Constant angle surfaces in 4-dimensional Minkowski space
aa r X i v : . [ m a t h . DG ] M a r CONSTANT ANGLE SURFACES IN 4-DIMENSIONALMINKOWSKI SPACE
PIERRE BAYARD, JUAN MONTERDE, RA ´UL C. VOLPE
Abstract.
We first define a complex angle between two oriented spacelikeplanes in 4-dimensional Minkowski space, and then study the constant anglesurfaces in that space, i.e. the oriented spacelike surfaces whose tangent planesform a constant complex angle with respect to a fixed spacelike plane. This no-tion is the natural Lorentzian analogue of the notion of constant angle surfacesin 4-dimensional Euclidean space. We prove that these surfaces have vanishingGauss and normal curvatures, obtain representation formulas for the constantangle surfaces with regular Gauss maps and construct constant angle surfacesusing PDE’s methods. We then describe their invariants of second order andshow that a surface with regular Gauss map and constant angle ψ = 0 [ π/ π/ π ] , with real or pure imaginary constant angle and describe theconstant angle surfaces in hyperspheres and lightcones. Keywords:
Minkowski space, spacelike surfaces, complex angle, constant angle sur-faces.
Introduction
A constant angle surface in R is a surface whose tangent planes have a constantangle with respect to some fixed direction in R . Constant angle surfaces in R have been studied in [4, 10]. This notion has then been extended to other geomet-ric contexts, especially to hypersurfaces in R n [4], surfaces in R [3] or surfaces in3-dimensional Minkowski space R , [8]. The aim of the paper is to introduce thenotion of constant angle surfaces in 4-dimensional Minkowski space R , , and studytheir main properties. We will first observe that a natural complex angle is definedbetween two oriented spacelike planes in R , , and then define a constant angle sur-face in R , as an oriented spacelike surface whose tangent planes form a constantcomplex angle with respect to some fixed oriented spacelike plane. This notion ap-pears to be the natural Lorentzian analogue of the notion of constant angle surfacein R , and also extends the definitions in the literature of constant angle surfacesin R and in R , . Let us note that the definition of the complex angle between twooriented spacelike planes seems to be new and might be of independent interest.We will suppose throughout the paper that the surfaces have regular Gauss map:under this additional assumption, we will obtain general representation formulas forthe constant angle surfaces in R , . These formulas rely on a representation formulafor surfaces with regular Gauss map and vanishing Gauss and normal curvaturesin R , given in [6], and reformulated in terms of spin geometry in [1]. With theseformulas at hand, the construction of a constant angle surface amounts to solving a PDE system. We will then describe the invariants of second order of the constantangle surfaces and prove that a surface with regular Gauss map and constant angle ψ = 0 [ π/
2] is never complete. Finally, we will show that a surface has constant an-gle π/ π ] if and only if it is a product of curves in orthogonal planes, we will studythe surfaces with real or pure imaginary constant angle and describe the constantangle surfaces in hyperspheres and lightcones. Additionally to the constructionsof the constant angle surfaces using PDE’s methods, we will give explicit concreteexamples of constant angle surfaces in the spirit of [9].The paper is organized as follows. We describe the Clifford algebra and the spingroup of R , in Section 1, we introduce the complex angle between two orientedspacelike planes of R , in Section 2 and the notion of constant angle surface in R , in Section 3. We then give general representation formulas and reduce theconstruction of these surfaces to the resolution of a PDE in Sections 4 and 5, andintroduce a frame adapted to a constant angle surface and describe the second orderinvariants of a constant angle surface in Section 6. We study the completenessof the constant angle surfaces in Section 7. We finally study the constant anglesurfaces which are product of plane curves in Section 8, surfaces with real or pureimaginary constant angle in Section 9 and constant angle surfaces in hyperspheresand lightcones in Section 10. Three appendices end the paper: we give in thefirst appendix an alternative construction and an elementary characterization ofthe complex angle between two oriented spacelike planes, we describe in the secondappendix the very special case of the surfaces of constant angle 0 [ π ] and we detailin the third appendix the computations leading to an explicit frame adapted to aconstant angle surface.1. Clifford algebra and spin group of R , The Minkowski space R , is the space R endowed with the metric g = − dx + dx + dx + dx . We recall here the description of the Clifford algebra and the spin group of R , using the complex quaternions, as introduced in [1], and refer to [5] for the generalbasic properties of the Clifford algebras and the spin groups. Let H C be the spaceof complex quaternions, defined by H C := { q q I + q J + q K, q , q , q , q ∈ C } where I, J, K are such that I = J = K = − , IJ = − JI = K. Using the Clifford map R , → H C (2)(1)( x , x , x , x ) (cid:18) ix x I + x J + x K − ix x I + x J + x K (cid:19) where H C (2) stands for the set of 2 × H C , the Clifford algebra of R , is Cl (1 ,
3) = (cid:26)(cid:18) a b b b b a (cid:19) , a, b ∈ H C (cid:27) ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 3 where, if ξ = q q I + q J + q K belongs to H C , we denote b ξ := q q I + q J + q K, the element of H C obtained by the usual complex conjugation of its coefficients.The Clifford sub-algebra of elements of even degree is(2) Cl (1 ,
3) = (cid:26)(cid:18) a b a (cid:19) , a ∈ H C (cid:27) ≃ H C . Let us consider the bilinear map H : H C × H C → C defined by(3) H ( ξ, ξ ′ ) = q q ′ + q q ′ + q q ′ + q q ′ where ξ = q q I + q J + q K and ξ ′ = q ′ q ′ I + q ′ J + q ′ K. It is C -bilinearfor the natural complex structure i on H C . We consider(4)
Spin (1 ,
3) := { q ∈ H C : H ( q, q ) = 1 } ⊂ Cl (1 , . Using the identification R , ≃ { ix x I + x J + x K, ( x , x , x , x ) ∈ R }≃ { ξ ∈ H C : ξ = − b ξ } , where, if ξ = q q I + q J + q K belongs to H C , we denote ξ = q − q I − q J − q K, we get the double coverΦ : Spin (1 , −→ SO (1 , q ( ξ ∈ R , q ξ b q − ∈ R , ) . Here and below SO (1 ,
3) stands for the component of the identity of the group ofLorentz transformations of R , . Complex angle between two oriented spacelike planes
We introduce here the complex angle between two oriented spacelike planes in R , . The definition relies on a model of the Grassmannian of the oriented spacelikeplanes in R , introduced in [1]: it is naturally a complex 2-sphere in C , and theintuition in Euclidean space R will then easily lead to the definition. Throughoutthe paper we will assume that R , is oriented by its canonical basis ( e o , e o , e o , e o ) , and that it is also oriented in time by e o : we will say that a timelike vector is future-oriented if its first component in the canonical basis is positive.2.1. The Grassmannian of the oriented spacelike 2-planes in R , . We con-sider Λ R , , the vector space of bivectors of R , endowed with its natural metric h ., . i (which has signature (3,3)). The Grassmannian of the oriented spacelike 2-planes in R , identifies with the submanifold of unit and simple bivectors Q = { η ∈ Λ R , : h η, η i = 1 , η ∧ η = 0 } . The Hodge ∗ operator Λ R , → Λ R , is defined by the relation(6) h∗ η, η ′ i = η ∧ η ′ for all η, η ′ ∈ Λ R , , where we identify Λ R , to R using the canonical volumeelement e o ∧ e o ∧ e o ∧ e o on R , . It satisfies ∗ = − id Λ R , and thus i := −∗ definesa complex structure on Λ R , . We also define(7) H ( η, η ′ ) = h η, η ′ i − i η ∧ η ′ ∈ C PIERRE BAYARD, JUAN MONTERDE, RA´UL C. VOLPE for all η, η ′ ∈ Λ R , . This is a C -bilinear map on Λ R , , and we have Q = { η ∈ Λ R , : H ( η, η ) = 1 } . The bivectors E = e o ∧ e o , E = e o ∧ e o , E = e o ∧ e o form a basis of Λ R , as a vector space over C ; this basis is such that H ( E i , E j ) = δ ij for all i, j. Since the Clifford map (1) identifies the vectors of R , with someelements of H C (2) , we can use this map and the usual matrix product to identifyΛ R , with elements of H C (2) . Recalling (2), these even elements may in turn beidentified with elements of H C . With these identifications, we easily get E = I, E = J, E = K and Λ R , = { Z I + Z J + Z K ∈ H C : ( Z , Z , Z ) ∈ C } ;moreover, the complex structure i and the quadratic map H defined above onΛ R , coincide with the natural complex structure i and the quadratic map H defined on H C , and(8) Q = { Z I + Z J + Z K : Z + Z + Z = 1 } = Spin (1 , ∩ ℑ m H C , where ℑ m H C stands for the linear space generated by I, J and K in H C . Let usfinally note that
Spin (1 ,
3) and Q identify respectively to the complex spheres S C := { Z Z I + Z J + Z K : Z + Z + Z + Z = 1 } and S C := { Z I + Z J + Z K : Z + Z + Z = 1 } (by (4) and (8)); we will consider below the bundle S C → S C (9) q q − Iq.
