PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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The Plancherel weights sum to one

Statement

For every $n \geq 0$ , $\sum_{λ ⊢ n} P_{n} (λ) = \frac{1}{n!} \sum_{λ ⊢ n} (f^{λ})^{2} = 1.$ Hence $P_{n}$ is a probability distribution on the finite set $Y_{n}$ (Finite probability spaces, outcome weights, events, and event probabilities): the weights are nonnegative and sum to one. In particular $\dim_{C} C [S_{n}] = n! = \sum_{λ ⊢ n} (f^{λ})^{2}$ is obtained in two independent ways.

Facts & Assumptions

Given: $n \geq 0$ ; the finite set $Y_{n}$ of partitions of $n$ and the weights $P_{n} (λ) = (f^{λ})^{2} / n!$ , where $f^{λ} = \dim_{C} S^{λ}$ is the number of standard $λ$ -tableaux and $P_{0} (\emptyset) = 1$ (The Plancherel measure on the partitions of $n$ ).

[F1]

If $G$ is a finite group and $k$ is algebraically closed with $char k ∤ ∣ G ∣$ , then there are finitely many irreducible representations $V_{1}, \dots, V_{r}$ of $G$ over $k$ , up to equivalence, and $k [G] ≅ V_{1}^{\oplus \dim V_{1}} \oplus \dots \oplus V_{r}^{\oplus \dim V_{r}}$ (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , there are finitely many irreducible representations, and each occurs in the regular representation with multiplicity equal to its degree).

[F2]

The character of $C [S_{n}]$ is $∣ S_{n} ∣$ at the identity and $0$ elsewhere, so $\dim_{C} C [S_{n}] = n!$ (The regular character is $∣ G ∣$ at $1$ and $0$ away from $1$ ).

[F3]

For $G = S_{n}$ over $C$ , the irreducible representations up to equivalence are exactly the Specht modules $S^{λ}$ , $λ ⊢ n$ (Specht modules classify the complex irreducibles of $S_{n}$ , Complex Specht modules are irreducible), and $\dim_{C} S^{λ} = f^{λ}$ (Standard polytabloids form a basis of a complex Specht module).

[F4]

$\sum_{λ ⊢ n} (f^{λ})^{2} = n!$ for every $n \geq 0$ , by the Robinson-Schensted count (The sum of squares of the standard tableau numbers, The Robinson-Schensted correspondence).

[F5]

A finite probability space is a finite set $Ω$ with weights $w \geq 0$ satisfying $\sum_{ω} w (ω) = 1$ (Finite probability spaces, outcome weights, events, and event probabilities).

Proof

technique · direct

1.1givenF1F2F3algebra

Regular-representation count: $C$ is algebraically closed of characteristic $0$ , and $∣ S_{n} ∣ = n! > 0$ , so $char C = 0$ does not divide $∣ S_{n} ∣$ , so [F1] applies to $G = S_{n}$ , $k = C$ : the regular representation is $C [S_{n}] ≅ ⨁_{i = 1}^{r} V_{i}^{\oplus \dim V_{i}}$ with $V_{1}, \dots, V_{r}$ representing the irreducible complex representations of $S_{n}$ up to equivalence. By [F3] this list is ${S^{λ} : λ ⊢ n}$ and $\dim_{C} S^{λ} = f^{λ}$ , so $C [S_{n}] ≅ ⨁_{λ ⊢ n} (S^{λ})^{\oplus f^{λ}}$ . Taking dimensions, which are additive over direct sums and multiplicative over direct powers, and using [F2] gives $n! = \dim_{C} C [S_{n}] = \sum_{λ ⊢ n} f^{λ} \cdot \dim_{C} S^{λ} = \sum_{λ ⊢ n} (f^{λ})^{2}$ .

1.2givenF4

Independent count: the same identity $\sum_{λ ⊢ n} (f^{λ})^{2} = n!$ is proved independently from the Robinson-Schensted bijection by [F4], so the two computations of $\dim_{C} C [S_{n}]$ agree without either appealing to the other.

2.1givenF5step 1.1step 1.2algebra

Normalization: dividing the identity of steps 1.1 and 1.2 by the positive integer $n!$ (for $n = 0$ both sides read $1 = 1$ and $P_{0} (\emptyset) = 1$ ) gives $\sum_{λ ⊢ n} P_{n} (λ) = 1$ . Each $P_{n} (λ) = (f^{λ})^{2} / n!$ is a quotient of a nonnegative integer by a positive integer, hence is $\geq 0$ , and $Y_{n}$ is finite (The Plancherel measure on the partitions of $n$ ); therefore $P_{n}$ , viewed as a function on the finite set $Y_{n}$ , satisfies both requirements of a finite probability space in [F5].

3.1givenstep 2.1algebra∎

Conclusion: the displayed normalization, the nonnegativity of the weights and the finite nonempty outcome set $Y_{n}$ are exactly the assertion that $P_{n}$ is a probability distribution on $Y_{n}$ ; the two independent evaluations computing $\dim_{C} C [S_{n}] = n!$ are steps 1.1 and 1.2. This holds for every $n \geq 0$ , including the degenerate case $n = 0$ with the single empty partition.

Depends on

Used by

Joint convergence in distribution and the normalized cycle-character observables Definition
The Plancherel measure on partitions of three Example
Plancherel expectations of the shifted character observables Proposition
Scaled Plancherel profile moments converge in probability Proposition
Plancherel Young diagrams converge to the limit shape Theorem

Dependency tree · two levels

61 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

Vladimir Ivanov and Grigori Olshanski, Kerov's central limit theorem for the Plancherel measure on Young diagrams, arXiv:math/0304010; survey-paper version in Symmetric Functions 2001, NATO Science Series II 74 (2002), 93-151 (standard reference, not scraped)
Dan Romik, The Surprising Mathematics of Longest Increasing Subsequences, Cambridge University Press 2015; author-hosted manuscript of 20 August 2014 (363 pp.) (standard reference, not scraped)