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

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[edit] 1 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

\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 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).

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

[edit] 2 References

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