I A sufficient condition for integrability of equation ##\nabla g=0##

[Jianbing_Shao]

In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this:
$$
\partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}.
$$
The integrability conditions for the existence of a global solution $F_{lj}$ is:
$$
R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0
$$
Then from the equation:
$$\nabla_b e_a= \Gamma^c_{ab} e_c$$
Using cartesian basis $e_I$ .$e_a(\mathbf{x}) = {e_a}^I(\mathbf{x}) e_I$. We get:
$$
\partial_a \left( {e_b}^I(\mathbf{x}) \right) = {\Gamma^c}_{ab}(\mathbf{x}){e_c}^I(\mathbf{x}) ~~~~~~~(1)
$$
Combine with definition:
$$ g_{ab} = {e_a}^I {e_b}^J \eta_{IJ} $$
Thenwe can get metric compatible equation:
$$
\partial_c g_{ab}-{\Gamma^e}_{ca}g_{e b}-{\Gamma^e}_{cb}g_{ae}=0~~~~~(2)
$$
From the theorem above, the integrability condition of equation (1) is $R({\Gamma^c}_{ab}(\mathbf{x}))=0$. Then we can find a global solution ${e_b}^I(\mathbf{x})$ satisfying the equation. So the metric field $g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$ is also a global solution to equation (2). And $g_{ab}$ can be a non-flat metric.

So we can draw a conclusion: non-flat metric $g_{ab}$ is compatible with a zero curvature connection ${\Gamma^c}_{ab}={e^c}_I \partial_a {e_b}^I$. Then how to explain this conclusion? A non-flat metric space can also have zero curvature?

 

[ergospherical]

Your connection $\Gamma^{c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$ is the teleparallel/Weitzenböck connection (see e.g. this Wikipedia article), which has no curvature but can have non-zero torsion.

For example, consider the 2-sphere with basis $(\hat{e}_{\theta}, \hat{e}_{\phi}) = (\partial_{\theta}, \tfrac{1}{\sin{\theta}} \partial_{\phi})$. The only non-zero Weitzenböck connection component is $\Gamma^{\phi}_{\theta \phi} = -\cot{\theta}$. In particular, note that $\Gamma^{\phi}_{\phi \theta} = 0$, hence the non-zero torsion. But the Riemann tensor does vanish. It's easy to compute here, because in 2 dimensions there is only one independent Riemann component to check.

(P.S. In this "teleparallel" idea, gravity arises conceptually because of the torsion rather than the curvature.)

 

[Jianbing_Shao]

Your connection $\Gamma^{c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$ is the teleparallel/Weitzenböck connection (see e.g. this Wikipedia article), which has no curvature but can have non-zero torsion.

From the theorem I mentioned we can get the same result. Especially if the basis field $e_a$ is a coordinate basis, then curvature and torsion tensor will be zero at the same time. Only when $e_a$ is a non-coordinate basis then the curvature is zero and torsion is non-zero. So I think the idea of teleparallel connection is closely related to non-coordinate basis.

And my question is: for zero-curvature teleparallel/Weitzenböck connections, what metric field the connection is compatible with? In GR, only mtric $\eta_{ab}$ is flat. So can we draw such a conclution: All teleparallel/Weitzenböck connections are compatible with $\eta_{ab}$?
Then from metric compatible equation:
$$
\partial_c g_{ab}-{\Gamma^e}_{ca}g_{e b}-{\Gamma^e}_{cb}g_{ae}=0
$$
If ${\Gamma^e}_{ca}$ in the differential equations are teleparallel/Weitzenböck connections. then solution of the differential equations only can be $\eta_{ab}$. Obviously it is not true. So teleparallel/Weitzenböck connections also can be compatible with non-flat metric field. then the meaning of flat seems to be not so clear.

 

[renormalize]

Especially if the basis field $e_a$ is a coordinate basis, then curvature and torsion tensor will be zero at the same time. Only when $e_a$ is a non-coordinate basis then the curvature is zero and torsion is non-zero. So I think the idea of teleparallel connection is closely related to non-coordinate basis.

That's simply not true. See the paper: The Geometrical Trinity of Gravity in which the authors derive the teleparallel equivalent of general relativity (TEGR) completely in a coordinate basis, and where they say:

 

[Jianbing_Shao]

That's simply not true. See the paper: The Geometrical Trinity of Gravity in which the authors derive the teleparallel equivalent of general relativity (TEGR) completely in a coordinate basis, and where they say:

If you introduce additional terms, then I am not doubt that you can get a non-zero torsion. it is not strange.
but if you didn't introduce any additional geometrical structure. Then I am sure I am not wrong.

So I am not so clear about your concept of non-metricity. To me, it seems to describe the connections which can not be compatible with any metric field. in other words, the metric compatible equation which contains non-metricity connection is not integrable. so we can't find a global solution to the equation.

