Skip to main content
10 events
when toggle format what by license comment
Oct 21, 2018 at 14:10 answer added John Shareshian timeline score: 3
Oct 21, 2018 at 13:14 comment added YCor A related paper (from a deleted answer) is The homology of "k-equal'' manifolds and related partition lattices. Adv. Math. 110 (1995), no. 2, 277–313. sciencedirect.com/science/article/pii/S0001870885710122 It computes the homology of $M_{n,k}\subset\mathbf{R}^n$, defined by the condition that no $k$-uple of coordinates is constant, so it's a bit distinct, but of the same flavor.
Oct 21, 2018 at 12:00 comment added YCor I guess $A(n,n-1)$ deformation retracts onto a graph but I don't have a proof now.
Oct 21, 2018 at 11:33 comment added YCor $A(n,3)$ is homeomorphic to the product of $\mathbf{R}$ with ($\mathbf{R}^{n-1}$ minus $2^{n-1}-1$ lines through zero), which itself is homeomorphic to the product of $\mathbf{R}^2$ with ($\mathbf{S}^{n-2}$ minus $2^{n}-2$ points); $\mathbf{S}^{n-2}$ minus $2^{n}-2$ points is homeomorphic to $\mathbf{R}^{n-2}$ minus $2^{n}-3$ points for $n\ge 2$. The homology is not hard to compute: for $n\ge 3$ it's the homology of a bunch of $2^n-3$ $(n-3)$-spheres.
Oct 21, 2018 at 10:59 comment added YCor $A(n,1)=\mathbf{R}^n$. $A(n,n)$ is homeomorphic to $\mathbf{R}^n\times\{1,\dots,n!\}$. As you say, $A(n,2)$ is homeomorphic to $\mathbf{R}^2\times\mathbf{S}^{n-2}$.
Oct 21, 2018 at 10:53 history edited YCor CC BY-SA 4.0
deleted 7 characters in body; edited tags
Oct 21, 2018 at 10:33 history edited user118503 CC BY-SA 4.0
deleted 35 characters in body; edited title
Oct 21, 2018 at 10:23 comment added user118503 @YCor: Yes, indeed. Maybe it's better to ask a new question
Oct 21, 2018 at 10:12 comment added user118503 @YCor: Thanks for the comment. I realized that I was thinking of tuples, but asking about finite sets. I will edit the question.
Oct 21, 2018 at 9:55 history asked user118503 CC BY-SA 4.0