This is a principal bundle of group S C = { cos A A I, A ∈ C } ⊂ S C acting on S C by multiplication on the left; it is equipped with a natural horizontaldistribution: a complex curve g : C → S C will be said to be horizontal if g ′ g − belongs to C J ⊕ C K. Definition of the complex angle.Definition 2.1.
Let p, q ∈ Q be two oriented spacelike planes of R , . The complexangle between p and q is the complex number ψ ∈ C such that H ( p, q ) = cos ψ. It is uniquely defined up to sign and the addition of π Z since cos : C → C issurjective and cos ψ = cos ψ ′ if and only if ψ = ± ψ ′ + 2 kπ, k ∈ Z . The complex angle between p and q ∈ Q is a kind of complex arc-length betweenthese two points in the complex sphere Q = S C : ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 5
Proposition 2.2.
Let p, q ∈ Q be two oriented spacelike planes of R , and ψ ∈ C be the complex angle between p and q. The following holds:a- If ψ = 0 [ π ] , there exists V ∈ T p Q such that H ( V, V ) = 1 and (10) q = cos ψ p + sin ψ V. b- If ψ = 0 [ π ] , there exists ξ ∈ T p Q such that H ( ξ, ξ ) = 0 and (11) q = ± p + ξ where the sign is positive if ψ = 0 [2 π ] and negative if ψ = π [2 π ] . Proof.
Since H ( p, q ) = cos ψ and H ( p, p ) = 1 , ξ := q − cos ψ p is such that H ( p, ξ ) = H ( p, q ) − cos ψ H ( p, p ) = 0i.e. ξ belongs to T p Q = { ξ ∈ I m H C : H ( p, ξ ) = 0 } . There are two cases:
Case 1: H ( ξ, ξ ) = 0 , which is equivalent to H ( q, q ) + cos ψ H ( p, p ) − ψ H ( p, q ) = 0 , which reads cos ψ = 1 since H ( p, p ) = H ( q, q ) = 1 and H ( p, q ) = cos ψ, i.e. ψ = 0 [ π ]; if γ ∈ C is such that γ = H ( ξ, ξ ) then γ = 0 and setting V = ξ/γ we obtain H ( V, V ) = 1 and q = cos ψ p + γ V. Since H ( p, p ) = H ( q, q ) = 1 and H ( p, V ) = 0 we deduce that 1 = cos ψ + γ , i.e. γ = ± sin ψ. Changing V by − V if necessary, we may suppose that γ = sin ψ, and (10) holds. Case 2: H ( ξ, ξ ) = 0 , which is equivalent to ψ = 0 [ π ]; we thus obtain ξ = q − ǫp with ǫ = 1 if ψ = 0 [2 π ] and ǫ = − ψ = π [2 π ] , and (11) follows. (cid:3) Remark 1.
The case ψ = 0 [ π ] has the following geometric meaning: by (11), thismeans that q = ± p + ξ for some ξ ∈ T p Q with H ( ξ, ξ ) = 0 , which holds if andonly if p and q belong to some degenerate hyperplane of R , : by (7), ξ ∈ T p Q and H ( ξ, ξ ) = 0 mean that ξ = u ∧ N for a unit vector u ∈ p and a null vector N normalto p, and the degenerate hyperplane containing p and q is p ⊕ R N. Note that thereare exactly two degenerate hyperplanes containing a given spacelike 2-plane p : thehyperplanes p ⊕ L and p ⊕ L ′ where L and L ′ are the two null lines in R , whichare normal to p. Interpretation of the complex angle in terms of two real angles.
Letus consider two oriented spacelike planes p, q ∈ Q and the complex angle ψ between p and q, and assume that ψ = 0 [ π ] . By Proposition 2.2, there exists V ∈ T p Q suchthat H ( V, V ) = 1 and q = cos ψ p + sin ψ V. By (7) the conditions V ∈ T p Q and H ( V, V ) = 1 mean that V is of the form u ∧ v for some unit spacelike vectors u and v belonging to p and p ⊥ , i.e. such that p = u ∧ u ⊥ and p ⊥ = v ∧ v ⊥ (where p ⊥ is the timelike plane orthogonal to p, with itsnatural orientation, and u ⊥ and v ⊥ are respectively unit spacelike and unit timelikevectors). Writing ψ = ψ + iψ and using the formulascos ψ = cos ψ cosh ψ − i sin ψ sinh ψ sin ψ = sin ψ cosh ψ + i cos ψ sinh ψ , PIERRE BAYARD, JUAN MONTERDE, RA´UL C. VOLPE a direct computation yields q = cos ψ u ∧ u ⊥ + sin ψ u ∧ v = (cosh ψ u + sinh ψ v ⊥ ) ∧ (cos ψ u ⊥ − sin ψ v );(12)the plane q is generated by the unit and orthogonal vectors cos ψ u ⊥ − sin ψ v andcosh ψ u + sinh ψ v ⊥ ; these vectors are determined by an euclidean rotation ofangle ψ in the spacelike plane generated by u ⊥ and v, and by a lorentzian rotationof angle ψ in the timelike plane generated by u and v ⊥ . Remark 2.
We deduce that the angle ψ = 0 [ π ] between the planes p and q isa real number (i.e. ψ = 0 ) if and only if there exists a spacelike hyperplanecontaining both p and q (by (12) this is the hyperplane p ⊕ R v ) and that ψ is apure imaginary complex number (i.e. ψ = 0 ) if and only there exists a timelikehyperplane containing p and q (this is the hyperplane p ⊕ R v ⊥ ). See also PropositionA.1 in Appendix A. First properties of a surface with constant angle
The Gauss map of a spacelike surface.
Let us consider an oriented space-like surface M in R . . We identify the oriented Gauss map of M with the map G : M → Q , x G ( x ) = u ∧ u , where ( u , u ) is a positively oriented orthonormal basis of T x M. We define the vectorial product of two vectors ξ, ξ ′ ∈ ℑ m H C by ξ × ξ ′ := 12 ( ξξ ′ − ξ ′ ξ ) ∈ ℑ m H C . We also define the mixed product of three vectors ξ, ξ ′ , ξ ′′ ∈ ℑ m H C by[ ξ, ξ ′ , ξ ′′ ] := H ( ξ × ξ ′ , ξ ′′ ) ∈ C . The mixed product is a complex volume form on ℑ m H C (i.e. with complex values, C -linear and skew-symmetric with respect to the three arguments); it induces anatural complex area form ω Q on Q by ω Q p ( ξ, ξ ′ ) := [ ξ, ξ ′ , p ]for all p ∈ Q and all ξ, ξ ′ ∈ T p Q . Note that ω Q p ( ξ, ξ ′ ) = 0 if and only if ξ and ξ ′ arelinearly dependent over C . We now recall the following expression for the pull-backby the Gauss map of the area form ω Q : Proposition 3.1. [1] If K and K N denote the Gauss and the normal curvaturesof M in R , , we have (13) G ∗ ω Q = ( K + iK N ) ω M , where ω M is the form of area of M. In particular, K = K N = 0 at x o ∈ M if andonly if the linear space dG x o ( T x o M ) belongs to some complex line in T G ( x o ) Q . As a consequence of the proposition, if K = K N = 0 and if G : M → Q isa regular map (i.e. if dG x is injective at every point x of M ), there is a uniquecomplex structure J on M such that dG x ( J u ) = i dG x ( u ) ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 7 for all x ∈ M and all u ∈ T x M. Indeed, for all x ∈ M, G ∗ ω Q x = ω Q ( dG x , dG x ) = 0in that case, and Im ( dG x ) is a complex line in T G ( x ) Q and we may set J u := dG x − ( i dG x ( u ))for all u ∈ T x M. The complex structure J coincides with the complex structureintroduced in [6]. Note that M cannot be compact under these hypotheses, since,on the Riemann surface ( M, J ) , the Gauss map G = G I + G J + G K is globallydefined, non-constant, and such that G , G and G are holomorphic functions.Thus, assuming moreover that M is simply connected, by the uniformization the-orem ( M, J ) is conformal to an open set of C , and thus admits a globally definedconformal parameter z = x + iy. Definition of a constant angle surface.Definition 3.2.
