AN EXPLICIT FORMULA FOR LOCAL

DENSITIES OF QUADRATIC FORMS

Tonghai Yang

0. Introduction.

Let S and T be two positive definite integral matrices. It is an ancient but still very challenging problem to determine how many times S can represent T, i.e., the number of integral matrices X with ^{t}XSX = T. However, Siegel proved in his celebrated paper ([Si]) that certain weighted averages of these numbers over the genus of S can be expressed as the Euler product of pure local data—confluent hypergeometric functions for p = ∞ and local densities α_{p}(T, S) for p < ∞ (see (1.5) for definition). This reduced a difficult global problem to pure local problems. Siegel himself extended this result, as did Weil and most recently Kudla and Rallis. They show that theta integrals equal the central value of Eisenstein series (see [We] or [KR] for example). When the central value of the Eisenstein series is forced to be zero by its functional equation, Kudla asserted and proved ([Ku1]) in a very interesting special case that the central derivative of the Eisenstein series is deeply connected with height pairing on Shimura varieties. One part of his proof depends on the explicit formula of local density α_{p}(T, S) for rank T = 1 or 2, which was only known in the ‘unramified case’, i.e., p does not divide 2 det S, ([Si], [Ki]), until very recently. Bruce Meyers ([Me]) extended the local density formula to the case that pS^{−1} is p-integral and p ≠ 2 (rank T ≤ 2). Local density was also used in Gross and Keating’s work on the interesection number of modular correspondence ([GK]). The purpose of this work is to give a formula for α_{p}(T, S) for arbitrary S and rank T ≤ 2, with one exception rank T = 2 = p. Our method is quite different from others in this subject. Instead of using reduction and functional equation to reduce the problem to a simpler one and then computing the simpler one, we relate the local density to a (Whittaker) integral and then compute the integral directly. The results may be used to complete Kudla’s work in the special case mentioned above. The local Whittaker integrals we computed are local factors of the Fourier Coefficients of Siegel Eisenstein series on Sp(n) and are of importance in their own right. It should be mentioned that functional equation in the general case is very complicated and is not relating S to itself. The reduction formula is also absent in general and is only true after a certain stage (see Theorem 3.3). We should

also mention that the case where S is hyperbolic, i.e., S = \frac{1}{2} ( \begin{matrix}0 & I_{r} \\ I_{r} & 0 \end{matrix} ), has been

extensively studied as related to the classical Siegel Eisenstein series. Most recently, Katsurada gave a beautiful formula in this case with both T and p arbitrary. We refer to [Kat] and its bibliography for further reference in this matter. On the other hand, the confluent hypergeometric functions, the local problem at p = ∞, have been extensively studied by Shimura ([Sh]).

Typeset by AMS-TEX

In section 1, we review the relation between local densities and local Whittaker functions and set up basic notations. In section 2, we record formulas for a couple of simple Gauss type integrals which will be used in sections 3 and 7. In sections 3 (p ≠ 2) and 4 (p = 2), we obtain the local density formula when T is a nonzero number, and derive some of its consequences. In particular, we generalize Siegel’s reduction formula and give an explicit formula for local Whittaker integrals in this case, which is also true for T = 0. In the appendix after section 4, we work out a special example which includes a unexpected simple relation between local Whittaker functions coming from split and nonsplit quaternion algebras. In section 5, we prove an integral transformation formula which will translate the local Whittaker integral problem to a problem on Gauss integral on GL_{2}(ℤ_{p}). In section 6, we give a complete solution to the Gauss integral on GL_{2}(ℤ_{p}). Results in both sections 5 and 6 should have independent interests. It is also interesting to note that the Gauss integral just mentioned is related to rational points of certain elliptic curves over the finite field 𝔽_{p} in some case (Proposition 6.5). In section 7, we obtain an explicit formula for α_{p}(T, S) in the case rank T = 2, p ≠ 2 and arbitrary S.

Acknowledgment This work was inspired by Kudla’s paper ([Ku1]) mentioned above. The author thanks him for the inspiration and his continuing advice and encouragement. The author also thanks G. Prasad and D. Zagier for their valuable help during the preparation of this paper. Finally, the author would like to thank the Max-Planck Institut für Mathematik at Bonn for its hospitality and for providing the excellent working environment. The main part of this work was done during his stay at the MPI in the summer of 1997.

1. Whittaker functionals and local densities.

In this section, we recall some basic facts on Whittaker functionals and local density and their relations and set up notations for this paper. We mainly follow Kudla ([Ku1, Appendix]).

Let (V, q) be a nondegerate quadratic space over ℚ_{p} of dimension m with quadratic form q(x) = \frac{1}{2}(x, x). Let

G = \{ \begin{matrix}Sp(n)(ℚ_{p}) & if m = even, \\ Mp(n)_{ℚ_{p}} & if m = odd. \end{matrix}

Here Mp(n)_{ℚ_{p}} is the metaplectic cover of Sp(n)(ℚ_{p}). We choose and fix an additive character of ℚ_{p} as follows

ψ(x) = e^{−2πiλ(x)}

with λ : ℚ_{p} ⟶ ℚ_{p}/ℤ_{p} ⟶ ℚ/ℤ. Then the reductive dual pair (O(V), Sp(n)) gives rise to a Weil representation ω = ω_{ψ,V} of G on S(V^{n}). We refer [Ku2] for explicit formula and refer ([Ho], [MVW]) for basic facts on reductive dual pair and Weil representation. Let

χ_{V} (a) = (a, (−1)^{\frac{m(m−1)}{2}} det V))_{p}

be the quadratic character of ℚ_{p}^{∗} associated to V, where (x, y)_{p} is the Hilbert symbol. Let P = NM be the Siegel parabolic subgroup of G where

N = \{n(b) = ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ) : ^{t}b = b\}

is the unipotent radical of P and

M = \{m(a) = ( \begin{matrix}a & 0 \\ 0 & ^{t}a^{−1} \end{matrix} ) : a ∈ GL_{n}(ℚ_{p})\}

is a Levi factor of P. Then ([Ku2])

(1.1)  ω(m(a))ϕ(x) = χ_{V,ψ}(a)|a|^{m/2}ϕ(xa)
ω(n(b))ϕ(x) = ψ(\frac{1}{2} tr(b(x, x)))ϕ(x)
ω(w)ϕ(x) = γ(V) ∫_{V^{n}} ϕ(y)ψ(− tr(x, y))d_{V} y

where

χ_{V,χ}(a) = \{ \begin{matrix}χ_{V} (det a) & if m = even, \\ χ_{V} (a)γ(det a, \frac{1}{2}ψ)^{−1} & if m = odd \end{matrix}

is a character of GL_{n}(ℚ_{p}) (and of P), w = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) and

γ(V) = ϵ(V)γ(\frac{1}{2}ψ)^{m}γ(det V, \frac{1}{2}ψ).

Here ϵ(V) is the Hasse invariant of V, γ(ψ) and γ(a, ψ) are local Weil indices ([We], [Rao]). The Haar measure d_{V} y on V is chosen to be self-dual with respect to the bilinear character ψ((x, y)), x, y ∈ V. The Haar measure on V^{n} is the product measure.

Let I_{n}(s, χ_{V}) = Ind_{P}^{G}χ_{V} | |^{s} be the degenerated principal series. Then the map

S(V^{n}) ⟶ I_{n}(s_{0}, χ_{V}), ϕ ↦ {g ↦ ω(g)ϕ(0)}

is G-intertwining, where s_{0} = \frac{m−n−1}{2} and S(V^{n}) is the space of locally constant functions on V^{n} with compact support. Given ϕ ∈ S(V^{n}), we associate a standard section Φ ∈ I_{n}(s, χ_{V}) via

Φ(g, s) = ω(g)ϕ(0)|a(g)|^{s},

where a(g) = det a if g = n(b)m(a)k with b ∈ Sym_{n}(ℚ_{p}), a ∈ GL_{n}(ℚ_{p}) and k ∈ Sp(n)(ℤ_{p}). We remark that a(g) is not well-defined but |a(g)| is. Given T ∈ Sym_{n}(ℚ_{p}), one defines the T -th Whittaker functional of ϕ (or Φ) by

W_{T} (g, s, ϕ) = ∫ _{Sym_{n}(ℚ_{p})} Φ(wn(b)g, s)ψ(− trT b)db.

Here db is the Haar measure on Sym_{n}(ℚ_{p}) so that meas( Sym_{n}(ℤ_{p})) = 1. One may assume g = 1 by replacing ϕ with ω(g)ϕ. Easy calculation gives

(1.3) W_{T} (1, s_{0}, ϕ) = γ(V) ∫ _{Sym_{n}(ℚ_{p})} ∫_{V^{n}} ϕ(x)ψ(\frac{1}{2} trb(x, x))ψ(− trT b) d_{V} x db.

¿From now on, we assume that T is nonsingular, i.e., det T ≠ 0. Define the Weighted local density for a datum (T, V, ϕ) by

α_{p}(T, V, ϕ) = lim_{t⟶∞} ∫\begin{matrix}x∈V^{n} \\ \frac{1}{2} (x,x)≡T mod p^{t} \end{matrix} ϕ(x)d_{V} x.

Then Kudla proved in [Ku1] that

(1.4)  W_{T} (1, s_{0}, ϕ) = γ(V)α_{p}(T, V, ϕ).

The following basic properties of weighted local densities follow easily from definition and (1.1).

Lemma 1.1. Write ϕ^{a}(x) = ϕ(xa) for a ∈ GL_{n}(ℚ_{p}). Then (1) α_{p}(T, V, ω(n(b))ϕ) = ψ( trT b)α_{p}(T, V, ϕ).

(2) α_{p}(T, V, ω(m(a))ϕ) = χ_{V,ψ}(a)|a|^{m/2}α_{p}(T, V, ϕ).

(3) For a ∈ ℚ_{p}^{∗}, one has

α(aT, aV, ϕ) = |a|^{− \frac{n(n+1)}{2}} α_{p}(T, V, ϕ).

Here aV means the same space V with quadratic form aq.

Two symmetric matrices T and T′ over ℚ_{p} of degree n are called ℤ_{p}-equivalent if there is a matrix a ∈ GL_{n}(ℤ_{p}) such that ^{t}aT a = T′. We often write T [a] for taT a.

Lemma 1.2. For a ∈ GL_{n}(ℤ_{p}), one has

α_{p}(T [a], V, ϕ) = α_{p}(T, V, ϕ^{a})

In particular, α_{p}(T, V, ϕ) is only dependent on the ℤ_{p}-equivalence class of T when ϕ is right GL_{n}(ℤ_{p})-invariant.

Now turn to the classical local density. Given two integral nondegenerated symmetric matrices S and T over ℤ_{p} of degree m and n respectively. We assume m > n ≥ 1. Then the local density of representing T by S is defined by Siegel ([Si]) as follows:

(1.5)  α_{p}(T, S) = lim_{t⟶∞}(p^{t})^{\frac{n(n+1)}{2} −mn}A_{t}(T, S),

where

A_{t}(T, S) = #{X ∈ M_{m,n}(ℤ_{p}/p^{t}) : S[X] ≡ T mod p^{t}}.

For a nondegenerate symmetric matrix S of degree m, we associate a quadratic space V = ℚ_{p}^{m} of columns of length m with quadratic form q(x) = \frac{1}{2}(x, x) = ^{t}xSx. Let L = ℤ_{p}^{m} be its standard lattice. We will use a different Haar measure on V instead of the self-dual one d_{V} x in this paper. On ℚ_{p}, we fix the Haar measure so that meas(ℤ_{p}) = 1. On V, we use the product measure dx. The difference between the product measure dx and the self-dual measure d_{V} x is a factor | det 2S|^{\frac{1}{2}} , that is

(1.6)  ∫_{V} ϕ(x)d_{V} x = | det 2S|^{\frac{1}{2}} ∫_{V} ϕ(x)dx, ϕ ∈ S(V).

Set

(1.7)  W (T, S) = ∫ _{Sym_{n}(ℚ_{p})} ∫_{L^{n}} ψ( trbq(x))ψ(− trT b) dx db.

Lemma 1.3. With notation as above, one has

α_{p}(T, S) = W (T, S) = | det S|^{\frac{−n}{2}} γ(V)^{−1}W_{T} (1, s_{0}, char(L)^{n}),

where s_{0} = \frac{m−n−1}{2}.

Proof. The last equality follows from (1.3), (1.6) and (1.7). So one has by (1.4) for t >> 0,

W (T, S) = p^{t \frac{n(n+1)}{2}} ∫\begin{matrix}X∈L^{n} \\ \frac{1}{2} (X,X)≡T mod p^{t} \end{matrix} dX

= p^{t \frac{n(n+1)}{2}} ∑_{\begin{matrix}X∈(L/p^{t}L)^{n} \\ \frac{1}{2} (X,X)≡T mod p^{t} \end{matrix}} ∫_{X+p^{t}L^{n}} dX

= p^{t \frac{n(n+1)}{2}} ∑_{\begin{matrix}X∈(L/p^{t}L)^{n} \\ \frac{1}{2} (X,X)≡T mod p^{t} \end{matrix}} p^{tmn} meas(L, dx)^{n}

= (p^{t})^{\frac{n(n+1)}{2} −mn} ∑_{\begin{matrix}X∈(L/p^{t}L)^{n} \\ S[X]≡T mod p^{t} \end{matrix}} 1

since meas(L, dx) = 1. This finishes the proof.

To compute the Whittaker function W_{T} (1, s, ( charL)^{n}), one can use Kudla’s trick ([Ku1, Appendix]) as follows.

Lemma 1.4. For any integer r ≥ 0, let S_{r} = S ⊥ \frac{1}{2} ( \begin{matrix}0 & I_{r} \\ I_{r} & 0 \end{matrix} ). Then

W_{T} (1, r + s_{0}, ( charL)^{n}) = | det 2S|^{\frac{n}{2}} γ(V)α_{p}(T, S_{r}).

It is known that W_{T} (1, s, ϕ) is a rational function of X = q^{−s}. So Lemma 1.4 gives a formula for W_{T} (1, s, ( charL)^{n}) once we know a formula for α_{p}(T, S).

By definition, W (T, S) is only dependent on the ℤ_{p}-equivalence class of S. By theory of quadratic forms, we may assume, for p ≠ 2, that S is ℤ_{p}-equivalent to

(1.8)  diag(ϵ_{1}p^{l_{1}}, ⋯, ϵ_{m}p^{l_{m}}), ϵ_{i} ∈ ℤ_{p}^{∗}, l_{1} ≤ l_{2} ≤ ⋯ ≤ l_{m} ∈ ℤ.

Notice that l_{m} is the smallest integer such that p^{l_{m}}S^{−1} is half-integral. We end this section with some more notation needed later. Denote for any b, c ∈ ℤ.

(1.9)  L(b) = {1 ≤ i ≤ m : b + l_{i} < 0}, l(b) = #L(b),
L(b, c) = {i ∈ L(b) : b + l_{i} ≡ k mod 2}, l(b, c) = #L(b, c).

Notice that l(b) = l(b, 0) + l(b, 1) and that l(b) = m when b + l_{m} < 0. Moreover, if b + l_{m} < 0, L(b, c) is only dependent on the parities of b and c. Denote

(1.10)  δ_{p} = \{ \begin{matrix}1 & if p ≡ 1 mod 4, \\ \sqrt{−1} & if p ≡ −1 mod 4. \end{matrix}

Finally, we write (\frac{α}{p}) for the Hilbert symbol (α, p)_{p} for α ∈ ℤ_{p} in this paper.

2. Some integrals.

In this section, we compute the following four integrals needed later.

(2.1)  I(t) = ∫_{ℤ_{p}} ψ(tx^{2})dx, t ∈ ℚ_{p}^{∗},
(2.2)  I^{∗}(t) = ∫_{ℤ_{p}^{∗}} ψ(tx^{2})dx, t ∈ ℚ_{p}^{∗},
(2.3)  J(βp^{b}) = ∫_{ℤ_{p}^{m}} ψ(βp^{b}q(x))dx, β ∈ ℤ_{p}^{∗}, b ∈ ℤ,
(2.4)  I(t, χ) = ∫_{ℤ_{p}^{∗}} χ(x)ψ(tx)dx, t ∈ ℚ_{p}^{∗}.

Here q(x) the quadratic form associated to S as in section 1, and χ is a character of ℚ_{p}^{∗}. Here the Haar measure on ℚ_{p}^{∗} is the restriction of the Haar measure on ℚ_{p}. We assume p ≠ 2 in this section.

Lemma 2.1. Let t = αp^{a} ∈ ℚ_{p}^{∗} with α ∈ ℤ_{p}^{∗} and a ∈ ℤ.

(1) Let γ_{a}(α) = 1 if a is even, and = δ_{p}(\frac{−α}{p}) if a is odd. Then

I(t) = \{ \begin{matrix}1 & if a ≥ 0, \\ p^{\frac{a}{2}} γ_{a}(α) & if a < 0. \end{matrix}

(2)

I^{∗}(t) = I(t) − p^{−1}I(p^{2}t) = \{ \begin{matrix}1 − p^{−1} & if a ≥ 0, \\ p^{− \frac{1}{2}} γ_{1}(α) − p^{−1} & if a = −1, \\ 0 & if a ≤ −2. \end{matrix}

In particular, I(t) = p^{−1}I(p^{2}t) when ord_{p}t ≤ −2.

