I've been doing some work with both Baez's Knots, gauge fields and gravity (1) and Gambini, Pullin's Loops, knots, gauge Theories and quantum gravity (2), lately.

I have basically two problems: I understand that, in the ADM formalism, the Lagrangian density for the cosmological term of Einstein equation is given by $$ L= q\Lambda \underline{N},$$ where $q$ is the determinant of the 3-metric, $\Lambda$ is the cosmological constant, and $\underline{N}$ is $q^\frac{-1}{2}$N (the lapse function). Also, that $$\tilde{E^i_a}=q^{\frac{1}{2}}E^i_a$$ are the densitized triads of the Ashtekar formalism. However, I don't get why $q$ can be given by the expression (7.53) from (2): $$q=\frac{1}{6}\underline{\epsilon_{abc}}\epsilon^{ijk}\tilde{E^a_i}\tilde{E^b_j}\tilde{E^c_K}. $$ Is there a way to obtain such expression?

The second problem is: after promoting the Ashtekar variables to operators ( $\hat{A^a_i}$ and $\hat{E^a_i}=\frac{\delta}{\delta A^a_i}$ ), it's can be shown, for the Chern-Simons state $$\psi_{\Lambda}= e^{-\frac{6}{\Lambda}S_{CS}},$$ with $S_{CS}$ being the Chern-Simons action $$S_{CS}= \int_{\Sigma} tr \,(A\wedge dA +\frac{2}{3}A\wedge A\wedge A),$$ that \begin{eqnarray*} \frac{\delta}{\delta A^i_a}\psi_\Lambda = \frac{3}{\Lambda}\overline{\epsilon^{abc}}F^i_{bc}\psi_\Lambda \\ \underline{\epsilon_{abc}}\frac{\delta}{\delta A^i_a}\psi_\Lambda = \frac{6}{\Lambda}F^i_{bc}\psi_\Lambda, \end{eqnarray*} which comes from expressions (7.70) and (7.71) from (2).

My problem is with the second line. Am I supposed to take

$$ \underline{\epsilon_{abc}}\overline{\epsilon^{abc}} = 2~?$$ Why would that be true?

Sorry for the lengthy post. I'd be glad if someone could help me with these.


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