Talk:Covariance and contravariance of vectors

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This article may be too technical for most readers to understand, and needs attention from an expert on its subject. Please expand it to make it accessible to non-experts, without removing the technical details.

Contents 1 Old comments 2 What 'contravariant' means 3 Example: covariant basis vectors in Euclidean R3 4 Usage in tensor analysis 5 Obtuse 6 Too Technical 6.1 replies to Ben 7 My attempt to explain 8 more technical 9 New Figure and Descriptions Added 10 Understanding 11 Merge 12 progress? 13 covectors: above or below? 14 the title is misleading in the first place

[edit] Old comments

Covariant components are mates to contravariant, but covariant transformations are ones that preserve certain invariants, the simplest one perhaps being "dot-products of covariant and contravariant components of the same vector. The distance function is preserved by covariant transformations, but its covariant components are not. The Riemann scalar is preserved by covariant transformations, but its covariant components are not. A merger is out of order.Pdn 04:18, 21 May 2005 (UTC)

I agree that there should be more than one article on this. Perhaps covariant transformation and covariant vector etc. I don't really like the naming now of simply covariant as it's an adjective and seems to preclude any covariant _____ articles. I'm also more than a little frustrated that the mergewith tags have been readded within a day without further discussion. --Laura Scudder | Talk 06:44, 22 May 2005 (UTC)

The conversatation has been moved to Wikipedia_talk:WikiProject_Mathematics. Please go there. linas 07:09, 22 May 2005 (UTC)

Well, long talk threads belong here, not there. I'm going to move this page to covariance and contravariance, since Laura's point about nouns is good (and covariance is taken by statistics right now). Covariant transformation seems to me too confusing a concept to be taken as primary. Charles Matthews 09:31, 27 May 2005 (UTC)

[edit] What 'contravariant' means

This section doesn't manage to be very clear. I know about tensors so this section which is only about Euclidean space should be a piece of cake, but it isn't. The only thing that I managed to distill is how contravariant components are formed and the picture makes it clear, although the wording is suboptimal and there is no caption, but the covariant case is neglected. Is there anything else that this section is trying to explain? Something about meaning maybe? Otherwise this section should be scrapped and a brand new section about how to take coordinates written. --MarSch 14:24, 24 Jun 2005 (UTC)

I agree. After half an hour of reading I was not able to determine what 'covariant' or 'contravariant' mean. There is a lot of information which would seem to suplement this knowledge, such as the relationship between the two concepts, etc. But there needs to be a simple intuitive explanation of the basic meaning, because without that all of this information is utterly useless. --Jack Senechal 08:58, 26 December 2005 (UTC)

[edit] Example: covariant basis vectors in Euclidean R3

Contravariant vector is not explained and neither is covariant vector and also not what the reciprocal system is. Then the section proceeds to talk of an unqualified vector v. Apparently it is possible to take the dot product of a vector with a contravariant vector and also of a vector and a covariant vector. How is that supposed to work? This section is extremely unclear to me. --MarSch 14:34, 24 Jun 2005 (UTC)

[edit] Usage in tensor analysis

First and third paragraphs have no information. Second paragraph explains about how a metric induces an isomorphism non-canonically, but is needlessly vague and wordy. --MarSch 14:38, 24 Jun 2005 (UTC)

[edit] Obtuse

The lay reader needs to walk away from this article with an intuitive understanding of what the covariant and contravariant components of a vector in arbitrary coordinates on an arbitrary manifold are.

Central to this understanding are the notion of a "geometric object", which is a collection of numbers, and a rule, such that if you know the numbers in one coordinate system, you can use the rule to find them in any other, and the notion that a person familiar with vector calculus in three dimensional Euclidian space will seek to make his life as uncomplicated as possible when dealing with curved coordinates in curved spaces.

Eschewing both unit and orthogonal basis vectors, he will seek to pick basis vectors such that the components of a vector have the simplest transformation rule possible, that of a differential, which transforms by multiplication by the matrix of partial derivatives of one set of coordinates with respect to the other. There are two fundamental ways of doing this, depending on whether you differentiate the old coordinates with respect to the new, or the new with respect to the old. These two choices correspond the geometric object being either a covariant or contravariant vector.

An important point here is while the components of the vector change with the choice of basis, the sum of the components times the basis vectors is an invariant independent of coordinate choice, which is the vector as an abstract mathematical object.

