# Existence of orthonormal frame

1. Oct 18, 2012

### semigroups

Let $$M$$ be a surface with Riemannian metric $$g$$. Recall that an orthonormal framing of $$M$$ is an ordered pair of vector fields $$(E_1,E_2)$$ such that $$g(E_i,E_j)=\delta_{ij}$$. Prove that an orthonormal framing exists iff $$M$$ is orientable and $$M$$ admits a nowhere vanishing vector field $$X$$.

Remark: It's obvious in $$\mathbb{R}^3$$, but how to formally justify it?
The definition for orientabily: $$M$$ is orientable if there exists an atlas $$(u_{\alpha},M_{\alpha})_{\alpha}$$ such that $$\mathrm{det}(\mathrm{d}(u_{\beta}\circ u_{\alpha}^{-1}))>0$$, for each $$(\alpha,\beta)$$ such that $$M_{\alpha} \cap M_{\beta} \neq \Phi$$

2. Oct 18, 2012

### quasar987

The idea here is that by definition, an orientation on M splits the basis at each tg space into 2 sets: the ones that are positively oriented and the ones that are not.

So start with a nowhere vanishing vector field X. You want to construct another one Y that is g-orthogonal to it. At each point, you have only two choices for Y, but one of those choices is canonical as only one of them makes (X,Y) a positively oriented basis.

Conversely, given a global orthonormal frame (X,Y), you get a canonical choice of orientation at each tg space: declare positively oriented those basis that share the same orientation as (X,Y).

I let you fill in the details.

3. Oct 20, 2012

### Bacle2

Still, if the surface S is parametrized in R^3 (which I assume is the case) and

orientable, doesn't if follow that S is either an embedded surface or a submanifold of

R^3. Then the tangent space T_pS to S at p is a subspace of the tangent space

T_pR^3 , so the result would follow. Right?