The Degenerate Ariki Categorification Theorem — Examples

The four monomials and the label sets. The finite alphabet is $I^{+}=\{0,1\}$ and each charge-zero factor is its one-box natural module. Hence its tensor-square basis is $M_{00},M_{01},M_{10},M_{11}$. Weakly increasing rows select $00,01,11$, and weakly decreasing rows select $00,10,11$, so these are respectively $\mathrm{S}\mathrm{t}\mathrm{d}$ and $\mathrm{R}\mathrm{e}\mathrm{v}$; the mixed Bruhat order is $01<10$. The actual coordinate weights are $2{\delta }_0,{\delta }_0+{\delta }_1,{\delta }_0+{\delta }_1,2{\delta }_1$; subtracting each from the vacuum weight $2{\delta }_0$ gives contents $0,{\alpha }_0,{\alpha }_0,2{\alpha }_0$. In particular $I_{{\alpha }_0}=\{(0)\}$ and $I_{2{\alpha }_0}=\{(0,0)\}$: residue sequences have length equal to tower degree, independently of the two tensor factors.

The transition vectors and the reciprocal convention. On this mixed space the ordinary Fock bar is $\psi M_{01}=M_{01}$ and $\psi M_{10}=M_{10}+(s^{-1}-s)M_{01}$, with coefficient bar $s\mapsto s^{-1}$. Direct substitution shows that $L_{10}=M_{10}-s{M}_{01}$ and $T_{10}=M_{10}+s^{-1}{M}_{01}$ are bar fixed in their respective positive- and negative-power triangular charts; the endpoints and $L_{01}$ equal their monomials. Thus $M_{10}=L_{10}+s{L}_{01}$, and the ambient inverse-chart coefficient is $d_{01,10}=s$. Its row gives $P_{01}=M_{01}+s{M}_{10}=s{T}_{10}$. Under $t=s^{-1}$ these same formulas become $L_{10}=M_{10}-t^{-1}{M}_{01}$, $T_{10}=M_{10}+t{M}_{01}$ and $P_{01}=M_{01}+t^{-1}{M}_{10}$; reciprocating the parameter does not transpose the inverse transition matrix.

The normalized projection. The source coproduct gives $F_0{M}_{00}=M_{10}+s^{-1}{M}_{01}=T_{10}$ and $F_0{T}_{10}=(s+s^{-1})M_{11}$. Hence $T_{10}$ lies in the ground-generated constituent and is fixed by its normalized projection. The ground highest vector is $M_{00}$; $E_0{M}_{01}=M_{00}$, so $M_{01}$ is not highest. The same coproduct gives $E_0{M}_{10}=s{M}_{00}$, $F_0{M}_{01}=M_{11}$ and $F_0{M}_{10}=s{M}_{11}$, so both $E_0$ and $F_0$ kill $L_{10}$. Its weight is zero for $h_0$, hence it spans a trivial constituent, complementary to the three-dimensional ground constituent since $T_{10}$ and $L_{10}$ are linearly independent. Projection onto the ground constituent along this line is a module map fixing the ground, so this directly constructs the normalized projection and its off-standard kernel $\mathbb{Q}(s)L_{10}$. Since $T_{10}-L_{10}=(s+s^{-1})M_{01}$, applying the projection gives $D_{01}=\pi M_{01}=T_{10}\mathrm{/}(s+s^{-1})$ and $\pi M_{10}=s{D}_{01}$. The endpoints are fixed, and $\pi L_{10}=0$. At $s=1$ this is the half-sum $(M_{01}+M_{10})\mathrm{/}2$. All divisions here take place over $\mathbb{Q}(s)$; they do not identify the projected costandard lattice with the ambient integral monomial lattice.

