LOCAL DENSITIES OF 2-ADIC QUADRATIC FORMS

Tonghai Yang

June 19, 2003

Abstract. In this paper, we give an explicit from formula for the local density number of representing a two by two 2-integral matrix T by a quadratic 2-integral lattice L over ℤ_{2}. The non-dyadic case was dealt in a previous paper. The special case when L is a (maximal) lattice in the space of trace zero elements in a quaternion algebra over ℚ_{2} yields a clean and interesting formula, which matches up perfectly with the non-dyadic case in terms of the Gross-Keating invariants. In this special case, The derivatives (when value is zero) matches up with certain local intersection numbers on Shimura curves.

1. Introduction and general set-up.

Local density of a quadratic form representing another quadratic form was first introduced by C. L. Siegel ([Si]) as the local building block of the Fourier coefficients of an Eisenstein series in 1935. Its connection with arithmetic geometry was discovered much later by Gross and Keating ([GK]), and by Kudla ([Ku]). It is very important to compute the local density and its derivative for arithmetic applications. We refer to [Ki], [HS], [Ka], [Ki], and [Ya] among others for introduction and reference on this subject. In fact, this paper is a complement to [Ya], where we considered the case p ≠ 2.

In general, let (L, q) be a non-degenerate even integral quadratic lattice over ℤ_{p} of rank n, i,e., q(x) ∈ ℤ_{p} for every x ∈ L. Let T be a m × m symmetric matrix over ℤ_{p}, we define the quadratic local density of L representing T as

(1.1)  W (T, L) = ∫ _{Sym_{m}(ℚ_{p})} ∫_{L^{m}} ψ( trbq(x))dxψ(− trT b)db.

Here the additive character ψ : ℚ_{p} ⟶ ℂ^{1} is given by ψ(x) = e^{−2πiλ(x)} where λℚ_{p}/ℤ_{p} ↪ ℚ/ℤ. In fact, the local density does not depend on the choice of

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Partially supported by a NSF grant DMS-0302043

ψ, we choose it for convenience. It depends of course on the choice of the Haar measures on L^{m} and Sym_{m}(ℚ_{p}). For any integer s ≥ 0, let L_{s} = L ⊕ H^{s} where H is the standard hyperplane H = ℤ_{p}^{2} with the quadratic form q(x_{1}, x_{2}) = x_{1}x_{2}. We define the quadratic local density polynomial W (X; T, L) via

(1.2)  W (p^{−s}; T, L) = W (T, L_{s}).

Let K be an open compact subgroup of GL_{m}(ℤ_{p}) acting on Sym_{m}(ℚ_{p}) via k.γ = kγ^{t}k. We define

(1.3)  A(γ, L) = ∫_{L^{m}} tr(γq(x))dx,

for γ ∈ Sym_{m}(ℚ_{p}), and

(1.4)  B(T, γ) = ∫_{K} ψ(− trk.T γ)dk.

Finally, for every γ ∈ Sym_{m}(ℚ_{p}) with det γ ≠ 0, there is a number C(γ) such that

(1.5)  C(γ) ∫_{K} ϕ(k.γ)dk = ∫_{O_{γ}} ϕ(x)dx

for every locally constant function ϕ on O_{γ}, where O_{γ} is the K-orbit of γ in Sym_{m}(ℚ_{p}). Here we define the Haar measure dk on GL_{m}(ℤ_{p}) as the restriction of the standard Haar measure on M_{m}(ℤ_{p}). In particular, vol(M_{m}(ℤ_{p})) = 1 and

(1.6)  vol( GL_{m}(ℤ_{p})) = p^{− \frac{m(m+1)}{2}} [ GL_{m}(ℤ_{p}) : K_{0}(p^{r})],

where

K_{0}(p^{r}) = {(g_{ij}) ∈ GL_{m}(ℤ_{p}) : g_{ij} ∈ 2^{r}ℤ_{p}, for i < j}.

Let A be a complete set of representatives of the K-orbits in Sym_{m}(ℚ_{p}) ∩ GL_{m}(ℚ_{p}), then we have as in [Ya, Section 7], and [HS]

(1.7)  W (T, L) = ∑_{γ∈A} A(γ, L)B(T, γ)C(γ).

As observed in [HS], the key is to choose K so that A is easy to describe and that A(γ, L), B(T, γ), and C(γ) are computable. Another problem is how to put things together to get a reasonable formula. When n = 1 or n = 2 but p ≠ 2, this was done by the author in [Ya]. Y. Hironaka and F. Sato extend the work to general n but p ≠ 2. The case p = 2 was left out in [Ya] and [HS] mainly because that A is more complicated, though completely determined by Watson in the case K = GL_{m}(ℤ_{2}) ([Wa]), and that putting things together might be too complicated.

In this paper, we deal with the case n = p = 2, and obtain explicit formulae for the local density polynomial W (X, T, L) in sections 2-4, according to the shape of the Jordan form of T (see (1.16)). They are complicated, I have to admit, and we have to use a computer program written by Song Guo to compute examples (http://www.math.wisc.edu/∼thyang/programs). However, in the case where B is a quaternion algebra over ℚ_{2} with maximal order O_{B}, and L is the lattice of trace zero elements in O_{B} with the reduced norm as its quadratic form, our formula for W (X, T, L) turns out to be surprisingly clean and simple in terms of the Gross-Keating invariants and to be of independent of the shape of the Jordan form of T (Theorems 5.5). In fact, it matches up with non-dyadic case ([Ya]) perfectly too (Theorem 5.7). From these formulae, it becomes clear (Corollaries 5.8 and 5.9), as first observed by Kudla ([Ku]), that when B = M_{2}(ℚ_{2}) and W (X, T, L) = 0, W′(1, T, L) is equal to the local intersection numbers computed by Gross and Keating ([GK, Proposition 5.4]). When B is a division algebra over ℚ_{2}, a similar identity is expected to be true as Kudla and Rapoport finish the geometric calculation soon ([KR2]). The same is known to be true for p ≠ 2 using the same method ([KR1]). Sections 6-9 are technical computations of the functions A(γ, L), B(T, γ), and C(γ). In Appendix A, we study the structure of U(1) in the 2-adic case for lack of literature reference,and give an explicit filtration of U(1) needed for computing C(γ). The ramified case was first computed by J. S. Li ([Li]). Appendix B is based on Kudla’s note and is essentially due to him. Here we computed the Gross-Keating invariants in terms of Jordan form. Both appendices should be of independent interest. In Appendix C, we take this opportunity to correct some minor mistakes in [Ya] and reformulate some of the results there in a more unified way.

Acknowledgement This work is motivated by the joint work with Kudla and Rapoport to relate the central derivative of a genus two Eisenstein series with some height pairing on a Shimura curve. I thank them for their encouragement and the joint work. The main formulae would be practically useless without the computer program written by Song Guo. The results in Section 5 were obtained using his program. I thank him for his generous help. I thank Kudla also for including his note here as an appendix and for suggesting me to use the Gross-Keating invariants. I thank J.S.Li for sharing his private note with me. Finally, I would like to thank the Max-Planck Institut für Mathematik at Bonn for its hospitality and for providing the excellent working environment. The final part of the work was done while I visit the MPIM at Bonn during the summer of 2003.

Now we set up the general framework and notation for the paper. Let c and r ≥ −2 be integers, we define the sets A_{r}(c) as follows. When r ≥ 3,

(1.8)  A_{r}(c) = {diag(β_{1}2^{c}, β_{3}2^{c+r}) : β_{1}, β_{2} ∈ (ℤ/8)^{∗}},

When r = 2, we define A_{2}(c) the same way except that β_{1} ∈ (ℤ/4)^{∗}, noticing that diag(1, 4β) is GL_{2}(ℤ_{2}) equivalent to diag(1 + 4β, 4 + β) in this case.

When r = 1, define

(1.9)  A_{1}(c) = {2^{c} diag(1, 2β), 2^{c} diag(−3, 2β), β ∈ (ℤ/8)^{∗}}.

Notice that diag(1, 2β) is GL_{2}(ℤ_{2}) equivalent to diag(1 + 2β, 2β(1 + 2β)).

When r = 0, we define A_{0}(c) = A_{0}^{+}(c) ∪ A_{0}^{−}(c) with

(1.10)  A_{0}^{+}(c) = \{±2^{c} ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) , ±2^{c} ( \begin{matrix}1 & 0 \\ 0 & −3 \end{matrix} )\},

and

(1.11)  A_{0}^{−}(c) = \{2^{c} ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) , 2^{c} ( \begin{matrix}1 & 0 \\ 0 & 3 \end{matrix} )\}.

Here we remark that diag(1, β) is GL_{2}(ℤ_{2}) equivalent to diag(β, 1) and to diag(1 + 4β, 4 + β).

We further define

(1.12)  A_{−1}(c) = \{2^{c} ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )\}, and A_{−2}(c) = \{2^{c} ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} )\}.

Then a theorem of Watson ([Wa, Theorem 3]) asserts that A = ∐ A_{r}(c), i.e., every 2 × 2 nonsingular symmetric matrix over ℚ_{2} is GL_{2}(ℤ_{2})-equivalent to exactly one matrix in ∐_{r≥−2,c∈ℤ} A_{r}(c). So one has

(1.13)  W (T, L) = ∑_{r≥−2,c∈ℤ} S_{r}(c, T, L),

with

(1.14)  S_{r}(c, T, L) = ∑_{γ∈A_{r}(c)} C(γ)A(γ, L)B(T, γ).

Let L = ℤ_{2}^{n} with quadratic form

(1.15)  q = ∑_{i=1}^{H} ϵ_{i}2^{l_{i}−1}x_{i}^{2} + ∑_{j=1}^{M} 2^{m_{j}}y_{j1}y_{j2} + ∑_{k=1}^{N} 2^{n_{k}}(z_{k1}^{2} + z_{k1}z_{k2} + z_{k2}^{2}).

Here H +2M +2N = n, ϵ_{i} ∈ ℤ_{2}^{∗}, and l_{i} ≥ 1, m_{j}, n_{k} ≥ 0 are integers. We reorder l_{i}, m_{j}, m_{j}, n_{k}, n_{k} into an increasing sequence 0 ≤ l′_{1} ≤ l′_{2} ≤ ⋯ ≤ l′_{n}. We choose the standard Haar measure on ℤ_{2}^{n} as the Haar measure on L. Let

(1.16)  T = 2^{a} ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , 2^{a} ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ) , or 2^{a} diag(α_{1}, α_{2}2^{t})

with a ≥ −1 in the first two cases and a, t ≥ 0 in the third case. Our purpose in this paper is to give an explicit formula for W (X, T, L) when L and T are given as in (1.15) and (1.16).

First notice that

(1.17) ∑_{c≥0,r≥−2} S_{r}(c, T, L) = ∫ _{Sym_{2}(ℤ_{2})} ∫_{L^{m}} ψ( trbq(x))dxψ(− trT b)db = 1

Secondly, one sees from definition that, for γ = A_{r}(−c) with c > 0,

(1.18)  A(γ, L_{s}) = A(γ, L)\{ \begin{matrix}X^{2c} & if r < 0, \\ X^{2c−r} & if 0 ≤ r < c, \\ X^{c} & if r ≥ c. \end{matrix}

Here X = 2^{−s}. We will write for convenience

S_{<a} = ∑_{r<c} S_{r} and S_{≥a} = ∑_{r≥a} S_{r}.

Combining (1.13)−(1.14) with (1.17)−(1.18), we obtain the following general proposition.

Proposition 1.1. Let the notation be as above. Then

W (X, T, L) = 1 + ∑_{c>0} S_{<0}(−c, T, L)X^{2c}

+ ∑_{c>0} ∑_{0≤r<c} S_{r}(−c, T, L)X^{2c−r} + ∑_{c>0} S_{≥c}(−c, T, L)X^{c}

= 1 + ∑_{c>0} R(c, T, L)X^{c}.

Here

R(c, T, L) = S_{≥c}(−c, T, L) + ∑_{0<l≤ \frac{m}{2}} S_{c−2l}(l − c, T, L) + S_{<0}(−\frac{c}{2}, T, L).

In this paper, we will use the following notations.

(1.19)  f_{1}(a, c) = \{ \begin{matrix}3 & if 1 ≤ c ≤ a + 1, \\ 1 & if c = a + 2, \\ 0 & if c > a + 2, \end{matrix}

(1.20)  f_{2}(a, c) = \{ \begin{matrix}1 & if 0 < c ≤ a + 1, \\ −1 & if c = a + 2, \\ 0 & if c > a + 2, \end{matrix}

(1.21)  g_{1}(c) = g_{1}(c, L) = ∑_{1≤i≤n} min(0, l′_{i} − c),

and

(1.22)  g_{2}(c) = g_{2}(c, L) = (−1)^{∑_{1≤i≤H} min(0,l_{i}−c)}.

We will prove the following lemma in Section 7 Lemma 1.2. One has

A(γ, L) = \{ \begin{matrix}2^{g_{1}(c)} & if γ ∈ A_{−1}(−c), \\ 2^{g_{1}(c)}g_{2}(c) & if γ ∈ A_{−2}(−c), \\ A(β_{1}2^{−c}, L)A(β_{2}2^{−c+r}, L) & if γ = 2^{−c} diag(β_{1}, 2^{r}β_{2}) ∈ A_{r}(−c), \end{matrix}

where

A(β2^{c}, L) = ∏_{1≤i≤H} I_{1}(ϵ_{i}β2^{c+l_{i}}) ∏_{1≤j≤M} 2^{min(0,c+m_{j})} ∏_{1≤k≤N}(−2)^{min(0,c+n_{k})},

where

I_{1}(α2^{c}) = ∫_{ℤ_{2}} ψ(α2^{c−1}x^{2})dx = \{ \begin{matrix}1 & if c > 0, \\ 0 & if c = 0, \\ 2^{\frac{c}{2}} ψ(\frac{α}{8})(2, α)^{c} & if c < 0. \end{matrix}

In particular, A(β2^{c}, L) = 1 whenever c ≥ 0. Lemma 1.3. Set

(1.)  Ã(c, L) = ∑_{β∈(ℤ/8)^{∗}} A(β2^{c}, L).

Then

2 ∑_{γ∈A_{0}^{−}(−c)} A(γ, L) + ∑_{γ∈A_{0}^{−}(−c)} A(γ, L) = (A(2^{−c}, L) + A(−2^{−c}, L)) Ã(−c, L)

and

∑_{γ∈A_{r}(−c)} A(γ, L) = \{ \begin{matrix}(A(2^{−c}, L) + A(−3 · 2^{−c}, L)) Ã(1 − c, L) & if r = 1, \\ (A(2^{−c}, L) + A(−2^{−c}, L)) Ã(2 − c, L) & if r = 2, \\ Ã(−c, L)Ã(r − c, L) & if r ≥ 3. \end{matrix}

Proof. The last equality follows from definitions of A_{r}(c) and Ã, and the fact

A( diag(γ_{1}, γ_{2}), L) = A(γ_{1}, L)A(γ_{2}, L).

The same reason gives

2 ∑_{γ∈A_{0}^{−}(−c)} A(γ, L) + ∑_{γ∈A_{0}^{−}(−c)} A(γ, L) − (A(2^{−c}, L) + A(−2^{−c}, L)) Ã(−c, L)

= A(2^{−c}, L)A(3 · 2^{−c}, L) − A(−2^{−c}, L)A(−3 · 2^{−c}, L). Lemma 1.2 implies

A(2^{−c}, L)A(3 · 2^{−c}, L) − A(−2^{−c}, L)A(−3 · 2^{−c}, L) = 0.

The following lemma will be proved in Section 9. Lemma 1.4. For γ ∈ A_{r}(−c), one has

C(γ) = 2^{3c−2}\{ \begin{matrix}\frac{1}{3} & if r = −2, \\ 1 & if r = −1, \\ 1 & if γ ∈ A_{0}^{−}(−c), \\ 2^{−1} & if γ ∈ A_{0}^{+}(−c), \\ 2^{−r−1} & if 1 ≤ r ≤ 2, \\ 2^{−r−2} & if r ≥ 3. \end{matrix}

2. The main formula I.

In this section, we consider the cases T = 2^{a} ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) ∈ A_{−1}(a) and T = 2^{a} ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ) ∈ A_{−2}(a) with a ≥ −1. The following lemma will be proved in Section 8.

8 TONGHAI YANG

Lemma 2.1. Let γ ∈ A_{r}(−c) with c > 0.

(1) When T ∈ A_{−2}(a) with a ≥ −1, one has

B(T, γ) = B(γ, T) = \{ \begin{matrix}\frac{3}{8}f_{2}(a, c) & if r ≠ 0, \\ \frac{3}{8} char(ℤ_{2})(2^{a−c+2}) & if r = 0. \end{matrix}

(2) When T ∈ A_{−1}(a) with a ≥ −1, one has

B(T, γ) = B(γ, T) = \{ \begin{matrix}\frac{3}{8}f_{2}(a, c) & if r < 0, \\ \frac{1}{8}f_{1}(a, c) & if r > 0, \\ \frac{1}{8}f_{1}(a, c)f_{2}(a, c) & if r = 0. \end{matrix}

Theorem 2.2. Let T = 2^{a} ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) with a ≥ −1. Then

W (X, T, L) = 1 + ∑_{0<c≤a+2} S_{<0}X^{2c}

+ ∑_{0<c≤a+2} ∑_{0≤r<c} S_{r}X^{2c−r} + ∑_{0<c≤a+2} S_{≥c}X^{c},

where

(2.1)  S_{<0} = 2^{3c−5+g_{1}(c)}(3 + g_{2}(c))f_{2}(a, c),
(2.2)  S_{0} = 2^{3c−6}f_{1}(a, c)f_{2}(a, c)Ã(−c, L) (A(2^{−c}, L) + A(−2^{−c}, L)) ,

and

(2.3) S_{r} = 2^{3c−6−r}f_{1}(a, c)Ã(r − c, L)\{ \begin{matrix}A(2^{−c}, L) + A(3 · 2^{−c}, L) & if r = 1, \\ A(2^{−c}, L) + A(−2^{−c}, L) & if r = 2, \\ \frac{1}{2} Ã(−c, L) & if r ≥ 3. \end{matrix}

and

(2.4)  S_{≥c} = \{ \begin{matrix}2^{2c−4}f_{1}(a, c)Ã(−c, L) & if c ≥ 3, \\ ∑_{c≤r<3} S_{r} + 2^{3c−7}f_{1}(a, c)Ã(−c, L) & if c < 3. \end{matrix}

Here f(r), f_{1}(a, c), and g_{i}(c) are given by formulae (1.19) − (1.22) above. S_{r} = S_{r}(−c, T, L).

Proof. It follows from Proposition 1.1, Lemmas 1.2-1.4 and 2.1, and formula (1.14), which we recall here:

S_{r}(−c, T, L) = ∑_{γ∈A_{r}(−c)} C(γ)B(T, γ)A(γ, L).

By Lemma 2.1, S_{r}(−c, T, L) = 0 unless c ≤ a + 2. Formula (2.1) follows from Lemmas 1.2, 1.4, and 2.1 immediately. So does (2.3) with r > 0, since C(γ) and B(T, γ) depend only on r. When r = 0, one has by Lemmas 1.2-1.4 and 2.1

S_{0}(−c, T, L) = ∑_{γ∈A_{0}^{−}(−c)} 2^{3c−2}A(γ, L)\frac{1}{8}f_{1}(a, c) + ∑_{γ∈A_{0}^{+}(−c)} 2^{3c−3}A(γ, L)\frac{1}{8}f_{1}(a, c)

= 2^{3c−6}f_{1}(a, c) (2 ∑_{γ∈A_{0}^{−}(−c)} A(γ, L) + ∑_{γ∈A_{0}^{+}(−c)} A(γ, L))

= 2^{3c−6}f_{1}(a, c)Ã(−c, T, L)(A(2^{−c}, L) + A(−2^{−c}, L))

as claimed by (2.2).

