Variations of Surfaces

Abstract

We derive the basics of Vector Calculus on surfaces and explore variations of surfaces. In particular, we compute the variational derivative of the area form $\det$ and of the shape operator A. We show that the critical points of the area functional are the surfaces with mean curvature $H=0$.

We derive the basics of Vector Calculus on surfaces and explore variations of surfaces. In particular, we compute the variational derivative of the area form $\det$ and of the shape operator A. We show that the critical points of the area functional are the surfaces with mean curvature $H=0$. If we constrain the enclosed volume, the critical points of area are the surface with constant mean curvature. These results mirror the situation for plane curves, where the analogous variational problems lead to straight lines ($\kappa =0$) or circles ($\kappa =\text{const}$).

1 Vector Calculus on Surfaces

Throughout this section, $M\subset \mathbb {R}^2$ is a Riemannian domain, $f\colon M\to \mathbb {R}^3$ a surface and $\langle \,,\rangle$ its induced metric. We will only use the area form $\det$, the $90^\circ$-rotation J and the Levi-Civita-connection $\nabla$, which by Theorems 6.22, 6.23 and 9.10 are already determined by the induced metric. This means that this section is dealing only with intrinsic geometry.

If $g\in C^\infty (M)$ is a smooth function, then for each $p\in M$ the restriction $(dg)|{ }_{T_pM}$ is a linear map on $T_pM$ and the restriction $\langle \,,\rangle |{ }_{T_pM\times T_pM}$ is a Euclidean scalar product. Therefore, there is a unique vector $Y(p)\in T_pM$ such that $dg(X)=\langle Y(p),X\rangle$ for all $X\in T_pM$. The smoothness of the vector field Y  defined in this way follows in the usual way, see for example the proof of Theorem 6.20. This leads us to the following:

Definition 12.1

For $g\in C^\infty (M)$ there is a unique vector field

$$\displaystyle \begin{aligned} \mathrm{grad}\,g\in\Gamma(TM)\end{aligned}$$

characterized by the fact that for all vector fields $X\in \Gamma (TM)$ we have

$$\displaystyle \begin{aligned} dg(X)=\langle \mathrm{grad}\,g,X\rangle.\end{aligned}$$

The vector field $\mathrm {grad}\,g\in \gamma (TM)$ is called the gradient of g.

So a function $g\in C^\infty (M)$ gives us a vector field $\mathrm {grad}\,g\in \Gamma (TM)$ (see Fig. 12.1). On the other hand, by taking the trace of the endomorphism field $\nabla Y$, a vector field $Y\in \Gamma (TM)$ gives us a function $\mathrm {div}\,Y\in C^\infty (M)$.

Fig. 12.1

The gradient vector field of the function $z\circ f$ (z being the third coordinate function on $\mathbb {R}^3$) for a surface $f\colon M\to \mathbb {R}^3$. On the left, the value of $z\circ f$ is indicated by color-coding

Definition 12.2

For a vector field $Y\in \Gamma (TM)$ the function

$$\displaystyle \begin{aligned} \mathrm{div}\,Y\colon M\to\mathbb{R},\ \mathrm{div}\,Y =\mathrm{tr}(\nabla Y)\end{aligned}$$

is called the divergence of Y .

The following theorem from Linear Algebra is useful for calculating the trace of an endomorphism field.

Theorem 12.3

Let W be a 2-dimensional vector space with a determinant function$\det$and$A\colon W\to W$a linear map. Then for any two vectors$X,Y \in W$we have

$$\displaystyle \begin{aligned} \det(AX,Y)-\det(AY,X)=\mathrm{tr}\,A \,\det(X,Y).\end{aligned}$$

Proof

If X and Y  are linearly dependent, both sides of the equation vanish. Otherwise, X and Y  form a basis of W and we can write

$$\displaystyle \begin{aligned} AX&=aX+cY\\ AY&=bX+dY.\end{aligned} $$

Our claim now follows from

$$\displaystyle \begin{aligned} \mathrm{tr}\,A=a+d.\end{aligned}$$

□

For the divergence of the product of a function and a vector field we have a Leibniz formula:

Theorem 12.4

For$g\in C^\infty (M)$and$Z\in \Gamma (TM)$we have

$$\displaystyle \begin{aligned} \mathrm{div}(gZ)=\langle \mathrm{grad}\,g,Z\rangle +g\,\mathrm{div}(Z).\end{aligned}$$