Can you give a rigorous definition of non-metricity.

 

[renormalize]

If you introduce additional terms, then I am not doubt that you can get a non-zero torsion. it is not strange.
but if you didn't introduce any additional geometrical structure. Then I am sure I am not wrong.

Please specify the "additional geometrical structure" to which you refer.

So I am not so clear about your concept of non-metricity.

Nowhere in my reply #4 is any mention non-metricity; it only involves curvature and torsion.

 

[Jianbing_Shao]

Please specify the "additional geometrical structure" to which you refer.

Nowhere in my reply #4 is any mention non-metricity; it only involves curvature and torsion.

In the paper you gave: He claimed:
The failure of the connection to be metric is encoded in the non-metricity
$$
Q_{\alpha\mu\nu}\equiv \nabla_\alpha g_{\mu\nu},
$$
In fact, I don't know why we should introduce the term $Q_{\alpha\mu\nu}$， Because this term destroy the compatibility relation between metric and connection. and what is the exact meaning of " failure of the connection to be metric".

 

[renormalize]

In fact, I don't know why we should introduce the term $Q_{\alpha\mu\nu}$， Because this term destroy the compatibility relation between metric and connection. and what is the exact meaning of " failure of the connection to be metric".

Metricity $Q_{\alpha\mu\nu}=0$ is a postulate (an assumption!) and holds in Riemannian spacetimes. But non-Riemannian spacetimes with $Q_{\alpha\mu\nu}\neq 0$ are possible and are perfectly consistent mathematically.

 

[Jianbing_Shao]

Metricity $Q_{\alpha\mu\nu}=0$ is a postulate (an assumption!) and holds in Riemannian spacetimes. But non-Riemannian spacetimes with $Q_{\alpha\mu\nu}\neq 0$ are possible and are perfectly consistent mathematically.

If $Q_{\alpha\mu\nu}= 0$, connection and metric field are compatible with each other, and if $Q_{\alpha\mu\nu}\neq 0$, it means the connection and metric field are not compatible with each other, But this can not rule out the possibilities that the connection can be compatible with another metric field. So can you prove that if we introduce the non-zero term $Q_{\alpha\mu\nu}$, then the connection can not be compatible with an arbitrary metric field?

 

[ergospherical]

From the theorem I mentioned we can get the same result. Especially if the basis field $e_a$ is a coordinate basis, then curvature and torsion tensor will be zero at the same time. Only when $e_a$ is a non-coordinate basis then the curvature is zero and torsion is non-zero. So I think the idea of teleparallel connection is closely related to non-coordinate basis.

And my question is: for zero-curvature teleparallel/Weitzenböck connections, what metric field the connection is compatible with? In GR, only mtric $\eta_{ab}$ is flat. So can we draw such a conclution: All teleparallel/Weitzenböck connections are compatible with $\eta_{ab}$?
Then from metric compatible equation:
$$
\partial_c g_{ab}-{\Gamma^e}_{ca}g_{e b}-{\Gamma^e}_{cb}g_{ae}=0
$$
If ${\Gamma^e}_{ca}$ in the differential equations are teleparallel/Weitzenböck connections. then solution of the differential equations only can be $\eta_{ab}$. Obviously it is not true. So teleparallel/Weitzenböck connections also can be compatible with non-flat metric field. then the meaning of flat seems to be not so clear.

Sorry for the delay - a few points:

1. The teleparallel connection ${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$ is flat in the sense that it has vanishing Riemann, $R= 0$. That is saying nothing about the $R$ of the Levi-Civita connection of $g$, which may well be nonzero (it is a different connection!).

Worth saying again - in this formulation, gravity is encoded in the torsion of a connection with zero curvature.

2. Metrics of the form $g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$ are compatible with the teleparallel connection. This is what follows from the definition of the teleparallel connection, $\partial_a {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$, since
\begin{align*}
\partial_c g_{ab} &= \eta_{IJ}\left[ (\partial_c {e_a}^{I} ) {e_b}^{J} + {e_a}^{I} (\partial_c {e_b}^{J})\right] \\
&= \eta_{IJ} \left[ {\Gamma}^d_{ca} {e_d}^I {e_b}^J + {\Gamma}^d_{cb} {e_a}^I {e_d}^J\right] \\
&= {\Gamma^d}_{ca} g_{db}+ {\Gamma^d}_{cb} g_{ad} \\
\end{align*}
which is the metric compatibility equation $\nabla_c g_{ab} = 0$.

 

[Jianbing_Shao]

Sorry for the delay - a few points:

1. The teleparallel connection ${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$ is flat in the sense that it has vanishing Riemann, $R= 0$. That is saying nothing about the $R$ of the Levi-Civita connection of $g$, which may well be nonzero (it is a different connection!).

