In three dimensions the Lorentz group is defined as

\begin{align} \text{O}(2,1)=\{\Lambda\in\text{GL}(3,\mathbb{R}),~~~ \eta=\Lambda^T\eta\Lambda,~~~ \text{det}(\Lambda)=\pm 1\},~~~\eta=\begin{pmatrix} -1&&0&&0\\0&&1&&0\\0&&0&&1\end{pmatrix}. \end{align} Is it true that when considered as a real Lie group it can be written as the union of four disconnected pieces; \begin{align} \text{O}(2,1)=\big\{\text{SO}^{\uparrow}(2,1),~\mathscr{P}\cdot\text{SO}^{\uparrow}(2,1),&~\mathscr{T}\cdot\text{SO}^{\uparrow}(2,1),~(\mathscr{P}\mathscr{T})\cdot\text{SO}^{\uparrow}(2,1)\big\},\tag{1}\\[5pt] \mathscr{P}=\begin{pmatrix} 1&&0&&0\\0&&-1&&0\\0&&0&&-1\end{pmatrix},&~~~\mathscr{T}=\begin{pmatrix} -1&&0&&0\\0&&1&&0\\0&&0&&1\end{pmatrix} \end{align} where \begin{align} \text{SO}^{\uparrow}(2,1)=\{\Lambda\in\text{O}(2,1),~~~\text{det}(\Lambda)=1,~~~\Lambda^0_{~0}\geq 1\}. \end{align} I ask this because in four dimensions it is clear that the parity operator $\mathscr{P}\notin \text{SO}^{\uparrow}(3,1)$ since it has determinant -1. However in three dimensions $\mathscr{P}\in\text{SO}^{\uparrow}(2,1)$ as it satisfies all of the above conditions. Does this mean that the Lorentz group in three dimensions is composed of not four but two disconnected pieces; \begin{align} \text{O}(2,1)=\big\{\text{SO}^{\uparrow}(2,1),~\mathscr{T}\cdot\text{SO}^{\uparrow}(2,1)\big\} \tag{2}\end{align} and $\text{SO}^{\uparrow}(2,1)$ is the component which is connected to the identity?

  • 2
    $\begingroup$ Yes, parity in two dimensions is just a rotation by 180 degrees, so it's connected to the identity. You can similarly construct parity in any even dimension as a series of such rotations. $\endgroup$ – knzhou Sep 28 '17 at 8:37
  • $\begingroup$ Thank you. I suppose by even dimensions you mean $n+1$ dimensions where n is even? $\endgroup$ – SigmaAlpha Sep 28 '17 at 8:46
  • $\begingroup$ Also, I cannot now say that we choose the branch with det$(\Lambda)=1$ because it preserves the spatial orientation (parity). So what is the physical justification in considering only Lorentz transformations with det$(\Lambda)=1$? Is the only justification that this is the component connected to the identity? $\endgroup$ – SigmaAlpha Sep 28 '17 at 10:05
  • $\begingroup$ Sure; also you could say it's the component that preserves the temporal direction. $\endgroup$ – knzhou Sep 28 '17 at 10:11
  • $\begingroup$ The Lorentz group still has 4 connected components in any spacetime dimension $\geq 2$. $\endgroup$ – Qmechanic Sep 28 '17 at 10:20

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