Talk:Christoffel symbols/Archive 1

Archive 1

Coordinate expression for curvature

I think this section is not directly relevant, it is ok to leave a coordinate description of curvature tensor, but defining Ricci curvature and Weyl curvature is too much. Tosha 09:35, 6 January 2006 (UTC)

A quick glance at the various curvature articles makes it clear that none(?) of them give coordinate expressions. As these are useful, and don't seem to be listed elsewhere, they can be left here. One possible idea would be to move them to an article Coordinate expression for curvature, but there doesn't seem to be a strong need. Anyone who would actually need to work with indexed version would end up looking at this article anyway. .. linas 15:20, 6 January 2006 (UTC)

I think it should be moved it to Curvature of Riemannian manifolds. Tosha 11:28, 9 January 2006 (UTC)

Most of what that article states is true for pseudo-Riemannian manifolds a well. Is there a better way of saying "Riemannian and pseudo-Riemanian", which makes for very long article titles and sentences? linas 16:56, 9 January 2006 (UTC)

Assuming no torsion?

Maybe I am naive, but I think that in order to derive the expression for the Christoffel symbols from the requirement that the covariant derivative of the metric vanishes, one has also to ASSUME that they form a torsion-free connection, i.e., that they are symmetric with respect to the last 2 indices. Can any real mathematician resolve this issue? 194.94.224.254 16:25, 24 January 2007 (UTC)

"Natural" basis

I would just like to comment that I do not think I was reverted appropriately. If someone would care to define what a "natural" basis is, then I will consider 86.101.197.154's version. Otherwise, it seems sort of silly to insist on a version of the lead in which some of the terms are misleading or undefined. I presume that 86.101.197.154 means "coordinate basis" (what the article later terms a holonomic basis). This is a more reasonable restriction, although some authors do not even require this much (and indeed it can occasionally be an unpleasant restriction). Since the sentence under dispute was about the most general sort of thing one can write down and still call a Christoffel symbol, using coordinate bases seems unnecessarily restrictive. I have, at any rate, replace the word "natural" with the more meaningful and accurate word "coordinate." Silly rabbit (talk) 13:37, 12 March 2008 (UTC)

I make it of model so please help me about this....... —Preceding unsigned comment added by 59.98.68.137 (talk) 07:58, 9 October 2008 (UTC)

Technical

I think I have nearly the required background to understand this article, but I really can't at the moment. A list of prerequisites, as described on Wikipedia:Make technical articles accessible, would be helpful. —BenFrantzDale

Yeah, me too. Can anyone make this article more accessible? —Preceding unsigned comment added by 70.64.188.255 (talk • contribs)

Off the top of my head I'd say that in order to understand this article you'd at least need Einstein notation and some understanding of covariance and contravariance and the metric tensor. The last two are linked to kinda early on in the article.

How do people here feel about using a line in the lead in the form of In order to understand them you need... like in Wikipedia:Make technical articles accessible#Articles that are unavoidably technical? — Laura Scudder ☎ 16:08, 25 August 2006 (UTC)

Looks like everyone agrees on this, but I thought I'd say it too: there's no way you could understand this page unless you already had an excellent understanding of Christoffel symbols. I'm not exactly a general audience (2nd year physics PhD) but even I find this page pretty worthless. Ckerr 11:16, 15 August 2007 (UTC)

Yeah, it's not really helpful right now. — Laura Scudder ☎ 13:53, 15 August 2007 (UTC)

I agree and disagree. It is not true one needs to understand the topic to actually understand the article. Indeed all one needs to know is the concept of covariant derivative. There are many such people that may understand that concept but not know how to manipulate the Christoffel symbols. Just like it is intuitively simple to understand the single variable derivative in introductory calculus but the quotient rule may seem abstruse. So the complaint is similar to saying quotient rule is opaque and one must already know it to find it useful. On the other hand, I have an idea why such a complaint is made. Often, physicist types learn such concepts in coordinate form. I notice several famous physicist treatises on such material define the covariant derivative in terms of Christoffel symbols. That would seemingly be why this article seems to presuppose knowledge of them.

