Specht Gram gcd detects p-regularity

For $j\ge 1$ let $P_j:=\{i:{\lambda }_i=j\}$ be the set of row indices of length $j$, and let $\mathrm{\Pi }:={\prod }_{j\ge 1}\mathrm{Sym}(P_j)$ be the finite group of all permutations of the rows of $[\lambda ]$ that preserve each row length. For $\pi =({\pi }_j{)}_j\in \mathrm{\Pi }$ and a tabloid $T$ define $T\star \pi$ by ${\text{row}}_i(T\star \pi ):={\text{row}}_{{\pi }_j(i)}(T)(i\in P_j)\mathrm{.}$ This is a right action of $\mathrm{\Pi }$ on tabloids: $(T\star \pi )\star \rho =T\star (\pi \rho )$ under the convention $(\pi \rho )(i)=\pi (\rho (i))$. It is free: if $T\star \pi =T$, then ${\text{row}}_{{\pi }_j(i)}(T)={\text{row}}_i(T)$ for all $i$, and distinct rows are disjoint nonempty sets, so ${\pi }_j(i)=i$ for all $i$. Therefore $\mathrm{\mid }\mathrm{\Pi }\mathrm{\mid }={\prod }_{j\ge 1}{z}_j!=L_{\lambda }$ and every orbit has $L_{\lambda }$ tabloids. Put $s_{\pi }:={\prod }_{j\ge 1}\mathrm{sgn}({\pi }_j{)}^j$.

Fix a $\lambda$-tableau $u$ and $\pi \in \mathrm{\Pi }$. Define a permutation ${\delta }_{\pi }\in S_n$ by ${\delta }_{\pi }(u(i,c)):=u({\pi }_j(i),c)(i\in P_j,1\le c\le j),$ which is well defined because the map $(i,c)\mapsto u(i,c)$ is a bijection from the cells of $[\lambda ]$ onto $\{1,\dots ,n\}$ by [F8] and because ${\pi }_j(i)\in P_j$, so that the cell $({\pi }_j(i),c)$ exists exactly when $c\le j$. For each column $c$ of $[\lambda ]$ the values $u(i,c)$ with ${\lambda }_i\ge c$ are permuted among themselves by ${\delta }_{\pi }$: indeed ${\delta }_{\pi }$ permutes, for each $j\ge c$, the set $E_{j,c}:=\{u(i,c):i\in P_j\}$ of the $z_j$ entries of column $c$ lying in rows of length $j$, and these sets partition the $c$-th column. Hence ${\delta }_{\pi }\in C_u$ by [F5]. Moreover $\mathrm{sgn}({\delta }_{\pi })={\prod }_{j\ge 1}\mathrm{sgn}({\pi }_j{)}^j=s_{\pi }$: the restriction of ${\delta }_{\pi }$ to $E_{j,c}$ corresponds to ${\pi }_j$ under the bijection $i\mapsto u(i,c)$, and the sets $E_{j,c}$ over all pairs $(j,c)$ with $c\le j$ are pairwise disjoint, so the signs multiply. Finally ${\delta }_{\pi }({\text{row}}_i(u))={\text{row}}_{{\pi }_j(i)}(u)$ for $i\in P_j$, since ${\delta }_{\pi }$ carries $u(i,c)$ to $u({\pi }_j(i),c)$ for every $c\le {\lambda }_i=j$.

Every integral polytabloid is an integral linear combination of the standard polytabloids, by [F1]. Fix an ordering $u_1,\dots ,u_d$ of the standard $\lambda$-tableaux and write $e_s={\sum }_i{a}_i{e}_{u_i}$ and $e_t={\sum }_j{b}_j{e}_{u_j}$ with integers $a_i,b_j$ and $G_{\lambda }=(G_{ij})=(\beta (e_{u_i},e_{u_j}))$. Bilinearity of $\beta$ gives $\beta (e_s,e_t)={\sum }_{i,j}{a}_i{b}_j{G}_{ij}$ for every pair of tableaux $s,t$. Hence the greatest common divisor of the entries $G_{ij}$ divides every pairing $\beta (e_s,e_t)$, while each $G_{ij}$ is itself one of the pairings appearing in the definition of $g_{\lambda }$; the two finite gcds therefore coincide, and $g_{\lambda }$ is the gcd of the entries of the integral Gram matrix $G_{\lambda }$.

