The degenerate cyclotomic Hecke center is the symmetric polynomial image

The associated graded. Give $x_i$ degree $1$ and the symmetric group degree $0$. By [F1] the transported relation has leading term $x_i^{\mathrm{\ell }}$ with all other dot terms of total degree below $\mathrm{\ell }$, and the bounded reduction sends monomials of dot degree $\le r$ to bounded monomials of dot degree $\le r$; independence of the bounded basis makes these the filtration steps $F_r$. The leading mixed relations give the variable permutation action of $S_d$, and the transported relations give $x_i^{\mathrm{\ell }}=0$ in the associated graded. Hence the natural surjection $F[X_1,\dots ,X_d]\mathrm{/}(X_1^{\mathrm{\ell }},\dots ,X_d^{\mathrm{\ell }})⋊F{S}_d⟶\mathrm{gr}{H}_d^{\mathrm{\Lambda }}$ is an isomorphism: the bounded monomials of the truncated algebra map degree-by-degree onto the basis of [F1].

Symmetric polynomials are central. For $x=x_i$, $y=x_{i+1}$, $s=s_i$ the relations of [F2] give, by induction on products, $sg(x,y)-g(y,x)s=\frac{g(y,x)-g(x,y)}{x-y}$ for every polynomial $g$; the quotient is a polynomial because the numerator vanishes at $x=y$. Applying this to a polynomial symmetric in $x,y$ shows that it commutes with $s$, and a globally symmetric polynomial commutes with every $s_i$ and every $x_j$, hence is central in $H_d^{\mathrm{\Lambda }}$. No division by an integer or by $x-y$ inside the algebra is used.

Cycle coefficients in the graded centralizer. In $B=F[X_1,\dots ,X_d]\mathrm{/}(X_1^{\mathrm{\ell }},\dots ,X_d^{\mathrm{\ell }})⋊F{S}_d$, put $h_r(I)=\sum X^b$ over $0\le b_i<\mathrm{\ell }$ and ${\sum }_{i\in I}{b}_i=(\mathrm{\mid }I\mathrm{\mid }-1)(\mathrm{\ell }-1)+r$, and $A^{(r)}=h_r(I)A$ for a cycle $A$ on $I$, $0\le r<\mathrm{\ell }$. For a coefficient $gA$ commuting with the $X_i$, the equations are $(X_i-X_{A(i)})g=0$. On one cycle they equate coefficients obtained by transferring one unit between consecutive coordinates, with out-of-range coefficients interpreted as zero. At fixed total degree, transfer units until the first $\mathrm{\mid }I\mathrm{\mid }-1$ coordinates are $\mathrm{\ell }-1$. If the total is $(\mathrm{\mid }I\mathrm{\mid }-1)(\mathrm{\ell }-1)+r$, every coefficient equals the coefficient at $(\mathrm{\ell }-1,\dots ,\mathrm{\ell }-1,r)$; if the total is smaller, the same transfers reach a zero coordinate followed by a transfer to exponent $-1$, and all coefficients vanish. Thus the solutions on that cycle have basis $h_0(I),\dots ,h_{\mathrm{\ell }-1}(I)$. For disjoint cycles apply these equations separately to each coefficient in the monomial basis of the complementary variables; this gives exactly the products of disjoint colored cycles as the polynomial centralizer basis. This argument also covers $\mathrm{\ell }=1$: the truncated polynomial ring is $F$, and its centralizer is $F{S}_d$.

Reduced-degree powers. Write $f(t)=t^{\mathrm{\ell }}+c_1{t}^{\mathrm{\ell }-1}+\cdots +c_{\mathrm{\ell }}$. The mixed relations give $x_{i+1}^{\mathrm{\ell }}=s_i{x}_i^{\mathrm{\ell }}{s}_i+{\sum }_{t=0}^{\mathrm{\ell }-1}{x}_i^t{x}_{i+1}^{\mathrm{\ell }-1-t}{s}_i$. Starting with $x_1^{\mathrm{\ell }}=-{\sum }_{a=1}^{\mathrm{\ell }}{c}_a{x}_1^{\mathrm{\ell }-a}$, induction on $i$ shows that $x_i^{\mathrm{\ell }}$ has filtration degree at most $\mathrm{\ell }-1$ and symbol $Y_i-c_1{X}_i^{\mathrm{\ell }-1}$ there, where $Y_i={\sum }_{j<i}(ji{)}^{(0)}$. For $k=p\mathrm{\ell }+r$, $0\le r<\mathrm{\ell }$, it follows that $x_i^k$ has filtration degree at most $p(\mathrm{\ell }-1)+r$ and symbol $(Y_i-c_1{X}_i^{\mathrm{\ell }-1}{)}^p{X}_i^r$. This is a reduced degree: the ordinary degree-$k$ symbol may be zero.

