Frobenius Characteristic and the Symmetric-Group Character Dictionary — Examples

The cycle types of $S_3$ are $(1^3),(2,1),(3)$, with $z_{(1^3)}=1^3\cdot 3!=6$, $z_{(2,1)}=2\cdot 1=2$ and $z_{(3)}=3$, so [F2] gives $h_1=p_1$, $h_2=\frac{p_1^2+p_2}{2}$ and $h_3=\frac{p_1^3+3p_1{p}_2+2p_3}{6}$.

The three relevant Schur functions are $s_{(3)}=h_3$, $s_{(2,1)}=h_2{h}_1-h_3$ and $s_{(1,1,1)}=e_3=\omega (h_3)$ by [F1] and [F3].

The numbers of standard tableaux are $f^{(3)}=1$, $f^{(2,1)}=2$ and $f^{(1,1,1)}=1$ by [F5].

Substituting step 1.1 into step 1.2: $s_{(3)}=\frac{p_1^3+3p_1{p}_2+2p_3}{6}$; $s_{(2,1)}=\frac{p_1^2+p_2}{2}{p}_1-\frac{p_1^3+3p_1{p}_2+2p_3}{6}=\frac{3p_1^3+3p_1{p}_2-p_1^3-3p_1{p}_2-2p_3}{6}=\frac{p_1^3-p_3}{3}$; and $s_{(1,1,1)}=\omega (h_3)=\frac{p_1^3-3p_1{p}_2+2p_3}{6}$.

By [F4], ${\chi }^{\lambda }(\rho )$ is the coefficient of $p_{\rho }\mathrm{/}{z}_{\rho }$ in the expansion of step 2.1; since $z_{(1^3)}=6$, $z_{(2,1)}=2$, $z_{(3)}=3$, the coefficients of $p_{(1^3)}\mathrm{/}6,p_{(2,1)}\mathrm{/}2,p_{(3)}\mathrm{/}3$ in $s_{(3)},s_{(2,1)},s_{(1,1,1)}$ are respectively $(1,1,1)$, $(2,0,-1)$ and $(1,-1,1)$.

The identity-column sum is $1\cdot 1+2\cdot 2+1\cdot 1=6=\mathrm{\mid }{S}_3\mathrm{\mid }$. The weighted squared row norms are $\frac{1}{6}+\frac{1}{2}+\frac{1}{3}=1$, $\frac{4}{6}+0+\frac{1}{3}=1$, and $\frac{1}{6}+\frac{1}{2}+\frac{1}{3}=1$ in the displayed row order. The weighted products of the three distinct row pairs are $\frac{2}{6}-\frac{1}{3}=0$, $\frac{1}{6}-\frac{1}{2}+\frac{1}{3}=0$, and $\frac{2}{6}-\frac{1}{3}=0$. The column squared norms are $1+4+1=6$, $1+0+1=2$, and $1+1+1=3$, and the distinct column products are $1-1=0$, $1-2+1=0$, and $1-1=0$. Thus the rows are orthonormal with weights $1\mathrm{/}{z}_{\rho }$, and the columns are orthogonal with squared norms $z_{\rho }$.

Listing the three $(2,1)$-tabloids by their two-element row: $\{12\}\mathrm{\mid }\{3\}$, $\{13\}\mathrm{\mid }\{2\}$, $\{23\}\mathrm{\mid }\{1\}$. The identity fixes all three; the transposition $(12)$ fixes exactly $\{12\}\mathrm{\mid }\{3\}$, because a fixed tabloid must have both its row sets $w$-invariant, and $(12)$ preserves $\{1,2\}$ and $\{3\}$ but moves $\{1,3\}$ to $\{2,3\}$ and $\{2,3\}$ to $\{1,3\}$; the $3$-cycle $(123)$ fixes none, since a row set of size $2$ is never invariant under a $3$-cycle. Hence ${\phi }^{(2,1)}=(3,1,0)$ on the cycle types $(1^3),(2,1),(3)$.

By [F1], $\mathrm{ch}({\phi }^{(2,1)})=h_2{h}_1$.

The Kostka numbers for $\mu =(2,1)$ are $K_{(3),(2,1)}=1$ (the single semistandard tableau $112$ of shape $(3)$) and $K_{(2,1),(2,1)}=1$ (the tableau with first row $11$ and second row $2$), while $K_{(1,1,1),(2,1)}=0$ both because two entries equal to $1$ would have to occur in the same column and because $(1,1,1)$ does not dominate $(2,1)$ [F4]; so Young's rule [F3] gives $M^{(2,1)}\cong S^{(3)}\oplus S^{(2,1)}$.

