Martin Traizet

On the genus of triply periodic minimal surfaces

we prove the existence of embedded minimal surfaces of arbitrary genus g ≥ 3 in any flat 3-torus. In fact we construct a sequence of such surfaces converging to a planar foliation of the 3-torus. In particular, the area of the surface can be chosen arbitrarily large.

Introduction

Triply periodic minimal surfaces in euclidean space are invariant by three independant translations. In the nineteenth century, five embedded triply periodic minimal surfaces were known to H.A. Schwarz and his school. In 1970, Alan Schoen described 12 further families of examples. The arguments were completed by H. Karcher who also proved the existence of many further examples.

If M is a triply periodic minimal surface and Λ is the lattice generated by its three periods, then M projects to a minimal surface in the flat 3-torus R 3 /Λ. Conversely, a (non-flat) minimal surface in R 3 /Λ lifts to a triply periodic minimal surface in R 3 .

A natural question is whether there exists non-flat minimal surfaces in any flat 3-torus. The examples constructed by H.A. Schwarz, A. Schoen and H. Karcher are very symmetric by construction, so they only construct examples in very particular 3-tori. In 1990, W. Meeks made the following conjecture ( [START_REF] Meeks | The theory of triply-periodic minimal surfaces[END_REF], conjecture 3.2) Conjecture 1 (Meeks) For any flat 3-torus R 3 /Λ and any integer g ≥ 3 there exist an embedded, orientable minimal surface of genus g in R 3 /Λ.

A related question asked by H. Karcher is the following ( [START_REF] Karcher | Construction of higher genus embedded minimal surfaces[END_REF], question 4)

Can triply periodic embedded minimal surfaces of arbitrary large genus exist in a nontrivial way -or does the genus stay bounded if one divides out by all translational symmetries ?

In this paper we prove that the conjecture is true and answer this question :

Theorem 1 For any flat 3-torus R 3 /Λ, and for any integer g ≥ 3, there exists a sequence of orientable, compact, embedded minimal surfaces (M n ) n∈N in R 3 /Λ which have genus g. The area of M n goes to infinity when n → ∞. Moreover, Figure 1: A minimal surface of genus 4 in the cubic 3-torus. One of the two planes is represented with dots for clarity.

if g ≥ 4, the only translation of R 3 /Λ which leaves M n invariant is the identity (so M n lifts to a triply periodic minimal surface in R 3 , whose lattice of periods is precisely Λ).

Here is what is known about this conjecture. It is known that a minimal surface of genus one in a flat 3-torus must be flat (a plane) and minimal surfaces of genus two in flat 3-tori do not exist ( [START_REF] Meeks | The theory of triply-periodic minimal surfaces[END_REF], corollary 3.1). W. Meeks has proved ( [START_REF] Meeks | The theory of triply-periodic minimal surfaces[END_REF], corollary 10.1) that the conjecture is true in the genus 3 case by using a min-max argument. Then by taking covers he concluded that the conjecture holds for arbitrary odd genus g ≥ 3. In other words, if Λ is a sub-lattice of Λ, a genus 3 minimal surface in R 3 /Λ lifts to an odd genus minimal surface in R 3 /Λ . The case of even genus remained open. Also, these examples do not answer the question of H. Karcher. Let us now explain the idea of the construction, which was suggested to the author by Antonio Ros. Consider a plane in space and its projection in the flat 3-torus R 3 /Λ. Depending on the plane, the area of the projection may be finite, or infinite. Choose the plane so that the area is very large. In the quotient, the plane will wrap around many time, and what we see locally is many parallel sheets very close to each other. Take another plane parallel to the first one. Our main result, Theorem 4, allows us to open small catenoidal necks between the two planes (on both sides), producing an embedded minimal surface in R 3 /Λ (see figure 1). Its genus is equal to the number of necks plus one.

It turns out that the placement of the necks is not arbitrary : they must satisfy a balancing condition which we will explain. One of our tasks in this paper is to give examples of balanced configurations of necks with arbitrary number of points.

The proof of the main theorem is in the spirit of recent gluing constructions [START_REF] Traizet | Adding handles to riemann's minimal surfaces[END_REF], [START_REF] Traizet | An embedded minimal surface with no symmetries[END_REF]. We use the Weierstrass Representation of minimal surfaces in its simplest form : consider a Riemann surface Σ and three holomorphic 1-forms φ 1 , φ 2 , φ 3 on Σ. Then X(z) = Re z z0 (φ 1 , φ 2 , φ 3 ) mod Λ defines a conformal minimal immersion X : Σ → R 3 /Λ provided the following conditions are satisfied ∀γ ∈ H 1 (Σ), Re

γ (φ 1 , φ 2 , φ 3 ) ∈ Λ φ 2 1 + φ 2 2 + φ 2 3 = 0 |φ 1 | 2 + |φ 2 | 2 + |φ 3 | 2 > 0
The first equation says that X is well defined and the second that it is conformal hence minimal since it is harmonic. The third one says that X is a regular immersion.

We define Σ by opening the nodes of a (singular) Riemann surface with nodes. We define φ 1 , φ 2 , φ 3 by prescribing their periods. We solve the above equations using the implicit function theorem.

The main difference with previous constructions is that in [START_REF] Traizet | Adding handles to riemann's minimal surfaces[END_REF] we used the Weierstrass representation in its classical form

X(z) = Re z z0 ( 1 2 (g -1 -g)dh, i 2 (g -1 + g)dh, dh)
where dh is a holomorphic 1-form and g is a meromorphic function on Σ (the Gauss map). Now on a high genus Riemann surface it is more natural to define holomorphic 1-forms (by prescribing periods) than meromorphic functions (where we have to face Abel's theorem). In fact in [START_REF] Traizet | Adding handles to riemann's minimal surfaces[END_REF] the Riemann surface Σ and the Gauss map g were defined at the same time, but that made the construction a little bit artificial. Also, the construction had some technical complications in the case where the Gauss map had multiple zeros or poles. It seems more natural to use φ 1 , φ 2 and φ 3 and avoid meromorphic functions completely.

The paper is organized as follows. In section 2 we explain the balancing condition and state our main theorem. In section 3 we prove the conjecture of Meeks. In section 4 we give examples of balanced configurations. In particular we prove the existence of triply periodic minimal surfaces with no symmetries beside translations. In section 5 we prove our main theorem.

Acknowledgements : This project was initiated when I visited Granada in 2003. It's a pleasure to thank Antonio Ros for the invitation and for suggesting that the conjecture of Meeks could be solved by desingularising a planar foliation of the 3-torus. I would also like to thank Laurent Mazet for carefuly reading the first draft of this paper, finding several mistakes and suggesting many improvements.

Main result

Simply periodic surfaces

Our main theorem in this paper is an adaptation, to the triply periodic case, of a construction of simply periodic minimal surfaces with ends asymptotic to horizontal planes [START_REF] Traizet | Adding handles to riemann's minimal surfaces[END_REF]. This construction may be described as follows.

Consider an infinite number of copies of the plane R 2 , labeled P k , k ∈ Z. In each plane P k , choose a finite number n k of points p k,i , 1 ≤ i ≤ n k . Identify, for each (k, i), the point p k,i in P k with the same point in P k+1 . This creates an abstract singular 2-manifold with double points (or nodes) which we call M 0 . Consider the map X 0 : M 0 → R 3 which sends each point (x, y) ∈ P k to the point (x, y, 0). We may think of X 0 as a minimal, isometric immersion of the singular 2-manifold M 0 into R 3 . We want to desingularise this singular minimal surface, namely : perturb M 0 into a regular 2-manifold M ε and X 0 into a minimal immersion X ε : M ε → R 3 . Intuitively, we do this by pushing slightly the planes away from each other and replace each double point by a small catenoid.

We assume the following periodicity : there exists an even integer N ≥ 2 and a vector T ∈ R 2 such that for all (k, i), p k+N,i = p k,i + T . We construct a family of minimal surfaces M ε which are periodic with period (T, ε), and have N ends asymptotic to horizontal planes in the quotient. (Here N must be even so that the quotient is orientable). We call the collection {p k,i } a simply periodic configuration. We call p k,i a point at level k of the configuration. Note however that all points p k,i are in R 2 . It will be convenient to identify R 2 with C and see each point p k,i as a complex number.

For the construction to work, the configuration must satisfy a balancing condition which we now explain. Let c k = 1/n k . Let ω k be the unique meromorphic 1-form on C ∪ {∞} with n k-1 simple poles at p k-1,i with residue c k-1 and n k simple poles at p k,i with residue -c k . (We assume that for each k, the points p k,i , p k-1,j are distinct). Let

F k,i = 1 2 Res p k,i ( ω k ) 2 dz + ( ω k+1 ) 2 dz
Explicitely,

ω k = n k-1 i=1 c k-1 z -p k-1,i dz - n k i=1 c k z -p k,i dz
which gives after a straightforward computation

F k,i = 2 n k j=1 j =i (c k ) 2 p k,i -p k,j - n k+1 j=1 c k c k+1 p k,i -p k+1,j - n k-1 j=1 c k c k-1 p k,i -p k-1,j
The analogy with 2-dimensional electrostatic forces suggests that we call F k,i a force. Each point p k,i interacts with the points p k,j at the same level and with points p k±1,j at the levels below and above it.

Definition 1 A simply periodic configuration is balanced if all forces F k,i are zero. It is non-degenerate if the differential of the map p → F has real co-rank 2, where p and F stand for the collection of p k,i and F k,i for 1 ≤ k ≤ N and

1 ≤ i ≤ n k .
The forces are clearly invariant by translation of all points, so vectors of the form (v, v, • • • , v), where v ∈ R 2 , are in the kernel of the differential. The nondegeneracy condition asks that these are the only ones. The following theorem is proven in [START_REF] Traizet | Adding handles to riemann's minimal surfaces[END_REF] Theorem 2 Given a balanced, non-degenerate, simply periodic configuration, there exists a smooth family of embedded simply periodic minimal surfaces M ε , for ε > 0 small enough, which have period (T, ε). The quotient has genus n -1, where n = n 1 + • • • + n N is the total number of necks, and has N horizontal planar ends. Moreover, M ε converges to the singular minimal surface M 0 when ε → 0. The last statement of the theorem should be understood as follows : if we see M ε as an abstract 2-manifold isometrically embedded in R 3 by the canonical injection X ε : M ε → R 3 , then the abstract manifold M ε converges to M 0 and X ε converges to X 0 on compact subsets of M 0 minus the double points.

Remark 1 In the sense of laminations, M ε converges to the lamination of R 3 by horizontal planes, with singular set {(p k,i , 0)}. Note however that this limit object does not contain enough information to reconstruct M ε . This is why we prefer to see the limit object M 0 as a minimal immersion of a singular 2manifold.

Section 2.1 of [START_REF] Traizet | Adding handles to riemann's minimal surfaces[END_REF] provides us with an explicit family of balanced configurations, depending on an integer parameter m ≥ 1.

Theorem 3

The following simply periodic configuration is balanced and nondegenerate :

N = 2, n 1 = m, n 2 = 1, p 1,j = cotan jπ m + 1 , p 2,1 = i and T = 2i (so p 0,1 = -i ).
For example, in the case m = 1, we have p k,1 = (k-1)i . The associated minimal surfaces M ε are the Riemann examples.

The case m = 4 is represented on figure 2. In the particular case of the configuration given by theorem 3, the corresponding minimal surfaces are hyperelliptic, so it is not too hard to write explicitely their Weierstrass representation and produce computer images, see figure 3.

Triply periodic surfaces

A natural, and interesting question, is whether we can carry a similar construction when there is an infinite number of points at each level, namely n k = ∞ for all k. It is not clear in general what the balancing condition should be, because if there is an infinite number of points, each force might be a diverging series.

In this paper we answer this question in the case where the points at each level are arranged in a doubly periodic way, namely they are invariant by two independant translations with vectors T 1 and T 2 . We assume that modulo T 1 and T 2 , there is a finite number n k of points at level k and we call them p k,i , 1 ≤ i ≤ n k . We see each point p k,i as an element of the torus T = R 2 /Γ, where Γ is the 2-dimensional lattice generated by T 1 and T 2 . We assume the following periodicity : there exists again an even integer N ≥ 2 and a vector T 3 ∈ R 2 (which replaces the vector T in the simply periodic case) so that p k+N,i = p k,i + T 3 . We call the collection {p k,i } a triply periodic configuration. We want to construct a family of triply periodic minimal surfaces with periods (T 1 , 0), (T 2 , 0) and (T 3 , ε).

Forces are defined as in the simply periodic case. Let again c k = 1/n k . Consider T as the genus one compact Riemann surface C /Γ. Let ω k be the meromorphic 1-form on T with n k-1 simple poles at p k-1,i with residue c k-1 , n k simple poles at p k,i with residue -c k , and pure imaginary periods. In other words,

Re T1 0 ω k = Re T2 0 ω k = 0.
The meromorphic 1-form ω k exists because the sum of the residues is zero, and the period condition makes it unique. Define forces as in the simply periodic case by

F k,i = 1 2 Res p k,i (ω k ) 2 dz + (ω k+1 ) 2 dz
The main difference with the simply periodic case is that forces cannot be computed as explicitly. They may be written in various ways in term of elliptic functions. We will do this in section 4.

Definition 2 A triply periodic configuration is balanced if all forces F k,i are zero. It is non-degenerate if the differential of the map p → F has real co-rank 2, where p and F are as in definition 1.

