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, 2019 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, 2019 at 11:04
  • $\begingroup$ If $\epsilon = \epsilon^{\alpha \beta}$, then $\epsilon^{-1} = ( \epsilon^{-1})_{\alpha \beta} = ( \epsilon^T)_{\alpha \beta}$. $\endgroup$
    – MBolin
    Aug 24, 2019 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, 2019 at 11:18

1 Answer 1


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 \tag{16} \label{16} \\ \varepsilon = \varepsilon^{\alpha\beta} &\implies \varepsilon^T = (\varepsilon^T)^{\beta\alpha} . \tag{17} \label{17} \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}.

Solving this conundrum requires a careful look at how matrix notation and index notation relate to each other. To that end, eqs. \eqref{16} and \eqref{17} are not quite correct. Focusing only on the first equality, a more rigorous statement would be $$ A = A_\alpha{}^\beta \; \pmb{e}^\alpha \otimes \pmb{e}_\beta , \tag{18} \label{18} $$ where $\pmb{e}^\alpha \otimes \pmb{e}_\beta$ is the basis of $A$. The basis vectors fulfill the following relations:* \begin{align} \pmb{e}_\alpha \cdot \pmb{e}^\beta &= \delta_\alpha^\beta \\ \pmb{e}^\alpha \cdot \pmb{e}^\beta &= \varepsilon^{\alpha\beta} \tag{19} \label{19} \\ \pmb{e}_\alpha \cdot \pmb{e}_\beta &= (\varepsilon^{-1})_{\alpha\beta} , \end{align} whereby e.g.** \begin{align} A \cdot B &= (A_\alpha{}^\beta \; \pmb{e}^\alpha \otimes \pmb{e}_\beta) \cdot (B_\mu{}^\nu \; \pmb{e}^\mu \otimes \pmb{e}_\nu) = A_\alpha{}^\beta B_\mu{}^\nu (\pmb{e}_\beta \cdot \pmb{e}^\mu) \; \pmb{e}^\alpha \otimes \pmb{e}_\nu \\ &= A_\alpha{}^\beta B_\mu{}^\nu (\delta_\beta^\mu) \; \pmb{e}^\alpha \otimes \pmb{e}_\nu = A_\alpha{}^\beta B_\beta{}^\nu \; \pmb{e}^\alpha \otimes \pmb{e}_\nu . \tag{20} \label{20} \end{align} With this knowledge we are finally in a position to clearly state how matrix and index notation relate to each other: while both notations keep the basis vectors implicit, tensor notation explicitly specifies the basis of components.

Unfortunately for the non-expert, the implicit handling of basis in matrix notation means that context becomes crucial when translating from matrix notation to index notation. Looking back at equation $(6)$, it can be seen from eq. \eqref{16} that eq. \eqref{13} actually should be understood to mean that $$ (\varepsilon^T)_{\alpha\beta} = (\varepsilon^{-1})_{\alpha\beta} . \tag{21} \label{21} $$ This is consistent with eq. \eqref{17}, as $(\varepsilon^T)^{\alpha\beta}=\varepsilon^{\beta\alpha}$ implies $$ (\varepsilon^T)_{\alpha\beta} = (\varepsilon^{-1})_{\alpha\mu} \, (\varepsilon^{-1})_{\beta\nu} \; \varepsilon^{\nu\mu} = (\varepsilon^{-1})_{\alpha\mu} \; \delta_\beta^\mu = (\varepsilon^{-1})_{\alpha\beta} . \tag{22} \label{22} $$

Similar comments apply to eqs. $(3)$ and \eqref{11}.

Edit: The last section (concerning the relation between index and matrix notation) initially contained some wrong statements, and have been edited accordingly. The moral I personally take from this is that while implicit notation can be blindingly obvious to the expert and the un-initiated, it can be devilishly abtruse to the intermediate student.$^\dagger$

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

**) To see why this sbould be the proper way to mix $\otimes$ and $\cdot$, consider the definition of the matrix product, or see this question over at math.stackexchange.

$\dagger$) Had I only known and focused on matrix notation, I would have seen no problems with the expressions from Scharf's book (and indeed, I now think there are none); it was my incomplete knowledge of index notation that sent me on this lengthy quest for answers. Incidently, this journey did not merly deepen my understanding of index notation, but also revealed that I did not understood matrix notation quite as well as I initially thought.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.