# Darboux Transforms of a Harmonic Inverse Mean Curvature Surface

Authors: Katsuhiro Moriya
Journal: Geometry
 (2013)

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## 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.


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## Body

## 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 aconstrained 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 to𝔼4 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.Aharmonic 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 map𝔣: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-formw 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.
LetL be a line bundle of a conformal map with mean curvature sphere S such that Lp≠{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 thatM 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.
LetL be a line bundle of a conformal map with mean curvature sphere S such that Lp≠{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
(1)𝔥=12(-μ+𝔣-NH-1).
Let M~ be the subset of M where 𝔥 is defined. The quaternionic vector space ℌ spanned by vL and vL𝔥¯ defines a Bäcklund transform of L on M~.We call the map𝔥¯ 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.
LetL be a line bundle of a conformal map with mean curvature sphere S such that Lp≠{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 𝔣^:=λ∞𝔥¯-1+𝔣 is a Darboux transform of 𝔣 on M~.An isothermic surface in𝔼4 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 map𝔣:M→ℍ is invariant under a subgroupℬ 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.
Letϕ(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 𝔣^:=λ∞λL-1+𝔣 is a classical Darboux transform of 𝔣 with dλ∞+d𝔣λL=0 and Bλ∞=λ∞∘τB  for every B∈ℛ. Then, 𝔣^ is an HIMC surface of revolution invariant under ℛ, and the function ϕ^(x) defined by (41) for 𝔣^ 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 thatM is a simply connected Riemann surface with complex structure JM 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, JMv) is a positive basis. We denote by TM the tangent bundle of M and by T*M the cotangent bundle of M.LetE be a vector space. We denote by Ωp(E) the set of p-forms on M with values in E. For a vector bundle F over M, we denote by Ωp(F) the set of p-forms on M with values in F. For ω∈Ω1(F), we define *ω by *ω=ω∘JM. We denote by E_ the trivial bundle over M. Then, Ωp(E) is naturally identified with Ωp(E_).We modelℝ4 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(a¯b)=2-1(a¯b+b¯a), and the norm of a∈ℍ is |a|:=(a¯a)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.Letω∈Ω1(ℍ). We define ωN∈Ω1(ℍ) and ωN∈Ω1(ℍ) by setting
(2)ωN:=12(ω-N*ω),ωN:=12(ω-*ωN).
Then, ω decomposes because ω=ωN+ω-N=ωN+ω-N. We see that *ωN=NωN and *ωN=ωN. Clearly, ω=ωN if and only if ω-N=0. Similarly, ω=ωN if and only if ω-N=0. The quaternionic conjugation provides an identity ωN¯=ω¯-N. It is known that ωN∧ωN=0 ([10], Proposition  16).LetE 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 byV 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 (v0*,v∞*) of V* such that
(3)v0*(v0)=1,v0*(v∞)=0,v∞*(v0)=0,v∞*(v∞)=1.
We define a map σ:V∖{v∞λ∣λ∈ℍ}→ℍ by
(4)σ(v)=(v∞*(v))(v0*(v))-1
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).Letf:M→ℙ(V)≅S4 be a map and L:=f*𝒯. Then, L is a quaternionic line subbundle of V_ such that Lp=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 mapf. We use the terminology of maps for line bundles.LetπV_/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:=πV_/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
(5)SL=L,∇S|L∈Ω1(L),*δ=πV_/LS∇|Ω0(L)=δS|Ω0(L).
The complex structure S is called the mean curvature sphere or the conformal Gauss map of L.LetL be a line bundle of a conformal map with mean curvature sphere S. We assume that Lp≠{v∞λ∣λ∈ℍ} for every p∈M. We define a map 𝔣:M→ℍ by 𝔣(p):=σ(Lp) for each p∈M. The section vL:=v0+v∞𝔣 of L does not vanish anywhere. We see that (v∞,vL) 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
(6)Sv∞=v∞N+vL(-H),SvL=vL(-R),N2=R2=-1,RH=HN,d𝔣H=(dN)N,Hd𝔣=(dR)-R.
Then, (df)-N=(df)R=0.Conversely, let𝔣: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 Lp={(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.LetL be a line bundle of a conformal map with mean curvature sphere S such that Lp≠{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-=-H-R. 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 defineA, Q∈Ω1(End(V_)) by
(7)2A=SdS+*dS,2Q=SdS-*dS.
