relating inverse metric to densitized triads

[jziprick]

Reading through an introductory Loop Quantum Gravity paper, I am given an induced Riemannian (space) metric:

\begin{equation}
q_{AB} = e^a_A e^b_B \delta_{ab}
\end{equation}

where $A = 1,2,3$ are covariant indices and $a = 1,2,3$ are internal indices of the triads $e^a_A$. The densitized triad is defined to be:

\begin{equation}
E^A_a := \frac{1}{2} \epsilon^{ABC}_{abc} e^b_B e^c_C.
\end{equation}

How do I determine the inverse metric $q^{AB}$ in terms of $E^A_a$? Must I guess the form and require that:

\begin{equation}
q^{AB}q_{BC} = \delta^A_C \ \ ?
\end{equation}

[jziprick]

Sorry about messing up the teX. Here it is again:

Reading through an introductory Loop Quantum Gravity paper, I am given an induced Riemannian (space) metric:


q_{AB} = e^a_A e^b_B \delta_{ab}


where A = 1,2,3 are covariant indices and a = 1,2,3 are internal indices of the triads e^a_A. The densitized triad is defined to be:


E^A_a := \frac{1}{2} \epsilon^{ABC}_{abc} e^b_B e^c_C.


How do I determine the inverse metric q^{AB} in terms of E^A_a? Must I guess the form and require that:


q^{AB}q_{BC} = \delta^A_C \ \ ?

[jziprick]

I figured it out, so I should post the solution. It becomes simple once you show that:


E^A_a = \sqrt{det(q)}e^A_a


where e^A_a is the inverse of e^a_A. Then clearly we have:


E^A_aE^B_b \delta^{ab} = det(q) q^{AB}


using the definition given in the original post for q_{AB}.