Volumes of Symmetric Spaces via Lattice Points

Henri Gillet and Daniel R. Grayson

Received: February 15, 2004 Revised: August 2, 2006

Communicated by Ulf Rehmann

Abstract. We show how to use elementary methods to compute the volume of Sl_{k} ℝ/ Sl_{k} ℤ. We compute the volumes of certain unbounded regions in Euclidean space by counting lattice points and then appeal to the machinery of Dirichlet series to get estimates of the growth rate of the number of lattice points appearing in the region as the lattice spacing decreases. We also present a proof of the closely related result that the Tamagawa number is 1.

2000 Mathematics Subject Classification: 11F06, 11H06, 11M45. Keywords and Phrases: special linear group, volume, lattice, Tauberian theorem, arithmetic group, Tamagawa number.

Introduction

In this paper we show how to use elementary methods to prove that the volume of Sl_{k} ℝ/ Sl_{k} ℤ is ζ(2)ζ(3) ⋯ ζ(k)/k; see Corollary 3.16. Using a version of reduction theory presented in this paper, we can compute the volumes of certain unbounded regions in Euclidean space by counting lattice points and then appeal to the machinery of Dirichlet series to get estimates of the growth rate of the number of lattice points appearing in the region as the lattice spacing decreases.

In section 4 we present a proof of the closely related result that the Tamagawa number of Sl_{k,ℚ} is 1 that is somewhat simpler and more arithmetic than Weil’s in [37]. His proof proceeds by induction on k and appeals to the Poisson summation formula, whereas the proof here brings to the forefront local versions (5) of the formula, one for each prime p, which help to illuminate the appearance of values of zeta functions in formulas for volumes.

The volume computation above is known; see, for example, [26] (with important corrections in [30]), formula (24) in [29], and Theorem 10.4 in [22]. The methods used in the computation of the volume of Sl_{k} ℝ/ Sl_{k} ℤ in the book [31, Lecture XV] have a different flavor from ours and do not involve counting lattice points. One positive point about the proof there is that it proceeds by induction on k, making clear how the factor ζ(k) enters in at k-th stage. See also [36, §14.12, formula (2)]. The proof offered there seems to have a gap which consists of assuming that a certain region (denoted by T there) is bounded, thereby allowing the application of [36, §14.4, Theorem 3]^{1}. The region in Example 2.7 below shows that filling the gap is not easy, hence if we want to compute the volume by counting lattice points, something like our use of reduction theory in Section 3 is needed.

An almost equivalent result was proved by Minkowski in formula (85.) of [16], where he computed the volume of SO(k)\ Sl_{k} ℝ/ Sl_{k} ℤ. The relationship between the two volume computations is made clear in the proof of [36, §14.12, Theorem 2].

Some of the techniques we use were known to Siegel, who used similar methods in his investigation of representability of integers by quadratic forms in [24, 25, 27]. See especially [25, Hilfssatz 6, p. 242], which is analogous to our Lemma 2.5 and the reduction theory of Section 3, where we show how to compute the volume of certain unbounded domains in Euclidean space by counting lattice points; see also the computations in [24, §9], which have the same general flavor as ours. See also [28, p. 581] where Siegel omits the laborious study, using reduction theory, of points at infinity; it is those details that concern us here.

We thank Harold Diamond for useful information about Dirichlet series and Ulf Rehmann for useful suggestions, advice related to Tamagawa numbers, and clarifications of Siegel’s work. We also thank the National Science Foundation for support provided by NSF grants DMS 01-00587 and 05-00762 (Gillet), and 99-70085 and 03-11378 (Grayson).

1 Counting with zeta functions

As in [8] we define the zeta function of a group G by summing over the subgroups H in G of finite index.

ζ(G, s) = ∑_{H⊆G}[G : H]^{−s}  (1)

Evidently, ζ(ℤ, s) = ζ(s) and the series converges for s > 1. For good groups G the number of subgroups of index at most T grows slowly enough as a function of T that ζ(G, s) will converge for s sufficiently large.

Let’s pick k ≥ 0 and compute ζ(ℤ^{k}, s). Any subgroup H of ℤ^{k} of finite index is isomorphic to ℤ^{k}; choosing such an isomorphism amounts to finding a matrix A : ℤ^{k} → ℤ^{k} whose determinant is nonzero and whose image is H.

1called Dirichlet’s Principle in [3, §5.1, Theorem 3]

Documenta Mathematica 11 (2006) 425–447

Any two matrices A, A′ with the same image H are related by an equation A′ = AS where S ∈ Gl_{k} ℤ.

Thus the terms in the sum defining ζ(ℤ^{k}, s) correspond to the orbits for the action of Gl_{k} ℤ via column operations on the set of k × k-matrices with integer entries and nonzero determinant. A unique representative from each orbit is provided by the matrices A that are in Hermite normal form (see [4, p. 66] or [19, II.6]), i.e., those matrices A with A_{ij} = 0 for i > j, A_{ii} > 0 for all i, and 0 ≤ A_{ij} < A_{ii} for i < j.

Let HNF be the set of integer k ×k matrices in Hermite normal form. Given positive integers n_{1}, …, n_{k}, consider the set of matrices A in HNF with A_{ii} = n_{i} for all i. The number of matrices in it is n_{1}^{k−1}n_{2}^{k−2} ⋯ n_{k−1}^{1}n_{k}^{0}. Using that, we compute formally as follows.

\begin{aligned}ζ(ℤ^{k}, s) & = ∑_{H⊆ℤ^{k}}[ℤ^{k} : H]^{−s} \\ & = ∑_{A∈HNF}(det A)^{−s} \\ & = ∑_{n_{1}>0,...,n_{k}>0}(n_{1}^{k−1}n_{2}^{k−2} ⋯ n_{k−1}^{1}n_{k}^{0})(n_{1} ⋯ n_{k})^{−s} \\ & = ∑_{n_{1}>0,...,n_{k}>0} n_{1}^{k−1−s}n_{2}^{k−2−s} ⋯ n_{k−1}^{1−s}n_{k}^{−s} \\ & = ∑_{n_{1}>0} n_{1}^{k−1−s} ∑_{n_{2}>0} n_{2}^{k−2−s} ⋯ ∑_{n_{k−1}>0} n_{k−1}^{1−s} ∑_{n_{k}>0} n_{k}^{−s} \\ & = ζ(s − k + 1)ζ(s − k + 2) ⋯ ζ(s − 1)ζ(s) \end{aligned}  (2)

The result ζ(s − k + 1)ζ(s − k + 2) ⋯ ζ(s − 1)ζ(s) is a product of Dirichlet series with positive coefficients that converge for s > k, and thus ζ(ℤ^{k}, s) also converges for s > k. This computation is old, and appears in various guises. See, for example: proof 2 of Proposition 1.1 in [8]; Lemma 10 in [15]; formula (1.1) in [32]; page 64 in [23]; formula (5) and the lines following it in [26], where the counting argument is attributed to Eisenstein, and its generalization to number rings is attributed to Hurwitz; and pages 37–38 in [37].