When r ≥ c ≥ 3, one has Ã(r − c, L) = 2^{2} by Lemmas 1.2-1.3. So for any r_{0} ≥ max(3, c), one has by (2.3)

S_{≥r_{0}} = 3^{3c−5}f_{1}(a, c)Ã(−c, L) ∑_{r≥r_{0}} 2^{−r}
= 2^{3c−4−r_{0}}f_{1}(a, c)Ã(−c, L).

Now (2.4) is clear.

Theorem 2.3. Let T = 2^{a} ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ) with a ≥ −1. Then

W (X, T, L) = 1 + ∑_{0<c≤a+2} S_{<0}X^{2c}

+ ∑_{0<c≤a+2} ∑_{0≤r<c} S_{r}X^{2c−r} + ∑_{0<c≤a+2} S_{≥c}X^{c},

where

(2.5)  S_{<0} = 2^{3c−5+g_{1}(c)}(3 + g_{2}(c))f_{2}(a, c),

and (2.6)

S_{r} = 3 · 2^{3c−6−r}Ã(r − c, L)\{ \begin{matrix}A(2^{−c}, L) + A(−2^{−c}, L) & if r = 0, \\ f_{2}(a, c)(A(2^{−c}, L) + A(3 · 2^{−c}, L)) & if r = 1, \\ f_{2}(a, c)(A(2^{−c}, L) + A(−2^{−c}, L)) & if r = 2, \\ \frac{1}{2}f_{2}(a, c)Ã(−c, L) & if r ≥ 3. \end{matrix}

and

(2.7)  S_{≥c} = \{ \begin{matrix}3 · 2^{2c−4}f_{2}(a, c)Ã(−c, L) & if c ≥ 3, \\ ∑_{c≤r<3} S_{r} + 3 · 2^{3c−7}f_{2}(a, c)Ã(−c, L) & if c < 3. \end{matrix}

Here f(r), f_{1}(a, c), and g_{i}(c) are given by formulae (1.19) − (1.22) above. S_{r} = S_{r}(−c, T, L).

The proof is the same as that of Theorem 2.2, and is left to the reader.

3. The Main Formula II.

In this section, we consider the case T = 2^{a} diag(α_{1}, α_{2}) ∈ A_{0}(a) with 0 ≤ a. Denote α = α_{1} + α_{2}. The following will be proved in Section 8.

Lemma 3.1. Let γ = 2^{−c} diag(β_{1}, β_{2}2^{r}) ∈ A_{r}(−c) with r ≥ 0 and c > 0.

(1) When r ≥ 3, B(T, γ) is independent of r or β_{2}, and is equal to 2^{−3}B(T, β_{1}2^{−c}), where

B(T, β2^{−c}) = \{ \begin{matrix}0 & if c > a + 3, \\ ψ(−\frac{αβ}{8} ) & if c = a + 3, \\ ψ(−\frac{αβ}{4} ) + ψ(−\frac{α_{1}β}{4} ) + ψ(−\frac{α_{2}β}{4} ) & if c = a + 2, \\ −1 & if c = a + 1, \\ 3 & if c ≤ a. \end{matrix}

(2) When r = 2,

B(T, γ) = 2^{−3}B(T, β_{1}2^{−c})\{ \begin{matrix}−1 & if c = a + 3, \\ 1 & if c ≠ a + 3. \end{matrix}

(3) When r = 1, one has

B(T, γ) = 2^{−3}\{ \begin{matrix}0 & if c > a + 3, \\ \frac{1}{2}ψ(−\frac{αβ_{1}}{8} ) (ψ(−\frac{α_{1}β_{2}}{4} ) + ψ(−\frac{α_{2}β−_{2}}{4} )) & if c = a + 3, \\ (−1)^{\frac{α_{1}−α_{2}}{2}} & if c = a + 2, \\ −1 & if c = a + 1, \\ 3 & if c ≤ a. \end{matrix}

(4) When r = 0, one has

B(T, γ) = \frac{1}{8}\{ \begin{matrix}0 & if c > a + 2, \\ ψ(−\frac{αβ_{1}}{4} ) − ψ(−\frac{α_{1}β_{1}}{4} ) − ψ(−\frac{α_{2}β_{1}}{4} ) & if c = a + 2, β_{2} ≡ β_{1} mod (4), \\ ψ(−\frac{(α_{1}−α_{2})β_{1}}{4} ) & if c = a + 2, β_{2} ≡ −β_{1} mod (4), \\ −1 & if c = a + 1, \\ 3 & if c ≤ a. \end{matrix}

In particular, when c ≤ a + 1, one has always

B(T, γ) = 2^{−3}f_{1}(a, c + 1)f_{2}(a, c + 1).

Theorem 3.2. For T = 2^{a} diag(α_{1}, α_{2}) with a ≥ 0, one has

W (X, T, L) = 1 + ∑_{0<c≤a+2} S_{<0}(−c, T, L)X^{2c} (3.1)

+ ∑_{0<c≤a+3} ∑_{0≤r<c} S_{r}(−c, T, L)X^{2c−r} + ∑_{0<c≤a+3} S_{≥c}(−c, T, L)X^{c}.

Here S_{<0}, S_{r}, and S_{≥c} are defined as follows. First,

(3.2) S_{<0}(−c, T, L) = \{ \begin{matrix}2^{3c−5+g_{1}(c)}(f_{1}(a, c)f_{2}(a, c) + g_{2}(c)) & if c ≤ a + 2, \\ 0 & if c > a + 2, \end{matrix}

Secondly,

(3.3)  S_{r}(−c, T, L) = ∑_{γ∈A_{r}(−c)} C(γ)B(T, γ)A(γ, L)

with A(γ, L), B(T, γ), and C(γ) given by Lemmas 1.2, 1.4, and 3.1. In particular, when 0 < c < a + 2, (3.4)

S_{r} = 2^{3c−6−r}f_{1}(a, c+1)f_{2}(a, c+1)Ã(r−c, L)\{ \begin{matrix}A(2^{−c}, L) + A(3 · 2^{−c}, L) & if r = 1, \\ A(2^{−c}, L) + A(−2^{−c}, L) & if r = 0, 2, \\ \frac{1}{2} Ã(−c, L) & if r ≥ 3. \end{matrix}

Finally,

(3.5)  S_{≥c}(−c, T, L) = 2^{2c−4} ∑_{β∈(ℤ/8)^{∗}} B(T, β2^{−c})A(β2^{−c}, L)

for c ≥ 3, and

(3.6) S_{≥c}(−c, T, L) = 2^{3c−7} ∑_{β∈(ℤ/8)^{∗}} B(T, β2^{−c})A(β2^{−c}, L)+ ∑_{c≤r<3} S_{r}(−c, T, L)

for 0 < c < 3.

The proof is the same as that of Theorem 4.4 below, and is left to the reader.

12 TONGHAI YANG

4. The Main formula III.

In this section, we consider the case T = 2^{a} diag(α_{1}, α_{2}2^{t}) = diag(α_{1}2^{a_{1}}, α_{2}2^{a_{2}}) with a = a_{1} ≥ 0 and t = a_{2} − a_{1} > 0. Corollaries 8.5 and 8.6 give

Lemma 4.1. When a_{2} > a_{1}, one has

B(T, γ) = B(γ, T) = B_{1}(γ, T) + B_{2}(γ, T) + B_{3}(γ, T),

where

B_{1}(γ, T) = \frac{1}{8} char(ℤ_{2})(2^{a_{1}+3−c})I_{1}(−2^{a_{2}+1−c}α_{2}β_{1})I_{1}(2^{a_{1}+3−c+r})
· ψ (−2^{a_{1}−c}α_{1}β_{1} − 2^{a_{2}−c+r}α_{2}β_{2})) ,
B_{2}(γ, T) = \frac{1}{8} char(ℤ_{2}^{2})(2^{a_{2}+3−c}, 2^{a_{1}+3−c+r})I_{1}(−2^{a_{1}+1−c}α_{1}β_{1})I_{1}(−2^{a_{2}+3−c+r})
· ψ(−2^{a_{2}−c}α_{2}β_{1} − 2^{a_{1}−c+r}α_{1}β_{2})

and

B_{3}(γ, T) = \frac{1}{8} char(ℤ_{2}^{2})(2^{a_{2}+3−c}, 2^{a_{1}+3−c+r})I_{1} (−2^{a_{1}+1−c}α_{1}β_{1} − 2^{a_{2}+1−c}α_{2}β_{1})
· ψ (−2^{a_{2}−c}α_{1}α_{2}αβ_{1} − 2^{a_{1}−c+r}α_{1}β_{2} − 2^{a_{2}−c+r}α_{2}β_{2}) .

Here α = α_{1} + 2^{t}α_{2}. In particular, B(T, γ) = 0 unless c ≤ a + t + 3 = a_{2} + 3. When 0 < c ≤ a + 1, one has

B(T, γ) = 2^{−3}f_{1}(a, c + 1)f_{2}(a, c + 1).

A couple of special cases are worth recording.

Corollary 4.2. Assume that t = a_{2} − a_{1} ≥ 3.

(1) When r = 0, then

B(T, γ) = \frac{1}{8} char(ℤ_{2})(2^{a_{1}+2−c})ψ(−2^{a_{1}−c}α_{1}β_{1})

+ \frac{1}{4} char(ℤ_{2})(2^{a_{1}+3−c})I_{1}(−2^{a_{1}+1−c}α_{1}β_{1})ψ(−2^{a_{2}−c}α_{2}β_{1} − 2^{a_{1}−c}α_{1}β_{2})

(2) When r ≥ 1, one has

B(T, γ) = \frac{1}{8} char(ℤ_{2})(2^{a_{1}+3−c})ψ(−2^{a_{1}−c}α_{1}β_{1})

+ \frac{1}{4} char(ℤ_{2}^{2})(2^{a_{2}+3−c}, 2^{a_{1}+3−c+r})I_{1}(−2^{a_{1}+1−c}α_{1}β_{1})

· ψ(−2^{a_{2}−c}α_{2}β_{1} − 2^{a_{1}−c+r}α_{1}β_{2})

Corollary 4.3. Assume t = a_{2} − a_{1} > 0 and r ≥ t + 3. Then B(T, γ) = \frac{1}{8}B(T, β_{1}2^{−c}) is independent of r or β_{2}, and

B(T, β2^{−c}) = char(ℤ_{2})(2^{a_{1}+3−c})I_{1}(−2^{a_{2}+1−c}α_{2}β)ψ(−2^{a_{1}−c}α_{1}β)
+ char(ℤ_{2})(2^{a_{2}+3−c})I_{1}(−2^{a_{1}+1−c}α_{1}β)ψ(−2^{a_{2}−c}α_{2}β)

+ char(ℤ_{2})(2^{a_{2}+3−c})I_{1}(−2^{a_{1}+1−c}αβ)ψ(−2^{a_{2}−c}α_{1}α_{2}αβ).

Here α = α_{1} + 2^{t}α_{2}.

Theorem 4.4. Let the notation be as above. Then

W (X, T, L) = 1 + ∑_{0<c≤a+2} S_{<0}(−c, T, L)X^{2c} (4.1)

+ ∑_{0<c≤a+t+3} ∑_{0≤r<c} S_{r}(−c, T, L)X^{2c−r} + ∑_{0<c≤a+t+3} S_{≥c}(−c, T, L)X^{c}.

Here S_{<0}, S_{r}, and S_{≥c} are given as follows. First,

(4.2)  S_{<0}(−c, T, L) = 2^{3c−5+g_{1}(c)}(f_{1}(a, c) + g_{2}(c)f_{2}(a, c)),

Secondly,

(4.3)  S_{r}(−c, T, L) = ∑_{γ∈A_{r}(−c)} C(γ)B(T, γ)A(γ, L)

with A(γ, L), B(T, γ), and C(γ) given in Lemmas 1.2, 1.4, and 4.1. In particular, when 0 < c ≤ a + 1, one has

(4.4) S_{r}(−c, T, L) = 2^{3c−r−6}f_{1}(a, c + 1)f_{2}(a, c + 1)Ã(r − c, L)

· \{ \begin{matrix}(A(2^{−c}, L) + A(−3 · 2^{−c}, L)) & if r = 1, \\ (A(2^{−c}, L) + A(−2^{−c}, L)) & if r = 0, 2, \\ \frac{1}{2} Ã(−c, L) & if r ≥ 3. \end{matrix}

Finally,

(4.5)  S_{≥c}(−c, T, L) = 2^{2c−4} ∑_{β∈(ℤ/8)^{∗}} B(T, β2^{−c})A(β2^{−c}, L)

for c ≥ t + 3, and

(4.6) S_{≥c}(−c, T, L) = ∑_{c≤r<t+3} S_{r}(−c) + 2^{3c−t−7} ∑_{β∈(ℤ/8)^{∗}} B(T, β2^{−c})A(β2^{−c}, L)

for 0 < c < t + 3. Here B(T, β2^{−c}) is given by Corollary 4.3.

Proof. (sketch) When r < 0, one has by Lemma 2.1 (switching roles of a and −c, and so on)

(4.7)  B(T, γ) = B(γ, T) = \{ \begin{matrix}\frac{3}{8}f_{2}(a, c) & if r = −2, \\ \frac{1}{8}f_{1}(a, c) & if r = −1. \end{matrix}

In particular, S_{<0}(−c, T, L) = 0 unless c ≤ a + 2. By Lemma 4.1, B(T, γ) = 0 and thus S_{r}(−c, T, L) = 0 unless c ≤ a + t + 3. By Proposition 1.1, this proves (4.1) with formulas (4.2) − (4.4) to be verified. (4.3) is a general formula given in (1.14). Combining (4.3), (4.7) with Lemmas 1.2 and 1.4, one proves (4.2). When 0 < c ≤ a + 1, one has by Lemmas 4.1, 1.4, and 1.3

S_{r}(−c, T, L) = 2^{−3}f_{1}(a, c + 1)f_{2}(a, c + 1) ∑_{γ∈A_{r}(−c)} C(γ)A(γ, L)

= 2^{3c−r−6}f_{1}(a, c + 1)f_{2}(a, c + 1)

· \{ \begin{matrix}2 ∑_{γ∈A_{0}^{−}(−c)} A(γ, L) + ∑_{γ∈A_{0}^{+}(−c)} A(γ, L) & if r = 0, \\ ∑_{γ∈A_{r}(−c)} A(γ, L) & if 1 ≤ r ≤ 2, \\ \frac{1}{2} ∑_{γ∈A_{r}(−c)} A(γ, L) & if r ≥ 3. \end{matrix}

= 2^{3c−r−6}f_{1}(a, c + 1)f_{2}(a, c + 1)Ã(r − c, L)

· \{ \begin{matrix}(A(2^{−c}, L) + A(−3 · 2^{−c}, L)) & if r = 1, \\ (A(2^{−c}, L) + A(−2^{−c}, L)) & if r = 0, 2, \\ \frac{1}{2} Ã(−c, L) & if r ≥ 3. \end{matrix}

This proves (4.4). Finally, when r ≥ t + 3, B(T, γ) = 2^{−3}B(T, β_{1}2^{−c}) and A(γ, L) = A(β_{1}2^{−c}, L) are independent of r or β_{2} by Corollary 4.3 and Lemma 1.2. So

S_{r}(−c, T, L) = 2^{3c−r−4}2^{−3} ∑_{γ∈A_{r}(−c)} B(T, β_{1}2^{−c})A(β_{1}2^{−c}, L)

= 2^{3c−r−5} ∑_{β∈(ℤ/8)^{∗}} B(T, β2^{−c})A(β2^{−c}, L).

So for any r_{0} ≥ t + 3, one has

S_{≥r_{0}}(−c, T, L) = 2^{3c−r_{0}−4} ∑_{β∈(ℤ/8)^{∗}} B(T, β2^{−c})A(β2^{−c}, L).

When c ≥ t + 3, taking r_{0} = c, ones gets (4.5). When c < t + 3, one obtains (4.6) by taking r_{0} = t + 3.

The main formulas in Section 2-4 are programmed by Song Guo using MAGMA, and can be downloaded from my web page http://www.math.wisc.edu/∼thyang/programs. One can get result with it for any given quadratic lattice of the form (1.15) and matrices of form (1.16). The next section is mainly a product of this program.

5. Examples.

Before going to the examples, we recall a proposition of Kudla. It tells when W (1, T, L) = 0, which is a good first test of whether a given formula for W (X, T, L) is wrong.

Let (V, q) be a quadratic (non-degenerate) space over a local field F of dimension n + 1. Let S be an associated symmetric matrix in the following sense that with respect to some isomorphism V ≅ F^{n+1}, one has

(5.1)  q(x) = ^{t}xSx.

Let T be a n × n symmetric matrix over F. Its associated quadratic space is V_{T} = F^{n} with quadratic form

q_{T} (x) = ^{t}xT x.

We say V represent T if there is x = ^{t}(x_{1}, ⋯, x_{n}) ∈ V^{n}, such that

(5.2)  \frac{1}{2}((x_{i}, x_{j})) = T

where (x, y) = q(x + y) − q(x) − q(y). We say that a lattice L represents T if x can be chosen in L^{n}.

It is known that W (1, T, L) = 0 if and only if L does not represent T. The following lemma is proved in [Ku, Proposition 1.3].

Lemma 5.1. Let the notation be as above, and let ϵ_{F} (V) = ϵ_{F} (S) be the Hasse invariant of V. Then V represents T if and only if

(5.3)  ϵ_{F} (V) = ϵ_{F} (T)(det T, − det S).

Now coming to our special examples. Let B be a quaternion algebra over ℚ_{2}, and let V be the trace zero elements in B with the reduced norm as the quadratic form. We use the superscript ra and s to stand for ramified and split case respectively when necessary. Notice that

ϵ_{2}(V) = \{ \begin{matrix}−1 & if V = V^{s}, \\ 1 & if V = V^{ra}. \end{matrix}

Set

μ_{2}(T) = −ϵ_{ℚ_{2}}(T)(det T, −1) (5.4)

= \{ \begin{matrix}(−1)^{a+1} & if T ∈ A_{−2}(a), \\ 1 & if T ∈ A_{−1}(a), \\ −(α_{1}, α_{2})(−1, α_{1}α_{2})(2, α_{1}α_{2})^{a}(2, α_{1})^{t} & if T = 2^{a} diag(α_{1}, 2^{t}α_{2}) ∈ A_{t}(a). \end{matrix}

Then Lemma 5.1 asserts that V^{s} represents T if and only if μ_{2}(T) = 1, and that V^{ra} represents T if and only if μ_{2}(T) = −1. In particular, for any lattice L of V^{s}, W (1, T, L) = 0 unless μ_{2}(T) = 1.

When B = M_{2}(ℚ_{2}), we take the following lattices of V for i ≥ 0:

(5.5)  L^{i} = \{X = ( \begin{matrix}a & b \\ 2^{i}c & −a \end{matrix} ) ∈ M_{2}(ℤ_{2}) ∩ V : a, b, c ∈ ℤ_{2}\},

with quadratic form

(5.6)  Q(X) = det X = −a^{2} − 2^{i}bc.

