Let $X$ be a non-compact manifold and let $C\_0(X)$ act on $L^2(X)$ by pointwise multiplication. 


We say $T\in\mathcal{B}(L^2(X))$ has *finite propagation* if there exists an $r>0$ such that: for all $f,g\in C\_c(X)$ with supports separated by a distance greater than $r$, we have $fTg=0$.


**Question:** Suppose $S\in\mathcal{B}(L^2(X))$ is a positive operator with finite propagation. Then does $S^{1/2}$, the unique positive square root of $S$, also have finite propagation?



Does the square root of a finite propagation operator have finite propagation?
Greg Kuperberg and I studied operators of this type in our AMS Memoir *A von Neumann Algebra Approach to Quantum Metrics*. We called them *finite displacement operators*.


The definition you give is ill-posed because it refers to distance and you have not specified a metric (unless "manifold" means "Riemannian manifold"?). However, I can give a counterexample on $L^2([0,\infty))$ with the standard distance (or on $L^2(\mathbb{R})$ if you like, simply by inclusion), which I think would cover any reasonable way of making the question precise.


It will suffice to find a positive operator whose displacement is arbitrarily large but whose square has displacement 1. Then you can put together a sequence of such things to get a positive operator with infinite displacement whose square has displacement 1. The examples I give can be modelled on $L^2([0,n))$, so after amalgamating them you would get an example on $L^2([0,1)) \oplus L^2([0,2)) \oplus \cdots \cong L^2([0,\infty))$ (with the standard distance).


I say "modelled on $L^2([0,n))$" because we can actually find an $n\times n$ matrix with displacement $n$ whose square has displacement $1$. You can then let this matrix act on $L^2([0,n)) = L^2([0,1)) \oplus \cdots \oplus L^2([n-1,n))$ in the obvious way.


The matrix counterexample will be constant on diagonals in the form $$A = \left[\begin{matrix}1&\epsilon&a\_2&a\_3&\cdots&a\_{n-1}\cr \epsilon&1&\epsilon&a\_2&\cdots&a\_{n-2}\cr a\_2&\epsilon&1&\epsilon&\cdots&a\_{n-3}\cr a\_3&a\_2&\epsilon&1&\cdots&a\_{n-4}\cr \vdots&\vdots&\vdots&\vdots&\ddots&\cr a\_{n-1}&a\_{n-2}&a\_{n-3}&a\_{n-4}&\cdots&1\end{matrix}\right].$$ Then $A^2$ will also be constant on diagonals and we want it to vanish off of the middle three diagonals (the main diagonal plus its immediate sub- and super-diagonals).


Since $A^2$ is constant on diagonals all we need to ensure is that its first column is nonzero only in the first two entries. Writing the formulas for the third, fourth, etc., entries, setting them equal to zero, and solving yields a formula for each $a\_k$ on the order of $\epsilon^k$. (This needs proof but I'm sure it's true.) If $\epsilon$ is small then the resulting matrix will be a small self-adjoint perturbation of the identity and hence will be positive.


