Talk:Covariant derivative/Archive 1

Archive 1

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)

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 field changes with under infinitesimal diffeomorphisms induced by a second vector field. The covariant derivative tells you how one vector field changes along the direction of a second vector. You use the first to see how a vector field changes under diffeomorphisms, and the second to see how a vector field 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)

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)

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)

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)

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:

$(X,j^{1}v)\in TM\times _{M}J^{1}(E)\mapsto \nabla _{X}j^{1}(v)\in E$

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)

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

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

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

"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 $DT$ (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)

Product rule?

I'm trying to grok Christoffel symbols and I feel like I'm missing a bit of insight... It looks to me that the following is an application of the product rule:

$\nabla _{\mathbf {v} }{\mathbf {u} }=\left(v^{i}u^{j}\Gamma ^{k}{}_{ij}+v^{i}{\partial u^{k} \over \partial x^{i}}\right){\mathbf {e} }_{k}=\left(u^{j}\left(v^{i}\Gamma ^{k}{}_{ij}\right)+v^{i}{\partial u^{k} \over \partial x^{i}}\right){\mathbf {e} }_{k}$

in that (in a handwavy sense) the total derivative with respect to the moving basis is the sum of u times the partial derivative of $\nabla$ wrt v plus $\nabla$ times the partial derivative of u. I'm not seeing the details, but this feels like it's more-or-less right. Is it? Could someone explain? In general, it would be very nice to have some diagrams and some more step-by-step details of how the covariant derivative appears in the example of cylindrical coordinates. —Ben FrantzDale (talk) 13:54, 8 December 2009 (UTC)

Slight incoherent Christoffel symbols index position

The article does not assume a symetric (torsion-free) connection, so it is not necessarily true that

$\Gamma ^{k}{}_{ij}=\Gamma ^{k}{}_{ji}$.

For example, in the section "coordinate description" the definition of the covariant derivative uses the "derivative index" of the Christoffel symbol to be the first lower index. The expression for the covariant derivative of a vector field is coherent with the definition used. On the other hand, the expressions for the covariant derivative of a dual field and for a general tensor field use the second lower index of the Christoffel symbol as the "derivative index".

My suggestion is to choose one convention and use it coherently. Personally, I'd rather use the second lower index of the Christoffel symbol as the "derivative index" because it maintains some coherence with the position of the index added by the semi-colon notation, but I have seen different texts using both of these conventions.

--Marcelo Roberto Jimenez (talk) 15:19, 15 July 2010 (UTC)

Wrong order?

Given a function $f\,$, the covariant derivative $\nabla _{\mathbf {v} }f$ coincides with the normal differentiation of a real function in the direction of the vector v, usually denoted by $f{\mathbf {v} }$ and by $df({\mathbf {v} })$.

The text is using the juxtaposition $f{\mathbf {v} }$ of a scalar field f and a vector v to denote multiplication with a scalar. I think the application of a vector (as a directional derivative operator) to a scalar field should be written in the opposite order, ${\mathbf {v} }f$. Otherwise there is no way to know which is meant. Cacadril (talk) 17:57, 10 September 2009 (UTC)

What is the rationale in using the "vector calculus notation" for vector fields anyway? In virtually every modern differential geometry book the tangent vectors are denoted by capital Roman letters. Expressions like Xf or X[f] are commonplace, but I don't remember seeing vf anywhere else. At least for me bold symbols for vectors suggest that one works in an Euclidean space, and then the directional derivative of a function along a vector field can be written as ${\boldsymbol {v}}\cdot \nabla f$. Lapasotka (talk) 21:35, 19 May 2011 (UTC)

Nonlinear covariant derivative

I have been thinking about writing an article which covers on a relatively high level (possibly nonlinear) Ehresmann-connections on slit tangent bundles TM\0. (I am cutting away the zero section to stay in the smooth category in the case of Finsler-metrics.) This is a rather peculiar topic because in this case it makes sense to talk about nonlinear covariant derivatives. Note that this would not make sense on general fibre bundles, because there is no linear structure that is needed to define a "covariant derivative", and an analogous definition of a nonlinear covariant derivative for a general vector bundle would not correspond to any Ehresmann connection. There is some material in double tangent bundle on this, but an individual article would be in place so that one can describe the relevant matter there (in particular the torsion and tension of a connection, which only make sense on TM\0.)

