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Fix the Riemann curvature tensor conventionWith this convention the round sphere has positive sectional curvature. Write $R(X,Y,Z,W)=g(R(X,Y)Z,W)$. We use skew-symmetry in the first and last pairs, pair interchange $R(X,Y,Z,W)=R(Z,W,X,Y)$, and the first Bianchi identity. These are the standard curvature symmetries of the Levi-Civita connection.

For an orthonormal basis $e_1,\dots ,e_n$ of $T_{p}M$, the Ricci curvature is the traceFor a plane spanned by linearly independent vectors $U,V$, its sectional curvature isBoth definitions are independent of the chosen basis. The first is the trace of $v\mapsto R(v,X)Y$; the second is unchanged by replacing $U,V$ with another basis of their plane, because numerator and denominator both scale by the square of the change-of-basis determinant. For $U=e_i,V=e_j$, $i\ne j$, the denominator is one. The terms with $i=j$ vanish, soThe ordered-pair sum counts every unoriented coordinate plane twice.

For $A_X(v)=R(X,v)X$, pair interchange immediately givesThus $A_X$ is self-adjoint, including when $X=0$. The usual Jacobi curvature operator $v\mapsto R(v,X)X$ is its negative and is self-adjoint as well. Keeping these two signs distinct will also fix the Jacobi field equation below.

The Bonnet-Myers theorem states that a connected, complete $n$-dimensional Riemannian manifold, $n\ge 2$, with $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$ has diameter at most $\pi /\sqrt{\kappa }$ and is compact. Completeness is required; a pointwise positive Ricci tensor without a uniform positive lower bound would not suffice for this stated conclusion.

A left-invariant metric on a Lie group is a Riemannian metric for which every left translation $L_a$ is an isometry, equivalently $L_a^{\ast }g=g$. It is determined by an inner product on the Lie algebra $\mathfrak{g}=T_{e}G$. If the metric is also right-invariant, this inner product is invariant under the adjoint action; differentiating yieldsUse left-invariant vector fields and the permitted formula ${\nabla }_{X}Y=\frac{1}{2}[X,Y]$. The Jacobi identity givesAdjoint skew-symmetry then impliesThis is the Ricci curvature of a bi-invariant Riemannian metric. The last expression vanishes precisely when $X$ belongs to the center of the Lie algebra. If that center is zero, it is positive for every nonzero $X$. Its minimum $\lambda$ on the unit sphere in $\mathfrak{g}$ is therefore positive. Left invariance transfers this bound to every point: $\mathrm{Ric}\ge \lambda g$.

Work in the identity component $G^0$, since ${\pi }_1(G,e)={\pi }_1(G^0,e)$. It is compact and hence complete. Give its universal cover ${\tilde{G}}^0$ the lifted metric. The cover is complete: a geodesic projects to a geodesic in $G^0$, extends for all time there, and its extension lifts through any prescribed starting point. The local isometry also preserves Ricci curvature, so the same uniform bound holds on the cover. Applying Bonnet-Myers with $\kappa =\lambda /(n-1)$ makes ${\tilde{G}}^0$ compact. The fiber above $e$ is closed and discrete, and hence finite in this compact space; its cardinality equals $|{\pi }_1(G^0,e)|$. Consequently ${\pi }_1(G,e)$ is finite. This is the finite fundamental group from a uniform positive Ricci bound. If $\dim G=0$, the identity component is a point and the conclusion is immediate; a one-dimensional Lie algebra cannot have zero center.

The ungauge-fixed Polyakov action has worldsheet diffeomorphism and Weyl transformation symmetries. They must survive quantization so that conformal gauge remains a valid gauge choice. In a curved target, the string nonlinear sigma model has a field-dependent coupling $g_{ab}(X)$. Its worldsheet Weyl anomaly is controlled by the renormalization of this coupling, the sigma-model beta function.

