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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.

A plane has both principal curvatures zero, hence $K=H=0$. A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm R^{-1}$, henceBecause the absolute value of mean curvature is preserved even if one reverses the chosen orientation, no Euclidean motion can carry any open piece of the cylinder to an open piece of the plane. Thus the condition $K\equiv 0$ does not characterize planar pieces.

There are analogous examples for both signs of constant Gaussian curvature. Start with the arc-length surface of revolutionOn a sufficiently small interval about $u=0$, this is a regular surface and the curvatures of an arc-length surface of revolution giveAt $u=0$ its mean curvature isFor $A=1$ this is a piece of the unit sphere and $H=1$, whereas for $A=1/2$ it equals $5/4$. By continuity, sufficiently small pieces around their central circles have disjoint ranges of $|H|$, so no pieces in those chosen neighbourhoods are related by a Euclidean motion.

For curvature minus one, useagain on a sufficiently small interval. ThenThe choices $A=1$ and $A=2$ give $H(0)=0$ and $H(0)=-3/4$. Restricting to small enough pieces again separates the ranges of $|H|$. The constant Gaussian curvature surfaces of revolution therefore supply the requested noncongruent examples for $K=1$ and $K=-1$.

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.

Choose a smooth unit normal $N$. For $v,w\in T_p\Sigma$, the first fundamental form and second fundamental form arewhere the shape operator is $S_p=-d{N}_p$. The principal curvatures are the eigenvalues ${\kappa }_1,{\kappa }_2$ of $S_p$; the mean curvature and Gaussian curvature areThe surface is minimal when $H=0$ everywhere.

For a local parametrization $X(u^1,u^2)$, the first fundamental form is the Euclidean inner product restricted to each tangent plane:Writing $u^1=u,u^2=v$, this is

After choosing an orientation, the Gauss map isThe second fundamental form is $II(v,w)=⟨S(v),w⟩$, where the shape operator is $S=-dN$. In coordinates,The Gaussian curvature and mean curvature are respectively the determinant and half the trace of $S$:Changing the chosen normal reverses $II$ and $H$ but leaves $K$ unchanged.

If a flat surface disc is replaced by a smooth disc agreeing with the original surface outside the same boundary, their boundary geodesic curvatures agree. Gauss-Bonnet theorem therefore forces the replacement’s total Gaussian curvature to be zero. If its curvature is everywhere nonnegative, continuity forces it to vanish everywhere. Thus a positive-curvature bulge cannot be glued this way without some negative curvature or a loss of smoothness.

If a compact regular surface lies in a closed Euclidean ball of radius $R$, maximize distance from the ball's centre. At a maximizing point, comparison of the second fundamental form with the tangent supporting sphere makes both principal curvatures have magnitude at least $R^{-1}$ and the same sign. The Gaussian curvature there is therefore at least $R^{-2}$.