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Motivation for this question:

Let $X$ be a vector field on a manifold $M$. Obviously the differential operator $D:C^{\infty}(M)\to C^{\infty}(M)$ with $D(f)=X.f$ is not an elliptic opetator when $\dim M>1$. Now consider a simple example: Put $M$ for $\mathbb{R}^2$, $X=x^2\partial_x+y^2 \partial_y$
and $D$ for the first order corresponding differential operator. Then if we dente by $\mathcal{F}$ the standard Fourier transform operator then the principal (last homogeneus) part of differential operator $\mathcal{F}^{-1} D \mathcal{F}$ is $xu_{xx} +yu_{yy}$. This operator is an elliptic operator if we restrict to the first positive quadrant $x>0, y>0$.This situation is discussed via some other quadratic system in Remark 2 and its consecutive example in page 5 of this note.

In this question we would like to globalize and generalize this situation.

Fourier transform on Riemannian manifolds and Lie groups

In this MSE post three methods of generalization of Fourier transform on a Riemannian manifold or a Lie group is discussed. So based on these definition we assume that the Fourier transform $\mathcal{F}$ is a linear isomorphism on $C^{\infty}(M)$ when $M$ is a Riemannian manifold.

Definition: A non-vanishing vector field $X$ with derivational operator $D$ on a Riemannian manifold $(M,g)$ or a Lie group $G$ is called Fourier elliptic vector field if the differential operator $\mathcal{F}^{-1}D\mathcal{F}$ gives us a global elliptic operator.

Question: What are some precise examples of both Riemannian manifold and Lie group cases which admit a Fourier elliptic vector field? Does every manifold $ M$ with a non-vanishing vector field $X$ admit a Riemannian metric such that the vector field $X$ would be a Fourier elliptic vector field?

Ali Taghavi
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    I wrote the linked MSE post. NONE of the three notions I described give you that the "Fourier transform is a linear isomorphism on $C^\infty(M)$". The closest case is that of the second notion, and even there the Fourier transform takes function spaces on a Lie group $G$ to functions defined on its dual group $\hat{G}$. For example, if you consider the group $\mathbb{T}^k$, its dual group is $\mathbb{Z}^k$. I am not sure how you intend to even define a differential operator on the dual group. – Willie Wong Apr 07 '20 at 01:29
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    Shouldn't the considered Fourier transform be defined on an analogue of the Schwartz class on $M$ to get an automorphism thereof? – Sylvain JULIEN Apr 07 '20 at 09:54
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    @SylvainJULIEN When $M$ is compact, there is no distinction between $C^\infty$ and $\mathcal{S}$; on the other hand, when $M$ is compact, most notions of frequencies will require the frequency space to be discrete and not a $\mathrm{dim}(M)$ dimensional manifold. For the case of Lie groups, it is also more convenient to formulate the general theory in terms of $L^2$ rather than $\mathcal{S}$: https://mathoverflow.net/questions/37021/is-fourier-analysis-a-special-case-of-representation-theory-or-an-analogue – Willie Wong Apr 07 '20 at 14:46
  • @SylvainJULIEN yes it was a typo. As you said it is Schartz space. I wrote in paragraph 3 of the arxived note I mentioned in the post. – Ali Taghavi Apr 07 '20 at 20:38
  • @WillieWong Thank you for your MO links. With help of these two MO and MSE link(your answer) I hope that one can introduce an abstract and global version I am searching for. To be honnest I am interested to find a generalization and globalization of this situation since more than 7 years ago. In fact after I posted that note to arxive, I realized that the resulting $\mathcal{F}^{-1}D\mathcal{F}$ is not GLOBALLY elliptic. So I wonder if there exist a vector field on the plane for which $D$ preserves the Schwartz class and $\mathcal{F}^{-1}D\mathcal{F}$ is globally elliptic – Ali Taghavi Apr 07 '20 at 20:45
  • For example does $D=cos x \partial_x +sin x \partial_y$ work? – Ali Taghavi Apr 07 '20 at 20:46
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    What exactly is the notion of elliptic you are working with? Your operator $F^{-1}D F$ for general vector fields $D$ (on $\mathbb{R}^2$) is not going to be a differential operator. I am not even sure it can in general be a pseudodifferential operator. – Willie Wong Apr 08 '20 at 02:20
  • @WillieWong A differential operator is a linear operator which decreases the support(according to Peetre theirem). So an elliptic one has an obvious definition. For polynomial vector field the resulting $F^{-1} D F$ is obviousely a differential operator. But as you said it is interesting to classify all $D$ whose $F^{-1}DF$ is restriction of a differential operatir from $C^{\infty}$ to Schwartz space. – Ali Taghavi Apr 08 '20 at 02:43
  • @WillieWong Now I realize what you say. but when the coefficients are polynomial functions the resulting $F^{-1}DF$ is a diff. operator again. Please read example in page 5 of this note(befor Remark 3) https://arxiv.org/pdf/1302.0001.pdf – Ali Taghavi Jan 22 '21 at 15:20
  • In this case: your $F^{-1} D F$ can be expressed as $P\ell$, where $\ell$ is a linear function on $\mathbb{R}^n$ and $P$ is a constant coefficient partial differential operator. Its classification is not too technical. – Willie Wong Jan 22 '21 at 16:33
  • @What is this linear functional for $D$ say $D=x^2\partial_x+y^2\partial_y+xy \Delta$? – Ali Taghavi Jan 22 '21 at 16:40
  • Delta is Laplacian – Ali Taghavi Jan 22 '21 at 16:41
  • The main aim of this question is to remedy the non ellipticity of certain linear operator associated to a derivational operator. – Ali Taghavi Jan 22 '21 at 16:42
  • In fact the non ellipticity which discussed in this post was the main motivation for our question: https://mathoverflow.net/questions/182415/elliptic-operators-corresponds-to-non-vanishing-vector-fields?rq=1 – Ali Taghavi Jan 22 '21 at 16:53

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