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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.

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 mentioned 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" description.

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 surface" of its "inverse" exists and provides a manifold $\hat M$ which can be compactified as a "Riemann surface" $M$ 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.

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.

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

This map is holomorphic, locally biholomorphic outside the diagonals ${r_j=r_i}$, $i\neq j$ (corresponding to multiple roots). This is a holomorphic covering, and sections gives the expected answer. Please notice that these section cannot be analytic at polynomials with multiple roots, you instead have a multivalued map. This map is well thought-of as an orbifold covering, I guess.

The topological structure of your covering is that of the complement of the hyperplane arrangement given by the diagonals, so the 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.

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 surface" of its "inverse" exists and provides a manifold $\hat M$ which can be compactified as a "Riemann surface" $M$ 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.