An oriented spacelike surface M is of constant angle with respectto a spacelike plane p o if the angle function ψ between p o and the tangent planes of M is constant. By Proposition 2.2 and Remark 1, if ψ = 0 [ π ] then the surface M, if it isconnected, belongs to a degenerate hyperplane p o ⊕ L where L is one of the twonull lines which are normal to p o , and nothing else can be said since reciprocallyan arbitrary surface in p o ⊕ L has constant angle ψ = 0 [ π ] . Details are given inthe Appendix B. Assuming thus that ψ = 0 [ π ] , this alternatively means that theGauss map image of M belongs to the complex circle of center p o and constantradius cos ψ in Q , i.e. G is of the form(14) G = cos ψ p o + sin ψ V for some function V : M → T p o Q such that H ( V, V ) = 1 (Proposition 2.2). Sincethe Gauss map image of a constant angle surface in R , is thus by definition acomplex curve, it is clear from (13) that such a surface has vanishing Gauss andnormal curvatures K = K N = 0 . It is known that constant angles surfaces in R have the same property [3].3.3. First examples: constant angle surfaces in a hyperplane. If M is aconstant angle surface of R , , of angle ψ ∈ C , it is clear from Remark 2 that if M belongs to a spacelike (resp. timelike) hyperplane of R , then ψ is a real (resp. pureimaginary) number. Since constant angle surfaces in R and in R , were studiedin [4, 10] and [8], we will be merely interested in the following in surfaces whichdo not belong to spacelike or timelike hyperplanes. We will see below examplesof constant angle surfaces belonging to degenerate hyperplanes of R , (the angleis ψ = 0 [ π ]), and, in contrast with the case ψ = 0 [ π ] , examples of real or pureimaginary constant angle surfaces which do not belong to any hyperplanes of R , . A new example.
Let us verify that the immersion(15) F ( x, y ) = e ax − by (cosh( x ) , sinh( x ) , cos( y ) , sin( y )) , ( x, y ) ∈ R defines a spacelike surface with constant angle. The associated tangent basis is ∂ x F = e ax − by ( a cosh( x ) + sinh( x ) , cosh( x ) + a sinh( x ) , a cos( y ) , a sin( y )) ,∂ y F = − e ax − by ( b cosh( x ) , b sinh( x ) , b cos( y ) + sin( y ) , − cos( y ) + b sin( y )) PIERRE BAYARD, JUAN MONTERDE, RA´UL C. VOLPE and the first fundamental form is e ax − by ) ( dx + dy ) . The immersion is thus spacelike, and z = x + iy is a conformal parameter. By adirect computation, its Gauss map reads G ( z ) := ∂ x F ∧ ∂ y F | ∂ x F ∧ ∂ y F | = ( a + ib ) E − (cosh z + ( a + ib ) sinh z ) E + i (( a + ib ) cosh z + sinh z ) E . Thus, the complex angle ψ between the tangent plane and the plane p o := E satisfies cos ψ := H ( G ( z ) , E ) = a + ib ;it is constant. Note that | F | = 0 , which means that the surface belongs to thelightcone at 0 . We may also verify by a direct computation that the mean curvaturevector ~H of the immersion is lightlike i.e. satisfies | ~H | = 0. Moreover, it is notdifficult to see that the surface belongs to a hyperplane if and only if a = ± b = 0 , i.e. ψ = 0 [ π ]; in that case it belongs in fact to the degenerate hyperplane x ± x = 1 . This in accordance with Remark 1. Writing ψ = ψ + iψ , ψ , ψ ∈ R , and since a = cos ψ cosh ψ and b = − sin ψ sinh ψ , we obtain for a ∈ ( − ,
1) (resp. a ∈ R \ [ − , b = 0 a surface with real(resp. pure imaginary) constant angle which does not belong to any hyperplane.Let us finally note that this example shows that there exist constant angle surfacesfor arbitrary values of the angle ψ. We will explain below how to systematicallyconstruct constant angle surfaces in R , . Representation of a surface with constant angle
In the sequel we will consider a complex circle in Q = S C , with center I andradius cos ψ, ψ = 0 [ π ] , and parametrized by(16) G ( z ) = cos ψ I + sin ψ J (cid:26) cos (cid:18) z sin ψ (cid:19) + sin (cid:18) z sin ψ (cid:19) I (cid:27) , z ∈ C . It is of the form (14) and such that H ( G ′ ( z ) , G ′ ( z )) = 4 for all z ∈ C . Recall theprincipal bundle S C → S C introduced in (9) and its natural horizontal distribution.Direct computations show the following: Lemma 4.1.
The function g : C → S C defined by g ( z ) = − cos ψ (cid:18) z tan ψ (cid:19) ψ (cid:18) z tan ψ (cid:19) I (17) + sin ψ (cid:18) z cot ψ (cid:19) J − sin ψ (cid:18) z cot ψ (cid:19) K is an horizontal lift of the function G defined in (16); it is such that H ( g ′ , g ′ ) = 1 and satisfies (18) g ′ g − = cos βJ + sin βK with (19) β = − z cot ψ. ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 9
Moreover, an horizontal lift of G is necessarily of the form g a := ag for someconstant a = cos( A )1 + sin( A ) I, A ∈ C , and satisfies g ′ a g − a = cos β a J + sin β a K with β a = β + 2 A. We now write the representation theorem for the flat surfaces with flat normalbundle and regular Gauss map in R , (Corollary 5 in [1], after [6]) in the contextof the prescribed Gauss map in the form (16). In the statement, the function g : C → S C is an horizontal lift of G and β : C → C is such that (18)-(19) hold. Theorem 1.
Let
U ⊂ C be a simply connected open set. If h , h : U → R are tworeal functions such that the vector fields α , α ∈ Γ( T U ) defined by α := ih cos β + h sin βα := ih sin β − h cos β are linearly independent at every point of U and satisfy [ α , α ] = 0 (as real vectorfields), then, setting (20) ξ := g − ( ω J + ω K ) b g where ω , ω : T U → R are the dual forms of α , α ∈ Γ( T U ) ,F = Z ξ : U → R , is a spacelike surface with constant angle ψ. Reciprocally, up to a rigid motion of R , , a spacelike surface of constant angle ψ and regular Gauss map may be locallywritten in that form. We also recall from [1] that the real functions h and h in the theorem are thecomponents of the mean curvature vector of the surface in a parallel frame normalto the surface, and the complex functions α , α are the expressions in z of a parallelframe tangent to the surface; moreover, these parallel frames are positively oriented,in space and in time. Remark 3.
Suppose that g : C → S C is an horizontal lift of G : C → S C as inLemma 4.1 and that β : C → C is such that (18) holds. For a ∈ S C , the function g a := ga : C → S C also satisfies (18), with the same function β . If h and h arereal functions as in the statement of Theorem 1, the corresponding forms ξ g and ξ g a are linked by ξ g a = a − ξ g b a : the immersions R ξ g and R ξ g a are thus congruent,i.e. differ one from the other by a rigid motion of R , (recall (5)). The representation theorem in terms of the metric
We aim to apply the representation theorem to construct all the constant anglesurfaces and give general explicit expressions in some special cases. In order todo this, we reformulate here the representation theorem (Theorem 1) using thecoefficients of the metric instead of the unknown functions h , h : the compatibilitycondition on these functions will then reduce to a hyperbolic PDE on the metriccoefficients, whose Cauchy problem is solvable. Determination of the metric.
Let us keep the notation of the previoussection and assume that the hypotheses of Theorem 1 hold. Writing β = u + iv, straightforward computations yield α = ( h sinh( v ) + h cosh( v )) sin u + i ( h cosh( v ) + h sinh( v )) cos u and α = − ( h sinh( v ) + h cosh( v )) cos u + i ( h cosh( v ) + h sinh( v )) sin u ;since α and α are everywhere independent vectors in R we have( h sinh( v ) + h cosh( v ))( h cosh( v ) + h sinh( v )) = 0 , and we may set µ and ν such that(21) 1 µ = h sinh( v ) + h cosh( v ) and 1 ν = h cosh( v ) + h sinh( v )and get the formulas(22) α = 1 µ sin( u ) + iν cos( u ) and α = − µ cos( u ) + iν sin( u ) . Since the tangent frame ( α , α ) is supposed to be orthonormal (by (20) the metricis ω + ω ), the metric reads(23) µ dx + ν dy . We write in the next lemma the condition [ α , α ] = 0 appearing in Theorem 1 interms of the metric coefficients µ, ν : Lemma 5.1.
