Say that you have two slow momenta \(P_{1}\) and \(P_{2}\) such that
$$ \Vert P_{1} \Vert^{2} = - m_{1}^{2}, \qquad \Vert P_{2} \Vert^{2} = - m_{2}^{2}. $$
That is,
$$ P_{1} = \begin{bmatrix} m_{1} \cosh{(\xi_{1})} & m_{1} \sinh{(\xi_{1})} \end{bmatrix}; $$
$$ P_{2} = \begin{bmatrix} m_{2} \cosh{(\xi_{2})} & m_{2} \sinh{(\xi_{2})} \end{bmatrix}. $$
You can introduce a 2-Mandelstam invariant:
$$ s_{12} = - \Vert P_{1} + P_{2} \Vert^{2}. $$
Now consider two null vectors \(N_{1}\) and \(N_{2}\):
$$ N_{1} = \begin{bmatrix} k_{1} & k_{1} \end{bmatrix}, \qquad N_{2} = \begin{bmatrix} k_{2} & -k_{2} \end{bmatrix}. $$
Note that
$$ N_{1} \cdot N_{2} = - 2 k_{1} k_{2}. $$
This means that the 2-Sudakov invariant \(S_{12}\) is given by
$$ S_{12} = - \Vert N_{1} + N_{2} \Vert^{2} = - 2 (N_{1} \cdot N_{2}) = 4 k_{1} k_{2}. $$
The Sudakov two-body decomposition is
$$ P_{1} = N_{1} + \frac{m_{1}^{2}}{S_{12}} N_{2}, \qquad P_{2} = N_{2} + \frac{m_{2}^{2}}{S_{12}} N_{1}. $$
From here it follows that
$$ m_{1} \cosh{(\xi_{1})} = k_{1} + \frac{m_{1}^{2}}{4 k_{1}}, \qquad m_{2} \cosh{(\xi_{2})} = k_{2} + \frac{m_{2}^{2}}{4 k_{2}}; $$
$$ m_{1} \sinh{(\xi_{1})} = k_{1} - \frac{m_{1}^{2}}{4 k_{1}}, \qquad m_{2} \sinh{(\xi_{2})} = -k_{2} + \frac{m_{2}^{2}}{4 k_{2}}. $$
Then
$$ k_{1} = \frac{1}{2} m_{1} \exp{(\xi_{1})}, \qquad k_{2} = \frac{1}{2} m_{2} \exp{(-\xi_{2})}. $$
This means that
$$ N_{1} \cdot N_{2} = - \frac{1}{2} m_{1} m_{2} \exp{(\xi_{1} - \xi_{2})} \quad \Longrightarrow \quad S_{12} = m_{1} m_{2} \exp{(\xi_{1} - \xi_{2})}. $$
Define the Regge-Sudakov invariant \(R_{12}\) as
$$ R_{12} = \frac{S_{12}}{m_{1} m_{2}} = \exp{(\xi_{1} - \xi_{2})}; $$
and the Regge-Mandelstam invariant \(r_{12}\) as
$$ r_{12} = - \frac{P_{1} \cdot P_{2}}{m_{1} m_{2}} = \cosh{(\xi_{1} - \xi_{2})}. $$
There is a relationship between \(r_{12}\) and \(R_{12}\):
$$ 2 r_{12} = R_{12} + \frac{1}{R_{12}} \quad \Longrightarrow \quad R_{12} = r_{12} + \sqrt{r_{12}^{2} - 1}. $$
There is also a relationship between \(r_{12}\) and \(s_{12}\):
$$ r_{12} = \frac{s_{12} - m_{1}^{2} - m_{2}^{2}}{2 m_{1} m_{2}}. $$
Combining these you can find an expression for \(R_{12}\) in terms of \(s_{12}\) and the masses.
The two Sudakov vectors introduced above form a linear basis for the whole set of energy-momentum vectors. That is, there is no transversal component to the Sudakov subspace. Any vector \(P\) can be written as
$$ P = \nu_{1} N_{1} + \nu_{2} N_{2}; $$
with
$$ \nu_{1} = \frac{N_{2} \cdot P}{N_{1} \cdot N_{2}}, \qquad \nu_{2} = \frac{N_{1} \cdot P}{N_{1} \cdot N_{2}}. $$
An important case is when \(P\) is slow and given by
$$ P = \begin{bmatrix} m \cosh{(\xi)} & m \sinh{(\xi)} \end{bmatrix}. $$
Then it follows that
$$ \nu_{1} = \frac{m \exp{(\xi)} }{2 k_{1}} = \left(\frac{m}{m_{1}} \right) \exp{(\xi - \xi_{1})} ; $$
$$ \nu_{2} = \frac{m}{2 k_{2} \exp{(\xi)} } = \left(\frac{m}{m_{2}} \right) \exp{(\xi_{2} - \xi)} . $$
These are the Sudakov components of \(P\). If you work with 2-to-\(N\) scattering, then all \(N\) outgoing energy-momentum vectors can be decomposed in the same way in terms of the two incoming energy-momentum vectors.
