For starters, in the context of the tangent space of a manifold in GR, we can derive that:
$$g'_{\mu \nu}=\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}g_{\rho \sigma} \ \ \ \ \ \ \ \ (1)$$
where of course $g$ is the metric tensor and where we have indicated with $'$ the objects in the new coordinate system.
From here we can derive that:
$$\partial '_\mu \cdot \partial '_\nu =\frac{\partial x^\rho}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}\partial _\rho \cdot \partial _\sigma \ \ \ \ \ \ \ \ (2)$$
where $\partial _\mu \ , \ \partial _\nu$ are the basis of the $\mathbb{M}^4$ tangent space of the manifold.
(we can derive this because the metric tensor is defined as $g_{\mu \nu}=\partial _\mu \cdot \partial _\nu$)
Then we can get:
$$\partial '_\mu=\frac{\partial x^\sigma}{\partial x'^\mu} \partial _\sigma \ \ \ \ \ \ \ \ (3)$$
and at last:
$$V'^\mu=\frac{\partial x'^\mu}{\partial x^\sigma}V^\sigma \ \ \ \ \ \ \ \ (4)$$
where $V$ is a vector of the tangent space.
Ok, the tedious part is over, as you can see the topic in which I am interested regards change of coordinates in the tangent space of a manifold. Regarding all the above I have a couple of questions:
- The vectors $x$ in the partial derivatives are part of the manifold or part of the tangent space of the manifold? (I strongly suspect that the first option is the correct one, but I am not completely sure)
- We can derive $(3)$ from $(2)$ thanks to the linearity of the scalar product?
- The formula $(4)$ for the change of coordinates should be general, so it should apply in the special case of a manifold equal to $\mathbb{M}^4$; in this case we should get the tensor corresponding to the usual Lorentz's Transformation: $$\Lambda =\begin{bmatrix}\gamma & -\beta \gamma & 0 & 0\\-\beta \gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$$ so we want to be able to prove that: $$\frac{\partial x'^\mu}{\partial x^\sigma}=\Lambda^\mu _\sigma$$ how should we do it?
