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Are the mapping class groups of $T^2 \times [0, 1$] and $T^2 \times S^1$ explicitly known?

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    $\begingroup$ Do your diffeomorphisms and isotopies fix the boundary pointwise or no? For the latter the answer is $GL_3 \Bbb Z$, due to Waldhausen, who gives the answer for all Haken manifolds. $\endgroup$ – Mike Miller Eismeier Jun 8 '20 at 23:11
  • $\begingroup$ Yes, for the former case, I am thinking about the diffeomorphisms which fix the boundary point-wise. $\endgroup$ – Luca Iliesiu Jun 9 '20 at 12:35
  • $\begingroup$ Then the mapping class group is $\Bbb Z^2$, generated by Dehn twists around each factor of the $T^2$ separately. This all requires a little bit of work but not too much from Waldhausen's result. In fact, you can compute using a result of Ivanov here the homotopy type of the entire diffeomorphism group; $GL_3(\Bbb Z) \rtimes T^3$ for the 3-torus, just $\Bbb Z^2$ for $T^2 \times I$ rel boundary, and $\Bbb Z/2 \times (GL_2(\Bbb Z) \rtimes T^2)$ for $T^2 \times I$. $\endgroup$ – Mike Miller Eismeier Jun 9 '20 at 14:38
  • $\begingroup$ Thanks a lot! What does the T^3 stand for in $GL_3(\mathbb Z) \rtimes T^3$? I think I might be misunderstanding the notation but you said that the mapping class group for the 3-torus is just $GL_3(\mathbb Z) $. Also, by ``rel boundary'' do you mean in the case in which you fix diffs on the boundary pointwise? $\endgroup$ – Luca Iliesiu Jun 9 '20 at 17:00
  • $\begingroup$ $T^3$ is the group given by the 3-torus, aka $\Bbb R^3/\Bbb Z^3$, and the action of $GL_3(\Bbb Z)$ descends from the action on $\Bbb R^3$. When you take connected components (the mapping class group) you are just left with $GL_3(\Bbb Z)$. I just thought I would mention these slightly stronger results as well which include information on the higher homotopy groups of Diff in addition to the mapping class group $\pi_0$ Diff, because they are accessible thanks to Ivanov. $\endgroup$ – Mike Miller Eismeier Jun 9 '20 at 17:16

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