The main result of the paper is the following Theorem 4 Given a balanced, non-degenerate triply periodic configuration, there exists a smooth family of embedded triply periodic minimal surfaces M ε , for ε > 0 small enough, which have period (T 1 , 0), (T 2 , 0) and (T 3 , ε). The quotient has genus n + 1, where n = n 1 + • • • + n N . Its area is close to N times the area of T. Moreover, M ε converges to M 0 when ε → 0, where M 0 is a singular minimal surface defined as in the beginning of section 2.1.

In fact, we construct a family of surfaces depending smoothly on ε, T 1 , T 2 and T 3 (in a neighborhood of the given values). This gives Corollary 1 Under the same hypothesis, there exists η > 0 such that the following is true : if V 1 , V 2 and V 3 are three independant vectors in space such that |V i -(T i , 0)| < η for i = 1, 2, 3, then there exists a triply periodic minimal surface with periods V 1 , V 2 , V 3 , which satisfies the same conclusion as the theorem.

Hence up to isometries, we construct a 6-dimensional family of surfaces. This is also the dimension of the space of flat 3-tori.

Proof : Simply consider the rotation with smallest angle which sends V 1 and V 2 to the horizontal plane.

Proof of the conjecture of Meeks

In section 4 we will prove Theorem 5 For any g ≥ 3, there exists T 1 , T 2 , T 3 , and a triply periodic, balanced, non-degenerate configuration with periods T 1 , T 2 , T 3 which has g -1 points, so the associated minimal surfaces have genus g.

By corollary 1, if V 1 , V 2 and V 3 are three independant vectors in space close to the horizontal vectors (T 1 , 0), (T 2 , 0), (T 3 , 0), there exists an embedded minimal surfaces of genus g with periods V 1 , V 2 , V 3 . It turns out that this already gives minimal surfaces of genus g in arbitrary flat 3-tori, up to scale ! To understand why, let us first consider an example.

Assume that T 1 = (1, 0), T 2 = (0, 1) and T 3 = (0, 0). How can we recover minimal surfaces of genus g in the cubic torus, namely with periods V 1 = (1, 0, 0), V 2 = (0, 1, 0), V 3 = (0, 0, 1) ?

Consider the following sequence of matrices

A n =   n 0 1 1 n 0 0 1 0  
Since this matrix has determinant 1, its columns form a basis of the cubic lattice.

On the other hand, when n → ∞, n -1 A n converges to the matrix with columns (T 1 , 0), (T 2 , 0) and (T 3 , 0). By corollary 1, for n large enough, there exists a minimal surface of genus g and periods the columns of n -1 A n . Scaling by n gives a sequence of minimal surfaces M n in the cubic torus.

In the general case, we need a result which we will prove in the next section. Let SL(3, Z) be the group of 3 × 3 matrices with integer coefficients and determinant one.

Proposition 1 Consider a matrix M ∈ M 3 (R) such that det(M ) = 0. Then there exists a sequence of matrices A n ∈ SL(3, Z) and a sequence of reals λ n such that λ n A n → M (so λ n → 0).
Using this proposition, we prove theorem 1. Let Λ be an arbitrary lattice in

R 3 . Let V 1 , V 2 , V 3 be a basis of Λ. We write [V 1 V 2 V 3 ] for the matrix whose columns are V 1 , V 2 , V 3 . Recall that SL(3, Z) acts on bases of Λ as follows : if A ∈ SL(3, Z), define V 1 , V 2 , V 3 by [V 1 V 2 V 3 ] = [V 1 V 2 V 3 ]A In other words, V j = i a ij V i . Then V 1 , V 2 , V 3 is
again a basis of Λ. Indeed, because A has integer coefficients, the lattice generated by V 1 , V 2 and V 3 is included in Λ. The converse is true because A is invertible. Let M ∈ M 3 (R) be the rank 2 matrix defined by

T 1 T 2 T 3 0 0 0 = [V 1 V 2 V 3 ]M By proposition 1 there exists a sequence A n ∈ SL(3, Z) and λ n ∈ R such that λ n → 0 and λ n A n → M . Define V n 1 , V n 2 , V n 3 by [V n 1 , V n 2 , V n 3 ] = λ n [V 1 , V 2 , V 3 ]A n Then V n 1 , V n 2 , V n 3 is a basis of the lattice λ n Λ and V n i → (T i , 0) when n → ∞, for i = 1, 2, 3.
By corollary 1, for n large enough, there exists an embedded triply periodic minimal surface of genus g with periods

V n 1 , V n 2 , V n 3 .
Scaling by 1/λ n , we have a sequence of embedded minimal surfaces of genus g in R 3 /Λ whose area goes to ∞. This proves theorem 1.

On singular matrices

We state a general result which we only use in the case of dimension d = 3. We use the following notations. Let M d (A) be the set of d × d matrices with coefficients in A, where A is Z, Q or R. Let N d (A) be the set of matrices in M d (A) whose determinant is zero. Let A d be the set of matrices M ∈ M d (R) such that there exists a sequence of matrices A n ∈ SL(d, Z) and a sequence of reals λ n such that λ n → 0 and

λ n A n → M . Proposition 2 A d = N d (R).
Proof : The inclusion ⊂ is clear. For the reverse inclusion, first observe the following facts.

(i) If M ∈ A d and P ∈ SL(d, Z) then P M ∈ A d and M P ∈ A d . (ii) If M ∈ A d and λ ∈ R then λM ∈ A d . (iii) A d is closed. (iv) N d (Q) is dense in N d (R).
By points (iii) and (iv), it suffices to prove that

N d (Q) ⊂ A d . By point (ii) it suffices to prove that N d (Z) ⊂ A d . Let M ∈ M d (Z)
. By standard theory of matrices with integer coefficients, M may be put into reduced form, namely

M = P    a 1 0 . . . 0 a d    Q where P, Q ∈ SL(d, Z) and a 1 , • • • a d ∈ Z.
Moreover, if r is the rank of M , then a i = 0 for i > r. Now assume that det M = 0 so a d = 0. By point (i) we only need to prove that the above diagonal matrix is in A d . When d = 3, which is the case we are interested in, simply write

  a 1 0 0 0 a 2 0 0 0 0   = lim n→∞ 1 n   a 1 n 1 0 0 a 2 n 1 1 0 0  
The matrix on the right has determinant 1. In the general case, write

     a 1 . . . 0 0 a d-1 0      = lim n→∞ 1 n       a 1 n 1 0 0 . . . . . . . . . a d-1 n 1 (-1) d+1 0 0       4 Examples

Preliminary observations

First observe that

F k,i = 1 2 Res p k,i N =1 (ω ) 2 dz because ω has no pole at p k,i if = k, k + 1.
Hence by the residue theorem, the forces sum up to zero :

N k=1 n k i=1 F k,i = 0
Also we have

F k,i = 1 4 Res p k,i (ω k -ω k+1 ) 2 dz To see this, observe that (ω k + ω k+1 ) 2 + (ω k -ω k+1 ) 2 = 2((ω k ) 2 + (ω k+1 ) 2 )
and ω k + ω k+1 has no pole at p k,i as the residues cancel.

Simplest configurations

Proposition 3 Take N = 2, n 1 = n 2 = 1. Consider a 2-dimensional lattice Γ = ZT 1 + ZT 2 and let T = C /Γ.
Consider some a ∈ T such that 0, a and -a are distinct points in T. The following triply periodic configuration is balanced : p 1,1 = 0, p 2,1 = a, T 3 = 2a (so p 0,1 = -a). The corresponding minimal surfaces have genus 3.

We will study non-degeneracy of this configuration in section 4.3.3. Observe that if we replace a by a + T /2 where T ∈ {T 1 , T 2 , T 1 + T 2 }, we get three other balanced configurations which have the same period T 3 modulo Γ.

Proof : The configuration is balanced by symmetry. Indeed, the meromorphic 1-form ω 1ω 2 has three simple poles at 0, ±a, with respective residues -2 and 1. Let σ(z) = -z. Then σ * (ω 1ω 2 ) has the same poles as ω 1ω 2 with the same residues, so their difference is a holomorphic 1-form on C /Γ with imaginary periods, so they are equal. In other words, the function (ω 1ω 2 )/dz is an odd function. Hence its square is an even function, so it has no residue at zero, so F 1,1 = 0. Since the forces sum up to zero, F 2,1 = 0, so the configuration is balanced.

Balanced configurations using Weierstrass ζ function

In this section we compute explicitly the forces in term of Weierstrass ζ function. This has several applications : first it may be used to compute the forces numerically and find numerical examples of balanced configurations. It will also give us an electrostatic interpretation of the forces. Finally, we use it to prove the existence of balanced, non-degenerate triply periodic configurations, as perturbations of simply periodic configurations.

Forces in term of Weierstrass ζ function

Given a lattice Γ = ZT 1 + ZT 2 in the complex plane, the Weierstrass ζ and ℘ functions are defined by

ζ(z) = 1 z + w∈Γ w =0 1 z -w + 1 w + z w 2 ℘(z) = 1 z 2 + w∈Γ w =0 1 (z -w) 2 - 1 w 2
We refer the reader to Ahlfors [START_REF] Ahlfors | Complex analysis[END_REF] for standard properties of these functions. The function ζ is odd, its derivative is -℘ and it is quasi-periodic, namely for

i = 1, 2 ζ(z + T i ) = ζ(z) + η i with η i = 2ζ(T i /2)
It has a simple pole at each point of the lattice Γ with residue 1. Hence we may write

ω k =   n k-1 j=1 c k-1 ζ(z -p k-1,j ) - n k j=1 c k ζ(z -p k,j ) + λ k   dz
where λ k is some constant. Indeed, the right term is well defined on T = C /Γ because the sum of the residues is zero, so the quasi-periodicities of ζ cancel, and two meromorphic 1-forms which have the same principal part differ by a holomorphic 1-form, so a constant times dz. The constant λ k is determined by the condition that the periods of ω k are imaginary. A straightforward computation gives

F k,i = n k j=1 j =i 2(c k ) 2 ζ(p k,i -p k,j ) - ± n k±1 j=1 c k c k±1 ζ(p k,i -p k±1,j ) + (λ k+1 -λ k )c k
Here the sum on ± means that there are two terms, one for the + sign and one for thesign. Let us now compute λ k . First observe that in our formula for ω k , we have chosen a representative in C for each point p k,i . The constant λ k depends on the choice of representatives. Let γ 1 and γ 2 be the standard generators of the homology of T (in other words, the homology class of the path from 0 to T 1 and 0 to

T 2 ). Given two points a, b in C , I claim that for i = 1, 2, γi ζ(z -a)dz -ζ(z -b)dz ≡ -η i (a -b) [2πi ]
Indeed, the integrant is a well defined meromorphic 1-form on T with two simple poles of residue ±1, so its periods are well defined mod 2πi . The formula clearly holds when a = b, and using the quasi-periodicity of the function ζ, both sides have the same derivative with respect to a, so the claim is true. Using this formula with a = p k-1,j or a = p k,j , and b = z 0 where z 0 is some fixed point, we obtain for i = 1, 2 (the z 0 terms cancel because the sum of the residues is zero)

Re γi ω k = Re   -η i   n k-1 j=1 c k-1 p k-1,j - n k j=1 c k p k,j   + λ k T i  
Let us define the center of mass µ k of the points at level k by

µ k = n k j=1 c k p k,j
Let (x k , y k ) ∈ R 2 be the coordinates of µ k in the basis T 1 , T 2 . Recall that η 1 and η 2 satisfy the Legendre relation ( [START_REF] Ahlfors | Complex analysis[END_REF], page 274)

η 1 T 2 -η 2 T 1 = 2πi
This gives for i = 1, 2

Re γi ω k = Re ((-η 1 (x k-1 -x k ) -η 2 (y k-1 -y k ) + λ k )T i ) Hence λ k = η 1 (x k-1 -x k ) + η 2 (y k-1 -y k )

Electrostatic interpretation of the forces

Assume that 2µ k = µ k-1 +µ k+1 for all k. This means that µ k+1 -µ k is constant, so the centers of mass µ k are regularly spaced. Then λ k+1 = λ k . Using the definition of the ζ function as a series, and after a tedious computation, we obtain the following formula for the force :

F k,i = n k j=1 j =i 2(c k ) 2 p k,i -p k,j - ± n k±1 j=1 c k c k±1 p k,i -p k±1,j + w∈Γ w =0   n k j=1 2(c k ) 2 p k,i -p k,j -w - ± n k±1 j=1 c k c k±1 p k,i -p k±1,j -w  
This formula may be interpreted as follows : this is the same formula as the force F k,i in the simply periodic case, except that there is an infinite number of terms : p k,i interacts with all the other points p ,j + w, except itself, namely for ( , j, w) = (k, i, 0). This provides an electrostatic interpretation of the force. Remark 2 It is not hard to see that the sum in the last formula converges, as written, if and only if the configuration satisfies 2µ k = µ k-1 + µ k+1 . This makes our hypothesis natural if we are to give an electrostatic interpretation of the forces.

The genus 3 case

We now return to the case N = 2, n 1 = n 2 = 1 which gives genus 3 surfaces. We may assume by translating the configuration that p 0,1 + p 2,1 = 0. Writing p 1,1 = xT 1 + yT 2 we obtain

F 1,1 = -ζ(xT 1 + yT 2 -T 3 /2) -ζ(xT 1 + yT 2 + T 3 /2) + 2xη 1 + 2yη 2
When x = y = 0, which is the configuration we considered in section 4.2, we have F 1,1 = 0 so the configuration is balanced, and using that the derivative of

ζ is -℘, ∂F 1,1 ∂x = 2℘(T 3 /2)T 1 + 2η 1 , ∂F 1,1 ∂y = 2℘(T 3 /2)T 2 + 2η 2
Given T 1 and T 2 , for generic values of T 3 , the differential of F 1,1 with respect to (x, y) has real rank 2 so the configuration is non-degenerate. But there is a real one dimensional family of values of T 3 for which the differential has rank one so the configuration is degenerate.

