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The first derivative of a scalar is a covector, soCommuting partial derivatives and the torsion-free connection give ${\phi }_{;lm}={\phi }_{;ml}$, the symmetric Hessian of a scalar field. Its next covariant derivative isIn particular, tensor index symmetrization over $l,m$ adds nothing: ${\phi }_{;(lm)n}={\phi }_{;lmn}$. This does not make the third index interchangeable with the first two. Applying the lower-index Ricci identity to the covector ${\phi }_{;l}$ yields the requested tensor index antisymmetrization:Parentheses and brackets include the usual averaging factors. To make the remaining symmetry explicit, let $S_{lmn}={\phi }_{;(lmn)}$. Since the first two derivative indices are already symmetric,The two commutator identities give ${\phi }_{;lnm}={\phi }_{;lmn}+R^p{}_{lmn}{\phi }_{;p}$ and ${\phi }_{;mnl}={\phi }_{;lmn}+R^p{}_{mln}{\phi }_{;p}$. Thus the third covariant derivatives of a scalar can also be writtenThe right-hand side displays exactly the obstruction to total symmetry. At a point where the scalar gradient vanishes, or where these curvature contractions vanish, the third derivative is totally symmetric.

For the final contraction use $R_{ab}=R^c{}_{acb}$. The arbitrary-valence formula givesThe Ricci tensor of the Levi-Civita connection is symmetric, whereas the last factor is antisymmetric. Their contraction is zero, provingNo symmetry of $T^{ik}$ was assumed. This double divergence of a contravariant tensor identity uses the metric-compatible connection of general relativity: zero torsion alone for an arbitrary affine connection does not guarantee a symmetric Ricci tensor and would not suffice for this final step.

A linear connection on a manifold is a rule $(X,Y)\mapsto {\nabla }_{X}Y$ for differentiating vector fields which is additive, is $C^{\infty }$-linear in $X$, is real-linear in $Y$, and obeys ${\nabla }_X(fY)=X(f)Y+f{\nabla }_{X}Y$. It extends to covector fields and arbitrary tensor fields by the Leibniz rule and compatibility with contraction. In coordinates,This is the covariant derivative; the affine connection terms correct the changes of the coordinate basis. A symmetric affine connection has zero torsion tensor, $T(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]=0$, or ${\Gamma }^c{}_{ab}={\Gamma }^c{}_{ba}$ in a coordinate basis. Symmetry does not imply metric compatibility; no metric is required for this question.

Additivity of each covariant derivative immediately gives additivity of ${\Delta }_{ab}$. Expanding the second derivative of a tensor product givesSubtract the expression with $a,b$ exchanged. The two mixed terms cancel, proving that the curvature commutator is a tensor derivation:For a scalar, ${\nabla }_a{\nabla }_{b}f={\partial }_a{\partial }_{b}f-{\Gamma }^c{}_{ab}{\partial }_{c}f$. Commuting partial derivatives and using the symmetric affine connection gives ${\Delta }_{ab}f=0$. Consequently ${\Delta }_{ab}(fU)=f{\Delta }_{ab}U$: the commutator is linear over smooth functions, not merely over constant scalars. Write $U=U^c{e}_c$ in a local basis. The tensor-product rule gives ${\Delta }_{ab}U=U^c{\Delta }_{ab}{e}_c$, showing that its value at a point depends only on $U$ at that point. Thus it defines the Riemann curvature tensor byThe antisymmetrized second covariant derivative is itself tensorial in the derivative indices, so this pointwise map transforms as a tensor. The formula or the defining commutator gives $R_{abc}{}^d=-R_{bac}{}^d$, hence $R_{abc}{}^d=R_{[ab]c}{}^d$ with normalized antisymmetrization.

Apply the commutator to the scalar contraction ${\omega }_c{U}^c$. Its action vanishes, soSince $U$ is arbitrary, ${\Delta }_{ab}{\omega }_c=-R_{abc}{}^d{\omega }_d$. Repeated use of the tensor-product rule gives one plus curvature action for an upper index and one minus action for each lower index. In particular,The type notation in the PDF means one contravariant and two covariant indices, not a fractional tensor order.

For the first Bianchi identity, put ${\omega }_c={\nabla }_{c}f$ and $H_{bc}={\nabla }_b{\nabla }_{c}f$. The symmetric Hessian of a scalar field satisfies $H_{bc}=H_{cb}$. Therefore the six third-derivative terms cancel inSubstituting the covector curvature action gives $(R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d){\nabla }_{d}f=0$. At a chosen point the gradient of a smooth scalar can be any covector, soThe second form follows from the already proved first-pair antisymmetry.

