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The identity matrix belongs to $O(1,3)$. If $A,B\in O(1,3)$, thenso the set is closed under multiplication. Taking determinants in $A^{T}MA=M$ gives $(\det A{)}^2=1$, so $A$ is invertible. Rearranging the defining identity givesIts inverse also preserves $M$, either directly or becauseThus the Lorentz group $O(1,3)$ is a group under matrix multiplication.

Invariance for every $x$ requiresThus $G=O(1,3)$, the Lorentz group; its identity component is $S{O}^{+}(1,3)$.

The Dirac spinor representation is the four-dimensional representation $(\frac{1}{2},0)\oplus (0,\frac{1}{2})$ of the connected Lorentz group. It is characterized byForone may takewith the signs of ${\Omega }_{\rho \sigma }$ fixed by the displayed vector-representation convention. This formula follows by differentiating the covariance relation and using the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$; moving both gamma indices upstairs changes the apparent parameter signs according to the metric.

Use the Weyl representation of the gamma matrices,

The Lorentz group consists of linear maps $x\mapsto \Lambda x$ satisfying ${\Lambda }^T\eta \Lambda =\eta$. The Poincaré group consists of affine isometries $x\mapsto \Lambda x+a$ and has multiplicationThus it is the semidirect product ${\mathbb{R}}^{1,3}⋊O(1,3)$, with the Lorentz group as the subgroup fixing the spacetime origin.

With $\eta =\mathrm{diag}(1,-1,-1)$, the defining condition for the $(1+2)$-dimensional Lorentz group isTaking determinants gives $(\det \Lambda {)}^2=1$, hence $\det \Lambda =\pm 1$. The norm of the zeroth column givesso $|{\Lambda }^0{}_0|\ge 1$. The Proper orthochronous Lorentz group is the component with $\det \Lambda =1$ and ${\Lambda }^0{}_0\ge 1$.

The Poincare group is the semidirect product of spacetime translations by the Lorentz group. It is the isometry group of Minkowski spacetime.

The proper orthochronous Lorentz group is the identity component of the Lorentz group. Its elements have determinant one and preserve the future light cone.

For $J=\mathrm{diag}(1,-1)$, the matrices satisfying $A^{T}JA=J$ and $A_{11}>0$ form the time-orientation-preserving Lorentz group in one spatial dimension. They are exactly $A=H(t)D_{\epsilon }$, where $H(t)=\left(\begin{array}{cc}\cosh t& \sinh t\\ \sinh t& \cosh t\end{array}\right)$, $D_{\epsilon }=\mathrm{diag}(1,\epsilon )$ and $\epsilon =\pm 1$. Preservation of the Minkowski metric makes the first column a future-pointing unit timelike vector; Minkowski inner product orthogonality determines the second up to sign. The identity $D_{\epsilon }H(s)=H(\epsilon s)D_{\epsilon }$ gives the multiplication law $(t,\epsilon )(s,\eta )=(t+\epsilon s,\epsilon \eta )$ and inverse $(-\epsilon t,\epsilon )$.