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We know that Mapping Class Group (MCG) of $S^2\times S^1$ is $Z_2$, generated by twists along the sphere (I am restricting to orientation preserving homeomorphisms). MCG does not act on the fundamental group of $S^2\times S^1$. So my question is, how does MCG act, if at all, on the cell complex triangulating $S^2\times S^1$?

Thank you

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(This is a long comment rather than a complete answer.) As Igor Rivin points out, the mapping class group is not ${\mathbb Z}_2$. There is another ${\mathbb Z}_2$ direct summand coming from a homeomorphism that reverses the orientation on both $S^1$ and $S^2$, hence preserves the orientation of $S^1\times S^2$. This homeomorphism can be taken to have order 2 and it seems likely that $S^1\times S^2$ can be triangulated so that the homeomorphism preserves this triangulation. On the other hand, it seems somewhat unlikely that the twist homeomorphism generating the other ${\mathbb Z}_2$ can be realized by a symmetry of a triangulation. This symmetry would have finite order, and the only reason the twist along $S^2$ has order 2 is that $\pi_1SO(3)$ is ${\mathbb Z}_2$, which depends on a well-known but slightly subtle deformation of a $4\pi$ rotation of space to the identity map, the so-called belt trick. In one lower dimension a Dehn twist on a nontrivial curve on a torus, for example, has infinite order in the mapping class group, so it cannot be realized by a symmetry of a fixed triangulation of the torus. It would be a little surprising if 3 dimensions was different in this regard.

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This seems to contradict Kawauchi which also has an extensive discussion.

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  • $\begingroup$ thank you for the correction. My question was ill posed on so many levels. :) $\endgroup$ – SKShukla Aug 11 '16 at 23:22

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