Intuitively, in three dimensional curvilinear coordinates, the covariant components of a vector correspond to using basis vectors which are tangents to the coordinate curves, and the contravariant components correspond to using basis vectors which are normals to the coordinate surfaces. If you perturb the coordinate grid lines, the covariant basis vectors will move with the grid lines, and the contravariant basis vectors will move in a contrary manner.

I really don't like the metaphor of drawing the vector on the manifold and projecting it onto the coordinates, because the vector lies in the tangent space to the manifold at a particular point, which is flat, and doesn't have physical extent on the manifold itself, although such handwaving will sometimes seem to give the right answer. Hermitian 04:58, 8 January 2006 (UTC)

I agree. Some new, more effective methods of communicating these important ideas are in order. Corax 22:14, 8 January 2006 (UTC)

Incidentally, note that Corax has never before made any contribution to any article on mathematics (at least under this name). User:Katefan0 and I both expected them of being sockpuppets, as they spend 99% of their time defending NAMBLA. Katefan even went to the length of having their IP addresses checked out, but it seems they're not SPs. However, I expect that Corax's endorsement of Hermitian's comment here may just be a case of the old buddy system, and can be ignored. Camillus (talk) 20:30, 9 January 2006 (UTC)

What does this have to do with adding clarity to the article on "Covariance and Contravariance?" Are you going to follow me around and troll all my posts because you don't like NAMBLA? (guffaw) Hermitian 21:14, 9 January 2006 (UTC)

[edit] Too Technical

I'm by no means new to vectors or tensors, but for the life of me I can't figure out what covariance or contravariance means from this article. I understand that a single (invariant) vector can be written in terms of any number of basis vectors, but I don't see how that relates with co- and contravariance. Could someone explain? —Ben FrantzDale 08:11, 26 March 2006 (UTC)

Let me take another crack at this. I still do not understand what this article is trying to say and it still does not seem to point me in the right direction to find out more. I have two fairly hazy hunches. (1) It seems that this topic deals with the difference between an abstract tensor—an invariant—and a "corporial" tensor in matrix form. It seems that there are two isomorphic numerical representations of a single invariant one being the covariant and the other the contrariant representation. (2) It seems that this topic deals with tensors expressed in terms of parametric coordinates. That is, if I have the vector (1,1) in cartesian space, I could also represent it in terms of another coordinate system with base vectors in the (√½,√½) and (−√½,√½) directions, in which case the same vector would be represented as (√½, 0). It is not clear to be if (1) or (2) or both of the above are right or are even close. Could someone explain? Am I on the right track? —Ben FrantzDale 05:20, 4 April 2006 (UTC)

Aside from the question of whether the article could or should be written more clearly, any discussion of covariance in multiple contexts is bound to be confusing. Physicists will use the term to mean that some mathematical entity is a well-defined geometric object, not an artifact of a particular coordinate choice. Physicists will also use the term in a closely related, but decidedly different, sense in discussing what they call tensors (which mathematicians call tensor fields). This latter use contrasts covariant with contravariant. Then we have mathematicians also talking about tensors, and about tensor fields, perhaps shifting meaning again. And the category theory mathematicians appropriate the same covariant/contravariant terminology for their own, quite different, use.

How do we know whether something is co- or anti-co-? Unfortunately, as with all binary choices, it's a matter of convention. (And often conventions are established too early. Thus electrons, the carriers of electric current, have negative charge; it's too late to call them positive!) Covariance can be extra challenging, because it involves multiple binary choices. Confusion is guaranteed.

In practice we are saved by working in only one context at a time, with only one convention, and with far less confusion. --KSmrq^T 06:56, 4 April 2006 (UTC)

Thanks for the reply. With regard to this article, the definitions I am most interested in understanding is the mathematical one (and perhaps the physics one as well). As the article stands I follow the first paragraph (which says very little) but get stuck at paragraph two:

In very general terms, duality interchanges covariance and contravariance, which is why these concepts occur together. For purposes of practical computation using matrices, the transpose relates two aspects (for example two sets of simultaneous equations). The case of a square matrix for which the transpose is also the inverse matrix, that is, an orthogonal matrix, is one in which covariance and contravariance can typically be treated on the same footing. This is of basic importance in the practical application of tensors.

The duality link isn't much help and the rest of the first sentence seems to say very little.

