Vector Fields in a Riemannian Manifold Suppose $(M,g)$ is a Riemannian manifold. 
Is there a way to classify manifolds where there exists a vector field that commutes with the laplace beltrami operator?
Thanks
 A: I give a geometric explanation of the calculations of Willie, which simultaneously elaborates the suggestion of Deane. 
The flow of a vector field commuting with Laplacian preserves the Laplacian  and therefore its symbol which is the inverse of the metric (i.e., $g^{ij}$). Then, the flow of the vector field preserves the metric and the vector field is Killing 
A related question is 
is $\nabla \cdot ( c^2 \nabla)$ a Laplace-Beltrami operator?
Added later: Just have seen that  Tobias Fritz  in his comment which appeared before my answer already had the same explanation. 
A: The commutator is easy enough to compute. 
Acting on scalars we have
$$ \begin{align}
[\triangle_g, \mathcal{L}_X] &= [g^{ab} \nabla_a \nabla_b ,\mathcal{L}_X] \\
&= [g^{ab} \nabla_a, \mathcal{L}_X]\nabla_b \\
&= {}^{(X,0)}\pi^{ab} \nabla_a \nabla_b + g^{ab} [\nabla_a, \mathcal{L}_X] \nabla_b \\
&= {}^{(X,0)}\pi^{ab} \nabla_a\nabla_b + g^{ab} {}^{(X,1)}\pi_{ab}^c \nabla_c
\end{align} $$
where the deformation tensors are
$$ {}^{(X,0)}\pi_{ab} = \mathcal{L}_X g_{ab} $$
and 
$$ {}^{(X,1)}\pi_{ab}^c = \frac12 g^{cd} \left( \nabla_a {}^{(X,0)}\pi_{bd} + \nabla_b {}^{(X,0)}\pi_{ad} - \nabla_d {}^{(X,0)}\pi_{ab} \right) $$
from this you see that for the commutator to vanish identically it is necessary and sufficient that $X$ is Killing. 
