I A question about metric compatibility equation

[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?

 

[Jianbing_Shao]

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.

Thanks a lot!
If we only take the rotation into account, it may be helpful, but it perhaps can not solve the problem.
For example, from the result I mentioned above: there exist infinite metric field different from flat metric $\eta$ compatible with a zero curvature connection. then here are some problems:
how to define a curved space? metric different with $\eta$? or the curvature is not zero? or if there exist a connection without a compatible metric field, then what does it mean?

 

[PeterDonis]

there exist infinite metric field different from flat metric $\eta$ compatible with a zero curvature connection

Since this is obviously false, your reasoning that leads to it must be wrong somewhere. I suggest taking a look at post #49.

 

[renormalize]

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.

Your answer is dancing around the mathematical inconsistency I pointed out in post #49. Below are the step-by-step details of your argument in post #47, as I understand them. I claim your deductive reasoning is manifestly self-contradictory:

1. Premise: $\gamma$ is an arbitrary, general connection $\Rightarrow Q\neq0$
2. $g(x)=P(x,x_0,\gamma)g(x_0)P^{T}(x,x_0,\gamma)$
3. $\partial_{\mu}g(x)=\partial_{\mu}P(x,x_0,\gamma)g(x_0)P^{T}(x,x_0,\gamma)+P(x,x_0,\gamma)g(x_0)\partial_{\mu}P^{T}(x,x_0,\gamma)$
4. $\partial_{\mu}P(x,x_0,\gamma)=\gamma_{\mu}(x)P(x,x_0,\gamma)$
5. $\partial_{\mu}g(x)=\gamma_{\mu}(x)P(x,x_0,\gamma)g(x_{0})P^{T}(x,x_0,\gamma)+P(x,x_0,\gamma)g(x_0)P^{T}(x,x_0,\gamma)\gamma^{T}_{\mu}(x)$
6. $\partial_{\mu}g(x)=\gamma_{\mu}(x)g(x)+g(x)\gamma^{T}_{\mu}(x)$
7. $\partial_{\mu}g(x)-\gamma_{\mu}(x)g(x)-g(x)\gamma^{T}_{\mu}(x)=0$
8. Conclusion: $\nabla_{\mu}g(x)=0\Rightarrow Q=0$ Contradiction!

I ask you to please clarify whether: 1) I have misunderstood your logic or made a mistake in one or more of my steps; if so could you please post your corrected version? Or, 2) you agree your logic is flawed and post #47 really proves nothing about the possible path-dependence of the metric tensor. Thanks!

 

[Jianbing_Shao]

Your answer is dancing around the mathematical inconsistency I pointed out in post #49. Below are the step-by-step details of your argument in post #47, as I understand them. I claim your deductive reasoning is manifestly self-contradictory:

1. Premise: $\gamma$ is an arbitrary, general connection $\Rightarrow Q\neq0$
2. $g(x)=P(x,x_0,\gamma)g(x_0)P^{T}(x,x_0,\gamma)$
3. $\partial_{\mu}g(x)=\partial_{\mu}P(x,x_0,\gamma)g(x_0)P^{T}(x,x_0,\gamma)+P(x,x_0,\gamma)g(x_0)\partial_{\mu}P^{T}(x,x_0,\gamma)$
4. $\partial_{\mu}P(x,x_0,\gamma)=\gamma_{\mu}(x)P(x,x_0,\gamma)$
5. $\partial_{\mu}g(x)=\gamma_{\mu}(x)P(x,x_0,\gamma)g(x_{0})P^{T}(x,x_0,\gamma)+P(x,x_0,\gamma)g(x_0)P^{T}(x,x_0,\gamma)\gamma^{T}_{\mu}(x)$
6. $\partial_{\mu}g(x)=\gamma_{\mu}(x)g(x)+g(x)\gamma^{T}_{\mu}(x)$
7. $\partial_{\mu}g(x)-\gamma_{\mu}(x)g(x)-g(x)\gamma^{T}_{\mu}(x)=0$
8. Conclusion: $\nabla_\mu g(x)=0\Rightarrow Q=0$ Contradiction!

I ask you to please clarify whether: 1) I have misunderstood your logic or made a mistake in one or more of my steps; if so could you please post your corrected version? Or, 2) you agree your logic is flawed and post #47 really proves nothing about the possible path-dependence of the metric tensor. Thanks!

