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Dec 2, 2019 at 16:58 vote accept user101010
Nov 4, 2019 at 18:16 answer added mme timeline score: 2
Oct 30, 2019 at 17:54 comment added Igor Belegradek The cover $S^3\to SO(3)$ is principal bundle with fiber $\mathbb Z_2$, so it is classified by maps into $B\mathbb Z_2=RP^\infty$.
Oct 30, 2019 at 17:27 history edited user101010 CC BY-SA 4.0
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Oct 30, 2019 at 17:27 comment added user101010 @IgorBelegradek Oh yes - orientable - thanks. What is the classifying map for 2-fold covers?
Oct 30, 2019 at 16:05 comment added Igor Belegradek For $TM$ to be trivial $M$ must be orientable. The classifying map for the 2-fold covering $S^3\to SO(3)$ induces an exact sequence of pointed sets $[M, S^3]\to [M, SO(3)]\to [M, RP^\infty]$. If $M$ is closed oriented, the degree identifies $[M, S^3]$ and $H^3(M)$, and also $[M, RP^\infty]=H^1(M;\mathbb Z_2)$ which are finitely generated abelian groups. It remains to see if the maps in the exact sequence are group homomorphisms.
Oct 30, 2019 at 11:00 comment added YCor The last sentence would better fit as a separate question.
Oct 30, 2019 at 10:57 history edited YCor
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Oct 30, 2019 at 9:49 history asked user101010 CC BY-SA 4.0