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For the standard catenoid, the two principal curvatures are opposite, so its mean curvature vanishes. Its Gaussian curvature is $K=-{\mathrm{sech}}^{4}u<0$, which also shows that no neighbourhood is locally isometric to a plane, cylinder, or regular part of a cone, all of which have zero Gaussian curvature.

Let $S\subset {\mathbb{R}}^3$ be a smooth surface. Suppose its Gaussian curvature tends to infinity in absolute value along every maximal geodesic with a finite affine-parameter endpoint. Then $S$ is an inextendible embedded surface: a smooth extension would let some geodesic reach the extension boundary in finite time while its curvature remained locally bounded.

An elliptic point of a surface is a point at which the Gaussian curvature is positive, equivalently where the two principal curvatures have the same nonzero sign.

For a proper Euclidean motion of Euclidean three-space $E(x)=Ax+b$, transport the unit normal by $\tilde{N}(E(p))=AN(p)$. The corresponding shape operators are orthogonally conjugate:Their determinant and trace, hence Gaussian curvature and mean curvature, are unchanged.

If a unit tangent makes angle $\theta$ with the first principal direction, the normal curvature is ${\kappa }_n={\kappa }_1{\cos }^2\theta +{\kappa }_2{\sin }^2\theta$. This follows by diagonalizing the second fundamental form relative to the first fundamental form. When the Gaussian curvature is nonnegative, $|{\kappa }_n|\le 2|H|$.

On a connected regular radial interval, trivial holonomy around circular fibres forces the continuous function $f^{\prime }$ to be integer-valued, hence constant. The resulting Riemannian metrics have $f(r)=nr+b$, with $n\in \mathbb{Z}$ and $f>0$ on the interval. Their Gaussian curvature is $-f^{\prime \prime }/f=0$. For $n\ne 0$, $\rho =f/|n|$ and $\theta =|n|\varphi$ give a local isometry to the Euclidean plane; for $n=0$, $r,b\varphi$ give a flat cylindrical chart. The polar developing map is injective on the regular domain when $|n|=1$, and is a multiple covering when $|n|>1$. Trivial holonomy therefore does not imply a global Euclidean chart. The slope is a discrete parameter, not an unrestricted real one.

The same contact does not order Gaussian curvature, because determinant is not monotone under addition of a positive-semidefinite matrix when the original Hessian is indefinite or negative definite. For example,satisfy $g\ge f$ and are tangent at zero, but $K_g(0)=1<4=K_f(0)$.

For a ring torus of radii $a>b>0$, the Gaussian curvature isIt is positive on the outer side, negative on the inner side and zero on the top and bottom circles. The sign contrast expresses two principal curvatures of the same sign outside and opposite signs inside.

For the regular $C^2$ ruled surface $\sigma (x,y)=\mathbf{a}(x)+y\mathbf{b}(x)$, put $W=|({\mathbf{a}}^{\prime }+y{\mathbf{b}}^{\prime })\times \mathbf{b}|>0$. The second-form coefficient in the ruling direction is zero, and the mixed coefficient is ${\mathbf{a}}^{\prime }\cdot (\mathbf{b}\times {\mathbf{b}}^{\prime })/W$. The determinant formula for Gaussian curvature therefore gives the displayed nonpositive expression. Its vanishing is equivalent to the zero scalar triple product, only at regular points.

A geodesic triangle is bounded by three geodesic segments. On a surface its angle excess is the integral of the Gaussian curvature over the triangle.

A hyperbolic paraboloid is a saddle-shaped quadric such as $z=xy$. Its graph parametrization $(x,y,xy)$ has Gaussian curvature $-1/(1+x^2+y^2{)}^2$ and is regular everywhere. It is not a hyperboloid: their quadratic equations have different forms.

A surface with a complete Riemannian metric of constant Gaussian curvature $-1$. Its local model is the Poincare half-plane model; the orientable case is a quotient by a free properly discontinuous group action of orientation-preserving isometries.

This metric pulls back the round spherical metric by $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$. It has unit Gaussian curvature and Ricci curvature equal to its metric, yet the meridian reaches a missing boundary in finite time. Every curve between $(0,0)$ and $(0,L)$ has length at least $|L|/\sqrt{2}$, so its diameter is infinite. It shows that the positive Ricci bound alone does not replace geodesic completeness in the Bonnet-Myers theorem.

Replacing the chosen unit normal $N$ by $-N$ replaces the shape operator $S$ by $-S$. On a surface, Gaussian curvature $K=\det S$ is unchanged, while mean curvature $H=\mathrm{tr}(S)/2$ changes sign.

An orthogonal transformation of a real inner-product space is a linear map $Q$ satisfying $Q^{T}Q=I$. It preserves inner products, lengths, angles, and therefore Gaussian curvature when applied to an embedded surface in Euclidean space.