Search Results

For the current version the basic argument still hold, using that $$ \| \nabla u\|_2 = \langle u, -\Delta u\rangle $$ and so behaves nicely for eigenfunctions of the Laplacian. …

Now as the principal part is the Laplacian and is invariant under the isometries, you can then conclude similarly that $c^i$ is a vector that is invariant under the action of the entire special orthogonal … Then quite clearly the second order operator $\mathcal{L}_X^2: f \mapsto X(X(f))$ commutes with the (one and only) Killing vector field, yet is not the Laplacian. …

So that when $X$ is Killing, automatically both the ${}^{(X,0)}\pi$ and ${}^{(X,1)}\pi$ vanish, and differentiation with $X$ commutes with the Laplacian. …

In fact, if you take the standard sphere $\mathbb{S}^n$, the Taylor expansion is precisely $r^2 + \ldots$ which suggests that the Laplacian is exactly $n\neq 0$. … The local expression of the Laplacian is $$ \frac{1}{\sqrt{|g|}} \partial_k g^{kl} \sqrt{|g|} \partial_l = g^{kl}\partial^2_{kl} + O(\partial) $$ where the $O(\partial)$ denotes terms that are first …