Talk:Levi-Civita connection

Stub history

Why did you kill this page? -- Anon

The problem is the article as it was written doesn't make sense. The text was: " The unique torsion free connection, preserving a riemannian metric". What does that mean? What does this relate to? Is it physics? biology? fiction? Remember you are writing for a non-specialist audience.

A quick Google tells me this is something to do with mathematics. The Anome gave a great example of how to write a good article on mathematical topics:

"(Mathematician) invented the concept of X in 18xx to represent (squeezy-pully-twisty things). A simple example, using modern notation is (example). (Explain notation). The idea has now been generalized to (stuff), which has uses in (other fields of math and science). The idea of X can be formalized as follows: let there be objects X such that (notation). Then (notation) (notation) (notation)..." (from this talk page)

If this article has nothing to do with mathematics, then that shows the problem even more clearly. In a case like this it is usually thought better to remove the article rather than have a confusing sub-stub. But much better is to improve the article, and if you can do that it would be great! I hope you stick around and give it a go. Regards -- sannse 19:30 26 Jun 2003 (UTC)

Formal definition

This section needs to be reviewed. With a firm background in vector calculus, I cannot interpret what the two equations are trying to express. Please explain where the RHS and LHS are coming from, and be more explicit with notation. I can only assume that X(g(Y,Z)) is supposed to mean the scalar multiplication of the vector field X with the inner product of the fields Y and Z, but this yields a vector quantity, and the RHS, which is the sum of inner products, would be a scalar. 98.235.167.113 (talk) 18:26, 8 March 2012 (UTC)[reply]

It is common for Xf to represent the Lie derivative of the function f by the vector field X (so that Xf is a function). This is not the same as the product of f and X, which would be written fX and would be a vector. Thus X(g(Y,Z)) means the Lie derivative of the function g(Y,Z), and the equations in this section make sense. — Preceding unsigned comment added by 83.104.131.53 (talk) 11:16, 3 May 2012 (UTC)[reply]

To say "It is common ..." is perhaps a bit strong, especially in this encyclopedic context. The notation strikes me as highly context-specific, and since the term "Lie derivative" does not occur once in the article, it is appropriate to add something like "... where Xf denotes the Lie derivative of ...". I would go so far as to say that a more explicit notation such as

{\mathcal {L}}_{\!X}g(Y,Z)

should be used, plus the mention that this represents the Lie derivative. — Quondum ^☏ 13:09, 3 May 2012 (UTC)[reply]

there is no need to introduce lie drivatives here. to make the notation more intuitive replace Xf by

\partial _{X}f

which is if necessary easily explained as df(X).Peter Grabs (talk) 06:28, 2 August 2018 (UTC)[reply]

I agree with your suggestion of simply using the partial derivative notation. Lie derivatives and Lie algebra, while sometimes encountered in general relativity, are not part of core formalism nor part of the notation common in modern physics literature, such as Physical Review D. Being an independent general relativist seeking a better understanding, I learned nothing from this presentation, alas. — Preceding unsigned comment added by 97.73.100.85 (talk) 21:56, 9 September 2024 (UTC)[reply]

The notation used in the article is standard in sources on this topic. In contrast, I have never seen the notation

\partial _{X}

used in this way. Tito Omburo (talk) 16:08, 10 September 2024 (UTC)[reply]

Derivative along curve

i suggest to replace this sections by something like this, because the given definition is not applicable for vectorfields along curves (as i know that notion):

The Levi-Civita connection (like any affine connection) also defines a derivative along curves, sometimes denoted by $D$ .

Given a smooth curve $\gamma$ on $(M,g)$ we consider vector fields along $\gamma$ , i.e. smooth sections $V\in \Gamma (\gamma ^{\ast }TM)$ of the pullback bundle $\gamma ^{\ast }TM$ , and define their derivative by

D_{t}V:=(\gamma ^{\ast }\nabla )_{1}V|_{t}

.

For $V=X\circ \gamma \in \Gamma (TM)\circ \gamma \subset \Gamma (\gamma ^{\ast }TM)$ this pullback connection specializes to

D_{t}V=\nabla _{{\dot {\gamma }}(t)}X|_{c(t)}

.

In particular, ${\dot {\gamma }}$ is a vector field along the curve $\gamma$ itself. If $D_{t}{\dot {\gamma }}$ vanishes, the curve is called a geodesic of the covariant derivative. If the covariant derivative is the Levi-Civita connection of a certain metric, then the geodesics for the connection are precisely those geodesics of the metric that are parametrised proportionally to their arc length.

