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Tetrads in General Relativity

II. Tetrad and Vierbein fields

Orthonormality Relations

   The Greek vierbein (or ‘curved’) indices can be raised or lowered by the metric, i.e. ${g^{\mu \nu }}$ or ${g_{\mu \nu }}$ , respectively. In contrast, the Latin (or ‘Lorentzian’ or ‘flat’) vierbein indices refer to the tangent space; they can be manipulated using ${\eta ^{ab}}$ or ${\eta _{ab}}$. For example:

2.10

\[{e^{\mu a}}: = {g^{\mu \nu }}e_\nu ^{\;{\kern 1pt} a}\qquad {e_{\nu a}}: = {\eta _{ab}}e_\nu ^{\;{\kern 1pt} b}\qquad {e^\mu }_a : = {\eta _{ab}}{g^{\mu \nu }}{e_\nu }^b \]

The vierbein field itself can be similarly manipulated: ${e^\nu }_a = {e^\mu }_a{e^\nu }_\mu $, since ${e^\nu }_\mu = \delta _\mu ^\nu $. [Wikipedia: Spin Connection]

   By appropriately raising and lowering indices and using (1.10), one can demonstrate that the vierbein fields defined by (2.6,7) satisfy the orthonormality relations

2.11

\[  {e^\mu }_a(x){e_\nu }^a(x) = \delta _\nu ^\mu {\quad}{e_\mu }^a(x){e^\mu }_b(x) = \delta _b^a  \]

confirming that ${e^\mu }_a$ is indeed the inverse of the vierbein ${e_\mu }^a$, and visa versa.

   It is noteworthy that the vierbeins serve double purposes:

  1. the vierbein serves as components of the coordinate basis vectors in terms of the orthonormal tetrad (2.6), and as components of the reciprocal tetrad in relation to the coordinate reciprocal basis.
  2. the inverse vierbein acts as the components of the orthonormal tetrad basis vectors in terms of the coordinate basis (2.7), and as components of the reciprocal coordinate basis in relation to the reciprocal tetrad basis;

   Formally, a linear operator can be assigned to map the coordinate frame into the orthonormal frame, expressed as:

2.12

\[ {{\mathbf{e}}_a} = {e^\mu }_a{{\mathbf{g}}_\mu }: = h({{\mathbf{g}}_\mu }) {\quad}{{\mathbf{g}}_\mu } = {e_\mu }^a{{\mathbf{e}}_a}: = {h^{ - 1}}({{\mathbf{e}}_a}){\text{ }}  \]

This operator represents the vierbein field. The assumption of invertibility of the map ensures that the adjoint map ${{\mathbf{e}}^a} = \bar h({{\mathbf{g}}^\mu })$ is also invertible.