Timeline for How many maximal length Bruhat paths from $u$ to $w$ can there be?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2018 at 1:28 | vote | accept | Matt Samuel | ||
| Nov 6, 2018 at 11:46 | answer | added | Nathan Reading | timeline score: 3 | |
| Nov 4, 2018 at 0:41 | comment | added | Matt Samuel | @Richard Thank you, I'll add that to my reading list. | |
| Nov 4, 2018 at 0:36 | comment | added | Richard Stanley | A kind of sequel to Stembridge's paper is my paper with Alex Postnikov at math.mit.edu/~rstan/papers/degschub.pdf. | |
| Nov 3, 2018 at 21:08 | comment | added | Matt Samuel | @Timothy It doesn't, but it's very closely related to what I'm working on, and it's definitely something I should familiarize myself with. Thanks. | |
| Nov 3, 2018 at 21:01 | comment | added | Timothy Chow | This may not answer your question, but there may be some useful information in John Stembridge's paper, "A weighted enumeration of maximal chains in the Bruhat order." | |
| Nov 3, 2018 at 20:58 | comment | added | Matt Samuel | @Sam The strong order. I'm aware the situation is much easier in weak order. | |
| Nov 3, 2018 at 20:56 | comment | added | Sam Hopkins | Is this the weak order or strong order? For weak order, the poset $[u,w]$ is isomorphic to $[e,wu^{-1}]$ (depending on convention I may be multiplying on wrong side), so any bounds for the $\ell(u)=0$ case carry over to general $u$ case. For strong order I guess the situation is different. | |
| Nov 3, 2018 at 20:37 | comment | added | Matt Samuel | @dhy I think I see what you mean. There are at most that many paths to all elements of length $\ell(u) $. | |
| Nov 3, 2018 at 20:18 | comment | added | dhy | I think the bound still applies - your objection implies that a stronger argument for a $(\ell(w)-\ell(u))!$ bound fails, but my bound only needs to fix an ordering for $w$ and not a distinguished subword corresponding to $u$. | |
| Nov 3, 2018 at 20:10 | comment | added | Matt Samuel | @dhy I don't think so, because there may be more than one subword for $u$. You would have to do something like multiply by the number of distinct subwords. | |
| Nov 3, 2018 at 20:09 | comment | added | dhy | A quick note: The same argument that you use in the $\ell(u)=0$ case gives a bound of $\ell(w)(\ell(w)-1)\cdots(\ell(u)+1)$ in general. | |
| Nov 3, 2018 at 19:36 | history | edited | Matt Samuel | CC BY-SA 4.0 |
added 178 characters in body; added 1 character in body
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| Nov 3, 2018 at 18:40 | history | asked | Matt Samuel | CC BY-SA 4.0 |