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Apr 13, 2017 at 12:19 history edited CommunityBot
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Oct 4, 2013 at 19:38 comment added James S. Cook where $\mathcal{H}$ is mostly a convenience since the direct product of $\mathbb{R}$ with itself is isomorphic. In the general case, the assumption of $\mathcal{A}$-differentiability necessitates $n^2-n$ Cauchy Riemann equations, so the systems of PDEs which this idea captures are probably only of a particularly special type.
Oct 4, 2013 at 19:35 comment added James S. Cook @WillieWong that might be a useful line of inquiry. Here's my thinking: if I can identify a certain PDE as allowing hyperbolic-calculus representation then it is likely possible to solve the problem in the hyperbolic notation. So, to make this interesting, I'd like to figure out how to take a given PDE and identify the algebra which might allow a simplification like the one illustrated in my post. Perhaps I should take a long look at Clifford Algebras. I know all the commutative, semi-simple algebras over $\mathbb{R}$ are direct sums of of $\mathbb{R}, \mathbb{C}$ and $\mathcal{H}$..
Oct 4, 2013 at 8:27 comment added Willie Wong The system $u_x = v_y$ and $u_y = v_x$ implies that $u,v$ solve the wave equation. The introduction of the hyperbolic numbers (which is the natural Clifford algebra associated to $\mathbb{R}^{1,1}$) allows you to factor the wave equation into the analogue of the Dirac equation you wrote down. So in the end you end up with $\partial w = w^2$ and $\bar{\partial} w = 0$ where $\partial$ is some notion of a Dirac operator. But all these is after-the-fact (and you of course recognize its similarity to the case of Complex Analysis), so I don't know if it helps you at all.
Oct 4, 2013 at 4:51 history asked James S. Cook CC BY-SA 3.0