For the second Bianchi identity, the same third-derivative cancellation can be organized as an operator identity. On vector-valued components set $D_a={\partial }_a+{\Gamma }_a$, with $({\Gamma }_a{)}^e{}_d={\Gamma }^e{}_{ad}$. Although a full second covariant derivative includes a term $-{\Gamma }^f{}_{ab}{D}_f$ for its lower derivative index, that term cancels on antisymmetrization because the affine connection is symmetric. Thus $[D_b,D_c]={\mathcal{R}}_{bc}$, where $({\mathcal{R}}_{bc}{)}^e{}_d=R_{bcd}{}^e$ acts by multiplication. Expanding the three nested operator commutators makes all six triple products cancel:Equivalently, the cyclic sum of ${\partial }_a{\mathcal{R}}_{bc}+[{\Gamma }_a,{\mathcal{R}}_{bc}]$ vanishes. These are the covariant derivative terms on the upper and last lower curvature indices. The two remaining lower indices add $-{\Gamma }^f{}_{ab}{R}_{fcd}{}^e-{\Gamma }^f{}_{ac}{R}_{bfd}{}^e$. In the cyclic sum these additions cancel in pairs, using ${\Gamma }^f{}_{ab}={\Gamma }^f{}_{ba}$ and $R_{fcd}{}^e=-R_{cfd}{}^e$. Hence the Bianchi identities from covariant derivative commutators giveBoth identities here apply to any torsion-free linear affine connection; the second derivation does not require choosing a metric affine connection.

For a scalar, ${\nabla }_a\varphi ={\partial }_a\varphi$, while its second covariant derivative is ${\nabla }_a{\nabla }_b\varphi ={\partial }_a{\partial }_b\varphi -{\Gamma }_{ab}^c{\partial }_c\varphi$. Partial derivatives commute and the Levi-Civita connection is torsion-free, ${\Gamma }_{ab}^c={\Gamma }_{ba}^c$, proving the asserted symmetry.

Write the Einstein equation without cosmological constant as $G_{ab}=\kappa T_{ab}$, $\kappa =8\pi G/c^4$. Tracing the given Ricci equation gives $R=g^{ab}{\partial }_a\varphi {\partial }_b\varphi$. HenceIn units $\kappa =1$ the prefactor is absent. Metric compatibility, the product rule and the just-proved symmetric Hessian of a scalar field giveThe contracted Bianchi identity makes the left side zero. Wherever the gradient covector is nonzero, at least one component is nonzero, forcing ${\nabla }_a{\nabla }^a\varphi =0$, even when that gradient is null.

Finally ${\Gamma }_{ab}^a={\partial }_b\log \sqrt{-g}$, obtained by differentiating the metric determinant. Thus for any vector $V$, ${\nabla }_a{V}^a=(\sqrt{-g}{)}^{-1}{\partial }_a(\sqrt{-g}{V}^a)$. Applying it to $V^a=g^{ab}{\partial }_b\varphi$ proves

The metric dependence of the minimally coupled scalar field Lagrangian is $\partial L/\partial g^{\alpha \beta }=-\frac{1}{2}{\nabla }_{\alpha }\varphi {\nabla }_{\beta }\varphi$. Part (ii) therefore gives the stress-energy tensorUse metric compatibility and the symmetric Hessian of a scalar field to computeThis is the scalar stress-energy divergence identity, an instance of the diffeomorphism Noether identity for a scalar field. On the scalar equation of motion it gives stress-energy conservation, $\boxed{{\nabla }_{\mu }{T}^{\mu }{}_{\alpha }=0}$.

PutUsing metric compatibility and the symmetric Hessian of a scalar field, the divergence of the proposed tensor isFor this to vanish for arbitrary solutions and local field data, the two coefficients must vanish. HenceThe resulting tensor is the Conserved Klein-Gordon scalar stress-energy ansatz.

For a scalar field, ${\nabla }_{\mu }\varphi ={\partial }_{\mu }\varphi$. Apply the product rule to the stress-energy tensor, using metric compatibility to pass the covariant derivative through $g^{\mu \nu }$:The middle terms cancel because the Levi-Civita connection gives a symmetric Hessian of a scalar field. Thus, with the sign conventions of this question, the scalar stress-energy divergence identity becomesThe last equality is the Klein-Gordon equation. This stress-energy conservation follows from the field equation and metric compatibility; it does not require imposing the Einstein field equations.

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.

Use the sequential convention ${\phi }_{;lmn}={\nabla }_n{\nabla }_m{\nabla }_l\phi$ and define $R$ by $X^a{}_{;mn}-X^a{}_{;nm}=R^a{}_{bmn}{X}^b$. The symmetric Hessian of a scalar field gives symmetry in $l,m$, but the curvature commutator on a covariant tensor gives ${\phi }_{;l[mn]}=-R^p{}_{lmn}{\phi }_{;p}/2$. Averaging the three distinct permutations yields ${\phi }_{;(lm)n}={\phi }_{;(lmn)}-(R^p{}_{lmn}+R^p{}_{mln}){\phi }_{;p}/3$. Thus a scalar's third covariant derivative is not generally totally symmetric.