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Within my limited experience, I have only known free groups to occur through two mechanisms: as fundamental groups of trees (graphs) and ping-pong. And sometimes only through one way: the fact that sufficiently high-powers of hyperbolic elements in a Gromov-hyperbolic group generate a free group arises via ping-pong. I know of no tree to represent the situation.

To the experts, the following question is surely either an obvious yes or no:

Can we explicitly describe all mechanisms by which finitely generated free subgroups of the Artin braid groups $B_n$ (for $n=2,3,\ldots$) arise ?

More specifically, seeing $B_n$ as the isotopy space of all $n$-pointed braids in the closed 2-disk, can we characterize all configurations generating the rank $k$ free group?

Much less specifically, do we know, say, that all finitely-generated free subgroups arise from ping-pong and can we describe all ping-pong games?

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Your final paragraph is a bit vague. What does it mean to say that a free subgroup "arises from ping-pong"? Ping-pong is a particular construction for free subgroups that depends on choices of various qualitative and quantitative parameters; are you asking whether for a given free subgroup those choices and parameters can all be set so as to produce that particular subgroup? – Lee Mosher Oct 27 '13 at 20:55
To be more precise, one can ask if every purely pseudo anosov free subgroup is Schottky. Then it is an open problem. – Misha Oct 27 '13 at 21:19

A theorem of Leininger-Margalit says that any two elements of the pure braid group either commute or generate a free group; see here. But I don't think there is any hope for classifying free subgroups that are generated by more than $2$ elements.

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Great reference, thanks. – J. Martel Oct 27 '13 at 23:50

This is a very extended comment to your question.

Your question does not have "obvious yes or no" answer; in part, this is because you asked 3 somewhat different questions:

  1. "Can one explicitly describe all finitely generated free subgroups..." In this form, the question has no answer since the word "explicit" is meaningless in this context.

  2. Your second question "can we explicitly enumerate/characterize all configurations which generate the rank k free group" is a bit better, since one can interpret it as a request for an algorithm that, given as its inputs tuples $(w_1,...,w_k)$ of words in the standard generators of $B_n$, would determine if they generate rank $k$ free subgroups of $B_n$. Existence of such an algorithm is unknown for $n\ge 6$. The reason is that $B_n$ contains $F_2\times F_2$ and, hence, also contains Mikhailova subgroups. In particular, it is undecidable if a given set of $k$ elements in $F_2\times F_2$ generates $F_2\times F_2$ or not.

  3. Lastly, you are asking something about ping-pong. What this "something" is, is very much unclear. Every free group $F_k$ has a ping-pong description (this is almost a tautology). On the other hand, the "collection" of such ping-pongs is not even a set! One can ask, however, a meaningful question restricting to "Schottky ping-pongs", by considering the action of $B_n$ on a certain topological sphere $S=S^{n-3}$, the space of projective classes of measured laminations on $n+1$ times punctured sphere. A Schottky ping-pong on $S$ is given by a collection of disjoint compact subsets $C_1, C_1',...,C_k, C_k'$ and elements $g_i\in B_n$, so that each $g_i$ sends the interior of $C_i$ homeomorphically to the exterior of $C_i'$. Then the subgroup $<g_1,...,g_k>\subset B_n$ is free of rank $k$ and is necessarily necessarily purely pseudo-Anosov; such subgroups are Schottky subgroups in $B_n$ by analogy with Schottky groups acting on hyperbolic spaces (and their ideal boundaries). It is also unknown if $B_n$ (or any mapping class group for this matter) contains a purely pseudo-Anosov free subgroup which is not Schottky. Note that already $PSL(2,C)$ contains free subgroups of rank 2 which are purely hyperbolic and not Schottky.

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A couple of remarks:

If a subgroup is torsion-free, and intersects the pure braid group $P_n$ in a free group, then the group is also free. So the question can be interpreted in $P_n$.

There is the Birman exact sequence $F_{n-1} \to P_n \to P_{n-1}$, which is obtained as deleting a puncture, and has kernel the fundamental group of an $n-1$-punctured plane. There is such a free subgroup for each puncture, and these subgroups are not ``quasiconvex" in any sense since they are normal. So I don't think there is some sort of ping-pong description of these from some action of $P_n$ on a nice space.

On the other hand, if the image of the subgroup generated by $g_1,\ldots,g_k$ in $P_{n-1}$ is free of rank $k$, then so is the subgroup of $P_n$. So I could imagine some inductive description of the free subgroups of $P_n$.

Suppose the image in $P_{n-1}$ is free. Then there is a Nielsen transformation of $g_1,\ldots,g_k$ such that $g_1,\ldots,g_j$ have image in $P_{n-1}$ generating a rank $j$ free subgroup, and $g_{j+1},\ldots,g_k$ have trivial image in $P_{n-1}$. Then the question is what is the extension of the free group $\langle g_1,\ldots,g_j\rangle$ by the normal subgroup of $\langle g_1,\ldots,g_n\rangle$ generated by $g_{j+1},\ldots, g_n$? I wouldn't expect a simple answer to this question though, so this is probably a difficult question. Also, the image in $P_{n-1}$ might not be free.

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I must admit i am reluctant to pursue the issue as an ``extension problem". Rather it seems all complication lies in determining how it arises that the intersections of a free subgroup $F$ of $P_n$ with the various kernels $F_{n-1}$ could all be nontrivial. – J. Martel Oct 28 '13 at 13:04

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