Fixed-tabloid reconstruction of the character table (Fox–Mullineux)

Fix $\lambda ,\nu ⊢n$. Evaluating the inner product by the class decomposition of $S_{\nu }$ and then applying Frobenius reciprocity [F4] yields ${\sum }_{\rho }{\chi }^{\lambda }(C_{\rho })a_{\nu \rho }={\sum }_{g\in S_{\nu }}{\chi }^{\lambda }(g)=\mathrm{\mid }{S}_{\nu }\mathrm{\mid }⟨{\chi }^{\lambda }{\mathrm{\mid }}_{S_{\nu }},1_{S_{\nu }}{⟩}_{S_{\nu }}=\mathrm{\mid }{S}_{\nu }\mathrm{\mid }⟨{\chi }^{\lambda },1_{S_{\nu }}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\uparrow }^{S_n}{⟩}_{S_n}$. By [F1] the induced character is the character of $M^{\nu }$, which by Young's rule [F2] is ${\sum }_{\mu }{K}_{\mu \nu }{\chi }^{\mu }$; orthonormality [F3] gives $⟨{\chi }^{\lambda },1_{S_{\nu }}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\uparrow }^{S_n}⟩=K_{\lambda \nu }$, hence $b_{\lambda \nu }=\mathrm{\mid }{S}_{\nu }\mathrm{\mid }{K}_{\lambda \nu }$. Since $K_{\lambda \nu }$ is a nonnegative integer vanishing unless $\lambda ⊵\nu$, with $K_{\nu \nu }=1$, and since $\mathrm{\mid }{S}_{\nu }\mathrm{\mid }={\prod }_i{\nu }_i!>0$, the matrix $B$ is a nonnegative integer matrix, upper triangular in the fixed order, with $b_{\nu \nu }=\mathrm{\mid }{S}_{\nu }\mathrm{\mid }>0$.

The entries $a_{\lambda \rho }=\mathrm{\mid }{S}_{\lambda }\cap C_{\rho }\mathrm{\mid }$ are nonnegative integers. If $a_{\lambda \rho }\mathrm{\ne }0$, choose $\sigma \in S_{\lambda }\cap C_{\rho }$: the cycles of $\sigma$ lie in the blocks of $S_{\lambda }$, so the parts of $\rho$ are split into groups whose sums are the parts ${\lambda }_1,{\lambda }_2,\dots$ of $\lambda$. For every $r\ge 1$ the $r$ largest parts of $\rho$ lie in at most $r$ distinct groups, so ${\sum }_{i\le r}{\rho }_i\le {\sum }_{i\le r}{\lambda }_i$ (the group sizes ${\lambda }_i$ decrease and the groups are labelled in that order), whence $\lambda ⊵\rho$; therefore $a_{\lambda \rho }=0$ unless $\lambda ⊵\rho$ and $A$ is upper triangular in the fixed order. For the diagonal, an element of $S_{\lambda }$ of cycle type $\lambda$ consists of exactly one cycle of length ${\lambda }_i$ on each block of size ${\lambda }_i$, and the number of ${\lambda }_i$-cycles on a ${\lambda }_i$-element block is $({\lambda }_i-1)!$, so $a_{\lambda \lambda }={\prod }_i({\lambda }_i-1)!>0$. By [F6] the determinant of the triangular matrix $A$ is the product of its diagonal, $\det A={\prod }_{\lambda ⊢n}{\prod }_i({\lambda }_i-1)!\mathrm{\ne }0$; a nonzero rational number is a unit of $\mathbb{Q}$, so $A$ is invertible over $\mathbb{Q}$ by [F6].

