Three linearly independent vector fields 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. 
 A: 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 $S^2\times S^1$.
A: 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.
