Talk:Covariant derivative

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Geometry guy 22:09, 14 April 2007 (UTC)

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Help with this template Comments: edit – history – watch – refresh Geometry guy 22:09, 14 April 2007 (UTC)

1 Metric compatible
2 Einstein notation
3 curvature, parallel transport, geodesics
4 ∇ vs D
5 A comment
6 Picture
7 Covariant derivative and Lie brackets.
8 difference with Koszul connection
9 Merged
10 Notations
11 Question
12 "Tensor"

[edit] Metric compatible

In the coordinate-specific section of this article, it is stated "By the way, this particular expression is equal to zero, because the covariant derivative of a function solely of the metric is always zero.". Presumably, this is something that can be shown for the covariant derivative induced from the metric?

[edit] Einstein notation

Don't you think it is better to switch to Einstein notation?

Tosha 13:35, 14 Jul 2004 (UTC)

I certainly agree -Lethe

[edit] curvature, parallel transport, geodesics

I removed the last two subsections (curvature and parallel transport-geodesic) they do not add anything to the correspondent articles and badly written.

The subsection on Levi-Civita connection is moved to Fundamental theorem of Riemannian geometry.

(I just realized that part of it can be used in lie bracket, but will do it next time) Tosha 20:47, 20 Jul 2004 (UTC)

[edit] ∇ vs D

I think to change from D to ∇, it will make it consistent with other articles. Tosha 13:24, 21 Jul 2004 (UTC)

I agree that we should stick to one standard. I'm going to edit the page for spelling/grammar in a moment... - Gauge 02:50, 3 Aug 2004 (UTC)

[edit] A comment

First i wish to thanks the autor of this article for the good work. My comment is concerned with the sentece: "Therefore, the covariant derivative is not a tensor." (in the notes section of General concepts); in fact i believe covariant derivative is a tensor - this is exactlly the reason such a object was "invented". Thanks and i really hope to read more things with your quality.

Whether it's a tensor depends on how you define a tensor. I think the most common definition is a multilinear map on the tangent/cotangent spaces of a manifold. The covariant derivative, not being linear in its direction argument, fails to meet this definition. This is the most common stance taken in the literature, and the one wikipedia should take, though I admit there are those authors who want to consider it a tensor. I would also say that the explanation here in the article about why it shouldn't be a tensor is not very clear and could be cleaned up a little. -lethe ^talk 17:55, July 26, 2005 (UTC)

Thanks for your answer Lethe. The way i like to talk about tensors (and i think we could call it the modern way) is the one you stated up here - a multilinear map on the tangent/cotangent spaces of a manifold. All modern literature define covariant derivative as not being a tensor as you very well pointed out. Old literature (tensors in the old way) define it to be a tensor (as you also said). Tensors are combined to form objects with meaning independent of coordinate charts - a very important feacture in physics. The object defined in wikipedia is not defined in this way and so has no geometrical value (in my opinion at the moment)- please note that i'm not trying to say it is not well defined :) The best books i have do it this way and i know it. I just want the opinion of someone more experimented than i am in the theme so i can understand why we should abdicate of the geometrical meaning of objects (maybe there is a new concept here that i didn´t figured it out). Thanks for your time.

I have before me as I type this the book Schaum's Outline Series : Vector Analysis SI (Metric) Edition And An Introduction To Tensor Analysis by Murray R. Spiegel ISBN 070843783. On pages 197 and 198, this book gives the coordinate form for the covariant derivative in terms of Christoffel symbols (using the older curly brace notation which is a better notation in some respects because it enmphasises that the Christoffel symbols do not transform as tensors). It then goes on to derive the transformation laws for the covariant derivatives of a (1,0) valent and a (0,1) valent tensor. These transformation laws match those for a (1,1) and a (0,2) valent tensor respectively. So I am puzzled somewhat by the above assertion that the covariant derivative is not a tensor, to put it mildly!

Perhaps someone might like to provide an explanation for the above? Calilasseia 03:33, 27 February 2006 (UTC)

What they're saying is that the symbol ∇ by itself is not a tensor. In other words, the function which assigns to a pair of vector fields v and w, the vector field ∇(v,w) = ∇_vw is not a bilinear map, since ∇(v, fw) ≠ f ∇(v, w). It satisfies a product rule instead. (The following is now an interesting, although unrelated rant on Lethe's comment.) Silly rabbit 22:23, 16 April 2007 (UTC)

~~I think what they're saying is that~~ the covariant derivative of a tensor is a tensor (period — that's why it was invented). But the Christoffel symbols are not tensors (at least not in the usual sense).

