Variation of the vielbein postulate

Question

The vielbein postulate is given by

$$\partial_\mu e_\nu{}^a + \omega_\mu{}^a{}_b e_\nu{}^b - \Gamma_{\mu\nu}^\rho e_\rho^a = 0 $$

whose anti-symmetric part, assuming the connection is a Levi-Civita connection, reads

$$2\partial_{[\mu} e_{\nu]} - 2\omega_{[\mu}{}^{ab}{} e_{\nu ]b} = 0 $$

The solution of this equation gives rise to an equation for the spin-connection $\omega_\mu{}^{ab}$ in terms of $e_\mu{}^{a}$.

As we use the "vielbein postulate", or the antisymmetric part of the vielbein postulate, we must be careful with the variation of this postulate. The variation of the postulate must also vanish otherwise it would yield a constraint. The transformation rules for the vielbein and the spin connection are

$$ \begin{align} \delta e_\mu{}^a &= \partial_\mu \xi^a + \omega_\mu{}^{ab} \xi_b - \lambda^{ab} e_{\mu b}\\ \delta \omega_\mu{}^{ab} &= \partial_\mu \lambda^{ab} + \omega_{\mu c}{}^{[a} \lambda^{b]c} \,. \end{align} $$

where $\xi_a$ and $\lambda_{ab}$ are the transformation parameters for boosts and Lorentz transformations. The transformation of the antisymmetrized vielbein postulate then reads

$$\xi_\nu R_{\rho\sigma}{}^{\mu\nu} = 0$$

provided that the vielbein postulate is satisfied. Here $R_{\mu\nu\rho\sigma}$ is the Riemann tensor. I have never seen such an identity. My question is: Has anyone ever seen something like this? Or am I missing something?

Answer 1

There is a lot of nonsense written about "vierbein postulate". This "thing" is not a postulate because a postulate can be accepted or rejected, but this formula is always true. It is simply the statement that the covariant derivative of the identity operator is zero. As such, is it is independent of any assumption of metricity or of torsion-freeness. It is simply a property of any connection on the tanget space.

Consider the vector-valued one form $$ {\bf E}= e^{*a}_\mu\, {\bf e}_a \otimes dx^\mu= {\bf e}_a\otimes {\bf e}^{*a} $$ which provides a map ${\bf E}:{\rm TM}\to {\rm TM}$. If ${\bf x}= x^b {\bf e}_b$ we have $$ {\bf E}({\bf x})= {\bf e}_a \otimes {\bf e}^{*a}({\bf x}) = {\bf e}_a x^a={\bf x}. $$ Thus ${\bf E}$, with components $e^{*a}_\mu$ is the identity map. In effect, however, ${\bf E}$ links two different interpretations of ``${\bf x}$'': the input ${\bf x}$ is a small displacement, starting at $p$, on the curved manifold M, while the output vector ${\bf x}$ is an element of the flat vector space that forms the fibre of TM at $p$.

Using the definitions of the connection in the vierbein frame and in the coordinate frame $$ \nabla_\mu {\bf e}_a = {\bf e}_b \,{\omega^{b}}_{a\mu}, \quad \nabla_\mu dx^\nu = -{ \Gamma^\nu}_{\lambda \mu}\, dx^\lambda $$ to ${\bf E} =e^{*a}_\nu\, {\bf e}_a \otimes dx^\nu$ we use Leibniz rule to compute
$$ \nabla_{\mu} {\bf E}=( \nabla_\mu e^{*a}_\nu) {\bf e}_a \otimes dx^\nu+e^{*a}_\nu(\nabla_\mu {\bf e}_a) \otimes dx^\nu+e^{*a}_\nu\, {\bf e}_a \otimes (\nabla_\mu dx^\nu)\\ = (\partial_\mu e^{*a}_\nu+ {\omega^a}_{b\mu} e^{*b}_{\nu}- {\Gamma^\lambda}_{\nu\mu}e^{*a}_\lambda)\, {\bf e}_a \otimes dx^\nu \nonumber $$ As ${\bf E}$ is an identity map, the property $\nabla_{\mu} {\bf E}=0$ is mandated by consistency, and we see that this condition is equivalent to one version of the ``tetrad postulate.''

