Calc. Var. (2008) 31:147–166 DOI 10.1007/s00526-007-0088-7

Calculus of Variations

Symmetric boundary values for the Dirichlet problem for harmonic maps from the disc into the 2-sphere Morgan Pierre

Received: 20 March 2006 / Accepted: 22 December 2006 / Published online: 9 October 2007 © Springer-Verlag 2007

Abstract Let Aut(D) denote the group of biholomorphic diffeormorphisms from the unit disc D onto itself and O(3) the group of orthogonal transformations of the unit sphere S2 . The existence of multiple solutions to the Dirichlet problem for harmonic maps from D into S2 is related to the symmetries (if any) of the boundary value γ : ∂D → S2 , by invariance of the Dirichlet energy under the action of Aut(D) × O(3). In this paper, we classify the stabilizers in Aut(D) × O(3) of boundary values in H 1/2 (S1 , S2 ) and C 0 (S1 , S2 ). We give two applications to the Dirichlet problem for harmonic maps. Mathematics Subject Classification (2000) 58E20 · 58G35

1 Introduction In this paper, we study the admissible symmetries for maps from ∂D into S2 , where D = {z ∈ C : |z| < 1} is the unit disc, ∂D = S1 its boundary, and S2 = {x ∈ R3 : |x|2 = 1} is the unit sphere in R3 . The symmetries that we take into consideration are, for ∂D, the restriction of conformal transformations from D onto D, and for S2 , the orthogonal transformations of S2 . The group of holomorphic or anti-holomorphic diffeormorphisms from D onto itself will be denoted Aut(D), and the group of the vectorial isometries of the euclidian space R3 , O(3). This study is motivated by the Dirichlet problem for harmonic maps from D into S2 , which can be formulated as follows: for a given boundary value γ : ∂D → S2 , find u : D → S2 such that

This work was partially supported by the CMLA, Ecole Normale Supérieure de Cachan, Cachan, France. M. Pierre (B) Laboratoire de Mathématiques, Université de Poitiers, Téléport 2 - BP 30179, Boulevard Marie et Pierre Curie, 86962 Futuroscope Cedex, France e-mail: [email protected]

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− ∆u = u|∇u|2 u=γ

in D,

on ∂D.

(1) (2)

If γ is constant, Lemaire [11] proved that the only solution u ∈ C ∞ (D, S2 ) to (1), (2) is the constant map. If γ ∈ C 1 (S1 , S2 ) is non constant, Brezis and Coron [5] showed that problem (1), (2) has at least two solutions, which are both minimizers of the Dirichlet energy
1 E(u) = |∇u|2 dxdy (3) 2 D

in their own (distinct) homotopy class (see also Jost [9]). The question of the existence of a minimizer of the Dirichlet energy in each homotopy class has been answered in [10,15,17] in terms of the existence of holomorphic or anti-holomorphic extensions of the boundary condition. Qing [14] proved the existence of multiple solutions to (1), (2) in one homotopy class for certain boundary values, by applying a Morse theory for harmonic maps. It is well-known that E, and hence Eq. (1), are invariant under the action of Aut(D) and O(3) (see for instance [11] and Sect. 3); the choice of Aut(D) and O(3) as groups of smooth diffeomorphisms which leave E invariant is even optimal [13]. Because of topological obstructions, a boundary value from ∂D to S2 which has symmetries with respect to these transformations cannot in general be extended in a given homotopy class to a map (harmonic or not) which has the same symmetries. This remark allowed us in [13] to build multiple minimizers in a same homotopy class for a large class of symmetric boundary values. The symmetries that we considered were essentially finite or continuous rotational symmetries. This result raises the question of finding all possible symmetries which lead to a topological obstruction. More generally, it indicates that the existence of multiple solutions to (1), (2) is strongly related to the symmetries of the problem. In Theorem 3, which is the main result of the paper, we shall see that the groups of rotations that we introduced in [13] are in fact, up to conjugacy, the only ones that occur in this problem — if one considers only orientation-preserving diffeomorphisms in Aut(D) and O(3). Theorem 3 also provides a classification of the boundary values which have an infinite symmetry group. The space H 1 (D, S2 ), defined by H 1 (D, S2 ) = {u ∈ H 1 (D, R3 ) : |u| = 1 a.e. in D}, is the natural space when using energy methods for problem (1), (2). The existence result of Brezis and Coron [5] is stated for any boundary value γ which is the trace on ∂D of a map u ∈ H 1 (D, S2 ). By density of C 1 (D, S2 ) into H 1 (D, S2 ) [5,16], every such map γ belongs to H 1/2 (S1 , S2 ) = {γ ∈ H 1/2 (S1 , R3 ) : |γ | = 1 a.e. in S1 }. Conversely, by a result of Bethuel and Demengel [3], we know now that every map in H 1/2 (S1 , S2 ) is the trace of a map u ∈ H 1 (D, S2 ). We shall therefore state the symmetry results in this paper for boundary values γ ∈ H 1/2 (S1 , S2 ). In fact, as it will be seen by the reader, these results also hold for every γ ∈ C 0 (S1 , S2 ), with similar and even easier proofs.

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We recall that H 1/2 (S1 , R3 ) = {γ ∈ L2 (S1 , R3 ) : |γ |H 1/2 (S1 ,R3 ) < ∞}, where the semi-norm is defined by (see for instance [4]): ⎛ ⎞1/2

2 |γ (x) − γ (y)| ⎜ ⎟ |γ |H 1/2 (S1 ,R3 ) = ⎝ dxdy⎠ . |x − y|2

(4)

(5)

S1 S1

In the right hand side, | · | is the euclidian norm in R2 and R3 . There are several equivalent definitions for the H 1/2 semi-norm, but with definition (5), | · | is clearly invariant by any isometry of R3 , and we shall see that it is also invariant by conformal transformation of D. This will be of great importance for us. The paper is structured as follows. In Sect. 2, we introduce some notations and well known results of group theory and complex analysis. In Sect. 3, we detail some obvious consequences of the Aut(D) × O(3) group action on the set H 1 (D, S2 ). The aim of Sect. 4 is to determine, in Theorem 3, the structure of stabilizers of boundary values in the subgroup Aut+ (D) × O+ (3) of Aut(D) × O(3) consisting of the orientation-preserving diffeomorphisms. We extend this result in Sect. 5 and Appendix A by giving the classification of stabilizers in Aut(D) × O(3). In the last section, we give two applications to the Dirichlet problem (1), (2).

2 Notations The elementary vocabulary concerning groups can be found for instance in [7]. If G is a group and H, K are two subgroups of G, we shall denote KH the set consisting of the elements of the form kh, k ∈ K, h ∈ H. A group generated by g will be denoted g. The index of H in G is the cardinal number of left cosets of H in G (i.e. sets of the form Hg, g ∈ G fixed) and is written [G : H]. If a group G acts on the left on a non-empty set X by (g, x)  → g · x (g ∈ G, x ∈ X), the stabilizer of x ∈ X is {g ∈ G : g · x = x}. If y ∈ X belongs to the orbit of x, i.e. y = g · x for some g ∈ G, then its stabilizer is conjugate to the stabilizer of x. ˆ = C ∪ {∞}, i.e. S2 C ˆ via It will be useful to consider S2 as the Riemann sphere C the stereographic projection ˆ Π : S2 → C (6) from the north pole. For any open subset Ω of S2 , we shall denote Aut(Ω) the group of diffeomorphisms from Ω onto Ω which are holomorphic or anti-holomorphic; Aut+ (Ω) will denote the subgroup of Aut(Ω) consisting of the the holomorphic diffeomorphisms, and Aut− (Ω) the subset of Aut(Ω) consisting of the anti-holomorphic diffeomorphisms. Obviously, we have Aut− (D) = Aut+ (D) ◦ (z  → z¯ ) = {z  → g(¯z) : g ∈ Aut+ (D)}.

