Principal curvatures as a self-adjoint problem

Try

At the origin of the graph z=(2x²+4xy−y²)/2, the gradient vanishes. With upward normal, the tangent metric is the identity and the shape operator is the Hessian matrix with rows (2,2) and (2,−1). Its eigenvalues are three and minus two, so K=−6 and H=1/2.

This convenient Hessian identification depends on the zero gradient and chosen normal. It is not a formula for the shape operator of an arbitrary graph point. In general the first and second fundamental forms must both be used.

Predict

Predict whether a symmetric coordinate matrix for II is automatically the matrix of S. Only in an orthonormal tangent basis. A nonorthogonal chart carries its own metric matrix g, and the operator matrix is g⁻¹b.

Predict what reversing the normal does to the eigenvalues. Both negate. Their half-sum H changes sign while their product K remains unchanged.

Experiment

The live explorer varies its own quadratic graph at the origin, separately from the fixed worked matrix above. Compare the directional normal curvature with the principal values. For the fixed example, independently verify each eigenvector equation by matrix multiplication.

Live experiment: curvature depends on direction

Study the graph z=(ax²+2bxy+cy²)/2 at its origin, with the upward normal and shape operator S=−dN. Its first fundamental form is the identity there, and its shape matrix is [[a,b],[b,c]]. Predict whether every tangent direction bends the same way before changing coefficients.

Coefficient a -2-1012Mixed coefficient b -101Coefficient c -2-1012Direction θ (degrees) 0153045607590120150180Reset experiment

For the unit tangent v=(cos θ,sin θ), normal curvature is vᵀSv=a cos²θ+2b sinθ cosθ+c sin²θ. At θ=45°, this is 0. The graph samples this exact formula at one-degree intervals; fixed axes make coefficient changes comparable.

Principal curvatures: -1 and 1. Gaussian curvature K=ac−b²=-1; mean curvature H=(a+c)/2=0. The origin is hyperbolic (saddle).

The eigenvalues solve (a−k)(c−k)−b²=0: k=H±√(((a−c)/2)²+b²). A direction for the larger curvature has angle 0° modulo 180°; the other principal direction is perpendicular. In principal coordinates this becomes Euler’s formula kₙ=k₁cos²φ+k₂sin²φ.

At the origin, the directional section has height z(tv)=kₙt²/2, so its second derivative equals the displayed normal curvature. Reversing the chosen normal negates both principal curvatures and H while preserving K. These formulas apply at this origin; away from it the first fundamental form is generally not the identity, so simply taking Hessian eigenvalues is incorrect.

Try a=c=1,b=0, then a=1,c=−1,b=0, then change b. Explain why zero curvature in one direction does not make a saddle flat. Values use consistent teaching length units; curvature has inverse-length units and K inverse-length-squared units. Settings are temporary.

Quantity in the fixed example	Value
Trace of S	1
Determinant of S	−6
Principal values	3 and −2
Mean curvature H	1/2

Check that the sum and product of the principal values reproduce the trace and determinant. A finite angle plot illustrates the directional quadratic form; the spectral argument below proves its extrema.

Understand

The shape operator is self-adjoint for the tangent inner product. In a coordinate basis, ⟨SX_i,X_j⟩=N·X_ij, symmetric in i,j. Bilinearity extends this symmetry to all tangent vectors. The spectral theorem therefore gives an orthonormal basis of principal directions and real eigenvalues k₁,k₂.

Let g be the first-form matrix and b the second-form matrix. By definition gS=b in coordinate components, so S=g⁻¹b. Principal components a solve ba=kga. Although g⁻¹b need not be symmetric in the ordinary matrix transpose sense, it is self-adjoint with respect to g. Treating it as a Euclidean symmetric matrix without changing basis can produce wrong directions.

Define Gaussian curvature K=det S=k₁k₂ and mean curvature H=(tr S)/2. At an umbilic point S is a scalar multiple of the identity, so every tangent direction is principal. Otherwise the two orthogonal principal lines are uniquely determined, up to order and sign.

Apply

For the fixed matrix, eigenvectors can be chosen along (2,1) for eigenvalue three and (−1,2) for minus two. Normalize them by √5. Their orthogonality and direct matrix multiplication independently verify the spectral calculation.

Repair “Negative mean curvature means a saddle.” Saddle behavior is detected by K<0, meaning opposite principal signs. H can be positive, negative or zero on such a point. Conversely reversing orientation changes H without changing the geometric saddle classification.

For optional depth, the spectral theorem used here is a finite-dimensional linear-algebra input. The geometric proof establishes the operator’s self-adjointness, which is the hypothesis needed to apply it.

Continue

Report g, b, normal orientation, principal directions and residual checks together. Continue to Euler and Meusnier to recover normal curvature in every tangent direction and relate it to the curvature of an actual space curve.