2. The finite general linear groups

In this section we mostly recall some relevant facts about the conjugacy classes and the character theory of $ G_n$ .

2·1. Partitions

An (integer) partition is a sequence $\lambda=(\lambda_1,\lambda_2,\dots)$ of nonnegative integers satisfying $\lambda_1\ge\lambda_2\ge\cdots$ . The set of partitions is denoted by Par. We often omit trailing zeros and write $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_k)$ if $\lambda_k>0$ and $\lambda_{k+1}=0$ . The size of $(\lambda_1,\lambda_2,\dots)$ is defined to be $\lvert\lambda\rvert=\lambda_1+\lambda_2+\cdots$ . If $\lvert\lambda\rvert=n$ , then we also say that $\lambda$ is a partition of n. We denote the unique partition of 0 by $\varnothing$ .

The Young diagram of a partition $(\lambda_1,\lambda_2,\dots,\lambda_k)$ of n is an array of n boxes with left-justified rows and top-justified columns, where row i contains $\lambda_i$ boxes. To each partition $\lambda$ belongs a conjugate partition $\lambda^{\prime}$ whose parts are the number of boxes in the columns of the Young diagram of $\lambda$ . For two partitions $\lambda=(\lambda_1,\lambda_2,\dots)$ and $\mu=(\mu_1,\mu_2,\dots)$ of the same size, we say that $\lambda$ dominates $\mu$ and write $\lambda\unrhd\mu$ if

\begin{equation*}\sum_{i=1}^k\lambda_i\ge\sum_{i=1}^k\mu_i\quad \text{for each $k\ge 1$}.\end{equation*}

This indeed defines a partial order on the set of partitions of a fixed size, which is called the dominance order.

2·2. Conjugacy classes

We shall now describe the conjugacy classes of $ G_n$ (see [ Reference Macdonald20 , chapter IV, section 3], for example). Let $\Phi$ be the set of monic irreducible polynomials in $\mathbb{F}_q[X]$ distinct from X. For $a\in\mathbb{F}_q^*$ (where $\mathbb{F}_q^*$ is the multiplicative group of $\mathbb{F}_q$ ), we shall often write a instead of $X-a$ when the meaning is clear from the context. We also write $\lvert f\rvert$ for the degree of $f\in\Phi$ . Let $\Lambda$ be the set of mappings $\underline{\lambda}\colon\Phi\to\text{Par}$ of finite support (with $\varnothing$ being the zero element in Par). We define the size of such a mapping to be

\begin{equation*}\lVert\underline{\lambda}\rVert=\sum_{f\in\Phi}\lvert\underline{\lambda}(f)\rvert\cdot\lvert f\rvert\end{equation*}

and put $\Lambda_n=\{\underline{\lambda}\in\Lambda\,:\,\lVert\underline{\lambda}\rVert=n\}$ . The companion matrix of $f\in\Phi$ with $f=X^d+f_{d-1}X^{d-1}+\cdots+f_1X+f_0$ is

\begin{equation*}C(f)=\begin{bmatrix} & \quad & \quad & \quad & \quad -f_0\\[4pt]1 & \quad & \quad & \quad & \quad -f_1\\[4pt] & \quad 1 & \quad & \quad & \quad -f_2\\[4pt] & \quad & \quad \ddots & \quad & \quad \vdots\\[4pt] & \quad & \quad & \quad 1 & \quad -f_{d-1}\\[4pt]\end{bmatrix}\in\mathbb{F}_q^{d\times d},\end{equation*}

(where blanks are filled with zeros). For $f\in\Phi$ of degree d and a positive integer k, we write

\begin{equation*}C(f,k)=\begin{bmatrix}C(f) & \quad I & \quad & \quad & \quad \\[4pt] & \quad C(f) & \quad I & \quad & \quad \\[4pt]& \quad& \quad\ddots& \quad\ddots\\[4pt] & \quad & \quad & \quad \ddots & \quad I\\[4pt] & \quad & \quad & \quad & \quad C(f)\end{bmatrix}\in\mathbb{F}_q^{kd\times kd},\end{equation*}

where I is an identity matrix of the appropriate size. For $f\in\Phi$ and $\sigma\in\text{Par}$ , we define $C(f,\sigma)$ to be the block diagonal matrix of order $\lvert\sigma\rvert\cdot\lvert f\rvert$ with blocks $C(f,\sigma_1),C(f,\sigma_2),\dots$ . Finally, with every $\underline{\sigma}\in\Lambda_n$ we associate the block diagonal matrix $R_{\underline{\sigma}}$ of order n whose blocks are $C(f,\underline{\sigma}(f))$ , where f ranges through the support of $\underline{\sigma}$ . Then every element g of $ G_n$ is conjugate to exactly one matrix $R_{\underline{\sigma}}$ for $\underline{\sigma}\in\Lambda_n$ , which is called the Jordan canonical form of g. Hence $\Lambda_n$ indexes the conjugacy classes of $ G_n$ ; we denote by $C_{\underline{\sigma}}$ the conjugacy class containing $R_{\underline{\sigma}}$ . The following result gives an explicit expression for the number of elements in $C_{\underline{\sigma}}$ .

Lemma 2·1 ([ Reference Stanley26 , theorem 1·10·7]). For each $\underline{\sigma}\in\Lambda_n$ , we have

\begin{equation*}\frac{\lvert G_n\rvert}{\lvert C_{\underline{\sigma}}\rvert}=\prod_{f\in\Phi} \prod_{i=1}^{\lvert\underline{\sigma}(f)\rvert} \prod_{j=1}^{m_i(\underline{\sigma}(f))} q^{\lvert f\rvert\, s_i\left(\underline{\sigma}(f)^{\prime}\right)} \left(1-q^{-\lvert f\rvert\,j}\right),\end{equation*}

where $m_i(\sigma)= \lvert\{j\ge 1\colon\sigma_j=i\}\rvert$ and $s_i(\sigma)=\sum_{j=1}^i \sigma_j$ for a partition $\sigma$ .

2·3. Parabolic induction

Recall that, given a finite group G, a subgroup H of G, and a class function $\phi$ on H, the induced class function $\text{Ind}_H^G(\phi)$ on G is given by

(2·1) \begin{equation}\text{Ind}_H^G(\phi)(g)=\frac{1}{\lvert H\rvert}\sum_{\substack{x\in G\\xgx^{-1}\in H}}\phi(xgx^{-1}). \end{equation}

The character theory of $ G_n$ crucially relies on the induction of characters from parabolic subgroups of $ G_n$ .

A composition is much like a partition, except that the parts do not need to be nonincreasing. Let $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_k)$ be a composition of n. Let $P_\lambda$ be the parabolic subgroup of $ G_n$ consisting of block upper-triangular matrices with block sizes $\lambda_1,\lambda_2,\dots,\lambda_k$ , namely

(2·2) \begin{equation}P_\lambda=\left\{\begin{bmatrix}g_1 & \quad * & \quad \cdots & \quad *\\[4pt] & \quad g_2 & \quad \cdots & \quad *\\[4pt] & \quad & \quad \ddots & \quad \vdots\\[4pt] & \quad & \quad & \quad g_k\\[4pt]\end{bmatrix}:g_i\in G_{\lambda_i}\right\}.\end{equation}

Let $\pi_i\,:\,P_\lambda\to G_{\lambda_i}$ be the mapping that projects to the ith diagonal block, so that

(2·3) \begin{equation}\pi_i\,:\,\begin{bmatrix}g_1 & \quad * & \quad \cdots & \quad *\\[4pt] & \quad g_2 & \quad \cdots & \quad *\\[4pt] & \quad & \quad \ddots & \quad \vdots\\[4pt] & \quad & \quad & \quad g_k\\[4pt]\end{bmatrix}\longmapsto g_i. \end{equation}

Let $\phi_i$ be a class function on $G_{\lambda_i}$ . Then

\begin{equation*}\prod_{i=1}^k(\phi_i\circ \pi_i)\end{equation*}

is a class function on $P_\lambda$ . We define the product $\phi_1\odot\phi_2\odot\cdots\odot\phi_k$ to be the induction of this class function to $ G_n$ , that is

(2·4) \begin{equation}\bigodot_{i=1}^k\phi_i=\text{Ind}_{P_\lambda}^{ G_n}\left(\prod_{i=1}^k(\phi_i\circ \pi_i)\right). \end{equation}

2·4. Character theory of $ G_n$

The complete set of complex irreducible characters has been obtained by Green [ Reference Green14 ]. A good treatment of this topic is also contained in [ Reference Macdonald20 , chapter IV]. The complex irreducible representations were obtained by Gelfand [ Reference Gelfand11 ] and the irreducible representations over fields of nondefining characteristic were obtained by James [ Reference James17 ]. The approach of [ Reference James17 ] is in fact very similar to the standard combinatorial approach to obtain the complex irreducible representations of the symmetric group (see [ Reference Sagan24 ], for example) and we mostly follow [ Reference James17 ] to recall some relevant background on the complex characters of $ G_n$ .

The irreducible characters of $ G_n$ are naturally indexed by $\Lambda_n$ and, for $\underline{\lambda}\in\Lambda_n$ , we denote by $\chi^{\underline{\lambda}}$ the corresponding irreducible character. We shall use the short-hand notation $\chi^{f\mapsto\lambda}$ for $\chi^{\underline{\lambda}}$ if $\underline{\lambda}$ is supported only on $f\in\Phi$ and $\underline{\lambda}(f)=\lambda$ . These are typically called the primary irreducible characters of $ G_n$ . It is well known (see [ Reference James17 , section 8], for example) that the irreducible characters of $ G_n$ satisfy

(2·5) \begin{equation}\chi^{\underline{\lambda}}=\bigodot_{f\in\Phi}\chi^{f\mapsto\underline{\lambda}(f)}. \end{equation}

In order to construct the primary irreducible characters, James [ Reference James17 ] constructs characters of $G_{dm}$ , denoted by $\xi^{f\mapsto\mu}$ , where $f\in\Phi$ has degree d and $\mu$ is a partition of m. Writing $\mu=(\mu_1,\mu_2,\dots,\mu_k)$ , these characters satisfy [ Reference James17 , (6·2)]

(2·6) \begin{equation}\xi^{f\mapsto\mu}=\bigodot_{i=1}^k\xi^{f\mapsto(\mu_i)} \end{equation}

and [ Reference James17 , (7·19)]

(2·7) \begin{equation}\xi^{f\mapsto\mu}=\sum_{\lambda}K_{\lambda\mu}\,\chi^{f\mapsto\lambda}, \end{equation}

where $\lambda$ ranges over the partitions of $\lvert\mu\rvert$ and $K_{\lambda\mu}$ is a Kostka number, which equals the number of semistandard Young tableaux of shape $\lambda$ and content $\mu$ . It is well known (see [ Reference Sagan24 , section 2·11], for example) that the Kostka numbers satisfy

(2·8) \begin{equation}\text{$K_{\mu\mu}=1$ and $K_{\lambda\mu}\ne 0\Rightarrow \lambda\unrhd\mu$}. \end{equation}

Conversely it is readily verified that there are integers $H_{\mu\lambda}$ satisfying$

(2·9) \begin{equation}\chi^{f\mapsto\lambda}=\sum_{\mu}H_{\mu\lambda}\,\xi^{f\mapsto\mu} \end{equation}

and

(2·10) \begin{equation}\text{$H_{\lambda\lambda}=1$ and $H_{\mu\lambda}\ne 0\Rightarrow \mu\unrhd\lambda$} \end{equation}

(see [ Reference Macdonald20 , p. 105], for example).

Now, for $\underline{\mu}\in\Lambda_n$ , we define the characters

(2·11) \begin{equation}\xi^{\underline{\mu}}=\bigodot_{f\in\Phi}\xi^{f\mapsto\underline{\mu}(f)}. \end{equation}

We denote by $\chi^{\underline{\lambda}}_{\underline{\sigma}}$ and $\xi^{\underline{\mu}}_{\underline{\sigma}}$ the characters $\chi^{\underline{\lambda}}$ and $\xi^{\underline{\mu}}$ , respectively, evaluated on the conjugacy class $C_{\underline{\sigma}}$ .

We now express $\xi^{\underline{\mu}}$ and $\chi^{\underline{\lambda}}$ in terms of each other. To do so, we define the shape of $\underline{\lambda}\in\Lambda_n$ to be the mapping $s\,:\,\Phi\to\mathbb{Z}$ given by $s(f)=\lvert\underline{\lambda}(f)\rvert$ for each $f\in\Phi$ . We write $\underline{\lambda}\sim\underline{\mu}$ if $\underline{\lambda},\underline{\mu}\in\Lambda_n$ have the same shape. Then $\sim$ is an equivalence relation on $\Lambda_n$ . For $\underline{\lambda},\underline{\mu}\in\Lambda_n$ with $\underline{\lambda}\sim\underline{\mu}$ , write

\begin{align*}K_{\underline{\lambda}\underline{\mu}}&=\prod_{f\in\Phi}K_{\underline{\lambda}(f)\underline{\mu}(f)},\\H_{\underline{\mu}\underline{\lambda}}&=\prod_{f\in\Phi}H_{\underline{\mu}(f)\underline{\lambda}(f)}.\end{align*}

We then find that

(2·12) \begin{align}\xi^{\underline{\mu}}&=\sum_{\underline{\lambda}\sim\underline{\mu}}K_{\underline{\lambda}\underline{\mu}}\;\chi^{\underline{\lambda}}\quad \text{for each $\underline{\mu}\in\Lambda_n$},\end{align}

(2·13) \begin{align}\chi^{\underline{\lambda}}&=\sum_{\underline{\mu}\sim\underline{\lambda}}H_{\underline{\mu}\underline{\lambda}}\;\xi^{\underline{\mu}}\quad \text{for each $\underline{\lambda}\in\Lambda_n$} .\end{align}

An explicit expression for the degree $\chi^{\underline{\lambda}}(1)$ (where 1 is the identity of $ G_n$ ) of $\chi^{\underline{\lambda}}$ is given by the so-called q-analog of the hook-length formula.