The second sentence (which isn't quite a sentence), with its talk of matrix transpose, seems to be talking about the rowspace and columnspace of a matrix.

Sentence three isn't clear. For example, what does "typically" mean? What does "the same footing"? mean? I'm inclined to think that this sentence means "when the transpose is the inverse of a matrix, xⁱ=x_i".

Sentence four doesn't add anything.

The third paragraph says a little bit more, but is similarly vague to the new reader, at least. Hopefully this disection will help someone who does know the subject find a place to start reworking the article. —Ben FrantzDale 12:44, 4 April 2006 (UTC)

[edit] replies to Ben

1. It seems that this topic deals with the difference between an abstract tensor—an invariant—and a "corporial" tensor in matrix form.

   Yes, the difference between the physicist's convention and the mathematician's convention depends on whether you want to know how components transform under passive transformations or how objects behave under active transformations.

2. It seems that there are two isomorphic numerical representations of a single invariant one being the covariant and the other the contrariant representation.

   There are only isomorphic representations if you have some isomorphism perhaps provided by a nondegenerate bilinear form, usually a metric tensor. The existence of such objects helps to conflate the notions of covariance and contravariance, and so when trying to understand the difference, prefer to consider spaces without such an isomorphism. Insist on strict separation in your mind of covariant and contravariant objects. This applies in either convention.

3. Hopefully this disection will help someone who does know the subject find a place to start reworking the article.

   I'm going to try to see what I can do with this article. In its current state, I have half a mind to just scrap the whole thing and start from scratch. -lethe ^{talk +} 15:55, 4 April 2006 (UTC)

A rewright or major overhaul sounds appropriate. I'd be happy to vet any new material for clarity.

Regarding 1 and looking at active and passive transformation and your section below, it sounds like the physicist's definition goes like this:

If I have a sheet of graph paper with a vector in the (1,0) direction drawn on it and that paper is aligned with a global coordinate system, if I physically rotate the sheet of paper by π/2 right hand up—an active transformation—then the vector will be covariant—it will varry with the transformation and now point in the (0,1) direction.

With the same setup, if I now rotate the global coordinate system by π/2 right hand up, the vector itself is unchanged but its representation in global coordinates is now (0, −1); that is, the vector transformed contravariantly with respect to the passive transformation.

Is that about right? —Ben FrantzDale 20:32, 4 April 2006 (UTC)

I think it's difficult to see the full picture if you only talk about components. Let's look at the components of the vector as well as the basis. Start with a basis e₁ and e₂, and components 1 and 0. So your vector is v = 1e₁+0e₂. Imagine those basis vectors attached to your sheet of paper. Rotate the sheet of paper, and e₁, e₂ and v all rotate as well. The new rotated vector is now v'= 1e'₁+0e'₂. Notice that the components didn't change: the vector is still (1,0). The components of the vector are just numbers, scalars, they do not transform. Vectors transform (covariantly), scalars do not. New basis vectors, same components, new vector.

Now imagine rotating your global coordinates, and re-writing the basis vectors in terms of the new coordinates. To get the same physical vector in terms of the new basis vectors, we change the components as well; the vector is v=0e'₁–1e'₂. The physical vector stays the same, but the components and the basis vectors transform (the vector covariantly, the components contravariantly). New components, new basis vectors, same resulting linear combination. -lethe ^{talk +} 21:16, 4 April 2006 (UTC)

I agree completely. Well put. I think we are making progress! —Ben FrantzDale 21:55, 4 April 2006 (UTC)

[edit] My attempt to explain

Covariance and contravariance is at its heart quite simple.

Covariant things in a space transform forward along transformations of the space, contravariant things transform backwards along tranformations of the space.

If you always work in a coordinate-independent way, things stay cleaner, and the situation is pretty simple. Vectors are covariant, functions and differential forms are contravariant (in apparent contradiction of the physicists' convention).

When you introduce local coordinates, you now have two kinds of transformations to consider: active transformations, which can move vectors and tensors around, and passive transformations, which change the basis, but leave other tensors fixed. (Active transformations are diffeomorphisms of manifolds, while passive transformations are transition functions on manifolds).

Under an active transformation, basis vectors, being vectors, are covariant, while their components, being scalars, do not change. Functions do not change, and the basis one forms change contravariantly while components do not change. The total object changes. This is the same statement as above, but viewed in local coordinates.