When B is ramified, i.e., a division algebra, let O_{B} be the maximal order of B and L^{ra} = V ∩ O_{B}. Then L^{ra} ≅ ℤ_{2}^{3} with quadratic form

(5.7)  Q(x) = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}.

The following three theorems are obtained by Song Guo’s computer program and tedious calculations.

Theorem 5.2. Let T = 2^{a} ( \begin{matrix}b & 1 \\ 1 & b \end{matrix} ) with a ≥ −1 and b = 0 or 2. Set

δ_{i} = \{ \begin{matrix}1 & if i = 0, [\frac{a}{2}] + 1, \\ −1 & if 0 < i < [\frac{a}{2}] + 1. \end{matrix}

(1) When a is even, then

\frac{W (X, T, L^{0})}{1 − 2^{−2}X^{2}} = ∑_{0≤i≤ \frac{a}{2}} 2^{i}(X^{2i} + μ_{2}(T)X^{2a+2−2i})

and

W (X, T, L^{ra}) = ∑_{0≤i≤ \frac{a}{2} +1} δ_{i}2^{i}(X^{2i} − μ_{2}(T)X^{2a+4−2i}).

(2) When a is odd,

\frac{W (X, T, L^{0})}{1 − 2^{−2}X^{2}} = ∑_{0≤i≤ \frac{a−1}{2}} 2^{i}(X^{2i} + μ_{2}(T)X^{2a+2−2i}) + 2^{\frac{a+1}{2}} X^{a+1},

and

W (X, T, L^{ra}) = 1 − μ_{2}(T)X^{2a+4} − ∑_{1≤i≤ \frac{a+1}{2}} 2^{i}(X^{2i} − μ_{2}(T)X^{2a+4−2i})

Theorem 5.3. Let T = 2^{a} diag(α_{1}, α_{2}) ∈ A_{0}(a) with α_{i} ∈ ℤ_{2}^{∗} and a ≥ 0. Then

\frac{W (X, T, L^{0})}{1 − 2^{−2}X^{2}} = ∑_{0≤i≤ \frac{a}{2}} 2^{i}(X^{2i} + μ_{2}(T)X^{2a+2−2i}),

and

W (X, T, L^{ra}) = ∑_{0≤i≤ \frac{a}{2} +1} δ_{i}2^{i}(X^{2i} − μ_{2}(T)X^{2a+4−2i}).

Here δ_{i} is as in Theorem 5.2.

Theorem 5.4. Let T = 2^{a} diag(α_{1}, 2^{t}α_{2}) ∈ A_{t}(a) with t ≥ 1 and a ≥ 0. Set v_{0} = (2, α_{1})_{2}.

(1) When a is odd or α_{1} ≡ 1 mod 4, one has Then

\frac{W (X, T, L^{0})}{1 − 2^{−2}X^{2}} = ∑_{0≤i≤ \frac{a}{2}} 2^{i}(X^{2i} + μ_{2}(T)X^{2a+t+2−2i}),

and

W (X, T, L^{ra}) = ∑_{0≤i≤ \frac{a}{2} +1} δ_{i}2^{i}(X^{2i} − μ_{2}(T)X^{2a+t+4−2i}).

(2) When a is even and α_{1} ≡ −1 mod 4, one has Then

\frac{W (X, T, L^{0})}{1 − 2^{−2}X^{2}} = ∑_{0≤i≤ \frac{a}{2}} 2^{i}(X^{2i} + μ_{2}(T)X^{2a+t+2−2i}) + 2^{\frac{a}{2} +1} ∑_{0≤k≤t−2} v_{0}^{k}X^{a+2+k},

and

W (X, T, L^{ra}) = 1 − μ_{2}(T)X^{2a+t+4} − ∑_{1≤i≤ \frac{a}{2} +1} 2^{i}(X^{2i} − μ_{2}(T)X^{2a+t+4−2i})
− v_{0}2^{\frac{a}{2} +2}(X^{a+3} − μ_{2}(T)X^{a+1+t}).

These formulae are almost the same. It can be unified using the Gross-Keating invariants of a quadratic lattice ([GK, Section 4]). In Appendix B, which is essentially due to Kudla, we recall the definition of the Gross-Keating invariants of a quadratic lattice and study the relation between the usual Jordan form and the Gross-Keating invariants when the rank is less than or equal to 3.

Theorem 5.5. Let the Gross-Keating invariants of diag(1, T) be (0, a_{2}, a_{3}), and let μ_{2}(T) be as given by (5.4). Set

δ_{i} = δ_{i}(T) = \{ \begin{matrix}1 & if i = 0, [\frac{a_{2}}{2} ] + 1, \\ −1 & if 0 < i < [\frac{a_{2}}{2} ] + 1. \end{matrix}

(1) When a_{2} is odd, one has

\frac{W (X, T, L^{0})}{1 − 2^{−2}X^{2}} = ∑_{0≤i≤ \frac{a_{2}−1}{2}} 2^{i}(X^{2i} + μ_{2}(T)X^{a_{2}+a_{3}−2i}),

and

W (X, T, L^{ra}) = ∑_{0≤i≤ \frac{a_{2}+1}{2}} δ_{i}2^{i}(X^{2i} − μ_{2}(T)X^{a_{2}+a_{3}−2i}),

(2) When a_{2} is even, one has

\frac{W (X, T, L^{0})}{1 − 2^{−2}X^{2}} = ∑_{0≤i≤ \frac{a_{2}}{2} −1} 2^{i}(X^{2i} + μ_{2}(T)X^{a_{2}+a_{3}−2i}) + 2^{\frac{a_{2}}{2}} ∑_{0≤k≤a_{3}−a_{2}}(ϵX)^{a_{2}+k},

and

W (X, T, L^{ra}) = ∑_{0≤i≤ \frac{a_{2}}{2}} δ_{i}2^{i}(X^{2i} − μ_{2}(T)X^{a_{2}+a_{3}+2−2i})

− ϵ2^{\frac{a_{2}}{2} +1}(X^{a_{2}+1} − μ_{2}(T)X^{a_{3}+1}).

Here ϵ = ϵ(T) is the Gross-Keating’s ϵ-constant given in [GK, page 236], see also Appendix B.

Remark 5.6. When T = 2^{a} diag(α_{1}, 2^{t}α_{2}) ∈ A_{t}(a) with t ≥ 2 and a even, ϵ(T) = (2, α_{1}) is the sign v_{0} in Theorem 5.4. In other cases, ϵ(T) does not make any difference in the formula.

Proof of Theorem 5.5 (sketch). (1) When T = 2^{a} ( \begin{matrix}b & 1 \\ 1 & b \end{matrix} ) ∈ A_{<0}(a) with

a ≥ −1 and b = 0, 2, the Gross-Keating invariants of diag(1, T) is (0, a+1, a+1) by Proposition B.4. When a is even a_{2} = a+1 is odd, Theorem 5.5(1) matches up perfectly with Theorem 5.2(1). When a is odd, a_{2} is even, and (ϵ_{v}X)^{a_{2}} = X^{a_{2}}. In this case, μ_{2}(T) = 1 and a_{2} = a_{3}, so ϵ2^{a_{2}/2+1}(X^{a_{2}} −μ_{2}(T)X^{a_{3}}) = 0, Theorem 5.5(2) matches up with Theorem 5.2(2) perfectly too.

(2) When T = 2^{a} diag(α_{1}, α_{2}) ∈ A_{0}(a), the Gross-Keating invariants of diag(1, T) is, by Proposition B.5,

(0, a_{2}, a_{3}) = \{ \begin{matrix}(0, a, a + 2) & if a is odd, \\ (0, a + 1, a + 1) & if a is even. \end{matrix}

So a_{2} is always odd, and Theorem 5.5(1) matches up perfectly with Theorem 5.3.

(3) When T = 2^{a} diag(α_{1}, 2α_{2}) ∈ A_{1}(a), the Gross-Keating invariants of diag(1, T) is, by Proposition B.5,

(0, a_{2}, a_{3}) = \{ \begin{matrix}(0, a, a + 3) & if a is odd, \\ (0, a + 1, a + 2) & if a is even. \end{matrix}

So a_{2} is always odd, and Theorem 5.5(1) matches up perfectly Theorem 5.4(1) perfectly. When a is even and α_{1} ≡ −1 mod 4, we now check that Theorem 5.5(1) matches with Theorem 5.4(2). When L = L^{0}, it is clear since t − 2 = −1 < 0. When L = L^{ra},

v_{0}μ_{2} = −(2, α_{1})(α_{1}, α_{2})(−1, α_{1}α_{2})(2, α_{2}) = −(−1, α_{1})(−α_{1}, α_{2}) = 1

since α_{1} ≡ −1 mod 4. So Theorem 5.4(2) gives

W (X, T, L^{ra}) = 1 − μ_{2}(T)X^{a_{2}+a_{3}+2} − ∑_{1≤i≤ \frac{a}{2}} 2^{i}(X^{2i} − μ_{2}(T)X^{a_{2}+a_{3}+2−2i})

− 2^{\frac{a}{2} +1}(X^{a+2} − μ_{2}(T)X^{a+3}) + 2^{\frac{a}{2} +2}(μ_{2}(T)X^{a+3} + X^{a+2})

= ∑_{0≤i≤ \frac{a_{2}+1}{2}} δ_{i}2^{i}(X^{2i} − μ_{2}(T)X^{a_{2}+a_{3}+2−2i}).

This matches up perfectly with Theorem 5.5(1).

(4) Finally, when T = 2^{a} diag(α_{1}, 2^{t}α_{2}) ∈ A_{t}(a) with t ≥ 2, the Gross-Keating invariants of diag(1, T) is, by Proposition B.5,

(0, a_{2}, a_{3}) = \{ \begin{matrix}(0, a, a + t + 2) & if a is odd, \\ (0, a + 1, a + t + 1) & if a is even and α_{1} ≡ 1 mod 4, \\ (0, a + 2, a + t) & if a is even and α_{1} ≡ 1 mod 4. \end{matrix}

Theorem 5.5 again matches up with Theorem 5.4 perfectly if we can verify v_{0} = ϵ, which is Remark 5.6. Indeed, assume that a_{2} is even, i.e., a is even and α_{1} ≡ −1 mod 4. Let e = {e_{1}, e_{2}, e_{3}} be a basis of L = ℤ_{2}^{3} with associated matrix diag(1, T), and let

(ψ_{1}, ψ_{2}, ψ_{3}) = (e_{1}, e_{2}, e_{3}) ( \begin{matrix}1 & 2^{\frac{β_{2}−β_{1}}{2}} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) .

Then ψ is an optimal basis of L by the proof of Lemma B.8. The subspace generated by ψ_{1} and ψ_{2}, is the same as that generated by e_{1} and e_{2}, and on which the quadratic form is Q(x_{1}e_{1} + x_{2}e_{2}) = x_{1}^{2} + 2^{a}ϵ_{1}x_{2}^{2}. This quadratic form is isotropic (i.e., ϵ = 1 if and only if (2, −ϵ_{1}) = (2, ϵ_{1}) = 1, i.e., v_{0} = 1. So ϵ = v_{0}.

It is interesting to compare Theorem 5.5 with [Ya, Propositions 8.6 and 8.7]. For a prime number p, odd or even, let μ_{p}(T) = ±1 according to whether

V_{p}^{s} = \{X = ( \begin{matrix}a & b \\ c & −a \end{matrix} ) ∈ M_{2}(ℚ_{p}) with quadratic form det represents T or not.

For p = 2, it is just our μ_{2}(T), and when p ≠ 2, it is the μ_{p}(T) in [Ku, Lemma 8.3] and

μ_{p}(T) = \{ \begin{matrix}v_{0}^{b} & if a is even, \\ v_{1} & if a is odd, \end{matrix}

in terms of the notation in [Ya, Propositions 8.6 and 8.7]. One sees that [Ya, Propositions 8.6 and 8.7] hold for p = 2 too if we replace a and b by a_{2} and a_{3}. We record the unified formula in a slightly different form as the the following theorem

Theorem 5.7. Assume that the Gross-Keating invariants of T is (0, a_{2}, a_{3}), and let μ_{p}(T) be as above. X = p^{−s} and

δ_{p,i} = δ_{p,i}(T) = \{ \begin{matrix}1 & if i = 0, [\frac{a_{2}}{2} ] + 1, \\ 1 − p & if 1 ≤ i ≤ [\frac{a_{2}}{2} ]. \end{matrix}

(1) When a_{2} is odd, one has

\frac{W (X, T, L^{0})}{1 − p^{−2}X^{2}} = ∑_{0≤i≤ \frac{a_{2}−1}{2}} p^{i}(X^{2i} + μ_{p}(T)X^{a_{2}+a_{3}−2i}),

and

W (X, T, L^{ra}) = ∑_{0≤i≤ \frac{a_{2}+1}{2}} δ_{p,i}p^{i}(X^{2i} − μ_{p}(T)X^{a_{2}+a_{3}−2i}),

(2) When a_{2} is even, one has

\frac{W (X, T, L^{0})}{1 − p^{−2}X^{2}} = ∑_{0≤i≤ \frac{a_{2}}{2} −1} p^{i}(X^{2i} + μ_{p}(T)X^{a_{2}+a_{3}−2i}) + p^{\frac{a_{2}}{2}} ∑_{0≤k≤a_{3}−a_{2}}(ϵX)^{a_{2}+k},

and

W (X, T, L^{ra}) = ∑_{0≤i≤ \frac{a_{2}}{2}} δ_{p,i}p^{i}(X^{2i} − μ_{p}(T)X^{a_{2}+a_{3}+2−2i})

− ϵp^{\frac{a_{2}}{2} +1}(X^{a_{2}+1} − μ_{p}(T)X^{a_{3}+1}).

Here ϵ = ϵ(T) is the Gross-Keating’s ϵ-constant given in [GK, page 236], see also Appendix B.

Because of this theorem, [Ya, Propositions 8.9] and [Ya, Theorem 8.10] are all true for p = 2, when one replaces a, b, and v_{0} by a_{2}, a_{3}, and ϵ respectively.

Corollary 5.8. Assume μ_{p}(T) = −1. Then W (1, T, L^{0}) = 0 and W (1, T, L^{1}) = 2(p + 1). Moreover,

\frac{W′(1, T, L^{0})}{1 − p^{−2}} = \{ \begin{matrix}− ∑_{0≤i≤ \frac{a_{2}−1}{2}} p^{i}(a_{2} + a_{3} − 4i) & if a_{2} is odd, \\ − ∑_{0≤i≤ \frac{a_{2}}{2} −1} p^{i}(a_{2} + a_{3} − 4i) − (a_{3} − a_{2} − 1)p^{\frac{a_{2}}{2}} & if a_{2} is even. \end{matrix}

In particular, when diag(1, T) is anisotropic,

W′(1, T, L^{0}) = −(1 − p^{−2})α_{p}( diag(1, T))

where α_{p}( diag(1, T)) is the local intersection number defined by Gross and Keating ([GK, Proposition 5.4]).

Corollary 5.9. Assume μ_{p}(T) = 1.

(1) One has W (1, T, L^{ra}) = 0, and

\frac{W (1, T, L^{0})}{1 − p^{−2}} = \{ \begin{matrix}2 ∑_{0≤i≤ \frac{a_{2}−1}{2}} p^{i} & if a_{2} is odd, \\ 2 ∑_{0≤i≤ \frac{a_{2}}{2} −1} p^{i} + p^{\frac{a_{2}}{2}} (1 + \frac{1+ϵ}{2}(a_{3} − a_{2})) & if a_{2} is even. \end{matrix}

(2) When a_{2} is odd, one has

W′(1, T, L^{ra}) = ∑_{0≤i≤ \frac{a_{2}}{2} +1} δ_{p,i}p^{i}(4i − a_{2} − a_{3} − 2)

= 4 ∑_{1≤i≤ \frac{a_{2}+1}{2}} p^{i} − (p + 1)(a_{2} + a_{3} + 2).

(3) When a_{2} is even, one has

W′(1, T, L^{ra}) = ∑_{0≤i≤ \frac{a_{2}}{2}} δ_{p,i}p^{i}(4i − a_{2} − a_{3} − 2) − ϵp^{\frac{a_{2}}{2} +1}(a_{2} − a_{3})

= 4 ∑_{1≤i≤ \frac{a_{2}}{2} +1} p^{i} + p^{\frac{a_{2}}{2} +1}(2 + (1 + ϵ)(a_{3} − a_{2})) − (p + 1)(a_{2} + a_{3} + 2).

(4) One has

W′(1, T, L^{ra}) = −(1 + p)(a_{2} + a_{3} + 2) + \frac{2p^{3}}{p^{2} − 1}W (1, T, L^{0}).

Encouraged by Theorem 5.7, we tried the case L = L^{1} and obtained the following theorem. It seems very interesting and possible to found a clean formula for L = L^{i} for any i ≥ 0.

Proposition 5.10. Let

L^{1} = \{( \begin{matrix}a & b \\ pc & −a \end{matrix} ) : a, b, c ∈ ℤ_{p}\}

with the determinant as the quadratic form. Let T be a p-integral 2×2 symmetric matrix over ℚ_{p} with the Gross-Keating invariants (0, a_{2}, a_{3}). Let ϵ and μ_{p}(T) be defined as above.

(1) When a_{2} is odd, one has

W (X, T, L^{1}) = 1 − μ_{p}(T)X^{a_{2}+a_{3}+2} + p^{\frac{a_{2}−1}{2}} ((p − 2)X^{a_{2}+1} + μ_{p}(T)pX^{a_{3}+1})

+ (p − 1) ∑_{1≤i≤ \frac{a_{2}−1}{2}} p^{i−1} ((p + 2)X^{2i} + μ_{p}(T)pX^{a_{2}+a_{3}+2−2i}) .

(2) When a_{2} is even, one has

W (X, T, L^{1}) = 1 − μ_{p}(T)X^{a_{2}+a_{3}+2} + p^{\frac{a_{2}}{2}} ∑_{a_{2}≤i≤a_{3}} C_{k}(ϵX)^{k+1}

+ (p − 1) ∑_{1≤i≤ \frac{a_{2}}{2}} p^{i−1} ((p + 2)X^{2i} + μ_{p}(T)pX^{a_{2}+a_{3}+2−2i})

with

C_{k} = \{ \begin{matrix}p & if k = a_{2}, \\ 2(p − 1) & if a_{2} < k < a_{3}, \\ p − 2 & if k = a_{3}. \end{matrix}

Proof. (sketch) When p ≠ 2, this follows from [Ya, Corollary 8.4] and [Ya, Proposition 8.7]. When p = 2, this follows from the Main formulae in Sections 2-4 plus tedious calculations, and can be verified via Song Guo’s computer program.

Theorem 5.7. Let L = L^{1} be given by (5.5), and let ϵ and μ_{p}(T) as in Theorem 5.5. Let (0, a_{2}, a_{3}) be the Gross-Keating invariants of diag(1, T).

(1) When a_{2} is odd, one has

W (X, T, L^{1}) = 1−μ_{2}(T)X^{a_{2}+a_{3}+2}+2μ_{2}(T)X^{a_{2}+a_{3}}+2 ∑_{1≤i≤ \frac{a_{2}−1}{2}} 2^{i}(X^{2i}+μ_{2}(T)X^{a_{2}+a_{3}−2i}).

