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Gauge transformation of Dirac equation in curved spacetime

For a research project I tried to explicitly show the invariance of the Dirac equation in curved space time under $\operatorname{Spin}(1,3) \cong SL(2,\mathbb C)$-gauge transformations.

I'm working with the spin-connection $\omega_{\mu}{}^{I}{}_{J}$ in the $\mathfrak{so}(1,3)$-matrix representation. Let's focus on a single Weyl spinor $\psi \in \mathbb C^2$. The equation I used for the gauge covariant derivative was

$$D \psi = \partial_\mu \psi - \omega_\mu{}^I{}_K \frac{1}{4} \eta^{KJ} \sigma_{[I} \bar \sigma_{J]} \psi$$

where $\sigma_I$ and $\bar \sigma_J$ come from the Dirac gamma matrices $$ \gamma^I= \begin{pmatrix} 0 & \sigma ^I\\ \bar \sigma^I&0 \end{pmatrix} $$

For the gauge transformations with $\mathcal M: \to SL(2,\mathbb C)$ I used $$ \tilde \psi(x) = S(x) \psi(x)$$ $$ \omega_\mu{}^{\tilde I}{}_{\tilde J}(x) = \Lambda^{\tilde I}{}_{I}[S(x)] \Lambda^{-1}{}^{ J}{}_{\tilde J}[S(x)] \omega_\mu{}^{I}{}_{J}(x) + \Lambda^{\tilde I}{}_{K}[S(x)] \partial_\mu \Lambda^{-1}{}^{ J}{}_{\tilde J}[S(x)]\,,$$ where the Lorentz transformation is implicitly defined by $$ \Lambda^{\tilde I}{}_{I}[S] \sigma_{\tilde I} = S \sigma_I S^\dagger\,$$ $$\Lambda^{\tilde I}{}_I[S] \bar \sigma_{\tilde I} = S^\dagger{}^{-1} \bar \sigma_I S^{-1}\,\tag{*}$$

Failed proof

I tried to prove the gauge invariance as follows:

\begin{align} \tilde D_\mu S \psi &= \partial_\mu (S \psi ) - \omega_\mu{}^{\tilde I}{}_{\tilde K} \frac{1}{4} \eta^{\tilde K \tilde J} \sigma_{[\tilde I}\bar \sigma_{\tilde J]} S \psi \nonumber\\ &= \partial_\mu (S \psi) - \Lambda^{\tilde I}{}_{I} \Lambda^{-1}{}^{ K}{}_{\tilde K} \omega_\mu{}^{I}{}_{K} \frac{1}{4} \eta^{\tilde K \tilde J} \sigma_{[\tilde I}\bar \sigma_{\tilde J]} S\psi % \nonumber\\ &- \Lambda^{\tilde I}{}_{L} \partial_\mu \Lambda^{-1}{}^{ L}{}_{\tilde K} \frac{1}{4} \eta^{\tilde K \tilde J} \sigma_{[\tilde I}\bar \sigma_{\tilde J]} S \psi\nonumber\\ &= (\partial_\mu S) \psi + S (\partial_\mu \psi)- \Lambda^{\tilde I}{}_{I} \eta^{KJ} \Lambda^{ \tilde J}{}_{J} \omega_\mu{}^{I}{}_{K} \frac{1}{4} \sigma_{[\tilde I}\bar \sigma_{\tilde J]} S\psi % \nonumber\\ &+ (\partial_\mu \Lambda^{\tilde I}{}_{L}) \eta^{LJ} \Lambda^{ \tilde J }{}_{J} \frac{1}{4} \sigma_{[\tilde I}\bar \sigma_{\tilde J]} S \psi\nonumber\\ &= (\partial_\mu S) \psi + S (\partial_\mu \psi)\nonumber\\ &- \eta^{KJ} \omega_\mu{}^{I}{}_{K} \frac{1}{8} \Lambda^{\tilde I}{}_{I} \sigma_{\tilde I} \Lambda^{ \tilde J}{}_{J}\bar \sigma_{\tilde J} S\psi \nonumber\\ % &+ \eta^{KJ} \omega_\mu{}^{I}{}_{K} \frac{1}{8} \Lambda^{ \tilde J}{}_{J} \sigma_{\tilde J} \Lambda^{\tilde I}{}_{I}\bar \sigma_{\tilde I} S\psi \nonumber\\ % % &+ \eta^{LJ} \frac{1}{8} \partial_\mu ( \Lambda^{\tilde I}{}_{L} \sigma_{\tilde I}) \Lambda^{ \tilde J }{}_{J}\bar \sigma_{\tilde J} S \psi\nonumber\\ &- \eta^{LJ} \frac{1}{8} \Lambda^{ \tilde J }{}_{J} \sigma_{\tilde J} \partial_\mu (\Lambda^{\tilde I}{}_{L}\bar\sigma_{\tilde I}) S\psi\nonumber\\ \end{align} Using $(*)$ I simplified to \begin{align} \tilde D_\mu S \psi &= (\partial_\mu S) \psi + S (\partial_\mu \psi)\nonumber\\ &- \eta^{KJ} \omega_\mu{}^{I}{}_{K} \frac{1}{4} S \sigma_{[ I} \bar \sigma_{ J]} \psi \nonumber\\ % % &+ \eta^{LJ} \frac{1}{8} \partial_\mu ( S \sigma_{L} S^\dagger) S^\dagger{}^{-1} \bar \sigma_{ J} S^{-1} S \psi\nonumber\\ &- \eta^{LJ} \frac{1}{8} S\sigma_{ J} S^\dagger \partial_\mu ( S^\dagger{}^ {-1} \bar \sigma_{L} S^{-1}) S\psi\nonumber\\ % % &= (\partial_\mu S) \psi+ S D_\mu \psi\nonumber\\ &+ \eta^{LJ} \frac{1}{8} (\partial_\mu S) \sigma_{L} S^\dagger S^\dagger{}^{-1} \bar \sigma_{ J} S^{-1} S \psi\nonumber\\ &+ \eta^{LJ} \frac{1}{8} S \sigma_{L}( \partial_\mu S^\dagger) S^\dagger{}^{-1} \bar \sigma_{ J} S^{-1} S \psi\nonumber\\ &- \eta^{LJ} \frac{1}{8} S\sigma_{ L} S^\dagger ( \partial_\mu S^\dagger{}^ {-1}) \bar \sigma_{J} S^{-1} S\psi\nonumber\\ &- \eta^{LJ} \frac{1}{8} S\sigma_{ J} S^\dagger S^\dagger{}^ {-1} \bar \sigma_{L} (\partial_\mu S^{-1}) S\psi\nonumber\\ &= (\partial_\mu S) \psi+ S D_\mu \psi\nonumber\\ &+ \eta^{LJ} \frac{1}{8} (\partial_\mu S) \sigma_{L} \bar \sigma_{ J} \psi\nonumber\\ &+ \eta^{LJ} \frac{1}{8} S\sigma_{ J} \bar \sigma_{L} S^{-1} (\partial_\mu S)\psi\,.\tag{**} \end{align} Then I used $\eta^{LJ} \sigma_{ J} \bar \sigma_{L} = - 4 \mathbb I$ to obtain \begin{align} \label{eq:stament_for_psi} \tilde D_\mu S \psi &= (\partial_\mu S) \psi+ S D_\mu \psi\nonumber\\ &- \frac{1}{2} (\partial_\mu S \psi)\nonumber\\ &- \frac{1}{2} (\partial_\mu S\psi)\nonumber\\ &=S D_\mu \psi\,. \end{align}

