Type-A Hecke Specht Modules and Cellular Bases — Examples

Substitution: in [F1], for $k=1$ the sum is empty, giving $L_1=0$; for $k=2$ it has the single term $q^{1-2}{T}_{(1,2)}=q^{-1}{T}_{(1,2)}$; for $k=3$ it has $q^{-1}{T}_{(2,3)}+q^{-2}{T}_{(1,3)}$; and for $k=4$ it has $q^{-1}{T}_{(3,4)}+q^{-2}{T}_{(2,4)}+q^{-3}{T}_{(1,4)}$, matching the displayed list.

Contents: reading the tableau $t$ with rows $(1,3,4)$ and $(2)$, the node of $1$ is $(1,1)$ with content $1-1=0$, the node of $2$ is $(2,1)$ with content $1-2=-1$, the node of $3$ is $(1,2)$ with content $2-1=1$ and the node of $4$ is $(1,3)$ with content $3-1=2$; by definition $[{\mathrm{content}}_t(k){]}_q$ is the q-integer of these contents.

Eigenvalues: by [F1], $[0{]}_q=0$, $[-1{]}_q=-q^{-1}[1{]}_q=-q^{-1}$, $[1{]}_q=1$ and $[2{]}_q=1+q$; substituting into the triangularity formula [F2] gives the stated diagonal eigenvalues $0,-1,1,2$ up to the q-powers $-q^{-1},1,1+q$. The correction terms are exactly the sum ${\sum }_{v⊳t}{a}_v{m}_v$ of [F2] and are not asserted to vanish outside the semisimple range.

Specialization: setting $q=1$ in the displayed formulas turns $L_k$ into ${\sum }_{j<k}(jk)=X_k$ by the specialization isomorphism of Hecke Jucys-Murphy elements and q-contents and [F1], giving $X_1=0$, $X_2=(12)$, $X_3=(13)+(23)$ and $X_4=(14)+(24)+(34)$; the eigenvalues $[m{]}_1=m$ become $0,-1,1,2$, the contents of $t$.

Contents of the two tableaux: for $s$ with rows $(1,2)$ and $(3)$ the nodes are $(1,1),(1,2),(2,1)$ for $1,2,3$, giving contents $0,1,-1$; for $t$ with rows $(1,3)$ and $(2)$ the nodes are $(1,1),(2,1),(1,2)$ for $1,2,3$, giving contents $0,-1,1$.

Collision: by [F1] and [F2], $[2{]}_{-1}=0$ and $[1{]}_{-1}=[-1{]}_{-1}=1$; hence $[{\mathrm{content}}_s(1){]}_q=0=[{\mathrm{content}}_t(1){]}_q$, $[{\mathrm{content}}_s(2){]}_q=1=[{\mathrm{content}}_t(2){]}_q$ and $[{\mathrm{content}}_s(3){]}_q=1=[{\mathrm{content}}_t(3){]}_q$, so the two q-content sequences coincide even though $s\mathrm{\ne }t$.

Interpolation and axial distance. The eigenvalue difference that would be needed to distinguish $s$ and $t$ at $k=2$ is $1-1=0$. Equal-value exclusion in a polynomial product does not make its literal denominators zero; it instead loses the factor that could distinguish these tuples. No polynomial in the JM operators can provide the two distinct joint-eigenvalue labels used in the separated-range construction. For the swap of $2$ and $3$, $\rho =1-(-1)=2$ and $[2{]}_{-1}=0$, so the displayed separated-range generator formula is undefined here.

Verify the failure of a JM-eigenbasis on this cell directly. Put $m_s=1+T_1$ and $m_t=(1+T_1)T_2$ modulo the higher row-shape ideal. The quadratic relation gives $m_s{T}_1=q{m}_s$. The higher row symmetrizer is $1+T_1+T_2+T_1{T}_2+T_2{T}_1+T_1{T}_2{T}_1$, so its vanishing in this cell gives $m_t{T}_1=-m_s-m_t$. At $q=-1$, since $L_2=q^{-1}{T}_1=-T_1$, it follows that $m_s{L}_2=m_s$ and $m_t{L}_2=m_s+m_t$. Thus $L_2-I$ is nonzero and $(L_2-I{)}^2=0$ on this two-dimensional cell. It is not diagonalizable over $\mathbb{C}$, so a simultaneous JM-eigenbasis cannot exist on this cell. Together with step 2.1 this supplies the claimed concrete boundary counterexample, without making a universal claim about all cells outside the separated range.

