# General Relativity - Confusion between choosing basis (orthonormal & coordinate) and coordinate transformations

I am reading the book 'Gravity' by Hartle and presently I am at the section discussing orthonormal and coordinate bases. I am confused about a few points I had read previously and can't exactly correlate with the above mentioned section.

Here the author says that at every point in spacetime I can define a basis that is orthonormal in the tangent space of that point. This would not necessarily be the coordinate basis which I would get from the line element. This would mean that the metric in the orthonormal basis becomes the flat spacetime metric at the point (from the definition of the components of the metric in terms of the dot product of basis vectors and the requirement of one timelike and three spacelike components). Now, I know that the way to locally transform the metric to the flat spacetime metric is to do a coordinate transformation on the line element that does the above, i.e to go to a local inertial frame for that point.

My question is - Is there any correlation between a coordinate transformation that transforms my line element to that of flat spacetime in the vicinity of a point to that of choosing an orthonormal coordinate system at that point? Is the latter something similar to choosing a coordinate system where my coordinate basis would align with the orthonormal basis at the point? So, I guess my question boils down to the two definitions of the metric - one from the dot product of basis vectors and the other from the line element. Are they completely equivalent?

My last question is - Is the coordinate transformation that takes me to a local inertial frame of a point unique? Can there be more than one coordinate transformations that take me to the local inertial frame of a point? Does this mean that at every point I can choose an orthonormal basis at every point in one and only one unique way?

I am sorry for not writing equations. I am very poor with Tex. I hope I have been able to put my question clearly.

• This would not necessarily be the coordinate basis which I would get from the line element. The coordinates don't come from the line element. You have to pick coordinates so you can express the line element. This would mean that the metric in the orthonormal basis becomes the flat spacetime metric at the point No, curvature is independent of the choice of coordinates. If the spacetime is curved, it's curved.
– user4552
Jul 8, 2019 at 17:08
• @Ben, I understand what you have said. But here's my point - Imagine spacetime that is curved. In some coordinate system I have a line element giving the distance between two arbitrarily close events at any point in spacetime. I have a coordinate system to label points which I use to write my line element. I can define basis vectors along those coordinate lines . This I call my coordinate basis. These are not orthonormal over the whole of spacetime. If they are at some point, spacetime is locally flat at that point, which must also be reflected in the line element at that point. Jul 9, 2019 at 10:26
• Continued - Now a consequence of spacetime being curved places the restriction that I cannot have a coordinate system whose basis vectors are orthonormal at every point in spacetime. This would mean spacetime is flat globally. Nor can I effect a transformation on the original line element that turns it into minkowski line element everywhere. But I can always choose an orthonormal basis at any specific point. My question is - Is this the same as effecting a transformation on the line element that turns it into flat at that point? Jul 9, 2019 at 10:45
• Continued - But an orthonormal basis can be chosen at a point in multiple ways. Simply rotating an orthonormal basis at a point gives me other basis. Does this mean that there are multiple transformations I can make on the line element to turn it into flat at that point? Jul 9, 2019 at 10:48
• Continued - In other words - are there multiple coordinate systems whose basis vectors at that point coincide with the different orientations of the orthonormal basis at that point? Please correct me if I have reasoned wrongly. Jul 9, 2019 at 10:56

First thing of note is that every frame (basis) at a single point is a coordinate frame. To see this, let $$\mathrm dx^\mu$$ be a coordinate cobasis, and let $$x$$ be a point of the manifold $$M$$. An arbitrary frame at $$x$$ can be expressed as $$\theta^a|_x=L^a_{\ \mu}\mathrm dx^\mu|_x,$$ where $$L^a_{\ \mu}$$ is a constant invertible matrix. But then if we define the new coordinates $$u^a=L^a_{\ \mu}x^\mu$$, then $$\mathrm du^a=L^a_{\ \mu}\mathrm dx^\mu=\theta^a,$$ where I was lazy in denoting it, but every expression is evaluated at $$x\in M$$.
The problem starts if one is considering frame fields, which are defined on an open domain. If $$\theta^a$$ is a coframe field on an open domain, then by Poincaré's lemma the necessary and sufficient condition for it to be a coordinate frame is that $$\mathrm d\theta^a=\mathrm d\theta^a_\nu\wedge\mathrm dx^\nu=\partial_\mu\theta^a_\nu\mathrm dx^\mu\wedge\mathrm dx^\nu=\frac{1}{2}\left( \partial_\mu\theta^a_\nu-\partial_\nu\theta^a_\mu\right)\mathrm dx^\mu\wedge\mathrm dx^\nu=0$$, which is a nontrivial integrability condition.
Even if this integrability condition is not satisfied, one can still write $$\theta^a|_x=\theta^a_\mu(x)\mathrm dx^\mu|_x$$ for some invertible matrix function $$\theta^a_\mu(x)$$, however in general there will be no such coordinate functions $$u^a(x)$$ such that $$\theta^a_\mu(x)=\partial_\mu u^a(x)$$.
Nontheless, for any $$x_0\in M$$ you may try to see $$\theta^a|_{x_0}$$ as a fixed cobasis at a point, and try the coordinate transformation $$u^a_{(x_0)}(x)=\theta^a_\mu(x_0)x^\mu$$, which will give at $$x_0$$, and only at $$x_0$$ $$\theta^a|_{x_0}=\mathrm du^a_{(x_0)}(x_0).$$