Fix a tableau $t$ and its row reversal $t^{\ast }(i,c)=t(i,{\lambda }_i+1-c)$. Suppose $T=\{\gamma t\}=\{\delta t^{\ast }\}$ with $\gamma \in C_t$ and $\delta \in C_{t^{\ast }}$. A row of $T$ of length $m$ contains one entry from each of columns $1,\dots ,m$ of $t$ and one from each of columns $1,\dots ,m$ of $t^{\ast }$. An entry originally in a row of length $j$ and column $c$ of $t$ lies in column $j+1-c$ of $t^{\ast }$. Take $m$ maximal among the row lengths still under consideration. The entry of a length-$m$ row of $T$ in $t$-column $m$ must come from an original length-$m$ row and occupies $t^{\ast }$-column $1$. Descending through $t$-columns $c=m-1,\dots ,1$, assume the preceding entries occupy $t^{\ast }$-columns $1,\dots ,m-c$. An entry in $t$-column $c$ from a shorter row has $t^{\ast }$-column $j+1-c\le m-c$, already occupied; thus it comes from a length-$m$ row and occupies $t^{\ast }$-column $m+1-c$. All length-$m$ rows of $T$ therefore use only entries from original length-$m$ rows, exhausting those entries. Remove these rows and repeat at the next largest length. Hence every row $i$ of $T$ contains only entries originally in rows of length ${\lambda }_i$. For $x=t(i,c)$, both $\gamma (x)$ and $\delta (x)$ belong to row $i$ of $T$. The former lies in $t$-column $c$; the latter lies in $t^{\ast }$-column ${\lambda }_i+1-c$, which is $t$-column $c$ among entries originally in rows of length ${\lambda }_i$. Since row $i$ of $T$ contains exactly one entry from that $t$-column, $\gamma (x)=\delta (x)$. Thus $\gamma =\delta$. Conversely, if $\gamma \in C_t\cap C_{t^{\ast }}$, the equal row sets of $t$ and $t^{\ast }$ give $\{\gamma t\}=\{\gamma t^{\ast }\}$. Consequently $\mathrm{supp}(e_t)\cap \mathrm{supp}(e_{t^{\ast }})=\{\{\gamma t\}:\gamma \in C_t\cap C_{t^{\ast }}\},$ and [F4] makes the coefficient of each common tabloid $\mathrm{sgn}(\gamma )$ in both polytabloids.

