My question is about 4-velocity but it is more general about my global comprehension of S.R.
In special relativity, we define the 4-velocity vector as ($\tau$ is the proper time) :
$$U=\frac{\partial x}{\partial \tau}$$
So, once I have chosen a frame $R$, it represent how the coordinates of the point change in $R$ when the proper time of the particle has changed of $d\tau$ : so the only frame dependence is on the upper part of the derivative, we always do the derivative according to $\tau$ no matter in which frame we are.
In my course, they say that this vector is "absolute" and doesn't depend on any frame.
But I have some questions about this.
In diff geometry, we define tensors as quantities that transform well, but here we have:
$$\partial_{\tau} x^{\beta}=\frac{\partial}{\partial \tau}\left(\frac{\partial x^{\beta}}{\partial y^\alpha}y^\alpha\right)=\frac{\partial^2 x^\beta}{\partial \tau \partial y^\alpha}y^\alpha+\frac{\partial x^{\beta}}{\partial y^\alpha}\frac{\partial y^\alpha}{\partial \tau}$$
So, the first term shouldn't be here to have a well defined quantity.
But here, we are focusing on inertial frame. So, they are linked with Lorentz boost that is a linear transformation. Thus the second derivative that is written above should be 0.
First question: From a math perspective, can we say that 4-velocity vector is indeed an absolute quantity because between inertial frames, the quantity transforms "well" as a tensor ?
Second question: in the course the teacher doesn't do such proof, he just says "we defined 4-velocity without referring a specific frame, thus it is a quantity independent of frames". I don't understand this, can we understand it is an absolute quantity without doing what I've written above?
Third question : In general relativity (that I just started to study), we are not focused on inertial frame only, we can do any change of coordinates. Thus is the 4-velocity still well defined ?