Parametrization vs. coordinate system

  • Thread starter lmedin02
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56
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I am reading Differential Topology by Guillemin and Pollack.

Definition: X in RN is a k-dimensional manifold if it is locally diffeomorphic to Rk.

Suppose U is an open subset of Rk and V is a neighborhood of a point x in X.
A diffeomorphism f:U->V is called a parametrization of the neighborhood V.

The inverse mapping f-1:V->U is called a coordinate system.

Why is f a parametrization and its inverse a coordinate system? How do these terms fit in the big picture of manifolds?

I understand that we are rewriting V in X into a new coordinate system in Rk, that is easier to work with as oppose to some abstract space X. I not to sure of how to interpret the parametrization f.
 
649
2
Think of it as follows:

By coordinate map one means a mapping that specifies coordinate values for each point in your manifold M. So it would go in the direction M -> R^m

Think of a path in M parametrized by a real number. It is represented as a mapping in the direction R -> M. Similarly, a parametrization of a surface in M could be a mapping R^2 -> M. A parametrization of a whole m-dimensional set in M would be a mapping of the type R^m -> M, exactly opposite to the coordinate mapping.

I didn't bother to specify subsets etc. above, just wanted to illustrate why a parametrization is a mapping from some R^m into M. But maybe you already understood this, and wondered about something else? Sorry about not using your terminology, I just wrote like I'm used to.

Torquil
 
56
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In my reading I got confuse with this definition:

Definition: [tex]X[/tex] in [tex]R^N[/tex] is a k-dimensional manifold if it is locally diffeomorphic to [tex]R^k[/tex]; meaning that each point [tex]x\in X[/tex] possesses a neighborhood [tex]V[/tex] in [tex]X[/tex] which is diffeomorphic to an open subset [tex]U[/tex] of [tex]R^k[/tex]. Is it necessary for [tex]U[/tex] to be open?

If [tex]U[/tex] was not open, we can use a local extension to define a smooth map?
 
489
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Neighborhoods are by definition open. It couldn't be homeomorphic to anything but an open subset of R^k.
 

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