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Under parallel transport, the gyroscope spin obeys $u^b{\nabla }_b{s}^a=0$, and its freely falling circular orbit is a geodesic, $u^b{\nabla }_b{u}^a=0$. Metric compatibility therefore givesThus $s^a{u}_a$ is constant along the orbit.

The first fundamental form of an embedded surface $S\subseteq {\mathbb{R}}^3$ is the inner product induced on each tangent plane:For a tangent vector field $V$ along a curve $\alpha$, its surface covariant derivative isthe tangential projection of its ordinary derivative. A curve is a geodesic when

A local isometry preserves the first fundamental form, so its Christoffel symbols and Levi--Civita covariant derivatives satisfyTaking $V={\alpha }^{\prime }$ shows that if $\alpha$ is geodesic, then $\varphi \circ \alpha$ is geodesic.

The converse is false. The dilationmaps every affinely parametrized straight-line geodesic to another such geodesic, butso it does not preserve the first fundamental form. This is the geodesic-preserving homothety that is not a local isometry.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

SetThe triangle inequality makes $f$ Lipschitz continuous and hence a continuous function. Since $\gamma$ is an isometric embedding,so $f(t)\to \infty$ as $|t|\to \infty$. The function is therefore coercive, and the extreme value theorem on a sufficiently large compact interval gives a minimizing parameter.

Suppose $t_1,t_2$ both minimize $f$, put $p=\gamma (t_1)$ and $q=\gamma (t_2)$, and letThe restriction of $\gamma$ between the two parameters is a geodesic from $p$ to $q$. Let $m$ be its midpoint. In the geodesic triangle with vertices $x,p,q$, the thin geodesic triangle condition gives a point $y$ on one of the other two sides with $d(m,y)\le \delta$. By symmetry suppose $y\in [x,p]$. ThensoBut $m$ lies on $\gamma (\mathbb{R})$ and $p$ is a closest point, so $R\le d(x,m)$. Therefore $L\le 4\delta$, which is stronger than the requiredThis is the closest point on a geodesic line in a hyperbolic metric space estimate.

The vector $T^a$ is the tangent to a reference member of a one-parameter family of affinely parametrized geodesics. The Jacobi field $Y^a$ is the infinitesimal connecting vector from that geodesic to a neighboring one at equal parameter. The geodesic deviation equation states that curvature determines their relative acceleration.

For $v\in T_{p}S$, let ${\gamma }_v$ be the maximal geodesic satisfyingThe exponential map ison the domain of the exponential mapThis is an open, star-shaped neighbourhood of zero. In local coordinates the geodesic equation is a smooth ordinary differential equation, so smooth dependence on its initial position and velocity proves that $(p,v)\mapsto {\exp }_p(v)$ is smooth on its domain.

For both requested phenomena consider the embedded surfacewith $p=(1,0,0)$. Its geodesics are Euclidean straight lines for as long as they remain in $S$. The initial vector $v=(-2,0,0)$ would givewhich reaches the missing origin at $t=1/2$. Thus $v\notin {\mathcal{D}}_p$, and the domain is not all of $T_{p}S$.

The point $q=(-1,0,0)$ is not in the image of ${\exp }_p$. Any geodesic from $p$ to $q$ would have to be their unique Euclidean straight line, which passes through the omitted origin. Hence this exponential map is also not surjective, as summarized by exponential map of the punctured plane.

For an affine parameter $\lambda$, the geodesic equation and the connection coefficients from part (a) givewhere dots denote derivatives with respect to $\lambda$.

On a line of constant $\varphi$, one has $\dot{\varphi }=0$. The equations reduce to $\ddot{r}=0$, so $r$ itself can be chosen as an affine parameter. Every such radial line is therefore a geodesic.

For a circle $r=r_0>0$, the first equation instead requiresBoth hyperbolic factors are positive when $r_0>0$, so $\dot{\varphi }=0$. This gives only a constant point, not a parametrized circle. Thus no circle of constant positive $r$ is a geodesic, in agreement with the coordinate geodesics of the hyperbolic polar metric.

An affinely parameterized geodesic $x^{\mu }(\lambda )$ satisfiesFor the conformally flat metric $g_{\mu \nu }={\phi }^2{\eta }_{\mu \nu }$, its Christoffel symbols areLet $u^{\mu }=d{x}^{\mu }/d\lambda$. Along a null geodesic, ${\eta }_{\alpha \beta }{u}^{\alpha }{u}^{\beta }=0$, so the equation reduces toThe acceleration is parallel to the tangent. A change of parameter removes it, leaving a straight null line of the Minkowski metric. Thus conformal rescaling changes the affine parameter but not the unparametrized light ray.

