Markovich and Saric proved the following remarkable theorem. Let $S$ be a compact surface of genus at least $2$ and let $MCG(S)=\pi_0(Homeo^{+}(S))$ be the mapping class group of $S$. There is then no right inverse to the natural projection $Homeo^{+}(S) \rightarrow MCG(S)$. This should be contrasted with the solution to the Nielsen realization problem by Kerckhoff, who proved that any finite subgroup of $MCG(S)$ can be realized by homeomorphisms of the surface.

My question is whether this is known for braid groups. Let me make this a little more precise. Let $X_n$ be a $2$-dimensional disc with $n$ punctures. By $Homeo^{+}(X_n)$, we mean homeomorphisms of $X_n$ that are the identity on the boundary and extend over the punctures (but are allowed to permute the punctures). The group $\pi_0(Homeo^{+}(X_n))$ is then the $n$-strand braid group $B_n$. My question is if it is known whether or not there is a right inverse to the natural projection $Homeo^{+}(X_n) \rightarrow B_n$.

Since $B_1=1$ and $B_2=\mathbb{Z}$, the first nontrivial case is $n=3$.

I would find it shocking if this is not known for $n=3$. The braid group then has two generators $a$ and $b$ and only one relation $aba=bab$. The problem is then asking whether or not we can find homeomorphisms $f,g \in Homeo^{+}(X_n)$ in the homotopy classes of $a$ and $b$ such that $fgf=gfg$. Surely this cannot be open!

  • $\begingroup$ The statement in your last paragraph is technically incorrect. For example, let $f=g$ then $fgf=gfg$ since they're both equal to $f^3$. The issue is finding homeomorphisms in the right homotopy class satisfying $fgf=gfg$. I believe it's an open problem for braid groups but I also think Markovic thinks his techniques should adapt to this situation without too much fighting -- as far as I know this is an interesting problem that hasn't been seriously attacked yet. $\endgroup$ – Ryan Budney Feb 15 '11 at 21:47
  • 1
    $\begingroup$ Ryan - Notice the phrase "in the homotopy classes of $a$ and $b$" in the last paragraph =). $\endgroup$ – Scott P Feb 15 '11 at 21:48
  • $\begingroup$ Ah, whoops. Missed that. FYI, there is a technique where one can realize the mapping class group of a hyperbolic surface as a subgroup of the group of homeomorphisms of the unit tangent bundle. I've seen this called "the Cheeger embedding" in a paper of Gromov's. The idea is to relate the unit tangent bundle to the configuration space of 3 points on the visual sphere/circle. $\endgroup$ – Ryan Budney Feb 15 '11 at 22:04
  • 1
    $\begingroup$ The cases with a complex 1-dimensional Teichmüller space, such as the 4-punctured sphere (i.e., 3 strand braid group), are easy to solve: these surface has a canonical Euclidean structure such that all homemorphisms are represented by affine maps, unique except in the case of $T^2$ where you can modify by translations. For the 3-strand braid group, you can use a regular tetrahedron for a model. Any braid gives a locally affine map of the tetrahedron to itself; this gives a lifting of the mapping class group to a group of homeomorphisms. $\endgroup$ – Bill Thurston Feb 15 '11 at 22:50
  • 1
    $\begingroup$ Note: for the 3 strand braid group, these generators are a: flip the three strands over, right over left (usually called Delta) b: take the rightmost strand and bring it to the left over the other two. It's easy to see that a^2 = b^3, and that by combining them, you can get any braid. In the hyperbolic plane, they correspond to rotating around the center of an edge and the center of a triangle in the regular ideal triangulation. $\endgroup$ – Bill Thurston Feb 16 '11 at 1:57

The problem is much easier, but still interesting and nontrivial, if you replace homeomorphisms by diffeomorphisms, so it might be worth to study this case. Logically, because any action by diffeomorphisms is an action by homeomorphisms and if no action by homeomorphisms exists, then certainly there cannot be any action by diffeomorphisms. But that is not the point.

