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Shape operator and Gauss-Kronecker curvature of a graph

Statement

Let $n \geq 2$ , let $U \subseteq R^{n - 1}$ be open, let $h \in C^{\infty} (U; R)$ , and let $S = {(y, h (y)) : y \in U} \subseteq R^{n}$ be the graph with the unit normal $ν (y) = (- \nabla h (y), 1) / \sqrt{1 + ∣ \nabla h (y) ∣^{2}}$ of positive last coordinate. At $p = (y, h (y))$ the shape operator $S_{ν}$ of $S$ with respect to $ν$ satisfies $\det S_{ν} = \det D^{2} h (y) / (1 + ∣ \nabla h (y) ∣^{2})^{(n + 1) / 2}$ . In particular the extrinsic Gaussian (Gauss-Kronecker) curvature of $S$ vanishes at $p$ if and only if $\det D^{2} h (y) = 0$ .

Facts & Assumptions

Given: $h \in C^{\infty} (U)$ , $X (y) = (y, h (y))$ , $b = \sqrt{1 + ∣ \nabla h ∣^{2}}$ , and $ν = (- \nabla h, 1) / b$ .

[F1]

The local Euclidean shape operator is $S_{ν} = - d ν$ , and it agrees with the usual hypersurface operator; curvature is its determinant. (Euclidean hypersurface normals, shape operators and curvature, Smooth Euclidean hypersurface graphs and compact localization)

[F2]

Determinants are multiplicative and invariant under change of basis. (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space)

[F3]

Mixed second partials agree. (Continuous mixed partials of order $k$ are invariant under permutations)

[F4]

The graph Gram determinant and surface density are $1 + ∣ \nabla h ∣^{2}$ and its square root. (Chart and partition independence of surface measure)

Proof

technique · direct; differentiate orthogonality and compute the determinant in the graph frame

1.1givenF1F3algebra

The columns $X_{j} = (e_{j}, h_{j})$ are independent, and their Gram matrix is $G = I + \nabla h \nabla h^{T}$ . The vector $ν$ is unit and perpendicular to each column. Hence $G$ is positive definite and this is the positive-last-coordinate normal. Differentiating $ν \cdot X_{k} = 0$ gives $⟨ S_{ν} X_{j}, X_{k} ⟩ = - \partial_{j} ν \cdot X_{k} = ν \cdot X_{j k} = h_{j k} / b$ . In particular the normal component of $X_{j k} = h_{j k} e_{n}$ is $(h_{j k} / b) ν$ ; the vertical vector $e_{n}$ itself need not be normal.

2.1F1F2F4step 1.1algebra∎

If $A$ is the matrix of $S_{ν}$ in the frame $X_{j}$ , the pairing in step 1.1 says $G A = D^{2} h / b$ . Consequently $A = G^{- 1} D^{2} h / b$ , and multiplicativity gives $\det S_{ν} = \det D^{2} h / (b^{n - 1} \det G) = \det D^{2} h / (1 + ∣ \nabla h ∣^{2})^{(n + 1) / 2}$ . The denominator is positive, so curvature vanishes exactly when the Hessian determinant vanishes. The Euclidean equivalence in [F1] identifies this determinant with the promised extrinsic Gaussian curvature.

Depends on

Used by

Dependency tree · two levels

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Sources

Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)