2.1 A palette of possibilities

To illustrate the range of possibilities that can arise, we will start this section with three examples, and then state our main results. We will elaborate on these examples in § 5.

Example 2.1 (Elliptic curve).

Let $a$ be a non-zero integer. Consider the $\mathbb{Z}$ -Lie algebra $\mathfrak{g}_{a}$ , introduced by Boston and Isaacs [Reference Boston and IsaacsBI04, § 3], which is spanned as a free $\mathbb{Z}$ -module by $\{v_{1},\ldots ,v_{9}\}$ , subject to the relations

$$\begin{eqnarray}[v_{1},v_{4}]=[v_{2},v_{5}]=[v_{3},v_{6}]=v_{7},\quad [v_{1},v_{5}]=[v_{2},v_{6}]=v_{8},\quad [v_{1},v_{6}]=av_{9},\quad [v_{2},v_{4}]=[v_{3},v_{4}]=v_{9}.\end{eqnarray}$$

All other brackets $[v_{i},v_{j}]$ with $i<j$ vanish. It will be shown in § 5.2 that if $p$ is a sufficiently large prime ( $p>1800$ will suffice) and $p$ does not divide $a$ , then

$$\begin{eqnarray}m_{\text{faithful}}(\exp (\mathfrak{g}_{a}\otimes _{\mathbb{Z}}\mathbb{F}_{p}))=3p^{2}.\end{eqnarray}$$

As we will see in the proof, the uniformity in $p$ is related to the fact that for such values of $p$ the cubic curve $Y^{2}=4aX^{3}+X^{2}-4X$ has a non-zero rational point over $\mathbb{F}_{p}$ . Note that in this example, aside from a finite set of primes, the value of $m_{\text{faithful}}(\exp (\mathfrak{g}_{a}\otimes _{\mathbb{Z}}\mathbb{F}_{p}))$ is given by one polynomial in $p$ .

Example 2.2 (Binary quadratic form).

Consider the $\mathbb{Z}$ -Lie algebra $\mathfrak{g}$ spanned as a free $\mathbb{Z}$ -module by $\{v_{1},\ldots ,v_{6}\}$ subject to the relations

$$\begin{eqnarray}[v_{1},v_{2}]=[v_{3},v_{4}]=v_{5},\quad [v_{1},v_{4}]=[v_{2},v_{3}]=v_{6},\end{eqnarray}$$

where all other commutators $[v_{i},v_{j}]$ with $i<j$ are defined to be $0$ . Then in § 5.3 we will show that for odd primes $p$ , the value of $m_{\text{faithful}}(\mathscr{G}_{p})$ is given by two different polynomials along two arithmetic progressions, namely,

$$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{p})=\left\{\begin{array}{@{}ll@{}}2p\quad & \text{if }p\equiv 1\;(\text{mod}\;4),\\ 2p^{2}\quad & \text{if }p\equiv 3\;(\text{mod}\;4).\end{array}\right.\end{eqnarray}$$

Put more formally, set

$$\begin{eqnarray}\mathscr{P}_{1}:=\{p\geqslant 3:p\equiv 1\;(\text{mod}\;4)\}\quad \text{and}\quad \mathscr{P}_{2}:=\{p\geqslant 3:p\equiv 3\;(\text{mod}\;4)\}.\end{eqnarray}$$

Also, set $g_{1}(T):=2T$ and $g_{2}(T):=2T^{2}$ . Then $m_{\text{faithful}}(\mathscr{G}_{p})=g_{i}(p)$ for all $p\in \mathscr{P}_{i}$ , $i=1,2$ .

As the next example shows, even this is not the end of the story.

Example 2.3 (Binary cubic form).

Consider the $\mathbb{Z}$ -Lie algebra $\mathfrak{g}$ spanned as a free $\mathbb{Z}$ -module by $\{v_{1},\ldots ,v_{8}\}$ with the following relations:

$$\begin{eqnarray}[v_{1},v_{4}]=[v_{2},v_{5}]=[v_{3},v_{6}]=v_{7},\quad [v_{1},v_{5}]=[v_{2},v_{6}]=[v_{3},v_{5}]=v_{8},\quad [v_{3},v_{4}]=-v_{8}.\end{eqnarray}$$

All other commutators $[v_{i},v_{j}]$ with $i<j$ vanish. Let $p$ be an odd prime. In § 5.4 we will show that

$$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{p})=\left\{\begin{array}{@{}ll@{}}p^{2}+p^{3}\quad & \displaystyle \text{if }\biggl(\frac{p}{23}\biggr)=-1,\\ 2p^{3}\quad & \text{if }p\text{ is represented by the form }2x^{2}+xy+3y^{2},\\ 2p^{2}\quad & \text{if }p\text{ is represented by the form }x^{2}+xy+6y^{2}\quad \text{or}\quad p=23,\end{array}\right.\end{eqnarray}$$

where $(\frac{\cdot }{p})$ is the Legendre symbol. The conditions defining this function split the set of prime numbers $p\geqslant 3$ into disjoint sets $\mathscr{P}_{1},\mathscr{P}_{2}$ and $\mathscr{P}_{3}$ . On each one of these sets, one of the polynomials $g_{1}(T)=T^{2}+T^{3}$ , $g_{2}(T)=2T^{3}$ and $g_{3}(T)=2T^{2}$ is applicable. It is worth mentioning that by Gauss genus theory, the sets $\mathscr{P}_{2}$ and $\mathscr{P}_{3}$ are not unions of arithmetic progressions (for example see [Reference KusabaKus67]).

Example 2.4 (Lee’s Lie algebra).

Consider the $\mathbb{Z}$ -Lie algebra $\mathfrak{g}$ spanned as a free $\mathbb{Z}$ -module by $\{v_{1},\ldots ,v_{8}\}$ with the following relations:

$$\begin{eqnarray}[v_{1},v_{4}]=[v_{2},v_{5}]=v_{6},\quad 2[v_{1},v_{5}]=[v_{3},v_{4}]=2v_{7},\quad [v_{2},v_{4}]=[v_{3},v_{5}]=v_{8}.\end{eqnarray}$$

All other commutators $[v_{i},v_{j}]$ with $i<j$ vanish. This Lie algebra was defined in [Reference LeeLee16]. Let $p$ be an odd prime. In § 5.5 we will show that

$$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{p})=\left\{\begin{array}{@{}ll@{}}p+2p^{2}\quad & \displaystyle \text{if }p\equiv 2\;(\text{mod}\;3)\text{ or }p=3,\\ 3p\quad & \text{if }p\equiv 1\;(\text{mod}\;3)\text{ and }p\text{ is represented by the form }x^{2}+27y^{2},\\ 3p^{2}\quad & \text{if }p\equiv 1\;(\text{mod}\;3)\text{ and }p\text{ is not represented by the form }x^{2}+27y^{2}.\end{array}\right.\end{eqnarray}$$

We remark that in [Reference LeeLee16], the author computes the order of the automorphism group of the Lie algebra $\mathfrak{g}_{p}$ . The formula obtained in [Reference LeeLee16] is given by different polynomials depending on the splitting of $\unicode[STIX]{x1D706}^{3}-2$ in $\mathbb{F}_{p}$ , and therefore its cases are parallel to the ones that appear above.

The important point to note here is that in each one of the above examples, the set of primes can be decomposed into finitely many arithmetically defined sets, such that the value of $m_{\text{faithful}}(\mathscr{G}_{p})$ on each one of these sets is given by a polynomial. Let us explain this in more detail. For a polynomial $g(T)\in \mathbb{Z}[T]$ , denote by ${\mathcal{V}}_{g}$ the set of primes $p$ for which the congruence $g(T)\equiv 0\;(\text{mod}\;p)$ has a solution. We will call ${\mathcal{V}}_{g}$ an elementary Frobenius set. Let $\mathbb{P}$ denote the set of prime numbers. By a Frobenius set we mean an element of the Boolean algebra inside the power set of $\mathbb{P}$ that is generated by elementary Frobenius sets. In other words, a Frobenius set is a finite union of sets of the form

$$\begin{eqnarray}{\mathcal{V}}_{g_{1}}\cap \cdots \cap {\mathcal{V}}_{g_{k}}\cap {\mathcal{V}}_{g_{k+1}}^{c}\cap \cdots \cap {\mathcal{V}}_{g_{l}}^{c}.\end{eqnarray}$$

We remark that every Frobenius set is a Frobenian set, as defined by Serre [Reference SerreSer12, § 3.3.1], but the converse does not hold. For more details about the connection between Frobenius and Frobenian sets, see [Reference LagariasLag83].

For $p>2$ the equation $x^{2}+1=0$ has a solution in $\mathbb{F}_{p}$ if and only if $p\equiv 1\;(\text{mod}\;4)$ . This shows that the sets appearing in Example 2.2 are Frobenius sets. One can see that the sets $\mathscr{P}_{i}$ in Example 2.3 are Frobenius sets as follows. First, using the quadratic reciprocity law one can easily verify that the set $\mathscr{P}_{1}$ consists of those primes $p\geqslant 3$ for which the equation $x^{2}+23$ has no solution in $\mathbb{F}_{p}$ . The other two parts are based on the less trivial fact that $p\geqslant 3$ can be represented by the quadratic form $a^{2}+ab+6b^{2}$ (respectively, $2a^{2}+ab+3b^{2}$ ) if and only if $p\not \in \mathscr{P}_{1}$ and $x^{3}-x-1$ has a solution (respectively, no solution) in $\mathbb{F}_{p}$ . Consequently, each $\mathscr{P}_{i}$ is a Frobenius set.

Let $\mathscr{P}$ be any set of primes. The Dirichlet density of $\mathscr{P}$ is defined by

$$\begin{eqnarray}d(\mathscr{P}):=\lim _{s\rightarrow 1^{+}}\frac{\mathop{\sum }_{p\in \mathscr{P}}p^{-s}}{\mathop{\sum }_{p\in \mathbb{P}}p^{-s}},\end{eqnarray}$$

if the limit exists. The Chebotarev density theorem implies that any infinite Frobenius set has a positive Dirichlet density. It follows from Dirichlet’s theorem on primes in arithmetic progressions that the Dirichlet density of the sets $\mathscr{P}_{i}$ in Example 2.2 is positive. For the sets $\mathscr{P}_{i}$ in Example 2.3, positivity of the Dirichlet density can be proved by class field theory. For more details we refer the reader to [Reference CoxCox13, Theorem 9.12].

With this preparation, we are now ready to state our first result.

Theorem 2.5. Let $\mathfrak{g}$ be a nilpotent $\mathbb{Z}$ -Lie algebra of nilpotency class $c$ which is finitely generated as an abelian group. Then there exist a partition $\mathscr{P}_{1},\ldots ,\mathscr{P}_{r}$ of the set of prime numbers larger than $c$ into Frobenius sets, and polynomials $g_{1}(T),\ldots ,g_{r}(T)$ with non-negative integer coefficients, depending only on $\mathfrak{g}$ , such that

$$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{p})=g_{i}(p)\quad \text{for all }p\in \mathscr{P}_{i},\end{eqnarray}$$

where $1\leqslant i\leqslant r$ .

The proof of this theorem relies on a theorem of Ax [Reference AxAx67] from model theory (see also van den Dries [Reference van den DriesVdDri91]), coupled with a parameterization of irreducible representations provided by the Kirillov machinery. As we shall see, studying $m_{\text{faithful}}(\mathscr{G}_{p})$ leads to questions related to the existence of rational points over finite fields of certain determinantal varieties associated with the commutator matrix. We remark that our proof is effective and we can use its method to describe the associated Frobenius sets and polynomials. We will demonstrate this by a detailed analysis of the above examples after we give the proof of Theorem 2.5.

