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Primitive tableau idempotents by Jucys-Murphy interpolation

Statement

Let $T$ be a standard tableau of size $n$ , let $μ$ be the shape of the restriction of $T$ to $[n - 1]$ , and let $A (μ)$ be the set of contents of the addable nodes of $μ$ . Set $P_{[1]} : = 1$ and define recursively, for $n \geq 2$ , $P_{T} : = P_{T ↓ [n - 1]} \prod_{\begin{array}{c} c \in A (μ) \\ c \neq c_{T} (n) \end{array}} \frac{X_{n} - c}{c_{T} (n) - c} \in C [S_{n}] .$ Then all displayed denominators are nonzero, $P_{T}$ is the rank-one idempotent projecting onto the Young line $C v_{T}$ , each $P_{T}$ is a polynomial in $X_{1}, \dots, X_{n}$ with rational coefficients, and the $P_{T}$ over the standard tableaux of size $n$ are pairwise orthogonal idempotents with $\sum_{T} P_{T} = 1$ .

Facts & Assumptions

Given: The chain $S_{1} \subset \dots \subset S_{n}$ , the Jucys-Murphy elements $X_{k}$ , and, for every standard tableau $T$ of size $n$ , the Young line $C v_{T} = im P_{T}$ with the idempotents $P_{T}$ of the diagonal-algebra theorem (The Gelfand-Tsetlin algebra is the diagonal algebra of the Young basis, The Jucys-Murphy elements of the symmetric group algebra).

[F1]

$G Z (n) = ⨁_{T} C P_{T}$ over the standard tableaux $T$ of size $n$ ; the $P_{T}$ are nonzero pairwise orthogonal idempotents with $\sum_{T} P_{T} = 1$ , and $C v_{T} = im P_{T}$ is one-dimensional (The Gelfand-Tsetlin algebra is the diagonal algebra of the Young basis).

[F2]

$X_{k} v_{T} = c_{T} (k) v_{T}$ for every standard tableau $T$ and every $k$ , with $c_{T} (k)$ the content of the node carrying $k$ (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]

For a partition $μ$ , the contents of distinct addable nodes are distinct, and the addable nodes of $μ$ are the nodes $(i, μ_{i} + 1)$ with $i = 1$ or $μ_{i - 1} > μ_{i}$ , together with the new-row node (Distinct addable nodes of a partition have distinct contents, Removable and addable nodes).

[F4]

The restriction of $T$ to $[n - 1]$ is a standard tableau of shape $μ$ that is obtained by deleting the node carrying $n$ ; conversely every standard tableau $S$ of size $n$ with $S ↓ [n - 1] = T ↓ [n - 1]$ is obtained by placing $n$ in an addable node of $μ$ (Tableaux and standard tableaux, Removable and addable nodes).

Proof

technique · induction

1.1F1givenbase

Base case. For $n = 1$ the only standard tableau is $[1]$ , the line is $C v_{[1]} = C \cdot 1$ , and $P_{[1]} : = 1$ is the rank-one projection onto it with rational (indeed integer) coefficients.

1.2F1givenih

Induction hypothesis. For every standard tableau $S$ of size $n - 1$ the recursively defined element $P_{S} \in C [S_{n - 1}]$ equals the idempotent $P_{S}$ of [F1], is the projection onto $C v_{S}$ , and is a polynomial in $X_{1}, \dots, X_{n - 1}$ with rational coefficients.

1.3F3F4givenalgebra

Let $T$ be a standard tableau of size $n$ and $μ : = shape (T ↓ [n - 1])$ . The set $A (μ)$ of contents of addable nodes is finite and $c_{T} (n) \in A (μ)$ : the node carrying $n$ is an addable node of $μ$ by [F4]. By [F3] the denominators $c_{T} (n) - c$ , $c \in A (μ)$ , $c \neq c_{T} (n)$ , are nonzero integers, so the displayed product is a well-defined element of $C [S_{n}]$ .

2.1step 1.2F1algebra

If $S$ is a standard tableau of size $n - 1$ with $S \neq T ↓ [n - 1]$ , then the factor $P_{T ↓ [n - 1]}$ acts as $0$ on the line $C v_{S}$ by step 1.2; hence $P_{T}$ acts as $0$ on every Young line of size $n$ whose restriction to $[n - 1]$ differs from $T ↓ [n - 1]$ .

