# Braid groups acting on CAT(0)-complexes

Does the braid group $B_n, n\ge 3$, act properly by isometries on a CAT(0) cube complex?

Update 1. During a recent talk of Nigel Higson in Pennstate Dmitri Burago asked whether the braid groups are a-T-menable. I seem to remember that somebody proved that they do act properly by isometries on CAT(0) cube complexes. That would imply a-T-menability by Niblo and Roller or Cherix, Martin and Valette. Hence the question. Note: no co-compactness required.

Update 2. It looks like my question is still an open problem for $n\gt 3$. I think my confusion came from terminology. Dan Farley proved that all braided diagram groups (including the R. Thompson group $V$) act property by isometries on CAT(0)-cube complexes. But braided diagram groups (defined by Victor Guba and myself) are not related to braid groups, at least not explicitly because wires there intersect and do not form braids. One can define the notion of "really braided" diagram groups where wires form braids, but I do not think Farley's method will work. So I got confused by my own terminology. By the way, I do not see an obvious reason that $B_n$ does not embed into $V$. $V$ is a big group with lots of complicated subgroups.

Update 3. As Bruce Hughes pointed out to me, even though the Haagerup property (a-T-menability) is unknown for $B_n, n\ge 4$, all forms of Baum-Connes conjecture have been proved for it by Thomas Schick.

Update 4 Concerning the question from Update 2. Collin Bleak proved that $V$ does not contain subgroups isomorphic to ${\mathbb Z}^2\ast {\mathbb Z}$. Does $B_n$ contain such subgroups?

• The Lawrence-Krammer representation preserves a sesquilinear form so that when you specialize the variables $q,t$ to unit complex numbers it becomes a Hermitian form. The signatures of the form can be a variety of numbers -- I haven't worked out the details (yet) but you can always realize positive-definate. For the 6-stranded braid group, you get signatures $-5, 1, 5, 7$ and $15$ (the rep has dimension $15$). So this gives faithful reps of the braid group on spheres and various level-surfaces of Hermitian forms of various signatures... This does not directly address your other restrictions – Ryan Budney Oct 19 '10 at 21:21
• I don't think this representation has any hope of preserving any kind of stratification of these hypersurfaces though. In the positive definate case my understanding is Stoimenow has proven this representation's image is dense in the corresponding orthogonal group $O_{n \choose 2}$. – Ryan Budney Oct 19 '10 at 21:43
• The question of a-T-menability of braid groups was actually asked first by Alain Valette in the book by Cherix, Cowling, Jolissaint, Julg and Valette in 2001. – user2412 Oct 20 '10 at 0:25
• That makes sense. As noted below, I just found this question in du Cornulier's thesis. He was a student of Valette. – HJRW Oct 20 '10 at 0:28
• It was proved by Haettel that, for $n\geq 4$, the braid group $B_n$ cannot act properly co-compactly on a CAT(0) cube complex: math.univ-montp2.fr/prepub/2015/2015_45%28Haettel%29.pdf – Alain Valette Dec 7 '18 at 14:30

As Sam Nead says, $B_n$ contains $\mathbb{Z}^2 * \mathbb{Z}$, and you can find an example the way he suggests.

If you'd like something much more explicit, you can simply take the first three standard generators $\sigma_1$, $\sigma_2$ and $\sigma_3$, and then it follows from the solution (by Crisp and Paris) of a conjecture of Tits that the subgroup $\langle \sigma_1^2, \sigma_2^2, \sigma_3^2 \rangle$ is isomorphic to $\mathbb{Z}^2 * \mathbb{Z}$.

Also, see this paper for more info and references about right angle Artin subgroups of mapping class groups.

• :) Glad to help. – Autumn Kent Oct 20 '10 at 15:56
• I only cannot understand why is it the same proof as Sam's. I think Crisp and Paris would be surprised to find out that their Inventiones paper reduces to one reference to Masur and Minsky. – Mark Sapir Oct 20 '10 at 17:59
• Oh, I didn't mean to suggest that they are the same. Getting the presentations for the groups generated by squares is much harder than getting presentations for groups generated by huge unknown powers. – Autumn Kent Oct 20 '10 at 19:03

This is an open problem. The answer is yes for $B_3$: it acts geometrically on a regular degree three tree crossed with the real line. I learned of this construction in an article by Tom Brady and Jon McCammond; specifically $B_3 \cong (x,a,b,c; x = ab = bc = ca)$, and the Cayley 2-complex can be given the structure of a square complex compatible with the natural action. For $B_4$, the group acts geometrically on a three dimensional CAT(0) complex whose cells are Euclidean tetrahedra isometric to the convex hull of $(0,0,0), (1,0,0), (1,1,0)$, and $(1,1,1)$; this complex does not seem to have an obvious structure of a cube complex. A similar construction can be made for all $n$. Jon McCammond and Tom Brady have an argument for showing $B_5$ is CAT(0). Woojung Choi, in her Ph.D. thesis supervised by McCammond, proved that these complexes are not CAT(0) (with this metric, anyhow) in sufficiently high dimensions.

As Henry points out, usually one asks for a co-compact action, but the natural question of whether there is a proper action on a cube complex is also open. I do not know of many obstructions to admitting such an action other than the result of Niblo and Roller which implies that infinite torsion free groups with property (T) do not have such actions. But braid groups do not have (T).