(3) For any β ∈ Z_{p}^{∗}, one has

∫_{β+pℤ_{p}} ψ(tx^{2})dx = p^{−1}ψ(tβ^{2}) char(p^{−1}ℤ_{p})(t).

Proof. (1). It is obvious for a ≥ 0. Assume a < 0. Let k = [\frac{−a+1}{2}] be the integral part of \frac{−a+1}{2}. Then

I(t) = ∑_{y∈ℤ_{p}/p^{k}} ∫_{y+p^{k}ℤ_{p}} ψ(tx^{2})dx

= ∑_{y∈ℤ_{p}/p^{k}} p^{−k}ψ(ty^{2}) ∫_{ℤ_{p}} ψ(2αp^{a+k}yx)dx

= p^{−k} ∑_{y∈ℤ_{p}/p^{k}} ψ(ty^{2}) char(p^{−a−k})(y)

When a is even, a = −2k and −a − k = k. So the only nonzero term in last integral is when y = 0. Therefore I(t) = p^{−k} = p^{a/2}. When a is odd, a = −2k + 1 and −a − k = k − 1. So

I(t) = p^{−k} ∑_{y∈ℤ_{p}/p} ψ(tp^{2k−2}y^{2})

= p^{−k} ∑_{y∈ℤ/p} e^{\frac{2πiαy^{2}}{p}}

= p^{−k}\sqrt{p}δ_{p}(\frac{−α}{p})

as expected, the final step is a well-known Gauss sum formula . Claim (2) follows from (1) easily. To prove (3), one first notices that

∫_{β+pℤ_{p}} ψ(tx^{2})dx = p^{−1}ψ(tβ^{2}) ∫_{ℤ_{p}} ψ(2βtpx)ψ(tp^{2}x^{2})dx.

Write t = αp^{a} with α ∈ ℤ_{p}^{∗} and a ∈ ℤ. When a ≥ −1,

∫_{ℤ_{p}} ψ(2βtpx)ψ(tp^{2}x^{2})dx = ∫_{ℤ_{p}} dx = 1

as expected. When a < −1, we have to prove that the integral vanishes. Substitute x by x + p^{−a−2}, one has

∫_{ℤ_{p}} ψ(2βtpx)ψ(tp^{2}x^{2})dx = ψ(2αβp^{−1}) ∫_{ℤ_{p}} ψ(2βtpx)ψ(tp^{2}x^{2})dx.

Since ψ(2αβp^{−1}) ≠ 1, one sees that the integral has to be zero.

As for integral (2.3), notice that q(x) = ∑ ϵ_{i}p^{l_{i}}x_{i}^{2} by (1.8). So Lemma 2.1 implies

Lemma 2.2. With notation as above, one has

J(βp^{b}) = p^{\frac{l(b)b}{2} + \frac{1}{2} ∑_{i∈L(b)} l_{i}}(−(\frac{β}{p})δ_{p})^{l(b,1)} ∏_{i∈L(b,1)}(\frac{ϵ_{i}}{p}).

In particular, J(βp^{b}) = 1 when b ≥ 0.

Corollary 2.3. Notation as above.

(1) J(βp^{b}) = (\frac{β}{p})^{l(b,1)}J(p^{b}).

(2) If b + l_{m} < 0 then

J(p^{b}) = p^{\frac{1}{2} (mb+∑ l_{i})}(−δ_{p})^{l(b,1)} ∏_{i∈L(b,1)}(\frac{ϵ_{i}}{p}).

Lemma 2.4. Let t = αp^{a} with α ∈ ℤ_{p}^{∗} and a ∈ ℤ, and let χ be a character of ℚ_{p} of conductor n(χ). Then

I(t, χ) = \{ \begin{matrix}1 − p^{−1} & if a ≥ 0 = n(χ), \\ p^{a}χ(−α)G(χ) & if n(χ) = −a, \\ 0 & otherwise. \end{matrix}

Here G(χ) is the Gauss sum of χ. In particular, one has

I(t, (\overline{p})) = p^{− \frac{1}{2}} δ_{p}(\frac{−α}{p}) char(p^{−1}ℤ_{p}^{∗})(t).

Proof. We may assume that a ≤ 0. First assume n(χ) = −a. Notice that χ can be viewed as a character of (ℤ/p^{n(χ)})^{∗}. So

I(t, χ) = ∑_{y∈(ℤ/p^{−a})} ∫_{y+p^{−a}ℤ_{p}} χ(x)ψ(tx) dx

= p^{a} ∑_{y∈(ℤ/p^{−a})^{∗}} χ(y)ψ(αp^{a}y)

= p^{a} ∑_{y∈(ℤ/p^{−a})^{∗}} χ(y)e^{2πiαyp^{a}}

= p^{a}χ(−α)G(χ)

as desired. Next we assume n(χ) > −a. In this case, one has

I(t, χ) = ∑_{y∈(ℤ/p^{−a})^{∗}} ∫_{y+p^{−a}ℤ_{p}} χ(x)ψ(tx) dx

= ∑_{y∈(ℤ/p^{−a})^{∗}} ψ(αp^{a}y)χ(y) ∫_{1+p^{−a}ℤ_{p}} χ(x)dx

= 0.

The case n(χ) < −a is similar.

3. The case n = 1 and p ≠ 2.

In this section, we assume that n = 1 and p ≠ 2 so that S is ℤ_{p}-equivalent to diag(ϵ_{1}p^{l_{1}}, ⋯, ϵ_{m}p^{l_{m}}) with ϵ_{i} ∈ ℤ_{p}^{∗} and l_{1} ≤ l_{2} ⋯ ≤ l_{m}. Set ϵ_{S} = ∏ ϵ_{i} ∈ ℤ_{p}^{∗}/ℤ_{p}^{∗2}. For any positive integer k, define

(3.1)  v_{k} = (\frac{−1}{p})^{[ \frac{l(−k,1)}{2} ]} ∏_{i∈L(−k,1)}(\frac{ϵ_{i}}{p}).

Here we recall l(−k, c) = #L(−k, c) for c = 0 or 1, and

L(−k, c) = {1 ≤ i ≤ m : l_{i} − k < 0 and l_{i} − k ≡ c mod 2}.

Denote

(3.2)  d(k) = k + \frac{1}{2} ∑_{l_{i}<k}(l_{i} − k).

Theorem 3.1. Let the notation be as above. Let t = αp^{a} with α ∈ ℤ_{p}^{∗} and a ∈ ℤ. Assume that p ≠ 2. Then α_{p}(t, S) = 0 unless a ≥ l_{1}. When a ≥ l_{1} one has

α_{p}(t, S) = p^{l_{1}} + (1 − p^{−1}) ∑_{\begin{matrix}l_{1}<k≤a \\ l(−k,1) even \end{matrix}} v_{k}p^{d(k)} + v_{a+1}f(t)p^{d(a+1)}

where

f(t) = \{ \begin{matrix}−\frac{1}{p} & if l(−a − 1, 1) is even, \\ (\frac{α}{p})\frac{1}{\sqrt{p}} & if l(−a − 1, 1) is odd. \end{matrix}

Proof. First notice that one has has by Lemma 1.1(3)

α_{p}(t, S) = p^{l}α_{p}(p^{−l}t, p^{−l}S).

It is not hard to see from this fact that it suffices to prove the formula for l_{1} = 0. In this case, it is obvious that α_{p}(t, S) = 0 when a < 0. Assume now a ≥ 0. By Lemmas 1.3 and 2.2, one has

α_{p}(t, S) = W (t, S)

= ∫_{ℚ_{p}} J(b)ψ(−tb)db

= 1 + ∑_{k>0} p^{k} ∫_{ℤ_{p}^{∗}} J(βp^{−k})ψ(−αβp^{a−k})dβ

= 1 + ∑_{\begin{matrix}k>0 \\ l(−k,1) even \end{matrix}} p^{d(k)}δ_{p}^{l(−k,1)} ∏_{i∈L(−k,1)}(\frac{ϵ_{i}}{p}) ∫_{ℤ_{p}^{∗}} ψ(−αβp^{a−k})dβ

+ ∑_{\begin{matrix}k>0 \\ l(−k,1) odd \end{matrix}} p^{d(k)}(−δ_{p})^{l(−k,1)} ∏_{i∈L(−k,1)}(\frac{ϵ_{i}}{p}) ∫_{ℤ_{p}^{∗}}(\frac{β}{p})ψ(−αβp^{a−k})dβ

Here δ_{p} is the number defined by (1.10). Notice that δ_{p}^{2} = (\frac{−1}{p}) and thus

v_{k} = \{ \begin{matrix}δ_{p}^{l(−k,1)} ∏_{i∈L(−k,1)}(\frac{ϵ_{i}}{p}) & if l(−k, 1) is even, \\ −δ_{p}^{l(−k,1)+1} ∏_{i∈L(−k,1)}(\frac{ϵ_{i}}{p}) & if l(−k, 1) is odd. \end{matrix}

So one has by Lemma 2.4

α_{p}(t, S) = 1 + ∑_{\begin{matrix}k>0 \\ l(−k,1) even \end{matrix}} v_{k}p^{d(k)}( char(p^{k}ℤ_{p}) − \frac{1}{p} char(p^{k−1}ℤ_{p}))(t)

+ ∑_{\begin{matrix}k>0 \\ l(−k,1) odd \end{matrix}} v_{k}p^{d(k)}(\frac{α}{p})p^{− \frac{1}{2}} char(p^{k−1}ℤ_{p}^{∗})(t)

= 1 + (1 − \frac{1}{p}) ∑_{\begin{matrix}k>0 \\ l(−k,1) even \end{matrix}} v_{k}p^{d(k)} + v_{a+1}f(t)p^{d(a+1)}.

When all l_{i} = 0, we recover a nice formula of Siegel ([Si])

Corollary 3.2. (Siegel) Assume p ≠ 2 and that S is unimodular and m > 1. Assume t = αp^{a} with a ∈ ℤ and α ∈ ℤ_{p}^{∗}. Then α_{p}(t, S) = 0 unless t is integral, i.e., a ≥ 0. In this case, let

ν = \{ \begin{matrix}((−1)^{\frac{m}{2}} det S, p)_{p} & if m is even, \\ ((−1)^{\frac{m−1}{2}} α det S, p)_{p} & if m is odd, \end{matrix}

and let

X = \{ \begin{matrix}p^{1− \frac{m}{2}} & if m is even, \\ p^{2−m} & if m is odd. \end{matrix}

Then

α_{p}(t, S) = \{ \begin{matrix}(1 − νp^{−1}X) ∑_{k=0}^{a}(νX)^{k} & if m is even, \\ (1 − p^{−1}X) ∑_{k=0}^{\frac{a−1}{2}} X^{k} & if m and a are odd, \\ (1 − p^{−1}X)(∑_{k=0}^{\frac{a}{2} −1} X^{k} + \frac{X^{\frac{a}{2}}}{1 − νp^{\frac{1−m}{2}}} ) & if m is odd and a is even. \end{matrix}

Proof. Since S is unimodular, S is ℤ_{p}-equivalent to diag(ϵ_{1}, ⋯, ϵ_{m}) with ϵ_{i} ∈ ℤ_{p}^{∗}. That is l_{i} = 0 for all i. So l(−k) = m for all k > 0, and l(−k, 1) = 0 or m depending on whether k is even or odd.

Case 1: m is even. In this case, v_{k} = ν^{k} and l(−k, 1) is always even. So

α_{p}(t, S) = 1 + (1 − \frac{1}{p}) ∑_{1≤k≤a} v_{k}p^{k(1− \frac{m}{2} )} − \frac{1}{p}v_{a+1}p^{(a+1)(1− \frac{m}{2} )}

= 1 + (1 − \frac{1}{p}) ∑_{1≤k≤a}(νX)^{k} − \frac{1}{p}(νX)^{a+1}

= (1 − νp^{−1}X) ∑_{0≤k≤a}(νX)^{k}

as expected.

Case 2: m is odd. In this case, l(−k, 1) is even if and only if k is even. One also has

v_{k} = \{ \begin{matrix}1 & if k is even, \\ ν(\frac{α}{p}) & if k is odd. \end{matrix}

When a is odd, one has

α_{p}(t, S) = 1 + (1 − \frac{1}{p}) ∑_{k=1}^{\frac{a−1}{2}} p^{2k(1− \frac{m}{2} )} − \frac{1}{p}p^{(a+1)(1− \frac{m}{2} )}

= (1 − p^{−1}X) ∑_{k=0}^{\frac{a−1}{2}} X^{k}

When a is even, a + 1 is odd.

α_{p}(t, S) = 1 + (1 − \frac{1}{p}) ∑_{k=1}^{\frac{a}{2}} p^{2k(1− \frac{m}{2} )} + (\frac{α}{p})\frac{1}{\sqrt{p}}ν(\frac{α}{p})p^{(a+1)(1− \frac{m}{2} )}

= ∑_{k=0}^{\frac{a}{2}} X^{k} − \frac{1}{p} ∑_{k=1}^{\frac{a}{2}} +νp^{\frac{1−m}{2}} X^{\frac{a}{2}}

= (1 − p^{−1}X) ∑_{k=0}^{\frac{a}{2} −1} X^{k} + X^{\frac{a}{2}} (1 + νp^{\frac{1−m}{2}} )

= (1 − p^{−1}X)(∑_{k=0}^{\frac{a}{2} −1} X^{k} + \frac{X^{\frac{a}{2}}}{1 − νp^{\frac{1−m}{2}}} )

as desired. Here we have used

1 − p^{−1}X = 1 − p^{1−m} = (1 + νp^{\frac{1−m}{2}} )(1 − νp^{\frac{1−m}{2}} ).

Theorem 3.3 (Induction formula). Let the notation be as in Theorem 3.1. Then for ord_{p}t ≥ l_{m} + 2, one has

α_{p}(tp^{2}, S) − α_{p}(t, S) = p^{2−m}(α_{p}(t, S) − α_{p}(tp^{−2}, S)).

Furthermore, if m is even, one has for ord_{p}t ≥ l_{m} + 1

α_{p}(tp, S) − α_{p}(t, S) = νp^{1− \frac{m}{2}} (α_{p}(t, S) − α_{p}(tp^{−1}, S)).

Here ν is the number defined in Corollary 3.2 with det S replaced by ϵ_{S} = ∏ ϵ_{i}.

Proof. Let t = αp^{a} with a ≥ l_{m}. First notice that d(k + 1) = d(k) + 1 − \frac{m}{2} for k > l_{m}. One has by Theorem 3.1

α_{p}(tp^{2}, S) − α_{p}(t, S)

= (1 − \frac{1}{p}) ∑_{\begin{matrix}a+1≤k≤a+2 \\ l(−k,1) even \end{matrix}} v_{k}p^{d(k)} + v_{a+3}f(tp^{2})p^{d(a+3)} − v_{a+1}f(t)p^{d(a+1)}

= p^{d(a+1)}\{(1 − \frac{1}{p}) ∑_{\begin{matrix}0≤k≤1 \\ l(−k−a−1,1) even \end{matrix}} v_{a+1+k}p^{k(1− \frac{m}{2} )}

+ v_{a+3}f(αp^{a+2})p^{2(1− \frac{m}{2} )} − v_{a+1}f(αp^{a})\}.

Notice that v_{k}, l(−k, 1), and f(αp^{k−1}) depend only on the parity of k when k > l_{m}. So the expression in the brace does not change when we replace a by a+2. Therefore

α_{p}(tp^{4}, S) − α_{p}(tp^{2}, S) = p^{2−m}(α_{p}(tp^{2}, S) − α_{p}(t, S))

when a = ord_{p}t ≥ l_{m}. This proves the first induction formula.

Now assume that m is even. Again write t = αp^{a} with a ≥ l_{m}. There are two cases. If l(−a − 1, 1) is even, so is l(−a − 2, 1) = m − l(−a − 1, 1) (thanks to a + 1 > l_{m}). So

α_{p}(tp, S) − α_{p}(t, S)

= (1 − \frac{1}{p})v_{a+1}p^{d(a+1)} − \frac{1}{p}v_{a+2}p^{d(a+2)} + \frac{1}{p}v_{a+1}p^{d(a+1)}

= v_{a+1}p^{d(a+1)}(1 − νp^{− \frac{m}{2}} ).

Here we have used the fact that when m is even

(3.3)  \frac{v_{a+1}}{v_{a+2}} = v_{a+1}v_{a+2} = ν.

Because of (3.3), it is now obvious that

α_{p}(tp^{2}, S) − α_{p}(tp, S) = νp^{1− \frac{m}{2}} (α_{p}(tp, S) − α_{p}(t, S))

for a ≥ l_{m}. If l(−a − 1, 1) is odd, so is l(−a − 2, 1) = m − l(−a − 1, 1). In this case Theorem 3.1 and Formula (3.3) imply

α_{p}(tp, S) − α_{p}(t, S) = v_{a+2}(\frac{α}{p})\frac{1}{\sqrt{p}}p^{d(a+2)} − v_{a+1}(\frac{α}{p})\frac{1}{\sqrt{p}}p^{d(a+1)}

= −v_{a+1}(\frac{α}{p})\frac{1}{\sqrt{p}}p^{d(a+1)}(1 − νp^{1− \frac{m}{2}} ).

