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.
Suggestion 1:
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1. Note that the dimensions must be equal.
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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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2. Use the homotopy classification of surfaces, so both surfaces must be $T^2$.
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3. Use the square model for the torus with edges identified, lying in a grid. Assume WLOG that the map preserves orientation (if not, then flip it). Homotope the map to the identity map.
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</wikitex>
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Suggestion 2:
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Do not use Suggestion 1.
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Revision as of 06:48, 10 January 2019

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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