maths2u
PhysicsMathsChemistryBiology
Tier
⌕ Search ⌘K
Theorem

Gauss's Theorema Egregium

T-107Home MU-306Threads space
Statement

Let \(M\) and \(\bar M\) be regular surfaces of class \(C^3\) in \(\mathbb{R}^3\), and let \(\varphi : U \to \bar U\) be a local isometry between an open neighbourhood \(U \subseteq M\) of a point \(p\) and an open neighbourhood \(\bar U \subseteq \bar M\) of \(\varphi(p)\); that is, \(\varphi\) is a diffeomorphism such that for every point \(q \in U\) and all tangent vectors \(X,Y \in T_qM\), \(\langle d\varphi_q(X), d\varphi_q(Y)\rangle = \langle X,Y\rangle\) — equivalently, \(\varphi\) preserves the first fundamental form \(I = E\,du^2+2F\,du\,dv+G\,dv^2\) when expressed in corresponding local coordinates. Let \(K_M(q)\) denote the Gaussian curvature of \(M\) at \(q\), defined extrinsically as the product of the principal curvatures, \(K_M(q)=\kappa_1(q)\kappa_2(q)=\det S_q\), where \(S_q = -dn_q\) is the shape operator of the unit normal field \(n\). Then \[ K_{\bar M}(\varphi(q)) = K_M(q) \qquad \text{for every } q \in U. \] In particular \(K\) is an intrinsic invariant: it can be computed from \(E,F,G\) and their first and second partial derivatives alone, with no reference to the embedding of \(M\) in \(\mathbb{R}^3\) or to \(L,M,N\) individually.

Why it matters

Gaussian curvature is defined via the shape operator, an object that only makes sense once a surface is sitting inside \(\mathbb{R}^3\) — it measures how the unit normal turns, so a priori it depends on the embedding, not just on the surface's internal geometry of lengths and angles. Gauss's theorem says this apparent dependence is an illusion for the product \(\kappa_1\kappa_2\): a two-dimensional being confined to the surface, able to measure only lengths of curves and angles between them, could in principle compute \(K\) without ever leaving the surface or knowing how (or whether) it curves through any ambient space.

This is the seed of Riemannian geometry: it licenses the idea of curvature as a property of an abstract metric space, detached from any embedding, and it is the reason surveyors and cartographers can (in principle) detect the curvature of the Earth using only measurements made on its surface.

