Consider a torus $T^n$. An differential operator $\mathcal{O}$ acts on differential forms on $T^n$. Let $R$ be a smooth vector field on $T^n$, whose orbits are dense on $T^n$ (for instance, irrational flow on $T^2$).


My question is: if $\mathcal{L}\_R$ and $ \mathcal{O}$ commutes, does it imply $\mathcal{O}$ also commutes with all $n$ generators of torus action (namely $\partial\_{\varphi\_i}$ in coordinate)?


I tested for example $n = 2$, and $\mathcal{O} = \mathcal{L}\_X$ for some smooth vector field $X$, and $R = p\partial\_{\varphi\_1} + q \partial\_{\varphi\_2}$ where $p$ and $q$ are co-irrational. Then $\mathcal{L}\_R\mathcal{O}=\mathcal{O}\mathcal{L}\_R$ implies $X$ can only has constant components and therefore must commute with $\partial\_{\varphi\_i}$. But I don't know how to generalize.


Thank you!


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**Edit**: as pointed out by Sergei Ivanov, the answer to original question is "No". Now let me put in a bit more assumptions. Put a standard flat metric on $T^n$, and let $R$ be also Killing, and $\partial\_{\varphi\_i}$ be generators of the isometry group. So I asked: does $\mathcal{O}$ commutes with the isometry group?



Does such an operator commutes with the whole torus action?
No. The assumption is coordinate-independent (i.e., preserved by self-diffeomorphisms) but the desired conclusion is not.


Begin with $R$ being the standard irrational flow and $\mathcal O$ a coordinate differentiation. Apply a generic self-diffeomorphism of the torus. The image of the coordinate field no longer commutes with the (new) coordinate ones but it commutes with the image of $R$.


