# Commutator with Dirichlet Laplacian

Consider the Dirichlet Laplacian $\Delta$ on a compact Riemannian manifold (with boundary). Consider the operator $T = \sqrt{-\Delta}$. My question is: is there any Leibniz/product rule? Can we say, for $u, v \in C^\infty_0(M)$, we have $$T(uv) = uT(v) + vT(u) ?$$ Otherwise, is there a nice'' estimate on the commutator $[u, T]$?

Also, if $T$ is a first order differential operator, then we have $$[u, [v, T]] = 0$$. Here $T$ is like a first order differential operator in a sense, though not quite. Would we still have the above result though?

• There is, in my opinion, no reason to expect such formulae. For example, your identity, if valid, would imply that a product of eigenfunctions is an eigenfunction, which is clearly false (say $M=(0,1)$). – Christian Remling Jul 14 '14 at 0:35
• @ChristianRemling: Technically, the identity as stated is only for $C^\infty_0$ functions, which the eigenfunctions are not, so we'd also need an approximation argument. – Nate Eldredge Jul 14 '14 at 0:45

$T$ is a first order pseudodifferential operator, and multiplication by a fixed function is a (pseudo)differential operator of order zero. So the identities you want hold modulo lower order terms. This follows by the asymptotic expansion of the composition of two pseudo differential operators that can be found in any book on pseudodifferential operators. I do, however, agree with Christian that it is unlikely that the exact identities hold. In fact, I believe that these identities come close to uniquely characterizing, among all first order pseudodifferential operators, first order differential operators with no zeroth order term.

• Are you sure $T$ is a pseudodifferential operator? I thought the presence of the boundary messes that up. It is possible I have a misconception. Could you please clarify? – Student Jul 14 '14 at 0:49
• It's certainly a pseudodifferential operator for functions that vanish on the boundary. – Deane Yang Jul 14 '14 at 4:35