Restore the Euclidean normalization $(4\pi {\alpha }^{\prime }{)}^{-1}\int g_{ab}(X)\partial X^a\partial X^b$. Use a covariant background field expansion of a string sigma model, with geodesic fluctuations about a slowly varying embedding. The quadratic fluctuation operator contains the target Riemann curvature tensor coupled to two background embedding derivatives. The ultraviolet coincident propagator is $⟨{\xi }^a{\xi }^b{⟩}_{\mathrm{UV}}=\alpha \prime g^{ab}\log (\Lambda /\mu )$: the kinetic operator supplies $2\pi \alpha \prime$, and the two-dimensional momentum integral supplies $(2\pi {)}^{-1}\log (\Lambda /\mu )$. Contracting the curvature vertex with this propagator gives a logarithmic metric counterterm proportional to $\alpha \prime R_{ab}$. The resulting one-loop metric counterterm givesThe trace of the worldsheet stress tensor contains this coefficient multiplying $\partial X^a\partial X^b$, together with the curvature anomaly controlled by total central charge. Thus, with zero antisymmetric background, constant dilaton, and the critical matter/ghost system,The central-charge condition also requires $d=26$ if there is no extra internal conformal theory. This is a leading-order equation: higher-curvature terms enter the beta function at higher orders in ${\alpha }^{\prime }$, so a Ricci-flat metric alone is not a general all-orders quantum consistency criterion.

To derive the local T-duality, work where the spacelike Killing vector has nonzero norm $V>0$. Use dimensionless coordinates and ${\alpha }^{\prime }=1$ units for the duality formulas. The isometry makes the action depend on $z$ only through its derivatives. Replace those derivatives by an independent worldsheet one-form $A_{\mu }$, and enforce its flatness with a Lagrange multiplier $\tilde{z}$. With ${\eta }_{\mu \nu }=\mathrm{diag}(-1,1)$ and ${ϵ}^{\tau \sigma }=1$, the relevant first-order worldsheet duality action isVarying $\tilde{z}$ sets $dA=0$. Locally $A=dz$, recovering the original theory. For the other elimination, integrate the multiplier term by parts. Varying $A_{\nu }$ gives $V{A}^{\nu }={ϵ}^{\mu \nu }{\partial }_{\mu }\tilde{z}$, orSubstituting into the entire first-order worldsheet duality action, including the multiplier term, givesThe multiplier contribution is necessary for the sign of the dual kinetic term. This is the first-order action for Abelian worldsheet duality, with Buscher rulesThe coordinate map is ${\partial }_{\tau }\tilde{z}=V{\partial }_{\sigma }z$, ${\partial }_{\sigma }\tilde{z}=V{\partial }_{\tau }z$. The original $z$ equation is precisely the integrability condition for $\tilde{z}$; conversely the original identity $d(dz)=0$ becomes the dual equation. Thus this transformation exchanges an equation of motion with the Bianchi identity for an Abelian p-form. It reverses one chiral derivative and preserves the other.

For full closed-string equivalence, the local derivation must include the global data. For a free compact circle isometry, the period of $\tilde{z}$ and the allowed gauge-field holonomies are chosen so that the Polyakov path integral exchanges momentum and winding modes. A circle of radius $R$ becomes a circle of radius ${\alpha }^{\prime }/R$ when dimensions are restored. A spacelike Killing field alone does not fix these periods or ensure a free global circle action; the global qualifications of Abelian T-duality are additional to the local field transformation.

At the quantum level integrating out $A_{\mu }$ also gives a regulated functional determinant. The Buscher dilaton shift isin the stated units and conventional Euclidean dilaton normalization. It generates the dilaton curvature coupling even though the original dilaton vanished. The dilaton curvature coupling is defined before fixing conformal gauge; its metric variation improves the worldsheet stress tensor even when the reference worldsheet is flat. As a normalization check, $e^{-2\tilde{\Phi }}\sqrt{|\tilde{g}|}=e^{-2\Phi }\sqrt{|g|}$ for this block-diagonal background. The local dual metric by itself is therefore only part of the equivalent quantum background.

With a nonconstant dilaton and no antisymmetric field, the leading metric-dilaton Weyl condition istogether with the scalar dilaton anomaly condition. It does not demand ${\tilde{R}}_{ab}=0$ separately. The new Ricci tensor can be nonzero while the Hessian of the shifted dilaton compensates it.

