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 in Finite group extensions and the Baum-Connes conjecture.
 Update 4  Concerning the question from Update 2. Collin Bleak and Olga Salazar-Diaz proved in Free products in R. Thompson's group V that $V$ does not contain subgroups isomorphic to ${\mathbb Z}^2\ast {\mathbb Z}$. Does $B_n$ contain such subgroups?
 A: 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 $\mathbb{Z}^2 * \mathbb{Z}$.
See The solution to a conjecture of Tits on the subgroup generated by the squares of the generators of an Artin group by Crisp and Paris.
Also, see Clay, Leininger and Mangahas' The geometry of right angled Artin subgroups of mapping class groups for more info and references about right angle Artin subgroups of mapping class groups.
A: 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).
A: 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. 
A: 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).