Lemma 2·2 ([ Reference Green14 , theorem 14]). We have

(2·14) \begin{equation}\frac{1}{\chi^{\underline{\lambda}}(1)}\;\prod_{i=1}^n(q^i-1)=\prod_{f\in\Phi}\frac{1}{q^{\lvert f\rvert b(\underline{\lambda}(f))}}\prod_{(i, j)\in\underline{\lambda}(f)}\left(q^{\lvert f\rvert h_{i, j}(\underline{\lambda}(f))}-1\right), \end{equation}

where, for each partition $\lambda=(\lambda_1,\lambda_2,\dots)$ ,

\begin{equation*}b(\lambda)=\sum_{i\ge 1}(i-1)\lambda_i\end{equation*}

and $h_{i, j}(\lambda)$ is the hook length of $\lambda$ at (i, j), namely

\begin{equation*}h_{i, j}(\lambda)=\lambda_i+\lambda^{\prime}_j-i-j+1\end{equation*}

and the corresponding product over (i, j) is over all boxes of the Young diagram of $\underline{\lambda}(f)$ .

It can be readily verified from Lemma 2·2 that the linear (degree-one) irreducible characters of $ G_n$ are precisely the primary characters $\chi^{f\mapsto(n)}$ , where $\lvert f\rvert=1$ . These are the only characters of $ G_n$ that we shall need explicitly. Let $\alpha$ be a generator of the multiplicative group $\mathbb{F}_q^*$ of $\mathbb{F}_q$ , let $\omega=\exp\!(2\pi \sqrt{-1}/(q-1))$ be a complex root of unity, and let $\theta\,:\,\mathbb{F}_q^*\to\mathbb{C}$ be the linear character of $\mathbb{F}_q^*$ given by $\theta\!\left(\alpha^i\right)=\omega^i$ . The following result is essentially given in [ Reference Green14 , pp. 415 and 444].

Lemma 2·3 ([ Reference Green14 ]). For all $g\in G_n$ , we have

\begin{equation*}\chi^{X-\alpha^i\mapsto(n)}(g)=\theta\!\left(\det(g)^i\right).\end{equation*}

In particular $\chi^{X-1\mapsto(n)}$ is the trivial character.

In what follows we consider certain characters of $G_n$ related to the permutation character of $G_n$ on the set of t-tuples of linearly independent elements of $\mathbb{F}_q^n$ . For $t\leq n$ , let $H_{n,t}$ be the stabiliser of a fixed t-tuple of linearly independent elements of $\mathbb{F}_q^n$ . We define $\zeta^{(t,i)}$ to be the character obtained by inducing the linear character

(2·15) \begin{equation}\begin{gathered}H_{n,t}\longrightarrow\mathbb{C}\\g\longmapsto \theta\!\left(\det(g)^i\right)\end{gathered}\end{equation}

to $ G_n$ . Then $\zeta^{(t,0)}$ is the permutation character of $ G_n$ on the set of t-tuples of linearly independent elements of $\mathbb{F}_q^n$ . These characters are related to each other in the following way.

Lemma 2·4. For each $g\in G_n$ , we have

\begin{equation*}\zeta^{(t,i)}(g)=\theta\!\left(\det(g)^i\right)\zeta^{(t,0)}(g).\end{equation*}

Proof. Since similar matrices have the same determinant, we find from (2·1) that

\begin{align*}\zeta^{(t,i)}(g)&=\frac{1}{\lvert H_{n,t}\rvert}\sum_{\substack{x\in G_n\\xgx^{-1}\in H_{n,t}}}\theta\!\left(\det(xgx^{-1})^i\right)\\&=\frac{1}{\lvert H_{n,t}\rvert}\sum_{\substack{x\in G_n\\xgx^{-1}\in H_{n,t}}}\theta\!\left(\det(g)^i\right)\\&=\theta\!\left(\det(g)^i\right)\zeta^{(t,0)}(g).\end{align*}

We shall also need the following information about the decomposition of $\zeta^{(t,i)}$ into irreducible characters of $ G_n$ .

Lemma 2·5. We have

\begin{equation*}\zeta^{(t,i)}=\sum_{\substack{\underline{\lambda}\in\Lambda_n}}m_{i,\underline{\lambda}}\,\chi^{\underline{\lambda}},\end{equation*}

where $m_{i,\underline{\lambda}}\ne 0$ if and only if $\underline{\lambda}\!\left(\alpha^i\right)_1\ge n-t$ .

Proof. We may choose $H_{n,t}$ to be

\begin{equation*}H_{n,t}=\left\{\begin{bmatrix}I & \quad *\\[4pt] & \quad g\end{bmatrix}\,:\,g\in G_{n-t}\right\},\end{equation*}

so that $H_{n,t}$ is a subgroup of the parabolic subgroup $P_{(t,n-t)}$ given in (2·2). Let $\pi_1$ and $\pi_2$ be the projections onto the diagonal blocks of orders t and $n-t$ , respectively, as given in (2·3). Using Lemma 2·3, the character (2·15) can be written as

(2·16) \begin{equation}(1\circ \pi_1)\,\left(\chi^{X-\alpha^i\mapsto (n-t)}\circ\pi_2\right). \end{equation}

where 1 is the trivial character of the trivial subgroup of $G_t$ . By Frobenius reciprocity, 1 induces on $G_t$ to the character

\begin{equation*}\sum_{\underline{\kappa}\in\Lambda_t}\chi^{\underline{\kappa}}(1)\,\chi^{\underline{\kappa}}.\end{equation*}

Since $P_{(t,n-t)}/H_{n,t}\cong G_t$ , it is then readily verified that (2·16) induces on $P_{(t,n-t)}$ to the character

\begin{equation*}\sum_{\underline{\kappa}\in\Lambda_t}\chi^{\underline{\kappa}}(1)\,(\chi^{\underline{\kappa}}\circ\pi_1)\,\left(\chi^{X-\alpha^i\mapsto (n-t)}\circ\pi_2\right).\end{equation*}

Hence, by transitivity of induction, we have

\begin{equation*}\zeta^{(t,i)}=\sum_{\underline{\kappa}\in\Lambda_t}\chi^{\underline{\kappa}}(1)\;\left(\chi^{\underline{\kappa}}\odot \chi^{X-\alpha^i\mapsto (n-t)}\right).\end{equation*}

It is well known [ Reference Macdonald20 , chapter IV, section 4] that, for each fixed $f\in\Phi$ , characters $\chi^{f\mapsto \lambda}$ form an algebra with multiplication $\odot$ that is isomorphic to the algebra of symmetric functions and the images of the characters $\chi^{f\mapsto \lambda}$ are the Schur functions. We then find from Pieri’s rule (see [ Reference Macdonald20 , chapter I, (5·16)], for example) that

\begin{equation*}\chi^{X-\alpha^i\mapsto\kappa}\odot\chi^{X-\alpha^i\mapsto(n-t)}=\sum_{\lambda}\chi^{X-\alpha^i\mapsto\lambda},\end{equation*}

where $\lambda$ runs through all partitions whose Young diagram is obtained from that of $\kappa$ by adding $n-t$ boxes, no two of which in the same column. Using (2·5) the statement of the lemma is then readily verified.

3. The Hoffman bound

Henceforth we use the following notation. For a field K and finite sets X and Y, we denote by K(X, Y) the set of $\lvert X\rvert\times\lvert Y\rvert$ matrices A with entries in K, where rows and columns are indexed by X and Y, respectively. For $x\in X$ and $y\in Y$ , the (x, y)-entry of A is written as A(x, y). If $\lvert Y\rvert=1$ , then we omit Y, so K(X) is the set of column vectors a indexed by X and, for $x\in X$ , the x-entry of a is written as a(x).

The adjacency matrix of a graph $\Gamma=(X,E)$ is the matrix $A\in\mathbb{R}(X,X)$ given by

\begin{equation*}A(x,y)=\begin{cases}1 & \text{for $\{x,y\}\in E$}\\0 & \text{otherwise}.\end{cases}\end{equation*}

Then A is a real symmetric matrix, which of course has an orthonormal system of $\lvert X\rvert$ eigenvectors forming a basis of $\mathbb{R}(X)$ . All eigenvalues of A are real and referred to as the eigenvalues of $\Gamma$ . Note that, if $\Gamma$ is d-regular, then d is an eigenvalue of $\Gamma$ and the all-ones vector is a corresponding eigenvector.

Our starting point arises from the following generalised versions of the Hoffman bound [ Reference Haemers15 ], stated and proved by Ellis, Friedgut, and Pilpel [ Reference Ellis, Friedgut and Pilpel7 , section 2·4] in the following form.

In order to study graphs induced by $ G_n$ and their eigenvalues, we shall bring the theory of association schemes into play. We refer to [ Reference Bannai and Ito4 ] and [ Reference Godsil and Meagher13 ] for background on association schemes. Every finite group gives rise to an association scheme (see [ Reference Bannai and Ito4 , section 2·7] or [ Reference Godsil and Meagher13 , section 3·3] for details). We shall recall relevant background about this association scheme and its symmetrisation for $ G_n$ .

For each $\underline{\sigma}\in\Lambda_n$ , we define $B_{\underline{\sigma}}=\mathbb{C}( G_n,G_n)$ by

\begin{equation*}B_{\underline{\sigma}}(x,y)=\begin{cases}1 & \text{for $x^{-1}y\in C_{\underline{\sigma}}$}\\0 & \text{otherwise}.\end{cases}\end{equation*}

The vector space generated by $\{B_{\underline{\sigma}}\,:\,\underline{\sigma}\in\Lambda_n\}$ over the complex numbers turns out to be a commutative matrix algebra $\mathbb{A}$ , which contains the identity and the all-ones matrix and is closed under conjugate transposition. The collection of zero-one matrices $B_{\underline{\sigma}}$ therefore defines an association scheme. Since $\mathbb{A}$ is commutative, it can be simultaneously diagonalised and therefore there exists a basis $\{F_{\underline{\lambda}}\,:\,\underline{\lambda}\in\Lambda_n\}$ of $\mathbb{A}$ consisting of primitive idempotent matrices. These matrices are given by [ Reference Bannai and Ito4 , theorem II·7·2]

(3·1) \begin{equation}F_{\underline{\lambda}}\,=\frac{\chi^{\underline{\lambda}}(1)}{\lvert G_n\rvert}\sum_{\underline{\sigma}\in\Lambda_n}\chi^{\underline{\lambda}}_{\underline{\sigma}}\,B_{\underline{\sigma}}. \end{equation}

Using the orthogonality of characters of the second kind, it is readily verified that

(3·2) \begin{equation}B_{\underline{\sigma}}=\sum_{\underline{\lambda}\in\Lambda_n}\frac{\lvert C_{\underline{\sigma}}\rvert}{\chi^{\underline{\lambda}}(1)}\,\overline{\chi}^{\underline{\lambda}}_{\underline{\sigma}}\,F_{\underline{\lambda}}, \end{equation}

where $\overline{\chi}^{\underline{\lambda}}$ is the character of $ G_n$ whose values at $g\in G_n$ are the complex conjugates of $\chi^{\underline{\lambda}}(g)$ .

For each $f\in\Phi$ , let $f^*\in\Phi$ be its reciprocal polynomial, namely the monic polynomial whose roots (in an algebraic closure of $\mathbb{F}_q$ ) are precisely the inverses of the roots of f. For each $\underline{\lambda}\in\Lambda_n$ , define $\underline{\lambda}^*$ to be the element of $\Lambda_n$ given by $\underline{\lambda}^*(f)=\underline{\lambda}(f^*)$ for all $f\in\Phi$ . We record the following lemma, in which we write $C^{-1}_{\underline{\sigma}}=\{g^{-1}\,:\,g\in C_{\underline{\sigma}}\}$ for $\underline{\sigma}\in\Lambda_n$ .

Proof. Statement (i) is a basic fact in linear algebra, (ii) is essentially [ Reference James17 , (7·32)], and (iii) can be deduced from (i) and (ii).

