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?

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

1 Answer 1


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, $\pi_3(G) = \Bbb Z$.

Now one has to use obstruction theory to see whether a global section exists.

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.

With the same argument, we can show that for 4 manifolds, we always have a non-trivial G-Bundle.


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