Limit Surfaces of Riemann Examples

David Hoffman, Wayne Rossman

1 Introduction

The only connected minimal surfaces foliated by circles and lines are domains on one of the following surfaces: the helicoid, the catenoid, the plane, and the examples of Riemann ([Ri] p329-33, [En] p403-6, [Ni] p85-6). All these surfaces are complete and embedded. Topologically they are planar domains: the helicoid is simply-connected, the catenoid is an annulus (conformally a twice-punctured sphere), and each Riemann example (see Figure 1) is conformal to the plane minus the points \{(n, 0), (\frac{1}{n}, 0) | n ∈ \} [HKR]. In this section we will show that the plane, helicoid, and catenoid arise naturally as limits of well-chosen and properly normalized sequences of Riemann examples. The local behavior of domains on Riemann examples accounts for the existence of these limits, thus allowing a change of topology in the limit surfaces.

Theorem 1.1 Any sequence of Riemann examples that converges to a surface in ^{3} converges either to a Riemann example, a helicoid, a catenoid, an infinite set of equally spaced parallel planes, or a single plane. Furthermore, there exist sequences of Riemann examples that converge to each of these possibilities.

In order to be precise, we state what we mean by convergence of surfaces in ^{3} in this context.

Definition 1.1 A sequence of surfaces {S_{i}}_{i=1}^{∞} converges as i → ∞ to a surface S in ^{3} if, for any compact region B ⊆ ^{3}, there exists an integer N_{B} such that for i > N_{B}, S_{i} ∩ B is a normal graph over S, and {S_{i} ∩ B}_{i=N_{B}}^{∞} converges to S in the C^{∞}-topology.

One can define analogously the convergence of a one-parameter family S_{t}, t ∈ , as t → t_{0}. Since this convergence is in the C^{∞}-topology, it requires that derivatives of all orders of the coordinate functions on the graphs converge to the corresponding derivatives of the coordinate functions on the limit-surface.

We observe that catenoids, helicoids, and planes are limit-surfaces of sequences of Riemann examples. Therefore the collection of connected embedded minimal surfaces foliated by circles and lines is itself connected in the topology associated with this convergence.

We wish to thank Rob Kusner, Pascal Romon, Ed Thayer, Johannes Nitsche and Harold Rosenberg for helpful conversations.

Figure 1: The Riemann Example R_{λ}, λ = 1

2 The Family of Riemann Examples

2.1 Definition of the Riemann Examples

We now describe properties of the Riemann examples, and we give a Weierstrass representation for these surfaces. Proofs of the statements in this section may be found in [HKR].

For any Riemann example R, there exists a translation T of ^{3} so that T: leaves R invariant; is orientation-preserving on R; generates the full cyclic orientation-preserving translation symmetry group of R. The quotient-surface R/T is a twice-punctured rectangular torus. Thus, Riemann examples can be parametrized via a Weierstrass representation on such a torus. Fix λ > 0. We first define M̄_{λ}:

M̄_{λ} = \{ (z, w) ∈ ( ∪ \{∞\})^{2} : w^{2} = z(z − λ)(z + \frac{1}{λ}) \} .  (2.1)

Let M_{λ} be the twice-punctured torus

M_{λ} = M̄_{λ} \ {(0, 0), (∞, ∞)} .  (2.2)

With the Weierstrass data

g(z, w) = z , η(z, w) = \frac{dz}{zw} ,  (2.3)

the Weierstrass representation for the Riemann examples R_{λ} is

R_{λ}(p) := Re ∫_{p_{0}}^{p} Φ_{(g,η)} dz , Φ_{(g,η)} := ( \frac{(1−z^{2})dz}{zw} ; \frac{i(1+z^{2})dz}{zw} ; \frac{2dz}{w} ) , p ∈ M_{λ} .  (2.4)

The family R_{λ} of Riemann examples is parametrized by λ ∈ (0, ∞).

The surface M_{λ} can be considered to be a double covering of the punctured z-plane, branched at λ and \frac{−1}{λ}. Consider a circle α in the z-plane with center \frac{−1}{2λ} and radius \frac{1}{2}(λ+ \frac{1}{λ}), and lift α to a closed curve α̂ in M_{λ}. The Weierstrass integral about α̂ results in a nonzero period-vector, and this is the translation T.

Letting ź = −z and ẃ = iw, we have

ẃ^{2} = ź(ź − \frac{1}{λ})(ź + λ) .

