I A question about metric compatibility equation

[Jianbing_Shao]

TL;DR Summary

metric compatibility equation, connection, constraint

We all know that The gradient of a scalar-valued function $f(x)$ in $IR^n$ is a vector field $V_\mu(x)=\partial_\mu f(x)$, Such a vector field is said to be conservative. Not all vector fields are conservative. A conservative vector field should meet certain constraints $curlV_\mu(x)=0$.

In the discussion of a vector field $V(x)$ , If we take the partial derivative of this vector field we can get $n$ matrix function $A_\mu(x)$.
$$
\partial_\mu V(x)=A_\mu(x)V(x)
$$If matrix functions $A_\mu(x)$ are continuously differentiable with respect to all variables. then we will find that only when$$
R_{\mu\nu}= \partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu , A_\nu]=0
$$ then there exist a matrix function $G(x)$ which satisfy: $A_\mu(x)=\partial_\mu G(x)G^{-1}(x)$.(we can find the proof of this conclusion in the book 'Product Integration, Its History And Applications'P41) also we can get: $V(x')=G(x')G^{-1}(x)V(x)$.

In general relativity, we calculate a Christoffel connection $\Gamma(x)$(torsion free)from a metric $g(x)$ using the metric compatibility equation $\nabla_\rho g_{\mu\nu}=0$. the metric compatible connection $\gamma(x)$ is not necessarily torsion free. From the discussion above, we can say that if the matric $g(x)$ is a matrix function, the metric compatible connection $\gamma(x)$ we get should also satisfy certain constraints. then what is it?

 

[vanhees71]

If you consider extensions of GR, where the affine connection is not torsion free, you need additional equations determining the torsion. The most natural theory is Einstein-Cartan theory, which you get from gauging Poincare symmetry of SR and include, e.g., Dirac-fermion fields. Then you get a connection that is metric compatible but necessarily with torsion. Treating the usual macroscopic case, where you have only scalar fields (to describe matter in terms of fluid dynamics) and a massless vector field (to describe electromagnetic fields), you get standard GR.

 

[Jianbing_Shao]

If you consider extensions of GR, where the affine connection is not torsion free, you need additional equations determining the torsion. The most natural theory is Einstein-Cartan theory, which you get from gauging Poincare symmetry of SR and include, e.g., Dirac-fermion fields. Then you get a connection that is metric compatible but necessarily with torsion. Treating the usual macroscopic case, where you have only scalar fields (to describe matter in terms of fluid dynamics) and a massless vector field (to describe electromagnetic fields), you get standard GR.

I am not discuss the connection with torsion, I only want to say that a metric compatible connection is not necessarily torsion free.

The problem here is just that if you try to calculate a connection from a metric, then the connection should meet certain constraints. just like you can get a vector field from a scalar field. the vector field should be conservative(curl of the vector field is 0). this constraint just comes from the fact that the nunber of components of a vector field is bigger than a scalar field., the nunber of components of a connection is bigger than a metric, so the connection we get from a metric should also meet some constraints..

so what is such a constraint.

 

[Jianbing_Shao]

If you consider extensions of GR, where the affine connection is not torsion free, you need additional equations determining the torsion. The most natural theory is Einstein-Cartan theory, which you get from gauging Poincare symmetry of SR and include, e.g., Dirac-fermion fields. Then you get a connection that is metric compatible but necessarily with torsion. Treating the usual macroscopic case, where you have only scalar fields (to describe matter in terms of fluid dynamics) and a massless vector field (to describe electromagnetic fields), you get standard GR.

Perhaps my question can be expressed in another way: from the example of the relationship between a conservative vector field which corresponds to a scalar field and an arbitrary vector field. I think an arbitrary connection field perhaps is not always a metric compatible connection.

So what is the distiction between a metric compatible connection and a non-metric compatible connection.

 

[Antarres]

Indeed, connection doesn't have to be torsion-free. However if it isn't torsion free, then it is an independent variable from the metric in theory. You can always put a connection with torsion in the form
$$\omega^{ij} = \tilde{\omega}^{ij} + K^{ij}$$
Where K is the contorsion tensor, dependent on torsion, and $\tilde{\omega}$ is the torsion-free connection of the same metric. So therefore you have a choice, either to have metric and connection as independent variables, or metric and torsion(because contorsion directly depends on torsion, and therefore you can find the connection if you know the torsion). So you cannot get connection directly from the metric in this case, is what I'm saying.

In this case you still have the metric compatibility condition. However, you can even choose to not have metric compatibility either. In that case, a new tensor will arise, similar to torsion called non-metricity, which will describe this deviation from metric compatibility, so you'll have another extra variable in the theory as well.

Edit: To add to the answer to your question if I understood you correctly, you want to know the restrictions that allow connection to be determined from the metric? In that case the answer is metric compatibility and the torsion zero constraint. With those two, you can derive the formula for connection that is determined completely by the metric, usually called Levi-Civita connection.

 

[Jianbing_Shao]

Edit: To add to the answer to your question if I understood you correctly, you want to know the restrictions that allow connection to be determined from the metric? In that case the answer is metric compatibility and the torsion zero constraint. With those two, you can derive the formula for connection that is determined completely by the metric, usually called Levi-Civita connection.

Yes ,it is just what I want to know,
From the discussion about the differential equations in mathematics, If the connection in the differential equations such as metric compatibility equation is arbitrary, we can not always find a metric (matrix function) then the pair (metric, connection) can satisfy the equation. So I want to know the restrictions on the connection that allow a connection to be metric copatible.

 

[renormalize]

So I want to know the restrictions on the connection that allow a connection to be metric copatible.

Have you read about the "nonmetricity tensor" ( https://en.wikipedia.org/wiki/Nonmetricity_tensor )?

 

[strangerep]

I am not discuss the connection with torsion, I only want to say that a metric compatible connection is not necessarily torsion free.

The problem here is just that if you try to calculate a connection from a metric, then the connection should meet certain constraints. just like you can get a vector field from a scalar field. the vector field should be conservative(curl of the vector field is 0). this constraint just comes from the fact that the nunber of components of a vector field is bigger than a scalar field., the nunber of components of a connection is bigger than a metric, so the connection we get from a metric should also meet some constraints..

A "metric-compatible connection" satisfies $$\nabla_{\!\lambda\,} g_{\mu\nu} ~=~ 0 ~.$$This equation is sufficient to determine the symmetric components of the connection. (Count the number of possible index values in the equation, and in the symmetric part of the connection.)

Try doing the usual computation by permutation of indices -- you'll find that the symmetric part of the connection contains extra terms involving the skewsymmetric part of connection. However, the skew part remains unspecified by this procedure. Something else must be introduced to specify the skew part.

 

[Jianbing_Shao]

Have you read about the "nonmetricity tensor" ( https://en.wikipedia.org/wiki/Nonmetricity_tensor )?

Thanks for your reply, I will read about it carefully.

 

[Jianbing_Shao]

A "metric-compatible connection" satisfies $$\nabla_{\!\lambda\,} g_{\mu\nu} ~=~ 0 ~.$$This equation is sufficient to determine the symmetric components of the connection. (Count the number of possible index values in the equation, and in the symmetric part of the connection.)

Try doing the usual computation by permutation of indices -- you'll find that the symmetric part of the connection contains extra terms involving the skewsymmetric part of connection. However, the skew part remains unspecified by this procedure. Something else must be introduced to specify the skew part.

Thanks for your reply.
From metric compatibility equation $$\nabla_\lambda g_{\mu\nu}=0$$ we can calculate a Levi-Civita connection, and the torsion part can not be totally determined by this equation.

Then does it means that an arbitrary connection can be expressed with a Levi-Civita connection and a torsion? Then for an arbitary connection we always can find a metric and if we substitude the two into the equation. the equation is true?

 

[Antarres]

I'm not sure you understood what I wrote in the first part of my reply. Like @strangerep mentioned, the metric compatibility equation determines the symmetric part of connection, and since Levi-Civita connection is by definition symmetric, it is always completely determined by this equation -given the metric-. If the connection isn't symmetric, then there is torsion present, and torsion influences connection, and like I wrote in the previous reply, any connection can be represented by a sum of Levi-Civita connection and torsion part(that is the contorsion tensor).

Obviously, if you somehow had torsion defined, and full connection defined, then in principle, knowing those two would be the same as knowing the Levi-Civita connection, so your question then becomes, if you know the Levi-Civita connection components, can you determine the metric? Sure, it's integrating a system of linear PDEs, but in practice you never have that. Sometimes that system is difficult, sometimes not so difficult, but your question then becomes, whether you can solve a linear system of PDEs. You can consult PDE theory about that. If you have both torsion, and the full connection(both symmetric and anti-symmetric part), that is metric compatible, you can solve for the metric.

