If $D$ is the unit circle in $\R^2$, then $\theta(z,u)=-(z^3u^{-2},z^2u^{-1})$, where $M=S^1$ is considered as a subset of the complex plane $\C$. Prove that the Haar measure on $\TT^2$ is $\theta$ invariant and that $\theta$ is not ergodic.
Since the outer normal equals the position $z=e^{it}$ we get for $u=e^{is}$:
$$
v
=u-2\Re(N\bar u)N
=u-(\bar zu+z\bar u)z
=-z^2\bar u
=-e^{i(2t-s)}
$$
and the next point of reflection $y=ze^{-i(\pi-2(t-s))}=-e^{i(3t-2s)}$, i.e.
$(y,v)=-(z^3u^{-2},z^2u^{-1})$.
2. The map $(z,u)=-(z^3u^{-2},z^2u^{-1})$ on $S^1\times S^1$ is on $\TT^2$ the composition of the homomorphism with matrix
$$
A\colon=\left(\begin{array}{cc}
3&-2\\
2&-1
\end{array}\right)
$$
and the translation $(s,t)\mapsto(s+\pi,t+\pi)$. Thus the Haar measure is invariant. The homomorphism corresponding to $A$ is not ergodic, since $A^t$ has eigen value $1$ with eigen vector $(1,-1)$, i.e. the function $t-s$ on $\TT^2$ is $u_A$ invariant, i.e. the function $(z,u)\mapsto zu^{-1}$ on $S^1\times S^1$ is invariant.
3. The function $(z,u)\mapsto zu^{-1}$ is also $\theta$ invariant. This simply says that the angle of $z$ and $u$ is invariant!