The condition [ α , α ] = 0 reads (24) ν ∂ y µ = − c µ ∂ x ν = c with (25) c = − sin(2 ψ )sin ( ψ ) + sinh ( ψ ) and c = − sinh(2 ψ )sin ( ψ ) + sinh ( ψ ) . Proof.
A straightforward computation yields[ α , α ] = (cid:18) ∂ y µ + ν ∂ x uµ ν (cid:19) ∂ x + (cid:18) − ∂ x ν + µ ∂ y uµν (cid:19) ∂ y , and [ α , α ] = 0 if and only if(26) ∂ y µ = − ν ∂ x u and ∂ x ν = µ ∂ y u. We have by definition u = Re( β ) = − z cot( ψ )) , which implies(27) ∂ x u = − ψ )) and ∂ y u = 2 Im(cot( ψ )) . Writing ψ = ψ + iψ , we easily get ∂ x u = c and ∂ y u = c where c and c aregiven by (25), and obtain from (26) the system (24). (cid:3) ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 11
Remark 4.
Computing the Christoffel symbols of the metric (23) and setting T := 1 µ ∂ x , T := 1 ν ∂ y it appears that (24) is equivalent to the equations (28) ∇ T = ( c dx + c dy ) T and ∇ T = − ( c dx + c dy ) T . Reformulation of the representation theorem.
We may then reformulateTheorem 1 as follows:
Theorem 2.
Let us assume that G : C → S C is given by (16), g : C → S C is anhorizontal lift of G and β = u + iv : C → C is such that (18) holds. If µ and ν are non-vanishing solutions of (24) on a simply connected open set U ⊂ C , then,setting (29) ω = sin( u ) µ dx + cos( u ) ν dy, ω = − cos( u ) µ dx + sin( u ) ν dy and (30) ξ := g − ( ω J + ω K ) b g, the formula F = Z ξ : U → R , defines a spacelike surface with constant angle ψ = ψ + iψ . Moreover, the metricis µ dx + ν dy . Reciprocally, up to a rigid motion of R , , a spacelike surface ofconstant angle ψ and regular Gauss map may be locally written in that form.Proof. In order to show that this is a reformulation of Theorem 1, we only observethat ω and ω are the dual forms of two independent vectors fields α and α ∈ Γ( T U ) such that [ α , α ] = 0 : the forms ω and ω are independent since µν = 0at every point, and their dual vectors fields α and α are given by (22); moreover,by Lemma 5.1 they are such that [ α , α ] = 0 since µ and ν are solutions of (24).Finally, the metric is ω + ω = µ dx + ν dy . (cid:3) Resolution of the system (24).
We now focus on the resolution of (24)and assume that ψ = 0 [ π/
2] or ψ = 0 so that c or c = 0 (if ψ = 0 [ π/
2] and ψ = 0 then ψ = 0 [ π ] or ψ = π/ π ] : the first case is studied in Appendix B andthe second case in Theorem 3 in Section 8 below). Let us first observe that theresolution of this system is then equivalent to the resolution of the single hyperbolicPDE(31) ∂ xy ζ = − c c ζ for ζ = µ or ν, which is a 1-dimensional Klein-Gordon equation. Indeed, if µ and ν satisfy (24) then they obviously also satisfy (31). Conversely, assuming firstthat c = 0 , if µ is a solution of (31) we obtain a solution µ, ν of (24) by setting ν := − c ∂ y µ, and, similarly, if c = 0 and ν is a solution of (31) we obtain asolution µ, ν of (24) by setting µ := c ∂ x ν. We finally note that we may solve aCauchy problem for (31): let us fix a smooth regular curve Γ = ( γ , γ ) in thecoordinate plane which does not intersect any line parallel to the coordinate axes in more than one point and let us consider two smooth functions f, g on Γ; thenthere exists a unique solution ζ of (31) such that(32) ( ζ | Γ = f∂ n ζ | Γ = g, where ∂ n denotes differentiation with respect to the direction normal to the curve.If the above geometric condition on Γ is not satisfied, the Cauchy problem is ingeneral insoluble. It thus appears that a general constant angle surface in R , locally depends on two arbitrary real functions of one real variable (the initialconditions f, g of the Cauchy problem (31)-(32) for µ or for ν ). Details on thisCauchy problem and its explicit resolution using a Bessel function may be found in[7, Chapter II].6. A frame adapted to a constant angle surface
With the last representation theorem and the explicit expression of the lift g ofthe Gauss map (Lemma 4.1), we can construct a special orthonormal frame adaptedto a given constant angle surface. We will first give a geometric construction of thisframe, and then its explicit expression. We finally use these results to obtain easilythe second order invariants of the surface.6.1. Geometric construction of an adapted frame.
Let us assume that theimmersion is given as in Theorem 2. We consider the orthonormal frame tangentto the immersion(33) T := 1 µ ∂ x F = 1 µ ξ ( ∂ x ) , T := 1 ν ∂ y F = 1 ν ξ ( ∂ y ) . Let us note that ( T , T ) is positively oriented: by the very definitions of ξ, ω and ω in Theorem 2 we have ∂ x F = ξ ( ∂ x )= g − ( ω ( ∂ x ) J + ω ( ∂ x ) K ) b g = µg − (sin u J − cos u K ) b g and similarly ∂ y F = νg − (cos u J + sin u K ) b g. We thus have, in H C , ∂ x F d ∂ y F = µνg − Ig = µνG, which implies that T · T = G, and thus that ( T , T ) is positively oriented.Let us recall that the curvature ellipse at a point x o ∈ M is the ellipse in thenormal plane { II ( w, w ) : w ∈ T x o M, | w | = 1 } ⊂ N x o M. Proposition 6.1.
The curvature ellipse is a segment [ µ N , ν N ] where N , N arenormal vectors such that | N | = −| N | = 1 and h N , N i = 0 . Moreover (34) II ( T , T ) = 2 µ N , II ( T , T ) = 2 ν N and II ( T , T ) = 0 . ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 13
The vector N is future-directed and ( N , N ) is a positively oriented basis of theplane normal to M. We thus obtain a natural moving frame ( N , N , T , T ) adapted to the constantangle surface. This frame is moreover positively oriented in R , and such that itsfirst vector is future-directed. Proof.
Since K N = 0 the curvature ellipse is a segment [ α, β ] ⊂ N x o M and since K = 0 , h α, β i = 0 (by the Gauss equation). Let us also note that α, β = 0 since II is not degenerate (if for instance α = 0 and w ∈ T x o M, | w | = 1 is such that II ( w, w ) = α, we would have II ( w, w ) = II ( w, w ⊥ ) = 0 (the curvature ellipse isa segment with extremal point II ( w, w )) and thus dG x o ( w ) = 0 , in contradictionwith G ′ ( x o ) = 0). Let us first show that II ( T , T ) = 0 . Differentiating G = T ∧ T , we easily get, for all w ∈ T x o M, (35) dG ( w ) = II ( T , w ) ∧ T + T ∧ II ( T , w ) . But we also have(36) dG ( w ) = G ′ w = 2 J (cid:26) − sin (cid:18) z sin ψ (cid:19) + cos (cid:18) z sin ψ (cid:19) I (cid:27) w. For w = T ≃ µ ∈ C we get H ( dG ( T ) , dG ( T )) = µ which is equivalent to(37) h dG ( T ) , dG ( T ) i = 4 µ and dG ( T ) ∧ dG ( T ) = 0 . In view of (35) with w = T , the second property reads(38) II ( T , T ) ∧ II ( T , T ) = 0 . Since II ( T , T ) belongs to the curvature ellipse [ α, β ] and II ( T , T ) is tangent tothe ellipse, we can write II ( T , T ) = α + λ ( β − α ) II ( T , T ) = λ ′ ( β − α )for some λ, λ ′ ∈ R , and (38) then implies λ ′ α ∧ β = 0 . This in turn implies λ ′ = 0 :by contradiction, if λ ′ = 0 we would obtain α ∧ β = 0 and since h α, β i = 0 , α and β would be collinear null vectors; the norm of dG ( T ) = α ∧ T + T ∧ λ ′ ( β − α )would then be zero, in contradiction with (37). Thus λ ′ = 0 and II ( T , T ) = 0 . Since II ( T , T ) = 0 we get that II ( T , T ) and II ( T , T ) are the extremal pointsof the curvature ellipse, and we assume that α = II ( T , T ) and β = II ( T , T ) . We deduce from (35) that dG ( T ) = α ∧ T and from (37) that | α | = µ . So thereexists a unit spacelike vector N such that α = µ N . Similarly, we obtain from (35)and (36) with w = T ≃ iν ∈ C that dG ( T ) = T ∧ β = iν G ′ , which implies that | β | = − ν , and thus that there exists a unit timelike vector N such that β = ν N . Let us finally show that ( N , N ) is a positively oriented basisof N x o M with N future-oriented: by (34), (35) and (36) with w := T ≃ µ and w := T ≃ iν we obtain2 N · T = G ′ and 2 T · N = iG ′ . Now, we have G ′ = 2 J (cid:18) − sin (cid:18) z sin ψ (cid:19) + cos (cid:18) z sin ψ (cid:19) I (cid:19) and thus G ′ = − H C ; this implies that, in H C ,N · N · T · T = i, which is also the canonical volume form e o · e o · e o · e o of R , . The basis ( N , N , T , T )is thus positively oriented in R , , and so is ( N , N ) in N x o M. The vector N isfuture-directed: we have ~H = 12 ( II ( T , T ) + II ( T , T ))= 1 µ N + 1 ν N = ( h sinh v + h cosh v ) N + ( h cosh v + h sinh v ) N by (21), i.e. ~H = h (sinh v N + cosh v N ) + h (cosh v N + sinh v N ) . Since h and h are by hypothesis the coordinates of ~H in a normal basis whichis positively oriented in space and in time, the vector sinh v N + cosh v N isfuture-directed, and so is N . This proves the proposition. (cid:3)
Explicit expression of the adapted frame.
