Abderemane Morame 
email: morame@math.univ-nantes.fr
  
Françoise Truc 
email: francoise.truc@ujf-grenoble.fr
  
Magnetic bottles on the Poincaré half-plane: spectral asymptotics

We consider a magnetic Laplacian -∆ A = (id + A) ⋆ (id + A) on the Poincaré upper-half plane H , when the magnetic field dA is infinite at the infinity such that -∆ A has pure discret spectrum. We give the asymptotic behavior of the counting function of the eigenvalues.

Introduction

In this paper we study the asymptotic distribution of large eigenvalues of magnetic bottles on the hyperbolic plane H. Magnetic bottles on H are Schrödinger operators of the form

-∆ A = y 2 (D x -A 1 ) 2 + y 2 (D y -A 2 ) 2 , (1.1)
where the magnetic field dA is infinite at the infinity . This property ensures that -∆ A has a compact resolvent. The precise formulation is given below.

In the Euclidean case the asymptotic distribution of large eigenvalues of magnetic bottles in R d has been given by Yves Colin de Verdière [Col], using partition in cubes and estimations for constant magnetic fields in the cubes. This method can still be used here, but cubes are replaced by rectangles adapted to the hyperbolic geometry and the formula we get is of the same type, taking into account the hyperbolic volume and the hyperbolic definition for magnetic fields.

The hyperbolic framework we recall below has been used mainly for studying the Maass Laplacian , which corresponds to the constant magnetic field case. This case has been studied by many authors (see [Gro], [Els], [Com] [D-I-M]). In the authors consider asymptotically constant magnetic fields and in they deal with Pauli operators. See also [Ike] for relationship between Maass Laplacian and Schrödinger operators with Morse potentials.

From an other point of view , the asymptotic distribution of large eigenvalues in the hyperbolic context has already been studied for Schrödinger operators (without magnetic field) (see ) . The method is based on Feynman-Kac representation of the heat kernel and the Tauberian theorem. As already mentioned our own method involves only min-max techniques so it does not require to study properties of the evolution semigroup. It is also local, so our result is valid for many surfaces of infinite area with fundamental domain H .

Let us now set up the hyperbolic framework of our problem.

In a connected and oriented Riemannian manifold (M, g) of dimension n , for any real one-form A on M , one can define the magnetic Laplacian

-∆ A = (i d + A) ⋆ (i d + A) , ( (i d + A)u = i du + uA , ∀ u ∈ C ∞ 0 (M) ) .
(1.2)

The magnetic field is the exact two-form ρ B = dA .

The two-form ρ B is associated with a linear operator B on the tangent space defined by

ρ B (X, Y ) = g(B.X, Y ) ; ∀ X , Y ∈ T M × T M . (1.3)
The magnetic intensity b is given by

b = 1 2 tr (B ⋆ B) 1/2 . (1.4)
Let us assume that dim(M) = 2, and denote by dv the Riemannian measure on M; then ρ B = b dv , with | b| = b . In this case, we can say that the magnetic field is constant iff b is constant. Now, we consider the case where M = H is the hyperbolic plane :

H = R×]0, +∞[ , g = dx 2 + dy 2 y 2 , A = A 1 (x, y) dx + A 2 (x, y) dy .
We will assume that

A j (x, y) ∈ C 2 (H; R) , ∀ j . (1.5) Let us define D x = 1 i ∂ x and D y = 1 i ∂ y . Then we have -∆ A = y 2 (D x -A 1 ) 2 + y 2 (D y -A 2 ) 2 , (1.6) b = y 2 (∂ x A 2 -∂ y A 1 ) b = | b| , and dv = y -2 dxdy .
It is well known that -∆ A is essentially self-adjoint on L 2 (H) , see for example [Shu].

As we are only interested on the spectrum of sp(-∆ A ), we will use that it is gauge invariant:

sp(-∆ A ) = sp(-∆ A+dϕ ) ; ∀ ϕ ∈ C 2 (H; R) .
(1.7)

For an operator H , sp(H) , sp es (H) , sp a (H) , sp d (H) and sp p (H) denote its spectrum, its essential part, its absolutely continuous part, its discret part and its ponctual part.

We will denote -∆ A by P (A) .

