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Geometry
Volume 2013 (2013), Article ID 902092, 9 pages
http://dx.doi.org/10.1155/2013/902092
Research Article

Darboux Transforms of a Harmonic Inverse Mean Curvature Surface

Division of Mathematics, Faculty of Pure and Applied Sciences, University of Tsukuba, 1-1-1 Tennodai, Tsukuba, Ibaraki 305-8571, Japan

Received 18 December 2012; Accepted 17 February 2013

Academic Editor: Manuel Sanchis

Copyright © 2013 Katsuhiro Moriya. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The notion of a generalized harmonic inverse mean curvature surface in the Euclidean four-space is introduced. A backward Bäcklund transform of a generalized harmonic inverse mean curvature surface is defined. A Darboux transform of a generalized harmonic inverse mean curvature surface is constructed by a backward Bäcklund transform. For a given isothermic harmonic inverse mean curvature surface, its classical Darboux transform is a harmonic inverse mean curvature surface. Then a transform of a solution to the Painlevé III equation in trigonometric form is defined by a classical Darboux transform of a harmonic inverse mean curvature surface of revolution.

1. Introduction

The theory of surfaces is connected with the theory of solitons through a compatibility condition of the Gauss-Weingarten equations. Bobenko [1] gave an outline of eight classes of surfaces in the three-dimensional Euclidean space in the formulation of the theory of solitons. They are minimal surfaces, surfaces of constant mean curvature, surfaces of constant positive Gaussian curvature, surfaces of constant negative Gaussian curvature, Bonnet surfaces, harmonic inverse mean curvature surfaces, Bianchi surfaces, and Bianchi surfaces of positive curvature. For the investigation of these surfaces, a matrices representation of quaternions is used to write their moving frames. Their moving frames are integrated by Sym's formula [2].

Quaternionic analysis by Pedit and Pinkall [3] is a technology to investigate surfaces in the Euclidean three- or four-space which are related to the soliton theory. In this theory, the Euclidean four-space is modeled on the set of all quaternions . A quaternionic line trivial bundle with a complex structure over a Riemann surface is associated with a conformal map from the Riemann surface to .

We can assume that a quaternionic line trivial bundle associated with a constrained Willmore surface in equips a harmonic complex structure [4]. If a constrained Willmore surface is neither minimal nor superconformal, then this complex structure defines a smooth family of flat connections on the line bundle. Then a holonomy spectral curve of a constrained Willmore torus is defined by a smooth family of holonomies of . The relation between a constrained Willmore torus and its holonomy spectral curve is discussed in detail in [4].

If a conformal map from a torus to is not a constrained Willmore torus, then the quaternionic line trivial bundle associated with the conformal map is accompanied with a nonharmonic complex structure. For a conformal map, its Darboux transform is defined. Then all Darboux transforms of a conformal map describe a multiplier spectral curve of a conformal map (see [46]). The property of a conformal map from a torus to is related to that of its multiplier spectral curve. Hence, a Darboux transform of a conformal map which is not a constrained Willmore is an important research subject.

A harmonic inverse mean curvature surface with a nonconstant mean curvature is a conformal map which is not a constrained Willmore. We abbreviate a harmonic inverse mean curvature surface as a HIMC surface. A HIMC surface is introduced by Bobenko [1]. A transform of a HIMC surface is defined in [7]. All isothermic HIMC surfaces are classified in terms of Painlevé transcendents in [8] (see [9]). We investigate a Darboux transform of a HIMC surface.

For a conformal map , a mean curvature sphere of is defined by (5). We define a function by (6). If up to translations in , then is real-valued and exactly a mean curvature function of with the induced metric on . Hence, an HIMC surface is a surface in such that is harmonic. We call a conformal map , such that is harmonic, a generalized harmonic inverse mean curvature surface and abbreviate it as a GHIMC surface (Definition 15). We advance the theory of GHIMC surfaces in a similar way to that of Willmore surfaces in the four-sphere in [10].

A one-form is constructed from a mean curvature sphere by (8). A constrained Willmore surface in is a conformal map such that there exists a one-form with is closed [4]. The one-form is closed if and only if a conformal map is Willmore [10]. On the other hand, we have the following.

