Image of a congruence subgroup is not always congruence

Let $G$ be a reductive algebraic group over $\mathbb{Q}$. Choose an embedding $G \hookrightarrow \mathrm{GL}_n$, and define

\[\Gamma_G(N) := G(\mathbb{Q}) \cap \{g \in \mathrm{GL}_n(\mathbb{Z}):g \equiv I_n \,\mathrm{mod}\,N\}.\]

Congruence subgroup of $G(\mathbb{Q})$ is any subgroup containing some $\Gamma(N)$ as a subgroup of finite index. Although $\Gamma_G(N)$ depends on embedding $G \hookrightarrow \mathrm{GL}_n$, being a congruence subgroup is not. In this post, we will show that the image of congruence subgroup need not to be congruence. More precisely, we give a solution for the Exercise 4.3 of Milne’s Introduction to Shimura Varieties:

The image in $\mathrm{PGL}_2(\mathbb{Q})$ of a congruence subgroup in $\mathrm{SL}_2(\mathbb{Q})$ need not to be congruence. In fact, the image of $\Gamma_{\mathrm{SL}_2}(8)$ in $\mathrm{PGL}_2(\mathbb{Q})$ is not congruence.

Proof. First, we embed $\mathrm{PGL}_2$ into $\mathrm{GL}_3$ using adjoint representation. $\mathrm{GL}_2$ acts on its Lie algebra $\mathfrak{gl}_2$ via conjugation, and the action preserves the subspace $\mathfrak{sl}_2$ (the set of traceless matrices), which has dimension 3. Choose a basis of $\mathfrak{sl}_2$ as

\[X = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \quad Y = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}, \quad Z = \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix},\]

then $g = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$ acts as

\[\begin{align*} gXg^{-1} &= \frac{1}{D} (a^2 X + ac Y + c^2 Z) \\ gYg^{-1} &= \frac{1}{D} (2ab X + (ad + bc) Y + 2cd Z) \\ gZg^{-1} &= \frac{1}{D} (b^2 X + bd Y + d^2 Z) \end{align*}\]

where $D = \det(g) = ad - bc$. Hence we obtain a map $\mathrm{GL}_2 \rightarrow \mathrm{GL}_3$ as

\[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \frac{1}{ad - bc} \begin{pmatrix} a^2 & 2ab & b^2 \\ ac & ad + bc & bd \\ c^2 & 2cd & d^2 \end{pmatrix},\]

and this induces an embedding $\iota: \mathrm{PGL}_2 \hookrightarrow \mathrm{GL}_3$, so we can define $\Gamma_{\mathrm{PGL}_2}(M)$ using this map. From now on, we will always consider representatives with integer entries. In fact, for $\bar{g} = \overline{\left(\begin{smallmatrix}a&b \\ c&d \end{smallmatrix}\right)}$ in $\mathrm{PGL}_2(\mathbb{Q})$, there is a unique representative $g$ with coprime integer entries (up to multiplication by $-I_2$), and $\iota(g) \in \mathrm{GL}_3(\mathbb{Z})$ implies $\det(g) = \pm 1$.

Now, we will show that $\alpha(\Gamma_{\mathrm{SL}_2}(8))$ is not a congruence subgroup, where $\alpha: \mathrm{SL}_2 \to \mathrm{PGL}_2$ is the natural map. In other words, we will show that for any $M \geq 1$, $\Gamma_{\mathrm{PGL}_2}(M) \not\subseteq \alpha(\Gamma_{\mathrm{SL}_2}(8))$.

First, assume that $8\nmid M$. We can easily check that $\overline{\left(\begin{smallmatrix} 1 & M \\ 0 & 1\end{smallmatrix}\right)}$ is in $\Gamma_{\mathrm{PGL}_2}(M)$, but not in the image of $\Gamma_{\mathrm{SL}_2}(8)$.

When $M$ is a multiple of 8, we will use the following lemma.

Lemma. Let $M$ be a positive integer divisible by $8$. Then $x^2 \equiv 1\,(\mathrm{mod}\,M)$ has a solution other than $\pm 1$ and not $1$ modulo $8$.

Let’s assume lemma for a moment and let $x$ as above. Then

\[g = \begin{pmatrix} 2x - x^3 & - \frac{(x^2 - 1)^2}{M} \\ M & x\end{pmatrix}\]

satisfies $\iota(\bar{g}) \equiv I_3 \,(\mathrm{mod}\,M)$, so $\bar{g} \in \Gamma_{\mathrm{PGL}_2}(M)$ but $\bar{g}$ is not in the image of $\Gamma_{\mathrm{SL}_2}(8)$ by Lemma.¹

Here’s a proof of the Lemma. Write $M = 2^k \cdot M_0$ where $k \geq 3$ and $2 \nmid M_0$. It is known that $5$ is an order $2^{k-2}$ element in $(\mathbb{Z}/2^k \mathbb{Z})^\times$ (one can check this by applying binomial theorem to $5 = 1 + 4$). Now let $x_0 = \pm 5^{2^{k-3}}$ where the sign is choosen as $x_0 \not \equiv 1 \,(\mathrm{mod}\,8)$. By Chinese remainder theorem, there exists $x$ such that

\[\begin{cases} x\equiv x_0 \,(\mathrm{mod}\, 2^k) \\ x \equiv 1 \, (\mathrm{mod}\, M_0) \end{cases}\]

and such $x$ satisfies the condition of lemma. $\square$

Tags: math