Unfortunately, I realized that I did a sign mistake when deriving the last line of $(**)$. Therefore, the whole calculation seems to fail.

Does anyone know, how to modify the Weyl equation in curved spacetime such that the transformation behaviour of $D\psi$ can be verified. Note, that I am interested in non-infinitesimal gauge transformations.

I looked much into the literature, but I couldn't find any solution.

Convention: I'm using $(-+++)$-signature following Wess and Bagger

Remark: I don't think this question is a duplicate. I am not asking just for the Dirac equation in curved spacetime. What I am looking for is a definition of the spin-covariant derivative consistent with the gauge transformations of the spin-connection.

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2 Answers 2

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The sign problem is caused by your definition of the gauge covariant derivative $$D \psi = \partial_\mu \psi - \omega_\mu{}^I{}_K \frac{1}{4} \eta^{KJ} \sigma_{[I} \bar \sigma_{J]} \psi$$ which should be $$D \psi = \partial_\mu \psi + \omega_\mu{}^I{}_K \frac{1}{4} \eta^{KJ} \sigma_{[I} \bar \sigma_{J]} \psi$$

Actually you are doing the proof in an unnecessaryly complicated way. Forget about the $\Lambda^{\tilde I}{}_{I}[S]$ as in $\Lambda^{\tilde I}{}_{I}[S] \sigma_{\tilde I} = S \sigma_I S^\dagger$. You can directly invoke the Lorentz gauge transformation property of spin-connection $\omega_\mu{} = \omega_\mu{}^I{}_K \frac{1}{4} \eta^{KJ} \sigma_{[I} \bar \sigma_{J]}$ as: $$ \omega_\mu{} \rightarrow S\omega_\mu{}S^{-1} + S\partial_\mu S^{-1} = S\omega_\mu{}S^{-1} - (\partial_\mu S) S^{-1} $$ where we have used the property $$S\partial_\mu S^{-1} = - (\partial_\mu S) S^{-1} $$ Therefore, gauge covariant derivative transforms as $$ D \psi = \partial_\mu \psi + \omega_\mu{} \psi \\ \rightarrow \partial_\mu (S\psi) + (S\omega_\mu S^{-1} - (\partial_\mu S) S^{-1}) (S\psi) \\ =S\partial_\mu \psi + (\partial_\mu S)\psi + S\omega_\mu \psi - (\partial_\mu S)\psi \\ = S D \psi $$

Well, if you prefer, you can of course expand $\omega_\mu{}$ into its components $\omega_\mu{}^I{}_K \frac{1}{4} \eta^{KJ} \sigma_{[I} \bar \sigma_{J]}$, and redo the proof in terms of its components. But the big picture is the same.

