In two-dimensional spacetime, an energy-momentum vector \(P\) has only two components:
$$ P = \begin{pmatrix} E & p \end{pmatrix}. $$
Given an energy-momentum vector \(P\), you can compute the \(\mathcal{T}\)-conjugate:
$$ \mathcal{T}(P) = \begin{pmatrix} -E & p \end{pmatrix}; $$
the \(\mathcal{P}\)-conjugate:
$$ \mathcal{P}(P) = \begin{pmatrix} E & -p \end{pmatrix}; $$
and the \(\mathcal{C}\)-conjugate:
$$ \mathcal{C}(P) = \begin{pmatrix} -E & -p \end{pmatrix}. $$
These three operations are involutions:
$$ \mathcal{T}(\mathcal{T}(P)) = P, \qquad \mathcal{P}(\mathcal{P}(P)) = P, \qquad \mathcal{C}(\mathcal{C}(P)) = P. $$
You also have:
$$ \mathcal{C}(P) = \mathcal{T}(\mathcal{P}(P)) = \mathcal{P}(\mathcal{T}(P)) = - P. $$
Let us look at the three kinds of energy-momentum vectors that you can have.
Slow Energy-Momenta
An energy-momentum vector \(P\) is slow if it satisfies the equation
$$ -m^{2} = \Vert P \Vert^{2} = -E^{2} + p^{2}, $$
with \(m > 0\). You can introduce a slow rapidity variable \(\xi\) via
$$ E = m \cosh{(\xi)}, \qquad p = m \sinh{(\xi)}. $$
Here \(- \infty < \xi < \infty\). In order to obtain all possible slow energy-momentum vectors, you also need to consider the \(\mathcal{T}\)-conjugates:
$$ \mathcal{T}(P) = \begin{pmatrix} -m \cosh{(\xi)} & m \sinh{(\xi)} \end{pmatrix}. $$
Note that the \(\mathcal{P}\)-conjugates correspond to changing the sign of the slow rapidity variable:
$$ \begin{pmatrix} m \cosh{(\xi)} & -m \sinh{(\xi)} \end{pmatrix} = \begin{pmatrix} m \cosh{(-\xi)} & m \sinh{(-\xi)} \end{pmatrix}. $$
The slow speed is defined as
$$ v^{2} \equiv \frac{p^{2}}{E^{2}} = \tanh^{2}{(\xi)}. $$
Since the hyperbolic tangent satisfies
$$ \tanh^{2}{(\xi)} < 1; $$
it follows that the slow speed is bounded from above:
$$ v^{2} < 1. $$
Null Energy-Momenta
An energy-momentum vector \(P\) is null if it satisfies the equation
$$ 0 = \Vert P \Vert^{2} = -E^{2} + p^{2}. $$
You can solve this via
$$ E = k, \qquad p = k; $$
with \(k > 0\). However, the three conjugates need to be included in order to obtain all possible null vectors:
$$ P, \qquad \mathcal{T}(P), \qquad \mathcal{P}(P), \qquad \mathcal{C}(P). $$
The null speed is defined in the same way as the slow speed:
$$ v^{2} = \frac{p^{2}}{E^{2}}; $$
but since \(p^{2} = E^{2}\), the null speed is fixed to a specific value:
$$ v^{2} = 1. $$
Fast Energy-Momenta
An energy-momentum vector \(P\) is fast if it satisfies the equation
$$ w^{2} = \Vert P \Vert^{2} = -E^{2} + p^{2}. $$
You can introduce a fast rapidity variable \(\rho\) via
$$ E = w \sinh{(\rho)}, \qquad p = w \cosh{(\rho)}. $$
Here \(-\infty < \rho < \infty\). In order to obtain all the fast energy-momentum vectors, you also need to consider the \(\mathcal{P}\)-conjugates:
$$ \mathcal{P}(P) = \begin{pmatrix} w \sinh{(\rho)} & -w \cosh{(\rho)} \end{pmatrix}. $$
Note that the \(\mathcal{T}\)-conjugates correspond to changing the sign of the fast rapidity variable:
$$ \begin{pmatrix} -w \sinh{(\rho)} & w \cosh{(\rho)} \end{pmatrix} = \begin{pmatrix} w \sinh{(-\rho)} & w \cosh{(-\rho)} \end{pmatrix}. $$
The fast speed is defined in the same way as the slow and null speeds:
$$ v^{2} = \frac{p^{2}}{E^{2}} = \coth^{2}{(\rho)}. $$
Since the hyperbolic cotangent satisfies
$$ \coth^{2}{(\rho)} > 1; $$
the fast speed is bounded from below:
$$ 1 < v^{2}. $$
Invariants
There are six kind of Lorentz invariants that you can construct.