One can also consider $m_{\text{faithful}}(\mathscr{G}_{q})$ for $q:=p^{f}$ when the prime $p$ is fixed and $f$ varies. Let $\mathscr{P}$ be one of the Frobenius sets in Theorem 2.5 with the associated polynomial $g(T)\in \mathbb{Z}[T]$ . The following example shows that it is not necessarily true that

$$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{q})=fg(q)\quad \text{for all }f\geqslant 1.\end{eqnarray}$$

Example 2.6. Take $\mathfrak{g}$ as in Example 2.2 and set $\mathscr{P}:=\mathscr{P}_{2}$ , where $\mathscr{P}_{2}$ is as in the same example. Recall that

$$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{p})=2p^{2}\quad \text{for all }p\in \mathscr{P}.\end{eqnarray}$$

However, in § 5.3 we shall demonstrate that

$$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{q})=\left\{\begin{array}{@{}ll@{}}2fq\quad & f\text{ is even},\\ 2fq^{2}\quad & f\text{is odd.}\end{array}\right.\end{eqnarray}$$

Nevertheless, we prove the following theorem which shows that in general, the behaviour of $m_{\text{faithful}}(\mathscr{G}_{q})$ for $q:=p^{f}$ with $p$ fixed is very similar to the above example.

The proof of Theorem 2.7 uses Dwork’s theorem on rationality of zeta functions of varieties and the Skolem–Mahler–Lech theorem.

At this point two remarks are in order. On the one hand, Theorems 2.5 and 2.7 set an upper limit on how complicated the value of $m_{\text{faithful}}(\mathscr{G}_{q})$ as a function of $p$ and $f$ can be. It would be desirable to know to what extent the collection of functions appearing in Theorem 2.5 can be realized as $m_{\text{faithful}}(\mathscr{G}_{q})$ for some Lie algebra $\mathfrak{g}$ . On the other hand, one may still hope that at least for a large class of naturally arising Lie algebras $\mathfrak{g}$ , the function $m_{\text{faithful}}(\mathscr{G}_{q})$ is given by a single polynomial. In the next section we address these two problems.

2.2 Pattern groups

Let us begin this section by introducing a large class of nilpotent groups which can be viewed as a generalization of the $n$ th Heisenberg group $\operatorname{Heis}_{2n+1}(\mathbb{F}_{q})$ defined in the introduction.

Set $[n]:=\{1,2,\ldots ,n\}$ , and let $([n],\prec )$ be a partially ordered set. Without loss of generality, we can assume that if $i\prec j$ then $i<j$ . To this partial order we assign the pattern Lie algebra

$$\begin{eqnarray}\mathfrak{g}_{\prec }:=\operatorname{Span}_{\mathbb{Z}}\{\mathbf{e}_{ij}:i\prec j\}\subseteq \mathfrak{gl}(n,\mathbb{Z}),\end{eqnarray}$$

where $\mathbf{e}_{ij}$ denotes the $n\times n$ matrix whose unique non-zero entry is a $1$ in the $(i,j)$ position.

Note that $\mathfrak{g}_{\prec }$ is nilpotent since the commutator relation

$$\begin{eqnarray}[\mathbf{e}_{ij},\mathbf{e}_{kl}]=\unicode[STIX]{x1D6FF}_{jk}\mathbf{e}_{il}-\unicode[STIX]{x1D6FF}_{li}\mathbf{e}_{kj}\end{eqnarray}$$

ensures that $\mathfrak{g}_{\prec }$ is a subalgebra of the Lie algebra $\mathfrak{u}_{n}$ of strictly upper-triangular $n$ by $n$ matrices. For instance, the $(2n+1)$ -dimensional Heisenberg Lie algebra corresponds to the partial order

(3) $$\begin{eqnarray}1\prec 2,3,\ldots ,n+1\prec n+2.\end{eqnarray}$$

For all $i\prec j$ define

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}(i,j):=\#\{k\in [n]:i\prec k\prec j\}.\end{eqnarray}$$

Moreover, the length of $([n],\prec )$ , denoted by $\unicode[STIX]{x1D706}_{\prec }$ , is defined to be the maximum value of $r$ such that there exists a chain $i_{0}\prec \cdots \prec i_{r}$ in $([n],\prec )$ . For instance the length of the partial order given in (3) is equal to $2$ . We remark that $\unicode[STIX]{x1D706}_{\prec }$ is equal to the nilpotency class of $\mathfrak{g}_{\prec }$ (see Lemma 4.1).

We can now state our theorem on the faithful dimension of pattern groups.

Theorem 2.10. Let $\prec$ be a partial order of length $\unicode[STIX]{x1D706}_{\prec }$ on the set $[n]$ , and let $q:=p^{f}$ where $p>\unicode[STIX]{x1D706}_{\prec }$ . Then

$$\begin{eqnarray}m_{\text{faithful}}(\exp (\mathfrak{g}_{\prec }\otimes _{\mathbb{Z}}\mathbb{F}_{q}))=\mathop{\sum }_{(i,j)\in I_{\text{ex}}}fq^{\unicode[STIX]{x1D6FC}(i,j)}.\end{eqnarray}$$

Theorem 2.10 generalizes (2), and its proof relies on Kirillov theory and more specifically on an explicit description of the size of coadjoint orbits in terms of combinatorial data of the partial order. Note that by Theorems 2.5 and 2.7, a priori there is no reason to expect the faithful dimension to be given by a single formula of the form $fg(p^{f})$ for a polynomial $g(T)$ . Theorem 2.10 is proved by a detailed analysis of the size of coadjoint orbits and a combinatorial lemma due to Rado and Horn.

Question 2.11. Let $F$ be a non-Archimedean local field with the ring of integers ${\mathcal{O}}$ , the unique maximal ideal $\mathfrak{p}$ and the associated residue field $\mathbb{F}_{q}$ . Is it true that $m_{\text{faithful}}(\exp (\mathfrak{g}_{\prec }\otimes _{\mathbb{Z}}{\mathcal{O}}/\mathfrak{p}^{n}))$ is given by the formula

(4) $$\begin{eqnarray}\mathop{\sum }_{\ell =0}^{\unicode[STIX]{x1D709}-1}\mathop{\sum }_{(i,j)\in I_{\text{ex}}}fq^{(n-\ell )\unicode[STIX]{x1D6FC}(i,j)},\quad \unicode[STIX]{x1D709}=\min \{n,e\},\end{eqnarray}$$

where $f$ is the absolute inertia degree and $e$ is the absolute ramification index of $F$ ? Formula (4) is suggested by (1) and Theorem 2.10.

An immediate consequence of Theorem 2.10 is the following corollary.

Corollary 2.12. For any non-zero polynomial $g(T)\in \mathbb{Z}[T]$ with non-negative coefficients, there exists a nilpotent $\mathbb{Z}$ -Lie algebra $\mathfrak{g}$ which is finitely generated as an abelian group such that

$$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{q})=fg(q),\end{eqnarray}$$

when $p\geqslant \deg g(T)+2$ and $f\geqslant 1$ .

2.3 Relatively free nilpotent groups

In this section we will turn to free objects in certain categories of nilpotent Lie algebras. Let $\mathfrak{f}_{n,c}:=\mathfrak{f}_{n,c}(\mathbb{Z})$ be the free nilpotent $\mathbb{Z}$ -Lie algebra on $n$ generators and of class $c$ ; it is defined to be the quotient algebra $\mathfrak{f}_{n}/\mathfrak{f}_{n}^{c+1}$ , where $\mathfrak{f}_{n}$ is the free $\mathbb{Z}$ -Lie algebra on $n$ generators, and $\mathfrak{f}_{n}^{c+1}$ denotes the $(c+1)$ th term in the lower central series of $\mathfrak{f}_{n}$ starting with $\mathfrak{f}_{n}^{1}=\mathfrak{f}_{n}$ . It is well known that the rank of the quotient $\mathfrak{f}_{n}^{c}/\mathfrak{f}_{n}^{c+1}$ (as a $\mathbb{Z}$ -module) is given by Witt’s formula

$$\begin{eqnarray}r_{n}(c):=\frac{1}{c}\mathop{\sum }_{d|c}\unicode[STIX]{x1D707}(d)n^{c/d},\end{eqnarray}$$

where $\unicode[STIX]{x1D707}$ is the Möbius function. Using the orbit method one can prove that

$$\begin{eqnarray}m_{\text{faithful}}(\exp (\mathfrak{f}_{n,c}\otimes _{\mathbb{Z}}\mathbb{F}_{q}))\geqslant r_{n}(c)fq.\end{eqnarray}$$

This lower bound is sharp for $\mathfrak{f}_{n,2}$ and $\mathfrak{f}_{n,3}$ .

The proofs of these results involve explicit computations with Hall bases and rely on a subtle combinatorial optimization.

Remark 2.14. For $2\leqslant c\leqslant 6$ , the value of $m_{\text{faithful}}(\exp (\mathfrak{f}_{2,c}\otimes _{\mathbb{Z}}\mathbb{F}_{p}))$ is given by Table 1.

Table 1.

The faithful dimension for 2-generated relatively free nilpotent groups of small nilpotency.

In § 7.4, we outline the computations that yield the values of $m_{\text{faithful}}(\exp (\mathfrak{f}_{2,c}\otimes _{\mathbb{Z}}\mathbb{F}_{p}))$ in Table 1. However, we are not able to obtain any general formula for $m_{\text{faithful}}(\exp (\mathfrak{f}_{2,c}\otimes _{\mathbb{Z}}\mathbb{F}_{p}))$ in terms of $c$ .

For a Lie algebra $\mathfrak{g}$ , let $(D^{k}\mathfrak{g})_{k\geqslant 0}$ be the derived series of $\mathfrak{g}$ . Thus $D^{0}\mathfrak{g}=\mathfrak{g}$ , $D\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$ , and $D^{k+1}\mathfrak{g}=[D^{k}\mathfrak{g},D^{k}\mathfrak{g}]$ for $k\geqslant 1$ . Note that $\mathfrak{g}/D^{2}\mathfrak{g}$ is the largest metabelian quotient of $\mathfrak{g}$ . When $\mathfrak{g}=\mathfrak{f}_{n,c}$ , this quotient is called the free metabelian Lie algebra of class $c$ on $n$ generators and will be denoted by $\mathfrak{m}_{n,c}$ .

In the course of the proof of Theorem 2.15 we will see that computing the faithful dimension of $\exp (\mathfrak{m}_{2,c}\otimes _{\mathbb{Z}}\mathbb{F}_{q})$ is linked to rational normal curves, that is, the image of the Veronese map given by

$$\begin{eqnarray}\unicode[STIX]{x1D708}_{c-2}:\mathbf{P}^{1}(\mathbb{F}_{q})\rightarrow \mathbf{P}^{c-2}(\mathbb{F}_{q}),\quad [X_{0}:X_{1}]\mapsto [X_{0}^{c-2}:X_{0}^{c-3}X_{1}:\cdots :X_{1}^{c-2}].\end{eqnarray}$$

This suggests that other tools (e.g. from the theory of determinantal varieties) might be relevant in the more general situation.

