This could be a long question - I apologise in advance!
I'm working over a real quadratic field $K$ (read indefinite quadratic form of signature $[1,1]$). Let $Q(x) = ax^2 + bxy + cy^2$ denote the quadratic form which has an associated matrix $A$, i.e. $$2 Q(x) = (x,y)^T A (x,y)$$
Then the group orthogonal to $Q(x)$ of determinant $1$ is given by $$SO(Q) = \left< \left( \begin{matrix} \cosh{x} & \sinh{x} \\ \sinh{x} & \cosh{x} \end{matrix} \right) , \left( \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right) \right> $$ for $x \in \mathbb{R}$. We'll take the connected component of this group as $$SO_0(Q) = \left< \left( \begin{matrix} \cosh{x} & \sinh{x} \\ \sinh{x} & \cosh{x} \end{matrix} \right) \right>$$
Define $Q_0(x) = x^2 - y^2$ and for $g_x \in SO_0(Q)$ we let $Q_x({v}) = Q_0(g_x^{-1} {v})$ for arbitrary $v$.
Define the theta function as \begin{equation*} \theta (\tau , x, L) = \sqrt{v} \sum_{{\lambda} \in L} e^{2 \pi i Q( {\lambda} ) u} e^{-2 \pi Q_x( {\lambda} ) v}. \end{equation*} where $L$ is a lattice that we sum over and $\tau = u + iv$.
I can show that \begin{equation*} \begin{split} \theta(\tau, \epsilon x, L) & = \theta(\tau, x, L), \end{split} \end{equation*} for $\epsilon \in SO_0(Q)$.
The automorphisms of a lattice $L$ are given by units of positive norm, $\mathcal{O}_K^{*,+} \subset SO_0(Q)$.
Now, my professor has said that we need to integrate away the symmetries of $\theta$. First question - why?. I understand that we basically get nothing from keeping the 'excess' info but there seems no need to integrate away.
Anyway, the story continues, and we looked at $$\begin{equation*} I(\tau, L) = \int\limits_{\mathcal{O}_K^{*,+} \backslash SO_0(Q)} \theta(\tau, x ,L) dx. \end{equation*}$$ again why specifically integrate over $\mathcal{O}_K^{*,+} \backslash SO_0(Q)$? Is it just to go over all known symmetries?
Next is the part I'm not convinced by:
$$\mathcal{O}_K^{*,+} \backslash SO_0(Q) \cong \log \epsilon \backslash \mathbb{R} \cong \mathbb{Z} \backslash \mathbb{R}$$ where $\epsilon$ is the fundamental unit of $K$. Is this correct, and if so, can anyone give the bijection(s)?
I have some further questions on this topic (namely I want to show that this is a Maass form of weight $0$ by computing the Fourier expansion) but this relies on the above to start! Perhaps I'll post a follow-up question in the future.