(7)

The group Aut(S2 ) can be identified with the group of Moebius transformations ˆ 2 [1]. It is well known that Aut+ (S2 ) is the family of transformations of the acting in R form  az + b a b TA : z  → with A = ∈ GL(2, C). (8) c d cz + d

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The map A  → TA from GL(2, C) onto Aut+ (S2 ) is a group homomorphism. Two matrices A and B (λ ∈ C∗ ) define the same orientation-preserving Moebius transformation if and only if B = λA for some λ ∈ C∗ . In particular, the following quantity, tr2 (g) =

tr2 (A) , det(A)

is well defined for every g ∈ Aut+ (S2 ), where A ∈ GL(2, C) is any matrix which represents g. The function tr2 is clearly invariant under conjugation. In fact, this function parametrizes the conjugacy classes in Aut+ (S2 ) [1]. We shall see in Theorem 1 that the same is almost true in Aut+ (D). Let H = {z ∈ C : Im z > 0} denote the open upper half-plane. The Moebius transformation z−i Φ0 : z  → (9) z+i maps bijectively H onto D, and R ∪ {∞} onto ∂D. In order to understand the elements of Aut+ (D), it will be useful to see D as H through Φ0 . It is well known [2] that Aut+ (H) = T(SL(2, R)), where T is the group homomorphism defined by (8). In particular, Aut+ (H) can be seen as a subgroup of Aut+ (S2 ). Using Φ0 , one finds that every g ∈ Aut+ (D) may be written g(z) = eiθ

z + z0 1 + z¯0 z

(θ ∈ R/2π Z, |z0 | < 1).

(10)

If g is defined by the right hand side in (10) for some θ ∈ R/2πZ and |z0 | < 1, we find tr2 (g) =

4 cos2 (θ/2) ∈ [0, +∞). 1 − |z0 |2

(11)

For d = 2 or 3, we will denote O(d) the group of vectorial isometries of the euclidian space Rd , O+ (d) the subgoup of O(d) consisting of the vectorial rotations of Rd ; O− (2) will be the subset of O(2) consisting of the reflections in a (vectorial) line. The space R3 has origin in O and axis Ox, Oy, Oz. For α ∈ R/2π Z, we shall write Rα the rotation of angle α around the z-axis in R3 , the orientation of the z-axis being the usual one; Rα will also denote the restriction of Rα to xOy R2 . We will often consider O(2) as a subset of Aut(D): Rα is naturally identified with z  → eiα z, and a reflection in a line with z  → eiβ z¯ for some β ∈ R/π Z. We denote DD ∈ O+ (3) the half-turn around the line D, SP ∈ O− (3) the reflection in the plane P , and sD ∈ O− (2) the reflection in the line D. The group O(3) can also be seen as a subgroup of Aut(S2 ); indeed, every T ∈ O+ (3) may be written T(z) =

αz + β ¯ + α¯ −βz

(α, β ∈ C, |α|2 + |β|2 = 1).

(12)

Finally, it will be useful to denote p1 : Aut(D) × O(3) → Aut(D) the projection onto the first component and p2 : Aut(D) × O(3) → O(3) the projection onto the second component. The maps p1 and p2 are both group homomorphisms. 3 The Aut(D) × O(3) group action: fundamental examples As pointed out in the introduction, the Dirichlet energy E (3) is invariant under the action of Aut(D) and O(3), in the sense that for all u ∈ H 1 (D, S2 ), g ∈ Aut(D) and

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151

T ∈ O(3), E(T ◦ u ◦ g−1 ) = E(u). Moreover, by (7) and (10), every g ∈ Aut(D) extends to a smooth diffeomorphism from a neighbourhood of D onto a neighbourhood of D, so for every T ∈ O(3), g ∈ Aut(D) and u ∈ L2 (D, R3 ), T ◦ u ◦ g−1 L2 (D) = u ◦ g−1 L2 (D) < +∞. As a consequence, the group Aut(D) × O(3) acts on the left on H 1 (D, S2 ) by (Aut(D) × O(3)) × H 1 (D, S2 ) → H 1 (D, S2 ), (g, T), u  → T ◦ u ◦ g−1 . In particular, Aut(D) × O(3) acts on the set of boundary values γ : ∂D → S2 which are traces of maps u ∈ H 1 (D, S2 ), i.e. the set H 1/2 (S1 , S2 ), by [3]. The stabilizer of a map γ ∈ H 1/2 (S1 , S2 ) in Aut(D) × O(3) is: S [γ ] = {(g, T) ∈ Aut(D) × O(3) : T ◦ γ ◦ g−1 = γ a.e. in S1 }.

(13)

In this notation, we use implicitely the fact that every g ∈ Aut(D) has a (unique) continuous extension to D, whose restriction to S1 = ∂D is a smooth (and even analytical) diffeomorphism from S1 onto S1 . By analycity, the restriction of g to ∂D defines uniquely g ∈ Aut(D). We shall say that γ ∈ H 1/2 (S1 , S2 ) is symmetric if its stabilizer is non trivial. Recall that a map u ∈ H 1 (D, S2 ) is harmonic if it satisfies (1) in the sense of distributions in D or equivalently, if it is a critical point of E in the following sense:  u + tϕ d , ∀ϕ ∈ Cc∞ (D, R3 ). (14) E dt |u + tϕ| |t=0 Thus, the group Aut(D) × O(3) acts on the set of solutions of (1) in H 1 (D, S2 ): for all (g, T) ∈ Aut(D) × O(3) and u ∈ H 1 (D, S2 ), u is harmonic ⇐⇒ T ◦ u ◦ g−1 is harmonic.

(15)

Hence the Dirichlet problem (1), (2) with boundary condition γ is equivalent to the Dirichlet problem with boundary condition T ◦γ ◦g−1 , for every (g, T) ∈ Aut(D)×O(3). Minimisation properties are also preserved. It is therefore tempting to try and classify the boundary conditions γ ∈ H 1/2 (S1 , S2 ) with respect to the Aut(D) × O(3) group action. This is a difficult question in general, but we shall see in Theorem 3 that every boundary value whose stabilizer has infinite cardinal is in the orbit of aeint for some a ∈ (0, 1], n ∈ N∗ . Another consequence of (15) is that if u is a solution to (1), (2) and if γ satisfies T ◦ γ ◦ g−1 = γ for some non-trivial (g, T) ∈ Aut(D) × O(3) then, because of non-uniqueness, T ◦ u ◦ g−1 may be a different solution to the same problem. In [13] we gave many examples of such situations (see below). In this paper, we seek more systematically the boundary values which present symmetries. Let n ∈ N∗ . The fundamental example of boundary value with continuous stabilizer is given by eit  → aeint (a ∈ C∗ ). We have (see [13] for details):



 S [eint ] = (eiα z, einα z) : α ∈ R ∪ (eiα z¯ , einα z¯ ) : α ∈ R



∪ (eiα z, einα z¯ −1 ) : α ∈ R ∪ (eiα z¯ , einα z−1 ) : α ∈ R , (16)

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or equivalently, {(Rα , Rnα ) : α ∈ R/2π Z}{(I, I), (I, SxOy ), (sOx , DOx ), (sOx , SxOz )}. Similarly, for a ∈ (0, +∞) \ {1},



 S [aeint ] = (eiα z, einα z), α ∈ R ∪ (eiα z¯ , einα z¯ ), α ∈ R , = {(Rα , Rnα ) : α ∈ R/2π Z}{(I, I), (sOx , SxOz )}. C∗ ,

(17)

Finally, if a ∈ is in the orbit of Notice also that for a > 0, the boundary values aeint , ae−int and 1/aeint belong to the same orbit. On the other hand, for a ∈ (0, 1], the images of ∂D by Π −1 (aeint ) are circles of S2 parallel to xOy with different radiuses, so the orbits of the boundary values eit  → aeint , a ∈ (0, 1] are disjoint. Let p ≥ 2 be an integer and n ∈ {0, . . . , p − 1}. A fundamental example of boundary value with discrete stabilizer is γp,n (eit ) = eint (eipt − 1). Since aeint

|a|eint .

ei2nπ/p γp,n (eit ) = γp,n (ei(t+2π/p) )

and

γp,n (eit ) = γp,n (e−it ),

S [γp,n ] contains (R2π/p , Rn2π/p ) and (sOx , SxOz ). Some additional computation (see

Proposition 4) will show that S [γp,n ] = (R2π/p , Rn2π/p ){(I, I), (sOx , SxOz )}.

