Isometric immersions into Lorentzian products

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Isometric immersions into Lorentzian products

Julien Roth

To cite this version:

Julien Roth. Isometric immersions into Lorentzian products. International Journal of Geometric

Methods in Modern Physics, World Scientific Publishing, 2011, 8 (6), pp 1-22. �hal-00528583�

JULIEN ROTH

Abstract. We give a necessary and sufficient condition for ann-dimensional Riemannian man- ifold to be isometrically immersed into one of the Lorentzian productsSⁿ×R1orHⁿ×R1. This condition is expressed in terms of its first and second fundamental forms, the tangent and nor- mal projections of the vectical vector field. As applications, we give an equivalent condition in a spinorial way and we deduce the existence of a one-parameter family of isometric maximal deformation of a given maximal surface obtained by rotating the shape operator.

1. Introduction

A fundamental question in the theory of submanifolds is to know when a (pseudo)-Riemannian manifold can be isometrically immersed in a given ambient space. In the case where the ambient space is a Riemannian space form, it is a well-known fact that the Gauss, Codazzi and Ricci equations are necessary and sufficient (see [4, 15]). In particular for immersions of codimension 1, the Ricci equation is trivial and the Gauss and Codazzi equations are equivalent to the existence of a local isometric immersion into the desired space form. This fact is also true for psuedo- Riemannian manifolds (see [12]). Namely, let (Mⁿ⁺¹, g) be an orientable psuedo-Riemannian manifold and (M, g) a hypersurface ofM. We denote by ∇and∇the Levi-Civita connections of (Mⁿ⁺¹, g) and (M, g) respectively and byR andR the associated curvature tensors. The shape operator of M associated with the normal unit vector field ν is the symmetric (1,1)-tensor S defined bySX =−∇ν. Then, it is well-known that the following equations holds for any vector fieldsX, Y, Z tangent toM:

(1) R(X, Y)Z =R(X, Y)Z+δ g(SX, Z)SY −g(SY, Z)SX , (2) ∇XSY − ∇YSX−S[X, Y] =R(X, Y)ν,

where δ= 1 isν is space-like andδ =−1 ifν is time-like. These two equations are respectively the Gauss and Codazzi equations. In the case where the ambient manifold is a space form of curvatureκ, they become

(3) R(X, Y)Z=κ g(X, Z)Y −g(Y, Z)X

+δ g(SX, Z)SY −g(SY, Z)SX ,

(4) ∇XSY − ∇YSX−S[X, Y] = 0.

Thus, the Gauss and Codazzi equations are defined intrinsically on M since they involve only the first and second fundamental forms. Moreover, it is well-known that these two equations are also sufficient conditions for (M, g) to be immersed locally and isometrically into the space form of curvatureκand compatible signature with given fundamental formsg andS.

In the general case, the Gauss and Codazzi equations are not intrinsic because of the curvature tensor of the ambient manifold. Nevertheless, in some special cases, for instance when the ambient space is a Riemannian product of a space form with a real lineMⁿ×Ror a 3-dimensional homoge- nous space, the Gauss and Codazzi equations can be expressed in an extrinsic way. Precisely, for a hypersurface ofMⁿ×R, the Gauss and Codazzi equations depend only on the first and second fundamental form, the projectionT of the vertical vector∂_tontoT M and its normal component f =h∂t, νi. In [5], Daniel proved that these two equations together with two more equations give

Date: October 22, 2010.

2000Mathematics Subject Classification. 53C27, 53C40, 53C80, 58C40.

Key words and phrases. Isometric Immersions, Gauss and Codazzi Equations, Maximal surfaces, Spinors.

1

a necessary and sufficient condition for a manifold M to be immersed locally and isometrically intoMⁿ×R. Namely, he showed the following

Theorem(Daniel [5]). Let(M, g)be an oriented, simply connected surface and∇its Riemannian connection. Let S be field of symmetric endomorphisms S_y :T_yM −→T_yM, T a vector field on M andf a smooth function on M, such that||T||²+f²= 1. If(M, g, S, T, f)satisfies the Gauss and Codazzi equations for Mⁿ(κ)×Rand fro all X ∈X(M),

∇XT =F SX and df(X) =−hSX, Ti, then, there exists an isometric immersion

F :M −→M²(κ)×R

so that the Weingarten operator of the immersion related to the normalν is dF◦S◦dF⁻¹

and such that

∂t=dF(T) +f ν.

Moreover, this immersion is unique up to a global isometry of Mⁿ(κ)×R which preserves the orientation ofR.

Remark 1. Note that contrary to the case of space forms, the Gauss and Codazzi equations are not sufficient here. Indeed, the compatiblity equations require two additional equtions coming form the fact that ∂_tis parallel.

The first aim of this paper is to give such a fundamental theorem of hypersyrfaces for the case of the Lorentzian productsMⁿ(κ)×R₁. We prove the following

Theorem 1. Let (M, g) be an oriented, simply connected Riemannian surface and ∇ its Rie- mannian connection . LetS be field of symmetric endomorphismsS_y:T_yM −→T_yM,T a vector field on M andf a smooth function onM, such that ||T||²−f²=−1. If (M, g, S, T, f) satisfies the Gauss and Codazzi equations forMⁿ(κ)×R₁ and the following equations

∇XT =f SX, df(X) =hSX, Ti, then, there exists an isometric immersion

F :M −→Mⁿ(κ)×R₁

so that the Weingarten operator of the immersion related to the time-like normal ν is dF◦S◦dF⁻¹

and such that

∂_t=dF(T) +f ν.

Moreover, this immersion is unique up to a global isometry of Mⁿ(κ)×R₁ which preserves the orientation ofRin the product.

For this, we use the fairly standard method based on differential forms first used by Cartan and after by Tenenblat [15] or Daniel [5] for instance. Then, from Theorem 1, we prove the existence of a one-parameter family of isometric maximal deformations of a given maximal surface into one of the above Lorentzian product.

Finally, we give a spinorial version of this theorem as in the Riemannian case [6, 11, 14] and generalzing the results of [8, 9] to 3-dimensional Lorentzian products.

2. Preliminaries

First of all, we fix our conventions for index and summation. Latin letters,i, j, k are integers between 1 and n whereas Greek letters denote integer between 0 and n+ 1. Therefore, for summations, we haveP

αuα=u0+· · ·un+1 andP

juj=u1+· · ·un.

