Finding the smooth inverse of a function

Question

Problem: The mapping $\phi: S^2 \longrightarrow S^2 $ by $$\phi(x,y,z)=(x\cos z+y\sin z,x\sin z-y \cos z,z)$$ is a diffeomorphism. Where $S^2$ is a unit sphere in $\mathbb{R}^3$.
I've already shown that $\phi$ is smooth and bijective. The only thing I can't find is $\phi^{-1}$ that is smooth.

Any help would be much appreciated!

Answer 1

To find $\phi^{-1}$ explicitly, let's define

$$ \begin{align*} a &= x \cos z + y \sin z \\ b &= x \sin z - y \cos z \\ c &= z \end{align*} $$

so that $\phi(x,y,z) = (a,b,c)$ and $\phi^{-1}(a,b,c) = (x,y,z)$. It just remains to write $x,y,z$ in terms of $a,b,c$.

Getting rid of $z$ is obvious:

$$ \begin{align*} a &= x \cos c + y \sin c \\ b &= x \sin c - y \cos c \end{align*} $$

To eliminate $x$, multiply the equations so both include the term $(x \cos c \sin c)$, then subtract:

$$ \begin{align*} a \sin c &= x \cos c \sin c + y \sin^2 c \\ b \cos c &= x \cos c \sin c - y \cos^2 c \\ a \sin c - b \cos c &= y (\sin^2 c + \cos^2 c) = y \end{align*}$$

Similarly eliminating $y$,

$$ \begin{align*} a \cos c &= x \cos^2 c + y \sin c \cos c \\ b \sin c &= x \sin^2 c - y \sin c \cos c \\ a \cos c + b \cos c &= x(\cos^2 c + \sin^2 c) = x \end{align*} $$

So finally,

$$ \phi^{-1}(a,b,c) = (a \cos c + b \sin c, a \sin c - b \cos c, c) $$

Look familiar? It happens that $\phi^{-1} = \phi$. This is because at each fixed value of $z=c$, the relationship between $(x,y)$ and $(a,b)$ describes a reflection of the $\mathbb{R}^2$ plane.

Answer 2

You can obtain the inverse $(x,y,z) = \phi^{-1}(X,Y,Z)$ as follows: $$x\cos z + y\sin z = X$$ $$-y\cos z+x\sin z = Y$$ $$ z = Z$$ The first two can be read as the complex multiplication $$ (x+iy)(\sin z+i\cos z) = Y+iX$$ which quickly (upon division) leads to $$ x=Y\sin Z +X\cos Z$$ $$ y=X\sin Z - Y\cos Z$$ Smoothness of $\phi^{-1}$ is then apparent.

Answer 3

Let $a=(x,y,z)\in S^2$. If we know that the derivative $d\phi_a: T_aS^2\to T_{\phi(a)}S^2$ is nonsingular, then we know that $\phi$ has a local inverse, defined and smooth in a neighborhood of $\phi(a)$.

Now, thinking for the moment of $\phi$ as a map on $\Bbb R^3$, we have $$d\phi_a = \begin{bmatrix} \cos z & \sin z & -x\sin z + y\cos z \\ \sin z & -\cos z & x\cos z+\sin z \\ 0 & 0 & 1\end{bmatrix}.$$ The determinant of $d\phi_a$ is $-1$ and hence $d\phi_a$ is invertible as a map $T_a \Bbb R^3\to T_{\phi(a)}\Bbb R^3$. It follows that the restriction of $d\phi_a$ to a map $T_a S^2\to T_{\phi(a)}S^2$ must be nonsingular (why?).

Therefore, it follows that $\phi$ has a local smooth inverse mapping a neighborhood of $\phi(a)\in S^2$ to $a\in S^2$. [If you haven't seen this application before, you can deduce it by using charts or parametrizations at $a$ and $\phi(a)$ and reducing to a question about the mapping of an open set in $\Bbb R^2$ to an open set in $\Bbb R^2$.]