Darboux Transforms of a Harmonic Inverse Mean Curvature Surface

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 𝔼³ 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 2×2 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 𝔼⁴ is modeled on the set of all quaternions ℍ. A quaternionic line trivial bundle with a complex structure over a Riemann surface is associated with a conformal map from the Riemann surface to 𝔼⁴.

We can assume that a quaternionic line trivial bundle associated with a constrained Willmore surface in 𝔼⁴ equips a harmonic complex structure [4]. If a constrained Willmore surface is neither minimal nor superconformal, then this complex structure defines a smooth family {∇^𝜇}_𝜇∈𝑆¹ 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 {∇^𝜇}_𝜇∈𝑆¹. 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 𝔼⁴ 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 [4–6]). The property of a conformal map from a torus to 𝔼⁴ 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 𝔣∶𝑀→ℍ, a mean curvature sphere of 𝔣 is defined by (5). We define a function 𝐻∶𝑀→ℍ by (6). If 𝔣(𝑀)⊂Imℍ up to translations in ℍ, then 𝐻 is real-valued and exactly a mean curvature function of 𝔣 with the induced metric on 𝑀. Hence, an HIMC surface is a surface in Imℍ such that 1/𝐻 is harmonic. We call a conformal map 𝔣∶𝑀→ℍ, such that 𝐻⁻¹ 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 𝑆⁴ in [10].

A one-form 𝑤 is constructed from a mean curvature sphere by (8). A constrained Willmore surface in 𝔼⁴ is a conformal map such that there exists a one-form 𝜂 with 𝑤+𝜂 is closed [4]. The one-form 𝑤 is closed if and only if a conformal map is Willmore [10]. On the other hand, we have the following.

Theorem 1. Let 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆 such that 𝐿_𝑝≠{𝑣_∞𝜆∣𝜆∈ℍ} for every 𝑝∈𝑀. Set 𝔣∶=𝜎(𝐿)∶𝑀→ℍ. Define a map 𝐻∶𝑀→ℍ by (6) and a one-form 𝑤 by (8). The map 𝔣 is a GHIMC surface if and only if there exists 𝑛∈ℍ with 𝑛≠1 such that 𝐻⁻¹∗𝑤𝐻⁻¹+𝑛∗𝑑𝐻⁻¹ is closed.

We assume that 𝑀 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 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆 such that 𝐿_𝑝≠{𝑣_∞𝜆∣𝜆∈ℍ} for every 𝑝∈𝑀. One assumes that 𝔣∶=𝜎(𝐿)∶𝑀→ℍ is a GHIMC surface. Define 𝐻∶𝑀→ℍ and 𝑁∶𝑀→𝑆² by (6). Let 𝜇∶𝑀→ℍ be a map such that 𝑑𝜇=∗𝑑𝐻⁻¹, and let 𝔥 be a map such that is defined by 1𝔥=2−𝜇+𝔣−𝑁𝐻⁻¹. (1) Let 𝑀^∼ be the subset of 𝑀 where 𝔥 is defined. The quaternionic vector space ℌ spanned by 𝑣_𝐿 and 𝑣_𝐿𝔥 defines a Bäcklund transform of 𝐿 on 𝑀^∼.

We call the map 𝔥 a backward Bäcklund transform of a GHIMC surface 𝔣. 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 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆 such that 𝐿_𝑝≠{𝑣_∞𝜆|𝜆∈ℍ} for every 𝑝∈𝑀. One assumes that 𝔣∶=𝜎(𝐿)∶𝑀→ℍ is a GHIMC surface. Let 𝔥∶𝑀^∼→ℍ be a map such that 𝔥 is a backward Bäcklund transform of 𝔣 on 𝑀^∼. Then, there exists a map 𝜆_∞∶𝑀^∼→ℍ such that 𝔣̂∶=𝜆_∞𝔥⁻¹+𝔣 is a Darboux transform of 𝔣 on 𝑀^∼.

An isothermic surface in 𝔼⁴ 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 𝔣∶𝑀→ℍ is invariant under a subgroup ℬ of Euclidean motions in 𝔼⁴, then there is a Darboux transform 𝑓 invariant under 𝐵 (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 𝔼³≅Imℍ invariant under rotations ℛ around the 𝑘-axis in Imℍ. Then, we obtain a transform of the solutions to the Painlevé III equation in trigonometric form as follows.

Theorem 5. Let 𝜙(𝑥) be a solution to the Painlevé III equation in trigonometric form (42) with 𝜙^′(𝑥)+2sin(𝜙(𝑥))≠0, and let 𝔣∶{𝑥+𝑦𝑖∈ℂ∣𝑥>0}→Imℍ be an HIMC surface of revolution defined by Theorem 20. Assume that 𝔣̂∶=𝜆_∞𝜆_𝐿⁻¹+𝔣 is a classical Darboux transform of 𝔣 with 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0 and 𝐵𝜆_∞=𝜆_∞∘𝜏_𝐵  for every 𝐵∈ℛ. Then, 𝔣̂ is an HIMC surface of revolution invariant under ℛ, and the function 𝜙(𝑥) defined by (41) for 𝔣̂ 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 𝑀 is a simply connected Riemann surface with complex structure 𝐽^𝑀 and maps, vector bundles, and sections are smooth. An orientation of 𝑀 is fixed so that, for any nonzero tangent vector 𝑣 of 𝑀, an ordered pair (𝑣, 𝐽^𝑀𝑣) is a positive basis. We denote by 𝑇𝑀 the tangent bundle of 𝑀 and by 𝑇^∗𝑀 the cotangent bundle of 𝑀.

Let 𝐸 be a vector space. We denote by Ω^𝑝(𝐸) the set of 𝑝-forms on 𝑀 with values in 𝐸. For a vector bundle 𝐹 over 𝑀, we denote by Ω^𝑝(𝐹) the set of 𝑝-forms on 𝑀 with values in 𝐹. For 𝜔∈Ω¹(𝐹), we define ∗𝜔 by ∗𝜔=𝜔∘𝐽^𝑀. We denote by 𝐸 the trivial bundle over 𝑀. Then, Ω^𝑝(𝐸) is naturally identified with Ω^𝑝(𝐸).

We model ℝ⁴ on the set of all quaternions ℍ and ℝ³ on the set of all purely imaginary quaternions Imℍ. For a quaternion 𝑎, we denote by 𝑎 the quaternionic conjugate of 𝑎. Set Re𝑎∶=(𝑎+𝑎)/2 and Im𝑎∶=(𝑎−𝑎)/2. Then, the inner product of 𝑎 and 𝑏∈ℝ⁴≅ℍ is ⟨𝑎,𝑏⟩=Re(𝑎𝑏)=2⁻¹(𝑎𝑏+𝑏𝑎), and the norm of 𝑎∈ℍ is |𝑎|∶=(𝑎𝑎)^1/2. We denote by Imℍ the set of all purely imaginary quaternions. If 𝑎, 𝑏∈Imℍ, then 𝑎𝑏 is the vector product 𝑎×𝑏. Let 𝑆² be the sphere of radius one centered at the origin in Imℍ. Then, 𝑆²={𝑎∈Imℍ∣𝑎²=−1}. Hence, 𝑆² is the set of all square roots of −1 in Imℍ.