(2) When a_{2} is even, one has

W (X, T, L^{1}) = 1 − μ_{2}(T)X^{a_{2}+a_{3}+2} + 2^{\frac{a_{2}}{2} +1} ∑_{a_{2}<k≤a_{3}}(ϵX)^{k}

+ ∑_{1≤i≤ \frac{a_{2}}{2}} 2^{i}(2X^{2i} + μ_{2}(T)X^{a_{2}+a_{3}+2−2i}).

LOCAL DENSITIES 23

6. Basic Integrals.

First we recall from [Ya, Lemmas 4.3 and 4.4] the following basic formulas for t = α2^{a} with α ∈ ℤ_{2}^{∗} and a ∈ ℤ

(6.1)  I_{1}(t) = ∫_{ℤ_{2}} ψ(\frac{1}{2}tx^{2})dx = \{ \begin{matrix}1 & if a > 0, \\ 0 & if a = 0, \\ 2^{\frac{a}{2}} ψ(\frac{α}{8})(2, α)^{a} & if a < 0, \end{matrix}

Alternatively,

(6.2)  I_{1}(α2^{a}) = \{ \begin{matrix}1 & if a > 0, \\ 0 & if a = 0, \\ \frac{a}{2}ψ(\frac{α}{8}) & if a < 0 is even, \\ \frac{a−1}{2}(1 + ψ(\frac{α}{4})) & if a < 0 is odd. \end{matrix}

(6.3)  I_{2}(t) = ∫_{ℤ_{2}^{2}} ψ(txy)dxdy = 2^{min(0,a)},

and

(6.4)  I_{3}(t) = ∫_{ℤ_{2}^{2}} ψ(t(x^{2} + xy + y^{2}))dxdy = (−2)^{min(0,a)}

Here (a, b) is the Hilbert symbol over ℚ_{2}.

The following two lemmas will be needed later. We leave the proof of Lemma 6.1 to the reader.

Lemma 6.1. For t ∈ ℚ_{2} and β ∈ ℚ_{2}, let

(6.5)  I_{1}(t, β) = ∫_{ℤ_{2}} ψ(t(x^{2} + βx))dx.

Then one has

I_{1}(t, β) = \{ \begin{matrix}ψ(−\frac{1}{4}tβ^{2})I_{1}(2t) & if β ∈ 2ℤ_{2}, \\ char(ℤ_{2})(2t) & if β ∈ ℤ_{2}^{∗}, \\ char(ℤ_{2})(βt) & if β ∉ ℤ_{2}. \end{matrix}

Lemma 6.2. For t ∈ ℚ_{2} and a, b ∈ \frac{1}{2}ℤ_{2}, let

(6.6)  I_{3}(t; a, b) = ∫_{a+ℤ_{2}} ∫_{b+ℤ_{2}} ψ(t(x^{2} + xy + y^{2}))dxdy.

Then

I_{3}(t; a, b) = \{ \begin{matrix}I_{3}(t) & if a, b ∈ ℤ_{2}, \\ char(2ℤ_{2})(t)ψ(\frac{t}{4}) & else. \end{matrix}

Proof. Of course, the formula depends only on a, b ∈ \frac{1}{2}ℤ/ℤ. The case a = b = 0 follows from the definition. When a = \frac{1}{2} and b = 0, one has, by Lemma 6.1

I_{3}(t; \frac{1}{2}, 0) = ∫_{\frac{1}{2} +ℤ_{2}} ψ(tx^{2})I_{1}(t, x)dx

= char(ℤ_{2})(\frac{1}{2}t)ψ(\frac{1}{4}t) ∫_{ℤ_{2}} ψ(t(x^{2} + x))dx

= char(2ℤ_{2})(t)ψ(\frac{1}{4}t).

Finally, one has by Lemma 6.1 again

I_{3}(t; \frac{1}{2}, \frac{1}{2}) = ∫_{\frac{1}{2} +ℤ_{2}} ψ(tx^{2})dx ∫_{\frac{1}{2} ℤ_{2}} ψ(t(xy + y^{2}))dy − I_{3}(t; \frac{1}{2}, 0)

= 2ψ(\frac{1}{4}t) ∫_{ℤ_{2}} ψ(t(x^{2} + x))I(\frac{t}{4}, 2x)dx − I_{3}(t; \frac{1}{2}, 0)

= 2 char(2ℤ_{2})(t)ψ(\frac{1}{4}t) − char(2ℤ_{2})(t)ψ(\frac{1}{4}t)

= char(2ℤ_{2})(t)ψ(\frac{1}{4}t).

7. The integral A(γ, L).

In this section, we will compute the integral

(7.1)  A(γ, L) = ∫_{L^{n}} ψ( trγq(x))dx,

in a general situation. Here γ ∈ Sym_{n}(ℚ_{2}) and (L, q) is a non-degenerate quadratic lattice over ℤ_{2}, and for x = (x_{1}, ⋯, x_{n}) ∈ L^{n}, q(x) = \frac{1}{2}((x_{i}, x_{j})) ∈ \frac{1}{2} Sym_{n}(ℤ_{2}). We only need the case n = 2 in this paper. We first record some useful easy facts.

Lemma 7.1. (1) A(k.γ, L) = A(γ, L) for any k ∈ GL_{n}(ℤ_{2}). Here k.γ = kγ^{t}k is the action of GL_{n} on Sym_{n}.

(2) A(γ, L_{1} ⊕ L_{2}) = A(γ, L_{1})A(γ, L_{2}).

(3) A( diag(γ_{1}, γ_{2}), L) = A(γ_{1}, L)A(γ_{2}, L).

(4) A(aγ, L) = A(γ, aL) for any a ∈ ℚ_{2}^{∗}. Here aL is the same lattice L with a new quadratic form q_{a}(x) = aq(x).

By this lemma and the classification of quadratic lattices, we need only to consider the following three cases for γ,

γ ∈ ℚ_{2}, γ = 2^{c} ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ), or γ = 2^{c} ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ),

and three cases for (L, q):

L_{1}(t) : L = ℤ_{2}, q(x) = \frac{1}{2}tx^{2},

L_{2}(t) : L = ℤ_{2}^{2}, q(x) = tx_{1}x_{2},

L_{3}(t) : L = ℤ_{2}^{2}, q(x) = t(x_{1}^{2} + x_{1}x_{2} + x_{2}^{2}).

The case γ ∈ ℚ_{2} was dealt with in [Ya, Section 4], we record it here for convenience.

Lemma 7.2. ([Ya, Lemmas 4.3 and 4.4]) For γ = β2^{c} and t = α2^{a} ∈ ℚ_{2}^{∗}, with α, β ∈ ℤ_{2}^{∗} and a, c ∈ ℤ, one has

A(γ, L_{i}(t)) = I_{i}(tγ) = \{ \begin{matrix}I_{1}(tγ) & if i = 1, \\ 2^{min(0,a+c)} & if i = 2, \\ (−2)^{min(0,a+c)} & if i = 3. \end{matrix}

Here I_{i}(t) are defined in Section 6.

Lemma 7.3. Let γ = 2^{c} ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) and t = α2^{a}. Then

A(γ, L_{i}(t)) = 2^{min(0,a+c) dim L_{i}(t)}

for 1 ≤ i ≤ 3.

Lemma 7.4. Let γ = 2^{c} ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ) and t = α2^{a}. Then

A(γ, L_{i}(t)) = (−2)^{min(0,a+c) dim L_{i}(t)}

26 TONGHAI YANG

for 1 ≤ i ≤ 3. Proof. We verify Lemma 7.4 and leave Lemma 7.3 to the reader. Set γ_{2} = 2^{a}. For z = ( \begin{matrix}x \\ y \end{matrix} ) ∈ L^{2}, one has

trγq(z) = γ_{2} ((x, x) + (x, y) + (y, y)))

and thus

(7.2)  A(γ, L) = ∫_{L^{2}} ψ (γ_{2} ((x, x) + (x, y) + (y, y)))) dxdy.

When L = L_{1}(t) = ℤ_{2}, one has (x, y) = txy, and A(γ, L_{1}(t)) = ∫_{ℤ_{2}^{2}} ψ(tγ_{2}(x^{2} + xy + y^{2}))dxdy = I_{3}(tγ_{2}).

When L = L_{2}(t) = ℤ_{2}^{2}, (x, y) = t(x_{1}y_{2} + x_{2}y_{1}) and

A(γ, L_{2}(t)) = ∫_{ℤ_{2}^{4}} ψ(tγ_{2}(2x_{1}x_{2} + x_{1}y_{2} + x_{2}y_{1} + 2y_{1}y_{2})dx_{1}dx_{2}dy_{1}dy_{2}. So

A(γ, L_{2}(t)) = ∫_{ℤ_{2}^{2}} char(ℤ_{2})(tγ_{2}(2x_{2} + y_{2})) char(ℤ_{2})(tγ_{2}(2y_{2} + x_{2}))dx_{2} dy_{2}
= ∫_{ℤ_{2}^{2}} char(ℤ_{2})(tγ_{2}y_{2}) char(ℤ_{2})(tγ_{2}x_{2})dx_{2} dy_{2}
= 2^{2 min(1,a+c)}.

Finally, when L = L_{3}(t), we rewrite (7.2) as

A(γ, L_{3}(t)) = ∫_{ℤ_{2}^{2}} ψ(\frac{3}{4}γ_{2}(y, y))dy ∫_{ℤ_{2}^{2}} ψ(γ_{2}(x + \frac{1}{2}y, x + \frac{1}{2}y)dx.

Substituting x′ = x + \frac{1}{2}y, and breaking the integral into 4 parts, one has

A(γ, L_{3}(t)) = ∑_{a,b∈ \frac{1}{2} ℤ/ℤ} ∫_{(2a+2ℤ_{2})×(2b+2ℤ_{2})} ψ(\frac{3}{4}γ_{2}(y, y))dy ∫_{(a+ℤ_{2})×(b+ℤ_{2})} ψ(γ_{2}(x′, x′))dx′

= \frac{1}{4} ∑_{a,b} I_{3}(6tγ_{2}; a, b)I_{3}(2tγ_{2}, a, b).

Now applying Lemma 6.2, one obtains

A(γ, L_{3}(t)) = \frac{1}{4}I_{3}(2tγ_{2})^{2} + \frac{3}{4} char(2ℤ_{2})(2tγ_{2})

= \{ \begin{matrix}\frac{1}{4} + \frac{3}{4} = 1 & if tγ_{2} ∈ ℤ_{2}, \\ \frac{1}{4}I_{3}(2tγ_{2})^{2} & if tγ_{2} ∉ ℤ_{2} \end{matrix}

= (−2)^{2 min(0,a+c)}.

Proof of Lemma 1.2. The case γ ∈ A_{r}(−c) with r ≥ 0 follows from Lemmas 7.1 and 7.2. We check the case r = −1 and omit the case r = −2 here. Indeed, Lemmas 7.1 and 7.3 imply

A(γ, L) = ∏_{1≤i≤H} A(γ, L_{1}(ϵ_{i}2^{l_{i}}) ∏_{1≤j≤M} A(γ, L_{2}(2^{m_{j}})) ∏_{1≤k≤N} A(γ, L_{3}(2^{n_{k}})
= 2^{∑_{i} min(0,l_{i}−c)}2^{2 ∑_{j} min(0,m_{j}−c)}2^{2 ∑_{k} min(0,n_{k}−c)}
= 2^{g_{1}(c)}.

8. The integral B(T, γ).

In this section, we compute the integral

(8.1)  B(T, γ) = ∫ _{GL_{2}(ℤ_{2})} ψ(− tr(k.T γ))dk = ∫ _{GL_{2}(ℤ_{2})} ψ(−T k.γ)dk.

Here T, γ ∈ Sym_{2}(ℚ_{2}) are symmetric matrices over ℚ_{2} of order 2, and we denote k.T = kT^{t}k for the action of GL_{2}(ℚ_{2}) on Sym_{2}(ℚ_{2}). Here dk is the restriction of the standard Haar measure on M_{2}(ℤ_{2}) to GL_{2}(ℤ_{2}). Clearly B(T, γ) are GL_{2}(ℤ_{2})-invariant in both variables in the sense

(8.2)  B(k.T, γ) = B(T, k.γ) = B(T, γ)

for every k ∈ GL_{2}(ℤ_{2}). It is also clear that

(8.3)  B(T, γ) = B(γ, T)

is symmetric on T and γ.

Let (8.4)

K = K_{1}(2) = \{k = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ GL_{2}(ℤ_{2}) : c ≡ 0 mod 2, a ≡ d ≡ 1 mod 2\}.

Then

(8.5)  GL_{2}(ℤ_{2}) = ∪Kg_{i}, g_{0} = 1, g_{1} = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , g_{2} = ( \begin{matrix}0 & 1 \\ 1 & 1 \end{matrix} ) .

28 TONGHAI YANG

Define

(8.6)  B_{1}(T, γ) = ∫_{K} ψ(−T k.γ)dk,

and

(8.7)  J(t, γ) = ∫_{ℤ_{2}^{∗}} ∫_{ℤ_{2}} ψ(t(x_{1}, x_{2})γ^{t}(x_{1}, x_{2}))dx_{1}dx_{2}
= ∫_{ℤ_{2}^{∗}} ∫_{ℤ_{2}} ψ(t(γ_{1}x_{1}^{2} + 2γ_{2}x_{1}x_{2} + γ_{3}x_{2}^{2}))dx_{1}dx_{2}.

Lemma 8.1. Let the notation be as above. Then

(8.8)  B(T, γ) = B_{1}(T, γ) + B_{1}(T, g_{1}.γ) + B_{1}(T, g_{2}.γ),

Lemma 8.2. Let T = diag(t_{1}, t_{3}), and γ = ( \begin{matrix}γ_{1} & γ_{2} \\ γ_{2} & γ_{3} \end{matrix} ). Then

B_{1}(T, γ) = \frac{1}{2}J(−t_{1}, γ)J(−t_{3}, ( \begin{matrix}γ_{3} & 2γ_{2} \\ 2γ_{2} & 4γ_{1} \end{matrix} )).

Proof. It follows from the formula

trT k.γ = t_{1}(γ_{1}k_{1}^{2} + 2γ_{2}k_{1}k_{2} + γ_{3}k_{2}^{2}) + t_{3}(γ_{3}k_{4}^{2} + 4γ_{2}k_{4}k_{3} + 4γ_{1}k_{3}^{2})

for k = ( \begin{matrix}k_{1} & k_{2} \\ 2k_{3} & k_{4} \end{matrix} ) ∈ K = K_{1}(2).

Lemma 8.3. Let the notation be as in Lemma 8.2. Then

J(t, γ) = \frac{1}{2}\{ \begin{matrix}char(ℤ_{2})(\frac{8t det γ}{γ_{3}} )I_{1}(2tγ_{3})ψ(\frac{t det γ}{γ_{3}} ) & if \frac{2γ_{2}}{γ_{3}} ∈ 2ℤ_{2}, \\ char(ℤ_{2}^{2})(8tγ_{1}, 2tγ_{3})ψ(tγ_{1}) & if \frac{2γ_{2}}{γ_{3}} ∈ ℤ_{2}^{∗}, \\ char(ℤ_{2}^{2})(8tγ_{1}, 2tγ_{2})ψ(tγ_{1}) & if \frac{2γ_{2}}{γ_{3}} ∉ ℤ_{2} \end{matrix}

Proof. Indeed, one has by definition

J(t, γ) = ∫_{ℤ_{2}^{∗}} ψ(tγ_{1}k_{1}^{2})dk_{1} ∫_{ℤ_{2}} ψ(tγ_{3} (k_{2}^{2} + \frac{2k_{1}γ_{2}}{γ_{3}} k_{2}) dk_{2}

= ∫_{ℤ_{2}^{∗}} ψ(tγ_{1}k_{1}^{2})I_{1}(tγ_{3}, \frac{2k_{1}γ_{2}}{γ_{3}} )dk_{1}.

LOCAL DENSITIES 29

When \frac{2γ_{2}}{γ_{3}} ∈ 2ℤ_{2}, one has by Lemma 6.1

J(t, γ) = I_{1}(2tγ_{3}) ∫_{1+2ℤ_{2}} ψ(\frac{t det γ}{γ_{3}} k_{1}^{2})dk_{1}

= \frac{1}{2}I_{1}(2tγ_{3})ψ(\frac{t det γ}{γ_{3}} ) ∫_{ℤ_{2}} ψ(\frac{4t det γ}{γ_{3}} (k_{1}^{2} + k_{1}))dk_{1}

= \frac{1}{2}I_{1}(2tγ_{3})ψ(\frac{t det γ}{γ_{3}} ) char(ℤ_{2})(\frac{8t det γ}{γ_{3}} )

as desired. When \frac{2γ_{2}}{γ_{3}} ∈ ℤ_{2}^{∗}, one has by Lemma 8.1 (notice that k_{1} ∈ ℤ_{2}^{∗})

J(t, γ) = char(ℤ_{2})(2tγ_{3}) ∫_{1+2ℤ_{2}} ψ(tγ_{1}k_{1}^{2})dk_{1}

= \frac{1}{2} char(ℤ_{2})(2tγ_{3})ψ(tγ_{1}) ∫_{ℤ_{2}} ψ(4tγ_{1}(k_{1}^{2} + k_{1}))dk_{1}

= \frac{1}{2} char(ℤ_{2})(2tγ_{3})ψ(tγ_{1}) char(ℤ_{2})(8tγ_{1})

as desired. The case \frac{2γ_{2}}{γ_{3}} ∉ ℤ_{2} is the same.

Now we have our first main formula of this section. Proposition 8.4. Assume that T = diag(t_{1}, t_{3}) with t_{3}/t_{1} ∈ ℤ_{2}. Then

(1) When \frac{γ_{2}}{γ_{3}} ∈ ℤ_{2}

8B_{1}(T, γ) = \{ \begin{matrix}char(ℤ_{2}^{2})(\frac{8t_{1} det γ}{γ_{3}} , \frac{8t_{3} det γ}{γ_{1}} )I_{1}(−2t_{1}γ_{3})I_{1}(−8t_{3}γ_{1}) \\ ·ψ(−\frac{t_{1} det γ}{γ_{3}} − \frac{t_{3} det γ}{γ_{1}} ) \\ char(ℤ_{2}^{3})(\frac{8t_{1} det γ}{γ_{3}} , 8t_{3}γ_{3}, 8t_{3}γ_{1})I_{1}(−2t_{1}γ_{3})ψ(−\frac{t_{1} det γ}{γ_{3}} − t_{3}γ_{3}) \\ char(ℤ_{2}^{3})(\frac{8t_{1} det γ}{γ_{3}} , 8t_{3}γ_{3}, 4t_{3}γ_{2})I_{1}(−2t_{1}γ_{3})ψ(−\frac{t_{1} det γ}{γ_{3}} − t_{3}γ_{3}) \end{matrix}

according to \frac{γ_{2}}{γ_{1}} ∈ 2ℤ_{2}, ℤ_{2}^{∗} or ∉ ℤ_{2}.

(2) When \frac{2γ_{2}}{γ_{3}} ∈ ℤ_{2}^{∗}, one has then

8B_{1}(T, γ) = \{ \begin{matrix}char(ℤ_{2}^{3})(8t_{1}γ_{1}, 2t_{1}γ_{3}, \frac{8t_{3} det γ}{γ_{1}} )I_{1}(−8t_{3}γ_{1})ψ(−t_{1}γ_{1} − \frac{t_{3} det γ}{γ_{1}} ) \\ char(ℤ_{2}^{2})(8t_{1}γ_{1}, 2t_{1}γ_{3})ψ(−t_{1}γ_{1} − t_{3}γ_{3}), \\ char(ℤ_{2}^{3})(8t_{1}γ_{1}, 2t_{1}γ_{3}, 4t_{3}γ_{2})ψ(−t_{1}γ_{1} − t_{3}γ_{3}), \end{matrix}

according to \frac{γ_{2}}{γ_{1}} ∈ 2ℤ_{2}, ℤ_{2}^{∗} or ∉ ℤ_{2}.

