Talk:Borel Conjecture for the 2-torus (Ex)

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(Created page with "<wikitex>; Suggestion: 1. Note that the dimensions must be equal. 2. Use the homotopy classification of surfaces, so both surfaces must be $T^2$. 3. Use the square model fo...")
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Suggestion:
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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.
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.
2. Use the homotopy classification of surfaces, so both surfaces must be $T^2$.
3. Use the square model for the torus with edges identified. Assume WLOG that the map preserves orientation. We'll homotope the map to the identity map. Homotope the boundaries to each other one at a time; this turns the interior into an open square. Then homotope the interior to itself.
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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 (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.


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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