Torsion-free connections with the same geodesic images need not coincide

The coefficients are symmetric in the lower indices, ${\mathrm{\Gamma }}^{\mathrm{\prime }a}{}_{bc}={\mathrm{\Gamma }}^{\mathrm{\prime }a}{}_{cb}$, so ${\mathrm{\nabla }}^{\mathrm{\prime }}$ is torsion-free by [L2]. It is not the zero connection: for instance ${\mathrm{\Gamma }}^{\mathrm{\prime }0}{}_{00}={\delta }^0{}_0{\delta }^0{}_0+{\delta }^0{}_0{\delta }^0{}_0=2\mathrm{\ne }0$, so ${\mathrm{\nabla }}^{\mathrm{\prime }}\mathrm{\ne }{\mathrm{\nabla }}_0$.

(Local ${\mathrm{\nabla }}^{\mathrm{\prime }}$-geodesic reparametrizations of straight lines.) Let $\gamma (t)=p+tv$ be a regular straight line and fix $t_0$ in its open parameter interval. For any $s_0$, choose an open interval $J$ about $s_0$ small enough that $1+2v^0(s-s_0)>0$ on $J$. If $v^0=0$, put $h(s)=t_0+(s-s_0)$. If $v^0\mathrm{\ne }0$, put $h(s):=t_0+\ln (1+2v^0(s-s_0))\mathrm{/}(2v^0)$. In either case, shrink $J$ if necessary so that $h(J)$ lies in the parameter interval of $\gamma$. Then $h$ is a $C^2$ diffeomorphism from $J$ onto its open image, with $h(s_0)=t_0$ and $h^{\mathrm{\prime }}\mathrm{\ne }0$; in the second case $h^{\mathrm{\prime }}=1\mathrm{/}(1+2v^0(s-s_0))$ and $h^{\mathrm{\prime }\mathrm{\prime }}=-2v^0(h^{\mathrm{\prime }}{)}^2$. The ${\mathrm{\nabla }}^{\mathrm{\prime }}$-geodesic equation for $\gamma \circ h$ is $(h^{\mathrm{\prime }\mathrm{\prime }}+2v^0(h^{\mathrm{\prime }}{)}^2)v^a=0$, so it is satisfied. This proves that every regular straight-line geodesic of ${\mathrm{\nabla }}_0$ is locally a reparametrized ${\mathrm{\nabla }}^{\mathrm{\prime }}$-geodesic.

(Every regular ${\mathrm{\nabla }}^{\mathrm{\prime }}$-geodesic is locally a reparametrized straight line.) Let $\sigma \mapsto x(\sigma )$ solve the ${\mathrm{\nabla }}^{\mathrm{\prime }}$-equation ${\ddot{x}}^a+2{\dot{x}}^0{\dot{x}}^a=0$ with $\dot{x}\mathrm{\ne }0$ on an interval. For $a=0$ this is ${\ddot{x}}^0=-2({\dot{x}}^0{)}^2$, so $y:={\dot{x}}^0$ solves $y^{\mathrm{\prime }}=-2y^2$. The set $A=\{\sigma :y(\sigma )=0\}$ is closed by continuity and open by uniqueness of solutions through $y({\sigma }_0)=0$ (the zero function is a solution, and [L4] gives uniqueness); hence on the connected parameter interval either $y\equiv 0$ or $y$ is nowhere zero. If $y\equiv 0$, the equations give ${\ddot{x}}^a=0$ for every $a$, so $x$ is an affine regular line. Otherwise, for $a=1,2,3$, $\frac{d}{d\sigma }\left({\dot{x}}^a\mathrm{/}{\dot{x}}^0\right)=({\ddot{x}}^a{\dot{x}}^0-{\dot{x}}^a{\ddot{x}}^0)\mathrm{/}({\dot{x}}^0{)}^2=0$, so $x^a=c^a{x}^0+d^a$ for constants $c^a,d^a$. Since $h(\sigma ):=x^0(\sigma )$ has nowhere-zero derivative, it is a $C^2$ local diffeomorphism; in the parameter $h$, the curve is the affine line $(h,c^{1}h+d^1,c^{2}h+d^2,c^{3}h+d^3)$, a regular ${\mathrm{\nabla }}_0$-geodesic by [L1]. Thus every regular ${\mathrm{\nabla }}^{\mathrm{\prime }}$-geodesic is locally a regular reparametrization of a ${\mathrm{\nabla }}_0$-geodesic.

${\mathrm{\nabla }}^{\mathrm{\prime }}$ is not $g$-compatible. For $(a,b,c)=(0,0,0)$ the identity of [L3] would read ${\mathrm{\partial }}_0{g}_{00}=2{\mathrm{\Gamma }}^{\mathrm{\prime }0}{}_{00}{g}_{00}$; the left side is $0$ because $g_{00}=-1$ is constant, while the right side is $2\cdot 2\cdot (-1)=-4\mathrm{\ne }0$.

Steps 1.2 and 1.3 show that ${\mathrm{\nabla }}_0$ and ${\mathrm{\nabla }}^{\mathrm{\prime }}$ have the same local unparametrized geodesic paths, while steps 1.1 and 2.1 show that ${\mathrm{\nabla }}^{\mathrm{\prime }}$ is a torsion-free connection distinct from ${\mathrm{\nabla }}_0$ and not $g$-compatible. Hence the displayed statement is refuted: torsion freedom plus equality of the local unparametrized paths do not determine the metric-selected connection, and metric compatibility is an independent condition.