Dominance unitriangularity of the symmetric-group decomposition matrix

By [F1] and [F8], $S_{\mathcal{O}}^{\lambda }=\mathcal{O}{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }$ is a free $\mathcal{O}$-module with the standard polytabloids as basis, it is stable under $S_n$, its reduction is $S_{\mathcal{O}}^{\lambda }\mathrm{/}\mathfrak{m}{S}_{\mathcal{O}}^{\lambda }\cong k{\otimes }_{\mathcal{O}}{S}_{\mathcal{O}}^{\lambda }\cong k{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }=S_k^{\lambda },$ and $K{\otimes }_{\mathcal{O}}{S}_{\mathcal{O}}^{\lambda }\cong K{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }=S_K^{\lambda }$. Thus $S_{\mathcal{O}}^{\lambda }$ is a stable $\mathcal{O}[S_n]$-lattice in $S_K^{\lambda }$ with reduction $S_k^{\lambda }$.

The matrix of the Hermitian product of [F3] in the standard basis of $S_{\mathbb{C}}^{\lambda }$ is $(⟨e_i,e_j⟩)=({\sum }_T{c}_i(T)\overline{c_j(T)})$, and since all tabloid coefficients of polytabloids are integers by [F1] this equals $({\sum }_T{c}_i(T)c_j(T))=G_{\lambda }$ by [F2]. By [F3] the restricted Hermitian form on $S_{\mathbb{C}}^{\lambda }$ is nondegenerate, so $G_{\lambda }$ is an invertible matrix over $\mathbb{C}$; since $G_{\lambda }$ has integer entries, $\det G_{\lambda }\mathrm{\ne }0$. As $K$ has characteristic $0$, the image of $\det G_{\lambda }$ in $K$ is nonzero, so the base-changed form ${\beta }_K$ has invertible Gram matrix on $S_K^{\lambda }$ and is nondegenerate there.

Let $D^{\nu }$ be a composition factor of $M_k^{\lambda }\mathrm{/}{S}_k^{\lambda }$. Then $D^{\nu }\mathrm{\ne }0$, so $\nu$ is $p$-regular by [F5]. Choose a composition series of $M_k^{\lambda }\mathrm{/}{S}_k^{\lambda }$; the factor $D^{\nu }$ is $N\mathrm{/}{N}^{\mathrm{\prime }}$ for submodules $N^{\mathrm{\prime }}\le N$ of $M_k^{\lambda }\mathrm{/}{S}_k^{\lambda }$. With $\pi :M_k^{\lambda }\twoheadrightarrow M_k^{\lambda }\mathrm{/}{S}_k^{\lambda }$ and $U:={\pi }^{-1}(N^{\mathrm{\prime }})\supseteq S_k^{\lambda }$ one has $M_k^{\lambda }\mathrm{/}U\cong (M_k^{\lambda }\mathrm{/}{S}_k^{\lambda })\mathrm{/}{N}^{\mathrm{\prime }}$, and $N\mathrm{/}{N}^{\mathrm{\prime }}\cong D^{\nu }$ is a nonzero submodule of that quotient; hence there is a nonzero $k[S_n]$-homomorphism $\psi :D^{\nu }\to M_k^{\lambda }\mathrm{/}U$. By [F6] with $(\nu ,\lambda )$ in place of its $(\lambda ,\mu )$ we get $\nu ⊵\lambda$, and if $\nu =\lambda$ then [F6] says $U$ does not contain $S_k^{\lambda }$, contrary to $U\supseteq S_k^{\lambda }$. Hence $\nu ⊳\lambda$: every composition factor of $M_k^{\lambda }\mathrm{/}{S}_k^{\lambda }$ is $D^{\nu }$ with $\nu$ strictly dominating $\lambda$.

By step 1.1 the stable lattice $S_{\mathcal{O}}^{\lambda }$ in $S_K^{\lambda }$ has reduction $S_k^{\lambda }$, so by [F8] the decomposition map sends $[S_K^{\lambda }]$ to $[S_k^{\lambda }]$. Since the classes of the simple modules form the integral basis of the modular Grothendieck group by [F7], and the expansion coefficients of $[S_k^{\lambda }]$ in that basis are the composition multiplicities by [F9], $d([S_K^{\lambda }])={\sum }_{\mu p\text{-regular}}[S_k^{\lambda }:D^{\mu }][D^{\mu }]={\sum }_{\mu p\text{-regular}}{d}_{\lambda \mu }[D^{\mu }]\mathrm{.}$ Hence the $d_{\lambda \mu }$ are exactly the decomposition numbers of the ordinary irreducible $S_K^{\lambda }$.

$S_K^{\lambda }$ is irreducible: if $0\mathrm{\ne }U\le S_K^{\lambda }$ is a proper submodule, then viewing $U$ inside $M_K^{\lambda }$ and applying the James submodule theorem [F4] gives $S_K^{\lambda }\le U$ (impossible) or $U\le (S_K^{\lambda }{)}^{\perp }$, and the latter forces $U\le S_K^{\lambda }\cap (S_K^{\lambda }{)}^{\perp }=0$ by the nondegeneracy of step 1.2, a contradiction. The same argument applies over any field extension $E\mathrm{/}K$: base change gives $E{\otimes }_K{S}_K^{\lambda }\cong E{\otimes }_{\mathbb{Z}}{S}_{\mathbb{Z}}^{\lambda }=S_E^{\lambda }$ by [F1], $\det G_{\lambda }\mathrm{\ne }0$ in $E$, and [F4] holds over $E$; so $S_E^{\lambda }$ is irreducible. Hence $S_K^{\lambda }$ is absolutely irreducible and is the ordinary irreducible attached to $\lambda$.

