The Geometry of Periodic Minimal Surfaces

William H. Meeks III^{∗} Mathematics Department University of Massachusetts

Amherst, MA 01003

Harold Rosenberg Mathematics Department Université de Paris 7, 75251 Paris, France

March 26, 1992

1 Introduction.

In this paper we shall demonstrate a surprising relationship between the topology of a properly embedded periodic minimal surface in ^{3} and its global geometry. We shall call a minimal surface periodic if it is connected and invariant under a group G of isometries that acts freely on ^{3}. We will analyze these surfaces by studying their quotient in ^{3}/G. We have already carried out this study for doubly-periodic minimal surfaces [16].

Recall that a surface has finite topology if it is homeomorphic to a closed surface with a finite number of points removed. Our main theorem is:

Theorem 1 A properly embedded minimal surface in a complete nonsimply connected flat three-manifold has finite total curvature if and only if it has finite topology.

∗The research described in this paper was supported by research grant DE-FG02-86ER25015 of the Applied Mathematical Science subprogram of Office of Energy Research, U.S. Department of Energy, and National Science Foundation grant DMS-8900285.

When the flat manifold is ^{3}, the existence of the helicoid (which has finite topology and infinite total curvature) demonstrates that the condition that N be nonsimply connected is a necessary one.

Theorem 1 has important topological and analytical consequences. One topological consequence is that a properly embedded orientable minimal surface of finite topology in an orientable flat nonsimply connected threemanifold always has an even number of ends or it is a plane (see Theorem 9 in Section 9).

A theorem of Huber [10] states that a complete Riemannian surface with nonpositive Gaussian curvature whose total curvature is finite must be conformally diffeomorphic to a closed Riemann surface punctured in a finite number of points. We will prove that a complete minimal surface of finite total curvature in a flat three-manifold can be described in terms of meromorphic data on its conformal compactification. We shall exploit these analytic conditions to prove the following uniqueness theorem.

Theorem 2 The plane and the helicoid are the only properly embedded simply connected minimal surfaces in ^{3} with infinite symmetry group.

In [16] we proved Theorem 1 in the case where the flat three-manifold N was isometric to the product × where is some flat torus. In fact we proved that a properly embedded minimal surface M in × has finite total curvature C(M) = 2πχ(M). It follows from the classification of flat threemanifolds that a flat, noncompact, nonsimply-connected three-manifold is finitely covered by × or by ^{3}/S_{θ} where S_{θ} is the right hand screw motion obtained by rotation around the positive x_{3}-axis by θ followed by a nontrivial translation along the x_{3}-axis. Thus, to prove Theorem 1, it remains to consider only the case where the manifold N is isometric to ^{3}/S_{θ} for some θ, 0 ≤ θ ≤ π. However, our proof of Theorem 1 will not actually depend on our previous theorem in the special case of × .

We have the following classification of the annular ends of the surfaces described in Theorem 1. As shown by work in [2, 3, 11, 12, 13], every ^{3}/S_{θ} has many examples with each possible end type.

Theorem 3 An annular end of a properly embedded minimal surface of finite topology in ^{3}/S_{θ} is asymptotic to a plane, a flat vertical annulus, or to an end of a helicoid (with horizontal limit tangent plane). If θ is nonzero and the end is asymptotic to a plane, then the plane is horizontal. If θ is irrational, then the end is not asymptotic to a flat vertical annulus.

The total curvature of minimal surfaces of finite topology in N = ^{3}/S_{θ} can be computed in terms of the winding numbers of its annular ends. Suppose A is the image of a proper embedding of the punctured disk D^{∗} in N. Let γ be the geodesic representing the image of the x_{3}-axis in N. After removing a compact neighborhood of ∂A, we may assume that A is disjoint from the ε-tubular neighborhood T of γ with boundary torus ∂T. The torus is obtained as a quotient by S_{θ} of the flat cylinder C of distance ε from the x_{3}-axis. A basis for π_{1}(∂T) is obtained from the quotient α of the oriented circle C ∩ ^{2} and the quotient β of the oriented right handed helical arc of least-length on C joining a point p with S_{θ}(p). The boundary curve of A is homotopic in N − γ to a unique element of π_{1}(∂T). Suppose ∂A is homotopic to nα + mβ. The winding number of the end of A is then defined to be \frac{1}{2π}|2π · n + m · θ|. If M is a complete embedded minimal surface of finite total curvature in ^{3}/S_{θ}, then define the total winding number of M to be the sum of the winding numbers of the ends of M. We let W (M) denote the total winding number of M.

Theorem 4 If M is a properly embedded minimal surface of finite topological type in ^{3}/S_{θ}, then the total curvature of M is

C(M) = 2π(χ(M) − W (M)).

When the ends are asymptotic to flat vertical annuli, this formula yields C(M) = 2πχ(M). When there are k planar ends, C(M) = 2π(χ(M) − k).

The paper is organized as follows. In Section 2 we develop the analytic theory of complete minimal surfaces of finite total curvature in ^{3}/S_{θ} when θ = 0 and in Section 3 we consider the case when θ ≠ 0. The main theorem of Section 3 is a Weierstrass-type analytic representation for a complete minimal surface M of finite total curvature in ^{3}/S_{θ}. In particular we show that these minimal surfaces are conformally equivalent to a closed Riemann surface \overline{M} punctured in a finite number of points and that the coordinates of M can be recovered from two meromorphic one-forms on \overline{M}. In Section 4 we characterize the asymptotic behavior of properly embedded minimal annuli of finite total curvature in ^{3}/S_{θ} and prove some global results on their geometry including the main reduction of the proof of Theorem 2 from Theorem 1. In Section 5 we prove a multi-valued version of Picard’s theorem that was used in the earlier Section 3. In Section 6 we prove that an annular end A of a properly embedded minimal surface in ^{3}/S_{θ} is trapped between two embedded minimal annuli of finite total curvature. In Sections 7 and 8 we use the result of Section 6 to show that A must have finite total curvature. The proof that A has finite total curvature breaks up into two cases depending on the asymptotic behavior of the finite total curvature annuli that trap it. This result on A proves Theorem 1. The remaining theorems are proved in Section 9.

We refer the reader to [3] and to [8] for related theoretical results.

2 Finite total curvature annular ends in R^{3}/T.

In this section we will analytically parametrize embedded finite total curvature ends A in N = ^{3}/T, where T is the group generated by translation by v ∈ ^{3}.

Let E denote a connected lifting of A to ^{3}. The Weierstrass data of E (Gauss map and holomorphic one-form) are invariant by T, hence, pass to a Gauss map on A and holomorphic one-form ω. As usual g denotes the composition of the Gauss map with stereographic projection to ∪ {∞}. Since A has finite total curvature, A is conformally the punctured disk D^{∗} = {z ∈ | 0 < |z| ≤ 1} and (g, ω) extend to meromorphic data at the origin. This last fact is well known, however in Section 3, we will prove a more general result.

Now, after a rotation of E in ^{3}, we can assume g(0) = 0, and after a conformal reparametrization (a subend of) of A we have:

g(z) = z^{p}, ω(z) = (\frac{c_{−q}}{z^{q}} + \frac{c_{−q+1}}{z^{q−1}} + ⋯) dz ,  (2.1)

where p ≥ 1.

The period vector is v = Re ∫_{S^{1}} ϕ, where

ϕ = (ϕ_{1}, ϕ_{2}, ϕ_{3}) = (\frac{(1 − g^{2})}{2}, \frac{i(1 + g^{2})}{2}, g)ω.

A (multi-valued) parametrization of E is given by x̃(z) = Re ∫^{z} ϕ. We have x̃(r, θ + 2nπ) = x̃(r, θ) + nv, z = re^{iθ}.

The case v = 0 is well known [23]. A is then an embedded finite total curvature annulus in ^{3} and is asymptotic to a plane or catenoid. Henceforth, assume v ≠ 0. The analysis of the type of A is determined by the order q of the pole of ω and the coefficient c_{−q}.

Theorem 5 Let (g, w) be as in equation 2.1 and assume v ≠ 0. The nature of the end A is determined by p and q as follows.

1. If q = 1, A is a Scherk-type end (this is made explicit shortly); 2. if q > 1, then A is embedded only if q = p + 1. When q > p + 1, the trace of A on a large cylinder is not embedded (this is Toubiana’s lemma). If q < p + 1, a translation by a large horizontal period will give a self-intersection point.

Proof. First suppose q = 1. We will see that A is a Scherk type end (e.g. the ends of Scherk’s surface g(z) = z, ω = i dz/(z^{4} − 1), on the sphere punctured at the four roots of unity) and converges to a flat annulus. Let a = c_{−q}. Then

g(z) = z^{p}, ω = (\frac{a}{z} + f(z)) dz, and

ϕ_{3}(z) = (az^{p−1} + h(z)) dz, f, h holomorphic in z ∈ D.

Hence ϕ_{3} is holomorphic at 0, v is a horizontal vector, and x_{3}(z) converges to a constant (which we take to be 0) as z → 0.

After a rotation about the x_{3}-axis, we can assume v = (−2π Im a, 0, 0), a ∈ i . Then

x_{1} = −\frac{1}{2}a_{0} arg (z) + O(1) ,

x_{2} = −\frac{1}{2}a_{0} ln |z| + O(1) ,

x_{3} = \frac{1}{p} Re (az^{p}) + O(1), a_{0} = Im (a) ,

x_{1} − i x_{2} = \frac{a}{2} log (z) + O(1) .

Here O(1) denotes a function continuous at z = 0. Henceforth, we will let O_{0}(1) denote such a function that vanishes at 0.

The image by x̃ of the line θ = 0, 0 < r ≤ 1, is a curve asymptotic to a line parallel to the x_{2}-axis: x_{3} = 0, x_{1} = c_{1} = lim_{r→0} x_{1}(r, 0). The image of the line θ = 2π, 0 < r ≤ 1, is the same curve translated by v. So a fundamental domain in ^{3} is a half band bounded by these two curves. The surface is asymptotic to the flat annulus x_{3} = 0, x_{2} > 0. If we think of v as vertical, then A is asymptotic to a vertical flat annulus.

Now suppose q > 1. E. Toubiana has proved that A embedded implies q ≤ p + 1 [26]. He proves this by showing the trace of A on a large cylinder centered at the x_{3}-axis, is not embedded when q > p + 1.

We now show that q = p + 1. Assume the contrary: q < p + 1. Then ϕ_{3}(z) is holomorphic at 0 and x_{3}(z) tends to a constant as z → 0, which we take to be zero. Notice that x_{3} changes sign on every circle |z| = r > 0, since it is harmonic on the disk. Also the period vector v is horizontal since the residue of ϕ_{3} is zero at 0. We have

(x_{1} + i x_{2})(z) = ∫ \overline{ω} − ∫ g^{2}ω = ∫ \overline{ω} + O(1) = \frac{1}{\overline{z}^{q−1}} (\frac{\overline{a}}{q − 1} + O_{0}(1)) ,

a = c_{−q}, R = \sqrt{x_{1}^{2} + x_{2}^{2}} = \frac{1}{r^{q−1}} (\frac{|a|}{q − 1} + O_{0}(1)) .

As z → 0 along a ray arg(z) = constant, (x_{1} + i x_{2})(z) is asymptotic to a straight line and X̃(z) is an embedded curve in ^{3} whose projection on the (x_{1}, x_{2})-plane is (x_{1} + i x_{2})(z) and whose x_{3}-coordinate tends to zero. For r fixed, r ≠ 0, the total change of the argument of (x_{1}, +i x_{2})(z), as z traverses once the circle |z| = r is (q − 1)2π + O_{0}(1), by the above formula for (x_{1}, +ix_{2})(z).

Now consider the surface M_{0} = X̃(0 ≤ arg(z) ≤ 4π, 0 < r ≤ 1). Let T_{R} denote the vertical cylinder of radius R, centered at the x_{3}-axis. For R_{∘} large, M_{0} projects surjectively onto the complement of the disk D_{R_{∘}} = {x_{1}^{2} + x_{2}^{2} ≤ R_{∘}^{2}}. (In fact, 4π can be replaced by 2π + ε for any ε > 0 for this projection to be surjective.)

Choose (r, θ) so that δ = x_{3}(X̃(r, θ)) > 0; for convenience, take θ = 0. Let α be the arc X̃(r, θ), 0 ≤ θ ≤ 2π; notice that the endpoints of α differ by v and so have the same x_{3}-coordinate. There are points on α above and below the (x_{1}, x_{2})-plane, since x_{3} changes sign on |z| = r. Let ε = 1/2 min{δ, | min x_{3}|α|}.

Now consider the surfaces M_{k} = X̃(2kπ ≤ arg(z) ≤ (2k+4)π, 0 < r ≤ 1). M_{k} is obtained from M_{0} by horizontal translation by 2kv. Let R_{1} > R_{∘} be chosen so that on the complement of T_{R_{1}}, M_{0} is at most a distance ε from the (x_{1}, x_{2})-plane.

Choose k so that kv > R_{1}. The arc α on M_{0} is translated horizontally to an arc α̃ on M_{k} by the translation by 2kv. Outside of T_{R_{1}}, M_{0} is at a height at most ε and M_{0} projects surjectively onto the complement of D_{R_{1}}, so M_{0} must intersect α̃. This contradicts A is embedded.

