Cyclotomic KLR Algebras and Highest-Weight Categorification — Examples

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.

Projective Hom self-duality. Evaluation ${\mathrm{HOM}}_B(B,B)\cong B$ gives $P^{\mathrm{\#}}={\mathrm{HOM}}_B(B,B{)}^{\psi }\cong B=P$ directly, the generator-fixing anti-involution producing the same left module; hence the canonical normalized projective is $P$ with no shift.

Vector duality and the shift. The functional $ϵ(x^a)=0$ for $a<\lambda -1$ and $ϵ(x^{\lambda -1})=1$ is the coefficient-of-$x^{\lambda -1}$ trace; it has degree $-2(\lambda -1)$ and its anti-diagonal pairing $ϵ(x^a{x}^b)={\delta }_{a+b,\lambda -1}$ is perfect over every field, so $D_{\psi }B\cong B\{-2(\lambda -1)\}$. Since $D_{\psi }(P\{s\})=D_{\psi }P\{-s\}$, vector self-duality of $P\{s\}$ solves $-2(\lambda -1)-s=s$, namely $s=-(\lambda -1)$, while projective Hom self-duality solves $-s=s$, namely $s=0$.

The general defect shift and the boundary. For a general normalized cyclotomic projective the same computation with the symmetric theorem gives $D_{\psi }P\cong P^{\mathrm{\#}}\{-2{\mathrm{def}}_{\mathrm{\Lambda }}(\beta )\}$, which is a comparison of the two dualities and does not redefine the projective normalization. At $\lambda =1$ the two centers agree; at $\lambda =2$ they differ by $1$ as computed; at $\lambda =0$ the defining relation is $1=0$ and the block vanishes, so there is no indecomposable projective here. The trace is never halved to pick the lower-basis class.

The word class. By [F1] the word projective decomposes as $(q+q^{-1})$ copies of $P_{i,2}$, so $[R^{\mathrm{\Lambda }}(2{\alpha }_i)]=(q+q^{-1})[P_{i,2}]$; under the K_0 isomorphism of the integral projective theorem this is the class of the ordinary product $F_i^2{v}_{2{\mathrm{\Lambda }}_1}$.

In the rank-one module $V(2{\mathrm{\Lambda }}_1)$, the Kashiwara string is $v,F_{i}v,F_i^{(2)}v$. These vectors generate its zero chart; they are bar fixed and generate the corresponding infinity chart. In weight $2{\alpha }_i$ the divided-power integral lattice is $A{F}_i^{(2)}v$: any divided word of that weight is an $A$-multiple of $F_i^{(2)}v$, since $F_i^{2}v=[2{]}_i{F}_i^{(2)}v$. Since an integral Laurent polynomial regular at both $0$ and infinity is an integer constant, the balanced intersection in that weight is $\mathbb{Z}{F}_i^{(2)}v$, and its positive string residue selects $F_i^{(2)}v$ as the lower global lift.

By [F2], $F_i^{2}v=[2{]}_i{F}_i^{(2)}v$. The coefficient $[2{]}_i=q+q^{-1}$ is a nonunit: evaluation at $q=1$ sends it to $2$, while an invertible Laurent polynomial must evaluate to a unit of $\mathbb{Z}$. Hence the word class cannot be the lower global vector identified in step 1.2.

This refutes the assertion that word-projective classes are automatically lower global vectors. Here the integral divided-power class $[P_{i,2}]$ gives $F_i^{(2)}v$ directly, whereas the regular word class has the nonunit coefficient $q+q^{-1}$. The finite example makes no general impossibility claim about extracting a basis from additional categorical or quantum-module structure.

The blocks. The rank-zero block is $k$ and the rank-one quotient is $k[x]\mathrm{/}(x^2)$ with basis $1,x$ and degrees $0,2$. Restriction of the vacuum has zero target. Thus the relevant functor values on the vacuum are $F_i^{\mathrm{\Lambda }}k=R^{\mathrm{\Lambda }}({\alpha }_i)$ and $F_i^{\mathrm{\Lambda }}{E}_i^{\mathrm{\Lambda }}k=0$.

The commutator. The relation of [F2] evaluated on the vacuum gives $E_i^{\mathrm{\Lambda }}{F}_i^{\mathrm{\Lambda }}k\cong k\oplus k\{2\}$, which is $R^{\mathrm{\Lambda }}({\alpha }_i)$ by [F1], so $q^{-2}{F}_i^{\mathrm{\Lambda }}{E}_i^{\mathrm{\Lambda }}\oplus \mathrm{I}\mathrm{d}\oplus q^2\mathrm{I}\mathrm{d}\cong E_i^{\mathrm{\Lambda }}{F}_i^{\mathrm{\Lambda }}$ holds at the first nontrivial weight.

The Grothendieck commutator. With the normalizations of [F3]: projective $E_i{F}_i[1_{\mathrm{\Lambda }}]=q^{-1}(1+q^2)[1_{\mathrm{\Lambda }}]=(q+q^{-1})[1_{\mathrm{\Lambda }}]$ and $F_i{E}_i[1_{\mathrm{\Lambda }}]=0$; KK module $F$ from the vacuum also carries $q^{-1}$ while KK $E$ is raw, and the canonical module action shifts $E$ and leaves $F$ raw, giving the same result in both module conventions. All agree with $(q^2-q^{-2})\mathrm{/}(q-q^{-1})$, the mixed relation evaluated at $K_i=q^2$. This verifies the commutator identity at the first nontrivial weight and is a finite rank-one computation using no choice.

The blocks and their composition structure. By [F1] the three nonzero blocks have graded dimensions $1$, $1+q^2$ and $(q+q^{-1}{)}^2$; the first regular module $B_1$ has two composition factors $S_1,S_1\{2\}$ with graded dimensions $1+q^2$, and the second block is the endomorphism matrix algebra of the graded column $S_2$, whose regular left module is the column tensored with its graded vector dual of dimension $q+q^{-1}$.

The Cartan pairing. The normalized cover-to-head Hom space is $k$ in degree zero and there are no other simple types in a block, so quotient-cover evaluation of [F2] identifies the tensor pairing with $T(P_r,S_s)={\delta }_{rs}$, with distinct roots paired by zero. The integral identification of [F3] maps $P_0,P_1,P_2$ to $v,Fv,F^{(2)}v$ and $[B_2]=(q+q^{-1})[P_2]$ to $F^{2}v$, and in $G_0$ the regular classes are $[B_1]=(1+q^2)[S_1]$ and $[B_2]=(q+q^{-1})[S_2]$.

The restricted dual. The canonical restricted-dual map sends $S_r$ to the form-dual of the corresponding divided lower vector, and since $T$ pairs $P_r$ with $S_s$ by ${\delta }_{rs}$ it does not identify $B_1$ with $S_1$; the scalar $q+q^{-1}$ is not removed by the pairing. This is a cover/simple pairing computation, not a claim that the Cartan matrix is all ones or that a trace removes a denominator, and it is a finite computation using no choice.