Remark 3 Using the above formula, we find numerically that there are other balanced configurations which are not as symmetric as the one we discussed in section 4.2. This confirms the already suspected fact that the space of genus 3 minimal surfaces in a 3-torus is quite intricate. We will not discuss the subject any further, as we are mainly interested in higher genus surfaces.

Triply periodic perturbation of a simply periodic configuration

In this section we start with a balanced, non-degenerate simply periodic configuration {p k,i } with period T (which cannot be zero, see proposition 2.4 in [START_REF] Traizet | Adding handles to riemann's minimal surfaces[END_REF]).

Consider two independant vectors T 1 and T 2 , let Γ be the lattice generated by T 1 , T 2 , and let T 3 = T . Given a real t = 0, we may consider the triply periodic configuration with periods T 1 /t, T 2 /t and T 3 , whose points are p k,i mod Γ/t. This configuration is of course not balanced anymore, so our goal is to perturb it into a balanced triply periodic configuration for t small enough. We write ζ(z, Γ/t) = 1 z and the limit is uniform with respect to z in compact subsets of C . Also

F k,i (t)
η i (Γ/t) = tη i (Γ), which gives λ k (t) = O(t 2 ). It follows from our formula in section 4.3.1 that lim t→0 F k,i (t) = F k,i
where F k,i is the force associated to the simply periodic configuration p k,i . Moreover, F k,i (t) extends real analytically to t = 0. The intuitive idea behind this formula, from the point of view of our electrostatic interpretation in section 4.3.2, is that when t → 0, T 1 /t and T 2 /t go to infinity, so p k,i does not interact anymore with the points p ,j + w/t, w = 0.

Theorem 6 Let {p 0 k,i } be a balanced, non-degenerate, simply periodic configuration with period T 3 . Fix some lattice Γ = ZT 1 + ZT 2 . Then for t close to 0, there exists analytic functions p k,i (t), such that p k,i (0) = p 0 k,i and for t = 0, the triply periodic configuration {p k,i (t) mod Γ/t} is balanced and non-degenerate.

Theorem 5 follows as a corollary of this theorem and theorem 3.

Proof : This is a quite straightforward application of the implicit function theorem. We fix p 1,1 = p 0 1,1 : this normalizes the translation invariance of the balancing condition. Let p be the collection of all variables p k,i , 1 ≤ k ≤ N , 1 ≤ i ≤ n k except p 1,1 , and let F(t, p) be the collection of the forces

F k,i (t) for 1 ≤ k ≤ N , 1 ≤ i ≤ n k except F 1,1 (t).
Then F is a real analytic map, F(0, p 0 ) = 0 and the partial differential of F with respect to p at (0, p 0 ) is an isomorphism. By the implicit function theorem, for t close to 0, there exists p(t) such that F(t, p(t)) = 0. If t = 0, the associated triply periodic configuration p k,i (t) mod Γ/t has all its forces zero except maybe F 1,1 (t). Since the sum of the forces is zero, F 1,1 (t) is also zero. Hence the configuration p k,i (t) mod Γ/t is balanced. It is also non-degenerate for t small enough by continuity.

Remark 4

In the simply periodic case, the force F k,i is a complex analytic function of the variables p ,j , so non-degeneracy is equivalent to the fact that the complex matrix of partial derivatives ∂ F k,i /∂p ,j has complex co-rank 1, which can be checked by computing a determinant.

In the triply periodic case, the force F k,i is only real analytic because of the period normalization. For this reason, non-degeneracy is much harder to deal with in this case. This makes the above theorem particularly valuable since non-degeneracy comes for free.

Surfaces with no symmetries

In the introduction of [START_REF] Hoffman | Some basic facts, old and new, about triply periodic embedded minimal surfaces[END_REF], D. Hoffman asked the following question (question 2) Does there exist a triply periodic embedded minimal surface with no symmetries other than translations ?

We give a positive answer to this question. First observe that for generic lattices Λ, the only symmetries of the torus R 3 /Λ are translations : x → x + v and symmetries about a point : x → -x + v. This means that in a generic flat 3-torus, we do not have to worry about planar symmetries since they are just impossible. Now consider the simply periodic configuration represented in figure 5. This configuration is obtained by combining two configurations of theorem 3, one with m = 2 and one with m = 3 scaled by 8/9 (the scale is chosen so that the forces at the points with n k = 1 cancel). It is proven in [START_REF] Traizet | Adding handles to riemann's minimal surfaces[END_REF], proposition 2.3, that configurations obtained this way are balanced and non-degenerate. The only symmetry of this configuration is the symmetry about the Ox 2 axis. Using theorem 5 to perturb this configuration into a triply periodic configuration, and arguing as in section 3, we have proven the existence, in any 3-torus R 3 /Λ, of a minimal surface whose only possible symmetry, if any, is a planar symmetry. Hence in a generic 3-torus, the surface will have no symmetry.

Balanced configurations using Weierstrass ℘ function

In this section we briefly explain how balanced configurations may be computed by solving purely algebraic (i.e. polynomial) equations. However, it does not seem possible to study non-degeneracy along these lines, so we won't go into very much details.

Consider for example the case where N = 2, n 1 = 2m is even and n 2 = 1 (which gives genus 2m + 2). We assume that the configuration is symmetric about the origin, namely p 1,i+m = -p 1,i for 1 ≤ i ≤ m, p 2,1 = T 3 /2 so p 0,1 = -T 3 /2. Then we can compute the forces in term of the Weierstrass ℘ function as follows : to simplify the notations let us write p i = p 1,i . Then

ω 1 -ω 2 = -2 n 1 m i=1 ℘ dz ℘ -℘(p i ) + ℘ dz ℘ -℘(T 3 /2)
Indeed, the right side has the right poles and residues and has imaginary periods. An easy computation gives

Res pi ℘ ℘ -℘(p i ) 2 = ℘ (p i ) ℘ (p i )
Recalling that (℘ ) 2 = 4℘ 3 -g 2 ℘-g 3 , where g 2 and g 3 are the modular invariants of C /Γ ([1] page 276), we obtain after elementary computations

F 1,i = ℘ (p i ) n 2 1   6℘ 2 (p i ) -g 2 /2 4℘ 3 (p i ) -g 2 ℘(p i ) -g 3 + j =i 2 ℘(p i ) -℘(p j ) - n 1 ℘(p i ) -℘(T 3 /2)  
By symmetry we have F 1,i+m = -F 1,i and F 2,1 = 0, so we have to solve the m equations F 1,1 = 0, • • • , F 1,m = 0. These are purely algebraic equations in the unknowns X i = ℘(p i ). For example, in the case m = 1, T 3 = 0, g 2 = 4 and g 3 = 0 (this is a square torus), we obtain the equation ℘ 2 (p 1 ) = 1/3. For modest values of m, the above system is easy to solve using for example Maple. This is how the configuration in figure 4 was computed.

5 Proof of Theorem 4

The initial Riemann surface with nodes Σ 0

For each k = 1, • • • , N , consider some complex parameter τ k such that Im(τ k ) = 0. Let T k be the genus one compact Riemann surface C /(Z + Zτ k ).

For

each k = 1, • • • , N and each i = 1, • • • , n k , consider a pair of points (a k,i , b k,i ) such that (a k,i , b k,i ) ∈ T k ×T k+1 if k < N , and (a N,i , b N,i ) ∈ T N ×T 1 .
(Here we assume that in each T k , the points a k,i , b k-1,j are distinct.)

Consider the disjoint union T 1 ∪• • •∪T N . Identify, for each pair of points, a k,i with b k,i to create a node (or double point). This creates a singular Riemann surface with nodes which we call Σ 0 . It depends on the complex parameters τ k , a k,i , b k,i .

Opening nodes

Following Fay [START_REF] Fay | Theta functions on riemann surfaces[END_REF] and Masur [START_REF] Masur | The extension of the weil petersson metric to the boundary of teichmuller space[END_REF] we define a (regular) Riemann surface Σ by "opening nodes".

Consider the local complex coordinates v k,i = za k,i in a neighborhood of a k,i in T k , and

w k,i = z -b k,i in a neighborhood of b k,i in T k+1 .
v k,i w k,i = t k,i
This identifies the two circles, with a Dehn twist of angle equal to the argument of t k,i , and defines a (possibly noded) Riemann surface Σ t which depends holomorphically on t (as well as on the other parameters τ k , a k,i , b k,i ).

When all the t k,i are nonzero, Σ t is a genuine compact Riemann surface (without nodes). From the topological point of view, its genus g is easily seen to be equal to n + 1, where n is the total number of nodes.

When t k,i = 0, this construction identifies the point a k,i with b k,i and Σ t has a node. In particular when t = 0, Σ t is the noded Riemann surface we started from.

Regular differentials

In this section, we explain all the tools we need to solve our problem. Following Masur [START_REF] Masur | The extension of the weil petersson metric to the boundary of teichmuller space[END_REF], we extend the notions of holomorphic 1-form and holomorphic quadratic differential to the case of Riemann surfaces with nodes. Consider a Riemann surface with nodes Σ where we see each node as a pair of points (a, b) which have been identified.

Definition 3 ([5]

) A regular q-differential ω on Σ is a form of type (q, 0) which is holomorphic outside the nodes and which, for each pair of points (a, b) which are identified to create a node, has two poles of order ≤ q at a and b. Moreover, the residues of ω at a and b must be opposite if q is odd and equal if q is even.

Here the residue of ω at some point p is the coefficient of dz q /z q in the expression of ω in term of a local coordinate z such that z(p) = 0.

When Σ has no nodes, a regular q-differential is just a holomorphic qdifferential. We are only interested in the cases q = 1 and q = 2 which correspond respectively to holomorphic 1-forms and holomorphic quadratic differentials.

We write Ω q (Σ) for the space of regular q-differentials on Σ. The space Ω 1 (Σ) has dimension g and Ω 2 (Σ) has dimension 3g -3, where g is the genus of Σ (provided g ≥ 2).

Moreover, these spaces depend holomorphically on parameters, in the following sense : proposition 4.1 of Masur [START_REF] Masur | The extension of the weil petersson metric to the boundary of teichmuller space[END_REF] says that there exists a basis ω 1,t , • • • , ω g,t of Ω 1 (Σ t ), which "depend holomorphically" on t in a neighborhood of 0. This is fundamental for the construction we have in mind, since we will apply the implicit function theorem at t = 0.

For our purpose, it will be enough to know that for any δ > 0, the restriction of ω j,t to the domain Ω δ defined by ∀k, ∀i, |v k,i | > δ, |w k,i | > δ, depends holomorphically on (z, t), z ∈ Ω δ and t in a neighborhood of 0. This makes sense because the domain Ω δ is independant of t provided t < δ 2 . There is a deeper way of expressing that ω t depends holomorphically on t [START_REF] Masur | The extension of the weil petersson metric to the boundary of teichmuller space[END_REF], but we will not need it.

The isomorphism between Ω 1 (Σ t ) and C g can be made explicit by computing periods as follows. Let A k , B k be the standard generators of the homology of the torus T k , namely A k is the straight path from 0 to 1 and B k from 0 to τ k . (Here we assume that A k and B k stay away from the nodes, taking different representatives of their homology class if necessary.) We write C(a k,i ) for the circle |za k,i | = in T k , where is some fixed small positive number, and

C(b k,i ) for the circle |z -b k,i | = in T k+1 . The circle C(a k,i ) is homologous in Σ t to the circle -C(b k,i ).
Again, when k = N , T N +1 should be understood as T 1 . More generally, we adopt the following cyclic convention : T k+N = T k , and in the same way,

a k+N,i = a k,i , b k+N,i = b k,i . Lemma 1 The map ω → ( A k ω 1≤k≤N , C(a k,i ) ω 1≤k≤N 2-δ k,1 ≤i≤n k ) is an isomorphism from Ω 1 (Σ t ) to C g .
In other words, we may define a regular 1-differential on Σ t by prescribing its periods on all indicated cycles. The condition 2δ k,1 ≤ i reads as i ≥ 2 if k = 1 and i ≥ 1 if k = 1. The total number of cycles is equal to the genus of Σ t .

Proof : One way to prove the lemma is to observe that the indicated set of cycles is the set of A-curves of a canonical homology basis. Here is another argument, which illustrates the use of Riemann surfaces with nodes, and will also work for the next lemma. We check that the statement is true in the case of the noded Riemann surface Σ 0 . By continuity, the statement will be true provided t is small enough, which is a weaker statement but enough for our needs.

We prove that the map is injective. Let ω be a regular 1-form on Σ 0 . In T k , ω has simple poles at a k,i and b k-1,i , and the residue of ω at b k-1,i is minus its residue at a k-1,i . By the residue theorem in

T k n k i=1 Res a k,i ω = - n k-1 i=1 Res b k-1,i ω = n k-1 i=1
Res a k-1,i ω so this sum is independant of k. When ω is in the kernel of the map of the lemma, this implies that all residues of ω are zero, so ω is in fact holomorphic on each T k , and has vanishing A k -period, so ω = 0.

Concerning regular quadratic differentials, proposition 5.1 of Masur [START_REF] Masur | The extension of the weil petersson metric to the boundary of teichmuller space[END_REF] says that there exists a basis ψ 1,t , • • • , ψ 3g-3,t of Ω 2 (Σ t ), which depends holomorphically on t.

