Gram determinant for the $2\to 3$ scattering cross-section I'll repeat the part of my previous question regarding the topic:
There is a book of Byckling and Kajantie, "Particle kinematics", discussing in particular (Chapter V) the kinematics of the $2\to 3$ cross-sections. It can be easily found in the internet. I just want to ask those who read it about the phase space evaluation.
I follow their idea of representing of the phase space
$$
d\Phi_{3} = \frac{d^{3}\mathbf p_{1}}{2E_{1}}\frac{d^{3}\mathbf p_{2}}{2E_{2}}\frac{d^{3}\mathbf p_{3}}{2E_{3}}\delta(k_{1}+k_{2}-p_{1}-p_{2}-p_{3})
$$ 
for the process $p_{a}+p_{b} \to p_{1}+p_{2}+p_{3}$ in terms of kinematic invariants 
$$
\tag 0 s_{1} = (p_{1}+p_{2})^{2},\quad s_{2} = (p_{2}+p_{3})^{2}, \quad t_{1} = (p_{a}-p_{1})^{2}, \quad t_{2} = (p_{b}-p_{3})^{2}
$$
(and one angle, which is typically trivial and therefore is irrelevant). In general, it is not hard to express all of the scalar products of 4-momentums through these invariants, and the $2\to 3$ cross-section is possible to write in the form
$$
\tag 1 \sigma  \sim \frac{1}{s}\int |M|^{2}d\Phi_{3} \sim \frac{1}{s^{2}}\int \frac{ds_{2}dt_{1}dt_{2}ds_{1}}{\sqrt{-\Delta_{4}}}|M(s,s_{1},s_{2},t_{1},t_{2})|^{2}, \quad s = (p_{a}+p_{b})^{2}
$$
where $M$ is the matrix element of the process, and $\Delta_{4}$ is the Gram determinant of 4 independent vectors combined from $p_{a},...,p_{3}$.
My question is following.
On p.138 there are discussed few forms of the Gram determinants, for example, $\Delta_{4}(k_{1},k_{2},p_{1},p_{3})$ and $\Delta_{4}(p_{2}+p_{3},k_{2},k_{1}+k_{2},p_{3})$. I don't understand whether they're equivalent. Precisely, suppose we initially fix the particle numeration and within this numeration define the kinematic invariants. May I then insert both the Gram invariants $\Delta_{4}(k_{1},k_{2},p_{1},p_{3})$ and $\Delta_{4}(p_{2}+p_{3},k_{2},k_{1}+k_{2},p_{3})$ into $(1)$?
 A: From your comment, iiuc, you are interested in the relation between $\Delta_4(p_1, p_2, p_3, p_4)$ and $\Delta_4(q_1, q_2, q_3, q_4)$ where $q_i = \sum_{i=1}^4 a_{ik} p_k$, as that would be a general change of basis. It is straightforward to show that
$$\Delta_4(q_1, q_2, q_3, q_4)=\det\left(A^TAD_4(p_1, p_2, p_3, p_4)\right),\tag{1} \label{cob}$$
where $A$ is the matrix $(a_{ij})_{1\le i,j\le 4}$, and denoting by $D(k_1, k_2, k_3, k_4)$ the matrix $\Delta_4(k_1, k_2, k_3, k_4)$ is the determinant of. Indeed
$$\begin{align}
D_4(q_1, q_2, q_3, q_4)_{ij} &= q_i\cdot q_j\\
&= \sum_{1\le k,l\le 4} a_{ik}a_{jl}p_k\cdot p_l
=\sum_{1\le k,l\le 4}A_{ik}D_4(p_1, p_2, p_3, p_4)_{kl}\left(A^T\right)_{lj}\\
&=\left(AD_4(p_1, p_2, p_3, p_4)A^T\right)_{ij}
\end{align}$$
Then we conclude using $\det AB = \det BA$ for any matrix $A$ and $B$.
$\square$
Therefore, the answer to your general question is "no": you can't just replace $\Delta_4(q_1, q_2, q_3, q_4)$ by $\Delta_4(p_1, p_2, p_3, p_4)$ in the general case. But if $A^TA$ happens to be the identity matrix, then the two Gram determinant are equal. The case where $A^TA$ is a multiple of the identity is also easy to handle. Anything more general, and you have to rely on the general formula (\ref{cob}).
