La maison d'Hades

It would be nice to be able to condense this condition. This opportunity might be given by visualizing cube A in a three-dimensional space. It is formed by the superposition of three elements of M(3, C) whose entries can generically be labeled as follows (arbitrary choice):

depth, row, column = 1, 2, 3

A^row_column-depth

With this convention, any node-based cube (3-3-3) in C is the superposition of three matrices:

 [A]_(depth)

So that:

• (i) by summing over the indices i from 1 to 3;
• (ii) noting the indices modulo 3;
• (iii) taking into account the antisymmetry on the lower indices,

...the condition can now be written as (the symbol Tr{...} denotes the trace of ...) :

∀ i = 1, 2, 3 :
( ∑_i Aⁱ₁₂ ). Tr{[A]₍₃₎}
+ ( ∑_i Aⁱ₂₃ ). Tr{[A]₍₁₎}
+ ( ∑_i Aⁱ₃₁ ). Tr{[A]₍₂₎}
=
∑_m (∑_i Aⁱ_m-m+1). Tr{[A]_(m+2)}
= 
0

A close observation of the deforming matrix [A] conventionally defined by:

[A]

=

... leads to the observation that each of the three sums:

∑_i Aⁱ_m-m+1is the sum of the entries of one of the three rows of the warping matrix [A].

Thus, by deciding to consider that any deforming matrix is the superposition of the dual representations in C³ of three vectors a_2.m+1, the condition transforming a cross product deformed by the matrix [A] into a Lie bracket deformed by this matrix is conventionally written as:

∑_m a^⊕_2m+1. Tr{[A]_(m+2)} = 0

This formulation is certainly concise, but at this stage, it is difficult to do anything with it.

There is another way to condense the condition concerning each component of the Jacobiator and I am going to present it.

It is however not entirely orthodox because it does not consider each component in isolation but the Jacobiator as a whole.

What follows can only be visualized on a computer and by activating the widget allowing the reading of equations (MathJax).

The Jacobiator of any element {a, b, c} in P{E(3, C) ; 3} for the cross product deformed by a matrix [A] in M(3, C) vanishes as soon as the following vector condition is verified:

{-[A].[J]^t._[J]Φ([A].| Ω >)}^⊕ = 0_{E(3, C)} 

This formula contains a Φ-type matrix mimicking the representation of an axial rotation. However, the matrix shown here differs completely from its little sister.
 
First, because it is no longer an element in M(3, C) but a matrix in M(3, E(3, C)). In other words, its entries are vectors with components in the commutative field of complex numbers rather than simply complex numbers.
 
Next, the vector X which is the argument of this Φ-type matrix is nothing other than the initial canonical basis Ω which is transformed by the deforming matrix [A].

Here, the basis is considered as any element in E³(3, C) but, at this stage of the explorations, nothing provides the assurance that the vector X represents a new basis.
 

Within the framework of this theory, a classical cross vector is also a cross product distorted by the generator of the cyclic group C6, i.e., the matrix [J]. By substituting the equality [A] = [J] into the condensed formula found above, it reduces to:

{_[J]Φ([J].| Ω >)}^⊕ = 0_{E(3, C)} 

It is then a tautology because the sum of the arguments of a matrix of type Φ is always zero. This proves once again that the classical cross product does indeed define a Lie bracket and that {E(3, C), ∧} is a Lie algebra.
 
Thus, although this second way of condensing the Jacobiator may seem unorthodox in the sense that it still leaves many doors open to unexplored subjects, it at least makes it possible to recover a known result. Which has the advantage of being reassuring.

© Thierry PERIAT

Page created on October 19, 2026

This page is a translation of the original French version