Talk:Borel Conjecture for the 2-torus (Ex)
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− | + | The selfhomotopy equivalences of an Eilenberg-Mac Lane space of type $(G,n)$ generally have $\pi_0 = Out(G)$, detected by the action of fundamental groups, and $\pi_n = C(G)$, with all other homotopy groups vanishing. In particular, $\pi_0(hAut(T^2)) = Gl_2(\mathbb Z)$ is detected by the action on $H_1(T^2)$ and all we have to do is produce a homeomorphism that acts by a given matrix $A \in Gl_2(\mathbb Z)$. But via the identification $T^2 = \mathbb R^2/\mathbb Z^2$ the action of $A$ on $\mathbb R^2$ produces such an element. | |
− | + | The same argument shows that the inclusion $Diff(T^n) \rightarrow hAut(T^n)$ splits on $pi_0$ In arbitrary dimensions. | |
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− | + | Note: as described by a (now erased) suggestion, there is a more elementary method. You can treat the map as a periodic map $\mathbb R^2 \to \mathbb R^2$ sending every lattice point to itself. This is because the image of the origin's generators also generate the target's homology. This map of $\mathbb R^2$ can be homotoped to the identity map linearly, producing a diffeomorphism. This also works for any $T^n$, as the above proof does. | |
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Latest revision as of 08:24, 10 January 2019
The selfhomotopy equivalences of an Eilenberg-Mac Lane space of type generally have , detected by the action of fundamental groups, and , with all other homotopy groups vanishing. In particular, is detected by the action on and all we have to do is produce a homeomorphism that acts by a given matrix . But via the identification the action of on produces such an element.
The same argument shows that the inclusion splits on In arbitrary dimensions.
Note: as described by a (now erased) suggestion, there is a more elementary method. You can treat the map as a periodic map sending every lattice point to itself. This is because the image of the origin's generators also generate the target's homology. This map of can be homotoped to the identity map linearly, producing a diffeomorphism. This also works for any , as the above proof does.