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For smooth spatial fields $E^i,C^i{}_j$, define a torsion-free connection on $\mathbb{R}\times {\mathbb{R}}^n$ by ${\Gamma }^i{}_{00}=-E^i$, ${\Gamma }^i{}_{0j}={\Gamma }^i{}_{j0}=-C^i{}_j/2$ and all other coefficients zero. The time geodesic equation is $\ddot{t}=0$. For $\dot{t}\ne 0$, use $t$ as affine parameter; the spatial geodesic equation becomes $d^2{x}^i/d{t}^2=E^i+C^i{}_{j}d{x}^j/dt$. In dimension three, $C^i{}_j=2{ϵ}^i{}_{kj}{B}^k$ realizes $\ddot{\mathbf{x}}=\mathbf{E}+2\mathbf{B}\times \dot{\mathbf{x}}$. Such a connection need not preserve a metric; a spacetime description of these trajectories only needs affine geometry.

The average of the two flat translation connections on a Lie group is a torsion-free connection. For left-invariant vector fields, the Jacobi identity gives $R(X,Y)Z=-\frac{1}{4}[[X,Y],Z]$ with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$. Contracting gives $\mathrm{Ric}(X,Y)=-\frac{1}{4}K(X,Y)$ for the Killing form $K$. It is flat exactly when all double Lie brackets vanish, including the case of a two-step nilpotent group. For a semisimple Lie algebra, $K$ is nondegenerate and supplies a bi-invariant indefinite metric whose Levi-Civita connection is ${\nabla }^0$.

The canonical vorticity two-form is ${\omega }_{ab}={\nabla }_a(ϵu_b)-{\nabla }_b(ϵu_a)$, with specific enthalpy per particle $ϵ$. Potential flow $ϵu_a={\nabla }_a\chi$ has $\omega =0$ for a torsion-free connection. Contracting this identity with $u^a$ yields $ϵu^a{\nabla }_a{u}_b={\nabla }_bϵ-u_b{u}^a{\nabla }_aϵ$, the isentropic relativistic Euler equation in signature $(+---)$. This canonical condition does not mean the unweighted four-velocity one-form is closed.

The spinor covariant derivative commutator is a differential two-form in its derivative indices, so the two-component spinor calculus decomposition defines symmetric curvature operators ${\Delta }_{AB}$ and ${\Delta }_{A^{\prime }{B}^{\prime }}$. With ${ϵ}^{AB}{ϵ}_{AB}=2$, the unprimed projection is ${\Delta }_{AB}={ϵ}^{A^{\prime }{B}^{\prime }}[{\nabla }_{A{A}^{\prime }},{\nabla }_{B{B}^{\prime }}]/2$, and the primed projection is analogous. For a torsion-free connection these are algebraic curvature derivations: they annihilate scalar functions and obey the Leibniz rule on tensor products.

For a local basis $e_a$ of the tangent bundle, an affine connection has components defined by ${\nabla }_{e_c}{e}_b={\Gamma }^a{}_{bc}{e}_a$. If $V=V^b{e}_b$, then $({\nabla }_{e_c}V{)}^a=e_c(V^a)+{\Gamma }^a{}_{bc}{V}^b$. The connection components depend on the chosen basis and have an inhomogeneous transformation law; they are not themselves a tensor field. In a coordinate basis, a torsion-free connection has symmetry in the two lower indices.

For a geodesic variation $F(s,\tau )$, the deviation vector $S=F_{\ast }{\partial }_s$ describes infinitesimal separation at equal parameter values. With $T=F_{\ast }{\partial }_{\tau }$, commuting parameter derivatives give $[S,T]=0$. A torsion-free connection then gives ${\nabla }_{T}S={\nabla }_{S}T$. For an affinely parametrized variation, geodesic deviation states ${\nabla }_T^{2}S=R(T,S)T$ with the convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$.

