vierbeins and spin connections as gauge fields of gravity


by jinbaw
Tags: connections, fields, gauge, gravity, spin, vierbeins
jinbaw
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#1
Jan13-11, 11:28 AM
P: 65
I was reading in a paper for Chamseddine the following:

"In the past many attempts were made to construct
gravity as a gauge theory of the Lorentz or Poincar6
groups in four dimensions . It later became clear
that if both the vierbein and the spin connection are
to be viewed as gauge fields, the Einstein action is
then only invariant under a constrained symmetry
where the torsion is set to zero."

I know that the first attempts to construct such a gauge theory were made by Utiyama and Kibble, but how does viewing the vierbein and spin connection as gauge fields lead to the idea that torsion is zero? Can someone give me some simple explanation for that? or may be advice me some reference to read? Thanks in advance.
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Mentz114
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#2
Jan13-11, 05:03 PM
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In the past many attempts were made to construct gravity as a gauge theory of the Lorentz or Poincar6groups in four dimensions . It later became clear that if both the vierbein and the spin connection are to be viewed as gauge fields, the Einstein action is
then only invariant under a constrained symmetry where the torsion is set to zero.
What that means is not clear to me either. Is he saying that if torsion is zero then 'the Einstein action is then only invariant ... ' ? Because I think that is true.

This paper is worth a look, although they don't address the torsion/no-torsion issue in great detail.

arXiv:gr-qc/9602013, "On the Gauge Aspects of Gravity", Gronwald & Hehl.
dextercioby
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#3
Jan13-11, 05:41 PM
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I don't have an expertise on this, but doesn't the normal (Palatini or Hilbert) action integral for gravity have the simple form it has only because the torsion is set to zero by hypothesis ? To me the answer is yes, so what the quoted author's saying is kinda trivial. Even in the Palatini formalism in the vierbein-spin-connection variant, the torsion is still zero, in the absence of a consistent coupling to some half-integral spinorial field.

Mentz114
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#4
Jan13-11, 06:38 PM
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vierbeins and spin connections as gauge fields of gravity


Quote Quote by bigubau
...only because the torsion is set to zero by hypothesis ?
If there's a metric, the connections are the Christoffel symbols,

[tex]
{\Gamma^m}_{ab} = \frac{1}{2}g^{mk}(g_{ak,b}+g_{bk,a}-g_{ab,k})
[/tex]

which are symmetric in a,b if the metric is symmetric. Does this ensure there's no torsion ?
jinbaw
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#5
Jan14-11, 12:46 AM
P: 65
Hi Mentz114

The Christoffel symbols take this form

[tex]

{\Gamma^m}_{ab} = \frac{1}{2}g^{mk}(g_{ak,b}+g_{bk,a}-g_{ab,k})

[/tex]

only when there is no torsion in the manifold, but i don't think this has anything to do with the existence or non-existence of a metric. I mean even when there is no torsion, these symbols could be written in terms of metric components and maybe other christoffel symbols, of course they won't be in the same form and maybe not even have in an explicit form.

Am I right?
dextercioby
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#6
Jan14-11, 02:40 AM
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To keep things rigorous, the Christoffel symbols always have that form and are noted with brackets, that's their definition, irrespective of curvature or torsion. However, the components of a general affine connection, which are always noted by [itex] \Gamma^{m}_{~ab} [/itex], contain other terms depending on the contortion tensor and the nonmetricity tensor.

For normal GR, there's no torsion, so the connection components are equal to the Christoffel symbols.
haushofer
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#7
Jan14-11, 02:41 AM
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Quote Quote by jinbaw View Post
I was reading in a paper for Chamseddine the following:

"In the past many attempts were made to construct
gravity as a gauge theory of the Lorentz or Poincar6
groups in four dimensions . It later became clear
that if both the vierbein and the spin connection are
to be viewed as gauge fields, the Einstein action is
then only invariant under a constrained symmetry
where the torsion is set to zero."

I know that the first attempts to construct such a gauge theory were made by Utiyama and Kibble, but how does viewing the vierbein and spin connection as gauge fields lead to the idea that torsion is zero? Can someone give me some simple explanation for that? or may be advice me some reference to read? Thanks in advance.
Gauging the Poincare group gives you the following:

* A vielbein and a spin connection, which are independent
* As gauge you have local translations, local rotations and general coordinate transformations

But you want the spin connection to become dependent (you don't want those extra degrees of freedom), and you want to get rid of the "local translations", which are not globally defined on a curved manifold.

