Complete stationary surfaces in \mathbb{R}^{4}_{1} with total curvature -\int K\mathrm{d}M=4\pi

# Complete stationary surfaces in R41 with total curvature −∫KdM=4π

Xiang Ma, Peng Wang

Abstract

Applying the general theory about complete spacelike stationary (i.e. zero mean curvature) surfaces in 4-dimensional Lorentz space , we classify those regular algebraic ones with total Gaussian curvature . Such surfaces must be oriented and be congruent to either the generalized catenoids or the generalized enneper surfaces. For non-orientable stationary surfaces, we consider the Weierstrass representation on the oriented double covering (of genus ) and generalize Meeks and Oliveira’s Möbius bands. The total Gaussian curvature are shown to be at least when is algebraic-type. We conjecture that there do not exist non-algebraic examples with .

Keywords: stationary surface, Weierstrass representation, finite total Gaussian curvature, singular end, non-orientable surfaces

MSC(2000):   53A10, 53C42, 53C45

## 1 Introduction

In a previous paper [5] we have generalized the classical theory about minimal surfaces in to zero mean curvature spacelike surfaces in 4-dimensional Lorentz space. Such an immersed surface , called a stationary surface (see [1] for related works before), admits a Weierstrass-type representation formula, which involves a pair of meromorphic functions (the Gauss maps) and a holomorphic -form (the height differential) on :

 x=2 Re∫(ϕ+ψ,−i(ϕ−ψ),1−ϕψ,1+ϕψ)dh.

Among complete examples, those with finite total curvature are most important, i.e., when the integral of the Gaussian curvature converges absolutely. For such surfaces, under mild assumptions we have established Gauss-Bonnet type formulas relating the total curvature with the Euler characteristic number of , the generalized multiplicities of each ends, the mapping degree of , and the indices of the so-called good singular ends:

 ∫MKdM=2π(2−2g−r−r∑j=1˜dj)=−2π(degϕ+degψ−∑j|indpj|)

On this foundation, here we go on to consider complete examples with total curvature , which is the smallest possible value among algebraic stationary surfaces. (Here we ingore the trivial case when is contained in a 3-dimensional degenerate subspace . The induced metric is flat in that case with total curvature . See Section 2.)

Recall that in , Osserman has shown that complete minimal surfaces with finite total curvature must be algebraic ones, i.e., they are given by meromorphic Weierstrass data over compact Riemann surfaces. In particular, immersed examples with are either the catenoid or the Enneper surface. (For other complete minimal surfaces in with small total curvature and the classification results, see [2, 6].)

These two classical examples have been generalized by us in [5] to stationary surfaces in (see Example 3.1 and 3.3). In this paper our main result is

Theorem A   Let be a complete, immersed, algebraic stationary surface with total curvature . Then it is either a generalized catenoid, or a generalized Enneper surface. In particular, there does not exist non-orientable examples with .

Compared to minimal surfaces in , here finite total Gaussian curvature (i.e., converges absolutely) still implies that is conformally equivalent to a compact Riemann surface with finite punctures . A main difference is that in our case, finite total curvature no longer implies algebraic-type. For counter-examples see Example 5.2 and Example 5.5. An interesting open problem is that whether there exist non-algebraic examples with . See discussions in Section 5.

Another new technical difficulty is that, to solve existence and uniqueness problems for complete stationary surfaces, now we must consider the following equation about complex variable :

 ϕ(z)=¯ψ(z), (1)

We have to show that there are no solutions to it for meromorphic functions with given algebraic forms and certain parameters on a compact Riemann surface (except at several points assigned to be good singular ends). This is because that on an immersed surface there must be (regularity condition). On the other hand, at one end where take the same value with equal multiplicities (bad singular end), the total curvature will diverge. Such a complex equation (1) involving both holomorphic and anti-holomorphic functions is quite unusual to the knowledge of the authors. Most of the time we have to deal with this problem by handwork combined with experience. See [5] or Appendix A for related discussions. Note that is a rare case where we overcome this difficulty easily, because this time , and this will never be equal to .

