Normalizers of the principal congruence subgroups in $\mathrm{GL}(n,\mathbf Q)$

Question

A question quite similar to this question. Let $n \geqslant 3$ and $m \geqslant 2$ be natural numbers and suppose that a matrix $A \in \mathrm{GL}(n,\mathbf Q)$ normalizes the principal congruence subgroup $\Gamma_n(m)$ of level $m$: $$ A\, \Gamma_n(m)\, A^{-1}=\Gamma_n(m). $$ What can be said about the matrix $A?$

A modification of Emerton's answer to the quoted question seems to imply (if I am not mistaken) that $A$ is of the form $1/d \cdot B,$ where $d$ is a natural number and all coefficients of a nonsingular matrix $B$ are of the form $p/m,$ where $p \in \mathbf Z.$ Any thoughts, and especially references, would be very much appreciated.

Answer 1

Proposition. For every $n\ge 0$ and $m\ge 1$ the normalizer of $\Gamma_n(m)$ in $\mathrm{GL}_n(\mathbf{Q})$ is $\mathbf{Q}^*\cdot\mathrm{GL}_n(\mathbf{Z})$.

Proof. One inclusion is clear so it is enough to prove the other one, by induction on $n$. The case $n\le 1$ is trivial. Suppose $n\ge 2$ and the result proved for smaller $n$. Since $\mathrm{GL}_n(\mathbf{Z})$ is transitive on the set of complete flags, we have $\mathrm{GL}_n(\mathbf{Q})=T_n(\mathbf{Q})\mathrm{GL}_n(\mathbf{Z})$. So it is enough to show that every $A\in T_n(\mathbf{Q})$ normalizing $\Gamma_n(m)$ has the required form.

We see that $Ae_{1n}(m)A^{-1}=e_{1n}(ma_{11}a_{nn}^{-1})$. So $a_{11}/a_{nn}$ is integral and since the same applies to $A^{-1}$ we deduce $a_{nn}=\pm a_{11}$.

The upper left $(n-1)\times (n-1)$ block of $A$ has the same property, so is in $\mathbf{Q}^*\cdot\mathrm{GL}_{n-1}(\mathbf{Z})$. Since its eigenvalues are rational, it follows that all its eigenvalues $a_{11},\dots,a_{n-1,n-1}$ are equal up to sign. Hence, after scalar multiplication, $A$ has only $\pm 1$ on the diagonal, and again by the size $n-1$ case applied to the upper-left and lower-right blocks, all its entries are integral, except maybe the $1,n$ entry. After right multiplication by an upper triangular matrix in $\mathrm{GL}_n(\mathbf{Z})$, we can then suppose $A=e_{1n}(s)$. Then we use that $Ae_{n1}(m)A^{-1}\in\Gamma_n(m)$: its $(1,1)$ entry is $1+ms$, so $s\in\mathbf{Z}$ and we are done.

Answer 2

Let me add a completely different proof (of a slightly stronger result). Unlike the previous one it is not self-contained linear algebra: it uses Zariski topology, being a lattice, Borel density etc.

Proposition: the normalizer $N$ of $\Gamma_n(m)$ in $\mathrm{GL}_n(\mathbf{R})$ is equal to $\mathbf{R}^*\cdot\mathrm{GL}_n(\mathbf{Z})$.

Let $F$ be the normalizer of $\Lambda=\Gamma_n(m)\cap\mathrm{SL}_n(\mathbf{Z})$ in $\mathrm{SL}_n(\mathbf{R}$. Let us first show that $F$ is reduced to $\mathrm{SL}_n(\mathbf{Z})$. Then since $\Lambda$ is discrete, $F^0$ centralizes it. But its centralizer, by an easy argument (e.g. using Borel density, or directly) is reduced to scalar matrices. So $F$ is discrete. Since $F$ contains the lattice $\Lambda$, it has to contain it with finite index. Hence $F$ preserves a finite index subgroup of $\mathbf{Z}^n$ (namely $\bigcap_{s\in F}s\mathbf{Z}^n$, which is a finite intersection since $s\mathbf{Z}^n$ only depends on the class of $s$ modulo $\Lambda$). Hence $F$ is conjugate to a subgroup of $\mathrm{SL}_n(\mathbf{Z})$, and in particular $\mathrm{covol}(F)\ge \mathrm{SL}_n(\mathbf{Z})$. But $\mathrm{SL}_n(\mathbf{Z})\subset F$, which implies $\mathrm{SL}_n(\mathbf{Z})=F$.

Now if $A\in\mathrm{GL}_n(\mathbf{R})$ normalizes $\Gamma_n(m)$, after multiplying by an integral reflection, $A$ has positive determinant and after multiplying by a scalar matrix, $A$ has determinant $1$, and then by the previous paragraph, $A$ is in $\mathrm{SL}_n(\mathbf{Z})$.