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here is this tricky identity to prove in an appendix of W.B. Supersymmetry and Supergravity that's driving me crazy. Some premises first:

This book use the van der Waerden's convention for spinor indices, where bispinors transforming in $\left(\frac{1}{2},0\right)$ and $\left(0,\frac{1}{2}\right)$ of SO(1,3) are indicated rispectively as: $$\delta\psi_\alpha=\frac{i}{2}{\left(\omega_{\mu\nu}\sigma^{\mu\nu}\right)_\alpha}^\beta\psi_\beta,\quad \delta\bar{\psi}^\dot{\alpha}=\frac{i}{2}{\left(\omega_{\mu\nu}\bar{\sigma}^{\mu\nu}\right)^\dot{\alpha}}_\dot{\beta}\bar{\psi}^\dot{\beta}$$ with ($\alpha=1,2$), and: $${{\sigma^{\mu\nu}}_\alpha}^\beta=\frac{1}{4}{\left(\sigma^\mu\bar{\sigma}^\nu-\sigma^\nu\bar{\sigma}^\nu\right)_\alpha}^\beta\\ {{\bar{\sigma}^{\mu\nu}}^\dot{\alpha}}_\dot{\beta}=\frac{1}{4}{\left(\bar{\sigma}^\mu\sigma^\nu-\bar{\sigma}^\nu\sigma^\nu\right)^\dot{\alpha}}_\dot{\beta}$$ The similarity transformation between the fundamental and anti-fundamental is given by: $$\psi^\alpha=\varepsilon^{\alpha\beta}\psi_\beta,\quad \bar{\psi}_\dot{\alpha}=\varepsilon_{\dot{\alpha}\dot{\beta}}\bar{\psi}^\dot{\beta}\\ {\bar{\sigma}^\mu}^{\dot{\alpha}\beta}=\varepsilon^{\beta\gamma}\varepsilon^{\dot{\alpha}\dot{\lambda}}\sigma^\mu_{\gamma\dot{\lambda}}\\ \sigma^\mu=(-\mathbb{I},\sigma^i)$$ where the $\sigma^i$ are Pauli's matrices and Levi-Civita's tensor $\varepsilon^{\alpha\beta}$ has signature $\varepsilon^{12}=\varepsilon_{21}=1$ so that $\varepsilon^{\alpha\beta}\varepsilon_{\beta\gamma}=\delta^\alpha_\gamma$.

Now the above mentioned identity is: $$\diamondsuit \quad \mathrm{Tr}\left(\sigma^{\mu\nu}\sigma^{\rho\tau}\right)=-\frac{1}{2}\left(\eta^{\mu\rho}\eta^{\nu\tau}-\eta^{\mu\tau}\eta^{\nu\rho}\right)-\frac{i}{2}\varepsilon^{\mu\nu\rho\tau}$$ where the signatures of the flat metric and Levi-Civita tensor are: $$\eta^{\mu\nu}=(-,+,+,+)\;,\quad \varepsilon^{0123}=-1$$ I understant why $\diamondsuit$ must have this form: $\eta^{\mu\nu}$ and $\varepsilon^{\mu\nu\rho\tau}$ are the only tensors with the correct index permutation structure and invariance under Lorentz, but can't obtain it more rigorously. Can you give me some help? Thank you in advance.

P.S. Position, geography of contraction and dotting of spinor indices are very important in this notation, with the only exception of $\varepsilon^{\alpha\beta}$ which always contract with the most left index.

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  • $\begingroup$ I do not know your exact circumstance, but in Penrose's notation if you have a basis of the two-spinor space, $\omicron^\bullet, \iota^\bullet: ~~\epsilon_{AB}\omicron^A\iota^B=\sqrt{1/2}$ then the Hermitian matrices over the two-spinors are a $(+---)$ Minkowski space, with Pauli matrices $$w^a=\omicron^A\omicron^{\dot A}+\iota^A\iota^{\dot A}\\x^a=\omicron^A\iota^{\dot A}+\iota^A\omicron^{\dot A}\\y^a=i~\omicron^A\iota^{\dot A}-i~\iota^A\omicron^{\dot A}\\z^a=\omicron^A\omicron^{\dot A}-\iota^A\iota^{\dot A}$$ being a 4-basis and $\eta^{ab}=\epsilon^{AB}\epsilon^{\dot A\dot B}$ the metric. $\endgroup$ – CR Drost Sep 28 '18 at 11:58
  • $\begingroup$ Well, that's true, and in van der Waerden's notation your suggestion can be translated in the map from Minkowsky space and the space of hermitian matrices whose determinant is conserved by SL(2,$\mathbb{C}$) similarity transformation (aka Lorentz) as:$$v^\mu\rightarrow v_{\alpha\dot{\alpha}}=v^\mu {\sigma_\mu}_{\alpha\dot{\alpha}}$$ but i don't get how can It be related to the identity that's obsessing me, sorry :P $\endgroup$ – Andrea Mosena Sep 28 '18 at 20:07
  • $\begingroup$ For a proof of this identity, consult "Supersymmetry: An Introduction with Conceptual and Calculational Details" by A. Wiedemann and Harald J. W. Mueller-Kirsten, page 84. $\endgroup$ – NormalsNotFar Sep 30 '18 at 13:03

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