The solution seems to be to give a very brief intuition on covariant differentiation and how the symbols arise. More than is currently in the lede. Other than that, there is nothing more that can be done without just duplicating material. --C S (talk) 08:27, 15 August 2008 (UTC)

OK, I added an intro paragraph that should make sense if you understand matrices and derivatives, and you realize that there's stuff beyond all that. Yes, it's not really a 'tensor' but that depends on whether you're a math major or physics major blah blah blah. And it applies to more than non-euclidian geometry, probably, but that's beyond the scope of the intro. and my understanding. OsamaBinLogin (talk) 23:56, 25 March 2009 (UTC)

Some errors in anholonomic expression

The anholonomic expression was written with coordinate derivatives, but there are no coordinate derivatives for an anholonomic basis. I changed them to subscript comma derivatives, which should work as shorthand for the Lie derivative, which is what you have to use. Furthermore, the commutation coefficient terms appear to have been incorrectly copied from Misner, Thorne and Wheeler, p. 210. Their convention for Christoffel symbols reverses the lower indices (p. 209), relative to what we defined here. I corrected the expression. If anyone is in doubt, I'd like to point out that it is consistent, for example, with Exercise 1. For another sanity check, compare the torsion now, with the torsion before. —Preceding unsigned comment added by Yzarc314 (talk • contribs) 15:50, 2 February 2009 (UTC)

A few hours later, SillyRabbit reversed the sign correction I made. Given the notation of this page, the correct sign should be +. Here is a proof:

As a special case of an anholonomic basis, I will use an orthonormal frame, so that $g_{ij}=\delta _{ij}$, the Kronecker delta function. The Christoffel symbols then reduce to $\Gamma ^{k}{}_{ij}={\frac {1}{2}}(c_{kij}+c_{kji}\pm c_{ijk})$, where the $\pm$ is the sign in dispute.

The Levi-Civita connection is torsion-free, so $\nabla _{i}e_{j}-\nabla _{j}e_{i}=[e_{i},e_{j}]$. In components, this is $(\Gamma ^{k}{}_{ij}-\Gamma ^{k}{}_{ji})e_{k}=c_{ij}{}^{k}e_{k}=c_{ijk}e_{k}$.

Substituting our expression for $\Gamma$ into the left hand side, we get $\Gamma ^{k}{}_{ij}-\Gamma ^{k}{}_{ji}={\frac {1}{2}}(\not c_{kij}+\not c_{kji}\pm c_{ijk}-\not c_{kji}-\not c_{kij}\mp c_{jik})=\pm c_{ijk}$. Therefore, $\pm =+$.

If you think I made a mistake somewhere, please let me know. In the meantime, I'm going to change it back. —Preceding unsigned comment added by Yzarc314 (talk • contribs) 20:41, 2 April 2009 (UTC)

Not tensor

The current intro claims that the Christoffel symbol is a 3-tensor. This is exactly wrong (as explained later in the article). Right? We want to say that it resembles a 3-tensor, in that it has n^3 coordinates, but is not a tensor. Mgnbar (talk) 04:09, 22 April 2009 (UTC)

Yes, that is right. I fix the main article.

TomyDuby (talk) 06:57, 22 April 2009 (UTC)

"Ugly"

Describing these symbols as "ugly" is subjective

You're right. I noticed this too and will change it. ---Mpatel (talk) 17:10, 3 November 2005 (UTC)

I was wondering why it is written $\Gamma ^{m}{}_{i\ell }$ instead of $\Gamma _{i\ell }^{m}$ in wikipedia - isn't the second one what we write on the board and on paper? Does this not show up correctly for some people? Orthografer 17:06, 1 August 2006 (UTC)

Actually this question just came up at Wikipedia talk:WikiProject Physics#Indices on Christoffel symbols. It turns out that they're staggered because the ordering is important when manipulating them. I'm hoping someone there comes up with a neater explanation than mine. — Laura Scudder ☎ 15:48, 25 August 2006 (UTC)

I propose to include the definition of $g^{ij}$ in the article, as it is for example in Fundamental theorem of Riemannian geometry. I've lost some time to find appropriate definition...