Let 𝜔∈Ω¹(ℍ). We define 𝜔_𝑁∈Ω¹(ℍ) and 𝜔^𝑁∈Ω¹(ℍ) by setting 𝜔_𝑁1∶=2(𝜔−𝑁∗𝜔),𝜔^𝑁1∶=2(𝜔−∗𝜔𝑁). (2) Then, 𝜔 decomposes because 𝜔=𝜔_𝑁+𝜔_−𝑁=𝜔^𝑁+𝜔^−𝑁. We see that ∗𝜔_𝑁=𝑁𝜔_𝑁 and ∗𝜔^𝑁=𝜔𝑁. Clearly, 𝜔=𝜔_𝑁 if and only if 𝜔_−𝑁=0. Similarly, 𝜔=𝜔^𝑁 if and only if 𝜔^−𝑁=0. The quaternionic conjugation provides an identity 𝜔_𝑁=𝜔^−𝑁. It is known that 𝜔^𝑁∧𝜔_𝑁=0 ([10], Proposition  16).

Let 𝐸 and 𝐹 be right quaternionic vector bundles over a manifold. We denote by Hom(𝐸,𝐹) the real vector bundle of quaternionic bundle homomorphism from 𝐸 to 𝐹 and denote by End(𝐸) the real vector bundle of quaternionic bundle endomorphisms of 𝐸. We denote the identity automorphism of 𝐸 by Id. We call 𝐽∈Ω⁰(End(𝐸)) a complex structure of 𝐸 if 𝐽²=−𝐼𝑑.

We denote by 𝑉 the right quaternionic vector space of dimension two and by ℙ(𝑉) the quaternionic projective space of dimension one. We model the conformal four-sphere 𝑆⁴ on ℙ(𝑉). Let 𝒯 be the tautological line bundle of ℙ(𝑉). We denote the trivial quaternionic vector bundle over ℙ(𝑉) of rank two by 𝒱. The tangent bundle 𝑇ℙ(𝑉) of ℙ(𝑉) is identified with a bundle Hom(𝒯,𝒱/𝒯).

We fix a basis (𝑣₀,𝑣_∞) of 𝑉. Let 𝑉^∗ be the dual vector space of 𝑉. There exists uniquely a basis (𝑣^∗₀,𝑣^∗_∞) of 𝑉^∗ such that 𝑣^∗₀𝑣₀=1,𝑣^∗₀𝑣_∞𝑣=0,_∗∞𝑣₀=0,𝑣^∗_∞𝑣_∞=1. (3) We define a map 𝜎∶𝑉⧵{𝑣_∞𝜆∣𝜆∈ℍ}→ℍ by 𝜎(𝑣)=𝑣^∗_∞(𝑣)𝑣^∗₀(𝑣)⁻¹ (4) for every 𝑣∈𝑉. The map 𝜎 induces a map from ℙ(𝑉)⧵{𝑣_∞𝜆|𝜆∈ℍ} to ℍ. This map is a stereographic projection of 𝑆⁴≅ℙ(𝑉) from {𝑣_∞𝜆∣𝜆∈ℍ}∈ℙ(𝑉).

Let 𝑓∶𝑀→ℙ(𝑉)≅𝑆⁴ be a map and 𝐿∶=𝑓^∗𝒯. Then, 𝐿 is a quaternionic line subbundle of 𝑉 such that 𝐿_𝑝=𝑓(𝑝) for every 𝑝∈𝑀. Conversely, if 𝐿 is a line subbundle of 𝑉, then there exists a map 𝑓∶𝑀→ℙ(𝑉) such that 𝐿=𝑓^∗𝒯. We call 𝐿 the line bundle of a conformal map 𝑓. We use the terminology of maps for line bundles.

Let 𝜋^𝑉/𝐿∶𝑉→𝑉/𝐿 be the projection. We consider 𝑣₀ and 𝑣_∞ as sections of 𝑉. Define a flat quaternionic connection ∇ on 𝑉 by ∇𝑣₀=∇𝑣_∞=0. Then, the differential map 𝑑𝑓∶𝑇𝑀→𝑇ℙ(𝑉) is identified with a one-form 𝛿∶=𝜋^𝑉/𝐿∇|_Ω⁰_(𝐿)∈Ω¹(Hom(𝐿,𝑉/𝐿)). A line bundle 𝐿 is called a line bundle of a conformal map if there exists a complex structure 𝑆∈Ω⁰(End(𝑉)) such that 𝑆𝐿=𝐿,∇𝑆|_𝐿∈Ω¹(𝐿),∗𝛿=𝜋^𝑉/𝐿𝑆∇|_Ω⁰_(𝐿)=𝛿𝑆|_Ω⁰_(𝐿). (5) The complex structure 𝑆 is called the mean curvature sphere or the conformal Gauss map of 𝐿.

Let 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆. We assume that 𝐿_𝑝≠{𝑣_∞𝜆∣𝜆∈ℍ} for every 𝑝∈𝑀. We define a map 𝔣∶𝑀→ℍ by 𝔣(𝑝)∶=𝜎(𝐿_𝑝) for each 𝑝∈𝑀. The section 𝑣_𝐿∶=𝑣₀+𝑣_∞𝔣 of 𝐿 does not vanish anywhere. We see that (𝑣_∞,𝑣_𝐿) is a frame of 𝑉. We call 𝑆 the mean curvature sphere of 𝔣. There exist 𝐻∶𝑀→ℍ, 𝑁∶𝑀→𝑆², and 𝑅∶𝑀→𝑆² such that 𝑆𝑣_∞=𝑣_∞𝑁+𝑣_𝐿(−𝐻),𝑆𝑣_𝐿=𝑣_𝐿(),𝑁−𝑅₂=𝑅²()=−1,𝑅𝐻=𝐻𝑁,𝑑𝔣𝐻=𝑑𝑁_𝑁,𝐻𝑑𝔣=(𝑑𝑅)^−𝑅. (6) Then, (𝑑𝑓)_−𝑁=(𝑑𝑓)^𝑅=0.

Conversely, let 𝔣∶𝑀→ℍ be a map with (𝑑𝔣)_−𝑁=(𝑑𝔣)^𝑅=0. Then, a map 𝐻∶𝑀→ℍ is defined by (6) except at the branch points or 𝔣. We define a line bundle 𝐿 by 𝐿_𝑝={(𝑣₀+𝑣_∞𝔣(𝑝))𝜆|𝜆∈ℍ} for each 𝑝∈𝑀. Then, 𝐿 is a line bundle of a conformal map with mean curvature sphere 𝑆 defined by (6) except at the branch points of 𝔣.