Proof

With the notation $G:=\mathrm {grad}\,g$ and with the help of Theorems 6.24 and 12.3, for $X,Y\in \Gamma (TM)$ we have

$$\displaystyle \begin{aligned} \mathrm{div}(gZ)\det(X,Y)&= \det(\nabla_X(gZ),Y)-\det(\nabla_Y(gZ),X) \\ &=\det(\langle X,G\rangle Z +g\nabla_X Z,Y ) -\det(\langle Y,G\rangle Z +g\nabla_YZ,X ) \\ &= -\det(\langle Y,G\rangle X-\langle X,G\rangle Y,Z)+g\,\mathrm{div}(Z)\det(X,Y) \\ &= (\det(-JG,Z)+g\,\mathrm{div}(Z))\det(X,Y)\\ &= (\langle G,Z\rangle +g\,\mathrm{div}(Z))\det(X,Y). \end{aligned} $$

□

Definition 12.5

The divergence of the gradient of a function $g\in C^\infty (M)$

$$\displaystyle \begin{aligned} \Delta g:= \mathrm{div}\, \mathrm{grad}\,g\end{aligned}$$

is called the Laplacian of g.

The divergence of a $90^\circ$-rotated gradient vanishes:

Theorem 12.6

For every$g\in C^\infty (M)$we have

$$\displaystyle \begin{aligned} \mathrm{div}(J\mathrm{grad}\,g)=0.\end{aligned}$$

Proof

Using again the notation $G:=\mathrm {grad}\,g$, by Theorems 9.6 and 12.3 we obtain

$$\displaystyle \begin{aligned} \mathrm{div}(J\mathrm{grad}\,g)\det(U,V) &= \det(\nabla_U(JG),V)-\det(\nabla_V(JG),U) \\&= \langle \nabla_V G,U\rangle -\langle\nabla_UG,V\rangle \\ &= d_V\langle G,U\rangle -d_U\langle G,V\rangle \\ &= d_Vd_U g -d_Ud_Vg\\&=0.\end{aligned} $$

□

The theorem below is a reformulation of Stokes Theorem in terms of vector fields instead of 1-forms. The integral

$$\displaystyle \begin{aligned} \int_{\partial M} g \,ds\end{aligned}$$

of a function $g\colon \partial M\to \mathbb {R}$ is defined in the same way as for total geodesic curvature—as the sum of integrals over the boundary loops.

Theorem 12.7 (Divergence Theorem)

Let$Y\in \Gamma (TM)$be a vector field and B the outward-pointing unit normal field on the boundary$\partial M$. Then

$$\displaystyle \begin{aligned} \int_M \mathrm{div}\,Y\,\det = \int_{\partial M} \langle Y,B\rangle \,ds.\end{aligned}$$

Proof

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

$$\displaystyle \begin{aligned} \omega(X)=\langle JY(p),X\rangle.\end{aligned}$$

Then, by Theorems 9.2, 9.4, 9.6 and Lemma 12.3,

$$\displaystyle \begin{aligned} d\omega(U,V)&=d_U\omega(V)-d_V\omega(U) \\ &= \langle J\nabla_UY,V\rangle +\langle JY,\nabla_UV\rangle -\langle J\nabla_VY,U\rangle -\langle JY,\nabla_VU\rangle \\&= \det(\nabla_UY,V)-\det(\nabla_VY,U) \\ &= \mathrm{tr}(\nabla Y)\det(U,V).\end{aligned} $$

Therefore $d\omega =\mathrm {div}\,Y \,\det .$ Using again the notation of the proof of Theorem 10.6 and applying Stokes Theorem 7.15 we obtain

$$\displaystyle \begin{aligned} \int_M \mathrm{div}\,Y\,\det &= \int_M d\omega \\&= \int_{\partial M}\omega \\&= \int_{\partial M} \langle JY,T\rangle ds \\&= \int_{\partial M} \langle Y,B\rangle ds.\end{aligned} $$

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2 One-Parameter Families of Surfaces

Throughout this chapter $M\subset \mathbb {R}^2$ will be a compact domain with smooth boundary and $[t_0,t_1]\subset \mathbb {R}$ a closed interval.

Definition 12.8

Let $g_t\colon M\to \mathbb {R}^n$ a smooth map, defined for each $t\in [t_0,t_1]$. Then the one-parameter family of maps $[t_0,t_1]\ni t\mapsto g_t$ is called smooth if the map

$$\displaystyle \begin{aligned} M\times [t_0,t_1]&\to \mathbb{R}^n,\ (p,t)\mapsto g_t(p)\end{aligned} $$

is smooth (as always, in the sense of Remark 1.2).

