Finite semisimple matrix density and the double commutant over Q

Let $U\subseteq W$ be a $G$-submodule and choose a rational linear projection $p:W\to U$ that is the identity on $U$. The average $P=\mathrm{\mid }G{\mathrm{\mid }}^{-1}{\sum }_{g\in G}gp{g}^{-1}$ has image in $U$, restricts to the identity on $U$, and satisfies $hP{h}^{-1}=P$ for every $h\in G$ by reindexing the finite sum. Thus $P$ is a $G$-linear projection and $\ker P$ complements $U$. Selecting a nonzero submodule of least positive rational dimension makes it simple; splitting it off and inducting on dimension writes $W={⨁}_{\lambda }{S}_{\lambda }^{m_{\lambda }}$ with finitely many pairwise nonisomorphic simple modules $S_{\lambda }$.

For a simple $S$, $D={\mathrm{End}}_B(S)$ is a division ring: a nonzero module map has zero kernel and nonzero image, so is bijective and its inverse is again a module map. Give $S$ the right $D^{\mathrm{o}\mathrm{p}}$-action $s\cdot d^{\mathrm{o}\mathrm{p}}=d(s)$. It has a finite basis over this division ring: a $D^{\mathrm{o}\mathrm{p}}$-independent list is rationally independent because $\mathbb{Q}{\mathrm{i}\mathrm{d}}_S\subseteq D$, so its length is bounded by ${\dim }_{\mathbb{Q}}S$; a list of maximal length spans. Write this basis as $v_1,\dots ,v_d$.

The evaluation image $I=\{(b{v}_1,\dots ,b{v}_d):b\in B\}\subseteq S^d$ is a $B$-submodule. If it were proper, [F1] would give a nonzero complement; projecting onto one simple summand of that complement and identifying it with $S$ gives a nonzero $B$-map $f:S^d\to S$ that vanishes on $I$. Every such map has the form $f(x_1,\dots ,x_d)={\sum }_j{d}_j(x_j)$ with $d_j\in D$. For $b\in B$, $0=f(b{v}_1,\dots ,b{v}_d)=b{\sum }_j{d}_j(v_j)$, so at $b=1$ one gets ${\sum }_j{v}_j\cdot d_j^{\mathrm{o}\mathrm{p}}=0$. The chosen right $D^{\mathrm{o}\mathrm{p}}$-basis forces every $d_j=0$, contradicting $f\mathrm{\ne }0$. Hence evaluation is onto. Any $D^{\mathrm{o}\mathrm{p}}$-linear endomorphism of $S$ is therefore the action of some $b\in B$: choose $b$ with the prescribed values on the basis, then both maps are $D^{\mathrm{o}\mathrm{p}}$-linear and agree on that basis. Thus the image of $B$ on $S$ is ${\mathrm{End}}_{D^{\mathrm{o}\mathrm{p}}}(S)$.

Apply the same evaluation argument simultaneously to one chosen $D_{\lambda }^{\mathrm{o}\mathrm{p}}$-basis of each $S_{\lambda }$. The target is $X={⨁}_{\lambda }{S}_{\lambda }^{d_{\lambda }}$. If the image of $B\to X$, $b\mapsto (b{v}_{\lambda ,j}{)}_{\lambda ,j}$, were proper, [F1] would provide a nonzero $B$-map $f:X\to S_{\mu }$ killing that image for some simple type $\mu$. A map from $S_{\lambda }$ to $S_{\mu }$ is zero for $\lambda \mathrm{\ne }\mu$, since a nonzero map between simple modules is an isomorphism; the remaining coordinates of $f$ are maps $d_j\in D_{\mu }$. Evaluating at $b=1$ contradicts the $D_{\mu }^{\mathrm{o}\mathrm{p}}$-independence of the chosen basis of $S_{\mu }$, exactly as in step 2.1. Therefore every tuple of target values on all these bases is attained by one $b$, and the image of $B$ on the distinct simple types is the full independent product ${\prod }_{\lambda }{\mathrm{End}}_{D_{\lambda }^{\mathrm{o}\mathrm{p}}}(S_{\lambda })$.

On $W_{\lambda }=S_{\lambda }^{m_{\lambda }}$ this product acts by the same full algebra $R_{\lambda }={\mathrm{End}}_{D_{\lambda }^{\mathrm{o}\mathrm{p}}}(S_{\lambda })$ on each copy. Its commutant on $S_{\lambda }$ is exactly $D_{\lambda }$: a map commutes with $R_{\lambda }$ iff it commutes with the image of $B$, which is equivalent to being in ${\mathrm{End}}_B(S_{\lambda })$. Hence ${\mathrm{End}}_B(W)={⨁}_{\lambda }{M}_{m_{\lambda }}(D_{\lambda })$. To take its commutant on $W_{\lambda }$, first commute with the diagonal and off-diagonal copy matrix units; this forces zero off-diagonal blocks and the same map $g:S_{\lambda }\to S_{\lambda }$ on every diagonal block. Commuting with the diagonal $D_{\lambda }$ maps is exactly the condition $g(s\cdot d^{\mathrm{o}\mathrm{p}})=g(s)\cdot d^{\mathrm{o}\mathrm{p}}$, so $g\in {\mathrm{End}}_{D_{\lambda }^{\mathrm{o}\mathrm{p}}}(S_{\lambda })=R_{\lambda }$. The independent central block projections eliminate maps between distinct $\lambda$. Thus the commutant of ${\mathrm{End}}_B(W)$ is precisely ${⨁}_{\lambda }{R}_{\lambda }$ acting diagonally on the multiplicity copies, which is the image of $B$ from step 3.1.