Rotating about the z-axis defines a smooth function on $S^{2}$

Question

I want to show that rotating about the z-axis defines a smooth function on $S^{2}$. To do this I used the function:

$f(x,y,z)=(x\cos(\theta)-y\sin(\theta),x\sin(\theta)+y\cos(\theta),z)$

where $\theta$ is the rotation angle. This function is smooth. Another portion of the question asks for calculations of the pushforward at different points but they all seem to have the same value so I lost confidence in my answer. Have I made a mistake? Is this the function I should be considering? I would be grateful for any help. Thank you in advance.

By pushforward I mean the derivative map between tagent spaces. I believe in John M. Lee's book it is denoted by $F_{*}$ and called the pushforward. — user99163, Oct 06 '13 at 19:23

score 0 · Answer 1 · answered Oct 08 '13 at 00:54

The function is correct.

the pushforward at different points but they all seem to have the same value

This can't possibly happen. The induced map $F_*$ at point $p$ (which I would call $D_pf$) goes from $T_p S^2$ to $T_{f(p)}S^2$. For different points $p$, these maps have disjoint domains, so they cannot be equal. They have disjoint ranges, too.

Perhaps what you mean is that by writing the linear map $F_*$ in spherical coordinates, you obtain the same matrix representation for it. That is perfectly fine, and to be expected. After all, $f$ plays remarkably well with the coordinates: one of them is fixed, and the other is increased by $\theta$.

Rotating about the z-axis defines a smooth function on $S^{2}$

1 Answers1