User hewhohungers - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:44:57Z http://mathoverflow.net/feeds/user/16486 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71629/function-theory-of-a-hyperbolic-variable Function theory of a hyperbolic variable HeWhoHungers 2011-07-30T01:04:21Z 2011-07-30T17:30:44Z <p>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 <a href="http://arxiv.org/PS_cache/math-ph/pdf/0507/0507053v2.pdf" rel="nofollow">http://arxiv.org/PS_cache/math-ph/pdf/0507/0507053v2.pdf</a>. 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.</p> <p>Particularly <a href="http://clifford-algebras.org/v8/81/MOTTER81.pdf" rel="nofollow">http://clifford-algebras.org/v8/81/MOTTER81.pdf</a> aksed for a hyperbolic equvalence of Cauchy integral formula. Thus two rather different answers were provided in <a href="http://arxiv.org/PS_cache/arxiv/pdf/0712/0712.0375v1.pdf" rel="nofollow">http://arxiv.org/PS_cache/arxiv/pdf/0712/0712.0375v1.pdf</a> and <a href="http://www.springerlink.com/content/kp44rl074g7187n2" rel="nofollow">http://www.springerlink.com/content/kp44rl074g7187n2</a> . Both articles mention immense applications that follow directly from their formulas, but I havn't been unable to find a single article discussing them.</p> <p>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!</p> http://mathoverflow.net/questions/70457/open-mapping-theorem-for-riemann-surfaces Open mapping theorem for Riemann surfaces HeWhoHungers 2011-07-15T19:26:22Z 2011-07-28T23:35:29Z <p>What restriction must one impose on a Riemann surface M in order for all biholomorphic $f:M\to\mathbb{C}$ to be open mappings, aka mappings of $M$ onto open subsets $f(M)\subset\mathbb{C}$?</p> http://mathoverflow.net/questions/70514/functions-anihilated-by-the-n-times-iterated-cauchy-riemann-operator Functions anihilated by the n-times iterated Cauchy-Riemann operator? HeWhoHungers 2011-07-16T18:43:18Z 2011-07-16T19:01:26Z <blockquote> <p><strong>Possible Duplicate:</strong><br> <a href="http://mathoverflow.net/questions/70515/n-times-iterated-cauchy-riemann-operator" rel="nofollow">n-times iterated Cauchy-Riemann operator</a> </p> </blockquote> <p>Is there a theory that deals with functions that sattisfy the following equation for $n\in\mathbb{N}$? $$\frac{\partial^n f}{\partial \bar z^n}=0$$ EDIT: I have posted the same question in another post also for the reasons stated bellow. Upon posting this question, my web browser suddenly froze and I didn't believe it got posted successively, therefor posting it once more. If such repetition goes against any site rules or regulations, as I would suspect it does, may somebody in power to do so please remove one of the posts. Thank you!</p> http://mathoverflow.net/questions/70462/range-of-biholomorphic-functions-in-mathbbc Range of biholomorphic functions in $\mathbb{C}$ HeWhoHungers 2011-07-15T19:56:47Z 2011-07-15T19:56:47Z <p>How could one prove that among the biholomorphic functions that map a fixed simply-connected open domain $D$ into open subsets of $\mathbb{C}$, there is such a biholomorphic map $f$ for each open simply-connected $U\subset\mathbb{C}$, such that $f(D)=U$? Also I would like to avoid the use of Riemann mapping theorem in such a proof.</p> http://mathoverflow.net/questions/71629/function-theory-of-a-hyperbolic-variable/71653#71653 Comment by HeWhoHungers HeWhoHungers 2011-07-30T15:01:22Z 2011-07-30T15:01:22Z 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 $\frac{1}{2}(\frac{\partial}{\partial x}+j\frac{\partial}{\partial y})$. http://mathoverflow.net/questions/71629/function-theory-of-a-hyperbolic-variable/71653#71653 Comment by HeWhoHungers HeWhoHungers 2011-07-30T14:58:04Z 2011-07-30T14:58:04Z 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. http://mathoverflow.net/questions/71629/function-theory-of-a-hyperbolic-variable/71653#71653 Comment by HeWhoHungers HeWhoHungers 2011-07-30T14:57:12Z 2011-07-30T14:57:12Z ...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. http://mathoverflow.net/questions/71629/function-theory-of-a-hyperbolic-variable/71653#71653 Comment by HeWhoHungers HeWhoHungers 2011-07-30T14:55:07Z 2011-07-30T14:55:07Z 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 $\frac{1}{2}\Big (\frac{\partial}{\partial x}-j{\partial}{\partial y}$