A semisimple diagonal subalgebra need not force same-label projective socles

The multiplication rules give basis $e_0,e_1,u,v$, with $u=e_{1}u{e}_0$, $v=e_{0}v{e}_1$, and all products of two arrows zero. The arrow span $J=\mathbb{C}u\oplus \mathbb{C}v$ is a two-sided ideal with $J^2=0$ and $B\mathrm{/}J\cong \mathbb{C}\oplus \mathbb{C}$. The diagonal subalgebra $A=\mathbb{C}{e}_0\oplus \mathbb{C}{e}_1\cong {\mathbb{C}}^2$ has regular left module $A=\mathbb{C}{e}_0\oplus \mathbb{C}{e}_1$, a direct sum of two one-dimensional simple modules, so it is semisimple under the Statement's convention.

First verify projectivity under [F1]. The regular left module $B$ is free of rank one: given a surjection $q:E\twoheadrightarrow M$ and a homomorphism $f:B\to M$, choose $e\in E$ with $q(e)=f(1)$ and define the lift $\tilde{f}(b)=be$. Thus $B$ is projective. Since $1=e_0+e_1$ and $e_0{e}_1=e_1{e}_0=0$, $B=B{e}_0\oplus B{e}_1$ as a left module. If $P$ is a direct summand of a projective module $B$, then for any $f:P\to M$, compose with the projection $B\to P$, lift that map through $q$ using projectivity of $B$, and restrict the lift to $P$; hence $P$ is projective. In particular $B{e}_0$ and $B{e}_1$ are projective. They have bases $B{e}_0=\mathbb{C}{e}_0\oplus \mathbb{C}u$ and $B{e}_1=\mathbb{C}{e}_1\oplus \mathbb{C}v$. Their top quotients are $B{e}_0\mathrm{/}\mathbb{C}u\cong S_0$ and $B{e}_1\mathrm{/}\mathbb{C}v\cong S_1$. Each quotient map is a projective cover: for $B{e}_0$, if a submodule $H$ satisfies $H+\mathbb{C}u=B{e}_0$, it contains some $a{e}_0+bu$ with $a\mathrm{\ne }0$; applying $e_0$ gives $a{e}_0\in H$, then $u=u{e}_0\in H$, so $H=B{e}_0$. The same argument with $e_1$ and $v=v{e}_1$ proves superfluity for the second kernel. The lines $\mathbb{C}u\cong S_1$ and $\mathbb{C}v\cong S_0$ are simple. Every vector $a{e}_0+bu\in B{e}_0$ with $a\mathrm{\ne }0$ generates $e_0$ and $u$, so no proper simple submodule of $B{e}_0$ is outside $\mathbb{C}u$; the analogous calculation in $B{e}_1$ shows every proper simple submodule lies in $\mathbb{C}v$. Hence [F3] gives $\mathrm{Soc}(B{e}_0)=S_1$ and $\mathrm{Soc}(B{e}_1)=S_0$, with the labels swapped.

The linear map $\ast$ reverses each defining multiplication relation: for example, $(e_{1}u{)}^{\ast }=v{e}_1=v=(u{)}^{\ast }$ and $(u{e}_0{)}^{\ast }=e_{0}v=v=(u{)}^{\ast }$, while $(uv{)}^{\ast }=vu=0$; the idempotent and unit relations are preserved as well. Hence it extends to the stated anti-involution. In the category of finite-dimensional $\mathbb{Z}$-graded $B$-modules and degree-preserving maps, define $D(M{)}_h=M_h^{\ast }$ with action $(b\cdot f)(m)=f(b^{\ast }m)$. Reversing multiplication verifies the action law. The functor is exact because finite-dimensional vector-space duality is exact, and the degree-preserving evaluation maps $M\to D^2(M)$ are natural isomorphisms, so $D$ is a contravariant duality. If $f\in D(M{)}_h$, then $u\cdot f$ is supported on $M_{h+1}$, because $v$ lowers its argument's degree; similarly $v\cdot f\in D(M{)}_{h-1}$. Thus the dual respects the assigned degree shifts. Each $S_i$ is concentrated in degree $0$ by the Statement's convention and fixed by the dual idempotent action, so $D(S_i)\cong S_i$. Exact degree-preserving duality fixing both simple labels therefore coexists with the swapped ungraded projective socles in Step 2.1. This algebraic model is not a VOA and makes no claim about any VOA's Zhu algebra.