Arakawa–Suzuki Functors and Higher-Level Schur–Weyl Duality — Examples

The permutation identities. Since $P_{ab}$ exchanges two tensor positions it is an involution, $P_{ab}^2=1$; and evaluating on the eight words gives the two braid-type identities used below, $P_{12}{P}_{02}=P_{01}{P}_{12}$ and $P_{02}{P}_{12}=P_{12}{P}_{01}$, together with their transposes $P_{01}{P}_{02}=P_{12}{P}_{01}$ and $P_{02}{P}_{01}=P_{01}{P}_{12}$ (both sides of each identity send a word to the same word).

The four relations. First $s_1^2=P_{12}^2=1$. Next, $[x_1,x_2]=P_{01}{P}_{02}-P_{02}{P}_{01}+P_{01}{P}_{12}-P_{12}{P}_{01}$, and the identities $P_{01}{P}_{02}=P_{12}{P}_{01}$, $P_{02}{P}_{01}=P_{01}{P}_{12}$ of step 1.1 cancel the four 3-cycle terms pairwise. Finally $s_1{x}_2-x_1{s}_1=P_{12}{P}_{02}+P_{12}^2-P_{01}{P}_{12}=P_{01}{P}_{12}+1-P_{01}{P}_{12}=1$ and, symmetrically, $x_2{s}_1-s_1{x}_1=P_{02}{P}_{12}+P_{12}^2-P_{12}{P}_{01}=P_{12}{P}_{01}+1-P_{12}{P}_{01}=1$. Hence all degenerate affine Hecke relations of [F2] hold in this specialization.

The explicit vector. For $v_1\otimes v_2\otimes v_1$: $s_1$ sends it to $v_1\otimes v_1\otimes v_2$, which $x_1=P_{01}$ fixes, so $x_1{s}_1(v_1\otimes v_2\otimes v_1)=v_1\otimes v_1\otimes v_2$; and $x_2=P_{02}+P_{12}$ sends it to $v_1\otimes v_2\otimes v_1+v_1\otimes v_1\otimes v_2$, which $s_1$ sends to $v_1\otimes v_1\otimes v_2+v_1\otimes v_2\otimes v_1$. Hence $(s_1{x}_2-x_1{s}_1)(v_1\otimes v_2\otimes v_1)=v_1\otimes v_2\otimes v_1$, the identity action on this vector.

The commutator caveat. On the same vector, $x_2{s}_1(v_1\otimes v_2\otimes v_1)=v_2\otimes v_1\otimes v_1+v_1\otimes v_2\otimes v_1$ and $s_1{x}_1(v_1\otimes v_2\otimes v_1)=v_2\otimes v_1\otimes v_1$, so $(x_2{s}_1-s_1{x}_1)(v_1\otimes v_2\otimes v_1)=v_1\otimes v_2\otimes v_1$, while $[s_1,x_2](v_1\otimes v_2\otimes v_1)=v_1\otimes v_1\otimes v_2-v_2\otimes v_1\otimes v_1$, which has no component along $v_1\otimes v_2\otimes v_1$. So $[s_1,x_2]\mathrm{\ne }1$: the affine cross relation relates $s_1{x}_2$ to $x_1{s}_1$, not to $x_2{s}_1$.

The operators on $V$. For $d=1$ the cross-column correction in [F1] is absent and ${\sum }_j(c_{\mathrm{c}\mathrm{o}\mathrm{l}(j)}+q_{\mathrm{c}\mathrm{o}\mathrm{l}(j)}-n)e_{j,j}=\mathrm{diag}(0,0,-1)$, so $x_1=E_{23}+\mathrm{diag}(0,0,-1)$: explicitly $x_1{v}_1=0$, $x_1{v}_2=0$ and $x_1{v}_3=v_2-v_3$. Hence $x_1(x_1+1)=0$, the eigenspaces of $x_1$ are $\mathbb{C}{v}_1\oplus \mathbb{C}{v}_2$ for the eigenvalue $0$ and $\mathbb{C}(v_3-v_2)$ for the eigenvalue $-1$, and the minimal polynomial is exactly $x(x+1)$: $x_1\mathrm{\ne }0$ since $x_1{v}_3=v_2-v_3\mathrm{\ne }0$, and $x_1+1\mathrm{\ne }0$ since $(x_1+1)v_1=v_1\mathrm{\ne }0$.

