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The Jucys-Murphy spectrum and projectors for S_3

Statement

The four standard tableaux of size three have content vectors $Cont = (0, 1, 2)$ for the row tableau of shape $(3)$ , $(0, 1, - 1)$ for the tableau $\begin{matrix} 1 & 2 \\ 3 \end{matrix}$ of shape $(2, 1)$ , $(0, - 1, 1)$ for $\begin{matrix} 1 & 3 \\ 2 \end{matrix}$ , and $(0, - 1, - 2)$ for the column tableau of shape $(1, 1, 1)$ . Writing $X_{2} = (1 2)$ and $X_{3} = (1 3) + (2 3)$ in $C [S_{3}]$ , the recursion of Primitive tableau idempotents by Jucys-Murphy interpolation gives the four primitive idempotents $P_{r o w} = \frac{(X_{2} + 1) (X_{3} + 1)}{6}, P_{\begin{matrix} 1 & 2 \\ 3 \end{matrix}} = \frac{(X_{2} + 1) (2 - X_{3})}{6}, P_{\begin{matrix} 1 & 3 \\ 2 \end{matrix}} = \frac{(1 - X_{2}) (X_{3} + 2)}{6}, P_{c o l} = \frac{(1 - X_{2}) (1 - X_{3})}{6},$ with denominators $2$ and $6$ ; they satisfy $P_{T}^{2} = P_{T}$ , $P_{T} P_{T^{'}} = 0$ for $T \neq T^{'}$ and $\sum_{T} P_{T} = 1$ , each $P_{T}$ acts on its own content vector by the identity and on the other content vectors by zero, and the sum of the two $(2, 1)$ -projectors is the central idempotent of the two-dimensional Specht module.

Facts & Assumptions

Given: The group algebra $C [S_{3}]$ with basis $1, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)$ and the elements $X_{2} = (1 2)$ , $X_{3} = (1 3) + (2 3)$ (The Jucys-Murphy elements of the symmetric group algebra).

[F1]

For the standard tableaux of size three the addable-content sets of the predecessor shapes are $A ((1)) = {1, - 1}$ , $A ((2)) = {2, - 1}$ and $A ((1, 1)) = {1, - 2}$ , and the recursion of the cited theorem gives the displayed four products; the denominators $c_{T} (n) - c$ are $\pm 2$ at step $n = 2$ and $\pm 3$ at step $n = 3$ ; their products give the denominator $6$ in the displayed formulas (Primitive tableau idempotents by Jucys-Murphy interpolation, Removable and addable nodes).

[F2]

$X_{k} v_{T} = c_{T} (k) v_{T}$ for the Young line of each standard tableau $T$ , and the four content vectors of size three are exactly the vectors satisfying conditions (1)-(3) (The joint spectrum of the Jucys-Murphy elements is the set of tableau content vectors, The content of a node and the content vector of a standard tableau).

[F3]

$G Z (3) = ⨁_{T} C P_{T}$ : the four lines are independent, the $P_{T}$ are pairwise orthogonal idempotents summing to $1$ , and the sum of the $P_{T}$ over the tableaux of a fixed shape is the corresponding central idempotent of $C [S_{3}]$ (The Gelfand-Tsetlin algebra is the diagonal algebra of the Young basis).

Proof

technique · direct

1.1givenalgebra

Multiplication in $S_{3}$ gives $X_{2}^{2} = 1$ , $X_{2} X_{3} = X_{3} X_{2} = (1 2 3) + (1 3 2)$ and $X_{3}^{2} = (1 2 3) + (1 3 2) + 2$ , whence also $(X_{2} X_{3})^{2} = X_{2} X_{3} + 2$ ; every product below is evaluated with these relations and the basis of the Given block.

1.2F2givenalgebra

Contents. The row tableau carries entries $1, 2, 3$ in the cells $(1, 1), (1, 2), (1, 3)$ of contents $0, 1, 2$ ; the tableau $\begin{matrix} 1 & 2 \\ 3 \end{matrix}$ carries them in $(1, 1), (1, 2), (2, 1)$ of contents $0, 1, - 1$ ; the tableau $\begin{matrix} 1 & 3 \\ 2 \end{matrix}$ in $(1, 1), (2, 1), (1, 2)$ of contents $0, - 1, 1$ ; and the column tableau in $(1, 1), (2, 1), (3, 1)$ of contents $0, - 1, - 2$ . This gives the four content vectors of the Statement.

