Equivariant Gauss sum of finite quadratic forms

TL;DR

This paper introduces an equivariant Gauss sum for finite quadratic forms, extending classical sums by incorporating group actions, with applications in modular form dimension formulas.

Contribution

It defines a new equivariant Gauss sum for finite quadratic forms and establishes arithmetic formulas for basic classes, linking to modular form theory.

Findings

Arithmetic formulas for basic quadratic forms

Invariant's role in vector-valued modular forms

Extension of classical Gauss sums to equivariant setting

Abstract

The classical quadratic Gauss sum can be thought of as an exponential sum attached to a quadratic form on a cyclic group. We introduce an equivariant version of Gauss sum for arbitrary finite quadratic forms, which is an exponential sum twisted by the action of the orthogonal group. We prove that simple arithmetic formulae hold for some basic classes of quadratic forms. In application, such invariant appears in the dimension formula for certain vector-valued modular forms.

Equations

$$G(A,\Gamma )=\sum _{[x]\in \Gamma \backslash A}\sum _{y\in \Gamma x}e((x,y)).$$

$$G(A,\mathrm{O}(A))=\prod _{p}G(A_p,\mathrm{O}(A_p)).$$

$$G(A,\mathrm{O}(A))=\{\begin{array}{ll}0,& p^k\equiv 3\text{mod}4,\\ {\left(\frac{a}{p}\right)}^k\sqrt{|A|},& p^k\equiv 1\text{mod}4.\end{array}$$

$$G(A,\mathrm{O}(A))=\{\begin{array}{ll}0,& p\equiv 3\text{mod}4,\\ 2{\left(\frac{2\delta }{p}\right)}^m\left(\frac{d(A)}{p}\right)\sqrt{|A|},& p\equiv 1\text{mod}4.\end{array}$$

$$G(A,\mathrm{O}(A))=G(A_1,\mathrm{O}(A_1))\cdot G(A_2,\mathrm{O}(A_2)).$$

$$G^{\prime }(A,\mathrm{O}(A))=\sum _{[x]\in \mathrm{O}(A)\backslash A}e(-q(x))\sum _{y\in \mathrm{O}(A)x}e((x,y)).$$

$$G^{\prime }(A,\mathrm{O}(A))=\frac{1}{2}\left(1+{\left(\frac{p}{3}\right)}^k\right){\left(\frac{2a}{p}\right)}^k\sqrt{|A|}\times \{\begin{array}{ll}1,& p^k\equiv 1\text{mod}4,\\ \sqrt{-1},& p^k\equiv 3\text{mod}4.\end{array}$$

$$G^{\prime }(A,\mathrm{O}(A))=\{\begin{array}{ll}0,& p\equiv 2\text{mod}3,\\ 2{\zeta }_{16}^{m^2(p-1{)}^2}{\left(\frac{2\delta }{p}\right)}^m\left(\frac{(-1{)}^{k}d(A)}{p}\right)\sqrt{|A|},& p\equiv 1\text{mod}3.\end{array}$$

$$G^{\prime }(A,\mathrm{O}(A))=G^{\prime }(A_1,\mathrm{O}(A_1))\cdot G^{\prime }(A_2,\mathrm{O}(A_2)).$$

$$(x,y)=q(x+y)-q(x)-q(y),x,y\in A,$$

$$G(A,\Gamma )=\sum _{[x]\in \Gamma \backslash A}\sum _{y\in \Gamma x}e((x,y)).$$

$$\sum _{y\in \Gamma x}e((\gamma x,y))=\sum _{y\in \Gamma x}e((x,{\gamma }^{-1}y))=\sum _{y\in \Gamma x}e((x,y)).$$

$$G^{\prime }(A,\Gamma )=\sum _{[x]\in \Gamma \backslash A}e(-q(x))\sum _{y\in \Gamma x}e((x,y)).$$

$$G(A,\{\mathrm{id}\})=\sum _{x\in A}e((x,x)),G^{\prime }(A,\{\mathrm{id}\})=\sum _{x\in A}e(q(x)),$$

$$⟨[x],[y]{⟩}_A=\sum _{y^{\prime }\in \mathrm{O}(A)y}e((x,y^{\prime })).$$

$$G(A,\mathrm{O}(A))=\sum _{[x]\in \mathrm{O}(A)\backslash A}⟨[x],[x]{⟩}_A,$$

$$G^{\prime }(A,\mathrm{O}(A))=\sum _{[x]\in \mathrm{O}(A)\backslash A}e(-q(x))⟨[x],[x]{⟩}_A.$$

$$G(A,\Gamma )=G(A_1,{\Gamma }_1)\cdot G(A_2,{\Gamma }_2),G^{\prime }(A,\Gamma )=G^{\prime }(A_1,{\Gamma }_1)\cdot G^{\prime }(A_2,{\Gamma }_2).$$

$$G(A,\Gamma )$$

$$G(A,\mathrm{O}(A))=\prod _{p}G(A_p,\mathrm{O}(A_p)),G^{\prime }(A,\mathrm{O}(A))=\prod _p{G}^{\prime }(A_p,\mathrm{O}(A_p)).$$

$$G(A,\mathrm{O}(A))=\{\begin{array}{ll}0,& p\equiv 3(4),k\text{odd},\\ {\left(\frac{a}{p}\right)}^k\sqrt{p^k},& \text{otherwise}.\end{array}$$

$$G(A,\mathrm{O}(A))=1+\frac{1}{2}\sum _{\stackrel{x\in \mathbb{Z}/p^k}{x\ne 0}}({\zeta }_{p^k}^{a{x}^2}+{\zeta }_{p^k}^{-a{x}^2})=\mathrm{Re}(G(A)).$$

$$G(A)=\left(\frac{a}{p^k}\right)\sqrt{p^k}\times \{\begin{array}{ll}1,& p^k\equiv 1(4),\\ \sqrt{-1},& p^k\equiv 3(4).\end{array}$$

$$A=\{0\}\cup (A_0\backslash \{0\})\cup \bigcup _{\mu \in {\mathbb{F}}_p^{\times }}{A}_{\mu }$$

$$x_{\mu }=u_1+(2^{-1}\mu )u_2\in U_{\mu }.$$

$$⟨\mu ,\lambda {⟩}_U=\sum _{x\in U_{\lambda }}{\zeta }_p^{(x_{\mu },x)}=\sum _{\nu \in {\mathbb{F}}_p^{\times }}{\zeta }_p^{2^{-1}(\mu \nu +\lambda {\nu }^{-1})}.$$

$$⟨\mu ,\lambda {⟩}_U=\{\begin{array}{ll}⟨U_{\mu }\backslash \{0\},U_{\lambda }⟩,& \lambda \ne 0,\\ ⟨U_{\mu }\backslash \{0\},U_0\backslash \{0\}⟩+1,& \lambda =0.\end{array}$$

$$G(U,\mathrm{O}(U))=\sum _{\mu \in {\mathbb{F}}_p}⟨\mu ,\mu {⟩}_U=\sum _{\nu \in {\mathbb{F}}_p^{\times }}\sum _{\mu \in {\mathbb{F}}_p}{\zeta }_p^{2^{-1}\mu (\nu +{\nu }^{-1})}.$$

$$G(A,\mathrm{O}(A))=\{\begin{array}{ll}0,& p\equiv 3(4),\\ 2{\left(\frac{2\delta }{p}\right)}^m\left(\frac{d(A)}{p}\right)\sqrt{|A|},& p\equiv 1(4).\end{array}$$

$$A_{\mu }=\underset{\lambda \in {\mathbb{F}}_p}{⨆}{U}_{\lambda }\times B_{\mu -\lambda }.$$

Full text

Supported by Grant-in-Aid for Scientific Research (S) 15H05738.