Let 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆 such that 𝐿_𝑝≠{𝑣_∞𝜆|𝜆∈ℍ} for every 𝑝∈𝑀. We induce a (singular) metric on 𝑀 by 𝔣=𝜎(𝐿). Then, 𝔣 is a conformal map except at the branch points of 𝔣. The map 𝔣 is called a conformal map with left normal 𝑁 and right normal 𝑅. Let ℋ be the mean curvature vector of 𝔣. Then, ℋ=−𝑁𝐻=−𝐻𝑅. The image of 𝔣 is contained in Imℍ up to translations in ℍ if and only if 𝑁=𝑅. If the image of 𝔣 is contained in Imℍ up to translations in ℍ, then 𝐻 is the mean curvature function of 𝔣.

We define 𝐴, 𝑄∈Ω¹(End(𝑉)) by 2𝐴=𝑆𝑑𝑆+∗𝑑𝑆,2𝑄=𝑆𝑑𝑆−∗𝑑𝑆. (7) The one-form 𝐴 and 𝑄 are called the Hopf fields of 𝐿. We have 4𝐴𝑣_∞=𝑣_𝐿(−∗𝑤),4𝐴𝑣_𝐿=𝑣_𝐿2𝑅(𝑑𝑅)_−𝑅,4𝑄𝑣_∞=𝑣_∞2𝑁(𝑑𝑁)_−𝑁+𝑣_𝐿∗(2𝑑𝐻−𝑤),4𝑄𝑣_𝐿=0,𝑤=𝑑𝐻+𝐻∗𝑑𝔣𝐻+𝑅∗𝑑𝐻−𝐻∗𝑑𝑁. (8)

For 𝐶∈End(𝑉), we set ⟨𝐶⟩∶=(1/8)tr_ℝ𝐶. Then, an indefinite scalar product ⟨,⟩ of End(𝑉) is defined by setting ⟨𝐶₁,𝐶₂⟩∶=⟨𝐶₁𝐶₂⟩ for 𝐶₁, 𝐶₂∈End(𝑉). The functional 𝑊(𝐿)1∶=𝜋_𝑀⟨𝐴∧∗𝐴⟩ (9) is called the Willmore functional of 𝐿. A conformal map whose line bundle is a critical line bundle of 𝑊 is called a Willmore surface. A critical line bundle is called a Willmore line bundle. It is known that a line bundle 𝐿 of a conformal map with 𝐿_𝑝≠{𝑣_∞𝜆∣𝜆∈ℍ} for every 𝑝∈𝑀 is Willmore if and only if 𝑑𝑤=0 ([10], Proposition  15).

The following lemma is proved in [10] implicitly.

Lemma 6. One has (2𝑑𝐻−𝑤)^−𝑁=0.

Proof. We have 𝑣_∞2𝑁(𝑑𝑁)_−𝑁+𝑣_𝐿∗(2𝑑𝐻−𝑤)=4𝑄𝑣_∞=−∗4𝑄𝑆𝑣_∞=−∗4𝑄𝑣_∞𝑁+𝑣_𝐿(−𝐻)=𝑣_∞−2(𝑑𝑁)_−𝑁+𝑣_𝐿(2𝑑𝐻−𝑤)𝑁=𝑣_∞2𝑁(𝑑𝑁)_−𝑁+𝑣_𝐿(2𝑑𝐻−𝑤)𝑁. (10) Hence, ∗(2𝑑𝐻−𝑤)=(2𝑑𝐻−𝑤)𝑁. Then, (2𝑑𝐻−𝑤)^−𝑁=0.

3. Transforms

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

Let 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆 such that 𝐿_𝑝≠{𝑣_∞𝜆∣𝜆∈ℍ} for every 𝑝∈𝑀. Set 𝔣∶=𝜎(𝐿)∶𝑀→ℍ and 𝑣_𝐿∶=𝑣₀+𝑣_∞𝔣. Define 𝑁∶𝑀→𝑆²,  𝑅∶𝑀→𝑆², and 𝐻∶𝑀→ℍ by (6). We have a splitting 𝑉=𝐿⊕(𝑉/𝐿). Let 𝜋^𝐿∶𝑉→𝐿 be the projection. We decompose ∇|_Ω⁰_(𝐿) by this splitting as ∇|_Ω⁰_(𝐿)=∇^𝐿+𝛿,∇^𝐿=𝜋^𝐿∇|_Ω⁰_(𝐿). (11) We see that ∇^𝐿 is a connection of 𝐿. Because ∇^𝐿𝑣_𝐿=𝜋^𝐿∇𝑣_𝐿=𝜋^𝐿𝑣_∞𝑑𝔣=0, the section 𝑣_𝐿 is a nowhere-vanishing parallel section of ∇^𝐿. Hence, ∇^𝐿 is a flat connection. Set 𝐷^𝐿∶=2⁻¹(∇^𝐿+𝑆∗∇^𝐿). The operator 𝐷^𝐿 is a quaternionic holomorphic structure of 𝐿 (see [11]). Because ∇^𝐿𝑣_𝐿=0, we have 𝐷^𝐿𝑣_𝐿=0.

Assume that 𝜆₀∶𝑀→ℍ and 𝜆₁∶𝑀→ℍ are maps such that 𝑣_𝐿𝜆₀ and 𝑣_𝐿𝜆₁ are linearly independent, 𝐷^𝐿(𝑣_𝐿𝜆₀)=𝐷^𝐿(𝑣_𝐿𝜆₁)=0 and |𝜆₀(𝑝)|+|𝜆₁(𝑝)|≠0 for any 𝑝∈𝑀. Let ℌ be the two-dimensional vector space spanned by 𝑣_𝐿𝜆₀ and 𝑣_𝐿𝜆₁. Then, ℌ is a trivial bundle over 𝑀 of rank two. Let ev∶ℌ→𝐿 be the evaluation map defined by ev𝑝,𝑣_𝐿𝜆₀𝜇₀+𝑣_𝐿𝜆₁𝜇₁=𝑝,𝑣_𝐿𝜆₀𝜇₀+𝜆₁𝜇₁𝑝. (12) The map ev∶ℌ→𝐿 is a surjective bundle homomorphism. Hence, we have an injective bundle homomorphism ev^∗∶𝐿⁻¹→ℌ^∗. Let 𝐿^∗ and ℌ^∗ be the dual bundles of 𝐿 and ℌ, respectively. Then, we have a line bundle ev^∗(𝐿^∗)⊂ℌ^∗ of a conformal map. The line bundle ev^∗(𝐿^∗) is called the Bäcklund transform of 𝐿 with respect to ℌ.

Lemma 7. A nonconstant map 𝜆∶𝑀→ℍ satisfies the equation (𝑑𝜆)_𝑅=0 if and only if the subspace spanned by 𝑣_𝐿 and 𝑣_𝐿𝜆 defines a Bäcklund transform of 𝐿.