All degrees and the cyclotomic quotient. By the minimal-polynomial clause of [F1], the operator $x_1$ on $V_c^{\otimes d}$ has minimal polynomial ${\prod }_j(x-a_j)=x(x+1)$ for every $d\ge 1$, so the tensor action factors through $H_d(\mathbf{a})$ with $\mathbf{a}=(0,-1)$, i.e. through $H_d(\mathrm{\Lambda })$ for $\mathrm{\Lambda }={\mathrm{\Lambda }}_0+{\mathrm{\Lambda }}_{-1}$; and by [F2] that quotient has dimension $2^{d}d!$. At $d=1$ the quotient is $\mathbb{C}[x]\mathrm{/}(x(x+1))\cong \mathbb{C}\times \mathbb{C}$, acting on $V$ through $x_1$ as the two-dimensional algebra of operators that are scalar on each of the two eigenspaces of step 1.1; hence the image is semisimple and $V$ is the direct sum of those two eigenspaces.

The Hecke action. By [F1] the generator $x_1$ acts as $0$ on $M\otimes V$, so the $H_1$-action is the evaluation action $x_1\mapsto 0$ (the quotient $H_1\mathrm{/}(x_1)=\mathbb{C}$ of The degenerate affine Hecke algebra specialized to the scalar $0$).

The first centralizer. Since $H_1$ acts by the scalars $\mathbb{C}$ on the four-dimensional space, ${\mathrm{End}}_{H_1}(M\otimes V)={\mathrm{End}}_{\mathbb{C}}({\mathbb{C}}^4)=M_4(\mathbb{C})$, of dimension $16$.

The image of the enveloping algebra. By [F2] the image of $U({\mathfrak{g}\mathfrak{l}}_2)$ in $\mathrm{End}(M\otimes V)$ is the subalgebra $1\otimes M_2(\mathbb{C})$, of dimension $4$. Hence the first double centralizer equality fails by a factor of $4$ in dimension.

The second centralizer. Writing an endomorphism as a $2\times 2$ block matrix indexed by the basis of $M$, commutation with every $1\otimes E_{ij}$ forces each block to be a scalar matrix. Thus the ${\mathfrak{g}\mathfrak{l}}_2$-endomorphisms are $M_2(\mathbb{C})\otimes 1$, of dimension $4$, while $\psi (H_1)=\mathbb{C}\cdot \mathrm{i}\mathrm{d}$ has dimension $1$; hence the second equality also fails. The missing hypothesis is the truncated/deformed setup: the double centralizer theorem applies to the finite-W-algebra tensor space $V_c^{\otimes d}$ over a pyramid where the Hecke action is the cyclotomic quotient action and the algebra on the opposite side is the finite W-algebra, not $U(\mathfrak{g})$ itself. Its associated graded image algebras are the corresponding images of $U({\mathfrak{g}}^e)$ and ${\mathbb{C}}_l[x_1,\dots ,x_d]⋊\mathbb{C}[S_d]$; this does not identify the deformed actions with the graded actions (The finite W and degenerate cyclotomic Hecke actions satisfy a double centralizer, The type-A finite W-algebra and the Whittaker category).

The case $d=1$. Here $B=\mathbb{C}[S_1]=\mathbb{C}$, so ${\mathrm{End}}_{{\mathfrak{g}\mathfrak{l}}_2}(V)=\mathbb{C}\cdot 1$ by Schur's lemma and the image ${\phi }_1(U({\mathfrak{g}\mathfrak{l}}_2))={\mathrm{End}}_{\mathbb{C}}(V)$ by [F4]; both Vust identities reduce to these two equal-dimensional statements.

The case $d=2$, second centralizer. By [F3], $V^{\otimes 2}=S^{2}V\oplus {\mathrm{\Lambda }}^{2}V$ with ${\mathrm{\Lambda }}^{2}V$ one-dimensional and $S^{2}V$ three-dimensional irreducible. Since the flip acts by $+1$ on $S^{2}V$ and by $-1$ on ${\mathrm{\Lambda }}^{2}V$, an endomorphism commuting with the $\mathbb{C}[S_2]$-action preserves each isotypic component, and its restrictions are arbitrary endomorphisms of the two (irreducible) components: ${\mathrm{End}}_{\mathbb{C}[S_2]}(V^{\otimes 2})=\mathrm{End}(S^{2}V)\oplus \mathrm{End}({\mathrm{\Lambda }}^{2}V)$, of dimension $9+1=10$.