Theorem 1. Let be a line bundle of a conformal map with mean curvature sphere such that for every . Set . Define a map by (6) and a one-form by (8). The map is a GHIMC surface if and only if there exists with such that is closed.

We assume that is simply connected. A backward Bäcklund transform of a Willmore surface in [10, 11] is a Bäcklund transform in [12]. Because a backward Bäcklund transform of a Willmore surface is a Willmore surface, a sequence of Willmore surfaces is obtained by repeating to take a backward Bäcklund transform (see [12]).

We define a counterpart of a backward Bäcklund transform of a Willmore surface in a GHIMC surface as follows.

Theorem 2. Let be a line bundle of a conformal map with mean curvature sphere such that for every . One assumes that is a GHIMC surface. Define and by (6). Let be a map such that , and let be a map such that is defined by Let be the subset of where is defined. The quaternionic vector space spanned by and defines a Bäcklund transform of on .

We call the map a backward Bäcklund transform of a GHIMC surface . It is unclear whether a backward Bäcklund transform of a GHIMC surface is a GHIMC surface. Hence, a sequence of GHIMC surfaces constructed by backward Bäcklund transforms is more involved.

A backward Bäcklund transform of a GHIMC surface gives a Darboux transform of a GHIMC surface as follows.

Theorem 3. Let be a line bundle of a conformal map with mean curvature sphere such that for every . One assumes that is a GHIMC surface. Let be a map such that is a backward Bäcklund transform of on . Then, there exists a map such that is a Darboux transform of on .

An isothermic surface in is a surface which has a curvature line coordinate around each nonumbilic point. For an isothermic surface, a classical Darboux transform is defined [13]. If an isothermic surface is a GHIMC surface or an HIMC surface, then we have the following.

Theorem 4. A classical Darboux transform of an isothermic GHIMC surface is a GHIMC surface. A classical Darboux transform of an isothermic HIMC surface is an HIMC surface.

We see that if a conformal map is invariant under a subgroup of Euclidean motions in , then there is a Darboux transform invariant under (Corollary 14). A surface of revolution is a surface invariant under a group of rotations around an axis. All HIMC surfaces of revolution are classified by the solutions to the Painlevé III equation in trigonometric form (42) (see Theorem 20). The formula in Theorem 20 gives an HIMC surface in invariant under rotations around the -axis in . Then, we obtain a transform of the solutions to the Painlevé III equation in trigonometric form as follows.

Theorem 5. Let be a solution to the Painlevé III equation in trigonometric form (42) with , and let be an HIMC surface of revolution defined by Theorem 20. Assume that is a classical Darboux transform of with and   for every . Then, is an HIMC surface of revolution invariant under , and the function defined by (41) for is a solution to the Painlevé III equation in trigonometric form (42).

2. Conformal Maps

We recall a conformal map and a line bundle associated with a conformal map [10].

Throughout this paper, we assume that is a simply connected Riemann surface with complex structure and maps, vector bundles, and sections are smooth. An orientation of is fixed so that, for any nonzero tangent vector of , an ordered pair (, ) is a positive basis. We denote by the tangent bundle of and by the cotangent bundle of .

Let be a vector space. We denote by the set of -forms on with values in . For a vector bundle over , we denote by the set of -forms on with values in . For , we define by . We denote by the trivial bundle over . Then, is naturally identified with .

We model on the set of all quaternions and on the set of all purely imaginary quaternions . For a quaternion , we denote by the quaternionic conjugate of . Set and . Then, the inner product of and is , and the norm of is . We denote by the set of all purely imaginary quaternions. If , , then is the vector product . Let be the sphere of radius one centered at the origin in . Then, . Hence, is the set of all square roots of in .

Let . We define and by setting Then, decomposes because . We see that and . Clearly, if and only if . Similarly, if and only if . The quaternionic conjugation provides an identity . It is known that ([10], Proposition  16).

Let and be right quaternionic vector bundles over a manifold. We denote by the real vector bundle of quaternionic bundle homomorphism from to and denote by the real vector bundle of quaternionic bundle endomorphisms of . We denote the identity automorphism of by . We call a complex structure of if .