In practice you never do that. You have field equations which are Einstein equations which are solved for the metric, but in that case the connection is already Levi-Civita. If torsion is present, then instead of one tensor equation(which is the Einstein equations), you also have another equation that determines torsion. That system is much more difficult to solve, as you can imagine, but by solving it, you find both connection and the metric, which are metric compatible(or torsion and metric, as I mentioned earlier).

 

[strangerep]

From metric compatibility equation $$\nabla_\lambda g_{\mu\nu}=0$$ we can calculate a Levi-Civita connection,

Not quite. You can calculate the symmetric part of the connection. (Did you attempt the exercise I suggested? I guess not, but you should. The calculation is... rather basic, alas.)

The Levi-Civita connection is
$$\frac12\, g^{\lambda\sigma} \Big( g_{\sigma\mu},_\nu + g_{\nu\sigma},_\mu
- g_{\mu\nu},_\sigma \Big) ~.$$ However, symmetric part of the most general metric-compatible connection is $$
\Gamma^\lambda_{~(\mu\nu)}
~=~ \frac12\, g^{\lambda\sigma} \Big( g_{\sigma\mu},_\nu + g_{\nu\sigma},_\mu
- g_{\mu\nu},_\sigma \Big)
~+ g^{\lambda\sigma} g_{\mu\alpha} \Gamma^\alpha_{~[\sigma\nu]}
~+ g^{\lambda\sigma} g_{\alpha\nu} \Gamma^\alpha_{~[\sigma\mu]} ~.$$ The torsion $\Gamma^\lambda_{~[\mu\nu]}$ is not determined by metric compatibility.

for an arbitrary connection we always can find a metric and if we substitude the two into the equation.

That would be an overdetermined system of partial differential equations (i.e., many more equations than unknowns). That system might, or might not, have a solution.

 

[Jianbing_Shao]

any connection can be represented by a sum of Levi-Civita connection and torsion part(that is the contorsion tensor).

Obviously, if you somehow had torsion defined, and full connection defined, then in principle, knowing those two would be the same as knowing the Levi-Civita connection, so your question then becomes, if you know the Levi-Civita connection components, can you determine the metric? Sure,

Thanks for your explanation.
I think my question perhaps can be expressed in such a way: Are all connections metric comatible? If I correctly understand your opinion, the conclusion is uncertain,

My discussion is based on the theory about PDEs in mathematics. If we start from a vector field $V(x)$, we can get a connection using equations:
$$
\partial_\mu V(x)=A_\mu(x)V(x)
$$
But from the equations above, we also can express vector using connection. In Sean carroll's book, he uses parallel propagrator $P(x.x_0)$(path ordered product) which is generated with $A_\mu(x)$ to describe the parallel transport of a vector according to the equation.

If given an arbitrary connection. the solution of the equation is $$V(x)=P(x.x_0)V(x_0)=P\exp\left(\int_x^{x_0}A^\mu dx^\mu\right)V(x_0)$$
and we can find that only if the $R(A)$ is zero, then $P(x.x_0)$ is path independent. so we can find a vector field can satisfy the equations. if $R(A)$ is not zero, then the equation only can describe a vector moving along a particular path. it is impossible to find a vector field which can satisfy the equation above,

Similarly we also can use parallel propagator to describe the metric from the metric compatibily equations. but the problem is if the curvature of the connection is not zero, the parallel propagrator we get is path dependent, then how to get a path independent metric using a path dependent parallel propagrator. In most cases I'm sure it is impossible. if it is possible then the connection should meet some constraints.

I'm not sure if the constraints I'm talking about is the same as you mentioned.

 

[haushofer]

A "metric-compatible connection" satisfies $$\nabla_{\!\lambda\,} g_{\mu\nu} ~=~ 0 ~.$$This equation is sufficient to determine the symmetric components of the connection. (Count the number of possible index values in the equation, and in the symmetric part of the connection.)

Try doing the usual computation by permutation of indices -- you'll find that the symmetric part of the connection contains extra terms involving the skewsymmetric part of connection. However, the skew part remains unspecified by this procedure. Something else must be introduced to specify the skew part.

In the gauge theory of the Poincare algebra this "something else" is the field strength of the spatial translations. Putting this to zero enables you to (1) solve for the spin connection and (2) to remove the local translations in the algebra in favour of general coordinate transformations. (See e.g. https://www.physicsforums.com/insights/general-relativity-gauge-theory/ ; maybe this adds something for TS)

 

[renormalize]

Similarly we also can use parallel propagator to describe the metric from the metric compatibily equations. but the problem is if the curvature of the connection is not zero, the parallel propagrator we get is path dependent, then how to get a path independent metric using a path dependent parallel propagrator. In most cases I'm sure it is impossible. if it is possible then the connection should meet some constraints.

Actually, there are no constraints. In any 4D spacetime with a symmetric metric $g$, the most general connection $\Gamma$ decomposes in a coordinate basis as:$$\Gamma{}^{\alpha}{}_{\mu\nu}=\left\{ _{\mu\nu}^{\alpha}\right\} +\frac{1}{2}\left(T^{\alpha}{}_{\mu\nu}+T_{\mu}{}^{\alpha}{}_{\nu}+T_{\nu}{}^{\alpha}{}_{\mu}\right)+\frac{1}{2}\left(Q^{\alpha}{}_{\mu\nu}-Q_{\mu}{}^{\alpha}{}_{\nu}-Q_{\nu}{}^{\alpha}{}_{\mu}\right)\qquad\text{(General connection; 64 components)}$$with:$$\left\{ _{\mu\nu}^{\alpha}\right\} \equiv\frac{1}{2}g^{\alpha\lambda}\left(g_{\lambda\mu,\nu}+g_{\lambda\nu,\mu}-g_{\mu\nu,\lambda}\right)\qquad\text{(Levi-Civita/Christoffel})$$and where:$$g_{\mu\nu}=g_{\nu\mu}\qquad\text{(Metric tensor; 10 components)}$$$$T^{\alpha}{}_{\mu\nu}\equiv\Gamma{}^{\alpha}{}_{\mu\nu}-\Gamma{}^{\alpha}{}_{\nu\mu}\qquad\text{(Torsion tensor; 24 components)}$$$$Q_{\alpha\mu\nu}\equiv\nabla_{\alpha}g_{\mu\nu}\qquad\text{(Nonmetricity tensor; 40 components)}$$Note that all 74 combined components of the metric and general connection $\left\{ g,\Gamma\right\}$ are expressed entirely in terms of the 3 tensor fields $\left\{ g,T,Q\right\}$. These fields may in general be freely specified (as long as the metric remains invertible) without encountering any inconsistencies or constraints to be satisfied. Of course, torsion and nonmetricity do lead to geometric differences compared to Riemannian spacetimes: parallelograms don't close and and vector lengths change under parallel transport:

(from https://www.mdpi.com/2218-1997/5/7/173/pdf )
But these differences can be useful for physics: one important historical example is the invention of gauge theory by Weyl in 1918 that is founded on nonmetricity, where he made $Q_{\alpha\mu\nu}=A_{\alpha}g_{\mu\nu}$ a function of the the electromagnetic vector potential $A_{\alpha}$.

 

[Jianbing_Shao]

Actually, there are no constraints. In any 4D spacetime with a symmetric metric $g$, the most general connection $\Gamma$ decomposes in a coordinate basis as:$$\Gamma{}^{\alpha}{}_{\mu\nu}=\left\{ _{\mu\nu}^{\alpha}\right\} +\frac{1}{2}\left(T^{\alpha}{}_{\mu\nu}+T_{\mu}{}^{\alpha}{}_{\nu}+T_{\nu}{}^{\alpha}{}_{\mu}\right)+\frac{1}{2}\left(Q^{\alpha}{}_{\mu\nu}-Q_{\mu}{}^{\alpha}{}_{\nu}-Q_{\nu}{}^{\alpha}{}_{\mu}\right)\qquad\text{(General connection; 64 components)}$$with:$$\left\{ _{\mu\nu}^{\alpha}\right\} \equiv\frac{1}{2}g^{\alpha\lambda}\left(g_{\lambda\mu,\nu}+g_{\lambda\nu,\mu}-g_{\mu\nu,\lambda}\right)\qquad\text{(Levi-Civita/Christoffel})$$and where:$$g_{\mu\nu}=g_{\nu\mu}\qquad\text{(Metric tensor; 10 components)}$$$$T^{\alpha}{}_{\mu\nu}\equiv\Gamma{}^{\alpha}{}_{\mu\nu}-\Gamma{}^{\alpha}{}_{\nu\mu}\qquad\text{(Torsion tensor; 24 components)}$$$$Q_{\alpha\mu\nu}\equiv\nabla_{\alpha}g_{\mu\nu}\qquad\text{(Nonmetricity tensor; 40 components)}$$Note that all 74 combined components of the metric and general connection $\left\{ g,\Gamma\right\}$ are expressed entirely in terms of the 3 tensor fields $\left\{ g,T,Q\right\}$. These fields may in general be freely specified (as long as the metric remains invertible) without encountering any inconsistencies or constraints to be satisfied. Of course, torsion and nonmetricity do lead to geometric differences compared to Riemannian spacetimes: parallelograms don't close and and vector lengths change under parallel transport:
View attachment 324962
(from https://www.mdpi.com/2218-1997/5/7/173/pdf )
But these differences can be useful for physics: one important historical example is the invention of gauge theory by Weyl in 1918 that is founded on nonmetricity, where he made $Q_{\alpha\mu\nu}=A_{\alpha}g_{\mu\nu}$ a function of the the electromagnetic vector potential $A_{\alpha}$.

