Matrices for the Weil representation

TL;DR

This paper revisits the matrix construction of the Weil representation, providing a detailed analysis and possibly new insights into its algebraic structure and applications.

Contribution

It offers a renewed examination of the matrix approach to the Weil representation, expanding on previous foundational work from 1972.

Findings

Clarifies the matrix construction of the Weil representation

Provides new insights into algebraic properties of the representation

Enhances understanding of applications in algebra and number theory

Abstract

These notes contain a revisit to the matrix construction of the Weil representation by the author in the Journal of Algebra in 1972.

Equations

$$(x{)}^{-1}\left(\sum _{v\in V}{\alpha }_y(v)\right)(x)=\sum _{v\in V}f(2v,x){\alpha }_y(v).$$

$$\sum _{x\in X}\psi (\mathrm{tr}(l(x)))$$

$$\sum _{l\in X^{l\mathrm{in}}}\psi (\mathrm{tr}(l(x)))$$

$$\rho =\sum _{\alpha \in GF(q)}\psi ({\alpha }^2)=\sum _{\xi \ne 0}\chi (\xi )\psi (\xi ),$$

$$\sum _{x\in X}\psi (Q(x))=\chi (Q){\rho }^{\mathrm{rank}Q}{q}^{\dim X-\mathrm{rank}Q}=\chi (Q)\left|X\right|{\delta }^{\dim X-\mathrm{rank}Q}{\rho }^{-\mathrm{rank}Q}.$$

$$\sum _x\psi (F(x))=\sum _x\psi (Q(x)+l(x))=\sum _{x_0}\psi (Q(x_0)+l(x_0))\sum _z\psi (l(z)).$$

$$\sum _x\psi (Q(x)+l(x))=\left|\mathrm{rad}Q\right|\sum _{x_0}\psi (Q(x_0)+l(x_0)).$$

$$\sum _{x_0}\psi (Q(x_0)+l(x_0))$$

$$\sum _x\psi (F(x))$$

$${\Theta }_g(x^{g-1},y^{g-1})=\phi (x^{g-1},y).$$

$$\frac{1}{2}(\phi (x^{g-1},y)-\phi (y^{g-1},x))$$

$$-\frac{1}{2}\phi (x^{g-1},y^{g-1})$$

$$B_g(x^{g-1},y^{g-1})$$

$$\tilde{s}(v)s=\sum _{u_1,u_2\in U}\psi (-Q(u_1))\psi (Q(u_2))f(u_1,v)f(u_1+v,u_2)(u_1+u_2+v).$$

$$\tilde{s}(v)s=\sum _{z\in U}\left\{\psi (Q(z)+\phi (v,z))\sum _{u_1\in U}\psi (2(\phi (u_1,v)-T(u_1,z)))\right\}(v+z).$$

$$T(u,z(v))=\phi (u,v)\text{ for all }u\in U,v\in V.$$

$$\tilde{s}(v)s=\left|U\right|\psi (Q(z(v)+\phi (v,z(v)))(v+z(v)).$$

$$\tilde{s}(v)s=\left|U\right|(v+z(v)).$$

$$Q(v^{g-1})=Q(z(v))=\phi (z(v),v)=\phi (v^{g-1},v)=\phi (v^g,v)=Q_g(v).$$

$$s(g)=\sum _{y\in V^{g-1}}\psi (Q_g(y))(y).$$

$$s(g)s(h)=\mu (g,h)s(gh)$$

$$s(g^{-1})=\sum _{y\in V^{g-1}}\psi (-Q_g(y))(y).$$

$$s(h{)}^{-1}s(g{)}^{-1}(x)s(g)s(h)=(x^{gh})=s(gh{)}^{-1}(x)s(gh).$$

$$Q_{g^{-1}}(v)=\phi (v^{g^{-1}},v)=\phi (v,v^g)=-\phi (v^g,v)=-Q_g(v).$$

$$Q_{h_{\gamma }}(v^{h-1})=\phi (v^{h_{\gamma }-1},v)=\phi (-{\gamma }^{-1}\phi (v,c)c,v)={\gamma }^{-1}\phi (v,c{)}^2.$$

$$(c)+(-c)=(0)+q^{-1}\sum _{\gamma \ne 0}(\psi (-\gamma )-1)s(h_{\gamma }).$$

$${\Theta }_{hk}(x,y)=\phi (u^{hk-1},v)=\phi (u^{h-1},v)=0,$$

$${\Theta }_{hk}(x_1,x_2)=\phi (u_1^{hk-1},u_2)=\phi (u_1^{h-1},u_2)={\Theta }_h(x_1,x_2),$$

$${\Theta }_{hk}(y_1,y_2)=\phi (v_1^{hk-1},v_2)=\phi (v_i^{k-1},v_2)={\Theta }_k(y_1,y_2).$$

$$\sum _{x\in V^{h-1}\cap V^{k-1}}\psi (Q_h(x)+Q_k(x)).$$

$$s(h)s(k)$$

Full text

Matrices for the Weil Representation

Harold N. Ward

Department of Mathematics

University of Virginia

Charlottesville, VA 22904

1 Introduction

These notes contain a revisit to the matrix construction of the Weil representation presented in [13] and [14]. Further consequences of that framework are presented.

2 Preliminaries and the symplectic algebra

As in [13], let $q$ be a power of the odd prime $p$, and let $V$ be a vector space of even dimension $2n$ over the finite field $GF(q)$, endowed with a nondegenerate symplectic form $\varphi$. The corresponding symplectic group will be denoted $\mathrm{Sp}(V)$. Let $K$ be $\mathbb{Q}(e^{2\pi i/p})$, the field of $p$-th roots of unity. Denote the trace function $GF(q)\rightarrow GF(p)$ by $\mathrm{tr}$: $\mathrm{tr}(\alpha)=\alpha+\alpha^{p}+\ldots+\alpha^{q/p}$. Then let $\psi$ be the canonical additive character of $GF(q)$ given by $\psi(\alpha)=e^{2\pi i/p\times\mathrm{tr}(\alpha)}$ [9, p. 190]. Finally, let $f:V\times V\rightarrow K$ be the function $f(x,y)=\psi(\varphi(x,y))$.

Lemma 2.1

*[13, Lemma 1.1] *For any $x,y,z$ in $V$ we have

1) $f(x+y,z)=f(x,z)f(y,z).$

2) $f(x,y)^{m}=f(mx,y)=f(x,my)$. Notice that $f(\alpha x,y)=f(x,\alpha y)$ for $\alpha\in GF(q).$

3) $f(y,x)=f(x,y)^{-1}.$

4) $f$ is nondegenerate: if $f(x,y)=1$ for all $y\in V$, then $x=0$.

Definition 2.2

The symplectic algebra of $V$ is the the twisted group ring $A$ of the additive group of $V$ over $K$, with the factor set $f$. Thus $A$ is a $K$-vector space with basis indexed by the members of $V$. The basis element corresponding to $x\in V$ is denoted $(x)$, and the multiplication rule is that $(x)(y)=f(x,y)(x+y)$.**

The following properties of $\mathcal{A}$ are straight-forward; some are in [13, Proposition 1.2].

Lemma 2.3

For the algebra $\mathcal{A}$ we have:

1) The identity is $(0).$

2) If $x\in V$, then $(x)$ is a unit and $(x)^{-1}=(-x)$.

3) If $x,y\in V$, then $(x)^{-1}(y)(x)=f(2y,x)(y)$.

4) The center of $\mathcal{A}$ is $K(0)$.

5) For $g\in\mathrm{Sp}(V)$, the map $\sum_{v\in V}\alpha_{v}(v)\rightarrow\sum_{v\in V}\alpha_{v}(v^{g})$ is a $K$-automorphism of $\mathcal{A}$. It will written as $\mathfrak{a}\rightarrow\mathfrak{a}^{g}$.

Proof. Only item 4 needs proof: certainly $K(0)$ is in the center of $\mathcal{A}$, by item 1. If $x\in V$, then

[TABLE]

Thus $\sum_{v\in V}\alpha_{y}(v)$ is in the center exactly when $f(2v,x)\alpha_{v}=\alpha_{v}$ for all $x$ and $v$. But if $v\neq 0$, we can choose $x$ to make $f(2v,x)\neq 1$, so that $\alpha_{v}$ must be 0.

