Surfaces and Riemannian Geometry

Abstract

The most simple quantity of a one-dimensional curve $\gamma \colon [a,b]\to \mathbb {R}^n$ is its speed $|\gamma '|\colon [a,b]\to \mathbb {R}$.

The most simple quantity of a one-dimensional curve $\gamma \colon [a,b]\to \mathbb {R}^n$ is its speed $|\gamma '|\colon [a,b]\to \mathbb {R}$. The goal of this chapter is to arrive at the analogous statement for a two-dimensional surface in $\mathbb {R}^n$. Our first task will be to replace the interval $[a,b]$ by a suitable domain of definition $M\subset \mathbb {R}^2$. For a surface $f\colon M\to \mathbb {R}^n$ the analog for the speed of a curve will be a Riemannian metric induced on M by f.

1 Surfaces in $\mathbb {R}^n$

Our investigations of curves in $\mathbb {R}^n$ will be the guideline when we now start to study surfaces. For the most part we will focus on surfaces in $\mathbb {R}^3$. We will study the curvature of surfaces and the analog of the length of a curve (obviously the area of a surface) as well as the analog of the total squared curvature (called the Willmore functional). We will study the critical points of the area under variations with support in the interior (these surfaces are called minimal surfaces) and of the Willmore functional. We will prove a famous result that concerns the surface analog of $\int _a^b \kappa \,ds$, the so-called Gauss-Bonnet theorem. We will investigate the analog (called the Euler characteristic) for the tangent winding number of a curve in $\mathbb {R}^2$.

In our discussion of (non-closed) curves $\gamma$ in $\mathbb {R}^n$, $\gamma$ was always defined on a closed interval $[a,b]$. It would have made little difference if $\gamma$ would have been defined on the finite union of pairwise disjoint intervals. In the case of surfaces, it will be useful to allow for such disconnected domains.

Definition 6.1

A subset $M\subset \mathbb {R}^2$ is called a connected compact domain with smooth boundary if

$$\displaystyle \begin{aligned} M=M_0\setminus\{\mathring{M}_1\cup\ldots \cup \mathring{M}_k\}\end{aligned}$$

where for each $j\in \{0,\ldots ,k\}$

$$\displaystyle \begin{aligned} M_j=\varphi_j(D)\end{aligned}$$

is the image of the unit disk

$$\displaystyle \begin{aligned} D:=\{p\in\mathbb{R}^2 \,\,|\,\, |p|\leq 1\}\end{aligned}$$

under a diffeomorphism

$$\displaystyle \begin{aligned} \varphi_j\colon D \to \mathbb{R}^2\end{aligned}$$

and the $M_j$ are pairwise disjoint and contained in the interior of $M_0$. A finite disjoint union of connected compact domains with smooth boundary is called a compact domain with smooth boundary (see Fig. 6.1).

Fig. 6.1

A compact domain with smooth boundary

Notation

Throughout the rest of the book, M will denote a compact domain with smooth boundary in $\mathbb {R}^2$.

In order to avoid having to mention regularity constantly, we include regularity in the definition of a surface (see Fig. 6.2):

Fig. 6.2

A surface $f\colon M\rightarrow \mathbb {R}^3$

Definition 6.2

A surface in $\mathbb {R}^n$ is a smooth map $f\colon M\to \mathbb {R}^n$ whose derivative $f'(p)$ is an $(n\times 2)$-matrix of rank 2 for all $p\in M$.

We will denote the coordinates in $\mathbb {R}^2$ by u and v. Partial derivatives with respect to u or v will be denoted by subscripts, so for a surface f in $\mathbb {R}^n$ the matrix-valued function $f'\colon M\to \mathbb {R}^{n\times 2}$ is of the form

$$\displaystyle \begin{aligned} f'=\begin{pmatrix} | & | \\ f_u & f_v \\ | & |\end{pmatrix}\end{aligned}$$

with $f_u(p),f_v(p)\in \mathbb {R}^n$ linearly independent for all $p\in M$. The following two definitions are special cases of the ones in Appendix A.1.

