What are Contravariant and Covariant Components of a Vector? Part 2

[AllOtherNamesTaken2]

Why is one set of coordinates contravariant and the other one covariant?

[Mathview]

@AllOtherNamesTaken2 Good question. The terms are somewhat arbitrary. Most current references don't bother explaining the term origin. i.e. Contra - variant to what? Co - variant with what else? The two types of vectors are defined via a coordinate transformation rules, or algebraically via the dual space concept. When changing coordinate systems the components of a covariant vector change by the factor (d x_old / d x_new). One could say it varies "with" x_old, so co-variant. More later.

[MrSlayer4ever]

Very useful videos.I was trying to understand how tensors work but it is difficult for me to study from books because there are some things that I don't understand like covariant and contravariant components of a vector.Finally I know what they are.I'm moving on to your videos about tensor algebra hoping that they will be as helpful as these were.Thanks.

[StudyAcademic]

I DON'T GET THAT FORMULA FOR VECTOR C in v and u coordinate system. Why did you use tan? how did you find upper componets?

[Mathview]

@StudyAcademic The formula can be found by doing trigonometry on the figure that shows the two coordinate systems. It's a little tricky, but you can get the tangent function and the rest from simple trig. on the drawing.

[yn30s]

Wow, that makes a lot of sense. I've been studying Reimannian metrics in detail so I actually know exactly what you're talking about. Of course Lee could have just indicated this relationship directly but I think he likes to torture you (which I must admit often does lead to a deeper appreciation even if it takes longer). Thanks so much for the communication, you have helped me to understand this a lot.

[Mathview]

BTW The notation that is more standard usage has the matrix of g with two lower indices as the metric tensor, and g with two upper indices as the inverse matrix. The calculations in the video are self consistent, though. I guess I should re-do this video sometime and fix-up this detail. Also, a good exercise is to derive this from scratch yourself on a blank sheet of paper. The trig calculation is all you need to do to get these formulas.

[scarface14441]

i know this is dumb but witch line is b im not sure

[Mathview]

@scarface14441 Ah..good Q. We have two coordinate systems in the plane. The oblique coordinates with non-orthogonal axes, and the regular old Cartesian coordinates, with orthogonal axes. In the Cartesian system the coordinates of the vector C are (a,b). So Cx = a, Cy = b. If we instead use the oblique coordinates, we can construct the covariant and contravariant components of C, as described in the video, which permits us to write the length of C, sqrt(a*a+b*b) as Csub1*Csup1 + Csub2*Csup2.

[yn30s]

I also know that there is a certain discrepancy between the classical and modern literature on this, having to do with how the prefixes "co" and "contra" are used. Like, "co" should mean "with," and "contra" should be "against," yet in differential geometry, pushing forward corresponds to contravariant tensors, while pulling back corresponds to covariant tensors ... or something like that. ...I feel like I can do the more complicated stuff, but somehow never fully comprehended the basics.

[yn30s]

Okay. (Thanks for responding by the way!) So let's say I'm at a fixed point p in a Reimannian manifold M. Would it make sense to say that my covariant coordinates are the ones I get from being locally homeomorphic to Rn (via some local coordinate chart), and that my contravariant coordinates are the ones from the induced basis on the local frame for the tangent bundle? (I know that the tangent bundle has by def. local trivializations, and that these with the local chart induce a local frame.)

[Mathview]

@yn30s Great comments! I agree, simple is better. So ... In any Euclidean Space, like the tangent space TMp at point p, we can define a new inner product ( x | y ) as (x | G | y ) where G is a symmetric positive definite matrix ( a well known Linear Algebra result.) The length of a vector X with this inner product is then (X|G|X) and we can consider the vectors (X| and G|X) as separate vectors, the co and contra vectors, and the matrix representation of G as the metric tensor.

[Mathview]

I should add that I have been describing vectors and tensors of classical physics where we have field quantities like the fluid flow velocity, the heat flux, or the electromagnetic field tensor where we have a physical interpretation. In pure mathematics we have more exotic objects like fiber bundles and "dual bundles" and their associated tangent spaces. In those examples the geometric interpretation can be a bit more difficult. More on this later. TY for these thought provoking comments.

[yn30s]

The covariant coordinates seem to depend on some knowledge of what is "actually" perpendicular to an axis, regardless of what the coordinate system says. How are we guaranteed any sense of this in an abstract space?

Secondly, I've been reading John Lee's DG books and he must have just assumed I already knew what covariant and contravariant mean because every time it comes up I feel lost. I had the impression it had to do with whether the vector acted on functions or on other vectors...?

