Observer Transformation in general relativity

Question

Let's say there is a particle moving with 4-velocity $U^{\mu}_{PA}$ in spacetime with respect to observer A and there is another observer B moving with velocity $U^{\mu}_{BA}$. In special relativity, if we wanted to find the velocity of particle with respect to observer B. We would just do Lorentz Transformation. However, in general relativity that is not possible. Sean Carroll in his textbook on general relativity suggests that relative velocity is an ill defined concept. In what context does he mean that?

1. Is it in general hard to define a 4 velocity of objects in curved spacetime
2. Or, the general coordinate transformation between observers is not well defined and we can only transform between observers that are close to each other. If so, then aren't we basically trying to setup local Minkowski spacetime (locally inertial frame) to perform the transformation and assuming that to the direction we take. Does this not imply that the particle whose velocity we wanna transform to another observer's frame of reference needs to be close.
3. What would happen if the observer A, observer B and particle are far away from each other so that we can't setup a locally inertial frame for all of them.

I would like to know the mathematical reasons behind why relative velocity is not well defined in General Relativity. Is it because in curved spacetime velocity of particle and velocity of observer B live in different tangent space.

Edit: I just found that

Gullstrand–Painlevé coordinates are a particular set of coordinates for the Schwarzschild metric – a solution to the Einstein field equations which describes a black hole. The ingoing coordinates are such that the time coordinate follows the proper time of a free-falling observer who starts from far away at zero velocity, and the spatial slices are flat.

But when we talk about Lorentz transformation we are talking about transformation that leaves the metric, as is. However these transformations change the metric tensor.

So, what do we truly mean by relative velocity is ill defined in general relativity? Does it have something to do with the fact that charts are local and velocities defined using coordinate could become ill defined.

Answer 1

You can always set up locally lorentz frames for any event. You can also define a 4 velocity for any particle trajectory, which takes values in the tangent space at that point. The issue, as you mention, is that they are in different points in space so the velocities live in different tangent spaces. In general, it only makes sense to compare two vectors at the same point. In order to compare vector spaces at different points, you need some invertible linear map between them to identify them. Which identification do you use? The answer is provided by parallel transport/connection. However! The resulting answer depends on the path you use to connect the two points, and curvature measures precisely the amount that these differ with the choice of two paths. The reason it works in special relativity is because the curvature is 0, so it doesn't matter what path we take, we get the same linear transformation.

Technically speaking, even nearby points can't really be compared because there's still multiple paths. You can cheat a bit and make things work either when spacetime is almost flat in the vicinity (as we do here on earth), or you can impose canonical paths e.g. if you're at a point $p$, you can compare with points close enough to $p$ s.t. there is a unique geodesic between them. The issue in this second one is you have to change what "close" means point to point (it'll be smaller at places with more curvature). It's probably better to just abandon the notion of relative velocity at different points.

I think for the last question, take a locally lorentz frame at each point along the free fall trajectory and use those coordinates? (However, these are not unique because of lorentz transformations)