Thanks!
I just start from an arbitrary connection which includes $Q\neq 0$.
Formula (1)~(7) just the procedure in which I'm trying to prove the metric generated with parallel prapagator can satify the metric compatibility equation. and we write the equation(7) in the form of components , then we get the usual metric compatibility equation we familiar.

In this procedure I can not find any reason to restrict that $Q$ should be zero. So I am confused about that why from $\nabla_\mu g(x)=0$ then we can draw a conclusion $Q=0$.

When I say we can not find a metric field compatible with a connection, I just mean that the metric field is defined on the whole space, If we confine the metric on a curve, then I am sure that we always can find a metric compatible with the connection. the metric can be expressed using formula(2),

The introduction of $Q$ can obviously make a connection more difficult to compatible with a global metric field. but it doesn't mean that if the connection compatible with a global metric field the $Q$ must be zero.

So I am not sure that the metric compatiblility equation can necesserily result in the conclusion $Q=0$.

 

[Jianbing_Shao]

Your answer is dancing around the mathematical inconsistency I pointed out in post #49. Below are the step-by-step details of your argument in post #47, as I understand them. I claim your deductive reasoning is manifestly self-contradictory:

1. Premise: $\gamma$ is an arbitrary, general connection $\Rightarrow Q\neq0$
2. $g(x)=P(x,x_0,\gamma)g(x_0)P^{T}(x,x_0,\gamma)$
3. $\partial_{\mu}g(x)=\partial_{\mu}P(x,x_0,\gamma)g(x_0)P^{T}(x,x_0,\gamma)+P(x,x_0,\gamma)g(x_0)\partial_{\mu}P^{T}(x,x_0,\gamma)$
4. $\partial_{\mu}P(x,x_0,\gamma)=\gamma_{\mu}(x)P(x,x_0,\gamma)$
5. $\partial_{\mu}g(x)=\gamma_{\mu}(x)P(x,x_0,\gamma)g(x_{0})P^{T}(x,x_0,\gamma)+P(x,x_0,\gamma)g(x_0)P^{T}(x,x_0,\gamma)\gamma^{T}_{\mu}(x)$
6. $\partial_{\mu}g(x)=\gamma_{\mu}(x)g(x)+g(x)\gamma^{T}_{\mu}(x)$
7. $\partial_{\mu}g(x)-\gamma_{\mu}(x)g(x)-g(x)\gamma^{T}_{\mu}(x)=0$
8. Conclusion: $\nabla_{\mu}g(x)=0\Rightarrow Q=0$ Contradiction!

I ask you to please clarify whether: 1) I have misunderstood your logic or made a mistake in one or more of my steps; if so could you please post your corrected version? Or, 2) you agree your logic is flawed and post #47 really proves nothing about the possible path-dependence of the metric tensor. Thanks!

So I can express my idea in such a way: From metric compatibility equation$\nabla_\mu g(x)=0$, if choose a particular path $l$, we can construct a metric confined on path $l$ $g(x)=P(x,x_0;l;\gamma)g(x_0)P^T(x,x_0;l;\gamma)$, here $g(x)$ and $\gamma$ are compatible with each other. Here $\gamma$ can be arbitrary.

If we try to find a global metric field compatible with the connection $\gamma$, To two arbitrary path $l$ and $l'$ connect point $x$ and $x_0$. we demand that:$$
P(x,x_0;l;\gamma)g(x_0)P^T(x,x_0;l;\gamma)=P(x,x_0;l';\gamma)g(x_0)P^T(x,x_0;l';\gamma)$$.
Then $g(x)$ can be a global metric field. this condition obviously put some restraints on the choose of the connection. and not all connection field can fulfill this condition.

 

[renormalize]

So I am not sure that the metric compatiblility equation can necesserily result in the conclusion $Q=0$.

How can you possibly be unsure of this? The definition of metric compatibility is: $\nabla_{\mu}g\equiv 0$. The definition of of nonmetricity is: $Q_{\mu}\equiv\nabla_{\mu}g$. So by definition, the statements "$\gamma_{\mu}$ is metric compatible" and "$Q_{\mu}\neq 0$" are mutually exclusive; i.e., they can never be true simultaneously. Your self-contradictory statement above seems to be based on your feelings about how things should work, or might work, instead of accepting the logical conclusion that your understanding of path-dependent metrics is simply wrong.

