Rim Hooks and the Murnaghan–Nakayama Rule — Examples

The class sizes and $M$. By [F1], $z_{\rho }={\prod }_i{i}^{m_i(\rho )}{m}_i(\rho )!$ evaluates to $5,4,6,6,8,12,120$ on the classes $(5),(4,1),(3,2),(3,1,1),(2,2,1),(2,1,1,1),(1^5)$: for instance $z_{(3,1,1)}=3\cdot 1^2\cdot 2!=6$ and $z_{(2,2,1)}=2^2\cdot 2!\cdot 1=8$. Hence $\mathrm{\mid }{C}_{\rho }\mathrm{\mid }=120\mathrm{/}{z}_{\rho }$ is $24,30,20,20,15,10,1$ and $M=\mathrm{diag}(5,4,6,6,8,12,120)$.

The matrix $A$. By [F2] the cycle types in $S_{\lambda }$ are obtained by combining the cycle types of the blocks. The single-block counts used are: in $S_1$, type $1$: $1$; in $S_2$, types $1^2,2$: $1,1$; in $S_3$, types $1^3,2\cdot 1,3$: $1,3,2$; in $S_4$, types $1^4,2\cdot 1^2,3\cdot 1,2^2,4$: $1,6,8,3,6$; and in $S_5$ the counts are the class sizes $1,10,15,20,20,30,24$ of types $1^5,2\cdot 1^3,2^2\cdot 1,3\cdot 1^2,3\cdot 2,4\cdot 1,5$. Combining block counts for each $\lambda$ (for example, in $S_{(3,2)}=S_3\times S_2$ the identity gives one element of type $1^5$; a $3$-cycle together with the identity gives the $2$ elements of type $3\cdot 1^2$; a $3$-cycle with the transposition gives the $2$ elements of type $3\cdot 2$; a transposition in $S_3$ combined with the identity, or the identity in $S_3$ combined with the transposition of $S_2$, gives the $3+1=4$ elements of type $2\cdot 1^3$; and a transposition in $S_3$ combined with the transposition in $S_2$ gives the $3$ elements of type $2^2\cdot 1$) yields exactly the displayed matrix $A$.

The matrix $B$. By [F3] the entry $b_{\lambda \nu }$ is $\mathrm{\mid }{S}_{\nu }\mathrm{\mid }$ times the number of semistandard tableaux of shape $\lambda$ and content $\nu$; the diagonal is $b_{\nu \nu }=\mathrm{\mid }{S}_{\nu }\mathrm{\mid }={\prod }_i{\nu }_i!=(120,24,12,6,4,2,1)$, and the displayed nonzero entries are confirmed by the corresponding counts: $b_{(4,1),(3,1,1)}=6\cdot 2=12$, because a semistandard tableau of shape $(4,1)$ and content $(3,1,1)$ must place $2$ or $3$ in the single lower cell and then is determined; $b_{(3,2),(2,2,1)}=4\cdot 2=8$ and $b_{(2,2,1),(2,1,1,1)}=2\cdot 2=4$ are checked the same way. The whole displayed matrix $B$ satisfies the Gram equation with the displayed $A$ and $M$: each entry of $B^{\mathsf{T}}B$ is the finite sum ${\sum }_{\lambda }{b}_{\lambda \nu }{b}_{\lambda \mu }$, and comparing with ${\sum }_{\rho }{z}_{\rho }{a}_{\nu \rho }{a}_{\mu \rho }$ gives, for example, $(B^{\mathsf{T}}B{)}_{(1^5),(1^5)}=1^2+4^2+5^2+6^2+5^2+4^2+1^2=120=z_{(1^5)}=(AM{A}^{\mathsf{T}}{)}_{(1^5),(1^5)}$ and $(B^{\mathsf{T}}B{)}_{(4,1),(3,2)}=24\cdot 12+24\cdot 12=576=6\cdot 8\cdot 2+8\cdot 3\cdot 3+12\cdot 6\cdot 4+120=(AM{A}^{\mathsf{T}}{)}_{(4,1),(3,2)}$. By the uniqueness clause of [F3] the nonnegative upper-triangular matrix with this Gram matrix and positive diagonal is $B$, so no other matrix can occur.

