$\def\FF{\mathbb{F}}\def\CC{\mathbb{C}}\def\QQ{\mathbb{Q}}\def\Sp{\text{Sp}}\def\SL{\text{SL}}\def\GL{\text{GL}}\def\PGL{\text{PGL}}$Let $p$ be an odd prime. The Weil representation is a $p^k$-dimensional complex representation of $\Sp_{2k}(\FF_p)$. If you read most descriptions of the Weil representation, they hit a key technical point: "linearizing the projective representation". They then usually apologize for how difficult this is and either (1) cite the details to some one else (2) perform intricate computations with generators and relations of $\Sp_{2k}(\FF_p)$ (3) perform computations in group cohomology or (4) invoke tools from algebraic geometry.

I think I have a way to linearize the projective representation which is completely elementary, and which allows me to write down the matrices of $\Sp_{2k}(\FF_p)$ in a completely elementary way. So I'm writing to ask if anyone has seen this, or if they know any reason it can't work.

Apologies for the length; it takes a while to lay out the notation.

The standard description: Let $L$ be a $k$-dimensional vector space over $\FF_p$, let $L^{\vee}$ be the dual space and let $V = L \oplus L^{\vee}$, equipped with a symplectic form $( \ , \ )$ in the usual way. Let $H$ be the Heisenberg group, which is a certain extension $1 \to \FF^+_p \to H \to V \to 1$. I'll denote $\FF^+_p$ as $Z$ when I am thinking of it as the center of $H$. The Heisenberg group comes with a natural action of $\Sp(V)$; I'll write it as $h \mapsto h^g$ for $h \in H$ and $g \in \Sp(V)$. It is important to know that $z^g=z$ for all $z \in Z$ and $g \in \Sp(V)$.

Let $S$ be the $p^k$-dimensional vector space of $\CC$ valued functions on $L$. There is a natural action of $H$ on $S$ called the "Schrodinger representation", which is characterized as the unique irreducible representation of $H$ where $c \in \FF_p \subset H$ acts by $\zeta^c$. We'll write $\rho_S : H \to \GL(S)$ for the Schrodinger representation. I'll give explicit matrix formulas for $\rho_S$ below.

Let $g \in \Sp(V)$. Then $h \mapsto \rho_S(h^g)$ is another representation of $H$, in which $Z$ acts by the same character, so this new representation is isomorphic to $S$. Thus, there is some matrix $\alpha(g)$, well defined up to scalar multiple, such that $\rho_S(h^g) = \alpha(g) \rho_s(h) \alpha(g)^{-1}$.

The issue with linearizing the projective representation What most sources now explain is that it is clear that $\alpha(g_1) \alpha(g_2) = \alpha(g_1 g_2)$ inside $\PGL(S)$, but that it is not clear that they can be lifted to matrices that obey this relation in $\GL(S)$. This is where the big tools come out.

How I want to do it My idea is to write down an explicit list $\Gamma$ of matrices in $\GL(S)$ which (1) form a subgroup and (2) normalize the image of $H$ in $\GL(S)$. Once I do this, $\Gamma$ will act on $H$ and hence will act on $H^{\text{ab}} \cong V$, and it is easy to show that this gives a map $\Gamma \to \Sp(V)$. I will then (3) show that this map is an isomorphism. In other words, my strategy is not to ask "given a matrix in $\Sp(V)$, how should it act on $S$?" but, rather, "what is a subgroup of $\GL(S)$ which normalizes $H$ and acts on $H$ in the right way?"

The Heisenberg representation in matrices First, let me tell you what the representation $\rho_S$ is. Remember that $S$ is the $\CC$-valued functions on $L$ so, for a matrix $M$ in $\GL(S)$, the rows and columns of $M$ are indexed by the elements of $L$. I'll generally denote them as $M_{yx}$, for $x$, $y \in L$.

