I am interested in a problem closely related to a problem stated by Jan Mycielski in his paper Can One Solve Equations in Group? (The American Mathematical Monthly, 1977, http://www.jstor.org/stable/2321255).

Let $p_1, q_1,...., p_m, q_m$ be fixed integers (both positive or negative). I am interested in the image of the mapping $SO(3)\times SO(3) \to SO(3)$ given by $(X, Y)\mapsto X^{p_1} Y^{q_1} X^{p_2} Y^{q_2}\cdots X^{p_m}Y^{q_m}$. By conjugacy and continuity, it is easy to see that the image is the set of all rotations by angles in $[0,\alpha]$ for some $\alpha$ that depends on $p_1, q_1,...., p_m, q_m$. Since $SO(3)$ contains a copy of the free group on two generators, $\alpha$ is always strictly positive.

Mycielski asked whether $\alpha$ is always equal to $\pi$. I am intersted to know whether $\alpha$ is always at least $\pi/2$.

Is there anything new to be said about this problem, or is it still wide open? I checked the papers citing Mycielski's paper, but none of them seem to have a solution.

  • 4
    $\begingroup$ @Watson, the interesting case is when the p's sum to zero and the q's sum to zero. $\endgroup$ Jan 27, 2016 at 6:42
  • 2
    $\begingroup$ If I remember correctly, Borel proved for any connected semisimple algebraic group that the image of this map is Zariski dense. Would not that settle your question, more or less? $\endgroup$ Jan 27, 2016 at 8:45
  • 1
    $\begingroup$ @VladimirDotsenko Borel's result does not help, since for every $\alpha>0$ the set of rotations of angle in $[0,\alpha]$ is Zariski dense. $\endgroup$
    – YCor
    Jan 27, 2016 at 10:11
  • 1
    $\begingroup$ @YCor : yes you're right of course. In the MO question you link, there is a paper of Andreas Thom mentioned in one of the comments (arxiv.org/abs/1003.4093), discussion of Remark 3.6 in that paper is relevant for the case of SO(3), and shows that probably there are more conjectures than actual results for this question. $\endgroup$ Jan 27, 2016 at 10:58
  • 2
    $\begingroup$ If $X$ is a rotation with angle $\beta\in (0,\pi/2)$ then $X^{\lceil \pi/(2\beta) \rceil}$ is a rotation with angle at least $\pi/2$ so $\alpha$ is at least $\pi/2$. $\endgroup$
    – user35593
    Jan 27, 2016 at 13:07

2 Answers 2


Let me collect a number of known results:

i) $\alpha$ can be arbitrarily small, see my paper

Andreas Thom, Convergent sequences in discrete groups, Canad. Math. Bull. 56 (2013), no. 2, 424–433.

ii) There is some interest in estimating how small $\alpha$ can be in terms of the word length, this has been studied in the paper above, but also in

Abdelrhman Elkasapy and Andreas Thom, On the length of the shortest non-trivial element in the derived and the lower central series. J. Group Theory 18 (2015), no. 5, 793–804.

iii) In many cases $\alpha= \pi$. This has been studied in

Abdelrhman Elkasapy and Andreas Thom, About Gotô's method showing surjectivity of word maps, Indiana Univ. Math. J. 63 (2014), no. 5, 1553–1565.

and a more recent preprint

Anton Klyachko and Andreas Thom, New topological methods to solve equations over groups, arXiv:1509.01376

iv) The shortest word that I know for which $\alpha< \pi/2$ is $w=[[[XY X,YXY],[XYX,Y]],[[XYYX,YXXY],[X,Y]]]$

  • $\begingroup$ (From the other post): in i): see esp. Corollary 3.3. In iv) convention is $[x,y]=xyx^{-1}y^{-1}$. $\endgroup$
    – YCor
    Jan 31, 2016 at 15:06

I'll post this as an answer so the question can be marked appropriately.

Corollary 3.3 in the paper Vladimir linked in the comments (arxiv.org/abs/1003.4093) says that $\alpha$ can be arbitrarily small.

  • 1
    $\begingroup$ Let me add that $w = [[[XYX,YXY],[XYX,Y]],[[XYYX,YXXY],[X,Y]]]$ is the shortest word that I know for which $\alpha< \pi/2$. $\endgroup$ Jan 31, 2016 at 14:29
  • $\begingroup$ $[a, b] := aba^{-1}b^{-1}$ $\endgroup$ Jan 31, 2016 at 14:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .