tracial triples Say that a triple of real numbers $(a,b,c)$ is a realizable triple if there are matrices $A,B\in SL_2(\mathbb{R})$ such that $tr (A)=a$, $tr (B)=b$, and $tr (AB)=c$. Question: what is the shape of the non-realizable set?
This is surely known, but I couldn't find an answer by myself, nor a reference on the www. It's easy to see that the set in question is non-empty (it contains the origin, for instance) and it lives in the 3-dimensional square $(-2,2)\times(-2,2)\times(-2,2)$. I'm guessing that the Fricke polynomial $a^2+b^2+c^2-abc$ should enter the picture as well.
Note that realizability is, well, a real problem but not a complex one.
 A: Let $x, y, z$ be traces of $A, B, AB$ respectively. Define
$$
k(x,y,z)= x^2+y^2 + z^2 -xyz -2.
$$
Then a triple of real traces $(x, y, z)$ is realizable in $SL(2,R)$, unless it is realizable in $SU(2)$, the latter happens if and only if $x, y, z\in [-2,2]$ and $k(x,y,z)\le 2$. See Goldman's paper "Topological components of spaces of representations", Inventiones, 1988. 
A: Thank you, Misha, for the answer. 
Goldman's proof is part of a bigger picture, and it is hard to rip it out of the context. Besides, it looks a bit too sophisticated for such a simple question. Helped by knowing what should happen, I eventually came up with an elementary and self-contained argument, though not particularly elegant. Feel free to improve.
Proposition (Goldman?)
A triple $(a,b,c)$ is not realizable iff $|a|,|b|,|c|<2$, $a^2+b^2+c^2−abc<4$.
(This set looks like a slightly inflated tetrahedron. A decent picture would be nice here. An indecent one even nicer.)
Proof.
If $|a|=2$, then $(a,b,c)$ is realizable:
\begin{align}
\begin{pmatrix} \pm 1 & \pm b-c
\newline
0 & \pm 1
\end{pmatrix}\begin{pmatrix} b & 1
\newline
-1 & 0
\end{pmatrix}=\begin{pmatrix} c & \pm 1
\newline
\mp 1 & 0
\end{pmatrix}.
\end{align}
Assume $|a|\neq 2$. Let $tr(A)=a$, $tr(B)=b$, $tr(AB)=c$; in particular, $A\neq \pm I$. Up to conjugating $A$ and $B$ by the same matrix, we may assume that $A$ takes the form
\begin{align}
A=\begin{pmatrix} a & \pm 1
\newline
\mp 1 & 0
\end{pmatrix}.\end{align}
Put
\begin{align} B=\begin{pmatrix} x & z
\newline
y & b-x
\end{pmatrix}
\end{align}
so $c=tr(AB)=ax\pm y\mp z$. In other words, $z=\pm ax +y\mp c$. The condition $\det B=1$ becomes
\begin{align}
x^2+y^2\pm axy -bx \mp cy+1=0
\end{align}
We can go backwards as well, so $(a,b,c)$ with $|a|\neq 2$ is realizable iff the above equation is solvable over the reals. The rest is even more basic high-school mathematics. Viewing $x$ as the main quadratic variable, the equation is not solvable iff the discriminant
\begin{align}
(a^2-4)y^2\pm 2(2c-ab)y+(b^2-4)
\end{align}
is negative for all real $y$. As $|a|\neq 2$, this happens precisely when $|a|<2$ and the discriminant
\begin{align}
16(a^2+b^2+c^2-abc-4)
\end{align}
is negative. For the symmetry's sake, observe that $|a|<2$ and $a^2+b^2+c^2−abc<4$ entail $|b|<2$ and $|c|<2$. 
