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A Riemannian metric is a smooth positive-definite symmetric inner product $g_x$ on every tangent space. A Levi-Civita connection is a connection on the tangent bundle which is torsion-free, ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$, and metric-compatible,Apply this identity with the cyclic triples $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and replace differences of covariant derivatives by Lie brackets. The result is the Koszul formulaIts right side depends only on $g$ and the vector fields; nondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely. This proves uniqueness.

For existence, in a coordinate chart defineThis is linear over smooth functions in $X$ and obeys the Leibniz rule in $Y$. Symmetry ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ proves zero torsion, and direct substitution giveswhich proves metric compatibility. On chart overlaps the two resulting connections satisfy the same intrinsic Koszul identity and therefore agree by uniqueness. They glue globally. Thus the existence and uniqueness of the Levi-Civita connection holds on every Riemannian manifold.

Fix the curvature conventionFor coordinate vectors $e_i$, define the Riemann curvature tensor components by $R_{ij,kl}=g(R(e_i,e_j)e_l,e_k)$. In these indices the Ricci and scalar contractions areExplicitly, $R_{ij,kl}=g_{ka}{R}^a{}_{lij}$, whereThese expressions define Ricci curvature and scalar curvature without ambiguity about which indices are contracted. With this convention a round unit $m$-sphere has $\mathrm{Ric}=(m-1)g$ and $s=m(m-1)$.

If $\mathrm{Ric}=\lambda g$ with a constant real $\lambda$, tracing gives $\boxed{s=m\lambda }$, so every Einstein metric has constant scalar curvature. The converse is not valid in arbitrary dimension. On $S^2\times S^1$ with the product Riemannian metric of the unit round sphere and a circle, the connection splits between factors. The mixed curvatures and circle curvature vanish, while the sphere has curvature one. HenceA scalar multiple of the whole metric would require $\lambda =1$ on the sphere directions and $\lambda =0$ on the circle direction, which is impossible. This proves that constant scalar curvature does not imply an Einstein metric, and supplies a counterexample to the unrestricted requested equivalence.

In dimension two, the curvature tensor is determined by the Gaussian curvature $K$:There the valid equivalence isIn dimension one both curvature and scalar curvature vanish identically. Without a two-dimensional restriction or another added geometric hypothesis, the general converse cannot be proved.

A Levi-Civita connection is a connection on the tangent bundle which is torsion-free and compatible with the Riemannian metric. Its defining identities areTo obtain uniqueness, write the metric identity for $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and use the torsion identity to interchange the derivatives. This gives the Koszul formulaThe metric is nondegenerate, so these inner products determine ${\nabla }_{X}Y$ uniquely.

For existence, let $K(X,Y,Z)$ denote the right-hand side. The bracket rules $[X,fY]=f[X,Y]+X(f)Y$ and $[fX,Y]=f[X,Y]-Y(f)X$ give, by cancellation of all extra derivatives,Thus the nondegenerate metric defines a unique smooth vector field ${\nabla }_{X}Y$ from $2g({\nabla }_{X}Y,Z)=K(X,Y,Z)$, and these identities give the connection rules. In coordinates the construction readsThe coefficients are smooth and symmetric in $i,j$, proving zero torsion. Substituting them giveswhich is exactly metric compatibility. Since the construction was given intrinsically by $K$, the local formulas agree on overlaps. This proves existence and uniqueness of the Levi-Civita connection on every Riemannian manifold.

Define the Riemann curvature tensor byThe connection rules make this expression tensorial in all three arguments. To fix the component convention, throughout this solution useWith this convention, the unit sphere has $R_{ijkl}=g_{ik}{g}_{jl}-g_{il}{g}_{jk}$ and positive Ricci curvature. In a coordinate frame, the components are obtained fromThe algebraic curvature symmetries aretogether with the first Bianchi identityThe first antisymmetry follows from the definition. Metric compatibility gives $g(R(X,Y)Z,W)+g(Z,R(X,Y)W)=0$, giving the second. Torsion-freeness reduces the cyclic sum to the Jacobi identity for Lie brackets of vector fields; pair interchange follows algebraically from these antisymmetries and the cyclic identity. These explain the identities rather than choosing unrelated sign conventions.

