Tensoriality of the curvature, its coordinate formula and its point-normal metric formula

(Function-linearity in the first slot.) Let $f$ be smooth. Using [L3] for the bracket and [L2] for the connection, $R(fX,Y)Z={\mathrm{\nabla }}_{fX}{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{fX}Z-{\mathrm{\nabla }}_{[fX,Y]}Z=f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_Y(f{\mathrm{\nabla }}_{X}Z)-{\mathrm{\nabla }}_{f[X,Y]-Y(f)X}Z=f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-Y(f){\mathrm{\nabla }}_{X}Z-f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-f{\mathrm{\nabla }}_{[X,Y]}Z+Y(f){\mathrm{\nabla }}_{X}Z=fR(X,Y)Z$, the two $Y(f){\mathrm{\nabla }}_{X}Z$ terms cancelling.

(Function-linearity in the third slot.) Expanding with [L2] and [L3], ${\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y(fZ)={\mathrm{\nabla }}_X(Y(f)Z+f{\mathrm{\nabla }}_{Y}Z)=X(Y(f))Z+Y(f){\mathrm{\nabla }}_{X}Z+X(f){\mathrm{\nabla }}_{Y}Z+f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z$, while ${\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X(fZ)=Y(X(f))Z+X(f){\mathrm{\nabla }}_{Y}Z+Y(f){\mathrm{\nabla }}_{X}Z+f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z$, and ${\mathrm{\nabla }}_{[X,Y]}(fZ)=[X,Y](f)Z+f{\mathrm{\nabla }}_{[X,Y]}Z$. Subtracting the second and third displays from the first, the $X(f)$, $Y(f)$ and $Z$-coefficient terms cancel in pairs and the remaining scalar coefficient is $X(Y(f))-Y(X(f))-[X,Y](f)=0$ by definition of the bracket, so $R(X,Y)(fZ)=fR(X,Y)Z$.

(Skew-symmetry and the second slot.) Directly from [L1] and $[Y,X]=-[X,Y]$, $R(Y,X)Z=-R(X,Y)Z$; combining this with step 1.1, $R(X,fY)Z=-R(fY,X)Z=-fR(Y,X)Z=fR(X,Y)Z$.

(Pointwise dependence.) Let $p\in M$ and let $(V,x)$ be a chart around $p$. By [L7], fields $X^{\mathrm{\prime }},Y^{\mathrm{\prime }},Z^{\mathrm{\prime }}$ with $X_p^{\mathrm{\prime }}=X_p$, $Y_p^{\mathrm{\prime }}=Y_p$, $Z_p^{\mathrm{\prime }}=Z_p$ have the form $X^{\mathrm{\prime }}=X+{\sum }_a{f}^a{\mathrm{\partial }}_a$, $Y^{\mathrm{\prime }}=Y+{\sum }_b{g}^b{\mathrm{\partial }}_b$, $Z^{\mathrm{\prime }}=Z+{\sum }_c{h}^c{\mathrm{\partial }}_c$ on a possibly smaller chart neighborhood, with $f^a(p)=g^b(p)=h^c(p)=0$. By additivity of $R$ in each field argument and the three function-linearity steps, expanding $R(X^{\mathrm{\prime }},Y^{\mathrm{\prime }},Z^{\mathrm{\prime }})$ yields $R(X,Y,Z)$ plus a finite sum of terms each carrying at least one of the smooth coefficients $f^a,g^b,h^c$ (a term with $k$ such factors expands, by steps 1.1, 2.1 and 1.2, into $k$-fold products of coefficients times $R$ evaluated on coordinate fields). Every such term vanishes at $p$ because all coefficients do, so $R(X^{\mathrm{\prime }},Y^{\mathrm{\prime }},Z^{\mathrm{\prime }}){\mathrm{\mid }}_p=R(X,Y,Z){\mathrm{\mid }}_p$: the value depends only on the pointwise data.

(Smooth tensor field.) By step 3.1 the assignment descends to a map $R_p:T_{p}M\times T_{p}M\times T_{p}M\to T_{p}M$, $(v,w,u)\mapsto R(X,Y)Z{\mathrm{\mid }}_p$ for any smooth local extensions, which is $\mathbb{R}$-multilinear because the field-level assignment is additive over $\mathbb{R}$ and function-linear. In a chart, $R({\mathrm{\partial }}_c,{\mathrm{\partial }}_d){\mathrm{\partial }}_b$ is a smooth field by [L1], so its coefficients $R^a{}_{bcd}$ in the coordinate frame are $C^{\mathrm{\infty }}$ functions on the chart domain. By [L9] these data constitute a smooth tensor field of type $(1,3)$ whose components are exactly $R^a{}_{bcd}$. This proves (a).