1. Introduction

In his study of the quadratic reciprocity law, Gauss introduced and evaluated the exponential sum $\sum_{x=0}^{p-1}e(ax^{2}/p)$ where $a\in{\mathbb{F}_{p}^{\times}}$, which is now called the Gauss sum. Here $e(z)={\rm exp}(2\pi iz)$ for $z\in{\mathbb{Q}/\mathbb{Z}}$. If we consider the ${\mathbb{Q}/\mathbb{Z}}$-valued quadratic form $q(x)=2^{-1}ax^{2}/p$ on the cyclic group $A={\mathbb{Z}}/p$ and the associated bilinear form $(x,y)=q(x+y)-q(x)-q(y)$, the Gauss sum can be written in the form $G(A)=\sum_{x\in A}e((x,x))$. Gauss evaluated such an exponential sum also for $A={\mathbb{Z}}/p^{k}$ (see [1]). If we consider $G(A)$ for general finite quadratic forms $A$, we have essentially no further problem concerning evaluation because the product formula $G(A_{1}\oplus A_{2})=G(A_{1})\cdot G(A_{2})$ holds and $A$ can be decomposed into cyclic forms as above and certain special forms on ${\mathbb{Z}}/2^{k}\oplus{\mathbb{Z}}/2^{k}$ (see [6]).

In this paper we introduce an equivariant version of $G(A)$ for a finite quadratic form $A=(A,q)$, which is a twist of $G(A)$ by the orthogonal group ${{\rm O}(A)}={\rm O}(A,q)$. Let $(\>,\>):A\times A\to{\mathbb{Q}/\mathbb{Z}}$ be the associated bilinear form as above. In general, for a subgroup $\Gamma$ of ${{\rm O}(A)}$, we define

[TABLE]

The classical Gauss sum $G(A)$ is the case $\Gamma=\{{\rm id}\}$. We are interested in the case $\Gamma={{\rm O}(A)}$, the motivation coming from certain modular forms (§6). If $A=\oplus_{p}A_{p}$ is the decomposition into $p$-parts, we have (Corollary 2.2)

[TABLE]

Hence we may restrict our attention to quadratic forms on $p$-groups.

Our main result is an arithmetic formula of $G(A,{{\rm O}(A)})$ for some basic quadratic forms, which has similar but different shape from that of $G(A)$. For simplicity we state the result only for $p>2$, referring to §4 for the case $p=2$. We write $(\frac{\cdot}{p})$ for the Legendre symbol.

Theorem 1.1** (§3).**

Let $p>2$.

$(1)$* Let $A$ be the symmetric bilinear form $(x,y)=axy/p^{k}$ on ${\mathbb{Z}}/p^{k}$ where $a\in{\mathbb{Z}_{p}^{\times}}$. Then*

[TABLE]

$(2)$* Let $A$ be a $p$-elementary form (quadratic space over ${\mathbb{F}_{p}}$) of dimension $m>1$. When $A$ contains an isotropic vector, then*

[TABLE]

Here $d(A)\in{\mathbb{F}_{p}^{\times}}/({\mathbb{F}_{p}^{\times}})^{2}$ is the discriminant of $A$, and $\delta\in{\mathbb{F}_{p}}$ is a square root of $-1$ which exists when $p\equiv 1$ mod $4$. When $A$ is the anisotropic plane, we have ${G(A,{\rm O}(A))}=(-1)^{(p+1)/2}p$.

$(3)$* Let $A=A_{1}\oplus A_{2}$ where $A_{1}\simeq{\mathbb{Z}}/p^{k}$ is a cyclic form as in $(1)$ with $k>1$ and $A_{2}$ is a $p$-elementary form as in $(2)$. Then*

[TABLE]

The product formula in $(3)$ is not trivial because ${{\rm O}(A)}$ does not preserve the decomposition $A=A_{1}\oplus A_{2}$ in general. When $p=2$, $k\leq 3$, this formula does not hold. The formula in $(2)$ already shows that the most naive product formula does not hold.

We were led to considering such an orbital exponential sum through the study of certain vector-valued modular forms. The Gauss sum ${G(A,{\rm O}(A))}$ appears in the dimension formula for them. We explain this in §6. In that formula also arises the following variant of ${G(A,{\rm O}(A))}$:

[TABLE]

We show that a similar arithmetic formula holds for ${G^{\prime}(A,{\rm O}(A))}$. We state the result only for $p>3$, referring to §5 for the case $p=2,3$.

Theorem 1.2** (§5).**

Let $p>3$.

(1) If $A$ is a cyclic form as in Theorem 1.1 (1), then

[TABLE]

(2) Let $A$ be a $p$-elementary form as in Theorem 1.1 (2). When $A$ contains an isotropic vector, we have

[TABLE]

Here $\zeta_{16}=e(1/16)$, $k=[m/2]$ and $\delta\in{\mathbb{F}_{p}}$ is a primitive $6$-th root of $1$ which exists when $p\equiv 1$ mod $3$. When $A$ is the anisotropic plane, we have ${G^{\prime}(A,{\rm O}(A))}=-\left(\frac{p}{3}\right)p$.

(3) If $A=A_{1}\oplus A_{2}$ is a quadratic form as in Theorem 1.1 (3), we have

[TABLE]

One finds that there are many cases of vanishing for both ${G(A,{\rm O}(A))}$ and ${G^{\prime}(A,{\rm O}(A))}$. One also finds that the absolute values of ${G(A,{\rm O}(A))}$ and ${G^{\prime}(A,{\rm O}(A))}$ are either [math] or $\sqrt{|A|}$ or $2\sqrt{|A|}$ in all cases we calculate.

Our evaluation of ${G(A,{\rm O}(A))}$ and ${G^{\prime}(A,{\rm O}(A))}$ is done by direct calculation based on the explicit description of the ${{\rm O}(A)}$-orbits. This seems to be difficult for more general quadratic forms. The author does not know if there is more systematic way to evaluate ${G(A,{\rm O}(A))}$ and ${G^{\prime}(A,{\rm O}(A))}$, and whether a uniform arithmetic formula holds for general finite quadratic forms.

2. Basic definitions

A finite quadratic form is a finite abelian group $A$ equipped with a ${\mathbb{Q}/\mathbb{Z}}$-valued quadratic form $q:A\to{\mathbb{Q}/\mathbb{Z}}$. This means that $q(ax)=a^{2}q(x)$ for $a\in{\mathbb{Z}}$ and $x\in A$, and that the ${\mathbb{Q}/\mathbb{Z}}$-valued pairing

[TABLE]

is symmetric bilinear. Unless stated otherwise, we assume that $(A,q)$ is nondegenerate, namely the bilinear form $(\,,\,)$ is nondegenerate. We often abbreviate $A=(A,q)$. The orthogonal direct sum of two finite quadratic forms $A_{1}$, $A_{2}$ is written as $A_{1}\oplus A_{2}$. A map $A\to A$ is called isometry if it is an isomorphism of abelian groups and preserves $q$. The group of isometries of $(A,q)$ is denoted by ${{\rm O}(A)}={\rm O}(A,q)$ and called the orthogonal group of $(A,q)$. Finite quadratic forms are also called finite quadratic modules in some literatures. A standard example is the discriminant form $L^{\vee}/L$ of an even lattice $L$, where the quadratic form $q$ is defined by $q(x+L)=(x,x)/2+{\mathbb{Z}}$ for $x\in L^{\vee}$. The associated bilinear form is the mod ${\mathbb{Z}}$ reduction of the pairing on the dual lattice $L^{\vee}$.

Let $A_{p}$ be the $p$-component of $A$. The canonical decomposition $A=\oplus_{p}A_{p}$ as an abelian group is automatically an orthogonal decomposition. Indecomposable quadratic forms on $p$-groups are classified by Wall [6]. When $p>2$, quadratic forms on $p$-groups can be reconstructed from the associated bilinear form by $q(x)=2^{-1}(x,x)$, as $2$ is invertible in ${\mathbb{Z}_{p}}$. We mainly consider the bilinear form $(\;,\;)$ in this case. The $2$-adic case is more subtle.