Worth saying again - in this formulation, gravity is encoded in the torsion of a connection with zero curvature.

2. Metrics of the form $g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$ are compatible with the teleparallel connection.

To a metrics $g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$, it can be compatible with Levi-Civita connection with non-zero curvature. but at the same time, it also can be compatible with the teleparallel connection with zero curvature, then why we call such a metric non-flat?

 

[ergospherical]

Right! Curvature is a property of the connection, not the metric. A metric is just a tensor field on the manifold (with the usual smooth, symmetric, positive-definite constraints). It has no intrinsic curvature.

When people talk about the 'curvature of a metric', they are typically referring to the curvature of the Levi-Civita connection that is uniquely determined by $g$.

 

[Jianbing_Shao]

Right! Curvature is a property of the connection, not the metric. A metric is just a tensor field on the manifold (with the usual smooth, symmetric, positive-definite constraints). It has no intrinsic curvature.

When people talk about the 'curvature of a metric', they are typically referring to the curvature of the Levi-Civita connection that is uniquely determined by $g$.

So in fact a metric field can be compatible with many different connections, and in them many have zero curvature, Then why a metric field can be compatible with many different connections?
And the same metric field, satisfy the same metric compatible equation. What is the difference between Levi-Civita connection and teleparallel connection?

 

[ergospherical]

Yes, in fact any two metric compatible connections differ by a "contortion" tensor $A_{ijk}$ which is antisymmetric in its last two indices (i.e. $A_{ijk} = -A_{ikj}$), thereby satisfying $g(A(X,Y),Z) = -g(A(X,Z), Y)$ for all vector fields $X,Y,Z$. If $\nabla$ is the Levi-Civita connection, then consider the new connection $\nabla' := \nabla + A$. Using the Leibniz rule, you can show that $({\nabla'}_{X}g)(Y,Z) = 0$ for any $X,Y,Z$ and hence that $\nabla'$ is also metric compatible.

[You can express the contortion tensor in terms of the torsion:
$A_{ijk} = \frac{1}{2}(T_{ijk} + T_{kij} - T_{jik})$]

 

[Jianbing_Shao]

Yes, in fact any two metric compatible connections differ by a "contortion" tensor $A_{ijk}$ which is antisymmetric in its last two indices (i.e. $A_{ijk} = -A_{ikj}$), thereby satisfying $g(A(X,Y),Z) = -g(A(X,Z), Y)$ for all vector fields $X,Y,Z$. If $\nabla$ is the Levi-Civita connection, then consider the new connection $\nabla' := \nabla + A$. Using the Leibniz rule, you can show that $({\nabla'}_{X}g)(Y,Z) = 0$ for any $X,Y,Z$ and hence that $\nabla'$ is also metric compatible.

From your description I still can’t find the difference between Levi-Civita connection and teleparallel connection,

If we start from a teleparallel connection ${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$, then from a given ${e^c}_{I}$ we can get a unique teleparallel connection ${\Gamma^c}_{ab}$, but from the definition of metric $g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$, If given an arbitrary group element $G_c^a \in SO(N)$, Then ${e_a}^{I}$ and ${e_c}^{I} G_c^a$ are compatible with the same metric field.
So all the connections compatible with the same metric field can be written in such a form:

$${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}+{e^d}_{I} G_d^c (\partial_a G_b^f ){e_f}^{I}$$

All these connections are compatible with the same metric field. The difference between then is ${e^d}_{I} G_d^c (\partial_a G_b^f ){e_f}^{I}$. Is this term equivalent to contorsion?

 

[Jianbing_Shao]

Yes, in fact any two metric compatible connections differ by a "contortion" tensor $A_{ijk}$ which is antisymmetric in its last two indices (i.e. $A_{ijk} = -A_{ikj}$), thereby satisfying $g(A(X,Y),Z) = -g(A(X,Z), Y)$ for all vector fields $X,Y,Z$. If $\nabla$ is the Levi-Civita connection, then consider the new connection $\nabla' := \nabla + A$. Using the Leibniz rule, you can show that $({\nabla'}_{X}g)(Y,Z) = 0$ for any $X,Y,Z$ and hence that $\nabla'$ is also metric compatible.

[You can express the contortion tensor in terms of the torsion:
$A_{ijk} = \frac{1}{2}(T_{ijk} + T_{kij} - T_{jik})$]

I have a question :
If all Levi-Civita connections compatible with a non-flat metric field only can have non-zero curvature. and we start from a Levi-Civita connection,then take the contortion tensor into account, Can we get all possible connections compatible with the same metric field? If we can, then how to get teleparallel connection with zero curvature from a Levi-Civita connection?