(Coordinate formula.) In the chart, [L3] gives $[{\mathrm{\partial }}_c,{\mathrm{\partial }}_d]=0$, so [L1] gives $R({\mathrm{\partial }}_c,{\mathrm{\partial }}_d){\mathrm{\partial }}_b={\mathrm{\nabla }}_{{\mathrm{\partial }}_c}{\mathrm{\nabla }}_{{\mathrm{\partial }}_d}{\mathrm{\partial }}_b-{\mathrm{\nabla }}_{{\mathrm{\partial }}_d}{\mathrm{\nabla }}_{{\mathrm{\partial }}_c}{\mathrm{\partial }}_b$. By [L4], ${\mathrm{\nabla }}_{{\mathrm{\partial }}_d}{\mathrm{\partial }}_b={\sum }_e{\mathrm{\Gamma }}^e{}_{db}{\mathrm{\partial }}_e$, so [L2] gives ${\mathrm{\nabla }}_{{\mathrm{\partial }}_c}\left({\sum }_e{\mathrm{\Gamma }}^e{}_{db}{\mathrm{\partial }}_e\right)={\sum }_e\left({\mathrm{\partial }}_c{\mathrm{\Gamma }}^e{}_{db}\right){\mathrm{\partial }}_e+{\sum }_{e,a}{\mathrm{\Gamma }}^e{}_{db}{\mathrm{\Gamma }}^a{}_{ce}{\mathrm{\partial }}_a$; exchanging $c,d$ and subtracting yields $R^a{}_{bcd}={\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{db}-{\mathrm{\partial }}_d{\mathrm{\Gamma }}^a{}_{cb}+{\sum }_e({\mathrm{\Gamma }}^a{}_{ce}{\mathrm{\Gamma }}^e{}_{db}-{\mathrm{\Gamma }}^a{}_{de}{\mathrm{\Gamma }}^e{}_{cb})$. This proves (b).

(Point-normal derivative of the Christoffel symbols.) In the point-normal chart of [L6], differentiate the Christoffel formula of [L5]. The derivative of the factor $g^{ae}$ multiplies the bracket $({\mathrm{\partial }}_b{g}_{ed}+{\mathrm{\partial }}_d{g}_{eb}-{\mathrm{\partial }}_e{g}_{db})$, which vanishes at $p$ because $\mathrm{\partial }g(p)=0$, so only the bracket's derivative survives: ${\mathrm{\partial }}_c{\mathrm{\Gamma }}^a{}_{db}(p)=\frac{1}{2}{g}^{ae}(p)({\mathrm{\partial }}_c{\mathrm{\partial }}_b{g}_{ed}+{\mathrm{\partial }}_c{\mathrm{\partial }}_d{g}_{eb}-{\mathrm{\partial }}_c{\mathrm{\partial }}_e{g}_{db})(p)$, and likewise ${\mathrm{\partial }}_d{\mathrm{\Gamma }}^a{}_{cb}(p)=\frac{1}{2}{g}^{ae}(p)({\mathrm{\partial }}_d{\mathrm{\partial }}_b{g}_{ec}+{\mathrm{\partial }}_d{\mathrm{\partial }}_c{g}_{eb}-{\mathrm{\partial }}_d{\mathrm{\partial }}_e{g}_{cb})(p)$. Since $\mathrm{\Gamma }(p)=0$, formula (b) of step 5.1 reduces at $p$ to the difference of these two derivatives, in which the terms ${\mathrm{\partial }}_c{\mathrm{\partial }}_d{g}_{eb}$ and ${\mathrm{\partial }}_d{\mathrm{\partial }}_c{g}_{eb}$ agree by [L8] and cancel: $R^a{}_{bcd}(p)=\frac{1}{2}{g}^{ae}(p)({\mathrm{\partial }}_b{\mathrm{\partial }}_c{g}_{ed}-{\mathrm{\partial }}_c{\mathrm{\partial }}_e{g}_{db}-{\mathrm{\partial }}_b{\mathrm{\partial }}_d{g}_{ec}+{\mathrm{\partial }}_d{\mathrm{\partial }}_e{g}_{cb})(p)$.

(Lowering and final form.) Lowering the first index with [L5], $R_{abcd}(p)=g_{ae}(p)R^e{}_{bcd}(p)=\frac{1}{2}({\mathrm{\partial }}_b{\mathrm{\partial }}_c{g}_{ad}-{\mathrm{\partial }}_c{\mathrm{\partial }}_a{g}_{db}-{\mathrm{\partial }}_b{\mathrm{\partial }}_d{g}_{ac}+{\mathrm{\partial }}_d{\mathrm{\partial }}_a{g}_{cb})(p)$. Using [L8] to commute the mixed partials and the symmetry $g_{db}=g_{bd}$, $g_{cb}=g_{bc}$ of the metric, this is $\frac{1}{2}(g_{ad,bc}+g_{bc,ad}-g_{bd,ac}-g_{ac,bd})(p)$, and raising back with $g^{ae}(p)$ recovers the equivalent raised display. This proves (c) and completes the proof.