Integral Specht Modules and Modular Simple Modules

By [F3] the tabloids $\gamma \cdot \{t\}$ with $\gamma \in C_t$ are pairwise distinct, so in the expansion $e_t={\sum }_{\gamma }\mathrm{sgn}(\gamma )\{\gamma \cdot t\}$ no tabloid occurs twice; every coefficient is $0$ or $\pm 1$, the coefficient of $\{t\}$ is $1$, and $e_t\mathrm{\ne }0$. In particular $e_t$ is a genuine nonzero integral vector of $M_{\mathbb{Z}}^{\lambda }$.

For $\sigma \in S_n$ the column sets satisfy $C_{\sigma \cdot t}=\sigma C_t{\sigma }^{-1}$, and the substitution $\gamma =\sigma \delta {\sigma }^{-1}$ gives $e_{\sigma \cdot t}={\sum }_{\delta \in C_t}\mathrm{sgn}(\delta )\{\sigma \delta \cdot t\}=\sigma \cdot e_t$. This identity of two integral vectors holds over $\mathbb{C}$ by [F8], and by the uniqueness of tabloid coordinates it therefore holds in $M_{\mathbb{Z}}^{\lambda }$. Since every $\lambda$-tableau is $\sigma \cdot t_0$ for the tableau $t_0$ that lists $1,\dots ,n$ along the rows of $[\lambda ]$, the lattice $S_{\mathbb{Z}}^{\lambda }$ is generated over $\mathbb{Z}[S_n]$ by the single polytabloid $e_{t_0}$.

The Garnir element of [F5] is an integral group-algebra element, and $G_{X,Y}{e}_t={\sum }_{g\in T}{\sum }_{\gamma \in C_t}\mathrm{sgn}(g)\mathrm{sgn}(\gamma )\{g\gamma \cdot t\}$ has integer coefficients in the tabloid basis. The identity $G_{X,Y}{e}_t=0$ holds in $M_{\mathbb{C}}^{\lambda }$ between two integral vectors, so by uniqueness of tabloid coordinates it holds in $M_{\mathbb{Z}}^{\lambda }$: the Garnir relation is integral.

For $\lambda =\mathrm{\varnothing }$ there is exactly one tabloid and one tableau, with $C_t=\{1\}$, ${\kappa }_t=1$ and $e_t=\{\mathrm{\varnothing }\}$; the empty tableau is standard by [F11]. Thus $M_{\mathbb{Z}}^{\mathrm{\varnothing }}=\mathbb{Z}\{\mathrm{\varnothing }\}\cong \mathbb{Z}$ and $S_{\mathbb{Z}}^{\mathrm{\varnothing }}=\mathbb{Z}\{\mathrm{\varnothing }\}$ has rank one with the empty polytabloid as basis.

Run the published spanning argument inside $M_{\mathbb{Z}}^{\lambda }$: for column-standard $s$ that is not standard, the integral Garnir relation of step 1.3 with the explicit transversal containing $1$ isolates the identity term and gives $e_s=-{\sum }_{A\mathrm{\ne }X}\mathrm{sgn}(g_A)e_{g_A\cdot s}$ with each $g_A$ an explicit product of disjoint swaps; sorting columns and using the covariance of step 1.2 rewrites each $e_{g_A\cdot s}$ as $\pm e_u$ with $u$ column-standard and $s\prec u$ in the finite order [F4]; and any tableau is taken to column-standard form by a column permutation, again by step 1.2. Reverse induction along [F4] therefore gives, for every $\lambda$-tableau $t$, an identity $e_t={\sum }_{u\text{ standard}}{c}_{tu}{e}_u$ with integer coefficients $c_{tu}$. Hence the standard polytabloids span $S_{\mathbb{Z}}^{\lambda }$ over $\mathbb{Z}$.

Order the standard $\lambda$-tableaux $t_1,\dots ,t_d$ so that $\{t_1\}<\cdots <\{t_d\}$ in the tabloid order of [F4], possible since distinct standard tableaux have distinct tabloids and the order is total. By [F7], $e_{t_i}=\{t_i\}+{\sum }_{T<\{t_i\}}{c}_{T}T$ with integer $c_T$, so the coefficient of $\{t_j\}$ in $e_{t_i}$ is $0$ for $j>i$ and $1$ for $j=i$. A relation ${\sum }_i{a}_i{e}_{t_i}=0$ with integral $a_i$ therefore forces $a_d=0$, then $a_{d-1}=0$, and so on; the standard polytabloids are $\mathbb{Z}$-linearly independent and, with step 2.1, form a $\mathbb{Z}$-basis of $S_{\mathbb{Z}}^{\lambda }$.

