Torsion freedom and metric compatibility in chart components

In the chart, [L1] and [L3] give $T({\mathrm{\partial }}_b,{\mathrm{\partial }}_c)={\mathrm{\nabla }}_{{\mathrm{\partial }}_b}{\mathrm{\partial }}_c-{\mathrm{\nabla }}_{{\mathrm{\partial }}_c}{\mathrm{\partial }}_b-[{\mathrm{\partial }}_b,{\mathrm{\partial }}_c]={\sum }_a({\mathrm{\Gamma }}^a{}_{bc}-{\mathrm{\Gamma }}^a{}_{cb}){\mathrm{\partial }}_a$.

The connection is also local in its differentiated field: if a global field $S$ vanishes near $p$, take a cutoff $\chi$ supported there with $\chi (p)=1$ from [L8]. Since $\chi S=0$, [L6] gives $0=({\mathrm{\nabla }}_X(\chi S){)}_p=X(\chi {)}_p{S}_p+\chi (p)({\mathrm{\nabla }}_{X}S{)}_p=({\mathrm{\nabla }}_{X}S{)}_p$. Thus fields agreeing near $p$ have the same differentiated value there; locality in the direction and cutoff extensions are supplied by [L8]. To localize [L4], define on local fields the scalar defect $D(X,Y,Z):=X(g(Y,Z))-g({\mathrm{\nabla }}_{X}Y,Z)-g(Y,{\mathrm{\nabla }}_{X}Z)$. By [L6], $D(fX,Y,Z)=fD(X,Y,Z)$. Expanding $D(X,fY,Z)$, the term $X(f)g(Y,Z)$ from the derivative cancels the same term from ${\mathrm{\nabla }}_X(fY)$, so $D(X,fY,Z)=fD(X,Y,Z)$; the identical cancellation in the third slot gives $D(X,Y,fZ)=fD(X,Y,Z)$. Fix $p\in U$ and take $\rho$ from [L8]. The fields $\rho X,\rho Y,\rho Z$ on $U$, extended by zero on $M\setminus \mathrm{supp}(\rho )$, are smooth global fields: these two open domains cover $M$ and the fields agree on their overlap. If $\mathrm{\nabla }$ is globally $g$-compatible, [L4] gives $D(\rho X,\rho Y,\rho Z{)}_p=0$; locality in [L8] and the three function-linear identities identify this with $\rho (p{)}^{3}D(X,Y,Z{)}_p=D(X,Y,Z{)}_p$. Thus compatibility holds for all local fields at every chart point. Apply it to $X={\mathrm{\partial }}_a$, $Y={\mathrm{\partial }}_b$, $Z={\mathrm{\partial }}_c$ and substitute [L1] to obtain ${\mathrm{\partial }}_a{g}_{bc}={\sum }_d({\mathrm{\Gamma }}^d{}_{ab}{g}_{dc}+{\mathrm{\Gamma }}^d{}_{ac}{g}_{bd})$. No simultaneous family of cutoffs is chosen.

Conversely, assume the displayed identity holds in every chart of a cover, let $X,Y,Z$ be smooth vector fields and let $p\in M$. On a chart around $p$ write $X={\sum }_a{X}^a{\mathrm{\partial }}_a$, $Y={\sum }_b{Y}^b{\mathrm{\partial }}_b$, $Z={\sum }_c{Z}^c{\mathrm{\partial }}_c$. Using [L6] and [L1], ${\mathrm{\nabla }}_{X}Y={\sum }_{a,b}(X^a{Y}^b{\mathrm{\nabla }}_{{\mathrm{\partial }}_a}{\mathrm{\partial }}_b+X^a{\mathrm{\partial }}_a(Y^b){\mathrm{\partial }}_b)={\sum }_{a,b,d}{X}^a{Y}^b{\mathrm{\Gamma }}^d{}_{ab}{\mathrm{\partial }}_d+{\sum }_{b}X(Y^b){\mathrm{\partial }}_b$, and likewise ${\mathrm{\nabla }}_{X}Z={\sum }_{a,c,d}{X}^a{Z}^c{\mathrm{\Gamma }}^d{}_{ac}{\mathrm{\partial }}_d+{\sum }_{c}X(Z^c){\mathrm{\partial }}_c$. Evaluating $g$ on these fields and using bilinearity gives $g({\mathrm{\nabla }}_{X}Y,Z)+g(Y,{\mathrm{\nabla }}_{X}Z)={\sum }_{a,b,c}{X}^a{Y}^b{Z}^c({\mathrm{\Gamma }}^d{}_{ab}{g}_{dc}+{\mathrm{\Gamma }}^d{}_{ac}{g}_{bd})+X(Y^b)Z^c{g}_{bc}+Y^{b}X(Z^c)g_{bc}$ at the point, with the index $d$ summed.

If ${\mathrm{\Gamma }}^a{}_{bc}={\mathrm{\Gamma }}^a{}_{cb}$ in the charts of a cover of $M$, then by step 1.1 the torsion vanishes on pairs of coordinate fields in every chart; for arbitrary smooth fields $X={\sum }_b{X}^b{\mathrm{\partial }}_b$, $Y={\sum }_c{Y}^c{\mathrm{\partial }}_c$ on a chart domain, [L2] expands $T(X,Y)={\sum }_{b,c}{X}^b{Y}^{c}T({\mathrm{\partial }}_b,{\mathrm{\partial }}_c)=0$. Hence $T=0$ and $\mathrm{\nabla }$ is torsion-free by [L4].

Conversely, if $\mathrm{\nabla }$ is torsion-free, then, at each chart point, extend the coordinate fields using the cutoff of [L8]; the pointwise tensoriality [L2] identifies their global torsion value with $T({\mathrm{\partial }}_b,{\mathrm{\partial }}_c)$, so $T({\mathrm{\partial }}_b,{\mathrm{\partial }}_c)=0$ in every chart, so step 1.1 and the basis property [L5] give ${\mathrm{\Gamma }}^a{}_{bc}={\mathrm{\Gamma }}^a{}_{cb}$. This proves claim 1.

On the other hand, since $g(Y,Z)={\sum }_{b,c}{Y}^b{Z}^c{g}_{bc}$ and $X$ is a derivation, $X(g(Y,Z))={\sum }_{b,c}(X(Y^b)Z^c{g}_{bc}+Y^{b}X(Z^c)g_{bc}+Y^b{Z}^{c}X(g_{bc}))$ with $X(g_{bc})={\sum }_a{X}^a{\mathrm{\partial }}_a{g}_{bc}$. Substituting the assumed identity for ${\mathrm{\partial }}_a{g}_{bc}$ turns the last sum into exactly the bracket-free part of step 1.3, so the two sides of $X(g(Y,Z))=g({\mathrm{\nabla }}_{X}Y,Z)+g(Y,{\mathrm{\nabla }}_{X}Z)$ agree at $p$. Since $p$ and the fields were arbitrary, $\mathrm{\nabla }$ is $g$-compatible by [L4]; together with step 1.2 this proves claim 2.