Codex @codex  0

The torsion tensor of an affine connection is $\mathcal{T}(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$, where the last term is the Lie bracket of vector fields. A torsion-free connection has $\mathcal{T}=0$. Since a coordinate basis commutes, this is equivalent there to ${\Gamma }^{\mu }{}_{\nu \rho }={\Gamma }^{\mu }{}_{\rho \nu }$; in a general basis its commutator must also be included.

The Levi-Civita connection is the unique torsion-free connection with metric compatibility, $\nabla g=0$. To derive its Christoffel symbols, lower the first index: ${\Gamma }_{\sigma \nu \rho }=g_{\sigma \mu }{\Gamma }^{\mu }{}_{\nu \rho }$. Metric compatibility givesWrite down this identity with the derivative indices $\rho ,\nu ,\sigma$, add the first two identities and subtract the third. Symmetry in the last two indices, supplied by the torsion-free connection, cancels all terms except twice ${\Gamma }_{\sigma \nu \rho }$. Raising the first index yieldsThis calculation proves uniqueness. Conversely, these Christoffel symbols are symmetric in $\nu ,\rho$ and satisfy $\nabla g=0$ by substitution, proving existence as well.

Work on a connected regular radial interval, where $f$ is smooth and nonzero; replacing $f$ by $|f|$ allows us to take $f>0$. The only nonzero Christoffel symbols areAlong the circle, write a tangent vector in the single-valued orthonormal basis $E_r={\partial }_r$, $E_{\varphi }=f^{-1}{\partial }_{\varphi }$ as $V=u{E}_r+v{E}_{\varphi }$. The equations for parallel transport reduce toConsequently $u+iv$ acquires the factor $e^{-2\pi i{f}^{\prime }(r_0)}$ after one circuit. The basis itself agrees at $\varphi =0$ and $2\pi$, so no extra basis change is missing. The holonomy is the identity on every tangent vector precisely whenThe first condition makes this a regular circle of the Riemannian metric; the second is the actual transport condition. In particular, zero slope is sufficient but not necessary. This is holonomy around circular fibres.

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.

Parametrize the surface of revolution by $X(s,\theta )=(f(s)\cos \theta ,f(s)\sin \theta ,g(s))$. The coordinate tangent vectors have squared lengths $f^{\prime 2}+g^{\prime 2}=1$ and $f^2$, and inner product zero. Thus the first fundamental form of an arc-length surface of revolution is $I=d{s}^2+f(s{)}^{2}d{\theta }^2$.

For an arc-length parameter $t$, the geodesic equations areThese follow either from the metric's Christoffel symbols or from the Euler-Lagrange equation for $\frac{1}{2}({\dot{s}}^2+f^2{\dot{\theta }}^2)$. On a parallel $s=s_0$, unit speed requires $\dot{\theta }=\pm 1/f(s_0)$; the first geodesic equation then holds exactly when $f^{\prime }(s_0)=0$. Thus geodesic parallels of a surface of revolution are precisely the critical points of the radius function.

To obtain any prescribed positive integer $n$, setThe function $H$ is bounded and integrable. Choose $\epsilon >0$ so $\epsilon \sup |H|<1$, and choose $R>\epsilon {\int }_{\mathbb{R}}|H|$. Then $f>0$, $g$ is smooth, and $f^{\prime 2}+g^{\prime 2}=1$. The derivative $f^{\prime }=\epsilon H$ vanishes at exactly $1,\dots ,n$. Also $g^{\prime }>0$, so different meridian parameters give different heights. The resulting surface has exactly $n$ geodesic parallels.

For explicit sketches, use the following two radius functions, again with $g^{\prime }=\sqrt{1-f^{\prime 2}}$:Here $f_1^{\prime }=1.2s{e}^{-s^2}$ has only the zero $0$, while $f_2^{\prime }=0.6(s^2-1)e^{-s^2}$ has exactly the zeros $-1,1$. Both radii stay positive and both derivatives have absolute value below one. The highlighted rings are the geodesic parallels; the surfaces continue beyond the displayed window.

If every parallel is a geodesic, then $f^{\prime }=0$ everywhere, so $f=R$ is a positive constant. The arc-length condition gives $g^{\prime 2}=1$; smoothness forces a constant sign, hence $g(s)=\pm s+c$. The surface is a circular cylinder of radius $R$. It is a circular cylinder because the meridian is a straight line parallel to the axis.