Lemma 2 For t close to 0, the map

ψ → L(ψ) = ( A k ψ dz 1≤k≤N , C(a k,i ) (z -a k,i )ψ dz 1≤k≤N 1≤i≤n k , C(a k,i ) ψ dz 1≤k≤N 2-δ k,1 ≤i≤n k , C(b k,i ) ψ dz 1≤k≤N 1+δ k,N ≤i≤n k ) is an isomorphism from Ω 2 (Σ t ) to C 3g-3 .
In this lemma, dz stands for the standard holomorphic 1-form on each T k , so dz is not globally defined on Σ t , which is why integrating ψ/dz on C(a k,i ) and C(b k,i ) gives independant results.

Proof : As in the previous lemma, it suffices to prove that the map is injective when t = 0. Consider an element ψ in the kernel. Recall that a regular quadratic differential on the noded Riemann surface Σ 0 has at most double poles at the nodes. The second summand of the map tells us that ψ has at most simple poles at all a k,i , hence at all b k,i (recall the definition of a regular quadratic differential). The third and fourth summands tell us that the only points where ψ might have poles are a k,1 for 2 ≤ k ≤ N and b N,1 . This means that ψ has at most one simple pole in each T k . Applying the residue theorem to the holomorphic 1-form ψ/dz in T k , we conclude that ψ has no poles at all so ψ is holomorphic in each T k . The first summand gives ψ = 0.

Let us now return to regular 1-forms. Let ω t be a regular 1-form on Σ t defined by prescribing periods (independant of t) on cycles as in lemma 1. We need to compute the derivative of ω t with respect to the parameters t k,i at t = 0.

Lemma 3

The derivative ∂ω t /∂t k,i at t = 0 is a meromorphic 1-form on Σ 0 which has double poles at a k,i and b k,i , is otherwise holomorphic, and has vanishing periods on all cycles of lemma 1. The principal part at the poles are

dz (z -a k,i ) 2 -1 2πi C(b k,i ) ω 0 z -b k,i at a k,i dz (z -b k,i ) 2 -1 2πi C(a k,i ) ω 0 z -a k,i at b k,i
Note that these conditions (principal parts + periods) determine a unique meromorphic 1-form.

Proof : First recall that when we say that ω t depends holomorphically on t, we only mean that its restriction to Ω δ depends holomorphically on (z, t). So the statement of the lemma is that the restriction to Ω δ of the derivative of ω t agrees with the indicated meromorphic 1-form.

To simplify the notations, write t = t k,i , v = v k,i and w = w k,i . Let us write the Laurent series of ω t in the annular region δ < |v| <

ω t = n∈Z c n (t)v n dv where c n (t) = 1 2πi |v|= ω t v n+1 Hence ∂ω t ∂t = n∈Z ∂c n (t) ∂t v n dv
The above series converges uniformly in the region δ < |v| < , so the series for n ≥ 0 converges uniformly in the disk |v| < and defines a holomorphic function. To compute the derivative of the coefficient c n for n < 0, first observe that c -1 is constant so its derivative is zero, and for n ≤ -2 and t = 0

c n (t) = -1 2πi |w|= ω t w n+1 t n+1 ∂c n (t) ∂t = n + 1 2πi |w|= ω t w n+1 t n+2 - 1 2πi |w|= ∂ω t ∂t w n+1 t n+1
Now ω t and its derivative are both uniformly bounded on the circle |w| = , so if we let t → 0, we see that for n ≤ -3, the derivative of c n (t) at t = 0 is zero, and for n = -2,

∂c -2 ∂t (0) = -1 2πi |w|= ω 0 w -1
This proves that the derivative of ω t with respect to t k,i has a double pole at a k,i with the indicated principal part. Entirely similar computations give that the derivative of ω t has a double pole at b k,i and no poles at the other nodes.

We need one more result to compute the integral of ω t on a path from the point a k,i + to the point b k,i + , which goes through the neck.

Lemma 4 The difference b k,i + a k,i + ω t -(log t k,i ) 1 2πi C(a k,i ) ω t
is a well defined analytic function of t k,i which extends analytically to t k,i = 0. B This lemma is proven in [START_REF] Traizet | An embedded minimal surface with no symmetries[END_REF], section 3.6, lemma 1, using a Laurent series to estimate ω t on the neck as in the proof of the previous lemma. Observe that the logarithm is not well defined because t k,i is a complex number, nor is the left integral because there is no canonical way to choose the integration path, but the multi-valuations cancel.

The Weierstrass data

Homology basis

Let A k,i be the homology class of the circle C(a k,i ) (positively oriented) in T k . For 1 ≤ k ≤ N and 2 ≤ i ≤ n k , let B k,i be a cycle as on figure 7, which intersects only A k,1 and A k,i , with respective intersection numbers -1 and +1. Let B be a cycle as on figure 7, which intersects all circles A k,1 with intersection numbers +1. Then the following set of cycles forms a homology basis of Σ t :

A 1,1 , B, A k,i , B k,i for 1 ≤ k ≤ N , 2 ≤ i ≤ n k , and A k , B k for 1 ≤ k ≤ N .
(Observe that if we replace B 1,i by B 1,i + B, then we have a canonical homology basis, namely each A-cycle intersects precisely one B-cycle with intersection number 1, and all other intersection numbers are zero).

The period problem

Without loss of generality we may assume that T 1 = (1, 0). We need to define three regular 1-forms φ 1 , φ 2 , φ 3 on Σ t such that Re (φ 1 , φ 2 , φ 3 ) on each cycle of the homology basis belongs to the lattice Λ generated by (T 1 , 0), (T 2 , 0) and (T 3 , ε). From the geometric picture of the surface we want to construct, we know in fact exactly what each period should be, namely

Re A k,i (φ 1 , φ 2 , φ 3 ) = 0

Parameters count

Let us count the number of parameters we have. Let n = n 1 + • • • + n N be the number of necks. We have the conformal parameters a k,i , b k,i , t k,i and τ k , for a total of 3n+N complex parameters. We have the period parameters α k,i; , α k; , for a total of 3(n -N + 1) + 3N = 3(n + 1) real parameters. Finally we have the real parameter x. If we see each complex parameter as two real parameters, this gives us a total of 9n + 2N + 4 real parameters.

Let us now count the number of equations we have to solve. We have the conformality equation φ 2 1 + φ 2 2 + φ 2 3 = 0, which amounts to 3g -3 = 3n complex equations. We have the period conditions on the cycles B k,i , B k and B, for a total of 3(n -N ) + 3N + 2 = 3n + 2 real equations. This gives us a total of 9n + 2 real equations.

So we see that we have 2N + 2 too many parameters, a lot more than expected. This is because the parameters are not independant and we can normalise the value of some of them. Indeed, if we multiply all parameters α k,i;3 and α k;3 by some λ and divide x by λ, we do not change the Weierstrass data, so we may impose one condition on these parameters. Also, if we translate all points a k,i and b k-1,j in T k by some fixed amount, we get an isomorphic Riemann surface Σ t , so we may fix the value of one point in each T k . So we have one real and N complex normalisations, for a total of 2N + 1 real normalisations.

Taking this into account, we have one more parameter than equation, so we expect to construct a one parameter family of minimal surfaces.

The conformality equations

Let Q = φ 2 1 + φ 2 2 + φ 2 3
. This is a regular quadratic differential on Σ t . To ensure that Q = 0 we will solve the system L(Q) = 0, where L is the linear operator defined in lemma 2. We start with the first three equations. We will solve the last one in section 5.7.

Proposition 4 For x close to 0, there exists (locally unique) values of the parameters α k;1 , α k;2 , α k,i;1 , α k,i;2 , t k,i (for all possible values of the indices k, i) such that the following equations are satisfied :

A k Q dz = 0 (1 ≤ k ≤ N ) C(a k,i ) (z -a k,i )Q dz = 0 (1 ≤ k ≤ N , 1 ≤ i ≤ n k ) C(a k,i ) Q dz = 0 (1 ≤ k ≤ N , 2 -δ k,1 ≤ i ≤ n k )
These values depend real analytically on X = x 2 (and on all other parameters a k,i , b k,i , τ k , α k,i;3 and α k;3 as well). Moreover, at the point x = 0, we have

t k,i = 0 , ∂t k,i ∂X = (α k,i;3 ) 2 4 α k;1 = 0 , α k;2 = (-1) k , α k,i;1 = α k,i;2 = 0 and for i ≥ 2 -δ k,1 , ∂ ∂X (α k,i;1 + (-1) k i α k,i;2 ) = -1 2 Res a k,i ( φ 3 ) 2 dz
Note that when the parameters have theses values, x = 0 implies t k,i (x) = 0 (for x small enough, provided α k,i;3 = 0) so Σ t(x) is a Riemann surface without nodes.

Proof : We use the implicit function theorem. First assume that x = 0, t k,i = 0, α k,i;1 = α k,i;2 = 0 (all indices k, i). Then φ 3 = 0 and φ 1 , φ 2 are holomorphic 1-forms on T k , so they are entirely determined by their period on A k , namely

φ 1 = (1 + i α k;1 )dz , φ 2 = i α k;2 dz φ 2 1 + φ 2 2 + φ 2 3 = (1 + 2i α k;1 -α 2 k;1 -α 2 k;
2 )dz 2 Hence we need α k;1 = 0 and α k;2 = ±1. The choice ±1 is determined by the orientation of the surface : φ 1 = dz, φ 2 = i dz gives the Weierstrass representation of a horizontal plane with normal (0, 0, -1) while φ 1 = dz, φ 2 = -i dz gives a horizontal plane with the opposite normal. Hence from the geometry of the surface we want to construct, the sign should alternate between consecutive T k . We choose α k;2 = (-1) k , which means that in T k , the normal will point down for k even and up for k odd. Then φ 1 = dz, φ 2 = (-1) k i dz in T k and Q = 0 so all equations are satisfied. We now compute the derivatives with respect to all parameters, except x, at this point. Let p be any of the parameters and γ be any cycle, then

∂ ∂p γ Q dz = γ 2φ 1 dz ∂φ 1 ∂p + 2φ 2 dz ∂φ 2 ∂p = γ 2 ∂φ 1 ∂p +2(-1) k i ∂φ 2 ∂p = 2 ∂ ∂p γ φ 1 +(-1) k i φ 2
This gives (recall the definition of these forms)

∂ ∂α k,i;1 C(a k,i ) Q dz = 4πi , ∂ ∂α k,i;2 C(a k,i ) Q dz = -4π(-1) k ∂ ∂α k;1 A k Q dz = 2i , ∂ ∂α k;2 A k Q dz = -2(-1) k
and all other partial derivatives of these functions are zero. By lemma 3, the derivatives of φ 1 and φ 2 with respect to the parameter t k,i have a double pole at a k,i . Since φ 1 = dz and φ 2 = (-1) k+1 i dz in T k+1 , their principal parts at a k,i are respectively -dz/(za k,i ) 2 and (-1) k i dz/(za k,i ) 2 . This gives

∂ ∂t k,i C(a k,i ) (z -a k,i )Q dz = C(a k,i ) (z -a k,i ) 2φ 1 dz ∂φ 1 ∂t k,i + 2φ 2 dz ∂φ 2 ∂t k,i = -8πi
The derivative of this function with respect to any of the other parameters is zero (because the derivatives of φ 1 and φ 2 have at most simple poles).

The first statement of the proposition now follows from the implicit function theorem. Note that the map is only real analytic because all the α parameters are real. Since (φ 3 ) 2 = X( φ 3 ) 2 , with X = x 2 , we have

∂ ∂X C(a k,i ) Q dz = C(a k,i ) ( φ 3 ) 2 dz = 2πi Res a k,i ( φ 3 ) 2 dz ∂ ∂X C(a k,i ) (z -a k,i )Q dz = C(a k,i ) (z -a k,i )( φ 3 ) 2 dz = 2πi (α k,i;3 ) 2
This gives the indicated derivatives of α k,i;1 , α k,i;2 and t k,i with respect to X.

5.6

The period problem

φ 3 = 0 (1 ≤ k ≤ N ) Re B k,i φ 3 = 0 (1 ≤ k ≤ N , 2 ≤ i ≤ n k ) n1 i=1 α 1,i;3 = -1
Moreover, when x = 0, we have α k,i;3 = -c k where c k = 1/n k . Finally, for x = 0 close to 0, we have

Re B φ 3 ∼ -log(x 2 )(c 1 + • • • + c N ) (3)
The third equation is a normalization, see section 5.4.4. The reason for the minus sign is that given the geometric picture of the surface we want to construct and the orientation of A k,i , we expect that the flux of each cycle A k,i points downward, so the imaginary part of the period of φ 3 is negative.

Proof : First observe that φ 3 depends linearly on the parameters α k,i;3 and α k;3 , so we have to solve linear equations : no need to use the implicit function theorem here ! The period of φ 3 on the cycle B k depends analytically on x because we can choose a representative of B k which stays away from the nodes. We represent B k,i by the composition of the following paths : The integral of φ 3 on the first and third paths depends analytically on x. We use lemma 4 to estimate the integral on the second and fourth paths.

φ 3 = (α k,i;3 -α k,1;3 ) log(x 2 ) + analytic
Hence the function (log(x 2 )) -1 Re B k,i φ 3 extends continuously to x = 0 with value α k,i;3α k,1;3 . Proposition 5 boils down to the following statement : Claim 1 When x = 0, the following linear map is an isomorphism

( α k,i;3 1≤k≤N 2-δ k,1 ≤i≤n k , α k;3 1≤k≤N ) → (α k,i;3 -α k,1;3 1≤k≤N 2≤i≤n k , Re B k φ 3 1≤k≤N , n1 i=1 α 1,i;3 )
Proof : The domain and target spaces have the same dimension. Using equation (1), it is straightforward to check that the kernel is zero.

