Is there a bi-4-vector representation of the Dirac gamma matrices and the spinor? I learned recently that if you have the Dirac spinor represented in the Weyl (chiral) basis $\Psi = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$, then given a Lorentz Transformation $\Lambda = exp[\frac{1}{2}\Omega_{\rho \sigma}M^{\rho\sigma}]$, the corresponding transformation $S[\Lambda]$ for $\Psi \rightarrow S[\Lambda]\Psi$ looks like $S[\Lambda] = exp[\frac{1}{2}\Omega_{\rho \sigma}S^{\rho\sigma}]$. In the chiral basis, this looks like each of the $\psi_L$ and $\psi_R$ transforming as the different spin-1/2 representations of $so(3,1)$.
Here, the $so(3,1)$ Lie algebra is represented by the standard Lorentz matrices $M^{\rho\sigma}$ for the $x^\mu$ transformation, and the other matrices $S^{\rho\sigma}$ are the algebra generated by the gamma matrices $S^{\rho\sigma} = \frac{1}{4}[\gamma^\rho,\gamma^\sigma]$.
My question is (in probably imprecise phrasing), is it possible to choose $\gamma^\mu$ to be 8x8 matrices in such a way that $S^{\rho\sigma} = \frac{1}{4}[\gamma^\rho,\gamma^\sigma] = M^{\rho\sigma} \oplus M^{\rho\sigma} $? i.e. can we pick $\gamma^\mu$ so that the chiral components $\psi_L, \psi_R$ each transform like 4-vectors would?
 A: *

*OP apparently wants to discuss reducible representations of the Clifford algebra $Cl(1,3;\mathbb{R})$. 

*Concretely, it seems that OP is asking about an 8-dimensional direct sum representation $$W~:=~V\oplus V$$ of 2 copies of the 4-dimensional Dirac spinor representation 
$$V~:=~(\frac{1}{2},0) \oplus (0,\frac{1}{2}).$$
See also this Phys.SE post.

*The $8\times 8$ gamma matrices 
$$\Gamma^{\mu} ~=~\begin{pmatrix} \gamma^{\mu} & 0 \cr 0 & \gamma^{\mu} \end{pmatrix}$$
in the $W$-representation are block matrices with 2 copies of $4\times 4$ gamma matrices in the $V$-representation.

*The $W$-representation of the Lorentz generators take a similar block diagonal form, cf. OP's last question (v1).
A: Without full appreciation of the gist of your question (8×8?), let me just review the chiral basis expressions for $\gamma_5$ and the Lorentz generators $S^{\mu\nu}$, which are both block diagonal with respect to the chiral projections, so they do not mix 
$\psi_L$ with $\psi_R$, unlike the γs:
$$\gamma^0 = \begin{pmatrix} 0 & I_2 \\ I_2 & 0 \end{pmatrix},\quad \gamma^k = \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix},\quad \gamma^5 = \begin{pmatrix} -I_2 & 0 \\ 0 & I_2 \end{pmatrix},$$
$$
S^{0j}  = \tfrac{1}{2}\begin{pmatrix} -\sigma^j & 0 \\ 0 & \sigma^j \end{pmatrix},  \qquad S^{jk}=\frac{i\epsilon^{kjm}}{2}  \begin{pmatrix} \sigma^m & 0 \\ 0 & \sigma^m \end{pmatrix} = i \epsilon^{kjm} \gamma_5 S^{0m}        .  $$
So,  generators (in the algebra),   manifestly constitute a reduced representation of non-interacting 2×2 blocks.
More formally, the reduced 2   ⊕ 2 rep is 
$$
2S^{0j}=(-\sigma^j)\oplus \sigma^j, \qquad 2S^{jk}=i\epsilon^{kjm} (\sigma^m\oplus \sigma^m). 
$$
