# Is the tetrad formalism in GR the same as using push-forwards and pull-backs in differential geometry?

I get that this is a math question; however I find the tetrad/verbein formulation of things to be much more known among physicists.

Looking at the metric tensor components $$g_{\mu\nu}$$ in terms of tetrads we have then:

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

Where here $$\eta_{ab}$$ are components of the Minkowski metric.

We can compare that with the induced metric $$g_{\mu\nu}$$ of some curved space embedded in another space as:

$$g_{\mu\nu}=\frac{\partial X^{a}}{\partial\phi^{\mu}}\frac{\partial X^{b}}{\partial\phi^{\nu}}g_{ab}$$

where $$g_{ab}$$ is the metric of the embedding space. this equation is referred to as the pullback by a diffeomorphism.

Or for another example we can take a vector on our coordinate basis (Greek indices) and write its components in terms of vector components in our Lorentzian basis (latin indices):

$$v^{\mu}=e_{a}^{\mu}v^{a}$$

The inverse tetrad then seems to be playing the part of a pushforward. Is that correct?

I'm mostly interested in writing the metric of a spacetime and associated quantities in terms of the embedding space ( say $$S^{3}\times\mathbb{R}\subset\mathbb{R}^{4,1}$$ for example) Can I then import the tools of the pushforward-pullback yet still utilize the tetrad formalism in GR (just using more general metrics than the Minkowskian)?

Note that because the embedding space is higher dimensional, the tetrads would no longer be square when viewed as matrices. I would very much appreciate input or books utilizing this approach.

Note: This question is closely related but much more general than another question here

Also similar but again, more general than a question here.

EDIT:

In response to an answer below I just wanted to be really specific. Looking again at the component form of the pullback operation:

$$g_{\mu\nu}=\frac{\partial\phi^{i}}{\partial x^{\mu}}\frac{\partial\phi^{j}}{\partial x^{\nu}}g_{ij}$$

where Greek indices indicate one space and latin the other. If we specify then that we're operating our pullback on Minkowski space we get:

$$g_{\mu\nu}=\frac{\partial\phi^{i}}{\partial x^{\mu}}\frac{\partial\phi^{j}}{\partial x^{\nu}}\eta_{ij}$$

Comparing this with the tetrad formulation we get:

$$g_{\mu\nu}=\frac{\partial\phi^{i}}{\partial x^{\mu}}\frac{\partial\phi^{j}}{\partial x^{\nu}}\eta_{ij}=e_{\mu}^{i}e_{\nu}^{j}\eta_{ij}$$

Which makes me think the tetrad formulation is just a particular choice of pullback. For different spacetime topologies it may make sense to take the pullback to some other manifold to take advantage of symmetries of the space.

I ask yet another, similarly themed question here.

• Not really, the concepts are different. The only similarity is that in both cases you use an invertible linear transformation. But the key point is that the tetrad does not have to be related to a coordinate system. Dec 19, 2020 at 4:16
• @Javier, I edited my question a bit, but the last section of the wiki article on the tetrad formalism seems to confirm my thoughts en.wikipedia.org/wiki/Tetrad_formalism Dec 20, 2020 at 9:19
• RE: EDIT II : Apologies to all for any messiness therein, I had to do some hasty copy, pasting and editing to make it applicable here from a project of mine. Dec 20, 2020 at 21:52

The tetrad is not a particular choice of pullback - if anything, it's the other way around. In general, given an arbitrary tetrad $$e^a{}_\mu$$ you cannot write

$$e^a{}_\mu = \frac{\partial \phi^a}{\partial x^\mu};$$

there's no function $$\phi$$ satisfying this. If there is, then yes, your tetrad vectors are the coordinate basis for an embedding, but in general that won't be the case. Read on for a more detailed explanation.

An embedding (or in general, a coordinate chart) defines a basis $$\{\mathbf{e}_\mu\}$$ at each point. If you have a coordinate chart and an embedding, the relation between the bases given by the two is

$$\mathbf{e}_\mu = E^a{}_\mu \mathbf{e}_a$$

$$g_{\mu\nu} = E^a{}_\mu E^b{}_\nu g_{ab},$$

where

$$E^a{}_\mu = \frac{\partial \phi^a}{\partial x^\mu}.$$

This has certain consequences: commutativity of partial derivatives implies that we must have $$\partial_\mu E^a{}_\nu = \partial_\nu E^a{}_\mu$$. In more abstract terms, the vectors of a coordinate basis Lie commute with each other.

Using a tetrad means replacing the $$E^a{}_\mu$$ in the above equations with an arbitrary (smooth, technically) matrix at each spacetime point (and replacing $$g_{ab}$$ with $$\eta_{ab}$$). And this is a much larger space of functions. You will no longer have Lie commutativity in general: there will be $$\phi$$ such that its corresponding $$E^a{}_\mu$$ is your tetrad $$e^a{}_\mu$$. You have chosen a basis at each point, but these bases don't fit together to make a coordinate system (or embedding, or diffeomorphism - it's all the same), unless all the Lie derivatives are zero.

The formulas are similar only in that they are a change of basis for the metric, but the tetrad encompasses a much larger class of bases, so it's definitely not the same as a pushforward.