 

[Jianbing_Shao]

If we start from a teleparallel connection ${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$, then from a given ${e^c}_{I}$ we can get a unique teleparallel connection ${\Gamma^c}_{ab}$, but from the definition of metric $g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$, If given an arbitrary group element $G_c^a \in SO(N)$, Then ${e_a}^{I}$ and ${e_c}^{I} G_c^a$ are compatible with the same metric field.
So all the connections compatible with the same metric field can be written in such a form:

In fact, if we start from a zero curvature teleparallel connection ${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$ and the invariance of metric field under the action of a $SO(N)$ group $G_c^a$. Then we can get all possible connections compatible with the same metric field.

Here the group action $G_c^a$ is not necessarily to be a group field. if it satisfy the equation:
$$\partial_a G_b^c=\gamma_{ab}^d G_d^c $$
and we also can find $\gamma_{ab}^d=-\gamma_{ad}^b$
Then all possible connections compatible with the same metric field is:
$${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}+{e^c}_{I} \gamma_{ab}^f{e_f}^{I}$$

Then if the curvature of connection $\gamma_{ab}^d$ is not zero, then $G_c^a$ changes path dependently, Then ${e_c}^{I} G_c^a$ also changes path dependently. So the connection $\{Gamma^c}_{ab}$ also have non-zero curvature. Then we can get non-zero curvature connections compatible with the same metric field.

And the difference between all these connections (have zero curvature or non-zero curvature) is just the term: $e^c}_{I} {\gamma'}_{ab}^f{e_f}^{I}$ and ${\gamma'}_{ab}^f$ satisfy: ${\gamma'}_{ab}^f=-{\gamma'}_{af}^b$.

If the two method are consistent then the term $e^c}_{I} {\gamma'}_{ab}^f{e_f}^{I}$ should be the same as the contorsion tensor. but it seems there exist some differences between them.

 

[Jianbing_Shao]

There are some mistakes, I made a revision.

If we start from a zero curvature teleparallel connection ${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$ and the invariance of metric field under the action of a $SO(N)$ group $G_c^a$. Then we can get all possible connections compatible with the same metric field.

Here the group action $G_c^a$ is not necessarily to be a group field. if it satisfy the equation:
$$\partial_a G_b^c=\gamma_{ab}^d G_d^c $$
and we also can find $\gamma_{ab}^d=-\gamma_{ad}^b$
Then all possible connections compatible with the same metric field is:
$${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}+{e^c}_{I} \gamma_{ab}^f{e_f}^{I}$$

Then if the curvature of connection $\gamma_{ab}^d$ is not zero, then $G_c^a$ changes path dependently, Then ${e_c}^{I} G_c^a$ also changes path dependently. So the connection ${\Gamma^c}_{ab}$ also have non-zero curvature. Then we can get non-zero curvature connections compatible with the same metric field.

And the difference between all these connections (have zero curvature or non-zero curvature) is just the term: ${e^c}_{I} {\gamma'}_{ab}^f{e_f}^{I}$ and ${\gamma'}_{ab}^f$ satisfy: ${\gamma'}_{ab}^f=-{\gamma'}_{af}^b$.

If the two method are consistent then the term ${e^c}_{I} {\gamma'}_{ab}^f{e_f}^{I}$ should be the same as the contorsion tensor. but it seems there exist some differences between them.

 

[Jianbing_Shao]

Sorry for the delay - a few points:

1. The teleparallel connection ${\Gamma^c}_{ab} = {e^c}_{I} \partial_a {e_b}^{I}$ is flat in the sense that it has vanishing Riemann, $R= 0$. That is saying nothing about the $R$ of the Levi-Civita connection of $g$, which may well be nonzero (it is a different connection!).

I read some papers about teleparallel connection. I have a question:

When we introduce spin connection $\omega^a_{\hphantom{a}\mu b}$. And get the full covariant derivative of the tetrad:
\begin{equation}
\partial_{\mu}e^a_{\nu}+\omega^a_{\mu b}e^b_{\nu}-\Gamma^{\alpha}_{\mu\nu}e^a_{\alpha}=0
\end{equation}

And tetrad $e’^a_{\nu}$ satisfy the equation:
\begin{equation}
\partial_{\mu}e’^a_{\nu}-\Gamma^{\alpha}_{\mu\nu}e’^a_{\alpha}=0
\end{equation}
If connection $\Gamma^{\alpha}_{\mu\nu}$ is same in the two equation. Then what is the difference between $e’^a_{\nu}$ and $e^a_{\nu}$?

 

[ergospherical]

I'm struggling a bit to follow your train of thought-

The tetrads $e'$ and $e$ are related by a (local) Lorentz transformation, i.e. $e' = \Lambda e$.
You mention the spin connection $\omega$. It too transforms, precisely because the tetrad postulate:

$\nabla_{\mu} {e^a}_{\nu} = \partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$

remains true in both tetrads. If you work it through, you'll find the transformation law $\omega' = \Lambda \omega \Lambda^{-1} + \Lambda d(\Lambda^{-1})$.

