3.6 Curvature of embedded surfaces

Given an embedded surface \(M\subset \mathbb{R}^{3},\) we may ask how we can define a notion of curvature at each point of \(M.\)

Consider a plane \(M=f^{-1}\left(\{0\}\right) \subset \mathbb{R}^3,\) where \(f : \mathbb{R}^3 \to \mathbb{R},\) \(p=(x,y,z) \mapsto Ax+By+Cz\) is a linear function and the constants \(A,B,C\) are not all zero. In this case a unit normal field is given by \[p\mapsto N(p)=\frac{1}{\sqrt{A^2+B^2+C^2}}\begin{pmatrix} A \\ B \\ C \end{pmatrix}_p.\] Observe that this unit normal field is constant when we forget about the basepoint. That is, writing \[N(p)=(\nu(p))_p\] for some map \(\nu : \mathbb{R}^3 \to M_{3,1}(\mathbb{R}),\) we have \[\nu(p)=\frac{1}{\sqrt{A^2+B^2+C^2}}\begin{pmatrix} A \\ B \\ C \end{pmatrix}\] so that \(\nu(p)\) is independent of \(p.\)

Intuitively a plane is a flat surface. In order to define a notion of curvature we can study how the unit normal field changes along an embedded surface. This leads to the notion of the shape operator.

Let \(M=f^{-1}\left(\{c\}\right)\) be an embedded surface and \(N : M \to TM^{\perp}\) a unit normal field. We may take \(N=\operatorname{grad}f/\Vert \operatorname{grad}f\Vert.\) By assumption \(\operatorname{grad}f(p)\) is non-zero for all points \(p \in M.\) Each point \(p\) of \(M\) admits an open neighbourhood on which \(\operatorname{grad}f(p)\) is also non-zero. This implies that \(N\) is well defined on an open subset \(U \subset \mathbb{R}^{3}\) which contains \(M.\) Again we write \[N(p)=\nu(p)_p\] for some function \(\nu : U \to M_{3,1}(\mathbb{R}).\) Explicitly we have \[\nu(p)=\begin{pmatrix} \nu_1(p) \\ \nu_2(p) \\\nu_{3}(p)\end{pmatrix}\] for real-valued functions \(\nu_i : U \to \mathbb{R}.\) Since \(N\) is a unit normal field, we have \[\sum_{i=1}^{3}\nu_i(p)^2=1\] for all \(p \in U.\) Taking the exterior derivative of this equation, we obtain \[\tag{3.2} 0=2\sum_{i=1}^{3}\nu_i(p)\mathrm{d}\nu_i|_{p}.\] Defining \[[\mathrm{d}\nu(\vec{v}_p)]_p:=\begin{pmatrix} \mathrm{d}\nu_1(\vec{v}_p) \\ \mathrm{d}\nu_2(\vec{v}_p) \\\mathrm{d}\nu_{3}(\vec{v}_p)\end{pmatrix}_p\] for all \(\vec{v}_p \in TU,\) (3.2) implies \[\langle N(p),[\mathrm{d}\nu(\vec{v}_p)]_p\rangle=0.\] Recall that \(T_pM^{\perp}\) is spanned by \(N(p),\) hence \([\mathrm{d}\nu(\vec{v}_p)]_p\) is an element of \(T_pM\) for all \(\vec{v}_p.\) In particular, for all \(p \in M\) we obtain a linear map \[S_p : T_p M \to T_p M, \qquad \vec{v}_p \mapsto [\mathrm{d}\nu(\vec{v}_p)]_p.\]

Here \(S^{2}\) denotes the \(2\)-sphere in \(M_{3,1}(\mathbb{R}),\) that is, those column vectors \(\vec{v}=(v_i)_{1\leqslant i\leqslant 3}\) so that \((v_1)^2+(v_2)^2+(v_3)^2=1.\)

Restricting \(\langle\cdot{,}\cdot\rangle_p\) to \(T_pM\subset T_p\mathbb{R}^{3},\) each tangent space of \(M\) is a Euclidean space and with respect to this Euclidean space structure we have the following important fact:

We thus obtain two symmetric bilinear forms on each tangent space:

We would like to have a geometric interpretation of \(\kappa(\vec{v}_p).\) By (3.4) the normal curvature of a graph \(\mathcal{G}_h\) at \((t,h(t))\) is given by (both for \(\vec{e}_p\) and for \(-\vec{e}_p\)) \[\frac{h^{\prime\prime}(t)}{(1+h^{\prime}(t)^2)^{3/2}}.\] This is precisely the signed curvature at \(t\) of the curve \(\gamma : I \to \mathbb{R}^2,\) \(t \mapsto (t,h(t)).\)

