# Derivation on 3-form Einstein equation from Palatini action

I was asked by a student a derivation problem on 3-form Einstein euqation obtained from Palatini action. He posted some pictures on the book he have been reading.

I derive firstly the gravitational part from Platini action which reads (for simplification I consider cosmological constant as zero so the term vanishes $\int\sqrt{-g}d^4x\;2\Lambda=0$)： \begin{align*} S_G&=\frac{1}{2\kappa}\int d^4x\,\sqrt{-g}R\xrightarrow{\epsilon_{IJKL}e^I\wedge e^J\wedge R^{KL}=-2|e|R\,dx^4} \frac{-1}{4\kappa}\int\epsilon_{IJKL}e^I\wedge e^J\wedge R^{KL}=\frac{-1}{4\kappa}\int \text{Tr}(e\wedge e\wedge R)\\ \delta S_G&=\frac{-1}{4\kappa}\delta\int\text{Tr}(e\wedge e\wedge R)=\frac{-1}{2\kappa}\int\text{Tr}(\delta e\wedge e\wedge R)+\frac{-1}{4\kappa}\int\text{Tr}(e\wedge e\wedge\delta R)\\ &(\text{Tr}(\delta e\wedge e\wedge R)+\text{Tr}(e\wedge\delta e\wedge R)=2\text{Tr}(\delta e\wedge e\wedge R)) \end{align*} These two parts can be calculated as \begin{align*} \int\text{Tr}(\delta e\wedge e\wedge R)&=\int\text{Tr}(e\wedge R\wedge\delta e)=\int\epsilon_{IJKL} e^I\wedge R^{JK}\wedge\delta e^L\\ \int\text{Tr}(e\wedge e\wedge\delta R)&=\int d\text{Tr}(e\wedge e\wedge\delta\omega)-2\int\text{Tr}(De\wedge e\wedge\delta\omega)=-2\int\text{Tr}(\mathscr{T}\wedge e\wedge\delta\omega)\\ &(R=d\omega+\omega\wedge\omega\;,\;De^I=de^I+\omega^I_J\wedge e^J=\mathscr{T}^I) \end{align*} Then come to matter field part： \begin{align*} \delta S_m&=\int d^4x\,\delta(\sqrt{-g}\mathcal{L}_m)\xrightarrow{-g=-\det(g_{\mu\nu})=\det(e)^2=|e|^2}\int d^4x\,\delta(|e|\mathcal{L}_m) =\int|e|d^4x\,\frac{1}{|e|}\delta(|e|\mathcal{L}_m)\\ &=\int\mathcal{V}\frac{1}{|e|}\delta(|e|\mathcal{L}_m)=\int*\left(\frac{1}{|e|}\delta(|e|\mathcal{L}_m)\right)\\ \\ \frac{1}{|e|}\delta(|e|\mathcal{L}_m)&=\frac{1}{\sqrt{-g}}\frac{\delta (\sqrt{-g}\mathcal {L}_m)}{\delta g_{\alpha\lambda}}\delta g_{\alpha\lambda}=\left(\frac{\delta\mathcal{L}_m}{\delta g_{\alpha\lambda}}-\frac{1}{2}g^{\alpha\lambda}\mathcal{L}_m\right)\delta g_{\alpha\lambda}=-\frac{1}{2}T^{\alpha\lambda}\delta g_{\alpha\lambda}\\ &=-\frac{1}{2}T^{\alpha\lambda}\eta_{NL}(e^{N}_\alpha +\delta^{\lambda}_\alpha \delta^{L}_{N} e_{\lambda}^{L})\delta e_{\lambda}^{L}=-\frac{1}{2}(T^{\alpha\lambda}e_{\alpha L}+T^{\lambda \lambda}e_{\lambda L})\delta e_{\lambda}^{L}=-T_L^\lambda\delta e_{\lambda}^{L}\\ \\ *(T_L^\lambda\delta e_{\lambda}^{L})&=\frac{|e|}{(4-0)!}T_L^\lambda\delta e_{\lambda}^{L}\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\wedge dx^\lambda=\frac{1}{4}T_L\wedge\delta e^L \end{align*} Above derivation I have used 4-form voulme and 3-form definition of energy-momentum tensor \begin{align*} \mathcal{V}&=|e|d^4x=\frac{1}{4!}\epsilon_{IJKL}e^I_\mu e^J_\nu e^K_\rho e^L_\lambda dx^\mu\wedge dx^\nu\wedge dx^\rho\wedge dx^\lambda=\frac{1}{4!}\epsilon_{IJKL} e^I\wedge e^J\wedge e^K\wedge e^L\\ T_L&=|e|T^\lambda_L\eta_\lambda=\frac{|e|}{3!}T^\lambda_L\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\\ \end{align*} When we condider within zero torsion spacetime ($\mathscr{T}=0$) then we will encounter the Einstein equation in 3-form： \begin{align*} \epsilon_{IJKL} e^I\wedge R^{JK}=\frac{\kappa}{2} T_L \end{align*} Which is $4\pi G$ gravitation coefficient rather then $2\pi G$ on the book.

