I have a question concerning the following paper: "The fundamental domain of the tree of GL(2) over the function field of an elliptic curve" by Shuzo Takahashi. (1993, Duke Math. J., Vol. 72, No. 1). Everyone familiar with the paper may skip to the part after the line.


**Notation**:
$k$ an arbitrary field (not necessarily finite)


$E:F(x,y) = 0$ an elliptic curve over $k$ with coordinate ring $k[E]$


$t := \frac{x}{y}$ local uniformizer at $\infty$, $k\_\infty := k((t))$


Embed $k[E] \hookrightarrow k\_\infty$ such that $\nu(x) = -2$, $\nu(y) = -3$ where $\nu$ is the order function of $k((t))$.


$\Theta\_\infty := k[[t]]$


$\Gamma := GL\_2(k[E])\\
K := GL\_2(\Theta\_\infty)\\
G := GL\_2(k\_\infty)\\
Z := Z(G) \text{ the center of } G$


We define a **tree** $\mathcal{T}$ with $vert(\mathcal{T}) := G/KZ$ and the following adjacency relation:


$gKZ \sim hKZ :\iff h^{-1}g = \begin{pmatrix} t & b \\ & 1 \end{pmatrix}$ (for $b \in k$) or $= \begin{pmatrix} t^{-1} & \\ & 1 \end{pmatrix}$ $\mod KZ$.


**Setting**
Consider the following vertices: 


$o := \begin{pmatrix} t & t^{-1} \\ & 1 \end{pmatrix}KZ\\
v(l) := \begin{pmatrix} t^2 & t^{-1} + lt \\ & 1\end{pmatrix}KZ \text{ for } l \in k$


All the $v(l)$ are neighbors of $o$. We are interested in the neighbors $w$ of $v(l)$ which are not $o$. These vertices are called the "successors of $v(l)$". ($o$ is called the "predecessor" of $v(l)$). We can reach them via the matrices $\begin{pmatrix} t & b \\ & 1 \end{pmatrix}$, i.e. $v(l)^{-1}w = \begin{pmatrix} t & b \\ & 1 \end{pmatrix} \mod KZ$. (the matrix $\begin{pmatrix} t^{-1} & \\ & 1 \end{pmatrix}$ lets us jump from any $v(l)$ back to $o$).




---


**The Problem / Question**


In the Proof of Theorem 4 of the paper (p. 93, first paragraph), Takahashi writes the following (I slightly rephrased it):


*"[...] it is easy to check that every successor of $v(l)$ is written as $\begin{pmatrix} t^3 & \frac{y-m}{x-l} \\ & 1 \end{pmatrix}$ for some $m \in k$."*


We can calculate the successors of $v(l)$ by using the adjacency relation:


$w$ is a successor of $v(l)$
$\iff$ $v(l)^{-1}w = \begin{pmatrix} t & b \\ & 1 \end{pmatrix} \mod KZ$
$\iff$ $w = v(l) \begin{pmatrix} t & b \\ & 1 \end{pmatrix} = \begin{pmatrix} t^3 & bt^2 + t^{-1} + lt \\ & 1 \end{pmatrix} \mod KZ$ (for some $b \in k$)


**Now I don't see any way of transforming this matrix to the one mentioned in the quote above.** I discussed this with a colleague and he suspects that the statement in the paper is in fact false.


**Bonus**
There is a proposition (Proposition 1 in the paper) which can be used to switch representatives of the cosets. It states:


*"For every $f\_1, f\_2, f\_1', f\_2'$, if $\nu(f\_1) = \nu(f\_1')$ and $\nu(f\_2 - f\_2') \geq \nu(f\_1)$, then $\begin{pmatrix} f\_1 & f\_2 \\ & 1 \end{pmatrix}KZ = \begin{pmatrix} f\_1' & f\_2' \\ & 1 \end{pmatrix}KZ$"*


However this proposition can not be applied here as $\nu(bt^2 + t^{-1} + lt - \frac{y-m}{x-l}) < 3 = \nu(t^3)$.




---


**On the answer of Matthias Wendt:**


Thanks again! I checked your approach for $l = 0$ and I think you are right: it works!


I tried it for $0 \not= l \in k$ and got the following. I denote by $T\_{\geq n}$ terms of order $\geq n$ and $a$ is defined by $x^{-1} = at^2 + T\_{\geq 3}$.


Expansion series for $\frac{y-m}{x-l}$:


$$
\frac{y-m}{x-l} = \frac{m-y}{l}\cdot\frac{1}{1-\frac{x}{l}} \\
= \frac{m-y}{l}\cdot \left(-\sum\limits\_{n=1}^\infty \frac{1}{(\frac{x}{l})^n}\right) \\
= \dots \text{ (use } y = \frac{x}{t} \text{)}\\
= (t^{-1} - m)\cdot \sum\_{n=0}^\infty l^n (\frac{1}{x})^n \quad\text{ (note index shift)} \\
= (t^{-1} -m) \cdot (1 + lat^2 + lT\_{\geq 3} + T\_{\geq 4})\\
= t^{-1} -m + lat -mlat^2 +T\_{\geq 3}
$$


We get:
$$
f = bt^2 + t^{-1} + \frac{y-m}{x-l}\\
= t^2(b + mla) - lat + m + T\_{\geq 3}
$$


and setting m := $-b(la)^{-1}$:
$f = -tla -b(la)^{-1} + T\_{\geq 3} \not= 0 + T\_{\geq 3}$. So it won't work for $l \not= 0$.


