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Aug 22, 2017 at 9:49 comment added Jianrong Li @darij grinberg, thank you very much. I corrected the theorem. In your example, if we take ${\bf i}=(4,1,2,1,5,3)$. Then we can take $r=2$.
Aug 22, 2017 at 9:48 history edited Jianrong Li CC BY-SA 3.0
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Aug 22, 2017 at 9:40 comment added Jianrong Li @Christian Stump, thank you very much. Are there some references about the results you mentioned?
Aug 21, 2017 at 10:24 comment added Christian Stump This is clearly wrong -- what you maybe want (to state / to use ?) is that inversion sets are closed in the sense that if $(i,j),(j,k)$ are inversions then so is $(i,k)$ and for any such reduced word $i = (i_1,\ldots,i_m)$, you have that $f_i(x) = (i,k)$ and $f_i(a) = (i,j), f_i(b) = (j,k)$ then $a < x < b$ or $b < x < a$.
Aug 21, 2017 at 10:19 comment added darij grinberg I don't think this can be true. You want to say that the inversions $(i,j),(i,k),(j,k)$ must be adjacent in the inversion word of every reduced sequence of $w$. I would be surprised if this holds for any reduced sequence of $w$; but definitely it cannot hold for every reduced sequence of $w$ ! (Try the element $s_1s_4s_2s_5s_1s$ in $S_6$, with the reduced sequence I just gave, and with $i=1, j=2, k=3$.)
Aug 21, 2017 at 9:53 history edited Jianrong Li CC BY-SA 3.0
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Aug 21, 2017 at 9:52 comment added Jianrong Li @Dirk Liebhold, thank you for your suggestions. I will edit the post.
Aug 21, 2017 at 9:16 comment added Dirk You should correct the theorem a little. For example, you have $f_i(r)$ twice, you should assume that $(i,j),(i,j),(j,k)$ are inversions, etc. Right now, your theorem is false in general, as one could simply pick $i < j < k$ such that $(i,j)$ is not an inversion of $w$...
Aug 21, 2017 at 8:37 history asked Jianrong Li CC BY-SA 3.0