Could anybody tell me where did I go wrong?

• If you tell us what book this comes from it will be much easier for us to help you. Aug 23, 2016 at 19:38
• @Blazej I have searched the key words inside the pictures and found that it should be C. Rovelli's book books.google.de/books/about/… Aug 23, 2016 at 19:41
• $\uparrow$ Which page? Apr 14, 2017 at 15:36
• @Qmechanic You could see it is §2.1 in that book. Apr 14, 2017 at 15:55

First, I found a mistake in my original derivation on the 4-form energy-momentum tnsor, lacking of other three terms, so it should be: \begin{align*} \frac{1}{|e|}\delta(|e|\mathcal{L}_m)&=-T^{\alpha\beta}\eta_{NL}(e^{N}_\alpha +\delta^{\beta}_\alpha \delta^{L}_{N} e_{\beta}^{L})\delta e_{\beta}^{L}=-T_L^\beta\delta e_{\beta}^{L}\\ *(T_L^\beta\delta e_{\beta}^{L})&=\frac{|e|}{(4-0)!}(T_L^{\underline\beta}\delta e_{\underline\beta}^{L})\epsilon_{\mu\nu\rho\lambda}dx^\underline\mu\wedge dx^\underline\nu\wedge dx^\underline\rho\wedge dx^\underline\lambda=4\cdot\frac{1}{4}T_L\wedge\delta e^L =T_L\wedge\delta e^L \end{align*} Hence the Einstein's field equation in such case should gain the form as $\epsilon_{IJKL} e^I\wedge R^{JK}=2\kappa T_L$, the coefficient is exactly the one in action $16\pi G$.
Second, We could check this by turning the 3-form back to normal tensor form: \begin{align*} \epsilon_{IJKL} e^I\wedge R^{JK}&=\frac{1}{2}\epsilon_{IJKL} e^I\wedge R^{JK}_{\;\;\;MN} e^M\wedge e^N =\frac{1}{2}\epsilon_{IJKL}\epsilon^{IMNE}R^{JK}_{\;\;\;MN}\eta_{EP}*e^P\\ &=-\frac{1}{2}(3!\delta^M_{[J}\delta_K^N\delta^E_{L]}R^{JK}_{\;\;\;MN}\eta_{EP}*e^P)=2(R^{PL}-\frac{1}{2}\eta_{PL}R)*e^P\\ &=\frac{|e|}{3!}2(R^{PL}-\frac{1}{2}\eta_{PL}R)e^P_\lambda\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\\ &=\frac{|e|}{3!}2(R^L_\lambda-\frac{1}{2}Re_\lambda^L)\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\\ T_L&=\frac{|e|}{3!}T^\lambda_L\epsilon_{\mu\nu\rho\lambda}dx^\mu\wedge dx^\nu\wedge dx^\rho\\ \end{align*} Then the equation could go back to the correct usual way $R^L_\lambda-\frac{1}{2}R e_\lambda^L=8\pi G\,T_\lambda^L$.