TITLE: Is the tangent bundle of $S^2 \times S^1$ trivial or not?
QUESTION [8 upvotes]: As the question title suggests, is the tangent bundle of $S^2 \times S^1$ trivial or not?
Progress: I suspect yes. If I could construct three independent vector fields, I would be done. But I'm not so sure how to do that. Could anyone help?

REPLY [2 votes]: A more explicit answer: identify $S^2\times\mathbb{R}$ with $\mathbb{R}^3\setminus\{0\}$ by the diffeomorphism $\psi(x,t):=2^t x$ (here we think $x\in S^2\subset\mathbb{R}^3$) and identify $S^1$ with $\mathbb{R}/\mathbb{Z}$.
Calling $\delta:\mathbb{R}^3\setminus\{0\}\to\mathbb{R}^3\setminus\{0\}$ the dilation by a factor $2$, i.e. $\delta(x):=2x$, it suffices to find three independent vector fields $X_1,X_2,X_3$ on $\mathbb{R}^3\setminus\{0\}$ s.t.
$$d\delta((X_i)_x)=(X_i)_{\delta(x)}\quad\quad(*)$$
(for $i=1,2,3$) because then you can define three independent vector fields $Y_1,Y_2,Y_3$ on $S^2\times S^1$ with the formula
$$(Y_i)_{(x,[t])}:=d(\pi\circ\psi^{-1})((X_i)_{\psi(x,t)})$$
(here $\pi:S^2\times\mathbb{R}\to S^2\times S^1$ is the product of $\mathrm{id}_{S^2}$ by the standard projection $\mathbb{R}\to S^1$). Note that this is a good definition: calling $\tau:S^2\times\mathbb{R}$, $\tau(x,t):=(x,t+1)$, it suffices to check that
$$ d(\pi\circ\psi^{-1})((X_i)_{\psi(x,t)})=d(\pi\circ\psi^{-1})((X_i)_{\psi\circ\tau(x,t)}). $$
But $\psi\circ\tau=\delta\circ\psi$, so you can rewrite the RHS as
$$ d(\pi\circ\psi^{-1})((X_i)_{\psi\circ\tau(x,t)})
=d((\pi\circ\tau^{-1})\circ\psi^{-1})((X_i)_{\delta\circ\psi(x,t)}) \\
=d(\pi\circ(\psi\circ\tau)^{-1})\circ d\delta((X_i)_{\psi(x,t)})
=d(\pi\circ(\delta\circ\psi)^{-1}\circ\delta)((X_i)_{\psi(x,t)})
=LHS $$
thanks to $(*)$. Besides checking these computations, I strongly suggest that you draw a picture and understand what is going on.
Now we have to build the vector fields $X_1,X_2,X_3$ s.t. $(*)$ holds, but this is very easy:
choose $(X_i)_x:=r(x)\frac{\partial}{\partial x^i}$ for $i=1,2,3$ (here $r:\mathbb{R}^3\setminus\{0\}\to\mathbb{R}$ is the distance from the origin).