Under a passive transformation, we need the product of the components with the basis to remain unchanged. The basis vectors, being vectors, still transform covariantly (covariantly is the only possibly way to change any vector under a smooth map unless it's invertible), so in order to keep the total object invariant, the components have to change contravariantly. Similarly, the basis of one forms change contravariantly (the only way they can) and to keep the total object invariant, the components have to change covariantly.

Thus the physicist, who thinks of the object solely in terms of its component and its behaviour under passive transformations, calls the vector contravariant and the one-form covariant, while the mathematician who thinks of the total object and how it behaves under active transformations (morphisms in the category of manifolds) calls the vector covariant and the one-form contravariant. -lethe ^{talk +} 15:36, 4 April 2006 (UTC)

That's a big help; the article is starting to make more sense. Is this related to the notion that the result of a cross product is a pseudovector or that a surface normal doesn't transform nicely under non-orthogonal transformations? —Ben FrantzDale 20:52, 4 April 2006 (UTC)

Well... if you'd like to think of it in terms of representation theory, for every representation of GL(n), there is a bundle whose vectors transform according to that rep. The tangent vectors transform according to the fundamental representation, covectors transform according to the dual representation, pseudovectors transform according to GL+(n) (the connected component of GL(n)), and if there is a metric tensor, all these groups can be reduced to the O(n) subgroup. To answer your question, contravariance versus covariance has some bearing on what representation you use, and so does pseudovectorness and orthogonality, but the notions are still independent. You can be a pseudovector or a pseudocovector, or a vector or covector, each is a different rep of GL(n). -lethe ^{talk +} 21:27, 4 April 2006 (UTC)

Wow. That was totally over my head. I wouldn't bother trying to explain that to me in any more depth right now. —Ben FrantzDale 21:55, 4 April 2006 (UTC)

Well, I was really stretching to figure out a way to relate pseudoness to covariance. A mention of represetnation theory might deserve only a footnote. -lethe ^{talk +} 22:06, 4 April 2006 (UTC)

It is possible to unify normals and vectors and dualities by embedding everything in Clifford algebra, but that's probably not the best step for you at this point in your learning. Let's try something different. A surface normal is part of a plane equation, typically written in matrix notation (using homogeneous coordinates) as 0 = G p. The G part is a "row vector", which is another way of saying a 1-form; the p part is a column vector, which is another way of saying just an ordinary vector. By a 1-form we mean a linear function from a vector space to real (or complex) numbers, the latter also consider a (trivial) vector space. The homogeneous coordinates are a nuisance required to handle an affine space with vector space tools. If we fix an origin and only consider planes through it, we can dispense with the extra component of homogeneous coordinates. Then G is precisely a normal vector.

A differential manifold, considered intrinsically, does not have a normal. "Normal" is a concept that only makes sense for a manifold embedded in an ambient space, like a spherical surface (a 2-manifold) embedded in 3D Euclidean space. Most of differential geometry as applied to general relativity has to work intrinsically. Most of our early intuition for it comes from the embedded examples.

So let's think about a simple smooth curving surface M embedded in 3D. Concentrate attention on a single point p∈M. Extrinsically, we have a unique plane tangent to the surface at that point. Intrinsically, we have a tangent space. This is a clever construct, but takes some practice to absorb. Consider all the smooth curves, C(t): R→M, passing through our point p. Switch attention to all the possible derivatives of those curves, dC/dt, at p. (To make it simple, parameterize the curves so that always C(0) = p.) This collection of derivatives becomes the tangent vector space, T_pM. For intuition, it helps to thing of the (extrinsic) tangent plane as a stand-in for the (intrinsic) tangent vector space, taking p as the origin.

When mathematicians speak of a tensor, we mean a function that takes some number of vectors and produces a real number, and that is separately linear in each vector argument. For example, the dot product of vectors u and v yields a real number that depends linearly on both u and v, so "dot product" is a "rank 2 tensor". Here "rank" refers to the number of vector arguments. More precisely, we have just defined a covariant tensor.

Linear forms over a vector space, such as the tangent space, comprise a vector space in their own right — a dual space. We can just as easily speak of multilinear functions, tensors, that take dual vectors (forms) as arguments. These are contravariant tensors. We can also have mixed tensors.