A Riemannian metric determines a unique torsion-free connection that is a metric connection. Metric compatibility and the torsion identity imply the Koszul formula, which proves uniqueness. Conversely the right-hand side of the Koszul formula is a smooth one-form in its last vector-field argument. The musical isomorphism defines $D_{X}Y$ from it; expanding the Lie bracket of vector fields verifies the connection rules, metric compatibility and zero torsion. Thus it constructs the Levi-Civita connection intrinsically.

For smooth stationary fields on ${\mathbb{R}}^3$, set ${\Gamma }^i{}_{tt}=-E^i$ and ${\Gamma }^i{}_{tj}={\Gamma }^i{}_{jt}={ϵ}^i{}_{jk}{B}^k$, with all other Christoffel symbols zero. This torsion-free connection makes the equation $\ddot{\mathbf{x}}=\mathbf{E}+2\mathbf{B}\times \dot{\mathbf{x}}$ a geodesic equation with time as affine parameter. Arbitrary smooth fields suffice; a metric realization imposes additional potential conditions.

For $\tilde{g}=e^{2\sigma }g$, the displayed correction is a symmetric tensor, so adding it to $\nabla$ gives a torsion-free connection. The two extra metric-compatibility pairings sum to $2d\sigma (X)g(Y,Z)$; this is exactly the derivative of the conformal factor. Uniqueness of the Levi-Civita connection proves the formula. It also yields the critical-point criterion for equal conformal connections.

For any torsion-free connection, expand the coordinate Lie derivative of a tensor field and cancel the symmetric connection terms to obtain the displayed identity. If the connection is metric-compatible, lowering the vector index commutes with the covariant derivative and it becomes ${\mathcal{L}}_{\xi }{g}_{ab}=2{\nabla }_{(a}{\xi }_{b)}$. Symmetry of the connection alone does not suffice: with $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the two expressions ${\mathcal{L}}_{\xi }g$ and $2{\nabla }_x{\xi }_x$ are $2e^{2x}$ and $4e^{2x}$, respectively.

For smooth spatial $\varphi$ and $g(c)=-c^2{e}^{2\varphi /c^2}d{t}^2+{\delta }_{ij}d{x}^{i}d{x}^j$, the only nonzero Christoffel symbols are ${\Gamma }^t{}_{ti}={\Gamma }^t{}_{it}=c^{-2}{\partial }_i\varphi$ and ${\Gamma }^i{}_{tt}=e^{2\varphi /c^2}{\partial }_i\varphi$. Consequently the Levi-Civita connection converges smoothly on compact sets as $c\to \infty$ although the metric itself diverges in these coordinates. The limiting torsion-free connection has only ${\Gamma }^i{}_{tt}={\partial }_i\varphi$ and gives $d^2{x}^i/d{t}^2=-{\partial }_i\varphi$ when $t$ is an affine parameter. With $R^a{}_{bcd}={\partial }_c{\Gamma }^a{}_{db}-{\partial }_d{\Gamma }^a{}_{cb}+\cdots$, its Ricci tensor is $(\Delta \varphi )dt\otimes dt$: quadratic connection terms vanish and $R^i{}_{tjt}={\partial }_j{\partial }_i\varphi$. Thus zero Ricci tensor is the Laplace equation, while full flatness would require the entire Hessian to vanish.

An affine connection on the tangent bundle of a smooth manifold assigns a vector field ${\nabla }_{X}Y$ to two vector fields, is $\mathbb{R}$-bilinear, and satisfiesThus its first slot is tensorial and its second obeys the Leibniz rule. Its action on covectors is defined by differentiating the pairing, and this extends it to tensor fields. A torsion-free connection has ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$.