The $\Gamma^{\sigma}_{\mu \nu}$ (with spacetime coordinate indices) don't change if you do a local Lorentz transformation of the tetrad (which affects the tetrad indices). You have just done a gauge transformation.

 

[Jianbing_Shao]

The tetrads $e'$ and $e$ are related by a (local) Lorentz transformation, i.e. $e' = \Lambda e$.
You mention the spin connection $\omega$. It too transforms, precisely because the tetrad postulate:

$\nabla_{\mu} {e^a}_{\nu} = \partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$

To the equation above, I have a question:
Because the equation above is a differential equations,, So even for equations like:
\begin{equation}
\partial_{\mu}e^a_{\nu}-\Gamma^{\alpha}_{\mu\nu}e^a_{\alpha}=0
\end{equation}
Not for all connections can we find a tetrad field which can satisfy the euqation.

Now if you introduce the term including the spin connection ${\omega^a}_{\mu b} {e^b}_{\nu}$.

$\nabla_{\mu} {e^a}_{\nu} = \partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$

So not for all connection ${\omega^a}_{\mu b}$ and $\Gamma^{\sigma}_{\mu \nu}$ there exists a tetrad field which can satisry the equation.
So I want to know what conditions should ${\omega^a}_{\mu b}$ and $\Gamma^{\sigma}_{\mu \nu}$ fulfill then such a tetrad field exist which can satisfy the equation.

 

[renormalize]

So I want to know what conditions should ${\omega^a}_{\mu b}$ and $\Gamma^{\sigma}_{\mu \nu}$ fulfill then such a tetrad field exist which can satisfy the equation.

Your equation is not an integrability condition that must be satisfied by $e,\omega,\Gamma$. Instead, it simply defines the spin connection. From page 5 of https://people.physics.illinois.edu/stone/torsion.pdf:

So you are free to select any arbitrary tetrad field $e$ and any connection $\Gamma$ (including one with arbitrary torsion and arbitrary nonmetricity) and use these equations to define a consistent spin connection $\omega$.

 

[Jianbing_Shao]

Your equation is not an integrability condition that must be satisfied by $e,\omega,\Gamma$. Instead, it simply defines the spin connection. From page 5 of https://people.physics.illinois.edu/stone/torsion.pdf:
View attachment 365038
So you are free to select any arbitrary tetrad field $e$ and any connection $\Gamma$ (including one with arbitrary torsion and arbitrary nonmetricity) and use these equations to define a consistent spin connection $\omega$.

When we say the teleparallel connection $\Gamma^{\alpha}_{\mu\nu} =e_a^{\alpha}\partial_{\mu}e^a_{\nu}$ is flat , it has vanishing Riemann, $R= 0$. Just because we demand that the tetrad $e^a_{\nu}$ must be a field.
But in such a equation:
\begin{equation}
\partial_{\mu}e^a_{\nu}-\Gamma^{\alpha}_{\mu\nu}e^a_{\alpha}=0
\end{equation}
The choose of connection is free. and if the curvature of the connection is not zero. then the tetrad $e^a_{\nu}$ is not a field. Because it changes path dependently.

So if we start from the equation:
$\partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$
The two connections $\omega,\Gamma$ can be defined independently, and ${e^a}_{\nu}$ is not always a field.
Only when ${e^a}_{\nu}$ is a tetrad field. Then the choose of $\omega,\Gamma$ is not free. Of course you can say we can express one in terms of the other.

But there is a question, If take the metric compatible condition into account. then what is the relation between the tetrad field $e$ and $\Gamma$ ?

 

[renormalize]

The two connections $\omega,\Gamma$ can be defined independently, and ${e^a}_{\nu}$ is not always a field.

Not according to the reference I posted above. Only the tetrad $e$ and connection $\Gamma$ can be independently and freely selected, from which the spin connection $\omega$ is then derived. Can you quote a credible reference that says otherwise?

 

[Jianbing_Shao]

Not according to the reference I posted above. Only the tetrad $e$ and connection $\Gamma$ can be independently and freely selected, from which the spin connection $\omega$ is then derived. Can you quote a credible reference that says otherwise?

I have read an interesting paper 'Conditions on a Connection to be a Metric Connection' by B. G. Schmidt(Commun. math. Phys. 29, 55—59 (1973)).
In this paper he discussed the integrability condition of metric compatible equation. He use holonomy group to tell us when a connection can be a metric connection. I don't know if the concept of metricity is equivalent to the concept ot metric connection? But he get an interesting result.

In fact we can use parallel propagator to calculate the holonomy group along a particular path. So if we try to discuss the integrability condition of differential equations like this:
$\partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$.
We can use parallel propagator. We can calculate the change of the tetrad along a
infinitesimal parallelogram using parallel propagator, if the tetrad is a field, then demand the change should be zero, then we can get integrability condition.
Here is an introduction of parallel propagator and some examples.

