La maison d'Hades

Let (E, +) be a set equipped with an internal additive operation. Let E^D be the set of D-tuples whose arguments are ordered elements of (E, +).

Let e = (e₀, ..., e_{D - 1}) be any element of E^D. The sum functor acts on it in such a way that:

e ∈ E^D→ e^⊕ = ∑_α e_α ∈ E

Let (E, +) be a set equipped with an internal additive operation. Let M(D, E) be the set of square matrices having D rows and D columns whose entries are elements of (E, +). Let [M] be any element of M(D, E). The sum functor acts on it in such a way that:

[M] = [M_αβ] ∈ M(D, E) → [M]^⊕ = ∑_α ∑_β M_αβ ∈ E

Let (E, +) be a set equipped with an internal additive operation. Let ⊞(D, E) be the set of cubes formed by the superposition of D elements in M(D, E). Let C be any element in ⊞(D, E). The sum functor acts on it in such a way that:

C = [c^χ_αβ] ∈ ⊞(D, E) → C^⊕ = ∑_α ∑_β ∑_χ c^χ_αβ ∈ E

Let (E, .) be a set equipped with an internal multiplicative operation. Let E^D be the set of D-tuples whose arguments are ordered elements of (E, +). Let e = (e₀, ..., e_{D - 1}) and f = (f₀, ..., f_{D - 1}) be any pair of elements in E^D. The left shift functor of e on f acts in such a way that:

(e, f) ∈ E^D × E^D

↓

 e • f = (e₀.f₀, ..., e_{D - 1}. f_{D - 1}) ∈ E^D

Sliding is an internal operation on E^D.

The principle explained through this example can easily be generalized to the elements of a vector space, a set of matrices, or a set of cubes. 

In the empty regions of the universe referred to a Minkowski geometry, it is possible to calculate the second invariant of an electromagnetic field by sliding its twice-covariant representation onto its twice-contravariant representation and then summing the result thus obtained.

The formalism of the demonstration forces me to use coding and you will therefore only be able to discover it by activating the widget.
 
I wish to calculate:

A) In a Minkowski metric with the (+ - - -) signature there is an equivalence between the classical tensor calculus and the "sliding-summation" calculation performed above regarding the second invariant of an electromagnetic field.
 
B) The "sliding-summation" calculation is extraordinarily more economical than the tensor calculus performed on the page "classical calculation of the second invariant of an electromagnetic field". Indeed, it requires only sixteen simple multiplications and one addition, i.e., 17 operations in total.

C) Potentially important for future developments, this result remains valid when the components of the electromagnetic field tensor are quaternions.

This section is devoted to the calculation of an expression concerning matrices, the formalism of which appears on various occasions in electromagnetic theory. I am referring to the set M-(4, C) of antisymmetric elements of M(4, C).

The objective is to generically calculate the expression:

∀ [M], [N] ∈ M-(4, C) : {[M].[N]}^⊕ = ?

The length of the demonstration once again prompts me to use coding. Activating the widget is therefore necessary to discover it.

Let apply this generic calculus to the specific case defined by: [M] = [F(2,0) and [N] = [F(0, 2)] when the firlds have components in R or C.

This mathematical excursion shows that there are physical circumstances making this expression equivalent to the second invariant of the electromagnetic field at hand.

Specifically, when the relationship below is verified:

½. {(E^⊕)^2 + (H^⊕)^2} = < H, H >

... then the sum of the entries of the matrix product between the twice-covariant version and the twice-contravariant version of the matrix representations of an electromagnetic field is equal to minus one times the value of the second invariant of this field when considered in an empty region.

Ψ = Inv2(E, H)

Ψ + {[F(2, 0)]. [F(0,2)]}^⊕ = 0

It is obvious that this algebraic calculation curiosity is not very economical in the number of operations to be performed; see the previous calculation method for comparison. However, it remains correct whenever the field has components in a commutative field.

Page first published on august 26, 2026