We denote by the right quaternionic vector space of dimension two and by the quaternionic projective space of dimension one. We model the conformal four-sphere on . Let be the tautological line bundle of . We denote the trivial quaternionic vector bundle over of rank two by . The tangent bundle of is identified with a bundle .

We fix a basis of . Let be the dual vector space of . There exists uniquely a basis of such that We define a map by for every . The map induces a map from to . This map is a stereographic projection of from .

Let be a map and . Then, is a quaternionic line subbundle of such that for every . Conversely, if is a line subbundle of , then there exists a map such that . We call the line bundle of a conformal map . We use the terminology of maps for line bundles.

Let be the projection. We consider and as sections of . Define a flat quaternionic connection on by . Then, the differential map is identified with a one-form . A line bundle is called a line bundle of a conformal map if there exists a complex structure such that The complex structure is called the mean curvature sphere or the conformal Gauss map of .

Let be a line bundle of a conformal map with mean curvature sphere . We assume that for every . We define a map by for each . The section of does not vanish anywhere. We see that is a frame of . We call the mean curvature sphere of . There exist , , and such that Then, .

Conversely, let be a map with . Then, a map is defined by (6) except at the branch points or . We define a line bundle by for each . Then, is a line bundle of a conformal map with mean curvature sphere defined by (6) except at the branch points of .

Let be a line bundle of a conformal map with mean curvature sphere such that for every . We induce a (singular) metric on by . Then, is a conformal map except at the branch points of . The map is called a conformal map with left normal and right normal . Let be the mean curvature vector of . Then, . The image of is contained in up to translations in if and only if . If the image of is contained in up to translations in , then is the mean curvature function of .

We define , by The one-form and are called the Hopf fields of . We have

For , we set . Then, an indefinite scalar product of is defined by setting for , . The functional is called the Willmore functional of . A conformal map whose line bundle is a critical line bundle of is called a Willmore surface. A critical line bundle is called a Willmore line bundle. It is known that a line bundle of a conformal map with for every is Willmore if and only if ([10], Proposition  15).

The following lemma is proved in [10] implicitly.

Lemma 6. One has .

Proof. We have Hence, . Then, .

3. Transforms

We recall Bäcklund transforms and Darboux transforms of a conformal map in [5, 11].

Let be a line bundle of a conformal map with mean curvature sphere such that for every . Set and . Define ,  , and by (6). We have a splitting . Let be the projection. We decompose by this splitting as We see that is a connection of . Because , the section is a nowhere-vanishing parallel section of . Hence, is a flat connection. Set . The operator is a quaternionic holomorphic structure of (see [11]). Because , we have .

Assume that and are maps such that and are linearly independent, and for any . Let be the two-dimensional vector space spanned by and . Then, is a trivial bundle over of rank two. Let be the evaluation map defined by The map is a surjective bundle homomorphism. Hence, we have an injective bundle homomorphism . Let and be the dual bundles of and , respectively. Then, we have a line bundle of a conformal map. The line bundle is called the Bäcklund transform of with respect to .

Lemma 7. A nonconstant map satisfies the equation if and only if the subspace spanned by and defines a Bäcklund transform of .

Proof. Recall that and that   is nowhere vanishing. For a nonconstant map , we have Hence, if and only if . Thus, and defines a Bäcklund transform if and only if is not constant and .

Lemma 8. Let and be maps satisfying the equation . Then a subspace of spanned by and defines a Bäcklund transform of .

Proof. We have . Hence . By Lemma 7, a subspace of spanned by and defines a Bäcklund transform of .

For , we denote by a lift of , that is, . Set The operator is a quaternionic holomorphic structure of . It is known that if , then there exists a unique lift such that . Then, a line bundle with for every away from zeros of is a line bundle of a conformal map. Set . Then, and . The line bundle of a conformal map is called a Darboux transform of with respect to . We call the map a Darboux transform of with respect to .

A Darboux transform is called classical if for each and is the mean curvature sphere of [13]. Let be the mean curvature sphere of , and set . If is a classical Darboux transform, then and share conformal curvature line parametrizations. The conformal maps and are called isothermic surfaces (see [13]).

The equation is related with a Darboux transform as follows.

