3
$\begingroup$

Let $F_n$ be a free group of rank n and consider all possible non-degenerate length functions on $F_n$.

Could I be directed to a reference that give two non-trivial non-power-conjugate elements $g,h \in F_n$ (i.e. no power of $g$ is conjugate to a power of $h$) such that there is some $K>1$ so that for any length function we have: $$ \|g\|_T \geq K \cdot \|h\|_T, $$ where $\|\cdot\|_T$ denote translation length for some $T \in \operatorname{CV}_n$? Obviously, we have this if $h^K=g$, but I specifically want a pair of elements that are not powers of one another.

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ It amounts to finding $g,h \in F_n$ and $K>1$ such that $\|g\| \geq K \cdot \|h\|$ for every free action on $F_n$ on a tree, right? If $g,h$ are two conjugates of some non-trivial element, what about $gh$ and $g^nh^n$? The point is that $g,h$ have the same translation length $\ell$, and, because the action is free, their axes must have intersection of diameter $\leq \ell$. So it should be possible to determine the axes and translation lengths of $gh$ and $g^nh^n$. (This is maybe a naive idea, I do not have the time to check the details right now.) $\endgroup$
    – AGenevois
    Commented Oct 9, 2021 at 6:15

1 Answer 1

Reset to default
3
$\begingroup$

After checking the details, here is a statement that seems to answer your question:

Fact: For every free action of $F_n$ on a real tree $T$ and for all non-commensurated elements $g,h \in F_n$, the inequality $$\|h^kg^k\|_T > \frac{1}{2} \|hg\|_T$$ holds for every $k \geq 2$.

This is an immediate consequence of the following observation:

Lemma: Let $T$ be a real tree and $g,h \in \mathrm{Isom}(T)$ two independent loxodromic isometries. Assume that $$\mathrm{diam}(\mathrm{Axis}(g) \cap \mathrm{Axis}(h)) < \|g\|,\|h\|.$$ Then $hg$ is also loxodromic and its translation length is

  • $\|h\|+ \|g\|$ if $\mathrm{diam}(\mathrm{Axis}(g) \cap \mathrm{Axis}(h))>0$ and if the translation directions of $g,h$ agree on the intersection $\mathrm{Axis}(g) \cap \mathrm{Axis}(h)$;
  • $\|h\|+\|g\| - 2 \cdot \mathrm{diam}(\mathrm{Axis}(g) \cap \mathrm{Axis}(h))$ if $\mathrm{diam}(\mathrm{Axis}(g) \cap \mathrm{Axis}(h))>0$ and if the translation directions of $g,h$ disagree on the intersection $\mathrm{Axis}(g) \cap \mathrm{Axis}(h)$;
  • $\|hg\|= \|h\|+\|g\| + 2 \cdot d(\mathrm{Axis}(h), \mathrm{Axis}(g))$ if $\mathrm{diam}(\mathrm{Axis}(g) \cap \mathrm{Axis}(h))=0$.

The proof is straightforward: it suffices to find a point $x \in T$ such that $hgx$ lies on $[x,(hg)^2x]$ and to compute the distance between $x$ and $hgx$. The different cases are illustrated by the following figures.

enter image description here

Improve this answer
$\endgroup$
5
  • $\begingroup$ I think we also need to specify that $h$ and $g$ have distinct axes in the statement of the fact. But yeah, that does the trick! $\endgroup$
    – NWMT
    Commented Oct 10, 2021 at 15:58
  • $\begingroup$ Right! I added the assumption, thank you. $\endgroup$
    – AGenevois
    Commented Oct 10, 2021 at 18:13
  • 1
    $\begingroup$ Very nice! The question asks for $K>1$, and this provides $K=1/2+\epsilon$, right? $\endgroup$
    – HJRW
    Commented Oct 11, 2021 at 10:07
  • 1
    $\begingroup$ @HJRW: Right, I did not pay attention to this condition... But then $\| (h^pg^p)^q\| > \frac{q}{2} \|hg\|$ for all $p,q \geq 2$. $\endgroup$
    – AGenevois
    Commented Oct 11, 2021 at 11:42
  • $\begingroup$ Good point! :)${}$ $\endgroup$
    – HJRW
    Commented Oct 11, 2021 at 13:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .