Length of a product of conjugates of an element in a free group

Question

Let $G$ be a free group generated by a set $S$. For $g\in G$, let $l(g)$ be the length of $g$ with respect to $S$.

Now for $a\in G$ and $g_1,\dotsc,g_n\in G$, let $$T=g_1^{-1}ag_1g_2^{-1}ag_2\dotsm g_n^{-1}ag_n.$$ We assume that $a$ is cyclically reduced; i.e., $l(a)$ is the minimum of the set $\{l(hah^{-1}): h\in G\}$.

What can we say about $l(T)$ in terms of $l(a)$? Is it known that $l(T)\geq l(a)$? (I conjectured this, but can't prove it.)

Background:

1. The length $l(a^n)$ in terms of $l(a)$ is easily determined. Products of conjugates of $a$ is a natural generalization of powers of $a$. Thus it's natural to ask the relation of the lengths with respect to this generalization.

2. I came up with this when studying orderability of groups.

3. A related result of Weinbaun (1972) tells us that $T$ can't be a subword of $g$. So for example, if $g=aba^{-1}b^{-1}$ then $T$ can't be $ab$. But his result doesn't rule out the possibility that, for instance, $T=ba$.

Answer 1

Let $S=\{s_1,s_2,s_3\}$.

Then $a=(s_1 s_2)^{-1}s_3(s_1s_2)$ has length $5$.

But $T=s_2as_2^{-1}=s_1^{-1}s_3s_1$ has length $3$.

Thus the conjecture about the lengths doesn't hold for the first power of $a$.

It you square $a$, you get $l(a^2)=l((s_1s_2)^{-1}s_3^2(s_1s_2))=6$.

But if $T=s_2^{-1}as_2(s_1s_2)a(s_1s_2)^{-1}=s_1^{-1}s_3s_1s_3$ has $L(T)=4$.

Thus this appears false. I guess the first observation already disproves it.

Answer 2

I'm sorry, I was completely wrong, please ignore this answer.
(I do find it reasonable that this post and this paper may be relevant, though).

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The correct formulation is with **cyclically reduced** length, and then the answer is **yes** (otherwise it's **no** as was commented).

Definition: The cyclically reduced length of a word $w$ as $|w|_c = \min\left\{\left|uwu^{-1}\right|: u\in F \right\}$.
Geometrically, this corresponds to erasing backtraces from the $w$-path in the Cayley graph $\textrm{Cay}\left(F/\langle\langle w \rangle\rangle, S\right)$ and then counting edges (instead of just counting edges).
A word $w$ is cyclically reduced if $|w|_c=|w|$.

Claim: $\left|\prod_{i=1}^n g_i^{-1} a g_i\right|_c \ge n\cdot |a|_c$.
(This answers your question positively in the case $|a|_c = |a|$).

Proof: Let's see both the geometric and algebraic interpretations.
There is a unique decomposition of the word $g_i = h_i g'_i$ such that $h_i^{-1} a h_i$ is cyclically reduced and $h_i$ is maximal among such words.
Geometrically, this means we decompose the traceback-erased $\rho$-shaped path $g_i^{-1}a g_i$ to its tail $g_i'^{\pm1}$ and cycle $h_i^{-1} a h_i$.
We may assume $h_i=1, g_i'=g_i$ (and in particular $a$ is cyclically reduced). Then all cancellations between $g_i^{-1}ag_i$ must occure between $g_i, g_{i+1}^{-1}$ only. Geometrically, the path $\prod_{i=1}^n g_i^{-1} a g_i $ is a tree with an $a$-cycle attached to every leaf. This finishes the proof.