It's metric tensor, there're already a couple of links to it in an article, as of now 212.75.204.52 (talk) 11:06, 12 July 2009 (UTC)

Force-free?

In the German wikipedia article use is made of the concept of a "force-free" particle, i.e one upon which no force is exerted other than gravity. This is not mentioned in the English article. Could someone explain why there is this difference in approach and what the English-language equivalent is for "force-free". Many thanks. --78.54.30.28 (talk) 22:17, 24 October 2009 (UTC)

I do not routinely look at the foreign language versions of articles. In particular, my understanding of German is weak (but better than my understanding of others, except French). So I cannot say why there is a similarity or difference between the articles.

The English expression for "force free" is "free falling".

The Christoffel symbol may be interpreted as the gravitational force field. However, due to the equivalence principle, one cannot make a coordinate-free separation of gravity from fictitious forces. That is, unlike other force fields, the gravitational force field is not a tensor. JRSpriggs (talk) 18:58, 25 October 2009 (UTC)

staggering indices

The Gamma symbols are not tensors, in the sense that they don't fit together to give a section of a tensor bundle built of out tangent and cotangent bundles. Therefore there is less rationale to stagger the indices, i.e. to leave a space below i before writing jk. We should either delete these spaces, or explain why they are needed. Tkuvho (talk) 14:54, 25 February 2010 (UTC)

I copy the earlier discussion from the physics archive: Tkuvho (talk) 14:59, 25 February 2010 (UTC)

A recent edit to Christoffel symbols replaced $\Gamma ^{m}{}_{i\ell }\,$ with $\Gamma _{i\ell }^{m}\,$ throughout the article. I seem to remember that there is actually a reason for the staggered indices, but it was some time ago that I looked into this and I never got it properly. Could somebody in the know please check, and if necessary revert while explaining the matter on the talk page? Cheers, Jitse Niesen (talk) 15:07, 25 August 2006 (UTC)

The ordering is important to the indices. So in a way the staggering is a way of remembering that

$g^{i\mu }g^{\ell \nu }\Gamma _{i\ell }^{m}=\Gamma ^{m\mu \nu }\neq \Gamma ^{\mu \nu m}$

There's probably a more fundamental way to say that, but we're delving into some old material here for me. — Laura Scudder ☎ 15:45, 25 August 2006 (UTC)

I agree that the changes Jitse noted should be reverted. ---CH 22:07, 25 August 2006 (UTC)

In most GR texts there is no space. The Christoffel symbol with all indices downstairs is defined by contracting with the metric tensor. You can avoid giving that definition by showing the space. But I think it is still necessary to write that in many books the symbol is written without the space and that it is the same as given in the article. Count Iblis 23:30, 25 August 2006 (UTC)

I have seen it both ways and I do not see the advantage of the longer form. There is no issue of ordering because the contravariant (upper) index is distinguished by the fact that it is upper rather than lower. I have never seen a Christoffel symbol with two or three upper indices. The only choice is one upper and two lower or all three lower. In the latter case, the first (leftmost) lower index corresponds to the upper index in the other case. Where is the problem? JRSpriggs 08:41, 26 August 2006 (UTC)

Well, it's just that one needs to define the symbol with all three indices downstairs by declaring that the upper index becomes the leftmost index when brought downstairs. That's not a problem, but one can avoid even mentioning this by indicating the space. Count Iblis 12:38, 26 August 2006 (UTC)

Note that the current version of the page does not have any Gamma symbols with the upper index lowered, so the advantage mentioned above is not relevant. Tkuvho (talk) 15:01, 25 February 2010 (UTC)

To Tkuvho: As one can see from the discussion which you copied over above, I agree with you. I have no objection to you removing the unnecessary spaces. JRSpriggs (talk) 02:09, 27 February 2010 (UTC)