Hypotheses
Regularity and smoothness: \(M\) is a regular surface of class \(C^3\).Regularity (\(r_u\times r_v\neq 0\)) is needed so that \(\{r_u,r_v\}\) spans the tangent plane and an orthonormal frame \(\{e_1,e_2,n\}\) exists; \(C^3\) is needed so that the shape operator is \(C^1\) and the connection form \(\omega_{12}\) built in the proof is differentiable, making \(d\omega_{12}\) meaningful. Drop regularity — e.g. at the apex of a cone, where \(r_u\times r_v \to 0\) — and the frame degenerates: curvature there is not a pointwise number at all but a concentrated (distributional) quantity, invisible to the smooth calculus used below. \(\varphi\) is a genuine isometry, i.e. it preserves the first fundamental form exactly.If \(\varphi\) is only required to preserve angles (conformal) or only areas (equiareal), the conclusion fails: stereographic projection from a sphere to a plane is conformal but not isometric, and the sphere has \(K=1/R^2\neq 0=K_{\text{plane}}\). The statement is local.\(\varphi\) need only be defined on a neighbourhood of \(p\); no global topological hypothesis (simple connectivity, compactness, orientability of the whole surface) is required. Strengthening to a *global* isometry adds nothing to this theorem, though it does matter for converses (Minding's theorem, discussed below), where reconstructing an isometry from matching curvature data is a genuinely global problem.
Proof
1
\{e_1,e_2\} := \text{Gram–Schmidt orthonormalisation of } \{r_u,r_v\}, \qquad n := e_1\times e_2
Regularity gives \(r_u,r_v\) linearly independent at every point of \(U\), so Gram–Schmidt (a purely algebraic, pointwise-smooth procedure since \(E,F,G\) are \(C^2\)) produces a smooth orthonormal tangent frame \(\{e_1,e_2\}\) with \(n\) the unit normal. A
2
dr = \theta^1 e_1+\theta^2 e_2, \qquad I = (\theta^1)^2+(\theta^2)^2
Define the dual coframe \(\theta^1,\theta^2\) by \(\theta^i(X)=\langle X,e_i\rangle\); since \(dr=r_u\,du+r_v\,dv\) is tangent to \(M\), it decomposes uniquely in the orthonormal basis \(\{e_1,e_2\}\), and \(\langle dr,dr\rangle = (\theta^1)^2+(\theta^2)^2\) recovers the first fundamental form. Crucially, \(\theta^1,\theta^2\) are algebraic functions of \(E,F,G\) only (via the Gram–Schmidt coefficients), so they are intrinsic data. A
3
d e_1 = \omega_{12}\,e_2+\omega_{13}\,n,\quad de_2=-\omega_{12}\,e_1+\omega_{23}\,n,\quad dn=-\omega_{13}\,e_1-\omega_{23}\,e_2
Differentiating \(\langle e_i,e_j\rangle=\delta_{ij}\) gives \(\langle de_i,e_j\rangle+\langle e_i,de_j\rangle=0\), so the matrix of connection 1-forms \((\omega_{ij})\) (writing \(de_i=\sum_j \omega_{ij}e_j\) in the frame \(\{e_1,e_2,n\}\)) is skew-symmetric: \(\omega_{ji}=-\omega_{ij}\), \(\omega_{ii}=0\). This is Cartan's moving-frame formalism applied to the orthonormal frame of Step 1. B
4
d\theta^1=\omega_{12}\wedge\theta^2, \qquad d\theta^2=-\omega_{12}\wedge\theta^1
Apply \(d\) to the identity \(dr=\theta^1e_1+\theta^2e_2\) of Step 2, using \(d^2r=0\) and the Leibniz rule \(d(\theta^ie_i)=d\theta^i\,e_i-\theta^i\wedge de_i\); substitute the Step 3 formulas for \(de_1,de_2\) and collect the coefficients of \(e_1,e_2\) (the coefficient of \(n\) instead gives \(\theta^1\wedge\omega_{13}+\theta^2\wedge\omega_{23}=0\), the Codazzi–Mainardi relation, not needed here). This is Cartan's first structure equation, the "torsion-free" condition for the Levi-Civita connection. B
5
\omega_{12} \text{ is the unique 1-form solving Step 4, and it is determined by } \theta^1,\theta^2 \text{ alone}
(Cartan's lemma / uniqueness of the Levi-Civita connection form.) If \(\omega,\omega'\) both satisfy \(d\theta^1=\omega\wedge\theta^2=\omega'\wedge\theta^2\) and \(d\theta^2=-\omega\wedge\theta^1=-\omega'\wedge\theta^1\), then \((\omega-\omega')\wedge\theta^1=(\omega-\omega')\wedge\theta^2=0\); since \(\theta^1,\theta^2\) are pointwise linearly independent 1-forms, this forces \(\omega-\omega'=0\). Because the *only* input to the equations of Step 4 is the coframe \(\theta^1,\theta^2\) — itself built purely from \(E,F,G\) in Step 2 — the solution \(\omega_{12}\) is computable from \(E,F,G\) and their derivatives with no reference to \(n\), \(L\), \(M\), or \(N\). This is the crux of the theorem. C
