Cores, Quotients, and Blocks of Symmetric Groups — Examples

Beads on the runners: reading positions $3a+i$ of $X=\{11,9,5,3,2,1\}$ gives runner $0$ the positions $9=3\cdot 3$, $3=3\cdot 1$, runner $1$ the position $1=3\cdot 0+1$, and runner $2$ the positions $11=3\cdot 3+2$, $5=3\cdot 1+2$, $2=3\cdot 0+2$. Hence the row-label sets are $X_0^{(3)}=\{3,1\}$, $X_1^{(3)}=\{0\}$, $X_2^{(3)}=\{3,1,0\}$, of counts $2,1,3$.

Top-packing moves the two beads of runner $0$ to rows $0,1$, the bead of runner $1$ to row $0$, and the three beads of runner $2$ to rows $0,1,2$, so $X^{(3)}=\{0,3\}\cup \{1\}\cup \{2,5,8\}=\{8,5,3,2,1,0\}$ and, by [F1] with the six beads listed in decreasing order, $C_3(\lambda )=P(X^{(3)})=(8-5,5-4,3-3,2-2,1-1,0-0)=(3,1)\mathrm{.}$

The normalisation $3\mid \mathrm{\mid }X\mathrm{\mid }=6$ holds, so the row-label sets of step 1.1 may be read as $1$-abaci: ${\lambda }^{(0)}=P(\{3,1\})=(2,1)$, ${\lambda }^{(1)}=P(\{0\})=\mathrm{\varnothing }$ and ${\lambda }^{(2)}=P(\{3,1,0\})=(1)$, so $Q_3(\lambda )=((2,1),\mathrm{\varnothing },(1))$ and $w_3(\lambda )=2+1+0+1=4$.

Four upward bead moves lead from $X$ to $X^{(3)}$: $11\mapsto 8$ on runner $2$ (the target $8\mathrm{\notin }X$), then $9\mapsto 6$ on runner $0$ ($6\mathrm{\notin }X$), then $3\mapsto 0$ on runner $0$ ($0\mathrm{\notin }X$), then $6\mapsto 3$ on runner $0$ (now $3\mathrm{\notin }X$); every move replaces a bead by a gap one row above it on its own runner, so by [F2] each is the removal of a rim hook of length $3$. In terms of [F1] the intermediate partitions are $(6,5,2,1^3)\to (4,4,2,1^3)\to (3,2,2,1^3)\to (3,2,2)\to (3,1),$ each step of size $3$ by [F4], and the terminal partition is $C_3(\lambda )=(3,1)$ as computed in step 1.2.

The size formula of [F5] reads $16=\mathrm{\mid }{C}_3(\lambda )\mathrm{\mid }+3w_3(\lambda )=4+3\cdot 4$, using steps 1.2 and 2.1; the fourth move of step 2.2 is the last of the $w_3(\lambda )=4$ removals of size $3$ exhibited there. The four hooks of $\lambda$ of $3$-divisible length, read from the hook-length diagram (rows $11,7,5,4,3,1$; $9,5,3,2,1$; $5,1$; $3$; $2$; $1$), are $H_{1,5}$ of length $3$, $H_{2,1}$ of length $9$, $H_{2,3}$ of length $3$ and $H_{4,1}$ of length $3$, and the bijection $f$ of [F5] sends them to the hooks $H_{1,1}({\lambda }^{(2)})$, $H_{1,1}({\lambda }^{(0)})$, $H_{1,2}({\lambda }^{(0)})$, $H_{2,1}({\lambda }^{(0)})$ of sizes $1,3,1,1$ of $(1)$, $(2,1)$, matching $\mathrm{\mid }H\mathrm{\mid }=3\mathrm{\mid }f(H)\mathrm{\mid }$ in each case; all assertions of the Example are thereby checked.

Beads on the runners: $7=3\cdot 2+1$, $5=3\cdot 1+2$, $3=3\cdot 1+0$, $1=3\cdot 0+1$, so the row-label sets are $X_0^{(3)}=\{1\}$ (bead $3$), $X_1^{(3)}=\{2,0\}$ (beads $7,1$) and $X_2^{(3)}=\{1\}$ (bead $5$). Top-packing puts the single beads of runners $0$ and $2$ in row $0$ and the two beads of runner $1$ in rows $0,1$, so $X^{(3)}=\{0\}\cup \{1,4\}\cup \{2\}=\{4,2,1,0\}$ and, listing the four beads in decreasing order, $C_3(\lambda )=P(X^{(3)})=(4-3,2-2,1-1,0-0)=(1)\mathrm{.}$

The moves of a single bead one row up are exactly $7\mapsto 4$ on runner $1$, $5\mapsto 2$ on runner $2$ and $3\mapsto 0$ on runner $0$, since $4,2,0\mathrm{\notin }X$; each target is a gap on the same runner as its bead, so by [F2] each move removes a rim hook of length $3$. In particular $X^{(3)}$ is reached after exactly these three moves, in any order.

