Gram invariants in momentum space

Given two vectors \(k_{1}\) and \(k_{2}\) in a $D$-dimensional euclidean space we can consider the parallelogram formed with them. The area of this parallelogram is given by the square root of the determinant of the Gram matrix \(G_{12}\):\[G_{12} = \begin{pmatrix} k_{1}^{2} & k_{1} \cdot k_{2} \\ k_{1} \cdot k_{2} & k_{2}^{2} \end{pmatrix} \quad \Longrightarrow \quad \det{(G_{12})} = k_{1}^{2} k_{2}^{2} - (k_{1} \cdot k_{2})^{2}\]We will denote the magnitude of each vector by \(|k_{i}| \equiv m_{i}\). Then it follows that\[\det{(G_{12})} = \left[m_{1}m_{2} - k_{1} \cdot k_{2} \right]\left[m_{1}m_{2} + k_{1} \cdot k_{2} \right]\]Introducing the invariant \(s_{12} \equiv (k_{1} + k_{2})^{2}\) we can write\[\det{(G_{12})} = \frac{1}{4} \left[ s_{12} - (m_{1} - m_{2})^{2} \right] \left[ (m_{1} + m_{2})^{2} - s_{12} \right]\]Hence, the area of the parallelogram generated by \(k_{1}\) and \(k_{2}\) is\[A_{12} = \frac{1}{2} \sqrt{\left[ s_{12} - (m_{1} - m_{2})^{2} \right] \left[ (m_{1} + m_{2})^{2} - s_{12} \right]}\]Note that this area is real in the domain\[(m_{1} - m_{2})^{2} \leq s_{12} \leq (m_{1} + m_{2})^{2}\]The notation that we have used is very suggestive. If the two vectors were momentum vectors, then \(m_{i}\) would correspond to masses and \(s_{12}\) would correspond to a Mandelstam invariant. Then \( (m_{1}+m_{2})^{2} \) is a mass threshold and \( (m_{1} - m_{2})^{2} \) a mass pseudothreshold.

We can study the case with three vectors. Now we have three masses and hence three three-body mass pseudothresholds:\[(m_{1} - m_{2} + m_{3})^{2} \qquad (m_{1} + m_{2} - m_{3})^{2} \qquad (m_{1} - m_{2} - m_{3})^{2}\]We also have three (two-body) Mandelstam invariants:\[s_{12} \equiv (k_{1} + k_{2})^{2} \qquad s_{23} \equiv (k_{2} + k_{3})^{2} \qquad s_{31} \equiv (k_{3} + k_{1})^{2}\]It might be convenient to introduce a three-body Mandelstam invariant:\[t_{123} \equiv (k_{1} + k_{2} + k_{3})^{2} = s_{12} + s_{23} + s_{31} - m_{1}^{2} - m_{2}^{2} - m_{3}^{2}\]Note that\[4 (m_{1}^{2} + m_{2}^{2} + m_{3}^{2}) = (m_{1} + m_{2} + m_{3})^{2} + (m_{1} - m_{2} + m_{3})^{2} +(m_{1} + m_{2} - m_{3})^{2} + (m_{1} - m_{2} - m_{3})^{2}\]Now the three vectors will form a parallelepiped whose volume is given by the square root of the Gram determinant:\[V_{123} \equiv \sqrt{\det{(G_{123})}}\]What is driving me nuts is how to write this determinant in a nice form that involves the above invariants. Naively we have\[\det{(G_{123})} = m_{1}^{2}m_{2}^{2}m_{3}^{2} - m_{1}^{2} (k_{2} \cdot k_{3})^{2} - m_{2}^{2}(k_{1} \cdot k_{3})^{2} - m_{3}^{2} (k_{1} \cdot k_{2})^{2} + 2 (k_{1} \cdot k_{2})(k_{2} \cdot k_{3})(k_{3} \cdot k_{1})\]The secret might be with traces.