
For each $t \in \R$, let $\phi_t$ denote the map of $\S^2$ into itself which is
defined by
$$
	\phi_t: \S^2 \to \S^2: \colvecthree{x}{y}{z} \mapsto
	\colvecthree{\cos(t)x - \sin(t) y}{\sin(t)x + \cos(t) y}{z}.
$$
Show that $\phi_t$ is a one-parameter group of diffeomorphisms, and compute and
graph the vector field $\vecv$ on $\S^2$ for which $\phi_t$ is the corresponding
flow.

Remark:  In defining the vector field $\vecv$, please specify how you are
viewing the tangent bundle of $\S^2$.

\emph{Solution.}  To show that $\phi_t$ is a one-parameter group of
diffeomorphisms, we need to show:
\begin{itemize}
\item Each $\phi_t$ is bijective on $\S^2$.
\item Each $\phi_t$ is smooth with smooth inverse.
\item For each $s$ and $t$ we have $\phi_s \circ \phi_t = \phi_{s+t}$.
\end{itemize}
First note that the map $\phi_t$ may be written in matrix form as
$$
	\colvecthree{x}{y}{z} \stackrel{\phi_t}{\to}
	\mxrsqthree
		{\cos(t)}{-\sin(t)}{0}
		{\sin(t)}{ \cos(t)}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z}.
$$
The matrix has determinant $\cos^2(t) + \sin^2(t)=1 \ne 0$ so it is invertible,
hence bijective.  In particular its inverse is
$$
	\colvecthree{x}{y}{z} \stackrel{\phi_t^{-1}}{\to}
	\mxrsqthree
		{\cos(t)}{\sin(t)}{0}
		{-\sin(t)}{\cos(t)}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z}.
$$

This much shows that $\phi_t$ is bijective on $\R^3$.  To show bijectivity on
$\S^2$, we need to show that $\phi_t(\vecq) \in \S^2$ for all $q \in \S^2$, and
likewise for $\phi_t^{-1}$.  Now, we've already shown that $\phi_t^{-1}$ is
rotation by $-t$ (since $\cos(-t)=\cos(t)$ and $\sin(-t)=\sin(t)$) so it
suffices to show this for $\phi_t$ only.  Namely, the coordinates of $(u, v, w)
= \phi_t(x,y,z)$ should satisfy $u^2+v^2+w^2=1$.  Check:
\begin{eqnarray*}
	u^2 + v^2 + w^2 &=& 
		(\cos(t)x - \sin(t)y)^2 + (\sin(t)x + \cos(t)y)^2 + z^2 \\
&&\\
	&=& \cos^2(t) x - \cos(t)\sin(t)xy + \sin^2(t)y^2 \\
	&+& \sin^2(t)x^2 + \cos(t)\sin(t)xy + \cos^2(t)y^2 \\
	&+& z^2 \\
&&\\
	&=& x^2 + y^2 + z^2 = 1.
\end{eqnarray*}

The transformation is linear, hence it is its own derivative and thus eminently
differentiable; likewise for the inverse.  (Remember we're checking smoothness
for fixed $t$.  Smoothness with respect to $t$ is a different question.) For
the composition property, we compute (using the sum formulas for sine and
cosine)
\begin{eqnarray*}
	\phi_{s+t}
	\colvecthree{x}{y}{z}
	%
	&=&
	\mxrsqthree
		{\cos(s+t)}{-\sin(s+t)}{0}
		{\sin(s+t)}{ \cos(s+t)}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z} \\
	%
	&=&
	\mxrsqthree
		{\cos(s)\cos(t)-\sin(s)\sin(t),}{-\sin(s)\cos(t)-\cos(s)\sin(t),}{0}
		{\sin(s)\cos(t)+\cos(s)\sin(t),}{ \cos(s)\cos(t)-\sin(s)\sin(t),}{0}
		{0,}{0,}{1}
	\colvecthree{x}{y}{z}.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
	\phi_s\circ\phi_t
	\colvecthree{x}{y}{z}
	&=&
	\mxrsqthree
		{\cos(s)}{-\sin(s)}{0}
		{\sin(s)}{ \cos(s)}{0}
		{0}{0}{1}
	\mxrsqthree
		{\cos(t)}{-\sin(t)}{0}
		{\sin(t)}{ \cos(t)}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z} \\
	&=&
	\mxrsqthree
		{\cos(s) \cos(t)-\sin(s)\sin(t),}{-\cos(s)\sin(t)-\sin(s)\cos(t),}{0}
		{\sin(s) \cos(t)+\cos(s)\sin(t),}{-\sin(s)\sin(t)+\cos(s)\cos(t),}{0}
		{0}{0}{1}
	\colvecthree{x}{y}{z}
\end{eqnarray*}
which is the same as desired.  This was the last item needed to prove
that the $\phi_t$'s form a one-parameter group of diffeomorphisms.

Recalling the necessary formula for the corresponding vector field, we have
\begin{eqnarray*}
	\vecv\pickpipe{\colvecthree{x}{y}{z}}
	&=& \frac{\D \phi_t}{\D t}\pickpipe{t=0}\colvecthree{x}{y}{z} \\
	&=& \frac{\D }{\D t}\pickpipe{t=0}
		\colvecthree{\cos(t)x - \sin(t) y}{\sin(t)x + \cos(t) y}{z} \\
	&=& \colvecthree{-\sin(t)x - \cos(t) y}{\cos(t)x - \sin(t) y}{0}
		\pickpipe{t=0} \\
	&=& \colvecthree{-y}{x}{0}\pickpipe{\colvecthree{x}{y}{z}} \\
	&=& -y\,\slashDx + x\,\slashDy.
\end{eqnarray*}
This is of course rotation about the $z$-axis.  [xxx insert figure here.] Here,
we are viewing the tangent bundle as directional derivatives.