Thanks a lot!
It is very interesting and I still have some questions here:
(1); In QFT, the connection can be expressed as: $\Gamma_\mu=A_\mu^i T^i$, So the decomposition of connection has any relationship with the classifying the connection with the generators of Lie group.

(2). You say 'Actually, there are no constraints'. Does it means that if we input an arbitrary connection into the metric compatibility equation, then will output a metric field.
then if the input is the general connection $\Gamma$, ouput should be $g_{\nu\mu}$.
if the input is $\left\{ _{\mu\nu}^{\alpha}\right\}$, then output $g'_{\nu\mu}$
if the input is $T^{\alpha}_{\mu\nu}$, then output $g''_{\nu\mu}$
if the input is $\left\{ _{\mu\nu}^{\alpha}\right\}+T^{\alpha}_{\mu\nu}$ ,then output is $g'''_{\nu\mu}$......
We can input different elements into the equation, then what is the difference between those output.

 

[Antarres]

From what @renormalize wrote in his post #15 you can see that a general connection depends on three tensors, metric, torsion and non-metricity. This connection is not a solution of metricity equation in general. Connection is a structure imposed on the manifold that in principle is imposed independently of the metric.

If you ask what kind of a metric you get if you define a connection using the metricity condition, the answer is, you find a metric that is compatible to this connection. In principle, a general connection does not satisfy the metricity condition. If you set in those equation that $Q_{\alpha\mu\nu} = 0$ you find a connection that is metric-compatible. Metric-compatibility is a condition that you impose on the metric and connection, it is not an equation that defines metric or connection.

If you impose metricity condition, that is, $\nabla_{\mu}g_{\nu\rho} = 0$, then you ask yourself, how do you get the metric from there. The point is that, in general, you don't get the metric from there. In order to get the metric, you also have to define the torsion tensor, because with only metricity in place, metric and torsion are two entities that independently govern the connection. Or you can reverse it and say that torsion and connection govern the metric. But connection by itself doesn't.

 

[jbergman]

From what @renormalize wrote in his post #15 you can see that a general connection depends on three tensors, metric, torsion and non-metricity. This connection is not a solution of metricity equation in general. Connection is a structure imposed on the manifold that in principle is imposed independently of the metric.

If you ask what kind of a metric you get if you define a connection using the metricity condition, the answer is, you find a metric that is compatible to this connection. In principle, a general connection does not satisfy the metricity condition. If you set in those equation that $Q_{\alpha\mu\nu} = 0$ you find a connection that is metric-compatible. Metric-compatibility is a condition that you impose on the metric and connection, it is not an equation that defines metric or connection.

If you impose metricity condition, that is, $\nabla_{\mu}g_{\nu\rho} = 0$, then you ask yourself, how do you get the metric from there. The point is that, in general, you don't get the metric from there. In order to get the metric, you also have to define the torsion tensor, because with only metricity in place, metric and torsion are two entities that independently govern the connection. Or you can reverse it and say that torsion and connection govern the metric. But connection by itself doesn't.

Another way to say this is that a connection and a metric on a manifold are two different types of structures. The metric computes the inner product of tangent vectors and a connection defines how you parallel transport vectors.

You can define a connection on manifold without a metric and vice versa. But, it's a nice property if you have defined both that when you parallel transport two tangent vectors, their inner product remains the same, but it isn't mandatory.

 

[renormalize]

...if the input is the general connection $\Gamma$, ouput should be $g_{\nu\mu}$.
if the input is $\left\{ _{\mu\nu}^{\alpha}\right\}$, then output $g'_{\nu\mu}$
if the input is $T^{\alpha}_{\mu\nu}$, then output $g''_{\nu\mu}$
if the input is $\left\{ _{\mu\nu}^{\alpha}\right\}+T^{\alpha}_{\mu\nu}$ ,then output is $g'''_{\nu\mu}$......
We can input different elements into the equation, then what is the difference between those output.

Getting the metric from the general connection requires integrating an involved system of first-order partial differential equations. But differentiation is much easier than integration! So instead, go the in opposite direction and consider the general connection $\Gamma$ to be output from the given input tensors $\left\{ g,T,Q\right\}$:$$\left\{ g,T,Q\right\} \Rightarrow\Gamma,\quad\left\{ g,0,0\right\} \Rightarrow\Gamma',\quad\left\{ g,T,0\right\} \Rightarrow\Gamma'',\quad\left\{ g,0,Q\right\} \Rightarrow\Gamma''',\quad\text{etc.}$$The differences between the various output connections can then be simply read-off from the first equation in post #15. Also note that we are free to set $g$ to be the flat metric $\eta$ and build connections strictly from the torsion $T$ and/or nonmetricity $Q$. Such connections give rise to various teleparallel (but flat) analogs of curved Riemannian spacetime.

 

[haushofer]

You can define a connection on manifold without a metric and vice versa. But, it's a nice property if you have defined both that when you parallel transport two tangent vectors, their inner product remains the same, but it isn't mandatory.

In my experience, textbooks on GR usually don't spend too much time on why you'd want on physical grounds that a connection is metric compatible.

 

[Jianbing_Shao]

Getting the metric from the general connection requires integrating an involved system of first-order partial differential equations. But differentiation is much easier than integration! So instead, go the in opposite direction and consider the general connection $\Gamma$ to be output from the given input tensors $\left\{ g,T,Q\right\}$:$$\left\{ g,T,Q\right\} \Rightarrow\Gamma,\quad\left\{ g,0,0\right\} \Rightarrow\Gamma',\quad\left\{ g,T,0\right\} \Rightarrow\Gamma'',\quad\left\{ g,0,Q\right\} \Rightarrow\Gamma''',\quad\text{etc.}$$The differences between the various output connections can then be simply read-off from the first equation in post #15. Also note that we are free to set $g$ to be the flat metric $\eta$ and build connections strictly from the torsion $T$ and/or nonmetricity $Q$. Such connections give rise to various teleparallel (but flat) analogs of curved Riemannian spacetime.

Thanks a lot:
In most cases differentiation is easier than integration, but in cases such as $expA(x)$($A(x)$ is a matrix function ), differentiation is also very complicate and in fact we probably can not get an exact result.

Although parallel propagator (it can be regard to be a integration of connection) is very complicate, but at first the differentiation of parallel propatator is very simple, although it is a product of exponents of matrix.

From parallel propagator we also can get property of path dependent.if the curvature of connection is not zero. just like the integral of a vector field whose curl is not zero will be path dependent. I think it is just why the reason why not all connections are metric compatible.

At last the parallel propagator of $P\exp\left(\int_x^{x_0}\left\{ _{\mu\nu}^{\alpha}\right\} dx^\mu\right)$ and $P\exp\left(\int_x^{x_0}\left(\left\{ _{\mu\nu}^{\alpha}\right\}+T^{\alpha}_{\mu\nu}\right) dx^\mu\right)$ is very complicate.
usually we can't find out the exact relationship between them.

So I think the integrational method perhaps is very important , and we also can use it to prove your result

 

[Jianbing_Shao]

In my experience, textbooks on GR usually don't spend too much time on why you'd want on physical grounds that a connection is metric compatible.

thanks a ltot.
Yes, but perhaps if we figure this out, I think it may help us to understand GR more clearly.

 

[Jianbing_Shao]

Another way to say this is that a connection and a metric on a manifold are two different types of structures. The metric computes the inner product of tangent vectors and a connection defines how you parallel transport vectors.

You can define a connection on manifold without a metric and vice versa. But, it's a nice property if you have defined both that when you parallel transport two tangent vectors, their inner product remains the same, but it isn't mandatory.

thanks a lot
Metric and connection are two different structures, and if we make it clear to what extent the two are connected from the metric compatibility equation I think perhaps it is very interesting.

 

[haushofer]

thanks a ltot.
Yes, but perhaps if we figure this out, I think it may help us to understand GR more clearly.

Two reasons I can think of:

- $\nabla_{\rho} g_{\mu\nu}=0$ is the covariantization of the condition $\partial_{\rho} \eta_{\mu\nu}=0$ on the Minkowski metric in Cartesian coordinates.

- Physical observables depend on inner products, e.g. the energy of a particle. Without metric compatibility, these observables would depend on the specific path (the "history") taken, which in a lot of cases is not desirable.

