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Let $D$ be the Levi-Civita connection of the Riemannian metric. We use the curvature sign specified in the question, the negative of the usual vector-bundle curvature convention in Question 4. Extend $D$ to the covariant exterior derivative on tangent-bundle-valued forms and define the curvature two-form by $R=-D^2$. In a local frame with connection matrix $A$, it is the endomorphism-valued two-form $-(dA+A\wedge A)$. The transformation law for $A$ makes this transform by conjugation, so it is an intrinsic tensor. It measures the infinitesimal failure of parallel derivatives to commute, with this chosen sign.

For a tangent-bundle-valued one-form $\eta$, the covariant exterior derivative isApply this to the one-form $DZ$, for which $(DZ)(Y)=D_{Y}Z$. This calculation proves, rather than just renames, the operator expressionIt is linear over smooth functions in $Z$: the extra first- and second-derivative terms in $R(X,Y)(fZ)$ cancel by $[X,Y]f=X(Yf)-Y(Xf)$. The connection's linearity in its vector-field argument similarly shows linearity over smooth functions in $X$ and $Y$. Thus $R(X,Y)$ is an endomorphism of the tangent bundle, not a differential operator on its sections, and $R$ is a well-defined endomorphism-valued two-form.

The torsion-free identity is $D_{X}Y-D_{Y}X=[X,Y]$. Expand the cyclic curvature sum and use this identity to obtainThe last equality is the Jacobi identity for the Lie bracket of vector fields. This proves the first Bianchi identity

Lower the output index with the metric and define the Riemann curvature tensor byThis is a covariant four-tensor, because the curvature endomorphism is tensorial in all three inputs. Skew-symmetry of the curvature two-form immediately gives $R_{ji,kl}=-R_{ij,kl}$.

Metric compatibility gives the second skew-symmetry. Apply the scalar operator $[X,Y]-XY+YX$, which is zero, to $g(Z,W)$. Expanding each derivative using $Dg=0$ cancels the mixed products of first derivatives and leavesThus $R_{ij,lk}=-R_{ij,kl}$.

For a direct proof of pair interchange, choose the geodesic coordinates of Question 5 at an arbitrary point $p$. There $\Gamma (p)=0$ and $\partial g(p)=0$. Write $D_{{\partial }_i}{\partial }_k={\Gamma }^a{}_{ik}{\partial }_a$. The operator formula gives, at $p$,The Levi-Civita formula is $g_{al}{\Gamma }^a{}_{jk}=\frac{1}{2}({\partial }_j{g}_{kl}+{\partial }_k{g}_{jl}-{\partial }_l{g}_{jk})$. Differentiating at $p$, the products involving first derivatives of the metric and Christoffel coefficients vanish. The terms ${\partial }_i{\partial }_j{g}_{kl}$ cancel, leavingHere a comma denotes ordinary coordinate derivatives at the normal-coordinate center. Swapping $(i,j)$ with $(k,l)$ leaves this expression unchanged, by symmetry of $g$ and commutation of its second partial derivatives. This is the pair symmetry from the normal-coordinate curvature formula. Since $p$ was arbitrary and the equality is tensorial, it holds in every frame. We have proved all the requested symmetries:Changing the overall curvature sign would change the displayed coordinate formula's sign, but would not change the first Bianchi identity or any of these algebraic symmetries.