Intereting Posts

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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($S^2 \times S^1$).

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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.

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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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