What are Contravariant and Covariant Components of a Vector? Part 1

A brief look at applications of differential geometry and the concept of contravariant and covariant components of a vector. It is shown that in the simple case of an oblique coordinate system in two dimensional Euclidean space the formula for the length requires covariant and contravariant components of a vector. The metric tensor is introduced and its components found using coordinate transformation matrices. Mysterious upper and lower vector indices are explained. More sophisticated algebraic operator versions of this concept, including the tangent and co-tangent dual, will be discussed in future videos.

Category:

Education

Tags:

• Contravariant
• Covariant
• Vector
• Differential
• Geometry
• Metric
• Tensor
• Euclidean
• Space

[mufc4everch]

Thanks, this is nice but you should try using different colours

[Mathview]

@mufc4everch TY for the note, in these early videos I purposely chose to use a single ink color for the screenboard. I find that too many colors, color for color sake, can be distracting. In more recent videos, like in Manifolds Part 1, I experimented with colors in diagrams. There the colors are keyed to distinct types of mathematical objects. Another use color is in text highlighting that guides the eye and connects the narration to specific text or equations on the screen.

[palui]

wouldn't the metric tensor need to be multiplied by 3 differentials since we're in 3-space? It seems like it would need dx dy dz. Can the metric tensor manipulate only dx and dy to move through 3-space? Not sure I understand.

[Mathview]

@palui Ah...yes. Good question. The metric tensor G can be used to get the length of a vector b like this... L^2 = G(b,b) where L^2 is the square length of the vector b. If you have two distinct vectors a, b then G(a,b) gives an inner product or dot product of the two vectors. This works for vectors in spaces of any dimension. In a d-dimensional space, the matrix form of the metric tensor becomes a d x d symmetric matrix and vectors have d components. Does that answer make sense?

[palui]

@Mathview So, in 3 space the metric tensor is a 3x3 matrix? But in your video you have ds^2 = g(ij) dxi dxj. So g(ij) would need to be 2x2. Still confused.

[Mathview]

@palui Yes... right on all points. In the x,y plane with Cartesian coordinates where x and y are orthogonal axes, the metric tensor matrix is just the 2x2 unit matrix. The indices (i, j) that show up in dx^i and g(i,j) range from 1 to d where d is the dimension of the vector space. Sometimes higher rank tensors are useful, for example the Riemann curvature tensor has rank four. So R(j,k,m,n) has 4 indices. If the space is three dimensional then the (j,k,m,n) indices range over 1,2,3.

[petermetric]

Enjoyed the animation but what about higher dimensions and also when 'orthogonal' has no meaning. i.e. a non-metric space

[PCGamerPortal]

@Mathview Hey, thanks for the reply! I am still very new to your channel so I haven't given myself the time to look at your newer material yet, but I know making videos is really tough at first so I'm sure you must have gotten a lot better by now. Thanks for doing what you do by the way.

[Mathview]

@PCGamerPortal Ah...yes I know what you mean. It's a tough choice sometimes, but perhaps the stuff that was edited out was way worse than a few bumps and clicks! Seriously, thanks for the comment. Back when we started there were lots of rough edges. With experience, and hard work the newer ones are a bit more ... polished? or at least not as rough. One issue is the headphone microphone seems to pick up alot of background sounds.

[PCGamerPortal]

you should try not to edit the audio, if you mess up just go with it. Otherwise the bumps and snips in the audio make it a bit hard to concentrate in my opinion.