 

[romsofia]

Section 4 of this paper might be useful: https://arxiv.org/pdf/1606.08756.pdf

 

[robphy]

thanks a ltot.
Yes, but perhaps if we figure this out, I think it may help us to understand GR more clearly.

This may be of interest:
an axiomatic approach
e.g.
"The geometry of free fall and light propagation" by Ehlers, Pirani, and Schild "EPS"
(editorial note by Trautman: https://link.springer.com/article/10.1007/s10714-012-1352-5 )

https://www.google.com/search?q=ehlers+pirani+schild

 

[Jianbing_Shao]

Two reasons I can think of:

- $\nabla_{\rho} g_{\mu\nu}=0$ is the covariantization of the condition $\partial_{\rho} \eta_{\mu\nu}=0$ on the Minkowski metric in Cartesian coordinates.

- Physical observables depend on inner products, e.g. the energy of a particle. Without metric compatibility, these observables would depend on the specific path (the "history") taken, which in a lot of cases is not desirable.

Thanks a lot:
Yes, if ’observables would depend on the specific path (the "history") taken, which in a lot of cases is not desirable‘。The property of depending on the specific path is path dependent.

As a scalar, if it's movement is path dependent, it perhaps a result of vector field with non-zero curl. If a vector's movement is path dependent, it depends on a connection field whose curvature is not zero.

And I also think that the metric compatibility eliminate some path dependence of a vector's movement in the space. So I think the metric compatibility is a constraint on the connection.
but I don't know how.

 

[Jianbing_Shao]

Getting the metric from the general connection requires integrating an involved system of first-order partial differential equations. But differentiation is much easier than integration! So instead, go the in opposite direction and consider the general connection $\Gamma$ to be output from the given input tensors $\left\{ g,T,Q\right\}$:$$\left\{ g,T,Q\right\} \Rightarrow\Gamma,\quad\left\{ g,0,0\right\} \Rightarrow\Gamma',\quad\left\{ g,T,0\right\} \Rightarrow\Gamma'',\quad\left\{ g,0,Q\right\} \Rightarrow\Gamma''',\quad\text{etc.}$$The differences between the various output connections can then be simply read-off from the first equation in post #15. Also note that we are free to set $g$ to be the flat metric $\eta$ and build connections strictly from the torsion $T$ and/or nonmetricity $Q$. Such connections give rise to various teleparallel (but flat) analogs of curved Riemannian spacetime.

I still have a question.
In GR, if we start from a flat metric $\eta_{\mu\nu}$, we will get a zero connection, then a zero curvature. But if we start from a zero-curvature connection. then it is not necessary to demand that the connection must be zero, then if we using such connection, from the metric compatibility equation, then the metric compatible to the connection usually is not $\eta_{\mu\nu}$, then I want to know that from a zero-curvature connection, what are the three parts of the general connection. and if the corresponding metric must be $\eta_{\mu\nu}$.

 

[renormalize]

In GR, if we start from a flat metric $\eta_{\mu\nu}$, we will get a zero connection, then a zero curvature. But if we start from a zero-curvature connection. then it is not necessary to demand that the connection must be zero, then if we using such connection, from the metric compatibility equation, then the metric compatible to the connection usually is not $\eta_{\mu\nu}$, then I want to know that from a zero-curvature connection, what are the three parts of the general connection. and if the corresponding metric must be $\eta_{\mu\nu}$.

Starting from the flat spacetime metric $\eta{_{\mu\nu}}$ (in cartesian coordinates) gives a zero Levi-Civita/Christoffel symbol, $\left\{ _{\mu\nu}^{\alpha}\right\}=0$, resulting in vanishing curvature $R^{\alpha}{}_{\beta\mu\nu}=0$. But even when spacetime is flat it can still carry a general connection $\Gamma$ containing torsion $T$ and/or nonmetricity $Q$ (see again the first equation of post #15). So the most general connection on flat spacetime in cartesian coordinates will be $\Gamma{}^{\alpha}{}_{\mu\nu}=\frac{1}{2}\left(T^{\alpha}{}_{\mu\nu}+T_{\mu}{}^{\alpha}{}_{\nu}+T_{\nu}{}^{\alpha}{}_{\mu}\right)+\frac{1}{2}\left(Q^{\alpha}{}_{\mu\nu}-Q_{\mu}{}^{\alpha}{}_{\nu}-Q_{\nu}{}^{\alpha}{}_{\mu}\right)$. And the flat metric $\eta$ is perfectly consistent with this general connection $\Gamma$ by its very construction, with $\nabla_{\alpha}\eta_{\mu\nu}=Q_{\alpha\mu\nu}$ (metric-noncompatible for $Q\neq0$, metric-compatible otherwise). So to answer your question: yes, in flat spacetime with a general connection $\Gamma$, the corresponding metric must be $\eta{_{\mu\nu}}$. How could it be otherwise? Flat spacetime is equivalent to a flat metric!

 

[Jianbing_Shao]

Starting from the flat spacetime metric $\eta{_{\mu\nu}}$ (in cartesian coordinates) gives a zero Levi-Civita/Christoffel symbol, $\left\{ _{\mu\nu}^{\alpha}\right\}=0$, resulting in vanishing curvature $R^{\alpha}{}_{\beta\mu\nu}=0$. But even when spacetime is flat it can still carry a general connection $\Gamma$ containing torsion $T$ and/or nonmetricity $Q$ (see again the first equation of post #15). So the most general connection on flat spacetime in cartesian coordinates will be $\Gamma{}^{\alpha}{}_{\mu\nu}=\frac{1}{2}\left(T^{\alpha}{}_{\mu\nu}+T_{\mu}{}^{\alpha}{}_{\nu}+T_{\nu}{}^{\alpha}{}_{\mu}\right)+\frac{1}{2}\left(Q^{\alpha}{}_{\mu\nu}-Q_{\mu}{}^{\alpha}{}_{\nu}-Q_{\nu}{}^{\alpha}{}_{\mu}\right)$. And the flat metric $\eta$ is perfectly consistent with this general connection $\Gamma$ by its very construction, with $\nabla_{\alpha}\eta_{\mu\nu}=Q_{\alpha\mu\nu}$ (metric-noncompatible for $Q\neq0$, metric-compatible otherwise). So to answer your question: yes, in flat spacetime with a general connection $\Gamma$, the corresponding metric must be $\eta{_{\mu\nu}}$. How could it be otherwise? Flat spacetime is equivalent to a flat metric!

Thanks for your reply:
Perhaps there is some difference about how to comprehend metric compatible. In my opinion, metric $g$ is compatible with $\Gamma$ just means that if we put $g$ and $\Gamma$ into metric compatibility equation. the equation is true.
So I think that it is possilbe that there exist a zero-curvature connection and a metric which is not $\eta$ can make metric compatibility equation work.

If I don't misunderstand your opinion. a general connection with zero Levi-Civita connection is compatible with flat metric $\eta$, To me this means that if we put the general connection and $\eta$ into the metric compatibility equation, the equation is true. Can you prove it?

 

[vanhees71]

I still have a question.
In GR, if we start from a flat metric $\eta_{\mu\nu}$, we will get a zero connection, then a zero curvature. But if we start from a zero-curvature connection. then it is not necessary to demand that the connection must be zero, then if we using such connection, from the metric compatibility equation, then the metric compatible to the connection usually is not $\eta_{\mu\nu}$, then I want to know that from a zero-curvature connection, what are the three parts of the general connection. and if the corresponding metric must be $\eta_{\mu\nu}$.

GR is based on the assumption that locally the laws of SR are valid, i.e., at each spacetime point you can always introduce a reference frame, such that the tangent space at this spacetime point is described as a Minkowski space. Together with the assumption of no torsion you necessarily get a pseudo-Riemannian spacetime manifold with Lorentzian signature of the fundamental form (i.e., (1,3) or (3,1) depending on your sign convention for the form).

 

[Jianbing_Shao]

GR is based on the assumption that locally the laws of SR are valid, i.e., at each spacetime point you can always introduce a reference frame, such that the tangent space at this spacetime point is described as a Minkowski space. Together with the assumption of no torsion you necessarily get a pseudo-Riemannian spacetime manifold with Lorentzian signature of the fundamental form (i.e., (1,3) or (3,1) depending on your sign convention for the form).

Thanks a lot:
Here my discussion just based on metric compatibility equation. I do not want to deny that if we put the flat metric $\eta$ and zero Levi-Civita connection into metric compatibility equation and the equation is true. but perhaps we can not negelect the possibilities that there can exist a non flat metric $g$ and a general connection whose curvature(I'm not sure if it is appropriate because in GR curvature only defined with Levi-Civita connection. ) is zero. In fact, in the discussion about differential equations in mathematics. such possibility do exist. and the non-zero curvature also indicate the property of path dependent.