(18)

We end this section by a first classification result. The following proposition shows that the orbits of aeit , a ∈ (0, 1] describe all the boundary conditions which are Moebius transformations. Proposition 1 Assume that the boundary condition γ : ∂D → S2 ⊂ R3 is (the restricaeit + b tion of) a Moebius transformation, i.e. Π ◦ γ (eit ) = it for a.e. t ∈ R/2π Z, with ce + d a, b, c, d ∈ C, ad − bc  = 0. There exist (g, T) ∈ Aut+ (D) × O+ (3) and α > 0 such that Π ◦ T ◦ γ ◦ g−1 (eit ) = αeit ,

for a.e. t ∈ R/2π Z.

In this proposition, we identify γ with the unique orientation-preserving Moebius ˆ. transformation which extends γ on C Proof First consider Π ◦ γ (∂D): it is the image by a Moebius transformation of a circle, so it is either a circle of C or a straight line [1], or equivalently, γ (∂D) is a circle of S2 . Let T be a rotation which maps the axis of γ (∂D) into the axis Oz, so that T ◦ γ (∂D) is a horizontal circle of S2 . Then, Π ◦ T ◦ γ (∂D) is a circle of radius R > 0 and center 0 in C. The map g1 = (1/R)Π ◦ T ◦ γ is an orientation-preserving Moebius transformation [recall (12)] which maps ∂D onto ∂D. By connexity, g1 (D) = D or ˆ \ D. In the first case, we set α = R and g = g1 , and in the second case, g1 (D) = C we set α = 1/R and g = (z  → 1/z) ◦ g1 . In both cases, g is an orientation-preserving Moebius transformation which maps D onto D, so g ∈ Aut+ (D). Summing up, we have proved that (1/α)Π ◦ T ◦ γ (z) = g(z), and the proof is complete. 4 Stabilizers in Aut+ (D) × O+ (3) The aim of this section is to determine the stabilizers G in Aut+ (D) × O+ (3) of every non constant boundary value γ ∈ H 1/2 (S1 , S2 ) (Theorem 3). We use two powerful results of complex analysis [1]. The proof is as follows:

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1. We recall in Theorem 1 the classification in conjugacy classes of elements in Aut+ (D): excepting I, Aut+ (D) contains only elliptic, parabolic and hyperbolic elements. 2. Using the invariance of the H 1/2 semi-norm (4) by conformal transformation of D (Lemma 1), we show in Lemma 2 that every element g ∈ p1 (G) is elliptic. A classical theorem of complex analysis (Theorem 2) asserts then that p1 (G) is conjugate to a rotation group. 3. We analyse the two easy cases where G contains an element of the form (I, T) (Proposition 2) or of the form (Rα , I) (Proposition 3); the first case, in particular, gives a strong relation between p1 (G) and p2 (G). This allows us to prove Theorem 3. The following theorem shows that tr2 parametrizes the conjugacy classes in Aut+ (D), with one exception. We recall that a point z ∈ S2 is a fixed point of g ∈ Aut+ (S2 ) if g(z) = z. Theorem 1 Let g ∈ Aut+ (D), g  = I. 1. If tr2 (g) ∈ [0, 4), then g is conjugate to a rotation z  → eiθ z (θ ∈ R/2π Z) in Aut+ (D); ˆ . Moreover, l ∈ ∂D and Φ −1 ◦g◦Φ0 2. if tr2 (g) = 4, then g has a unique fixed point l in C 0 + is conjugate to a translation z  → z ± 1 in Aut (H); ˆ . Moreover, l+ , l− ∈ 3. if tr2 (g) ∈ (4, +∞), then g has exactly two fixed points l+ , l− in C −1 ∂D and Φ0 ◦ g ◦ Φ0 is conjugate to a transformation z  → λz (0 < λ < 1) in Aut+ (H). This result is well-known in complex analysis [1]. In case (1), we say that g is elliptic, in case (2), we say that g is parabolic and in case (3), we say that g is hyperbolic. Notice that the two parabolic transformations z  → z + 1 and z  → z − 1 are not conjugated in Aut+ (H). For the reader’s convenience, we sketch an elementary proof. z + z0 ∈ Aut+ (D), g  = I, with θ ∈ R/2π Z, |z0 | < 1. If z0 = 0, 1 + z¯0 z g(z) = eiθ z is a rotation and tr2 (g) = 4 cos2 (θ/2) ∈ [0, 4). Now assume z0  = 0. Then ∞ is not a fixed point of g. For all l ∈ C, Proof Let g(z) = eiθ

g(l) = l ⇐⇒ z¯0 l2 + (1 − eiθ )l − eiθ z0 = 0.

(19)

Computing, ∆ = (1 − + = − Assume Then, by (11), |z0 |2 − sin2 (θ/2) < 0, so g has two distinct fixed points in C,    ieiθ/2 sin θ/2 ± sin2 θ/2 − |z0 |2 . l± = z¯0 eiθ )2

4eiθ

4eiθ [|z0 |2

sin2 (θ/2)].

tr2 (g)

∈ [0, 4).

Choosing θ ∈ [0, 2π], we have sin(θ/2) > |z0 | > 0, so l+ , l−  = 0 and l+ , l− have the ¯ . Moreover, |l+ | ≥ |l− | and since same argument. By (19), |l+ l− | = 1 so l− = 1/l+ l+  = l− , we have |l+ | > 1 > |l− |. Now, let h(z) =

z − l−

¯z 1 − l−

∈ Aut+ (D).

Computing, we find h ◦ g(z) = qh(z) with |q| = 1 so h ◦ g ◦ h−1 is a rotation, and this concludes the first case. The two other cases can be treated similarly. The following technical lemma shows the invariance of the H 1/2 (S1 ) semi-norm under conformal transformation of D. This is not too much a surprise in view of the invariance of E, the H 1 (D) semi-norm.