Thus we may consider the map (z, w) → (ź, ẃ) to be a conformal diffeomorphism between M_{λ} and M_{\frac{1}{λ}}.

Because the metric on R_{λ} ([Os] p65, [HKR]) is

ds^{2} = \frac{(1 + |z|^{2})^{2}}{4|z|^{2}|w|^{2}} |dz|^{2} ,

the maps (z, w) → (z̄, w̄) and (z, w) → (z̄, −w̄) both restrict to isometries of M_{λ} in the induced metric. The set A = {(t, w) ∈ M_{λ} | t ∈ }, which consists of the union of their fixed-point sets, is mapped by the Weierstrass integral to geodesics in R_{λ}. It is clear from the Weierstrass integral that the intervals in A of the form t ≤ \frac{−1}{λ} or 0 < t ≤ λ represent lines in R_{λ}, while intervals in A of the form \frac{−1}{λ} ≤ t < 0 or t ≥ λ represent planar geodesics in R_{λ}. Planar geodesics must be principal curves.

The surfaces R_{λ} have the following additional properties:

1) R_{λ} is foliated by circles and lines in horizontal planes. All the lines in R_{λ} are the image under the Weierstrass representation of points in A such that t ≤ \frac{−1}{λ} or 0 < t ≤ λ, which is the fixed-point set of the map (z, w(z)) → (z̄, −w̄(z)).

2) R_{λ} has an infinite number of equally spaced horizontal flat ends. These ends correspond to the punctures of M̄_{λ} at z = 0, ∞.

3) R_{λ} is invariant under reflection in a plane parallel to the (x_{1}, x_{3})-plane. The intersection of this plane with R_{λ} consists of the image under the Weierstrass representation of points in A such that \frac{−1}{λ} ≤ z < 0 or z ≥ λ, which is the fixed-point set of the map (z, w(z)) → (z̄, w̄(z)).

4) R_{λ} is invariant under rotation about horizontal lines that are perpendicular to the (x_{1}, x_{3})-plane, and meet the surface orthogonally at the points where g(z) = ±i. The map (z, w) → (\frac{−1}{z}, \frac{w}{z^{2}} ) restricted to M_{λ} represents this rotation.

We will consider a Riemann example to be any surface that is the image of one of these surfaces R_{λ} under an isometry or homothety of ^{3}.

2.2 Conjugate Pairs of Riemann Examples

Proposition 2.1 For any positive λ ≠ 1, the Riemann examples R_{λ} and R_{\frac{1}{λ}} are conjugate but not congruent. R_{1} is self-conjugate.

Proof. Choose λ > 1. The conjugate surface of R_{λ} has Weierstrass data

g = z ,

iη = \frac{idz}{zw} = \frac{dz}{(−z)(iw)} .

The Weierstrass data for the conjugate surface, expressed in terms of (ź, ẃ) = (−z, iw) on M_{\frac{1}{λ}}, is

g̃(ź, ẃ) = −ź ,

η̃(ź, ẃ) = \frac{−dź}{źẃ} .

Because

Φ_{(g̃,η̃)} = ( \begin{matrix}−1 & 0 & 0 \\ 0 & −1 & 0 \\ 0 & 0 & 1 \end{matrix} ) Φ_{(g,η)} ,

it follows from equation (2.4) that the minimal surface in ^{3} with this Weierstrass data is the image of R_{\frac{1}{λ}} under a rotation by π about a vertical axis. Hence R_{λ} and R_{\frac{1}{λ}} are conjugate surfaces.

We use the Weierstrass data to show that R_{λ} and R_{\frac{1}{λ}} cannot be congruent for λ ≠ 1. Consider the points on R_{λ} where g = ±1. As shown in the previous Section 2.1 (Property 1), when λ > 1, these points lie on the lines in R_{λ}. By contrast, the points of R_{\frac{1}{λ}} where g(z) = ±1 lie on the planar geodesics in the vertical plane of reflective symmetry of the surface. Suppose there exists a symmetry of ^{3} taking R_{λ} to R_{\frac{1}{λ}}, then this symmetry must take lines to lines, ends to ends, and planar geodesics to planar geodesics. Since the lines in these two surfaces are all parallel to the x_{2}-axis and the ends are horizontal, the symmetry of ^{3} must be of the form

( \begin{matrix}±1 & 0 & 0 \\ 0 & ±1 & 0 \\ 0 & 0 & ±1 \end{matrix} ) .

