\section{Studying $K$-spherical representations via Hecke algebras (Notes by Sabine Lang)}
For the time being, we can take $G$ any locally profinite group, and $K$ any compact open subgroup of $G$.

Recall that in finite group theory, one studies representations of $G$ by viewing them as $\cmplx[G]$-representations. We can identify $\cmplx[G]\iso\text{Maps}(G,\cmplx)$, which is an identification of $\cmplx$-algebras given by $$\sum_{g\in G}a_gg\to(f:G \to \cmplx, f(g)=a_g).$$ The multiplication on $\text{Maps}(G,\cmplx)$ is the convolution of functions, i.e., for $f_1,f_2:G \to \cmplx$, we have $(f_1*f_2)(g)=\sum_{x\in G}f_1(x)f_2(x^{-1}g)$. Now we will mimic this for locally profinite groups.

\begin{defn}
 The Hecke algebra $\mc{H}(G)$ of $G$ is the $\cmplx$-algebra of functions $f:G\to\cmplx$ that are locally constant with compact support. We write $\mc{H}(G)=\mc{C}_c^{\infty}(G)$. The multiplication is given by $$(f_1*f_2)(g)=\int_Gf_1(x)f_2(x^{-1}g)d\mu_G(x),\text{ for all } g\in G, f_1,f_2 \in \mc{C}_c^{\infty}(G),$$ where $\mu_G$ is a Haar measure on $G$ (more details on Haar measure will come later in the notes).
\end{defn}

\begin{rmk}
 For any $f\in \mc{C}_c^{\infty}(G)$, there exists a compact open subgroup $K < G$ such that $f$ is $K$-bi-invariant (i.e., $f$ factors through $K\backslash G/K$).
\end{rmk}
\proof
Because $f$ is locally constant, for every $x\in G$ there exists a neighborhood $U_x$ of $x$ such that $f\mid_{U_x}=f(x)$. Since $G$ is locally profinite, there exists a compact open subgroup $K_x$ such that $xK_x \subset U_x$ and $K_xx \subset U_x$ (we can take $\widetilde{K_x}$ with $\widetilde{K_x}x\subset U_x$, $\overline{K_x}$ with $x\overline{K_x}\subset U_x$ and we define $K_x=\widetilde{K_x}\cap\overline{K_x}$). Write $G$ as $G=\cup_{x \in G}xK_x=\cup_{x \in G} K_x x$. Since $f$ has compact support, we can extract finite subcovers of the support of $f$, so $\text{supp}(f)=\cup_{i=1,\dots,n}x_iK_{x_i}\cap \text{supp}(f)=\cup_{j=1,\dots,m}K_{y_j}y_j\cap \text{supp}(f)$. By choosing $K=\cap_{i=1,\dots,n} K_{x_i} \bigcap \cap_{j=1, \ldots, m} K_{y_j}$, we observe that $f$ is $K$-bi-invariant.
\endproof

From the remark, we can write $\mc{H}(G)=\cup_K\mc{H}(G,K)$ where we take the union over all the compact open subgroups $K$ of $G$. Here $\mc{H}(G,K)$ (also written $\mc{H}_K$) denotes the subalgebra (check this!) of $K$-bi-invariant functions in $\mc{H}(G)$.

\subsection{Overview of the translation from smooth $G$-representations to smooth $\mc{H}(G)$-modules}
We will prove the two following results (smooth $\mc{H}(G)$-modules will be defined later in the notes):

\begin{enumerate}
 \item There is an equivalence of categories $$\mc{M}(G) \to \{\text{smooth }\mc{H}(G)\text{-modules}\}.$$
 \item For all $K$, there is a bijection $$\quotient{\{\text{Irreducible smooth }\\G\text{-rep. }\pi \text{ with }\pi^K\neq 0\}}{\iso}\to\quotient{\{\text{Simple }\mc{H}_K\text{-modules}\}}{\iso}.$$
\end{enumerate}