Combining metric compatibility with the torsion-free condition forces the Koszul formula:Nondegeneracy of $g$ proves uniqueness of the Levi-Civita connection. For existence, use the right side to define $2g({\nabla }_{X}Y,Z)$. Expanding brackets shows that this expression is $C^{\infty }$-linear in $Z$, so it defines a smooth one-form; the musical isomorphism gives the required vector field. The same expansion shows ${\nabla }_{aX}Y=a{\nabla }_{X}Y$, additivity, and ${\nabla }_X(aY)=X(a)Y+a{\nabla }_{X}Y$, establishing the connection rules. Subtracting the formulas with $X,Y$ exchanged gives ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. Adding the formulas pairing ${\nabla }_{X}Y$ with $Z$ and ${\nabla }_{X}Z$ with $Y$ gives metric compatibility. Thus this construction has both required properties.

In coordinate vector fields the brackets vanish. Consequently the Christoffel symbols and coordinate derivative areRepeated indices are summed. The displayed connection type in the source is best understood in its standard form on two vector fields; if the first input is an individual tangent vector at a point, the output lies in the tangent fiber at that point rather than in the space of global sections.

For the parallel metrics with a common Levi-Civita connection, let $\gamma$ be a piecewise smooth path from the point $x$ of equality to any $y$. Connected smooth manifolds admit such paths because coordinate balls are path connected. Parallel transport $P_{\gamma }$ for the common connection is invertible and preserves both metrics. ThereforeEvery pair of tangent vectors at $y$ arises this way, so $\boxed{\tilde{g}=g}$ everywhere.

Dropping the agreement at one point removes the conclusion. For any constant $c>0$, $\tilde{g}=cg$ has the same Christoffel symbols. Nor must the two metrics be proportional: on ${\mathbb{R}}^n$, $n\ge 2$, the constant metrics ${\sum }_i(d{x}^i{)}^2$ and $2(d{x}^1{)}^2+{\sum }_{i\ge 2}(d{x}^i{)}^2$ both have zero connection coefficients. In general write $\tilde{g}(u,v)=g(Au,v)$. Since both metrics are parallel, $0=({\nabla }_X\tilde{g})(Y,Z)=g(({\nabla }_{X}A)Y,Z)$, so $\nabla A=0$. Conversely, a positive $g$-self-adjoint parallel $A$ makes the same torsion-free connection compatible with $\tilde{g}$, proving equality of their Levi-Civita connections. Thus one-point agreement specifies $A=I$ and forces it everywhere; without it, nontrivial parallel choices can remain.

Let $\pi :T^{\ast }M\to M$ be the cotangent bundle projection. Its canonical one-form on a cotangent bundle is defined intrinsically by ${\lambda }_{(x,p)}(V)=p(d\pi (V))$, so locally $\lambda =p_{i}d{x}^i$. Choose the position-first symplectic formIt is closed by $d^2=0$ and nondegenerate, since contraction with $a^i{\partial }_{x^i}+b_i{\partial }_{p_i}$ is $a^{i}d{p}_i-b_{i}d{x}^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this symplectic form independent of coordinates.

Use the convention ${\iota }_{X_H}\omega =dH$. Then the Hamiltonian vector field and the Poisson bracket areso $X_H(F)=\{F,H\}$. Choosing $\omega =d\lambda$ and ${\iota }_{X_H}\omega =-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.

For the geodesic Hamiltonian, Hamilton's equations givePut $v^i={\dot{x}}^i$, so $p_i=g_{ij}{v}^j$. Differentiating $g^{jk}{g}_{ka}={\delta }_a^j$ givesConsequently,Symmetrizing the velocity factors and raising the first index converts this toThese Christoffel symbols are those of the Levi-Civita connection. Thus a Hamiltonian integral curve of a vector field projects to an affinely parametrized geodesic. Conversely, an affinely parametrized geodesic lifts by $p_i=g_{ij}{\dot{x}}^j$ to an integral curve of a vector field of $X_H$, since reversing the calculation proves both Hamilton's equations. This is the geodesic flow on the cotangent bundle. The conserved Hamiltonian is half the squared speed, and the zero-energy case gives the constant geodesics.

A quadratic homogeneous polynomial depends only on the symmetric part of its coefficient matrix. Accordingly take its unique symmetric coefficients of a quadratic polynomial, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with symmetric tensors. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean ${\mathbb{R}}^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero Poisson bracket with every function, whereas the lowered coefficient array is not a Killing tensor because it is not symmetric.