Therefore points of R_{λ} where g = ±1 must be mapped to points of R_{\frac{1}{λ}} where g = ±1. This implies that the points of R_{λ} and R_{\frac{1}{λ}} where g = ±1 are the points where the lines and planar geodesics in the surface intersect i.e. where g = ±λ^{±1}. But we are assuming λ ≠ 1. Thus R_{λ} and R_{\frac{1}{λ}} are not congruent. □

Remark. One checks that the mapping (z, w) → (z′, w′) from M_{λ} to M_{\frac{1}{λ}} has the property that on fundamental cycles it preserves the periods (one of which is always zero) of (2.4). Hence this mapping induces an isometry of R_{λ} and R_{\frac{1}{λ}}. □

Upon considering the behavior of R_{λ} and R_{\frac{1}{λ}} as λ → 0, one finds that R_{λ} and R_{\frac{1}{λ}} do not have limits that are regular surfaces. However, if these surfaces are normalized appropriately they converge to a catenoid and its conjugate surface, the helicoid, which is consistent with Proposition 2.1. We shall prove this in the next section.

3 Proof of the Main Result

To find a sequence of Riemann examples that converges to a catenoid or helicoid, we renormalize the surfaces R_{λ} as follows. Begin the integration in the Weierstrass representation at the point z_{0} = 1. If λ > 1, rescale by \sqrt{λ}; if λ < 1, rescale by \frac{1}{\sqrt{λ}}. We name these normalized Riemann examples Ŕ_{λ}. Note that the rescaling is accomplished by multiplying η in the Weierstrass representation by \sqrt{λ} and \frac{1}{\sqrt{λ}}, respectively. Thus, the Weierstrass representation for Ŕ_{λ} is

g = z , η = \frac{\sqrt{λ}dz}{zw} , λ ≥ 1 ,

g = z , η = \frac{dz}{\sqrt{λ}zw} , λ ≤ 1 .

One can check as in Proposition 2.1 that Ŕ_{λ} and Ŕ_{\frac{1}{λ}} are conjugate.

Let K denote Gauss curvature.

Lemma 3.1 There exists a universal bound c such that |K| ≤ c on Ŕ_{λ} for all λ ∈ (0, ∞).

Proof. With Weierstrass data given locally as g and η = fdz, the Gauss curvature of a minimal surface is ([Os], p76)

K = −(\frac{4|g′|}{|f|(1 + |g|^{2})^{2}} )^{2} .  (3.5)

In the case of Ŕ_{λ} a computation using equations (2.1) and (3.5) yields

|K| = 16λ^{α} \frac{|z − λ||z + λ^{−1}|}{|z|(|z| + |z|^{−1})^{4}} .

where α = 1 if λ ≤ 1, and α = −1 if λ ≥ 1. If λ ≤ 1, then

|K| ≤ 16\frac{(|z| + 1)λ(|z| + λ^{−1})}{|z|(|z| + |z|^{−1})^{4}} ≤ 16\frac{(|z| + 1)^{2}}{|z|(|z| + |z|^{−1})^{4}} ≤ 4 .

Similarly, if λ ≥ 1, then

|K| ≤ 16\frac{λ^{−1}(|z| + λ)(|z| + 1)}{|z|(|z| + |z|^{−1})^{4}} ≤ 16\frac{(|z| + 1)^{2}}{|z|(|z| + |z|^{−1})^{4}} ≤ 4 .

Thus, |K| has a universal upper bound. □

Remark. More calculation shows that |K| on R_{λ} is maximized at a fixed point of one of the symmetries of the surface. Thus K is maximized in one of the following three places: somewhere along a straight line in R_{λ}; somewhere along a planar geodesic in the plane of reflective symmetry parallel to {x_{2} = 0}; or at the points where g(z) = ±i, which are the fixed points of the normal rotation. At the points where z = g(z) = ±i, the value of |K| on R_{λ} is λ + \frac{1}{λ}. Hence on Ŕ_{λ} the value is 1 + \frac{1}{λ^{2}} (resp. 1 + λ^{2}) when λ > 1(resp. when λ < 1).

