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

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.