A generalised Gauss circle problem and integrated density of states

Counting lattice points inside a ball of large radius in Euclidean space is a classical problem in analytic number theory, dating back to Gauss. We propose a variation on this problem: studying the asymptotics of the measure of an integer lattice of affine planes inside a ball. The first term is the volume of the ball; we study the size of the remainder term. While the classical problem is equivalent to counting eigenvalues of the Laplace operator on the torus, our variation corresponds to the integrated density of states of the Laplace operator on the product of a torus with Euclidean space. The asymptotics we obtain are then used to compute the density of states of the magnetic Schroedinger operator.


Introduction and Main results
The first problem we are considering in this paper has several equivalent formulations.
1.1. Number theoretic formulation. For ρ > 0 and k ∈ R d , let B(ρ; k) be the ball of radius ρ centered at k. Let S(ρ; k) be the number of integer points inside the disk B(ρ, k) ⊂ R 2 . The classical Gauss Circle Problem consists in estimating the remainder term (1.1)R(ρ; 0) = S(ρ; 0) − πρ 2 Hardy and (Edmund) Landau have found lower bounds for this problem, while the current best upper bound is given by Huxley in [Hux]. This problem has also been studied for balls of dimension higher than two, see e.g. [Go], and it is well-known that averaging over the radius of the ball improves regularity of the remainder.
In this paper, we consider a variation on this problem: we estimate the measure of the intersection of affine planes sitting on integer coordinates with balls of large radius in R d . More precisely, put and let B d (ρ, k) be a ball in R d of radius ρ centred at (k, 0) ∈ R k × R l , where k + l = d. Denote by S(ρ; k; d, k) the l-dimensional volume of the set B d (ρ, k) ∩ A k . This quantity has been studied in [KY] for general open sets rather than balls. However, in our case we are left only with their leading term (1.3) S(ρ; k; d, k) = ω l γ∈Z k |γ−k|<ρ (ρ 2 − |γ − k| 2 ) l/2 , where ω d is the volume of the unit ball in R d . One can see that the integral of R(ρ, k) over k ∈ T k = R k /Z k , is the same as the remainder term where H L is the restriction of H to the cube [0, L] d with appropriate self-adjoint boundary conditions andÑ (λ, H L ) is the counting functions of the (discrete) eigenvalues of H L . While this formulation of the IDS is important for Theorem 1.5, for periodic V we use an useful equivalent definition. Following [RS], we express H as a direct integral Then, one can express N (λ; H) in terms of the counting functions of the fibre operators H(k): where N (λ, H(k)) is the eigenvalue counting function of H(k). Remarkably, despite the fact that the asymptotic behaviour of N (λ, H(k)) for fixed k and λ → ∞ is very irregular (so that even the precise size of the remainder is unknown), integration over all quasimomenta k ∈ T d := R d /Z d makes things extremely regular, so that there exists a complete asymptotic expansion of N (λ) in powers of λ as λ → ∞, [PS1], [PS2]. Here, we have denoted is the volume of the unit ball in R d . The question we want to study is what would happen if, instead of integrating against all quasimomenta, we integrate over a subset of them, say over an affine plane. We write k = (k 1 , k 2 ), where k 1 ∈ T k , k 2 ∈ T l and define the partial density of states (PDS) as Our aim is to investigate the asymptotic behaviour of the PDS as λ → ∞. Obviously, the regularity at infinity will be improving as l increases and so the larger l is, the more asymptotic terms we are likely to obtain. This asymptotic problem can be treated in two steps: Step 1. Obtain the asymptotic behaviour of the PDS for unperturbed operator H 0 := −∆. More precisely, we want to obtain as good an estimate on (1.12) R 0 (λ; k 1 ; k, d) := N 0 p (λ; k 1 ; d, k) − C d λ d/2 as possible (of course, superscript 0 refers to the fact that we are dealing with the case V = 0). A simple calculation shows that if l = d, then R 0 (λ; k 1 ; l, d) = 0, so this step is trivial when dealing with the IDS. In the case of l < d this step becomes quite non-trivial and interesting. Once we have performed this step, we can move to Step 2. Compute (or estimate) the difference (1.13) N p (λ; k 1 ; d, k) − N 0 p (λ; k 1 ; d, k) and try to obtain as many asymptotic terms of it as possible. It follows from a simple computation that (1.14) hence the main aim of this paper deals with the first step of this programme; we intend to perform the second step in a separate publication.
1.3. Second spectral theoretic formulation. Consider the operatorH = −∆+ V acting on T l × R k with a smooth potentialṼ : T l × R k → R. We assume that, as a function on R k ,Ṽ is periodic with the lattice of periods (2πZ) k . Then it is easy to see that (1.15) N (λ;H) = (2π) l N p (λ; 0; d, k), that is to say that the integrated density of states equals the partial density of states up to a constant. If we consider a more general (but also less natural) operator H k1 , the domain of which consists of functions on T l × R k which become periodic after multiplication by e ik1x1 , then the IDS ofH k1 equals, again up to the same constant, N p (λ; k 1 ; d, k).

