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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, $T=ba$.

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  • $\begingroup$ Did you mean $G=F$? $\endgroup$ Sep 26 at 21:15

2 Answers 2

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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.

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  • $\begingroup$ You are right. I forgot to mention that the $a$ should be cyclically reduced. $\endgroup$
    – user45392
    Sep 28 at 12:51
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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).


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.

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  • $\begingroup$ I do not see why this argument is correct: for example, if $F=\langle x,y \rangle$ and $a=x$, then the first copy of $a$ cancels in the reduction of $T=a \cdot x^{-1}y a y^{-1} x$. The correct proof would have to use the fact that $T$ is the product of conjugates of $a$ only (and no conjugates of $a^{-1}$ are involved). Indeed, otherwise we can take $a=xy^{-n}$, which has length $n+1$, and $T=xyx^{-1}y^{-1}=a \cdot y a^{-1} y^{-1}$, which has length $4$. $\endgroup$ Sep 27 at 16:15
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    $\begingroup$ Your claimed formula is not correct. Consider in the free group with generators $x$ and $y$. Consider $a=xyx^{-1}y^{-1}$ which is cyclically reduced, the length of $a$ is 4. Let $a'=yx^{-1}y^{-1}x$, it's a conjugate of $a$. But the length of $aa'$ is 6, cyclically reduced length of $aa'$ is 4$. $\endgroup$
    – user45392
    Sep 28 at 12:40
  • $\begingroup$ You may want to modify your claimed inequality to $|\prod_{i=1}^ng_i^{-1}ag_i|_c\geq|a|_c$ (by deleting the $n$ multiplier on the RHS), which looks stronger than my conjecture, but equivalent. But I don't see how your argument proves this either. After you wrinting $g_i=h_ig_i'$ such that $a_i:=h_i^{-1}ah_i$ is still cyclically reduced and $l(h_i)$ is maximal, you write the product to the form $T=\prod_{i=1}^n g_i'^{-1}a_ig_i$. Now the difficulty is that there might be cancellations among $a_i$ and $a_{i+1}$. (It's relatively easy to control the number of such cancellations when $n=2$.) $\endgroup$
    – user45392
    Sep 28 at 13:09
  • $\begingroup$ Sorry, in my example above, the reduced length of $aa'$ is still 6, not 4. (In fact, because conjugate is allowed, to prove $|T|_c\geq |a|_c$ is equivalent to prove $|T|\geq|a|$, if $a$ is cyclically reduced. ) $\endgroup$
    – user45392
    Sep 29 at 2:08

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