12
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ Since you are assuming the 'vielbein postulate', you can simply determine $\omega$ as a function of $e$, $\omega(e)$ and substitute it everywhere in the action. Then vary w.r.t. $e$ only. $\endgroup$
    – Prahar
    Feb 6, 2017 at 17:59
  • $\begingroup$ Sure, but "assuming" the vielbein postulate should not come at a price. But varying the vielbein postulate tells me that if I assume the postulate, then the contraction of the Riemann tensor with a killing vector should vanish, which is something I have never heard of. $\endgroup$
    – John Doe
    Feb 6, 2017 at 21:34
  • 1
    $\begingroup$ I guess these variation formulas of $e$ and $\omega$ came from Poincare gauge theory of gravity, right? $\endgroup$ Feb 18, 2017 at 12:36
  • $\begingroup$ Yes, that's exactly true. The transformation rules can be found by using the [P,M] and [M,M] commutators. $\endgroup$
    – John Doe
    Feb 18, 2017 at 19:45
  • $\begingroup$ I realize I'm several years too late and no one will probably read this comment. But, conceptually, it seems like you are taking an equation that transforms as a tensor under both local Lorentz transformations and diffeomorphisms, applying infintesimal local Lorentz an diffeomorphisms, and finding the equation is not invariant. This can't be right. One thing that jumps out is that if $\xi$ is the parameter of an infintesimal diff, it should have a $\mu$ index, not an $a$ index. $\endgroup$
    – Andrew
    Aug 22, 2021 at 0:19

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I know that it "must be" wrong! But for the postulate to add nothing new, the postulate should transform to itself, so I can safely set it to zero. My question is why? Why do I get that when I vary the "vielbein postulate"? What could possibly be wrong with this simple calculation? The reason why I also provided the transformation rules is that so someone can tell me what I do wrong in the transformation of the postulate. If you simply vary it with the provided transformation rules, you'll see that $\xi_{\nu} R^{\mu\nu\rho\sigma}= 0$ must be satisfied for the vielbein postulate to be valid $\endgroup$
    – John Doe
    Feb 13, 2017 at 10:50
  • $\begingroup$ @John I edited to explain that given the transformation laws I too get your result (provided the caveat outlined in the edit is fulfilled). Do you want to help me understand where they come from? Is there some restriction on the Killing vector $\xi_a$ (I have concluded that it is a Killing vector based on your comment to the OP)? $\endgroup$ Feb 13, 2017 at 23:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.