Covector fields - did I get them wrong?

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In summary, the covector field integrated over the path is still coordinate independent, but the length you measure will be different depending on the coordinates you choose.
  • #1
Kontilera
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Hello!
I am a bit confused about how I can use covector fields on a differentiable manifold.
John M. Lee writes that they can be integrated in a coordinate independent way so I thought that the covector fields could give me a coordinate independent way of calculating distance over a manifold.

Lets say we are working in R^3. This means that if I have a curve [tex]\gamma: I \rightarrow \mathbb{R}^3[/tex] I can measure how far it stretches in the y-direction by doing the integral,
[tex] \int_\gamma dy .[/tex]
If we change coordinates my covector field, [tex]\omega = dy[/tex] gets pullbacked to [tex]\omega' = dy/dy' dy'[/tex] and we get,
[tex] \int_\gamma \frac{dy}{dy'} dy' .[/tex]
It seems coordinate independent in this sense but what if we would have started with the coordinates dy' form the beginning?
Then we would have arrived at:
[tex] \int_\gamma dy' .[/tex]
Which gives another value right?
What have I missed in this subject? :/
Thanks so much,
All the best!
/ Kontilera
 
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  • #2
The point is that the integral of the form w is independant of the coordinate system used to compute its integral.

Here, you start with the form w = dy. Then you change the cordinates and remark that it is written (dy/dy')dy' in the new coordinates. Then you are confused by the fact that the integral of dy' will give something different. But it's not the same form! In the y' cordinates, the form w is (dy/dy')dy', not dy'.

Hope this helps.
 
  • #3
Thanks quasar! It helped. Could you agree with me on this:
When doing an integral over a certain covector field and path it doesn't matter in which coordinate system you start as long as the path and the field are constructed with regard to this coordinatesystem.
For example, if you start to draw R^2 by two orthogonal lines without any scale, it doesn't matter how big you make your coordinate grid, the result of the integral is, trivially, invariant of your size.

The transformation matrix is needed when you change coordinates in that sense that the calculation is taking place in one coordinate system and the geometrical objects such as the path (for example the unit circle) and the covector field are constructed from another coordinate system.

This makes sense for me at least. :)
All the best!
 
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  • #4
This sounds correct. :)
 
  • #5
Howdy,

It looks like your question about changing coordinates was answered, but your comment about lengths was unclear, so I thought I'd add something.

You won't get lengths from covector fields, rather you'll get net changes. E.g. the integral of dy over your curve returns the net change in the y direction. If you wanted total change in the y direction, you would need to integrate |dy|, which is not strictly speaking a differential form.

To measure length, you need a formula for ds. For example to calculate euclidean length there are many possible formulas:
[itex] ds = \sqrt{dx^2+dy^2} [/itex]
[itex] ds=\sqrt{dx^2+x^2dy^2}[/itex]
The specific formula depends on the coordinates you choose. The first corresponds to cartesian coordinates of the plane. The second to polar coords. The formula for the element of arclength is called a riemannian metric.

|dy| is like a degenerate metric: [itex] |dy|=\sqrt{0x^2+y^2}[/itex]
 
  • #6
Thanks! Yeah, net change is a better way to put it. :)
Another question that popped up was if it is misleading that we write integration as:
[tex]\int_S f(x,y)\,dxdy \quad ?[/tex]
After all, we are integrating over differential forms, ie:
[tex]\int_S f(x,y)\,dxdy = \int_S f(x,y)\,dx\wedge dy[/tex]
but juxtaposition of dxdy is used for the symmetric product so we should have:
[tex]dxdy = dx \wedge dy[/tex]
Which is not true since the left hand side is symmetric and the right hand side is alternating. :grumpy:
 
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  • #7
Both notations make sense and are used for good reason.

The dxdy notation is measure theoretic and is the more fundamental. It has nothing to do with tensors. It is shorthand for the product measure dA and by Fubini's theorem it makes sense to write it as dxdy or dydx.

The differential form is a completely different object. By definition, its integrals are the same up to a sign that indicates orientation. You could define this integral directly if you were doing a Riemann type integral, but if you want it to be defined Lebesgue style, it just makes more sense to essentially define it as a signed version of the dxdy (measure theoretic) integral.

You just have to tell from context whether dx or dy is referring to Lebesgue measure or a differential form. And often it doesn't really matter anyway.
 

1. What is a covector field?

A covector field is a mathematical concept used in vector calculus to represent a quantity that has both direction and magnitude at each point in space. It is a collection of covectors, which are linear functions that map vectors to real numbers.

2. How is a covector field different from a vector field?

A vector field assigns a vector to each point in space, while a covector field assigns a covector to each point in space. The main difference is that vectors represent physical quantities with both direction and magnitude, while covectors represent dual quantities that can act on vectors to produce a scalar value.

3. How are covector fields used in physics?

Covector fields are used in physics to describe a variety of physical quantities, such as electric and magnetic fields, stress and strain in materials, and fluid flow. They are also used in the formulation of mathematical models and equations in fields such as electromagnetism and fluid dynamics.

4. What are some common applications of covector fields?

Covector fields have many applications in mathematics, physics, and engineering. They are used in differential geometry, where they are used to define tangent spaces and differential forms. In physics, they are used to describe the physical quantities mentioned above. In engineering, they are used to model and analyze complex systems such as fluid flow and heat transfer.

5. What is the relationship between a vector field and its corresponding covector field?

Every vector field has a corresponding covector field, and vice versa. This relationship is known as the duality principle. The covector field acts on the vector field to produce a scalar value, known as the dot product. This duality relationship is fundamental in many areas of mathematics and physics, and allows for the formulation of elegant and powerful mathematical models.

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