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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
Jan 24, 2020 at 14:28 vote accept Meths
Jan 24, 2020 at 1:37 answer added Ty Ghaswala timeline score: 3
Jan 23, 2020 at 15:20 comment added Meths @MarkSapir That's unfortunate - the description I gave was intended to match that of the Definition given in Section 2.2 of the linked paper. I believe $\text{Mot}_{n}(M)$ is then $\pi_{1}(C_{n}(M)/S_{n})$ (which the above discussion helped me understand). According to the paper, this is indeed $B_{n}(M)$. I guess that just leaves question 2(ii).
Jan 23, 2020 at 15:10 comment added user6976 After the clarification, the question has become less clear.
Jan 23, 2020 at 14:42 comment added Meths Ah, I see. That makes sense if $T$ is an unordered set (which is implicit in your description). So these $n$ curves in $M$ just look like one loop in $C_{n}(M)$. If $T$ was ordered, then you'd only get a loop for the trivial permutation? I guess there are some subtleties about $M$ being path-connected meaning $C_{n}(M)$ being path-connected too (for $T$ unordered), allowing us to forget about the basepoint.
Jan 23, 2020 at 14:07 history edited Meths CC BY-SA 4.0
Clarification 2
Jan 23, 2020 at 14:03 comment added John Klein Isn't your description just an equivalence class of path $\gamma: [0,1] \to C_n(M)$ such that $\gamma(0) = T = \gamma(1)$, where $C_n(M)$ is the space of subsets of cardinality $n = |T|$?
Jan 23, 2020 at 14:00 comment added John Klein What you wrote in your post is still not clear: you have a curve in $M$ and you also have one in $M\times [0,1]$. Please commit and make it precise.
Jan 23, 2020 at 13:58 comment added Meths *I would identify $π_1(C_n(M))$ with the homotopy equivalence classes of curves that fix $\gamma_i(0)=\gamma_i(1)$ in my description.
Jan 23, 2020 at 13:39 comment added Meths I've clarified what I meant in the post now. @JohnKlein I don't have a clear understanding of why $\pi_{1}(C_{n}(M))$ doesn't restrict to the trivial permutation?
Jan 23, 2020 at 13:27 history edited Meths CC BY-SA 4.0
Clarifying "homotopy classes of paths that permute these points".
Jan 23, 2020 at 10:04 comment added user6976 @Lee: The OP probably meant a path in the configuration space of $n$-tuples of points. Of course he needs to make ir clear.
Jan 23, 2020 at 7:40 review Close votes
Jan 31, 2020 at 3:05
Jan 23, 2020 at 3:48 comment added Lee Mosher Without further care in formulating definitions, I do not know what it might mean for a path to permute points. I presume by a "path" you mean a continuous function $f : [0,1] \to M$, in which case "permuting points" is not something a path ordinarily does.
Jan 23, 2020 at 3:21 history rollback user6976
Rollback to Revision 1
Jan 23, 2020 at 2:15 comment added John Klein Questions about the definition. Let $T\subset M$ be a finite subset of cardinality $n$.This gives a basepoint for the configuration space of $n$ unordered points in $M$. Call this configuration space $C_n(M)$. Then it seems to me what you are describing is $\pi_0$ of the loop space $\Omega C_n(M)$. Is that right? If so, then your group is nothing more than $\pi_1(C_n(M))$. Right?
Jan 23, 2020 at 1:10 history edited YCor
edited tags
Jan 23, 2020 at 1:03 history asked Meths CC BY-SA 4.0