# Relating inverse metric to densitized triads

1. Oct 13, 2009

### jziprick

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 \ \ ?

Last edited: Oct 13, 2009
2. Oct 18, 2009

### 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:

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

3. Oct 19, 2009

### 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 $$q_{AB}$$.