# Operators of finite propagation

In the starting post of this series we considered infinite band matrices (with uniformly bounded entries) acting on infinite vectors and asked at the end the question how to determine whether a given matrix, which is not a band matrix, can be approximated by such. Today we provide the setup in order to answer this question properly in future posts. The notions introduced in this post are fundamental to large scale geometry.

We consider a metric space $$(X,d)$$. On it we consider the space $$\ell^2(X)$$ of all square-summable sequences indexed by points in $$X$$. The metric $$d$$ on $$X$$ comes into play by the following definition:

A linear operator $$T$$ acting on $$\ell^2(X)$$ is said to have finite propagation, if there exists an $$R > 0$$ such that for all $$v \in \ell^2(X)$$ we have $\mathrm{supp}(Tv) \subset \mathrm{Neigh}(\mathrm{supp}(v),R),$ where $$\mathrm{Neigh}(\mathrm{supp}(v),R)$$ is the neighbourhood of distance $$R$$ around $$\mathrm{supp}(v)$$, i.e., all points in $$X$$ of distance at most $$R$$ to $$\mathrm{supp}(v)$$.

If $$(X,d)$$ is the set $$\mathbb{Z}$$ equipped with the usual distance, then operators of finite propagation are exactly the infinite band matrices. The smallest possible value of $$R$$ in the above definition is called the propagation of $$T$$.

In the following, given a linear operator $$T$$ on $$\ell^2(X)$$, its entry $$T_{x,y}$$ is defined as $$(T\delta_y)(x)$$, where $$\delta_y \in \ell^2(X)$$ denotes the sequence which has only one non-zero entry, namely $$1$$ on the point $$y$$.

In the starting post we noticed that if the entries of an infinite band matrix are uniformly bounded (in absolute value), then the matrix defines a bounded operator on $$\ell^2(\mathbb{Z})$$ with norm bound $$\|T\| \le M\cdot K$$, where $$M$$ is the thickness of the band of non-zero entries and $$K$$ is the uniform upper bound on the absolute values of the entries. But on a general metric space $$(X,d)$$, if we are given an operator of finite propagation with uniformly bounded entries it does not necessarily define a bounded operator on $$\ell^2(X)$$. Morally, the reason for this is that the $$M$$ in the estimate is actually not the thickness of the band, but it is the number of elements in a row (or column) of the band. And on a general metric space this might be infinite. Hence the following definition:

The metric space $$(X,d)$$ is said to have bounded geometry, if for every $$R>0$$ we have $\sup_{x\in X} \#\mathrm{Neigh}(x,R) < \infty.$

Note that a metric space of bounded geometry is necessarily discrete, since every ball of finite radius must have only finitely many elements.

The following basic estimate is now immediate:

Let $$(X,d)$$ be a metric space of bounded geometry and let $$T$$ be a linear operator of finite propagation on $$\ell^2(X)$$. If the entries of $$T$$ are uniformly bounded, then $$T$$ defines a bounded operator on $$\ell^2(X)$$ with $\|T\| \le \sup_{x\in X} \#\mathrm{Neigh}(x,\mathrm{prop}(T)) \cdot \sup_{x,y \in X} |T_{x,y}|,$ where $$\mathrm{prop}(T)$$ denotes the propagation of $$T$$.

Note that the converse to this result is obvious: if $$T$$ is a bounded operator on $$\ell^2(X),$$ then its entries must be uniformly bounded: $$|T_{x,y}| \le \|T\|$$.

In this generalized setup, the original question of approximating infinite matrices by band matrices becomes the question of investigating for a metric space $$(X,d)$$ of bounded geometry the operator norm closure of all bounded, linear operators of finite propagation on $$\ell^2(X)$$. We will start this investigation in a future post.