In chapter 34 of his Quantum Field theory handbook, Srednicki discusses invariants of the Lorentz group and how they appear in the decomposition in irreducible representations of Lorentz tensors. As one example [Eq. 34.32], the fully antisymmetric Levi-Civita tensor $\epsilon^{\mu\nu\rho\sigma}$ appears in the direct product of

$$(2,2) \otimes (2,2) \otimes (2,2)\otimes (2,2)~=~(1,1)_A \oplus \ldots \tag{34.32} $$

Here, $(2,2)$ denotes the irreducible representation corresponding to the fourvector (or equivalently a tensor with one left-handed and one right-handed spinor index) and $(1,1)$ is the singlet.
Is there a straightforward way to see why there can be a singlet there that is antisymmetric? The singlet I can understand of course from the direct product rules of the $SU(2)$ group representations, but not the antisymmetric nature.

Am familiar with all the basics of Lie groups, $SU(N)$, Young tableaux and such, but fail to see how it translates to the case of the Lorentz group here. I guess my more general question would be how can one determine the (anti)symmetric nature of the irreducible representations in the decomposition of a general Lorentz group tensor.


I) Here is one derivation. Let us first consider the undotted 2-component left Weyl spinor. The decomposition of the 4th tensor power in terms of irreducible representations/Young diagrams reads: $$\begin{array}{rcccccc} \underline{\large\bf 2}^{\otimes 4} &\cong& 2~\underline{\large\bf 1}&\oplus& 3~\underline{\large\bf 3}&\oplus& \underline{\large\bf 5}\cr\cr [~~]^{\otimes 4} &\cong& 2~\begin{array}{rl} [~~]&[~~]\cr [~~] &[~~]\end{array}&\oplus& 3~\begin{array}{rcl} [~~]&[~~]&[~~]\cr [~~] \end{array}&\oplus& \begin{array}{rccl} [~~]& [~~] & [~~] & [~~] \end{array} \end{array}$$ Interestingly, there are 2 singlets! If we define the shorthand notation $$I~:=~\epsilon^{\alpha\beta}\epsilon^{\gamma\delta}, \qquad II~:=~\epsilon^{\alpha\gamma}\epsilon^{\beta\delta}, \qquad III~:=~\epsilon^{\alpha\delta}\epsilon^{\beta\gamma}, $$ there is a Fierz-like identity $$I-II+III~=~0.$$ Two independent tensors that each correspond to the singlet Young diagram $$\begin{array}{rl} [~~]&[~~]\cr [~~] &[~~]\end{array}$$ are $$ A~:=~~\epsilon^{\alpha\gamma}\epsilon^{\beta\delta}+\epsilon^{\alpha\delta}\epsilon^{\beta\gamma}~=~II+III,$$ and $$ B~:=~\epsilon^{\alpha\beta}\epsilon^{\gamma\delta}-\epsilon^{\alpha\delta}\epsilon^{\beta\gamma} ~=~I-III~=~II-2III.$$ The tensor $A$ is manifestly symmetric under $\alpha\leftrightarrow \beta$ and $\gamma\leftrightarrow \delta$. The tensor $B$ is manifestly symmetric under $\alpha\leftrightarrow \gamma$ and $\beta\leftrightarrow \delta$.

II) There is a similar story for the dotted 2-component right Weyl spinor. So there are altogether $2\times 2= 4$ singlets on the rhs. of Srednicki's eq. (34.32)! It turns out that the antisymmetric singlet $$(\underline{\large\bf 1},\underline{\large\bf 1})_A,$$ which is totally antisymmetric under permutation of spinor indices pairs $$ (\alpha,\dot{\alpha})\quad\leftrightarrow \quad (\beta,\dot{\beta})\quad\leftrightarrow \quad (\gamma,\dot{\gamma})\quad\leftrightarrow \quad (\delta,\dot{\delta}), $$ corresponds (up to an overall normalization) to the following linear combination of tensors $$ A\dot{B}-B\dot{A}, $$ in a hopefully obvious notation.

  • $\begingroup$ Correction to the answer (v2): Replace the word Fierz-like identity with Schouten identity. $\endgroup$ – Qmechanic Feb 9 '17 at 20:47

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