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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.

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.

Let ${\gamma }_X(s)$ be the affinely parametrized geodesic of the affine connection with ${\gamma }_X(0)=p$ and ${\dot{\gamma }}_X(0)=X$. For fixed $t$, the curve $\eta (u)={\gamma }_X(tu)$ has initial velocity $tX$. The geodesic equation is homogeneous of degree two in the velocity, so its left-hand side for $\eta$ is $t^2$ times the left-hand side for ${\gamma }_X$, and is zero. Uniqueness of the geodesic initial value problem gives $\eta ={\gamma }_{tX}$.

By the definition of the affine exponential map, rescaling the initial velocity rescales affine parameter:At $t=0$ both sides equal $p$. The statement is made where the geodesics exist; without geodesic completeness, the affine exponential map is defined only on a suitable neighbourhood of zero in $T_{p}M$, not necessarily all of $T_{p}M$. An affine parameter distance need not be a metric length.

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.

There is also an explicit converse using null geodesics. For any physical trajectory, setThis makes its five-dimensional tangent null. The conserved momentum of the cyclic coordinate $t$ becomesThe quantity in parentheses is conserved, because its derivative is $\dot{\mathbf{x}}\cdot (2\mathbf{B}\times \dot{\mathbf{x}})=0$. Hence the $t$ equation, as well as the spatial and $u$ Euler-Lagrange equations, is satisfied. Every trajectory admits a null geodesic lift with $p_u=1$, and every such lift projects to the required trajectory.

Finally, a time-independent gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ is absorbed by $u\mapsto u+2\chi$. The one-form $du-2A_{i}d{x}^i-Udt$ and the five-dimensional metric are unchanged. These gauge transformations of an Eisenhart-Duval lift alter the reduced Lagrangian only by the total derivative $-2d\chi /dt$.