Physlib.Relativity.LorentzGroup.Proper

Let $\mathcal{L}_d$ denote the Lorentz group acting on a $(1+d)$-dimensional Minkowski space. For any element $\Lambda \in \mathcal{L}_d$, the determinant of its matrix representation is either $1$ or $-1$. That is, $$\det(\Lambda) = 1 \quad \text{or} \quad \det(\Lambda) = -1.$$

The definition provides a topological space structure on the group $\mathbb{Z}_2$ (the cyclic group of order 2), where the topology assigned is the discrete topology.

The topological space defined on the group $\mathbb{Z}_2$ (the cyclic group of order 2) is a discrete space.

The cyclic group of order 2, $\mathbb{Z}_2$, is a topological group when equipped with the discrete topology. This implies that the group multiplication and inversion operations are continuous maps.

The function is a continuous map $f: \{-1, 1\} \to \mathbb{Z}_2$, where the domain $\{-1, 1\}$ is a subset of the real numbers $\mathbb{R}$ and $\mathbb{Z}_2$ is the cyclic group of order 2. The map is defined such that $f(1)$ is the identity element of $\mathbb{Z}_2$ (corresponding to $0 \in \mathbb{Z}/2\mathbb{Z}$) and $f(-1)$ is the non-identity element of $\mathbb{Z}_2$ (corresponding to $1 \in \mathbb{Z}/2\mathbb{Z}$).

The continuous map $\det: \mathcal{L}_d \to \mathbb{Z}_2$ from the Lorentz group $\mathcal{L}_d$ to the cyclic group of order 2. For a Lorentz matrix $\Lambda$, it maps the determinant $\det(\Lambda) \in \{-1, 1\}$ to the corresponding element in $\mathbb{Z}_2$, where $\det(\Lambda) = 1$ corresponds to the identity element and $\det(\Lambda) = -1$ corresponds to the non-identity element.

For any element $\Lambda$ in the Lorentz group $\mathcal{L}_d$, the continuous determinant map $\det: \mathcal{L}_d \to \mathbb{Z}_2$ evaluates to the identity element of $\mathbb{Z}_2$ if and only if the determinant of the matrix $\Lambda$ is $1$.

For any element $\Lambda$ in the Lorentz group $\mathcal{L}_d$ acting on $(1+d)$-dimensional Minkowski space, the continuous determinant map $\det_{\text{cont}}: \mathcal{L}_d \to \mathbb{Z}_2$ evaluates to the non-identity element of $\mathbb{Z}_2$ if and only if the determinant of the matrix representation of $\Lambda$ is $-1$.

For any elements $\Lambda$ and $\Lambda'$ in the Lorentz group $\mathcal{L}_d$, the values of the continuous determinant map $\det_{\text{cont}}: \mathcal{L}_d \to \mathbb{Z}_2$ are equal if and only if the determinants of their matrix representations are equal. That is, $$\det_{\text{cont}}(\Lambda) = \det_{\text{cont}}(\Lambda') \iff \det(\Lambda) = \det(\Lambda').$$

The group homomorphism $\det: \mathcal{L}_d \to \mathbb{Z}_2$ from the Lorentz group $\mathcal{L}_d$ to the cyclic group of order 2. For any Lorentz transformation $\Lambda \in \mathcal{L}_d$, the map sends $\Lambda$ to its determinant $\det(\Lambda) \in \{1, -1\}$, represented as an element of $\mathbb{Z}_2$. Specifically, a determinant of $1$ corresponds to the identity element of $\mathbb{Z}_2$, and a determinant of $-1$ corresponds to the non-identity element.

Let $\mathcal{L}_d$ be the $d$-dimensional Lorentz group and $\mathbb{Z}_2$ be the cyclic group of order 2 equipped with the discrete topology. The determinant representation $\text{detRep}: \mathcal{L}_d \to \mathbb{Z}_2$, which maps a Lorentz transformation $\Lambda$ to its determinant $\det(\Lambda) \in \{1, -1\}$ represented in $\mathbb{Z}_2$, is continuous.

For any two Lorentz transformations $\Lambda$ and $\Lambda'$ in the Lorentz group $\mathcal{L}_d$ (the group of linear transformations preserving the Lorentz metric in $d$ dimensions), if $\Lambda'$ belongs to the same connected component as $\Lambda$, then their determinants are equal: $\det(\Lambda) = \det(\Lambda')$.

Let $\mathcal{L}_d$ be the Lorentz group in $d$ dimensions. For any two Lorentz transformations $\Lambda, \Lambda' \in \mathcal{L}_d$, if $\Lambda'$ belongs to the same connected component as $\Lambda$, then their images under the determinant representation $\text{detRep}: \mathcal{L}_d \to \mathbb{Z}_2$ are equal, i.e., $\text{detRep}(\Lambda) = \text{detRep}(\Lambda')$.

For any two Lorentz transformations $\Lambda$ and $\Lambda'$ in the $d$-dimensional Lorentz group, if $\Lambda$ and $\Lambda'$ are joined by a path, then their determinants are equal: $\det(\Lambda) = \det(\Lambda')$.

For a Lorentz transformation $\Lambda$ in the Lorentz group of dimension $d$, the property $\text{IsProper}(\Lambda)$ holds if the determinant of the matrix representation of $\Lambda$ is equal to $1$, denoted as $\det(\Lambda) = 1$.

For a Lorentz transformation $\Lambda$ in the Lorentz group of dimension $d$, the property $\text{IsProper}(\Lambda)$, which is defined by the condition that the determinant of the matrix representation of $\Lambda$ is equal to $1$ ($\det(\Lambda) = 1$), is a decidable predicate.

For any two Lorentz transformations $\Lambda$ and $\Lambda'$ in the Lorentz group of dimension $d$, if $\Lambda$ and $\Lambda'$ are proper transformations (meaning $\det \Lambda = 1$ and $\det \Lambda' = 1$), then their product $\Lambda \Lambda'$ is also a proper Lorentz transformation.

For any Lorentz transformation $\Lambda$ in the Lorentz group $\mathcal{L}_d$ of dimension $d$, $\Lambda$ is a proper Lorentz transformation ($\text{IsProper } \Lambda$) if and only if its image under the determinant representation $\text{detRep}: \mathcal{L}_d \to \mathbb{Z}_2$ is the identity element $1$.

The identity element of the Lorentz group in $d$ dimensions is a proper Lorentz transformation (i.e., its determinant is equal to $1$).

For any two Lorentz transformations $\Lambda$ and $\Lambda'$ in the Lorentz group of dimension $d$, if $\Lambda'$ belongs to the same connected component as $\Lambda$, then $\Lambda$ is a proper Lorentz transformation if and only if $\Lambda'$ is a proper Lorentz transformation. A Lorentz transformation is called proper if its determinant is equal to $1$.