Let $\Gamma$ be a subgroup of ${{\rm O}(A)}$. We define the equivariant Gauss sum of $(A,\Gamma)$ as the exponential sum

[TABLE]

This is well-defined: if we use $\gamma x$ in place of $x$ where $\gamma\in\Gamma$, then

[TABLE]

Similarly, we define the equivariant Gauss sum of the second kind by

[TABLE]

When $\Gamma$ is trivial,

[TABLE]

are the classical quadratic Gauss sum. We will write $G(A)=G(A,\{{\rm id}\})$ and $G^{\prime}(A)=G^{\prime}(A,\{{\rm id}\})$. The evaluation of $G(A)$ and $G^{\prime}(A)$ is well-known: see [1]. We especially have the Milgram formula $G^{\prime}(A)=e(\sigma(A)/8)\sqrt{|A|}$, where $\sigma(A)\in{\mathbb{Z}}/8{\mathbb{Z}}$ is the signature of $A$.

Our object of study is the case $\Gamma={{\rm O}(A)}$. In the rest of this paper, for two ${{\rm O}(A)}$-orbits $[x],[y]\in{{\rm O}(A)}\backslash A$ we will write

[TABLE]

(We often omit $A$ in the subscript.) This sum does not depend on the choice of an element $x$ from $[x]$, as checked above. In particular, since $-{\rm id}\in{{\rm O}(A)}$, we see that $\langle[x],[y]\rangle$ is a real number. Note that $\langle[x],[y]\rangle\neq\langle[y],[x]\rangle$ in general. The equivariant Gauss sums can be written as

[TABLE]

[TABLE]

Then ${G(A,{\rm O}(A))}$ is a real algebraic integer.

We first localize the equivariant Gauss sum to each prime.

Lemma 2.1**.**

Let $A$ be of the form $A=A_{1}\oplus A_{2}$ and assume that $\Gamma\subset{{\rm O}(A)}$ can be decomposed as $\Gamma=\Gamma_{1}\times\Gamma_{2}$ such that $\Gamma_{i}\subset{\rm O}(A_{i})$. Then

[TABLE]

Proof.

For $x=(x_{1},x_{2})\in A_{1}\oplus A_{2}$ the orbit $\Gamma x$ is decomposed as $\Gamma x=(\Gamma_{1}x_{1})\times(\Gamma_{2}x_{2})$. Hence

[TABLE]

The case of $G^{\prime}(A,\Gamma)$ is similar. ∎

For $\Gamma={{\rm O}(A)}$ we have the canonical decomposition ${{\rm O}(A)}=\prod_{p}{\rm O}(A_{p})$. Hence we obtain

Corollary 2.2**.**

We have

[TABLE]

The evaluation of ${G(A,{\rm O}(A))}$ and ${G^{\prime}(A,{\rm O}(A))}$ is thus reduced to the case of quadratic forms on $p$-groups. For ${G(A,{\rm O}(A))}$ we study the case $p>2$ in §3 and the case $p=2$ in §4. We study ${G^{\prime}(A,{\rm O}(A))}$ in §5.

3. Nondyadic case

Let $p>2$. Let $A$ be an abelian $p$-group endowed with a ${\mathbb{Q}/\mathbb{Z}}$-valued symmetric bilinear form $(\>,\>)$. In this section we evaluate ${G(A,{\rm O}(A))}$ for the following quadratic forms: (1) cyclic forms, (2) $p$-elementary forms (quadratic spaces over ${\mathbb{F}_{p}}$), and (3) direct sum $A=A_{1}\oplus A_{2}$ where $A_{i}$ is as in $(i)$. We write $A(\varepsilon)$ for the scaling of $A$ by $\varepsilon\in{\mathbb{Z}}_{p}^{\times}$. The symmetric bilinear form on ${\mathbb{Z}}/p^{k}$ defined by $(x,y)=axy/p^{k}$, $a\in{\mathbb{Z}_{p}^{\times}}$, is denoted by $A_{p^{k},a}$. We write $\left(\frac{\cdot}{p}\right)$ for the Legendre symbol. We also write $\zeta_{p^{k}}=e(1/p^{k})$.

3.1. Cyclic forms

Proposition 3.1**.**

Let $A=A_{p^{k},a}$ with $p>2$. Then

[TABLE]

Proof.

The orthogonal group ${{\rm O}(A)}$ consists of $\pm{\rm id}$, and $-{\rm id}$ fixes no nonzero element. Hence

[TABLE]

By [1] Theorem 1.5.2 we have

[TABLE]

So ${\rm Re}(G(A))=0$ when $p^{k}\equiv 3\;(4)$, and ${\rm Re}(G(A))=\left(\frac{a}{p}\right)^{k}\sqrt{p^{k}}$ otherwise. ∎

3.2. $p$-elementary forms

Let $A$ be a symmetric form on a $p$-elementary group with $p>2$. We can naturally view $A$ as a quadratic space over ${\mathbb{F}_{p}}$. By Witt’s extension theorem ([4] Theorem 2.44), two nonzero vectors of $A$ are ${{\rm O}(A)}$-equivalent if and only if they have the same norm. For $\mu\in{\mathbb{F}_{p}}$ let $A_{\mu}$ be the subset of $A$ of vectors $x$ such that $(x,x)=\mu$. Then

[TABLE]

is the ${{\rm O}(A)}$-orbit decomposition of $A$.

We first consider the case $A$ has a nonzero isotropic vector. In that case $A$ contains as a direct summand the hyperbolic plane $U$, namely the symmetric form on ${\mathbb{F}_{p}}\oplus{\mathbb{F}_{p}}$ given by the Gram matrix $\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$. As the first step we calculate the case $A=U$.

Lemma 3.2**.**

*We have $G(U,{\rm O}(U))=(1+(-1)^{(p-1)/2})p$. *

Proof.

Let $u_{1},u_{2}$ be the standard basis of $U$. Then $U_{0}={\mathbb{F}_{p}}u_{1}\cup{\mathbb{F}_{p}}u_{2}$, and for $\mu\neq 0$ we have $U_{\mu}=\{\nu u_{1}+\nu^{-1}(2^{-1}\mu)u_{2}\;|\;\nu\in{\mathbb{F}_{p}^{\times}}\}$. As a reference vector in $U_{\mu}$ for $\mu\in{\mathbb{F}_{p}}$ we take

[TABLE]

We write for $\mu,\lambda\in{\mathbb{F}_{p}}$

[TABLE]

We have

[TABLE]

The second equality of (3.1) holds even when $\lambda=0$ (both sides equal to $-1$ when $\mu\neq 0$, and to $p-1$ when $\mu=0$). This expression shows that $\langle\mu,\lambda\rangle_{U}=\langle\lambda,\mu\rangle_{U}$.

By definition we have

[TABLE]

When $p\equiv 1\;(4)$, the equation $\nu+\nu^{-1}=0$ has two solutions, namely the square roots of $-1$, hence $G(U,{\rm O}(U))=2p$. When $p\equiv 3\;(4)$, we have $\nu+\nu^{-1}\neq 0$ for any $\nu\in{\mathbb{F}_{p}^{\times}}$. Thus $G(U,{\rm O}(U))=0$ in this case. ∎

We consider the general case.

Proposition 3.3**.**

Let $A$ be an $m$-dimensional quadratic space over ${\mathbb{F}_{p}}$ with $p>2$ that contains a nonzero isotropic vector. Let $d(A)\in{\mathbb{F}_{p}^{\times}}/({\mathbb{F}_{p}^{\times}})^{2}$ be the discriminant of $A$. When $p\equiv 1\;(4)$, we choose $\delta\in{\mathbb{F}_{p}}$ with $\delta^{2}=-1$. Then

[TABLE]

Proof.

