Principal Series Representations of GL N over a Finite Field

Multiplication of $A$ is continuous by [L1], and limits in $A$ are unique because $A$ is separated by [F2]. A sequence is Cauchy precisely when for every $n$ its terms eventually have a fixed residue modulo $I^{n}A$; completeness then gives its unique limit with those eventual residues. In particular a sequence with $s_m\in I^{m}A$ tends to zero, and a series with its $m$-th term in $I^{m}A$ has Cauchy partial sums, whose limit agrees with each partial sum modulo the ideal containing its tail.

Let $c_m:=\left(\genfrac{}{}{0px}{}{-1\mathrm{/}2}{m}\right)\in \mathbb{C}$ for $m\ge 0$. If $\epsilon \in IA$ then ${\epsilon }^m\in I^{m}A$ for every $m$, so the partial sums $S_N:={\sum }_{m=0}^N{c}_m{\epsilon }^m$ satisfy $S_{N^{\mathrm{\prime }}}-S_N\in I^{N+1}A$ whenever $N^{\mathrm{\prime }}\ge N$: the series converges, and its limit $f(\epsilon ):={\sum }_{m\ge 0}{c}_m{\epsilon }^m$ satisfies $f(\epsilon )\equiv S_N(\mathrm{m}\mathrm{o}\mathrm{d}{I}^{N+1}A)$ for every $N$ by step 1.1. To justify the formal identity, put $f(u)={\sum }_{m\ge 0}{c}_m{u}^m$. The binomial coefficients satisfy $2(m+1)c_{m+1}=-(2m+1)c_m$, so $2(1+u)f^{\mathrm{\prime }}(u)+f(u)=0$. Consequently the formal derivative of $(1+u)f(u{)}^2$ is zero, and its constant term is $1$; over $\mathbb{C}$ this gives $(1+u)f(u{)}^2=1$. Thus the identity $(1+u)({\sum }_{m=0}^N{c}_m{u}^m{)}^2=1+{\sum }_{m>N}{d}_m{u}^m$ of truncated formal power series over $\mathbb{C}$ shows $(1+\epsilon )S_N^2\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{I}^{N+1}A)$; passing to limits using the continuity of multiplication and the uniqueness of limits gives $(1+\epsilon )f(\epsilon {)}^2=1$.

Let $e\in A$ be an idempotent and put $B:=(1-e)A(1-e)$, with unit $1-e$. Then $B$ is an associative $R$-algebra with unit $1-e$ and scalar map $r\mapsto r(1-e)$, finitely generated as an $R$-module, and $I^{n}B=B\cap I^{n}A$ for every $n$: the inclusion $I^{n}B\subseteq B\cap I^{n}A$ is clear, and for $b\in B\cap I^{n}A$ the computation $\pi (b)=b$, where $\pi (a):=(1-e)a(1-e)$ is the $R$-linear idempotent projection onto $B$, exhibits $b\in \pi (I^{n}A)\subseteq I^{n}B$. The projection $\pi$ satisfies $\pi (I^{n}A)\subseteq I^{n}B\subseteq I^{n}A$, hence is continuous, so it induces an idempotent endomorphism $\mathrm{\Pi }$ of the completion of [F2]; its image is exactly with the maps induced by $\pi$. Since agrees with $\mathrm{\Pi }$ on $B$ and $\mathrm{\Pi }$ is the identity on $B$, the map is an isomorphism: it is injective by the separatedness of $A$, and surjective onto the image of $\mathrm{\Pi }$. Thus $B$ is $I$-adically complete.

Put $y:=2x-1$ and $\epsilon :=y^2-1=4(x^2-x)\in IA$. Since $\epsilon$ is a polynomial in $y$, the element $z:=yf(\epsilon )$ of step 2.1 satisfies $z^2=y^{2}f(\epsilon {)}^2=(1+\epsilon )f(\epsilon {)}^2=1.$ Hence $e:=\frac{1}{2}(1+z)$ satisfies $e^2=\frac{1}{4}(1+2z+z^2)=e$, and $e-x=\frac{1}{2}(z-y)=\frac{1}{2}y(f(\epsilon )-1)\in A\cdot IA\subseteq IA$, because $f(\epsilon )-1={\sum }_{m\ge 1}{c}_m{\epsilon }^m\in IA$. Any idempotent $\bar{x}\in A\mathrm{/}IA$ has a representative $x\in A$ with $x^2-x\in IA$, so the preceding construction produces an idempotent $e$ of $A$ with image $\bar{x}$: assertion (1) holds.