Check the boundary case $n=e=2$. The row and column tableaux have contents $(0,1)$ and $(0,-1)$, whose q-content tuples both equal $(0,1)$ at $q=-1$ by [F2]. These are all standard tableaux of size two, so the theorem's global sets are ${\mathcal{R}}_1=\{0\}$ and ${\mathcal{R}}_2=\{1\}$. For either label every inner product in its Lagrange formula is empty, giving $F_{\mathrm{r}\mathrm{o}\mathrm{w}}=F_{\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{n}}=1$. Their product is $1\mathrm{\ne }0$ in the nonzero algebra $H_{\mathbb{C},-1}(S_2)$, which has basis $1,T_1$. Hence they are not orthogonal, disproving replacement of $e>n$ by $e\ge n$ in the global theorem. This supplements the same-shape collision and Jordan block at size three.

Cell $(2,1)$: $S_{(2,1)}=⟨s_1⟩$, so $m_{(2,1)}=1+T_1$; the two standard tableaux of shape $(2,1)$ are $t=\begin{array}{cc}1& 2\\ 3\end{array}$ (row reading word $123$, so $d(t)=1$) and $u=\begin{array}{cc}1& 3\\ 2\end{array}$ (row reading word $132$, so $d(u)=s_2$). Hence $m_{tt}=1\cdot (1+T_1)\cdot 1=1+T_1$, $m_{tu}=(1+T_1)T_2$, $m_{ut}=T_2^{\ast }(1+T_1)=T_2(1+T_1)$ and $m_{uu}=T_2^{\ast }(1+T_1)T_2=T_2(1+T_1)T_2$, using $T_2^{\ast }=T_2$ and $m_{(2,1)}^{\ast }=m_{(2,1)}$.

Cells $(3)$ and $(1^3)$: $S_{(3)}=S_3$ has six elements with reduced expressions $1,s_1,s_2,s_1{s}_2,s_2{s}_1,s_1{s}_2{s}_1$, so $m_{(3)}=1+T_1+T_2+T_1{T}_2+T_2{T}_1+T_1{T}_2{T}_1=T_{d(s)}^{\ast }{m}_{(3)}{T}_{d(s)}$ for the unique standard tableau; $S_{(1^3)}$ is trivial, so its group algebra sum is the identity, $m_{(1^3)}=1$, and $m_{vv}=1$.

Count and anti-involution: the three cells contribute $1^2+2^2+1^2=6=3!=\mathrm{\mid }{S}_3\mathrm{\mid }$ elements, in agreement with $H$ being free of rank $6$; and $m_{tu}^{\ast }=m_{ut}$ is the instance of the general identity $m_{st}^{\ast }=m_{ts}$ of [F1], the middle cell being the one with two indices.

Specialization: setting $q=1$ turns $T_w$ into $w$, so the displayed elements become the Murphy basis of $R[S_3]$; this is the $n=3$ case of [F3].

Check the lift. Localization makes $\widehat{q}$ a unit since its residue is $-1$, so the exact PID hypotheses of [F1] hold; the generic factorial is nonzero. The identities $[2]=1+\widehat{q}$, $[4]=(1+\widehat{q})(1+{\widehat{q}}^2)$ and evaluation of the odd quantum integers at $-1$ prove [F2]. These values refer to this lift, not an arbitrary field or lift with quantum characteristic $2$.

Compute the balanced symbols explicitly. For $(3,2)$, $\beta =(6,5,3,1,0)$ and its hook array is $\begin{array}{ccc}4& 3& 1\\ 2& 1\end{array}$. The pair $(1,1),(1,2)$ gives $(9,2,3,1,0)$, which sorts to the beta sequence of $(1^5)$ with sign $-1$ and coefficient $v(4)-v(3)=1$. The pair $(1,1),(1,3)$ gives $(7,5,2,1,0)$, shape $(2,2,1)$ with sign $+1$ and coefficient $v(4)-v(1)=1$. The pair $(2,1),(2,2)$ gives $(7,4,3,1,0)$, shape $(3,1^2)$ with sign $+1$ and coefficient $v(2)-v(1)=1$. The remaining pair has repeated beta entries and vanishes. Thus the first displayed Weyl identity follows by direct substitution.

