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Let $\lambda$ be a partition of $ n$. Let $ SYT(\lambda) $ denote the set of standard Young tableaux of shape $ \lambda $.

For $ i = 1, \dots, n $, let me define permutations $ S_i $ of the set $ SYT(\lambda) $. Let $ S_n(T) $ be the Schutzenberger involution of $ T $. For $ i < n $, let $ S_i(T) $ be the $i$th "partial" Schutzenberger involution. By this, I mean that we fix the part of $ T $ containing $ i+1, \dots, n $ and perform Schutzenberger involution of the part of $ T $ containing $ 1, \dots, i $.

Berenstein and Kirillov studied these permutations, in their article "Groups generated by involutions, Gelfand-Tsetlin patterns and combinatorics of Young tableaux" (available at math.uoregon.edu/~arkadiy/bk1.pdf). They prove that these permutations give rise to the action of a certain group, called $ G_n $, on $ SYT(\lambda) $.

Question

Does the group $ G_n$ act transitively on $ SYT(\lambda)$? In other words, given two standard Young tableaux can I turn one into the other by applying a sequence of partial Schutzenberger involutions?

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    $\begingroup$ Apologies for commenting on a question from long ago, but it's related to something I am now thinking about. In the BK paper they show that the group generated by these partial Schutzenberger involutions, which they denote $q_i$, is the same as the group generated by the Bender-Knuth involutions, which they denote $t_i$. But acting on SYT these $t_i$ act very simply: they just swap $i$ and $i+1$ if they are non-adjacent. Then isn't it quite easy to see that the $t_i$ act transitively on the set of SYT of shape $\lambda$? $\endgroup$ Nov 22, 2021 at 14:29
  • $\begingroup$ Why is it so easy to see that the $t_i$ act transitively? $\endgroup$ Dec 5, 2021 at 21:46
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    $\begingroup$ I think this is clear by induction. It suffices to show that I can e.g. bring $n$ to the highest outer corner box in the shape. If it's already there, great. If not, well then restrict to the tableau with entries $1,\ldots,n-1$, and by induction I can bring $n-1$ to the right box, which then I can finally swap with $n$. In fact the claim should hold more generally for BK involutions acting on the linear extensions of any finite poset. $\endgroup$ Dec 5, 2021 at 21:49
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    $\begingroup$ For an explicit statement of the transitivity of the Bender-Knuth involutions in the more general context of arbitrary (finite) posets, see Proposition 1.3 of doi.org/10.5070/C61055363. $\endgroup$ Dec 11, 2021 at 22:07

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Yes. It has been a while since I did this calculation but I think this is the gist of it:

Let $ s_{1q} $ be the permutation induced by the partial Schützenberger involution $ S_q(\cdot) $. Then define the permutations $$ s_{pq} := s_{1q} s_{1(q-p+1)} s_{1q}. $$ The $Q$-symbol of the RSK correspondence gives a bijection between words of length $n$ with shape $\lambda$ and $SYT(\lambda)$. One can check that on words the operators $s_{pq}$ when $p-q = 2$ induce the Knuth moves - they are in fact just the cactus operators, hence the choice of notation. Since the Knuth moves act transitively so does the group $G_n$.

You can see this nicely using Speyer's cylindrical growth diagrams defined here.

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    $\begingroup$ Sorry, I know this is a bit vague. I am in the process of writing some results related to this down as part of my phd thesis. I'd be happy to add to it once I have the story straight in my head again! $\endgroup$
    – Noah White
    Dec 12, 2014 at 14:17
  • $\begingroup$ Thanks, this is quite interesting. I would be happy to learn more about your PhD thesis. Please send it to me when it is complete. $\endgroup$ Dec 14, 2014 at 19:03
  • $\begingroup$ No worries. I will! $\endgroup$
    – Noah White
    Dec 15, 2014 at 15:01

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