Obtaining curved space Dirac equation from action (tetrad formalism)

Question

I'm reading the book Covariant Loop quantum gravity by C.Rovelli where in 3.2 the action of a dirac fermion is presented in the tetrad formalism:

$$S= int bar{psi} gamma^{I} Dpsi wedge e^J wedge e^{K} wedge e^{L} epsilon_{IJKL} $$

where $D = D_{mu} dx^{mu}$

I can't obtain from this action the typical action in terms of the coordinates. Could anyone give some guidance on how to proceed?

Cham · Answer

Here's a standard formulation of the Dirac field lagrangian (where $hbar = 1  = c$, and $bar{Psi} equiv Psi^{dagger} , gamma^0$, as usual) :
begin{equation}tag{1}
mathscr{L}_{textsf{D}} = i , frac{1}{2} big( bar{Psi} : Gamma^{mu} , (, D_{mu} , Psi ,) - (, D{_{mu}} , bar{Psi} ,) , Gamma^{mu} , Psi , big) - m , bar{Psi} , Psi.
end{equation}
Here, $D_{mu}$ is the covariant derivative of the Dirac field, under local Lorentz transformations and under arbitrary coordinates transformations.  Neglecting the electromagnetic and Yang-Mills fields for simplicity :
begin{equation}tag{2}
D_{mu} , Psi equiv partial_{mu} , Psi + frac{1}{2} : omega_{mu}^{; ab} : M_{ab} , Psi,
end{equation}
where $M_{ab}$ is the set of the Lorentz group generators :
begin{equation}tag{3}
M_{ab} = i , frac{1}{2} ( gamma_a , gamma_b - gamma_b , gamma_a),
end{equation}
and $omega_{mu}^{; ab}$ are the spin-connection coefficients.  Those can be defined from the tetrad field.  Under some coordinates $x^{mu}$, the tetrad is $boldsymbol{e}^a equiv e_{mu}^a(x) , boldsymbol{d}x^{mu}$ and can be found from the metric :
begin{equation}tag{4}
dboldsymbol{s}^2 = g_{mu nu} : boldsymbol{d}x^{mu} otimes boldsymbol{d}x^{nu} equiv eta_{ab} : e_{mu}^a(x) , e_{nu}^b(x) : boldsymbol{d}x^{mu} otimes boldsymbol{d}x^{nu} equiv eta_{ab} : boldsymbol{e}^a otimes boldsymbol{e}^b.
end{equation}
Then the "generalized" gamma matrices are
begin{equation}tag{5}
Gamma^{mu}(x) equiv gamma^a : e_a^{mu}(x),
end{equation}
and the spin connection coefficients are given by this formula :
begin{equation}tag{6}
omega_{mu ; b}^{; a}(x) = e_{lambda}^a : nabla_{mu} , e_b^{lambda}.
end{equation}
Using the lagrangian density (1) above in the Euler-Lagrange equation gives the usual Dirac equation, in a gravitational field (i.e. curved spacetime).  The calculations are laborious.

Is that what you are looking for ?

mike stone · Answer

After you have written $gamma^I= e^I_mu gamma^mu$ you can use ${bf e}^I= e_mu^I dx^mu$ to write
$$
epsilon_{IJKL} {bf e}^Iwedge {bf e}^Jwedge {bf e}^Kwedge{bf e}^L = sqrt{g}d^4x
$$
and so get the usual coordinate action.
$$
S= int  bar psi gamma^mu nabla_mu psi sqrt{g}d^dx. 
$$

Obtaining curved space Dirac equation from action (tetrad formalism)

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