Verify the two zero forms directly at $q=-1$. For each $\eta \in \{(3,2),(3,1,1)\}$, write $v_j$ for the Murphy vector whose row-reading word is, respectively, $12345,12435,12534,13425,13524$ for $j=0,\dots ,4$; the shape $(3,1,1)$ has the additional word $14523$, giving $v_5$. The row action gives $v_0{T}_1=v_0{T}_2=-v_0$, $v_1{T}_2=v_3$ and $v_2{T}_2=v_4$. The first-column Garnir belt between the first two rows gives $v_3{T}_1=-v_0-v_1-v_3$: its row-standard terms have lower first entry $1,2,3,4$, and its other side has shape $(4,1)$, strictly above either $\eta$. Also $v_3{T}_4=v_4$, $v_1{T}_4=v_2$ and $v_0{T}_4=-v_0$. The last identity is the row action for $(3,2)$; for $(3,1,1)$ it follows from $x_{(3,1,1)}(1+T_4)=x_{(3,2)}$, a higher-shape element. Since $T_1{T}_4=T_4{T}_1$, these identities give $v_4{T}_1=v_0-v_2-v_4$. These are finite row-action and Garnir calculations from the supplied integral Murphy construction.

Put $r_j=⟨v_0,v_j⟩$. Self-adjointness of $T_1,T_2$ and $v_0{T}_i=-v_0$ imply $⟨v_0,v_j{T}_i⟩=-r_j$. The four identities of step 1.3 consequently give $r_1=-r_0$, $r_2=r_0$, $r_3=-r_1$ and $r_4=-r_2$. But $r_0=[\eta ]{!}_{-1}=0$, since $x_{\eta }^2=[\eta ]{!}_q{x}_{\eta }$ and $[3]{!}_{-1}$ contains $[2{]}_{-1}=0$. Thus $r_0,\dots ,r_4$ are zero for both shapes.

Compute the smaller rows rather than assuming them. The same calculation for $(3,1^2)$ gives ${\sum }_{i>0}[W^{(3,1^2)}(i)]=[W^{(2,2,1)}]+[W^{(1^5)}]$: the pairs with hook lengths $(5,2)$ and $(2,1)$ give those two terms, while the $(5,1)$ coefficient is zero. For $(2,2,1)$, the only nonzero symbols arise from pairs $(4,2)$ and $(3,1)$, both of valuation difference zero. For $(2,1^3)$ the sole pair is $(5,1)$, also zero; $(1^5)$ has no pair. Hence these three Weyl modules have zero positive sums and zero radicals, so they equal their simple heads. For $(3,1^2)$ each displayed simple occurs once in the positive sum; because its first submodule is the radical and later ones lie inside it, both radical multiplicities are exactly $1$. Its class is consequently $[L^{(3,1^2)}]+[L^{(2,2,1)}]+[L^{(1^5)}]$. Substitution into step 1.2 now gives $[L^{(3,1^2)}]+2[L^{(2,2,1)}]$, proving the stated preliminary multiplicity conclusions without reading a signed individual Weyl coefficient as a simple multiplicity.

For the remaining vector of $(3,1,1)$, put $x=x_{(3,1,1)}$. Its distinguished inverse word is $T_3{T}_4{T}_2{T}_3$. Expanding the finite row symmetrizers with $T_i^2=-2T_i-1$ and the braid relations gives the exact identity $x{T}_3{T}_4{T}_2{T}_{3}x=x_{(5)}-(1+T_4)x_{(4,1)}(1+T_4)-x_{(3,2)}\mathrm{.}$ Every term on the right belongs to the higher dominance ideal for $(3,1,1)$, so the cell-product coefficient is $r_5=0$. Together with step 2.1, this puts the canonical vector $v_0$ in the form radical of each cell. That radical is a submodule, and $v_{0}H$ is the entire Murphy cell, since its standard basis is $x_{\eta }{T}_{d(t)}$ modulo the higher ideal. Hence both forms are identically zero. Their Schur-simple images are zero by the exact radical-corner identity in [F3]. This uses no general Gram-entry divisor or finite-$e$ classification.