Assertion (2) follows by finite iteration. Given orthogonal idempotents ${\bar{x}}_1,\dots ,{\bar{x}}_k$ in $A\mathrm{/}IA$, choose representatives $x_1,\dots ,x_k$; they satisfy $x_i^2-x_i\in IA$ and $x_i{x}_j\in IA$ for $i\mathrm{\ne }j$. Suppose $e_1,\dots ,e_j\in A$ are pairwise orthogonal idempotents with $e_i\equiv x_i(\mathrm{m}\mathrm{o}\mathrm{d}IA)$ for $i\le j$, put $E_j:=e_1+\cdots +e_j$ and $B_j:=(1-E_j)A(1-E_j)$, and set $X_j:=(1-E_j)x_{j+1}(1-E_j)\in B_j$. Expanding, $X_j^2-X_j=(1-E_j)(x_{j+1}^2-x_{j+1}-x_{j+1}{E}_j{x}_{j+1})(1-E_j)$, and both $x_{j+1}^2-x_{j+1}$ and $x_{j+1}{E}_j{x}_{j+1}\equiv {\sum }_{i\le j}{x}_{j+1}{x}_i{x}_{j+1}\equiv 0$ lie in $IA$, so $X_j^2-X_j\in (1-E_j)IA(1-E_j)\subseteq I{B}_j$. As $B_j$ is complete by step 2.2, step 3.1 applied in $B_j$ supplies an idempotent $e_{j+1}\in B_j$ with $e_{j+1}\equiv X_j(\mathrm{m}\mathrm{o}\mathrm{d}I{B}_j)$; then $e_{j+1}$ is orthogonal to $e_1,\dots ,e_j$, and $e_{j+1}\equiv X_j\equiv x_{j+1}(\mathrm{m}\mathrm{o}\mathrm{d}IA)$ because $(1-E_j)$ is congruent to $1-{\sum }_{i\le j}{x}_i$ modulo $IA$. Starting from $j=0$, where $B_0=A$ and $X_0=x_1$, this yields orthogonal idempotents with the required congruences. If ${\bar{x}}_1+\cdots +{\bar{x}}_k=1$, the last lift may be replaced by $e_k:=1-e_1-\cdots -e_{k-1}$: it is idempotent, orthogonal to $e_1,\dots ,e_{k-1}$, and congruent to $1-{\sum }_{i<k}{\bar{x}}_i={\bar{x}}_k$ modulo $IA$.

Left multiplication is $\mathbb{C}$-linear and satisfies $L_{a+\lambda b}=L_a+\lambda L_b$ and $L_{ab}=L_a\circ L_b$, since $cx$ is linear in $c$ with $x$ fixed and $(ab)x=a(bx)$; hence $(\cdot ,\cdot )$ is $\mathbb{C}$-bilinear. Moreover $(ab,c)=\mathrm{tr}(L_{(ab)c})=\mathrm{tr}(L_{a(bc)})=(a,bc)$, and symmetry of the form follows from $(a,b)=\mathrm{tr}(L_a{L}_b)=\mathrm{tr}(L_b{L}_a)=(b,a)$, where the middle equality is the cyclic property of the matrix trace applied to the matrices of $L_a$ and $L_b$ and their composite, with the matrix of a composite given by the product and the trace read in one basis by [F1]. This proves assertion (1).