Definition 6.3

A map $f\colon M\to \mathbb {R}^n$ is called smooth if it is of the form $f=\tilde {f}|{ }_{M}$ for some smooth map $\tilde {f}\colon U\to \mathbb {R}^n$ where $U\subset \mathbb {R}^2$ is an open set that contains M.

It is easy to check that even at boundary points $p\in M$ the Jacobian matrix $\tilde {f}'(p)$ only depends on f, not on the specific way in which $\tilde {f}$ extends f. We therefore can safely define $f'(p):=\tilde {f}\,\!'(p)$.

Definition 6.4

If $M,\tilde {M}\subset \mathbb {R}^2$ are two compact domains with smooth boundary, a bijective map $\varphi \colon M \to \tilde {M}$ is called a diffeomorphism if both $\varphi$ and $\varphi ^{-1}$ are smooth. A diffeomorphism $\varphi$ is called orientation-preserving if $\det \varphi '(p)>0$ for all $p\in M$.

Definition 6.5

If $f\colon M\to \mathbb {R}^n$ and $\tilde {f}\colon \tilde {M}\to \mathbb {R}^n$ are two surfaces, then $\tilde {f}$ is called an (orientation-preserving) reparametrization of f if there is an (orientation-preserving) diffeomorphism $\varphi \colon \tilde {M}\to M$ such that

$$\displaystyle \begin{aligned} \tilde{f}=f\circ \varphi,\end{aligned}$$

(see Fig. 6.3).

Fig. 6.3

A reparametrization $\tilde f$ of a surface f

As in the case of curves in the plane, it is not difficult to check that reparametrization (as well as orientation-preserving reparametrization) defines an equivalence relation on the set of surfaces in $\mathbb {R}^n$. Although we will not formalize this, we are only interested in properties of surfaces that are invariant under orientation-preserving reparametrization, so the real objects of our study are the equivalence classes of surfaces under reparametrization.

2 Tangent Spaces and Derivatives

Let $M\subset \mathbb {R}^2$ be a compact domain with smooth boundary and $f\colon M\to \mathbb {R}^k$ a smooth map. Then the directional derivative of f at a point $p\in M$ in the direction of a vector $\hat {X}\in \mathbb {R}^2$ is given by

$$\displaystyle \begin{aligned} df(p,\hat{X}):=f'(p)\hat{X}.\end{aligned}$$

This means that all these directional derivatives are encoded in a map $df\colon M\times \mathbb {R}^2\to \mathbb {R}^k$:

Definition 6.6

For a point $p\in M$, a tangent vector to M at p is a pair $X=(p,\hat {X})$ where $\hat {X}\in \mathbb {R}^2$. The set

$$\displaystyle \begin{aligned} T_pM=\{p\}\times \mathbb{R}^2\end{aligned}$$

of all these tangent vectors is called the tangent space to M at p. We make each $T_pM$ into a two-dimensional real vector space by defining for $X=(p,\hat {X}), Y=(p,\hat {Y})$ and $\lambda \in \mathbb {R}$

$$\displaystyle \begin{aligned} X+Y &=(p,\hat{X}+\hat{Y})\\ \lambda X&=(p,\lambda \hat{X}).\end{aligned} $$

The union $TM=M\times \mathbb {R}^2$ of all these tangent spaces is called the tangent bundle of M. The map

$$\displaystyle \begin{aligned} \pi\colon TM &\to M ,\ (p,\hat{X}) \mapsto p\end{aligned} $$

is called the projection map of the tangent bundle.

One immediate benefit of this definition is a more concise notation for derivatives:

Definition 6.7

For a smooth map $f\colon M\to \mathbb {R}^n$ we define the derivative$df\colon TM \to \mathbb {R}^n$ of f by setting for $X\in TM, X=(p,\hat {X})$

$$\displaystyle \begin{aligned} df(X)=f'(p)\hat{X},\end{aligned}$$

(see Fig. 6.4).

Fig. 6.4

Two tangent vectors $X,Y\in T_pM$ and their image under df

The restriction of df to each tangent space $T_pM$ is a linear map from $T_pM$ to $\mathbb {R}^n$.

Definition 6.8

A smooth map $X\colon M \to TM$ is called a vector field if $\pi \circ X= \mathrm {id}_M$, which means that $X(p)\in T_pM$ for all $p\in M$.

