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Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\mathrm{grad}$. A Riemannian submersion is a surjective smooth submersion $\pi :(M,g)\to (N,h)$ for which, at every $p$, the restriction of $d{\pi }_p$ to $H_p=(\ker d{\pi }_p{)}^{\perp }$ is a linear isometry onto $T_{\pi (p)}N$. The spaces $V_p=\ker d{\pi }_p$ and $H_p$ are its vertical and horizontal spaces. Its fibres are totally geodesic submanifolds precisely when ${\nabla }_{U}V$ is vertical for vertical vector fields $U,V$: their second fundamental form vanishes. Equivalently, a geodesic initially tangent to a fibre remains in that fibre while defined.

For a smooth $f:N\to \mathbb{R}$, its basic function $F=f\circ \pi$ is constant along each fibre. The Riemannian gradient of $F$ is the horizontal lift of a vector field through a submersion of ${\mathrm{grad}}_{N}f$, sincefor horizontal $X$, and $dF(U)=0$ for vertical $U$. In particular no derivative of $f$ in a vertical direction occurs.

Here is the needed connection fact, which also follows directly from the Koszul formula: for horizontal lifts $X,Y$ of vector fields $\bar{X},\bar{Y}$ on $N$, the horizontal component of ${\nabla }_{X}Y$ projects to ${\nabla }_{\bar{X}}^N\bar{Y}$. To see this, pair the Koszul formula with a third horizontal lift $Z$. The horizontal inner products are pulled back from $N$, and the horizontal components of their Lie brackets of vector fields project to the brackets on $N$. Thus all six terms are the pullbacks of the corresponding terms on $N$.

Choose an adapted Riemannian orthonormal frame $X_1,\dots ,X_n,U_1,\dots ,U_r$, with the $X_a$ horizontal lifts. Using $\mathrm{Hess}F(A,A)=A(AF)-({\nabla }_{A}A)F$, the connection fact givesFor a vertical $U_b$, both $U_{b}F=0$ and $({\nabla }_{U_b}{U}_b)F=0$, the latter because the fibres are totally geodesic submanifolds. Taking the negative metric trace of the Riemannian Hessian therefore proves the basic-function Laplacian identityThis identity is local and does not require compactness. Vanishing mean curvature of the fibres would already suffice; total geodesicity makes each vertical summand vanish separately.

For the discrete eigenspace assertion, assume the two Riemannian manifolds are closed manifolds. Without a discrete spectral realization, an unrestricted noncompact version need not have an eigenbasis. The projections of the Riemannian product $M\times N$ are Riemannian submersions with totally geodesic submanifolds as fibres. Its Levi-Civita connection splits into the two factor connections. Consequently its positive Laplace-Beltrami operator isThe cross term in the product rule for the positive Laplace-Beltrami operator is zero because the two factor Riemannian gradients are orthogonal.

We use the standard compact elliptic compact elliptic spectral theorem: the positive Laplace-Beltrami operator on a closed manifold is self-adjoint, has compact resolvent, and has a complete orthonormal eigenbasis of smooth eigenfunctions, with finite-dimensional eigenspaces and eigenvalues tending to infinity. Let ${\Delta }_M{u}_i={\mu }_i{u}_i$ and ${\Delta }_N{v}_j={\nu }_j{v}_j$. Fubini's theorem and completeness on each factor show that $u_i(x)v_j(y)$ form a complete orthonormal basis of $L^2(M\times N)$. For example, a function orthogonal to all these products has, for each $i$, zero $u_i$-coefficient as an $L^2(N)$ function, hence is zero.

The displayed operator identity makes $u_i{v}_j$ an eigenfunction with eigenvalue ${\mu }_i+{\nu }_j$. Conversely, if $\Delta w=\lambda w$, self-adjointness of the positive Laplace-Beltrami operator givesAll other coefficients vanish. Only finitely many pairs can have ${\mu }_i+{\nu }_j=\lambda$, since both spectra are nonnegative and have finitely many eigenvalues below any fixed bound. Thus the product Laplacian eigenspace decomposition isThe tensor product summands are mutually orthogonal; their elements are actual smooth eigenfunctions, so this is an equality of eigenspaces, not just a formal expansion.

Write $R=a+b\cos u>0$. The tangent vectors areTheir squared lengths and inner product give the first fundamental formThe unit normal determined by this ordered parametrization isTaking inner products of $\nu$ with ${\sigma }_{uu}$, ${\sigma }_{uv}$ and ${\sigma }_{vv}$ yields $e=b$, $f=0$, $g=R\cos u$. Consequently the second fundamental form of a ring torus isThis normal points into the tube. Choosing the outward unit normal reverses both coefficients and gives $-II$; the choice must therefore accompany an answer for the second fundamental form.

