Physlib.Relativity.LorentzAlgebra.ExponentialMap

Let $A$ be an element of the Lorentz algebra $\mathfrak{so}(1, 3)$ and let $\eta$ denote the Minkowski metric tensor. The transpose of the matrix representation of $A$, denoted $A^T$, satisfies the relation $A^T = -\eta A \eta$.

Let $A$ be an element of the Lorentz algebra $\mathfrak{so}(1, 3)$ and let $\eta$ denote the Minkowski metric matrix. Then the matrix exponential of the transpose of $A$ satisfies the identity: $$\exp(A^T) = \eta \exp(-A) \eta$$

Let $A$ be an element of the Lorentz algebra $\mathfrak{so}(1, 3)$. Then the matrix exponential of $A$, denoted $\exp(A)$, is an element of the Lorentz group $O(1, 3)$.

Given that \( \mathbb{K} \) is a uniform space, the space of \( m \times n \) matrices over \( \mathbb{K} \) is also equipped with a uniform space structure.

For any real matrix $A$ with rows and columns indexed by $\text{Fin } 1 \oplus \text{Fin } 3$, the trace of $A$ equals the sum of its diagonal elements, i.e., $\operatorname{trace}(A) = \sum_i A_{ii}$.

For any element $A$ of the Lorentz algebra $\mathfrak{so}(1, 3)$ (the Lie algebra of the Lorentz group in 1+3 Minkowski space), the trace of its matrix representation is zero, i.e., $\operatorname{tr}(A) = 0$.

For any semiring $R$, finite types $n$ and $\iota$, a bijection $e : n \simeq \iota$, and an $n \times n$ matrix $A$ over $R$, the trace of the matrix obtained by reindexing the rows and columns of $A$ using the inverse bijection $e^{-1}$ is equal to the trace of $A$.

For any real or complex-like field $k$, finite types $n$ and $\iota$, a bijection $e : n \simeq \iota$, and an $n \times n$ matrix $A$ over $k$, the matrix exponential of the matrix obtained by reindexing the rows and columns of $A$ using the inverse bijection $e^{-1}$ is equal to the matrix obtained by reindexing the rows and columns of the matrix exponential of $A$ using $e$.

For any element $A$ of the Lorentz algebra $\mathfrak{so}(1, 3)$, the matrix exponential $\exp(A)$ is a proper Lorentz transformation, meaning its determinant is equal to $1$.

For any element $A$ of the Lorentz algebra $\mathfrak{so}(1, 3)$, its matrix exponential $\exp(A)$ is orthochronous. Specifically, the $(0,0)$-component of the resulting Lorentz transformation matrix satisfies $(\exp A)_{00} \geq 1$.

For any element $A$ of the Lorentz algebra $\mathfrak{so}(1, 3)$, its matrix exponential $\exp(A)$ is an element of the restricted Lorentz group $SO^+(1, 3)$.