Another possibility is just that for some general connections we even can not find any metric field who can make the equation work. so this is just why I say perhaps not all connections are metric compatible.

 

[vanhees71]

Standard GR simply assumes that spacetime is a pseudo-Riemannian manifold with the uniquely defined torsion-free metric-compatible connection.

If you just "gauge" the Poincare invariance of Minkowski space, you are lead to Einstein-Cartan theory, i.e., a spacetime manifold with torsion as soon as you want to have matter fields with spin. For scalar fields + em. fields (U(1) gauge fields) the torsion turns out to be 0 in this approach, and you get standard GR. For a good introductory overview on this approach, see

P. Ramond, Field Theory: A Modern Primer,
Addison-Wesley, Redwood City, Calif., 2 edn. (1989).

 

[Jianbing_Shao]

Standard GR simply assumes that spacetime is a pseudo-Riemannian manifold with the uniquely defined torsion-free metric-compatible connection.

If you just "gauge" the Poincare invariance of Minkowski space, you are lead to Einstein-Cartan theory, i.e., a spacetime manifold with torsion as soon as you want to have matter fields with spin. For scalar fields + em. fields (U(1) gauge fields) the torsion turns out to be 0 in this approach, and you get standard GR. For a good introductory overview on this approach, see

P. Ramond, Field Theory: A Modern Primer,
Addison-Wesley, Redwood City, Calif., 2 edn. (1989).

Thanks!
From metric compatibility equation, we can use parallel propagator(defined in Sean Carroll' book) to express the metric with connection.
Then the difference between a metric compatible with a torsion free connection and the metric compatible with the same torsion free connection plus a torsion. in other words if the torsion part have effects on the metric compatible with the connection.

My original problem is just to find out that if an arbitrary connection have a metric compatible with it then we put them into the equation can make the equation work and if such connections do exist then why.

 

[renormalize]

To me this means that if we put the general connection and $\eta$ into the metric compatibility equation, the equation is true. Can you prove it?

Sorry, but I am unclear on what you are asking for mathematically. Rather than guess, can you please post the tensor equation (or equations) that you need to see proved?

 

[jbergman]

Thanks!
From metric compatibility equation, we can use parallel propagator(defined in Sean Carroll' book) to express the metric with connection.
Then the difference between a metric compatible with a torsion free connection and the metric compatible with the same torsion free connection plus a torsion. in other words if the torsion part have effects on the metric compatible with the connection.

My original problem is just to find out that if an arbitrary connection have a metric compatible with it then we put them into the equation can make the equation work and if such connections do exist then why.

So just so I understand, you want to show the following, given an arbitrary connection, does there exist a metric compatible with it and what does it look like? You don't care about the signature of the resulting metric, whether it is lorentzian or Riemannian?

 

[Jianbing_Shao]

Sorry, but I am unclear on what you are asking for mathematically. Rather than guess, can you please post the tensor equation (or equations) that you need to see proved?

Thanks:
So the question is : from the metric compatibility equation
$$\nabla_\rho g_{\mu\nu}=g_{\mu\nu,\rho}-g_{\lambda \nu}\gamma^\lambda_{\rho\mu}-g_{\mu\lambda}\gamma^\lambda_{\rho\nu}=0$$
If we choose an arbitrary connection $\gamma^\lambda_{\rho\mu}(x)$, then if we always can find a metric field $g_{\mu\nu}(x)$ which can make this equation work.

 

[Jianbing_Shao]

So just so I understand, you want to show the following, given an arbitrary connection, does there exist a metric compatible with it and what does it look like? You don't care about the signature of the resulting metric, whether it is lorentzian or Riemannian?

Thanks!
Yes it is just what I want to say:
Just because I from the discussion about partial differential equations, I find that in most cases, a global solution does not exist, we only can get a solution confined on a curve. So I doubt that to metric compatibility equation, we will get a similar result.
So if there exist some connections can not compatible with a metric field, then what is the difference between these connections and those compatible with metric.

 

[robphy]

You might find the papers of Graham Hall (U. Aberdeen) interesting
https://www.abdn.ac.uk/ncs/profiles/g.hall#research
https://scholar.google.com/scholar?q=graham+hall++connection+curvature+metric&btnG=
For example,
https://arxiv.org/abs/gr-qc/0509067

On the compatibility of Lorentz-metrics with linear connections on 4-dimensional manifolds

G.S. Hall, D.P. Lonie
This paper considers 4-dimensional manifolds upon which there is a Lorentz metric, h, and a symmetric connection and which are originally assumed unrelated. It then derives sufficient conditions on the metric and connection (expressed through the curvature tensor) for the connection to be the Levi-Civita connection of some (local) Lorentz metric, g, and calculates the relationship between g and h. Some examples are provided which help to assess the strength of the sufficient conditions derived.

 

[renormalize]

So the question is : from the metric compatibility equation
$$\nabla_\rho g_{\mu\nu}=g_{\mu\nu,\rho}-g_{\lambda \nu}\gamma^\lambda_{\rho\mu}-g_{\mu\lambda}\gamma^\lambda_{\rho\nu}=0$$
If we choose an arbitrary connection $\gamma^\lambda_{\rho\mu}(x)$, then if we always can find a metric field $g_{\mu\nu}(x)$ which can make this equation work.

Yes, you can always make this equation work!

Using your metric-compatibility equation written in all-covariant (lower) indices, I begin by cyclically permuting those indices to get 3 versions of the same equation:$$0=g_{\mu\nu,\rho}-\gamma{}_{\nu\rho\mu}-\gamma{}_{\mu\rho\nu}\:,\qquad0=g_{\rho\mu,\nu}-\gamma{}_{\mu\nu\rho}-\gamma{}_{\rho\nu\mu}\:,\qquad0=g_{\nu\rho,\mu}-\gamma{}_{\rho\mu\nu}-\gamma{}_{\nu\mu\rho}\tag{1a,b,c}$$From these, I can form the linear-combination (eq.(1b)+eq.(1c)-eq.(1a))/2 and proceed to solve for the unique symmetric-piece of the general metric-compatible connection $\gamma$:$$0=\frac{1}{2}\left(g_{\rho\mu,\nu}+g_{\nu\rho,\mu}-g_{\mu\nu,\rho}\right)-\frac{1}{2}\left(\gamma{}_{\mu\nu\rho}+\gamma{}_{\rho\mu\nu}-\gamma{}_{\nu\rho\mu}\right)-\frac{1}{2}\left(\gamma{}_{\rho\nu\mu}+\gamma{}_{\nu\mu\rho}-\gamma{}_{\mu\rho\nu}\right)$$$$=\left\{ \rho,\mu\nu\right\} -\frac{1}{2}\left(\gamma{}_{\rho\mu\nu}+\gamma{}_{\rho\nu\mu}\right)+\frac{1}{2}\left(\gamma{}_{\mu\rho\nu}-\gamma{}_{\mu\nu\rho}\right)+\frac{1}{2}\left(\gamma{}_{\nu\rho\mu}-\gamma{}_{\nu\mu\rho}\right)$$$$=\left\{ \rho,\mu\nu\right\} -\gamma{}_{\rho\{\mu\nu\}}+\frac{1}{2}\left(T_{\mu\rho\nu}+T_{\nu\rho\mu}\right)$$or:$$\gamma{}_{\rho\{\mu\nu\}}=\left\{ \rho,\mu\nu\right\} +\frac{1}{2}\left(T_{\mu\rho\nu}+T_{\nu\rho\mu}\right)\tag{2}$$Adding the identity $\gamma{}_{\rho[\mu\nu]}\equiv\frac{1}{2}T_{\rho\mu\nu}$ to eq.(2) gives finally the most general metric-compatible connection:$$\gamma{}_{\rho\mu\nu}=\left\{ \rho,\mu\nu\right\} +\frac{1}{2}\left(T_{\rho\mu\nu}+T_{\mu\rho\nu}+T_{\nu\rho\mu}\right)\tag{3}$$(Note that this is the same as the first equation in post #15 for the case of vanishing nonmetricity $Q$.)

Since $\left\{g,T\right\}$ are already arbitrary fields, you cannot write down a more general metric-compatible connection than that given by eq.(3)! The point here is that, starting from your condition for metric-compatibility, we derive and exhibit in eq.(3) the unique form of the most general compatible connection $\gamma$, strictly using tensor algebra (no integration or differentiation!). That means there is no need to "find a metric field...that can make this work": it works for all metric fields $g$ and all torsion fields $T$ by construction. That's the meaning of the statement that one may arbitrarily specify both the connection $\gamma$ (or equivalently $\left\{T,Q\right\}$) and the metric $g$ on any spacetime, with no worry that this could somehow lead to an inconsistency or restrictive condition involving the metric and connection.

 

[jbergman]

Yes, you can always make this equation work!

