General relationship between braid groups and mapping class groups

I just finished correcting my answer on visualizing braid groups as fundamental groups of configuration spaces, and in the process became interested in the other pictorial definition of the braid group $B_n$, namely as the mapping class group of the $n$-punctured closed unit disk.

Some definitions. If $X$ is a topological space, let $F_n(X)$ be the subspace of $X^n$ consisting of tuples with distinct coordinates, on which the symmetric group $S_n$ acts by permuting coordinates, and then define the quotient $SF_n(X):=F_n(X)/S_n$. The braid group of $X$ is defined to be the fundamental group $B_n(X):=\pi_1(SF_n(X))$. Note $B_n=B_n(\Bbb C)$ is the usual braid group.

The automorphism group ${\rm Aut}(X)$ is the group of homeomorphisms $X\to X$ which fix its boundary $\partial X$ pointwise. Denote by ${\rm Aut}_0(X)$ those automorphisms which are isotopic to the identity map. The mapping class group is ${\rm Mod}(X):={\rm Aut}(X)/{\rm Aut}_0(X)$.

Denote by $\overline{{\Bbb D}^2}$ the closed unit disk. Then $B_n(\Bbb C)\cong {\rm Mod}(\overline{\Bbb D^2}-\{x_1,\cdots,x_n\})$ for any choice of $n$ distinct points. It seems obvious to me that $B_n(\Bbb C)=B_n(\overline{\Bbb D^2})$, which inspires the question:

• For what kinds of surfaces/manifolds does $B_n({\cal S})={\rm Mod}({\cal S}-\{x_1,\cdots,x_n\})$?

And this leads to the larger question:

• What is the general relationship between braid groups and mapping class groups?

Sorry if these facts are well-known somewhere. (I am pretty new to homotopy theory and algebraic topology in general too, so I might be a bit slow. It's possible I am biting off more than I am supposed to be chewing.) Here is my likely invalid argument:

Claim. ${\rm Mod}({\cal S}-\{x_1,\cdots,x_n\})=B_n({\cal S})$ for "nice" spaces $\cal S$.

"Proof". Assume ${\rm Aut}(S)$ acts transitively on $n$-subsets of ${\cal S}$. (Intuitively it feels like this should follow automatically from $\cal S$ being homogeneous, i.e. ${\rm Aut}({\cal S})$ only acting transitively on $\cal S$ itself, by cordoning off a nbhd around any $n$ points, but I haven't tried proving it.) We should be able to identify ${\rm Aut}({\cal S}-\{x_1,\cdots,x_n\})$ with ${\rm Stab}_{{\rm Aut}({\cal S})}(\{x_1,\cdots,x_n\})$ the setwise (not pointwise) stabilizer. By the orbit-stabilizer theorem, we have

$${\rm Aut}(S)/{\rm Stab}_{{\rm Aut}(\cal S)}(\{x_1,\cdots,x_n\})\cong SF_n({\cal S})$$

So to get $B_n({\cal S})$ we apply $\pi_1$ to the left side. Here I invoke a lemma:

Lemma. If $G$ is connected, simply connected, $H$ a subgroup, and $H^\circ$ the connected component of the identity in $H$, then $\pi_1(G/H)\cong H/H^\circ$ via $[\gamma]\mapsto\gamma(1)H^\circ$.

If we apply with $G={\rm Aut}({\cal S})$ and $H={\rm Stab}_{{\rm Aut}({\cal S})}(\{x_1,\cdots,x_n\})$ then $H/H^\circ$ should be the identified group ${\rm Aut}({\cal S}-\{x_1,\cdots,x_n\})$ modulo isotopy, i.e. ${\rm Mod}({\cal S}-\{x_1,\cdots,x_n\})$, no?

My argument must have issues, because the claim doesn't work for ${\cal S}=\Bbb C$: as I understand it, we instead have that $\Bbb C\cong\Bbb S^2-\{\rm pt\}$ and so ${\rm Mod}({\Bbb C}-\{x_1,\cdots,x_n\})$ is a subgroup of $B_{n+1}(\Bbb S^2)$ with elements having a marked string always trivial, and this seems like a different group.

Let $\mathcal S$ be a compact surface, possibly with boundary. Let $\text{Homeo}^+(\mathcal S)$ refer to the group of orientation-preserving homeomorphisms that fix the boundary pointwise with the compact-open topology. Throwing an $n$ in there means we add $n$ marked points in the interior, which I prefer over deleting points. (I'd be worried about the validity of any particular version of this result if we delete points instead of marking them.) I don't know the correct topology on $\text{Homeo}^+$ if you use noncompact surfaces.

Then the correct general statement of your dream is that $$\text{Homeo}^+(\mathcal S,n) \to \text{Homeo}^+(\mathcal S) \to SF_n(\mathcal S)$$ is a fiber bundle. The tools you need to analyze this are the homotopy long exact sequence, the Earle-Eels theorem (you can find a statement and proof in Appendix B here, and that $\text{Homeo}^+(D^2)$ is contractible.

