Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 7 / 4 / b / Solution

For each rotation, unitarity givesAveraging and using self-adjointness of $Q$ yields the Dirichlet form of the Kac collision operatorThe velocity integral is necessary: its omission from the printed right-hand side would leave a function of $\mathbf{v}$ rather than a scalar. This identity applies to every $L^2$ function, with complex modulus when necessary, and is nonnegative.

If $(I-Q)F=0$, every nonnegative angular integral is zero, so $\Vert U_{ij,\theta }F-F{\Vert }_2=0$ for almost every angle. Strong continuity in angle extends equality to every angle. The coordinate-plane Givens rotations generate the special orthogonal group $SO(N)$, hence $F$ is invariant in $L^2$ under every element of this group. To identify its shape rigorously despite almost-everywhere representatives, average over the normalized Haar measure of $SO(N)$. This averaging leaves $F$ unchanged, while transitivity of the rotation group on each sphere makes the average a radial function. Thus $F(\mathbf{v})=\varphi (|\mathbf{v}|)$ almost everywhere.

Conversely, every radial function is fixed by every coordinate-plane rotation, and so by $Q$. ThereforeThis is the radial kernel of the Kac collision operator. The rotation correction is essential to this conclusion: with the literal printed map, even $F(\mathbf{v})=e^{-|\mathbf{v}{|}^2}$ in dimension two is not fixed. At $\mathbf{v}=(0,1)$ its printed-map angular average is the average of $e^{-{\sin }^2\theta }$, strictly greater than its value $e^{-1}$.