Jucys–Murphy Elements and Seminormal Forms

Let $\lambda ⊢n$ with Young diagram $[\lambda ]$ (Partitions, English diagrams, and conjugation), so that the nodes of $[\lambda ]$ are the pairs $(r,c)$ with $r,c\ge 1$ and $c\le {\lambda }_r$ (rows numbered downward, columns rightward).

The content of a node $(r,c)\in [\lambda ]$ is $c(r,c):=c-r,$ the column index minus the row index. The node $(r,c)$ therefore has content $t$ exactly when it lies on the diagonal $c-r=t$; the content of $(1,1)$ is $0$, contents increase by $1$ along a row and decrease by $1$ down a column.

Now let $T$ be a standard tableau of shape $\lambda$ (Tableaux and standard tableaux). For $1\le k\le n$ let $(r_k,c_k)$ be the node of $[\lambda ]$ carrying the entry $k$, so that $T(r_k,c_k)=k$. The content vector of $T$ is $\mathrm{Cont}(T):=(c_T(1),\dots ,c_T(n))\in {\mathbb{Z}}^n,c_T(k):=c(r_k,c_k)=c_k-r_k,$ the list of contents of the nodes read in the order of the entries $1,2,\dots ,n$. For the empty tableau of shape $\mathrm{\varnothing }$ the content vector is the empty vector, the unique element of ${\mathbb{Z}}^0$.

• Well-definedness. A tableau $T$ of shape $\lambda$ is a bijection $[\lambda ]\to \{1,\dots ,n\}$, so each $k\in \{1,\dots ,n\}$ occupies exactly one node $(r_k,c_k)$ and the integers $c_T(k)=c_k-r_k$ are determined by $T$. The map $T\mapsto \mathrm{Cont}(T)$ uses the fixed row and column coordinates and the labels of $T$: it depends only on which entry stands in which node.

• The entries of the content vector are the contents of the shape. Since $T$ is a bijection onto the $n$ nodes of $[\lambda ]$, the multiset of entries of $\mathrm{Cont}(T)$ equals the multiset of node contents $\{c(x):x\in [\lambda ]\}$, independent of $T$. In particular two standard tableaux of the same shape have content vectors that differ by a permutation of their entries, and the multiset of contents of a tableau is an invariant of its shape.

• Two extreme examples. The row tableau of shape $(n)$ carries $k$ in $(1,k)$, so its content vector is $(0,1,2,\dots ,n-1)$; the column tableau of shape $(1^n)$ carries $k$ in $(k,1)$, so its content vector is $(0,-1,-2,\dots ,-(n-1))$.

• First entries. The entry $1$ always occupies the node $(1,1)$, because every other node has a node weakly to its left and weakly above it, whose entry would be smaller. Hence $c_T(1)=0$ for every nonempty standard tableau $T$. Similarly, $c_T(k)$ need not be monotone in $k$: for the tableau $\begin{array}{cc}1& 2\\ 3\end{array}$ of shape $(2,1)$ one has $\mathrm{Cont}=(0,1,-1)$.

For $n\ge 1$ and $1\le m\le n$ let $Z(\mathbb{C}[S_m])$ be the centre of the complex group algebra $\mathbb{C}[S_m]$ (The center $Z(k[G])$ of the group algebra, The group ring $R[G]$ of finitely supported formal $R$-linear combinations of group elements). We use the standard chain of symmetric groups $S_1\subset S_2\subset \cdots \subset S_n,S_m=\mathrm{Sym}(\{1,\dots ,m\}),$ in which $S_m$ is the subgroup of $S_n$ fixing every element of $\{m+1,\dots ,n\}$ pointwise; thus $\mathbb{C}[S_m]$ is the subalgebra of $\mathbb{C}[S_n]$ spanned by $S_m$, and $Z(\mathbb{C}[S_1])=\mathbb{C}\cdot 1$. Each centre $Z(\mathbb{C}[S_m])$ is thereby regarded as a subalgebra of $\mathbb{C}[S_n]$.

