# Vielbeins as gauge fields of local translations

## Main Question or Discussion Point

Hi,

I have a question about gravity.

I think most of you know that we can obtain Einstein gravity by gauging the Poincaré algebra and imposing constraints. The Poincaré algebra consists of {P,M}. P describes translations, and M describes Lorentz rotations.

Gauging M gives us the so-called spin connection. The gauge field of the local translations is often taken (e.g. in a lot of supergravity texts) to be the vielbein e,

$$\eta_{ab} e_{\mu}{}^a e_{\nu}{}^b = g_{\mu\nu}$$

What is the precise reason that this identification is justified? What do these "local translations" (I regard them as abstract internal transformations a la Yang-Mills) precisely have to do with the metric? Apart from the index structure I'm not really sure why this choice is justified.

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Hi,

I have a question about gravity.

I think most of you know that we can obtain Einstein gravity by gauging the Poincaré algebra and imposing constraints. The Poincaré algebra consists of {P,M}. P describes translations, and M describes Lorentz rotations.

Gauging M gives us the so-called spin connection. The gauge field of the local translations is often taken (e.g. in a lot of supergravity texts) to be the vielbein e,

$$\eta_{ab} e_{\mu}{}^a e_{\nu}{}^b = g_{\mu\nu}$$

What is the precise reason that this identification is justified? What do these "local translations" (I regard them as abstract internal transformations a la Yang-Mills) precisely have to do with the metric? Apart from the index structure I'm not really sure why this choice is justified.
The local translations are eliminated in this analysis since tangent space has a preferred origin. You must ask yourself, what do you translate? It is merely an analogy because the degrees of freedom match ...

Careful

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qsa
http://en.wikipedia.org/wiki/Cartan_formalism_(physics [Broken])

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That last article is interesting, but a quick glance (I'll read it carefully soon) still doesn't answer my question.

I know the local translations are removed by a curvature constraint, but I still don't see why the vielbein is a proper gauge field for these "local translations", even though they are "replaced" by diffeomorphisms later on.

I know the local translations are removed by a curvature constraint, but I still don't see why the vielbein is a proper gauge field for these "local translations", even though they are "replaced" by diffeomorphisms later on.
I think I answered your question. Try to construct the local action of your translations: there is none. The vielbein only feels coordinate and Lorentz transformations.

I'm sorry, but I still don't see your point. Could you be more elaborate on that? :)

I'm sorry, but I still don't see your point. Could you be more elaborate on that? :)
Usually, if we speak about about generators of a symmetry (such as translation symmetry) we do one of the following:
(a) we give a geometrical action of the symmetry in terms of the spacetime coordinates or fields
(b) we calculate (in classical physics) the Poisson bracket algebra of the generators of the symmetry group
Usually, both are equivalent. Now, you say you have a local representation of the Poincare group; I told you that the Lorentz group works on the vielbein e^a_{\mu} and spin connection in the $a$ indices. Furthermore, I asked you how the translation part of the Poincare group acts (and gave you the answer it doesn't act at all).

Careful

Usually, if we speak about about generators of a symmetry (such as translation symmetry) we do one of the following:
(a) we give a geometrical action of the symmetry in terms of the spacetime coordinates or fields
(b) we calculate (in classical physics) the Poisson bracket algebra of the generators of the symmetry group
Usually, both are equivalent. Now, you say you have a local representation of the Poincare group; I told you that the Lorentz group works on the vielbein e^a_{\mu} and spin connection in the $a$ indices. Furthermore, I asked you how the translation part of the Poincare group acts (and gave you the answer it doesn't act at all).

Careful
But that's AFTER the constraint R(P)=0. Before the constraint I can realize the local translations as

$$\delta_P(\zeta^c) e_{\mu}{}^a = \partial_{\mu}\zeta^a - \omega_{\mu}{}^{ab}\zeta^b$$

and

$$\delta_P(\zeta^c) \omega_{\mu}{}^{ab} = 0$$

according to the Poincare algebra. I would say that you should have a reason to identify the gauge field belonging to local translations as the vielbein BEFORE you start imposing curvature constraints, right?

