Let $\Gamma(N)$ denote the kernel of the reduction map $\text{SL}_2(\mathbb{Z})\rightarrow\text{SL}_2(\mathbb{Z}/N\mathbb{Z})$
Let $p$ be a prime that is $1$ mod $N$, and let $M$ be the set of matrices
$$\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$$ with $a,b,c,d\in\mathbb{Z}$, $ad-bc = p$, and congruent to the identity mod $N$.
$M$ has the property that $\Gamma(N)M\Gamma(N) = M$.
In Drinfeld's paper "Two Theorems on Modular Curves", he claims that assuming for every $m \in M$, there is an $h\in\Gamma(N)$ such that $m(\infty) = h(\infty)$, THEN for any $m\in M$ and $\gamma\in\text{SL}_2(\mathbb{Z})$, there is some $h\in\Gamma(N)$ such that $m\gamma(\infty) = \gamma h(\infty)$, where here we view all matrices as acting on the compactified upper half plane via fractional linear transformations.
This is equivalent to saying that there is some $h$ such that the left column of $m\gamma$ agrees with the left column of $\gamma h$.
Does anyone see why this is true?
thanks,
- will
Let us write $$ N = 3, p = 7. $$
$$ m = \left[\begin{array}{ll}4 & 3 \\ 3 & 4\end{array}\right].$$ $$ \gamma = \left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right].$$ $$ m \gamma = \left[\begin{array}{ll}7 & 3 \\ 7 & 4\end{array}\right].$$ It follows that $$ h = \left[\begin{array}{ll}7 & x \\ 0 & y\end{array}\right]$$ and it is not possible for $h$ to have determinant $1.$