I Metric compatibility? Why is it an additional property?

[George Keeling]

TL;DR Summary

Metric compatibility? Why is it an additional property?

In chapter 3 of Sean Carroll's Introduction to General Relativity he 'makes the demand' of metric compatibility of a connection that $\nabla_\mu g_{\lambda\nu}=0$. Metric compatibility becomes a phrase that is used frequently. However metric compatibility seems to arise naturally. One only has to require that the covariant derivative is a tensor (!) and that it obeys the Leibniz rule:$$
\nabla\left(T\ \bigotimes\ \ S\right)=\left(\nabla T\right)\ \ \bigotimes\ \ S+T\ \bigotimes\ \ \left(\nabla S\right)
$$If the covariant derivative is a tensor then we could say that $\ \nabla_\mu V^\nu\equiv\ \left(\nabla V\right)_\mu^{\ \ \ \nu}$ or $\nabla V$ is the tensor and $\mu,\nu$ are its indices. In that case we must have$$
g_{\lambda\nu}\nabla_\mu V^\lambda=g_{\lambda\nu}\left(\nabla V\right)_\mu^{\ \ \ \lambda}=\left(\nabla V\right)_{\mu\nu}^{\ \ \ }=\nabla_\mu V_\nu
$$
By the Leibniz rule we have $$
\mathrm{\nabla}_\mu V_\nu=\nabla_\mu\left(g_{\lambda\nu}V^\lambda\right)=V^\lambda\nabla_\mu g_{\lambda\nu}+g_{\lambda\nu}\nabla_\mu V^\lambda
$$Picking bits out of those we get$$
V^\lambda\nabla_\mu g_{\lambda\nu}=0
$$Since that holds for any $V^\lambda$ we must have$$
\nabla_\mu g_{\lambda\nu}=0
$$We can call that metric compatibility but why say it is an additional property of a connection?
This harks back back to an answer StevenDaryl :

Here's the way I understand #3.

$g^{\lambda \nu} (\nabla_\mu T_{\nu \lambda \rho}) = \nabla_\mu (g^{\lambda \nu} T_{\nu \lambda \rho})$

So 'rule 3' of covariant derivatives is that they are tensors.

 

[PAllen]

Requiring that a metric be available to relate Contravectors and covectors, and applying that to covariant derivative is just another way of requiring metric compatibility. In general, a connection can define a covariant derivative with all the expected derivative properties in a context with no metric provided at all, in which there is no way to raise and lower indices. In fact, for a given topolological space, you can provide many connections as well as a metric. Only one connection will work with the metric as you outline - the metric compatible one.

 

[PAllen]

Just to put some 'measure' on this. I believe that you could find a measure on the set of all symmetric connections possible on some topological space such the the subset compatible with any metric that can be placed the space is of measure 0. Further, for any given metric, there will only be be one compatible connection (up to coordinate transformations). Note that, in general, there are unaccountably many metrics that can be placed on a given topological space.

 

[George Keeling]

I think that answers my question.

Requiring that a metric be available to relate Contravectors and covectors, and applying that to covariant derivative is just another way of requiring metric compatibility.

All the manifolds I have met have a metric with the property of raising and lowering indices. When I look up topological space I see that " manifolds and metric spaces, are specializations of topological spaces". I'm not sure why Carroll went so theoretical here or if it is necessary in this book. I have been warned. Thanks!

 

[vanhees71]

Only one connection will work with the metric as you outline - the metric compatible one.

Only one torsion-free connection. Otherwise there are many metric-compatible connections for a given metric.

 

[PAllen]

Only one torsion-free connection. Otherwise there are many metric-compatible connections for a given metric.

I had previously said I was restricting to symmetric connection, which is the same thing.

 

[pervect]

Consider the 2 sphere. With the Levi-Civita connection, geodesics on the sphere are great circles. But suppose one wants to plot the course of a naval vessel on a spherical ocean, and one wants to plot courses of constant heading rather than courses of shortest distance. These are known as rhumb lines, and also as loxodromes.

With the proper non Levi-Civita connection, rhumb lines are geodesics (and great circles are not). With this alternate connection, a rhumb line is the curve that parallel transports its tangent vector.

This may be some what of a "toy" example, but there are serious theories (such as Einstein-Cartan theory https://en.wikipedia.org/wiki/Einstein–Cartan_theory) which do not use the Levi-Civita connection. Einstein-Cartan theory has fewer problems with spin 1/2 particles than GR does, but under normal conditions the predictions aren't that different from GR. Under extreme conditions (like the interior of black holes), the two theories make different predictiotns.

However, for learning GR, it's sufficient to deal only with metric compatible connections, which makes life a lot simpler. I'm rather fond of Einstein-Cartan theory in the abstract, but not fond enough to want to learn how to deal with torsion.