Lorentz generators $J_i,K_i$ and Casimir invariants in bispinor representation Lorentz generators satisfy the Lie algebra
$$[J_i,J_j]=i\epsilon_{ij}^kJ_k, ~~~~[J_i,K_j]=i\epsilon_{ij}^kK_k, ~~~~[K_i,K_j]=-i\epsilon_{ij}^kJ_k.$$
Now, define $$A_i=\frac{J_i+iK_i}{2},~~~~B_i=\frac{J_i-iK_i}{2},$$ and we can easily prove that
$$[A_i,A_j]=i\epsilon_{ij}^kA_k\, ,[B_i,B_j]=i\epsilon_{ij}^kB_k\, ,[A_i,B_j]=0.$$
We also see that the $\{M^{\mu\nu}\}$ Lie algebra is isomorphic to two $SU(2)$ Lie algebras and they have the Casimir invariants $A^2, B^2$.
Now, I do not understand what is the meaning of

$A^2$ and $B^2$ commuting the spinor representation provided by $\sum^{\mu\nu}=\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}]$ evaluate the two casimir invariant.

Any help is highly appreciated.
 A: This is a direct plugin, and I fear I am doing your homework for you, denying you the opportunity to figure it out. Rather than doing it, I'll demonstrate to you how  more efficiently you can get the answer by the tensor product representation of Dirac matrices as 2×2 matrices tensored with each other,
$$
 \gamma^0 = \sigma^3 \otimes I ,  \qquad \gamma^j = i\sigma^2 \otimes \sigma^j ,
$$
so, by sleepwalker's plugin, you find
$$
-\Sigma^{0j}=K_j= -\frac{i}{2} \sigma^1\otimes \sigma^j, \qquad -\frac{\epsilon ^{jkl}}{2}\Sigma^{kl}=J_j= -\tfrac{1}{2} I\otimes \sigma^j, ~~\leadsto \\
A_j=-P_A \otimes \frac{\sigma^j}{2}, \qquad B_j=-P_B \otimes \frac{\sigma^j}{2},
$$
where, of course, the projectors to either Kronecker factor are
$$
P_A= (I- \sigma^1)/2, \Longrightarrow P_A^2= P_A, \\
P_B= (I+ \sigma^1)/2, \Longrightarrow P_B^2= P_B\\
P_A P_B=P_B P_A=0. 
$$
It is then evident that
$$
\vec A \cdot \vec  A = P_A \otimes \vec \sigma \cdot \vec \sigma /4 = \tfrac{3}{4} P_A\otimes I,\\
\vec B \cdot \vec  B = P_B \otimes \vec \sigma \cdot \vec \sigma /4 =\tfrac{3}{4} P_B\otimes I, 
$$
your desired 4×4 matrices.
