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The global Gauss-Bonnet theorem for a compact oriented surface without boundary isFor a connected closed orientable surface of genus $g$, this is

No. The missing point is not assumed to complete $S$ as a smooth surface, so the hypotheses of the global Gauss-Bonnet theorem do not apply to $S\cup \{p\}$.

For a concrete counterexample, rotate a smooth simple arc $(r(s),z(s))$, parametrized by arc length for $0\le s<L$, about the $z$-axis. Arrange that it starts smoothly on the axis,stays in $r>0$ for $0<s<L$, and converges to another point of the axis as $s\uparrow L$, withThe endpoint is omitted. The resulting surface of revolution is diffeomorphic to ${\mathbb{S}}^2$ minus one point, and adjoining its limiting endpoint makes the set compact.

For an arc-length surface of revolution,Therefore the total Gaussian curvature of a punctured surface with a singular compactification point iswhich can take values different from $4\pi$. Compactness of the one-point closure alone therefore does not determine the integral.

A geodesic triangulation of a closed smooth surface decomposes it into finitely many triangular discs whose edges are geodesic arcs and whose intersections are common faces. Its Euler number isThe Gauss-Bonnet theorem for a closed surface statesFor any triangulation of a closed surface, each triangular face has three edges and every edge belongs to two faces, so $3F=2E$. The sum of all vertex valencies is $2E$. Hence the average valency isThe torus has Euler characteristic $\chi =V-E+F=0$. Using $F=2E/3$ gives $V-E/3=0$, so $E=3V$ and

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

Since $K\ge 0$, Gauss-Bonnet gives $4\pi (1-g)\ge 0$, hence $g\le 1$. If $g=1$, the integral of the continuous nonnegative function $K$ is zero, so $K$ vanishes everywhere.

That is impossible for a compact regular surface in ${\mathbb{R}}^3$. Choose $p\in S$ maximizing the Euclidean distance from the origin, after translating the origin off the surface if necessary. The sphere centered at the origin through $p$ is a supporting sphere. The supporting-sphere curvature bound shows that both principal curvatures at $p$ have the same sign and nonzero magnitude, so $K(p)>0$. Therefore $g\ne 1$, and $g=0$. The classification theorem for surfaces now shows that $S$ is diffeomorphic to the sphere.

False with the $\mathrm{limsup}$ printed in the paper. Let the even-indexed surfaces be one fixed unit sphere and the odd-indexed surfaces one fixed ring torus. Thenbut infinitely many surfaces are tori. The two fixed surfaces lie in one ball and have uniformly bounded area. Their injectivity radii have a positive common lower bound, and compactness gives a common bound on $|\nabla K|$, hence on every radial derivative $|{\partial }_{r}K|$. Thus this alternating sequence satisfies all four additional conditions as well.

The intended statement becomes true if $\mathrm{limsup}$ is replaced byHere is the proof. Put ${\delta }_n=\max (0,-{\inf }_{S_n}{K}_n)$, so ${\delta }_n\to 0$. Choose $p_n$ maximizing distance from the centre of the fixed containing ball. The supporting-sphere curvature bound givesWrite $K_0=K_n(p_n)$ and $j(r,\theta )=\sqrt{G(r,\theta )}$ in geodesic polar coordinates. Choose a fixed sufficiently small $c>0$ and set ${\rho }_n=c{K}_0^{-1/2}$. Conditions (3) and (4) allow $c$ to be chosen so that ${\rho }_n\le {ϵ}_0$ andThe Jacobi equation $j_{rr}=-K_{n}j$, with $j(0)=0$ and $j_r(0)=1$, and the Sturm comparison theorem then give $j(r,\theta )\ge c_{1}r$ on this ball for a constant $c_1>0$ independent of $n$. ConsequentlySince $K_n\ge -{\delta }_n$ elsewhere and $\mathrm{Area}(S_n)\le M$,for all sufficiently large $n$. Gauss-Bonnet forces $\chi (S_n)>0$, and the classification theorem for surfaces makes each such $S_n$ 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 $K<0$ there is at most one. Two essential simple closed geodesics cannot form a geodesic bigon by Gauss-Bonnet theorem, so they are disjoint; the annulus between them would have $\chi =0$ and geodesic boundary, forcing $\int K=0$, a contradiction. Thus multiple geodesics in the flat case are disjoint, while in the positive example they need not be.

The geodesic curvature is the tangential component of $\alpha \prime \prime$, namely $k_g=⟨\alpha \prime \prime ,N\times \alpha \prime ⟩$ up to orientation; a curve is geodesic when $k_g=0$. Gauss-Bonnet theorem for a compact region says $\int K+{\int }_{\partial }{k}_g+\sum \text{exterior angles}=2\pi \chi$.

For $K=0$, a flat cylinder has infinitely many parallel closed geodesics.

