Covariant and Contravariant: What Are the Differences in Differential Geometry?

[gvk]

Let me see if I understand. You are saying that a coordinate system, say in 3d, defines not only

basis tangent vectors $$\mathbf{e}_1 ,\mathbf{e}_2,\mathbf{e}_3$$, but basis cotangent vectors or one-forms

$$\acute{E}_1,\acute{E}_2,\acute{E}_3$$,... .

Actually, the right way to think is the following:
the vector is an invariant object which does not depend on any coordinate system at all!
This point of view was formulated above
"COORDINATES ArE NOT a natural property of vectors"

Vectors (and tensors) exist without any coordinate system. The coordinate systems are only needed to describe them in a proper manner.

The final physical results do not depend on the choise of coords.
If space has a metric, which, by the way, also does not depend on the coordinate systems, there is no difference between covarient and
contravarient (or tangent and cotangent) vectors-tensors.
It is the same vector(tensor).
In fact, you confirm this using the same notation $$/omega$$ for covariant and contravariant vectors. Right?

 

[ObsessiveMathsFreak]

OK, I think I've got the representation of a vector part now.

If you consider a vector w as a tangent to some curve s(u,v) in, let's say a 2d plane, parameterised by coordinates u and v, then
$$\mathbf{w} = \frac{d \mathbf{s}}{dt} = \frac{\partial \mathbf{s}}{\partial u} \frac{du}{dt} + \frac{\partial \mathbf{s}}{\partial v} \frac{dv}{dt}$$

But $$\frac{du}{dt}$$ and $$\frac{dv}{dt}$$ can themselves be represented by (excuse the lapse into cartesians for a moment)

$$\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} = \nabla u \cdot (\frac{dx}{dt},\frac{dy}{dt})$$

$$\frac{dv}{dt} = \frac{\partial v}{\partial x} \frac{dx}{dt} + \frac{\partial v}{\partial y} \frac{dy}{dt} = \nabla v \cdot (\frac{dx}{dt},\frac{dy}{dt})$$

But since $$(\frac{dx}{dt},\frac{dy}{dt})=\frac{d \mathbf{s}}{dt}=\mathbf{w}$$, what we can actually say is that

$$\frac{du}{dt}=\nabla u \cdot \mathbf{w}$$
$$\frac{dv}{dt}=\nabla v \cdot \mathbf{w}$$

But, if we want to avoid using the dot product and cartesian gradients, instead of using $$\nabla u \ \ \nabla v$$, we can define the one forms $$d\acute{u}\ \ \ d\acute{v}$$ so that $$\nabla u \cdot \mathbf{w} = d\acute{u}(\mathbf{w})$$, etc.

These one forms have representations like $$d\acute{u} = \frac{\partial u}{\partial x} d\acute{x} + \frac{\partial u}{\partial y} d\acute{y}$$, in the caresian coordinate system, but of course, like vectors, one-forms have no preferred coordinate system! We can just visualise them as a field of gradient lines! So we just refer to the objects $$d\acute{u}\ \ \ d\acute{v}$$, which operate on vectors to give a number and so the representation of the vector becomes.

$$\mathbf{w}= d\acute{u}(\mathbf{w})\frac{\partial \mathbf{s}}{\partial u} + d\acute{v}(\mathbf{w})\frac{\partial \mathbf{s}}{\partial v}$$

And letting the partial derivatives of s be our basis vectors we would have.

$$\mathbf{w}= d\acute{u}(\mathbf{w})\mathbf{e}_u + d\acute{v}(\mathbf{w})\mathbf{e}_v$$

So, this concept of the coordinates being one forms seems clear to me now.

However I'm still not solid on what all this talk of covariant and contravariant vectors is. Is this a misnomer? Should we really be speaking of contravariant vectors and covariant one-forms?

 

[gvk]

However I'm still not solid on what all this talk of covariant and contravariant vectors is. Is this a misnomer? Should we really be speaking of contravariant vectors and covariant one-forms?

No, it is not misnomer. Covariant and contravariant are different if you don't have a metric in your space. Go through another thread in this forum:

https://www.physicsforums.com/showthread.php?p=413246#post413246

and may be it helps.