Note that in using Liebniz we must remember that the array of numbers $e^{*a}_\nu$ are functions and so their derivative is an ordinary partial derivative.

Similarly, If we expand ${\bf E}$ as $$ {\bf E} =e_a^\nu \partial_\nu \otimes {\bf e}^{*a} ) $$ we get $$ \nabla_\mu {\bf E}= (\partial_\mu e_a^\nu- {e}_a^\nu {\omega^a}_{b\mu} + e_{a}^\lambda{\Gamma^\nu}_{\lambda\mu})( { \partial_\nu} \otimes {\bf e}^{*a}), $$ or $$ \partial_\mu e_a^\nu- {e}_a^\nu {\omega^a}_{b\mu} +e_{a}^\lambda {\Gamma^\nu}_{\lambda\mu}=0, $$ which is the second version of the ``tetrad postulate.''

As long as one remembers that the this vierbein "thing" is an identity, and that there is no more difference between ${\omega^{b}}_{a\mu}$ and ${ \Gamma^\nu}_{\lambda \mu}$ than there is between two sets of ${\Gamma^\nu}_{\lambda \mu}$'s in different coordinate systems, then you will not get confused.

Answer 2

If I understand your notation correctly, your vielbein postulate is simply an expression of the metric compatability of the connection in that the connection on the frame bundle is compatible with the connection on the bundle of orthonormal frames.

By this I mean that if $e^a = e_\nu{}^adx^\nu$ labels co-tetrad one-forms such that $e_\nu{}^a$ are its component functions in the holonomic (coordinate) frame, that $\Gamma^\rho_{\mu\nu}$ are the holonomic connection coefficients, and that $\omega^a{}_b = \omega_\mu{}^a{}_bdx^\mu$ labels the connection forms of the tetrad, then $$\nabla e^a = -\omega_\mu{}^a{}_bdx^\mu \otimes e^b = -\omega_\mu{}^a{}_b e_\nu{}^bdx^\mu \otimes dx^\nu, \tag{1}$$ by definition, but in the holonomic frame: $$ \nabla e^a = \nabla \left(e_\nu{}^a dx^\nu\right) = \left(\partial_\mu e_\nu{}^a - \Gamma^\rho_{\mu\nu}e_\rho{}^a \right)dx^\mu \otimes dx^\nu, \tag{2} $$ and since the connections are compatible we must have $(1) = (2)$. This is not a postulate, but an identity, and yields exactly your "vielbein postulate."

In the antisymmetric expression you seem to have (erroneously) changed sign on the term belonging to the connection on orthonormal frames, but otherwise it is an expression of the same compatability, which is commonly expressed as the first Cartan equation: $$ de^a = e^b \wedge \omega^a{}_b. $$

Both expressions can thus be seen to be identities for metric compatible connections. I also think it is unwise to label $\omega_\mu{}^{ab}$ a "spin" connection, as it is a connection on the bundle of orthonormal frames (compatible with that of the frame bundle), not on the bundle of spin frames.

From this it is clear that the identity $$\xi_\nu R_{\rho\sigma}{}^{\mu\nu} = 0 \tag{0}$$ must be wrong, since the "vielbein postulate" adds nothing new, and $(0)$ is not true in general.

This paper goes through significant effort to explain the details of the relation between the connections in relation to the "vielbein postulate," and is published in International Journal of Modern Physics D, so should count as reputable, but might be a bit snarky at times.

EDIT: Using the transformation laws given in the OP, but with $$ \delta\omega_\mu{}^{ab} = \partial_\mu\lambda^{ab} + 2 \omega_{\mu c}{}^{[a}\lambda^{b]c}, $$ I too end up with $(0)$. Without the additional factor of 2 on the latter term, the result is different. However, without knowing where the transformation laws come from, it's difficult to say more. It might be worth nothing is that $(0)$ is equivalent to $$ \xi^\mu{}_{;[\rho\sigma]} = 0, \tag{3} $$ by the Ricci identity. As before, if $\xi^\mu$ is an arbitrary Killing vector, then this must be wrong.