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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.

Differentiate the parametrized surface. The coefficients of the first fundamental form simplify toIf $c={\sigma }_u\times {\sigma }_v$, then $|c{|}^2=E$. MoreoverThus the second-fundamental-form coefficients satisfy $M=1/\sqrt{E}$ and $N=0$. The Gaussian curvature is consequently

For a local parametrization $\varphi (u,v)$, write the first fundamental form and second fundamental form asThe Gaussian curvature is

To prove the Theorema Egregium, set $g_{ij}=⟨{\varphi }_i,{\varphi }_j⟩$ and decomposeDifferentiating $g_{ij}$ and solving for the tangential coefficients givesso every Christoffel symbol depends only on $E,F,G$ and their first derivatives. Equality of mixed third derivatives yields the Gauss equationThe Riemann curvature tensor is formed from first derivatives and quadratic products of the Christoffel symbols, and hence from the first and second derivatives of $E,F,G$. Consequentlydepends only on the first fundamental form.

Suppose that a connected smooth surface $\tilde{S}\subset {\mathbb{R}}^3$ properly contains $S$. Since $S$ and $\tilde{S}$ have the same dimension, $S$ is open in $\tilde{S}$. Choose a point of its relative boundary. Inside a sufficiently small compact normal neighbourhood in $\tilde{S}$, choose $q\in S$ and a closest point $p\in \tilde{S}\setminus S$. A minimizing $\tilde{S}$-geodesic from $q$ to $p$ stays in $S$ before its endpoint, since encountering the complement earlier would contradict the choice of $p$.

The restriction of this curve is consequently a geodesic of $S$ with a finite maximal endpoint at $p$. On the other hand, the Gaussian curvature of $S$ is the restriction of the smooth Gaussian curvature of $\tilde{S}$, so it remains bounded near $p$. This contradicts condition $(\ast )$. Henceso $S$ is an inextendible embedded surface. This is the curvature-blowup criterion for inextendibility of a surface.

A smooth surface in ${\mathbb{R}}^3$ is a subset locally parametrized by a smooth map of two variables whose derivative has rank two and which is a homeomorphism onto its image.

Writing $\rho =\sqrt{x^2+y^2}$, the first surface equation becomesIt is the torus of major radius $R=2\sqrt{2}$ and minor radius $r=1$, with smooth parametrizationSince $R>r$, this is an embedding. The Gaussian curvature of a torus is

For the second surface, use the orthogonal transformationIts equation becomes the same torus equation $(\sqrt{X^2+Y^2}-2\sqrt{2}{)}^2+Z^2=1$. Orthogonal transformations are Euclidean isometries and preserve Gaussian curvature. Curvature vanishes where $\cos v=0$, equivalently $Z=\pm 1$. In the original coordinates the zero-curvature points are therefore exactly

Away from a zero of $\varphi$, the first Bogomolny vortex equation givesSubstitution into the second equation, with ${\nabla }^2=4{\partial }_z{\partial }_{\bar{z}}$, givesNow set $\tilde{\Omega }=\Omega |\varphi {|}^2$. The Gaussian curvature formula givesTherefore $(\tilde{K}+1/2)\tilde{\Omega }=(K+1/2)\Omega$ implieswhich is exactly the vortex equation.

The metricis the pullback of the $n=1$ Poincare disc model metric under $w=z^n$. It consequently has $K=-1/2$ away from the origin. Comparing it with $d{s}_1^2$ gives the Witten hyperbolic vortexPutting $n=N+1$ gives winding number $N$. A gauge choice with positive radial factor isand then

A regular surface is a minimal surface when its mean curvature vanishes identically, equivalently when its two principal curvatures sum to zero everywhere.

The catenoidis an example. In part (a), take $f(v)=\cosh v$ and $g(v)=v$, so $q=\cosh v$. The formula there givesA plane and a circular cylinder have zero Gaussian curvature, as does a flat cone away from its vertex. Because a local isometry preserves Gaussian curvature, the catenoid is minimal and is not locally isometric to any of those surfaces; this is the catenoid is a minimal surface.

