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definition

Lorentz Algebra

The Lorentz algebra as a Lie subalgebra over $\mathbb{R}$ of the space of real matrices indexed by $\text{Fin } 1 \oplus \text{Fin } 3$ .

theorem

$A^T \eta = -\eta A$ for $A \in \mathfrak{so}(1, 3)$

#transpose_eta

For any element $A$ in the Lorentz algebra $\mathfrak{so}(1, 3)$ , the matrix representation of $A$ satisfies the relation $A^T \eta = -\eta A$ , where $A^T$ is the transpose of $A$ and $\eta$ is the Minkowski metric tensor.

theorem

$A^\top \eta = -\eta A$ implies $A \in \mathfrak{so}(1,3)$

#mem_of_transpose_eta_eq_eta_mul_self

Let $A$ be a real matrix indexed by $\text{Fin } 1 \oplus \text{Fin } 3$ . If $A$ satisfies the condition $A^\top \eta = -\eta A$ , where $\eta$ is the Minkowski metric tensor, then $A$ is an element of the Lorentz algebra $\mathfrak{so}(1,3)$ .

theorem

$A \in \mathfrak{so}(1, 3) \iff A^\top \eta = -\eta A$

#mem_iff

Let $A$ be a real matrix indexed by $\text{Fin } 1 \oplus \text{Fin } 3$ . Then $A$ is an element of the Lorentz algebra $\mathfrak{so}(1, 3)$ if and only if it satisfies the condition $A^\top \eta = -\eta A$ , where $\eta$ is the Minkowski metric tensor.

theorem

$A \in \mathfrak{so}(1, 3) \iff A = -\eta A^\top \eta$

#mem_iff'

Let $A$ be a real matrix indexed by $\text{Fin } 1 \oplus \text{Fin } 3$ . Then $A$ is an element of the Lorentz algebra $\mathfrak{so}(1, 3)$ if and only if it satisfies the condition $A = -\eta A^\top \eta$ , where $\eta$ is the Minkowski metric tensor and $A^\top$ is the transpose of $A$ .

theorem

Diagonal entries of $\Lambda \in \mathfrak{so}(1, 3)$ are zero

#diag_comp

Let $\Lambda$ be a matrix in the Lorentz algebra $\mathfrak{so}(1, 3)$ . For any index $\mu \in \text{Fin } 1 \oplus \text{Fin } 3$ , the diagonal entry of $\Lambda$ is zero, i.e., $\Lambda_{\mu\mu} = 0$ .

theorem

$\Lambda_{i,0} = \Lambda_{0,i}$ for $\Lambda \in \mathfrak{so}(1, 3)$

#time_comps

Let $\Lambda$ be a matrix in the Lorentz algebra $\mathfrak{so}(1, 3)$ , which is indexed by the set $\{0\} \cup \{1, 2, 3\}$ where $0$ represents the temporal dimension and $\{1, 2, 3\}$ represent the spatial dimensions. For any spatial index $i \in \{1, 2, 3\}$ , the mixed temporal-spatial entries of the matrix are equal, i.e., $\Lambda_{i,0} = \Lambda_{0,i}$ .

theorem

$\Lambda_{ij} = -\Lambda_{ji}$ for spatial indices $i, j$ of $\Lambda \in \mathfrak{so}(1, 3)$

#space_comps

Let $\Lambda$ be a matrix in the Lorentz algebra $\mathfrak{so}(1, 3)$ , where the indices consist of a temporal dimension and three spatial dimensions $\{1, 2, 3\}$ . For any spatial indices $i, j \in \{1, 2, 3\}$ , the spatial components of the matrix are antisymmetric, satisfying $\Lambda_{ij} = -\Lambda_{ji}$ .

Physlib.Relativity.LorentzAlgebra.Basic

Lorentz Algebra

ATη=−ηAA^T \eta = -\eta AATη=−ηA for A∈so(1,3)A \in \mathfrak{so}(1, 3)A∈so(1,3)

A⊤η=−ηAA^\top \eta = -\eta AA⊤η=−ηA implies A∈so(1,3)A \in \mathfrak{so}(1,3)A∈so(1,3)

A∈so(1,3) ⟺ A⊤η=−ηAA \in \mathfrak{so}(1, 3) \iff A^\top \eta = -\eta AA∈so(1,3)⟺A⊤η=−ηA

A∈so(1,3) ⟺ A=−ηA⊤ηA \in \mathfrak{so}(1, 3) \iff A = -\eta A^\top \etaA∈so(1,3)⟺A=−ηA⊤η

Diagonal entries of Λ∈so(1,3)\Lambda \in \mathfrak{so}(1, 3)Λ∈so(1,3) are zero

Λi,0=Λ0,i\Lambda_{i,0} = \Lambda_{0,i}Λi,0​=Λ0,i​ for Λ∈so(1,3)\Lambda \in \mathfrak{so}(1, 3)Λ∈so(1,3)

Λij=−Λji\Lambda_{ij} = -\Lambda_{ji}Λij​=−Λji​ for spatial indices i,ji, ji,j of Λ∈so(1,3)\Lambda \in \mathfrak{so}(1, 3)Λ∈so(1,3)

$A^T \eta = -\eta A$ for $A \in \mathfrak{so}(1, 3)$

$A^\top \eta = -\eta A$ implies $A \in \mathfrak{so}(1,3)$

$A \in \mathfrak{so}(1, 3) \iff A^\top \eta = -\eta A$

$A \in \mathfrak{so}(1, 3) \iff A = -\eta A^\top \eta$

Diagonal entries of $\Lambda \in \mathfrak{so}(1, 3)$ are zero

$\Lambda_{i,0} = \Lambda_{0,i}$ for $\Lambda \in \mathfrak{so}(1, 3)$

$\Lambda_{ij} = -\Lambda_{ji}$ for spatial indices $i, j$ of $\Lambda \in \mathfrak{so}(1, 3)$