The one-form A and Q are called the Hopf fields of L. We have
(8)4Av∞=vL(-*w),4AvL=vL2R(dR)-R,4Qv∞=v∞2N(dN)-N+vL*(2dH-w),4QvL=0,w=dH+H*d𝔣H+R*dH-H*dN.ForC∈End(V), we set 〈C〉:=(1/8)trℝC. Then, an indefinite scalar product 〈,〉 of End(V) is defined by setting 〈C1,C2〉:=〈C1C2〉 for C1, C2∈End(V). The functional
(9)W(L):=1π∫M‍〈A∧*A〉
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 Lp≠{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(10)v∞2N(dN)-N+vL*(2dH-w)=4Qv∞=-*4QSv∞=-*4Q(v∞N+vL(-H))=(v∞(-2(dN)-N)+vL(2dH-w))N=v∞2N(dN)-N+vL(2dH-w)N.
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].LetL be a line bundle of a conformal map with mean curvature sphere S such that Lp≠{v∞λ∣λ∈ℍ} for every p∈M. Set 𝔣:=σ(L):M→ℍ and vL:=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 πL:V_→L be the projection. We decompose ∇|Ω0(L) by this splitting as
(11)∇|Ω0(L)=∇L+δ,∇L=πL∇|Ω0(L).
We see that ∇L is a connection of L. Because ∇LvL=πL∇vL=πLv∞d𝔣=0, the section vL is a nowhere-vanishing parallel section of ∇L. Hence, ∇L is a flat connection. Set DL:=2-1(∇L+S*∇L). The operator DL is a quaternionic holomorphic structure of L (see [11]). Because ∇LvL=0, we have DLvL=0.Assume thatλ0:M→ℍ and λ1:M→ℍ are maps such that vLλ0 and vLλ1 are linearly independent, DL(vLλ0)=DL(vLλ1)=0 and |λ0(p)|+|λ1(p)|≠0 for any p∈M. Let ℌ be the two-dimensional vector space spanned by vLλ0 and vLλ1. Then, ℌ_ is a trivial bundle over M of rank two. Let ev:ℌ_→L be the evaluation map defined by
(12)ev(p,vLλ0μ0+vLλ1μ1)=(p,(vL(λ0μ0+λ1μ1))(p)).
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 mapλ:M→ℍ satisfies the equation (dλ)R=0 if and only if the subspace spanned by vL and vLλ defines a Bäcklund transform of L.Proof.
Recall thatDLvL=0 and that  vL is nowhere vanishing. For a nonconstant map λ:M→ℍ, we have
(13)2DLvLλ=∇L(vLλ)+S*∇L(vLλ)=vL2(dλ)R.
Hence, DL(vLλ)=0 if and only if (dλ)R=0. Thus, vL and vLλ defines a Bäcklund transform if and only if λ is not constant and (dλ)R=0.Lemma 8.
Letλ∞:M→ℍ and λL:M→ℍ be maps satisfying the equation dλ∞+d𝔣λL=0. Then a subspace of Ω0(L) spanned by vL and vLλL defines a Bäcklund transform of L.Proof.
We have0=d(dλ∞)=d(-d𝔣λL)=d𝔣∧dλL. Hence (dλL)R=0. By Lemma 7, a subspace of Ω0(L) spanned by vL and vLλL defines a Bäcklund transform of L.Forϕ∈Ω0(V_/L), we denote by ϕ~∈Ω0(V_) a lift of ϕ, that is, πV_/Lϕ~=ϕ. Set
(14)DV_/Lϕ:=πV_/L(12(∇ϕ~+S*∇ϕ~)).
The operator DV_/L is a quaternionic holomorphic structure of V_/L. It is known that if DV_/Lϕ=0, then there exists a unique lift ϕ^ such that πV_/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^:=πV_/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 𝔣^=σ(L^) a Darboux transform of 𝔣 with respect to ϕ.A Darboux transformL^ is called classical if Lp≠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 𝔣^:=σ(L^). If L^ is a classical Darboux transform, then 𝔣 and 𝔣^ share conformal curvature line parametrizations. The conformal maps 𝔣 and 𝔣^ are called isothermic surfaces (see [13]).The equationdλ∞+d𝔣λL=0 is related with a Darboux transform as follows.Lemma 9.