Let $\Omega_n$ be the subset of $\Lambda_n$ that contains all $\underline{\lambda}\in\Lambda_n$ satisfying $\underline{\lambda}=\underline{\lambda}^*$ and precisely one of $\underline{\lambda}$ or $\underline{\lambda}^*$ for all $\underline{\lambda}\in\Lambda_n$ satisfying $\underline{\lambda}\ne\underline{\lambda}^*$ . For $\underline{\lambda}\in\Omega_n$ , we define the character

\begin{equation*}\psi^{\underline{\lambda}}=\begin{cases}\chi^{\underline{\lambda}} & \text{for $\underline{\lambda}=\underline{\lambda}^*$}\\\chi^{\underline{\lambda}}+\chi^{\underline{\lambda}^*} & \text{otherwise},\end{cases}\end{equation*}

and, for $\underline{\sigma}\in\Omega_n$ , we define $D_{\underline{\sigma}}=C_{\underline{\sigma}}\cup C_{\underline{\sigma}^*}$ . Lemma 3·2 implies that $\psi^{\underline{\lambda}}$ is constant on $D_{\underline{\sigma}}$ . We write

(3·3) \begin{equation}\psi^{\underline{\lambda}}_{\underline{\sigma}}=\psi^{\underline{\lambda}}(g),\quad \text{where} \, g \, \text{is an arbitrary element of $D_{\underline{\sigma}}$}. \end{equation}

For $\underline{\sigma},\underline{\lambda}\in\Omega_n$ , write

(3·4) \begin{equation}A_{\underline{\sigma}}=\begin{cases}B_{\underline{\sigma}} & \text{for $\underline{\sigma}=\underline{\sigma}^*$}\\B_{\underline{\sigma}}+B_{\underline{\sigma}^*} & \text{otherwise}\end{cases}\quad \text{and}\quad E_{\underline{\lambda}}=\begin{cases}F_{\underline{\lambda}} & \text{for $\underline{\lambda}=\underline{\lambda}^*$}\\F_{\underline{\lambda}}+F_{\underline{\lambda}^*} & \text{otherwise}.\end{cases}\end{equation}

Note that $A_{\underline{\sigma}}$ is symmetric, so all of its eigenvalues are real, and that $E_{\underline{\lambda}}$ has only real entries. Let $V_{\underline{\lambda}}$ be the column span over the reals of $E_{\underline{\lambda}}$ and, for $\underline{\sigma},\underline{\lambda}\in\Omega_n$ , write

(3·5) \begin{equation}P\!\left(\underline{\lambda},\underline{\sigma}\right)=\frac{\lvert D_{\underline{\sigma}}\rvert}{\psi^{\underline{\lambda}}(1)}\;\psi^{\underline{\lambda}}_{\underline{\sigma}}. \end{equation}

The following lemma, containing essentially standard results, will be crucial in the following.

Lemma 3·3. We have the following orthogonal direct sum decomposition

\begin{equation*}\mathbb{R}(G_n)=\bigoplus_{\underline{\lambda}\in\Omega_n}V_{\underline{\lambda}}.\end{equation*}

Moreover, for all $\underline{\sigma},\underline{\lambda}\in\Omega_n$ , every element of $V_{\underline{\lambda}}$ is an eigenvector of $A_{\underline{\sigma}}$ and the corresponding eigenvalue is $P\!\left(\underline{\lambda},\underline{\sigma}\right)$ .

Proof. Since $F_{\underline{\lambda}}$ is a primitive idempotent in $\mathbb{C}(G_n,G_n)$ for each $\underline{\lambda}\in\Lambda_n$ , it is readily verified that $E_{\underline{\lambda}}$ is a primitive idempotent in $\mathbb{R}(G_n,G_n)$ for each $\underline{\lambda}\in\Omega_n$ . Therefore the $E_{\underline{\lambda}}$ are pairwise orthogonal, namely we have $E_{\underline{\lambda}}E_{\underline{\mu}}=\delta_{\underline{\lambda}\underline{\mu}}E_{\underline{\lambda}}$ for all $\underline{\lambda},\underline{\mu}\in\Omega_n$ . Since $E_{\underline{\lambda}}$ is idempotent, $\text{rk}(E_{\underline{\lambda}})$ is just the trace of $E_{\underline{\lambda}}$ . It follows from (3·1) that the trace of $F_{\underline{\lambda}}$ equals $\chi^{\underline{\lambda}}(1)^2$ . Hence we have

\begin{equation*}\sum_{\underline{\lambda}\in\Omega_n}\dim V_{\underline{\lambda}}=\sum_{\underline{\lambda}\in\Omega_n}\text{rk}(E_{\underline{\lambda}})=\sum_{\underline{\lambda}\in\Lambda_n}\chi^{\underline{\lambda}}(1)^2=\lvert G_n\rvert\end{equation*}

by standard properties of the degrees of irreducible characters. This proves the first statement. We have $\chi^{\underline{\lambda}}(1)=\chi^{\underline{\lambda}^*}(1)$ by Lemma 2·2, from which together with (3·2) and Lemma 3·2 it is readily verified that

\begin{equation*}A_{\underline{\sigma}}=\sum_{\underline{\lambda}\in\Omega_n}P\!\left(\underline{\lambda},\underline{\sigma}\right)\,E_{\underline{\lambda}}.\end{equation*}

Since the $E_{\underline{\lambda}}$ are pairwise orthogonal, we obtain the second statement.

In fact the proof of Lemma 3·3 shows that $\{A_{\underline{\sigma}}\,:\,\underline{\sigma}\in\Omega_n\}$ is a symmetric association scheme with primitive idempotents given by $\{E_{\underline{\lambda}}\,:\,\underline{\lambda}\in\Omega_n\}$ . However we will not exploit this further.

Note that $A_{\underline{\sigma}}$ is the adjacency matrix of a $\lvert D_{\underline{\sigma}}\rvert$ -regular graph for each $\underline{\sigma}\in\Omega_n$ , except for $\underline{\sigma}$ given by $\underline{\sigma}(1)=(1^n)$ , and that $P\!\left(\underline{\lambda},\underline{\sigma}\right)=\lvert D_{\underline{\sigma}}\rvert$ if $\underline{\lambda}\in\Omega_n$ is given by $X-1\mapsto (n)$ .

The strategy to prove Theorems 1·1 and 1·2 is as follows (Theorems 1·5 and 1·6 will be proved using slight modifications). We call an element $x\in G$ a t-derangement if there is no t-tuple of linearly independent elements of $\mathbb{F}_q^n$ that is fixed by x. Equivalently $x\in G_n$ is a t-derangement if $\text{rk}(x-I)>n-t$ . It is readily verified that either all elements of $D_{\underline{\sigma}}$ are t-derangements or none of them. We wish to identify an appropriate subset $\Sigma$ of $\Omega_n$ such that $D_{\underline{\sigma}}$ consists of t-derangements for all $\underline{\sigma}\in\Sigma$ and then apply Proposition 3·1 to the graph $\Gamma$ with adjacency matrix $\sum_{\underline{\sigma}\in\Sigma}A_{\underline{\sigma}}$ and $\lvert D_{\underline{\sigma}}\rvert$ -regular spanning subgraphs $\Gamma_{\underline{\sigma}}$ having adjacency matrix $A_{\underline{\sigma}}$ for $\underline{\sigma}\in\Sigma$ . In view of Lemma 3·3, we wish to construct some $w\in\mathbb{R}(\Sigma)$ such that both the minimum value and the negative of the second-largest absolute value over all $\underline{\lambda}\in\Omega_n$ of

(3·6) \begin{equation}\sum_{\underline{\sigma}\in\Sigma}w(\underline{\sigma})P\!\left(\underline{\lambda},\underline{\sigma}\right) \end{equation}

equals

(3·7) \begin{equation}\eta=-\frac{1}{(q^n-1)(q^n-q)\cdots\left(q^n-q^{t-1}\right)-1} \end{equation}

and such that w is normalised in the sense that (3·6) equals 1 if $\psi^{\underline{\lambda}}$ is the trivial character (or equivalently $\underline{\lambda}\in\Omega_n$ is given by $X-1\mapsto (n)$ ). This will ensure that Proposition 3·1 will give the bounds of Theorems 1·1 and 1·2.

4. An invertible matrix

This section contains some key preparations for our main proofs. We first identify relevant conjugacy classes of $G_n$ whose elements are either t-derangements or do not fix a t-space. We then use these conjugacy classes to identify a matrix related to the character table of $G_n$ . A key step is to show that this matrix is invertible.

We call an element of $G_n$ regular elliptic if its characteristic polynomial is irreducible. The following lemma shows that regular elliptic elements in $ G_n$ play the role of an n-cycle in the symmetric group $S_n$ .

Note that, for each $f\in\Phi$ of degree d, its companion matrix satisfies $\det\!\left(C_f\right)=({-}1)^df(0)$ . It is well known [ Reference Hansen and Mullen16 ] that, for each $a\in\mathbb{F}_q^*$ , there exists an irreducible polynomial $f\in\mathbb{F}_q[x]$ of degree d such that $f(0)=a$ . Hence we can always find a polynomial in $\Phi$ with prescribed degree and prescribed nonzero determinant of its companion matrix. Also note that, for each $f\in\Phi$ , we have $f(0)f^*(0)=1$ and therefore

\begin{equation*}\det\!\left(C_f\right)\det\!\left(C_{f^*}\right)=1.\end{equation*}

We now continue to use $\alpha$ to denote a fixed generator of $\mathbb{F}_q^*$ . For all integers $\ell,j$ satisfying $0\leq \ell<n$ and $0\leq j\leq q-2$ , we fix an irreducible polynomial $h_{\ell,j}\in\Phi$ of degree $n-\ell$ such that its companion matrix has determinant $\alpha^j$ and such that $h_{\ell,j}^*=h_{\ell,-j}$ . We define

\begin{equation*}\Sigma_{\ell,j}=\{\text{$\underline{\sigma}\in\Lambda_n\,:\,\underline{\sigma}(h_{\ell,j})=(1)$}\}.\end{equation*}

and

\begin{equation*}\Sigma_\ell=\bigcup_{j=0}^{q-2}\Sigma_{\ell,j}\quad \text{and}\quad \Sigma_{\le t}=\bigcup_{\ell=0}^t\Sigma_\ell.\end{equation*}

Note that, for each $\underline{\sigma}\in\Sigma_{\le t-1}$ , the conjugacy class $C_{\underline{\sigma}}$ consists of elements that do not fix a t-space of $\mathbb{F}_q^n$ . In addition, for each $\underline{\sigma}\in\Sigma_t$ with the $q-1$ exceptions $\underline{\sigma}\in\Sigma_t$ satisfying $\underline{\sigma}(X-1)=(1^t)$ , the conjugacy class $C_{\underline{\sigma}}$ consists of elements that do not fix a t-space pointwise. Next we define

\begin{equation*}\Pi_{k,i}=\{\underline{\lambda}\in\Lambda_n\,:\,\underline{\lambda}\!\left(\alpha^i\right)_1=n-k\}.\end{equation*}

and

\begin{equation*}\Pi_k=\bigcup_{i=0}^{q-2}\Pi_{k,i}\quad \text{and}\quad \Pi_{\le t}=\bigcup_{k=0}^t\Pi_k.\end{equation*}

Note that, for $k<n/2$ , we have $\lvert\Pi_{k,i}\rvert=\lvert\Sigma_{k,i}\rvert$ and $\lvert\Omega_n\cap\Pi_{k,i}\rvert=\lvert\Omega_n\cap\Sigma_{k,i}\rvert$ .

We define $Q\in\mathbb{R}(\Omega_n,\Omega_n)$ by

\begin{equation*}Q\!\left(\underline{\lambda},\underline{\sigma}\right)=\psi^{\underline{\lambda}}_{\underline{\sigma}}\quad \text{for each $\underline{\lambda},\underline{\sigma}\in\Omega_n$}\end{equation*}

and let $Q_t$ be the restriction of Q to $\mathbb{R}(\Omega_n\cap\Pi_{\le t},\Omega_n\cap\Sigma_{\le t})$ . We emphasise that $Q_t$ is a square matrix. A key step in our proof is the following proposition.

In the remainder of this section we essentially only prove Proposition 4·2. The reader who is interested in maintaining the flow of the proof of our main results may wish to skip to the next section at first reading.

We define $R\in\mathbb{C}(\Lambda_n,\Lambda_n)$ by

\begin{equation*}R\!\left(\underline{\lambda},\underline{\sigma}\right)=\chi^{\underline{\lambda}}_{\underline{\sigma}}\quad \text{for each $\underline{\lambda},\underline{\sigma}\in\Lambda_n$}\end{equation*}

and let $R_t$ be the restriction of R to $\mathbb{C}(\Pi_{\le t},\Sigma_{\le t})$ . We shall prove a counterpart of Proposition 4·2 for the matrix $R_t$ .

Note that $Q_t$ is obtained from $R_t$ by first applying elementary row operations, then deleting some rows, and then (in view of (3·3)) deleting duplicate columns. Hence Proposition 4·2 follows from Proposition 4·3.