Slow and Slow
If \(P_{1}\) and \(P_{2}\) are both slow and given by
$$ P_{1} = \begin{pmatrix} m_{1} \cosh{(\xi_{1})} & m_{1} \sinh{(\xi_{1})} \end{pmatrix}, \qquad P_{2} = \begin{pmatrix} m_{2} \cosh{(\xi_{2})} & m_{2} \sinh{(\xi_{2})} \end{pmatrix} $$
with both \(m_{1} > 0\) and \(m_{2} > 0\), then the inner product is given by
$$ P_{1} \cdot P_{2} = -E_{1} E_{2} + p_{1} p_{2} = -m_{1} m_{2} \left(\cosh{(\xi_{1})} \cosh{(\xi_{2})} - \sinh{(\xi_{1})} \sinh{(\xi_{2})} \right). $$
This is equivalent to
$$ P_{1} \cdot P_{2} = -m_{1} m_{2} \cosh{(\xi_{1} - \xi_{2})}. $$
Since \(m_{1}\) and \(m_{2}\) are positive, and the hyperbolic cosine is greater or equal to 1, you have
$$ -\frac{P_{1} \cdot P_{2}}{m_{1} m_{2}} \geq 1. $$
Note that
$$ P_{1} \cdot \mathcal{P}(P_{2}) = \mathcal{P}(P_{1}) \cdot P_{2} = -E_{1}E_{2} - p_{1}p_{2} = -m_{1} m_{2} \cosh{(\xi_{1} + \xi_{2})}; $$
as expected from \(\mathcal{P}\)-conjugation changing the sign of the slow rapidity. Furthermore,
$$ P_{1} \cdot \mathcal{T}(P_{2}) = \mathcal{T}(P_{1}) \cdot P_{2} = E_{1}E_{2} + p_{1}p_{2} = m_{1} m_{2} \cosh{(\xi_{1} + \xi_{2})}; $$
and also
$$ P_{1} \cdot \mathcal{C}(P_{2}) = \mathcal{C}(P_{1}) \cdot P_{2} = - P_{1} \cdot P_{2} = m_{1} m_{2} \cosh{(\xi_{1} - \xi_{2})}. $$
Null and Null
If \(P_{1}\) and \(P_{2}\) are both null and given by
$$ P_{1} = \begin{pmatrix} k_{1} & k_{1} \end{pmatrix}, \qquad P_{2} = \begin{pmatrix} k_{2} & k_{2} \end{pmatrix}; $$
with both \(k_{1} > 0\) and \(k_{2} > 0\), then the inner product is
$$ P_{1} \cdot P_{2} = -E_{1} E_{2} + p_{1} p_{2} = -k_{1} k_{2} + k_{1} k_{2} = 0. $$
However, you have
$$ P_{1} \cdot \mathcal{T}(P_{2}) = \mathcal{T}(P_{1}) \cdot P_{2} = E_{1} E_{2} + p_{1} p_{2} = 2 k_{1} k_{2}; $$
$$ P_{1} \cdot \mathcal{P}(P_{2}) = \mathcal{P}(P_{1}) \cdot P_{2} = -E_{1} E_{2} - p_{1} p_{2} = -2 k_{1} k_{2}; $$
$$ P_{1} \cdot \mathcal{C}(P_{2}) = \mathcal{C}(P_{1}) \cdot P_{2} = E_{1} E_{2} - p_{1} p_{2} = 0. $$
Note that both non-zero invariants are factorizable.
