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For the unit-speed curve, the normal curvature is ${\kappa }_n=t^{\prime }\cdot n$, and the geodesic curvature is ${\kappa }_g=t^{\prime }\cdot (n\times t)$. Differentiating $\gamma =r(u,v)$ twice and taking the normal component kills the terms in $r_u,r_v$, givingAn affinely parametrized geodesic has zero tangential acceleration, or equivalently its covariant derivative of velocity is zero. For unit speed, $t^{\prime }\cdot t=0$, and $t,n\times t,n$ form an orthonormal frame. Therefore $t^{\prime }={\kappa }_{n}n+{\kappa }_g(n\times t)$, so zero tangential acceleration is exactly ${\kappa }_g=0$.

For the sphere, choose the inward normal so that the second fundamental form is $I/a$. Both principal curvatures are $1/a$, giving $K=1/a^2$; the outward choice reverses both signs and gives the same product.

The octant has area $4\pi a^2/8=\pi a^2/2$, so its curvature integral is $\pi /2$. Its sides are great-circle geodesics, hence have zero geodesic curvature. Each interior angle is $\pi /2$, so each exterior angle is also $\pi /2$. Thereforeverifying the polygon formula directly.

Use the unit normal from part (c), and orient ${\gamma }_a$ by increasing $v$. Write $T={\gamma }_a^{\prime }/|{\gamma }_a^{\prime }|$ for its unit tangent vector. The signed geodesic curvature convention used below iswhere $s$ is arc length. The second equality follows by differentiating $T$: the term parallel to ${\gamma }_a^{\prime }$ disappears in the scalar triple product. It makes the correction for a parameter which is not arc-length parametrization explicit.

Put $C=\cos a$, $S=\sin a$, $D=1+S^2$, and $G=3+2C$. The speed is the constant $\sqrt{G}$. At $v=0$,soTaking its dot product with $N=(C,S,-S)/\sqrt{D}$ yields $-S(3+2C)/\sqrt{D}=-SG/\sqrt{D}$. ThereforeThe screw motion of part (b) makes this independent of $v$. Thus ${\gamma }_a$ is a geodesic precisely when $a$ is an integer multiple of $\pi$. Reversing the direction of traversal or the unit normal changes the sign of signed geodesic curvature; its magnitude is the absolute value of the displayed expression. The signed convention is needed in part (e).

First interpret the surface integral through the given parametric surface, counting multiplicity. SetIts derivative isUsing part (c), the fundamental theorem of calculus givesOn the other hand part (d), including its arc length factor, givesand the corresponding expression for $b$ is $-(d-c)J(b)$. Subtraction provesIf $0<b-a<2\pi$, the parametric surface is injective on the rectangle, so this is exactly the stated surface integral over its image. It also remains the ordinary image integral at $b-a=2\pi$, since the repeated seam has zero surface area.

The assertion for arbitrary $a<b$ needs a multiplicity convention or the restriction $b-a\le 2\pi$. This omission is present in the native PDF. For example, take $a=0$, $b=5\pi /2$, $c=0$, $d=1$. The image is the entire cylindrical band $0\le z\le 1$, since $u$ already covers a full turn. Its ordinary surface integral of Gaussian curvature isBut the difference of the two signed geodesic curvature integrals is $J(5\pi /2)-J(0)=1/\sqrt{2}$. Thus ordinary integration over the image set makes the unrestricted assertion false. The boxed parameter integral is valid for all $a<b$ and $c<d$ and is the appropriate corrected statement. This distinction is the multiplicity of a parametrized surface integral.

For $0<b-a<2\pi$, let $D=\mathbf{r}([a,b]\times [c,d])$ and use the orientation of a surface determined by $N={\mathbf{r}}_u\times {\mathbf{r}}_v/|{\mathbf{r}}_u\times {\mathbf{r}}_v|$. Its positive boundary orientation follows the bottom side in increasing $u$, the right side in increasing $v$, the top side in decreasing $u$, and the left side in decreasing $v$. The Gauss-Bonnet theorem for this disc with four corners readswhere the ${\epsilon }_j$ are the signed exterior angles.

The bottom and top sides have cancelling geodesic curvature integrals: a screw motion takes one to the other with opposite boundary orientation. Explicitly, along increasing $u$ at fixed $v$, the speed is one and the signed geodesic curvature is $-\sin u/\sqrt{1+{\sin }^{2}u}$ on both sides. The right and left sides contribute, respectively, the integral along ${\gamma }_b$ and minus the integral along ${\gamma }_a$.

