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A parametrization is $X(\theta ,z)=(\cosh z\cos \theta ,\cosh z\sin \theta ,z)$. It gives a homeomorphism $S^1\times \mathbb{R}\to S$, with inverse obtained from the height and angular position. Its first fundamental form isFor a surface of revolution with radius $r(z)$, the Gaussian curvature is $-r^{\prime \prime }/[r(1+r^{\prime 2}{)}^2]$. Thus $\boxed{K=-{\mathrm{sech}}^{4}z<0}$. The waist $z=0$ is a closed geodesic: ${\Gamma }_{\theta \theta }^z=-\tanh z$ vanishes there, and $\theta$ can be parametrized at constant speed.

For the uniqueness of a closed geodesic on a negatively curved cylinder, use Gauss-Bonnet theorem with piecewise geodesic boundary:A geodesic disk with smooth boundary would require curvature integral $2\pi$. A disk bounded by a geodesic monogon would require $\pi +\alpha >0$, and a geodesic bigon would require $\alpha +\beta >0$. All are impossible when $K<0$.

Here is the topological reduction needed to apply this observation without assuming simplicity. Lift a closed geodesic to the universal covering plane of the cylinder. A self-intersection of its lift supplies an innermost monogon by the Jordan curve theorem, so the lift is embedded. A contractible closed geodesic would similarly supply a geodesic disk or monogon, so its winding number is nonzero. Its lift is consequently a proper line, invariant under a nonzero power of the deck translation. Two distinct translates of that line cannot intersect: periodicity would repeat an intersection, and innermost arcs between intersections would bound a bigon. The disjoint translates are ordered transversely, with the deck translation preserving that order. A finite cyclic permutation of this ordered family is impossible, so the lift is invariant under the primitive deck translation too. Thus the original geodesic has a simple image and is at most a repeated traversal of it.

Likewise two distinct simple essential closed geodesics cannot intersect: their periodic lifts would produce a bigon. If disjoint, the Jordan curve theorem on the cylinder says they bound a compact annulus. Its two boundaries have zero geodesic curvature and $\chi =0$, so Gauss-Bonnet gives $\int KdA=0$, again impossible. Therefore there is at most one closed geodesic image. Reparametrizations, reversed orientation and multiple traversals are understood to describe the same geometric geodesic.

For a regular parameter $t$, the curvature and torsion of a space curve areThe torsion of a space curve requires $\kappa \ne 0$. Reparameterize by arc length $s$ and set $T={\alpha }_s$, $N=T_s/\kappa$, $B=T\times N$. Differentiating the pairwise inner products of this orthonormal Frenet frame shows that its derivative matrix is skew symmetric. Since $T_s=\kappa N$, defining $\tau =N_s\cdot B$ gives the Frenet-Serret formulasThe fundamental theorem of regular space curves says that prescribed smooth $\kappa >0$ and smooth $\tau$ on an interval determine a unit-speed curve uniquely up to an orientation-preserving Euclidean rigid motion. Solve this frame's linear differential equation with an initial oriented orthonormal frame, then integrate ${\alpha }_s=T$; uniqueness proves the assertion.

Under the intended convention that the shared parameter $s$ is arc length, equality on an open interval gives equal position and equal Frenet frames at one point. Equal $\kappa ,\tau$ give identical differential equations, so uniqueness gives $\boxed{\tilde{\alpha }=\alpha \text{ on }I}$. The printed “regular smooth” hypothesis alone does not establish this convention. Without common arc length, an orientation-preserving smooth change of speed that is the identity on $J$ but not elsewhere reparameterizes a circle into a counterexample: both curves have $\kappa =1$, $\tau =0$ and agree on $J$, but differ elsewhere as parameterized curves.

For an oriented surface with unit normal $n$, define normal curvature ${\kappa }_n=T_s\cdot n$ and geodesic curvature ${\kappa }_g=T_s\cdot (n\times T)$. ThusA unit-speed curve is a geodesic exactly when ${\kappa }_g=0$. If also ${\kappa }_n\equiv 0$, then $T_s=0$, so it is a straight line. A vertical ruling of a circular cylinder is such a geodesic, although the cylinder contains no open piece of a plane. A nonconstant regular straight line cannot be a simple closed curve. Hence no planar patch is required, and such a geodesic cannot be simple and closed.