The supported blocks. Set $x=x_1$, $y=x_2$, $u=s_1$. The defining relations of The degenerate cyclotomic Hecke algebra give $u^2=1$, $[x,y]=0$, $ux=yu-1$, $uy=xu+1$ and $x^2=0$ in the relevant degree; from $u{x}^2=0$ and the cross relations one obtains $y^{2}u=x+y$, hence $y^2=(x+y)u$, and reducing the relation $y^2=(x+y)u$ decreases the $y$-exponent while killing $x$-exponents at least $2$, so the eight words $1,x,y,xy,u,xu,yu,xyu$ span the degree-$2$ algebra. Map these generators to $(N,-N,-J)$ on $M_2(\mathbb{C})$ with $N=E_{12}$ and $J$ the swap matrix, to $(ϵ,1+ϵ,1)$ on $B_{+}=\mathbb{C}[ϵ]\mathrm{/}{ϵ}^2$, and to $(ϵ,-1+ϵ,-1)$ on $B_{-}=\mathbb{C}[ϵ]\mathrm{/}{ϵ}^2$; direct matrix computation verifies $u^2=1$, $[x,y]=0$, $x^2=0$, $ux=yu-1$ and $uy=xu+1$ in all three factors. The central element $z=x+y$ has images $0$, $1+2ϵ$ and $-1+2ϵ$, and the Hermite polynomials $E_0=(1-z^2{)}^2$, $E_{\pm }=(1-E_0\pm (3z-z^3)\mathrm{/}2)\mathrm{/}2$ map to the three separate identities; each factor is generated by its images of $x$ and $u$, so the map from the span of the eight words onto the eight-dimensional product is surjective and hence, by the spanning bound, an isomorphism. The $I=\{0\}$ spectral content cut keeps the factor $M_2(\mathbb{C})$ at content $2{\alpha }_0$ and discards the adjacent-content factors, so the supported degree-$2$ block is $M_2(\mathbb{C})$ with simple dimension $2$; degree one is $\mathbb{C}[x]\mathrm{/}{x}^2$ with simple dimension $1$ at content ${\alpha }_0$, and degree zero is $\mathbb{C}$. To verify vanishing independently of categorification, suppose an all-zero-residue supported module exists in degree at least three. Its restriction to the first two strands lies in the displayed $M_2$ factor, hence $x_2=-x_1$ and $x_1^2=0$. Since $s_2$ commutes with $x_1$, it also commutes with $x_2$. The affine recursion gives $x_3=s_2{x}_2{s}_2+s_2=x_2+s_2$. But a commuting sum of a square-zero operator and an involution has generalized eigenvalues only $1,-1$: on the two $s_2$ eigenspaces it is $\pm 1+x_2$, each invertible with inverse $\pm 1-x_2$. This contradicts the nilpotence required by residue zero for $x_3$. Every nonzero finite-dimensional algebra has its nonzero regular module, so no such supported algebra exists in degree three or higher.

The decomposition matrix by intrinsic modules. The intrinsic affine-induced Specht presentation in Schur functor images of standard, simple, projective and tilting objects gives, for either single-box label $01$ or $10$, the degree-one module $H_1\mathrm{/}{H}_{1}x=\mathbb{C}$; the two different singleton components have the same charge zero. For $11$, with one box in each component, it gives $H_2\mathrm{/}(H_{2}x+H_{2}y)$, since the inducing algebra is the polynomial algebra on the two strand variables and both act as zero on its one-dimensional inducing module. In the three factors of step 2.2, $x,y$ generate the whole left ideal in $B_{+}$ and $B_{-}$ because $y-x=\pm 1$; in the matrix factor $y=-x=-N$, so the quotient is $M_2\mathrm{/}{M}_{2}N$. Here $M_{2}N$ consists of matrices with first column zero, and the first-column map identifies the quotient with the defining simple ${\mathbb{C}}^2$. The degree-zero Specht is $\mathbb{C}$. Thus each of these four Spechts is the indicated simple, directly giving the displayed $3\times 4$ multiplicity matrix. Independently step 1.2 gives $d_{01,10}(1)=1$ and identity on standard columns, so these actual multiplicities agree with [F1]. There is one simple in each of the three supported blocks by step 2.2, so there is no additional simple $D_{10}$. The standard square submatrix is the identity; the rectangular matrix is not asserted invertible.