If $\hat {X}\colon M\to \mathbb {R}^2$ is a smooth map, then the assignment

$$\displaystyle \begin{aligned} X\colon M &\to TM,\ X(p) = (p,\hat{X}(p))\end{aligned} $$

is a smooth vector field on M and all smooth vector fields are obtained in this way. Here is some convenient notation:

Definition 6.9

• (i) The vector space of all smooth functions $f\colon M\to \mathbb {R}$ is denoted by $C^\infty (M)$.

• (ii) The vector space of all smooth functions $f\colon M\to \mathbb {R}^n$ is denoted by $C^\infty (M,\mathbb {R}^n)$.

• (iii) The vector space of all smooth vector fields on M is denoted by $\Gamma (TM)$.

As is known from calculus class, for a smooth map $f\colon M\to \mathbb {R}^n$ the vector $f'(p)\hat {X}\in \mathbb {R}^n$ can also be interpreted as the directional derivative of f at p in the direction of the vector $\hat {X}\in \mathbb {R}^2$. With this in mind we define the directional derivative of $f\in C^\infty (M,\mathbb {R}^n)$ in the direction of a vector field $X\in \Gamma (TM)$ by

$$\displaystyle \begin{aligned} (d_Xf)(p):=d_{X(p)}f=df(X(p)).\end{aligned}$$

Definition 6.10

The coordinate vector fields$U,V\in \Gamma (TM)$ are defined as

$$\displaystyle \begin{aligned} U(p) = \left(p,\left(\begin{array}{c} 1 \\0\end{array}\right)\right), \qquad V(p)=\left(p,\left(\begin{array}{c} 0 \\1\end{array}\right)\right).\end{aligned}$$

The directional derivatives in the direction of U or V  are just partial derivatives:

$$\displaystyle \begin{aligned} d_U f=f_u, \qquad d_Vf=f_v.\end{aligned}$$

Definition 6.11

Let $M,\tilde {M}\subset \mathbb {R}^2$ be two compact domains with smooth boundary and $\varphi \colon \tilde {M}\to M$ a diffeomorphism. Then we define

$$\displaystyle \begin{aligned} d\varphi\colon T\tilde{M}&\to TM,\ d\varphi(X)=\left(\varphi(p),\varphi'(p)\hat{X}\right) \quad \text{for} \quad X=\left(p,\hat{X}\right).\end{aligned} $$

Remark 6.12

Note that for $X\in T_p\tilde {M}$ the vector $d\varphi (X)$ is an element of $T_{\varphi (p)}M$, while for a surface $f\colon M\to \mathbb {R}^n$ the vector $df(X)$ is just an element of $\mathbb {R}^n$, not an element of something like $T_{f(p)}\mathbb {R}^n$. We are relying here on the fact that in our situation we can naturally identify all such tangent spaces $T_{q}\mathbb {R}^n$ with $\mathbb {R}^n$ itself. This mild context-dependency of notation should lead to no confusion. It is very useful and common in Differential Geometry.

With this notation in place, the chain rule now emerges in its most elegant form:

Theorem 6.13

• (i) Suppose$\tilde {f}\colon \tilde {M}\to \mathbb {R}^n$is a reparametrization of the surface$f\colon M\to \mathbb {R}^n$, i.e.$\tilde {f}=f\circ \varphi$for some diffeomorphism$\varphi \colon \tilde {M}\to M$. Then

  $$\displaystyle \begin{aligned} d\tilde{f}=df\circ d\varphi.\end{aligned}$$

• (ii) If$M,\tilde {M},\hat {M}\subset \mathbb {R}^2$are compact domains with smooth boundary and$\varphi \colon \tilde {M}\to M$and$\tilde {\varphi }\colon \hat {M}\to \tilde {M}$are diffeomorphisms, then

  $$\displaystyle \begin{aligned} d(\varphi \circ \tilde{\varphi})=d\varphi \circ d\tilde{\varphi}.\end{aligned}$$

Proof

The proof just involves spelling out our definitions and applying the ordinary chain rule. □

