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I am browsing through the book "Abel’s Theorem in Problems and Solutions" which outlines a topological proof of the Abel-Ruffini theorem (due to V.Arnold).

For a polynomial $p$, one investigates the monodromy groups of the Riemann surface of $\omega(z)$, where $p(w(z))+z=0$. Theorem 11 in section 2.13 then says:

If the multi-valued function $\omega(z)$ is representable by radicals, its monodromy group is solvable.

Remark 1 in the corresponding section says that one is also allowed to use, besides radicals, for example all analytic single-valued functions ($e.g. \sin{z}$) and also $\log{z}$. Unfortunately, I find no further explanation in the book of how this works. Is there a reference where this is outlined in more detail?

In this setting, as indicated in this post, there is a connection/isomorphism bewteen the monodromy group and Galois groups. Can one conlude with this "isomorphism" that: If the Galois group is not solvable, the roots of $p(x)$ are not expressable in radicals and trigonometric functions?

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2 Answers 2

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Let $F=\Bbb{C}(z)$ and a polynomial $g\in F[X]$, factorize $g(X)=\prod_j (X-w_j)$ and $K=F(w_1,\ldots,w_d)$ its splitting field.

You are looking at $g=p(X)+z$ with $p\in \Bbb{C}[X]$.

Then it is one of the main theorem of Galois theory that $w_j$ is radical over $F$ iff $Gal(K/F)$ is solvable.

Radical means that $w_j$ has an expression which is a tree of additions, multiplications, quotients and $n$-th roots of elements of $F$.

In every case, away from the finitely many poles and branch points located at the zeros/poles of the coefficients of $g$ and $Disc(g)\in F$ then $w_j$ is an analytic function of $z$. When $g$ is not solvable then this analytic function has no simple expression (this happens for most polynomial of degree $\ge 5$).

There are no trigonometric functions, $\exp $ and $\log$ in there: only algebraic functions.

Next, $Gal(K/F)$ is the monodromy group of $w_1$. Take a base point $z_0\in \Bbb{C}$ where all the $w_j=w_j(z)$ are analytic, then $\gamma \in Gal(K/F)$ can be represented as a closed curve $z_0\to z_0$ and $\gamma(w_j)$ is the function analytic near $z_0$ obtained by continuing $w_j$ analytically along $\gamma$.

This is because if you take $u_1,\ldots,u_M$ all the functions analytic near $z_0$ obtained by analytic continuation of $w_1$, then (since the analytic continuation commutes with the algebraic operations) all of them are roots of $g$, and $\prod_{m=1}^M (X-u_m(z))$ is a polynomial whose coefficients are locally analytic away from finitely many poles and branch points, and those coefficients stay the same under analytic continuation along closed-loops. Together with the polynomial growth at $\infty$ and near the branch points this implies that the coefficients of $\prod_{m=1}^M (X-u_m(z))$ are meromorphic on the Riemann sphere: they are in $F$. Whence $g=\prod_{m=1}^M (X-u_m(z))$.

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  • $\begingroup$ Thank you for your answer. "When g is not solvable then this analytic function has no simple expression." Here by "simple expression" you mean an expression in terms of elementary functions (en.wikipedia.org/wiki/Elementary_function)? $\endgroup$
    – Breaking M
    Feb 8, 2020 at 20:04
  • $\begingroup$ Yes, any kind of expression simpler than "it is a root of that irreducible polynomial". Often there are special functions generating the roots of families of polynomials, for example the non-solvable quintic has its roots generated by elliptic functions, but this is far from a simpler expression. $\endgroup$
    – reuns
    Feb 9, 2020 at 13:05
  • $\begingroup$ Thanks, that's what I was thinking. And why does "g is not solvable" imply that no such simple expression exists? As far as I can see, Galois theory rules out a radical expression, but how does one extend this result to the other elementary functions? $\endgroup$
    – Breaking M
    Feb 9, 2020 at 17:34
  • $\begingroup$ The trigonometric functions are defined by differential equations, not quite related to algebraic functions (even if there are a lot of algebraic relations between those solutions of differential equations, which is why the quintic has its roots given by inverting an elliptic function). $\endgroup$
    – reuns
    Feb 9, 2020 at 17:41
  • $\begingroup$ Alright, if "g is not solvable", then it is possible to express roots by elliptic functions, even though they cannot be expressed by means of radicals. I'm sorry, but I still don't get it: Why we can, based on the non-solvability of g, exclude that a root-formula exists if we are allowed to use, resides radicals, also elementary functions (exp, log, tan)? $\endgroup$
    – Breaking M
    Feb 9, 2020 at 19:32
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Can one conlude with this "isomorphism" that: If the Galois group is not solvable, the roots of p(x) are not expressable in radicals and trigonometric functions?

Yes, one can conclude that.

The references you are looking for are the publications of Askold Khovanskii.

[Khovanskii 2014]:
"Vladimir Igorevich Arnold discovered that many classical questions in mathematics are unsolvable for topological reasons. In particular, he showed that a generic algebraic equation of degree 5 or higher is unsolvable by radicals precisely for topological reasons. Developing Arnold’s approach, I constructed in the early 1970s a one-dimensional version of topological Galois theory. According to this theory, the way the Riemann surface of an analytic function covers the plane of complex numbers can obstruct the representability of this function by explicit formulas. The strongest known results on the unexpressibility of functions by explicit formulas have been obtained in this way.
...
The monodromy group of an algebraic function is isomorphic to the Galois group of the associated extension of the field of rational functions. Therefore, the monodromy group is responsible for the representability of an algebraic function by radicals. However, not only algebraic functions have a monodromy group. It is defined for the logarithm, arctangent, and many other functions for which the Galois group does not make sense. It is thus natural to try using the monodromy group for these functions instead of the Galois group to prove that they do not belong to a certain Liouville class. This particular approach is implemented in one-dimensional topological Galois theory ..."

Khovanskii:
"If the monodromy group of an algebraic function is solvable one can represent it via radicals. But if it is unsolvable one cannot represent it by a formula which involves meromorphic functions and elementary functions and uses integration, composition and meromorphic operations."

[Khovanskii 2021]:
"Theorem 12. If the monodromy group of an algebraic function is unsolvable then one can not represent it by a formula which involves meromorphic functions and elementary functions and uses integration, composition and meromorphic operations."
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[Khovanskii 2014] Khovanskii, A.: Topological Galois Theory - Solvability and Unsolvability of Equations in Finite Terms. Springer 2014

[Khovanskii 2019] Khovanskii, A.: One dimensional topological Galois theory. 2019

[Khovanskii 2021] Topological Galois Theory - Slides 2021, University Toronto

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