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.stack.imgur.com/QSC0S.png

[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.

Edit 03.12.17

I have made some progress in finding a method that gives the linear transformation for which the multiplication law has the simplest possible form. For the degree-3-case the result above is reproduced.

For the degree-4-case: Until now I could only handle the special (algebraically irrelevant) case $d=1$, for which the multiplication is given by $$ \left( \begin{array}{c} x_1 \\ y_1 \\ z_1 \\ w_1 \end{array} \right) \cdot \left( \begin{array}{c} x_2 \\ y_2 \\ z_2 \\w_2 \end{array} \right) = \left( \begin{array}{c}x_1x_2 + (y_1w_2+ z_1z_2 + w_1y_2)\\ x_1y_2 + y_1x_2 + (z_1w_2 +w_1z_2) \\ x_1z_2 + y_1y_2 + z_1x_2 + (w_1w_2) \\ x_1w_2 + y_1z_2+z_1y_2+w_1x_2 \end{array} \right).$$ (The brakets indicate where for general $d$ a factor $d$ occurs.)

Applying the linear transformation $$ \left( \begin{array}{c} x' \\ y' \\ z' \\w' \end{array} \right) = \left( \begin{array}{ccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \\w \end{array} \right) $$ gives the group multiplication law in the new coordinates as

$$ \left( \begin{array}{c} x'_1 \\ y'_1 \\ z'_1 \\w'_1 \end{array} \right) \cdot \left( \begin{array}{c} x'_2 \\ y'_2 \\ z'_2 \\ w'_2 \end{array} \right) = \left( \begin{array}{c} x'_1x'_2 \\ y'_1y'_2 \\ z'_1z'_2 - w'_1w'_2 \\ z'_1w'_2 + z'_2w'_1 \end{array} \right),$$ i.e. a simple multiplication in $x'$- and $y'$-directions und complex multiplication (i.e. addition of angles and multiplication of modulus) in the $z'-w'$-plane.

I assume/hope that my method also works for general $d$, but could not yet verify it due to very cumbersome expressions. I will try for special values of $d$ numerically.

Edit 05.12.17

Following the hint given by Lee Mosher, I have had a look into Stewart/Tall, Algebraic Number Theory and Fermat's Last Theorem, 3rd edition. Chapter 8 contains more or less what I was looking for. (Lesson learnt for me: Obviously the goal of transforming the multiplication law in the most simple form is far easier achievable by means of algebraic number theory than with my approach of (multi)linear algebra (sort of generalized eigenwert equation for 3-tensors), but it was good to see it worked, at least for low degrees.)

  • 3
    $\begingroup$ Have you looked at the Higher Composition Laws papers of Manjul Bhargava? $\endgroup$ – Stanley Yao Xiao Nov 26 '17 at 22:35
  • $\begingroup$ @Stanley Yao Xiao. Many thanks for pointing this out! I will try to have a look at these papers, probably "Higher composition laws I", Annals of Mathematics, 159 (2004), 217-250 is a good starting point. $\endgroup$ – Andreas Rüdinger Nov 26 '17 at 22:54
  • 2
    $\begingroup$ The geometric properties of norm 1 surfaces, and their interaction with multiplication in the number ring and the number field, are the underlying ideas behind the proof of the Dirichlet units theorem. I like how the book by Stewart and Tall presents these ideas, but I don't know very many textbooks in this field, and I suspect you should be able to find these ideas in any algebraic number theory book $\endgroup$ – Lee Mosher Dec 3 '17 at 21:08
  • $\begingroup$ @Lee Mosher. Many thanks! I have had a look into Stewart/Tall, Algebraic Number Theory and Fermat's Last Theorem, 3rd edition. Chapter 8 contains more or less what I was looking for. I will study it in more detail. Many thanks again. $\endgroup$ – Andreas Rüdinger Dec 5 '17 at 21:50

My rep. is not high enough to leave a comment so this has to go here. I comment that I have done a detailed study of matrices similar to those you mention for cubic fields only. A small part of that will be in: A cubic generalization of Brahmagupta's identity, J. Ramanujan Math. Soc., 32, (2017) no. 4, 327-337. Also, see : http://mathoverflow.net/questions/61859/, which is related. bf Edit 29/11/17: I can give a partial answer to the question in the following: Let $$N_{\mathcal{C}}^{(\alpha )} = \left( \begin{array}{ccc} u & -a d y & -a d x-b d y \\ x & u-b x-c y & -c x-d y \\ y & a x & u-c y \\ \end{array} \right) ,$$ where $\mathcal{C}(x, y) = (a, b, c, d)$ is an index form for the cubic field $K = \mathbb{Q}(\delta )$, $\delta \in \mathbb{R} $ satisfies $\mathcal{C}(\delta , 1) = 0$, and $u,x,y$ are coefficients of Belabas' integral basis $\{ 1, \theta , \phi \}$, $\theta = a \delta $, $\phi = a \delta^2 + b \delta $,and $\alpha = u + x \theta + y \phi $. With $\alpha \in K$ or the ring of integers, the matrices commute with one another, their determinant is the norm, their sum can be used to produce the sum $\alpha_1 + \alpha_2$, the trace of the matrix is the trace of $\alpha $, the inverse of the matrix can be used to produce the inverse of the $\alpha \in K$. One extracts the first column and builds $u + x \theta + y \phi \in K$ from it. Now the proofs of all of this should work if instead of letting $u,x,y \in \mathbb{Q}$, we let $u, x, y \in F$, another field for which we know the associated matrices, e.g. the quadratic field $\mathbb{Q}(\sqrt{D})$. From that it should be possible to find matrices that do the same job as these do for the appropriate extension field, e.g. the normal closure of $K$.

  • $\begingroup$ Many thanks for having added details to your answer. I'm still struggling to understand how exactly I can apply your approach to my problem, but will think about it. $\endgroup$ – Andreas Rüdinger Nov 29 '17 at 20:01

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.