3 Riemannian Domains

When it comes to investigating the geometry of a surface $f\colon M\to \mathbb {R}^n$, the geometry of M as it sits in $\mathbb {R}^2$ is completely irrelevant. Things like the length of a vector or the angle between vectors should be computed in the target space $\mathbb {R}^n$ of f, not in $\mathbb {R}^2$. Accordingly, we endow each tangent space $T_pM$ with its own private Euclidean scalar product by defining

$$\displaystyle \begin{aligned} \langle\,,\rangle_f\colon \bigcup_{p\in M} (T_pM\times T_pM) &\to \mathbb{R},\ \langle X,Y \rangle_f = \langle df(X),df(Y)\rangle.\end{aligned} $$

It is easy to check that for each $p\in M$ the restriction of $\langle \,,\rangle _f$ to $T_pM\times T_pM$ is indeed a positive definite scalar product on $T_pM$. With respect to this scalar product, $X\in T_pM$ is a unit vector if and only if $df(X)\in \mathbb {R}^n$ is a unit vector.

Definition 6.14

$\langle \,,\rangle _f$ as defined above is called the metric on M induced by f.

Remark 6.15

In older texts the induced metric is often called the first fundamental form. We will not use this terminology.

In general, objects like $\langle \,,\rangle _f$ are interesting even when they are not induced by a map $f\colon M\to \mathbb {R}^n$. That is, $\langle \,,\rangle _f$ has the properties of a scalar product between two vectors, which allows us to measure lengths an angles. However, one can freely choose other ways to define such a metric, without explicit reference to a surface f, as long as the scalar product properties are satisfied.

Definition 6.16

Let M be a compact domain with smooth boundary in $\mathbb {R}^2$. Then:

• (i) A map

  $$\displaystyle \begin{aligned} \langle\,,\rangle\colon \bigcup_{p\in M} (T_pM\times T_pM) \to \mathbb{R}\end{aligned}$$

  is called a Riemannian metric on M if for each $p\in M$ the restriction of $\langle \,,\rangle$ to $T_pM\times T_pM$ is a positive definite scalar product and for any two smooth vector fields $X,Y\in \Gamma (TM)$ the function

  $$\displaystyle \begin{aligned} \langle X,Y\rangle\colon M\to \mathbb{R}\end{aligned}$$

  is smooth.

• (ii) M together with a Riemannian metric $\langle \,,\rangle$ on M is called a Riemannian domain.

Remark 6.17

For brevity of the notation we will usually omit the index $(\cdot )_f$ even for Riemannian metrics which are induced by some $f\colon M\to \mathbb {R}^3$. The inserted vectors should provide enough context to avoid confusion.

A Riemannian metric $\langle \,,\rangle$ gives rise to a function

$$\displaystyle \begin{aligned} |\cdot|\colon TM &\to \mathbb{R},\ |X|=\sqrt{\langle X,X\rangle}.\end{aligned} $$

The restriction of $|\cdot |$ to each tangent space is indeed a norm on $T_pM$. One should note that the coordinate vector fields U and V  are not necessarily orthonormal with respect to this induced metric. In fact, this is only true for special surfaces. We will elaborate more on this in Sect. 6.5.

Example 6.18

The norm corresponding to the metric $\langle \,,\rangle _{\iota }$ induced by the inclusion map

$$\displaystyle \begin{aligned} \iota\colon M &\to \mathbb{R}^2,\ (u,v) \mapsto \begin{pmatrix}u \\ v\end{pmatrix}\end{aligned} $$

satisfies

$$\displaystyle \begin{aligned} |\cdot|{}_{\iota}^2=du^2+dv^2.\end{aligned}$$

The above equation should be read as a literal equality of two functions on TM.

4 Linear Algebra on Riemannian Domains

Even in the absence of a Riemannian metric, each single tangent space $T_pM$ is a playing field for Linear Algebra.

