An angle-doubling trick of Kirillov and Berenstein [closed]

Kirillov and Berenstein, in their article "Groups generated by involutions, Gelfand-Tsetlin patterns and combinatorics of Young tableaux" (available at math.uoregon.edu/~arkadiy/bk1.pdf), present a construction that, starting from involutions $t_1,t_2,\dots,t_n$ satisfying $(t_i t_j)^2 = 1$ for $|i-j|>1$ and $(t_i t_j)^6 = 1$ for $|i-j|=1$, create new involutions $s_1,s_2,\dots$ satisfying $(s_i s_j)^2 = 1$ for $|i-j|>1$ and $(s_i s_j)^3 = 1$ for $|i-j|=1$; specifically, they define $s_i = q_i t_1 q_i^{-1}$ with $q_i = (t_1)(t_2 t_1) \cdots (t_i t_{i-1} \cdots t_2 t_1)$.

Unfortunately K&B do not prove directly that their construction has the desired properties, but rather derive it from properties of a more complicated construction. (See page 18: "Part 10 of Theorem 1.1 follows from part 20 and the equality (1.18)." Theorem 1.1 itself appears on page 16.)

Can anyone give a self-contained explanation of why this trick works?

(Is it a special case of a more general trick that, given a Coxeter group, constructs a subgroup whose diagram is similar to that of the original, but simpler beause some of the Coxeter exponents are replaced by proper divisors of those exponents?)

I have verified the claim up through $(s_2 s_3)^3 = 1$, but even proving $(s_3 s_4)^3 = 1$ is giving me trouble.

closed as off-topic by Stefan Kohl, Steven Sam, David Loeffler, S. Carnahan♦Aug 28 '14 at 22:11

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• Are you sure they actually claim this? The equation (1.18) is nowhere in your axioms. (Disclaimer: I have read even less of the paper than you.) – darij grinberg Aug 26 '14 at 19:14
• It is indeed possible that K&B's proof of the Theorem 1.1 makes use of properties of the $t_i$'s that I haven't included. (This would account for why I haven't been able to prove $(s_3 s_4)^3 = 1$ using just the Coxeter relations.) I don't understand the article well enough to assess this. So I guess I should have started the original post by saying "It seems to me that Kirillov and Berenstein..." – James Propp Aug 26 '14 at 20:18
• It seems that (as Darij speculated) Kirillov and Berenstein didn't do what I said they did; their arguments use other properties of the $t_i$'s. To see this, let $t_1$, $t_2$, and $t_3$ be the permutations of $\{1,2,\dots,7\}$ with respective cycle-decompositions $(25)(46)$, $(13)(27)$, and $(26)(45)$. Then the $t_i$'s satisfy the hypotheses of my claim but the $s_i$'s don't satisfy the conclusions of my claim. This leaves me with only the not-very-MathOverflow-ish question "Can someone explain to me what Kirillov and Berenstein are doing in their construction of the $s_i$'s?" – James Propp Aug 27 '14 at 3:41
• If anything, MathOverflow could profit from more of the "please explain/simplify the proof in that paper" type of questions. But I guess I'm not in the majority with this. – darij grinberg Aug 27 '14 at 10:45
• I think this would be quite a nice question, if the construction would actually work in the claimed generality. But as James himself has noticed, it doesn't. What remains is merely a not self-contained question of the type "can someone explain what Kirillov and Berenstein are doing in their paper?". Therefore I am voting to close. – Stefan Kohl Aug 27 '14 at 11:07