3.1 Central characters of faithful representations of $p$ -groups

Let $A$ be a finite abelian group. We denote the minimal number of generators of $A$ by $d(A)$ . For an exact sequence of finite abelian groups $0\rightarrow A_{1}\rightarrow A\rightarrow A_{2}\rightarrow 0$ , the numbers $d(A)$ , $d(A_{1})$ and $d(A_{2})$ satisfy the inequalities

(5) $$\begin{eqnarray}\max \{d(A_{i}):i=1,2\}\leqslant d(A)\leqslant d(A_{1})+d(A_{2}).\end{eqnarray}$$

The number of invariant factors of $A$ will be denoted by $d^{\prime }(A)$ . It can be easily seen from elementary divisor theory that $d^{\prime }(A)=d(A)$ . Evidently $m_{\text{faithful}}(A)\leqslant d^{\prime }(A)$ . Now for a given faithful representation $\unicode[STIX]{x1D70C}:A\rightarrow \operatorname{GL}_{m}(\mathbb{C})$ , by decomposing $\unicode[STIX]{x1D70C}$ into irreducible components and applying (5) we obtain $d(A)=d^{\prime }(A)\leqslant m_{\text{faithful}}(A)$ . This implies that

$$\begin{eqnarray}m_{\text{faithful}}(A)=d(A)=d^{\prime }(A).\end{eqnarray}$$

In particular, we obtain the following lemma.

Lemma 3.1. For a finite abelian $p$ -group $A$ ,

$$\begin{eqnarray}d(A)=d^{\prime }(A)=m_{\text{faithful}}(A)=\dim _{\mathbb{Z}/p\mathbb{Z}}(A\otimes _{\mathbb{Z}}\mathbb{Z}/p\mathbb{Z})=\dim _{\mathbb{Z}/p\mathbb{Z}}(\unicode[STIX]{x1D6FA}_{1}(A)).\end{eqnarray}$$

Let $E$ be a finite elementary abelian $p$ -group equipped with the canonical $\mathbb{Z}/p\mathbb{Z}$ -vector space structure. Every one-dimensional representation $\unicode[STIX]{x1D712}:E\rightarrow \mathbb{C}^{\ast }$ factors uniquely as $\unicode[STIX]{x1D712}=\unicode[STIX]{x1D716}\circ \unicode[STIX]{x1D712}_{\circ }$ , where $\unicode[STIX]{x1D712}_{\circ }\in \operatorname{Hom}(E,\mathbb{Z}/p\mathbb{Z})$ and the embedding $\unicode[STIX]{x1D716}:\mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{C}^{\ast }$ is defined by

$$\begin{eqnarray}\unicode[STIX]{x1D716}(x+p\mathbb{Z})=\exp ((2\unicode[STIX]{x1D70B}ix)/p).\end{eqnarray}$$

Hence the $\mathbb{Z}/p\mathbb{Z}$ -linear map

(6) $$\begin{eqnarray}\widehat{E}\rightarrow \operatorname{Hom}(E,\mathbb{Z}/p\mathbb{Z}),\quad \unicode[STIX]{x1D712}\mapsto \unicode[STIX]{x1D712}_{\circ },\end{eqnarray}$$

provides an isomorphism of $\mathbb{Z}/p\mathbb{Z}$ -vector spaces between $\widehat{E}$ and $\operatorname{Hom}(E,\mathbb{Z}/p\mathbb{Z})$ . Now, let $\text{G}$ be a finite $p$ -group. Applying (6) to $\unicode[STIX]{x1D6FA}_{1}(\text{Z}(G))$ , we obtain the $\mathbb{Z}/p\mathbb{Z}$ -isomorphism

(7) $$\begin{eqnarray}\operatorname{Hom}(\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\text{G})),\mathbb{C}^{\ast })\rightarrow \operatorname{Hom}(\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\text{G})),\mathbb{Z}/p\mathbb{Z}).\end{eqnarray}$$

Hereafter the $\mathbb{Z}/p\mathbb{Z}$ -vector space $\text{Hom}(\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\text{G})),\mathbb{C}^{\ast })$ will be denoted by $\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\text{G}))$ .

We recall the following simple lemma.

The following observation, due to Meyer and Reichstein [Reference Meyer and ReichsteinMR10], will play a crucial role in computing the faithful dimension of $p$ -groups.

Lemma 3.4. Let $\text{G}$ be a finite $p$ -group and let $(\unicode[STIX]{x1D70C}_{i},V_{i})_{1\leqslant i\leqslant n}$ be a family of irreducible representations of $\text{G}$ with central characters $\unicode[STIX]{x1D712}_{i}$ . Assume that the set of characters

$$\begin{eqnarray}\{{\unicode[STIX]{x1D712}_{i}}_{|_{\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\text{G}))}}:1\leqslant i\leqslant n\}\end{eqnarray}$$

spans $\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\text{G}))$ . Then $\bigoplus _{1\leqslant i\leqslant n}\unicode[STIX]{x1D70C}_{i}$ is a faithful representation of $\text{G}$ .

Proof. Since the set $\{{\unicode[STIX]{x1D712}_{i}}_{|_{\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\text{G}))}}:1\leqslant i\leqslant n\}$ spans $\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\text{G}))$ , from the $\mathbb{Z}/p\mathbb{Z}$ -isomorphism (7) and Lemma 3.3 we obtain

$$\begin{eqnarray}\mathop{\bigcap }_{i=1}^{n}\ker {\unicode[STIX]{x1D712}_{i}}_{|_{\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\text{G}))}}=\{\mathbf{1}\}.\end{eqnarray}$$

Hence $\bigoplus _{1\leqslant i\leqslant n}\unicode[STIX]{x1D70C}_{i}$ is a faithful representation of $\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\text{G}))$ . Remark 3.2 implies that $\bigoplus _{1\leqslant i\leqslant n}\unicode[STIX]{x1D70C}_{i}$ is a faithful representation of $\text{G}$ .◻

Lemma 3.5. Let $\text{G}$ be a finite $p$ -group and let $\unicode[STIX]{x1D70C}$ be a faithful representation of $\text{G}$ with the smallest possible dimension. Then $\unicode[STIX]{x1D70C}$ decomposes as a direct sum of exactly $r:=d(\text{Z}(\text{G}))$ irreducible representations

$$\begin{eqnarray}\unicode[STIX]{x1D70C}=\unicode[STIX]{x1D70C}_{1}\oplus \cdots \oplus \unicode[STIX]{x1D70C}_{r}.\end{eqnarray}$$

Therefore the set of central characters $\{{\unicode[STIX]{x1D712}_{i}}_{|_{\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\text{G}))}}:1\leqslant i\leqslant r\}$ is a basis of $\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\text{G}))$ .

Proof. Let $\unicode[STIX]{x1D70C}=\bigoplus _{1\leqslant i\leqslant n}\unicode[STIX]{x1D70C}_{i}$ be the decomposition of $\unicode[STIX]{x1D70C}$ , and let $\unicode[STIX]{x1D712}_{i}$ , $1\leqslant i\leqslant n$ , denote the central character of $\unicode[STIX]{x1D70C}_{i}$ . Since $\unicode[STIX]{x1D70C}$ is faithful and $r=d(\text{Z}(\text{G}))$ , it follows that $n\geqslant r$ . Furthermore, faithfulness of $\unicode[STIX]{x1D70C}$ also implies $\bigcap _{i=1}^{n}\ker \unicode[STIX]{x1D712}_{i}=\{\mathbf{1}\}$ . Hence, from Lemma 3.3, Lemma 3.4, and the minimality of $\dim (\unicode[STIX]{x1D70C})$ it follows that $n=r$ and also that the set

$$\begin{eqnarray}\{{\unicode[STIX]{x1D712}_{i}}_{|_{\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\text{G}))}}:1\leqslant i\leqslant r\}\end{eqnarray}$$

is a basis of $\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\text{G}))$ .◻

3.2 Kirillov’s orbit method

The orbit method was introduced by Kirillov [Reference KirillovKir62] to study unitary representations of simply connected nilpotent Lie groups. For such a group $\text{G}$ with Lie algebra $\mathfrak{g}$ , this method provides an explicit bijection between the unitary dual $\widehat{\text{G}}$ of $\text{G}$ , and the set $\operatorname{Hom}^{\text{cont}}(\mathfrak{g},\mathbb{C}^{\ast })/\text{G}$ of orbits of the induced action of $\text{G}$ on $\operatorname{Hom}^{\text{cont}}(\mathfrak{g},\mathbb{C}^{\ast })$ , called the coadjoint orbits. Since Kirillov’s work, this method has been extended to study representations of nilpotent groups in other contexts. Relevant to this work is the version applicable to finite $p$ -groups, which we now briefly explain. For more details we refer the reader to [Reference Boyarchenko and SabitovaBS08]. Let $\text{G}$ be a $p$ -group of nilpotency class $c<p$ . By the Lazard correspondence, there exists a unique finite $\mathbb{Z}$ -Lie algebra $\mathfrak{g}:=\text{Lie}(\text{G})$ of cardinality $|\text{G}|$ and nilpotency class $c$ such that $\text{G}\cong \exp (\mathfrak{g})$ . Note that in the definition of $\exp (\mathfrak{g})$ the underlying set of the group is $\mathfrak{g}$ and the multiplication law is defined by the Campbell–Baker–Hausdorff formula. Usually we identify the underlying sets of the group $\text{G}$ and the Lie algebra $\mathfrak{g}$ . A simple application of the Campbell–Baker–Hausdorff formula shows that in this identification the centre of $\mathfrak{g}$ (as a Lie algebra) will be mapped onto the centre of $\text{G}$ as a group.

Consider now the coadjoint action of $\text{G}$ on $\widehat{\mathfrak{g}}:=\operatorname{Hom}_{\mathbb{Z}}(\mathfrak{g},\mathbb{C}^{\ast })$ , defined by

$$\begin{eqnarray}\unicode[STIX]{x1D703}^{x}(y):=\unicode[STIX]{x1D703}\biggl(\mathop{\sum }_{n=0}^{c}\frac{\operatorname{ad}_{x}^{n}(y)}{n!}\biggr),\end{eqnarray}$$

where $x,y\in \mathfrak{g}$ and $\unicode[STIX]{x1D703}\in \widehat{\mathfrak{g}}$ . Note that since $p>c$ , the sum is well defined.

Theorem 3.6. Assume that $p\geqslant 3$ and let $\text{G}$ be a $p$ -group of nilpotency class $c<p$ . Furthermore, assume that $\mathfrak{g}=\text{Lie}(\text{G})$ . Then there exists a bijection between $\text{G}$ -orbits $\unicode[STIX]{x1D6E9}\subseteq \widehat{\mathfrak{g}}$ and irreducible representations $\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D6E9}}\in \widehat{\text{G}}$ such that Kirillov’s character formula holds:

$$\begin{eqnarray}\unicode[STIX]{x1D712}_{\unicode[STIX]{x1D6E9}}(x):=\unicode[STIX]{x1D712}_{\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D6E9}}}(x)=|\unicode[STIX]{x1D6E9}|^{-1/2}\mathop{\sum }_{\unicode[STIX]{x1D703}\in \unicode[STIX]{x1D6E9}}\unicode[STIX]{x1D703}(x).\end{eqnarray}$$

Let $\unicode[STIX]{x1D6E9}\subset \widehat{\mathfrak{g}}$ be the orbit of $\unicode[STIX]{x1D703}_{0}\in \widehat{\mathfrak{g}}$ . Then from Kirillov’s character formula we see that the central character of $\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D6E9}}$ is ${\unicode[STIX]{x1D703}_{0}}_{|_{\text{Z}(\mathfrak{g})}}$ and

(8) $$\begin{eqnarray}\dim (\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D6E9}})=|\unicode[STIX]{x1D6E9}|^{1/2}=[\mathfrak{g}:\text{Stab}_{\text{G}}(\unicode[STIX]{x1D703}_{0})]^{1/2}.\end{eqnarray}$$

Proposition 3.8. The stabilizer of $\unicode[STIX]{x1D703}_{0}$ is given by

(9) $$\begin{eqnarray}\text{Stab}_{\text{G}}(\unicode[STIX]{x1D703}_{0})=\{x\in \mathfrak{g}:\unicode[STIX]{x1D703}_{0}([x,y])=1\;\forall y\in \mathfrak{g}\}.\end{eqnarray}$$

