Skip to main content
2 of 6
edited tags
YCor
  • 63.9k
  • 5
  • 187
  • 286

Permuting $n$ points in a $2$-manifold

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

  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][1]

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). [1]: https://arxiv.org/abs/1302.6536

Meths
  • 309
  • 1
  • 11