The solution to both problems is to put the torsion (or, the curvature of translations!) to zero, R(P)=0. Only then you can write down the Hilbert action, which is invariant under local rotations and general coordinate transformations, and obtain Einstein's theory.
jinbaw
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#8
Jan14-11, 03:50 AM
P: 65
Quote Quote by haushofer View Post
Gauging the Poincare group gives you the following:

* A vielbein and a spin connection, which are independent
* As gauge you have local translations, local rotations and general coordinate transformations

But you want the spin connection to become dependent (you don't want those extra degrees of freedom), and you want to get rid of the "local translations", which are not globally defined on a curved manifold.

The solution to both problems is to put the torsion (or, the curvature of translations!) to zero, R(P)=0. Only then you can write down the Hilbert action, which is invariant under local rotations and general coordinate transformations, and obtain Einstein's theory.
I do understand that setting the torsion to zero will allow us to write the spin connection in terms of the veirbeins which makes them dependent, but how does this let us get rid of local translations?
haushofer
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#9
Jan14-11, 06:26 AM
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Quote Quote by jinbaw View Post
I do understand that setting the torsion to zero will allow us to write the spin connection in terms of the veirbeins which makes them dependent, but how does this let us get rid of local translations?
This depends heavily on a relation between gauge transformations, curvatures and general coordinate transformations, which goes for every closed Lie algebra which is gauged:

[tex]
\delta_{gct}(\xi^{\lambda})B_{\mu}^a = \xi^{\lambda}R_{\lambda\mu}^a(B) + \sum_{\{b\}}\delta(\xi^{\lambda}B_{\lambda}^{b})B_{\mu}^a
[/tex]

Here, B is a gauge field, and the sum b is over all the gauge transformations on the field B, with the specific gauge parameter mentioned. If you're not convinced, plug in the explicit definitions of curvatures and gauge transformations for a general algebra,


[tex]
\delta(\epsilon)B_{\mu}^a = \partial_{\mu}\epsilon^a - \epsilon^b B_{\mu}^c f_{bc}^a
[/tex]

and

[tex]
R_{\mu\nu}^a (B) = 2\partial_{[\mu}B_{\nu]}^a + B_{\mu}^b B_{\nu}^c f_{bc}^a
[/tex]

Now, apply this to the Poincare algebra.This gives

[tex]
\xi^{\lambda}\partial_{\lambda}e_{\mu}^a + \partial_{\mu}\xi^{\lambda}e_{\lambda}^a & = \xi^{\lambda}R_{\lambda\mu}^a (P) + \delta_{P}(\xi^{\lambda}e_{\lambda}^b)e_{\mu}^a + \delta_M(\xi^{\lambda}\omega_{\lambda}^{ef})e_{\mu}^a
[/tex]
and

[tex]
\xi^{\lambda}\partial_{\lambda}\omega_{\mu}^{ab} + \partial_{\mu}\xi^{\lambda}\omega_{\lambda}^{ab} & = \xi^{\lambda}R_{\lambda\mu}^{ab} (M) + \delta_{P}(\xi^{\lambda}e_{\lambda}^b)\omega_{\mu}^{ab} + \delta_M(\xi^{\lambda}\omega_{\lambda}^{ef})\omega_{\mu}^{ab}

[/tex]

Now, putting R(P)=0 does two things for you:
[*] You can rewrite a P-transformation on the vielbein as a gct plus a local Lorentz transformation; so it can be eliminated, and
[*] The spin connection becomes dependent

The first step is achieved by choosing a very specific vector xi for your general coordinate transformations, namely

[tex]
\xi^{\lambda} = e^{\lambda}{}_a \zeta^a
[/tex]

where zeta parametrizes the local P transformation.

Ofcourse, now the spin connection depends on the vielbein, potentially you could ruin its transformation properties! However, you can directly check (but this is quite tedious!) that the dependent spin connection still transforms in the same way under local Lorentz transformations as before; a hint of this is given by the fact that according to the algebra, the R(P)=0 constraint is invariant under Local Lorentz transformations.

But, the R(P)=0 constraint is NOT invariant under local P transformations! The dependent spin connection now suddenly does transform under P-transformations according to the algebra, and in a very nasty way. However, this poses no problem; the P-transformation can be "removed" from the vielbein (it is a combination of a gct and a local Lorentz transformation), so you don't care about this.
jinbaw
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#10
Jan15-11, 03:33 PM
P: 65
Thanks haushoffer! :) that was of good help!
haushofer
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#11
Jan17-11, 05:04 AM
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P: 869
Glad I could help. It's important, because in a lot of SUGRA texts people motivate SUGRA by stating that gauging the SUSY-transformations, one obtains local translations, which seem to indicate that a theory with local SUSY should also somehow incorporate diffeomorphisms as symmetry transformations, hence gravity.

But of course, diffeomorphisms are not local translations. You need to impose constraints in order to do so, which is non-trivial.


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