We only give here results ofcalculations, and refer to Appendix C for more details. Direct computations usingthe special lift (17) of the Gauss map and the representation formula (29)-(30) givethe following explicit formulas: T = ( − sinh ( ψ ) cosh ( ϕ ) , − sinh ( ψ ) sinh ( ϕ ) , cosh ( ψ ) sin ( ϕ ) , cosh ( ψ ) cos ( ϕ ))and T = (sin ( ψ ) sinh ( ϕ ) , sin ( ψ ) cosh ( ϕ ) , − cos ( ψ ) cos ( ϕ ) , cos ( ψ ) sin ( ϕ ))where ψ = ψ + iψ and ϕ := 2 z sin ψ = ϕ + iϕ . Let us note that (28) and (34) imply that(39) dT = ( c T + 2 N ) dx + c T dy, dT = − c T dx + ( − c T + 2 N ) dy which may naturally also be obtained by direct computations. Similarly, we alsohave N = (cos( ψ ) sinh( ϕ ) , cos( ψ ) cosh( ϕ ) , sin( ψ ) cos( ϕ ) , − sin( ψ ) sin( ϕ )) ,N = (cosh( ψ ) cosh( ϕ ) , cosh( ψ ) sinh( ϕ ) , − sinh( ψ ) sin( ϕ ) , − sinh( ψ ) cos( ϕ )) , (40) dN = ( − T + c N ) dx − c N dy, dN = c N dx + (2 T − c N ) dy, and thus(41) ∇ ′ N = ( c dx − c dy ) N , ∇ ′ N = ( c dx − c dy ) N . It appears on these formulas that the special frame ( T , T , N , N ) only dependson the constant angle ψ and the value of the parameter z. We have by (33)(42) F = Z µT dx + νT dy, ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 15 and the immersion with constant angle ψ with respect to e o ∧ e o is entirely deter-mined by µ and ν, in accordance with Theorem 2.6.3. Second order invariants of a constant angle surface.
Let us fix a point x o ∈ M and consider the quadratic form δ : T x o M → R defined by δ := 12 dG x o ∧ dG x o where Λ R , is naturally identified with R as in Section 2.1. Let us recall from[2] that δ determines the asymptotic directions of the surface at x o , and also thefourth numerical invariant of the surface at that point∆ := disc δ (the other three invariants are K, K N and | ~H | ). We recall that a non-zero vector w ∈ T x o M is said to be an asymptotic direction of M at x o if δ ( w ) = 0; the existenceof a pair of asymptotic directions is thus determined by the sign of ∆ . In the nextlemma we compute δ and the invariant ∆ for a constant angle surface: Lemma 6.2.
The matrix of δ in the orthonormal basis ( T , T ) of T x o M is M at ( δ, ( T , T )) = 2 µν (cid:18) (cid:19) . In particular ∆ = − µ ν and T , T are the asymptotic directions of M at x o . Proof.
By (34) and (35), we have dG ( T ) = II ( T , T ) ∧ T = 2 µ N ∧ T and dG ( T ) = T ∧ II ( T , T ) = 2 ν T ∧ N . Thus δ ( T , T ) = δ ( T , T ) = 0 and δ ( T , T ) = 2 µν N ∧ N ∧ T ∧ T ≃ µν since ( N , N , T , T ) is a positively oriented and orthonormal basis of R , with N timelike and future-oriented. (cid:3) Let us also mention that the mean curvature vector of a constant angle surfaceis given by ~H = 12 ( II ( T , T ) + II ( T , T )) = 1 µ N + 1 ν N and thus that(43) | ~H | = 1 µ − ν . We thus have the following
Proposition 6.3.
The four invariants of M are K = K N = 0 , | ~H | = 1 µ − ν and ∆ = − µ ν . Incompleteness of the constant angle surfaces.
Proposition 7.1.
An oriented spacelike surface in R , with regular Gauss mapand constant angle ψ = 0 [ π/ is not complete.Proof. Recalling (25), the property ψ = 0 [ π/
2] is equivalent to c or c = 0 . Letus first assume that c = 0 . Since h µ T , ∂x i = 1 and h µ T , ∂y i = 0 , the gradient of the function x on U is ∇ x = µ T . Let us consider its norm f ( t ) := |∇ x | = | µ | − along an integral curve of T . By the first equation in (24) it satisfies( f ) ′ = − µ − dµ ( T ) = 2 c µ − = 2 c f . Since f does not vanish, this equation implies that the flow of T cannot be definedfor all t ∈ R , and thus that the surface is not complete. If c = 0 and c = 0 , weanalogously consider g ( t ) = |∇ y | = | ν | − along an integral curve of T . It satisfies ( g ) ′ = 2 c g , which also implies that thesurface is not complete. (cid:3) Remark 5.
The surface H ( r ) × S ( r ) ⊂ R , × R = R , with H ( r ) := { ( x , x ) ∈ R , : x − x = r , x > } , more generally a product of two regular and complete curves γ × γ ⊂ R , × R ( γ spacelike), is a complete spacelike surface with regular Gauss map and constantangle ψ = π/ π ] . In fact, in view of the proposition and of Theorem 3 below, upto a congruence, all the complete surfaces in R , with constant complex angle andregular Gauss map are of that form. Characterization of constant angle surfaces which are productof plane curves
We show here that the surfaces of constant angle π/ π ] are product of planecurves. More precisely, we have the following result: Theorem 3.
A surface has constant angle ψ = π/ π ] with respect to a spacelikeplane p o if and only if it is a product γ × γ of curves in the perpendicular planes p o and p ⊥ o . Proof.
We assume that p o = e o ∧ e o . Since ψ = π/ π ] and ψ = 0 , we have c = c = 0 (see (25)) and (24) implies that µ only depends on x, and ν onlydepends on y ; moreover, the explicit formulas for T and T in Section 6.2 read T = (0 , , sin 2 x, cos 2 x ) and T = ( − sinh 2 y, cosh 2 y, , . Formula (42) gives the result. Reciprocally, for a product γ × γ in R × R , where γ is a spacelike curve, setting p o = R × { } , T = γ ′ / | γ ′ | and T = γ ′ / | γ ′ | theangle ψ between the surface γ × γ and p o is by definition such thatcos ψ = H ( p o , T ∧ T ) = h p o , T ∧ T i + i p o ∧ T ∧ T = 0;this implies that ψ = π/ π ] . (cid:3) ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 17
Since surfaces with constant angle ψ = 0 [ π ] were described in Remark 1, thisresult completes the description of the surfaces with constant angle ψ = 0 [ π/ . Characterization of constant angle surfaces with real or pureimaginary constant angle
We describe in the following theorem the spacelike surfaces with regular Gaussmap and real or pure imaginary constant angle in R , . Let us recall that a holonomytube over a spacelike curve γ ∈ R , is a surface obtained by the normal paralleltransport along γ of some curve c initially given in a fixed hyperplane normal to γ : c is the curve of the starting points of the tube. Theorem 4.