Main results.
Our first main result is as follows: Theorem 1.1. The error term R(ρ; k 1 ) satisfies the asymptotic estimates We do not pretend that all of these estimates are optimal, but some of them are, as can be seen from the following result: Theorem 1.3. For k > 1 and ρ sufficiently large, there exists a positive constant C d,k and k 1 ∈ O such that where > 0 is arbitrary. When d ≡ 1 mod 4, the lower bound R(ρ; k 1 ; d, k) ≥ C d,k ρ d−1 2 holds for k = 1.
In particular, this theorem means that for 1 ≤ k < d+1 2 and d ≡ 1 mod 4, we cannot get improvements on the upper bounds found in Theorem 1.1. It also means that for d ≡ 1 mod 4, k = 1, we cannot get improvements in the exponent.
Remark 1.4. It seems interesting that, after we have integrated N (λ; H(k)) (d − 1)/2 times, additional integrations do not improve the remainder estimate, until we perform the last (d-th) integration, which makes the remainder equal zero.
Open problem. The results in [Go] imply that for k = d, our upper bound is not optimal, but as d → ∞, our upper bound converges to the optimal one. Hence we may ask what is the optimal upper bound for k ≥ d+1 2 . 1.5. Operators with constant magnetic field. Another type of problems we consider in this paper is the asymptotic behaviour of the density of states of the (Lev) Landau Hamiltonian (Schrödinger operator with constant magnetic field).
Let D j = −i ∂ ∂xj . Then we define the Landau Hamiltonian H d as the operator acting in R d whose action is given by: Of course, only operators H 2 and H 3 make real physical sense, but for the sake of completeness we will deal with all dimensions.
Let Ω d (ρ) for d ≥ 2 be the parabolic domain in R d given by Defining P (ρ; d, k) analogously to S(ρ; 0; d, k), that is, The IDS N (λ; H d ) is related to P (ρ; d, k) by the following proposition.
Proposition 1.5. Let H d be the d-dimensional Landau Hamiltonian. Then, its integrated density of states is given by This proposition is particularly useful because we get an asymptotic expression for P (ρ; d, k), via the next theorem.
we obtain the following theorem.
Corollary 1.7. The integrated density of states of the Landau Hamiltonian on R 3 admits the asymptotic expansion The rest of the paper is organised as follows: in Section 2 we formulate several results which will be used in the proof of the main theorems, but we will postpone their proofs until Section 6. In Section 3 we prove the upper bounds in the Laplace case, and in Section 4 we obtain lower bounds. Finally, in Section 5 we deal with the magnetic case.

