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
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3$\begingroup$ Clearly Killing vector fields have this property. But in general, I guess that one can think of such vector fields as generators of "isospectralities" (instead of isometries). What would be a nice example of a manifold with such a vector field that is not Killing? $\endgroup$– Tobias FritzCommented Aug 21, 2015 at 0:37
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5$\begingroup$ Actually I'm not sure that my statement on "isospectralities" makes sense: the principal symbol of the Laplace operator is the metric itself. So doesn't this imply that every vector field that commutes with the Laplacian must be Killing? $\endgroup$– Tobias FritzCommented Aug 21, 2015 at 3:48
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$\begingroup$ That is a good point but then should we consider the principal symbol of the commutator? That would be the poisson bracket of the two principal symbols $\endgroup$– AliCommented Aug 21, 2015 at 4:13
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2 Answers
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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.
answered Aug 21, 2015 at 7:50
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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.
answered Aug 21, 2015 at 4:28
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$\begingroup$ It seems to me that there should be a calculation-free argument. Something based on the fact that the Laplacian uniquely determines the metric. So if the Laplacian commutes with the vector field, then so does the metric. $\endgroup$ Commented Aug 21, 2015 at 5:02
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1$\begingroup$ @DeaneYang: yes. But then you need the fact that the Laplacian uniquely determines the metric. Which you get by looking at its principal part, and then we come back to Tobias' comments. (In any case, if you are talking about my tendency to provide lowbrow answers on MO, I am definitely guilty as charged. :-p) $\endgroup$ Commented Aug 21, 2015 at 5:17
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$\begingroup$ Willie, I certainly wasn't knocking your "lowbrow" answer. That's the best kind, especially for anyone learning things the first time. I was just wondering about a more precise way to express the fact that we all knew in advance that the vector field had to be Killing. $\endgroup$ Commented Aug 21, 2015 at 15:31
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$\begingroup$ @WillieWong Is there a Riemannian metric on $\mathbb{R}^2 $ or the punctured plane such that the vander pol vector field would be a Killing vector field. In this case the space of Harmonic maps would be invariant under flow then one can think to compute the algebraic index of such operator (to have some dynamical information on X) $\endgroup$ Commented Jun 5, 2017 at 20:29
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1$\begingroup$ @AliTaghavi: the flow of a Killing vector field is a one-parameter family of isometries. So it is volume preserving. $\endgroup$ Commented Jun 6, 2017 at 19:57