Fast and Fast
If \(P_{1}\) and \(P_{2}\) are both fast and given by
$$ P_{1} = \begin{pmatrix} w_{1} \sinh{(\rho_{1})} & w_{1} \cosh{(\rho_{1})} \end{pmatrix}, \qquad P_{2} = \begin{pmatrix} w_{2} \sinh{(\rho_{2})} & w_{2} \cosh{(\rho_{2})} \end{pmatrix} $$
with both \(w_{1} > 0\) and \(w_{2} > 0\), then the inner product is given by
$$ P_{1} \cdot P_{2} = -E_{1} E_{2} + p_{1} p_{2} = -w_{1} w_{2} \left(\sinh{(\rho_{1})} \sinh{(\rho_{2})} - \cosh{(\rho_{1})} \cosh{(\rho_{2})} \right). $$
This is equivalent to
$$ P_{1} \cdot P_{2} = w_{1} w_{2} \cosh{(\rho_{1} - \rho_{2})}. $$
Since \(w_{1}\) and \(w_{2}\) are positive, and the hyperbolic cosine is greater or equal to 1, you have
$$ \frac{P_{1} \cdot P_{2}}{w_{1} w_{2}} \geq 1. $$
Note that
$$ P_{1} \cdot \mathcal{T}(P_{2}) = \mathcal{T}(P_{1}) \cdot P_{2} = E_{1}E_{2} + p_{1}p_{2} = w_{1} w_{2} \cosh{(\rho_{1} + \rho_{2})}; $$
as expected from \(\mathcal{T}\)-conjugation changing the sign of the fast rapidity. Furthermore,
$$ P_{1} \cdot \mathcal{P}(P_{2}) = \mathcal{P}(P_{1}) \cdot P_{2} = -E_{1}E_{2} - p_{1}p_{2} = -w_{1} w_{2} \cosh{(\rho_{1} + \rho_{2})}; $$
and also
$$ P_{1} \cdot \mathcal{C}(P_{2}) = \mathcal{C}(P_{1}) \cdot P_{2} = - P_{1} \cdot P_{2} = -w_{1} w_{2} \cosh{(\rho_{1} - \rho_{2})}. $$
Slow and Null
If \(P_{1}\) is slow and given by
$$ P_{1} = \begin{pmatrix} m_{1} \cosh{(\xi_{1})} & m_{1} \sinh{(\xi_{1})} \end{pmatrix}; $$
and \(P_{2}\) is null and given by
$$ P_{2} = \begin{pmatrix} k_{2} & k_{2} \end{pmatrix}; $$
with both \(m_{1} > 0\) and \(k_{2} > 0\), then the inner product is
$$ P_{1} \cdot P_{2} = -E_{1} E_{2} + p_{1} p_{2} = -m_{1} k_{2} (\cosh{(\xi_{1})} - \sinh{(\xi_{1})}) .$$
This is equivalent to
$$ P_{1} \cdot P_{2} = -m_{1} k_{2} \exp{(-\xi_{1})}. $$
Since \(m_{1}\) and \(k_{2}\) are positive, and the exponential function is never negative, you have
$$ -\frac{P_{1} \cdot P_{2}}{m_{1} k_{2}} \geq 0 $$
Note that
$$ P_{1} \cdot \mathcal{T}(P_{2}) = \mathcal{T}(P_{1}) \cdot P_{2} = E_{1} E_{2} + p_{1} p_{2} = m_{1} k_{2} \exp{(\xi_{1})}. $$
Both of these inner products are factorizable.
