# Math Help - inverse function of polynomials

1. ## inverse function of polynomials

Hi all,

I have a polynomial p(z), say of degree n; view it as an n-to-1 map of the complex Riemann sphere to itself. Can I always construct an inverse which is defined on large piece of the complex plane? of course I can define the inverse locally around each point where $p'(z)\ne 0$, but I want global inverses

example: z^2, its inverse is $\sqrt z$ which we can define e.g. on $C\setminus [\infty,0]$

I hope to get that for any polynomial there exists a cut (or a few) of the complex plane such that the inverse is defined there globally

any ideas will be appreciated...

2. Originally Posted by choovuck
Hi all,

I have a polynomial p(z), say of degree n; view it as an n-to-1 map of the complex Riemann sphere to itself. Can I always construct an inverse which is defined on large piece of the complex plane? of course I can define the inverse locally around each point where $p'(z)\ne 0$, but I want global inverses

example: z^2, its inverse is $\sqrt z$ which we can define e.g. on $C\setminus [\infty,0]$

I hope to get that for any polynomial there exists a cut (or a few) of the complex plane such that the inverse is defined there globally

any ideas will be appreciated...
Hi,

I don't have a full proof yet, but I'm almost sure the following holds:

Let $C=\{P(z)|P'(z)=0\}$. On any simply-connected open subset of $\mathbb{C}\setminus C$, it is possible to define an inverse of $P$.

In terms of "slit plane", the inverse may thus be defined on the whole plane except for a simple path joining all the points of $C$ and going to infinity. The slit could more generally be any "planar tree" containing the elements of $C$ and $\infty$.

I have a few ideas for a proof but there are still parts missing. I'll post it if I can complete it.

3. thanks. that's what I've been getting too. it looks like all we need is the monodromy theorem (Monodromy theorem - Wikipedia, the free encyclopedia), with only exception that here we can define f in a small neighborhood of each point rather than along every path... I feel this should be easy to deduce but I'm not getting it yet

4. Originally Posted by choovuck
thanks. that's what I've been getting too. it looks like all we need is the monodromy theorem (Monodromy theorem - Wikipedia, the free encyclopedia), with only exception that here we can define f in a small neighborhood of each point rather than along every path... I feel this should be easy to deduce but I'm not getting it yet
Thanks, this theorem was a part I missed.

Meanwhile, I've found a wonderful course on Complex Variables by Joseph L. Taylor from University of Utah, which provides a well-written answer.

Look at Chapter 7. The result is Theorem 7.3.4. Using analytic continuation along curves, and the monodromy theorem, this theorem allows to lift an analytic function through an analytic covering, on a simply-connected subset of $\mathbb{C}$. Explanation below.

In our case, the polynomial $P$ is a ( $n$-sheeted) analytic covering from $V=\mathbb{C}\setminus\{z|P'(z)=0\}$ to $W=\mathbb{C}\setminus\{P(z)|P'(z)=0\}$, which means that for any point $z\in W$, there is an open neighbourhood $A$ of $z$ in $W$ and $n$ disjoint open subsets $B_1,\ldots,B_n$ of $V$ such that, for $i=1,\ldots,n$, $P$ is a bijective map from $B_i$ to $A$. This is a consequence of the inversion theorem applied at every pre-image of $z$ (with $A$ being the intersection of the neighbourhoods provided by each pre-image).

Theorem 7.3.4 thus says that for any analytic map $f:U\to W$ where $U$ is simply connected, it is possible to find an analytic map $g:U\to V$ such that $f(z)=P(g(z))$, i.e. we can choose an element $g(z)\in P^{-1}(f(z))$ in a "coherent" way on all $U$.

The answer to your question is obtained as a particular example if you consider $f:W\to W$ to be the identity map.

The proof does not get any simpler in this case (simply because the general case is a direct consequence of this one), and anyway the proof is neatly written, I read it in detail, it is pretty natural.

I wish you'll enjoy the reading.