Using your metric-compatibility equation written in all-covariant (lower) indices, I begin by cyclically permuting those indices to get 3 versions of the same equation:$$0=g_{\mu\nu,\rho}-\gamma{}_{\nu\rho\mu}-\gamma{}_{\mu\rho\nu}\:,\qquad0=g_{\rho\mu,\nu}-\gamma{}_{\mu\nu\rho}-\gamma{}_{\rho\nu\mu}\:,\qquad0=g_{\nu\rho,\mu}-\gamma{}_{\rho\mu\nu}-\gamma{}_{\nu\mu\rho}\tag{1a,b,c}$$From these, I can form the linear-combination (eq.(1b)+eq.(1c)-eq.(1a))/2 and proceed to solve for the unique symmetric-piece of the general metric-compatible connection $\gamma$:$$0=\frac{1}{2}\left(g_{\rho\mu,\nu}+g_{\nu\rho,\mu}-g_{\mu\nu,\rho}\right)-\frac{1}{2}\left(\gamma{}_{\mu\nu\rho}+\gamma{}_{\rho\mu\nu}-\gamma{}_{\nu\rho\mu}\right)-\frac{1}{2}\left(\gamma{}_{\rho\nu\mu}+\gamma{}_{\nu\mu\rho}-\gamma{}_{\mu\rho\nu}\right)$$$$=\left\{ \rho,\mu\nu\right\} -\frac{1}{2}\left(\gamma{}_{\rho\mu\nu}+\gamma{}_{\rho\nu\mu}\right)+\frac{1}{2}\left(\gamma{}_{\mu\rho\nu}-\gamma{}_{\mu\nu\rho}\right)+\frac{1}{2}\left(\gamma{}_{\nu\rho\mu}-\gamma{}_{\nu\mu\rho}\right)$$$$=\left\{ \rho,\mu\nu\right\} -\gamma{}_{\rho\{\mu\nu\}}+\frac{1}{2}\left(T_{\mu\rho\nu}+T_{\nu\rho\mu}\right)$$or:$$\gamma{}_{\rho\{\mu\nu\}}=\left\{ \rho,\mu\nu\right\} +\frac{1}{2}\left(T_{\mu\rho\nu}+T_{\nu\rho\mu}\right)\tag{2}$$Adding the identity $\gamma{}_{\rho[\mu\nu]}\equiv\frac{1}{2}T_{\rho\mu\nu}$ to eq.(2) gives finally the most general metric-compatible connection:$$\gamma{}_{\rho\mu\nu}=\left\{ \rho,\mu\nu\right\} +\frac{1}{2}\left(T_{\rho\mu\nu}+T_{\mu\rho\nu}+T_{\nu\rho\mu}\right)\tag{3}$$(Note that this is the same as the first equation in post #15 for the case of vanishing nonmetricity $Q$.)

Since $\left\{g,T\right\}$ are already arbitrary fields, you cannot write down a more general metric-compatible connection than that given by eq.(3)! The point here is that, starting from your condition for metric-compatibility, we derive and exhibit in eq.(3) the unique form of the most general compatible connection $\gamma$, strictly using tensor algebra (no integration or differentiation!). That means there is no need to "find a metric field...that can make this work": it works for all metric fields $g$ and all torsion fields $T$ by construction. That's the meaning of the statement that one may arbitrarily specify both the connection $\gamma$ (or equivalently $\left\{T,Q\right\}$) and the metric $g$ on any spacetime, with no worry that this could somehow lead to an inconsistency or restrictive condition involving the metric and connection.

It looks like you solved for $\gamma$ given $g$. I think he was asking to go the other way, i.e., solve for $g$ given $\gamma$.

 

[renormalize]

It looks like you solved for $\gamma$ given $g$. I think he was asking to go the other way, i.e., solve for $g$ given $\gamma$.

I understand that's what the OP would like to see. But they cannot hope to solve for the metric $g$ from the metric-compatibility condition $\nabla_{\rho}g_{\mu\nu}\equiv g_{\mu\nu,\rho}-g_{\lambda\nu}\gamma_{\rho\mu}^{\lambda}-g_{\mu\lambda}\gamma_{\rho\nu}^{\lambda}=0$ because, in spite of its appearance, it is not a partial differential equation for $g$!

I prove this by defining:$$\Delta{}^{\lambda}{}_{\mu\nu}\equiv\gamma{}^{\lambda}{}_{\mu\nu}-\left\{ _{\mu\nu}^{\lambda}\right\}$$ to be the difference between the general connection and the Christoffel symbol, so I can write the connection as:$$\gamma{}^{\lambda}{}_{\mu\nu}=\left\{ _{\mu\nu}^{\lambda}\right\} +\Delta{}^{\lambda}{}_{\mu\nu}$$This represents a fully-general decomposition of the connection since $\Delta$ remains completely arbitrary to this point. I now insert this decomposition into the metric-compatability condition, written in the form $g_{\mu\nu,\rho}=\gamma{}_{\nu\rho\mu}+\gamma{}_{\mu\rho\nu}$, to find:$$g_{\mu\nu,\rho}=\left\{ \nu,\rho\mu\right\} +\Delta_{\nu\rho\mu}+\left\{ \mu,\rho\nu\right\} +\Delta_{\mu\rho\nu}$$$$=\frac{1}{2}\left(g_{\nu\rho,\mu}+g_{\nu\mu,\rho}-g_{\rho\mu,\nu}\right)+\frac{1}{2}\left(g_{\mu\rho,\nu}+g_{\mu\nu,\rho}-g_{\rho\nu,\mu}\right)+\Delta_{\nu\rho\mu}+\Delta_{\mu\rho\nu}$$$$=g_{\mu\nu,\rho}+\Delta_{\nu\rho\mu}+\Delta_{\mu\rho\nu}$$That is:$$0=\Delta_{\nu\rho\mu}+\Delta_{\mu\rho\nu}\tag{1}$$Thus, all derivatives of $g$ disappear from the compatibility condition, and it collapses to just a symmetry condition for $\Delta$! By index manipulations analogous to those in post #40, the general solution of (1) is then:$$\Delta_{\rho\mu\nu}=\frac{1}{2}\left(T_{\rho\mu\nu}+T_{\mu\rho\nu}+T_{\nu\rho\mu}\right)$$where $T_{\rho\mu\nu}\equiv\Delta_{\rho\mu\nu}-\Delta_{\rho\nu\mu}$ is the torsion tensor.

Bottom line: the metric-compatibility condition is not a differential equation for the metric $g$. It is merely an algebraic condition imposed on the components of the difference $\Delta$ between the general connection $\gamma$ and the Christoffel symbol.

 

[Jianbing_Shao]

I understand that's what the OP would like to see. But they cannot hope to solve for the metric $g$ from the metric-compatibility condition $\nabla_{\rho}g_{\mu\nu}\equiv g_{\mu\nu,\rho}-g_{\lambda\nu}\gamma_{\rho\mu}^{\lambda}-g_{\mu\lambda}\gamma_{\rho\nu}^{\lambda}=0$ because, in spite of its appearance, it is not a partial differential equation for $g$!

Thanks a lot!
But I can not agree with your conclusion. Sean Carroll using parallel propagator $$P(x,x_0;l;\gamma)= P\exp\left(\int_x^{x_0}-\gamma^\nu_{\mu\rho}(x) dx^\rho\right)$$
to describe the parallel trsport of a vector along a particular path $l$. Simillarly we also can use parallel propagator $P(x,x_0;l;\gamma)$ generated with connection $\gamma^\nu_{\mu\rho}(x)$ to describe the metric $g$:
Because a metric can be difined with a basis $e$, $g=e e^T$. and the parallel transport of a basis $e(x_0)$ can be described using parallel propagator.
$$e(x)=P(x,x_0;l;\gamma)e(x_0)$$
so:
$$g(x)=P(x,x_0;l;\gamma)g(x_0)P^T(x,x_0;l;\gamma)$$
This expression is right because if you differentiate this formula, you can find that we can get metric compatibility equation.
So the problem here is that the parallel propagator $P(x,x_0;l;\gamma)$ is path dependent if the curvature of the connection $\gamma^\nu_{\mu\rho}(x)$ is not zero, then the movement of metric in most cases is also path dependent, this means that we probably can not get a metric field.
So just like the situations in a scalar field, you differentiate a scalar field only can get a zero curl vector field, when you integrate a non-zero curl vector field, you can find a property of path dependence so can not find a scalar field corresponding to a non-zero curl vector field. Only when we try to describe a scalar field using the integration of a vector field, then we can find the property of path dependence.

 

[Jianbing_Shao]

It looks like you solved for $\gamma$ given $g$. I think he was asking to go the other way, i.e., solve for $g$ given $\gamma$.

Thanks!
Here the logic is we can not get a field just result from path dependence, So we should express $g$ with $\gamma$ and prove that $g$ is path indenpendent for an arbitrary $\gamma$.

