Cyclotomic KLR quotients in ranks zero, one and two

The empty word gives $R^{\mathrm{\Lambda }}(0)=k$. One strand has no crossings, so its presentation is $k[x]\mathrm{/}{x}^{{\lambda }_i}$ with $\mathrm{deg}x=2d_i$. For ${\lambda }_i>0$ the monomials $1,x,\dots ,x^{{\lambda }_i-1}$ give the stated Hilbert series. The unique graded simple is $k$ at degree zero, and the regular module is its indecomposable projective cover. Evaluation identifies ${\mathrm{HOM}}_B(B,B{)}^{\psi }$ with $B$, so this is the projective Hom-self-dual normalization, without a vector-dual centering shift. The coefficient-of-$x^{{\lambda }_i-1}$ functional pairs the monomial basis anti-diagonally with unit entries; it is a nondegenerate symmetric Frobenius functional of degree $-2d_i({\lambda }_i-1)$. If ${\lambda }_i=0$, its defining relation is $1=0$, so the block is zero and has no simple; if ${\lambda }_i=1$, it is $k$.

For $n\ge 1$, put $S=S_n$ and $C=S[z]\mathrm{/}(z^n-e_1{z}^{n-1}+\cdots +(-1{)}^n{e}_n)$, embedded in $P_n$ by $z=x_1$. Let $e_a^{\mathrm{\prime }}$ be the elementary symmetric functions of $x_2,\dots ,x_n$. The identities $e_a=e_a^{\mathrm{\prime }}+z{e}_{a-1}^{\mathrm{\prime }}$, $e_0^{\mathrm{\prime }}=1$, express the $e_a^{\mathrm{\prime }}$ recursively in $z,e_1,\dots ,e_a$; the final $e_n^{\mathrm{\prime }}=0$ is precisely the displayed monic relation. Hence $C=k[x_1,e_1^{\mathrm{\prime }},\dots ,e_{n-1}^{\mathrm{\prime }}]$, and the Artin basis for the remaining $n-1$ variables makes $P_n$ free of rank $(n-1)!$ over $C$. On its $S$-basis $1,z,\dots ,z^{n-1}$, multiplication by $z^{{\lambda }_i}$ is repeated $(n-1)!$ times in multiplication by $x_1^{{\lambda }_i}$ on $P_n$. In a full matrix algebra over $S$, left and right matrix units isolate each matrix entry and insert it into any chosen entry. Therefore its two-sided ideal is the matrix ideal of exactly those scalar entries, so the cyclotomic quotient is the polynomial-column matrix algebra over their quotient of $S$.

Define $ϵ:C\to S$ to be the coefficient of $z^{n-1}$ in the power basis. The matrix $ϵ(z^{a+b})$ for $0\le a,b<n$ is zero for $a+b<n-1$ and one for $a+b=n-1$, hence anti-triangular with unit anti-diagonal and invertible over $S$. Its dual basis is therefore integral over $S$. Pairing multiplication by $z^{{\lambda }_i}$ with this invertible trace matrix shows that its entry ideal is exactly the ideal generated by $ϵ(z^{{\lambda }_i+a+b})$, $0\le a,b<n$. Put $h_r=0$ for $r<0$ and $h_0=1$; the monic equation gives the recurrence $h_r=e_1{h}_{r-1}-e_2{h}_{r-2}+\cdots +(-1{)}^{n-1}{e}_n{h}_{r-n}$. Induction on the power gives $ϵ(z^{n-1+r})=h_r$, including the stated initial values. All indices in the entry list are between ${\lambda }_i-n+1$ and ${\lambda }_i+n-1$. The first consecutive $n$ indices $h_{{\lambda }_i-n+1},\dots ,h_{{\lambda }_i}$ occur, and the recurrence expresses every higher one using them. Thus the exact corner ideal is $(h_{{\lambda }_i-n+1},\dots ,h_{{\lambda }_i})$.

If $n>{\lambda }_i$, the consecutive list contains $h_0=1$, and the quotient is zero. If $1\le n\le {\lambda }_i$, every index in that list is positive, so the ideal is contained in the augmentation ideal of $S$ and the corner quotient is nonzero. It is finite dimensional by F2 and the full matrix description of step 1.2. Its degree-zero part is $k$, and its positive-degree ideal is nilpotent because the graded vector space is finite and positively graded. It is therefore graded local with exactly one simple up to shift, its augmentation quotient. Matrix equivalence gives exactly one graded simple column in each nonzero root block. For $n=0$ there is the vacuum, so the total number of normalized rank-one simples is exactly ${\lambda }_i+1$.

The augmentation simple column is $P_n{\otimes }_{S}k$, the ordinary coinvariant polynomial representation with its Artin basis, centered by $-d_i\left(\genfrac{}{}{0px}{}{n}{2}\right)$. Its Hilbert series is $q_i^{-\left(\genfrac{}{}{0px}{}{n}{2}\right)}{\prod }_{r=1}^n(1+q_i^2+\cdots +q_i^{2(r-1)})=[n{]}_{q_i}!$. The integral perfect Artin trace in F1 pairs complementary degrees, so this centering makes its generator-fixing finite vector dual isomorphic to itself. This yields the unique normalized simple in every nonzero block. The same matrix model retains the symmetric corner of step 2.1 and its polynomial-column projectives, proving all ranks of part (c), rather than identifying the whole algebra with a commutative quotient.

At $n={\lambda }_i=2$, the corner is $k[e_1,e_2]\mathrm{/}(h_1,h_2)=k$, since $h_1=e_1$, $h_2=e_1^2-e_2$. The polynomial column is $C_2=k[x_1,x_2]\mathrm{/}(x_1+x_2,x_1{x}_2)$, with basis $1,x_1$ and $x_2=-x_1$, $x_1^2=0$, so its centered column $S_2=C_2\{-d_i\}$ has Hilbert series $q_i^{-1}+q_i$. The block is ${\mathrm{End}}_k(S_2)$; its unique simple column is also its normalized indecomposable projective. The regular left module is $S_2\otimes S_2^{\ast }$, with the second factor giving shifts $\pm d_i$, and thus its class is $[2{]}_{q_i}[P_2]$ in projective $K_0$ and $[2{]}_{q_i}[S_2]$ in module $G_0$. The matrix identification and these formulas hold in characteristic two as well. A one-dimensional zero-dot module would violate the equal-colour mixed relation with unit right side, so no such simple exists.

Finally the coprime polynomials $u,u-1$ give $k[u]\mathrm{/}(u(u-1))=k\oplus k$, with two ordinary simples at root ${\alpha }_i$. This is the ungraded monic deformation: ordinary finiteness and symmetry retain their stated scope, but it is not the standard homogeneous graded rank-one highest-weight identification. Parts (a)–(d) follow from steps 1.1–5.1 with arbitrary symmetrizer and every field, and part (e) from this finite coprime-factor calculation.