# General Relativity: change of coordinates in tangent space

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:

1. 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)
2. We can derive $$(3)$$ from $$(2)$$ thanks to the linearity of the scalar product?
3. 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?

1. The numbers $$x^\mu$$ which appear in the partial derivatives are not vectors (or the components of vectors) at all, they are the coordinates in the two coordinate charts.
3. If your coordinate transformation is a simple Lorentz transformation on Minkowski space, then $$x'^\mu = \Lambda^\mu_{\ \ \sigma} x^\sigma$$, so the derivative is trivial.