2.1. The compatiblity equations. Let (Mⁿ, g) be an oriented Riemannian hypersurface of Mⁿ(κ)×R₁ with normal unit vector fieldν. We denote by∂_t the vertical vector ofMⁿ(κ)×R₁, that is, the unit vector field giving the orientation ofR₁ inMⁿ(κ)×R₁. The projection of∂t on T M is denoted byT and its normal part isf. Therefore, we have∂t=T+f ν, withf =−h∂t, νi sincehν, νi=−1. We can compute the curvature tensor ofMⁿ(κ)×R₁ for vector fields tangent toM. We have the following:

Proposition 2.1. For any vector fields X, Y, Z, W ∈Γ(T M), we have R(X, Y)Z, W

= κ hX, Zi hY, Wi − hY, Zi hX, Wi

+hY, Ti hW, Ti hX, Zi −εhX, Ti hZ, Ti hY, Wi

− hX, Ti hW, Ti hY, Zi+εhY, Ti hZ, Ti hX, Wi

and

R(X, Y)ν, Z

= −f κ hY, Wi hX, Ti − hX, Wi hY, Ti Proof : LetX, Y, Z, W ∈Γ(T M). We can write these vector fields as follows:

X =Xe+x∂t, Y =Ye+y∂t, Z=Ze+z∂t, W =Wf+w∂t, withX,e Y ,e ZeandWftangent toMⁿ(κ) andx, y, zandwsome real-valued functions.

Since,∂tis parallel, we have R(X, Y)Z, W

=

RMⁿ(X,e Ye)Z,e fW

= κ

hX,e ZeihY ,e Wfi − hY ,e Zihe X,e fWi

= κ

hX−x∂t, Z−z∂tihY −y∂t, W−w∂ti

−hY −y∂_t, Z−z∂_tihX−x∂_t, W−w∂_ti

Moreover, since the vector fields X, Y, Z and W are tangent to M and h∂t, ∂_ti = −1, we have x=− hX, Ti, y =− hY, Ti, z =− hZ, Tiand w=− hW, Ti. So, a straightforward computation yields the first identity.

For the second identity, we denote, ν =νe+n∂t with eν tangent to Mⁿ(κ). Clearly, we see that n=− hν, ∂ti=f. So, we have

R(X, Y)ν, W

=

RMⁿ(X,e Ye)eν,Wf

= κ

hX,e eνihY ,e Wfi − hY ,e νihe X,e Wfi

= κ

hX−x∂t, ν−f ∂tihY −y∂t, W−w∂ti

−hY −y∂_t, ν−f ∂_tihX−x∂_t, W−w∂_ti

which gives easily the wanted identity.

Now, using again the fact that the vector field∂_tis parallel, we get the following identities.

Proposition 2.2. For any X∈X(M), we have (1) ∇XT =f SX,

(2) df(X) =hSX, Ti.

Proof : Since∂t=T+f ν is parallel, we have

0 = ∇X∂_t=∇XT+df(X)ν+f∇Xν

= ∇XT−

∇XT, ν

ν+df(X)ν+f∇Xν

But, by definition, ∇Xν = −SX, so by identification of normal and tangential parts, we have

∇XT =f SX anddf(X) =

∇XT, ν

=−

∇Xν, T

=hT, SXi. This concludes the proof.

2.2. Moving frames. The main ingeredient to prove Theorem 1 is the technic of moving frames.

This tool is fairly standard and one can refer for instance to [16, 15, 13, 5].

Let (Mⁿ, g) be a Riemannian manifold, ∇its Levi-Civita connection and Rits curvature tensor.

We consider a field of symmetric operatorsSonM. Let{e1, . . . , en}be a local orthonormal frame onM and{ω¹,· · ·, ωⁿ} its dual basis,i.,e.,ωⁱ(ek) =δ_kⁱ. We also defineωⁿ⁺¹ = 0 and the forms ω_jⁱ,ω_jⁿ⁺¹,ω_n+1ⁱ andωⁿ⁺¹_n+1 by

ω_jⁱ(ek) =h∇eke_j, e_ii, ωⁿ⁺¹_j (ek) =hS(ek), eji, ω_n+1^j =−ω_jⁿ⁺¹ andω_n+1ⁿ⁺¹= 0.

Obviously, we have

∇eke_j =X

i

ωⁱ_j(ek)ei and Se_k=X

j

ωⁿ⁺¹_j (ek)ej.

Finally, we denote Rⁱ_klj = hR(ek, e_l)ej, e_ii. Then, we have the following well-known structure forumlas.

Proposition 2.3. We have

(5) dωⁱ+X

a

ω_aⁱ ∧ω^a= 0,

(6) X

a

ωⁿ⁺¹_a ∧ω^a= 0,

(7) dωⁱ_j+X

a

ω_aⁱ ∧ω^a_j =−1 2

X

k

X

l

Rⁱ_kljω^k∧ω^l.

(8) dωⁿ⁺¹_j +X

a

ω_aⁿ⁺¹∧ω^a_j =−1 2

X

k

X

l

h∇ekSe_l− ∇elSe_k−S[ek, e_l], ejiω^k∧ω^l. These formulas are well-known. One can see [5] for a complete proof of these proposition.

2.3. Other facts about Mⁿ(κ)×R₁ and their hypersurfaces. In all this section, we will consider (Mⁿ, g) a hypersurface of Mⁿ(κ)×R₁. I We denote by R^p

1 the p-dimensional Lorentz space, that is,R^p endowed with the metric

(dx0)²+· · ·+ (dxp−2)²−(dx^p−1)².

andR^p₂thep-dimensional pseudo-Euclidean space of signature (p−2,2), that is,R^pendowed with the metric

−(dx0)²+ (dx1)²+· · ·+ (dx^p−2)²−(dx^p−1)². We have the following inclusions

Sⁿ×R₁⊂Rⁿ⁺¹×R₁=Rⁿ⁺²₁ and Hⁿ×R₁⊂Rⁿ⁺¹₁ ×R₁=Rⁿ⁺²₂ ,

whereSⁿ andHⁿ are the respectively the sphere and the hyperbolic space of dimensionndefined by

Sⁿ={(x⁰,· · · , xⁿ)∈Rⁿ⁺¹|(x⁰)²+· · ·+ (xⁿ)²= 1}, and Hⁿ={(x⁰,· · ·, xⁿ)∈Rⁿ⁺¹₁ | −(x⁰)²+· · ·+ (xⁿ)²=−1; x⁰>0}.

