# Noncoordinate basis vector fields

1. Oct 27, 2006

### desic

I'm self-studying Schutz Geometric methods of mathematical physics, having problems with ex. 2.1. page 44.

r=cos(theta)x+sin(theta)y
theta=-sin(theta)x+cos(theta)y

show this is non-coordinate basis, i.e. show commutator non-zero.

I try to apply his formula 2.7, assuming

V1=cos(theta), V2=sin(theta)
W1=-sin(theta), W2=cos(theta)
x(r)=r cos(theta)
y(r)=r sin(theta)
x(theta)=cos(theta)
y(theta)=sin(theta)

I believe the component of x should be (sin(theta))/r, however I get (sin(theta) - r sin(theta))/r.

would appreciate any help

2. Dec 1, 2006

### Chris Hillman

Vector fields

Hi, desic,

Vector fields, flows (in the sense of manifold theory), and first order linear homogeneous partial differential operators, are all tighly inter-related concepts, and it helps to able to freely intertranslate between these various representations.

E.g. in $$\mathbold{R}^2$$ "translation along x" and "rotation around the origin" can be written
$\vec{T} = \partial_x$
$\vec{R} = \partial_\theta = -y \, \partial_x + x \, \partial_y$
Now you can just compute [tex] \vec{T} \vec{R} f - \vec{R} \vec{T} f[itex] for an undetermined function f, from which you can read off the commutator.

See Olver, Equivalence, Invariants, and Symmetry for more about the virtues of this style of thinking/computing.

Chris Hillman

Last edited: Dec 1, 2006