8
$\begingroup$

The Set of Complex numbers is a field as well as a nice topological space homeomorphic to $\mathbb{R}^2$. But why such a particular interest for this space? For instance what is more special about it than any other $\mathbb{R}^n$? I agree that each function on a complex domain will look like a single variable function and may have a nice definition for a derivative with respect to that single variable where as in $\mathbb{R}^3$ we cannot(?) have such a single variable to differentiate a function with.Particularly Holomorphic functions are as many times differentiable as we want. But is that all? I believe there is more to it but couldn't just get a picture of it. Any insights? Thanks

  • 0
    @Dinesh: Dear Dinesh, please refer to this thread: http://mathoverflow.net/questions/3819/why-do-functions-in-complex-analysis-behave-so-well-as-opposed-to-functions-in2011-09-06

2 Answers 2

4

I'll post this as an answer since I'm not allowed to comment.

As a response to your comment "I never knew multiplicative structure of C was responsible for holomorphicity", you have to use the field structure of the complex numbers to define the derivative via limits of quotients.

You might also wonder whether one can define such derivatives via quotiens in other $\mathbb{R}^n$ for n>2. However, in order to define a quotient one need an algebra structure such that any nonzero element has an inverse, ie, a division algebra structure. Due to classical theorems of Frobenius and Hurwitz this forces n=4 or 8 and corresponding algebra is not commutative (or even associative in the case n=8), which is undesirable if one wants do do calculus.

  • 1
    The Wikipedia articles are well written: [Frobenius Theorem](http://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)) and [Hurwitz Theorem](http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(normed_division_algebras))2011-09-12
3

I think you got the main point: The definition of complex differentiable (holomorphic) functions uses the multiplicative structure of $\mathbb{C}$ in an essential way. It's not just a differentiable function from $\mathbb{R}^2 \to \mathbb{R}^2$. There are no higher-dimensional field extensions of $\mathbb{R}$, so we don't have an analog in higher dimensions.

  • 0
    You can define holomorphicity without the multiplicative structure but it isn't _nearly_ as natural.2011-09-05