We keep the notation in the proof of Lemma 3.2. We choose a splitting $A=U\oplus B$, which gives the partition

[TABLE]

As a reference vector in $A_{\mu}$ we use $(x_{\mu},0)\in U_{\mu}\times B_{0}$. Then

[TABLE]

by this partition. The formula of $|B_{\alpha}|$ can be found in [4] §2.8. It depends on the parity of $m={\dim}(A)$.

(1) Let $m=2k$ be even. We write $\delta_{A}=\left(\frac{(-1)^{k}d(A)}{p}\right)$. Then $\delta_{A}=1$ if and only if $B\simeq U^{k-1}$. By [4] Theorem 2.59 we have

[TABLE]

In particular, $|B_{\alpha}|$ for $\alpha\neq 0$ is independent of $\alpha$ and hence

[TABLE]

Since

[TABLE]

we have

[TABLE]

and the proposition follows from Lemma 3.2.

(2) Let $m=2k+1$ be odd. We put $\tilde{d}(A)=(-1)^{k}d(A)$. Then $B\simeq U^{k-1}\oplus\langle\tilde{d}(A)\rangle$. By [4] Theorem 2.60 we have for $\alpha\in{\mathbb{F}_{p}}$

[TABLE]

This depends on the class of $\alpha$ in ${\mathbb{F}_{p}}/({\mathbb{F}_{p}^{\times}})^{2}=\{\bar{0},\bar{1},\bar{\varepsilon}\}$ where $\varepsilon\in{\mathbb{F}_{p}^{\times}}$ is a nonsquare. If we write

[TABLE]

[TABLE]

we obtain from (3.3) and (3.5)

[TABLE]

We shall show that

[TABLE]

from which the proposition follows. In case $p\equiv 3\;(4)$, since $-1$ is nonsquare, switching $\mu$ and $\lambda$ gives

[TABLE]

In case $p\equiv 1\;(4)$, writing $\mu=\lambda+\alpha^{2}$ for $(\mu,\lambda)\in S$, we have

[TABLE]

The square roots $\pm\delta$ of $-1$ are the solutions of $\nu+\nu^{-1}=0$. We thus obtain

[TABLE]

∎

It remains to consider the anisotropic case. Since the $1$-dimensional case is covered in §3.1, what remains is the anisotropic plane.

Proposition 3.4**.**

Let $A$ be the anisotropic plane over ${\mathbb{F}_{p}}$ with $p>2$. Then ${G(A,{\rm O}(A))}=(-1)^{(p+1)/2}p$.

Proof.

Fix a nonsquare $\varepsilon\in{\mathbb{F}_{p}^{\times}}$. Since the scaling $A(\varepsilon)$ is isometric to $A$ itself, there exists an isomorphism $j:A\to A$ of abelian groups such that $(j(x),j(y))=\varepsilon(x,y)$ for every $x,y\in A$. For $\lambda\in{\mathbb{F}_{p}^{\times}}$ we write

[TABLE]

and set

[TABLE]

$\tilde{A}$ is a double covering of $A\backslash\{0\}$, with the covering transformation $(\lambda,x)\mapsto(-\lambda,x)$ which switches $A_{\lambda}^{\pm}$ and $A_{-\lambda}^{\pm}$. On $\tilde{A}$ we have an ${\mathbb{F}_{p}^{\times}}$-action defined by $\alpha\cdot(\lambda,x)=(\alpha\lambda,\alpha x)$. We can choose a reference point $(\lambda,x_{\lambda}^{\pm})\in A_{\lambda}^{\pm}$ for each $\lambda$ so that $x_{\alpha\lambda}^{\pm}=\alpha x_{\lambda}^{\pm}$ and $x_{\lambda}^{-}=j(x_{\lambda}^{+})$. If we consider the mapping

[TABLE]

we have

[TABLE]

Claim 3.5*.*

The fibers of $\varphi$ over ${\mathbb{F}_{p}^{\times}}\subset{\mathbb{F}_{p}}$ have constant cardinality.

Proof.

We let ${\mathbb{F}_{p}^{\times}}$ act on ${\mathbb{F}_{p}}$ by weight $2$, i.e., $\alpha(t)=\alpha^{2}t$. By our choice of $x_{\lambda}^{\pm}$, the map $\varphi$ is ${\mathbb{F}_{p}^{\times}}$-equivariant. Hence its fibers have constant cardinality over $({\mathbb{F}_{p}^{\times}})^{2}$ and over $\varepsilon({\mathbb{F}_{p}^{\times}})^{2}$ respectively. We shall show that

[TABLE]

which would then imply $|\varphi^{-1}(({\mathbb{F}_{p}^{\times}})^{2})|=|\varphi^{-1}(\varepsilon({\mathbb{F}_{p}^{\times}})^{2})|$. We consider the map $j:A^{+}\to A^{-}$ defined by $(\lambda,x)\to(\lambda,j(x))$. Since $x_{\lambda}^{-}=j(x_{\lambda}^{+})$, we have $\varphi(j(\lambda,x))=\varepsilon\cdot\varphi(\lambda,x)$ for $(\lambda,x)\in A^{+}$. Hence

[TABLE]

[TABLE]

[TABLE]

Since $j:A^{+}\to A^{-}$ is bijective, the three inclusions are all equality. ∎

By this claim we have

[TABLE]

In the last equality we used

[TABLE]

We are thus reduced to calculating $|\varphi^{-1}(0)|$.

For each $\lambda\neq 0$, $\varphi^{-1}(0)\cap A_{\lambda}^{\pm}$ is identified with the set of vectors in $(x_{\lambda}^{\pm})^{\perp}\cap A$ having the same norm as $x_{\lambda}^{\pm}$. This set is non-empty if and only if $d(A)=-\varepsilon$ is a square, i.e., $\left(\frac{-1}{p}\right)=-1$. In that case it consists of two elements. It follows that

[TABLE]

Therefore ${G(A,{\rm O}(A))}=-\left(\frac{-1}{p}\right)p$. ∎

The above method can be extended to general $p$-elementary forms of even dimension. The result agrees with Proposition 3.3, of course.

3.3. Product formula

Let $p>2$. Let $A$ be the direct sum $A=A_{p^{k},a}\oplus B$ where $k>1$ and $B$ is $p$-elementary. We show that a product formula holds for ${G(A,{\rm O}(A))}$. This is not trivial as ${{\rm O}(A)}$ does not preserve the decomposition in general. We first describe the orthogonal group ${{\rm O}(A)}$, then classify the ${{\rm O}(A)}$-orbits, and finally calculate the Gauss sum.

Let $e$ be the standard generator of $A_{p^{k},a}\simeq{\mathbb{Z}}/p^{k}$. Using $e$, we can express an isomorphism $A\to A$ of abelian groups in the matrix form

[TABLE]

We define a subgroup of ${{\rm O}(A)}$ isomorphic to $B\rtimes{{\rm O}(B)}$ as follows. For $(v,g)\in B\rtimes{{\rm O}(B)}$ we define $x_{v}\in({\mathbb{Z}}/p^{k})^{\times}$ and $f_{v,g}:B\to{\mathbb{Z}}/p$ by

[TABLE]

[TABLE]

where we view $p(v,v),p(g^{-1}(v),?)\in{\mathbb{Z}}/p$. We define $\gamma_{v,g}:A\to A$ by

[TABLE]

It is easy to check that $\gamma_{v,g}$ is an isometry and that the group

[TABLE]

is isomorphic to the semiproduct $B\rtimes{{\rm O}(B)}$.

Proposition 3.6**.**

We have ${{\rm O}(A)}=\{\pm 1\}\times\Gamma$.

Proof.

The isometry condition for the matrix (3.7) is

[TABLE]

[TABLE]

[TABLE]

The solutions of (3.8) are explicitly given by $x=\pm x_{v}$. Then $f$ is uniquely determined from $v,x$ and $g$ by (3.9). ∎

We consider the following subsets of $A$:

[TABLE]

We have $A=A_{0}\cup A_{1}$. Each $A_{0}$, $A_{1}$ is preserved by ${{\rm O}(A)}$.

