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Given $n$ points on a connected $2$-manifold $M$, I'd like to consider the homotopy classes of paths that "permute" these points.

Edit (Clarifying what I mean by this):

Given a set of $n$ distinct points $T=\{x_{1},\ldots,x_{n}\}\subset M$, to each point we assign a continuous simple curve $\gamma_{i}:[0,1]\to M$ such that $\gamma_{i}(0)=x_{i}, \gamma_{i}(1) \in T$ and $\gamma_{i}(s)\neq\gamma_{j}(s)$ for all $s\in[0,1]$ (where $i\neq j$). I'd like to consider the homotopy classes of all such possible curves.

  1. It seems obvious that these homotopy classes should constitute the elements of a group. Is that right? If so, what's the name of this group? I'm inclined to simply call this the motion group $\text{Mot}_{n}(M)$ of the $n$ points on $M$. Does this coincide with the mapping class group of $n$ points in $M$? Also, do I need any more restrictions on $M$? If so, why?

  • E.g. considering $3$-space for a moment, it is obvious that $\text{Mot}_{n}(\mathbb{R}^{3})\cong S_{n}$ (where $S_{n}$ is the permutation group).

  • It is also obvious that $\text{Mot}_{n}(\mathbb{D}_{2})\cong \text{Mot}_{n}(\mathbb{R}^{2})\cong B_{n}$, where $\mathbb{D}_{2}$ is the $2$-disk with boundary and $B_{n}$ is the braid group.

  1. Consider a presentation of $\text{Mot}_{n}(M)$ with relations $R$.
    (i) Is it true that $\text{Mot}_{n}(M)\cong B_{n}(M)$, where $B_{n}(M)$ is the surface braid group for $M$?
    (ii) Under what conditions will it be true that the generator relations $G$ of $B_{n}$ will be a subset of $R$?

For instance, I'm sure that $\text{Mot}_{n}(S^{2})\cong B_{n}(S^{2})$ : in which case, we do have $G\subset R$ (in fact, $B_{n}(S^{2})$ is a quotient of $B_{n}$).

Relevant Resources:

A survey of surface braid groups and the lower algebraic K-theory of their group rings

I think my notion of $\text{Mot}_{n}(M)$ coincides with the Definition in Section 2.2 of the above paper. If so, then the answer to 2(i) is yes (according to the paper). Building on this, I believe Theorems 12 and 13 of Bellingeri (for $m=0$) may provide a partial answer to 2(ii).

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    $\begingroup$ 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? $\endgroup$
    – John Klein
    Commented Jan 23, 2020 at 2:15
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    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Commented Jan 23, 2020 at 3:48
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    $\begingroup$ @Lee: The OP probably meant a path in the configuration space of $n$-tuples of points. Of course he needs to make ir clear. $\endgroup$
    – user6976
    Commented Jan 23, 2020 at 10:04
  • $\begingroup$ 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? $\endgroup$
    – Meths
    Commented Jan 23, 2020 at 13:39
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    $\begingroup$ 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 $n = |T|$? $\endgroup$
    – John Klein
    Commented Jan 23, 2020 at 14:03

1 Answer 1

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The comments seem to have answered questions 1 and 2i, to show that the group $\operatorname{Mot}_n(M)$ is indeed the surface braid group $B_n(M)$.

To answer 2ii, consider a disk $D \subset M$ such that $T \subset D$. Then the inclusion map $D \hookrightarrow M$ induces a homomorphism $B_n \to B_n(M)$, so the relations in $B_n$ always hold in $B_n(M)$.

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  • $\begingroup$ Simple enough ! $\endgroup$
    – Meths
    Commented Jan 24, 2020 at 14:17
  • $\begingroup$ Bump... quick question: for your (induced homomorphism) argument to be true, doesn't this also mean there has to be a continuous map from the (unordered) configuration space of $D$ to that of $D\subset M$? That doesn't seem immediately obvious to me. Sorry for the posthumous enquiry! $\endgroup$
    – Meths
    Commented Feb 22, 2020 at 11:45
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    $\begingroup$ It's not immediately obvious, but It does induce a continuous map between unordered configuration spaces. One way to see this is to first check it on the ordered configuration spaces, and then make sure the induced inclusion respects the quotient maps. However, it's clearer if you use the viewpoint you presented in your original question, of looking at elements of the surface braid group as a collection of paths on $D$ or $M$. Then a collection of paths in $D$ is clearly a collection of paths in $M$ once you include $D$ into $M$. $\endgroup$
    – Ty Ghaswala
    Commented Feb 24, 2020 at 3:56
  • $\begingroup$ Got it, thanks for your help! $\endgroup$
    – Meths
    Commented Feb 24, 2020 at 5:53

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