The Laplacian Commutes with Isometries

A short proof using naturality of the exterior derivative and Hodge star.

This short note proves the following theorem relating the laplacian on a Riemannian manifold and its isometries:

Let $(M,g)$ be a Riemannian manifold and $\Phi\colon M\to M$ an isometry. Then for any $F\colon M\to\RR$ $\Delta(F\circ \Phi)=(\Delta F)\circ \Phi$

Here’s the main idea: the composition $F\circ \Phi$ is the pullback of $F$ by $\Phi$ , $F\circ\Phi = \Phi^\ast F$ And the laplacian is a repeated composition of hodge stars and exterior derivatives. So, we just need to commute each one with the pullback, one at a time. We prove each of these possible as a lemma below:

Exterior Derivative Commutes with Pullbacks

Let $\alpha$ be a $k$ -form on $M$ and $\Phi\colon M\to M$ . Then $d(\Phi^\ast \alpha)=\Phi^\ast(d\alpha)$

Note this first result is completely general and does not care that $\Phi$ is an isometry (as one expects, since $d$ does not know about the metric structure). However the hodge dual is defined in terms of the metric (indirectly through the volume form) so the isometry is crucial below.

Hodge Star Commutes with Pullbacks

Let $(M,g)$ be a Riemannian manifold and $\Phi$ an orientation preserving isometry of $M$ . Then if $\alpha$ is a $k$ -form, $\star (\Phi^\ast \alpha)=\Phi^\ast(\star \alpha)$

Now we can prove the main theorem by a short computation:

\begin{align} \Delta(F\circ \Phi) &=\Delta(\Phi^\ast F)\\ &= \star d\star d(\Phi^\ast F)\\ &=\star d\star \Phi^\ast (d F)\\ &= \star d\,\Phi^\ast(\star dF)\\ &= \star\, \Phi^\ast(d\star dF)\\ &= \Phi^\ast(\star d\star d F)\\ &= \Phi^\ast \Delta F\\ &= (\Delta F)\circ \Phi \end{align}

This computation goes through exactly if $\Phi$ is orientation preserving. If it is orientation reversing, then we need to modify the argument for the Hodge star, and see that in fact commuting introduces a minus sign. But the Laplacian uses the hodge star twice, so these cancel and the final result holds for any isometry of $M$ .