These are some general notations we shall use: as in [14], $\chi$ is the quadratic character on $GF(q)$, and $\delta=\chi(-1)=(-1)^{(q-1)/2}$. If $U$ is a subspace of $V$, $\mathcal{A}(U)$ is the $K$-subalgebra generated by the $(u)$, $u\in U$. If $Y$ is a subspace of $X$, then $Y^{F}=\left\{x|F(y,x)=0\text{ for all }y\in Y\right\}$ and ${}^{F}Y=\left\{x|F(x,y)=0\text{ for all }y\in Y\right\}$. (However, we shall write $Y^{\perp}$ when $F=\varphi$.) The form $F$ is nondegenerate if $X^{F}=\left\{0\right\}$, and that is equivalent to $\det F\neq 0$. If $F$ is nondegenerate, we also have ${}^{F}X=\left\{0\right\}$. For $A$ an additive group, $A^{\emptyset}$ is the set of nonzero members of $A$. If the size of a square matrix needs indicating, it is put on as a subscript. Finally, if $\mathfrak{m}$ is a linear transformation on a vector space $X$, then $X^{\mathfrak{m}}=\mathop{\mathrm{I}m}\mathfrak{m}$ and $X_{\mathfrak{m}}=\ker(\mathfrak{m}-1)$, the fixed-point subspace of $\mathfrak{m}$.

Remark 2.4

Here are some facts that will be needed in the sequel:**

The trace form $(\alpha,\beta)\rightarrow\mathrm{tr}(\alpha\beta)$ on $GF(q)$ is nondegenerate [9, Theorem 5.7]. It follows from this that if $X$ is a finite-dimensional vector space over $GF(q)$, then the functions $x\rightarrow\psi(\mathrm{tr}(l(x)))$, where $l\in X^{\mathrm{lin}}$, the space of $GF(q)$-linear functionals on $X$, give the characters of the additive group of $X$. The standard character summations are

[TABLE] 2. 2.

Let

[TABLE]

as in [14, Section 2]. Then $\rho^{2}=\delta q$. Moreover, for $\gamma\neq 0$, $\sum_{\alpha\in GF(q)}\psi(\gamma\alpha^{2})=\chi(\gamma)\rho$. If $Q$ is a quadratic form on $X$, then

[TABLE]

Here $\chi(Q)$ means $\chi(\det(Q^{\prime}))$, where $Q^{\prime}$ is the quadratic form induced by $Q$ on $X/\mathrm{rad}Q$, and we take $\chi(0)=1$. This classical result can be proved as follows: square the sum defining $\rho$ to get $\rho^{2}=\sum_{\alpha,\beta\in GF(q)}\psi(\alpha^{2}+\beta^{2})$. By [9, Theorem 6.26], the number of solutions to $\zeta=\alpha^{2}+\beta^{2}$ is $q-\delta$ if $\zeta\neq 0$ and $q-\delta+q\delta$ if $\zeta=0$. Thus $\rho^{2}=(q-\delta)\sum_{\zeta\in GF(q)}\psi(\zeta)+q\delta\psi(0)$. As $\psi$ is a nontrivial character, the sum is 0, so that $\rho^{2}=q\delta\psi(0)=q\delta$. For the second statement, $\sum_{\alpha\in GF(q)}\psi(\gamma\alpha^{2})=\sum_{\alpha\in GF(q)}\psi(\alpha^{2})$ if $\gamma$ is a square. If not, $\sum_{\alpha\in GF(q)}\psi(\alpha^{2})+\sum_{\alpha\in GF(q)}\psi(\gamma\alpha^{2})=2\sum_{\zeta\in GF(q)}\psi(\zeta)=0$, making $\sum_{\alpha\in GF(q)}\psi(\gamma\alpha^{2})=-\sum_{\alpha\in GF(q)}\psi(\alpha^{2})=\chi(\gamma)\rho$. The quadratic form result comes from taking an orthogonal basis for $Q$ on $X$ and using the previous sums. 3. 3.

Continuing, let $F(x)=Q(x)+l(x)$, where $l\in X^{\mathrm{lin}}$. We wish to evaluate $\sum_{x}\psi(F(x))$. Write $X=X_{0}+\mathrm{rad}Q$ for some subspace $X_{0}$ complementary to $\mathrm{rad}Q$, and suppose first that $l$ is not identically [math] on $\mathrm{rad}Q$. Putting $x=x_{0}+z$, with $x_{0}\in X_{0}$ and $z\in\mathrm{rad}Q$, we have

[TABLE]

By item 1, the second sum is [math], so that $\sum_{x}\psi(F(x))=0$. On the other hand, if $l$ is [math] on $\mathrm{rad}Q$, then

[TABLE]

If $B$ is the polarization of $Q$, we can find $y_{0}\in X_{0}$ with $l(x)=2B(x,y_{0})$ for all $x$. Then $Q(x_{0})+l(x_{0})=Q(x_{0}+y_{0})-Q(y_{0})$. Thus

[TABLE]

since $\dim X_{0}=\mathrm{rank}Q$. Thus in this case

[TABLE]

These remarks imply the following:

Lemma 2.5

If $X$ is a finite-dimensional vector space over $GF(q)$, and $F(x)=Q(x)+l(x)$, where $Q$ is a quadratic form on $X$ and $l\in X^{\mathrm{lin}}$ (that is, $F$ is a polynomial of degree at most 2 on $X$, with $F(0)=0$), then $\left|\sum_{x}\psi(F(x))\right|\leq\left|X\right|$. Equality holds if and only if $F$ is identically 0.

Corollary 2.6

If $\psi(F(x))=1$ for all $x$, then $F$ is identically 0.

3 The symplectic group in $\mathcal{A}$

In this section we produce the embedding of $\mathrm{Sp}(V)$ that is the main ingredient in the treatment of the Weil representation in [13]. However, several other useful results are proved in its development. First we recall the adjoint map $\mathrm{ad}$ for $\varphi$: it is an involutory antiautomorphism $\mathfrak{m}\rightarrow\mathfrak{m}^{\mathrm{ad}}$ of $\mathrm{End}_{GF(q)}(V)$ with the property that $\varphi(x^{\mathfrak{m}},y)=\varphi(x,y^{\mathfrak{m}^{\mathrm{ad}}})$ for all $x,y\in V$ [5, Section 36.3]. The members of $\mathrm{Sp}(V)$ are the invertible $\mathfrak{m}$ for which $\mathfrak{m}^{\mathrm{ad}}=\mathfrak{m}^{-1}$.

Lemma 3.1

If $\mathfrak{m}\in\mathrm{End}_{GF(q)}(V)$, then $\ker\mathfrak{m}=\mathop{\mathrm{I}m}(\mathfrak{m}^{\mathrm{ad}})^{\perp}$.

Proof. The equality follows from this sequence of equivalent statements: $x\in\mathop{\mathrm{I}m}(\mathfrak{m}^{\mathrm{ad}})^{\perp}$; $\varphi(x,y^{\mathfrak{m}^{\mathrm{ad}}})=0$ for all $y$; $\varphi(x^{\mathfrak{m}},y)=0$ for all $y$; $x^{\mathfrak{m}}=0$, from the nondegeneracy of $\varphi$; $x\in\ker\mathfrak{m}$.

Applying this to $\mathfrak{m}=g-1$, $g\in\mathrm{Sp}(V)$, we have

Corollary 3.2

If $g\in\mathrm{Sp}(V)$, then $V_{g}=(V^{g-1})^{\perp}$.

Definition 3.3

Let $g\in Sp(V)$. Following Thomas [11] and Wall [12], we introduce the theta form $\Theta_{g}$ on $V^{g-1}$, defined by

[TABLE]

Lemma 3.4

[12, Lemma 1.1.1 and equation (1.1.3)]* The form $\Theta_{g}$ is well-defined and nondegenerate. Its skew-symmetric part is $-\varphi/2$ (on $V^{g-1}$), and its symmetric part $B_{g}$ is given by $B_{g}(x^{g-1},y^{g-1})=(\varphi(x^{g},y)+\varphi(y^{g},x))/2$. The corresponding quadratic form $Q_{g}$ has $Q_{g}(x^{g-1})=B_{g}(x^{g-1},x^{g-1})=\varphi(x^{g},x)$.*

Proof. If $y_{1}^{g-1}=y_{2}^{g-1}$, then $y_{1}-y_{2}\in V_{g}$, and $\varphi(x^{g-1},y_{1}-y_{2})=0$, by Corollary 3.2. Thus $\varphi(x^{g-1},y_{1})=\varphi(x^{g-1},y_{2})$, as needed to make $\Theta_{g}$ well-defined. Then if $\Theta_{g}(x^{g-1},y^{g-1})=\varphi(x^{g-1},y)=0$ for all $x$, $y$ must be in $(V^{g-1})^{\perp}$ and hence be a fixed vector of $g$, whereupon $y^{g-1}=0$. This shows that $\det\Theta_{g}\neq 0$ (relative to any basis of $V^{g-1}$).

