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

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## 1 Problem

Let $\sigma$$== Problem == ; Let \sigma be a generator of \mathbb{Z}/ 4\mathbb{Z} and consider the free action of \Z/4 on S^2 \times S^2 defined by \sigma(x, y) = (y, -x), \text{ where } (x,y)\in S^2 \times S^2. Let M := S^2\times S^2/ \langle \sigma \rangle be the quotient of S^2 \times S^2 obtained from this free action. To understand the structure of this quotient, first, notice that \sigma^2 restricted to the diagonal copy of S^2 \subset S^2 \times S^2 is the antipodal map. So the diagonal projects down to the projective plane \mathbb{R}P^2 inside the quotient. Denote a normal disk bundle neighbourhood of this projective plane by N. Off the diagonal, the structure of S^2 \times S^2/\langle \sigma \rangle is that of a mapping cylinder. Namely, the mapping cylinder of the double cover of the lens space L(8,1) by the lens space L(4, 1). So N \cup \rm{MCyl} (L(4, 1) \to L(8, 1)) is a model for the quotient M = S^2\times S^2/ \langle \sigma \rangle. Modifying that mapping cylinder by taking the double covering L(4, 1) \to L(8, 3), it can be shown that N \cup \rm{MCyl}(L(4, 1) \to L(8, 1)) and N \cup \rm{MCyl} (L(4, 1) \to L(8, 3)) are homotopy equivalent. In {{cite|Hambleton&Hillmann2017}} it is shown that there are at most four topological manifolds in this homotopy type, half of which are stably smoothable. ''Question'': Are N \cup \rm{MCyl}(L(4, 1) \to L(8, 1)) and N \cup \rm{MCyl}(L(4, 1) \to L(8, 3)) diffeomorphic? This question was posed by Jonathan Hillmann at the [[:Category:MATRIX 2019 Interactions|MATRIX meeting on Interactions between high and low dimensional topology.]] == References == {{#RefList:}} [[Category:Problems]] [[Category:Questions]] [[Category:Research questions]]\sigma$ be a generator of $\mathbb{Z}/ 4\mathbb{Z}$$\mathbb{Z}/ 4\mathbb{Z}$ and consider the free action of $\Z/4$$\Z/4$ on $S^2 \times S^2$$S^2 \times S^2$ defined by

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

Let $M := S^2\times S^2/ \langle \sigma \rangle$$M := S^2\times S^2/ \langle \sigma \rangle$ be the quotient of $S^2 \times S^2$$S^2 \times S^2$ obtained from this free action.

To understand the structure of this quotient, first, notice that $\sigma^2$$\sigma^2$ restricted to the diagonal copy of $S^2 \subset S^2 \times S^2$$S^2 \subset S^2 \times S^2$ is the antipodal map.

So the diagonal projects down to the projective plane $\mathbb{R}P^2$$\mathbb{R}P^2$ inside the quotient. Denote a normal disk bundle neighbourhood of this projective plane by $N$$N$.

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

So
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$N \cup \rm{MCyl} (L(4, 1) \to L(8, 1))$ is a model for the quotient $M = S^2\times S^2/ \langle \sigma \rangle$$M = S^2\times S^2/ \langle \sigma \rangle$. Modifying that mapping cylinder by taking the double covering $L(4, 1) \to L(8, 3)$$L(4, 1) \to L(8, 3)$, it can be shown that
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$N \cup \rm{MCyl}(L(4, 1) \to L(8, 1))$ and
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$N \cup \rm{MCyl} (L(4, 1) \to L(8, 3))$ are homotopy equivalent.

In [Hambleton&Hillmann2017] it is shown that there are at most four topological manifolds in this homotopy type, half of which are stably smoothable.

Question: Are
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$N \cup \rm{MCyl}(L(4, 1) \to L(8, 1))$ and
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$N \cup \rm{MCyl}(L(4, 1) \to L(8, 3))$ diffeomorphic?

This question was posed by Jonathan Hillmann at the MATRIX meeting on Interactions between high and low dimensional topology.

## 2 References

• [Hambleton&Hillmann2017] I. Hambleton and J. Hillmann, Quotients of $S^2 \times S^2$$S^2 \times S^2$. Available at the arXiv:17172.04572.