Codazzi and compatible surface data

Try

Suppose someone prescribes the Euclidean metric du²+dv² and second form 2du²+3dv² on a patch. The metric has intrinsic K=0, but the proposed shape operator has determinant six. No surface can realize both forms because the Gauss equation would require zero to equal six.

Compatibility is not merely symmetry of the second form. Derivatives of the bending coefficients must also agree with the metric connection. Those differential restrictions are the Codazzi equations.

Predict

Predict whether g=du²+dv² and b=kdu², with constant k, are compatible. Their determinant is zero, matching the flat metric, and constant coefficients satisfy the derivative conditions. A cylinder with suitable orientation realizes them locally.

Predict whether a local reconstruction theorem guarantees a globally embedded surface. Integrating compatible data produces a local immersion; global identifications, monodromy and self-intersections remain separate issues.

Experiment

Compare candidate data.

First form g	Second form b	Compatibility result
Identity metric	Zero	Plane
Identity metric	diag(k,0), constant k	Cylinder or plane
Identity metric	diag(2,3)	Fails Gauss
Identity metric	diag(k(v),0)	Codazzi requires k′(v)=0

Derive the last row directly: with zero Christoffel symbols, one Codazzi component becomes ∂v b_uu=∂u b_vu=0. Matching determinant zero was necessary but not sufficient.

Predict whether two rectangle paths agree

Prescribed dataConstant curvature, sheared cylinderTransverse curvature k+v, no shear Base curvature k00.512 Shear c (ignored in incompatible case)-101 Endpoint u00.250.51 Endpoint v00.250.51Reset frame experiment

Fixed xz projection; both endpoints have y=v. Navy dot/solid normal: u then v. Red square/dashed normal: v then u. Normal segments represent length0.3 in physical coordinates; unequal x/z display scales are labeled and angles should be read from the numbers. The gray curve appears only for compatible cylinder data, not for a nonexistent incompatible surface.

Effective shear 1; metric [1, 1; 1, 2], prescribed second form [1, 1; 1, 1]. Codazzi residual ∂v b_uu−∂u b_vu = 0.

u-then-v position (0.841471, 0.500000, 0.459698), tangent (0.540302, 0.00000, 0.841471), normal (-0.841471, 0.00000, 0.540302). v-then-u position (0.841471, 0.500000, 0.459698), tangent (0.540302, 0.00000, 0.841471), normal (-0.841471, 0.00000, 0.540302).

Frame rotation angles: 1.00000 and 1.00000 radians; difference 0.00000. These compatible data reconstruct a local immersed cylinder, or a plane when k=0.

Repair the claim

Zero determinant of the second form satisfies the flat Gauss condition but does not check Codazzi. For b=(k+v)du², one derivative is1 instead of0. Integrating around different rectangle paths exposes this failure. For constant k and shear c, the mixed second-form term kc is required; omitting it changes the prescribed geometry.

Closed-form bounded teaching examples; no general PDE solver, animation or saved state. The proof below supplies the existence/uniqueness argument. Retain your prediction and changed-case calculation in written practice.

Understand

The normal component of the same ambient flatness identity used for Gauss gives (∇_Xb)(Y,Z)=(∇_Yb)(X,Z). Here (∇_Xb)(Y,Z)=X(b(Y,Z))−b(∇_XY,Z)−b(Y,∇_XZ). To see the terms, expand D_X(D_YZ): differentiating b(Y,Z)N contributes X(b(Y,Z)) in the normal direction, and differentiating ∇_YZ contributes b(X,∇_YZ). Subtract the swapped expression and the bracket term. Torsion-free cancellation organizes the result into the two covariant derivatives. This proves Codazzi for every smooth surface patch.

An immediate application concerns an everywhere umbilic connected surface with S=λ identity. Operator Codazzi gives X(λ)Y=Y(λ)X. Choose independent tangent directions to conclude both derivatives of λ vanish. Hence λ is constant. If λ=0, N is constant and the surface lies locally in a plane. If λ≠0, differentiating X+N/λ gives zero, so the surface lies locally on a sphere centered at that constant vector.