The skew-symmetric part of $\Theta_{g}$ has the values

[TABLE]

But

[TABLE]

the same thing.

For the symmetric part $B_{g}$ of $\Theta_{g}$, we have

[TABLE]

Theorem 3.5

[12, Theorem 1.1.2]* Let $U$ be a subspace of $V$ endowed with a nondegenerate bilinear form $T$ whose skew-symmetric part is $-\varphi/2$ on $U$. Then if $B=T+\varphi/2$ and $Q(u)=B(u,u)$, $u\in U$, there is an element $g\in\mathrm{Sp}(V)$ for which $U=V^{g-1}$ and $\Theta_{g}=T$. If we start with $h\in\mathrm{Sp}(V)$ and take $U=V^{h-1}$ and $T=\Theta_{h}$, then $g=h$. Consequently, the original $g$ is unique.*

Proof. Let $s=\sum\psi(Q(u))(u)$ and $\widetilde{s}=\sum\psi(-Q(u))(u)$, both sums over $u\in U$. For $v\in V$,

[TABLE]

Put $u_{2}=z-u_{1}$ and use $T(u_{1},z)=B(u_{1},z)-\varphi(u_{1},z)/2$ to get

[TABLE]

Now $u\rightarrow\varphi(u,v)-T(u,z)$ is a linear functional on $U$. So if it is not identically 0, the inner sum (a character sum) is 0. Since $T$ is nondegenerate, there is a unique $z=z(v)$ making $\varphi(u,v)-T(u,z(v))=0$:

[TABLE]

The map $v\rightarrow z(v)$ is linear. Thus

[TABLE]

Moreover, $T(z(v),z(v))=\varphi(z(v),v)$. That is, $Q(z(v))+\varphi(v,z(v))=0$, and $Q(z(v))=\varphi(z(v),v)$. So

[TABLE]

From $v=0$ we get $\widetilde{s}s=\left|U\right|(0)$, so that $s$ is invertible and $s^{-1}=\left|U\right|^{-1}\widetilde{s}$. Consequently $g:v\rightarrow v+z(v)$ is nonsingular.

Since $\mathfrak{a}\rightarrow s^{-1}\mathfrak{a}s$ is an automorphism of $\mathcal{A}$, $f(v_{1}^{g},v_{2}^{g})=f(v_{1},v_{2})$ for all $v_{1},v_{2}$. In Lemma 2.5, let $X=V\oplus V$ and $F(v_{1},v_{2})=\varphi(v_{1}^{g},v_{2}^{g})-\varphi(v_{1},v_{2})$. Then $\psi(F(v_{1},v_{2})=1$. By Corollary 2.6, $F(v_{1},v_{2})=0$, so that $\varphi(v_{1}^{g},v_{2}^{g})=\varphi(v_{1},v_{2})$, wherewith $g\in\mathrm{Sp}(V)$. As $z(v)=v^{g-1}$,

[TABLE]

Thus $Q=Q_{g}$, and so $B=B_{g}$ and $T=\Theta_{g}$. Certainly $V^{g-1}\subseteq U$, since $z(v)\in V^{g-1}$. But given $z\in U$, there is a $v$ with $T(z,u)=\varphi(u,v)$ for all $u\in U$, since $\varphi$ is nonsingular. Then $z=z(v)$, and $U\subseteq V^{g-1}$. So $U=V^{g-1}$.

Finally, let $U=V^{h-1}$ and $T=\Theta_{h}$, for some $h\in\mathrm{Sp}(V)$. Then on the one hand, $\Theta_{h}(u,z(v))=\varphi(u,v)$ for all $u\in U$ and all $v\in V$, by (3.1). On the other hand, $\Theta_{h}(u,v^{h-1})=\varphi(u,v)$. Since $\Theta_{h}$ is nondegenerate, it must be that $z(v)=v^{h-1}$, so that $v^{h}=v+z(v)=v^{g}$. Thus $g=h$.

Corollary 3.6

*[13, Proposition 2.1] *For $g\in\mathrm{Sp}(V)$ there is a unique invertible element $s(g)\in\mathcal{A}$ with $(0)$-coefficient 1, for which $s(g)^{-1}(x)s(g)=(x^{g})$ for all $x\in V$. One has

[TABLE]

If $g,h\in\mathrm{Sp}(V)$, then

[TABLE]

for some nonzero $\mu(g,h)$ of $K$. (This factor is the inverse of the “$\mu(g,h)$” in [13].) The map $g\rightarrow s(g)$ is injective, and

[TABLE]

Proof. The existence, invertibility, and formula for $s(g)$ follow from Theorem 3.5. If $s_{1}$ and $s_{2}$ are two members of $\mathcal{A}$ for which $s_{1}^{-1}(x)s_{1}=s_{2}^{-1}(x)s_{2}$ for all $x$, then $s_{1}s_{2}^{-1}$ commutes with all $(x)$, so that $s_{1}=\zeta s_{2}$ for some nonzero $\zeta\in K$, by Lemma 2.3. If the $(0)$-coefficients of $s_{1}$ and $s_{2}$ are both 1, then $\zeta=1$ and $s_{1}=s_{2}$, giving the uniqueness.

For $g,h\in\mathrm{Sp}(V)$,

[TABLE]

Thus again $s(g)s(h)$ must be a scalar multiple of $s(gh)$, and that multiple is defined to be $\mu(g,h)$. As the $(0)$-coefficient of $s(gh)$ is 1, we can find $\mu(g,h)$ by computing the $(0)$-coefficient in $s(g)s(h)$. Since $f(u,v)=f(-u,-v)$, we may write $s(h)=\sum_{y=z^{h}-z}f(z^{h},z)(-y)$. Then the $(0)$-coefficient in $s(g)s(h)$ is $\sum f(x^{g},x)f(z^{h},z)$, the sum over $y\in V^{g-1}\cap V^{h-1}$ with $y=x^{g}-x=z^{h}-z$. This coefficient is $\sum_{y\in V^{g-1}\cap V^{h-1}}\psi(Q_{g}(y)+Q_{h}(y))$. Here $Q_{g}(y)+Q_{h}(y)$ is a quadratic form on $V^{g-1}\cap V^{h-1}$, by Lemma 3.4, so that item 2 of Remark 2.4 implies that $\mu(g,h)$ is a power of $\rho$, up to sign.

That $g\rightarrow s(g)$ is injective just follows from the fact that $s(g)^{-1}(x)s(g)=(x^{g})$, which determines $g$. Since $s(g^{-1})s(g)$ and $\widetilde{s}s(g)$ are scalar multiples of $(0)$, where $\widetilde{s}=\sum_{\unit{y}\in V^{g-1}}\psi(-Q_{g}(u)(u)$ from the proof of Theorem 3.5, both $s(g^{-1})$ and $\widetilde{s}$ are scalar multiples of $s(g)^{-1}$ and so of each other. As both have $(0)$-coefficient 1, it must be that $s(g^{-1})=\widetilde{s}$. Note that

[TABLE]

Proposition 3.7

Let $g\in\mathrm{Sp}(V)$ be an involution, with eigenspaces $E_{\varepsilon}=\left\{v|v^{g}=\varepsilon v\right\}$, $\varepsilon=\pm 1$. Then $Q_{g}=0$ and $s(g)=\sum_{x\in E_{-1}}(x)$. If $g\in\mathrm{Sp}(V)$ and $Q_{g}=0$, then $g$ is an involution.

Proof. We have $V=E_{1}\oplus E_{-1}$, an orthogonal direct sum. For $v=v_{1}+v_{-1}$, with $v_{\varepsilon}\in E_{\varepsilon}$, $v^{g-1}=2v_{-1}=-v_{-1}^{g-1}$. Thus $V^{g-1}=E_{-1}$, and $Q_{g}(v^{g-1})=\varphi(2v_{-1},v)=0$. Then $s(g)=\sum_{x\in E_{-1}}(x)$.

If $Q_{g}=0$ for some $g\in\mathrm{Sp}(V)$, then $\Theta_{g}=-\varphi/2$ on $V^{g-1}$, by Lemma 3.4, so that $\varphi$ is nonsingular on $V^{g-1}$, since $\Theta_{g}$ is. We have both $s(g)=\sum_{x\in V^{g-1}}(x)$ and $s(g^{-1})=\sum_{x\in V^{g-1}}(x)$, from Corollary 3.6. Thus $g=g^{-1}$ and $g$ is an involution.