Remark 5

The function 1/ log(x 2 ) does not extend to a smooth function at 0, so the solutions to our equations are not smooth functions of x. To deal with this problem, we write x = exp(-1/ξ 2 ), which extends to a smooth function at ξ = 0. Then 1/ log(x 2 ) = -1 2 ξ 2 , so our equations and their solutions depend smoothly on the auxiliary variable ξ in a neighborhood of 0. Note however that we (sadly) leave the realm of analytic functions, but there seems to be no way to avoid this ! 5.6.2 Solution of the period problem for φ 1 and φ 2 For = 1, 2, 3, write T = (T ;1 , T ,2 , 0). Proposition 6 Assume the parameters t k,i , α k; and α k,i; ( = 1, 2, 3) are given as functions of x by propositions 4 and 5. For x close to 0, there exists (locally unique) values of the parameters τ k and b k,i (all indices) such that the following equations are satisfied

Re

B k φ = T 2; (1 ≤ k ≤ N , = 1, 2) Re B k,i φ = 0 (1 ≤ k ≤ N , 2 ≤ i ≤ n k , = 1, 2) Re B φ = T 3; ( = 1, 2)
Moreover, when x = 0, we have

τ k = T 2 if k is odd and τ k = T 2 if k is even, b k,i = a k,i if 1 ≤ k ≤ N -1, and b N,i = a N,i -T 3 .
Proof : We first solve the equations when x = 0 and then conclude using the implicit function theorem. When x = 0, we have φ 1 = dz and φ 2 = (-1) k i dz in T k . Hence

T 2;1 = Re B k φ 1 = Re τ k 0 dz = Re(τ k ) T 2;2 = Re B k φ 2 = Re τ k 0 (-1) k i dz = (-1) k+1 Im(τ k ) This determines τ k . Then 0 = Re B k,i φ 1 = Re a k,i a k,1 dz + b k,1 b k,i dz = Re(a k,i -a k,1 + b k,1 -b k,i ) 0 = Re B k,i φ 2 = Re a k,i a k,1 (-1) k i dz + b k,1 b k,i (-1) k+1 i dz = (-1) k+1 Im(a k,i -a k,1 -b k,1 + b k,i ) This gives a k,i -a k,1 = b k,i -b k,1 (1 ≤ k ≤ N , 2 ≤ i ≤ n k ) (4)
As explained in section 5.4.4, we may normalise translation by fixing one point in each

T k . For k = 1, • • • , N -1, we fix the value of b k,1 in T k+1 by asking that b k,1 = a k,1 . This normalizes translation in T 2 , • • • , T N . (We will normalise translation in T 1 later). Equation (4) gives b k,i = a k,i for 1 ≤ k ≤ N -1.
Computations entirely similar to the previous ones give

T 3;1 = Re B φ 1 = Re a1,1 bN,1 dz + N -1 k=1 a k+1,1 b k,1 dz = Re(a N,1 -b N,1 ) T 3;2 = Re B φ 2 = Im(-a N,1 -b N,1 )
This gives b N,1 = a N,1 -T 3 . Equation (4) gives b N,i = a N,i -T 3 .

To apply the implicit function theorem, we need to study the smoothness of the above functions. The integrals on the cycles B k are clearly analytic functions of x. By proposition 4, the periods of φ 1 and φ 2 on the circles C(a k,i ), namely the parameters α k,i;1 and α k,i;2 , are of order x 2 . By lemma 4, their periods on the cycles B k,i are analytic functions plus terms of order x 2 log t k,i = O(x 2 log(x 2 )). This extends to a smooth function of the auxiliary parameter ξ at ξ = 0, see remark 5. Hence the periods are smooth functions of ξ and the other parameters. The partial differential with respect to the parameters b k,i and τ k is clearly an isomorphism. The proposition then follows from the implicit function theorem at ξ = 0.

The balancing condition

It remains to solve the last equation in lemma 2.

Proposition 7 Assume the parameters t k,i , α k;1 , α k;2 , α k,i;1 and α k,i;2 are given as analytic functions of X = x 2 by proposition 4. Let

f k,i = 1 2πi C(b k,i ) Q dz (1 ≤ k ≤ N , 1 + δ k,N ≤ i ≤ n k )
Then f k,i = x -2 f k,i extends analytically to x = 0 and its value at x = 0 is

f k,i (0) = R k,i if 1 ≤ k ≤ N , 2 -δ k,1 ≤ i ≤ n k f k,1 (0) = R k,1 + k-1 =1 n i=1 ϕ k-(R ,i ) if 2 ≤ k ≤ N -1
where ϕ(z) = z denotes conjugation in C and

R k,i = ϕ Res a k,i ( φ 3 ) 2 dz + Res b k,i ( φ 3 ) 2 dz
Proof : If f (z) is an analytical function of z such that f (0) = 0 then f (z)/z extends analytically to z = 0 with value f (0). So all we have to do is to compute the derivative of f k,i with respect to X at X = 0. We have

∂f k,i ∂α k,i;1 = -2 , ∂f k,i ∂α k,i;2 = 2(-1) k i
and the derivatives of f k,i with respect to the other parameters (except X) are zero. Hence, by the chain rule,

∂f k,i ∂X = -2 ∂α k,i;1 ∂X + 2(-1) k i ∂α k,i;2 ∂X + Res b k,i ( φ 3 ) 2 dz
When 2δ k,1 ≤ i ≤ n k , using the last formula in proposition 4, this gives ∂f k,i /∂X = R k,i . To compute the derivative of f k,1 for 2 ≤ k ≤ N -1, we use equation ( 1) to obtain (the α terms cancel)

n k i=1 ∂f k,i ∂X - n k-1 i=1 ϕ ∂f k-1,i ∂X = n k i=1 Res b k,i ( φ 3 ) 2 dz - n k-1 i=1 ϕ Res b k-1,i ( φ 3 ) 2 dz = n k i=1 Res b k,i ( φ 3 ) 2 dz + n k i=1 ϕ Res a k,i ( φ 3 ) 2 dz = n k i=1 R k,i
In the second line, we have used the residue theorem in T k . The result follows by induction on k.

Assume the parameters t k,i , α k; , α k,i; ( = 1, 2, 3) and b k,i are given as functions of x and the remaining parameters (namely a k,i ) by proposition 4, 5 and 6.

Proposition 8 For x small enough, there exists (locally unique) values of the parameters a k,i such that f k,i = 0 for 1

≤ k ≤ N , 1 + δ k,N ≤ i ≤ n k .
This determines the value of all parameters as functions of x so that the period problem is solved and φ 2 1 + φ 2 2 + φ 2 3 = 0. The solution depends smoothly on the auxiliary parameter ξ (see remark 5). We also still have T 1 , T 2 and T 3 as free parameters. The solution depends smoothly on these parameters.

Proof : First assume that x = 0. Given the value of τ k found in proposition 6, we have T k = ϕ k+1 (T) where T = C /Γ and ϕ(z) = z. Let a k,i = ϕ k+1 (p k,i ) where p k,i is the given configuration. Then from proposition 6, b k,i = ϕ k (p k,i ) for 1 ≤ k ≤ N -1 and b N,i = p N,i -T 3 = p 0,i . It follows that

φ 3 = ω k in T k if k odd ϕ * ω k in T k if k even
where ω k is the meromorphic 1-form on T defined in section 2.2. (Indeed, these forms are holomorphic and have the same poles and residues, and both have imaginary periods). This implies that

R k,i = 2ϕ k (F k,i )
where F k,i is the force defined in the section 2.2. Hence since the configuration is balanced, we have f k,i = 0. We may normalise translation in T 1 by fixing the position of p 1,1 . Then since the configuration is non-degenerate, it follows from elementary linear algebra that the partial differential of ( f k,i : 1 ≤ k ≤ N , 1 + δ k,N ≤ i ≤ n k ) with respect to (p ,j : 1 ≤ k ≤ N , 1 + δ k,1 ≤ i ≤ n k ) at x = 0 is an (R linear) isomorphism. The conclusion follows by the implicit function theorem again.

Embeddedness

At this point, we have constructed a one parameter family of conformal, minimal immersions from Σ t(x) to R 3 /Λ x , x in a neighborhood of 0, where Λ x is the lattice generated by the periods (T 1 , 0), (T 2 , 0), (T 3 , ε(x)) and by equation (3) in proposition 5, ε(x) -x 2 log(x 2 )(c

1 + • • • + c N )
Here we assume that x > 0, as changing x into -x only changes the sign of the third coordinate. We need to prove that the immersion is regular (free of branch points) and an embedding. Given a small δ > 0, let Ω k,δ be the set of points in T k which are at distance greater than δ from the points a k,i and b k-1,i (all indices i), and

Ω δ = Ω 1,δ ∪ • • • ∪ Ω N,δ .
The immersion is regular if

|φ 1 | 2 + |φ 2 | 2 + |φ 3 | 2 > 0.
It is straightforward to check that this is true on Ω δ because we know explicitly the limits of these forms on this domain. The problem is to prove that this holds on the necks. We prove that φ 3 has no zeroes on the necks. The zeros of φ 3 are the same as the zeros of φ 3 . When x = 0, φ 3 has n k + n k-1 poles in T k , so it has n k + n k-1 zeros, which lie in Ω k,δ provided δ is small enough. By continuity, for x close to 0, φ 3 has n k + n k-1 zeros in Ω k,δ . This gives a total of 2n zeros in Ω δ , where n = n 1 + • • • + n N . But if x = 0, φ 3 is a holomorphic 1-form on a compact Riemann surface of genus g = n + 1 so it has 2g -2 = 2n zeros. This means that φ 3 has no further zeros and proves that the surface is regular.

We now prove embeddedness. Let X x = (X 1 x , X 2 x , X 3 x ) be the immersion given by the Weierstrass representation we have constructed, and let X x = (X 1

x , X 2 x , x -1 X 3 x ). We will prove that the image of X x is embedded, and compute explicitely its limit (after suitable vertical translation) when x → 0.

First observe that on each domain Ω k,δ , the Gauss map converges to (0, 0, -1) if k is even and (0, 0, 1) if k is odd, hence its image is locally a graph. Then for z ∈ Ω k,δ we have, up to translation, lim where f k (z) = Re z ω k . This is a well defined function of z ∈ T because the residues of ω k are real. It has logarithmic singularities at the poles of ω k . Hence the image X x (Ω δ ) converges (up to translation) when x → 0 to the graph of f k on T minus disks of radius δ around the singularities. So it is included in a slab whose width is bounded by some constant C(δ). By our computation in section 5.6.1, the distance between consecutive slabs has order O(log(x 2 )), so these slabs are disjoint for x small enough.

It remains to understand the behavior of X x on the necks. There exists c > 0 such that for each k, the horizontal sections x 3 = c (resp. x 3 = -c) of the graph of f k consists of n k (resp. n k-1 ) disjoint convex curves. Hence for x small enough, we may find numbers δ + k and δ - k (depending on x), with δ - k < δ + k < δ - k+1 , such that the intersection of the surface with the slab δ - k < x 3 < δ + k is bounded by convex horizontal curves, and is a graph, and the intersection with the slab δ + k < x 3 < δ - k+1 consists of n k annuli, each bounded by two horizontal convex curves. By a theorem of Schiffman [START_REF] Shiffman | On surfaces of stationary area bounded by two circles, or convex curves, in parallel planes[END_REF], a minimal annulus bounded by two horizontal convex curves is fibered by horizontal convex curves. It follows that the surface is embedded. This concludes the proof of Theorem 4.

Figure 2 :

 2 Figure 2: A simply periodic configuration with N = 2, n 1 = 4 and n 2 = 1. Points at level 0, 1 and 2 are represented respectively by a black dot, a star, and a white dot. The configuration extends periodically.

Figure 3 :

 3 Figure 3: An element in the corresponding family of minimal surfaces. Computer image made by the author using J. Hoffman MESH software.

Figure 4 :

 4 Figure 4: A numerically computed triply periodic configuration with N = 2, n 1 = 4 and n 2 = 1, obtained by perturbation of the simply periodic configuration of figure 2.

Figure 5 :

 5 Figure 5: A simply periodic configuration with N = 4, n 1 = 2, n 2 = 1, n 3 = 3 and n 4 = 1.

Figure 6 :

 6 Figure 6: Left : definition of Σ t in the case N = 2, n 1 = 3, n 2 = 1. Right : topological model of Σ t . The top circle is identified with the bottom one.

Figure 7 :

 7 Figure 7: Homology basis (genus 5)

  1. a path from a k,1 + to a k,i + in T k , (depending continuously on parameters and avoiding nodes), 2. a path from v k,i = to w k,i = (going through the neck), 3. a path from b k,i + to b k,1 + in T k+1 , 4. a path from w k,1 = to v k,1 = .

x→0 X 1 x

 1 (z) + i X 2 x (z) = Re z dz + i Re z (-1) k i dz = ϕ k+1 (z)where as before ϕ(z) = z, andlim x→0 X 3 x (z) = Re z (ϕ k+1 ) * ω k = f k (ϕ k+1 (z))

  5.6.1 Solution of the Period problem for φ 3Proposition 5 Assume the parameters t k,i are given as functions of x by proposition 4. Then for x close to 0, there exists (locally unique) values of the parameters α k;3 , α k,i;3 (all indices) such that the following equations are satisfied

	Re
	B k

Re

The period problem for the A-curves will be automatically solved by definition of φ 1 , φ 2 , φ 3 . The period problem for the B-curves is a system of equations we will have to solve. It is not necessary to solve the last equation Re B φ 3 = ε, we will use it to define ε.

