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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.

A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm 1/R$, according to orientation. Hence $K=0$ and $|H|=1/(2R)$.

This is the circular generating-curve case of a Cylinder; the curvature formulas here refer to its lateral surface.

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

Curvature depresses the equilibrium melting temperature of solid convex into liquid. Here $\mathcal{K}$ is the sum of principal curvatures, equal to twice a common convention for mean curvature, and $T_m$ is absolute temperature. The coefficient uses solid-liquid surface energy, mass density and latent heat. This stabilizes fine corrugations of a solidification front.

Put $q=(f^{\prime 2}+g^{\prime 2}{)}^{1/2}$ for the profile parametrizationWith unit normal $N=(-g^{\prime }\cos u,-g^{\prime }\sin u,f^{\prime })/q$, the principal curvatures areConsequently $H=(k_1+k_2)/2$ and $K=k_1{k}_2$.

The differential geometry of smooth surfaces studies their induced Riemannian metrics and intrinsic or extrinsic curvature. The first fundamental form measures lengths, while the second fundamental form and shape operator describe bending in an ambient Euclidean space. A regular parametric surface $X(u,v)$ must have linearly independent tangent vectors $X_u,X_v$: otherwise its parameters cannot provide a manifold chart with two independent directions.

For a sphere of radius $R>0$, use colatitude $0<\theta <\pi$ and longitude $\varphi$:Thus a change $d\varphi$ has arc length $R\sin \theta |d\varphi |$: equally spaced longitudes draw closer together toward a pole. The coordinate degeneracy at the poles belongs to this parametrization, not to the smooth surface.

Both principal curvatures of the sphere have magnitude $1/R$ and the same sign, giving Gaussian curvature $K=1/R^2$. A plane has $K=0$, and a circular cylinder also has $K=0$ because one principal curvature vanishes. These three familiar shapes show why visible bending alone cannot measure intrinsic Gaussian curvature.

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.

Rolling a sheet into a circular cylinder bends it without changing its internal lengths and angles. Fitting that same sheet smoothly onto a sphere requires distortion of its first fundamental form. Gaussian curvature distinguishes these situations: it measures the intrinsic curvature of a smooth surface, even though its simplest definition uses how the surface bends in Euclidean space.

The Gaussian curvature at a point is the determinant of the shape operator, or the product of the two principal curvatures:Here $I=Ed{u}^2+2Fdudv+Gd{v}^2$ is the first fundamental form and $II=ed{u}^2+2fdudv+gd{v}^2$ is the second fundamental form, with $S=-dN$. Regularity makes $EG-F^2>0$. Reversing the unit normal negates $e,f,g$ and both principal curvatures, leaving Gaussian curvature unchanged. ^[ref]

The sign describes bending at the point: positive Gaussian curvature means the principal curvatures have the same nonzero sign, as at an elliptic point; negative Gaussian curvature means opposite signs, as on a saddle; zero Gaussian curvature means at least one principal curvature vanishes. Zero Gaussian curvature at a single point does not imply that a neighborhood is a plane. ^[ref]

Four worked models make the signs concrete:

The patches use a common length unit, with $R=1$ for the sphere and circular cylinder. The marked red and blue curves have principal directions at the black point; the arrow is the chosen unit normal. For physical lengths, principal curvature and mean curvature have units of inverse length, while Gaussian curvature has units of inverse squared length. Rescaling a surface by $\lambda >0$ divides its principal curvatures by $\lambda$ and its Gaussian curvature by ${\lambda }^2$. Writing the saddle as $h=(x^2-y^2)/(2L)$ restores an explicit length scale: $K(0)=-1/L^2$.

The Theorema Egregium says that Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. A local isometry therefore preserves Gaussian curvature, even if the shape operator and the principal curvatures change. ^[ref]

Here is the outline, rather than a full proof. The Gauss formula splits second derivatives of a parametric surface into tangential Christoffel symbols and the second fundamental form. The Christoffel symbols are determined by the first fundamental form and its first derivatives. Commuting the next partial derivatives gives the Gauss equation, which identifies $\det S$ with the intrinsic sectional curvature computed from the Riemann curvature tensor. Thus the apparently extrinsic product $k_1{k}_2$ is determined by the Riemannian metric. ^[ref]

One can see the distinction directly by rolling a plane strip with coordinates $(u,v)$ into the circular cylinder $X(u,v)=(R\cos (u/R),R\sin (u/R),v)$. Both have $I=d{u}^2+d{v}^2$, so the map is a local isometry, and is one-to-one on an open strip of width less than $2\pi R$. Extending $u$ over all real numbers wraps around periodically; it does not give a global isometry between the entire plane and the entire circular cylinder. A nonempty open patch of a sphere cannot be flattened by a local isometry into a plane, because its Gaussian curvature $1/R^2$ cannot become zero. The obstruction concerns preservation of the first fundamental form, rather than a particular drawing or choice of seam.