We now prove Proposition 4·3. We let $S\in\mathbb{C}(\Lambda_n,\Lambda_n)$ be the matrix defined by

(4·1) \begin{equation}S(\underline{\mu},\underline{\sigma})=\xi^{\underline{\mu}}_{\underline{\sigma}}\quad \text{for each $\underline{\mu},\underline{\sigma}\in\Lambda_n$} \end{equation}

and let $S_t$ be the restriction of S to $\mathbb{C}(\Pi_{\le t},\Sigma_{\le t})$ . Now recall the equivalence relation $\sim$ on $\Lambda_n$ and the numbers $K_{\underline{\lambda}\underline{\mu}}$ from Section 2·4. Define $T\in\mathbb{C}(\Lambda_n,\Lambda_n)$ by

\begin{equation*}T(\underline{\mu},\underline{\lambda})=\begin{cases}K_{\underline{\lambda}\underline{\mu}} & \text{for $\underline{\lambda}\sim\underline{\mu}$}\\[4pt]0 & \text{otherwise}\end{cases}\end{equation*}

and let $T_t$ be the restriction of T to $\mathbb{C}(\Pi_{\le t},\Pi_{\le t})$ . We first prove the following.

Proof. From (2·12) we have $S=TR$ and T is block diagonal, where the blocks are induced by the equivalence classes under $\sim$ . Each diagonal block corresponds to one equivalence class. If $s\,:\,\Phi\to\mathbb{Z}$ is the shape of such an equivalence class, then the corresponding block can be written as a Kronecker product

\begin{equation*}\bigotimes_{f\in\Phi}K^{(s(f))},\end{equation*}

where $K^{(m)}\in\mathbb{C}(\text{Par}_m,\text{Par}_m)$ is a Kostka matrix given by $K^{(m)}(\mu,\lambda)=K_{\lambda\mu}$ with the convention $K^{(0)}=(1)$ and $\text{Par}_m$ is the set of partitions of m. By (2·8) the Kostka matrices are invertible. Hence T is a block-diagonal matrix whose blocks are Kronecker products of matrices of full rank and so T itself has full rank. This proves (i).

From (2·8) we find that $S_t=T_tR_t$ . Note that $T_t$ is still block diagonal with one diagonal block for each equivalence class of $\Lambda_n$ under $\sim$ whose shape $s\,:\,\Phi\to\mathbb{Z}$ satisfies $s\!\left(\alpha^i\right)\ge n-t$ for some i. The corresponding block can be written as

(4·2) \begin{equation}\tilde K^{(s\!\left(\alpha^i\right))}\otimes\bigotimes_{f\in\Phi\setminus\{\alpha^i\}}K^{(s(f))}, \end{equation}

where $\tilde K^{(s\!\left(\alpha^i\right))}$ is the matrix $K^{(s\!\left(\alpha^i\right))}$ restricted to partitions $\lambda$ of $s\!\left(\alpha^i\right)$ satisfying

\begin{equation*}\lambda\unrhd \left(n-t,1^{s\!\left(\alpha^i\right)-(n-t)}\right).\end{equation*}

From (2·8) we find that, after a suitable ordering of rows and columns, all matrices occuring in the Kronecker product (4·2) are upper-triangular with ones on the diagonal. Again $T_t$ is a block-diagonal matrix whose blocks are Kronecker products of matrices of full rank and so $T_t$ itself has full rank.

From the proof of [ Reference Ellis, Friedgut and Pilpel7 , theorem 20] we know that $\tilde K^{(s\!\left(\alpha^i\right))}$ is independent of n. Moreover all other matrices occuring in the Kronecker product (4·2) are also independent of n. Hence $T_t$ itself is also independent of n. This proves (ii).

Next we shall show that the matrix $S_t$ has full rank. Recall that, for a composition $\lambda$ , we denote by $P_{\lambda}$ the parabolic subgroup of $G_{\lvert\lambda\rvert}$ given in (2·2). We start with the following lemma.

Lemma 4·5. Let m and n be positive integers satisfying $m<n$ and let $\phi$ and $\psi$ be class functions of $G_m$ and $G_n$ , respectively. Let $\pi_1\,:\,P_{(m,n)}\to G_m$ and $\pi_2\,:\,P_{(m,n)}\to G_n$ be the natural projections onto the corresponding diagonal blocks. Let $g\in P_{(m,n)}$ be such that $\pi_2(g)$ is regular elliptic. Then we have

\begin{equation*}(\phi\odot \psi)(g)=\phi(\pi_1(g))\,\psi(\pi_2(g)).\end{equation*}

Proof. From (2·4) we have

(4·3) \begin{equation}(\phi\odot \psi)(g)=\frac{1}{\lvert P_{(m,n)}\rvert}\sum_{\substack{x\in G_{m+n}\\xgx^{-1}\in P_{(m,n)}}}\phi\!\left(\pi_1\!\left(xgx^{-1}\right)\right)\psi\!\left(\pi_2\!\left(xgx^{-1}\right)\right). \end{equation}

Since $\pi_2(g)$ is regular elliptic and $m<n$ , we find from Lemma 4·1 that g stabilises a unique m-dimensional subspace U of $\mathbb{F}_q^{m+n}$ . Hence the number of $x\in G_{m+n}$ such that $xgx^{-1}\in P_{(m,n)}$ is the number of ordered bases $\{u_1,\dots,u_m,w_1,\dots,w_n\}$ of $\mathbb{F}_q^{m+n}$ such that $\{u_1,\dots,u_m\}$ spans U. This number equals $\lvert P_{(m,n)}\rvert$ . Since $xgx^{-1}\in P_{(m,n)}$ for each $x\in P_{(m,n)}$ , we conclude that

\begin{equation*}\left\{x\in G_{m+n}\,:\,xgx^{-1}\in P_{(m,n)}\right\}=P_{(m,n)}.\end{equation*}

Since $\pi_i(xgx^{-1})$ is conjugate to $\pi_i(g)$ for each $i\in\{1,2\}$ and each $x\in P_{(m,n)}$ , the statement of the lemma follows from (4·3).

We use Lemma 4·5 to prove the following lemma on the structure of the matrix S.

Lemma 4·6. Let $k,\ell$ be integers satisfying $0\leq k,\ell<n/2$ and let $\underline{\mu}\in\Pi_{k,i}$ and $\underline{\sigma}\in\Sigma_{\ell,j}$ . If $k>\ell$ , then we have $\xi^{\underline{\mu}}_{\underline{\sigma}}=0$ . For $k\leq \ell$ , let $\nu$ be the partition obtained from $\underline{\mu}(X-\alpha^i)$ by replacing the part $n-k$ by $\ell-k$ and define $\underline{\nu},\underline{\tau}\in\Lambda_\ell$ by

\begin{equation*}\underline{\nu}(f)=\begin{cases}\nu & \text{for $f=X-\alpha^i$}\\\underline{\mu}(f) & \text{otherwise}\end{cases}\quad \text{and}\quad \underline{\tau}(f)=\begin{cases}\varnothing & \text{for $f=h_{\ell,j}$}\\\underline{\sigma}(f) & \text{otherwise}.\end{cases}\end{equation*}

If $k\leq \ell$ , then we have $\xi^{\underline{\mu}}_{\underline{\sigma}}=\xi^{\underline{\nu}}_{\underline{\tau}}\,\omega^{ij}$ .

Proof. Let $g\in C_{\underline{\sigma}}$ . Define $\underline{\kappa}\in\Lambda_k$ by

\begin{equation*}\underline{\kappa}(f)=\begin{cases}\left(\underline{\mu}\!\left(\alpha^i\right)_2,\underline{\mu}\!\left(\alpha^i\right)_3,\dots\right) & \text{for $f=X-\alpha^i$}\\[4pt]\underline{\mu}(f) & \text{otherwise},\end{cases}\end{equation*}

so that by (2·6) and (2·11)

(4·4) \begin{equation}\xi^{\underline{\mu}}=\xi^{\underline{\kappa}}\odot\xi^{X-\alpha^i\mapsto (n-k)}. \\\end{equation}

For $\xi^{\underline{\mu}}(g)$ to be nonzero, g must be conjugate to an element of $P_{(k,n-k)}$ . Each such element fixes a k-dimensional subspace of $\mathbb{F}_q^n$ . If $k>\ell$ , then by Lemma 4·1, g fixes no k-dimensional subspace of $\mathbb{F}_q^n$ and hence $\xi^{\underline{\mu}}(g)=0$ .

Henceforth assume that $k\leq \ell$ . We shall frequently use $\xi^{f\mapsto (m)}=\chi^{f\mapsto (m)}$ , which follows from (2·7) and (2·8). Since $k\leq \ell$ we have

(4·5) \begin{equation}\xi^{\underline{\nu}}=\xi^{\underline{\kappa}}\odot\xi^{X-\alpha^i\mapsto (\ell-k)}. \end{equation}

Write

\begin{equation*}E=\bigcup_{\substack{\underline{\rho}\in\Lambda_{n-k}\\\underline{\rho}(h_{\ell,j})=(1)}} C_{\underline{\rho}}.\end{equation*}

We claim that

(4·6) \begin{equation}\xi^{X-\alpha^i\mapsto (n-k)}(e)=\left(\xi^{X-\alpha^i\mapsto (\ell-k)}\odot \xi^{X-\alpha^i\mapsto (n-\ell)}\right)(e)\quad \text{for each $e\in E$}. \end{equation}

Indeed, each $e\in E$ is conjugate to an element of $P_{(\ell-k,n-\ell)}$ with blocks $e_1\in G_{\ell-k}$ and $e_2\in G_{n-\ell}$ on the main diagonal, where $e_2$ is regular elliptic. Hence we find from Lemma 2·3 that, for each $e\in E$ , the left-hand side of (4·6) equals

\begin{align*}\theta\!\left(\det(e)^i\right)&=\theta\!\left(\det(e_1)^i\right)\cdot \theta\!\left(\det(e_2)^i\right)\\&=\xi^{X-\alpha^i\mapsto (\ell-k)}(e_1)\cdot \xi^{X-\alpha^i\mapsto (n-\ell)}(e_2),\end{align*}

which by Lemma 4·5 equals the right hand side of (4·6). From (4·4) we have

\begin{equation*}\xi^{\underline{\mu}}(g)=\frac{1}{\lvert P_{(k,n-k)}\rvert}\sum_{\substack{x\in G_n\\xgx^{-1}\in P_{(k,n-k)}}}\xi^{\underline{\kappa}}\!\left(\pi_1\!\left(xgx^{-1}\right)\right)\xi^{X-\alpha^i\mapsto (n-k)}\left(\pi_2\!\left(xgx^{-1}\right)\right),\end{equation*}

where $\pi_1\,:\,P_{(k,n-k)}\to G_{k}$ and $\pi_2\,:\,P_{(k,n-k)}\to G_{n-k}$ are the natural projections onto the diagonal blocks. Since $k,\ell<n/2$ , Lemma 4·1 implies that each $\pi_2(xgx^{-1})$ occuring in the summation is forced to lie inside E. Hence by subsequent applications of (4·4), (4·6), and (4·5) we then find that

\begin{align*}\xi^{\underline{\mu}}(g)&=\left(\xi^{\underline{\kappa}}\odot \xi^{X-\alpha^i\mapsto (n-k)}\right)(g)\\&=\left(\xi^{\underline{\kappa}}\odot\xi^{X-\alpha^i\mapsto (\ell-k)}\odot \xi^{X-\alpha^i\mapsto (n-\ell)}\right)(g)\\&=\left(\xi^{\underline{\nu}}\odot \xi^{X-\alpha^i\mapsto (n-\ell)}\right)(g).\end{align*}

Without loss of generality, we may assume that $g\in P_{(\ell,n-\ell)}$ and that the diagonal blocks of g are $g_1$ and $g_2$ , where $g_1\in C_{\underline{\tau}}$ and $g_2$ is the companion matrix of $h_{\ell,j}$ . Since $g_2$ is regular elliptic, we may apply Lemma 4·5 once more to obtain

\begin{equation*}\xi^{\underline{\mu}}(g)=\xi^{\underline{\nu}}(g_1)\,\xi^{X-\alpha^i\mapsto (n-\ell)}(g_2).\end{equation*}

Since $g_1\in C_{\underline{\tau}}$ , we have $\xi^{\underline{\nu}}(g_1)=\xi^{\underline{\nu}}_{\underline{\tau}}$ , and since $g_2$ is the companion matrix of $h_{\ell,j}$ , we find from Lemma 2·3 that

\begin{equation*}\xi^{X-\alpha^i\mapsto (n-\ell)}(g_2)=\theta\!\left(\det(g_2)^i\right)=\omega^{ij}.\end{equation*}

Hence we obtain $\xi^{\underline{\mu}}(g)=\xi^{\underline{\nu}}_{\underline{\tau}}\,\omega^{ij}$ , as required.

We can now prove the required property of the matrix $S_t$ .

Proof. To indicate dependence on n, write $S^{(n)}$ for the matrix S given in (4·1) and $S^{(n)}_t$ for the corresponding restricted matrix $S_t$ . Let $n>2t$ . From Lemma 4·6 we find that all entries in $S^{(n)}_t$ are independent of n, which proves the second statement of the lemma.

