A previous answer nicely show the relationship between the Pauli transfer matrix (PTM) and the Bloch representation of a quantum channel that acts on single-qubit density matrix. In short, given a quantum channel $\mathcal{E}$, the PTM is defined as $R_{i, j} = \frac{1}{2^n}Tr[P_{i}\mathcal{E}(P_{j})]$, where the $P's$ are the n-qubit Pauli basis (from $P_{0}$ up to $P_{4^n - 1}$). For a single qubit, $n = 1$, and these are just the three Pauli matrices + identity. For the 2-qubit case, $n = 2$, and these are all the tensor products of the former + identity (total of 16). So far it all make sense.
From the PTM one can infer the Bloch representation of a quantum channel. Given that any single-qubit channel can be written as $\mathcal{E}(\rho) = \frac{1}{2}(I + (M \vec{r} + \vec{t}) \cdot \sigma)$, where $\vec{r}$ is the Bloch vector of the state $\rho = \frac{1}{2}(I + \vec{r} \cdot \sigma) $. The matrix M and vector $\vec{t}$ can be obtained from the single-qubit $4 \times 4$ PTM as the bottom $3 \times 3$ matrix and the $3 \times 1$ vector of the left column, i.e.: PTM = $\pmatrix{1 \ \ \ 0 \\ \vec{t} \ \ \ M}$. The physical interpretation is immediate here, as $\vec{r}$ is a vector on the Bloch sphere, so one can either describe how a channel acts on a density matrix $\rho$, or, equivalently, describe the transformation of the vector on the Bloch sphere. It also quite nice, as it provides a state-independent Euclidian terms which can be used for several norm calculations.
My question is: What about the 2-qubit case? The PTM is well defined for this case. Then, is it true to say that, in a similar fashion, the Bloch representation is $(M, \vec{t})$, where now M is the bottom $15 \times 15$ matrix, and $\vec{t}$ is the left $15 \times 1$ column? And, if so, what does it mean? What is the 2-qubit equivalent geometrical interpretation to the 1-qubit case?