Slow and Fast
If \(P_{1}\) is slow and given by
$$ P_{1} = \begin{pmatrix} m_{1} \cosh{(\xi_{1})} & m_{1} \sinh{(\xi_{1})} \end{pmatrix} ; $$
and \(P_{2}\) is fast and given by
$$ P_{2} = \begin{pmatrix} w_{2} \sinh{(\rho_{2})} & w_{2} \cosh{(\rho_{2})} \end{pmatrix} ; $$
with \(m_{1} > 0\) and \(w_{2} > 0\), then the inner product is given by
$$ P_{1} \cdot P_{2} = -E_{1} E_{2} + p_{1} p_{2} = -m_{1}w_{2}(\cosh{(\xi_{1})} \sinh{(\rho_{2})} - \sinh{(\xi_{1})} \cosh{(\rho_{2})}). $$
This is equivalent to
$$ P_{1} \cdot P_{2} = m_{1} w_{2} \sinh{(\xi_{1} - \rho_{2})}. $$
Since \(m_{1}\) and \(w_{2}\) are positive, and the hyperbolic sine takes all values in the real line, you have
$$ -\infty < \frac{P_{1} \cdot P_{2}}{m_{1} w_{2}} < \infty. $$
Note that
$$ P_{1} \cdot \mathcal{T}(P_{2}) = \mathcal{T}(P_{1}) \cdot P_{2} = E_{1} E_{2} + p_{1} p_{2} = m_{1} w_{2} \sinh{(\xi_{1} + \rho_{2})}; $$
$$ P_{1} \cdot \mathcal{P}(P_{2}) = \mathcal{P}(P_{1}) \cdot P_{2} = -E_{1} E_{2} - p_{1} p_{2} = -m_{1} w_{2} \sinh{(\xi_{1} + \rho_{2})}; $$
$$ P_{1} \cdot \mathcal{C}(P_{2}) = \mathcal{C}(P_{1}) \cdot P_{2} = E_{1} E_{2} - p_{1} p_{2} = -m_{1} w_{2} \sinh{(\xi_{1} - \rho_{2})}. $$
Null and Fast
If \(P_{1}\) is null and given by
$$ P_{1} = \begin{pmatrix} k_{1} & k_{1} \end{pmatrix}; $$
and \(P_{2}\) is fast and given by
$$ P_{2} = \begin{pmatrix} w_{2} \sinh{(\rho_{2})} & w_{2} \cosh{(\rho_{2})} \end{pmatrix}; $$
with \(k_{1} > 0\) and \(w_{2} > 0\), then the inner product is
$$ P_{1} \cdot P_{2} = -E_{1} E_{2} + p_{1} p_{2} = - k_{1} w_{2} (\sinh{(\rho_{2})} - \cosh{(\rho_{2})}) = k_{1} w_{2} \exp{(-\rho_{2})}. $$
Since \(k_{1}\) and \(w_{2}\) are positive, and the exponential function is never negative, you have
$$ \frac{P_{1} \cdot P_{2}}{k_{1} w_{2}} \geq 0. $$
Note that
$$ P_{1} \cdot \mathcal{T}(P_{2}) = \mathcal{T}(P_{1}) \cdot P_{2} = E_{1} E_{2} + p_{1} p_{2} = k_{1} w_{2} \exp{(\rho_{2})}; $$
$$ P_{1} \cdot \mathcal{P}(P_{2}) = \mathcal{P}(P_{1}) \cdot P_{2} = -E_{1} E_{2} - p_{1} p_{2} = -k_{1} w_{2} \exp{(\rho_{2})}; $$
$$ P_{1} \cdot \mathcal{C}(P_{2}) = \mathcal{C}(P_{1}) \cdot P_{2} = E_{1} E_{2} - p_{1} p_{2} = -k_{1} w_{2} \exp{(-\rho_{2})}. $$
All of these inner products are factorizable.
