KLR Algebras, Induction, and Quantum-Serre Categorification — Examples

(a) In weight ${\alpha }_i$ the only sequence is $(i)$, and there are no crossings, so the presentation reduces to the single generator $x_1$ with relation (2) $x_{1}e(i)=e(i)x_1$; hence $R({\alpha }_i)=k[x_1]e(i)$ with basis $\{x_1^a\}$, $\mathrm{deg}{x}_1=2d_i$ by The KLR grading and the diagrammatic multiplication, and $\mathrm{g}\mathrm{d}\mathrm{i}\mathrm{m}R({\alpha }_i)={\sum }_{a\ge 0}{q}^{2d_{i}a}=(1-q^{2d_i}{)}^{-1}$. A graded simple is cyclic and bounded below; the positive-degree central $x_1$ cannot act bijectively on it, so simplicity forces $x_{1}S=0$. Thus its unique graded simple is $k=R({\alpha }_i)\mathrm{/}(x_1)$ up to shift; $R({\alpha }_i)$ is free of rank one over itself, hence projective, indecomposable (its endomorphism ring is $k[x_1]$, whose only idempotents are $0,1$), and its quotient onto $k$ is a graded projective cover: if a graded submodule $U$ satisfies $U+(x_1)=R$, comparison in degree zero gives $1\in U$, hence $U=R$; the pairing value is the normalization of The KLR Cartan (graded Hom) pairing on the projective Grothendieck group: $P_i^{\psi }{\otimes }_{R({\alpha }_i)}{P}_i\cong k[x_1]$ as a graded space.

(b) In weight $2{\alpha }_i$ there is one sequence $(ii)$ and one crossing ${\tau }_1$, with relations (4) and (6) specialized to $Q_{ii}=0$: ${\tau }_1^2=0$ and, on the equal-colour component, ${\tau }_1{x}_1=x_2{\tau }_1+e(ii)$, ${\tau }_1{x}_2=x_1{\tau }_1-e(ii)$. The PBW basis [F1] is $\{x_1^a{x}_2^b,x_1^a{x}_2^b{\tau }_1{\}}_{a,b\ge 0}$, which is free of rank $2$ over $k[x_1,x_2]$; the graded dimension is the sum of the two geometric series $\frac{1}{(1-q^{2d_i}{)}^2}+\frac{q^{-2d_i}}{(1-q^{2d_i}{)}^2}=\frac{1+q^{-2d_i}}{(1-q^{2d_i}{)}^2}$ because ${\tau }_1$ has degree $-b_{ii}=-2d_i$. For the idempotent: $e_{i,2}^2=x_1{\tau }_1{x}_1{\tau }_1=x_1(x_2{\tau }_1+e(ii)){\tau }_1=x_1{x}_2{\tau }_1^2+x_1{\tau }_1=x_1{\tau }_1=e_{i,2}$ using ${\tau }_1^2=0$, and $e_{i,2}\mathrm{\ne }0$ by the PBW basis; its degree is $2d_i+(-2d_i)=0$.

(b, continued) By [F2] with $n=2$ the element $e_{i,2}=x_1{\tau }_1$ is the divided-power idempotent, $P_{i,2}=R(2{\alpha }_i)\psi (e_{i,2})\{-d_i\}$ is the unique indecomposable projective summand up to shift, and $[P_{i,2}]=\gamma ({\theta }_i^{(2)})$; the shift here is $-\left(\genfrac{}{}{0px}{}{2}{2}\right)d_i=-d_i$.

(Two-strand basis and dimension.) In weight ${\alpha }_i+{\alpha }_j$ the sequences are $ij$ and $ji$, and $S_2=\{1,s_1\}$, so the PBW basis [F3] consists of the four families $x_1^a{x}_2^{b}e(ij)$, $x_1^a{x}_2^b{\tau }_{1}e(ij)$, $x_1^a{x}_2^{b}e(ji)$, $x_1^a{x}_2^b{\tau }_{1}e(ji)$, which is rank $4$ over $k[x_1,x_2]$; grouping by idempotent gives rank $2$ over $k[x_1,x_2]e(ij)\oplus k[x_1,x_2]e(ji)$. The graded dimension is the two geometric series for the dots in degrees $2d_i,2d_j$ in each corner plus the crossing contributions: in the $ij$-corner, ${\tau }_{1}e(ij)$ has degree $-b_{ij}$, and in the $ji$-corner ${\tau }_{1}e(ji)$ has degree $-b_{ji}=-b_{ij}$, giving $\frac{1+q^{-b_{ij}}}{(1-q^{2d_i})(1-q^{2d_j})}$ in each corner, total $\frac{2(1+q^{-b_{ij}})}{(1-q^{2d_i})(1-q^{2d_j})}$.

