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I'm studying the formalism of gravity with torsion, the Einstein-Cartan (EC) theory, and i've encountered this book by H. Kleinert "Gauge fields in condensed matter", in which he derives the basic framework of EC theory [in part 4].

When defining the basic differential geometric quantities, he does it all in term of a tetrad basis. Naively speaking the tetrad is defined as a connection between the Minkowski space and the curved one, and in fact he uses the tetrad to define define both torsion and curvature of the affine connection (also defined in terms of tetrad) and of course the metric. Namely the curvature is defined by

$$R_{\mu \nu \lambda \rho} = e_{a \rho} ( \partial_{\mu}\partial_{\nu} - \partial_{\nu}\partial_{\mu} ) e^{a}_{\lambda}. (1)$$

Later when talking about spinors in a curved manifold he invokes the notion of a local basis field, the vierbein, $h_{\alpha \mu}$, that satisfies the same properties of the tetrad, such as

$$ g_{\mu \nu} = e_{a \mu} e^{a}_{\nu} (2.1)$$ $$ g_{\mu \nu} = h_{a \mu} h^{a}_{\nu}.(2.2)$$

Also the tetrad and the vierbein components are related via a Lorentz transformation. For me the problem arises when he says that if we want to define the curvature tensor in terms of the vierbein field, we would always get zero, that is:

$$R_{\mu \nu \lambda \rho} = h_{\alpha \rho} ( \partial_{\mu}\partial_{\nu} - \partial_{\nu}\partial_{\mu} ) h^{\alpha}_{\lambda} = 0. (3)$$

I understand that they are different objects, and in this language the tetrad is just a coordinate transformation that does not obey the integrability condition and the vierbein connects the manifold with the tangent plane at each point.

But in practice how do I tell them apart? I mean, given a metric, how do i obtain the tetrad and vierbein such that they satisfy the integrability conditions (1) and (3)?

EDIT: as suggested, here are the definitions the book gives of tetrad and local basis.
notation:
${\alpha, \beta, \gamma, \delta...}$ = indices in the tangent plane

${\mu, \nu, \rho...}$ = spacetime indices
latin letters = minkowski indices
The vierbein field is defined as a "differential coordinate transformation" $$dx^\alpha = dx^\mu h^\alpha_\mu,$$ but it is worth noting that the integrability condition (3) is part of the definition.

The tetrad field is presented as: $$e_\mu = e^a_\mu e_a = \frac{\partial x^a}{\partial x^\mu}e_a$$

where the coordinate transformation $x^a = x^a(x^\mu)$ satisfy the relation $$(\partial_\mu \partial_\nu - \partial_\nu \partial_\mu)x^a \neq 0.$$ In this sense that i said the tetrad is defined as a connection between minkowski spacetime to the curved one. The thing I'm calling vierbein are also said to be non-holonomic coordinates. The whole thing is that they obey the same properties (except for the integrability condition) and that confuses me on how to tell them apart in a practical situation.

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  • $\begingroup$ What do you mean? Tetrad and vierbein are different names for the same thing. $\endgroup$ – Qmechanic May 26 at 2:03
  • $\begingroup$ @Qmechanic that's the whole point of my question hahah. They are mostly defined as having the same components. But this book defines them as different objects (reasonable) and address different integrability conditions for them. $\endgroup$ – Marcos Vinicius May 26 at 2:11
  • $\begingroup$ That seems to be non-standard terminology. $\endgroup$ – Qmechanic May 26 at 2:37
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    $\begingroup$ This question might be clearer if you stated the author's definitions of the tetrad and the vierbein. $\endgroup$ – G. Smith May 26 at 6:15
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    $\begingroup$ There's also a third possibility - and you don't need to know German or Greek - just English - an orthogonal field frame. Or in the spirit of Cartan, a moving orthogonal field frame. $\endgroup$ – Cinaed Simson May 27 at 17:48
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In the physics literature there is a lot of notational confusion caused by failing to distinguish between geometric tensor objects and their numerical components. Let start with a manifold $M$ equiped with a local coordinate system $x^\mu$. We define define a vector field to be an object like ${\bf V}=V^\mu(x) {\boldsymbol \partial}_\mu$ where the $${\boldsymbol \partial}_\mu\equiv \frac{\partial}{\partial x^\mu} $$ are the basis vectors of the tangent space $T(m)_x$ at the point $x$. The $V^\mu$ are componts of the basis indepenedent object ${\bf V}$. Now we can introduce a set of vector fields ${\bf e}_a= e_a^\mu(x) {\boldsymbol\partial }_\mu$ and if they are linearly independent and span $T(M)_x$ at all $x$ we can use them as a basis and write ${\bf V}= V^a(x){\bf e}_a$. The $V^a$ are the components of ${\bf V}$ with respect to this frame field. In four dimensions relativists call the ${\bf e}_a$ a vierbein from the German vier=four, bein=leg. In three dimensions we would have a dreibein and an one diemsnion an einbein. In general we have a vielbein (many legs).