Assume $A$ is semisimple. If $A=0$ then the empty product gives assertion (3) and the form is nondegenerate vacuously. If $A\mathrm{\ne }0$, Wedderburn-Artin gives a ring isomorphism $A\cong {\prod }_{i=1}^r{\mathrm{M}}_{n_i}(D_i)$ with division rings $D_i$ and $n_i\ge 1$ by [F2]; since $A$ is a $\mathbb{C}$-algebra, the scalar copy of $\mathbb{C}$ lies in the centre of each factor, so each $D_i$ is a finite-dimensional division algebra over $\mathbb{C}$. For $d\in D_i$ the left multiplication $L_d$ on the nonzero finite-dimensional $\mathbb{C}$-space $D_i$ has an eigenvalue $\lambda \in \mathbb{C}$ by [F2], and $L_{d-\lambda }=L_d-\lambda$ is then not injective, so $d-\lambda$, being either $0$ or a unit of the division ring $D_i$, must be $0$; hence $d\in \mathbb{C}$ and $D_i=\mathbb{C}$. Thus $A\cong {\prod }_{i=1}^r{\mathrm{M}}_{d_i}(\mathbb{C})$, and the simple left $A$-modules are the column modules ${\mathbb{C}}^{d_i}$ by [F2]. For $Z\in {\mathrm{M}}_d(\mathbb{C})$ and the basis $\{E_{kl}\}$ of matrix units, $L_Z(E_{kl})=Z{E}_{kl}={\sum }_i{z}_{ik}{E}_{il}$ has no diagonal contribution from the summand indexed by $(k,l)$ other than the $E_{kl}$-coefficient $z_{kk}{E}_{kl}$, so $\mathrm{tr}(L_Z)={\sum }_{k,l}{z}_{kk}=d\mathrm{tr}(Z)$. Hence, under the isomorphism, $(X,Y)={\sum }_i{d}_i\mathrm{tr}(X_i{Y}_i)$; if the first argument is orthogonal to the whole factor $i$, testing against all matrix units $E_{kl}$ of that factor shows $X_i=0$, with $d_i\mathrm{\ne }0$ in $\mathbb{C}$; therefore the form is nondegenerate. This proves the reverse implication of assertion (2) and, with the identification of the simple modules, assertion (3).

Assume conversely that $(\cdot ,\cdot )$ is nondegenerate. Let $J:=J(A)$, which is a two-sided ideal with $A\mathrm{/}J$ semisimple and which is nilpotent by [F3]. For $x\in J$ and $b\in A$ associativity puts $xb\in J$, so $(xb{)}^m=0$ for some $m\ge 1$ and $L_{xb}^m=L_{(xb{)}^m}=0$ by the product rule of step 1.1, that is, $L_{xb}$ is nilpotent; by [F1] it has a strictly upper triangular matrix in some ordered basis, so $(x,b)=\mathrm{tr}(L_{xb})=0$. Since $b\in A$ was arbitrary and the form is nondegenerate, $x=0$. Hence $J=0$ and $A=A\mathrm{/}J$ is semisimple, which is the forward implication of assertion (2).

Assertion (1) is step 1.1, the two directions of assertion (2) are steps 1.2 and 2.1, and assertion (3) is the structure statement proved in step 1.2, including the case $A=0$ as the empty product. All objects constructed are determined by the cited decomposition theorems and by explicit basis computations; no choice principle is invoked, so the item is choice-free.

Write $M={⨁}_{i=1}^r{M}_i$ with $M_i\cong V_i^{\oplus m_i}$ for each $i$, the isotypic decomposition supplied by [F2]. For $j\mathrm{\ne }i$ Schur's lemma over $\mathbb{C}$ gives ${\mathrm{Hom}}_A(V_j,M_i)=0$, because a nonzero such map would be an isomorphism from the simple module $V_j$ onto a simple submodule of $M_i$, which is a direct sum of copies of $V_i$; and ${\mathrm{Hom}}_A(V_i,M_i)\cong {\mathrm{Hom}}_A(V_i,V_i{)}^{m_i}$ because the finite direct sum $M_i$ is a biproduct, with the components read off by the inclusions and projections of the $m_i$ copies of $V_i$. With $D_i:={\mathrm{End}}_A(V_i)$ a division ring by [F1], this identifies ${\mathrm{Hom}}_A(V_i,M)$ with $D_i^{m_i}$ as a $\mathbb{C}$-vector space.