We determine the intersection $C_t\cap C_{t^{\ast }}$. First let $\gamma \in C_t\cap C_{t^{\ast }}$ and let $j\ge 1$. For a value $x=t(i,c)$ with ${\lambda }_i=j$, the value $x$ lies in the $t^{\ast }$-column ${\lambda }_i+1-c$, so $\gamma (x)$, being in $C_{t^{\ast }}$, lies in that same $t^{\ast }$-column; say $\gamma (x)=t(i^{\mathrm{\prime }\mathrm{\prime }},{\lambda }_{i^{\mathrm{\prime }\mathrm{\prime }}}+1-({\lambda }_i+1-c))=t(i^{\mathrm{\prime }\mathrm{\prime }},c+{\lambda }_{i^{\mathrm{\prime }\mathrm{\prime }}}-{\lambda }_i)$ for some $i^{\mathrm{\prime }\mathrm{\prime }}$ with ${\lambda }_{i^{\mathrm{\prime }\mathrm{\prime }}}\ge {\lambda }_i+1-c$. On the other hand $\gamma \in C_t$ means $\gamma (x)=t(i^{\mathrm{\prime }},c)$ for some row $i^{\mathrm{\prime }}$ by [F5]. Comparing the two descriptions cell by cell gives $i^{\mathrm{\prime }\mathrm{\prime }}=i^{\mathrm{\prime }}$ and $c+{\lambda }_{i^{\mathrm{\prime }\mathrm{\prime }}}-{\lambda }_i=c$, that is, ${\lambda }_{i^{\mathrm{\prime }}}={\lambda }_i=j$. Therefore $\gamma (x)$ lies in a row of length $j$ for every $x$ in a row of length $j$; since $\gamma$ is bijective, $\gamma (R_j)=R_j$ for every $j$, where $R_j:={\bigcup }_{i\in P_j}{\text{row}}_i(t)$. Second, conversely, suppose $\gamma \in C_t$ satisfies $\gamma (R_j)=R_j$ for every $j$. Let $c\ge 1$ and let $x=t(i,{\lambda }_i+1-c)$ be an entry of the $t^{\ast }$-column $c$, so ${\lambda }_i\ge c$ and $x\in R_{{\lambda }_i}$. Then $\gamma (x)\in R_{{\lambda }_i}$ and $\gamma \in C_t$ preserve the $t$-column of $x$, which is ${\lambda }_i+1-c$; hence $\gamma (x)=t(i^{\mathrm{\prime }},{\lambda }_i+1-c)$ with ${\lambda }_{i^{\mathrm{\prime }}}={\lambda }_i$, that is, $\gamma (x)=t(i^{\mathrm{\prime }},{\lambda }_{i^{\mathrm{\prime }}}+1-c)$, an entry of the $t^{\ast }$-column $c$. Thus $\gamma \in C_{t^{\ast }}$. This proves $C_t\cap C_{t^{\ast }}=\{\gamma \in C_t:\gamma (R_j)=R_j\text{ for every }j\}\mathrm{.}$

Let $u$ be a $\lambda$-tableau, $\gamma \in C_u$ and $\pi \in \mathrm{\Pi }$. By step 1.2, for $i\in P_j$, ${\text{row}}_i(\gamma {\delta }_{\pi }\cdot u)=\gamma ({\text{row}}_{{\pi }_j(i)}(u))={\text{row}}_{{\pi }_j(i)}(\gamma \cdot u)={\text{row}}_i(\{\gamma \cdot u\}\star \pi ),$ so $\{\gamma \cdot u\}\star \pi =\{\gamma {\delta }_{\pi }\cdot u\}$. By [F4] the coefficient changes by $\mathrm{sgn}({\delta }_{\pi })=s_{\pi }$. Applying ${\pi }^{-1}$ gives the converse for support. Thus, writing $c_u(T)$ for the coefficient of $T$ in $e_u$, $c_u(T\star \pi )=s_{\pi }{c}_u(T)(T\text{ any tabloid},\pi \in \mathrm{\Pi }),$ including when both coefficients vanish.

Such a $\gamma$ is exactly a choice, for every pair $(j,c)$ with $c\le j$, of an arbitrary permutation of the $z_j$ values $E_{j,c}=\{t(i,c):i\in P_j\}$ that column $c$ of $[\lambda ]$ receives from the rows of length $j$, the choices for the finitely many pairs $(j,c)$ being independent; these permutations determine $\gamma$ and lie in $C_t$ because the sets $E_{j,c}$ partition the value sets of the columns, and they satisfy $\gamma (R_j)=R_j$ and hence $\gamma \in C_{t^{\ast }}$ by step 1.5. Therefore $\mathrm{\mid }{C}_t\cap C_{t^{\ast }}\mathrm{\mid }={\prod }_{j\ge 1}{\prod }_{c=1}^j{z}_j!={\prod }_{j\ge 1}(z_j!{)}^j=U_{\lambda }\mathrm{.}$