If mapsλ∞:M→ℍ and λL:M→ℍ satisfy the equation dλ∞+d𝔣λL=0, then 𝔣^=λ∞λL-1+𝔣 is a Darboux transform of 𝔣 with respect to v∞λ∞, and 𝔣¯ is a Darboux transform of 𝔣^¯ with respect to v∞λL-1¯.Proof.
We identifyV_/L with kerv0*_={v∞λ∣λ∈ℍ}_. For v∞λ∞∈Ω0(kerv0*_)≅Ω0(V_/L) with λ∞:M→ℍ, we have
(15)2DV_/Lv∞λ∞=πV_/L(v∞dλ∞+Sv∞*dλ∞)=πV_/L(v∞(dλ∞+N*dλ∞))=v∞2(dλ∞)-N.
Hence, DV_/Lv∞λ∞=0 if and only if (dλ∞)-N=0.
A lift ofϕ=v∞λ∞ is written as ϕ~=v∞λ∞+vLλL for λL:M→ℍ. Because σ(Lp)=𝔣(p) for every p∈M, we have
(16)πV/L∇ϕ~=πV/L(v∞dλ∞+v∞d𝔣λL+vLdλL)=v∞(dλ∞+d𝔣λL).
Hence, ϕ~=v∞λ∞+vLλL satisfies the equation πV/L∇ϕ~=0 if and only if dλ∞+d𝔣λL=0. Thus, if maps λ∞:M→ℍ and λL:M→ℍ satisfy the equation dλ∞+d𝔣λL=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
(17)ϕ~=v∞λ∞+vLλL=v0λL+v∞(λ∞+𝔣λL),
we have σ(L^)=(λ∞+𝔣λL)λL-1=λ∞λL-1+𝔣=𝔣^.
We haved𝔣^=λ∞dλL-1. Hence, dλL-1¯+d𝔣^¯(-λ∞-1¯)=0. Then, λL-1¯(-λ∞¯)+𝔣^¯=𝔣¯ is a Darboux transform of 𝔣^¯ with respect to v∞λL-1¯.We have the following by straightforward calculation.Lemma 10.
DefinevL^, T∞, TL, N^, R^, and H^ by
(18)vL^=v0+v∞𝔣^=v∞T∞+vLTL,S^v∞=v∞N^+vL^(-H^),S^vL^=vL^(-R^),N^2=R^2=-1,R^H^=H^N^,d𝔣^H^=(dN^)N^,H^d𝔣^=(dR^)-R^.
Then,
(19)TL=1,T∞=𝔣^-𝔣,NT∞+T∞HT∞+T∞R=0,N^-N=(𝔣^-𝔣)H,R^-R=H^(𝔣^-𝔣).
The map 𝔣 is an isothermic surface, and 𝔣^ is a classical Darboux transform if and only if R^=-T∞-1NT∞. If a map 𝔣 is an isothermic surface and 𝔣^ is a classical Darboux transform, then H^=H.Proof.
Because(20)v∞T∞+vLTL=v∞T∞+(v0+v∞𝔣)TL=v0TL+v∞(T∞+𝔣TL),
we have TL=1 and T∞=𝔣^-𝔣.
Because(21)SL^=L^,SvL^=S(v∞T∞+vL)=(v∞N+vL(-H))T∞+vL(-R)=v∞NT∞+vL(-HT∞-R)=v∞NT∞+(-v∞T∞+vL^)(-HT∞-R)=v∞(NT∞+T∞HT∞+T∞R)+vL^(-HT∞-R),
we have NT∞+T∞HT∞+T∞R=0. Because
(22)-HT∞-R=-T∞-1(T∞HT∞+T∞R)=T∞-1NT∞,
we have SvL^=vL^T∞-1NT∞.
Because*δ^=S^δ^=Sδ^, we have
(23)πV_/L^S^v∞=πV_/L^Sv∞,Sv∞=v∞N+vL(-H)=v∞N+(-v∞T∞+vL^)(-H)=v∞(N+T∞H)+vL^(-H),S^v∞=v∞N^+vL^(-H^).
