I'm trying to read about Mumford curves. I've barely begun and I've already encountered a stumbling block. I'm sure this is probably a basic question that an expert could resolve quickly. I would very much appreciate help on this.

Let $k$ be a $p$-adic field (a finite extension of $\mathbb{Q}_p)$, and let $K$ be a complete and algebraically closed field containing $k$.

Let $\Gamma\le PGL_2(k)$ be a discontinuous subgroup. This means, that:

  1. For all $x\in\mathbb{P}^1(K)$, The closure of the orbit $\Gamma x$ is a compact subset of $\mathbb{P}^1(K)$.

  2. There exists a point $x\in\mathbb{P}^1(K)$ which is not a limit point of $\Gamma$. Ie, for any $y\in\mathbb{P}^1(K)$ and any nonrepeating sequence $\{\gamma_n\}_{n\ge 1}\subset\Gamma$, $\lim \gamma_n(y)\ne x$.

In particular, property (2) implies that any discontinuous subgroup is discrete as a subgroup of $PGL_2(k)$.

Now let $\Gamma$ be such a discontinuous subgroup, and suppose that $\infty$ is not a limit point of $\Gamma$. Then, for any infinite sequence of $\Gamma$, by (1), there exists a subsequence $\gamma_n = \begin{bmatrix}a_n & b_n\\c_n & d_n\end{bmatrix}$ such that the sequences of points (in $\mathbb{P}^1(K)$) $$\gamma_n(\infty) = \frac{a_n}{c_n},\qquad \gamma_n(0) = \frac{b_n}{d_n},\qquad\text{and}\quad-\gamma_n^{-1}(\infty) = \frac{d_n}{c_n}$$ are convergent. Thus, we may write $$\lim_{n\rightarrow\infty}\begin{bmatrix}\frac{a_n}{c_n} & \frac{b_n}{c_n}\\1 & \frac{d_n}{c_n}\end{bmatrix} = \begin{bmatrix}a & b\\1 & d\end{bmatrix}$$

On page 7 in "Schottky Groups and Mumford Curves" (Lecture notes in mathematics volume 817), they assert that

$$\text{"From the discreteness of $\Gamma$, it follows that $ad = b$. Furthermore, for $q\in\mathbb{P}^1(K)$},$$ $$\text{we find $\lim_{n\rightarrow\infty} \gamma_n(q) = a$ unless $q = -d$ and the sequence $\frac{d_n}{c_n}$ is constant."}$$

Can someone explain why the quoted text above follows from discreteness? Certainly $\begin{bmatrix}a & b\\1 & d\end{bmatrix}$ is a limit point of $\Gamma$, and hence it cannot be in $\Gamma$, but I don't see why that implies $ad = b$ (ie, the limit has determinant 0). For example, I don't see anything stopping the matrix from being an elliptic element of infinite order (no discrete subgroup can contain such an element).

The second statement in the quoted text is even more mysterious.

  • 2
    $\begingroup$ The first fact is indeed because by assumption you have an injective sequence in a discrete subgroup, and discrete subgroups are closed. If the matrix $M=[[a,b],[1,d]]$ were invertible, then the convergence of the sequence to the image of $M$ in $PGL_n$ would hold, contradicting that no injective sequence in a discrete subgroup can converge. $\endgroup$
    – YCor
    Commented Oct 19, 2016 at 3:37
  • $\begingroup$ @YCor Hmm, why must discrete subgroups be closed? This is certainly not true for $\mathbb{R}$ - for example $(\{1/n : n\ge 1\}\cup\mathbb{N})\subset\mathbb{R}$ seems to be a discrete multiplicative subgroup which is not closed in $\mathbb{R}$... $\endgroup$ Commented Oct 19, 2016 at 3:42
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    $\begingroup$ I'm saying that discrete subgroups of topological groups are closed (exercise). Here you're taking a discrete subset of a topological monoid ($\mathbf{R}$ with multiplication). $\endgroup$
    – YCor
    Commented Oct 19, 2016 at 3:44
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    $\begingroup$ $\gamma_n(q)=(a_nq+b_n)/(c_nq+d_n)$. After modding out both the numerator and the denominator by $c_n$, they converge to $a(q+d)$ and $q+d$. So if $q\neq -d$, this $\gamma_n(q)$ indeed converges to $a$. $\endgroup$
    – YCor
    Commented Oct 19, 2016 at 3:47
  • 1
    $\begingroup$ The numerator converges to $aq+b=aq+ad=a(q+d)$. $\endgroup$
    – YCor
    Commented Oct 19, 2016 at 3:59

1 Answer 1


There is a surjective homomorphism from $PGL(2,k)$ onto $k^*/(k^*)^2$ whose kernel is precisely the image (i.e. $SL(2,k)/\{\pm 1\}$) of $SL(2,k)$ in the group $PGL(2,k)$.. The group $k^*/(k^*)^2$ is a finite group since $k$ is a finite extension of $\mathbb{Q}_p$. Hence the $SL(2)$ image has finite index. By replacing the discrete subgroup $\Gamma $ by its intersection with the subgroup $SL(2,k)/\{\pm 1\}$, we may assume that all the matrices in $\Gamma $ have determinant one.

Since $a_n/c_n$ and $d_n/c_n$ converge, it follows that if the sequence $c_n$ were bounded, $a_n,d_n$ are also bounded, and hence $b_n/d_n$ is also bounded; this cannot happen, since that means that $\gamma _n$ converges. Hence $c_n$ tends to infinity in absolute value. But $ad-b$ is the limit of $\frac{a_nd_n-b_nc_n}{c_n^2}=\frac{1}{c_n^2}$ and the latter limit is zero.

Next, for a general point $q\in \mathbb{P}^1(K)$, we have $\gamma_n(q)=\frac{a_nq+b_n}{c_nq+d_n}$ and since $a_n/c_n$ and $d_n/c_n$ tends to $a,d$ respectively, we have $\gamma _n(q)$ tends to $\frac{aq+b}{q+d}$ (divide by $c_n$). Write $b=ad$ and you see that $\frac{aq+b}{q+d}=a$.

  • $\begingroup$ I was typing my answer and did not realize that YCor has answered this in the comments during that period. $\endgroup$ Commented Oct 19, 2016 at 4:06
  • $\begingroup$ No problem! I like your approach as well! Though, in the second paragraph did you mean to say that "since $b_n/c_n$ and $d_n/c_n$ converge..."? Also, I'm guessing YCor won't write an actual answer so I'll probably end up accepting your answer anyway. $\endgroup$ Commented Oct 19, 2016 at 4:12
  • $\begingroup$ Also, I'm a little confused about why $\gamma_n$ can't converge? I thought the whole point is that $\gamma_n$ does converge...? $\endgroup$ Commented Oct 19, 2016 at 4:13
  • $\begingroup$ the elements $\gamma _n$ are elements of the discrete group. If they converge, then the sequence would be constant (well, a subsequence would be constant) $\endgroup$ Commented Oct 19, 2016 at 4:16
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    $\begingroup$ if $c_n$ is bounded, all the numbers $a_n,b_n,c_n,d_n$ are bounded; that is ruled out, since that would mean ($SL(2,k)$ being a closed subset of the vector space $M_2(k)$), that every bounded sequence lies in a compact set in $M_2(k)$ (an easy version of the Heine Borel property) and hence has a convergent subsequence in $M_2(k)$ and hence in $SL(2,k)$. $\endgroup$ Commented Oct 19, 2016 at 4:42

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