When physicists speak of a tensor, the meaning shifts. No longer are we talking about objects defined on a single vector space, T_pM, and its dual. Now we are talking about an object defined across all the tangent spaces, TM (and their duals). Mathematicians prefer to call this a tensor field, a (continuous) mapping from points p∈M to (mathematicians') tensors. We do insist on continuity, thus the rank and covariance/contravariance are the same at every point.

Because physics is based on experiment, the question naturally arises if some experimentally or theoretically discovered mapping from points to functions, defined in terms of coordinates, has intrinsic meaning. Does it fit the mathematicians' definition of a tensor field? If it does, old physics terminology proclaims this thing "covariant", a third meaning. Sigh.

Does this help? --KSmrq^T 00:21, 5 April 2006 (UTC)

[edit] more technical

I've started a rewrite of this article, wherein I hope to make the difference between the different conventions for co- and contravariant completely transparent. Unfortunately, as I write it, I find I'm writing an article that is far more technical than the existing article, rather than one that is less technical. It is important that we have an article which is mostly comprehensible to, say, someone encountering special relativity for the first time. This person may not have the slightest idea what a manifold or a topological space is. So this is actually a hard article to write. -lethe ^{talk +} 16:39, 4 April 2006 (UTC)

Sounds good. Where's the draft? I'd say write what comes naturally and then we can work on translating it to something more reasonable for beginners. The more I look at the current version of the article the more I think the problem isn't that it's too technical but that it's just incoherent. I might have a chance of understanding a fully-technical article by following enough links. —Ben FrantzDale 20:36, 4 April 2006 (UTC)

The draft is at User:Lethe/covariance, but be warned, I'm not sure if it's a suitable direction for an article on this subject, and even if it is, it's still very much a work in progress. -lethe ^{talk +} 21:19, 4 April 2006 (UTC)

It looks like a good start. I'll watch it. —Ben FrantzDale 21:55, 4 April 2006 (UTC)

What this article seriously needs is a split, either internally or externally. By simultaneously trying to discuss all the variations, it only succeeds in causing confusion. We need clear separation between physics, tensor mathematics, category theory, geometric objects, and coordinate systems. Instead we have a witches' brew, with a little eye of newt here and wool of bat there. Yuck. --KSmrq^T 23:08, 4 April 2006 (UTC)

I would really like it if this could all be covered clearly in one article. Treating covariance in physics and mathematics as separate and different notions I think makes the concepts harder to understand. I'm not sure it can be done well, but I'm going to try. Give me some time to see what I can come up, and then let's decide whether we need to split. -lethe ^{talk +} 23:45, 4 April 2006 (UTC)

As a relative layperson (someone who shouldn't encounter covariance/contravariance for about 8 months on eir mathematics course) I've encountered both manifolds and topological spaces. Is it really likely that such a technically advanced notation will attract the interest of laypeople? 12:30, 16 May 2007 (UTC)

[edit] New Figure and Descriptions Added

Dear All, it seems apparent that the meanings of contravariance and covariance were not clear to most readers. Therefore I've added another figure showing how the component-wise basis vector terms are oriented in a 2D curvilinear and non-orthogonal grid to represent some generic vector. I am also going to add some bits on the transformation properties of the basis vectors. It may be a little rough at first, so I definitely invite anyone to clean up the notation and presentation...

...OK, I've put a ton of new stuff in there, but it is still not entirely complete. —Preceding unsigned comment added by Hernlund (talk • contribs)

Please sign your messages with "~~~~". Also please put new messages at the BOTTOM of the talk page, not at the top, in accordance with accepted practice. Thanks for your contribution to the article. JRSpriggs 06:59, 10 November 2006 (UTC)

With regard to the figure, why is tangent to the line labeled x¹ = α? Shouldn't it be tangent to the line labeled x² = β? I have a similar comment about ; I think it should be tangent to x¹ = α instead of x² = β. ChrisChiasson 12:33, 7 January 2007 (UTC)

Those vectors point in the direction that the corresponding variable changes. The lines you are referring to are lines along which the variable does NOT change. So e₁ points in a direction in which x₁ changes and all other variables (including x₂) do not change. JRSpriggs 13:17, 7 January 2007 (UTC)

So the contravariant unit vectors are actually orthogonal to all other coordinate axes (besides their own)? ChrisChiasson 01:42, 9 January 2007 (UTC)