Hence, N^-N=(𝔣^-𝔣)H.
By Lemma9, a conformal map 𝔣¯ is a Darboux transform of 𝔣^¯. Because d𝔣¯(-H¯)=(dR)R and d𝔣^¯(-H^¯)=(dR^)R^, we have R-R^=(𝔣¯-𝔣^¯)(-H^¯).
Then,R^-R=H^(𝔣^-𝔣).
We assume that𝔣 is an isothermic surface and a Darboux transform L^ of L is classical.
We have(24)δ^SvL^=δ^vL^T∞-1NT∞=v∞df^T∞-1NT∞,δ^S^vL^=δ^vL^(-R^)=v∞df^(-R^).
Hence, R^=-T∞-1NT∞.
Conversely, ifR^=-T∞-1NT∞, then, *δ^=δ^S^=δ^S. Because L^ is a Darboux transform of L, we have S^δ^=Sδ^. Hence, L^ is a classical Darboux transform of L.
If𝔣 is an isothermic surface and 𝔣^ is a classical Darboux transform, then 𝔣 is a classical Darboux transform of 𝔣^. Hence N-N^=(𝔣-𝔣^)H^. Then H^=H.A pair of two conformal maps𝔣 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 mapsλ∞:M→ℍ, λL:M→ℍ, 𝔣:M→ℍ, and 𝔤:M→ℍ satisfy the equation dλ∞+d𝔣λL=0 and dλL+d𝔤λ∞=0. Set 𝔣^:=λ∞λL-1+𝔣 and 𝔤^:=λLλ∞-1+𝔤. If dλ∞∧dλL-1=dλL∧dλ∞-1=0, then d𝔣∧d𝔤=d𝔤∧d𝔣=0 and d𝔣^∧d𝔤^=d𝔤^∧d𝔣^=0. The conformal maps 𝔣^ and 𝔤^ are classical Darboux transforms of 𝔣 and 𝔤, respectively.Proof.
We have(25)d𝔣∧d𝔤=dλ∞λL-1∧dλLλ∞-1=-dλ∞∧dλL-1λLλ∞-1=0,d𝔤∧d𝔣=dλLλ∞-1∧dλ∞λL-1=-dλL∧dλ∞-1λ∞λL-1=0.
Because d𝔣^=λ∞dλL-1 and d𝔤^=λLdλ∞-1, we have
(26)d𝔣^∧d𝔤^=λ∞dλL-1∧λLdλ∞-1=-λ∞λL-1dλL∧dλ∞-1=0,d𝔤^∧d𝔣^=λLdλ∞-1∧λ∞dλL-1=-λLλ∞-1dλ∞∧dλL-1=0.
We assume that (d𝔣)-N=0 and (d𝔣^)R^=0. Because d𝔣=dλ∞λL-1, d𝔣^=λ∞dλL-1, and d𝔣^∧λLdλ∞-1=-d𝔣^∧λLλ∞-1dλ∞λ∞-1=0, we have R^=-λLλ∞-1Nλ∞λL-1=-T∞-1NT∞. Hence, 𝔣^ is a classical Darboux transform of 𝔣.We see that a pair of an isothermic surface in𝔼3 and its classical Darboux transform form a Bonnet pair in [15] up to similarities in 𝔼3.For an isothermic surface in𝔼3, we have the following.Corollary 12.
One assumes thatL^ 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  𝔣^ is contained in Imℍ up to translations in ℍ.Proof.
Assume that the image of𝔣 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,
(27)N^=N+(𝔣^-𝔣)H,R^=R+H^(𝔣^-𝔣)=N+(𝔣^-𝔣)H=N^.
Then, the image of  𝔣^ is contained in Imℍ up to translations in ℍ.
The converse is trivial.LetB: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.
Let𝔣:M→ℍ be a conformal map and 𝔣^=λ∞λL-1+𝔣 a Darboux transform of  𝔣 with respect to v∞λ∞ such that dλ∞+d𝔣λL=0. Then, B𝔣^ is a Darboux transform of B𝔣 with respect to v∞Bλ∞.Proof.
We have(28)B𝔣^=Bλ∞sλL-1s-1+B𝔣.
