2 deleted 3 characters in body; deleted 3 characters in body

Following notions from [1], call a set of elements $g_1, \dots, g_k \in G = SU(2)$ Diophantine if it satisfies the following property: there exists a constant $D$ such that for every word $W_m$ of length $m$ in $g_1, \dots, g_k$, if $W_m \neq \pm e$ (identity element), then $\Vert W_m \pm e\Vert \geq D^{-m}$. It can be proved that elements with algebraic entries are Diophantine.

Bourgain and Gambrud prove that a certain property (spectral gap) holds for $g_1, \dots g_k$ that: a) generate a free subgroup of $SU(2)$, b) are Diophantine. The first property is generic in measure. The second is not known to be generic in measure, but sets of Diophantine elements are dense in $G^k$ (since all rational-entry elements are). Does it follow that set of elements that a) generate free subgroup, b) are Diophantine, is dense in $G^{k}$?

Maybe it's almost obvious, since the set of elements generating a free subgroup is a complement of a countable union of codimension one varietiesproper subvarieties, but somehow one must rule out the possibility that being Diophantine and freeness are "anticorrelated" (a priori, we only know that elements with algebraic elements are Diophantine, and this is a countable set). Or maybe in fact Diophantine elements form a much larger set?

[1] Jean Bourgain, Alex Gamburd, "On the spectral gap for finitely-generated subgroups of SU(2)"

1

# Diophantine elements in SU(2)

Following notions from [1], call a set of elements $g_1, \dots, g_k \in G = SU(2)$ Diophantine if it satisfies the following property: there exists a constant $D$ such that for every word $W_m$ of length $m$ in $g_1, \dots, g_k$, if $W_m \neq \pm e$ (identity element), then $\Vert W_m \pm e\Vert \geq D^{-m}$. It can be proved that elements with algebraic entries are Diophantine.

Bourgain and Gambrud prove that a certain property (spectral gap) holds for $g_1, \dots g_k$ that: a) generate a free subgroup of $SU(2)$, b) are Diophantine. The first property is generic in measure. The second is not known to be generic in measure, but sets of Diophantine elements are dense in $G^k$ (since all rational-entry elements are). Does it follow that set of elements that a) generate free subgroup, b) are Diophantine, is dense in $G^{k}$?

Maybe it's almost obvious, since the set of elements generating a free subgroup is a complement of a countable union of codimension one varieties, but somehow one must rule out the possibility that being Diophantine and freeness are "anticorrelated" (a priori, we only know that elements with algebraic elements are Diophantine, and this is a countable set). Or maybe in fact Diophantine elements form a much larger set?

[1] Jean Bourgain, Alex Gamburd, "On the spectral gap for finitely-generated subgroups of SU(2)"