The Gaussian curvature is unchanged by that reversal:Since the denominator is positive, $K>0$ on the outside of the torus, where $\cos u>0$, and $K<0$ on its inner side, where $\cos u<0$. It vanishes on the two circles $u=\pi /2,3\pi /2$. This is the Gaussian curvature of a ring torus.

Write derivatives with respect to $u$. The arc-length condition is $f^{\prime 2}+g^{\prime 2}=1$. The tangent vectors areTheir scalar products give the first fundamental formChoose the unit normal vector $N=(-g^{\prime }\cos v,-g^{\prime }\sin v,f^{\prime })$, since ${\sigma }_u\times {\sigma }_v=fN$ and $f>0$. Scalar products of $N$ with the second partial derivatives giveThe mixed coefficient is zero. Reversing $N$ reverses the second fundamental form but not the Gaussian curvature. Its formula isDifferentiating the arc-length identity gives $f^{\prime }{f}^{\prime \prime }+g^{\prime }{g}^{\prime \prime }=0$. Substitute it directly, without dividing by $g^{\prime }$ (which may vanish), to obtain

At a regular point, choose the unit normal $\mathbf{n}=({\sigma }_x\times {\sigma }_y)/|{\sigma }_x\times {\sigma }_y|$. The first fundamental form is the induced squared lengthThe second fundamental form isThe Gaussian curvature is the determinant of the shape operator, $K=(eg-f^2)/(EG-F^2)$, independent of the normal orientation.

For the given graph, ${\sigma }_x=(1,0,y)$ and ${\sigma }_y=(0,1,x)$, so with $S=1+x^2+y^2$,ThereforeThe surface is regular everywhere. The printed description “hyperboloid” is a naming error: this graph is a hyperbolic paraboloid, and the computation uses the actual parametrization.

Assume the parametrization is $C^2$ and regular, so its curvature is defined. Write $\mathbf{v}={\mathbf{a}}^{\prime }+y{\mathbf{b}}^{\prime }$ and $W=|\mathbf{v}\times \mathbf{b}|>0$. Since ${\sigma }_x=\mathbf{v}$ and ${\sigma }_y=\mathbf{b}$, its first fundamental form isFor the unit normal $\mathbf{n}=(\mathbf{v}\times \mathbf{b})/W$, the mixed coefficient of the second fundamental form iswhere the term containing ${\mathbf{b}}^{\prime }$ twice vanishes. Also $g=\mathbf{n}\cdot {\sigma }_{yy}=0$, while $EG-F^2=W^2$. ThusThe Gaussian curvature vanishes exactly when the displayed scalar triple product is zero. This is the Gaussian curvature of a ruled surface criterion; at a singular point with $W=0$, curvature and the claimed equivalence are not defined.

For a local parametrization $\varphi (u,v)$, write the first fundamental form and second fundamental form asThe Gaussian curvature is

To prove the Theorema Egregium, set $g_{ij}=⟨{\varphi }_i,{\varphi }_j⟩$ and decomposeDifferentiating $g_{ij}$ and solving for the tangential coefficients givesso every Christoffel symbol depends only on $E,F,G$ and their first derivatives. Equality of mixed third derivatives yields the Gauss equationThe Riemann curvature tensor is formed from first derivatives and quadratic products of the Christoffel symbols, and hence from the first and second derivatives of $E,F,G$. Consequentlydepends only on the first fundamental form.

Take $n^a{n}_a=-1$. The first fundamental form, or induced metric, is the spatial projection tensorWith the convention used in the question, define the second fundamental form byLocally $n_a$ is proportional to the gradient of a defining function for the hypersurface, so $h_a{}^c{h}_b{}^d{\nabla }_{[c}{n}_{d]}=0$ and $K_{ab}=K_{ba}$. Both derivative indices are projected tangentially. If two extensions of $n$ agree on $\Sigma$, their difference vanishes there, as does its tangential derivative, so $K_{ab}$ is independent of the extension.

WriteThe regularity of the profile makes $q>0$. The coordinate tangent vectors areso the first fundamental form of a surface of revolution has coefficientsChoose the unit normalTaking scalar products with the second derivatives of $\varphi$ gives the second fundamental form coefficientsThus the coordinate directions are principal directions, with principal curvaturesThe curvatures of a parametrized surface of revolution are thereforeReversing the orientation of a surface reverses the sign of $H$ but leaves $K$ unchanged.

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 a chosen unit normal field $N$, the second fundamental form isIn a local parametrization $\varphi$, its coefficients, up to the equivalent simultaneous sign convention, areThe Gaussian curvature is the determinant of the shape operator, henceTheorema Egregium states that $K$ is determined entirely by the first fundamental form. In particular, it is intrinsic and is preserved by every local isometry.

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.