Unequal-Parameter Hecke Cells and Representations: Examples and Counterexamples

Since $m(s,t)=4$ is even, the weight extension criterion [F1] permits $L(s)=1,L(t)=2$. Put $r=st$. The presentation gives $t=sr$ and $srs=ts=r^{-1}$. Hence every word is $r^j$ or $r^{j}s$ for some $j\in \{0,1,2,3\}$, so $\mathrm{\mid }W\mathrm{\mid }\le 8$. Map $s$ to reflection across the horizontal axis and $t$ to reflection across the diagonal $y=x$. Their product is a rotation by $90$ degrees, so these transformations generate all eight symmetries of the square and define a surjection from $W$ onto that group. Thus $W$ is that dihedral group. In particular $1,s,t,st,ts$ are distinct; $s,t$ have length one and $st,ts$ length two. The reduced-subword theorem [F1] then gives the lower intervals below $st$ and $ts$ as $\{1,s,t,st\}$ and $\{1,s,t,ts\}$.

By [F3], $C_s=T_s+v^{-1}$ and $C_t=T_t+v^{-2}$. Expanding in the standard basis [F2] gives $C_s{C}_t=T_{st}+v^{-2}{T}_s+v^{-1}{T}_t+v^{-3}$. Each lower term is indexed by a strict Bruhat subword from step 1.1 and has negative exponent. Since $C_s,C_t$ are bar invariant, their product is bar invariant; uniqueness in [F3] identifies it with $C_{st}$. Reversing the product gives $C_t{C}_s=T_{ts}+v^{-1}{T}_t+v^{-2}{T}_s+v^{-3}=C_{ts}$.

The quadratic relation gives $C_s^2=T_s^2+2v^{-1}{T}_s+v^{-2}=(v+v^{-1})T_s+1+v^{-2}=(v+v^{-1})C_s$. Using $L(t)=2$ gives $C_t^2=(v^2+v^{-2})C_t$. Also $C_s{C}_1=C_s$, $C_t{C}_1=C_t$, $C_t{C}_s=C_{ts}$, and $C_s{C}_t=C_{st}$. By [F4], these products give the four directed transitions displayed in the Statement and diagram. The proof establishes only these edges and makes no claim that the list is exhaustive.

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.

The generic algebra has two distinct roots $v,-v^{-1}$ over $\mathbb{Q}(v)$, so it is split with its two displayed one-dimensional characters. The KL basis is $C_1=1$, $C_s=T_s+v^{-1}$, and $C_s^2=(v+v^{-1})C_s$ by expansion of the quadratic relation. Therefore $a(1)=0$, $a(s)=1$, and $\tau (C_1)=1$, $\tau (C_s)=v^{-1}$ give $\mathcal{D}=\{1,s\}$ and $n_1=n_s=1$. The only nonzero $\gamma$ are ${\gamma }_{1,1,1}={\gamma }_{s,s,s}=1$.

P2, P3, P5, P6, P7 and P8 follow immediately from these two diagonal $\gamma$ entries and $s^2=1$. P4 holds for the only products $C_1{C}_w=C_w=C_w{C}_1$ and $C_s^2=(v+v^{-1})C_s$. For P15′, the condition $a(w)=a(y)$ forces $w=y$, since the two values are distinct. Its left side is then ${\delta }_{w,x^{\mathrm{\prime }}}{h}_{x,w,w}$ and its right side is ${\delta }_{y,x^{\mathrm{\prime }}}{h}_{x,w,y}$, identical because $w=y$. Thus every required P property is checked, not imported from a general cell theorem.

For the two characters write $\tau =c_{+}^{-1}{\chi }_{+}+c_{-}^{-1}{\chi }_{-}$. Evaluating on $1$ and $T_s$ gives $c_{+}^{-1}+c_{-}^{-1}=1$ and $v{c}_{+}^{-1}-v^{-1}{c}_{-}^{-1}=0$. Hence $c_{+}=1+v^2$ and $c_{-}=1+v^{-2}$. Their lowest terms give $(f_{+},{\alpha }_{+})=(1,0)$ and $(f_{-},{\alpha }_{-})=(1,1)$ as in [F2]. Both specialize to zero at $v=i$, whereas the leading integers stay nonzero.

At $v=i$ both roots equal $i$, so $H_k=\mathbb{C}[T_s]\mathrm{/}((T_s-i{)}^2)$. Put $\epsilon =T_s-i$. The element $\epsilon$ is nonzero and square-zero, since $1,T_s$ form the rank-two basis. On a simple module, the nilpotent ideal $(\epsilon )$ must act by zero: otherwise its image would be the whole simple and iteration would contradict its nilpotence. Thus the quotient $H_k\mathrm{/}(\epsilon )=\mathbb{C}$ gives the unique simple $M$. Both global free rank-one character models reduce to it, so the decomposition column is $(1,1{)}^t$. Its unique least Schur exponent is zero, with multiplicity one, selecting exactly the first label. These are also precisely the axioms and uniqueness conclusion in [F3]. This verifies concretely that full-Schur vanishing is compatible with the general nonsemisimple basic-set theorem.