Euclidean contraction identities for the Levi-Civita tensor

i. Overview

This file proves the "epsilon-epsilon" contraction identities for the rank-four Levi-Civita tensor leviCivita (notation ε4) in d = 3, stated in terms of the standard-basis components of ε4 itself (realLorentzTensor.leviCivita_basis_repr_apply).

The underlying facts about the generalizedKroneckerDelta alone, with no tensor content — lives in Physlib.Mathematics.KroneckerDelta.Contraction, next to the definition of generalizedKroneckerDelta. Here we specialise those facts to the components of ε4, where (ε4)_b = (Tensor.basis _).repr ε4 b is the standard-basis component of ε4, an integer Levi-Civita symbol carried to the reals, and the sums run over the remaining (uncontracted) component slots.

ii. Key results

leviCivita_symbol_contract_zero : ∑_b (ε4)_b · (ε4)_b = 24 (full Euclidean contraction).

leviCivita_symbol_contract_one : ∑_h (ε4)_{a,h} · (ε4)_{b,h} = 6 · δ[a,b].

leviCivita_symbol_contract_two : ∑_h (ε4)_{r,s,h} · (ε4)_{t,w,h} = 2 · (δ[r,t]·δ[s,w] - δ[r,w]·δ[s,t]).

iii. Table of contents

A. The combinatorial bridge lemma

B. Euclidean epsilon-epsilon contraction identities

iv. References

A. Euclidean epsilon-epsilon contraction identities

Full Euclidean Levi-Civita contraction ∑_b (ε4)_b · (ε4)_b = 24 at the symbol level: summing the square of every standard-basis component of ε4 over all four Fin 4 index slots, paired naively (no metric), counts the 4! = 24 permutations. The Lorentz contraction ε^{μνρσ} ε_{μνρσ} lowers one factor with η and equals -24 instead.

Triple Euclidean Levi-Civita contraction ∑_h (ε4)_{a,h} · (ε4)_{b,h} = 6 · δ[a,b] at the symbol level: contracting three of the four Fin 4 component slots of ε4 with the naive Kronecker pairing leaves one free pair a, b and the factor 3! = 6. The Lorentz form carries an extra det η = -1.

Double Euclidean Levi-Civita contraction ∑_h (ε4)_{r,s,h} · (ε4)_{t,w,h} = 2 · (δ[r,t]·δ[s,w] - δ[r,w]·δ[s,t]) at the symbol level: contracting two of the four Fin 4 component slots of ε4 with the naive Kronecker pairing leaves two free pairs and the factor 2! = 2. The Lorentz form carries an extra det η = -1.