First removal order: move $7\mapsto 4$, then $5\mapsto 2$, then $3\mapsto 0$, giving the $\beta$-sets $\{7,5,3,1\}\to \{5,4,3,1\}\to \{4,3,2,1\}\to \{4,2,1,0\}$ and, by [F1], the partitions $(4,3,2,1)\to (2,2,2,1)\to (1,1,1,1)\to (1),$ each step of size $3$ by [F2] and [F3] and terminating at $C_3(\lambda )=(1)$ by step 1.1.

Second removal order: move $3\mapsto 0$, then $7\mapsto 4$, then $5\mapsto 2$, giving $\{7,5,3,1\}\to \{7,5,1,0\}\to \{5,4,1,0\}\to \{4,2,1,0\}$ and the partitions $(4,3,2,1)\to (4,3)\to (2,2)\to (1),$ again three steps of size $3$ ending at $(1)$; both orders are maximal, since the terminal configuration $\{4,2,1,0\}$ has no bead with a gap one row above it on its runner, so no rim $3$-hook remains.

The size formula of [F4] reads $10=\mathrm{\mid }{C}_3(\lambda )\mathrm{\mid }+3w_3(\lambda )=1+3\cdot 3$, using step 1.1 and the three steps of either order; the partition $(1)$ has the single hook length $1$, so indeed no hook of length $3$ survives, and the two orders agree on the terminal partition as asserted.

The partition $(5)$. Its hook-length diagram is the single row $5,4,3,2,1$, so by [F1] the unique hook of length $3$ is $H_{1,3}(5)$, a horizontal rim hook of size $3$ spanning the nodes $(1,3),(1,4),(1,5)$ whose removal leaves the partition $(2)$; hence $(5)$ has exactly one removable $3$-hook. Its $\beta$-set is $X_{\lambda }=\{b_1\}=\{5\}$, a single bead on runner $2$ at row $1$; top-packing slides it to row $0$, so $X^{(3)}=\{2\}$ and $C_3((5))=P(\{2\})=(2)$, a $3$-core since $2<3$ and $(2)$ has hook lengths $2,1$.

The partition $(4,1)$. Its hook-length diagram is $5,3,2,1$ in the first row and $1$ in the second, so by [F1] the unique hook of length $3$ is $H_{1,2}(4,1)$, whose rim hook is the horizontal strip $(1,2),(1,3),(1,4)$ with complement $(1,1)$; hence $(4,1)$ also has exactly one removable $3$-hook. Its $\beta$-set is $X_{\mu }=\{b_1,b_2\}=\{5,1\}$, with $5$ on runner $2$ at row $1$ and $1$ on runner $1$ at row $0$; top-packing slides the bead $5$ to the position $2$, so $X^{(3)}=\{2,1\}$ and $C_3((4,1))=P(\{2,1\})=(1,1)$, a $3$-core since $(1,1)$ has hook lengths $2,1$.

The two partitions therefore have the same number (one) of removable $3$-hooks, but their $3$-cores are the distinct partitions $(2)$ and $(1,1)$ of size $2$; by [F3] the Specht modules $S_k^{(5)}$ and $S_k^{(4,1)}$ lie in different blocks of $k[S_5]$ for every splitting field $k$ of characteristic $3$. The top-packed abacus test of [F2] distinguishes the two configurations (one bead on runner $2$ versus beads on runners $2$ and $1$), while the local removable-hook count does not, so no such count can replace the core test.

The labels and their cores. The hook-length diagrams of the three partitions of $3$ are $3,2,1$ for $(3)$; $3,1$ in the first row and $1$ in the second for $(2,1)$; and $3,2,1$ vertically for $(1,1,1)$. In each case there is exactly one hook of length $3$ and removing its rim hook leaves the empty partition, so all three partitions have $3$-core $\mathrm{\varnothing }$; conversely any partition of $3$ with core $\mathrm{\varnothing }$ is one of these, since a partition whose core has size $3$ is its own core, and $\mathrm{\varnothing }$ has size $0=3-3$. Dominance orders them as $(3)⊳(2,1)⊳(1,1,1)$.

Compute the Weyl matrix directly. First work over ${\mathbb{F}}_3$ with lift $R={\mathbb{Z}}_{(3)}$, $\widehat{q}=1$, so the only nonzero hook valuation at rank three is $v([3{]}_1)=1$. The conjugate beta sequence of $(3)$ is $(3,2,1)$. Its hook pairs $(3,2)$ and $(3,1)$ give the balanced sequences $(5,0,1)$ and $(4,2,0)$, of sorted shapes $(1,1,1)$ with sign $-1$ and $(2,1)$ with sign $+1$, respectively. The other pair has coefficient zero. Thus its positive sum is $[W^{(2,1)}]-[W^{(1,1,1)}]$. For $(2,1)$, beta $(4,2,0)$ and its only nonzero pair $(3,1)$ give $(5,1,0)$, shape $(1,1,1)$ with sign $+1$, so its positive sum is $[W^{(1,1,1)}]$. The column shape has no row pair and zero positive sum, hence $W^{(1,1,1)}=L^{(1,1,1)}$. Positivity and $W(1)=\mathrm{rad}W$ then show that $W^{(2,1)}(1)=L^{(1,1,1)}$ and every higher submodule is zero, giving $[W^{(2,1)}]=[L^{(2,1)}]+[L^{(1,1,1)}]$. Substitute into the first sum to get $[L^{(2,1)}]$; the same argument gives $[W^{(3)}]=[L^{(3)}]+[L^{(2,1)}]$. This proves exactly the displayed complete Weyl matrix. Scalar extension to any original field $F$ of characteristic three preserves the Weyl modules/forms and the composition series, since all Schur simple heads are absolutely simple by [F2].

