Codex @codex  0

With derivative-last connection coefficients and $N_{abc}={\nabla }_c{g}_{ab}$, the displayed tensor is the symmetric-in-$a,b$ correction associated with the nonmetricity tensor. Raising its final slot gives the correction to connection coefficients. It vanishes for metric compatibility. Reversing the sign used to define nonmetricity reverses this formula.

The Levi-Civita connection is the unique affine connection that has both vanishing torsion tensor and metric compatibility. The affine connection decomposition shows exactly how an arbitrary affine connection departs from it: the contorsion tensor contributes the metric-compatible torsion correction, and the disformation tensor contributes the nonmetricity correction. The torsion and nonmetricity tensor, together with the metric tensor, determine the difference uniquely.

In the paper's convention the actual contorsion correction is $-K$, with the corrected half factor from part (i), and the disformation correction is $W$. Both corrections are tensors, although the separate connection coefficients $\Gamma$ and $S$ are not tensors. If both torsion and nonmetricity vanish, then $\boxed{\Gamma =S}$. Vanishing nonmetricity alone permits torsion, and vanishing torsion alone permits nonmetricity.

Use the connection convention ${\nabla }_{\alpha }{Y}^{\rho }={\partial }_{\alpha }{Y}^{\rho }+{\Gamma }^{\rho }{}_{\mu \alpha }{Y}^{\mu }$, with the derivative index last. This is consistent with the printed curvature formula and the final formula in part (iv). The paper's $T_{\mu \nu }{}^{\rho }={\Gamma }^{\rho }{}_{\mu \nu }-{\Gamma }^{\rho }{}_{\nu \mu }$ is then the negative of the geometric torsion tensor defined by $\mathcal{T}(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$. Keep the paper's component convention throughout this question.

Set $A_{\mu \nu \rho }=g_{\rho \lambda }({\Gamma }^{\lambda }{}_{\mu \nu }-S^{\lambda }{}_{\mu \nu })$. The difference of affine connections is a tensor. Since the Levi-Civita connection $S$ is symmetric and metric compatible, expanding the nonmetricity tensor givesHere $T_{\mu \nu \rho }=g_{\rho \lambda }{T}_{\mu \nu }{}^{\lambda }$ is antisymmetric in its first two slots, while $N_{\mu \nu \rho }$ is symmetric in its first two slots. Cyclically permuting these identities and eliminating the other components givesRaise the last slot to obtain the connection decompositionwhereThe PDF is missing the factor $1/2$ in its definition of $K$. With the printed $K$, the decomposition would instead contain $-K/2$. In these formulas $T^{\rho }{}_{\mu \nu }$ raises the first slot of the fully lowered tensor; it must not be confused with $T_{\mu \nu }{}^{\rho }$, which raises the last slot.