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For a minimal surface, the shape operator has eigenvalues $k,-k$, so $⟨dN(v),dN(w)⟩=k^2⟨v,w⟩$. Conversely this metric identity only implies that the two principal curvatures have equal squares. A sphere has equal nonzero principal curvatures and satisfies the identity while having nonzero mean curvature. The distinction is between opposite and equal signs, not equality of magnitudes.

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.

The Bishop-Gromov inequality forces every volume ratio to equal one if the asymptotic ratio is one. Equality in the radial Jacobian comparison forces the radial operator $A(X)={\nabla }_X{\partial }_r$ of each distance sphere to be $r^{-1}I$ (it is the negative of the outward shape operator in the convention $S=-\nabla \nu$), by the traced radial Riccati equation for distance spheres and equality in Cauchy-Schwarz inequality. Any finite cut time would remove a positive cone of radial volume at a later radius, contradicting equality. Thus the Riemannian exponential map is global and one-to-one, and the polar metric is $d{r}^2+r^2{g}_{S^{n-1}}$, the Euclidean metric.

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.

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.

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.

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.

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.

Let $i:M\hookrightarrow N$ be an embedded submanifold. In the general terminology, a vector field along a map $i$ is a smooth section $X\in \Gamma (i^{\ast }TN)$, so $X_p\in T_{p}N$. A local extension of a vector field along a submanifold near $p$ is a vector field $\tilde{X}$ on an open neighbourhood in $N$ with $\tilde{X}{|}_M=X$ on that neighbourhood. Choose adapted coordinates $(x^1,\dots ,x^m,y^{m+1},\dots ,y^n)$ in which $M$ is given by $y=0$. Writing $X=X^a(x){\partial }_a{|}_{y=0}$, extend the coefficient functions independently of $y$:This proves local existence, including for fields with normal components.

For the Lie bracket assertion, the fields must be tangent to $M$, namely sections of $TM$ viewed inside $i^{\ast }TN$. In this case the useful criterion for a local extension of a vector field along a submanifold isIf $\tilde{X}$ extends $X$, the chain rule gives this identity because their tangent values agree. Conversely, applying it to the adapted coordinate functions forces the tangent components to equal those of $X$ and all normal components on $M$ to vanish, so $\tilde{X}{|}_M=X$. The criterion also shows that functions vanishing on $M$ are differentiated to functions vanishing on $M$ by $\tilde{X}$.

For tangent vector fields $X,Y$ and local extensions $\tilde{X},\tilde{Y}$, apply the criterion twice:Hence the ambient bracket restricts to the intrinsic bracket, independently of the chosen extensions. Tangency is essential. For example, on the $x$-axis in ${\mathbb{R}}^2$, let $X={\partial }_y{|}_M$ and $Y=0{|}_M$. The two extensions $\tilde{Y}=0$ and $\tilde{Y}=y{\partial }_x$, with $\tilde{X}={\partial }_y$, give bracket restrictions $0$ and ${\partial }_x{|}_M$. Thus general sections with normal components do not possess the extension-independent Lie bracket asserted for tangent fields.

For the hypersurface $M^n\subset {\mathbb{R}}^{n+1}$, choose a smooth unit normal $\nu$ locally. A global choice is needed only if one wants a globally signed shape operator; an arbitrary hypersurface need not come equipped with such a choice. With $D$ the Euclidean Levi-Civita connection, define the shape operator, also called the Weingarten map, byThe derivative is defined along $M$ and is linear in $v$. Since $v⟨\nu ,\nu ⟩=2⟨D_v\nu ,\nu ⟩=0$, its value is tangent to $M$, so this is an endomorphism of the tangent space. Differentiating $⟨\nu ,Y⟩=0$ gives, for tangent vector fields $X,Y$,This is the scalar second fundamental form for the chosen normal. The Euclidean connection on a vector bundle has zero torsion tensor, giving $D_{X}Y-D_{Y}X=[X,Y]$. The bracket is tangent to $M$ by the previous argument; thereforeConsequently $\boxed{⟨S_{p}v,w⟩=⟨v,S_{p}w⟩}$ for every $v,w\in T_{p}M$: the shape operator is self-adjoint. Reversing $\nu$ reverses $S$ and $h$ but preserves this self-adjointness.

