Talk:Covariance and contravariance of vectors

Notation burdened with crap.

Definition section would benefit hugely from being written with some sort of coherent standard of notation. I think we are probably allowed to assume that the reader knows what a linear transformation, basis, and coordinate vector with respect to a basis are, or certainly is capable of clicking links. There's no need to be summing over indices in both superscripts and subscripts and what the hell is this needless reference to the basis in brackets doing anyway? — Preceding unsigned comment added by 50.141.31.25 (talk) 21:47, 19 November 2015 (UTC)

1. I don't see any place the notation is incoherent. The components of a vector depend on the choice of basis, and the notation indicates clearly shows the basis. This dependence is appropriately emphasized throughout the section ("coherently"). In fact, I would say that the whole definition section is scrupulously coherent. A transformation of the basis is also shown clearly in the notation, and it is easy to show how this changes the vector and covector. Another way to indicate this dependence is using parentheses as opposed to square brackets (see, for example, Raymond Wells, Differential analysis on complex manifolds), with subscripts (e.g., Kobayashi and Nomizu, Foundations of differential geometry), or just with some ad hoc method (like "Let $v^{i}$ be the components of v in the basis f, and ${\tilde {v}}^{i}$ be the components of v in the basis ${\tilde {\mathbf {f} }}$"). It seems here better to use a notation where the dependence on the basis is explicitly emphasized, because that's what the article is really about.

2. "I think we are probably allowed to assume that the reader knows what a linear transformation, basis, and coordinate vector with respect to a basis are, or certainly is capable of clicking links." I agree with this, which is why the article doesn't define these things, except to write them down and fix the notation as it's used in the article, which is just a best practice for writing clear mathematics and not intended to be a substitute for knowledge that can be gleaned from clicking links. Also, linear transformations are not even mentioned until the very last section of the article. And the article doesn't define those either.

3. As for why the Einstein summation convention is not used in the definition section, it seems likely that some readers of the article will not have seen it before. It does no harm to include the summation sign. It's not like we are going to run out of ink, or expressions are going to become tangled messes of indices there. Covariance and contravariance should arguably be understood before the Einstein summation convention anyway, so presenting the additional of abstraction doesn't seem like there is any clear benefit to readers. Sławomir
Biały 22:41, 19 November 2015 (UTC)

It might be beneficial to change notation to such that vectors are bold-face lower-case letter, transformations are bold-face capital letters, while standard italics are used for scalars only. This convention is widespread in continuum mechanics, and, I believe, used in most elementary undergraduate courses. 130.209.157.54 (talk) 13:22, 14 January 2019 (UTC)

Standard definition apparent inconsistency

Could someone please explain the fallacy in the following?

Suppose that in $\mathbf {R} ^{3}$ we change from a basis $B$ in units of meters to a basis ${\overline {B}}$ in units of centimeters. That is,

$\mathbf {\overline {b}} ^{i}=\left({\frac {1}{100}}\right)\mathbf {b} ^{i},\quad (1\leq i\leq 3).$

Then the coordinate transformation is

${\overline {x}}^{i}=(100)x^{i},\quad (1\leq i\leq 3),$ 1

and the coordinate transformation in matrix form is

$\mathbf {A} ={\begin{pmatrix}100&0&0\\0&100&0\\0&0&100\end{pmatrix}}.$ 2

A position vector

$\mathbf {r} =3\mathbf {b} ^{1}+4\mathbf {b} ^{2}+12\mathbf {b} ^{3}$ 3

with magnitude $|\mathbf {r} |=13{\text{ m}}$ is transformed to

$\mathbf {\overline {r}} =300\mathbf {\overline {b}} ^{1}+400\mathbf {\overline {b}} ^{2}+1200\mathbf {\overline {b}} ^{3}$ 4

with magnitude $|\mathbf {\overline {r}} |=1300{\text{ cm}}.$

Suppose that a scalar function $f(x^{i})=-2{x}^{1}$ is defined in a region containing r. Then,