The case $d=2$, first centralizer. The image of $U({\mathfrak{g}\mathfrak{l}}_2)$ in $\mathrm{End}(V^{\otimes 2})$ is the diagonal action on the two isotypic components, which by the mutual-centralizer statement of [F3] is the block-diagonal algebra $\mathrm{End}(S^{2}V)\oplus \mathrm{End}({\mathrm{\Lambda }}^{2}V)$, also of dimension $10$; hence ${\phi }_2(U({\mathfrak{g}\mathfrak{l}}_2))={\mathrm{End}}_{\mathbb{C}[S_2]}(V^{\otimes 2})$. The other direction, ${\mathrm{End}}_{{\mathfrak{g}\mathfrak{l}}_2}(V^{\otimes 2})=\mathbb{C}[S_2]$, follows because ${\psi }_2$ is injective ($\dim \mathbb{C}[S_2]=2$ and the flip is not a scalar on $V^{\otimes 2}$) and $\dim {\mathrm{End}}_{{\mathfrak{g}\mathfrak{l}}_2}(V^{\otimes 2})=2$ by Schur's lemma applied to the two nonisomorphic components.

Range check. For $d\le 2=q_l$ the stable-range hypothesis $q_l\ge d$ holds, and for $d=3$ it fails; this example therefore realizes the theorem exactly in its stable range and shows that the level-one case is the classical Schur-Weyl duality.

By [F2], the tensor space contains the trivial $S_3$ module with multiplicity four, the standard module with multiplicity two, and no sign module. The decomposition [F3] therefore makes the group-algebra image exactly $\mathbb{C}\times M_2(\mathbb{C})$, of dimension five. By classical Schur-Weyl duality [F2] this is ${\mathrm{End}}_{{\mathfrak{g}\mathfrak{l}}_2}(V^{\otimes 3})$, whereas $\dim \mathbb{C}[S_3]=6$.

The killed sign block. By [F3] the block decomposition of $\mathbb{C}[S_3]$ is $\mathbb{C}\times \mathbb{C}\times M_2(\mathbb{C})$; the one-dimensional sign block is spanned by the antisymmetrizer $a$ of [F4], which acts on $V^{\otimes 3}$ by $0$. Hence the kernel of ${\psi }_3$ contains the sign block; since the image has dimension $5$ by step 1.1, the kernel has dimension $1$ and is exactly the sign block. So ${\psi }_3$ kills precisely the sign block and is injective on the direct sum of the trivial and standard blocks.

The block idempotent corner. Let $f$ be the central idempotent of the trivial and standard blocks, so that $\mathbb{C}[S_3]f=\mathbb{C}[S_3]f^2$ is the corresponding two-sided ideal; its corner $f\mathbb{C}[S_3]f$ has dimension $1+4=5$ by [F3]. Multiplication by $f$, $x\mapsto fxf$, is an algebra surjection $\mathbb{C}[S_3]\twoheadrightarrow f\mathbb{C}[S_3]f$ whose kernel contains the sign block and has dimension $6-5=1$, hence is exactly the sign block; therefore $\mathbb{C}[S_3]\mathrm{/}\text{\hspace{0.17em}⁣}\ker {\psi }_3\cong f\mathbb{C}[S_3]f$ and the image $H_3(\lambda ,c)={\psi }_3(\mathbb{C}[S_3])$ is isomorphic to the block idempotent corner $f\mathbb{C}[S_3]f$, which is the central-corner form $f{H}_3(\mathbf{a})f$ of the Definition.

The stable-range check. Here $q_l=q_1=2$ while $d=3$, so the stable range fails and the image is not the full algebra $H_3(\mathbf{a})=\mathbb{C}[S_3]$ (which has dimension $6$); the corner of step 3.1 realizes the boundary case of the Definition, and the missing dimension $1$ is precisely the killed sign block.