Curvature

Abstract

From this chapter on we will focus attention on surfaces $f\colon M \to \mathbb {R}^3$. The most fundamental tool for analysing such a surface is its unit normal field $N\colon M\to S^2$ which is a map to the unit sphere $S^2\subset \mathbb {R}^3$.

From this chapter on we will focus attention on surfaces $f\colon M \to \mathbb {R}^3$. The most fundamental tool for analysing such a surface is its unit normal field $N\colon M\to S^2$ which is a map to the unit sphere $S^2\subset \mathbb {R}^3$. The derivative of N reveals information about the curvature of f. In particular, the area covered by N on $S^2$ provides us with a geometric interpretation of the so-called Gaussian curvature of f.

1 Unit Normal of a Surface in $\mathbb {R}^3$

Most of the material in the Chaps. 6 and 7 was concerned with the intrinsic geometry of Riemannian domains or with surfaces in $\mathbb {R}^n$. From now on we will focus on surfaces $f\colon M\to \mathbb {R}^3$.

Definition 8.1

Let $M\subset \mathbb {R}^2$ be a domain with smooth boundary and $f\colon M\to \mathbb {R}^3$ a surface. Then there is a unique smooth map $N\colon M\to \mathbb {R}^3$ with $\langle N,N \rangle =1$ such that

• (i) For all $p\in M$ and all $X\in T_pM$ we have

  $$\displaystyle \begin{aligned} \langle N(p),df(X)\rangle =0.\end{aligned}$$

• (ii) For all $p\in M$ and every positively oriented basis $X,Y$ of $T_pM$ we have

  $$\displaystyle \begin{aligned} \det(N(p),df(X),df(Y))>0.\end{aligned}$$

N is called the unit normal of f (see Fig. 8.1).

Fig. 8.1

The normal vector $N(p)$ of a surface f at a point p

In terms of the coordinate vector fields U and V  we can express N as

$$\displaystyle \begin{aligned} N=\frac{f_u\times f_v}{|f_u\times f_v|}.\end{aligned}$$

Theorem 8.2

For all$p\in M$and all$X,Y\in T_pM$we have

$$\displaystyle \begin{aligned} df(JX)&=N(p)\times df(X) \\ {\det}_f(X,Y)&=\det(N(p),df(X),df(Y)).\end{aligned} $$

For the area of f we get

$$\displaystyle \begin{aligned} \mathcal{A}(f)=\int_M {\det}_f =\int_M {\det}_f(U,V) =\int_M \det(N,f_u,f_v)=\int_M |f_u\times f_v|.\end{aligned}$$

Similar as for a surface $f\colon M\to \mathbb {R}^3$, we can consider the derivative dN of the unit normal $N\colon M\to \mathbb {R}^3$. In the case of plane curves the derivative of the normal N gave us the curvature $\kappa$ via the equation

$$\displaystyle \begin{aligned} N'=\kappa \gamma'.\end{aligned}$$

In order to find the analogous equation for surfaces, let us consider a vector field $X\in \Gamma (TM)$ and take the derivative in the direction of X of the equation $1=\langle N,N\rangle$:

$$\displaystyle \begin{aligned} 0=d_X\langle N,N\rangle=2\langle d_XN,N\rangle.\end{aligned}$$

This means that for all $X\in T_pM$ the vector $dN(X)$ lies in the image of the restriction of df to $T_pM$. Therefore, there is a vector $Y\in T_pM$ such that $dN(X)=df(Y)$. Obviously, the dependence of Y  on X is linear, so there is a linear map $A_p\colon T_pM\to T_pM$ such that for all $X\in T_pM$ we have

$$\displaystyle \begin{aligned} dN(X)=df(AX).\end{aligned}$$

We leave it to the reader to check that A is a smooth endomorphism field on M.

Definition 8.3

The smooth endomorphism field A is called the shape operator of f.

