# Why can Principal $G$ Bundles be Trivialized when $G = SU(N)$?

Reading about TQFT one usually comes about the fact that over 3-manifolds, Simply Connect Lie Group-bundles can be trivialized, yet it is a bit hard to find a clear answer online. Why is that the case?

• Here is a math stack exchange answer on the topic that effectively says the same, just goes a bit more in depth. Commented Nov 25, 2023 at 7:19

A Principal G-Bundle $$\pi: P\rightarrow M$$ is said to be trivial if it is isomorphic to $$M \times G$$, which means that a global section exists.
In the case of a simply connected Lie Group G, its fundamental group and second homotopy group are trivial, and unless they it is trivial, $$\pi_3(G) = \Bbb Z$$.
Specifically we care about the classes $$o_j \in H^{j+1}(M,\pi_j(G))$$. If o is 0 for all j's then we have a global section. Since we care about 3 manifolds, only j=0,1,2 are of any concern to us, but since all the corresponding homotopy groups are trivial, we see that indeed all the $$o_j$$ are 0, and thus we have a trivial bundle.