Proposition 3.8

Let $h_{\gamma}$ be the transvection $v\rightarrow v-\gamma^{-1}\varphi(v,c)c$. Then $s(h_{\gamma})=\sum_{\zeta\in GF(q)}\psi(\gamma\zeta^{2})(\zeta c)$.**

Proof. Here $v^{h_{\gamma}-1}=-\gamma^{-1}\varphi(v,c)c$, so that $V^{h_{\gamma}-1}=GF(q)c$. Furthermore,

[TABLE]

Thus with $v^{h_{\gamma}-1}=\zeta c$, $Q_{h_{\gamma}}(\zeta c)=\gamma\zeta^{2}$, and $s(h_{\gamma})=\sum_{\zeta}\psi(\gamma\zeta^{2})(\zeta c)$.

Corollary 3.9

[13, Proposition 2.2]* The centralizer $\mathcal{C}$ of $j$ is the $K$-span of the $s(g)$, $g\in\mathrm{Sp}(V)$.*

Proof. As in [13],

[TABLE]

Proposition 3.10

Let $h,k\in\mathrm{Sp}(V)$. If $V^{h-1}\cap V^{k-1}=\left\{0\right\}$, that is, the sum $V^{h-1}+V^{k-1}$ is direct, then $s(h)s(k)=s(hk)$ and $\mu(h,k)=1$. Conversely, if $\mu(h,k)=1$, then $V^{h-1}\cap V^{k-1}=\left\{0\right\}$. When this happens, $V^{hk-1}=V^{h-1}\oplus V^{k-1}$ and $\dim V^{hk-1}=\dim V^{h-1}+\dim V^{k-1}$. Moreover, $\Theta_{hk}(V^{h-1},V^{k-1})=0$, and with a basis for $V^{hk-1}$ adapted to the direct sum, the matrix $\left[\Theta_{hk}\right]$ for $\Theta_{hk}$ is \left[\begin{array}[]{cc}\left[\Theta_{h}\right]&0\\ \ast&\left[\Theta_{k}\right]\end{array}\right].

Proof. We have $v^{hk-1}=v^{h-1}+v^{h(k-1)}$, showing that $V^{hk-1}\subseteq$ $V^{h-1}+V^{k-1}$. Suppose that $V^{h-1}\cap V^{k-1}=\left\{0\right\}$. Then in writing out the product $s(h)s(k)$, there is no collection of terms with the same basis element $(v)$. Thus $s(hk)=s(h)s(k)$ and $\mu(h,k)=1$. Each term in $s(h)s(k)$ appears as a product in only one way. So $V^{hk-1}=V^{h-1}+V^{k-1}$ and the sum must be direct. If $x=u^{hk-1}\in V^{hk-1}$ and $x\in V^{h-1}$ too, then in the direct sum $u^{h(k-1)}=0$ and $x=u^{h-1}$. Likewise, if $y=v^{hk-1}$ and $y\in V^{k-1}$, then $v^{h-1}=0$, so that $v\in V_{h}$. Thus

[TABLE]

from the $\varphi$-orthogonality of $V^{h-1}$ and $V_{h}$.

With $x_{i}=u_{i}^{hk-1}$, where $u_{i}^{h(k-1)}=0$ and $x_{i}=u_{i}^{h-1}$, we have

[TABLE]

making $\Theta_{hk}|V^{h-1}=\Theta_{h}$. Similarly, with $y_{i}=v_{i}^{hk-1}\in V^{k-1}$, where $v_{i}^{h-1}=0$ and $y_{i}=v_{i}^{k-1}$,

[TABLE]

Hence $\Theta_{hk}|V^{k-1}=\Theta_{k}$. All this implies that by taking a basis for $V^{hk-1}$ that is the union of one in $V^{h-1}$ followed by one in $V^{k-1}$, a matrix for $\Theta_{hk}$ can be arranged as displayed.

From the proof of Corollary 3.6, the $(0)$-coefficient in $s(h)s(k)$ is

[TABLE]

As observed there, $Q_{h}+Q_{k}$ is a quadratic form on $V^{h-1}\cap V^{k-1}$. If its rank is $r$, then $r\leq\dim(V^{h-1}\cap V^{k-1})$, and the sum, up to sign, is $\rho^{r}q^{\dim(V^{h-1}\cap V^{k-1})-r}=\pm\rho^{2\dim(V^{h-1}\cap V^{k-1})-r}$, by Remark 2.4. That must be $\mu(h,k)$, and the only way it can be 1 is that $\dim(V^{h-1}\cap V^{k-1})=0$.

Corollary 3.11

Suppose that $g\in\mathrm{Sp}(V)$ and that $V^{g-1}=X\oplus Y$, both terms nonzero. Suppose also that $\Theta_{g}(X,Y)=0$, so that $\Theta_{g}|X$ and $\Theta_{g}|Y$ are nondegenerate. Let $h,k\in\mathrm{Sp}(V)$ with $X=V^{h-1},Y=V^{k-1}$, and $\Theta_{g}|X=\Theta_{h},\Theta_{g}|Y=\Theta_{k}$, following Theorem 3.5. Then $g=hk$.

Proof. We have $s(h)=\sum_{x\in X}\psi(Q_{h}(x))(x)$ and $s(k)=\sum_{y\in Y}\psi(Q_{k}(y))(y)$. Thus

[TABLE]

The argument of $\psi$ here is

[TABLE]

and since $\Theta_{g}=B_{g}-\varphi/2$, it is $Q_{g}(x+y)-2\Theta_{g}(x,y)=Q_{g}(x+y)$. Thus $s(h)s(k)=\sum_{x\in X,y\in Y}\psi(Q_{g}(x+y))=s(g)$. Since $s(h)s(k)=s(hk)$, from $V^{h-1}\cap V^{k-1}=\left\{0\right\}$, $g=hk$.

Remark 3.12

*By the arguments for $s(g)$ we have been using, if $g=g_{1}\cdots g_{r}$ and $\dim V^{g-1}=\sum_{i=1}^{r}\dim V^{g_{i}-1}$, then $s(g)=s(g_{1})\cdots s(g_{r})$ and $V^{g-1}=V^{g_{1}-1}\oplus\cdots\oplus V^{g_{r}-1}$. Moreover, $\left[\Theta_{g}\right]$ *is lower block-triangular, with $\left[\Theta_{g_{1}}\right],\ldots,\left[\Theta_{g_{r}}\right]$ on the diagonal.

4 $\mathcal{A}$ as a matrix ring

For any commutative ring $S$ we let $\mathcal{M}_{m}(S)$ be the $m\times m$ matrix ring over $S$. The goal of this section is to show that the symplectic algebra $\mathcal{A}$ is isomorphic to $\mathcal{M}_{q^{n}}(K)$. We prove this by creating a set of matrix units that will be of use later on. The flow of the argument is quite standard.

Let $W$ and $W^{\ast}$ be two maximal isotropic (Lagrangian) subspaces of $V$ that are complementary, so that $V=W\oplus W^{\ast}$, an internal direct sum. Then $\varphi$ is identically [math] on both $W$ and $W^{\ast}$, and $\varphi$ sets up a nondegenerate pairing between $W$ and $W^{\ast}$. The subalgebras $\mathcal{A}(W)$ and $\mathcal{A}(W^{\ast})$ are simply the group algebras for the two subspaces as additive groups. Moreover, the characters of the additive group of $W$ are produced by the maps $w\rightarrow f(w,a)$, $a\in W^{\ast}$. Set $e_{0}=q^{-n}\sum_{w\in W}(w)$. If $x\in V$, then

[TABLE]

As $w\rightarrow f(w,x)$ is a character of $W$, $\sum_{w\in W}f(w,x)=q^{n}$ if $x\in W$, and $\sum_{w\in W}f(w,x)=0$ otherwise. Thus

[TABLE]

Now let $e_{ab}=(-a)e_{0}(b)$, where $a,b\in W^{\ast}$ (we may separate the two subscripts by a comma when more complicated expressions appear). Then $e_{ab}e_{cd}=(-a)e_{0}(b-c)e_{0}(d)$, and by (4.1), this is 0 if $b\neq c$, and $e_{ad}$ if $b=c$. Thus the $e_{ab}$ multiply like matrix units. As customary, put $e_{a}=e_{aa}$, so that $e_{0}$ retains its original meaning. These $e_{a}$ are orthogonal idempotents in $\mathcal{A(W)}$. By item 3 of Lemma 2.3, $e_{a}=q^{-n}\sum_{w\in W}f(w,2a)(w)$. Then

[TABLE]

The inner sum is 0 if $w\neq 0$, and $q^{n}$ if $w=0$. So $\sum_{a\in W^{\ast}}e_{a}=(0)$. Furthermore,

[TABLE]

Thus

[TABLE]

Hence $e_{a}(x)e_{b}\in Ke_{ab}$ for all $x$, so that $e_{a}\mathcal{A}e_{b}=Ke_{ab}$. Then for $\mathfrak{a}\in\mathcal{A}$,

[TABLE]

with the $\alpha_{ab}\in K$. As the number of the $e_{ab}$ is $q^{2n}$, which is $\dim_{K}\mathcal{A}$, the $e_{ab}$ form a $K$-basis of $\mathcal{A}$. Thus we have

Theorem 4.1

The algebra $\mathcal{A}$ is indeed isomorphic to $\mathcal{M}_{q^{n}}(K)$.