Let $P:{\mathbb{Z}}^d\to M_{\mathbb{Z}}^{\lambda }$ send the $i$-th standard basis vector to $e_{t_i}$, so that $\mathrm{im}P=S_{\mathbb{Z}}^{\lambda }$ by step 3.1, and let $\pi :M_{\mathbb{Z}}^{\lambda }\to {\mathbb{Z}}^d$ take tabloid coordinates at $\{t_1\},\dots ,\{t_d\}$. By step 3.1 the matrix of $\pi \circ P$ is upper unitriangular with integer entries, hence invertible over $\mathbb{Z}$ by integer back-substitution, and $\rho :=(\pi \circ P{)}^{-1}\circ \pi$ satisfies $\rho \circ P=\mathrm{i}\mathrm{d}$. Thus $P$ is injective and split, $S_{\mathbb{Z}}^{\lambda }$ is a direct summand of the free $\mathbb{Z}$-module $M_{\mathbb{Z}}^{\lambda }$, and $M_{\mathbb{Z}}^{\lambda }\mathrm{/}{S}_{\mathbb{Z}}^{\lambda }$ is free: the lattice is saturated.

Let $R$ be a commutative ring and use the identification $R{\otimes }_{\mathbb{Z}}{M}_{\mathbb{Z}}^{\lambda }\cong M_R^{\lambda }$ of [F10]. The map ${\mathrm{i}\mathrm{d}}_R\otimes P$ has image $R{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }$ and ${\mathrm{i}\mathrm{d}}_R\otimes \rho$ is a left inverse of it, so ${\mathrm{i}\mathrm{d}}_R\otimes P$ is injective; hence the canonical map $R{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }\to M_R^{\lambda }$ is injective onto the $R$-span of the vectors $1\otimes e_{t_i}$, which is the $R$-span of the standard polytabloids. By the identity of step 2.1 every polytabloid is an integral combination of standard ones, so this span is exactly $S_R^{\lambda }$; therefore $S_R^{\lambda }\cong R{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }$ for every commutative ring $R$, including rings of prime characteristic.

Taking $R=\mathbb{Z}$ in step 5.1 returns $S_{\mathbb{Z}}^{\lambda }$, and taking $\lambda =\mathrm{\varnothing }$ returns the rank-one lattice $\mathbb{Z}\{\mathrm{\varnothing }\}$ of step 1.4; the standard-polytabloid basis is step 3.1 and saturation is step 4.1. This proves the three asserted properties for all $n\ge 0$, including $n=0$.

The prescription $\beta (T,U)={\delta }_{TU}$ on the tabloid basis extends uniquely to a $\mathbb{Z}$-bilinear map by [F1], and the formula $\beta ({\sum }_T{a}_{T}T,{\sum }_U{b}_{U}U)={\sum }_T{a}_T{b}_T$ is finite because only finitely many coefficients are nonzero. If $\beta (x,U)=0$ for every tabloid $U$, then taking $U$ to be a tabloid occurring in $x$ gives $a_U=0$, so all coefficients vanish and $x=0$; the form is nondegenerate.

For $\sigma \in S_n$, [F3] says that $T\mapsto \sigma T$ is a bijection of the tabloid set, so $\beta (\sigma T,\sigma U)={\delta }_{\sigma T,\sigma U}={\delta }_{TU}=\beta (T,U)$ on basis vectors, and bilinearity extends the identity to all $x,y$.

Every entry $\beta (e_s,e_t)$ is a finite sum ${\sum }_T{c}_s(T)c_t(T)$ of products of tabloid coefficients, each of which lies in $\{0,1,-1\}$ by [F4]; hence $\beta (e_s,e_t)\in \mathbb{Z}$ and, using the standard basis [F2], $G_{\lambda }$ is a matrix over $\mathbb{Z}$. Since $\beta$ is symmetric, so is $G_{\lambda }$.