In [8], Meeks initiated the study of complete non-orientable minimal surfaces in . Such surfaces are represented on its oriented double covering space, and the example with least possible total curvature was constructed (Meeks’ Möbius strip). Here we generalize this theory to non-orientable stationary surfaces in (Section 4). A key result is the following lower bound estimation of the total curvature which helps to establish Theorem A above.

Theorem B   Given a non-orientable surface whose double covering space   has genus and finite many ends, for any complete algebraic stationary immersion with finite total curvature there must be

We conjecture that is the best lower bound which could always been attained. Note that this agrees with the estimation for non-orientable minimal surfaces in , and the conjecture is still open even in that special case [7].

We organize this paper as below. In Section 2 we review the basic theory about stationary surfaces in . The orientable case and non-orientable case are discussed separately in Section 3 and 4. In Section 5 we give non-algebraic examples with small total curvature. The proofs to several technical lemmas are left to Appendix A and B.

Acknowledgement   We thank two colleagues of the first author at Peking University, Professor Fan Ding for providing the proof to the topological Theorem 7.1 in Appendix B, and Professor Bican Xia for verifying Lemma 6.1 in Appendix A using a computational method developed by him before. We also thank the encouragement of Professor Changping Wang. This work is supported by the Project 10901006 of National Natural Science Foundation of China.

## 2 Preliminary

Let be an oriented complete spacelike surface in 4-dimensional Lorentz space. The Lorentz inner product is given by

 ⟨x,x⟩=x21+x22+x23−x24.

We will briefly review the basic facts and global results established in [5] about such surfaces with zero mean curvature (called stationary surfaces).

Let be the induced Riemannian metric on with respect to a local complex coordinate . Hence

 ⟨xz,xz⟩=0,  ⟨xz,x¯z⟩=12e2ω.

Choose null vectors in the normal plane at each point such that

 ⟨y,y⟩=⟨y∗,y∗⟩=0,  ⟨y,y∗⟩=1,  det{xu,xv,y,y∗}>0 .

Such frames are determined up to scaling

 {y, y∗}→{λy, λ−1y∗} (2)

for some non-zero real-valued function . After projection, we obtain two well-defined maps (independent to the scaling (2))

 [y], [y∗]:M→S2≅{[v]∈RP3|⟨v,v⟩=0}.

The target space is usually called the projective light-cone, which is well-known to be homeomorphic to the 2-sphere. By analogy to , we call them Gauss maps of the spacelike surface in .

The surface has zero mean curvature if, and only if, are conformal mappings (yet they induce opposite orientations on ). Since , we may represent them locally by a pair of holomorphic and anti-holomorphic functions . The Weierstrass-type representation of stationary surface is given by [5]:

 x=2 Re∫(ϕ+ψ,−i(ϕ−ψ),1−ϕψ,1+ϕψ)dh (3)

in terms of two meromorphic functions and a holomorphic -form locally. We call the Gauss maps of and the height differential.

###### Remark 2.1.

When , by (3) we obtain a minimal surface in , or a maximal surface in . This recovers the Weierstrass representation in these classical cases. When or is constant, we get a zero mean curvature spacelike surface in the 3-space with an induced degenerate inner product, which is essentially the graph of a harmonic function on complex plane .

Convention: In this paper, we always assume that neither of is a constant unless it is stated otherwise. According to the remark above, we have ruled out the trivial case of stationary surfaces in . (According to (7) below, such surfaces have flat metrics and zero total Gaussian curvature.)

###### Remark 2.2.

The induced action of a Lorentz orthogonal transformation of on the projective light-cone is nothing but a Möbius transformation on , or equivalently, a fractional linear transformation on given by with . The Gauss maps and the height differential transform as below:

 ϕ⇒aϕ+bcϕ+d ,  ψ⇒¯aψ+¯b¯cψ+¯d ,  dh⇒(cϕ+d)(¯cψ+¯d)dh . (4)

This is repeatedly used in Section 2 and Section 3 to simplify or to normalize the representation of examples.

