# Symmetry of Plancherel measure (for $S_n$)

For each $n \geq 1$ consider the reverse lexicographical order on the set $P(n)$ of partitions of $n$. Example for $n=7$:

$$\begin{pmatrix} \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \hline 7 & 6 & 5 & 5 & 4 & 4 & 4 & 3 & 3 & 3 & 3 & 2 & 2 & 2 & 1 \\ 0 & 1 & 2 & 1 & 3 & 2 & 1 & 3 & 2 & 2 & 1 & 2 & 2 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 2 & 1 & 1 & 2 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline \end{pmatrix}$$

Thus a partition of $n$, equivalently a Young diagram, is a column $j$ of the table above. For each $j$ let $Y_n(j)$ be the number of standard Young tableaux of shape $j$. Then, as shown by the figures below, the sequence $Y_n(1), \ldots, Y_{n}(T(n))$ is not symmetric when $n \geq 6$. Equivalently the Plancherel probability measure on $P(n)$ is not symmetric for this ordering.

However it seems that symmetry occurs for another ordering of $P(n)$. Is it true and if so what is the order making symmetry ?

## EDIT

I think I'm wrong: for $n=8$ there's no possible symmetry (unless my graph is wrong). I'm going to delete this question soon.

## EDIT 2

Sorry I have not deleted this post yet because I have in mind a variant of the question but I don't have the time to develop it now (shortly, I am under the impression that the bar chart is "almost symmetric" for $n \geq 8$.)

If you want a total ordering that is reversed by transposition, then your best bet is to extend the dominance ordering so as to preserve the anti-isomorphism property. This means breaking ties for the incomparable pairs in the dominance ordering. The smallest such pairs are for $n=6$, namely $([4,1,1],[3,3])$ and the transposed pair $([3,1,1,1],[2,2,2])$. Since these pairs are separated in the dominance ordering by the partition $[3,2,1]$, an extension is possible. But incomparable pairs get more numerous quickly as $n$ grows. For $n=7$ one can still arrange things symmetrically around the unique self-dual partition $[4,1,1,1]$, but for $n=8$ there are two (necessarily incomparable) self-dual partitions $[4,2,1,1]$ and $[3,3,2]$. They have different numbers of tableaux ($90$ and $42$, respectively) which numbers then have odd multiplicity in the list of the $Y_n(j)$ (in fact both are unique), which means that for a symmetric arrangement both these numbers need to be at the central position in the list, which obviously that cannot.
This type of problem rapidly gets worse as $n$ increases.
• Thank you (though I will need some time to understand your answer). About by 2nd edit, I am under the impression that for $n \geq 8$ it is possible to have a symmetric bar chart up to one point (i.e. we allow to remove one point). But I need to check again with the computer. May 22, 2013 at 10:05
• As my answer indicates, the obstruction for symmetry comes from the self-conjugate partitions (whose number is the same as that of partitions into distinct odd parts). Supposing all self-conjugate partitions have different numbers of tableaux (likely) then allowing $m$ asymmetries you can accommodate up to $2m+1$ self-conjugate partitions. The smallest number with $4$ self-conjugate partitions is $n=15$: $[8,1^{(7)}], [6,3,3,1,1,1], [5,4,3,2,1], [4,4,4,3]$; indeed you will need $2$ asymmetries for this case. May 22, 2013 at 11:39