Geodesics, proper time and parallel transport of the flat model

The connection $\mathrm{\nabla }$ has $\mathrm{\Gamma }\equiv 0$, so it is torsion-free; and the compatibility identity ${\mathrm{\partial }}_a{g}_{bc}={\sum }_d({\mathrm{\Gamma }}^d{}_{ab}{g}_{dc}+{\mathrm{\Gamma }}^d{}_{ac}{g}_{bd})$ holds because both sides are $0$ with constant metric components and vanishing coefficients. By [L2] it is the metric-selected connection of $g$.

Substituting $\mathrm{\Gamma }=0$ into [L3], the geodesic equation is ${\ddot{x}}^k=0$, whose solutions are exactly the affine lines $t\mapsto p+tv$ on parameter intervals; by [L5] the maximal geodesic through $(p,v)$ is this line extended to all of $\mathbb{R}$.

If $V={\sum }_a{V}^a{\mathrm{\partial }}_a$ along a curve, then by [L3] $D_{t}V=0$ if and only if ${\mathrm{\partial }}_t{V}^a=0$ for all $a$, that is, the components are constant; so parallel sections are exactly the constant-component fields along the curve and $P_{\gamma }$ is the identity identification of fibres in the identity chart. Hence transport preserves $g$ and is independent of the path (it depends only on the endpoints), by [L2] and the component computation.

Let $\gamma (t)=p+tv$ with $g(v,v)<0$ and $g(v,{\mathrm{\partial }}_0)<0$; then $\gamma$ is a future timelike worldline and its density is the constant $\sqrt{-g(v,v)}$. By [L4] the proper-time increment over $[t_0,t]$ is $\sqrt{-g(v,v)}(t-t_0)\mathrm{/}c$, and the four-velocity is the constant vector $U=cv\mathrm{/}\sqrt{-g(v,v)}$ with $g(U,U)=-c^2$. In particular, for the future timelike unit vector $u$ with $g(u,u)=-c^2$ the worldline $\tau \mapsto p+\tau u$ is parametrized by proper time up to its origin.

By step 2.1, the geodesic through $p$ with initial tangent $v$ is $t\mapsto p+tv$ on all of $\mathbb{R}$, so $1\in I_{p,v}$ for every $v$ and ${\exp }_p(v)=p+v$ on all of $T_{p}M$. Differentiating $v\mapsto p+v$ at $0$ gives the identity, and the chart $y:=x-x(p)$ centered at $p$ is the identity chart translated; in it $g_{ab}=\mathrm{diag}(-1,1,1,1)$ identically, so $g_{ab}(p)={\eta }_{ab}$ and ${\mathrm{\partial }}_a{g}_{bc}(p)=0$ hold exactly, not merely at first order.