Auxilliary results
In order to prove Theorem 1.1, it will be useful to give an alternate expression for S(ρ, d, k). Let us define the function χ : R k → R as We can then observe that We would like to use Poisson's summation formula with f = χ. This will allow us to get upper bounds for all k 1 ∈ O, from the relation For the rest of this section, we therefore consider k 1 = 0, and it will be seen in the proof of Lemma 2.2 that this assumption is made without loss of generality. Unfortunately, Equation (2.3) holds only for f ∈ C ∞ , so we need to smooth out χ. To do so, we will consider its convolution with a Friederich mollifier Ψ . Hence, setting χ = Ψ * χ we get that Theorem 1.1 follows from two lemmas. The first one finds asymptotic upper and lower bounds for S: for all x ∈ R k . Immediately, if we define we get that Since χ ± are smooth functions, we can use Poisson's summation formula to compute the asymptotic expansion of S ± . The second lemma therefore gives the asymptotic expansion ofχ(ξ).
We will postpone the proof of these lemmas until Section 6.
3. Proof of Theorem 1.1 In this section, we prove Theorem 1.1 using both Lemmas 2.2 and 2.1. We have that Let us therefore find asymptotic expansions on S ± . We shall split those computations in two cases : whether k ≥ (d + 1)/2 or k < (d + 1)/2 3.1. Case 1. Here, we assume that k ≥ (d + 1)/2. Let us find asymptotic expansions on S ± . Since χ is a smooth compactly supported function of x, we may use Poisson's summation formula (2.3) to obtain Since we have that we get, assuming 1/ρ, that This is equivalent to Observe thatΨ(ξ) = O(|ξ| j ) for any j whenever |ξ| > 1 and bounded for |ξ| ≤ 1. Recall from Lemma 2.2 that χ(ξ) = O(|ξ| −(d+1)/2 ). Hence, choosing j = d−2k−1 2 , the last summand in (3.5) can be split into two terms, becoming The first sum can be estimated by The second sum can be estimated by (3.8) One can notice that the asymptotic estimate in ρ we obtain for both summands is the same whenever k > (d + 1)/2. Furthermore, when equality holds, the polynomial component is the same. Therefore, we have to choose = ρ −j such that We can remark that in the case where k = (d + 1)/2, we can choose = ρ −k , yielding the announced result. In general, the best choice of j is when This is achieved exactly when This gives us the announced asymptotic estimates when k ≥ (d + 1)/2, that is

Case 2.
We now assume that k < (d + 1)/2. In this case, we have that the sum converges with Ψ = O(1). Hence, the asymptotic expansion for S ± simplifies to (3.13) Since that last sum converges, we see that choosing = ρ −(d+1)/2 satisfies Theorem 1.1. We also see that choosing smaller does not improve the estimate.
Note that Equation (2.4) ensures that these estimates hold for all k 1 ∈ O.

