Assume that $M$ is a compact Riemannian manifold whose Laplacian is denoted by $\Delta$. Assume that the Euler characteristic of $M$ is zero. Does $M$ admit a non vanishing vector field $X$ which satisfy $$(\*) \qquad \Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$$


What can be said about the structure of the Lie algebra of all vector fields $X$ with the property $(\*)$?


**As a second question:** Every vector field $X$ on $M$ defines a second order differential operator on $C^{\infty}(M)$ with $$D(f)=\Delta(X.f)-X.\Delta(f)$$


This is a second order operator since the third order terms cancel each others.


What is the principal symbol of this operator , precisely? Can this PDE be an elliptic operator when $M$ is a compact manifold?(I mean:is there an example of this situation in compact case?) Does every compact manifold admit a vector field $X$ for which this PDE would be an elliptic operator?What would be a dynamical interpretation for the index of this PDE. [This is a dynamical motivation for the later question.](https://mathoverflow.net/questions/182415/elliptic-operators-corresponds-to-non-vanishing-vector-fields).



On equation $\Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$ on a Riemannian manifold
The following formula is known among the experts but hard to find in the literature, so I figure I will document it here. Throughout $(M,g)$ denote an arbitrary *pseudo-Riemannian* manifold, and $\nabla$ its Levi-Civita connection. 



> 
> **Definition** Given a vector field $X$, its corresponding *0th order deformation tensor* is defined to be ${}^{(X,0)}\pi := \mathcal{L}\_X g$, where $\mathcal{L}\_X$ is Lie differentiation with respect to $X$.  
> 
>  The corresponding *1st order deformation tensor* is defined using a formula similar to that of Christoffel symbols:
>  $$ {}^{(X,1)}\pi\_{ab}{}^c := \frac12 g^{cd} \left[ \nabla\_a ( {}^{(X,0)}\pi\_{bd}) + \nabla\_b ({}^{(X,0)}\pi\_{ad}) - \nabla\_d ({}^{(X,0)}\pi\_{ab}) \right] $$
> 
> 
> **Lemma** Let $\Xi$ be an arbitrary $k$-covariant tensor field. And let $X$ be a vector field. The following formula holds for the commutation:
>  $$ [ \nabla\_a, \mathcal{L}\_X ] \Xi\_{b\_1\cdots b\_k} = \sum\_{j = 1}^k {}^{(X,1)}\pi\_{a b\_j}{}^c \Xi\_{b\_1 \cdots b\_{j-1} c b\_{j+1} \cdots b\_k} $$
> 
> 
> 


With the aid of these formulas, we have immediately that, writing $\triangle\_g$ for the Laplace-Beltrami operator, first


$$ [ \nabla\_X, \triangle\_g] f = [\mathcal{L}\_X, \triangle\_g ] f $$


because Lie derivation and covariant differentiation act identically on scalars, and then


$$ [\mathcal{L}\_X, g^{ab}\nabla\_a\nabla\_b] f = \mathcal{L}\_X (g^{ab}) \nabla^a \nabla\_b f + g^{ab} [\mathcal{L}\_X, \nabla\_a] \nabla\_b f + g^{ab} \nabla\_a [\mathcal{L}\_X, \nabla\_b ]f $$


The first factor we can compute to get


$$ \mathcal{L}\_X(g^{ab}) = - {}^{(X,0)}\pi^{ab} $$


using that $g^{ab} g\_{bc} = \delta^a\_c$. The third factor vanishes because Lie differentiation commutes with exterior differentiation. And we use our Lemma for the second term. We get, finally


$$ [\nabla\_X, \triangle\_g] f = - {}^{(X,0)}\pi^{ab} \nabla\_a\nabla\_b f - g^{ab} ~{}^{(X,1)}\pi\_{ab}{}^c \nabla\_c f. $$


Remarks: 


* Notice that the first order deformation tensor is defined in terms of the 0th order one. So that when $X$ is Killing, automatically both the ${}^{(X,0)}\pi$ and ${}^{(X,1)}\pi$ vanish, and differentiation with $X$ commutes with the Laplacian.
* In the case ${}^{(X,0)}\pi = \phi g$ for some scalar function $\phi$ (so $X$ is conformally Killing), one can check that the formula reduces to [the one I gave in a comment above](https://mathoverflow.net/questions/334060/on-equation-delta-circ-partial-partial-x-partial-partial-x-circ-delta#comment835997_334130).
* When the function $\phi$ in the previous item is a non-zero constant (which some people refer to as $X$ being a *homothetic vector field*) one gets the special case
$$ [ \nabla\_X, \triangle\_g] f = \phi \triangle\_g f $$