My concern is how to incorporate this article to the other connection-related articles. Mathematically it should be the most general case of "a covariant derivative on a smooth manifold", but it is still quite marginal research interest and probably not very well known outside Finsler-community. Especially I would like to hear opinions from the people with a background in general relativity or continuum mechanics. I have understood that there has been some interest in Finslerian space-times and elastic materials, which definitely require this kind of machinery. Lapasotka (talk) 23:17, 6 October 2010 (UTC)

Any objections on incorporating the general definition into this article?

A covariant derivative along X∈TM is a mapping D_X:Γ(TM)→Γ(TM) in the module of smooth vector fields on M such that

• $D_{X}(\alpha Y+\beta Z)\,=\,\alpha D_{X}Y+\beta D_{X}Z$ for every $\alpha ,\beta \in \mathbb {R}$.
• $D_{X}(fY)\,=\,X[f]Y+fD_{X}Y$ for every $f\in C^{\infty }(M)$.

I also suggest switching to the standard notation for vector fields in differential geometry. Lapasotka (talk) 21:53, 19 May 2011 (UTC)

I'd like to see a separate article on the topic of nonlinear covariant derivatives. I'm interested in the topic, but I don't think it should supercede the linear case here. I'm also interested in the invariants that can be constructed from such an object (curvature, torsion, etc.) I think there is ample scope for a new article that could focus on nonlinear connections in TM\{0}. Sławomir Biały (talk) 21:00, 3 July 2011 (UTC)

Isometrically embedded

From the lead: "In the special case of a manifold isometrically embedded into a higher dimensional Euclidean space..." Shouldn't we say Riemannian manifold there? I'd be bold and do it myself even though I'm pretty new to WP math editing, but I'm learning that the lead of a math article is usually the most contentious part, so I thought I'd better post here first. — ChalkboardCowboy^[T] 13:26, 12 July 2012 (UTC)

What other kind of differentiable manifold can you isometrically embed in Euclidean space? Cloudswrest (talk) 15:00, 12 July 2012 (UTC)

This "formal definition" is as wrong as it gets.

See http://fr.wiki.x.io/wiki/D%C3%A9riv%C3%A9e_covariante#D.C3.A9finition_formelle for the correct definition. --92.204.8.118 (talk) 19:49, 1 October 2012 (UTC)

Note that one of the errors was mentioned two years ago: http://en.wiki.x.io/wiki/Talk:Covariant_derivative#Formal_definition_typo.3F but nobody fucking cares. --92.204.8.118 (talk) 20:01, 1 October 2012 (UTC)

The section you deleted was correct. I reverted your deletion. If you still maintain that it is wrong, please explain why here before you delete it again. JRSpriggs (talk) 21:42, 1 October 2012 (UTC)

Add Images

Can anyone add some images to illustrate these definitions in article? Thank!113.160.85.34 (talk) 12:58, 2 January 2013 (UTC)

I'm not a good graphic artist, but here is a suggestion for a picture. Draw a circle on a flat sheet of paper. Travel around the circle at a constant speed. The derivative of your velocity, your acceleration vector, always points radially inward. Roll this sheet of paper into a cylinder. Now the (Euclidean) derivative of your velocity has a component that sometimes points inward toward the axis of the cylinder depending on whether you're near a solstice or an equinox. This is the (Euclidean) Normal component. The Covariant derivative component is the component parallel to the cylinder's surface, and is the same as that before you rolled the sheet into a cylinder. Cloudswrest (talk) 19:37, 2 January 2013 (UTC)

I'll give it a shot, but I'll need some time, as I'm currently busy with other stuff and I also need to read a bit more on the subject before I feel comfortable about approaching it. Thanks for the heads up, Cloudswrest. — Kieff | Talk 16:41, 4 January 2013 (UTC)

I removed weasle words tag in Introduction section

I don't understand the rationale for the tag and there is no discussion by whomever added it. Until clarification is added, it serves no purpose - imho.173.189.74.254 (talk) 20:57, 30 August 2014 (UTC)

It was a [which] tag. It was immediately followed by what it was asking for, so its removal was appropriate. —Quondum 21:17, 30 August 2014 (UTC)

For covector fields, please a reference

The formula given for the covariant derivative of covector field, where can I corroborate? kmath (talk) 02:30, 19 December 2014 (UTC)

Formal definition typo?