Proof. One inclusion is obvious. For the inclusion $\subseteq$ note that

$$\begin{eqnarray}\text{Stab}_{\text{G}}(\unicode[STIX]{x1D703}_{0})=\{x\in \mathfrak{g}:\unicode[STIX]{x1D703}_{0}^{x}(y)=\unicode[STIX]{x1D703}_{0}(y),\forall y\in \mathfrak{g}\}=\bigg\{x\in \mathfrak{g}:\unicode[STIX]{x1D703}_{0}\biggl(\mathop{\sum }_{n=0}^{c}\frac{\operatorname{ad}_{x}^{n}(y)}{n!}\biggr)=\unicode[STIX]{x1D703}_{0}(y),\forall y\in \mathfrak{g}\bigg\}.\end{eqnarray}$$

Fix $x\in \text{Stab}_{\text{G}}(\unicode[STIX]{x1D703}_{0})$ . Then

(10) $$\begin{eqnarray}\unicode[STIX]{x1D703}_{0}\biggl(\mathop{\sum }_{n=1}^{c}\frac{\operatorname{ad}_{x}^{n}(y)}{n!}\biggr)=0\quad \text{for all }y\in \mathfrak{g}.\end{eqnarray}$$

Choose an arbitrary element $y\in \mathfrak{g}^{c-1}$ . Since $\mathfrak{g}^{c+1}=0$ , it follows from (10) that $\unicode[STIX]{x1D703}_{0}(\operatorname{ad}_{x}(y))=0$ . Next choose an arbitrary $y\in \mathfrak{g}^{c-2}$ , and note that

$$\begin{eqnarray}\mathop{\sum }_{n=1}^{c}\frac{\operatorname{ad}_{x}^{n}(y)}{n!}=\operatorname{ad}_{x}(y)+\frac{\operatorname{ad}_{x}^{2}(y)}{2}.\end{eqnarray}$$

In light of the previous step, $\unicode[STIX]{x1D703}_{0}(\operatorname{ad}_{x}(y))=0$ . Continuing this process, the claim follows for all $y\in \mathfrak{g}$ .◻

Let us now illustrate the power of the orbit method by showing how it can be used to compute the faithful dimension of certain 2-step nilpotent groups.

Proof. Let $\unicode[STIX]{x1D70C}$ be a faithful representation of $\text{G}$ of minimal dimension. Since the centre of $\text{G}$ is cyclic, it follows from Lemma 3.5 that $\unicode[STIX]{x1D70C}$ is irreducible. By the Lazard correspondence $\text{G}=\exp (\mathfrak{g})$ , where $\mathfrak{g}$ is a finite Lie algebra of class $2$ . It suffices to show that $\dim \unicode[STIX]{x1D70C}=\sqrt{[\mathfrak{g}:\text{Z}(\mathfrak{g})]}$ . Thanks to the orbit method, there exists $\unicode[STIX]{x1D703}_{0}\in \widehat{\mathfrak{g}}$ such that

$$\begin{eqnarray}\unicode[STIX]{x1D712}_{\unicode[STIX]{x1D70C}}(x)=\frac{1}{\sqrt{|\unicode[STIX]{x1D6E9}|}}\mathop{\sum }_{\unicode[STIX]{x1D703}\in \unicode[STIX]{x1D6E9}}\unicode[STIX]{x1D703}(x),\end{eqnarray}$$

where $\unicode[STIX]{x1D6E9}$ is the $\text{G}$ -orbit of $\unicode[STIX]{x1D703}_{0}$ . Therefore

$$\begin{eqnarray}\dim \unicode[STIX]{x1D70C}=\sqrt{|\unicode[STIX]{x1D6E9}|}=\sqrt{[\mathfrak{g}:\text{Stab}_{\text{G}}(\unicode[STIX]{x1D703}_{0})]}.\end{eqnarray}$$

Recall that we identify $\text{Z}(\text{G})$ with $\text{Z}(\mathfrak{g})$ . The restriction to $\text{Z}(\mathfrak{g})$ of the central character of $\unicode[STIX]{x1D70C}$ is $\unicode[STIX]{x1D703}_{0}$ , and so $\unicode[STIX]{x1D703}_{0}:\text{Z}(\mathfrak{g})\rightarrow \mathbb{C}^{\ast }$ is faithful. Now let $x\in \text{Stab}_{\text{G}}(\unicode[STIX]{x1D703}_{0})$ . Then $\unicode[STIX]{x1D703}_{0}([x,y])=1$ for all $y\in \mathfrak{g}$ , so that $[x,y]=0$ . Consequently, $\text{Stab}_{\text{G}}(\unicode[STIX]{x1D703}_{0})=\text{Z}(\mathfrak{g})$ . This completes the proof.◻

The class of $p$ -groups covered by Proposition 3.9 includes all the extra special $p$ -groups for $p\geqslant 3$ ; a $p$ -group $\text{G}$ is called extra special when its centre $\text{Z}(\text{G})$ has $p$ elements and $\text{G}/\text{Z}(\text{G})$ is an elementary abelian $p$ -group. It is well known that an extra special $p$ -group has order $p^{2n+1}$ for some positive integer $n$ . Thus the faithful dimension of $\text{G}$ is $p^{n}$ .

4.1 Proof of Theorem 2.10

Write $m:=\#I_{\text{ex}}$ and $n:=fm$ . As before, we identify $\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\mathfrak{g}_{q}))$ with $\bigoplus _{(i,j)\in I_{\text{ex}}}\mathbb{F}_{q}$ which has dimension $n$ as an $\mathbb{F}_{p}$ -vector space. Let $\unicode[STIX]{x1D70C}$ be a faithful representation of $\mathscr{G}_{q}$ with the smallest possible dimension. We will show that the dimension of $\unicode[STIX]{x1D70C}$ is bounded from below by the right-hand side of (16). Using Lemma 3.5 we can decompose $\unicode[STIX]{x1D70C}$ as a direct sum of $n$ irreducible representations, each of which obtained via the orbit method as described above. Hence, we can write

$$\begin{eqnarray}\unicode[STIX]{x1D70C}=\bigoplus _{k=1}^{n}\unicode[STIX]{x1D70C}_{\mathbf{a}_{k}},\end{eqnarray}$$

with vectors $\mathbf{a}_{k}$ given by

$$\begin{eqnarray}\mathbf{a}_{k}=(a_{st}(k))_{s\prec t}\in \bigoplus _{s\prec t}\mathbb{F}_{q}.\end{eqnarray}$$

Since the central character of $\unicode[STIX]{x1D70C}_{\mathbf{a}_{k}}$ is the restriction of $\unicode[STIX]{x1D713}_{\mathbf{a}_{k}}$ , Lemma 3.5 implies that the set $\{\unicode[STIX]{x1D713}_{\mathbf{a}_{k}}:1\leqslant k\leqslant n\}$ is a basis of $\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\mathscr{G}_{q}))$ and therefore the set

$$\begin{eqnarray}\{(a_{ij}(k))_{(i,j)\in I_{\text{ex}}}:1\leqslant k\leqslant n\}\end{eqnarray}$$

is a basis of the $\mathbb{F}_{p}$ -vector space $\bigoplus _{(i,j)\in I_{\text{ex}}}\mathbb{F}_{q}$ . At this point we need a combinatorial lemma whose proof relies on a theorem of Rado and Horn.

We will use the following theorem of Rado and Horn [Reference HornHor55]. The proof of this theorem itself is based on Hall’s marriage theorem and ideas from matroid theory. We refer the reader to [Reference BollobásBol86, § 18] for more details.

Proof of Lemma 4.6.

We apply Theorem 4.7 with $k=f$ to the set of vectors $S$ . Consider an arbitrary set $J\subseteq \{1,\ldots ,mf\}$ and let $d=\dim _{\mathbb{F}_{q}}\text{Span}_{\mathbb{F}_{q}}\{v_{j}:j\in J\}$ . Then

$$\begin{eqnarray}\#\text{Span}_{\mathbb{F}_{q}}\{v_{j}:j\in J\}=q^{d}=p^{fd}.\end{eqnarray}$$

Clearly $\text{Span}_{\mathbb{F}_{p}}\{v_{j}:j\in J\}\subseteq \text{Span}_{\mathbb{F}_{q}}\{v_{j}:j\in J\}$ , and since $\{v_{j}:j\in J\}$ is linearly independent over $\mathbb{F}_{p}$ we obtain

$$\begin{eqnarray}p^{\#J}=\#\text{Span}_{\mathbb{ F}_{p}}\{v_{j}:j\in J\}\leqslant \#\text{Span}_{\mathbb{F}_{q}}\{v_{j}:j\in J\}=p^{fd}.\end{eqnarray}$$

It follows from the above inequality that

$$\begin{eqnarray}\#J\leqslant f\dim _{\mathbb{F}_{q}}\text{Span}_{\mathbb{F}_{q}}\{v_{j}:j\in J\}.\end{eqnarray}$$

Thus by the Rado–Horn theorem, the set $\{1,\ldots ,fm\}$ can be partitioned into ${\mathcal{A}}_{1},\ldots ,{\mathcal{A}}_{f}$ such that each of the sets $\{v_{\ell }:\ell \in {\mathcal{A}}_{i}\}$ is linearly independent over $\mathbb{F}_{q}$ . Note that $\#{\mathcal{A}}_{i}\leqslant m$ since $\dim _{\mathbb{F}_{q}}V=m$ . But the ${\mathcal{A}}_{i}$ partition $\{1,\ldots ,mf\}$ and so ${\mathcal{A}}_{i}$ has the size $m$ which implies that $\{v_{\ell }:\ell \in {\mathcal{A}}_{i}\}$ is a basis of $V$ over $\mathbb{F}_{q}$ .◻

We return to the proof of Theorem 2.10. Set $V=\bigoplus _{(i,j)\in I_{\text{ex}}}\mathbb{F}_{q}$ , which is an $\mathbb{F}_{p}$ -vector space of dimension $n=mf$ . Also set $v_{k}=(a_{ij}(k))_{(i,j)\in I_{\text{ex}}}\in V$ . Recall that $S=\{v_{k}:1\leqslant k\leqslant n\}$ is a basis of $V$ as an $\mathbb{F}_{p}$ -vector space and so by Lemma 4.6 there exist $f$ disjoint sets $S_{1},\ldots ,S_{f}$ , each of size $m$ , such that each $S_{\ell }$ is a basis of $V$ as an $\mathbb{F}_{q}$ -vector space. For $1\leqslant \ell \leqslant f$ , let $A_{\ell }$ denote an $m\times m$ matrix whose rows are elements of $S_{\ell }$ . Note that $A_{\ell }$ is invertible, since $S_{\ell }$ is a basis of $V$ as an $\mathbb{F}_{q}$ -vector space. Using the Leibniz expansion of the determinant of $A$ , we can assume that up to a permutation of the rows, all of the diagonal entries of $A_{\ell }$ are non-zero. Thus Proposition 4.3 implies that

$$\begin{eqnarray}\mathop{\sum }_{(i,j)\in I_{\text{ex}}}q^{\unicode[STIX]{x1D6FC}(i,j)}\leqslant \mathop{\sum }_{\mathbf{a}_{k}\in S_{\ell }}\dim \unicode[STIX]{x1D70C}_{\mathbf{a}_{k}}\quad \text{for }1\leqslant \ell \leqslant f.\end{eqnarray}$$

Summing over all $\ell$ , we obtain

$$\begin{eqnarray}f\mathop{\sum }_{(i,j)\in I_{\text{ex}}}q^{\unicode[STIX]{x1D6FC}(i,j)}\leqslant \dim \unicode[STIX]{x1D70C},\end{eqnarray}$$

which finishes the proof.

