I'm currently reading the textbook "Finite Quantum Electrodynamics" by Günter Scharf, but I find myself stuck already on page 24.


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:

  1. 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 $$ A^* = \varepsilon^{-1} (A^{-1})^\dagger \varepsilon . \tag{5} \label{5} $$

  2. 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 ...


  1. Are my derivations correct?

  2. Is Scharf's result correct, and if so, how does it squares with my derivations, provided they too are correct?

  3. 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 $\varepsilon^T = (\varepsilon^T)^{\alpha\beta}$.

  • $\begingroup$ From the expression you gave for $\epsilon$, you do have $\epsilon^{-1} = \epsilon^T$. $\endgroup$ – MBolin Aug 24 at 10:54
  • $\begingroup$ @MBolin: As far as I can see, this depends on the multiplication rules of $\varepsilon^T$, i.e. the correct way to "give indices to" $\varepsilon^T$. Numerically (i.e. component-wise) they are of course identical. $\endgroup$ – B. Bergtun Aug 24 at 11:04
  • $\begingroup$ If $\epsilon = \epsilon^{\alpha \beta}$, then $\epsilon^{-1} = ( \epsilon^{-1})_{\alpha \beta} = ( \epsilon^T)_{\alpha \beta}$. $\endgroup$ – MBolin Aug 24 at 11:11
  • $\begingroup$ @MBolin: It's obvious that $\varepsilon=\varepsilon^{\alpha\beta} \implies \varepsilon^{-1}=(\varepsilon^{-1})_{\alpha\beta}$, but why isn't $\varepsilon^T=(\varepsilon^T)^{\alpha\beta}$? Specifically, I don't see how equation \eqref{9} can be correct if $\varepsilon^T=(\varepsilon^T)_{\alpha\beta}$—the indices does not add up. $\endgroup$ – B. Bergtun Aug 24 at 11:18

Here is my own attempt at an answer:

All the derivations are correct. To see this, let $$ A^{-1}\equiv \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} , \tag{10} \label{10} $$ whereby \begin{align} \varepsilon (A^{-1})^T \varepsilon^{-1} &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} \\ &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} \begin{pmatrix} b & -a \\ d & -c \\ \end{pmatrix} \\ &= \begin{pmatrix} d & -c \\ -b & a \\ \end{pmatrix}, \tag{11} \label{11} \end{align} while \begin{align} \varepsilon^{-1} (A^{-1})^T \varepsilon &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix} \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} \\ &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} \begin{pmatrix} -b & a \\ -d & c \\ \end{pmatrix} \\ &= \begin{pmatrix} d & -c \\ -b & a \\ \end{pmatrix} = \varepsilon (A^{-1})^T \varepsilon^{-1}. \tag{12} \label{12} \end{align}

And yes, this does imply that $$ \varepsilon^T = \varepsilon^{-1}. \tag{13} \label{13} $$

As to the indices, the wikipedia article on the transpose of a linear map states that if $f : V \to W$ is a linear map, then the transpose of $f$ is defined to be $$ {}^t f : W^* \to V^* , \tag{14} \label{14} $$ where $V^*, W^*$ are the dual spaces of $V$ and $W$, respectively.

Hence, using that for any finite vector space $V$ one can always set $V^{**} = V$, one sees that \begin{align} A : V^* \to V^* &\implies A^T : V \to V \\ \varepsilon : V^* \to V &\implies \varepsilon^T : V^* \to V, \tag{15} \label{15} \end{align} i.e. \begin{align} A = A_\alpha{}^\beta &\implies A^T = (A^T)^\beta{}_\alpha \\ \varepsilon = \varepsilon^{\alpha\beta} &\implies \varepsilon^T = (\varepsilon^T)^{\beta\alpha} . \tag{16} \label{16} \end{align}

This index placement is in nice agreement with eqs. $(2)$ and $(4)$ to $(9)$. However, it is seemingly in conflict with eq. $(3)$ , and also eqs. \eqref{11} and \eqref{13}, as we appear to be rising and lowering indices twice!

This conundrum is solved by a rigorous treatment of how matrix notation relates to index notation. Specifically, eq. \eqref{16} is not quite correct. Focusing only on the first equality, a more rigorous statement would be $$ A = A_\alpha{}^\beta \; \pmb{e}^\alpha \otimes \pmb{e}_\beta , \tag{17} \label{17} $$ where $\pmb{e}^\alpha \otimes \pmb{e}_\beta$ is the basis of $A$. Hence, eq. \eqref{13} merely states that $$ (\varepsilon^T)^{\alpha\beta} \; \pmb{e}_\alpha \otimes \pmb{e}_\beta = (\varepsilon^{-1})_{\alpha\beta} \; \pmb{e}^\alpha \otimes \pmb{e}^\beta , \tag{18} \label{18} $$ and not that $(\varepsilon^T)^{\alpha\beta} = (\varepsilon^{-1})_{\alpha\beta}$.

As for eqs. $(3)$ and \eqref{11}, these are straightforward if one employs the convention $\varepsilon_{\alpha\beta} = \varepsilon^{\alpha\beta}$,* since then the relevant part simply reads $$ A_\alpha{}^\beta = \varepsilon_{\alpha\gamma} \ ((A^{-1})^T)^\gamma{}_\delta \; (\varepsilon^{-1})^{\delta\beta} . \tag{19} \label{19} $$ If one instead employs the convention $\varepsilon_{\alpha\beta} = -\varepsilon^{\alpha\beta}$, eq. \eqref{19} is still a valid representation, since $\varepsilon_{\alpha\beta} = -\varepsilon^{\alpha\beta} \implies (\varepsilon^{-1})^{\alpha\beta} = -(\varepsilon^{-1})_{\alpha\beta}$. Alternatively, one could write the less visually appealing $$ A_\alpha{}^\beta = \varepsilon_{\gamma\alpha} \ ((A^{-1})^T)^\gamma{}_\delta \; (\varepsilon^{-1})^{\beta\delta} , \tag{20} \label{20} $$ since $\varepsilon_{\alpha\beta} = -\varepsilon^{\alpha\beta} \iff \varepsilon_{\beta\alpha} = \varepsilon^{\alpha\beta}$.

All this is not quite to say that "indices can be moved around freely"—instead, the relations** \begin{align} \pmb{e}_\alpha \cdot \pmb{e}^\beta &= \delta_\alpha^\beta \\ \pmb{e}^\alpha \cdot \pmb{e}^\beta &= \varepsilon^{\alpha\beta} \tag{21} \label{21} \\ \pmb{e}_\alpha \cdot \pmb{e}_\beta &= (\varepsilon^{-1})_{\alpha\beta} , \end{align} can be used to "project out" meaningful index expressions from matrix expressions.

*) From here on out, all unbalanced index equations are to be understood in a numerical sense only. Thus, $\varepsilon_{\alpha\beta} = \varepsilon^{\alpha\beta}$ is shorthand for $\varepsilon_{12} = \varepsilon^{12} = 1$, and so on.

**) Adapted from "Gravity: An Introduction to Einstein's General Relativity" by James B. Hartle (2003), page 425.


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