I think there is a typo in the "Formal definition" section. Under "Vector fields", v is a vector, and w is a vector field, yet, in the first rule, their linear combination (fv+gw) is taken. (Nota bene: I'm trying to learn all this from the article itself, so unfortunately, I can't correct it myself, since I don't know what should be there. I just noticed the apparent inconsistency.) 92.249.133.91 (talk) 22:01, 17 August 2010 (UTC)

I fixed it. - Patrick (talk) 13:28, 5 April 2015 (UTC)

Here is another comment:

It is not clear what (formal) language is used in the formal definition. The first sentence of the subsection "Functions" in section "Formal definition" appears to contain a nonstandard "the [...] the" construction in the English language. If this is an informal definition one could just remove the first of the two "the" in the construction.

Clarification please

You have "Incidentally, this particular expression is equal to zero, because the covariant derivative of a function solely of the metric is always zero." That is not incidental: it is vitally important. Do you mean $\nabla _{c}{\sqrt {-g}}=({\sqrt {-g}})_{;c}=0$ ? If so, make that clear in the equation! Also, if I am interpreting this statement correctly, you also mean $\nabla _{c}f(g_{\mu \nu })=(f(g_{\mu \nu }))_{;c}=0,$, right? If so, please include that also. TonyMath (talk) 14:46, 28 April 2015 (UTC)

As stated, the statement was nonsensical. I have removed the bit about densities; this is not directly relevant to the covariant derivative, but relates to the generalization as applied to a tensor density, and IMO, it should be confined to that article. —Quondum 16:48, 28 April 2015 (UTC)

What do you mean, nonsensical?

The covariant derivative is the same as the partial derivative in a reference frame where $g_{\mu \nu ,\sigma }=0$. In such a reference frame, the determinant of g is constant to order 1 as are any functions of it.

Therefor, I have reverted. JRSpriggs (talk) 04:28, 29 April 2015 (UTC)

@JRSpriggs: I do not believe that this inclusion is appropriate in this context. The article is about the covariant derivative of tensors, not of every mathematical object that you choose to generalize the concept to. Also, the statement "this particular expression is equal to zero, because the covariant derivative of a function solely of the metric is always zero" is downright wrong, at least if it is referring to the previous expression. Please do not introduce tensor densities in to every article on tensors; in this article, the most that is warranted is a "see also", with the covariant derivative of a tensor density to be explained in a article dedicated to tensor densities. —Quondum 04:38, 29 April 2015 (UTC)

No. It is not wrong.

$0=({\sqrt {-g}})_{;c}=({\sqrt {-g}})_{,c}-{\sqrt {-g}}\,\Gamma ^{d}{}_{dc}$

and thus

$({\sqrt {-g}})_{,c}={\sqrt {-g}}\,\Gamma ^{d}{}_{dc}$

are true facts which I have seen in at least two books on GR. JRSpriggs (talk) 05:02, 29 April 2015 (UTC)

That might be the case if the context is stipulated, but which is missing here. "the covariant derivative of a function solely of the metric is always zero" is not true if considered in terms of components of a metric representation; there are many unmentioned conditions to be applied to make this condition true, such as tracking of what the weight of the function is to even make sense of the derivative, not stated; it is not any function, but only functions that are densities to which this statement applies. What value does this add to the article? —Quondum 06:02, 29 April 2015 (UTC)

I have seen $({\sqrt {-g}})_{;c}=0$ mentioned in the context of General Relativity which tells me that somehow it should be indicated in the article. I have been told that the covariant derivative of a 2nd rank tensor components can be expressed in terms of an ordinary derivative using the determinant of the metric as $t_{;c}^{ij}=({\sqrt {-g}})^{-1}*({\sqrt {-g}}*t^{ij})_{,c}$. Can anyone confirm? TonyMath (talk) 06:28, 29 April 2015 (UTC)

The expression ${\sqrt {-g}}$ is not a tensor scalar, and so its covariant derivative is not defined in this context. If, however, one considers it as a tensor density of the appropriate weight, and uses the definition of a covariant derivative as extended to cover tensor densities (and this definition seems to require information about the weight in its definition), and provided the object t has the correct weight, and the coordinate system is suitably restricted, and the basis is a coordinate basis ... then maybe the equation you give holds. But it is almost certainly not true of a general tensor or even tensor density t^ij. Tensor densities get used in the context of general relativity, but not every calculation tool should be mentioned in an article. It clutters and confuses, especially when assumptions are made but not stated, as is the case here. —Quondum 14:54, 29 April 2015 (UTC)

What is a non-tensor scalar? (I have forgotten most about this topic.) YohanN7 (talk) 15:46, 29 April 2015 (UTC)

Guess the answer is here. Such things should probably not be called scalars at all in this context because they do change under transformations. YohanN7 (talk) 16:06, 29 April 2015 (UTC)

I thought I'd weight in. JRSpriggs is correct, but it seems that the text concerning how to differentiate densities is not ideal. More generally, one can compute the Lie derivative of a density along the flow of any vector field. (The case of the covariant derivative is obtained by specializing to a vector field obtained in a natural way from the exponential map.)

