# What are the Voronoi cones in 4 variables?

Question: What are the top dimensional cones of the 2nd Voronoi decomposition of the space of positive definite forms in $4$ variables?

The 2nd Voronoi decomposition of the cone of positive definite in $g$ variables is a $\operatorname{GL}_{g}(\mathbb{Z})$-invariant decomposition into convex cones. When $g=4$, there are suppose to be 3 top-dimensional cones up to $\operatorname{GL}_{4}$-equivalence. One cone is the principal cone spanned by the forms $$\{ x_i^2, (x_i-x_j)^{2} \}$$ (or rather the intersection of their span with the cone of positive definite forms). The other two cones can be taken to be cones subdividing the cone spanned by $$\{ x_i^2, (x_1-x_3)^2, (x_1-x_4)^2, (x_2-x_3)^2, (x_2-x_4)^2, (x_3-x_4)^2, (x_1+x_2-x_3)^2, (x_1+x_2-x_4)^2, (x_1+x_2-x_3-x_4)^{2} \}$$ (or rather the intersection of their span with the cone of positive definite forms).

But which two cones? What is the subdivision?

P.S. Recall the definition of the 2nd Voronoi decomposition. Suppose that we are given a positive definite quadratic form $q$ in $g$ variables. If $v \in \mathbb{R}^{g}$ is a vector, then let $w_1, \dots, w_n \in \mathbb{Z}^{g}$ be the lattice points nearest to $v$ with respect to the metric defined by $q$ and define their convex hull to be $D(a)$. The $D(a)$'s form a polyhedreal decomposition of $\mathbb{R}^{g}$ called the Delony decomposition.

We say that two forms $q_1$ and $q_2$ are Voronoi-equivalent if their define the same Delony decomposition of $\mathbb{R}^{g}$. The closures of the resulting equivalence classes are the cones defining the 2nd Voronoi decomposition of the cone of positive definite forms.

P.P.S.: I am tagging this as algebraic geometry because this fan comes up in the study of moduli of abelian varieties.

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Maybe you'll find Lee/Szcarba's paper "On the homology of $K_3 \mathbb{Z}$ and $K_4 \mathbb{Z}$" useful, or Paul Gunnell's appendix in Stein's AMS book "Modular Forms: A computational approach", or any of the Gunnells/McConnell/Ash papers. Or ask any one of them. –  J. Martel Jul 1 '13 at 5:54