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The map $T(x)=1/x-\lfloor 1/x\rfloor$ on $(0,1]$, with $T(0)=0$. Its digit $N(x)=\lfloor 1/x\rfloor$ gives the successive partial quotients $N(T^{j}x)$ of a continued fraction. This is distinct from the geometric Gauss map assigning surface normals.

The Gauss map sends each point of an oriented surface to its chosen unit normal $N(p)$. Both $T_{p}S$ and $T_{N(p)}{S}^2$ are the plane orthogonal to $N(p)$, so the derivative is an endomorphism of that plane. In local coordinates $r(u,v)$, differentiating $⟨N,r_j⟩=0$ gives $⟨N_i,r_j⟩=-⟨N,r_{ij}⟩$. Since $r_{ij}=r_{ji}$, this equals $⟨N_j,r_i⟩$. Bilinearity then gives $⟨dN(v),w⟩=⟨v,dN(w)⟩$ for all tangent vectors. The derivative of the Gauss map is self-adjoint.

If the Gauss map is a diffeomorphism onto $S^2$, the surface is compact and $dN$ is invertible everywhere. Its Gaussian curvature $K=\det (dN)$ is continuous and nowhere zero, so has constant sign on the connected surface. To determine that sign, maximize a height function $r\mapsto ⟨r,a⟩$. At a maximum, $a$ is parallel to the normal and the Hessian of the height function is semidefinite. Thus the second fundamental form is semidefinite, up to a sign. Its determinant is nonnegative and cannot be zero because $dN$ is invertible. Hence $K>0$ at that point and therefore everywhere.

The converse is false under the stated hypotheses. An open hemisphere of the unit sphere is a connected oriented surface with $K=1$, but its outward Gauss map has only that hemisphere as its image, and is not a diffeomorphism onto all of $S^2$. The missing compactness hypothesis matters.

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.

Parametrize the torus by $F(u,v)=((2+\cos u)\cos v,(2+\cos u)\sin v,\sin u)$. Its outward Gauss map isThe derivative $N_u$ has unit length, while $N_v$ has length $|\cos u|$ and is perpendicular to $N_u$. Thus rank drops precisely at $u=\pi /2$ or $3\pi /2$. These are the two circles of height $z=1$ and $z=-1$ and radius two, and their images are the north and south poles. Therefore the critical values are $(0,0,\pm 1)$.

For any nonpolar target point, its third coordinate determines two values of $u$, one with positive and one with negative $\cos u$. For each, the longitude fixes a unique $v$ modulo $2\pi$, with a shift of $\pi$ when $\cos u$ changes sign. Thus every regular value has exactly two preimages, one on the outer and one on the inner side of the torus. ConsequentlyThis explicit two-sheet description justifies the count without relying on a sketch.

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.

The Gauss map of an oriented surface is its smooth unit normal $N:S\to S^2$. With the convention that the shape operator is $A_p=-d{N}_p$, the second fundamental form is ${\mathrm{II}}_p(v,w)=⟨A_{p}v,w⟩$ and the normal curvature in a nonzero direction $w$ is $k_n(w)=\mathrm{II}(w,w)/⟨w,w⟩$. The two principal curvatures are the real eigenvalues $k_1,k_2$ of this self-adjoint operator; the mean curvature is $H=(k_1+k_2)/2$.

For directions successively spaced around the tangent plane, write their angles to a principal direction as ${\theta }_j={\theta }_0+(j-1)\pi /m$. The Euler formula for normal curvature givesThe geometric series ${\sum }_{j=0}^{m-1}{e}^{2\pi ij/m}=0$ for $m\ge 2$ makes the cosine sum zero. ThereforeHere equal angular spacing is understood in the usual directed sense. If the printed consecutive angles were only unsigned, without requiring the same rotational sense, the assertion would be false: for $m=3$ the directions $0,\pi /3,0$ satisfy those unsigned angle conditions but do not generally give $3H$.

A minimal surface has $H=0$, so at each point $k_2=-k_1$. In an orthonormal principal basis, $A=\mathrm{diag}(k_1,-k_1)$ and $A^2=k_1^{2}I$. Hencewhich proves the identity with $\boxed{\mu (p)=k_1^2=-K(p)\ge 0}$, where $K$ is the Gaussian curvature.

The converse is false. The identity only forces $k_1^2=k_2^2$, which allows either $k_2=-k_1$ or $k_2=k_1$. A sphere of radius $R$ has $dN=I/R$ for its outward normal, so it satisfies the identity with $\mu =R^{-2}$, yet $H=-1/R\ne 0$ under the chosen convention. Thus a conformal Gauss map can also arise from an umbilic nonminimal surface.

For an oriented smooth embedded surface, the Gauss map sends each point to its chosen unit normal. This surface is the ring torus with major radius two and minor radius one. Its outward Gauss map isOn the $y=0$ cross-section, the normal points radially away from the center of each generating circle, which gives this formula for $u=0,\pi$ and then rotational symmetry gives it for all $u$.

