Young Diagrams Tableaux and Permutation Modules — Examples

The six lists are complete: a partition of $n\le 5$ whose largest part is $a$ is exactly a partition of $n-a$ with all parts at most $a$, with the part $a$ adjoined, so running over $a=n,n-1,\dots ,1$ recovers each list, and for $n=5$ this gives $a=5$: $(5)$; $a=4$: $(4,1)$; $a=3$: $(3,2)$ and $(3,1,1)$; $a=2$: $(2,2,1)$ and $(2,1,1,1)$; $a=1$: $(1,1,1,1,1)$, exactly the seven partitions displayed, with the same recursion for $n\le 4$.

Each displayed conjugate is read off as the column-height sequence of the diagram: $(4,1)$ has column heights $2,1,1,1$, so $(4,1{)}^{\mathrm{\prime }}=(2,1,1,1)$; $(3,2)$ has column heights $2,2,1$, so $(3,2{)}^{\mathrm{\prime }}=(2,2,1)$; $(3,1,1)$ has column heights $3,1,1$ and is self-conjugate; $(3,1)$ transposes to $(2,1,1)$ and $(2,2)$ to itself, matching the listed pairs, and taking column heights twice returns the original partition as in [F1].

For $n=5$, each listed consecutive pair is comparable, by the prefix sums of the two partitions: $(5)⊳(4,1)$ since $5>4$; $(4,1)⊳(3,2)$ since $4>3$; $(3,2)⊳(3,1,1)$ since $3=3$ and $5>4$; $(3,1,1)⊳(2,2,1)$ since $3>2$; $(2,2,1)⊳(2,1,1,1)$ since $2=2$, $4>3$; and $(2,1,1,1)⊳(1,1,1,1,1)$ since $2>1$.

The same computation for $n=4$ gives the chain $(4)⊳(3,1)⊳(2,2)⊳(2,1,1)⊳(1,1,1,1)$: the prefix sums compare as $4>3$; then $3>2$; then $2=2$ and $4>3$; then $2>1$, while the partitions of $n\le 3$ form the chains $(3)⊳(2,1)⊳(1,1,1)$, $(2)⊳(1,1)$ and the single partitions of $n\le 1$.

At $n=6$ the partition $(4,1,1)$ has prefix sums $4,5,6$ and $(3,3)$ has prefix sums $3,6,6$, so $4>3$ rules out $(3,3)⊵(4,1,1)$ while $5<6$ rules out $(4,1,1)⊵(3,3)$: the two are incomparable, and taking conjugates gives $(4,1,1{)}^{\mathrm{\prime }}=(3,1,1,1)$ and $(3,3{)}^{\mathrm{\prime }}=(2,2,2)$, whose prefix sums $3,4,5,6$ and $2,4,6,6$ also cross, as [F3] requires.

Steps 1.3 and 1.4 exhibit a chain through all partitions of each $n\le 5$, so any two partitions of the same $n\le 5$ are comparable by transitivity of the partial order $⊵$; together with step 1.5, which exhibits an incomparable pair of partitions of $6$, dominance is total exactly through size five and the first incomparable pair occurs at $n=6$. ∎

For $\lambda =(3,3,1)$ the row lengths are $3,3,1$, so the row endpoints are $(1,3),(2,3),(3,1)$; by [F1] the removable ones are those with ${\lambda }_i>{\lambda }_{i+1}$, and with ${\lambda }_4:=0$ these are $(2,3)$, because $3>1$, and $(3,1)$, because $1>0$, while ${\lambda }_1={\lambda }_2$ excludes $(1,3)$: deleting $(1,3)$ would leave the row lengths $2,3,1$, which are not weakly decreasing, so this row endpoint is not removable. Deleting the two removable nodes instead gives the partitions $(3,2,1)$ and $(3,3)$ of size $6$.

For $\lambda =(3,3,1)$ the addable points of the form $(i,{\lambda }_i+1)$ are $(1,4)$, which is in the first row, and $(3,2)$, because ${\lambda }_2=3>{\lambda }_3=1$, while $(2,4)$ fails the test ${\lambda }_1>{\lambda }_2$ since $3=3$; the point $(4,1)$ opens a new row and is addable by [F1]. Inserting these three points gives the partitions $(4,3,1)$ from $(1,4)$, $(3,3,2)$ from $(3,2)$ and $(3,3,1,1)$ from $(4,1)$.

For $\mu =(3,2,1)$ the row lengths are strictly decreasing, $3>2>1>0$, so by [F1] all three row endpoints are removable: $(1,3)$ with deletion $(2,2,1)$, $(2,2)$ with deletion $(3,1,1)$, and $(3,1)$ with deletion $(3,2)$, each of size $5$.

For $\mu =(3,2,1)$ the addable points are $(1,4)$ in the first row, $(2,3)$ because ${\mu }_1=3>{\mu }_2=2$, $(3,2)$ because ${\mu }_2=2>{\mu }_3=1$, and $(4,1)$ opening a new row, and no other point of the form $(i,{\mu }_i+1)$ passes the test of [F1]; inserting $(2,3)$ and $(3,2)$ gives the partitions $(3,3,1)$ and $(3,2,2)$ of size $7$.

For $\mathrm{\varnothing }$ the diagram has no nodes, so $\mathrm{Rem}(\mathrm{\varnothing })=\mathrm{\varnothing }$; a point $(i,j)$ outside the empty diagram leaves a Young diagram after insertion only for $(i,j)=(1,1)$, since the diagrams $\{(i,j)\}$ with $j\ge 2$ or $i\ge 2$ are not left-justified, so $\mathrm{Add}(\mathrm{\varnothing })=\{(1,1)\}$ and the resulting partition is $(1)$, in agreement with [F1].