For $\lambda \in L$, $\lambda^{\vee} \in L^{\vee}$ and $c \in \FF_p$, the corresponding matrix in $\GL(S)$ is $$\rho_S(\lambda, \lambda^{\vee}, c)_{yx} = \begin{cases} \zeta^{\lambda^{\vee}(x)+c} & y=x+\lambda \\ 0 & \text{otherwise} . \end{cases}.$$

The Weil representation I will now give a similar description of a list of matrices normalizing $H$. Each matrix will be indexed by the following data: (1) A vector space $R \subseteq L \oplus L$ and (2) a quadratic form $q$ on $R$. These will obey conditions, to be described later. Define $$K(R,q)_{yx} = \begin{cases} \zeta^{q((x,y))} & (x,y) \in R \\ 0 & \text{otherwise} . \end{cases}.$$

Remark: The following may give some intuition. If $R = L \oplus L$ and $q(x_1, x_2, \ldots, x_g, y_1, y_2, \ldots, y_g) = \sum x_j y_j$, then this is the finite Fourier transform. If $R$ is the graph of some isomorphism $\phi : L \to L$, and $q=0$, then $\left[ \begin{smallmatrix} \phi & 0 \\ 0 & \phi^{-T} \end{smallmatrix} \right]$ is in $\Sp(V)$, and this is the standard description of how such a matrix acts in the Weil representation. If $R$ is the diagonal $\{ (x,x) : x \in L \}$, and $q$ is a quadratic form on $L$, then we can think of $q$ as a self-adjoint map $L \to L^{\vee}$; then $\left[ \begin{smallmatrix} 1 & 0 \\ q & 1 \end{smallmatrix} \right]$ is in $\Sp(V)$, and this is the standard description of how such a matrix acts in the Weil representation.

I have found the following criterion for when such a matrix is invertible:

Lemma If $K(R,q)$ is invertible, then the projections of $R$ onto $L \oplus 0$ and $0 \oplus L$ must both be surjective. (In particular, $\dim R \geq \dim L$.) Given $R$ such that these projections are surjective, define $X = R \cap (L \oplus 0)$ and $Y = R \cap (0 \oplus L)$. For such an $R$ and any $q$, the formula $\langle x,y \rangle = q((x,y)) - q((x,0)) - q((0,y))$ defines a bilinear pairing between $X$ and $Y$. The matrix $K(R,q)$ is invertible if and only if this pairing is nondegenerate. If so, then $$\det K(R,q) = \pm (p^{\ast})^{(\dim L) (\dim R - \dim L)/2} \ \text{where}\ p^{\ast} = (-1)^{(p-1)/2}.$$

Remark If $\dim R = \dim L$, then the condition that $R$ surjects onto $L \oplus 0$ and onto $0 \oplus L$ just says that $R$ is the graph of an isomorphism $\phi: L \to L$. In this case, $X=Y=0$, so the condition on $q$ is automatic. This corresponds to the representation theory of the subgroup $\left[ \begin{smallmatrix} \phi & 0 \\ \ast & \phi^{-T} \end{smallmatrix} \right]$ in $\Sp(V)$.

For $(R,q)$ such that $K(R,q)$ is invertible, set $$\gamma(R,q) = \tfrac{\pm 1}{(p^{\ast})^{(\dim R-\dim L)/2}} K(R,q)$$ where the $\pm 1$ is chosen to make the determinant $1$. Since $p$ is odd, we know that $\dim S = |L|$ is odd, and this therefore defines a unique sign. (This is the key step which has no analogue for the real symplectic group; there is no determinant for operators on an infinite dimensional Hilbert space.)

Our group $\Gamma$ will be the collection of $\gamma(R,q)$, for $R$ and $q$ obeying the condition of the lemma.

Incidentally, it is easy to describe the group $\Gamma H$: One just allows $R$ to be an affine linear space and takes $q$ an inhomogenous polynomial function of degree $\leq 2$ on $R$. When $k=1$, I believe this is the full normalizer of $H$ in $\SL_p(\CC)$; when $k>1$, I think it is the normalizer of $H$ in $\SL_{p^k}(\QQ(\zeta))$.