It is tempting to try to interpret \(\kappa(\vec{v}_p)\) as a signed curvature of a plane curve as well. To this end write \[p=\begin{pmatrix} x_1 \\ x_2 \\ x_{3}\end{pmatrix} \qquad \text{and} \qquad N(p)=\begin{pmatrix} v_1 \\ v_2 \\ v_{3}\end{pmatrix}_p\qquad \text{and}\qquad \vec{v}_p=\begin{pmatrix} w_1 \\ w_2\\ w_{3}\end{pmatrix}_p\] and consider the affine \(2\)-plane \(U_{\vec{v}_p}\subset \mathbb{R}^{3}\) passing through \(p\) and which is spanned by \(N(p)\) and \(\vec{v}_p\) \[U_{\vec{v}_p}:=\{(x_1+s_1v_1+s_2w_1,x_2+s_1v_2+s_2w_2, x_{3}+s_1v_{3}+s_2w_{3}) \,|\, s_1,s_2 \in \mathbb{R}\}\] The intersection of this affine \(2\)-plane with \(M\) is a plane curve. Let \(\epsilon>0\) and \(\gamma : (-\epsilon,\epsilon) \to U_{\vec{v}_p}\cap M\) be a smooth unit speed curve with \(\gamma(0)=p\) and \(\dot{\gamma}(0)=\vec{v}_p\) which is contained in the intersection \(U_{\vec{v}_p}\cap M.\) In order to apply the definition of the signed curvature of a plane curve, we choose a vector space isomorphism \(\Psi : U_{\vec{v}_p} \to \mathbb{R}^2\) and compute the signed curvature of the curve \(\delta:=\Psi \circ \gamma : I \to \mathbb{R}^2\) at \(t=0.\) Let \(\Psi : U_{\vec{v}_p} \to \mathbb{R}^2\) be the vector space isomorphism so that \[\Psi\left((x_1+s_1v_1+s_2w_1,x_2+s_1v_2+s_2w_2, x_{3}+s_1v_{3}+s_2w_{3})\right)=(s_1,s_2)\] for all \(s_1,s_2 \in \mathbb{R}.\) Observe that \[\Psi_*(N(p))=\vec{X} \qquad \text{and} \qquad \Psi_*(\vec{v}_p)=\vec{Y},\] where here we simplify notation and write \(\vec{X},\vec{Y}\) for the standard basis of \(T_{0_{\mathbb{R}^2}}\mathbb{R}^2\) \[\vec{X}=\begin{pmatrix} 1 \\ 0 \end{pmatrix}_{0_{\mathbb{R}^2}}, \qquad \vec{Y}=\begin{pmatrix} 0 \\ 1 \end{pmatrix}_{0_{\mathbb{R}^2}}.\] By definition, the signed curvature of \(\delta\) at \(t=0\) is the real number \(\kappa_0\) so that \[\ddot{\delta}(0)=\kappa_0J(\dot{\delta}(0)),\] where \(J : T_{0_{\mathbb{R}^2}}\mathbb{R}^2 \to T_{0_{\mathbb{R}^2}}\mathbb{R}^2\) is the unique linear map satisfying \[J(\vec{X})=\vec{Y} \qquad \text{and} \qquad J(\vec{Y})=-\vec{X}.\] Let \(\langle\!\langle\cdot{,}\cdot\rangle\!\rangle\) denote the inner product on \(T_{0_{\mathbb{R}^2}}\mathbb{R}^2.\) Since \(N(p),\vec{v}_p\) are orthonormal vectors in \(T_p\mathbb{R}^{3}\) and \(\vec{X},\vec{Y}\) are orthonormal vectors in \(T_{0_{\mathbb{R}^2}}\mathbb{R}^2,\) it follows that \[\Psi_* : \operatorname{span}\{\vec{v}_p,N(p)\} \to T_{0_{\mathbb{R}^2}}\mathbb{R}^2\] is an orthogonal transformation. Using this fact we compute \[\begin{aligned} \langle\ddot{\gamma}(0),N(p)\rangle&=\langle\!\langle \Psi_*(\ddot{\gamma}(0)),\Psi_*(N(p))\rangle\!\rangle=\langle\!\langle \ddot{\delta}(0),\vec{X}\rangle\!\rangle=-\kappa_0\langle\!\langle J(\dot{\delta}(0),J(\vec{Y}))\\ &=-\kappa_0\langle\!\langle \dot{\delta}(0),\vec{Y}\rangle\!\rangle=-\kappa_0\langle\!\langle \Psi_*(\dot{\gamma}(0)),\vec{Y}\rangle\!\rangle\\ &=-\kappa_0\langle\!\langle \Psi_*(\vec{v}_p),\vec{Y}\rangle\!\rangle=-\kappa_0\langle\!\langle \vec{Y},\vec{Y}\rangle\!\rangle=-\kappa_0. \end{aligned}\] We also have:

In summary we see that \[\kappa(\vec{v}_p)=-\langle \ddot{\gamma}(0),N(p)\rangle=\kappa_0\] where \(\kappa_0\) is the signed curvature at \(p\) of the curve cut out of \(M\) by the affine \(2\)-plane \(U_{\vec{v}_p}\) passing through \(p\) and which is spanned by \(N(p)\) and \(\vec{v}_p.\) Notice that \(\kappa(\vec{v}_p)\) depends on the choice of unit normal vector field \(N.\) Reversing the sign of \(N(p)\) reverses the sign of \(\kappa(\vec{v}_p).\)

Since \(S_p : T_pM \to T_pM\) is self-adjoint for all \(p \in M,\) the Spectral Theorem from M06 Linear Algebra II implies that \(T_pM\) admits an ordered orthonormal basis \((\vec{v}_p,\vec{w}_p)\) consisting of eigenvectors of \(S_p.\)

The average trace of the shape operator is known as the mean curvature and the determinant of the shape operator as the Gauss curvature:

Using the principal curvatures we can classify the points of an embedded surface into different types:

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