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Edit: As requested, I give here the proof of the statement about $t$. I believe that this statement is due in some form to Fricke in the nineteenth century:

If $t:G \rightarrow K$ isthe trace of a representation $r: G \rightarrow {\rm{SL}}_2(K)$, then one has $t(gh)+t(gh^{-1}) = t(g)t(h)$ for all $g,h$ in $G$ (since this formula holds when $t$ is the trace, and $g,h$ are matrices in ${\rm SL}_2(K)$, as easily checked). Now this formula, and an easy induction on the lengthof a word $g$ in $G$ shows that $t(g)$ can be expressed as a polynomial (with integral coefficients) in $t(u)$, $t(v)$ and $t(uv)$. Hence a $t$ satisfying (1) and (2) is unique if it exists.

For the existence, let $U$ and $V$ be two matrices in ${\rm SL}_2(\mathbb{C})$ such that tr$U$,tr$V$, and tr$UV$ are algebraically independent (this is really easy). Then see $\mathbb{Z}[X,Y,Z]$ as a subring of $\mathbb{C}$ by sending $X$ on tr$U$, $Y$ on tr$V$, $Z$ on tr$UV$.Let $r$ be the representation $G \rightarrow \rm{SL}_2(\mathbb{C})$ sending $u$ on $U$ and $v$ on $V$, and $t=$tr $r$. Then by definition $t$ satisfies (2), with $K=\mathbb{C}$, one has $t(u)=X$, $t(v)=Y$, $t(uv)=Z$ by construction and $t$ takes value in the subring $\mathbb{Z}[X,Y,Z]$ of $\mathbb{C}$ by what we have said earlier. Hence the existence of $t$.

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# Which polynomials are Fricke polynomials ?

Let me recall the definition which seems the most standard of Fricke polynomials. Let $G$ be the free group with two generators $u,v$. It is not very hard to prove that there exists a unique application $t : G \rightarrow \mathbb{Z}[X,Y,Z]$ such that

(1) $t(u)=X,t(v)=Y,t(uv)=Z$,

(2) $t$ is the character (i.e. the trace) of a representation $r : G \rightarrow \rm{SL}_2(K)$ where $K$ is some field containing $\mathbb{Z}[X,Y,Z]$.

A Fricke polynomial is just an element of $\mathbb{Z}[X,Y,Z]$ which is in the mage of $t$.

What is the image of $t$ ?