To show that $S^{(n)}_t$ is invertible, we view $S^{(n)}_t$ as a block matrix, where the blocks are indexed by $\Pi_k$ and $\Sigma_\ell$ for $k,\ell\in\{0,1,\dots,t\}$ . Let $B_{k,\ell}$ be the block corresponding to $\Pi_k$ and $\Sigma_\ell$ . Lemma 4·6 implies that $B_{k,\ell}$ is zero for $k>\ell$ and, for $0\leq k\leq t$ , the block $B_{kk}$ is the Kronecker product of $S^{(k)}$ and the Vandermonde matrix $(\omega^{ij})_{0\leq i, j\leq q-2}$ . Since the character table of irreducible characters of every finite group is invertible, Lemma 4·4 implies that $S^{(k)}$ is invertible and so $B_{kk}$ is invertible. Hence $S^{(n)}_t$ is block upper-triangular and all diagonal blocks are invertible. Therefore $S^{(n)}_t$ itself is invertible.

Finally, by combining Lemmas 4·4 and 4·7, we obtain a proof of Proposition 4·3.

5. Proof of Theorems 1·1 and 1·2

Now recall the definition (3·5) of the eigenvalues $P\!\left(\underline{\lambda},\underline{\sigma}\right)$ and the definition (3·7) of the prescribed extremal eigenvalue $\eta$ . As a first step in constructing the required weight function w occuring in (3·6), we prove the following result.

Proposition 5·1. Let n and t be positive integers satisfying $n>2t$ . Then there exists $w\in\mathbb{R}(\Omega_n\cap\Sigma_{\le t})$ such that $w(\underline{\sigma})=0$ for $\underline{\sigma}(1)=(1^t)$ and

(5·1) \begin{equation}\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}w(\underline{\sigma})P\!\left(\underline{\lambda},\underline{\sigma}\right)=\begin{cases}1 & \text{for $\underline{\lambda}\in\Omega_n\cap\Pi_{0,0}$}\\\eta & \text{for $\underline{\lambda}\in\Omega_n\cap\Pi_{k,0}$ and $1\leq k\leq t$}\\0 & \text{for $\underline{\lambda}\in\Omega_n\cap\Pi_{k,i}$ and $0\leq k\leq t$ and $1\leq i\leq q-2$}\end{cases} \end{equation}

and

(5·2) \begin{equation}\lvert w(\underline{\sigma})\rvert\le \frac{\gamma_t}{\lvert D_{\underline{\sigma}}\rvert}\quad \text{for all $\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}$} \end{equation}

for some constant $\gamma_t$ depending only on t.

Proof. From Proposition 4·2 we know that $Q_t$ has full rank. In view of (3·5) there exists a unique $w\in\mathbb{R}(\Omega_n\cap\Sigma_{\le t})$ satisfying (5·1).

We now show that $w(\underline{\sigma})=0$ for the $\lfloor q/2\rfloor+1$ elements $\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}$ satisfying $\underline{\sigma}(1)=(1^t)$ . Without loss of generality we may assume that $\Omega_n$ contains $X-\alpha^i$ and $h_{t,j}$ for all $i, j=0,1,\dots,\lfloor q/2\rfloor$ . Accordingly we define $\underline{\sigma}_j\in\Sigma_{t,j}$ by $\underline{\sigma}_j(1)=(1^t)$ for $j=0,1,\dots,\lfloor q/2\rfloor$ . Recall the definition of the character $\zeta^{(t,i)}$ from Section 2·4 and write $\zeta^{(t,i)}_{\underline{\sigma}}$ for this character evaluated on the conjugacy class $C_{\underline{\sigma}}$ . We evaluate the sum

(5·3) \begin{equation}S_i=\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}w(\underline{\sigma})\lvert D_{\underline{\sigma}}\rvert\,\left(\zeta^{(t,i)}_{\underline{\sigma}}+\zeta^{(t,-i)}_{\underline{\sigma}}\right) \end{equation}

in two ways. Since $\zeta^{(t,0)}$ is the permutation character on the set of t-tuples of linearly independent elements of $\mathbb{F}_q^n$ , we find by Lemma 2·4 that the summand in (5·3) is nonzero only when the elements of $C_{\underline{\sigma}}$ fix a t-tuple of linearly independent elements of $\mathbb{F}_q^n$ , hence only when $\underline{\sigma}=\underline{\sigma}_j$ for some j. By the definition of $\underline{\sigma}_j$ , each element in $C_{\underline{\sigma}_j}$ has determinant $\alpha^j$ . Hence by applying Lemma 2·4 twice we obtain

\begin{equation*}\zeta^{(t,i)}_{\underline{\sigma}_j}=\omega^{ij}\zeta^{(t,0)}_{\underline{\sigma}_j}=\omega^{ij}\zeta^{(t,0)}_{\underline{\sigma}_0}\end{equation*}

and therefore

(5·4) \begin{equation}S_i=2\zeta^{(t,0)}_{\underline{\sigma}_0}\sum_{j=0}^{\lfloor q/2\rfloor}w\!\left(\underline{\sigma}_j\right)\,\lvert D_{\underline{\sigma}_j}\rvert\,\cos\bigg(\frac{2\pi ij}{q-1}\bigg). \end{equation}

On the other hand, since $\zeta^{(t,i)}+\zeta^{(t,-i)}$ is a real-valued class function, we find from Lemma 3·2 that it is a linear combination of $\psi^{\underline{\lambda}}$ for $\underline{\lambda}\in\Omega_n$ . Hence by Lemma 2·5 there exists numbers $n_{i,\underline{\lambda}}$ such that

\begin{equation*}\zeta^{(t,i)}_{\underline{\sigma}}+\zeta^{(t,-i)}_{\underline{\sigma}}=\sum_{\substack{\underline{\lambda}\in\Omega_n\\\underline{\lambda}\!\left(\alpha^i\right)_1\ge n-t}}n_{i,\underline{\lambda}}\,\psi^{\underline{\lambda}}_{\underline{\sigma}}\end{equation*}

and hence

(5·5) \begin{equation}S_i=\sum_{\substack{\underline{\lambda}\in\Omega_n\\\underline{\lambda}\!\left(\alpha^i\right)_1\ge n-t}}n_{i,\underline{\lambda}}\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}w(\underline{\sigma})\lvert D_{\underline{\sigma}}\rvert\,\psi^{\underline{\lambda}}_{\underline{\sigma}}. \end{equation}

Since (5·1) holds, we conclude that $S_i=0$ for each i satisfying $1\leq i\leq \lfloor q/2\rfloor$ . Since $\zeta^{(t,0)}$ is a permutation character, it contains the trivial character with multiplicity 1 (this can be seen by Frobenius reciprocity, for example). Hence we have $n_{0,\underline{\lambda}}=2$ for $\underline{\lambda}\in\Omega_n$ satisfying $\underline{\lambda}(1)=(n)$ . We therefore find from (5·5) and (5·1) that

\begin{equation*}S_0=2+\eta\sum_{\substack{\underline{\lambda}\in\Omega_n\\n-t\leq \underline{\lambda}(1)_1<n}}n_{0,\underline{\lambda}}\,\psi^{\underline{\lambda}}(1)=2+2\eta\left(\zeta^{(t,0)}(1)-1\right).\end{equation*}

Since $\zeta^{(t,0)}(1)$ equals the number of t-tuples of linearly independent elements of $\mathbb{F}_q^n$ , we have

(5·6) \begin{equation}\zeta^{(t,0)}(1)=(q^n-1)(q^n-q)\cdots\left(q^n-q^{t-1}\right). \end{equation}

Therefore $S_0=0$ and so $S_i=0$ for each i satisfying $0\leq i\leq \lfloor q/2\rfloor$ . Since each element of $C_{\underline{\sigma}_0}$ fixes a t-tuple of linearly independent elements of $\mathbb{F}_q^n$ , we have $\zeta^{(t,0)}_{\underline{\sigma}_0}\ne 0$ . Thus (5·4) implies

\begin{equation*}\sum_{j=0}^{\lfloor q/2\rfloor}w\!\left(\underline{\sigma}_j\right)\,\lvert D_{\underline{\sigma}_j}\rvert\,\,\cos\bigg(\frac{2\pi ij}{q-1}\bigg)=0\quad \text{for each} \, i \, \text{satisfying $0\leq i\leq \lfloor q/2\rfloor$}\end{equation*}

and it is readily verified, using that $(\omega^{ij})_{0\leq i, j<q-1}$ is a Vandermonde matrix, that this in turn implies that $w\!\left(\underline{\sigma}_j\right)=0$ for all j satisfying $0\leq j\leq \lfloor q/2\rfloor$ , as required.

Now, for each $\underline{\lambda}\in\Omega_n$ satisfying $n-t\leq \underline{\lambda}(1)_1<n$ , we find from Lemma 2·5 that

\begin{equation*}\lvert\eta\rvert\, \psi^{\underline{\lambda}}(1)\le\lvert\eta\rvert\,\left(\zeta^{(t,0)}(1)-1\right)=1,\end{equation*}

using (5·6). Since $\psi^{\underline{\lambda}}(1)=\chi^{\underline{\lambda}}(1)=1$ for $\underline{\lambda}\in\Pi_{0,0}$ , we conclude from (5·1) that

\begin{equation*}\Bigg\lvert\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}w(\underline{\sigma})\lvert D_{\underline{\sigma}}\rvert\,\psi^{\underline{\lambda}}_{\underline{\sigma}}\Bigg\rvert\le 1 \quad \text{for each $\underline{\lambda}\in\Omega_n\cap\Pi_{\le t}$}.\end{equation*}

By Lemma 4·2 all entries of $Q_t$ (which are precisely the values of $\psi^{\underline{\lambda}}_{\underline{\sigma}}$ occuring in the sum) are independent of n and so are uniformly bounded by some value only depending on t. The same also holds for the inverse of $Q_t$ , which establishes (5·2).

In what follows we treat the remaining eigenvalues.

Lemma 5·2. Let n and t be positive integers satisfying $n>2t$ and let $w\in\mathbb{R}(\Omega_n\cap\Sigma_{\le t})$ be such that

\begin{equation*}\lvert w(\underline{\sigma})\rvert\le \frac{\gamma_t}{\lvert D_{\underline{\sigma}}\rvert}\quad \text{for all $\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}$}\end{equation*}

for some constant $\gamma_t$ depending only on t. Then

\begin{equation*}\Bigg\lvert\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}w(\underline{\sigma})P\!\left(\underline{\lambda},\underline{\sigma}\right)\Bigg\rvert \lt \lvert\eta\rvert\quad \text{for all $\underline{\lambda}\in\Omega_n\setminus\Pi_{\le t}$},\end{equation*}

provided that n is sufficiently large compared to t.

In the proof of the lemma we use the usual scalar product on class functions of $G_n$ , which is given by

(5·7) \begin{equation}\langle\chi,\psi\rangle=\frac{1}{\lvert G_n\rvert}\sum_{g\in G_n}\chi(g)\overline{\psi(g)}, \end{equation}

where $\chi,\psi$ are class functions of $G_n$ .

Proof of Lemma 5·2. By the definition (3·5) of $P\!\left(\underline{\lambda},\underline{\sigma}\right)$ and (3·3) we have

(5·8) \begin{equation}P\!\left(\underline{\lambda},\underline{\sigma}\right)=\frac{\lvert G_n\rvert}{\psi^{\underline{\lambda}}(1)}\,\big\langle\psi^{\underline{\lambda}},1_{D_{\underline{\sigma}}}\big\rangle. \end{equation}

Since $\chi^{\underline{\lambda}}$ is irreducible, we have $\langle\psi^{\underline{\lambda}},\psi^{\underline{\lambda}}\rangle=1$ or 2 and therefore we obtain, by an application of the Cauchy–Schwarz inequality,

\begin{equation*}\big\lvert\big\langle\psi^{\underline{\lambda}},1_{D_{\underline{\sigma}}}\big\rangle\big\rvert\le \sqrt{2\,\big\langle1_{D_{\underline{\sigma}}},1_{D_{\underline{\sigma}}}\big\rangle}=\sqrt\frac{2\lvert D_{\underline{\sigma}}\rvert}{\lvert G_n\rvert}.\end{equation*}

From (5·8) and our hypothesis on w we then find that

\begin{align*}\Bigg\lvert\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}w(\underline{\sigma})P\!\left(\underline{\lambda},\underline{\sigma}\right)\Bigg\rvert&\le \sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}\lvert w(\underline{\sigma})\rvert\,\lvert P\!\left(\underline{\lambda},\underline{\sigma}\right)\rvert\\[4pt]&\le\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}\frac{\gamma_t}{\lvert D_{\underline{\sigma}}\rvert}\,\frac{\lvert G_n\rvert}{\psi^{\underline{\lambda}}(1)}\sqrt{\frac{2\lvert D_{\underline{\sigma}}\rvert}{\lvert G_n\rvert}}\\[4pt]&\le\frac{\gamma_t\,\lvert\Sigma_{\le t}\rvert}{\psi^{\underline{\lambda}}(1)}\max_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}\sqrt{\frac{2\lvert G_n\rvert}{\lvert D_{\underline{\sigma}}\rvert}}\\[4pt]&\le\frac{\gamma_t\,\lvert\Sigma_{\le t}\rvert}{\chi^{\underline{\lambda}}(1)}\max_{\underline{\sigma}\in\Sigma_{\le t}}\sqrt{\frac{2\lvert G_n\rvert}{\lvert C_{\underline{\sigma}}\rvert}}.\end{align*}

Note that $\lvert\Sigma_{\le t}\rvert$ is independent of n. Using Lemmas 7·1 and 7·2, to be stated and proved in Section 7, we find that there is a constant $\gamma^{\prime}_t$ , depending only on t, such that

\begin{equation*}\Bigg\lvert\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}}w(\underline{\sigma})P\!\left(\underline{\lambda},\underline{\sigma}\right)\Bigg\rvert\le \frac{\gamma^{\prime}_t}{q^{n/2}}\,\frac{1}{q^{nt}}\end{equation*}

for all $\underline{\lambda}\in\Omega_n\setminus\Pi_{\le t}$ and all sufficiently large n. The right-hand side is certainly strictly smaller than $1/q^{nt}$ for all sufficiently large n and the proof is completed by noting that $\lvert\eta\rvert>1/q^{nt}$ .

