# The holomorphic version of Galois theory

We identify the space of polynomials of degree n with $\mathbb{C}^{n+1}-\mathbb{C}^{n}$, that is an $n+1$ tuple $(a_{n},a_{n-1},\ldots,a_{0})$ with $a_{n} \neq 0$ is identified with $p(z)=a_{n}z^{n} +\ldots a_{1}z+a_{0}$. In this question we search for a holomorphic representation for the roots of P. That is a holomorphic maps which send each P to its roots. Motivating by the special case n=2 and the "Radical formula" for the roots, it is natural to search for an appropriate version of "Riemann surface of radical-type functions". So here is our question, precisely:

Question:

Does there exist a $n+1$ dimensional complex manifold M with a covering space structure $\pi :M \rightarrow \mathbb{C}^{n+1}-\mathbb{C}^{n}$ and a holomorphic function $f:M \rightarrow \mathbb{C}^{n}$, such that for every $\tilde{P} \in M$ with $\pi(\tilde{P})=P$, all n roots of $P$ is arranged in the n_tuple $f(\tilde{P})$?

Remark:
Since "Galois Theory " is an obstruction for existence of a radical (algebraic) formulation for the roots, it was natural that we search for a holomorphic analogy

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The thing you are asking was much studied in connection with Hilbert Problem 13. The roots of a polynomial of degree exactly $d$ form an unordered $d$-tuple. The set of unordered $d$-tuples is called the configuration space. It is the factor of $C^d$ (or $P^d$) over the action of permutation group. It is equivalent to the space of polynomials of degree exactly $d$ modulo multilication by a non-zero constant. One recent reference is http://arxiv.org/pdf/math/0403120v3, and it contains many other references. The version of Hilbert problem 13 asks whether this function, mapping a polynomial to its roots, can be represented as a composition of functions of fewer number of variables.

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Say $a_n=1$. You can obtain this map as a section of the map $\sigma$ sending the $n$-tuple of the roots $(r_1,\ldots,r_n)$ to the coefficients of the polynomial using the symmetric polynomials, corresponding to the equality $$\prod _{j=1}^n (z-r_j) = \sum_{j=0}^n a_j z^j$$

This map is holomorphic, locally biholomorphic outside the union $\Delta$ of the diagonals ${r_j=r_i}$, $i\neq j$ (corresponding to multiple roots). $\sigma$ is a holomorphic covering of degree $n!$ outside $\Delta$, and the Riemann manifold of its "inverse" exists and provides a manifold $\hat M$ which can be compactified as a Riemann manifold $M$ for "$\sigma^{-1}$" (since $\sigma$ is polynomial).

The topological structure of the covering space $\hat M$ is that of the complement of the hyperplanes arrangement given by $\Delta$, so its fundamental group will be a braid group.

As you mentionned what you can write down is limited by Galois theory, so you're not going to have anything "explicit" starting from degree 5, so I'm afraid you'll have to be satisfied with the above "inverse" describtion.

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Thank you.puting $a_{n}=1$ we simplify the question as follows: we identify all monic polynomial of degree n with $\mathbb{C}^{n}$. We search for covering space $M \rightarrow \mathbb{C}^{n}$. It seems that I donot underestand something. Does your covering map cover wholle C^n? could you please more explain? –  Ali Taghavi Dec 11 '13 at 12:41
According to the original formulation of my question we search for complex manifold M (Not orbifold)So do you think that the answer of my question is "No"? I mean that is it possible to prove that there is no a complex manifold with those properies? –  Ali Taghavi Dec 11 '13 at 13:02
I think that the radical formula has a Riemann surface so this Riemann surface is the desired M –  Ali Taghavi Dec 11 '13 at 13:12
@AliTaghavi: See edited answer. I removed all my comments since they were incorporated in the edit. –  Loïc Teyssier Dec 11 '13 at 13:25
There is no any obstruction for compactification to obtain a complex manifold again? There is no any obstruction for holomorphic extension? I would appreciate if you explain more clear. Where you used the polynomial ness of $\sigma$. Could you please give me a reference for some background? –  Ali Taghavi Dec 11 '13 at 15:01