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The Jucys-Murphy elements commute pairwise

Statement

For all $1 \leq i, j \leq n$ one has $X_{i} X_{j} = X_{j} X_{i}$ in $Z [S_{n}]$ , 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.

Facts & Assumptions

Given: An integer $n \geq 1$ , the Jucys-Murphy elements $X_{1}, \dots, X_{n} \in Z [S_{n}]$ , and for each $k$ the transposition sum $T_{k} = \sum_{1 \leq i < j \leq k} (i j)$ (The Jucys-Murphy elements of the symmetric group algebra).

[F1]

$X_{1} = 0$ , $X_{k} = T_{k} - T_{k - 1}$ for $2 \leq k \leq n$ , and therefore $\sum_{k = 2}^{n} X_{k} = T_{n} - T_{1} = T_{n}$ ; for $m \leq n$ the element $X_{k}$ of $Z [S_{m}]$ maps to $X_{k}$ under the inclusion $Z [S_{m}] ↪ Z [S_{n}]$ , and the identity $Z [S_{n}] \to R [S_{n}]$ of the base change is a unital ring homomorphism sending $X_{k}$ to $X_{k}$ (The Jucys-Murphy elements of the symmetric group algebra).

[F2]

For $h \in S_{n}$ and distinct $i, j \in {1, \dots, n}$ one has $h (i j) h^{- 1} = (h (i) h (j))$ (Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$ ).

Proof

technique · induction on $n$

1.1givenbase

Base case. For $n = 1$ the only element is $X_{1} = 0$ ; for $n = 2$ the elements are $X_{1} = 0$ and $X_{2} = (1 2)$ . In both cases every pair among $X_{1}, \dots, X_{n}$ consists of two commuting elements, namely $0$ or the single element $X_{2}$ .

1.2givenihF1

Induction hypothesis. Let $n \geq 3$ and assume that $X_{2}, \dots, X_{n - 1}$ commute pairwise in $Z [S_{n - 1}]$ . By [F1] the inclusion of group rings is a unital ring homomorphism carrying these elements to the corresponding elements of $Z [S_{n}]$ , so $X_{2}, \dots, X_{n - 1}$ commute pairwise in $Z [S_{n}]$ as well.

1.3F2givenalgebra

The element $T_{n} = \sum_{1 \leq i < j \leq n} (i j)$ is central in $Z [S_{n}]$ . Indeed, for $h \in S_{n}$ , [F2] gives $h (i j) h^{- 1} = (h (i) h (j))$ for every pair $i < j$ , so $h T_{n} h^{- 1} = \sum_{i < j} (h (i) h (j)) = \sum_{i < j} (i j) = T_{n}$ , because ${i, j} \mapsto {h (i), h (j)}$ is a bijection of the set of $2$ -element subsets of ${1, \dots, n}$ . Thus $h T_{n} = T_{n} h$ for every $h \in S_{n}$ , and extending by linearity over the basis $S_{n}$ gives $z T_{n} = T_{n} z$ for every $z \in Z [S_{n}]$ .

2.1step 1.2step 1.3F1algebra

For $2 \leq j \leq n - 1$ compare the two expansions of $X_{j} T_{n}$ . On the one hand $X_{j} T_{n} = T_{n} X_{j}$ by step 1.3; on the other hand, using [F1] and the induction hypothesis of step 1.2, $X_{j} T_{n} = X_{j} (\sum_{k = 2}^{n - 1} X_{k} + X_{n}) = \sum_{k = 2}^{n - 1} X_{j} X_{k} + X_{j} X_{n} = \sum_{k = 2}^{n - 1} X_{k} X_{j} + X_{j} X_{n}$ , while $T_{n} X_{j} = (\sum_{k = 2}^{n - 1} X_{k} + X_{n}) X_{j} = \sum_{k = 2}^{n - 1} X_{k} X_{j} + X_{n} X_{j}$ . Subtracting the common term $\sum_{k = 2}^{n - 1} X_{k} X_{j}$ gives $X_{j} X_{n} = X_{n} X_{j}$ . Together with the induction hypothesis and the base case this covers every pair, so all of $X_{1}, \dots, X_{n}$ commute pairwise in $Z [S_{n}]$ .

3.1step 2.1F1discharge-induction: step 1.1∎

Finally let $R$ be a commutative ring. The base change $Z [S_{n}] \to R [S_{n}]$ is a unital ring homomorphism and carries $X_{k}$ to $X_{k}$ for every $k$ by [F1]; applying it to the identity $X_{i} X_{j} = X_{j} X_{i}$ of step 2.1 gives $X_{i} X_{j} = X_{j} X_{i}$ in $R [S_{n}]$ . Hence the subalgebra generated by the $X_{k}$ is commutative over any commutative ring.

Remarks

Integrality and no choice. The argument takes place entirely in $Z [S_{n}]$ 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 $Z [S_{k}]$ , since conjugation by $S_{k}$ permutes the transpositions of $S_{k}$ . This is the only property of $T_{n}$ used above.

Depends on

Used by

Elementary symmetric Jucys-Murphy evaluations are cycle-count class sums Theorem

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Garsia, Young Seminormal Representation, Murphy Elements and Content Evaluations, UCSD lecture notes (2003), section 3, printed pp. 18-25, Theorem 3.1(a) (standard reference, not scraped)
Okounkov-Vershik, A New Approach to the Representation Theory of the Symmetric Groups, Selecta Math. (N.S.) 2 (1996) 581-605; complete arXiv repost math/0503040, equation (2.1) and its following commutativity observation, printed p. 10 (standard reference, not scraped)