2 The constant magnetic Laplacian on the hyperbolic plane

In this section, we explain how to get the well-known properties of the spectrum of a constant magnetic Laplacian on the hyperbolic plane. The original study was done by J. Elstrodt in [Els]. We consider the case where y

2 (∂ x A 2 (x, y) -∂ y A 1 (x, y)) is constant.
We choose a gauge such that A 2 = 0 , so A 1 (x, y) = ±by -1 . We can assume that A 1 (x, y) = by -1 , even if we change x into -x , which is a unitary operator on L 2 (H) . The operator we are interested in is

-∆ A b = y 2 (D x -by -1 ) 2 + y 2 D 2 y , with b ≥ 0 constant. (2.1)
Let U be the unitary operator

U : L 2 (H) → L 2 (R × R * + ) , Uf = y -1 f ; (2.2) R × R *
+ is endowed with the standard Lebesgue measure dxdy . Then

P b = U(-∆ A b )U ⋆ = (D x -by -1 )y 2 (D x -by -1 ) + D y y 2 D y . (2.3)
Using partial Fourier transform we get that sp

(P b ) = ξ∈R sp(P b (ξ)) ,
where P b (ξ) is the self-adjoint operator on L 2 (R + ) defined by

P b (ξ)f = (yξ -b) 2 f (y) + D y (y 2 D y f )(y) ; ∀ f ∈ C ∞ 0 (R + ) .
(2.4)

When ξ > 0 , by scaling, y → ξ -1 y , we get that sp(P b (ξ)) = sp(P b (1)) , (if ξ > 0) .

In the same way, we get that

sp(P b (ξ)) = sp(P b (-1)) , if ξ < 0 .
It is easy to see that sp For the proof, we use the method of . We define

K b = y -b -1 -iyD y ; so K ⋆ b = y -b + iyD y . (2.6) Then K ⋆ b K b = P b (1) + b and K b K ⋆ b = P b+1 (1) -b -1 . (2.7) When b > 1/2 , we define ϕ b (y) = 2 b-1/2 Γ(2b -1) y b-1 e -y , ϕ b ∈ Ker(K ⋆ b-1 ) , (2.8) ϕ b is the ground state of P b (1) : P b (1)ϕ b = bϕ b . As K b (P b (1) + 2b + 1) = P b+1 (1)K b and K -1 b f (y) = y -b-1 e y +∞ y s b e -s f (s)ds ; ∀ f ∈ [ϕ b+1 ] ⊥ ; we get that, if µ + 2b + 1 < b 2 + 1 4 , then µ ∈ sp d (P b (1)) ⇒ µ + 2b + 1 ∈ sp d (P b+1 (1))
,

and if λ -2b -1 ≥ b , λ ∈ sp d (P b+1 (1)) \ {b + 1} ⇒ λ -2b -1 ∈ sp d (P b ) .
One gets the well-known following theorem:

Theorem 2.1 The spectrum of P b (±1) is formed by its absolutely continuous part and its discret part, and

sp(P b (-1)) = sp ac (P b (-1)) = sp ac (P b (1)) = [b 2 + 1 4 , +∞[ sp(P b (1)) = sp ac (P b (1)) , if b ≤ 1 2 sp d (P b (1)) = {(2j + 1)b -j(j + 1) ; j ∈ N , j < b - 1 2 } if b > 1 2 . Corollary 2.2 The spectrum of -∆ A b is essential: sp(-∆ A b ) = sp es (-∆ A b ) .
Its absolutely continuous part is given by sp ac (-

∆ A b ) = [b 2 + 1 4 , +∞[ . The remaining part of its spectrum is empty if 0 ≤ b ≤ 1/2 ,
otherwise it is formed by a finite number of eigenvalues of infinite multiplicity given by

sp p (-∆ A b ) = {(2j + 1)b -j(j + 1) ; j ∈ N , j < b - 1 2 } , (if 1 2 < b .)
3 The case of a magnetic bottle (with compact resolvent)

The following theorem deals with the case of a magnetic field which fulfills magnetic bottles type assumptions. and if ∃ C 0 > 0 such that, for any vector field X on H ,

|X b| ≤ C 0 (| b| + 1) g(X, X) ; (3.2) then P (A) = -∆ A has a compact resolvent.
(d(x, y) denotes the hyperbolic distance of (x, y) to (0, 1) ) .

Proof : The standard proof for elliptic operators on the flat R n can be applied using the estimate given by the following Lemma.