So one has again by (3.3)

α_{p}(tp^{2}, S) − α_{p}(tp, S) = νp^{1− \frac{m}{2}} (α_{p}(tp, S) − α_{p}(t, S))

for a = ord_{p}t ≥ l_{m}. This proves the second induction formula.

Theorem 3.4. (1) For any ϕ ∈ S(V), g ∈ G and t ≠ 0, W_{t}(g, s + s_{0}, ϕ) is a polynomial of X = p^{−s} over ℚ_{p}.

(2) Suppose ϕ = char(L) for a lattice L of V, let S = diag(ϵ_{1}p^{l_{1}}, ⋯, ϵ_{m}p^{l_{m}}) be the matrix of L with respect to a suitable basis with l_{1} ≤ ⋯ ≤ l_{m} as explained in section 1. Then W_{t}(1, s+s_{0}, char(L)) is a polynomial of X of degree max(0, ord_{p}t+ 1). More precisely, for t = αp^{a} with a ≥ min(0, l_{1}), one has

(3.4) C^{−1}W_{t}(1, s + s_{0}, char(L))

= 1 + (1 − p^{−1}) ∑_{0<k≤min(l_{1},a)} p^{k}X^{k}

+ (1 − p^{−1}) ∑_{\begin{matrix}min(l_{1},a)<k≤a \\ l(−k,1) even \end{matrix}} v_{k}p^{d(k)}X^{k} + v_{a+1}f(t)p^{d(a+1)}X^{a+1}

for l_{1} ≥ 0, and

C^{−1}W_{t}(1, s + s_{0}, char(L)) (3.5)

= p^{l_{1}} + (1 − p^{−1}) ∑_{\begin{matrix}l_{1}<k≤a \\ l(−k,1) even \end{matrix}} v_{k}p^{d(k)}X^{max(k,0)} + v_{a+1}f(t)p^{d(a+1)}X^{max(a+1,0)}

for l_{1} < 0, where C = | det S|^{\frac{1}{2}} γ(V) is a constant.

Proof. Since W_{t}(g, s, ϕ) = W_{t}(1, s, ω(g)ϕ) and every ϕ ∈ S(V) is a linear combination of characteristic functions char(L) of lattices, it is enough to prove the formulas (3.4) and (3.5). For any large integer r > 0, one has by Lemmas 1.3 and (1.4)

W_{t}(1, r + s_{0}, char(L)) = Cα_{p}(t, S_{r})

where

S_{r} = S ⊥ ( \begin{matrix}0 & \frac{1}{2}I_{r} \\ \frac{1}{2}I_{r} & 0 \end{matrix} ) .

Now (3.4) and (3.5) follow from Theorem 3.1 easily. For example, if l_{1} < 0, let l_{r}(−k, 1) and d_{r}(k) stand for the numbers l(−k, 1) and d(k) for the matrix S_{r} respectively. Then

l_{r}(−k, 1) = \{ \begin{matrix}l(−k, 1) & if k ≤ 0, \\ l(−k, 1) + 2r & if k > 0, \end{matrix}

and

d_{r}(k) = \{ \begin{matrix}d(k) & if k ≤ 0, \\ d(k) − kr & if k > 0. \end{matrix}

In particular, l_{r}(−k, 1) is even if and only if l(−k, 1) is even. Also notice that v_{k} and f(t) do not change. Applying, one obtains formula (3.5). Formula (3.4) is similar and left to the reader.

Remark 3.5. The formulas in Theorem 3.4 are also true for t = 0, i.e., a = ∞. In this case, W_{0}(1, s + s_{0}, charL) is a rational function of X and has a pole at s = s_{0}. In fact, Theorem 3.1 is also true for t = 0 for m > n+1. We will give an interesting example involving quaternion algebras in the appendix after section 4.

4. The case n = 1 and p = 2.

In this section, we consider the case n = 1 and p = 2. By [Ca, Lemma 8.4.1], a symmetric matrix (nonsingular) over ℚ_{2} is ℤ_{2} equivalent to

(4.1) diag(ϵ_{1}2^{l_{1}}, ⋯, ϵ_{L}2^{l_{L}}) ⊕ (⊕_{i=1}^{M}ϵ′_{i}2^{m_{i}} ( \begin{matrix}0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{matrix} )) ⊕ (⊕_{j=1}^{N}ϵ′′_{j}2^{n_{j}} ( \begin{matrix}1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{matrix} ))

where ϵ_{h}, ϵ′_{i}, ϵ′′_{j} ∈ ℤ_{2}^{∗}, l_{h}, m_{i}, and n_{j} are all integers. Let l be the smallest integer among all l_{h}, m_{i}, and n_{j}. Of course, L + 2M + 2N = m is the degree of S. The corresponding lattice has a basis over which the corresponding quadratic form is given by

(4.2)  q = ∑_{h=1}^{L} ϵ_{h}2^{l_{h}}x_{i}^{2} + ∑_{i=1}^{M} ϵ′_{i}2^{m_{i}}y_{i1}y_{i2} + ∑_{j=1}^{N} ϵ′′_{j}2^{n_{j}}(z_{j1}^{2} + z_{j1}z_{j2} + z_{j2}^{2}).

For each integer k > 0 and t = α2^{a} with α ∈ ℤ_{2}^{∗} and a ∈ ℤ, we denote

p(k) = (−1)^{∑_{n_{j} <k}(n_{j}−k)}
(4.3)  ϵ(k) = ∏_{h∈L(−k+1,1)} ϵ_{h}
μ_{k}(t) = α2^{a−k+3} − ∑_{l_{h}<k−1} ϵ_{h}

and

(4.4)  d(k) = k + \frac{1}{2} ∑_{l_{h}<k−1}(l_{h} − k + 1) + ∑_{m_{i}<k}(m_{i} − k) + ∑_{n_{j}<k}(n_{j} − k).

Here

L(b, c) = {1 ≤ h ≤ L : l_{h} + b < 0 & l_{h} + b ≡ c mod 2}

and l(b, c) = #L(b, c) for any integers. We also define the Kronecker symbol for x ∈ ℚ_{2}

(4.5)  (\frac{2}{x}) = \{ \begin{matrix}(2, x)_{2} & if x ∈ ℤ_{2}^{∗}, \\ 0 & otherwise. \end{matrix}

Finally, we define δ(k) = 0 or 1 depending on whether there exists l_{h} = k − 1 or not.

Theorem 4.1. With notation as above, one has α_{2}(α2^{a}, S) = 0 unless a ≥ l. Furthermore, for a ≥ l one has

α_{2}(α2^{a}, S) = p^{l} + ∑_{\begin{matrix}l<k≤a+3 \\ l(−k+1,1) odd \end{matrix}} δ(k)p(k) (\frac{2}{μϵ(k)}) 2^{d(k)− \frac{3}{2}}

+ ∑_{\begin{matrix}l<k≤a+3 \\ l(−k+1,1) even \end{matrix}} δ(k)p(k) (\frac{2}{ϵ(k)}) 2^{d(k)−1}ψ(\frac{μ}{8}) char(4ℤ_{2}))(μ).

Here we write μ = μ_{k}(t) to save space.

We remark that the terms in the formula with k ≤ a are independent of a. From this it is not difficult to derive a inductive formula for p = 2 similar to Theorem 3.3. However, even when m is even, one does not have a reduction formula for α_{2}(2t, S)−α_{2}(t, S) similar to that of Theorem 3.3 since there is not a good reduction formula for μ_{k}(t). In other words, we only have

Corollary 4.2. Let the notation be as in Theorem 4.1. Then for a ≥ 4+max(l_{h}, m_{i}, n_{j}), one has

α_{2}(4t, S) − α_{2}(t, S) = 2^{2−m} (α_{2}(t, S) − α_{2}(\frac{1}{4}t, S)) .

In order to prove Theorem 4.1 we need two lemmas. The first one is similar to Lemmas 2.1 and 2.4 and the proof is left to the reader.

Lemma 4.3. (1) For α ∈ ℤ_{2}^{∗} and a ∈ ℤ, one has

∫_{ℤ_{2}} ψ(α2^{a}x^{2})dx

= \{ \begin{matrix}1 & if a ≥ 0 \\ 0 & if a = −1 \\ 2^{\frac{a+1}{2}} ψ(\frac{α}{8})(\frac{2}{α})^{a+1} & if a < −1. \end{matrix}

LOCAL DENSITIES 15

(2) Let χ be a character of ℚ_{2}^{∗} with conductor n(χ). Then for α and a as above, one has

∫_{ℤ_{2}^{∗}} χ(x)ψ(α2^{a}x)dx

= \{ \begin{matrix}\frac{1}{2} & if n(χ) = 0 & a ≥ 0 \\ −\frac{1}{2} & if n(χ) = 0 & a = −1 \\ 2^{a}χ(−α)G(χ) & if n(χ) = −a \\ 0 & otherwise. \end{matrix}

In particular, one has

∫_{ℤ_{2}^{∗}}(\frac{2}{x})ψ(α2^{a}x)dx = \frac{1}{2\sqrt{2}}(\frac{2}{−α})δ_{−a,3}.

Here δ_{m,n} is the usual Kronecker symbol. Lemma 4.4. For t = α2^{a} with α ∈ ℤ_{2}^{∗} and a ∈ ℤ, one has

(4.6)  ∫_{ℤ_{2}^{2}} ψ(ty_{1}y_{2})dy_{1} dy_{2} = min(1, 2^{a})

and ∫_{ℤ_{2}^{2}} ψ(t(z_{1}^{2} + z_{1}z_{2} + z_{2}^{2})dz_{1}dz_{2} (4.7)

= \{ \begin{matrix}1 & if a ≥ 0 \\ 2^{a}(−1)^{a} & if a < 0. \end{matrix}

Proof. Only (4.7) needs some verification for a < 0. Set k = [\frac{−a+1}{2}]. Then ∫_{ℤ_{2}} ψ(t(z_{1}z_{2} + z_{2}^{2}))dz_{2}

= ∑_{x∈ℤ/2^{k}} 2^{−k}ψ(t(z_{1}x + x^{2})) ∫_{ℤ_{2}} ψ(α2^{a+k}(z_{1} + 2x)z_{2})dz_{2}

= 2^{−k} ∑_{x∈ℤ/2^{k}} ψ(t(z_{1}x + x^{2})) char(−2x + 2^{−a−k}ℤ_{2})(z_{1}).

So ∫_{ℤ_{2}^{2}} ψ(t(z_{1}^{2} + z_{1}z_{2} + z_{2}^{2}))dz_{1}dz_{2}

= 2^{−k} ∑_{x∈ℤ/2^{k}} ∫_{−2x+2^{−a−k}ℤ_{2}} ψ(t(z_{1}^{2} + z_{1}x + x^{2}))dz_{1}

= 2^{a} ∑_{x∈ℤ/2^{k}} ψ(3tx^{2}) ∫_{ℤ_{2}} ψ(−3xα2^{−k}z_{1})ψ(α2^{−2k−a}z_{1}^{2})dz_{1}.

When a = −2k is even, the last integral is

∫_{ℤ_{2}} ψ(−3xα2^{−k}z_{1})dz_{1} = char(2^{k}ℤ_{2})(x).

This proves the even case. When a = −2k + 1 is odd, the last integral is

∫_{ℤ_{2}} ψ(−3xα2^{−k}z_{1})ψ(\frac{α}{2}z_{1}^{2})dz_{1}

= ∫_{2ℤ_{2}} ψ(−3xα2^{−k}z_{1})dz_{1} − ∫_{1+2ℤ_{2}} ψ(−xα2^{−k}z_{1})dz_{1}

= \frac{1}{2}(1 − ψ(−3xα2^{−k})) char(2^{k−1}ℤ_{2})(x)

= char(2^{k−1}ℤ_{2}^{∗})(x).

Therefore

∫_{ℤ_{2}^{2}} ψ(t(z_{1}^{2} + z_{1}z_{2} + z_{2}^{2}))dz_{1}dz_{2}

= 2^{a}ψ(3α2^{a}2^{2k−2})

= 2^{a}ψ(\frac{3α}{2}) = −2^{a}.

Proof of Theorem 4.1 Now the proof of Theorem 4.1 becomes a formality. Just as in the proof of Theorem 3.1, we may assume that l = 0. Obviously, α_{2}(t, S) = 0 for a < 0. Assume a ≥ 0. By Lemma 1.3, one has for t = α2^{a}

α_{2}(t, S))

= ∫_{ℚ_{2}} ∫_{ℤ_{2}^{m}} ψ(bq(x))dxψ(−tb)db

= 1 + ∑_{k=1}^{∞} 2^{k} ∫_{ℤ_{2}^{∗}} J_{1}(β2^{−k})J_{2}(β2^{−k})J_{3}(β2^{−k})ψ(−αβ2^{a−k})dβ

where

J_{1}(b) = ∏_{h=1}^{L} ∫_{ℤ_{2}} ψ(ϵ_{h}2^{l_{h}}bx_{h}^{2})dx

J_{2}(b) = ∏_{i=1}^{M} ∫_{ℤ_{2}^{2}} ψ(ϵ_{i}2^{m_{i}}by_{i1}y_{i2})dy_{i1}dy_{i2}

J_{3}(b) = ∏_{j=1}^{N} ∫_{ℤ_{2}^{2}} ψ(ϵ_{j}2^{n_{j}}b(z_{j1}^{2} + z_{j1}z_{j2} + z_{j2}^{2}))dz_{j1}dz_{j2}.

By Lemma 4.3, one has

J_{1}(β2^{−k}) = δ(k)(\frac{2}{ϵ(k)})2^{∑^{l_{h}<k−1} \frac{l_{h}−k+1}{2}} ∏_{l_{h}<k−1} ψ(\frac{ϵ_{h}β}{8} )(\frac{2}{α})^{l(−k+1,1)}.

By Lemma 4.4, one has

J_{2}(β2^{−k}) = 2^{∑_{m_{i}<k} m_{i}−k}

and

J_{3}(β2^{−k}) = p(k)2^{∑_{n_{j} <k} n_{j}−k}.

Therefore

α_{2}(α2^{a}, S)

= 1 + ∑_{k=1}^{∞} δ(k)p(k)(\frac{2}{ϵ(k)})2^{d(k)} ∫_{ℤ_{2}^{∗}}(\frac{2}{α})^{l(−k+1,1)}ψ(\frac{−μ_{k}(t)}{8}β)dβ.

The last integral is always zero for k > a + 3 since μ_{k}(t) ∉ ℤ_{2} in this case.

When l(−k + 1, 1) is odd, the last integral is equal to 2^{− \frac{3}{2}} (\frac{2}{μ_{k}(t)}) by Lemma 4.3 and our notation (4.5). When l(−k + 1, 1) is even, the last integral is equal to

∫_{ℤ_{2}^{∗}} ψ(\frac{−μ_{k}(t)}{8}β)dβ = \frac{1}{2}ψ(−\frac{μ_{k}(t)}{8}) ∫_{ℤ_{2}} ψ(\frac{−μ_{k}(t)}{4}β)dβ

= \frac{1}{2}ψ(\frac{μ_{k}(t)}{8}) char(4ℤ_{2})(μ).

Now Theorem 4.1 follows easily.

Just as Theorem 3.4, the following theorem follows easily from Theorem 4.1.

Theorem 4.5. Let L be a lattice of V with matrix S given by (4.1) (with respect to a suitable basis). Let t = α2^{a}. Then W_{t}(1, s + s_{0}, L) = 0 unless a ≥ min(0, l). Set C = | det 2S|^{\frac{1}{2}} γ(V) and X = 2^{−s}.

(1) When l ≥ 0, one has

C^{−1}W_{t}(1, s + s_{0}, char(L))

= 1 + ∑_{\begin{matrix}l<k≤a+3 \\ l(−k+1,1) odd \end{matrix}} δ(k)p(k) (\frac{2}{μϵ(k)}) 2^{d(k)− \frac{3}{2}} X^{k}

+ ∑_{\begin{matrix}0<k≤a+3 \\ l(−k+1,1) even \end{matrix}} δ(k)p(k) (\frac{2}{ϵ(k)}) 2^{d(k)−1}ψ(\frac{μ}{8}) char(4ℤ_{2}))(μ)X^{k}.

where μ = μ_{k}(t). Moreover, the k-th term with 0 < k ≤ min(l, a) is just 2^{k−1}X^{k}. (2) When l < 0, one has

C^{−1}W_{t}(1, s + s_{0}, char(L))

= p^{l} + ∑_{\begin{matrix}l<k≤a+3,k≠0 \\ l(−k+1,1) odd \end{matrix}} δ(k)p(k) (\frac{2}{μϵ(k)}) 2^{d(k)− \frac{3}{2}} X^{max(0,k)}

+ ∑_{\begin{matrix}l<k≤a+3,k≠0 \\ l(−k+1,1) even \end{matrix}} δ(k)p(k) (\frac{2}{ϵ(k)}) 2^{d(k)−1}ψ(\frac{μ}{8}) char(4ℤ_{2}))(μ)X^{max(0,k)}

Here μ = μ_{k}(t) as above.

(3) For any ϕ ∈ S(V), g ∈ G and t ≠ 0, W_{t}(g, s + s_{0}, ϕ) is a polynomial of X = 2^{−s} over ℚ_{2}.