Theorem 8.4

The shape operator A is a self-adjoint endomorphism field with respect to the induced metric, i.e. for all$X,Y\in \Gamma (TM)$we have

$$\displaystyle \begin{aligned} \langle AX,Y\rangle = \langle X,AY\rangle.\end{aligned}$$

Proof

Since at each point $p\in M$ the two vectors $U(p),V(p)$ form a basis of $T_pM$, it is sufficient to prove the theorem in the special case $X=U, Y=V$. Using the fact that

$$\displaystyle \begin{aligned} \langle N,df(U)\rangle=\langle N,df(V)\rangle=0\end{aligned}$$

we obtain

$$\displaystyle \begin{aligned} \langle AU,V\rangle &= \langle df(AU),df(V)\rangle \\&= \langle dN(U),df(V)\rangle \\&=d_U\langle N,df(V)\rangle -\langle N,d_Udf(V)\rangle \\&=-\langle N,f_{vu}\rangle\\&=-\langle N,f_{uv}\rangle \\&= d_V\langle N,df(U)\rangle -\langle N,d_Vdf(U)\rangle \\&= \langle dN(V),df(U)\rangle\\&= \langle df(AV),df(U)\rangle \\&= \langle AV,U\rangle, \end{aligned} $$

where we used that the partial derivatives commute, i.e. $f_{uv}=f_{vu}$. □

2 Curvature of a Surface

The shape operator A of a surface $f\colon M\to \mathbb {R}^3$ captures all the information about how the surface is curved. In fact it measures deviation from being planar:

Theorem 8.5

Let$M\subset \mathbb {R}^2$be a connected compact domain with smooth boundary and$f\colon M\to \mathbb {R}^3$a surface with shape operator A. Then A vanishes identically if and only if there is a plane$E\subset \mathbb {R}^3$with$f(M)\subset E$.

Proof

If $f(M)\subset E$ with

$$\displaystyle \begin{aligned} E=\{\mathbf{p}\in \mathbb{R}^3\,\,|\,\, \langle \hat{N},\mathbf{p}\rangle =c\}\end{aligned}$$

for some unit vector $\hat {N}\in \mathbb {R}^3$ and $c\in \mathbb {R}$, then

$$\displaystyle \begin{aligned} \langle \hat{N},df(X)\rangle =d_X\langle \hat{N},f\rangle=0\end{aligned}$$

for all $X\in TM$, so the unit normal of f satisfies $N(p)=\pm \hat {N}$ for all $p\in M$. In particular, $dN=0$ and therefore $A=0$.

Conversely, by the connectedness of M, $A=0$ implies that N is constant, i.e. $N(p)=\hat {N}$ for some $\hat {N}\in \mathbb {R}^3$ and all $p\in M$. Then $d\langle \hat {N},f\rangle =0$ and (by the connectedness of M) there is $c\in \mathbb {R}$ such that $\langle \hat {N},f(p)\rangle = c$ for all $p\in M$. □

At a given point, a surface can be curved by a different amount in different directions. We call a vector $X\in TM$ a direction if $\langle X,X\rangle =1$.

Definition 8.6

For a direction $X\in TM$ we define the directional curvature$\kappa (X)$ of f in the direction of X as

$$\displaystyle \begin{aligned} \kappa(X):= \langle AX,X\rangle.\end{aligned}$$

If $X_1,X_2$ is an orthonormal basis of $T_pM$ then we can parametrize all unit vectors in $T_pM$ as

$$\displaystyle \begin{aligned} X(\theta)=\cos \theta \,X_1 + \sin \theta \,X_2.\end{aligned}$$

Figure 8.2 contains a plot of the function $\theta \mapsto \kappa (X(\theta ))$.

Fig. 8.2

For every unit vector $X\in T_pM$ a surface $f\colon M\to \mathbb {R}^3$ has a different directional curvature

By Theorem 8.4, for all $p\in M$ the linear map

$$\displaystyle \begin{aligned} A_p:=A|{}_{T_pM}\colon T_pM \to T_pM\end{aligned}$$

is self-adjoint, so there is an orthonormal basis $X_1,X_2$ in $T_pM$ such that $X_1$ and $X_2$ are eigenvectors of $A_p$:

$$\displaystyle \begin{aligned} AX_1 &= \kappa_1(p)X_1 \\ AX_2 &= \kappa_2(p)X_2.\end{aligned} $$

If we assume $\kappa _1\geq \kappa _2$ the eigenvalue functions $\kappa _1,\kappa _2\colon M \to \mathbb {R}$ are well-defined and continuous. They arise from solving the characteristic equation of $A_p$, in which a square root is involved. This means that in general (if there are points where $\kappa _1(p)$ and $\kappa _2(p)$ coincide) they are not smooth functions.

