The equal-coordinate rank-one principal series of GL_2

By [F1] the dimension is ${\dim }_{\mathbb{C}}I(\chi )=[M:B_2]=q+1$, and by [F2] the $M$-module $I(\chi )$ and each of its submodules are semisimple.

Define $f_0(g):=a(\det g{)}^{-1}$ for $g\in M$. Then $f_0\in I(\chi )$: for $b\in B_2$, using multiplicativity of $\det$ and [F8], $f_0(gb)=a(\det g{)}^{-1}a(\det b{)}^{-1}=\tilde{\chi }(b{)}^{-1}{f}_0(g)$ since $\tilde{\chi }(b)=a(\det b)$. The line $\mathbb{C}V:=\mathbb{C}{f}_0$ is stable under $M$ because $\det$ is multiplicative: $(g\cdot f_0)(x)=a(\det (g^{-1}x){)}^{-1}=a(\det g)a(\det x{)}^{-1}=a(\det g)f_0(x)$, so $V\cong a\circ \det$ is a one-dimensional submodule of $I(\chi )$, and for $a=1$ it is spanned by the constant function $f_0=1$.

For $\chi =(a,a)$ the stabiliser is $W_{\chi }=S_2$, so [F3] gives ${\dim }_{\mathbb{C}}{\mathrm{End}}_M(I(\chi ))=2$.

We show that $V$ occurs in $I(\chi )$ with multiplicity exactly one, that it has a complement $W$ with ${\dim }_{\mathbb{C}}W=q$, and that $W$ is simple. If $V$ occurred at least twice, then by semisimplicity (step 1.1) $I(\chi )$ would contain $V\oplus V$ as a direct summand, and every endomorphism of $V\oplus V$ extended by zero would be an $M$-endomorphism of $I(\chi )$, so ${\mathrm{End}}_M(I(\chi ))$ would contain ${\mathrm{M}}_2(\mathbb{C})$ and have dimension at least $4$, contradicting step 1.3. Hence $V$ occurs with multiplicity one. By semisimplicity $I(\chi )=V\oplus W$ for some submodule $W$, necessarily nonzero because $\dim I(\chi )=q+1\ge 3$ and $\dim V=1$. No simple constituent of $W$ is isomorphic to $V$, since that would again give multiplicity at least two; hence ${\mathrm{Hom}}_M(V,W)={\mathrm{Hom}}_M(W,V)=0$ by [F4], and restriction gives ${\mathrm{End}}_M(I(\chi ))\cong {\mathrm{End}}_M(V)\oplus {\mathrm{End}}_M(W)$. Since ${\mathrm{End}}_M(V)=\mathbb{C}$ and $\dim {\mathrm{End}}_M(I(\chi ))=2$ by step 1.3, we get ${\dim }_{\mathbb{C}}{\mathrm{End}}_M(W)=1$. If $W$ had a nonzero proper submodule, Maschke would give a nontrivial invariant splitting of $W$. Its projection would be an idempotent in the one-dimensional algebra ${\mathrm{End}}_M(W)=\mathbb{C}{\mathrm{i}\mathrm{d}}_W$ different from $0$ and ${\mathrm{i}\mathrm{d}}_W$, which is impossible. Thus $W$ is simple. Therefore $I(\chi )\cong V\oplus W$ with $V\cong ̸W$ simple, each of multiplicity one, and ${\mathrm{End}}_M(I(\chi ))\cong \mathbb{C}\oplus \mathbb{C}$; in particular $V$ is the unique one-dimensional constituent, since $\dim W=q\mathrm{\ne }1$.

Let $f\in I(\chi )$ be fixed by $B_2$. Then $f(b^{-1}g)=f(g)$ for all $b\in B_2$ and $g\in M$, so $f$ is left $B_2$-invariant; taking $g=1$ and using covariance gives $f(b)=f(1\cdot b)=\tilde{\chi }(b{)}^{-1}f(1)$, while left invariance gives $f(b)=f(1)$, so $(\tilde{\chi }(b{)}^{-1}-1)f(1)=0$ for all $b\in B_2$. If $a\mathrm{\ne }1$, then $\tilde{\chi }\mathrm{\ne }1$: choosing $t\in T_2$ with $\chi (t)\mathrm{\ne }1$ (possible since $a$ is a nontrivial character of ${\mathbb{F}}_q^{\times }$) gives $b\in B_2$ with $\tilde{\chi }(b)\mathrm{\ne }1$, so $f(1)=0$; then $f=0$ on $B_2$ by the formula, and for $g\in B_2\dot{s}{B}_2$ writing $g=b_1\dot{s}{b}_2$ by [F9], left invariance and right covariance give $f(g)=\tilde{\chi }(b_2{)}^{-1}f(\dot{s})$, while applying covariance to $g=\dot{s}$ and $b={\dot{s}}^{-1}t\dot{s}\in B_2$ gives $f(\dot{s})=f(t\dot{s})=\tilde{\chi }({\dot{s}}^{-1}t\dot{s}{)}^{-1}f(\dot{s})=\chi (t{)}^{-1}f(\dot{s})$, so $f(\dot{s})=0$ and $f=0$. Hence $I(\chi )$ has no nonzero $B_2$-fixed vector when $a\mathrm{\ne }1$. For $a=1$ the constant function is fixed by $B_2$, since $\tilde{\chi }=1$.