Definition 6.19

A smooth map

$$\displaystyle \begin{aligned} A\colon TM\to TM\end{aligned}$$

is called an endomorphism field if its restriction to each tangent space $T_pM$ is a linear map

$$\displaystyle \begin{aligned} A_p\colon T_pM \to T_pM.\end{aligned}$$

Remember that $TM=M\times \mathbb {R}^2$, so it is clear what we mean by a smooth map from TM to TM. In particular, it is clear what we mean by a smooth endomorphism field. For every smooth endomorphism field A there are smooth functions $a,b,c,d\colon M\to \mathbb {R}$ such that

$$\displaystyle \begin{aligned} AU &= a\cdot U+c\cdot V \\ AV&=b\cdot U+d\cdot V.\end{aligned} $$

This means that a smooth endomorphism field basically is the same thing as a smooth map

$$\displaystyle \begin{aligned} p\mapsto \begin{pmatrix}a&b\\c&d\end{pmatrix}\in \mathbb{R}^{2\times 2}.\end{aligned}$$

We always have the identity map as a canonical endomorphism field:

$$\displaystyle \begin{aligned} I\colon TM &\to TM,\ IX =X \quad \mbox{for all } X\in TM.\end{aligned} $$

If A is an arbitrary smooth endomorphism field on M, for each $p\in M$ we can take the determinant or trace of the restriction of A to $T_pM$ and obtain smooth functions

$$\displaystyle \begin{aligned} \det A, \mathrm{tr} \,A\colon M\to \mathbb{R}.\end{aligned}$$

In the presence of a Riemannian metric we can define the adjoint of an endomorphism field:

Theorem 6.20

Let$\langle \,,\rangle$be a Riemannian metric on M and A a smooth endomorphism field on M. Then there is a unique smooth endomorphism field$A^*$on M such that for all vector fields$X,Y\in \Gamma (TM)$we have

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

Proof

By definition of a Riemannian metric, the map

$$\displaystyle \begin{aligned} G\colon M&\to \mathbb{R}^{2\times 2},\ G=\begin{pmatrix}\langle U,U\rangle & \langle U,V\rangle \\ \langle V,U\rangle &\langle V,V\rangle\end{pmatrix}\end{aligned} $$

is smooth and the matrix $G(p)$ is invertible for all $p\in M$. Now one can check that, given an endormorphism field A such that

$$\displaystyle \begin{aligned} AU&=aU+cV \\ AV&=bU+dV\end{aligned} $$

the endomorphism field $A^*$ defined by

$$\displaystyle \begin{aligned} A^*U = \tilde{a}\cdot U +\tilde{c}\cdot V \\ A^*V = \tilde{b}\cdot U +\tilde{d}\cdot V \end{aligned} $$

with

$$\displaystyle \begin{aligned} \begin{pmatrix}\tilde{a}&\tilde{c}\\\tilde{b}&\tilde{d}\end{pmatrix}= G^{-1}\begin{pmatrix}a&b\\c&d\end{pmatrix}^T G\end{aligned}$$

is smooth and satisfies the desired identity. The uniqueness part of the claim is straightforward. □

The endomorphism field I defined above is self-adjoint, which means $I^*=I$. The only structure on M we want to inherit from $\mathbb {R}^2$ is the notion of orientation:

Definition 6.21

Two vectors

$$\displaystyle \begin{aligned} X=(p,\hat{X}) \,\,,\,\,Y=(p,\hat{Y})\in T_pM\end{aligned}$$

are said to form a positively oriented basis of $T_pM$ if $\hat {X},\hat {Y}\in \mathbb {R}^2$ are a positively oriented basis of $\mathbb {R}^2$, i.e. $\det _{\mathbb {R}^2}(\hat X, \hat Y)>0$.

Each tangent space $T_pM$ of a Riemannian domain comes with its own determinant form:

Theorem 6.22

Let$\langle \,,\rangle$be a Riemannian metric on M. Then there is a unique map

$$\displaystyle \begin{aligned} \det\colon \bigcup_{p\in M} (T_pM\times T_pM) \to \mathbb{R}\end{aligned}$$

such that for every$p\in M$the restriction

$$\displaystyle \begin{aligned} \det|{}_{T_pM\times T_pM}\end{aligned}$$

is a skew-symmetric bilinear form such that

$$\displaystyle \begin{aligned} \det(X,Y)=1\end{aligned}$$

for every positively oriented orthonormal basis of$T_pM$. The map$\det$is called thearea formof the Riemannian domain$(M,\langle \,,\rangle )$.