2.1F1step 1.2algebra

The recursion of [F1] at $n = 2$ gives $P_{[1 2]} = (X_{2} + 1) / 2$ and $P_{\begin{matrix} 1 \\ 2 \end{matrix}} = (1 - X_{2}) / 2$ , because the two addable contents of $(1)$ are $1$ and $- 1$ ; at $n = 3$ the factors are $(X_{3} + 1) / 3$ and $(2 - X_{3}) / 3$ over the shape $(2)$ , and $(X_{3} + 2) / 3$ and $(1 - X_{3}) / 3$ over the shape $(1, 1)$ . Multiplying gives exactly the four displayed elements, with denominators $2$ and $6$ .

3.1step 2.1algebra

Expanded in the basis of the Given block, the four elements are $P_{r o w} = \frac{1}{6} (1 + (1 2) + (1 3) + (2 3) + (1 2 3) + (1 3 2)),$ $P_{\begin{matrix} 1 & 2 \\ 3 \end{matrix}} = \frac{1}{3} \cdot 1 + \frac{1}{3} (1 2) - \frac{1}{6} (1 3) - \frac{1}{6} (2 3) - \frac{1}{6} (1 2 3) - \frac{1}{6} (1 3 2),$ $P_{\begin{matrix} 1 & 3 \\ 2 \end{matrix}} = \frac{1}{3} \cdot 1 - \frac{1}{3} (1 2) + \frac{1}{6} (1 3) + \frac{1}{6} (2 3) - \frac{1}{6} (1 2 3) - \frac{1}{6} (1 3 2),$ $P_{c o l} = \frac{1}{6} (1 - (1 2) - (1 3) - (2 3) + (1 2 3) + (1 3 2)) .$

4.1step 1.1step 3.1F3algebra

Idempotence and orthogonality. Squaring each of the four elements of step 3.1 with the relations of step 1.1 returns the same element, and each product of two distinct ones vanishes; equivalently, the four elements are the orthogonal rank-one projections of [F3], and their sum is $\frac{1}{6} + \frac{1}{3} + \frac{1}{3} + \frac{1}{6} = 1,$ using the coefficient sums in step 3.1. This verifies all algebraic assertions about $P_{T}$ .

5.1F2F3step 1.2step 4.1algebra

Action on the spectrum. By [F3] each $P_{T}$ is the projection onto the line $C v_{T}$ , so $P_{T}$ acts by $1$ on the content-vector eigenvector of $T$ and by $0$ on the eigenvectors of the other tableaux, whose content vectors are the four listed in step 1.2.

5.2step 3.1step 4.1F3algebra

The sum of the two $(2, 1)$ -projectors is $P_{\begin{matrix} 1 & 2 \\ 3 \end{matrix}} + P_{\begin{matrix} 1 & 3 \\ 2 \end{matrix}} = \frac{2}{3} \cdot 1 - \frac{1}{3} (1 2 3) - \frac{1}{3} (1 3 2)$ , which is central in $C [S_{3}]$ and, by [F3], is the central idempotent of the two-dimensional Specht module $V^{(2, 1)}$ ; the remaining two projectors are the central idempotents of the one-dimensional modules $V^{(3)}$ and $V^{(1, 1, 1)}$ .

6.1F1step 2.1step 5.2∎

The example exhibits the interpolation formula at the smallest nontrivial size: the recursion inverts only $2$ at step $n = 2$ and $3$ at step $n = 3$ , giving the denominator $6$ in the products, and the failure of the formula in characteristic $2$ is recorded separately in Ordinary Jucys-Murphy projection formulas do not survive content collision.

Remarks

The four spectra. The content vectors $(0, 1, 2), (0, 1, - 1), (0, - 1, 1), (0, - 1, - 2)$ are the four vectors satisfying conditions (1)-(3) for $n = 3$ ; the first and last belong to the one-dimensional modules of the trivial and sign representations.
The central idempotent. $\frac{2}{3} \cdot 1 - \frac{1}{3} (1 2 3) - \frac{1}{3} (1 3 2)$ has the standard form $(\dim V / ∣ G ∣) \sum_{g} χ (g^{- 1}) g$ of the central idempotent attached to $V^{(2, 1)}$ .

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Sources

Garsia, Young Seminormal Representation, Murphy Elements and Content Evaluations, UCSD lecture notes (2003), Theorems 3.4-3.5, printed pp. 22-24 (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, sections 5-6, printed pp. 17-24 (standard reference, not scraped)