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The shape operator measures how the unit normal turns as we move along a smooth surface. A plane has a constant unit normal, whereas the outward unit normal of a sphere turns in every direction in the tangent plane. Keeping the direction of this change, rather than just its magnitude, lets one recover the principal curvatures and their principal directions.

Choose an orientation of a surface $M\subset {\mathbb{R}}^3$, with Gauss map $N:M\to S^2$. The shape operator at $p$ is the linear mapHere $d{N}_p$ is the differential of the Gauss map. Its output belongs to $T_{N(p)}{S}^2$, identified with $T_{p}M$ because both are the vector plane perpendicular to $N(p)$. The minus sign is our convention throughout this branch. ^[ref]

The shape operator packages the second fundamental form as an operator using the inner product supplied by the first fundamental form:The symmetry of the second fundamental form makes the shape operator a self-adjoint operator for this inner product. Thus an orthonormal basis of principal directions diagonalizes it. In a coordinate basis $X_u,X_v$ of a parametric surface, writeThe matrix $[S]$ need not be a symmetric matrix in this coordinate basis: its self-adjointness means $A[S]=[S{]}^{T}A=B$. Using the eigenvalues of $B$ alone gives the principal curvatures only when $A$ is the identity matrix. ^[ref]

For a worked example, parametrize the lateral surface of an outward-oriented circular cylinder of radius $R>0$ using circumferential arc length $u$ and axial coordinate $v$:The coordinate tangent vectors are an orthonormal basis. Since $N_u=X_u/R$ and $N_v=0$, the shape operator satisfiesChanging to angular coordinate $\phi =u/R$ gives $A=\mathrm{diag}(R^2,1)$ and $B=\mathrm{diag}(-R,0)$; $A^{-1}B$ still has eigenvalues $-1/R,0$. This is why the first fundamental form must participate in the calculation.

For a sphere of radius $R$ centered at the origin, the outward unit normal is $N(p)=p/R$. Hence $S_{p}v=-v/R$, so both principal curvatures are $-1/R$. For a plane, $dN=0$ and the shape operator vanishes. Reversing the unit normal negates the shape operator and both principal curvatures; the Gaussian curvature remains unchanged while the mean curvature changes sign, as in orientation reversal of surface curvature.

For a normal vector $\xi$, this identity defines the corresponding shape operator using the vector-valued second fundamental form. It is self-adjoint because of the symmetry of the second fundamental form. Differentiating $⟨\xi ,Y⟩=0$ shows that $A_{\xi }X=-(D_X\xi {)}^{\top }$. In higher codimension there is a different operator for each normal direction.

A smooth embedded surface in ${\mathbb{R}}^3$ is a two-dimensional smooth manifold locally represented by an injective regular parametrized surface $\sigma :U\subset {\mathbb{R}}^2\to {\mathbb{R}}^3$, with ${\sigma }_u\times {\sigma }_v\ne 0$. The first fundamental form measures lengths and angles, and a choice of unit normal determines the second fundamental form, which measures bending. A surface with boundary additionally uses local half-plane charts along its boundary.

If a compact regular surface lies in a closed Euclidean ball of radius $R$, maximize distance from the ball's centre. At a maximizing point, comparison of the second fundamental form with the tangent supporting sphere makes both principal curvatures have magnitude at least $R^{-1}$ and the same sign. The Gaussian curvature there is therefore at least $R^{-2}$.

The second fundamental form of an embedded submanifold of a Riemannian manifold is symmetric. Its ambient Levi-Civita connection is a torsion-free connection, soThe Lie bracket of vector fields tangent to the submanifold is tangent and therefore has zero normal component. This proof applies to a curved ambient manifold as well as Euclidean space.

For a nondegenerate induced metric tensor, a totally geodesic submanifold has vanishing second fundamental form; equivalently, every ambient geodesic initially tangent to it remains in it. This extends the notion of a totally geodesic hypersurface to arbitrary codimension.

Differentiate the orthogonality of the unit normal to the two independent coordinate tangent vectors. If the second fundamental form vanishes, each derivative of the normal is orthogonal to both tangent vectors; differentiating its unit length also makes it orthogonal to the normal. Thus both normal derivatives vanish. On a connected parameter domain the normal is constant, and differentiating the position's scalar product with that constant normal proves the surface lies in a plane.