###### Theorem 2.3.

[5] Given holomorphic -form and meromorphic functions globally defined on a Riemann surface . Suppose they satisfy the regularity condition 1),2) and period conditions 3) as below:

1) on and their poles do not coincide;

2) The zeros of coincide with the poles of or with the same order;

3) Along any closed path the periods satisfy

 ∮γϕdh=−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∮γψdh,   (horizontal period condition) (5)
 Re∮γdh=Re∮γϕψdh=0.   (vertical period condition) (6)

Then (3) defines a stationary surface .

Conversely, any stationary surface can be represented as (3) in terms of such and over a (necessarily non-compact) Riemann surface .

The structure equations and the integrability conditions are given in [5]. An extremely important corollary is the formula below for the total Gaussian and normal curvature over a compact stationary surface with boundary :

 ∫M(−K+iK⊥)dM=2i∫Mϕz¯ψ¯z(ϕ−¯ψ)2dz∧d¯z=−2i∫∂Mϕzϕ−¯ψdz=−2i∫∂M¯ψ¯zϕ−¯ψd¯z. (7)

At one end with , the integral of total curvature above will become an improper integral. An important observation in [5] is that this improper integral converges absolutely only for a special class of such ends.

###### Definition 2.4.

Suppose is an annular end of a regular stationary surface (with boundary) whose Gauss maps and extend to meromorphic functions on the unit disk . It is called a regular end when

 ϕ(0)≠¯ψ(0).   (Thus ϕ(z)≠¯ψ(z), ∀ z∈D.)

It is a singular end if where the value could be finite or .

When the multiplicities of and at are equal, we call a bad singular end. Otherwise it is a good singular end.

###### Proposition 2.5.

[5] A singular end of a stationary surface is good if and only if the curvature integral (7) converges absolutely around this end.

For a good singular end we introduced the following definition of its index.

###### Definition 2.6.

[5] Suppose is an isolated zero of in ’s neighborhood , where holomorphic functions and take the value with multiplicity and , respectively. The index of at (when are both holomorphic at ) is

 indp(ϕ−¯ψ)≜12πi∮∂Dpdln(ϕ−¯ψ)={m,mn. (8)

The absolute index of at is

 ind+p(ϕ−¯ψ)≜∣∣indp(ϕ−¯ψ)∣∣. (9)

For a regular end our index is still meaningful with For convenience we also introduce

 ind1,0≜12(ind++ind),   ind0,1≜12(ind+−ind), (10)

which are always non-negative.

Note that our definition of index of is invariant under the action of fractional linear transformation (4). So it is well-defined for a good singular end of a stationary surface. In particular, we can always assume that our singular ends do not coincide with poles of ; hence the definition above is valid.

A stationary surface in is called an algebraic stationary surface if there exists a compact Riemann surface with such that is a vector valued meromorphic form defined on . In other words, the Gauss map and height differential extend to meromorphic functions/forms on . For this surface class we have established Gauss-Bonnet type formulas involving the indices of the good singular ends.

###### Theorem 2.7 ([5]).

For a complete algebraic stationary surface given by (3) in terms of without bad singular ends, the total Gaussian curvature and total normal curvature are related with the indices at the ends (singular or regular) by the following formulas:

 ∫MK⊥dM =0 , (11) ∫MKdM =−4π(degϕ−∑jind1,0(ϕ−¯ψ)) (12) =−4π(degψ−∑jind0,1(ϕ−¯ψ)), (13)

From (12)(13) we have equivalent identities:

 ∑jindpj(ϕ−¯ψ)=degϕ−degψ . (14) ∫MKdM=−2π(degϕ+degψ−∑jind+pj(ϕ−¯ψ)) . (15)
###### Definition 2.8.

The multiplicity of a regular or singular end for a stationary surface in is defined to be

 ˜dj=dj−ind+pj,

where is equal to the order of the pole of  at .

###### Theorem 2.9 (Generalized Jorge-Meeks formula [5]).

Given an algebraic stationary surface with only regular or good singular ends . Let be the genus of compact Riemann surface , the number of ends, and the multiplicity of . We have

 ∫MKdM=2π(2−2g−r−r∑j=1˜dj) ,  ∫MK⊥dM=0 . (16)
###### Proposition 2.10 ([5]).

