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For a manifold $M$, define the mapping class group $Mod(M)$ to be the set of self-diffeomorphisms of $M$, modulo isotopy. In symbols, $Mod(M) = \pi_0 Diff(M)$. Of course, every self-diffeomorphism gives an automorphism of $\pi_1(M)$, well-defined up to conjugacy because of basepoint issues. Thus we have a homomorphism $\sigma: Mod(M) \to Out ( \pi_1 M)$.

When $M$ is a surface, the Dehn-Nielsen-Baer theorem says $\sigma: Mod(M) \to Out ( \pi_1 M)$ is an isomorphism. My question is: what can be said in higher dimensions?

Assuming $M$ is a $K(\pi, 1)$, one can identify $Out ( \pi_1 M)$ with the set of homotopy classes of homotopy equivalences of $M$. Through this lens, injectivity of $\sigma$ is the question of whether two homotopic diffeomorphisms need to be isotopic. Surjectivity of $\sigma$ is the question of whether every homotopy equivalence is homotopic to a self-diffeomorphism.

From a naive point of view, both injectivity and surjectivity seem hard. What is known about them?

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see my answer to this question: mathoverflow.net/questions/66459/… –  Ian Agol May 30 '11 at 19:28
    
Thanks, Ian. Both the references you mention (Gabai and Farrell-Jones) are for hyperbolic manifolds. Looking at citations of Farrell-Jones gives me the impression that very little is known beyond the hyperbolic world. Is that accurate? –  Dave Futer May 30 '11 at 19:44

2 Answers 2

up vote 3 down vote accepted

In general it's not injective, nor surjective. It's just a map.

For example, if $M = S^n$, the sphere. $\pi_0 Diff(M)$ is the group of homotopy $(n+1)$-spheres, but $Out(\pi_1 M)$ is trivial. So you have surjectivity but not injectivity provided exotic spheres exist in that dimension. There's lots of much more complicated examples of this type, for example $M = S^1 \times D^n$, this goes back to work of Farrell, Hatcher, Quillen, Igusa.

I think you mean to formulate the question a little differently since manifolds can have orientiation-reversing diffeomorphisms and you rarely "capture" that by $Out(\pi_1)$. For example, $\pi_0 Diff(S^2) \simeq \mathbb Z_2$ yet $Out(\pi_1 S^2)$ is trivial.

But similarly, $\pi_0 Diff(S^2 \times S^1) \simeq \mathbb Z_2^2$ and $Out(\pi_1 S^2 \times S^1)$ is $\mathbb Z_2$. But even in the orientation-preserving case, there's no isomorphism.

edit: okay if you're specifically interested in $K(\pi,1)$ spaces, there's the computation of Hatcher -- $\pi_0 Diff( (S^1)^n)$ is an extension of $GL_n \mathbb Z$ by:

$$ \mathbb Z_2^\infty\oplus\binom n2\mathbb Z_2\oplus\sum_{i=0}^n\binom n i\Gamma_{i+1} $$

That's an infinite direct-sum of $\mathbb Z_2$'s together with a direct sum of many groups of exotic spheres. See the Wikipedia page.

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Thanks, Ryan. Of course, for spheres (or other simply connected examples), $Out(\pi_1)$ is going to be trivial. That's why I was trying to look at $K(\pi, 1)$'s... –  Dave Futer May 30 '11 at 19:51
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FYI: I suspect that infinite direct sum of $\mathbb Z_2$'s has a very nice geometric interpretation. That's one of my current research projects. –  Ryan Budney May 30 '11 at 20:32

For $M$ oriented closed hyperbolic, I guess the surjectivity follow from Mostow's rigidity. In fact, $Out(\pi_1(M))\cong Isom(M)$ so the map $\sigma$ should be a projection $Mod^{\pm}(M)\to Isom(M)$ in disguise.

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