Because dλ∞+d𝔣λL=0, we have
(29)d(Bλ∞)+d(B𝔣)sλLs-1=0.
Hence, B𝔣^ is a Darboux transform of B𝔣 with respect to v∞Bλ∞.For a subgroupℬ of Euclidean motions of 𝔼n   (n=3,4), we say that a map 𝔣:M→𝔼n is invariant underℬ if there exists a subgroup {τB}B∈ℬ of holomorphic transformations of M such that B𝔣=𝔣∘τB for any B∈ℬ.Corollary 14.
Letℬ 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 τB:M→M be a holomorphic transformation of M such that B𝔣=𝔣∘τB for each B∈ℬ. Assume that λ∞:M→ℍ and λL:M→ℍ are maps such that dλ∞+d𝔣λL=0 and Bλ∞=λ∞∘τB for every B∈ℬ. Set 𝔣^:=λ∞λL-1+𝔣. Then, 𝔣^ is a Darboux transform of 𝔣 invariant under ℬ.Proof.
LetB∈ℬ. Then, there exists r, s∈ℍ with |r|=|s|=1 such that Ba=ras-1 for any a∈ℍ. We have
(30)B(dλ∞+d𝔣λL)=d(Bλ∞)+d(B𝔣)sλLs-1=0,τB*(dλ∞+d𝔣λL)=d(λ∞∘τB)+(d(𝔣∘τB))λL∘τB=d(Bλ∞)+(d(B𝔣))λL∘τB=0.
Hence, λL∘τB=sλLs-1. By the definition, the map 𝔣^ is a Darboux transform of 𝔣 with respect to v∞λ∞. We have
(31)B𝔣^=B(λ∞λL-1+𝔣)=(Bλ∞)sλL-1s-1+B𝔣=(λ∞∘τB)(λL∘τB)-1+𝔣∘τB=𝔣^∘τB,
for each B∈ℬ. Hence, 𝔣^ is invariant under B.


## 4. GHIMC Surfaces

A conformal map𝔣: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.LetL be a line bundle of a conformal map with mean curvature sphere S such that Lp≠{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 map𝔣 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 surface𝔣 whose image is contained in Imℍ up to translations in ℍ is exactly an HIMC surface.Proof of Theorem4.
LetL be a line bundle of a conformal map with mean curvature sphere S such that Lp≠{v∞λ∣λ∈ℍ} for every p∈M, and let  L^ be a Darboux transform of L. Set 𝔣:=σ(L) and 𝔣^:=σ(L^). We assume that the Darboux transform 𝔣^ 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 𝔣^ is.
If𝔣 is an HIMC surface, then the image of 𝔣^ is contained in Imℍ up to translations in ℍ by Corollary 12. Hence, 𝔣^ is an HIMC surface. Similarly, if 𝔣^ 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-formA, 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 Theorem1.
We have(32)-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+d(𝔣-NH-1).
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 Theorem1 in a different way. Set vL:=v0+v∞𝔣. Then, (v∞,vL) is a frame of V_. Set ψ:=-𝔣v0*+v∞*. Then, (ψ,v0*) is the dual frame of (v∞,vL).Corollary 16.
LetL be a line bundle of a conformal map with mean curvature sphere S such that Lp≠{v∞λ|λ∈ℍ} for every p∈M. Set 𝔣:=σ(L):M→ℍ, vL:=v0+v∞𝔣, and ψ:=-𝔣v0*+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(33)4H-1ψ(Qv∞)H-1=-2*dH-1-H-1*wH-1.
Hence, the corollary follows by Theorem 1.Let𝔟:M→ℍ be a map such that d𝔟=w/2-dH. If 𝔣 is Willmore, then the subspace spanned by vL and vL𝔟 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 Theorem2.
DefineQ by (7). Then,
(34)2d𝔥=-dμ+d(𝔣-NH-1)=-2*dH-1-H-1*wH-1=4H-1ψ(Qv∞)H-1.
We have ψ(Qv∞)=*(2dH-w). By Lemma 6, we have (ψ(Qv∞))-N=0. Hence,
(35)*(H-1ψ(Qv∞)H-1)=H-1*(ψ(Qv∞))H-1=H-1ψ(Qv∞)NH-1=H-1ψ(Qv∞)H-1R.