Assume that $g$ is a smooth nondegenerate metric tensor. Metric compatibility and a vanishing torsion tensor imply the Koszul formula,The right-hand side is determined by $g$ and the Lie bracket of vector fields. Nondegeneracy determines ${\nabla }_{X}Y$ uniquely; conversely this formula defines an affine connection with both required properties, proving existence as well as uniqueness. In a coordinate frame the brackets vanish. Combining the three differentiated metric tensor identities givesThis is the Levi-Civita connection. Positive definiteness is unnecessary: the argument also works for a pseudo-Riemannian metric.

For the other affine connection define $S(X,Y)={\bar{\nabla }}_{X}Y-{\nabla }_{X}Y$. The derivative of $f$ cancels between the two Leibniz rules, so $S$ is $C^{\infty }$-linear in both slots. The difference of affine connections is a tensor: here $S$ is a section of $TM\otimes T^{\ast }M\otimes T^{\ast }M$ with components ${\bar{\Gamma }}^a{}_{bc}-{\Gamma }^a{}_{bc}$. Both affine connections are torsion-free, so $S(X,Y)=S(Y,X)$.

Agreement of geodesics means agreement of their unparametrized curves; different affine parameters are allowed. For every nonzero tangent vector $v$, existence of a geodesic with that initial velocity implies that $S(v,v)$ is parallel to $v$. The symmetric bilinear diagonal-parallel lemma now determines $S$. Choose a basis $e_i$ and write $S(e_i,e_i)={\alpha }_i{e}_i$. Applying the parallelism condition to $e_i+e_j$ and $e_i-e_j$ givesWith $V(e_i)={\alpha }_i/2$, bilinearity yieldsThis is projective equivalence of affine connections. It also covers dimension one. For two Levi-Civita connections, the projective covector from metric volume densities gives the explicit answerIndeed ${\Gamma }^a{}_{ac}={\partial }_c\log \sqrt{|\det g|}$. The determinant ratio is a scalar, so this expression is a genuine covector, although either determinant alone is coordinate-dependent. This determinant expression requires nondegeneracy of $\bar{g}$; the difference-tensor and earlier trace formula require only the two torsion-free affine connections.

Conversely, the displayed tensor gives ${\bar{\nabla }}_{\dot{\gamma }}\dot{\gamma }=2V(\dot{\gamma })\dot{\gamma }$ along an affinely parametrized $\nabla$-geodesic. Choosing a new parameter $t(\tau )$ satisfying $t^{\prime \prime }/t^{\prime }=2V(\dot{\gamma })$ removes that tangential acceleration. This geodesic reparametrization under projective equivalence proves sufficiency. If agreement were required with the very same affine parameter, polarization would instead force $S=0$ and $V=0$.

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.

Use sequential semicolon indices: $X^a{}_{;kl}={\nabla }_l({\nabla }_k{X}^a)$, where the second covariant derivative also differentiates the lower index of ${\nabla }_k{X}^a$. For a torsion-free connection, direct expansion givesSubtract the expression with $k,l$ exchanged. Commuting partial derivatives cancel the first term; the first-derivative terms cancel in pairs; symmetry of the lower Christoffel symbol indices cancels both terms containing ${\Gamma }^p{}_{kl}$. What remains isThis is the Ricci identity with the curvature convention printed on the PDF's first page. In terms of derivative operators it is $[{\nabla }_l,{\nabla }_k]X^a=R^a{}_{bkl}{X}^b$; writing the operator order explicitly avoids reversing the sign.

The lower-index rule follows by applying the commutator to the scalar contraction $Y_a{X}^a$. Its second derivatives commute, so the product rule implies $Y_{a;kl}-Y_{a;lk}=-R^p{}_{akl}{Y}_p$. The derivative commutator is a derivation on tensor products, since the cross terms cancel. It therefore acts once on every index, with a positive sign on each upper index and a negative sign on each lower index. For an arbitrary tensor of valence $(r,s)$,Local sums of decomposable tensors establish the formula for every tensor field, rather than only for a single tensor product.