Lemma 3.2 For any ǫ ∈]0, 1[ , there exists C ǫ > 0 s.t. ∀ f ∈ C ∞ 0 (H) , H b|f | 2 dv ≤ (1 + ǫ 2 ) -∆ A f |f L 2 (H) + C ǫ f L 2 (H) .
For the proof, one can use the unitary operator defined in 2.2

U : L 2 (H) → L 2 (R × R * + ) , Uf (x, y) = y -1 f (x, y) . We get that UP (A)U ⋆ = y 2 (D x -A 1 ) 2 + y(D y -A 2 ) 2 y .
In this form, we can write

UP (A)U ⋆ = K ⋆ K + b = K ⋆ K -b with K = y(D x -A 1 ) -i(D y -A 2 )y and K = y(D x -A 1 ) + i(D y -A 2 )y . So ± b ≤ UP (A)U ⋆ .
We cover R × R * + by two open sets O 0 , O 1 , such that O 0 is bounded and y and 1/y are bounded on O 0 , and 1 ≤ b on O 1 .

Taking an associated partition of unity χ j , (j = 0, 1) , and using that ± b ≤ UP (A)U ⋆ , we get

R×R * + b|χ 1 f | 2 dxdy ≤ R×R * + UP (A)U ⋆ (χ 1 f )χ 1 f dxdy .
The Lemma comes easily from this estimate.

4 Spectral asymptotics for magnetic bottles

The main theorem

For a self-adjoint operator P , and for any real λ ≤ inf sp es (P ) , we denote by N(λ; P ) the number of eigenvalues of P , (counted with their multiplicity), which are in ] -∞, λ[ . Theorem 4.1 Under the assumptions of Theorem 3.1, P (A) = -∆ A has a compact resolvent and for any δ ∈ ] 1 3 , 2 5 [ , in (4.15), there exists a constant C > 0 such that

1 2π H (1- C (b(m) + 1) (2-5δ)/2 )b(m) +∞ k=0 [λ(1-Cλ -3δ+1 )- 1 4 -(2k+1)b(m)] 0 + dv ≤ N(λ, -∆ A ) ≤ (4.1) 1 2π H (1+ C (b(m) + 1) (2-5δ)/2 )b(m) +∞ k=0 [λ(1+Cλ -3δ+1 )- 1 4 -(2k+1)b(m)] 0 + dv [ρ] 0 + is the Heaviside function: [ρ] 0 + = 1 , if ρ > 0 0 , if ρ ≤ 0 .
This result can be compared to the one obtained in [Col] . The difference between the two results is the additional term -1 4 , which comes from the geometry of the problem . It becomes really significant in the following Corollary 4.2 Under the assumptions of Theorem 3.1 and if the function

ω(µ) = H [µ -b(m)] 0 + dv satisfies ∃ C 1 > 0 s.t. ∀ µ > C 1 , ∀ τ ∈ ]0, 1[ , ω ((1 + τ ) µ) -ω(µ) ≤ C 1 τ ω(µ) , (4.2) then N(λ; -∆ A ) ∼ 1 2π H b(m) k∈N [λ - 1 4 -(2k + 1)b(m)] 0 + dv . (4.3)
The assumption (4.2) is satisfied when ω(λ) ∼ αλ k ln j (λ) when λ → +∞ , with k > 0 , or k = 0 and j > 0 .

For example this allows us to consider magnetic fields of the type b(x, y) = x y 2j + g(y) , with j ∈ N ⋆ and g(y) = p 1 (y) + p 2 (1/y),

where p 1 (s) and p 2 (s) are, for large s, polynomial functions of order ≥ 1 . In this case ω(λ) ∼ αλ 1 2j ln(λ) when λ → +∞ . For the proof of Theorem 4.1, we will establish some transformations, prove some technical lemmas and then use the minimax technique on quadratic forms as in Colin de Verdière's result to get successively a lower bound and an upper bound for N(λ; -∆ A ).

Technical transformations

Change of variables

Let us consider the diffeomorphism φ : R 2 → H , (x, y) = φ(x, t) := (x, e t ) which induces a unitary operator

U : L 2 (H; dv) → L 2 (R 2 ; dxdt) ( Uf )(x, t) := e -t/2 f (x, e t ) for any f ∈ L 2 (H). U maps C ∞ 0 (H) onto C ∞ 0 (R 2
) and the inverse U -1 is given by ( U -1 g)(x, y) := y 1/2 g(x, ln y) for each g ∈ L 2 (R 2 ).