This is a vector field on $\R^3$.  [xxx to do:  convert to graph coordinates, or spherical.]

\begin{center}* * *\end{center}

\textbf{Variation:}  Let $\omega = x\,dy\wedge dz +
y\,dz\wedge dx + z\, dx\wedge dy$.  Compute $\phi_t^*(\omega)$.

\emph{Solution.}  Let $c=\cos(t)$ and $s=\sin(t)$.  Then, replacing $x$ with
$cx-sy$ and $y$ with $sx+cy$, we have
\begin{eqnarray*}
	\phi_t^*(\omega)
		&=& (cx-sy)\, d(sx+cy)\wedge dz \\
		&+& (sx+cy)\, dz\wedge d(cx-sy) \\
		&+& z\, d(cx-sy)\wedge d(sx+cy) \\
	&&\\
		&=& (cx-sy)\, (sdx\wedge dz +c dy\wedge dz) \\
		&+& (sx+cy)\, (c dz\wedge dx - s dz \wedge dy) \\
		&+& z\, (cdx-sdy)\wedge (sdx+cdy) \\
	&&\\
		&=& (cx-sy)\, (-sdz\wedge dx +c dy\wedge dz) \\
		&+& (sx+cy)\, (c dz\wedge dx + s dy \wedge dz) \\
		&+& z\,(c^2+s^2) dx \wedge dy \\
	&&\\
		&=& -cs\,x\, dz\wedge dx
		+c^2\,x\, dy \wedge dz
		+s^2\,y\, dz \wedge dx
		-cs\,y\, dy\wedge dz \\
	&+& cs\,x\, dz\wedge dx
		+s^2\,x\, dy \wedge dz
		+c^2\,y\, dz \wedge dx
		+cs\,y\, dy\wedge dz \\
		&+& z\, dx \wedge dy \\
	&&\\
		&=& x\, dy \wedge dz +y\, dz \wedge dx + z\, dx \wedge dy \\
	&=& \omega.
\end{eqnarray*}
Note that this is a bit messy, only because I chose a form with three terms in
it.

\textbf{Variation:}  Show that the two-form $\omega =
x\,dy\wedge dz + y\,dz\wedge dx + z\, dx\wedge dy$ is invariant under the flow
of $\phi_t$.

\emph{Solution.}  One option is to do what we just did.  Another option is to
compute the Lie derivative [xxx for prolrev, xref forward or reorg] of $\omega$
in the direction of $\vecv$ and [xxx].  Since
$$
	\vecv = -y\, \slashDx + x\, \slashDy,
$$
we have
$$
	L_\vecv(x) = -y, \quad
	L_\vecv(y) =  x, \quad
	L_\vecv(z) =  0.
$$
Recall that $L_\vecv$ is a derivation, i.e. it follows the product rule through
wedges.  Also, it commutes with $d$ for exact forms.  For the first term of
$\omega$, namely $x\, dy\wedge dz$, we then have
\begin{eqnarray*}
	L_\vecv(x\, dy \wedge dz)
		&=& L_\vecv(x) dy \wedge dz
		\;+\; x \, d(L_\vecv(y)) \wedge dz
		\;+\; x \, dy \wedge d(L_\vecv(z)) \\
	%
		&=& -y\, dy \wedge dz
		\;+\; x\, dx \wedge dz
		\;+\; x \, dy \wedge 0 \\
	%
		&=& -y\, dy \wedge dz \;-\; x\, dz \wedge dx.
\end{eqnarray*}
For the second term:
\begin{eqnarray*}
	L_\vecv(y\, dz \wedge dx)
		&=& L_\vecv(y) dz \wedge dx
		\;+\; y \, d(L_\vecv(z)) \wedge dx
		\;+\; y \, dz \wedge d(L_\vecv(x)) \\
	%
		&=& x\, dz \wedge dx
		\;+\; y \, 0 \wedge dx
		\;-\; y \, dz \wedge dy \\
	%
		&=& x\, dz \wedge dx \;+\; y \, dy \wedge dz.
\end{eqnarray*}
For the third term:
\begin{eqnarray*}
	L_\vecv(z\, dx \wedge dy)
		&=& L_\vecv(z) dx \wedge dy
		\;+\; z \, d(L_\vecv(x)) \wedge dy
		\;+\; z \, dx \wedge d(L_\vecv(y)) \\
	%
		&=& 0\, dx \wedge dy
		\;-\; z \, dy \wedge dy
		\;+\; z \, dx \wedge dx \\
	%
		&=& 0.
\end{eqnarray*}
Combining these, we have
\begin{eqnarray*}
	L_\vecv(\omega)
	&=& -y\, dy \wedge dz \;-\; x\, dz \wedge dx \\
	&+& x\, dz \wedge dx \;+\; y \, dy \wedge dz \\
	&=& 0.
\end{eqnarray*}
which shows that $\omega$ is invariant under the flow of $\phi_t$.

[xxx do it again using the magic formula.  Much nicer.]