6
\omega_{13}=h_{11}\theta^1+h_{12}\theta^2, \quad \omega_{23}=h_{21}\theta^1+h_{22}\theta^2, \quad h_{12}=h_{21}
\(\omega_{13},\omega_{23}\) are 1-forms vanishing on nothing outside the tangent directions, hence expand in the coframe \(\{\theta^1,\theta^2\}\) with coefficients \(h_{ij}=\langle S(e_i),e_j\rangle\) the components of the shape operator \(S=-dn\) in the frame \(\{e_1,e_2\}\) (matching \(\omega_{13}(e_j) = \langle de_1,\cdot\rangle\) type identities via \(dn=-\omega_{13}e_1-\omega_{23}e_2\)). Symmetry \(h_{12}=h_{21}\) follows because \(\langle n_u,r_v\rangle = -\langle n,r_{uv}\rangle = \langle n_v,r_u\rangle\), using \(r_{uv}=r_{vu}\) (Clairaut's theorem, valid since \(r\in C^3\)); this is the standard self-adjointness of the shape operator with respect to the first fundamental form. A
7
d\omega_{12} = -\,\omega_{13}\wedge\omega_{23}
Apply \(d\) to \(de_1=\omega_{12}e_2+\omega_{13}n\) and use \(d(de_1)=0\) together with Step 3's formulas for \(de_2,dn\); expanding and using \(\alpha\wedge\alpha=0\) for any 1-form \(\alpha\), the coefficient of \(e_2\) in \(0=d(de_1)\) gives exactly this identity. This is Cartan's second structure equation for a surface in \(\mathbb{R}^3\). C
8
\omega_{13}\wedge\omega_{23} = (h_{11}h_{22}-h_{12}h_{21})\,\theta^1\wedge\theta^2 = K\,\theta^1\wedge\theta^2
Substitute Step 6 into the wedge product and use bilinearity together with \(\theta^i\wedge\theta^i=0\), \(\theta^2\wedge\theta^1=-\theta^1\wedge\theta^2\): \((h_{11}\theta^1+h_{12}\theta^2)\wedge(h_{21}\theta^1+h_{22}\theta^2)=(h_{11}h_{22}-h_{12}h_{21})\theta^1\wedge\theta^2\). Since \(\{e_1,e_2\}\) is orthonormal, the matrix \((h_{ij})\) *is* the matrix of \(S\) in an orthonormal basis, so its determinant is exactly \(\kappa_1\kappa_2=K\) by the extrinsic definition of \(K\) recalled in the Statement. A
9
d\omega_{12} = -K\,\theta^1\wedge\theta^2 \quad\Longrightarrow\quad K = -\,\frac{d\omega_{12}}{\theta^1\wedge\theta^2}
Combine Steps 7 and 8. The right-hand side depends only on \(\omega_{12}\) and \(\theta^1\wedge\theta^2\), both of which were shown in Steps 2 and 5 to be determined by \(E,F,G\) alone. (Formally: \(\theta^1\wedge\theta^2\) is the area form \(\sqrt{EG-F^2}\,du\wedge dv\), also visibly intrinsic.) No reference to \(n\), \(L\), \(M\), \(N\), or the embedding survives on the right. B
10
\varphi \text{ intertwines } (\theta^1,\theta^2,\omega_{12}) \text{ on } U \text{ with } (\bar\theta^1,\bar\theta^2,\bar\omega_{12}) \text{ on } \bar U
Because \(\varphi\) preserves the first fundamental form, in matching coordinates \(\bar E=E,\bar F=F,\bar G=G\); the Gram–Schmidt construction of Step 1 depends only on \(E,F,G\), so \(\varphi\) carries the coframe \(\theta^1,\theta^2\) exactly onto \(\bar\theta^1,\bar\theta^2\), and by the uniqueness of Step 5, \(\varphi\) carries \(\omega_{12}\) onto \(\bar\omega_{12}\) (uniqueness of solutions of a system that \(\varphi^*\) manifestly preserves, since \(\varphi^*\) commutes with \(d\) and \(\wedge\)). B
11
K_{\bar M}(\varphi(q)) = -\frac{d\bar\omega_{12}}{\bar\theta^1\wedge\bar\theta^2}\bigg|_{\varphi(q)} = -\frac{d\omega_{12}}{\theta^1\wedge\theta^2}\bigg|_{q} = K_M(q)
Apply Step 9 on \(\bar M\) at \(\varphi(q)\), pull the identity of the right-hand terms back through \(\varphi\) using Step 10 (pullback commutes with \(d\) and preserves the ratio of 2-forms since \(\varphi^*(\bar\theta^1\wedge\bar\theta^2)=\theta^1\wedge\theta^2\)), then apply Step 9 on \(M\) at \(q\). A
Result
K \text{ depends only on } E,F,G \text{ and their derivatives: local isometries preserve } K.