30 TONGHAI YANG

(3) When \frac{2γ_{2}}{γ_{3}} ∉ ℤ_{2}, one has

8B_{1}(T, γ) = \{ \begin{matrix}char(ℤ_{2}^{3})(8t_{1}γ_{1}, 2t_{1}γ_{2}, \frac{8t_{3} det γ}{γ_{1}} )I_{1}(−8t_{3}γ_{1})ψ(−t_{1}γ_{1} − \frac{t_{3} det γ}{γ_{1}} ), \\ char(ℤ_{2}^{3})(8t_{1}γ_{1}, 2t_{1}γ_{2}, 8t_{3}γ_{3})ψ(−t_{1}γ_{1} − t_{3}γ_{3}), \\ char(ℤ_{2}^{3})(8t_{1}γ_{1}, 2t_{1}γ_{2}, 8t_{3}γ_{3})ψ(−t_{1}γ_{1} − t_{3}γ_{3}), \end{matrix}

according to \frac{γ_{2}}{γ_{1}} ∈ 2ℤ_{2}, ℤ_{2}^{∗} or ∉ ℤ_{2}. Corollary 8.5. Let T = diag(t_{1}, t_{3}) with t_{3}/t_{1} ∈ ℤ_{2}, and γ = diag(γ_{1}, γ_{3})

with γ_{3}/γ_{1} ∈ ℤ_{2}. Let g_{1} = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) and g_{2} = ( \begin{matrix}0 & 1 \\ 1 & 1 \end{matrix} ) as in (8.5).

(1) One has

B_{1}(T, γ) = \frac{1}{8} char(ℤ_{2})(8t_{1}γ_{1})I_{1}(−2t_{1}γ_{3})I_{1}(−8t_{3}γ_{1})ψ(−t_{1}γ_{1} − t_{3}γ_{3}),

(2) One has

B_{1}(T, g_{1}.γ) = \frac{1}{8} char(ℤ_{2}^{2})(8t_{1}γ_{3}, 8t_{3}γ_{1})I_{1}(−2t_{1}γ_{1})I_{1}(−8t_{3}γ_{3})ψ(−t_{1}γ_{3} − t_{3}γ_{1}).

(3) When γ_{3}/γ_{1} ∈ 2ℤ_{2}, one has

B_{1}(T, g_{2}.γ) = \frac{1}{8} char(ℤ_{2}^{2})(8t_{1}γ_{3}, 8t_{3}γ_{1})I_{1}(−2t_{1}(γ_{1}+γ_{3}))ψ(−\frac{t_{1}γ_{1}γ_{3}}{γ_{1} + γ_{3}} −t_{3}(γ_{1}+γ_{3})).

In particular, if c_{1} ≤ c_{3} − 3, then

B_{1}(T, g_{2}.γ) = \frac{1}{8} char(ℤ_{2}^{2})(8t_{1}γ_{3}, 8t_{3}γ_{1})I_{1}(−2t_{1}γ_{1})ψ(−t_{1}γ_{3} − t_{3}γ_{1}).

(4) When γ_{3}/γ_{1} = u ∈ ℤ_{2}^{∗}, one has

B_{1}(T, g_{2}.γ) = \{ \begin{matrix}\frac{1}{8} char(ℤ_{2})(4t_{1}γ_{1})ψ(−t_{1}γ_{1} − 2t_{3}γ_{1}) & if u ≡ 1 mod 4, \\ \frac{1}{8} char(ℤ_{2})(2t_{1}γ_{1})ψ(−t_{1}γ_{1}) & if u ≡ −1 mod 4. \end{matrix}

Proof. We verify (3) and (4), and leave (1) and (2) to the reader. Write

g_{2}.γ = ( \begin{matrix}γ̃_{1} & γ̃_{2} \\ γ̃_{2} & γ̃_{3} \end{matrix} )

with γ̃_{1} = γ̃_{2} = γ_{3} and γ̃_{3} = γ_{1} + γ_{3}.

When c_{1} < c_{3}, one has \frac{2γ̃_{2}}{γ̃_{3}} ∈ 2ℤ_{2}, and \frac{γ̃_{2}}{γ̃_{1}} = 1 ∈ ℤ_{2}^{∗}. So Proposition 8.4 gives

8B_{1}(T, g_{2}.γ) = char(ℤ_{2}^{3})(\frac{8t_{1}γ_{1}γ_{3}}{γ̃_{3}} , 8t_{3}γ̃_{3}, 8t_{3}γ̃_{1})I_{1}(−t_{1}γ̃_{3})ψ (−\frac{t_{1}γ_{1}γ_{3}}{γ̃_{3}} − t_{3}γ̃_{3})

= char(ℤ_{2}^{2})(8t_{1}γ_{3}, 8t_{3}γ_{1})I_{1}(−t_{1}(γ_{1} + γ_{3}))ψ (−\frac{t_{1}γ_{1}γ_{3}}{γ_{1} + γ_{3}} − t_{3}γ̃_{3}) .

When γ_{3} = γ_{1}u with u ≡ 1 mod 4, one has γ̃_{3} = γ_{1}(1 + u) and \frac{2γ̃_{2}}{γ̃_{3}} = \frac{2}{1+u} ∈ ℤ_{2}^{∗}. One still has \frac{γ̃_{2}}{γ̃_{1}} = 1 ∈ ℤ_{2}^{∗}, and one has then by Proposition 8.4

8B_{1}(T, g_{2}.γ) = char(ℤ_{2}^{3})(8t_{1}γ_{3}, 2t_{1}γ_{1}(1 + u), 8t_{3}γ_{1}(1 + u))ψ(−t_{1}γ_{3} − t_{3}(t_{1} + γ_{3}))
= char(4t_{1}γ_{1})ψ(−t_{1}γ_{1} − 2t_{3}γ_{1}).

Here we have used the fact t_{i}γ_{3} ≡ t_{i}γ_{1} mod 4t_{1}γ_{1}.

When γ_{3} = γ_{1}u with u ≡ −1 mod 4, one has γ̃_{3} = γ_{1}(1 + u) and \frac{2γ̃_{2}}{γ̃_{3}} = \frac{2}{1+u} ∉ ℤ_{2}. One still has \frac{γ̃_{2}}{γ̃_{1}} = 1 ∈ ℤ_{2}^{∗}, and one has then by Proposition 8.4

8B_{1}(T, g_{2}.γ) = char(ℤ_{2}^{3})(8t_{1}γ_{3}, 2t_{1}γ_{3}, 8t_{3}γ_{1}(1 + u))ψ(−t_{1}γ_{3} − t_{3}(t_{1} + γ_{3}))
= char(ℤ_{2})(2t_{1}γ_{1})ψ(t_{1}γ_{1}).

Proof of Lemma 3.1: In our new notation, t_{1} = α_{1}2^{a}, t_{3} = α_{2}2^{a}, γ_{1} = β_{1}2^{−c}, and γ_{3} = β_{2}2^{r−c}.

We first look at the case r ≥ 1. We have by Lemma 6.1

B(T, γ) = B(γ, T) = B_{1}(γ, T) + B_{2}(γ, T) + B_{3}(γ, T)

where B_{2}(γ, T) = B_{1}(γ, g_{1}.T), and B_{3}(γ, T) = B_{1}(γ, g_{2}.T). Applying Corollary 8.5, one has

8B_{1}(γ, T) = char(ℤ_{2})(2^{a+3−c})I_{1}(−2^{a+1−c}α_{2}β_{1})ψ(−2^{a−c}α_{1}β_{1} − 2^{a−c+r}α_{2}β_{2})

= \{ \begin{matrix}0 & if c > a + 3, \\ I_{1}(−\frac{α_{2}β_{1}}{4} )ψ(−\frac{α_{1}β_{1}+2^{r}α_{2}β_{2}}{8} ) & if c = a + 3, \\ I_{1}(−\frac{α_{2}β_{1}}{2} )ψ(−\frac{α_{1}β_{1}+2^{r}α_{2}β_{2}}{4} ) & if c = a + 2, \\ 0 & if c = a + 1, \\ 1 & if c ≤ a, \end{matrix}

8B_{2}(γ, T) = char(ℤ_{2})(2^{a+3−c})I_{1}(−2^{a+1−c}α_{1}β_{1})ψ(−2^{a−c}α_{2}β_{1} − 2^{a−c+r}α_{1}β_{2})

= \{ \begin{matrix}0 & if c > a + 3, \\ I_{1}(−\frac{α_{1}β_{1}}{4} )ψ(−\frac{α_{2}β_{1}+2^{r}α_{1}β_{2}}{8} ) & if c = a + 3, \\ I_{1}(−\frac{α_{1}β_{1}}{2} )ψ(−\frac{α_{2}β_{1}+2^{r}α_{1}β_{2}}{4} ) & if c = a + 2, \\ 0 & if c = a + 1, \\ 1 & if c ≤ a, \end{matrix}

and

8B_{3}(γ, T) = \{ \begin{matrix}char(ℤ_{2})(2^{a+2−c})ψ(−2^{a−c}α_{1}β_{1} − 2^{a+1−c+r}α_{1}β_{2}) & if α_{2} ≡ α_{1} mod 4, \\ char(ℤ_{2})(2^{a+2−1})ψ(2^{a−c}α_{1}β_{1}) & if α_{2} ≡ −α_{1} mod 4 \end{matrix}

When c = a + 3, one has then I_{1}(\frac{b}{4}) = \frac{1}{2}ψ(\frac{b}{8}) for any b ∈ ℤ_{2}^{∗} and thus

8B(T, γ) = \frac{1}{2}ψ(−\frac{α_{2}β_{1}}{8} )ψ(−\frac{α_{1}β_{1} + 2^{r}α_{2}β_{2}}{8} ) + \frac{1}{2}ψ(−\frac{α_{1}β_{1}}{8} )ψ(−\frac{α_{2}β_{1} + 2^{r}α_{1}β_{2}}{8} )

= \frac{1}{2}ψ(−\frac{αβ_{1}}{8} ) (ψ(−2^{r−3}α_{1}β_{2}) + ψ(−2^{r−3}α_{2}β_{2})) .

When c = a + 2, I_{1}(\frac{b}{2}) = \frac{1}{2}(1 + ψ(\frac{b}{4})) for every b ∈ ℤ_{2}^{∗}. So

8(B_{1} + B_{2}) = \frac{1}{2} (1 + ψ(−\frac{α_{2}β_{1}}{4} )) ψ(−\frac{α_{1}β_{1} + 2^{r}α_{2}β_{2}}{4} )

+ \frac{1}{2} (1 + ψ(−\frac{α_{1}β_{1}}{4} )) ψ(−\frac{α_{2}β_{1} + 2^{r}α_{1}β_{2}}{4} )

= ψ(2^{r−2}) (ψ(−\frac{αβ_{1}}{4} ) + \frac{1}{2}ψ(−\frac{α_{1}β_{1}}{4} ) + \frac{1}{2}ψ(−\frac{α_{2}β_{1}}{4} )) .

Here we have used the fact ψ(2^{r−2}b) = ψ(2^{r−2}) for r ≥ 1. If α_{2} ≡ α_{1} mod 4, then

8B(T, γ) = ψ(2^{r−2}) (ψ(−\frac{αβ_{1}}{4} ) + \frac{1}{2}ψ(−\frac{α_{1}β_{1}}{4} ) + \frac{1}{2}ψ(−\frac{α_{2}β_{1}}{4} )) + ψ(−\frac{α_{1}β_{1}}{4} )

= \{ \begin{matrix}1 & if r = 1, \\ −1 + 2ψ(−\frac{α_{1}β_{1}}{4} ) & if r ≥ 2. \end{matrix}

= −ψ(2^{r−2}) + (ψ(2^{r−2}) + 1)ψ(−\frac{α_{1}β_{1}}{4} )

If α_{2} ≡ −α_{1} mod 4, then

8B(T, γ) = 8(B_{1} + B_{2}) = ψ(2^{r−2}).

Notice that, in both cases, one has

8B(T, γ) = \{ \begin{matrix}ψ(−\frac{αβ_{1}}{4} ) + ψ(−\frac{α_{1}β_{1}}{4} ) + ψ(−\frac{α_{2}β_{1}}{4} ) & if r ≥ 2, \\ (−1)^{\frac{α_{1}−α_{2}}{2}} & if r = 1. \end{matrix}

The case c ≤ a + 1 is obvious. This proves (1) − (3) of Lemma 3.1. For the case r = 0, we use

B(T, γ) = B_{1}(T, γ) + B_{2}(T, γ) + B_{3}(T, γ).

LOCAL DENSITIES 33

By Corollary 8.5, one has

8B_{1}(T, γ) = char(ℤ_{2})(2^{a+3−c})I_{1}(−2^{a+1−c}α_{1}β_{2})
· I_{1}(−2^{a+3−c}α_{2}β_{1})ψ(−2^{a−c}(α_{1}β_{1} + α_{2}β_{2}))

= \{ \begin{matrix}0 & if c > a + 2, \\ I_{1}(−\frac{α_{1}β_{2}}{2} )ψ(−\frac{α_{1}β_{1}+α_{2}β_{2}}{4} ) & if c = a + 2, \\ 0 & if c = a + 1, \\ 1 & if c ≤ a, \end{matrix}

8B_{2}(T, γ) = char(ℤ_{2})(2^{a+3−c})I_{1}(−2^{a+1−c}α_{1}β_{1})

· I_{1}(−2^{a+3−c}α_{2}β_{2})ψ(−2^{a−c}(α_{1}β_{2} + α_{2}β_{1}))

= \{ \begin{matrix}0 & if c > a + 2, \\ I_{1}(−\frac{α_{1}β_{1}}{2} )ψ(−\frac{α_{1}β_{2}+α_{2}β_{1}}{4} ) & if c = a + 2, \\ 0 & if c = a + 1, \\ 1 & if c ≤ a, \end{matrix}

and

8B_{3}(T, γ) = \{ \begin{matrix}cha(ℤ_{2})(2^{a+2−c})ψ(−2^{a−c}α_{1}β_{1} − 2^{a+1−c}α_{2}β_{1}) & if β_{2} ≡ β_{1} mod 4 \\ cha(ℤ_{2})(2^{a+1−c})ψ(−2^{a−c}α_{1}β_{1}) & if β_{2} ≡ −β_{1} mod 4 \end{matrix}

= \{ \begin{matrix}0 & if c > a + 2, \\ −ψ(−\frac{α_{1}β_{1}}{4} ) & if c = a + 2, β_{2} ≡ β_{1} mod 4, \\ 0 & if c = a + 2, β_{2} ≡ −β_{1} mod 4, \\ −1 & if c = a + 1, \\ 1 & if c ≤ a. \end{matrix}

Since I_{1}(\frac{b}{2}) = \frac{1}{2}(1+ψ(\frac{b}{2})) for b ∈ ℤ_{2}^{∗}, one has when c = a+2 and β_{2} ≡ β_{1} mod 4

8B(T, γ) = 2I_{1}(−\frac{α_{1}β_{1}}{2} )ψ(−\frac{αβ_{1}}{)} − ψ(−\frac{α_{1}β_{1}}{4} )

= ψ(−\frac{αβ_{1}}{4} ) − ψ(−\frac{α_{2}β_{1}}{4} ) − ψ(−\frac{α_{1}β_{1}}{4} )

as claimed. When c = a + 2 and β_{2} ≡ −β_{1} mod 4, one has

8B(T, γ) = \frac{1}{2} (1 + ψ(\frac{α_{1}β_{1}}{4} ) ψ(\frac{(α_{2} − α_{1})β_{1}}{4} ) + \frac{1}{2} (1 + ψ(−\frac{α_{1}β_{1}}{4} ) ψ(\frac{(α_{1} − α_{2})β_{1}}{4} )

= ψ(\frac{(α_{2} − α_{1})β_{1}}{4} )

as claimed.

Proof of Lemma 4.1: The proof of Lemma 4.1 is again a routine calculation from Corollary 8.5 and is left to the reader. Proposition 8.4 also gives

34 TONGHAI YANG

Corollary 8.6. Let T = diag(t_{1}, t_{3}) with t_{3}/t_{1} ∈ ℤ_{2}. Let g_{1} = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) and

g_{2} = ( \begin{matrix}0 & 1 \\ 1 & 1 \end{matrix} ) as in (8.5).

(1) When γ = γ_{2} ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ), one has

B_{1}(T, γ) = B_{1}(T, g_{1}.γ) = \frac{1}{8} char(ℤ_{2})(2t_{1}γ_{2}),

and

B_{1}(T, g_{2}.γ) = \frac{1}{8} char(ℤ_{2})(4t_{1}γ_{2})ψ(−2t_{3}γ_{2}).

(2) When γ = γ_{2} ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ), one has

B_{1}(T, γ) = B_{1}(T, g_{1}.γ) = \frac{1}{8} char(ℤ_{2})(4t_{1}γ_{2})ψ(−2(t_{1} + t_{3})γ_{2}),

and

B_{1}(T, g_{2}.γ) = \frac{1}{8} char(ℤ_{2})(4t_{1}γ_{2})ψ(−2(t_{1} + 3t_{3})γ_{2}).

Proposition 8.7. Let g_{i} be as in (8.5) with i = 0, 1, 2. Let T = t ( \begin{matrix}d_{1} & 1 \\ 1 & d_{1} \end{matrix} ), and γ = γ_{2} ( \begin{matrix}d_{2} & 1 \\ 1 & d_{2} \end{matrix} ), with d_{j} = 0, 2. Then one has

B_{1}(T, g_{i}.γ) = \frac{1}{8} char(ℤ_{2})(4tγ_{2})ψ(2tγ_{2}).

Proof. First assume T = t ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ). We will prove a slightly stronger result. For a general γ = ( \begin{matrix}γ_{1} & γ_{2} \\ γ_{2} & γ_{3} \end{matrix} ) and k = ( \begin{matrix}k_{1} & k_{2} \\ k_{3} & k_{4} \end{matrix} ), one has

trT k.γ = 2t(γ_{1}k_{3} + γ_{2}k_{4})k_{1} + 2t(γ_{2}k_{3} + γ_{3}k_{4})k_{2}.

So

B_{1}(−T, γ) = ∫_{2ℤ_{2}×ℤ_{2}^{∗}} dk_{3}dk_{4} ∫_{1+2ℤ_{2}} ψ(2t(γ_{1}k_{3} + γ_{2}k_{4})k_{1})dk_{1}
· ∫_{ℤ_{2}} ψ(2t(γ_{2}k_{3} + γ_{3}k_{4})k_{2})dk_{2}

= \frac{1}{2} ∫_{2ℤ_{2}×ℤ_{2}^{∗}} ψ(2tγ_{1}k_{3} + 2tγ_{2}k_{4}) char(ℤ_{2})(4t(γ_{1}k_{3} + γ_{2}k_{4}))

· char(ℤ_{2})(2t(γ_{2}k_{3} + γ_{3}k_{4}))dk_{3}dk_{4}.

LOCAL DENSITIES 35

If γ_{1}/γ_{2} ∈ ℤ_{2}, then k_{3} ∈ 2ℤ_{2} and k_{4} ∈ ℤ_{2}^{∗} imply

char(ℤ_{2})(4t(γ_{1}k_{3} + γ_{2}k_{4})) = char(ℤ_{2})(4tγ_{2}).