Remark 12.9

The variable t is also referred to as the time.

Given a smooth one-parameter family

$$\displaystyle \begin{aligned} t\mapsto (g_t\colon M\to \mathbb{R}^n),\quad t\in [t_0,t_1]\end{aligned}$$

of maps and a vector field $X\in \Gamma (TM)$, also

$$\displaystyle \begin{aligned} t\mapsto d_Xg_t\end{aligned}$$

is a smooth one-parameter family of maps $d_Xg_t\colon M\to \mathbb {R}^n$. The same holds for ) where ) is defined as

The following fact will be used many times in upcoming chapters:

Theorem 12.10

For a smooth one-parameter family of maps$t\mapsto g_t$from M to$\mathbb {R}^n$, the directional derivative in the direction of a vector field$X\in \Gamma (TM)$commutes with the time derivative:

Proof

In the special case where X is one of the coordinate vector fields U and V , this is just the fact that partial derivatives of the smooth map $(p,t)\mapsto g_t(p)$ commute. In the general case, we can write

$$\displaystyle \begin{aligned} X=a\,U+b\,V\end{aligned}$$

where $a,b\in C^\infty (M)$ are independent of t. Then

□

Definition 12.11

A smooth one-parameter family $t\mapsto g_t$ of maps from M to $\mathbb {R}^n$ is called a variation of a smooth map$g\colon M\to \mathbb {R}^n$ if

$$\displaystyle \begin{aligned} t_0 < 0 < t_1\end{aligned}$$

and

$$\displaystyle \begin{aligned} g_0=g.\end{aligned}$$

In this context, we will also use the notation

One should compare the arguments below with our reasoning in Sect. 2.4.

Definition 12.12

A variation of a surface$f\colon M\to \mathbb {R}^n$ is a smooth one-parameter family of surfaces

$$\displaystyle \begin{aligned} f_t\colon M\to \mathbb{R}^n, \quad t\in [-\epsilon,\epsilon]\end{aligned}$$

such that

$$\displaystyle \begin{aligned} f_0=f.\end{aligned}$$

The map ) defined as

is called the variational vector field of the variation $t\mapsto f_t$.

Definition 12.13

Let $M\subset \mathbb {R}^2$ be a compact domain with smooth boundary. Suppose we have a way to assign to each surface $f\colon M\to \mathbb {R}^n$ a real number $\mathcal {E}(f)$. Then $\mathcal {E}$ is called a smoothfunctional if for every smooth one-parameter family

$$\displaystyle \begin{aligned} t\mapsto f_t, \quad t\in [t_0,t_1]\end{aligned}$$

of surfaces $f\colon M\to \mathbb {R}^n$ the function

$$\displaystyle \begin{aligned} [t_0,t_1]\to \mathbb{R},\ t\mapsto \mathcal{E}(f_t)\end{aligned} $$

is smooth.

In many circumstances, we want to consider only variations of $f\colon M\to \mathbb {R}^n$ that keep the surface fixed near the boundary $\partial M$:

Definition 12.14

Let $M\subset \mathbb {R}^2$ be a compact domain with smooth boundary and $f\colon M\to \mathbb {R}^n$ a surface. Then a variation

$$\displaystyle \begin{aligned} t\mapsto f_t,\quad t\in [-\epsilon,\epsilon]\end{aligned}$$

of f is said to have support in the interior of M if there is a compact set $M_0\subset \mathring {M}$ such that for all $p\in M, p\notin M_0$ we have

$$\displaystyle \begin{aligned} f_t(p)=f(p)\quad \text{for all}\quad t\in [-\epsilon,\epsilon].\end{aligned}$$

Definition 12.15

Let $M\subset \mathbb {R}^2$ be a compact domain with smooth boundary and $\mathcal {E}$ be a smooth functional defined on the space of surfaces $f\colon M\to \mathbb {R}^n$. Then a surface $f\colon M\to \mathbb {R}^n$ is called a critical point of $\mathcal {E}$ if for all variations $t\mapsto f_t$ of f with support in the interior of M we have

$$\displaystyle \begin{aligned} \left.\frac{d}{dt}\right|{}_{t=0} \mathcal{E}(f_t)=0.\end{aligned}$$

Definition 12.15 spells out the notion of an equilibrium of a variational energy $\mathcal {E}$, to which we will refer to in later sections. Moreover, one should note that, as already explained in the beginning of Sect. 2.4, we will work with a definition of a critical point under constraints that is slightly stronger than the standard one.