Lemma 1.1. #{H ⊆ ℤ^{k} | [ℤ^{k} : H] ≤ T } ∼ ζ(2)ζ(3) ⋯ ζ(k)T^{k}/k for k ≥ 1.

The right hand side is interpreted as T when k = 1. The notation f(T) ∼ g(T) means that lim_{T →∞} f(T)/g(T) = 1.

Proof. We give two proofs.

The first one is more elementary, and was told to us by Harold Diamond. Writing ζ(s−k+1) = ∑ n^{k−1}n^{−s} and letting B(T) = ∑_{n≤T} n^{k−1} be the corresponding coefficient summatory function we see that B(T) = T^{k}/k + O(T^{k−1}). If k ≥ 3 we may apply Theorem A.2 to show that the coefficient summatory function for the Dirichlet series ζ(s)ζ(s−k+1) behaves as ζ(k)T^{k}/k+O(T^{k−1}).

Applying it several more times shows that the coefficient summatory function for the Dirichlet series ζ(s)ζ(s − 1) ⋯ ζ(s − k + 3)ζ(s − k + 1) behaves as ζ(k)ζ(k − 1) ⋯ ζ(3)T^{k}/k + O(T^{k−1}). Applying it one more time we see that the coefficient summatory function for ζ(ℤ^{k}, s) = ζ(s) ⋯ ζ(s − k + 1) behaves as ζ(k)ζ(k − 1) ⋯ ζ(2)T^{k}/k + O(T^{k−1} log T), which in turn implies the result.

The second proof is less elementary, since it uses a Tauberian theorem. From (2) we know that the rightmost (simple) pole of ζ(ℤ^{k}, s) occurs at s = k, that the residue there is the product ζ(2)ζ(3) ⋯ ζ(k), and that Theorem A.4 can be applied to get the result.

Now we point out a weaker version of lemma 1.1 whose proof is even more elementary.

Lemma 1.2. If T > 0 then #{H ⊆ ℤ^{k} | [ℤ^{k} : H] ≤ T } ≤ T^{k}.

Proof. As above, we obtain the following formula.

\begin{aligned}#\{H ⊆ ℤ^{k} | [ℤ^{k} : H] ≤ T \} & = #\{A ∈ HNF | det A ≤ T \} \\ & = ∑_{{n_{1}>0,...,n_{k}>0 \atop n_{1}·····n_{k}≤T}} n_{1}^{k−1}n_{2}^{k−2} ⋯ n_{k−1}^{1}n_{k}^{0} \end{aligned}

We use it to prove the desired inequality by induction on k, the case k = 0 being clear.

#\{H ⊆ ℤ^{k} | [ℤ^{k} : H] ≤ T \} = ∑_{n_{1}=1}^{⌊T ⌋} n_{1}^{k−1} ∑_{{n_{2}>0,...,n_{k}>0 \atop n_{2}···n_{k}≤T /n_{1}}} n_{2}^{k−2} ⋯ n_{k−1}^{1}n_{k}^{0}

= ∑_{n_{1}=1}^{⌊T ⌋} n_{1}^{k−1} · #{H ⊆ ℤ^{k−1} | [ℤ^{k−1} : H] ≤ T/n_{1}}

≤ ∑_{n_{1}=1}^{⌊T ⌋} n_{1}^{k−1}(T/n_{1})^{k−1} [by induction on k]

= ∑_{n_{1}=1}^{⌊T ⌋} T^{k−1} = ⌊T ⌋ · T^{k−1} ≤ T^{k}

2 Volumes

Recall that a bounded subset U of Euclidean space ℝ^{k} is said to have Jordan content if its volume can be approximated arbitrarily well by unions of boxes contained in it or by unions of boxes containing it, or in other words, that the the characteristic function χ_{U} is Riemann integrable. Equivalently, the boundary ∂U of U has (Lebesgue) measure zero (see [21, Theorem 105.2, Lemma 105.2, and the discussion above it]). If U is a possibly unbounded subset of ℝ^{k} whose boundary has measure zero, its intersection with any ball will have Jordan content.

Now let’s consider the Lie group G = Sl_{k} ℝ as a subspace of the Euclidean space M_{k}ℝ of k × k matrices. Siegel defines a Haar measure on G as follows (see page 341 of [29]). Let E be a subset of G. Letting I = [0, 1] be the unit interval and considering a number T > 0, we may consider the following cones.

\begin{aligned}I · E & = \{t · B | B ∈ E, 0 ≤ t ≤ 1\} \\ T · I · E & = \{t · B | B ∈ E, 0 ≤ t ≤ T \} \\ ℝ^{+} · E & = \{t · B | B ∈ E, 0 ≤ t\} \end{aligned}

Observe that if B ∈ T · I · E, then 0 ≤ det B ≤ T^{k}.

Definition 2.1. We say that E is measurable if I · E is, and in that case we define μ_{∞}(E) = vol(I · E) ∈ [0, ∞].

The Jacobian of left or right multiplication by a matrix γ on M_{k}ℝ is (det B)^{k}, so for γ ∈ Sl_{k} ℝ volume is preserved. Thus the measure is invariant under G, by multiplication on either side. According to Siegel, the introduction of such invariant measures on Lie groups goes back to Hurwitz (see [10, p. 546] or [9]).

Let F ⊆ G be the fundamental domain for the action of Γ = Sl_{k} ℤ on the right of G presented in [15, section 7]; it’s an elementary construction of a fundamental domain which is a Borel set without resorting to Minkowski’s reduction theory. In each orbit they choose the element which is closest to the identity matrix in the standard Euclidean norm on M_{k}ℝ ≅ ℝ^{k^{2}}, and ties are broken by ordering M_{k}ℝ lexicographically. This set F is the union of an open subset of G (consisting of those matrices with no ties) and a countable number of sets of measure zero.

The intersection of T · I · F with a ball has Jordan content. To establish that, it is enough to show that the measure of the boundary ∂F in G is zero. Suppose g ∈ ∂F. Then it is a limit of points g_{i} ∉ F, each of which has another point g_{i}h_{i} in its orbit which is at least as close to 1. Here h_{i} is in Sl_{k}(ℤ) and is not 1. The sequence i ↦ g_{i}h_{i} is bounded, and thus so is the sequence h_{i}; since Sl_{k}(ℤ) is discrete, that implies that h_{i} takes only a finite number of values. So we may assume h_{i} = h is independent of i, and is not 1. By continuity, gh is at least as close to 1 as g is. Now g is also a limit of points f_{i} in F, each of which has f_{i}h not closer to 1 than f_{i} is. Hence gh is not closer to 1 than g is, by continuity. Combining, we see that gh and g are equidistant from 1. The locus of points g in Sl_{k}(ℝ) such that gh and g are equidistant from 1 is given by the vanishing of a nonzero quadratic polynomial, hence has measure zero. The boundary ∂F is contained in a countable number of such sets, because Sl_{k}(ℤ) is countable, hence has measure zero, too.

