Is the tangent bundle of $S^2 \times S^1$ trivial or not?

Question

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?

Answer 1

It's true, more generally, that any product of spheres has trivial tangent bundle, as long as (at least) one of them has odd dimension. To prove this requires some yoga with vector bundles.

However, yes, you can explicitly write down three linearly independent vector fields on $S^2\times S^1$. Here's a hint to get you started: For any vector $v\in\Bbb R^3$, consider its projection onto $T_pS^2$. Can you figure out how to assign a vector in $T_q S^1$?

Answer 2

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).

Answer 3

Just a note on terminology. A manifold with trivial tangent bundle is called parallelizable, so your question can be rephrased as "Is $S^2\times S^1$ parallelizable?"

As has already been pointed out, the answer is yes. In fact, every closed orientable three-dimensional manifold is parallelizable. If you are trying to explicitly write down three linearly independent vector fields on such a manifold (e.g. $S^2\times S^1$), this fact (and its proof) will not help.

Another general result about parallelizable manifolds which includes $S^2\times S^1$ is the one mentioned in Ted Shifrin's answer: a (non-trivial) product of spheres is parallelizable if and only if at least one of the spheres is odd-dimensional. Again, the proof of this fact will not help you to construct any vector fields. However, there is a paper by Parton, Explicit parallelizations on products of spheres, which does exactly this. In particular, section $2$ deals with the case $S^m\times S^1$.