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For $n\ne 0$, put $\rho =f(r)/|n|$ and $\theta =|n|\varphi$. Then $d{\rho }^2=d{r}^2$ andThese give a local isometry wherever $\rho >0$. For $n=0$, $f=b>0$, and the local coordinates $X=r$, $Y=b\varphi$ similarly give the Euclidean plane metric tensor. Equivalently, the Gaussian curvature of the original Riemannian metric is $-f^{\prime \prime }/f=0$ throughout this family.

A local isometry need not be a global isometry. If $|n|=1$, the displayed polar map is injective modulo $\varphi \sim \varphi +2\pi$ and identifies the regular domain with a Euclidean annulus, possibly a punctured disc or the punctured Euclidean plane, according to the radial interval. If $|n|>1$, it is a $|n|$-fold covering of that annulus: distinct angles differing by $2\pi /|n|$ have the same image. This developing map cannot be made injective by a different choice of Euclidean coordinates, since two local Euclidean isometries on a connected flat domain differ by a rigid motion once their value and differential are fixed at one point. For $n=0$, the domain is a flat cylinder; its circular curves are closed geodesics, whereas the Euclidean plane has no nonconstant closed geodesic.

All these metrics are locally Euclidean; they are not all globally Euclidean. Even the regular $|n|=1$ domain is an annular region rather than the entire Euclidean plane. A regular axis can be added without a conical singularity only when $|n|=1$; with the full radial range this completion becomes the Euclidean plane. Trivial holonomy alone therefore does not supply a global injective Euclidean chart.

Write $R=a+b\cos u>0$. The tangent vectors areTheir squared lengths and inner product give the first fundamental formThe unit normal determined by this ordered parametrization isTaking inner products of $\nu$ with ${\sigma }_{uu}$, ${\sigma }_{uv}$ and ${\sigma }_{vv}$ yields $e=b$, $f=0$, $g=R\cos u$. Consequently the second fundamental form of a ring torus isThis normal points into the tube. Choosing the outward unit normal reverses both coefficients and gives $-II$; the choice must therefore accompany an answer for the second fundamental form.

The Gaussian curvature is unchanged by that reversal:Since the denominator is positive, $K>0$ on the outside of the torus, where $\cos u>0$, and $K<0$ on its inner side, where $\cos u<0$. It vanishes on the two circles $u=\pi /2,3\pi /2$. This is the Gaussian curvature of a ring torus.

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.

The Poincare disc model is the open unit disc $D=\{z=x+iy:|z|<1\}$ equipped with the Riemannian metricThis normalization gives Gaussian curvature $-1$. In particular, a Euclidean curve has hyperbolic length $\int 2|dz|/(1-|z{|}^2)$; the boundary circle is infinitely far away in hyperbolic distance.

Write derivatives with respect to $u$. The arc-length condition is $f^{\prime 2}+g^{\prime 2}=1$. The tangent vectors areTheir scalar products give the first fundamental formChoose the unit normal vector $N=(-g^{\prime }\cos v,-g^{\prime }\sin v,f^{\prime })$, since ${\sigma }_u\times {\sigma }_v=fN$ and $f>0$. Scalar products of $N$ with the second partial derivatives giveThe mixed coefficient is zero. Reversing $N$ reverses the second fundamental form but not the Gaussian curvature. Its formula isDifferentiating the arc-length identity gives $f^{\prime }{f}^{\prime \prime }+g^{\prime }{g}^{\prime \prime }=0$. Substitute it directly, without dividing by $g^{\prime }$ (which may vanish), to obtain

The generating circle has $f^{\prime 2}+g^{\prime 2}=1$, and its revolution is a torus. From part (a), $K=\cos u/(2+\cos u)$ and the area element of a surface is $dA=fdudv=(2+\cos u)dudv$. ThereforeThe positive Gaussian curvature on the outer side exactly cancels the negative curvature on the inner side. Equivalently, the Gauss-Bonnet theorem gives ${\int }_{S}KdA=2\pi \chi (T^2)=0$ because the Euler characteristic of the torus is zero.

Constant Gaussian curvature one gives $f^{\prime \prime }+f=0$, so, after a shift of the arc-length parameter, $f=C\cos u$ with $C>0$ on an interval contained in $(-\pi /2,\pi /2)$. To close the surface by adding only two points, the limiting parallels at both ends must collapse to points on the axis. If an end had limiting radius $f>0$, it would contribute a whole missing circle, which two added points cannot supply. Thus the parameter interval is the full interval between consecutive zeros, and its two ends are $u=\pm \pi /2$.

At a smooth axis point the tangent plane of a surface of revolution is invariant under all rotations about the axis, so it must be horizontal. The unit meridian tangent must consequently have $g^{\prime }\to 0$ and $|f^{\prime }|\to 1$ there. But $|f^{\prime }(\pm \pi /2)|=C$, so smoothness forces $C=1$. The arc-length identity also illustrates the alternatives: $C>1$ prevents reaching either zero with real $g^{\prime }$, while $0<C<1$ gives a nonhorizontal limiting meridian and a conical singularity.

Now $g^{\prime 2}=1-f^{\prime 2}={\cos }^{2}u$. On the connected interior, $g^{\prime }$ has one fixed sign because $\cos u>0$. Therefore $g=g_0\pm \sin u$, andThe revolved surface is the unit sphere centered at $(0,0,g_0)$, with only its two poles $(0,0,g_0\pm 1)$ omitted. Those points are antipodal. This proves the claim using smooth endpoint criterion for a surface of revolution, without integrating a general square root.