I also have no objection, as I already stated at Talk:Geodesic, although I don't completely agree with the reasons stated there. Sławomir Biały (talk) 03:36, 28 February 2010 (UTC)

first kind vs second kind symmetric vs second kind asymmetric

Anyone using these in practice needs to know which is which. Gamma is very standard for the second kind, and the symmetric definition seems to be the most frequent in modern usage, but we still have to be careful. People should still be aware of older usages. Boud (talk) 13:34, 27 December 2010 (UTC)

See Christoffel symbols#Change of variable. The way I see it, the essential feature of Christoffel symbols is that they satisfy

${\overline {\Gamma _{ij}^{k}}}={\frac {\partial x^{p}}{\partial y^{i}}}\,{\frac {\partial x^{q}}{\partial y^{j}}}\,\Gamma _{pq}^{r}\,{\frac {\partial y^{k}}{\partial x^{r}}}+{\frac {\partial y^{k}}{\partial x^{m}}}\,{\frac {\partial ^{2}x^{m}}{\partial y^{i}\partial y^{j}}}\,.$

This allows them to be used to cancel out undesirable terms in the transformation laws of the partial derivatives of tensors, i.e. to define a covariant derivative — an adjusted partial derivative which is a tensor. Only one form of the Christoffel symbol is needed for that purpose. And for simplicity, we should assume that its properties are like those of the term

${\frac {\partial y^{k}}{\partial x^{m}}}\,{\frac {\partial ^{2}x^{m}}{\partial y^{i}\partial y^{j}}}\,.$

To wit, it should be symmetrical in the lower indices and, for each event, it should be zero in some coordinate system. Fortunately, we can choose that coordinate system to be the one, at that event, in which the partial derivatives of the metric tensor are zero, thereby making zero the covariant derivative of the metric everywhere in all coordinate systems. JRSpriggs (talk) 17:51, 27 December 2010 (UTC)

This Is the (Interpersonal) Reason the Article is Unclear

I just noticed my 10-Feb addition was removed for being "redundant." Some of us would call this rephrasing of the symbols in what are called "words" an explanation. It is such "redundancies" that make obscure mathematics clearer for the rest of us, who aren't mathematicians and don't speak mathematese. What remains clear after the undo is that the mathematicians taking charge of this article don't understand people. If you're aiming the article exclusively at the select few who already understand mathematics, you couldn't be doing a better job. JKeck (talk) 23:45, 17 February 2011 (UTC)

For reference, the text that was removed starts with "The interpretation" below:

The Christoffel symbols of the second kind, using the definition symmetric in i and j - ref name="wolfram2ndkind" -, $\Gamma _{ij}^{k}$ (sometimes Γ^k_ij ) are defined as the unique coefficients such that the equation

$\nabla _{i}e_{j}=\Gamma _{ij}^{k}e_{k}$

holds, where $\nabla _{i}$ is the Levi-Civita connection on M taken in the coordinate direction e_i. The interpretation of $\Gamma _{ij}^{k}$ is that it is the kth component of $\partial e_{i}/\partial x^{j}$; the indices denote the basis vector being differentiated (i), the coordinate with respect to which it is being differentiated (j), and the component of the resulting derivative vector (k). - ref -Bernard F. Schutz, A first course in general relativity (Cambridge Univ. Press, 1990), 135.-ref -

I tend to agree with JKeck. I'd phrase this a bit differently, but a plain-English description would be helpful. I think JKeck's is a good start. —Ben FrantzDale (talk) 15:53, 18 February 2011 (UTC)

My derivation

I struggled a lot to understand what's the point of this symbol. After struggling for a week, the symbols come up out of the blue, when I tried to calculate the acceleration felt by the particle which is moving on a path.

I come up with this down to Earth derivation:

(I'll use a bit relaxed form of the Einstein summation notation: if an index appears on all sides of the equation, it's kept; if not, it's summed.)

Let's assume you have a manifold immersed in Euclidean space, defined by this: $y_{i}(x^{j})$ ($i$ functions, each taking $j$ parameters, upper index is not a power as you know from Einstein's tensor notation.) For example a sphere in 3 dimensions (3 ambient coordinates), whose points defined by latitude and longitude (2 local coordinates).

