How to define the square root of $1-\Delta $?

Question

If $M$ is a Riemannian manifold with $\Delta $ its Laplacian, how can we define $(1-\Delta)^{1/2}$?

The book I am reading says that $(1-\Delta)^{1/2}$ is an invertible first-order pseudo-differential operator with the inverse $(1-\Delta)^{-1/2}$. Naively, I try to construct $(1+|\xi|^2)^{1/2}$ as its principal symbol, but I cannot find an operator whose square is $(1-\Delta)$.

Another possible way to define $(1-\Delta)^{1/2}$ is to use the spectral decomposition of $1-\Delta$. If $\{\phi_i\}$ is an orthonormal basis of $L^2(M)$ such that $(1-\Delta)\phi_i=\lambda_i\phi_i$, then we should have $(1-\Delta)^{1/2}\phi_i=\lambda_i^{1/2}\phi_i$. But I cannot see why this is a pseudo-differential operator since I cannot check the definition locally.

Answer 1

There is a general result of Seeley which states that if $A$ is an elliptic, selfadjoint positive scalar pseudo-differential operator of order $k$ on a compact Riemann manifold, then for any $\newcommand{\bR}{\mathbb{R}}$$s\in\bR$ the operator $A^s$ defined by functional calculus is in fact described by an elliptic selfadoint pseudo-differential operator of order $ks$. This is by no means an obvious result.

The special case you are interested in is simpler and for a proof I refer to page 298 in the book

M.Taylor: Pseudodifferential operators, Princeton University Press, 1981.

Answer 2

You are done once you know that you have a functional calulus for the Laplace-Beltrami operator on $M$. For this, show that it is self-adjoint and has nonpositive spectrum (there are various ways to do it, using different prerequisites). Once you have that, you define the mentioned operator via functional calculus.

\Edit: I forgot that you also wanted to know that it is a pseudo-differential operator, which is not as obvious.

Answer 3

Here is a functional calculus that comes in handy very frequently: in more generality, if $A$ is a sectorial operator with angle $\omega$ on a Banach space $X$ and $0 < \alpha < 1$, then $$A^\alpha u = \frac{\text{sin}\alpha\pi}{\pi}\int^\infty_0t^{\alpha - 1}(t + A)^{-1}Au dt$$ for all $u$ in the domain of $A$.

In particular, it is known that on a Hilbert space $H$, a self-adjoint dissipative operator $A : \mathcal{D}(A) \subset H \longrightarrow H$ is sectorial with $\omega = 0$. See the online notes by M. Haase for all the relevant definitions and more.