Recall that $V_{\underline{\lambda}}$ is the column span of $E_{\underline{\lambda}}$ . Define

\begin{equation*}U_t=\sum_{\substack{\underline{\lambda}\in\Omega_n\\\underline{\lambda}(1)_1\ge n-t}} V_{\underline{\lambda}}.\end{equation*}

Now we obtain the following.

Proof. As explained at the end of Section 3, we apply Proposition 3·1 to the graph with adjacency matrix

\begin{equation*}\sum_{\substack{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}\\\underline{\sigma}(1)\ne (1^t)}}A_{\underline{\sigma}}\end{equation*}

and the $\lvert D_{\underline{\sigma}}\rvert$ -regular spanning subgraphs with adjacency matrix $A_{\underline{\sigma}}$ for those $\underline{\sigma}$ occuring in the above set union. Since none of the elements in $D_{\underline{\sigma}}$ for such $\underline{\sigma}$ fix a t-space pointwise, every t-intersecting set in $G_n$ is an independent set in this graph. Recall from Lemma 3·3 that every element of $V_{\underline{\lambda}}$ is an eigenvector of $A_{\underline{\sigma}}$ with eigenvalue $P\!\left(\underline{\lambda},\underline{\sigma}\right)$ . Let $w\in\mathbb{R}(\Omega_n\cap\Sigma_{\le t})$ be the vector given by Proposition 5·1 and write

\begin{equation*}P(\underline{\lambda})=\sum_{\substack{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t}\\\underline{\sigma}(1)\ne (1^t)}}w(\underline{\sigma})P\!\left(\underline{\lambda},\underline{\sigma}\right).\end{equation*}

Proposition 5·1 and Lemma 5·2 imply that, for all sufficiently large n, we have

\begin{equation*}P(\underline{\lambda})=\begin{cases}1 & \text{for $\underline{\lambda}(1)_1=n$}\\\eta & \text{for $n-t\leq \underline{\lambda}(1)_1<n$}\\\end{cases}\end{equation*}

and $\lvert P(\underline{\lambda})\rvert<\lvert\eta\rvert$ for $\underline{\lambda}(1)_1<n-t$ . Hence, writing $\underline{\lambda}_0$ for $X-1\mapsto (n)$ , we have $P(\underline{\lambda}_0)=1$ and

\begin{equation*}\eta=\min_{\underline{\lambda}\ne \underline{\lambda}_0}P(\underline{\lambda})\quad \text{and}\quad \lvert\eta\rvert=\max_{\underline{\lambda}\ne \underline{\lambda}_0}\,\lvert P(\underline{\lambda})\rvert.\end{equation*}

Then the required result follows from Proposition 3·1 and the decomposition of $\mathbb{R}(G_n)$ given in Lemma 3·3.

Our proof of Theorems 1·1 and 1·2 is completed by the following result.

Proof. Let $\mathcal{A}_t$ be the set of t-tuples of linearly independent elements of $\mathbb{F}_q^n$ . Define the incidence matrix $M_t\in\mathbb{C}( G_n,\mathcal{A}_t\times\mathcal{A}_t)$ of elements of $ G_n$ versus t-cosets by

\begin{equation*}M_t(x,(u,v))=\begin{cases}1 & \text{for $xu=v$}\\0 & \text{otherwise},\end{cases}\end{equation*}

so that the columns of $M_t$ are precisely the characteristic vectors of the t-cosets. Let $\zeta^t=\zeta^{(t,0)}$ be the permutation character of the set of t-tuples of linearly independent elements of $\mathbb{F}_q^n$ and define $C_t\in\mathbb{C}( G_n, G_n)$ by

\begin{equation*}C_t(x,y)=\zeta^t\left(x^{-1}y\right).\end{equation*}

Denoting by $\unicode{x1D7D9}_{xu=v}$ the indicator of the event that $x\in G_n$ maps u to v, we have

\begin{align*}\left(M_tM_t^T\right)(x,y)&=\sum_{u,v}M_t(x,(u,v))M_t(y,(u,v))\\&=\sum_{u,v}\unicode{x1D7D9}_{xu=v}\unicode{x1D7D9}_{yu=v}\\&=\sum_{u}\unicode{x1D7D9}_{xu=yu}\\&=\sum_{u}\unicode{x1D7D9}_{x^{-1}yu=u}\\&=\zeta^t\left(x^{-1}y\right)=C_t(x,y).\end{align*}

Hence we have $C_t=M_tM_t^T$ and so the column span of $C_t$ equals the column span of $M_t$ or equivalently the span of the characteristic vectors of the t-cosets.

From Lemma 2·5 we have

\begin{equation*}\zeta^t=\sum_{\substack{\underline{\lambda}\in\Lambda_n\\\underline{\lambda}(1)_1\ge n-t}} m_{\underline{\lambda}}\,\chi^{\underline{\lambda}}\end{equation*}

for some integers $m_{\underline{\lambda}}$ satisfying $m_{\underline{\lambda}}\ne 0$ for each $\underline{\lambda}$ occuring in the summation. Since $\zeta^t$ is real-valued, we find by Lemma 3·2 that $m_{\underline{\lambda}^*}=m_{\underline{\lambda}}$ and therefore have

(5·9) \begin{equation}\zeta^t=\sum_{\substack{\underline{\lambda}\in\Omega_n\\\underline{\lambda}(1)_1\ge n-t}} m_{\underline{\lambda}}\,\psi^{\underline{\lambda}}. \end{equation}

Lemma 2·2 implies that $\chi^{\underline{\lambda}}(1)=\chi^{\underline{\lambda}^*}(1)$ . We therefore obtain from (3·4) and (3·1) that

\begin{equation*}E_{\underline{\lambda}}(x,y)=\frac{\chi^{\underline{\lambda}}(1)}{\lvert G_n\rvert}\psi^{\underline{\lambda}}\left(x^{-1}y\right)\end{equation*}

and thus find from (5·9) that

(5·10) \begin{equation}C_t=\lvert G_n\rvert\sum_{\substack{\underline{\lambda}\in\Lambda_n\\\underline{\lambda}(1)_1\ge n-t}}\frac{m_{\underline{\lambda}}}{\chi^{\underline{\lambda}}(1)}\,E_{\underline{\lambda}}. \end{equation}

Hence the column span of $C_t$ is contained in $U_t$ . Conversely, let v be a column of $E_{\underline{\kappa}}$ for some $\underline{\kappa}\in\Omega_n$ satisfying $\underline{\kappa}(1)_1\ge n-t$ . Since $E_{\underline{\lambda}}$ is idempotent, we have $E_{\underline{\lambda}}v=v$ for $\underline{\kappa}=\underline{\lambda}$ and Lemma 3·3 implies $E_{\underline{\lambda}}v=0$ for $\underline{\kappa}\ne \underline{\lambda}$ . Hence from (5·10) we find that

\begin{equation*}C_tv=\lvert G_n\rvert\frac{m_{\underline{\kappa}}}{\chi^{\underline{\kappa}}(1)}\,v,\end{equation*}

and, since $m_{\underline{\kappa}}\ne 0$ , we conclude that v is in the column span of $C_t$ . This completes the proof.

6. Proof of Theorems 1·5 and 1·6

Our proofs of Theorems 1·5 and 1·6 follow along similar lines as those in the previous section and therefore our proofs will be less detailed.

Since the parabolic subgroup $P_{(t,n-t)}$ is the stabiliser of a t-space of $\mathbb{F}_q^n$ , the character $\xi^{X-1\mapsto (n-t,t)}$ is the permutation character of the set of t-spaces of $\mathbb{F}_q^n$ . From (2·7) we obtain its decomposition

(6·1) \begin{equation}\xi^{X-1\mapsto (n-t,t)}=\sum_{s=0}^t\chi^{X-1\mapsto (n-s,s)}. \end{equation}

Let $\genfrac{[}{]}{0pt}{}{{n}}{{k}}_q$ denote the q-binomial coefficient, which counts the number of k-spaces of $\mathbb{F}_q^n$ . Then we have

(6·2) \begin{equation}\xi^{X-1\mapsto (n-t,t)}(1)=\genfrac{[}{]}{0pt}{}{{n}}{{t}}_q, \end{equation}

and so (6·1) implies that

(6·3) \begin{equation}\chi^{X-1\mapsto (n-s,s)}(1)=\genfrac{[}{]}{0pt}{}{{n}}{{s}}_q-\genfrac{[}{]}{0pt}{}{{n}}{{s-1}}_q. \end{equation}

Also note that $\psi^{X-1\mapsto \lambda}=\chi^{X-1\mapsto \lambda}$ for all partitions $\lambda$ . Throughout this section, we define

\begin{align*}\varepsilon = -\frac{1}{\genfrac{[}{]}{0pt}{}{{n}}{{t}}_q-1},\end{align*}

which will be our prescribed extremal eigenvalue.

We begin with the following counterpart of Proposition 5·1.

Proposition 6·1. Let n and t be positive integers satisfying $n>2t$ . Then there exists $w\in\mathbb{R}(\Omega_n\cap\Sigma_{\le t-1})$ such that

(6·4) \begin{equation}\sum_{\underline{\sigma}\in\Omega_n \cap \Sigma_{\le t-1}} w(\underline{\sigma}) P\!\left(\underline{\lambda},\underline{\sigma}\right)=\begin{cases}1 & \text{for $\underline{\lambda}(1)=(n)$},\\\varepsilon & \text{for $\underline{\lambda}(1)=(n-s,s)$ with $1\leq s\leq t$},\\0 & \text{for $\underline{\lambda} \in \Omega_n \cap \Pi_{\le t-1}$, where}\\&\quad \, \underline{\lambda}(1)\ne (n-s,s) \text{with} \, 0 \, \leq s\leq t-1\end{cases} \end{equation}

and

(6·5) \begin{equation}\lvert w(\underline{\sigma})\rvert \le \frac{\gamma_t}{\lvert D_{\underline{\sigma}}\rvert} \quad \text{for all $\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t-1}$} \end{equation}

for some constant $\gamma_t$ depending only on t.

Proof. From Lemma 4·2 we know that $Q_{t-1}$ has full rank. In view of (3·5) there exists a unique $w\in\mathbb{R}(\Omega_n\cap\Sigma_{\le t-1})$ satisfying (6·4) except for $\underline{\lambda}$ of the form $\underline{\lambda}(1)=(n-t,t)$ .

Next we show that (6·4) also holds when $\underline{\lambda}(1)=(n-t,t)$ . By Lemma 4·6 we have

$\xi^{X-1\mapsto (n-t,t)}_{\underline{\sigma}}=0$ for each $\underline{\sigma}\in\Sigma_{\le t-1}$ . Hence we have

(6·6) \begin{align}0&=\sum_{\underline{\sigma}\in\Omega_n \cap \Sigma_{\le t-1}} w(\underline{\sigma}) \lvert D_{\underline{\sigma}}\rvert\xi^{X-1\mapsto (n-t,t)}_{\underline{\sigma}} \nonumber\\&=\sum_{s=0}^t\sum_{\underline{\sigma}\in\Omega_n \cap \Sigma_{\le t-1}} w(\underline{\sigma}) \lvert D_{\underline{\sigma}}\rvert\chi^{X-1\mapsto (n-s,s)}_{\underline{\sigma}}, \end{align}

using (6·1). Since (6·4) holds with the only exception $\underline{\lambda}(1)=(n-t,t)$ , the inner sum equals 1 for $s=0$ and $\varepsilon\,\chi^{X-1\mapsto (n-s,s)}(1)$ for each s satisfying $1\leq s\leq t-1$ . Assuming that this is true also for $s=t$ and using (6·3), the right-hand side of (6·6) is indeed

\begin{equation*}1+\varepsilon\,\sum_{s=1}^t\bigg(\genfrac{[}{]}{0pt}{}{{n}}{{s}}_q-\genfrac{[}{]}{0pt}{}{{n}}{{s-1}}_q\bigg)=1+\varepsilon\,\bigg(\genfrac{[}{]}{0pt}{}{{n}}{{t}}_q-1\bigg)=0.\end{equation*}

Hence (6·4) also holds when $\underline{\lambda}(1)=(n-t,t)$ .