The point is that one can use cohomology theory (which is impossible in the homeomorphism case, see below). The case of closed surfaces and diffeomorphisms was solved by Morita around 1985, and here is a punchline.

Consider a closed oriented $n$-manifold $M$ and the orientation-preserving diffeomorphism group $Diff(M)$. Let $Diff(M)^{\delta}$ be the diffeomorphism group with the discrete topology. There is an induced map $B Diff(M)^{\delta} \to B Diff(M)$, inducing a map in rational cohomology. We now describe cohomology classes that automatically lie in the kernel. Let $f:A Diff(M) \to B Diff(M)$ be the universal $M$-bundle. It has a vertical tangent bundle $T \to A Diff(M) $ of the universal $M$-bundle. It has Pontrjagin classes $p_i \in H^{4i}$, and with the help of the Gysin map, you produce classes $f_! (q(p_1,p_2,..))$ for each polynomial in the Pontrjagin classes (let us call them Mumford-Morita-Miller or MMM-classes).

Now let $X$ be a smooth manifold and $X \to B Diff (M)^{\delta}$ a map, classifying a manifold bundle $E \to X$. Since the structural group is discrete, there would be a foliation of $E$ by leaves that are transverse to the fibres.The normal bundle of the foliation is the vertical tangent bundle.

But there is a theorem of Bott. He showed that the Pontrjagin classes of the normal bundle of a foliation are zero in a range of degrees (I forgot the range, but it is sufficient to guarantee that in the case of surfaces $p_{1}^{k}$ is zero for $k \geq 2$). This relies on Chern-Weil theory. Upshot of the discussion: in a range of degrees, the MMM-classes map to zero in $H^* (B Diff(M)^{\delta})$.

Now assume that there is a section $s: MCG(M) \to Diff(M)$. It would, clearly, factor through $Diff(M)^{\delta}$. In particular, a portion of the MMM-classes map to zero under the section.

In the surface case, there is a theorem by Earle and Eells, saying that the components of $Diff$ are contractible. Consequence: the induced map $Bs: B MCG (M) \to B Diff (M)$ is a homotopy equivalence, but it maps many MMM-classes to zero. This is a contradiction once you know that enough MMM classes are nonzero. This is precisely what Morita did.

The question for homeomorphisms cannot be attacked using these methods, because of the following, really awesome, theorem by Dusa Mc Duff (I think, first stated by Bill Thurston, and big portions of the proof were done by Mather and Segal). It says that for each (maybe PL is necessary) manifold $M$, the natural map $B Homeo(M)^{\delta} \to B Homeo (M)$ is a homology equivalence!! So you will never get a contradiction in that way.

Now if you wish to study the Braid group case (and diffeomorphisms), then some of these things will work (and may be easier): The components off the corresponding diffeomorphism groups are contractible, the (rational) cohomology of the braid groups is explicitly known; and it should be relatively easy to figure out the Pontrjagin classes of the vertical tangent bundle. Once some of them are nontrivial, you are done.

  • 5
    $\begingroup$ It's also true that for G = biLipschitz homeomorphisms, $BG_\delta \rightarrow BG$ induces an isomorphism in homology --- this was the form I first used to show that every subbundle of the tangent bundle of a smooth manifold is homotopic to the tangent bundle of a foliation with $C^\infty$-smooth leaves but only biLipshcitz holonomy. For every surface bundle over a CW complex, there is another CW complex mapping to it and inducing isomorphism in homology with arbitrary coefficients (can be a module over pi_1(range) so the pullback bundle has a flat, bilipschitz connection. Almost a lifting. $\endgroup$ – Bill Thurston Feb 16 '11 at 2:13

Hot off the presses: a new preprint by Nick Salter and Bena Tshishiku proves that the braid groups cannot be realized by diffeomorphisms for $n \geq 5$.


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.