Levi-Civita Connection on an Eleven-Dimensional Lorentzian Manifold

For smooth local fields define $K(X,Y,Z):=Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)+g([X,Y],Z)+g([Z,X],Y)-g([Y,Z],X)\mathrm{.}$ Using the bracket rules in [F3] and symmetry of $g$, the derivative-of-$f$ terms in the three expansions are respectively $Y(f)(g(Z,X)-g(X,Z))+Z(f)(g(X,Y)-g(Y,X)),$ $X(f)(g(Y,Z)-g(Z,Y))+Y(f)(g(Z,X)-g(X,Z)),$ $2X(f)g(Y,Z)+Z(f)(g(Y,X)-g(X,Y))\mathrm{.}$ The first two expressions vanish by symmetry, while the third leaves $2X(f)g(Y,Z)$. Therefore $K(fX,Y,Z)=fK(X,Y,Z),K(X,Y,fZ)=fK(X,Y,Z),K(X,fY,Z)=fK(X,Y,Z)+2X(f)g(Y,Z)\mathrm{.}$ In particular, for each fixed $X,Y$, $K(X,Y,\cdot )$ is a smooth one-form.

Since $g$ is nondegenerate, its flat map $v\mapsto g(v,\cdot )$ is an isomorphism on every tangent space. Define ${\mathrm{\nabla }}_{X}Y$ uniquely by $2g({\mathrm{\nabla }}_{X}Y,Z)=K(X,Y,Z)$ for every $Z$. In a chart, applying this to coordinate fields and using [F3] gives $2g({\mathrm{\nabla }}_{{\mathrm{\partial }}_M}{\mathrm{\partial }}_N,{\mathrm{\partial }}_Q)={\mathrm{\partial }}_M{g}_{NQ}+{\mathrm{\partial }}_N{g}_{MQ}-{\mathrm{\partial }}_Q{g}_{MN},$ so the coefficients are exactly those in the Statement. They are smooth by [F4]. The pointwise definition uses only the global tensors, fields and bracket, so it agrees on chart overlaps and defines a global smooth vector field.

Additivity and real homogeneity of $\mathrm{\nabla }$ follow from those of $K$. The first identity in step 1.1 gives ${\mathrm{\nabla }}_{fX}Y=f{\mathrm{\nabla }}_{X}Y$; the third gives $2g({\mathrm{\nabla }}_X(fY),Z)=fK(X,Y,Z)+2X(f)g(Y,Z)=2g(f{\mathrm{\nabla }}_{X}Y+X(f)Y,Z)\mathrm{.}$ Nondegeneracy of $g$ yields ${\mathrm{\nabla }}_X(fY)=f{\mathrm{\nabla }}_{X}Y+X(f)Y$, so $\mathrm{\nabla }$ is an affine connection by [F2].

Subtracting the two Koszul expressions and using symmetry of $g$ and skew-symmetry of the bracket gives $K(X,Y,Z)-K(Y,X,Z)=2g([X,Y],Z)\mathrm{.}$ Thus $2g({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y],Z)=0$ for every $Z$, and nondegeneracy gives zero torsion. Adding the expressions with $Y,Z$ interchanged gives $K(X,Y,Z)+K(X,Z,Y)=2Xg(Y,Z),$ so the defining identity implies metric compatibility.

If ${\mathrm{\nabla }}^{\mathrm{\prime }}$ is any torsion-free metric-compatible affine connection, substitute compatibility and torsion freedom into the six terms of $K$. The derivative terms become $g({\mathrm{\nabla }}_X^{\mathrm{\prime }}Y,Z)+g(Y,{\mathrm{\nabla }}_X^{\mathrm{\prime }}Z)+g({\mathrm{\nabla }}_Y^{\mathrm{\prime }}Z,X)+g(Z,{\mathrm{\nabla }}_Y^{\mathrm{\prime }}X)-g({\mathrm{\nabla }}_Z^{\mathrm{\prime }}X,Y)-g(X,{\mathrm{\nabla }}_Z^{\mathrm{\prime }}Y),$ and the bracket terms become $g({\mathrm{\nabla }}_X^{\mathrm{\prime }}Y,Z)-g({\mathrm{\nabla }}_Y^{\mathrm{\prime }}X,Z)+g({\mathrm{\nabla }}_Z^{\mathrm{\prime }}X,Y)-g({\mathrm{\nabla }}_X^{\mathrm{\prime }}Z,Y)-g({\mathrm{\nabla }}_Y^{\mathrm{\prime }}Z,X)+g({\mathrm{\nabla }}_Z^{\mathrm{\prime }}Y,X)\mathrm{.}$ Pairings are symmetric, so all terms cancel except $2g({\mathrm{\nabla }}_X^{\mathrm{\prime }}Y,Z)$. Hence $K(X,Y,Z)=2g({\mathrm{\nabla }}_X^{\mathrm{\prime }}Y,Z)$. The defining identity for $\mathrm{\nabla }$ and nondegeneracy imply ${\mathrm{\nabla }}_X^{\mathrm{\prime }}Y={\mathrm{\nabla }}_{X}Y$ for all fields. This proves uniqueness.