It remains to prove (6·5). For each s satisfying $1\leq s\leq t$ , we find from (6·1) that

\begin{equation*}\lvert\varepsilon\rvert\, \chi^{X-1\mapsto(n-s,s)}(1)\le\lvert\varepsilon\rvert\left(\xi^{X-1\mapsto (n-t,t)}(1)-1\right)=1,\end{equation*}

using (6·2). Since $\chi^{X-1\mapsto (n)}(1)=1$ , we conclude from (6·4) that

\begin{equation*}\Bigg\lvert\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t-1}}w(\underline{\sigma})\lvert D_{\underline{\sigma}}\rvert\,\psi^{\underline{\lambda}}_{\underline{\sigma}}\Bigg\rvert\le 1 \quad \text{for each $\underline{\lambda}\in\Omega_n\cap\Pi_{\le t-1}$}.\end{equation*}

By Lemma 4·2 all entries of $Q_{t-1}$ are independent of n and so are uniformly bounded by some value only depending on t. The same also holds for the inverse of $Q_t$ , which establishes (6·5).

The bound (6·5) and Lemma 5·2 ensure that the right-hand side of (6·4) is small in modulus for each $\underline{\lambda}\in\Omega_n\setminus\Pi_t$ . It therefore remains to deal with the case that $\underline{\lambda}\in\Omega_n\cap\Pi_t$ except for $\underline{\lambda}\in\Omega_n$ given by $\underline{\lambda}(1)=(n-t,t)$ , which is the subject of the following lemma.

Lemma 6·2. Let $w\in\mathbb{R}(\Omega_n\cap\Sigma_{\le t-1})$ be given in Proposition 6·1 (so that $n>2t$ ). Then, for all $\underline{\lambda}\in \Omega_n\cap \Pi_t$ with $\underline{\lambda}(1)\ne (n-t,t)$ , we have

\begin{align*}\bigg\lvert\sum_{\underline{\sigma} \in \Omega_n \cap \Sigma_{\le t-1}} w(\underline{\sigma}) P\!\left(\underline{\lambda},\underline{\sigma}\right)\bigg\rvert<\lvert\varepsilon\rvert,\end{align*}

provided that n is sufficiently large compared to t.

Proof. By slight abuse of notation, we view w as an element of $\mathbb{R}(G_n)$ by setting $w(x)=0$ if $x\not\in\Omega_n\cap\Sigma_{\le t-1}$ and $w(x)=w(\underline{\sigma})$ if $x\in\Omega_n\cap\Sigma_{\le t-1}$ and $x\in D_{\underline{\sigma}}$ . Recalling the scalar product on class functions of $G_n$ from (5·7), the statement of the lemma is equivalent to

(6·7) \begin{equation}\frac{\lvert G_n\rvert}{\psi^{\underline{\lambda}}(1)}\;\big\lvert\big\langle w,\psi^{\underline{\lambda}}\big\rangle\big\rvert<\lvert\varepsilon\rvert \end{equation}

for all $\underline{\lambda}\in \Omega_n\cap \Pi_t$ with $\underline{\lambda}(1)\ne (n-t,t)$ , provided that n is sufficiently large compared to t.

Pick $\underline{\lambda}\in\Omega_n\cap \Pi_t$ such that $\underline{\lambda}(1)\ne(n-t,t)$ . Then $\underline{\lambda}\!\left(\alpha^i\right)_1=n-t$ for some i. First assume that $\lvert\underline{\lambda}(1)\rvert\ne n$ . Denoting by $\text{Re}\ x$ the real part of a complex number x, we find from Lemma 3·2 and (2·13) that

\begin{equation*}\frac{1}{2}\big\lvert\big\langle w,\psi^{\underline{\lambda}}\big\rangle\big\rvert\le \lvert\text{Re}\;\langle w,\chi^{\underline{\lambda}}\rangle\rvert=\bigg\lvert\sum_{\underline{\mu}\sim\underline{\lambda}}H_{\underline{\mu}\underline{\lambda}}\,\text{Re}\;\big\langle w,\xi^{\underline{\mu}}\big\rangle\bigg\rvert.\end{equation*}

Lemma 4·6 implies that $\xi^{\underline{\mu}}_{\underline{\sigma}}=0$ for each $\underline{\mu}\not\in\Pi_{\le t-1}$ and each $\underline{\sigma}\in\Sigma_{\le t-1}$ . For $\underline{\mu}\in\Lambda_n$ , we have

\begin{align*}\text{Re}\;\big\langle w,\xi^{\underline{\mu}}\big\rangle&=\sum_{\underline{\kappa}\sim\underline{\mu}}K_{\underline{\kappa}\underline{\mu}}\,\text{Re}\;\big\langle w,\chi^{\underline{\kappa}}\big\rangle.\end{align*}

By (2·8), the summation can be taken over all $\underline{\kappa}$ such that $\underline{\kappa}\!\left(\alpha^i\right)\unrhd\underline{\mu}\!\left(\alpha^i\right)$ . Hence if $\mu\in\Pi_{\le t-1}$ , then $\underline{\kappa}\in\Pi_{\le t-1}$ . By the assumed properties of w given in Proposition 6·1, we have $\langle w,\psi^{\underline{\kappa}}\rangle=0$ for each $\underline{\kappa}\in\Omega_n\cap\Pi_{\le t-1}$ satisfying $\lvert\underline{\kappa}(1)\rvert\ne n$ . Since $\lvert\underline{\lambda}(1)\rvert\ne n$ we conclude that $\langle w,\psi^{\underline{\lambda}}\rangle=0$ .

Now assume that $\lvert\underline{\lambda}(1)\rvert=n$ and write $\underline{\lambda}(1)=\lambda$ . From (2·9) and (2·10) we have

\begin{equation*}\big\langle w,\psi^{X-1\mapsto\lambda}\big\rangle=\sum_{\substack{\mu\unrhd\lambda\\\mu_1>n-t}}H_{\mu\lambda}\,\big\langle w,\xi^{X-1\mapsto\mu}\big\rangle,\end{equation*}

since by Lemma 4·6 in the case $\mu_1=n-t$ we have $\xi^{X-1\mapsto\mu}_{\underline{\sigma}}=0$ for each $\underline{\sigma}\in\Sigma_{\le t-1}$ . From (2·7) and (2·8) we then find that$

(6·8) \begin{align}\big\langle w,\psi^{X-1\mapsto\lambda}\big\rangle&=\sum_{\substack{\mu\unrhd\lambda\\\mu_1>n-t}}H_{\mu\lambda}\,\sum_{\kappa\unrhd\mu}K_{\kappa\mu}\,\big\langle w,\psi^{X-1\mapsto\kappa}\big\rangle\nonumber \\&=\frac{1}{\lvert G_n\rvert}\sum_{\substack{\mu\unrhd\lambda\\\mu_1>n-t}}H_{\mu\lambda}+\sum_{\substack{\mu\unrhd\lambda\\\mu_1>n-t}}H_{\mu\lambda}\sum_{(n)\rhd\kappa\unrhd\mu}K_{\kappa\mu}\,\big\langle w,\psi^{X-1\mapsto\kappa}\big\rangle, \end{align}

using that $\lvert G_n\rvert\,\langle w,\psi^{X-1\mapsto (n)}\rangle=1$ by the assumed properties of w given in Proposition 6·1 and $K_{(n)\mu}=1$ for each partition $\mu$ of n. We first show that the first sum is zero. We have

(6·9) \begin{equation}\sum_{\substack{\mu\unrhd\lambda\\\mu_1>n-t}}H_{\mu\lambda}=\sum_{\substack{\mu\unrhd\lambda}}K_{(n)\mu}H_{\mu\lambda}-\sum_{\substack{\mu\unrhd\lambda}}K_{(n-t,t)\mu}H_{\mu\lambda}, \end{equation}

using that $\lambda_1=n-t$ and that, for each partition $\mu$ of n, we have

\begin{equation*}K_{(n-t,t)\mu}=\begin{cases}1 & \text{for $\mu_1=n-t$}\\0 & \text{for $\mu_1>n-t$}.\end{cases}\end{equation*}

It is readily verified that

(6·10) \begin{equation}\sum_{\mu\unrhd\lambda} K_{\kappa\mu}H_{\mu\lambda}=\delta_{\kappa\lambda}. \end{equation}

Since $\lambda$ is neither (n) nor $(n-t,t)$ , we conclude that (6·9) equals zero. Hence (6·8) becomes

(6·11) \begin{equation}\big\langle w,\psi^{X-1\mapsto\lambda}\big\rangle=\sum_{\substack{\mu\unrhd\lambda\\\mu_1>n-t}}H_{\mu\lambda}\sum_{(n)\rhd\kappa\unrhd\mu}K_{\kappa\mu}\,\big\langle w,\psi^{X-1\mapsto\kappa}\big\rangle .\end{equation}

By the assumed properties of w given in Proposition 6·1, the inner summand is nonzero only when $\kappa=(n-s,s)$ for some s satisfying $1\leq s\leq t-1$ . In particular, for $\kappa$ of this form, Proposition 6·1 and (6·3) give

\begin{equation*}\lvert G_n\rvert\,\big\lvert\big\langle w,\psi^{X-1\mapsto \kappa}\big\rangle\big\rvert=\frac{\genfrac{[}{]}{0pt}{}{{n}}{{s}}_q-\genfrac{[}{]}{0pt}{}{{n}}{{s-1}}_q}{\genfrac{[}{]}{0pt}{}{{n}}{{t}}_q-1}\le \frac{\genfrac{[}{]}{0pt}{}{{n}}{{t-1}}_q}{\genfrac{[}{]}{0pt}{}{{n}}{{t}}_q}=\frac{q^t-1}{q^{n-t+1}-1}\le \frac{q^{2t-1}}{q^n}.\end{equation*}

By Lemma 4·4 the Kostka numbers $K_{\kappa\mu}$ occuring in (6·11) are independent of n and it is readily verified from (6·10) that the numbers $H_{\mu\lambda}$ occuring in (6·11) are also independent of n. Moreover the number of summands in (6·11) is also independent of n. From Lemma 7·2, to be stated and proved in Section 7, we have $\psi^{X-1\mapsto\lambda}(1)\ge \delta_{t-1}\,q^{nt}$ for some constant $\delta_{t-1}$ only depending on t. Hence there is a constant $c_t$ , depending only on t, such that

\begin{equation*}\frac{\lvert G_n\rvert}{\psi^{X-1\mapsto\lambda}(1)}\,\big\lvert\big\langle w,\psi^{X-1\mapsto\lambda}\big\rangle\big\rvert\le \frac{c_t}{q^{n(t+1)}}.\end{equation*}

Since $\lvert\varepsilon\rvert>1/q^{nt}$ , this shows that (6·7) holds provided that n is sufficiently large compared to t.

Recall that $V_{\underline{\lambda}}$ is the column span of $E_{\underline{\lambda}}$ . Define

\begin{equation*}W_t=\sum_{\substack{\underline{\lambda}\in\Omega_n\\\underline{\lambda}(1)\unrhd (n-t,t)}} V_{\underline{\lambda}}.\end{equation*}

Now we obtain the following.

Theorem 6·3. Let t be a positive integer. Then, for all sufficiently large n, the following holds:

(i) every t-space-intersecting set Y in $ G_n$ satisfies

\begin{equation*}\lvert Y\rvert\le \Bigg[\prod_{i=0}^{t-1}\left(q^t-q^i\right)\Bigg]\Bigg[\prod_{i=t}^{n-1}\left(q^n-q^i\right)\Bigg]\end{equation*}

and, in case of equality, we have $1_Y\in W_t$ ;

(ii) every pair of t-space-cross-intersecting sets Y, Z in $ G_n$ satisfies

\begin{equation*}\sqrt{\lvert Y\rvert\cdot\lvert Z\rvert}\le \Bigg[\prod_{i=0}^{t-1}\left(q^t-q^i\right)\Bigg]\Bigg[\prod_{i=t}^{n-1}\left(q^n-q^i\right)\Bigg]\end{equation*}

and, in case of equality, we have $1_Y,1_Z\in W_t$ .

Proof. We apply Proposition 3·1 to the graph with adjacency matrix

\begin{equation*}\sum_{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t-1}}A_{\underline{\sigma}}\end{equation*}

and the $\lvert D_{\underline{\sigma}}\rvert$ -regular spanning subgraphs with adjacency matrix $A_{\underline{\sigma}}$ for those $\underline{\sigma}$ occuring in the above set union. Every t-space-intersecting set in $G_n$ is an independent set in this graph. Let $w\in\mathbb{R}(\Omega_n\cap\Sigma_{\le t-1})$ be given by Proposition 6·1 and write

\begin{equation*}P(\underline{\lambda})=\sum_{\substack{\underline{\sigma}\in\Omega_n\cap\Sigma_{\le t-1}}}w(\underline{\sigma})P\!\left(\underline{\lambda},\underline{\sigma}\right).\end{equation*}

Proposition 6·1 and Lemmas 5·2 and 6·2 imply that, for all sufficiently large n, we have

\begin{equation*}P(\underline{\lambda})=\begin{cases}1 & \text{for $\underline{\lambda}(1)=(n)$}\\\varepsilon & \text{for $\underline{\lambda}(1)=(n-s,s)$ with $1\leq s\leq t$}\\\end{cases}\end{equation*}

and $\lvert P(\underline{\lambda})\rvert<\lvert\varepsilon\rvert$ for $\underline{\lambda}(1)\ne (n-s,s)$ with some s satisfying $0\leq s\leq t$ . Hence, writing $\underline{\lambda}_0$ for $X-1\mapsto (n)$ , we have $P(\underline{\lambda}_0)=1$ and

\begin{equation*}\varepsilon=\min_{\underline{\lambda}\ne \underline{\lambda}_0}P(\underline{\lambda})\quad \text{and}\quad \lvert\varepsilon\rvert=\max_{\underline{\lambda}\ne \underline{\lambda}_0}\,\lvert P(\underline{\lambda})\rvert.\end{equation*}

Then the required result follows from Proposition 3·1 and the decomposition of $\mathbb{R}(G_n)$ given in Lemma 3·3.