By [F5] the dictionary example gives $s_{(3)}+s_{(2,1)}=\frac{p_1^3+3p_1{p}_2+2p_3}{6}+\frac{p_1^3-p_3}{3}=\frac{3p_1^3+3p_1{p}_2}{6}=\frac{p_1^3+p_1{p}_2}{2}$; on the other hand $h_2=\frac{p_1^2+p_2}{2}$, so $h_2{h}_1=\frac{(p_1^2+p_2)p_1}{2}=\frac{p_1^3+p_1{p}_2}{2}$. Therefore $h_2{h}_1=s_{(3)}+s_{(2,1)}$, and step 1.2 identifies this with $\mathrm{ch}({\phi }^{(2,1)})$.

Adding the values of ${\chi }^{(3)}$ and ${\chi }^{(2,1)}$ from [F5] gives $(1,1,1)+(2,0,-1)=(3,1,0)$ on the cycle types $(1^3),(2,1),(3)$, exactly the values computed for ${\phi }^{(2,1)}$ in step 1.1.

The two routes agree: directly, ${\phi }^{(2,1)}=(3,1,0)=1\cdot {\chi }^{(3)}+1\cdot {\chi }^{(2,1)}$ by step 2.2, matching the Young's-rule decomposition $M^{(2,1)}\cong S^{(3)}\oplus S^{(2,1)}$ of step 1.3; symmetrically, $\mathrm{ch}({\phi }^{(2,1)})=h_2{h}_1=s_{(3)}+s_{(2,1)}$ by steps 1.2 and 2.1, matching the dictionary table of [F5].

The cycle types of $S_4$ together with their centralizer orders are $(1^4)\text{\hspace{0.17em}⁣}:24$, $(2,1,1)\text{\hspace{0.17em}⁣}:4$, $(2,2)\text{\hspace{0.17em}⁣}:8$, $(3,1)\text{\hspace{0.17em}⁣}:3$ and $(4)\text{\hspace{0.17em}⁣}:4$; hence by [F2] and [F3], $h_4=\frac{p_1^4+6p_1^2{p}_2+8p_1{p}_3+3p_2^2+6p_4}{24}$ and $e_4=\frac{p_1^4-6p_1^2{p}_2+8p_1{p}_3+3p_2^2-6p_4}{24}$.

With $h_1=p_1$ and $h_3=\frac{p_1^3+3p_1{p}_2+2p_3}{6}$ and $e_3=\omega (h_3)=\frac{p_1^3-3p_1{p}_2+2p_3}{6}$, [F1] gives $s_{(3,1)}=h_3{h}_1-h_4$ and $s_{(2,1,1)}=e_3{e}_1-e_4$.

Substituting step 1.1 into step 1.2: $s_{(3,1)}=\frac{p_1^4+3p_1^2{p}_2+2p_1{p}_3}{6}-\frac{p_1^4+6p_1^2{p}_2+8p_1{p}_3+3p_2^2+6p_4}{24}=\frac{p_1^4+2p_1^2{p}_2-p_2^2-2p_4}{8}$, and $s_{(2,1,1)}=\frac{p_1^4-3p_1^2{p}_2+2p_1{p}_3}{6}-\frac{p_1^4-6p_1^2{p}_2+8p_1{p}_3+3p_2^2-6p_4}{24}=\frac{p_1^4-2p_1^2{p}_2-p_2^2+2p_4}{8}$.

By [F4] the values ${\chi }^{\lambda }(\rho )$ are the coefficients of $p_{\rho }\mathrm{/}{z}_{\rho }$ in step 2.1; with the centralizer orders of step 1.1 this gives ${\chi }^{(3,1)}=(3,1,-1,0,-1)$ and ${\chi }^{(2,1,1)}=(3,-1,-1,0,1)$ on the cycle types $(1^4),(2,1,1),(2,2),(3,1),(4)$.

