$G$-composition length of representation

Question

I do not understand what is meant by $G$-composition length. The textbook I'm using - The Local Langlands for GL(2) - makes the following statement.

Let $\chi=\chi_1 \otimes \chi_2$ be a character of the maximal non-split torus $T$ of $G=GL_2(F)$ where $F$ is a non-Archemidean local field of characteristic $0$. Inflate $\chi$ to the Borel subgroup $B$, and form its smooth induction $(\Sigma,X)=\text{Ind}_B^G \chi$. If $(\Sigma,X)$ is reducible, then $X$ has $G$-composition length $2$. One composition factor of $X$ has dimension 1, the other has infinite dimension.

The following I know so far. There is a canonical map $\alpha_\chi:\text{Ind}_B^G \chi \rightarrow \mathbb{C}, f \mapsto f(1);$ we set $V=\ker \alpha_\chi$. This space carries a representation of $B$. It's known that $X/V$ is isomorphic to the $T$-representation $(\chi,\mathbb{C})$. The space $V(N)$ is a linear subspace of $V$ spanned by elements of the form $v-\Sigma(n)v,\ v\in V, n \in N.$ It's known $V/V(N)$ is isomorphic to the $T$-representation $(\chi^\omega \otimes \delta_B^{-1},\mathbb{C})$, where $\chi^\omega(t)=\chi(\omega t\omega)$ for the permutation matrix $\omega=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$, and $\delta_B$ is the module of $B$. Furthermore, we prove $V(N)$ is irreducible as a representation of a subgroup of $B$, so irreducible as a representation of $B$.

Am I to understand the following then: we have the inclusions $X \supset V \supset V(N)$ as representations of $B$. There are no (non-trivial) $B$-subspaces of $V(N)$ by irreduciblity so the chain of inclusions stops there. Would one say $X$ has a $B$-composition length of $2$ or $3$? And do the composition factors refer to $V$ and $V(N)$, or $X/V$ and $V/V(N)$.

Lastly, how does one pass from $B$-composition length to $G$-composition length since neither $V$ nor $V(N)$ are representations of $G$.

Answer 1

What I have found:

In general, given a ring $R$ and a $R$-module $V$, a finite composition series of $V$ of length $N-1$ consists of a filtration \begin{equation*} 0=V_1\subset V_2 \subset \dots \subset V_{N-1} \subset V_N = V \end{equation*} where $V_i/V_{i-1}$ is simple for each $i$. The quotients $V_i/V_{i-1}$ are referred to as the composition factors.

Thus, $X \supset V \supset V(N) \supset 0$ is a $B$-composition series of length $3.$ Two composition factors - $X/V$ and $V/V(N)$ - are of dimension 1; the other $V(N)/0$ is of infinite dimension.

Answer 2

This is explained in these notes of Murnaghan (just search for "length"). The general statement is that principal series representations have length at most the order of the Weyl group. In your case the Weyl group has length 2, so any reducible principal series representation must have a composition series of length exactly two. The key idea is to look at Jacquet modules.