Timeline for Counting chains of inclusions
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 1, 2014 at 12:30 | comment | added | darij grinberg | @MikeShulman: done so. | |
| Jun 1, 2014 at 12:30 | answer | added | darij grinberg | timeline score: 4 | |
| Jun 1, 2014 at 4:16 | vote | accept | Mike Shulman | ||
| Jun 1, 2014 at 4:14 | vote | accept | Mike Shulman | ||
| Jun 1, 2014 at 4:16 | |||||
| May 31, 2014 at 23:35 | comment | added | Mike Shulman | @KarolSzumiło, you're probably right. | |
| May 31, 2014 at 22:18 | comment | added | Mike Shulman | @darijgrinberg, that looks like an answer; why don't you post it as one? | |
| May 30, 2014 at 19:25 | answer | added | Ira Gessel | timeline score: 7 | |
| May 30, 2014 at 18:02 | answer | added | Martin Tancer | timeline score: 6 | |
| May 30, 2014 at 17:59 | comment | added | darij grinberg | (The $k = 0$ addend, of course, is only relevant for $n = 0$. It is more important that your sign is different because of the $k$ shift.) The Möbius function of the Boolean lattice is easily computed, as it is multiplicative and the Boolean lattice is a product of $2$-chains. | |
| May 30, 2014 at 17:55 | comment | added | darij grinberg | Hall's formula for the Möbius function of a finite bounded ranked poset says that any such poset $P$ satisfies $\mu\left(P\right) = \sum_{k\geq 0} \left(-1\right)^k \left(\text{number of chains } 0 = x_0 < x_1 < \cdots < x_k = 1 \text{ in } P\right)$, where $0$ and $1$ denote the lower bound and the upper bound of $P$. In your case, the poset is the Boolean lattice, but notice that your $g\left(n,k\right)$ counts chains $0 = x_0 < x_1 < \cdots < x_{k+1} = 1$ rather than $0 = x_0 < x_1 < \cdots < x_k = 1$ so you are skipping the $k = 0$ addend of Hall's formula. | |
| May 30, 2014 at 17:43 | comment | added | Karol Szumiło | This not an answer to the question as you posed it, but I suspect that your "category-theoretic argument involving traces of geometric realizations in derivators" reduces to a (probably) simpler homotopy theoretic proof. Namely, $g(n,k)$ is the number of $(k-1)$-simplices in the barycentric subdivision of the boundary of the standard $(n-1)$-simplex. Hence your sum (up to the $k=0$ summand and up to sign) computes the Euler characteristic of the $(n-2)$-sphere. | |
| May 30, 2014 at 17:06 | history | asked | Mike Shulman | CC BY-SA 3.0 |