That was long for a Mathoverflow post, but compared to any papers I have seen addressing the linearization issue, it is pretty short and much more explicit! The remaining tasks are to check of points (1), (2) and (3) in the "how I want to do it" paragraph. All of these are on the level of exercises.

So, has anyone seen this? Or, I suppose, does anyone have a reason to think I've screwed up?

  • 1
    $\begingroup$ @SamHopkins Thanks. I noticed that and decided it wasn't worth fixing but, now that you mention it, I have. Trying to use $k$ for the rank of the symplectic group now. $\endgroup$ Oct 14, 2022 at 2:44
  • 1
    $\begingroup$ For the $\Gamma$ you have constructed, why is the map $\Gamma \rightarrow Sp(V)$ an isomorphism? $\endgroup$
    – spin
    Oct 14, 2022 at 6:00
  • 2
    $\begingroup$ Your approach is more geometric than most I have seen (but I am an algebraist, after all), and is very clean. The idea of using unimodular matrices to induce the right automorphisms( especially automorphisms of order prime to $p$ in this context) has occurred incharacter theory and representation theory of finite groups to define "canonical"extensions which impose some rigidity which is useful in trying to extend further. This observation may have been made first by Marty Isaacs. For odd $p$, a key point is that $\endgroup$ Oct 14, 2022 at 20:56
  • 1
    $\begingroup$ @L Spice: that ${\rm Sp}(2n,p)$ is generated by elements of order prime to $p$ ( except when $p= 3$ and $n=1$), so it suffices to understand how automorphims of order prime to $p$ are represented. ( I had completed the comment, but was chided by the system for taking more than 5 minutes to edit,then ran out of steam. I may also complete a post I started to write ,if I get to the point that I think it sheds any new light. $\endgroup$ Oct 14, 2022 at 22:56
  • 3
    $\begingroup$ Have you compared your approach to that in "B. Bolt, T. G. Room and G. E. Wall, On the Clifford collineation, transform and similarity groups. I, J. Austral. Math. Soc., 2 (1961-62), 60-79." ? $\endgroup$
    – spin
    Oct 15, 2022 at 7:09

2 Answers 2


Just a comment, here are some original references for the construction of the Weil representation.

[1] B. Bolt, T. G. Room and G. E. Wall, On the Clifford collineation, transform and similarity groups. I, J. Austral. Math. Soc., 2 (1961-62), 60-79. DOI

[2] I. M. Isaacs, Characters of solvable and symplectic groups, Amer. J. Math., 95 (1973), 594-635. DOI

[3] R. E. Howe, On the characters of Weil’s representations, Trans. Amer. Math. Soc., 177 (1973), 287-298. DOI

[4] P. Gérardin, Weil representations associated to finite fields, J. Algebra, 46 (1977), 54-101. DOI

[5] H. N. Ward, Representations of symplectic groups, J. Algebra, 20 (1972), 182-195. DOI


I will attempt to write up a representation-theoretic argument influenced by papers I have seen (and sometimes written) over the years, making use of automorphisms of extraspecial groups. It may come under the umbrella of proofs which you have seen, and consider more complicated than the one you have posted (this is certainly more algebraic and less geometric than yours, but I think it points out some underlying ideas common to both approaches). I do not give all details,and while I make mention of relevant background results from time to time, the treatment is essentially self-contained (apart from the use of the simplicity of ${\rm Sp}(2n,p)$).

If we take an extraspecial group $E$ of order $p^{2n+1}$ and exponent $p$, where $p$ is an odd prime (and we take $p >3$ if $n = 1$, that case needing individual treatment, which we don't pursue here), we may proceed as follows.

The group $E$ has an explicit monomial (and, in particular, unitary) complex irreducible representation of degree $p^{n}$ which is induced from a one-dimensional representation of a maximal Abelian (and normal) subgroup $A$ of $E$. The irreducible character $\chi$ afforded vanishes outside $Z(E)$,and $Z(E)$ has order $p$. This induced representation,say $\rho$ gives an embedding of $E$ into $T = {\rm SU}(p^{n},\mathbb{C})$.