Moreover, when 4tγ_{2} ∈ ℤ_{2}, one has

char(ℤ_{2})(2t(γ_{2}k_{2} + γ_{3}k_{3})) = char(ℤ_{2})(2tγ_{3}).

So

B_{1}(T, γ) = \frac{1}{2} char(ℤ_{2}^{2})(4tγ_{2}, 2tγ_{3}) ∫_{2ℤ_{2}×ℤ_{2}^{∗}} ψ(2tγ_{1}k_{3} + 2tγ_{2}k_{4})dk_{3} dk_{4}

= \frac{1}{8} char(ℤ_{2}^{2})(4tγ_{2}, 2tγ_{3})ψ(2tγ_{2})

So we have proved

Lemma 8.8. Let T = t ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ), and γ = ( \begin{matrix}γ_{1} & γ_{2} \\ γ_{2} & γ_{3} \end{matrix} ) with γ_{1}/γ_{2} ∈ ℤ_{2}. Then

B_{1}(T, γ) = \frac{1}{8} char(ℤ_{2}^{2})(4tγ_{2}, 2tγ_{3})ψ(2tγ_{2}).

Now the case T = t ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) follows from this lemma immediately. The case T = ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ) follows from the following lemma easily.

Lemma 8.9. Let T = ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ), and γ = ( \begin{matrix}γ_{1} & γ_{2} \\ γ_{2} & γ_{3} \end{matrix} ) with γ_{1}/γ_{2}, γ_{3}/γ_{2} ∈ 2ℤ_{2}.

Then

B_{1}(T, γ) = \frac{1}{8} char(ℤ_{2})(4tγ_{2})ψ(2tγ_{2}).

Proof. One checks that

trT k.γ = 2tγ_{1} (k_{1}^{2} + (k_{3} + \frac{2γ_{2}}{γ_{1}} k_{2} + \frac{γ_{2}}{γ_{1}} k_{4})k_{1}) + Q

with

Q = 2γ_{3}k_{4}^{2} + 4γ_{2}k_{3}k_{4} + 2γ_{3}k_{2}k_{4} + 2γ_{2}k_{2}k_{3} + 2γ_{1}k_{3}^{2} + 2γ_{3}k_{2}^{2}.

Now Lemma 1.1 implies

∫_{ℤ_{2}^{∗}} ψ(2tγ_{1} (k_{1}^{2} + (k_{3} + \frac{2γ_{2}}{γ_{1}} k_{2} + \frac{γ_{2}}{γ_{1}} k_{4})k_{1}))dk_{1}

= char(ℤ_{2})(2tγ_{1} (k_{3} + \frac{2γ_{2}}{γ_{1}} k_{2} + \frac{γ_{2}}{γ_{1}} k_{4})) − \frac{1}{2} char(ℤ_{2})(4tγ_{1} (k_{3} + \frac{2γ_{2}}{γ_{1}} k_{2} + \frac{γ_{2}}{γ_{1}} k_{4}))

= char(ℤ_{2})(2tγ_{2}) − \frac{1}{2} char(ℤ_{2})(4tγ_{2})

= \frac{1}{2} char(ℤ_{2})(4tγ_{2})ψ(2tγ_{2}).

On the other hand, when 4tγ_{2} ∈ ℤ_{2}, k_{3} ∈ 2ℤ_{2}, one has 2tγ_{3} ∈ Z and thus Q ∈ Z. So

B_{1}(−T, γ) = ∫ ψ(Q)dk_{2}dk_{3}dk_{4} ∫_{ℤ_{2}^{∗}} ψ(2tγ_{1} (k_{1}^{2} + (k_{3} + \frac{2γ_{2}}{γ_{1}} k_{2} + \frac{γ_{2}}{γ_{1}} k_{4})k_{1}))dk_{1}

= \frac{1}{2} char(ℤ_{2})(4tγ_{2})ψ(2tγ_{2}) ∫ ψ(Q)dk_{2}dk_{3}dk_{4}

= \frac{1}{2} char(ℤ_{2})(4tγ_{2})ψ(2tγ_{2}) ∫_{ℤ_{2}×2ℤ_{2}×ℤ_{2}^{∗}} dk_{2}dk_{3}dk_{4}

= \frac{1}{8} char(ℤ_{2})(4tγ_{2})ψ(2tγ_{2}).

Proof of Lemma 2.1 Now Lemma 2.1 follows from Corollary 8.6 and Proposition 8.7. We check the case T ∈ A_{−1}(a) with and leave the other case to the reader. When r < 0, Proposition 8.7 asserts

B(T, γ) = B_{1}(T, γ) + B_{1}(T, g_{1}.γ) + B_{1}(T, g_{2}.γ)

= \frac{3}{8} char(ℤ_{2})(2^{a+2−c})ψ(2^{a+1−c})

= \frac{3}{8}f_{2}(a, c).

When r ≥ 0, Corollary 8.6 implies

B(T, γ) = B(γ, T)

= B_{1}(γ, T) + B_{1}(γ, g_{1}.T) + B_{1}(γ, g_{2}.T)

= \frac{1}{4} char(ℤ_{2})(2^{a+1−c}) + \frac{1}{8} char(ℤ_{2})(2^{a+2−c})ψ(2^{a+1−c+r})

= \{ \begin{matrix}\frac{1}{8}f_{1}(a, c)f_{2}(a, c) & if r = 0, \\ \frac{1}{8}f_{1}(a, c) & if r > 0. \end{matrix}

9. The constant C(γ).

In this section, we compute the constant C(γ) in (1.5) when n = 2 and prove Lemma 1.4. We start with some general calculation related with C(γ). Recall that C(γ) is defined via

(9.1)  C(γ) ∫ _{GL_{n}(ℤ_{2})} ϕ(k.γ)dk = ∫_{O_{γ}} ϕ(t) dt

for every locally constant function ϕ on O_{γ}. For a positive integer m > 0, Let O_{m,γ} be the (1+2^{m}M_{n}(ℤ_{2}))-orbit of γ, and let O_{m}(γ) = (1+2^{m}M_{n}(ℤ_{2}))∩O(γ) be the stabilizer of γ in 1 + 2^{m}M_{n}(ℤ_{2}). Let ϕ be the characteristic function of O_{m,γ}, then (9.1) gives

C(γ)vol((1 + 2^{m}M_{n}(ℤ_{2}))O(γ)) = vol(O_{m,γ}).

So we have

(9.2)  C(γ) = \frac{2^{mn^{2}}vol(O_{m,γ})}{[O(γ) : O_{m}(γ)]}.

Let

γ = diag(ϵ_{1}2^{l_{1}}, ⋯, ϵ_{H}2^{l_{H}} , 2^{m_{1}}A, ⋯, 2^{m_{M}} A, 2^{n_{1}}B, ⋯, 2^{n_{N}} B),

with ϵ_{i} ∈ ℤ_{2}^{∗}, l_{i}, m_{j}, n_{k} ∈ ℤ_{2}, and H + 2M + 2N = n, and

A = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , B = ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ) .

We reorder l_{i}, m_{j}, m_{j}, n_{k}, n_{k} into an increasing sequence l′_{1} ≤ l′_{2} ≤ ⋯ ≤ l′_{n}. We assume that γ is invertible, i.e, l′_{n} < ∞ (Notice that these numbers are not the same as those in Section 1).

Lemma 9.1. Let the notation be as above. Then one has when m >> 0

vol(O_{m,γ}) = 2^{− \frac{mn(n+1)}{2} −n} ∏_{1≤i≤n} 2^{−(n−i+1)l′_{i}}.

Proof. A typical element in O_{m,γ} is

α = (α_{ij}) = (1 + 2^{m}k).γ = γ + 2^{m}(kγ + γ^{t}k) + 2^{2m}k.γ

with k = (k_{ij}) ∈ M_{n}(ℤ_{2}). We choose m > 0 so that m + l′_{1} ≥ l′_{n} + 4. Then we have

α_{ij} − γ_{ij} ≡ 2^{m} ∑_{a=1}^{n} k_{ia}(γ_{ja} + k_{ja}γ_{ia}) mod 2^{m+l′_{n}+4}.

From this, one can easily check that α_{ij} −γ_{ij} runs over the set 2^{m+δ_{ij}+l_{ij}}M_{n}(ℤ_{2}), where δ_{ij} is the usual Kronecker symbol, and l_{ij} are given as follows.

l_{ii} = \{ \begin{matrix}l_{i} & if 1 ≤ i ≤ H, \\ m_{j} & if 1 ≤ j = [\frac{i−H+1}{2}] ≤ M, \\ n_{k} & if 1 ≤ k = [\frac{i−H+1}{2}] − M ≤ N, \end{matrix}

and l_{ij} = min(l_{ii}, l_{jj}). So

vol(O_{m,γ}) = 2^{− \frac{mn(n+1)}{2} −n} ∏_{1≤i≤j≤n} 2^{−l_{ij}}.

Clearly, l′_{i} appears as l_{ij} with multiplicity (n − i + 1), and so

∏_{1≤i≤j≤n} 2^{−l_{ij}} = ∏_{1≤i≤n} 2^{−(n+1−i)l′_{i}}.

Next we compute [O(γ) : O_{m}(γ)] when γ is of rank 2. There are three cases.

Lemma 9.2. Let γ = diag(γ_{1}, γ_{3}) with γ_{3}/γ_{1} ∈ ℤ_{2}. Let r = ord_{2}(γ_{3}/γ_{1}). Then

[O(γ) : O_{m}(γ)] = \{ \begin{matrix}2^{m+2} & if c ≥ 3, \\ 2^{m+1} & if 1 ≤ r ≤ 2, \\ 2^{m+1} & if r = 0, γ_{3}/γ_{1} ≡ 1 mod 4, \\ 2^{m} & if r = 0, γ_{3}/γ_{1} ≡ −1 mod 4. \end{matrix}

Proof. Let Δ = −γ_{3}/γ_{1}. It is easy to check that [O(γ) : SO(γ)] = 2 with

SO(γ) = \{k ∈ O(γ) : det k = 1\} = \{( \begin{matrix}k_{1} & k_{2} \\ Δk_{2} & k_{1} \end{matrix} ) : k_{1}^{2} − Δk_{2}^{2} = 1, k_{1}, k_{2} ∈ ℤ_{2}\}.

Let K = ℚ_{2}(\sqrt{Δ}) be the quadratic ètale extension of ℚ_{2} (i.e., K = ℚ_{2} × ℚ_{2} is allowed). Let G = K^{1} be the norm one subgroup of K^{∗} in the nonsplit case and G = {(x, x^{−1}) : x ∈ ℤ_{2}^{∗}} in the split case. Write Δ = 2^{2a}Δ_{0} with ord_{2}Δ_{0} = 0 or 1. As in the Appendix, we set

G_{n} := {g = a + bδ_{0} ∈ G : b ∈ 2^{n}ℤ_{2}, a − 1 ∈ 2^{n}Δ_{0}ℤ_{2}}.

Here δ_{0} = \sqrt{Δ}_{0} in the nonsplit case and (\sqrt{Δ}_{0}, −\sqrt{Δ}_{0}) in the split case. Let δ = 2^{a}δ_{0} = \sqrt{Δ}. Then we have a group isomorphism

SO(γ) ⟶ {±1}G_{a}, k ↦ g = k_{1} + δk_{2},

under which O_{m}(γ) maps onto G_{a+m}. When r = ord_{2}(γ_{3}/γ_{1}) ≥ 3, either a ≥ 2 or a = 1 and ord_{2}Δ_{0} = 1. In either case, one has −1 ∉ G_{a} and Propositions A.8 in the Appendix A gives

[SO(γ) : O_{m}(γ)] = 2[G_{a} : G_{a+m}] = 2^{m+1}.

When 1 ≤ r ≤ 2, a = 1 with ord_{2}Δ_{0} = 0 or a = 1 with ord_{2}δ_{0} = 1. In either case, −1 ∈ G_{a}, and

[SO(γ) : O_{m}(γ)] = [G_{a} : G_{a+m}] = 2^{m}.

Finally, when r = 0, one has a = 0 and Δ_{0} = Δ = −γ_{3}/γ_{1} ∈ ℤ_{2}^{∗}. So −1 ∈ G_{0}. Proposition A.8 in the Appendix A implies

[SO(γ) : O_{m}(γ)] = [G_{a} : G_{a+m}] = \{ \begin{matrix}2^{m} & if Δ_{0} ≡ −1 mod 4, \\ 2^{m−1} & if Δ_{0} ≡ 1 mod 4. \end{matrix}

This proves the Lemma.

Lemma 9.3. When γ = 2^{c} ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) ≠ 0, one has

[O(γ) : O_{m}(γ)] = 2^{m}.

Proof. Let SO(γ) be the special orthogonal group associated to γ as in the proof of Lemma 9.2. One has again [O(γ) : SO(γ)] = 2. It is easy to check that

ℤ_{2}^{∗} ⟶ SO(γ), α ↦ diag(α, α^{−1})

is a group isomorphism, which maps 1 + 2^{m}ℤ_{2} onto O_{m}(γ). From this, it is easy to see that

[SO(γ) : O_{m}(γ)] = 2^{m−1}.

Lemma 9.4. Let γ = 2^{c} ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ). Then

[O(γ) : O_{m}(γ)] = 3 · 2^{m}

for m ≥ 1.

Proof. Again we have [O(γ); SO(γ)] = 2. Notice that for k ∈ SL_{2}(ℤ_{2}), k.γ = γ if and only if

( \begin{matrix}k_{1} & k_{2} \\ k_{3} & k_{4} \end{matrix} ) ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ) = ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ) ( \begin{matrix}k_{4} & −k_{3} \\ −k_{2} & k_{1} \end{matrix} ) ,

that is,

(9.3)  k_{1}^{2} + k_{1}k_{2} + k_{2}^{2} = 1, k_{3} = −k_{2}, k_{4} = k_{1} + k_{2}.

Let K = ℚ_{2}(\sqrt{−3}), and G = K^{1} is the subgroup of norm one elements in K^{∗}. Then

SO(γ) ⟶ G, ( \begin{matrix}k_{1} & k_{2} \\ k_{3} & k_{4} \end{matrix} ) ↦ k_{1} + \frac{−1 + \sqrt{−3}}{2}k_{2} = (k_{1} − \frac{1}{2}k_{2}) + \frac{1}{2}k_{2}\sqrt{−3}

is a group isomorphism, under which O_{m}(γ) maps onto G_{m}. Indeed, one can

verify using (9.3) that if ( \begin{matrix}k_{1} & k_{2} \\ k_{3} & k_{4} \end{matrix} ) ∈ SO_{m}(γ), then k_{2} ∈ 2^{m+1}ℤ_{2}. Now Propo-

sition A.8 in the Appendix A implies

[SO(γ) : O_{m}(γ)] = [G : G_{m}] = 3 · 2^{m−1}

for m ≥ 2.

Proof of Lemma 1.4 Now Lemma 1.4 follows from (9.2) and Lemmas 9.1-9.4 immediately.

Appendix A: Structure of U(1) over a 2-adic field.

In this appendix, we study the structure of U(1) over a 2-adic field, which is needed in Section 9.

Let F be a finite field extension of ℚ_{2} with ring of integers O and a uniformizer π. Let K = F (δ) be a quadratic field extension of F with Δ = δ^{2} ∈ F^{∗}. We assume that ν(Δ) = 0 or 1, where ν is the 2-adic valuation on F such that v(π) = 1. Let

(A.1)  r = ν(2), t = max{n : Δ − x^{2} ∈ π^{n}O for some x ∈ O^{∗}}.

By Hensel’s lemma, one can show t ≤ 2r (otherwise Δ ∈ F^{∗2}). Let

(A.2)  G = K^{1} = {g ∈ K : gḡ = 1}

be the group of norm one elements in K. Here ḡ = a − bδ if g = a + bδ. We denote for any integer n

(A.3)  G_{n} = {g = a + bδ ∈ G : b ∈ π^{n}O, and a − 1 ∈ π^{n}ΔO}.

Clearly, G_{n} is a subgroup of G when n ≥ 0, we will later prove that G_{n} is always a group.

The next two simple lemmas will be fundamental for our calculation.

Lemma A.1. Let the notation be as above. Then the map

(A.4)  g : F ⋃\{∞\} ⟶ G, x ↦ g(x) = \frac{δ + x}{δ − x} = a(x) + b(x)δ

is a bijection of sets with g(x)^{−1} = g(−x). Here

(A.5) a(x) − 1 = \frac{2x^{2}}{Δ − x^{2}} , b(x) = \frac{2x}{Δ − x^{2}} , and a(x) + 1 = \frac{2Δ}{Δ − x^{2}} ,

Moreover, if we write

(A.6)  g(x_{1})g(x_{2})^{−1} = g(x_{1})g(−x_{2}) = g(x_{3}),

then

(A.7)  a(x_{3}) − 1 = \frac{2Δ(x_{1} − x_{2})^{2}}{(Δ − x_{1}^{2})(Δ − x_{2}^{2})}, b(x_{3}) = \frac{2(x_{1} − x_{2})(Δ − x_{1}x_{2})}{(Δ − x_{1}^{2})(Δ − x_{2}^{2})}.

Lemma A.2. Let x ↦ g(x) = a(x) + b(x)delta the bijection defined in Lemma A.1.

(1) When n ≥ 1, one has

x ∈ π^{n}O^{∗} ⇔ \{ \begin{matrix}a(x) − 1 ∈ π^{r+2n}Δ^{−1}O^{∗}, \\ b(x) ∈ π^{r+n}Δ^{−1}O^{∗}, \\ a(x) + 1 ∈ π^{r}ΔO^{∗}.. \end{matrix}

(2) When n ≥ 1, one has

x ∈ π^{−n}O^{∗} ⇔ \{ \begin{matrix}a(x) − 1 ∈ π^{r}Δ^{−1}O^{∗}, \\ b(x) ∈ π^{r+n}O^{∗}, \\ a(x) + 1 ∈ π^{r+2n}ΔO^{∗}. \end{matrix}

(3) When ν(Δ) = 1, one has

x ∈ O^{∗} ⇔ \{ \begin{matrix}a(x) − 1 ∈ π^{r}O^{∗} \\ b(x) ∈ π^{r}O^{∗}, \\ a(x) + 1 ∈ π^{r+1}O^{∗}. \end{matrix}

(4) When ν(Δ) = 0, one has, for x ∈ O^{∗} and 0 ≤ c ≤ t,

Δ − x^{2} ∈ π^{c}O^{∗} ⇔ \{ \begin{matrix}a(x) ± 1 ∈ π^{r−c}O^{∗}, \\ b(x) ∈ π^{r−c}O^{∗}. \end{matrix}

Here t is the integer defined in (A.1)

Proof. The “⇒” parts follow from Lemma A.1 directly. Since the conditions on either side are mutually exclusive, the “⇐” parts follow from the “⇒” parts.

The following two propositions follow from Lemma A.2 directly. I learnt Proposition A.3 first from J.S. Li’s private note ([Li]).

Proposition A.3 (Li). Assume first that ν(Δ) = 1. Then G = G_{r−1} and for n ≥ r

G_{n} = {g(x) : x ∈ π^{n−r+1}O} = {g = a + bδ : b ∈ π^{n}O, a − 1 ∈ π^{2n−r+1}O}.

Moreover, one has

(1) [G : G_{r}] = 2.

(2) [G_{n} : G_{n+1}] = q for n ≥ r.

