It is relatively easy to show that the Laplacian

$$ \Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} $$

Is the unique second order linear differential operator that is invariant under rotations in the sense that

$$ \Delta (f(R\mathbf{x})) = (\Delta f)(R\mathbf{x}). $$

The way I remember proving this was to write down a general

$$ D = \sum_i a_i \frac{\partial}{\partial x_i} + \sum_{ij} b_{ij} \frac{\partial^2}{\partial x_i \partial x_j}, $$

and then demand the invariance property. A multiple of the Laplacian will then fall out.

I am wondering if there is a way to general all such operators (up to a certain degree). Are they all powers of the Laplacian? What happens for the vector Laplacian?

  • 1
    $\begingroup$ The premise of the question is wrong: there are many more such rotationally order 2 invariant operators, even modulo first order operators. For example, consider $r^2\Delta$. Were you thinking of constant coefficient differential operators? If so, then the first fundamental theorem of invariant theory for the orthogonal group ${\rm O}_n$ shows that indeed every constant coefficient differential operator on $K^n$ ($K$ a field) is a polynomial in $\Delta$. There is an analogous description, due to Roger Howe, for the case of ${\rm O}_n$-invariant polynomial coefficient differential operators. $\endgroup$ May 3, 2019 at 4:39

3 Answers 3


They are not all powers of the Laplacian. For example, if a function $g$ is invariant under rotation, then the 0th order differential operator $D_g(f) = g f$ is invariant under rotation. There is also the (first-order) radial derivative operator $$ R(f) = xf_x + yf_y\,. $$ Any (non-commutative) polynomial in $D_g$, $R$ and $\Delta$ will be a rotationally invariant differential operator, but these are not independent. For example, we have the identity $$ \left[\Delta, D_{x^2+y^2}\right] = D_4 + 4R, $$ so $R$ is already in the ring generated by $\Delta$ and the $D_g$ where $g$ varies through rotationally invariant functions.

Added comment: Of course, I should have mentioned the other first order linear rotationally invariant operator, the angular derivative operator: $$ A(f) = x\,f_y - y\,f_x\,, $$ with the identity $$ R^2 + A^2 = D_{x^2+y^2}\,\Delta $$


Polynomials in the Laplacian are invariant under all isometries, not just rotations. This characterizes them by e.g. Helgason (1959, Thm 11) $=$ (1962, Prop. X.2.10) $=$ (1984, Prop. II.4.11), or Folland (1995, Thm 2.1).


More a comment than an answer - since $\nabla $ transforms as a vector, you could construct scalars by contracting an arbitrary number (say $n$) of $\nabla $s with an arbitrary rank-$n$ tensor. So it seems that to generate all operators in question, you'd at least have to generate all tensors on your underlying space (which implicitly seems to be 2-dimensional?). You'd discard some, such as antisymmetric of rank greater than 2, since the contraction would vanish, but it still seems there is a large number of possibilities. Perhaps some irreducibility specification is needed to make the question sharper?


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .