Asymptotic induction identifies leading modular simples

Every simple $H$-module $M$ is finite-dimensional, since for $0\mathrm{\ne }m\in M$ the map $H\to M$, $h\mapsto hm$, is surjective. Since $D_1=1$, the maximum $a=a_M$ exists; $I_{a+1}M=0$ and $I_{a}M\mathrm{\ne }0$. Set $T(M)=B_a{\otimes }_{H}M$. Its map $\pi :T(M)\to M$, $[f]\otimes m\mapsto fm$, is well-defined: changing $f$ by $I_{a+1}$ changes no value, and the tensor balancing relation is respected. It is $H$-linear by multiplication and nonzero by $I_{a}M\mathrm{\ne }0$, so it is surjective.

If $x={\sum }_j[f_j]\otimes m_j\in \ker \pi$ and $g\in I_a$, then $gx={\sum }_j[g{f}_j]\otimes m_j=[g]\otimes {\sum }_j{f}_j{m}_j=0$ by tensor balancing. Thus $I_a$ annihilates $\ker \pi$. Every simple factor $M^{\mathrm{\prime }}$ of that kernel has $a_{M^{\mathrm{\prime }}}<a$, because all $D_w$ with $a(w)\ge a$ belong to $I_a$. In particular $[\ast T(M):M]=1$. Every $J_k$-simple factor of $T(M)$ has degree $a$, by the degree support of $B_a$.

Decompose the semisimple $J_k$-module $T(M)={⨁}_E{V}_E^{m_E}$. Restriction preserves the direct sum. The equality $[\ast T(M):M]=1$ implies that exactly one occurring label $E$ has $M$ as a factor of $\ast V_E$, and both $m_E$ and $[\ast V_E:M]$ equal one. Define this label to be $E(M)$. Its degree is ${\alpha }_{E(M)}=a_M$. The direct-sum inclusion of this $\ast V_E$ into $\ast T(M)$ and step 2.1 show that all its other simple factors have strictly smaller $a$-value. If $E(M)=E(M^{\mathrm{\prime }})$, differing $a$-values would contradict their common degree; at equal degree the unique factor just proved forces $M\cong M^{\mathrm{\prime }}$. This proves the assignment's injectivity.

Put $\mathrm{\Delta }={\mathrm{End}}_H(M)$. A nonzero endomorphism of the simple finite-dimensional module is bijective, so $\mathrm{\Delta }$ is a division $k$-algebra. It acts on $T(M)$ by $1\otimes d$, commuting with $J_k$. Consequently it preserves every $J_k$-isotypic component. The $E(M)$-component has multiplicity one, so restriction gives a unital homomorphism $\mathrm{\Delta }\to {\mathrm{End}}_{J_k}(V_{E(M)})=k$. The equality on the right follows from the full matrix block of the split algebra $J_k$. The homomorphism's kernel is zero because $\mathrm{\Delta }$ is a division algebra, and its image contains all scalar maps. Thus $\mathrm{\Delta }=k$.

We verify explicitly that $\mathrm{\Delta }=k$ implies absolute simplicity here. A finite direct sum of copies of a simple module is semisimple: by induction, every submodule has a complement. Indeed, in $S\oplus U$, split its intersection with $U$ by the induction hypothesis, quotient that intersection, and consider its projection to $S$; this is zero or $S$, and in the latter case the remaining submodule is the graph of a homomorphism from $S$ into the remaining summand of $U$. That graph has the remaining $U$ as a complement. Now choose a $k$-basis $m_1,\dots ,m_r$ of $M$. The image of $H\to M^r$, $h\mapsto (h{m}_1,\dots ,h{m}_r)$, is a submodule. If it were proper, semisimplicity would give a nonzero map $M^r\to M$ vanishing on that image. Its components belong to ${\mathrm{End}}_H(M)=k$, so they are scalars $c_i$, not all zero. Evaluating at $h=1$ would give ${\sum }_i{c}_i{m}_i=0$, contrary to basis independence. Thus the map is surjective. Every prescribed list of images of the basis is realized by some $h$, proving that $H\to {\mathrm{End}}_k(M)$ is surjective. After any field extension, its image remains the full matrix algebra acting on the extended vector space; its standard module is simple. Hence every $M$ is absolutely simple, and $H$ is split.