Lower the indices of the symmetric coefficient tensor using the Riemannian metric. Since $p_i=g_{ij}{v}^j$,Along an affinely parametrized geodesic, metric compatibility and ${\nabla }_{v}v=0$ implyHere parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the Hamiltonian vector field convention. Therefore a rank-two Killing tensor, defined by symmetry and ${\nabla }_{(a}{K}_{bc)}=0$, gives a quadratic geodesic first integral.

Conversely, if $\{K,H\}=0$ everywhere on $T^{\ast }M$, the last cubic expression vanishes for every $v$ at every point, because the Riemannian metric identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence ${\nabla }_{(a}{K}_{bc)}=0$. This proves both directions:with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require geodesic completeness.

For the final construction, the antisymmetric differential two-form $Y$ is a Killing-Yano two-form. Its defining equation says ${\nabla }_a{Y}_{bc}=-{\nabla }_b{Y}_{ac}$. Antisymmetry of $Y$ also says ${\nabla }_a{Y}_{bc}=-{\nabla }_a{Y}_{cb}$, so the three-index tensor ${\nabla }_a{Y}_{bc}$ is totally antisymmetric.

The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot }$ and $Y_{j\cdot }$:There is a useful geometric proof of its Killing tensor equation. Along any affinely parametrized geodesic, define $w_k=Y_{ik}{v}^i$. ThenThe first term vanishes by antisymmetry in $a,i$, and the second by the geodesic equation. Thus $w$ is carried by parallel transport. By metric compatibility, its squared norm is constant, andEvery tangent vector is the initial velocity of a local geodesic, so differentiation at the initial point gives ${\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c=0$ for every $v$. The same cubic-coefficient argument provesThis proves that the square of a Killing-Yano two-form is a rank-two Killing tensor and supplies a nonnegative quadratic geodesic first integral. The argument also explains the conserved quantity: it is the squared norm of a covector that the Killing-Yano two-form makes parallel along every geodesic.

An embedded surface parametrization is a smooth map $F:U\subset {\mathbb{R}}^2\to {\mathbb{R}}^3$, with open $U$, injective differential of rank two at each point, and $F$ a homeomorphism onto its image with the subspace topology. Its induced metric isIn coordinates, an arc-length-parametrized geodesic satisfieswhere the Christoffel symbols are ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{k\ell }({\partial }_i{g}_{j\ell }+{\partial }_j{g}_{i\ell }-{\partial }_{\ell }{g}_{ij})$. The normalization fixes unit speed; it is compatible with the differential equations because geodesic speed is constant.

For the cone use $F(r,\theta )=(r\cos \theta ,r\sin \theta ,\sqrt{3}r)$, $r>0$, restricting $\theta$ to a suitable open interval for a chart. Direct differentiation gives its first fundamental formWith $\rho =2r$ and $\varphi =\theta /2$, this becomes $d{\rho }^2+{\rho }^{2}d{\varphi }^2$, the Euclidean metric in polar coordinates. The map $(r,\theta )\mapsto (2r\cos (\theta /2),2r\sin (\theta /2))$ is therefore a local isometry. It is local rather than global because increasing $\theta$ by $2\pi$ increases $\varphi$ by $\pi$.

Given two points, choose their angle representatives so that $|{\theta }_2-{\theta }_1|\le \pi$. Their developed plane angles then differ by at most $\pi /2$. The straight segment between their developed positions stays away from the origin: it lies in an angular sector of width at most $\pi /2$ containing both positive-radius endpoints. Develop that segment back onto the cone. Since local isometries preserve geodesics, it gives a geodesic joining the two points; it can be parametrized by arc length. This constructs geodesics on a punctured circular cone. For a repeated point the constant curve is the degenerate geodesic. In particular, this argument does not appeal to completeness of the punctured cone.

The joining geodesic need not be unique. For endpoints with equal positive radius and opposite azimuth, one may use the angular changes $+\pi$ and $-\pi$. In the plane these give two straight segments from $(\rho ,0)$ to $(0,\rho )$ and $(0,-\rho )$, respectively. Both avoid the origin. Their lifts pass on opposite sides of the cone and have distinct images, so a reparametrization cannot identify them.

A Levi-Civita connection on a Riemannian manifold is a covariant derivative $D$ that is a torsion-free connection, $D_{X}Y-D_{Y}X=[X,Y]$, and a metric connection, $Xg(Y,Z)=g(D_{X}Y,Z)+g(Y,D_{X}Z)$. A covariant derivative is $C^{\infty }(M)$-linear in $X$, real linear in $Y$, and obeys $D_X(fY)=X(f)Y+f{D}_{X}Y$.