Definition 8.7

For $p\in M$ the numbers $\kappa _1(p)$ and $\kappa _2(p)$ are called the principal curvatures of f at p. A vector $X\in T_pM$ with $\langle X,X\rangle =1$ is called a principal direction corresponding to the principal curvature $\kappa _j$ if

$$\displaystyle \begin{aligned} AX=\kappa_j(p)X.\end{aligned}$$

If we parametrize directions $X(\theta )$ at p based on principal directions $X_1,X_2$ as above we obtain

$$\displaystyle \begin{aligned} \kappa(\theta)&=\langle A(\cos \theta \,X_1 + \sin \theta \,X_2),\cos \theta \,X_1 + \sin \theta \,X_2\rangle \\&=\kappa_1(p)\cos^2\theta + \kappa_2(p)\sin^2 \theta\\&=\frac{\kappa_1(p)+\kappa_2(p)}{2}+ \frac{\kappa_1(p)-\kappa_2(p)}{2} \cos{}(2\theta).\end{aligned} $$

Definition 8.8

The mean value

$$\displaystyle \begin{aligned} H(p):=\frac{1}{2\pi}\int_0^{2\pi}\kappa(\theta)d\theta\end{aligned}$$

is called the mean curvature of f at the point p.

We have

$$\displaystyle \begin{aligned} H(p)=\frac{\kappa_1(p)+\kappa_2(p)}{2}=\frac{1}{2}\mathrm{tr}(A_p),\end{aligned}$$

so the function $H\colon M\to \mathbb {R}$ is smooth.

Definition 8.9

The smooth function

$$\displaystyle \begin{aligned} K\colon M \to \mathbb{R},\ K(p)=\det A_p= \kappa_1(p)\kappa_2(p)\end{aligned} $$

is called the Gaussian curvature of f.

If $K(p)>0$ then the directional curvatures at p are either all positive or all negative. In the first case, the surface looks convex when viewed from “outside” (when we think of N as pointing “outward”). Otherwise it looks concave. Figure 8.3 shows surfaces whose Gaussian curvature is positive everywhere on M.

Fig. 8.3

Three surfaces with positive Gaussian curvature

If $K(p)<0$ Then the surface bends towards $N(p)$ is some directions and away from N in other directions. Figure 8.4 shows surfaces whose Gaussian curvature is negative everywhere on M.

Fig. 8.4

Three surfaces with negative Gaussian curvature

Points where the principal curvatures coincide (and therefore all directions are principal directions) are special and we give them a name:

Definition 8.10

A point $p\in M$ is called an umbilic point of the surface f if at p the surface has the same curvature in all directions, i.e. for all directions $X\in T_pM$ we have

$$\displaystyle \begin{aligned} \kappa(X)=H(p).\end{aligned}$$

The most interesting theorems in Differential Geometry lead from local assumptions (curvature properties at each given point) to conclusions about global shape. Here is our first theorem of this kind in the context of surfaces:

Definition 8.11

A subset $S\subset \mathbb {R}^3$ of the form

$$\displaystyle \begin{aligned} S=\{\mathbf{p}\in \mathbb{R}^3\,\,|\,\, \langle \mathbf{p}-\mathbf{m}, \mathbf{p}-\mathbf{m}\rangle =r^2\}\end{aligned}$$

with $\mathbf {m}\in \mathbb {R}^3$ and $r>0$ is called a round sphere.

Theorem 8.12 (Umbilic Point Theorem)

Let$M\subset \mathbb {R}^2$be a connected compact domain with smooth boundary and$f\colon M\to \mathbb {R}^3$a surface. Then the following are equivalent:

• (i) All points$p\in M$are umbilic points.

• (ii) Either$f(M)\subset E$for some plane$E\subset \mathbb {R}^3$or$f(M)\subset S$for some round sphere

  $$\displaystyle \begin{aligned} S=\left\{\mathbf{p}\in \mathbb{R}^3\,\,|\,\, \langle \mathbf{p}-\mathbf{m},\mathbf{p}-\mathbf{m}\rangle=r^2\right\}.\end{aligned}$$

  with center$\mathbf {m}$and radius$r>0$.

Proof

If $f(M)$ is contained in a plane, we already know that $A=0$ and therefore all points are umbilic points. If $f(M)$ is contained in a round sphere, then there is a point $\mathbf {m}\in \mathbb {R}^3$ and a radius $r>0$ such that

$$\displaystyle \begin{aligned} \langle f-\mathbf{m},f-\mathbf{m}\rangle=r^2.\end{aligned}$$

Clearly then, $f-\mathbf {m}\neq 0$ for all $p\in M$. Differentiating the above equation reveals that for all $p\in M$ and all $X\in T_pM$ we have

$$\displaystyle \begin{aligned} \langle df(X),f-\mathbf{m}\rangle=0.\end{aligned}$$