Proposition A.4. Assume Δ ∈ O^{∗} and n > r. Then

(1)  G_{n} = g(π^{n−r}O) = {g = a + bδ : b ∈ π^{n}O, a − 1 ∈ π^{2n−r}O}

is a subgroup of G, and

(2)  [G_{n} : G_{n+1}] = q.

From now on, we assume Δ ∈ O^{∗} and n ⟶ r. Recall that t is the largest integer such that

Δ = x_{0}^{2} mod π^{t}

has a solution x_{0} ∈ O^{∗}. We fix such an x_{0}. Depending on whether t is an even or odd number, we write in this appendix t = 2c or t = 2c + 1. Recall that t ≤ 2r.

LOCAL DENSITIES 43

Lemma A.5. Assume that Δ ∈ O^{∗}. Then G = G_{r−t}, and G_{n} is a subgroup of G for every integer n. Moreover,

(1) For 0 ≤ α ≤ c, one has

G_{r−2α} = {g(x) : ν(x − x_{0}) ≤ α}.

Moreover, G_{r−2α−1} = G_{r−2α} if α < c.

(2) For 0 < α ≤ c and 2α < t, one has

G_{r−2α} − G_{r−2α+1} = {g(x) : x − x_{0} ∈ π^{α}O^{∗}}.

(3) When t = 2c, one has

G − G_{r−2c+1} = {g(x) : x − x_{0} ∈ π^{c}O}.

(4) When t = 2c + 1, one has

G − G_{r−2c} = {g(x) : x − x_{0} ∈ π^{c+1}O}.

Proof. By Lemma A.2, one has

g(x) ∈ \{ \begin{matrix}G_{r+1} & if x ∈ πO, \\ G_{r} & if x ∉ O, \\ G_{r−β} − G_{r−β−1} & if x ∈ O^{∗} and ν(Δ − x^{2}) = β ≥ 0. \end{matrix}

In particular, g(x) ∈ G_{r} for all x ∉ O^{∗}.

For any x ∈ O^{∗}, write

(A.9)  x = x_{0} + π^{α}y

with α ≥ 0 and y ∈ O^{∗}. If α > c, then

x^{2} ≡ x_{0}^{2} ≡ Δ mod π^{t}.

By the definition of t, one has then ν(Δ − x^{2}) = t and thus g(x) ∈ G_{r−t}. If α ≤ c and 2α < t, then

x^{2} − Δ = π^{2α} (\frac{x_{0}^{2} − Δ}{π^{2α}} + \frac{2}{π^{α}} x_{0}y + y^{2}) ∈ π^{2α}O^{∗}.

So Lemma A.2 implies that

(A.10)  g(x) ∈ G_{r−2α} − G_{r−2α+1},

whenever α ≤ c and 2α < t. This proves

G = G_{r−t}, G_{r−2α} = G_{r−2α−1}, 0 ≤ α < c

and claim (2). Notice that ν(x − x_{0}) ≤ 0 ≤ α when x ∉ O^{∗}. This, together with (A.10), implies claim (1). Claim (3) is the same as Claim (2) except for the fact that g(x) ∈ G even when ν(x − x_{0}) > c. Claim (4) follows from Claim (1). Now we check G_{n} is a group. We may assume n = r − 2α with 0 ≤ α ≤ c. By induction, we may assume ν(x_{1} − x_{0}) = α, i.e, g(x_{1}) ∈ G_{r−2α} − G_{r−2α+1}, and g(x_{2}) ∈ G_{r−2α}. Finally, when 0 ≤ ν(x_{2} −x_{0}) = β ≤ α, one has Δ−x_{2}^{2} ∈ π^{2β}O^{∗},

x_{1} ≡ x_{2} mod π^{β}, and Δ − x_{1}x_{2} ≡ Δ − x_{2}^{2} ≡ 0 mod π^{β}.

Write g(x_{1})g(x_{2})^{−1} = g(x_{3}) as in Lemma A.1, then

a(x_{3}) − 1 = \frac{2Δ(x_{1} − x_{2})^{2}}{(Δ − x_{1}^{2})(Δ − x_{2}^{2})} ∈ π^{r+2β−2α−2β}O = π^{r−2α}O,

and

b(x_{3}) = \frac{2(x_{1} − x_{2})(Δ − x_{1}x_{2})}{(Δ − x_{1}^{2})(Δ − x_{2}^{2})} ∈ π^{r+2β−2α−2β}O =∈ π^{r−2α}O.

So g(x_{3}) ∈ G_{r−2α}. When x_{2} ∉ O, one has −g(x_{2}) ∈ G_{r+1}, and then

−g(x_{1})g(x_{2})^{−1} = −g(x_{3}) ∈ G_{r−2α}

by what has just been proved. In summary, G_{r−2α} is a subgroup of G.

Proposition A.6. Assume that Δ ∈ O^{∗}, and let t = 2c or 2c + 1 as before. Then

(1) [G_{r} : G_{r+1}] = 2(q − 1).

(2) When 0 < α ≤ c and 2α < t, one has

[G_{r−2α} : G_{r−2α+1}] = [G_{r−2α} : G_{r−2α+2}] = q.

(3) When t = 2c, one has [G : G_{r−2c+1}] = [G : G_{r−2c+2}] = q + 1.

(4) When t = 2c + 1, one has [G : G_{r−2c}] = 2.

Proof. By Lemma A.5, one has

G_{r} = {g(x) : ν(x − x_{0}) ≤ 0} = {g(x) : x ∉ O or x − x_{0} ∈ O^{∗}}.

Let H = {g(x) : x − x_{0} ∈ O^{∗}}, then G_{r} = H ∪ (−H), and H is a subgroup of G. For x_{1}, x_{2} ∈ x_{0} + O^{∗}, one has

Δ − x_{1}^{2} ∈ O^{∗}, Δ − x_{2}^{2} ∈ O^{∗}.

Write g(x_{1})g(x_{2})^{−1} = g(x_{3}) as in Lemma A.1, one has then that

a(x_{3}) − 1 = \frac{2Δ(x_{1} − x_{2})^{2}}{(Δ − x_{1}^{2})(Δ − x_{2}^{2})} ∈ π^{r+1}O

if and only if x_{1} ≡ x_{2} mod π. One also has

b(x_{3}) = \frac{2(x_{1} − x_{2})(Δ − x_{1}x_{2})}{(Δ − x_{1}^{2})(Δ − x_{2}^{2})} ∈ π^{r+1}

when x_{1} ≡ x_{2} mod π. So g(x_{1}) ∈ g(x_{2})G_{r+1} if and only if x_{1} ≡ x_{2} mod π, and thus x ↦ g(x) induces a bijection between {x ∈ O/π, x ≠ x_{0}} and H/G_{r+1}. So [H : G_{r+1}] = (q − 1) and [G_{r} : G_{r+1}] = 2(q − 1). This proves (1). When 0 < α ≤ c and 2α < t, similar argument shows that a ↦ g(x_{0} + π^{α}a) induces a bijection between (O/π)^{∗} and (G_{r−2r} − G_{r−2α+1})/G_{r−2α+1}. So

[G_{r−2α} : G_{r−2α+1}] = 1 + #(O/π)^{∗} = q.

This proves (2). Claim (3) is the same as Claim (2) except for the fact that a = 0 is allowed (i.e., g(x_{0}) ∈ G − G_{r−2c+1}). For claim (4), if g(x) ∈ G − G_{r−2c}, one has x − x_{0} ∈ π^{c+1} by Lemma A.5. So Formula (A.7) implies that g(x)g(x_{0})^{−1} ∈ G_{r−2c}, and thus [G : G_{r−2c}] = 2.

In Section 9, we also need the split case Δ ∈ 𝔽^{∗2}. In this case, K = F ⊕F is a quadratic etale extension of F, and δ = (\sqrt{Δ}, −\sqrt{Δ}) ∈ ℤ_{2}^{∗} for a fixed square-root \sqrt{Δ} of Δ in F. We identify K^{1} = \{(x, x^{−1}) : x ∈ F^{∗}\} ≅ F^{∗}. We take in this case

G = \{(a + b\sqrt{Δ}, a − b\sqrt{Δ}) ∈ K^{1} : a + b\sqrt{Δ} ∈ O^{∗}\} ≅ O^{∗},

and

G_{n} = \{(a + b\sqrt{Δ}, a − b\sqrt{Δ}) ∈ G : b ∈ π^{n}O, a − 1 ∈ π^{n}O\} ≅ 1 + π^{n}O

as in (A.3). Then we have

Proposition A.7. In the split case, one has

[G : G_{1}] = q − 1, and [G_{n} : G_{n+1}] = q, for n ≥ 1.

The following special case will be used in Section 9.

46 TONGHAI YANG

Proposition A.8. Let F = ℚ_{2}, and K = F (\sqrt{Δ}) with ord_{2}Δ = 0 or 1.

(1) When K/ℚ_{2} is a ramified field extension, one has G = G_{0}, and [G_{n} : G_{n+1}] = 2 for every n ≥ 0.

(2) When K/ℚ_{2} is an inert or split extension, one has

[G_{n} : G_{n+1}] = \{ \begin{matrix}2 & if n ≥ 1, \\ 1 & if n = 0, \end{matrix}

and

[G : G_{0}] = \{ \begin{matrix}3 & if inert, \\ 1 & if split. \end{matrix}

Appendix B. The Gross-Keating invariants.

This appendix is essentially due to Steve Kudla, who is very generous to allow it be included here.

Gross and Keating define invariants of a ℤ_{p}–lattice L with a quadratic form Q as follows. Let

(x, y) = Q(x + y) − Q(x) − Q(y)

be the associated bilinear form. Then Q(x) = \frac{1}{2}(x, x). For any ℤ_{p}–basis ψ = (ψ_{1}, ψ_{2}, …, ψ_{n}) of L, write

(B.1)  Q(∑_{i} x_{i}ψ_{i}) = ∑_{i≤j} q_{ij} x_{i}x_{j} = xT^{t}x.

Here the q_{ij} ∈ ℤ_{p}, x = (x_{1}, ⋯, x_{n}) ∈ ℤ_{p}^{n}, and T = (t_{ij}) = (\frac{1}{2}(ψ_{i}, ψ_{j})) is a symmetric matrix over \frac{1}{2}ℤ_{p} with t_{ii} ∈ ℤ_{p}. We have

q_{ij} = \{ \begin{matrix}2t_{ij} & if i < j \\ t_{ii} & if i = j, \end{matrix}

for the q_{ij} defined by (1). Note that t_{ij} ∈ \frac{1}{2}ℤ_{p}. Let

S(ψ) = { y = (y_{1}, …, y_{n}) ∈ ℤ^{n} | y_{1} ≤y_{2} ≤ ⋯ ≤ y_{n} (B.2)

and \frac{1}{2}(y_{i} + y_{j}) ≤ o(q_{ij}), ∀i ≤ j \}.

Here o(q_{ij}) = ord_{p}(q_{ij}). Let

S = ∪_{ψ} S(ψ).

Finally, define

a_{1} = max_{y∈S}{y_{1}}
(B.3)  a_{2} = max_{y=(a_{1},y_{2},...)∈S}{y_{2}}
…
a_{n} = max_{y=(a_{1},...,a_{n−1},y_{n})∈S}{y_{n}}

If (a_{1}, …, a_{n}) ∈ S(ψ) we will say that ψ is optimal; such a basis exists, by the definition. It should be pointed out however that it happens sometimes a_{i} < o(Q(ψ_{i}) even when ψ is optimal.

Actually, Gross and Keating consider the case n = 3. In that case, they further define a sign ϵ = ϵ(L)—the Gross-Keating constant—when a_{1} ≡ a_{2} mod (2) and a_{2} < a_{3} as follows. For ψ optimal, let ϵ = +1 if the span of ψ_{1} and ψ_{2} is isotropic and −1 if it is anisotropic. We want to have an explicit expression for these invariants when a Jordan decomposition of the lattice L is given. When p ≠ 2, it is a fact that if ψ is a basis with

(B.4)  T (ψ) = diag(ϵ_{1}p^{α_{1}}, …, ϵ_{n}p^{α_{n}}),

where α_{1} ≤ ⋯ ≤ α_{n} and the ϵ_{i}’s are units, then

(B.5)  (a_{1}, …, a_{n}) = (α_{1}, …, α_{n}).

So, the interesting case is when p = 2.

Following Kitaoka ([Ki]), for a lattice L, let

n(L) = ideal in ℤ_{p} generated by { Q(x) | x ∈ L },

and

(B.6)  s(L) = \frac{1}{2}(L, L).

Then

(B.7)  o(n(L)) = min{ o(q_{ij}) },

and

(B.8)  o(s(L)) = min{ o(t_{ij}) }.

In particular, for n > 1,

(B.9)  2s(L) ⊂ n(L) ⊂ s(L).

Lemma B.1. For any L, a_{1} = o(n(L)).

Proof. Let α = o(n(L)). Then, for any basis ψ, (α, …, α) ∈ S(ψ), and hence 𝔞 ≤ a_{1}. On the other hand, if ψ is an optimal basis, then for all i ≤ j,

a_{1} ≤ \frac{1}{2}(a_{i} + a_{j}) ≤ o(q_{ij}),

so that a_{1} ≤ α.

For r = 1, …, n, let ∧^{r}(L) be the rth exterior product of the lattice L. It is the ℤ_{p} submodule spanned by the vectors

ψ_{i} = ψ_{i_{1}} ∧ … ψ_{i_{r}}

where i = (i_{1}, …, i_{r}) with i_{1} < i_{2} ⋯ < i_{r}. Define a symmetric bilinear form B_{r} on ∧^{r}(L) by

B_{r}(ψ_{i}, ψ_{j}) := det ( B(ψ_{i_{a}}, ψ_{j_{b}}) ) = det(T (ψ)_{ij}) = det. of the ij th minor of T (ψ),

where a indexes the row and b indexes the column. Note that, for r = 1, this agrees with Kitaoka’s notation. The values of B_{r} can lie in 2^{−r}ℤ_{p}. In any case, the values

ρ_{r}(L) := min_{i,j}(o(B_{r}(ψ_{i}, ψ_{j}))) = s(∧^{r}(L))

depend only on L, and ρ_{1}(L) = o(S(L)). Also, ρ_{n}(L) = det(T (ψ)).

Lemma B.2. (i) For any n, a_{1} + a_{2} ≤ ρ_{2}(L) + 2. (ii) For n = 3,

a_{1} + a_{2} + a_{3} ≤ o(det(T)) + 2.

Proof. Suppose that ψ is an optimal basis, so that, entrywise:

( \begin{matrix}a_{1} & \frac{1}{2}(a_{1} + a_{2}) − 1 & … \\ & a_{2} & … \\ & … & … \end{matrix} ) ≤ ( \begin{matrix}o(t_{11}) & o(t_{12}) & … \\ & o(t_{22}) & … \\ & … & … \end{matrix} )

Now consider the quantities o(det(T (ψ)_{ij}). For example, if r = 2, then, for a principal minor, we have

o(t_{ii}t_{jj} − t_{ij}^{2}) ≥ min{o(t_{ii}) + o(t_{jj}), 2o(t_{ij})}.

But

a_{i} + a_{j} ≤ o(t_{ii}) + o(t_{jj}),  and  a_{i} + a_{j} − 2 ≤ 2o(t_{ij}).

so that

a_{i} + a_{j} ≤ o(t_{ii}t_{jj} − t_{ij}^{2}) + 2,

For a nonprincipal minor,

o(t_{i_{1}j_{1}}t_{i_{2}j_{2}} − t_{i_{2}j_{1}}t_{i_{1}j_{2}}) ≥ min{o(t_{i_{1}j_{1}}) + o(t_{i_{2}j_{2}}), o(t_{i_{2}j_{1}}) + o(t_{i_{1}j_{2}})},

o(t_{i_{1}j_{1}}) + o(t_{i_{2}j_{2}}) ≥ \frac{1}{2}(a_{i_{1}} + a_{j_{1}} + a_{i_{2}} + a_{j_{2}}) − 2,

and

o(t_{i_{1}j_{2}}) + o(t_{i_{1}j_{2}}) ≥ \frac{1}{2}(a_{i_{1}} + a_{j_{1}} + a_{i_{2}} + a_{j_{2}}) − 2.

Thus,

a_{1} + a_{2} ≤ ρ_{2}(L) + 2.

Note, for example, that when n = 2, the extreme case a_{1} + a_{2} = o(det(T)) + 2 does occur.

For r = 3, we have, say for n = 3,

o(det(T)) ≥ min{o(t_{11}) + o(t_{22}) + o(t_{33}), o(t_{12}) + o(t_{23}) + o(t_{13}) + 1,

o(t_{11}) + 2o(t_{23}), o(t_{22}) + 2o(t_{13}), o(t_{33}) + 2o(t_{12}) }

≥ a_{1} + a_{2} + a_{2} − 2.

Proposition B.3. Assume n = 2. Let L be an ℤ_{2}-integral quadratic lattice of rank 2 and let T be the matrix of L with respect to a basis e = (e_{1}, e_{2}) of L.

(1) When T = diag(ϵ_{1}p^{α}, ϵ_{2}p^{β}), where 0 ≤ α ≤ β, one has n(L) = s(L) and

(a_{1}, a_{2}) = \{ \begin{matrix}(α, β) & if α ≢ β mod 2, \\ (α, β + 1) & if α ≡ β mod 2 and ϵ_{1} ≡ ϵ_{2} mod 4, \\ (α, β + 2) & if α ≡ β mod 2 and ϵ_{1} ≡ −ϵ_{2} mod 4. \end{matrix}

Note that the basis ψ which diagonalizes Q is not optimal in the last two cases.

(2) When

T (ψ) = 2^{α} ( \begin{matrix}a & 1 \\ 1 & a \end{matrix} ) , a = 0, 2,

one has

(a_{1}, a_{2}) = (α + 1, α + 1).

Proof. (1) In this case, clearly n(L) = s(L) = 2^{α}. So a_{1} = α. By Lemma B.2, one has then a_{1} + a_{2} ≤ α + β + 2 and so a_{2} ≤ β + 2. Since (α, β) ∈ S(e), one has a_{2} ≥ β.

First assume that β ≢ α mod 2. If a_{2} > β, let ψ = e(α_{ij}) be an optimal basis, then

β < a_{2} ≤ o(Q(ψ_{2})) = ϵ_{1}2^{α}α_{12}^{2} + ϵ_{2}2^{β}α_{22}^{2}

implies that α_{12}, α_{22} ∈ 2ℤ_{2}, and thus (α_{ij}) ∉ GL_{2}(ℤ_{2}), a contradiction. So a_{2} = β.

Now assume β ≡ α mod 2. If ϵ_{1} + ϵ_{2} ≡ 0 mod 4, choose

(ψ_{1}, ψ_{2}) = (e_{1}, e_{2}) ( \begin{matrix}1 & 2^{\frac{β−α}{2}} \\ 0 & 1 \end{matrix} ) .

Then Q(ψ_{2}) = 2^{β}(ϵ_{1} + ϵ_{2}) ∈ 2^{β+2}ℤ_{2}, and q_{12} = (ψ_{1}, ψ_{2}) = 2^{β+1}ϵ_{1}). So (α, β + 2) ∈ S(ψ), and (a_{1}, a_{2}) = (α, β + 2) as claimed. On the other hand, if ϵ_{1} ≡ ϵ_{2} mod 4, i.e., ϵ_{1} + ϵ_{2} ≡ 2 mod 4, the same basis ψ gives (α, β + 1) ∈ S(ψ). Furthermore, for any optimal basis ψ = e(α_{ij}) with GL_{2}(ℤ_{2}), at least one of α_{12} or α_{22} is a unit, one has then

a_{2} ≤ o(Q(ψ_{2})) = o(ϵ_{1}2^{α}α_{12}^{2} + ϵ_{2}2^{β}α_{22}^{2}) ≤ β + 1.

