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 $$q_{AB}$$.