Thus, we set Eⁿ⁺² =Rⁿ⁺²

1 ifκ= 1 andEⁿ⁺² =Rⁿ⁺²

2 if κ=−1. With these notation, we have Mⁿ(κ)×R₁∈Eⁿ⁺².

Now, let (Mⁿ, g) a Riemannian hypersurface ofMⁿ(κ)×R₁. We denote respectively by∇,∇and

∇ the Levi-Civita connections of M, Mⁿ(κ)×R₁ and Eⁿ⁺². We consider ν the unit normal to Mⁿ(κ)×R₁into Eⁿ⁺², that is,

ν(x) = (x⁰, x¹,· · · , xⁿ,0),

and ν the unit normal to M into Mⁿ(κ)×R₁. We denote by S and S the associated shape operators, that is,S is defined by

∇XY =∇XY+< SX, Y > ν,

i.e.,SX=−κ∇Xν. Note that the signκcomes for the fact that we have< ν, ν >=κ.

Let{e1, . . . , e_n}be a local orthonormal frame onM,en+1=ν ande0=ν. We consider the forms ω_jⁱ,ω_i^j,ω_n+1ⁱ andωⁿ⁺¹_n+1 as above and in addition, we set

ω⁰_γ(ek) =< Sek, e_γ >, and ω₀^γ =−κω_γ⁰. SinceS_X =−κ∇Xν =κ(−X− hX, ∂ti∂_t), we have

ω⁰_γ(ek) =−κhek, e_γi+κεhek, ∂_ti heγ, ∂_ti. Hence, by definition, we have∇e^ke_β=P

αω^α_β(ek)eα.

Now, let{E0,· · ·, En+1}be a frame ofEⁿ⁺²so thathE0, E0i=κ,hEj, Eji= 1 forj∈ {1,· · ·, n}

and En+1 = ∂t (hence, hEn+1, En+1i = −1). We denote by A ∈ Mn+2(R) the matrix whose columns are the coordinates of the vectorseβ in the frame {E0,· · · , En+1},i.e.,eβ=P

αA^α_βEα. Hence, we have

∇ekeβ=X

α

dA^α_βEα. But, on the other hand, we have

∇ekeβ=X

α,γ

ω_β^γ(ek)A^α_γEα, which means thatA⁻¹dA= Ω, with Ω = (ω_β^α).

We setG= diag(κ,1,· · ·,1,−1)∈ Mn+2(R). We define the two following sets S(Eⁿ⁺²) =

Z ∈ Mn+2(R)|^tZGZ =G,detZ= 1 , and

s(Eⁿ⁺²) =

H ∈ Mn+2(R)|^tHG+GH = 0, ∼=so(Eⁿ⁺²).

Moreover, we setS⁺(Eⁿ⁺²) the connected component of the identity inS(Eⁿ⁺²).

Thus, one can see easily thatA∈ S⁺(Eⁿ⁺²) and Ω∈s(Eⁿ⁺²).

3. The fundamental theorem of hypersurfaces

3.1. The compatibility equations. Let (Mⁿ, g) be a simply connected Riemannian manifold with a field of symmetric operatorsS, a vector fieldT and a smooth functionf onM. From the section of preliminaries, we introduce the following definition.

Definition 3.1. We say that (M, g, S, T, f)satisfies the compatibility equation fro Mⁿ(κ)×R₁ if

||T||²−f²=−1and for any X, Y, Z, W ∈X(M),

R(X, Y)Z, = hSY, ZiX− hSX, ZiY +κ hX, ZiY − hY, ZiX (9)

+κ

hY, Ti hX, ZiT− hX, Ti hZ, TiY +hX, Ti hY, ZiT+hY, Ti hZ, TiX

∇XSY − ∇YSX−S[X, Y] = −f κ hX, TiY − hY, TiX (10)

∇XT =f SX.

(11)

df(X) =hSX, Ti (12)

Remark 2. Note that by differentiating ||T||²−f² =−1, we see that (11) implies (12)at any point wheref does not vanish.

Thus, Theorem 1 says that these compatibilty equations are a necessarey and sufficient condition to the existence of a local isometric immersion intoMⁿ(κ)×R₁.

3.2. Proof of Theorem 1. In order to simplify and without lost of generality, we can assume thatκ= 1 orκ=−1. We defineωⁱ,ωⁿ⁺¹,ω_jⁱ,ω_jⁿ⁺¹,ω_n+1ⁱ andω_n+1ⁿ⁺¹ as above and we set

Eⁿ⁺²=

Rⁿ⁺²

1 ifκ >0, Rⁿ⁺²

2 ifκ <0.

Let{e0,· · ·, en+1 =∂t}be the canonical frame of Ewith |e0,|² =κ,|ei|²= 1 for i∈ {1,· · ·, n}

and|en+1|²=−1.

We also consider the following functionsT^k=< T, ek >,Tⁿ⁺¹=−f andT⁰= 0, and the following forms: ω_j⁰(ek) = κ(T^jT^k+δ_j^k), ω_n+1⁰ (ek) =−κf T^k, ω₀ⁱ =−κω_i⁰, ω₀ⁿ⁺¹ =−κω_n+1⁰ and ω⁰₀ = 0.

Finally, we setσ(X) =< T, X >, that is,σ=T^kω^k and Ω = (ω^α_β)∈ Mn+2(R).

Lemma 3.2. We have

dσ= 0.

Proof : LetX, Y ∈Γ(T M), we have

dσ(X, Y) = X(σ(Y))−Y(σ(X))−σ([X, Y])

= <∇XT, Y >−<∇YT, X >

= −f < SX, Y >+f < SY, X >

= 0,

becauseS is symmetric.

Lemma 3.3. We have

dT^α=X

β

T^βω^β_α.

Proof : Forα= 0, we haveT⁰= 0 and then dT⁰= 0. Moreover, we have X

β

T^βω₀^β(ek) = X

j

T^jω^j₀(ek) +Tⁿ⁺¹ω₀ⁿ⁺¹(ek)

= X

j

κ²(T^jT^k+δ^j_k)T^j+κ²f Tⁿ⁺¹T^k

= X

j

(κ²T^jT^kT^j) +κ²T^k−κ²f²T^k

= κ²T^k X

j

(T^j)²+ 1−f²

= 0.

Indeed, we have,P

j(T^j)²−f²=|T|²−f²=−1.