Consequently Nordstrom theory of gravitation predictsfor light passing a star. This contradicts the observed nonzero deflection and the successful general-relativistic value $4GM/(b{c}^2)$.

Because $S_1$ and $S_2$ are isometric,The two induced orientations on their common boundary $\gamma$ are opposite. Ifis computed with the boundary orientation from $S_1$, then its value for $S_2$ is $-B$. Applying the Gauss-Bonnet theorem to the two halves givesThe equalities above imply $B=0$. This is the isometric halves give zero total boundary geodesic curvature argument.

Every ambient isometry ${\varphi }_{x,y}$ preserves the magnitude of geodesic curvature and sends $x$ to $y$. Hence $|k_g|$ is constant along the connected curve $\gamma$ by the transitive curve symmetry makes geodesic-curvature magnitude constant. If that constant were positive, continuity would force the signed $k_g$ to have one fixed sign, making its integral nonzero. Therefore $k_g\equiv 0$. A curve has zero geodesic curvature exactly when it is a geodesic, so $\gamma$ is geodesic.

For local coordinates $x^1,\dots ,x^n$, a curve is an affinely parametrized geodesic exactly when it satisfies the geodesic equationwhere the Christoffel symbols of the Levi-Civita connection are

A horizontal lift of $\gamma$ from $p$ is a curve $\tilde{\gamma }$ in $E$ projecting to $\gamma$, starting at $p$, and tangent to the horizontal distribution of the connection. In a local frame write $\tilde{\gamma }(t)=(\gamma (t),v(t))$. Horizontality is the linear ordinary differential equationwhose initial-value theorem gives local existence and uniqueness; successive trivializations continue the lift.

A geodesic satisfies ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$, andfor the geodesic with initial velocity $v$. Since $d({\exp }_p{)}_0=1$, the inverse function theorem makes ${\exp }_p$ a diffeomorphism near zero; its inverse gives normal coordinates. A geodesic sphere is ${\exp }_p(\{v:|v|=r\})$ inside such a normal neighborhood.

The Gauss lemma statesFor the variation ${\gamma }_s(t)={\exp }_p(t(v+sw))$, let $J={\partial }_s{\gamma }_s{|}_{s=0}$. The coordinate vector fields commute, so metric compatibility and constant geodesic speed giveSince $J(0)=0$, evaluation at $t=1$ proves the formula. In particular radial and spherical directions are orthogonal.

Affinely parametrized geodesics are the critical curves of the geodesic LagrangianFor $w=1+r^2$ and $w^{\prime }=2r$, the nonzero Christoffel symbols, up to symmetry in the lower indices, areThey follow either from the Euler-Lagrange equations or directly from the Levi-Civita connection formula.

A unit-speed surface curve $\alpha$ is a geodesic when its acceleration is normal to the surface, equivalently when its covariant acceleration vanishes. In coordinates $u^1,u^2$, the Christoffel symbols of the Levi-Civita connection areThe coordinate functions of a geodesic satisfy the geodesic equation

An orientation of $S$ chooses a unit normal at every point. The Gauss map isFor the unit tangent $T=\dot{\alpha }$, the surface covariant derivative is the tangential projectionThe signed geodesic curvature isBecause $|T|=1$, $D_{s}T$ is perpendicular to $T$ and henceConsequently ${\kappa }_g=0$ means $D_{s}T=0$: the tangent is parallel along $\alpha$, so $\alpha$ is a geodesic.

For a connected Riemannian manifold $(M,g)$, its Riemannian distance iswhere the infimum is over the piecewise smooth curves from $p$ to $q$. Connectedness of a smooth manifold implies path connectedness, so this set of curves is nonempty.

The Gauss lemma says that the differential of ${\exp }_p$ preserves the radial inner product: for $v,w\in T_{p}M$,Consequently radial geodesics from $p$ are orthogonal to the images of tangent vectors to spheres centred at the origin in $T_{p}M$. In a sufficiently small normal neighbourhood of $p$, this impliesevery competing curve has length at least the total variation of its radial coordinate, and the radial geodesic has that length.