The Gelfand-Tsetlin algebra of the chain is $\mathrm{G}\mathrm{Z}(n):=⟨Z(\mathbb{C}[S_1]),Z(\mathbb{C}[S_2]),\dots ,Z(\mathbb{C}[S_n])⟩\subseteq \mathbb{C}[S_n],$ the unital subalgebra of $\mathbb{C}[S_n]$ generated by all these centres.

• Well-definedness. Each $Z(\mathbb{C}[S_m])$ is a $\mathbb{C}$-subalgebra of $\mathbb{C}[S_m]$, hence, via the inclusion $\mathbb{C}[S_m]\subseteq \mathbb{C}[S_n]$, a subset of $\mathbb{C}[S_n]$ closed under addition, multiplication and scalar multiplication; the generated subalgebra is therefore a unital subalgebra of $\mathbb{C}[S_n]$. It is finite-dimensional because $\mathbb{C}[S_n]$ is: each $\mathbb{C}[S_m]$ is a finite-dimensional semisimple algebra (If $\mathrm{char}k\nmid \mathrm{\mid }G\mathrm{\mid }$, then $k[G]$ is a semisimple ring), so its centre is finite-dimensional as well.

• Commutativity. Every element of $Z(\mathbb{C}[S_m])$ is central in $\mathbb{C}[S_m]$. If $m\le k$ and $z\in Z(\mathbb{C}[S_m])$, $w\in Z(\mathbb{C}[S_k])$, then $\mathbb{C}[S_m]\subseteq \mathbb{C}[S_k]$ and $w$ commutes with every element of $\mathbb{C}[S_k]$, in particular with $z$; hence $zw=wz$. Thus all the generating centres commute with one another, and $\mathrm{G}\mathrm{Z}(n)$ is a commutative subalgebra of $\mathbb{C}[S_n]$. Only this centrality, not semisimplicity, is used.

• The top centre is included. The term $m=n$ shows $Z(\mathbb{C}[S_n])\subseteq \mathrm{G}\mathrm{Z}(n)$; in fact the centre of $\mathbb{C}[S_n]$ is the algebra generated by its class sums, and the other centres are generated by the class sums of the subgroups, so $\mathrm{G}\mathrm{Z}(n)$ is generated by class sums of the chain together with $\mathbb{C}[S_1]=\mathbb{C}$.

• The algebra used below. $\mathrm{G}\mathrm{Z}(n)$ is the diagonal algebra of the Young basis: The Gelfand-Tsetlin algebra is the diagonal algebra of the Young basis identifies it with ${⨁}_T\mathbb{C}{P}_T$ over the standard tableaux $T$ of size $n$, and The Jucys-Murphy elements generate the Gelfand-Tsetlin algebra shows that it is also the algebra generated by the Jucys-Murphy elements.

Let $n\ge 1$ and let $R$ be a commutative ring. In the group ring $R[S_n]$ of the symmetric group on $\{1,\dots ,n\}$ (Partitions, English diagrams, and conjugation, The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition, The group ring $R[G]$ of finitely supported formal $R$-linear combinations of group elements), where $(jk)$ denotes the transposition of the distinct entries $j,k\in \{1,\dots ,n\}$ (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition), the Jucys-Murphy elements are $X_1:=0,X_k:={\sum }_{j=1}^{k-1}(jk)\in R[S_n](2\le k\le n)\mathrm{.}$ Each $X_k$ is a finite sum of group elements with coefficient $1$, so the formula already defines an element of the integral group ring $\mathbb{Z}[S_n]$, and its image under the base change $\mathbb{Z}[S_n]\to R[S_n]$ is the displayed element (Restriction of scalars and extension of scalars $S{\otimes }_{R}M$ along a ring homomorphism $R\to S$); for $1\le k\le m\le n$ the element $X_k$ of $R[S_m]$, computed in the subgroup $S_m=\mathrm{Sym}(\{1,\dots ,m\})\subseteq S_n$ of permutations fixing $m+1,\dots ,n$, maps to $X_k$ under the inclusion $S_m\hookrightarrow S_n$. Equivalently $X_k=T_k-T_{k-1}$, where $T_k={\sum }_{1\le i<j\le k}(ij)$ is the sum of all transpositions in $S_k$.