Lemma 3.7**.**

(1) The subset $(({\mathbb{Z}}/p^{k})^{\times}e)\times\{0\}$ of $A_{0}$ is a representative for $\Gamma\backslash A_{0}$. For $x\in({\mathbb{Z}}/p^{k})^{\times}$ we have

[TABLE]

(2) The $\Gamma$-orbit of $xe+w\in A_{1}$, where $x\in p({\mathbb{Z}}/p^{k-1})$ and $w\in B$, is

[TABLE]

Proof.

(1) Let $x\in({\mathbb{Z}}/p^{k})^{\times}$. The description (3.10) of the orbit is apparent. It implies that $x^{\prime}e\not\in\Gamma(xe)$ if $x\neq x^{\prime}\in({\mathbb{Z}}/p^{k})^{\times}$. Since $|\Gamma(xe)|=|B|$, then $\Gamma(({\mathbb{Z}}/p^{k})^{\times}e)=A_{0}$. The assertion (2) follows from

[TABLE]

for $x\in p({\mathbb{Z}}/p^{k-1})$ and $w^{\prime}\in{{\rm O}(B)}w\subset B$. ∎

Before calculating ${G(A,{\rm O}(A))}$, let us prepare a general formula.

Lemma 3.8**.**

Let $p>2$, $a\in{\mathbb{Z}_{p}^{\times}}$ and suppose that $k\geq 2$. Then

[TABLE]

Proof.

If we write $S=(({\mathbb{Z}}/p^{k})^{\times})^{2}\subset{\mathbb{Z}}/p^{k}$, the right hand side is written as $2\sum_{y\in S}\zeta_{p^{k}}^{ay}$. By the local square theorem, we have the additive action of $p({\mathbb{Z}}/p^{k-1})$ on $S$. On each orbit, say $S_{i}=y_{i}+p({\mathbb{Z}}/p^{k-1})$, we have

[TABLE]

Since $\zeta_{p^{k-1}}^{a}\neq 1$, this sum is equal to [math]. ∎

We are now ready to calculate the Gauss sum.

Proposition 3.9**.**

Let $A$ be the direct sum $A=C\oplus B$ where $C=A_{p^{k},a}$ with $p>2$, $k>1$ and $B$ is $p$-elementary. Then

[TABLE]

Proof.

We take the sum over each stratum $A_{0},A_{1}$. We first consider $A_{0}$. By Lemma 3.7 (1), we can take $(({\mathbb{Z}}/p^{k})^{\times}/-1)e$ as reference points of ${{\rm O}(A)}\backslash A_{0}$. For $x\in({\mathbb{Z}}/p^{k})^{\times}$ we have by (3.10)

[TABLE]

Hence

[TABLE]

by Lemma 3.8. We next consider the stratum $A_{1}$. For $xe+w\in A_{1}$ we have

[TABLE]

by Lemma 3.7 (2), where $[xe]\in C/-1$ and $[w]\in B/{{\rm O}(B)}$. Since $\langle[x^{\prime}e],[x^{\prime}e]\rangle_{C}=\langle[xe],[xe]\rangle_{C}$ for $x^{\prime}\in x+p^{k-1}({\mathbb{Z}}/p)$, we have

[TABLE]

Here the second equality is a consequence of Lemma 3.8. ∎

4. $2$-adic case

In this section we study quadratic forms $A=(A,q)$ on $2$-groups. Let $(\>,\>):A\times A\to{\mathbb{Q}/\mathbb{Z}}$ be the associated bilinear form. For $x\in A$, $(x,x)=2q(x)$ is well-defined as an element of ${\mathbb{Q}}/2{\mathbb{Z}}$, not just of ${\mathbb{Q}/\mathbb{Z}}$. We work with this refined symmetric bilinear form, from which the quadratic form $q$ can be recovered. This is one of the differences with the case $p>2$. Classification of quadratic forms on $2$-groups is more complicated ([6]). Furthermore, the local square theorem is now $({\mathbb{Z}}_{2}^{\times})^{2}=1+8{\mathbb{Z}}_{2}$, hence ${\mathbb{Z}}_{2}^{\times}/({\mathbb{Z}}_{2}^{\times})^{2}=({\mathbb{Z}}/8)^{\times}$. Note also that the square $x^{2}$ of $x\in{\mathbb{Z}}/2^{k}$ can be defined as an element of ${\mathbb{Z}}/2^{k+1}$ and hence $x^{2}/2$ is well-defined as an element of $\frac{1}{2}{\mathbb{Z}}/2^{k}$.

For an odd number $a$ we write $A_{2^{k},a}$ for the quadratic form $(x,y)=axy/2^{k}$ on ${\mathbb{Z}}/2^{k}$. ($ax^{2}/2^{k}$ is considered as an element of $2^{-k}{\mathbb{Z}}/2{\mathbb{Z}}$ as noted above.) We denote by $U,V$ the quadratic forms on $({\mathbb{Z}}/2)^{\oplus 2}$ expressed by the Gram matrices

[TABLE]

respectively.

4.1. Cyclic forms

Proposition 4.1**.**

Let $A=A_{2^{k},a}$. When $k=1$, we have ${G(A,{\rm O}(A))}=0$. When $k>1$, we have

[TABLE]

Proof.

The case $k=1$ is clear. Let $k>1$. We have ${{\rm O}(A)}=\{\pm 1\}$, and $-1$ has two fixed points (the zero element and the unique order $2$ element). Then

[TABLE]

We have $G(A)=\left(\frac{2}{a}\right)^{k}(1+\sqrt{-1}^{a})\sqrt{2^{k}}$ by [1] Proposition 1.5.3. ∎

4.2. $2$-elementary forms

Let $A$ be a quadratic form on a 2-elementary group. By the nondegeneracy there exists a unique element $x_{A}\in A$ such that $(x,x_{A})=(x,x)$ mod ${\mathbb{Z}}$ for every $x\in A$. This element $x_{A}$ is called the characteristic element of $A$. We denote $A_{+}=x_{A}^{\perp}$, the group of elements of $A$ whose norm is in ${\mathbb{Z}}$. When $x_{A}=0$, namely $A=A_{+}$, then $A$ is said to be special. In that case we have $A\simeq U^{m}$ or $A\simeq U^{m-1}\oplus V$. We set $\delta_{A}$ by $\delta_{A}=1$ in the first case and $\delta_{A}=-1$ in the second case. When $A$ is non-special, it is isometric to $A^{\prime}\oplus A^{\prime\prime}$ where $A^{\prime}$ is special and $A^{\prime\prime}\simeq(A_{2,1})^{i}\oplus(A_{2,-1})^{j}$ with $1\leq i+j\leq 2$ (see [6]). When $i+j=1$, $x_{A}$ is the generator of $A^{\prime\prime}$. When $i+j=2$, $x_{A}$ is the unique nonzero element of $A^{\prime\prime}$ with $(x_{A},x_{A})\in{\mathbb{Z}}$, and generates the radical of $A_{+}$.

By definition the characteristic element $x_{A}$ is invariant under the action of ${{\rm O}(A)}$. Except this, description of ${{\rm O}(A)}$-orbits is the same as the nondyadic case. We use the following $2$-adic versions of Witt cancelation.

Lemma 4.2**.**

$(1)$* Let $A$ be a $2$-elementary form and $A_{1},A_{2}\subset A$ be nondegenerate special forms. If $A_{1}\simeq A_{2}$, then $A_{1}^{\perp}\simeq A_{2}^{\perp}$. In particular, there exists an isometry $A\to A$ which maps $A_{1}$ to $A_{2}$.*

$(2)$* Let $A_{1},A_{2}$ be non-special $2$-elementary forms and let $a=1$ or $-1$. If $A_{1}\oplus A_{2,a}\simeq A_{2}\oplus A_{2,a}$, then $A_{1}\simeq A_{2}$.*

Proof.