${\frac {\partial {f}}{\partial {{x}^{1}}}}=-2,{\frac {\partial {f}}{\partial {{x}^{2}}}}={\frac {\partial {f}}{\partial {{x}^{3}}}}=0,$ 5

the gradient vector is

$\mathbf {\nabla {f}} ={\frac {\partial {f}}{\partial {{x}^{1}}}}\mathbf {b} ^{1}+{\frac {\partial {f}}{\partial {{x}^{2}}}}\mathbf {b} ^{2}+{\frac {\partial {f}}{\partial {{x}^{3}}}}\mathbf {b} ^{3}=-2\mathbf {b} ^{1},$ 6

and the gradient vector at r,

$\mathbf {\nabla {f}} (\mathbf {r} )=-2\mathbf {b} ^{1},$ 7

with magnitude $|\mathbf {\nabla {f}} (\mathbf {r} )|=-2{\text{ m}}^{-1}$, is transformed to

$\mathbf {\overline {\nabla {f}}} (\mathbf {\overline {r}} )=-0.02\mathbf {\overline {b}} ^{1},$ 8

with magnitude $|\mathbf {\overline {\nabla {f}}} (\mathbf {\overline {r}} )|=-0.02{\text{ cm}}^{-1}.$

While the position vector r and the gradient vector $\mathbf {\nabla {f}} (\mathbf {r} )$ are invariant, the components of the position vector in equations (3) and (4) transform inversely as the basis vectors, and the components of the gradient vector in equations (7) and (8) transform directly as the basis vectors.

Now, most textbooks agree that a vector u is a contravariant vector if its components $(T^{i})$ and $({\overline {T}}^{i})$ relative to the respective coordinate systems $(x^{i})$ and $({\overline {x}}^{i})$ transform according to

${\overline {T}}^{i}=T^{r}{\frac {\partial {{\overline {x}}^{i}}}{\partial {x^{r}}}},\quad (1\leq i\leq 3)$ 9

and that a vector w is a covariant vector if its components $(T_{i})$ and $({\overline {T}}_{i})$ relative to the respective coordinate systems $(x^{i})$ and $({\overline {x}}^{i})$ transform according to

${\overline {T}}_{i}=T_{r}{\frac {\partial {x^{r}}}{\partial {{\overline {x}}^{i}}}},\quad (1\leq i\leq 3),$ 10

but equations (4) and (8) seem to indicate the exact opposite of equations (9) and (10). Why?—Anita5192 (talk) 17:44, 24 November 2018 (UTC)

There is an error in your comment right at the beginning. So I stopped reading it at that point. If $x^{i}$ is the position in meters and ${\overline {x}}^{i}$ is the position in centimeters, then the conversion is just:

${\overline {x}}^{i}=100\cdot x^{i},\quad (1\leq i\leq 3).$

OK? JRSpriggs (talk) 22:46, 24 November 2018 (UTC)

The $x^{i}$, with $|x^{i}|=1{\text{ m}}$, are the old basis vectors; the ${\overline {x}}^{i}$, with $|{\overline {x}}^{i}|=0.01{\text{ m}}=1{\text{ cm}}$ are the new basis vectors.—Anita5192 (talk) 03:17, 25 November 2018 (UTC)

Maybe this helps:

Let's have two ordered bases of a 3-D vector space over $\mathbb {R} ,$ consisting of three orthogonal vectors each

$\{e_{i}\mid i\in \{1,2,3\}\}\quad$ and $\quad \{{\bar {e}}_{i}\mid i\in \{1,2,3\}\}\quad$ with

$||e_{i}||=1{\text{m}}\quad$ and $\quad ||{\bar {e}}_{i}||=1{\text{cm}}\quad$ for $i\in \{1,2,3\}.$ (To be honest, I am unsure how to formally conflate a multiplicative group of units with the norm of the vector space)

Further, let (for simplicity, and satisfying the above)