(Case $a_{ij}=0$.) Here $Q_{ij}=1$ is a nonzero constant with $Q_{ji}=1$, so [F1] gives ${\tau }_1^{2}e(ij)=e(ij)$ and ${\tau }_1^{2}e(ji)=e(ji)$, i.e. ${\tau }_1^2=e(ij)+e(ji)$, and the dot slides ${\tau }_1{x}_1=x_2{\tau }_1$, ${\tau }_1{x}_2=x_1{\tau }_1$ (distinct colours). Right multiplication by ${\tau }_1$ is an isomorphism of graded left modules $R({\alpha }_i+{\alpha }_j)e(ij)\to R({\alpha }_i+{\alpha }_j)e(ji)$ with inverse right multiplication by ${\tau }_1$ on the other side: both composites are ${\tau }_1^2=\mathrm{i}\mathrm{d}$ on the respective corners, and the map is degree-zero because $\mathrm{deg}{\tau }_1=-b_{ij}=0$. Hence $[P_{ij}]=[R({\alpha }_i+{\alpha }_j)e(ij)]=[R({\alpha }_i+{\alpha }_j)e(ji)]=[P_{ji}]$ in $K_0(R)$.

(Case $a_{ij}=-1$.) With $P_{ij}=u+v$, $P_{ji}=1$ we have $P_{ij}(u,v)P_{ji}(v,u)=(u+v)\cdot 1=Q_{ij}(u,v)$, a valid factorization with both factors nonzero, so [F2] applies: ${\tau }_1(fe(ij))=(x_1+x_2)f(x_2,x_1)e(ji)$ and ${\tau }_1(fe(ji))=1\cdot f(x_2,x_1)e(ij)$. Applying twice, ${\tau }_1^2(fe(ij))=(x_1+x_2)f$ and ${\tau }_1^2(fe(ji))=(x_1+x_2)f$, so ${\tau }_1^{2}e(\nu )=Q_{{\nu }_1{\nu }_2}(x_1,x_2)e(\nu )$ on both summands, as required by [F1]. The freeness statements are step 1.1. For the Serre relation, apply [F4] with $a_{ij}=-1$: the sum runs over $a+b=2$ with signs $+,-,+$, i.e. $[P_{i,2}][P_j]-[P_{i,1}][P_j][P_{i,1}]+[P_j][P_{i,2}]=0$, which in the divided-word notation is $[P_{i(2)j}]-[P_{iji}]+[P_{ji(2)}]=0$, i.e. $[P_{iji}]=[P_{i(2)j}]+[P_{ji(2)}]$.

(a) Specialize relation (4) of [F1]. On an equal-colour pair ${\nu }_r={\nu }_{r+1}$ one has $Q_{{\nu }_r{\nu }_{r+1}}=Q_{ii}=0$ or $Q_{jj}=0$, so ${\tau }_r^{2}e(\nu )=0$. On $(i,j)$ one has $Q_{ij}(x_r,x_{r+1})=-(x_r-x_{r+1}{)}^2$, and on $(j,i)$ the symmetric formula $Q_{ji}(u,v)=-(u-v{)}^2$ gives the same expression, so ${\tau }_r^{2}e(\nu )=-(x_r-x_{r+1}{)}^{2}e(\nu )$ in both orientations.

(b) On $\nu =(i,j,i)$ the outer colours $i,i$ agree and the middle colour differs, so relation (7) of [F1] applies with $a=1$: the left side is ${\tau }_2{\tau }_1{\tau }_{2}e(\nu )-{\tau }_1{\tau }_2{\tau }_{1}e(\nu )$ in the sign normalization fixed there (relation (6) of the definition, in which ${\tau }_1{x}_1-x_2{\tau }_1=e(\nu )$ on the equal-colour component), and the right side is $(Q_{ij}(x_1,x_2)-Q_{ij}(x_3,x_2))\mathrm{/}(x_3-x_1)$. Substituting $Q_{ij}(u,v)=-(u-v{)}^2$ gives $((x_3-x_2{)}^2-(x_1-x_2{)}^2)\mathrm{/}(x_3-x_1)=(x_3-x_1)(x_3+x_1-2x_2)\mathrm{/}(x_3-x_1)=x_1+x_3-2x_2$, a polynomial. The element $(x_1+x_3-2x_2)e(\nu )$ is nonzero by [F3], since its distinct dot monomials have coefficients $1,1,-2$, and the first two remain nonzero in every characteristic.