Finally let the principal curvatures be ${\kappa }_1,{\kappa }_2$. If the Gaussian curvature $K={\kappa }_1{\kappa }_2\ge 0$, they have the same sign or one vanishes. Euler formula for normal curvature makes ${\kappa }_n$ a weighted average of them. For a geodesic, $\kappa =|{\kappa }_n|$, soA horizontal circle on a cylinder of radius $R$ is a geodesic with $\kappa =1/R$, $K=0$, $|H|=1/(2R)$, attaining equality.

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.

For a geodesic triangle with interior angles $\alpha ,\beta ,\gamma$, the local Gauss-Bonnet theorem isbecause its geodesic sides have zero geodesic curvature.

Triangulate a closed oriented surface into $F$ geodesic triangles, with $E$ edges and $V$ vertices. Summing the local formula, the angles around each vertex total $2\pi$, soSince every triangular face has three edges and every edge belongs to two faces, $3F=2E$. Hencewhich is the global Gauss-Bonnet theorem.

For the sphere $S_r$, the unit normal is $N(p)=p/r$. Its shape operator is, up to the conventional sign, $(1/r)I$ on each tangent plane. Both principal curvatures therefore have magnitude $1/r$, and the Gaussian curvature is

An octant has one eighth of the sphere's area:Thus ${\int }_{T}KdA=\pi /2$. Its three great-circle sides meet at three right angles, soThe two sides of the local Gauss-Bonnet formula agree directly.

For a positively oriented local parametrization $X(u,v)$ with first fundamental formthe area element of a surface isUnder an orientation-preserving coordinate change, the Jacobian from $du\wedge dv$ cancels the inverse Jacobian in the square root of the metric determinant. Hence these local expressions agree and define a global two-form.

The Euler characteristic may be defined from any finite triangulation byor equivalently by the alternating sum of the dimensions of the rational homology groups. Subdivision leaves $V-E+F$ unchanged, and the homological formula shows that it is a topological invariant, so the definition does not depend on the triangulation.

Give the boundary its induced orientation and parametrize it by arc length. If $T$ is its unit tangent and $N$ is the chosen unit normal to the surface, its signed geodesic curvature iswhere $D_s$ is the surface covariant derivative. The Gauss-Bonnet theorem for a compact oriented surface with smooth boundary and no corners iswhere $K$ is the Gaussian curvature.

For a unit-speed space curve, setThese form the Frenet trihedron. With the signed torsion convention compatible with the question, the Frenet-Serret formulas areFor an oriented surface with unit normal $N$, the signed geodesic curvature is

On the unit sphere $N=\alpha$. Since $\alpha \cdot t=0$, differentiation givesWrite $\alpha =An+Bb$. Then $A=-1/\kappa$, and differentiating this expression and comparing the $n$ coefficient with $\dot{\alpha }=t$ givesThus, wherever the displayed quantities are defined,Finally,which is the stated curvature decomposition for a spherical curve.

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.

Suppose that two closed geodesics ${\gamma }_1$ and ${\gamma }_2$ were disjoint. Under the usual convention that a closed geodesic curve is simple, the Jordan curve theorem on the sphere shows that they bound an annulus $A\subset S$. Its Euler characteristic is zero. Give $A$ the induced orientation and apply the Gauss-Bonnet theorem:Both boundary components are geodesics, so their geodesic curvature vanishes. It follows thatThis is impossible because $K>0$ everywhere and $A$ has positive area. Hence the two closed geodesics must intersect, as summarized by the intersection of closed geodesics on a positively curved sphere.

For unit speed on an oriented surface,These coefficients are normal and geodesic curvature; the curve is geodesic iff ${\kappa }_g=0$.

If a flat surface disc is replaced by a smooth disc agreeing with the original surface outside the same boundary, their boundary geodesic curvatures agree. Gauss-Bonnet theorem therefore forces the replacement’s total Gaussian curvature to be zero. If its curvature is everywhere nonnegative, continuity forces it to vanish everywhere. Thus a positive-curvature bulge cannot be glued this way without some negative curvature or a loss of smoothness.