${\bar {e}}_{i}={\frac {1}{100}}e_{i},\quad$ again for $i\in \{1,2,3\}.$

This yields the matrix $A$ from above. Now let's have a vector, and its decompositions wrt the two bases (adopting Gaussian summation convention)

$r=r^{i}e_{i}={\bar {r}}^{i}{\bar {e}}_{i}.$

For the last equalities to hold, the scalar coefficients must satisfy

${\bar {r}}^{i}=100r^{i}\quad$ for $i\in \{1,2,3\}.$

This yields the relation for coefficients, given by JRSpriggs above. Maybe it's all about the vectors, formed from the coefficients suitably arranged in tuples (and freely identified with abstract vectors) that vary contravariantly? (I'm struggling myself.)

OK? Purgy (talk) 13:14, 25 November 2018 (UTC)

This makes perfect sense to me, and I believe this is what I wrote in equations (1,2,3,4), but I do not see how equations (9,10) agree with this. For example, equation (1) seems to imply that for the basis vectors, $x^{i}$ and ${\overline {x}}^{i}$,

${\frac {\partial {{\overline {x}}^{i}}}{\partial {x^{i}}}}={\frac {1}{100}},\quad (1\leq i\leq 3)$ 11

(the other partial derivatives are zero), but then equation (9) for the components $(T^{i})$ and $({\overline {T}}^{i})$ of $(\mathbf {r} )$ and $(\mathbf {\overline {r}} )$ would be

${\overline {T}}^{i}=T^{i}{\frac {1}{100}},\quad (1\leq i\leq 3),$ 12

in contradiction to equation (4).

I have faith that equations (9,10) in textbooks are somehow correct, but because they look incorrect to me, I evidently have a misconception somewhere. I am trying to determine where and what that misconception is—Anita5192 (talk) 20:27, 25 November 2018 (UTC)

I do not know what you mean by equation (2) because you do not explain how you are using A. Equation (3) is obvious bull-shit. It should be

$x^{1}=3;x^{2}=4;x^{3}=12.$

Again I stopped reading at that point until you clean up your act. You must explain what all the variables in your equations mean or you will never get your head straight. JRSpriggs (talk) 01:58, 26 November 2018 (UTC)

I regret having mistaken the question. I thought to perceive a mix up of scalar coefficients and vectors, wherefore I dismissed the $x^{i}$s, seemingly used for both kinds, and introduced the $e_{i}$s for vectors and the $r^{i}$s for scalars, and wrote that boring triviality. Maybe, I tripped over the possibility of having abstract indices to denote vectors, but again, I have no formal conception for these. Purgy (talk) 07:08, 26 November 2018 (UTC)

I think I see where some of my misconceptions were. Correct me if I am wrong. (By chance, I discovered something in one of my books that clued me in, although it was not obvious.) Most textbooks that I have seen for linear algebra and tensors do not make clear the distinction between basis vectors, coordinate vectors, coordinates, and components. I tend to agree with Granzer92 (talk) in the previous section, that the Introduction did not make this clear either, and still does not. I have edited my original post above. Now equations (4) and (8) seem to agree with equations (9) and (10). Please let me know if I have this correct now.–Anita5192 (talk) 22:56, 27 November 2018 (UTC)

I agree to the reservations addressed by Granzer92, and admit not being a jour with JRSpriggs improvements. To my impression, much of this potential of confusion is caused by the wide spread, convenient notation of a basis-transformation, involving -say- "v-vectors" with bases-vectors as coordinates, and matrices of scalars as linear maps, without (sufficiently) emphasizing that the "v-vectors" live in "another" space than the basis-vectors, and that these spaces just share their scalars. The didactic migration from coordinate-bound to coordinate-free has not fully matured, yet, imho. Purgy (talk) 08:49, 28 November 2018 (UTC)