Put $\theta (r)=\mathrm{Vol}B(p,r)/({\omega }_n{r}^n)$. The Bishop-Gromov inequality says that $\theta$ is nonincreasing and at most one, while smoothness gives $\theta (r)\to 1$ as $r\downarrow 0$. The prescribed limit at infinity is also one. Hence $\theta (r)=1$ for every $r>0$.

Here is how equality gives the global Euclidean rigidity of maximal asymptotic volume ratio. Assume first $n\ge 2$. In geodesic polar coordinates centered at $p$, let $c(\xi )$ be the cut time along unit direction $\xi$, and $J(r,\xi )$ the positive radial volume density before that time. The comparison giving the Bishop-Gromov inequality says $J(r,\xi )\le r^{n-1}$, andNo cut time can be finite. If $c({\xi }_0)<t$, then the radial segment at time $t$ is not minimizing, so $d(p,{\exp }_p(t{\xi }_0))<t$. This strict inequality persists for nearby directions by continuity, and their cut times are also at most $t$. At any radius $R>t$, that open set of directions omits all radial volume from $t$ to $R$. Even the maximal density $r^{n-1}$ cannot then give Euclidean volume, contradicting $\theta (R)=1$. Thus $c(\xi )=\infty$ for all directions. Equality of all ball volumes and continuity now force $J(r,\xi )=r^{n-1}$ everywhere.

Let $A(r)X={\nabla }_X{\partial }_r$ on the distance sphere. This is the negative of the outward shape operator in the convention $S=-\nabla \nu$. Put $h=\mathrm{tr}A={\partial }_r\log J$. The traced radial Riccati equation for distance spheres isSince $h=(n-1)/r$, it givesBut $\mathrm{tr}(A^2)\ge h^2/(n-1)=(n-1)/r^2$ by the Cauchy-Schwarz inequality, and the Ricci term is nonnegative. Equality therefore holds in both, so $A=r^{-1}I$. Writing the polar metric as $d{r}^2+g_r$, its evolution is ${\partial }_r{g}_r=2g_r/r$. Its small-radius Euclidean limit then givesThe Riemannian exponential map is onto by the Hopf-Rinow theorem, has no conjugate points, and is injective because there is no finite cut time. Explicitly, two distinct minimizing radial geodesics reaching the same point would, after extending one past that point, create a length-minimizing broken curve with a genuine corner; smoothing the corner shortens it. That contradicts global minimization of every radial segment. Thus ${\exp }_p$ is a global diffeomorphism, and the displayed metric provesIn dimension one, a connected complete manifold without boundary is a line or a circle. The circle has limiting volume ratio zero, so the hypothesis leaves only the Euclidean line.

At $P=\sigma (u,v)$, regularity makes ${\sigma }_u,{\sigma }_v$ linearly independent. The tangent space is $T_{P}U=\mathrm{span}\{{\sigma }_u,{\sigma }_v\}$, and an oriented unit normal vector isReversing its sign chooses the opposite orientation. The first fundamental form is the squared length of a tangent displacement:where $E={\sigma }_u\cdot {\sigma }_u$, $F={\sigma }_u\cdot {\sigma }_v$, $G={\sigma }_v\cdot {\sigma }_v$. Its determinant $EG-F^2=|{\sigma }_u\times {\sigma }_v{|}^2$ is positive.

Use $N_{\mathrm{II}}={\sigma }_{vv}\cdot \mathbf{n}$ to distinguish the scalar coefficient of the second fundamental form from the normal vector. Since $\mathbf{n}\cdot \mathbf{n}=1$, its derivatives are perpendicular to it and hence tangent. Write ${\mathbf{n}}_u=a{\sigma }_u+b{\sigma }_v$ and ${\mathbf{n}}_v=c{\sigma }_u+d{\sigma }_v$. Differentiating $\mathbf{n}\cdot {\sigma }_u=\mathbf{n}\cdot {\sigma }_v=0$ givesSubstitution of the tangent expansions provesThe shape operator is $-d\mathbf{n}$; its matrix in the parametrized tangent basis is the negative transpose of the coefficient matrix shown. Its determinant, the product of the principal curvatures, is thereforethe Gaussian curvature. Transposition and the two minus signs do not change this determinant.

