Projective and simple dual bases in a degenerate cyclotomic block

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.