2.2step 1.2F2F4algebra

Now let $S$ be a standard tableau of size $n$ with $S ↓ [n - 1] = T ↓ [n - 1]$ , and let $x$ be the addable node of $μ$ carrying $n$ in $S$ , so that $c_{S} (n) = c (x)$ by [F4] and [F2]. Then $P_{T ↓ [n - 1]}$ acts as the identity on $C v_{S}$ by step 1.2, and the interpolation factor acts on $C v_{S}$ by the scalar $\prod_{\begin{array}{c} c \in A (μ) \\ c \neq c_{T} (n) \end{array}} \frac{c_{S} (n) - c}{c_{T} (n) - c},$ because $X_{n}$ acts on $C v_{S}$ by $c_{S} (n)$ by [F2].

2.3step 1.2step 1.3algebra

Polynomial form. By step 1.2 the factor $P_{T ↓ [n - 1]}$ is a polynomial in $X_{1}, \dots, X_{n - 1}$ with rational coefficients; each factor $(X_{n} - c) / (c_{T} (n) - c)$ is a polynomial in $X_{n}$ with rational coefficients because $c_{T} (n) - c \in Z ∖ {0}$ by step 1.3; hence $P_{T}$ is a polynomial in $X_{1}, \dots, X_{n}$ with rational coefficients.

3.1F3step 1.3step 2.2algebra

The scalar of step 2.2 equals $1$ when $S = T$ , and equals $0$ when $S \neq T$ : if $S = T$ then $c_{S} (n) = c_{T} (n)$ and every factor is $(c_{T} (n) - c) / (c_{T} (n) - c) = 1$ ; if $S \neq T$ then $S$ places $n$ in a different addable node of $μ$ , so $c_{S} (n) \neq c_{T} (n)$ by [F3], and the factor with $c = c_{S} (n)$ has numerator $0$ while the denominator is nonzero by step 1.3.

4.1step 2.1step 3.1F1algebra

Steps 2.1-2.3 show that $P_{T}$ acts as the identity on the line $C v_{T}$ and as $0$ on every other Young line of size $n$ . Since $G Z (n) = ⨁_{S} C P_{S}$ is the algebra of operators diagonal in the Young basis by [F1], the element $P_{T}$ equals the idempotent $P_{T}$ of [F1], namely the rank-one projection onto $C v_{T}$ . In particular $P_{T}^{2} = P_{T}$ and $P_{T} \neq 0$ .

5.1step 4.1F1

Orthogonality and the partition of unity are inherited from [F1]: for $T \neq T^{'}$ the idempotents $P_{T}, P_{T^{'}}$ of [F1] multiply to $0$ , and $\sum_{T} P_{T} = 1$ over the standard tableaux of size $n$ .

6.1step 1.1step 4.1step 5.1step 2.3discharge-induction∎

Steps 1.1, 2.1-2.3, 3.1 and 4.1 are the base, the successor and the conclusions of an induction on $n$ ; therefore the recursive formula defines the tableau idempotents for every $n$ , with all properties asserted.

Remarks

Interpolation at the spectrum. The factor is the Lagrange polynomial that takes the value $1$ at the content $c_{T} (n)$ and vanishes at the other contents in $A (μ)$ ; the contents in $A (μ)$ are pairwise distinct by [F3], and they are the eigenvalues of $X_{n}$ on the lines over $T ↓ [n - 1]$ by [F2]. This is Garsia's recursion for the seminormal units, stated here for the tableau idempotents of the diagonal algebra.
Integrality fails only at the denominators. Over $Z$ the formula must be cleared of denominators; over $C$ the rational coefficients are harmless. The first nontrivial denominator occurs for $n = 2$ : the addable contents of $(1)$ are $1, - 1$ , giving the projectors $(1 + X_{2}) / 2$ and $(1 - X_{2}) / 2$ .
Choice. The recursion selects no object: the addable node carrying $n$ in a given tableau is determined by that tableau, and the idempotents are built from the fixed chain.

Depends on

Used by

Ordinary Jucys-Murphy projection formulas do not survive content collision Counterexample
The Jucys-Murphy spectrum and projectors for S₃ Example
The Jucys-Murphy elements generate the Gelfand-Tsetlin algebra Theorem

Dependency tree · two levels

21 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), 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)