Orthogonality does not enter into contravariant vectors. Consider the fact that the x-axis in the x-y-plane is given by the equation y=0. Similarly the y-axis is x=0. The contravariant e^x vector at a point is directed along the constant-y line through the point. Whereas, the covariant e_x vector is perpendicular to the constant-x line through the point. In this plane they appear to be the same vector. JRSpriggs 07:57, 9 January 2007 (UTC)

"Whereas, the covariant e_x vector is perpendicular to the constant-x line through the point." - I'm lost :-[ Does this mean the picture should have e₁ perpendicular to x¹? ChrisChiasson 23:20, 9 January 2007 (UTC)

Sorry, I was the one who was confused. I am not used to thinking about the basis vectors. Frankly, I do not find them helpful. So I was reversing contravariant and covariant in my mind and in my last message. The picture is correct. The covariant e_x vector at a point is directed along the constant-y line through the point. Whereas, the contravariant e^x vector is perpendicular to the constant-x line through the point.
So the answer to your question "So the contravariant unit vectors are actually orthogonal to all other coordinate axes (besides their own)?" is Yes.
In most of the books I have read on this subject and in my own thinking, the sequence of components (coefficients) IS a contravariant vector. And the sequence of components IS a covariant vector. The basis vectors vary the opposite way from their associated components, i.e. contravariant components go with covariant basis vectors and vice-versa. JRSpriggs 10:26, 10 January 2007 (UTC)

It would be informative to see unit vector field plots of e₁, e₂, e¹, and e², along with streamlines (defining coordinate axes?) (created by solving the differential equations defining the vector field and the initial conditions from selecting an initial point) and constant coordinate lines (e.g. x² = 15). I would create them, but I think I would do it incorrectly because I don't really understand general curvilinear coordinates. ChrisChiasson 18:50, 12 January 2007 (UTC)

The way that I think of them is (as I wrote in the article some time back)

A contravariant vector is one which transforms like , where are the coordinates of a particle at its proper time . A covariant vector is one which transforms like , where is a scalar field.

That is a contravariant vector is like a little arrow, e.g. velocity. A covariant vector is like a gradient which could be better visualized as a set of level lines rather than as an arrow. The arrows shown in these diagrams for the covariant vectors are just the normal vectors which are actually contravariant vectors perpendicular to the level lines. What is being hidden here is the metric tensor which is being used implicitly to convert between covariant and contravariant. JRSpriggs 08:21, 13 January 2007 (UTC)

[edit] Understanding

I think I finally get it. Too many years thinking in an orthonormal basis rotted my mind; this isn't too complicated. Here's the beginning of the explanation I needed. If it's accurate, I'll incorporate it (with diagrams) into the article.

When vectors are written in terms of an orthonormal basis, the covariant and contravariant basis vectors align, so the distinction between the two is unnecessary. In an orthonormal basis, the components of a vector are determined by dotting the vector with each basis vector, so a vector, x, that points up and to the right with a length of is the vector (1,1) in the usual basis. We will first consider basis vectors of different lengths, then consider non-orthogonal basis vectors.

Consider two basis vectors, e₁ points right and is two units long and e₂ is one unit long and points up [image]. By inspection, x written in this coordinate system is (1/2,1). This was determined by projecting x onto these two directions, but also scaling so that . That is, for orthogonal basis vectors that are not necessarily unit length,

This vector, , is the contravariant basis vector.

Now consider the case of a skiewed coordinate system such that is a unit vector to the right but . Obviously we expect x represented in this coordinate system to have coordinates (0,1). Again, projecting onto each vector does not give the coordinates of x. The correct vectors to project onto—the correct linear functionals—are the contravariant basis vectors [image].

That's a start. Feel free to rephrase the above. —Ben FrantzDale 18:03, 10 May 2007 (UTC)

This makes a _lot_ more sense now. Huzzah for concrete examples! —Preceding unsigned comment added by 98.203.237.75 (talk) 09:28, 7 January 2008 (UTC)

I think the image is useful but could be improved if the contravariant and covariant basis vecotors were in different colours. —Preceding unsigned comment added by 137.205.125.78 (talk) 12:02, 23 April 2008 (UTC)

[edit] Merge

I re-added the proposal to merge with Covariant transformation. This page and that may both deserve to exist, but at the moment, the two seem redundant. If they aren't merged completely, they should be refactored so that one page (this one?) is an introduction and the other goes into detail. —Ben FrantzDale 13:26, 11 May 2007 (UTC)

[edit] progress?

