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I will start by recalling some general points to properly frame the discussion beginning on this page. The theory of (E) Question begins with the approach widely explained in the introduction to this chapter.
As already explained on other pages, the deformed cross product can be understood as a tensor product deformed by a cube (3-3-3) of which the lower indices of the nodes are endowed with an antisymmetry property; review the page dealing with deforming cubes.
Aχαβ + Aχβα = 0 ⇔ A ∈ ⊞-(D, K)
On the other hand, it has not yet been explained on the page defining the various types of deformed products (tensorial, alternating, and vector) whether deformed vector products could sometimes be deformed Lie products. I am therefore going to explore this aspect here.
In this theory, a deformed vector product is by definition half of an alternating deformed tensor product built on a cube whose lower indices satisfy the so-called antisymmetry property.
It effectively becomes a deformed Lie product when it equips the space with a Lie algebra structure; that is, when the following three properties are satisfied:
∀ a, b, c ∈ E(D, K)
∀ A ∈ ⊞-(D, K)
[a, a]A = 0
[a, b]A + [b, a]A = 0
J[...,...]{a, b, c}
=
[a, [b, c]A]A + [b, [c, a]A]A + [c, [a, b]A]A
=
0
Scientific literature refers to the third condition by stating that the Jacobi identity, for the Lie bracket of any subset of E(D, K) whose cardinality is three, is zero.
Since a deformed cross product is equal to half of a deformed alternating tensor product built on an antisymmetric cube, the one-half factor plays no structural role and the definition of a deformed Lie product can be rewritten using the symbolism of deformed alternating tensor products.
∀ a, b, c ∈ E(D, K)
∀ A ∈ ⊞-(D, K)
∧A(a, a) = 0
∧A(a, b) + ∧A(b, a) = 0
∧A(a, ∧A(b, c)) + ∧A(b, ∧A(c, a)) + ∧A(c, ∧A(a, b)) = 0
Furthermore, still by definition:
∧A(a, b)
=
2. ∑χ {∑α < β ∑β Aχαβ. (aα. bβ - aβ. bα)}.eχ
It easily follows that the first property is always satisfied. The same holds true for the second as long as K is a set equipped with a commutative product:
∧A(a, b) + ∧A(b, a)
=
2. ∑χ {∑α < β ∑β Aχαβ. (aα. bβ - aβ. bα)
+ (bα. aβ - bβ. aα)}.eχ
=
0
The characterization of the third property requires much more work and holds a few surprises that will prove to be interesting.
The detailed calculations in this section are not accessible on mobile phones.
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The calculations presented in extenso below concern only three-dimensional spaces (D = 3).
The deforming cubes A are therefore represented by arbitrary elements [A] in M(3, K).
Let the three double-products be:
The Jacobian is:
The formalism of the Jacobiator for the case where D = 3 immediately demonstrates the existence of two particular subsets of situations that allow for its cancellation.
Subset 1: It contains the situations for which, independently of the arguments a, b and c, the entries of the respective deforming matrices respect the condition:
∀ a, b, c ∈ E(3, K)
A ∈ ⊞-(3, K)
∀ i = 1, 2, 3 :
(Ai12. A223 + Ai12. A113 + Ai13. A323)
=
(Ai23. A313 + Ai23. A212 + Ai13. A112)
Subset 2: It contains the situations for which, regardless of the deforming matrices, each triplet {a, b, c} respects the conditions:
< a, b ∧ c > = 0K
< b, c ∧ a > = 0K
< c, a ∧ b > = 0K
Each of these two particular subsets is the subject of a detailed separate development on another page.
In three-dimensional vector spaces (D = 3) built on commutative fields (for example real numbers and complex numbers), it is possible to deform classical cross products; see the previous page.
These deformed cross products become deformed Lie products under certain conditions which can be divided into at least two subsets of particular situations ( see above).
© Thierry PERIAT - Page created on the 18th of October 2026