Definition 12.16

Let $M\subset \mathbb {R}^2$ be a compact domain with smooth boundary, $f\colon M\to \mathbb {R}^n$ a surface and $\mathcal {E},\tilde {\mathcal {E}}$ two smooth functionals on the space of all surfaces $\tilde {f}\colon M\to \mathbb {R}^n$. Then f is called a critical point of $\mathcal {E}$under the constraint of fixed $\tilde {\mathcal {E}}$ if for all variations $t\mapsto f_t$ of f with support in the interior of M

$$\displaystyle \begin{aligned} \left.\frac{d}{dt}\right|{}_{t=0} \tilde{\mathcal{E}}=0\end{aligned}$$

implies

$$\displaystyle \begin{aligned} \left.\frac{d}{dt}\right|{}_{t=0} \mathcal{E}=0.\end{aligned}$$

Using the Linear Algebra Therorem 2.21 in the same way as we used it in Sect. 2.4, we obtain

Theorem 12.17

Let$M\subset \mathbb {R}^2$be a compact domain with smooth boundary and$\mathcal {E},\tilde {\mathcal {E}}$two smooth functionals on the space of all surfaces$f\colon M\to \mathbb {R}^n$. Suppose we have a way to associate to each surface$f\colon M\to \mathbb {R}^n$smooth maps

$$\displaystyle \begin{aligned} G_f,\widetilde{G}_f\colon M\to \mathbb{R}^n\end{aligned}$$

such that for all variations$t\mapsto f_t$of f with support in the interior of M we have

Then f is a critical point of$\mathcal {E}$under the constraint of fixed$\tilde {\mathcal {E}}$if and only if there is a constant$\lambda \in \mathbb {R}$such that

$$\displaystyle \begin{aligned} G_f=\lambda \widetilde{G}_f.\end{aligned}$$

For reasons already explained in Sect. 2.4, we call $\lambda$ a Lagrange multiplier for the constraint of fixed $\tilde {\mathcal {E}}$.

3 Variation of Curvature

Given a smooth variation $t\mapsto f_t$ of a surface f, we are mainly interested in the time derivative at time zero of quantities like the area form $\det _t$ or the shape operator $A_t$ associated with the surfaces $f_t$. In situations where it clear with which variation $t\mapsto f_t$ we are dealing, we will usually drop the index zero when we mean the time derivative at time zero. So, for example, we will write

Theorem 12.18

Let$f\colon M\to \mathbb {R}^3$be a surface with unit normal N, shape operator A and Levi-Civita connection$\nabla$. Let$t\mapsto f_t$be a variation of f whose variational vector field

is described in terms of a function$\phi \in C^\infty (M)$and a vector field$Z\in \Gamma (TM)$. Denote by$N_t$and$A_t$the unit normals and the shape operators of the surfaces$f_t$. Define vector fields$G,W\in \Gamma (TM)$as

$$\displaystyle \begin{aligned} G&:=\mathrm{grad}\,\phi\\ W &:=AZ-G.\end{aligned} $$

Then

Proof

The proof of the first equation is straightforward:

Differentiating $\langle N,df(X)\rangle =0$ with respect to time we obtain

This holds for all $X\in TM$ and this implies the second equation. For $X,Y\in T_pM$ we know that ) (which is orthogonal to N), $df(X)$ and $df(Y)$ are linearly dependent. Using this and Theorem 12.3 we obtain

This proves the third equation. For the fourth equation, consider the directional derivative of the second equation in the direction of X and make use of the first:

The normal part of this equation is satisfied automatically. The tangential part, together with the Codazzi equation (Theorem 9.7) gives us

This proves the fourth equation. For the fifth we take the trace of the fourth and multiply by $\frac {1}{2}$. The last two terms in the fourth equation do not contribute because we see here the commutator of two endomorphisms A and $\nabla Z$, which always has zero trace. Regarding the first term, one can verify (for example by taking the directional derivative of the equation in Theorem 12.3 in the direction of Z) that indeed for any endomorphism field $\tilde A$

$$\displaystyle \begin{aligned} \mathrm{tr}(\nabla_Z\tilde A)=d_Z(\mathrm{tr}\,\tilde A).\end{aligned}$$

Finally, by diagonalizing A one can easily check the equality

$$\displaystyle \begin{aligned} \frac{1}{2}\mathrm{tr}\,A^2 = 2H^2-K.\end{aligned}$$

□

4 Variation of Area

Variations of surfaces (as defined in Definition 12.12) are needed in order to define and determine those surfaces that represent equilibria of geometrically interesting variational functionals.

