# Does reversal of one spatial direction count as a discrete Lorentz transformation?

A transformation $\Lambda$ is a Lorentz transformation if it satisfies $\Lambda^T g \Lambda = g$, for the flat metric $g = \left( \begin{array}{cccc} 1 &&& \\ & -1 &&& \\ &&-1&& \\ &&&-1 \end{array} \right)$

In addition to rotations and boosts, time reversal ($T: t \rightarrow -t$) and parity ($P: x_i \rightarrow -x_i$ for all spatial coordinates) are singled-out as discrete Lorentz transformations. I find that reversal of a single spatial coordinate ($x_1 \rightarrow -x_1$, all others unchanged) also satisfies the definition of a Lorentz transformation. So the question is: why isn't it considered as another discrete transformation, alongside $T$ and $P$?

• It can be composed out of the parity and a rotation. – Meng Cheng Jun 6 '15 at 22:09

1. Recall that the Lorentz group $G=O(3,1)$ has 4 connected components $$G~=~G_0 ~\cup~ P\cdot G_0 ~\cup~ T\cdot G_0 ~\cup~ PT\cdot G_0.$$ Here the connected components $G_0$ that contains the identity is the restricted Lorentz group $G_0=SO^+(3,1)$.

2. It is straightforward to see that $G_0$ is a normal subgroup of $G$. Therefore the quotient $G/G_0$ is a group.

3. For the Lorentz group $G=O(3,1)$, the quotient $G/G_0$ is isomorphic to the Klein Vierergruppe $V\cong \mathbb{Z}_2\times \mathbb{Z}_2$, which is generated by two elements.

4. Moreover, the quotient $G/G_0$ is discrete. Formally speaking, the elements of $G/G_0$ constitute the 'discrete Lorentz transformations'. In practice, one often picks a representative for each equivalence class, and called these representatives the 'discrete Lorentz transformations', with the implicit understanding that one might as well pick other representatives, that deviate with elements of $G_0$.

5. Returning to OP's example, the difference between $P$ and the mirror reflection $x^1 \mapsto -x^1$ in the $(x^2,x^3)$-plane is a $\pi$-rotation around the $x^1$-axis, which belongs to $G_0$, cf. above comment by Meng Cheng.

• From what you say, is it correct to conclude that the mirror reflection is not called a "discrete" transformation because it requires a rotation? I tried searching for the definition of a "discrete transformation" but only the discrete Fourier transform shows up. – yjc Jun 7 '15 at 0:52
• I updated the answer. – Qmechanic Jun 7 '15 at 9:32

A single transformation is not strictly speaking discrete, but a group of transformations can be. Every transformation of finite order (i.e. $\Lambda^n = I$ for some $n$) generates a discrete group of transformations, so elements of finite order are sometimes called discrete transformations.

As Qmechanic explains, $T$ and $P$ generate a subgroup of coset representatives of the connected components. There are many other coset representatives though. Why $T$ and $P$ get a special treatment is mainly for aesthetic and/or physical reasons: if we call $S$ the transformation mapping $x\mapsto -x$ (and leaving $t$, $y$ and $z$ untouched), then we could also write

$$O(3,1)~=~SO^+(3,1) ~\cup~ S\cdot SO^+(3,1)~\cup~ T\cdot SO^+(3,1) ~\cup~ ST\cdot SO^+(3,1).$$

Mathematically this is fine, but aesthetically it is less nice because this doesn't treat the spatial dimensions in a symmetrical way, whereas the group itself does.