Although the coordinate sides are generally not perpendicular, the corner contributions still sum to $2\pi$. Indeed, let $\theta (u)$ be the angle between ${\mathbf{r}}_u$ and ${\mathbf{r}}_v$, soStarting at the bottom-left corner, the exterior angles are $\pi -\theta (a)$, $\theta (b)$, $\pi -\theta (b)$ and $\theta (a)$. Their sum is $2\pi$. Cancelling this from the Gauss-Bonnet theorem leaveswhich is the identity in part (e). For a rectangle covering more than one turn, apply the same argument on the parameter rectangle with its induced Riemannian metric, or partition it into narrower rectangles and cancel the artificial internal boundary contributions. This gives the version counting multiplicity, as required by the correction in part (e).

For an oriented surface normal $N$ and a unit-speed curve tangent $T$, take signed geodesic curvature $k_g=(dT/ds)\cdot (T\times N)$. This convention uses boundary orientation with the region on the right; reversing to the usual left-hand boundary convention reverses both orientation and the corresponding curvature convention. With positive orientation consistent this way, the Gauss-Bonnet theorem for a region $D$ with piecewise smooth boundary isFor a disk with smooth boundary there are no corner terms and $\chi (D)=1$. Stating the signed convention is necessary to reproduce the sign in part (ii).

Let $J$ be rotation by ninety degrees in the oriented tangent plane, $Jv=N\times v$. In local coordinates seek a curve withThis smooth second-order ordinary differential equation, written as a first-order system for $(x,\dot{x})$, has a unique local solution with the specified initial position and velocity. Its speed is constant, sinceThe initial speed is one, so the parameter is arc-length and the signed geodesic curvature is $⟨{\nabla }_s\dot{\gamma },J\dot{\gamma }⟩=\lambda$. Conversely every requested curve satisfies this system, which proves the claimed uniqueness as well as existence on some $(-ϵ,ϵ)$.

If two simple closed geodesics were disjoint on a sphere-like surface, they would bound an annulus. The Gauss-Bonnet theorem applied to it gives ${\int }_{A}KdA+{\int }_{\partial A}{k}_{g}ds=2\pi \chi (A)=0$. Both boundary geodesic curvatures vanish, but positive Gaussian curvature makes the area integral strictly positive, a contradiction. Hence the two geodesics must intersect.

The assertion fails for nonzero constant signed geodesic curvature. On the unit round sphere a latitude of colatitude $\theta$, oriented in increasing longitude, has $k_g=\cot \theta$. The latitude of colatitude $\pi -\theta$ has the opposite value; reversing its orientation gives the same value as the first. Choosing $0<\theta <\pi /2$ produces two disjoint smooth simple closed curves with the same positive nonzero curvature. Reversing both handles negative values, and $\theta$ can be chosen for any specified nonzero $\lambda$. Thus constant nonzero geodesic curvature does not force intersection.

For a regular parametrized surface $\sigma (u,v)$, the first fundamental form is the induced metricChoose the unit normal $N=({\sigma }_u\times {\sigma }_v)/|{\sigma }_u\times {\sigma }_v|$. With the convention $I{I}_{ij}=N\cdot {\sigma }_{ij}$, the second fundamental form isThe Gaussian curvature, the product of the principal curvatures, isFor a compact smooth embedded surface without boundary, the Gauss-Bonnet theorem giveswhere $e(S)$ is the Euler characteristic. The no-boundary condition is needed for this form; a smooth embedded surface with boundary has the additional boundary geodesic curvature integral.

For the surface of revolution, useThe parameters are periodic, with local open parameter rectangles providing the coordinate charts. Put $R=r+a\sin u$ and $D=\sqrt{a^2{\cos }^{2}u+b^2{\sin }^{2}u}$. Since $r>a>0$ and $b>0$, both $R$ and $D$ are positive. The tangent vectors areTheir inner products are $E=D^2$, $F=0$, and $G=R^2$. ThusTheir cross product is $R(b\sin u\cos v,b\sin u\sin v,a\cos u)$, so its normalization gives the outward unit normalNext,Taking their inner products with $N$ yields $e=-ab/D$, $f=0$, and $g=-bR\sin u/D$. Hence the fundamental forms of an elliptic ring torus giveChoosing the inward unit normal changes both displayed coefficients of $II$ to their negatives and leaves $I$ and $K$ unchanged. As a consistency check, $K=a{b}^2\sin u/(R{D}^4)$; its integral over the torus is zero, since $KdA=a{b}^2\sin ududv/D^3$ cancels under $u\mapsto u+\pi$. This agrees with $e(S)=0$ in the Gauss-Bonnet theorem.

