# Does a coordinate chart have a smooth orthonormal frame w.r.t. a semi-Riemannian metric?

If the metric is definite, then the answer is yes: the coordinate basis vectors form a smooth local frame, and each basis vector has non-zero length by definiteness, and one can run the Gram-Schmidt algorithm for frames to get a smooth orthonormal frame. (See Lemma 8.13 in Lee's "Introduction to Smooth Manifolds".)

However, if the metric is indefinite, then the existence of null vectors is the main difficulty. At each point $p$, one can find a nondegenerate basis (see this answer for the definition) for $T_p M$, but it's not clear to me that one can do so smoothly. e.g. the method in this answer has multiple cases depending on which elements are non-zero, which a priori doesn't seem smooth.

Another approach would be to find a nondegenerate basis at a given point, use Gram-Schmidt to orthonormalize it, and then parallel-transport it to the rest of the coordinate chart. This works only if the chart domain is contractible. One can deduce that the chart domain is smoothly contractible to some point $p_0$, then for each point $p$ one can smoothly map to a smooth path from $p_0$ to $p$. Then one can find a nondegenerate basis of $T_{p_0} M$ and orthonormalize it, and parallel transport that to the rest of the coordinate chart from the smooth paths. Then this yields a smooth frame, since everything involved is smooth.

Then the above can be extended to unions of contractible domains, subsets of such unions, etc.

So my question boils down to:

1) Can one find a smooth local frame that reduces to a nondegenerate basis at each point? Or, can one tweak the method above to find a nondegenerate basis such that it varies smoothly with the metric?

2) Or, can one extend the above parallel transport method to any open set?

• If you looking for a local orthonormal frame, i've made some construction by modifying Gram-Schmidt method here – Kelvin Lois Aug 13 at 1:18

To answer question (1): Not necessarily. Here's a counterexample: On $M = \mathbb R^2\smallsetminus\{(0,0)\}$, define $$g = x\, dx^2 + 2 y \, dx\,dy - x\, dy^2.$$ The identity map is a smooth coordinate chart on all of $M$; but there's no global nondegenerate frame on $M$. The problem is that $M$ is not time-orientable: If you declare $\partial/\partial y$ to be the future time direction at the point $(1,0)$, and then try to extend it continuously around the unit circle, any extension comes back past-oriented.
Even if you assume time-orientability, it might not be possible. For example, on $\Bbb R^+ \times \Bbb S^n$, let $g$ be the Lorentz metric $$g = - dt^2 + \mathring g,$$ where $\mathring g$ is the round metric on the sphere. This manifold is diffeomorphic to $\Bbb R^{n+1}\smallsetminus\{0\}$, so it has a global coordinate chart. If there were a global nondegenerate frame $(v_0,v_1,\dots,v_n)$, then we could replace $v_0$ by $\partial/\partial t$ and still have a nondegenerate frame (exercise), and applying Gram-Schmidt to that frame would yield a global frame $(\partial/\partial t, E_1,\dots,E_n)$ in which $(E_1,\dots,E_n)$ restricts to a global frame on $\Bbb S^n$. Unless $n=1$, $3$, or $7$, this is impossible.
• These examples are very helpful. Thanks again Professor! One small fix: I believe you want $\Bbb R^+ \times \Bbb S^n$ instead of $\Bbb R \times \Bbb S^n$. – Fred Akalin Jan 31 at 5:58