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Lemma 1 Let γ ∈ H 1/2 (S1 , R3 ). For every Φ : H → D which is a biholomorphic diffeomorphism from H onto D,

S1

S1

|γ (x ) − γ (y )|2   dx dy = |x − y |2



R R

|γ ◦ Φ(x) − γ ◦ Φ(y)|2 dxdy |x − y|2

(20)

where | · | is the Euclidian norm in R, R2 or R3 . By (9) and (10) Φ has a unique continuous extension from H ∪ {∞} onto D. This extension maps homeomorphically ∂ H ∪ {∞} onto ∂D and is also denoted Φ in this lemma. Proof Let γ ∈ H 1/2 (∂D, R3 ) and let Iγ denote the left hand side of (20). First assume 2 iz + 1 for all x ∈ R, the change of variable that Φ(z) = . Since |Φ  (x)| = z+i 1 + x2 x = Φ(x), y = Φ(y) in Iγ yields

Iγ = R R

|γ ◦ Φ(x) − γ ◦ Φ(y)|2 4 dxdy. |Φ(x) − Φ(y)|2 (1 + x2 )(1 + y2 )

It is now easy to check (see for instance [2]) that Φ −1 restricted to ∂D = S1 is the stereographic projection from the north pole from S1 onto R ∪ {∞} = ∂ H ∪ {∞} and that |Φ(x) − Φ(y)|2 =

4|x − y|2 , (1 + x2 )(1 + y2 )

∀x, y ∈ R.

iz + 1 . z+i Now, since Aut+ (H) = T(SL(2, R)), any biholomorphic transformation from H onto D can be written      iz + 1 az + b a b z → ◦ z → , ∈ SL(2, R). c d z+i cz + d This concludes the proof for the case Φ(z) =

We only need to check that the right hand side in (20) is invariant under transformaaz + b with a, b, c, d ∈ R, ad − bc = 1 be such a tions g ∈ T(SL(2, R)). Let g(z) = cz + d   transformation. Letting x = g(x ) y = g(y ) in the right hand side of (20), we obtain, since |g (x)| = (cx + d)−2 for all x ∈ R,

Iγ = R R

|γ ◦ Φ ◦ g(x ) − γ ◦ Φ ◦ g(y )|2 dx dy .   2  |g(x ) − g(y )| |cx + d|2 |cy + d|2

A straightforward computation yields g(x ) − g(y ) =

(cx

x − y , + d)(cy + d)

∀x , y ∈ R,

Symmetric boundary values from S1 to S2

so



Iγ = R R

155

|γ ◦ Φ ◦ g(x ) − γ ◦ Φ ◦ g(y )|2   dx dy , |x − y |2

and this concludes the proof. The following result is the key of the proof of Theorem 3. Lemma 2 Let γ ∈ H 1/2 (∂D, S2 ), g ∈ Aut+ (D) and T ∈ O(3) such that T ◦ γ = γ ◦ g a.e. in ∂D. If γ is not constant on ∂D, then g is elliptic or I. Proof First assume by contradiction that g is hyperbolic. By Theorem 1, there exists h ∈ Aut+ (H) such that h ◦ Φ0−1 ◦ g ◦ Φ0 ◦ h−1 (z) = λz = g1 (z), 0 < λ < 1, where Φ0 : H → D is defined by (9). The function Φ0 ◦ h−1 is a smooth diffeomorphism from R onto S1 \ {l}, where l = Φ0 ◦ h−1 (∞). Hence, the function γ1 = γ ◦ Φ0 ◦ h−1 satisfies T ◦ γ 1 = γ1 ◦ g 1

a.e. in R,

(21)

1/2

and γ1 ∈ Hloc (R). In particular, for every A > 0, we have

|γ1 (x) − γ1 (y)|2 dxdy < +∞. Iγ1 (A) = |x − y|2 [−A,A] [−A,A]

The change of variable Iγ1 (A) =

x

= g1 (x) = λx in the above integral yields

|γ1 ◦ g1−1 (x ) − γ1 ◦ g1−1 (y )|2   dx dy . |x − y |2

[−λA,λA] [−λA,λA]

Using (21) and T ∈ O(3), Iγ1 (A) = Iγ1 (λA). Iterating,

|γ1 (x) − γ1 (y)|2 dxdy, Iγ1 (A) = Iγ1 (λn A) = |x − y|2 [−λn A,λn A] [−λn A,λn A]

for every n ∈ N. Since 0 < λ < 1, the right hand side tends to 0 as n → +∞, so Iγ1 (A) = 0 for every A > 0. This proves that γ1 is constant on R, which contradicts the assumption on γ . If we assume that g is parabolic, we find similarly, using the invariance of the H 1/2 -semi-norm by translations z  → z ± 1, that γ is constant. The proof is complete. We shall use a theorem which is well known in the context of complex analysis [1,12]: Theorem 2 Let G be a subgroup of Aut+ (D). If G \ {I} contains only elliptic elements, then G is conjugate to a rotation group in Aut+ (D). For the reader’s convenience and also for later purpose, we shall make an elementary proof, following [12]. This theorem is an immediate consequence of the following lemma, together with Theorem 1. Lemma 3 Let R : z  → eiα z, α ∈ (0, 2π) be a rotation and let g ∈ Aut+ (D), g  = I be elliptic. If [R, g] = RgR−1 g−1 is elliptic or I, then g is a rotation.

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Proof By Theorem 1, tr2 ([R, g]) ∈ [0, 4). Letting g(z) = eiθ

z + z0 (θ ∈ [0, 2π), |z0 | < 1), z¯0 z + 1

we compute: −z + eiθ z0 , z¯0 z − eiθ e−iα z + z0 , RgR−1 (z) = ei(α+θ ) z¯0 e−iα z + 1 (|z0 |2 − e−iα )z + z0 eiθ (e−iα − 1) , [R, g] = ei(α+θ ) z¯0 (1 − e−iα )z + |z0 |2 ei(θ −α) − eiθ g−1 (z) =

and we find tr2 ([R, g]) = 4

(1 − |z0 |2 cos α)2 . (1 − |z0 |2 )2

The assumption implies tr2 ([R, ϕ]) ≤ 4. Since |z0 | < 1, this is equivalent to 1 − |z0 |2 cos α ≤ 1 − |z0 |2 ⇐⇒ |z0 |2 (1 − cos α) ≤ 0, Thus |z0 | = 0 or cos α = 1. This last case is impossible by assumption, so z0 = 0, and the proof is complete. The following proposition analyses the case where S [γ ] contains some (I, T). Proposition 2 Let γ be a non constant map in L1 (S1 , R3 ) and T ∈ O(3) such that T ◦ γ = γ a.e. in S1 . Then T = I or T is a symmetry with respect to a (vectorial) plane P and γ (x) ∈ P for a.e. x ∈ S1 . In particular, ∀T1 , T2 ∈ O+ (3),

T1 ◦ γ = T2 ◦ γ

a.e. in S1 ⇒ T1 = T2 .

Proof If γ is continuous, the conclusion is obvious (use T(γ (x)) = γ (x) for all x ∈ S1 ). In the general case, use a mollifier. The following proposition studies the case where S [γ ] contains some (Rα , I). Proposition 3 Let γ be a non constant map in L1 (S1 , R3 ) and α ∈ R/2π Z such that γ ◦ Rα = γ a.e. in S1 . Then Rα is of finite order, or equivalently, α ∈ 2π Q. Proof Let γ = (γ 1 , γ 2 , γ 3 ). For j = 1, 2, 3, we have γ j ◦ Rα = γ j a.e. in S1 . The proof is immediate by identifying the Fourier components of t  → γ j (eit ) and t  → γ j (ei(α+t) ). We are now ready to prove our main result. Theorem 3 Let γ ∈ H 1/2 (S1 , S2 ) be non constant. Then the stabilizer G of γ in Aut+ (D) × O+ (3) is conjugate in Aut+ (D) × O+ (3) to a rotation group of the form (R2π/p , Rn2π/p ) for a unique p ∈ N∗ and a unique n ∈ (Z/pZ)/{−1, +1} (finite case) or of the form {(Rα , Rnα ) : α ∈ R} for a unique n ∈ N∗ (infinite case). In the latter case, there exist (g, T) ∈ Aut+ (D) × O+ (3) and a ∈ C∗ such that Π ◦ T ◦ γ ◦ g(eit ) = aeint for a.e. t ∈ R/2π Z. The same result holds if γ ∈ C 0 (S1 , S2 ). The proof is similar. By the remark in Sect. 3, we may also choose a ∈ (0, 1] in this theorem.