We can play the same game with the cotangent spaces $T^*(M)_x$ with coordinate basis ${\bf dx}^\mu$ dual to the ${\boldsymbol\partial}_\mu$ (Most people don't use boldface here, just writing $dx^\mu$ but I am doing so to distinguish tensor objects from their components). A covariant vector field ${\bf A}$(a section of $T^*(M)$) can be expandend as ${\bf A}= A_\mu(x) {\bf dx}^\mu$. If we introduce a co-frame ${\bf e}^{*a}= e^{*a}_\mu {\bf dx}^\mu$ dual to the ${\bf e}_a$ (this means that ${\bf e}^{*a}({\bf e}_b)= \delta^a_b$), then we have ${\bf A}= A_a(x) {\bf e}^{*a}$, where $A_a= e_a^\mu A_\mu$.

In casual use people call the arrys of numbers $e^\mu_a$ or the $e^{*a}_\mu$ "vierbeins". They also often don't bother to put the star on $e^{*a}_\mu$ just writing $e^a_\mu$ and trusting that the Greek and Roman letters serve to distinguish between the components of the frame and the co-frame. This is OK in many places but is $e^3_2$ a component of a frame or a coframe?

Using the co-frame one can write the metric ${\bf g}$ as either $$ {\bf g}= g_{\mu\nu}(x) {\bf dx}^\mu\otimes {\bf dx}^\nu $$ or $$ {\bf g}= \eta_{ab} {\bf e}^{*a}\otimes {\bf e}^{*n}. $$ It is common to choose the coframe so that the the $\eta_{ab}$'s become constants. The inner product ${\bf g}({\bf V},{\bf W})$ evaluates to either $g_{\mu\nu}V^\mu W^\nu$ or $\eta_{ab}V^aW^b$ when one usues ${\bf dx}^\mu({\boldsymbol \partial}_\nu)= \delta^\mu_\nu$ or ${\bf e}^{*a}({\bf e}_b)= \delta^a_b$.

From what you say, it seems as though Hagen K treats the numerical array $e^\mu_a$ as if the were components of a tensor and feels free to raise and lower induces using the metric. These numbers are a set of change-of-basis coefficients and not the components of a tensor, however, and this abuse of notation poses serious risks. I recommend not to do it.

Now to EC theory: In the teleparallel EC formalism we simply choose frame ${\bf e}_a$ and declare it to be everywhere parallel. "Everwhere parallel" means that the "spin connection" coefficients ${\omega^b}_{a\mu}$ defined by
$$ \nabla_X {\bf e}_a= {\bf e}_b {\omega^b}_{a\mu} X^\mu $$ are all zero.