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  • $\begingroup$ Thanks a lot for your answer. Let me clarify a few things: I see that gauge invariance is proven easier if you work in the spin representation $\omega_\mu \in \mathfrak{sl}(2,\mathbb C)$. The reason I worked with the connection in $\mathfrak{so}(1,3)$ representation was, that I also want to consider the covariant derivative of tensor with frame indices $T_{IJ}^{KL}$. $\endgroup$ Commented Oct 2 at 7:45
  • $\begingroup$ So want I am looking for is an explicit representation of $L: \mathfrak{so}(1,3) \stackrel{\cong}{\to} \mathfrak{sl}(2,\mathbb C)$ such that $$\Lambda^{\tilde I}{}_{K} \partial_\mu \Lambda^{-1}{}^{K}{}_{\tilde J} L_{\tilde I}{}^{\tilde J} = S (\partial_\mu S^{-1})\,.$$ In your answer, it is taken for granted that $$L_I{}^J := \frac{1}{4} \eta^{JK} \sigma_{[I} \bar\sigma_{K]}$$ satisfies the condition above. However, this is what I wanted to prove but failed. $\endgroup$ Commented Oct 2 at 7:57
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I managed to find an answer to the question.

Firstly, the overall sign $\pm \frac{1}{4} \eta^{KJ} \sigma_{[I} \bar \sigma_{J]} \psi$ depends on the signature of $\eta$, since it must match the sign of $\eta^{LJ} \sigma_{ J} \bar \sigma_{L} = \pm 4 \mathbb I$.

This origin of the sign disagreement with @MadMax .

As explained in the question, the calculation works perfectly up to the line

\begin{align} \tilde D_\mu S \psi &= (\partial_\mu S) \psi+ S D_\mu \psi\nonumber\\ &+ \eta^{LJ} \frac{1}{8} (\partial_\mu S) \sigma_{L} S^\dagger S^\dagger{}^{-1} \bar \sigma_{ J} S^{-1} S \psi\nonumber\\ &+ \eta^{LJ} \frac{1}{8} S \sigma_{L}( \partial_\mu S^\dagger) S^\dagger{}^{-1} \bar \sigma_{ J} S^{-1} S \psi\nonumber\\ &- \eta^{LJ} \frac{1}{8} S\sigma_{ L} S^\dagger ( \partial_\mu S^\dagger{}^ {-1}) \bar \sigma_{J} S^{-1} S\psi\nonumber\\ &- \eta^{LJ} \frac{1}{8} S\sigma_{ J} S^\dagger S^\dagger{}^ {-1} \bar \sigma_{L} (\partial_\mu S^{-1}) S\psi\nonumber\\ \end{align}

Solution: Where I need to show that the second plus the third line vanish. The error is to think that they need to cancel each other. However, one can show that they vanish individually.

Let's work in a specific representaiton of $\sigma_I$ and $\bar \sigma_I$. The representation of Wess and Bagger is given by $$ (\sigma_I) = \left( \begin{pmatrix} 1 &0\\ 0&1 \end{pmatrix}, \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}, \begin{pmatrix} 0 &-i\\ i&0 \end{pmatrix}, \begin{pmatrix} 1 &0\\ 0&-1 \end{pmatrix} \right) $$ and $\bar \sigma_0 = \sigma_0$, $\bar \sigma_{1,2,3} = - \sigma_{1,2,3}$. Since $S \partial_\mu S^{-1}$ lies in the Lie-algebra $\mathfrak{sl}(2,\mathbb C)$ it follows that $\operatorname{tr}(S^\dagger \partial_\mu S^{-1}{}^\dagger) =0$. Therefore, it is sufficient to show, that $$ \eta^{IJ} \sigma_I A \bar \sigma_J = 0 $$ if $A$ is trace free. Let $A:= \begin{pmatrix} a & b\\ c& -a \end{pmatrix}$. We then have \begin{align} \eta^{IJ} \sigma_I A \bar \sigma_J &= - \sigma_0 A \sigma_0 - \sigma_1 A \sigma_1 - \sigma_2 A \sigma_2- \sigma_3 A \sigma_3\\ &= -\begin{pmatrix} a & b \\ c & -a \end{pmatrix} -\begin{pmatrix} -a & c \\ b & a \end{pmatrix} -\begin{pmatrix} -a & -c \\ -b & a \end{pmatrix} -\begin{pmatrix} a & -b \\ -c & -a \end{pmatrix}\\ &= \begin{pmatrix} 0&0\\ 0&0 \end{pmatrix}\,. \end{align}

From this one gets that the second and third line vanish. Therefore, the whole calculation works.

Caveat: I do not know if this is independent of the representation of the Infeld-Van der Waerden symbols.

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