Supply the actual dimension argument. The regular-weight Gram product for the Murphy cell $(2,2,1)$ is ${\widehat{q}}^7\frac{[4{]}_{\widehat{q}}}{[2{]}_{\widehat{q}}}[3{]}_{\widehat{q}}^4={\widehat{q}}^7(1+{\widehat{q}}^2)[3{]}_{\widehat{q}}^4,$ since its two signed symbols have shapes $(1^5)$ and $(2,1^3)$ with standard-tableau counts $1$ and $4$. The five standardizations have lengths $0,1,1,2,3$, giving exponent $7$. At $\widehat{q}=-1$ this determinant is $-2\mathrm{\ne }0$. Thus $S_{\mathrm{M}}^{(2,2,1)}=D_{\mathrm{M}}^{(2,2,1)}$ is simple of dimension $5$, counted by its five standard tableaux. Both Schur simples $L^{(3,1^2)}$ and $L^{(3,2)}$ are killed by the local zero-form calculation of step 3.1 and the exact corner/form identity [F3]. Applying the functor to the positive sum of step 2.2 therefore gives ${\sum }_{i>0}[S_{\mathrm{M}}^{(3,2)}(i)]=2[D_{\mathrm{M}}^{(2,2,1)}]$. The form of $(3,2)$ is identically zero by step 3.1, so its first submodule is the whole five-dimensional cell. Every composition factor of that cell must appear in the positive sum, hence must be this five-dimensional simple. Dimension forces exactly one factor, proving $S_{\mathrm{M}}^{(3,2)}\cong D_{\mathrm{M}}^{(2,2,1)}$. Exact projection of a composition series, with distinct nonzero simple images as in [F3], gives $d_{(3,2),(2,2,1)}=1$.

A further consistency check is the regular-weight Gram product ${\widehat{q}}^8[3{]}_{\widehat{q}}[4{]}_{\widehat{q}}^4[2{]}_{\widehat{q}}^6$ for $(3,2)$, whose valuation is $10$, agreeing with two five-dimensional positive Specht submodules. This is a check of the complete balanced formula; it is not the unsupported full-hook product used in the former classification proof.

Translate simultaneously by $\theta$ and conjugation. The actual ordinary/Murphy module-and-form comparison sends the isomorphism of step 4.1 to $S_{\mathrm{D}\mathrm{J}}^{(2,2,1)}\cong D_{\mathrm{D}\mathrm{J}}^{(3,2)}$. The partition $(2,2,1)$ is not regular, so its own cell-form quotient is zero; this does not say its actual head is zero, since the module itself is simple. The dominance inequality $(3,2)⊵(2,2,1)$ agrees with the ordinary-convention unitriangular bound. This proves every claimed numeric conclusion.

The abstract instance: in $A=F[x]\mathrm{/}(x^2)$ with $c_{11}=1$, $c_{22}=x$ and $2⊳1$, the product $c_{22}{c}_{22}=x^2=0$ is congruent to $0\cdot c_{22}$ modulo $A^{⊳2}=0$, so the axiom holds with zero form value; the products $c_{11}(a+bx)=a+bx\equiv a{c}_{11}(\mathrm{m}\mathrm{o}\mathrm{d}(x))$ and $c_{22}(a+bx)=ax=a{c}_{22}$ confirm the triangularity. Hence $C^2$ has identically zero form, $\mathrm{r}\mathrm{a}\mathrm{d}{C}^2=C^2$ and $D^2=0$; the module $C^2$ itself is one-dimensional with $a+bx$ acting as $a$, hence simple with zero Jacobson radical and head $C^2\mathrm{\ne }0$. Moreover $a+bx$ is a unit exactly when $a\mathrm{\ne }0$: in that case its inverse is $a^{-1}-b{a}^{-2}x$, while every element of $(x)$ squares to zero. Thus $(x)$ is the unique maximal ideal, so $\mathrm{rad}A=(x)\mathrm{\ne }0$. The regular module is not semisimple: its submodule $(x)$ has no complementary ideal, since any nonzero ideal meeting $A\setminus (x)$ contains a unit and equals $A$.