Let be a regular or a good singular end which is further assumed to be complete at . Then its multiplicity satisfies

###### Corollary 2.11 (The Chern-Osserman type inequality [5]).

Let be an algebraic stationary surface without bad singular ends, . Then

 ∫KdM≤2π(χ(M)−r)=4π(1−g−r). (17)
###### Corollary 2.12 (Quantization of total Gaussian curvature [5]).

Under the same assumptions of the theorem above, when are not constants (equivalently, when is not a flat surface in ), there is always

 −∫MKdM=4πk≥4π,

where is a positive integer.

## 3 Orientable case and examples with −∫KdM=4π

This section is dedicated to the classification of complete stationary surfaces immersed in with finite Gaussian curvature which are orientable and of algebraic type.

Under our hypothesis, the generalized Jorge-Meeks formula (16) yields

 r+∑˜dj+2g=4, (18)

and the index formulas (12)(13) read

 degϕ−∑ind1,0=1,  degψ−∑ind0,1=1. (19)

Since , and for any end, there must be , and we need only to consider five cases separately as below.

Case 1: (torus with one end).

Since there is only one end, at least one of the indices is zero. By (19) we know either or is a meromorphic function of degree . Yet this contradicts the well-known fact that over a torus there do not exist such functions. So we rule out this possibility.

Case 2: and the unique end is regular.

Such examples exist and they are generalization of the classical Enneper surface.

###### Example 3.1 (The generalized Enneper surfaces).

This is given by

 ϕ=z, ψ=cz , dh=s⋅zdz, (20)

or

 ϕ=z+1, ψ=cz , dh=s⋅zdz, (21)

over with complex parameters . has no singular points if and only if the parameter is not zero or positive real numbers in (20), or

 c1−c22+14<0 (22)

in (21). When in (20) we obtain the Enneper surface in .

Indeed they are all examples in Case 2 according to the following result in [5].

###### Theorem 3.2.

[5] A complete immersed algebraic stationary surface in with and one regular end is a generalized Enneper surface.

Case 3: with a good singular end.

Suppose there exists such an example. Without loss of generality we assume that the singular end has positive index. Since , by definition we know that at the function takes the value with multiplicity at least . On the other hand, and , which is a contradiction to the observation above. Hence such examples do not exist.

Case 4: and both ends are regular.

The classical catenoid is one of such examples. The generalization in is

###### Example 3.3 (The generalized catenoids).

This is defined over with

 ϕ=z+t, ψ=−1z−t, dh=sz−tz2dz.     (−1

When , it is the classical catenoid in .

###### Theorem 3.4.

[5] A complete immersed algebraic stationary surface in with total curvature and two regular ends is a generalized catenoid.

Case 5: with at least one good singular ends.

This is the most difficult case in our discussion. We will show step by step that there are no such examples.

First, assume there is such a surface. We assert that it must have two singular ends whose indices have opposite signs. Otherwise, if there is only one good singular end which might be assumed to have positive index, similar to the discussion in Case 3 we can show and has multiplicity greater than at the end, which is a contradiction. In the same way we can rule out the possibility that both ends are singular with the same signs.

Second, without loss of generality we may suppose and the good singular ends are and , with . By (19), . If , by definition we know has multiplicity at least at where must has higher multiplicity, which is impossible since . If , by definition and (19) we know , which contradicts the requirement that must has multiplicity greater than at the first end . In summary there must be

 ind0=m≥1,  ind∞=−m,  degϕ=degψ=m+1≥2. (24)

We observe that . Otherwise, since are both singular ends, there must be . Because is a good singular end and , has multiplicity at least at and multiplicity at . This is impossible when .

This observation enables us to make the following normalization. Without loss of generality, suppose . Since meromorphic functions must be rational functions satisfying restrictions (24), we know

 ϕ(z)=zm(z−a),  ψ(z)=zm+1z−b,  dh=ρz−bzkdz, (25)

where are arbitrary nonzero complex parameters. Note that takes the form as above because is regular at . On the other hand, at the ends and it should satisfy according to Proposition 2.10, which implies by the definition of .

