I am studying co- and contra- variant vectors and I found
the video at https://www.youtube.com/watch?v=8vBfTyBPu-4
very useful. It discusses the slanted coordinate system where the X, Y axes
are at an angle of α. One can get the components of **v** either by dropping perpendiculars to the axes (v_{i}) or by
dropping a line parallel to the other axis (v^{i}). These give correct results
for the norm v_{i} v^{i} and the dot product v_{i} w^{i}.
(I have not shown **w**). So the v^{i}
are called contravariant and the v_{i} are called covariant. According to
the video Dirac thought this was a great example.

But both v

_{i} and v

^{i} contra-vary with a change of scale of the basis
vectors. This contradicts some definitions of contravariant and covariant components,
e.g. this one on Wikipedia.
These definitions say that covariant components co-vary with a change of scale.

Is there a simple resolution to this apparent contradiction?

Submitted as comment on video and

on Physics Forums at

https://www.physicsforums.com/threads/covariant-coordinates-dont-co-vary.959888/
Orodruin answered on PF seven hours later with "The covariant and contravariant components belong to different basis vectors." Cryo came up with something a bit more complicated and I followed Orodruin. So here's how it goes:

To get

**v** with the covariant coordinates we must have

**v** = v_{x}**e**_{x} + v_{y}**e**_{y} (1)

I have drawn them in on the diagram and **e**_{x} is clearly smaller than **e**^{x}, which I have conveniently drawn so that v_{x}
= 1. We must also have

v_{x}**e**_{x} = v^{x}**e**^{x} (2)

and similarly

v_{y}**e**_{y}
= v^{y}**e**^{y} (3)

which give us

**v** = v^{x}**e**^{x} + v^{y}**e**^{y} (4)

which is as it should be.

(3) gives us

v_{y} = **e**^{y}(**e**_{y})^{-1}v^{y} (5)

where (**e**_{y})^{-1}
is the inverse of **e**_{y}: or **e**_{y}(**e**_{y})^{-1} = 1.

(5) shows us that v_{y} does indeed vary with **e**^{y} with a strange constant
of proportionality (**e**_{y})^{-1}v^{y}.

##
Back to the drawing board

Orodruin did not like that at his #5 and then gave me a tip at #7 that the dual (or covariant) bases are given by ê

^{a} = ∇x

^{a}. I will use ê for basis vectors in future, following Carroll. From the formula I could calculate the covariant basis vectors in terms of the Cartesian basis vectors because I already had the inverse metric of the covariant system. This was calculated from the transformations of the systems to and from Cartesian.

The answer came:

and draw them properly (x = 1, y = 2)

From (NF12) it is clear that ê

^{x} is π/2 - α below the Cartesian X axis. Therefore the angle between ê

^{y} and ê

^{x} is a right angle, and the projected line from v

^{y }is parallel to ê

^{x} as we should have guessed and Orodruin intimated. Likewise ê

^{y} is parallel to the projected line from v

_{x} and always on the Cartesian Y axis. In addition |ê

^{i} | > |ê

_{i} |.

Getting back to the original question, what happens to v

_{x} if we double the contravariant basis ê

_{x} to ê'

_{x}? We could leave v

_{x}, ê

^{x}, v

_{y}, ê

^{y} alone and they would still give

**v**. We could double ê

^{x} and halve v

_{x}. Neither option would co-vary. Both would give incorrect values for |

**v**|. We need to find the metric of the primed system so we can just calculate the primed covariant coordinates and bases. We also cannot use the previous technique (calculating from the transformations of the systems to and from Cartesian), because we don't know how to transform primed covariant coordinates to Cartesian - we don't know what the primed coordinates are.

Therefore we use the pullback operator which we met two posts ago in

Commentary on Appendix A: Mapping S2 and R3.

If we have our contravariant primed bases, with scale factors

* a,b so*
*ê'*_{x} = aê_{x} , ê'_{y} = bê_{y}

which give

*v'*^{x} = v^{x} ⁄ a , v'^{y} =v^{y} ⁄ b

Turning all the handles we end up with

*v'*_{x} = a^{2}v'^{x} + abv'^{y}cosα

*v'*_{y} =b^{2}v'^{y} + abv'^{x}cosα

Since

* v' *^{i} decrease as

*a, b* and

*v'*_{i} increase as

*a*^{2} ,b^{2} ,ab, we can safely say that v'

_{i} increase as ê

_{i} .

**They co-vary**. It also looks as if they will 'compensate' for

*v' *^{i} , giving correct |

**v**| and

**v.w**. We would just have to follow the trail.

The full details of all the above are contained in sections 5b and Note F of

Commentary 1.1 Tensors matrices and indexes.pdf