Examples of smooth functionals of surfaces are the Willmore functional (to be introduced in Sect. 13.1) and the cone volume that will be defined in Sect. 12.5. In this chapter we will focus on the area functional

$$\displaystyle \begin{aligned} \mathcal{A}(f)= \int_M {\det}_f.\end{aligned}$$

Theorem 12.19 (First Variation Formula of Area)

As in Theorem12.18, suppose the variational vector field of a variation$t\mapsto f_t$of a surface$f\colon M\to \mathbb {R}^3$is written as

with$\phi \in C^\infty (M)$and$Z\in \Gamma (TM)$. Then

$$\displaystyle \begin{aligned} \left.\frac{d}{dt}\right|{}_{t=0} \mathcal{A}(f_t)=2\int_M \phi H \det+\int_{\partial M} \langle Z,B\rangle \,ds\end{aligned}$$

where B is the outward pointing unit normal on$\partial M$.

Proof

By Theorem 12.18 and the Divergence Theorem 12.7,

□

Definition 12.20

A surface $f\colon M\to \mathbb {R}^3$ is called a minimalsurface if it is a critical point of the area functional $\mathcal {A}$.

Figure 12.2 shows a minimal surface whose six boundary curves are all mapped onto prescribed circles. In fact, it is here a solution of the so-called Plateau problem, which means that it minimizes area among all surfaces whose boundary is mapped onto a prescribed set of curves.

Fig. 12.2

The Schwarz-P minimal surface

Remark 12.21

The Plateau problem was first solved by Jesse Douglas [12] and Tibor Rado [32] independently.

Theorem 12.22

A surface$f\colon M\to \mathbb {R}^3$is a minimal surface if and only if is mean curvature H vanishes.

Proof

If $H=0$ and $t\mapsto f_t$ is a variation of f with support in the interior of M, then Z vanishes near the boundary of M and by Theorem 12.19 the variation of area is zero. Conversely, suppose that f is a minimal surface but there is a point $p\in M$ for which $H(p)\neq 0$. Then there is such a p also in the interior of M, so we assume $p\in \mathring {M}$. Let us treat the case $H(p)>0$, the case $H(p)<0$ being similar. Then we can construct a bump function $g\in C^\infty (M)$ such that g vanishes outside of a compact set contained in the interior of M and

$$\displaystyle \begin{aligned} g(p)&=1 \\ H(q)\leq 0 &\implies g(q)=0.\end{aligned} $$

Then, for small enough $\epsilon >0$,

$$\displaystyle \begin{aligned} t &\mapsto f_t,\quad t\in[-\epsilon,\epsilon] \\ f_t&=f+t\cdot g\cdot N\end{aligned} $$

(N being the unit normal of f) will be a smooth variation of f with support in the interior of M and

$$\displaystyle \begin{aligned} \left.\frac{d}{dt}\right|{}_{t=0} \mathcal{A}(f_t)=\int_M gH>0,\end{aligned}$$

which contradicts our assumption that f is minimal. □

As the reader may verify, the Enneper surfaces defined in Sect. 6.5 have mean curvature $H=0$, so by Theorem 12.22 they are minimal surfaces. Figure 12.3 shows one of these Enneper surfaces.

Fig. 12.3

An Enneper surface is a minimal surface

5 Variation of Volume

Definition 12.23

Let $M\subset \mathbb {R}^2$ be a compact domain with smooth boundary and $f\colon M\to \mathbb {R}^3$ a surface. Then the cone volume of f is defined as

$$\displaystyle \begin{aligned} \mathcal{V}(f)=\frac{1}{3}\int_M \det(f,f_u,f_v).\end{aligned}$$

$\mathcal {V}(f)$ can be interpreted as the volume covered by the map

$$\displaystyle \begin{aligned} F\colon[0,1]\times M&\to \mathbb{R}^3,\ F(s,p)=s\cdot f(p).\end{aligned} $$