For a direct example, take flat polar coordinates away from the origin, with $d{s}^2=d{r}^2+r^{2}d{z}^2$ and flat spectator directions. The polar-coordinate T-dual background isHereBoth metric equations vanish. In critical dimension the scalar equation also holds: $4|\tilde{\nabla }\tilde{\Phi }{|}^2-4\tilde{\square }\tilde{\Phi }-\tilde{R}=4/r^2-8/r^2+4/r^2=0$. The example is local on $r>0$; the circle degenerates at the excluded origin.

Nonzero dual Ricci curvature is consistent because quantum T-duality transforms the dilaton as well as the metric. The displayed curvature equations and classical Buscher rules are interpreted at their stated leading derivative order; higher-order renormalized descriptions require the corresponding corrections and field-redefinition conventions. A constant $V$ is a special case in which the dual metric may also remain Ricci-flat.

Use precisely the stated convention for the Riemann curvature tensor, and let $D={\nabla }_T$. The geodesic equation says ${\nabla }_{T}T=0$ because proper time is an affine parameter. The commuting variation and the torsion-free connection giveThus the geodesic deviation equation isHere $(R(X,Y)Z{)}^a=R^a{}_{bcd}{Z}^b{X}^c{Y}^d$. Declaring this index convention fixes the sign: writing $R(S,T)T$ instead would require a minus sign. A deviation vector satisfying this equation is a Jacobi field.

Contracting the assumed Riemann curvature tensor gives the Ricci tensorThe contracted Bianchi identity and metric compatibility now implyThus ${\nabla }_{b}R=0$, so the scalar curvature is constant on each connected component of spacetime. The sectional curvature is consequently the constant $\kappa =R/12$. This is the four-dimensional case of the Schur theorem in pseudo-Riemannian geometry.

Keep only first order in the metric perturbation and use the Minkowski metric to raise and lower indices; set $G=c=1$. The Levi-Civita connection hasThe quadratic products of Christoffel symbols in the Riemann curvature tensor can be dropped at this order. Contracting the remaining derivatives giveswhere $\square ={\eta }^{\rho \sigma }{\partial }_{\rho }{\partial }_{\sigma }=-{\partial }_t^2+{\nabla }^2$ is the d'Alembert operator. In four dimensions the trace-reversed metric perturbation satisfies $\bar{h}=-h$ and $h_{\mu \nu }={\bar{h}}_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }\bar{h}$. Writing the Linearized Einstein equations in terms of it givesIn Lorenz gauge in linearized gravity, the first, second and fourth terms vanish. Therefore $G_{\mu \nu }^{(1)}=8\pi T_{\mu \nu }$ becomesThe divergence of this equation also gives ${\partial }^{\mu }{T}_{\mu \nu }=0$, the leading-order stress-energy conservation required by the contracted Bianchi identity.

The Equivalence principle permits Riemann normal coordinates at any event, in which the first derivatives of the metric tensor, and hence the Christoffel symbols, vanish. A putative universal local gravitational stress-energy tensor constructed from the metric and its first derivatives would vanish in that freely falling description. If it were a genuine tensor, it would then vanish in every description, even where other coordinates assign nonzero gravitational energy density. The gravitational field therefore does not have a local stress-energy tensor analogous to that of matter, with coordinate-independent energy density and the usual energy-balance interpretation.

Gravitational energy is not represented by a universal local covariant stress-energy tensor. This does not assert that all local curvature tensors vanish or are forbidden: the Riemann curvature tensor measures tidal gravity, and higher-curvature tensors can be constructed. They do not provide the sought ordinary local gravitational energy density. Boundary charges, quantities associated with a background, and suitable averaged wave energies have different definitions and evade that local requirement. This is the absence of a local gravitational stress-energy tensor.

Extract the Riemann curvature tensor from ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ using the curvature 2-forms just computed. Contraction $R_{bd}=R^a{}_{bad}$ gives the diagonal Ricci tensor in the orthonormal coframe:For example, the $00$ component receives $a^2$ from each of the $1$ and $2$ directions and $q$ from direction $3$. The signs in the spatial components reflect the same Lorentzian index lowering used for the connection 1-forms.