The pullback of a density f under a diffeomorphism φ is defined so that $\int _{U}\phi ^{*}f=\int _{\phi (U)}f$ for all open sets U. So, if $\phi _{t}$ is a one-parameter group of diffeomorphisms with generator $X={\frac {d}{dt}}\phi _{t}|_{t=0}$, the Lie derivative along X is the unique density satisfying $\int _{U}L_{X}f=\lim _{t\to 0}t^{-1}\int _{U\cap \phi _{t}(U)}(\phi _{t}^{*}f-f)$.

Now, when working with densities, it is typical to work in coordinates. This is an association of a scalar function $f[\mathbf {e} ]$ to every basis e of the tangent space, together with the relation under change of basis that $f[\mathbf {e} A]=|\det A|f[\mathbf {e} ]$ for any change of basis matrix A in GL(n). In particular, when $\mathbf {e} =(\partial _{1},\dots ,\partial _{n})$ is a coordinate basis, the product of $f[\mathbf {e} ]$ with the Lebesgue measure in the coordinate system, $f[\mathbf {e} ]|dx^{1}\dots dx^{n}|$, is actually invariant under changes to the basis (by the transformation law for f[e], together the Jacobian change of variables theorem applied to the Lebesgue measure).

The Lie derivative of the density represented by f[e] along a vector field X is then $X(f[\mathbf {e} ])+\partial _{i}X^{i}$. (This last term is "div X" in the coordinate system, which admits the usual interpretation as the infinitesimal change in the volume of the coordinate box $|dx^{1}\cdots dx^{n}|$ under the flow of X.) Sławomir Biały (talk) 00:03, 30 April 2015 (UTC)

Aren't you missing the wood for the trees? Is "the covariant derivative of a function solely of the metric is always zero" is applied correctly? g is not a function solely of the metric; it is also a function of the basis (as you describe your f). Do you think that the covariant derivative as defined in this article applies without generalization to tensor densities (if not, this generalization must be defined, and indeed tensor densities must be defined), and do you support its inclusion here rather than dealing with it at that its mention here rather than simply in the Tensor density article? —Quondum 03:25, 30 April 2015 (UTC)

Instead of suggesting that the covariant derivative of a tensor density belongs in tensor density, you started out by saying that something was "nonsensical" and removed a chunk of text from the article. What do you expect? You are the one not sticking to "the point" as you call it, because you argue about technicalities (and JR and S had to go long stretches to explain them to you). YohanN7 (talk) 09:52, 30 April 2015 (UTC)

It could belong either here or there. The articles name is the generic Covariant derivative YohanN7 (talk) 09:52, 30 April 2015 (UTC)

How about I try to break it down a bit:

• The name of an article does not give its scope (you might want to check article naming guidelines on this). We must determine the scope, for example as in Tensor having been chosen to cover the real finite-dimensional case. Any generalizations belong in duly labelled generalization section, duly describing the direction of generalization, and not unintroduced as though they are part of the main topic. I have already indicated that the scope of this article, as it stands, is the covariant derivative of (real, finite-dimensional) tensors. If we wish to generalize it to tensor densities, that is a decision for consensus to determine. I would oppose, as I would argue that this generalization belongs in a separate article (either in Tensor density or its own article) or in a generalization section of this article, and we need an article about the covariant derivative of tensors without the confusion of densities in the mix.
• I objected to a nonsensical statement. That statement would allow me to conclude that the covariant derivative of 1+g is defined and is zero, but this expression is not even a tensor density and thus has no covariant derivative, even under the generalization. That it is nonsensical is a good argument for its removal and non-re-inclusion, at least until a sensible version is produced. Don't misread me as saying that the covariant derivative (of tensor densities) of sqrt(g) is not zero.
• There is insufficient context and definition given to allow interpretation of the statements. The covariant derivative (as used in the statement) is not defined prior to its use, or anywhere other than implied in the link tensor density. This means that the piece is out of place.