MENTOR NOTE: non-peer reviewed reference.

 

[Jianbing_Shao]

Metrics of the form $g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$ are compatible with the teleparallel connection. This is what follows from the definition of the teleparallel connection, $\partial_a {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$, since
\begin{align*}
\partial_c g_{ab} &= \eta_{IJ}\left[ (\partial_c {e_a}^{I} ) {e_b}^{J} + {e_a}^{I} (\partial_c {e_b}^{J})\right] \\
&= \eta_{IJ} \left[ {\Gamma}^d_{ca} {e_d}^I {e_b}^J + {\Gamma}^d_{cb} {e_a}^I {e_d}^J\right] \\
&= {\Gamma^d}_{ca} g_{db}+ {\Gamma^d}_{cb} g_{ad} \\
\end{align*}
which is the metric compatibility equation $\nabla_c g_{ab} = 0$.

Because Levi-Civita connection is also compatible with metric. So it also satisfy the equation:
$\partial_a {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$,
And the curvature of Levi-Civita connection is not zero. So ${e_b}^{I}$ can not be a tetrad field.
Then how to get a metric field from such a tetrad ${e_b}^{I}$ using formula:
$g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$

 

[renormalize]

I have read an interesting paper 'Conditions on a Connection to be a Metric Connection' by B. G. Schmidt(Commun. math. Phys. 29, 55—59 (1973)).

How is this paper relevant to your claim in post #23?

So if we start from the equation:
$\partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$
The two connections $\omega,\Gamma$ can be defined independently, and ${e^a}_{\nu}$ is not always a field.
Only when ${e^a}_{\nu}$ is a tetrad field. Then the choose of $\omega,\Gamma$ is not free.

The tetrad $e$ and spin-connection $\omega$ appear nowhere in the Schmidt paper. And your later GitHub reference is not refereed so it's disallowed on Physics Forums.
My citation in post #22 explicitly states that the equation:$$\partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$$is nothing more than the definition of $\omega$ in terms of the arbitrary chosen fields $e,\Gamma$. (Equivalently, it can serve to define $\Gamma$ in terms of arbitrary $e,\omega$.)
You on the other hand wish to discuss something different: choosing the two connections $\Gamma,\omega$ independently, from which you then derive a path-dependent tetrad $e$. But this puts the burden on you to cite a credible published reference that implements this approach and demonstrates that it leads to a consistent theory of gravity in spacetime. If you cannot produce such a reference, you are proposing a personal theory that is off limits in Physics Forums.

 

[Jianbing_Shao]

How is this paper relevant to your claim in post #23?

Metric compatible equations and the equations you gave are two differential equations , In fact your equations is more complex, The integrability condition of metric compatible equation is hard to get a result, So if you demand the tetrad be a field in your equation, Then the result should be more complex.

My citation in post #22 explicitly states that the equation:$$\partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$$is nothing more than the definition of $\omega$ in terms of the arbitrary chosen fields $e,\Gamma$. (Equivalently, it can serve to define $\Gamma$ in terms of arbitrary $e,\omega$.)

If you demand that the tetrad is a field, then the field can be describe with a zero curvature teleparallel connection. When you introduce an extra connection $\omega$. Then the relation between $\omega$ and $\Gamma$ is determined by the tetrad field, In other words, you use two types of connections to describe the same tetrad field. So now $\omega$ itself can not tell us anyting about the tetrad field. So what physics the spin connection $\omega$ can tell us?

You on the other hand wish to discuss something different: choosing the two connections $\Gamma,\omega$ independently, from which you then derive a path-dependent tetrad ##e#

I wnat to use tetrad to find out how to get a non-zero curvature connection from a metric field.

To the metric compatible equations. if the metric is a field, then the paths between two points have no distiction. but to a non-zero curvature connection, the paths are different. Then I wander where is the path dependent property comes from? So I doubt in fact we can not get a non-zero curvature connection purely from a metric field.

 

[PeterDonis]

I doubt in fact we can not get a non-zero curvature connection purely from a metric field.

The Levi-Civita connection, which will have nonzero curvature in any metric that is not the flat Minkowski metric, can be derived purely from the metric and its first derivatives. Doesn't that contradict the claim of yours just quoted above?

 

[Jianbing_Shao]

The Levi-Civita connection, which will have nonzero curvature in any metric that is not the flat Minkowski metric, can be derived purely from the metric and its first derivatives. Doesn't that contradict the claim of yours just quoted above?

Yes, I really think there is contradiction here, In the paper I quoted, He claimed that the holonomy group generated from a metric connection(I think it is not necessarily to be Levi-Civita connection, but all possibe connections compatible with the metric field.) should be a rotation group, and the metric is rotation invariant.
Here is just where the problem is, A metric field is rotation invariant. then do you think we can get a connection from such a metric field, then the connection can tell us how the tetrad rotate when moves along a colsed path? is it possible ?

