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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?

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    $\begingroup$ 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. $\endgroup$
    – Willie Wong
    Commented Apr 7, 2020 at 1:29
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    $\begingroup$ Shouldn't the considered Fourier transform be defined on an analogue of the Schwartz class on $M$ to get an automorphism thereof? $\endgroup$
    – Sylvain JULIEN
    Commented Apr 7, 2020 at 9:54
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    $\begingroup$ @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 $\mathcal{S}$: mathoverflow.net/questions/37021/… $\endgroup$
    – Willie Wong
    Commented Apr 7, 2020 at 14:46
  • $\begingroup$ @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. $\endgroup$
    – Ali Taghavi
    Commented Apr 7, 2020 at 20:38
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    $\begingroup$ 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. $\endgroup$
    – Willie Wong
    Commented Apr 8, 2020 at 2:20

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