Now we can make this more precise. First, we will do some preliminaries on integration. Recall that any locally compact (Hausdorff) topological group has a unique (up to scaling) left- (or right-) invariant Haar measure, i.e., a functional $I:\mc{C}_c(G)\to \reals$ such that 
\begin{enumerate}
 \item $I(f)\geq 0$ for $f\geq 0,$
 \item $I(\lambda_g \cdot f)=I(f)$ for all $g\in G$ and $f \in \mc{C}_c(G)$, where $\lambda_g$ denotes the left-regular representation of $G$ on $\mc{C}_c(G)$.
\end{enumerate}
We can rewrite the second condition as $\int_Gf(g^{-1}x)d\mu(x)=\int_Gf(x)d\mu(x).$ We also introduce the notation $\rho_g$ for the right-regular representation of $G$ on $\mc{C}_c(G)$.

But with $G$ locally profinite, we just use the following bare-hands approach : a left-invariant Haar measure will be such a functional $I$ defined on $\mc{C}_c^{\infty}(G)$ (and not the whole space $\mc{C}_c(G)$), and it is constructed as follows:

\begin{itemize}
 \item Fix $K<G$ a compact open subgroup.
 \item Declare $I(\mathbbm{1}_K)=1$ for $\mathbbm{1}_K$ the characteristic function of $K$ (normalization).
 \item For any $K'\subset K$, write $K$ as a finite disjoint union $K=\sqcup a_iK'$ and the left-invariance forces $I(\mathbbm{1}_{K'})=\dfrac{1}{[K:K']}$. Likewise, left-invariance forces $I(\mathbbm{1}_{gK'})=\dfrac{1}{[K:K']}$ for all $g \in G$ and for all such $K'$.
 \item Extend linearly.
\end{itemize}

This will force a unique construction of a left-invariant functional $I:\mc{C}_c^{\infty}(G)\to \reals$.

\subsection{Relation between left and right Haar measures}
Let $I$ be a left-invariant Haar measure, written as $I(f)=\int_Gf(g)d\mu_G(g)$. Then for any $g \in G$, we consider $I(\rho_g\cdot f)=\int_Gf(xg)d\mu_G(x)$. Since left and right regular representations commute, the map that sends $f$ to $I(\rho_g\cdot f)$ is also a left-invariant Haar measure. By uniqueness (up to scaling) of a left-invariant Haar measure on $G$, there exists $\delta_G(g)\in \reals_{>0}$ such that $\delta_G(g)I(\rho_g\cdot f)=I(f)$ for all $f$.  

We can check that $\delta_G:G\to\reals_{>0}$ is a smooth character : we apply it to $g_1\cdot g_2$ as follows :
$$I(f)=\delta_G(g_1\cdot g_2)I(\rho_{g_1\cdot g_2}f)=\delta_G(g_1\cdot g_2)I(\rho_{g_1}\rho_{g_2}f)$$
$$=\delta_G(g_1\cdot g_2)\delta_G(g_1)^{-1}I(\rho_{g_2}f)=\delta_G(g_1\cdot g_2)\delta_G(g_1)^{-1}\delta_G(g_2)^{-1}I(f),$$ so we obtain $\delta_G(g_1\cdot g_2)=\delta_G(g_1)\delta_G(g_2).$ To check that $\delta$ is smooth, fix any compact open subgroup $K$. For $k \in K$, 
\[
I(\mathbbm{1}_K)=I(\rho_k \mathbbm{1}_K)= \delta_G(k)^{-1} I(\mathbbm{1}_K),
\]
so $\delta_G$ is trivial on $K$ (note that if we take a subgroup $K$ of $G$ which is not compact open, then $\mathbbm{1}_K$ is not an element of $\mc{C}^{\infty}_c(G)$, so this argument does not apply). We call $\delta_G$ the modulus character of $G$.