So (a_{1}, a_{2}) = (α, β).

(2) In the second case, clearly n(L) = 2^{α+1} = 2s(L). So a_{1} = α + 1 by Lemma B.1. Lemma B.2 implies then a_{2} = β + 1 too.

Next, we consider the case n = 3.

Proposition B.4. Let

T = diag(ϵ_{1}2^{α}, 2^{β} ( \begin{matrix}a & 1 \\ 1 & a \end{matrix} )) with a = 0, 2.

Then

(a_{1}, a_{2}, a_{3}) = \{ \begin{matrix}(α, β + 1, β + 1) & if α ≤ β + 1, \\ (β + 1, β + 1, α) & if α > β + 1. \end{matrix}

Proof. First assume that α ≤ β + 1. Then n(L) = s(L) = 2^{α}, and thus a_{1} = α by Lemma B.1. Notice that (α, β + 1β + 1) ∈ S(e), where e is a basis of L with respect to which the matrix of L is T. So Lemma B.2 implies (a_{1}, a_{2}, a_{3}) = (α, β + 1, β + 1). Now assume that α > β + 1, then n(L) = 2^{β+1} = 2s(L) and a_{1} = β + 1. This implies a_{2} ≥ β + 1. Lemma B.2 implies then (a_{1}, a_{2}, a_{3}) = (β + 1, β + 1, α).

Proposition B.5. T = diag(ϵ_{1}2^{β_{1}}, ϵ_{2}2^{β_{2}}, ϵ_{3}2^{β_{3}}) with β_{1} ≤ β_{2} ≤ β_{3}.

(1) When β_{2} ≢ β_{1} mod 2, one has (a_{1}, a_{2}, a_{3}) = (β_{1}, β_{2}, β_{3} + 2).

(2) When β_{2} ≡ β_{1} mod 2 and β_{3} ≤ β_{2} + 1, one has (a_{1}, a_{2}, a_{3}) = (β_{1}, β_{2} + 1, β_{3} + 1).

(3) When β_{2} ≡ β_{1} mod 2, and β_{3} ≥ β_{2} + 2, one has

(a_{1}, a_{2}, a_{3}) = \{ \begin{matrix}(β_{1}, β_{2} + 1, β_{3} + 1) & if ϵ_{1} ≡ ϵ_{2} mod 4, \\ (β_{1}, β_{2} + 2, β_{3}) & if ϵ_{1} ≡ −ϵ_{2} mod 4. \end{matrix}

For clarity, we divide the proof into several lemmas.

Lemma B.6. When β_{2} ≢ β_{1} mod 2, one has (a_{1}, a_{2}, a_{3}) = (β_{1}, β_{2}, β_{3} + 2).

Proof. We first check that (β_{1}, β_{2}, β_{3}+2) ∈ S(ψ) for some basis ψ. When β_{3}−β_{1} is odd, set

(ψ_{1}, ψ_{2}, ψ_{3}) = (e_{1}, e_{2}, e_{3}) ( \begin{matrix}1 & 0 & 2^{\frac{β_{3}−β_{1}+1}{2}} α_{13} \\ 0 & 1 & 2^{\frac{β_{3}−β_{2}}{2}} β_{1}_{23} \\ 0 & 0 & 1 \end{matrix} )

Then q_{33} = Q(ψ_{3}) = 2^{β_{3}}(2ϵ_{1}α_{13}^{2} + ϵ_{2}α_{23}^{2} + ϵ_{3}). Clearly, there are α_{13}, α_{23} ∈ ℤ_{2} such that o(q_{33}) ≥ β_{3} + 2. For this choice, it is easy to see that (β_{1}, β_{2}, β_{3} + 2) ∈ S(ψ). When β_{3} − β_{1} is even, β_{3} − β_{2} has to be odd. The same argument with the role of β_{1} and β_{2} switched works. Now, by Lemma B.2, it suffices to prove a_{2} = β_{2}. Clearly, a_{2} ≥ β_{2}. If a_{2} > β_{2}, let ψ = (ψ_{1}, ψ_{2}, ψ_{3}) be an optimal basis with ψ = e(α_{ij}) and (α_{ij}) ∈ GL_{2}(ℤ_{2}). However

β_{1} < β_{2} < a_{j} ≤ o(Q(ψ_{j})) = o(ϵ_{1}2^{β_{1}}α_{1j}^{2} + ϵ_{2}2^{β_{2}}α_{2j}^{2} + ϵ_{3}2^{β_{3}}α_{3j}^{2})

for j = 2, 3 implies that α_{ij} ∈ 2ℤ_{2} for i = 1, 2 and j = 2, 3. This implies (α_{ij}) ∉ GL_{2}(ℤ_{2}).

Lemma B.7. When β_{2} ≡ β_{1} mod 2 and β_{3} ≤ β_{2} + 1, one has (a_{1}, a_{2}, a_{3}) = (β_{1}, β_{2} + 1, β_{3} + 1).

Proof. When β_{3} = β_{2}, take

(ψ_{1}, ψ_{2}, ψ_{3}) = (e_{1}, e_{2}, e_{3}) ( \begin{matrix}1 & 2^{\frac{β_{2}−β_{1}}{2}} & 2^{\frac{β_{2}−β_{1}}{2}} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) .

Then (β_{1}, β_{2} + 1, β_{2} + 1) ∈ S(ψ). By Lemma B.2, this has to be (a_{1}, a_{2}, a_{3}).

When β_{3} = β_{2} + 1 and ϵ_{1} + ϵ_{2} ≡ 2 mod 4, take

(ψ_{1}, ψ_{2}, ψ_{3}) = (e_{1}, e_{2}, e_{3}) ( \begin{matrix}1 & 2^{\frac{β_{2}−β_{1}}{2}} & 2^{\frac{β_{2}−β_{1}}{2}} \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ) .

Then (β_{1}, β_{2} + 1, β_{3} + 1) ∈ S(ψ). By Lemma B.2, this has to be (a_{1}, a_{2}, a_{3}).

Finally, when β_{3} = β_{2} + 1 and ϵ_{1} + ϵ_{2} ≡ 0 mod 4, take

(ψ_{1}, ψ_{2}, ψ_{3}) = (e_{1}, e_{2}, e_{3}) ( \begin{matrix}1 & 2^{\frac{β_{2}−β_{1}}{2}} & 2^{\frac{β_{2}−β_{1}}{2}} \\ 0 & 1 & 1 \\ 0 & 1 & 2 \end{matrix} ) .

Then (β_{1}, β_{2} + 1, β_{3} + 1) ∈ S(ψ). By Lemma B.2, this has to be (a_{1}, a_{2}, a_{3}).

Lemma B.8. Assume β_{2} ≡ β_{1} mod 2, and β_{3} ≥ β_{2} + 2. Then

(a_{1}, a_{2}, a_{3}) = \{ \begin{matrix}(β_{1}, β_{2} + 1, β_{3} + 1) & if ϵ_{1} ≡ ϵ_{2} mod 4, \\ (β_{1}, β_{2} + 2, β_{3}) & if ϵ_{1} ≡ −ϵ_{2} mod 4. \end{matrix}

Proof. First assume that ϵ_{1} ≡ −ϵ_{2} mod 4, i.e., ϵ_{1} + ϵ_{2} ≡ 0 mod 4. Set

(ψ_{1}, ψ_{2}, ψ_{3}) = (e_{1}, e_{2}, e_{3}) ( \begin{matrix}1 & 2^{\frac{β_{2}−β_{1}}{2}} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) .

Then (β_{1}, β_{2} + 2, β_{3}) ∈ S(ψ). We claim that a_{3} ≥ β_{3}. This, together with Lemma B.2, implies (a_{1}, a_{2}, a_{3}) = (β_{1}, β_{2} + 2, β_{3}).

To prove a_{3} ≥ β_{3}, we may assume that β_{2} < a_{2} ≤ β_{3}. Let ψ = e(α_{ij}) be an optimal basis. If α_{22} ∈ ℤ_{2}^{∗}, then (a_{1}, a_{2}, β_{3}) ∈ S({ψ_{1}, ψ_{2}, e_{3}}) and thus a_{3} ≥ β_{3}. If α_{23} ∈ ℤ_{2}^{∗}, then (a_{1}, a_{2}, β_{3}) ∈ S({ψ_{1}, ψ_{3}, e_{3}}) and thus a_{3} ≥ β_{3}. Indeed,

\frac{1}{2}(a_{1} + a_{2}) ≤ \frac{1}{2}(a_{1} + a_{3}) ≤ o((ψ_{1}, ψ_{3})),

\frac{1}{2}(a_{1} + β_{3}) ≤ o((ψ_{1}, e_{3})) = β_{3} + o(α_{31}),

and

\frac{1}{2}(a_{2} + β_{3}) ≤ β_{3} ≤ o((ψ_{3}, e_{3})) = β_{3} + o(α_{33}).

Clearly one has also

a_{1} ≤ o(Q(ψ_{1})), a_{2} ≤ a_{3} ≤ o(Q(ψ_{3})), , β_{3} = o(Q(e_{3})).

Finally, if α_{22}, α_{23} ∈ 2ℤ_{2}, then α_{12}, α_{13} ∈ 2ℤ_{2} due to the fact o(Q(ψ_{2})) ≥ a_{2} > β_{1} and o(Q(ψ_{3})) > β_{1}. This implies that (α_{ij}) ∉ GL_{2}(ℤ_{2}), a contradiction.

Now assume that ϵ_{1} ≡ ϵ_{2} mod 4, i.e., ϵ_{1} + ϵ_{2} ≡ 2 mod 4. Then one easily check (β_{1}, β_{2} + 1, β_{3}) ∈ S(ψ), where ψ is the basis just given. So a_{2} ≥ β_{2} + 1. In fact, we can do better. If β_{3} ≡ β_{1} mod 2, choose

(ψ_{1}, ψ_{2}, ψ_{3}) = (e_{1}, e_{2}, e_{3}) ( \begin{matrix}1 & 2^{\frac{β_{2}−β_{1}}{2}} & 2^{\frac{β_{3}−β_{1}}{2}} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) .

Then (β_{1}, β_{2} + 1, β_{3} + 1) ∈ S(ψ). If β_{3} ≢ β_{1} mod 2, choose

(ψ_{1}, ψ_{2}, ψ_{3}) = (e_{1}, e_{2}, e_{3}) ( \begin{matrix}1 & 2^{\frac{β_{2}−β_{1}}{2}} & 2^{\frac{β_{3}−β_{1}−1}{2}} \\ 0 & 1 & 2^{\frac{β_{3}−β_{2}−1}{2}} \\ 0 & 0 & 1 \end{matrix} ) .

Then (β_{1}, β_{2} + 1, β_{3} + 1) ∈ S(ψ). This, together with Lemma B.2, implies that (a_{1}, a_{2}, a_{3}) = (β_{1}, β_{2} + 1, β_{3} + 1) as claimed if we can prove a_{2} = β_{2} + 1.

If a_{2} ≥ β_{2} + 2, let ψ = e(α_{ij}) be an optimal basis for it. Then

Q(ψ_{2}) = ϵ_{1}2^{β_{1}}α_{12}^{2} + ϵ_{2}2^{β_{2}}α_{22}^{2} + ϵ_{3}2^{β_{3}}α_{32}^{2} ∈ 2^{β_{2}+2}ℤ_{2},

and so α_{12} = 2^{\frac{β_{2}−β_{1}}{2}} α̃_{12} with α̃_{12} ∈ ℤ_{2}, and

ϵ_{1}α̃_{12}^{2} + ϵ_{2}α_{22}^{2} + ϵ_{3}2^{β_{3}−β_{2}}α_{32}^{2} ≡ 0 mod 4.

So ϵ_{1}α̃_{12}^{2}+ϵ_{2}α_{22}^{2} ≡ 0 mod 4. Since ϵ_{1}+ϵ_{2} ≡ 2 mod 4, this implies α̃_{12}, α_{22} ∈ 2ℤ_{2}. For the same reason, o(Q(ψ_{3})) ≥ a_{3} ≥ a_{2} ≥ β_{2} + 2 implies α_{13}α_{23} ∈ 2ℤ_{2}. However, this implies (α_{ij}) ∉ GL_{2}(ℤ_{2}), a contradiction. Therefore a_{2} = β_{2} + 1, which proves the proposition.

Appendix C: Correction and reformulation in non-dyadic case.

In [Ya], we deal with p-adic local density in the cases n = 1 and n = 2 ≠ p. In this appendix, we correct some typos and reformulate some formulae. The notation α(X, T, S) there is the same as W (X, T, L) in this paper, where S is the matrix of L.

C1. We assume that n = 2 and p ≠ 2. [Ya, Propositions 8.6 and 8.7] is reformulated as Theorem 5.7. In fact, [Ya, Prop 8.7] has a minor problem when a = 0 which is corrected in the new version. A computer program for computing the local density when p ≠ 2 using [Ya, Theorem 7.1] was written by Song Guo and is available in my web page

http://www.math.wisc.edu/∼ thyang/programs.

There are a couple of typos in the formulaes leading to [Ya, Theorem 7.1]—the main formula. Using the notation in [Ya], the correct ones are the following.

When a = b,

I_{2,2} = v(a+1)δ^{+}(a+1)((\frac{−α_{1}α_{2}}{p})−p^{−1}) ∑_{0<k≤a} δ^{+}(k)v(k)p^{a+d(a+1)+d(k)}X^{a+1+k}.

When a < b, I_{2,3} missed a factor X^{k_{1}+k_{2}} and should be

I_{2,3} = (1 − p^{−1}) ∑_{0<k_{2}<a+1<k_{1}≤b} v(k_{1})v(k_{2})δ^{+}(k_{2})g(k_{1})p^{\frac{a+k_{1}}{2} +d(k_{1})+d(k_{2})}X^{k_{1}+k_{2}}.

Of course, the p-adic case with p ≠ 2 can also be computed in this paper’s notation. The relation with [Ya, Theorem 7.1] is as follows.

Proposition C1. One has for T = diag(α_{1}p^{a}, α_{2}p^{b}) ∈ A_{t}(a) with t = b − a ≥ 0,

W (X, T, L) = 1 + ∑_{0<c≤b+1} ∑_{0≤r<c} S_{r}(−c, T, L)X^{2c−r} + ∑_{0<c≤b+1} S_{≥c}(−c, T, L)X^{c}.

Here S_{r} and S_{≥c} are given as follows.

(1) When a = b, i.e., t = 0, one has

∑_{0<c≤a} S_{0}(−c, T, L)X^{2c} = (1 − p^{−1})I_{2,8},
S_{0}(−a − 1, T, L)X^{2a+2} = I_{2,6},
∑_{0<r<c≤a} S_{r}(−c, T, L)X^{2c−r} = (1 − p^{−1})I_{2,1},
∑_{0<r<c=a+1} S_{r}(−c, T, L)X^{2c−r} = (1 − p^{−1})I_{2,2},
∑_{0<c≤a} S_{≥c}(−c, T, L)X^{c} = I_{1,1},
S_{≥a+1}(−a − 1, T, L)X^{a+1} = I_{1,2}.

(2) When a < b, i.e., t > 0, one has

∑_{0<c≤a} S_{0}(−c, T, L)X^{2c} = (1 − p^{−1})I_{2,8},
S_{0}(−a − 1, T, L)X^{2a+2} = (1 − p^{−1})I_{2,7},
S_{r}(−c, T, L) = 0 if r < c − a − 1,
∑_{0<r<c≤a} S_{r}(−c, T, L)X^{2c−r} = (1 − p^{−1})I_{2,1},
∑_{0<r<c=a+1} S_{r}(−c, T, L)X^{2c−r} = (1 − p^{−1})I_{2,2},
∑_{a+1<c≤b} ∑_{c−a≤r<c} S_{r}(−c, T, L)X^{2c−r} = (1 − p^{−1})I_{2,3},
∑_{a+1<c≤b} ∑_{0<r≤c−a−1} S_{r}(−c, T, L)X^{2c−r} = (1 − p^{−1})I_{2,5},
∑_{c−a≤r<c=b+1} S_{r}(−c, T, L)X^{2c−r} = (1 − p^{−1})I_{2,4},
∑_{0<r≤b−a} S_{r}(−b − 1, T, L)X^{2b+2−r} = I_{2,6},
∑_{0<c≤a} S_{≥c}(−c, T, L)X^{c} = I_{1,1},
S_{≥a+1}(−a − 1, T, L)X^{c} = I_{1,2},
∑_{a+1<c≤b} S_{≥c}(−c, T, L)X^{c} = I_{1,3},
S_{≥b+1}(−b − 1, T, L)X^{b+1} = I_{1,4}.

C2. Now we assume n = 1 and p is any prime, including 2. In [KRY, Proposition 13.4], we reformulate [Ya, Proposition 8.3] uniformly in terms of local L-function. Here we reformulate it in terms of the Gross-Keating invariants uniformly. There are quite a few obvious typos in [Ya, Proposition 8.3]. We thank J.K. Yu for pointing them and other minor mistakes in [Ya] out.

Proposition C2. Let L^{0} = \{X = ( \begin{matrix}a & b \\ c & −a \end{matrix} ) ∈ M_{2}(ℤ_{p})\} with quadratic form

q(X) = −κ det X, where κ ∈ ℤ_{p}^{∗}. For t = αp^{a} ∈ ℚ_{p}^{∗} with α ∈ ℤ_{p}^{∗} and a ∈ ℤ_{≥0}. Let (0, a_{2}) be the Gross-Keating invariant of diag(−κ, t). Let μ_{p} = (κα, p).

(1) When a_{2} is odd, one has

\frac{W (X, t, L^{0})}{1 − p^{−2}X^{2}} = ∑_{0≤k≤ \frac{a_{2}−1}{2}}(p^{−1}X^{2})^{k}.

(2) When a_{2} is even, one has

W (X, t, L^{0}) = (1−p^{−2}X^{2}) ∑_{0≤k≤ \frac{a_{2}}{2}} (p^{−1}X^{2})^{k} −μ_{p}p^{− \frac{a_{2}}{2} −1}(X^{a_{2}+1} −μ_{p}p^{−1}X^{a_{2}+2}).

It might be interesting to notice that −κ is det L^{0}. In complement to the split case, we also record the corresponding formula for the ramified case L = L^{ra} (with the quadratic form twisted by −κ as above.

When a_{2} is odd, one has

W (X, t, L^{ra}) = ∑_{0≤k≤ \frac{a_{2}+1}{2}} δ_{p,i}(p^{−1}X^{2})^{k}.

Here and in the next formula,

δ_{p,i} = \{ \begin{matrix}1 & if i = 0, [\frac{a_{2}}{2} ] + 1, \\ 1 − p & if 1 ≤ i ≤ [\frac{a_{2}}{2} ]. \end{matrix}

When a_{2} is even, one has

W (T, t, L^{ra}) = ∑_{0≤k≤ \frac{a_{2}}{2}} δ_{p,i}(p^{−1}X^{2})^{k} − μ_{p}p^{− \frac{a_{2}}{2} −1}X^{a_{2}+1}.

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Department of Mathematics, University of Wisconsin, Madison, WI 53706

   E-mail address: thyang@math.wisc.edu