Forα=n+ 1, we have dTⁿ⁺¹(ek) = df(ek) =−< Sek, T >. On the other hand, we have X

β

T^βω_n+1^β (ek) = X

j

T^jω_n+1^j (ek) +T^βωⁿ⁺¹_n+1(ek)

= −X

j

T^jω_jⁿ⁺¹(ek)

= −< Se_k, T >

= dTⁿ⁺¹(ek)

Finally, forj∈ {1,· · · , n},

dT^j(ek) = ek < T, ej>

= <∇ekT, e_j>+< T,∇eke_j>

= −f < Sek, e_j >+X

i

< Tⁱe_i,∇eke_j >

= −f ωⁿ⁺¹_j (ek) +X

i

Tⁱω_jⁱ(ek)

= Tⁿ⁺¹ω_jⁿ⁺¹(ek) +X

i

Tⁱωⁱ_j(ek).

This achieves the proof.

Lemma 3.4. We have

dΩ + Ω∧Ω = 0.

Proof : We set Φ =dΩ + Ω∧Ω. We recall that Ω = (ω_β^α). Then, from (7), we deduce immediately Φ^j_i =ω_n+1ⁱ ∧ω_jⁿ⁺¹+ωⁱ₀∧ω_j⁰−1

2 X

k,l

Rⁱ_kljω^k∧ω^l.

But, the Gauss equation (9) is satisfied, that is,

Rⁱ_klj =Rⁱ_klj+ω_jⁿ⁺¹+∧ωⁿ⁺¹_i (ek, e_l), with

Rⁱ_klj =κ(δ_j^kδ^l_i−δ^l_jδ^k_i +T^lTⁱδ^k_j +T^kT^jδ^l_i−T^kTⁱδ^l_j−T^lT^jδ^k_i).

Moreover, a straightforward computation givesRⁱ_klj =ω₀ⁱ∧ω⁰_j(ek, e_l). Thus, from these last four relations, we deduce that Φⁱ_j= 0 for any i, j∈ {1,· · · , n}.

Now, from (8), we have Φⁿ⁺¹_j = 1

2 X

k,l

h∇ekSe_l− ∇elSe_k−S[ek, e_l], ejiω^k∧ω^l−ω₀ⁿ⁺¹∧ω_n+1⁰ . Since the Coddazzi equation is satisfied, we have

<∇ekSe_l− ∇elSe_k−S[ek, e_l], ej > = −f κ T^kδ^j_l −T^lδ_k^j

= −κ T^lTⁿ⁺¹δ_k^j−T^kTⁿ⁺¹δ^j_l

Moreover, a simple computation yieldsω₀ⁿ⁺¹∧ω⁰_n+1(ek, el) =κ(T^kTⁿ⁺¹δ_j^l−T^lTⁿ⁺¹δ_j^k). Thus, we deduce Φⁿ⁺¹_j = 0.

Now, we will show that Φ⁰_n+1= 0. First, we have by definitionω⁰_j =κ(T^jσ+ω^j). By Lemma 3.2, we know that dσ= 0, so we deduce

dω⁰_j =κ(dT^j∧σ+ dω^j) =κdT^j∧σ−1κX

k

ω^j_k∧ω^k,

where we have used (5). Thus, we get Φ⁰_j(ep, e_q) = dω_j⁰(ep, e_q) +X

k

ω⁰_k∧ω^k_j(ep, e_q) +ω_n+1⁰ ∧ω_jⁿ⁺¹(ep, e_q)

= κ dT^j(ep)σ(eq)−dT^j(eq)σ(ep)−ω_q^j(ep) +ω^j_p(eq) +κ

T^pX

k

T^kω_j^k(eq)−T^qX

k

T^kω_j^k(ep) +ω_j^p(eq)−ω_j^q(ep) +κ T^pTⁿ⁺¹ωⁿ⁺¹_j (eq)−T^qTⁿ⁺¹ωⁿ⁺¹_j (ep)

Sinceσ(ep) =T^p andσ(eq) =T^q, we conclude by Lemma 3.3 withα=j that Φ⁰_j = 0.

Finally, we will prove that Φ⁰_n+1 = 0. For this, we recall thatω⁰_n+1 =κTⁿ⁺¹σ, so that, dω_n+1⁰ = κdTⁿ⁺¹σ. Therefore, we get

Φ⁰_n+1(ep, e_q) = dω_n+1⁰ (ep, e_q) +X

k

ω⁰_k∧^k_n+1(ep, e_q)

= κ T^qdTⁿ⁺¹(ep)−T^pdTⁿ⁺¹(eq) +κ

T^pX

k

T^kω_n+1^k (eq)−T^qX

k

T^kω^k_n+1(ep) +κ −ω_n+1^p (eq) +ω_n+1^q (ep)

.

Using the fact thatS is symmetric and Lemma 3.3 withα=n+ 1, we conclude that Φ⁰_n+1= 0.

To finish, it is obvious from the definition that Φ⁰₀= Φⁿ⁺¹_n+1= 0, and we achieve the proof by noting

that Φⁱ_n+1=−Φⁿ⁺¹_i .

Let x ∈ M, we define Z(x) as the set of matrices Z ∈ S⁺(Eⁿ⁺²) such that the coefficients of the last line ofZ areZ_βⁿ⁺¹=T^β(x). Then,Z(x) is a manifold of dimension ⁿ⁽ⁿ⁺¹⁾₂ . Indeed, this comes from the fact that the map

Fe : S⁺(Eⁿ⁺²) −→ S(Eⁿ⁺²) Z 7−→ (Z_βⁿ⁺¹)β

withS(Eⁿ⁺²) ={X∈Eⁿ⁺²| < X, X >=−1} is a submersion.

Now, we prove the following

Proposition 3.5. Assume that the compatibility equations forMⁿ(κ)×R₁are fulfiled. Letx0∈M andA0∈ Z(x0). Then, there exists a neighbourhoodU0 ofx0∈M and a unique map A:U0−→

S⁺(Eⁿ⁺²)so that

A⁻¹dA= Ω, A(x0) =A0 and ∀x∈U0, A(x)∈ Z(x).

Proof : LetU a neighbourhood ofx0 inM. We set F=

(x, Z)∈U× S⁺(Eⁿ⁺²)|Z ∈ Z(x) .

Obviously,F is a manifold of dimensionn+ⁿ⁽ⁿ⁺¹⁾₂ and the tangent space of F is given by T(x,Z)F=n

(u, ξ)∈TxU⊕TZS⁺(Eⁿ⁺²)|ξⁿ⁺¹_β = (dT^β)x(u)o .