Define the Ricci curvature by the tensor contractionThis is a bilinear form, since $R$ is tensorial and trace is linear. At a point choose an orthonormal basis. Pair interchange givesso Ricci curvature is symmetric.

For the final determination in dimension three, an algebraic curvature tensor is determined by its entries in the three unordered index pairs $12,13,23$: antisymmetry removes repeated indices, and pair interchange makes the resulting $3\times 3$ array symmetric. It therefore has at most six independent entries. WriteThe diagonal Ricci components giveso, for example, $K_{12}=({\mathrm{Ric}}_{11}+{\mathrm{Ric}}_{22}-{\mathrm{Ric}}_{33})/2$, and the other two follow cyclically. The three off-diagonal entries are also recovered explicitly:Thus every independent curvature entry is determined by Ricci; this proves injectivity, not merely a dimension count.

An invariant formula making the reconstruction explicit is, with $S={\mathrm{tr}}_g\mathrm{Ric}$,To verify it, the right-hand side has the two antisymmetries, pair interchange and the cyclic identity by symmetry of $g$ and Ricci. Contracting its first and third indices in dimension three givesSubtract it from the original tensor. The difference has zero Ricci contraction, and the six recovered-entry formulas above force it to vanish. This completes the proof of three-dimensional curvature from the Ricci tensor in arbitrary coordinates.

Let $D$ be the ambient Levi-Civita connection and $\nabla$ the one for the induced metric on $M$. The Gauss formula iswhere the second fundamental form $II(X,Y)=(D_{X}Y{)}^{\perp }$ is normal. The tangential part of $D$ is metric-compatible and torsion-free, so the existence and uniqueness of the Levi-Civita connection identifies it with $\nabla$.

For a normal field $\xi$, the Weingarten formula defines the shape operator in a normal direction $A_{\xi }$ and the normal connection byDifferentiate $⟨\xi ,Y⟩=0$ using metric compatibility to getThe symmetry of the second fundamental form follows from the ambient torsion-free condition:since the Lie bracket of vector fields tangent to $M$ is tangent. Thereforeso every normal-direction shape operator is self-adjoint.

The paper uses $R(X,Y)=D_{[X,Y]}-[D_X,D_Y]$, so use the Gauss equation with reversed curvature convention. Applying the Gauss formula and Weingarten formula twice givesSubtracting in the stated curvature order yieldsPair with a tangent field $W$ and use the defining relation for the shape operator in a normal direction:This is the Gauss equation in the supplied sign convention. If one defines curvature by $[D_X,D_Y]-D_{[X,Y]}$ instead, the two quadratic terms have the opposite signs.

For an orthonormal frame $X,Y$ on a surface, positive Gaussian curvature with the convention here is $K=⟨R^M(X,Y)X,Y⟩$. If the ambient space is Euclidean and $\xi$ is a unit normal, the equation becomesThe Koszul formula determines $\nabla$ from the induced metric, and hence determines its Riemann curvature tensor and $K$. Therefore the Gaussian curvature depends only on that metric, even though its expression as a product of principal curvatures appears extrinsic. This is Theorema Egregium.

The Levi-Civita connection is the unique affine connection which is both a metric connection and a torsion-free connection. Uniqueness is encoded by the Koszul formulaIt follows by adding the metric-compatibility identities with derivative fields $X,Y$, subtracting the one with derivative field $Z$, and replacing differences of covariant derivatives by Lie brackets. Nondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely from all its pairings with $Z$. The existence and uniqueness of the Levi-Civita connection supplies existence; no existence proof is needed here.

The product Riemannian metric isIt is smooth and symmetric. If $(u_1,u_2)\ne 0$, at least one component is nonzero and contributes a strictly positive squared norm; the other contributes a nonnegative term. Thus $g$ is positive definite.

Let the factor connections be Levi-Civita connections. The torsion form of the product affine connection vanishes on two lifted fields from one factor because it equals the factor torsion; on opposite-factor fields it vanishes because both covariant derivatives and the Lie bracket of vector fields vanish. By tensoriality, torsion vanishes for all fields.