Conclusion. Steps 1.1-3.1 verify the transition, projection and decomposition statements, including the inverse-chart entry $d_{01,10}(1)=1$ in place of any negative transposed-inverse value, the identity on standard columns and the absence of an off-standard simple $D_{10}$; step 2.2 verifies the block dimensions $1,1,2$ at contents $0,{\alpha }_0,2{\alpha }_0$ of the supported tower and the vanishing of higher supported degrees. The two further adjacent-residue degree-$2$ blocks of the full algebra are outside this supported window and are treated in the companion showcase; no unprojected monomial is used as a simple class.

The degree-one block. For $I=\{0\}$, level $2$ and charge $(0,0)$ the degree-one cyclotomic quotient is $H_1^{(0,0)}=\mathbb{C}[x_1]\mathrm{/}(x_1^2)$ by the level-two instance of The degenerate cyclotomic Hecke algebra, and it is already supported, so $B=\mathbb{C}[x]\mathrm{/}{x}^2$ with $x=x_1$. Its radical is $(x)$ and its quotient is $\mathbb{C}$, so $B$ is local; the unique simple is $D_{01}=\mathbb{C}$ with $x\mapsto 0$, the regular module $Y_{01}=B$ is its projective cover, and the regular composition series $0\subset (x)\subset B$ has successive quotients $D_{01},D_{01}$, giving $[Y_{01}]=2[D_{01}]$ in $G_0$. Every $B$-linear map $B\to D_{01}$ is determined by the image of $1$, so $\dim \mathrm{Hom}(Y_{01},D_{01})=1$.

The symmetric form on $B$ and self-duality. The coefficient functional $\tau (a+bx)=b$ is a symmetrizing trace on $B$: it is linear, $\tau (uv)=\tau (vu)$ for all $u,v$ (both sides pick out the $x$-coefficient of the product), and its radical is zero because $\tau (x)\mathrm{\ne }0$, so the associated form $⟨u,v⟩=\tau (uv)$ has matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ in the basis $(1,x)$, which is nonsingular. The anti-involution $\ast$ of The degenerate cyclotomic Hecke algebra that fixes $x$ is the identity of $B$ here, so the form is generator-fixing, and it exhibits an isomorphism $Y_{01}\cong Y_{01}^{⊛}$: the dual of the regular module with respect to the trace. This is the actual symmetric-algebra proof of cover self-duality, not an inference from self-duality of the head.