Thus, q = p + 1 and x_{3} = α_{0} ln |z| + β arg (z) + O(1), α_{0}, β ∈ . As before, we have R = \sqrt{x_{1}^{2} + x_{2}^{2}} = 1/r^{q−1} (|a|/(q − 1) + O_{0}(1)) so x_{3} = α ln R + β arg(z) + O(1), α ∈ . (Note that α ≠ α_{0}.)

The trace of E on the cylinder S_{R} is converging to the helix α · ln R + β · arg(z), for 0 < arg(z) < 2π, hence E is embedded for R sufficiently large. Notice that the period vector v need not be vertical: e.g. g(z) = z, ω = (i/z^{2} + 1/z) dz. This is an embedded helicoidal type end with a non vertical axis v. If both α and β are non zero, then the helicoidal end has a logarithmic growth as R → ∞, given by α; just as the usual catenoid where β = 0. We shall see later that if the end A is part of a properly embedded surface M of finite topology and β ≠ 0, then there is no logarithmic growth (α = 0), v is vertical, and so A is asymptotic to a helicoid (see Theorem 3 in the Introduction).

In summary, the embedded ends in ^{3}/T are planar type (asymptotic to flat annuli), catenoid type, or helicoidal type. The latter ends are helicoidal ends which may have logarithmic growth and possibly an axis not orthogonal to the end. □

3 Finite total curvature annular ends in R^{3}/S_{θ} and their Weierstrass Representation.

Let A be a finite total curvature minimal annulus, embedded in N = ^{3}/S_{θ} where 0 < θ < 2π. In this section we will derive meromorphic data on the disk that parametrizes A, and describes its aymptotic behavior at ∞.

We take S_{θ} to be a translation along the x_{3}-axis followed by rotation by θ about the x_{3}-axis. Since A has finite total curvature, A is conformally the punctured disk D^{∗}. We no longer have a single valued Gauss map g on A; g is a multi-valued meromorphic map on D^{∗} whose values differ by multiplication by λ^{m}, λ = e^{2π i θ}. To see this, let E be a connected lifting of the universal covering space of A to ^{3}. The Gaussian image of the normal vector to E at p ∈ E and the image of the normal vector to E at S_{θ}(p), differ by rotation about the x_{3}-axis by θ. Hence, the stereographic projections of these vectors on the sphere, differ by rotation by θ in , i.e., by multiplication by λ = e^{2π i θ}.

Lifting g to the Riemann surface of g (i.e., the covering Riemann surface where g is defined), we have a well defined meromorphic map g̃, on the half plane H = {x ≤ 0}, satisfying g̃(z + 2πmi) = λ^{m}g̃(z), for z ∈ H. Then g = g̃ (exp^{−1}).

We wish to show that A has a limiting tangent plane at ∞, i.e., g extends continuously to 0 (even though g is multi-valued). This will follow from the fact that the area of the spherical image of g (i.e. a single valued branch of g on the slit punctured disk D′) is finite (see Theorem 6 below). We are grateful to Dennis Sullivan for explaining the length-area inequality of conformal maps which is used repeatedly in the proof of Theorem 6.

Theorem 6 Let g be a multi-valued meromorphic map on D^{∗}, g = g̃ (exp^{−1}), with g̃ (z + 2π i) = λg̃ (z), for z ∈ H, and some λ, |λ| = 1. If Area(g(D′)) is finite, then g extends continuously to 0.

The proof of Theorem 6 will be postponed to Section 5.

Now we shall use Theorem 6 to obtain a Weierstrass representation on the disk D, for finite total curvature annuli A in N = ^{3}/S_{θ}. We use the notation of Section 2. By Theorem 6, the multi-valued g extends continuously to 0 and since the limiting value is fixed by multiplication by λ and λ ≠ 1, the limiting value is 0 or ∞; so we can assume g(0) = 0. Write λ = e^{2π i a} with 0 < a < 1. Since g̃ (z + 2π i) = λg̃(z), the map z^{−a}g(z) is indeed single valued on D^{∗}. Furthermore, z^{1−a}g(z) is bounded in a neighborhood of 0, hence g(z) = z^{a−1}h(z) where h is holomorphic in a neighborhood of 0. Hence, dg/g is a well defined meromorphic one-form on D^{∗}, and 0 is a removable singularity. The multi-valued g on D, is obtained from this form by g(z) = exp(∫ dg/g).

Next notice that ϕ_{3} is a well defined holomorphic form on A. To see this let x(u, v) be local conformal coordinates about a point p ∈ E. Then x̃ (u, v) = S_{θ} (x (u, v)) are local coordinates about S_{θ}(p) and x̃_{3} (u, v) = x_{3} (u, v) + t_{0}, t_{0} the vertical translation component of S_{θ}. Hence,

ϕ̃_{3} (u, v) = \frac{∂x̃_{3}}{∂u} − i \frac{∂x̃_{3}}{∂v} = \frac{∂x_{3}}{∂u} − i \frac{∂x_{3}}{∂v} = ϕ_{3} (u, v).

Denote ϕ_{3} by η. We claim 0 is a removable singularity of η: the metric on A is given by

ds = \frac{1}{2}(|g| + \frac{1}{|g|}) |η|,

(|g| is well defined on A), and since the metric is regular and complete, η ≠ 0 on A, and for γ a path on A tending to 0, we have ∫_{γ} ds = ∞. Since g(0) = 0 and g is continuous at 0, this implies ∫_{γ} |η|/|g| = ∞. Now there is an integer m such that |g(z)| > |z|^{m} for |z| small (|g(z)| is of the order |z|^{a+n} for 0 < a < 1 and n ≥ 1 an integer, so m = 2n works). Then |η|/|g| < |η|/|z^{m}|, hence ∫_{γ} |η|/|z^{m}| = ∞ for every path γ in D^{∗} tending to 0. This implies 0 is a removable singularity of η/z^{m}, hence of η too, [19].

We take as Weierstrass data on A the pair (dg/g, η); these forms are meromorphic at the puncture and A is obtained from this data by the formula g = exp (∫ dg/g),

x(z) = Re ∫ (\frac{1}{2g} − \frac{g}{2}, \frac{i}{2g} + \frac{ig}{2} , 1) η .

In particular, we have proved:

Theorem 7 Let M be a complete finite total curvature minimal surface in 3/S_{θ}. Then there exists a conformal compactification \overline{M} of M, and meromorphic forms (dg/g, η) on \overline{M}, such that M is parametrized by

x(z) = Re ∫ (\frac{1}{2g} − \frac{g}{2}, \frac{i}{2g} + \frac{ig}{2} , 1) η

where g = exp( ∫ \frac{dg}{g}) .

Remark 3.1 H. Karcher has given many new examples of such M with this data [11].

4 Some global properties of finite total curvature M in N = R^{3}/S_{θ}.

Let M be a properly embedded minimal surface in N of finite total curvature. Since the lift of M to ^{3} is orientable (since it is embedded) and M is invariant under S_{θ}, S_{θ}^{2} acts on the lifted surface in an orientation preserving manner. Hence, after lifting to a two-sheeted covering space, we can assume M is orientable. We know M is conformally equivalent to a compact Riemann surface \overline{M} punctured at a finite number of points. A neighborhood of a puncture in M is an embedded annulus of finite total curvature, hence the previous section applies and (dg/g, η) is meromorphic on \overline{M}; in particular, each of the ends of M has a limiting normal vector.

Proposition 4.1 Let A be a finite total curvature embedded minimal annulus (homeomorphic to S^{1} × [0, ∞)) in ^{3}/S_{θ}, with limiting normal vector g(0). Then A is asymptotic to a plane, a catenoid, a flat annulus, or a helicoidal-catenoid type end. This means:

i) If θ ≠ 0, then g(0) is parallel to the axis of translation of S_{θ} (the x_{3}-axis) and there are real numbers α, β such that (we assume A parametrized by D^{∗})

x_{3}(z) = α ln R + β arg (z) + O(1),

where R = \frac{1}{|z|^{q+a}} (c + O(1)),

q an integer ≥ 1, 2π a = θ, and c a real constant. If α = β = 0, this is a planar end; if β = 0 and α ≠ 0, a catenoid type end; if β ≠ 0, α = 0, a helicoidal type end. And if α ≠ 0, β ≠ 0, we call this a helicoidal-catenoid type end.

ii) If θ = 0 and g(0) is not orthogonal to the axis of S_{θ}, then the same statement for x_{3}(z) as in (i) holds, where x_{3} is the coordinate parallel to g(0). If g(0) is orthogonal to the axis of S_{θ}, then A is a Scherk type end, asymptotic to a flat annulus.

Definition 4.1 An annular end as in Proposition 4.1 will be called a standard end.

Proof of Proposition 4.1. Assume first, that θ ≠ 0, so that the limiting normal vector to A at infinity is parallel to the axis of translation of S_{θ}, i.e. the x_{3}-axis. Then we can assume g(0) = 0, A is parametrized by the punctured disk D^{∗}. Let T_{R} be the torus tubular neighborhood of radius R, the radius R cylinder centered at the x_{3}-axis modulo S_{θ}. For R large, A intersects ∂T_{R} transversally in a simple closed curve A(R). We have x_{3} = Re ∫ η and Toubiana’s lemma (more precisely: the proof of Toubiana’s lemma [26]) implies that η has at most a pole of order one at 0, since A(R) is embedded.

First suppose η is holomorphic at 0. Then x_{3}(z) tends to a constant as z → 0, x_{3} is a well defined function on A, and A lifts to an embedded annular end in ^{3}. A is then a planar end, asymptotic to a horizontal plane.

Now suppose η has a pole of order one. Then

x_{1} − i x_{2} = ∫ \frac{η}{g} + O(1),

R = \frac{1}{r^{q+a}} (c + O(1)),

where c is a real constant, q an integer greater than or equal to one and a = \frac{θ}{2π}. Thus,

x_{3} = α log R + β arg (z) + O(1),

for some real constants α, β. For R large, A(R) is approximately the curve α · log R + β · arg (z). This is a horizontal circle for β = 0 and a helix for β ≠ 0. When β = 0, A is a catenoid type end, x_{3} = α · log R + O(1). When β ≠ 0, A is a helicoid type end with a logarithmic growth rate term; a (rather clumsy) appropriate name for these ends is helicoidal–catenoid type end.

When θ = 0, so S_{θ} is translation by a vector v, the asymptotic behavior of A was analyzed in Section 2. We chose as x_{3}-axis the limiting normal to the end and we found A was also a planar, catenoid or helicoidal–catenoid type end. We remark, that one also had x_{3} = α · log R + β · arg(z) + O(1) except when the translation vector v was orthogonal to g(0) (this was the case q = 1, a Scherk type end). □

Now we return to our globally embedded M in ^{3}/S_{θ}. We will now prove that all the ends of M are of the same type, with the same coefficients α and |β| of ln R and arg(z). First assume that the translation vector v is vertical. If the α’s of two ends of M were distinct, then the distance between the lifted trace curves A(R) to ^{3}, would tend to ∞ as R → ∞. So for some value of R, they would intersect on the torus ∂T_{R}. Hence M would not be embedded. If two |β|’s were not the same, then one would have helices of different slope on ∂T_{R} and they would intersect. Notice the β’s can be of opposite sign and equal (as for the standard helicoid). As θ increases from 0 to 2π, β > 0 yields a helix with x_{3}-coordinate increasing and β < 0 a helix going down, i.e., x_{3} decreasing. If v is not vertical, let T_{R} denote the torus tubular neighborhood of radius R around a translation axis of M. Note that if α ≠ 0, v is not horizontal. The above discussion shows that if the α′s of two ends of M were distinct, then the trace curves A(R) on ∂T_{R} would intersect for some large values of R. If two |β|′s were not the same, then the trace curves are not homotopic on ∂T_{R}/S_{θ} and so must intersect.

Proposition 4.2 Let A_{1}, …, A_{n} be the ends of M and (α, β_{i}) the coefficients of log R and arg(z) at each end. Then ∑_{i=1}^{n} β_{i} = 0 and α = 0. In particular, there are an even number of ends when β_{i} ≠ 0 for all i.

Proof. Let M_{0} = M −∪_{i=1}^{n} Int(A_{i}); M_{0} is a compact surface with one boundary component ∂A_{i} coming from each A_{i}. Consider the holomorphic form η on M_{0}. We have:

0 = ∫_{M_{0}} dη = ∫_{∂M_{0}} η = ∑_{i=1}^{n} ∫_{∂A_{i}} η = 2πi ∑_{i−1}^{n} c_{i},

where at the end A_{i}, η(z) = (c_{i}/z + O(1)) dz. Since x_{3} = Re ∫ η, we have α = Re(c_{i}) and β_{i} = − Im(c_{i}). Hence, α = 0 and ∑_{i=1}^{n} β_{i} = 0. □

E. Toubiana proved that an embedded minimal two punctured sphere in ^{3}/T, T a translation, is a helicoid, provided the total curvature is finite [26]. The question naturally arises whether this remains true in screw motion spaces. The answer is affirmative.