Using lemma 1, we define three regular 1-forms φ 1 , φ 2 and φ 3 by prescribing the following periods :

All the α numbers in the right hand side are real parameters. The reason for the 2πi factor is that we will interpret the parameters α k,i; as residues. We define φ 3 = x φ 3 , where x is a real number in a neighborhood of 0. When x = 0, φ 3 = 0 so the surface will be locally a flat horizontal plane : this is the limit case. When x = 0, we want to adjust the other parameters, so that the period problem is solved and φ 2 1 + φ 2 2 + φ 2 3 = 0. We will do this using the implicit function theorem at x = 0. We will prove regularity and embeddedness in section 5.8.

Note that we do not prescribe the periods of φ 1 , φ 2 , φ 3 on the circles C(a 2,1 ), • • • , C(a N,1 ). It will be convenient to define

These can be expressed as functions of the other parameters :

To see this, let ω be a holomorphic 1-form on Σ t . Using Cauchy theorem in the domain of T k bounded by the circles C(a k,i ) and C(b k-1,i ), we have


Sebastien Elis

Hayden-William Courtland

Yingjie Wu

Hui Sun

Valerie Williams

Karl Jepsen

Shoshana Yakar

Elevated levels of serum IGF-1 restore peak bone properties in the absence of tissue IGF-1 and enhance bone properties in its presence Elevated level of endocrine IGF-1 restore the severe bone phenotype of IGF-1 null mice at 8 weeks of age. Nevertheless, autocrine/paracine IGF-1 levels are important to establish bone size early postnatally

The growth hormone (GH)/Insulin-like growth factor-1 (IGF-1) axis has major effects on skeletal growth and integrity. Studies of skeletal structure in the IGF-1 and GH receptor (GHR) null mice showed decreases in bone length, total cross-sectional bone area and cortical bone area. However, the IGF-1 null mice exhibit greater impairment in bone accretion than GHR or GH null mice, suggesting GH-independent effect of IGF-1 during postnatal growth. Moreover, while GH deficiency in human and mice does not affect birth size, mutations in the igf-1 or the igf-1 receptor (igf-1r) genes lead to severe intrauterine and postnatal growth retardation. IGF-1 acts on the growing skeleton in an endocrine and autocrine/paracrine manner. In our previous study with the liver-IGF-1 deficient (LID) mice, which exhibit 75% reductions in serum IGF-1 and normal skeletal IGF-1 expression, we found that serum IGF-1 regulates periosteal bone growth, determines bone size and by inference bone strength.

This study aimed to test whether elevated levels of IGF-1 in serum can rescue the severe growth phenotype of IGF-1 null mice.

We performed longitudinal analyses of three mouse models: 1) control mice, which express normal levels of autocrine/paracrine and endocrine IGF-1; 2) Hepatic IGF-1 transgenic (HIT) mice, which express normal levels of autocrine/paracrine IGF-1 but overexpress endocrine IGF-1 and 3) IGF-1 null mice, which do not express autocrine/paracrine IGF-1 but overexpress endocrine IGF-1 (KO-HIT) Increased level of IGF-1 lead to a bigger bone and more robust phenotype in HIT mice, by affecting bone size (Tt.Ar, Ct.Ar, Le) but also mineralization (increased TMD). Those modifications lead to stronger bones, as showed by the mechanical testing (stiffer bones and increased max load). HIT mice reach a plateau of the morphological bone parameters at 8 weeks old. Nevertheless, mineralization and thus strength, continue to increase till 16 weeks old. KO-HIT mice exhibit a delay in bone growth, as they have smaller bones at 4 weeks old (Tt.Ar, Ct.Ar, Ma.Ar, Le) and thus, probably weaker bones as suggested by their slenderness and J o (10 and 23% decrease, respectively). Increased endocrine IGF-1 lead to a catch up of the bone at 8 weeks old on most of the morphological parameters. This is explained by an increase in growth rate observed in both HIT in KO-HIT. KO-HIT mice, as oppose to HIT mice and control, keep a high growth rate between 8 and 16 weeks of age, so that they reach an intermediate phenotype between control and HIT by 16 weeks of age. High level of IGF-1 can lead to increase in mineralization but only if the autacrine/paracrine IGF-1 is at a normal level, as suggested by the absence of difference in TMD between KO-HIT and control mice. The longitudinal modification of the bone also suggest that increased endocrine IGF-1 lead to an increase in size of the bone.

ACKNOLEDGMENTS : National Institutes of Health NIH AR0549-19-01

  KO-HIT are smaller, but then they caught up body weight, length and bone length by 8 w At 4w, all KO-HIT bone compartment are smaller, but then they caught up by 8w, and have increased Ct.Ar and Ct.Th by 16w Between 4 and 8w and between 8 and 16w, KO-HIT have higher growth rate of Tt.Ar and Ct.Ar than control. KO-HIT is the only group that keep a high growth rate between 8 and 16w.Slenderness is defined by Tt.Ar/Le, such as less slender bones (more robust bones) have greater TtAr/Le ratio. The polar moment of inertia (Jo) is a measure of tissue distribution from the center of the bones and gives and indication of resisteance to bending. Tissue mineral density (TMD), which is the average mineral value of the bone voxels only and is expressed in hydroxyapatite (HA) density equivalents.


Rouba Borghol
email: borghol@univ-tours.fr

Laurent Véron
email: veronl@lmpt.univ-tours.fr

Boundary singularities of N -harmonic functions *

Introduction

Let Ω be a domain is R N (N ≥ 2) with a C 2 compact boundary ∂Ω.

A function u ∈ W 1,p loc (Ω) is p-harmonic if Ω |Du| p-2 Du, Dφ dx = 0 (1.1)
for any φ ∈ C 1 0 (Ω). Such functions are locally C 1,α for some α ∈ (0, 1). In the case p = N , the function u is called N -harmonic. The N -harmonic functions play an important role as a natural extension of classical harmonic functions. They also appear in the theory of bounded distortion mappings [START_REF] Rešetnjak | Spatial mappings with bounded distortion (Russian)[END_REF]. One of the main properties of the class of N -harmonic functions is its invariance by conformal transformations of the space R N . This article is devoted to the study of N -harmonic functions which admit an isolated boundary singularity. More precisely, let a ∈ ∂Ω and u ∈ W 1,N loc (Ω) ∩ C(Ω \ {a}) be a N -harmonic function vanishing on ∂Ω \ {a}, then u may develop a singularity at the point a. Our goal is to show the existence of such singular solutions, and then to classify all the positive N -harmonic functions with a boundary isolated singularity. We denote by n a the outward normal unit vector to Ω at a The main result we prove are presented below:

There exists a unique positive N -harmonic function u = u 1,a in Ω, vanishing on ∂Ω \ {a} such that lim

x → a x-a |x-a| → σ |x -a| u(x) = -σ, n a (1.2)
uniformly on S N -1 ∩ Ω = {σ ∈ S N -1 : σ, n a < 0}.

The functions u 1,a plays a fundamental role in the description of all the positive singular N -harmonic functions since we the next result holds Let u be a positive N -harmonic function in Ω, vanishing on ∂Ω \ {a}. Then there exists k ≥ 0 such that u = ku 1,a .

(1.3)

When u is no longer assumed to be positive we obtain some classification results provided its growth is limited as shows the following

Let u be a N -harmonic function in Ω, vanishing on ∂Ω \ {a} and verifying

|u| ≤ M u 1,a ,
for some M ≥ 0. Then there exists k ∈ R such that u = ku 1,a .

(1.4)

In the last section we give a process to construct p-harmonic regular functions (p > 1) or N -harmonic singular functions as product of one variable functions. Starting from the existence of p-harmonic functions in the plane under the form u(x) = u(r, σ) = r β ω(θ) (see [START_REF] Krol | The behavior of the solutions of a certain quasilinear equation near zero cusps of the boundary[END_REF]), our method, by induction on N , allows us to produce separable solutions of the spherical p-harmonic spectral equation

-div σ β 2 v 2 + |∇ σ v| 2 (p-2)/2 ∇ σ v = λ N,β β 2 v 2 + |∇ σ v| 2 (p-2)/2 v. (1.5) on S N -1 , where λ N,β = β (N -1 + (β -1)(p -1)
). This equation equation is naturally associated to the existence of p-harmonic functions under the form u(x) = |x| β v(x/ |x|). As a consequence, we express p-harmonic functions under the form of a product of N -explicit functions of one real variable. If we represent R N as the set of {x = (x 1 , ..., x N )} where

x 1 = r sin θ N -1 sin θ N -2 ... sin θ 2 sin θ 1 , x 2 = r sin θ N -1 sin θ N -2 ... sin θ 2 cos θ 1 , ..., x N -1 = r sin θ N -1 cos θ N -2 and x N = r cos θ N -1 with θ 1 ∈ [0, 2π] and θ k ∈ [0, π],
for k = 2, ..., N -1, then, for any integer k the function

u(x) = (r sin θ N -1 sin θ N -2 ... sin θ 2 ) β k ω k (θ 1 ) (1.6)
is p-harmonic in R N , in which expression β k > 1 is an algebraic number depending on k and ω k is a π/k-antiperiodic solutions of a completely integrable homogeneous differential equation. Moreover N -harmonic singular functions are also obtained under the form

u(x) = r -β k (sin θ N -1 sin θ N -2 ... sin θ 2 ) β k ω k (θ 1 ). (1.7)
Our paper is organized as follows: 1-Introduction. 2-Construction of fundamental singular N -harmonic functions. 3-The classification theorem. 4-Separable solutions of the p-harmonic spectral problem.

Construction of fundamental singular N -harmonic functions

We denote by H N the group of conformal transformations in R N . This group is generated by homothethies, inversion and isometries. Our first result is classical, but we repeat the proof for the sake on completeness.

Proposition 2.1 Let u be a N -harmonic function in a domain G ⊂ R N and h ∈ H N . Then u h = u • h is N -harmonic in h -1 (G).
Proof. Because for any p > 1 the class of p-harmonic functions is invariant by homothethies and isometries, it is sufficient to prove the result if h is the inversion I 1 0 with center the origin in R N and power 1. We set y = I 1 0 (x) and v(y) = u(x). For any i = 1, ..., N

u xi (x) = j δ ij |x| -2 -2 |x| -4 x i x j v yj (y). Then |Du| 2 (x) = |x| -4 |Dv| 2 (y) = |y| 4 |Dv| 2 (y).
If φ is a test function, we denote similarly ψ(y) = φ(x), thus

Du, Dφ = |x| -4 Dv, Dψ = |y| 4 Dv, Dψ ,
and

G |Du| N -2 Du, Dφ dx = I 1 0 (G) |y| 2N |Dv| N -2 Dv, Dψ DI 1 0 dy Because DI 1 0 = |det(∂x i /∂y j )| = |y| -2N
, the result follows.

Proposition 2.2 Let N ≥ 2, B = B 1 (0) and a ∈ ∂B. Then there exists a unique positive N -harmonic function U i in B which vanishes on ∂B \ {a} and satisfies

U i (x) = 1 -|x| |x -a| 2 (1 + •(1)) as x → a. (2.1)
Proof. We first observe that the coordinates functions x i are N -harmonic and positive in the half-space H i = {x ∈ R N : x i > 0} and vanishes on ∂H i . Therefore, the functions χ i (x) = x i / |x| 2 are also N -harmonic and singular at 0. Without loss of generality we can assume that a is the origin of coordinates, and that B is the ball with radius 1 and center (-1, 0, ..., 0). Let ω be the point with coordinates (-2, 0, ..., 0). By the inversion I 4 ω , a is invariant and B is transformed into the half space H 1 . Since χ 1 is N -harmonic in H 1 , the function

x → χ 1 • I 4 ω (x) = - |x| 2 + 2x 1 2 |x| 2
is N -harmonic and positive in B = {x : |x| 2 + 2x 1 < 0}, vanishes on ∂B and is singular at

x = 0. If we set x ′ 1 = x 1 + 1, x ′ i = x i for i = 2, ..., N and U i (x ′ ) = χ 1 • I 4 ω (x)
, then the x ′ coordinates of a are (1, 0, ..., 0) and

U i (x ′ ) = 1 -|x ′ | 2 2 |x ′ -a| 2 = 1 -|x ′ | |x ′ -a| 2 (1 + •(1)) as x ′ → a.
Let Ũ i be another positive N -harmonic function in B which verifies (2.1 ) and vanishes on ∂B \ {a}. Thus, for any δ > 0, (1 + δ) Ũ i , is positive, N -harmonic, and U i -(1 + δ) Ũ i is negative near a. By the maximum principle, U i ≤ (1 + δ) Ũ i . Letting δ → 0, and permuting U i and Ũ i yields Ũ i = U i .

By performing the inversion I 1 0 , we derive the dual result

Proposition 2.3 Let N ≥ 2, G = B c 1 (0)
and a ∈ ∂B. Then there exists a unique positive N -harmonic function U e in G which vanishes on ∂B \ {a} and satisfies

U e (x) = •(ln |x|) as |x| → ∞, (2.2)
and

U e (x) = |x| -1 |x -a| 2 (1 + •(1)) as x → a. (2.3)
Proof.

The assumption (2.2 ) implies that the function

U = U e • I 1 0 , which is N -harmonic in B \ {0} verifies U (x) = •(ln(1/ |x|)) near 0.
By [START_REF] Serrin | Local behaviour of solutions of quasilinear equations[END_REF], 0 is a removable singularity and thus U can be extended as a positive N -harmonic function in B which satisfies (2.1 ). This implies the claim.