We can define a path in local coordinates, my making it a function of time. So we have this:

$y_{i}(x^{j}(t))$

The tangent vector (aka it's velocity in ambient coordinates can be obtained taking the first derivative):

${\frac {dy_{i}}{dt}}={\frac {\partial y_{i}}{\partial x^{j}}}{\frac {dx^{j}}{dt}}$

(Using the rules of the total derivative)

To get the acceleration we'll need to differentiate again:

${\frac {d^{2}y_{i}}{dt^{2}}}={\frac {\partial y_{i}}{\partial x^{j}}}{\frac {d^{2}x^{j}}{dt^{2}}}+{\frac {dx^{j}}{dt}}{\frac {dx^{k}}{dt}}{\frac {\partial ^{2}y_{i}}{\partial x^{j}\partial x^{k}}}$

The particle cannot feel the acceleration from the extra dimensions, only its projection on the tangent space. So we'll need to project this acceleration vector into the tangent space. Let's call the acceleration vector $V_{i}$. It's composition of the required tangent vector $p$ and a normal component $W$ (in ambient coordinates):

$V_{i}=p_{i}+W_{i}$

We'll need to express this projected tangent vector in term of local coordinates:

$V_{i}=p^{j}{\frac {\partial y_{i}}{\partial x^{j}}}+W_{i}$

We can do this if we express this tangent vector using a path: $x^{j}(t)=p^{j}t+m^{j}$, where $p$ is set of local coordinates, $m$ is the point on the manifold the particle is at. Then the first derivation of this yields substitution above.

The $W$ is perpendicular to all tangent vectors so dotting both sides of the equation with one, we can eliminate it. I've chosen these tangent vectors to correspond the basis of the local coordinate system (so one of the coordinates is 1, the others are zero). So we have $k$ equations now:

$s_{k}=V_{i}{\frac {\partial y_{i}}{\partial x^{k}}}=p^{j}{\frac {\partial y_{i}}{\partial x^{j}}}{\frac {\partial y_{i}}{\partial x^{k}}}$

This way the metric appears on the right side:

$s_{k}=V_{i}{\frac {\partial y_{i}}{\partial x^{k}}}=p^{j}g_{jk}$

And using metric's inverse we can get our projected tangent vector:

$p^{j}=g^{jk}V_{i}{\frac {\partial y_{i}}{\partial x^{k}}}$

Now let's substitute back the acceleration vector we got before:

$p^{j}=g^{jk}\left({\frac {\partial y_{i}}{\partial x^{l}}}{\frac {d^{2}x^{l}}{dt^{2}}}+{\frac {dx^{l}}{dt}}{\frac {dx^{m}}{dt}}{\frac {\partial ^{2}y_{i}}{\partial x^{l}\partial x^{m}}}\right){\frac {\partial y_{i}}{\partial x^{k}}}$

Eliminating the parenthesis:

$p^{j}=g^{jk}{\frac {\partial y_{i}}{\partial x^{k}}}{\frac {\partial y_{i}}{\partial x^{l}}}{\frac {d^{2}x^{l}}{dt^{2}}}+g^{jk}{\frac {\partial ^{2}y_{i}}{\partial x^{l}\partial x^{m}}}{\frac {\partial y_{i}}{\partial x^{k}}}{\frac {dx^{l}}{dt}}{\frac {dx^{m}}{dt}}$

Now that inverse metric multiplied by partial derivatives thing in the right term is the Christoffel symbol of the second kind (to see how it corresponds to the metric, see it here), while the left term can also be simplified:

$p^{j}=g^{jk}g_{kl}{\frac {d^{2}x^{l}}{dt^{2}}}+\Gamma _{lm}^{j}{\frac {dx^{l}}{dt}}{\frac {dx^{m}}{dt}}$ =

$p^{j}=\delta _{l}^{j}{\frac {d^{2}x^{l}}{dt^{2}}}+\Gamma _{lm}^{j}{\frac {dx^{l}}{dt}}{\frac {dx^{m}}{dt}}$ =

$p^{j}={\frac {d^{2}x^{j}}{dt^{2}}}+\Gamma _{lm}^{j}{\frac {dx^{l}}{dt}}{\frac {dx^{m}}{dt}}$

For example on a sphere, if you are forced to move along a latitude circle with 1 radians per sec, you will feel $\mathrm {sin} (x)\,\mathrm {cos} (x)$ radians/s² acceleration towards the poles, where $x$ is your latitude.

