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Relating inverse metric to densitized triads

  1. Oct 13, 2009 #1
    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: Oct 13, 2009
  2. jcsd
  3. Oct 18, 2009 #2
    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:

    [tex]
    q_{AB} = e^a_A e^b_B \delta_{ab}
    [/tex]

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

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

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

    [tex]
    q^{AB}q_{BC} = \delta^A_C \ \ ?
    [/tex]
     
  4. Oct 19, 2009 #3
    I figured it out, so I should post the solution. It becomes simple once you show that:

    [tex]
    E^A_a = \sqrt{det(q)}e^A_a
    [/tex]

    where [tex]e^A_a[/tex] is the inverse of [tex]e^a_A[/tex]. Then clearly we have:

    [tex]
    E^A_aE^B_b \delta^{ab} = det(q) q^{AB}
    [/tex]

    using the definition given in the original post for [tex]q_{AB}[/tex].
     
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