Continue the immersion: path independence is the global question

Suppose the stated C² positive metric and C¹ symmetric second form are prescribed on a connected, simply connected open domain U in R² and satisfy Gauss–Codazzi everywhere. The rectangle construction extends to an immersion on all of U, unique up to one rigid motion with the normal convention fixed. Here is the continuation argument, separate from any assertion of injectivity.

Continue a chosen initial frame and position along a path by covering its compact image with finitely many small rectangles contained in U. On overlaps, the local uniqueness proof says that equal frame and position at one point give the same continued solution on the connected overlapping component. Thus subdividing the path or changing the rectangle cover has no effect. Linear ODE solutions exist along the entire compact path: its coefficients are bounded on a compact neighborhood, and the exponential norm bound used in the local proof prevents finite-path blowup.

If two paths with the same endpoints are homotopic with endpoints fixed, cover the image of their homotopy square by finitely many such rectangles. The Lebesgue-number argument subdivides the square into a grid whose small cells map into individual rectangles. Replace one edge route around a cell by the other. Local path independence makes the endpoint frame and position unchanged. Sweeping through the finitely many cells shows that the two boundary paths give identical endpoint data. A continuous homotopy can be used here because each image cell lies in a rectangle; piecewise smooth paths in those rectangles suffice for continuation. Simply connectedness makes any two endpoint-fixed paths homotopic, so the global frame and position are well defined. Their local expressions are the already constructed C² immersion, and uniqueness follows along paths from the initial data.

On a non-simply-connected domain, returning around a loop can change the frame by a rotation or the position by a translation. All such rigid-motion return data must be trivial for that continued immersion to descend from the universal cover. Vanishing local curvature defect is not itself this global condition.

Exact examples separating three claims

On the cylinder’s universal coordinate domain R², put

r(u,v)=(sin u, v, 1−cos u).

Direct differentiation gives g=du²+dv² and, with the chosen normal (−sin u,0,cos u), b=du². The Gauss–Codazzi equations hold globally. The map is a regular immersion but not injective because r(u+2π,v)=r(u,v). It descends to an embedding of the quotient (R/2πZ)×R as a cylinder. Thus simply connected data give a global immersion, not necessarily a global embedding of their original parameter domain.

For a different failure, give the cylinder abstractly the same flat metric but prescribe b=0. Locally the reconstruction is a plane with r(u,v)=(u,v,0). Continuing once around its u-period produces translation (2π,0,0), so no globally single-valued immersion with these two forms descends to that cylinder. This is positional monodromy even though the frame returns unchanged. It is not a contradiction to local existence or to the simply connected theorem.

Embedding criterion. An injective immersion that is proper is an embedding. Locally the constant-rank theorem gives an embedded coordinate patch. Globally properness makes the map closed: near a target limit point, intersect with a compact closed ball and use compactness of its inverse image to extract a source limit. A continuous injective closed map has continuous inverse onto its image. Without properness, additional global reasoning is needed; a local rank calculation alone does not establish inverse continuity.

Changed-case task. Replace sin u and cos u by sin(2u)/2 and cos(2u)/2, keeping the third coordinate (1−cos(2u))/2. Calculate g,b, the period and which quotient makes the cylinder map one-to-one. Then state why changing only b from 0 to 2 du² changes the return data without changing the flat metric.

Feedback and proof criteria

The u-derivative is (cos 2u,0,sin 2u), so g=du²+dv². The unit normal is (−sin 2u,0,cos 2u), and rᵤᵤ·N=2, giving b=2 du². The period is π, so quotienting u by πZ gives the one-to-one cylinder parametrization; quotienting only by 2πZ would still wrap twice. For b=0 the same metric gives translational planar monodromy on either periodic quotient. The metric alone does not determine the immersion: the second form and global return data matter. Verify rank, normal convention and quotient explicitly rather than equating equal curvature with congruence.

