I'm reading the review by H. K. Dreiner, H. E. Haber and S. P. Martin (arXiv:0812.1594) about the two-component spinor formalism. There are some identities and notational conventions which lead to some confusion on my part after I when them.
Definitions / Conventions
Since this topic is very prone to mistakes from mixing different conventions, I'll try to collect the relevant definitions below. If in doubt, I'd like to stick to the conventions of the paper mentioned above.
The authors define the anti-symmetric symbol as (Eq. (2.19)) $\epsilon^{12} = -\epsilon^{21} = \epsilon_{21} = -\epsilon_{12} = 1$ and $\epsilon^{\dot{\alpha}\dot{\beta}} = (\epsilon^{\alpha\beta})^*$, $\epsilon_{\dot{\alpha}\dot{\beta}} = (\epsilon_{\alpha\beta})^*$, i.e. the numerical values for dotted and undotted indices are identical.
With these objects, we can raise and lower spinor indices (Eq. (2.20)) $$ \psi_\alpha = \epsilon_{\alpha\beta} \psi^\beta \,, \quad \psi^\alpha = \epsilon^{\alpha\beta} \psi_\beta \,, \quad \psi^\dagger_{\dot{\alpha}} = \epsilon_{\dot{\alpha}\dot{\beta}} \psi^{\dagger\dot{\beta}} \,, \quad \psi^{\dagger\dot{\alpha}} = \epsilon^{\dot{\alpha}\dot{\beta}} \psi^\dagger_{\dot{\beta}} $$ and similar for higher rank objects (Eq. (2.21)): $$ A^{\gamma\delta} = \epsilon^{\gamma\alpha} \epsilon^{\delta\beta} A_{\alpha\beta} \,, \qquad A_{\gamma\delta} = \epsilon_{\gamma\alpha} \epsilon_{\delta\beta} A^{\alpha\beta} \,. $$
The Pauli matrices are defined as (Eqs. (2.27), (2.28)) $$ (\sigma^\mu)_{\alpha\dot{\beta}} = (\mathbb{1}_{2\times2},\vec{\sigma}) \,, \qquad (\bar{\sigma}^\mu)^{\dot{\alpha}\beta} = (\mathbb{1}_{2\times2},-\vec{\sigma}) \,. $$ With them we can define (Eqs. (2.71), (2.72)) $$ (\sigma^{\mu\nu})_\alpha{}^\beta = \frac{i}{4} \left(\sigma^\mu_{\alpha\dot{\gamma}} \bar{\sigma}^{\nu\dot{\gamma}\beta}-\sigma^\nu_{\alpha\dot{\gamma}} \bar{\sigma}^{\mu\dot{\gamma}\beta}\right) \,, \qquad (\bar{\sigma}^{\mu\nu})^{\dot{\alpha}}{}_{\dot{\beta}} = \frac{i}{4} \left(\bar{\sigma}^{\mu\dot{\alpha}\gamma}\sigma^\nu_{\gamma\dot{\beta}}-\bar{\sigma}^{\nu\dot{\alpha}\gamma}\sigma^\mu_{\gamma\dot{\beta}}\right) \,. $$ for which the following identities hold (Eq. (2.77)) $$ (\sigma^{\mu\nu})_\alpha{}^\beta = \epsilon_{\alpha\tau} \epsilon^{\beta\gamma} (\sigma^{\mu\nu})_\gamma{}^\tau, \qquad (\bar{\sigma}^{\mu\nu})^{\dot{\alpha}}{}_{\dot{\beta}} = \epsilon^{\dot{\alpha}\dot{\tau}} \epsilon_{\dot{\beta}\dot{\gamma}} (\bar{\sigma}^{\mu\nu})^{\dot{\gamma}}{}_{\dot{\tau}} \,. \label{eq:id}\tag{1} $$ I've verified these identities by inserting the explicit representations for the Pauli matrices and performing all sums.
Moreover, they discuss the correspondence between their component notation and objects with two spinor indices viewed as matrices. Around Eqs. (2.33) and (2.34) they write $$ (V^T)_{\alpha\dot{\beta}} = V_{\beta\dot{\alpha}} \,, \quad (V^*)_{\dot{\alpha}\beta} = (V_{\alpha\dot{\beta}})^* \,, \quad (V^\dagger)_{\alpha\dot{\beta}} = (V_{\beta\dot{\alpha}})^* = (V^*)_{\dot{\beta}\alpha} \,, \label{eq:mat}\tag{2} $$ $$ (W^T)_\alpha{}^\beta = W^\alpha{}_\beta \,, \quad (W^*)_{\dot{\alpha}}{}^{\dot{\beta}} = (W_\alpha{}^\beta)^* \,, \quad (W^\dagger)^{\dot{\beta}}{}_{\dot{\alpha}} = (W_\alpha{}^\beta)^* = (W^*)_{\dot{\alpha}}{}^{\dot{\beta}} \,. \label{eq:mat2}\tag{3} $$
Questions
If I use the index raising and lowering properties of the $\epsilon$ symbol on the identities (\ref{eq:id}) I get, e.g. in the undotted case, $$ (\sigma^{\mu\nu})_\alpha{}^\beta = (\sigma^{\mu\nu})^\beta{}_\alpha \,, $$ which I would interpret, using the conventions (\ref{eq:mat2}), as $$ (\sigma^{\mu\nu})_\alpha{}^\beta = \left((\sigma^{\mu\nu})^T\right)_\alpha{}^\beta \,, $$ i.e. $\sigma^{\mu\nu} = (\sigma^{\mu\nu})^T$ in matrix notation. As one can quickly verify using explicit representations for the Pauli matrices, this does not hold (check, e.g., $(\mu,\nu) = (0,2)$ or $(1,3)$). My question is now: Where do I misinterpret the conventions or make a mistake?
When I try to combine the notation for transposition and complex conjugation for the mixed undotted/dotted case in (\ref{eq:mat}), I don't get the same expression for hermitian conjugation as given there: $$ (V^\dagger)_{\alpha\dot{\beta}} = \left((V^*)^T\right)_{\alpha\dot{\beta}} = (V^*)_{\beta\dot{\alpha}} = (V_{\dot{\beta}\alpha})^* \,. $$ This seems to be exactly the opposite concerning the dottedness compared to what (\ref{eq:mat}) claims. Did I overlook anything?
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