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Proof Let G1 = p1 (G) be the projection of G onto the first component, i.e. G1 = {g ∈ Aut+ (D) : ∃T ∈ O+ (3), T ◦ γ = γ ◦ g a.e. in ∂D}. By Lemma 2, every element in G1 is elliptic or I. By Theorem 2, G1 is conjugate in Aut+ (D) to a rotation group, and we may assume without loss of generality that G1 ⊂ {Rα : α ∈ R/2π Z}. First case: G1 is finite. Then G1 = R2π/p  for some p ∈ N∗ (we may have p = 1); p, being the order of G1 , is unique. By Proposition 2, there is a unique T ∈ O+ (3) such p that T ◦ γ = γ ◦ R2π/p . We have T p ◦ γ = γ ◦ R2π/p = γ , and by Proposition 2 again, p T = I, so T is a rotation of angle 2nπ/p for some n ∈ Z/pZ, and we may assume without loss of generality that T = R2nπ/p . We have (R2π/p , T) ⊂ G. Conversely, for every (g, T  ) ∈ G, g = Rk2π/p for some k ∈ Z/pZ, and T  ◦ γ = γ ◦ Rk2π/p = T k ◦ γ , so T  = T k . Thus G = (R2π/p , R2nπ/p ). The integer n is defined by the angle of T, which is unique up to the sign (for every θ , the rotations Rθ and R−θ are conjugate in O+ (3), since R−θ = DOx Rθ DOx ); T itself is uniquely defined by R2π/p , whose choice is unique up to the sign of the angle; so n is unique in the set (Z/pZ)/{−1, +1} , and this concludes the finite case.

Second case: G1 is infinite. We know that G is a subgroup of R/2π Z × O+ (3). The map + 2 1 2 F : (α, T)  → T ◦ γ ◦ R−1 α from R/2π Z × O (3) into L (S , S ) is clearly continuous, so G = F −1 ({γ }) is closed in R/2π Z × O+ (3), hence compact and G1 = p1 (G) also. Thus G1 = {Rα : α ∈ R/2π Z} R/2π Z. Choose α0 ∈ (0, 2π) \ 2π Q and let T0 ∈ O+ (3) be the unique rotation such that T0 ◦ γ = γ ◦ Rα0 . Then T0 is conjugate in O+ (3) to some Rθ0 (θ0 ∈ R/2π Z), and we may assume without loss of generality that T0 = Rθ0 . Now let (Rα , T) ∈ G, α ∈ R/2π Z. There exists a sequence of integers (mk ) such that m m m m Rα0k → Rα and Rθ0k → Rθ in R/2π Z. For every mk , Rθ0k ◦ γ = γ ◦ Rα0k a.e. in S1 . Passing to the limit, Rθ ◦ γ = γ ◦ Rα = T ◦ γ a.e. in S1 . Thus T = Rθ for a (unique) θ ∈ R/2π Z. Now consider the map f : α → θ from R/2π Z into R/2π Z where θ is defined above (for a given Rα , Rθ is the unique rotation in O+ (3) such that Rθ ◦ γ = γ ◦ Rα ). This map is a continuous group homomorphism. Equivalently, there exists n ∈ Z such that f (α) = nα for all α ∈ R/2π Z (n is the degree of f ). Since Rθ is conjugate to R−θ in O+ (3), we may assume n ≥ 0. This proves that G is conjugate to (Rα , Rnα ) : α ∈ R/2π Z}. In other words, for every α ∈ R/2π Z, Rnα ◦ γ (x) = γ ◦ Rα (x) for a.e. x ∈ S1 . We fix again α = α0 ∈ (0, 2π) \ 2π Q in the latter equation. By Proposition 3, n  = 0. Now we want to see that Π ◦ γ (eit ) = aeint for some a ∈ C∗ . The proof is similar to that of Proposition 3: using the decomposition γ = (γ 1 , γ 2 , γ 3 ) and the Fourier decomposition of γ3 and γh = γ1 + iγ2 , we obtain γ3 = constant  = ±1, and γh (eit ) = cn eint , cn int cn ∈ C∗ . Thus Π ◦ γ (eit ) = e a.e. in S1 , and the proof is complete. 1 − γ3 5 Examples of stabilizers in Aut(D) × O(3) With a little more work, it is possible to obtain the complete list of stabilizers in Aut(D) × O(3). The reader will find it at the end of this paper in Appendix A. In this section, we shall only give the basic idea and some examples.

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We introduce a new notation. For an element (g, T) ∈ Aut(D) × O(3), σ1 (g, T) ∈ {−1, +1} is the sign of g (i.e. σ1 = +1 is g if orientation-preserving and σ1 = −1 otherwise), σ2 (g, T) ∈ {−1, +1} is the determinant of T, and σ (g, T) = (σ1 (g, T), σ2 (g, T)) ∈ {−1, +1}2 . The maps σ1 , σ2 , σ1 σ2 and σ are group homomorphisms; σ|G denotes the restriction of σ to a subgroup G of Aut(D) × O(3). Now let G be the stabilizer in Aut(D) × O(3) of a non constant boundary value γ . The fundamental idea for the computation of G is that Ker σ|G is the stabilizer of γ in Aut+ (D) × O+ (3). Since the homomorphism σ : G → {−1, +1}2 induces a bijection from G/Ker σ|G onto σ (G), the index [G : Ker σ|G ] is either 1, 2 or 4. In particular, G is infinite if and only if Ker σ|G is infinite. This remark, together with Theorem 3 and (16) and (17), gives us all the stabilizers in Aut(D) × O(3) whose cardinal is infinite. As an application, we compute the fundamental example of discrete symmetries given in Sect. 3. Proposition 4 Let p ≥ 2, n ∈ {0, . . . , p − 1} and γp,n (eit ) = eint (eipt − 1). Then the stabilizer S [γn,p ] of γp,n is given by (18). Proof Clearly (see Sect. 3), (R2π/p , Rn2π/p ){(I, I), (sOx , SxOz )} ⊂ S [γn,p ] =: G. In particular, p1 (G) contains R2π/p . By Lemmas 2 and 3,   p1 (G) ∩ Aut+ (D) ⊂ O+ (2).   Now, let (R, T) ∈ G ∩ O+ (2) × O+ (3) . Writing (recall (12)) R(z) = eiα z and T(z) =

αz + β , |α|2 + |β|2 = 1, ¯ + α¯ −βz

we see that T ◦ γ (eit ) = γ ◦ Rα (eit ) for all t if and only if (by computation) (R, T) ∈ (R2π/p , Rn2π/p ). In other words, Ker σ|G = (R2π/p , Rn2π/p ). Similarly, by computa  tion, we obtain G ∩ O+ (2) × O− (3) = ∅. Thus, [G : Ker σ|G ] = 2 and the proposition is proved. In Proposition 5 which follows, we propose a general method to build a boundary value with given (admissible) stabilizer G. The computations concerning p1 (G) are based on Lemmas 2 and 3 and on the following two lemmas. The computations concerning p2 (G) are based on the classification of finite groups in O(3) and are postponed to Appendix A. Lemma 4 Let g ∈ Aut− (D). If g2 ∈ O+ (2), then g2 = I.If, furthermore, (Rα g)2 ∈ O+ (2) for some α ∈ (0, 2π), then g ∈ O− (2). z¯ + z0 Proof By (7) and (10), g(z) = eiθ for some θ ∈ R/2π Z and |z0 | < 1. Comput1 + z¯0 z¯ ing, g2 (z) =

(1 + eiθ z20 )z + z¯0 + eiθ z0 1 + z¯0 2 e−iθ + (z0 + z¯0 e−iθ )z

.