Proof. Recall that 𝐷^𝐿𝑣_𝐿=0 and that  𝑣_𝐿 is nowhere vanishing. For a nonconstant map 𝜆∶𝑀→ℍ, we have 2𝐷^𝐿𝑣_𝐿𝜆=∇^𝐿𝑣_𝐿𝜆+𝑆∗∇^𝐿𝑣_𝐿𝜆=𝑣_𝐿2(𝑑𝜆)_𝑅. (13) Hence, 𝐷^𝐿(𝑣_𝐿𝜆)=0 if and only if (𝑑𝜆)_𝑅=0. Thus, 𝑣_𝐿 and 𝑣_𝐿𝜆 defines a Bäcklund transform if and only if 𝜆 is not constant and (𝑑𝜆)_𝑅=0.

Lemma 8. Let 𝜆_∞∶𝑀→ℍ and 𝜆_𝐿∶𝑀→ℍ be maps satisfying the equation 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0. Then a subspace of Ω⁰(𝐿) spanned by 𝑣_𝐿 and 𝑣_𝐿𝜆_𝐿 defines a Bäcklund transform of 𝐿.

Proof. We have 0=𝑑(𝑑𝜆_∞)=𝑑(−𝑑𝔣𝜆_𝐿)=𝑑𝔣∧𝑑𝜆_𝐿. Hence (𝑑𝜆_𝐿)_𝑅=0. By Lemma 7, a subspace of Ω⁰(𝐿) spanned by 𝑣_𝐿 and 𝑣_𝐿𝜆_𝐿 defines a Bäcklund transform of 𝐿.

For 𝜙∈Ω⁰(𝑉/𝐿), we denote by 𝜙^∼∈Ω⁰(𝑉) a lift of 𝜙, that is, 𝜋^𝑉∼/𝐿𝜙=𝜙. Set 𝐷^𝑉/𝐿𝜙∶=𝜋^𝑉/𝐿12∇𝜙^∼+𝑆∗∇𝜙^∼. (14) The operator 𝐷^𝑉/𝐿 is a quaternionic holomorphic structure of 𝑉/𝐿. It is known that if 𝐷^𝑉/𝐿𝜙=0, then there exists a unique lift 𝜙 such that 𝜋^𝑉/𝐿∇𝜙=0. Then, a line bundle 𝐿 with 𝐿_𝑝={𝜙(𝑝)𝜆|𝜆∈ℍ} for every 𝑝∈𝑀 away from zeros of 𝜙 is a line bundle of a conformal map. Set 𝛿∶=𝜋^𝑉/𝐿∇. Then, 𝑆𝐿=𝐿 and ∗𝛿=𝑆𝛿. The line bundle 𝐿 of a conformal map is called a Darboux transform of 𝐿 with respect to 𝜙. We call the map 𝔣̂=𝜎(𝐿) a Darboux transform of 𝔣 with respect to 𝜙.

A Darboux transform 𝐿 is called classical if 𝐿_𝑝≠𝐿_𝑝 for each 𝑝∈𝑀 and 𝑆 is the mean curvature sphere of 𝐿 [13]. Let 𝑆 be the mean curvature sphere of 𝐿, and set 𝔣̂∶=𝜎(𝐿). If 𝐿 is a classical Darboux transform, then 𝔣 and 𝔣̂ share conformal curvature line parametrizations. The conformal maps 𝔣 and 𝔣̂ are called isothermic surfaces (see [13]).

The equation 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0 is related with a Darboux transform as follows.

Lemma 9. If maps 𝜆_∞∶𝑀→ℍ and 𝜆_𝐿∶𝑀→ℍ satisfy the equation 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0, then 𝔣̂=𝜆_∞𝜆_𝐿⁻¹+𝔣 is a Darboux transform of 𝔣 with respect to 𝑣_∞𝜆_∞, and 𝔣 is a Darboux transform of 𝔣̂ with respect to 𝑣_∞𝜆_𝐿⁻¹.

Proof. We identify 𝑉/𝐿 with ker𝑣^∗₀={𝑣_∞𝜆∣𝜆∈ℍ}. For 𝑣_∞𝜆_∞∈Ω⁰(ker𝑣^∗₀)≅Ω⁰(𝑉/𝐿) with 𝜆_∞∶𝑀→ℍ, we have 2𝐷^𝑉/𝐿𝑣_∞𝜆_∞=𝜋^𝑉/𝐿𝑣_∞𝑑𝜆_∞+𝑆𝑣_∞∗𝑑𝜆_∞=𝜋^𝑉/𝐿𝑣_∞𝑑𝜆_∞+𝑁∗𝑑𝜆_∞=𝑣_∞2𝑑𝜆_∞−𝑁. (15) Hence, 𝐷^𝑉/𝐿𝑣_∞𝜆_∞=0 if and only if (𝑑𝜆_∞)_−𝑁=0.
A lift of 𝜙=𝑣_∞𝜆_∞ is written as 𝜙^∼=𝑣_∞𝜆_∞+𝑣_𝐿𝜆_𝐿 for 𝜆_𝐿∶𝑀→ℍ. Because 𝜎(𝐿_𝑝)=𝔣(𝑝) for every 𝑝∈𝑀, we have 𝜋^𝑉/𝐿∇𝜙^∼=𝜋^𝑉/𝐿𝑣_∞𝑑𝜆_∞+𝑣_∞𝑑𝔣𝜆_𝐿+𝑣_𝐿𝑑𝜆_𝐿=𝑣_∞𝑑𝜆_∞+𝑑𝔣𝜆_𝐿. (16) Hence, 𝜙^∼=𝑣_∞𝜆_∞+𝑣_𝐿𝜆_𝐿 satisfies the equation 𝜋^𝑉/𝐿∇𝜙^∼=0 if and only if 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0. Thus, if maps 𝜆_∞∶𝑀→ℍ and 𝜆_𝐿∶𝑀→ℍ satisfy the equation 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0, then a line bundle 𝐿 with 𝐿_𝑝={𝜙^∼(𝑝)𝜆∣𝜆∈ℍ} for every 𝑝∈𝑀 away from zero of 𝜙^∼ is a Darboux transform of 𝐿 with respect to 𝑣_∞𝜆_∞. Because 𝜙^∼=𝑣_∞𝜆_∞+𝑣_𝐿𝜆_𝐿=𝑣₀𝜆_𝐿+𝑣_∞𝜆_∞+𝔣𝜆_𝐿, (17) we have 𝜎(𝐿)=(𝜆_∞+𝔣𝜆_𝐿)𝜆_𝐿⁻¹=𝜆_∞𝜆_𝐿⁻¹+𝔣=𝔣̂.
We have 𝑑𝔣̂=𝜆_∞𝑑𝜆_𝐿⁻¹. Hence, 𝑑𝜆_𝐿⁻¹+𝑑𝔣̂(−𝜆_∞⁻¹)=0. Then, 𝜆_𝐿⁻¹(−𝜆_∞)+𝔣̂=𝔣 is a Darboux transform of 𝔣̂ with respect to 𝑣_∞𝜆_𝐿⁻¹.

We have the following by straightforward calculation.