Check metric compatibility on lifted fields as well. When the two paired fields are from one factor, their metric pairing depends only on that factor. Differentiation along that factor gives the usual factor compatibility identity, and differentiation along the other factor gives zero on both sides. When the paired fields are from different factors, their pairing is identically zero and their derivatives stay in their original summands, so both sides are again zero. Since $\nabla g$ has tensoriality, these checks imply $\nabla g=0$ globally.

The existence and uniqueness of the Levi-Civita connection therefore givesEquivalently, applying the Christoffel symbol formula to the block metric gives the factor coefficients and zero mixed coefficients, because each factor metric is independent of the other coordinates.

Put $u=t-z$, $\ell =du=e^0-e^3$, $p=A^{\prime }/A$, $q=B^{\prime }/B$, with primes denoting $d/du$. The exterior derivatives are $d{e}^0=d{e}^3=0$, $d{e}^1=p\ell \wedge e^1$, $d{e}^2=q\ell \wedge e^2$. For a Levi-Civita connection, the connection 1-forms obey both Cartan's first structure equation and metric compatibility. With lower indices ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$, the result isAll other lower-index forms vanish or follow from ${\omega }_{ab}=-{\omega }_{ba}$. Raising the first time index changes its sign: for example ${\omega }^0{}_1={\omega }^1{}_0=p{e}^1$, whereas ${\omega }^1{}_3=-p{e}^1$ and ${\omega }^3{}_1=p{e}^1$.

For direct verification, ${\omega }^1{}_0\wedge e^0+{\omega }^1{}_3\wedge e^3=-p\ell \wedge e^1=-d{e}^1$, and the same holds for index two. For indices zero and three each term wedges $e^1$ or $e^2$ with itself, so it is zero. These forms are metric-compatible, and uniqueness follows from the existence and uniqueness of the Levi-Civita connection. The printed uniqueness statement implicitly includes metric compatibility: the torsion equation alone permits adding nonzero forms $C^a{}_{bc}{e}^c$ with $C^a{}_{bc}=C^a{}_{cb}$, since their wedge with $e^b$ vanishes.

The Levi-Civita connection of a Riemannian manifold $(M,g)$ is the unique connection on a vector bundle on $TM$ that is torsion-free and compatible with the Riemannian metric. Any such connection must satisfy the Koszul formulaNondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely from the right-hand side. Conversely, define $\nabla$ by this formula. Direct substitution shows that it is $C^{\infty }$-linear in $X$, satisfies the Leibniz rule in $Y$, preserves $g$, and obeys ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. It is therefore a torsion-free metric connection. This proves the existence and uniqueness of the Levi-Civita connection.

For the metric vector bundle $E$, choose the stated orthonormal local frame $(e_1,\dots ,e_m)$ and write $d_A{e}_j=A^i{}_j{e}_i$. Since $⟨e_i,e_j⟩={\delta }_{ij}$, metric compatibility givesHence the connection matrix in an orthonormal frame is skew-symmetric:

On an oriented Riemannian $d$-manifold, the Hodge star operator is defined byIt is an orthogonal map $\ast :{\Lambda }^r{T}_x^{\ast }M\to {\Lambda }^{d-r}{T}_x^{\ast }M$ and satisfies ${\ast }^2=(-1{)}^{r(d-r)}$. In dimension $d=2n$ on middle-degree forms, the Adjoint of the Hodge star on middle-degree forms isThus it is self-adjoint when $n$ is even, but it is not always self-adjoint. For $n=1$ on the oriented Euclidean plane,so its matrix in the orthonormal basis $(dx,dy)$ is skew-adjoint.

The Laplace-Beltrami operator on differential forms is the Hodge Laplacianwhere the codifferential $\delta$ is the formal $L^2$ adjoint of the exterior derivative. On a compact manifold without boundary, if $\Delta \omega =\lambda \omega$ for a nonzero differential form $\omega$, then integration by parts givesTherefore the nonnegativity of the Hodge Laplacian yields