We denote by ρ(x) the signed distance from x to ∂Ω. Since ∂Ω is C 2 , there exists

β 0 > 0 such that if x ∈ R N verifies -β 0 ≤ ρ(x) ≤ β 0 , there exists a unique ξ x ∈ ∂Ω such that |x -ξ x | = | ρ(x)|. Furthermore, if ν ξx is the outward unit vector to ∂Ω at ξ x , x = ξ x -ρ(x)ν ξx .
In particular ξ x -ρ(x)ν ξx and ξ x + ρ(x)ν ξx have the same orthogonal projection ξ x onto ∂Ω.

Let T β0 (Ω) = {x ∈ R N : -β 0 ≤ ρ(x) ≤ β 0 }, then the mapping Π : [-β 0 , β 0 ] × ∂Ω → T β0 (Ω) defined by Π(ρ, ξ) = ξ -ρν(ξ) is a C 2 diffeomorphism. Moreover DΠ(0, ξ)(1, e) = e -ν ξ for any e belonging to the tangent space T ξ (∂Ω) to ∂Ω at ξ. If x ∈ T β0 (Ω), we define the reflection of x through ∂Ω by ψ(x) = ξ x + ρ(x)ν ξx . Clearly ψ is an involutive diffeomorphism from Ω ∩ T β0 (Ω) to Ω c ∩ T β0 (Ω), and Dψ(x) = I for any x ∈ ∂Ω. If a function v is defined in Ω ∩ T β0 (Ω), we define ṽ in T β0 (Ω) by ṽ(x) = v(x) if x ∈ Ω ∩ T β0 (Ω) -v • ψ(x) if x ∈ Ω c ∩ T β0 (Ω). (2.4) Lemma 2.4 Assume that 0 ∈ ∂Ω. Let v ∈ C 1,α (Ω ∩ T β0 (Ω) \ {0}) be a solution of (1.1 ) in Ω ∩ T β0 (Ω) vanishing on ∂Ω \ {0}. Then ṽ ∈ C 1,α (T β (Ω) \ {0}) is solution of a quasilinear equation j ∂ ∂x j Ãj (x, Dṽ) = 0 (2.5) in T β (Ω) \ {0}
where the Ãj are C 1 functions defined in T β (Ω) where they verify

                 (i) Ãj (x, 0) = 0 (ii) i,j ∂ Ãj ∂η i (x, η)ξ i ξ j ≥ Γ |η| p-2 |ξ| 2 (iii) i,j ∂ Ãj ∂η i (x, η) ≤ Γ |η| p-2 (2.6) for all x ∈ T β (Ω) \ {0} for some β ∈ (0, β 0 ], η ∈ R N , ξ ∈ R N and some Γ > 0.
Proof. The assumptions (2.6 ) implies that weak solutions of (2.5 ) are C 1,α , for some α > 0 [START_REF] Tolksdorff | Regularity for a more general class of quasilinear elliptic equations[END_REF] and satisfy the standard a priori estimates. As it is defined the function ṽ is clearly

C 1 in T β0 (Ω) \ {0}. Writing Dv(x) = -D(ṽ • ψ(x)) = -Dψ(x)(Dṽ(ψ(x))) and x = ψ(x) = ψ -1 (x) Ω∩T β (Ω) |Dv| p-2 Dv.Dζdx = Ω c ∩T β (Ω)
|Dψ(Dṽ)| p-2 Dψ(Dṽ).Dψ(Dζ) |Dψ| dx.

But

Dψ(Dṽ).Dψ(Dζ)

= k i ∂ψ i ∂x k ∂ṽ ∂x i   j ∂ψ j ∂x k ∂ζ ∂x j   = j   i,k ∂ψ i ∂x k ∂ψ j ∂x k ∂ṽ ∂x i   ∂ζ ∂x j .
We set b(x) = |Dψ|,

A j (x, η) = |Dψ| |Dψ(η)| p-2 i k ∂ψ i ∂x k ∂ψ j ∂x k η i , (2.7)
and

A(x, η) = (A 1 (x, η), ..., A N (x, η)) = |Dψ| |Dψ(η)| p-2 (Dψ) t Dψ(η). (2.8)
For any ξ ∈ ∂Ω, the mapping Dψ ∂Ω (ξ) is the symmetry with respect to the hyperplane T ξ (∂Ω) tangent to ∂Ω at ξ, so |Dψ(ξ)| = 1. Inasmuch Dψ is continuous, a lengthy but standard computation leads to the existence of some β ∈ (0, β 0 ] such that (2.6 ) holds in T β (Ω)∩Ω c . If we define à to be |η| p-2 η on T β (Ω)∩Ω and A on T β (Ω)∩Ω c , then inequalities (2.6 ) are satisfied in T β (Ω).

These three results allows us to prove our main result Theorem 2.5 Let Ω be an open subset of R N with a compact C 2 boundary, ρ(x) = dist (x, ∂Ω) and a ∈ ∂Ω. Then there exists one and only one positive N -harmonic function u in Ω, vanishing on ∂Ω \ {a} verifying

lim x → a x-a |x-a| → σ |x -a| u(x) = -σ, n a (2.9)
uniformly on S N -1 ∩ Ω, and

u(x) = •(ln |x|)) as |x| → ∞, (2.10)
if Ω is not bounded.

Proof. Uniqueness follows from (2.9 ) by the same technique as in the previous propositions.

Step 1 (Existence). If Ω is not bounded, we perform an inversion

I |m-a| 2 m
with center some m ∈ Ω. Because of (2.10 ), the new function u

• I |m-a| 2 m is N -harmonic in Ω ′ = I |m-a| 2 m
(Ω) and satisfies (2.9 ). Thus we are reduced to the case were Ω is bounded. Since Ω is C 2 , it satisfies the interior and exterior sphere condition at a. By dilating Ω, we can assume that the exterior and interior tangent spheres at a have radius 1. We denote them by B 1 (ω e ) and B 1 (ω i ), their respective centers being ω i = an a and ω e = a + n a . We set V i (x) = U i (x -ω i ) and V e (x) = U e (x -ω e ) where U i and U e are the two singular Nharmonic functions described in Proposition 2.2 and Proposition 2.3, respectively in B 1 (ω i ) and B c 1 (ω e ), with singularity at point a. For ǫ > 0, we put

Ω ǫ = Ω \ B ǫ (a), Σ ǫ = Ω ∩ ∂B ǫ (a) and ∂ * Ω ǫ = ∂Ω ∩ B c ǫ (a). Let u ǫ be the solution of    div(|Du ǫ | N -2 Du ǫ ) = 0 in Ω ǫ u ǫ = 0 on ∂ * Ω ǫ u ǫ = V e on Σ ǫ . (2.11)
This solution is obtained classicaly by minimisation of a convex functional over a class of functions with prescribed boudary value on ∂Ω ǫ . For any x ∈ B 1 (ω i ), there holds

dist (x, ∂B 1 (ω e )) = |x -ω e | -1 ≥ dist (x, ∂Ω) ≥ dist (x, ∂B 1 (ω i )) = 1 -x -ω i . thus V i (x) ≤ V e (x) ∀x ∈ B 1 (ω i ),
by using (2.1 ), (2.3 ) and the maximum principle. Therefore

V i (x) ≤ u ǫ (x) ≤ V e (x) ∀x ∈ B 1 (ω i ) ∩ Ω ǫ and u ǫ (x) ≤ V e (x) ∀x ∈ Ω ǫ .
Finally, for 0 < ǫ ′ < ǫ, u ǫ ′ Σǫ ≤ V e Σǫ = u ǫ Σǫ . Thus

u ǫ ′ (x) ≤ u ǫ (x) ∀x ∈ Ω ǫ .
The sequence {u ǫ } is increasing with ǫ. By classical a priori estimates concerning quasilinear equations, it converges to some positive N -harmonic function u in Ω which vanishes on ∂Ω \ {a} and verifies

V i (x) ≤ u(x) ∀x ∈ B 1 (ω i ), and u(x) ≤ U e (x) ∀x ∈ Ω. This implies 1 -|x -ω i | 2 2 |x -a| 2 ≤ u(x) ∀x ∈ B 1 (ω i ), (2.12
)

u(x) ≤ |x -ω e | 2 -1 2 |x -a| 2 ∀x ∈ Ω, (2.13)
By scaling we can prove the following estimate

u(x) ≤ C ρ(x) |x -a| 2 ∀x ∈ Ω. (2.14)
for some C > 0: for simplicity we can assume that a is the origin of coordinates and, for r > 0 set u r (y) = u(ry). Clearly u r is N -harmonic in Ω/r and max{|Du r (y)| :

y ∈ Ω/r ∩ (B 3/2 \ B 2/3 )} ≤ C max{|u r (z)| : z ∈ Ω/r ∩ (B 2 \ B 1/2 )},
where C, which depends on the curvature of ∂Ω/r, remains bounded as long as r ≤ 1. Since Du r (y) = rDu(ry), we obtain by taking ry = x, |y| = 1 and using (2.13 ) with general a,

|Du(x)| ≤ C |x -a| -2
. By the mean value theorem, since u vanishes on ∂Ω \ {a}, (2.14 ) holds.

Step 2. In order to give a simple proof of the estimate (2.9 ), we fix the origin of coordinates at a = 0 and the normal outward unit vector at a to be -e N . If ũ is the extension of u by reflection through ∂Ω, it statisfies (2.5 ) in T β (Ω)\{0} (see lemma 2.4 ). For r > 0, set ũr (x) = rũ(rx). Then ũr is solution of For 0 < a < b fixed and for some 0

< r 0 ≤ 1 the spherical shall Γ a,b = {x ∈ R N : a ≤ |x| ≤ b} is included into T β/r (Ω/r
) for all 0 < r ≤ r 0 . By the classical regularity theory for quasilinear equations [START_REF] Tolksdorff | Regularity for a more general class of quasilinear elliptic equations[END_REF] and lemma 2.4 , there holds

Dũ r C α (Γ 2/3,3/2 ) ≤ C r ũr L ∞ (Γ 1/2,2 ) ,
where C r remains bounded because r ≤ 1. By Ascoli's theorem, (2.12 ) and (2.14 ), ũr (x) converges to x N |x| -2 in the C 1 (Γ 2/3,3/2 )-topology. This implies in particular that r 2 Dũ(rx) converges uniformly in Γ 2/3,3/2 to -2x N |x| -4 x + |x| -2 e N . Using the expression of Dũ in spherical coordinates we obtain r 2 ũr i -rũ φ e + r sin φ ∇ σ ′ ũ → -2σ N i + e N uniformly on S N -1 as r → 0, where cos φ = x N |x| -1 , i = x/|x|, e is derived from x/|x| by a rotation with angle π/2 in the plane 0, x, N (N being the North pole), and ∇ σ ′ is the covariant gradient on S N -2 . Inasmuch i, e and ∇ σ ′ are orthogonal, the components of e N are cos φ, sin φ and 0, thus

rũ φ (r, σ ′ , φ) → -sin φ as r → 0. Since ũ(r, σ ′ , φ) = φ π/2 ũφ (r, σ ′ , θ)dθ,
the previous convergence estimate establishes (2.9 ).

Definition 2.6

We shall denote by u 1,a the unique positive N -harmonic function satisfying (2.9 ), and call it the fundamental solution with a point singularity at a.

The classification theorem

In this section we characterize all the positive N -harmonic functions vanishing on the boundary of a domain except one point. The next statement is an immediate consequence of Theorem 2.5 and [2, Th. 2.11].

Theorem 3.1 . Let Ω be a bounded domain with a C 2 boundary and a ∈ ∂Ω. If u is a positive N -harmonic function in Ω vanishing on ∂Ω \ {a}, there exists M ≥ 0 such that

u(x) ≤ M u 1,a (x) ∀x ∈ Ω (3.1)
In the next theorem, which extends [2, Th. 2.13], we characterize all the signed Nharmonic functions with a moderate growth near the singular point. Proof. We define k as the minimum of the ℓ such that u ≤ ℓu 1,a in Ω. Without any loss of generality we can assume k > 0. Then either the tangency of the graphs of the functions u and ku 1,a is achieved in Ω \ {a}, or it is achieved asymptotically at the singular point a. In the first case we considered two sub-cases: (i) The coincidence set G of u and ku 1,a has a connected component ω isolated in Ω. In this case there exists a smooth domain U such that ω ⊂ U and δ > 0 such that ku 1,a -u ≥ δ on ∂U. The maximum principle implies that ku 1,a -u ≥ δ in U, a contradiction. (ii) In the second sub-case any connected component ω of the coincidence set touches ∂Ω\{a}, or the two graphs admits a tangency point on ∂Ω \ {a}. If m ∈ ω ∩ ∂Ω \ {a} or is such a tangency point, the regularity theory implies ∂u(m)/∂n m = ku 1,a (m)/∂n m . By Hopf boundary lemma, u 1,a (m)/∂n m < 0. By the mean value theorem, the function w = ku 1,a -u satisfies an equation

Lw = 0 (3.2)
which is elliptic and non degenerate near m (see [START_REF] Friedman | Singular solutions of some quasilinear elliptic equations[END_REF], [START_REF] Kichenassamy | Singular solutions of the p-Laplace equation[END_REF]), it follows that w vanishes in a neighborhood of m and the two graphs cannot be tangent only on ∂Ω \ {a}. Assuming that ω = Ω, let x 0 ∈ Ω \ ω such that dist (x 0 , ω) = r 0 < ρ(x 0 ) = dist (x 0 , ∂Ω, and let y 0 ∈ ω be such that |x 0 -y 0 | = r 0 . Since u 1,a has at most a finite number of critical points, we can choose x 0 such that y 0 is not one of these critical points. By assumption w = ku 1,a -u is positive in B r0 (x 0 ) and vanishes at a boundary point y 0 . Since the equations are not degenerate at y 0 there holds k∂u 1,a (y 0 )/∂ν -∂u(y 0 )/∂ν < 0 where ν = (y 0 -x 0 )/r 0 , which contradicts the fact that the two graphs are tangent at y 0 .