Sorry that I've combined what I consider several reasons, each of which stands on its own, for removing the disputed part. I hope I have adequately separated them here. —Quondum 14:25, 30 April 2015 (UTC)

Sorry I stated my earlier result incorrectly, it was a actually a covariant 'divergence' and given in 2D i.e. $t_{;j}^{ij}=({\sqrt {-g}})^{-1}*({\sqrt {-g}}*t^{ij})_{,j}$ but I could not tell you about the general conditions. I want to know if it holds in 3D. I could use a reference if anyone has seen this. TonyMath (talk) 18:01, 30 April 2015 (UTC)

Is t assumed to be skew? Otherwise, there is a Christoffel term $\Gamma _{jk}^{i}t^{jk}$ that isn't likely to disappear. Sławomir Biały (talk) 22:33, 30 April 2015 (UTC)

As a matter of fact $t^{ii}=0$ for both the 2D and 3D cases I am dealing with.04:15, 1 May 2015 (UTC)

I am sorry that I have not been able to participate in this discussion for two days due to the failure of my Internet connection. Of course, the statement that the covariant derivative of any function of the metric is zero assumes that the covariant derivative of the differentiable function in question is defined, otherwise it is not applicable. This fact is a simple consequence of the chain rule for differentiation. In a reference frame where the partial derivative of the metric is zero (i.e. the Christoffel symbols are zero), we have

${\frac {\partial f(g_{\mu \nu })}{\partial x^{\sigma }}}={\frac {\partial f(g_{\mu \nu })}{\partial g_{\mu \nu }}}g_{\mu \nu ,\sigma }={\frac {\partial f(g_{\mu \nu })}{\partial g_{\mu \nu }}}0=0\,.$

OK? JRSpriggs (talk) 02:04, 3 May 2015 (UTC)

Ok but assume I am dumb and/or pedantic. How many steps/definitions between that and $\nabla _{c}f(g_{\mu \nu })=(f(g_{\mu \nu }))_{;c}=0,$ if there are any?TonyMath (talk) 08:39, 7 May 2015 (UTC)

The gravitation and Cosmology lecture notes of Christian von Schultz - Tuesday 2008-11-04 do mention that "The metric is covariantly constant" i.e. $D_{\mu }g_{\nu \lambda }=0$ as written in their notation. Still need a reference to my covariant divergence question though. TonyMath (talk) 09:08, 7 May 2015 (UTC)

Although the article seems not to mention this fact, one way of defining the covariant derivative is as the tensor which agrees with the partial derivative in a locally inertial frame of reference. So one could compute it by: (1) transforming to an inertial frame of reference (one where the partial of the metric is zero), (2) taking the partial derivative of the tensor and assuming the result is a tensor, then (3) transforming back to the original coordinate system. JRSpriggs (talk) 20:28, 7 May 2015 (UTC)

Wouldn't the appropriate place for this be at the ref desk? —Quondum 20:42, 7 May 2015 (UTC)

Maybe so but these identities are so useful, this article should give at least a few. About that question $t_{;j}^{ij}=({\sqrt {-g}})^{-1}*({\sqrt {-g}}*t^{ij})_{,j}$, it holds indeed for skew-symmetric tensors $t^{ij}$ as indicated in Integration in General Relativity by Andrew DeBenedictis, but by using the Maple computer algebra system, I found it it is true for a symmetric 2D tensor $t^{ij}$ for an isotropic metric i.e. $g_{ij}=f(x,y,z)*\delta _{ij}$. This is used in a paper by Yale, Mann and Ohta in 2+1 gravitational cosmology. In 3D, it almost holds true for a symmetric traceless tensor where $t^{ii}=0$ but there is a multiplicative factor of 5/3 that has to be dealt with. I cannot find this result in any book but it is so useful to me. No ref desk has this invaluable info (sigh!). TonyMath (talk) 15:58, 9 May 2015 (UTC)

I meant that you should ask at WP:Reference desk/Mathematics as the appropriate forum in WP for asking this. You seem to be thinking of something else. You'll get essentially the same crowd of editors responding, and if it becomes clear that there is something not included in an article that should be, it could then trigger a discussion on the relevant article's talk page; don't assume ahead of time that it should be this one. As I've tried to suggest, I do not consider this article as the appropriate home for a discourse on the manipulation of tensor densities. It might belong in other pages though, such as on the page Tensor density. That page already contains far more than this page does on the subject of the covariant derivative of a tensor density. —Quondum 16:59, 9 May 2015 (UTC)