The Christoffel symbols are tensors in an "unusual" sense though. They satisfy a second order transformation law, which means that they are tensors on a higher order jet bundle. I lean towards this interpretation more than the conventional dogmatic one. Philosophically, they are tensors because of the idea that tensors should correspond to natural objects — objects which can be written down generically. A favorite quote of mine, due to Salomon Bochner is: "A tensor is something with indices." Indeed!

Yet another way in which the Christoffel symbols can be treated as tensors (in the ordinary sense) is as a tensor-valued function depending on an initial choice of connection (not the L-C connection: usually the flat connection for a particular coordinate system). With this interpretation, the Christoffel symbols are

$\Gamma(\nabla)(X,Y) = \nabla_X^{LC}Y - \nabla_XY$

This transforms as a tensor in its arguments X, Y. Silly rabbit 22:13, 16 April 2007 (UTC)

[edit] Picture

It is not clear for me why they removed the picture, I do not think it was copyrighted, anyway they could say in advance and not on the image page, but here. BTW the other one is also mared for delition! Tosha 17:42, 12 October 2005 (UTC)

[edit] Covariant derivative and Lie brackets.

The Wikipedia states that the Lie derivative is a gadget that allows us to derive a vector field in the direction of another vector field. This (very good) entry about convariant derivaties also state that the covariant derivatives are gadgets that allow us to derive object (and, in particular, vector fields) along a given vector field. I suggest then that there should be some remark about the difference between the Lie derivative and the convariant derivative (in the paricular case where we are deriving vector fields.)

Let me try to be a tad more clear: Problem: I want to define what is the derivative of a vector field Y along a vector field X. Solution: (a) take the Lie derivative [Y,X] or (b) take a covariant derivative $\nabla_X Y$

Question: what is the difference between method (a) and (b)? Since method (b) depends on a choice of a connection, and method (a) is there for you always, why even bother about (b)?

The Lie derivative tells you how one vector changes with under infinitesimal diffeomorphisms induced by a second vector. The covariant derivative tells you how one vector changes along the direction of a second vector. You use the first to see how a vector changes under diffeomorphisms, and the second to see how a vector changes under parallel transport. -lethe ^talk 04:26, 24 January 2006 (UTC)

I think I can phrase what bothers me in the following way: given X, Y vector fields on a manifold M, and a fixed connection on it, we can calculate [X,Y] and , $\nabla_Y X$ at, say, a point p. If y is a solution curve of Y on M such that y(0) = p then we can use both

[X, Y] p

and $\nabla_{Y \ p} X$ to approximate the value of X at a point y(t), t > 0. Both of these approximations should be the "best" possible linear approximations. To this last phrase it should be added that the Lie derivative is the best linear approx. if we use the local diffeomorphisms generated by Y to compare the tangent spaces at y(0) and y(t), and the covariant derivative is the best linear approx. we can get if we use the connection to compare the vector spaces at y(0) and y(t) [i.e. if we use the parallel transport induced by the connection]. However, in any case, both are telling me how one vector is changing in the direction of another. At least, by using this linear approx. intrerpreation, this is how I understand this. —The preceding unsigned comment was added by 201.51.232.183 (talk • contribs) .

According to your problem, the best solution if you want to see how Y changes along the (flow of the) vector field X, use the Lie derivative (and this change is fixed by the flow). If you want to see how Y changes along an arbitrary curve which heads out in the direction of vector X_p, use the covariant derivative (and this change is arbitrary). All curves heading in the same direction give the same linear approximation for the parallel transport of vector field Y, but the linear approximation of the pushforward of Y along different vector fields X can be different, even if they match at p. I've added text to that affect to the article, I'd appreciate your feedback. -lethe ^{talk +} 09:10, 7 March 2006 (UTC)

[edit] difference with Koszul connection

I was wondering whether there is any difference between a Koszul connection and a covariant derivative on a vector bundle. I couldn't distill any from the articles. --MarSch 09:55, 29 June 2006 (UTC)

In the covariant derivative article, all the bundles involved are tensor bundles. Silly rabbit 13:54, 29 June 2006 (UTC)

Stated another way, the covariant derivative is defined by generalized Christoffel symbols which are chosen so that the derivative is covariant with respect to coordinate transitions. For a Koszul connection, no such covariance is required; in fact, it may not even make sense at all. If we happen to be working over a tensor bundle, then the Koszul connection gives covariance for free, in a sense.