%Since $\delta_G$ is continuous with image in $\reals_{>0}$ and the only compact subgroup of $\reals_{>0}$ is the trivial group, we see that $\delta_G$ is trivial on any compact open subgroup of $G$. Note that if we take a subgroup $K$ of $G$ which is not a compact open subgroup of $G$, then $\mathbbm{1}_K$ is not an element of $\mc{C}_c(G)$.

\begin{exmpl}
\begin{itemize}
 \item  As we have just seen, if $G$ is compact, then $\delta_G$ is trivial.
  \item If $G=G(F)$ is reductive, then $\delta_G$ is trivial.
\item A typical example with $\delta_G$ non-trivial arises when we take $G$ to be a Borel subgroup.
\end{itemize}
\end{exmpl}

Moreover, $\delta_G$ measures the difference between the left- and right-invariant Haar measures. Define $I'(f)=I(\delta_G^{-1}\cdot f)$. Then $I'$ is a right-invariant Haar measure : 

$$I'(\rho_g \cdot f)=I(\delta_G^{-1}\cdot \rho_g\cdot f)=I(\delta_G(g)\cdot \rho_g(\delta_G^{-1}\cdot f))=\delta_G(g)I(\rho_g(\delta_G^{-1}\cdot f))=I(\delta_G^{-1}\cdot f)=I'(f).$$

Now we return to the setting of interest with $F$ a non-archimedean local field, $\mc{O}$ its ring of integers, $G=G(F)$ and $K=G(\mc{O})$. One of our aims is to classify the $K$-spherical representations of $G$.

How? One way, as in Theorem \ref{ups}, is to understand how to find all $K$-spherical representations inside Ind$_B^G(\chi)$ for $\chi:T(F)/T(\mc{O})\to\cmplx^{\times}$. We will take a less explicit approach to the classification, but one that arises directly from the Satake isomorphism. We first point out one more thing about unramified principal series that was omitted last time.

\begin{rmk}
 (A subtlety arising from the failure of unimodularity of $B$.) We saw that Ind$_B^G(\chi)^K$ is $1$-dimensional in this case but Ind$_B^G(\chi)$ is not necessarily irreducible. For example, for $G=\gl{2}{F}$, we have the short exact sequence 
$$0 \to \mathbbm{1}\to \text{ Ind}_B^G(\mathbbm{1})\to\text{ St }\to 0,$$ which does not split, and when we dualize it we get 
$$0 \to \text{ St } \to \langdual{\text{ Ind}_B^G(\mathbbm{1})}\to\mathbbm{1}\to 0,$$
where St is the Steinberg representation. Here we do NOT have $\langdual{\text{ Ind}_B^G(\mathbbm{1})}\iso \text{ Ind}_B^G(\mathbbm{1})$. We have to be careful : in general, $\langdual{\text{ Ind}_B^G(\chi)}\iso \text{ Ind}_B^G(\chi^{-1}\cdot \delta_B^{-1})$.

Pursuing this, we get that Ind$_B^G(\chi)$ is irreducible either when $\chi_1=\chi_2$ or when $\chi_1=\chi_2\cdot \mid\hspace*{2mm}\mid^2,$ where $\chi=(\chi_1,\chi_2)$. We have a $G$-equivariant pairing
$$\text{Ind}_B^G(\chi)\times\text{ Ind}_B^G(\chi^{-1}\delta_B^{-1})\to \cmplx$$ given by $$(f_1,f_2)\mapsto \int_{B\backslash G}f_1(g)f_2(g)d\mu_{B\backslash G}(g),$$
which implies that $\langdual{\text{ Ind}_B^G(\chi)}\iso \text{ Ind}_B^G(\chi^{-1}\cdot \delta_B^{-1})$. The subtelty here is that we can only define a right-invariant measure $\mu_{B\backslash G}$ on the set $\{f:G \to \cmplx \mid f(bg)=\delta_B(b)^{-1}f(g), f$  locally constant with compact support$\}.$
\end{rmk}

\begin{rmk}
 Here $\langdual{(\hspace{2mm})}$ means the smooth dual, i.e., the smooth vectors in the algebraic dual.
\end{rmk}

Now, we need to learn how to integrate over $B$ for $B$ a Borel subgroup. (Note that the following discussion works equally well for a parabolic $P=M \ltimes N$ instead of $B$.)