The pairing $(x+S_k^{\lambda },y)\mapsto {\beta }_k(x,y)$ from $(M_k^{\lambda }\mathrm{/}{S}_k^{\lambda })\times S_k^{\lambda \perp }$ to $k$ is well defined because ${\beta }_k(S_k^{\lambda },S_k^{\lambda \perp })=0$, and it is nondegenerate: on the right, ${\beta }_k(M_k^{\lambda },y)=0$ forces $y=0$ by nondegeneracy of ${\beta }_k$ from [F2]; on the left, $(S_k^{\lambda \perp }{)}^{\perp }=S_k^{\lambda }$ because $\dim W^{\perp }=\dim M_k^{\lambda }-\dim W$ for a nondegenerate form and $\dim M_k^{\lambda }-\dim S_k^{\lambda \perp }=\dim S_k^{\lambda }$. Hence $\mathrm{\Phi }:S_k^{\lambda \perp }\to (M_k^{\lambda }\mathrm{/}{S}_k^{\lambda }{)}^{\ast }$, $\mathrm{\Phi }(y)={\beta }_k(\cdot ,y)$, is an isomorphism of $k[S_n]$-modules, equivariant by the invariance of ${\beta }_k$ in [F2]. Dualizing a composition series $0=M_0<\cdots <M_r=M_k^{\lambda }\mathrm{/}{S}_k^{\lambda }$ gives exact sequences $0\to (M_i\mathrm{/}{M}_{i-1}{)}^{\ast }\to M_i^{\ast }\to M_{i-1}^{\ast }$ and, by induction on $i$, the composition factors of $(M_k^{\lambda }\mathrm{/}{S}_k^{\lambda }{)}^{\ast }$ are the duals of those of $M_k^{\lambda }\mathrm{/}{S}_k^{\lambda }$ with the same multiplicities. Each $D^{\nu }$ is self-dual by [F5], so by step 1.3 every composition factor of $S_k^{\lambda \perp }$ is $D^{\nu }$ with $\nu ⊳\lambda$.

The chain $0\subseteq R^{\lambda }=S_k^{\lambda }\cap S_k^{\lambda \perp }\subseteq S_k^{\lambda }\subseteq M_k^{\lambda }$ is a chain of $k[S_n]$-submodules, and $S_k^{\lambda }\mathrm{/}{R}^{\lambda }=D^{\lambda }$ if $\lambda$ is $p$-regular, and $D^{\lambda }=0$ otherwise, by [F5]. By additivity of composition multiplicities [F9] over this chain, every composition factor of $S_k^{\lambda }$ is a composition factor of $R^{\lambda }$ or of $S_k^{\lambda }\mathrm{/}{R}^{\lambda }$; the factors of $R^{\lambda }$ are among those of $S_k^{\lambda \perp }$, hence have the form $D^{\nu }$ with $\nu ⊳\lambda$ by step 2.3. Consequently: (i) every composition factor of $S_k^{\lambda }$ is $D^{\mu }$ with $\mu ⊵\lambda$; and (ii) if $\lambda$ is $p$-regular then $[S_k^{\lambda }:D^{\lambda }]=1$, since the quotient $S_k^{\lambda }\mathrm{/}{R}^{\lambda }$ contributes exactly one copy of $D^{\lambda }$ and no factor of $R^{\lambda }$ is $D^{\lambda }$ (those have $\nu ⊳\lambda$), while if $\lambda$ is not $p$-regular then $D^{\lambda }=0$. With step 2.1 this is assertion 1 and assertion 2.

Let $\mu ⊵\lambda$ with $\mu \mathrm{\ne }\lambda$ and let $r$ be the least index with ${\mu }_r\mathrm{\ne }{\lambda }_r$ (sequences padded by zeros). The first $r-1$ partial sums of $\mu$ and $\lambda$ agree, so if ${\mu }_r<{\lambda }_r$ the $r$-th partial sum of $\mu$ would be strictly smaller than that of $\lambda$, contradicting $\mu ⊵\lambda$; hence ${\mu }_r>{\lambda }_r$ and $\mu$ is strictly larger than $\lambda$ in decreasing lexicographic order by [F10]. Therefore, for $p$-regular $\lambda$, a nonzero $d_{\lambda \mu }$ forces $\mu =\lambda$ or $\mu >\lambda$ lexicographically. Listing the $p$-regular partitions in decreasing lexicographic order as rows (in a block placed first) and as columns, all nonzero entries of the leading $p$-regular square block lie on or below the diagonal, and the diagonal entries equal $1$ by step 3.1. This is assertion 3.

Assertions 1, 2 and 3 are steps 3.1, 3.1 and 4.1; the decomposition-number identification of the $d_{\lambda \mu }$ is step 2.1, and the irreducibility of the ordinary modules $S_K^{\lambda }$ is step 2.2. For $n=0$ there is one partition $\mathrm{\varnothing }$, which is $p$-regular, $S_k^{\mathrm{\varnothing }}\cong k$ is the trivial module and $d_{\mathrm{\varnothing }\mathrm{\varnothing }}=1$, so the statements hold with a $1\times 1$ block. The theorem gives only dominance constraints: it does not compute the off-diagonal entries $d_{\lambda \mu }$ with $\mu ⊳\lambda$, which depend on $p$. No step divides by $p$ or by a group order, none uses positivity of the modular form (positivity is used only over $\mathbb{C}$ in step 1.2 to see that $G_{\lambda }$ is nonsingular), and characteristic $2$ is included.