We remark that the formulas in Theorem 4.5 is also true for t = 0, i.e., a = ∞. In this case, W_{0}(1, s+s_{0}, char(L)) is a rational function of X instead of a polynomial. One has μ_{k}(0) = ϵ_{2}(k) for all k.

Appendix: An example.

We work out an interesting example in this appendix for future application. Let p < ∞ be any prime number. Let B be a quaternion algebra over ℚ_{p}, and let R be a maximal order of B. Let κ ∈ ℚ_{p}^{∗}. Let V = V_{κ} = {x ∈ B : trx = 0} be with quadratic form q(x) = −κN(x), and let L = R ∩ V be a lattice of V. We use superscript ‘s’ or ‘r’ to refer to the split or ramified case. For example, B^{s} stands for the split quaternion algebra over ℚ_{p}. Denote l = ord_{p}κ and assume l = 0 or 1. t ∈ ℚ_{p}^{∗}, we write t = αp^{a+l} with α ∈ ℤ_{p}^{∗} and a ∈ ℤ. Put

(A1)  C = C(κ) = p^{− \frac{3}{2} l− \frac{1}{2} δ_{p,2}}(κ, −1)_{p}γ(−κ, \frac{1}{2}ψ)γ(\frac{1}{2}ψ)^{3},

and

(A2)  v(a) = v(t, κ) = \{ \begin{matrix}(ακp^{−l}, δ_{p,2} − p)_{p} & if a is even, \\ −1 & if a is odd. \end{matrix}

Here δ_{m,n} is the usual Kronecker symbol. Notice that ακp^{−1} is just the ‘unit’ part of tκ. We also define

(A3)  δ_{8}(x) = ψ(\frac{x}{8}) char(4ℤ_{2})(x)

= \{ \begin{matrix}1 & if x ≡ 0 mod 8, \\ −1 & if x ≡ 4 mod 8, \\ 0 & otherwise. \end{matrix}

Finally, we associate a polynomial F (x) = F_{t,κ}(x) to the pair (t, κ) as follows. When p = 2 and a = ord_{p}t − ord_{p}k is even,

(A4)  F (x) = \frac{1}{2} ∑_{k=1}^{\frac{a}{2}} 2^{−k}x^{2k} + v(a)2^{− \frac{a}{2} −2}x^{a+2} + δ_{8}(α − κp^{−l})2^{− \frac{a}{2} −2}x^{a+3},

and otherwise,

(A5)  F (x) = (1 − p^{−1}) ∑_{k=1}^{[ \frac{a}{2} ]} p^{−k}x^{2k} + v(a)p^{−[ \frac{a+1}{2} ]−1}X^{a+1}.

Then calculation using Theorems 3.4 and 4.5 will give the following theorem.

Theorem A1. Let p be any prime number and let the notation be as above. Set X = p^{−s}. Then W_{t}(1, s + \frac{1}{2}, charL) = 0 unless t ∈ ℤ_{p}. Moreover,

(1) When a ≥ 0, one has

C^{−1}W_{t}(1, s + \frac{1}{2}, charL)

= \{ \begin{matrix}1 + p^{l}X^{l}((1 − p^{−1})l + F (X)) & if B = B^{s}, \\ −p^{−1} + p^{l}X^{l}(−p^{−1}(1 − p^{−1})l + F (X)) & if B = B^{r}. \end{matrix}

(2) When l = 1 and a = −1 (so t ∈ ℤ_{p}^{∗}), one has

C^{−1}W_{t}(1, s + \frac{1}{2}, charL) = (1 − X)\{ \begin{matrix}1 & if B = B^{s}, \\ −\frac{1}{p} & if B = B^{r}. \end{matrix}

Corollary A2. With notation as in Theorem 4.6, one has for a ≥ 0

W_{t}(1, s + \frac{1}{2}, charL^{s}) − W_{t}(1, s + \frac{1}{2}, charL^{r}) = C(1 − p^{−1}) + C(1 − p^{−2})lp^{l}X^{l}.

In particular, when l = 0, i.e., κ ∈ ℤ_{p}^{∗}, the difference is a constant.

As mentioned before, the results here are also true for t = 0 with a = ∞, i.e., with F (x) replaced by a simpler rational function \frac{p^{−1}(1−p^{−1})x^{2}}{1−p^{−1}x^{2}} . More precisely, we have

Corollary A3. With notation as above, one has

W_{0}(1, s + \frac{1}{2}, charL)

= \{ \begin{matrix}\frac{1+lp^{l}(1−p^{−1})X^{l}−p^{−1}X^{2}+p^{l−1}(1−p^{−1})(1−l)X^{2+l}}{1−p^{−1}X^{2}} & if B = B^{s}, \\ \frac{−1}{p} \frac{1+lp^{l}(1−p^{−1})X^{l}−p^{−1}X^{2}−p^{l}(1−p^{−1})(1+lp^{−1})X^{2+l}}{1−p^{−1}X^{2}} & if B = B^{r}. \end{matrix}

We divide the proof of Theorem A1 into two parts and four cases.

Part 1 We first assume that p ≠ 2. We will verify the case B = B^{r} in detail and leave the case B = B^{s} to the reader where the matrix of L^{s} is ℤ_{p}-equivalent to κ diag(1, 1, −1). We use the notation in section 3. When B is a division quaternion algebra, the matrix of L is ℤ_{p}-equivalent to S = κ diag(β, p, −βp) where β ∈ ℤ_{p}^{∗} is a non-square modulo p. So we have l_{1} = l, l_{2} = l_{3} = l + 1,

l(−k − l, 1) = \{ \begin{matrix}1 & if k > 0 is odd, \\ 2 & if k > 0 is even, \end{matrix}

and

v_{k+l} = \{ \begin{matrix}−(κp^{−1}, p)_{p} & if k > 0 is odd, \\ −1 & if k > 0 is even. \end{matrix}

Moreover, one has d(k + l) = −\frac{1}{2}k + l + 1 for k > 0, and

f(αp^{a+l}) = \{ \begin{matrix}−\frac{1}{p} & if a ≥ 0 is odd, \\ (\frac{α}{p})\frac{1}{\sqrt{p}} & if a ≥ 0 is even. \end{matrix}

Finally, the constant in Theorem 3.4 is C_{L} = −p^{−1}C(κ) in this case. Now all we need to do is to plug all the data to the formula in Theorem 3.4. The case l = 1 and a = −1 is obvious. When a ≥ 0 is odd, one has

− pC^{−1}W_{t}(1, s + \frac{1}{2}, charL^{r})

= 1 + (1 − p^{−1})lp^{l}X^{l} + (1 − p^{−1}) ∑_{0<k≤a, even} −p^{− \frac{1}{2} k+l+1}X^{k}

+ \frac{1}{p}p^{− \frac{1}{2} (a+1)+l+1}X^{a+1+l}

= 1 + p^{l+1}X^{l}(lp^{−1}(1 − p^{−1}) − F (X))

Multiplying both sides by −p^{−1}, one gets the desired formula. When a ≥ 0 is even, one has

− pC^{−1}W_{t}(1, s + \frac{1}{2}, charL^{r})

= 1 + (1 − p^{−1})lp^{l}X^{l} + (1 − p^{−1}) ∑_{0<k≤a, even} −p^{− \frac{1}{2} k+l+1}X^{k}

− v(a)\frac{1}{\sqrt{p}}d^{− \frac{a+1}{2} +l+1}X^{a+1+l}

= 1 + p^{l+1}X^{l}(lp^{−1}(1 − p^{−1}) − F (X))

Multiplying both sides by −p^{−1}, one gets the desired formula again.

Part 2 Now we assume that p = 2. First we change the definition of F (x) a little and rephrase the theorem in a slightly better way, taking advantage of the fact 1 − \frac{1}{2} = \frac{1}{2}. Let

(E2)  F_{1}(x) = ∑_{k=1}^{[ \frac{a}{2} ]} 2^{−k}x^{2k} − 2^{− \frac{a+1}{2}} x^{a+1}

when a > 0 is odd, and

(E3)  F_{1}(x) = ∑_{k=1}^{[ \frac{a}{2} ]} 2^{−k}x^{2k} + (\frac{−1}{ακp^{−l}} ) 2^{− \frac{a+2}{2}} x^{a+2} + δ_{8}(α − κ′)2^{− \frac{a+2}{2}} X^{a+3}

when a ≥ 0 is even. Then F_{1}(x) = 2F (x).

Theorem A4. W_{t}(1, s + \frac{1}{2}, charL) = 0 unless t ∈ ℤ_{2}. Put X = 2^{−s}.

(1) When l = 1 and a = −1 (so t ∈ ℤ_{2}^{∗}), one has

C^{−1}W_{t}(1, s + \frac{1}{2}, charL) = (1 − X)\{ \begin{matrix}1 & if B = B^{s}, \\ −\frac{1}{2} & if B = B^{r}. \end{matrix}

(2) When a ≥ 0, one has

C^{−1}W_{t}(1, s + \frac{1}{2}, charL)

= \{ \begin{matrix}1 + 2^{l−1}X^{l}(l + F_{1}(X)) & if B = B^{s}, \\ −\frac{1}{2} + 2^{l−1}X^{l}(−\frac{1}{2}l + F_{1}(X)) & if B = B^{r}. \end{matrix}

Proof. First we assume that B = B^{s} is split. We may identify B with M_{2}(ℚ_{2}) in

such a way that R = M_{2}(ℤ_{2}). So for x = ( \begin{matrix}x_{1} & x_{2} \\ x_{3} & −x_{1} \end{matrix} ), one has q(x) = κ(x_{1}^{2}+x_{2}x_{3}).

In the notation of section 4, we have L = M = 1, l_{1} = m_{1} = l, and N = 0. Now the case l = 1 and a = −1 can be verified without any difficulty from Theorem 4.5. We assume a ≥ 0 from now on. Write κ′ = κp^{−l}. Then δ(k + l) = 0 for k = 1 and = 1 otherwise. One has also p(k + l) = 1. Furthermore

l(−k − l + 1, 1) = \{ \begin{matrix}1 & if k > 0 is even, \\ 0 & if k > 0 is odd, \end{matrix}
ϵ_{1}(k + l) = \{ \begin{matrix}κ′ & if k > 0 is even, \\ 1 & if k > 0 is odd, \end{matrix}

and μ_{k+l}(t) = αp^{a−k+3} − κ′ for k > 0. So l(−k − l + 1, 1) is odd if and only if k > 0 is even. When k > 0 is even, we compute (μ = μ_{k+l}(t) for short)

(\frac{2}{μϵ_{1}(k + l)}) = \{ \begin{matrix}1 & if 0 < k ≤ a, \\ −1 & if k = a + 1, \\ (\frac{−1}{ακ′} ) & if k = a + 2, \\ 0 & if k = a + 3. \end{matrix}

Here we have used the fact that (\frac{2}{4x−1}) = −1 and (\frac{2}{2x−1}) = (\frac{−1}{x}) for any x ∈ ℤ_{2}^{∗}, where (\frac{−1}{x}) = (−1, x)_{2}. Moreover,

char(4ℤ_{2})(μ) = \{ \begin{matrix}0 & if 0 < k ≤ a + 2, \\ char(4ℤ_{2})(α − κ′) & if k = a + 3. \end{matrix}

Finally, we have d(k + l) = −\frac{1}{2}k + l + \frac{1}{2} for k > 0, and that the constant C_{L} = | det 2S|^{\frac{1}{2}} γ(V) in Theorem 4.5 is equal to C(κ) defined by (A1). Now all we need to do it to plug in all the data to the formula in Theorem 4.5, with a replaced by a + l. When a > 0 is odd, a + 3 is even, one has

C^{−1}W_{t}(1, s + \frac{1}{2}, charL)

= 1 + 2^{l−1}X^{l}δ_{l,1} + ∑ ; even^{0<k≤a} 2^{− \frac{1}{2} +k−1}X^{k+l} − 2^{− \frac{a+1}{2} +l−1}X^{a+1+l}

= 1 + 2^{l−1}X^{l}(l + F (X)).

Here we have used the trivial fact that δ_{l,1} = l for l = 0 or 1. When a ≥ 0 is even, a + 3 is odd. One has in this case

C^{−1}W_{t}(1, s + \frac{1}{2}, charL)

= 1 + 2^{l−1}X^{l}l + ∑ ; even^{0<k≤a} 2^{− \frac{1}{2} k+l−1}X^{k+l} + (\frac{−1}{ακ′})2^{− \frac{a+2}{2} +l−1}X^{a+2+l}

+ ψ(α − κ′) char(4ℤ_{2})(α − κ′)2^{− \frac{a+2}{2} +l−1}X^{a+3+l}

= 1 + 2^{l−1}X^{l}(l + F_{1}(X)).

This proves the split case.

Finally we assume that B = B^{r} is a division quaternion algebra over ℚ_{2}. The following easy lemma is well known and we include a proof here for completeness..

Lemma A5. There is a ℤ_{2} basis {v_{1}, v_{2}, v_{3}} of L with respect to which the quadratic form q is given by

(A6)  q(∑ x_{i}v_{i}) = κ(−3x_{1}^{2} + 2(x_{2}^{2} + x_{2}x_{3} + x_{3}^{2})).

Proof. We may assume that κ = 1, i.e., q(x) = −N(x) is the negative reduced norm. Since every quadratic extension of ℚ_{2} splits in B, there is α ∈ B such that α^{2} = −3. There exists β ∈ B^{∗} such that β^{2} ∈ ℚ_{2} and βk = k̄β for every k ∈ ℚ_{2}(α). Since ℚ_{2}(α) is unramified over ℚ_{2}, every units in ℤ_{2} is a norm from ℚ_{2}(α). So replacing β by βk for some k ∈ ℚ_{2}(α) if necessary, we may assume β^{2} = 2. Set γ = αβ. The condition βα = −αβ implies that γ^{2} = 6, αγ = −γα, and βγ = −γβ. Put v_{1} = α, v_{2} = \frac{1}{2}(β + γ), and v_{3} = \frac{1}{2}(β − γ). We claim that \{v_{1}, v_{2}, v_{3}\} is what we sought. First v_{1}^{2} = −3 and v_{2}^{2} = v_{3}^{2} = 2 imply v_{i} ∈ L. Direct computation shows that the quadratic form q(x) has the form (A6) with respect to {v_{1}, v_{2}, v_{3}}. It remains to check that v_{i} generate L as a ℤ_{2} module. Given x ∈ L, we write x = ∑ x_{i}v_{i} with x_{i} ∈ ℚ_{2}, we need to prove x_{i} ∈ ℤ_{2}. Taking norm, one finds

−Nx = −3x_{1}^{2} + 2(x_{2}^{2} + x_{2}x_{3} + x_{3}^{2}) ∈ ℤ_{2}.

It is easy to see that ord_{2}x_{2} = ord_{2}x_{3} if either of the two is not in ℤ_{2}. Assume m = ord_{2}x_{2} = ord_{2}x_{3}, and write x_{i} = 2^{m}y_{i} for some y_{i} ∈ ℤ_{2}^{∗}, i = 2, 3. So

−Nx = −3x_{1}^{2} + 2^{2m+1}(y_{2}^{2} + y_{2}y_{3} + y_{3}^{2}) ∈ ℤ_{2}.

Notice that y_{2}^{2} +y_{2}y_{3} +y_{3}^{2} is a unit. So one must have that x_{1} ∈ ℤ_{2} and 2m+1 ≥ 0, i.e., m ≥ 0. This proves x_{i} ∈ ℤ_{2}.

Now turn back to the proof of Theorem A4. Let {v_{1}, v_{2}, v_{3}} be a basis of L given by Lemma A5. In the notation of section 4, one has L = N = 1, l_{1} = l, n_{1} = l + 1, and M = 0. We again verify the case a ≥ 0 and leave the other case to the reader. First, as in the split case, δ(k + l) = 0 for k = 1 and = 1 otherwise. One also has

l(−k − l + 1, 1) = \{ \begin{matrix}1 & if k > 0 is even, \\ 0 & if k > 0 is odd, \end{matrix}

is odd if and only if k > 0 is even. However,

ϵ_{1}(k + l) = \{ \begin{matrix}−3κ′ & if k > 0 is even, \\ 1 & if k > 0 is odd, \end{matrix}

and μ_{k+l}(t) = αp^{a−k+3} + 3κ′ for k > 0. When k > 0 is even, we have as in the split case

(\frac{2}{μϵ_{1}(k + l)}) = \{ \begin{matrix}1 & if 0 < k ≤ a, \\ −1 & if k = a + 1, \\ (\frac{−1}{ακ′} ) & if k = a + 2, \\ 0 & if k = a + 3, \end{matrix}

and

char(4ℤ_{2})(μ) = \{ \begin{matrix}0 & if 0 < k ≤ a + 2, \\ char(4ℤ_{2})(α + 3κ′) & if k = a + 3. \end{matrix}

Moreover, we have p(k +l) = (−1)^{k−1}, and d(k +l) = −\frac{1}{2}k +l+ \frac{3}{2} for k > 0. Finally the constant C_{L} = −\frac{1}{2}C(κ). Now one can obtain the desired formula by doing the same calculation as in the split case, plus the following identity:

δ_{8}(α + 3κ′) = −δ_{8}(α − κ′).

We leave the detail to the reader.

