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 $\mathbb{Q}[\sqrt[3]{d}]$, 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 
[![https://i.sstatic.net/QSC0S.png][1]][1] 

[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. 

[![https://i.sstatic.net/mCbWR.png][2]][2] 

[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: 

[![https://i.sstatic.net/suz3l.png][3]][3]

Any comments or hints are welcome. 


  [1]: https://i.sstatic.net/QSC0S.png
  [2]: https://i.sstatic.net/mCbWR.png
  [3]: https://i.sstatic.net/suz3l.png