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Consider the Minkowski 2D metric $\eta = \text{diag}(-1, 1)$. The Lorentz group is the set of matrices such that, for a transformation $\Lambda$, we get

$$\eta = \Lambda^T \eta \Lambda$$

This means

\begin{eqnarray} \Lambda^\dagger \eta \Lambda &=& \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}\\ &=& \begin{pmatrix} c^2 - a^2 & cd - ab \\ cd - ab & d^2 - b^2 \end{pmatrix} \end{eqnarray}

So that we get the equalities

\begin{eqnarray} cd &=& ab\\ c^2 &=& a^2 - 1\\ d^2 &=& 1 + b^2 \end{eqnarray}

We can replace $c$ with, let's say at random, $\sinh(u)$, as it is a bijection on $\mathbb{R}$, so that

\begin{eqnarray} \sinh^2(u) + 1 &=& a^2 \end{eqnarray}

meaning that $a = \pm \cosh(u)$. Similarly with $b = \sinh(v)$, we get

$$d^2 = 1 + \sinh^2(v)$$

so $d = \pm \cosh(v)$. Now there are four possible cases to treat $cd = ab$, depending on the signs of $a$ and $d$ : $++$, $+-$, $-+$ and $--$.

For $++$ :

\begin{eqnarray} cd - ab &=& \sinh(u) \cosh(v) - \cosh(u) \sinh(v)\\ &=& \sinh(u - v) \end{eqnarray}

Meaning $u = v$, this is just the orthochronous oriented Lorentz transform,

\begin{eqnarray} \Lambda = \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix} \end{eqnarray}

For $+-$ :

\begin{eqnarray} cd - ab &=& -(\sinh(u) \cosh(v) + \cosh(u) \sinh(v))\\ &=& -\sinh(u + v) \end{eqnarray}

Meaning that $u = -v$. From the parity properties of hyperbolic functions, that is

\begin{eqnarray} \Lambda = \begin{pmatrix} \cosh(u) & -\sinh(u) \\ \sinh(u) & -\cosh(u) \end{pmatrix} \end{eqnarray}

For $-+$ :

\begin{eqnarray} cd - ab &=& \sinh(u) \cosh(v) + \cosh(u) \sinh(v)\\ &=& \sinh(u + v) \end{eqnarray}

Meaning that $u = -v$.

\begin{eqnarray} \Lambda = \begin{pmatrix} -\cosh(u) & -\sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix} \end{eqnarray}

with similar proofs, the rest will be

\begin{eqnarray} \Lambda = \begin{pmatrix} -\cosh(u) & \sinh(u) \\ \sinh(u) & -\cosh(u) \end{pmatrix} \end{eqnarray}

Now the first matrix is entirely fine as it is the $SO^\uparrow(2)$ matrix, and the second one is a Lorentz transformation with parity reversal. But the rest doesn't seem to correspond to the usual matrices, which are the Lorentz transform composed with parity and time reversal :

\begin{eqnarray} \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -\cosh(u) & \sinh(u) \\ -\sinh(u) & \cosh(u) \end{pmatrix} \end{eqnarray}

\begin{eqnarray} \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \cosh(u) & -\sinh(u) \\ \sinh(u) & -\cosh(u) \end{pmatrix} \end{eqnarray}

\begin{eqnarray} \begin{pmatrix} \cosh(u) & \sinh(u) \\ \sinh(u) & \cosh(u) \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} -\cosh(u) & -\sinh(u) \\ -\sinh(u) & -\cosh(u) \end{pmatrix} \end{eqnarray}

where is the error here?

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  • $\begingroup$ Maybe because the sign of $u$ is not determined, so that your third Lorentz matrix is in fact the same as the first of the last three matrices, with $u\rightarrow-u$. And similarly your fourth Lorentz matrix is just the last of the last three matrices. $\endgroup$ – chaostang Aug 9 at 11:17
  • $\begingroup$ True, flipping the signs of the last two would give back all the relevant matrices $\endgroup$ – Slereah Aug 9 at 11:18

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