The curve is symmetric in both coordinate axes and under interchange of $x$ and $z$. Its shape, including its eight inflection points, is:

Writing $r^2=x^2+y^2$, the surface of revolution has equationLetIts gradient isIf this vanished, then $r^2\in \{0,1\}$ and $z^2\in \{0,1\}$, but at such a point $(r^2-1{)}^2+(z^2-1{)}^2\le 2$, not five. Hence zero is a regular value, and the regular level set theorem shows that $S=F^{-1}(0)$ is a smooth embedded surface.

For a surface of revolution, the two principal curvatures are the curvature of the meridian and the normal curvature of a parallel. Along $r=1$,so the normal is vertical in the meridian plane and the parallel principal curvature vanishes. Therefore the Gaussian curvature, the product of the principal curvatures, is zero there.

These are not the only zero-curvature points. The meridian itself has inflection points, at which its principal curvature vanishes. Setting $u=r^2$ and $v=z^2$, their positive squared coordinates satisfyBesides interchanging $u$ and $v$, the solution is approximatelyEach corresponding meridian inflection point sweeps out another circle on $S$, so $r=1$ does not exhaust the zero set of the Gaussian curvature.

For an oriented smooth embedded surface, the Gauss map sends each point to its chosen unit normal. This surface is the ring torus with major radius two and minor radius one. Its outward Gauss map isOn the $y=0$ cross-section, the normal points radially away from the center of each generating circle, which gives this formula for $u=0,\pi$ and then rotational symmetry gives it for all $u$.

The upper hemisphere condition is $\sin v>0$, hence $0<v<\pi$. For this torus, the Gaussian curvature and area element areTherefore the positive outer and negative inner contributions cancel:

Put $h(r,\theta )=\sqrt{G(r,\theta )}$. The Jacobi equation in geodesic polar coordinates and the initial conditions from part (b) areInside a geodesic polar coordinate ball, $h>0$. If $K\le 0$, then $h_{rr}\ge 0$, so $h_r\ge 1$ and therefore $h(r,\theta )\ge r$. The Riemannian area element is $hdrd\theta$, whenceThus nonpositive Gaussian curvature makes such a ball at least as large as the Euclidean disk of the same radius.

For a compact oriented regular surface $S\subset {\mathbb{R}}^3$ without boundary, the Gauss-Bonnet theorem statesHere $K=k_1{k}_2$ is the Gaussian curvature, the product of the two principal curvatures; $dA$ is the area element induced by the first fundamental form; and $\chi (S)$ is the Euler characteristic. If $S$ is connected and has genus $g$, the classification theorem for surfaces gives $\chi (S)=2-2g$, so

Take the ring torus with minor radius $1$ and major radius $R_n=n+2$. With angular coordinate $\theta$ around the generating circle, its Gaussian curvature isand henceThus the displayed $\mathrm{limsup}$ condition holds, while every $S_n$ has genus one and is not diffeomorphic to the sphere.

For an oriented smooth surface $S$, the Gauss map sends $p$ to the chosen unit normal vector $N(p)\in S^2$. Since $|N{|}^2=1$, differentiation shows that $D{N}_p(X)$ is perpendicular to $N(p)$ and hence lies in $T_{p}S$.

In a local parametrization $\varphi (u,v)$ with $n=N\circ \varphi$, differentiatinggivesThus the bilinear form $(X,Y)\mapsto D{N}_p(X)\cdot Y$ is symmetric, so $D{N}_p$ is self-adjoint. The Gaussian curvature isWriting the coefficients of the first fundamental form asand those of the second fundamental form asone obtains

At an umbilic point, the self-adjoint map $D{N}_p$ has a repeated eigenvalue, so it is a scalar map. If every point is umbilic, there is a function $\lambda$ withEquality of mixed partial derivatives givesThe two tangent vectors are linearly independent, hence ${\lambda }_u={\lambda }_v=0$. Since ${\mathbb{R}}^2$ is connected, $\lambda$ is constant.

If $\lambda =0$, then $n$ is constant andso $S$ lies in a plane. If $\lambda \ne 0$, thenThus $n-\lambda \varphi =c$ is constant, andTherefore $S$ lies in a sphere of radius $1/|\lambda |$. This proves that the surface is part of a plane or part of a sphere.