The assertion (1) is proved in [5] Proposition 1.5. We consider (2) in case ${\dim}(A_{1})={\dim}(A_{2})$ is odd. Using (1) for $U^{k}\hookrightarrow A_{i}$, we may assume that $A_{1}$ and $A_{2}$ are one of

[TABLE]

When $A_{1}\not\simeq A_{2}$, we see that $A_{1}\oplus A_{2,a}$ and $A_{2}\oplus A_{2,a}$ have different representation numbers by direct calculation, hence cannot be isometric. The case ${\dim}(A_{1})={\dim}(A_{2})$ even is similar. ∎

Proposition 4.3**.**

Let $A$ be a finite quadratic form on a 2-elementary group and $x,y\in A$ be nonzero, non-characteristic elements. The elements $x$ and $y$ are ${{\rm O}(A)}$-equivalent if and only if they have the same norm in ${\mathbb{Q}}/2{\mathbb{Z}}$.

Proof.

It suffices to prove the ”if” direction. We first consider the case $(x,x)\not\in{\mathbb{Z}}$. Since $\langle x\rangle$ and $\langle y\rangle$ are nondegenerate, we have the orthogonal decomposition $A=\langle x\rangle\oplus x^{\perp}=\langle y\rangle\oplus y^{\perp}$. Since $x$ and $y$ are non-characteristic, $x^{\perp}$ and $y^{\perp}$ are non-special. Then $x^{\perp}\simeq y^{\perp}$ by Lemma 4.2 (2), so $x$ and $y$ are ${{\rm O}(A)}$-equivalent.

Next let $(x,x)\in{\mathbb{Z}}$. We assume $A_{+}/{\rm rad}(A_{+})\not\simeq V$: this exceptional case can be treated directly. We shall show that there exists an embedding $U\hookrightarrow A$ whose image contains $x$. The same also applies to $y$, and then our assertion follows from Lemma 4.2 (1). Now since $x$ is nonzero and non-characteristic, it is not contained in the radical of $A_{+}$. Hence we can find an element $x^{\prime}\in A_{+}$ such that $(x,x^{\prime})=1/2$. If either $x$ or $x^{\prime}$ has norm [math], then $\langle x,x^{\prime}\rangle\simeq U$. If both $x$ and $x^{\prime}$ have norm $1$, then $\langle x,x^{\prime}\rangle\simeq V$. By our assumption $\langle x,x^{\prime}\rangle^{\perp}\cap A_{+}$ contains $U$ or $V$, so it contains a norm $1$ vector $x^{\prime\prime}$. Then $\langle x,x^{\prime}+x^{\prime\prime}\rangle\simeq U$. ∎

For $\mu\in 2^{-1}{\mathbb{Z}}/2{\mathbb{Z}}$ we write $A_{\mu}\subset A$ for the subset of vectors of norm $\mu$. Then the ${{\rm O}(A)}$-orbit decomposition of $A$ is

[TABLE]

We first consider the case $A$ contains $U$.

Proposition 4.4**.**

Let $A$ be a 2-elementary form that contains $U$. Then

[TABLE]

Proof.

We write $A=U\oplus B$ and let $u_{1},u_{2}$ be the standard basis of $U$. In case $A$ is non-special, we can also write $A=A_{2,1}\oplus A_{2,-1}\oplus C$ because $U\oplus A_{2,\pm 1}\simeq(A_{2,\pm 1})^{2}\oplus A_{2,\mp 1}$. We denote by $u_{\pm}$ the generator of $A_{2,\pm 1}$. As a reference vector in $A_{\mu}$ we can take $x_{\mu}=u_{1}$, $u_{1}+u_{2}$, $u_{\pm}$ for $\mu=0,1,\pm 1/2$ respectively. We have the simple expression

[TABLE]

because $(x_{\lambda},x_{A})=(x_{A},x_{A})$ for $\lambda=(x_{A},x_{A})$. Since we have the partitions

[TABLE]

[TABLE]

we obtain

[TABLE]

[TABLE]

Similarly, we have

[TABLE]

[TABLE]

and hence

[TABLE]

Hence ${G(A,{\rm O}(A))}=0$ when $A$ is non-special, and ${G(A,{\rm O}(A))}=|A_{0}|-|A_{1}|$ when $A$ is special. In the latter case we have $|A_{0}|-|A_{1}|=\delta_{A}\sqrt{|A|}$ by induction on ${\dim}(A)/2$. ∎

The remaining cases are covered by the following.

Proposition 4.5**.**

We have

[TABLE]

Proof.

This can be checked directly. ∎

4.3. Product formula

Let $A$ be the direct sum $A=A_{2^{k},a}\oplus B$ such that $k\geq 2$ and $B$ is $2$-elementary. We show that a product formula like Proposition 3.9 holds when $k\geq 4$, but not always when $k\leq 3$. As in (3.7), we express an isomorphism $A\to A$ in the form $\begin{pmatrix}x&2^{k-1}f\\ v&g\end{pmatrix}$. The isometry condition is

[TABLE]

[TABLE]

[TABLE]

where $2(v,v)\in{\mathbb{Z}}/4$ and $2(v,g(?)),f(?)\in{\mathbb{Z}}/2$. The condition (4.6) becomes $g\in{\rm O}(B)$ when $k\geq 3$. When $k\geq 4$, the equation (4.4) always has two solutions, one of which is given by

[TABLE]

In case $k=3$, by the local square theorem, (4.4) has a solution if and only if $(v,v)\in{\mathbb{Z}}$. In that case, a solution is given by $x_{v}=1+4(v,v)$.

Let $k\geq 4$. For $(v,g)\in B\rtimes{{\rm O}(B)}$ we define the isometry $\gamma_{v,g}$ of $A$ by $\gamma_{v,g}=\begin{pmatrix}x_{v}&2^{k-1}f_{v,g}\\ v&g\end{pmatrix}$ where $f_{v,g}$ is defined from $v$ and $g$ by (4.5). We put

[TABLE]

We have ${{\rm O}(A)}=\{\pm 1\}\times\Gamma$ by the same argument as in Proposition 3.6, and the assertion of Lemma 3.7 still holds. Also Lemma 3.8 holds in $k\geq 4$ by the local square theorem in $p=2$:

[TABLE]

Hence the argument of §3.3 works in $p=2$, $k\geq 4$:

Proposition 4.6**.**

Let $A$ be the direct sum $A=C\oplus B$ such that $C=A_{2^{k},a}$ with $k\geq 4$ and $B$ is $2$-elementary. Then

[TABLE]

We consider the case $k=2,3$.

Proposition 4.7**.**

Let $A=C\oplus B$ where $C=A_{8,a}$ and $B$ is $2$-elementary. Then

[TABLE]

By §4.2, the product formula does not hold when $B=V$, $(A_{2,a})^{2}$, $(A_{2,a})^{4}$.

Proof.

We replace (4.7) by the group $\Gamma_{+}=B_{+}\rtimes{\rm O}(B)$ where $B_{+}\subset B$ is the subgroup of elements of norm $\in{\mathbb{Z}}$. Then ${{\rm O}(A)}=\{\pm 1\}\times\Gamma_{+}$. Let

[TABLE]

which are preserved by ${{\rm O}(A)}$. If $xe+w\in A_{1}$, then

[TABLE]

where $2(B_{+},w)$ is a subgroup of ${\mathbb{Z}}/2{\mathbb{Z}}$. Hence we have

[TABLE]

where $\sum_{2C/-1}\langle[z],[z]\rangle_{C}=0$ by direct calculation. For $A_{0+}$, the coset ${{\rm O}(A)}\backslash A_{0+}$ consists of $[e]$ and $[3e]$. Since

[TABLE]

for $x=1,3$, we obtain

[TABLE]

As shown in the proof of Proposition 4.4, we have $|B_{0}|-|B_{1}|=G(B,{\rm O}(B))$ when $B$ is special and $B\neq V$, which proves the proposition in this case. The case $B=V$ is immediate. Suppose that $B$ is non-special. If we choose $w_{0}\in B\backslash B_{+}$, then ${{\rm O}(A)}\backslash A_{0-}$ consists of $[e+w_{0}]$ and $[3e+w_{0}]$. For $x=1,3$ we have

[TABLE]

It follows that

[TABLE]

Therefore ${G(A,{\rm O}(A))}=0$ in this case. ∎

Proposition 4.8**.**

Let $A=A_{4,a}\oplus B$ where $B$ is $2$-elementary. Then

[TABLE]

Proof.

