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What can one say about two closed oriented 3-manifolds $M_1$ and $M_2$ such that $S^2 \times M_1$ is diffeomorphic to $S^2 \times M_2$?

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Closed, aspherical 3-manifolds with isomorphic fundamental groups are diffeomorphic (by Waldhausen, Mostow, Scott and Perelman, I think). – HJRW Jun 8 '13 at 9:08
In the case when the 3-manifolds are lens spaces some partial classification is due to Sadeeb Ottenburger, see, and I am not sure whether there is a clean simple statement in this case. If memeory serves, he can handle all fundamental groups $\mathbb Z_r$ where $r$ is coprime to $6$. The proofs are surgery theoretic. – Igor Belegradek Jun 8 '13 at 16:18
I suspect the torsion will still distinguish them. The Reidemeister torsion distinguishes Lens spaces, and is roughly a determinant of the cellular chain complex of the universal cover as a group-ring module. When you cross with $S^2$, the chain complex changes by a copy of the chain complex and a shift by 2. I suspect the Reidemeister torsion will be determined by that of the Lens space, but one would have to go through the details to check this. – Ian Agol Jun 12 '13 at 4:59
@Agol: After checking some details, I agree that the Reidemeister torsion distinguishes $L_1\times S^2$ and $L_2\times S^2$ if $L_1$ and $L_2$ are non-diffeomorphic 3-dimensional lens spaces. In brief: (1) Multiplying by $S^2$ squares the Reidemeister torsion (see theorem B in section 4 of Milnor's "Two complexes which are homeomorphic but combinatorially distinct"). (2) The Reidemeister torsion of the lens space $L(m,n)$ is $(1-\zeta)(1-\zeta^a)$ mod $\{\pm \zeta^k\}$, where $\zeta$ is a primitive $m$-th root of unity, and $a n \equiv 1$ mod $m$. (to be continued) – Ricardo Andrade Jun 14 '13 at 4:26
(continued) See the notes at for the calculation. (3) Finally, if $L(m,n)\times S^2$ has the same Reidemeister torsion as $L(m,p)\times S^2$ (with respect to some isomorphism of the fundamental groups given by multiplication by $r$ in $\mathbb{Z}/m$), then $A = \frac{(1-\zeta^r)(1-\zeta^{ar})}{(1-\zeta)(1-\zeta^b)}$ is a $2m$-th root of unity, by (1) and (2). Since $A\in\mathbb{Q}[\zeta]$, if $A$ is a root of unity at all, then it is actually of the form $\pm \zeta^k$, implying that $L(m,n)$ and $L(m,p)$ already had equal Reidemeister torsions. – Ricardo Andrade Jun 14 '13 at 4:58

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