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2 deleted trivially false first part of the question

# Inverses ofinfinitematrices(ormaybe:inverses in convolution algebras)

Suppose $A$,$B$ are infinite matrices (with entries in a field) such that $AB=I$. Suppose further that $A$,$B$ are both row-column-finite: every row of $A$ and $B$ contains at most finitely many nonzero entries, and the same holds of the columns. Does it follow that $BA=I$?

Since this may well be too general to be true, let me say where this is coming from.

Let $G$ be a locally compact totally disconnected group, and to make life easy let's suppose its Haar measure is bi-invariant. Let $C_c(G)$ be the space of locally constant complex functions on $G$ with compact support, which forms an algebra under convolution. Suppose $e \in C_c(G)$ is an idempotent, so that $H = eC_c(G)e$ is an algebra with identity. Is it now true that if $f \star g = e$ in $H$, then $g \star f = e$ as well?

Again this

This may be too general to be true, so to be more specific: suppose $K < G$ is a compact subgroup such that $K\backslash G/K$ is countable, suppose $\phi : G \to \mathbf{C}^{\times}$ is a character, and suppose the idempotent $e$ is the function with support in $K$ such that $e(x) = \phi(x^{-1})/\mu(K)$ for $x \in K$. Now is it true that if $f \star g = e$ in $H$ then $g \star f = e$ as well?

[I am reading something that claims some $f$ is a unit but then checks it by checking the existence of $g$ such that $f \star g = e$. So really, the question is whether it follows by some general business that $g \star f = e$, or whether one has to do another computation to check the other direction. (And if the former, how general is the general business?) direction.]

1

# Inverses of infinite matrices (or maybe: inverses in convolution algebras)

Suppose $A$,$B$ are infinite matrices (with entries in a field) such that $AB=I$. Suppose further that $A$,$B$ are both row-column-finite: every row of $A$ and $B$ contains at most finitely many nonzero entries, and the same holds of the columns. Does it follow that $BA=I$?

Since this may well be too general to be true, let me say where this is coming from. Let $G$ be a locally compact totally disconnected group, and to make life easy let's suppose its Haar measure is bi-invariant. Let $C_c(G)$ be the space of locally constant complex functions on $G$ with compact support, which forms an algebra under convolution. Suppose $e \in C_c(G)$ is an idempotent, so that $H = eC_c(G)e$ is an algebra with identity. Is it now true that if $f \star g = e$ in $H$, then $g \star f = e$ as well?

Again this may be too general to be true, so to be more specific: suppose $K < G$ is a compact subgroup such that $K\backslash G/K$ is countable, suppose $\phi : G \to \mathbf{C}^{\times}$ is a character, and suppose the idempotent $e$ is the function with support in $K$ such that $e(x) = \phi(x^{-1})/\mu(K)$ for $x \in K$. Now is it true that if $f \star g = e$ in $H$ then $g \star f = e$ as well?

[I am reading something that claims some $f$ is a unit but then checks it by checking the existence of $g$ such that $f \star g = e$. So really, the question is whether it follows by some general business that $g \star f = e$, or whether one has to do another computation to check the other direction. (And if the former, how general is the general business?) ]