What is global Lorentz transformation and what is local Lorentz transformation? I will consider $\textbf{spacetime}$  as $(M,\eta)$ where $M$ is a four dimensional $\textbf{manifold}$ and $\eta$ the metric which in this coordinates
 $$
\begin{align*}
  x \colon M &\longrightarrow \mathbb{R}^4\\
  p &\mapsto x(p)=(x_0,x_1,x_2,x_3).
\end{align*}
$$ 
 is given by 
$$\eta=dx^0\otimes dx^0-dx^1\otimes dx^2-dx^2\otimes dx^1-dx^3\otimes dx^3 \tag1$$ 
An $\textbf{observer}$ is a worldline $\gamma$ with together with a choice of basis
$
O=v_{\gamma,\gamma(\lambda)} \equiv e_0(\lambda) , e_1(\lambda), e_2(\lambda), e_3(\lambda)
$
of each $T_{\gamma(\lambda)}M$ where the observer worldline passes, if 
$$
\eta(e_a(\lambda), e_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab} \tag2
$$
$v_{\gamma,\gamma(\lambda)}$  is the tangent vector of the the curve $\gamma$ at the point $\gamma(\lambda)$
In text books i've found three definition of $\textbf{Lorentz transformation} \quad \Lambda$ 


*

*$\Lambda \colon \mathbb{R}^4 \longrightarrow \mathbb{R}^4$ is a group of coordinate transformations that leave eq.1 in the same form ,that is $\Lambda \cdot x(p)=y(p)=(y_0,y_1,y_2,y_3)$ such that in this coordinate  $$\eta=dy^0\otimes dy^0-dy^1\otimes dy^2-dy^2\otimes dy^1-dy^3\otimes dy^3 $$ 

*$\Lambda \colon M \longrightarrow M$ a Spacetime diffeomorphism such that $\Lambda_* \eta=\eta$ where $\Lambda_* \eta$ is the pullback of the metric $\eta$

*$\Lambda \colon T_pM \longrightarrow T_pM$ such that $\Lambda O=O'=e'_0(\lambda) , e'_1(\lambda), e'_2(\lambda), e'_3(\lambda)$ satisfy the eq.2 that is 
$$
\eta(e'_a(\lambda), e'_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab} 
$$
My question is which is of these transformation is global Lorentz transformation and which is local? 
 A: The three definitions are the same. They are ways of saying the same thing. Since you have a manifold $(M,\eta)$ this is a flat, Minkowski, spacetime. The Lorentz transformation is global on Minkowski spacetime.
In a curved spacetime the metric is usually denoted $g$, rather than $\eta$. $g$ is, in general, a function of time and position. At each point there is a Minkowski tangent space, meaning that the manifold is locally Minkowski (to the accuracy of measurement) and that local Lorentz transformations can be applied within a neighbourhood of each point.
A: The basic definition of the Lorentz transformation is, given a vector space $V$ equipped with a Minkowski metric $\eta$, it is the group that leaves the norm invariant (in other words, definition 1).
In a spacetime, every tangent space can be considered as a copy of Minkowski space, ie for every $p \in M$, $T_p M \cong V$, since by Sylvester's law, we can always put the metric tensor $g_p$ at that point in the appropriate form by a change of basis (that basis being an orthonormal basis $\{ e_\mu \}$). Then at each point, you can perform a Lorentz transform of that basis. 
Diffeomorphisms on a manifold are a rather large class, but there exists a subset of diffeomorphisms such that, if $\phi \in \mathrm{Diff}(M)$, the pushforward on a vector  at $p$, $\phi^* v$, corresponds to a Lorentz transform. Just by having say
\begin{equation}
\frac{\partial x^\mu(p)}{\partial y^\nu(p)} \in \mathrm{SO}(3,1)
\end{equation}
Locally, in the Riemann/Fermi coordinates, this is roughly equivalent to a Lorentz transform, since the normal neighbourhood is diffeomorphic to $T_pM$. 