 

[Jianbing_Shao]

My citation in post #22 explicitly states that the equation:$$\partial_{\mu} {e^a}_{\nu} + {\omega^a}_{\mu b} {e^b}_{\nu} - \Gamma^{\sigma}_{\mu \nu} {e^a}_{\sigma} = 0$$is nothing more than the definition of $\omega$ in terms of the arbitrary chosen fields $e,\Gamma$. (Equivalently, it can serve to define $\Gamma$ in terms of arbitrary $e,\omega$.)

In your definition, the fields $e,\Gamma$ is chosen independently. But if we demand that the connection $\Gamma$ to be the Levi-Civita connection compatible with the metric field defined using tetrad field $e$. Then what is the property of spin connection $\omega$.

 

[PeterDonis]

I really think there is contradiction here

You missed my point. My statement in post #29, which is a completely uncontroversial fact, contradicts your claim that I quoted. That means your claim is wrong. It doesn't mean there's a contradiction in the literature. There isn't.

 

[renormalize]

But if we demand that the connection $\Gamma$ to be the Levi-Civita connection compatible with the metric field defined using tetrad field $e$. Then what is the property of spin connection $\omega$.

Just look at Wikiedia: https://en.wikipedia.org/wiki/Spin_connection!

So for the Levi-Civita connection $\left\{ _{\mu\nu}^{\alpha}\right\}$, the spin-connection $\omega$ is expressed uniquely and unambiguously in terms of $e$ and its first derivatives. For more general connections $\Gamma_{\mu\nu}^{\alpha}\,$, $\omega$ will involve the torsion $T$ and nonmetricity $Q$ tensors as well.

 

[Jianbing_Shao]

You missed my point. My statement in post #29, which is a completely uncontroversial fact, contradicts your claim that I quoted. That means your claim is wrong. It doesn't mean there's a contradiction in the literature. There isn't.

Then return to the question we have discussed. From a non-zero curvature Levi-Civita connection and metric compatible equation, then how to get a metric field. and why the path dependent property disspeared?
A right theory can answer any question we ask.

 

[renormalize]

From a non-zero curvature Levi-Civita connection and metric compatible equation, then how to get a metric field. and why the path dependent property disspeared?

You avoided my request above for a reference, so I request again: please supply a credible published reference that explicitly discusses the possibility, existence and consistency of a path-dependent metric/tetrad in 4D spacetime. Otherwise you waste everyone's time trying to debate an ill-conceived personal theory.

 

[Jianbing_Shao]

You missed my point. My statement in post #29, which is a completely uncontroversial fact, contradicts your claim that I quoted. That means your claim is wrong. It doesn't mean there's a contradiction in the literature. There isn't.

Another problem ,Do you think a non trivial metric field can also be compatible with a zero curvature connection. they call it teleparallel connection. If it is true. then why we must believe we can use it to tell if a metric space is curved.

 

[Jianbing_Shao]

Just look at Wikiedia: https://en.wikipedia.org/wiki/Spin_connection!
View attachment 365158
So for the Levi-Civita connection $\left\{ _{\mu\nu}^{\alpha}\right\}$, the spin-connection $\omega$ is expressed uniquely and unambiguously in terms of $e$ and its first derivatives. For more general connections $\Gamma_{\mu\nu}^{\alpha}\,$, $\omega$ will involve the torsion $T$ and nonmetricity $Q$ tensors as well.

Simply from a tetrad field , then we can calculate the difference between teleparallel connection and Levi-Civita connection:
If we define:
$\Gamma_{abc} = \frac{1}{2}(g_{ab,c} + g_{ac,b} - g_{bc,a} + c_{abc} + c_{acb} - c_{bca})$
and
$g^{ab} = {e^a}_I {e^b}_J \eta^{IJ}$,

${c_{ab}}^c e_c = [e_a, e_b]$
We can get:
$\Gamma_{abc} = \frac{1}{2} ( {{e_a}^I}_{,c} e_{bI} + {{e_b}^I}_{,c} e_{aI} + {{e_a}^I}_{,b} e_{cI} + {{e_c}^I}_{,b} e_{aI} - {{e_b}^I}_{,a} e_{cI} - {{e_c}^I}_{,a} e_{bI} + {{e_b}^I}_{,a} e_{cI} - {{e_a}^I}_{,b} e_{cI} + {{e_c}^I}_{,a} e_{bI} - {{e_a}^I}_{,c} e_{bI} - {{e_c}^I}_{,b} e_{aI} + {{e_b}^I}_{,c} e_{aI} )$
At last we can find that:
$\Gamma_{abc} = e_{aI} {{e_b}^I}_{,c}$
So the difference is very clear:
$c_{abc} + c_{acb} - c_{bca})$.
Then is the tern equal the spin connection? and if the tetrad $e_a$ field is a coordinate basis, then spin connection is zero?