Reading. Although Gaussian curvature is defined by watching how the surface's normal vector twists in the surrounding space — an extrinsic, embedding-dependent recipe — its value is secretly a function of measurements a surface-bound observer can make: lengths of curves and angles between them. Bend a sheet without stretching or tearing it (an isometric deformation) and \(K\) is unchanged at every corresponding point, even though the shape in space, the normal direction, and the individual principal curvatures \(\kappa_1,\kappa_2\) all typically change.

Scope. Applies to any \(C^3\) regular surface and any local isometry between (open subsets of) two such surfaces; it is a pointwise, local statement and requires no global hypotheses. It does not assert anything about mean curvature \(H=\tfrac12(\kappa_1+\kappa_2)\), which is genuinely extrinsic, nor does it by itself produce an isometry from matching curvature data (see Corollaries).

Corollaries & converses
  • Bending invariance. \(K\) is invariant under any isometric bending of a surface; \(\kappa_1,\kappa_2\) individually and \(H\) are not — this pins down exactly which curvature-type quantities are "shape of the abstract surface" versus "shape of the embedding."
  • No perfect flat maps. Since the round sphere has \(K=1/R^2 \gt 0\) everywhere and the plane has \(K=0\), no local isometry from any open piece of a sphere to the plane can exist — hence no flat map of any region of the Earth can be exactly distance-preserving (a rigorous form of the classical cartography obstruction).
  • Developable surfaces. A surface with \(K\equiv 0\) is a *necessary* condition for local isometry to the plane; the converse direction (constructing the isometry) is the separate content of Minding's theorem for constant-curvature surfaces, not of Theorema Egregium itself.
  • Converse in general: false. Two surfaces need not be locally isometric merely because they attain the same numerical values of \(K\) somewhere, or even because their curvature functions have the same range. See Problem 5 and "Fails without" for why matching \(K\) pointwise under an arbitrary correspondence is not sufficient.
  • Converse for constant curvature: true (Minding's theorem, cited not proved here). Any two \(C^3\) surfaces with the same constant Gaussian curvature \(K_0\) are locally isometric to each other; this is the genuine converse partner to Theorema Egregium, but it requires \(K\) constant, not merely matching at a point.
Fails without
  • Drop "isometry", keep "conformal" (angle-preserving) or "equiareal" (area-preserving). Stereographic projection \(\sigma\) from the unit sphere minus a pole to the plane is conformal (it preserves angles) but stretches lengths radially; the sphere has \(K\equiv 1\) while the plane has \(K\equiv 0\), so \(K_{\text{plane}}(\sigma(p))\neq K_{\text{sphere}}(p)\). An equiareal projection (e.g. Lambert's) similarly preserves \(dA\) but not \(ds\), and again fails to match curvature. The theorem genuinely needs the *full* first fundamental form preserved, not a weaker invariant derived from it.
  • Drop regularity/smoothness at a point. The cone \(z=\sqrt{x^2+y^2}\) is flat (\(K=0\)) and locally isometric to a planar sector at every regular point away from the apex — consistent with the theorem. At the apex itself the surface is not \(C^1\) (let alone \(C^3\)), the frame \(\{e_1,e_2,n\}\) of Step 1 is undefined, and curvature is not a smooth pointwise quantity there at all: it becomes a concentrated "angle defect" (the cone's total turning deficit), governed by a distributional generalisation (Alexandrov geometry) that lies outside the reach of this proof.
Common errors
  • Assuming mean curvature \(H\) is also intrinsic by analogy with \(K\): the plane (\(H=0\)) rolled isometrically into a cylinder of radius \(R\) has \(H=1/(2R)\neq 0\) — only the *product* \(\kappa_1\kappa_2\), not the sum, survives bending.
  • Reading the theorem as biconditional: concluding two surfaces are isometric because their Gaussian curvatures agree at corresponding points (or even everywhere), rather than the correct one-way implication isometry \(\Rightarrow\) equal \(K\).
  • Conflating "isometric" (same first fundamental form / same lengths of curves) with "congruent" (identical shape in \(\mathbb{R}^3\), related by a rigid motion) — leading to the false belief that isometric surfaces must look alike as embedded objects.
  • Using the orthogonal-coordinate curvature formula \(K=-\frac{1}{2\sqrt{EG}}\big[\partial_u(G_u/\sqrt{EG})+\partial_v(E_v/\sqrt{EG})\big]\) in coordinates where \(F\neq 0\), forgetting it was derived under \(F=0\).
  • Believing "intrinsic" means "computable without ever choosing an embedding": in practice \(E,F,G\) are usually *obtained* via some parametrisation \(r(u,v)\subset\mathbb{R}^3\); the theorem's content is that the resulting number \(K\) does not depend on *which* embedding produced those \(E,F,G\), not that no embedding is ever used.
Discussion