After fixing the form of , we verify the period conditions. It is easy to see that the vertical period conditions are satisfied. The horizontal period conditions are satisfied if and only if In summary, such examples have Weierstrass data

 ϕ(z)=zm(z−a),  ψ(z)=zm+1z−b,  dh=ρz−bzm+2dz, (26)

with parameters

 m≥1, a,b,ρ∈C∖0, a+b=−¯ρ/ρ. (27)

If we can find nonzero parameters as above so that the regularity condition holds true for any , then new examples with are found. But according to Lemma 6.1 in Appendix A, for any given nonzero parameters there always exist nonzero solutions to the equation for given above. We conclude that there exist no examples in Case 5, The proof to the following theorem has been finished.

###### Theorem 3.5.

Complete regular algebraic stationary surfaces with are either the generalized catenoids or the generalized Enneper surfaces under the assumption that is orientable.

Another interesting observation is that if we make change of variables in (25), and choose the power to be a even number suitably, then the period conditions always hold true and we don’t need the restriction in (27). In this situation, if parameters is chosen suitably, the regularity condition is satisfied. See Lemma 6.3. In this way we find a complete, immersed stationary surface in , yet with total curvature . See the example below (which has appeared in [5]).

###### Example 3.6 (Genus zero, two good singular ends and ∫MKdM=−8π).
 M=C∖{0}, ϕ=w2(w2−a), ψ=w4w2−a, dh=w2−aw4dw.  (a∈C∖{0})

The regularity, completeness and period conditions are satisfied when is a sufficiently large positive real number (e.g. ). For the proof of regularity, see Lemma 6.3.

## 4 Non-orientable stationary surfaces and examples

In this section we will consider non-orientable algebraic stationary surfaces and show that the total curvature of them is always greater than . For this purpose we need to consider their oriented double covering surface , and characterize the Weierestrass data over . This is a natural extension of Meeks’ characterization of non-orientable minimal surfaces in [8].

### 4.1 Representation of non-orientable stationary surfaces

###### Theorem 4.1.

Let be a Riemann surface with an anti-holomorphic involution (i.e., a conformal automorphism of reversing the orientation) without fixed points. Let be a set of Weierstrass data on such that

 ϕ∘I=¯ψ,  ψ∘I=¯ϕ,  I∗dh=¯¯¯¯¯¯dh, (28)

which satisfy the regularity and period conditions as well. Then they determine a non-orientable stationary surface

 M=˜M/{id,I}→R41

by the Weierstrass representation formula (3).

Conversely, any non-orientable stationary surface could be constructed in this way.

###### Remark 4.2.

Geometrically, (28) is the consequence of reversing the orientation of the tangent plane by , and reversing the induced orientation of the normal plane by interchanging the lightlike normal directions .

###### Proof to Theorem 4.1.

We prove the converse first. It is well-known that any non-orientable surface has a orientable two-sheeted covering surface with an orientation-reversing homeomorphism , and is realized as the quotient surface

 M=˜M/Z2=˜M/{id,I}.

Denote the quotient map. Notice that is endowed with the complex structure induced from the metric. When is a local complex coordinate over a domain which projects to one-to-one, is also a coordinate over compatible with the orientation on .

Consider the stationary surface . In the chart we have

 ~xzdz=(ϕ+ψ,−i(ϕ−ψ),1−ϕψ,1+ϕψ)dh .

Then in the corresponding chart , consider and we have

 ~x∗wdw=(¯ϕ+¯ψ,i(¯ϕ−¯ψ),1−¯ϕ¯ψ,1+¯ϕ¯ψ)d¯h .

This implies (28).

Now we prove the first part. If as described in the theorem and satisfy condition (28) as well as the regularity and period conditions, then the integral along any path yields two stationary surfaces

 ~x = 2 Re∫γ(ϕ+ψ,−i(ϕ−ψ),1−ϕψ,1+ϕψ)dh, ~x∘I = ~x∗ = 2 Re∫γ(¯ψ+¯ϕ,−i(¯ψ−¯ϕ),1−¯ψ¯ϕ,1+¯ψ¯ϕ)d¯h .

