12
$\begingroup$

Let $M$ be a smooth manifold, like $\mathbb{R}^n$ for instance. The existence of a parametrix for an operator $P$ on $C^\infty(M)$ in any reasonable pseudodifferential calculus implies that $P$ is hypoelliptic. Is there a converse to this statement?

The converse would have to take place in some very general setting that encompasses all possible pseudodifferential calculi. Here's how I'd like to phrase it...

An operator $P$ on $C^\infty(M)$ is "very regular" if its Schwartz kernel has the following two properties:

  • (1) $p$ is properly supported and semiregular in both variables, ie, $p \in (C^\infty(M) \hat\otimes \mathcal{E}'(M)) \cap (\mathcal{E}'(M) \hat\otimes C^\infty(M))$,

  • (2) $p$ is equal to a smooth function off the diagonal.

The first condition means $P$ maps each of $C_c^\infty(M)$, $C^\infty(M)$, $\mathcal{E}'(M)$ and $\mathcal{D}'(M)$ to itself. The second means $P$ is pseudo-local.

Suppose a very regular operator $P$ is hypoelliptic, in the sense that every preimage of a smooth function is smooth. Does this mean that there is a very regular $Q$ which is a parametrix in the sense that $PQ-I$ and $QP-I$ are smoothing operators?

$\endgroup$
1
  • $\begingroup$ As Erik Van Erp pointed out to me, the answer seems to be no based on the answer to <u>this question</u>, since the property of having a very regular parametrix is invariant under transpose, while the property of being hypoelliptic is apparently not. Unfortunately, I don't know how to prove the pertinent claim in the answer given to that question, and didn't get a response there (it's a very old thread). If anyone can help out, we'll have a solution. $\endgroup$ Jul 27, 2015 at 10:19

1 Answer 1

3
$\begingroup$

In the case of hypoelliptic operators with constant coefficients, you have a characterization found by L.Hörmander, which implies that you do have a pseudo-differential parametrix with a symbol in a class $S^{-m}_{\rho,\delta}$.

For hypoelliptic operators with variable coefficients, this is a different story: although the hypoelliptic operator in two dimensions $$ D_x^2+x^2 D_y^2, $$ does have a pseudo-differential parametrix, it is not so easy to put it in a classical framework of pseudo-differential operators. Moreover, the operator in three dimensions $$ D_x^2+x^4(D_y+x D_z)^2, $$ is hypoelliptic, but is not likely to have a parametrix with a symbol in any $\rho, \delta$ class and one can even doubt that a parametrix could live in a general class of pseudo-differential operators such as the ones created by R.Beals & C.Fefferman or L.Hörmander.

$\endgroup$
2
  • $\begingroup$ This is fascinating. What are you basing your comments on? Why shouldn't we expect a nice parametrix for the second operator, for instance? Any references or hints would be appreciated. $\endgroup$ Jun 25, 2021 at 5:49
  • $\begingroup$ @Bob Yuncken If you take a look at the "lifting" procedure due to Rothschild \& Stein to prove the optimal hypoellipticity of the second operator, you will see that this is very far from the explicit construction of a parametrix. Also if you look at Chapter 27 on Subellipticity in L. Hörmander's ALPDO volume IV, you will check that the subellipticity of $D_x+i x^2(D_y+ c\vert D_z\vert)$ requires the full strength of the whole chapter which does not lead to a parametrix construction. $\endgroup$
    – Bazin
    Jun 26, 2021 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.