In one of my revisions of Koszul connections, I tried to make this sharper. I said something to the effect of: "The notion of covariance only makes sense for a Koszul connection if it is associated to a Cartan connection." But this opens a whole other can of worms: "equivariance" versus covariance. I decided in favor of removing the statement, because a priori there is no group of transformations in the Koszul point of view (hence no kind of "variance" -- whatever that means). One can start trying to impose with extra structures, but that should probably come later in the article. Silly rabbit 14:10, 29 June 2006 (UTC)

You seem to be implying that a Koszul connection can be defined over a non-vector bundle, although all of the article text deals with vector bundles. Or perhaps I should understand this as: 1) def of cov. der. as collection of Christoffel symbols, 2) def. of Koszul connection as a map between certain vector bundles, 3) theorem that these are the same. This would contradict a statement from the lead that the Koszul connection generalizes the covariant derivative. --MarSch 07:57, 2 July 2006 (UTC)

Did I make such an implication somewhere? There are vector bundles which aren't tensor bundles in the sense of classical tensor analysis. Silly rabbit 19:04, 2 July 2006 (UTC)

Incidentally, I would actually prefer to have covariant differentiation defined in terms of Christoffel symbols equipped with a transformation law which makes the operator covariant under coordinate transformations. In these terms, it becomes quite clear how the two notions differ. A Koszul connection will then determine a covariant derivative (in this sense) if, and only if, the bundle for the connection is the tangent (or cotangent) bundle. There are other, more general notions of covariance which may be able to capture Koszul connections associated to any Cartan connection as well as connections subordinate to a pseudogroup structure. But covariance certainly cannot be meaningfully applied to a connection in a vector bundle which is in no way tied down to some kind of frames on the manifold itself. This is one reason why, if you go back and read papers prior to Koszul's 1950 paper, you find that mathematicians practically have to bend over backwards just to define a connection in a general vector bundle. Weyl (and a lot of the earlier work of Chern) uses gauges and gauge transformations, which is a pretty inelegant way of doing things. Unless you want to perform some nitty-gritty calculations, which is one reason physicists still use this formalism. Silly rabbit 19:43, 2 July 2006 (UTC)

And before anyone cries neologism, this is actually a fairly standard distinction among connection-theorists per se. Cf. with Hermann's appendix to Cartan (1984) Geometry of Riemannian Spaces, Math Sci Press, Tr. James Glazebrook, and with Spivak, M. (1999) A Comprehensive Introduction to Differential Geometry Volume 2, Publish or Perish. Cartan himself is careful to avoid discussion of covariant differentiation in anything but the classical context (hence his locution absolute differentiation in, well, just about any Cartan paper about connections). Nevertheless, when a distinction is not important, one tends to conflate the two meanings of Koszul and covariant. But: (1) a distinction must be of the utmost importance in an encyclopedia, and (2) the conflation is the result of an "abuse of language" anyway. Silly rabbit 20:02, 2 July 2006 (UTC)

[edit] Merged

I've merged Koszul connection here. I don't like the fact that a redirect, such as

# REDIRECT [[Covariant derivative#Koszul connection]]

isn't able to focus the point to a specific section of the article. Is there a way to do this? Silly rabbit 17:23, 29 June 2006 (UTC)

[edit] Notations

I feel confused about the notations of $\frac{\partial u^i}{\partial x^j}$ and $\frac{\partial}{\partial x^j} = e_j$ . The former seams to denote the derivation of (a coordinate of) a vector field with (omitted) arguments from a particular, choosen coordinate system (R^n), whereas the later is a purely symbolic notation of a vector of the tangent spaces at each point. So $e_j u_i \neq \frac{\partial u^i}{\partial x^j}$ .