The degree-two block and its simple. With $x=x_1$, $y=x_2$, $u=s_1$ the relations give $u^2=1$, $[x,y]=0$, $ux=yu-1$, $uy=xu+1$, $x^2=0$, and from $u{x}^2=0$ one gets $y^2=(x+y)u$, so the eight words $1,x,y,xy,u,xu,yu,xyu$ span the degree-two algebra; the models $(N,-N,-J)$ on $M_2(\mathbb{C})$ with $N=E_{12}$ and $J$ the swap matrix, $(ϵ,1+ϵ,1)$ and $(ϵ,-1+ϵ,-1)$ on $\mathbb{C}[ϵ]\mathrm{/}{ϵ}^2$ satisfy all relations, and $z=x+y$ is central with images $0,1+2ϵ,-1+2ϵ$; the Hermite polynomials $E_0=(1-z^2{)}^2$, $E_{\pm }=(1-E_0\pm (3z-z^3)\mathrm{/}2)\mathrm{/}2$ map to the separate factor identities. Each factor is generated by its images of $x,u$, so the homomorphism is onto the eight-dimensional product; the eight-word spanning bound proves that the span of the eight words is isomorphic to $M_2(\mathbb{C})\oplus B_{+}\oplus B_{-}$ with $B_{\pm }=\mathbb{C}[ϵ]\mathrm{/}{ϵ}^2$. The supported content cut at $2{\alpha }_0$ keeps the factor $M_2(\mathbb{C})$; its simple module ${\mathbb{C}}^2$ has Gram matrix $J$, since $N^{t}J=JN$ and $(-J{)}^{t}J=J(-J)$, and it is self-dual and projective, so its projective cover is itself and its endomorphism ring is $\mathbb{C}$; degree $0$ contributes $\mathbb{C}$, and Hom-spaces between distinct content blocks vanish because they are different blocks. In particular the one-dimensional Fock weight vector $M_{11}$ is not the two-dimensional module $D_{11}$, and the degree-two projective cover has vector-space dimension $2$ and composition length $1$. Higher supported degrees vanish directly: in an all-zero-residue module on at least three strands, restriction to strands $1,2$ forces $x_2=-x_1$ and $x_1^2=0$ by the matrix factor. Since $s_2$ commutes with $x_1$, it commutes with $x_2$, and the recursion gives $x_3=x_2+s_2$. On either $s_2$ eigenspace this is $x_2\pm 1$, which is invertible, contradicting nilpotence of $x_3$ at residue zero. Thus such a nonzero module cannot exist; applying this to the regular module proves the supported algebra is zero in every degree at least three.

The projected dual bases at parameter one. In the tensor square at parameter one, $f{M}_{00}=M_{01}+M_{10}$, $f(M_{01}+M_{10})=2M_{11}$, and $e$ reverses these moves with coefficients $2,1$. Hence the ground constituent has basis $P_{00},P_{01},P_{11}$ as displayed, and its complementary trivial line is $M_{10}-M_{01}$. Projection along this line gives $D_{01}=\pi M_{01}=\frac{1}{2}(M_{01}+M_{10})$ and fixes the endpoints. The normalized form is determined by $C_0(M_{00},M_{00})=1$, $e\mathrm{/}f$ adjunction and orthogonality of distinct weights: $C_0(P_{01},P_{01})=C_0(M_{00},e{P}_{01})=2$, and $2C_0(P_{11},P_{11})=C_0(P_{01},e{P}_{11})=2$. Thus the mixed vector is $P_{01}=M_{01}+M_{10}$ of $C_0$-norm $2$, so the correct field dual is $D_{01}=\frac{1}{2}(M_{01}+M_{10})$, and the endpoint vectors satisfy $P_{00}=D_{00}=M_{00}$ and $P_{11}=D_{11}=M_{11}$. The Gram matrix in the $D$-basis is $\mathrm{diag}(1,\frac{1}{2},1)$, and multiplying its first-argument basis by $\mathrm{diag}(1,2,1)$ gives the identity matrix, which is the pairing $C_0(P_A,D_B)={\delta }_{AB}$; this directly computes the parameter-one projected-simple dictionary of The dual-canonical, canonical and quasi-canonical Fock bases and Simples, projectives and tiltings map to the three distinguished Fock bases, not an identification of $L_{01}=M_{01}$ with the simple.

