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It seems similar questions have been asked about these notes by Leo Goldmakher (linked below) which give a topological proof by Arnold of the insolvability of the quintic by radicals. I think I understand everything except for how to prove exercise 1 on p. 4.

For a bit of context, $\mathcal{F}_n$ is used to denote the space of complex polynomials (with leading coefficient equal to 1) without repeated roots. So a polynomial is of the form $z^n+a_{n-1}z^{n-1}+...+a_1z + a_0$; it is determined by the coefficients and so such polynomials may be regarded as a subset of $\mathbb{C}^n$. The roots may be regarded as tuples in $\mathbb{C}^n$.

For example, if we take $n=2$, we have some quadratics and there is a quadratic formula which gives us roots. However, if you move the coefficients in a large loop (so take a loop in $\mathcal{F}_2$) and keep track of the roots, it's possible to permute the roots. This means that the quadratic formula does not always map a loop of $\mathcal{F}_2$ to a loop in $\mathbb{C}^2$. This is, of course, because square roots aren't really continuous functions on $\mathbb{C}$. We need branch cuts and the theory of Riemann surfaces originates at this point.

Now, here is the exercise.

Exercise 1: Suppose $\gamma_1$ and $\gamma_2$ are two loops based at the same point in $\mathcal{F}_n$, and pick any continuous function $f: \mathcal{F}_n \to \mathbb{C}$. Then for any $\alpha \in \mathbb{Q}$ the image of $f(p)^\alpha$ as $p$ traverses the commutator loop $[\gamma_1, \gamma_2]$ is also a loop in $\mathbb{C}$.

This exercise seems to be saying there is something special about commutators; I don't think the statement is true for arbitrary loops and it seems to me that the $\alpha$ exponent should give rise to the same issues as we have with square roots not being continuous. So I'm somewhat at a lost on how to prove the statement though I imagine that it should be rather elementary. Any help is appreciated.

https://web.williams.edu/Mathematics/lg5/394/ArnoldQuintic.pdf

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    $\begingroup$ There's a really excellent video about Arnold's proof here (by math.SE user not all wrong, by the way). Maybe that can help you sort it out? $\endgroup$ Oct 16, 2021 at 13:35
  • $\begingroup$ @HansLundmark Thanks for sharing the video. I am still curious about this particular exercise though and I don't think the video discusses it. $\endgroup$
    – inkievoyd
    Oct 17, 2021 at 15:05
  • $\begingroup$ @inkievoyd: Start at 22:47 of the video; the climax (and essentially a restatement of the counting argument you give below) is at 26:53 or so. The video calls the function $r$ instead of $f$, but the idea seems the same. $\endgroup$
    – Brian Tung
    Dec 17, 2021 at 0:21

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I think one argument goes like this. We can just focus on $\alpha = 1/n$ and consider loops $\gamma_1,\gamma_2$ such that $f(\gamma_1),f(\gamma_2)$ are loops around zero. Otherwise, the $1/n$ isn't an issue.

WLOG, let's suppose that we may begin the image of the loops at $1 \in \mathbb{C}$. Then $f(\gamma_1)^{1/n}$ must have endpoint on some $n$th root of unity, say $e^{2\pi ki/n}$ for some $0 \leq k \leq n-1$. Similarly, $f(\gamma_2)^{1/n}$ ends at some other $n$th root of unity, say $e^{2\pi \ell i/n}$. Then the path traced out by $f([\gamma_1,\gamma_2])^{1/n}$ goes $k$ steps, then $\ell$ steps, then back $k$, then back $\ell$. This brings us back to $1 \in \mathbb{C}$ and hence, is a loop..

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  • $\begingroup$ Oh, that's clever. Ultimately, it's a counting argument... $\endgroup$ Oct 17, 2021 at 23:58

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