The following conjecture has been verified numerically: The absolute value of the Gaussian curvature on R_{λ} is maximized at the points where z = g(z) = ±i. This implies that the optimal value in Lemma 3.1 is c = 2. □

Proposition 3.1 There exist sequences of Riemann examples that converge to a helicoid and sequences that converge to a catenoid.

Proof. We will prove this proposition by showing that the limit, as λ → 0, of Ŕ_{λ} is a catenoid, and the limit, as λ → ∞, of Ŕ_{λ} is a helicoid (see Figures 2 through 4). We consider first the case λ → 0.

For a large positive number L, let A_{L} be the annular ring \{z ∈ : \frac{1}{L} < |z| < L\}. One can think of z in (2.1) as a map from M_{λ} to \ {0}. Then for λ sufficiently small, z^{−1}(A_{L}) consists of two disjoint sets in M_{λ}. Choose either one and name it Á_{L,λ}. Since z parametrizes Á_{L,λ}, we shall refer to points in Á_{L,λ} by their z coordinates.

Let B_{r} be a ball centered at the origin in ^{3} with radius r. Let S_{r} be the horizontal slab {(x_{1}, x_{2}, x_{3}) ∈ ^{3} | − r ≤ x_{3} ≤ r}, with boundary planes x_{3} = ±r.

Let C be the catenoid obtained by using the Weierstrass data

g(z) = z , η(z) = \frac{dz}{z^{2}} ,

and integrating over \ {0} with base point z_{0} = 1. The Weierstrass representation for the catenoid C is

C(p) := Re ∫_{1}^{p} ( \frac{(1−z^{2})dz}{z^{2}} ; \frac{i(1+z^{2})dz}{z^{2}} ; \frac{2dz}{z} ) , p ∈ \ \{0\} .

Let Ĉ_{L} be the image of the restriction of C(p) to A_{L}. By examination of the third coordinate of the Weierstrass representation, we see that Ĉ_{L} = C ∩ S_{2 ln L}. Choose a large value for r, and choose L > e^{r} so that C ∩ S_{2r} ⊆ Ĉ_{L}.

For λ < 1, the Weierstrass representation for Ŕ_{λ} is

Ŕ_{λ}(p) := Re ∫_{1}^{p} ( \frac{(1−z^{2})dz}{\sqrt{λ}zw} ; \frac{i(1+z^{2})dz}{\sqrt{λ}zw} ; \frac{2dz}{\sqrt{λ}w} ) = C(p) + Re ∫_{1}^{p} ( \begin{matrix}f_{0,λ}(z)\frac{(1−z^{2})dz}{z^{2}} \\ f_{0,λ}(z)\frac{i(1+z^{2})dz}{z^{2}} \\ f_{0,λ}(z)\frac{2dz}{z} \end{matrix} ) ,  (3.6)

Figure 2: Portions of Riemann Examples Ŕ_{λ}, λ = 0.1 (left), λ = 0.5 (right) Figure 3: Portions of Riemann Examples Ŕ_{λ}, λ = 1.0 (left), λ = 3.0 (right) Figure 4: Portions of Riemann Examples Ŕ_{λ}, λ = 5.0 (left), λ = 10.0 (right) where

f_{0,λ}(z) = \begin{matrix}√ & \overline{z} \\ √ & \overline{(z − λ)(λz + 1)} \end{matrix} − 1 .

If λ is close to zero, then the Weierstrass integral Ŕ_{λ} along any closed curve in Á_{L,λ} is zero. Therefore the Weierstrass integral depends only on the endpoints of a path in Á_{L,λ}. All paths from z_{0} = 1 to p ∈ A_{L} may be chosen to be of length less than cL, where c is some fixed constant.

Note that the functions

\frac{1 − z^{2}}{z^{2}} , \frac{1 + z^{2}}{z^{2}} , \frac{2}{z}

are bounded on A_{L}. Also, for all z ∈ A_{L},

lim_{λ→0} |f_{0,λ}(z)| = 0 .

It follows that the last integral in equation (3.6) becomes arbitrarily small, uniformly on Á_{L,λ}, as λ → 0. In particular, for all p ∈ A_{L}, lim_{λ→0} Ŕ_{λ}(p) = C(p).

Let R̂_{λ} be the image of Á_{L,λ} under the map Ŕ_{λ}(p). It follows that R̂_{λ} is a small perturbation of Ĉ_{L} for λ close to zero. We conclude that Ŕ_{λ} ∩ B_{r} ⊆ R̂_{λ}. Thus to examine the behavior of Ŕ_{λ}, as λ → 0, in arbitrary compact regions of ^{3}, it is enough to examine R̂_{λ} in B_{r} for arbitrarily large fixed r.

