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It is stated in various places that, in the euclidean space $\mathbb{R}^n$, a differential operator which commutes with every isometries can be expressed as a polynomial of the Laplacian on $\mathbb{R}^n$.

This is for instance stated in this question in the broader context of Riemannian geometry.

I can't find a proof of this fact. Would you be able to point me to a document that proves it and/or to give me the key ingredients of the proof?

Arctic Char
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Weier
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  • @MarianoSuárez-Álvarez I was thinking, is there a way to use the Fourier transform (then restricting the set of functions on which we assume that the Laplacian and isometries are acting) to turn this into a form that would be provable using algebraic considerations? (e.g. we would require a polynomial P corresponding to the operator to depend only on $x^2 + y^2 + z^2$) – Weier Nov 17 '23 at 11:48
  • Are you assuming constant coefficients? – Deane Nov 17 '23 at 15:29
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    @Deane I actually realized I made this assumption in my answer, but the result should hold for coefficients as functions. I don't know if I can get the full result using my method... – Weier Nov 17 '23 at 15:37
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    @Weier the way you reduce to constant coefficients is that translations are also isometries of $\Bbb{R}^n$, so commuting with translations implies the coefficients must be constant (actually it’s equivalent). – peek-a-boo Nov 17 '23 at 16:14
  • @peek-a-boo Indeed... and that still works if I consider the full Poincaré group (the Lorentzian case seems a bit more tricky). – Weier Nov 17 '23 at 16:34

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(Analysis is not my thing but after giving it some thought I think something like this will work — please correct me if I'm wrong)

Let $D = P(D^1, ..., D^n)$ be a differential operator commuting with isometries.

In particular $D(f \circ g) = Df \circ g$ (*) for any Schwartz function $f$ and isometry $g$.

Taking the Fourier transform $\mathfrak{F}$ we get (using the scaling property of the Fourier transform mentioned here for linear operators — I am omitting the determinant which will factor out):

$$ \mathfrak{F}(D(f \circ g)) = P \cdot \mathfrak{F}(f \circ g) = P \cdot \mathfrak{F}(f) \circ g^{-1} \\ \mathfrak{F}(Df \circ g) = F(D f) \circ g^{-1} = (P \cdot \mathfrak{F}(f)) \circ g^{-1} $$

By (*) we get $P = P \circ g^{-1}$ (considering a $f$ such that the support of $\mathfrak{F}(f)$ is the whole real line).

The only polynomials $P(X_1, ..., X_n)$ which are invariant by all isometries are those which are functions of ${X_1}^2 + ... + {X_n}^2$ (we can show that it factors as $P - \lambda = Q \cdot ({X_1}^2 + ... + {X_n}^2)$ for some scalar $\lambda$ then the same for $Q$ etc. until we reach the degree of $P$).

Note that this should also work in the case of the Lorentz group and and the d'Alembert operator.

Weier
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