Without ambiguity, we denote byZ the projectionU× S⁺(Eⁿ⁺²)−→ S⁺(Eⁿ⁺²), and we consider onF the matrix of 1-forms Θ =Z⁻¹dZ−Ω. Thus, for (x, Z)∈ F and (u, ξ)∈T(x,Z)F, we have Θ(x,Z)(u, ξ) =Z⁻¹ξ−Ωx(u). Finally, we setD(x, Z) = ker Θ(x,Z).

Claim: For any (x, Z)∈ F,D(x, Z) has dimensionn.

Indeed, first, we note that Θ belongs tos(Eⁿ⁺²) as Ω andZ⁻¹dZ. In addition, we have by Lemma 3.3

(ZΘ)ⁿ⁺¹_β = dZ_βⁿ⁺¹−X

γ

Z_γⁿ⁺¹ω_β^γ= dT^β−X

γ

Z_γⁿ⁺¹ω_β^γ = 0.

Hence, Θ(x,Z)has values in the spaceH=n

H ∈s(Eⁿ⁺²)|(ZH)ⁿ⁺¹_β = 0o

. But,His of dimension

n(n+1)

2 since the mapF defined above is a submersion andH ∈ Hif and only ifZH ∈ker (dF)Z. Moreover, {(0, ZH)|H ∈ H} lies into T(x,Z)F and the restriction of Θ(x,Z) to this subspace is just (0, ZH) 7−→ H. Hence, the map Θ(x,Z) is onto H, therefore, its rank is ⁿ⁽ⁿ⁺¹⁾₂ and so the dimension of its kernel isn, which proves the claim.

Claim: The distributionDis integrable.

By Lemma 3.4, we have

dΘ = −Z⁻¹dZ∧Z⁻¹dZ−dΩ

= −(Θ + Ω)∧(Θ + Ω)−dΩ

= −Θ∧Θ−Θ∧Ω−Ω∧Θ.

From this, we deduce that ifξ₁, ξ₂∈ D, then dΘ(ξ1, ξ₂) = 0. So, we deduce that Θ([ξ1, ξ2]) =ξ1·Θ(ξ2)−ξ2·Θ(ξ1)−dΘ(ξ1, ξ2) = 0.

So this proves the claim since by Frobenius theorem,Dis integrable.

Now, les A be the integral manifold through (x0, A0). If ξ ∈ T(x0,A0)S⁺(Eⁿ⁺²) is such that (0, ξ)∈ D(x0, A0) = TA0A, thenA⁻¹₀ = Θ_x0,A0)(0, ξ) = 0. Soξ= 0 and so

T(x0,A0)A ∩ {0} ×TA⁰S⁺(Eⁿ⁺²)

={(0,0)}.

Therefore,Ais locally the graph of a functionA:U₀−→ S⁺(Eⁿ⁺²), whereU₀is a neighbourhood ofx0 inU. Moreover, by construction, this map is unique and satisfies the properties of Proposi-

tion 3.5.

Now, we can prove Theorem 1 using this propostion.

Proof of Theorem 1:

Let x0 ∈ M, A ∈ Z(x0) and t0 ∈ R. Let {e1, . . . , en} be a local orthonromal frame ofM in a neighbourhood ofx0. By Proposition 3.5, there exists a unique mapAU0−→ S⁺(Eⁿ⁺²) satisfying

A⁻¹dA= Ω, A(x0) =A0 and ∀x∈U0, A(x)∈ Z(x).

For anyx∈U0, we setF⁰(x) =A⁰₀(x),Fⁱ(x) =Aⁱ₀(x) and we considerFⁿ⁺¹ the unique function so that dFⁿ⁺¹=εσ and Fⁿ⁺¹(x0) =t0. Note thatFⁿ⁺¹ exists because dσ= 0 andU0 is simply connected. Hence, this define a mapF :U0−→Eⁿ⁺².

Claim: The mapF satisfies all the properties of Theorem 1.

First, we show that the image of U0 lies into Mⁿ(κ)×R₁. Indeed, since Aⁿ⁺¹₀ = T⁰ = 0 and A∈SO⁺(Eⁿ⁺¹), we have,

κ(F⁰)²+ (F¹)²+ Xn i=2

(Fⁱ)²=κ.

Therefore, (F0,· · ·, F_n) lies intoMⁿ(κ) and the values ofF lie intoMⁿ(κ)×R₁. Now, since dA=AΩ, we have for α∈ {0,· · · , n+ 1}, we have

dF^α(ek) = X

j

A^α_jω₀^j(ek) +A^α_n+1ω₀ⁿ⁺¹(ek)

= X

j

A^α_j(−δ_j^k−T^jT^k)−A^α_n+1Tⁿ⁺¹T^k

= −A^α_k −T^kX

β

A^α_βAⁿ⁺¹_β

= −A^α_k and

dFⁿ⁺¹(ek) =−σ(ek) =−T^k=−Aⁿ⁺¹_k .

Hence, dF(ek) is the opposite thek-th column of the matrixA. So, sinceAis invertible, dF has rankn. ThusF is an immersion. Moreover,A∈ S⁺(Eⁿ⁺²), then<dF(ep),dF(eq)>=δ_q^p andF is an isometry.

The columns ofA(x) give a direct orthornormal frame ofEⁿ⁺¹, and columns 1 to ngive a direct orthonromal frame of TF(x)F(M). The column 0 is the projection ofF(x) onMⁿ× {0},i.e., the

unit normal ν(F(x)) toMⁿ(κ)×R₁ at the point F(x). Therefore, the (n+ 1)-th column is the unit normalν(F(x)) toF(M) inMⁿ(κ)×R₁at thr point F(x).

Now, we setXj= dF(ej) and we have

<dXj(Xk), ν > = X

α

dA^α_j(ek)A^α_n+1

= X

α,β

A^α_βA^α_n+1ω^β_j(ek)

= ω_jⁿ⁺¹(ek) =< Sek, ej>,

that is, the shape operator of F(M) in Mⁿ(κ)×R₁ is dF ◦S◦dF⁻¹. Finally, the last line of A gives the coordinates of the vertical vector ∂t in the orthornormal frame {ν, X1,· · · , Xn, ν}.

Hence, sinceA(x)∈ Z(x) forx∈U₁, we have

∂t=X

j

T^jXj+Tⁿ⁺¹ν = dF(T)−f ν.