We claim that the surface Ŕ_{λ} ∩ B_{r} is a graph over a subset of the catenoid C, for λ close to 0. By Lemma 3.1, the Gaussian curvature K is uniformly bounded for all Ŕ_{λ}. This implies that the normal curvatures are also uniformly founded. This in turn implies Ŕ_{λ} ∩B_{r} is a union of graphs over a subset of the catenoid C. Since the surface Ŕ_{λ} is foliated by lines and circles in horizontal planes, there is a single graph.

We now show that Ŕ_{λ} ∩ B_{r} converges to C ∩ B_{r}, in the C^{∞}-topology, as λ → 0. Because the curvature of Ŕ_{λ} is bounded uniformly in λ, it follows from standard elliptic theory that the convergence is C^{∞}. (Curvature bounds give C^{1} estimates.) However, we can give in our specific case a direct proof. Let p ∈ C ∩ B_{r}. Let z_{p} ∈ A_{L} be the point such that C(z_{p}) = p, and let B_{h}(z_{p}) ⊆ A_{L} be a ball of radius h about z_{p}. Note that Ŕ_{λ}(B_{h}(z_{p})) consists of disks in ^{3} converging in the C^{0}-norm, as λ → 0, to a disk containing p and lying on the catenoid C. Let x_{i,λ}(z) be the i’th coordinate function of Ŕ_{λ}, let x_{i}(z) be the i’th coordinate function on C, and let f_{λ}(z) = x_{i,λ}(z) − x_{i}(z). Note that

max_{w∈B_{h}(z_{p})} |f_{λ}(w)| → 0 , as λ → 0 .

Since f_{λ}(z) is harmonic, it is the real part of a holomorphic function F_{λ}(z), and F_{λ}(z) can be chosen so that

max_{w∈B_{h}(z_{p})} |F_{λ}(w)| → 0 , as λ → 0 .

Assuming z ∈ B_{\frac{h}{2}} (z_{p}), and using the Cauchy integral formula, we have

|f_{λ}^{(k)}(z)| ≤ |F_{λ}^{(k)}(z)| = |\frac{k!}{2πi} ∫_{∂B_{h}(z_{p})} \frac{F (ξ)}{(ξ − z)^{k+1}} dξ|

≤ \frac{k!}{2π} (\frac{2}{h})^{k+1} 2πh max_{w∈B_{h}(z_{p})} |F_{λ}(w)| ,

and so we have that |f_{λ}^{(k)}(z)| → 0 ∀k, ∀z ∈ B_{\frac{h}{2}} (z_{p}), as λ → 0. Therefore |f_{λ}^{(k)}(z)| → 0 ∀k, ∀z ∈ A_{L}, and the convergence is in the C^{∞}-topology in B_{r}.

Since r is arbitrary, we have completed the proof that Ŕ_{λ} converges to the catenoid C as λ → 0.

It is now intuitively clear that the surfaces Ŕ_{λ} converge to a helicoid for the following reason: as λ → ∞, their conjugate surfaces Ŕ_{\frac{1}{λ}} converge to a catenoid. One can give an explicit proof of this, similar to the proof just given for the case λ → 0. However, there are some differences in the proof of the case λ → ∞. For λ close to ∞, the representation for Ŕ_{λ} is

Ŕ_{λ}(p) := Re ∫_{1}^{p} ( \frac{\sqrt{λ}(1−z^{2})dz}{zw} ; \frac{i\sqrt{λ}(1+z^{2})dz}{zw} ; \frac{2\sqrt{λ}dz}{w} ) =

− Re ∫_{1}^{p} ( \frac{i(1−z^{2})dz}{z^{2}} ; \frac{−(1+z^{2})dz}{z^{2}} ; \frac{2idz}{z} ) + Re ∫_{1}^{p} ( \begin{matrix}f_{∞,λ}(z)\frac{i(1−z^{2})dz}{z^{2}} \\ f_{∞,λ}(z)\frac{−(1+z^{2})dz}{z^{2}} \\ f_{∞,λ}(z)\frac{2idz}{z} \end{matrix} ) ,  (3.7)

where

f_{∞,λ}(z) = 1 − \begin{matrix}√ & \overline{z} \\ √ & \overline{(1 − \frac{z}{λ})(z + \frac{1}{λ})} \end{matrix} .

