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What is an example of a manifold $M$ with $\dim(M)>1$ whose Lie algebra $\chi^{\infty}(M)$ of smooth vector fields admit an elliptic operator $D:\chi^{\infty}(M)\to \chi^{\infty}(M)$ such that $D$ is a Lie algebra derivation on $\chi^{\infty}(M)$?

Does every manifold admit such an operator?

Is there a Riemannian manifold for which the Laplace operator $D=\Delta$, naturally defined on $\chi^{\infty}(M)\simeq \Omega^1(M)$, would be a derivation of $\chi^{\infty}(M)$?

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First a classical result about elliptic operators: in dimension $\ge 3$ the order of elliptic operators is even. In dimension 2, say in $\mathbb R^2$ you have elliptic vector fields such as $$ \bar \partial=\frac12(\partial_x+i\partial_y),\quad \partial=\frac12(\partial_x-i\partial_y). $$ Of course vector fields are derivations and I suspect that, since a derivation on a manifold must be a differential operator, that differential operator must be first-order from Leibniz' formula and thus should be a vector field since no mutiplication by a (non-zero) function is a derivation.

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  • $\begingroup$ Thanks for your answer. Is there a complete classification of all derivations of $\chi^{\infty}(M)$?(as a vector field analogy of the fact that every derivation of $C^{\infty}(M)$, the space of smooth functions on $M$ is in the form $f\mapsto X.f$ for a unique vector field on $M$? what is your argument that every derivation on the space of vector fields is of first order?What is your argument that every elliptic operator on the sections of a vector bundle of rank greater than 1 is off even degree? $\endgroup$
    – Ali Taghavi
    Commented Apr 27, 2019 at 18:20
  • $\begingroup$ Finally is not the operator $(u,v,w)\mapsto (u_x-v_y,u_y+v_x,w)$ counted as an elliptic operator? $\endgroup$
    – Ali Taghavi
    Commented Apr 27, 2019 at 18:24
  • $\begingroup$ Do not you think that the classical result you mentioned is about "functions" not vector fields?what is a proof of that classical result(or a reference, please)? $\endgroup$
    – Ali Taghavi
    Commented Apr 27, 2019 at 18:30
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I would like to apply the hints presented in the answer by Bazin to give the following answer.

Every derivation of $\chi^{\infty}(M)$ is inner and it is obvious that every inner derivation is a non elliptic operator. The reason is that every non vanishing vector field is locally look like $\partial/\partial_x$, whose corresponding adjoint operator is obviousely non elliptic.

So the non ellipticity of a derivation operator lies in its adjoint-ness not merely on its order.It is first order but this order situation is not enough to conclude it is non elliptic. In fact there is an example of a first order elliptic pde in dim 3 as follows: $$D(u,v, w)= (u_x-v_y, u_y+v_x, w)$$

This operator $D$ satisfies the definition of an elliptic PDE but is of first order.

One can construct a first order elliptic PDE in dimension 4 without a term of order 0. Construct a linear PDE on $\chi^{\infty}(\mathbb{R}^4)$ whose principal symbol is the $4\times 4$ matrix representation of quaterniouns $$\xi_1+\xi_2 i + \xi_3 j+\xi_4 k$$ where $(\xi_1,\xi_2,\xi_3,\xi_4)$ is a cotangent vector.

So it seems that the first line of the answer by Bazin does not work.

Remark: The above linked paper contains a reference to a result by F.Takens that every derivation of $\chi^{\infty}(M)$ is an inner derivation.

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