The axioms $d_g(p,q)\ge 0$, symmetry, and the triangle inequality follow directly from length and concatenation. Certainly $d_g(p,p)=0$. If $q\ne p$, choose a normal ball $B_g(p,r)$ that does not contain $q$. Every curve from $p$ to $q$ first meets its boundary, and its initial part has length at least $r$ by the Gauss lemma. Hence $d_g(p,q)\ge r>0$. Thus $d_g(p,q)=0$ if and only if $p=q$.

The metric $g$ is geodesically complete when every maximal affinely parametrized geodesic is defined on all of $\mathbb{R}$; equivalently, ${\exp }_p$ is defined on every $T_{p}M$ for every $p\in M$.

The Hopf-Rinow theorem says that for a connected Riemannian manifold, the following are equivalent: geodesic completeness; completeness of the Riemannian distance $d_g$; compactness of every closed bounded subset; and the existence, between every two points, of a length-minimizing geodesic. It is enough in the exponential-map formulation that ${\exp }_p$ be defined on all of $T_{p}M$ for one point $p$.

Let $\gamma :[0,\ell ]\to M$ be a unit-speed geodesic, let $F(s,t)$ be a smooth variation with fixed endpoints, and letbe its variation vector field. Then $V(0)=V(\ell )=0$. If $V^{\perp }=V-⟨V,T⟩T$ is its component normal to $\gamma$, the second variation of Riemannian arc length isHere $D_t={\nabla }_T$ is the covariant derivative along $\gamma$, $R$ is the Riemann curvature tensor, and $I$ is the Riemannian index form. Fixed endpoints remove the boundary term. The normal projection removes a tangential change of parametrization, which does not change length to second order.

The Bonnet-Myers theorem states that if a complete connected $n$-dimensional Riemannian manifold satisfiesfor some $k>0$, thenIn particular, $M$ is compact and has finite fundamental group.

By the Hopf-Rinow theorem, points $p,q\in M$ are joined by a unit-speed length-minimizing geodesic $\gamma :[0,\ell ]\to M$. Choose a parallel orthonormal frame $E_1,\dots ,E_{n-1}$ normal to $T=\dot{\gamma }$ and setThe endpoint-vanishing fields $V_i$ arise from fixed-endpoint variations. Since $\gamma$ minimizes length, its Riemannian index form is nonnegative on each $V_i$. Summing the second variation of Riemannian arc length givesUsing the Ricci curvature bound and integrating ${\sin }^2$ and ${\cos }^2$ yieldsso $\ell \le \pi /\sqrt{k}$. Taking the supremum over $p,q$ proves the diameter bound. Hopf-Rinow now makes the closed bounded space $M$ compact. Finally, the same bound applies to the complete universal cover; a compact universal cover has finite fibres over $M$, so ${\pi }_1(M)$ is finite.

A line in a Riemannian manifold is a unit-speed geodesic $\gamma :\mathbb{R}\to M$ that minimizes globally:for all $s,t\in \mathbb{R}$. A connected noncompact manifold is disconnected at infinity if some compact set $K$ has a complement with at least two unbounded connected components.

Choose points $p_j$ and $q_j$ in two such components withThe Hopf-Rinow theorem supplies a length-minimizing geodesic ${\gamma }_j$ from $p_j$ to $q_j$. Its image must meet $K$, since otherwise it would connect the two different components of $M\setminus K$. Reparametrize so that ${\gamma }_j(0)=x_j\in K$. After taking a subsequence, compactness gives $x_j\to x\in K$ and the unit tangent vectors ${\dot{\gamma }}_j(0)$ converge to some unit $v\in T_{x}M$.

Both endpoint parameters tend to infinity because their distances from $K$ do. Smooth dependence of geodesics on initial data therefore makes ${\gamma }_j$ converge on every compact parameter interval to the complete geodesicEvery finite segment of every ${\gamma }_j$ minimizes length. Passing to the limit gives $d_g(\gamma (s),\gamma (t))=|s-t|$, so $\gamma$ is a line.

A Killing vector field $K$ obeys ${\mathcal{L}}_{K}g=0$, or equivalently ${\nabla }_{(a}{K}_{b)}=0$. Along an affinely parametrized geodesic with tangent $U$,The second term vanishes by the geodesic equation, while the first contracts the symmetric tensor $U^a{U}^b$ with the antisymmetric part selected by the Killing equation. Thus $g(K,U)$ is a geodesic conserved quantity from a Killing vector. In Minkowski spacetime, translational Killing fields give conserved energy-momentum and rotational or boost Killing fields give the corresponding angular-momentum and boost charges; the same construction works for any spacetime isometry.