We remark that HNF contains a unique representative for each orbit of the action of Sl_{k} ℤ on {A ∈ M_{k}ℤ | det A > 0}. The same is true for ℝ^{+} · F.

Restricting our attention to matrices B with det B ≤ T^{k} we see that #(T · I · F ∩ M_{k}ℤ) = #{A ∈ HNF | det A ≤ T^{k}}.

Warning: HNF is not contained in ℝ^{+} · F. To convince yourself of this,

consider the matrix A = (\begin{matrix}5 & −8 \\ 3 & 5 \end{matrix}) of determinant 49. Column operations

with integer coefficients reduce it to B = (\begin{matrix}49 & 18 \\ 0 & 1 \end{matrix}), but (1/7)A is closer to

the identity matrix than (1/7)B is, so B ∈ HNF, but B ∉ ℝ^{+} · F.

We want to approximate the volume of T ·I ·F by counting the lattice points it contains, i.e., by using the number #(T · I · F ∩ M_{k}ℤ), at least when T is large. Alternatively, we may use #(I · F ∩ r · M_{k}ℤ), when r is small.

Definition 2.2. Suppose U is a subset of ℝ^{n}. Let

N_{r}(U) = r^{n} · #{U ∩ r · ℤ^{n}}

and let

μ_{ℤ}(U) = lim_{r→0} N_{r}(U),

if the limit exists, possibly equal to +∞. An equation involving μ_{ℤ}(U) is to be regarded as true only if the limit exists.

Lemma 2.3. μ_{ℤ}(I · F) = ζ(2)ζ(3) ⋯ ζ(k)/k

Proof. We replace r above with 1/T:

μ_{ℤ}(I · F) = lim_{T →∞} T^{−k^{2}} · #(T · I · F ∩ M_{k}ℤ)

= lim_{T →∞} T^{−k^{2}} · #{A ∈ HNF | det A ≤ T^{k}}

= lim_{T →∞} T^{−k^{2}} · #{H ⊆ ℤ^{k} | [ℤ^{k} : H] ≤ T^{k}}

= ζ(2)ζ(3) ⋯ ζ(k)/k [using lemma 1.1]

Lemma 2.4. If U is a bounded subset of ℝ^{n} with Jordan content, then μ_{ℤ}(U) = vol U.

Proof. Subdivide ℝ^{n} into cubes of width r (and of volume r^{n}) centered at the points of rℤ^{n}. The number #{U ∩ r · ℤ^{n}} lies between the number of cubes contained in U and the number of cubes meeting U, so r^{n} · #{U ∩ r · ℤ^{n}} is captured between the total volume of the cubes contained in U and the total volume of the cubes meeting U, hence approaches the same limit those two quantities do, namely vol U.

Lemma 2.5. Let B_{R} be the ball of radius R > 0 centered at the origin, and let U be a subset of ℝ^{n} whose boundary has measure zero.

1. For all R, the quantity μ_{ℤ}(U) exists if and only if μ_{ℤ}(U −B_{R}) exists, and in that case, μ_{ℤ}(U) = vol(U ∩ B_{R}) + μ_{ℤ}(U − B_{R}).

2. If μ_{ℤ}(U) exists then μ_{ℤ}(U) = vol(U) + lim_{R→∞} μ_{ℤ}(U − B_{R}).

3. If vol(U) = +∞, then μ_{ℤ}(U) = +∞.

4. If lim_{R→∞} lim sup_{r→0} N_{r}(U − B_{R}) = 0, then μ_{ℤ}(U) = vol(U).

Proof. Writing U = (U ∩ B_{R}) ∪ (U − B_{R}) we have

N_{r}(U) = N_{r}(U ∩ B_{R}) + N_{r}(U − B_{R}).

For each R > 0, the set U ∩ B_{R} is a bounded set with Jordan content, and thus lemma 2.4 applies to it. We deduce that

lim inf_{r→0} N_{r}(U) = vol(U ∩ B_{R}) + lim inf_{r→0} N_{r}(U − B_{R})

and

lim sup_{r→0} N_{r}(U) = vol(U ∩ B_{R}) + lim sup_{r→0} N_{r}(U − B_{R}),

from which we can deduce (1), because vol(U ∩ B_{R}) < ∞. We deduce (2) from (1) by taking limits. Letting R → ∞ in the equalities above we see that

lim inf_{r→0} N_{r}(U) = vol(U) + lim_{R→∞} lim inf_{r→0} N_{r}(U − B_{R})

and

lim sup_{r→0} N_{r}(U) = vol(U) + lim_{R→∞} lim sup_{r→0} N_{r}(U − B_{R}),

in which some of the terms might be +∞. Now (3) follows from lim inf_{r→0} N_{r}(U) ≥ vol(U), and (4) follows because if

lim_{R→∞} lim sup_{r→0} N_{r}(U − B_{R}) = 0,

then

lim_{R→∞} lim inf_{r→0} N_{r}(U − B_{R}) = 0

also.

Lemma 2.6. If U is a subset of ℝ^{n} whose boundary has measure zero, and μ_{ℤ}(U) = vol(U), then vol(T · U) ∼ #(T · U ∩ ℤ^{n}) as T → ∞.

Proof. The statement follows immediately from the definitions.

Care is required in trying to compute the volume of I · F by counting lattice

points in it, for it is not a bounded set (even for k = 2, because (\begin{matrix}a & 0 \\ 0 & 1/a \end{matrix}) ∈ F).

Example 2.7. It’s easy to construct an unbounded region where counting lattice points does not determine the volume, by concentrating infinitely many very thin spikes along rays of rational slope with small numerator and denominator. Consider, for example, a bounded region B in ℝ^{2} with Jordan content and nonzero area v = vol B, for which (by Lemma 2.4) μ_{ℤ}B = vol B. Start by replacing B by its intersection B′ with the lines through the origin of rational (or infinite) slope – this doesn’t change the value of μ_{ℤ}, because every lattice point is contained in a line of rational slope, but now the boundary ∂B′ does not have measure zero. To repair that, we enumerate the lines M_{1}, M_{2}, … through the origin of rational slope, and for each i = 1, 2, 3, … we replace R_{i} = B ∩ M_{i} by a suitably scaled and rotated version L_{i} of it contained in the line N_{i} of slope i through the origin, with scaling factor chosen precisely so L_{i} intersects each r · ℤ^{2} in the same number of points as does R_{i}, for every r > 0. The scaling factor is the ratio of the lengths of the shortest lattice points in the lines M_{i} and N_{i}. The union L = ⋃ L_{i} has μ_{ℤ}L = μ_{ℤ}B = v ≠ 0, but it and its boundary have measure zero.

3 Reduction Theory

In this section we apply reduction theory to show that the volume of I · F can be computed by counting lattice points.

We introduce a few basic notions about lattices. For a more leisurely introduction see [7].

Definition 3.1. A lattice is a free abelian group L of finite rank equipped with an inner product on the vector space L ⊗ ℝ.

We will regard ℤ^{k} or one of its subgroups as a lattice by endowing it with the standard inner product on ℝ^{k}.

