A notational confusion in a Bell-type inequality In the tripartite Bell-type inequality know as Svetlichny inequality, given in this (freely available) article. The quantity $M_{ijk} = Tr [\rho(\sigma_i \otimes \sigma_j \otimes \sigma_k)]$, $i,j,k\in \{1,2,3 \}$, where $\sigma_i$ represents Pauli matrix, is defined below equation 6.
My question: After this quantity, they define the matrix  $M_{j, ik}$, but it is not clear to me how this matrix is related to $M_{ijk}$. In other words, how do I obtain $M_{j,ik}$ from $M_{ijk}$?
 A: I think they mean a $3\times9$ matrix like
$$M=(M_{j,ik})=\left(
\begin{matrix}
M_{111}&M_{112}&M_{113}&M_{121}&M_{122}&M_{123}&M_{131}&M_{132}&M_{133}\\
M_{211}&M_{212}&M_{213}&M_{221}&M_{222}&M_{223}&M_{231}&M_{232}&M_{233}\\
M_{311}&M_{312}&M_{313}&M_{321}&M_{322}&M_{323}&M_{331}&M_{332}&M_{333}\\
\end{matrix}
\right),$$
that is, the first index is the row and the other two are the $3^2=9$ possible 2-element permutations with repetition over $\{1,2,3\}$. This makes sense given the sentence after Eq. $(6)$:

where $λ_1$ is the maximal singular value of the matrix $M = (M_{j,ik})$, [...] and
the two degenerate nine-dimensional singular vectors corresponding to $λ_1$ take the form [...]

(boldface mine).
The singular values $\lambda$ of $M_{3\times 9}$ are the $3$ diagonal values of $\Sigma_{3\times 9}$, where $M=U_{3\times 3}\Sigma_{3\times 9} V_{9\times 9}$ is the singular value decomposition (SVD) of $M$. The singular vectors $\mathbf v$ are the first $3$ columns of $V$, which are in fact 9-dimensional, and satisfy $$M\mathbf{v}=\lambda\mathbf{u},$$ with $\mathbf{u}$ the $3$ columns of $U$. For reference, read the introduction to the SVD article on Wikipedia and also its section 3.1.
