I got interested in the question of possible geometric interpretations of the multiplication in algebraic number fields of degree $>2$ (with application to multiplication of units in the ring of algebraic integers of that field), inspired by a talk of Franz Lemmermeyer in honor of Peter Roquette's 90th anniversary some weeks ago.
First, let's consider a pure cubic field $\mathbb{Q}[\sqrt[3]{d}]$$\mathbb{Q}(\sqrt[3]{d})$, an element $(x,y,z)^{\top} = x + d^{1/3}y + d^{2/3}z$ with norm $N((x,y,z)^{\top}) = x^3 + dy^3 + d^2z^3 - 3dxyz$ and multiplication
$$ \left( \begin{array}{c} x_1 \\ y_1 \\ z_1 \end{array} \right) \cdot \left( \begin{array}{c} x_2 \\ y_2 \\ z_2 \end{array} \right) = \left( \begin{array}{c}x_1x_2 + d(y_1z_2+z_1y_2)\\ x_1y_2 + y_1x_2 + dz_1z_2 \\ x_1z_2 + y_1y_2 + z_1x_2 \end{array} \right).$$
Due to
$$
\begin{aligned}
& x^3 + dy^3 + d^2z^3 - 3dxyz = \\
& (x + d^{1/3}y + d^{2/3}z)(x^2 + d^{2/3}y^2 + d^{4/3}z^2 - d^{1/3}xy - d^{2/3}xz - dyz) = 1
\end{aligned}
$$
one realizes that the norm-1-surface is funnel-shaped with an asymptotic plane $x + d^{1/3}y + d^{2/3}z = 0$ and the opening of the funnel showing in direction $(d^{2/3}, d^{1/3}, 1)^{\top}$. The angle between the normal of the plane and the funnel direction is $\arccos(3d^{2/3}/(1+d^{2/3}+d^{4/3})$. Cutting the surface with planes parallel to the asymptotic plane yields ellipses.
See figure
[Norm-1-surface for $d=2$ shown with $(1,1,1)$ (red) and $(1,1,1) \cdot (1,1,1) = (5,4,3)$ (green). The green line indicates the normal of the asymptotic plane, the (nearly not visible) blue line the funnel direction.]
For the number field irrelevant, geometrically yet reasonable case $d=1$ the norm-1-surface gives a rotationally symmetric funnel (cf.
[Norm-1-surface for $d=1$ shown with powers (multiples) of a point, spiraling around the funnel]) and the group multiplication is just multiplication of the height above the asymptotic plane and complex multiplication (i.e. addition of angles) in the plane perpendicular (parallel to asymptotic plane).
Interestingly this is also valid for the case of general $d$:
Applying the linear transformation $$
\left( \begin{array}{c} x' \\ y' \\ z' \end{array} \right) =
\left(
\begin{array}{ccc}
1 & d^{1/3} & d^{2/3} \\
1 & -\frac{1}{2}d^{1/3} & -\frac{1}{2}{d^{2/3}} \\
0 & \frac{1}{2} \sqrt{3} d^{1/3} & -\frac{1}{2} \sqrt{3} d^{2/3} \\
\end{array}
\right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right)
$$
gives the group multiplication law in the new coordinates as
$$ \left( \begin{array}{c} x'_1 \\ y'_1 \\ z'_1 \end{array} \right) \cdot \left( \begin{array}{c} x'_2 \\ y'_2 \\ z'_2 \end{array} \right) = \left( \begin{array}{c} x'_1x'_2 \\ y'_1y'_2 - z'_1z'_2 \\ y'_2z'_1 + y'_1z'_2 \end{array} \right),$$
i.e. a simple multiplication in $x'$-direction und complex multiplication (i.e. addition of angles and multiplication of modulus) in the $y'-z'$-plane.
I don't know but would assume that this has been remarked before (I'm not at all an expert in algebraic number theory and don't have an overview of the literature.) I found it pretty interesting that by a pure linear transformation one can bring the group multiplication in such a simple form and one thus has a kind of geometric interpretation of units in the corresponding ring.
My question is whether this observation could be generalized to higher degree ($>3$) algebraic fields, e.g. quartic or quintic fields.
For the cubic field I found the linear transformation above rather easily by hand and a bit of Mathematica calculation.
But the more proper (and for higher degrees, thus higher dimensions necessary) way to proceed would be some sort of normal form theory of 3-tensors, which I'm not familiar with and which I could not find easily.
Here my thoughts: If the multiplication of the algebraic numbers in vector components (generalization of the multiplication given above) is given by a 3-tensor $M$, i.e. $(x_1 \cdot x_2)_k = \sum_{i,j=1}^{\text{degree of the field}} M_{ijk}x_{1i} x_{2j}$, then by applying a linear transformation $x_i = \sum B_{ii'} x'_{i'}$ one gets $M'_{i'j'k'} = \sum_{ijk}B^{-1}_{k'k}M_{ijk}B_{ii'}B_{jj'}$, and we are looking for $B$ such that $M'$ gets as simple as possible (ideally diagonal or with $2\times2$-blocks, or even $4\times 4$-blocks and quaternion multiplication(?); for degree three we find one one-dimensional and one two-dimensional block, see above the multiplication law in the prime coordinates).
As I have it on my hand, here a 3d slice of the norm-1-surface for a pure cubic field of degree 4:
Any comments or hints are welcome.