Definition 3.2. If L is a lattice, then a sublattice L′ ⊆ L is a subgroup with the induced inner product. The quotient L/L′, if it’s torsion free, is made into a lattice by equipping it with the inner product on the orthogonal complement of L′.

There’s a way to handle lattices with torsion, but we won’t need them.

Definition 3.3. If L is a lattice, then covol L denotes the volume of a fundamental domain for L acting on L ⊗ ℝ.

The covolume can be computed as | det(θv_{1}, ⋯, θv_{k})|, where θ : L⊗ℝ → ℝ^{k} is an isometry, {v_{1}, …, v_{k}} is a basis of L, and (θv_{1}, …, θv_{k}) denotes the matrix whose i-th column is θv_{i}. We have the identity covol(L) = covol(L′) · covol(L/L′) when L/L′ is torsion free.

If L is a subgroup of ℤ^{k} of finite index, then covol L = [ℤ^{k} : L].

Definition 3.4. If L is a nonzero lattice, then min L denotes the smallest length of a nonzero vector in L.

If L is a lattice of rank 1, then min L = covol L.

Proposition 3.5. For any natural number k > 0, there is a constant c such that for any S ≥ 1 and for any T > 0 the following inequality holds.

cS^{−k}T^{k^{2}} ≥ #{L ⊆ ℤ^{k} | [ℤ^{k} : L] ≤ T^{k} and min L ≤ T/S}.

Proof. For k = 1 we may take c = 2, so assume k ≥ 2. Letting N be the number of these lattices L, we bound N by picking within each L a nonzero vector v of minimal length, and counting the pairs (v, L) instead. For each v occurring in such pair we write v in the form v = n_{1}v_{1} where n_{1} ∈ ℕ and v_{1} is a primitive vector of ℤ^{k}, and then we extend {v_{1}} to a basis B = {v_{1}, …, v_{k}} of ℤ^{k}. We count the lattices L occurring in such pairs with v by putting a basis C for L into Hermite normal form with respect to B, i.e., it will have the form C = {n_{1}v_{1}, A_{12}v_{1} + n_{2}v_{2}, …, A_{1k}v_{1} + ⋯ + A_{k−1,k}v_{k−1} + n_{k}v_{k}}, with n_{i} > 0 and 0 ≤ A_{ij} < n_{i}. Notice that n_{1} has been determined in the previous step by the choice of v. The number of vectors v ∈ ℤ^{k} satisfying ∥v∥ ≤ T/S is bounded by a number of the form c(T/S)^{k}; for c we may take a large enough multiple of the volume of the unit ball. With notation as above, and counting the bases for C in Hermite normal form as before, we see that

N ≤ ∑_{∥v∥≤T/S} ∑_{{n_{2}>0,...,n_{k}>0 \atop n_{1}···n_{k}≤T^{k}}} n_{1}^{k−1}n_{2}^{k−2} ⋯ n_{k−1}^{1}n_{k}^{0}

= ∑_{∥v∥≤T/S} n_{1}^{k−1} ∑_{{n_{2}>0,...,n_{k}>0 \atop n_{2}···n_{k}≤T^{k}/n_{1}}} n_{2}^{k−2} ⋯ n_{k−1}^{1}n_{k}^{0}

= ∑_{∥v∥≤T/S} n_{1}^{k−1} · #{H ⊆ ℤ^{k−1} | [ℤ^{k−1} : H] ≤ T^{k}/n_{1}}

≤ ∑_{∥v∥≤T/S} n_{1}^{k−1}(T^{k}/n_{1})^{k−1} [by Lemma 1.2]

= ∑_{∥v∥≤T/S} T^{k(k−1)}

≤ c(T/S)^{k}T^{k(k−1)}

= cS^{−k}T^{k^{2}}.

Corollary 3.6. The following equality holds.

0 = lim_{S→∞} lim sup_{T →∞} T^{−k^{2}} · #{L ⊆ ℤ^{k} | [ℤ^{k} : L] ≤ T^{k} and min L ≤ T/S}

The following two lemmas are standard facts. Compare them, for example, with [2, 1.4 and 1.5].

Lemma 3.7. Let L be a lattice and let v ∈ L be a primitive vector. Let L̄ = L/ℤv, let w̄ ∈ L̄ be any vector, and let w ∈ L be a vector of minimal length among all those that project to w̄. Then ∥w∥^{2} ≤ ∥w̄∥^{2} + (1/4)∥v∥^{2}.

Proof. The vectors w and w ± v project to w̄, so ∥w∥^{2} ≤ ∥w ± v∥^{2} = ∥w∥^{2} + ∥v∥^{2} ± 2⟨w, v⟩, and thus |⟨w, v⟩| ≤ (1/2)∥v∥^{2}. We see then that

∥w̄∥^{2} = ∥w − \frac{⟨w, v⟩}{∥v∥^{2}}v∥^{2}

= ∥w∥^{2} − \frac{⟨w, v⟩^{2}}{∥v∥^{2}}

≥ ∥w∥^{2} − \frac{1}{4}∥v∥^{2}.

Lemma 3.8. Let L be a lattice of rank 2 with a nonzero vector v ∈ L of minimal length. Let L′ = ℤv and L′′ = L/L′. Then covol L′′ ≥ (\sqrt{3}/2) covol L′.

Proof. Let w̄ ∈ L′′ be a nonzero vector of minimal length, and lift it to a vector w ∈ L of minimal length among possible liftings. By lemma 3.7 ∥w∥^{2} ≤ ∥w̄∥^{2}+(1/4)∥v∥^{2}. Combining that with ∥v∥^{2} ≤ ∥w∥^{2} we deduce that covol L′′ = ∥w̄∥ ≥ (\sqrt{3}/2)∥v∥ = (\sqrt{3}/2) covol L′.

Definition 3.9. If L is a lattice, then minbasis L denotes the smallest value possible for (∥v_{1}∥^{2} + ⋯ + ∥v_{k}∥^{2})^{1/2}, where {v_{1}, …, v_{k}} is a basis of L.

Proposition 3.10. Given k ∈ ℕ and S ≥ 1, for all R ≫ 0, for all T > 0, and for all lattices L of rank k with covol L ≤ T^{k}, if minbasis L ≥ RT then min L ≤ T/S.

Proof. We show instead the contrapositive: provided covol L ≤ T^{k}, if min L > T/S then minbasis L < RT. There is an obvious procedure for producing an economical basis of a lattice L, namely: we let v_{1} be a nonzero vector in L of minimal length; we let v_{2} be a vector in L of minimal length among those projecting onto a nonzero vector in L/(ℤv_{1}) of minimal length; we let v_{3} be a vector in L of minimal length among those projecting onto a vector in L/(ℤv_{1}) of minimal length among those projecting onto a nonzero vector in L/(ℤv_{1} + ℤv_{2}) of minimal length; and so on. A vector of minimal length is primitive, so one can show by induction that the quotient group L/(ℤv_{1} + ⋯ + ℤv_{i}) is torsion free; the case where i = k tells us that L = ℤv_{1} + ⋯ + ℤv_{k}. Let L_{i} = ℤv_{1}+⋯+ℤv_{i}, and let α_{i} = covol(L_{i}/L_{i−1}), so that α_{1} = ∥v_{1}∥ = min L > T/S.