The upper hemisphere condition is $\sin v>0$, hence $0<v<\pi$. For this torus, the Gaussian curvature and area element areTherefore the positive outer and negative inner contributions cancel:

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 an oriented regular surface with unit normal $N$, the Gauss map sends each point $p$ to $N(p)\in S^2$. The outward normal to the catenoid isIts image is the open southern hemisphere, apart from its limiting equator and south pole. The Gauss map is one-to-one after taking $v$ modulo $2\pi$ and reverses orientation. Since the Jacobian of the Gauss map is the Gaussian curvature, its signed spherical area is the total Gaussian curvature:Equivalently, $KdA=-{\mathrm{sech}}^{2}ududv$, whose integral over $u>0$ and $0\le v<2\pi$ is $-2\pi$.

The Gauss map sends $p$ to the chosen unit normal $N(p)$. Differentiating $⟨N,N⟩=1$ shows that $d{N}_p$ maps $T_{p}S$ into itself. The shape operator $S_p=-d{N}_p$ is self-adjoint because its associated second fundamental form is symmetric. Its eigenvalues $k_1,k_2$ are the principal curvatures. A point is umbilical when $k_1=k_2$, and the surface is minimal when its mean curvature $H=(k_1+k_2)/2$ vanishes everywhere.

In an orthonormal principal basis, $d{N}_p$ has diagonal matrix $\mathrm{diag}(-k_1,-k_2)$. It is conformal exactly whenwhich is equivalent to $k_1^2=k_2^2\ne 0$. At a nonumbilical point this says $k_1=-k_2$, exactly the condition $H=0$. Hence, when there are no umbilical points, the surface is minimal if and only if its Gauss map is conformal.

Choose the orientation of a surface represented by a smooth Gauss map $N:S\to {\mathbb{S}}^2$. The shape operator at $p$ isIf its principal curvatures are $k_1,k_2$, then by definition

Let $E(x)=Ax+b$ be a proper Euclidean motion of Euclidean three-space, so $A\in SO(3)$. Orient $\tilde{S}=E(S)$ by transporting the normal:The chain rule gives the orthogonal conjugacyThe determinant and trace are invariant under similarity, so the Euclidean invariance of the shape operator yields

If the surface orientation is reversed, then $N$ and $S_p$ are multiplied by $-1$. Therefore the orientation reversal of surface curvature says

Use the global parametrizationIts tangent vectors areandThus $X$ is a regular parametrization and the graph is a smooth surface.

The first fundamental form isChoosing the upward orientation, the Gauss map is

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.

An orientation of $S$ chooses a unit normal at every point. The Gauss map isFor the unit tangent $T=\dot{\alpha }$, the surface covariant derivative is the tangential projectionThe signed geodesic curvature isBecause $|T|=1$, $D_{s}T$ is perpendicular to $T$ and henceConsequently ${\kappa }_g=0$ means $D_{s}T=0$: the tangent is parallel along $\alpha$, so $\alpha$ is a geodesic.

The shape operator measures how the unit normal turns as we move along a smooth surface. A plane has a constant unit normal, whereas the outward unit normal of a sphere turns in every direction in the tangent plane. Keeping the direction of this change, rather than just its magnitude, lets one recover the principal curvatures and their principal directions.

Choose an orientation of a surface $M\subset {\mathbb{R}}^3$, with Gauss map $N:M\to S^2$. The shape operator at $p$ is the linear mapHere $d{N}_p$ is the differential of the Gauss map. Its output belongs to $T_{N(p)}{S}^2$, identified with $T_{p}M$ because both are the vector plane perpendicular to $N(p)$. The minus sign is our convention throughout this branch. ^[ref]

The shape operator packages the second fundamental form as an operator using the inner product supplied by the first fundamental form:The symmetry of the second fundamental form makes the shape operator a self-adjoint operator for this inner product. Thus an orthonormal basis of principal directions diagonalizes it. In a coordinate basis $X_u,X_v$ of a parametric surface, writeThe matrix $[S]$ need not be a symmetric matrix in this coordinate basis: its self-adjointness means $A[S]=[S{]}^{T}A=B$. Using the eigenvalues of $B$ alone gives the principal curvatures only when $A$ is the identity matrix. ^[ref]

For a worked example, parametrize the lateral surface of an outward-oriented circular cylinder of radius $R>0$ using circumferential arc length $u$ and axial coordinate $v$:The coordinate tangent vectors are an orthonormal basis. Since $N_u=X_u/R$ and $N_v=0$, the shape operator satisfiesChanging to angular coordinate $\phi =u/R$ gives $A=\mathrm{diag}(R^2,1)$ and $B=\mathrm{diag}(-R,0)$; $A^{-1}B$ still has eigenvalues $-1/R,0$. This is why the first fundamental form must participate in the calculation.

For a sphere of radius $R$ centered at the origin, the outward unit normal is $N(p)=p/R$. Hence $S_{p}v=-v/R$, so both principal curvatures are $-1/R$. For a plane, $dN=0$ and the shape operator vanishes. Reversing the unit normal negates the shape operator and both principal curvatures; the Gaussian curvature remains unchanged while the mean curvature changes sign, as in orientation reversal of surface curvature.