Multiplying pointwise by the sign values $\mathrm{sgn}(\rho )=(-1{)}^{4-\mathrm{\ell }(\rho )}$, namely $(+1,-1,+1,+1,-1)$, turns the values of step 3.1 into $(3\cdot 1,1\cdot (-1),(-1)\cdot 1,0\cdot 1,(-1)\cdot (-1))=(3,-1,-1,0,1)$, which are exactly the values of ${\chi }^{(2,1,1)}$; this agrees with the module isomorphism $S^{(3,1)}\otimes \mathrm{sgn}\cong S^{(2,1,1)}$ and with $\omega (s_{(3,1)})=s_{(2,1,1)}$ of [F3].

The subgroup $H=S_1\times S_1$ is the trivial subgroup of $S_2$, and ${\mathbf{1}}_1⊠{\mathbf{1}}_1$ is the trivial character of $H$. By [F2], ${\mathrm{Ind}}_H^{S_2}({\mathbf{1}}_1⊠{\mathbf{1}}_1)(1)=\frac{1}{1}{\sum }_{x\in S_2}1=2$, while for the transposition $\tau$ the condition $x^{-1}\tau x\in H=\{1\}$ fails for every $x$, so $\mathrm{Ind}(\tau )=0$.

Since ${\mathbf{1}}_1$ is the trivial character of $S_1$, whose only cycle type is $(1)$ with $z_{(1)}=1$, [F3] gives $\mathrm{ch}({\mathbf{1}}_1)=p_1$; hence $\mathrm{ch}({\mathbf{1}}_1\circ {\mathbf{1}}_1)=\mathrm{ch}({\mathbf{1}}_1{)}^2=p_1^2$.

By [F7], $h_2=\frac{p_1^2+p_2}{2}$, so $e_2=\omega (h_2)=\frac{p_1^2-p_2}{2}$ by [F6], and therefore $h_2+e_2=p_1^2$.

The tensor product of two $1$-dimensional complex representations of $S_1$ is a $1$-dimensional complex representation of $S_1$ with character the product of the two characters, so it lies in $\mathrm{c}\mathrm{f}(S_1)$ and has degree $1$; the value of the product of two trivial characters at the identity is $1$.

By [F5] and step 1.3 applied to the identities $s_{(2)}=h_2,s_{(1,1)}=e_2$, one has $s_{(2)}+s_{(1,1)}=h_2+e_2=p_1^2$.

By [F4] and [F3], $\mathrm{ch}({\chi }^{(2)})=s_{(2)}=h_2=\frac{p_1^2+p_2}{2}$ and $\mathrm{ch}({\chi }^{(1,1)})=s_{(1,1)}=e_2=\frac{p_1^2-p_2}{2}$; reading the coefficients of $p_{\rho }\mathrm{/}{z}_{\rho }$ in these expansions with $z_{(1,1)}=z_{(2)}=2$ gives ${\chi }^{(2)}=(1,1)$ and ${\chi }^{(1,1)}=(1,-1)$ on the cycle types $(1^2),(2)$.

Steps 1.2 and 2.1 give $\mathrm{ch}({\mathbf{1}}_1\circ {\mathbf{1}}_1)=p_1^2=s_{(2)}+s_{(1,1)}$ in ${\mathrm{\Lambda }}^2$.

Step 3.1 and step 2.2 show that the character of the induced module ${\mathbf{1}}_1\circ {\mathbf{1}}_1$ equals ${\chi }^{(2)}+{\chi }^{(1,1)}$, namely $(2,0)$ as computed in step 1.1; by [F9] the induced module is isomorphic to $S^{(2)}\oplus S^{(1,1)}$, so the outer coefficient $c_{(1),(1)}^{(2)}=1$ is the multiplicity of $S^{(2)}$ in a module induced from $S_1\times S_1$ to $S_2$.

The two sides are therefore objects attached to different symmetric groups: the outer product ${\mathbf{1}}_1\circ {\mathbf{1}}_1$ is an element of $R(S_2)$, namely the degree-$2$ character $(2,0)$ with $\mathrm{ch}({\mathbf{1}}_1\circ {\mathbf{1}}_1)=p_1^2$, while the tensor product of the two one-dimensional trivial representations of $S_1$ is an element of $R(S_1)$ of degree $1$ with value $1$ at the identity [step 1.4]; a class function on $S_2$ with value $2$ at $1$ and a class function on the one-element group $S_1$ cannot be the same function, and an outer product coefficient records a multiplicity for representations of $S_m\times S_n$ inducing to $S_{m+n}$, not a tensor-product multiplicity inside one $R(S_n)$. This refutes the identified claim.