Applying Lemma 3.8 to the rank 2 lattice L_{i}/L_{i−2} shows that α_{i} ≥ Aα_{i−1}, where A = \sqrt{3}/2, and repeated application of Lemma 3.7 shows that ∥v_{i}∥^{2} ≤ α_{i}^{2} + (1/4)(α_{i−1}^{2} + ⋯ + α_{1}^{2}), so of course ∥v_{i}∥^{2} ≤ (1/4)(α_{k}^{2} + ⋯ + α_{i+1}^{2}) + α_{i}^{2} + (1/4)(α_{i−1}^{2} + ⋯ + α_{1}^{2}). We deduce that

minbasis L ≤ (∑_{i=1}^{k} ∥v_{i}∥^{2})^{1/2} ≤ (\frac{k + 3}{4} ∑ α_{i}^{2})^{1/2}.  (3)

Going a bit further, we see that

T^{k} ≥ covol L
= α_{1} ⋯ α_{k}
≥ A^{0+1+2+···+(i−2)}α_{1}^{i−1} · A^{0+1+2+···+(k−i)}α_{i}^{k−i+1}
> c_{1}(T/S)^{i−1}α_{i}^{k−i+1}

where c_{1} is some constant depending on S which we may take to be independent of i. Dividing through by T^{i−1} we get T^{k−i+1} > c_{2}α_{i}^{k−i+1}, from which we deduce that T > c_{3}α_{i}, where c_{2} and c_{3} are new constants (depending only on S). Combining these latter inequalities for each i, we find that (((k + 3)/4) ∑ α_{i}^{2})^{1/2} < RT, where R is a new constant (depending only on S); combining that with (3) yields the result.

Corollary 3.11. The following equality holds.

0 = lim_{R→∞} lim sup_{T →∞} T^{−k^{2}} · #{L ⊆ ℤ^{k} | [ℤ^{k} : L] ≤ T^{k} and minbasis L ≥ RT }

Proof. Combine (3.6) and (3.10).

If in the definition of our fundamental domain F we had taken the smallest element of each orbit, rather than the one nearest to 1, we would have been almost done now. The next lemma takes care of that discrepancy.

Definition 3.12. If L is a (discrete) lattice of rank k in ℝ^{k}, then size L denotes the value of (∥w_{1}∥^{2} + ⋯ + ∥w_{k}∥^{2})^{1/2}, where {w_{1}, …, w_{k}} is the (unique) basis of L satisfying (w_{1}, …, w_{k}) ∈ ℝ^{+} · F.

Lemma 3.13. For any (discrete) lattice L ⊆ ℝ^{k} of rank k the inequalities

minbasis L ≤ size L ≤ minbasis L + 2\sqrt{k}(covol L)^{1/k}

hold.

Proof. Let {v_{1}, …, v_{k}} be the basis envisaged in the definition of minbasis L, let {w_{1}, …, w_{k}} be the basis of L envisaged the definition of size L, and let U = (covol L)^{1/k} = (det(v_{1}, …, v_{k}))^{1/k} = (det(w_{1}, …, w_{k}))^{1/k}. The following chain of inequalities gives the result.

minbasis L = ∥(v_{1}, …, v_{k})∥ ≤ size L

= ∥(w_{1}, …, w_{k})∥ ≤ ∥(w_{1}, …, w_{k}) − U · 1_{k}∥ + U\sqrt{k}

≤ ∥(v_{1}, …, v_{k}) − U · 1_{k}∥ + U\sqrt{k}

≤ ∥(v_{1}, …, v_{k})∥ + 2U\sqrt{k} = minbasis L + 2U\sqrt{k} Corollary 3.14. The following equality holds.

0 = lim_{Q→∞} lim sup_{T →∞} T^{−k^{2}} · #{L ⊆ ℤ^{k} | [ℤ^{k} : L] ≤ T^{k} and size L ≥ QT }

Proof. It follows from (3.13) that given R > 0, for all Q ≫ 0 (namely Q ≥ R + 2\sqrt{k}) if covol L ≤ T^{k} and size L ≥ QT then minbasis L ≥ RT. Now apply (3.11).

Theorem 3.15. vol(I · F) = μ_{ℤ}(I · F).

Proof. Observe that #{L ⊆ ℤ^{k} | covol L ≤ T^{k} and size L ≥ QT } = #((T · I · F −B_{QT})∩M_{k}ℤ) = #((I ·F −B_{Q})∩T^{−1}M_{k}ℤ), so replacing 1/T by r, Corollary 3.14 implies that lim_{Q→∞} lim sup_{r→0} N_{r}(I · F − B_{Q}) = 0, which allows us to apply Lemma 2.5 (4).

The theorem allows us to compute the volume of F arithmetically, simultaneously showing it’s finite.

Corollary 3.16. μ_{∞}(G/Γ) = ζ(2)ζ(3) ⋯ ζ(k)/k

Proof. Combine the theorem with lemma 2.3 as follows.

μ_{∞}(G/Γ) = μ_{∞}(F) = vol(I · F) = μ_{ℤ}(I · F) = ζ(2)ζ(3) ⋯ ζ(k)/k

Remark 3.17. Theorem 10.4 in [22] states that the volume of G/Γ is ζ(2)ζ(3) ⋯ ζ(k)\sqrt{k}. The difference arises from a different choice of Haar measure on G. Theirs assigns volume \sqrt{k} to 𝔰𝔩_{k}(ℝ)/ 𝔰𝔩_{k}(ℤ), whereas ours assigns volume 1/k to it, as we see in formula (14) below. The ambiguity is unavoidable, because there is no canonical choice of Haar measure. (The Tamagawa number resolves that ambiguity.)

4 p-adic volumes

In this section we reformulate the computation of the volume of G/Γ to yield a natural and informative computation of the Tamagawa number of Sl_{k}. We are interested in the form of the proof, not its length, so we incorporate the proofs of (3.16) and (2) rather than their statements. The standard source for information about p-adic measures and Tamagawa measures is Chapter II of [37], and the proof we simplify occurs there in sections 3.1 through 3.4. See also [11] and [20].

We let μ_{p} denote the standard translation invariant measure on ℚ_{p} normalized so that μ_{p}(ℤ_{p}) = 1. Let μ_{p} also denote the product measure on the ring of k by k matrices, M_{k}(ℚ_{p}). Observe that μ_{p}(M_{k}(ℤ_{p})) = 1.

For x ∈ ℚ_{p}, let |x|_{p} denote the standard valuation normalized so that |p|_{p} = 1/p

If A ∈ M_{k}(ℚ_{p}) and U ⊆ ℚ_{p}^{k}, then μ_{p}(A · U) = | det A|_{p} · μ_{p}(U). (To prove this, first diagonalize A using row and column operations, and then assume that U is a cube.) It follows that if V ⊆ M_{k}(ℚ_{p}), then μ_{p}(A · V) = | det A|_{p}^{k} · μ_{p}(V).