Theorem 8 (Toubiana’s theorem in ^{3}/S_{θ}.) Let M be a complete embedded minimal annulus of finite, nonzero, total curvature in ^{3}/S_{θ}. Then M is a helicoid.

Proof. M is conformally − {0} and of finite total curvature. We have the meromorphic data (dg/g, η) on the Riemann sphere, that parametrizes M. We write η = g ω,

g(z)z^{−a} = \frac{P (z)}{Q(z)}, ω(z)z^{a} = c \frac{Q^{2}(z)}{z^{n}} dz,

were 0 < a < 1, P and Q polynomials, relatively prime. Parametrize M so that the normal vectors are vertical at the punctures. We can assume g(0) = 0, so Q(0) ≠ 0, (we know the limiting normal vectors are vertical).

By Corollary 1 in [9], a nontotally geodesic properly embedded orientable surface in a flat orientable three-manifold N must separate N, so since g(0) = 0, we have g(∞) = ∞. That is, M ∩ ∂T_{R} consists of two embedded curves that separate ∂T_{R} into 2 components, and the normal vector to M in N always points into the same component of N − M. Now Toubiana’s lemma [26] implies that if P (z) = z^{m} P_{1}(z), m ≥ 0, then n ≤ m + 1 ≤ p + 1, p = deg P.

Consider the end of M, where g(∞) = ∞. We rotate ^{3} by the matrix

( \begin{matrix}−1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & −1 \end{matrix} )

so that the (g̃, ω̃) at the rotated end satisfy:

g̃ = \frac{ϕ̃_{3}}{ω̃} = −\frac{1}{g}, ω̃ = ϕ̃_{1} − i ϕ̃_{2} = g^{2}ω.

Then

g̃(z) z^{a} = \frac{Q(z)}{P (z)}, ω̃(z) z^{−a} = c \frac{P^{2}(z)}{z^{n}} dz,

and ∞ is a zero of order deg P − deg Q of g̃. Also (P^{2}(z)/z^{n}) dz has a pole at ∞ of order 2 · deg P + 2 − n, so by Toubiana’s lemma:

2 · deg P + 2 − n ≤ deg P − deg Q + 1.

Thus,

deg P + deg Q + 1 ≤ n ≤ m + 1 ≤ deg P + 1.

Hence deg Q = 0 and P (z) = c_{1}z^{m}, m = n − 1, c_{1} ∈ . This gives

z^{−a}g(z) = z^{n−1}, z^{a}ω = \frac{dz}{z^{n}} ,

which is a helicoid. □

5 A multi-valued Picard’s Theorem.

We now prove the Theorem 6 stated in Section 3. The length-area inequality will be applied to on the one hand to the circles of an annulus and also to the radial lines of an annulus.

Let D′ be the punctured disk D^{∗}, slit along θ = 0. Suppose g is meromorphic on D′ and g(r, 0) = λ g(r, 2π) for 0 < r ≤ 1. Let C(r) = {z ∈ D′ | |z| = r}, and ℓ(r) = the length of g(C(r)). Then

ℓ(r) = ∫_{0}^{2π} | dg(T_{r}) | dθ = r ∫_{0}^{2π} \sqrt{J(g(z))} dθ,

where T_{r} = r(cos θ, sin θ), and J(g) is the Jacobian of g.

We have: ℓ(r)^{2}/r^{2} ≤ (∫_{0}^{2π} J(g) dθ) · (2π), so integrating with respect to r dr we obtain:

∫_{0}^{r} \frac{ℓ(r)^{2}}{r} dr ≤ 2π ∫_{0}^{r} ∫_{0}^{2π} J(g) r dr dθ ≤ 2π (Area(g(D′))) < ∞.

Hence there exists a sequence r_{n} → 0 such that ℓ(r_{n}) → 0. Let B(n) be the curve g(C(r_{n})). First consider the case λ = 1 (this means g is single valued on D^{∗}) so each B(n) is a closed curve, and ℓ(B(n)) → 0 as n → ∞.

We show that all the B(n) accumulate at the same point. This shows g extends continuously to zero since by the open mapping property of g, the annulus between C(r_{n}) and C(r_{n+1}) gets sent close to this accumulation point as well (otherwise the image of this annulus would cover almost the whole sphere).

Assume, on the contrary, that B(n_{2i}) accumulates at p and B(n_{2i+1}) accumulates at q, with p ≠ q. Then the annulus between C(r(n_{2i})) and C(r(n_{2i+1})) gets sent to the region of the sphere between B(n_{2i}) and B(n_{2i+1}), by the open mapping property of g. It follows that the spherical image of this annulus has area at least 2π. Since this holds for an infinite sequence of annuli, tending to 0, the area of the image of g would be infinite. Thus p = q and the theorem is proved when λ = 1.

Now suppose λ ≠ 1. The endpoints of B(n) differ by multiplication by λ and since the lengths of the B(n) tend to zero, the only possible accumulation points of the B(n) are 0 and ∞.

First suppose the B(n) accumulate at both 0 and ∞. We will show this is impossible by showing the image of D′ by g would have infinite area. So assume the sequence r_{1} > r_{2} > ⋯ > r_{n} > ⋯, satisfies: B(n_{2i}) tends to 0, B(n_{2i+1}) tends to ∞ and r_{i} → 0. We will derive a contradiction by showing the image by g of the annular region between C(r_{2i}) and C(r_{2i+1}) has definite spherical area, thus the area of the image of g would be infinite.

For notational convenience, let r_{1} = r_{2i}, r_{2} = r_{2i+1}, C_{1} = C(r_{2i}), C_{2} = C(r_{2i+1}), and let B_{1} = g(C_{1}) be a short curve near 0 and B_{2} = g(C_{2}) be a short curve near ∞. Let F be the annulus on the Riemann sphere S^{2} bounded by the tropic of cancer (= F_{2}) and the tropic of capricorn (= F_{1}).

B_{1} is in the disk on S^{2} below F_{1} and B_{2} in the disk above F_{2}.

For each θ, 0 ≤ θ ≤ 2π, let r_{1}(θ), r_{2}(θ) be chosen in [r_{1}, r_{2}] so that for r_{1}(θ) ≤ r ≤ r_{2}(θ), g(r, θ) ∈ F, and g(r_{1}(θ), θ) ∈ F_{1}, g(r_{2}(θ), θ) ∈ F_{2}. This is possible since g(r_{1}, θ) ∈ B_{1} and g(r_{2}, θ) ∈ B_{2}.

Let α_{θ}(r) = r(cos θ, sin θ) and L_{θ} = length (g(α_{θ})) for r_{1}(θ) ≤ r ≤ r_{2}(θ). We have L_{θ} ≥ π/4 for each θ.

Now do a length-area calculation:

L_{θ} = ∫_{r_{1}(θ)}^{r_{2}(θ)} |dg(αθ^{′})| dr = ∫_{r_{1}(θ)}^{r_{2}(θ)} \sqrt{J(g)} dr = ∫_{r_{1}(θ)}^{r_{2}(θ)} \sqrt{r}\sqrt{J} \frac{1}{\sqrt{r}} dr,

L_{θ}^{2} ≤ (∫_{r_{1}(θ)}^{r_{2}(θ)} J(g)r dr)(∫_{r_{1}(θ)}^{r_{2}(θ)} \frac{dr}{r}) ≤ (∫_{r_{1}(θ)}^{r_{2}(θ)} J(g)r dr) × K,

K = sup_{θ} ln(\frac{r_{2}(θ)}{r_{1}(θ)}) (the sup exists since ln(\frac{r_{2}(θ)}{r_{1}(θ)}) ≤ ln(\frac{r_{2}}{r_{1}} )).

Now integrate with respect to θ:

∫_{0}^{2π} L_{θ}^{2} dθ ≤ K ∫_{0}^{2π} ∫_{r_{1}(θ)}^{r_{2}(θ)} J(g)r dr dθ

≤ KArea (g(D0^{′})),

where D0^{′} is the slit annulus between C(r_{1}) and C(r_{2}). Since L_{θ} ≥ \frac{π}{4}, this yields

\frac{π^{3}}{8K} ≤ Area(g(D0^{′})).

Now our previous length-area calculation for ℓ_{r} = length(g(|z| = r)) yielded:

(\frac{ℓ_{r}}{r})^{2} ≤ 2π ∫_{0}^{2π} J(g) dθ.

Hence for θ fixed:

∫_{r_{1}(θ)}^{r_{2}(θ)}(\frac{ℓ_{r}}{r})^{2}r dr ≤ 2π ∫_{r_{1}(θ)}^{r_{2}(θ)} ∫_{0}^{2π} J(g)r dr dθ ≤ 2π · Area (g(D0^{′})).

Now, for r_{1}(θ) ≤ r ≤ r_{2}(θ), g(r, θ) ∈ F, so there exists a constant c such that l_{r} ≥ c > 0 (the endpoints of g(|z| = r) differ by multiplication by λ, so in F their distance is uniformly bounded from below). Thus the last integral inequality yields

c^{2} ln(\frac{r_{2}(θ)}{r_{1}(θ)}) ≤ 2π · Area (g(D0^{′})).

This holds for all θ so:

c^{2}K ≤ 2π · Area (g(D0^{′})).

Multiply this with the inequality π^{3}/8K ≤ Area (g(D0^{′})), to get:

Area(g(D0^{′})) ≥ \frac{π · c}{4},

and this contradicts our finite area hypothesis and completes the proof of Theorem 1 in the case that the B(n) accumulate at both 0 and ∞.

It remains to consider the case when all the B(n) accumulate at one of the points, 0 say. If g is not continuous at 0, then there is a sequence x_{n} → 0 with g(x_{n}) → q, and q ≠ 0. Each x_{n} is in an annulus A_{n}, bounded by circles C(r_{n}), C(r_{n+1}), which get sent by g to short curves near 0. We can suppose q is on the boundary of a disk E (which we take to be of radius one for convenience) centered at 0, and all the circles in ∂A_{n} get sent by g into the disk centered at 0 of radius \frac{1}{4}. Let F be the annulus in E bounded by the circle F_{1}, of radius \frac{1}{2}, and the circle F_{2} of radius \frac{3}{4}.

Fix an annulus A_{n} and for notational convenience let r_{1} be its inner radius and r_{2} its outer radius, C_{1} = C(r_{1}), C_{2} = C(r_{2}), B_{1} = g(C_{1}), B_{2} = g(C_{2}). Let g be well defined on A_{n} slit along θ = θ_{0}. Notice that if g(|z| = r) is contained in a disk of radius R centered at 0, then any other determination of the multi-valued g has the same property, since two determinations differ by multiplication by λ^{m}, for some m, and |λ| = 1.

Now x_{n} ∈ A_{n} has polar coordinates (r(x_{n}), θ(x_{n})). Recall that g(x_{n}) converges to q, so we can suppose g(x_{n}) is not in the disk D(\frac{3}{4}), of radius \frac{3}{4}, centered at 0. Consider the image by g of the radial segment α(θ(x_{n})) in A_{n} joining (r_{1}, θ(x_{n})) to (r_{2}, θ(x_{n})). Since the extremities of this segment get sent to points in D(\frac{1}{4}), and g(x_{n}) ∉ D(\frac{3}{4}), there are r_{1}(θ(x_{n})), r_{2}(θ(x_{n})), such that:

r_{1} < r_{1}(θ(x_{n})) < r_{2}(θ(x_{n})) < r_{2},
g(r_{2}(θ_{n}), θ(x_{n})) ∈ F_{2}, g(r_{1}(θ(x_{n})), θ(x_{n})) ∈ F_{1},

and

g(r, θ(x_{n})) ∈ F, for r_{1}(θ(x_{n})) ≤ r ≤ r_{2}(θ(x_{n})).

Observe now, that for any θ between θ_{0} and θ_{0} + 2π, the same property holds, i.e., there are r_{1}(θ), r_{2}(θ) such that:

r_{1} < r_{1}(θ) < r_{2}(θ) < r_{2}, g(r_{2}(θ), θ) ∈ F_{2}, g(r_{1}(θ), θ) ∈ F_{1},

and

g(r, θ) ∈ F for r_{1}(θ) ≤ r ≤ r_{2}(θ).

For if this failed to hold, then for some θ, the image by g of the radial segment α_{θ}(r) = r(cos θ, sin θ), r_{1} ≤ r ≤ r_{2}, would be contained entirely in D(\frac{3}{4}). Then let D_{1} be the disk obtained from A_{n} by cutting A_{n} along α_{θ}, and let g_{1} be a single-valued branch of g on D_{1}. The boundary of D_{1} gets sent into D(\frac{3}{4}) by hypothesis, so D_{1} gets sent into D(\frac{3}{4}) or S^{2} − D(\frac{3}{4}) is in the image of D_{1} by the open mapping theorem. In the latter case, the image has area at least 2π. In the former case, we conclude all the determinations of g on A_{n} get sent into D(\frac{3}{4}) and this contradicts g(x_{n}) ∉ D(\frac{3}{4}). So for each θ, we have r_{1}(θ), r_{2}(θ) as desired.

Now do the two length-area calculations, just as in the case when 0 and ∞ were accumulation points of the B(n). The same reasoning shows there is a constant C > 0 such that Area(g(A_{n})) ≥ C for all n. This completes the proof of Theorem 6. □

6 The trapping lemma for embedded minimal annuli.

Before stating the Trapping Lemma, we prove a topological property for properly embedded surfaces in N = ^{3}/S_{θ}.