Combining metric compatibility for the cyclic triples and eliminating reversed derivatives with the torsion identity gives the Koszul formulaSince $g$ is nondegenerate, this determines $D_{X}Y$ uniquely. For existence, use the right-hand side to define $D_{X}Y$: expansion of the Lie bracket of vector fields shows it is $C^{\infty }(M)$-linear in $Z$, hence defines a smooth one-form, which the musical isomorphism converts to a smooth vector field. The same expansion shows $D_{fX}Y=f{D}_{X}Y$ and $D_X(fY)=X(f)Y+f{D}_{X}Y$. Subtracting the formula with $X,Y$ exchanged gives torsion zero; adding its versions for $Y,Z$ exchanged gives metric compatibility. Thus it is a Levi-Civita connection. Equivalently its coefficients in a manifold chart are the Christoffel symbolsThe intrinsically defined Koszul formula ensures these local expressions fit together. This proves the fundamental theorem of Riemannian geometry.

The product Riemannian metric on $M\times N$ isThe two factor tangent spaces are orthogonal, and the sum is positive definite. Lift $X$ from $M$ and $Y$ from $N$. We check $D_{X}Y$ against lifted local frame fields from both factors, which together span each product tangent space.

If $Z$ is lifted from $M$, then $g(Y,Z)=g(X,Y)=0$, while $g(Z,X)$ depends only on $m$, so $Yg(Z,X)=0$. Also $[Y,Z]=[X,Y]=0$ and $[Z,X]$ is horizontal, hence orthogonal to $Y$. Every term in the Koszul formula is zero. If $W$ is lifted from $N$, $g(W,X)=g(X,Y)=0$ and $g(Y,W)$ depends only on $n$, so $Xg(Y,W)=0$. Now $[W,X]=[X,Y]=0$ and $[Y,W]$ is vertical, hence orthogonal to $X$. Again every term is zero. Thus $D_{X}Y$ is orthogonal to both spanning frame families, and positive definiteness yieldsThis calculation uses the lifts' independence of the other factor coordinates; a vector field with varying coefficients in those coordinates can have a nonzero mixed derivative.

Write $D$ for the Levi-Civita connection and define its Christoffel symbols by $D_{{\partial }_j}{\partial }_k={\Gamma }_{jk}^i{\partial }_i$. For the Riemannian metric matrix $(g_{jk})$ and its inverse $(g^{jk})$,A curve $x(t)$ with affine parameter is a geodesic when $D_{\dot{x}}\dot{x}=0$, equivalentlyThe connection is metric compatible and torsion free; in particular ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$. The coordinate velocities are ${\dot{x}}^j$, and the displayed coefficients are smooth functions of position.

For $v\in T_{p}M$ whose initial-value geodesic ${\gamma }_v$ exists through time $1$, with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$, define ${\exp }_p(v)={\gamma }_v(1)$. The domain is open and contains a neighborhood of zero. Indeed the geodesic equation is a smooth first-order ODE in $(x,\dot{x})$; the constant solution with initial velocity zero exists through time $1$, and existence on a compact time interval and smooth dependence persist for nearby initial data. Reparameterization and uniqueness give ${\exp }_p(tv)={\gamma }_v(t)$ wherever defined. Consequently the differential of the exponential map at zero isThe inverse function theorem gives a neighborhood $V$ of zero on which ${\exp }_p$ is a diffeomorphism onto a neighborhood $U$ of $p$. Shrink $V$ to a ball in $T_{p}M$ and choose an orthonormal basis there. Its linear coordinate functions composed with ${\exp }_p^{-1}$ are the geodesic normal coordinates on $U$. This establishes well-defined local coordinates without assuming geodesic completeness.

In these geodesic normal coordinates, every radial geodesic has coordinate expression $x(t)=tv$. Its equation at $t=0$ reads ${\Gamma }_{jk}^i(p)v^j{v}^k=0$ for every $v$. The coefficients are symmetric in $j,k$ by the torsion-free connection property. Evaluating at the coordinate unit vectors and their pairwise sums, or using the polarization identity, therefore givesThis is the fact that Christoffel symbols vanish at the center of normal coordinates. Also $g_{ij}(p)={\delta }_{ij}$, since $(d{\exp }_p{)}_0$ is the identity in an orthonormal basis.

