Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields

The functional $K$ is additive in each of its three arguments, and expanding its six terms with [L2] gives the three scalar rules $K(fX,Y,Z)=fK(X,Y,Z),K(X,Y,fZ)=fK(X,Y,Z),$ $K(X,fY,Z)=fK(X,Y,Z)+2X(f)g(Y,Z)$ for every smooth function $f$: in the first, the terms containing $Y(f)$ and $Z(f)$ produced by $[fX,Y]$ and $[fZ,X]$ cancel against the derivative-of-$f$ terms in $Y(g(Z,fX))$ and $-Z(g(fX,Y))$; in the second, the derivative terms cancel in pairs by symmetry of $g$; in the third, the terms $X(f)g(Y,Z)$ occur twice and all other derivative terms cancel.

Work with smooth local fields on each open chart domain and its overlaps. For fixed smooth $X,Y$, the rule of step 1.1 and bilinearity of $g$ show that the equation $2g(V,Z)=K(X,Y,Z)$ for all smooth $Z$ is equivalent, in a chart, to the single set of identities $2g(V,{\mathrm{\partial }}_c)=K(X,Y,{\mathrm{\partial }}_c)$ for $c=0,1,2,3$: writing $Z={\sum }_c{Z}^c{\mathrm{\partial }}_c$ on the chart gives $K(X,Y,Z)={\sum }_c{Z}^{c}K(X,Y,{\mathrm{\partial }}_c)$.

In each smooth chart $(U,x)$ put $V_U:=\frac{1}{2}{\sum }_{a,b}{g}^{ab}K(X,Y,{\mathrm{\partial }}_b){\mathrm{\partial }}_a$, a smooth field on $U$; by step 2.1 and $g_{ac}{g}^{ab}={\delta }_c{}^b$ it satisfies $2g(V_U,{\mathrm{\partial }}_c)=2{\sum }_b{g}_{ac}{g}^{ab}\frac{1}{2}K(X,Y,{\mathrm{\partial }}_b)=K(X,Y,{\mathrm{\partial }}_c)$ for every $c$, hence $2g(V_U,Z)=K(X,Y,Z)$ for every smooth $Z$ on $U$. If $(U,x)$ and $(U^{\mathrm{\prime }},x^{\mathrm{\prime }})$ overlap, then $g(V_U-V_{U^{\mathrm{\prime }}},Z)=0$ for every smooth $Z$ there, so $V_U=V_{U^{\mathrm{\prime }}}$ pointwise by nondegeneracy; the local fields therefore patch to one smooth vector field ${\mathrm{\nabla }}_{X}Y$ on $M$ with $2g({\mathrm{\nabla }}_{X}Y,Z)=K(X,Y,Z)$ for all $Z$.

$\mathrm{\nabla }$ is an affine connection. Additivity and real homogeneity in $X$ and in $Y$ hold termwise for $K$; scalar linearity in $X$ is $K(fX,Y,Z)=fK(X,Y,Z)$; and $2g({\mathrm{\nabla }}_X(fY),Z)=K(X,fY,Z)=fK(X,Y,Z)+2X(f)g(Y,Z)=2g(X(f)Y+f{\mathrm{\nabla }}_{X}Y,Z)$ for all $Z$, so ${\mathrm{\nabla }}_X(fY)=X(f)Y+f{\mathrm{\nabla }}_{X}Y$ by nondegeneracy.

Torsion vanishes. Subtracting the definitions of $K(X,Y,Z)$ and $K(Y,X,Z)$ and using symmetry of $g$ and skew-symmetry of the bracket cancels the $X,Y,Z$-derivative terms and the four bracket terms except $g([X,Y],Z)-g([Y,X],Z)=2g([X,Y],Z)$; hence $2g({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y],Z)=K(X,Y,Z)-K(Y,X,Z)-2g([X,Y],Z)=0$ for all $Z$, and $T(X,Y)={\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y]=0$ by nondegeneracy.

Metric compatibility holds. Adding $K(X,Y,Z)$ and $K(X,Z,Y)$ cancels the $Y$- and $Z$-derivative terms by symmetry of $g$ and cancels the bracket terms by skew-symmetry, leaving $K(X,Y,Z)+K(X,Z,Y)=2X(g(Y,Z))$; dividing by two and using the defining property of $\mathrm{\nabla }$ gives $X(g(Y,Z))=g({\mathrm{\nabla }}_{X}Y,Z)+g(Y,{\mathrm{\nabla }}_{X}Z)$.

Any torsion-free $g$-compatible connection ${\mathrm{\nabla }}^{\mathrm{\prime }}$ satisfies $2g({\mathrm{\nabla }}_X^{\mathrm{\prime }}Y,Z)=K(X,Y,Z)$. Indeed, compatibility applied to the three pairings and torsion freedom written as $[X,Y]={\mathrm{\nabla }}_X^{\mathrm{\prime }}Y-{\mathrm{\nabla }}_Y^{\mathrm{\prime }}X$, $[Z,X]={\mathrm{\nabla }}_Z^{\mathrm{\prime }}X-{\mathrm{\nabla }}_X^{\mathrm{\prime }}Z$, $[Y,Z]={\mathrm{\nabla }}_Y^{\mathrm{\prime }}Z-{\mathrm{\nabla }}_Z^{\mathrm{\prime }}Y$ turn the six terms of $K(X,Y,Z)$ into ${\mathrm{\nabla }}^{\mathrm{\prime }}$-terms in which $g({\mathrm{\nabla }}_X^{\mathrm{\prime }}Y,Z)$ occurs twice and every other term cancels in pairs. Hence ${\mathrm{\nabla }}^{\mathrm{\prime }}=\mathrm{\nabla }$ by the displayed defining property and nondegeneracy, proving claim (b) and the uniqueness assertion of (a).

