Let $\Gamma_g=< a_1,...,a_g,b_1,...,b_g | \prod_{i=1}^g [a_i,b_i]>$ (a surface group). What is known about maximal free subgroups of $\Gamma_g$ for $g>1.$ (I.e. free subgroups which are not properly embedded into any other free subgroup)?

For example,

  1. Is $<a_1,...,a_g,b_1,...,b_{g-1},b_g^2>$ free? Maximal?
  2. What are the lower, upper bounds on ranks of the maximal free subgroups?
  3. Can the free subgroups of the minimal rank be classified somehow?
  • $\begingroup$ In 3, do you mean "the maximal free subgroups of minimal rank"? $\endgroup$
    – Lee Mosher
    Commented Nov 30, 2013 at 23:08
  • $\begingroup$ @Lee: There must be the minimum rank, $m(g)$, of such groups. Question 3 asks if the maximal free subgroups of rank $m(g)$ can be classified somehow. $\endgroup$
    – Adam
    Commented Nov 30, 2013 at 23:15

1 Answer 1

  1. The subgroup is free, and therefore not maximal (see 2.). To see this, the subgroup generated by all the generators except $b_g$ is the fundamental group of a subsurface obtained by cutting along a curve (dual to $b_g$). This surface lifts to a 2-fold cover dual to the curve, and the subgroup is obtained by adding another element corresponding to $b_g^2$, so it has infinite index in the 2-fold cover (since its rank is too small to generate).
  2. Maximal rank free subgroups (i.e. subgroups of infinite index) must be infinitely generated. More generally, this holds for lattices in $PSL_2(\mathbb{C})$ (the argument they gives works for $PSL_2(\mathbb{R})$, and is even easier, since finitely-generated subgroups are geometrically finite).

  3. Since maximal free subgroups have infinite rank, I don't think there will be a nice description of them. They can't be normal, since the only possible quotient groups could have no proper non-trivial subgroups, so must be finite cyclic groups of prime order, a contradiction.


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