5.1 Proof of Theorem 5.3

In this section we assume that $p$ is chosen as in Theorem 5.3. By abuse of notation, we denote the images in $\mathfrak{g}_{q}$ of the $\mathbf{v}_{i}$ , the $\mathbf{w}_{i}$ and the $\mathbf{z}_{i}$ that are chosen above by the same letters. Let $\unicode[STIX]{x1D713}:\mathbb{F}_{q}\rightarrow \mathbb{C}^{\ast }$ be the primitive additive character defined in § 4. Choose a basis

$$\begin{eqnarray}\{\mathbf{u}_{1}+(\text{Z}(\mathfrak{g}_{q})+\mathfrak{g}_{q}^{\prime }),\ldots ,\mathbf{u}_{l_{3}}+(\text{Z}(\mathfrak{g}_{q})+\mathfrak{g}_{q}^{\prime })\}\end{eqnarray}$$

of $\mathfrak{g}_{q}/(\text{Z}(\mathfrak{g}_{q})+\mathfrak{g}_{q}^{\prime })$ . Since $p>C_{2}$ , the set

$$\begin{eqnarray}\{\mathbf{w}_{1},\ldots ,\mathbf{w}_{l_{1}},\mathbf{w}_{l_{1}+1},\ldots ,\mathbf{w}_{m},\mathbf{z}_{1},\ldots ,\mathbf{z}_{l_{2}},\mathbf{u}_{1},\ldots ,\mathbf{u}_{l_{3}}\}\end{eqnarray}$$

is a basis of $\mathfrak{g}_{q}$ . For

(19) $$\begin{eqnarray}\mathbf{a}=(a_{1},\ldots ,a_{l_{1}},a_{l_{1}+1},\ldots ,a_{m},b_{1},\ldots ,b_{l_{2}},c_{1},\ldots ,c_{l_{3}})\in \mathbb{F}_{q}^{m+l_{2}+l_{3}},\end{eqnarray}$$

let $\unicode[STIX]{x1D713}_{\mathbf{a}}\in \widehat{\mathfrak{g}}_{q}$ be defined by

$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D713}_{\mathbf{a}}\biggl(\mathop{\sum }_{i=1}^{l_{1}}w_{i}\mathbf{w}_{i}+\mathop{\sum }_{i=l_{1}+1}^{m}w_{i}\mathbf{w}_{i}+\mathop{\sum }_{i=1}^{l_{2}}z_{i}\mathbf{z}_{i}+\mathop{\sum }_{i=1}^{l_{3}}u_{i}\mathbf{u}_{i}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad :=\unicode[STIX]{x1D713}\biggl(\mathop{\sum }_{i=1}^{l_{1}}a_{i}w_{i}+\mathop{\sum }_{i=l_{1}+1}^{m}a_{i}w_{i}+\mathop{\sum }_{i=1}^{l_{2}}b_{i}z_{i}+\mathop{\sum }_{i=1}^{l_{3}}c_{i}u_{i}\biggr).\nonumber\end{eqnarray}$$

The assignment $\mathbf{a}\mapsto \unicode[STIX]{x1D713}_{\mathbf{a}}$ identifies $\widehat{\mathfrak{g}}_{q}$ with $\mathbb{F}_{q}^{m+l_{2}+l_{3}}$ . For $\mathbf{a}$ as in (19), we write $\mathbf{a}=(\mathbf{a}^{\prime },\mathbf{a}^{\prime \prime },\mathbf{b},\mathbf{c})$ , where $\mathbf{a}^{\prime }\in \mathbb{F}_{q}^{l_{1}}$ , $\mathbf{a}^{\prime \prime }\in \mathbb{F}_{q}^{m-\ell _{1}}$ , $\mathbf{b}\in \mathbb{F}_{q}^{l_{2}}$ and $\mathbf{c}\in \mathbb{F}_{q}^{l_{3}}$ , and define the projection maps

$$\begin{eqnarray}\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{1}:\mathbb{F}_{q}^{m+l_{2}+l_{3}}\longrightarrow \mathbb{F}_{q}^{m+l_{2}},\quad \mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{2}:\mathbb{F}_{q}^{m+l_{2}+l_{3}}\longrightarrow \mathbb{F}_{q}^{l_{1}+l_{2}},\quad \mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}:\mathbb{F}_{q}^{m+l_{2}+l_{3}}\longrightarrow \mathbb{F}_{q}^{m},\end{eqnarray}$$

by $\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{1}(\mathbf{a})=(\mathbf{a}^{\prime },\mathbf{a}^{\prime \prime },\mathbf{b})$ , $\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{2}(\mathbf{a})=(\mathbf{a}^{\prime },\mathbf{b})$ , and $\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}(\mathbf{a})=(\mathbf{a}^{\prime },\mathbf{a}^{\prime \prime })$ .

In the rest of this section, we identify $\text{Z}(\mathscr{G}_{q})$ with $\text{Z}(\mathfrak{g}_{q})$ . Let $\unicode[STIX]{x1D70C}$ be an irreducible representation of $\mathscr{G}_{q}$ . By the orbit method, $\unicode[STIX]{x1D70C}$ is obtained from a character $\unicode[STIX]{x1D703}\in \widehat{\mathfrak{g}}_{q}$ , whose restriction to $\text{Z}(\mathfrak{g}_{q})$ coincides with the central character of $\unicode[STIX]{x1D70C}$ . Assume that $\unicode[STIX]{x1D703}=\unicode[STIX]{x1D713}_{\mathbf{a}}$ for some $\mathbf{a}\in \mathbb{F}_{q}^{m+l_{2}+l_{3}}$ , whose entries are indexed as in (19). Our next goal is to prove that

(20) $$\begin{eqnarray}\dim \unicode[STIX]{x1D70C}=q^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(a_{1},\ldots ,a_{m}))/2}=q^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}(\mathbf{a})))/2}.\end{eqnarray}$$

The proof is similar to the argument of [Reference O’Brien and VollO’BV15, Lemma 3.3], but for the reader’s convenience we provide some details. Proposition 3.8 implies that $\text{Z}(\mathfrak{g}_{q})\subseteq \text{Stab}_{\mathscr{G}_{q}}(\unicode[STIX]{x1D703})$ . Since $p>C_{3}$ , it follows that $K$ is invertible in $\mathbb{F}_{q}$ , so that after tensoring by $\mathbb{F}_{q}$ the $\mathbf{v}_{i}$ form a basis of $\mathfrak{g}_{q}/\text{Z}(\mathfrak{g}_{q})$ . For $x=\sum _{i=1}^{n}x_{i}\mathbf{v}_{i}\in \text{Stab}_{\mathscr{G}_{q}}(\unicode[STIX]{x1D703})$ and $y=\sum _{i=1}^{n}y_{i}\mathbf{v}_{i}\in \mathfrak{g}_{q}/\text{Z}(\mathfrak{g}_{q})$ we have $\unicode[STIX]{x1D703}([x,y])=1$ . From (17) and the fact that $p>C_{1}$ it follows that

$$\begin{eqnarray}\unicode[STIX]{x1D713}\biggl(\mathop{\sum }_{i=1}^{n}\biggl(\mathop{\sum }_{1\leqslant r<i}\mathop{\sum }_{k=1}^{m}a_{k}\unicode[STIX]{x1D706}_{ri}^{k}x_{r}-\mathop{\sum }_{i<s\leqslant n}\mathop{\sum }_{k=1}^{m}a_{k}\unicode[STIX]{x1D706}_{is}^{k}x_{s}\biggr)y_{i}\biggr)=\unicode[STIX]{x1D713}_{\mathbf{a}}\biggl(\mathop{\sum }_{1\leqslant i<j\leqslant n}\mathop{\sum }_{k=1}^{m}\unicode[STIX]{x1D706}_{ij}^{k}(x_{i}y_{j}-x_{j}y_{i})\mathbf{w}_{k}\biggr)=1.\end{eqnarray}$$

Since $\unicode[STIX]{x1D713}$ is a primitive character, it follows that $\text{Stab}_{\mathscr{G}_{q}}(\unicode[STIX]{x1D703})/\text{Z}(\mathfrak{g}_{q})$ is defined by the linear equations

$$\begin{eqnarray}\mathop{\sum }_{i<s\leqslant n}\mathop{\sum }_{k=1}^{m}a_{k}\unicode[STIX]{x1D706}_{is}^{k}x_{s}-\mathop{\sum }_{1\leqslant r<i}\mathop{\sum }_{k=1}^{m}a_{k}\unicode[STIX]{x1D706}_{ri}^{k}x_{r}=0,\quad 1\leqslant i\leqslant n.\end{eqnarray}$$

Consequently, $x=\sum _{i=1}^{n}x_{i}\mathbf{v}_{i}\in \text{Stab}_{\mathscr{G}_{q}}(\unicode[STIX]{x1D703})/\text{Z}(\mathfrak{g}_{q})$ if and only if $(x_{1},\ldots ,x_{n})\in \ker F_{\mathfrak{g}}(a_{1},\ldots ,a_{m})$ . The last statement implies that $\#\text{Stab}_{\mathscr{G}_{q}}(\unicode[STIX]{x1D703})=q^{\dim _{\mathbb{F}_{q}}(\mathfrak{g}_{q})-\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(a_{1},\ldots ,a_{m}))}$ . Equality (20) now follows from (8).

Definition 5.5 (Admissible sets of vectors).

A set of vectors

$$\begin{eqnarray}\{\mathbf{a}_{\ell }\in \mathbb{F}_{q}^{m+l_{2}+l_{3}}:1\leqslant \ell \leqslant (l_{1}+l_{2})f\}\end{eqnarray}$$

is called an admissible set of vectors if $\{\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{2}(\mathbf{a}_{\ell }):1\leqslant \ell \leqslant (l_{1}+l_{2})f\}$ is a basis of the $\mathbb{F}_{p}$ -vector space $\mathbb{F}_{q}^{l_{1}+l_{2}}$ .