Some special cases:

When $\mathcal S = D^2$, you immediately obtain that $B_n = \text{MCG}(\mathcal S,n)$.

Ignore the issues with topologizing $\text{Homeo}^+(\mathbb R^2)$. The trick with your proof is that this is not actually simply connected! $\text{Homeo}^+(\mathbb R^2)$ should be the same as $\text{Homeo}^+(S^2,1)$ (send every homeomorphism to its compactification). This is homotopy equivalent to $SO(2) = S^1$.

When $\mathcal S = \Sigma_{g,k}$, a genus $g$ surface with $k$ boundary components, and either $g \geq 2$ or $k \geq 1$, applying Earle-Eels you obtain the exact sequence $$1 \to B_n(\mathcal S) \to \text{MCG}(\mathcal S,n) \to \text{MCG}(\mathcal S) \to 1.$$

When $\mathcal S = S^2$, using Smale's theorem that $\text{Homeo}^+(S^2) \simeq SO(3)$, and using Earle-Eels you can prove that the components of $\text{Homeo}^+(S^2,n)$ are contractible, so you get a short exact sequence $$1 \to \mathbb Z/2 \to B_n(S^2) \to \text{MCG}(S^2,n) \to 1.$$

When $\mathcal S = T^2$, $\text{Homeo}^+(T^2) \simeq SL_2(\mathbb Z) \times T^2$. If you can get some control on $\text{Homeo}^+(T^2,n)$ you should get something interesting here, but a couple mindless attempts didn't work. (Idea: work with Diff instead so that at each marked point you can 'pull apart' the marked points and obtain diffeomorphisms of $T^2$ minus some open discs without control on the way the diffeomorphisms behave on the boundary. This space might be assailable with EE.)

The fiber sequence above generalizes perfectly well to higher-dimensional manifolds. You might enjoy playing with it for 3-manifolds, when $\text{Homeo}^+(M)$ is known in many examples. $S^2 \times S^1$ might be fun.

You would probably enjoy Farb's primer on mapping class groups. The fiber sequence above just comes from modifying his proof of Birman's exact sequence to having $n$ points instead of one.

• Whenever you've got $H \to G \to G/H$, my favorite way to check that it's a fiber bundle is that there's a local cross-section $G/H \to G$ at every point (this automatically gives you a local trivialization!) I think this trick is due to Palais. See Theorem A of "Local Triviality of the Restriction Map for Embeddings". Use this to prove the above thing is a fiber bundle. – user98602 Jul 3 '15 at 0:25
• Ah... is rotating the plane $360$ degrees not a contractible loop in ${\rm Homeo}^+(\Bbb C)$? If so, I see why that wouldn't rear its head for the closed unit disk. Thanks for the answer, and glad to know that my reasoning was (mostly) correct. BTW where do you know of Homeo groups of spheres, planes and tori from? (And does $\simeq$ mean homotopy equivalence?) I indeed have Farb checked out from the library, but just reading the first chapter I realized I should study the classification of surfaces and more hyperbolic geometry first, so that's what I've been doing a little of this summer. – anon Jul 3 '15 at 14:18
• Anyway, why is ${\rm Homeo}^+({\cal S})$ being a ${\rm Homeo}^+({\cal S},n)$-bundle on $SF_n({\cal S})$ the correct general version of my dream? I haven't been able to mentally picture such a bundle, and I've never handled bundles before. [My intuition was that paths in configuration space "come from" paths in ${\rm Homeo}^+(\cal S)$ starting $\rm id$ that braid the $n$ points over time, hence have an endpoint in ${\rm Homeo}^+({\cal S},n)$, and when I noticed ${\rm Homeo}^+({\cal S},n)$ was a stabilizer my algebraic instinct to use orbit-stabilizer kicked in.] – anon Jul 3 '15 at 14:18
• 1) That's correct. Of course, if we didn't fix the boundary of the disc pointwise, we're still in trouble because the same rotation works again. The homeomorphism group of $S^2$ is Smale's theorem; I don't know who the torus one is attributed to; the plane you get by fiddling with Smale's theorem a bit. I didn't learn these from a book, just from random findings and from conversations with people. – user98602 Jul 3 '15 at 14:33
• 2) The point is more that the long exact sequence it gives you is the precise version of the relationship you're trying to find between $B_n(\mathcal S)$ and $\text{MCG}(\mathcal S,n)$, which isn't quite so simple as you'd hoped. It also tells you about the higher homotopy of $SF_n(\mathcal S)$; if you can prove some sort of Earle-Eeles theorem for marked-point homeomorphism groups, you should be able to show its higher homotopy groups vanish (for a wide class of surfaces $\Sigma$). – user98602 Jul 3 '15 at 14:36