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Does there exist a prime 3-manifold such that its mapping class group has an abelian representation in which the 2$\pi$ rotation is represented by -1?

In detail: Let $M$ be a closed orientable prime 3-manifold. Let $D_F(M,p)$ be the group of diffeomorphisms of $M$ that fix a point $p$ of $M$ and a frame there. Define the mapping class group (MCG) of $M$ to be the zeroth homotopy group of $D_F(M,p)$. Then the $2\pi$ rotation is an element of MCG that is the equivalence class of the following diffeo, $R_{2\pi}$: Consider a coordinate ball of radius 2, $B2$, centred on $p$. $R_{2\pi}$ fixes the ball of radius 1, $B_1$, centred on $p$ and everything outside the sphere of radius 2. In between $B_1$ and $B_2$ the $R_{2\pi}$ maps $(x,y,z)\rightarrow(x\cos\theta+y\sin\theta,y\cos\theta−x\sin\theta,z)$ where $\theta$ is a function of $r=\sqrt(x^2+y^2+z^2)$ which increases smoothly and monotonically from 0 to 2$\pi$ as $r$ increases from 1 to 2. The square of the 2$\pi$ rotation is the identity in MCG. A manifold is spinorial if $\[R_{2\pi}\]$ is non-trivial in MCG.

Background motivation:

This question is interesting because of the possibility that fermions can be built on non-trivial spatial topology. $M$ is the manifold of a 3-D spatial hypersurface in spacetime. The fixed point is the point at infinity (where the metric is asymptotically flat) and fixing a frame there has the same effect on $\pi_0(D_F)$ as requiring some falloff conditions on the diffeomorphisms at infinity or requiring them to be the identity outside some ball. The configuration space of General Relativity in this asymptotically flat setting is (space of asymptotically flat metrics on $M$)/$D_F$ and its first homotopy group is isomorphic to (what I called above) MCG, see http://arxiv.org/abs/math-ph/0606066 (I know it is not the usual definition of MCG). The quantum state, on canonical quantisation of General Relativity, carries a unitary irreducible representation (UIR) of the MCG and different choices of UIR give different physics. Prime 3-manifolds are potentially candidates for elementary particles built from pure geometry: topological geons (this is speculative!). A prime 3-manifold can be the basis for a spinorial particle (i.e. spin 1/2, spin 3/2 ....) if $R_{2\pi}$ is nontrivial. Because particles must be able to be pair produced and annihilated, topology change must be allowed in the theory which means that the theory should be quantised in a sum-over-histories framework rather than a canonical quantisation framework. Within the sum-over-histories framework it is challenging to realise nonabelian reps of MCG. Abelian reps on the other hand are more easily accommodated by attaching phases to topologically distinct sectors of the path integral. Moreover certain rules that would result in a spin-statistics correlation for topological geons would also force the reps to be abelian, http://arxiv.org/abs/gr-qc/9609064 (hence the need for abelian reps). However, if there were no spinorial primes with abelian reps this would rule out spinorial geons and therefore fermions.

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What do you mean by 'the' $2\pi$ rotation? –  HJRW May 16 '11 at 12:21
    
Please edit the question statement to put in the clarifications. You're also presumably assuming that the manifold is spin, and looking at the spin mapping class group. –  Dylan Thurston May 16 '11 at 13:25
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1. This is not the usual definition of MCG. 2. By the belt trick, the $2\pi$ rotation squares to be the identity. 3. The $2\pi$ rotation is central. 4. I don't know what you mean when you say "abelian representation" or "represented by -1". –  Sam Nead May 16 '11 at 13:28
    
Even with the edit, "represented by -1" is not clear. Are you asking for this element of the mapping class group to map onto the generator of an infinite cyclic group? That's how I read your question. –  Ryan Budney May 16 '11 at 19:33
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1 Answer

up vote 8 down vote accepted

Yes, there exists such a manifold $M$. This follows if there exists aspherical $M$ with $Diff(M)\simeq 0$ (contractible) and $H^2(M;\mathbb{Z}/2\mathbb{Z})=0$. I claim there exists such manifolds. Let's see why such $M$ suffice.

There are two fibrations: $$ D_F(M,p) \to Diff(M,p) \to GL(3,\mathbb{R})$$ and $$ Diff(M,p) \to Diff(M) \to M.$$

The first comes from considering the derivative at $p$ of a diffeomorphism fixing $p$. The second comes from considering the image of $p$ under a diffeomorphism of $M$.

From the long exact sequence of homotopy for a fibration (e.g. 4.41 Hatcher), we have the exact sequence $$0=\pi_2(M) \to \pi_1(Diff(M,p)) \to \pi_1(Diff(M))=0 \to \pi_1(M)\to $$ $$\pi_0(Diff(M,p))\to \pi_0(Diff(M))=0.$$ We see that $\pi_0(Diff(M,p))=\pi_1(M)$, and $\pi_1(Diff(M,p))=0$. Similarly, $$0=\pi_1(Diff(M,p))\to \mathbb{Z}/2\mathbb{Z}=\pi_1(GL(3,\mathbb{R})) \to \pi_0(D_F(M,p))\to \pi_0(Diff(M,p))\to 0 .$$

From the second sequence, we see that $\pi_0(D_F(M,p))$ is therefore a $\mathbb{Z}/2\mathbb{Z}$ extension of $\pi_1(M)$. These are classified by $H^2(\pi_1(M);\mathbb{Z}/2\mathbb{Z})=H^2(M;\mathbb{Z}/2\mathbb{Z})=0$ (since any such extension is central). Thus, it is a trivial extension, so $\pi_0(D_F(M,p))=\mathbb{Z}/2\mathbb{Z}\times \pi_1(M)$. The $\mathbb{Z}/2\mathbb{Z}$ factor is generated by $[R_{2\pi}]$ in your notation, so clearly the desired abelian representation exists.

Now we need to see that a closed orientable aspherical manifold $M$ with $Diff(M)\simeq 0$ and $H^2(M;\mathbb{Z}/2)=0$ exists. In fact, we may assume $M$ is a hyperbolic homology sphere. For example, take a hyperbolic knot complement with trivial isometry group. Perform Dehn filling of slope $1/k$ for $k$ large to get a closed hyperbolic manifold. This manifold will be an aspherical homology sphere and will have trivial isometry group for $k>>0$ (this follows from Thurston's Dehn surgery theorem and the Margulis lemma, since the isometries must preserve the short core geodesic of the Dehn surgery). By a result of Gabai, $Diff(M)$ will be contractible.

One can work in somewhat greater generality with hyperbolic 3-manifolds $M$ which have the property that $H^2(Aut(\pi_1(M));\mathbb{Z}/2\mathbb{Z})=0$, since the first exact sequence implies that $\pi_0(Diff(M,p))=Aut(\pi_1(M))$. I'm not sure what happens if $H^2(Aut(\pi_1(M));\mathbb{Z}/2\mathbb{Z})\neq 0$; I assumed it $=0$ as a convenient way to see that the group $\pi_0(D_F(M,p))$ splits.

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Thank you for your answer. I'm still reading through it to try and understand it all but it looks like it's what I'm looking for. Do you mean that $\pi_0(D_F(M,p))$ (as opposed to $\pi_0(Diff(M,p))$) is a $\mathbb{Z}/2\mathbb{Z}$ extension of $\pi_1(M)$? –  Dionigi Benincasa May 17 '11 at 15:33
    
Thanks, I fixed it. –  Ian Agol May 17 '11 at 16:20
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