Multiplying with this vector again with metric again you'll get $r^{2}\mathrm {sin} (x)\,\mathrm {cos} (x)$ which gives you the acceleration in "real" units in m/s².

Just my 2 cents. This might not be correct or precise enough, but it should give the idea.

Calmarius (talk) 10:41, 31 August 2013 (UTC)

I've only skimmed this, but here are a couple of comments. First, you seem to be endowing the manifold with a metric by embedding the manifold in an ambient Euclidean space. This is not wrong, but the article currently treats manifolds intrinsically, rather than as subsets of Euclidean space. Actually, that entire issue is left to other articles. Second, Wikipedia forbids original research. So we don't get to use your derivation in the article, whether it's right or not. Mgnbar (talk) 11:48, 31 August 2013 (UTC)

I'm not intended to add it to the article as it is, just put it here to help the next guy who comes here to complain, that he cannot get the point. Also derivations are not original research. I immersed the manifold in the Euclidean space to make it and the notation look familiar. But just as many textbooks and in the article, we can omit the $y_{i}$ and play with the differential operators alone, but that looks very weird to me... Calmarius (talk) 17:51, 2 September 2013 (UTC)

Thanks for quoting the relevant part of the policy. I didn't know that it was written like that. The breadth of the language surprises me. It seems to allow arbitrarily difficult mathematical theorem proving. Mgnbar (talk) 22:26, 2 September 2013 (UTC)

From Christoffel symbols to metric tensor

Is there a formula to go from Christoffel symbols to metric tensor? Jackzhp (talk) 17:08, 27 November 2013 (UTC)

While we wait for someone else to give you a better answer: If I recall correctly, not all connections arise from metrics, so no such formula can exist in general. But, for connections that do arise from metrics, you might have more luck at Levi-Civita connection? Mgnbar (talk) 17:54, 27 November 2013 (UTC)

Not exactly. If you are given the value of the metric on a hyper-surface (co-dimension 1), then you can use the Christoffel symbols to calculate it elsewhere. From Christoffel symbols#Christoffel symbols of the second kind (symmetric definition), we get

$0={\frac {\partial g_{ik}}{\partial x^{\ell }}}-g_{mk}\Gamma ^{m}{}_{i\ell }-g_{im}\Gamma ^{m}{}_{k\ell }$

and thus

${\frac {\partial g_{ik}}{\partial x^{\ell }}}=g_{mk}\Gamma ^{m}{}_{i\ell }+g_{im}\Gamma ^{m}{}_{k\ell }\,.$

Using this we can integrate from the hyper-surface outward to get the metric throughout spacetime (assuming the Christoffel symbols are available everywhere).

With a little more work, you should be able to start from the metric at one event to get it on a line, then on a plane, then on a hyper-surface, then everywhere. JRSpriggs (talk) 06:28, 29 November 2013 (UTC)

Covariant derivatives of tensors

I don't understand the notation $V^{m}$ for a vector field in this section. The notation is very unclear; what is the index m? 129.97.134.116 (talk) 02:50, 2 March 2014 (UTC)

The text is referring to a vector field V. A vector field has one upper index, which the text is denoting m, to prepare you (the reader) for what follows. But technically the vector field is just V, and the index is unnecessary, so I'll remove it. Mgnbar (talk) 03:12, 2 March 2014 (UTC)

I put it back. A vector is not one number, but a one-dimensional array of real numbers. Each element of the array is associated with one of the n dimensions of the underlying tangent space of the manifold. The index m=1,2,...,n is a variable which selects one of those elements. JRSpriggs (talk) 15:42, 2 March 2014 (UTC)

I don't think anyone in this discussion thinks that a vector is a single number. A vector is also not an array of numbers (see Tensor), but that is not the crux of our disagreement...