Note that the first integral in the above sum in equation (3.7) is the Weierstrass integral for a helicoid, since Weierstrass data for a helicoid is

g(z) = z , η(z) = \frac{idz}{z^{2}} .

If λ is large, then the vertical distance between adjacent planar ends of Ŕ_{λ} is approximately 2π. This can be verified by integrating the third function in the integrand of the Weierstrass representation from +1 to −1 along the half of the unit circle lying in the upper half of the complex plane. This curve is z = e^{it}, 0 ≤ t ≤ π. The distance between adjacent planar ends for large λ is

Re ∫_{0}^{π} \frac{2\sqrt{λ}ie^{it}dt}{\sqrt{e^{it}(e^{it} − λ)(e^{it} + \frac{1}{λ})}} ≈ Re ∫_{0}^{π} \frac{2\sqrt{λ}idt}{\sqrt{−λ}} = 2π .

The major difference from the proof in the case λ → 0 is this: When λ is close to ∞, then a homologically nontrivial loop in Á_{L,λ} has a nonzero real period with respect to the Weierstrass integral. Now one may only assume that a path from z = 1 to z = p in A_{L} has length less than cnL, where n is the number of times that the path wraps about the origin, and c is some fixed constant. Since the distance between adjacent ends is approximately 2π, it follows that if the wrapping number n about the origin of a path from 1 to p is sufficiently large, then z = p will be mapped by the Weierstrass integral to a point outside B_{r}. Thus, we may assume there is an upper bound N for |n|, depending only on r. Hence, for large λ, we may assume that any path from 1 to p has length less than ćL, where ć = cN.

The other parts of the proof of the case λ → ∞ transfer directly from the proof of the case λ → 0. □

Corollary 3.1 As λ → ∞, the radius of any level-circle of Ŕ_{λ} diverges to infinity.

Corollary 3.2 The planar curve described by the set of centers of the horizontal circles that foliate Ŕ_{λ} has the property that its maximum curvature approaches zero as λ → ∞.

It remains to determine what other surfaces are limits of Riemann examples. Three possibilities are clear: a single flat plane; an infinite number of equally-spaced flat planes; and another Riemann example. The following proposition states that these are the only other possibilities, completing the proof of Theorem 1.1.

Proposition 3.2 Any convergent sequence of Riemann examples converges to one of the following surfaces:

a Riemann example;

a helicoid;

a catenoid;

a single flat plane;

an infinite number of equally-spaced flat planes.

Proof. Let {R_{j}}_{j=1}^{∞} be a sequence of Riemann examples converging to a surface S. From the definition of convergence, S is clearly minimal, properly embedded, and complete. Without loss of generality we may assume that the limiting normals v⃗_{j} at the ends of the surfaces R_{j} converge in S^{2} to a vertical vector. With this normalization, the surfaces R_{j} are foliated by circles and lines lying in planes that are becoming horizontal as j → ∞. These almosthorizontal planes intersect R_{j} in connected sets, and these connected sets each consist of a single circle or a single line. Thus every curve S ∩ {x_{3} = c} is the limit set of a sequence of circles and/or lines. It follows that S ∩{x_{3} = c} is either a circle, a line, the empty set or the plane {x_{3} = c}. Recall from the introduction that this implies that each component of S is either a helicoid, a catenoid, a Riemann example or a plane. Moreover: the helicoid and catenoid must have vertical axes; the plane and the ends of the Riemann example must be horizontal. If S has more than one component, they must all be horizontal planes (because S is fibred by circles or lines, one in each horizontal plane). Suppose S consists of two or more horizontal planes. Since the points on the surfaces R_{j} where the Gauss map is vertical must approach the ends of R_{j} as j → ∞, it is the equally spaced-ends of the surfaces R_{j} that converge to S. Therefore S is an infinite collection of equally-spaced planes. □

References

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[Ni] Johannes C. C. Nitsche. Lectures on Minimal Surfaces Vol. 1. Cambridge University Press, New York, NY, 1989.

[Os] Robert Osserman. A Survey of Minimal Surfaces. Dover Publications, Inc., New York, NY, 1986.

[Ri] B. Riemann. Gesammelte mathematische Werke. 2nd edition, B. G. Teubner, Leipzig, 1892.