Orbit sums and colored-cycle multiplication. Conjugation permutes the basis of step 2.1, so its fixed vectors have basis the orbit sums $z_{\lambda }$, indexed by colored cycle types, including trivial color-zero fixed points to fill rank $d$. No averaging is used. If $\mathrm{\mid }I\cap J\mathrm{\mid }=c>0$, direct multiplication gives $h_r(I)h_s(J)={\mathrm{\ell }}^{c-1}{h}_{r+s+(c-1)(\mathrm{\ell }-1)}(I\cup J)$. For $c=1$, fixing the exponents away from the common variable fixes its exponent uniquely and gives the asserted coefficient one. For $c=2$, the basic repeated-edge computation is $h_0(i,j{)}^2=\mathrm{\ell }{X}_i^{\mathrm{\ell }-1}{X}_j^{\mathrm{\ell }-1}$: only equal complementary exponents survive, with exactly $\mathrm{\ell }$ choices. Joining the remaining common variables successively reduces the general case to these two computations. Consequently disjoint colored cycles commute; two cycles meeting at one point multiply to their merged cycle of color $r+s$; when they meet in at least two points the product vanishes unless $r=s=0$, and otherwise is a scalar times disjoint cycles all of color $\mathrm{\ell }-1$. Indeed the displayed exponent equals the maximal degree on the union when $c=2$, and exceeds it when $c>2$ and $\mathrm{\ell }>1$; when $\mathrm{\ell }=1$ every polynomial coefficient is one and the assertion reduces to ordinary cycle multiplication. Iterating gives disjoint-cycle reductions, never increasing the union of supports.

The star-cycle expansion. Let $y_i(p\mathrm{\ell }+r)$ be the sum of $(j_1\cdots j_{p}i{)}^{(r)}$ over ordered, distinct $j_1,\dots ,j_p<i$, and let $y_i(0)=1$. Expanding $Y_i^p$, sequences with distinct leaves give precisely $y_i(p\mathrm{\ell })$: the product of the star transpositions is one cycle, with a unique ordered leaf sequence for each such cycle. Any repeated leaf uses at most $p$ vertices including $i$; at its first repetition the intersecting cycles have at least two common vertices, so step 3.1 turns the result into saturated (color $\mathrm{\ell }-1$) disjoint cycles or zero. Further star factors preserve saturation by the one-point rule, or vanish by the multiple-point rule. The term $X_i^{\mathrm{\ell }-1}$ is itself a saturated singleton containing $i$; choosing it anywhere in $(Y_i-c_1{X}_i^{\mathrm{\ell }-1}{)}^p$ has the same consequence and uses at most $p$ vertices. Thus this power equals $y_i(p\mathrm{\ell })$ plus saturated disjoint-cycle products on at most $p$ vertices, always with $i$ in their support. Multiplication by $X_i^r$ gives exactly $y_i(p\mathrm{\ell }+r)$ when $r>0$, because it raises the color of the distinct-leaf cycle to $r$, while every saturated error containing $i$ vanishes. (For $\mathrm{\ell }=1$ there is no $r>0$ case.) This proves the complete power-symbol formula, including $p=0$.

Triangular symmetric lifts. Index colored cycle types by partitions $\mu$ obtained by replacing each cycle of length $a$ and color $r$ by $k=(a-1)\mathrm{\ell }+r$ and discarding zero parts. These are exactly the partitions satisfying $N(\mu )=\mathrm{length}(\mu )+{\sum }_j\lfloor {\mu }_j\mathrm{/}\mathrm{\ell }\rfloor \le d$; the inverse adds $d-N(\mu )$ trivial fixed points. Put $D(\mu )={\sum }_j({\mu }_j-\lfloor {\mu }_j\mathrm{/}\mathrm{\ell }\rfloor )$ and $p_{\mu }={\sum }_{\nu \sim (\mu ,0,\dots ,0)}{x}_1^{{\nu }_1}\cdots x_d^{{\nu }_d}$, summing each distinct exponent tuple once. It is central by step 1.2 and lies in filtration degree $D(\mu )$ by step 4.1. Its symbol at that degree is $z_{\mu }$ plus orbit sums with strictly smaller $N$: each product of the leading star cycles on disjoint supports has the prescribed type, and its coefficient is one, since each cycle's largest vertex uniquely chooses its star hub, its ordered other vertices uniquely choose its star expansion, and the unordered exponent tuple chooses each assignment only once, even for repeated parts. Overlapping supports strictly decrease the number of used vertices, as does every error from step 4.1; disjoint-cycle reduction preserves this bound. The entire symbol is central, so the remainder is a combination of orbit sums by step 3.1, with smaller $N$ and the same homogeneous degree. Hence, ordering the finite colored types by $N$, the symbols of the $p_{\mu }$ form a unit triangular change of the orbit-sum basis, degree by degree.

Filtration descent. If $z\in Z(H_d^{\mathrm{\Lambda }})$ has filtration degree $D$, its degree-$D$ symbol lies in $Z(B)$. By step 5.1 it is a linear combination of degree-$D$ symbols of central symmetric evaluations $p_{\mu }$; subtract that same combination in $H_d^{\mathrm{\Lambda }}$ to lower the filtration degree. Repeating terminates because degrees are nonnegative; in degree zero the same triangular basis applies, and the next filtration step is zero. Thus every central element is a symmetric-polynomial evaluation, and step 1.2 proves the reverse inclusion. Rank zero was handled separately. No division by $\mathrm{\ell }$, $d!$ or the field characteristic occurs.