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.



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