The quadratic form related to the operator P (A) = -∆ A is given, for any u ∈ L 2 (H), by

q(u) := H 2 |y(D x -A 1 )u| 2 + |y(D y -A 2 )u| 2 dxdy y 2 = R 2 |e t (D x -Ã1 )u(φ)| 2 + |e t (e -t D t -Ã2 )u(φ)| 2 e -t dxdt = R 2 |e t/2 (D x -Ã1 )u(φ)| 2 + |e t/2 (e -t D t -Ã2 )u(φ)| 2 dxdt with Ãi (x, t) := A i (x, e t ) , i = 1, 2 .
After defining w := Uu, the preceding form becomes

q(w) := R 2 |e t (D x -Ã1 )w| 2 + |(e -t/2 D t e t/2 -e t Ã2 )w| 2 dxdt so P ( Ã) := UP (A) U -1 = e 2t (D x -Ã1 ) 2 + (D t -e t Ã2 ) 2 + 1/4.

Gauge

We want to work with a gauge such that A 2 = 0. Since

b = y 2 (∂ x A 2 -∂ y A 1 )
we can take

A 1 (x, y) = - y 1 b(x, s) s 2 ds which gives Ã1 (x, t) := - e t 1 b(x, s)
s 2 ds (4.4)

and

P ( Ã) = e 2t D x + e t 1 b(x, s) s 2 ds 2 + D 2 t + 1/4.
The associated quadratic form is

q Ã(w) = R 2 |e t (D x -Ã1 )w| 2 + |D t w| 2 + 1/4|w| 2 dxdt .
An application of the assumption (3.2) is the following Lemma.

Lemma 4.3 For any a > 0 and any ε 0 > 0 small enough

(ε 0 < min{ 1 2 , 1 C 0 (a + 1) }) , there exists C 1 > 1 such that , if (x 0 , y 0 ) ∈ H and b(x 0 , y 0 ) > 1 , then 1 C 1 b(x 0 , y 0 ) ≤ b(x, y) ≤ C 1 b(x 0 , y 0 ) ; ∀ (x, y) ∈ Ω(x 0 , y 0 , a, ε 0 )
where Ω(x 0 , y 0 , a, ε 0

) := {(x, y) / |x -x 0 | ≤ aε 0 y 0 , |y -y 0 | ≤ ε 0 y 0 } .
The proof comes directly from the assumption (3.2). Performing Taylor expansion , we get Let a 0 > 1 be given.

| b(x, y) -b(x 0 , y 0 )| ≤ (|x -x 0 | + |y -y 0 |) sup z∈Ω (|∂ x b(z)| + |∂ y b(z)|) so | b(x, y) -b(x 0 , y 0 )| ≤ ε 0 C 0 (a +
Any nonnegative constant depending only on a 0 , will be denoted invariably C .

Let X 0 = (x 0 , t 0 ) ∈ R 2 such that b(z 0 ) > 1 ; (z 0 = (x 0 , e t 0 ) ) ; |X 0 | can be very large.

Let us choose ε 0 ∈]0, 1[ , ε 0 can be very small. For a ∈] 1 a 0 , a 0 ] , let

K := X 0 + K 0 , K 0 = ] -ε 0 a e t 0 2 , ε 0 a e t 0 2 [ × ] - ε 0 2 , ε 0 2 [ . (4.5)
We consider the Dirichlet operator P K ( Ã) on K associated to the quadratic form

q à K (w) = K |e t (D x -Ã1 )w| 2 + |D t w| 2 + 1/4|w| 2 dxdt ∀ w ∈ W 1 0 (K) .
We are interested only by the spectrum of P K ( Ã) . It is gauge invariant, sp(P K ( Ã)) = sp(P K ( Ã + ∇ϕ)) , (4.6) so by taking ϕ(x, t) = - Let us define the magnetic potential related to a constant magnetic field

A 0 (x, t) = (A 0 1 , 0) with A 0 1 := -(t -t 0 ) e -t 0 b(x 0 , e t 0 ) . (4.7)
We want to compare N(λ; P K ( Ã)) to N(λ; P 0 K (A 0 )) for λ >> 1 , where P 0 K (A 0 ) is the Dirichlet operator on K, associated to the quadratic form

q A 0 ,0 K (w) = K |e t 0 (D x -A 0 1 )w| 2 + |D t w| 2 + 1/4|w| 2 dxdt ∀ w ∈ W 1 0 (K) .
We begin with comparing the associated magnetic potentials.

Lemma 4.4 Under the above assumptions there exists a constant C , depending only on a 0 in (4.5), such that for any (x, t) ∈ K : = -e t 0 (e t-t 0 -1) b(x, e τ ) e 2τ (4.9)

| Ã1 (x, t) -A 0 1 (x, t)| ≤ C ε 2 0 e -t
Writing A := e t 0 b(x, e τ ) e 2τ -b(x 0 , e t 0 ) e t 0 we get from the definition 4.7

| Ã1 (x, t) -A 0 1 (x, t)| ≤ C|t -t 0 ||A| ≤ C ε 0 |A| .
But from the lemma 4.3, we get the following estimate for any (x, τ ) ∈ K :

|A| = |e t 0 b(x, e τ ) e 2τ - b(x 0 , e t 0 ) e t 0 | ≤ C ε 0 e -t 0 b(x 0 , e t 0 ) .

