# Parametrization vs. coordinate system

1. Feb 5, 2010

### lmedin02

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.

2. Feb 5, 2010

### torquil

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

3. Feb 5, 2010

### lmedin02

In my reading I got confuse with this definition:

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

If $$U$$ was not open, we can use a local extension to define a smooth map?

4. Feb 6, 2010

### zhentil

Neighborhoods are by definition open. It couldn't be homeomorphic to anything but an open subset of R^k.