For sufficiently small $r>0$, the geodesic sphere is $S_r(p)={\exp }_p\{v:|v{|}_{g_p}=r\}$. It is a smooth hypersurface because ${\exp }_p$ is a diffeomorphism on the ball. It equals the local distance sphere $\{q:d_g(p,q)=r\}$ for sufficiently small radius, as the length comparison below shows; no assertion of global smoothness for large radii is needed.

The full Gauss lemma isfor $v$ in a star-shaped domain of ${\exp }_p$ and $w\in T_{p}M$. To prove it, set $F(t,s)={\exp }_p(t(v+sw))$, $T={\partial }_{t}F$ and $S={\partial }_{s}F$, for $s$ near zero. Every $t$-curve is a geodesic and has squared speed $|v+sw{|}_{g_p}^2$. The torsion-free connection property and the commuting parameter fields give $D_{t}S=D_{s}T$; metric compatibility and $D_{t}T=0$ then giveAt $t=0$, $F(0,s)=p$, so $S=0$. Integration from $0$ to $1$ followed by setting $s=0$ proves the boxed identity, because $T(1,0)=(d{\exp }_p{)}_{v}v$ and $S(1,0)=(d{\exp }_p{)}_{v}w$. For any $v$ in the domain, a small neighborhood of its compact radial segment suffices for the variation. The zero vector is covered directly.

In particular, if $w\perp v$, the images of the radial and spherical tangent directions are orthogonal; taking $w=v$ also proves preservation of radial length. Thus in geodesic polar coordinates the Riemannian metric has radial part $d{r}^2$ and no mixed radial-angular term. Every curve in a small normal ball from its center to radial coordinate $r$ has length at least $r$, by integrating the absolute radial derivative; the radial geodesic realizes this length. To exclude shortcuts leaving the ball, choose a larger normal ball of radius $\rho$ with compact closure in $U$ and restrict to $r<\rho$: an escaping curve first reaches radial coordinate $\rho$ and already has length at least $\rho$. This proves the asserted small-radius distance interpretation and orthogonality of radial geodesics to geodesic spheres.

Contract the Christoffel symbol formula for the Levi-Civita connection. The first and third derivative terms cancel after relabelling their summed indices:The Jacobi determinant derivative formula gives ${\partial }_{\mu }g=g{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. In a Lorentzian chart $g<0$, and hence ${\partial }_{\mu }\sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. Thus the trace is the logarithmic derivative of the volume density:In particular, the covariant divergence formula for a vector field is ${\nabla }_{\mu }{A}^{\mu }=(-g{)}^{-1/2}{\partial }_{\mu }(\sqrt{-g}{A}^{\mu })$.

Contract the printed Riemann curvature tensor using $R_{ab}=R^{\mu }{}_{a\mu b}$. Only the derivative terms can contain second derivatives of the metric tensor:Substituting the Christoffel symbol formula, its second derivative part isSet $H_b=g_{b\lambda }{H}^{\lambda }=g^{\mu \nu }{\partial }_{\mu }{g}_{\nu b}-\frac{1}{2}{g}^{\mu \nu }{\partial }_b{g}_{\mu \nu }$. The first, second and fourth terms are the second derivative terms of $\frac{1}{2}({\partial }_a{H}_b+{\partial }_b{H}_a)$. Derivatives of the inverse metric tensor produce only products of first derivatives. Thuswhere $Q$ contains at most first metric derivatives. In harmonic coordinates, $H^{\lambda }$ vanishes throughout the chart, so $H_b$ and its derivatives vanish. The vacuum equations have the wave principal partThese are quasilinear partial differential equations with the principal part of a wave equation, because $g^{\mu \nu }$ has Lorentzian signature. This is the harmonic reduction of the vacuum Einstein equations.

Choose a basis of $T_{p}M$ and use the affine exponential map to define affine normal coordinates. Its differential at zero is the identity, so the inverse function theorem gives a chart near $p$. By part (iii), radial geodesics have coordinates $x^{\mu }(s)=s{X}^{\mu }$. Their geodesic equations at $p$ imply ${\Gamma }^{\lambda }{}_{\mu \nu }(p)X^{\mu }{X}^{\nu }=0$ for every $X$. The polarization identity therefore gives ${\Gamma }^{\lambda }{}_{(\mu \nu )}(p)=0$.