Remarks on notation

In the equations that follow Einstein Summation Notation is used extensively. This notation requires that any term in which an upper index has a matching lower index be interpreted to be a sum of the term on all possible (matched) values of the index. The range of the indices may be implicit from context and not specified explicitly. It is generally easy to tell the difference between a superscript being used as an index, and one meaning power (square, cube, etc.) from context.
In the following examples, the index runs from 1 to 3:
Z^k = A_k + Xⁱ_i means Z^k = A_k + X¹₁ + X²₂ + X³₃
Q^m_j¹R^j means Q^m₁¹R¹ + Q^m₂¹R² + Q^m₃¹R³
Please add something like that. Also, I find some of the other notation elusive. Is the reader coming to this article supposed to know what <•;•> means? Really? You shouldn't preach to the choir. 173.189.77.81 (talk) 12:15, 19 January 2016 (UTC)

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

It appears that the Einstein notation is being used (now), if I understand correctly. I do not understand why the editor(s) chose to implement it and NOT explicitly acknowledge/mention/note the fact BEFORE (as is blatantly obvious) its first use. Perhaps he believes the reader can't possibly interpret XⁱY_i any other way? Wow. I suggest the following be added in its own section just prior to Informal definition... section.

You're welcome to go ahead and add that. I don't think anyone would object. Sławomir
Biały 12:59, 19 January 2016 (UTC)

Small change in the (local coordinate) formular for tensor fields

I swapped the indices "c" and "d" for the Christoffel symbols for the vector part because the connection is not necessarily torsion-free. I do not know the case of a tensor density. — Preceding unsigned comment added by 93.207.2.74 (talk) 22:37, 16 March 2016 (UTC)

Very poor choice of notation

The subsection titled Vector fields contains this passage:

"A covariant derivative $\nabla$ at a point p in a smooth manifold assigns a tangent vector $(\nabla _{\mathbf {v} }\mathbf {u} )_{p}$ to each pair $(\mathbf {u} ,\mathbf {v} )$, consisting of a tangent vector v at p and vector field u defined in a neighborhood of p, such that the following properties hold (for any vectors v, x and y at p, vector fields u and w defined in a neighborhood of p, scalar values g and h at p, and scalar function f defined in a neighborhood of p):

1. $\left(\nabla _{\mathbf {v} }\mathbf {u} \right)_{p}$ is linear in $\mathbf {v}$ so

   $\left(\nabla _{g\mathbf {x} +h\mathbf {y} }\mathbf {u} \right)_{p}=\left(\nabla _{\mathbf {x} }\mathbf {u} \right)_{p}g+\left(\nabla _{\mathbf {y} }\mathbf {u} \right)_{p}h$

2. $\left(\nabla _{\mathbf {v} }\mathbf {u} \right)_{p}$ is additive in $\mathbf {u}$ so:

   $\left(\nabla _{\mathbf {v} }\left[\mathbf {u} +\mathbf {w} \right]\right)_{p}=\left(\nabla _{\mathbf {v} }\mathbf {u} \right)_{p}+\left(\nabla _{\mathbf {v} }\mathbf {w} \right)_{p}$

3. $(\nabla _{\mathbf {v} }\mathbf {u} )_{p}$ obeys the product rule; i.e., where $\nabla _{\mathbf {v} }f$ is defined above,

   $\left(\nabla _{\mathbf {v} }\left[f\mathbf {u} \right]\right)_{p}=f(p)\left(\nabla _{\mathbf {v} }\mathbf {u} )_{p}+(\nabla _{\mathbf {v} }f\right)_{p}\mathbf {u} _{p}$."

This could hardly be written more confusingly.

First of all, using g and h "scalar values" is already confusing for two reasons: 1) The letters g and h in the context of differential geometry are almost always reserved for functions, not scalars; and 2) For people coming to this article to try to understand covariant derivative, the scalars should be real numbers and the phrase "real numbers" is much clearer than the phrase "scalar values".