Proof

The vector fields

$$\displaystyle \begin{aligned} X&:=\frac{U}{\sqrt{\langle U,U\rangle}} \\ Y&:= \frac{\langle U,U\rangle V-\langle V,U\rangle U}{\sqrt{\langle U,U\rangle}\sqrt{\langle U,U\rangle \langle V,V \rangle -\langle U,V\rangle^2}}\end{aligned} $$

are orthonormal at each point $p\in M$. Therefore, the function $\det$ we are looking for has to satisfy

$$\displaystyle \begin{aligned} 1=\det(X,Y)=\frac{\det(U,V)}{\sqrt{\langle U,U,\rangle \langle V,V \rangle -\langle U,V\rangle^2}}.\end{aligned} $$

The skew-symmetric bilinear forms on $T_pM$ form a 1-dimensional vector space, so there is a unique such form $\det$ for which

$$\displaystyle \begin{aligned} \det(U,V)=\sqrt{\langle U,U\rangle \langle V,V \rangle -\langle U,V\rangle^2}\, ,\end{aligned}$$

where the equality has to be read point-wise. This form already satisfies $\det (X,Y)=1$. On the other hand, every positively oriented orthonormal basis of $T_pM$ is of the form

$$\displaystyle \begin{aligned} \tilde{X}&=\cos \alpha \,X(p)-\sin\alpha \,Y(p) \\ \tilde{Y}&=\sin \alpha \,X(p) +\cos\alpha\, Y(p)\end{aligned} $$

for some $\alpha \in \mathbb {R}$. Therefore, we also have $\det (\tilde {X},\tilde {Y})=1$. □

Theorem 6.23

Let$\langle \,,\rangle$be a Riemannian metric on M and$\det$the area form defined in Theorem6.22. Then there is a unique endomorphism field J on M such that for all$p\in M$and all$X,Y\in T_pM$we have

$$\displaystyle \begin{aligned} \langle JX,Y\rangle =\det(X,Y).\end{aligned}$$

In terms of the coordinate vector fields, J is given by

$$\displaystyle \begin{aligned} JZ=\frac{\langle U,Z\rangle V -\langle V,Z\rangle U}{\sqrt{\langle U,U\rangle \langle V,V \rangle -\langle U,V\rangle^2}}.\end{aligned}$$

Hence J is smooth. For every positively oriented orthonormal basis of $T_pM$ we have

$$\displaystyle \begin{aligned} JX&=Y \\ JY&=-X.\end{aligned} $$

So, J operates in each tangent space $T_pM$ as the $90^\circ$-rotation in the positive sense.

The proof is straightforward and left to the reader. The theorem will be useful in several instances:

Theorem 6.24

For vectors$X,Y,Z\in T_pM$the following identity holds:

$$\displaystyle \begin{aligned} \langle X,Z\rangle Y -\langle Y,Z\rangle X = \det(X,Y)JZ.\end{aligned}$$

Proof

Both sides of the claimed identity are linear in X and Y  and by Theorem 6.23 the formula that we want to prove is true whenever $X,Y\in \{U(p),V(p)\}$. Since $U(p)$ and $V(p)$ form a basis of $T_pM$, the claimed identity then is true for all $X,Y\in T_pM$. □

5 Isometric surfaces

Definition 6.25

Two surfaces $f,\tilde {f}\colon M\to \mathbb {R}^n$ are called isometric if they induce the same Riemannian metric $\langle \,,\rangle$ on M.

Note that f and $\tilde {f}$ are isometric if and only if

$$\displaystyle \begin{aligned} \langle f_u, f_u \rangle &= \langle \tilde{f}_u, \tilde{f}_u\rangle \\ \langle f_u, f_v \rangle &= \langle \tilde{f}_u, \tilde{f}_v\rangle \\ \langle f_v, f_v \rangle &= \langle \tilde{f}_v, \tilde{f}_v\rangle. \end{aligned} $$

The physical intuition concerning isometries is as follows: The deformation of the surface f to the surface $\tilde {f}$ involves only bending, without any intrinsic deformation such as stretching within the surface. In the nineteenth century, geometers liked to demonstrate this using leather patches. By methods known for example to shoemakers, a leather patch can be brought into any initial shape. After the initial preparation, the leather can still be bent easily, but it will not allow stretching. In Fig. 6.5, the initial shape f of the patch is a planar ring. This ring can easily be fitted to a cone, assuming a shape $\tilde {f}$.