A surface with real constant angle ψ = ψ ∈ R (resp. pure imaginaryconstant angle ψ = iψ ∈ i R ) with respect to a spacelike plane p o is a holonomy tubeover a plane curve γ ∈ p o . Moreover, if c ∈ N m o γ is the curve of the starting pointsof the tube, then c is an helix curve in N m o γ ≃ R , with respect to a spacelike(resp. timelike) direction.Proof. Let us fix m o = ( x o , y o ) ∈ U ⊂ R and consider the curves γ ( x ) := F ( x, y o )and c ( y ) := F ( x o , y ) . We assume that p o is the plane { } × R ⊂ R , × R = R , . The curve γ is everywhere tangent to T = (0 , , sin (cid:18) x sin ψ (cid:19) , cos (cid:18) x sin ψ (cid:19) ) , by the expression of T with ψ = 0 in Section 6.2, and is thus a curve in p o . Thecurve c belongs to the hyperplane normal to the curve γ at x o . Indeed, h c ( y ) − γ ( x o ) , γ ′ ( x o ) i = h F ( x o , y ) − F ( x o , y o ) , ∂ x F ( x o , y o ) i = h Z yy o ∂ y F ( x o , t ) dt, ∂ x F ( x o , y o ) i = Z yy o µ ( x o , y o ) ν ( x o , t ) h T ( x o , t ) , T ( x o , y o ) i , which is zero since T ( x o , y o ) = T ( x o , t ) is orthogonal to T ( x o , t ) ( T ( x o , t ) doesnot depend on t, by the expression of T above). Finally, if we fix y = y , the curve x F ( x, y ) may be regarded as a normal section of γ ; it is parallel since ∂ x F ( x, y ) = µ ( x, y ) T ( x, y ) = µ ( x, y ) T ( x, y o )is tangent to γ at x. The curve c is an helix in N m o γ : its unit tangent is y T ( x o , y ) = (sin ψ sinh ϕ , sin ψ cosh ϕ , − cos ψ cos ϕ , cos ψ sin ϕ ) , and if ~A is the fixed direction (0 , , − cos ϕ , sin ϕ ) in N m o γ we have h T ( x o , y ) , ~A i = cos ψ . The curve c is thus a constant angle curve in R , with respect to the spacelikedirection ~A ; the constant angle is ψ . The proof for ψ = iψ ∈ i R is analogous. (cid:3) Characterization of constant angle surfaces in hyperspheres andlightcones
The immersion in the adapted orthonormal frame.
Let us write theimmersion of a constant angle surface in the form(44) F = f T + ˜ f T + g N + ˜ g N where ( T , T , N , N ) is the frame adapted to the surface introduced in Section 6and f, ˜ f , g and ˜ g are smooth real functions of the variables x and y. Formulas (28)and (34) imply the following:
Theorem 5.
We assume that ψ = 0 [ π/ and ψ = 0 (i.e. c , c = 0 ) andsuppose that µ and ν are solutions of (24) . Then the immersion reads (45) F = f T + ∂ y fc T + g N + ∂ y gc N where f and g are solutions of (46) ∂ x f = µ + (4 + c ) g − ∂ yy g , ∂ y f = c ( c g − ∂ yy g )2 c ,∂ x g = − c ν + ( c − f + ∂ yy f , ∂ y g = c ν − c ( c f + ∂ yy f )2 c ,∂ xy f = − c c f, ∂ xy g = − c c g. Reciprocally, given two functions f, g solving this PDE system for µ and ν solutionsof (24) , the immersion F given by (45) is a spacelike immersion in R , of constantcomplex angle. The advantage of this formulation lies in the fact that the solutions of this systemdirectly give the immersion; in the previous formulations, Theorems 1 and 2, a lastintegration was still required to obtain the immersion F from the 1-form ξ. Proof. If F is an immersion of constant angle ψ and metric µ dx + ν dy , we definethe functions f = h F, T i , ˜ f = h F, T i , g = h F, N i and ˜ g = −h F, N i ;they are such that (44) holds. Using (28) and (34) we compute df = h dF, T i + h F, dT i = h µT dx + νT dy, T i + h F, ( c dx + c dy ) T + 2 N dx i = ( µ + c ˜ f + 2 g ) dx + c ˜ f dy and similarly d ˜ f = − c f dx + ( ν − c f − g ) dy, dg = ( − f − c ˜ g ) dx + c ˜ gdy and d ˜ g = − c gdx + ( c g − f ) dy. This implies that ˜ f = ∂ y fc , ˜ g = ∂ y gc and f, g satisfy (46).Reciprocally, if f, g are solutions of (46), straightforward computations using(39) and (40) show that the function F defined by (45) satisfies ∂ x F = µT and ∂ y F = νT ; it is thus an immersion with Gauss map G = T ∧ T , and, since ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 19 H ( I, G ) = cos ψ (by the explicit formulas for T and T in Section 6.2), it is ofconstant angle ψ with respect to p := I. (cid:3) Remark 6.
The system (46) is in fact equivalent to the smaller system formed bythe first four equations and one of the last two equations: the sixth equation mayindeed be easily obtained from these five equations.
Description of the constant angle surfaces in hyperspheres andlightcones.
We determine here the constant angle surfaces in hyperspheres andlightcones, i.e., up to translations, immersions of constant angle and constant norm.
Corollary 1.
Keeping the notation introduced above, we have the following:1. The constant angle immersion F belongs to a hypersphere if and only if µ = r e c y − c x + r e − ( c y − c x ) and ν = − r e c y − c x + r e − ( c y − c x ) for some constants r , r = 0 . In that case, and up to a translation, we have | F | = r r .
2. The constant angle immersion F belongs to a lightcone if and only if (47) µ = re ǫ ( c y − c x ) and ν = − ǫre ǫ ( c y − c x ) for some constant r = 0 and ǫ = ± .Moreover, if the immersion belongs to a hypersphere or a lightcone it can be written,up to a translation, in the form F = − µN + νN where N , N ∈ R , are the unit orthogonal vector fields introduced in Section 6.Proof. From Theorem 5, we can write the immersion F in the form F = f T + ∂ y fc T + g N + ∂ y gc N with f and g solutions of (46). But F has constant norm if and only if h F, F x i = h F, F y i = 0, that is f = 0. The function g is thus a solution of µ + (4 + c ) g − ∂ yy g , c g − ∂ yy g,∂ x g = − c ν, ∂ y g = c ν, ∂ xy g = − c c g. Using (24) we have ∂ y g = c ν = − ∂ y µ g ( x, y ) = − µ ( x, y )2 + t ( y )for some function t ; but the last condition now reads ∂ xy g = − ∂ xy µ = 12 c c µ − c c t, and since ∂ xy µ = − c c µ we get t = 0. With g = − µ ∂ yy µ = c µ, ∂ x µ = c ν, ∂ y µ = − c ν, ∂ xy µ = − c c µ whose solutions are µ = r e c y − c x + r e − ( c y − c x ) and ν = − r e c y − c x + r e − ( c y − c x ) where r and r are real numbers. We thus obtain F = gN + ∂ y gc N = − µ N − ∂ y µ c N = − µN + νN | F | = µ − ν r r . Finally, the immersion belongs to the lightcone at the origin if and only if r or r = 0 , which implies the last claim in the statement. (cid:3) Remark 7.
The second part of the corollary in particular shows that the surfacesin (15) are, up to a congruence and scaling, the unique surfaces with constant anglein a lightcone.
Corollary 2.
Let us assume that M is a spacelike surface of constant angle ψ =0 [ π/ . Then the following properties are equivalent:a) ~H ∈ Γ( N M ) is parallel;b) ~H is everywhere lightlike;c) the parameter z = x + iy ∈ U is conformal, i.e. µ = ν on U . If one of these properties occurs then the surface belongs to a lightcone and is, upto a congruence, of the form (15).Proof.
Since β = u + iv is an holomorphic function and by (26) we have ∂ y v = ∂ x u = − ν ∂ y µ. Recalling (21) we deduce that − ∂ y µµ = ∂ y h sinh v + ∂ y h cosh v + ( h cosh v + h sinh v ) ∂ y v = ∂ y h sinh v + ∂ y h cosh v − ν ∂ y µ and using (43) that(48) − | ~H | ∂ y µ = ∂ y h sinh v + ∂ y h cosh v. We similarly have − ∂ x νν = ∂ x h sinh v + ∂ x h cosh v + ( h cosh v + h sinh v ) ∂ x v = ∂ x h sinh v + ∂ x h cosh v − µ ∂ x ν and thus(49) | ~H | ∂ x ν = ∂ x h cosh v + ∂ x h sinh v. ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 21
Since, by (26) and (27), we have1 ν ∂ y µ + i µ ∂ x ν = − ∂ x u + i∂ y u = 2 cot ψ, (48) and (49) yield2 | ~H | cot ψ = − ν ( ∂ y h sinh v + ∂ y h cosh v ) + i µ ( ∂ x h cosh v + ∂ x h sinh v ) . Since h and h are the components of the mean curvature vector ~H in a parallelframe normal to the surface, this formula implies the equivalence between (a) and(b). The equivalence between (b) and (c) is a consequence of (43). If now theproperties (a)-(c) hold then µ = ν and the system (24) easily implies that µ, ν areof the form (47). Corollary 1 then gives the result. (cid:3) Examples.