Applying step 1.2 to each term of ${\kappa }_t$ gives $\beta ({\kappa }_{t}x,y)={\sum }_{\gamma \in C_t}\mathrm{sgn}(\gamma )\beta (\gamma x,y)={\sum }_{\gamma }\mathrm{sgn}(\gamma )\beta (x,{\gamma }^{-1}y)$; reindexing $\gamma \mapsto {\gamma }^{-1}$ and using $\mathrm{sgn}({\gamma }^{-1})=\mathrm{sgn}(\gamma )$ this is $\beta (x,{\kappa }_{t}y)$, so ${\kappa }_t$ is self-adjoint for $\beta$.

Let $R$ be a commutative ring. Under the identification of [F1], the $R$-bilinear form ${\beta }_R$ with orthonormal tabloid basis sends $1\otimes x,1\otimes y$ to $\beta (x,y)$ times $1_R$, by expanding $x$ and $y$ in the tabloid basis; hence ${\beta }_R$ is the scalar extension of $\beta$, and the matrix of ${\beta }_R$ in the standard basis $1\otimes e_{s_i}$ is the scalar extension of $G_{\lambda }$.

Over the real field, $x={\sum }_T{a}_{T}T$ with $a_T\in \mathbb{R}$ satisfies $\beta (x,x)={\sum }_T{a}_T^2\ge 0$, with equality only when all $a_T=0$, i.e. $x=0$; this is the positivity that holds on real vectors and it is the same quantity as the Hermitian form of [F5] on real coefficients, while the conjugate-linearity of [F5] and the positivity have no counterpart over a general commutative ring, and in particular cannot be reduced modulo a prime. The self-adjointness of step 2.1 and the integrality and scalar extension of steps 1.3 and 2.2 complete the asserted definition.

The field $k$ is the residue field of the splitting $p$-modular system fixed above, hence has characteristic $p$ and is a splitting field for $S_n$ and all its subgroups by [F1]. Thus $M_k^{\lambda }$ is a finite-dimensional $k$-vector space with the tabloids as basis, and $S_k^{\lambda }$ is the $k$-span of the polytabloids with the standard polytabloids as basis by [F2]. The form ${\beta }_k$ of [F3] is nondegenerate and $S_n$-invariant.

If $U\subseteq M_k^{\lambda }$ is an $S_n$-submodule, then $U^{\perp }$ is an $S_n$-submodule: for $x\in U^{\perp }$, $u\in U$ and $\sigma \in S_n$, invariance gives ${\beta }_k(\sigma x,u)={\beta }_k(x,{\sigma }^{-1}u)=0$ because ${\sigma }^{-1}u\in U$, so $\sigma x\in U^{\perp }$.

Since $S_k^{\lambda }$ is an $S_n$-submodule by [F2] and [F3], step 1.2 shows that $(S_k^{\lambda }{)}^{\perp }$ is an $S_n$-submodule, hence so is $R^{\lambda }=S_k^{\lambda }\cap (S_k^{\lambda }{)}^{\perp }$; the quotient $D^{\lambda }=S_k^{\lambda }\mathrm{/}{R}^{\lambda }$ is therefore a $k[S_n]$-module, and $D^{\lambda }=0$ holds exactly when $S_k^{\lambda }\subseteq (S_k^{\lambda }{)}^{\perp }$.

Let $\phi :S_k^{\lambda }\to (S_k^{\lambda }{)}^{\ast }$ be $\phi (v)={\beta }_k(v,\cdot ){\mathrm{\mid }}_{S_k^{\lambda }}$. Its kernel is exactly $R^{\lambda }$, and in the standard basis of $S_k^{\lambda }$ paired with its dual basis the matrix of $\phi$ is $\overline{G_{\lambda }}$, so by [F6] the rank of $\phi$ equals ${\mathrm{rank}}_k\overline{G_{\lambda }}$. Rank-nullity [F5] gives ${\dim }_k{S}_k^{\lambda }={\mathrm{rank}}_k\overline{G_{\lambda }}+{\dim }_k{R}^{\lambda }$, hence ${\dim }_k{D}^{\lambda }={\mathrm{rank}}_k\overline{G_{\lambda }}$.

