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For $v\in T_{p}S$ sufficiently small, let ${\gamma }_v$ be the geodesic with ${\gamma }_v(0)=p$, ${\dot{\gamma }}_v(0)=v$, and define ${\exp }_p(v)={\gamma }_v(1)$. The domain consists more generally of initial velocities whose geodesic exists through time one. Uniqueness and affine rescaling give ${\exp }_p(tv)={\gamma }_v(t)$ near zero, soThus the differential is the identity under the natural identification of tangent spaces, and the inverse function theorem makes the exponential map a local diffeomorphism near zero.

A punctured plane has geodesics reaching its missing point in finite time, so its exponential map need not be global; adding the point remedies this example. But an extension cannot always remedy the failure. On the punctured circular cone $S=\{(r\cos \theta ,r\sin \theta ,r):r>0\}$, radial generators are geodesics and reach the missing point in finite intrinsic length. Any globally defined extension would have to include their limiting point $(0,0,0)$. The cone's tangent planes have different limits as $\theta$ varies; no smooth embedded surface through that point can contain this punctured circular cone. Hence no such extension exists for this $S$.

For an oriented surface and a closed topological disc $D$ with smooth positively oriented boundary, the Gauss-Bonnet theorem with boundary saysHere $k_g$ is signed geodesic curvature with the disc on the left. A smooth boundary has no corner terms; piecewise smooth boundaries require exterior-angle terms.

For the flat disc in the problem this gives ${\int }_{\partial D}{k}_{g}ds=2\pi$. Conditions (i) and (ii) mean that the proposed replacement disc has the same boundary and agrees with the original smooth surface from the outside. Smoothness therefore gives the same tangent planes, metric derivatives and boundary geodesic curvature on the replacement, with consistent orientation. Applying Gauss-Bonnet to that disc forcesThus these gluing conditions fix the total curvature of any smooth replacement disc, regardless of its detailed interior shape.

The nonnegative continuous Gaussian curvature on the compact replacement disc, together with its zero integral established by Gauss-Bonnet, forces it to vanish at every point of that disc. Outside it the replacement surface agrees with the original flat surface, so its curvature also vanishes there; smoothness covers the common boundary. Consequently no point of the replacement surface can have positive Gaussian curvature. Therefore no surface can satisfy all three conditions. A disc with some positive curvature could only have compensating negative curvature or alter the smooth boundary gluing, both excluded here.

Choose an orthonormal basis $(e_1,e_2)$ of $T_{p}S$ and write $e(\theta )=\cos \theta e_1+\sin \theta e_2$. On a punctured disc on which the exponential map is a diffeomorphism, the mapdefines geodesic polar coordinates centred at $p$. The curves $r\mapsto X(r,\theta )$ are unit-speed geodesics, so the first coefficient of the first fundamental form isThe Gauss lemma says that radial and angular coordinate vectors are orthogonal, hence

The angular variation $J=X_{\theta }$ is a Jacobi field along each radial geodesic, with initial dataSince $G=⟨X_{\theta },X_{\theta }⟩=|J{|}^2$, smooth dependence of the geodesic flow givesConsequentlyas required.

The Poincare half-plane model iswith the orientation inherited from the complex plane. Forone hasThese identities show directly that ${\gamma }^{\ast }d{s}^2=d{s}^2$. The map is holomorphic with nonzero derivative, so it preserves orientation. The matrices $I$ and $-I$ induce the same map, giving the action of PSL2(R).

Conversely, let $F$ be an orientation-preserving isometry. The PSL2(R) action is transitive on $\mathbb{H}$, so compose $F$ with an element taking $F(i)$ back to $i$. The resulting isometry fixes $i$ and acts on $T_i\mathbb{H}$ by an orientation-preserving orthogonal map, hence a rotation. The stabilizer of $i$ in PSL2(R),realizes every such tangent rotation. An isometry is determined by its value and differential at one point because it preserves geodesics and the exponential map. Thus the composed isometry belongs to PSL2(R), and so does $F$.