-- Peter Grabs (talk) 19:17, 29 April 2012 (UTC)[reply]

Connection preserves metric

The expression $\nabla g=0$ does not make sense to me. Isn't the property of metric preservation exactly $Xg(Y,Z)=g(\nabla _{X}Y,Z)+g(Y,\nabla _{X}Z)$ ? — Preceding unsigned comment added by 2003:6B:A1E:4801:4000:2B7E:EC01:A50 (talk) 10:06, 25 May 2015 (UTC)[reply]

if you have a notion of derivation for functions (

\nabla _{X}f=\partial _{X}f=df(X)

) and vectors (

\nabla _{X}Y

, e.g. Levi Civita of TM), you can carry it over to arbitrary tensors like

g\in \Gamma (T^{*}M\otimes T^{*}M)

by requiring the leibniz rule, i.e.

\nabla _{X}(g(Y,Z))=(\nabla _{X}g)(Y,Z)+g(\nabla _{X}Y,Z)+g(Y,\nabla _{X}Z)

, this defines

\nabla g

and lets you read off the equivalence of the two definitions of metricity.Peter Grabs (talk) 06:56, 2 August 2018 (UTC)[reply]

Animation of parallel transport

The animation in the section on parallel transport indicates that it shows parallel transport of a vector under two different metrics. However, the metrics are simply both the flat metric, but in Cartesian and polar coordinates. I'm not sure how the right-side animation is supposed to be interpreted. If the axes still represent Cartesian coordinates $x,y$ then the animation is clearly wrong as parallel transport should be identical to that in the left image. If the coordinate axes are for $r,\theta$ really, then the parallel transport curve (unit circle) passes through the coordinate singularity at $r=0$ and secondly the length of the vector is not preserved with respect to the metric. Jeldering (talk) 22:18, 20 January 2016 (UTC)[reply]

This turned out to be an erroneous anonymous edit that I reverted. Still, I think this section is not clear and needs rewriting/improvement. I may have a go at it if I find time. Jeldering (talk) 14:18, 21 January 2016 (UTC)[reply]

I spent a long time thinking about this section and it still confuses me. Part of the problem is that we are not used to "see" non-euclidean metrics on the plane. Also, if I'm not mistaken, both metrics should be the same in S^1, which is the situation shown here. I would appreciate it if someone took the time to explain in more detail (possibly in charts) how the different parallel transports arise. Emilparole (talk) 13:08, 20 February 2018 (UTC)[reply]

Some other people edited this page and screwed up the metrics again. I've tried my best, with the limited time and understanding I currently have, to make it at least a little bit clearer that $dx^{2}+dy^{2}=dr^{2}+r^{2}d\theta ^{2}$ and that the second transport is for a completely different metric. It still needs a rewrite, though. 73.114.132.157 (talk) 02:56, 18 July 2018 (UTC)[reply]

Parallel transport example is wrong

First the whole discussion about a singularity in the origin is besides the point, because both have this singularity, the big scary formula although not wrong is distracting and pointless and should be removed. The error is in the claim that both connections are "different" Levi-Civita connections, the one on the left is indeed a Levi-Civita connection but the one on the right is NOT! (this error is also present on the file page of the animation.) The animation on the right has a connection but it's not a Levi-Civita connection, i.e. it's non-symmetrical in the last two indices. The connection can be found by using the general formula for the Connection coefficients in a nonholonomic basis using the formula given in this article will result in all zero coefficients and parallel transport that does not change the vector and so identical to the animation on the left. --DelftUser (talk) 15:24, 25 February 2020 (UTC)[reply]

The metric used in the right image is wrong! the right image does not use the coordinates basis and so the metric can't be written as $ds^{2}=dr^{2}+d\theta ^{2}$ . Here is how the title & captions of the two animations should be:

Parallel transports under different connections

This transport is in the coordinates basis:

{\partial  \over \partial r},{\partial  \over \partial \theta }

with the Levi-Civita connection of the metric

ds^{2}=dr^{2}+r^{2}d\theta ^{2}

.

This transport is in the natural local tetrad:

e_{(r)},e_{(\theta )}

and the associated Ricci Rotation Coefficients.