Lemma 10. Define 𝑣𝐿, 𝑇_∞, 𝑇_𝐿, 𝑁, 𝑅, and 𝐻 by 𝑣𝐿=𝑣₀+𝑣_∞𝔣̂=𝑣_∞𝑇_∞+𝑣_𝐿𝑇_𝐿,𝑆𝑣_∞=𝑣_∞𝐿−𝐻,𝐿𝐿−𝑅,𝑁𝑁+𝑣𝑆𝑣=𝑣₂=𝑅²𝑅𝐻𝑑𝔣̂𝑑𝑁𝑁,̂𝑑𝑅=−1,𝐻=𝑁,𝐻=𝐻𝑑𝔣=⁻𝑅. (18) Then, 𝑇_𝐿=1,𝑇_∞=𝔣̂−𝔣,𝑁𝑇_∞+𝑇_∞𝐻𝑇_∞+𝑇_∞̂𝐻̂.𝑅=0,𝑁−𝑁=𝔣−𝔣𝐻,𝑅−𝑅=𝔣−𝔣 (19) The map 𝔣 is an isothermic surface, and 𝔣̂ is a classical Darboux transform if and only if 𝑅=−𝑇_∞⁻¹𝑁𝑇_∞. If a map 𝔣 is an isothermic surface and 𝔣̂ is a classical Darboux transform, then 𝐻=𝐻.

Proof. Because 𝑣_∞𝑇_∞+𝑣_𝐿𝑇_𝐿=𝑣_∞𝑇_∞+𝑣₀+𝑣_∞𝔣𝑇_𝐿=𝑣₀𝑇_𝐿+𝑣_∞𝑇_∞+𝔣𝑇_𝐿, (20) we have 𝑇_𝐿=1 and 𝑇_∞=𝔣̂−𝔣.
Because 𝑆𝐿𝑣𝐿=𝐿,𝑆𝑣=𝑆_∞𝑇_∞+𝑣_𝐿=𝑣_∞𝑁+𝑣_𝐿(−𝐻)𝑇_∞+𝑣_𝐿(−𝑅)=𝑣_∞𝑁𝑇_∞+𝑣_𝐿−𝐻𝑇_∞−𝑅=𝑣_∞𝑁𝑇_∞+−𝑣_∞𝑇_∞𝐿+𝑣−𝐻𝑇_∞−𝑅=𝑣_∞𝑁𝑇_∞+𝑇_∞𝐻𝑇_∞+𝑇_∞𝑅𝐿+𝑣−𝐻𝑇_∞−𝑅, (21) we have 𝑁𝑇_∞+𝑇_∞𝐻𝑇_∞+𝑇_∞𝑅=0. Because −𝐻𝑇_∞−𝑅=−𝑇_∞⁻¹𝑇_∞𝐻𝑇_∞+𝑇_∞𝑅=𝑇_∞⁻¹𝑁𝑇_∞, (22) we have 𝐿𝐿𝑆𝑣=𝑣𝑇_∞⁻¹𝑁𝑇_∞.
Because ∗𝛿=𝑆𝛿=𝑆𝛿, we have 𝜋^𝑉/𝐿𝑆𝑣_∞=𝜋^𝑉/𝐿𝑆𝑣_∞,𝑆𝑣_∞=𝑣_∞𝑁+𝑣_𝐿(−𝐻)=𝑣_∞𝑁+−𝑣_∞𝑇_∞𝐿+𝑣(−𝐻)=𝑣_∞𝑁+𝑇_∞𝐻𝐿+𝑣(−𝐻),𝑆𝑣_∞=𝑣_∞𝐿𝑁+𝑣−𝐻. (23) Hence, 𝑁−𝑁=(𝔣̂−𝔣)𝐻.
By Lemma 9, a conformal map 𝔣 is a Darboux transform of 𝔣̂. Because 𝑑𝔣(−𝐻)=(𝑑𝑅)_𝑅 and 𝑑𝔣̂(−𝑅𝐻)=(𝑑𝑅), we have 𝑅−𝑅=(𝔣−𝔣)̂(−𝐻).
Then, 𝑅−𝑅=𝐻(𝔣̂−𝔣).
We assume that 𝔣 is an isothermic surface and a Darboux transform 𝐿 of 𝐿 is classical.
We have 𝐿=𝐿𝛿𝑆𝑣𝛿𝑣𝑇_∞⁻¹𝑁𝑇_∞=𝑣_∞𝑑𝑓𝑇_∞⁻¹𝑁𝑇_∞,𝛿𝐿=𝐿𝑆𝑣𝛿𝑣−𝑅=𝑣_∞𝑑𝑓−𝑅. (24) Hence, 𝑅=−𝑇_∞⁻¹𝑁𝑇_∞.
Conversely, if 𝑅=−𝑇_∞⁻¹𝑁𝑇_∞, then, ∗𝛿=𝛿𝑆=𝛿𝑆. Because 𝐿 is a Darboux transform of 𝐿, we have 𝑆𝛿=𝑆𝛿. Hence, 𝐿 is a classical Darboux transform of 𝐿.
If 𝔣 is an isothermic surface and 𝔣̂ is a classical Darboux transform, then 𝔣 is a classical Darboux transform of 𝔣̂. Hence 𝑁−𝑁=(𝔣−𝔣)̂𝐻. Then 𝐻=𝐻.

A pair of two conformal maps 𝔣 and 𝔤 is called a Christoffel pair in [14] if 𝑑𝔣∧𝑑𝔤=𝑑𝔤∧𝑑𝔣=0. It is known that if a pair of  𝔣 and 𝔤 is a Christoffel pair, then maps 𝔣 and 𝔤 are isothermic. If a map 𝔣 is isothermic, then there exists a map 𝔤 such that a pair of 𝔣 and 𝔤 is a Christoffel pair (see [14]).

Corollary 11. One assumes that conformal maps 𝜆_∞∶𝑀→ℍ, 𝜆_𝐿∶𝑀→ℍ, 𝔣∶𝑀→ℍ, and 𝔤∶𝑀→ℍ satisfy the equation 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0 and 𝑑𝜆_𝐿+𝑑𝔤𝜆_∞=0. Set 𝔣̂∶=𝜆_∞𝜆_𝐿⁻¹+𝔣 and 𝔤∶=𝜆_𝐿𝜆_∞⁻¹+𝔤. If 𝑑𝜆_∞∧𝑑𝜆_𝐿⁻¹=𝑑𝜆_𝐿∧𝑑𝜆_∞⁻¹=0, then 𝑑𝔣∧𝑑𝔤=𝑑𝔤∧𝑑𝔣=0 and 𝑑𝔣̂∧𝑑𝔤=𝑑𝔤∧𝑑𝔣̂=0. The conformal maps 𝔣̂ and 𝔤 are classical Darboux transforms of 𝔣 and 𝔤, respectively.