At a regular point, choose the unit normal $\mathbf{n}=({\sigma }_x\times {\sigma }_y)/|{\sigma }_x\times {\sigma }_y|$. The first fundamental form is the induced squared lengthThe second fundamental form isThe Gaussian curvature is the determinant of the shape operator, $K=(eg-f^2)/(EG-F^2)$, independent of the normal orientation.

For the given graph, ${\sigma }_x=(1,0,y)$ and ${\sigma }_y=(0,1,x)$, so with $S=1+x^2+y^2$,ThereforeThe surface is regular everywhere. The printed description “hyperboloid” is a naming error: this graph is a hyperbolic paraboloid, and the computation uses the actual parametrization.

Assume the parametrization is $C^2$ and regular, so its curvature is defined. Write $\mathbf{v}={\mathbf{a}}^{\prime }+y{\mathbf{b}}^{\prime }$ and $W=|\mathbf{v}\times \mathbf{b}|>0$. Since ${\sigma }_x=\mathbf{v}$ and ${\sigma }_y=\mathbf{b}$, its first fundamental form isFor the unit normal $\mathbf{n}=(\mathbf{v}\times \mathbf{b})/W$, the mixed coefficient of the second fundamental form iswhere the term containing ${\mathbf{b}}^{\prime }$ twice vanishes. Also $g=\mathbf{n}\cdot {\sigma }_{yy}=0$, while $EG-F^2=W^2$. ThusThe Gaussian curvature vanishes exactly when the displayed scalar triple product is zero. This is the Gaussian curvature of a ruled surface criterion; at a singular point with $W=0$, curvature and the claimed equivalence are not defined.

Use the Gaussian curvature $-1$ normalizationof the Poincare half-plane model. This is a complete conformal metric, invariant under real Möbius transformations. Its quotient by the free effective principal congruence subgroup is again complete: a geodesic lifts to the complete covering plane and extends there for all time. Transporting the quotient metric through the biholomorphism supplies the required complete conformal metric.

A nonconstant closed geodesic corresponds to a hyperbolic Möbius transformation $A$ of $\Gamma (2)$, and its hyperbolic translation length is $2\mathrm{arcosh}(|\mathrm{tr}A|/2)$. Indeed its eigenvalues have magnitudes $\lambda ,{\lambda }^{-1}$ with $\lambda >1$, and its axis quotient has length $2\log \lambda$. For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in $\Gamma (2)$, $bc\equiv 0(\mathrm{mod}4)$ implies $ad\equiv 1(\mathrm{mod}4)$. The odd numbers $a,d$ therefore have the same residue modulo four, andA hyperbolic Möbius transformation must have $|\mathrm{tr}A|>2$, so its smallest possible absolute trace is six. It is attained byConsequentlyThis element is primitive: a proper power would have a shorter root represented by a hyperbolic Möbius transformation, contradicting the trace bound. No essential simple closed geodesic exists on the three-punctured sphere, since every essential simple loop is peripheral and corresponds to a parabolic Möbius transformation. The minimizing closed geodesic is therefore nonsimple. Finally, the PDF leaves the numerical Gaussian curvature unspecified: the displayed length uses Gaussian curvature $-1$; for Gaussian curvature $-{\kappa }^2$, it is divided by $\kappa$.

For fixed $g\ge 2$, Mumford's compactness theorem states thatHere ${\mathcal{M}}_g$ is the unmarked moduli space of Riemann surfaces, with its usual topology, and the hyperbolic systole is the shortest nonconstant closed hyperbolic geodesic, in Gaussian curvature $-1$. Equivalently, a subset of ${\mathcal{M}}_g$ is relatively compact exactly when its hyperbolic systoles have a common positive lower bound.

We use two standard structural results. The Bers pants decomposition theorem provides a constant $B_g$ such that every closed genus-$g$ hyperbolic surface has a pants decomposition with all $3g-3$ cuff lengths at most $B_g$. For a fixed topological pants decomposition, the Fenchel–Nielsen coordinates identify Teichmüller space withThe length coordinates are ${\ell }_i$, and the twist coordinates ${\tau }_i$ are measured in length units: a full Dehn twist changes ${\tau }_i$ by ${\ell }_i$. Reconstruction from these coordinates is continuous; locally the marked metrics can be chosen to vary smoothly on a fixed reference surface, and the quotient by the mapping class group is the Hausdorff moduli space of Riemann surfaces.

Take any sequence in ${\mathcal{M}}_g^{\epsilon }$. The pants decompositions supplied by the Bers pants decomposition theorem have cuff lengths in $[\epsilon ,B_g]$. There are finitely many topological types of pants decomposition: their dual graphs have $2g-2$ vertices and $3g-3$ edges, with loops and multiple edges allowed, giving finitely many finite graphs. Choose a subsequence of one type, and choose markings carrying each decomposition to a fixed reference one. Compose these markings with Dehn twists so that $0\le {\tau }_i\le {\ell }_i$. The resulting points of Teichmüller space lie in the compact boxThey therefore have a convergent subsequence inside Teichmüller space; its continuous projection gives a convergent subsequence in the moduli space of Riemann surfaces. If $\epsilon >B_g$ the thick set is empty, which is already compact. Equivalently, using all the finitely many reference decompositions gives a finite union of compact projected boxes containing the whole thick set.

Finally the hyperbolic systole is continuous. Nearby marked hyperbolic surfaces admit metric comparisons with bi-Lipschitz distortion tending to one; the length of every loop, and hence the infimum over all essential loops, obeys the same multiplicative comparison. Therefore ${\mathcal{M}}_g^{\epsilon }$ is closed in that compact union and is compact. A compact subset has a positive minimum hyperbolic systole, proving the converse characterization of relative compactness. This theorem concerns the unmarked quotient: repeated Dehn twists can give an unbounded sequence in Teichmüller space while leaving the underlying surface and its hyperbolic systole unchanged.

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.