Nonvanishing vector field on an odd sphere

Question

This is an exercise I am somewhat confused about. Here $X$ looks like a vector field on $\mathbb{R}^{2n}$, not $S^{2n-1}$. Then how should I interpret $X$ to make it a vector field on the sphere? Could anyone please explain?

Answer 1

For every $p\in S^{2n-1}$, $X(p)\in T_p S^{2n-1}$ (you have to prove this), so $X_{\vert S^{2n-1}}$ is a vector field on the sphere. What is meant by "$X$ is a vector field on $S^{2n-1}$" is "$X$ is a vector field on $S^{2n-1}$ when it is restricted to the $S^{2n-1}$".

Answer 2

The sphere $S^n$ is an imbedded submanifold of $R^{n+1}$ under the inclusion map $i$. Hence, at every point $p\in S^n$, the differential $di$ establishes a one-to-one correspondence between the tangent space $T_pS^n$ and a subspace of $T_p\mathbb R^{n+1}\equiv \mathbb R^{n+1}$. Suppose that $\gamma(t)=(x_1(t),...,x_{n+1}(t))$ is an arbitrary curve passing through $\gamma(0)$ in $S^n$, so that $\gamma’(0)$ is an arbitrary tangent vector to $S^n$, and that $x_1^2(t)+...+x_{n+1}^2(t)=1$ implies that the dot product $(x_1’(0),...,x_{n+1}’(0)).(x_1(0),...,x_{n+1}(0))=0$. On the other hand, if $y=(y_1,...,y_{n+1})$ is an arbitrary element in $\mathbb R^{n+1}$ whose dot product with a point in the sphere is $0$, then $y=di(\gamma’(0))$ for some $\gamma(t)$ in the sphere.

Using the above considerations, you easily see why the restriction of your vector field to the sphere is a vector field on $S^n$.