5. A proposition.

The purpose of this section is to prove a result needed for the main calculation in section 7. We fix an element u ∈ ℤ_{p}^{∗} such that (\frac{u}{p}) = −1. Let H = Sym_{2}(ℚ_{p}) ∩ GL_{2}(ℚ_{p}) be the set of nonsingular symmetric matrices over ℚ_{p}, and let GL_{2}(ℤ_{p}) acts on H via g.X = ^{t}gXg. By a well-known theorem ([Ca, Theorem 8.3.1]), the following is a complete representative set of the GL_{2}(ℤ_{p})-orbits in H

(5.1)  p^{b} ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) , p^{b} ( \begin{matrix}1 & 0 \\ 0 & u \end{matrix} ) , and ( \begin{matrix}u_{1}p^{b_{1}} & 0 \\ 0 & u_{2}p^{b_{2}} \end{matrix} )

with b_{1} < b_{2} and b ∈ ℤ and u_{i} = 1 or u. Recall also that every GL_{2}(ℤ_{p})-orbit is open in H since two quadratic forms over ℚ_{p} sufficiently close p-adically are GL_{2}(ℤ_{p})-equivalent ( [Ca, chapter 8]).

As before, we choose on ℚ_{p}the Haar measure dx so that meas(ℤ_{p}) = 1. On sym-

metric matrices, we choose the Haar measure dx = dx_{1}dx_{2}dx_{3} for x = ( \begin{matrix}x_{1} & x_{3} \\ x_{3} & x_{2} \end{matrix} ).

On GL_{2}(ℚ_{p}) we choose the Haar measure dx = dx_{1}dx_{2}dx_{3}dx_{4}. In particular, the measure of GL_{2}(ℤ_{p}) is (1 − p^{−1})^{2}(1 + p^{−1}) instead of 1.

Proposition 5.1. Let f be a locally constant bounded function on Sym_{2}(ℚ_{p}) such that f ∈ L^{1}( Sym_{2}(ℚ_{p}). Then

∫ _{Sym_{2}(ℚ_{p})} f(x)dx

= ∑_{b∈ℤ} p^{−3b} (\frac{1}{2(1 − (\frac{−1}{p})p^{−1})} ∫ _{GL_{2}(ℤ_{p})} f(p^{b} ^{t}xx)dx

+\frac{1}{2(1 + (\frac{−1}{p})p^{−1})} ∫ _{GL_{2}(ℤ_{p})} f(p^{b} ^{t}x diag(1, u)x)dx)

+ \frac{1}{4} ∑_{b_{1}<b_{2},u_{i}=1 or u} p^{−2b_{1}−b_{2}} ∫ _{GL_{2}(ℤ_{p})} f(^{t}x diag(u_{1}p^{b_{1}}, u_{2}p^{b_{2}})x)dx

Proof. Since Sym_{2}(ℚ_{p}) − H is a closed subvariety of codimension 1, one has

∫ _{Sym_{2}(ℚ_{p})} f(x)dx = ∫_{H} f(x)dx

= ∑ ∫_{𝕆} f(x)dx. (5.2)

Here the sum runs over all GL_{2}(ℤ_{p})-orbits 𝕆 of H. For each matrix diag(t_{1}, t_{2}), let 𝕆(t_{1}, t_{2}) be the GL_{2}(ℤ_{p})-orbit of diag(t_{1}, t_{2}), and let O(t_{1}, t_{2}) be the stabilizer of diag(t_{1}, t_{2}) in GL_{2}(ℤ_{p}). Let dh be the Haar measure on O(t_{1}, t_{2}) with total measure 1. Since the Haar measure on H is GL_{2}(ℤ_{p}) invariant, there is a constant C = C(t_{1}, t_{2}) such that for every locally constant function ϕ in GL_{2}(ℤ_{p}), one has

(5.3)  C ∫ _{GL_{2}(ℤ_{p})} f(x)dx = ∫_{𝕆(t_{1},t_{2})} ∫_{O(t_{1},t_{2})} f(hx)dhdx.

24 TONGHAI YANG

Take ϕ = char(1 + p^{n}M_{2}(ℤ_{p})) for sufficiently large integer n. Then the left hand side of (5.3) is Cp^{−4n}, and the right hand side is equal to

meas(O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p}))) · meas(𝕆_{n}(t_{1}, t_{2})).

Here 𝕆_{n}(t_{1}, t_{2}) is the (1 + p^{n}M_{2}(ℤ_{p}))-orbit of diag(t_{1}, t_{2}). Since

t(1 + p^{n} ( \begin{matrix}x_{1} & x_{2} \\ x_{3} & x_{4} \end{matrix} )) diag(t_{1}, t_{2})(1 + p^{n} ( \begin{matrix}x_{1} & x_{2} \\ x_{3} & x_{4} \end{matrix} ))

= ( \begin{matrix}t_{1} + 2t_{1}p^{n}x_{1} & t_{1}p^{n}x_{2} + t_{2}p^{n}x_{3} \\ t_{1}p^{n}x_{2} + t_{2}p^{n}x_{3} & t_{2} + 2t_{2}p^{n}x_{4} \end{matrix} ) p^{2n} ( \begin{matrix}∗ & ∗ \\ ∗ & ∗ \end{matrix} ) ,

one has

meas(𝕆_{n}(t_{1}, t_{2})) = p^{−3n−2b_{1}−b_{2}}.

On the other hand,

meas(O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p}))) = \frac{1}{[O(t_{1}, t_{2}) : O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p}))]}.

Applying the lemma below, one has then

(5.4)  C(t_{1}, t_{2}) = p^{−2b_{1}−b_{2}}μ(t_{1}, t_{2})

where

(5.5)  μ(t_{1}, t_{2}) = \{ \begin{matrix}\frac{1}{4} & if b_{1} < b_{2}, \\ \frac{1}{2(1 − (\frac{−u_{1}u_{2}}{p})p^{−1})} & if b_{1} = b_{2}. \end{matrix}

(t_{i} = u_{i}p^{b_{i}}). Applying (5.3) − (5.5) to (5.2), one proves the proposition. Lemma 5.2. One has

[O(t_{1}, t_{2}) : O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p}))] = μ(t_{1}, t_{2})^{−1}p^{n}.

Proof. Direct computation gives

O(t_{1}, t_{2}) = SO(t_{1}, t_{2}) ∝ {±1}

where

SO(t_{1}, t_{2}) = \{( \begin{matrix}x_{1} & x_{2} \\ −\frac{t_{1}}{t_{2}} x_{2} & x_{1} \end{matrix} ) : x_{i} ∈ ℤ_{p}, \frac{t_{1}}{t_{2}} x_{2} ∈ ℤ_{p}, x_{1}^{2} + \frac{t_{1}}{t_{2}} x_{2}^{2} = 1\}

≅ \{(x_{1}, x_{2}) ∈ ℤ_{p}^{2} : x ∈ \frac{t_{2}}{t_{1}} ℤ_{p}, x_{1}^{2} + \frac{t_{1}}{t_{2}} x_{2}^{2} = 1\}

and O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p}))

≅ \{(x_{1}, x_{2}) ∈ ℤ_{p}^{2} : x_{1} ≡ 1 mod p^{n}, x_{2} ≡ 0 mod p^{n+b_{2}−b_{1}}, x_{1}^{2} + \frac{t_{1}}{t_{2}} x_{2}^{2} = 1\}.

When b_{1} < b_{2}, the map (x_{1}, x_{2}) ↦ x_{2} is a two-to-one map from SO(t_{1}, t_{2}) onto p^{b_{2}−b_{1}}ℤ_{p} and a one-to-one correspondence from O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p})) onto p^{n+b_{2}−b_{1}}ℤ_{p}. This proves the case b_{1} < b_{2}.

When b_{1} = b_{2}, \frac{t_{1}}{t_{2}} = \frac{u_{1}}{u_{2}} ∈ ℤ_{p}^{∗}. Let α = \sqrt{−\frac{u_{1}}{u_{2}}} . If α ∈ ℤ_{p}^{∗}, i.e., (\frac{−u_{1}u_{2}}{p}) = 1,

then (x_{1}, x_{2}) ↦ x_{1} + αx_{2} is an isomorphism between SO(t_{1}, t_{2}) to ℤ_{p}^{∗} and maps O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p})) onto 1 + p^{n}ℤ_{p}. This implies

[SO(t_{1}, t_{2}) : O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p}))] = [ℤ_{p}^{∗} : 1 + p^{n}ℤ_{p}] = p^{n}(1 − p^{−1}).

If α ∉ ℤ_{p}^{∗}, i.e., (\frac{−u_{1}u_{2}}{p}) = −1, then K = ℚ_{p}(α) is a quadratic extension. Let K^{1} be the norm-1 subgroup of K and let

K^{1}(n) = {z ∈ K^{1} : z ≡ 1 mod p^{n}}.

Then the map (x_{1}, x_{2}) ↦ x_{1} + αx_{2} gives an isomorphism from SO(t_{1}, t_{2}) onto K^{1} and from O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p})) onto K^{1}(n). So in this case, one has

[SO(t_{1}, t_{2}) : O(t_{1}, t_{2}) ∩ (1 + p^{n}M_{2}(ℤ_{p}))] = [K^{1} : K^{1}(n)] = p^{n}(1 + p^{−1}).

This finishes the proof of the lemma.

6. Gauss integrals over GL_{2}.

For any matrix M = ( \begin{matrix}t_{1} & t_{2} \\ t_{3} & t_{4} \end{matrix} ) ∈ M_{2}(ℚ_{p}), we compute the following Gauss

integral in this section.

(6.1)  I^{∗}(t_{1}, t_{2}, t_{3}, t_{4}) = I(M) = ∫ _{GL_{2}(ℤ_{p})} ψ(∑ t_{i}x_{i}^{2})dx_{1}dx_{2}dx_{3}dx_{4}.

Here dx_{i} is the Haar measure on ℚ_{p} so that meas(ℤ_{p}) = 1. So

meas( GL_{2}(ℤ_{p})) = (1 + p^{−1})(1 − p^{−1})^{2}.

The integral is a generalization of the integral (2.2) to GL_{2}, and will be needed in next section. It also seems to be of independent interest. We will use either of the two notations whenever convenient. The following lemma is obvious from the definition.

Lemma 6.1. (1) The value of the integral I^{∗}(t_{1}, t_{2}, t_{3}, t_{4}) does not change under any of the following operations: switching t_{1} with t_{4}, switching t_{2} with t_{3}, or switching the pair (t_{1}, t_{4}) with the pair (t_{2}, t_{3}).

(2) For any α ∈ ℤ_{p}^{∗}, one has

I(α^{2}t_{1}, α^{2}t_{2}, t_{3}, t_{4}) = I(t_{1}, t_{2}, t_{3}, t_{4}).

In other words, multiplying a column or a row of M by a square does not change the value of I(M).

Lemma 6.2. (1)

I^{∗}(t_{1}, t_{2}, t_{3}, t_{4})

= I^{∗}(t_{1})I^{∗}(t_{4})I^{∗}(t_{2})I(t_{3}) + I^{∗}(t_{1})I(t_{4})I^{∗}(t_{2})I^{∗}(t_{3}) + p^{−1}I^{∗}(t_{1})I^{∗}(t_{4})I(p^{2}t_{2})I(t_{3})

+ p^{−1}I(p^{2}t_{1})I(t_{4})I^{∗}(t_{2})I^{∗}(t_{3}) − I^{∗}(t_{1})I^{∗}(t_{4})I^{∗}(t_{2})I^{∗}(t_{3})

− p^{−4} char((p^{−1}ℤ_{p})^{4})(t_{1}, t_{2}, t_{3}, t_{4}) ∑_{x_{i}∈𝔽_{p}^{∗},x_{1}x_{4}=x_{2}x_{3}} ψ(∑ t_{1}x_{i}^{2}).

Here 𝔽_{p} is the finite field of p elements, I(t) and I^{∗}(t) are integrals given by (2.1) and (2.2).

(2) For α_{i} ∈ ℤ_{p}, one has

I^{∗}(α_{1}p^{−1}, α_{2}p^{−1}, α_{3}p^{−1}, α_{4}p^{−1})

= p^{−4} ∑_{X∈ GL_{2}(𝔽_{p})} e_{p}(− ∑_{i=1}^{4}(α_{i}x_{i}^{2})).

Here e_{p}(x) = e^{\frac{2πix}{p}}, and X = ( \begin{matrix}x_{1} & x_{2} \\ x_{3} & x_{4} \end{matrix} ).

Proof. We first remark that for x ∈ ℤ_{p} or x ∈ 𝔽_{p}, we choose a rational integer n ≡ x mod p, then e_{p}(x) = e_{p}(n) is independent of the choice of n and is what we mean in (2). By the definition of ψ, one has ψ(xp^{−1}) = e_{p}(−x) for x ∈ ℤ_{p}. Now (2) is obvious.

To prove (1), first notice that GL_{2}(ℤ_{p}) ⊂ ℤ_{p}^{4} is the disjoint union of the subsets (coordinates are in the order (x_{1}, x_{4}, x_{2}, x_{3}))

pℤ_{p} × ℤ_{p} × ℤ_{p}^{∗} × ℤ_{p}^{∗}, ℤ_{p}^{∗} × pℤ_{p} × ℤ_{p}^{∗} × ℤ_{p}^{∗}, ℤ_{p}^{∗} × ℤ_{p}^{∗} × pℤ_{p} × ℤ_{p},

and

{(x_{1}, x_{4}, x_{2}, x_{3}) ∈ ℤ_{p}^{∗3} × ℤ_{p} : x_{3} ≢ x_{2}^{−1}x_{1}x_{4} mod p}.

So

I^{∗}(t_{1}, t_{2}, t_{3}, t_{4})
= p^{−1}I(p^{2}t_{1})I(t_{4})I^{∗}(t_{2})I^{∗}(t_{3}) + p^{−1}I^{∗}(t_{1})I(p^{2}t_{4})I^{∗}(t_{2})I^{∗}(t_{3})
+ p^{−1}I^{∗}(t_{1})I^{∗}(t_{4})I(p^{2}t_{2})I(t_{3}) + I^{∗}(t_{1})I^{∗}(t_{4})I^{∗}(t_{2})I(t_{3})
− ∫_{ℤ_{p}^{∗3}} ψ(t_{1}x_{1}^{2} + t_{2}x_{2}^{2} + t_{4}x_{4}^{2})dx_{1}dx_{2}d_{4} ∫_{x_{3}≡x_{2}^{−1}x_{1}x_{4} mod p} ψ(t_{3}x_{3}^{2})dx_{3}.

Since p^{−1}I(p^{2}t_{4}) = I(t_{4}) − I^{∗}(t_{4}), it suffices to show that

∫_{ℤ_{p}^{∗3}} ψ(t_{1}x_{1}^{2} + t_{2}x_{2}^{2} + t_{4}x_{4}^{2})dx_{1}dx_{2}d_{4} ∫_{x_{3}≡x_{2}^{−1}x_{1}x_{4} mod p} ψ(t_{3}x_{3}^{2})dx_{3}

(6.2) = p^{−4} char((p^{−1}ℤ_{p})^{4})(t_{1}, t_{2}, t_{3}, t_{4}) ∑_{x_{i}∈𝔽_{p}^{∗},x_{1}x_{4}=x_{2}x_{3}} ψ(∑ t_{1}x_{i}^{2}).

By Lemma 2.1, the left hand side of (6.2) is

p^{−1} char(p^{−1}ℤ_{p})(t_{3}) ∫_{ℤ_{p}^{∗3}} ψ(t_{1}x_{1}^{2})ψ(t_{2}x_{2}^{2})ψ(t_{4}x_{4}^{2})ψ(t_{3}x_{2}^{−2}x_{1}^{2}x_{4}^{2})dx_{1}dx_{2}dx_{4}.

This is zero unless t_{1}, t_{4} ∈ p^{−1}ℤ_{p} since it contains a factor I^{∗}(t_{1} + t_{3}x_{2}^{−1}x_{2}^{2}) resp. I^{∗}(t_{4} + t_{3}x_{2}^{−2}x_{1}^{2}). The same proof as in Lemma 2.1(3) would make the integral zero unless t_{2} ∈ p^{−1}ℤ_{p} (under the condition t_{3} ∈ p^{−1}ℤ_{p}). On the other hand, if t_{1}, ..., t_{4} ∈ p^{−1}ℤ_{p}, then the integral equals

p^{−1} ∑_{x_{1},x_{2},x_{4}∈𝔽_{p}^{∗}} ψ(t_{1}x_{1}^{2})ψ(t_{2}x_{2}^{2})ψ(t_{4}x_{4}^{2})ψ(t_{3}x_{2}^{−2}x_{1}^{2}x_{4}^{2}) ∫_{(1+pℤ_{p})^{3}} dy

= p^{−4} ∑_{x_{i}∈𝔽_{p}^{∗},x_{1}x_{4}=x_{2}x_{3}} ψ(∑ t_{i}x_{i}^{2}).

This finishes the proof of Lemma 6.2.

To obtain a more precise formula for I^{∗}(t_{1}, t_{2}, t_{3}, t_{4}), write t_{i} = α_{i}p^{a_{i}} with α_{i} ∈ ℤ_{p}^{∗} and a_{i} ∈ ℤ. We may and will assume that a_{1} ≤ a_{2}, a_{4}, and a_{2} ≤ a_{3} without loss of generality in the rest of this section (see Lemma 6.1).