We now use the previous results to give new simple explicit ex-amples of constant angle surfaces. Since examples of constant angle surfaces inlightcones were already given in Section 3.4, we will focus on examples in hyper-spheres, and on examples which do not belong to hyperspheres nor to lightcones.All the examples arise from the application of Theorem 5 and its first corollary: westart with a given solution µ of (31), we then find two functions f, g solving (46),and we finally obtain the immersion F from the general form (45). For the sake ofbrevity we only write the final expressions.10.3.1. Immersions in hyperspheres: µ = 2 sinh( c y − c x ) . As proved in Corollary1, the metric coefficient µ has to be of the form µ = r e c y − c x + r e − ( c y − c x ) . If we take r = − r = 1 we get µ = 2 sinh( c y − c x ) and then ν = 2 cosh( c x + c y ).We assume that ( x, y ) ∈ R is such that c y = c x, so that µ = 0 . The immersionis thus F = − sinh( c y − c x ) N + cosh( c y − c x ) N , whose norm is | F | = − Immersions which are not in hyperspheres nor in lightcones: µ = sin( c y + c x ) . For this µ , solution to (31), we have ν = − cos( c x + c y ) . Assuming thatsin( c y + c x ) = 0 and cos( c x + c y ) = 0 (for suitable ( x, y ) in R ), the functions f and g are f ( x, y ) = cos( c x + c y ) , g ( x, y ) = − sin( c x + c y ) . Thus, the immersion reads F = cos( c x + c y ) c T − sin( c x + c y ) c T − sin( c x + c y ) N − cos( c x + c y ) N = − c νT − c µT − µN + νN , whose norm is | F | = ( c + 1) ν + ( c − µ = ( c + 1) cos ( c x + c y ) + ( c −
1) sin ( c x + c y ) . Another example: µ = ( c y − c x )(sin( c y + c x ) + cos( c y + c x )) . For this µ we get ν = (1 − c y + c x ) cos( c y + c x ) − (1 + c y − c x ) sin( c y + c x ) . We assume that ( x, y ) belongs to an open subset of R such that µν = 0. It appearsthat the functions f ( x, y ) = c ((2 + c − c )( c y − c x )(cos( c y + c x ) + sin( c y + c x ))+( c + c )(cos( c y + c x ) − sin( c y + c x ))) ,g ( x, y ) = (2 + c − c )( c y − c x )(cos( c y + c x ) − sin( c y + c x )) − ( c + c )(cos( c y + c x ) + sin( c y + c x ))are solutions of (46), which defines the immersion by (45). Appendix A. Angles and orthogonal projections
A.1.
An alternative construction of the complex angle.
We give here an-other construction of the complex angle between two oriented and spacelike planes p and q in R , . Let us denote by π : q → p and π ′ : q → p ⊥ the restrictions to q of the orthogonal projections R , → p and R , → p ⊥ , and consider the quadraticforms Q and Q ′ defined on q by Q ( x ) = h π ( x ) , π ( x ) i and Q ′ ( x ) = h π ′ ( x ) , π ′ ( x ) i for all x ∈ q. They are linked by the relation Q ( x ) + Q ′ ( x ) = | x | for all x ∈ q, andthus satisfy(50) Q ( x ) + Q ′ ( x ) = 1for all x ∈ q, | x | = 1 . There exists a positive orthonormal basis ( u, u ⊥ ) of q suchthat Q ( u ) = max x ∈ q, | x | =1 Q ( x ) , Q ( u ⊥ ) = min x ∈ q, | x | =1 Q ( x ) . By (50) this basis is also such that Q ′ ( u ) = min x ∈ q, | x | =1 Q ′ ( x ) , Q ′ ( u ⊥ ) = max x ∈ q, | x | =1 Q ′ ( x ) , and satisfies h π ( u ) , π ( u ⊥ ) i = 0 and h π ′ ( u ) , π ′ ( u ⊥ ) i = 0 . We need to divide the discussion into three main cases, according to the dimensionof the range of π ′ : Case 1: rank ( π ′ ) =2. This condition means that p ⊕ q = R , . Since p ⊥ is timelike, Q ′ has then signature (1 , . The relation (50) then implies that π ′ ( u ) is timelike, Q ( u ) > Q ( u ⊥ ) <
1. We choose moreover u such that π ′ ( u ) is future-directed:the basis ( u, u ⊥ ) is then uniquely defined. We then consider the positively orientedand orthonormal basis ( e , e , e , e ) of R , such that e is future-directed, ( e , e )is a positive basis of p and π ′ ( u ) = a e , π ′ ( u ⊥ ) = a e , π ( u ) = a e , π ( u ⊥ ) = a e for some constants a , a , a , a such that a , a ≥ . Since u = a e + a e and u ⊥ = a e + a e we have | u | = 1 = − a + a and | u ⊥ | = 1 = a + a and we may set ψ , ψ ∈ R , ψ ≥ , such that a = sinh ψ , a = cosh ψ , a = − sin ψ and a = cos ψ . Since Q ′ ( u ) = − a , Q ′ ( u ⊥ ) = a and the signature of Q ′ is (1 , , we in fact have a a = 0 , that is ψ > ψ = 0 [ π ] . With these definitions, we have(51) u = sinh ψ e + cosh ψ e and u ⊥ = − sin ψ e + cos ψ e which easily yields for ψ := ψ + iψ q = u ∧ u ⊥ = cos ψ p + sin ψ V for V = e ∧ e . Case 2: rank ( π ′ ) = 1 . In that case Im ( π ′ ) is a line, which may be timelike,spacelike or lightlike: • Q ′ has signature (0 , , i.e. Im ( π ′ ) is a timelike line: q belongs to the timelikehyperplane p ⊕ Im ( π ′ ) . We may then follow the lines of the previous case: the basis( u, u ⊥ ) is such that Q ( u ) > Q ( u ⊥ ) = 1 , and we may suppose that π ′ ( u ) isfuture-directed. We then define the basis ( e , e , e , e ) of R , and the angles ψ and ψ as above, and observe that ψ = 0 [ π ] and ψ > Q ′ ( u ⊥ ) = a = 0 heresince Q ( u ⊥ ) = 1 and by (50)). The angles ψ and ψ have the following inter-pretations: the planes p and q are two oriented planes in the timelike hyperplane p ⊕ Im ( π ′ ); this hyperplane is naturally oriented by the orientation of p and thefuture-orientation of Im ( π ′ ); the lines normal to the oriented planes p and q arethus also oriented, and ψ = 0 [2 π ] if these lines are both future-oriented, and ψ = π [2 π ] in the other case. The angle ψ is the measure of the lorentzian anglebetween the future-directed lines normal to p and q. • Q ′ has signature (1 , , i.e. Im ( π ′ ) is a spacelike line: q belongs to the spacelikehyperplane p ⊕ Im ( π ′ ) . The basis ( u, u ⊥ ) is such that Q ( u ) = 1 and Q ( u ⊥ ) < , i.e. Q ′ ( u ) = 0 and Q ′ ( u ⊥ ) > . It is no more possible to suppose here that π ′ ( u ) is time-like (it is the vector zero!), i.e. u is only defined up to sign. Nevertheless, choosingfor e the future-directed unit vector orthogonal to p ⊕ R π ′ ( u ⊥ ) we may constructas above a unique positively oriented and orthonormal basis ( e , e , e , e ) of R . adapted to π ′ ( u ⊥ ) , π ( u ) and π ( u ⊥ ) . Defining ψ and ψ as above we easily see that ψ = 0 and ψ is defined up to sign and a multiple of 2 π : it is thus representedby a unique real number in [0 , π ] . This angle has the following interpretation: theplanes p and q are two oriented planes in the spacelike hyperplane p ⊕ Im ( π ′ ) whichis naturally oriented by the election of its future-directed normal direction in R , and the canonical orientation of R , ; ψ ∈ [0 , π ] is a measure of the angle of theoriented lines normal to these planes in p ⊕ Im ( π ′ ) . • Q ′ is zero, i.e. Im ( π ′ ) is a lightlike line: q belongs to the degenerate hyperplane p ⊕ Im ( π ′ ) . In that case we set ψ = 0 [ π ] and ψ = 0: more precisely, a givenorientation of Im ( π ′ ) induces an orientation of p ⊕ Im ( π ′ ) , and we set ψ = 0 [2 π ]if it coincides with the sum of the orientations of q and Im ( π ′ ) , and ψ = π [2 π ] inthe other case. Case 3: rank ( π ′ ) =0. In that case q = ± p, i.e. the planes p and q coincidein R , , with the same orientation ( q = p ) or the opposite orientation ( q = − p ); we set ψ = 0 and ψ = 0 [2 π ] if q = p or ψ = π [2 π ] if q = − p. It is not difficult to verify that in all the cases described above the complex angle ψ := ψ + iψ coincides with the complex angle ψ defined in Section 2.2, if wemoreover choose the sign of ψ in Definition 2.1 such that ψ := ℑ m ψ ≥ . We haveobtained the following additional informations concerning the relative position of q and p in terms of the angle ψ = ψ + iψ : Proposition A.1. • q does not belong to any hyperplane containing p if andonly if ψ = 0 [ π ] and ψ > • q = ± p belongs to a spacelike (resp. timelike) hyperplane containing p ifand only if ψ = 0 [ π ] and ψ = 0 (resp. ψ = 0 [ π ] and ψ > ); • q belongs to a null hyperplane containing p if and only if ψ = 0 [ π ] and ψ = 0 . A.2.
