Action of braid groups on regular trees

Question

Question:

Are there any well known actions of braid groups on trees? For example is there some action of a braid group $ B_n $ on a $ p $ regular tree for some $ p $ such that the action is transitive on vertices and also transitive on directed edges?

Context:

We often understand a group by its actions. For a finite group or compact group, the theory of its unitary representations is very well behaved. Even for a general reductive algebraic group its representation theory is well behaved.

For finite groups we can also learn things by thinking of the finite group acting as automorphisms of a graph.

I'm not used to working with infinite groups that aren't algebraic groups over some field. Braid groups are an example of infinite groups that aren't algebraic groups over any field. Although there is a theory of unitary representations of Braid groups (e.g. certain values of the Burau representation and Lawrence–Krammer representation) the image of the Braid group is not a closed subgroup of the unitary group and so it is a bit beyond my intuition.

Since the Braid group is an infinite "discrete" group in my intuition it might be related to the automorphisms of an infinite tree.

Cross-posted from MSE https://math.stackexchange.com/questions/4729118/action-of-braid-groups-on-regular-trees

Answer 1

In the article A group theoretic criterion for property FA, Culler and Vogtmann notice that

for $n \geq 5$, the braid group $B_n$ has property $A \mathbb{R}$,

meaning that every non-trivial (i.e. without a global fixed point) action of $B_n$ on a tree has an invariant bi-infinite geodesic. As a consequence, the only regular tree on which $B_n$ can act transitively is the line. The proof given in the paper, available online, is short and elementary.

For n=3: About $B_3= \langle \sigma_1, \sigma_2 \mid \sigma_1 \sigma_2 \sigma_1 = \sigma_2 \sigma_1 \sigma_2 \rangle$, one can notice that the relation is equivalent to $(\sigma_1 \sigma_2)^3= (\sigma_1 \sigma_2 \sigma_1)^2$. Thus, setting $a:=\sigma_1 \sigma_2$ and $b:= \sigma_1 \sigma_2 \sigma_1$, one finds the alternative presentation $\langle a,b \mid a^3=b^2 \rangle$ of $B_3$. This implies that $B_3$ is a non-trivial amalgamated sum $\mathbb{Z} \underset{\mathbb{Z}}{\ast} \mathbb{Z}$. As such, it admits an edge-transitive action on a tree.

For n=4: As mentioned by Ishan Banerjee in the comments, there is a surjective morphism $B_4 \twoheadrightarrow B_3$, so the previous action of $B_3$ on a tree yields a similar action of $B_4$ (with an infinite kernel though). The morphism consists in taking the presentation $$\langle \sigma_1, \sigma_2, \sigma_3 \mid [\sigma_1, \sigma_3]=1,\ \sigma_1 \sigma_2 \sigma_1 = \sigma_2 \sigma_1 \sigma_2,\ \sigma_2 \sigma_3 \sigma_2 = \sigma_3 \sigma_2 \sigma_3 \rangle$$ of $B_4$ and setting $\sigma_1= \sigma_3$.