We write $B$ as the semidirect product $B=T\ltimes N$, which is isomorphic to $T\times N$ as a topological space. But $T$ and $N$ are both unimodular (i.e., $\delta_T$ and $\delta_N$ are trivial). Fix both left- and right-invariant Haar measures $I_N$ and $I_T$ on $\mc{C}_c^{\infty}(N)$ and $\mc{C}_c^{\infty}(T)$. We use them to define a left-invariant Haar measure in $B$ by 
$I_B(f)=$``$\int_Bf(b)db=\int_T(\int_Nf(tn)dn)dt$''. The following commutative diagram makes sense of this notation :

\begin{center}
\begin{tikzcd}
\mc{C}_c^{\infty}(T)\otimes \mc{C}_c^{\infty}(N) \ar[r,"I_N"] \arrow[d, "\iso"]
& \mc{C}_c^{\infty}(T) \arrow[d,"I_T"] \\
\mc{C}_c^{\infty}(B) \arrow[r,"I_B"]
& \cmplx \\
\end{tikzcd} 
\end{center}

and the isomorphism $$\mc{C}_c^{\infty}(T)\otimes \mc{C}_c^{\infty}(N) \to \mc{C}_c^{\infty}(B)$$ is given by 
$$\phi \otimes \psi \mapsto (b=tn\mapsto \phi(t)\psi(n)).$$

We observe that $I_B$ is left-invariant. It suffices to show that $I_B(\lambda_{(t_0,n_0)^{-1}}\phi\otimes \psi )=I_B(\phi \otimes \psi),$ for all $\phi \in \mc{C}_c^{\infty}(T)$ and $\psi \in \mc{C}_c^{\infty}(N).$ Analyzing $\lambda_{(t_0,n_0)^{-1}}\phi\otimes \psi$, we get
$$(\lambda_{(t_0,n_0)^{-1}}\phi\otimes \psi)(t,n)=(\phi \otimes \psi)((t_0,n_0)\cdot (t,n))=(\phi \otimes \psi)(t_0t,t^{-1}n_0tn)=\phi(t_0t)\psi( t^{-1}n_0tn).$$ Using the fact that $t^{-1}n_0t$ is an element of $N$, and the invariance of both $I_N$ and $I_T$, we obtain $I_B(\lambda_{(t_0,n_0)^{-1}}\phi\otimes \psi )=I_B(\phi \otimes \psi).$

We want to use this definition of the left-invariant measure $I_B$ to calculate the corresponding modulus character $\delta_B$. Recall that it is defined by $\delta_B(b)I_B(\rho_b\cdot f)=I_B(f).$ Fix now $b_0=t_0n_0.$ We compute
$$I_B(\rho_{b_0}\cdot f)=\int_T\int_Nf(tnt_0n_0)dndt=\int_T\int_Nf(tt_0t_0^{-1}nt_0n_0)dndt,$$ and writing $f=\phi\otimes \psi$ we get
$$I_B(\rho_{b_0}\cdot f)=\int_T\phi(tt_0)dt\int_N\psi(t_0^{-1}nt_0n_0)dn=I_T(\phi)\int_N\psi(t_0^{-1}nt_0)dn,$$ using the fact that both $I_T$ and $I_N$ are unimodular. Note that we cannot use the invariance of $I_N$ to simplify $\int_N\psi(t_0^{-1}nt_0)dn$ since we are multiplying by elements of $T$, and not elements of $N$. 

\begin{rmk}
 For a fixed $t_0\in T$, the functional sending $\psi$ to $\int_N\psi(t_0^{-1}nt_0)dn$ is a Haar measure on $N$. By unicity (up to scaling) of a Haar measure, we obtain $\int_N\psi(t_0^{-1}nt_0)dn=cst(t_0)I_N(\psi).$ Here $cst(t_0)$ does not depend on $\psi$, so we can choose any $\psi$ we like to compute this quantity. 
\end{rmk}

We observe that $I_B(\rho_{b_0}\cdot f)=I_T(\phi)cst(t_0)I_N(\psi)=cst(t_0)I_B(f)$, so $\delta_B(b_0)=cst(t_0)^{-1}$.

\begin{exmpl}
 Take $G=\gl{2}{F}$, $t=\bp t_1 & \\ & t_2 \ep$ and $\psi=\mathbbm{1}_{N(\mc{O})}$. In this case, $N(\mc{O})=\{\bp 1 & x \\ 0 & 1\ep\mid x\in \mc{O}\}$. We may assume the normalization $I_N(\mathbbm{1}_{N(\mc{O})})=1.$ We want to compute $\int_{G(F)}\mathbbm{1}_{N(\mc{O})}(\bp1 & x\frac{t_2}{t_1} \\ 0 & 1\ep)dx=\int_{F}\mathbbm{1}_{\frac{t_1}{t_2}\mc{O}}(x)dx.$ So the question is : given that $\mc{O}$ has volume $1$, what volume does $\frac{t_1}{t_2}\mc{O}$ have?

Answer : $|\frac{t_1}{t_2}|$, where $|\hspace{2mm}|$ is normalized as $(\#\frac{\mc{O}}{\pi\mc{O}})^{-\nu(\frac{t_1}{t_2})}$ with $\nu : \mc{O} \twoheadrightarrow \ints.$ So in this case, $\delta_B(\bp t_1 & * \\ 0 & t_2 \ep)=|\frac{t_1}{t_2}|^{-1}=|\frac{t_2}{t_1}|.$
\end{exmpl}

More generally, for a group $G$, we obtain $\delta_B(t)=|\det(\text{Ad}(t)\mid_{\mf{n}})|^{-1}.$

\subsection{Back to Hecke algebras}
For now, $G$ can be any locally profinite group. Recall : $$\mc{H}(G)=\mc{C}_c^{\infty}(G)=\cup_{K}\mc{H}(G,K),$$ where the union is taken over the compact open subgroups $K$ of $G$, and $\mc{H}(G,K)=\mc{H}_K$ is the subspace of $K$-bi-invariant functions. We observe that $\mc{H}_K$ is a subalgebra of $\mc{H}(G)$, and $\mc{H}_K=e_K*\mc{H}(G)*e_K$ where $e_K=\frac{\mathbbm{1}_K}{\text{vol}(K)}$. We can check that the $e_K$ are idempotents : for $g\in G$, we have
$$(e_K*e_K)(g)=\int_Ge_K(x)e_K(x^{-1}g)d\mu_G(x)=\frac{1}{\text{vol}(K)}\int_Ke_K(x^{-1}g)d\mu_G.$$ Since we consider now $x \in K$, this expression is zero if $x^{-1}g\notin K$, i.e., if $g \notin K$. In the case where $g \in K$, we have 
$$(e_K*e_K)(g)=\frac{1}{\text{vol}(K)}\int_Ke_K(x^{-1}g)d\mu_G=(\frac{1}{\text{vol}(K)})^2\text{vol}(K)=\frac{1}{\text{vol}(K)},$$ so we conclude that $e_K*e_K=\frac{\mathbbm{1}_K}{\text{vol}(K)}=e_K.$

\subsection{Construction of the equivalence $\mc{M}(G)\to\{\text{ smooth }\mc{H}(G)-\text{modules}\}$}
Let $(\pi,V)\in \mc{M}(G)$. Let $f \in \mc{H}(G)$, and let $v\in V$. We define an action of $\mc{H}(G)$ on $V$ by 
$$\pi(f)\cdot v=\int_Gf(g)\pi(g)(v)dg.$$

\begin{ex}
The expression $\pi(f)\cdot v$ can be rewritten as a finite sum (Hint : use the fact that $f$ is locally constant with compact support, and that $\pi$ is a smooth representation.)
\end{ex}

\begin{ex}
 This defines an action. Note that there is no unit in $\mc{H}(G)$, so there is no unit condition to check : we only want the associativity. So we need to check that $\pi(f_1*f_2)\cdot v=\pi(f_1)\cdot(\pi(f_2)\cdot v),$ which can be done by developing both sides explicitely and using the left-invariance of the Haar measure on $G$.
\end{ex}

What kinds of $\mc{H}(G)$-modules do we obtain?

\begin{defn}
 An $\mc{H}(G)$-module $V$ is smooth if for all $v \in V$, there exists a compact open subgroup $K$ of $G$ such that $e_K\cdot v=v$.
\end{defn}

\begin{prop}
 The map $\mc{M}(G)\to\{\text{ smooth }\mc{H}(G)-\text{modules}\}$ is an equivalence of categories.
\end{prop}

\begin{proof}
 First, we need to check that $(\pi,V)\in \mc{M}(G)$ gives a smooth $\mc{H}(G)$-module: we know that for each $v \in V$, there exists a compact open subgroup $K$ of $G$ with $\pi^K\neq 0$. Then we can check that $\pi(e_K)\cdot v=\int_G \frac{\mathbbm{1}_K}{\text{vol}(K)}(g)\pi(g)(v)dg=\int_K\frac{1}{\text{vol}(K)}\pi(g)(v)dg=\pi(g)(v)=v$ since $g\in K$ from the previous equality and $v$ is an element of $\pi^K$.

Then, we need to give a quasi-inverse map. Given a smooth $\mc{H}(G)$-module $V$, let $v\in V$ and $g \in G$. We have to define $\pi(g).$ Choose any $K$ such that $e_K\cdot v=v$ (such a $K$ exists by smoothness of $V$). Then we set $\pi(g)\cdot v =e_{gK}\cdot v$. We need to check that this is well-defined.

Let $K'$ be another subgroup with $e_{K'}\cdot v=v$. We can reduce to the case where $K'\subset K$. Now if $K'$ is a subgroup of $K$, we have 
$$(e_{gK'}*e_K)(y)=\int_G\frac{\mathbbm{1}_{gK'}(x)}{\text{vol}(gK')}\frac{\mathbbm{1}_K(x^{-1}y)}{\text{vol}(K)}dx=
\int_{gK'}\frac{\mathbbm{1}_K(x^{-1}y)}{\text{vol}(gK')\text{vol}(K)}dx,$$ so we have here $x^{-1}\in K'g^{-1}$ and we get
$$(e_{gK'}*e_K)(y)=0 \text{ if }y \notin gK,$$
$$(e_{gK'}*e_K)(y)=\frac{1}{\text{vol}(K)}=\frac{1}{\text{vol}(gK)} \text{ if }y \in gK,$$
so we have shown that $e_{gK'}*e_K=e_{gK}.$

In particular, $e_{gK'}\cdot v=(e_{gK'}*e_K)\cdot v=e_{gK}\cdot v,$ since $e_K\cdot v=v$ by choice of $K$. So this is a well-defined action, and checking the equivalence is now easy.
\end{proof}

Our next goal is to show that we have the following bijection, for any $K$:
$$\quotient{\{\text{Irreducible smooth }\\G\text{-rep. }(\pi,V) \text{ with }\pi^K\neq 0\}}{\iso}\to\quotient{\{\text{Irreducible }\mc{H}_K\text{-modules}\}}{\iso}$$
induced by
$$V \mapsto V^K.$$

\begin{lem}
 For all $(\pi,V) \in \mc{M}(G),$ $\pi(e_K):V\to V^K$ is a projection. In particular, $\mc{H}_K=e_K*\mc{H}(G)*e_K$ acts on $V^K$. 
\end{lem}

\begin{proof}
 We already saw that $e_K$ is an idempotent. The next exercises shows that the image of $V$ by $\pi(e_K)$ lies indeed in $V^K$ and concludes the proof.
\end{proof}

\begin{ex}
 Check that for all $k\in K$, we have $\pi(k)\cdot\pi(e_K)(v)=\pi(e_K)(v).$ (Hint: use a change of variable.)
\end{ex}

\begin{prop}
The map sending $V$ to $V^K$ is a bijection on isomorphism classes of irreducible objects, as claimed.
\end{prop}

\begin{proof}
 First, we check that if $(\pi,V)$ is irreducible, then $V^K$ is an irreducible $\mc{H}_K$-module. Since we consider our map as  $$\quotient{\{\text{Irreducible smooth }\\G\text{-rep. }(\pi,V) \text{ with }\pi^K\neq 0\}}{\iso}\to\quotient{\{\text{Irreducible }\mc{H}_K\text{-modules}\}}{\iso},$$
then $\pi^K\neq0$. Suppose $0\subsetneq M \subsetneq V^K$ is a proper $\mc{H}_K$-submodule. Then by definition of $M\subset V^K$ we have $M=\pi(e_K)M$, and $\mc{H}(G)M=V$ by the previous equivalence and the irreducibility of $V$. So we get
$$V^K=\pi(e_K)\cdot V=\pi(e_K)\mc{H}(G)\cdot M=\pi(e_K)\mc{H}(G)\pi(e_K)\cdot M=\mc{H}_K\cdot M=M,$$ which is a contradiction since we assumed $M \subsetneq V^K$. This implies that $V^K$ is irreducible.

Now we construct a quasi-inverse. Given an irreducible $\mc{H}_K$-module $M$, then there exists an ideal $I$ of $\mc{H}_K$ such that $M\iso \quotient{\mc{H}_K}{I}$. We now consider $I\subset \mc{H}_K\subset \mc{H}(G)$ and form the $\mc{H}(G)$-submodule generated by $I$ inside $\mc{H}(G)$, call it $\widetilde{I}$. We also denote by $\widetilde{\mc{H}_K}$ the $\mc{H}(G)$-module generated by $\mc{H}_K$. The quotient $\quotient{\widetilde{\mc{H}_K}}{\widetilde{I}}$ is a $\mc{H}(G)$-module but it might be too big. Using Zorn's lemma, we can take $V$ to be any non-zero irreducible quotient, i.e., $\quotient{\widetilde{\mc{H}_K}}{\widetilde{I}}\twoheadrightarrow V.$

First, we check that $V^K$ is non-zero : we apply $(\hspace{2mm})^K$ to the short exact sequence 
$$0 \to U \to \quotient{\widetilde{\mc{H}_K}}{\widetilde{I}} \to V \to 0$$
and by exactness of $(\hspace{2mm})^K$ we have the short exact sequence
$$0 \to U^K \to (\quotient{\widetilde{\mc{H}_K}}{\widetilde{I}})^K \to V^K \to 0.$$
But we know that $(\quotient{\widetilde{\mc{H}_K}}{\widetilde{I}})^K=\quotient{\mc{H}_K}{I}=M$ by definition. So the exact sequence is
$$0 \to U^K \to M \to V^K \to 0.$$ If $V^K=0$ we obtain $U^K=M$, and therefore $U$ generates $\quotient{\widetilde{\mc{H}_K}}{\widetilde{I}}$. But this implies that $V=0$, which is a contradiction. So we have $V^K\neq 0$ and by irreducibility of $M$ we see that $V^K=M$ as desired. 
\end{proof}

\subsection{A couple of loose ends}
For $G$ a locally profinite group, if $G/K$ is countable for some compact open $K$ then :
\begin{enumerate}
 \item Any smooth $(\pi,V)\in\mc{M}(G)$ has countable dimension.
\item Schur's lemme holds : if $(\pi,V)$ is irreducible, then $\into_G(\pi)=\cmplx.$
\end{enumerate}

\begin{proof}
 We will leave $(1)$ as an exercise and prove $(2)$. Since $\pi$ is irreducible, $\into_G(\pi)$ is a divison algebra over $\cmplx$, not a priori finite dimensional. Fix $v\in V\backslash\{0\}$. Any $\phi \in \into_G(\pi)$ is then determined by $\phi(v)$ since $v$ generates $(\pi,V)$. So by $(1)$, $\into_G(\pi)$ has countable dimension. If $\phi \notin \cmplx$, then we consider (note that $\cmplx(\phi)$ is a field since $\into_G(\pi)$ is a division algebra)
$$\{\frac{1}{\phi-a} \mid a \in \cmplx\}\subset \cmplx(\phi).$$
All the elements $\frac{1}{\phi-a}$ are linearly independant over $\cmplx$, and there are uncountably many of these. So $\into_G(\pi)$ would have uncountable dimension, which is a contradiction. This shows that $\into_G(\pi)=\cmplx.$
\end{proof}
