I understand the Lorentz group have 4 connected components, one of them being the proper orthochronous group $SO(1,3)^{+}$, where its matrix elements have determinant +1 and the component at column $0$ and row $0$ is $\geq1$ (i.e. $\Lambda_0^0\geq1$). The other three components will be given by:

{$T(SO(1,3)^{+})$ , $P(SO(1,3)^{+})$, $T(P(SO(1,3)^{+}))$} where $T$ is a linear transformation that changes the sign of $\Lambda_0^0$ of a Lorentz transformation $\Lambda$, and $P$ is a linear transformation that changes the sign of the determinant of a lorentz transformation. If we consider:


$$\Lambda_{T}=\begin{pmatrix}-1&&&\\&1&&\\&&1&\\&&&1\end{pmatrix}.$$ I thought $T$ and $P$ would be defined by: $T(A) = \Lambda_TA$, $P(A) = \Lambda_PA$

I was trying to verify this was true but most authors seem to exclude stating them more explicitly. I found in Wikipedia that these transformations actually behave as an adjoint action $Ad_{\Lambda_P}$, you can check here. It seems they are defined by:

$T(A) = \Lambda_T^{-1}A\Lambda_T$, $P(A) = \Lambda_P^{-1}A\Lambda_P$ , for $A$ in $SO(1,3)^{+}$

Is this true? If so, what is the intuition behind them having to be similarity transformations? why can't they just be a single matrix multiplication (from the left) with the given $\Lambda_T$ and $\Lambda_P$ as I thought?

I was looking for a more reputable source that stated P and T more explicitly, if you can give one I would appreciate it.


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  • $\begingroup$ Who is A ? An element of the full Lorentz group, or a geometrical object subject to a full Lorentz group transformation? @Qmechanic addresses the second possibility. $\endgroup$ – DanielC Dec 5 '18 at 13:09

It seems that OP is essentially asking

How do a physical quantity $A$ transforms under the Lorentz group $G=O(3,1)$, in particular wrt. the discrete generators $P$ and $T$?

Brief answer: That depends on the physical quantity $A$, and which $G$-representation it transforms under. To mention one simple example: A pseudoscalar is a quantity that changes sign under $P$, while a true scalar does not.


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