We identify $\mathcal{A}$ with $\mathcal{M}_{q^{n}}(K)$ by this isomorphism, that is, by means of the matrix units $e_{ab}$. If we wish to emphasize that we are regarding $\mathfrak{a\in}\mathcal{A}$ as a matrix, we shall write $[\mathfrak{a}]$.

Remark 4.2

*The trace of the map $\mathfrak{b}\rightarrow\mathfrak{ab}$ *is $q^{n}\mathrm{tr}\mathfrak{a}$, with $\mathfrak{a,b}\in\mathcal{A}$. If $x\neq 0$, $\mathfrak{b}\rightarrow(x)\mathfrak{b}$ has trace [math], so that $\mathrm{tr}[(x)]=0$. Thus

[TABLE]

Remark 4.3

Suppose that $\mathcal{M}$ is a matrix algebra over a field and that $\mathcal{M}\emph{\ }$has two sets of matrix units, the $e_{ab}$ and the $e_{ab}^{\prime}$, indexed by the same finite set. Then by [8, Theorem 3, p. 59], there is an invertible $u$ in $\mathcal{M}$ with $e_{ab}^{\prime}=u^{-1}e_{ab}u$, for all $a,b$. It follows that traces and determinants of members of $\mathcal{M}$ computed from the two sets of units are the same.

5 The Weil representation

Continue now with the set-up and notation of Section 4. We have $s(-1)=\sum_{v\in V}(v)$, by Proposition 3.7. As in [13], we set $j=q^{-n}s(-1)$, an involution. To determine $[j]$, observe that

[TABLE]

Once again, the inner sum is 0 if $u\notin W$ and $q^{n}$ if $u\in W$. So

[TABLE]

Thus

[TABLE]

This last is 0 unless $b=-a$, when it is just $e_{a,-a}$.

Partition $W^{\ast}$ as $P\cup-P\cup\{0\}$ in some way, and on ordering $P$, order $-P$ so that $-a<-b$ in $-P$ when $a<b$ in $P$. Then label rows and columns with the members of $W^{\ast}$ following that arrangement. This makes

[TABLE]

where $I$ is the identity matrix of size $(q^{n}-1)/2$ and the various 0s have appropriately matching sizes. Direct computations give the following result:

Proposition 5.1

Let $\mathcal{C}$ be the centralizer of $j$ in $\mathcal{A}$. Then the members of $\mathcal{C}$ are the matrices of the form

[TABLE]

the blocks matching those of $\left[j\right]$. The map given by

[TABLE]

provides an isomorphism of $\mathcal{C}$ with the direct sum $\mathcal{M}_{(q^{n}-1)/2}(K)\oplus\mathcal{M}_{(q^{n}+1)/2}(K)$.

The representation $g\rightarrow s(g)$ is a projective representation of $\mathrm{Sp}(V)$, with factor set $\mu$. A key result in any development of the Weil representation is that it is equivalent to an ordinary representation. Thus we seek a function $\eta:\mathrm{Sp}(V)\rightarrow K$ such that with $t(g)=\eta(g)s(g)$, the representation $g\rightarrow t(g)$ is ordinary; that is, $\mu(g,h)=\eta(gh)\eta(g)^{-1}\eta(h)^{-1}$, $\mu$ as in (3.3). Now

[TABLE]

Let $\flat$ be the homomorphism

[TABLE]

of $\mathcal{C}$ onto $\mathcal{M}_{(q^{n}-1)/2}(K)$, as inferred from Proposition 5.1. Then

[TABLE]

Consequently:

Lemma 5.2

If we define

[TABLE]

we have the needed equation $\mu(g,h)=\eta(gh)\eta(g)^{-1}\eta(h)^{-1}$.

Remark 5.3

The comments in Remark 4.3 show that $\eta$ is well-defined: it does not depend on the choices made in obtaining the matrix realization of $A$. In fact, if we also have $\mu(g,h)=\eta^{\prime}(gh)\eta^{\prime}(g)^{-1}\eta^{\prime}(h)^{-1}$, then $\eta^{\prime}\eta^{-1}$ is a homomorphism of $Sp(V)$ to $C^{\varnothing}$. As $Sp(V)^{\prime}=Sp(V)$ except when $n=1$ and $q$ (odd) is 3, $\eta^{\prime}=\eta$ with that one exception.**

For example, $j^{\flat}=-I_{(q^{n}-1)/2}$, so that $\det(j^{\flat})=(-1)^{(q^{n}-1)/2}=\det(j)$. Then

[TABLE]

Since $(q^{n}-1)/2\equiv n(q-1)/2\,(\bmod\ 2)$, $(-1)^{(q^{n}-1)/2}=(-1)^{n(q-1)/2}=\delta^{n}$. As $s(-1)=q^{n}j$, we get

[TABLE]

Consequently,

[TABLE]

Definition 5.4

The representation $\mathcal{W}$ of $Sp(V)$ given by $g\rightarrow t(g)$, with degree $q^{n}$, is the Weil representation. By Proposition 5.1, $\mathcal{W}$ has the two constituents $\mathcal{W}_{-}:g\rightarrow t(g)^{\flat}$ and $\mathcal{W}_{+}:g\rightarrow t(g)^{\sharp}$, of degrees $(q^{n}-1)/2$ and $(q^{n}+1)/2$, respectively. Since $C$ is spanned by the $t(g)$, from Corollary 3.9, both of these representations are irreducible. We denote the characters of these three representations by $\omega$, $\omega_{-}$, and $\omega_{+}$, so that $\omega=\omega_{-}+\omega_{+}$.**

6 The Weil character

Before examining the Weil character in detail, we record this implication of the matrix picture above:

Proposition 6.1

For the Weil characters,

[TABLE]

Thus

[TABLE]

Proof. If t(g)=\left[\begin{array}[]{ccc}A&B&\mathbf{b}\\ B&A&\mathbf{b}\\ \mathbf{a}&\mathbf{a}&\alpha\end{array}\right], then $t(g)^{\flat}=A-B$ and

[TABLE]

Thus

[TABLE]

To begin with, from $t(g)=\eta(g)s(g)$ and $\mathrm{tr}s(g)=q^{n}$, we have for the Weil character $\omega$,

[TABLE]

For example,

[TABLE]

To obtain formulas for $\omega$, we start with any $g\in\mathrm{Sp}(V)$ for which $W^{g}=W$. Let $g^{\ast}$ be the map induced by $g$ on $W^{\ast}$: if $a\in W^{\ast}$, $W+a^{g^{\ast}}=W+a^{g}$. For $[s(g)]$, we have $e_{0}^{g}=e_{0}$, so $e_{0}s(g)e_{a}=s(g)e_{0}e_{a}=0$ unless $a=0$; similarly $e_{a}s(g)e_{0}=0$ if $a\neq 0$. Thus

[TABLE]

Here

[TABLE]

the last from (4.1). Thus

[TABLE]

We have $e_{a}^{g}=(-a^{g})e_{0}(a^{g})$. But $a^{g}=w+a^{g^{\ast}}$ for some $w\in W$, so

[TABLE]

Therefore

[TABLE]

Thus with $\left[s(g)\right]=\left[\alpha_{ab}\right]$,

[TABLE]

This can be nonzero only when $b=a^{g^{\ast}}$. Therefore $[s(g)]$ is monomial, and it follows that both $A+B$ and $A-B$ are monomial. Moreover, $\det(A+B)=(-1)^{\beta}\det(A-B)$, where $\beta$ is the number of nonzero entries in $B$. This is the number of members $a$ of $P$ for which $a^{g^{\ast}}\in-P$. The sign $(-1)^{\beta}$ is evaluated in the next lemma.