In the coordinate frame we define the Christoffel symbols in the usual way as the derivative the coordinate frame basis vectors
$$ \nabla_X {\boldsymbol \partial}_\mu = {\Gamma^\lambda}_{\mu\nu} X^\nu{\boldsymbol \partial}_\lambda. $$ Using ${\boldsymbol \partial}_\mu = e^{*a}_\mu {\bf e}_a$ we evaluate $$ \nabla_X {\boldsymbol \partial}_\mu = (\nabla_X e^{*a}_\mu) {\bf e}_a+ e^{*a}_\mu\nabla_X {\bf e}_a\\ = (X^\nu \partial_\nu e^{*a}_\mu) {\bf e}_a+0\\ = e^{*\lambda}_a (X^\nu \partial_\nu e^{*a}_\mu){\boldsymbol\partial}_\lambda $$ to read off that $$ {\Gamma^\lambda}_{\mu \nu}= e^\lambda_a\partial_\nu e^{*a}_\mu. $$ In this calculation we have used that the covaraint derivative is a derivation (obeys Leibnitz rule) and observed that the $e^{*a}_{\mu}$ are just numbers so $\nabla_X e^{*a}_{\mu}\equiv X^\nu \partial_\nu e^{*a}_{\mu}$.

In my option this teleparallel game is a total waste of time. We are obscuring what is going on by making a completely arbitrary choice of the ${\bf e}_a$ that we are going to call parallel.

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  • $\begingroup$ I feel it is important to note that either the vierbein and tetrad, in the language of the book, have inverse and are defined as different objects. The problem is how to distinguish them in a practical situation since they must differ by the integrability condition posed. about the EC: yeah, in teleparallell they obey the integrability condition because the curvature vanishes. I dont know what you mean by "EC is simply a confusing rewriting of standard GR", i'm sorry. $\endgroup$ – Marcos Vinicius May 26 at 14:06
  • $\begingroup$ I'll amend my answer. $\endgroup$ – mike stone May 26 at 16:41
  • $\begingroup$ sorry, but I still do not understand how to tell the fields apart, or if there is a way to tell them to be the same ignoring the integrability conditions. In the beggining of my question i pointed that I was studying the EC theory, i suppose you're talking about teleparalelism, which is not the same thing. EC theory, in general, consider both curvature AND torsion as non-vanishing. $\endgroup$ – Marcos Vinicius May 26 at 23:39
  • $\begingroup$ The vierbein and the tetrad are just different names for the same thing. I do not have have Keinert's book, but I expect that he wrote different chapters at different times and so used different names for the same object. Your other equations do not make sense, so I think you are misunderstanding something. I suggest that you read some other source on EC. $\endgroup$ – mike stone May 27 at 12:24
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This answer only addresses the integrability condition.

It doesn't address the calculation of Jacobi identity, the Riemann curvature tensor or the Second Bianchi identity.

I'm assuming a Riemann-Cartan space-time, namely, $Dg=0$ and $T\neq0$, where $D$ is the covariant derivative (typically denoted by $\nabla$) using an affine connection, and $g$ is the metric tensor which is assumed compatible with the connection. Since $T$ is the non-zero it excludes the Levi-Civita connection.

And I'm using Latin letters where $a,b,c,d$ for the field frame indices and $i,j,k,l$ for the space-time indices.

If you start with a metric, then you need to ensure the connection is compatible with the metric, i.e., $Dg=0$.

Otherwise, the first thing you will need is a connection $1$-form, namely

$$\Gamma^a_{\;b}=\Gamma^a_{ib}dx^i.$$

Then the curvature matrix based on the connection is defined as

$$\Omega\equiv d\Gamma + \Gamma\wedge\Gamma$$

or in component form, $$\Omega^a_{\;b}=\Gamma^{a}_{\;b}+ \Gamma^{a}_{\;c}\wedge\Gamma^{c}_{\;b} $$ and is called the structure equation or the curvature $2$-form - which can be shown to transform as a tensor.

The exterior derivative of the structure equation is the integrable condition

$$d\Omega=\Omega \wedge \Gamma-\Gamma \wedge \Omega$$

where the commutator of covariant derivatives based on the affine connections must satisfy the Jacobi identity.

The structure equation can be written in terms of the Riemann curvature tensor

$$\Omega^a_{\;b}=\frac{1}{2} R^a_{bjk} dx^j\wedge dx^k$$

noting $R^a_{bjk}$ is a $(1,3)$ tensor and $\Omega^a_{\;b}$ is a $(1,1)$ tensor.

In this case, the exterior derivative of the structure equation must satisfy the Second Bianchi identity.

This is a typical pattern of Cartan - differential forms with tensor valued coefficients.

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