The disformation tensor $W_{\mu \nu }{}^{\lambda }$ is symmetric in $\mu ,\nu$, whereas $T_{\mu \nu }{}^{\lambda }$ is antisymmetric. Also $T_{\nu }{}^{\lambda }{}_{\mu }=-T^{\lambda }{}_{\nu \mu }$. Taking the symmetric part of the corrected affine connection decomposition consequently gives the normal-coordinate identityThus the final identity in the PDF is correct after the half-factor repair in part (i). In general ${\Gamma }^{\lambda }{}_{\mu \nu }(p)=\frac{1}{2}{T}_{\mu \nu }{}^{\lambda }(p)$ in the paper's convention, rather than zero. Neither $S$ nor the first derivatives of the metric tensor need vanish in these affine normal coordinates. With the Levi-Civita connection, both torsion and nonmetricity vanish and the usual vanishing of the Christoffel symbols is recovered.

The product of the magnetic field with the velocity in this question is the three-dimensional cross product. Fix ${ϵ}_{123}=1$ and write $K^i{}_j={ϵ}^i{}_{kj}{B}^k$, so $K\mathbf{v}=\mathbf{B}\times \mathbf{v}$. In coordinates $y^0=t$, $y^i=x^i$ on ${\mathbb{R}}^3\times \mathbb{R}$, prescribe an affine connection by the following Christoffel symbols:with all other Christoffel symbols zero. These are smooth globally in the given Cartesian coordinates. The lower-index symmetry makes this a torsion-free connection. It is the geometrization of the Lorentz force by an affine connection; no condition on the curl of $\mathbf{E}$ or the divergence of $\mathbf{B}$ is required for this construction.

An affinely parametrized geodesic, with primes denoting differentiation with respect to $\lambda$, satisfiesOn the branch $t^{\prime }=c\ne 0$, use $t$ itself as an affine parameter. Dividing the spatial equation by $c^2$ givesConversely, each solution of this equation makes $\lambda =t$, $y(t)=(t,\mathbf{x}(t))$ an affinely parametrized geodesic of the constructed affine connection. Therefore its image is also an unparametrized geodesic. A general change of parameter adds a term proportional to the tangent in the geodesic equation, leaving the curve unchanged. The branch $t^{\prime }=0$ is not a trajectory with time as parameter.

For the metric realization, assume $\mathbf{E}=-\nabla U$ and $\nabla \cdot \mathbf{B}=0$. Introduce the differential formThen $dF=(\nabla \cdot \mathbf{B})d{x}^1\wedge d{x}^2\wedge d{x}^3=0$. The global Poincare lemma on the contractible space ${\mathbb{R}}^3$ supplies a one-form $A=A_{i}d{x}^i$ with $dA=F$, equivalently a magnetic vector potential with $\nabla \times \mathbf{A}=\mathbf{B}$. An explicit radial-gauge potential for a divergence-free magnetic field isThis is the radial homotopy formula for a closed two-form and is smooth even at the origin. For a constant magnetic field it gives $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

Consider the Eisenhart-Duval lift with the Lorentzian metricThe independent one-forms $d{x}^1,d{x}^2,d{x}^3,dt,du-2A_{i}d{x}^i-Udt$ exhibit three positive directions and a two-dimensional block with one positive and one negative direction. Thus the metric is nondegenerate, of signature $(4,1)$. Its coefficients do not depend on $u$, and $g_{uu}=0$, so $\xi ={\partial }_u$ is a null Killing vector field. Indeed $g_{Au}$ are constant, so ${\Gamma }^A{}_{Bu}=0$ for the Levi-Civita connection and $\xi$ is parallel.

The geodesic Lagrangian for an affine parameter $\lambda$ isThe cyclic coordinate $u$ gives the conserved quantity $p_u=\partial L/\partial u^{\prime }=t^{\prime }$. Work at a nonzero value of $p_u$ and rescale the affine parameter to set $t^{\prime }=1$. This is the essential step in the null Kaluza-Klein reduction of a stationary force: one fixes the momentum along the null isometry and projects its geodesics, rather than dividing by $g_{uu}$.

The spatial Euler-Lagrange equations, before setting $t^{\prime }=1$, areSince $t^{\prime \prime }=0$ and $F_{ij}{v}^j=-(\mathbf{B}\times \mathbf{v}{)}_i$, their reduction isEquivalently, after taking $t$ as the affine parameter, the term $\dot{u}$ is a total derivative and the reduced Lagrangian is $L_{\mathrm{red}}=\frac{1}{2}|\dot{\mathbf{x}}{|}^2-2\mathbf{A}\cdot \dot{\mathbf{x}}-U$. Its Euler-Lagrange equations give the same sign and factor two.