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If a unit tangent makes angle $\theta$ with the first principal direction, the normal curvature is ${\kappa }_n={\kappa }_1{\cos }^2\theta +{\kappa }_2{\sin }^2\theta$. This follows by diagonalizing the second fundamental form relative to the first fundamental form. When the Gaussian curvature is nonnegative, $|{\kappa }_n|\le 2|H|$.

The second fundamental form records the normal component of second-order bending of the surface. For a surface curve $\gamma (t)=\sigma (u(t),v(t))$, the terms containing $u^{\prime \prime }$ and $v^{\prime \prime }$ are tangent and thereforeThus on a unit tangent direction it gives the normal curvature; for a general coordinate direction the normal curvature is the ratio of the second to the first fundamental form. Equivalently the quadratic part of displacement out of the tangent plane is one half of this form. Reversing the chosen unit normal changes its sign.

Now suppose $L=M=N=0$. Differentiate $\mathbf{N}\cdot {\sigma }_u=\mathbf{N}\cdot {\sigma }_v=0$. This givesAlso differentiation of $\mathbf{N}\cdot \mathbf{N}=1$ gives ${\mathbf{N}}_u\cdot \mathbf{N}={\mathbf{N}}_v\cdot \mathbf{N}=0$. The parametrization is regular, so ${\sigma }_u,{\sigma }_v,\mathbf{N}$ form a basis of three-dimensional space. Both normal derivatives are orthogonal to this basis and hence vanish:The parameter ball is connected; integrating these derivatives along a straight segment between parameter points shows that $\mathbf{N}={\mathbf{N}}_0$ is constant. Finally both derivatives of $\sigma \cdot {\mathbf{N}}_0$ vanish, because ${\sigma }_u,{\sigma }_v$ are tangent. Thus the zero second fundamental form implies planar image statement isThe PDF has ${\mathbf{N}}_v$ in the second derivative equality; the apparent ${\mathbf{N}}_t$ in the TeX is not a distinct coordinate.

The Gauss map of an oriented surface is its smooth unit normal $N:S\to S^2$. With the convention that the shape operator is $A_p=-d{N}_p$, the second fundamental form is ${\mathrm{II}}_p(v,w)=⟨A_{p}v,w⟩$ and the normal curvature in a nonzero direction $w$ is $k_n(w)=\mathrm{II}(w,w)/⟨w,w⟩$. The two principal curvatures are the real eigenvalues $k_1,k_2$ of this self-adjoint operator; the mean curvature is $H=(k_1+k_2)/2$.

For directions successively spaced around the tangent plane, write their angles to a principal direction as ${\theta }_j={\theta }_0+(j-1)\pi /m$. The Euler formula for normal curvature givesThe geometric series ${\sum }_{j=0}^{m-1}{e}^{2\pi ij/m}=0$ for $m\ge 2$ makes the cosine sum zero. ThereforeHere equal angular spacing is understood in the usual directed sense. If the printed consecutive angles were only unsigned, without requiring the same rotational sense, the assertion would be false: for $m=3$ the directions $0,\pi /3,0$ satisfy those unsigned angle conditions but do not generally give $3H$.

A minimal surface has $H=0$, so at each point $k_2=-k_1$. In an orthonormal principal basis, $A=\mathrm{diag}(k_1,-k_1)$ and $A^2=k_1^{2}I$. Hencewhich proves the identity with $\boxed{\mu (p)=k_1^2=-K(p)\ge 0}$, where $K$ is the Gaussian curvature.

The converse is false. The identity only forces $k_1^2=k_2^2$, which allows either $k_2=-k_1$ or $k_2=k_1$. A sphere of radius $R$ has $dN=I/R$ for its outward normal, so it satisfies the identity with $\mu =R^{-2}$, yet $H=-1/R\ne 0$ under the chosen convention. Thus a conformal Gauss map can also arise from an umbilic nonminimal surface.

For a regular parameter $t$, the curvature and torsion of a space curve areThe torsion of a space curve requires $\kappa \ne 0$. Reparameterize by arc length $s$ and set $T={\alpha }_s$, $N=T_s/\kappa$, $B=T\times N$. Differentiating the pairwise inner products of this orthonormal Frenet frame shows that its derivative matrix is skew symmetric. Since $T_s=\kappa N$, defining $\tau =N_s\cdot B$ gives the Frenet-Serret formulasThe fundamental theorem of regular space curves says that prescribed smooth $\kappa >0$ and smooth $\tau$ on an interval determine a unit-speed curve uniquely up to an orientation-preserving Euclidean rigid motion. Solve this frame's linear differential equation with an initial oriented orthonormal frame, then integrate ${\alpha }_s=T$; uniqueness proves the assertion.

