 Solution

At $P=\sigma (u,v)$, regularity makes ${\sigma }_u,{\sigma }_v$ linearly independent. The tangent space is $T_{P}U=\mathrm{span}\{{\sigma }_u,{\sigma }_v\}$, and an oriented unit normal vector isReversing its sign chooses the opposite orientation. The first fundamental form is the squared length of a tangent displacement:where $E={\sigma }_u\cdot {\sigma }_u$, $F={\sigma }_u\cdot {\sigma }_v$, $G={\sigma }_v\cdot {\sigma }_v$. Its determinant $EG-F^2=|{\sigma }_u\times {\sigma }_v{|}^2$ is positive.

Use $N_{\mathrm{II}}={\sigma }_{vv}\cdot \mathbf{n}$ to distinguish the scalar coefficient of the second fundamental form from the normal vector. Since $\mathbf{n}\cdot \mathbf{n}=1$, its derivatives are perpendicular to it and hence tangent. Write ${\mathbf{n}}_u=a{\sigma }_u+b{\sigma }_v$ and ${\mathbf{n}}_v=c{\sigma }_u+d{\sigma }_v$. Differentiating $\mathbf{n}\cdot {\sigma }_u=\mathbf{n}\cdot {\sigma }_v=0$ givesSubstitution of the tangent expansions provesThe shape operator is $-d\mathbf{n}$; its matrix in the parametrized tangent basis is the negative transpose of the coefficient matrix shown. Its determinant, the product of the principal curvatures, is thereforethe Gaussian curvature. Transposition and the two minus signs do not change this determinant.