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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?

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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 $.

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  • $\begingroup$ For the last sentence- explicitly how? $\endgroup$
    – Kanghun Kim
    Commented Mar 5, 2023 at 4:42
  • $\begingroup$ @KanghunKim - I'm not aware of a reasonably simple explicit expression for a general number of dimensions, but I'll note that the counting of the number of parameters works out: A general $SO(N)$ matrix depends on $N(N-1)/2$ parameters, and the antisymmetric rank-2 tensor in $N$ dimensions has $N(N-1)/2$ independent components. $\endgroup$
    – Michael Engelhardt
    Commented Mar 5, 2023 at 6:55
  • $\begingroup$ I attempted to compute the exponential of the antisymmetric tensor operator using the Cauchy contour integral formula along with the resolvent; Bing Chat simply refuses to do it and manually doing it results in a complicated series-integral that seems impossible to re-express in analytic (let alone closed) form. Maybe I should not have diagonalised so(N). $\endgroup$
    – Kanghun Kim
    Commented Jan 16 at 13:28

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