First of all, wasn't sure what could be a good title for this question. If mods think of a better name, pls feel free to change it...

Let $F$ be a (non-abelian) free group of finite rank, a vector $\textbf{a} = \left( a_1,\ldots , a_{n+1} \right)$ of non-trivial elements in $F$, and a non-trivial element $u\in F$ (which we can assume to be root (i.e. not a proper power)).

Define $$W(X, \textbf{a}) = a_1 X^{-1} u^{k_1} X a_2 X^{-1} u^{k_1} X a_2 \ldots a_n X^{-1} u^{k_n} X a_{n+1} ,$$

where $k_i \in \mathbb{Z}$.

Immediately notice that when $V$ and $T$ are from the same conjugacy class of $u$ (i.e. when $V^{-1}uV=T^{-1}uT$, or equivalently $T=u^{\alpha}V$ for some $\alpha\in \mathbb{Z}$) then $W(T, \textbf{a}) = W(V, \textbf{a})$.

$\textit{Question}$: How many different elements $V \in F$ satisfy $W(V, \textbf{a}) = u^{l}$ for some power $l\in \mathbb{Z}$, where by different we mean elements from different conjugacy classes of $u$. In other words, what is the maximal number of different (in the above sense) solutions of the equation $$ \[ W(X, \textbf{a}), u \] = 1 \ \ ?$$

$\textit{Wishful guess}$: My guess is that the answer is at most $n$.

$\textit{Remark.}$ I am looking for an answer that is a function of $n$ only.

The reason for the above remark is that it is known that the number of "solutions" (irreducible algebraic components) is finite, however this number depends on the coefficients (i.e. elements $a_1, \ldots, a_{n+1}, u$).

Now goes the proof for the case of $n=1$. Suppose that there are two elements $W$ and $V$ such that $$a_1 V^{-1} u^{k} V a_2 = u^{l} \ \ \text{and} \ \ a_1 W^{-1} u^{k} W a_2 = u^{m}$$ for some $l,m\in \mathbb{Z}$.

Three cases: 1) $l=m$. We have then $V^{-1} u^{k} V = W^{-1} u^{k} W $, so that $V$ and $W$ are from the same conjugacy class of $u$.

2) $|l-m| \geq 2$. We have then $$a_1 V^{-1} u^{k} V W^{-1} u^{-k} W a_1^{-1} = u^{l-m}.$$ There are two ways to proceed here. First one is to notice that the left-hand side can be written as a product of two powers so that $$\widetilde{V}^{k} \widetilde{W}^{-k} = u^{l-m}$$, and use a result that says that if $x,y,z\in F$ and $x^p y^q = z^r$ where $p, q, r \geq 2$ then $x,y$ and $z$ pairwise commute. This implies that $ \widetilde{V}$, $\widetilde{W}$, and $u$ commute. Then it is easy to show that $u=1$.

Another way to show that is (as suggested by Henry (HW) in a related question here ) is to represent the left side as a commutator, so that $$\[ VW^{-1},\ u^k \]^{Wa_1^{-1}} = u^{l-m} ,$$ and then either use an old fact that in a free group non-trivial commutator can not be a proper power or a big gun result like Duncan-Howie which implies the same since scl is always at least $1\over 2$.

So that the case $|l-m| \geq 2$ is not possible as it implies $u=1$.

3) $|l-m|=1$. Let's assume $l=m+1$. We have then

$$\[ VW^{-1},\ u^k \]^{Wa_1^{-1}} = u.$$

Here comes a fun part. Element $u$ then clearly belongs to the derived subgroup $F'=F_1$. But then, by the above equation, since $u^k$ obviously belongs to $F'$ as well, the left side of the equation belongs to the second element $\[F, F_1\] = F_2$ of the lower central series of $F$. By induction,

$$u\in \cap_{i=1}^{\infty} F_{i}, $$

where $F_i$ is the i-th element of the lower central series. But the intersection of the lower central series of a free group of finite rank is trivial. Hence $u=1$.

I am not able to generalize the approach or apply similar ideas to the general case of any $n$.

Just wanted to hear any suggestions on how to attack this problem, as well as thoughts on whether it is hopelessly complex, or (who knows!) trivial :)



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