To see this we decompose A in 3 parts

A 1 = e t 0 -2τ ( b(x, e τ )b(x 0 , e τ ))

A 2 = e t 0 -2τ ( b(x 0 , e τ )b(x 0 , e t 0 ))

A 3 = e t 0 b(x 0 , e t 0 ) ( 1 e 2τ - 1 e 2t 0 )
According to the assumption (3.2) and to the Lemma 4.3 we have

( b(x, e τ ) -b(x 0 , e τ )) ≤ e t 1 -t 0 b(x 0 , e t 0 ) , so |A 1 | ≤ e -t 0 C|x -x 0 |e -t 0 b(x 0 , e t 0 ) ≤ Caε 0 e -t 0 b(x 0 , e t 0 ) , |A 2 | ≤ e -t 0 C|e τ -e t 0 |e -t 0 b(x 0 , e t 0 ) ≤ Ce -t 0 |τ -t 0 | b(x 0 , e t 0 ) ≤ Cε 0 e -t 0 b(x 0 , e t 0 ) ,
The third term is also bounded by the same expression

|A 3 | ≤ C ε 0 e -t 0 b(x 0 , e t 0 )
, so we finished the proof.

Quadratic forms on K

Let us define

q Ã,0 K (w) := K |e t 0 (D x -Ã1 )w| 2 + |D t w| 2 + 1/4|w| 2 dxdt , ∀w ∈ W 1 0 (K) .
Lemma 4.5 There exists a constant C depending only on a 0 of (4.5), s.t.

(1

-ε 0 C) q Ã,0 K (w) ≤ q à K (w) ≤ (1 + ε 0 C) q Ã,0 K (w) .
Proof : Write

q à K (w) = K e 2(t-t 0 ) |e t 0 (D x -Ã1 )w| 2 + |D t w| 2 dxdt| + 1/4|w| 2 dxdt and use that |t -t 0 | ≤ 1 in K .
Lemma 4.6 There exists a constant C depending only on a 0 of (4.5), such that, for any τ ∈ ]0, 1[ , (with z 0 = (x 0 , e t 0 ) ) ,

(1τ 2 ) q A 0 ,0 K (w) + (1 -

1 τ 2 ) C ε 4 0 b 2 (z 0 ) w 2 ≤ q Ã,0 K (w) ≤ (1 + τ 2 ) q A 0 ,0 K (w) + (1 + 1 τ 2 ) C ε 4 0 b 2 (z 0 ) w 2 .
Proof : This is a straightforward application of lemma 4.4, when we write e t 0 (D x -Ã1 )w = e t 0 (D x -A 0 1 )we t 0 ( Ã1 -A 0 1 )w .

Spectral asymptotics for a rectangle.

An immediate application of Theorem A.2 in the appendix is the following Lemma.

Lemma 4.7 For any real λ ,

N(λ, P 0 K (A 0 )) ≤ |K|b(x 0 , e t 0 ) 2πe t 0 +∞ k=0 [λ - 1 4 -(2k + 1)b(x 0 , e t 0 )] 0 + . (4.10)
Moreover, there exists a constant C 0 depending only on a 0 of (4.5), such that, if

ε -2 0 /C 0 ≤ b(x 0 , e t 0 ) ≤ λ , then ∀ τ ∈]0, 1[ , (1-τ ) 2 |K|b(x 0 , e t 0 ) 2πe t 0 +∞ k=0 [λ- 1 4 - C 0 (τ ε 0 ) 2 -(2k+1)b(x 0 , e t 0 )] 0 + ≤ N(λ, P 0 K (A 0 )) .
(4.11)

Proof. Change variables (x, t) → (ξ, θ) = (e -t 0 (xx 0 ), tt 0 ) and apply (A.3) to get (4.10), and (A.4) to get (4.11).

Taking into account (4.6), Lemmas 4.5 -4.7, we get the following proposition.