Primary comparison: Sasaki, A global formulation of the fundamental theorem of the theory of surfaces, inspected 30 October 2026. The continuation proof here is for smooth data on a simply connected planar domain, not a claim about arbitrary low-regularity manifolds or global isometric embedding of every metric.

Apply

Construct the local surface instead of invoking integrability

Work on an open rectangle about (0,0), shrinking it when needed. Assume g is a C² positive-definite symmetric2×2 matrix field and b is a C¹ symmetric2×2 field. We will construct a local immersion realizing them when their Gauss–Codazzi compatibility holds. Regularity, positivity and locality are hypotheses, not conclusions inferred from one cylinder. Use the normal convention b_ij=r_ij·N and shape operator S=−dN=g⁻¹b.

Let Γ_u have columns (Γ^1_11,Γ^2_11) and (Γ^1_12,Γ^2_12); let Γ_v have columns (Γ^1_12,Γ^2_12) and (Γ^1_22,Γ^2_22). All Christoffel symbols come from the usual metric formula. Let b_u=(b11,b12)ᵀ, b_v=(b12,b22)ᵀ and S_u=g⁻¹b_u, S_v=g⁻¹b_v. The prospective coordinate frame F has columns r_u,r_v,N; it is generally not orthonormal. Define

$$A=\begin{pmatrix}\Gamma_u&-S_u\\b_u^T&0\end{pmatrix},\qquad B=\begin{pmatrix}\Gamma_v&-S_v\\b_v^T&0\end{pmatrix},\qquad F_u=FA,\quad F_v=FB.$$

Mixed differentiation requires B_u−A_v+AB−BA=0. The sign follows by directly comparing (F_u)_v=FBA+FA_v and (F_v)_u=FAB+FB_u; it depends on our convention that frame vectors are columns and coefficients multiply on the right.

The top-left block of B_u−A_v+[A,B] is

$$\partial_u\Gamma_v-\partial_v\Gamma_u+[\Gamma_u,\Gamma_v]-S_u b_v^T+S_v b_u^T.$$

Its vanishing is the Gauss curvature identity with this connection convention. The bottom-left block is ∂u b_vᵀ−∂v b_uᵀ+b_uᵀΓ_v−b_vᵀΓ_u, the two Codazzi equations. The bottom-right block vanishes because b_uᵀg⁻¹b_v=b_vᵀg⁻¹b_u. The top-right block is the metric-raised Codazzi equation: differentiate g⁻¹b, using ∂i g=Γ_iᵀg+gΓ_i and Γ_u’s second column equal to Γ_v’s first. Thus Gauss and Codazzi give the entire matrix compatibility, not merely equality of the two determinants. The matrices also satisfy A e2=B e1 by symmetry of b and of the Christoffel lower indices.

Solve along two directions and control the defect

Choose F0 with positive determinant and F0ᵀF0=diag(g(0,0),1), for example by a positive Gram factor. First solve F_u(u,0)=F(u,0)A(u,0), F(0,0)=F0. For each u, solve F_v(u,v)=F(u,v)B(u,v) starting from that horizontal solution. These are finite linear ODEs, covered by the Picard construction. On a compact smaller rectangle the coefficients are bounded, so successive local solutions extend across it; the exponential norm estimate prevents finite-interval blowup.

We need parameter differentiation, not just pointwise existence. Write the vertical integral equation F(u,v)=F(u,0)+∫ from0 to v of F(u,t)B(u,t)dt. Difference quotients in u satisfy a linear integral equation with coefficient B and forcing converging to F B_u, because B is C¹. Subtract the candidate derivative equation; iterating the resulting integral inequality bounds the error by its vanishing forcing error times an exponential of the coefficient bound. This is the same factorial-series estimate used in Picard iteration. Hence F_u exists continuously and (F_u)_v=F_u B+F B_u. Negative v is handled by reversing the interval. This justifies the following calculation rather than assuming commuting unknown partial derivatives.