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Thus, if g2 (z) = eiα z for some α  ∈ R/2π Z, then z¯0 + z0 eiθ = 0. This identity and z¯ + z0 the above formula yield g2 (z) = z. If, furthermore, Rα g(z) = ei(α+θ ) satisfies 1 + z¯0 z¯ (Rα g)2 ∈ O+ (2) then, by the previous computation, z¯0 + z0 ei(α+θ ) = 0. This, together with z¯0 + z0 eiθ = 0 and α ∈ (0, 2π), implies z0 = 0, so g(z) = eiθ z¯ and the proof is complete. Lemma 5 Let z1 , z2 and z3 be three distinct points in ∂D. If g ∈ Aut(D) fixes z1 , z2 and z3 , then g = I. Proof The statement is clear, knowing that the unique map in Aut+ (S2 ) which fixes z1 , z2 and z3 is I [1], and noticing that z  → z¯ −1 , which belongs to Aut− (S2 ) \ Aut− (D), also fixes z1 , z2 and z3 . As a consequence, we have: Proposition 5 For every p ∈ N∗ and every n ∈ {0, . . . , p − 1}, there exists a map γ ∈ C 0 (S1 , S2 ) ∩ H 1/2 (S1 , S2 ) such that its stabilizer G in Aut(D) × O(3) is equal to (R2π/p , R2nπ/p ). Proof Assume p ≥ 2. We first choose       γ˜ ∈ C 0 [0, 2π/p], S2 ∩ C 1 (0, 2π/p), S2 ∩ H 1/2 (0, 2π/p), S2 √

i t such that γ˜ (2π/p) = R2nπ/p γ˜ (0) and limt→0+ |∇ γ˜ (t)| = +∞ (for instance γ˜ (t)  =e near t = 0). We also build γ˜ such that, for some t0 ∈ (0, 2π/p) \ {π/p}, γ˜ ∈ C 2 (0, t0 ) ∪  (t0 , 2π/p), S2 but γ˜ is not C 2 near t0 . We also assume that γ˜ does not take all its values in a plane. Then, we extend γ˜ on [0, 2π] by defining    2kπ 2(k + 1)π 2kπ k for t ∈ , , k = 1, . . . , p − 1. γ˜ (t) = R2nπ/p ◦ γ˜ t − p p p

Finally, we set γ (eit ) = γ˜ (t). The map γ has the required stabilizer G. Indeed, it is clear by construction that γ ∈ C 0 (S1 , S2 ) ∩ H 1/2 (S1 , S2 ) and that G contains (R2π/p , R2nπ/p ). By Lemmas 2 and 3, every element g ∈ Aut+ (D) ∩ p1 (G) belongs to O+ (2), and then, by Lemma 4, every element g ∈ Aut− (D) ∩ p1 (G) belongs to O− (2) (recall p ≥ 2). In other words, p1 (G) ⊂ O(2). Notice now that for every (g, T) ∈ Aut(D) × O(3) and for every t ∈ R/2π Z, the map T ◦ γ ◦ g−1 has, near the point γ (eit ), the same regularity as γ near the point eit . Thus, every g ∈ p1 (G) leaves both set of points {1, R2π/p (1), . . . , R2(p−1)π/p (1)}

and

{eit0 , R2π/p eit0 , . . . , R2(p−1)π/p (eit0 )}

globally invariant. The only elements in p1 (G) ⊂ O(2) that satisfy such a requirement are the rotations in R2π/p . Hence, p1 (G) = R2π/p  and, by Theorem 3, Ker σ|G = (R2π/p , R2nπ/p ) and [G : (R2π/p , R2nπ/p )] = 1 or 2. This index cannot be 2, by application of Lemma 9, otherwise γ would take its values in a plane. Thus, G = (R2π/p , R2nπ/p ) and the proof is complete when p ≥ 2. When p = 1, we use a similar construction, together with Lemma 5 (see Proposition 6 below for details).

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6 Two applications to the Dirichlet problem The following proposition shows that there exist multiple solutions to the Dirichlet problem in a same homotopy class which are not due to symmetry. Proposition 6 There exists a non symmetric γ ∈ C 2 (S1 , S2 ) such that the corresponding Dirichlet problem (1), (2) has at least two distinct solutions which belong to a same homotopy class. “Non symmetric” means that the stabilizer of γ in Aut(D) × O(3) is {(I, I)}. Proof Let α ∈ (0, 1). By a result of Qing [14], for any perturbation γ ∈ C 1,α (S1 , S2 ) of λ¯z2n (n ∈ N∗ , 0 < λ < 1) which is small enough in C 1,α (S1 )-norm, there exists in the homotopy class E1 (with the notations of Qing) more than one solution to the Dirichlet problem. Choose three distinct points z1 , z2 , z3 in ∂D. It is clearly possible to build a map γ ∈ C 2 (S1 , S2 ) ∩ C ∞ (S1 \ {z1 , z2 , z3 }, S2 ) which has regularity C j+1 but not C j+2 at point zj , j = 1, 2, 3, which does not take all its values in a plane, and which is a small enough perturbation of λ¯z2n in C 1,α (S1 )-norm. Let G = S [γ ]. Every g ∈ p1 (G) fixes the three points z1 , z2 , z3 and therefore g = I, by Lemma 5. By Proposition 2, G = {(I, I)}. The following result generalizes a remark in [13]. Theorem 4 Let a > 0, n ∈ N∗ . If u ∈ H 1 (D, S2 ) is a solution of the Dirichlet problem (1), (2) with γ (eit ) = Π −1 (aeint ), Π ◦ u different from z  → azn and z  → a¯z−n , then its stabilizer in Aut(D) × O(3) is finite. In particular, there exists p ∈ N∗ such that the maps uα = Rnα ◦ u ◦ R−α ,

α ∈ [0, 2π/p)

define a continuum of distinct solutions to (1), (2) with the same boundary value γ (eit ) = Π −1 (aeint ). For n = 1, a > 0, it is still an open problem (see [4]) to find a solution to (1), (2) with γ (eit ) = Π −1 (aeint ), different from z  → azn and z  → a¯z−n (for n ≥ 2, the result in [5] provides one). Proof Let u ∈ H 1 (D, S2 ) be a solution of (1), (2) with γ (eit ) = Π −1 (aeint ). By the regularity result of Hélein [8], u is smooth up to the boundary. Its stabilizer G in Aut(D) × O(3) is obviously a subgroup of S [γ ]. As in the proof of Theorem 6, we have [G : Ker σ|G ] = 1, 2 or 4. It is therefore sufficient to prove that Ker σ|G is finite. By (16), (17), Ker σ|G is a subgroup of {(Rθ , Rnθ ), θ ∈ R/2π Z}. Since Ker σ|G is closed, it is sufficient to prove that Ker σ|G is different from this group. Now, assume by contradiction that Ker σ|G = {(Rθ , Rnθ ), θ ∈ R/2π Z} ⇐⇒ Rnθ ◦ u = u ◦ Rθ ,

∀θ ∈ R/2π Z.