But that's AFTER the constraint R(P)=0. Before the constraint I can realize the local translations as

$$\delta_P(\zeta^c) e_{\mu}{}^a = \partial_{\mu}\zeta^a - \omega_{\mu}{}^{ab}\zeta^b$$

and

$$\delta_P(\zeta^c) \omega_{\mu}{}^{ab} = 0$$

according to the Poincare algebra. I would say that you should have a reason to identify the gauge field belonging to local translations as the vielbein BEFORE you start imposing curvature constraints, right?
Sure, you can do that for infinitesimal translations; the difference with the local Lorentz transformations is that here the entire group is represented. In the case of the translations this is only so for the Lie algebra. That is why I said that the origin is kept fixed (or better only ''infinitesimally'' shifted ). You might now want to figure out why this implies the vielbein cannot be regarded as a gauge field; actually the action you wrote down is just given by the Lie derivative and therefore no gauge terms are developped.

Careful

Sure, you can do that for infinitesimal translations; the difference with the local Lorentz transformations is that here the entire group is represented. In the case of the translations this is only so for the Lie algebra. That is why I said that the origin is kept fixed (or better only ''infinitesimally'' shifted ).
Ah, ok, because the Lie algebra constitutes the origin of the Lie group as a manifold!

actually the action you wrote down is just given by the Lie derivative and therefore no gauge terms are developped.
What do you mean? The action of local P-translations which I wrote down, $\delta_P$, are not Lie-derivatives. A Lie derivative on the vielbein would be given by

$$\delta(\xi) e_{\mu}{}^a = \xi^{\rho}\partial_{\rho}e_{\mu}{}^a - \partial_{\mu}\xi^{\rho}e_{\rho}{}^a$$

Only after the R(P)=0 constraint can one rewrite the local P-translation as a Lie derivative on the vielbein (and a local Lorentz rotation).

What do you mean? The action of local P-translations which I wrote down, $\delta_P$, are not Lie-derivatives. A Lie derivative on the vielbein would be given by

$$\delta(\xi) e_{\mu}{}^a = \xi^{\rho}\partial_{\rho}e_{\mu}{}^a - \partial_{\mu}\xi^{\rho}e_{\rho}{}^a$$
Sure, but that is the same expression I believe. First of all you mix spacetime indices with vector indices here; if you consistently relate spacetime indices to vector indices by means of the vielbein, both expressions coincide. I think we are quibelling over something very simple; wether you prefer the vielbein and spin connection as dynamical variables or only the vielbein. The classical theories are fully equivalent, the quantum theories might not be so.

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No, after the manipulation you mention a local translation can be written as

$$\delta_P(\xi^{\lambda}} e_{\lambda}^b) e_{\mu}{}^a = \delta_{gct}(\xi^{\lambda}) e_{\mu}{}^a - \xi^{\lambda}\omega_{\lambda}^{ab}(e, \partial e)e_{\mu}^b$$

after the R(P)=0 constraint; without this constraint R(P) would appear on the right hand side and the spin connection would still be independent. So if one chooses the zeta parameter of the P-transformations as

$$\zeta^a = \xi^{\lambda} e_{\lambda}{}^a$$

one sees that this particular P-transformation is a combination of a Local LT and a gct. Not just a gct.

This is how in "gravity as a gauged Poincare algebra" the local P-translations are "removed" (they are just a combination of the transformations which already acted on the vielbein, and thus can be discarded).

No, after the manipulation you mention a local translation can be written as

$$\delta_P(\xi^{\lambda}} e_{\lambda}^b) e_{\mu}{}^a = \delta_{gct}(\xi^{\lambda}) e_{\mu}{}^a - \xi^{\lambda}\omega_{\lambda}^{ab}(e, \partial e)e_{\mu}^b$$

after the R(P)=0 constraint; without this constraint R(P) would appear on the right hand side and the spin connection would still be independent.
As I said in my previous message, I think we are quibbeling over something as simple as whether the spin connection is a variable independent of the vielbein or not. This choice of variables is completely irrelevant classically , it depends upon how one looks at the action. Let me stress this that these gauge transformations generated by the Lie algebra of translations only have meaning on-shell , that is after the constraints have been implemented. Prior to that, there is no significance to it. So my point of view (which was just inspired by the geometrical form of the action) actually also holds in the case one wishes to see the spin connection as independent variable, at least classically it is so.

I see why you were confused about the fact I said it simply is the lie derivative; actually this is also a matter of convention. The ordinary Lie derivative doesn't make sense on objects with a Lorentz index, that's why the more mathematically inclined just redefine it to include the spin connection, so that it is ok. What you do is subdivide something that makes sense in two parts which don't; that's something I don't do, I directly think in those terms which are natural.

Careful