The character table. Multiplying $B$ on the right by the inverse transpose of $A$ gives the displayed $C$; equivalently, $C{A}^{\mathsf{T}}=B$, and this is the identity ${\sum }_{\rho }{\chi }^{\lambda }(C_{\rho })a_{\nu \rho }=b_{\lambda \nu }$ of the theorem. Two full rows of the product check it: the row $(5)$ consists of ones, and ${\sum }_{\rho }{a}_{\nu \rho }=\mathrm{\mid }{S}_{\nu }\mathrm{\mid }=b_{(5),\nu }$ for every $\nu$; for the row $(3,2)=(0,-1,1,-1,1,1,5)$ one gets ${\sum }_{\rho }{\chi }^{(3,2)}(C_{\rho })a_{(3,2),\rho }=0\cdot 0+(-1)\cdot 0+1\cdot 2+(-1)\cdot 2+1\cdot 3+1\cdot 4+5\cdot 1=12=b_{(3,2),(3,2)}$ and ${\sum }_{\rho }{\chi }^{(3,2)}(C_{\rho })a_{(4,1),\rho }=(-1)\cdot 6+(-1)\cdot 8+1\cdot 3+1\cdot 6+5\cdot 1=0=b_{(3,2),(4,1)}$; the remaining rows are verified by the same finite sums from the displayed $A$ and $B$. By the theorem's determinant formula, $\det C={\prod }_{\lambda ⊢5}{\prod }_i{\lambda }_i=5\cdot 4\cdot 6\cdot 3\cdot 4\cdot 2\cdot 1=2880$, the product of all parts of all partitions of $5$.

The orthogonality spot check of the example statement: with the class sizes of step 1.1 and the rows of $C$, $⟨{\chi }^{(3,2)},{\chi }^{(2,2,1)}⟩=\frac{1}{120}(24\cdot 0\cdot 0+30\cdot (-1)\cdot 1+20\cdot 1\cdot (-1)+20\cdot (-1)\cdot (-1)+15\cdot 1\cdot 1+10\cdot 1\cdot (-1)+1\cdot 5\cdot 5)=\frac{1}{120}(0-30-20+20+15-10+25)=0$, and $⟨{\chi }^{(3,2)},{\chi }^{(3,2)}⟩=\frac{1}{120}(30\cdot 1+20\cdot 1+20\cdot 1+15\cdot 1+10\cdot 1+25\cdot 1)=1$; the entries are real, so the conjugate in [F4] plays no role. Thus the displayed table is orthonormal on the checked pairs, as a character table must be.

A partition $\lambda ⊢8$ contains $(3,2)$ exactly when ${\lambda }_1\ge 3$ and ${\lambda }_2\ge 2$, leaving the eleven candidates $(6,2),(5,2,1),(4,2,2),(4,2,1,1),(3,2,2,1),(3,2,1,1,1),(5,3),(4,3,1),(3,3,2),(3,3,1,1),(4,4)$; among their skew differences $\lambda \mathrm{/}(3,2)$, which all have $3$ nodes, exactly four are edge-connected and contain no $2\times 2$ block, namely the horizontal strip $\{(1,4),(1,5),(1,6)\}$ with $c=3$ columns in $\lambda =(6,2)$, the corners $\{(1,4),(2,3),(2,4)\}$ and $\{(3,1),(3,2),(4,1)\}$ with $c=2$ in $\lambda =(4,4)$ and $\lambda =(3,2,2,1)$, and the vertical strip $\{(3,1),(4,1),(5,1)\}$ with $c=1$ in $\lambda =(3,2,1,1,1)$; the other seven differences are disconnected, for instance $\lambda =(5,3)$, where the node $(2,3)$ is not adjacent to $(1,4)$ or $(1,5)$. The heights are $3-c$, namely $0,1,1,2$, so the four signs $+,-,-,+$ and the displayed expression for ${\mathrm{\Psi }}_{(3,2)}$ follow.

The four dimensions of [F5] give ${\mathrm{\Psi }}_{(3,2)}(1)=20-14-70+64=0$; equivalently ${\mathrm{\Psi }}_{(3,2)}={\chi }^{(3,2)}\circ {\psi }_3$ by [F1] and ${\psi }_3(1)=0$ because the identity of $S_3$ is not a $3$-cycle, so the degree vanishes.

On the class of $8$-cycles each of the four shapes contains a $2\times 2$ block (rows $1$ and $2$, columns $1$ and $2$), so none is a hook and ${\chi }^{\lambda }=0$ on that class for all four $\lambda$ by the full-cycle lemma [F4]; hence ${\mathrm{\Psi }}_{(3,2)}$ vanishes there.

On the class $(4,4)$ the rule [F3] gives: ${\chi }^{(6,2)}=0$, since the unique $4$-strip takes $(6,2)$ to $(2,2)$ with height $0$ and $(2,2)$ has no $4$-strip; ${\chi }^{(4,4)}=+1-(-1)=2$ from the strips to $(4)$ (height $0$) and to $(3,1)$ (height $1$), with ${\chi }^{(4)}(4)=1$ and ${\chi }^{(3,1)}(4)=-1$; ${\chi }^{(3,2,2,1)}=(-1)+(-1)=-2$ from the strips to $(3,1)$ and to $(1^4)$, both of height $2$, with ${\chi }^{(3,1)}(4)={\chi }^{(1^4)}(4)=-1$; and ${\chi }^{(3,2,1,1,1)}=0$ because $(3,2,1,1,1)$ has no $4$-strip. Hence ${\mathrm{\Psi }}_{(3,2)}=+0-2-(-2)+0=0$ on the class $(4,4)$, whose cycle type has no part $3$.

