Given some function $f \in L^2(\mathbb{R}^2)$, I'm interested in finding a positive semi-definite differential operator $\mathcal P: L^2(\mathbb{R}^2) \rightarrow L^2(\mathbb{R}^2)$ that is quadratic in $f$ and invariant under the the action of $\textrm{SL}{(2, \mathbb{R})},$ such that $\forall A \in \textrm{SL}(2, \mathbb{R})$ and $\forall {\bf x} \in \mathbb{R}^2,$ $$ {\mathcal P} f(A {\bf x}) = [{\mathcal P} f] (A {\bf x} ).$$

After thinking for some time, I've come up with two operators that are invariant and P.S.D, but not quadratic.

For example, suppose we consider the operator $${\mathcal P} = \left( \frac{\partial^2}{\partial x^2} \frac{\partial^2}{\partial y^2} - \left[\frac{\partial^2}{\partial x y}\right] \right)^2,$$ which is the squared determinant of the Hessian. It's clear that it is P.S.D. and invariant under transformations in $\textrm{SL}(2, \mathbb{R})$, though it is quartic in $f$.

Furthermore, letting $H$ denote the Hessian and $J \in \textrm{SO}(2)$ be a rotation by $90^\circ$, the operator $$ {\mathcal P} = \left(\nabla^T J^T H \ J \ \nabla\right)^2,$$ is also invariant and P.S.D., but is not quadratic.

I'm asking this question in the hope that someone might know of such a quadratic P.S.D. differential operator that is invariant under $\textrm{SL}(2, \mathbb{R})$ (if it even exists) or be able to point me toward a with a few other ideas I could try.

Some possibly related question(s):

Projective invariant differential operator

Classification of diffeomorphisms by association of differentials with Lie groups

Proof that $a\nabla u = b u$ is the only homogenous second order 2D PDE unchanged/invariant by rotation

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    Also advertising this question in [the Pearl Dive](https://chat.stackexchange.com/rooms/102837/pearl-dive). I am not competent to judge whether this question is easy, difficult or near impossible to answer. If you are such a viewer, please comment. We need more competent people sifting the pearls from the sand. – Jyrki Lahtonen Aug 22 '20 at 06:20
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    Not that I know an answer off-hand, but you should clarify if you are looking for a continuous operator and if the operator should be defined on the entire $L^2$ (seems unlikely given your examples) or only on a dense subset of it, what exactly do you mean by "quadratic PSD operator" (there are two possible interpretations), etc. As for references, Helgason wrote a couple of books on invariant differential operators. – Moishe Kohan Aug 23 '20 at 16:37
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    Tommym, can you comment on the point raised by @MoisheKohan, please? If you don't want to, I guess the answerers are free to interpret the question any way they see fit. For my part I will say that any sensible interpretation is eligible for the bounty. – Jyrki Lahtonen Aug 27 '20 at 14:43
  • My apologies, I've had a busy few days. @MoisheKohan I'm looking for a continuous (operator defined over the entirety of $L^2$. By quadratic I mean that each term is at most of degree two in the partial derivatives of $f$, but these derivatives can be of any order. By P.S.D. I mean that ${\mathcal P} f \geq 0, \ \forall f \in L^2(\mathbb{R}^2)$. – tommym Aug 27 '20 at 15:47
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    Then I have no idea what the question is: partial derivatives applied to $L^2$ functions are no longer functions, but distributions. – Moishe Kohan Aug 27 '20 at 16:13

1 Answers1


I will take your question seriously but not literally since you are asking for a differential operator $D: L^2({\mathbb R}^2)\to L^2({\mathbb R}^2)$ and there are no differential operators $D$ of order $>0$ which take all $L^2$-functions to $L^2$-functions: You would need distributions as values of $D$.

Thus, I will assume that $L^2({\mathbb R}^2)$ in your question means $C^\infty({\mathbb R}^2)$. Then you get your example:

Take $Z=x\frac{\partial}{\partial x} + y \frac{\partial}{\partial y}$: As a vector-field, it sends each point with coordinates $(x,y)$ to vector with coordinates $(x,y)$, this is why $Z$ is invariant under the action of $GL(2, {\mathbb R})$. Then take $D=Z\otimes Z$. As a differential operator, it acts on smooth functions by $$ D: f\mapsto (x\frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y})^2. $$
One can prove that among strictly 1st order strictly quadratic PSD differential operators, up to scalar, this is the only one which is $SL(2, {\mathbb R})$-invariant.

The same works in higher dimensions as well, your $GL(n, {\mathbb R})$-invariant differential operator will be $$ Z\otimes Z, Z=\sum_{i=1}^n x_i\frac{\partial}{\partial x_i}. $$

If I were to take your question literally but not seriously, my answer would be $$ D: L^2({\mathbb R}^2)\to L^2({\mathbb R}^2), D(f)=a f^2 $$ where $a\ge 0$ is a fixed constant. Such $D$ is a PSD, continuous, quadratic differential operator of order 0.

Moishe Kohan
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  • I was half expecting something built from the Casimir operator to pop out (now that I had time to think a bit more). It's linear though. Of course, you're right about this not being defined on all of $L^2$ in the usual sense. I was worried about that :-) – Jyrki Lahtonen Aug 28 '20 at 20:19
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    @JyrkiLahtonen: Casimir was my first thought as well, until there was a clarification of what PSD means. Needless to say, that definition was not the one I expected. Another thing: Casimir is a central element in the universal enveloping algebra, which would correspond to the problem finding biinvariant differential operators on the Lie group $G$ itself. But the question is (mostly) about left-invariant operators on $G/U$ ($U$ is the upper unipotent) and this is a different problem. Thinking about elements of UEA commuting with the Lie algebra of $U$ is how I got the answer. – Moishe Kohan Aug 28 '20 at 20:25
  • @JyrkiLahtonen in case you are still awake https://math.stackexchange.com/questions/3806488/strange-notation-related-to-continued-fractions-in-finnish-language – Will Jagy Aug 28 '20 at 21:06
  • Moishe Kohan, I generally approve of your approach to the site. If you see something *pearly* (the meaning of that has not converged yet) on the site, feel free to bring it up/endorse it in the Pearl Dive. Usually I'm looking for something that is not straight out of ten textbooks and/or has a cute answer. This particular bounty was a bit of a dud due to the flaws in the question, but I'm not gonna worry about that :-) Your specialties have been a bit underrepresented there. – Jyrki Lahtonen Aug 29 '20 at 04:59
  • (cont'd) The general idea of the Pearl Dive was outlined [here](https://math.meta.stackexchange.com/q/31105/11619), but it is still in its infancy. No pressure, of course, but if you happen to run into something that "makes your day" on this site... – Jyrki Lahtonen Aug 29 '20 at 05:02
  • @WillJagy See above. Applies to you as well. – Jyrki Lahtonen Aug 29 '20 at 05:02
  • @JyrkiLahtonen: Thank you, I will keep this in mind. For the current question, I was unsure if I should answer it or to vote for closure as unclear, once the bounty is over. On the balance, I decided that it was interesting enough to be answered. – Moishe Kohan Aug 29 '20 at 20:52