An orthonormal tetrad, or vierbein, is a local basis $e_I{}^a$ of tangent vectors together with inverse coframe $e_a{}^I$, satisfying $g_{ab}{e}_I{}^a{e}_J{}^b={\eta }_{IJ}$ and $g_{ab}=e_a{}^I{e}_b{}^J{\eta }_{IJ}$. Capital indices label the locally orthonormal Lorentz frame; lower-case indices label spacetime coordinates. Changing the frame by a local Lorentz transformation leaves the metric unchanged. A spinor description uses a compatible local spin frame and the torsion-free spin connection.

Let fixed flat matrices satisfy $\{{\Gamma }_I,{\Gamma }_J\}=2{\eta }_{IJ}1$. The curved spacetime gamma matrices are ${\gamma }_a=e_a{}^I{\Gamma }_I$ and ${\gamma }^a=e_I{}^a{\Gamma }^I$. ThereforeThe identity matrix is four by four for a four-dimensional Dirac spinor. The combined tensor/spin covariant derivative makes ${\nabla }_a{\gamma }_b=0$, the Clifford version of the tetrad postulate.

Fix the curvature slots by $R_{abcd}=g_{ae}{R}^e{}_{bcd}$, with $[{\nabla }_c,{\nabla }_d]v^a=R^a{}_{bcd}{v}^b$, and $R=g^{ac}{g}^{bd}{R}_{abcd}$. The torsion-free spin connection then has $[{\nabla }_a,{\nabla }_b]\psi =\frac{1}{4}{R}_{abcd}{\gamma }^c{\gamma }^d\psi$. These choices fix the signs in the requested identities.

Write ${\gamma }^{ab}={\gamma }^{[a}{\gamma }^{b]}$ and similarly for four indices. Clifford anticommutation givesOn contraction with the Riemann curvature tensor, the four-form term vanishes by the first Bianchi identity. The first and last two-form terms vanish by pair antisymmetry; the remaining two-form terms are symmetric Ricci contractions against antisymmetric gamma matrices and vanish too. The scalar terms are zero, minus R and minus R respectively. Thus the Riemann tensor Clifford contraction isThis algebraic result is independent of Lorentz signature when the contraction convention is held fixed.

Apply the Dirac operator twice, using covariant constancy of the gamma matrices. Define the rough wave operator by $\square =g^{ab}{\nabla }_a{\nabla }_b$, with the full connection acting on the intermediate covector-spinor derivative. Splitting the product into symmetric and antisymmetric parts givesConsequently the Lichnerowicz spinor-square formula and the assumed massless Dirac equation imply $\boxed{(\square -R/4)\psi =0}$. Torsion would modify the commutator and introduce extra terms, so the torsion-free connection is part of this argument.

For a scalar, the Lie derivative of a function is its directional derivative:The Lie derivative of a vector field is their Lie bracket of vector fields:For a torsion-free connection,The last term vanishes because the connection coefficients are symmetric in $b,c$. This proves ${\mathcal{L}}_{\xi }X={\nabla }_{\xi }X-{\nabla }_X\xi$.

For the metric tensor statements use the Levi-Civita connection, which also has metric compatibility. The tensor derivation property of the Lie derivative of a tensor field givesExpanding the first term using $\nabla g=0$ and using the vector identity cancels the derivatives of $X$ and $Y$. What remains isSince $X,Y$ are arbitrary, the Killing equation isMetric compatibility is needed here, even though symmetry alone sufficed for the vector identity. In general the Lie derivative of a metric with nonmetricity contains a term ${\xi }^c{\nabla }_c{g}_{ab}$. For example, with the metric tensor $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the Lie derivative of a tensor field is $2e^{2x}$, whereas $2{\nabla }_x{\xi }_x=4e^{2x}$. The usual metric connection interpretation of the question supplies the required compatibility.

