# Transitivity for Schutzenberger involutions on standard Young tableaux

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?

• 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 $\lambda$? Nov 22, 2021 at 14:29
• Why is it so easy to see that the $t_i$ act transitively? Dec 5, 2021 at 21:46
• 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. Dec 5, 2021 at 21:49
• 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. Dec 11, 2021 at 22:07

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 $G_n$.