Therefore, at each $p\in M$ the unit normal of f must be given by

$$\displaystyle \begin{aligned} N(p)=\pm\frac{1}{r}(f(p)-\mathbf{m}).\end{aligned}$$

By the connectedness of M this implies

$$\displaystyle \begin{aligned} N=\pm \frac{1}{r}(f-\mathbf{m})\end{aligned}$$

and therefore all points are umbilic points:

$$\displaystyle \begin{aligned} dN=\pm \frac{1}{r}df.\end{aligned}$$

Conversely, assume that all points $p\in M$ are umbilic points of f. Then

$$\displaystyle \begin{aligned} H_vf_u+Hf_{uv}=N_{uv}=N_{vu}= H_uf_v +Hf_{vu} \end{aligned} $$

and therefore $H_vf_u-H_uf_v=0.$ By the connectedness of M, this means that H is constant. In the case $H=0$ we have $A=0$ and by Theorem 8.5 we know that $f(M)$ is contained in a plane. Otherwise, there is a constant $r>0$ such that

$$\displaystyle \begin{aligned} H=\pm \frac{1}{r}.\end{aligned}$$

The function

$$\displaystyle \begin{aligned} \mathbf{m}\colon M \to \mathbb{R}^3,\ \mathbf{m}(p)=f(p)\pm r N(p)\end{aligned} $$

then satifies $d\mathbf {m}=0$ and, by the connectedness of M, must be constant. This means that $f(M)$ lies on a sphere around $\mathbf {m}$ with radius r. □

3 Area of Maps Into the Plane or the Sphere

Recall the second formula from Theorem 8.2: The area form $\det$ of a surface $f\colon M\to \mathbb {R}^3$ with unit normal N is given on $X,Y\in T_pM$ by

$$\displaystyle \begin{aligned} {\det}_f(X,Y)=\det(N(p),df(X),df(Y)).\end{aligned}$$

There are situations where we know what N should be, even if f is not a surface but just a smooth map whose derivative $d_pf\colon T_pM\to \mathbb {R}^3$ might fail to have a two-dimensional image for some $p\in M$: Define the Euclidean plane$E^2$ as the subset of $\mathbb {R}^3$ where the third component is zero. Then at any point $\mathbf {p}\in E^2$ we consider the third basis vector ${\mathbf {e}}_3$ as the unit normal vector of $E^2$ at $\mathbf {p}$. Define the unit two-sphere $S^2$ as the set of all $\mathbf {p}\in \mathbb {R}^3$ with $|\mathbf {p}|=1$. Then at any point $\mathbf {p}\in S^2$ we consider $\mathbf {p}$ itself as the unit normal vector of $S^2$ at $\mathbf {p}$.

Definition 8.13

Let $M\subset \mathbb {R}^2$ be a compact domain with smooth boundary and g a smooth map defined on M with values in either $E^2$ or $S^2$. Then we define the covered area form$\sigma _g\in \Omega ^2(M)$ on $X,Y\in T_pM$ as follows:

• (i) For a smooth map $g\colon M\to E^2$ we define

  $$\displaystyle \begin{aligned} \sigma_g(X,Y)=\det\left({\mathbf{e}}_3,dg(X),dg(Y)\right).\end{aligned}$$

• (ii) For a smooth map $g\colon M\to S^2$ we define

  $$\displaystyle \begin{aligned} \sigma_g(X,Y)=\det\left(g(p),dg(X),dg(Y)\right).\end{aligned}$$

If we identify $E^2$ with $\mathbb {R}^2$ in the obvious way and use the standard determinant $\det$ on $\mathbb {R}^2$, the first part of the above definition becomes:

Definition 8.14

Let $M\subset \mathbb {R}^2$ be a compact domain with smooth boundary and $g\colon M\to \mathbb {R}^2$ a smooth map. Then we define the covered area form$\sigma _g\in \Omega ^2(M)$ on $X,Y\in T_pM$ as

$$\displaystyle \begin{aligned} \sigma_g(X,Y)=\det(dg(X),dg(Y)).\end{aligned}$$

Definition 8.15

Let $M\subset \mathbb {R}^2$ be a compact domain with smooth boundary and g a smooth map defined on M with values in either $E^2$, $S^2$ or $\mathbb {R}^2$. Then we define the area covered by a mapg as

$$\displaystyle \begin{aligned} \int_M \sigma_g.\end{aligned}$$

Given a smooth map $g\colon M\to \mathbb {R}^2$ from the unit disk M into $\mathbb {R}^2$, we obtain a loop $\gamma \colon [0,2\pi ]\to \mathbb {R}^2$ defined by

$$\displaystyle \begin{aligned} \gamma(x)=g\left(\!\begin{pmatrix}\cos x\\ \sin x\end{pmatrix}\!\right).\end{aligned}$$

