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If G is a simply connected Lie Group then why is every G-bundle over an orientable 3-manifold trivial? (Why is orientability important?)

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Think 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 $f$. –  Ryan Budney Aug 24 '10 at 17:40
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I 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? –  Jason DeVito Aug 24 '10 at 17:51
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So 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. –  Ryan Budney Aug 24 '10 at 19:55

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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.

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