The representation $\rho$ of $E$ in fact extends uniquely to a homomorphism of the semidirect product $E.{\rm SL}(2n,p)$ into $ T = {\rm SU}(p^{n},\mathbb{C})$, as we will outline here, summarizing and amalgamating the results and methods of many authors, as referenced,for example, in @spin's answer . Notice that $N_{T}(E\rho)$ is a finite group (from now on, for ease of notation, we will identify $E$ with $E\rho$ if there is no danger of ambiguity). For $C_{T}(E\rho)$ has order $p$, and $N_{T}(E\rho)/C_{T}(E\rho)$ is isomorphic to a subgroup of the finite group ${\rm Aut}(E).$

The group $N_{T}(E)/E$ is isomorphic to a subgroup of ${\rm Sp}(2n,p)$ ( and is in fact all of ${\rm Sp}(2n,p),$ as will be seen).

Since ${\rm Sp}(2n,p)$ is a non-Abelian simple group (remember the excluded case!),it is generated by its elements of order prime to $p$). This simplifying remark (no pun intended) is not essential, but it makes the exposition easier if we assume it.

The idea of the construction to follow is that for any automorphism $\sigma$ of order prime to $p$ of $E$, there is a unique way to extend the representation $\rho$ of $E$ to $E \langle \sigma \rangle $ in such a way that $\sigma \rho \in {\rm SU}(p^{n},\mathbb{C})$ ,and in fact in that case, $\sigma \rho$ has trace $\epsilon(\sigma) p^{d(\sigma)}$, where $\epsilon(\sigma)$ is a unique sign forced by determinant one,and $|C_{E}(\sigma)| = p^{2d(\sigma)+1}.$

As I have said in comments, this uniqueness of the unimodular extension to $E\langle \sigma \rangle$ (as far as I know), goes back to Marty Isaacs,( see the answer by @spin), and the fact that the character values of the unique extension are as stated was known to Isaacs. The character values can also be calculated using Glauberman correspondence (given the unimodularity).

If we accept this uniqueness for the moment, we see that $\langle E\rho, \sigma \rho$ : $\sigma\in {\rm Sp}(2n,p)$ is $p$-regular $\rangle$ must be all of $N_{T}(E\rho)$ , and that $N_{T}(E)/E$ is (isomorphic to) all of ${\rm Sp}(2n,p)$ we have made use of the fact that ${\rm Sp}(2n,p)$ is generated by its elements of order prime to $p$ (also known as $p$-regular elements)

Notice, in particular, that in the case that $\sigma$ is the central element of order $2$ in ${\rm Sp}(2n,p)$, which fixes $Z(E)$ and induces elementwise inversion on $E/Z(E)$, we see that $\sigma \rho$ has trace $\epsilon,$ where $p \equiv \epsilon$ (mod $4$) and $\epsilon = \pm 1$.

This gives that $N_{T}(E) = EC_{T}(\sigma)^{\prime}$ for this choice of $\sigma$, since we have $N_{T}(E) \cap C_{T}(\sigma) \cong Z(E) \times {\rm Sp}(2n,p)$ ( here,the superscript used denotes taking the derived subgroup, as usual).

The uniqueness of $\sigma\rho$ for a chosen general automorphism $\sigma \in {\rm Sp}(2n,p)$ of order $h$, coprime to $p$, can be seen as follows, which is a (familiar) explicit way to define a matrix with the correct intertwining properties ( unique up to scalars, due to irreducibility of the representation of $E$, as noted in the question).

We define the matrix $M(\sigma)$ via $M(\sigma) = \sum_{u \in E/C_{E}(\sigma)} (u^{-1}u^{\sigma})\rho$ ( which is also $|C_{E}(\sigma)|^{-1} \sum_{u \in E}(u^{-1}u^{\sigma})\rho.$

The latter expression makes it clear that we have $(v\rho)^{-1}M(\sigma) (v^{\sigma}\rho) = M(\sigma)$ for each $v \in E$, so that $M(\sigma)^{-1}(v\rho) M(\sigma) = (v^{\sigma} \rho)$ for each $v \in E$, so conjugation by the matrix $M(\sigma)$ has the same effect on $E\rho$ as $\sigma$ does on $E$.