Now let $\tilde{\unicode[STIX]{x1D70C}}$ be a faithful representation of $\mathscr{G}_{q}$ with the smallest possible dimension. Note that the dimension of $\unicode[STIX]{x1D6FA}_{1}(\text{Z}(\mathfrak{g}_{q}))=\text{Z}(\mathfrak{g}_{q})$ over $\mathbb{F}_{p}$ is $(l_{1}+l_{2})f$ and $\text{Z}(\mathscr{G}_{q})\cong \text{Z}(\mathfrak{g}_{q})$ . Therefore

$$\begin{eqnarray}\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\mathscr{G}_{q}))\cong \widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\mathfrak{g}_{q}))\cong \bigoplus _{\ell =1}^{l_{1}+l_{2}}\mathbb{F}_{q}.\end{eqnarray}$$

Thus by Lemma 3.5 the representation $\tilde{\unicode[STIX]{x1D70C}}$ decomposes into $(l_{1}+l_{2})f$ irreducible representations

$$\begin{eqnarray}\tilde{\unicode[STIX]{x1D70C}}=\bigoplus _{\ell =1}^{(l_{1}+l_{2})f}\unicode[STIX]{x1D70C}_{\mathbf{a}_{\ell }},\quad \mathbf{a}_{\ell }\in \mathbb{F}_{q}^{m+l_{2}+l_{3}},\end{eqnarray}$$

where the representation $\unicode[STIX]{x1D70C}_{\mathbf{a}_{\ell }}$ is obtained by $\unicode[STIX]{x1D713}_{\mathbf{a}_{\ell }}\in \widehat{\mathfrak{g}}_{q}$ . Since the restriction of $\unicode[STIX]{x1D713}_{\mathbf{a}_{\ell }}$ is the central character of $\unicode[STIX]{x1D70C}_{\mathbf{a}_{\ell }}$ , it follows from Lemma 3.5 that the set

$$\begin{eqnarray}\{\unicode[STIX]{x1D713}_{\mathbf{a}_{\ell }}|_{\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\mathfrak{g}_{q}))}:1\leqslant \ell \leqslant (l_{1}+l_{2})f\}\end{eqnarray}$$

is a basis of $\widehat{\unicode[STIX]{x1D6FA}}_{1}(\text{Z}(\mathfrak{g}_{q}))$ and therefore the set

$$\begin{eqnarray}\{\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{2}(\mathbf{a}_{\ell }):1\leqslant \ell \leqslant (l_{1}+l_{2})f\}\end{eqnarray}$$

is a basis of the $\mathbb{F}_{p}$ -vector space $\mathbb{F}_{q}^{l_{1}+l_{2}}$ . To summarize, we have proven that for each faithful representation $\tilde{\unicode[STIX]{x1D70C}}$ with the smallest possible dimension we can find an admissible set of vectors

$$\begin{eqnarray}\{\mathbf{a}_{\ell }\in \mathbb{F}_{q}^{m+l_{2}+l_{3}}:1\leqslant \ell \leqslant (l_{1}+l_{2})f\}\end{eqnarray}$$

such that

(21) $$\begin{eqnarray}\dim (\tilde{\unicode[STIX]{x1D70C}})=\mathop{\sum }_{\ell =1}^{(l_{1}+l_{2})f}q^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}(\mathbf{a})))/2}.\end{eqnarray}$$

Conversely, let $\{\mathbf{a}_{\ell }\in \mathbb{F}_{q}^{m+l_{2}+l_{3}}:1\leqslant \ell \leqslant (l_{1}+l_{2})f\}$ be an admissible set of vectors. Then by Lemma 3.4, we can construct a faithful representation $\tilde{\unicode[STIX]{x1D70C}}$ , not necessarily of minimal dimension, such that its dimension is equal to (21). In the definition of admissible vectors we considered $\mathbb{F}_{q}^{l_{1}+l_{2}}$ as an $\mathbb{F}_{p}$ -vector space. We now consider $\mathbb{F}_{q}^{l_{1}+l_{2}}$ as an $\mathbb{F}_{q}$ -vector space and define the following notion.

Definition 5.6 (Regular sets of vectors).

A set of vectors

$$\begin{eqnarray}\{\mathbf{a}_{\ell }\in \mathbb{F}_{q}^{m+l_{2}+l_{3}}:1\leqslant \ell \leqslant l_{1}+l_{2}\},\end{eqnarray}$$

is called a regular set of vectors if the set $\{\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{2}(\mathbf{a}_{\ell }):1\leqslant \ell \leqslant (l_{1}+l_{2})\}$ is a basis of the $\mathbb{F}_{q}$ -vector space $\mathbb{F}_{q}^{l_{1}+l_{2}}$ .

We now claim that

(22) $$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{q})=\min \bigg\{\mathop{\sum }_{\ell =1}^{l_{1}+l_{2}}fq^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}(\mathbf{a}_{\ell })))/2}:\{\mathbf{a}_{\ell }\in \mathbb{F}_{q}^{m+l_{2}+l_{3}}\}_{\ell =1}^{l_{1}+l_{2}}\text{ is a regular set}\bigg\}.\end{eqnarray}$$

Let $\{\unicode[STIX]{x1D714}_{1},\ldots ,\unicode[STIX]{x1D714}_{f}\}$ be a basis of $\mathbb{F}_{q}$ over $\mathbb{F}_{p}$ and let

$$\begin{eqnarray}\{\mathbf{a}_{\ell }\in \mathbb{F}_{q}^{m+l_{2}+l_{3}}:1\leqslant \ell \leqslant l_{1}+l_{2}\}\end{eqnarray}$$

be a regular set of vectors that minimizes (22). Clearly $\{\unicode[STIX]{x1D714}_{i}\mathbf{a}_{\ell }:1\leqslant i\leqslant f,1\leqslant \ell \leqslant l_{1}+l_{2}\}$ is an admissible set of vectors and

$$\begin{eqnarray}\mathop{\sum }_{\ell =1}^{l_{1}+l_{2}}fq^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}(\mathbf{a}_{\ell })))/2}=\mathop{\sum }_{i=1}^{f}\mathop{\sum }_{\ell =1}^{l_{1}+l_{2}}q^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}(\unicode[STIX]{x1D714}_{i}\mathbf{a}_{\ell })))/2}\geqslant m_{\text{faithful}}(\mathscr{G}_{q}).\end{eqnarray}$$

Conversely, let $\tilde{\unicode[STIX]{x1D70C}}$ be a faithful representation with the smallest possible dimension. From the above discussion we obtain an admissible set of vectors

$$\begin{eqnarray}\{\mathbf{a}_{\ell }\in \mathbb{F}_{q}^{m+l_{2}+l_{3}}:1\leqslant \ell \leqslant (l_{1}+l_{2})f\}.\end{eqnarray}$$

From Lemma 4.6 this set can be partitioned into $f$ sets ${\mathcal{B}}_{i}$ in which each ${\mathcal{B}}_{i}$ is a regular set of vectors. Without loss of generality assume that

$$\begin{eqnarray}\mathop{\sum }_{\mathbf{a}_{\ell }\in {\mathcal{B}}_{1}}q^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}(\mathbf{a}_{\ell })))/2}\leqslant \mathop{\sum }_{\mathbf{a}_{\ell }\in {\mathcal{B}}_{i}}q^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}(\mathbf{a}_{\ell })))/2}\quad 2\leqslant i\leqslant f.\end{eqnarray}$$

Thus

$$\begin{eqnarray}\mathop{\sum }_{\mathbf{a}_{\ell }\in {\mathcal{B}}_{1}}fq^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{3}(\mathbf{a}_{\ell })))/2}\leqslant \dim (\tilde{\unicode[STIX]{x1D70C}}),\end{eqnarray}$$

which proves the claim.

We are now ready to finish the proof of Theorem 5.3. Let $\{\mathbf{a}_{\ell }\in \mathbb{F}_{q}^{m+l_{2}+l_{3}}:1\leqslant \ell \leqslant l_{1}+l_{2}\}$ be a regular set of vectors. Let $A\in \text{GL}_{l_{1}+l_{2}}(\mathbb{F}_{q})$ be the matrix whose rows are

$$\begin{eqnarray}\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{2}(\mathbf{a}_{1}),\ldots ,\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{2}(\mathbf{a}_{l_{1}+l_{2}}).\end{eqnarray}$$

Therefore

$$\begin{eqnarray}A=\left(\begin{array}{@{}cccccc@{}}a_{11} & \cdots \, & a_{1l_{1}} & b_{11} & \cdots \, & b_{1l_{2}}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{l_{1}1} & \cdots \, & a_{l_{1}l_{1}} & b_{l_{1}1} & \cdots \, & b_{l_{1}l_{2}}\\ a_{(l_{1}+1)1} & \cdots \, & a_{(l_{1}+1)l_{1}} & b_{(l_{1}+1)1} & \cdots \, & b_{(l_{1}+1)l_{2}}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{(l_{1}+l_{2})1} & \cdots \, & a_{(l_{1}+l_{2})l_{1}} & b_{(l_{1}+l_{2})1} & \cdots \, & b_{(l_{1}+l_{2})l_{2}}\end{array}\right)\!,\end{eqnarray}$$

where $\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{2}(\mathbf{a}_{i})=(a_{i1},\ldots ,a_{il_{1}},b_{i1},\ldots ,b_{il_{2}})$ . The first $l_{1}$ columns of $A$ are linearly independent over $\mathbb{F}_{q}$ , and therefore we can find an invertible $l_{1}\times l_{1}$ submatrix of the first $l_{1}$ columns. By possibly permuting the rows of $A$ we can assume that this submatrix lies at the intersection of the first $l_{1}$ rows and $l_{1}$ columns of $A$ . It is clear that

$$\begin{eqnarray}\mathop{\sum }_{\ell =1}^{l_{1}+l_{2}}q^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(a_{\ell 1},\ldots ,a_{\ell m}))/2}\geqslant \mathop{\sum }_{\ell =1}^{l_{1}}q^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(a_{\ell 1},\ldots ,a_{\ell m}))/2}+l_{2}.\end{eqnarray}$$

From this and (22) we conclude that

(23) $$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{q})\geqslant \min \left\{\mathop{\sum }_{\ell =1}^{l_{1}}fq^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(a_{\ell 1},\ldots ,a_{\ell m}))/2}:\left(\begin{array}{@{}ccc@{}}a_{11} & \cdots \, & a_{1l_{1}}\\ \vdots & \ddots & \vdots \\ a_{l_{1}1} & \cdots \, & a_{l_{1}l_{1}}\end{array}\right)\in \operatorname{GL}_{l_{1}}(\mathbb{F}_{q})\right\}+fl_{2}.\end{eqnarray}$$

Conversely, let

$$\begin{eqnarray}\{a_{\ell (l_{1}+1)},\ldots ,a_{\ell m}\}_{1\leqslant \ell \leqslant l_{1}}\end{eqnarray}$$

be an arbitrary set of elements of $\mathbb{F}_{q}$ and let

$$\begin{eqnarray}B:=\left(\begin{array}{@{}ccc@{}}a_{11} & \cdots \, & a_{1l_{1}}\\ \vdots & \ddots & \vdots \\ a_{l_{1}1} & \cdots \, & a_{l_{1}l_{1}}\end{array}\right)\in \operatorname{GL}_{l_{1}}(\mathbb{F}_{q})\end{eqnarray}$$

be an arbitrary invertible matrix. Then the rows of the matrix

$$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}B & 0\\ 0 & I_{l_{1}\times l_{1}}\end{array}\right)\in \operatorname{GL}_{l_{1}+l_{2}}(\mathbb{F}_{q})\end{eqnarray}$$

are projections (under $\mathsf{p}\mathsf{r}\mathsf{o}\mathsf{j}_{2}$ ) of a regular set of vectors in $\mathbb{F}_{q}^{m+l_{2}+l_{3}}$ . Similar to the proof of (22), using this regular set of vectors we can construct a faithful representation of $\mathscr{G}_{q}$ of dimension

$$\begin{eqnarray}\mathop{\sum }_{\ell =1}^{l_{1}}fq^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(a_{\ell 1},\ldots ,a_{\ell m}))/2}+fl_{2}.\end{eqnarray}$$

From this we conclude that

(24) $$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{q})\leqslant \min \left\{\mathop{\sum }_{\ell =1}^{l_{1}}fq^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(a_{\ell 1},\ldots ,a_{\ell m}))/2}:\left(\begin{array}{@{}ccc@{}}a_{11} & \cdots \, & a_{1l_{1}}\\ \vdots & \ddots & \vdots \\ a_{l_{1}1} & \cdots \, & a_{l_{1}l_{1}}\end{array}\right)\in \operatorname{GL}_{l_{1}}(\mathbb{F}_{q})\right\}+fl_{2}.\end{eqnarray}$$

Therefore, by combining (23) and (24) we obtain Theorem 5.3.

We now address Examples 2.1, 2.2, 2.3, and 2.6 in detail. By a straightforward calculation, one can verify that in all of these examples $C_{1}=C_{2}=C_{3}=1$ , and hence Theorem 5.3 is applicable for $p>2$ .