A vector field is a single mathematical object, that is represented by a capital Roman letter in this article. For example, see the preceding section "Relationship to index-free notation", which begins with the text "Let X and Y be vector fields with components Xⁱ and Y^k". We should have similar text in this section. Mgnbar (talk) 19:44, 2 March 2014 (UTC)

If IP user 129.97.134.116 understood what you and I understand, he would not have asked his question. Instead of answering the question, you chose to sweep the issue under the rug by removing the index. You may say that the vector is a single object, but that is not helpful. It is more useful to think of it as a function from indices to components; this function can be represented as an array. If one allowed for changing the event where the tensor is located or the coordinate system in which it is represented, then its functional dependence would be even more complicated. Saying that a tensor is something else than an array is merely an exercise in mindless abstraction like Plato's world of Forms. JRSpriggs (talk) 05:04, 3 March 2014 (UTC)

At this point I do not accept your interpretation of the IP editor's complaint. He could easily have been confused by the fact that V is a vector field, but V¹ (or V^m for any particular value of m) is not. Your conceptualization of a vector field as a function from indices to components is fine with me, but the name of this function is V, and its value at index m is V^m.

What do you think of the compromise, implicitly proposed above, in which we state that V is a vector field with coordinates V^m? It seems a harmless way of resolving this small dispute. Mgnbar (talk) 06:12, 3 March 2014 (UTC)

In a context such as this (i.e. in a reference), it would be best having more explicit wording of this nature. The shorthand phrasing "the vector V^m" is for initiates, not for someone such as the typical reader who is less familiar with the typical conventions. One could also use the shorter phrase "vector components V^m". The whole article (and many on related subjects) should probably be copyedited with this in mind. Care should also be taken if multiple notations are mixed. —Quondum 06:26, 3 March 2014 (UTC)

To 129.97.134.116: You should also see Ricci calculus. JRSpriggs (talk) 08:55, 4 March 2014 (UTC)

Possible misdirected link for "are well behaved"

Under "Definition in Euclidean space", 5th line from the bottom of that section, the link at the end of the parenthetical phrase "(as long as the manifold and coordinate system are well behaved)" appears to jump to "https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives" rather than a page on what "well behaved" means in the context of Christoffel symbols. As I am an expert on neither Christoffel symbols nor Wikipedia edits, I feel it best to leave any decisions and/or changes in more experienced hands. BornRightTheFirstTime (talk) 02:53, 21 December 2018 (UTC)

Definition about connection coefficient in nonholonomic basis

I'm studying connection coefficient with a MIT physics textbook (http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/1211.2.K.pdf) and found the definition of connection coefficient in this book is very different from that of here. Definition there is almost same to the Ricci rotation coefficient here. But I found a inconsistent point in THIS article that when you substitute

$\omega ^{k}{}_{ij}:={\mathbf {u} }^{k}\cdot \left(\nabla _{j}{\mathbf {u} }_{i}\right)\,$

into

$\nabla _{\mathbf {u} _{i}}\mathbf {u} _{j}=\omega ^{k}{}_{ij}\mathbf {u} _{k}$,

you find that order of index, to say that order of i and j, is inversed. 

So I think

$\nabla _{\mathbf {u} _{i}}\mathbf {u} _{j}=\omega ^{k}{}_{ji}\mathbf {u} _{k}$

would the correct definition, and is consistent with my textbook's definition. (or we can inverse the order of indexes in the second eq.) But as a physics graduate with not through mathematical knowledge I'm not sure so I want someone to confirm this. Thanks. — Preceding unsigned comment added by 137.229.78.205 (talk) 04:59, 28 August 2017 (UTC)

Transformation law for change of variables

Sunlitsky (talk · contribs) said in an edit summary "The christoffel symbol in previous edit was not wrong, but can be reduced to the form presented in the current edit.". This is incorrect. You would have to change the sign since you are taking the second derivative of the inverse function. JRSpriggs (talk) 11:33, 10 January 2019 (UTC)