Proposition 4.8 There exists a constant C 1 > 1 depending only on a 0 of (4.5), such that for any ε 0 ∈ ]0, 1/(2C 1 )[ , for any real λ > 1 and for any

η ∈]0, 1/2[ , N(λ, P K ( Ã)) ≤ |K|b(x 0 , e t 0 ) 2πe t 0 +∞ k=0 [Λ M (λ) - 1 4 -(2k + 1)b(x 0 , e t 0 )] 0 + , (4.12) with Λ M (λ) = (1 -η 2 ) -1 [ λ 1 -ε 0 C 1 + ε 4 0 η 2 C 1 b 2 (x 0 , e t 0 )
] . Moreover, there exists a constant C 0 depending only on a 0 of (4.5), such that for any ε 0 ∈ ]0, 1/(2C 1 )[ , for any real λ > 1 and for any η ∈]0,

1/2[ , if ε -2 0 /C 0 ≤ b(x 0 , e t 0 ) , then ∀ τ ∈]0, 1[ , (1-τ ) 2 |K|b(x 0 , e t 0 ) 2πe t 0 +∞ k=0 [Λ m (λ)- 1 4 - C 0 (τ ε 0 ) 2 -(2k+1)b(x 0 , e t 0 )] 0 + ≤ N(λ, P K ( Ã)) , (4.13) with Λ m (λ) = (1 + η 2 ) -1 [ λ 1 + ε 0 C 1 - ε 4 0 η 2 C 1 b 2 (x 0 , e t 0 )] . Without the condition that ǫ -2 0 /C 0 ≤ b(x 0 , e t 0 )
, we have in the same way that

|K|b(x 0 , e t 0 ) 4πe t 0 [Λ m (λ) - 1 4 -C 0 ( √ λ(1 + b(x 0 , e t 0 )) 1/2 ] ≤ N(λ, P K ( Ã)) . (4.14)
For the lower bound (4.14), use the same method as for (4.13), by using the lower bound in (A.5) instead of (A.4).

4.4 Lower bound and upper bound for the N (λ; -∆ A )

A partition adapted to b

Let a 0 and δ 0 be given such

1 < a 0 and δ 0 ∈ ] 1 3 , 2 5 [ . (4.15)
For any α ∈ Z 2 , we denote the rectangle

K(α) = ] - e α 2 2 + e α 2 α 1 , e α 2 α 1 + e α 2 2 [×] - 1 2 + α 2 , α 2 + 1 2 [ . (4.16) So R 2 = ∪ α K(α) and K(α) ∩ K(β) = ∅ if α = β .
Taking into account Lemma 4.3, each K(α) can be parted, (if necessary), into M(α) rectangles:

K(α) = ∪ M (α) j=1 K α,j , K α,j = ]- ǫ α,j e t α,j 2 +x α,j , x α,j + ǫ α,j e t α,j 2 [×]- ǫ α,j 2 +t α,j , t α,j + ǫ α,j 2 [ , (4.17) 
with

1 a 0 (1 + b δ 0 (x α,j , e t α,j ) ) ≤ ǫ α,j ≤ a 0 (1 + b δ 0 (x α,j , e t α,j ) ) , (4.18) 
and such that K α,k ∩ K α,j = ∅ if k = j .

We will denote Γ = {(α, j) ; α ∈ Z 2 , j ∈ {1, . . . , M(α)}} , X γ = (x γ , t γ ) the center of K γ , (γ ∈ Γ) , and z γ = (x γ , e tγ ) .

The lower bound estimate

Proposition 4.9 Under the assumptions of Theorem 3.1 and on δ 0 in (4.15), there exists a constant C 0 > 0 such that

1 2π H (1- C 0 (b(m) + 1) (2-5δ 0 )/2 )b(m) +∞ k=0 [λ(1-C 0 λ -3δ 0 +1 )- 1 4 -(2k+1)b(m)] 0 + dv ≤ N(λ, -∆ A ) . (4.19)
Proof. Any constant depending only on the assumptions will be denoted invariably by C .

As R 2 = γ∈Γ K γ , and

K γ ∩ K ρ = ∅ if γ = ρ ,
we get that γ∈Γ N(λ, P Kγ ( Ã)) ≤ N(λ, -∆ A ) .

For large |γ| , we use the lower bound estimate (4.13) , with

η 2 = ǫ γ = b -δ 0 (z γ ) and τ = b -(5δ 0 -2)/2 (z 0 ) ,
and also the fact that on

K γ , |e -t b(x, e t ) -e -tγ b(z γ )| ≤ ǫ γ C .
For small |γ| , (even the γ such that b(z γ ) ≤ λ 1-2(3δ 

N(λ, -∆ A ) ≤ (4.20) 1 2π H (1+ C 0 (b(m) + 1) (2-5δ 0 )/2 )b(m) +∞ k=0 [λ(1+C 0 λ -3δ 0 +1 )- 1 4 -(2k+1)b(m)] 0 + dv .
Proof. Any constant depending only on the assumptions will be denoted invariably by C . We keep the partition and the notation used in the lower bound.