For the Riemann curvature tensor calculation, keep the original PDF's convention. With $R_{abcd}=g_{ae}{R}^e{}_{bcd}$ and $R_{abc}{}^d=g^{de}{R}_{abce}$, it meansThus the commutator on a covector has the opposite sign. Write $K_{abc}={\xi }_{a;bc}={\nabla }_c{\nabla }_b{\xi }_a$. Taking a covariant derivative of the Killing equation makes $K$ antisymmetric in its first two indices, while the Ricci identity givesCombining this with the first-two-index antisymmetry gives the three relationsAdd the first and third and subtract the second. Pair symmetry and the first Bianchi identity imply $R_{adbc}+R_{bdca}-R_{cdab}=-2R_{abcd}$. ConsequentlyThis is the second covariant derivative of a Killing vector in the paper's convention; its sign is tied to the declared curvature-sign convention in Killing derivative identities, not to raising a particular index.

Set $L_{ab}={\xi }_{a;b}$. Along any smooth curve with tangent vector $V$, the chain rule and the just-proved identity give the Killing transport equationsChoose a parallel-propagated orthonormal frame along the curve. These become a homogeneous linear system of ordinary differential equations for the components of $(\xi ,L)$. Uniqueness of its initial value problem implies that zero initial data remain zero. Any point in the same connected component as $P$ can be joined to $P$ by a piecewise smooth curve. Therefore zero value and zero first derivative at $P$ force the Killing field and its derivative to vanish throughout that component, with no completeness assumption. On a manifold with one connected component this is the requested global result. Connectedness cannot be dropped: on two disjoint flat spacetimes, take the zero field on one and a nonzero translation on the other.

On a pseudo-Riemannian manifold of dimension $n$ with one connected component, the map $\xi \mapsto ({\xi }_a(P),L_{ab}(P))$ is therefore injective. There are $n$ independent components of the value and $n(n-1)/2$ of its antisymmetric derivative. The dimension bound for Killing vector fields isIt is attained in flat space by ${\xi }_a=A_a+B_{ab}{x}^b$ with constant antisymmetric $B$, giving translations and infinitesimal rotations or Lorentz transformations. The four-dimensional maximum is 10. For a disconnected manifold the bound applies componentwise; there is no uniform global bound depending only on $n$ if the number of components is unrestricted.

Use the Levi-Civita connection and the Riemann curvature tensor conventionfor orthonormal $T,E$. Thus the round sphere has positive sectional curvature. Work on a fixed compact parameter interval $[a,b]$ inside the given open interval; this keeps all differentiations and endpoint evaluations well-defined. The base geodesic is regular and has constant speed, so a constant rescaling of $t$ makes it unit-speed. Put $T={\partial }_t\varphi$, $J={\partial }_s\varphi$, $D_t={\nabla }_T$, $D_s={\nabla }_J$ and evaluate at $s=0$. The torsion-free connection gives $D_{s}T=D_{t}J$ because the two parameter derivatives commute. The geodesic equation is $D_{t}T=0$.

Differentiating the length of a curve once and twice, using metric compatibility, givesandLet $A=D_{s}J$. The curvature commutator gives $D_s{D}_{t}J=D_{t}A+\mathcal{R}(J,T)J$, and the skew-adjoint curvature symmetry gives $⟨\mathcal{R}(J,T)J,T⟩=-⟨\mathcal{R}(J,T)T,J⟩$. Integration by parts therefore proves the moving-endpoint second variation of Riemannian arc length:Equivalently, with $W=J-⟨J,T⟩T$, the integral is $I(W,W)$, the Riemannian index form. The endpoints of this geodesic variation generally move, so the displayed boundary term must be retained. For a base speed $c$ instead of one, the boundary and integral have an overall factor $1/c$, and the subtracted square inside the integral is $⟨D_{t}J,T{⟩}^2/c^2$. Returning to the original parameter, every ruling has constant speed, so $L_{[a,b]}(s)=(b-a)|{\partial }_t\varphi (0,s)|$ and the full open-interval length is $2\delta |{\partial }_t\varphi (0,s)|$. ConsequentlyThe preceding curvature formula, with the base-speed factor $1/c$ when necessary, therefore evaluates the original full length derivative using any compact subinterval. No extension to the missing endpoints is required.

