Murphy elements and Murphy tableau elements

Let $R$ be a commutative ring, $q\in R^{\times }$, and $H=H_{R,q}(S_n)$ with standard basis $\{T_w\}$ and star anti-automorphism $\ast$ (The generic type-A Hecke algebra, The standard basis of the generic type-A Hecke algebra, The star anti-automorphism of the type-A Hecke algebra). For a composition (in particular a partition) $\mu$ of $n$ put $m_{\mu }:={\sum }_{w\in S_{\mu }}{T}_w\in H$, the row symmetrizer of the standard Young subgroup $S_{\mu }$ (Young subgroups, tabloids, and permutation modules); the sum has $\mathrm{\mid }{S}_{\mu }\mathrm{\mid }={\prod }_i{\mu }_i!$ terms.

A tableau is row standard when its entries increase along each row. Use the right tableau action $t\cdot w:=w^{-1}\cdot t$, where the left action applies $w$ to the entries. For a row-standard $\mu$-tableau $s$, let $d(s)$ be the unique permutation with $s=t^{\mu }\cdot d(s)$, where $t^{\mu }$ is filled by $1,2,\dots ,n$ in row-major order. Explicitly, if $s$ carries entry $a$ in the $p$-th node in that order, then $d(s)(a)=p$: $d(s)$ is the inverse of the row-reading permutation. The action on tableaux is free; the stabilizer of the row tabloid of $t^{\mu }$ is $S_{\mu }$ (Row and column stabilizers).

Write $D_{\mu }$ for these distinguished permutations. Row-standardness says that $d^{-1}$ is increasing on each canonical row block. The coset $S_{\mu }d$ consists exactly of the tableaux with the same row sets, so sorting each row produces its unique representative in $D_{\mu }$. Moreover, $\mathrm{\ell }(wd)=\mathrm{\ell }(w)+\mathrm{\ell }(d)(w\in S_{\mu },d\in D_{\mu })\mathrm{.}$ To verify this, count inversions of $(wd{)}^{-1}=d^{-1}{w}^{-1}$. Pairs of labels within one canonical block contribute exactly the inversions of $w^{-1}$, since $d^{-1}$ is increasing there. For labels in different blocks, $w^{-1}$ permutes the labels within their respective blocks, so the total number of cross-block inversions equals that of $d^{-1}$. The two contributions give the displayed equality, using inversion invariance of length. It also proves minimality and uniqueness in $S_{\mu }d$. These conventions extend to compositions by using their consecutive row blocks. For row-standard $\mu$-tableaux $s,t$ define the Murphy element $m_{st}:=T_{d(s)}^{\ast }{m}_{\mu }{T}_{d(t)}\in H,$ and for a standard $\lambda$-tableau $t$ write the diagonal element as $m_{tt}=T_{d(t)}^{\ast }{m}_{\lambda }{T}_{d(t)}$; the notation $m_t$ is reserved for a cell-module vector, rather than an element of $H$. Then $m_{st}^{\ast }=m_{ts}$ for all row-standard $s,t$ of the same shape: applying the anti-automorphism, which fixes $m_{\mu }$ because $m_{\mu }^{\ast }={\sum }_{w\in S_{\mu }}{T}_{w^{-1}}=m_{\mu }$, gives $m_{st}^{\ast }=(T_{d(s)}^{\ast }{m}_{\mu }{T}_{d(t)}{)}^{\ast }=T_{d(t)}^{\ast }{m}_{\mu }^{\ast }{T}_{d(s)}^{\ast \ast }=T_{d(t)}^{\ast }{m}_{\mu }{T}_{d(s)}=m_{ts}\mathrm{.}$ Finally, $m_{\mu }{T}_w=q^{\mathrm{\ell }(w)}{m}_{\mu }$ for every $w\in S_{\mu }$: it suffices to prove this for $w=s_i$ a simple transposition with $i,i+1$ in the same block of $\mu$, since any $w\in S_{\mu }\setminus \{1\}$ has a reduced expression $w=w^{\mathrm{\prime }}{s}_i$ with $\mathrm{\ell }(w)=\mathrm{\ell }(w^{\mathrm{\prime }})+1$ and the claim follows by induction, $m_{\mu }{T}_w=m_{\mu }{T}_{w^{\mathrm{\prime }}}{T}_i=q^{\mathrm{\ell }(w^{\mathrm{\prime }})}{m}_{\mu }{T}_i=q^{\mathrm{\ell }(w)}{m}_{\mu }$. For such $i$ the right multiplication $v\mapsto v{s}_i$ is an involution of the finite group $S_{\mu }$, and pairing the summands of $m_{\mu }{T}_i={\sum }_{v\in S_{\mu }}{T}_v{T}_i$ in twos gives, in the two cases of the standard-basis multiplication rule $T_v{T}_i=T_{v{s}_i}$ if $\mathrm{\ell }(v{s}_i)=\mathrm{\ell }(v)+1$ and $T_v{T}_i=(q-1)T_v+q{T}_{v{s}_i}$ if $\mathrm{\ell }(v{s}_i)=\mathrm{\ell }(v)-1$, $T_v{T}_i+T_{v{s}_i}{T}_i=q(T_v+T_{v{s}_i}),$ so summing over the $\frac{1}{2}\mathrm{\mid }{S}_{\mu }\mathrm{\mid }$ pairs yields $m_{\mu }{T}_i=q{m}_{\mu }$.