An elementary characterization of the two real angles.
We still sup-pose that p and q are oriented spacelike planes in R , and we consider their realangles ψ , ψ constructed above.The following descriptions of ψ and ψ are verysimilar to the elementary definitions of the two principal angles between two ori-ented planes in R (see for example [3]). Proposition A.2.
We have cosh ψ = sup x ∈ q, y ∈ p, | x | = | y | =1 h x, y i . Moreover, if u ∈ q and v ∈ p, | u | = | v | = 1 are such that cosh ψ = h u, v i , then cos ψ = h u ⊥ , v ⊥ i where u ⊥ and v ⊥ are unit vectors in q and p such that ( u, u ⊥ ) and ( v, v ⊥ ) arepositive orthonormal bases of q and p. Proof.
We assume for simplicity that rank( π ′ )=2 and keep the notations introducedin the previous section to study that case. If x ∈ q and y ∈ p are unit vectors, wehave h x, y i = h π ( x ) , y i ≤ | π ( x ) | ≤ sup x ∈ q, | x | =1 | π ( x ) | = cosh ψ . Moreover, the equality holds for x = u = sinh ψ e + cosh ψ e and y = v = e where ( e , e , e , e ) is the basis defined above (Case 1). This proves the first claim.Since u ⊥ = − sin ψ e + cos ψ e (by (51)) and v ⊥ = e the last claim readilyfollows. (cid:3) Appendix B. Characterization of surfaces of constant angle ψ = 0 [ π ]We study here the very special case of the surfaces with regular Gauss map andconstant angle ψ = 0 [ π ] with respect to some given spacelike plane p o . The resultis the following:
Proposition B.1.
A connected and spacelike surface with regular Gauss map hasa constant angle ψ = 0 [ π ] with respect to a fixed spacelike plane p o if and only if itbelongs to a affine hyperplane x o + p o ⊕ L where x o is a point of R , and L is oneof the two null lines of the timelike plane p ⊥ o . ONSTANT ANGLE SURFACES IN 4-DIMENSIONAL MINKOWSKI SPACE 25
Proof.
We only prove the ”only if” direction (the converse is trivial). Let us fixa positive orthonormal basis ( u, u ⊥ ) of p o and a positive basis ( N, N ′ ) of p ⊥ o suchthat h N, N i = h N ′ , N ′ i = 0 and h N, N ′ i = − , and set ξ = u ∧ N and ξ ′ = u ⊥ ∧ N ′ . ( ξ, ξ ′ ) form a basis of the C -linear space T p o Q such that H ( ξ, ξ ) = H ( ξ ′ , ξ ′ ) = 0 and H ( ξ, ξ ′ ) = i. Recall the definition of the complex parameter z at the end of Section 3.1. Since H ( G ( z ) , p o ) = cos ψ = 1 we have H ( G ( z ) − p o , p o ) = H ( G ( z ) , p o ) − H ( p o , p o ) = 1 − H ( G ( z ) − p o , G ( z ) − p o ) = H ( G ( z ) , G ( z )) + H ( p o , p o ) − H ( G ( z ) , p o )= 1 + 1 −
2= 0 , which means that G ( z ) − p o belongs to one of the two complex lines C ξ and C ξ ′ of T p o Q . Thus G ′ ( z ) also belongs C ξ ∪ C ξ ′ and by continuity G ′ ( z ) belongs to C ξ forall value of z, or G ′ ( z ) belongs to C ξ ′ for all value of z ( G ′ does not vanish since G is assumed to be regular!). If G ′ belongs to C ξ then G belongs to p o + C ξ, whichmeans that all the tangent planes of M belongs the hyperplane p o ⊕ L with L = R N. This finally easily implies that M belongs to an affine hyperplane x o + p o ⊕ L . (cid:3) Appendix C. Details for the computations of the adapted frame
Let us briefly describe here the computations leading to the explicit formulas forthe special frame ( T , T , N , N ) adapted to a complex angle surface. RecallingTheorem 2, we have T := 1 µ ∂ x F = 1 µ ξ ( ∂ x ) = 1 µ g − ( ω ( ∂ x ) J + ω ( ∂ x ) K ) b g = g − (sin( u ) J − cos( u ) K ) b g, where g, b g and u are given by the formulas g − = ¯ g = − cos ψ (cid:18) z tan ψ (cid:19) − cos ψ (cid:18) z tan ψ (cid:19) I − sin ψ (cid:18) z cot ψ (cid:19) J + sin ψ (cid:18) z cot ψ (cid:19) K, b g = − cos ¯ ψ (cid:18) ¯ z tan ¯ ψ (cid:19) ψ (cid:18) ¯ z tan ¯ ψ (cid:19) I + sin ¯ ψ (cid:18) ¯ z cot ¯ ψ (cid:19) J − sin ¯ ψ (cid:18) ¯ z cot ¯ ψ (cid:19) K and u = − z cot( ψ )) . Direct and long computations (or the use of a software of symbolic computation)lead to the formula T = − sin (cid:18) ψ − ¯ ψ (cid:19) cos (cid:0) z csc ψ − ¯ z csc ¯ ψ (cid:1) (cid:18) ψ − ¯ ψ (cid:19) sin (cid:0) z csc ψ − ¯ z csc ¯ ψ (cid:1) I + cos (cid:18) ψ − ¯ ψ (cid:19) sin (cid:0) z csc ψ + ¯ z csc ¯ ψ (cid:1) J + cos (cid:18) ψ − ¯ ψ (cid:19) cos (cid:0) z csc ψ + ¯ z csc ¯ ψ (cid:1) K where csc( ψ ) = 1 / sin ψ. Writing ψ = ψ + iψ and z/ sin( ψ ) = ϕ + iϕ , we finallyobtain T = − i sinh ψ cosh ϕ − sinh ψ sinh ϕ I + cosh ψ sin ϕ J + cosh ψ cos ϕ K which is the expression given in Section 6.2. Similar computations give the expres-sion of T . We now explain how to determine N and N . On one side, we have by(34) and (35) dG ( T ) = II ( T , T ) ∧ T = 2 µ N ∧ T and dG ( T ) = T ∧ II ( T , T ) = 2 ν T ∧ N . On the other side, since dG = G ′ dz we have dG ( T ) = 1 µ dG ( ∂ x ) = 1 µ G ′ = 2 µ J {− sin ϕ + cos ϕI } and dG ( T ) = 1 ν dG ( ∂ y ) = iν G ′ = 2 iν J {− sin ϕ + cos ϕI } . We deduce that, in H C , N = − J {− sin ϕ + cos ϕI } T and N = iT J {− sin ϕ + cos ϕI } . The expressions of N and N then follow from the expressions of T and T , anddirect computations. Acknowledgments:
P. Bayard was supported by the project PAPIIT-UNAMIA106218, J. Monterde was partially supported by the Spanish Ministry of Econ-omy and Competitiveness DGICYT grant MTM2015-64013 and R. C. Volpe wassupported by the Generalitat Valenciana VALi+D grant ACIF/2016/342.
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