Physlib.Relativity.Tensors.LeviCivita.Contractions

The sum of the squares of all components of the 4-dimensional Euclidean Levi-Civita symbol $\epsilon_{i_0 i_1 i_2 i_3}$ over all possible indices $i_0, i_1, i_2, i_3 \in \{0, 1, 2, 3\}$ is equal to 24: $$\sum_{i_0, i_1, i_2, i_3 = 0}^3 \epsilon_{i_0 i_1 i_2 i_3}^2 = 24$$ The Levi-Civita symbol $\epsilon_{i_0 i_1 i_2 i_3}$ is defined to be the sign of the permutation $(i_0, i_1, i_2, i_3)$ if the indices are a permutation of $(0, 1, 2, 3)$, and zero otherwise. This result represents the full Euclidean contraction of the tensor components, effectively counting the $4! = 24$ possible permutations.

Let $\epsilon_{ijkl}$ denote the 4-dimensional Euclidean Levi-Civita symbol for indices in $\{0, 1, 2, 3\}$, and let $\delta_{ab}$ denote the Kronecker delta. For any indices $a, b \in \{0, 1, 2, 3\}$, the triple contraction over the remaining three indices is given by: $$\sum_{h_1, h_2, h_3 = 0}^3 \epsilon_{a h_1 h_2 h_3} \epsilon_{b h_1 h_2 h_3} = 6 \delta_{ab}$$

For any indices $r, s, t, w \in \{0, 1, 2, 3\}$, the double contraction of the 4-dimensional Euclidean Levi-Civita symbol $\epsilon_{ijkl}$ over its last two indices satisfies: $$\sum_{h_1, h_2 = 0}^{3} \epsilon_{r s h_1 h_2} \epsilon_{t w h_1 h_2} = 2 (\delta_{rt}\delta_{sw} - \delta_{rw}\delta_{st})$$ where $\delta_{ij}$ denotes the Kronecker delta.

In 4-dimensional Minkowski spacetime, let $\epsilon^{\mu\nu\rho\sigma}$ be the rank-4 contravariant Levi-Civita tensor (with $\epsilon^{0123} = 1$). For any multi-index $b = (\mu, \nu, \rho, \sigma)$, the components of the covariant Levi-Civita tensor $\epsilon_{\mu\nu\rho\sigma}$, obtained by lowering all four indices using the Minkowski metric, are the negative of the corresponding contravariant components: $$(\epsilon_{\mu\nu\rho\sigma})_b = -(\epsilon^{\mu\nu\rho\sigma})_b$$

In four-dimensional spacetime (corresponding to $d=3$ spatial dimensions), the sum of the squares of the standard-basis components of the rank-four contravariant Levi-Civita tensor $\epsilon^{\mu\nu\rho\sigma}$ is $24$: $$\sum_{\mu, \nu, \rho, \sigma = 0}^3 (\epsilon^{\mu\nu\rho\sigma})^2 = 24$$ where the sum is taken over all multi-indices $b = (\mu, \nu, \rho, \sigma)$ in the standard basis.

In 4-dimensional spacetime (corresponding to $d=3$ spatial dimensions), let $\epsilon^{\mu\nu\rho\sigma}$ denote the components of the rank-four contravariant Levi-Civita tensor $\epsilon^4$. For any indices $a, b \in \{0, 1, 2, 3\}$, the contraction of the first three components of the Levi-Civita symbol is given by: $$\sum_{h_0, h_1, h_2 = 0}^{3} \epsilon^{h_0 h_1 h_2 a} \epsilon^{h_0 h_1 h_2 b} = 6 \delta_{ab}$$ where $\delta_{ab}$ is the Kronecker delta, which equals 1 if $a = b$ and 0 otherwise.

In 4-dimensional spacetime ($d=3$), let $\epsilon^{\mu\nu\rho\sigma}$ denote the rank-four Levi-Civita tensor. For any component index $b = (\sigma, \tau)$, the $(\sigma, \tau)$-th component of the tensor contraction of two Levi-Civita tensors over three indices, $\epsilon^{\mu\nu\rho\sigma} \epsilon_{\mu\nu\rho\tau}$, is given by the sum: $$\left[ \epsilon^{\mu\nu\rho\sigma} \epsilon_{\mu\nu\rho\tau} \right]_{\sigma\tau} = \sum_{\mu, \nu, \rho} \epsilon^{\mu\nu\rho\sigma} \epsilon_{\mu\nu\rho\tau}$$ where the indices $\mu, \nu, \rho$ are summed over the set $\{0, 1, 2, 3\}$.

The full contraction of the rank-4 Levi-Civita tensor $\epsilon^{\mu \nu \rho \sigma}$ with its fully lowered form $\epsilon_{\mu \nu \rho \sigma}$ in 4-dimensional spacetime ($d=3$ spatial dimensions) is equal to the sum over all multi-indices $b$ of the products of their corresponding standard-basis components. Specifically, $$(\epsilon^{\mu \nu \rho \sigma} \epsilon_{\mu \nu \rho \sigma}) = \sum_{b} \epsilon_{b} \cdot (\epsilon^*)_{b}$$ where $\epsilon_{b}$ denotes the standard-basis component of the Levi-Civita tensor for the multi-index $b$, and $(\epsilon^*)_{b}$ denotes the component of its fully lowered form.