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

(Difference between revisions)
Jump to: navigation, search
m (Problem)
m
Line 1: Line 1:
<!-- 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 ==
== Problem ==
<wikitex>;
<wikitex>;
Let $\sigma$ be the generator of the free action of $\mathbb{Z}/ 4\mathbb{Z}$ on $S^2 \times S^2$ determined by
+
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
$$\sigma(x, y) = (y, -x), \text{ where } (x,y)\in S^2 \times S^2.$$
$$\sigma(x, y) = (y, -x), \text{ where } (x,y)\in S^2 \times S^2.$$
There is a (unique) geometric quotient $M = S^2\times S^2/ \langle \sigma \rangle$ obtained
+
Let $M := S^2\times S^2/ \langle \sigma \rangle$ be the quotient of $S^2 \times S^2$ obtained from this free action.
by this free action.
+
To understand the structure of this quotient, first, notice that $\sigma^2$ is just the antipodal map on the diagonal.
+
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 disk bundle neighbourhood of this projective plane by $N$.
+
+
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 $L(4, 1)$.
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 $L(4, 1)$.
Line 27: Line 13:
So $N \cup \rm{MCyl} (L(4, 1) \to L(8, 1))$ is homotopy equivalent to the quotient $M = S^2\times S^2/ \langle \sigma \rangle$.
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.
+
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.
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.

Revision as of 06:02, 8 January 2019

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).

So
Tex syntax error
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
Tex syntax error
and
Tex syntax error
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
Tex syntax error
and
Tex syntax error
are homeomorphic or even diffeomorphic.

2 References

Personal tools
Namespaces
Variants
Actions
Navigation
Interaction
Toolbox