Relating inverse metric to densitized triads

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The discussion focuses on deriving the inverse metric \( q^{AB} \) in terms of the densitized triad \( E^A_a \) within the context of Loop Quantum Gravity. The induced Riemannian metric is defined as \( q_{AB} = e^a_A e^b_B \delta_{ab} \). The solution reveals that \( E^A_a = \sqrt{\text{det}(q)} e^A_a \), leading to the relationship \( E^A_a E^B_b \delta^{ab} = \text{det}(q) q^{AB} \). This establishes a clear connection between the densitized triad and the inverse metric.

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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}
 
Last edited:
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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:

<br /> q_{AB} = e^a_A e^b_B \delta_{ab}<br />

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:

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

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

<br /> q^{AB}q_{BC} = \delta^A_C \ \ ?<br />
 
I figured it out, so I should post the solution. It becomes simple once you show that:

<br /> E^A_a = \sqrt{det(q)}e^A_a<br />

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

<br /> E^A_aE^B_b \delta^{ab} = det(q) q^{AB}<br />

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

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