# Action of braid groups on regular trees

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.

• I should point out that the existence of a non-elementary action on a finite degree tree is much stronger than that of a non-elementary action on an arbitrary tree. (And assuming regular of finite degree is even more restrictive.)
– YCor
Commented Jul 21, 2023 at 8:05
• If you don't care about the tree being regular, every finitely generated residually finite group (for example braid groups) admits a faithful action on a rooted tree: math.tamu.edu/~grigorch/publications/freevsnonfree.pdf. You take a nested sequence of finite index subgroups intersecting trivially, and their cosets forms a tree. For braid groups I think you can probably choose all the subgroups in such a way that the vertices of the tree all have the same degree except the root, so it's a "regular rooted tree". Commented Jul 21, 2023 at 10:57
• You wrote: Since the Braid group is an infinite "discrete" group in my intuition it might be related to the automorphisms of an infinite tree. Within the class of infinite discrete groups, there is a special subclass FA consisting of groups such that every action on a tree has a global fixed point (see the paper mentioned in the answer below). It is a rich an interesting class, and its complement is a rich and interesting class. So for a general infinite discrete group, there is no particular reason to expect it to lie in FA or its complement. Commented Jul 24, 2023 at 13:49

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 $$\sigma_1= \sigma_3$$.

• Do you know why it's false for $n=2,3$? they just say (p682) "It can be seen from the presentation that each of the groups $B_3$ and $B_4$ admits a non-trivial action on a tree with no invariant line."
– YCor
Commented Jul 21, 2023 at 11:33
• @YCor: I added an explanation about $B_3$. For $B_4$, I need to think a bit. Commented Jul 21, 2023 at 12:10
• B_4 admits a surjective map to B_3. Commented Jul 21, 2023 at 18:45
• So the idea is that $B_3$ has an action on a 3-regular tree and $B_3$ is a quotient of $B_4$ (basically for the same reason $S_3$ is a quotient of $S_4$) so the action of $B_3$ on a 3-regular tree can be lifted to an action of $B_4$? Whereas for any other $B_n, n \neq 3,4$ the only action is on the trivial tree (line). Commented Jul 25, 2023 at 19:41
• The action of a group $G$ on a tree $T$ can be equivalently described as a morphism $G \to \mathrm{Isom}(T)$. Consequently, for all group $H$ and morphism $H \to G$, you naturally get an action of $H$ on $T$ through the morphism $H \to G \to \mathrm{Isom}(T)$. Commented Jul 26, 2023 at 4:50