The categorified residue operators. In these all-zero-residue blocks the residue projectors are the identities, so the bimodules are the explicit tower-inclusion bimodules. Induction from degree $0$ to the degree-one block gives the regular module of $B$; the supported induction from $B$ to the degree-two block $M_2(\mathbb{C})$ sends the simple $\mathbb{C}$ to ${\mathbb{C}}^2$ and the regular module $B$ to $M_2(\mathbb{C})$, which is two copies of the simple ${\mathbb{C}}^2$; explicitly $M_2{\otimes }_B\mathbb{C}=M_2\mathrm{/}{M}_{2}N$ has dimension two and is the unique matrix-factor simple, whereas $M_2{\otimes }_{B}B=M_2$; restriction of ${\mathbb{C}}^2$ to $B$ is the regular Jordan module $B$ itself (the generator $x$ of $B$ acts on ${\mathbb{C}}^2$ as $N$ with $N^2=0$ and $\dim \ker N=1$), and restriction of $D_{01}=\mathbb{C}$ to degree zero is the one-dimensional $\mathbb{C}$-module. The right $B$-module $M_2$ is free of rank two: each row is a regular $B$-module for right multiplication by $N$, so these induction functors are exact directly. These explicit bimodules realize the residue functors of The degenerate cyclotomic Hecke tower and its residue functors and yield the displayed columns $f{D}_{00}=2D_{01}$, $f{D}_{01}=D_{11}$, $f{D}_{11}=0$ and $e{D}_{00}=0$, $e{D}_{01}=D_{00}$, $e{D}_{11}=2D_{01}$; equivalently, in the $D$-basis, $f$ has columns $(0,2,0{)}^{\mathsf{T}},(0,0,1{)}^{\mathsf{T}},0$ and $e$ has columns $0,(1,0,0{)}^{\mathsf{T}},(0,2,0{)}^{\mathsf{T}}$, so $[e,f]=ef-fe$ is $\mathrm{diag}(2,0,-2)$ in the $s{l}_2$-normalization of Divided-power commutation and the simple $U_{q_i}({\mathfrak{s}\mathfrak{l}}_2)$ string modules.

The normalized finite identifications. On projective classes the same bimodules give $f[Y_{00}]=[Y_{01}]$, $f[Y_{01}]=2[Y_{11}]$, $e[Y_{01}]=2[Y_{00}]$ and $e[Y_{11}]=[Y_{01}]$. On simple classes they give step 4.1. Therefore the maps $[Y_A]\mapsto P_A$ and $[D_A]\mapsto D_A$ intertwine both $e$ and $f$ with the explicit Fock matrices of step 3.1 and send the degree-zero class to the normalized ground. Over $\mathbb{Q}$ each module is ground generated, so these normalized intertwining maps are unique; consequently they agree with any named categorification identification with that normalization, rather than assuming a general bridge to establish this finite instance. Each cover/simple Hom space within a block has dimension one, by step 1.1 and the scalar endomorphism ring in step 2.2, while different degrees give zero Hom in the tower direct sum. Exactness of Hom from a projective makes the pairing additive on short exact sequences, so the pairing matrix is the identity on these bases. Its vector realization is precisely $C_0(P_A,D_B)={\delta }_{AB}$ from step 3.1, and the composition Cartan matrix is $\mathrm{diag}(1,2,1)$.

The index-two lattice and its cyclicity. The divided powers $f^{(2)}=f^2\mathrm{/}2$ and $e^{(2)}=e^2\mathrm{/}2$ give $f^{(2)}{D}_{00}=D_{11}$ and $e^{(2)}{D}_{11}=D_{00}$, and the displayed matrices preserve evenness of the $D_{01}$-coordinate whenever the input has that coordinate even, so the lattice $L_{\mathbb{Z}}=\mathbb{Z}{D}_{00}\oplus 2\mathbb{Z}{D}_{01}\oplus \mathbb{Z}{D}_{11}$ is stable under $e,f$, their divided powers and the toral operators at weights $2,0,-2$. It is generated from the ground because $f{D}_{00}=2D_{01}$ and $f^{(2)}{D}_{00}=D_{11}$. Its elements have $D_{01}$-coordinate even, so it does not contain $D_{01}$; hence it is a proper integral submodule containing the ground, of index $2$ in the full costandard lattice, and $D_{01}$ is not in the integral ground-generated span. Over $\mathbb{Q}$ division by $2$ is available, so the ground vector generates the whole three-dimensional costandard space; the standard lattice is by step 5.1 the lattice with basis $(P_{00},P_{01},P_{11})$, which is integrally generated by its ground under the divided powers and pairs perfectly with the full costandard lattice by step 3.1. This proves that integral costandard generation fails while the full integral duality and the standard divided-power statement survive.