Corollary 6.3. With notation and assumption as above, one has

I^{∗}(t_{1}, t_{2}, t_{3}, t_{4}) = \{ \begin{matrix}(1 − p^{−1})^{2}(1 + p^{−1}) & if a_{1} ≥ 0, \\ I(t_{1})I(t_{4})I^{∗}(t_{2})I^{∗}(t_{3}) & if a_{1} ≤ −2. \end{matrix}

In particular, if a_{1}, a_{2} ≤ −2, then I^{∗}(t_{1}, t_{2}, t_{3}, t_{4}) = 0.

Proof. When a_{1} ≥ 0, one has a_{i} ≥ 0 for 1 ≤ i ≤ 4. So

I^{∗}(t_{1}, t_{2}, t_{3}, t_{4}) = ∫ _{GL_{2}(ℤ_{p})} dx_{1}dx_{2}dx_{3}dx_{4}
= p^{−4}# GL_{2}(𝔽_{p})
= (1 − p^{−1})^{2}(1 + p^{−1}).

On the other hand, if a_{1} ≤ −2 then I^{∗}(t_{1}) = 0 by Lemma 2.1. So Lemma 6.2 implies

I^{∗}(t_{1}, t_{2}, t_{3}, t_{4}) = p^{−1}I(p^{2}t_{1})I(t_{4})I^{∗}(t_{2})I^{∗}(t_{3}) = I(t_{1})I(t_{4})I^{∗}(t_{2})I^{∗}(t_{3}).

The case a_{1} = −1 is a little bit more complicated and more interesting. We first assume at least one of a_{i}, say a_{4} is nonnegative. Then (assuming a_{4} ≥ 0)

∑_{x_{i}∈𝔽_{p}^{∗},x_{1}x_{4}=x_{2}x_{3}} ψ(∑ t_{i}x_{i}^{2})

= ( ∑_{x_{1}∈𝔽_{p}^{∗}} ψ(t_{1}x_{1}^{2}))( ∑_{x_{2}∈𝔽_{p}^{∗}} ψ(t_{2}x_{2}^{2}))( ∑_{x_{3}∈𝔽_{p}^{∗}} ψ(t_{3}x_{3}^{2}))

= p^{3}I^{∗}(t_{1})I^{∗}(t_{2})I^{∗}(t_{3})

Also notice that I(t_{4}) = 1 and I^{∗}(t_{4}) = 1 − p^{−1} in this case. Now Lemmas 6.2 and 2.1 give Lemma 6.4. Let the notation be as above. Assume that a_{1} = −1 and a_{4} ≥ 0. Then

I^{∗}(t_{1}, t_{2}, t_{3}, t_{4}) = (1 − p^{−1})I^{∗}(t_{1})I(t_{2})I(t_{3}) + p^{−1}I^{∗}(t_{2})I^{∗}(t_{3}).

More precisely, one has

I^{∗}(t_{1}, t_{2}, t_{3}, t_{4})

= \{ \begin{matrix}(1 − p^{−1}) (\frac{δ_{p}}{\sqrt{p}}(\frac{−α_{1}}{p}) − \frac{1}{p^{2}} ) & if a_{2}, a_{3} ≥ 0, \\ p^{−1}(1 − p^{−1}) ((\frac{−α_{1}α_{2}}{p}) − \frac{1}{p}) & if a_{2} = −1, a_{3} ≥ 0, \\ p^{−3}(1 + (\frac{−α_{2}α_{3}}{p})) + \frac{δ_{p}}{p\sqrt{p}}(\frac{α_{1}α_{2}α_{3}}{p}) \\ −\frac{δ_{p}}{p^{2}\sqrt{p}} ((\frac{α_{1}α_{2}α_{3}}{p}) + (\frac{−α_{2}}{p}) + (\frac{−α_{3}}{p})) & if a_{2} = a_{3} = −1. \end{matrix}

Finally, it comes to the case where a_{1} = a_{2} = a_{3} = a_{4} = −1. By Lemma 6.2, it reduces to compute the following Gauss sum

(6.3)  S(M) = ∑_{X∈ GL_{2}(𝔽_{p})} e_{p}(∑ α_{i}x_{i}^{2})

where M = ( \begin{matrix}α_{1} & α_{2} \\ α_{3} & α_{4} \end{matrix} ) ∈ M_{2}(𝔽_{p}), X = ( \begin{matrix}x_{1} & x_{2} \\ x_{3} & x_{4} \end{matrix} ), and e_{p}(x) = e^{\frac{2πix}{p}}. We also

write S(α_{1}, α_{2}, α_{3}, α_{4}) for S(M) when it is more convenient. One has

(6.4)  I^{∗}(α_{1}p^{−1}, α_{2}p^{−1}, α_{3}p^{−1}, α_{4}p^{−1}) = p^{−4}S(−α_{1}, −α_{2}, −α_{3}, −α_{4})

for α_{i} ∈ ℤ_{p}. Here we also use α_{i} for its image in 𝔽_{p}. When some of α_{i} = 0, the sum (or the integral) is given by Lemma 6.4. The following lemma was proved with the help of Don Zagier. I thank him for his help and for allowing me to publish it here.

Lemma 6.5. Let M = ( \begin{matrix}α_{1} & α_{2} \\ α_{3} & α_{4} \end{matrix} ) with α_{i} ∈ 𝔽_{p}^{∗}.

(1) When det M = 0, i.e, M = ( \begin{matrix}α_{1} & α_{2} \\ βα_{1} & βα_{2} \end{matrix} ) with some β ∈ 𝔽_{p}^{∗}, one has

S(M) = −p^{2}((\frac{−β}{p}) + (\frac{−α_{1}α_{2}}{p}))(1 − (\frac{−α_{1}α_{2}}{p})p^{−1}).

(2) When det M ≠ 0, let λ = \frac{α_{2}α_{3}}{α_{1}α_{4}} , and let

E_{λ} : y^{2} = x(x − 1)(x − λ)

be the corresponding elliptic curve over 𝔽_{p}. Let

a_{p}(λ) = a_{p}(E_{λ}) = #E_{λ}(𝔽_{p}) − p − 1.

be the trace of the Frobenious on E_{λ}. Then

S(M) = (\frac{∏ α_{i}}{p})p^{2} − (\frac{α_{1}α_{4}}{p})pa_{p}(λ) + p

− p\sqrt{p}δ_{p}(\frac{det M}{p}) ((\frac{α_{1}}{p}) + (\frac{−α_{2}}{p}) + (\frac{−α_{3}}{p}) + (\frac{α_{4}}{p})) .

LOCAL DENSITIES 29

Here δ_{p} is the number defined by (1.10), i.e., it is 1 or i depending on whether p ≡ 1 mod 4 or p ≡ −1 mod 4. Proof. By the classical Gauss sum formula, one has

S(M) = ∑_{X∈M_{2}(𝔽_{p})} e_{p}(∑ α_{i}x_{i}^{2}) − ∑_{det X=0} e_{p}(∑ α_{i}x_{i}^{2})

= (\frac{∏ α_{i}}{p})p^{2} − ∑_{det X=0} e_{p}(∑ α_{i}x_{i}^{2})

Since det X = 0 means that either X = ( \begin{matrix}a & b \\ μa & μb \end{matrix} ) with a, b, μ ∈ 𝔽_{p} or X = ( \begin{matrix}0 & 0 \\ c & d \end{matrix} ) with X = 0 being counted p + 1 times, one has

∑_{det X=0} e_{p}(∑ α_{i}x_{i}^{2}) = ∑_{μ} ∑_{a} e_{p}((α_{1} + α_{3}μ^{2})a^{2}) ∑_{b} e_{p}((α_{2} + α_{4}μ^{2})b^{2})

+ ∑_{c,d} e_{p}(α_{3}c^{2} + α_{4}d^{2}) − p.

Applying the classical Gauss sum formula to the second term, one obtains

S(M) = (\frac{∏ α_{i}}{p})p^{2} + p − p(\frac{−α_{3}α_{4}}{p})

− ∑_{μ} ∑_{a} e_{p}((α_{1} + α_{3}μ^{2})a^{2}) ∑_{b} e_{p}((α_{2} + α_{4}μ^{2})b^{2}) (6.5)

Notice that

(6.6) ∑_{a} e_{p}((α_{1} +α_{3}μ^{2})a^{2}) = \{ \begin{matrix}p & if (\frac{−α_{1}α_{3}}{p}) = 1&μ = ±\sqrt{−\frac{α_{1}}{α_{3}}} , \\ \sqrt{p}δ_{p}(\frac{α_{1}+α_{3}μ^{2}}{p}) & otherwise. \end{matrix}

Case 1 When det M = 0, \frac{−α_{1}}{α_{3}} = \frac{−α_{2}}{α_{4}} . In this case, the roots μ = ±\sqrt{\frac{−α_{1}}{α_{3}}} and μ = ±\sqrt{\frac{−α_{2}}{α_{4}}} are the same if they exist in 𝔽_{p}. Let β = \frac{α_{3}}{α_{1}} , then one has by (6.6)

∑_{μ} ∑_{a} e_{p}((α_{1} + α_{3}μ^{2})a^{2}) ∑_{b} e_{p}((α_{2} + α_{4}μ^{2})b^{2})

= ∑_{μ} p(\frac{−1}{p})(\frac{α_{1} + α_{3}μ^{2}}{p})(\frac{α_{2} + α_{4}μ^{2}}{p}) + (1 + (\frac{−β}{p}))p^{2}

= (\frac{−α_{1}α_{2}}{p})p ∑_{μ}(\frac{1 + βμ^{2}}{p})^{2} + (1 + (\frac{−β}{p}))p^{2}

= (\frac{−α_{1}α_{2}}{p})p(p − 1 − (\frac{−β}{p})) + (1 + (\frac{−β}{p}))p^{2}

= (1 + (\frac{−β}{p}) + (\frac{−α_{1}α_{2}}{p}))p^{2} − (\frac{−α_{1}α_{2}}{p})(1 + (\frac{−β}{p}))p.

Plugging this into (6.5), one proves (1).

Case 2 When det M ≠ 0, the roots μ = ±\sqrt{\frac{−α_{1}}{α_{3}}} and μ = ±\sqrt{\frac{−α_{2}}{α_{4}}} are different if they exist in 𝔽_{p}. So

∑_{μ} ∑_{a} e_{p}((α_{1} + α_{3}μ^{2})a^{2}) ∑_{b} e_{p}((α_{2} + α_{4}μ^{2})b^{2})

= ∑_{μ} p(\frac{−1}{p})(\frac{α_{1} + α_{3}μ^{2}}{p})(\frac{α_{2} + α_{4}μ^{2}}{p})

+ (1 + (\frac{−α_{1}α_{3}}{p}))p ∑_{b} e_{p}((α_{2} − \frac{α_{1}α_{4}}{α_{3}} )b^{2})

+ (1 + (\frac{−α_{2}α_{4}}{p}))p ∑_{a} e_{p}((α_{1} − \frac{α_{2}α_{3}}{α_{4}} )a^{2})

= (\frac{−α_{3}α_{4}}{p})p ∑_{x∈𝔽_{p}}(\frac{x + α_{1}α_{3}^{−1}}{p})(\frac{x + α_{2}α_{4}^{−1}}{p})(1 + (\frac{x}{p}))

+ ((\frac{α_{1}}{p}) + (\frac{−α_{3}}{p}))(\frac{det M}{p})p\sqrt{p}δ_{p} + ((\frac{−α_{2}}{p}) + (\frac{α_{4}}{p}))(\frac{det M}{p})p\sqrt{p}δ_{p}.

So

∑_{μ} ∑_{a} e_{p}((α_{1} + α_{3}μ^{2})a^{2}) ∑_{b} e_{p}((α_{2} + α_{4}μ^{2})b^{2})

= (\frac{−α_{3}α_{4}}{p})p ∑_{x}(\frac{(x + α_{1}α_{3}^{−1})(x + α_{2}α_{4}^{−1})}{p})

+ (\frac{−α_{3}α_{4}}{p})p ∑_{x}(\frac{x(x + α_{1}α_{3}^{−1})(x + α_{2}α_{4}^{−1})}{p})

+ p\sqrt{p}δ_{p}(\frac{det M}{p})((\frac{α_{1}}{p}) + (\frac{−α_{2}}{p}) + (\frac{−α_{3}}{p}) + (\frac{α_{4}}{p})).

Given a polynomial f(x) ∈ 𝔽_{p}[x], let C_{f}^{a} be the affine curve defined by y^{2} = f(x). Since, for a fixed x ∈ 𝔽_{p}, the equation y^{2} = f(x) has 1 + (\frac{f(x)}{p}) solutions in 𝔽_{p}, one has

(6.7)  ∑_{x∈𝔽_{p}}(\frac{f(x)}{p}) = #C_{f}^{a}(𝔽_{p}) − p.

It is easy to see from this that

(6.8)  ∑_{x∈𝔽_{p}}(\frac{x^{2} + ax + b}{p}) = \{ \begin{matrix}p − 1 & if a^{2} − 4b = 0, \\ −1 & if a^{2} − 4b ≠ 0. \end{matrix}

In particular, one has (det M ≠ 0)

∑_{x}(\frac{(x + α_{1}α_{3}^{−1})(x + α_{2}α_{4}^{−1})}{p}) = −1.

Next, let E_{M} be the elliptic curve defined by

E_{M} : y^{2} = x(x + α_{1}α_{3}^{−1})(x + α_{2}α_{4}^{−1}).

Notice that E_{M} has one extra point (infinity) to its affine points. So (6.7) gives

(\frac{x(x + α_{1}α_{3}^{−1})(x + α_{2}α_{4}^{−1})}{p}) = a_{p}(E_{M}).

So

∑_{μ} ∑_{a} e_{p}((α_{1} + α_{3}μ^{2})a^{2}) ∑_{b} e_{p}((α_{2} + α_{4}μ^{2})b^{2})

= −(\frac{−α_{3}α_{4}}{p})p + (\frac{−α_{3}α_{4}}{p})pa_{p}(E_{M})

+ p\sqrt{p}δ_{p}(\frac{det M}{p})((\frac{α_{1}}{p}) + (\frac{−α_{2}}{p}) + (\frac{−α_{3}}{p}) + (\frac{α_{4}}{p})).

Plugging this into (6.5) and comparing the result with the desired formula in (2) leads us to verify

(6.9)  a_{p}(E_{M}) = (\frac{−α_{1}α_{3}}{p})a_{p}(λ).

Notice that the map x ↦ −α_{1}α_{3}^{−1}x and y ↦ (−α_{1}α_{3}^{−1})^{2}y gives a bijection between E′(𝔽_{p}) and E_{M}(𝔽_{p}), where

E′ : −α_{1}α_{3}^{−1}y^{2} = x(x − 1)(x − λ)

is the quadratic twist of E_{λ} by −α_{1}α_{3}^{−1} and λ = \frac{α_{2}α_{3}}{α_{1}α_{4}} . So (6.9) follows from the general fact that

(6.10)  a_{p}(E^{d}) = (\frac{d}{p})a_{p}(E)

for any elliptic curve E over 𝔽_{p} and any quadratic twist E^{d} with d ∈ 𝔽_{p}^{∗}. Indeed, if d is a square, E^{d} is isomorphic to E over 𝔽_{p} and (\frac{d}{p}) = 1. If d is not a square, one has (\frac{d}{p}) = −1. On the other hand, we may assume that E is defined by an equation y^{2} = f(x) with a nonsigular cubic polynomial f(x) over 𝔽_{p} (p ≠ 2). Then E^{d} is given by dy^{2} = f(x). So (6.7) implies

a_{p}(E^{d}) + a_{p}(E) = ∑_{x}(\frac{f(x)}{p}) + ∑_{x}(\frac{df(x)}{p}) = 0.

Corollary 6.6. (1) For α_{i}, β ∈ ℤ_{p}^{∗}, one has

I^{∗}(α_{1}p^{−1}, α_{2}p^{−1}, α_{1}βp^{−1}, α_{2}βp^{−1})

= −p^{−2}((\frac{−β}{p}) + (\frac{−α_{1}α_{2}}{p}))(1 − (\frac{−α_{1}α_{2}}{p})p^{−1}).

(2) For α_{i} ∈ ℤ_{p}^{∗} with α_{1}α_{4} ≠ α_{2}α_{3}, one has

I^{∗}(α_{1}p^{−1}, α_{2}p^{−1}, α_{3}p^{−1}, α_{4}p^{−1})

= (\frac{∏ α_{i}}{p})p^{−2} + p^{−3} − (\frac{α_{1}α_{4}}{p})p^{−3}a_{p}(λ)

− p^{−3}\sqrt{p}δ_{p}(\frac{α_{1}α_{4} − α_{2}α_{3}}{p}) ((\frac{−α_{1}}{p}) + (\frac{α_{2}}{p}) + (\frac{α_{3}}{p}) + (\frac{−α_{4}}{p})) .

7. The case n = 2 and p ≠ 2.

In the section, we will obtainan explicit formula of local density for n = 2 and p ≠ 2.. We first recall some notations for the convenience of the reader. Let S be a nondegenerate symmetric matrix over ℚ_{p} of rank m > 2. Then S is ℤ_{p} equivalent to diag(ϵ_{1}p^{l_{1}}, ⋯, ϵ_{m}p^{l_{m}}) with ϵ_{i} ∈ ℤ_{p}^{∗} and l_{1} ≤ l_{2} ⋯ ≤ l_{m}. Denote, for integers b and c,

(7.1) L(b, c) = {1 ≤ i ≤ m : b + l_{i} < 0 and b + l_{i} ≡ k mod 2}, l(b, c) = #L(b, c).