Consider Gl_{k}(ℤ_{p}) as an open subset of M_{k}(ℤ_{p}). The following computation occurs on page 31 of [37].

\begin{aligned}μ_{p}(Gl_{k}(ℤ_{p})) & = #(Gl_{k}(𝔽_{p}))/p^{k^{2}} \\ & = (p^{k} − 1)(p^{k} − p) ⋯ (p^{k} − p^{k−1})/p^{k^{2}} \\ & = (1 − p^{−k})(1 − p^{−k+1}) ⋯ (1 − p^{−1}) \end{aligned}  (4)

Weil considers the open set M_{k}(ℤ_{p})^{∗} = {A ∈ M_{k}(ℤ_{p}) | det A ≠ 0}.

Lemma 4.1. μ_{p}(M_{k}(ℤ_{p})^{∗}) = 1

Proof. Let Z = M_{k}(ℤ_{p}) \ M_{k}(ℤ_{p})^{∗} be the set of singular matrices. If A ∈ Z, then one of the columns of A is a linear combination of the others. (This depends on ℤ_{p} being a discrete valuation ring – take any linear dependency with coefficients in ℚ_{p} and multiply the coefficients by a suitable power of p to put all of them in ℤ_{p}, with at least one of them being invertible.) For each n ≥ 0 we can get an upper bound for the number of equivalence classes of elements of Z modulo p^{n} by enumerating the possibly dependent columns, the possible vectors in the other columns, and the possible coefficients in the linear combination: μ_{p}(Z) ≤ lim_{n→∞} k · (p^{nk})^{k−1} · (p^{n})^{k−1}/(p^{n})^{k^{2}} = lim_{n→∞} k · p^{−n} = 0.

We call rank k submodules J of ℤ_{p}^{k} lattices. To each A ∈ M_{k}(ℤ_{p})^{∗} we associate the lattice J = Aℤ_{p}^{k} ⊆ ℤ_{p}^{k}. This sets up a bijection between the lattices J and the orbits of Gl_{k}(ℤ_{p}) acting on M_{k}(ℤ_{p})^{∗}. The measure of the orbit corresponding to J is μ_{p}(A · Gl_{k}(ℤ_{p})) = |det A|_{p}^{k} · μ_{p}(Gl_{k}(ℤ_{p})) = [ℤ_{p}^{k} : J]^{−k} · μ_{p}(Gl_{k}(ℤ_{p})). Now we sum over the orbits.

\begin{aligned}1 & = μ_{p}(M_{k}(ℤ_{p})^{∗}) \\ & = ∑_{J}([ℤ_{p}^{k} : J]^{−k} · μ_{p}(Gl_{k}(ℤ_{p}))) \\ & = (∑_{J}[ℤ_{p}^{k} : J]^{−k}) · μ_{p}(Gl_{k}(ℤ_{p})) \end{aligned}  (5)

An alternative way to prove (5) would be to use the local analogue of (2), which holds and asserts that ∑_{J}[ℤ_{p}^{k} : J]^{−s} = (1 − p^{k−1−s})^{−1}(1 − p^{k−2−s})^{−1} ⋯ (1 − p^{−s})^{−1}; we could substitute k for s and compare with the number in (4). The approach via lemma 4.1 and (5) is preferable because M_{k}(ℤ_{p})^{∗} provides natural glue that makes the computation seem more natural.

The product ∏_{p} μ_{p}(Gl_{k}(ℤ_{p})) doesn’t converge because ∏_{p}(1 − p^{−1}) doesn’t converge, so consider the following formula instead.

1 = ((1 − p^{−1}) ∑_{J}[ℤ_{p}^{k} : J]^{−k}) · ((1 − p^{−1})^{−1}μ_{p}(Gl_{k}(ℤ_{p})))

Now we can multiply these formulas together.

1 = (∏_{p}(1 − p^{−1}) ∑_{J}[ℤ_{p}^{k} : J]^{−k}) · ∏_{p}((1 − p^{−1})^{−1}μ_{p}(Gl_{k}(ℤ_{p}))) (6)

We’ve parenthesized the formula above so it has one factor for each place of ℚ, and now we connect each of them with a volume involving Sl_{k} at that place.

We use the Haar measure on Sl_{k}(ℤ_{p}) normalized to have total volume

# Sl_{k}(𝔽_{p})/p^{dim Sl^{k}}.

The normalization anticipates (13), which shows how a gauge form could be used to construct the measure, or alternatively, it ensures that the exact sequence 1 → Sl_{k}(ℤ_{p}) → Gl_{k}(ℤ_{p}) → ℤ_{p}^{×} → 1 of groups leads to the desired assertion μ_{p}(Gl_{k}(ℤ_{p})) = μ_{p}(ℤ_{p}^{×}) · μ_{p}(Sl_{k}(ℤ_{p})) about multiplicativity of measures. We rewrite the factor of the right hand side of (6) corresponding to the prime p as follows.

\begin{aligned}(1 − p^{−1})^{−1}μ_{p}(Gl_{k}(ℤ_{p})) & = μ_{p}(ℤ_{p}^{×})^{−1} · μ_{p}(Gl_{k}(ℤ_{p})) \\ & = μ_{p}(Sl_{k}(ℤ_{p})). \end{aligned}  (7)

To evaluate the left hand factor of the right hand side of (6), we insert the complex variable s. Because the ring ℤ is a principal ideal domain, any finitely generated sub-ℤ-module H ⊆ ℤ^{k} is free. Hence a lattice H ⊆ ℤ^{k} is determined freely by its localizations H_{p} = H ⊗_{ℤ} ℤ_{p} ⊆ ℤ_{p}^{k} (where H_{p} = ℤ_{p}^{k} for all but finitely many p), and its index is given by the formula

[ℤ^{k} : H] = ∏_{p}[ℤ_{p}^{k} : H_{p}],  (8)

Volumes of Symmetric Spaces via Lattice Points 439

in which only a finite number of terms are not equal to 1.

res_{s=k}ζ(ℤ^{k}, s)

= res_{s=k} ∑_{H}[ℤ^{k} : H]^{−s} [by (1)]

= lim_{s→k+} ζ(s − k + 1)^{−1} · ∑_{H}[ℤ^{k} : H]^{−s}

= lim_{s→k+}(ζ(s − k + 1)^{−1}( ∑_{H⊆ℤ^{k}} ∏_{p}[ℤ_{p}^{k} : H_{p}]^{−s})) [by (8)]

= lim_{s→k+}(ζ(s − k + 1)^{−1}(∏_{p} ∑_{J⊆ℤ_{p}^{k}}[ℤ_{p}^{k} : J]^{−s})) [positive terms]

= lim_{s→k+} ∏_{p}((1 − p^{−s+k−1}) ∑_{J}[ℤ_{p}^{k} : J]^{−s})

= ∏_{p}(1 − p^{−1}) ∑_{J}[ℤ_{p}^{k} : J]^{−k}

(9)

Starting again we get the following chain of equalities.

res_{s=k} ζ(ℤ^{k}, s) = res_{s=k} ζ(s − k + 1)ζ(s − k + 2) ⋯ ζ(s − 1)ζ(s)

= ζ(2) ⋯ ζ(k − 1)ζ(k)

= k · lim_{T →∞} T^{−k}#{H ⊆ ℤ^{k} | [ℤ^{k} : H] ≤ T } [by 1.1]

= k · lim_{T →∞} T^{−k^{2}}#{H ⊆ ℤ^{k} | [ℤ^{k} : H] ≤ T^{k}}

= k · lim_{T →∞} T^{−k^{2}}#{A ∈ HNF | det A ≤ T^{k}}

= k · μ_{ℤ}(I · F) [by definition 2.2]

= k · μ_{∞}(Sl_{k}(ℝ)/ Sl_{k}(ℤ)) [by 3.15 and 2.1]

(10)

Combining (9) and (10) we get the following equation.

