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How would I find out if T($S^2 \times S^1$) is trivial or not?

Using the hairy ball theorem I can show that T($S^2$) is not trivial, and it is straight forward to show that T($S^1$) is trivial.

But I have no idea about T($S^2 \times S^1$).

I know that T($S^2 \times S^1$)$\cong$ T($S^2)\times$T($S^1$). I tried to say something about restricting a global frame to $S^2$ and deriving a contradiction, but I couldn't get anywhere.

Bates
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  • Think normal bundles. –  Aug 24 '14 at 18:33
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    I don't see how this helps. Sorry, I am really having a lot of trouble understanding this. Can you give me more. – Bates Aug 24 '14 at 18:36
  • Do you agree that the sum of the tangent bundle on $S^2$ and the normal bundle on $S^2$ is trivial? If so, find a way to get the normal bundle in there from what you've got! If not... convince yourself. :) –  Aug 24 '14 at 18:37
  • Even if I believe that the sum is trivial, (I'm not convinced of that yet) how does this help me solve the problem? Is there some way I can relate the tangent bundle of the product to the sum of the bundles on $S^2$? – Bates Aug 24 '14 at 19:00
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    Even more: all oriented 3-manifolds are parallelizable, see http://math.stackexchange.com/questions/46297/which-manifolds-are-parallelizable. – Moishe Kohan Aug 24 '14 at 19:41
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    @Georges: Mike is referring to the normal bundle of $S^2$ in $\Bbb R^3$. The big hint is $T(S^1)$ is trivial so $T(S^2\times S^1)\cong T(S^2)\oplus\epsilon^1$, where $\epsilon^1$ is the trivial line bundle on $S^2$. ... But I will comment that it's not difficult to write down explicitly three everywhere linearly independent vector fields on $S^2\times S^1$, once again using the standard embedding of $S^2\subset\Bbb R^3$. – Ted Shifrin Aug 24 '14 at 20:04
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    Dear @Ted: yes, I know the idea but you are equating a bundle on $S^2\times S^1$ with a bundle on $S^2$ in your isomorphism. That abuse of language may be confusing for a beginner. I encourage you to write a complete answer, with the suitable pull-backs explicitly displayed, and I will gladly upvote you (and Bates would probably accept your answer...) . – Georges Elencwajg Aug 24 '14 at 20:21
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    I think it would help me a lot if someone wrote a complete answer. – Bates Aug 24 '14 at 20:21
  • Sorry for confusing others with my comments. –  Aug 24 '14 at 20:35
  • @TedShifrin: Thanks for the enlightenment. – Bombyx mori Aug 24 '14 at 20:41

1 Answers1

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Let $p_1, p_2$ be the respective projection onto $S^1$ and $S^2$. Canonically we have $T(S^1 \times S^2) \cong p_1^{-1}T(S^1) \oplus p_2^{-1}T(S^2)$. As the tangent bundle of the circle is trivial, the former is isomorphic the trivial line bundle on $S^1 \times S^2$. Let $\xi$ be the trivial line bundle on $S^2$. Then this is again isomorphic to $p_2^{-1}T(S^2) \oplus p_2^{-1}(\xi) \cong p_2^{-1}(T(S^2) \oplus \xi)$. Consider $S^2$ as embedded in $\Bbb R^3$ in the standard manner; then it has trivial normal bundle (because there is a nonvanishing section of it), and $T(S^2) \oplus N(S^2) \cong S^2 \times \Bbb R^3$. So $$p_2^{-1}(T(S^2) \oplus \xi) \cong p_2^{-1}(T(S^2) \oplus N(S^2)) \cong p_2^{-1}(S^2 \times \Bbb R^3) \cong S^2 \times S^1 \times \Bbb R^3.$$ So the tangent bundle of $S^2 \times S^1$ is trivial.

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    A similar argument (you have to futz to get a copy of the trivial line bundle) works to show that $S^n \times S^k$ has trivial tangent bundle for $n$ odd. –  Aug 24 '14 at 20:45
  • +1 for the answer and the interesting comment (in which I made acquaintance with the verb "futz"). Did you learn about these parallelizable products of spheres in Hirsch ? I did. – Georges Elencwajg Aug 24 '14 at 20:50
  • @GeorgesElencwajg It's an exercise in Bredon's "Topology and Geometry" - before the author has introduced Whitney sums! (It's marked with a symbol that means 'come back later'.) –  Aug 24 '14 at 21:00
  • Thanks for the explanation, Mike. – Georges Elencwajg Aug 24 '14 at 21:17
  • @GeorgesElencwajg And thank you for your kind comments. –  Aug 24 '14 at 21:26