We consider the covering of R 2 by open rectangles:

R 2 = γ∈Γ K γ , K γ = X γ + (1/τ γ )(K γ -X γ ) , with τ γ ∈ b -(5δ 0 -2)/2 (z γ )[a -1 0 , a 0 ] . Then there exists a partition of unity (χ γ (x, t)) satisfying        γ χ 2 γ = 1 support(χ γ ) ⊂ K γ |D x χ γ | ≤ C/(e tγ ǫ γ τ γ ) |D t χ γ | ≤ C/(ǫ γ τ γ )        (4.21)
Proof: Its comes easily from THEOREME 1.6 of [Dem].

Let us sketch a proof. By scaling and dividing λ by b, we need only to establish the formula when b = 1 .

We take

Ω R = (] -R 2 , R 2 [) 2
, the Dirichlet boundary conditions on x = ± R 2 and the Floquet conditions: e ixy/2 u(x, y) is R-periodic in y .

As u → ue ixy/2 is a unitary operator, by performing this gauge transform,

H Ω R 0 becomes H 0 = D 2 x + (D y + x) 2 , for the Dirichlet boundary conditions on x = ± R
2 and the periodic ones on y = ± R 2 . H Ω R 0 and H 0 have the same spectrum. Using discret Fourier expansion, we get that N(λ, H

Ω R 0 ) = k∈Z N(λ, H k,R ) , where H k,R is the Dirichlet operator on I R =] -R 2 , R 2 [ , associated to the har- monic oscillator D 2 x + ( 2kπ R + x) 2 . As N(λ, H k,R ) = 0 when |k| > 1 2π (R √ λ + R 2 2 ) , we get N (λ, H 0 ) ≤ ♯{n ∈ N; 2n + 1 < λ} × lim R→∞ 1 πR 2 (R √ λ + R 2 2 ) , or equivalently N (λ, H 0 ) ≤ 1 2π ♯{n ∈ N; 2n + 1 < λ} .
Now, for any fixed ǫ ∈]0, 1[ , ( for example ǫ = 1/ √ R ), and for any k such that |k| ≤ (1ǫ) R 2 4π , the exponential decreasing of the eigenfunctions of the harmonic oscillator on R 2 leads to

N(λ, H k,R ) ≥ ♯{n ∈ N; 2n + 1 < λ - C λ ǫ 2 R 2 }
, where C λ depends only on λ .

To see this, with ǫ chosen as previously and R >> λ + 1, just use the fact that, for any u

∈ χ(4x/ √ R)E λ (H k,∞ )[L 2 (R)] , R/2 -R/2 H k,R uu dx ≤ (λ + C/R) R/2 -R/2 |u| 2 dx . H k,∞ denotes the harmonic oscillator D 2 x + ( 2kπ R + x) 2 on L 2 (R) , χ is a cut-off function, supported in [-1, 1] and equal to 1 in [-1/2, 1/2] , and E λ (H k,∞ ) denotes the spectral projection on ]-∞, λ[ of the self-adjoint operator H k,∞ .
Then, with the same ǫ and using the left-hand side continuity of the function λ → ♯{n ∈ N; 2n + 1 < λ} , we get also that

N (λ, H 0 ) ≥ 1 2π ♯{n ∈ N; 2n + 1 < λ} .
A.2 Eigenvalues estimate in the euclidian rectangle for a constant magnetic field

Let us consider the Dirichlet problem H Ω R D,b associated to the Schrödinger operator with a constant magnetic field The classical Weyl estimate is the following. There exists a constant C 0 > 0 such that, if 0 ≤ b ≤ λ and (C 0 √ b) -1 ≤ R j for j = 1, 2 , then .5) Proof: The upper bound (A.3) and the lower bound (A.4) come from the density of state using the same proof as in Colin de Verdière paper [Col].