Gauss proved this result in his 1827 Disquisitiones generales circa superficies curvas, and himself called it egregium — "outstanding" or "remarkable" — precisely because the definition of \(K=\kappa_1\kappa_2\) he had been using gives every appearance of being an extrinsic quantity, tied to how the surface sits in space, and yet turns out to be computable from the metric alone. Gauss's own motivation was geodetic: he had spent years on the triangulation survey of the Kingdom of Hanover, and the question of how curvature of the Earth could be detected from surface measurements (rather than from outside) was not idle.

The theorem is the historical hinge between classical surface theory and Riemannian geometry. Once curvature is known to be a function of the metric tensor alone, Riemann (1854) could define curvature for abstract \(n\)-dimensional spaces equipped only with a metric, with no ambient Euclidean space required at all — the sectional curvature of Riemannian geometry is the direct descendant of \(K\) here, and the moving-frame proof given above (Cartan's structure equations) is exactly the machinery that generalises cleanest to higher dimensions.

It also underlies the Gauss–Bonnet theorem, \(\iint_D K\,dA + \int_{\partial D}\kappa_g\,ds = 2\pi\chi(D)\): because \(K\,dA\) is built from purely intrinsic data (Step 9's \(-d\omega_{12}\), integrated), the left-hand side is meaningful for an abstract Riemannian surface, and the theorem becomes a genuine bridge between local geometry and global topology (\(\chi\)) rather than a fact about a particular embedding.

A subtlety worth flagging: the proof shows \(\omega_{12}\) is determined by \(\theta^1,\theta^2\) via a *first-order linear system* (Step 4), so \(\omega_{12}\) involves one derivative of the metric, and \(K=-d\omega_{12}/(\theta^1\wedge\theta^2)\) involves a second. This is exactly mirrored in coordinate formulas for \(K\) (e.g. the Brioschi determinant formula), which always involve \(E,F,G\) and their derivatives up to second order — never third or higher — reflecting that curvature is a second-order differential invariant of the metric, the simplest one that is not itself linear in the metric's derivatives.

