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Let $T$ be the solid 2-torus and let $\sim$ be the equivalence relation on $T$ generated by the relation $\{(\alpha,\beta) \sim (\beta,\alpha) \mid \alpha, \beta \in S^1\}$ on the boundary $\partial T=S^1\times S^1$. What kind of space is $T/{\sim}$?

I believe that it is $S^3$, but I can't prove it.

I posted this question already here:

https://math.stackexchange.com/questions/4339019/identifying-boundary-of-solid-torus-with-itself-by-swapping-coordinates

Unfortunately, I didn't get a satisfying answer. Not sure if it's because it is harder than I thought or actually nobody cares. Maybe both :)

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  • $\begingroup$ But the meridian of the torus is contractible in the solid torus and represents a generator for the fundamental group of the Mobius strip. $\endgroup$
    – IJL
    Dec 22, 2021 at 17:34
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    $\begingroup$ Why is your quotient space a manifold? $\endgroup$
    – Tim Campion ♦
    Dec 22, 2021 at 20:02
  • $\begingroup$ @TimCampion oh good question.. I don't know and I will think about it. But I'm not sure if this issue is serious enough to unaccept IJLs answer $\endgroup$
    – Bipolar Minds
    Dec 22, 2021 at 20:58
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    $\begingroup$ @TimCampion The only issues would be on the diagonal. Pick a tubular neighborhood of the diagonal of $S^1 \times S^1$ such that the involution acts as negation on the fibers. Now pick a point $p$ in the diagonal. We can write down a nice half open neighborhood of $p$ by taking a small open neighborhood of $p$ in the diagonal, taking the restriction of the tubular neighborhood, and then taking the collar on that in the solid torus. By construction, the gluing can be viewed as gluing $\mathbb{R}_p \times [0,1)$ along negation on $\mathbb{R}$, parametrized by the points in the open around p. $\endgroup$
    – Connor Malin
    Dec 22, 2021 at 22:13
  • $\begingroup$ @TimCampion It is geometrically clear that these 2 dimensional gluings give $\mathbb{R}^2$ (it's like taping the edge of a sheet of paper to itself). So when we parametrize by the open neighborhood of $p$ in the diagonal, it becomes a $\mathbb{R}^3$. $\endgroup$
    – Connor Malin
    Dec 22, 2021 at 22:15

3 Answers 3

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Your quotient manifold is homeomorphic to $S^3$. I know this because it is a closed 3-manifold and its fundamental group is trivial, so I'm quoting the Poincare conjecture/Perelmann's theorem. The fundamental group can be seen to be trivial because there is a CW-decomposition of the (2-manifold) torus with one 0-cell, three 1-cells and two 2-cells that is respected by your involution, where the 1-cells are the longitude, meridian and diagonal of the torus, and the two 2-cells are triangles. The involution swaps the longitude and meridian while fixing the diagonal pointwise, and swaps the two 2-cells. The quotient of the torus by the involution has one vertex, two 1-cells and one 2-cell. The fundamental group of this space is infinite cyclic, generated by the loop around the 1-cell that started life as both the longitude and meridian of the torus.
This quotient space can be viewed as a copy of the Mobius strip, as in Taras Banakh's comment.

But now think about how the insides fits in: a CW-structure on the solid torus can be made by adding in one more 2-cell, attached around the meridian of the torus, and one 3-cell. The extra 2-cell wraps around the generator for the fundamental group of the piece discussed in the previous paragraph. So the fundamental group of the overall space is trivial.

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  • $\begingroup$ Thx for this nice answer! $\endgroup$
    – Bipolar Minds
    Dec 22, 2021 at 18:46
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I believe that your quotient space can be seen as the quotient of the $3$-sphere $S^3$ in $\mathbf C^2$ by the action of complex conjugation. The $3$-sphere $S^3$ can be identified with the join $S^1\star iS^1$ of the real circle $S^1=S^3\cap \mathbf R^2$ with the imaginary circle $iS^1=S^3\cap i\mathbf R^2$. Since complex conjugation acts trivially on $S^1$ and antipodally on $iS^1$, the quotient space is again homeomorphic to a join $S^1\star S^1$ of $2$ circles, hence is a $3$-sphere.

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    $\begingroup$ Thx for this nice answer! $\endgroup$
    – Bipolar Minds
    Dec 22, 2021 at 18:46
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As yet another proof: let $M$ be the quotient of $\partial T$. Let $U$ be a small closed regular neighbourhood of $M$. Thus $U$ is a solid torus. Let $V$ be the closure of the complement of $U$. So $V$ is $T$, minus an open regular neighbourhood of its boundary. So $V$ is another solid torus. Let $S$ be the intersection of $U$ and $V$. So $S$ is a (genus one) Heegaard surface for the three-manifold. It is an “exercise” to show that the boundaries of the meridian disk in $U$ and $V$, respectively, meet once. Thus $S$ is the standard Heegaard splitting of the three-sphere.

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