If $x\in A$ is of order $4$, then for all $y\in{{\rm O}(A)}x$ we have $(x,y)\equiv\pm 1/4$ mod ${\mathbb{Z}}$ and $y\neq-y$. Then

[TABLE]

hence ${G(A,{\rm O}(A))}$ reduces to

[TABLE]

When $B=A_{2,1}\oplus A_{2,-1}$, the assertion follows by direct calculation. When $B$ is anisotropic, the decomposition $A=A_{4,a}\oplus B$ is canonical, so the assertion follows from Lemma 2.1 and the results of §4.1 and §4.2. We consider the general case $B=B^{\prime}\oplus B^{\prime\prime}$ where $B^{\prime}\neq 0$ is special and $B^{\prime\prime}=(A_{2,1})^{i}\oplus(A_{2,-1})^{j}$ with $0\leq i+j\leq 2$. We will show that ${G(A,{\rm O}(A))}=0$ in this case. Let $x_{B}\in B^{\prime\prime}$ be the characteristic element of $B$. Let $e\in A_{4,a}$ be a generator. We have the exceptional elements [math], $2e$, $x_{B}$, $2e+x_{B}$ fixed by ${{\rm O}(A)}$. ($x_{B}$ can be [math].) Extending Lemma 4.3, we can show that other two elements of order $2$ in $A$ are ${{\rm O}(A)}$-equivalent if and only if they have the same norm. The possibilities for $x_{B}$ are: (1) $x_{B}=0$, (2) $x_{B}\neq 0$, $(x_{B},x_{B})=0$, (3) $(x_{B},x_{B})=1$, and (4) $(x_{B},x_{B})=\pm 1/2$.

We consider only the case (2): other cases can be calculated similarly. If $x\in A$ has norm [math] with $x\neq 0,x_{B}$, then

[TABLE]

Hence

[TABLE]

Similarly, for other orbits $[x]$, we have

[TABLE]

The contribution from the exceptional orbits [math], $2e$, $x_{B}$, $2e+x_{B}$ is $1+1+1+1=4$, hence ${G(A,{\rm O}(A))}=0$. ∎

5. Equivariant Gauss sum of the second kind

In this section we evaluate the equivariant Gauss sum ${G^{\prime}(A,{\rm O}(A))}$ of the second kind for quadratic forms $A$ as in the previous sections. Some part is similar to the case of ${G(A,{\rm O}(A))}$, so we omit the detail there.

5.1. Cyclic forms

Proposition 5.1**.**

Let $A=A_{p^{k},a}$ with $p\geq 2$ and $a\in{\mathbb{Z}_{p}^{\times}}/({\mathbb{Z}_{p}^{\times}})^{2}$.

$(1)$* If $p>3$, then*

[TABLE]

$(2)$* For $p=2,3$ we have*

[TABLE]

Proof.

When $p>2$, we have

[TABLE]

When $p>3$, the assertion follows from the formula of $G(A(\varepsilon))$ ([1] Theorem 1.5.2) and the quadratic reciprocity $(\frac{-3}{p})=(\frac{p}{3})$. When $p=3$, $A(-6)\simeq A(3)$ is degenerate with kernel $3^{k-1}A\simeq{\mathbb{Z}}/3$, so we have

[TABLE]

When $p=2$, $k\geq 3$, we have

[TABLE]

The case $p=2$, $k\leq 2$ can be calculated directly. ∎

5.2. $p$-elementary forms

We first consider the case $p>2$. We begin with the hyperbolic plane $U$.

Lemma 5.2**.**

Let $p>2$. We have $G^{\prime}(U,{\rm O}(U))=\left(1+\left(\frac{p}{3}\right)\right)p$.

Proof.

As in (3.2) we have

[TABLE]

When $p\neq 3$, the equation $-1+\nu+\nu^{-1}=0$ has (two) solutions in ${\mathbb{F}_{p}^{\times}}$ if and only if $\left(\frac{-3}{p}\right)=1$, and we have $\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)$. When $p=3$, this equation has the unique solution $\nu=-1$. ∎

Proposition 5.3**.**

Let $A$ be a quadratic space over ${\mathbb{F}_{p}}$ with $p>2$ of dimension $m$ and discriminant $d(A)$. We write $k=[m/2]$. When $p\equiv 1$ mod $3$, choose a solution $\delta\in{\mathbb{F}_{p}}$ of $\delta^{2}-\delta+1=0$. ($\delta$ is a primitive $6$-th root of $1$ in ${\mathbb{F}_{p}}$.)

1) When $A$ contains an isotropic vector, then

[TABLE]

(2) When $A$ is the anisotropic plane, then ${G^{\prime}(A,{\rm O}(A))}=-\left(\frac{p}{3}\right)p$.

Proof.

(1) We reuse the notation in the proof of Proposition 3.3. When $m=2k$ is even, we have as in (3.3) and (3.5)

[TABLE]

In the third equality we used $\sum_{\mu,\lambda}\zeta_{p}^{-2^{-1}\mu}\langle\mu,\lambda\rangle_{U}=0$. The proposition follows from Lemma 5.2.

Next let $m=2k+1$ be odd. We set

[TABLE]

Then ${G^{\prime}(A,{\rm O}(A))}$ is given by

[TABLE]

So it suffices to calculate $F^{\prime}$. We have

[TABLE]

When $p\equiv 2\;(3)$, we have $\nu+\nu^{-1}\neq 1$ for any $\nu\in{\mathbb{F}_{p}^{\times}}$ so that $F^{\prime}=0$. When $p\equiv 1\;(3)$, we have two solutions $\delta,\delta^{-1}$ of $\nu+\nu^{-1}=1$. Then $F^{\prime}=p(G(A_{p,-2\delta})-1)$. Finally, when $p=3$, $\nu+\nu^{-1}=1$ has the unique solution $\nu=-1$ and thus $2F^{\prime}=3(G(A_{3,-1})-1)$. It is now straightforward to finish the calculation of ${G^{\prime}(A,{\rm O}(A))}$.

(2) We keep the notation in the proof of Proposition 3.4. Instead of $\varphi$, we consider the map

[TABLE]

Then

[TABLE]

As in Claim 3.5, $\varphi^{\prime}$ has constant fiber cardinality over ${\mathbb{F}_{p}^{\times}}\subset{\mathbb{F}_{p}}$. Hence

[TABLE]

If $\varphi^{\prime}(\lambda,x)=0$, we can write $x=(x_{\lambda}^{\pm}+y)/2$ for some $y\in(x_{\lambda}^{\pm})^{\perp}$. Hence

[TABLE]

When $p>3$, this set is non-empty (consisting of two points) if and only if $3\in d(A)\cdot({\mathbb{F}_{p}^{\times}})^{2}$, namely $-3$ is nonsquare. It follows that

[TABLE]

When $p=3$, $(\varphi^{\prime})^{-1}(0)\cap A_{\lambda}^{\pm}$ consists of one point and hence $|(\varphi^{\prime})^{-1}(0)|=2(p-1)$. ∎

Proposition 5.4**.**

Let $A$ be a 2-elementary form of dimension $m>1$. Then

[TABLE]

Proof.