Now, we finish the proof by showing that the local immersionFis unique up to a global isometry of Mⁿ(κ)×R₁. For this, letFe:U2−→Mⁿ(κ)×R₁be another immersion satisfying all the properties of the theorem, with U2 a simply connected neighbourhood of x0 included in U0 and (Xeβ) the associated frame, that is,Xej =Fe(ej),Xen+1is the unit normal ofF(Me ) intoMⁿ(κ)×R₁ andXe0

is the unit normal toMⁿ(κ)×R₁ intoEⁿ⁺². LetAebe the matrix of the coordinates of the vectors Xeβ in the frame {Eα}. Obviously, up to a direct isometry of Mⁿ(κ)×R₁, we can assume that F(x0) =Fe(x0) and the frame (Xβ) and (Xe_β) coincide at the pointx0 and henceA0(x0) =A(xe 0).

Note that this isometry fixes∂_tsince theT^α are the same. Moreover, these two matrices satisfy all the properties of Proposition 3.5, so by uniqueness of the solution in Proposition 3.5, we have A0(x) =A(x) for alle x∈M. Hence, by construction ofF andFe from A andA, we deduce thate F =FeonU2.

Now, we will prove that F can be extended in a unique way toM. For his, we consider x1 ∈M and a curve Γ : [0,1]−→M so that Γ(0) =x0 and Γ(1) =x1. Then, for each point Γ(t), there exists a neighbourhood of Γ(t) such that there exists an isometric immersion of this neighbouhood intoMⁿ(κ)×R₁satisfying the properties of the theorem. From this family of neighbourhood, we can extract a finite subsequence (V0,· · ·, Vp) covering Γ withV0=U0. Hence, by uniqueness, we can extendF to Vp and define F(x1).We conclude by noting that sinceM is simply connected, the value ofF(x1) does not depend on the choice of the curve Γ.

4. Application to maximal surfaces: the associate family

We deduce a first application of Theorem 1, namely the existence of a one-parameter family of maximal conformal immersion associated to a given maximal surface. This result is analogous to the result of B. Daniel for minimal surfaces into Riemannian productsMⁿ(κ)×R. First, we give this elementary proposition.

Proposition 4.1. Assume that(M², g, S, T, f)satisfies the compatibility equations forM²(κ)×R₁ and that S is trace-free. Forθ∈R, we define

S_θ=e^θJS= cos(θ)S+ sin(θ)J S, Sθ=e^θJT = cos(θ)S+ sin(θ)J T.

ThenSθis symmetric, trace-free,||Tθ||²−f²=−1, and(M², g, Sθ, Tθ, f)satisfies the compatibility equations for M²(κ)×R₁.

Remark 3. Note that Sθ andTθ are obtained fromS andT by a rotation of angle θ.

Proof : The fact thatSθis symmetric is just coming from the definition and a simple computation shows thatSθ is trace-free. Moreover since< T, J T >= 0, we deduce immediately that||Tθ||=

||T||. Sincee^JS commutes with∇X, the conditions (11) and (12) are fulfiled forSθandTθ. Now,

we will prove that the Gauss and Codazzi equations are satified. First, we remark that on surface, the Gauss equation becomes

K=−det (S) +κ(−1− ||T||²) =−det (S)−κf². Since det (e^θJ) = 1, we have det (Sθ) = det (S) and then the Gauss equation.

From this, we deduce the existence of the associated familiy.

Theorem 2. Let (M², g)be a simply connected Riemannian surface and F :M −→M²(κ)×R₁ a conformal maximal immersion with induced normal ν. Let S be the symmetric operator on M induced by the shape operator ofF. LetT be the vector field onM so thatdF(T)is the projection of ∂tonT(F(M)), andf =−< ν, ∂t>.

Let x0∈M. Then, there exists a unique family(Fθ)θ∈Rof conformal maximal immersions ofM intoMⁿ(κ)×R₁ such that

(1) F_θ(x0) =F(x0)and(dFθ)x⁰ = (dF)x⁰,

(2) the metrics induces onM byF andF_θ are the same,

(3) the symmetric operator onM induce by the shape operator ofFθ ise^θJS, (4) ∂t= dFθ(e^θJT) +f νθ, whereνθ is the unit normal to Fθ.

Moreover,F₀=F and the family(Fθ)θ∈R is continous with respect toθ.

Proof : Letg be the metric onM induced by F. Thus, (M, g, S, T, f) satisfies the compatibily equations forM²(κ)×R₁. So, by Proposition 4.1, (M, g, Sθ, Tθ, f) also satisfy these compatibility equations. Therefore, by Theorem 1, there exists a unique isometric immersionFθ satisfying the properties of the theorem. By uniqueness , it is clear thatF0 =F. Moreover, From the proof of Theorem 1, (M, g, Sθ, Tθ, f) defines for anyθa matrix of 1-forms Ωθand a matrix of functionsAθ

withA⁻¹_θ dAθ= Ωθ. From the definition of Ωθ, we get the coninuity of Ωθ and hence, ofA_θ . By the construction ofFθ using onlyAθ, we deduce thatFθis also continuous.

5. Spinorial characterization of surfaces intoM²(κ)×R₁

The second application of the main result is the spinorial characterization of surfaces into M²(κ)×R₁. It is now well-known that a necessary and sufficient condition to get an isometric immersion of a surface into Euclidean 3-space is existence of a special spinor field calledrestricted Killing spinor field (see [6, 11]). These results are the geometrically invariant version of previ- ous work on the spinorial Weierstrass representation by U. Abresch, D. Sullivan, R. Kusner, N.

Schmidt, and many others (see [7]). This representation was expressed by T. Friedrich ([6]) for surfaces inR³and then extended to other 3-dimensional Riemannian manifolds ([11, 14]). In a re- cent work, we have treated this question for surfaces of arbitrary signature in pseudo-Riemannian 3-dimensional space forms [9].

More generally, the restriction ϕof a parallel spinor field onRⁿ⁺¹ to an oriented Riemannian hypersurfaceMⁿ is a solution of a restricted Killing equation

(13) ∇^ΣM_X ϕ=−1

2γ^M(S(X))ϕ,

whereγ^M and∇^ΣM are respectively the Clifford multiplication and the spin connection onMⁿ, andS is the Weingarten tensor of the immersion. Conversely, Friedrich proves in [6] that, in the two dimensional case, if there exists a restricted Killing spinor field satisfying equation (13), where Sis an arbitrary field of symmetric endomorphisms ofT M, thenSsatisfies the Gauss and Codazzi equations of hypersurface theory and is consequently the Weingarten tensor of a local isometric immersion of M intoR³. Moreover, in this case, the solutionϕof the restricted Killing equation is equivalently a solution of the Dirac equation

(14) Dϕ=Hϕ,

where|ϕ|is constant and H is a real-valued function.

