Gauss–Codazzi equations

Remarks

Let $D$ be the flat connection of $\R^3$, given by directional derivatives. The Gauss formula $D_X Y = \nabla_X Y + I\!I(X, Y) N$ and the Weingarten formula $D_X N = -S X$ split $D$ into tangent and normal parts. Since $D$ has zero curvature, the tangent part of $D_X D_Y Z - D_Y D_X Z - D_{[X, Y]} Z = 0$ gives the Gauss equation, and the normal part gives the Codazzi–Mainardi equation. Taking $X, Y$ orthonormal and $Z = Y$, $W = X$ in the Gauss equation gives $\ip{R(X, Y) Y}{X} = \det S = K$, the Gaussian curvature. Example: for the sphere of radius $r$ with the inner normal, $S = \frac{1}{r} \id$, so $(\nabla_X S) Y = \frac{1}{r} (\nabla_X Y - \nabla_X Y) = 0$ and both sides of the Codazzi equation vanish, while the Gauss equation gives sectional curvature $1/r^2$.

Going deeper

For a student

The Gauss equation contains the Theorema Egregium: the left side depends only on the metric, so $K = \det S$ does too. Conversely, Bonnet's theorem says that on a simply connected domain, a metric and a symmetric form satisfying these equations come from a surface in $\R^3$, unique up to rigid motion. This is the surface analogue of the fundamental theorem of space curves. The same equations hold for hypersurfaces of $\R^{n+1}$, and with extra terms for submanifolds of a general Riemannian manifold.