Talk:Covariance and contravariance of vectors

message: 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)
thread_title:  Introduction for the layman 

message: 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)
replies: text: 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)
replies: text: 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)

text: I back my position up with the reference Gravitation (book) page 59. JRSpriggs (talk) 05:12, 15 July 2020 (UTC)
replies: text: I meant that if a gradient g is seen as a linear form, say for instance from \R^3 to \R, and M \in GL_3(\R) (resp. N \in GL_1(\R)) is an invertible matrix corresponding to a change of basis B \rightarrow B' in \R^3 (resp. a change C \rightarrow C' in \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)

text: I find it 1) not likely that the author of this bit is right and everyone else who writes on the topic is wrong, and 2) pretty obvious that gradient is covariant. If the temperature of a fluid drops by 1 degree per meter (a temperature gradient), then it's going to drop by ten degrees per ten meters, so if you change your unit of measure from one metre to 10 metres, the temperature gradient (change in degrees per unit of distance) likewise multiplies by 10. If the author of the "actually, gradient is contravariant" bit would care to plug that simple example into their \nabla f = \mathbf{e}_i (\nabla f)^i=\mathbf{e}_i \delta^{i}_j (\nabla f)^j=\mathbf{e}_i (T)_k^i(T^{-1})_j^k (\nabla f)^j\ , and explain why they get a different result, perhaps this would be helpful. 203.13.3.93 (talk) 03:38, 29 January 2024 (UTC)
replies: text: There are two different conventions for cpvariant versus contravariant; one gives primacy to covectors (diiferential forms) and on gives primacy to vectors. The gradient of a scalar field is a covector fiels; if vctors are covariant then it is contravariant and if vectors are contravariant then it is covatiant. I don't recall which convention the "Texas Telephone Directory" (MTW) uses. -- Shmuel (Seymour J.) Metz Username:Chatul (talk) 15:15, 29 January 2024 (UTC)

text: I think a few comments here have confusion based on the fact that the gradient of a function on R3 can be interpreted equally well either as a covariant or contravariant vector, because R3 has a canonical Riemannian metric. On a more general space, the derivative of a function is covariant (i.e. a 1-form) and if there is a metric it can be converted, in a way depending on the particular metric, to contravariant (i.e. a tangent vector). In common usage the former is called the differential of the function and the latter is called the gradient. Gumshoe2 (talk) 15:55, 29 January 2024 (UTC)
thread_title:  Covariance of gradient 

message: 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)
replies: text: OK. Go ahead and delete that sentence. JRSpriggs (talk) 14:47, 28 August 2020 (UTC)
thread_title:  Example of vector which is not contravariant is arguably not a vector 

message: 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)
thread_title:  Geometrical construction of coordinates 

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

text: What about scalar densities (or if you refer to work from a concrete example, 'density' as in kg/m^3)? These are rank-zero tensors (or tensor fields) but nevertheless they are covariant. Can't recall the name, but one text I have read suggests that you can also have a 'scalar capacity', which can also be a field, which is a contravariant scalar.  — Preceding unsigned comment added by 203.13.3.89 (talk) 03:26, 29 January 2024 (UTC)
thread_title:  Mis-use of the word "scalar" 

message: 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)
thread_title:  Vectors vs. Covectors 

message: I am adding this (3/26/2022) without reading the below because I have a comment on the definition section. We have  with each of the Xis being basis a vector. I was going to add a remark that said this makes f a matrix and indeed it is used as a matrix in the unnumbered equation v=f v[f]. I decided not
to add that comment as it contradicts Note 1.  In the note preceding Eq 1 it says "regarding f as a row vector whose entries are the elements of the basis". Maybe I am naive but I find this pretty confusing as I've always considered matrices and vectors to be different objects. If f is to be considered a row vector but with each element a basis vector instead of just a number (as one usually thinks of vectors) this needs to be explained. If the note is incorrect then it should be removed. — Preceding unsigned comment added by  76.113.29.12 (talk • contribs) 14:45, 26 March 2022 (UTC)
replies: text: I removed the template. It is sometimes useful when dealing with matrices to consider a matrix as a column vector of row vectors or a row vector of column vectors. In this sense, a vector is just a list of some object even though it's entries are not strictly from a field. I tend to agree with your level of scrutiny, which if applied would require we don't use the word vector (see here). The wording of vector could be changed to that of list, but I don't think that would be a helpful change and actually could be a little detrimental, so I am leaving it. 69.166.46.156 (talk) 19:35, 25 April 2022 (UTC)
thread_title: Inconsistency of note 1 with the main text

message: Should the article discuss relative vectors, whose transformation includes a power of the transformation determinant as a factor? In more modern language, these are tensor products of vectors with tensor densities, or for orientable manifolds, liftings of line bundles. -- Shmuel (Seymour J.) Metz Username:Chatul (talk) 14:31, 13 November 2023 (UTC)
replies: text: This seems like a good place for it.  Note that in tensor there is some content on tensor densities which uses notation compatible with this article.  Tito Omburo (talk) 22:31, 15 March 2024 (UTC)
thread_title:  Relative vectors?