Let T be a 2 × 2 nondegenerate symmetric matrix over ℚ_{p}, ℤ_{p}-equivalent to diag(α_{1}p^{a_{1}}, α_{2}p^{a_{2}}) with α_{i} ∈ ℤ_{p}^{∗}, and a_{1} ≤ a_{2}. When both S and T are ℤ_{p}- integral, the local density α_{p}(T, S) is defined as

(7.2)  α_{p}(T, S) = lim_{t⟶∞}(p^{t})^{3−2m}A_{t}(T, S),

where

A_{t}(T, S) = #{X ∈ M_{m,2}(ℤ_{p}/p^{t}) : S[X] ≡ T mod p^{t}}.

In general, we associate S to a ℤ_{p}-lattice L = ℤ_{p}^{m} with the quadratic form q(x) = \frac{1}{2}(x, x) = ^{t}xSx. Define

(7.3)  W (T, S) = ∫ _{Sym_{2}(ℚ_{p})} ∫_{L^{2}} ψ( trbq(x))ψ(− trT b) dx db.

where ψ is a canonical additive character of ℚ_{p} given by ψ(x) = e^{−2πix′} where x′ is a rational number with x′ − x ∈ ℤ_{p}. Then Lemma 1.3 gives

(7.4)  α_{p}(T, S) = W (T, S)

when S and T are integral. We use this equation as the definition of α_{p}(T, S) when either S or T is not integral. It is known (Lemma 1.2) that α_{p}(T, S) depends only on the ℤ_{p}-equivalence classes of S and T. For this reason, we sometime write α_{p}(α_{1}p^{a_{1}}, α_{2}p^{a_{2}}, S) for α_{p}(T, S). As in section 5, we fix a unit u ∈ ℤ_{p}^{∗} with (\frac{u}{p}) = −1.

Denote

(7.5)  d(k) = k + \frac{1}{2} ∑_{l_{i}<k}(l_{i} − k),

v_{k} = (\frac{−1}{p})^{[ \frac{l(−k,1)}{2} ]} ∏_{i∈L(−k,1)}(\frac{ϵ_{i}}{p})

as before, and

(7.6)  δ^{±}(k) = \frac{1 ± (−1)^{l(−k,1)}}{2} .

Note that δ^{+}(k) (resp. δ^{−}(k) ) is 1 or 0 depending on whether l(−k, 1) is even (resp. odd) or odd (resp. even). Define

f_{1}(αp^{a}) = \{ \begin{matrix}−\frac{1}{p} & if l(−1 − a, 1) is even, \\ (\frac{α}{p})\frac{1}{\sqrt{p}} & if l(−1 − a, 1) is odd, \end{matrix}

(7.7)  f_{2}(αp^{a}) = \{ \begin{matrix}−\frac{1}{p} & if l(−1 − a, 1) is odd, \\ (\frac{−α}{p})\frac{1}{\sqrt{p}} & if l(−1 − a, 1) is even. \end{matrix}

LOCAL DENSITIES 33

We also define

(7.8)  g(k) = \{ \begin{matrix}δ^{+}(k) & if a_{1} − k is even, \\ (\frac{α_{1}}{p})δ^{−}(k) & if a_{1} − k is odd. \end{matrix}

Theorem 7.1. Assume a_{1} = a_{2} = a and α_{i} ∈ ℤ_{p}^{∗}. Then the local density α_{p}(α_{1}p^{a}, α_{2}p^{a}, S) = 0 unless a ≥ l_{1}. When a ≥ l_{1}, one has

α_{p}(α_{1}p^{a}, α_{2}p^{a}, S)

= p^{3l_{1}} + (1 − p^{−2})p^{l_{1}} ∑_{\begin{matrix}0<k≤a \\ l(−k,1) even \end{matrix}} v_{k}p^{k+d(k)}

+ δ^{+}(1 + a)v_{1+a}p^{l_{1}}((\frac{−α_{1}α_{2}}{p}) − p^{−1})p^{a+d(1+a)}

+ (1 − p^{−1}) ∑_{0<k≤a}(δ^{+}(k) + p^{−1}δ^{−}(k))p^{k+2d(k)}

+ (1 − p^{−1})^{2}(1 + p^{−1}) ∑_{\begin{matrix}0<k_{2}<k_{1}≤a \\ l(−k_{i},1) even \end{matrix}} v_{k_{1}}v_{k_{2}}p^{k_{1}+d(k_{1})+d(k_{2})}

+ δ^{+}(1 + a)v_{1+a}((\frac{−α_{1}α_{2}}{p}) − p^{−1})(1 − p^{−1}) ∑_{\begin{matrix}0<k≤a \\ l(−k,1) even \end{matrix}} v_{k}p^{a+d(1+a)+d(k)}

− ((\frac{−α_{1}α_{2}}{p})δ^{+}(1 + a) + δ^{−}(1 + a))p^{a−1+2d(1+a)}.

Theorem 7.2. Assume a_{1} < a_{2} and α_{i} ∈ ℤ_{p}^{∗}. Then α_{p}(α_{1}p^{a_{1}}, α_{2}p^{a_{2}}, S) = 0 unless a_{1} ≥ l_{1}. When a_{1} ≥ l_{1}, one has

α_{p}(α_{1}p^{a_{1}}, α_{2}p^{a_{2}}, S)

= p^{3l_{1}} + (1 − p^{−2})p^{l_{1}} ∑_{l_{1}<k≤a_{1}} v_{k}δ^{+}(k)p^{k+d(k)}

+ v_{1+a_{1}}p^{l_{1}} ((\frac{α_{1}}{p})\frac{1}{\sqrt{p}}δ^{−}(1 + a_{1}) − p^{−2}δ^{+}(1 + a_{1})) p^{1+a_{1}+d(1+a_{1})}

+ (1 − p^{−1}) ∑_{l_{1}<k≤a_{1}}(δ^{+}(k) + p^{−1}δ^{−}(k))p^{k+2d(k)}

+ (1 − p^{−1})δ^{−}(1 + a_{1})p^{a_{1}+2d(1+a_{1})}

+ (1 − p^{−1})^{2}(1 + p^{−1}) ∑_{l_{1}<k_{2}<k_{1}≤a_{1}} v_{k_{1}}v_{k_{2}}δ^{+}(k_{1})δ^{+}(k_{2})p^{k_{1}+d(k_{1})+d(k_{2})}

+ (1 − p^{−1}) ((\frac{α_{1}}{p})\frac{1}{\sqrt{p}}δ^{−}(1 + a_{1}) − p^{−2}δ^{+}(1 + a_{1}))

· ∑_{l_{1}<k≤a_{1}} v_{1+a_{1}}v_{k}δ^{+}(k)p^{1+a_{1}+d(1+a_{1})+d(k)}

+ (1 − p^{−1})p^{l_{1}} ∑_{1+a_{1}<k<1+a_{2}} v_{k}g(k)p^{\frac{a_{1}+k}{2} +d(k)}

+ (1 − p^{−1})^{2} ∑_{l_{1}<k_{2}<1+a_{1}<k_{1}<1+a_{2}} v_{k_{1}}v_{k_{2}}δ^{+}(k_{2})g(k_{1})p^{\frac{a_{1}+k}{2} +d(k_{1})+d(k_{2})}

+ (1 − p^{−1})v_{1+a_{1}}f_{1}(α_{1}p^{a_{1}}) ∑_{1+a_{1}<k<1+a_{2}} v_{k}g(k)p^{\frac{a_{1}+k}{2} +d(1+a_{1})+d(k)}

+ v_{1+a_{2}}p^{\frac{1+a_{1}+a_{2}}{2} +d(1+a_{2})}G(a_{1}, a_{2}).

Here

G(a_{1}, a_{2}) = p^{l_{1}}f_{1}(α_{2}p^{a_{2}}) + v_{1+a_{1}}f_{1}(α_{1}p^{a_{1}})f_{1}(α_{2}p^{a_{2}})p^{d(1+a_{1})}

+ (1 − p^{−1}) ∑_{l_{1}<k≤a_{1}} v_{k}δ^{+}(k)f_{1}(α_{2}p^{a_{2}})p^{d(k)}

when a_{1} + a_{2} is odd, and

G(a_{1}, a_{2}) = (\frac{α_{1}}{p})p^{l_{1}}f_{2}(α_{2}p^{a_{2}}) + (\frac{α_{1}}{p})v_{1+a_{1}}f_{1}(α_{1}p^{a_{1}})f_{2}(α_{2}p^{a_{2}})p^{d(1+a_{1})}

+ (1 − p^{−1})(\frac{α_{1}}{p}) ∑_{l_{1}<k≤a_{1}} v_{k}δ^{+}(k)f_{2}(α_{2}p^{a_{2}})p^{d(k)}

when a_{1} + a_{2} is even.

Remark 7.3. When S is unimodular, one can verify (by tedious calculation) that our formulas coincide with those of Kitaoka’s ([Ki, Theorem 2]). We will give a ‘shorter’ version of Theorem at the end of this section.

Corollary 7.4. Let V be a nondegerated quadratic space over ℚ_{p} of dimension m > 2. Let ϕ ∈ S(V^{2}) be a locally constant function on V^{2} with compact support. Let T be a nonzero 2 × 2 symmetric matrix over ℚ_{p}. Then the local Whittaker function W_{T} (g, s + s_{0}, ϕ), defined in section 1, is a polynomial of X = p^{−s} for any g ∈ G, where s_{0} = \frac{m−3}{2} and G = Sp_{2}(ℚ_{p}) or Mp_{2,ℚ_{p}} depending on whether m is even or odd.

Proof. When T is non-degenerate, it follows from Theorems 7.1 and 7.2 and the proof of Theorem 3.4. When T has rank one, standard calculation reduces the Whittaker function into Whittaker functions on Sp_{1} or Mp_{1}, where we can apply Theorem 3.4.

The rest of this section is to prove Theorems 7.1 and 7.2.

Lemma 7.5. With notation as above, one has

(7.9) α_{p}(T, S) = ∑_{k∈ℤ} p^{3k}J(p^{−k})^{2}R(k) + ∑_{k_{1}>k_{2}} p^{2k_{1}+k_{2}}J(p^{−k_{1}})J(p^{−k_{2}})R(k_{1}, k_{2}).

Here

R(k) = \frac{1}{2(1 − (\frac{−1}{p})p^{−1})}I^{∗}(−α_{1}p^{a_{1}−k}, −α_{1}p^{a_{1}−k}, −α_{2}p^{a_{2}−k}, −α_{2}p^{a_{2}−k})

+ \frac{(−1)^{l(−k,1)}}{2(1 + (\frac{−1}{p})p^{−1})}I^{∗}(−α_{1}p^{a_{1}−k}, −α_{1}up^{a_{1}−k}, −α_{2}p^{a_{2}−k}, −α_{2}up^{a_{2}−k}),

and

R(k_{1}, k_{2}) = \frac{1}{4} ∑_{β_{1},β_{2}=1 or u}(\frac{β_{1}}{p})^{l(−k_{1},1)}(\frac{β_{2}}{p})^{l(−k_{2},1)}
· I^{∗}(−α_{1}β_{1}p^{a_{1}−k_{1}}, −α_{1}β_{2}p^{a_{1}−k_{2}}, −α_{2}β_{1}p^{a_{2}−k_{1}}, −α_{2}β_{2}p^{a_{2}−k_{2}}).

Proof. By Proposition 5.1 and (7.4), one has

α_{p}(T, S) = ∑_{b∈ℤ} p^{3b} (\frac{1}{2(1 − (\frac{−1}{p})p^{−1})}K(p^{b}, p^{b}) + \frac{1}{2(1 + (\frac{−1}{p})p^{−1})}K(p^{b}, up^{b}))

+ \frac{1}{4} ∑_{\begin{matrix}b_{1}<b_{2} \\ β_{i}=1 or u \end{matrix}} p^{−2b_{1}−b_{2}}K(β_{1}p^{b_{1}}, β_{2}p^{b_{2}})

Where

K(t_{1}, t_{2}) = ∫ _{GL_{2}(ℤ_{p})} J(t_{1}, t_{2})ψ(− trT ^{t}g diag(t_{1}, t_{2})g)dg

and

J(t_{1}, t_{2}) = ∫_{(ℤ_{p}^{m})^{2}} ψ(\frac{1}{2} tr^{t}g ( \begin{matrix}t_{1} & 0 \\ 0 & t_{2} \end{matrix} ) g ( \begin{matrix}(x, x) & (x, y) \\ (x, y) & (y, y) \end{matrix} )) dxdy.

A substitution of ( \begin{matrix}x \\ y \end{matrix} ) by g ( \begin{matrix}x \\ y \end{matrix} ) gives

J(t_{1}, t_{2}) = ∫_{(ℤ_{p}^{m})^{2}} ψ(\frac{1}{2} tr ( \begin{matrix}t_{1} & 0 \\ 0 & t_{2} \end{matrix} ) ( \begin{matrix}(x, x) & (x, y) \\ (x, y) & (y, y) \end{matrix} ))
= ∫_{(ℤ_{p}^{m})^{2}} ψ(t_{1}q(x))ψ(t_{2}q(y)) dx dy
= J(t_{1})J(t_{2}).

In particular, J(t_{1}, t_{2}) does not depend on g. Direct calculation gives

∫ _{GL_{2}(ℤ_{p})} ψ(− trT ^{t}g diag(t_{1}, t_{2})g)dg
= I^{∗}(−α_{1}p^{a_{1}}t_{1}, −α_{1}p^{a_{1}}t_{2}, −α_{2}p^{a_{2}}t_{1}, −α_{2}p^{a_{2}}t_{2}).

Finally, J(βp^{b}) = (\frac{−1}{p})^{l(b,1)}J(p^{b}) by Corollary 2.3. Now putting pieces together, and replacing b_{i} by −k_{i}, one proves the lemma.

Lemma 7.6. Let the notation be as above.

(1) One has

J(p^{−k})^{2} = p^{∑_{l_{i}<k}(l_{i}−k)}(\frac{−1}{p})^{l(−k,1)}.

(2) One has

(\frac{−1}{p})^{l(−k,1)}R(k)

= \{ \begin{matrix}(1 − p^{−1})(δ^{+}(k) + p^{−1}δ^{−}(k)) & if k ≤ a_{1}, \\ p^{−1}(1 − p^{−1})δ^{−}(k) & if k = 1 + a_{1}&a_{1} < a_{2}, \\ −p^{−2}((\frac{−α_{1}α_{2}}{p})δ^{+}(k) + δ^{−}(k)) & if k = 1 + a_{1}&a_{1} = a_{2}, \\ 0 & if k > 1 + a_{1}. \end{matrix}

Proof. Claim (1) follows from Lemma 2.2 directly. When k ≤ a_{1}, one has by Corollary 6.3

I^{∗}(−α_{1}p^{a_{1}−k}, −βα_{1}p^{a_{1}−k}, −α_{2}p^{a_{2}−k}, −α_{2}βp^{a_{2}−k}) = (1 − p^{−1})^{2}(1 + p^{−1}).

So

R(k) = (1 − p^{−1})^{2}(1 + p^{−1})\{\frac{1}{2(1 − (\frac{−1}{p})p^{−1})} + \frac{(−1)^{l(−k,1)}}{2(1 + (\frac{−1}{p})p^{−1})}\}

= (1 − p^{−1})(δ^{+}(k) + (\frac{−1}{p})p^{−1}δ^{−}(k)).

Notice that

(7.10)  (\frac{−1}{p})^{l(−k,1)}(xδ^{+}(k) + y(\frac{−1}{p})δ^{−}(k)) = xδ^{+}(k) + yδ^{−}(k)

for any number x and y. So

(\frac{−1}{p})^{l(−k,1)}R(k) = (1 − p^{−1})(δ^{+}(k) + p^{−1}δ^{−}(k)).

When k = 1 + a_{1} and a_{1} < a_{2}, one has a_{2} − k ≥ 0. In this case, Corollary 6.4 implies

I^{∗}(−α_{1}p^{a_{1}−k}, −βα_{1}p^{a_{1}−k}, −α_{2}p^{a_{2}−k}, −βα_{2}p^{a_{2}−k})

= p^{−1}(1 − p^{−1})(\frac{−β}{p})\{1 − (\frac{−β}{p})p^{−1}\}.

Now simple calculation (using (7.10)) gives the desired formula for this case. The case where k = 1 + a_{1} and a_{1} = a_{2} follows similarly from Corollary 6.6 and Lemma 6.1

Finally, when k > 1 + a_{1}, a_{1} + b ≤ −2. So one has by Corollary 6.3

I^{∗}(−α_{1}p^{a_{1}−k}, −βα_{1}p^{a_{1}−k}, −α_{2}p^{a_{2}−k}, −α_{2}βp^{a_{2}−k}) = 0.

So R(k) = 0.

Next come to the second sum in (7.9). First we make a renormalization of the factors involved. Set for an integer k

θ(k) = \{ \begin{matrix}1 & if l(−k, 1) is even, \\ δ_{p} & if l(−k, 1) is odd. \end{matrix}

Then the following trivial observation will be frequently used in the rest of the paper.