 

[renormalize]

If we try to find a global metric field compatible with the connection $\gamma$, To two arbitrary path $l$ and $l'$ connect point $x$ and $x_0$. we demand that:$$
P(x,x_0;l;\gamma)g(x_0)P^T(x,x_0;l;\gamma)=P(x,x_0;l';\gamma)g(x_0)P^T(x,x_0;l';\gamma)$$Then $g(x)$ can be a global metric field. this condition obviously put some restraints on the choose of the connection. and not all connection field can fulfill this condition.

No, the only condition that the connection needs to satisfy is metric compatibility. Compatibility guarantees that, unlike every other parallel-transported tensor, the metric tensor remains path-independent under transport. That's what I proved in post #46. I ask that you please re-read that post and take the effort to understand it. Realizing that the metric is the only inherently path-independent tensor-field should eliminate your concerns that motivated this thread. (Or if you find an error in that proof, please do let me know!)

 

[Jianbing_Shao]

No, the only condition that the connection needs to satisfy is metric compatibility. Compatibility guarantees that, unlike every other parallel-transported tensor, the metric tensor remains path-independent under transport. That's what I proved in post #46. I ask that you please re-read that post and take the effort to understand it. Realizing that the metric is the only inherently path-independent tensor-field should eliminate your concerns that motivated this thread. (Or if you find an error in that proof, please do let me know!)

Thanks a lot!
logically if we want to prove a property of path independence, we can prove that if we move along different paths we can get the same result, that is just what I am doing.

Now if we start from your point, when you say the connection satisfy metric compatibility. compatibility guarantees the metric tensor path independent, then the connection can be an arbitrary connection? It seems that the problem is still unsettled. The problem is that when you define metric compatibility you think we already have a metric field, then it is not metric compatibility guarntee the path independence of the metric, but you have a path independent metric field at first.

If you are sure that an arbitrary connection can compatible with a metric field defined in the whole space. as I noted above you obviously can prove that:
$$P(x,x_0;l;\gamma)g(x_0)P^T(x,x_0;l;\gamma))=P(x,x_0;l';\gamma)g(x_0)P^T(x,x_0;l';\gamma))$$
In fact I have constructed a global metric field from a connection. and It seemed that the path independence is related to the cuvature of the connection. and In this question I think perhaps there is no need to classify different kind of connection.

 

[renormalize]

If you are sure that an arbitrary connection can compatible with a metric field defined in the whole space. as I noted above you obviously can prove that:
$$P(x,x_0;l;\gamma)g(x_0)P^T(x,x_0;l;\gamma))=P(x,x_0;l';\gamma)g(x_0)P^T(x,x_0;l';\gamma))$$
In fact I have constructed a global metric field from a connection. and It seemed that the path independence is related to the cuvature of the connection. and In this question I think perhaps there is no need to classify different kind of connection.

Did you read and absorb post #46, as I asked you too? It proves that:

• All tensor fields, including the metric, are path-dependent under parallel transport for a general connection that includes nonmetricity.
• But if the nonmetricity vanishes (i.e., the connection is metric compatible, but can still include torsion) the metric becomes the one and only tensor field that is path independent under transport, even in curved spacetime. All other tensor fields remain path-dependent.

 

[Jianbing_Shao]

Did you read and absorb post #46, as I asked you too? It proves that:

• All tensor fields, including the metric, are path-dependent under parallel transport for a general connection that includes nonmetricity.
• But if the nonmetricity vanishes (i.e., the connection is metric compatible, but can still include torsion) the metric becomes the one and only tensor field that is path independent under transport, even in curved spacetime. All other tensor fields remain path-dependent.

Thanks!
Of course a connection includes monmetricity is a general connection, then I agree on your first conclusion. It is just what I mean, perhaps the difference between us is that you think the path dependence of the metric comes from nonmetricity. I think it perhaps a conbination of rotation invariance of metric and curvature determine the property of path dependence of metric.

Because for a connection $exp(\gamma_\mu dx^\mu)$ is an infinitesimal action. So can you show what kind of action $exp(Q_\mu dx^\mu)$ is?

About your conclusion I have a question: the first one is it seems that you think that a connection includes nonmetricity makes the transportation of tensor path dependent? Does it means that only when connection is zero then the the transportation of tensor can be path independent..