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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!


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?

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up vote 3 down vote accepted

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 $R$.

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Thanks for your answer. You are right. But my question was over-general. What if I add a standard flat metric on $T^n$, and let $R$ be Killing and $\partial_{\varphi_i}$ be the generators of the isometry group. And then let $\mathcal{O}$ commutes with $\mathcal{L}_R$. Does these new assumptions make any difference? – Lelouch Nov 8 '13 at 17:38
If $R$ has constant coordinates, then yes. Commuting with a vector field is the same thing as commuting with its flow. And since the orbit is dense, translations by $R$ can approximate any translation. Hence the operator commutes with translations (but not with the full isometry group, which includes other connected components). – Sergei Ivanov Nov 8 '13 at 18:24

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