(7.11)  θ(k)^{−1}(xδ^{+}(k) + yδ_{p}δ^{−}(k)) = xδ^{+}(k) + yδ^{−}(k),

where x and y are any complex numbers and k is any integer. Denote

(7.12)  R^{∗}(k_{1}, k_{2}) = (θ(k_{1})θ(k_{2}))^{−1}R(k_{1}, k_{2})

and

J^{∗}(k_{1}, k_{2}) = θ(k_{1})θ(k_{2})J(p^{−k_{1}})J(p^{−k_{2}}).

Then

(7.13)  J^{∗}(k_{1}, k_{2}) = v_{k_{1}}v_{k_{2}}p^{\frac{1}{2} [∑_{l_{i}<k_{1}} (l_{i}−k_{1})+∑_{l_{i}<k_{2}} (l_{i}−k_{2})]}

where v_{k} is defined by (7.5), and

(7.14)  J^{∗}(k_{1}, k_{2})R^{∗}(k_{1}, k_{2}) = J(p^{−k_{1}})J(p^{−k_{2}})R(k_{1}, k_{2}).

The following lemma follows directly from Corollary 6.3.

Lemma 7.7. (1) If k_{2} < k_{1} ≤ a_{1}, then

R^{∗}(k_{1}, k_{2}) = (1 − p^{−1})^{2}(1 + p^{−1})δ^{+}(k_{1})δ^{+}(k_{2}).

(2) If k_{1} > 1 + a_{2} or k_{2} > 1 + a_{1}, then R^{∗}(k_{1}, k_{2}) = 0.

Lemma 7.8. (1) When k_{1} = 1 + a_{1} > k_{2} and a_{1} < a_{2}, one has

R^{∗}(k_{1}, k_{2}) = (1 − p^{−1})δ^{+}(k_{2}) ((\frac{α_{1}}{p})\frac{1}{\sqrt{p}}δ^{−}(k_{1}) − \frac{1}{p^{2}} δ^{+}(k_{1})) .

(1′) When k_{1} = 1 + a_{1} > k_{2} and a_{1} = a_{2}, one has

R^{∗}(k_{1}, k_{2}) = p^{−1}(1 − p^{−1}) ((\frac{−α_{1}α_{2}}{p}) − \frac{1}{p}) δ^{+}(k_{1})δ^{+}(k_{2}).

(2) When 1 + a_{2} > k_{1} > 1 + a_{1} > k_{2}, one has

R^{∗}(k_{1}, k_{2}) = (1 − p^{−1})^{2}p^{\frac{a_{1}−k_{1}}{2}} δ^{+}(k_{2})g(k)

(3) When 1 + a_{2} > b_{1} > 1 + a_{1} = k_{2}, one has

R^{∗}(k_{1}, k_{2}) = (1 − p^{−1})p^{\frac{a_{1}−k_{1}}{2}} f_{1}(α_{1}p^{a_{1}})g(k)

(4) When k_{1} = 1 + a_{2} > 1 + a_{1} > k_{2}, one has

R^{∗}(k_{1}, k_{2}) = (1 − p^{−1})p^{\frac{a_{1}−a_{2}−1}{2}} δ^{+}(k_{2})

· \{ \begin{matrix}f_{1}(α_{2}p^{a_{2}}) & if a_{1} − a_{2} is odd, \\ (\frac{α_{1}}{p})f_{2}(α_{2}p^{a_{2}}) & if a_{1} − a_{2} is even. \end{matrix}

(5) When k_{1} = 1 + a_{2} > 1 + a_{1} = k_{2}, one has

R^{∗}(k_{1}, k_{2}) = (1 − p^{−1})p^{\frac{a_{1}−a_{2}−1}{2}} f_{1}(α_{1}p^{a_{1}})

· \{ \begin{matrix}f_{1}(α_{2}p^{a_{2}}) & if a_{1} − a_{2} is odd, \\ (\frac{α_{1}}{p})f_{2}(α_{2}p^{a_{2}}) & if a_{1} − a_{2} is even. \end{matrix}

Proof. The proof is case by case verification using results in section 6. To save space, we write

I(β_{1}, β_{2}) = I^{∗}(−α_{1}β_{1}p^{a_{1}−k_{1}}, −α_{1}α_{2}p^{a_{1}−k_{2}}, −α_{2}β_{1}p^{a_{2}−k_{1}}, −α_{2}β_{2}p^{a_{2}−k_{2}})

temporarily.

When k_{1} = 1 + a_{1} > k_{2} and a_{1} < a_{2}, one has a_{1} − k_{1} = −1, a_{1} − k_{2} ≥ 0, a_{2} − k_{1} ≥ 0, and a_{2} − k_{2} ≥ 0. So Lemma 2.1 and Corollary 6.4 imply

I(β_{1}, β_{2}) = (1 − p^{−1})I^{∗}(−α_{1}β_{1}p^{−1}) + p^{−1}(1 − p^{−1})^{2}

= (1 − p^{−1})(δ_{p}(\frac{α_{1}β_{1}}{p})\frac{1}{\sqrt{p}} − \frac{1}{p^{2}} ).

Since (\frac{u}{p}) = −1, one has then

R(k_{1}, k_{2}) = \frac{1}{4} ∑_{β_{i}=1 or u}(\frac{β_{1}}{p})^{l(−k_{1},1)}(\frac{β_{2}}{p})^{l(−k_{2},1)}I(β_{1}, β_{2})

= (1 − p^{−1})δ^{+}(k_{2}) ((\frac{α_{1}}{p})\frac{1}{\sqrt{p}}δ_{p}δ^{−}(k_{1}) − \frac{1}{p^{2}} δ^{+}(k_{1})) .

Applying (7.11), one proves (1).

When k_{1} = 1 + a_{1} > k_{2} and a_{1} = a_{2}, one has a_{2} − k_{1} = a_{1} − k_{1} = −1, and a_{1} − k_{2} = a_{2} − k_{2} ≥ 0. So Lemma 2.1 and Corollary 6.4 imply

I(β_{1}, β_{2}) = (1 − p^{−1})I^{∗}(−α_{1}β_{1}p^{−1})I(−α_{2}β_{1}p^{−1}) + p^{−1}(1 − p^{−1})I^{∗}(−α_{2}β_{1}p^{−1})

= p^{−1}(1 − p^{−1})((\frac{−α_{1}α_{2}}{p}) − p^{−1})

Here we have used δ^{2} = (\frac{−1}{p}). Plugging this into R^{∗}(b_{1}, b_{2}) and applying (7.11), one obtains (1′).

When 1 + a_{2} > k_{1} > 1 + a_{1} > k_{2}, one has a_{1} − k_{1} ≤ −2, a_{1} − k_{2} ≥ 0, a_{2} − k_{1} ≥ 0, and a_{2} − k_{2} ≥ 0. So Lemma 2.1 and Corollary 6.3 imply

I(β_{1}, β_{2}) = (1 − p^{−1})^{2}I(−α_{1}β_{1}p^{a_{1}−k_{1}})

= (1 − p^{−1})^{2}p^{\frac{a_{1}−k_{1}}{2}} \{ \begin{matrix}1 & if a_{1} − k_{1} is even, \\ δ_{p}(\frac{α_{1}β_{1}}{p}) & if a_{1} − k_{1} is odd. \end{matrix}

Plugging this into R^{∗}(b_{1}, b_{2}) and applying (7.11), one obtains (2).

When 1+a_{2} > k_{1} > 1+a_{1} = k_{2}, one has a_{1} −k_{1} ≤ −2, a_{1} −k_{2} = −1, a_{2} −k_{1} ≥ 0, and a_{2} − k_{2} ≥ 0. So Lemma 2.1 and Corollary 6.3 imply

I(β_{1}, β_{2}) = (1 − p^{−1})I(−α_{1}β_{1}p^{a_{1}−k_{1}})I^{∗}(−α_{1}β_{2}p^{−1})

= (1 − p^{−1})p^{\frac{a_{1}−k_{1}}{2}} (δ_{p}(\frac{α_{1}β_{2}}{p})\frac{1}{\sqrt{p}} − \frac{1}{p})

· \{ \begin{matrix}1 & if a_{1} − k_{1} is even, \\ δ_{p}(\frac{α_{1}β_{1}}{p}) & if a_{1} − k_{1} is odd. \end{matrix}

Plugging this into R^{∗}(k_{1}, k_{2}) and applying (7.11), one obtains (3).

When 1+a_{2} = k_{1} > 1+a_{1} > k_{2}, one has a_{1} −k_{1} ≤ −2, a_{2} −k_{1} = −1, a_{1} −k_{2} ≥ 0, and a_{2} − k_{2} ≥ 0. So Lemma 2.1 and Corollary 6.3 imply

I(β_{1}, β_{2}) = (1 − p^{−1})I(−α_{1}β_{1}p^{a_{1}−k_{1}})I^{∗}(−α_{2}β_{1}p^{−1})

= (1 − p^{−1})p^{\frac{a_{1}−a_{2}−1}{2}} (δ_{p}(\frac{α_{2}β_{1}}{p})\frac{1}{\sqrt{p}} − \frac{1}{p})

· \{ \begin{matrix}1 & if a_{1} − k_{1} is even \\ δ_{p}(\frac{α_{1}β_{1}}{p}) & if a_{1} − k_{1} is odd \end{matrix}

= (1 − p^{−1})p^{\frac{a_{1}−a_{2}−1}{2}} \{ \begin{matrix}δ_{p}(\frac{α_{2}β_{1}}{p})\frac{1}{\sqrt{p}} − \frac{1}{p} & if a_{1} + a_{2} is odd, \\ (\frac{α_{1}}{p})\{(\frac{−α_{2}}{p})\frac{1}{\sqrt{p}} − δ_{p}(\frac{β_{1}}{p})\frac{1}{p}\} & if a_{1} − a_{2} is even. \end{matrix}

Here we have used δ_{p}^{2} = (\frac{−1}{p}). Plugging this into R^{∗}(k_{1}, k_{2}) and applying (7.11), one obtains (4).

Finally, when 1 + a_{2} = k_{1} > 1 + a_{1} = k_{2}, one has a_{1} − k_{1} ≤ −2, a_{2} − k_{1} = −1, a_{1} − k_{2} = −1, and a_{2} − k_{2} ≥ 0. So Lemma 2.1 and Corollary 6.3 imply

I(β_{1}, β_{2}) = I(−α_{1}β_{1}p^{a_{1}−k_{1}})I^{∗}(−α_{2}β_{1}p^{−1})I^{∗}(−α_{1}β_{2}p^{−1})

= p^{\frac{a_{1}−a_{2}−1}{2}} (δ_{p}(\frac{α_{1}β_{2}}{p})\frac{1}{\sqrt{p}} − \frac{1}{p})

· \{ \begin{matrix}δ_{p}(\frac{α_{2}β_{1}}{p})\frac{1}{\sqrt{p}} − \frac{1}{p} & if a_{1} + a_{2} is odd \\ (\frac{α_{1}}{p})\{(\frac{−α_{2}}{p})\frac{1}{\sqrt{p}} − δ_{p}(\frac{β_{1}}{p})\frac{1}{p}\} & if a_{1} − a_{2} is even. \end{matrix}

Plugging this into R^{∗}(k_{1}, k_{2}) and applying (7.11), one obtains (5).

Proof of Theorem 7.1 Now the proof of Theorem 7.1 becomes tedious but easy calculation. We may assume l_{1} = 0 by shifting. By Lemma 7.6, one has

(7.15) ∑_{k∈ℤ} p^{3k}J(p^{−k})^{2}R(k)

= (1 − p^{−1}) ∑_{k≤0} p^{3k} + (1 − p^{−1}) ∑_{0<k≤a} p^{k+2d(k)}(δ^{+}(k) + p^{−1}δ^{−}(k)

− p^{−1+a+2d(1+a)}((\frac{−α_{1}α_{2}}{p})δ^{+}(1 + a) + δ^{−}(1 + a))

= (1 − p^{−1})(1 − p^{−3})^{−1} + (1 − p^{−1}) ∑_{0<k≤a} p^{k+2d(k)}(δ^{+}(k) + p^{−1}δ^{−}(k))

− p^{−1+a+2d(1+a)}((\frac{−α_{1}α_{2}}{p})δ^{+}(1 + a) + δ^{−}(1 + a)).

By Lemma 7.7 and (7.13) − (7.14), one has

∑_{k_{1}>k_{2}} p^{2k_{1}+k_{2}}J(p^{−k_{1}})J(p^{−k_{2}})R(k_{1}, k_{2})

= ∑_{k_{2}<k_{1}} v_{k_{1}}v_{k_{2}}p^{k_{1}+d(k_{1})+d(k_{2})}R^{∗}(k_{1}, k_{2})

= ∑_{k_{2}<k_{1}≤0} + ∑_{k_{2}≤0<k_{1}≤a} + ∑_{k_{2}≤0, k_{1}=a+1} + ∑_{0<k_{2}<k_{1}≤a} + ∑_{0<k_{2}≤a<k_{1}=a+1}

where every sum is on v_{k_{1}}v_{k_{2}}p^{k_{1}+d(k_{1})+d(k_{2})}R^{∗}(k_{1}, k_{2}). By Lemmas 7.7, one has

∑_{k_{2}<k_{1}≤0} v_{k_{1}}v_{k_{1}}v_{k_{2}}p^{k_{1}+d(k_{1})+d(k_{2})}R^{∗}(k_{1}, k_{2})

= (1 − p^{−1})^{2}(1 + p^{−1}) ∑_{k_{2}<k_{1}≤0} p^{2k_{1}+k_{2}}

= p^{−1}(1 − p^{−1})(1 + p^{−1})(1 − p^{−3})^{−1}.

Notice that the sum of this term and the term (1 − p^{−1})(1 − p^{−3})^{−1} in the sum just computed is one. By Lemmas 7.7, one has

∑_{k_{2}≤0<k_{1}≤a} v_{k_{2}}p^{k_{1}+d(k_{1})+d(k_{2})}R^{∗}(k_{1}, k_{2})

= ∑_{0<k_{1}≤a} v_{k_{1}}p^{k_{1}+d(k_{1})}δ^{+}(k_{1})(1 − p^{−1})^{2}(1 + p^{−1}) ∑_{k_{2}≤0} p^{k_{2}}

= (1 − p^{−2}) ∑_{\begin{matrix}0<k≤a \\ l(−k,1) even \end{matrix}} v_{k}p^{k+d(k)}

Similar calculation using Lemmas 7.8(1′) and 7.7 gives

∑_{k_{2}≤0, k_{1}=a+1} v_{k_{1}}v_{k_{2}}p^{k_{1}+d(k_{1})+d(k_{2})}R^{∗}(k_{1}, k_{2})

= δ^{+}(1 + a)v_{1+a}((\frac{−α_{1}α_{2}}{p}) − p^{−1})p^{a+d(1+a)},

and

∑_{0<k_{2}<k_{1}≤a} v_{k_{1}}v_{k_{2}}p^{k_{1}+d(k_{1})+d(k_{2})}R^{∗}(k_{1}, k_{2})

= (1 − p^{−1})^{2}(1 + p^{−1}) ∑_{\begin{matrix}0<k_{2}<k_{1}≤a \\ l(−k_{i},1) even \end{matrix}} v_{k_{1}}v_{k_{2}}p^{k_{1}+d(k_{1})+d(k_{2})},

and

∑_{0<k_{2}≤a<k_{1}=a+1} v_{k_{1}}v_{k_{2}}p^{k_{1}+d(k_{1})+d(k_{2})}R^{∗}(k_{1}, k_{2})

= δ^{+}(1 + a)v_{1+a}((\frac{−α_{1}α_{2}}{p}) − p^{−1})(1 − p^{−1}) ∑_{\begin{matrix}0<k≤a \\ l(−k,1) even \end{matrix}} v_{k}p^{a+d(1+a)+d(k)}.

Putting pieces together, one proves Theorem 7.1.

The proof of Theorem 7.2 is similar. A ‘shorter’ version of Theorem 7.2 is

Theorem 7.2′. Let the notation be as in Theorem 7.2. Then

α_{p}(α_{1}p^{a_{1}}, α_{2}p^{a_{2}}, S) = p^{3l_{1}} + (1 − p^{−1})δ^{−}(1 + a_{1})p^{a_{1}−1+2d(a_{1}+1)}

+ (1 − p^{−1}) ∑_{l_{1}<k≤a_{1}}(δ^{+}(k) + p^{−1}δ^{−}(k))p^{k+2d(k)}

+ (1 − p^{−1})^{−1}p^{l_{1}} ∑_{l_{1}<k_{1}≤1+a_{2}} p^{2k_{1}}J^{∗}(k_{1}, 0)R^{∗}(k_{1}, 0)

+ ∑_{\begin{matrix}0<k_{1}≤1+a_{2} \\ 0<k_{2}≤1+a_{1} \\ k_{2}<k_{1} \end{matrix}} p^{2k_{1}+k_{2}}J^{∗}(k_{1}, k_{2})R^{∗}(k_{1}, k_{2})

where J^{∗}(k_{1}, k_{2}) is given by (7.13) and R^{∗}(k_{1}, k_{2}) is given by Lemma 7.9.

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Department of Mathematics, University of Michigan, Ann Arbor, MI 48109

   E-mail address: thyang@math.lsa.umich.edu