 

[weirdoguy]

A right theory can answer any question we ask.

Can you answer questions you've been asked?

 

[PeterDonis]

From a non-zero curvature Levi-Civita connection and metric compatible equation, then how to get a metric field

Um, solve the differential equation? If you know the connection is the Levi-Civita connection for some metric, then you have a differential equation for the metric in terms of the connection.

 

[PeterDonis]

Do you think a non trivial metric field can also be compatible with a zero curvature connection. they call it teleparallel connection.

Mathematically, sure. But math isn't the same as physics.

If it is true. then why we must believe we can use it to tell if a metric space is curved.

In physics, we pick a specific connection for physical reasons, regardless of the fact that it's only one of many connections we could use mathematically. In General Relativity, we use the Levi-Civita connection for the metric, and the curvature of that connection, to tell whether spacetime is physically curved, and we derive the physical properties we associate with gravity from the curvature. The fact that there are other connections we could use is irrelevant, because we don't use them.

In teleparallel gravity, which is a different theory from GR, the properties we associate with gravity are derived from the torsion, not the curvature. So the fact that the connection used in teleparallel gravity has zero curvature doesn't matter, because we aren't deriving the physical properties of gravity from the curvature. We're deriving them from the torsion. As @ergospherical told you way back in post #2 of this thread.

 

[Jianbing_Shao]

Um, solve the differential equation?

Yes, I think we should treat the metric compatible equation as differential equations. To do so we at least can understand two problems:
1, Why a connection can be compatible with many metric fields, Because the different initial conditions we choose.
2, Why a metric field can be compatible with many connections, Because the inner-symmetric structure of the metric field. From the symmetric group of metric we can know what is the difference between the different connections.

If you know the connection is the Levi-Civita connection for some metric, then you have a differential equation for the metric in terms of the connection.

If we know how to use the connection to describe the mtric field. then we can know the physical meaning of Levi-Civita connection and metric field better.

 

[PeterDonis]

Why a connection can be compatible with many metric fields, Because the different initial conditions we choose.

No, because "compatible" isn't just one thing. "Compatible" if the connection is the Levi-Civita connection gives you one equation that relates the metric and the connection. But any other connection compatible with the same metric will have a different equation that relates the metric and the connection. It has nothing to do with different initial conditions; it has to do with different equations relating the metric and the connection.

Why a metric field can be compatible with many connections, Because the inner-symmetric structure of the metric field.

No, because, as above, "compatible" isn't just one thing. The answer here is the same as the answer above.

From the symmetric group of metric we can know what is the difference between the different connections.

How? Do you have a reference?

 

[Jianbing_Shao]

No, because "compatible" isn't just one thing. "Compatible" if the connection is the Levi-Civita connection gives you one equation that relates the metric and the connection. But any other connection compatible with the same metric will have a different equation that relates the metric and the connection. It has nothing to do with different initial conditions; it has to do with different equations relating the metric and the connection.

Because compatible is defined with metric compatible equation, and this equation is different with the normal differential equation in $IR^n$. But we can analyze the equation in such a way. At first we can write the metric compatible equation in the combination of two equations:
$\partial_a {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$ or $e_a^J\partial_J {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$.
and
$g_{ab} = \eta_{IJ} {e_a}^{I} {e_b}^{J}$.

Then I am confused by such a problem: Does the operator $\partial_a$ associate with a coordinate system. or at least is defined with a non-coordinate basis field. What is your opinion on this problem?

 

[PeterDonis]

compatible is defined with metric compatible equation

Which "metric compatible equation:"? There isn't just one. There are as many different metric compatible equations as there are connections. But you don't even seem to recognize this.

Does the operator $\partial_a$ associate with a coordinate system

Of course. That's what it means. It's a partial derivative with respect to the coordinates.

at least is defined with a non-coordinate basis field.

What "non-coordinate basis field"? There isn't one as far as the operator $\partial_a$ is concerned.

What is your opinion on this problem?

My opinion is that there is no problem at all, just your continued misunderstandings.

 

[Jianbing_Shao]

Which "metric compatible equation:"? There isn't just one. There are as many different metric compatible equations as there are connections. But you don't even seem to recognize this.

Of course. That's what it means. It's a partial derivative with respect to the coordinates.

So $e_a^J$ should be a tetrad field. then from the equation:
$\partial_a {e_b}^{I} = {\Gamma^c}_{ab} {e_c}^{I}$
If $e_a^J$ is a tetrad field, then the connection can be written as:
${\Gamma^c}_{ab}={e^c}_{I}\partial_a {e_b}^{I}$
It is a teleparallel connection, and the curvature of the connection is zero. is it right?