Conclusion. Steps 1.1 and 2.1 verify the degree-one cover, its composition length against its Hom dimension, and the symmetric-form self-duality; step 2.2 and step 3.1 verify the degree-two block, the projected dual bases and the pairing identity; step 4.1 verifies the residue operators, their commutator and the categorified $[Y_{01}]=2[D_{01}]$; step 6.1 verifies the exact index-two lattice, its stability, its failure to contain $D_{01}$ and the rational cyclicity. No step identifies the simple image with the unprojected monomial $L_{01}$, and no self-duality is inferred from the head alone.

The eight ambient labels. There is one bipartition of size zero, two of size one and five of size two: $(\mathrm{\varnothing },\mathrm{\varnothing })$; $(\mathrm{\varnothing },1),(1,\mathrm{\varnothing })$; and $(1,1),(\mathrm{\varnothing },2),(2,\mathrm{\varnothing }),(\mathrm{\varnothing },11),(11,\mathrm{\varnothing })$. The ground-difference column rule at charge $(0,0)$ gives respectively $OO,OU,UO,UU,OR,RO,OC,CO$, so $C0$ has the vertical partition in the first component and $0C$ in the second. Subtracting the ground weights gives degree $0$ for $00$, ${\alpha }_0$ for $01,10$, $2{\alpha }_0$ for $11$, ${\alpha }_0+{\alpha }_1$ for $0R,R0$, and ${\alpha }_0+{\alpha }_{-1}$ for $0C,C0$. Strict columns with weakly increasing rows select $00,01,11,0R,0C$; reverse-standard rows select $00,10,11,R0,C0$. Reading the two columns and inserting in this small window gives ${10}^{\downarrow }=01$, $R{0}^{\downarrow }=0R$, $C{0}^{\downarrow }=0C$ and fixes $00,11$. Component reversal and partition transpose give precisely the displayed ambient permutation.

The full height-two window. With $x=x_1$, $y=x_2$, $u=s_1$ the relations give $u^2=1$, $[x,y]=0$, $ux=yu-1$, $uy=xu+1$ and $x^2=0$; from $u{x}^2=0$ one derives $y^2=(x+y)u$, and reducing by this relation decreases the $y$-exponent while killing $x$-exponents at least two, so the eight words $1,x,y,xy,u,xu,yu,xyu$ span the degree-two algebra. The three models $(N,-N,-J)$ on $M_2(\mathbb{C})$ with $N=E_{12}$ and $J$ the swap matrix, $(ϵ,1+ϵ,1)$ on $B_{+}$, and $(ϵ,-1+ϵ,-1)$ on $B_{-}$ satisfy the relations by direct matrix computation; the central element $z=x+y$ has images $0$, $1+2ϵ$ and $-1+2ϵ$, and the Hermite polynomials $E_0=(1-z^2{)}^2$, $E_{\pm }=(1-E_0\pm (3z-z^3)\mathrm{/}2)\mathrm{/}2$ map to the three separate identities. Each factor is generated by its images of $x$ and $u$, so the map is surjective onto the eight-dimensional product and the spanning bound makes it an isomorphism; hence there are no other height-two blocks and $H_0,H_1$ are the direct factors listed. The two-dimensional simple of $M_2(\mathbb{C})$ carries the generator-fixed contravariant form $J$, since $N^{t}J=JN$ and $(-J{)}^{t}J=J(-J)$; sign is inner on $M_2(\mathbb{C})$ via conjugation by $K=\mathrm{diag}(1,-1)$ (which satisfies $KN=-NK$ and $KJ=-JK$), so the unique simple and its cover are fixed there, while on $B_{\pm }$ sign identifies $ϵ$ with $-ϵ$ and exchanges the two factors, fixing the unique objects of those contents as classes.