A Characterization of Invertible Trace Maps Associated with a Substitution

Abstract

Let $F=\langle a,b\rangle$ be the free group generated by $a,b$. Let $\phi\in\mathrm{Hom}(F,SL(2,\mathbf{C}))$ be a homomorphism from $F$ to $SL(2,\mathbf{C})$. Define $T(\phi)=(\mathrm{tr}\,\phi(a),\mathrm{tr}\,\phi(b),\mathrm{tr}\,\phi(ab))$, where $\mathrm{tr}\,A$ stands for the trace of the matrix $A$. Let $\sigma\in\mathrm{Aut}F$. Then from [2, 12, 4], there exists a unique polynomial map $\Phi_{\sigma}\in(\mathbf{Z}[x,y,x])^3$, such that \[ \mathrm{tr}\,\phi(\sigma(a)),\mathrm{tr}\,\phi(\sigma(b)),\mathrm{tr}\,\phi(\sigma(ab)))=\Phi_{\sigma}(\mathrm{tr}\,\phi(a),\mathrm{tr}\,\phi(b),\mathrm{tr}\,\phi(ab)) \] with $x=\mathrm{tr}\,\phi(a),y=\mathrm{tr}\,\phi(b),z=\mathrm{tr}\,\phi(ab)$, and there exists a unique polynomial $Q_{\sigma}$, such that $\lambda\circ\Phi_{\sigma}=\lambda\cdot Q_{\sigma}$, where $\lambda(x,y,z)=x^2+y^2+z^2-xyz-4$. In this paper, we will show that $\sigma\in\mathrm{Aut}F$ if and only if $Q_{\sigma}(2,2,z)\equiv 1$, and that this result cannot be improved.