(22)

The map u = (u1 , u2 , u3 ) is completely defined by its values on the segment {(r, 0) : 0 < r ≤ 1}. The PDE (1) can be replaced by a system of differential equations. In the

Symmetric boundary values from S1 to S2

161

following lemma, we solve this system explicitly (more precisely, we solve ∆u∧u = 0), and using the boundary condition, we show that the only solutions are azn and a¯z−n . This will conclude the proof. Lemma 6 Let a > 0, n ∈ N∗ . If u ∈ C ∞ (D, S2 ) is a solution of the Dirichlet problem (1), (2) with γ (eit ) = Π −1 (aeint ), and u satisfies (22), then Π ◦ u(z) = azn or a¯z−n . Proof Let (r, θ ) be the polar coordinates on D and λ, ϕ ∈ [0, π] the spherical coordinates in S2 (λ is the longitude with of u respect to xOz and ϕ is the angle of u with the Oz axis). The space R3 is oriented by {O, e1 , e2 , e3 }. We write u(r, 0) = (sin ϕ(r) cos λ(r), sin ϕ(r) sin λ(r), cos ϕ(r)), = sin ϕ(r)e(λ(r)) + cos ϕ(r)e3 ,

(23)

with e(λ) = cos λe1 + sin λe2 . More precisely, ϕ(r) = arccos(u3 (r, 0)), so that ϕ ∈

C 0 ([0, 1], [0, π]). By assumption, u(1, 0) = Π −1 (a)  = (0, 0, ±1), so ϕ(1)  ∈ {0, π}. Define

 r0 = sup r ∈ [0, 1) : sin ϕ(r) = 0

(24)

(notice that by (22), u(0, 0) is the north pole or the south pole, so the set on the right hand side contains 0). Then r0 ∈ [0, 1) and ϕ is smooth on (r0 , 1]. The S1 -valued map r  → (u1 (r, 0), u2 (r, 0))/ sin ϕ(r) is well-defined, smooth on (r0 , 1] and has value 1 at r = 1: it has a unique smooth lifting λ : (r0 , 1] → R such that λ(1) = 0, and this is our choice of λ. The computations that follow are valid for r ∈ (r0 , 1]. By (1), ∆u ∧ u = 0. By assumption (22), u(r, θ ) = Rnθ ◦ u(r, 0)∀r, θ . Computing with (23), and using the expression of ∆u in polar coordinates, we find ∆u ∧ u in the basis {Rnθ e(λ), Rnθ e(λ + π/2), e3 }, and we obtain the system 2ϕ  cos2 ϕλ + cos ϕ sin ϕλ + (1/r) sin ϕ cos ϕλ = 0, 

2



2

2

(25)

ϕ − sin ϕ cos ϕλ + (1/r)ϕ − (n /r ) sin ϕ cos ϕ = 0,

(26)

−2ϕ  cos ϕ sin ϕλ − sin2 ϕλ − (1/r) sin2 ϕλ = 0.

(27)

Since sin ϕ  = 0, (27) is equivalent to 2ϕ  cos ϕλ + sin ϕλ + (1/r) sin ϕλ = 0

(28)

[notice that (25) is also proportional to (28)]. Thus, λ = A/(r sin2 ϕ) with A ∈ R constant. Using this in (26), we find ϕ  − A2

sin ϕ cos ϕ + (1/r)ϕ  − (n2 /r2 ) sin ϕ cos ϕ = 0. r2 sin4 ϕ

Multiplying by 2r2 ϕ  and integrating, we obtain r2 ϕ 2 = −

A2 + n2 sin2 ϕ + B. sin2 ϕ

If A  = 0, then the right hand side tends to −∞ as r tends to r0 , r > r0 . This is not possible, because the left hand side is ≥ 0. So A = 0, λ is constant, and r2 ϕ 2 = n2 sin2 ϕ + B

on (r0 , 1].

(29)

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Taking the limit in (29) as r → r0 , r > r0 , we have B ≥ 0. By continuity, ϕ  has a constant sign on (r0 , 1], and   n2 sin2 ϕ + B n2 sin2 ϕ + B   ϕ = (30) or ϕ = − on (r0 , 1]. r r Assume by contradiction that r0 > 0. Then, on the disc D(r0 ) = {z ∈ C : |z| < r0 }, u satisfies (1) with the constant boundary condition u = (0, 0, cos ϕ(r0 )) on ∂D(r0 ). By Lemaire’s theorem [11], u is constant on D(r0 ). In particular, √ ∂r u(r0 , 0) = 0, so by (23) ϕ  (r0 ) = 0. This, together with (30), yields ϕ  (r0 ) = 0 = B/r0 , i.e. B = 0. But for B = 0, (30) becomes ϕ =

n sin ϕ r

or

ϕ = −

n sin ϕ on (r0 , 1], r

(31)

whose solutions are ϕ = 2 arctan(A+ rn ) and ϕ = 2 arctan(A− r−n ), respectively; the constants A+ and A− in R are uniquely defined by the initial condition ϕ(1) (in fact, A+ = A− = a). This contradicts r0 > 0 [recall that r0 is defined by (24)]. Notice that these solutions are well known: they correspond to the maps z  → a¯z−n and z  → azn , respectively. Now we know that r0 = 0 in (29). But E(u) < ∞ implies B = 0. By the same computation as previously, Π ◦ u(z) = azn or a¯z−n . This concludes the proof. A Stabilizers in Aut(D) × O(3) The aim of this appendix is to classify the finite stabilizers in Aut(D) × O(3). We use the notations σ , σ1 and σ2 introduced in Sect. 5. The proof is divided as follows: 1. 2.

3.

We prove that every stabilizer in Aut(D) × O(3) is conjugate to a subgroup of O(2) × O(3) (Theorem 5). In Lemma 9, we look for the stabilizers G such that [G : Ker σ|G ] = 2. The proof uses the fact that all finite subgroups of O(3) are known [18]; only a few among them are possible for p2 (G) (Lemma 8). Using [G : Ker σ|G ] = 1, 2 or 4, Lemma 9 and Proposition 2, we obtain a list of finite groups. Some of them cannot be stabilizers, by an additional property, Lemma 10. We start with the following

Theorem 5 Let γ ∈ H 1/2 (S1 , S2 ) be non constant. Then the stabilizer G of γ in Aut(D) × O(3) is conjugate [in Aut(D) × O(3)] to a subgroup of O(2) × O(3). Proof First, by Lemma 2 and Theorem 2 again, p1 (G) ∩ Aut+ (D) is conjugate to a rotation group, so we may assume without loss of generality that p1 (G) ∩ Aut+ (D) ⊂ O+ (2). If p1 (G) ∩ Aut+ (D)  = {I}, there exists α ∈ (0, 2π) such that Rα ∈ p1 (G), and by application of Lemma 4, for every g ∈ p1 (G) ∩ Aut− (D) we have g ∈ O− (2). If p1 (G) ∩ Aut+ (D) = {I} then p1 (G) ∩ Aut− (D) contains at most one element which is an involution, and thus conjugate to an orthogonal reflection in a line, by Lemma 7. This concludes the proof. For the proof of the following lemma, see for instance [1].

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Lemma 7 If g ∈ Aut− (D) is an involution, i.e. g2 = I, then g is conjugate in Aut(D) to an element of O− (2). In fact, the involutions in Aut− (D) are either reflection in lines or reflection in circles orthogonal to ∂D [1]. If G is a finite stabilizer, then G2 = p2 (G) is a finite subgroup of O(3) with a special structure, by Theorem 3. From the well known classification of the finite subgroups of O(3) (see for instance [18]), we deduce the following useful Lemma 8 Let G2 be a finite subgroup of O(3) which contains R2π/q  (q ∈ N∗ ) as a finite subgroup of index 2. Then G2 is conjugate in O(3) to one of the following groups: Rπ/q ,

R2π/q DOx ,

R2π/q −I,

R2π/q SxOz ,

−IRπ/q .

More precisely, if q ≥ 2, there exists α ∈ R/2π Z such that Rα G2 R−α is equal to one of the above groups. The following result gives a partial list of finite stabilizers. Lemma 9 Let p ∈ N∗ , n ∈ N ∩ [0, p/2], and G be a finite subgroup of O(2) × O(3) which satisfies the following conditions: (i) (ii) (iii) (iv)

Ker σ|G = (R2π/p , R2nπ/p ); Ker σ|G has index 2 in G; For all T ∈ O+ (3), (I, T) ∈ G ⇒ T = I; If there exists T ∈ O− (3) such that (I, T) ∈ G, then T is a reflection in a vectorial plane P and R2nπ/p (P ) = P .