(c) In the word ${\tau }_2{\tau }_1{\tau }_{2}e(\nu )$ the factors are applied right to left: the rightmost ${\tau }_2$ acts on the pair of positions $(2,3)$ of $\nu =(i,j,i)$, which carries colours $(j,i)$, so its degree is $-b_{ji}=2$; the middle ${\tau }_1$ then acts on the pair of positions $(1,2)$ of the permuted sequence $(i,i,j)$, which carries equal colours $(i,i)$, so its degree is $-b_{ii}=-2$; the leftmost ${\tau }_2$ acts on the pair of positions $(2,3)$ of $(i,i,j)$, which carries colours $(i,j)$, so its degree is $-b_{ij}=2$. The composite has degree $2+(-2)+2=2$.

The right-hand side $x_1+x_3-2x_2$ is a sum of degree-$2$ monomials, so the correction in (b) is homogeneous of degree $2=-2b_{ij}-2d_i$ (with $b_{ij}=-2$, $d_i=1$); on a simple edge $a_{ij}=-1$ with $d_i=d_j=1$ the analogous correction is the degree-$0$ constant $Q_{ij}(x_1,x_2)-Q_{ij}(x_3,x_2)$ divided by $x_3-x_1$, so the doubled Cartan entry is exactly what raises the degree of the correction by $2$.

(a) Using [F1], $e_{i,2}^2=x_1\tau x_1\tau =x_1(x_2\tau +e)\tau =x_1{x}_2{\tau }^2+x_1\tau =x_1\tau =e_{i,2}$ because ${\tau }^2=0$, and $e_{i,2}\mathrm{\ne }0$ because $x_1\tau$ is a basis monomial; its degree is $2d_i+(-2d_i)=0$. Likewise $\psi (e_{i,2}{)}^2=\tau x_1\tau x_1=(x_2\tau +e)\tau x_1=x_2{\tau }^2{x}_1+\tau x_1=\tau x_1$, so $\psi (e_{i,2})$ is an idempotent of degree $0$. The left ideal $R(2{\alpha }_i)\psi (e_{i,2})$ is the polynomial representation: the map $a\psi (e_{i,2})\mapsto a\cdot 1$ is surjective onto $k[x_1,x_2]$ because $\psi (e_{i,2})1=1$, and its inverse sends $f$ to $f\tau x_1$; the two maps are inverse because $\psi (e_{i,2})$ is the rank-one projection exhibiting $N{H}_2={\mathrm{End}}_{k[x_1,x_2{]}^{S_2}}(k[x_1,x_2])$, so the module is indecomposable projective and its homogeneous endomorphism algebra is the symmetric polynomial ring $k[x_1,x_2{]}^{S_2}$, whose degree-zero part is $k$. As a projective with simple head it is the projective cover of the unique graded simple column $k[x_1,x_2]\mathrm{/}(x_1+x_2,x_1{x}_2)$, of dimension $2$, up to shift. The positive symmetric-center ideal gives its simple quotient; its kernel is graded superfluous by the least-degree argument, or by the matrix equivalence with the connected positively graded center; the displayed shift $-d_i$ is the one fixed in [F2] and does not affect these statements except by transport.

(b) The endomorphism ring computation of step 1.1 shows $R(2{\alpha }_i)\cong {\mathrm{End}}_{k[x_1,x_2{]}^{S_2}}(k[x_1,x_2])$ as a graded module over itself, and this is the direct sum of $2=[2{]}_i!{\mathrm{\mid }}_{q_i=1}$ shifted copies of the polynomial column $P_{i,2}$: the polynomial ring is free of rank $2$ over its symmetric subring with basis $\{1,x_1\}$ (or the Artin monomials $\{1,x_1\}$ for $m=2$), so the duals of the two basis monomials have degrees $0,-2d_i$, so the regular module is $V\oplus V\{-2d_i\}$ for the polynomial column $V$. Since $P_{i,2}=V\{-d_i\}$, this is $P_{i,2}\{d_i\}\oplus P_{i,2}\{-d_i\}$. By [F4] the decomposition into indecomposables is unique up to order and shift, so $P_{i,2}$ is the unique indecomposable projective summand up to shift. The graded multiplicity in the normalization of [F2] is $[2{]}_{q_i}!=[1{]}_{q_i}[2{]}_{q_i}=1\cdot (q_i+q_i^{-1})$, the value recorded in the statement, and at $q_i=1$ it is the ungraded multiplicity $2$; the two polynomial columns carry the permutation shifts of the two Artin monomials and are absorbed by this normalization.

(c) The class identification $[P_{i,2}]=\gamma ({\theta }_i^{(2)})$ is [F2] with $n=2$. The self-pairing is the rank-$n$ value of [F3] at $n=2$: $([P_{i,2}],[P_{i,2}])={\prod }_{a=1}^2(1-q_i^{2a}{)}^{-1}=(1-q_i^2{)}^{-1}(1-q_i^4{)}^{-1}$, which is the Lusztig-normalized value $B_L({\theta }_i^{(2)},{\theta }_i^{(2)})$ by the form comparison.