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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.

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    $\begingroup$ If I understood correctly the problem is asking you to find the spinor rep of the two Casimir invariants. $\Sigma^{\mu\nu}$ are the Lorentz generators for the $\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)$ rep. $\endgroup$ Jan 30, 2021 at 8:11
  • $\begingroup$ Actually, I was confuse with language and now it make sense. Thank you. $\endgroup$
    – user286848
    Jan 30, 2021 at 19:00

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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.

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    $\begingroup$ I think you are using $\{\gamma^{\mu},\gamma^{\nu}\}=2\eta^{\mu\nu}$ and signature of the metric is $\eta=diag(+,-,-,-)$. I have done this question explicitly by using first principle and now I learnt a new easy and elegant method. Thank you very much and appreciate that. $\endgroup$
    – user286848
    Jan 30, 2021 at 19:07
  • $\begingroup$ Yes; delighted you may track down conventions... I've long decoupled from that.... $\endgroup$ Jan 30, 2021 at 19:14

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