Talk:Levi-Civita connection

message: :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)
replies: text: 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)
replies: text: 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)
replies: text:  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)
thread_title: Formal definition

message: 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_tV:=(\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_tV = \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)
thread_title:  Derivative along curve 

message: The expression \nabla g=0 does not make sense to me. Isn't the property of metric preservation exactly X g(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ)?  — Preceding unsigned comment added by 2003:6B:A1E:4801:4000:2B7E:EC01:A50 (talk) 10:06, 25 May 2015 (UTC)
replies: text: 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)
thread_title:  Connection preserves metric 

message: 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)
thread_title:  Animation of parallel transport 

message: 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)
replies: text: 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)
thread_title:  Parallel transport example is wrong