construct a smooth non-vanishing tangent vector field

Question

Fix $m\in \mathbb{N}$. Construct a smooth non-vanishing tangent vector field on $S^{2m-1}$

My attempt

Let $M\subseteq \mathbb{R}^k$ be a manifold. A smooth tangent vector field on $M$ is a smooth function $V:M \rightarrow \mathbb{R}^k$ such that $V(x) ≡ Vx\in TxM$ fir all $x\in M$.

Smooth non-vanishing means the derivatives won't vanish I am assuming

I don't really know how to do this...

Answer 1

It is very classical.

Consider $S^{2n-1}$ as the unit sphere in $\bf C^n$ with its standard hermitian norm ; it has equation $ 1= \sum _{i=1}^n \vert z_i\vert ^2$.

If $z=(z_1,...,z_n)\in S$, then $V(z)=iz$ is orthogonoal to $z$, and therefore tangent to $S$.

proof : the curve $t\to \exp it. z$ belongs to $S$ and its derivative at $t=0$ is $iz$.

The field $V(z)=iz$ is therefore a nowhere vanishing field on $S$.