Then G is conjugated in O(2) × O(3) to one of the following (different) groups: 1.

(Rπ/p , SxOy Rnπ/p ), (Rπ/p , −IRnπ/p ), (R2π/p , R2nπ/p )(I, SxOy ), or

2.

(R2π/p , R2nπ/p )(I, SxOz )

(R2π/p , R2nπ/p )(sOx , DOx ), or (R2π/p , R2nπ/p )(sOx , I)

3.

if n = p/2; if n = 0, p/2;

(R2π/p , R2nπ/p )(sOx , SxOz ), or

(R2π/p , R2nπ/p )(sOx , −I)

if n = 0, p/2.

¯ T) ¯ ∈ G \ (R2π/p , R2nπ/p ). By (i), (ii) we have Proof Let (R, ¯ T). ¯ G = (R2π/p , R2nπ/p ) ∪ (R2π/p , R2nπ/p )(R,

(32)

The map σ|G : G → {−1, +1}2 induces a bijection from G/(Ker σ|G ) onto σ|G (G). Thus σ|G (G) has cardinal 2, and there are three possibilities: σ|G (G) \ {(1, 1)} = {(+1, −1)}, {(−1, +1)} or {(−1, −1)}. First case: σ|G (G) \ {(1, 1)} = {(+1, −1)}. If we assume p1 (G)  = R2π/p , then p1 (G) is a subgroup of O+ (2) of order 2p, i.e. p1 (G) = Rπ/p . Let T ∈ O− (3) such that (Rπ/p , T) ∈ G. By condition (iii), such a T is unique and (Rπ/p , T)2 = (R2π/p , T 2 ) = (R2π/p , R2nπ/p ), i.e. T 2 = R2nπ/p . If n  = 0, T = −IRnπ/p or T = SxOy Rnπ/p . If ¯ T) ¯ = (Rπ/p , T) in (32), we obtain n = 0, T is conjugate to −I or SxOy . Choosing (R, the first two groups of the list. Otherwise, p1 (G) = R2π/p , and there exists T ∈

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¯ T) ¯ ∈ G \ (R2π/p , R2nπ/p ) and consider O− (3) such that (I, T) ∈ G (choose any (R, −1 −1 n ¯ ¯ ¯ ¯ (R, T)(R , (R ) ) ∈ G). By condition (iv), T is a reflection in a plane; applying Lemma 8 to G2 = p2 (G) = R2nπ/p  ∪ R2nπ/p T, we may assume without loss of generality that T = SxOy or T = SxOz . If T = SxOz , condition (iv) implies 2nπ/p = 0 or π [2π], i.e. n = 0 or p/2. We find the last two groups of the first case — notice that the case n = 0, T = SxOz is conjugate to the case n = 0, T = SxOy , so we keep only the latter in our list of different groups. In the second case σ|G (G) \ {(1, 1)} = {(−1, +1)}, we find similarly the groups listed in the second case of the lemma. The third case σ|G (G) \ {(1, 1)} = {(−1, −1)} gives us the third case of the lemma. In fact, the last group in the above list cannot be the stabilizer of a boundary condition γ ∈ H 1/2 (S1 , S2 ), as shown by the following Lemma 10 Let γ ∈ H 1/2 (S1 , S2 ), q ∈ N∗ and T, T  ∈ O(3). If T ◦ γ = γ ◦ R2π/q

and

T  ◦ γ = γ ◦ sOx a.e. in S1 ,

(33)

there exists v ∈ S2 such that Tv = T  v. In particular, choosing q = 1, T = I and T  = −I, we find that (sOx , −I) cannot belong to the stabilizer of any map γ ∈ H 1/2 (S1 , S2 ). Proof If γ is continuous, the conclusion is obvious by letting v = γ (e−iπ/q ). In the general case, following Brezis [4] and Brezis and Nirenberg [6], we let γ˜ (t) = γ (eit ) and define for ε > 0 and for t ∈ R,
1 γ˜ε (t) = γ˜ (y)dy. 2ε (t−ε,t+ε)

The map γ˜ε is 2π-periodic and continuous. By (33), T ◦ γ˜ε (t) = γ˜ε (t + 2π/q)

and

T  ◦ γ˜ε (t) = γ˜ε (−t),

∀t ∈ R.

Since γ belongs to VMO(S1 , R3 ), the space of functions with “vanishing mean oscillations”, |γ˜ε (t)| → 1, uniformly in t ∈ [−π, π] (see [4] for details). Hence, for ε > 0 γ˜ε (−π/q) small enough, γ˜ε (−π/q)  = 0, and we can choose v = . |γ˜ε (−π/q)| We can now state: Theorem 6 Let γ ∈ H 1/2 (S1 , S2 ) ∪ C 0 (S1 , S2 ). If the stabilizer G of γ in Aut(D) × O(3) is finite, there exist a unique p ∈ N∗ and a unique n ∈ (Z/pZ)/{−1, +1} such that G is conjugate in Aut(D) × O(3) to one of the following different groups: (R2π/p , R2nπ/p ); one of the groups listed in Lemma 9, excepting the last one (i.e. the one which contains (sOx , −I)); (Rπ/p , SxOy Rnπ/p )(sOx , DOx ); (Rπ/p , SxOy )(sOx , I) with n = 0 [p]; (Rπ/p , −IRnπ/p )(sOx , DOx ); (R2π/p , R2nπ/p ){(I, I), (I, SxOy ), (sOx , DOx ), (sOx , SxOz )};

Symmetric boundary values from S1 to S2

165

(R2π/p , R2nπ/p ){(I, I), (I, SxOy ), (sOx , I), (sOx , SxOy )} with n = 0, p/2 [p]; (R2π/p , Rπ ){(I, I), (I, SxOz ), (sOx , DOx ), (sOx , SxOy )} with n = p/2 [p]; (R2π/p , Rπ ){(I, I), (I, SxOz ), (sOx , DOy ), (sOx , SyOz )} with n = p/2 [p]; Conversely, for every p ∈ N∗ , n ∈ Z/pZ, and for every group G of this list, there exists γ ∈ H 1/2 (S1 , S2 ) ∩ C 0 (S1 , S2 ) whose stabilizer in Aut(D) × O(3) is equal to G. Proof Since G is finite, γ is non constant. By the proofs of Theorems 3 and 5, G is (conjugate to) a finite subgroup of O(2)×O(3) such that Ker σ|G = R2π/p , R2nπ/p  for a unique p ∈ N∗ and a unique n ∈ (Z/pZ)/{−1, +1}. The homomorphism σ|G : G → {−1, +1}2 induces an isomorphism from G/(Ker σ|G ) onto σ|G (G), so [G : Ker σ|G ] = 1, 2, or 4. If [G : Ker σ|G ] = 1, we have G = (R2π/p , R2nπ/p ). If [G : Ker σ|G ] = 2, G satisfies the four conditions of Lemma 9 (conditions (iii) and (iv) are both a consequence of Proposition 2), so G is conjugate to one of the groups listed in Lemma 9; the one which contains (sOx , −I) has to be excluded by Lemma 10. If [G : Ker σ|G ] = 4, then Ker σ1|G satisfies the conditions of Lemma 9 and is conjugate to one of the four groups in the first case of Lemma 9. By adapting the proof of Lemma 9 to this case, and using Lemma 10, we find all the finite groups listed in the theorem. Conversely, for every group G in the list, it is possible to build by the method given in Proposition 5 a boundary value γ ∈ C 0 (S1 , S2 ) ∩ H 1/2 (S1 , S2 ) whose stabilizer is G. The details are left to the reader. Acknowledgements I would like to thank F. Bosio for the reference [1] and B. Merlet for his help on the proof of Lemma 2.

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