this is my first question here, so I hope I am doing it right. :)
I'm currently reading a paper about the tree of GL_2 over a discretely valued field (similarly to Serre). Here's the setting:
$k$ an arbitrary field
$t$ local parameter (to a valuation $\nu$)
$\Theta_\infty := k[[t]]$ ($=$ the valuation ring to $\nu$)
$G := GL_2(k((t))), K := GL_2(\Theta_\infty), Z := $ the centre of $G$
Let $G/KZ$ be the vertex set of the tree. (the adjacency relation doesn't play a role for my question) Then two vertices $gKZ$ and $hKZ$ are equal iff $h^{-1}g \in KZ$. However, I have two given vertices of which ones I am sure that they are not equal (they are adjacent):
\begin{align*} o &:= \begin{pmatrix} t & t^{-1} \\ & 1 \end{pmatrix}KZ \\ v &:= \begin{pmatrix} t^2 & t^{-1} + lt \\ & 1 \end{pmatrix}KZ \text{ (for some $l \in k$).} \end{align*}
I calculate $$ o^{-1}v = \begin{pmatrix} t & l \\ & 1 \end{pmatrix} $$
But this matrix is in $GL_2(\Theta_\infty) = K$, isn't it? The $t$-valuations of the entries are: $\begin{pmatrix} = 1 & = 0 \\ = \infty & = 0 \end{pmatrix}$. It would follow that $oKZ = vKZ$ which can't be.
What am I doing wrong?
Okay, that took me too long. The point is that $o^{-1}\nu$ is not in $K$, because its inverse is not a matrix with entries in $\Theta_\infty$.