A circular cylinder of radius $\rho$ has induced metric locally equal to $d{u}^2+d{z}^2$ with $u=\rho \phi$. Its Riemann curvature tensor therefore vanishes, despite its nonzero extrinsic curvature. With outward unit normal and convention $K_{ij}=e_i^a{e}_j^b{\nabla }_a{n}_b$, its trace is $1/\rho$ and its principal curvatures are $1/\rho$ and zero. Its averaged mean curvature is $1/(2\rho )$. The angular identification changes global topology, not local flatness.

Mean curvature measures the average signed bending of a smooth surface. Unlike Gaussian curvature, it adds the two principal curvatures rather than multiplying them. A circular cylinder therefore has nonzero mean curvature despite zero Gaussian curvature, whereas a catenoid has zero mean curvature despite negative Gaussian curvature. The cancellation on a catenoid explains how a curved minimal surface can be stationary for surface area.

Choose a unit normal $N$ and the shape operator $S=-dN$, as in the second fundamental form convention $II(v,w)=⟨Sv,w⟩$. For a smooth surface in Euclidean space of dimension three, define the signed scalar mean curvature byHere $E,F,G$ are the coefficients of the first fundamental form, and $e,f,g$ those of the second fundamental form. The factor $1/2$ makes $H$ an average; the scalar trace convention instead calls $k_1+k_2=2H$ the mean curvature. Authors who use $S=+dN$ reverse both signs. A formula involving mean curvature is therefore incomplete unless its shape operator, unit normal, and averaging convention are specified.

In orientation reversal of surface curvature, both principal curvatures and $H$ change sign, while Gaussian curvature and the mean curvature vector do not. The units of mean curvature are inverse length, compared with inverse squared length for Gaussian curvature. The distinction matters in soap film pressure balance: pressure couples to the sum of the signed principal curvatures.

Scale the catenoid by its waist radius $a>0$ and choose a unit normal with outward radial component:The hyperbolic cosine controls the profile $r(z)=a\cosh (z/a)$. Direct differentiation yields the first fundamental form and second fundamental formThe coordinate tangent vectors are orthogonal, so the meridional and circumferential principal curvatures areThe opposite principal curvatures cancel at every point, not just at the waist. This is precisely why the catenoid is a minimal surface, while its nonzero second fundamental form and negative Gaussian curvature show that it is not a plane.

For a sphere of radius $R>0$, its outward unit normal is $N(p)=p/R$. Applying the shape operator gives $S(v)=-v/R$, so the two principal curvatures coincide:For the lateral surface of a circular cylinder with outward unit normal, moving around a circle turns $N$ at rate $1/R$, while moving along the axis leaves $N$ constant. Thus the circumferential and axial principal curvatures giveHere $K$ is Gaussian curvature. The circular cylinder bends in only one principal direction, so its mean curvature is half that of the equally sized sphere. These formulas concern the smooth lateral surface, excluding any edges where end caps meet it.

The common length scale is $\ell$, equal to the radius $R$ for the sphere and circular cylinder, and to the waist radius $a$ for the catenoid. The mean curvature marker is halfway between the two principal curvature markers. Gaussian curvature has a different unit and is listed separately beneath each panel.

A minimal surface has vanishing mean curvature vector. For a smooth surface in three-dimensional Euclidean space, this means $H=0$, equivalently $k_1+k_2=0$. Thus Gaussian curvature is $K=-k_1^2\le 0$, but neither principal curvature needs to vanish. A plane and a catenoid both satisfy the condition, with very different Gaussian curvatures.

By the first variation of area formula, a minimal surface is a stationary point of surface area for every smooth variation that fixes its boundary, or has compact support in its interior. The word “minimal” specifies this differential equation; it does not guarantee a global minimum of surface area, or even a local minimum. A stable minimal surface imposes a further condition on the second variation.

For $0<A<1$, rotate the unit-speed profilenear $s=0$. Since $K=-Y^{\prime \prime }/Y$, the resulting regular surface of revolution has $K=1$. At $s=0$ its principal curvatures are $A$ and $A^{-1}$, so the point is not umbilical when $A\ne 1$ and no neighbourhood of it lies on a sphere.

A parallel surface is a constant-distance normal displacement $P_d=P+dn$ of a regular surface with unit normal vector $n$. With shape operator $W=-Dn$, its differential is $(I-dW)DP$; it can become singular where $1-d{\kappa }_i=0$ for a principal curvature ${\kappa }_i$.

