Three linearly independent vector fields

Question

How can one find three linearly independent vector fields on $S^1\times S^2$? I know that $S^1\times S^2 \cong SO_3( \mathbb{R})$, i.e. the set of orthogonal $3 \times 3$ matrices with determinant $1$, which is a Lie group and thus parallelizable. I am, however, interested in an explicit form for three vector fields.

Answer 1

Let $Z$ be a nontrivial vector field on $S^1$ and $X, Y$ linearly independent vector fields on $S^2$. The three vector fields $(Z, 0), (0, X), (0, Y)$ on $S^1 \times S^2$ are linearly independent. This means: in the tangent space at any $(p, q)\in S^1 \times S^2$, the vectors $(Z_p,0), (0, X_p), (0, Y_p)$ are linearly independent.

It's more difficult to give formulas for three vector fields whose values at each point are linearly independent. But the Cartesian product of any finite number of spheres of various dimensions is parallelizable provided one of the spheres is odd-dimensional.

Answer 2

Draw a picture of $S^2 \times [0,1]$ as a thickened sphere in $\mathbb{R}^3$ where we think of the outside sphere as $S^2\times \{1\}$ and the inside as $S^2\times \{0\}$. We will view $S^2\times S^1$ as a quotient of this space where we identify $(p,0)$ with $(p,1)$.

Now, at each point, just draw the usual $x,y,z$ arrows. This defines a nonvanishing smooth vector field on $S^2\times [0,1]$. The identification made on the boundary maps the usual $x,y,z$ arrows to the usual $x,y,z$ arrows, so this descends to a non-vanishing vector field on $S^2\times S^1$.