In the case of immersions of Riemannian surfaces into a Lorentzian space two spinor fields are needed, as we will see in Theorem 3.

5.1. Basic facts about spin geometry of hypersurfaces. Here, we recall the basics of spin geometry. For more details, one can refer to [10, 2, 1] for instance.

Let (M^p,q, g),p+q= 2, be an oriented pseudo-Riemannian surface of arbitrary signature isometri- cally immersed into a three-dimensional pseudo-Riemannian spin manifold (N^r,s, g). We introduce the parameterδas follows: δ=iif the immersion is timelike andδ= 1 if the immersion is space- like. Let ν be a unit vector normal to M. The fact that M is oriented implies that M carries a spin structure induced from the spin structure ofN and we have the following identification of the spinor bundles and Clifford multiplications:

ΣN|M ≡ΣM.

X·ϕ|M = δX•ν•ϕ)|M,

where ·and• are the Clifford multiplications, respectively onM andN. Moreover, we have the following well-known spinorial Gauss formula

(15) ∇Xϕ=∇Xϕ+δ

2SX·ϕ,

whereS is the shape operator of the immersion. Finally, we recall the Ricci identity onM

(16) R(e1, e2)ϕ= 1

2ε1ε2R1221e1·e2·ϕ, wheree1, e2 is a local orthonormal frame of M andεj=g(ej, ej).

The complex volume element on the surface depends on the signature and is defined by ω_p,q^C =i^q+1e1·e2.

Obviously ω_p,q^C 2

= 1 independently of the signature and the action of ω^C splits ΣM into two eigenspaces Σ^±M of real dimension 2. Therefore, a spinor fieldϕcan be written asϕ=ϕ⁺+ϕ⁻ withω^C·ϕ^±=±ϕ^±. Finally, we denoteϕ=ω^C·ϕ=ϕ⁺−ϕ⁻.

5.2. Special spinor fields.

5.2.1. Special spinor fields onM²(κ)×R₁. First, we construct particular spinor fields onM²(κ)×R₁ by lifting a Killing spinor ofM²(κ) to the productM²(κ)×R₁. For this, we will consider some particular and interesting sections, precisely the sections which do not depend ont.

We have the spinorial Gauss formula,

∇Xϕ=∇Xϕ+ i

2γ(SX)γ(ν)ϕ,

where ∇is the spinorial connection on M²(κ)×R₁, the spinorial connection onM²(κ) is ∇, the Clifford multiplication onM²(κ)×R₁ is γ andS the shape operator of the immersion ofM²(κ) into M²(κ)×R₁. Since M²(κ) is totally geodesic in the productM²(κ)×R₁, we get by taking ϕ_t=ϕ0a Killing spinor on M²(κ),i.e. ∇Xϕ0=ηγ^M2(X)ϕ0:

∇Xϕ = ηγ^M2(X)ϕ=iηγ(X)γ(∂t)ϕ

On the other hand, the complexe volume formωC=−ie1·e2·∂tacts as identity, so we have γ(e1)γ(∂t)ϕ=iγ(e2)ϕ, andγ(e2)γ(∂t)ϕ=−iγ(e1)ϕ.

So we deduce that

(17)













∇e1ϕ=iηγ(e2)ϕ,

∇e2ϕ=−iηγ(e1)ϕ,

∇∂tϕ= 0.

These particular spinor fields are the analogue of Killing spinor field for space forms and will be play a important role in the sequel.

5.2.2. Restriction to a surface. Now, let (M, g) be a surface ofM²(κ)×R, oriented by ν. Since M is oriented, it could be equiped with a spin structure induce from the spin structure ofM²×R. Moreover, as we saw, we have the following identification between the spinor bundles

Σ (M²×R)|N ∼= ΣM,

and the spinorial Gauss forumla (15) gives the relation between the spinorial connctions ofM and M²×R. For anyX∈X(M) and anyψ∈Γ(Σ (M²×R)), we have

∇XψM = ∇X ψM

+ i

2γ^M(SX)ψM.

If we use this forumla for the particular spinor field onM²×Rgiven by (17), we get

∇Xϕ = ηγ(Xt)γ(∂t)ϕ− i

2γ^M(SX)ϕ, whereX_tis the part ofX tangent toM², that is,

X_t=X− hX, ∂ti∂_t=X− hX, TiT−fhX, Tiν.

So, we deduce that

∇Xϕ = iηγ(X)γ(∂t)ϕ−iηhX, ∂tiγ(∂t)γ(∂t)ϕ− i

2γ^M(SX)ϕ

= iηγ(X)γ(∂t)ϕ+iηhX, ∂tiϕ− i

2γ^M(SX)ϕ

= iηγ(X)γ(T)ϕ+iηf γ(X)γ(ν)ϕ+ηhX, Tiϕ− i

2γ^M(SX)ϕ.

On the other hand, if we denote by ω = e1 ·

Me2 the real volume element on M, we have the following well-known relations (see [9])

γ(X) =−γ^M(X)γ^N(ω), γ(ν) =−iγ^M(ω).

By using these two identities, the fact thatω²=−1 and thatω anti-commuts with vector fields tangent toM, we get

∇Xϕ = iηγ^M(X)γ^M(ω)γ^M(T)γ^M(ω)ϕ+iηhX, Tiϕ−ηf γ^M(X)γ^M(ω)γ^M(ω)ϕ−i

2γ^M(SX)ϕ Now, we can rewrite this equation in an intrinsic way as follows

∇Xϕ = iηX·T ·ϕ+ηf X·ϕ+iηhX, Tiϕ− i

2SX·ϕ.

(18)

where “·” stands for the Clifford multiplication onM. 5.3. A spinorial version of the fundamental theorem.