The map $\tau (z)=-\overline{z}$ is an orientation-reversing isometry. Composing any orientation-reversing isometry with $\tau$ gives an orientation-preserving one, so

A hyperbolic line is a vertical Euclidean line or a semicircle orthogonal to the real axis. Its hyperbolic reflection ${\sigma }_{\ell }$ is the unique orientation-reversing isometry that fixes every point of $\ell$. If $\ell ,{\ell }^{\prime }$ meet at angle $\theta$, thenis the hyperbolic rotation about $A$ through angle $2\theta$. The generators satisfyIf $\theta /\pi =p/q$ in lowest terms, then $\rho$ has order $q$ and the generated group is the finite dihedral group of order $2q$. If $\theta /\pi$ is irrational, $\rho$ has infinite order and the generated group is the infinite dihedral group. In the degenerate case $\ell ={\ell }^{\prime }$, the group has order two.

Suppose, for contradiction, that every future null geodesic orthogonal to the compact trapped surface $T$ extends beyond $L=A/|{\theta }_0|$. Both families begin with expansion at most ${\theta }_0<0$, so part d gives a point conjugate to $T$ on every generator by affine length $L$. A generator of the achronal boundary ${\dot{J}}^{+}(T)$ cannot remain on that boundary beyond its first conjugate point, because afterward it can be deformed to a timelike curve from $T$.

The two bundles of initial null directions over compact $T$, restricted to $0\le \lambda \le L$, form a compact set. Their image under the geodesic exponential map contains all of ${\dot{J}}^{+}(T)$, so this achronal boundary is compact. Project it along a complete timelike flow onto the noncompact Cauchy hypersurface $\Sigma$. The projection is both open and closed in the connected Cauchy surface, hence would be all of $\Sigma$; compactness of the source would then make $\Sigma$ compact, a contradiction.

Therefore at least one orthogonal future null geodesic cannot extend to affine length $L$. Its maximal future development is future-inextendible with total affine lengthwhich is the incompleteness conclusion of the Penrose singularity theorem.

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 $v$ near $0\in T_{p}M$, let ${\gamma }_v$ be the unique geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The exponential map is ${\exp }_p(v)={\gamma }_v(1)$.

The constant initial velocity $0$ gives ${\exp }_p(0)=p$. Varying the initial velocity through $sv$ yields ${\exp }_p(sv)={\gamma }_v(s)$, soThus $(d{\exp }_p{)}_0$ is the identity. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighborhood of $0$ onto a neighborhood of $p$. Coordinates from an orthonormal basis of $T_{p}M$ are the geodesic normal coordinates.

For $v$ in a sufficiently small neighbourhood of $0\in T_{p}S$, let ${\gamma }_v$ be the geodesic satisfying ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The exponential map at $p$ isChoose an oriented orthonormal basis of $T_{p}S$, write $e(\theta )=(\cos \theta ,\sin \theta )$, and define geodesic polar coordinates by

The Gauss lemma says that radial and angular coordinate curves are orthogonal. Since $r\mapsto \varphi (r,\theta )$ is a unit-speed geodesic,Moreover, torsion-freeness of the surface connection gives $D_r{\varphi }_{\theta }=D_{\theta }{\varphi }_r$, and thereforeAt $r=0$, ${\varphi }_{\theta }=0$, so $⟨{\varphi }_r,{\varphi }_{\theta }⟩=0$ for every sufficiently small $r$. Thus the first fundamental form isand $\sqrt{G}/r\to 1$ as $r\downarrow 0$ because $d({\exp }_p{)}_0$ is the identity.

The Cheeger-Gromoll splitting theorem states that a complete connected Riemannian manifold with nonnegative Ricci curvature that contains a line in a Riemannian manifold is isometric to a Riemannian product

The Hadamard-Cartan theorem states that if a complete simply connected Riemannian manifold has nonpositive sectional curvature, then for every point $p$ its exponential mapis a diffeomorphism. In particular, the manifold is diffeomorphic to Euclidean space and is contractible.

For a connected Riemannian manifold, two-point homogeneity is equivalent to the isometry group being transitive on the unit tangent bundle. One direction follows by taking short equal-length radial geodesic segments: the Gauss lemma identifies their distance, and injectivity of the exponential map identifies the initial directions after the endpoints are matched. Conversely, unit tangent transitivity implies point homogeneity, hence completeness. The Hopf-Rinow theorem supplies minimizing geodesics for arbitrary equal-distance pairs. Matching their initial unit tangent vectors and using uniqueness of the geodesic equation matches their other endpoints.