I was just checking back on this page, to which I had added the transformation laws and the figure on a non-orthogonal grid some time ago. I was curious whether people were understanding the difference between covariant and contravariant better after this addition, and it seems there might still be some confusion, though matters may have improved a bit after that. Another way to look at this is constrasting how the differentials vs. the gradient vectors of a coordinate system transform...one gets the same results as the contravariant and covariant laws in these scenarios as well. I'd like to know if there is anything else I can contribute...drop a note here, and I'll read it later on and work on it over time, as I'm a rather busy fellow. Cheers! Hernlund (talk) 03:33, 22 November 2007 (UTC)

To be honest, I still don't really understand. On the other hand, I don't use tensors often. However, I have never had any formal training with them. Please don't feel obligated to improve the article; only do it if you want to - and thanks! ChrisChiasson (talk) 14:41, 22 November 2007 (UTC)

[edit] covectors: above or below?

It seems that amateurs are trying to bring more confusion between covariant and contravariant issues but just as a mnemonic device remember that coordinated functions are indexed above x^μ and then differential forms also dx^μ (reason: coordinated functions are linear and also its differentials are indexed above), and as everybody knows in electromagnetism: dA = F which in components means

for the covector potential and the bivector force , in an holonomic frame. Conclusion, this article sucks!... 'cuz (further) it brings more confusion to the non-experts, besides there, in Eric Weinstein stuff, it is wrong... it is better to check against Misner-Thorne-Wheeler's Gravitation--kiddo (talk) 04:28, 10 March 2008 (UTC)

[edit] the title is misleading in the first place

Dear Wikipeditors,

First of all the title is misleading since it concerns not only "vectors" but also "covectors", both of which are a special case of tensors, namely 1-tensors. This should also be made clear to the reader in the introduction of the article. The first question is whether or not there should be a separate article concerning this special case. In my opinion the whole matter should be dealt within the article about tensor. The reason is that co- and contravariance aren't really concepts themselves. However "covariant 1-tensors" and "contravariant 1-tensors" are. I think this is one of the reasons why students in general are confused about these things. If everything is stated in a precise and unambiguous manner it shouldnt really be confusing: A contravariant 1-tensor (or (1,0) tensor) corresponds to a vector. For physicists such a tensor x is a set of tuples a=(a_1,...,a_n) for EACH basis. However, one is not allowed to chose these tuples independently. The extra condition is that a_1 = M a_2 if a_1 and a_2 are the tuples of x with respect to two given bases and M is the change of coordinates matrix. Hence this notion is really the same as the notion of vector, only that we specify the coordinates with respect to all bases simultaneously. A covaraint 1-tensor is (or (0,1) tensor) has a different definition and corresponds to a covector (i.e. a linear transformation V \to k). Similarly this is a set of coordinates for each basis. The condition in this case will instead be a_1 = (M^T)^{-1} a_2. The reason one uses this definition is that the tuples should correspond to the coordinates of covector with respect to all possible choices of basis. (See below).

There is also a "subarticle" Classical_treatment_of_tensors in the article tensor which should be expanded. There it would be a good idea to start of with 1-tensors before proceeding to the general case. As I said, I think the whole matter must be discussed within the setting of tensors.

Furthermore, I think that the notion "components" of vectors is also presented in a confusing manner, and should be properly defined. The argument for why a covector transforms by the inverse transpose is completely adequate, and possibly incorrect. A proper explation only takes a few lines:

If two bases B_1 and B_2 are given and [x]_1 and [x]_2 are the corresponding coordinates of x, let f be covector. Denote the coordinates of f with respect to the two bases similarily. We then have by definition (1) [f]_1^T [x]_1 = [f]_2^T [x]_2 and (2) [x]_2 = M[x]_1 for all x. We now want to compute the matrix that transforms [f]_1 into [f]_2. Subsituting (2) into (1) yields [f]_1^T [x]_1 = [f]_2^T M [x]_1, which implies [f]_1^T = [f]_2^T M. This gives that indeed [f]_2 = (M^T)^{-1}[f]_1.

Yours sincerely, Zariski —Preceding unsigned comment added by Zariski (talk • contribs) 19:03, 19 March 2008 (UTC)