Lemma 6.2

Let $Z$ be a vector space of dimension $n>0$ over $GF(q)$, $q$ odd. Partition $Z$ into $P\cup-P\cup\{0\}$, as above. Let $\sigma(g)=(-1)^{\left|P^{g}\cap(-P)\right|}$ for $g$ a nonsingular linear transformation on $Z$. Then

[TABLE]

Proof. Index $q^{n}-1$ independent variables $X_{z}$ by the nonzero members of $Z$, and in the $\mathbb{Z}$-ring of polynomials $R$ in the $X_{z}$, let $\pi=\mathop{\displaystyle\prod}\limits_{z\in P}(X_{z}-X_{-z})$. Let $g$ act on $R$ by $X_{z}\longrightarrow X_{z^{g}}$. Then $\pi^{g}=\sigma(g)\pi$, and it follows that $g\longrightarrow\sigma(g)$ is a homomorphism from $GL(Z)$ to $\left\langle-1\right\rangle$. If $\sigma$ is nontrivial, then it must be that $\sigma(g)=\chi(\det g)$, since $GL(Z)/GL(Z)^{\prime}$ is cyclic of order $q-1$ [1, Theorem 4.7].

For the nontriviality of $\sigma$, observe first that if $P^{\prime}\cup-P^{\prime}\cup\{0\}$ is a second partition of $Z$ and $\pi^{\prime}$ the corresponding product, then $\pi^{\prime}=\varepsilon\pi$ for some sign, $\varepsilon$. It follows that $\pi^{\prime g}=\sigma(g)\pi^{\prime}$, so that $\sigma(g)$ does not depend on the partition used. Now realize $Z$ as $GF(q^{n})$ and let $\alpha$ be a primitive element. Let $P=\left\{1,\alpha,\ldots,\alpha^{(q^{n}-3)/2}\right\}$; $\alpha^{(q^{n}-1)/2}=-1$. Then $g:z\longrightarrow\alpha z$ has $\left|P^{g}\cap(-P)\right|=1$, $\sigma(g)=-1$, and $\sigma$ is indeed nontrivial. By way of corroboration, this $g$ has determinant $\alpha+\alpha^{q}+\ldots+\alpha^{q^{n-1}}=\alpha^{(q^{n}-1)/(q-1)}$ (the norm of $\alpha$), a primitive element of $GF(q)$. Thus $\chi(\alpha^{(q^{n}-1)/(q-1)})=-1=\sigma(g)$.

Now the matrix \left[\begin{array}[]{ccc}A&B&0\\ B&A&0\\ 0&0&\alpha\end{array}\right] is equivalent to \left[\begin{array}[]{ccc}A-B&B-A&0\\ 0&A+B&0\\ 0&0&\alpha\end{array}\right], with no scalings involved. Thus

[TABLE]

by the lemma. From $\eta(g)=(\det s(g)^{\flat})^{2}(\det s(g))^{-1}$ (equation (5.2)) and $\det s(g)^{\flat}=\det(A-B)$, we conclude:

Theorem 6.3

If $g\in\mathrm{Sp}(V)$ and $W^{g}=W$, then

[TABLE]

Corollary 6.4

Suppose that $g\in\mathrm{Sp}(V)$ is an involution. Let $E_{\varepsilon}$ be the eigenspace of $g$ for eigenvalue $\varepsilon$ and let $\dim E_{-1}=2m$. Then $\eta(g)=\delta^{m}q^{-m}$ and $\omega(g)=\delta^{m}q^{n-m}$.

Proof. As in Proposition 3.7, $\varphi$ is nonsingular on $E_{-1}$, so that $\dim E_{-1}$ is indeed even. From Remark 4.2 we can tailor the choice of $W$ and $W^{\ast}$ to $g$. Thus we set up $W$ and $W^{\ast}$ by taking decompositions $E_{\varepsilon}=W_{\varepsilon}\oplus W_{\varepsilon}^{\ast}$, with $W=W_{1}\oplus W_{-1}$, $W^{\ast}=W_{1}^{\ast}\oplus W_{-1}^{\ast}$. Here $\dim W_{-1}=\dim W_{-1}^{\ast}=m$, and $\dim W_{1}=\dim W_{1}^{\ast}=n-m$. We have $V^{g-1}=E_{-1}$, and $Q_{g}=0$. Thus in $\left[s(g)\right]$, $\alpha=\left|W\cap V^{g-1}\right|=\left|W_{-1}\right|=q^{m}$. On $W^{\ast}$, $a\rightarrow a^{g^{\ast}}$ has fixed point space $W_{1}^{\ast}$ and so $(q^{n}-q^{n-m})/2=q^{n-m}(q^{m}-1)/2$ two-cycles. Hence the sign of $\pi_{g}$ is $(-1)^{(q^{m}-1)/2}$. As noted before, this is $\delta^{m}$. Thus $\eta(g)=\delta^{m}q^{-m}$. Once again, we get $\eta(-1)=\delta^{n}q^{-n}$.

Corollary 6.5

For such an involution $g$,

[TABLE]

Proof. Here $-g$ is also an involution with $\dim E_{-1}=2n-2m$, and Proposition 6.1 applies.

Corollary 6.6

Let $h$ be the transvection $v\rightarrow v-\gamma^{-1}\varphi(v,c)c$, where $\gamma\neq 0$ and $c\neq 0$. Then $\eta(h)=\rho^{-1}\chi(\gamma)$ and $\omega(h)=q^{n}\rho^{-1}\chi(\gamma)$.

Proof. Now choose $W$ so that $c\in W$. From Proposition 3.8,

[TABLE]

By Remark 2.4, this is $\chi(\gamma)\rho$. On $W^{\ast}$, $h^{\ast}$ is the identity. Thus $\eta(h)=\rho^{-1}\chi(\gamma)$.

With these results in hand, we can now prove a version of the formula in [11, Remark 1.3].

Theorem 6.7

Let $g\in\mathrm{Sp}(V)$ with theta form $\Theta_{g}$. Then

[TABLE]

Proof. Recall that $\chi(\Theta_{g})$ means $\chi(\det\Theta_{g})$, $\det\Theta_{g}$ computed relative to any basis. The two examples we have so far corroborate the formula: for an involution $g$, $V^{g-1}=V_{-1}$, and with $\dim V_{-1}=2m$, Corollary 6.4 gives $\eta(g)=\delta^{m}q^{-m}$. As we saw in Proposition 3.7, $Q_{g}=0$ and $\Theta_{g}=-\varphi/2$, making $\det\Theta_{g}=(-1/2)^{2m}$ and $\chi(\Theta_{g})=1$. Moreover, $q=\delta\rho^{2}$ makes $\eta(g)=\delta^{m}\delta^{m}\rho^{-2m}=\rho^{-r}$, as asserted. Similarly, for $h$ the transvection $h:v\rightarrow v-\gamma^{-1}\varphi(v,c)c$, $V^{h-1}=GF(q)c$, and $\Theta_{h}(\xi c,\zeta c)=\xi\zeta\gamma$, with $\det\Theta_{h}=\gamma$, all by Proposition 3.8. From Corollary 6.6, $\eta(h)=\rho^{-1}\chi(\gamma)$, again in agreement.

We can thus induct on $\dim V^{g-1}$. Suppose that $V^{h-1}\cap V^{k-1}=\left\{0\right\}$ for $h,k\in\mathrm{Sp}(V)$, neither $h$ nor $k$ the identity. Then Proposition 3.10 implies that $\det\Theta_{hk}=\det\Theta_{h}\det\Theta_{k}$. Because $\mu(h,k)=1$, it follows that $\eta(hk)=\eta(h)\eta(k)$, and by induction,

[TABLE]

But $\dim V^{hk-1}=\dim V^{h-1}+\dim V^{k-1}$, as shown in the proof of Proposition 3.10, and then $\eta(hk)=\rho^{-\dim V^{hk-1}}\chi(\Theta_{hk})$ indeed.

