TITLE: On equation $\Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$ on a Riemannian manifold
QUESTION [6 upvotes]: 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..

REPLY [6 votes]: 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. 
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
$$ [ \nabla_X, \triangle_g] f = \phi \triangle_g f $$