References

Co-/contra- variance of 'vectors' (per se) or of their components

It is my understanding that a vector, per se (i.e. not its representation), is neither covariant or contravariant. Instead covariance or contravariance refers to a given representation of the vector, depending on the basis being used. For example, one can write the same vector in either manner as, ${\vec {A}}=a^{i}e_{i}=a_{i}e^{i}$. I think this point (in addition to the variance topic as a whole) can be both subtle and confusing to people first learning these topics, and thus the terminology should be used very carefully, consistently, and rigorously. I tried to change some of the terminology in the article to say, "vectors with covariant components" instead of "covariant vectors" (for example), but this has been reversed as inaccurate. So I wanted to open a discussion in case I am mistaken or others disagree. @JRSpriggs. Zhermes (talk) 16:45, 2 January 2020 (UTC)

I have several textbooks that describe covariance and contravariance of tensors. Most of them refer to the tensors themselves as having these properties; however, some of them indicate that these properties apply only to the components and that the tensors are invariant. I am inclined to agree with the latter, that is, that only the components change—not the tensors. I think we should point out in the article two things: 1. what we think is the correct parlance, and 2. the fact that some textbooks say otherwise. I can supply citations.—Anita5192 (talk) 17:54, 2 January 2020 (UTC)

The key point to understand is that the distinction between "covariant" and "contravariant" only makes sense for vector fields in the context of a preferred tangent bundle over an underlying manifold.

Otherwise, all the structures could just be called simply "vectors" and the choice of one basis or another would be completely arbitrary.

So, restricting ourselves to structures built from the tangent bundle, the vectors in the tangent bundle itself are called "contravariant", and the vectors in the dual or cotangent bundle are called "covariant". Tensor products of tangent and cotangent bundles may have both covariant and contravariant properties (depending on the index considered). JRSpriggs (talk) 02:13, 3 January 2020 (UTC)

I'm afraid that in a strict sense, there are two distinct meanings in play here, and unless we distinguish them, confusion will ensue. One is a description of components, and mentioned above, and the meaning used by JRSpriggs essentially means "is an element of the (co)tangent bundle". The former makes sense in the context of a vector space and its dual space, independently of whether these are associated with a manifold. It is unfortunate that texts often conflate a tensor with its components with respect to a basis. I expect that the interpretation of a "contravariant vector" as being "an element of the tangent bundle" arises from exactly this conflation, and should be avoided (there is a better term: a "vector field") in favour of restricting its use to describing the behaviour of components with respect to changes of a basis. If in doubt about the confusion implicit in the mixed use of the terms, consider this conundrum: the set of components of a vector is contravariant (transform as the inverse of the transformation of the basis), whereas the basis (a tuple of vectors) is covariant (transform as the transformation of the basis by definition, in the first sense), yet we would describe the elements of the basis as being contravariant (in the sense of belonging to the tangent space), making them simultaneously "covariant" and "contravariant". Let's not build such confusion into the article, and restrict its meaning to (the no doubt original) sense. —Quondum 18:06, 31 March 2020 (UTC)

I had precisely the same question, that I opened in math stackexchange. I haven't got any satisfactory answer yet. I hope somebody will pick this thread and continue the discussion. https://math.stackexchange.com/questions/4297246/conceptual-difference-between-covariant-and-contravariant-tensors — Preceding unsigned comment added by 99.62.6.96 (talk) 05:24, 29 December 2021 (UTC)

Introduction for the layman

I would add a paragraph to the introduction, which is easier to understand. Maybe something like this:

contravariant vectors

Let's say we have three base vectors a, b and c. Each one has the length 1 meter. d = (3,2,4) is a vector (≙ (3m,2m,4m)).If we now double the length of every base vector, so that |a| = |b| = |c| = 2m, then d must be (1.5, 1, 2) using the new a, b, c basis.(but d would still be (3m,2m,4m))

covariant vectors

Let's say f is a scalar function and the base of the coordinate system is a, b and c. And suppose |a| = |b| = |c| = 1 meter. Then suppose that ∇f=(2,5,9); so that the slope in x-direction is 2/ meter (2 per meter). If we now double the length of each base vector, so that |a| = |b| = |c| = 2m, ∇f becomes (4,10,18). The slope in x-direction is the same: 4/2m = 2/m.