H 0 = (D x -b y 2 ) 2 + (D y + b x 2 ) 2 ; (b > 0 is a constant), in a rectangle Ω R =] -R 1 2 , R 1 2 [×] -R 2 2 , R 2 2 [ ; R = (R 1 , R 2 ) ∈ (R ⋆ + ) 2 .
|Ω R | 4π λ[1 -C 0 √ b √ λ ] ≤ N(λ, H Ω R D,b ) ≤ |Ω R | 4π λ[1 + C 0 √ b √ λ ] . (A
We sketch the proof of the lower bound (A.4). x-x j 2 ) 2 , where (x j , y j ) is the center of Ω(j) . Now, it is easy to get the uniform Weyl formula: ∃C 0 > 0 s.t. ∀ j + 1, . . . , M ,

|Ω(j)| 4π λ b [1 -C 0 √ b √ λ ] ≤ N( λ b , H Ω(j) D,1 ) ≤ |Ω(j)| 4π λ b [1 + C 0 √ b √ λ ] .
To be convinced, see that ∃ C 0 > 0 s.t. ∀ τ ∈]0, 1] , ∀ u ∈ W 1 0 (Ω(j) ),

(1 -τ 2 ) ∇u 2 - C 0 τ 2 u 2 ≤ (D -A(j))u 2 ≤ (1 + τ 2 ) ∇u 2 + C 0 τ 2 u 2 ,
and take τ = 

  es (P b (±1)) = b 2 + sp ac (P 0 (1)) = sp ac (P b (±1)) , and, (see for example the exercise I6 p. 1573 in [Du-Sc]), sp(P b (-1)) = sp ac (P b (-1)) = [b 2 + 1 4 , +∞[ = sp ac (P b (1)) . (2.5) P b (1) may have some eigenvalues in [b, b 2 + 1 4 [ .

Theorem 3. 1

 1 Under the assumptions (1.5) and (1.6), if b(x, y) → +∞ as d(x, y) → +∞ , (3.1)

  in a suitable rectangle in R 2

x 0 A

 0 1 (s, t 0 )ds , we can assume that Ã1 (x, t) := -

Theorem A. 2

 2 The Colin de Verdière upper bound holds for any λ > 1 :N(λ, H Ω R D,b ) ≤ b|Ω R | 2π ♯{n ∈ N ; (2n + 1)b < λ} . (A.3)For the lower bound, we will need to precise Colin de Verdière's one as follows.There exists a constantC 0 > 0 s.t., if 0 < b < λ and 1 ≤ √ b min R j , then ∀ ǫ ∈]0, 1] , (1ǫ) 2 b|Ω R | 2π ♯{n ∈ N ; (2n + 1)b < λ -C 0 (ǫ min R i ) 2 } ≤ N(λ, H Ω R D,b ) . (A.4) 

  We set : R(b) = R √ b . By scaling, we change Ω R , b and λ into Ω R(b) , 1 and λ/b . Then we take a large rectangle Ω R(b),M,ǫ = M j=1 Ω(j, ǫ) where (D -A(j)) 2 = (D x -y-y j 2 ) 2 + (D y +

  So we get the lower bound of (A.5). We get in the same way the upper bound by considering the Neumann operatorsH As in Theorem A.2, λb 2 ≤ ♯ {n ∈ N ; (2n + 1)b < λ} ≤ λ + b 2 ,so the upper bound (A.3) is sharp compared to the one in (A.5). The lower bound (A.4) is sharp, compared to the one in (A.5), when ǫ < (b/λ) .

  0 b(x 0 , e t 0 ) .

	Proof : As	Ã1 (x, t) := -	e t e t 0	b(x, s) s 2 ds ,	(4.8)
	there exists τ = τ (x) ∈]t 0 , t[ such that		
	Ã1 (x, t) = -(e t -e t 0 )	b(x, e τ ) e 2τ	

Keywords : spectral asymptotics, magnetic bottles, hyperbolic plane , minimax principle.

We write

Thus on K γ , V (x, t) ≤ C/(ǫ γ τ γ ) 2 , and it follows easily from the min-max principle that

Then we get (4.20) from (4.22), as for (4.19), but using only (4.12), (instead of (4.13) and (4.14) ).

APPENDIX A Constant magnetic laplacian on the flat plane

A.1 The density of states for the euclidian constant magnetic field

Let us consider on L 2 (R 2 ) the Schrödinger operator with a constant magnetic field

The density of states of H 0 , N (λ, H 0 ) , is defined, (see [D-I-M]), by

Theorem A.1 The Colin de Verdière formula holds for any λ > 0 :

We consider the large rectangle Ω R(b),M = M j=1 Ω(j) where Ω(j) = z j + Ω R(b) .

So there exists a constant C 0 and a partition of unity (χ j ) s.t.

support(χ j ) ⊂ Ω(j) ,

.

We can write, for any

so we get that for any real µ ,

.

By the density of states formula, we have lim We change gauge in each Ω(j) in order to consider H Ω(j) D,1 as the operator