Proof. We have 𝑑𝔣∧𝑑𝔤=𝑑𝜆_∞𝜆_𝐿⁻¹∧𝑑𝜆_𝐿𝜆_∞⁻¹=−𝑑𝜆_∞∧𝑑𝜆_𝐿⁻¹𝜆_𝐿𝜆_∞⁻¹=0,𝑑𝔤∧𝑑𝔣=𝑑𝜆_𝐿𝜆_∞⁻¹∧𝑑𝜆_∞𝜆_𝐿⁻¹=−𝑑𝜆_𝐿∧𝑑𝜆_∞⁻¹𝜆_∞𝜆_𝐿⁻¹=0. (25) Because 𝑑𝔣̂=𝜆_∞𝑑𝜆_𝐿⁻¹ and 𝑑𝔤=𝜆_𝐿𝑑𝜆_∞⁻¹, we have 𝑑𝔣̂∧𝑑𝔤=𝜆_∞𝑑𝜆_𝐿⁻¹∧𝜆_𝐿𝑑𝜆_∞⁻¹=−𝜆_∞𝜆_𝐿⁻¹𝑑𝜆_𝐿∧𝑑𝜆_∞⁻¹𝑑̂=0,𝔤∧𝑑𝔣=𝜆_𝐿𝑑𝜆_∞⁻¹∧𝜆_∞𝑑𝜆_𝐿⁻¹=−𝜆_𝐿𝜆_∞⁻¹𝑑𝜆_∞∧𝑑𝜆_𝐿⁻¹=0. (26) We assume that (𝑑𝔣)_−𝑁=0 and (𝑑𝔣)̂𝑅=0. Because 𝑑𝔣=𝑑𝜆_∞𝜆_𝐿⁻¹, 𝑑𝔣̂=𝜆_∞𝑑𝜆_𝐿⁻¹, and 𝑑𝔣̂∧𝜆_𝐿𝑑𝜆_∞⁻¹=−𝑑𝔣̂∧𝜆_𝐿𝜆_∞⁻¹𝑑𝜆_∞𝜆_∞⁻¹=0, we have 𝑅=−𝜆_𝐿𝜆_∞⁻¹𝑁𝜆_∞𝜆_𝐿⁻¹=−𝑇_∞⁻¹𝑁𝑇_∞. Hence, 𝔣̂ is a classical Darboux transform of 𝔣.

We see that a pair of an isothermic surface in 𝔼³ and its classical Darboux transform form a Bonnet pair in [15] up to similarities in 𝔼³.

For an isothermic surface in 𝔼³, we have the following.

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

Proof. Assume that the image of 𝔣 is contained in Imℍ up to translations in ℍ. Then, 𝑁=𝑅 and 𝐻 is real-valued. By Lemma 10, we have 𝐻=𝐻. Hence, 𝐻 is real-valued. Then, ̂𝐻̂̂𝑁=𝑁+𝔣−𝔣𝐻,𝑅=𝑅+𝔣−𝔣=𝑁+𝔣−𝔣𝐻=𝑁. (27) Then, the image of  𝔣̂ is contained in Imℍ up to translations in ℍ.
The converse is trivial.

Let 𝐵∶ℍ→ℍ be an Euclidean motion of 𝔼⁴≅ℍ fixing the origin. Then, there exists 𝑟, 𝑠∈ℍ with |𝑟|=|𝑠|=1 such that 𝐵𝑎=𝑟𝑎𝑠⁻¹ for any 𝑎∈ℍ. If 𝑟=𝑠, then 𝐵 is an Euclidean motion of 𝔼⁴≅ℍ fixing ℝ⊂ℍ and Imℍ.

Lemma 13. Let 𝔣∶𝑀→ℍ be a conformal map and 𝔣̂=𝜆_∞𝜆_𝐿⁻¹+𝔣 a Darboux transform of  𝔣 with respect to 𝑣_∞𝜆_∞ such that 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0. Then, 𝐵𝔣̂ is a Darboux transform of 𝐵𝔣 with respect to 𝑣_∞𝐵𝜆_∞.

Proof. We have 𝐵𝔣̂=𝐵𝜆_∞𝑠𝜆_𝐿⁻¹𝑠⁻¹+𝐵𝔣. (28) Because 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0, we have 𝑑𝐵𝜆_∞+𝑑(𝐵𝔣)𝑠𝜆_𝐿𝑠⁻¹=0. (29) Hence, 𝐵𝔣̂ is a Darboux transform of 𝐵𝔣 with respect to 𝑣_∞𝐵𝜆_∞.

For a subgroup ℬ of Euclidean motions of 𝔼^𝑛   (𝑛=3,4), we say that a map 𝔣∶𝑀→𝔼^𝑛 is invariant under ℬ if there exists a subgroup {𝜏_𝐵}_𝐵∈ℬ of holomorphic transformations of 𝑀 such that 𝐵𝔣=𝔣∘𝜏_𝐵 for any 𝐵∈ℬ.

Corollary 14. Let ℬ be a subgroup of Euclidean motions of 𝔼⁴≅ℍ such that each element of ℬ fixes the origin of ℍ, and let 𝔣∶𝑀→ℍ be a conformal map invariant under ℬ. Let 𝜏_𝐵∶𝑀→𝑀 be a holomorphic transformation of 𝑀 such that 𝐵𝔣=𝔣∘𝜏_𝐵 for each 𝐵∈ℬ. Assume that 𝜆_∞∶𝑀→ℍ and 𝜆_𝐿∶𝑀→ℍ are maps such that 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0 and 𝐵𝜆_∞=𝜆_∞∘𝜏_𝐵 for every 𝐵∈ℬ. Set 𝔣̂∶=𝜆_∞𝜆_𝐿⁻¹+𝔣. Then, 𝔣̂ is a Darboux transform of 𝔣 invariant under ℬ.

Proof. Let 𝐵∈ℬ. Then, there exists 𝑟, 𝑠∈ℍ with |𝑟|=|𝑠|=1 such that 𝐵𝑎=𝑟𝑎𝑠⁻¹ for any 𝑎∈ℍ. We have 𝐵𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=𝑑𝐵𝜆_∞+𝑑(𝐵𝔣)𝑠𝜆_𝐿𝑠⁻¹𝜏=0,_∗𝐵𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=𝑑𝜆_∞∘𝜏_𝐵+𝑑𝔣∘𝜏_𝐵𝜆_𝐿∘𝜏_𝐵=𝑑𝐵𝜆_∞+(𝑑(𝐵𝔣))𝜆_𝐿∘𝜏_𝐵=0. (30) Hence, 𝜆_𝐿∘𝜏_𝐵=𝑠𝜆_𝐿𝑠⁻¹. By the definition, the map 𝔣̂ is a Darboux transform of 𝔣 with respect to 𝑣_∞𝜆_∞. We have 𝐵𝔣̂=𝐵𝜆_∞𝜆_𝐿⁻¹+𝔣=𝐵𝜆_∞𝑠𝜆_𝐿⁻¹𝑠⁻¹=𝜆+𝐵𝔣_∞∘𝜏_𝐵𝜆_𝐿∘𝜏_𝐵⁻¹+𝔣∘𝜏_𝐵=𝔣̂∘𝜏_𝐵, (31) for each 𝐵∈ℬ. Hence, 𝔣̂ is invariant under 𝐵.