For $\nu ,\mu ⊢n$, step 1.1 gives $(B^{\mathsf{T}}B{)}_{\nu \mu }={\sum }_{\lambda }{b}_{\lambda \nu }{b}_{\lambda \mu }=\mathrm{\mid }{S}_{\nu }\mathrm{\mid }\mathrm{\mid }{S}_{\mu }\mathrm{\mid }{\sum }_{\lambda }{K}_{\lambda \nu }{K}_{\lambda \mu }$. By Young's rule [F2] the characters of $M^{\nu }$ and $M^{\mu }$ are ${\sum }_{\lambda }{K}_{\lambda \nu }{\chi }^{\lambda }$ and ${\sum }_{\lambda }{K}_{\lambda \mu }{\chi }^{\lambda }$, so orthonormality [F3] gives ${\sum }_{\lambda }{K}_{\lambda \nu }{K}_{\lambda \mu }=⟨1_{S_{\nu }}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\uparrow }^{S_n},1_{S_{\mu }}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\uparrow }^{S_n}{⟩}_{S_n}$. For $g$ of cycle type $\rho$, Frobenius' formula [F4] and the change-of-variable count $\mathrm{\mid }\{x\in S_n:x^{-1}gx=h\}\mathrm{\mid }=z_{\rho }$ for each $h\in C_{\rho }$ give $(1_{S_{\nu }}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\uparrow }^{S_n})(g)=z_{\rho }{a}_{\nu \rho }\mathrm{/}\mathrm{\mid }{S}_{\nu }\mathrm{\mid }$. Hence $⟨1_{S_{\nu }}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\uparrow }^{S_n},1_{S_{\mu }}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\uparrow }^{S_n}⟩=\frac{1}{n!}{\sum }_{\rho }\mathrm{\mid }{C}_{\rho }\mathrm{\mid }\frac{z_{\rho }{a}_{\nu \rho }}{\mathrm{\mid }{S}_{\nu }\mathrm{\mid }}\frac{z_{\rho }{a}_{\mu \rho }}{\mathrm{\mid }{S}_{\mu }\mathrm{\mid }}=\frac{1}{\mathrm{\mid }{S}_{\nu }\mathrm{\mid }\mathrm{\mid }{S}_{\mu }\mathrm{\mid }}{\sum }_{\rho }{z}_{\rho }{a}_{\nu \rho }{a}_{\mu \rho }$, where $\mathrm{\mid }{C}_{\rho }\mathrm{\mid }=n!\mathrm{/}{z}_{\rho }$ by [F5]. Multiplying by $\mathrm{\mid }{S}_{\nu }\mathrm{\mid }\mathrm{\mid }{S}_{\mu }\mathrm{\mid }$ gives $(B^{\mathsf{T}}B{)}_{\nu \mu }={\sum }_{\rho }{z}_{\rho }{a}_{\nu \rho }{a}_{\mu \rho }=(AM{A}^{\mathsf{T}}{)}_{\nu \mu }$, the last equality by the definitions of $A$, $M$ and matrix multiplication.

Index the fixed total order by $1,\dots ,q$ and put $G:=AM{A}^{\mathsf{T}}=B^{\mathsf{T}}B$. For any upper-triangular real matrix $U$ with positive diagonal and $U^{\mathsf{T}}U=G$, process columns $j=1,\dots ,q$, and in column $j$ process rows $i=1,\dots ,j$. If $i<j$, the equation $G_{ij}={\sum }_{k=1}^i{u}_{ki}{u}_{kj}$ determines $u_{ij}=(G_{ij}-{\sum }_{k<i}{u}_{ki}{u}_{kj})\mathrm{/}{u}_{ii}$: column $i$ and the earlier rows of column $j$ are already determined, and $u_{ii}>0$. For $i=j$, $G_{jj}={\sum }_{k<j}{u}_{kj}^2+u_{jj}^2$ determines the unique positive $u_{jj}$. Induction therefore gives at most one such $U$, and step 2.1 supplies $U=B$, whose diagonal is $\mathrm{\mid }{S}_{\nu }\mathrm{\mid }$ by step 1.1. This proves the asserted uniqueness even among all upper-triangular real matrices with positive diagonal; no dominance sparsity is assumed of a competitor.

By definition $B=C{A}^{\mathsf{T}}$, and $A^{\mathsf{T}}$ is invertible over $\mathbb{Q}$ by step 1.2, so multiplying the identity on the right by $(A^{\mathsf{T}}{)}^{-1}$ gives $C=B(A^{\mathsf{T}}{)}^{-1}$: the character table is recovered from $A$ alone (which gives $M$ and the Gram matrix $AM{A}^{\mathsf{T}}$) together with the Gram equation (which determines $B$ by step 3.1). For the determinant, $B$ and $A^{\mathsf{T}}$ are triangular, so their determinants are the products of their diagonals by [F6]: $\det B={\prod }_{\lambda ⊢n}{b}_{\lambda \lambda }={\prod }_{\lambda ⊢n}\mathrm{\mid }{S}_{\lambda }\mathrm{\mid }={\prod }_{\lambda ⊢n}{\prod }_i{\lambda }_i!$, and $\det (A^{\mathsf{T}})=\det A={\prod }_{\lambda ⊢n}{\prod }_i({\lambda }_i-1)!$ by step 1.2. Therefore $\det C=\det B\cdot \det ((A^{\mathsf{T}}{)}^{-1})=\det B\mathrm{/}\det (A^{\mathsf{T}})={\prod }_{\lambda ⊢n}{\prod }_i\frac{{\lambda }_i!}{({\lambda }_i-1)!}={\prod }_{\lambda ⊢n}{\prod }_i{\lambda }_i$, which is the product of all parts of all partitions of $n$.