A characteristic-two specialization collapses four weighted cell labels

Since the only finite edge has even label $2$, the assignments $L(s)=1,L(t)=2$ extend to a weight function by [F2]. The relation $(st{)}^2=1$ gives $st=ts$, so every element is one of $1,s,t,st$. The map to $(\mathbb{Z}\mathrm{/}2\mathbb{Z}{)}^2$ sending $s,t$ to the two standard basis vectors separates these four elements; their lengths are $0,1,1,2$.

In $k={\mathbb{F}}_2$, both $3$ and $3^2$ reduce to $1$, and their negatives and inverses also reduce to $1$. Put $x=T_s-1$ and $y=T_t-1$ in $H_k$. The specialized quadratic relations and characteristic two give $x^2=y^2=0$, and $x,y$ commute, so the generators define a surjection $k[x,y]\mathrm{/}(x^2,y^2)\to H_k$. Its source has basis $1,x,y,xy$, while [F3] and the four-element group from step 1.1 give $H_k$ dimension four with standard basis $1,T_s,T_t,T_{st}$; since $T_s=1+x$ and $T_t=1+y$, the map is surjective between four-dimensional spaces and is an isomorphism. The ideal $\mathfrak{r}=(x,y)$ satisfies ${\mathfrak{r}}^3=0$ and $H_k\mathrm{/}\mathfrak{r}\cong k$. If $M$ is a simple $H_k$-module, then $\mathfrak{r}M$ is a submodule; if it equalled $M$, nilpotence would imply $M={\mathfrak{r}}^{3}M=0$. Thus $\mathfrak{r}M=0$, so $M$ is the unique one-dimensional simple module of $H_k\mathrm{/}\mathfrak{r}$.

Let $C_1=1$, $C_s=T_s+v^{-1}$, and $C_t=T_t+v^{-2}$. The quadratic relations give $C_s^2=(v+v^{-1})C_s$ and $C_t^2=(v^2+v^{-2})C_t$. Since $s,t$ commute, $C_{st}:=C_s{C}_t=C_t{C}_s=T_{st}+v^{-2}{T}_s+v^{-1}{T}_t+v^{-3}$. The standard basis and Bruhat subword property in [F1,F3] show triangularity with negative lower coefficients; bar invariance and uniqueness in [F3] identify the KL elements. The complete left products are $C_s{C}_1=C_s$, $C_t{C}_1=C_t$, $C_s{C}_s=(v+v^{-1})C_s$, $C_t{C}_s=C_{st}$, $C_s{C}_t=C_{st}$, $C_t{C}_t=(v^2+v^{-2})C_t$, $C_s{C}_{st}=C_s^2{C}_t=(v+v^{-1})C_{st}$, and $C_t{C}_{st}=C_s{C}_t^2=(v^2+v^{-2})C_{st}$. Right products agree because the algebra is commutative. Hence the only non-loop arrows are $1\to s$, $1\to t$, $s\to st$, and $t\to st$, so all four cells are singletons. In the $\{1\}$-quotient $C_s,C_t$ vanish, giving $(T_s,T_t)=(-v^{-1},-v^{-2})$; in the $\{s\}$-quotient $C_s^2$ gives $T_s=v$ and $C_t{C}_s=C_{st}$ vanishes, giving $T_t=-v^{-2}$; in the $\{t\}$-quotient $C_s{C}_t=C_{st}$ vanishes, giving $T_s=-v^{-1}$, and $C_t^2$ gives $T_t=v^2$; in the $\{st\}$-quotient the last two products give $(T_s,T_t)=(v,v^2)$. By [F4] these are well-defined cell modules, with exactly the four action pairs in the counterexample.

At $v=3$, the two possible $T_s$-eigenvalues $3,-3^{-1}$ are distinct in $K$, since $3+3^{-1}=10\mathrm{/}3\mathrm{\ne }0$. The two possible $T_t$-eigenvalues $3^2,-3^{-2}$ are distinct since $3^2+3^{-2}=82\mathrm{/}9\mathrm{\ne }0$. Thus the four pairs listed in this counterexample label four distinct one-dimensional simple $H_K$-modules. Each has an $O$-stable lattice $\mathrm{\Lambda }=O$, because all four eigenvalues lie in $O$.

Each of the four lattices from step 3.1 reduces to the one-dimensional character $T_s=T_t=1$, which is the unique simple module found in step 2.1. Each reduction is itself simple, so its composition multiplicity is one. Hence the decomposition matrix has four rows and one column, all entries equal to $1$, by [F5]. The four generic labels have collapsed to one special simple label, proving the claimed failure of universal label preservation. This calculation does not decide whether some subset satisfies the canonical-basic-set conditions.