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Nov 10, 2015 at 16:44 comment added Christian Stump (It might be better to move this into an email conversation.)
Nov 10, 2015 at 16:43 comment added Christian Stump I also wouldn't know why the pieces should be balls or spheres, I'd rather expect not to be (I will try to construct a counterexample tonight). Finally, the "exists" means I wonder whether one can decompose an Artin subword complex $(Q,w)$ into subword complexes of the form $(Q,w1,S1) \sqcup ... \sqcup (Q,wm,Sm)$. Every element in $(Q,w)$ is in one of the pieces $(Q,wi,Si)$, but I would actually hope that the complete pieces are present.
Nov 10, 2015 at 16:40 comment added Christian Stump There is no reduction for Artin words. I believe that the greedy product only depends on a word up to braid relations (as I think it only depends on the "colored inversion sequence" r1,s1(r2),s1s2(r3),... for a word s1,s2,s3,... with corresponding simple roots r1,r2,r3,...).
Nov 10, 2015 at 15:53 comment added Allen Knutson So in this theory, you look at subwords that are Artin-reduced but not necessarily Coxeter-reduced? Is it clear that they should all have the same greedy product? I guess you can always decompose according to (greedy product, skip locations), but maybe the pieces won't be balls? I don't understand what "exists" means.
Nov 10, 2015 at 12:19 comment added Christian Stump Do you happen to see whether the decomposition in #2 always exists?
Nov 10, 2015 at 12:18 comment added Christian Stump You can also do that if you prefer. But since $(Q \setminus(S \setminus T),w,T)$ is "contained" in $(Q,w,T)$, you can as well send it to $(Q,w,T)$. I wanted to use the latter to emphasize that you can embedd everything into $(Q,w,\{\})$, and for an element in $(Q,w,\{\})$, you can recover all $S$ mapping onto this element. (Or am I overseeing some obvious problem?)
Nov 10, 2015 at 11:04 comment added Allen Knutson Should that map in #3 be to $(Q\setminus (S\setminus T), w, T)$?
Nov 10, 2015 at 10:10 history answered Christian Stump CC BY-SA 3.0