Is there a way to put the formulas clear even for those who don't know already? 84.160.245.126 13:57, 19 November 2006 (UTC)

Give an eye to [1]. Also, any vector is a derivation y any derivation is a vector. This a two ways to see those objets greets--kiddo 21:00, 19 November 2006 (UTC)

[edit] Question

In which bundle does a connection on some bundle E take values? Another question: what is an integrable connection? Thanks. Jakob.scholbach 15:43, 10 April 2007 (UTC)

The first question may be ambiguous, but one answer would be the bundle Hom(TM,E). The second question is easy: an integrable connection is the same thing as a flat connection, i.e., a connection whose curvature is identically zero. Geometry guy 21:13, 10 April 2007 (UTC)

For the first question, the most obvious approach is to take the definition of a connection defined on the sheaf Γ(E):

$(X,v)\in TM\times_M \Gamma(E) \mapsto \nabla_X v \in \Gamma(E)$

and linearize in v to obtain a mapping of vector bundles:

$(X,j^1v)\in TM\times_M J^1(E) \mapsto \nabla_X j^1(v) \in E$

Taking into account the linearity gives ∇ ∈ T^*M ⊗ J¹(E)^* ⊗ E. Of course, not everything in this bundle comes from a connection: you also need ∇ to be correctly coupled to the exterior derivative. In fact, connections do not form a vector bundle in themselves since they aren't closed under linear combinations, although they are closed under affine combinations. So connections define an affine subbundle of T^*M ⊗ J¹(E)^* ⊗ E.

Furthermore, if the connection acts on a particular (1-jet of a) section v of E, then you do indeed get something in Hom(TM, E). Physicists, and some mathematicians, make this clear notationally by using abstract index notation. An expression such as

∇_a v^B (which should really be written (∇_a v)^B to indicate dependence on the derivative of v)

indicates an operation which accepts a vector in the tangent bundle TM (carrying lowercase latin indices a,b,...) to produce an element of the bundle E (whose indices are A, B, ...).

Sometimes a more useful answer to your question, though, is to consider the pullback of the connection to the total space of E. In this situation, a connection is a section of Ω¹(E,TE). Regarded as a bundle mapping TE → TE over E, the connection is required to be a projection onto the kernel of π_* where π : E → M is the submersion of the bundle. (Note that projections are also closed under affine combinations.) This is more in the spirit of Ehresmann connections.

In reference to G.G.'s answer to your second question, have a look at the second example in integrability condition. Granted this is for the curvature of a Riemannian manifold, but essentially the same approach works to show that a flat connection locally admits a parallel frame. (I.e., it is integrable in the sense of Frobenius.) Silly rabbit 21:47, 16 April 2007 (UTC)

[edit] "Tensor"

The Remarks section says:

The covariant derivative can be described by a "tensor" in a fixed coordinate chart, but it is not a true tensor in the sense that it is not invariant under coordinate changes.

I'm a bit confused here. The very reason for introducing a covariant derivative is that it is covariant (under coordinate changes), right? More precisely, if $T$ is a tensor field of type (r,s), then $D T$ (defined as $DT(X) = \nabla_X T$ ) should be a tensor of type (r,s+1). Cf. also the EOM entry. I think that remark needs some correction (or should be removed). --84.130.78.245 (talk) 15:06, 5 March 2008 (UTC)

Yes, the statement is confusing. It means that the covariant derivative of the elements of the coordinate frame do not transform tensorially. Symbolically, the indexed quantity defined in a coordinate system xⁱ by

$\nabla_{\partial/\partial x^i} \frac{\partial}{\partial x^j} = \sum_{k=1}^n\Gamma_{ij}^k\frac{\partial}{\partial x^k}$

does not transform tensorially under changes of the coordinate system. It probably should be reworded somehow. Silly rabbit (talk) 15:15, 5 March 2008 (UTC)

"Although Christoffel symbols are written in the same notation as tensors with index notation, they are not tensors. Indeed, they do not transform like tensors under a change of coordinates". Perhaps this link will help. --Ancheta Wis (talk) 18:04, 5 March 2008 (UTC)

Yes, that seems to fit much better. Perhaps one should even extend it, saying: "Although Christoffel symbols and partial derivatives are written in the same notation [...] neither of them are tensors." And this would probably go into the section "Coordinate description". --84.130.69.214 (talk) 21:45, 5 March 2008 (UTC)

In this vein, I propose striking the problematic sentence from the Remarks section. --Ancheta Wis (talk) 10:25, 6 March 2008 (UTC)