Prove radial vector field is not tangent to $S^n$ rigorously

Question

Well, this seems to be a very trivial fact that the radial vector field $x^i\frac{\partial}{\partial x^i} $is an outward pointing vector field on $S^n$, but after some thinking, I’m not sure how can I rigorously prove this. The first thought comes into my mind is to check this is true for the north pole, and then using symmetric argument to conclude the result, but how can I claim this symmetry argument rigorously? The second thought is to prove this by using coordinate chart, but I’m not sure what should I calculate here. Any hints?

Answer 1

It's just the usual calculus argument. If you accept the fact that $S^n$ is defined by the equation $f(x)=\|x\|^2-1=0$, then $df_p(v)=0$ for any $v$ in the tangent space at $p\in S^n$. But $df_p(v) = 2\langle \sum x^i\frac{\partial}{\partial x^i},v\rangle$, so the given vector is in fact orthogonal to the tangent space. (This is the same thing as applying the vector field to $f$ and getting $0$.)

Now, what does outward-pointing mean? Has the assignment given an orientation on the sphere? Of course, we can all look and see it points radially outward, but you need to provide an orientation for the sphere.