// Target: geometry-and-topology.bigb

For a regular parametric surface $r(u,v)$, setThe first fundamental form is the induced Riemannian metricThe second fundamental form, for this choice of unit normal vector, isThe Gaussian curvature is the product of the principal curvatures, equivalently the determinant of the shape operator:Reversing the normal reverses the second fundamental form and both principal curvatures, while leaving $K$ unchanged.

For a regular parametric surface $\mathbf{r}(u,v)$, its tangent vectors ${\mathbf{r}}_u,{\mathbf{r}}_v$ are linearly independent. The first fundamental form is the induced inner product on the tangent plane, writtenIt measures squared lengths and angles; its determinant satisfies $EG-F^2=|{\mathbf{r}}_u\times {\mathbf{r}}_v{|}^2>0$. Choose the unit normalThe second fundamental form, with shape operator $S=-dN$, isThe subscript distinguishes the coefficient $g_2$ from a genus or a Riemannian metric. In these coordinates the shape operator has matrix $I^{-1}II$, so the Gaussian curvature, the product of its two principal curvatures, isReversing $N$ reverses $II$ and both principal curvatures, leaving the Gaussian curvature unchanged.

Use the cylindrical coordinate system basis ${\mathbf{e}}_r,{\mathbf{e}}_{\phi },{\mathbf{e}}_z$. The meridional unit tangent vector and outward unit normal vector consistent with the given geometry areAn surface of revolution has $\theta$ independent of $\phi$. ConsequentlyApply the product rule to the vector force balance:Equating the three independent components to zero, and using $r_s=\cos \theta$, givesThus, away from the rotation axis where $r>0$, the scalar mechanical equilibrium equations areThe two geometric factors ${\theta }_s$ and $\sin \theta /r$ are the signed meridional and azimuthal principal curvatures in this convention. The normal equation is the Young–Laplace equation generalized to unequal principal membrane tensions.

Let $D$ be the ambient Levi-Civita connection and $\nabla$ the one for the induced metric on $M$. The Gauss formula iswhere the second fundamental form $II(X,Y)=(D_{X}Y{)}^{\perp }$ is normal. The tangential part of $D$ is metric-compatible and torsion-free, so the existence and uniqueness of the Levi-Civita connection identifies it with $\nabla$.

For a normal field $\xi$, the Weingarten formula defines the shape operator in a normal direction $A_{\xi }$ and the normal connection byDifferentiate $⟨\xi ,Y⟩=0$ using metric compatibility to getThe symmetry of the second fundamental form follows from the ambient torsion-free condition:since the Lie bracket of vector fields tangent to $M$ is tangent. Thereforeso every normal-direction shape operator is self-adjoint.

The paper uses $R(X,Y)=D_{[X,Y]}-[D_X,D_Y]$, so use the Gauss equation with reversed curvature convention. Applying the Gauss formula and Weingarten formula twice givesSubtracting in the stated curvature order yieldsPair with a tangent field $W$ and use the defining relation for the shape operator in a normal direction:This is the Gauss equation in the supplied sign convention. If one defines curvature by $[D_X,D_Y]-D_{[X,Y]}$ instead, the two quadratic terms have the opposite signs.

For an orthonormal frame $X,Y$ on a surface, positive Gaussian curvature with the convention here is $K=⟨R^M(X,Y)X,Y⟩$. If the ambient space is Euclidean and $\xi$ is a unit normal, the equation becomesThe Koszul formula determines $\nabla$ from the induced metric, and hence determines its Riemann curvature tensor and $K$. Therefore the Gaussian curvature depends only on that metric, even though its expression as a product of principal curvatures appears extrinsic. This is Theorema Egregium.

Evaluate all quantities at $(u_0,v_0)$ and assume a regular surface point, $S_u\times S_v\ne 0$. The tangent plane passes through $S_0=S(u_0,v_0)$ and is spanned by $S_u,S_v$. ThusChoose the unit normal $n=(S_u\times S_v)/\Vert S_u\times S_v\Vert$. The first fundamental form has coefficientswhile the second fundamental form has coefficientsTo derive the curvature formula, differentiate $n\cdot S_u=n\cdot S_v=0$. This gives $-n_u\cdot S_u=e$, $-n_u\cdot S_v=f$, $-n_v\cdot S_u=f$ and $-n_v\cdot S_v=g$. Therefore the shape operator $-dn$, expressed in the tangent basis, isIts eigenvalues are the principal curvatures. Their product, the Gaussian curvature, is its determinant:Regularity ensures $EG-F^2=\Vert S_u\times S_v{\Vert }^2>0$. Reversing $n$ changes the signs of $e,f,g$ and both principal curvatures but leaves $K$ unchanged. At a singular parameter point this quotient and the cross-product normal vector are undefined; an alternative regular chart or separate geometric analysis is required.