∏_{p}(1 − p^{−1}) ∑_{J}[ℤ_{p}^{k} : J]^{−k} = k · μ_{∞}(Sl_{k}(ℝ)/ Sl_{k}(ℤ))  (11)

We combine (6), (7) and (11) to obtain the following equation.

1 = k · μ_{∞}(Sl_{k}(ℝ)/ Sl_{k}(ℤ)) · ∏_{p} μ_{p}(Sl_{k}(ℤ_{p}))  (12)

To relate this to the Tamagawa number we have to introduce a gauge form ω on the algebraic group Sl_{k} over ℚ, invariant by left translations, as in sections 2.2.2 and 2.4 of [37]. We can even get gauge forms over ℤ. Let X be a generic

Documenta Mathematica 11 (2006) 425–447

element of Gl_{k}. The entries of the matrix X^{−1}dX provide a basis for the 1forms invariant by left translation on Gl_{k}. On Sl_{k} we see that tr(X^{−1}dX) = d(det X) = 0, so omitting the element in the (n, n) spot will provide a basis of the invariant forms on Sl_{k}. We let ω be the exterior product of these forms. Just as in the proof of Theorem 2.2.5 in [37] we obtain the following equality.

∫_{Sl_{k}(ℤ_{p})} ω_{p} = μ_{p}(Sl_{k}(ℤ_{p}))  (13)

The measure ω_{p} is defined in [37, 2.2.1] in a neighborhood of a point P by writing ω = f dx_{1} ∧ ⋯ ∧ dx_{n} and setting ω_{p} = |f(P)|_{p}(dx_{1})_{p} … (dx_{n})_{p}, where (dx_{i})_{p} is the Haar measure on ℚ_{p} normalized so that ∫_{ℤ_{p}}(dx_{i})_{p} = 1, and |c|_{p} is the p-adic valuation normalized so that d(cx)_{p} = |c|_{p}(dx)_{p}.

Now we want to determine the constant that relates our original Haar measure μ_{∞} on Sl_{k}(ℝ) to the one determined by ω_{∞}. For this purpose, it will suffice to evaluate both measures on the infinitesimal parallelepiped B in Sl_{k}(ℝ) centered at the identity matrix and spanned by the tangent vectors εe_{ij} for i ≠ j and ε(e_{ii} − e_{kk}) for i < k. Here ε is an infinitesimal number, and e_{ij} is the matrix with a 1 in position (i, j) and zeroes elsewhere. For the purpose of this computation, we may even take ε = 1. We remark that B is a fundamental domain for 𝔰𝔩_{k}(ℤ) acting on the Lie algebra 𝔰𝔩_{k}(ℝ). We compute easily that ∫_{B} ω_{∞} = 1 and

μ_{∞}(B) = vol(I · B)
= (1/k^{2}) · | det(e_{11} − e_{kk}, ⋯, e_{k−1,k−1} − e_{kk}, ∑ e_{ii})|
= (1/k^{2}) · | det(e_{11} − e_{kk}, ⋯, e_{k−1,k−1} − e_{kk}, ke_{kk})|
= (1/k^{2}) · | det(e_{11}, ⋯, e_{k−1,k−1}, ke_{kk})|
= 1/k

(14)

We obtain the following equation.

μ_{∞}(Sl_{k}(ℝ)/ Sl_{k}(ℤ)) = \frac{1}{k} ∫_{Sl_{k}(ℝ)/ Sl_{k}(ℤ)} ω_{∞}  (15)

See [36, §14.12, (3)] for an essentially equivalent proof of this equation. We may now rewrite (12) as follows.

1 = ∫_{Sl_{k}(ℝ)/ Sl_{k}(ℤ)} ω_{∞} · ∏_{p} ∫_{Sl_{k}(ℤ_{p})} ω_{p}  (16)

(If done earlier, this computation would have justified normalizing μ_{∞} differently.)

The Tamagawa number τ(Sl_{k,ℚ}) = ∫_{Sl_{k}(𝔸_{ℚ})/ Sl_{k}(ℚ)} ω is the same as the right hand side of (16) because F × ∏_{p} Sl_{k}(ℤ_{p}) is a fundamental domain for the action of Sl_{k}(ℚ) on Sl_{k}(𝔸_{ℚ}). Thus τ(Sl_{k,ℚ}) = 1. This was originally proved by Weil in Theorem 3.3.1 of [37]. See also [14], [12], and [36, §14.11, Corollary to Langlands’ Theorem].

See also [33, §8] for an explanation that Siegel’s measure formula amounts to the first determination that τ(SO) = 2.

A Dirichlet series

Theorem A.1. Suppose we are given a Dirichlet series f(s) := ∑_{n=1}^{∞} a_{n}n^{−s} with nonnegative coefficients. Let A(T) := ∑_{n≤T} a_{n}. If A(T) = O(T^{k}) as T → ∞, then ∑_{n=T}^{∞} a_{n}n^{−s} = O(T^{k−s}) as T → ∞, and thus f(s) converges for all complex numbers s with Re s > k.

Proof. Write σ = Re s and assume σ > k. We estimate the tail of the series as follows.

∑_{n=T}^{∞} a_{n}n^{−s} = ∫_{T}^{∞} x^{−s} dA(x)

= x^{−s}A(x)]_{T}^{∞} − ∫_{T}^{∞} A(x) d(x^{−s})
= x^{−s}A(x)]_{T}^{∞} + s ∫_{T}^{∞} x^{−s−1}A(x) dx
= O(x^{k−σ})]_{T}^{∞} + s ∫_{T}^{∞} x^{−s−1}O(x^{k}) dx
= O(T^{k−σ}) + s ∫_{T}^{∞} O(x^{k−σ−1}) dx
= O(T^{k−σ})

Theorem A.2. Suppose we are given two Dirichlet series

f(s) := ∑_{n=1}^{∞} a_{n}n^{−s} g(s) := ∑_{n=1}^{∞} b_{n}n^{−s}

with nonnegative coefficients and corresponding coefficient summatory functions

A(T) := ∑_{n≤T} a_{n} B(T) := ∑_{n≤T} b_{n}

Assume that A(T) = O(T^{i}) and B(T) = cT^{k} + O(T^{j}), where i ≤ j < k. Let h(s) := f(s)g(s) = ∑_{n=1}^{∞} c_{n}n^{−s}, and let C(T) := ∑_{n≤T} c_{n}. Then C(T) = cf(k)T^{k} + O(T^{j} log T) if i = j, and C(T) = cf(k)T^{k} + O(T^{j}) if i < j.