Lemma 6.1 Let M be a properly embedded surface in N that separates N and has at least one annular end. Then there is a finite covering p: Ñ → N such that p^{−1}(M) has more than one end.

Remark 6.1 If M is an orientable minimal surface, not a plane, then M separates N (cf. Corollary 1 in [9]).

Proof. Let A be an annular end of M. If π_{1}(A) → π_{1}(N) is not an isomorphism, then one can clearly lift M to a covering space so that the lifted surface has more than one end. So suppose it is an isomorphism. Choose R so that ∂T_{R} intersects A transversally; ∂T_{R} is the torus in N which is all points a distance R from the x_{3}-axis. Also suppose ∂A ⊂ T_{R}. Since ∂A ⊂ T_{R} and the end of A lies outside T_{R}, A ∩ ∂T_{R} contains an odd number of simple closed curves, each a generator of π_{1}(∂T_{R}), and perhaps some null homotopic cycles. In particular, there is a cycle β on ∂T_{R} whose intersection number with A is odd. Since M separates N, the intersection number of β and M is zero. Thus M must have other ends. □

Lemma 6.2 (The Trapping Lemma) Let T_{R} denote the image in N = 3/S_{θ} of the solid vertical cylinder of radius R in ^{3} around the x_{3}-axis.

Suppose A is an annular end of a properly embedded minimal surface in N with more than one end. Then for some R > 0, there exist two disjoint standard ends E_{1}, E_{2} (hence of the same type, see Definition 4.1) that satisfy:

1. (E_{1} ∪ E_{2}) ∩ T_{R} = ∂E_{1} ∪ ∂E_{2} ⊂ ∂T_{R};

2. E_{1} ∪ E_{2} separates N − T_{R} into 2 components C_{1}, C_{2} ;

3. A has an annular end A′ ⊂ A with A′ ⊂ C_{1} or A′ ⊂ C_{2} .

Proof. If A has finite total curvature, we have shown in Proposition 4.1 that A is asymptotic to a standard end E. A vertical translation is well defined in N and so E_{1} and E_{2} can be obtained by small vertical (up and down) translations of E. Assume now that A has infinite total curvature. By Remark 6.1, M separates N into two components whose closures we denote by C and C′. Since ∂A is not homologous to zero in M, it can not be homologous to zero (with _{2}-coefficients) in both C and C′, (since H_{2}(N) = 0). Assume that ∂A represents a nontrivial class in C.

By [6]), a stable minimal surface with compact boundary in a flat orientable three-manifold has finite total curvature. In particular A contains a compact subdomain that is unstable. After replacing A by a subend that is disjoint from this unstable compact domain, we may assume that ∂A disconnects ∂C into two unstable minimal surfaces.

Choose an exhaustion F_{1} ⊂ F_{2} … of A by smooth compact subannuli with ∂A ⊂ ∂F_{1}. Since ∂C has nonnegative mean curvature, every smooth 1-cycle Γ in C that bounds in C is the boundary of an embedded least-area surface in C (see Theorem 1 in [18] and also [24]). In particular ∂F_{i} is the boundary of a smooth embedded surface Σ_{i} in C that is least area and _{2}- homologous to F_{i} rel(∂F_{i}). Since C is orientable and F_{i}∪Σ_{i} is a _{2}-boundary in C, Σ_{i} is orientable. The usual compactness and regularity theorems for least-area surfaces (see [24]) imply that a subsequence of the Σ_{i} converge to a least-area orientable surface Σ ⊂ C with ∂Σ = ∂A. Since Σ is stable and both components of ∂C − ∂A are unstable, all three are different so the maximum principle implies that Σ ∩ ∂C = ∂Σ.

The surface Σ separates C into two components; let C̃ be the component containing A. Let Ã be a proper annular subend of A. As for A in C solve the Plateau problem for ∂Ã in C̃ to obtain a stable minimal surface Σ̃ in C̃ with boundary ∂Ã. Clearly Σ̃ is disjoint from Σ. Since Σ and Σ̃ are stable, they have finite total curvature. Let E′_{1}, E′_{2} be annular ends of Σ, Σ̃, respectively. Since E′_{1} and E′_{2} are standard and disjoint, Proposition 4.1 implies that for R large, E_{i} = E′_{i} ∩ (N − Int(T_{R})) for i = 1, 2, are disjoint standard ends. It follows directly from the asymptotic properties of standard ends in N that E_{1} ∪ E_{2} separates N − T_{R} into two components C_{1} and C_{2}. Since A is proper and disjoint from E_{1} ∪ E_{2}, it has an annular end representative A′ with A′ ⊂ C_{1} or A′ ⊂ C_{2}. This completes the proof of the lemma. □

7 Trapped minimal annuli in a wedge.

Throughout this section A will denote a properly embedded minimal annulus in N = ^{3}/T where T is a vertical translation. Let S be the flat vertical annulus with boundary whose inverse image in ^{3} is a vertical half plane with boundary the x_{3}-axis. Let γ = ∂S be the quotient of the x_{3}-axis in N. A wedge is a region between two such vertical annuli and whose interior angle is less than π.

Lemma 7.1 If A is contained in a wedge, then A has finite total curvature.

Proof. Let Δ denote a wedge and suppose A ⊂ Δ. We will prove that the Gauss map on a subend of A misses a curve of values.

Consider a family F_{t} 0 < t < ∞ of parallel vertical flat annuli of distance t from γ (the image of the x_{3}-axis in N) and whose intersection with Δ gives rise to a foliation of Δ − γ by parallel compact flat annuli. Let h: Δ → ^{+} denote the level set function of this foliation and note that h|A is a proper harmonic function on A. It is known (see for example Lemma 1 in [16]) that A contains an end A′ that can be parametrized by D^{∗} = {z ∈ | 0 < |z| ≤ 1} and that h|D^{∗} = K ln |z| + K′ for some constants K and K′. In particular h|D^{∗} has no critical points, so the foliation F_{t} is always transverse to A′. Hence the normal vector field to F_{t} is never normal to A′.

Note that the Gauss map of A′ and its stereographic projection to ∪{∞} gives rise to a holomorphic map g : D^{∗} → ∪ {∞} that misses the 2 normal vectors of F_{t}. Since the integral of the Gaussian curvature on A′ equals the negative of the area (counted with multiplicity) of the Gauss map of A′, Picard’s theorem shows that at most 2 values of g can be taken finitely often. However, by changing the angle of F_{t} slightly, the above argument shows that A′ has an end A′′ such that g|A′′ omits 2 new values. As we already observed this possibility contradicts Picard’s theorem and the infinite total curvature assumption on A. □

Lemma 7.2 If S is a vertical annulus in N and A∩S = ∅, then A has finite total curvature.

Proof. Choose S_{1}, S_{2} so that S ∪ S_{1}, S ∪ S_{2}, S_{1} ∪ S_{2} are congruent wedges (cut a pie in three equal pieces). Since A is disjoint from S, we can translate A away from γ in the direction parallel to S, until ∂A is contained in the interior of the wedge W whose boundary is S_{1} ∪S_{2}. We can also assume that A intersects ∂W transversally.

We will show that A ∩ S_{1} or A ∩ S_{2} has a noncompact component. If A is disjoint from ∂W, then A ⊂ W and Lemma 7.1 shows A has finite total curvature. Hence A must intersect one of the vertical faces, S_{1} or S_{2}, of ∂W. Suppose A ∩ S_{1} is nonempty. If S_{1} ∩ A has more than one component that is compact, there would be a compact domain Σ ⊂ A with ∂Σ ⊂ S_{1}. But then there would be an interior point on Σ of maximal distance from the complete vertical annulus containing S_{1}. The existence of such a point on Σ contradicts the maximum principle. Thus, S_{1} ∩ A consists of at most one closed curve and this curve does not bound a disk on A. On the other hand if S_{1} ∩ A consists of a single compact component that is a homotopically nontrivial curve, then the end of A with boundary this curve is contained in one of the convex wedges which is impossible by Lemma 7.1. Hence we are left with the possibility that A ∩ S_{1} has a noncompact component.

Consider any noncompact proper curve in A ∩ S_{1}. This curve separates A into two components, C_{1}, C_{2}, where C_{1} is simply connected. Our earlier remarks show that C_{1} ∩ (S_{1} ∪ S_{2}) contains no compact components. We now check that C_{1} intersects W or one of the adjacent wedges W_{1} in a component C that is simply connected and has its entire boundary in S_{1} or in S_{2}. This statement is clear if C_{1} ∩ S_{2} = ∅. But if C_{1} ∩ S_{2} ≠ ∅, then C_{1} intersects the other wedge W_{2}. In this case any component C of C_{1} ∩ W_{2} suffices.

The following assertion proves that C can not exist; a contradiction from which the lemma follows.

Assertion 7.1 Suppose X is a wedge in N. If C is a properly embedded simply connected minimal surface in X with boundary in the interior of one of the faces of ∂X, then C is contained in ∂X.

Remark 7.1 One should note that a properly embedded simply connected minimal surface C in a wedge X̃ in ^{3} rather than in N whose boundary is in the interior of one of the faces of ∂X̃, is not in general contained in ∂X̃. For example, an end of one of Scherk’s surfaces, asymptotic to a half plane, can be chosen in a wedge with its boundary in the boundary of the wedge.

Proof of Assertion 7.1. Let F_{1} and F_{2} be the faces of X and assume F_{1} ∩ F_{2} = γ. Suppose ∂C ⊂ F_{1} and C is not contained in F_{1}. If F_{3} is a vertical annulus in X with ∂F_{3} = γ and F_{3} ∩ C = ∅, then we can replace X by the smaller wedge with faces F_{3}, F_{1}. This replacement shows that we may assume that X is minimal in the sense that if F is a flat annulus in X with ∂F = γ and F ∩ C = ∅, then F = F_{2}. If the angle between F_{1} and F_{2} is greater than \frac{π}{4}, then choose an F_{3} in X that is transverse to C and that makes an angle at most \frac{π}{4} with F_{2}. In this case replace C by a simply connected component in the subwedge of X bounded by F_{2} ∪ F_{3}. Clearly to derive a contradiction it suffices to prove that this component is contained in the flat annulus F_{3}, since F_{3} was choosen to be transverse to C. Thus we may assume that the angle between F_{1} and F_{2} is less than \frac{π}{4}.

For visual convenience we now change coordinates so that γ corresponds to the image of the x_{1}-axis, F_{1} is horizontal with nonnegative x_{2}-coordinate and F_{2} is a graph over F_{1}. If C were a graph over F_{1}, then it is not difficult to prove that C is contained in F_{1}. (When C is a graph we shall find a curve on C, which is a graph over part of γ of slope less than one and rising arbitrarily high. Since the length of γ is bounded, this is impossible.) We essentially reduce the general case to this graph case. (See Figure 1.)

[figure: t; F_{1}; p; x_{3}; x_{1}; γ; F_{2}; V_{t}; W_{t}; Σ_{t}; E_{t}; C]

Figure 1:

Arbitrarily choose a point p ∈ ∂C. Let V_{t} denote the compact vertical annulus in X whose x_{2}-coordinate is t. Suppose t is large enough so that the x_{2}-coordinate of p is less than t and suppose that V_{t} is transverse to C. If V_{t} ∩ C contained a simple closed curve, then this curve would bound a disk in C. Then this disk would be contained in V_{t} by the maximum principle, an impossibility. Hence V_{t} ∩ C consists of arcs whose boundary points lie in F_{1}. Let E_{t} ⊂ C denote the compact disk component of C − V_{t} that contains p. E_{t} separates the compact region of X bounded by V_{t} into two components where we denote by W_{t} the closure of the component containing F_{2}. The Geometric Dehn’s Lemma in [18] implies that ∂E_{t} is the boundary of an embedded disk Σ_{t} ⊂ W_{t} of least area in W_{t}. (If a Jordan curve on the boundary of a mean-convex domain is homotopically trivial in the domain, then it spans an embedded minimal disk.)

Remark 7.2 The maximum principle implies that either Σ_{t} equals E_{t} or Σ_{t} ∩ ∂W_{t} = ∂E_{t}. If Σ_{t} ⊄ ∂W_{t}, then Σ_{t} ∪ E_{t} is an embedded sphere in W_{t} which must bound a ball in W_{t}. In particular, any arc in W that joins a point of E_{t} to F_{2} must intersect Σ_{t}. We will have further use of this remark.

Choose t_{∘} such that V_{2t_{∘}} is transverse to C and V_{t} is transverse to Σ_{2t_{∘}} where t is approximately t_{∘}. Let D_{t} = Σ_{2t_{∘}} ∩ W_{t}. In this way, for most values of t, we produce a collection of disks D_{t} that are stable in W_{t}.

Let θ denote the angle between F_{1} and F_{2}. For η, 0 < η ≤ θ, let F_{η} denote the flat annulus in W with ∂F_{η} = γ and such that the angle between F_{2} and F_{η} is η. By our earlier choice of F_{2}, F_{η} ∩ C ≠ ∅ for η ≤ θ. Let W (η) denote the wedge between F_{η} and F_{2}.