In a chart, the brackets of coordinate fields vanish by [L2], so evaluating $K$ on coordinate fields with [L2] and [L4] gives $K({\mathrm{\partial }}_b,{\mathrm{\partial }}_c,{\mathrm{\partial }}_a)={\mathrm{\partial }}_b{g}_{ca}+{\mathrm{\partial }}_c{g}_{ba}-{\mathrm{\partial }}_a{g}_{bc}$; substituting ${\mathrm{\nabla }}_{{\mathrm{\partial }}_b}{\mathrm{\partial }}_c={\sum }_e{\mathrm{\Gamma }}^e{}_{bc}{\mathrm{\partial }}_e$ into step 3.1 gives ${\sum }_e{g}_{ea}{\mathrm{\Gamma }}^e{}_{bc}=\frac{1}{2}({\mathrm{\partial }}_b{g}_{ca}+{\mathrm{\partial }}_c{g}_{ba}-{\mathrm{\partial }}_a{g}_{bc})$ and multiplying by $g^{ad}$ gives the displayed Christoffel formula, whose right side is smooth by [L5].

(Dual and tensor components.) The dual connection of [L6] is a connection on $T^{\ast }M$, and its components are $({\mathrm{\nabla }}_a^{\ast }\omega {)}_b={\mathrm{\partial }}_a{\omega }_b-{\sum }_d{\mathrm{\Gamma }}^d{}_{ab}{\omega }_d$: with $\omega ={\sum }_b{\omega }_{b}d{x}^b$, the characterizing identity gives $({\mathrm{\nabla }}_{{\mathrm{\partial }}_a}^{\ast }\omega )({\mathrm{\partial }}_b)={\mathrm{\partial }}_a{\omega }_b-\omega ({\mathrm{\nabla }}_{{\mathrm{\partial }}_a}{\mathrm{\partial }}_b)={\mathrm{\partial }}_a{\omega }_b-{\sum }_d{\mathrm{\Gamma }}^d{}_{ab}{\omega }_d$.

The product connection of [L6] extends $\mathrm{\nabla }$ and ${\mathrm{\nabla }}^{\ast }$ to all tensor bundles and satisfies the Leibniz rule on tensor products. Its two elementary values are ${\mathrm{\nabla }}_{{\mathrm{\partial }}_c}{\mathrm{\partial }}_b={\sum }_d{\mathrm{\Gamma }}^d{}_{bc}{\mathrm{\partial }}_d$ and ${\mathrm{\nabla }}_{{\mathrm{\partial }}_c}d{x}^a=-{\sum }_d{\mathrm{\Gamma }}^a{}_{dc}d{x}^d$: the second follows from step 5.1 and $d{x}^a({\mathrm{\partial }}_b)={\delta }^a{}_b$. Expanding a smooth tensor field $T$ in the coordinate product frame as $T=\sum T^{a_1\cdots a_r}{}_{b_1\cdots b_s}{\mathrm{\partial }}_{a_1}\otimes \cdots \otimes d{x}^{b_1}\otimes \cdots$ and applying the Leibniz rule, exactly one $+\mathrm{\Gamma }$ term appears for each upper slot and one $-\mathrm{\Gamma }$ term for each lower slot, and the coefficient derivative ${\mathrm{\partial }}_c{T}^{\cdots }$ appears once; this yields the displayed component formula, the sum over $d$ replacing the slot index.

The induced derivative commutes with contraction. Let $T$ be of type $(r,s)$ with $r,s\ge 1$, contract one upper slot $a_i$ with one lower slot $b_j$, and contract the derivative in the same slots. The $+\mathrm{\Gamma }$ term for the upper slot contributes ${\sum }_d{\mathrm{\Gamma }}^d{}_{ec}{T}^{\cdots e\cdots }{}_{\cdots d\cdots }$ and the $-\mathrm{\Gamma }$ term for the lower slot contributes $-{\sum }_d{\mathrm{\Gamma }}^e{}_{dc}{T}^{\cdots d\cdots }{}_{\cdots e\cdots }$; relabelling the dummy indices $(d,e)\mapsto (e,d)$ in the second sum makes the two contributions equal and opposite, while the remaining terms are exactly the derivative of the contracted components by step 6.1. Summing over $d$ is a finite index sum, so both orders give the same smooth function.

The metric and inverse metric are annihilated. By step 6.1 with $T=g$, writing the two lower slots explicitly, $g_{ab;c}={\mathrm{\partial }}_c{g}_{ab}-{\sum }_d{\mathrm{\Gamma }}^d{}_{ac}{g}_{db}-{\sum }_d{\mathrm{\Gamma }}^d{}_{bc}{g}_{ad}=0$ by the compatibility identity of [L5]. Differentiating $g^{ad}{g}_{db}={\delta }^a{}_b$ in the direction ${\mathrm{\partial }}_c$ and substituting the compatibility identity gives ${\mathrm{\partial }}_c{g}^{af}+{\sum }_d{\mathrm{\Gamma }}^f{}_{dc}{g}^{ad}+{\sum }_d{\mathrm{\Gamma }}^a{}_{dc}{g}^{df}=0$, which is $g^{af}{}_{;c}=0$ by step 6.1. This proves claim (c) and completes the proof.