Fig. 6.5

The maps f and $\tilde f$ are isometric, as the flat leather patch f can be placed onto a cone without tearing or stretching. It can even slide freely on the cone

It is clear that one can slide the ring freely around on the cone, in all directions. Few surfaces have the property that one can take a piece of the surface and slide it without distortion or stretching around on the surface. For example, the leather patch on the surface in Fig. 6.6 is clearly stuck in place. One famous surface on which such a patch can freely slide, already known in the nineteenth century and a popular tool for the leather demonstration, is the pseudosphere, that can be parametrized as follows:

Fig. 6.6

The leather patch on this dodecahedron is stuck in place

Suppose that $M\subset \{(u,v)\in \mathbb {R}^2\,\,|\,\,v>1\}$ and define

$$\displaystyle \begin{aligned} f\colon M\to \mathbb{R}^3,\ f(u,v)=\left(\begin{array}{c}\frac{\cos{}(u)}{v}\\ \frac{\sin{}(u)}{v} \\ \log\left(\sqrt{ v^2-1}+ v\right)-\frac{\sqrt{v^2-1}}{v}\end{array}\right).\end{aligned} $$

Choose $\lambda >1$ and $\mu \in \mathbb {R}$ and define $\tilde {f}\colon M\to \mathbb {R}^3$ by $\tilde {f}(u,v)=f(\lambda u+\mu , \lambda v)$.

We leave it to the reader to verify

$$\displaystyle \begin{aligned} \langle f_u,f_v\rangle = \langle \tilde{f}_u,\tilde{f}_v\rangle &=0 \\ \langle f_u,f_u\rangle =\langle f_v,f_v \rangle =\langle \tilde{f}_u,\tilde{f}_u\rangle=\langle \tilde{f}_v,\tilde{f}_v\rangle &= \frac{1}{v^2}\end{aligned} $$

and that therefore f and $\tilde {f}$ are isometric (see Fig. 6.7).

Fig. 6.7

A leather patch fitted to the pseudosphere (at first in one place) is able to slide around freely and even rotate freely

Here is another example, which will also be of interest later when we study minimal surfaces: Suppose that $M\subset \{(u,v)\in \mathbb {R}^2\,\,|\,\,u>0\}$. Let $k,\ell \in \mathbb {Z}$ be two integers with $k+\ell \neq -1$. Then define the Enneper surface$f\colon M\to \mathbb {R}^3$ by

$$\displaystyle \begin{aligned} f(u,v)=\left(\begin{array}{c}u^{2k+1}\frac{\cos{}((2k+1)v)}{2k+1}-u^{2\ell+1}\frac{\cos{}((2\ell+1)v)}{2\ell+1} \\ u^{2k+1}\frac{\sin{}((2k+1)v)}{2k+1}+u^{2\ell+1}\frac{\sin{}((2\ell+1)v)}{2\ell+1} \\ 2u^{k+\ell+1}\frac{\cos{}((k+\ell+1)v)}{k+\ell+1} \end{array}\right).\end{aligned}$$

Again, we leave it to the reader to verify that

$$\displaystyle \begin{aligned} \langle f_u, f_v\rangle &=0 \\ |f_u(u,v)|& = u^{2k}+u^{2\ell} \\|f_v(u,v)|&=u^{2k+1}+u^{2\ell+1}\end{aligned} $$

and that for arbitrary $\lambda \in \mathbb {R}$ the analogous formulas also hold for

$$\displaystyle \begin{aligned} \tilde{f}\colon M &\to \mathbb{R}^3 \\ \tilde{f}(u,v) &= f(u,v+\lambda).\end{aligned} $$

This means that also here f and $\tilde {f}$ are isometric, so a leather patch has at least one degree of freedom to slide on the surface without stretching (see Fig. 6.8).

Fig. 6.8

A leather patch fitted to the Enneper surface is able to slide around, but only in one direction. It cannot rotate