# How to vary fermion action in the index-free Clifford notation with respect to spin-connection?

In ref (1), it is claimed that the Dirac action (2.30) $$S_D \sim \int ( \overline \psi \star eee D \psi + \overline {D\psi} \star eee \psi)$$ becomes $$\delta_\omega S_D \sim \int - \{\psi\overline \psi, \star eee\} \;\delta \omega$$ where varying the action with respect to the spin connection $\omega$. I don't understand why we must rearrange the terms in action this way.

References

(1) A.C.Randono, 'In Search of Quantum de Sitter Space: Generalizing the Kodama State' link

$$D_\mu \psi^A = \partial_\mu \psi^A - \frac i 2 \omega_\mu{}^{IJ}(\mathcal J_{IJ})^A{}_B \, \psi^B\;,$$ where $\mathcal J_{IJ}\equiv - \frac i 4 [\gamma_I, \gamma_J]$. Or we may simply write \begin{eqnarray} D_\mu \psi^A &=&\partial_\mu\psi^A - \frac 1 8 \omega_\mu{}^{IJ}[\gamma_I,\gamma_J]^A{}_B\, \psi^B\;,\\ &=& \partial_\mu\psi^A - \frac 1 4 \omega_\mu{}^{IJ}(\gamma_I,\gamma_J)^A{}_B\, \psi^B\;. \end{eqnarray}
\begin{eqnarray} \delta_\omega(D_\mu \psi^A ) &=&- \frac 1 4 \delta(\omega_\mu{}^{IJ})(\gamma_I,\gamma_J)^A{}_B\, \psi^B\;,\\ &\equiv& - \delta(\omega_\mu{}^{IJ})^A{}_B\, \psi^B\;, \end{eqnarray} also, since we use the real representation of $\gamma$'s (In this thesis it use $\{\gamma_I,\gamma_J\}=+2\eta_{IJ}$ with $\eta=diag(-1,1,1,1)$ ) $$\delta_\omega(\overline{D_\mu \psi_A} ) = - \delta(\omega_\mu{}^{IJ})^B{}_A\,\overline{ \psi_B}\;.$$ In shorthand notation we write \begin{eqnarray} \delta_\omega \, D\psi &=& -\delta\omega \,\psi \;,\\ \delta_\omega \,\overline {D\psi} &=& - \overline {\psi}\,\delta\omega \; \end{eqnarray} Note that to see $\omega$ in Clifford basis we try \begin{eqnarray} \omega= \frac 1 4 \omega_{IJ}\gamma^I \gamma^J ,\; \omega \wedge \omega &=& \frac 1 {16} \omega_{IJ} \wedge \omega_{KL} \gamma^I < \gamma^J,\gamma^K> \gamma^L\;,\\ &=& \frac 1 {4} \omega_{IK} \wedge \omega^K{}_{L} \gamma^I \gamma^L\;.\\ F = \mathrm{d} \omega+ \omega \wedge \omega &=& \frac 1 4 \mathrm{d}\omega_{IJ}\gamma^I \gamma^J + \frac 1 {4} \omega_{IK} \wedge \omega^K{}_{L} \gamma^I \gamma^L\;,\\ &=&\frac 1 4 \Big[ \mathrm{d}\omega_{IJ}+ \omega_{IK} \wedge \omega^K{}_{J} \Big] \gamma^I \gamma^J \equiv \frac 1 4 F_{IJ} \gamma^I \gamma^J\;. \end{eqnarray} Which consistence to $F$ in the thesis. \begin{eqnarray} S &\sim& \int\Big( \overline \psi_A (\star e e e)^A{}_B (D\psi)^B + (\overline{D\psi})_A (\star eee)^A{}_B \psi^B \Big)\;,\\ \delta_\omega S &\sim& \int \Big( \overline \psi_A (\star e e e)^A{}_B (-\delta \omega\,\psi)^B + (-\overline{\psi}\delta \omega)_A (\star eee)^A{}_B \psi^B \Big)\;,\\ &=&\int - \overline \psi_A (\star e e e)^A{}_B \delta \omega^B{}_C\,\psi^C -\overline{\psi}_C\delta \omega^C{}_A (\star eee)^A{}_B \psi^B \;,\\ &=& \int - \psi^C \overline \psi_A (\star e e e)^A{}_B \delta \omega^B{}_C\,- (\star eee)^A{}_B \psi^B \overline{\psi}_C\delta \omega^C{}_A\;,\\ (&\equiv& \int - \psi^C \otimes \overline \psi_A (\star e e e)^A{}_B \delta \omega^B{}_C\,- (\star eee)^A{}_B \psi^B \otimes\overline{\psi}_C\delta \omega^C{}_A)\;,\\ &=& \int - \psi \overline \psi (\star e e e) \delta \omega\,- (\star eee)\psi \overline{\psi}\delta \omega\;,\\ &=& -\int \Big\{\psi \overline \psi ,(\star e e e) \Big\}\delta \omega\, \end{eqnarray}
$$\delta_\omega S_D \sim \int Tr ( \overline \psi \star eee (-\delta \omega) \psi + (-\delta \omega \overline {\psi}) \star eee \psi)\\ =\int Tr ( -\overline \psi \star eee \psi - \overline {\psi} \star eee \psi)\delta \omega\\$$ By using $Tr(ABC)=Tr(CAB)=Tr(BCA)$ we can write $$S_D \sim \int Tr ( - \star eee \psi \overline \psi- \psi\overline {\psi} \star eee )\delta \omega\\$$ This is what you want.
• I think $\delta \omega$ cannot move freely to the right. It's a clifford algebra vulued in this case. – Saksith Jaksri Feb 22 '17 at 23:56