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.

Parametrize the surface by $r(x,y)=(x,y,xy)$. Its tangent vectors are $r_x=(1,0,y)$ and $r_y=(0,1,x)$. Choose the upward unit normalThe first fundamental form coefficients areSince $r_{xx}=r_{yy}=0$ and $r_{xy}=(0,0,1)$, the second fundamental form coefficients are $e=g=0$ and $f=(1+x^2+y^2{)}^{-1/2}$. The determinant of the shape operator gives the Gaussian curvature:Changing the normal's sign changes both second-form factors and leaves this determinant unchanged. Although the question calls the surface a hyperboloid, $z=xy$ is a hyperbolic paraboloid; the calculation uses the stated surface equation.

At a non-planar point of a minimal surface, the shape operator is not zero. Minimality gives nonzero opposite principal curvatures, so $K<0$ nearby and $dN$ is invertible. The inverse function theorem makes the Gauss map a local diffeomorphism onto a patch of the sphere; part (ii) makes it conformal.

Choose stereographic coordinates $(u,v)$ on that sphere patch. Their metric is $4(d{u}^2+d{v}^2)/(1+u^2+v^2{)}^2$. Pulling them back by $N$, the conformal identity $N^{\ast }{g}_{S^2}=\lambda g_S$ givesThe positive coefficient shows that these are local isothermal coordinates. No general isothermal-coordinate existence theorem is needed: the nondegenerate conformal Gauss map supplies them explicitly.

Let $A=-dN$ be the self-adjoint shape operator. In an orthonormal principal basis, $⟨dN(v),dN(w)⟩=⟨Av,Aw⟩$, with diagonal coefficients $k_1^2,k_2^2$. The Gauss map is conformal with nonzero factor exactly when $k_1^2=k_2^2=\lambda >0$.

If the surface is minimal and has no umbilics, $k_2=-k_1$ and $k_1\ne 0$, so this equality holds. Conversely equality of the squares gives either $k_1=k_2$ or $k_1=-k_2$. The first possibility is excluded by the no-umbilic hypothesis. Hence $H=0$, proving minimality is equivalent to conformality of the Gauss map under the stated hypothesis. The factor is automatically positive because it equates positive definite inner products. The hypothesis is essential: a sphere has conformal Gauss map and nonzero mean curvature.

For a regular parametric surface $P(u,v)$, form $N=P_u\times P_v$, $n=N/\Vert N\Vert$ and define the offset evaluatorIt is already a parametric surface with the same parameter domain, even if its coordinates no longer belong to the original polynomial or rational patch family. Point interrogation requires evaluations of $P,P_u,P_v$ and one normalization. For tangent or normal vector interrogation, differentiate the evaluator. For example,and similarly for $v$. The offset normal vector is the normalized cross product $(P_d{)}_u\times (P_d{)}_v$ wherever it is nonzero. These derivatives also support intersection, projection and curvature calculations by the usual methods for parametric surfaces. For a variable normal offset, add $d_{u}n$ and $d_{v}n$ to the corresponding tangent derivatives.

With shape operator $W=-Dn$, the constant-distance offset differential is $D{P}_d=(I-dW)DP$. Along a principal direction its factor is $1-d{\kappa }_i$, where ${\kappa }_i$ is a principal curvature. Consequently regular interrogation fails at $d{\kappa }_i=1$; on regular pieces the offset principal curvatures are ${\kappa }_i/(1-d{\kappa }_i)$ with the continued normal orientation. Global self-intersections can occur even when both local factors are nonzero. Evaluate and differentiate the base surface and its normalized normal; an exact conversion back to the original patch type is unnecessary.

For an oriented surface with unit normal $N$, the Gauss map sends $p$ to $N(p)\in S^2$. The shape operator is $-dN$ on the tangent plane; its eigenvalues are the principal curvatures $k_1,k_2$. The Gaussian curvature is $K=k_1{k}_2$ and the mean curvature is $H=(k_1+k_2)/2$. Reversing orientation reverses both principal curvatures and $H$, but preserves $K$. The Theorema Egregium says that Gaussian curvature is determined by the first fundamental form, so is invariant under local isometries.