Understanding Adelic Hilbert modular forms. I am reading about adelic Hilbert modular forms. Let us fix notation as in Shih.
I am having difficulty understanding the definition of Hecke operators. These are defined on the group of cusps $C_\mathfrak n$ of $K(\mathfrak n )$ as follows.
First, we say that for each prime $\mathfrak p_v$ we have \begin{equation} K_1(\mathfrak n)\begin{pmatrix} \varpi _v & 0 \\ 0 & 1 \end{pmatrix} K_1 (\mathfrak n) = \sqcup _i \gamma _i K_1(\mathfrak n ) =\sqcup _j K_1(\mathfrak n)\beta _j  \quad (1)\end{equation}  and then define $T(\mathfrak q ).c = \sum _i c\gamma _i$.
Where does $(1)$ come from? (this could probably be Bruhat decomposition but at this moment I do not know almost anything about it) Similar equation appears with $\beta _j$ replaced with $\gamma _i$. At another place there is similar equation
\begin{equation} K_1(\mathfrak np^r)\begin{pmatrix} \varpi _v & 0 \\ 0 & 1 \end{pmatrix} K_1 (\mathfrak np^r) = \sqcup _{u\in \mathcal O_v /\varpi _v} \begin{pmatrix} \varpi _v & u \\ 0 & 1 \end{pmatrix} K_1(\mathfrak np^r )  \quad (2)\end{equation}
Any help is appreciated and feel free to provide any reference. Please let me know if should provide more background or/and notation.
 A: The first decomposition is nothing as fancy as the Bruhat decomposition, it's just the observation that $K_1(\mathfrak n)\begin{pmatrix} \varpi _v & 0 \\ 0 & 1 \end{pmatrix} K_1 (\mathfrak n)$ carries a left and right action of $K_1(\mathfrak n)$ and may be decomposed into orbits for each of those actions. The second quality is more interesting, it amounts to giving explicit representatives for the orbits of the right action.
So let's prove this decomposition. The only place where something interesting happens is $v$, so it suffices to show that
$$K_{1,v}(\mathfrak{n}_vp)\begin{pmatrix} \varpi _v & 0 \\ 0 & 1 \end{pmatrix} K_{1,v} (\mathfrak{n}_vp)= \bigsqcup _{u\in \mathcal O_v /\varpi _v} \begin{pmatrix} \varpi _v & u \\ 0 & 1 \end{pmatrix} K_{1,v}(\mathfrak n_vp^r )$$ inside the ring $\mathrm{M}_{2\times 2}(\mathcal O_v)$. Let's just set $I:=n_vp^r$. Fix a system $U$ of representatives for $\mathcal O_v/\varpi_v$. The reduction homomorphism $\mathcal O_v \to \mathcal O_v/I$ induces a reduction homomorphism $M_{2\times 2}(\mathcal O_v) \to M_{2\times 2}(\mathcal O_v/I)$.
The inclusion$K_{1,v}(I)\begin{pmatrix} \varpi _v & 0 \\ 0 & 1 \end{pmatrix} K_{1,v} (I)\supseteq \bigcup _{u\in \mathcal O_v /\varpi _v} \begin{pmatrix} \varpi _v & u \\ 0 & 1 \end{pmatrix} K_{1,v}(I)$ follows by writing $\begin{pmatrix}\varpi & u \\ 0 & 1 \end{pmatrix}$ as $\begin{pmatrix}1 & u \\ 0 & 1 \end{pmatrix}\begin{pmatrix}\varpi & 0 \\ 0 & 1 \end{pmatrix}$.
Working mod $I$, we need to consider products of the form $\begin{pmatrix}a & b \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \varpi _v & 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix}a\varpi_v & b \\ 0 & 1 \end{pmatrix}$ und their orbit under right multiplication by matrices of the form $\begin{pmatrix}c & d \\ 0 & 1 \end{pmatrix}$.
We necessarily have that $a$ is a unit, so we can multiply from the right by $\begin{pmatrix}a^{-1} & 0 \\ 0 & 1 \end{pmatrix}$ to obtain $\begin{pmatrix}\varpi_v & b \\ 0 & 1 \end{pmatrix}$. Now let $u \in U$ such that $b\equiv u \pmod{\varpi_v}$ which means that we can write $u=b+c\varpi_v$ for some $c$. Now right multiply the matrix $\begin{pmatrix}\varpi_v & b \\ 0 & 1 \end{pmatrix}$ by $\begin{pmatrix}1 & c \\ 0 & 1 \end{pmatrix}$ to obtain $\begin{pmatrix}\varpi_v & u \\ 0 & 1 \end{pmatrix}$, as required.
To show uniqueness, suppose that for $u,u \in U$, we have $\begin{pmatrix}\varpi_v & u \\ 0 & 1 \end{pmatrix}K_{1,v}(I)=\begin{pmatrix}\varpi_v & u' \\ 0 & 1 \end{pmatrix}K_{1,v}(I)$ this means that $K_{1,v}$ contains
$\begin{pmatrix}\varpi_v & u \\ 0 & 1 \end{pmatrix}^{-1}\begin{pmatrix}\varpi_v & u \\ 0 & 1 \end{pmatrix}=\begin{pmatrix}1 & \frac{u'-u}{\varpi_v} \\ 0 & 1 \end{pmatrix}$
which is only the case if $u=u'$.