 

[jbergman]

Thanks a lot!
But I can not agree with your conclusion. Sean Carroll using parallel propagator $$P(x,x_0;l;\gamma)= P\exp\left(\int_x^{x_0}-\gamma^\nu_{\mu\rho}(x) dx^\rho\right)$$
to describe the parallel trsport of a vector along a particular path $l$. Simillarly we also can use parallel propagator $P(x,x_0;l;\gamma)$ generated with connection $\gamma^\nu_{\mu\rho}(x)$ to describe the metric $g$:
Because a metric can be difined with a basis $e$, $g=e e^T$. and the parallel transport of a basis $e(x_0)$ can be described using parallel propagator.
$$e(x)=P(x,x_0;l;\gamma)e(x_0)$$
so:
$$g(x)=P(x,x_0;l;\gamma)g(x_0)P^T(x,x_0;l;\gamma)$$
This expression is right because if you differentiate this formula, you can find that we can get metric compatibility equation.
So the problem here is that the parallel propagator $P(x,x_0;l;\gamma)$ is path dependent if the curvature of the connection $\gamma^\nu_{\mu\rho}(x)$ is not zero, then the movement of metric in most cases is also path dependent, this means that we probably can not get a metric field.
So just like the situations in a scalar field, you differentiate a scalar field only can get a zero curl vector field, when you integrate a non-zero curl vector field, you can find a property of path dependence so can not find a scalar field corresponding to a non-zero curl vector field. Only when we try to describe a scalar field using the integration of a vector field, then we can find the property of path dependence.

Thanks a lot!
But I can not agree with your conclusion. Sean Carroll using parallel propagator $$P(x,x_0;l;\gamma)= P\exp\left(\int_x^{x_0}-\gamma^\nu_{\mu\rho}(x) dx^\rho\right)$$
to describe the parallel trsport of a vector along a particular path $l$. Simillarly we also can use parallel propagator $P(x,x_0;l;\gamma)$ generated with connection $\gamma^\nu_{\mu\rho}(x)$ to describe the metric $g$:
Because a metric can be difined with a basis $e$, $g=e e^T$. and the parallel transport of a basis $e(x_0)$ can be described using parallel propagator.
$$e(x)=P(x,x_0;l;\gamma)e(x_0)$$
so:
$$g(x)=P(x,x_0;l;\gamma)g(x_0)P^T(x,x_0;l;\gamma)$$
This expression is right because if you differentiate this formula, you can find that we can get metric compatibility equation.
So the problem here is that the parallel propagator $P(x,x_0;l;\gamma)$ is path dependent if the curvature of the connection $\gamma^\nu_{\mu\rho}(x)$ is not zero, then the movement of metric in most cases is also path dependent, this means that we probably can not get a metric field.
So just like the situations in a scalar field, you differentiate a scalar field only can get a zero curl vector field, when you integrate a non-zero curl vector field, you can find a property of path dependence so can not find a scalar field corresponding to a non-zero curl vector field. Only when we try to describe a scalar field using the integration of a vector field, then we can find the property of path dependence.

I'm not sure this is a fruitful approach. We know that curvature in general is not an obstacle to defining a metric so building a metric with this propagator seems to only work in spaces without curvature.

 

[renormalize]

So the problem here is that the parallel propagator $P(x,x_0;l;\gamma)$ is path dependent if the curvature of the connection $\gamma^\nu_{\mu\rho}(x)$ is not zero, then the movement of metric in most cases is also path dependent, this means that we probably can not get a metric field.

Your assertion here is simply false. If a metric-compatible connection $\Gamma$ could somehow lead to a path-dependent metric tensor $g$, don't you think that fact would rate at least a mention in every textbook on general relativity? Have you made any attempt to calculate this yourself? It's not hard to do for infinitesimal transport, as I now demonstrate.

As you noted, parallel transport of a contravariant vector field $V^{\mu}$ along a spacetime curve $\mathcal{C}(\lambda)$ between the points $\lambda_{0}$ and $\lambda$ can be expressed as $$\overset{*}{V^{\mu}}\left(\lambda\right)=P^{\mu}{}_{\nu}\left(\lambda,\lambda_{0}\right)V^{\nu}\left(\lambda_{0}\right)\tag{1})$$where the "parallel propagator" $P$ is defined as the path-ordered exponential of the line-integral of minus the connection $\Gamma$ (see Carroll eq.(3.45) in 3.Curvature):$$P^{\mu}{}_{\nu}\left(\lambda,\lambda_{0}\right)\equiv\mathcal{P}\exp\left(-\intop_{\lambda_{0}}^{\lambda}\Gamma^{\mu}{}_{\sigma\nu}\left(\eta\right)\frac{dx^{\sigma}}{d\eta}d\eta\right)\tag{2}$$ I will focus on parallel transport along an infinitesimal displacement $dx$, for which eq.(2) reduces to:$$P^{\mu}{}_{\nu}\left(x+dx,x\right)=\delta_{\nu}^{\mu}-\Gamma^{\mu}{}_{\sigma\nu}\left(x\right)dx^{\sigma}+\mathcal{O}\left(dx^{2}\right)\tag{3}$$so that (1) becomes:$$ \overset{*}{V^{\mu}}\left(x+dx\right)=V^{\mu}\left(x\right)-\Gamma^{\mu}{}_{\sigma\nu}\left(x\right)V^{\nu}\left(x\right)dx^{\sigma}+\mathcal{O}\left(dx^{2}\right)\tag{4}$$Now compare this result to the Taylor expansion of the actual vector field $V$ at $x+dx$:$$ V^{\mu}\left(x+dx\right)=V^{\mu}\left(x\right)+\frac{\partial V^{\mu}\left(x\right)}{\partial x^{\sigma}}dx^{\sigma}+\mathcal{O}\left(dx^{2}\right)\tag{5}$$Subtracting (4) from (5) defines the difference $\delta_{P}V$ between the actual and the parallel-transported vector fields:$$ \delta_{P}V^{\mu}\left(x\right)\equiv V^{\mu}\left(x+dx\right)-\overset{*}{V^{\mu}}\left(x+dx\right)=\left(\frac{\partial V^{\mu}\left(x\right)}{\partial x^{\sigma}}+\Gamma^{\mu}{}_{\sigma\nu}\left(x\right)V^{\nu}\left(x\right)\right)dx^{\sigma}+\mathcal{O}\left(dx^{2}\right)\tag{6}$$or:$$ \delta_{P}V^{\mu}\left(x\right)=\nabla_{\sigma}V^{\mu}\left(x\right)dx^{\sigma}+\mathcal{O}\left(dx^{2}\right)\tag{7}$$This difference accumulates to $\Sigma_{i}\delta_{P_{i}}V$ as $V$ is parallel-transported along multiple, linked infinitesimal displacements. If those displacements form a closed path such as a parallelogram, it’s well-known that $\Sigma_{i}\delta_{P_{i}}V\propto RVdA$, where $R$ is the curvature tensor and $dA$ is the infinitesimal area enclosed by the path. Parallel transport of a vector $V$ between any two points in curved spacetime does indeed depend on the particular path linking those points. This fact evidently motivates your claim that the metric tensor itself is similarly path-dependent.

But that claim is provably wrong. The contravariant metric tensor $g$ parallel-transports according to $\overset{*}{g^{\mu\nu}}\left(\lambda\right)=P^{\mu}{}_{\alpha}\left(\lambda,\lambda_{0}\right)g^{\alpha\beta}\left(\lambda_{0}\right)P^{\nu}{}_{\beta}\left(\lambda,\lambda_{0}\right)$. Using (3), infinitesimal-transport of $g$ from $x$ to $x+dx$ therefore takes the form:$$\overset{*}{g^{\mu\nu}}\left(x+dx\right)=g^{\mu\nu}\left(x\right)-\Gamma^{\mu}{}_{\sigma\alpha}\left(x\right)g^{\alpha\nu}\left(x\right)dx^{\sigma}-\Gamma^{\nu}{}_{\sigma\alpha}\left(x\right)g^{\mu\alpha}\left(x\right)dx^{\sigma}+\mathcal{O}\left(dx^{2}\right)\tag{8}$$whereas the Taylor expansion of the actual metric tensor at $x+dx$ is:$$g^{\mu\nu}\left(x+dx\right)=g^{\mu\nu}\left(x\right)+\frac{\partial g^{\mu\nu}\left(x\right)}{\partial x^{\sigma}}dx^{\sigma}+\mathcal{O}\left(dx^{2}\right)\tag{9}$$Subtracting (8) from (9) gives $\delta_{P}g$:$$\delta_{P}g^{\mu\nu}\left(x\right)\equiv g^{\mu\nu}\left(x+dx\right)-\overset{*}{g^{\mu\nu}}\left(x+dx\right)=\left(\frac{\partial g^{\mu\nu}\left(x\right)}{\partial x^{\sigma}}+\Gamma^{\mu}{}_{\sigma\alpha}\left(x\right)g^{\alpha\nu}\left(x\right)+\Gamma^{\nu}{}_{\sigma\alpha}\left(x\right)g^{\mu\alpha}\left(x\right)\right)dx^{\sigma}+\mathcal{O}\left(dx^{2}\right)$$$$=\nabla_{\sigma}g^{\mu\nu}\left(x\right)dx^{\sigma}+\mathcal{O}\left(dx^{2}\right)=0+ \mathcal{O}\left(dx^{2}\right)\tag{10}$$ for a metric-compatible $\Gamma$. So to first-order in $dx$, $\delta_{P}g=0$, or equivalently: $$g^{\mu\nu}\left(x+dx\right)=\overset{*}{g^{\mu\nu}}\left(x+dx\right)\tag{11}$$That is, the actual metric $g$ and the transported metric $\overset{*}{g}$ are the same. When the connection is metric compatible, the metric tensor (unlike every other tensor field) parallel transports into itself under infinitesimal displacements! And around an infinitesimal parallelogram, $\Sigma_{i}\delta_{P_{i}}g=\Sigma_{i}0=0$ irrespective of the curvature $R$.

