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I know almost nothing about this field, so my following question may be stupid.

We know in a free group F, if $ax^{-1}a^{-1}x=1$, then $x=t^{k}, a=t^{l}$ for some t.

Now consider a more general equation, $a_1x^{-1}a_1^{-1}a_2xa_2^{-1}...a_{2n-1}x^{-1}a_{2n-1}^{-1}a_{2n}xa_{2n}^{-1}=1$.

I guess this equation has only obvious solutions, i.e., this equation splits into n equations like $a_ix^{-1}a_i^{-1}a_{i+1}xa_{i+1}^{-1}=1$ (of course, there are other ways to split the equation.) How to prove it has no "nontrivial" solutions?

It seems Wicks forms might work, but I think it is too powerful to deal with this problem. I feel this should be a consequence of a classical result.

I would appreciate if anyone can give me hints or references.

Thank you.

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  • $\begingroup$ Assuming you really mean to keep $x$ fixed, you are working within a subgroup generated by finitely many conjugates of $x.$ A subgroup of a free group is free, so I think that should take you a long way (bearing in mind that some of the conjugates may be the same element in different guises). $\endgroup$
    – Geoff Robinson
    May 18, 2014 at 9:39
  • $\begingroup$ Any relationship between those $a_i$'s? $\endgroup$
    – Alexey Kvashchuk
    May 18, 2014 at 16:23
  • $\begingroup$ Yes， x is a fixed element. no relations between any $a_i$ and $a_j$. $\endgroup$
    – guest123
    May 19, 2014 at 1:25
  • $\begingroup$ How about $ax^{-1}a^{-1}bxb^{-1}cx^{-1}c^{-1}x=1$, $a,b,c,x\in $ a free group of rank 3? Is there any easy way to solve this equation? $\endgroup$
    – guest123
    May 19, 2014 at 1:32
  • $\begingroup$ Whether you are solving in a free group of rank 3 or 4 is immaterial. It is equally difficult. Also, in your general equation you have even number of coefficients, but in the example you have here you have 3 coefficients. Finally, if you consider this example equation, then all I can show is that there are no line solutions (unless <a,b,c> is not of rank 3 i.e. there are some relationships between the coefficients). On the other hand this equation is consistent since x=1 is a solution. Whether there are any other solutions I don't know. $\endgroup$
    – Alexey Kvashchuk
    May 19, 2014 at 14:02

2 Answers 2

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This is not easy. The complete modern description of the solution sets (Lyndon original result) was refined, for example, in the paper by Remeslennikov and Chiswell, and then a bit later by Myasnikov et. al.

The general result states that the solution set of an arbitrary one-variable equation can be represented as a sum of irreducible components which are a) points, b) lines, and c) shifted lines. Point is a set $V=\{ a\}$, line is a centralizer of some element i.e. $V=C(a)$, and shifted line is a set of the form $V=C(a)b$.

This paper ( http://arxiv.org/pdf/math/0607176.pdf ) also provides an algorithm for finding the solutions.

For your equation, all I can say is that if it has a line as a solution then it is the centralizer of one of the coefficients.

Finding point solutions is the hardest part.

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  • $\begingroup$ Thanks you, Alexey. Now I realize I underestimated the difficulty of this kind of problems. It might have some nontrivial solutions probably.. $\endgroup$
    – guest123
    May 19, 2014 at 1:28
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This seems to be resolved completely in R. C. Lyndon's 1960 paper.

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  • $\begingroup$ Thank you, Igor. I am reading this paper. It seems this paper provides a solution to the general equations. Is there any statement concerning this specific kind of equation (consisting of conjugates of $x$ and its inverse )? $\endgroup$
    – guest123
    May 19, 2014 at 1:24
  • $\begingroup$ @guest123 it seems that on the very first page Lyndon's very first lemma is very similar to your conjecture, so I am assuming that this case should be tractable, though not being an expert in this specific field, I might be missing a subtlety. $\endgroup$
    – Igor Rivin
    May 19, 2014 at 2:21
  • $\begingroup$ HI, Igor: the first page of this paper is introduction; do you mean proposition 1 on the 2nd page? $\endgroup$
    – guest123
    May 19, 2014 at 3:29
  • $\begingroup$ @guest123 yes, that's what I meant, sorry. $\endgroup$
    – Igor Rivin
    May 19, 2014 at 3:31
  • $\begingroup$ I have no clue how to apply that proposition. Even in the cas e $n=2$, $a_1x^{-1}a_1^{-1}a_2xa_2^{-1}a_3x^{-1}a_3^{-1}x=1$, there are too many possiblities... $\endgroup$
    – guest123
    May 19, 2014 at 4:06

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