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.



Discontinuous subgroups of $PGL_2(\mathbb{Q}_p)$
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 $\frac{aq+b}{q+d}=a$.


