Extrinsic Curvature and the Weingarten Map

Extrinsic curvature measures how the tangent spaces of a non-null hypersurface turn within the ambient spacetime. It can be read from the derivative of the unit normal, from the normal part of the ambient derivative of tangent vectors, or from the change of the induced metric along the normal. These equivalent definitions agree only after one sign convention is fixed.

Unlike intrinsic curvature, it depends on a coorientation and on the embedding. Nevertheless it is a genuine tensor on the hypersurface: changing ambient coordinates does not create or remove bending. Its trace and trace-free part separate isotropic normal change of the induced volume from anisotropic bending, much as expansion and shear separate a congruence deformation.

Required background. Affine Connections and Covariant Derivatives supplies covariant differentiation, and The Levi–Civita Connection supplies metric compatibility; The Riemann Curvature Tensor supplies the curvature convention used by the embedding identities; Pseudo-Riemannian Metrics and Signature supplies unit normals and induced signatures. Helpful background. Calculus and Linear Algebra for Relativity supplies symmetric maps and eigenvalues; Exterior Algebra and Differential Forms supplies orientation algebra, while Orientation, Densities, and Volume Elements supplies induced measures.

Second fundamental form and shape map

Throughout this chapter, for a unit normal $n^\mu$ with $n^2=\epsilon=\pm1$, define

$$K_{\mu\nu} =-h_\mu{}^\alpha h_\nu{}^\beta\nabla_\alpha n_\beta.$$

This is the ADM/Gourgoulhon sign convention. For tangent vectors $X$ and $Y$,

$$K(X,Y)=-g(\nabla_Xn,Y).$$

The Weingarten map, or shape operator, is

$$S(X)=-\nabla_Xn.$$

The normalization of $n$ implies $\nabla_Xn$ is tangent, so $K(X,Y)=h(SX,Y)$. Hypersurface orthogonality and the torsion-free connection imply that $K$ is symmetric. Its eigenvalues with respect to $h$ are the principal curvatures when that spectral language is appropriate.

Metric compatibility also gives

$$g(\nabla_XY,n)=K(X,Y),$$

and hence the Gauss and Weingarten formulas

$$\nabla_XY=D_XY+\epsilon K(X,Y)n, \qquad \nabla_Xn=-S(X).$$

These formulas keep the sign $\epsilon$ associated with extracting a normal component separate from the chosen minus sign in $K$.

Normal evolution of the metric

Extend the unit normal to a neighborhood while preserving its norm. For tangent arguments,

$$(\mathcal L_n g)(X,Y) =g(\nabla_Xn,Y)+g(X,\nabla_Yn).$$

Symmetry of $K$ therefore yields

$$K_{\mu\nu} =-\frac12 h_\mu{}^\alpha h_\nu{}^\beta \mathcal L_n h_{\alpha\beta}.$$

The projection is important if the extension away from the hypersurface is arbitrary. The trace

$$K=h^{\mu\nu}K_{\mu\nu}$$

is minus the normal logarithmic rate of the induced volume element. This statement uses the declared convention; sources using the opposite definition reverse every displayed $K$.

For a spacelike foliation with evolution vector $t^\mu=Nn^\mu+\beta^\mu$,

$$K_{ij} =-\frac1{2N} \left(\partial_t h_{ij}-\mathcal L_\beta h_{ij}\right).$$

This kinematical identity reconstructs extrinsic curvature from lapse, shift, and metric change. It is not an Einstein evolution equation.

Principal curvature and Gaussian normal distance

Where the induced metric is Riemannian, the self-adjoint shape map can be diagonalized in an orthonormal basis. Its eigenvalues are principal curvatures, their sum is the trace of the extrinsic curvature, and their product combinations enter the intrinsic curvature through the Gauss relation. For a timelike hypersurface the induced metric is indefinite, so real diagonalization is not automatic; tensor formulas remain valid even when Euclidean spectral intuition fails.

Near a non-null hypersurface, one may sometimes launch geodesics along the normal and use signed proper distance or proper time as a coordinate. In such Gaussian normal coordinates the shift vanishes and the normal coordinate has unit lapse in magnitude. The extrinsic curvature then directly gives minus one half of the normal derivative of the tangential metric. This construction is local: normal geodesics may intersect, and the coordinate system can fail at a focal point even while the hypersurface and ambient geometry remain smooth.

The acceleration of the chosen normal extension should not be confused with extrinsic curvature. The vector $\nabla_n n$ describes how normal integral curves bend. The tensor $K(X,Y)$ differentiates the normal in tangent directions. Gaussian normal extension sets the former to zero locally, but the latter can remain nonzero, as the sphere and Milne examples show.

Orientation and elementary embeddings

Reversing the coorientation sends $n\mapsto-n$ and therefore

$$K_{\mu\nu}\mapsto-K_{\mu\nu}, \qquad S\mapsto-S,$$

while the induced metric is unchanged. Any quoted sign for a boundary or slice is incomplete until the normal direction is stated.

A plane in Euclidean space has constant normal, hence $K=0$. For a sphere of radius $R$ with outward unit normal, $\nabla_Xn=X/R$, so the convention here gives

$$K(X,Y)=-\frac1R h(X,Y).$$

With the inward normal the sign reverses. The sphere is therefore a quick check both of magnitude and coorientation.

For a constant-time slice of Minkowski spacetime with future normal $n=\partial_t$, $K_{ij}=0$. For Milne slices,

$$ds^2=-d\tau^2+\tau^2\gamma_{ij}dx^idx^j,$$

the lapse is one and shift zero, so

$$K_{ij}=-\frac12\partial_\tau h_{ij} =-\frac1\tau h_{ij}.$$

The ambient spacetime is flat even though these hyperboloidal slices have nonzero extrinsic curvature. These equivalent definitions and checks align with Gourgoulhon 2007, §§2.3.4–2.5, pp. 23–37.

Common pitfalls

Extrinsic curvature is convention sensitive. Declare its sign before using a source or comparing formulas. Do not repair a discrepancy by changing only one Gauss–Codazzi term.

Nonzero extrinsic curvature does not imply ambient curvature. The round sphere, cylinder, and Milne slices all embed nontrivially in flat ambient spaces.

Null boundaries need separate data. Their normals cannot be normalized, and their null second fundamental forms carry boost weight. They are not obtained by setting $\epsilon=0$ here.

Exercises

Compute the extrinsic curvature of the timelike cylinder $r=R$ in Minkowski spacetime using its outward normal.

Solution

In coordinates $(t,r,\theta,\phi)$ the outward normal is $n=\partial_r$, with $\epsilon=+1$. The induced metric is

$$h=-dt^2+R^2d\Omega^2.$$

Because the normal flow changes only the angular metric,

$$K=-\frac12\mathcal L_n h =-R\,d\Omega^2.$$

Thus the time principal curvature is zero and the two angular principal curvatures are $-1/R$. Reversing the normal reverses all three values.

References

• Gourgoulhon, Éric. 3+1 Formalism and Bases of Numerical Relativity. Lecture notes, 2007. arXiv:gr-qc/0703035.