Shape operator
= Shape operator
{title2=$S_p$}
= Weingarten map
{c}
{synonym}
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 map>
$$
S_p=-dN_p:T_pM\to T_pM.
$$
Here $dN_p$ is the <differential> of the <Gauss map>. Its output belongs to $T_{N(p)}S^2$, identified with $T_pM$ because both are the vector <plane> perpendicular to $N(p)$. The minus sign is our convention throughout this branch. https://ghomi.math.gatech.edu/Classes/Math497C/LectureNotes9.pdf[Ghomi, Lecture Notes 9, §2.2: Gauss map and shape operator]{ref}
The <shape operator> packages the <second fundamental form> as an operator using the <inner product> supplied by the <first fundamental form>:
$$
II(v,w)=I(S_pv,w).
$$
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>, write
$$
A=\begin{pmatrix}E&F\\F&G\end{pmatrix},\qquad B=\begin{pmatrix}e&f\\f&g\end{pmatrix},\qquad [S]=A^{-1}B.
$$
The <matrix> $[S]$ need not be a <symmetric matrix> in this coordinate <basis>: its <self-adjointness> means $A[S]=[S]^TA=B$. Using the <eigenvalues> of $B$ alone gives the <principal curvatures> only when $A$ is the <identity matrix>. https://people.maths.ox.ac.uk/ritter/geometry-of-surfaces/ritter-geom-surf-67.pdf[Ritter, Geometry of Surfaces, §12.1 and Theorem 12.8: self-adjointness and coordinate matrices]{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 $v$:
$$
X(u,v)=(R\cos(u/R),R\sin(u/R),v),\qquad N(u,v)=(\cos(u/R),\sin(u/R),0).
$$
The coordinate <tangent vectors> are an <orthonormal basis>. Since $N_u=X_u/R$ and $N_v=0$, the <shape operator> satisfies
$$
S(X_u)=-\frac1R X_u,\qquad S(X_v)=0,\qquad A=\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad B=\begin{pmatrix}-1/R&0\\0&0\end{pmatrix}.
$$
Changing to angular coordinate $\varphi=u/R$ gives $A=\operatorname{diag}(R^2,1)$ and $B=\operatorname{diag}(-R,0)$; $A^{-1}B$ 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 $S_pv=-v/R$, 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>.