Each $D_i$ is a finite-dimensional $\mathbb{C}$-division algebra: it is a $\mathbb{C}$-algebra because $A$ is one and the action is $\mathbb{C}$-linear, and it is finite-dimensional because $V_i$ is a quotient of the finite-dimensional algebra $A$, hence finite-dimensional over $\mathbb{C}$. For $0\mathrm{\ne }T\in D_i$ the left multiplication $L_T:D_i\to D_i$ is a $\mathbb{C}$-linear endomorphism of a nonzero finite-dimensional $\mathbb{C}$-space, so it has an eigenvalue $\lambda$ by [F4]; then $T-\lambda$ is not invertible, so $T=\lambda$ and $T\in \mathbb{C}$; hence $D_i=\mathbb{C}$. In particular ${\dim }_{\mathbb{C}}{\mathrm{Hom}}_A(V_i,M)=m_i$ by step 1.1.

The matrix description of endomorphisms of the finite direct sum $M={⨁}_i{M}_i$ from [F3], together with the vanishing ${\mathrm{Hom}}_A(V_j,M_i)=0$ for $j\mathrm{\ne }i$ of step 1.1, gives a $\mathbb{C}$-algebra isomorphism $E\cong {\prod }_{i=1}^r{\mathrm{End}}_A(M_i)$; applying the matrix description again inside each isotypic block, and $D_i=\mathbb{C}$ from step 2.1, gives ${\mathrm{End}}_A(M_i)\cong {\mathrm{M}}_{m_i}({\mathrm{End}}_A(V_i))={\mathrm{M}}_{m_i}(\mathbb{C})$, so $E\cong {\prod }_i{\mathrm{M}}_{m_i}(\mathbb{C})$. If $r=0$ then $M=0$, $E=0$ is semisimple by the definition of Semisimple modules as direct sums of simple modules read for the zero ring, and the statement is the empty product; if $r\ge 1$ then $E\mathrm{\ne }0$ and Wedderburn-Artin [F5] applied to the product of matrix rings over the division rings $\mathbb{C}$ shows that $E$ is semisimple. This proves assertion (1).

The space $S_i={\mathrm{Hom}}_A(V_i,M)$ is a left $E$-module by postcomposition $(e\cdot f)(v)=e(f(v))$, since a composite of $A$-linear maps is $A$-linear, and by step 1.1 it is $\mathbb{C}$-linearly isomorphic to ${\mathrm{End}}_A(V_i{)}^{m_i}={\mathbb{C}}^{m_i}$. Under the isomorphism $E\cong {\prod }_i{\mathrm{M}}_{m_i}(\mathbb{C})$ of step 3.1 and this identification, an endomorphism $e=(e^{(1)},\dots ,e^{(r)})$ acts on the $i$-th component of $M$ through the matrix $e^{(i)}$ acting on ${\mathbb{C}}^{m_i}$ by matrix multiplication; hence $S_i$ is precisely the natural column module ${\mathbb{C}}^{m_i}$ of the $i$-th matrix factor of $E$, a simple $E$-module, and by [F5] these column modules represent all isomorphism classes of simple $E$-modules, one per factor. Therefore $i\mapsto S_i$ is a bijection onto those classes and ${\dim }_{\mathbb{C}}{S}_i=m_i$: assertions (2) and (3).

Assertion (1) is step 3.1, and assertions (2) and (3) are step 4.1 together with the identification ${\dim }_{\mathbb{C}}{\mathrm{Hom}}_A(V_i,M)=m_i$ of step 2.1; this proves everything asserted for $E={\mathrm{End}}_A(M)$. If $E^{\mathrm{\prime }}$ is semisimple and $\phi :E^{\mathrm{\prime }}\to E$ is an algebra isomorphism, the map $S\mapsto S$ with $E^{\mathrm{\prime }}$-action $e^{\mathrm{\prime }}\cdot f:=\phi (e^{\mathrm{\prime }})\cdot f$ identifies the simple $E^{\mathrm{\prime }}$-modules with the simple $E$-modules and preserves dimensions, so the indexing and multiplicities transport along $\phi$; the same argument applies to any abstract isomorphism onto ${\mathrm{End}}_A(M)$. Every step used only finite-dimensional $\mathbb{C}$-linear algebra, the cited structure theorems and explicit component projections, so no choice principle is used.