If we assign the same initial value, then after either integration above we get the same result, because they are the real parts of a holomorphic vector-valued function and its complex conjugate. So and are mapped to the same point in , yet with opposite induced orientations on the same surface. After taking quotient we get a stationary immersion of the non-orientable into . This finishes the proof. ∎∎

As an application of this theorem, we give a natural generalization of Meeks and Oliveira’s construction of minimal Möbius strip.

###### Example 4.3 (Generalization in R41 of Meeks’ minimal Möbius strip).

This is defined on with involution , and the Weierstrass data be

 ϕ=z−λz−¯λ⋅z2m,  ψ=1+¯λz1+λz⋅1z2m,  dh=i(z−¯λ)(1+λz)z2dz, (29)

where is a complex parameter satisfing , and the integer .

###### Remark 4.4.

When we have , and the example above is equivalent to Oliveira’s examples in [9]. (Meeks’ example [8] corresponds to the case .) Otherwise this is a full map in . Furthermore, for fixed these examples are not congruent to each other unless the values of the parameter are the same or differ by complex conjugation, because the cross ratio

 cr(0,∞;λ,¯λ)=λ¯λ

between the zeros and poles in the normal form of is an invariant.

###### Proposition 4.5.

Example 4.3 is a complete immersed stationary Möbius strip with a regular end and total Gaussian curvature .

###### Proof.

We start from a general case, a Möbius strip with

 ˜M=C∖{0},  I:z→−1/¯z,  ϕ(z)=az+bcz+d⋅z2m.   (a,b,c,d∈C,ad−bc≠0)

To satisfy condition (28), there should be

 ψ=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ϕ(−1/¯z)=¯bz−¯a¯dz−¯c⋅1z2m .

The surface is regular outside the ends . Together with , this implies

 dh=i(cz+d)(¯dz−¯c)z2dz

up to multiplication by a real constant. Under these conditions it is easy to verify that the metric is complete.

Next, let us check the period conditions. The horizontal periods vanish automatically since has no residues at and . The vertical periods must vanish, hence Without loss of generality we may write

 ϕ=z−λz−¯λ⋅z2m,   |λ|=1.

To simplify to this form we have utilized the freedom to change complex coordinate by and the (fractional) linear transformation induced from the Lorentz transformation of (see (4)).

We are left to verify over . (At the ends it is obviously true. So they are regular ends.) Suppose form some . Substitute the expressions of into it. We obtain

 |z|4m=(z−¯λ)(−1/¯z−λ)(z−λ)(−1/¯z−¯λ)=cr(z,−1/¯z;¯λ,λ).

Since the cross ratio at the right hand side takes a real value, four points are located on a circle in the complex plane .

We assert that this circle could not be identical to the unit circle. (Otherwise and the cross ratio above is . This holds true only if , which is impossible, or , which has been ruled out in Example 4.3.)

Circle intersects the unit circle at and . Observe that any circle passing through will intersect the unit circle at an antipodal point pair. (Because under the inverse of the standard stereographic projection, correspond to two antipodal points on , and the unit circle corresponds to the equator. Any circle passing through the inverse images of on will intersect the equator again at two antipodal points. After taking stereographic projection back to we get the conclusion.) As a consequence, . But this time the aforementioned cross ratio could only take value as a negative real number (because on circle , must be separated by ). This contradiction finishes our proof. ∎∎

When this example has smallest possible total curvature among non-orientable algebraic stationary surfaces. (Note that the classical Henneberg surface in has total curvature , yet with four branch points.) This conclusion is the corollary of a series of propositions below.

### 4.2 Non-orientable stationary surfaces of least total curvature

In general we are interested in finding least possible total curvature for non-orientable stationary surfaces of a given topological type. This is motivated by discussions of F. Martin in [7]. Compared with minimal surfaces in , this general case looks even more interesting (at least to the authors).

As a consequence of Theorem 4.1, for a complete non-orientable stationary surface with double covering of genus with ends, there must be ; the index formula (15) as well as the Jorge-Meeks formula (16) implies

 −∫MK=2π(degϕ−r∑j=1|indpj|)=2π(g+r−1+r∑j=1~dj) . (30)

Because and , we know

 −∫MK=≥2π(g+1).

