# What is the topological/smooth mapping class group of an n-dimensional torus?

The page "mapping class groups" on wikipedia says the topological MCG of T^n is GL(n,Z), but does anyone know a reference? Also, is the smooth MCG of T^n known?

• If you were to take the mapping class group to be $\pi_0 HomotopyEquiv(M)$, i.e. the path-components of the homotopy equivalences of the manifold $M$, then provided $M$ is a $K(\pi,1)$, the mapping class group would be $Out(\pi_1 M)$. So provided $\pi_1 M$ is abelian, this is just $Aut(\pi_1 M)$. Aug 18, 2010 at 0:01
• I'm using $MCG(M)$ in the sense of either $\pi_0 Homeo(M)$ or $\pi_0 Diff(M)$, but probably it's your version that is meant by claim on wikipedia. Aug 18, 2010 at 0:11
• I patched-up the Wikipedia page a little. Still could use work. Thanks Gogolev and Liu. Aug 18, 2010 at 1:10
• For $n=3$ this result is due to Hatcher: dx.doi.org/10.1016/0040-9383(76)90027-6 Aug 18, 2010 at 20:45

• Indeed, $MCG(\mathbb T^n)=GL(n,\mathbb Z)$ in dimension $n<4$, but it is not simple. In dimension 2 it was first proved by Earle and Eells using complex analysis.[Edit: As Allen Hatcher points out this was known for a long time, Earle and Eells prove much stronger statement: $\mathbb T^2$ is deformation retraction of $Diff_0(\mathbb T^2)$]
• This is not correct in dimension >4, the MCG is semidirect product of $GL(n,\mathbb Z)$ with another (non-finitely generated group). Let me just quote Hatcher:
If $n\ge 5$ then there are split exact sequences $$0\to \mathbb Z_2^\infty\to\pi_0(Top(\mathbb T^n))\to GL(n,\mathbb Z)\to 0$$ $$0\to \mathbb Z_2^\infty\oplus\binom n2\mathbb Z_2\to\pi_0(PL(\mathbb T^n))\to GL(n,\mathbb Z)\to 0$$ $$0\to \mathbb Z_2^\infty\oplus\binom n2\mathbb Z_2\oplus\sum_{i=0}^n\binom n i\Gamma_{i+1}\to\pi_0(Diff(\mathbb T^n))\to GL(n,\mathbb Z)\to 0$$ where $\Gamma_i$ are Kervaire-Milnor finite abelian groups of homotopy spheres, $\mathbb Z_2$ is just the group of order 2 and $\mathbb Z_2^\infty$ is the group of finite strings.
• I doubt there's much known about $\pi_0 Diff( (S^1)^4 )$. Aug 17, 2010 at 23:56