Did I do something wrong?



Switching representatives of right coset in a paper on fundamental domain of tree of GL(2)
Thinking about my comment again, maybe my remark on necessity of using left multiplication with $\Gamma$ was too hasty.


First, some general comment on the coset change in the specified situation.
Assume that there is a matrix $A\in KZ$ such that 
$$
\left(\begin{array}{cc}
t^3&bt^2+t^{-1}+lt\\0&1
\end{array}\right)=
\left(\begin{array}{cc}
t^3&\frac{y-m}{x-l}\\0&1
\end{array}\right)\cdot A
$$
Then, working out the matrix multiplication, it is clear that
$A=e\_{12}(f)$ with  

$$
ft^3=bt^2+t^{-1}+lt-\frac{y-m}{x-l}, f\in \Theta\_\infty.
$$
Rephrasing, the right coset representatives can be changed if and only if
$bt^2+t^{-1}+lt$ and $\frac{y-m}{x-l}$ agree up to
$t^3\Theta\_\infty$. The right cosets are the same if and only if
Takahashi's Proposition 1 can be applied to prove it.


There are situations where that is actually possible (I think): let's look at the simplest example, which is $l=0$. In
this case, we can start the Laurent series expansion at infinity:
$\frac{y-m}{x}=t^{-1}-\frac{m}{x}$. As 
$x^{-1}=at^2$ with $a\neq 0$ up to terms of order $\geq 3$, we choose
$m=-ba^{-1}$ and find that 
$$
\frac{y-m}{x}=t^{-1}+bt^2+ft^3
$$
for some $f\in\Theta\_\infty$. In this case, it is in fact possible to
change the right 
coset representative as required in Takahashi's paper. 


[Added later:] In the general case, we make the following Laurent series expansion (correcting some mistakes in the reformulated question): 
$$
\frac{y-m}{x-l}=\frac{m-y}{l}\cdot\frac{1}{1-\frac{x}{l}}=
\frac{m-y}{l}\left(-\sum\_{n=1}^\infty\frac{1}{\left(\frac{x}{l}\right)^n}\right)
=$$
$$
=\frac{y-m}{l}\left(\sum\_{n=1}^\infty\frac{l^n}{x^n}\right)=\left(t^{-1}-\frac{m}{x}\right)\sum\_{n=0}^\infty\frac{l^n}{x^n}.$$


Now we write $x^{-1}=a\_2t^2+a\_3t^3+T\_{\geq 4}$, and expand the above product:
$$
\left(t^{-1}-\frac{m}{x}\right)\sum\_{n=0}^\infty\frac{l^n}{x^n}=\left(t^{-1}-\frac{m}{x}\right)\left(1+l\left(a\_2t^2+a\_3t^3+T\_{\geq 4}\right)+R\_{\geq 4}\right)=\left(t^{-1}+la\_2t+la\_3t^2+S'\_{\geq 3}\right)-m\left(a\_2t^2+S''\_{\geq 3}\right)
$$
with all the $T\_{\geq 4}, R\_{\geq 4},S'\_{\geq 3},S''\_{\geq 3}$ power series of the respective valuation. 
We collect the terms of order $\leq 2$ to get the following equality modulo $t^3\Theta\_\infty$:
$$
\frac{y-m}{x-l}=t^{-1}+la\_2t+\left(la\_3-ma\_2\right)t^2.
$$
I guess it is not a problem to choose $x$ such that $a\_2=1$, and then the above allows to set $m=la\_3-b$. This should show that also in the case $l\neq 0$, it is always possible to change the coset representative as required in Takahashi's paper. 


Maybe the argument Takahashi had in mind was to see the case $l=0$ and then apply Proposition 6 of the paper to reduce the general case to the special case $l=0$.


As an aside, maybe a bonus: if you are not necessarily interested in finding an explicitly embedded fundamental domain, you may get the computation of $\Gamma\backslash \mathcal{T}$ via Atiyah's classification of vector bundles on the elliptic curves. Originally, Atiyah's classification is for $k$ algebraically closed, but this has been generalized by Pumplün to arbitrary base fields as in Takahashi's setting. Then you are looking at the classification of vector bundles on the curve $\overline{E}=E\cup\{\infty\}$ which are trivial on $E$. The central point $o$ corresponds to the unique stable bundle with determinant at infinity. The points $v(l)$ form the $\mathbb{P}^1$ of semistable bundles of rank $2$ and degree $0$. All other points are direct sums of suitable line bundles. The stabilizer of the vertex in the tree corresponds to the automorphism group of the vector bundle (all this is in Serre's book on trees). You can work out the action of the automorphism groups on the links, compute the quotients of links modulo the automorphism group actions and glue these local models together - you get exactly Takahashi's description of $\Gamma\backslash\mathcal{T}$. I find this way of argument a bit simpler than the explicit calculations in Takahashi's paper (this way I worked out the next case, the quotient of the two-dimensional building modulo $GL\_3(k[E])$).