We now apply the curvature estimates of Schoen [22]: the Gaussian curvature at a point on a stable orientable minimal surface Σ in a flat threemanifold is bounded from above in absolute value by c/d^{2} where c is a constant independent of Σ and d is the intrinsic distance from the point to ∂Σ. We will now show that these curvature estimates imply: For all δ > 0 there exists a positive ε(δ) such that if η < ε(δ) and Σ is a stable orientable minimal surface in W with boundary in F_{1}, then the normal line L_{q} to Σ at q ∈ Σ ∩ W (η) makes an angle less than δ from the normal line to F_{η}. (This means that Σ ∩ W (η) is almost parallel to F_{η}.)

We now give the proof of the above implication. Suppose that the implication were false. Then there exists a δ > 0, a positive sequence of numbers η_{i} → 0, and sequence of stable orientable minimal surfaces Σ_{i} in W, ∂Σ_{i} ⊂ F_{1} with points q_{i} ∈ Σ_{i} ∩ W (η_{i}) such that the angle L_{q_{i}} makes with the normal to F_{η_{i}} is always greater than δ. Similar statements hold by lifting everything to ^{3} so that the inverse image of W is a wedge between two half planes. We will use the same notation for the lifted surfaces and subsets in ^{3}. Since we now consider W (η) to be contained in ^{3}, it is invariant under homothety.

After a homothety, we may assume that q_{i} ∈ Σ_{i} has x_{2}-coordinate equal to 1. Since η_{i} → 0, the distance between q_{i} and F_{2} goes to zero as i → ∞ and distance of q_{i} to ∂Σ_{i} is greater than a fixed constant. The second fundamental form of Σ_{i} is uniformly bounded in geodesic coordinate systems of some fixed radius by Schoen’s estimates, and we can choose these coordinate systems to be graphs. Since q_{i} is converging to F_{2} and Σ_{i} is disjoint from F_{2}, it is evident that the angle between L_{q_{i}} and the normal vector to F_{2} is converging to zero. But the normal vector of F_{η_{i}} is converging to the normal vector of F_{2}. This contradicts the assumption that the angle between L_{q_{i}} and the normal line of F_{η_{i}} is greater than δ for all i. This contradiction proves the implication. Henceforth, we will work in N instead of ^{3}.

Since ∂Σ_{2t_{∘}} ⊂ F_{1} ∪ V_{2t_{∘}} and t is approximately t_{∘}, these same curvature estimates imply that for all δ > 0 there exists a positive ε(δ) such that if η < ε(δ) and q ∈ D_{t} ∩ W (η), then the angle between L_{q} and the normal line to F_{η} is less than δ.

Now choose δ = \frac{π}{4} and fix η, η < ε(δ). Recall C ∩ F_{τ} ≠ ∅ for all τ < θ. If C ∩ W (η) stays a bounded distance from F_{η}, then C ∩ W (\frac{η}{2}) is compact which is impossible by the maximum principle. Hence we can pick a point q ∈ C ∩ W (η) such that the distance of q to F_{η} is greater than the length of γ. Choose t large enough so that q ∈ E_{t}. In particular q ∈ E_{2t}. Let ℓ denote the line segment joining q to F_{2} and that is orthogonal to F_{η}. By Remark 7.2, ℓ must intersect D_{t} in a point q_{1} whose distance from F_{η} is greater than the length of γ.

Let H be the flat annulus in W (η) containing ℓ and whose boundary consists of a circle on F_{2} parallel to γ and a circle on F_{η}. We can assume that H is transverse to D_{t}. Let α be the component of H ∩ D_{t} that contains the point q_{1}. Change coordinates in W (η) (by rotation around γ) so that F_{η} is horizontal and H is vertical. Since δ < \frac{π}{4}, the normal line to D_{t} along α makes an angle of less than \frac{π}{4} with the (new) vertical. Hence the slope of the tangent line along α is always less than one. If α were a closed curve, then it would bound a disk on D_{t} and this disk would be contained in H by the maximum principle, an impossibility. Hence, α is an arc with two boundary points on F_{η}. Since α is embedded, it is a graph over the circle H ∩F_{η}. Since the slope of α is less than 1, its maximum height can be at most the length of H ∩ F_{η} which equals the length of γ. But q_{1} ∈ α has height greater than the length of γ. This contradiction completes the proof of Assertion 7.1. As remarked before, the assertion proves Lemma 7.2. □

Lemma 7.3 If A is trapped between standard ends that are Scherk type ends, then A has finite total curvature.

Proof. By Lemma 7.2 we need only show that A is disjoint from some vertical flat annulus S. Suppose that A is trapped between standard ends that are Scherk type ends E_{1}, E_{2}. In this case E_{1} is asymptotic to a flat vertical annulus S_{1} and E_{2} is asymptotic to a flat vertical annulus S_{2}. If S_{1} and S_{2} are disjoint, then we can clearly find the desired annulus S, so we must show that S_{1} and S_{2} are disjoint. Assume on the contrary that S_{1} ∩ S_{2} ≠ ∅. Since vertical flat annuli in N intersect in a compact set or the intersection contains a noncompact subannulus, we can take S_{1} = S_{2}.

At this point, we could appeal to the maximum principle at infinity given in [17], which implies E_{1} and E_{2} can not be asymptotic at infinity and be disjoint. However, a direct proof is rather easy here so we proceed with the proof.

Since E_{1}, E_{2} are standard ends, we can choose them, by replacing them by subends, so that they can be expressed as graphs over S_{1} tending to zero. Without loss of generality we may assume that S_{1} is the flat vertical annulus whose inverse image in ^{3} is the half plane with boundary the x_{3}- axis and containing the positive x_{2}-axis. Assume that the x_{1}-coordinate of E_{1} is greater than the x_{1}-coordinate of E_{2}. Choose a small ε > 0 and ε < dist(∂E_{1}, ∂E_{2}), and so that E′_{1} = E_{1} + (−ε, 0, 0) intersects E_{2} transversely. By the classical maximum principle, E′_{1} ∩ E_{2} does not contain a component that bounds a compact subdomain of E′_{1} or of E_{2}. Hence there are annular subends Ẽ_{1} ⊂ E′_{1}−E_{2} and Ẽ_{2} ⊂ E_{2}−E′_{1} with common boundary curve γ. Let n⃗_{i} denote the inward pointing unit conormal vector field along the boundary of the surface Ẽ_{i}. Since the x_{1}-coordinate of Ẽ_{1} is less than the x_{1}-coordinate of Ẽ_{2}, grad(x_{1}|∂Ẽ_{1}) · n⃗_{1} < grad(x_{1}|∂Ẽ_{2}) · n⃗_{2}. Integrating this inequality, we obtain inequalities on the fluxes of grad(x_{1}) across the common boundary of Ẽ_{1} and Ẽ_{2}:

a_{1} = ∫_{∂_{E_{1}}} grad(x_{1}|Ẽ_{1}) · n⃗_{1} < a_{2} = ∫_{∂_{E_{2}}} grad(x_{1}|Ẽ_{2}) · n⃗_{2}.

Since the coordinate functions of a minimal surface in ^{3}/T are harmonic, the divergence theorem implies that the flux of a harmonic function is constant on homologous cycles. Now by Proposition 4.1, grad(x_{1}|Ẽ_{1}) converges uniformly to zero as x_{2} tends to infinity. The number a_{1} equals the flux of x_{1} across a cycle on Ẽ_{1} defined by x_{2} equals constant, the constant arbitrarily large. Since these cycles are of bounded length, it follows that a_{1} (and a_{2} by the same reasoning) is zero, a contradiction. □

8 The proof of Theorem 1.

Suppose M is a properly embedded minimal surface in a complete flat nonsimply connected three-manifold N and suppose that M has finite topology. We shall prove that M has finite total curvature. Since N is finitely covered by a flat three-torus, × , or by ^{3}/S_{θ}, we may assume, after lifting M to a covering space, that N is × or ^{3}/S_{θ} where θ = 0 or θ is irrational (clearly M is compact if one is in a flat three-torus). Now it suffices to prove each annular end A has finite total curvature. The theorem in × can be reduced to the theorem in ^{3}/S_{θ} since if A is an annular end in × , then π_{1}(A) is contained in an infinite cyclic subgroup of π_{1}( ). Hence, one can lift A to a covering space (S^{1} × ) × of × so that the lifted A is an annular end of a properly embedded minimal surface M̃ that covers M.

By Remark 6.1, Lemma 6.1, and the Trapping Lemma, we can assume (after passing to a finite covering) that each annular end of M (or M̃) is trapped between two standard ends in N. The Trapping Lemma (Lemma 6.2) shows that we may assume that A is trapped between 2 standard ends that are helicoidal, planar or are Scherk type ends. Note that Scherk ends can only occur for θ rational. Lemma 7.3 shows that Theorem 1 is true when A is trapped between Scherk type ends.

To complete the proof of Theorem 1, we must prove that an embedded annular end A that is trapped between two helicoidal (perhaps with logarithmic growth) or planar standard ends, E and D, is of finite total curvature.

Assume the limiting normal vector to E is vertical and E is never vertical. We will show that there exists an annular subend A′ of A that never has a vertical tangent plane. Then a lifting of A′ to ^{3} has the same property so the lifting is stable (its Gaussian image is contained in a hemisphere) Hence A is stable also and thus has finite total curvature. The construction of A′ will involve interesting geometric constructions and occupy all of Section 8.

Let B = {(x_{1}, x_{2}) | x_{1}^{2} + x_{2}^{2} ≥ 1} and i: B → E be a parametrization of E, sending the circles of radius R in B to E ∩ ∂T_{R}, i.e., the helices of E for large R. We work in the manifold W = B × with the flat metric induced by the submersion: (x_{1}, x_{2}, t) ↦ i(x_{1}, x_{2}) + (0, 0, t) ∈ N.

Observe that this metric on W is asymptotic to the product flat metric (a flat metric on B with ) since the metric on the end E is asymptotically flat. Notice that the region trapping A in N, i.e., the region bounded by E, D and a large compact cylinder (parallel to the period vector v) lifts isometrically to W and A is in this region as well. Clearly vertical lines in N correspond to vertical lines in W, hence it suffices to find a never vertical subannulus in the lifting of A to W. Henceforth, we shall work in W. For simplicity, assume that the trapping region lifts to a subdomain of B × [0, 1] with E lifting to B × {0}. Let B_{t} ⊂ W be B × {t}.

Proposition 8.1 There exists a compact subset K_{1} of W (1) = B×[0, 1] such that for any other compact subset K_{2} containing K_{1}, the following statement holds: For every x ∈ W (1), sufficiently far from K_{2}, and for every vertical plane P_{x} at x, there is a foliation F of a neighborhood of W (1) − K_{1}, such that:

1. the leaves of F are compact minimal annuli F_{t}, 0 ≤ t < ∞, with one boundary component in B_{0}, and the other boundary component in B_{1};

2. x ∈ F_{1} and the tangent plane of F_{1} at x is P_{x};

3. F_{1} ⊂ W (1) − K_{2};

4. F_{0} ⊂ K_{1}.

Remark 8.1 In fact K_{1} will be the compact region of W (1) bounded by a stable minimal annulus whose boundary consists of a circle of radius R in B_{0} and its parallel translate to B_{1}, for some large R. Outside of some larger compact set, the leaves of F will be compact annuli with boundary; circles of larger radius, one in B_{0} and the other its parallel translate to B_{1}.

Before proving Proposition 8.1, we show why it implies A contains a subannulus that is never vertical.

Let h: W (1) → [0, ∞) be a proper function. Let T_{1} be a regular value of h such that K_{1} ∪ ∂A ⊂ h^{−1}[0, T_{1}]. Choose T_{2} > T_{1} a regular value, such that the component of A∩h^{−1}[0, T_{2}] that contains ∂A, also contains A∩h^{−1}[0, T_{1}]. Let K_{2} = h^{−1}[0, T_{2}]. Note that any compact subdomain of A whose boundary is contained in W (1) − K_{2}, is disjoint from K_{1}.

Now apply Proposition 8.1 to K_{1} and K_{2}. If A were vertical at points arbitrarily far from ∂A, then there is an x ∈ A sufficiently far from ∂A to which we apply the proposition. Let F be the foliation, F_{1} the minimal annulus in W (1) − K_{2} such that F_{1} is tangent to A at x. Then A ∩ F_{1} is a compact singular one cycle in A (a singularity at x) and A − F_{1} contains a component Δ with compact closure, with boundary in F_{1} ⊂ W (1) − K_{2}. By our choice of K_{1}, K_{2}, we have Δ disjoint from K_{1}. Now there is a largest (or smallest) value t such that Δ ∩ F_{t} is nonempty. At such a point Δ is on one side of F_{t} and this contradicts the maximum principle.

Hence it remains to prove Proposition 8.1. This will be carried out in a series of lemmas.

Let C be a catenoid in ^{3} with waist circle at height t = \frac{1}{2} and let S_{0}, S_{1} be the circles of C at heights 0 and 1 respectively; each of radius R_{∘}. Assume R_{∘} is sufficiently large so that the angle that the normal vector to C, along S_{0} ∪ S_{1}, makes with the horizontal is less than π/8. We will work in regions of W where the vertical cylinder of height three and radius 3R_{∘} isometrically embeds in ^{3}. Then the catenoid C isometrically embeds in this region of W as a vertical stable catenoid.

