I am trying to understand the proof of the following lemma, which comes from section 9.3 of The Local Langlands Conjecture for GL(2). In this question, $T$ is the subgroup of $GL_2(F),$ $F$ a non-archimedean local field, of the form $\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ with $a,b \in F^{\times}$.
Lemma: Let $(\sigma,U)$ be a smooth representation of $T.$ As usual, $(\Sigma,X)=\text{Ind}_B^G\sigma.$ There is an exact sequence of representations of $T$:
$$\require{AMScd}
\begin{CD}
0 @>{}>> \sigma^{\omega} \otimes \delta_B^{-1} @>{}>> \Sigma_N @>{\alpha_{\sigma}}>> \sigma @>{}>> 0.
\end{CD}$$
The proof proceeds as follows. We can view $\sigma$ as a representation of $B$, the standard Borel subgroup of $G$, by letting it act trivially on $N$, the unipotent radical of $B$. Recall that the space $X$ consists of $G$-smooth functions $f:G \rightarrow U$ satisfying $f(bg)=\sigma(b)f(g)$ for all $b \in B, g \in G.$ There is a canonical $B$-homomorphism $\alpha_\sigma: \text{Ind}_B^G \sigma \rightarrow U$ given by $f \mapsto f(1)$. Set $V=\ker(\alpha_\sigma)$. Thus $V$ provides a smooth representation of $B$ (this I understand and can show) and there is an exact sequence:
$$\require{AMScd}
\begin{CD}
0 @>{}>> V_N @>{}>> X_N @>{\alpha_{\sigma}}>> U @>{}>> 0.
\end{CD}$$
(1): since $V$ in not a representation of $G$ one cannot form the Jacquet module of $V$ at $N$. Given this, I'm interpreting $V_N$ to be the kernel of the map $X_N \rightarrow U,$ as this is the only other thing sensible interpretation I can come up with. Is it correct?
(2): the map $\alpha_\sigma: X_N \rightarrow U$ is a $T$-homomorphism, induced by $\alpha_\sigma: X \rightarrow U$. How does one show it, or $\alpha_\sigma:X \rightarrow U$ is surjective?
Best Answer
(1): $V$ is a representation of $B$, so it makes sense to take the $N$-coinvariants which are denoted by $V_N$.
(2): One can show that $\alpha_\sigma:\mathrm{Ind}^G_H \sigma \to W, f \mapsto f(1)$ is surjective for a closed subgroup $H$ of a locally profinite group $G$ and a smooth $H$-representation $(W,\sigma)$ (cf. 2.4)
To prove this, take $w \in W$. Then there is some open compact subgroup $K'$ of $H$ that fixes $w$. So $K'$ is an open neighborhood in $H$ of the identity element $e$. By definition of the subspace topology, there is some open subset $U$ of $G$ such that $H \cap U=K'$. As $G$ is locally profinite, one can take $K$ to be an open compact subgroup of $G$ contained in $U$, then we have $H \cap K \subset K'$. Now the set of right $K$-cosets $G/K$ is discrete and it has a left $H$-action. We may write a $H$-stable decomposition: $$G/K=HK/K \sqcup X$$ where $X \subset G/K$ is some $H$-stable subset. Now simply define $\tilde{f}:G/K\to \Bbb C$ by setting $\tilde{f}(x)=0$ for $x \in X$ and $\tilde{f}(hk)=\sigma(h)w$ for $h \in H, k \in K$. Note that this is well-defined because $H \cap K\subset K'$ fixes $w$. Because $G/K$ is discrete, $\tilde{f}$ is automatically continuous. Furthermore, by construction $\tilde{f}$ satisfies $\tilde{f}(hg)=\sigma(h)\tilde{f}(g)$ for any $h \in H, g\in G$. Define $f:G \to \Bbb C$ by composing $\tilde{f}$ with the projection $G \to G/K$. Then from the definition and properties of $\tilde{f}$, it is easily verified that $f \in \mathrm{Ind}_H^G\sigma$ and $f(1)=w$.