A better estimation is given in the following proposition.

###### Proposition 4.6.

Given a non-orientable surface whose double covering space   has genus and finite many punctures, there does not exist complete algebraic stationary immersion with total Gaussian curvature . In other words, under our assumptions there must be

 −∫MKdM≥2π(g+2). (31)
###### Proof.

Consider the lift of , i.e., . Since the immersion is algebraic and , it has finite many regular or good singular ends, and the total number is a even number ( is the number of ends of ). By the modified Jorge-Meeks formula (30) and , a Chern-Osserman type inequality is obtained:

 −∫MKdM≥2π(g+1).

Suppose the equality is achieved. Then there must be two ends for and . Both of them are regular ends or good singular ends at the same time. We will show that in either case there will be a contradiction.

Case 1: regular end(s). The multiplicity , and for the end has a pole of order . In a local coordinate chart with we write out the Laurent expansion of :

 ~xz=1z2 v2+1z v1+(holomorphic part).

Since this is a regular end, is an isotropic vector whose real and imaginary parts span a 2-dimensional spacelike subspace. is a real vector orthogonal to by the period condition and . Thus in there exist a constant non-zero real vector .

At the other end with local coordinate , because , we know the same is orthogonal to the principal part of the Laurent series. Thus is a holomorphic -form, and is a harmonic function defined on the whole compact Riemann surface. It must be a constant; hence as well as is contained in a 3-dimensional subspace of .

Yet this is impossible. Since in or there exist no immersed spacelike non-oriented surfaces. The possibility of could be ruled out by Schoen’s famous result [11] that any complete, connected, oriented minimal surface in with two embedded ends is congruent to the catenoid. (Alternatively, we may argue by the maximal principle once again. Since the unique end of is an embedded end in , which is either a catenoid end or a planar end, one can choose the coordinate of suitably so that the height function is bounded from below over the whole . Such a harmonic function must be a constant, and . Contradiction.)

Case 2: good singular end(s). At the good singular end , without loss of generality, suppose it has and . Then , and has a pole of order at . There always exists a suitable local coordinate such that and

 dh=dzzm+2,  ϕ(z)=a0zm+a1zm+1+O(zm+2),  ψ(z)=b1zm+1+O(zm+2).

By (3) we know

 ~xzdz = = dzzm+2 ⎛⎜ ⎜ ⎜⎝0011⎞⎟ ⎟ ⎟⎠+dzz2 ⎛⎜ ⎜ ⎜⎝a0−ia000⎞⎟ ⎟ ⎟⎠+dzz ⎛⎜ ⎜ ⎜⎝a1+b1−i(a1−b1)00⎞⎟ ⎟ ⎟⎠+(holomorphic part).

Take . We can argue as in case 1 to show that is a harmonic function defined on the whole compact Riemann surface, hence be a constant. (The key point is that has only one end.) Thus as well as is contained in an affine space (orthogonal to ). Yet this is also impossible for a non-oriented spacelike surface. ∎∎

We will show that the lower bound could be improved to be , the same as the case for non-orientable minimal surfaces in .

###### Theorem 4.7.

Given a non-orientable surface whose double covering space   has genus and finite many punctures, there does not exist complete algebraic stationary immersion with total Gaussian curvature . In other words, under our assumptions there must be

 −∫MKdM≥2π(g+3). (32)
###### Proof.

As in the proof to Proposition 4.6, consider the lift of , i.e., . Suppose the lower bound is attained. Then has two ends with by (30). By symmetry, both of them are regular or good singular ends at the same time. Each possibility is ruled out using different arguments.

When both ends are good singular ends, we use the same argument as in Case 2 of Proposition 4.6. At the good singular end , without loss of generality, suppose it has and . Then , and has a pole of order at . There always exists a suitable local coordinate such that and

 dh=dzzm+3,  ϕ(z)=a0zm+a1zm+1+O(zm+2),  ψ(z)=b1zm+1+O(zm+2).

By the Weierstrass representation formula we know

 ~xz =