5.2 Details for Example 2.1

From the defining bracket relations we deduce that $\mathfrak{g}_{a}^{\prime }=\text{Z}(\mathfrak{g}_{a})=\operatorname{Span}_{\mathbb{Z}}\{v_{7},v_{8},v_{9}\}$ and so $\mathfrak{g}_{a}$ is a 2-step nilpotent Lie algebra. From the relations we also obtain the following commutator matrix:

$$\begin{eqnarray}F_{\mathfrak{g}}(T_{1},T_{2},T_{3})=\left(\begin{array}{@{}cc@{}}0 & M\\ -M^{\text{tr}} & 0\end{array}\right)\!,\quad M:=M(T_{1},T_{2},T_{3})=\left(\begin{array}{@{}ccc@{}}T_{1} & T_{2} & aT_{3}\\ T_{3} & T_{1} & T_{2}\\ T_{3} & 0 & T_{1}\end{array}\right)\!.\end{eqnarray}$$

Observe that the determinant of $M$ is

$$\begin{eqnarray}g(T_{1},T_{2},T_{3}):=T_{3}T_{2}^{2}+T_{1}^{3}-T_{1}T_{2}T_{3}-aT_{1}T_{3}^{2}.\end{eqnarray}$$

By Theorem 5.3 the faithful dimension of $\mathscr{G}_{a,p}=\exp (\mathfrak{g}_{a}\otimes _{\mathbb{Z}}\mathbb{F}_{p})$ , for $p\geqslant 3$ , is the minimum value of

(25) $$\begin{eqnarray}p^{\text{rk}_{\mathbb{F}_{p}}(M(x_{11},x_{12},x_{13}))}+p^{\text{rk}_{\mathbb{F}_{p}}(M(x_{21},x_{22},x_{23}))}+p^{\text{rk}_{\mathbb{F}_{p}}(M(x_{31},x_{32},x_{33}))}\end{eqnarray}$$

subject to the condition

(26) $$\begin{eqnarray}\left(\begin{array}{@{}ccc@{}}x_{11} & x_{12} & x_{13}\\ x_{21} & x_{22} & x_{23}\\ x_{31} & x_{32} & x_{33}\end{array}\right)\in \operatorname{GL}_{3}(\mathbb{F}_{p}).\end{eqnarray}$$

Let $p$ be a prime not dividing $a$ . Computing all $2\times 2$ minors of $M$ shows that

$$\begin{eqnarray}2\leqslant \text{rk}_{\mathbb{F}_{p}}(M(x,y,z))\leqslant 3\end{eqnarray}$$

unless $x=y=z=0$ . Let us consider the question of existence of a vector $(x,y,z)\in \mathbb{F}_{p}^{3}$ such that $x\neq 0$ and

(27) $$\begin{eqnarray}g(x,y,z)=y^{2}z+x^{3}-xyz-axz^{2}=0.\end{eqnarray}$$

Obviously we should take $z\neq 0$ Picking $x=1$ , we obtain the equation $zy^{2}-zy+(1-az^{2})=0$ , whose discriminant with respect to $y$ is equal to $4az^{3}+z^{2}-4z$ . Thus to solve (27) it suffices to show that for any non-zero $a\in \mathbb{Z}$ the curve $Y^{2}=4aX^{3}+X^{2}-4X$ has a rational point in $\mathbb{F}_{p}$ with $X\neq 0$ . This can be done by noticing that $Y^{2}-4aX^{3}-X^{2}+4X$ is absolutely irreducible [Reference SchmidtSch76, Corollary, p. 13] and thus by Hasse’s bound on the number of $\mathbb{F}_{p}$ -points of elliptic curves [Reference SchmidtSch76, Theorem 2A, p. 10] one can verify that such a point exists for $p>1800$ . Let $(1,y,z)$ be a solution of (27). Then the vectors $(0,1,0)$ and $(0,0,1)$ and $(1,y,z)$ satisfy (26), and minimize (25). Thus the faithful dimension of $\mathscr{G}_{a,p}$ is equal to $3p^{2}$ .

5.3 Details for Examples 2.2 and 2.6

We discuss these examples together by proving the following formula:

$$\begin{eqnarray}m_{\text{faithful}}(\mathscr{G}_{q})=\left\{\begin{array}{@{}ll@{}}2fq\quad & \text{if }p\equiv 1\;(\text{mod}\;4),\\ 2fq\quad & \text{if }p\equiv 3\;(\text{mod}\;4)\text{ and }f\text{ is even},\\ 2fq^{2}\quad & \text{if }p\equiv 3\;(\text{mod}\;4)\text{ and }f\text{ is odd}.\end{array}\right.\end{eqnarray}$$

The commutator relations imply that $\mathfrak{g}^{\prime }=\text{Z}(\mathfrak{g})=\operatorname{Span}_{\mathbb{Z}}\{v_{5},v_{6}\}$ , and that $\mathfrak{g}$ is a 2-step nilpotent Lie algebra. The commutator matrix of $\mathfrak{g}$ can be easily seen to be

$$\begin{eqnarray}F_{\mathfrak{g}}(T_{1},T_{2})=\left(\begin{array}{@{}cccc@{}}0 & T_{1} & 0 & T_{2}\\ -T_{1} & 0 & T_{2} & 0\\ 0 & -T_{2} & 0 & T_{1}\\ -T_{2} & 0 & -T_{1} & 0\end{array}\right)\!.\end{eqnarray}$$

Note that $\det F_{\mathfrak{g}}(T_{1},T_{2})=(T_{1}^{2}+T_{2}^{2})^{2}$ . By Theorem 5.3 the faithful dimension of $\mathscr{G}_{q}$ is given by

$$\begin{eqnarray}\min \left\{fq^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(x_{11},x_{12}))/2}+fq^{\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(x_{21},x_{22}))/2}:\left(\begin{array}{@{}cc@{}}x_{11} & x_{12}\\ x_{21} & x_{22}\end{array}\right)\in \operatorname{GL}_{2}(\mathbb{F}_{q})\right\}.\end{eqnarray}$$

For $p\equiv 1\;(\text{mod}\;4)$ , let $\unicode[STIX]{x1D6FC}$ denote a square root of $-1$ in $\mathbb{F}_{q}$ . Then the trivial lower bound $2fq$ can be realized by the choice of vectors $(\unicode[STIX]{x1D6FC},1)$ and $(-\unicode[STIX]{x1D6FC},1)$ . We now consider the case $p\equiv 3\;(\text{mod}\;4)$ . For these primes, observe that $-1$ is a square in $\mathbb{F}_{q}$ if and only if $f$ is even. By the above argument the faithful dimension of $\mathscr{G}_{q}$ is $2fq$ when $f$ is even. Now suppose that $f$ is odd. Then $-1$ is not a square in $\mathbb{F}_{q}$ , and therefore $\det F_{\mathfrak{g}}(a_{1},a_{2})\neq 0$ for all non-zero vectors $(a_{1},a_{2})\in \mathbb{F}_{q}^{2}$ . This implies that

$$\begin{eqnarray}\text{rk}_{\mathbb{F}_{q}}(F_{\mathfrak{g}}(a_{1},a_{2}))=4\quad \text{for all }0\neq (a_{1},a_{2})\in \mathbb{F}_{q}^{2}.\end{eqnarray}$$

Therefore the faithful dimension of $\mathscr{G}_{q}$ is at least $2fq^{2}$ , which can be realized by the standard basis.

5.4 Details for Example 2.3

The commutator relations imply that $\mathfrak{g}^{\prime }=\text{Z}(\mathfrak{g})=\operatorname{Span}_{\mathbb{Z}}\{v_{7},v_{8}\}$ , implying that $\mathfrak{g}$ is a 2-step nilpotent Lie algebra with the commutator matrix given by

(28) $$\begin{eqnarray}F_{\mathfrak{g}}(T_{1},T_{2})=\left(\begin{array}{@{}cc@{}}0 & M\\ -M^{\text{tr}} & 0\end{array}\right)\quad \text{where }M:=M(T_{1},T_{2})=\left(\begin{array}{@{}ccc@{}}T_{1} & T_{2} & 0\\ 0 & T_{1} & T_{2}\\ -T_{2} & T_{2} & T_{1}\end{array}\right)\!.\end{eqnarray}$$

By Theorem 5.3 the faithful dimension of $\mathscr{G}_{p}$ is given by

$$\begin{eqnarray}\min \left\{p^{\text{rk}_{\mathbb{F}_{p}}(M(x_{11},x_{12}))}+p^{\text{rk}_{\mathbb{F}_{p}}(M(x_{21},x_{22}))}:\left(\begin{array}{@{}cc@{}}x_{11} & x_{12}\\ x_{21} & x_{22}\end{array}\right)\in \operatorname{GL}_{2}(\mathbb{F}_{p})\right\}.\end{eqnarray}$$

A simple inspection of $2\times 2$ minors of $M$ shows that for a non-zero vector $(T_{1},T_{2})$ , the matrix $M(T_{1},T_{2})$ has rank at least $2$ , implying $m_{\text{faithful}}(\mathscr{G}_{p})\geqslant 2p^{2}$ . Note that $\det M=T_{1}^{3}-T_{1}T_{2}^{2}-T_{2}^{3}$ is the homogenization of the polynomial $T_{1}^{3}-T_{1}-1$ . This leads us to consider the number of roots of $f(T)=T^{3}-T-1$ over $\mathbb{F}_{p}$ .

At this point we will make a digression and consider the more general question of determining the number of roots of a given integer polynomial over finite fields. Let $C_{f}$ be the companion matrix of a given polynomial $f(T)\in \text{Z}[T]$ , and set $M(T_{1},T_{2})=T_{1}I_{d\times d}-T_{2}C_{f}$ , where $d:=\deg f(T)$ . The determinant of $M(T_{1},T_{2})$ is the homogenization of $f(T)$ . This construction leads to a general collection of interesting examples of 2-step nilpotent Lie algebras with commutator matrix as in (28). We refer the reader to Serre’s book [Reference SerreSer12, § 2.1.2] or his beautiful paper [Reference SerreSer03] for more details on what follows.

Let $f(T)\in \mathbb{Z}[T]$ be a monic integer polynomial. The discriminant of $f(T)$ is defined to be

$$\begin{eqnarray}\text{Disc}_{f}=\unicode[STIX]{x1D6E5}_{f}^{2},\quad \unicode[STIX]{x1D6E5}_{f}=\mathop{\prod }_{1\leqslant i<j\leqslant n}(\unicode[STIX]{x1D6FC}_{i}-\unicode[STIX]{x1D6FC}_{j}),\end{eqnarray}$$

where $\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{n}$ are the roots of $f(T)$ in an algebraic closure of $\mathbb{Q}$ . Note that since $f(T)$ is a monic polynomial, the discriminant $\text{Disc}_{f}$ is in $\mathbb{Z}$ . Henceforth, $p$ will denote an odd prime which does not divide $\text{Disc}_{f}$ . Denote the reduction of $f(T)$ modulo $p$ by $\bar{f}$ . Then the roots of $\bar{f}(T)$ are also simple. Define ${\mathcal{O}}_{f}=\mathbb{Z}[\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{n}]$ and let $\mathfrak{p}$ be a prime ideal of ${\mathcal{O}}_{f}$ such that $\mathfrak{p}\cap \mathbb{Z}=p\mathbb{Z}$ . Such an ideal exists since ${\mathcal{O}}_{f}$ is integral over $\mathbb{Z}$ . For such a prime we can define a unique element in the Galois group of $f$ , which is called the Frobenius automorphism.

For a monic integer polynomial $f(T)\in \mathbb{Z}[T]$ define

$$\begin{eqnarray}N_{f}(p):=\#\{a\in \mathbb{F}_{p}:f(a)=0\}.\end{eqnarray}$$

Theorem 5.7 shows that $N_{f}(p)$ also counts the number of fixed points of $\unicode[STIX]{x1D70E}_{\mathfrak{p}}$ permuting the roots of $f$ .

