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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.

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

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.

For a coordinate curve $v=v_0$, the length between $u_1$ and $u_2$ iswhere $E$ is a coefficient of the first fundamental form. Thus $\partial E/\partial v=0$ makes the two $u$-directed sides of every coordinate rectangle equally long. Similarly,so $\partial G/\partial u=0$ makes the two $v$-directed sides equally long.

Conversely, suppose every pair of opposite sides has equal length. Fix $v_0,v_1$. Equality of the first pair for every interval $[u_1,u_2]$ givesThe fundamental theorem of calculus implies that the continuous integrand vanishes, so $E$ is independent of $v$. The same argument shows that $G$ is independent of $u$. Hence the stated equal-length property is equivalent to $E_v=G_u=0$.

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.

Take the circular cylinderIts first fundamental form is $R^{2}d{\theta }^2+d{z}^2$. The universal covering parametrization ${\mathbb{R}}^2\to S$ is therefore a local isometry from the Euclidean plane, after scaling the first coordinate. By geodesic preservation by a local isometry, every geodesic on the cylinder lifts to a straight line defined for all $t\in \mathbb{R}$. Thus the cylinder is geodesically complete. It is noncompact because $z$ is unbounded, and it is not a plane because one of its principal curvatures is $1/R$.

Take $n^a{n}_a=-1$. The first fundamental form, or induced metric, is the spatial projection tensorWith the convention used in the question, define the second fundamental form byLocally $n_a$ is proportional to the gradient of a defining function for the hypersurface, so $h_a{}^c{h}_b{}^d{\nabla }_{[c}{n}_{d]}=0$ and $K_{ab}=K_{ba}$. Both derivative indices are projected tangentially. If two extensions of $n$ agree on $\Sigma$, their difference vanishes there, as does its tangential derivative, so $K_{ab}$ is independent of the extension.

Choose an orthonormal basis $(e_1,e_2)$ of $T_{p}S$ and write $e(\theta )=\cos \theta e_1+\sin \theta e_2$. On a punctured disc on which the exponential map is a diffeomorphism, the mapdefines geodesic polar coordinates centred at $p$. The curves $r\mapsto X(r,\theta )$ are unit-speed geodesics, so the first coefficient of the first fundamental form isThe Gauss lemma says that radial and angular coordinate vectors are orthogonal, hence

The angular variation $J=X_{\theta }$ is a Jacobi field along each radial geodesic, with initial dataSince $G=⟨X_{\theta },X_{\theta }⟩=|J{|}^2$, smooth dependence of the geodesic flow givesConsequentlyas required.

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

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 smooth curve $\gamma (t)$ on a Riemannian surface, its energy of a curve isChoose a local parameterization $\mathbf{X}(u,v)$ and write $\gamma (t)=\mathbf{X}(u(t),v(t))$. With coefficients of the first fundamental formthe Lagrangian isThe Euler-Lagrange equations are equivalentlyAfter multiplying by the inverse metric these become the geodesic equationwhere the Christoffel symbols are determined by $E,F,G$.

If a straight line segment lies in the surface, parameterize it byThen $\ddot{\gamma }=0$, so its acceleration has zero tangential component. The two displayed equations hold directly, and the segment is a geodesic.

For the one-sheeted hyperboloidputTwo distinct ruling lines through $P$ areIndeed, the direction $d_{\pm }$ satisfiesso substitution shows that every point of the line lies in $H$. These are geodesics by the straight-line argument.

A third geodesic is the meridian through $P$. Choose $s_0$ with $\sinh s_0=z_0$; thenIts acceleration is ${\gamma }_m$, which is normal to $H$, so it is a geodesic. If $z_0\ne 0$, these give the required three distinct subsets.

If $z_0=0$, there is also the equatorial circle. Writing $P=(\cos {\theta }_0,\sin {\theta }_0,0)$,Its acceleration is normal to $H$ along $z=0$, so it is a fourth geodesic distinct from the meridian and the two rulings.

Finally, write $\rho =\sqrt{x^2+y^2}=\sqrt{1+z^2}$. Clairaut's relation gives the conserved quantityChoose initial data in $z>0$ withSince $|\sin \psi |\le 1$, every point of the resulting geodesic satisfies $\rho \ge c$, and thereforeContinuity keeps the geodesic in the component $z>0$, and the stated completeness assumption defines it for every real time. The continuum of choices of $c$ supplies infinitely many such geodesics.

For each $p\in S$, the first fundamental form is defined without coordinates bythe restriction of the Euclidean inner product to the tangent space of $S$.

If $\varphi (u,v)$ is a local parametrization andthenTaking their dot product giveswhere

Since $z>0$, the cone has the parametrizationIts coordinate derivatives satisfyso its first fundamental form is

On a sufficiently narrow angular patch, defineand map the cone to the plane point with polar coordinates $(R,\Theta )$. The Euclidean metric pulls back asThe map is therefore a local isometry from a circular cone to the plane, proving that $\Sigma$ is locally isometric to the Euclidean plane.

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

The first fundamental form of an embedded surface $S\subseteq {\mathbb{R}}^3$ is the inner product induced on each tangent plane:For a tangent vector field $V$ along a curve $\alpha$, its surface covariant derivative isthe tangential projection of its ordinary derivative. A curve is a geodesic when

A local isometry preserves the first fundamental form, so its Christoffel symbols and Levi--Civita covariant derivatives satisfyTaking $V={\alpha }^{\prime }$ shows that if $\alpha$ is geodesic, then $\varphi \circ \alpha$ is geodesic.

The converse is false. The dilationmaps every affinely parametrized straight-line geodesic to another such geodesic, butso it does not preserve the first fundamental form. This is the geodesic-preserving homothety that is not a local isometry.

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.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

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.