According to the following link, page 248, the braid group modulo its center is isomorphic to the mapping class group of the $N$-times punctured plane, i.e. $B_N/Z(B_N)\cong M_N(\mathcal(R)^2)$. Could anybody explain to me, how it is related to the mapping class group of the punctured $2$-sphere, please?


The key theorem is the following.

The braid group $B_n$ is isomorphic to the mapping class group $$ M_n(D^2) \;=\; \mathrm{MCG}\bigl(D^2 - \{p_1,\ldots,p_n\},\partial D^2\bigr) $$ of a disk with $n$ punctures rel the boundary.

Here "rel the boundary" means that all homeomorphisms and isotopies must fix the boundary of the disk pointwise.

The idea of this theorem is that a homeomorphism of the disk "moves" the $n$ punctures, and you can keep track of this motion using a braid.

More precisely, given a homeomorphism $h$ of $D^2 - \{p_1,\ldots,p_n\}$ that fixes $\partial D^2$ pointwise, we can extend $h$ to all of $D^2$ by filling in the punctures. Now, every homeomorphism of $D^2$ that fixes $\partial D^2$ pointwise is isotopic rel $\partial D^2$ to the identity. Therefore, there exists some isotopy $h_t\,$ rel $\partial D^2$ for which $h_0$ is the identity map on $D^2$ and $h_1=h$. Then each puncture $p_i$ follows a path $h_t(p_i)$ under this isotopy, and the union of the graphs $$ \bigcup_{i=1}^n \{(h_t(p_i),t) \mid t\in[0,1]\} $$ is a braid embedded in the solid cylinder $D^2 \times [0,1]$.

In more concrete terms, the mapping class group $M_n(D^2)$ is generated by half Dehn twists around pairs $\{p_i,p_{i+1}\}$ of punctures, and these correspond to half-twists of adjacent pairs of strands in the braid group.

In any case, the theorem about a punctured sphere is the following.

Let $M$ be the mapping class group of an $(n+1)$-punctured sphere. Let $p$ be one of the punctures, and let $M_p$ be the subgroup of $M$ consisting of mapping classes that fix $p$. Then $M_p$ is isomorphic to $B_n / Z(B_n)$, where $Z(B_n)$ is the center of $B_n$.

Before explaining this theorem, I should mention that the center $Z(B_n)$ of $B_n$ is infinite cyclic, and is generated by a full $360^\circ$ twist of the $n$ strands. As an element of $M_n(D^2)$, this corresponds to a Dehn twist around a curve that lies close to the boundary of the disk.

Now, there is an obvious homomorphism $M_n(D^2) \to M_p$, obtained by mapping the boundary of the disk to a small circle around the puncture $p$. Since $M_n(D^2)$ is isomorphic to $B_n$, this gives a homomorphism $B_n \to M_p$. Now, it should not be too hard to see that:

  1. This homomorphism is onto, and

  2. The center $Z(B_n)$ lies in the kernel of this homomorphism.

The explanation for (2) is that the generator for the center is mapping to a Dehn twist along a small circle surrounding $p$, which is clearly isotopic to the identity.

It should also make sense intuitively that $Z(B_n)$ would be the entire kernel of this homeomorphism, and thus $M_p$ is isomorphic to $B_n / Z(B_n)$.

  • $\begingroup$ Thank you Jim! Just one tiny additional question: The braid group that you use here is the one defined one the plane/disc, right? I ask, since there is also the spherical braid group and $M_N(S^2)\cong B_N(S^2)/\mathbb(Z)_2$. $\endgroup$ – Hamurabi Dec 27 '13 at 10:11
  • $\begingroup$ Right. I'm using $B_n$ to mean the usual braid group on $n$ strands, i.e. the fundamental group of the configuration space of $n$ unlabeled points in the plane. $\endgroup$ – Jim Belk Dec 27 '13 at 17:04
  • $\begingroup$ Jim, could you also comment on the relation of all that to the mcg of the plane? How do I use Birman's exact sequence here? Thank you! $\endgroup$ – Hamurabi Dec 29 '13 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.