# References about mapping between special linear groups, dimension two to three

Live and learn: this mapping is given on page 301 of CASSELS, although the version he writes would fit my 2 by 2 matrix $$W = \left( \begin{array}{rr} \alpha & \gamma \\ \beta & \delta \end{array} \right),$$ which is the transpose of what I use below. He also points out that we are talking about the automorphism group of $f_0(\vec{x}) = xz - y^2,$ and that the mapping is the spin representation of $O^+_{\mathbb Q}$ from chapters 10 and 11. Finally, on page 303, (5.19), he says clearly that any isotropic ternary form is related, there is a nonzero integer $m$ and a nonsingular integral matrix $M$ with $m f(x) = f_0(Mx).$ This goes back to Fricke and Klein (1897), pages 507-508.

I've gotten a lot of mileage out of the group isomorphism described below, from $PSL_2 \mathbb Z$ to $PSL_3 \mathbb Z.$ Very handy. I found this earliest in Fricke and Klein (1897) in a few places, later in Magnus, Zagier, other places. My impression is that this also has a physics name; Twistor Map; Bratwurst Map; Spinor Map; Snickerdoodle Map; not sure. There could be Fuchsian groups involved. Or worse.

Anyway, looking for (easily found) places that say firmly that this is an injective group homomorphism, and confirm that the whole thing is pretty.

Given an element of the modular group $$W = \left( \begin{array}{rr} \alpha & \beta \\ \gamma & \delta \end{array} \right),$$ so that $$\alpha \delta - \beta \gamma = 1,$$ then construct $$M = \left( \begin{array}{ccc} \alpha^2 & 2 \alpha \beta & \beta^2 \\ \alpha \gamma & (\alpha \delta + \beta \gamma) & \beta \delta \\ \gamma^2 & 2 \gamma \delta & \delta^2 \end{array} \right).$$

Oh, $\det M = (\alpha \delta - \beta \gamma)^3,$ but I already required that to be $1.$ The characteristic polynomial of $M$ is $$\lambda^3 - \left( (\alpha + \delta)^2 - 1 \right) \lambda^2 + \left( (\alpha + \delta)^2 - 1 \right) \lambda - 1 = (\lambda - 1) \left( \lambda^2 - \left( (\alpha + \delta)^2 - 2 \right) \lambda + 1 \right)$$ Part of this is the claim that $1$ is an eigenvalue of $M.$ Here is an eigenvector with eigenvalue $1:$ $$X = \left( \begin{array}{c} 2 \beta \\ \delta - \alpha \\ - 2 \gamma \end{array} \right),$$ with $MX=X,$ and $X$ is nonzero unless $W = \pm I.$

There are two reasons I like it: first, given the Hessian matrix $H$ of the indefinite ternary quadratic form $y^2 - z x,$ we get $$M^T H M = H.$$ Second, writing a binary quadratic form as a row of three coefficients $R = \langle A,B,C \rangle,$ the coefficient vector of the form transformed by the 2 by 2 matrix $A$ is just $RM.$ $$\langle A,B,C \rangle \left( \begin{array}{ccc} \alpha^2 & 2 \alpha \beta & \beta^2 \\ \alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta \\ \gamma^2 & 2 \gamma \delta & \delta^2 \end{array} \right) =$$ $$\langle A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2 \rangle$$ Here $\langle A,B,C \rangle$ refers to $$f(x,y) = A x^2 + B xy + C y^2.$$ Compare $W^T QW,$ where $$Q = \left( \begin{array}{rr} 2A & B \\ B & 2C \end{array} \right).$$

• I think this is the so-called symmetric square representation. Namely, $SL_2(\mathbb Z)$ acts naturally on polynomials in $2$-variables with $\mathbb Z$-coefficients, i.e. the ring $\mathbb Z[x,y]$, and preserves degree of homogeneous polynomials. The action on the homogeneous linear forms is the standard matrix action of $SL_2$ on 2-dim'l vectors, and the action on homogeneous quadratic forms gives the three dimensional representation that you've written down. – tracing Aug 23 '16 at 22:07
• @tracing thank you. I edited that in, the action on binary quadratic forms is half the reason this is useful to me. Combined with the stuff about $y^2 - zx,$ this is the reason that an indefinite ternary quadratic form that possesses a nonzero integer null vector can have all of them parametrized by a finite number of mappings, of the same type that give all primitive Pythagorean triples. Usually more than one such recipe. – Will Jagy Aug 23 '16 at 22:32