Steps 2.1, 2.2 and 2.3 verify the vanishing predicted by [F1] on the identity class, the class of $8$-cycles and the class $(4,4)$, all of which are classes of $S_8$ without a $3$-cycle, confirming the displayed value ${\mathrm{\Psi }}_{(3,2)}=0$ there.

Row $(4)$: for each part $r$ of the cycle type the unique removable $r$-strip of the row $(4)$ is horizontal and has height $0$, leaving the row $(4-r)$, so iterating the rule [F1] gives ${\chi }^{(4)}=1$ on all five classes. Row $(1^4)$ is the sign character: the unique $r$-strip of the column is vertical of height $r-1$, so iterating [F1] gives ${\chi }^{(1^4)}(\rho )={\prod }_i(-1{)}^{r_i-1}$ for the parts $r_i$ of the cycle type, that is the values $1,-1,1,1,-1$ on $(1^4),(2,1,1),(2,2),(3,1),(4)$.

Row $(3,1)$: by [F2] the dimension is $4!\mathrm{/}(4\cdot 2\cdot 1\cdot 1)=3$; on $(2,1,1)$ the only $2$-strip removes the domino $\{(1,2),(1,3)\}$ and leaves $(1,1)$ with height $0$, so the value is ${\chi }^{(1,1)}(1,1)=1$; on $(2,2)$ the same strip gives ${\chi }^{(1,1)}(2)=-1$; on $(3,1)$ the hook lengths $4,2,1,1$ contain no $3$, so there is no $3$-strip and the value is $0$; on $(4)$ the whole diagram $(3,1)$ is the unique $4$-strip, of height $1$, so the value is $-{\chi }^{\mathrm{\varnothing }}=-1$. This gives the row $3,1,-1,0,-1$.

Row $(2,2)$: the two $2$-strips are the bottom row to $(2)$ (height $0$) and the right column to $(1,1)$ (height $1$), so on $(2,1,1)$ the value is ${\chi }^{(2)}(1,1)-{\chi }^{(1,1)}(1,1)=1-1=0$, on $(2,2)$ it is ${\chi }^{(2)}(2)-{\chi }^{(1,1)}(2)=1-(-1)=2$; on $(3,1)$ the unique $3$-strip (leaving the single node $(1)$) has height $1$ and gives $-{\chi }^{(1)}(1)=-1$; on $(4)$ the hook lengths $3,2,2,1$ contain no $4$, so the value is $0$; the dimension is $2$ by [F2]. Row $(2,1,1)$: the unique $2$-strip removes the vertical domino $\{(2,1),(3,1)\}$ and leaves $(2)$ with height $1$, so the values on $(2,1,1)$ and $(2,2)$ are $-{\chi }^{(2)}(1,1)=-1$ and $-{\chi }^{(2)}(2)=-1$; there is no $3$-strip because the hook lengths are $4,2,1,1$; the whole diagram is the unique $4$-strip, of height $2$, giving ${\chi }^{\mathrm{\varnothing }}=1$ on $(4)$; the dimension is $3$ by [F2]. This gives the rows $2,0,2,-1,0$ and $3,-1,-1,0,1$.

With the class sizes $1,6,3,8,6$ [F3] and the standard inner product [F4], the five rows have norms $\frac{1}{24}(1+6+3+8+6)=1$ for $(4)$ and $(1^4)$, $\frac{1}{24}(9+6+3+0+6)=1$ for $(3,1)$ and $(2,1,1)$, and $\frac{1}{24}(4+0+12+8+0)=1$ for $(2,2)$; pairwise orthogonality holds as in the displayed computation $⟨{\chi }^{(3,1)},{\chi }^{(2,2)}⟩=\frac{1}{24}(6-6)=0$, and likewise $⟨{\chi }^{(2,2)},{\chi }^{(2,1,1)}⟩=\frac{1}{24}(1\cdot 2\cdot 3+6\cdot 0\cdot (-1)+3\cdot 2\cdot (-1)+8\cdot (-1)\cdot 0+6\cdot 0\cdot 1)=0$, the remaining pairs checking in the same way. Hence the five rows computed by the recursion are exactly the character table of $S_4$ displayed in the statement, with the rows orthonormal with respect to the class sizes of [F3].