If G is a simply connected Lie Group then why is every G-bundle over an orientable 3-manifold trivial? (Why is orientability important?)
Principal bundles on 3-manifolds
8
$\begingroup$
differential-geometry
-
2Think about what the clutching maps are like. This is technically simplest in the classifying space formalism for bundles, your bundle is the pull-back of its classifying map $f:M\to BG$. Your assumptions are saying that $BG$ has trivial $\pi_1$ and $\pi_2$, but all Lie groups have trivial $\pi_2$ so $BG$ has trivial $\pi_3$, so there's the standard obstruction-theory construction of a null-homotopy of $f$. – 2010-08-24
-
2I wanted to add that orientability of a 3-manifold is enough to guarantee that the tangent bundle is trivial. Maybe that's why you're thinking you need orientability? – 2010-08-24
-
2So this argument shows that a principle $G$-bundle over a space $X$ is trivial provided $X$ is a CW-complex of dimension at most $3$, and $G$ is simply-connected. – 2010-08-24
1 Answers
2
Are you sure that orientability is necessary? The result is proved in lemma 4.1.1. here and I do not see where orientability is used.