Plancherel Measure and Asymptotic Young Diagrams — Examples

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 $P_3(3)=1\mathrm{/}6$, $P_3(2,1)=4\mathrm{/}6$ and $P_3(1^3)=1\mathrm{/}6$, whereas the uniform weights on the three partitions would be $1\mathrm{/}3$ each. Since $4\mathrm{/}6\mathrm{\ne }1\mathrm{/}6$, the Plancherel measure on $Y_3$ is not the uniform measure, refuting uniformity already at $n=3$.

The general case: let $n\ge 3$. 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 $N=n\ge 3$ one has $f^{(n-1,1)}=n-1>1$. Hence $P_n(n)=1\mathrm{/}n!$ while $P_n(n-1,1)=(n-1{)}^2\mathrm{/}n!>1\mathrm{/}n!$. 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 $Y_n$.

Conclusion: the two partitions $(n)$ and $(n-1,1)$ of $n$ carry distinct Plancherel weights for every $n\ge 3$, 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 $P_3(2,1)=4\mathrm{/}6>1\mathrm{/}6=P_3(3)$.

The three shapes: by [F1] the partitions of $3$ are exactly $(3)$, $(2,1)$ and $(1^3)$, and [F2] and [F3] give $f^{(3)}=1$, $f^{(2,1)}=2$ and $f^{(1^3)}=1$; in particular each $f^{\lambda }$ is a positive integer and the standard-tableau counts are as displayed.

The weights: by definition of the Plancherel measure $P_3(\lambda )=(f^{\lambda }{)}^2\mathrm{/}3!$ with $3!=6$, step 1.1 gives $P_3(3)=1\mathrm{/}6$, $P_3(2,1)=4\mathrm{/}6=2\mathrm{/}3$ and $P_3(1^3)=1\mathrm{/}6$; summing, $1\mathrm{/}6+4\mathrm{/}6+1\mathrm{/}6=6\mathrm{/}6=1$.

Conclusion: the computed weights are the values displayed in the Example, their sum is one as predicted by [F4], and the middle shape carries four times the weight of either extreme shape, so $n=3$ is the smallest case exhibiting non-uniformity of the Plancherel weights.

Explicit insertions: applying [F1] to each word, one letter at a time, gives the following tableaux (written as the list of their rows): $123⇝((1,2,3))$, shape $(3)$; $132$: $(1)$, then $(1,3)$, then $2$ bumps the $3$, giving rows $(1,2)$ and $(3)$, shape $(2,1)$; $213$: $(2)$, then $1$ bumps $2$ giving rows $(1)$ and $(2)$, then $3$ is appended in the first row, shape $(2,1)$; $231$: $(2)$, then $(2,3)$, then $1$ bumps $2$, giving rows $(1,3)$ and $(2)$, shape $(2,1)$; $312$: $(3)$, then $1$ bumps $3$ giving rows $(1)$ and $(3)$, then $2$ is appended in the first row, shape $(2,1)$; $321$: $(3)$, then $2$ bumps $3$, giving rows $(2)$ and $(3)$, then $1$ bumps $2$ and the expelled $2$ bumps $3$, giving rows $(1)$, $(2)$, $(3)$, shape $(1^3)$. Each bumping step is the deterministic rule of [F1] applied to the displayed entries.

Frequencies: by [F2] the shapes recorded in step 1.1 are the Robinson-Schensted shapes of the six permutations, so among the six words the shape $(3)$ occurs once, $(2,1)$ four times and $(1^3)$ once; with the uniform weight $1\mathrm{/}6$ on each permutation the frequencies are $1\mathrm{/}6,4\mathrm{/}6,1\mathrm{/}6$.

Comparison: the frequencies of step 2.1 are exactly the Plancherel weights $P_3(3)=1\mathrm{/}6$, $P_3(2,1)=4\mathrm{/}6$ and $P_3(1^3)=1\mathrm{/}6$ of [F3], confirming the law of the shape of a uniform permutation at $n=3$; every step was a finite computation with integer entries, and no choice principle was used.