Because the basis vectors are different it's wrong to write the metric with ( $dr,d\theta$ ), so this is my fix; I'll wait a while for comments before changing the article.--DelftUser (talk) 09:31, 29 February 2020 (UTC)[reply]

The example is correct, interpreting r and theta as functions on R^2. The former defines the standard metric on R^2 and the latter defines an incomplete metric on R^2-{(0,0)} which is isometric to the standard circular cone. I think it should be explained more clearly though, and there is probably a better example to use. Gumshoe2 (talk) 23:04, 5 January 2022 (UTC)[reply]

The Asymmetric Metric

Consider an asymmetric metric, as it relates to Relativity. In Einstein's The Meaning of Relativity, 5th Edition, 1953, on page 71, equations 69 and 70 show the Christoffel symbols, the Levi-Civita connection being equation 70. I propose some alternative definitions by swapping the indices of the 2nd term on the right of equation 69:

${[(g)ij,k]\equiv {\frac {1}{2}}(g_{ik,j}+g_{kj,i}-g_{ij,k})}$

The swapping makes no difference for a symmetric metric, but better serves the case for an asymmetric metric (Notice a new notation, where this alternate Christoffel symbol of the 1st kind is now an operator, on a rank 2 covariant tensor). For the case of the symmetric metric, the normal results ensue. When $t$ is nondegenerate I propose the following connection:

${\Gamma ^{k}{}_{ij}\equiv s^{mk}[(s)ij,m]+t^{mk}[(t)ij,m],(RP,1Feb1993)}$

where $g_{ij}=s_{ij}+t_{ij}$ , $s_{ij}=s_{ji}$ , $t_{ij}=-t_{ji}$ , and similarly for the associated contravariant metric.

For the case of an asymmetric metric these alternative definitions provide for Torsion with metricity (the covariant derivative of the metric is identically zero), and the Bianchi identity remains valid.

Regards, Ric Peregrino Ric.Peregrino (talk) 20:42, 18 April 2024 (UTC)[reply]

In an attempt to further develop the results of an asymmetric metric towards a potentially modified Einstein tensor, it seems the Bianchi identity doesn't simplify the same way. Taking a step back, I had started studying the remaining symmetries of the resulting curvature tensor in an effort to arrive at a better result for such an Einstein tensor. As you may have noticed, this was decades ago.

I've shared this topic here, in an effort to get someone who might understand what I'm trying to propose with an asymmetric metric, the alternative connection which reduces to the Levi-Civita connection with a symmetric metric, and the ensuing results, if any, towards an alternative Einstein tensor. Hopefully this makes sense to someone and they might comment. Or perhaps this has already been dealt with, in which case some pointers would be appreciated.

I have more information with some details of my claims, which I'd gladly share upon request.

At this time, there's been some feedback, and I'm no longer considering posting such a chapter to the article.

Regards, Ric Ric.Peregrino (talk) 02:56, 19 April 2024 (UTC)[reply]

Please don't edit the article, this isn't the right place for original research. Regardless, I believe your formula for Γ^k_ij doesn't make sense since t is generally not nondegenerate, so t^mk is meaningless. (Certainly this is at least the case in the most important special case, t=0.)

There is some work out there on nonsymmetric metrics, notably by Einstein. Gumshoe2 (talk) 17:30, 28 April 2024 (UTC)[reply]

Regarding that work by Einstein, it seems to suggest that if |t|!=0 at any point in spacetime, it's nonzero every where and when. So to be more precise, I should state the formula only applies if |t|!=0, and else if |t|=0 revert to Levi-Civita formula. I think they further arrive at there being only 2 unique components of t in 4d spacetime, and arrive at a complicated set of formula for the connection which I may or may not be able to eventually understand.

The result I provided seems simply stated, and may or may not be valid, and if valid may or may not be equivalent to Einstein's quoted work. I hope I can find out.

I'd like to ask you, where might be a good place to discuss this topic? Perhaps here in this talk is ok, at least to start? I may poke around different topics and see if another one makes more sense... Ric.Peregrino (talk) 20:00, 28 April 2024 (UTC)[reply]

Wikipedia isn't a good place for such discussion, since the focus here is on writing exposition and very much against doing novel work. I would suggest math stackexchange or math overflow. Gumshoe2 (talk) 20:17, 28 April 2024 (UTC)[reply]

Roger wilco, and thanks for the response.

Thanks so much for that reference! Looks like I have some homework.

Indeed t^mk is meaningless if t_ij=0, but I suggest the redefinition of the Christoffel symbol of the 1st kind, results in the ensuing connection resolving to the Levi-Civita connection.

Cheers,

Ric Ric.Peregrino (talk) 17:51, 28 April 2024 (UTC)[reply]