$ to be trivial, let $s_n$ be a global section. Can you extend $s_n$ to a global section $s_{n+1}$ of $P_{n+1}$? Consider a $(n+1)$face $F$ and its border $\partial F$. The desired section is already defined along the border $\partial F$ (it's the restriction of $s_n$ to $\partial F$), so you have what is essentially a (continuous) map $c_{\partial F}$ from $\mathbb{S}^n$ (the border $\partial F$) to $G$, and you want to extend it to a map from the euclidean $(n+1)$unit ball $\mathbb{B}^{n+1}$ (which would be $F$) to $G$. This problem is equivalent to nulhomotopy of the map $c_{\partial F}$, and you thus find that $P_{n+1}$ is trivial $\mathrm{iff}$ all such maps are nullhomotopic. In particular, if $\pi_n (G)$ is trivial, the above problem is always solvable.
You might want to have a look at Homotopy groups of Lie groups if you are unfamiliar with the fact that the second fundamental group of a Lie group is trivial. This fact implies, as Thomas Nikolaus has already pointed out, that a principal $G$ bundle over a surface or $3$ manifold with simply connected $G$ is always trivial. Another good reference is Milnor and Stasheff's excellent book "Characteristic Classes", where everything I list is explained in better and shorter form.