# 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 proper 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)"

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Perhaps one could start with a single free subgroup of SU(2) with algebraic coefficients (which can be constructed by a ping-pong argument in some p-adic topology, as in the standard demonstration of the Banach-Tarski paradox), and work with subgroups of that free subgroup (which are automatically themselves free, by the Nelson-Schrier theorem). As the free subgroup is dense in the Archimedean topology, I would imagine that the set of independent generators g_1,...,g_k in it would also be dense. –  Terry Tao Mar 2 '12 at 16:22

Tao's argument essentially does the job. However, I would start with a cocompact arithmetic lattice $\Gamma$ in $SL(2, {\mathbb R})$ (which is, hence, not commensurable, up to conjugation, to $SL(2, {\mathbb Z})$). Then every Galois conjugation $\sigma(\Gamma)$ of $\Gamma$ will be a subgroup of $SU(2)$ (essentially, by the definition of arithmeticity). The group $\sigma(\Gamma)$ is isomorphic to $\Gamma$. Thus, every free subgroup $F$ in $\Gamma$ yields a free subgroup $F':=\sigma(F)$ of $SU(2)$ whose matrix entries are contained in a fixed number field $K$. Thus, $F'$ will be dense in $SU(2)$. In particular, every $k$ elements $g_1,...,g_k\in SU(2)$ can be approximated (arbitrarily well) by elements $f_1,...,f_k\in F'$. The elements $f_1,...,f_k$ will generate a free subgroup $F''$ in $SU(2)$ and the homomorphisms $\psi: F_k\to F''$ sending free generators $x_i$ to the elements $f_i$, will approximate the given homomorphism $\phi: F_k\to SU(2)$ sending $x_i$ to $g_i$. This answers your original question. Note that, a priori, rank of $\psi(F_k)$ is less than $k$.
One can do better and replace $f_i\in F$ by elements $h_i\in F$ so that the elements $h_i\in F$ generate a free subgroup of rank $k$ and the homomorphisms $\eta: x_i\to \sigma(h_i)$ still approximate $\phi$. You would need the following:
1. If $H\subset F$ is a finitely-generated subgroup of infinite index then every element $g\in SU(2)$ can be approximated arbitrarily well by elements $\sigma(f)$, where $f\in F$ is such that the cyclic group $\langle f\rangle$ generated by $f$ intersects $H$ only at the identity.
2. Let $H\subset F$ be a finitely-generated subgroup, $\hat{h}\in F$ is such that group $\langle \hat{h}\rangle$ intersects $H$ only at the identity. Then for all sufficiently high powers $h:=(\hat{h})^m$, the group generated by $H$ and $h$ is the free product $H \star {\mathbb Z}$ and has infinite index in $F$.
Given this, you construct elements $h_i\in F$ inductively, taking $\hat{h}_{i}$ in $F- \langle h_1,...,h_{i-1}\rangle$ (so that $\sigma(h_i)$ is close to $g_i\in SU(2)$) and then replace this $\hat{h}_i$ with its high power $h_i$ so that $\sigma(h_i)$ is still close to $g_i$ and the group generated by $h_1,...,h_i$ is free of rank $i$.