Hence, (d𝔥)-R=0. Then, (d𝔥¯)R=0. By Lemma 7, the subspace spanned by vL and vL𝔥¯ 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.
LetL be a line bundle of a conformal map with mean curvature sphere S such that Lp≠{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
(36)𝔥=12(-μ+𝔣-NH-1).
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.
LetB: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 that𝔣: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 sphereS~ of B𝔣 defines the functions N~:M→S2, R~:M→S2, and H~:M→ℍ by
(37)vL~=v0+v∞B𝔣,S~v∞=v∞N~+vL~(-H~),S~vL~=vL~(-R~),N~2=R~2=-1,R~H~=H~N~,d(B𝔣)H~=(dN~)N~,H~d(B𝔣)=(dR~)-R~.
Then, N~=rNr-1 and R~=sRs-1. Let ℋ~ be the mean curvature vector of B𝔣. Then, ℋ~=rℋs-1. Hence,
(38)H~=-ℋ~¯N~=-rℋs-1¯rNr-1=-sℋ¯Nr-1=-sHr-1.
Set μ~:=rμs-1. Then, dμ~=*dH~-1=rdμs-1. Hence,
(39)B(2𝔥)=r(-μ+𝔣-NH-1)s-1+t=-rμs-1+r𝔣s-1-rNr-1rH-1s-1+t=-μ~+B𝔣-N~H~-1.
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 Theorem3.
BecausevL and vL𝔥¯ define a Bäcklund transform of L by Theorem 2, there exists a map λ∞:M→ℍ such that dλ∞+d𝔣𝔥¯=0, and the map 𝔣^=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. Letℛ 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.
Let𝔣:M→ℍ be an isothermic GHIMC surface invariant under ℛ, and let {τB}B∈ℛ be the subgroup of holomorphic transformations of M such that B𝔣=𝔣∘τB for each B∈ℛ. Assume that λ∞:M→ℍ and λL:M→ℍ are maps such that dλ∞+d𝔣λL=0, 𝔣^:=λ∞λL-1+𝔣 is a classical Darboux transform and Bλ∞=λ∞∘τB  for every B∈ℛ. Then 𝔣^ is a GHIMC surface invariant under ℛ. If 𝔣:M→Imℍ is an isothermic HIMC surface invariant under ℛ, then 𝔣^ is a HIMC surface invariant under ℛ.Proof.
By Corollary14, the map 𝔣^ is invariant under ℛ. By Theorem 4, the map 𝔣^ is a GHIMC surface. By Theorem 4, if 𝔣 is an HIMC surface, then 𝔣^ 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]).
If𝔣:ℂ→Imℍ with the induced metric on ℂ is a surface of revolution, then it has a parametrization
(40)𝔣(x+yi)=eu(x)/2cos(ay)ai+eu(x)/2sin(ay)aj+c(x)ak,u:ℝ⟶ℝ,c:ℝ⟶ℝ,a∈ℝ∖{0}.
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
(41)sin(ϕ(x))=12c′(x)e-u(x)/2,cos(ϕ(x))=14u′(x)
is a solution to the Painlevé III equation in trigonometric form
(42)x[ϕ′′(x)-2sin(2ϕ(x))+ϕ′(x)+2sin(ϕ(x))]=0.
Conversely, letϕ(x) be an arbitrary solution to (42) with ϕ′(x)+2sin(ϕ(x))≠0. Then, the map 𝔣:ℂ→Imℍ defined by (40) and
(43)eu(x)=x2(ϕ′(x)+2sin(ϕ(x)))24,c(x)=-x24{(ϕ′(x))2-4[sin(ϕ(x))]2}
is an HIMC surface of revolution.Then, we have a transform of a solution to the Painlevé III equation in trigonometric form.Proof of Theorem5. An HIMC surface 𝔣 in Theorem 20 is invariant under ℛ. Because a surface of revolution is an isothermic surface, a classical Darboux transform 𝔣^ is an HIMC surface. An HIMC surface 𝔣^ is invariant under ℛ by Corollary 19. The function ϕ^(x) is defined by (41) for 𝔣^, and it is a solution to the the Painlevé III equation in trigonometric form (42) by Theorem 20.



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