I am studying representations of the free group on $2$ generators $F_2$ into the group of isometries of the hyperbolic $n$-space $\mathrm{Isom}(\mathbb{H}^n)$, which involves looking at the right angled hexagon given by two generators $a$ and $b$ (obtained by first drawing the axis of $a$, $b$ and $ab$, and then by drawing the common perpendicular of each couple of lines, except if they intersect). I am in a setting where I can make sure that these right-angled hexagons (RAH) make sense and can always be built (all primitive elements are sent to loxodromic isometries, and the representation is irreducible).
I am interested in a geometric property of representations that I know to be true if $n\leq 3$, namely that the distance between two vertices of a given RAH (obtained from a basis of $F_2$) that both are on the axis of a primitive element $u$ is half its translation length. For convenience, let us denote by $d_{\mathcal{H}}(u)$ the given side length, and by $l(u)$ the translation length, so that our property is $d_\mathcal{H}(u) = \frac12l(u)$.
The only proof that I know of in dimension $3$ is to write the isometries as the product of the half-turns around the common perpendiculars of the RAH, which comes down to factorizing the associated representation of $F_2$ by a representation of $W_3^2=F(p,q,r)/<p^2,q^2,r^2>$. In bigger dimensions, I know that not all representations can be factorized through $W_3^2$, so that this proof can't apply, but it still shows that there are some representations having this property.
I would like to know whether all representations have this property for arbitrary $n$, and if not, which representations have it (are the one that factor through $W_3^2$ the only ones ?). Even a counterexample would be great !
All I have managed to prove for now is that if you extend a RAH given by $a$,$b$ and $ab$ (say $\mathcal{H}$) with the one given by $a$, $b$ and $ba$ (say $\mathcal{H}'$), then $d_\mathcal{H}(a)+d_\mathcal{H'}(a) = l(a)$, so it remains to show (if it is true…) that $d_\mathcal{H}(a) = d_\mathcal{H'}(a)$. The proof is quite simple: $a$ sends the axis of $ba$ onto the axis of $ab$, so that the common perpendicular with $ab$ is the image by $a$ of the common perpendicular with $ba$.
It seems to me that this is a problem where everything can be computed (except I don't know how to compute the axis of common perpendiculars), so that it may be a classical result I am not aware of.