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For any vector field $Y$, apply the Leibniz rule to the scalar $\omega (Y)$:Using ${\mathcal{L}}_{V}Y=[V,Y]$ and expanding the Lie bracket in a coordinate chart leavesThe connection terms cancel because the Levi-Civita connection is torsion-free. Applying this formula to each slot of the metric tensor and using metric compatibility givesEquivalently, one may prove both identities at a point in normal coordinates; since both sides are tensors, the result then holds in every coordinate system.

Varying the Ricci tensor and contracting givesSubstitution of part (i), followed by metric compatibility, reduces the divergence toThus the metric variation of scalar curvature isand consequently

For a covariant rank-two tensor,Impose metric compatibility $\nabla g=0$, write this equation and its two cyclic permutations in $\alpha ,\mu ,\nu$, and use the assumed torsion-free symmetry of the connection. Solving gives the unique Levi--Civita connection

If $S$ is a holomorphic subbundle of a Hermitian vector bundle $E$, its induced Chern connection is $D_S={\pi }_S{D}_E$ on sections of $S$. Orthogonal projection preserves metric compatibility, and holomorphicity of $S$ preserves the required $(0,1)$ operator. Uniqueness of the Chern connection therefore identifies the projected connection.

For a minimally coupled scalar field with a Levi-Civita connection, metric compatibility and the symmetric Hessian of a scalar field give the displayed off-shell identity. It is a special case of the diffeomorphism Noether identity for a scalar field, and gives stress-energy conservation when the scalar equation is satisfied. No gravitational field equation is needed.

With signature $+---$ and $n^{\mu }{n}_{\mu }=1$, the spatial projection tensor is $P^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }-n^{\mu }{n}_{\nu }$. Projecting the spacetime metric tensor gives $q_{\mu \nu }=g_{\mu \nu }-n_{\mu }{n}_{\nu }$, which is negative definite on spatial vectors. The positive spatial induced metric is $h_{\mu \nu }=-q_{\mu \nu }$. A plus sign in $g_{\mu \nu }+n_{\mu }{n}_{\nu }$ belongs to a different normal/signature convention and is not transverse in this one. Fully projected covariant derivatives of either spatial metric vanish by metric compatibility.

The square is a rank-two Killing tensor. It is symmetric because it pairs the covectors $Y_{a\cdot }$ and $Y_{b\cdot }$ using the metric. Along every affinely parametrized geodesic, its contraction $K(v,v)$ is the squared norm of the carried by parallel transport covector $Y(v,\cdot )$. Metric compatibility makes that norm constant, proving the Killing tensor equation. On a Riemannian manifold this tensor is positive semidefinite.