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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.

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Could you please write down next time the polynomial equations you get in terms of the entries of A and B? I change the notation $$ det(A) = ad-bc = 1 , \; det (B) = a'd'-b'c', \; tr (A)=a+d = \alpha, \; tr(B) = a'+d'= \beta, \; tr AB = aa'+dd'+bc'+b'c.$$ Where does the Fricke Polynomial turn up? –  Marc Palm Apr 26 '13 at 6:49
    
$trAB=aa′+dd′+bc′+b′c$ –  Marc Palm Apr 26 '13 at 6:49
    
$$trAB=aa'+dd'+bc'+b'c = \delta$$ –  Marc Palm Apr 26 '13 at 6:50
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In case you are interested: The seemingly algebraic answer has a geometric interpretation. Namely, these are necessary and sufficient conditions for nonexistence of a spherical triangle with the given side-lengths, provided you use logs of eigenvalues instead of traces. –  Misha Apr 28 '13 at 20:39

2 Answers 2

up vote 3 down vote accepted

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.

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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 $|c|<2$.

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The above argument has the graciousness of a geriatric turtle's gait, but here's a redeeming feature: it can be used to show that, over finite fields of characteristic not equal to 2, all triples are realizable. In fact, if I am to believe my doodles, non-realizability over any field of characteristic not equal to 2 could be characterized as Hilbert symbol conditions on a,b,c's. –  BN2 Apr 29 '13 at 9:12

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