In the case $a=1$ the submodule $V$ of step 1.2 is the trivial module spanned by the constants, and we define the Steinberg representation by $\mathrm{St}:=W$ for the decomposition $I(1)=V\oplus W$ of step 2.1; it is a simple module of dimension $q$. We claim that for every character $a$ of ${\mathbb{F}}_q^{\times }$ there is an isomorphism of $M$-modules $I(\chi )\cong I(1)\otimes (a\circ \det ),\chi =(a,a),$ where $a\circ \det$ is the one-dimensional $M$-module $g\mapsto a(\det g)$. Let $\psi :=a\circ \det$ and let $W_{\psi }=\mathbb{C}z$ be the one-dimensional space on which $M$ acts by $\psi$; define $\mathrm{\Phi }:I(1)\otimes W_{\psi }\to I(\chi )$ by $\mathrm{\Phi }(f\otimes z)(x):=f(x)\psi (x{)}^{-1}$. This is well defined and lands in $I(\chi )$ because $f(xb)=f(x)$ and $\psi (xb{)}^{-1}=\psi (x{)}^{-1}\psi (b{)}^{-1}$, while $\tilde{\chi }(b)=\psi (b)$ by [F8]; it is $M$-equivariant because $\mathrm{\Phi }(g\cdot (f\otimes z))(x)=f(g^{-1}x)\psi (g)\psi (x{)}^{-1}=(\mathrm{\Phi }(f\otimes z))(g^{-1}x)$, using the diagonal action of [F7]; and its inverse sends $F\in I(\chi )$ to the function $x\mapsto F(x)\psi (x)$ tensored with $z$, which is right $B_2$-invariant. Thus $\mathrm{\Phi }$ is an isomorphism. Tensoring the decomposition $I(1)=V\oplus \mathrm{St}$ of step 2.1 with the one-dimensional module $a\circ \det$ and applying $\mathrm{\Phi }$ gives $I(\chi )\cong (a\circ \det )\oplus (\mathrm{St}\otimes (a\circ \det )),$ where $\mathrm{\Phi }(V\otimes W_{\psi })=\mathbb{C}\cdot (a^{-1}\circ \det )$ is the one-dimensional constituent $a\circ \det$ of step 1.2 and $\mathrm{St}\otimes (a\circ \det )$ is simple of dimension $q$; the two summands are non-isomorphic because their dimensions $1$ and $q\ge 2$ differ, and each occurs with multiplicity one.

Take $a=1$, so that $I(1)=\mathbb{C}\cdot 1\oplus \mathrm{St}$ by step 3.1, and recall $B_s=R_{T_s}$ under the identification $\mathbb{C}[M]e_B\cong I(1)$ of [F5] and [F6]. The vector ${\sum }_{g\in M}g$ satisfies $({\sum }_{g}g)e_B={\sum }_{g}g$ because right multiplication by $B_2$ permutes $M$, and it corresponds to the constant function under [F5]; since ${\sum }_{g}g$ is invariant under left multiplication by $M$, it spans the trivial constituent. Applying $B_s$ gives $B_s({\sum }_{g}g)=({\sum }_{g}g)T_s=q({\sum }_{g}g)e_B\dot{s}{e}_B=q({\sum }_{g}g)\dot{s}{e}_B=q({\sum }_{g}g)e_B=q{\sum }_{g}g$, using $\dot{s}$ permuting $M$ on the left and $({\sum }_{g}g)e_B={\sum }_{g}g$; hence $B_s$ acts on the trivial constituent by $q$. The corner anti-isomorphism $h\mapsto R_h$ transfers the polynomial identity of [F6] to $B_s^2=(q-1)B_s+q\mathrm{i}\mathrm{d}$. The projections $(B_s+\mathrm{i}\mathrm{d})\mathrm{/}(q+1)$ and $(q\mathrm{i}\mathrm{d}-B_s)\mathrm{/}(q+1)$ split $I(1)$ into its $q$ and $-1$ eigenspaces. They are $M$-equivariant and preserve $\mathrm{St}$: maps from this nontrivial simple module to the trivial summand vanish by [F4]. Simplicity of $\mathrm{St}$ therefore makes $B_s$ scalar on it, with value $q$ or $-1$. The eigenvalue cannot be $q$: if $B_s=q\mathrm{i}\mathrm{d}$ on $\mathrm{St}$ as well, then $B_s=q\mathrm{i}\mathrm{d}$ on $I(1)$, but $B_s(e_B)=e_B{T}_s=T_s{e}_B=q{e}_B\dot{s}{e}_B{e}_B=q{e}_B\dot{s}{e}_B=T_s$ and $T_s,e_B$ are distinct basis elements of $H$, so $B_s(e_B)\mathrm{\ne }q{e}_B$. Hence $B_s$ acts on $\mathrm{St}$ by $-1$, and the displayed quadratic identity $B_s^2=(q-1)B_s+q{\mathrm{i}\mathrm{d}}_{I(\chi )}$ holds.

Steps 1.1 and 2.1 give the dimension, the multiplicity-one splitting into two non-isomorphic simple constituents and ${\mathrm{End}}_M(I(\chi ))\cong \mathbb{C}\oplus \mathbb{C}$; step 3.1 identifies the constituents as $a\circ \det$ and $\mathrm{St}\otimes (a\circ \det )$ with $\mathrm{St}$ of dimension $q$ and shows uniqueness of the one-dimensional constituent; steps 2.2 and 4.1 give the fixed-vector statement and the action of $B_s$ in the spherical case. All modules are finite-dimensional over $\mathbb{C}$, all decompositions are finite, and no choice principle is used.