Extrinsic curvature ($K$)

For a non-null hypersurface in a pseudo-Riemannian manifold, choose a unit normal vector $n$ and the ambient Levi-Civita connection $\nabla$. One sign convention for its extrinsic curvature isfor tangent vector fields $X,Y$. Reversing the normal reverses $K$; some authors use the overall negative convention.

Differentiating $g(n,Y)=0$ gives $K(X,Y)=-g(n,{\nabla }_{X}Y)$. The torsion-free property and tangency of $[X,Y]$ show that $K$ is a symmetric bilinear form. It measures the normal part of ambient acceleration along the hypersurface.

For a sphere of radius $R$ in Euclidean SPACE, the outward unit normal is $n(x)=x/R$, so ${\nabla }_{X}n=X/R$ and $K=g{|}_{TS}/R$. A plane has constant normal and $K=0$. The latter is the local totally geodesic case.

A hypersurface is totally geodesic when its intrinsic geodesics are also ambient geodesics. Equivalently, for a non-null hypersurface with the induced metric, its extrinsic curvature vanishes. The normal part of the ambient covariant derivative along a curve of the tangent is the second fundamental form, so its vanishing identifies the two geodesic equations.

This is a local statement: an ambient geodesic starting tangent to the hypersurface stays in it for a sufficiently short interval. Global containment needs the relevant completeness or closedness hypothesis. An open line segment inside the Euclidean plane has zero extrinsic curvature, but an ambient straight line leaves the segment in finite time. The whole line is totally geodesic and complete. A great circle on a round sphere is another complete example.