• Well-definedness. The sum defining $X_k$ has the $k-1$ terms $(1k),\dots ,(k-1k)$, each of which is an element of the subgroup $S_k\subseteq S_n$, hence a basis element of the free $R$-module $R[S_n]$; a finite sum of basis elements is an element of $R[S_n]$ independent of any ordering of the summands. The formula uses only the labels $1,\dots ,k$, so it is preserved by the inclusion $S_m\hookrightarrow S_n$ for $k\le m\le n$ and by the base change $\mathbb{Z}[S_n]\to R[S_n]$. The group-ring multiplication is bilinear and sends basis elements $g,h$ to $gh$ (The group ring $R[G]$ is a unital $R$-algebra with basis $G$, and each $g\in G$ is a unit of $R[G]$). Thus the subgroup inclusions and the coefficient map ${\sum }_g{a}_{g}g\mapsto {\sum }_g(a_g{1}_R)g$ preserve products and the identity.

• The identity $X_k=T_k-T_{k-1}$. The transpositions of $S_k$ are $(ij)$ with $1\le i<j\le k$; those with $j<k$ are exactly the transpositions of $S_{k-1}$, and the remaining ones are $(ik)$ with $i<k$. Hence $T_k-T_{k-1}={\sum }_{j=1}^{k-1}(jk)=X_k$. For $k=2$ this reads $X_2=(12)=T_2-T_1$, where $T_1=0$. Since each $T_k$ is a sum of all transpositions of the subgroup $S_k$, it is a sum of full conjugacy classes of $S_k$.

• Integral normalization. Every coefficient in $X_k$ is $1$, not $\pm 1$ or a fraction; no characteristic is inverted, so $X_k$ is defined over $\mathbb{Z}$ and over every commutative ring. This is the normalization used throughout this page; the spectral statements below specialize the coefficient ring to $\mathbb{C}$, but no integral identity of this page uses division.

• Centrality. The elements $X_k$ need not be central in $R[S_n]$. The element $X_1=0$ is always central, and for $n=2$ the algebra $R[S_2]$ is commutative, so $X_2$ is central as well. For $n\ge 3$ and $R\mathrm{\ne }0$, conjugation by $(23)$ sends $X_2=(12)$ to $(13)\mathrm{\ne }{X}_2$: these are distinct basis elements and $1_R\mathrm{\ne }0$. Thus $X_2$ is not central in this case. Pairwise commutativity of the $X_k$ is proved in The Jucys-Murphy elements commute pairwise.

Let $n\ge 1$, let $S_{n-1}\le S_n$ be the stabilizer of $n$, and let $g\in S_n$. There is $h\in S_{n-1}$ with $h^2=1$ (the identity is allowed) and $hg{h}^{-1}=g^{-1}$; here such a self-inverse permutation is called an involution. In particular every element of $S_n$ is conjugate to its inverse by an element of $S_{n-1}$.

• The source's shorter argument is incomplete as printed. The cited source proves the fact by deleting the letter $n$ from $g$ and choosing an element $h\in S_{n-1}$ that conjugates the deletion $g^{\mathrm{\prime }}$ to $g^{\mathrm{\prime }-1}$; it then asserts that such an $h$ realizes $g^{-1}=hg{h}^{-1}$. That step is not correct for an arbitrary such $h$: for $g=(1234)$ and $h=(23)$ one has $h(123)h^{-1}=(132)=(123{)}^{-1}$, while $hg{h}^{-1}=(1324)\mathrm{\ne }{g}^{-1}=(1432)$. The construction above chooses the explicit cycle-reversing involution, which does satisfy $hg{h}^{-1}=g^{-1}$; only that corrected construction is used later, in The centralizer of $\mathbb{C}[S_{n-1}]$ in $\mathbb{C}[S_n]$ is commutative.

