# Coordinate-independent basis

1. Jul 14, 2014

### befj0001

Suppose you have a differentiable manifold where at each point you have attached a set of basis vectors X_1,X_2,...,X_n. One thing that I don't have clear is the difference between a coordinate basis and a non-coordiante basis. I've been told that there is a way to check if the set of basis vectors is coordinate independent. One do this by simply taking the commutator between any pair of basis vectors and if the commutator is zero then my basis is not coordinate dependent. But why is that?

2. Jul 14, 2014

### Matterwave

Given a manifold $\mathcal{M}$ of dimension n, one can always construct on any open patch $U$ a set of coordinate mappings $\phi:U\rightarrow\mathbb{R}^n$. The set of coordinates $\{x_i\}$ can then be used to construct a set of coordinate basis vectors $\{\partial_{x_i}\}$.

A coordinate basis (or sometimes called a holonomic basis) is one in which one can find a set of coordinates $\{y_i\}$ such that in this coordinate system my basis is the set $\{\partial_{y_i}\}$. A non coordinate basis is a basis in which one cannot find such a coordinate system.

Because of the equality of mixed partial derivatives, it is easy to see, then, since: $[\partial_{x_i},\partial_{y_j}]=\partial_{x_i}\partial_{y_j}-\partial_{y_j}\partial_{x_i}=0$ that a 0 Lie bracket (what you term commutator) is a necessary condition that my basis is a holonomic one. That this condition is also sufficient needs to be proved. Unfortunately, I cannot recall the proof off the top of my head, but the proof should be found in any standard text on differential geometry.

Last edited by a moderator: Jul 14, 2014
3. Jul 14, 2014

### befj0001

So, given a manifold we can always define a coordinate system such that the Lie bracket is zero in that system. But I'm confused about this sentence:

"A non coordinate basis is a basis in which one cannot find such a coordinate system."

How can one not find such a coordinate system if "one can always find such a coordinate system" whenever the underlying topology is a manifold? It seems to contradict the definition.

Last edited by a moderator: Jul 14, 2014
4. Jul 14, 2014

### Matterwave

Sorry if my post was unclear. One can always find a coordinate system on a patch of a manifold. From this coordinate system, one can find a set of coordinate bases.

But it doesn't mean that given any basis, I can find a coordinate system for which that basis is a coordinate basis. Usual examples of non coordinate bases are orthonormal bases.

Perhaps it was confusing for me to introduce the notion that a coordinate patch can always be found.

Given a coordinate system I can construct a set of coordinate bases. Given a basis I may or may not be able to construct a coordinate system for which that basis is the coordinate basis. Is that clear?

5. Jul 14, 2014

### stevendaryl

Staff Emeritus
Here's an example of the distinction:

In polar coordinates, we have the coordinate basis $e_r, e_\theta$ that has the nice property:

If $\mathcal{P}$ is a point with coordinates $(r,\theta)$, then $\mathcal{P}+\alpha e_r$ is a point with coordinates $(r + \alpha, \theta)$ and $\mathcal{P}+ \beta e_\theta$ is a point with coordinates $(r, \theta + \beta)$. That's the property that $e_r, e_\theta$ are coordinate bases for $r, \theta$.

A related basis that is often used with polar coordinates is $\hat{r}, \hat{\theta}$. This is nice for a different reason: $|\hat{r}| = |\hat{\theta}| = 1$. This is not a coordinate basis.

These two bases are related (I think) by:

$\hat{r} = e_r$
$\hat{\theta} = e_\theta/r$

6. Jul 16, 2014

### befj0001

Ok, I think I understand!

So what happens if we consider for example a sphere in R^3 ? There is no coordinate system that covers the whole sphere, but two coordinate systems are enough and we can associate a basis to each one of these coordinate systems. But I suppose I can find just one basis for the entire sphere, for instance, the "spherical directions" would do.

Am I correct?

7. Jul 16, 2014

### George Jones

Staff Emeritus
No. You can't comb a hairy ball.

8. Jul 16, 2014

### WannabeNewton

As George mentioned, you cannot find such a basis. Manifolds which admit global frames are called parallelizable. A 2-sphere is not parallelizable as a consequence of the hairy ball theorem: http://en.wikipedia.org/wiki/Hairy_ball_theorem

9. Jul 16, 2014

### George Jones

Staff Emeritus
Another interesting result:

Let M be a non-compact 4-dimensional spacetime (Lorentzian manifold). There is a set of four continuous orthonormal vector fields (there exists a global tetrad; M is parallelizable) defined on all of M if and only if M admits a spinor structure.

All compact Lorentzian manifolds have closed timelike curves, so non-compactness seems physically reasonable.