Here the “volume covered” should not be understood as the volume of the image $F([0,1]\times M)$, but rather in the spirit of Theorem 8.17. At first sight, the cone volume does not look like an honorable geometric functional. For example, the version $\tilde {f}=f+\mathbf {a}$ of f that has been translated by a vector $\mathbf {a}\in \mathbb {R}^3$ in general does not have the same cone volume as f. On the other hand, for closed surfaces the cone volume is invariant under translations:

Theorem 12.24

If$(f,\rho )$is an oriented closed surface (Definition11.1) and$\mathbf {a}\in \mathbb {R}^3$, then

$$\displaystyle \begin{aligned} \mathcal{V}(f+\mathbf{a})=\mathcal{V}(f).\end{aligned}$$

Proof

Define a 1-form $\omega \in \Omega ^1(M)$ by

$$\displaystyle \begin{aligned} \omega(X)=\frac{1}{6}\det(\mathbf{a},f,df(X)).\end{aligned}$$

Then

$$\displaystyle \begin{aligned} d\omega(U,V)=\frac{1}{6}(\det(\mathbf{a},f,f_v)_u-\det(\mathbf{a},f,f_u)_v =\frac{1}{3}\det(\mathbf{a},f_u,f_v)\end{aligned} $$

and therefore, by Stokes Theorem 7.15,

$$\displaystyle \begin{aligned} \mathcal{V}(f+\mathbf{a})-\mathcal{V}(f)= \int_M d\omega = \int_{\partial M} \omega =0.\end{aligned} $$

The last equality follows from the fact that $(f,\rho )$ is oriented, and therefore the integrals of $\omega$ over the various boundary curves of M cancel in pairs. □

Moreover, by almost the same reasoning as in the above proof one can show:

Theorem 12.25

Let$M\subset \mathbb {R}^2$be a compact domain with smooth boundary,

$$\displaystyle \begin{aligned} t\mapsto f_t,\quad t\in [-\epsilon,\epsilon]\end{aligned}$$

a variation with support in the interior of M of a surface$f\colon M\to \mathbb {R}^3$and$\mathbf {a}\in \mathbb {R}^3$. Then

$$\displaystyle \begin{aligned} \left.\frac{d}{dt}\right|{}_{t=0} \mathcal{V}(f_t+\mathbf{a})=\left.\frac{d}{dt}\right|{}_{t=0} \mathcal{V}(f_t).\end{aligned}$$

Theorem 12.25 implies that for the purposes of variational calculus the cone volume $\mathcal {V}$ behaves in the same way as a translationally invariant functional (see Fig. 12.4).

Fig. 12.4

The cone volume of f(left) and of a variation $f_t$ of f(right)

We can view df as an $\mathbb {R}^3$-valued 1-form on M. Given smooth maps ) we then obtain a scalar valued 1-form

Theorem 12.26 (First Variation of Cone Volume)

Let$f\colon M\to \mathbb {R}^3$be a surface. Then for every variation$t\mapsto f_t$of f we have

Proof

We have

□

It is easy to see that, on its own, the cone volume functional does not have any critical points. However, we can use it in the context of variational problems under a volume constraint. Here is our first application of Theorem 12.17:

Theorem 12.27

Let$M\subset \mathbb {R}^2$be a compact domain with smooth boundary. Then a surface$f\colon M\to \mathbb {R}^3$is a critical point of the area$\mathcal {A}$under the constraint of fixed cone volume$\mathcal {V}$if and only if the mean curvature H of f is constant.

Proof

By Theorems 12.19, 12.26, and 12.17, f is a critical point of area under fixed cone volume if and only if there is a constant $\lambda \in \mathbb {R}$ such that

$$\displaystyle \begin{aligned} HN=\lambda N.\end{aligned}$$

□

The surface in Fig. 12.5 minimizes area among all surfaces that are bounded by the same six circles as the first surface shown in Sect. 12.4 and have a certain prescribed volume:

Fig. 12.5

A surface with the same boundary as the surface in Fig. 12.2. It is a critical point of the area functional under the constraint of having a prescribed cone volume

Remark 12.28

In 1984 Henry Wente found a counterexample to a conjecture by Heinz Hopf which stated that every closed surface in $\mathbb {R}^3$ with constant mean curvature is round sphere [43]. In Fig. 12.6 it is shown how the Wente torus can be build from a fundamental piece.

Fig. 12.6

A Wente torus—a closed surface of genus $g=1$ with constant mean curvature $H\neq 0$

Nevertheless, the conjecture is true if one demands that the surface is embedded in $\mathbb {R}^3$, or has genus $g=0$. These results are due to Alexandrov [1] and Hopf [17].