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.


  • 3
    $\begingroup$ 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

1 Answer 1


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..

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.