The exceptional cases can be checked directly. Let $A$ contain $U$. We have as in (4.1)

[TABLE]

By (4.2) and (4.3) we obtain ${G^{\prime}(A,{\rm O}(A))}=0$. ∎

5.3. Product formula

Proposition 5.5**.**

Let $p\geq 2$ and $A$ be the direct sum $A=C\oplus B$ where $C=A_{p^{k},a}$ with $k>1$ and $B$ is $p$-elementary. Then

[TABLE]

Proof.

We first consider the case $p>2$. We use the notation in §3.3. For $[xe]\in{{\rm O}(A)}\backslash A_{0}$, $x\in({\mathbb{Z}}/p^{k})^{\times}$, we have by (3.3)

[TABLE]

If $p>3$, we obtain

[TABLE]

by Lemma 3.8. When $p=3$, we have

[TABLE]

This is equal to [math] when $k>2$, and to $G(B(2))\cdot 3\zeta_{3}^{a}$ when $k=2$. For the stratum $A_{1}$ we have by (3.12)

[TABLE]

The second term is $G^{\prime}(B,{\rm O}(B))$. When $(p,k)\neq(3,2)$, the first term is equal to $G^{\prime}(C,{\rm O}(C))$ by Lemma 3.8. Let $(p,k)=(3,2)$. Then the first term is equals to $3$. If $B$ contains $U$, we may use Proposition 5.3 to obtain

[TABLE]

If $B$ is anisotropic, the decomposition $A=C\oplus B$ is canonical and we can use Lemma 2.1.

Next let $p=2$. When $k\geq 4$, the above calculation is still valid. When $k=3$, we have as in the proof of Proposition 4.7

[TABLE]

[TABLE]

and similarly $\sum_{{{\rm O}(A)}\backslash A_{0-}}e(-q(x))\langle[x],[x]\rangle=0$. When $k=2$, we have $G^{\prime}(C,{\rm O}(C))=0$ by Proposition 5.1. We shall show that ${G^{\prime}(A,{\rm O}(A))}=0$. As in the proof of Proposition 4.8, ${G^{\prime}(A,{\rm O}(A))}$ can be reduced to

[TABLE]

and we may assume that $B^{\prime}\neq 0$. When $x_{B}$ is nonzero and isotropic, (4.8) and (4.9) show that the contribution from orbits other than [math], $x_{B}$, $2e$ and $x_{B}+2e$ cancels to [math]. The contribution from those elements also amounts to [math], so we have ${G^{\prime}(A,{\rm O}(A))}=0$. The case of other $x_{B}$ is similar. ∎

Remark 5.6*.*

We also checked that ${G^{\prime}(A,{\rm O}(A))}=0$ for $A=A_{2^{k},a}\oplus A_{4,b}$ and $A=A_{2^{k},a}\oplus A_{4,b}\oplus A_{2,c}$ where $k\geq 2$. Since $G^{\prime}(A_{4,b},{\rm O}(A_{4,b}))=0$ by Proposition 5.1, the product formula also holds in this case.

6. Modular forms for the Weil representation

In this section we explain that the equivariant Gauss sums ${G(A,{\rm O}(A))}$ and ${G^{\prime}(A,{\rm O}(A))}$ appear in the dimension formula for certain vector-valued modular forms. This was our original motivation to study them.

Let $A=(A,q)$ be a finite quadratic form. We write $\sigma(A)\in{\mathbb{Z}}/8$ for its signature. We recall the Weil representation associated to $A$ (see [2]). Let ${{\rm Mp}_{2}(\mathbb{Z})}$ be the metaplectic double cover of ${{\rm SL}_{2}(\mathbb{Z})}$. Its elements are pairs $(M,\phi(\tau))$ where $M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in{{\rm SL}_{2}(\mathbb{Z})}$ and $\phi(\tau)$ is a holomorphic function on the upper half plane such that $\phi(\tau)^{2}=c\tau+d$. The group ${{\rm Mp}_{2}(\mathbb{Z})}$ is generated by the elements

[TABLE]

Its center is cyclic of order $4$ generated by

[TABLE]

Let ${\mathbb{C}}A$ be the group algebra over $A$. We denote by $\mathbf{e}_{x}\in{\mathbb{C}}A$ the standard basis vector corresponding to $x\in A$. The Weil representation $\rho_{A}$ is the unitary representation of ${{\rm Mp}_{2}(\mathbb{Z})}$ on ${\mathbb{C}}A$ defined by

[TABLE]

The orthogonal group ${{\rm O}(A)}={\rm O}(A,q)$ acts on ${\mathbb{C}}A$ by permuting the basis vectors ${{\mathbf{e}}}_{x}$, namely $\gamma({{\mathbf{e}}}_{x})={{\mathbf{e}}}_{\gamma x}$ for $\gamma\in{{\rm O}(A)}$. This action commutes with $\rho_{A}$:

[TABLE]

Let $l\in\frac{1}{2}{\mathbb{Z}}$. A ${\mathbb{C}}A$-valued holomorphic function $f(\tau)$ on the upper half plane is called a modular form of type $\rho_{A}$ and weight $l$ if it satisfies

[TABLE]

and is holomorphic at the cusp. We write $M_{l}(\rho_{A})$ for the space of such modular forms. The group ${{\rm O}(A)}$ acts on $M_{l}(\rho_{A})$ naturally through its action on ${\mathbb{C}}A$. Our purpose is to compute the dimension of the invariant part $M_{l}(\rho_{A})^{{{\rm O}(A)}}$. This is related to constructing modular forms for the full orthogonal group of an even lattice by means of lifting [2]. Let $({\mathbb{C}}A)^{{{\rm O}(A)}}$ be the ${{\rm O}(A)}$-invariant subspace of ${\mathbb{C}}A$. As its basis we can take

[TABLE]

Since the $\rho_{A}$-action commutes with the ${{\rm O}(A)}$-action, it preserves $({\mathbb{C}}A)^{{{\rm O}(A)}}$. We write $\rho_{A}^{inv}$ for the restriction of $\rho_{A}$ on $({\mathbb{C}}A)^{{{\rm O}(A)}}$. Then

[TABLE]

so the problem is to compute the dimension of the space of $\rho_{A}^{inv}$-valued modular forms. We write

[TABLE]

If $\lambda\in{\mathbb{Q}}/{\mathbb{Z}}$, we denote by $\alpha(\lambda)$ its representative in $[0,1)$. We set

[TABLE]

Proposition 6.1**.**

Let $l\equiv\sigma(A)/2$ mod $2{\mathbb{Z}}$ with $l\geq 2$. Then

[TABLE]

When $l\not\equiv\sigma(A)/2$ mod $2{\mathbb{Z}}$, we have $M_{l}(\rho_{A}^{inv})=\{0\}$.

Proof.

Since $\rho_{A}(Z)$ maps ${{\mathbf{e}}}_{x}$ to $\sqrt{-1}^{-\sigma(A)}{{\mathbf{e}}}_{-x}$, we have

[TABLE]

The invariance under $Z$ means $\sqrt{-1}^{2l-\sigma(A)}=1$, so we must have $2l-\sigma(A)\in 4{\mathbb{Z}}$. In general, for a representation $\rho$ of ${{\rm Mp}_{2}(\mathbb{Z})}$ such that $\rho(Z)$ acts by the scalar multiplication by $\sqrt{-1}^{2l}$, a general dimension formula for $\rho$-valued modular forms of weight $l$ is given by Skoruppa (see, e.g., [3] p. 129). In the present case, it takes the form

[TABLE]

We shall show that

[TABLE]

We use the natural basis $v_{[x]}$ of $({\mathbb{C}}A)^{{{\rm O}(A)}}$ to compute the traces. Then

[TABLE]

This gives the trace formula for $\rho_{A}^{inv}(S)$. For $ST$ we have

[TABLE]

and hence

[TABLE]

This gives the trace formula for $\rho_{A}^{inv}(ST)$. ∎