Parametrize a regular curve by arclength $s$, write its unit tangent as $T$, and choose an oriented surface unit normal $N$. Its signed geodesic curvature isIt vanishes exactly when the acceleration of the unit-speed curve is normal to the surface, equivalently when the curve is a geodesic with arclength parametrization. For an arbitrary regular parametrization, this means the curve becomes a geodesic after reparametrization.

The global Gauss-Bonnet theorem for a compact oriented surface with smooth boundary, positively oriented with the surface on the left, isFor piecewise smooth boundary, add the signed exterior turning angles at its corners to the left side.

A simple closed curve on the sphere bounds two topological disks. Equal areas give area $2\pi$ for each, since the unit sphere has area $4\pi$. Apply Gauss-Bonnet theorem to either disk $D$, with $K=1$ and $\chi (D)=1$:A continuous geodesic curvature that were strictly positive everywhere, or strictly negative everywhere, would have a correspondingly strict sign for this integral. Therefore neither everywhere strictly positive nor everywhere strictly negative geodesic curvature is possible. Orientation reversal changes its sign but not this conclusion.

A geodesic is a curve with zero covariant acceleration in an affine parameter. For a unit-speed curve on an oriented surface with unit normal $N$, its geodesic curvature is $k_g=⟨{\gamma }^{\prime \prime },N\times {\gamma }^{\prime }⟩$: it measures the tangent-plane component of acceleration perpendicular to the tangent. A unit-speed curve is geodesic exactly when this component vanishes.

The exponential map ${\exp }_p(v)$ follows for unit time the geodesic starting at $p$ with initial velocity $v$, whenever it is defined. Since ${\exp }_p(tv)={\gamma }_v(t)$, its derivative at zero is $\boxed{d({\exp }_p{)}_0={\mathrm{id}}_{T_{p}S}}$. Choose an oriented orthonormal basis of $T_{p}S$ and let $v(\theta )$ be its unit circle. Geodesic polar coordinates are $F(r,\theta )={\exp }_p(rv(\theta ))$ in a normal neighborhood, with the usual angular singularity at $r=0$.

Put $T=F_r$ and $J=F_{\theta }$. Radial curves are unit-speed geodesics, so $E=⟨T,T⟩=1$ and ${\nabla }_{r}T=0$. The torsion-free connection gives ${\nabla }_{r}J={\nabla }_{\theta }T$, and thereforeSince $J(0,\theta )=0$, this proves $F=⟨T,J⟩=0$, the radial form of the Gauss lemma. Moreover $J(r,\theta )=r{v}^{\prime }(\theta )+O(r^2)$, with $|v^{\prime }|=1$, by the derivative of the exponential map. Hence $G=|J{|}^2=r^2+O(r^3)$, givingThe last derivative is the one-sided radial limit.

The global Gauss-Bonnet theorem for a compact oriented surface with piecewise smooth boundary isHere $K$ is Gaussian curvature, the boundary is positively oriented with the surface on its left, ${\epsilon }_j$ are signed exterior turning angles at corners, and $\chi$ is the Euler characteristic. Smooth geodesic boundary components contribute neither curvature nor corner terms.

A contractible simple closed geodesic on the cylinder would bound a disc, giving $\int KdA=2\pi$, impossible when $K<0$. Thus any such geodesic is an essential loop. If two distinct essential simple loops on an annulus intersect, the elementary bigon criterion gives two subarcs bounding an innermost disc: lifting the homotopic loops to the covering strip and choosing an innermost region supplies such a bigon. Distinct geodesics cannot have tangential intersections by uniqueness of the geodesic initial-value problem, so its two interior angles $\alpha ,\beta$ are positive. Gauss-Bonnet on this geodesic bigon gives $\int KdA=\alpha +\beta >0$, another contradiction. Therefore two distinct simple closed geodesics would be disjoint. They would then bound a compact annulus with geodesic boundary and Euler characteristic zero, forcing $\int KdA=0$, again impossible. There is at most one simple closed geodesic. Completeness of the noncompact surface is not needed for these compact-region arguments.

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.