Let $T_{\tau}:=\{\begin{pmatrix}a&b\tau\\b&a\end{pmatrix}:a,b\in\mathbb{Q}_p,a^2-b^2\tau\neq0\}$, where $\tau\in\mathbb{Q}_p^{\star}$ with $\tau\notin(\mathbb{Q}_p^{\star})^2$.
Author claimed that $T_{\tau}/Z$ is compact, where $Z:=\{\begin{pmatrix}r&0\\0&r\end{pmatrix}:r\in\mathbb{Q}_p^{\star}\}$, i.e $Z$ is a center of $GL_2(\mathbb{Q}_p)$.
To prove this, author showed that any element $t$ in $T_{\tau}$ can be written in the form $t=\begin{pmatrix}p^n&0\\0&p^n\end{pmatrix}\cdot k$, $k\in T_{\tau}\cap GL_2(\mathbb{Z}_p)$ under the assumtion $\tau\in \mathbb{Z}_p^{\star}$ and $p>2$.
Then, author says that this completes the proof that $T_{\tau}/Z$ is compact.
I understand how $t$ can be written in the above form but why this implies that $T_{\tau}/Z$ is compact?
Also, author called $T_{\tau}$ as torus. Is there any reason to this?
Full arguments can be found in [Dorian goldfeld, automorphic representations and l-functions for the general linear group: volume 1, page 189]
Thank you.
To put this into a possibly enlightening context, notice that for any field $K$ and any $\tau \in K^* \setminus (K^*)^2$, we have $a^2-b^2\tau =0$ for $a,b \in K$ if and only if $a=b=0$, so the algebra of matrices $\pmatrix{a&b\tau \\b&a}$ identifies (via $\mapsto a +b\sqrt{\tau}$) with the field extension $L_\tau:=K(\sqrt{\tau})$, your $T_\tau$ with its multiplicative group $L_\tau^*$, and the determinant $a^2-b^2\tau$ of such a matrix with the norm map $N_\tau:L \rightarrow K$ which induces a homomorphism of multiplicative groups $L_\tau^* \rightarrow K^*$.
Calling $U_\tau$ the kernel of this homomorphism, i.e. the multiplicative group of norm-$1$-elements of $L_\tau$, we further always have a short exact sequence
$$ 1\rightarrow U_\tau \rightarrow L^*_\tau \rightarrow N_\tau(L_\tau^*) \rightarrow 1$$
Now in the case of $K$ being a local field, that norm-$1$-group $U_\tau$ is always compact. In fact, in the archimedean case $K=\mathbb R$, wlog $\tau =-1$, this sequence becomes
$$ 1\rightarrow e^{i\mathbb R} \rightarrow \mathbb C^* \rightarrow (\mathbb R_{>0}, \cdot) \rightarrow 1$$
which is well-known to split i.e. give the standard "polar coordinates" $z = r e^{i\phi}$ for complex numbers, with $U_\tau$ being the unit circle i.e. the original torus $\simeq S^1$.
In the non-archimedean case, it's well-known that the absolute value of $K$ extends essentially uniquely to $L$ via $| \cdot |_L := \sqrt{N(\cdot)}$. In the case you specialize to, namely $K=\mathbb Q_p$ with $p \neq 2$ and $|\tau|=1$, you are looking at the unramified quadratic extension and the exact sequence becomes
$$ 1\rightarrow \{x \in \mathcal{O}_L^\times:N(x)=1\} \rightarrow L^* \rightarrow p^{2\mathbb Z} \cdot \mathbb Z_p^* \rightarrow 1$$
which looks non-split, at least including the group on the right back into the middle certainly does not provide a splitting.
That being said, the construction you cite is kind of "orthogonal" to that one and looks at the exact sequence
$$ 1\rightarrow K^* \rightarrow L^*_\tau \rightarrow L_\tau^*/ K^* \rightarrow 1$$
The quotient on the right is what you are interested in. As orangeskid notes in their answer, the underlying topological space of that quotient is $P^1(K)$, the one-dimensional projective space over $K$ (this forgets a lot of structure, as it does not even need $L$ to be a field, just a two-dimensional vector space over $K$). Already in the case $K=\mathbb R$ this quotient is not actually the full unit circle $e^{i\mathbb R}$, but its quotient modulo $\{\pm 1\}$ (which however topologically is still the circle $S^1$).
In your case, the product decomposition from your source certainly shows (as per KCd's comment) that the quotient is compact. But in fact I was not joking when I said this sequence is orthogonal to the other one, as we always have the following commutative diagram with exact rows and columns: $\require{AMScd}$ \begin{CD} @. 1 @. 1 @. 1@. \\ @. @VVV @VVV @VVV \\ 1 @>>> \pm 1 @>>> K^* @>()^2>> (K^*)^2=N_\tau(K^*) @>>> 1\\ @. @VVV @VVV @VV a V \\ 1 @>>> U_\tau @>>> L_\tau^* @>N_\tau>> N_\tau(L_\tau^*)@>>> 1\\ @. @VVV @VVV @VVV \\ 1 @>>> U_\tau/\pm 1 @>b>> L_\tau^*/K^* @>>> N_\tau(L_\tau^*)/N_\tau(K^*)@>>> 1\\ @. @VVV @VVV @VVV \\ @. 1 @. 1 @.1@.\\ \end{CD}
The map $b$ is an isomorphism if and only if the includion $a$ is an equality, which is the case for $K=\mathbb R$, but not for other local fields. However, for $p$-adic fields with $p \neq 2$, the bottom right object is just of size $2$ (and even for $p=2$ it is finite), so the object we are interested in (the middle of the last row) is "up to finite obstacles" the torus $U_\tau$. (In particular, it is as compact as $U_\tau$ is).