Secondly: In the first equation, equation 1., it is a very bad idea to place the so-called scalar values on the right side of the expressions $\left(\nabla _{\mathbf {x} }\mathbf {u} \right)_{p}$ and $\left(\nabla _{\mathbf {y} }\mathbf {u} \right)_{p}$. For, each of these represents a vector field which can act on a function to its right ... as well as a constant function represented by a "scalar value".

Thirdly: In equation 3., the expression $\left(\nabla _{\mathbf {v} }f\right)_{p}\mathbf {u} _{p}$ would be much clearer if instead of $\left(\nabla _{\mathbf {v} }f\right)_{p}$ it read $\mathbf {v} \left(f\right)_{p}$. The action of a vector, at a point, on a function (as a differential operator) is much better understood than covariant derivative ... which people come to this article in order to understand.50.205.142.35 (talk) 19:52, 13 November 2019 (UTC)

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?

I think you can use differentiation of a composite function. D(f(g(x))=f’(g(x))•g’(x). If g(x) is the metric then it’s covariant derivative is zero and therefore by the chain rule any function solely of it is zero because the derivative of the metric tensor is 0 and it is at the end of the chain.

You have to know that the covariant derivative follows the chain rule and that the covariant derivative of the metric tensor is always zero. Justintruth (talk) 01:56, 1 August 2020 (UTC)

inconsistent notation of connection coefficients in the article

In the "Informal definition..." the first lower index "i" of $\Gamma^k_{ij}$ denotes the direction of the derivative, while in the section "Coordinate description" the second lower index "j" of $\Gamma^k_{ij}$ plays that role. Personally, I would prever to use the first lower index for that purpose as it is done in the informal definition (and many authors agree), but at the very least the article should be self-consistent. - Peter Schupp 91.7.253.201 (talk) 11:25, 15 February 2023 (UTC)

not all vectors are covariant

Wiki nowadays says "A vector may be described as a list of numbers in terms of a basis, but as a geometrical object a vector retains its own identity regardless of how one chooses to describe it in a basis. This persistence of identity is reflected in the fact that when a vector is written in one basis, and then the basis is changed, the components of the vector transform according to a change of basis formula. Such a transformation law is known as a covariant transformation."

This is not correct, not all vectors are covariant. In fact, position vector, the most basic one, is usually contravariant.

A vector (i.e. a first order tensor) is neither covariant nor contravariant, it is invariant; that is the whole point of tensors. As you say the vector is independent of the basis you describe it in. You can describe the vector with contravariant coordinates in a covariant basis, or with covariant coordinates in a contravariant basis. The reason that basis vectors "change" under a coordinate transformation is that you switched to a different set of basis vectors; the original basis vectors themselves didn't change. Also, calling a set of coordinates by themselves a vector (what you probably mean by a "position vector"), while quite common especially in physics, in my opinion just confuses everyone (in the context of differential geometry at least; in euclidean geometry you can mostly get away with it). -P

92.34.140.23 (talk) 10:14, 9 June 2022 (UTC) P

Don't basis vectors always transform covariantly by definition? And a vector can then either transform with the same transformation map or its inverse i.e. either covariantly or contravariantly? Codeinpappi (talk) 12:56, 1 May 2023 (UTC)

Recently added citation of writing of Niccolai, Edoardo

I recommend removal of the citation to a writing of Niccolai, Edoardo (recently added by JeppOne (talk · contribs)), unless it can be supported as reliable and noteworthy.

Please note that there are similar discussions at Talk:Bounded mean oscillation § Recently added citation of writing of Niccolai, Edoardo, Talk:Einstein–Cartan theory § Recently added citation of writing of Niccolai, Edoardo, and Talk:Covariant derivative § Recently added citation of writing of Niccolai, Edoardo. There is also discontinued discussion at User talk:JeppOne § Articles by Niccolai, Edoardo. —Quantling (talk | contribs) 13:41, 1 May 2023 (UTC)

I'm just trying to let people know that there are some important writings by my professor. That's all. And I do it for the love of science. Maybe you should inquire about the person in question before "triggering" an unfair censorial judgment, which is to the detriment of the knowledge itself.

Thank you very much JeppOne (talk) 11:57, 2 May 2023 (UTC)

Hyphen in "coordinate-free language"

I edited the article to add a hyphen because

• "Coordinate free language" (no hyphen) means a "free language" that is of type "coordinate"
• "Coordinate-free language" (with hyphen) means a "language" that is of type "coordinate free"

and we mean the latter. —Quantling (talk | contribs) 14:24, 26 July 2023 (UTC)