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The Plancherel measure is not uniform on partitions

Statement refuted

For every $n \geq 3$ , the Plancherel measure on $Y_{n}$ is uniform on the partitions of $n$ .

Facts & Assumptions

Given: the Plancherel weights $P_{n} (λ) = (f^{λ})^{2} / n!$ on the partitions $λ ⊢ n$ , where $f^{λ} = \dim_{C} S^{λ}$ is the number of standard $λ$ -tableaux (The Plancherel measure on the partitions of $n$ , Standard polytabloids form a basis of a complex Specht module).

[F1]

For every $n \geq 0$ and $λ ⊢ n$ , $f^{λ} = n! / \prod_{x \in [λ]} h (x)$ , the empty product being $1$ ; in particular $f^{(n)} = 1$ for $n \geq 1$ and $f^{(1^{n})} = 1$ (The hook length formula).

[F2]

For every integer $N \geq 2$ , the hook shape $(N - 1, 1)$ has first-row hook lengths $N, N - 2, N - 3, \dots, 1$ and second-row hook length $1$ , so its hook product is $N (N - 2)!$ and $f^{(N - 1, 1)} = N! / (N (N - 2)!) = N - 1$ (The hook length formula).

[F3]

Partitions of a fixed integer are the weakly decreasing positive sequences summing to it; $(1^{3})$ denotes the column $(1, 1, 1)$ (Partitions, English diagrams, and conjugation).

Counterexample

The Plancherel measure on the partitions of $3$ is not the uniform measure on the three partitions $(3), (2, 1), (1^{3})$ : the middle shape has four times the weight of either extreme shape, since $P_{3} (2, 1) = 4 / 6$ while $P_{3} (3) = P_{3} (1^{3}) = 1 / 6$ . More generally, uniform weights on $Y_{n}$ are not the Plancherel weights for any $n \geq 3$ .

Proof technique: direct.

1.1givenF1F2F3algebra

The case $n = 3$ : the partitions of $3$ are $(3), (2, 1), (1^{3})$ by [F3]; [F1] gives $f^{(3)} = 1$ and $f^{(1^{3})} = 1$ , and [F2] with size $3$ gives $f^{(2, 1)} = 3 - 1 = 2$ . Hence the Plancherel weights of order $3$ are $P_{3} (3) = 1 / 6$ , $P_{3} (2, 1) = 4 / 6$ and $P_{3} (1^{3}) = 1 / 6$ , whereas the uniform weights on the three partitions would be $1 / 3$ each. Since $4 / 6 \neq 1 / 6$ , the Plancherel measure on $Y_{3}$ is not the uniform measure, refuting uniformity already at $n = 3$ .

1.2givenF1F2algebra

The general case: let $n \geq 3$ . The shapes $(n)$ and $(n - 1, 1)$ are partitions of $n$ (Partitions, English diagrams, and conjugation); by [F1] $f^{(n)} = 1$ , and by [F2] with $N = n \geq 3$ one has $f^{(n - 1, 1)} = n - 1 > 1$ . Hence $P_{n} (n) = 1 / n!$ while $P_{n} (n - 1, 1) = (n - 1)^{2} / n! > 1 / n!$ . Two partitions of $Y_{n}$ therefore carry distinct Plancherel weights, so the weights on $Y_{n}$ are not all equal, and in particular $P_{n}$ is not the uniform distribution on $Y_{n}$ .

2.1givenstep 1.1step 1.2algebra∎

Conclusion: the two partitions $(n)$ and $(n - 1, 1)$ of $n$ carry distinct Plancherel weights for every $n \geq 3$ , so no uniform probability distribution on $Y_{n}$ can coincide with $P_{n}$ ; the computation of step 1.1 is the case $n = 3$ in which the failure is witnessed explicitly by $P_{3} (2, 1) = 4 / 6 > 1 / 6 = P_{3} (3)$ .

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Sources

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)