in these constraints; to take just one example, follows from and . One could in fact specify $\Sigma^{{\operatorname{I}}}_n$ using just three of these nine constraints if desired. However, this redundancy will be useful in the sequel, as we will be taking full advantage of all nine of these identities.

The identities - form an algebraic set that can be parameterised (perhaps up to some bounded multiplicity) by fixing three of the six coordinates $a,b,c,d,e,f$ and solving for the other three coordinates. For instance, using the coordinates $a,c,d$, one easily verifies that $$\Sigma^{{\operatorname{I}}}_n = \left\{ (a, \frac{na+c}{4acd-n}, c, d, \frac{4a^2d+1}{4acd-n}, 4acd-n): a,c,d \in {\mathbb{C}}^3; 4acd \neq n \right\}$$ and similarly for the other $\binom{6}{3}-1 = 14$ choices of three coordinates; we omit the elementary but tedious computations. Thus we see that $\Sigma^{{\operatorname{I}}}_n$ is a three-dimensional algebraic variety. From we see that the map $$\pi^{{\operatorname{I}}}_n: (a,b,c,d,e,f) \mapsto (abdn, acd, bcd)$$ maps $\Sigma^{{\operatorname{I}}}_n$ to $S_n$. After quotienting out by the dilation symmetry $$\label{dilation}
(a,b,c,d,e,f) \mapsto (\lambda a, \lambda b, \lambda c, \lambda^{-2} d, e, f)$$ of $\Sigma^I_n$, this map is injective.

If $n$ is a natural number, then $\pi^I_n$ clearly maps ${\mathbb{N}}$-points of $\Sigma^I_n$ to ${\mathbb{N}}$-points of $S_n$, and if $c$ is coprime 