Let L_{0} be a smooth simple closed curve in B_{0} and L_{1} the vertical translate of L_{0} to B_{1}. For each x ∈ L_{0}, we consider the two catenoids C_{x}, parallel translations of C, which contain x and x + (0, 0, 1) in their boundary and whose tangent line γ_{x} to ∂C_{x} at x is the horizontal projection of the tangent line of L_{0}.

Figure 2:

Let ℓ_{x} be the horizontal line of length 2R_{0}, centered at x and normal to γ_{x} at x. Let W_{x} be the vertical strip over ℓ_{x} of height one; W_{x} is a rectangle of base ℓ_{x} and side one.

Let β_{x} = W_{x} ∩ C_{x}; β_{x} is a Jordan curve, smooth except at x and x + (0, 0, 1). β_{x} consists of two meridian curves on C_{x} (one on each catenoid of C_{x}) joining x to x + (0, 0, 1). Clearly β_{x} bounds a disk in W_{x}.

Now define the torus barrier T = T (L_{0}) to be ⋃_{x∈L_{0}}β_{x}. In general, T is neither embedded nor a barrier, however, if it is, then we have the following lemma.

Lemma 8.1 Suppose the torus barrier of L_{0} is embedded and mean convex (i.e. the mean curvature vector of T − (L_{0} ∪ L_{1}) points into the solid torus S bounded by T). Then L_{0} ∪ L_{1} is the boundary of a stable embedded minimal annulus in S, and any embedded minimal annulus in S with boundary L_{0}∪L_{1} is stable.

Remark 8.2 We will use this lemma to construct the foliation of Proposition 8.1. We will construct a foliation of B_{0} by simple closed curves L_{0}(s), 1 ≤ s < ∞, and the vertical translation L_{1}(s) will foliate B_{1}. Lemma 8.1 will be used to show that L_{0}(s) ∪L_{1}(s) bounds a unique stable minimal annulus and these annuli will foliate a region of W. The difficulty in this construction is to appropriately construct the foliation L_{0}(s) in order to obtain a vertical tangent plane of the annulus A as defined in Proposition 8.1.

Proof. The angle of ∂S − T along L_{0} ∪ L_{1} is always less than π so by [18], ∂S is an appropriate barrier for solving the Plateau problem in S. Let Σ be a least-area annulus with ∂Σ = L_{0} ∪ L_{1}(L_{0} is homotopic to L_{1} in S and L_{0} is not null homotopic in S). By the Geometric Dehn’s Lemma [18], Σ is embedded.

It remains now to show an embedded minimal annulus Σ in S with boundary L_{0} ∪ L_{1} is stable. Since the vertical makes sense in W, the angle the normal vector to Σ makes with (0,0,1) is a well defined function on Σ, hence log |g| is a well defined harmonic map on Σ (where Σ is not horizontal); g the Gauss map. Notice that |g| is well defined even though g is multi-valued.

Now if Σ is never horizontal, then the angle between the normal lines to Σ and the vertical vary between 3π/8 and 5π/8. To see this, notice that the normal lines along ∂Σ have this property because Σ is between two catenoids with normal lines making angles with the horizontal at most π/8. Also log |g| is harmonic on Σ, hence the maximum and minimum values are assumed on ∂Σ.

Let Σ̃ be a connected lifting of Σ to ^{3} and let D ⊂ Σ̃ be a compact domain. The Gauss map of Σ̃ (hence of D as well) takes its values in a band about the equator whose maximum angle with the equator is π/8. In particular, the area of the spherical image of D (not counted with multiplicity) is less than 2π. By the theorem of Barbosa-Do Carmo, D is stable, hence Σ̃ too [1]. Since Σ̃ covers Σ, Σ is stable as well (see [4] or [7]).

It remains to show Σ is never horizontal. The proof will use a winding number argument. Let S̃ be the universal covering space of S and Σ̃ ⊂ S̃ the lifting of Σ. On Σ̃ we have a well defined meromorphic Gauss map g. Let α be an embedded arc on Σ̃ joining a point x ∈ L̃_{0} to x+(0, 0, 1) ∈ L̃_{1}, disjoint from the zeros and poles of g. Let σ be a nontrivial covering transformation of Σ̃. Now consider the disk D bounded by α, σ(α), an arc ℓ_{0} on L̃_{0} and its’ parallel arc ℓ_{1} on L̃_{1}. We claim that for any y ∈ L̃_{0}, g(y) and g(y + (0, 0, 1)) have arguments whose difference is less than π. To see this, observe that Σ̃ separates S̃ into two components and the normal vector of Σ̃ points into the same component. Call this component B and let M = ∂B − Σ̃. At y ∈ L̃_{0}, we have a horizontal unit vector v(y) that is normal to L̃_{0} at y and the scalar product of v(y) with the exterior normal (to B) of M at y is positive. Notice that v(y + (0, 0, 1)) is the parallel translation of v(y) to y + (0, 0, 1).

The angle between Σ̃ and M at y is less than π/8 and g(y) and v(y) are both orthogonal to L̃_{0}, so the scalar product of g(y) and v(y) is positive. Similarly g(y + (0, 0, 1)) and v(y) have a positive scalar product. Hence g(y) and g(y + (0, 0, 1)) lie in the same open hemisphere so their arguments (thought of as complex numbers after stereographic projection) differ by at most π.

The Gauss map g is never 0 or ∞ on ∂D. Observe that g restricted to ∂D has degree zero, thought of as a map into ^{∗} = − {0}. To see this one calculates the winding number. For z ∈ α, g(z) and g(σ(z)) differ by rotation by a fixed θ_{0} (θ_{0} is a multiple of the angle of the flat structure, 3/S_{θ}, on W). So the total change of the argument arising by traversing α and then –σ(α) is zero. For y ∈ ℓ_{0}, the argument of g(y) and g(y + (0, 0, 1)) differ by less than π. Hence the total change of argument as one traverses ℓ_{0} and then –ℓ_{1} is less than π. Consequently the total winding number (in absolute value) is less than π, hence zero. Now g is conformal so every value has positive degree. Hence g misses 0 and ∞ and Σ is never horizontal. This completes the proof of Lemma 8.1. □

In the following lemma we will give a natural condition on a curve L_{0} ⊂ B_{0} which ensures that the torus barrier is, in fact, embedded.

Note that for each x ∈ W sufficiently far from ∂W, the vertical cylinder V_{x} of radius 3R_{∘}, centered at the vertical line through x, is embedded in W.

Let D_{x} ⊂ V_{x} be the horizontal disk of radius 3R_{∘} centered at x. (Note that D_{x} is not part of a B_{t}.)

Lemma 8.2 Let L_{0} ⊂ B_{0} be a smooth simple closed curve. For x ∈ L_{0}, suppose the vertical projection Γ_{x} of L_{0}∩V_{x} to D_{x} satisfies: the disks bounded by the circles in D_{x}, of radius R_{∘} and tangent to Γ_{x} at x (one on each side of Γ_{x}), intersect Γ_{x} at x only. Then the torus barrier of L_{0} is embedded.

Proof. For each x ∈ L_{0}, let ℓ_{x} denote the normal line to Γ_{x} at x, centered at x and of length 2R_{∘}. Observe that if x ≠ y, x, y ∈ L_{0}, then the vertical strips W_{x} and W_{y} over ℓ_{x} and ℓ_{y} are disjoint. For if they intersect, then ℓ_{y} ⊂ V_{x} and the vertical projection ℓ̃_{y} of ℓ_{y} onto D_{x}, must intersect ℓ_{x} at a point z. Let a and b be the points of ℓ_{x} and ℓ̃_{y} in Figure 3.

[figure: a; b; z; Π(y); x; ℓ_{x}; ℓ̃_{y}]

Figure 3:

Let Π denote orthogonal projection onto D_{x}. Assuming d(z, b) < d(z, a), we have: d(a, z) + d(z, Π(y)) = R_{∘}, d(b, Π(y)) ≤ d(b, z) + d(z, Π(y)), hence d(b, Π(y)) < R_{∘}. But then Π(y) is in the disk of radius R_{∘} centered at b and Π(y) ∈ Γ_{x}. This contradicts our hypothesis and hence the vertical strips over the ℓ_{x} are pairwise disjoint.

Now consider the construction of our torus barrier using the catenoids C_{x}, for x ∈ L_{0}. Each C_{x} (recall there are two catenoids in C_{x}) intersects the vertical strip over ℓ_{x} in an embedded curve β_{x}, that bounds a disk in this strip. The union of these curves is the embedded torus barrier. □

Remark 8.3 We claim that if L_{0} is close to a horizontal plane P and if the injectivity radius of L_{0} is large, then the hypothesis of lemma 8.2 are satisfied hence the torus barrier of L_{0} is embedded. More precisely there is a C > 0 and ϵ > 0 such that if the injectivity radius of L_{0} is greater than C and if L_{0} is ϵ − C^{2} close to P (i.e. the distance of L_{0} to P is less than ϵ and the curvature and torsion of L_{0} are less than ϵ) then the hypothesis of lemma 8.2 are satisfied.

To see this, for x ∈ L_{0}, consider the solid vertical cylinders V_{x}, of radius R_{0}, tangent to Γ_{x} at x (there are two of these cylinders). We choose ϵ small enough so that the osculating plane of L_{0} is always within π/4 of the horizontal. L_{0} does not enter V_{x} at x, otherwise the curvature of L_{0} at x would be greater than the curvature of the helice on ∂V_{x}, making a constant angle π/4 with the horizontal. Since the curvature of this helice only depends on R_{0}, ϵ can be chosen so this is impossible.

Now choose C large enough so that the tubular neighborhood of L_{0}, of radius 2R_{0} along a fixed (small) arc on L_{0}, centered at x, contains V_{x}. We need work with V_{x} of height R_{0} (since L_{0} is within R_{0} of P) so such a choice of C is possible. Then the only point of L_{0} in V_{x} is x and the hypothesis of lemma 8.2 are satisfied.

Now when will the torus barrier T of L_{0} be mean convex. We claim that if L_{0} is close to a plane curve (in the C^{2}-topology) and if the curvature of L_{0} is small enough then T is mean convex. Consider first, a plane curve L_{0}. For x ∈ L_{0}, γ_{x} has two smooth arcs, each joining x to x+(0, 0, 1). One is an outside arc a(x) and the other an inside arc b(x); i.e., b(x) is on the side of L_{0} to which L_{0} is curving at x. Clearly, along the outer arc a(x), the mean curvature vector is pointing inside T; T is even locally convex along a(x). At a point y on b(x), T will be mean convex if one can find two orthogonal directions such that the sum of the normal curvatures in these directions has the right sign, the same sign as that of the normal curvature of b(x) at y. Clearly if the curvature of L_{0} at x is sufficiently smaller than the curvature of b(x) at y (both in absolute value) then this will be satisfied. Also one can bound the mean curvature away from zero by choosing the curvature of L_{0} small.

Now if L̃_{0} is C^{2}-close to a horizontal curve L_{0} and if the curvature of L̃_{0} is sufficiently small, then T (L̃_{0}) will be mean convex as well. We saw in the last paragraph that bounding the mean curvature of T (L_{0}) away from zero only depended on the curvature of L_{0} being small.

We shall say a curve L_{0} is R_{0}-admissible, if T (L_{0}) is embedded and mean convex. In the sequel we shall work in regions of W in the complement of the tubular neighborhood of radius R about the period vector. As R → ∞, the metric in this exterior domain converges to the flat product metric on B × . The curves L_{0} we will work with will be in the exterior domain and contained in B_{0} = B × (0). By choosing the injectivity radius of L_{0} large, and R large, we will have L_{0} a R_{0}-admissible curve.

Lemma 8.3 Let L_{0}(s) be a family of simple closed curves in B_{0}, 0 ≤ s ≤ 1, that foliate an annulus of B_{0}. Let L_{1}(s) denote the foliation in B_{1} obtained by vertical translation of the curves L_{0}(s). For each s, assume the torus barrier T (s) of Lemma 8.1, defined by L_{0}(s) ∪ L_{1}(s), is embedded and mean convex. Let S_{0} be a stable minimal annulus in T (0) with boundary L_{0}(0) ∪ L_{1}(0).

Then there is a foliation S(s) by stable minimal annuli satisfying:

1. S(0) = S_{0},

2. ∂S(s) = L_{0}(s) ∪ L_{1}(s),

3. S(s) is in T (s).

Proof. Let s be between 0 and 1. First observe that if Σ is a minimal annulus in T (s) with boundary L_{0}(s) ∪ L_{1}(s), then along L_{0}(s) ∪ L_{1}(s), Σ makes a strictly positive angle with T (s), and the interior of Σ is in the interior of T (s). This follows from the boundary maximum principle and the maximum principle.

Now for s near 0, L_{0}(s) ∪ L_{1}(s) bounds a stable minimal annulus S(s), since a stable minimal surface varies smoothly with a smooth change of boundary data. This type of result can be found in [25] or [27]. By our previous paragraph and since the T (s) vary smoothly, S(s) is contained in T (s) for s near 0. Since the variation vector field is a Jacobi field that is never zero on the boundary (L_{0}(0) ∪ L_{1}(0)), by stability (the index theorem) it can not vanish inside. Hence, the family of surfaces that one obtains by moving along the variation vector field at time s is indeed a foliation for s near zero.