4. GHIMC Surfaces

A conformal map 𝔣∶𝑀→Imℍ is called a harmonic inverse mean curvature surface if 𝑑∗𝑑𝐻⁻¹=0 [1]. We introduce the notion of a generalized harmonic inverse mean curvature surface and define its transform.

Let 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆 such that 𝐿_𝑝≠{𝑣_∞𝜆∣𝜆∈ℍ} for every 𝑝∈𝑀. Set 𝔣∶=𝜎(𝐿)∶𝑀→ℍ. Define 𝐻∶𝑀→ℍ by (6).

Definition 15. One calls the map 𝔣 a generalized harmonic inverse mean curvature surface if 𝑑∗𝑑𝐻⁻¹=0.

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 𝔣 whose image is contained in Imℍ up to translations in ℍ is exactly an HIMC surface.

Proof of Theorem 4. Let 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆 such that 𝐿_𝑝≠{𝑣_∞𝜆∣𝜆∈ℍ} for every 𝑝∈𝑀, and let  𝐿 be a Darboux transform of 𝐿. Set 𝔣∶=𝜎(𝐿) and 𝔣̂∶=𝜎(𝐿). We assume that the Darboux transform 𝔣̂ is classical. Define 𝐻∶𝑀→ℍ by (6) and 𝐻∶𝑀→ℍ by (18). By Lemma 10, we have 𝐻=𝐻. Hence, 𝑑∗𝑑𝐻=0 if and only if 𝑑∗𝑑𝐻=0. Hence, 𝔣 is a GHIMC surface if and only if 𝔣̂ is.
If 𝔣 is an HIMC surface, then the image of 𝔣̂ is contained in Imℍ up to translations in ℍ by Corollary 12. Hence, 𝔣̂ is an HIMC surface. Similarly, if 𝔣̂ is a HIMC surface, then, 𝔣 is a HIMC surface.

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

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

Proof of Theorem 1. We have −𝐻⁻¹∗𝑤𝐻⁻¹=𝐻⁻¹×(−∗𝑑𝐻+𝐻𝑑𝔣𝐻+𝑅𝑑𝐻−𝐻𝑑𝑁)𝐻⁻¹=∗𝑑𝐻⁻¹+𝑑𝔣+𝐻⁻¹𝑅𝑑𝐻𝐻⁻¹−𝑑𝑁𝐻⁻¹=∗𝑑𝐻⁻¹+𝑑𝔣+𝑁𝐻⁻¹𝑑𝐻𝐻⁻¹−𝑑𝑁𝐻⁻¹=∗𝑑𝐻⁻¹+𝑑𝔣−𝑁𝑑𝐻⁻¹−𝑑𝑁𝐻⁻¹=∗𝑑𝐻⁻¹+𝑑𝔣−𝑁𝐻⁻¹. (32) Then, 𝐻⁻¹∗𝑤𝐻⁻¹+∗𝑑𝐻⁻¹ is closed. Hence, 𝐿 is a GHIMC surface if and only if there exists 𝑛∈ℍ with 𝑛≠1 such that 𝐻⁻¹∗𝑤𝐻⁻¹+𝑛∗𝑑𝐻⁻¹ is closed.

We explain Theorem 1 in a different way. Set 𝑣_𝐿∶=𝑣₀+𝑣_∞𝔣. Then, (𝑣_∞,𝑣_𝐿) is a frame of 𝑉. Set 𝜓∶=−𝔣𝑣^∗₀+𝑣^∗_∞. Then, (𝜓,𝑣^∗₀) is the dual frame of (𝑣_∞,𝑣_𝐿).

Corollary 16. Let 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆 such that 𝐿_𝑝≠{𝑣_∞𝜆|𝜆∈ℍ} for every 𝑝∈𝑀. Set 𝔣∶=𝜎(𝐿)∶𝑀→ℍ, 𝑣_𝐿∶=𝑣₀+𝑣_∞𝔣, and 𝜓∶=−𝔣𝑣^∗₀+𝑣^∗_∞. Define a map 𝐻∶𝑀→ℍ by (6), a one-form 𝑄 by (7), and a one-form 𝑤 by (8). The map 𝔣∶𝑀→ℍ is a GHIMC surface if and only if 𝐻⁻¹𝜓(𝑄𝑣_∞)𝐻⁻¹ is closed.

Proof. We have 4𝐻⁻¹𝜓𝑄𝑣_∞𝐻⁻¹=−2∗𝑑𝐻⁻¹−𝐻⁻¹∗𝑤𝐻⁻¹. (33) Hence, the corollary follows by Theorem 1.

Let 𝔟∶𝑀→ℍ be a map such that 𝑑𝔟=𝑤/2−𝑑𝐻. If 𝔣 is Willmore, then the subspace spanned by 𝑣_𝐿 and 𝑣_𝐿𝔟 defines a Bäcklund transform of 𝐿 and the map 𝔟 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 𝑄 by (7). Then, 2𝑑𝔥=−𝑑𝜇+𝑑𝔣−𝑁𝐻⁻¹=−2∗𝑑𝐻⁻¹−𝐻⁻¹∗𝑤𝐻⁻¹=4𝐻⁻¹𝜓𝑄𝑣_∞𝐻⁻¹. (34) We have 𝜓(𝑄𝑣_∞)=∗(2𝑑𝐻−𝑤). By Lemma 6, we have (𝜓(𝑄𝑣_∞))^−𝑁=0. Hence, ∗𝐻⁻¹𝜓𝑄𝑣_∞𝐻⁻¹=𝐻⁻¹∗𝜓𝑄𝑣_∞𝐻⁻¹=𝐻⁻¹𝜓𝑄𝑣_∞𝑁𝐻⁻¹=𝐻⁻¹𝜓𝑄𝑣_∞𝐻⁻¹𝑅. (35) Hence, (𝑑𝔥)^−𝑅=0. Then, (𝑑𝔥)_𝑅=0. By Lemma 7, the subspace spanned by 𝑣_𝐿 and 𝑣_𝐿𝔥 defines a Bäcklund transform of 𝐿.

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 𝐿 be a line bundle of a conformal map with mean curvature sphere 𝑆 such that 𝐿_𝑝≠{𝑣_∞𝜆|𝜆∈ℍ} for every 𝑝∈𝑀. One assumes that 𝔣∶=𝜎(𝐿) is a GHIMC surface. Define 𝐻∶𝑀→ℍ and 𝑁∶𝑀→𝑆² by (6). Let 𝜇∶𝑀→ℍ be a map such that 𝑑𝜇=∗𝑑𝐻⁻¹, and let 𝔥∶𝑀→ℍ be a map defined by 1𝔥=2−𝜇+𝔣−𝑁𝐻⁻¹. (36) Then one calls the map 𝔥 the backward Bäcklund transform of 𝔣.
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 𝐵∶ℍ→ℍ be an Euclidean motion of 𝔼⁴≅ℍ. Then, there exists 𝑟, 𝑠∈ℍ with |𝑟|=|𝑠|=1 and 𝑡∈ℍ such that 𝐵𝑎=𝑟𝑎𝑠⁻¹+𝑡 for any 𝑎∈ℍ.

