= Solution
Choose the <orientation of a surface> represented by a smooth <Gauss map> $N:S\to\mathbb S^2$. The <shape operator> at $p$ is
$$
S_p=-dN_p:T_pS\longrightarrow T_pS.
$$
If its <principal curvature>[principal curvatures] are $k_1,k_2$, then by definition
$$
\boxed{K(p)=\det S_p=k_1k_2,
\qquad
H(p)=\frac12\operatorname{tr}S_p=\frac{k_1+k_2}{2}.}
$$
Let $E(x)=Ax+b$ be a <proper Euclidean motion of Euclidean three-space>, so $A\in SO(3)$. Orient $\widetilde S=E(S)$ by transporting the normal:
$$
\widetilde N(E(p))=AN(p).
$$
The <chain rule> gives the orthogonal conjugacy
$$
\widetilde S_{E(p)}
=A S_pA^{-1}.
$$
The <determinant> and <matrix trace>[trace] are invariant under <similar matrix>[similarity], so the <Euclidean invariance of the shape operator> yields
$$
\boxed{\widetilde K(E(p))=K(p),
\qquad
\widetilde H(E(p))=H(p).}
$$
If the surface orientation is reversed, then $N$ and $S_p$ are multiplied by $-1$. Therefore the <orientation reversal of surface curvature> says
$$
\boxed{K\text{ is independent of orientation},
\qquad H\text{ changes sign}.}
$$
Back to article page