Next we are reduced to the case where the graphs of u and ku 1,a are separated in Ω and asymptotically tangent at the singular point a. There exists a sequence {ξ n } ⊂ Ω such that lim n→∞ u(ξ n )/u 1,a (ξ n ) = k. We set |x n -a| = r n , u n (y) = r n u(a + r n y) and v n (y) = r n u 1,a (a + r n y). Both u n and v n are N -harmonic in Ω n = (Ω -a)/r n . The functions u n and v n are locally uniformly bounded in Ω n \ {0}. It follows, by using classical regularity results, that, there exists sub-sequences, such that {u n k } and {v n k } converge respectively to U and V in the C 1 loc -topology of Ω n k \ {0}. The functions U and V are N -harmonic in H ≈ R N + = {x = (x 1 , x 2 , ..., x N ) : x N > 0} and vanish on ∂H \ {0}. Since it can be assumed that (ξ n k -a)/r n k → ξ, there holds U ≤ kV in H, U (ξ) = kV (ξ), if ξ ∈ H, and ∂U (ξ)/∂x N = k∂V (ξ)/∂x N > 0, if ξ ∈ ∂H (notice that |ξ| = 1). If ξ ∈ ∂H, Hopf lemma applies to V at ξ and, using the same linearization with the linear operator L as in the previous proof, it yields to U = kV . If ξ ∈ H, we use the fact that |Du 1,a (x)| ≥ β > 0 for |x -a| ≤ α for some β, α > 0. Thus |Dv n (ξ)| ≥ β. The non-degeneracy of V and the strong maximum principle lead again to U = kV . Whatever is the position of ξ, the equality between U and kV and the convergence in C 1 loc leads to the fact that for any ǫ > 0 there exists

n ǫ ∈ N such that n ≥ n ǫ implies (k -ǫ)u 1,a (x) ≤ u(x) ≤ (k + ǫ)u 1,a (x) ∀x ∈ Ω ∩ ∂B rn (a).
By the comparison principle between N -harmonic functions this inequality holds true in Ω \ ∂B rn (a). Since r n → 0 and ǫ is arbitrary, this ends the proof.

Remark.

The assumption that u 1,a has only isolated critical points in Ω is clearly satisfied in the case of a ball, a half-space or the complementary of a ball where no critical point exists. It is likely that this assumption always holds but we cannot prove it. However the Hopf maximum principle for p-harmonic functions (see [START_REF] Tolksdorff | On the Dirichlet problem for quasilinear equations in domains with conical boundary points[END_REF]) implies that u 1,a cannot have local extremum in Ω.

Separable solutions of the p-harmonic spectral problem

In this section we present a technique for constructing signed N -harmonic functions, regular or singular, as a product of functions depending only on one real variable. Some of the results were sketched in [START_REF] Véron | Singularities of some quasilinear equations, Nonlinear diffusion equations and their equilibrium states[END_REF]. The starting point is the result of Krol [START_REF] Krol | The behavior of the solutions of a certain quasilinear equation near zero cusps of the boundary[END_REF] dealing with the existence of 2-dimensional separable p-harmonic functions (the construction of singular separable p-harmonic functions was performed in [START_REF] Kichenassamy | Singular solutions of the p-Laplace equation[END_REF]).

Theorem 4.1 (Krol) Let p > 1. For any positive integer k there exists a unique β k > 0 and ω k : R → R, with least antiperiod π/k, of class C ∞ such that

u k (x) = |x| β k ω k (x/ |x|) (4.1) is p-harmonic in R 2 ; β k is the unique root ≥ 1 of (2k -1)X 2 - pk 2 + (p -2)(2k -1) p -1 X + k 2 = 0. (4.2)
(β k , ω k ) is unique up to translation and homothety over ω k .

This result is obtained by solving the homogeneous differential equation satisfied by ω k = ω:

-β 2 ω 2 + ω 2 θ (p-2)/2 ω θ θ = β (1 + (β -1)(p -1)) β 2 ω 2 + ω 2 θ (p-2)/2 ω. (4.3)
In the particular case k = 1, then β 1 = 1 and ω 1 (θ) = sin θ. For the other values of k the β k are algebraic numbers and the ω k are not trigonometric functions, except if p = 2. More generally, if one looks for p-harmonic functions in R N \ {0} under the form u

(x) = u(r, σ) = r β v(σ), r = |x| > 0, σ = x/ |x| ∈ S N -1 , one obtains that v verifies -div σ β 2 v 2 + |∇ σ v| 2 (p-2)/2 ∇ σ v = λ N,β β 2 v 2 + |∇ σ v| 2 (p-2)/2 v (4.4)
on S N -1 , where λ N,β = β (N -1 + (β -1)(p -1)) and div σ and ∇ σ are respectively the divergence and the gradient operators on S N -1 (endowed with the Riemaniann structure induced by the imbedding of the sphere into R N ). This equation, called the spherical p-harmonic spectral problem, is the natural generalization of the spectral problem of the Laplace-Beltrami operator on S N -1 . Since it does not correspond to a variational form (except if p = 2), it is difficult to obtain solutions. In the range of 1 < p ≤ N -1, Krol proved in [START_REF] Krol | The behavior of the solutions of a certain quasilinear equation near zero cusps of the boundary[END_REF] the existence of solutions of (4.4 ), not on the whole sphere, but on a spherical cap (which reduced (4.4 ) to an non-autonomous nonlinear second order differential equation).

His methods combined ODE estimates and shooting arguments. Later on, Tolksdorf [START_REF] Tolksdorff | On the Dirichlet problem for quasilinear equations in domains with conical boundary points[END_REF] introduced an entirely new method for proving the existence of solutions on any C 2 spherical domain S, with Dirichlet boundary conditions. Only the case β > 0 was treated in [START_REF] Tolksdorff | On the Dirichlet problem for quasilinear equations in domains with conical boundary points[END_REF], and, by a small adaptation of Tolksdorf approach, the case β > 0 was considered in [START_REF] Véron | Singularities of some quasilinear equations, Nonlinear diffusion equations and their equilibrium states[END_REF]. We develop below a method which allows to express solutions as product of explicit one variable functions.

The 3-D case

Let (r, θ, φ) ∈ (0, ∞) × [0, 2π] × [0, π] be the spherical coordinates in R 3    x 1 = r cos θ sin φ x 2 = r sin θ sin φ x 3 = r cos φ Then (4.4 ) turns into - 1 sin φ sin φ β 2 v 2 + v 2 φ + v 2 θ sin 2 φ (p-2)/2 v φ φ - 1 sin 2 φ β 2 v 2 + v 2 φ + v 2 θ sin 2 φ (p-2)/2 v θ θ = β (2 + (β -1)(p -1)) β 2 v 2 + v 2 φ + v 2
(x) = u(r, θ, φ) = r β k (sin φ) β k ω k (θ) (4.7)
where β k and ω k are as in Theorem 4.1.

In the case p = 3 we can use the conformal invariance of the 3-harmonic equation in R 3 to derive Theorem 4.3 Assume p = N = 3. Then for any positive integer k there exists a pharmonic function u in R 3 \ {0} under the form

u(x) = u(r, θ, φ) = r -β k (sin φ) β k ω k (θ) (4.8)
where β k and ω k are as in Theorem 4.1 with p = 3.

As a consequence of Theorem 4.3 we obtain signed 3-harmonic functions under the form (4.7 ) in the half space R 3 + = {x : x 2 > 0}, vanishing on ∂R 3 + \ {0}, with a singularity at x = 0. They correspond to even integers k. The extension to general smooth domains Ω is a deep chalenge. In the particular case k = 1, we have seen that β 1 = 1 and ω 1 (θ) = sin θ = x 2 , that we already know.

The general case

We assume that N > 3 and write the spherical coordinates in R N under the form

x = (r, σ) ∈ (0, ∞) × S N -1 = (r, sin φ σ ′ , cos φ) : σ ′ ∈ S N -2 , φ ∈ [0, π] .
(4.9)

The main result concerning separable p-harmonic functions is the following. where β k is the unique root ≥ 1 of (4.2 ) and w is solution of (4.15 ) with β = β k . Furthermore, if p = N there exists a singular N -harmonic function under the form

u(x) = u(r, σ ′ , φ) = r -β k (sin φ) β k w(σ ′ ). (4.11)
Proof. We first recall (see [START_REF] Vilenkin | Fonctions spéciales et théorie de la représentation des groupes[END_REF] for details) that the SO(N ) invariant unit measure on S N -1 is dσ = a N sin N -2 φ dσ ′ for some a N > 0, and

∇ σ v = -v φ e + 1 sin φ ∇ σ ′ v.
where e is derived from x/ |x| by the rotation of center 0 angle π/2 in the plane going thru 0, x/ |x| and the north pole. The weak formulation of (4.4 ) expresses as

π 0 S N -2 β 2 v 2 + v 2 φ + 1 sin 2 φ |∇ σ ′ v| 2 (p-2)/2 v φ ζ φ + 1 sin 2 φ ∇ σ ′ v.∇ σ ′ ζ sin N -2 φ dσ ′ dφ = λ N,β π 0 S N -2 β 2 v 2 + v 2 φ + 1 sin 2 φ |∇ σ ′ v|
∇ σ ′ v = λ N,β β 2 v 2 + v 2 φ + 1 sin 2 φ |∇ σ ′ v| 2 (p-2)/2 v (4.13)
where div σ ′ is the divergence operator acting on vector fields on S N -2 . We look again for p-harmonic functions under the form u(r, σ) = u(r, σ ′ , φ) = r β v(σ ′ , φ) = r β sin β φ w(σ ′ ). on S N -2 , which is the desired induction.

In order to be more precise, we can completely represent the preceding solutions by introducing the generalized Euler angles in R N = {x = (x 1 , ..., x N )}

              
x 1 = r sin θ N -1 sin θ N -2 ... sin θ 2 sin θ 1 x 2 = r sin θ N -1 sin θ N -2 ... sin θ 2 cos θ 1 . . . x N -1 = r sin θ N -1 cos θ N -2 x N = r cos θ N -1 (4.16) where θ 1 ∈ [0, 2π] and θ k ∈ [0, π], for k = 2, ..., N -1. Notice that θ N -1 is the variable φ in the representation (4.9 ). The above theorem combined with the induction process yields to the following.

  Ãj (rx, Dũ r ) = 0(2.15) in T β/r (Ω/r)\{0}. By the construction of Ãj (x, η), we can note thatlim r→0 Ãj (rx, η) = |η| p-2 η j , ∀η ∈ R N .Furthermore, for any x ∈ T β (Ω)\{0}, ρ(x) = ρ(ψ(x)) and c |x| ≤ |ψ(x)| ≤ c -1 |x| for some c > 0, the estimate (2.14 ) holds if u is replaced by ũr , Ω by T β/r (Ω/r) and ρ(x) by ρ r (x) := dist(x, Ω/r) i.e.|ũ r (x)| ≤ C|x| -2 ρ r (x) ∀x ∈ T β/r (Ω/r).

Theorem 3 . 2 .

 32 Let Ω be a bounded domain with a C 2 boundary and a ∈ ∂Ω. Assume that u 1,a has only a finite number of critical points in Ω. If u is a N -harmonic function in Ω vanishing on ∂Ω \ {a} verifying |u(x)| ≤ M u 1,a (x) for some M > 0 and any x ∈ Ω, there exists k ∈ [-M, M ] such that u = ku 1,a .

5 ) 1 Theorem 4 . 2

 5142 We look for a function v under the form v(θ, φ) = (sin φ) β ω(θ) (4.6) It follows that ω satisfies the same equation (4.3 ). The next result follows immediately from Theorem 4.Assume N = 3 and p > 1. Then for any positive integer k there exists a p-harmonic function u in R 3 under the form u

Theorem 4 . 4

 44 Let N > 3 and p > 1. For any positive integer k there exists p-harmonic functions in R N under the form u(x) = u(r, σ ′ , φ) = (r sin φ) β k w(σ ′ ).(4.10)

2 =× β 2 w 2 + 2 ∇ 2 ∇ σ ′ w Finally w satisfies -div σ ′ β 2 w 2 +

 22222 (sin φ) (β-1)(p-2) β 2 w 2 + |∇ σ ′ w| 2 (p-2)/2 , sin φ) (β-1)(p-1)-1 (N -2 + (β -1)(p -1)) -(N -1 + (β -1)(p -1)) sin 2 φ |∇ σ ′ w| 2 (p-2σ ′ v = (sin φ) (β-1)(p-1)-1 div σ ′ β 2 w 2 + |∇ σ ′ w| 2 (p-2)/|∇ σ ′ w| 2 (p-2)/2 ∇ σ ′ w = λ N -1,β β 2 w 2 + |∇ σ ′ w|

Theorem 4 . 5

 45 Let N > 3 and p > 1. For any positive integer k there exists p-harmonic functions in R N under the formu(x) = (r sin θ N -1 sin θ N -2 ... sin θ 2 ) β k ω k (θ 1 ) (4.17)where (β k , ω k ) are obtained in Theorem 4.1. Furthermore, if p = N there exists a singular N -harmonic function under the formu(x) = r -β k (sin θ N -1 sin θ N -2 ... sin θ 2 ) β k ω k (θ 1). (4.18)