Conversely, every loop $\gamma \colon [0,2\pi ]\to \mathbb {R}^2$ arises in this way:

Theorem 8.16

Let$\gamma \colon \mathbb {R}\to \mathbb {R}^2$be a loop and$M\subset \mathbb {R}^2$the unit disk. Then there is a smooth map$g\colon M\to \mathbb {R}^2$such that the Fig.8.5becomes a commutative diagram, i.e.$\gamma =g\circ s$.

Fig. 8.5

By Theorem 8.17, the area covered by the map g equals the sector area of the boundary loop $\gamma$, i.e. the blue region is counted positively, the orange region is counted negatively and the region with mixed color is not counted at all

Proof

Let $\varphi \colon [0,1]\to \mathbb {R}$ be a smooth function such that

$$\displaystyle \begin{aligned} \varphi(1)&=1 \\ \varphi(x)&=0 \qquad \text{for}\quad x<\frac{1}{3}.\end{aligned} $$

Then we can define $g\colon M\to \mathbb {R}^2$ as the unique map such that for all $r\in [0,1]$ and all $t\in \mathbb {R}$ we have

$$\displaystyle \begin{aligned} g\left(r\begin{pmatrix}\cos t\\ \sin t\end{pmatrix}\right)=\varphi(r)\gamma(t).\end{aligned}$$

□

Theorem 8.17

Let$\gamma \colon [0,2\pi ]\to \mathbb {R}^2$be a loop,$M\subset \mathbb {R}^2$the unit disk and$g\colon M\to \mathbb {R}^2$any smooth map such that Fig.8.5is a commutative diagram. Then the area covered by g equals the sector area of$\gamma$.

Proof

Define a 1-form $\omega \in \Omega ^1(M)$ by setting for $X\in T_pM$

$$\displaystyle \begin{aligned} \omega(X)=\frac{1}{2}\det(g(p),dg(X)).\end{aligned}$$

Then

$$\displaystyle \begin{aligned} 2d\omega(U,V)&=d_U\det(g,dg(V))-d_V\det(g,dg(U))\\&= \det(d_U g,d_V g)+\det(g,d_U d_V g)\\ &\quad -\det(d_V g,d_U g) -\det(g,d_V d_U g) \\&=2\sigma_g(U,V).\end{aligned} $$

Our claim now follows from Stokes Theorem. With

$$\displaystyle \begin{aligned} \tilde{\gamma}(t):=\begin{pmatrix}\cos t\\ \sin t\end{pmatrix}\end{aligned}$$

we obtain

$$\displaystyle \begin{aligned} \frac{1}{2}\int_0^{2\pi}\det(\gamma,\gamma') &= \frac{1}{2}\int_0^{2\pi} \det(g\circ \tilde{\gamma},(g'\circ \tilde{\gamma})\tilde{\gamma}')\\&=\int_{[0,2\pi]}\tilde{\gamma}^*\omega \\&=\int_{\partial M} \omega \\ &=\int_M d\omega \\ &= \int_M \sigma_g.\end{aligned} $$

□

By Definition 8.13, we obtain a similar interpretation for the area covered by a map $g\colon M\to S^2$. For us, the most important case is $g=N$ where N is the unit normal of a surface $f\colon M\to \mathbb {R}^3$ (see Fig. 8.6):

Fig. 8.6

If the Gaussian curvature K is positive in some subregion $\tilde {M}\subset M$, the normal map N will be orientation-preserving in $\tilde {M}$(left). If K is negative, N will be orientation-reversing (middle). On the right we see a situation where K changes sign in $\tilde {M}$

Theorem 8.18

Let$f\colon M\to \mathbb {R}^3$be a surface with unit normal N and Gaussian curvature K. Then the covered area form of N is

$$\displaystyle \begin{aligned} \sigma_N=K\,{\det}_f.\end{aligned}$$

Proof

For vector fields $X,Y\in \Gamma (TM)$ we have

$$\displaystyle \begin{aligned} \sigma_N(X,Y)&=\det(N,dN(X),dN(Y))\\ &= \det(N,df(AX),df(AY)) \\&= {\det}_f(AX,AY)\\&= \det A \,{\det}_f(X,Y) =K \,{\det}_f(X,Y).\end{aligned} $$

□