I'm currently reading the textbook "Finite Quantum Electrodynamics" by Günter Scharf, but I find myself stuck already on page 24.
Background
Scharf introduces the index-raising symbol (spinor metric) $$ \epsilon^{\alpha\beta}= \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}, \tag{1} \label{1} $$ before writing the equation $$ A^T \epsilon A = \epsilon, \tag{2} \label{2} $$ where $A = A_\alpha{}^\beta \in SL(2, \mathbb{C})$. He states that this equation can be proven "simply by multiplying the $2 \times 2$-matrices", and therefore I assume that $\varepsilon = \epsilon^{\alpha\beta}$, since this makes eq. \eqref{2} straightforward.
The problem
Immediately thereafter, he claims that by multiplying eq. \eqref{2} with $A^{-1} \varepsilon^{-1}$ and "taking the adjoint" one obtains $$ A^* = \varepsilon (A^{-1})^\dagger \varepsilon^{-1} . \tag{3} \label{3} $$ I see two ways to try and derive this result, but both fails:
Start by multiplying eq. \eqref{2} with $\varepsilon^{-1} (A^T)^{-1} = \varepsilon^{-1} (A^{-1})^T$ from the left, obtaining $$ A = \varepsilon^{-1} (A^{-1})^T \varepsilon , \tag{4} \label{4} $$ whereby $$ A^* = \varepsilon^{-1} (A^{-1})^\dagger \varepsilon . \tag{5} \label{5} $$
Alternatively, following Scharf's approach, I obtain \begin{align} A^T &= \varepsilon A^{-1} \varepsilon^{-1} \\ \implies \quad A &= (\varepsilon A^{-1} \varepsilon^{-1})^T \\ &= (\varepsilon^{-1})^T (A^{-1})^T \varepsilon^T \\ &= (\varepsilon^T)^{-1} (A^{-1})^T \varepsilon^T. \tag{6} \label{6} \end{align} This can be shown to be equal to eq. \eqref{4} by using that $\varepsilon^{-1} \varepsilon^T = (\varepsilon^T)^{-1} \varepsilon = -I_2$, whereby \begin{align} A &= (\varepsilon^T)^{-1} (A^{-1})^T \varepsilon^T \\ &= \varepsilon^{-1} \varepsilon^T (\varepsilon^T)^{-1} (A^{-1})^T \varepsilon^T (\varepsilon^T)^{-1} \varepsilon \\ &= \varepsilon^{-1} (A^{-1})^T \varepsilon . \tag{7} \label{7} \end{align}
My two derivations thus seems to be consistent with each other. In order to arrive at eq. \eqref{3}, I seem to be forced to assume that $\varepsilon^T = \varepsilon^{-1}$ whereby $(\varepsilon^{-1})^T = (\varepsilon^T)^T = \varepsilon$, but the horror arises when I try to translate these expressions into index notation ...
Questions
Are my derivations correct?
Is Scharf's result correct, and if so, how does it squares with my derivations, provided they too are correct?
As eluded to above, I'm really struggling with spinor index notation. If $A = A_\alpha{}^\beta$ and $\varepsilon = \varepsilon^{\alpha\beta}$, how does one go about giving indices to $A^T$ and $\varepsilon^T$? In particular, it seems natural that $$ \varepsilon^{-1} = (\varepsilon^{-1})_{\alpha\beta} \qquad \text{and} \qquad \varepsilon^T = (\varepsilon^T)^{\alpha\beta}, \tag{8} \label{8} $$ but given that $(\varepsilon^{-1})_{\alpha\beta}$ and $(\varepsilon^T)^{\alpha\beta}$ are numerically the same—both equal $\varepsilon^{\beta\alpha}$—is it most correct to consider them unequal (because they behave differently with regards to multiplication), or equal (implying that the indices can be moved around freely to obtain sensible expressions)?
Extra information
Scharf seemingly does not explicitly specify $\varepsilon_{\alpha\beta}$. I'm aware some authors use $\varepsilon_{\alpha\beta} = -\varepsilon^{\alpha\beta}$ while others prefer $\varepsilon_{\alpha\beta} = \varepsilon^{\alpha\beta}$ (c.f. Moniz, 2010, pages 216–217). Thus, please be clear on which convention you are using if it makes any difference for the answer.
An additional reason for assuming that $\varepsilon^T$ are given the indices $(\varepsilon^T)^{\alpha\beta}$ comes on page 27 of Scharf's text, where he writes the derivation $$ \partial^{\alpha \dot{\beta}} = \varepsilon^{\alpha\gamma} \varepsilon^{\dot{\beta}\dot{\delta}} \partial_{\gamma \dot{\delta}} = (\varepsilon \partial \varepsilon^T)^{\alpha \dot{\beta}}, \tag{9} \label{9} $$ an equation which to my eye doesn't makes much sense unless $\varepsilon^T = (\varepsilon^T)^{\alpha\beta}$.