Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I've found quite a number of articles on the basics of function theory in one hyperbolic (split-complex, dual, duplex, motro,..) variable, perhaps the most notable being http://arxiv.org/PS_cache/math-ph/pdf/0507/0507053v2.pdf. What is covered in this and the other articles are more or less only the counterparts of most elementary topics in complex analysis. Are there really no deeper results in this theory or are they just so hard to find? If so, some links or summaries of results would be dearly appriciated.

Particularly http://clifford-algebras.org/v8/81/MOTTER81.pdf aksed for a hyperbolic equvalence of Cauchy integral formula. Thus two rather different answers were provided in http://arxiv.org/PS_cache/arxiv/pdf/0712/0712.0375v1.pdf and http://www.springerlink.com/content/kp44rl074g7187n2 . Both articles mention immense applications that follow directly from their formulas, but I havn't been unable to find a single article discussing them.

And it is quite apparent that hyperbolic Cauchy-like formulas don't yield the percise same results as in complex analysis since it is easy to show not only that hyperbolic holomorphic functions are not allways analytic, but that they need not be even $C^2$! So could somebody explain what those mentioned direct implications of hyperbolic Cauchy formulas and also explain why are there two formulas to begin with? Thank you!

share|improve this question
    
You would probably like to read this article on "double numbers" (yet another name for hyperbolic numbers) researchgate.net/publication/… I found it useful as it has considerably more care than the article you linked here. In particular, it is important to note $||zw|| < \sqrt{2}||z|| ||w||$ indicating Euclidean length and hyperbolic algebra product. Furthermore, a vastly more general version of this was set out by Ketchum in his 1928 paper (which I'm currently trying to believe). In any event, I share your pain. –  James S. Cook Aug 31 at 18:26
    
In particular, Eq. 5.89 on page 19 of the article arxiv.org/PS_cache/math-ph/pdf/0507/0507053v2.pdf assumes that $||zw||=||z||||w||$ to prove the $ML$-theorem... I think that can be repaired with the insertion of some $\sqrt{2}$'s at the appropriate points. –  James S. Cook Aug 31 at 18:30

1 Answer 1

The problem is that "adjoining" a "number" $j$ to $\mathbb R$ such that $j^2=+1$ rather than $i^2=-1$ gives, by Sun-Ze's theorem, $$ \mathbb R[j] \approx \mathbb R[x]/\langle x^2-1\rangle \approx \mathbb R[x]/\langle x-1\rangle \oplus \mathbb R[x]/\langle x+1\rangle \approx \mathbb R\oplus \mathbb R $$ In particular, there are $0$-divisors, such as $(0,1)\cdot (1,0)=(0,0)$. A corresponding change of coordinates gives a analogue of the Cauchy-Riemann operator: just $\frac{\partial}{\partial x}\pm \frac{\partial}{\partial y}$.

Notably, these two operators are the factors of the one (spatial) dimensional wave equation, whose analytic features/failings caused so much consternation/interest pre-1800, namely, any function of the form $F(x,y)=f(x-y)$ is apparently annihilated by $\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$, even when $f$ is not as smooth as one might think it ought to be.

This is in extreme contrast to the $i^2=-1$ situation, where the Cauchy-Riemann operator's vanishing gives a definite constraint (a.k.a. "ellipticity").

Edit: [thx for helpful edit-corrections...] If one uses the "usual" norm/length in the denominator, in the definition of "derivative", this does avoid zero-divisors, but changes the thing to being something else entirely, I think.

The fact that $\mathbb R$ with $j$ such that $j^2=+1$ adjoined is not a field is inescapable, and quite unlike the case of complex numbers. On another hand, it is certainly true that Clifford analysis is useful, if not quite in this fashion. Yes, "Dirac operators" have many roles, as factoring second-order differential operators into first-order. Yes, the Laplacian in $\mathbb R^2$ factors into Cauchy-Riemann and its conjugate, and the one-dimensional wave equation (as noted above) factors into two "real" linear operators. This is the goal, actually. For higher dimensions, these operators cannot factor into scalar differential operators, but Clifford-algebra-valued ones. Nevertheless, I don't think it's quite that the underlying "scalars" are made more exotic, but, rather, are enhanced by allowing operator-valued functions.

share|improve this answer
    
The structure $\mathbb{R}[x]/(x-1)$ corresponds to $z=x+jy$ for $x,y$ real. In this notation zero divisors are percisely those numbers with $x^2-y^2=(x+y)(x-y)=0$. Than you can construct a basis alternative to $\{1,j\}$ from zero divisors $\{(1+j)/2,(1-j)/2\}$ so that $$z=x+jy=\frac{1+j}{2}(x+y)+\frac{x-y}{2}(x-y)$$ Now in this form, the CR operator indeed gets the form $\frac{1+j}{2}\frac{\partial}{\partial (x-y)}+\frac{1+j}{2}\frac{\partial}{\partial (x-y)}$ which can easily be proven equivalent to the form $\frac{1}{2}\Big (\frac{\partial}{\partial x}-j{\partial}{\partial y}$ –  HeWhoHungers Jul 30 '11 at 14:55
    
...to the form $\frac{1}{2}(\frac{\partial}{\partial x}-j\frac{partial}{\partial y})$. Your account is incorrect because of the following fact: consider the operator $$\frac{1+j}{2}\frac{\partial}{\partial (x+y)}+\frac{1-j}{2}\frac{\partial}{\partial (x-y)}$$ The function annihilated by it are the hyperbolic equivalent of antiholomorphic. –  HeWhoHungers Jul 30 '11 at 14:57
    
Also if one defines differentiability for functions in $\mathbb{R}[j]$ as the existance of such $f'(z)$ for every direction $$\lim_{h\to 0}\frac{f(z+h)-f(z)-f'(z)h}{\lVert h\rVert}=0$$ Here $\lVert h\rVert$ is the standard euclidean norm on $\mathbb{R}^2$, using which prevents the problem of zero devisors with the definition of derivatives. Notice that the condition that $f'(z)$ exists is the same as requiering CR operator to annihilate the function at $z$, simmilarly as in complex analisys. –  HeWhoHungers Jul 30 '11 at 14:58
    
It does however not follow that such functions need be analytic or even $C^{2}$. Therefore the hyperbolic CR operator perhaps doesn not impose as much structure on functions it annihilates as does its complex counterpart, but it gives some structure none the less. **please forgive me for consistantly failing to write this down correctly: what I wanted to say was that the CR operator is equal to $\frac{1}{2}(\frac{\partial}{\partial x}+j\frac{\partial}{\partial y})$. –  HeWhoHungers Jul 30 '11 at 15:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.