Lower bounds
Let us first follow the argument given in [DT] for d = k = 2. The beginning of the argument is the same, which we add for completeness. Since R(ρ; k 1 ) is periodic in k 1 with respect to Γ, we can compute its Fourier coefficients, obtaining (4.1) for C, c positive constants whose value can change throughout. Whenever d ≡ 1 mod 4, we have that hence in that case, fixing γ ∈ Γ, we conclude that there exists r * such that for all We conclude that whenever d ≡ 1 mod 4, The remaining case, that is when d ≡ 1 mod 4 is more subtle. We will use results found in [PSo][Theorem 3.1, Lemma 3.3]. Indeed, from Equation (4.3), we have From Lemma 3.3 in [PSo], we know that, if k ≥ 2, for all > 0, there exists ρ 0 > 0 and α ∈ (0, 1/2) such that for all ρ > ρ 0 there exists γ ∈ Γ such that |γ| < (2πρ) and the distance from 2ργ to an integer is greater than α. Choosing such a γ bounds cos(2πρ|γ| − π/2) away from 0, and we get that For H 2 , we can write the solutions as u(x 1 , x 2 ) = e 2πin L x1 f (x 2 ), which reduces the problem to solving the eigenvalue problem This is a shifted quantum harmonic oscillator. We have that σ(H 2 ) = {2j + 1 : j ∈ N}, each with infinite multiplicity. It is a standard computation, see e.g. [Nak], that for λ ≥ 1, and 0 otherwise. Extending the methods of [Nak] to higher dimensions, it is again a simple computation to show that for λ ≥ 1, and, more generally, that Thus, from the definition of P (ρ; d, k), we have indeed that
By Theorem 1.1, we have Comparing with the integral, we get that for all X as defined above, For any d, we can use the Euler-Maclaurin formula : for any integer p ≥ 1, where B k is the kth Bernoulli number. Note that for integer a, Hence, by the Euler-Maclaurin formula, we get that Obviously, when d is odd, this last sum is actually finite and the error term 0. When ρ is not an integer, we can write ρ = a + τ , where τ is the fractional part. In that case, using the Euler-Maclaurin formula again, we get Let us observe that where q(τ ) = O(1) is a polynomial expression in τ . This computation holds whenever (d − 1)/2 − 2n − 1 > 0, after which point the term in τ gets more important than the term in ρ. Hence, we obtain the asymptotic expansion When k = 1, we already have that X(ρ) = 0. Therefore, we have that from which we recover a (quite sharp) asymptotic integrated density of states for the magnetic hamiltonian H d+1 . Let us combine equations (5.6) and (5.11). When k = d, we get that the error term from X is greater than d−1 2 , and as such, When k > d+2 2 , we get that P (ρ; d, k) Finally, when k ≤ d+2 2 , we get that P (ρ; d, k) where δ = 1 if k = d+2 2 and 0 otherwise.

Proofs of auxiliary results
6.1. Smoothing of the cut-off function. Let us define a smooth, even bump function ψ in C ∞ c (R), supported in [−1, 1], such that the integral where V k−1 is the area of the unit sphere in R k . Using this function, we can define the radial bump function Ψ on R k , of total mass 1 to be given by Let Ψ := Ψ 1 and χ (x) = Ψ (x) * χ(x). Its Fourier transform is given by Let χ + and χ − be defined on R k by We can now proceed with the proof of Lemma 2.1.
Therefore, it only remains to show that (6.6) holds for all x ∈ R k . First note that if |x| ≥ 1 + , both sides are 0. We shall split the remaining cases in |x| ≤ and < |x| < 1 + .
In the case where < |x| ≤ 1 + , we need to show that It is equivalent to show that 1 − (|x| − ) 2 ≤ (1 + ) 2 − |x| 2 . This is the case if (6.10) Since the last line is true by hypothesis, we can conclude that the left-hand side inequality of (2.7) is true.
In order to get an upper bound on χ(x), we proceed in a similar fashion, averaging χ (x) on a ball of radius around x, which yields (6.11) otherwise.
As we did before, it suffices to show that (6.12) Notice that the left hand side of that equation is 0 whenever |x| ≥ 1 − . Like before, we see that is equivalent to |x| < 1 − . This concludes the proof. 6.2. Fourier transform of χ.
Using identities of the Gamma function, we get that (6.16) ω l 2 d B( d + 1 2 , d + 1 2 ) = π d/2 Γ( d 2 + 1) = ω d , which is the desired value. If k = 2, then the Fourier transform is given by (6.17) Working in polar coordinates, we get that which is the desired result. [GR][Eq. 8.411, 6.567 and 8.451] were used respectively for an integral formula for the Bessel function, its integral, and its asymptotic expansion.
using [GR][Eq. 8.411] in the first line, which is the desired result. Additionnally, we have that (6.22) .
Using identities of the Gamma function, we get that (6.23) χ(0)ω d−k = ω d which is once again the desired value. One can note that in each of those cases, we ignored the trigonometric term to get an upper bound, considering it to be 1. Hence, since translation by k 1 is simply multiplication by a complex exponential in Equation (2.3), it can be ignored in just the same fashion. This completes the proof of Lemma 2.2.