For a one-dimensional $f(x)$, the dilation operator $f(ax)$ can be expressed as $\exp(g(D))f(x)$, where $g$ is a closed-form function. This is easily checked by e.g. formal Taylor series expansion.


However, it is not clear what to do in the multidimensional case- even if one restricts $U$ to be orthogonal with determinent 1, all attempts to expand $f(xU)$ result in a Taylor series in $xU$, which cannot be simplified further without explicitly expanding it in terms of $x$, a very complicated process. I'm not a big fan of Occam's razor, but I believe there should be a multidimensional $g(D)$ that works even in the multidimensional case. What should I do here?



Differential form of the multidimensional "orthogonal dilation" operator
In 3 dimensions, rotations, i.e., transformations corresponding to orthogonal $U$ with determinant 1, are generated by the (orbital) angular momentum operator $\vec{L} $ with components $L\_i =-i \epsilon\_{ijk} x\_j \,\partial / \partial x\_k $. By Euler's rotation theorem, any given such transformation can be effected by rotating around a specific axis $\vec{e} $ by an angle $\alpha $. Then, the desired rotation operator is
$$
\exp \left(-i \alpha \ \vec{e} \cdot \vec{L} \right) \ .
$$
In other than 3 dimensions, there isn't, of course, such an intuitive description in terms of a vector axis and an angle, but the modification is purely on the level of the rotation theorem -- once this is adapted, one will still then generate the rotations using the antisymmetric tensor operator $x\_j \,\partial / \partial x\_k - x\_k \,\partial / \partial x\_j $.


