INTERSECTION THEOREMS FOR FINITE GENERAL LINEAR GROUPS

NEW

Problem: For a fixed prime power $q$, let $G_n=\mathrm{GL}_n(\mathbb F_q)$. Two matrices $x,y\in G_n$ are $t$-intersecting if

$$\dim\ker(x-y)\ge t$$

equivalently they agree on $t$ linearly independent vectors. Sets $Y,Z\subseteq G_n$ are $t$-cross-intersecting if every $(y,z)\in Y\times Z$ is $t$-intersecting. A $t$-coset is a set

$$\{g\in G_n: gu_i=v_i\ (1\le i\le t)\},$$

where $u_1,\dots,u_t$ and $v_1,\dots,v_t$ are linearly independent. Its size is

$$M=\prod_{i=t}^{n-1}(q^n-q^i).$$

The reconstructed conjecture is: if $n$ is sufficiently large compared with $t$, and $Y,Z\subseteq G_n$ are $t$-cross-intersecting with equality in Ernst–Schmidt Theorem 1.2,

$$\sqrt{|Y||Z|}=M,$$

then $Y=Z$, and $Y$ or $Y^T$ is a $t$-coset. Here $Y^T=\{y^T:y\in Y\}$. The phrase “sizes meet the bound” is interpreted as equality in Theorem 1.2.

Result: The conjecture is true.

Let

$$N=\prod_{i=0}^{t-1}(q^n-q^i),\qquad |G_n|=MN.$$

In the proof of Ernst–Schmidt Theorem 5.3, for all sufficiently large $n$, there is a real symmetric weighted adjacency operator $A$, supported only on non-$t$-intersecting pairs, such that with respect to the normalized inner product on $\mathbb R(G_n)$,

• $A\mathbf 1=\mathbf 1$;
• on the relevant nontrivial eigenspace the eigenvalue is

$$\eta=-\frac1{N-1};$$

• every other nontrivial eigenvalue has absolute value strictly less than $\rho=1/(N-1)$.

Let $f=1_Y$, $g=1_Z$, and write

$$f=a\mathbf 1+f_0,\qquad g=b\mathbf 1+g_0,$$

with $f_0,g_0\perp \mathbf 1$. Since $Y,Z$ are cross-intersecting and $A$ is supported on non-$t$-intersecting pairs,

$$0=\langle f,Ag\rangle=ab+\langle f_0,Ag_0\rangle .$$

Equality in the product bound gives

$$ab=\frac{|Y||Z|}{|G_n|^2}=\frac{M^2}{M^2N^2}=\frac1{N^2}.$$

By the spectral bound,

$$ab\le \rho\|f_0\|\|g_0\| =\frac1{N-1}\sqrt{a(1-a)b(1-b)}.$$

Substituting $ab=N^{-2}$ gives

$$(1-a)(1-b)\ge \left(1-\frac1N\right)^2.$$

But $ab=N^{-2}$, so AM–GM gives $a+b\ge 2/N$, hence

$$(1-a)(1-b)=1-a-b+\frac1{N^2}\le \left(1-\frac1N\right)^2.$$

Thus equality holds throughout, so $a=b=1/N$, i.e.

$$|Y|=|Z|=M.$$

Moreover equality in the spectral inequality forces $f_0,g_0$ to lie in the $\eta=-1/(N-1)$ eigenspace, and equality in Cauchy–Schwarz gives $f_0=g_0$. Hence $f=g$, so $Y=Z$.

Now $Y=Z$ and cross-intersection imply that $Y$ is $t$-intersecting. Since $|Y|=M$, the Ellis–Kindler–Lifshitz equality theorem for forbidden intersections in $\mathrm{GL}_n(\mathbb F_q)$ applies: for sufficiently large $n$, an extremal $t$-intersecting family in $G_n$ is either a $t$-coset or the transpose of one. Therefore $Y$ or $Y^T$ is a $t$-coset.

This proves Conjecture 1.4.

Citation: The spectral operator and eigenvalue facts are from Ernst–Schmidt, Intersection theorems for finite general linear groups, Math. Proc. Cambridge Philos. Soc. 175 (2023), 129–160, especially Theorem 5.3 and its proof. The extremal $t$-intersecting classification is from Ellis–Kindler–Lifshitz, Forbidden intersection problems for families of linear maps, Discrete Analysis 2023:19, arXiv:2208.04674.