We may note that $M(\sigma)$ really is invertible as follows. The only time that $(u^{-1}u^{\sigma})\rho$ can have non-zero trace is when $u^{-1}u^{\sigma} \in Z(E)$,in which case ${\rm trace}(u^{\sigma}\rho) = \lambda {\rm trace}(u\rho)$ for some $p$-th root of unity $\lambda$.

Since $\sigma$ has order prime to $p$, this is easily seen to force $\lambda = 1$, so we need $u^{\sigma} = u$ in that case. It follows then that ${\rm trace}(M(\sigma) ) = p^{n} = \chi(1).$ In particular, $M(\sigma)$ is not nilpotent.

Since $\sigma$ is an automorphism of order $h$ coprime to $p$, it follows that $M(\sigma)^{h}$ is a non-zero scalar matrix. Hence we may take a scalar multiple of $T(\sigma)$ of multiplicative order $h$ (in fact an easy character-theoretic argument tells us that $M(\sigma) = \epsilon(\sigma)p^{n-d(\sigma)} F(\sigma)$, where $F(\sigma)$ is a matrix of order $h$ and determinant one, so the unique way to extend the (special) unitary representation $\rho$ of $E$ to $E\rangle \sigma \rangle$ is to set $\sigma \rho = F(\sigma).$ This is because we find that ${\rm trace}(M(\sigma)) = [E:C_{E}(\sigma)] \frac{| {\rm trace}(F(\sigma)|^{2}} {\chi(1)},$ while we know that $M(\sigma)$ has trace $p^{n}.$

While it might appear that the necessary choice of scalar is not explicit, and outside our control, the scalar is in fact uniquely recoverable from the representation-theoretic and character-theoretic information we have available.

The choice of $\sigma \rho$ as $F(\sigma)$ tells us that we have extended the character $\chi$ to $E\langle \sigma\rangle$ in such a way that $M(\sigma) = [E:C_{E}(\sigma)] \frac{\overline{\chi(\sigma)}}{\chi(1)}(\sigma \rho)$, and also that $|\chi(\sigma)|^{2} = \frac{|C_{E}(\sigma)|}{p}.$ In other words, the explicit multiple of $M(\sigma)$ need to obtain a matrix $F(\sigma)$ of finite order $h$ is determined up to a scalar (of absolute value one) multiple, and forcing the determinant to be one fixes the choice of scalar.

It is a small exercise to see that general theory and the fact that $E\rho$ consists of unitary matrices (and is irreducible) forces $F(\sigma)$ to be unitary (given that it has finite order).

Hence the uniquess of the extension of $\rho$ to a unimodular representation of $E\langle \sigma \rangle$ is established whenever $\sigma \in {\rm Sp}(2n,p)$ is $p$-regular, and the demonstration that $N_{T}(E\rho)/E\rho$ is isomorphic to ${\rm Sp}(2n,p)$ is complete.

Further edit: I realise that the fact that $p$-regular elements of ${\rm Sp}(2n,p)$ have rational trace in the above representation is not really fully explained. This can be explained at the level of characters, by arguments which I think originate with Isaacs.

At the level of characters, the irreducible character $\chi$ of $E$ has been shown to have a unique extension to an irreducible character of $E{\rm Sp}(2n,p)$ (which we still call $\chi$) such that the well- defined linear character ${\rm det} \chi$ is the trivial character.

Now we note that if $\tau$ is a Galois automorphism (of the appropriate cyclotomic field) which fixes $p$-power roots of unity, then the irreducible character $\chi^{\tau}$ has the same properties as $\chi$, so must be equal to $\chi$ by the uniqueness of $\chi$. Then we see by Galois theory that $\chi(g) \in \mathbb{Q}$ whenever $g$ is $p$-regular.


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