I A question about metric compatibility equation

[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..

 

[renormalize]

I think it perhaps a conbination of rotation invariance of metric and curvature determine the property of path dependence of metric.

It's frustrating when you continue to make these statements about what you "think". What matters is what you can prove! The whole point of my post #46 is to demonstrate (prove!) that the metric $g$ is unchanged by parallel-transport through an infinitesimal-distance along a path by any metric-compatible connection $\Gamma$ (see eq.(10)), regardless of the curvature or torsion. So I must cease commenting in this thread unless you do me the courtesy of one of the following:

• Prove using equations (not words) that "curvature determine the property of path dependence of metric", or
• Prove using equations (not words) that there is an error in post #46 that renders eq.(10) invalid, or
• Acknowledge that eq.(10) is valid and proves that $g$ is invariant when parallel-transported by a metric-compatible $\Gamma$.

 

[berkeman]

Thread closed for Moderation...

 

[PeterDonis]

After moderator review, this thread will remain closed as the OP question has been sufficiently addressed.