Gauss-Bonnet theorem gives the area of a curvature-$-1$ triangle as $\pi -(\alpha +\beta +\gamma )$. Split the regular polygon into $2n$ congruent right triangles. If $\theta$ is a polygon half-angle, the right-triangle identity and part (a) giveThe polygon area is $(n-2)\pi -2n\theta$. Rewriting $\theta =\pi /2-{\cot }^{-1}(\tan \theta )$ yields

In the Poincare half-plane model, the Riemannian metricassigns to a smooth curve $\gamma (t)=(x(t),y(t))$ the lengthThe metric is conformal to the Euclidean metric, so hyperbolic and Euclidean angles agree. Its area element is

The two geodesics from $e^{i\alpha }$ and $e^{i\beta }$ to infinity are the vertical lines $x=\cos \alpha$ and $x=\cos \beta$, while the third side is the unit semicircle. The triangle lies above that semicircle, so its area isIts interior angles are $\alpha$, $\pi -\beta$, and zero at the ideal vertex, and henceTriangulating a geodesic polygon with $n$ sides into $n-2$ triangles givesthe area of a hyperbolic geodesic polygon and the polygonal Gauss-Bonnet theorem.

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.

Because $S_1$ and $S_2$ are isometric,The two induced orientations on their common boundary $\gamma$ are opposite. Ifis computed with the boundary orientation from $S_1$, then its value for $S_2$ is $-B$. Applying the Gauss-Bonnet theorem to the two halves givesThe equalities above imply $B=0$. This is the isometric halves give zero total boundary geodesic curvature argument.

Every ambient isometry ${\varphi }_{x,y}$ preserves the magnitude of geodesic curvature and sends $x$ to $y$. Hence $|k_g|$ is constant along the connected curve $\gamma$ by the transitive curve symmetry makes geodesic-curvature magnitude constant. If that constant were positive, continuity would force the signed $k_g$ to have one fixed sign, making its integral nonzero. Therefore $k_g\equiv 0$. A curve has zero geodesic curvature exactly when it is a geodesic, so $\gamma$ is geodesic.

After the $2n$ sides are identified in pairs, the polygonal-schema Euler count gives one face, $n$ edges and some number $V\ge 1$ of vertices. The Euler characteristic of a closed orientable genus-$g$ surface is $2-2g$, so

For genus two, take a regular octagon in the Poincare disc model and identify each side with its opposite side, with the orientation reversed along the boundary. One cyclic labelling isThe quotient has $V=1$, $E=4$, and $F=1$, hence Euler characteristic $-2$ and genus two. All eight vertices become one point, so smoothness requires their angles to sum to $2\pi$. Each interior angle is consequentlyEquivalently, the hyperbolic polygon area formula shows that the regular hyperbolic octagon fundamental polygon has areaas required by the Gauss-Bonnet theorem for a genus-two surface of curvature $-1$.

The Polyakov action contains the dilaton couplingFor a constant dilaton ${\Phi }_0$, the Gauss-Bonnet theorem gives $S_{\Phi }={\Phi }_0\chi (\Sigma )$, so the path-integral weight contributes $e^{-{\Phi }_0\chi }=g_s^{-\chi }$ with string coupling $g_s=e^{{\Phi }_0}$. A connected closed oriented genus-$g$ worldsheet has $\chi =2-2g$. Including one conventional factor of $g_s$ for each of $n$ external closed-string vertices gives the string genus expansionFor four external states on the sphere, this is $g_s^2$.

For a positively oriented simple closed curve bounding a disc $D$ on the unit sphere, the local Gauss-Bonnet theorem givesHere $K=1$, soAt fixed length $L$, maximizing the enclosed area $A$ is therefore exactly the same as minimizing ${\int }_0^L{\kappa }_{g}ds$.

A hyperbolic triangle is bounded by three hyperbolic geodesic segments or rays. Its vertices may lie in the hyperbolic plane; an ideal vertex is their endpoint on the boundary at infinity. Every ideal angle is zero. Gauss-Bonnet theorem with curvature $-1$ givesso an all-ideal triangle has area $\pi$.

For fixed admissible angles, hyperbolic trigonometry determines all three side lengths from the angles, for exampleThus two such triangles are congruent. An orientation-preserving isometry can send one chosen vertex and oriented tangent to the corresponding data of the other, and the determined side lengths and angles then send the entire triangle to it. Since the orientation-preserving isometry group is $PS{L}_2(\mathbb{R})$, represented by $S{L}_2(\mathbb{R})$, the stated action is transitive.

Let $M$ be a compact oriented surface with piecewise smooth, positively oriented boundary. If ${\theta }_j$ are the signed exterior turning angles at its corners, the global Gauss-Bonnet theorem with boundary isFor a smooth boundary the corner sum is absent, and for a closed surface both boundary terms are absent.

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.

A line segment has zero normal curvature in its tangent direction, impossible when Gaussian curvature is positive because both principal curvatures have the same nonzero sign. Two disjoint closed geodesics would bound an annulus. Applying Gauss-Bonnet theorem, its geodesic boundary contributes zero and $\chi =0$, but $\int KdA>0$, a contradiction.