Differentiating $D_{t}T=0$ in $s$ yields the Jacobi field equationThe same ruling curves are geodesics of the surface with its induced metric, because the tangential part of their ambient acceleration is zero. Write its second fundamental form as $II$ and its induced Levi-Civita connection as $D^N$. In particular $II(T,T)=0$. The scalar $f=⟨J,T⟩$ satisfies $f^{\prime \prime }=0$, by the Jacobi field equation, so $W=J-fT$ is a Jacobi field both intrinsically and in the ambient manifold. In a regular surface chart $h=|W|>0$, and $E=W/h$ is the perpendicular unit tangent. Along a two-dimensional geodesic, $D_t^{N}E=0$: its derivative is orthogonal to both $E$ and the parallel field $T$. The intrinsic Jacobi field equation consequently givesFor the ambient derivative, the Gauss formula gives $D_{t}W=h^{\prime }E+hII(T,E)$. Applying the same norm differentiation used above to $h=|W|$ givesComparing these identities proves the curvature comparison for a geodesically ruled surface:This is also the Gauss equation in a curved ambient manifold with $II(T,T)=0$. It applies locally to each regular patch of the surface.

For strict inequality, take the Euclidean embedded submanifold parametrized by $\varphi (t,s)=(t,s,ts)$. Each $t$-curve is a straight-line geodesic. Its first fundamental form has coefficients $E_0=1+s^2$, $F_0=st$, $G_0=1+t^2$; the normal $(-s,-t,1)/\sqrt{1+t^2+s^2}$ gives second fundamental form coefficients $e=0$, $f_0=1/\sqrt{1+t^2+s^2}$, $g_0=0$. HenceThus equality can fail everywhere, although every ruling is an ambient geodesic.

The length differentiation requires a nonconstant base ruling; a regular parametrized surface patch guarantees this. If “image of a smooth map” is read literally without a rank assumption, a degenerate member can fail even to have a first length derivative. For example, in the flat cylinder $\mathbb{R}\times S^1$, the map $\varphi (t,s)=(st,e^{is})$ for $|t|<\delta$, $|s|<4\pi$ has geodesic rulings and a two-dimensional open image: every nonzero-$s$ point has full-rank parametrization, and the sole zero-$s$ image point also has a full-rank preimage at $s=2\pi ,t=0$. Yet $L({\gamma }_s)=2\delta |s|$ is not differentiable at zero. The variation formula therefore applies to nonconstant members, as usual. The curvature inequality still holds on the whole image: regular values are dense by Sard theorem, and both curvatures extend continuously to the remaining points.

A Riemannian isometry $F:(M,g)\to (N,h)$ is a diffeomorphism satisfying $F^{\ast }h=g$. A local isometry is a local diffeomorphism with the same pullback identity; equivalently $d{F}_p$ is an isometric linear isomorphism for every $p$. An isometric immersion into a higher-dimensional manifold does not satisfy this local-diffeomorphism requirement.

Let ${\varphi }_t$ be the local flow of the generating vector field $K$. The identity ${\varphi }_t^{\ast }g=g$ differentiates to ${\mathcal{L}}_{K}g=0$, where $\mathcal{L}$ is the Lie derivative of a tensor field. For arbitrary vector fields $X,Y$, metric compatibility and the torsion-free connection giveThis proves the Killing equation. Conversely, a smooth Killing field has a local flow, andSince ${\varphi }_0$ is the identity, ${\varphi }_t^{\ast }g=g$ wherever the flow is defined. Its inverse is ${\varphi }_{-t}$ locally, so the local flow consists of local isometries. Neither direction assumes that $K$ is complete.