Lemma 18. Assume that 𝔣∶𝑀→ℍ is a GHIMC surface with mean curvature sphere 𝑆. Define 𝑁∶𝑀→𝑆² and 𝐻∶𝑀→ℍ by (6). Let 𝔥=(−𝜇+𝔣−𝑁𝐻⁻¹)/2 be a backward Bäcklund transform of 𝔣 with 𝑑𝜇=∗𝑑𝐻⁻¹. Then, 𝐵(2𝔥)/2 is a backward Bäcklund transform of 𝐵𝔣.

Proof. The mean curvature sphere 𝑆^∼ of 𝐵𝔣 defines the functions 𝑁^∼∶𝑀→𝑆², 𝑅^∼∶𝑀→𝑆², and 𝐻^∼∶𝑀→ℍ by 𝑣^∼_𝐿=𝑣₀+𝑣_∞𝐵𝔣,_∼𝑆𝑣_∞=𝑣𝑁_∞^∼+𝑣^∼_𝐿−𝐻^∼,𝑆^∼𝑣^∼_𝐿=𝑣^∼_𝐿−𝑅^∼,_∼𝑁²=𝑅^∼2=−1,𝑅^∼𝐻^∼=𝐻^∼𝑁^∼,𝑑(𝐵𝔣)𝐻^∼=𝑑𝑁^∼∼_𝑁,𝐻^∼𝑑(𝐵𝔣)=𝑑𝑅^{∼−∼𝑅}. (37) Then, 𝑁^∼=𝑟𝑁𝑟⁻¹ and 𝑅^∼=𝑠𝑅𝑠⁻¹. Let ℋ^∼ be the mean curvature vector of 𝐵𝔣. Then, ℋ^∼=𝑟ℋ𝑠⁻¹. Hence, 𝐻^∼=−ℋ^∼𝑁^∼=−𝑟ℋ𝑠⁻¹𝑟𝑁𝑟⁻¹=−𝑠ℋ𝑁𝑟⁻¹=−𝑠𝐻𝑟⁻¹. (38) Set 𝜇^∼∶=𝑟𝜇𝑠⁻¹. Then, 𝑑𝜇^∼=∗𝑑𝐻^∼−1=𝑟𝑑𝜇𝑠⁻¹. Hence, 𝐵(2𝔥)=𝑟−𝜇+𝔣−𝑁𝐻⁻¹𝑠⁻¹+𝑡=−𝑟𝜇𝑠⁻¹+𝑟𝔣𝑠⁻¹−𝑟𝑁𝑟⁻¹𝑟𝐻⁻¹𝑠⁻¹+𝑡=−𝜇^∼+𝐵𝔣−𝑁^∼𝐻^∼−1. (39) Then, 𝐵(2𝔥)/2 is a backward Bäcklund transform of 𝐵𝔣.

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

Proof of Theorem 3. Because 𝑣_𝐿 and 𝑣_𝐿𝔥 define a Bäcklund transform of 𝐿 by Theorem 2, there exists a map 𝜆_∞∶𝑀→ℍ such that 𝑑𝜆_∞+𝑑𝔣𝔥=0, and the map 𝔣̂=𝜆_∞𝔥⁻¹+𝔣 is a Darboux transform of 𝔣 with respect to 𝑣_∞𝜆_∞ by Lemma 9.

We consider an HIMC surface of revolution. Let ℛ be a group consisting of all the elements of the rotation group of ℍ which fix the origin and a plane containing the origin, 1 and 𝑘. A conformal map 𝔣∶𝑀→Imℍ invariant under ℛ is called a surface of revolution. It is known that a surface of revolution is an isothermic surface.

Corollary 19. Let 𝔣∶𝑀→ℍ be an isothermic GHIMC surface invariant under ℛ, and let {𝜏_𝐵}_𝐵∈ℛ be the subgroup of holomorphic transformations of 𝑀 such that 𝐵𝔣=𝔣∘𝜏_𝐵 for each 𝐵∈ℛ. Assume that 𝜆_∞∶𝑀→ℍ and 𝜆_𝐿∶𝑀→ℍ are maps such that 𝑑𝜆_∞+𝑑𝔣𝜆_𝐿=0, 𝔣̂∶=𝜆_∞𝜆_𝐿⁻¹+𝔣 is a classical Darboux transform and 𝐵𝜆_∞=𝜆_∞∘𝜏_𝐵  for every 𝐵∈ℛ. Then 𝔣̂ is a GHIMC surface invariant under ℛ. If 𝔣∶𝑀→Imℍ is an isothermic HIMC surface invariant under ℛ, then 𝔣̂ is a HIMC surface invariant under ℛ.

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

All HIMC surfaces of revolution are classified in [8]. Let (𝑥,𝑦) be a coordinate of ℝ² such that 𝑥+𝑦𝑖 is the standard holomorphic coordinate ℂ.

Theorem 20 (see [8, Lemma 2, Theorem 1]). If 𝔣∶ℂ→Imℍ with the induced metric on ℂ is a surface of revolution, then it has a parametrization 𝔣=𝑒𝑥+𝑦𝑖^𝑢(𝑥)/2cos𝑎𝑦𝑎𝑒𝑖+^𝑢(𝑥)/2sin𝑎𝑦𝑎𝑐(𝑥)𝑗+𝑎{0}.𝑘,𝑢∶ℝ⟶ℝ,𝑐∶ℝ⟶ℝ,𝑎∈ℝ⧵ (40) If 𝔣∶{𝑥+𝑦𝑖∈ℂ∣𝑥>0}→Imℍ with the above parametrization as an HIMC surface, then a real-valued function 𝜙∶ℝ→ℝ defined by 𝜙(𝑥=1sin)2𝑐^′(𝑥)𝑒^{−𝑢(𝑥)/2}𝜙(𝑥=1,cos)4𝑢^′(𝑥) (41) is a solution to the Painlevé III equation in trigonometric form 𝑥𝜙^′′(𝑥)−2sin2𝜙(𝑥)+𝜙^′(𝑥)+2sin𝜙(𝑥)=0. (42)
Conversely, let 𝜙(𝑥) be an arbitrary solution to (42) with 𝜙^′(𝑥)+2sin(𝜙(𝑥))≠0. Then, the map 𝔣∶ℂ→Imℍ defined by (40) and 𝑒^𝑢(𝑥)=𝑥²𝜙^′(𝑥)+2sin𝜙(𝑥)²4,𝑐(𝑥)𝑥=−²4𝜙^′(𝑥)²−4sin𝜙(𝑥)² (43) 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 𝔣 in Theorem 20 is invariant under ℛ. Because a surface of revolution is an isothermic surface, a classical Darboux transform 𝔣̂ is an HIMC surface. An HIMC surface 𝔣̂ is invariant under ℛ by Corollary 19. The function 𝜙(𝑥) is defined by (41) for 𝔣̂, 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.