Lemma 9. If maps and satisfy the equation , then is a Darboux transform of with respect to , and is a Darboux transform of with respect to .

Proof. We identify with . For with , we have Hence, if and only if .
A lift of is written as for . Because for every , we have Hence, satisfies the equation if and only if . Thus, if maps and satisfy the equation , then a line bundle with for every away from zero of is a Darboux transform of with respect to . Because we have .
We have . Hence, . Then, is a Darboux transform of with respect to .

We have the following by straightforward calculation.

Lemma 10. Define , , , , , and by Then, The map is an isothermic surface, and is a classical Darboux transform if and only if . If a map is an isothermic surface and is a classical Darboux transform, then .

Proof. Because we have and .
Because we have . Because we have .
Because , we have Hence, .
By Lemma 9, a conformal map is a Darboux transform of . Because and , we have .
Then, .
We assume that is an isothermic surface and a Darboux transform of is classical.
We have Hence, .
Conversely, if , then, . Because is a Darboux transform of , we have . Hence, is a classical Darboux transform of .
If is an isothermic surface and is a classical Darboux transform, then is a classical Darboux transform of . Hence . Then .

A pair of two conformal maps and is called a Christoffel pair in [14] if . It is known that if a pair of   and is a Christoffel pair, then maps and are isothermic. If a map is isothermic, then there exists a map such that a pair of and is a Christoffel pair (see [14]).

Corollary 11. One assumes that conformal maps , , , and satisfy the equation and . Set and . If , then and . The conformal maps and are classical Darboux transforms of and , respectively.

Proof. We have Because and , we have We assume that and . Because , , and , we have . Hence, is a classical Darboux transform of .

We see that a pair of an isothermic surface in and its classical Darboux transform form a Bonnet pair in [15] up to similarities in .

For an isothermic surface in , we have the following.

Corollary 12. One assumes that is a classical Darboux transform of a line bundle of an isothermic surface. The image of is contained in up to translations in if and only if the image of   is contained in up to translations in .

Proof. Assume that the image of is contained in up to translations in . Then, and is real-valued. By Lemma 10, we have . Hence, is real-valued. Then, Then, the image of   is contained in up to translations in .
The converse is trivial.

Let be an Euclidean motion of fixing the origin. Then, there exists , with such that for any . If , then is an Euclidean motion of fixing and .

Lemma 13. Let be a conformal map and a Darboux transform of   with respect to such that . Then, is a Darboux transform of with respect to .

Proof. We have Because , we have Hence, is a Darboux transform of with respect to .

For a subgroup of Euclidean motions of    , we say that a map is invariant under if there exists a subgroup of holomorphic transformations of such that for any .

Corollary 14. Let be a subgroup of Euclidean motions of such that each element of fixes the origin of , and let be a conformal map invariant under . Let be a holomorphic transformation of such that for each . Assume that and are maps such that and for every . Set . Then, is a Darboux transform of invariant under .

Proof. Let . Then, there exists , with such that for any . We have Hence, . By the definition, the map is a Darboux transform of with respect to . We have for each . Hence, is invariant under .

4. GHIMC Surfaces

A conformal map is called a harmonic inverse mean curvature surface if [1]. We introduce the notion of a generalized harmonic inverse mean curvature surface and define its transform.

Let be a line bundle of a conformal map with mean curvature sphere such that for every . Set . Define by (6).

Definition 15. One calls the map a generalized harmonic inverse mean curvature surface if .

We abbreviate a harmonic inverse mean curvature surface as an HIMC surface and a generalized harmonic inverse mean curvature surface as a GHIMC surface. A GHIMC surface whose image is contained in up to translations in is exactly an HIMC surface.

Proof of Theorem 4. Let be a line bundle of a conformal map with mean curvature sphere such that for every , and let   be a Darboux transform of . Set and . We assume that the Darboux transform is classical. Define by (6) and by (18). By Lemma 10, we have . Hence, if and only if . Hence, is a GHIMC surface if and only if is.
If is an HIMC surface, then the image of is contained in up to translations in by Corollary 12. Hence, is an HIMC surface. Similarly, if is a HIMC surface, then, is a HIMC surface.

