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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

tommym
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    Also advertising this question in the 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

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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, 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