# Does Every Coordinate System Admit Local Othornormal Basis Vectors

1. Apr 30, 2008

### friend

I wonder if there are coordinate systems that gobally curve and twist and turn and curl, that do NOT admit local orthonormal basis. I know that the Gram-Schmidt procedure converts ANY set of linear independent vectors into an orthnormal set that can be used as local basis vectors. And I assume ANY coordinate system, no matter how it globally
twists and turn, has local basis vectors that are at least linear independent, is this right? This sounds like ANY coordinate system, no matter how it may gobally twist and turn and curve, etc., must necessarily admit local orthnormal basis vectors? Is this right?
Thanks.

2. Apr 30, 2008

### shoehorn

In general you cannot define a global coordinate system on a manifold.

Before you can have an orthonormal basis you need something with which to measure orthonormality. In effect, you need a metric. Given a chart $(U,\varphi)$ in a neighborhood of a point $p\in M$, such that $\varphi(p)=x^\mu$, we have a coordinate basis of $T_pM$

$$\left\{\frac{\partial}{\partial x^\mu}\right\} = \{\partial_\mu\}.$$

In general a coordinate basis will not be orthonormal, i.e.,

$$g(\partial_\mu,\partial_\nu) \ne \delta_{\mu\nu}[/itex] If, in addition to the chart, you have a metric on $M$, you can form an orthonormal basis [tex]\{e_a\}=\{e^\mu_a \partial_\mu\}, \qquad e^\mu_a\in\textrm{GL}(m,\mathbb{R})$$

The orthonormality of this basis simply means that

$$g(e_a,e_b) = e^\mu_a e^\nu_b g_{\mu\nu} = \delta_{ab}$$

Therefore, given any admissable coordinate system and a metric, you can construct a (pseudo-)orthonormal basis by a $\textrm{GL}(m,\mathbb{R})$ rotation of the coordinate basis.

Last edited: Apr 30, 2008
3. Apr 30, 2008

### gel

What does a basis of a coordinate system mean? Do you mean a basis for the tangent space of a manifold? That wouldn't have anything to do with the coordinate system.

4. May 1, 2008

### friend

I know that for ALL manifolds, you can impose coordinates on patches of it, and you can construct an orthonormal basis on the tangent space at each point of the manifold, right?

But I think I want to be more general than manifolds. Are there any other structures on which coordinates can be imposed, which have a metric to determine distance, and which may not admit local orthonormal basis at each point? Or is it true that ANY coordinate sytem imosed on any structure (manifold or otherwise) always necessarily admits local orthonormal basis vectors at each point? I'm thinking that at each point the lines constructed by only one coordinate parameter varying at a time have tangent lines at each point. If we consider all the unit vectors tangent to each coordinate line at a point, do they at least always form a linear independent set of vectors? And can an orthonormal basis be constructed of these? Thanks.

Last edited: May 1, 2008
5. May 1, 2008

### shoehorn

It would be terribly helpful if you'd take the time to read the replies. For instance, I've already pointed out that to have an orthonormal basis of $T_pM$ you need a metric. The situation is as follows:

• If you do not have a metric on a manifold, you cannot construct an orthonormal basis since you have no means of "measuring" the orthonormality of a basis.
• If you do have a metric on a manifold, an orthonormal basis (or, more generally, a pseudo-orthonormal basis) can always be constructed by an appropriate $\textrm{GL}(m,\mathbb{F})$-rotation of a coordinate basis, where $\mathbb{F}$ is some field to which $M$ is locally homeomorphic.

6. May 1, 2008

### friend

So I guess the question becomes then, can there be a metric apart from some kind of manifold? I already know that all manifolds can have a orthonormal basis in their tangent spaces. But can there be structures other than manifolds that can have a metric defined on it so that questions of orthogonality can be determined? Thanks.