- #1
Kontilera
- 179
- 24
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
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