# Conjugation Curvature

Recently I saw some papers on the arXiv on conjugation curvature of finitely generated groups (also called medium-scale curvature, transportation curvature, metric Ricci curvature or comparison curvature for Cayley graphs). I got a bit interested in it and so decided to write up a short post about it.

Let $$G$$ be a finitely generated group and let $$S$$ be a chosen finite generating set for it. We assume that $$S = S^{-1}$$ and that $$S$$ does not contain the identity element $$e$$of $$G$$. For a group element $$g$$ we define $\mathrm{GenCon}(g) := \frac{1}{|S|}\sum_{a \in S} |a^{-1}ga|\,,$ where $$|-|$$ denotes the word length derived from the generating set $$S$$. The value $$\mathrm{GenCon}(g)$$ is the average word length of a conjugate of $$g$$ by a generator. We then define conjugation curvature at $$g$$ as $\kappa(g) := \frac{|g| – \mathrm{GenCon}(g)}{|g|}\,.$ So negative conjugation curvature, for example, occurs at group elements whose conjugates by generators have on average longer word lengths.

Let us collect some basic properties of conjugation curvature from a preprint by Bar-Natan, Duchin and Kropholler (arXiv:1712.02484):

• Central elements have vanishing conjugation curvature. Especially, abelian groups have identically zero conjugation curvature at all group elements.
• Conjugation curvature, even its sign, depends on the chosen generating set.
• If $$w$$ is a group element such that $$(e,w,w^2,w^3,\ldots)$$ is a geodesic in $$G$$, which means that $$|w^k| = |w|^k$$ for every natural number $$k$$, then $$\kappa(w) \le 0$$.
• In the free group $$F_n$$ on $$n$$ generators ($$n \ge 2$$ and with respect to the standard generating set, every non-identity group element has negative conjugation curvature.
• Generalizing the examples of abelian groups and of free groups, a group element $$g$$ of a right-angled Artin group (RAAG) has with respect to the natural group presentation of the RAAG vanishing conjugation curvature if it is central and negative conjugation curvature otherwise.

Because of the second bullet point above, conjugation curvature is not a large scale invariant. The latter would mean that it is a quasi-isometry invariant and especially, independent of the chosen generating set.

Since there are several classes of groups that come with naturally given generating sets, it is still worthwhile to investigate this notion. For example, the solvable Baumslag-Solitar groups are defined as $\mathrm{BS}(1,n) := \langle a,t \mid t a t^{-1} = a^n\rangle\,.$ For these groups Taback and Walker computed the conjugation curvature of certain elements and showed that there is a positive density of elements of positive, negative and of zero conjugation curvature (arXiv:2006.14525).

Up to now we stated computations of conjugation curvature for various groups. One can to a certain extend also go the other way around and derive conclusions about the group from the information of its conjugation curvature. Let me mention just three examples, the first two are taken from the first mentioned preprint of Bar-Natan, Duchin and Kropholler and the third is from a preprint by Nguyen-Wang (arXiv:1910.05822):

• If the conjugation curvature is exactly zero for all elements of the group, then the group is virtually abelian.
• If the conjugation curvature is positive for all elements outside of a large ball in the group, then the group is finite.
• If the conjugation curvature is negative for all elements outside of a large ball in the group, then the group has exponential growth.