[Mathview]

@yn30s Wow Great discussion on this topic TY for posting. All this index calculus takes place on a Riemann Manifold which is a Metric Space with a coordinate system labeling the points. So we have points and coordinates but no Vectors, so far. Where do we get Vectors? Riemann manifolds have another property. They are "locally Euclidean." Meaning that epsilon neighborhoods of points are Euclidean (vector) Spaces. At small scale the space becomes flat and we have Euclidean Vectors locally.

[Mathview]

@yn30s Where do we get co-and contra- coordinates? After all, a Manifold is a Metric Space with No Vectors No inner product No angles, just points and coordinates. Angles do sneak in if we look at an epsilon neighborhood of a point P in M. If we start at P, the coordinate curves look like straight lines for short distances from P, i.e. in the neighborhood of P. So we are allowed to construct a Euclidean space at every P in M. These Tangent Spaces are complete Vector Spaces same dimension as M.

[Mathview]

@yn30s Where do we get co- and contra- coordinates? (cont.) So we have the Tangent Space at a point P in M. The Tangent Space Tp is flat and the lines tangent to the coordinate curves at P form a natural coordinate system in Tp. The differentials of the coordinate curves at P form a basis for the space. The angles between the differentials do not have to be right angles. So in general we have oblique coordinates in Tp. That's why we need co- and contra- basis for Components of Vectors Tp.

[spiritwalkerXXl]

TY for the answer! ok, so from what you say there is no intrinsically covariant or contravariant vector, it depends on the choice of the coordinates for the representation, but most books define covariant and contravariant vectors (tensors in general) depending on the way they transform under a change of coordinates, it seems a different concept of covariant and contravariant compared to what you explain.

[Mathview]

@spiritwalkerXXl Ah...can be confusing in some older texts. Consider this: Let A^k be contravariant components of the vector A, and A_n be covariant components (downstairs index.) The metric tensor g is used to "raise or lower indices" so we have A_n = g_n,k A^k (sum over k.) g converts "contra to co." If we change coordinates, the transformations for the A_n's comes from transforming both A^k AND the g_n,k. That's why the transform for A_n looks different from the transform of the A^k.

[Mathview]

Ooops... typo... last sentence in third comment below should end with "... by raising or lowering the index with the metric tensor."

[Mathview]

I would say that in casual conversation we talk about covariant and contravariant vectors as though they are different. They do seem different if you are doing a given calculation using index calculus. But strictly speaking the difference comes from the coordinate systems. Q: An abstract vector, or in physics, a vector field, exists mathematically in a tangent bundle on a Riemann manifold. Does the tangent bundle have a geometric or physical reality?

[Mathview]

More: We have two distinct coordinate systems on the manifold, covariant and contravariant coordinates. X with lower index and X with upper index. A given geometric point of a Riemann manifold has two coordinate representations, but they label the same geometric location in the manifold. It is always possible to convert from one representation to the other. Excellent question, glad you posted it. Please let me know if this explanation is on the mark, raises more questions, or is not clear.

[Mathview]

TY for posting an excellent question! Let me get directly to your question. The difference between covariant and contravariant representations of an abstract vector is analogous to a geometric Euclidean vector represented by Cartesian (x,y,z) or Sperical polar coordinates (r,theta, phi). Same vector, different coordinates. One hint that covariant vectors are related to contravariant vectors is that we can convert one into the other by raising or lower with the index metric tensor.

[spiritwalkerXXl]

Excelent video! BTW there is something that I still don´t understand, the relation of what you explain in the video with a covariant or contravariant VECTOR ( the usual example for a covariant vector is the gradient) it seems that a vector is intrinsically covariant or contravariant (depending on the way they transform under a change of coordinates) and is not related with what you have explained, that every vector have covariant and contravariant components

[nementerc]

Looking at your video, I finally see clearly the difference between covariant and contravariant vectors, If the same applies to tensors, then I believe I finally can visualize what general relativists have been writing all along. Thanks for shinning the light.

[aparsloe1]

I'm very grateful for this, I'm doing relativity with the OU and they don't explain this anywhere as clearly as this. Thanks

[KikiManini]

Am I write in saying that the covariant components (c_u and c_v) cannot be attached to the basis vectors u and v to construct the vector c? If so does that mean when we express the vector c using the notation (c_u,c_v) that we are not expressing it as the sum of a set of basis vectors?