Common misconception. Some students infer from "curvature is intrinsic" that a surface's *shape* is somehow also intrinsic, or that isometric surfaces are indistinguishable as objects. Neither is true: the cylinder and the plane are isometric yet look nothing alike, and infinitely many non-congruent embeddings can realise the same abstract intrinsic metric (bend a sheet of paper any way you like without stretching it). What is forced to agree across all of them is only \(K\), not the embedding itself.

Worked examples
1
\text{No exact flat map of a sphere exists.}
Set-up: let \(S_R^2\) be a sphere of radius \(R\), so \(K_{S_R^2}\equiv 1/R^2\); let \(\Pi\) be the plane, \(K_\Pi\equiv 0\). Suppose, for contradiction, a local isometry \(\varphi: U\subset S_R^2\to \Pi\) existed on some open \(U\). A
2
K_\Pi(\varphi(q)) = K_{S_R^2}(q) = 1/R^2 \quad \text{for all } q\in U
Apply the theorem to \(\varphi\). A
3
1/R^2 = 0 \quad \text{(contradiction, since } R \lt \infty)
But \(K_\Pi\equiv 0\) identically on the plane, contradicting Step 2. A
\text{No open subset of a sphere admits a distance-preserving flat map.}

Reading. Every world map necessarily distorts either distances, angles, or areas (usually all three to some degree) — this is not a failure of cartographic technique but a theorem.

1
\text{Catenoid } X(u,v)=(\cosh u\cos v,\ \cosh u\sin v,\ u) \ \text{and helicoid } Y(u,v)=(u\cos v,\ u\sin v,\ v)
Compute first fundamental forms directly: for \(X\), \(X_u=(\sinh u\cos v,\sinh u\sin v,1)\), \(X_v=(-\cosh u\sin v,\cosh u\cos v,0)\), giving \(E=\cosh^2u,\ F=0,\ G=\cosh^2u\). A
2
\text{Reparametrise the helicoid by } u=\sinh\bar u: \quad \bar E=\bar G=\cosh^2\bar u,\ \bar F=0
For \(Y\), \(E_Y=1,\ F_Y=0,\ G_Y=u^2+1\). Substituting \(u=\sinh\bar u\) (\(du=\cosh\bar u\,d\bar u\)) rescales the first coefficient to \(E_Y(du/d\bar u)^2=\cosh^2\bar u\) and \(G_Y=u^2+1=\sinh^2\bar u+1=\cosh^2\bar u\); \(F\) is unaffected since it was already \(0\). The two metrics now agree exactly as functions of \((\bar u,v)\): \(ds^2=\cosh^2\bar u\,(d\bar u^2+dv^2)\) on both surfaces. B
3
(\bar u,v)\mapsto X(\bar u,v) \text{ is a local isometry catenoid} \leftrightarrow \text{helicoid}
Two surfaces parametrised so that \(E,F,G\) agree identically as functions of the same coordinates are, by definition, related by a local isometry (the coordinate map itself). A
4
K = -\frac{1}{2\lambda}\Delta(\ln\lambda), \qquad \lambda=\cosh^2\bar u \ \ (\text{conformal metric } ds^2=\lambda(d\bar u^2+dv^2))
For a conformal coframe \(\theta^1=\sqrt\lambda\,d\bar u,\ \theta^2=\sqrt\lambda\,dv\), solving Step 4–5 of the Proof explicitly gives \(\omega_{12}=-\tfrac12\lambda_v/\lambda\,d\bar u+\tfrac12\lambda_{\bar u}/\lambda\,dv\), and Step 9 (\(K=-d\omega_{12}/\theta^1\wedge\theta^2\)) reduces after direct computation to Liouville's formula \(K=-\Delta(\ln\lambda)/(2\lambda)\), a special case of the general machinery proved above. C
5
\ln\lambda = 2\ln\cosh\bar u, \quad \partial_{\bar u}^2(\ln\lambda)=2\,\mathrm{sech}^2\bar u,\quad \partial_v^2(\ln\lambda)=0
Direct differentiation: \(\partial_{\bar u}\ln\cosh\bar u=\tanh\bar u\), \(\partial_{\bar u}^2\ln\cosh\bar u=\mathrm{sech}^2\bar u\). A
6
K = -\frac{2\,\mathrm{sech}^2\bar u}{2\cosh^2\bar u} = -\frac{1}{\cosh^4\bar u}
Substitute into Step 4's formula. A
K_{\text{catenoid}}(u,v) = K_{\text{helicoid}}(\bar u,v) = -\frac{1}{\cosh^4 u}, \ u=\sinh\bar u

