On algebras of harmonic quaternion fields in ${\mathbb R}^3$

TL;DR

This paper extends a classical complex analysis result to three dimensions, characterizing harmonic quaternion fields in the unit ball and describing their multiplicative linear functionals as point evaluations.

Contribution

It introduces a 3D analog of the classical disk algebra characterization, defining harmonic quaternion fields and their characters, and proves a homeomorphism with the unit ball.

Findings

The set of quaternionic characters corresponds exactly to point evaluations within the ball.

Each commutative subalgebra is isometrically isomorphic to the disk algebra.

The space of H-characters is homeomorphic to the unit ball.

Abstract

Let ${\mathscr A}(D)$ be an algebra of functions continuous in the disk $D=\{z\in{\mathbb C}\,|\,\,\,|z|\leqslant 1\}$ and {\it holomorphic} into $D$. The well-known fact is that the set ${\mathscr M}$ of its characters (homomorphisms ${\mathscr A}(D)\to\mathbb C$) is exhausted by the Dirac measures $\{\delta_{z_0}\,|\,\,z_0\in D\}$ and a homeomorphism ${\mathscr M}\cong D$ holds. We present a 3d analog of this classical result as follows. Let $B=\{x\in{\mathbb R}^3\,|\,\,|x|\leqslant 1\}$. A quaternion field is a pair $p=\{\alpha,u\}$ of a function $\alpha$ and vector field $u$ in the ball $B$. A field $p$ is {\it harmonic} if $\alpha, u$ are continuous in $B$ and $\nabla\alpha={\rm rot\,}u,\,{\rm div\,}u=0$ holds into $B$. The space ${\mathscr Q}(B)$ of such fields is not an algebra w.r.t. the relevant (point-wise quaternion) multiplication. However, it contains the commutative algebras ${\mathscr A}_\omega(B)=\{p\in{\mathscr Q}(B)\,|\,\,\nabla_\omega\alpha=0,\,\nabla_\omega u=0\}\,\,(\omega\in S^2)$, each ${\mathscr A}_\omega(B)$ being isometrically isomorphic to ${\mathscr A}(D)$. This enables one to introduce a set ${\mathscr M}^{\mathbb H}$ of the $\mathbb H$-valued linear functionals on ${\mathscr Q}(B)$ ({\it $\mathbb H$-characters}), which are multiplicative on each ${\mathscr A}_\omega(B)$, and prove that ${\mathscr M}^{\mathbb H}=\{\delta^{\mathbb H}_{x_0}\,|\,\,x_0\in B\}\cong B$, where $\delta^{\mathbb H}_{x_0}(p)=p(x_0)$.