[Mathview]

@KikiManini Thank you for this question, it's a good one. To get a Vector C in component form as a sum of basis vectors times components, we must use two distinct basis vector systems. They are called Covariant Basis and Contravariant Basis systems, or a Dual Basis. Matching the covariant basis with the covariant components allows them to be Vector added ( e.g. parallelogram construction) to get C. There are some animations of this in the Playlist Differential Geometry 1. Great Q!

[KikiManini]

I'm currently studying relativistic electromagnetism as a part of a course and our lecturer did not explain any of this with the clarity you have. Thank you very much!

[Mathview]

Hi All. If I missed your comment or your comment didnt get posted, please try to post it on the channel comments setion of the home page. It seems some comments might be getting lost, maybe a software glitch from Google YouTube, not sure.

[Mathview]

@maxbob427 Ah yes! Good question. If you can wait a few days, there is a new video almost finished that revisits this topic using projection operators and tangent spaces. I hope to include your comment and some discussion. Posted is an animation of co/con basis vectors in E^2 for non-orthogonal (oblique) coordinate systems. There is an interesting scale change for the oblique case. Also distinguishes basis SEQUENCES from basis SETS, order implies handedness. TY for the comment.

[mathphys75]

very good very helpful, clear, thank you!

[johnswburg]

First I would like to tell you that your videos on the covariant/contravariant/metric tensor are great.....I'm doing research in convex membrane finite elements dealing with line stretch along a convex surfaces in a reference and deformed pressure domain...This really brings light to the physical meaning of the metric tensor...keep up the great work!

[Mathview]

@johnswburg Ah yes! Convex membrane finite elements. Are you looking at the equililbrium solution or time dependent? And you get such nice pictures using visualization tools. Matlab? or? Great comment and encouragement. TY

[johnswburg]

@Mathview a static analysis through various pressures...using tracking markers for stretch values on photographs through various pressure states (photogrammetry)...then taking advantage of static determinacy when solving for a stress in a convex membrane (if it has a known internal pressure and a thin wall)....joining these two parts of data a stress/stretch graph can be made for every data point on a nonlinear heterogeneous membrane (biotissue)....But yes lots of matlab

[IQ20000Berta]

So easy to understand thanks for your great work

[robkim55]

Nice explanation of covariant and contravariant components

[ciaoidea]

clear explanation tnx a lot! helpful video.

[ayanmoitra1]

Very helpful

[fix520]

nice, very easy to folo. I understand it somwhat plan to keep learning!

[mva286]

how can we calculate the covariant and contravariant components of a vector w.r.t a Gaussian (curvilinear) coordinate system?

[Johanadrian]

amazing!! I dont even have this in any class, but Im browsing through more and more videos. Nice way to spend a friday night ;)

[davepinsent]

Thank you for that clear explanation! I've been struggling to understand contravariant and covariant vectors for weeks now. I'll be sure to check out your other videos on differential geometry.

[elliottc2001]

this is awesome

[Kashrlyyk]

The vector c in the oblique coordinates picture and the vector c in the triangle are the same. So no problem in adding the triangle to the obligue coordinate system. You literally see that a is bigger than c^u. In order to see how much we need to split up the c^v vector into two vectors: one parallel to the x and one parallel to the y axis of the cartesian coordinate-system. c^v is parallel to v so alpha is the angle between v and u and between c^v and u.

[Kashrlyyk]

Then use pythagoras and this gives you the first of the equations in my post above.

For the second equation we use pythagoras too. "c^v" is the long side and "b" and "c^v * cos(alpha)" are the two short sides. This gives you the first part of the second equation: (c^v)^2 = (c^v *cos(alpha))^2 + b^2 => (1-cos(alpha)^2)*(c^v)^2=b^2 => (sin(alpha))^2 * (c^v)^2 = b^2 => c^v = b/sin(alpha)

[xslaya22x]

Thank you so much for posting this. I have only been presented with this material in an abstract format, and the examples you have shown make the ideas much easier to understand.

[ofeyofey]

Can't see connection between triangle abc and the oblique coordinate system. Therefore I can't see why c^u=a-b/tan(alpha).

[Kashrlyyk]

Draw the cartesian coordinate system and then add the oblique coordinate system, c^u, c^v and the triangle abc to that. It is obvious then that a is longer than c^u. You get: c^u = a - c^v * cos(alpha).

(c^v)^2 = (c^v *cos(alpha))^2 + b^2 => c^v = b/sin(alpha). QED.

[volintine]

can u explain in detail plz?

[Kashrlyyk]

If the answer above is not enough just holler.

[ofeyofey]

Remember 'Co below' for covariant index is below and opposite, raised index for contravariant index.