Induction in Stages: $\textrm{Ind}_H^G = \textrm{Ind}_N^G \textrm{Ind}_H^N$

Question

$\DeclareMathOperator{\Ind}{Ind}$ Let $H \subseteq N \subseteq G$ be closed subgroups of a td-type group $G$. Then $\Ind_H^G$ and $\Ind_N^G \circ \Ind_H^N$ are additive functors from the category of smooth representations of $H$ to the category of smooth representations of $G$. I'm trying to understand why these functors are naturally isomorphic. To a smooth representation $(\sigma,W)$ of $H$, we need to define a $G$-linear isomorphism

$$T_{\sigma}: \Ind_N^G \Ind_H^N(\sigma) \rightarrow \Ind_H^G(\sigma)$$

which is natural in $\sigma$.

Answer 1

$\DeclareMathOperator{\Ind}{Ind}$We are fixing a representation space $(\pi,W)$ of $H$.

Fix $F \in \Ind_N^G \Ind_H^N(\sigma)$. To each $g \in G$, we have a function $F(g): N \rightarrow W$ such that $F(g)(hn) = \sigma(h) F(g)(n)$ for all $h \in H, n \in N$, and such that $F(g)(nk) = F(g)(n)$ for all $n \in N$ and all $k$ in some open compact subgroup of $N$, depending on $F$ and $g$.

We also have that $F(ng)(n') = F(g)(n'n)$ for all $g \in G, n,n' \in N$. Finally, $G$ acts by right translation on these functions $F$: $g \cdot F(g') := F(g'g)$, and for each $g \in G$ there exists an open compact subgroup $K$ of $G$ such that the functions $F(gk)$ and $F(g)$ are equal.

Define $T_{\sigma}(F): G \rightarrow W$ by $T_{\sigma}(F)(g) = F(g)(1_N)$. This defines our map $T_{\sigma}: \Ind_N^G \Ind_H^N(\sigma) \rightarrow \Ind_H^G(\sigma)$. There are several things to check:

1 . $T_{\sigma}(F)$ lies in $\Ind_H^N(\sigma)$.

First, for any $h \in H$ and $g \in G$, we have

$$T_{\sigma}(F)(hg) = F(hg)(1_N) = F(g)(h) = F(g)(h1_N) = \sigma(h)F(g)(1_N) = \sigma(h)T_{\sigma}(F)(g)$$

Next, for each $g \in G$, there exists an open compact subgroup $K$ of $G$ such that $F(gk)$ and $F(g)$ are the same function $N \rightarrow W$. In particular, $T_{\sigma}(gk) = F(gk)(1_N) = F(g)(1_N) = T_{\sigma}(g)$.

2 . $T_{\sigma}$ is a $G$-map.

It is immediate that $T_{\sigma}$ is a linear transformation, so we just need to show that $T_{\sigma}$ preserves the action of $G$. For $g, g' \in G$, and $F \in \Ind_N^G \Ind_H^N(\sigma)$ we have

$$T_{\sigma}(g \cdot F)(g') = (g \cdot F)(g')(1_N) = F(g'g)(1_N) = T_{\sigma}(F)(g'g) = g \cdot T_{\sigma}(F)(g)$$

So $T_{\sigma}(g \cdot F) = g \cdot T_{\sigma}(F)$.

3 . $T_{\sigma}$ is a bijection.

For injectivity, suppose $T_{\sigma}(F)$ is the zero map. This means that for all $g \in G$, $F(g)(1_N)$ is zero in $W$. But for any $n \in N$ and any $g \in G$, we have $0 = F(ng)(1_N) = F(g)(n)$, so this says that for each $g \in G$, $F(g): N \rightarrow W$ is the zero function. Hence $F = 0$.

For surjectivity, let $f \in \Ind_H^G(\sigma)$. For each $g \in G$, define a function $F(g): N \rightarrow W$ by $F(g)(n) = f(ng)$. If we can show that $F(g)$ lies in $\Ind_H^N(\sigma)$, and that moreover the function $F: G \rightarrow \Ind_H^N(\sigma)$ itself lies in $\Ind_N^G \Ind_H^N(\sigma)$, then we will have

$$T_{\sigma}(F)(g) = F(g)(1_N) = f(g)$$

and hence $T_{\sigma}(F) = f$, as required.

First, to show $F(g)$ lies in $\Ind_H^N(\sigma)$ for a fixed $g$. For $h \in H, n \in N$, we have

$$F(g)(hn) = f(hng) = \sigma(h) f(ng) = \sigma(h)F(g)(n)$$

To finish showing that $F(g)$ lies in $\Ind_H^N(\sigma)$, we need to find an open compact subgroup of $N$ for which $F(g)$ is right invariant. Let $K$ be an open compact subgroup of $G$ such that $f$ is right $K$-invariant. Then $N \cap gKg^{-1}$ is a compact open subgroup of $N$, and for $n \in N, k \in N \cap gKg^{-1}$, we have

$$F(g)(nk) = f(nkg) = f(ngg^{-1}kg) = f(ng) = F(g)(n)$$

Finally, we need to show that $F: G \rightarrow \textrm{Ind}_H^N(\sigma)$ lies in $\Ind_N^G \Ind_H^N(\sigma)$.

We first show that $F$ appropriately distributes the action of $N$, i.e. $F(ng)(n') = F(g)(n'n)$ for all $g \in G, n, n' \in N$. This is immediate, since $F(ng)(n') = f(n'ng) = F(g)(n'n)$.

Next, we show that $F$ is right invariant for some open compact subgroup of $G$. An open compact subgroup $K$ of $G$ such that $f$ is right $K$-invariant will do:

$$F(gk)(n) = f(ngk) = f(ng) = F(g)(n)$$

4 . $T_{\sigma}$ is natural.

Let $\phi:(W,\sigma) \rightarrow (U,\tau)$ be a morphism of smooth $H$-representations.

This is a nasty definition check. Let $F \in \Ind_N^G \Ind_H^N(\sigma)$.

First, we have to look at $(\Ind_N^G \Ind_H^N(\phi))(F)$, which is an element of $\Ind_N^G \Ind_H^N(\tau)$. This is by definition a function from $G$ to $\Ind_H^N(\tau)$. If we evaluate this function at a given $g$, this is by definition the element $\Ind_H^N(\phi)(F(g))$, which is a function from $N$ to $U$. If we further evaluate this function at a given $n$, then this is by definition $\phi(F(g)(n))$. Thus

$$T_{\tau} \circ [\Ind_N^G \Ind_H^N(\phi)](F)(g) = [\Ind_N^G \Ind_H^N(\phi)]F(g)(1_N) = \phi(F(g)(1_N))$$

On the other hand, we have $\Ind_H^G(\phi) \circ T_{\sigma}(F)(g) = \phi(T_{\sigma}(F)(g)) = \phi(F(g)(1_N))$. This establishes the naturality.