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Will Jagy
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I wouldn't call it GL, it is the orthogonal group of the lattice we are discussing. References, as i said, include Lattices and Codes by W. Ebeling, Rational Quadratic Forms by Cassels, these two being available and inexpensive.

We double the quadratic form to get an integral Gram matrix:

$$ G \; = \; \left( \begin{array}{rrr} 2 & 1 & 0 \\\ 1 & 2 & 0 \\\ 0 & 0 & -2 \end{array} \right). $$

Your solutions, the light cone, are column vectors $v$ such that $ v^T G v = 0.$ A root, since we have doubled everything to get an "even" lattice, is a vector $r$ with norm 2, $ r^T G r = 2.$ The general definition of reflection (Cassels calls this a symmetry, page 19) in any vector $w$ is that $$ x \mapsto \; \; x \; - \; \frac{2 \, x^T G w}{w^T G w} \; w.$$ As a result, when we take $w$ to be a root, the factors of 2 cancel and we are taking lattice points to other lattice points. In your original form, a root $(x,y,z)$ solves $x^2 + x y + y^2 = 1 + z^2.$ Then the reflection in the root is just a linear map, determinant $-1,$ and is therefore given by a square matrix with respect to the original basis. Finally, the reflection is an isometry, part of the orthogonal group of the quadratic form, and if we call the matix $A,$ it solves $$A^T G A = G.$$

So far I have found two reflections. Taking the root $(-3,7,6)^T$ gives the reflection $$ A_1 \; = \; \left( \begin{array}{rrr} 4 & 33 & -36 \\\ -7 & -76 & 84 \\\ -6 & -66 & 73 \end{array} \right). $$

If we take the initial triple in the light cone to be $c = (3,5,7)^T,$ we get $A_1 c = (-75,187,163)^T$ which is indeed a new solution.

Taking the root $(5,15,18)^T$ gives the reflection $$ A_2 \; = \; \left( \begin{array}{rrr} -124 & -175 & 180 \\\ -375 & -524 & 540 \\\ -450 & -630 & 649 \end{array} \right). $$ we get $A_2 c = (13,35,43)^T$

Ian Agol would know how many reflections are necessary. Nothing really wrong with finding too many.

Will Jagy
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