It remains to show the set of s for which the foliation exists is closed. So assume the foliation exists and satisfies 1, 2 and 3 for s < τ. We know that T (s) converges to T (τ). A subsequence S(s_{n}) converges to a minimal annulus S(τ) in T (τ). By Lemma 8.1, S(τ) is stable. By the openness property, S(τ) is part of a foliation near S(τ). Since S(s_{n}) converge to S(τ), the maximum principle implies S(s_{n}) must be a leaf of this foliation for n large. Hence S(s) converges to S(τ) as s → τ. This proves Lemma 8.3. □

Before proving Proposition 8.1, we will describe the idea of the proof in the special case when E is a flat horizontal annulus. The general situation is a metric perturbation of this case near infinity but it will help the reader to consider the special case of a flat annulus first.

So assume E = {(x, y, z) | z = 0, x^{2} + y^{2} ≥ 1} and W = E × . Suppose there exists a sequence p_{n} ∈ E × [0, 1] diverging to ∞, and a sequence of vertical planes P_{n} at p_{n}. Here is a resumé of what we shall do next. We construct an R_{∘}-admissible curve L as follows. Let Γ be a planar convex curve as in Lemma 8.4. We take a long arc on Γ, centered at q, and join it to a convex curve as in Figure 4.

We translate L in E_{0} so that q is near p_{n}, and then rotate about the vertical line through p_{n}, so that the new curve L so obtained satisfies: L∪L_{1} bounds a stable minimal annulus F_{1} and F_{1} is tangent to P (n) at p_{n}. (In fact one will be obliged to move F_{1} vertically to realize this tangency, since p_{n} is not necessarily at height z = \frac{1}{2}.)

Next construct a foliation in E_{0} by R_{∘}-admissible Jordan curves L_{0}(s), 0 ≤ s < ∞, such that L_{0}(0) is the circle of radius 2 centered at the origin, L_{0}(1) is the L we constructed above and L_{0}(s) for large s is also a circle. Now apply Lemma 8.3 where S_{0} is a catenoid. Observe that F_{1} is necessarily a leaf of the foliation given by Lemma 8.3. This follows from the maximum principle. This foliation contradicts our assumption so Proposition 8.1 follows.

We remark that the boundary planes we worked with in the above argument were E_{0} and E_{1}. In fact, when we prove Proposition 8.1, we will need to work at heights, such as −1 and 2, to acquire the tangency at p_{n}. Also, our construction of L must be done with great care since the metric on E is not flat in the general case. Instead it is asymptotic to a flat metric which is why we will construct a sequence of foliations F(n), working at p_{n} when n is large. This is the end of the resumé.

Now we continue with the proof of the case E a flat annulus; i.e., we shall make precise the previous resumé!

We now need a technical lemma which is not difficult to prove but is necessary for our proof.

Lemma 8.4 There is a planar curve Γ contained in the positive quadrant of the (x, y) plane having the following properties:

1. Γ is convex, asymptotic to the x and y axis and invariant under (x, y) → (y, x);

2. Γ is R_{∘}-admissible, i.e., Γ and its parallel translate Γ_{1} to E_{1} define an embedded envelope E(Γ). Topologically E(Γ) is S^{1} × , and E(Γ) is defined as in Lemma 8.1 using the catenoids C_{x};

3. Γ∪Γ_{1} bound a unique, area minimizing strip M(Γ), contained in E(Γ);

4. M(Γ) is invariant by reflection in E_{1/2} and by reflection in the vertical plane y = x.

Notice that Property 4 implies that M(Γ) is vertical along E_{1/2}, (cf. Figure 5).

[figure: E_{1/2} ∩ M(Γ)] Figure 5:

Proof. Let A be the infinite strip {0 ≤ z ≤ 1, y = −x}. Γ is constructed as a graph over one of the boundary components of A so as to satisfy conditions 1 and 2. Then Γ_{1} is the same graph over the other component of ∂A. It is known that every continuous function on ∂A extends to a solution of the minimal surface equation in A, so M(Γ) is the graph of this solution [5]. Using catenoids as barriers above and below the graph of Γ ∪ Γ_{1} it’s easy to see that M(Γ) is contained in the torus barrier T (Γ). To see that M(Γ) is unique in T (Γ), one reasons as follows. Let M be any other minimal surface in T (Γ) with boundary Γ∪Γ_{1}. A straightforward application of the Alexandrov reflection principle, using planes parallel to A, shows M is also a graph over A (see [23] for this type of argument). Now one has two minimal graphs over A, with the same boundary values and whose difference is bounded. It is known that this implies they are equal [5]. Since minimal graphs are area minimizing, we have proved Property 3. Property 4 follows from unicity. □

We will need to fix a base point q on Γ. We take q to be the intersection of Γ with y = x.

Remark 8.4 Lemma 8.4 also holds in a sector whose angle is almost Π; so the curvature of Γ can be as small as desired.

Replacing p_{n} by a subsequence, we can assume the vertical cylinder V_{3n}(p_{n}), of radius 3n and centered at the vertical line through p_{n}, embeds in W (2) = E × [−1, 2].

Translate V_{3n}(p_{n}) horizontally and rotate so that the plane P_{n} becomes the plane y = −x and p_{n} is on the z-axis. Let f_{n} denote this rigid motion of V_{3n}(p_{n}) and let V_{R} be the vertical cylinder of radius R centered at the z-axis.

We construct a foliation G of the planes {z = −1} ∪ {z = 2} as follows. Foliate z = 0 by translating Γ along the line y = x. G is obtained by parallel translation (vertically) of this foliation to the planes {z = −1} ∪ {z = 2}. Note that G is the boundary of a foliation of ^{2} × [−1, 2] by minimal strips {M(Γ)}, parallel translates of a fixed M(Γ). This M(Γ) is the same as in Lemma 8.4 except that the planes z = 0 and z = 1 have become z = −1 and z = 2, respectively.

We construct a foliation G(n) of part of {z = −1} ∪ {z = 2} by Jordan curves as follows. In the top and bottom of V_{2n}(p_{n}), G(n) is the foliation by arcs f_{n}^{∗}(G). In the complement of the top and bottom of V_{3n}(p_{n}), G(n) is the foliation by circles centered at the z-axis. In the top and bottom of V_{3n}(p_{n}) − V_{2n}(p_{n}), G(n) is a foliation by arcs so that the resulting foliation is a foliation by R_{∘}-admissible Jordan curves; Figure 6. It is not hard to show G(n) exists for n large. We choose the innermost circle of G(n) to be of radius ten; this guarantees that the stable catenoid, whose boundary is these circles of radius ten, has a waist circle of radius greater than one.

We need to choose the R_{∘} of the last paragraph so the stable vertical catenoid bounded by two circles of radius R_{∘}, and of height 3, exists, and makes an angle less than π/8 with the horizontal along its boundary. Henceforth, we work with this value of R_{∘}.

Now apply Lemma 8.3 to G(n); there is a foliation F(n) by stable minimal annuli F_{t}, 0 ≤ t < ∞, inducing G(n) on the boundary. Notice that each F_{t} intersects each horizontal plane in a simple closed curve, since the foliation of F_{t} induced by the horizontal planes can only have hyperbolic singularities (maximum principle), so it has no singularities.

We will prove the following assertion:

Assertion 8.1 Let q_{n} be the vertical projection of p_{n} onto E_{1/2}. The trace of F(n) on the intersection of the plane P_{n} with the vertical cylinder V_{n}(q_{n}), has a unique singularity, q̂_{n}, that is near q_{n}, for n large.

Before proving this we will explain how this assertion completes the proof of Proposition 8.1 in the special case E is flat. First we arrange so that q̂_{n} is on the same vertical line as q_{n}. To do this, one redoes the construction of the foliation G starting with a curve Γ(s) in z = 0 where Γ(s) is the curve Γ translated a distance s along the tangent line to Γ at q, −1 ≤ s ≤ 1. Then the foliation F_{n}(s) so obtained will yield a unique point q̂_{n}(s), which is the singularity the trace foliation F_{n}(s) induces on P_{n} ∩ V_{n}(p_{n}). This point q̂_{n}(s) is near q_{n}(s), for n large, where q_{n}(s) is the translation of q_{n} a distance s along P_{n} ∩E_{1/2}. Since F_{n}(s) varies continuously with s, the points q̂_{n}(s) vary continuously with s. So for an appropriate choice of s, q̂_{n}(s) and q_{n} will be on the same vertical line. For convenience, assume s = 0, and the leaf of q̂_{n}(s) is labelled F_{1}.

Once q̂_{n} and q_{n} are on the same vertical line, one does a vertical translation of F_{n} to make q̂_{n} coincide with p_{n}. It’s easy to see the translated foliation has a trace on W (1) = E × [0, 1] as desired. K_{1} can be chosen to be the vertical cylinder of radius ten, centered at the z-axis, intersected with W (1). The points p_{n} diverge so one constructs the foliation of Jordan curves ∂F_{n} so that the leaves passing through the top of V_{n}(p_{n}) are outside K_{2} and their torus barriers are also outside of K_{2}. This guarantees that F_{1} ⊂ W (1) − K_{2}. Proof of Assertion 8.1. Let R_{1} be the rectangle P_{n} ∩ V_{2n}(p_{n}). For points x on R = P_{n} ∩ V_{n}(p_{n}), far enough from the vertical line through q_{n}, the leaf F_{t}(x) of x is in the torus barrier T of its’ boundary curves. By construction T then intersects R_{1} in disks D_{1}, D_{2} each of whose boundary is smooth except along two points, one on the top of R_{1}, the other on the bottom. Clearly F_{t}(x) intersects D_{1} ∪ D_{2} transversally and joins the top of R_{1} to the bottom of R_{1}. This shows the trace of F(n) on R has at least two nonsingular leaves as in Figure 7.

[figure: q_{n}^{−}; Figure 7:; q̂_{n}; R; q_{n}^{+}; γ_{1}; γ_{2}; q_{n}; x]

The trace foliation near the segments γ_{1} ∪ γ_{2} is as in Figure 7, since P_{n} is transverse to the foliation ∂F_{n} except at q_{n}^{+} and q_{n}^{−}. This boundary data and the fact that minimal surfaces must have hyperbolic contact with R guarantees that there is exactly one singularity q̂_{n} of the trace foliation. It remains to prove q̂_{n} is near q_{n} for n large.

To show q̂_{n} is near q_{n} we will show the foliation H(n) ∩ V_{10} converges to the foliation {M(Γ)} ∩ V_{10}, where H(n) is the image of F(n) by f_{n}. Clearly, the torus barriers of the upper and lower trace leaves of H(n) converge to the corresponding torus barriers of the curves of G. This will imply the leaves of H(n) converge to the foliation, extending G, whose leaves are the parallel translates of M(Γ). Since H(n) induces a foliation of a convex compact region of ^{3} by compact minimal leaves, each leaf is area minimizing relative to its boundary. Hence, any subsequence of the leaves of H(n) whose boundaries converge to a leaf L_{−1} ∪ L_{2} of G contains a convergent subsequence, that converges to a minimal surface which is contained in the torus barrier of L_{−1} ∪ L_{2}. We showed earlier that there is a unique such surface in this torus barrier (our graph argument and the Alexandrov reflection principle), and it is a translate of M(Γ). Hence, the leaves of H(n) converge to parallel translates of M(Γ). The Gauss map of M(Γ) is injective in a neighborhood of q_{n} so the same is true for nearby minimal surfaces. It is now clear that q̂_{n} is near q_{n} as n → ∞. This proves Assertion 8.1 and completes the proof of Proposition 8.1 when E is a flat annulus. □

Proof of Proposition 8.1. Now we work in W (2) = B × [−1, 2], with the flat metric induced from the map (x, t) ↦ i(x)+(0, 0, t). The torus ∂T_{R} (the boundary of an R-tubular neighborhood about the period vector v) intersects E in a simple closed curve C(R), for R large, whose geodesic curvature tends to 0 as R → ∞. By choosing a subend of E, we can assume E is foliated by the C(R), 1 ≤ R < ∞, (we work from R = 1 for notational convenience). We can also assume that for R ≥ 10, the vertical translates C_{−1}(R), C_{2}(R) of C(R) to B_{−1} and B_{2} are R_{∘}-admissible. Finally, assume C(10) has the property that the torus barrier of C_{−2}(10) ∪ C_{1}(10) is contained in T_{20}. Then define K_{1} to be T_{20} ∩ W (1). We will show that for every divergent sequence of points p_{n} ∈ W (1) and vertical planes P_{n} at p_{n}, there is a foliation F_{n} for n large, satisfying the conclusion of 8.1.

Consider such a sequence of planes P_{n} and points p_{n}. Replacing p_{n} by a subsequence, we can assume the vertical cylinder V_{3n}(p_{n}) in W (2), of radius 3n and centered at the vertical line through p_{n}, embeds in W (2), and isometrically embeds in ^{3}.