Proof. The basic idea for this proof was told to us by Harold Diamond.

Observe that Theorem A.1 ensures that f(k) converges. Let’s fix the notation β(T) = O(γ(T)) to mean that there is a constant C so that |β(T)| ≤ Cγ(T) for all T ∈ [1, ∞), and simultaneously replace O(T^{j} log T) in the statement by O(T^{j}(1 + log T)) in order to avoid the zero of log T at T = 1. We will use the notation in an infinite sum only with a uniform value of the implicit constant C.

We examine C(T) as follows.

C(T) = ∑_{n≤T} c_{n} = ∑_{n≤T} ∑_{pq=n} a_{p}b_{q} = ∑_{pq≤n} a_{p}b_{q}

= ∑_{p≤T} a_{p} ∑_{q≤T/p} b_{q} = ∑_{p≤T} a_{p}B(T/p)

= ∑_{p≤T} a_{p}{c(T/p)^{k} + O((T/p)^{j})}

= cT^{k} ∑_{p≤T} a_{p}p^{−k} + O(T^{j}) ∑_{p≤T} a_{p}p^{−j}

= cT^{k}{f(k) + O(T^{i−k})} + O(T^{j}) ∑_{p≤T} a_{p}p^{−j}

= cf(k)T^{k} + O(T^{i}) + O(T^{j}) ∑_{p≤T} a_{p}p^{−j}

If i < j then ∑_{p≤T} a_{p}p^{−j} ≤ f(j) = O(1). Alternatively, if i = j, then

∑_{p≤T} a_{p}p^{−j} = ∑_{p≤T} a_{p}p^{−i} = ∫_{1−}^{T} p^{−i} d(A(p))

= p^{−i}A(p)]_{1−}^{T} − ∫_{1−}^{T} A(p) d(p^{−i})
= T^{−i}A(T) + i ∫_{1−}^{T} A(p)p^{−i−1} dp
= O(1) + O(∫_{1−}^{T} p^{−1} dp) = O(1 + log T)

In both cases the result follows.

The proof of the following “Abelian” theorem for generalized Dirichlet series is elementary.

Theorem A.3. Suppose we are given numbers R, k ≥ 1, and 1 ≤ λ_{1} ≤ λ_{2} ≤ ⋯ → ∞. Suppose that

N(T) := ∑_{λ_{n}≤T} 1 = (R + o(1))\frac{T^{k}}{k} (T → ∞)

for some number R. Then the generalized Dirichlet series ψ(s) := ∑ λ_{n}^{−s} converges for all real numbers s > k, and lim_{s→k+}(s − k)ψ(s) = R.

Proof. In the case R ≠ 0, the proof can be obtained by adapting the argument in the last part of the proof of [3, Chapter 5, Section 1, Theorem 3]: roughly, one reduces to the case where k = 1 by a simple change of variables, shows λ_{n} ∼ n/R, uses that to compare a tail of ∑ λ_{n}^{−s} to a tail of ζ(s) = ∑ n^{−s}, and then uses lim_{s→1+}(s − 1)ζ(s) = 1.

Alternatively, one can refer to [34, Theorem 10, p. 114] for the statement about convergence, and then to [34, Theorem 2, p. 219] for the statement about the limit. Actually, those two theorems are concerned with Dirichlet series of the form F (s) = ∑ a_{n}n^{−s}, but the first step there is to consider the growth rate of ∑_{n≤x} a_{n} as x → ∞. Essentially the same proof works for F (s) = ψ(s) by considering the growth rate of N(x) instead.

The result also follows from the following estimate, provided to us by Harold Diamond. Assume s > k.

ψ(s) := ∑ λ_{n}^{−s}

= ∫_{1−}^{∞} x^{−s} dN(x)

= x^{−s}N(x)]_{1−}^{∞} + s ∫_{1}^{∞} x^{−s−1}N(x) dx

= O(x^{k−s})]^{∞} + s ∫_{1}^{∞} x^{−s−1}(R + o(1))\frac{x^{k}}{k} dx (x → ∞)

= \frac{s(R + o(1))}{k} ∫_{1}^{∞} x^{−s−1+k} dx (s → k+)

= \frac{s(R + o(1))}{k(s − k)} (s → k+)

Notice the shift in the meaning of o(1) from one line to the next, verified by writing ∫_{1}^{∞} = ∫_{1}^{b} + ∫_{b}^{∞} and letting b go to ∞; it turns out that for sufficiently small ϵ the major contribution to ∫_{1}^{∞} x^{−1−ϵ} dx comes from ∫_{b}^{∞} x^{−1−ϵ} dx.

The following Wiener-Ikehara “Tauberian” theorem is a converse to the previous theorem, but the proof is much harder.

Theorem A.4. Suppose we are given numbers R > 0, k > 0, 1 ≤ λ_{1} ≤ λ_{2} ≤ ⋯ → ∞, and nonnegative numbers a_{1}, a_{2}, . . .. Suppose that the Dirichlet series ψ(s) = ∑ a_{n}λ_{n}^{−s} converges for all complex numbers with Re s > k, and that the function ψ(s) − R/(s − k) can be extended to a function defined and continuous for Re s ≥ k. Then

∑_{λ_{n}≤T} a_{n} ∼ RT^{k}/k.

Proof. Replacing s by ks allows us to reduce to the case where k = 1, which can be deduced directly from the Landau-Ikehara Theorem in [1], from Theorem 2.2 on p. 93 of [35], from Theorem 1 on p. 464 of [17], or from Theorem 1 on p. 534 of [18]. See also Theorem 17 on p. 130 of [40] for the case where λ_{n} = n, which suffices for our purposes. A weaker prototype of this theorem was first proved by Landau in 1909 [13, §241]. Other relevant papers include [39], [6], and [5]. See also Bateman’s discussion in [13, Appendix, page 931] and the good exposition of Abelian and Tauberian theorems in chapter 5 of [38].

References

[1] S. Bochner. Ein Satz von Landau und Ikehara. Math. Z., 37:1–9, 1933.

[2] Armand Borel. Introduction aux groupes arithmétiques. Publications de l’Institut de Mathématique de l’Université de Strasbourg, XV. Actualités Scientifiques et Industrielles, No. 1341. Hermann, Paris, 1969.

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Henri Gillet University of Illinois at Chicago http://www.math.uic.edu/˜ henri/ henri@math.uic.edu Daniel R. Grayson University of Illinois

at Urbana-Champaign http://www.math.uiuc.edu/˜ dan/ dan@math.uiuc.edu 448

Documenta Mathematica 11 (2006)