Comparing a Willmore conformal map with a GHIMC surface, we advance the theory of GHIMC surfaces.

Define one-form , , and by (7) and (8). The map is Willmore if and only if is closed (see [10]). Theorem 1 is a counterpart in GHIMC surfaces.

Proof of Theorem 1. We have Then, is closed. Hence, is a GHIMC surface if and only if there exists with such that is closed.

We explain Theorem 1 in a different way. Set . Then, is a frame of . Set . Then, is the dual frame of .

Corollary 16. Let be a line bundle of a conformal map with mean curvature sphere such that for every . Set , , and . Define a map by (6), a one-form by (7), and a one-form by (8). The map is a GHIMC surface if and only if is closed.

Proof. We have Hence, the corollary follows by Theorem 1.

Let be a map such that . If is Willmore, then the subspace spanned by and defines a Bäcklund transform of and the map is called a backward Bäcklund transform (see [10]). Theorem 2 is a counterpart of a backward Bäcklund transform of a Willmore surface in GHIMC surfaces.

Proof of Theorem 2. Define by (7). Then, We have . By Lemma 6, we have . Hence, Hence, . Then, . By Lemma 7, the subspace spanned by and defines a Bäcklund transform of .

As a counterpart in a GHIMC surface, we introduce the notion of a backward Bäcklund transform of a GHIMC surface as follows.

Definition 17. Let be a line bundle of a conformal map with mean curvature sphere such that for every . One assumes that is a GHIMC surface. Define and by (6). Let be a map such that , and let be a map defined by Then one calls the map the backward Bäcklund transform of .
A backward Bäcklund transform of a Willmore surface is Willmore. On the other hand, it is unclear whether a backward Bäcklund transform of a GHIMC surface is a GHIMC surface.
Let be an Euclidean motion of . Then, there exists , with and such that for any .

Lemma 18. Assume that is a GHIMC surface with mean curvature sphere . Define and by (6). Let be a backward Bäcklund transform of with . Then, is a backward Bäcklund transform of .

Proof. The mean curvature sphere of defines the functions , , and by Then, and . Let be the mean curvature vector of . Then, . Hence, Set . Then, . Hence, Then, is a backward Bäcklund transform of .

We can construct a Darboux transform of a GHIMC surface from the backward Bäcklund transform.

Proof of Theorem 3. Because and define a Bäcklund transform of by Theorem 2, there exists a map such that , and the map is a Darboux transform of with respect to by Lemma 9.

We consider an HIMC surface of revolution. Let be a group consisting of all the elements of the rotation group of which fix the origin and a plane containing the origin, and . A conformal map invariant under is called a surface of revolution. It is known that a surface of revolution is an isothermic surface.

Corollary 19. Let be an isothermic GHIMC surface invariant under , and let be the subgroup of holomorphic transformations of such that for each . Assume that and are maps such that , is a classical Darboux transform and   for every . Then is a GHIMC surface invariant under . If is an isothermic HIMC surface invariant under , then is a HIMC surface invariant under .

Proof. By Corollary 14, the map is invariant under . By Theorem 4, the map is a GHIMC surface. By Theorem 4, if is an HIMC surface, then is a HIMC surface.

All HIMC surfaces of revolution are classified in [8]. Let be a coordinate of such that is the standard holomorphic coordinate .

Theorem 20 (see [8, Lemma 2, Theorem 1]). If with the induced metric on is a surface of revolution, then it has a parametrization If with the above parametrization as an HIMC surface, then a real-valued function defined by is a solution to the Painlevé III equation in trigonometric form
Conversely, let be an arbitrary solution to (42) with . Then, the map defined by (40) and is an HIMC surface of revolution.

Then, we have a transform of a solution to the Painlevé III equation in trigonometric form.

Proof of Theorem 5. An HIMC surface in Theorem 20 is invariant under . Because a surface of revolution is an isothermic surface, a classical Darboux transform is an HIMC surface. An HIMC surface is invariant under by Corollary 19. The function is defined by (41) for , and it is a solution to the the Painlevé III equation in trigonometric form (42) by Theorem 20.

Acknowledgment

This work is supported by JSPS KAKENHI Grant no. 22540064.

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