Non-orientable quotients of the product of two 2-spheres by Z/4Z

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Let $<!-- COMMENT: To achieve a unified layout, along with using the template below, please OBSERVE the following: besides, $...$ and $$...$$, you should use two environments: - For statements like Theorem, Lemma, Definition etc., use e.g. {{beginthm|Theorem 1|(Milnor)}} ... ... ... {{endthm}}. - For references, use e.g. {{cite|Milnor1958b}}. END OF COMMENT --> {{Stub}} == Problem == <wikitex>; Let $\sigma$ be the generator of the free action of $\mathbb{Z}/ 4\mathbb{Z}$ on $S^2 \times S^2$ determined by $$\sigma(x, y) = (y, -x), \text{ where } (x,y)\in S^2 \times S^2.$$ There is a (unique) geometric quotient $M = S^2\times S^2/ \langle \sigma \rangle$ obtained by this free action. To understand the structure of this quotient, first, notice that $\sigma^2$ is just the antipodal map on the diagonal. So the diagonal projects down to the projective plane $\mathbb{R}P^2$ inside the quotient. Denote a disk bundle neighbourhood of this projective plane by $N$. Off the diagonal, the structure of $S^2 \times S^2/\langle \sigma \rangle$ is that of a mapping cylinder. Namely, the mapping cylinder of the double cover of the lens space $L(8,1)$ by the lens space $L(4, 1)$. So $N \cup \rm{MCyl} (L(4, 1) \to L(8, 1))$ is homotopy equivalent to the quotient $M = S^2\times S^2/ \langle \sigma \rangle$. Modifying that mapping cylinder by taking the map to $L(8, 3)$, it can be shown that $N \cup \rm{MCyl}(L(4, 1) \to L(8, 1))$ and $N \cup \rm{MCyl} (L(4, 1) \to L(8, 3))$ have the same homotopy type. In \url{https://arxiv.org/pdf/1712.04572.pdf}, it is shown that there are exactly four topological manifolds in this homotopy type, two of which are smoothable and two which have non-trivial Kirby-Siebenmann invariant. The question is if $N \cup \rm{MCyl}(L(4, 1) \to L(8, 1))$ and $N \cup \rm{MCyl}(L(4, 1) \to L(8, 3))$ are homeomorphic or even diffeomorphic. </wikitex> == References == {{#RefList:}} <!-- == External links == * The Wikipedia page about [[Wikipedia:Page_name|link text]]. --> <!-- Please modify these headings or choose other headings according to your needs. --> [[Category:Questions]] [[Category:Research questions]]\sigma$ be the generator of the free action of $\mathbb{Z}/ 4\mathbb{Z}$ on $S^2 \times S^2$ determined by

$$\displaystyle \sigma(x, y) = (y, -x), \text{ where } (x,y)\in S^2 \times S^2.$$

There is a (unique) geometric quotient $M = S^2\times S^2/ \langle \sigma \rangle$ obtained by this free action.

To understand the structure of this quotient, first, notice that $\sigma^2$ is just the antipodal map on the diagonal. So the diagonal projects down to the projective plane $\mathbb{R}P^2$ inside the quotient. Denote a disk bundle neighbourhood of this projective plane by $N$.

Off the diagonal, the structure of $S^2 \times S^2/\langle \sigma \rangle$ is that of a mapping cylinder. Namely, the mapping cylinder of the double cover of the lens space $L(8,1)$ by the lens space $L(4, 1)$.

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In \url{https://arxiv.org/pdf/1712.04572.pdf}, it is shown that there are exactly four topological manifolds in this homotopy type, two of which are smoothable and two which have non-trivial Kirby-Siebenmann invariant.

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2 References