• I am not asking for a co-compact action. Example are the R. Thompson groups $F, T, V$ and other f.g. diagram groups (Farley). They act on properly by isometries on CAT(0)-cubical complexes but are not CAT(0)-groups. – Mark Sapir Oct 19 '10 at 23:06
• Robert, the Brody-McCammond construction is quite old. $B_3$ is the fundamental group of the trefoil complement in $S^3$. The trefoil complement is a once-punctured torus bundle over the circle, so it's universal cover has the homotopy-type of your complex, simply by lifting the standard cell decomposition of the punctured torus bundle. – Ryan Budney Oct 19 '10 at 23:14
• Yes, that is a good point; and I'm sure they knew this too! But the novelty of their construction is that it generalizes in an obvious way to many spherical Artin groups. The Artin group of type $I(m)$ has a presentation $(x, a_1, \dots, a_m; x = a_1 a_2 = a_2 a_3 = \dots = a_m a_1)$, and the Cayley 2-complex is CAT(0). In general, one considers the universal cover of the presentation complex (with additional higher dimensional cells attached) arising from a "dual braid monoid"; this construction can be adapted to work for any Artin group. – Robert Bell Oct 19 '10 at 23:41

Here's question 2.16 of Bestvina's problem list:

Q 2.16. (Ruth Charney) Are all [finite type] Artin groups CAT(0)?

The answer is yes for small numbers of generators by the work of Krammer, Tom Brady, Jon McCammond, Robert Bell. The question is open even for braid groups.

So it seems that this is may be an open question.

EDIT: I notice that you only ask for a proper action by isometries, whereas Bestvina may require that the action of a 'CAT(0) group' be cocompact.

FURTHER UPDATE: On page 102 of Yves de Cornulier's thesis (2005), he asserts that it's unknown whether braid groups of at least 4 strings are Haagerup (a-T-menable).

• Yes, the difference is "co-compact". I do not ask for that. See my comment below. – Mark Sapir Oct 19 '10 at 23:26
• @Henry: OK, perhaps Yves knows. I will ask him. Thanks! – Mark Sapir Oct 20 '10 at 2:16

Update 4: Does $B_n$ contain a copy of $\mathbb{Z}^2 \ast \mathbb{Z}$?

Answer: Definitely. The Abelian factor is generated by a pair of (powers of) disjoint twists while the cyclic factor can be generated by a (power of a) Dehn twist with support curve "sufficiently distant" from the first two curves, in the curve complex. The proof is the bounded geodesic image theorem of Masur and Minsky.

There is probably an easier way, as well, using a pA map for the cyclic factor.

Edit: I've now spent considerable time thinking about this problem and I think that a very "light" proof is possible. It is enough to use subsurface projection, the fact that the curve complex has infinite diameter (due to Kobayashi and Feng Luo), and Behrstock's lemma on projection to overlapping surfaces. Now, Behrstock's proof used the hierarchy machine, but Mangahas has written up a very pretty (with pictures!) and elementary proof as Lemma 2.13 in her paper "A recipe for short-word pseudo-Anosovs".

Now to sketch an answer. Fix $S$, a surface. Let $0 << m << M$ be correctly chosen constants. Suppose that $X, Y$ are disjoint annuli in $S$ and $Z$ is an annulus so that $d_S(X,Z) > M$. Let $x, y, z$ be the $M$-th powers of the Dehn twists about the corresponding core curves. Let $U_r$ be the set of curves $\alpha \in C(S)$ so that $d_Z(\alpha, z^r(X)) \leq m$. Similarly, let $V_{p,q}$ be the set of curves $\alpha \in C(S)$ so that $d_X(\alpha, x^p(Z))$ and $d_Y(\alpha, y^q(Z))$ are both bounded by $m$.

Note that if $r \neq s$ then $U_r \cap U_s = \emptyset$ and the same holds for the sets $V_{p,q}$ and pairs of vectors in $\mathbb{Z}^2$. By Behrstock's lemma we find that if $r \neq 0$ then $U_r \subset V_{0,0}$. Similarly, if $(p,q) \neq (0,0)$ then $V_{p,q} \subset U_0$. We also have $z^r(U_0) \subset U_r$ and also $x^py^q(V_{0,0}) \subset V_{p,q}$. (This last because subsurface projection to $\Sigma \subset S$ commutes with $\rm{Mod}(\Sigma) \subset \rm{Mod}(S)$.)

Let $w$ be any word in $x, y, z$ and their inverses which is a minimal product of syllables. Here, a word $u$ is a syllable if $u = z^r$ for $r \in \mathbb{Z} - 0$ or $u = x^p y^q$ for $(p, q) \in \mathbb{Z}^2 - (0,0)$. Let $\beta$ be a midpoint of the geodesic $[X, Z] \subset C(S)$. So $\beta \in U_0 \cap V_{0,0}$. We can now follow the orbit of $\beta$ and induct on the syllable length of $w$ to obtain the result.

I hope that this is helpful and I apologize if I made this sound "easy" by my earlier comments. Also, I'll say again that this proof is very "light" and does not use anything difficult from Masur and Minsky.

• @Sam: Proof? A very similar argument "works" for $V$ as well. – Mark Sapir Oct 20 '10 at 13:24
• @Sam: Here is an argument that "works" for $V$. Take two elements of $F\subset V$ with very small disjoint supports. These generate ${\mathbb Z}^2$. For a generator of the free cyclic factor, take any sufficiently complicated element of $V$ whose support contains the supports of the first two elements. The proof is using the rotation number quasi-morphism of Ghys and Sergiescu. There is probably an easier proof using the results of Brin and others about the dynamics of the action of $V$ on $S^1$. The problem of course is that the argument does not work, and the result is false. – Mark Sapir Oct 20 '10 at 14:10