It's known that it's possible to map a 4-vector $x^\mu=(t,x,y,z)$, here i use $c=1$, into a 2x2 hermitian matrix as linear combination of Pauli matrices, thus the mapping $x^\mu \leftrightarrow X$. Where $\mathcal{H}^{2x2}$ is the set of 2x2 hermitian matrices. $$X=t\sigma^0+x\sigma^x+y\sigma^y+z\sigma^z=\begin{pmatrix} t+z & x-iy \\\ x+iy & t-z\end{pmatrix}=X^\dagger \in \mathcal{H}^{2x2}$$ that transform in the following way under $M\in SL(2,C)$ $$MXM^\dagger=X'$$ preserving the Lorentz metric because $\det M=1$ implies $x_\mu x^\mu=x'_\mu x'^\mu=t^2-x^2-y^2-z^2$.
Now, the Pauli 4-vector is a vector where the components are Pauli matrices and it transform as a 4-vector under proper orthocronous transformations $\sigma '^\mu=\Lambda^\mu_\nu \sigma^\nu$ $$\sigma^\mu=(I_2,\vec{\sigma})=(\sigma^0,\vec{\sigma})=(\sigma^0,\sigma^x,\sigma^y,\sigma^z)$$ it's correct to build the following matrix as it's done for the generic 4-vector identifying the matrix elements in terms of the components of the Pauli 4-vector? $$\Sigma=\begin{pmatrix} \sigma^0+\sigma^z & \sigma^x-i\sigma^y \\\ \sigma^x+i\sigma^y & \sigma^0-\sigma^z\end{pmatrix}=\Sigma^\dagger$$
If yes, what kind of quantity is $\Sigma$ and how could i write the starting relation to build it? i.e. for a generic vector i can write $X=t\sigma^0+x\sigma^x+y\sigma^y+z\sigma^z$ but what about the matrix corresponding to the Pauli 4-vector? The latter is a 4-vetor of course but its components are matrices instead of real or complex numbers. I would write $\Sigma=\sigma^0\cdot\sigma^0+\sigma^x\cdot\sigma^x+\sigma^y\cdot\sigma^y+\sigma^z\cdot\sigma^z$ but with the row column product $\cdot$ is meaningless because the square of each Pauli matrix is the identity 2x2 matrix $I_2$. It could be true using the tensor product?