For a null vector $k$ in this frame, $(k^0{)}^2=(k^1{)}^2+(k^2{)}^2+(k^3{)}^2$. ThereforeThe Einstein field equations imply $8\pi T_{ab}{k}^a{k}^b=R_{ab}{k}^a{k}^b$, since the metric term vanishes for a null vector; an included cosmological-constant term would also vanish. Choose $k^a=(1,0,0,1)$ and apply the null energy condition. It follows thatConversely, this inequality makes the same contraction nonnegative for every null vector, so it is exactly the null energy condition for a planar warped spacetime when its matter stress-energy tensor is defined by the Einstein field equations. Null vectors tangent to the planar slices saturate the condition. As checks, constant $A$ gives flat spacetime and $A=e^{bz}$ gives constant negative sectional curvature with $a^{\prime }=0$, also saturating it.

Fix the curvature conventionTo prove tensoriality, use $[hX,Y]=h[X,Y]-Y(h)X$ and the connection rules. The two $Y(h){\nabla }_{X}Z$ terms cancel, giving $R(hX,Y)Z=hR(X,Y)Z$. Antisymmetry in $X,Y$ gives linearity over smooth functions in the second input. Expanding the third input givesIt is therefore a smooth tensor of type $(1,3)$. Lowering the output with $g$ gives the type $(0,4)$ Riemann curvature tensor $R_4(X,Y,Z,W)=g(R(X,Y)Z,W)$.

For an independent pair $X,Y$, define sectional curvature byMetric compatibility gives $g(R(X,Y)Z,W)=-g(R(X,Y)W,Z)$ by applying $XY-YX-[X,Y]$ to $g(Z,W)$. Together with antisymmetry in $X,Y$, this shows that replacing the pair by $(aX+bY,cX+dY)$ multiplies both numerator and denominator by $(ad-bc{)}^2$. Thus the value depends only on the plane. Define Ricci curvature by $\mathrm{Ric}(Y,Z)={\sum }_{i}g(R(e_i,Y)Z,e_i)$ for any orthonormal basis; a trace is independent of the orthonormal basis.

For the curvature of the round unit sphere, the outward unit normal is the position vector $p$. The tangential projection of ambient differentiation is torsion-free and has metric compatibility, so uniqueness identifies it with $\nabla$. Ambient differentiation $D$ satisfies $D_{X}p=X$ andsince differentiating $g(Y,p)=0$ gives its normal component. The ambient curvature is zero. Take tangential components of $D_X{D}_{Y}Z-D_Y{D}_{X}Z-D_{[X,Y]}Z=0$ to obtainHenceEvery two-plane has sectional curvature one. Tracing the first formula gives $\mathrm{Ric}(Y,Z)=ng(Y,Z)-g(Y,Z)$, and consequentlyThus the sphere is an Einstein manifold. When $n=1$ there are no tangent two-planes, the curvature tensor is zero and the same Ricci formula gives zero. The declared slot convention fixes all signs.

Adopt the printed Riemann curvature tensor convention and contract $R_{ab}=R^c{}_{acb}$. At an arbitrary point choose normal coordinates for the original metric tensor. There the original connection coefficients vanish. Varying the coordinate curvature formula leavesAt that point these partial derivatives equal the covariant derivatives of the tensor $C$. Both sides are tensors, so the result is valid in every coordinate system:Contracting its first and third indices proves the Palatini identityThe normal-coordinate argument is applied independently at every point; it does not assume a flat background or set derivatives of the original connection to zero.

The two coordinate tangent vectors of the embedding areTheir Euclidean inner products give the pullback of a Riemannian metric, equivalently the induced metric:Locally set $u=\rho \phi$. The line element becomes $d{u}^2+d{\hat{z}}^2$, so the induced Riemann curvature tensor vanishes identically. This is the intrinsic flatness of a circular cylinder. Its bending in the ambient space is extrinsic curvature, not intrinsic curvature; periodicity of the angular coordinate does not change the local flatness.