Reading. The catenoid and helicoid, wildly different-looking surfaces (one compact-waisted and rotationally symmetric, one an unbounded spiral ramp), are two different embeddings of the same abstract Riemannian surface; Theorema Egregium forces their curvatures to agree pointwise under the isometry, and direct computation confirms it.

Scope. The identification is a genuine isometric bending: physically, the catenoid can be continuously "unrolled" into the helicoid (through the associate family of minimal surfaces) while preserving all intrinsic lengths at every stage.

Problems
  1. Verify Theorema Egregium directly for the plane and the cylinder: parametrise the cylinder of radius \(a\) as \(r(u,v)=(a\cos(u/a),a\sin(u/a),v)\), show \(r\) is a local isometry from the plane (with coordinates \(u,v\)), and compute \(K\) on each side independently via \(K=(LN-M^2)/(EG-F^2)\).
    SolutionFor the plane with coordinates \((u,v)\), \(E=G=1,F=0\); since it is a plane, \(L=M=N=0\) trivially (or: it embeds with zero shape operator), so \(K_{\text{plane}}=0\). For the cylinder, \(r_u=(-\sin(u/a),\cos(u/a),0),\ r_v=(0,0,1)\), giving \(E=1,F=0,G=1\) — identical to the plane's first fundamental form in these coordinates, so \(r\) is a local isometry by definition. The unit normal is \(n=(-\cos(u/a),-\sin(u/a),0)\); \(r_{uu}=(-\cos(u/a)/a,-\sin(u/a)/a,0)\), so \(L=\langle r_{uu},n\rangle=1/a\); \(r_{uv}=r_{vv}=0\) so \(M=N=0\). Hence \(K_{\text{cyl}}=(LN-M^2)/(EG-F^2)=(0-0)/1=0\). Both give \(K=0\), consistent with the theorem, even though \(L\) itself is \(1/a\neq 0\) — the individual second fundamental form coefficients are not preserved, only the combination \(K\).
  2. A surface of revolution with unit-speed profile curve has first fundamental form \(ds^2=du^2+f(u)^2dv^2\) for some \(f(u)\gt 0\). Using the orthogonal-coordinates formula \(K=-\frac{1}{2\sqrt{EG}}[\partial_u(G_u/\sqrt{EG})+\partial_v(E_v/\sqrt{EG})]\) with \(E=1,G=f^2\), show \(K=-f''/f\), then find \(f\) (up to the obvious symmetries) so that \(K\equiv 1/R^2\), and identify the surface.
    SolutionWith \(E=1,G=f^2,F=0\): \(\sqrt{EG}=f\), \(G_u=2ff'\), \(E_v=0\). Then \(\partial_u(G_u/\sqrt{EG})=\partial_u(2ff'/f)=\partial_u(2f')=2f''\), and the \(v\)-term vanishes. So \(K=-\frac{1}{2f}(2f'')=-f''/f\). Setting \(-f''/f=1/R^2\) gives \(f''+f/R^2=0\), with general solution \(f(u)=A\cos(u/R)+B\sin(u/R)\); the boundary condition \(f(0)=0,f'(0)=1\) appropriate to a smooth axis point gives \(f(u)=R\sin(u/R)\), the profile of a sphere of radius \(R\) (with \(u\) the arclength from the pole) — confirming \(K\equiv 1/R^2\) for the sphere via purely intrinsic data.
  3. Use the plane-vs-cylinder isometry of Problem 1 to show explicitly that mean curvature \(H=\tfrac12(\kappa_1+\kappa_2)\) is *not* preserved by local isometries, even though \(K\) is — i.e. Theorema Egregium does not generalise from \(K=\kappa_1\kappa_2\) to \(H\).
