Curvature survives a first-order normal chart: an explicit metric, its normal form and its deviation field

(First-order normal form.) With $g_{11}=1+(x^0{)}^2$ and all other components constant, $g(0)=\eta$ and ${\mathrm{\partial }}_a{g}_{bc}(0)=0$ because ${\mathrm{\partial }}_0{g}_{11}=2x^0$ vanishes at $0$ and no other component depends on the coordinates. By [L1] the Christoffel symbols at $0$ are linear combinations of the first derivatives $\mathrm{\partial }g(0)$, so they vanish as well; the identity chart is point-normal at $0$ by [L3].

(Christoffel symbols.) By [L1], the only nonzero first derivative of a metric component is ${\mathrm{\partial }}_0{g}_{11}=2x^0$, so a symbol can be nonzero only if one of its indices is a derivative index $0$ and another index refers to the component $g_{11}$. Checking the three slots: ${\mathrm{\Gamma }}^1{}_{10}={\mathrm{\Gamma }}^1{}_{01}=\frac{1}{2}{g}^{11}{\mathrm{\partial }}_0{g}_{11}=\frac{1}{2}(1+(x^0{)}^2{)}^{-1}(2x^0)=\frac{x^0}{1+(x^0{)}^2}$, and ${\mathrm{\Gamma }}^0{}_{11}=\frac{1}{2}{g}^{00}(-{\mathrm{\partial }}_0{g}_{11})=\frac{1}{2}(-1)(-2x^0)=x^0$; all remaining symbols have a factor ${\mathrm{\partial }}_a{g}_{bc}$ with $a\mathrm{\ne }0$ or $g_{bc}$ one of the constant components and vanish.

(Curvature at the origin.) By steps 1.1 and 1.2 the chart is point-normal at $0$, so [L2] applies at that point. For $(a,b,c,d)=(0,1,0,1)$ the four terms are $g_{01,10}+g_{10,01}-g_{11,00}-g_{00,11}$, of which only $-g_{11,00}(0)=-{\mathrm{\partial }}_0{\mathrm{\partial }}_0(1+(x^0{)}^2){\mathrm{\mid }}_0=-2$ survives, so $R_{0101}(0)=-1$; interchanging the pairs gives $R_{1010}(0)=-1$ as well, and the last-pair skew gives $R_{0110}(0)=+1$. In particular $R(0)\mathrm{\ne }0$, although all Christoffel symbols vanish at $0$.

(An explicit geodesic family.) Let $\mathrm{\Phi }(s,r):=(s,r,0,0)$. By step 1.2, ${\mathrm{\Gamma }}^a{}_{00}=0$ identically, so by [L5] each curve $s\mapsto \mathrm{\Phi }(s,r)$ is a geodesic with affine parameter $s$; its tangent $T={\mathrm{\partial }}_0$ satisfies $g(T,T)=-1$. The variation field is $J={\mathrm{\partial }}_1$.

(The curvature term.) By [L6] and step 1.2, $R^1{}_{001}={\mathrm{\partial }}_0{\mathrm{\Gamma }}^1{}_{10}-{\mathrm{\partial }}_1{\mathrm{\Gamma }}^1{}_{00}+{\mathrm{\Gamma }}^1{}_{0e}{\mathrm{\Gamma }}^e{}_{01}-{\mathrm{\Gamma }}^1{}_{1e}{\mathrm{\Gamma }}^e{}_{00}=\frac{1-s^2}{(1+s^2{)}^2}+0+\frac{s^2}{(1+s^2{)}^2}-0=\frac{1}{(1+s^2{)}^2}$, all other components $R^a{}_{001}$ vanishing because their Christoffel combinations vanish. Hence $R(T,J)T=R^1{}_{001}{\mathrm{\partial }}_1=\frac{1}{(1+s^2{)}^2}{\mathrm{\partial }}_1$.

(The variation field and its derivative.) By [L1], $D_{s}J={\mathrm{\nabla }}_{{\mathrm{\partial }}_0}{\mathrm{\partial }}_1={\sum }_a{\mathrm{\Gamma }}^a{}_{10}{\mathrm{\partial }}_a=\frac{x^0}{1+(x^0{)}^2}{\mathrm{\partial }}_1$, and differentiating along the curve, $D_s^{2}J=\frac{d}{ds}\left(\frac{s}{1+s^2}\right){\mathrm{\partial }}_1+\frac{s}{1+s^2}{\mathrm{\nabla }}_{{\mathrm{\partial }}_0}{\mathrm{\partial }}_1=\frac{1-s^2}{(1+s^2{)}^2}{\mathrm{\partial }}_1+\frac{s^2}{(1+s^2{)}^2}{\mathrm{\partial }}_1=\frac{1}{(1+s^2{)}^2}{\mathrm{\partial }}_1$.

(Verification of the deviation equation.) Steps 3.1 and 2.3 give $D_s^{2}J=R(T,J)T$ for every $s$, in accordance with [L4]. At $s=0$ the relative acceleration is ${\mathrm{\partial }}_1\mathrm{\ne }0$, while every Christoffel symbol of the chart vanishes there: the first-order normal form of the metric cannot remove the tidal (curvature) effect, and the deviation equation detects it at the very event where the connection symbols are zero. This completes the verification.