To obtain such a factoring, we may assume that a given $g$ is neither an involution nor a transvection, for both of which the formula is already correct. Thus the quadratic form $Q_{g}$ on $V^{g-1}$ is not identically 0, by Proposition 3.7. Take $c\in V^{g-1}$ with $Q_{g}(c)\neq 0$. Put $X=\left\langle c\right\rangle$ and $Y=\left\langle c\right\rangle^{\Theta_{g}}$; $Y\neq\left\{0)\right\}$. Corollary 3.11 applies, since $Q_{g}(c)\neq 0$ implies that $X\cap Y=\left\{0)\right\}$. Then $g=hk$, where $X=V^{h-1}$ and $Y=V^{k-1}$, and neither $h$ nor $k$ is 1. The product result (6.6) now gives the asserted formula for $g$.

7 Computations

7.1 Scaling

Let $\nu$ be a nonsquare in $GF(q)$ and let $\varphi^{\prime}=\nu\varphi$. Construct the algebra $\mathcal{A}^{\prime}$ with the procedure used for $\mathcal{A}$ but with $\varphi^{\prime}$ in place of $\varphi$. Label corresponding ingredients for $\mathcal{A}^{\prime}$ with the dash. The symplectic group for $\varphi^{\prime}$ is the same as that for $\varphi$. But the form in Definition 3.3 changes to $\Theta_{g}^{\prime}=\nu\Theta_{g}$. Then $\chi(\Theta_{g}^{\prime})=(-1)^{\dim V^{g-1}}\chi(\Theta_{g})$, and

[TABLE]

from Theorem 6.7. That is, $\eta^{\prime}$ comes from $\eta$ by changing $\rho$ to $-\rho$. The same holds for $\omega^{\prime}$ from $\omega$. As pointed out when this scaling is discussed in [14, Section 3], $-\rho$ may not be an algebraic conjugate of $\rho$.

7.2 Permuting $V$

Proposition 7.1

(compare [4, Theorem 4.4(c)]) The homomorphism $g\rightarrow t(g^{-1})^{T}\otimes t(g)\in\mathcal{M}_{q^{2n}}(K)$ is the permutation representation of $\mathrm{Sp}(V)$ on $V$. Consequently $\omega(g^{-1})\omega(g)=\left|V_{g}\right|$.

Proof. This is immediate from the fact that $t(g^{-1})(v)t(g)=(v^{g})$ for $v\in V$.

This leads to a corroboration of Theorem 6.7: we have

[TABLE]

and

[TABLE]

Now $v^{g-1}=-v^{g(g^{-1}-1)}$ implies that $V^{g-1}=V^{g^{-1}-1}$. If $u,v\in V^{g-1}$ and $v=y^{g-1}=(-y^{g})^{g^{-1}-1}$, then $\Theta_{g}(u,v)=\varphi(u,y)$ and $\Theta_{g^{-1}}(u,v)=-\varphi(u,y^{g})$. Thus $\Theta_{g^{-1}}(u,v)=-\Theta_{g}(u,v^{g})$. Hence $\det\Theta_{g^{-1}}=(-1)^{\dim V^{g-1}}\det(g|V^{g-1})\det\Theta_{g}$. But since $v^{g}\equiv v(\bmod\ V^{g-1})$ and $\det g=1$, we have $\det(g|V^{g-1})=1$. So $\det\Theta_{g^{-1}}=(-1)^{\dim V^{g-1}}\det\Theta_{g}$. Then

[TABLE]

Therefore

[TABLE]

as $V_{g}=(V^{g-1})^{\perp}$.

7.3 Embedding

From the “internal” formula for $\eta$ in Theorem 6.7, we can prove the following: let $V=U\oplus U^{\prime}$, with $U^{\prime}=U^{\perp}$; $\varphi$ is nondegenerate on $U$ and $U^{\prime}$. Then $\mathrm{Sp}(U)\times\mathrm{Sp}(U^{\prime})$ embeds into $\mathrm{Sp}(V)$: if $g\in\mathrm{Sp}(U)$ and $g^{\prime}\in\mathrm{Sp}(U^{\prime})$, then $(u+u^{\prime})^{g\times g^{\prime}}=u^{g}+(u^{\prime})^{g^{\prime}}$, where $u\in U$ and $u^{\prime}\in U^{\prime}$. We have $V^{g\times g^{\prime}-1}=U^{g-1}\oplus(U^{\prime})^{g^{\prime}-1}$. and $\Theta_{g\times g^{\prime}}(u_{1}+u_{1}^{\prime},u_{2}+u_{2}^{\prime})=(\Theta_{U})_{g}(u_{1},u_{2})(\Theta_{U^{\prime}})_{g^{\prime}}(u_{1}^{\prime},u_{2}^{\prime})$, the $\Theta$s on the right taken for $U$ and $U^{\prime}$. Then from the formula, $\eta(g\times g^{\prime})=\eta_{U}(g)\eta_{U^{\prime}}(g^{\prime})$ and $\omega(g\times g^{\prime})=\omega_{U}(g)\omega_{U^{\prime}}(g^{\prime})$.

7.4 Eigenvalue restrictions and semisimplicity

Proposition 7.2

If no eigenvalue of $g$ is 1, that is, $g-1$ is invertible, then

[TABLE]

Proof. For such a $g$, $V^{g-1}=V$. With $y_{2}=x_{2}^{g-1}$, we have $\Theta_{g}(y_{1},y_{2})=\varphi(y_{1},y_{2}^{(g-1)^{-1}})$. In matrix terms, $\Theta_{g}(y_{1},y_{2})=y_{1}\Phi(y_{2}(g-1)^{-1})^{T}=y_{1}\Phi(g-1)^{-T}y_{2}^{T}$, $\Phi$ being the matrix for $\varphi$. Thus $\det\Theta_{g}=\det(\Phi(g-1)^{-T})=\det\Phi\det(g-1)^{-1}$. But relative to a symplectic basis, $\det\Phi=1$. Hence $\chi(\Theta_{g})=\chi(\det(g-1))$. So

[TABLE]

from which the stated formula follows. Once again, $\eta(-1)=\delta^{n}q^{-n}$, because $\chi(\det(-2I_{2n}))=\chi(-2)^{2n}=1$.

Corollary 7.3

Let $h$ be the usual transvection $v\rightarrow v-\gamma^{-1}\varphi(v,c)c$, where $\gamma\neq 0$ and $c\neq 0$. Then

[TABLE]

Proof. All eigenvalues of $h$ are 1, so those of $-h$ are $-1$. Thus $\det(-h-1)=(-1)^{2n}=1$, and $\omega(-h)=\delta^{n}$. The formula follows from Propositions 6.1 and 7.2.

Corollary 7.4

If $g\in\mathrm{Sp}(V)$ is semisimple, then $\omega(g)=\omega^{\prime}(g)$.

Proof. Here $\omega^{\prime}$ is the Weil character for $\mathcal{A}^{\prime}$ from Subsection 7.1, and the restriction on $g$ means that $g$ is diagonalizable in some extension of $GF(q)$. If $E$ is the eigenspace of $g$ for eigenvalue 1, then the semisimplicity implies that $\varphi$ is nonsingular on $E$ and $g|E$ is the identity. With the decomposition $V=E\oplus E^{\perp}$, the results in Subsection 7.3 apply, $g$ being identified with $(g|E)\times(g|E^{\perp})$. For $g|E^{\perp}$, Proposition 7.2 gives $\omega_{E^{\perp}}(g|E^{\perp})=\delta^{n-m}\chi(\det(g|E^{\perp}-1))$, where $\dim E=2m$. As $\omega_{E}(g|E)=q^{m}$, we get $\omega(g)=q^{m}\delta^{n-m}\chi(\det(g|E^{\perp}-1))$. There being no $\rho$ here, we also have the same formula for $\omega^{\prime}(g)$, and $\omega^{\prime}(g)=\omega(g)$.

7.5 Constructions

One way to set up $V$ and $\varphi$ is to take $V=GF(q^{2n})$ and to let $\varphi(x,y)=\mathrm{tr}(\varepsilon xy^{q^{n}})$, where the trace is from $GF(q^{2n})$ to $GF(q)$ and $\varepsilon^{q^{n}}=-\varepsilon$. We have

[TABLE]

as needed, and the nondegeneracy of $\varphi$ follows from the nondegeneracy of the trace form itself. If $z\in V$ and $z^{q^{n}+1}=1$, then the multiplication $g:x\rightarrow zx$ is a member of $Sp(V)$. With $z\neq 1$, Proposition 7.2 applies. As in Lemma 6.2, the determinant of $x\rightarrow ax$ is the norm $a^{(q^{2n}-1)/(q-1)}$ of $a$. Moreover, $\chi(\alpha)=\alpha^{(q-1)/2}$ for $\alpha\in GF(q)$. Thus

[TABLE]

(Once again, we use $(-1)^{(q^{n}-1)/2}=\delta^{n}$.) Now $z\rightarrow z^{(q^{n}+1)/2}$ is the “quadratic character” $\chi_{Z}$ on the cyclic subgroup $Z$ of order $q^{n}+1$ in $GF(q^{2n})^{\emptyset}$: $\chi_{Z}(z)=1$ if $z$ is a square in $Z$, $-1$ if not. So $\chi(\det(g-1))=-\delta^{n}\chi_{Z}(z)$, and $\omega(g)=\delta^{n}\chi(\det(g-1))=-\chi_{Z}(z)$.