Isometrically embed V_{3n}(p_{n}) in ^{3} so that the vertical line through p_{n} goes to the z-axis and the origin corresponds to a point q_{n} on B_{1/2}, and P_{n} goes to the vertical plane through the line y = −x. Let V_{3n} denote the image cylinder of radius 3n in ^{3}. Consider the foliation of the (x, y)-plane by the parallel translates of Γ along y = x. Notice that y = −x is tangent to a leaf of this foliation at (0, 0). Pull back this foliation, to the top and bottom disks of V_{3n} by vertical projection onto the (x, y)-plane. Now consider the induced foliation of the top and bottom of V_{2n}(p_{n}), back in W (2).

Let D_{r} be the bottom disk of B_{−1} ∩V_{r}(p_{n}) for r ≤ 3n. Now foliate B_{−1} as follows: in D_{2n} we take the above induced foliation of the previous paragraph. In B_{−1} − D_{3n} we take the foliation by the curves C_{−1}(R), 10 ≤ R < ∞, and then fill in the foliation in D_{3n} − D_{2n} so that each leaf L of this foliation, is a simple closed curve, and L together with its vertical translation to B_{2} is R_{∘}-admissible.

Then by Lemma 8.3, there is a foliation F(n) in W (2) by stable minimal annuli F_{t}, 0 ≤ t < ∞, inducing the previous foliation in B_{−1} ∪ B_{2}. Notice that each F_{s} intersects each B_{t} in a simple closed curve since the induced foliation of F_{s} can have only hyperbolic singularities.

Now the top and bottom of V_{2n}(p_{n}) converge, in the isometric embedding into ^{3} to horizontal planes at heights 2 and −1 respectively, as n → ∞. This is because the geometry of a standard end E converges to the Euclidean metric near infinity. Therefore the foliation F(n) in V (n), when viewed in ^{3} under the isometric embedding, converges to the foliation given by parallel translates by M(Γ). The same argument as in our special case E flat, proves the above assertion.

It now follows that there is a unique point q̂_{n} ∈ V_{2n}(p_{n}), near q_{n}, such that the leaf of F(n), through q̂_{n}, is tangent to P_{n}. Just as in the special case E flat, we can modify F(n) in V_{2n}(p_{n}) to make q̂_{n} move horizontally. So one can assume q̂_{n} and p_{n} are on the same vertical line and have distance less than one. Vertical translation is well defined in W so one can vertically translate the foliation F(n), taking q̂_{n} to p_{n}, and this new foliation (intersected with W (1)) satisfies the conclusions of Proposition 8.1. We have now completed the proof of Proposition 8.1 and hence of Theorem 1. □

Remark 8.5 In [20] Rosenberg and Toubiana constructed a properly immersed minimal annulus A in ^{3} with proper x_{1}-coordinate function and infinite total curvature. Projecting A into ^{3}/S_{θ} yields a properly immersed minimal annulus. Thus, the embeddedness assumption in Theorem 1 is a necessary one for proving the finite total curvature property.

A second remark is that a properly embedded minimal annulus A with compact boundary in ^{3}/S_{θ} has finite total curvature. The proof of Theorem 1 shows that this is the case if A is the end of a properly embedded minimal surface, since in this case A can be trapped between standard ends. The trapping argument can be generalized using the technical results in [17] to show that a general A can be trapped, thereby proving A has finite total curvature.

9 Applications of Theorem 1.

In this section we shall give the proofs of the remaining theorems. The proofs of these theorems are based on Theorem 1 and the results of Sections 2-4 on the geometry of properly embedded minimal surfaces of finite total curvature. Proof of Theorem 2. Suppose M is a properly embedded simply connected minimal surface in ^{3} with infinite symmetry group Sym(M) ⊂ Sym( ^{3}).

If M is not the helicoid, then its symmetry group is a discrete subgroup of Sym( ^{3}). Every discrete infinite subgroup of Sym( ^{3}) contains a screw motion (which may be a translation) and so, after a possible rigid motion of M, we may assume that M is invariant under a screw motion S_{θ}. Since S_{θ} acts freely and properly discontinuously on ^{3}, M/S_{θ} is a properly embedded minimal annulus in ^{3}/S_{θ}. By Theorem 1 M has finite total curvature and so Theorem 8 implies M is a helicoid or a flat plane. □

Proof of Theorem 3. Let M ⊂ ^{3}/S_{θ} be such a surface. By lifting to a 2-sheeted cover of ^{3}/S_{θ}, we may assume that M is orientable. Theorem 1 implies M has finite total curvature and Propositions 4.1 and 4.2 show that an end of M must be asymptotic to a plane, a vertical flat annulus, a helicoid or else θ = 0 and the end is a nonhorizontal helicoidal type end. We now prove that the last case can not occur. For convenience rotate the surface so that the normal vector on a punctured disk neighborhood D^{∗} of the end is vertical and let T denote the translation by the period vector v of the lifted surface in ^{3}. Note v has a nonzero horizontal component.

Let \overline{g}: D → ∪ \{∞\} be the extension of g|D^{∗} to the origin. We may assume that g(0) = 0. In this case we have by the work in Section 2 that

g(z) = z^{p} , ω(z) = (\frac{c_{p+1}}{z^{p+1}} + \frac{c_{p}}{z^{p}} + ⋯)dz

where c_{p+1} ∈ and c_{1} ∈ i . It remains to prove that c_{1} = 0, i.e. the translation vector v is actually vertical, a contradiction. Consider the extended map \overline{g} : \overline{M} → S^{2} = ∪ \{∞\} to the conformal compactification across the puncture points P ⊂ \overline{M}, corresponding to the ends of M. Since M is embedded with an even number of helicoidal type ends, we see that \overline{g}(P) = \{0, ∞\}. Since M has helicoidal ends and it is embedded, the map ĝ : ^{∗} → ^{3}/T, defined by ĝ(z) = ∑_{p∈g^{−1}(z)} p, where the sum is taken in the abelian group ^{3}/T and taken with multiplicity, is a constant, or a complete branched minimal immersion. (See [16, 21] for these properties of ĝ.)

We claim that ĝ is the helicoid. To see this, recall that the helicoidal type ends of M, with the same limiting normal value, have the same coefficients c_{p+1} in ω (Proposition 4.2). Notice that c_{p+1} = −iβ ≠ 0, β given by Proposition 4.1. To obtain the “sum surface” ĝ, one adds all points with the same g values. Parametrize the ends of M with the same limiting normal values by

g(z) = z^{p}, ω(z) = (\frac{c}{z^{p+1}} + o(z^{−p}))dz.

Then the Weierstrass Representation of the sum of these ends is given by

ĝ(η) = η, ω̂(η) = ω̂_{1}(η) + … + ω̂_{ℓ}(η),

where ω̂_{i}(η) is the 1-form of the “sum surface” of the i’th end A_{i}; A_{1}, …, A_{ℓ} the ends with the same limiting normal.

We calculate ω̂_{i}(η): For fixed η let z be a p’th root of η. So z, jz, …, j^{p−1}z are all the p’th roots of η, where j is a nontrivial p’th root of unity. Then

ω̂_{i}(η) = c_{p+1} ∑_{k=0}^{p−1}(\frac{1}{(zj^{k})^{p+1}} + o(z^{−p}))d(j^{k}z) = \frac{c_{p+1}}{z^{p+1}} (∑_{k=0}^{p−1} j^{−pk} + o(z^{−p}))d(z) =

(\frac{(p − 1)c_{p+1}}{z^{p+1}} + o(z^{−p}))dz = [(\frac{p − 1}{p})(c_{p+1})\frac{1}{η^{2}} + o(η^{−1})]dη.

Since the c_{p+1′s} are the same at A_{1}, …, A_{ℓ}, this shows

ω̂(η) = (\frac{ℓ(p − 1)c_{p+1}}{p})[\frac{1}{η^{2}} + o(η^{−1})]dη.

Now to obtain the ω̂ of the sum surface, one must also add the ω′s at the points of M having η as normal value, which are not on the ends of M. These ω′s are holomorphic forms at these points hence their sum as well. It follows ω̂ = (c/η^{2} + o(η^{−1}))dη and the ends of ĝ are helicoidal, c some constant.

Now ω̂ is a meromorphic form on S^{2} with a double pole at zero and regular at infinity. Hence ω̂(η) = c dη/η^{2}. Thus ĝ is an associate surface of a genuine helicoid. Since the coefficient of 1/η^{2} in ω̂ is purely imaginary, the surface is a genuine helicoid. But the translation vector of the helicoid is vertical whereas the vector v has a horizontal component. This contradiction completes the proof of the first part of Theorem 3.

If θ ≠ 0, it is clear that the planar ends are horizontal and when θ is irrational ^{3}/S_{θ} does not contain vertical flat annuli. This completes the proof of the theorem. □

Remark 9.1 The above argument proves that if M is a properly embedded, finite topology, minimal surface in ^{3}/T, and M has helicoidal ends, then the sum surface M + M is a genuine helicoid. It’s not hard to see (by a similar argument) that if the ends are planar, then M + M is a point. If M has four Scherk type ends, then M + M is a Scherk surface.

Theorem 9 A properly embedded orientable minimal surface of finite topology in an orientable flat nonsimply connected three-manifold has an even number of ends or it is a plane.

Proof. By Theorem 1, the minimal surface M ⊂ N has finite total curvature. If the manifold N is isometric to × , then Theorem 3.1 in [16] states that M has an even number of ends. If N is compact, then M is closed and has zero ends, an even number. The only other possibility is that N is isometric to some ^{3}/S_{θ}. If M has helicoidal type ends, then M has an even number of ends by Proposition 4.2. If M is not a plane, it must separate N by Remark 6.1. If the ends of M are planar, then the work in Section 2 shows that for large R the vertical torus ∂T_{R} of radius R centered along the x_{3}-axis intersects M in a family of parallel simple closed curves, one for each end of M. Since M separates N, the number of curves in ∂T_{R} ∩ M is even. The remaining case is when the ends of M are Scherk type ends. For large R, ∂T_{R} ∩ M consists of parallel almost-vertical, simple, closed curves, one for each end. As in the previous case, this implies M has an even number of ends. □

Proof of Theorem 4. Suppose that M ⊂ N = ^{3}/S_{θ} is a properly embedded minimal surface of finite topology. By Theorem 1, M has finite total curvature and by Theorem 3 the ends of M are asymptotic to parallel planes, vertical flat annuli or to ends of parallel helicoids in N. We will implicity use the analytic results of Sections 2–4.

Let M_{R} = T_{R} ∩ M and note that for R large Int(M_{R}) is homeomorphic to M_{R} and M − Int(M_{R}) consists of the annular ends of M. Hence χ(M_{R}) = χ(M).

By Gauss-Bonnet, the total curvature of M_{R} is

C(M_{R}) = ∫_{M_{R}} K dA = 2πχ(M) − ∫_{∂M_{R}} κ_{g}

where K is the Gaussian curvature and κ_{g} is the geodesic curvature of ∂M_{R}. Since C(M) = lim_{R→∞} C(M_{R}), the theorem will follow by showing that the total geodesic curvature of ∂M_{R} converges to 2π · W (M) as R → ∞, where W (M) is the total winding number of M.

First consider the case when the ends of M are planar. In this case it is clear that the geodesic curvature of a component δ_{R} of ∂M_{R} converges to 2π. It is equally clear that δ_{R} is homotopically trivial in N since it lifts to 3. Recall the definition of the curves α and β used in defining the winding number of an end of M. Define the related curves α_{R} = ∂T_{R} ∩ ^{2} and β_{R} on ∂T_{R}. Since δ_{R} is homotopically trivial in N, δ_{R} ⊂ ∂T_{R} is homotopic to α_{R} +0·β_{R}. Hence, the winding number of the end associated to δ_{R} is 1 which proves the formula when the ends of M are planar.

Suppose R is large and the ends of M are asymptotic to flat vertical annuli. Let A_{R} be a component of M − M_{R}. Since A_{R} is proper, θ must be a rational multiple of 2π. Suppose θ = 2π · \frac{m}{n} where m and n are relatively prime. As R → ∞ the curve δ_{R} approximates a geodesic of M_{R}, almost vertical in N. On the other hand it is clear that the absolute value of the intersection number of δ_{R} with α_{R} is n and with β_{R} is m. Since the sign of δ_{R}∩α_{R} is the negative of the sign of δ_{R}∩β_{R}, δ_{R} is homotopic to ±(mα_{R}−nβ_{R}). By definition, the winding number of A_{R} is |2π · m − n · 2π · \frac{m}{n}| = 0. Thus, the total curvature formula holds when M has Scherk type ends.

Now consider the case where the ends of M are helicoidal. Let A_{R} be a component of M −M_{R}. Note that ∂A_{R} approximates a helix h_{R} on ∂T_{R} when R is large. Since the limiting normal vector to A_{R} is vertical, the geodesic curvature of ∂A_{R} converges to the curvature of h_{R} as R → ∞. It remains to calculate the winding number and the total curvature of h_{R} as R → ∞. The helix γ_{R} is homotopic to nα_{R} + mβ_{R} for some relatively prime integers n, m. It is geometrically evident that the total curvature of γ_{R} converges to |2πn + m · θ|. This observation completes the proof of Theorem 4. □

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