CaffeineWitcher (talk) 12:59, 23 May 2020 (UTC)

Covariance of gradient

I am a bit confused : this article takes the gradient to be a prime example of a "covariant vector", but the Gradient article claims that it is a contravariant vector. Which is correct? (Sorry if this is the wrong place to ask) --93.25.93.82 (talk) 19:56, 7 July 2020 (UTC)

After thinking a little, I think I understand where my confusion comes from: the gradient of a function is covariant with respect to the input basis, but contravariant with respect to the output basis. Is this a valid interpretation? If it is, maybe this could be clarified in the lead of this article. --93.25.93.82 (talk) 20:08, 7 July 2020 (UTC)

I do not understand your second comment. The gradient article is wrong because it is choosing the primary meaning as the one which includes the metric (i.e. the dot product) rather than the one which is merely the derivative. JRSpriggs (talk) 07:31, 9 July 2020 (UTC)

I back my position up with the reference Gravitation (book) page 59. JRSpriggs (talk) 05:12, 15 July 2020 (UTC)

I meant that if a gradient $g$ is seen as a linear form, say for instance from $\mathbb {R} ^{3}$ to $\mathbb {R}$, and $M\in GL_{3}(\mathbb {R} )$ (resp. $N\in GL_{1}(\mathbb {R} )$) is an invertible matrix corresponding to a change of basis $B\rightarrow B'$ in $\mathbb {R} ^{3}$ (resp. a change $C\rightarrow C'$ in $\mathbb {R}$), then $g$ will be expressed in the bases $B'$ and $C'$ as $N^{-1}gM$ (assuming of course that $g$ was originally expressed as a row matrix in the bases $B$ and $C$). — Preceding unsigned comment added by 93.25.93.82 (talk) 13:03, 10 July 2020 (UTC)

Example of vector which is not contravariant is arguably not a vector

Currently: "On the other hand, for instance, a triple consisting of the length, width, and height of a rectangular box could make up the three components of an abstract vector, but this vector would not be contravariant, since a change in coordinates on the space does not change the box's length, width, and height: instead these are scalars." I think this is not helpful since there is no obvious meaning to the vector space of box [length, width, height]. What do vector scaling and vector addition correspond to? I suggest this example be removed. Intellec7 (talk) 05:19, 28 August 2020 (UTC)

OK. Go ahead and delete that sentence. JRSpriggs (talk) 14:47, 28 August 2020 (UTC)

Geometrical construction of coordinates

If you have a basis $\mathbf {e} ^{i}$ in a three dimensional Euclidean space you can construct the coordinates of a given point by drawing lines through the point parallel to each basis vector. Those lines will intersect with each other and the distance of the intersection point from the origin divided by the length of the corresponding basis vector gives you the covariant components $q_{i}$. But how do you construct the coordinates in respect to the dual basis $\mathbf {e} _{i}$ ???? 2003:E7:2F3B:B0D8:11C:4108:841E:BA24 (talk) 07:00, 17 April 2021 (UTC)

Mis-use of the word "scalar"

The word "scalar" is used in the page to mean something that multiplies a unit vector. But a scalar is supposed to be coordinate-free or gauge-invariant or invariant to changes of coordinates. The quantities that multiply unit vectors do not have this property, because the unit vectors change (and hence the components of the vectors change) as you change coordinates. --David W. Hogg (talk) 12:58, 16 May 2021 (UTC)

I tried to fix that. Let me know if there is still a problem. JRSpriggs (talk) 18:26, 16 May 2021 (UTC)

Vectors vs. Covectors

I have some "issues" with the fundamental jargon used in this article. I would like to help improve it; but I don't want to start an editing war. So, let me test the waters with a few comments:

Vectors and covectors are not the same thing. And they are both invariant under (proper, invertible) linear transformation. (Note: this is the passive/alias viewpoint of transformations). It is a semantic error to say that a vector is contravariant (or that a covector is covariant). The co/vectors, and tensors in general, are invariant under linear transformations. If the space has a metric, then one can "convert" a vector into a covector, and vice versa. But the existence of a metric is not obligatory and, in fact, confuses people into thinking that vectors and covectors are fungible. They are not. Vectors are linear approximations to the curves defined by the intersection of $n-1$ coordinate functions; covectors are linear functional approximations to the level (hyper)surfaces of a single coordinate function. The figure at the top of the article kind of hints at this, but then garbles the message by overlaying arrow quantities with level-surface quantities on the right-hand side. Too bad. Remove those blue arrows and you'd have a right proud representation of covectors in a cobasis.

Co/contra-variance is a property of the components and of the basis elements. For a vector $(\mathbf {v} )$, the components $(v^{i})$ are contravariant and the basis vectors $(\mathbf {e} _{i})$ are covariant; for a covector $({\boldsymbol {\omega }})$, the components $(\omega _{i})$ are covariant and the basis covectors $(\mathbf {e} ^{i})$ are contravariant. In either case, when you contract the components with the basis—one of which is covariant and the other contravariant—then you get an invariant quantity, as required of a tensor.

I won't try to define/defend here (yet;-) what co/contra-variant mean. (Spoiler alert: contravariant quantities transform as the Jacobian of a coordinate transformation; covariant quantities transform as the inverse Jacobian; this seems backwards to what I, for one, would expect from the concepts of co- and contra-; but it is what it is!)

To whomever has purview over this article: if these comments make sense and seem worth the trouble of editing the article, please let me know and I'll try to collaborate. On the other hand, if you think these are distinctions without a difference—or worse, misguided—then I'll stand down.--ScriboErgoSum (talk) 08:25, 15 November 2021 (UTC)

NPOV

The article describes covariance and contravariance in terms of coordinates and components, a perspective that is rather dated. The terms have a meaning independent of any choice of basis or coordinates, and the article should reflect that. There is a lot of variation in the literature, but essentially there are three styles:

1. Define the tangent and cotangent bundles independently, then prove duality.
   1. Authors often define tangent vectors as equivalence classes of curves through a point.
   2. Authors often define cotangent vectors in the language of germs.
2. Define the tangent bundle, define the cotangent space at a point as the dual of the tangent space at that point and then define the cotangent bundle.
3. Define the cotangent bundle, define the tangent space at a point as the dual of the cotangent space at that point and then define the tangent bundle.

Tensors can then be defined as either tensor products or as multilinear maps. It is common to just classify tensors by the covariant and contravariant ranks, but if there is a (pseudo)metric involved then order matters because of raising and lowering of indexes. --Shmuel (Seymour J.) Metz Username:Chatul (talk) 11:32, 1 March 2022 (UTC)

This article mostly describes tensors; what you are describing is tensor field. But taking your point, there should be some section for the tensor product language on this page. Even so, there is no need to choose one way over another. (I object to the idea that bases and coordinates are dated.) Gumshoe2 (talk) 07:26, 13 March 2022 (UTC)

There is a large amount of material in the Tensors article that does not belong there, but should rather be in Tensor fields. The distinction between covariant and contravariant is really only relevant for tensor fields on a manifold, but as long as it is in this article the text should reflect its character.

I didn't say that coordinates and bases are dated, but rather that defining, e.g., tensor, covariant, in terms of coordinates and bases rather than intrinsically is dated. --Shmuel (Seymour J.) Metz Username:Chatul (talk) 12:51, 13 March 2022 (UTC)

I agree that the tensor product definition should be given here, but I think it should be given together with the present one. Why do you say that co(/ntra)variance is only relevant for tensor fields? There is of course a distinction between an element of a vector space and an element of the dual vector space, for instance. Gumshoe2 (talk) 17:37, 13 March 2022 (UTC)