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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 $n \geq 8$.)

enter image description here

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Your question is related to the fact that the lexicographic ordering of partitions is not reversed by taking the transpose of all partitions. (Taking transpose does not affect the number of standard Young tableaux.) The important partial odering by dominance does have the property that transposition is an anti-isomorphism.

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.

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  • $\begingroup$ 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. $\endgroup$ May 22, 2013 at 10:05
  • $\begingroup$ 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. $\endgroup$ May 22, 2013 at 11:39
  • $\begingroup$ Thank you Marc. Your answers provide more information than I expected. $\endgroup$ Jul 3, 2013 at 20:52

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