Let ${\mathcal{O}}_{E}$ be the ring of integers of $E$ and $\mathfrak{P}$ be a prime ideal of ${\mathcal{O}}_{E}$ such that $\mathfrak{P}\cap \mathbb{Z}=p\mathbb{Z}$ , where $p$ does not divide the discriminant of $E$ . Then, as above, one can prove the existence of a unique automorphism $\unicode[STIX]{x1D70E}_{\mathfrak{P}}\in \text{Gal}(E/\mathbb{Q})$ such that $\unicode[STIX]{x1D70E}_{\mathfrak{P}}(\unicode[STIX]{x1D6FC})\equiv \unicode[STIX]{x1D6FC}^{p}\;(\text{mod}\;\mathfrak{P})$ for all $\unicode[STIX]{x1D6FC}\in {\mathcal{O}}_{E}$ . Let $\mathfrak{P}\cap {\mathcal{O}}_{f}=\mathfrak{p}$ . Since the elements of $\text{Gal}(E/\mathbb{Q})$ are uniquely determined by their restrictions to ${\mathcal{O}}_{f}$ , we have $\unicode[STIX]{x1D70E}_{\mathfrak{P}}=\unicode[STIX]{x1D70E}_{\mathfrak{p}}$ . The automorphism $\unicode[STIX]{x1D70E}_{\mathfrak{p}}$ is called the Frobenius automorphism and it describes the splitting behaviour of the prime $p$ . It is well known that $p$ splits completely in $E$ if and only if $\unicode[STIX]{x1D70E}_{\mathfrak{p}}$ is the identity element. Now let $\mathfrak{P}$ and $\mathfrak{P}^{\prime }$ be two primes in ${\mathcal{O}}_{E}$ lying above the rational prime $p$ . One can show (see [Reference NeukirchNeu99, § 9] for more details) that there exists $\unicode[STIX]{x1D70F}\in \text{Gal}(E/\mathbb{Q})$ such that $\unicode[STIX]{x1D70F}\unicode[STIX]{x1D70E}_{\mathfrak{P}}\unicode[STIX]{x1D70F}^{-1}=\unicode[STIX]{x1D70E}_{\mathfrak{P}^{\prime }}$ . This implies that the conjugacy class of $\unicode[STIX]{x1D70E}_{\mathfrak{P}}$ is independent of the choice of $\mathfrak{P}$ . Let us now turn to the question of computing $N_{f}(p)$ when $f(T)$ is a cubic polynomial. The following proposition relates the Legendre symbol of the discriminant of $f$ to the number of the irreducible factors of $\bar{f}$ .

Proof. Continuing to use the same notation as before, we denote the splitting field of $f$ by $E$ and its ring of integers by ${\mathcal{O}}_{E}$ . Set $D=\text{Disc}_{f}$ and set $K=\mathbb{Q}(\sqrt{D})$ which is a subfield of $E$ . Let $\mathfrak{p}$ be a prime in ${\mathcal{O}}_{E}$ lying over $p$ and write ${\wp}=\mathfrak{p}\cap K$ . Then $\unicode[STIX]{x1D70E}_{{\wp}}:=\unicode[STIX]{x1D70E}_{\mathfrak{p}|_{K}}$ is the Frobenius automorphism assigned to ${\wp}$ in $K/\mathbb{Q}$ . Suppose $\unicode[STIX]{x1D70E}_{\mathfrak{p}}$ is an even permutation. Then $\unicode[STIX]{x1D70E}_{\mathfrak{p}}(\unicode[STIX]{x1D6E5}_{f})=\unicode[STIX]{x1D6E5}_{f}$ and so $\unicode[STIX]{x1D70E}_{{\wp}}$ is trivial over $K$ , which implies that $(\frac{D}{p})=1$ . If, on the other hand, $\unicode[STIX]{x1D70E}_{\mathfrak{p}}$ is an odd permutation, then $\unicode[STIX]{x1D70E}_{\mathfrak{p}}(\unicode[STIX]{x1D6E5}_{f})=-\unicode[STIX]{x1D6E5}_{f}$ , and $\unicode[STIX]{x1D70E}_{{\wp}}$ is not-trivial, implying that $(\frac{D}{p})=-1$ . We have thus shown that $\text{sgn}(\unicode[STIX]{x1D70E}_{\mathfrak{p}})=(\frac{D}{p})$ . Let $n_{i}=\deg f_{i}$ . Viewing $\unicode[STIX]{x1D70E}_{\mathfrak{p}}$ as a permutation of the roots of $f$ , from Theorem 5.7, we obtain

$$\begin{eqnarray}\text{sgn}(\unicode[STIX]{x1D70E}_{\mathfrak{p}})=(-1)^{\mathop{\sum }_{i=1}^{g}(n_{i}-1)}=(-1)^{n-g},\end{eqnarray}$$

since $\sum _{i=1}^{g}n_{i}=n$ . This finishes the proof.◻

As an application we obtain the following result.

Corollary 5.9. Let $f(T)\in \mathbb{Z}[T]$ be an irreducible monic cubic polynomial with discriminant $D$ , and suppose $p$ is an odd prime which does not divide $D$ . Then

$$\begin{eqnarray}N_{f}(p)=\left\{\begin{array}{@{}ll@{}}0\text{ or }3\quad & \displaystyle \text{if }\biggl(\frac{D}{p}\biggr)=1,\\ 1\quad & \displaystyle \text{if }\biggl(\frac{D}{p}\biggr)=-1.\end{array}\right.\end{eqnarray}$$

We now turn to the special case $f(T)=T^{3}-T-1$ . The discriminant of $f$ is $-23$ and then by the quadratic reciprocity we deduce that for $p\neq 23$

$$\begin{eqnarray}N_{f}(p)=\left\{\begin{array}{@{}ll@{}}0\text{ or }3\quad & \displaystyle \text{if }\biggl(\frac{p}{23}\biggr)=1,\\ 1\quad & \displaystyle \text{if }\biggl(\frac{p}{23}\biggr)=-1.\end{array}\right.\end{eqnarray}$$

When $(\frac{p}{23})=1$ , we will need the reduction theory of binary quadratic forms to determine $N_{f}(p)$ . We refer the reader to [Reference FlathFla89, ch. 2, § 8] for more details. Let $\unicode[STIX]{x1D6E5}<0$ be an integer and assume that $\unicode[STIX]{x1D6E5}\equiv 0,1\;(\text{mod}\;4)$ . The modular group $\text{SL}_{2}(\mathbb{Z})$ acts on

$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{\unicode[STIX]{x1D6E5}}:=\{g(x,y)=ax^{2}+bxy+cy^{2}:a,b,c\in \mathbb{Z},a>0,\gcd (a,b,c)=1,b^{2}-4ac=\unicode[STIX]{x1D6E5}\},\end{eqnarray}$$

by linear change of variables. By the reduction theory of positive definite integral binary quadratic forms (for example see [Reference CoxCox13, Theorem 3.9]), the number $h(\unicode[STIX]{x1D6E5})$ of $\text{SL}_{2}(\mathbb{Z})$ -orbits is finite. This number is called the class number of $\unicode[STIX]{x1D6E5}$ . In the case at hand, we have $h(-23)=3$ , and $\text{SL}_{2}(\mathbb{Z})$ -orbits of $\unicode[STIX]{x1D6F4}_{-23}$ are represented by the forms $x^{2}+xy+6y^{2}$ and $2x^{2}\pm xy+3y^{2}$ . Note that $2x^{2}+xy+3y^{2}$ and $2x^{2}-xy+3y^{2}$ are $\text{GL}_{2}(\mathbb{Z})$ -equivalent and thus represent the same set of integers. It is easy to show that $(\frac{p}{23})=1$ , if and only if $p$ is represented by exactly one of the form $x^{2}+xy+6y^{2}$ or $2x^{2}+xy+3y^{2}$ (see [Reference FlathFla89, Proposition 10.2]). Let $L$ be the cubic extension of $\mathbb{Q}$ obtained by adding a root of $f(T)$ and set $K=\mathbb{Q}(\sqrt{-23})$ .

Note that $\text{Gal}(E/\mathbb{Q})=S_{3}$ . For $p\neq 23$ , set $p{\mathcal{O}}_{K}=\mathfrak{p}\overline{\mathfrak{p}}$ . Since the class number of $K$ is $3$ and $E$ is unramified over $K$ , $E$ is the Hilbert class field of $K$ , i.e. the maximal unramified abelian extension of $K$ . From this we can conclude that $\mathfrak{p}$ splits completely in $E$ if and only if $\mathfrak{p}$ is a principal ideal [Reference CoxCox13, Corollary 5.25]. Moreover, note that the ring of integers of the quadratic extension $K$ is $\mathbb{Z}[(1+\sqrt{-23})/2]$ and so $\mathfrak{p}$ is a principal ideal if and only if $p=x^{2}+xy+6y^{2}$ . Putting all these together, we conclude that $p=x^{2}+xy+6y^{2}$ if and only if $p$ splits completely in $E$ . This means that $T^{3}-T-1$ has three roots in $\mathbb{F}_{p}$ if and only if $p=x^{2}+xy+6y^{2}$ . This also shows that $p=2x^{2}+xy+3y^{2}$ if and only if $T^{3}-T-1$ has no root in $\mathbb{F}_{p}$ . Consequently,

(29) $$\begin{eqnarray}N_{f}(p)=\left\{\begin{array}{@{}ll@{}}1\quad & \displaystyle \text{if }\biggl(\frac{p}{23}\biggr)=-1,\\ 0\quad & \text{if }p\text{ is of the form }2x^{2}+xy+3y^{2},\\ 3\quad & \text{if }p\text{ is of the form }x^{2}+xy+6y^{2}.\end{array}\right.\end{eqnarray}$$

Let $N_{\mathbf{X}}(p)$ denote the number of rational points of the projective variety $\mathbf{X}:=T_{1}^{3}-T_{1}T_{2}^{2}-T_{2}^{3}=0$ in $\mathbf{P}^{1}(\mathbb{F}_{p})$ . Then from (29), for all $p\neq 23$ we obtain

$$\begin{eqnarray}N_{\mathbf{X}}(p)=\left\{\begin{array}{@{}ll@{}}1\quad & \displaystyle \text{if }\biggl(\frac{p}{23}\biggr)=-1,\\ 0\quad & \text{if }p\text{ is of the form }2x^{2}+xy+3y^{2},\\ 3\quad & \text{if }p\text{ is of the form }x^{2}+xy+6y^{2}.\end{array}\right.\end{eqnarray}$$

When $(\frac{p}{23})=-1$ , we have $N_{\mathbf{X}}(p)=1$ and hence the faithful dimension of $\mathscr{G}_{p}$ equals $p^{2}+p^{3}$ . When $p$ is of the form $2x^{2}+xy+3y^{2}$ then $N_{\mathbf{X}}(p)=0$ and so the minimum is exactly $2p^{3}$ . This implies that the faithful dimension is $2p^{3}$ . In the remaining case, we can find distinct points $(x_{11},x_{12})$ and $(x_{21},x_{22})$ in $\mathbf{X}$ . Since $M(x_{11},x_{12})$ and $M(x_{21},x_{22})$ have both rank $2$ , it follows that in this case the faithful dimension is $2p^{2}$ . Moreover, $T_{1}^{3}-T_{1}-1$ has a double root and a simple root in $\mathbb{F}_{23}$ . Thus the same argument shows that in this case the faithful dimension is $2(23)^{2}$ .