Determine the specific corner images directly. The group row symmetrizer $x_{(3)}={\sum }_{w\in S_3}w$ satisfies $x_{(3)}^2=6x_{(3)}$, so its rank-one Murphy form is zero in characteristic three. In the rank-one column cell the generator is $1$ and its square coefficient is $1$; the higher ideal contains $1+T_1$ and $1+T_2$ (the latter is a conjugate of the former), so this cell and its nonzero form quotient are the sign module. For the middle cell put $x=1+T_1$, $c_0=x$ and $c_1=x{T}_2$ modulo the higher row ideal. Its row symmetrizer satisfies $x_{(3)}=x+x{T}_2+x{T}_2{T}_1$, so $c_1{T}_1=-c_0-c_1$ while $c_0{T}_1=c_0$; and $T_2^2=1$ gives $c_0{T}_2=c_1$, $c_1{T}_2=c_0$. Thus its generator matrices are $M_1=\left(\begin{array}{cc}1& 0\\ -1& -1\end{array}\right)$ and $M_2=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The independent cell products $x^2=2x$, $x{T}_{2}x=x_{(3)}-x\equiv -x$ and $x{T}_2^{2}x=2x$ give $G=\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 2& 2\end{array}\right)$ over $F$. Its radical is the line $(1,-1)$; modulo it both generators act by $1$, so $D_{\mathrm{M}}^{(2,1)}$ is the nonzero trivial module. These two simple images are distinct since $1\mathrm{\ne }-1$. The exact corner/form identity [F3] therefore kills precisely $L^{(3)}$ among the three labels. Projecting the Weyl matrix just proved gives all displayed Murphy Specht rows over the original $F$, without any general finite-$e$ nonzero-label criterion.

Dipper-James rows and the direct group computation. Conjugating the partitions $(3)\leftrightarrow (1,1,1)$, $(2,1)\leftrightarrow (2,1)$ and applying the signed $\theta$-pullback to every Hecke module, the rows of step 3.1 become $[S^{(1,1,1)}]=[D^{(2,1)}]$, $[S^{(2,1)}]=[D^{(2,1)}]+[D^{(3)}]$ and $[S^{(3)}]=[D^{(3)}]$ in the $3$-regular labels, that is, the matrix $\left(\begin{array}{cc}1& 0\\ 1& 1\\ 0& 1\end{array}\right)$ with rows $(3),(2,1),(1,1,1)$ and columns $(3),(2,1)$. For the independent direct check, construct the tabloid module over $k$ using [F5]. Every middle-shape polytabloid is a difference $v_i-v_j$, and these span the two independent vectors $e_t,e_u$ displayed in [F5], so they are a basis of $S_k^{(2,1)}$. A one-row polytabloid is its unique tabloid, giving the trivial module; a one-column polytabloid is the nonzero alternating sum of all six distinct tabloids, giving the sign module. Put $w:=e_t+e_u=v_3-v_1+v_2-v_1=v_1+v_2+v_3$ in characteristic $3$: then $w\mathrm{\ne }0$ spans a one-dimensional submodule, since $\sigma \cdot v_i=v_{\sigma (i)}$ gives $\sigma \cdot w=w$ for every $\sigma \in S_3$, so it realises the trivial module; and in the quotient, $(12)\cdot e_t+e_t=(v_3-v_2)+(v_3-v_1)=2v_3-v_1-v_2=3v_3-w\equiv 0$ shows that $(12)$ acts by $-1$. All transpositions are conjugate and act on a one-dimensional quotient, so they all act by $-1$ and the quotient is sign. The orthonormal tabloid form has middle Gram matrix $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ in $(e_t,e_u)$; its characteristic-three kernel is $k(e_t+e_u)=kw$. Thus the James quotient $D^{(2,1)}$ is precisely this sign quotient, while the row form has norm $1$ and $D^{(3)}$ is trivial. With $S^{(3)}=\mathrm{triv}$, $S^{(1,1,1)}=\mathrm{sgn}$ by [F5], and the just-computed $D^{(3)}=\mathrm{triv}$, $D^{(2,1)}=\mathrm{sgn}$ (the complete list is [F4]), the direct rows of $S^{(3)},S^{(2,1)},S^{(1,1,1)}$ are $(1,0),(1,1),(0,1)$, matching the matrix obtained from the Schur functor entry by entry.