The braid groups satisfy a number of properties that one would expect of a hyperbolic group, liking having a solvable word problem, and having exponential growth. Are the braid groups hyperbolic groups? If not, is there any obvious property of hyperbolic groups showing that they are not?
Is the braid group hyperbolic
braid-groupsgeometric-group-theorygroup-theoryhyperbolic-groups
Related Solutions
A similar question was asked on MathOverflow recently. The answer of Genevois is pretty comprehensive.
It does not follow that for groups which are not finitely generated the notion of hyperbolicity makes sense.
What follows, instead, is that for any pair $(G,S)$ such that $G$ is a group and $S$ is a generating set, the notion of hyperbolicity is well-defined: the Cayley graph $\Gamma(G,S)$ is certainly defined (and it is connected), and hyperbolicity of $\Gamma(G,S)$ certainly makes sense.
The trouble is that without any control on the cardinality of $S$, it is possible to have a group $G$ and two generating sets $S_1,S_2$ such that $\Gamma(G,S_1)$ is hyperbolic whereas $\Gamma(G,S_2)$ is not.
In fact, here's an example: $G = \mathbb Z^2$ with $S_1 = \{(\pm 1,0),(0,\pm 1)\}$ and with $S_2 = \mathbb Z^2$. The Cayley graph $\Gamma(G,S_1)$ is not hyperbolic whereas $\Gamma(G,S_2)$ is bounded and is therefore hyperbolic. That's an example where $G$ is in fact finitely generated, but one can easily cook up examples where $G$ is not finitely generated, e.g. $G = \mathbb Z^\infty$ with the "standard" generating set $S_1 = \{\pm e_i \mid i \ge 1\}$ is not hyperbolic, but with the generating set $S_2=G$ it is hyperbolic.
What happens in the finitely generated case is that if $S_1,S_2$ are two finite generating sets for the same group $G$, then $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ are quasi-isometric to each other, and hyperbolicity is a quasi-isometry invariant, so $\Gamma(G,S_1)$ is hyperbolic if and only if $\Gamma(G,S_2)$ is hypebolic. Thus one says that a finitely generated group $G$ is hyperbolic if and only if some (hence any) finite generating set $S$ results in a hyperbolic Cayley graph $(G,S)$.
The summary statement is that hyperbolicity of a finitely generated group is well-defined independent of the choice of finite generating set, but take away those bold faced words and the statement becomes false.
ADDED: In answer to a followup question in the comments, every Gromov hyperbolic space (i.e. every geodesic metric space satisfying the thin triangle property) has a Gromov boundary.
And, given a group $G$ and two generating sets $S_1$, $S_2$, if $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ are both hyperbolic, then both of them do have a Gromov boundary. And if $S_1,S_2$ are both finite then one can use the quasi-isometry between $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ to deduce that their Gromov boundaries are homeomorphic; so in the finitely generated case you are free to speak about THE (homeomorphism class of the) Gromov boundary.
However, without the assumption that $S_1,S_2$ are both finite, it need not be true that the Gromov of $\Gamma(G,S_1)$ is homeomorphic to the Gromov boundary of $\Gamma(G,S_2)$.
For a counterexample, take your favorite infinite, finitely generated, hyperbolic group $G$. With any finite generating set $S_1$, the Gromov boundary is nonempty. With the infinite generating set $S_2=G$, the Gromov boundary is empty.
Anyway, it follows that to speak of THE Gromov boundary of $G$ is nonsensical in the infinitely generated case.
Best Answer
Outside of the one and two strand case they are not hyperbolic. One obstruction is that braid groups on $n>2$ strands, $B_n$, contain $\mathbb Z^2$ subgroups which can not happen in hyperbolic groups.
First note that $B_3$ is a subgroup of $B_n$ for $n>2$. Order the strands and name them $b_1,b_2,b_3,...,b_n$. Let $\sigma$ be the element which braids $b_1$ around $b_2,b_3$ where $b_1$ goes around the back of those two and then cross in front and let $\rho$ be the element which braids $b_2,b_3$ (basically an element in $B_2$). These two elements "obviously" commute so we get $\langle \sigma, \rho \rangle \cong \mathbb{Z}^2$ as a subgroup of $B_n$ for $n>2$ (if you want a formal proof there is enough detail in A Primer on Mapping Class Groups by Farb and Margalit).
In the case of $B_1,B_2$ we have the trivial group and $\mathbb{Z}$ respectively and these are hyperbolic but not in an "interesting" way.
You mention that one of your motivations is that there do seems to be some similarities to hyperbolic groups and it turns out this is a really good observation which gets into some interesting math. I won't be giving much details but it seems good to mention some of these ideas.
There is a different way to describe braid groups and it is as mapping class groups of punctured disk (basically the group of topological symmetries up to a natural identification). Very roughly you can think about punctures moving around each other in the disk as braids moving around each other. If you "plot" this with a time axis you can see the punctures moving "drawing" the braid.
This becomes important because now you get tools from surface theory and their mapping class groups. The mapping class group acts naturally on something called the curve complex which, somewhat surprisingly, is $\delta$-hyperbolic, proved by Masur and Minsky, and infinite diameter (you can choose $\delta =19$). Lots of the hyperbolic behavior can actually be seen in this action on the curve complex. Further, Masur and Minsky developed a way of studying the geometry of mapping class groups through the curve complex of the surface and curve complexes of all its (essential) subsurfaces by "piecing together" the geometric information of the curve complexes in some consistent way.