The upshot is: the metric $g$ is never path-dependent for an any connection $\Gamma$ that it is compatible with that metric ($\nabla_{\sigma}g^{\mu\nu}$=0).

 

[Jianbing_Shao]

Your assertion here is simply false. If a metric-compatible connection $\Gamma$ could somehow lead to a path-dependent metric tensor $g$, don't you think that fact would rate at least a mention in every textbook on general relativity? Have you made any attempt to calculate this yourself? It's not hard to do for infinitesimal transport, as I now demonstrate.

}\right)\tag{7}$$This difference accumulates to $\Sigma_{i}\delta_{P_{i}}V$ as $V$ is parallel-transported along multiple, linked infinitesimal displacements. If those displacements form a closed path such as a parallelogram, it’s well-known that $\Sigma_{i}\delta_{P_{i}}V\propto RVdA$, where $R$ is the curvature tensor and $dA$ is the infinitesimal area enclosed by the path. Parallel transport of a vector $V$ between any two points in curved spacetime does indeed depend on the particular path linking those points. This fact evidently motivates your claim that the metric tensor itself is similarly path-dependent.
The upshot is: the metric $g$ is never path-dependent for an any connection $\Gamma$ that it is compatible with that metric ($\nabla_{\sigma}g^{\mu\nu}$=0).

Thanks!
So at first we all acknowledge that just as Sean Carroll pointed the parallel propagator $P(x,x_0)$ can solve the parallel transport equation. then if the curvature of the connection is not zero, the parallel propagator $P(x,x_0)$ is path dependent,

it is just because this that I try to find out if there exist connections which can make the movement of a metric is path dependent, But if you only start from a metric field, because a field just implicitly means that the change of the metric is path independent. so if you demand that $(8)=(9)$, it will put some restrictions on the choose of the connection.

So I start from parallel propagator generated with an arbitrary connection, and the metric can be expressed as: $g(x)=P(x,x_0)g(x_0)P^T(x,x_0)$, conbine with $\partial_\mu P(x,x_0)=\gamma_\mu P(x,x_0)$. we can get:
$$\partial_\rho g(x)= \partial_\rho P(x,x_0)g(x_0)P^T(x,x_0)+P(x,x_0)g(x_0)\partial_\rho P^T(x,x_0)$$
$$=\gamma_\rho g(x)+(\gamma_\rho g(x))^T$$
This is just the metric compatibility equation. So $\gamma_\mu$ is compatible with metric $g(x)$. because the parallel propagator $P(x,x_0)$ is possibly path dependent, so why the metric defined with $P(x,x_0)$ can not be path dependent.

I also can give an example from which we can get a strange result. If a connection is defined in such a way: $\gamma_\mu= \partial_\mu G(x)G^{-1}(x)$ ,($G(x)$ is an invertible matrix function ) , then an easy calculation can reveal that the curvature of $\gamma_\mu$ is zero. and the parallel propagator is very simple $P(x,x_0)=G(x)G^{-1}(x_0)$, it is path independent. so we can get a metric field:
$$g(x)=G(x)G^{-1}(x_0)g(x_0)(G^{-1})^T(x_0)G^T(x)$$
it obviously is not always a flat metric. so it seems that there are many non-flat metric fields who is compatible with a zero curvature connection. This result seems to be very strange in GR. but I can not find any problem in the calculations, So where is the problem?

 

[jbergman]

Thanks!
So at first we all acknowledge that just as Sean Carroll pointed the parallel propagator $P(x,x_0)$ can solve the parallel transport equation. then if the curvature of the connection is not zero, the parallel propagator $P(x,x_0)$ is path dependent,

it is just because this that I try to find out if there exist connections which can make the movement of a metric is path dependent, But if you only start from a metric field, because a field just implicitly means that the change of the metric is path independent. so if you demand that $(8)=(9)$, it will put some restrictions on the choose of the connection.

So I start from parallel propagator generated with an arbitrary connection, and the metric can be expressed as: $g(x)=P(x,x_0)g(x_0)P^T(x,x_0)$, conbine with $\partial_\mu P(x,x_0)=\gamma_\mu P(x,x_0)$. we can get:
$$\partial_\rho g(x)= \partial_\rho P(x,x_0)g(x_0)P^T(x,x_0)+P(x,x_0)g(x_0)\partial_\rho P^T(x,x_0)$$
$$=\gamma_\rho g(x)+(\gamma_\rho g(x))^T$$
This is just the metric compatibility equation. So $\gamma_\mu$ is compatible with metric $g(x)$. because the parallel propagator $P(x,x_0)$ is possibly path dependent, so why the metric defined with $P(x,x_0)$ can not be path dependent.

I also can give an example from which we can get a strange result. If a connection is defined in such a way: $\gamma_\mu= \partial_\mu G(x)G^{-1}(x)$ ,($G(x)$ is an invertible matrix function ) , then an easy calculation can reveal that the curvature of $\gamma_\mu$ is zero. and the parallel propagator is very simple $P(x,x_0)=G(x)G^{-1}(x_0)$, it is path independent. so we can get a metric field:
$$g(x)=G(x)G^{-1}(x_0)g(x_0)(G^{-1})^T(x_0)G^T(x)$$
it obviously is not always a flat metric. so it seems that there are many non-flat metric fields who is compatible with a zero curvature connection. This result seems to be very strange in GR. but I can not find any problem in the calculations, So where is the problem?

Your questions are interesting because you are making me think about these questions in new ways. With a metric compatible connection, I think, parallel transport can change the direction of vectors along different paths but not lengths! So maybe that is the answer to your riddle.

But as @renormalize said there exists connections from which you can't construct metric compatible connections because they don't satisfied a required algebraic relationship.

I'm too lazy to try and work out the details via the propagator.

 

[renormalize]

So I start from parallel propagator generated with an arbitrary connection, and the metric can be expressed as: $g(x)=P(x,x_0)g(x_0)P^T(x,x_0)$, conbine with $\partial_\mu P(x,x_0)=\gamma_\mu P(x,x_0)$. we can get:
$$\partial_\rho g(x)= \partial_\rho P(x,x_0)g(x_0)P^T(x,x_0)+P(x,x_0)g(x_0)\partial_\rho P^T(x,x_0)$$
$$=\gamma_\rho g(x)+(\gamma_\rho g(x))^T$$
This is just the metric compatibility equation. So $\gamma_\mu$ is compatible with metric $g(x)$. because the parallel propagator $P(x,x_0)$ is possibly path dependent, so why the metric defined with $P(x,x_0)$ can not be path dependent.

Do you see the inconsistency here? First you say: "So I start from parallel propagator generated with an arbitrary connection...", yet you conclude with: "So $\gamma_\mu$ is compatible with metric $g(x)$." But a truly arbitrary connection $\gamma_\mu$ is not metric-compatible because it can include nonmetricity $Q$. There is a flaw in your derivation: can you spot it?

 

[Jianbing_Shao]

Do you see the inconsistency here? First you say: "So I start from parallel propagator generated with an arbitrary connection...", yet you conclude with: "So $\gamma_\mu$ is compatible with metric $g(x)$." But a truly arbitrary connection $\gamma_\mu$ is not metric-compatible because it can include nonmetricity $Q$. There is a flaw in your derivation: can you spot it?

Thanks!
Just because the parallel propagator is also called path ordered product, So when we talked about the parallel propagator, a particular path $l$ have been chosen at first. when I say "So $\gamma_\mu$ is compatible with metric $g(x)$." I mean that metric $g(x)$ only well defined on the path we choose. it is not a global metric field.

About the general connection, to change a vector is just to change it's direction and length. so only with rotation and nonmetricity $Q$ we can fully describe the change of a vector. and the torsion part is different from the other two, it is result from the lack of a global coordinate system, so it is hard for us to define a closed parallelograms. Is it right?