• +1 Of course you are right for the partial derivative, If you can address my last question edit though I'll accept your answer. The last edit is sort of why I asked this question in the first place. Dec 20, 2020 at 20:35
• I suppose its more of a generalized "covariantized" pullback that I am trying to identify with the tetrads Dec 20, 2020 at 21:26
• @R.Rankin It doesn't matter if the target manifold has some additional group structure, if the you get a basis from pulling back a diffeomorphism, that's a coordinate basis, and it Lie commutes. I still don't think it has anything to do with a tetrad, but that's as far as I think I can answer anyway - maybe your second edit should be a separate question, written a bit more clearly (that much math is unnecessary IMO) and emphasizing that you're asking about a Lie group, not just any manifold. Dec 22, 2020 at 13:25
• $$g_{\mu\nu}=\left(D_{\mu}\phi^{j}\right)g_{ij}\left(D_{\nu}\phi^{i}\right)$$ with: $D_{\mu}=\partial_{\mu}+A_{\mu}$ It is important to note the connection takes values in the group preserving the quadratic form on the \mathit{embedding} space (Latin indices), not our spacetime itself. For the pullback to Minkowski space, then we get: $$g_{\mu\nu}=\left(D_{\mu}\phi^{j}\right)\eta_{ij}\left(D_{\nu}\phi^{i}\right)$$ with: $D_{\mu}=\partial_{\mu}+A_{\mu}$ With $A\in SO(3,1)$ Then clearly Latin basis (both vector and 1-form) CANNOT Lie commute Dec 26, 2020 at 1:03
• @R.Rankin To answer properly, I'd have to go review some of the theory of Lie groups and algebras, invariant vector fields, and so on. I cannot promise to do that, so I can't really promise a more complete answer either. But I'm pretty sure the key lies in the distinction between the group elements and tangent vectors. A coordinate basis in a manifold always Lie commutes (in the sense of the Lie derivative); if you have something that doesn't commute (like group elements), it must be another operation. Dec 26, 2020 at 19:58

The tetrad formalism is not related to the pullback operation. The similarity you point out has more to do with the statement that both $$g_{\mu\nu}=\frac{\partial X^a}{\partial x^\mu}\frac{\partial X^b}{\partial x^\nu}g_{ab}$$ and $$g_{\mu\nu}=e^a_\mu e^b_\nu \eta_{ab}$$ transform covariantly under diffeomorphisms (which is a pullback of the spacetime to itself via some diffeomorphism, if you prefer that language).

Let me also point out that the purpose of the tetrad formalism is to separate the local Lorentz group from the action of diffeomorphisms on the fields of the theory. This is particularly important in the case of fermions (fields transforming under a spinnor representation of the Lorentz group). Since they transform as a half-integer representation, we need the tetrads to couple them covariantly to other fields.

The tetrads also have the nice advantage of making gravity more like gauge theories based on internal symmetries as it essentially turns the local Lorentz symmetry into something which acts on "internal" indices (separated from diffeomorphisms), though it's still not a Yang-Mills theory.

Necessarily then, the machinery of principle bundles come along with tetrads, something which is not seen by considering pullbacks.

EDIT: Let me also point out in response to the edit to the question some additional points. The tetrad formalism demands (in all instances I'm aware of) that the internal indices $$a,b,\ldots$$ run over a range the same dimension as the spacetime. If it were the case that the tetrads were nothing more than the pullback from some embedding space, it would need to be the case that this embedding space is the same dimension as the manifold in question and hence is diffeomorphic to it. Of course, this is not generically the case as many manifolds will have a non-trivial lower bound on the dimension of spaces they can be embedded in.

I would then also point out that because the tetrads are defined only to be sections of some principle bundle, they need not be globally defined and hence cannot come from a globally defined diffeomorphism, as required to make a (globally defined) pullback.

If you would like to consider only locally defined pullbacks, I would argue that you will find essentially every object to be described by stuff that looks like "locally defined pullbacks." But this is actually kind of trivial because every manifold is, by definition locally isomorphic to some part of $$\mathbb{R}^n$$ (or Minkowski space if that's what your signature dictates). With this in mind, all coordinate expressions will look and feel like "pullbacks" precisely because they are in this definitional sense.

Tetrads, however, are well-defined as coordinate-independent objects.

• +1 I am definitely interested in globally defined pullbacks/mappings. instead of an embedding space, I could say map a space to it's homogenous isotropic equivalent, then topologically flat space would go over to minkowskian. and something like a closed-cyclic FLRW universe would map to $S^{3}XS^{1}=SU(2)XU(1)$ Dec 21, 2020 at 18:32
Echoing @javier's observation, to put it in the context of differential forms, if we regard the components of the pullback operation $$E^i_{\mu}=\frac{\partial\phi^{i}}{\partial x^{\mu}}$$ as components of 1-form $$E^i$$, then $$E^i$$ is exact $$E^i = d\phi^i$$ therefore, $$E^i$$ is closed $$dE^i=0$$ because $$d^2 = 0$$
On the other hand, for the tetrad/verbein 1-form $$e^i$$, in general it's not closed $$de^i \neq 0$$