Our proof of Theorems 1·5 and 1·6 is completed by the following result.

Proof. The proof is almost identical to that of Theorem 5·4 with $\mathcal{A}_t$ replaced by the set of t-spaces and $\zeta^t$ replaced by the permutation character $\xi^{X-1\mapsto(n-t,t)}$ of t-spaces and the decomposition of $\zeta^t$ replaced by the decomposition given in (6·1).

7. Estimates on conjugacy class sizes and character degrees

In this section we provide bounds on the size of certain conjugacy classes and degrees of certain irreducible characters of $ G_n$ , which are used in the proof of Lemma 5·2.

Lemma 7·1. Let n and t be positive integers satisfying $n>2t$ and let $\underline{\sigma}\in\Sigma_{\le t}$ . Then we have

\begin{equation*}\frac{\lvert G_n\rvert}{\lvert C_{\underline{\sigma}}\rvert}\le q^{t^5} q^n.\end{equation*}

Proof. From Lemma 2·1 we find that (with the same notation as in Lemma 2·1)

(7·1) \begin{equation}\frac{\lvert G_n\rvert}{\lvert C_{\underline{\sigma}}\rvert}\le\prod_{f\in\Phi} \prod_{i=1}^{\lvert\underline{\sigma}(f)\rvert} q^{\lvert f\rvert\, s_i\left(\underline{\sigma}(f)^{\prime}\right)\,m_i(\underline{\sigma}(f))}. \end{equation}

Since $\underline{\sigma}\in\Sigma_{\le t}$ and $t<n/2$ , there is exactly one polynomial $h\in\Phi$ of degree at least $n-t$ in the support of $\underline{\sigma}$ . This polynomial must satisfy $\underline{\sigma}(h)=(1)$ and the corresponding factor in (7·1) is at most $q^n$ . There are at most t other polynomials in the support of $\underline{\sigma}$ . Each such polynomial f has degree at most t and satisfies $\lvert\underline{\sigma}(f)\rvert\le t$ and hence the corresponding factor in (7·1) has a crude upper bound of $q^{t^4}$ . As there are at most t such factors, the proof is completed.

Lemma 7·2. Let t be a positive integer. Then there is a constant $\delta_t$ such that, for all sufficiently large n and for all $\underline{\lambda}\in\Lambda_n\setminus \Pi_{\le t}$ , we have

\begin{equation*}\chi^{\underline{\lambda}}(1)\ge \delta_t\,q^{n(t+1)}.\end{equation*}

Proof. Let $\underline{\lambda}\in\Lambda_n\setminus \Pi_{\le t}$ . Using elementary calculus we find that

\begin{equation*}1-x\ge 4^{-x}\quad \text{for $0\leq x\leq 1/2$}\end{equation*}

and therefore

\begin{equation*}\frac{\prod_{i=1}^n(q^i-1)}{q^{\frac{1}{2}n(n+1)}}=\prod_{i=1}^n\bigg(1-\frac{1}{q^i}\bigg)\ge \prod_{i=1}^n\bigg(1-\frac{1}{2^i}\bigg)\ge \prod_{i=1}^n4^{-1/2^i}\ge\prod_{i=1}^\infty4^{-1/2^i}=\frac{1}{4}.\end{equation*}

Substitute into (2·14) of Lemma 2·2 to give

(7·2) \begin{equation}\frac{1}{\chi^{\underline{\lambda}}(1)}\le 4q^{N(\underline{\lambda})-M(\underline{\lambda})-\frac{1}{2}n(n+1)}, \end{equation}

where

\begin{align*}N(\underline{\lambda})&=\sum_{f\in\Phi}\lvert f\rvert\sum_{(i, j)\in\underline{\lambda}(f)}h_{i, j}(\underline{\lambda}(f)),\\M(\underline{\lambda})&=\sum_{f\in\Phi}\lvert f\rvert\,b(\underline{\lambda}(f))\end{align*}

and b and $h_{i, j}$ are as in Lemma 2·2. Note that for each partition $\lambda$ , we have

(7·3) \begin{equation}\sum_{(i, j)\in\lambda}h_{i, j}(\lambda)\le \sum_{k=1}^{\lvert\lambda\rvert}k=\frac{1}{2}\lvert\lambda\rvert(\lvert\lambda\rvert+1). \end{equation}

First assume that there exists a polynomial $h\in\Phi$ such that $\lvert h\rvert=1$ and $\underline{\lambda}(h)^{\prime}_1\ge n-t$ . In this case we have

\begin{equation*}M(\underline{\lambda})\ge b(\underline{\lambda}(h))\ge \sum_{k=1}^{n-t-1}k=\frac{1}{2}(n-t)(n-t-1)\end{equation*}

and by (7·3)

\begin{align*}N(\underline{\lambda})&\le \frac{1}{2}\sum_{f\in\Phi}\lvert f\rvert\lvert\underline{\lambda}(f)\rvert(\lvert\underline{\lambda}(f)\rvert+1)\\&\le \frac{n+1}{2}\sum_{f\in\Phi}\lvert f\rvert\lvert\underline{\lambda}(f)\rvert\\&=\frac{n(n+1)}{2}.\end{align*}

Therefore (7·2) implies that

\begin{equation*}\frac{1}{\chi^{\underline{\lambda}}(1)}\le 4q^{-\frac{1}{2}(n-t)(n-t-1)},\end{equation*}

so that we have $\chi^{\underline{\lambda}}(1)\ge q^{n(t+1)}$ for all sufficiently large n by very crude estimates.

Hence we can assume that $\underline{\lambda}(f)^{\prime}_1\leq n-t-1$ and $\underline{\lambda}(f)_1\leq n-t-1$ for all $f\in\Phi$ satisfying $\lvert f\rvert=1$ . Note that the second assumption is implied by the hypothesis $\underline{\lambda}\not\in\Pi_{\le t}$ . In what follows we use the trivial bound $M(\underline{\lambda})\ge 0$ . We distinguish two cases.

In the first case we assume that $\lvert\underline{\lambda}(f)\rvert\le n-t-1$ for all $f\in\Phi$ satisfying $\lvert f\rvert=1$ . Let $\ell$ be the maximum of $\lvert\underline{\lambda}(f)\rvert$ over all $f\in\Phi$ satisfying $\lvert f\rvert=1$ , hence $\ell\leq n-t-1$ . By (7·3) we have

\begin{align*}N(\underline{\lambda})&\le\frac{1}{2}\sum_{f\in\Phi}\lvert f\rvert\lvert\underline{\lambda}(f)\rvert(\lvert\underline{\lambda}(f)\rvert+1)\\&=\frac{n}{2}+\frac{1}{2}\sum_{f\in\Phi}\lvert f\rvert\lvert\underline{\lambda}(f)\rvert^2.\end{align*}

If $\ell\leq n/2$ , then we have $\lvert\underline{\lambda}(f)\rvert\le n/2$ for all $f\in\Phi$ and so $N(\underline{\lambda})\le n^2/4+n/2$ . From (7·2) we then find that $\chi^{\underline{\lambda}}(1)\ge q^{n(t+1)}$ for all sufficiently large n, again by very crude estimates. If $\ell>n/2$ , then

\begin{align*}N(\underline{\lambda})&\le \frac{1}{2}\left(n+\ell^2+(n-\ell)^2\right)\\&\le \frac{1}{2}\left(n+(n-t-1)^2+(t+1)^2\right)\\&=\frac{n^2+n}{2}-n(t+1)+(t+1)^2,\end{align*}

where we have used that $x^2+(n-x)^2$ is increasing for $x\ge n/2$ . Hence in this case we obtain $1/\chi^{\underline{\lambda}}(1)\le 4q^{-n(t+1)+(t+1)^2}$ by (7·2).

In the remaining case we assume that there exists $h\in\Phi$ such that $\lvert h\rvert=1$ and $\lvert\underline{\lambda}(h)\rvert\ge n-t$ . Recall that we also assume that $\underline{\lambda}(h)_1\leq n-t-1$ and $\underline{\lambda}(h)^{\prime}_1\leq n-t-1$ . Since $N(\underline{\lambda})$ depends only on the hook lengths of $\underline{\lambda}(f)$ for $f\in\Phi$ , we may replace $\underline{\lambda}(h)$ by its conjugate $\underline{\lambda}(h)^{\prime}$ . Assuming that n is sufficiently large, namely $n\ge (t+2)^2$ , we have $\underline{\lambda}(h)_1\ge t+2$ or $\underline{\lambda}(h)^{\prime}_1\ge t+2$ and we assume without loss of generality that $\underline{\lambda}(h)_1\ge t+2$ . Write $\underline{\lambda}(h)_1=n-r$ , so that our assumptions imply $t+1\leq r\leq n-t-2$ . Then, writing $s=\lvert\underline{\lambda}(h)\rvert$ , there exist nonnegative integers $c_j$ satisfying

\begin{equation*}\sum_{j=1}^{n-r}h_{1j}(\underline{\lambda}(h))=\sum_{j=1}^{n-r}(j+c_j),\quad \text{where}\quad \sum_{j=1}^{n-r}c_j=s-(n-r).\end{equation*}

Hence

\begin{equation*}\sum_{j=1}^{n-r}h_{1j}(\underline{\lambda}(h))=\binom{n-r+1}{2}+(s-n+r).\end{equation*}

Application of (7·3) with $\lambda=(\underline{\lambda}(h)_2,\underline{\lambda}(h)_3,\dots)$ gives

\begin{align*}\sum_{(i, j)\in\underline{\lambda}(h)}h_{i, j}(\underline{\lambda}(h))&\le \binom{s-n+r+1}{2}+\binom{n-r+1}{2}+(s-n+r)\\&=\frac{s^2}{2}+\frac{3s}{2}+n^2-sn-n+r(r-(2n-s-1))\\&\le \frac{s^2}{2}+\frac{3s}{2}+n^2-sn-n+(t+1)((t+1)-(2n-s-1)),\end{align*}

since the term depending on r is maximised for $r=t+1$ over the interval $[t+1,n-t-2]$ . This last expression equals

\begin{equation*}\frac{s}{2}+\frac{1}{2}s(s-2(n-t-2))+n^2-n+(t+1)((t+1)-(2n-1)).\end{equation*}

The second summand is increasing for $s\ge n-t$ and so is at most $\frac{1}{2}n(n-2(n-t-2))$ . Hence we obtain

\begin{equation*}\sum_{(i, j)\in\underline{\lambda}(h)}h_{i, j}(\underline{\lambda}(h))\le \frac{s}{2}+\frac{n^2}{2}-n(t+1)+(t+1)(t+2).\end{equation*}

Invoking (7·3) once more, we obtain

\begin{equation*}N(\underline{\lambda})\le \sum_{(i, j)\in\underline{\lambda}(h)}h_{ij}+\frac{1}{2}\sum_{\substack{f\in\Phi\\f\ne h}}\lvert f\rvert\lvert\underline{\lambda}(f)\rvert(\lvert\underline{\lambda}(f)\rvert+1).\end{equation*}

We have

\begin{equation*}\frac{s}{2}+\frac{1}{2}\sum_{\substack{f\in\Phi\\f\ne h}}\lvert f\rvert\lvert\underline{\lambda}(f)\rvert=\frac{1}{2}\sum_{f\in\Phi}\lvert f\rvert\lvert\underline{\lambda}(f)\rvert=\frac{n}{2}\end{equation*}

and

\begin{equation*}\frac{1}{2}\sum_{\substack{f\in\Phi\\f\ne h}}\lvert f\rvert\lvert\underline{\lambda}(f)\rvert^2\leq \frac{1}{2}\left(\sum_{\substack{f\in\Phi\\f\ne h}}\lvert f\rvert\lvert\underline{\lambda}(f)\rvert\right)^2\leq \frac{t^2}{2}.\end{equation*}

Collecting all terms, we find that

\begin{equation*}N(\underline{\lambda})\le \frac{n(n+1)}{2}-n(t+1)+(t+1)(t+2)+\frac{t^2}{2}.\end{equation*}

From (7·2) we then obtain

\begin{equation*}\frac{1}{\chi^{\underline{\lambda}}(1)}\le 4q^{-n(t+1)+(t+1)(t+2)+\frac{1}{2}t^2},\end{equation*}

which completes the proof.