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 chosen unit normal field $N$, the second fundamental form isIn a local parametrization $\varphi$, its coefficients, up to the equivalent simultaneous sign convention, areThe Gaussian curvature is the determinant of the shape operator, henceTheorema Egregium states that $K$ is determined entirely by the first fundamental form. In particular, it is intrinsic and is preserved by every local isometry.

Withthe coordinate tangent vectors satisfyThus the first fundamental form isGeometrically, $S$ is the half of a helicoid with positive radius, winding upwards indefinitely.

Consider the upper half of the catenoid, parametrized as the surface of revolutionIt has the same first fundamental form, so $(u,v)\mapsto Y(u,v)$ defines a local isometry. By the invariance of Gaussian curvature under local isometry, the curvatures agree; directly, the conformal metric above gives

For an oriented regular surface with unit normal $N$, the Gauss map sends each point $p$ to $N(p)\in S^2$. The outward normal to the catenoid isIts image is the open southern hemisphere, apart from its limiting equator and south pole. The Gauss map is one-to-one after taking $v$ modulo $2\pi$ and reverses orientation. Since the Jacobian of the Gauss map is the Gaussian curvature, its signed spherical area is the total Gaussian curvature:Equivalently, $KdA=-{\mathrm{sech}}^{2}ududv$, whose integral over $u>0$ and $0\le v<2\pi$ is $-2\pi$.

The Gaussian curvature $K$ is the determinant of the shape operator, equivalently the product of the two principal curvatures. Yes,This is the Theorema Egregium: Gaussian curvature depends only on the first fundamental form, which a local isometry preserves.

The tangent vectors ofareHence the first fundamental form has coefficientsPutThe upward unit normal is $N=(-f_u,-f_v,1)/W$, so the second fundamental form has coefficientsThus the two forms are

The Gaussian curvature isSince $EG-F^2=W^2$, the graph formula isas in Gaussian curvature of a graph surface.

For the final claim, fix a point of $\gamma$ and make a rigid motion taking $P$ to the plane $z=0$. The common tangent plane is horizontal, so locally $\Sigma$ is the graph $z=f(u,v)$. Along the projected curve $c(s)$ of tangency,Differentiating the second identity givesBecause $c$ is a smooth curve, $c^{\prime }(s)\ne 0$, so the Hessian is singular. Its determinant is zero, and the graph formula gives $K=0$ at every point of $\gamma$. This is tangency to a plane along a curve forces zero Gaussian curvature.

The tangent planecuts the hyperboloid inso its intersection is the pair of straight linesTheir accelerations vanish, hence are normal to the surface, so both lines are geodesics. Their tangent vectors at their intersection $(1,0,0)$ are $(0,1,1)$ and $(0,-1,1)$, whose inner product is zero. They therefore intersect at a right angle, giving the second case of plane-section geodesics of the unit one-sheet hyperboloid.

There are also geodesics entirely contained in $z>0$. In the coordinatesthe metric isFor a unit-speed geodesic, the Clairaut first integral for a surface of revolution is $c=(1+z^2)\dot{\varphi }$, and the constant-speed equation becomesChoose $c>1$ and initial height $z_0=\sqrt{c^2-1}$. The geodesic initially tangent to the parallel has $\dot{z}=0$, and the displayed identity prevents it from entering $0<z<z_0$. Since the parallel at $z_0>0$ is not itself a geodesic, the curve turns there and otherwise has $z>z_0$. Hence it remains entirely in $z>0$; this is a geodesic trapped in one half of the unit one-sheet hyperboloid.

Finally, the waistis a geodesic and is preserved setwise by every isometry of the hyperboloid. Indeed, its Gaussian curvature isThe value $K=-1$ occurs exactly at $z=0$. Since Gaussian curvature is intrinsic and therefore preserved by isometries, every isometry preserves the waist. Thus the answer to the final question is also yes, with the isometry-invariant waist geodesic of the unit one-sheet hyperboloid as an example.