    SolutionFrom Problem 1, on the plane \(L=M=N=0\) so \(\kappa_1=\kappa_2=0\) and \(H_{\text{plane}}=0\). On the cylinder, with \(E=1,F=0,G=1,L=1/a,M=N=0\), the principal curvatures solve \(\det(S-\kappa I)=0\); since the shape operator here is already diagonal with entries \(L/E=1/a\) and \(N/G=0\), \(\kappa_1=1/a,\kappa_2=0\), giving \(H_{\text{cyl}}=1/(2a)\neq 0=H_{\text{plane}}\). The isometry of Problem 1 preserves \(K=\kappa_1\kappa_2=0\) on both sides (trivially, since one factor is always \(0\) here) but does not preserve \(H\), confirming \(H\) is extrinsic.
  4. Let \(r(x,y)=(x,y,xy)\) (a Monge patch for the saddle \(z=xy\)). Compute \(K\) at the origin using \(E=1+f_x^2,F=f_xf_y,G=1+f_y^2,\ L=f_{xx}/W,\ M=f_{xy}/W,\ N=f_{yy}/W\) with \(W=\sqrt{1+f_x^2+f_y^2}\), and interpret the sign.
    SolutionWith \(f=xy\): \(f_x=y,f_y=x,f_{xx}=0,f_{xy}=1,f_{yy}=0\). At the origin \(f_x=f_y=0\), so \(E=G=1,F=0,W=1\); hence \(L=0,\ M=1,\ N=0\). Then \(K=(LN-M^2)/(EG-F^2)=(0-1)/(1-0)=-1 \lt 0\). The negative sign confirms the origin is a saddle point: the surface curves oppositely in the two principal directions (concave up along \(y=x\), concave down along \(y=-x\)), consistent with \(z=xy\) locally resembling \(\tfrac12(u^2-v^2)\) after a \(45^\circ\) rotation.
  5. Explain, using Theorema Egregium, why finding two points \(p\) on an ellipsoid (not a sphere) and \(q\) on a sphere with \(K(p)=K(q)\) numerically is not enough to conclude a local isometry taking a neighbourhood of \(p\) to a neighbourhood of \(q\) could exist.
    SolutionAn ellipsoid that is not a sphere has non-constant Gaussian curvature: on any neighbourhood of any point \(p\), \(K\) takes more than one value (this follows since the principal curvatures vary continuously and are not both constant unless the surface is a sphere or plane, a standard rigidity fact). Suppose a local isometry \(\varphi\) existed from a neighbourhood of \(p\) to a neighbourhood of \(q\) on the sphere with \(K(p)=K(q)=1/R^2\). By Theorema Egregium, \(K_{\text{sphere}}(\varphi(p'))=K_{\text{ellipsoid}}(p')\) for every \(p'\) near \(p\). But \(K_{\text{sphere}}\equiv 1/R^2\) is constant on the whole sphere, forcing \(K_{\text{ellipsoid}}\) to be constant \((=1/R^2)\) throughout the neighbourhood of \(p\) — contradicting non-constancy of \(K\) near \(p\) on a genuine (non-spherical) ellipsoid. So matching \(K\) at the single points \(p,q\) is far from sufficient; the theorem constrains the whole local behaviour of \(K\) under any hypothetical isometry, not just its value at one point, which is exactly why the converse of Theorema Egregium fails in general and needs the much stronger hypothesis of *constant* curvature (Minding's theorem) to succeed.