The construction involves no choices of lifts: $M_k^{\lambda }$, $S_k^{\lambda }$ and ${\beta }_k$ are obtained from the integral objects by scalar extension, and by [F2] every element of $S_k^{\lambda }$ is a $k$-combination of the standard polytabloids, so the descriptions of $R^{\lambda }$, $D^{\lambda }$ and ${\dim }_k{D}^{\lambda }$ depend only on $\lambda$, $p$ and $k$.

By step 2.1 the form radical is an $S_n$-submodule and $D^{\lambda }$ is defined for every $\lambda$, possibly zero; by step 2.2 its dimension is the rank of the reduced Gram matrix. The identification $R^{\lambda }=\mathrm{rad}(S_k^{\lambda })$ with the radical of [F4] is asserted only later, for the nonzero case; nothing in this definition presupposes it.

For $\gamma ,\delta \in C_t$ the sign is multiplicative, so in the group algebra over any ring ${\kappa }_t\gamma =\mathrm{sgn}(\gamma ){\kappa }_t$ and in particular ${\kappa }_t^2=\mathrm{\mid }{C}_t\mathrm{\mid }{\kappa }_t$; and ${\kappa }_t$ acts $F$-linearly on $M_F^{\lambda }$ through the group action. Moreover the coefficient of $\{t\}$ in $e_t$ is $1$ by [F2], so $e_t\mathrm{\ne }0$ over $F$.

Let $\{s\}$ be a $\mu$-tabloid whose row contains two entries $x,y$ lying in one column of $t$, so that $\tau =(xy)\in C_t$. Writing $Z$ for a set of left coset representatives of $⟨\tau ⟩$ in $C_t$ gives the integral group-algebra identity ${\kappa }_t={\sum }_{z\in Z}\mathrm{sgn}(z)z(1-\tau )$; since $\tau \in R_s$ fixes the tabloid $\{s\}$ by [F3], applying this to $\{s\}$ gives ${\kappa }_t\{s\}={\sum }_{z\in Z}\mathrm{sgn}(z)(z\{s\}-z\tau \{s\})=0$. This is an identity between integral vectors, so it holds in $M_{\mathbb{Z}}^{\mu }$ and hence over $F$: the collision criterion of [F4] is field-independent.

Suppose ${\kappa }_t\{s\}\mathrm{\ne }0$ for a $\mu$-tabloid $\{s\}$. Then step 1.2 shows no row of $\{s\}$ contains two entries from one column of $t$, i.e. every row of a representing tableau meets every column of $t$ in at most one entry; by [F5] this gives $\lambda ⊵\mu$, and when $\lambda =\mu$ it gives $\rho \in R_s$, $\gamma \in C_t$ with $\rho \cdot s=\gamma \cdot t$. In the equal-shape case ${\kappa }_t\{s\}={\kappa }_t\{\rho \cdot s\}={\kappa }_t\gamma \cdot \{t\}=\mathrm{sgn}(\gamma ){\kappa }_t\{t\}=\mathrm{sgn}(\gamma )e_t$ by step 1.1.

Every element of $M_F^{\lambda }$ is an $F$-combination of $\lambda$-tabloids, and by step 2.1 each ${\kappa }_t\{s\}$ is either $0$ or $\pm e_t$; hence ${\kappa }_t{M}_F^{\lambda }\subseteq F{e}_t$. Since $e_t={\kappa }_t\{t\}\mathrm{\ne }0$ lies in the image, ${\kappa }_t{M}_F^{\lambda }=F{e}_t$, as asserted.

If ${\kappa }_t{M}_F^{\mu }\mathrm{\ne }0$, some $\mu$-tabloid $\{s\}$ satisfies ${\kappa }_t\{s\}\mathrm{\ne }0$, so step 2.1 gives $\lambda ⊵\mu$ in the order of [F6].

Over $\mathbb{C}$ the conclusions of steps 3.1 and 3.2 are exactly the published statements [F7]; the present proof rederives them over an arbitrary field from the integral collision identity of step 1.2 and the combinatorial lemma [F5], both of which involve only coefficients $0,\pm 1$, so the argument applies in characteristic two. For $n=0$ the empty tableau has ${\kappa }_t=1$ and $e_t=\{\mathrm{\varnothing }\}\mathrm{\ne }0$, the only partition is $\mathrm{\varnothing }$ with $\mathrm{\varnothing }⊵\mathrm{\varnothing }$, and the three displayed claims hold.