Define E=F_u−FA. It vanishes on v=0. Differentiate the constructed solution:

$$E_v=EB+F(AB+B_u-BA-A_v)=EB.$$

The final equality uses compatibility. Uniqueness for this homogeneous linear ODE with zero initial data gives E=0. Thus the constructed frame satisfies both equations. C¹ coefficients then make F C². This is the local integrability mechanism; no separate Frobenius theorem is being invoked.

Preserve the metric and integrate position

Set H=diag(g,1). Metric compatibility gives H_u=AᵀH+HA and H_v=BᵀH+HB. The actual Gram matrix G=FᵀF satisfies the same equations along the constructed horizontal and vertical paths, with the same initial value. Linear ODE uniqueness gives G=H. Therefore the first two frame columns are independent, the third is a unit normal orthogonal to them, and det F stays positive. This is why the resulting position map will be regular rather than merely a formal solution.

Write F1,F2,F3 for its columns. Because A e2=B e1, ∂v F1=F B e1=F A e2=∂u F2. Given a starting position r0, define

$$r(u,v)=r_0+\int_0^u F_1(s,0)\,ds+\int_0^v F_2(u,t)\,dt.$$

Differentiate to get r_v=F2. For r_u, replace ∂u F2 in the second integral by ∂v F1; the fundamental theorem of calculus yields r_u=F1. This proves exactness of the position differential directly on the rectangle. The induced first form is g, the oriented normal is F3, and the normal coefficients of F_u and F_v are exactly b. Thus r is the required local immersion.

Two positively oriented initial frames with the same Gram matrix differ by a fixed rotation Q. Linear ODE uniqueness makes their entire reconstructed frames differ by Q; integrating position adds only a fixed translation. Therefore the local immersion is unique up to an orientation-preserving rigid motion with the fixed normal convention. Reversing the chosen normal changes the sign of b. Neither this rectangle proof nor the formulas assert global injectivity, completeness or absence of monodromy on a larger domain.

A nonorthogonal worked reconstruction

Take constant k and shear c, set s=u+cv, and prescribe

$$g=\begin{pmatrix}1&c\\c&1+c^2\end{pmatrix},\qquad b=k\begin{pmatrix}1&c\\c&c^2\end{pmatrix}.$$

The metric is positive definite with determinant1, and all Christoffel symbols vanish. The shape matrix is k times [[1,c],[0,0]], with principal curvatures k and0. The frame matrix A has rows (0,0,−k), (0,0,0), (k,kc,0), and B=cA. Constant coefficients commute, so compatibility holds. For k≠0 a reconstruction is r=(sin(ks)/k, v, (1−cos(ks))/k); at k=0 its continuous limit is (s,v,0). Its normal is (−sin(ks),0,cos(ks)), consistent with r_u×r_v. Direct differentiation verifies both forms, including the mixed term kc.

Now remove the shear and prescribe g=I, b=(k+v)du². Gauss still gives determinant0, but Codazzi requires ∂v b11=0 and instead gives1. The two frame paths in the widget rotate by ku and (k+v)u respectively, differing by uv. Their pathwise position integrals differ too. They do not define two valid reconstructions of the same data; they diagnose the missing compatibility. At u=0 or v=0 they can coincide despite the nonzero defect elsewhere.

Primary reference: Ghomi, Curves and Surfaces Lecture Notes12, November2004, which states the coordinate-frame equations, compatibility and fundamental theorem. This lesson supplies the rectangle ODE/defect construction with explicit regularity and orientation conventions.

Repair “Prescribing K and H determines the surface.” Even their pointwise values do not supply the full metric, principal-direction field or compatibility derivatives needed by reconstruction.

Continue

When proposing surface data, check positivity, symmetry, Gauss and Codazzi before invoking reconstruction. Continue to Jacobi fields to see intrinsic curvature control the separation of nearby geodesics.