If $z\neq-1$, we can apply this to $g^{\prime}=-g$ and get

[TABLE]

So $\omega(-g)=\delta^{n}\chi_{Z}(z)$. If $g$ (and $z$) has order $o(g)$, then $\chi_{Z}(z)=(-1)^{(q^{n}+1)/o(g)}$. Thus for $z\neq\pm 1$,

[TABLE]

by 6.1.

A second construction for $V$ and $\varphi$ takes $V=GF(q^{n})\oplus GF(q^{n})$, with $\varphi((x,y),(x^{\prime},y^{\prime}))=\mathrm{tr}(xy^{\prime}-yx^{\prime})$, the trace now from $GF(q^{n})$ to $GF(q)$. This time the map $g:(x,y)\rightarrow(zx,z^{-1}y)$, for $z\in GF(q^{n})^{\emptyset}$, is in $Sp(V)$. Taking $z\neq 1$, we have

[TABLE]

Then

[TABLE]

with $\chi_{Z}$ now the quadratic character on $GF(q^{n})$. Thus $\omega(g)=\delta^{n}\chi(\det(g-1))=\chi_{Z}(z)$. Similarly, if $z\neq-1$ and $g^{\prime}=-g$, $\chi(\det(g^{\prime}-1))=\delta^{n}\chi_{Z}(-z)=\chi_{Z}(z)$, and $\omega(-g)=\delta^{n}\chi_{Z}(z)$. So this time, if $z\neq\pm 1$, $\omega_{-}(g)=\chi_{Z}(z)(1-\delta^{n}\times\delta^{n})/2=0$.

7.6 Some characters of $\mathrm{SL}(2,q)$

With the various character values in hand, we can write out the part of the character table for $\mathrm{Sp}(2,q)(\simeq\mathrm{SL}(2,q))$ that involves $\omega,\omega_{-}$, and $\omega_{+}$(see the table in [3, Section 38], for example, based on work of I. Schur). Let \left[\begin{array}[]{cc}0&1\\ -1&0\end{array}\right] be the matrix for $\varphi$. The transvection $h:v\rightarrow v-\gamma^{-1}\varphi(v,(0,1))(0,1)$ we have been using has matrix \left[\begin{array}[]{cc}1&-\gamma^{-1}\\ 0&1\end{array}\right]. From Corollary 6.6, $\omega(h)=\delta\rho\chi(\gamma)$, and by 7.3, $\omega(-h)=\delta$.

[TABLE]

Using $\mathcal{A}^{\prime}$ from Subsection 7.1, we can add two more irreducible characters $\omega_{-}^{\prime}$ and $\omega_{+}^{\prime}$ to this table, with the same rows as for $\omega_{-}$ and $\omega_{+}$ but with $-\rho$ in place of $\rho$.

8 Factorizations

The embedding of $\mathrm{Sp}(V)$ into $\mathcal{A}$ can be used to study factorizations of group elements. The best known is the fact that every member of $\mathrm{Sp}(V)$ is a product of transvections. Initially one shows that the transvections generate $\mathrm{Sp}(V)$ without paying attention to how many transvections are needed in a product for a given group member [10, Theorem 8.5]. That number was examined by Dieudonné [2], and we shall deal with it in this section.

Theorem 8.1

[2, Theorem 2]* If $g\in\mathrm{Sp}(V)$ and $g$ is not an involution, then $g$ is the product of $\dim V^{g-1}$ transvections. If $g$ is an involution, then $g$ is the product of $\dim V^{g-1}+1$ transvections. As usual, $1$ is declared to be an empty product.*

Proof. If $g=g_{1}\cdots g_{r}$ with transvections $g_{i}$, then $V^{g-1}\subseteq V^{g_{1}-1}+\cdots+V^{g_{r}-1}$. Since $\dim V^{g_{i}-1}=1$, by Proposition 3.8, $\dim V^{g-1}\leq r$; so there must be at least $\dim V^{g-1}$ transvections in the product. Suppose that $g$ is not an involution. Then $Q_{g}$ is not identically 0, by Proposition 3.7. When $\dim V^{g-1}=1$, $g$ is already a transvection. Thus we can assume that $\dim V^{g-1}>1$. If there is $c\in V^{g-1}$ for which $Q_{g}(c)=\gamma\neq 0$ and for which $Q_{g}$ is not identically 0 on $Y=\left\langle c\right\rangle^{\Theta_{g}}$, then with $X=GF(q)c$ in Corollary 3.11, $g=hk$, where $h$ is the transvection $v\rightarrow v-\gamma^{-1}\varphi(v,c)c$ and $Y=V^{k-1}$. Here $k$ is also not an involution since $Q_{k}=Q_{g}|Y$ is not identically 0. Induction then implies that $k$ is the product of $\dim Y=\dim V^{g-1}-1$ transvections, making $g$ the product of $\dim V^{g-1}$ transvections.

For the needed $c$, note that the map $\left\langle c\right\rangle\rightarrow\left\langle c\right\rangle^{\Theta_{g}}$ is one-to-one between lines and hyperplanes of $V^{g-1}$, because $\Theta_{g}$ is nondegenerate. Suppose that there were three different hyperplanes in $V^{g-1}$ that are totally singular for $Q_{g}$. Then the intersection of any two of them must be the radical, $\mathrm{rad}Q_{g}$, of $Q_{g}$. The induced quadratic form on the two-dimensional space $V^{g-1}/\mathrm{rad}Q_{g}$ would then have at least three singular lines. However, nonsingular planes have either none or two [10, pp. 138–139]. Thus in general, $V^{g-1}$ has at most two $Q_{g}$-totally singular hyperplanes. But since $\dim V^{g-1}>1$, there are at least three lines $\left\langle c\right\rangle$ with $Q_{g}(c)\neq 0$, as is easy to see. Therefore we can find the required $c$ with $Q_{g}(c)\neq 0$ and $Q_{g}$ not identically 0 on $\left\langle c\right\rangle^{\Theta_{g}}$.

Now let $g$ be an involution, with $\dim V^{g-1}=2l$. It cannot be that $g=g_{1}\cdots g_{2l}$ for transvections $g_{i}$. For if so, it must be that $V^{g-1}=V^{g_{1}-1}\oplus\cdots\oplus V^{g_{2l}-1}$, by the dimensions. If we set $h=g_{1}$ and $k=g_{2}\cdots g_{2l}$ in Remark 3.12, then $\Theta_{g}|V^{g_{1}-1}=\Theta_{g_{1}}$. But the left is 0, since $Q_{g}=0$ and $\Theta_{g}=-\varphi/2$ on $V^{g-1}$. Yet the right is not 0, by Proposition 3.8. Thus a transvection product for $g$ must have more than $2l$ factors.

If $c\in V^{g-1}$, $c\neq 0$, then $c^{g}=-c$. Let $h$ be the transvection $v\rightarrow$. Then $gh=hg$: for the successive images of $v$ by $gh$ are $v\rightarrow v^{g}\rightarrow v^{g}-\alpha\varphi(v^{g},c)c$; and by $hg$ are $v\rightarrow v-\alpha\varphi(v,c)c\rightarrow v^{g}+\alpha\varphi(v,c)c$. But this last is indeed $v^{g}+\alpha\varphi(v^{g},c^{g})c=v^{g}-\alpha\varphi(v^{g},c)c$. Thus $(gh^{-1})^{2}=h^{-2}$ and $gh^{-1}$ is not an involution. So it is a product $g_{1}\cdots g_{r}$ of transvections, with$r\leq\dim V^{g-1}$, since $V^{gh-1}\subseteq V^{g-1}$. Then $g=hg_{1}\cdots g_{r}$, and now it must be that $r=\dim V^{g-1}=2l$. Notice the freedom of choice for $h$.