• No choice. For each $g$ the involution $h$ is given by explicit formulas on the finitely many cycles of $g$, so the statement is proved without any selection principle, and the argument is integral and characteristic-free: it uses only the group structure of $S_n$.

For all $1\le i,j\le n$ one has $X_i{X}_j=X_j{X}_i$ in $\mathbb{Z}[S_n]$, hence in $R[S_n]$ for every commutative ring $R$; that is, the Jucys-Murphy elements commute pairwise, and the subalgebra they generate is commutative.

• Integrality and no choice. The argument takes place entirely in $\mathbb{Z}[S_n]$ and uses only bilinear expansion in the group basis and the bijection $\{i,j\}\mapsto \{h(i),h(j)\}$ on two-element subsets. No characteristic is inverted, no module is selected and no choice principle is used; the base-change sentence is the only place a general ring appears.

• The centrality of $T_k$ in the subgroup. The same computation with $n$ replaced by $k$ shows that each $T_k$ is central in $\mathbb{Z}[S_k]$, since conjugation by $S_k$ permutes the transpositions of $S_k$. This is the only property of $T_n$ used above.

Let $\lambda ,\mu ⊢n$. If the multiset of contents $\{c(x):x\in [\lambda ]\}$ equals the multiset $\{c(y):y\in [\mu ]\}$, then $\lambda =\mu$.

• Why the diagonal data are Frobenius coordinates. The numbers $a_1>\cdots >a_d$ and $b_1>\cdots >b_d$ are the arm and leg lengths of the diagonal nodes, the Frobenius coordinates of $\nu$; the formula of step 1.3 is the usual reconstruction of a partition from them. The lemma says that the content multiset, which records the $a_i$ and $b_j$ through the diagonal counts of steps 1.1 and 1.2, is equivalent to that data.

• Sharper statement. The proof shows the two multisets $A(\nu )$ and $B(\nu )$ separately, not merely their union; both are needed, since the nonnegative and negative contents determine the arms and the legs respectively.

Let $\lambda$ be a partition. If $x\mathrm{\ne }y$ are addable nodes of $[\lambda ]$, then $c(x)\mathrm{\ne }c(y)$; equivalently the content map is injective on the set $\mathrm{Add}(\lambda )$ of addable nodes.

• The content is the addable-node coordinate. For an addable node $(i,{\lambda }_i+1)$ the content ${\lambda }_i+1-i$ is the integer at which the interpolation factors of the projector recursion of Primitive tableau idempotents by Jucys-Murphy interpolation are evaluated; the lemma is what makes all their denominators nonzero.

• Empty partition. For $\lambda =\mathrm{\varnothing }$ the only addable node is $(1,1)$, of content $0$, so injectivity is vacuous there; the argument above applies with $k=0$ and $I=\{1\}$.

Let $n\ge 2$ and put $s_i:=(ii+1)$ for $1\le i\le n-1$, the Coxeter generators of $S_n$. Then in $\mathbb{Z}[S_n]$, hence in $R[S_n]$ for every commutative ring $R$: (a) $s_i{X}_j=X_j{s}_i$ whenever $j\mathrm{\notin }\{i,i+1\}$; (b) $s_i{X}_i+1=X_{i+1}{s}_i$; equivalently $s_i{X}_i{s}_i+s_i=X_{i+1}$ and $s_i{X}_{i+1}=X_i{s}_i+1$. In particular the subalgebra generated by $s_i,X_i,X_{i+1}$ satisfies the local $H(2)$ relations $s_i^2=1$, $X_i{X}_{i+1}=X_{i+1}{X}_i$, $s_i{X}_i+1=X_{i+1}{s}_i$.