Degenerate parameters $Q_{ij}$ destroy the faithful polynomial-representation route

(a) Define a $k$-algebra $E$ with $k$-basis $\{x_1^a{x}_2^{b}e(ij),x_1^a{x}_2^{b}e(ji),x_1^a{x}_2^b\tau e(ij),x_1^a{x}_2^b\tau e(ji){\}}_{a,b\ge 0}$ and multiplication determined by: $x_1,x_2$ commuting with each other and with the idempotents; $\tau e(ij)=e(ji)\tau$, $\tau e(ji)=e(ij)\tau$; ${\tau }^2=0$; and moving a dot across a crossing permutes the two variables. Associativity holds because the dot permutation is an algebra automorphism of $k[x_1,x_2]$, the idempotents are orthogonal, and every product containing two consecutive crossings or an idempotent mismatch is zero. The presentation of $R({\alpha }_i+{\alpha }_j)$ with $Q=0$ is satisfied by this model, so the universal property gives a surjection $R({\alpha }_i+{\alpha }_j)\twoheadrightarrow E$, while the same Coxeter/dot reduction that proves spanning [F2] shows the model basis spans $R({\alpha }_i+{\alpha }_j)$; a linear dependence among those spanning elements in $R$ would map to a dependence among basis elements of $E$, hence they are independent and the PBW basis of the statement holds for $Q=0$. This proves (a) without applying the nondegenerate PBW theorem outside its hypotheses.

(b) A product $P_{ij}(u,v)P_{ji}(v,u)=0$ in the domain $k[u,v]$ forces one factor to be zero, so no factorization has both factors nonzero; taking $P_{ij}=0$, $P_{ji}=1$ and applying the formula of Faithful polynomial representation of a KLR algebra defines a legitimate action: on the $ij$ summand the crossing operator vanishes, on the $ji$ summand it is the swap. It kills the element ${\tau }_{1}e(ij)$, which is nonzero by the basis of step 1.1, so this action (and by the same argument every factorized action, since at least one corner factor vanishes) is not faithful; this is exactly the failure of the faithfulness hypothesis [F3].

(c) For the non-hermitian pair, associativity of the word ${\tau }_1{\tau }_1{\tau }_1$ on $e(ij)$ computed in the two groupings gives, using relation (4) on the $ij$ corner for the right grouping and on the $ji$ corner after passing through ${\tau }_1$ for the left grouping, $(Q_{ij}(x_2,x_1)-Q_{ji}(x_1,x_2)){\tau }_{1}e(ij)=0$; the dot slides for distinct colours are used to move dots across the crossings in the outer factors. Substituting $Q_{ij}(u,v)=u+2v$ and $Q_{ji}(u,v)=u+v$ gives $(x_2+2x_1)-(x_1+x_2)=x_1$, so $x_1{\tau }_{1}e(ij)=0$; applying the dot slide ${\tau }_1{x}_2=x_1{\tau }_1$ on the $ij$ corner gives ${\tau }_1{x}_{2}e(ij)=x_1{\tau }_{1}e(ij)=0$. The element $x_1{\tau }_{1}e(ij)$ belongs to the spanning family $\{x_1^a{x}_2^b{\tau }_1^{\epsilon }e(\nu )\}$ and is nonzero in the free algebra of the presentation, so that spanning family is not linearly independent and the graded PBW claim fails; equivalently, Rouquier’s PBW criterion shows that PBW is equivalent to the hermitian symmetry $Q_{ij}(u,v)=Q_{ji}(v,u)$, which the pair violates, and the failure is a linear dependence rather than a degree mismatch.