Let $C$ be a $\mathrm{\Pi }$-orbit with base point $T_0$. By step 1.1 the map $\pi \mapsto T_0\star \pi$ is a bijection $\mathrm{\Pi }\to C$, and by step 2.1, for any two polytabloids $e_s,e_t$, ${\sum }_{T\in C}{c}_s(T)c_t(T)={\sum }_{\pi \in \mathrm{\Pi }}{c}_s(T_0\star \pi )c_t(T_0\star \pi )={\sum }_{\pi \in \mathrm{\Pi }}{s}_{\pi }^2{c}_s(T_0)c_t(T_0)=L_{\lambda }{c}_s(T_0)c_t(T_0)\mathrm{.}$ Summing over all orbits gives $\beta (e_s,e_t)=L_{\lambda }{N}_{s,t}$ for an integer $N_{s,t}$; hence $L_{\lambda }\mid g_{\lambda }$.

Combining steps 1.4 and 2.2, each of the $U_{\lambda }$ common tabloids contributes $\mathrm{sgn}(\gamma {)}^2=1$ to the pairing, so $\beta (e_t,e_{t^{\ast }})=U_{\lambda }\mathrm{.}$ Since $U_{\lambda }$ is one of the pairings whose positive gcd is $g_{\lambda }$, the gcd divides it: $g_{\lambda }\mid U_{\lambda }$. With step 3.1 this gives $L_{\lambda }\mid g_{\lambda }\mid U_{\lambda }$, assertions 1 and 2 of the statement.

Let $p$ be a prime. A prime divides the factorial $z_j!$ if and only if $z_j\ge p$. Hence $p\mid L_{\lambda }$ if and only if $z_j\ge p$ for some $j$, and the same equivalence holds for the product $U_{\lambda }={\prod }_j(z_j!{)}^j$; the two products therefore have the same prime divisors. By step 4.1, $p\mid g_{\lambda }$ if and only if $p\mid L_{\lambda }$, that is, if and only if $z_j\ge p$ for some $j$; by [F6] this is exactly the failure of $p$-regularity of $\lambda$. Therefore $g_{\lambda }$ is nonzero modulo $p$ if and only if $\lambda$ is $p$-regular, assertion 3.

It remains to verify the field relation of assertion 4. Let $F$ be a field and let $M_F^{\lambda }$ be the field-valued tabloid module, with ${\beta }_F$ the scalar extension of $\beta$ and with the same symbols ${\kappa }_t,e_t$. By [F7] the image of ${\kappa }_t$ on $M_F^{\lambda }$ is the line $F{e}_t$, so ${\kappa }_t\cdot e_{t^{\ast }}=h{e}_t$ for a unique $h\in F$. Using [F2], the normalization ${\beta }_F(e_t,\{t\})=1$ (the coefficient of $\{t\}$ in $e_t$ is $1$ by [F4]), and ${\kappa }_t\{t\}=e_t$, we compute $h=h{\beta }_F(e_t,\{t\})={\beta }_F(h{e}_t,\{t\})={\beta }_F({\kappa }_t{e}_{t^{\ast }},\{t\})={\beta }_F(e_{t^{\ast }},{\kappa }_t\{t\})={\beta }_F(e_{t^{\ast }},e_t)=U_{\lambda }\cdot 1_F,$ where the last equality is the base change of the integral identity of step 4.1. Hence ${\kappa }_t\cdot e_{t^{\ast }}=U_{\lambda }{e}_t$ in $M_F^{\lambda }$, over every field and in particular in every prime characteristic, with no division by a group order.

Finally take $\lambda =\mathrm{\varnothing }$ and $n=0$. There is exactly one tabloid, exactly one tableau, and $C_t=\{1\}$, so $e_t$ is the unique basis vector and $\beta (e_t,e_t)=1$; the products $L_{\mathrm{\varnothing }}$ and $U_{\mathrm{\varnothing }}$ are empty products equal to $1$, and the empty partition is $p$-regular for every prime $p$ by [F6]. Thus $L_{\mathrm{\varnothing }}=U_{\mathrm{\varnothing }}=g_{\mathrm{\varnothing }}=1$ and all four assertions hold in this case.