Theorem 3. Let (M, g) be a connected, oriented and simply connectd Riemannian surface. Let T be a vector field, f and H two real functions onM satisfying f²− ||T||²=−1. The following three data are equivalent:

i) There exist two non-trivial spinor fieldsϕ1 andϕ2, orthogonal and solution of the equation

∇Xϕ=−i

2SX·ϕ+iηX·T·ϕ+ηf X·ϕ+iηhX, Tiϕ, whereS satisfies

∇XT =f SX, and df(X) =hSX, Ti.

ii) There exists an isometric immersionF fromM intoM²(κ)×R₁ of mean curvatureH, such that the shape operator related to the time-like normalν is given by

dF◦S◦dF⁻¹ and such that

∂t=dF(T) +f ν.

5.4. Restricted Killing fields and compatiblity equations. In a first time, we show that the existence of a restricteded Killing spinor implies the Gauss and Codazzi equations. For this, let (M, g) be an oriented surface with a non-trivial spinor field solution of the equation (18). We will see that the integrability conditions for this equation are precisely the Gauss and Codazzi equations.

Proposition 5.1. Let (M, g) be an oriented surface with ϕ solution of (18) and such that the equations (11) and (12) are satisfied. Then, we have

R1212+ det (S)−κf²

ω·ϕ=i d^∇S(e1, e2) +κf J(T)

·ϕ,

Proof: The proof of this proposition is based on the computation of the spinorial curvature applied to the spinor fieldϕ. We recall that the spinorial curvature is defined as follows

R(X, Y)ϕ=∇X∇Yϕ− ∇Y∇Xϕ− ∇[X,Y]ϕ,

for X, Y ∈ X(M). Here, we do the computations for ϕ. It is clear that the same computation for ψ with only some changes of sign yields the result. Using the expression given by (18), the equations (11) and (12), the fact that ω² = −1 and that ω anticommuts with the vector fields tangent toM, we get

∇X∇Yϕ = −η

2f Y ·AX·ϕ

| {z }

α1(X,Y)

−η²Y ·T·X·T·ϕ

| {z }

α2(X,Y)

+η²f Y ·T·X·ϕ

| {z }

−α3(X,Y)

+η

2Y ·T·AX·ϕ

| {z }

−α4(X,Y)

−ηhAX, TiY ·ϕ

| {z }

−α5(X,Y)

+η²f Y ·X·T ·ϕ

| {z }

−α6(X,Y)

−η²hX, TiY ·T·ϕ

| {z }

α⁷(X,Y)

−η²f²Y ·X·ϕ

| {z }

−α8(X,Y)

+η²fhX, TiY ·ϕ

| {z }

−α9(X,Y)

−iηf Y ·AX·ϕ

| {z }

−α10(X,Y)

+iηfhY, AXiϕ

| {z }

α11(X,Y)

−η²hY, TiX·T·ϕ

| {z }

α12(X,Y)

+η²fhY, TiX·ϕ

| {z }

−α13(X,Y)

−η²hX, Ti hY, Tiϕ

| {z }

α14(X,Y)

+η

2hY, TiAX·ϕ

| {z }

−α15(X,Y)

−i

2∇X(AY)·ϕ

| {z }

−α16(X,Y)

+η

2AY ·X·T·ϕ

| {z }

−α17(X,Y)

−η

2f AY ·X·ϕ

| {z }

−α18(X,Y)

+η

2hX, TiAY ·ϕ

| {z }

−α19(X,Y)

−1

4AY ·AX·ϕ

| {z }

−α20(X,Y)

+iη∇XY ·T·ϕ

| {z }

α21(X,Y)

−iηf∇XY ·ϕ

| {z }

α22(X,Y)

+iηh∇XY, Tiϕ

| {z }

α²³(X,Y)

That is,

∇X∇Yϕ= X23 i=1

α_i(X, Y).

Obviously, by symmetry, we have

∇Y∇Xϕ= X23 i=1

α_i(Y, X).

On the other hand, we have

∇[X,Y]ϕ = iη[X, Y]·T·ϕ

| {z }

β1([X,Y])

−iηf[X, Y]·ϕ

| {z }

−β2([X,Y])

+iηh[X, Y], Tiϕ

| {z }

β3([X,Y])

−i

2A[X, Y]·ϕ

| {z }

−β4([X,Y])

.

Since the connection∇ is torsion-free, we get∇XY − ∇YX−[X, Y] = 0, which implies that α21(X, Y)−α21(Y, X)−β1([X, Y]) = 0,

α22(X, Y)−α22(Y, X)−β2([X, Y]) = 0, α23(X, Y)−α23(Y, X)−β3([X, Y]) = 0.

Moreover, by symmetry, we have

α11(X, Y)−α11(Y, X) = 0 and

α₁₄(X, Y)−α₁₄(Y, X) = 0.

Other terms vanishes by symmetry. Namely,

α1(X, Y) +α10(X, Y) +α18(X, Y)−α1(Y, X)−α10(Y, X)−α18(Y, X) = 0, α3(X, Y) +α6(X, Y)−α3(Y, X)−α6(Y, X) = 0,

and

α4(X, Y) +α5(X, Y) +α15(X, Y) +α17(X, Y) +α19(X, Y)

−α4(Y, X)−α5(Y, X)−α15(Y, X)−α17(X, Y)−α19(X, Y) = 0.

The termsα2,α7,α8et α12 can be combined. Indeed, if we set α=α2+α7+α8+α12, then

α(X, Y)−α(Y, X) = η²h

f²(Y ·X−X·Y) +Y ·T·X·T −X·T·Y ·Ti

·ϕ

= η²h

f²(Y ·X−X·Y) +||T||²(Y ·X−X·Y)i

·ϕ

−2η²(hX, TiY ·T− hY, TiX·T)·ϕ, by takingX andY as an orthonormal frame{e1, e2}, we have

α(e1, e2)−α(e2, e1) = 2η²h

−T1e2(T1e1+T2e2) +T2e2(T1e1+T2e2)i

·ϕ

−2η²e1·e2·ϕ

= −2η²ω·ϕ+ 2η² T₁²ω·+T1T2−T1T2+T₂²ω·

ϕ

= −2η²f²ω·ϕ.

Always forX =e1 andY =e2,

α9(e1, e2) +α13(e1, e2)−α9(e2, e1)−α13(e2, e1) = 2T·ω·ϕ=−2J(T)·ϕ, whereJ is the rotation of positive angle ^π₂ onT M. Finally, we get

R(e1, e2)ϕ = 1

4(Ae2·Ae1−Ae1·Ae2)·ϕ−2η²f²ω·ϕ

−1

2d^∇A(e1, e2)·ϕ−2J(T)·ϕ.