Under the intended convention that the shared parameter $s$ is arc length, equality on an open interval gives equal position and equal Frenet frames at one point. Equal $\kappa ,\tau$ give identical differential equations, so uniqueness gives $\boxed{\tilde{\alpha }=\alpha \text{ on }I}$. The printed “regular smooth” hypothesis alone does not establish this convention. Without common arc length, an orientation-preserving smooth change of speed that is the identity on $J$ but not elsewhere reparameterizes a circle into a counterexample: both curves have $\kappa =1$, $\tau =0$ and agree on $J$, but differ elsewhere as parameterized curves.

For an oriented surface with unit normal $n$, define normal curvature ${\kappa }_n=T_s\cdot n$ and geodesic curvature ${\kappa }_g=T_s\cdot (n\times T)$. ThusA unit-speed curve is a geodesic exactly when ${\kappa }_g=0$. If also ${\kappa }_n\equiv 0$, then $T_s=0$, so it is a straight line. A vertical ruling of a circular cylinder is such a geodesic, although the cylinder contains no open piece of a plane. A nonconstant regular straight line cannot be a simple closed curve. Hence no planar patch is required, and such a geodesic cannot be simple and closed.

Finally let the principal curvatures be ${\kappa }_1,{\kappa }_2$. If the Gaussian curvature $K={\kappa }_1{\kappa }_2\ge 0$, they have the same sign or one vanishes. Euler formula for normal curvature makes ${\kappa }_n$ a weighted average of them. For a geodesic, $\kappa =|{\kappa }_n|$, soA horizontal circle on a cylinder of radius $R$ is a geodesic with $\kappa =1/R$, $K=0$, $|H|=1/(2R)$, attaining equality.

Suppose $S$ lies in the closed ball of radius $R$ centred at $c$. By compactness, the continuous function $q\mapsto |q-c|$ attains its maximum $r\le R$ at some $p\in S$. The sphere of radius $r$ about $c$ is a supporting sphere tangent to $S$ at $p$.

Put $N=(p-c)/r$. For any unit vector $v\in T_{p}S$, choose a surface curve $\alpha$ with $\alpha (0)=p$ and ${\alpha }^{\prime }(0)=v$. Since $|\alpha (t)-c{|}^2$ has a local maximum at zero,The scalar $⟨{\alpha }^{\prime \prime }(0),N⟩$ is the normal curvature in direction $v$, so every normal curvature is at most $-1/r$ with this choice of unit normal. Applying this to principal directions shows that both principal curvatures satisfy $k_i\le -1/r$. ThereforeThis proves both assertions, including the existence of an elliptic point on every compact regular surface.

At a point of a smooth surface, a normal section of a surface can bend by different amounts depending on its tangent vector. The principal curvatures are the largest and smallest signed normal curvatures among unit tangent vectors. Equivalently they are the two eigenvalues $k_1,k_2$ of the shape operator, with an orthonormal basis $e_1,e_2$ of corresponding principal directions. Here the labels follow the selected directions, without requiring $k_1\le k_2$. ^[ref]

For a unit tangent vector making angle $\theta$ with $e_1$, Euler formula for normal curvature givesThe normal curvature is a weighted average of the principal curvatures; the extremes occur in the principal directions. If the principal curvatures coincide, every direction in the tangent plane is a principal direction: this is an umbilical point. The Gaussian curvature and mean curvature summarize the pair by

For an outward-oriented sphere of radius $R$, both principal curvatures are $-1/R$, so the normal curvature is $-1/R$ in every direction. For an outward-oriented circular cylinder, choose $e_1$ around the circumference and $e_2$ along the axis. The principal curvatures are $-1/R,0$, giving $k_n(\theta )=-{\cos }^2\theta /R$; halfway between the principal directions, $k_n(\pi /4)=-1/(2R)$.

At the origin of the upward-oriented hyperbolic paraboloid $z=(x^2-y^2)/2$, the principal directions are the coordinate axes and the principal curvatures are $1,-1$ in the chosen length unit. Thus $k_n(\theta )=\cos (2\theta )$, with zero normal curvature at $\theta =\pi /4,3\pi /4$. Zero normal curvature in these directions coexists with Gaussian curvature $-1$: the shape operator is not zero. The mean curvature is zero at this point; this does not say that the whole hyperbolic paraboloid is a minimal surface.

Each graph uses angle $\theta$ in the tangent plane, measured from the indicated first principal direction. The sphere and circular cylinder have $R=1$; the saddle graph refers only to the origin. Reversing the unit normal reflects each normal curvature graph through zero, negates both principal curvatures and the mean curvature, and preserves their product, the Gaussian curvature.

A principal direction is an eigenvector of the shape operator. The normal curvature in a unit principal direction is the corresponding principal curvature.

A supporting sphere for an embedded surface at $p$ is a sphere tangent at $p$ such that the surface lies locally on one side of the sphere. Comparing second fundamental forms at the tangency controls the surface's normal curvature.