Put $H(X,Y)={\nabla }_X({\nabla }_{Y}K)-{\nabla }_{{\nabla }_{X}Y}K$ and $B(X,Y,Z)=⟨H(X,Y),Z⟩$. These expressions are tensorial in $X,Y,Z$. Take the covariant derivative in direction $X$ of the Killing equation, subtracting the terms from differentiating $Y$ and $Z$. It follows thatWith the curvature convention $\mathcal{R}$ of Solution 1, the torsion-free connection also givesDefine $C(X,Y,Z)=⟨\mathcal{R}(X,K)Y,Z⟩$. It is skew in $Y,Z$ by skew-adjointness of the Riemann curvature tensor. The first Bianchi identity givesso $C$ has the same antisymmetrization in $X,Y$ as $B$. Consequently $D=B-C$ is symmetric in its first two slots and skew in its last two. These symmetries force it to vanish:We have proved the second covariant derivative of a Killing vector identity, including its sign:The printed identity uses the opposite curvature-sign convention in Killing derivative identities. If $R=-\mathcal{R}$, namely $R(X,Y)Z={\nabla }_Y{\nabla }_{X}Z-{\nabla }_X{\nabla }_{Y}Z+{\nabla }_{[X,Y]}Z$, the same result reads ${\nabla }_{X,Y}^{2}K=R(K,X)Y$, exactly as printed. The convention must change with the sign; under $\mathcal{R}$, writing $\mathcal{R}(K,X)Y$ would be incorrect.

Finally let $H_0=K-\tilde{K}$ have zero value and zero first covariant derivative at $p$. Along any geodesic issuing from $p$, the identity just proved saysThus $H_0$ is a Jacobi field with zero initial value and derivative. Uniqueness for this linear ordinary differential equation implies $H_0=0$ on each such geodesic, hence on a convex normal neighborhood of $p$. Consider the set of points where both $H_0$ and $\nabla H_0$ vanish. It is closed by continuity and open by this same local argument; on the resulting neighborhood $H_0$ vanishes identically, so its derivative does too. It is nonempty, and $M$ is connected, therefore it is all of $M$. This proves $K=\tilde{K}$ everywhere and the general statement that a Killing field is determined by its value and first covariant derivative.

For a Lie group $G$, Left translation on a Lie group and right translation on a Lie group are the diffeomorphisms $L_g(h)=gh$ and $R_g(h)=hg$. Their differentials identify the tangent space at the identity with every tangent space. Translating a basis $e_a$ of the Lie algebra gives global left-invariant vector fields $E_a(g)=(d{L}_g{)}_e{e}_a$ and right-invariant vector fields $F_a(g)=(d{R}_g{)}_e{e}_a$. Left and right translations commute. If $[e_a,e_b]=c^c{}_{ab}{e}_c$, thenThe opposite sign for the right-invariant frame is essential.

Define an affine connection ${\nabla }^{-}$ by declaring the $E_a$ parallel. Explicitly, ${\nabla }_X^{-}(f^a{E}_a)=X(f^a)E_a$. Its curvature form of a connection vanishes, since successive derivatives of the component functions satisfy the vector-field commutator identity. Its torsion tensor isSimilarly, declaring the $F_a$ parallel gives another flat affine connection ${\nabla }^{+}$, with $T^{+}(F_a,F_b)=-[F_a,F_b]$. In a left-invariant frame the latter connection obeysTo see this, express $E_a=F_{{\mathrm{Ad}}_g{e}_a}$. Along $g\exp (t{e}_b)$, differentiation of ${\mathrm{Ad}}_{g\exp (t{e}_b)}{e}_a$ gives ${\mathrm{Ad}}_g[e_b,e_a]$, producing the displayed bracket. Consequently $T^{+}(X,Y)=[X,Y]$ for left-invariant $X,Y$. These are the flat translation connections on a Lie group. They have opposite torsion; if $G$ is Abelian, both torsion tensors vanish and the two connections coincide. The phrase “with torsion” does not mean nonzero torsion for Abelian groups.