The three shapes are thus completely described: $\mathrm{Rem}(3,3,1)=\{(2,3),(3,1)\}$ with the non-removable row endpoint $(1,3)$, $\mathrm{Rem}(3,2,1)=\{(1,3),(2,2),(3,1)\}$, $\mathrm{Rem}(\mathrm{\varnothing })=\mathrm{\varnothing }$, together with the addable sets $\mathrm{Add}(3,3,1)=\{(1,4),(3,2),(4,1)\}$, $\mathrm{Add}(3,2,1)=\{(1,4),(2,3),(3,2),(4,1)\}$ and $\mathrm{Add}(\mathrm{\varnothing })=\{(1,1)\}$. ∎

For $\lambda =(n)$ the blocks of [F1] are the single block $\{1,\dots ,n\}$, so $S_{(n)}=S_n$, and every $\lambda$-tableau has the single row set $\{1,\dots ,n\}$; hence any two $\lambda$-tableaux are row equivalent, there is exactly one tabloid, and $M^{(n)}$ is one-dimensional with that tabloid as basis.

For $\lambda =(1^n)$ the blocks are the $n$ singletons $\{1\},\dots ,\{n\}$, so $S_{(1^n)}=\{1\}$, and the row set of row $i$ of a tableau is the singleton carrying its single entry, so two tableaux are row equivalent exactly when they agree entrywise and each tabloid is a singleton class; a $\lambda$-tableau is a bijection from the $n$ boxes onto $\{1,\dots ,n\}$, so there are $n!$ of them, and the map $\sigma \mapsto \sigma \cdot t_0$ from $S_n$ to the tableaux is a bijection, being injective because $t_0$ is surjective and surjective because $\sigma (t_0(i,j)):=t(i,j)$ defines a permutation, whence $\sigma \mapsto \{\sigma \cdot t_0\}$ is a bijection onto the tabloids with $\tau \cdot \{\sigma \cdot t_0\}=\{\tau \sigma \cdot t_0\}$ for all $\tau \in S_n$.

For $\lambda =(n)$ and any $\sigma \in S_n$ the tableau $\sigma \cdot t_0$ again has the single row set $\{1,\dots ,n\}$, so $\sigma \cdot \{t_0\}=\{t_0\}$: every $\sigma$ acts as the identity on the one-dimensional $M^{(n)}$, which by [F3] is the trivial representation, in agreement with [F2] because $S_n\mathrm{/}{S}_n$ is a single coset.

Transporting the basis along the bijection of step 1.2 turns the action of [F1] into left multiplication on $\mathbb{C}[S_n]$, which by [F3] is the regular representation, so $M^{(1^n)}\cong \mathbb{C}[S_n]$; the coset description of [F2] reduces to the same statement because $S_{(1^n)}=\{1\}$ and the cosets $\sigma \{1\}$ are the singletons of $S_n$.

For $n=1$ the two partitions coincide, $(1)=(1^1)$, and both conclusions above give the same one-dimensional module; for $n=0$ the only partition is $\mathrm{\varnothing }$, whose tabloid set consists of the single empty tabloid, so $M^{\mathrm{\varnothing }}=\mathbb{C}$ is one-dimensional with trivial action and coincides with the regular module $\mathbb{C}[S_0]$ of the trivial group $S_0=\{1\}$. Thus $M^{(n)}$ is the trivial module and $M^{(1^n)}$ the regular module for every $n\ge 1$, and for $n=0$ both descriptions give the same one-dimensional module. ∎

For shape $(2,1)$ and content $(2,1)$ the filling carries two $1$'s and one $2$, so the single entry $2$ occupies one of the three boxes; placing it in the first box of the top row gives $21\mathrm{/}1$, whose top row violates weak increase since $2>1$; placing it in the second box of the top row gives $12\mathrm{/}1$, whose column has entries $1,1$ and violates strict increase; placing it in the bottom box gives $11\mathrm{/}2$, whose rows are weakly increasing and whose column entries $1<2$ strictly increase, and no other filling is available, so $K_{(2,1),(2,1)}=1$, realized by $11\mathrm{/}2$.

For shape $(2,1)$ and content $(1,1,1)$ the fillings are the bijections of the three boxes onto $\{1,2,3\}$, six in all; each of $12\mathrm{/}3$ and $13\mathrm{/}2$ is semistandard, since its rows are weakly increasing and its column entries are $1<3$ and $1<2$, while $21\mathrm{/}3$, $31\mathrm{/}2$ and $32\mathrm{/}1$ have a top row that is not weakly increasing and $23\mathrm{/}1$ has column entries $2,1$, which are not strictly increasing, and hence $K_{(2,1),(1,1,1)}=2$, in agreement with $K_{\lambda ,(1^n)}=f^{\lambda }$ and $f^{(2,1)}=2$.

For shape $(3)$ the diagram is a single row, whose weak increase forces the entries to be sorted, so the filling of content $(2,1)$ must be $1,1,2$, that is $112$, and this filling is semistandard because a one-row diagram has no column condition; hence $K_{(3),(2,1)}=1$.

For shape $(1,1,1)$ the diagram is a single column of three boxes, and strict increase down the column forces the three entries to be pairwise distinct; the content $(2,1)$ supplies only the two distinct labels $1$ and $2$, with $1$ repeated twice, so no filling of that content is semistandard and $K_{(1,1,1),(2,1)}=0$.

Steps 1.1, 1.2, 1.3 and 1.4 compute the four displayed numbers $1,2,1,0$, by checking all three content fillings in step 1.1 and all six in step 1.2, and by using the row and column conditions in steps 1.3 and 1.4 to leave respectively one and no semistandard fillings. Each of the latter two shapes has three fillings of content $(2,1)$ before imposing those conditions. ∎