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Antisymmetric tensor of rank two can be connected with spinor formalism by the formula $$ M_{\mu \nu} = \frac{1}{2}(\sigma_{\mu \nu})^{\alpha \beta}h_{(\alpha \beta )} - \frac{1}{2}(\sigma_{\mu \nu})^{\dot {\alpha} \dot {\beta} }h_{(\dot {\alpha} \dot {\beta} )}, $$ where $$ h_{(\alpha \beta )} = (\sigma^{\mu \nu})_{\alpha \beta}M_{\mu \nu}, \quad h_{(\dot {\alpha} \dot {\beta} )} = -(\tilde {\sigma}^{\mu \nu})_{\dot {\alpha }\dot {\beta }}M_{\mu \nu} \qquad (.1) $$ are an irreducible spinor representations (for other definitions look here).

With generator $J_{\mu \nu}$ of the Lorentz group and corresponding irreducible representation $T(g) = e^{\frac{i}{2}\omega^{\mu \nu}J_{\mu \nu}}$, by rewriting antisymmetric tensor $\omega^{\mu \nu}$ with using spinor formalism we can get

$$ T(g) = e^{\frac{i}{2}\left(\omega^{(ab)}J_{(ab)} + \omega^{(\dot {a}\dot {b})}J_{(\dot {a}\dot {b})}\right)}, $$ where (compare with $(.1)$) $$ J_{(ab)} = \frac{1}{2}(\sigma^{\mu \nu})_{a b}J_{\mu \nu}, \quad J_{(\dot {a} \dot {b} )} = -\frac{1}{2}(\tilde {\sigma}^{\mu \nu})_{\dot {a }\dot {b}}J_{\mu \nu}, $$ so the Lorentz group is generated by two symmetrical spinor tensors.

I got commutation relations of these tensors: $$ [J_{(\dot {a} \dot {b})}, J_{(\dot {c} \dot {d})}] = \frac{i}{2}\left( \varepsilon_{\dot {a}\dot {c}}J_{(\dot {b} \dot {d})} + \varepsilon_{\dot {b} \dot {d}}J_{(\dot {a} \dot {c})} + \varepsilon_{\dot {a} \dot {d}}J_{(\dot {b} \dot {c})} + \varepsilon_{\dot {b} \dot {c}}J_{(\dot {a} \dot {d})}\right), $$ $$ [J_{(a b)}, J_{(c d)}] = \frac{i}{2}\left( \varepsilon_{ac}J_{(bd)} + \varepsilon_{bd}J_{(ac)} + \varepsilon_{ad}J_{(bc)} + \varepsilon_{bc}J_{(ad)}\right). $$ But commutator $[J_{(a b)}, J_{(\dot {c} \dot {d})}]$ isn't equal to zero, against expectations. It's equal to $$ [J_{(a b)}, J_{(\dot {c} \dot {d})}] = -\frac{i}{8}\left( (\sigma^{\beta})_{b\dot {c}}(\sigma^{\nu})_{a\dot {d}} + (\sigma^{\beta })_{b \dot {d}}(\sigma^{\nu})_{a \dot {c}} + (\sigma^{\beta })_{a \dot {c}}(\sigma^{\nu})_{b \dot {d}} + (\sigma^{\beta})_{a \dot {d}}(\sigma^{\nu})_{b \dot {c}}\right)J_{\beta \nu}, $$ which isn't zero (look here).

Should it be so?

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Is representation T(g) Dirac representation ? –  user10001 Aug 29 '13 at 5:07
    
@user10001 . What do you mean? I'm only know that Dirac representation is one of the representations of Dirac matrices. –  PhysiXxx Aug 29 '13 at 8:06
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sorry i didn't read it carefully. T(g) is usual vector representation and you are talking about its splitting into two su(2)'s right? By Dirac representation i meant 4d complex spinor representation which is direct sum of right and left Weyl. –  user10001 Aug 29 '13 at 14:19
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up vote 1 down vote accepted

Your last expression is equals to zero, because the first term is symmetric in $\beta$ and $\nu$, while $J_{\beta\nu}$ is antisymmetric in $\beta$ and $\nu$.

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Thank you! You helped me very much. I forgot about $J_{\beta \nu}$. –  PhysiXxx Aug 29 '13 at 9:13
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