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Let $R\subset\mathbb{C}$ be a subring, and let $A,B\in SL(2,\mathbb{C})$ be matrices such that $A,B,AB$ all have trace in $R$.

For which $R$ can we then deduce that $A,B$ are simultaneously conjugate to a pair of matrices in $SL(2,R)$?

In particular I'm wondering about $R = \mathbb{Z}, \mathbb{Q}$, or a general number field.

(Each of $A,B$ is conjugate to a matrix in $SL(2,R)$, see the comments.)

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    $\begingroup$ 2: never (unless $R=\mathbf{C}$): pick $t\notin R$, define $A=e_{12}(t)$ and $B=e_{21}(1)$; then $AB$ has trace $t+2$. So $A,B$ have characteristic polynomial $(x-1)^2$ which is in $R[x]$, but $AB$ has trace $t+2$ so $A,B$ are not simultaneously conjugate into $SL_2(R)$. $\endgroup$
    – YCor
    Commented Oct 29, 2016 at 0:20
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    $\begingroup$ 1: always: if $A$ is a scalar matrix (scalar $t$), its trace is $2t$ so $t$ has to be in the ring. If $A$ is not scalar, it's conjugated to its companion matrix, whose entries are $0,1$ or coefficients of the characteristic polynomial. $\endgroup$
    – YCor
    Commented Oct 29, 2016 at 0:25
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    $\begingroup$ @YCor (my motivation is to see if the character variety of homorphisms $F_2\rightarrow SL(2,\mathbb{C})$ of commutator trace $k$ given by the equation $x^2 + y^2 + z^2 - xyz-2 = k$, where $x,y,z$ are the traces of $A,B,AB$ respectively, is functorial in the ring of coefficients of $SL_2$) $\endgroup$
    – stupid_question_bot
    Commented Oct 29, 2016 at 1:33
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    $\begingroup$ Yes assuming that the trace of $AB$ is in $R$ changes the answer, then there's a positive result. $\endgroup$
    – YCor
    Commented Oct 29, 2016 at 1:40
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    $\begingroup$ The functions $tr(A)$, $tr(B)$, and $tr(AB)$ generate the ring of $GL_2$-invariant functions on $Hom(F_2, SL_2)$, so in that sense they completely determine the pair $(A,B)$ up to conjugacy. A good reference is the book by Brumfiel and Hilden, "SL_2 representations of finitely presented groups." They work over $\mathbb Z$ or $\mathbb Z[1/2]$ for the whole book. $\endgroup$
    – Peter Samuelson
    Commented Oct 29, 2016 at 9:28

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