# Quotient of solid torus by swapping coordinates on boundary

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 :)

• But the meridian of the torus is contractible in the solid torus and represents a generator for the fundamental group of the Mobius strip.
– IJL
Dec 22, 2021 at 17:34
• Why is your quotient space a manifold? Dec 22, 2021 at 20:02
• @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 Dec 22, 2021 at 20:58
• @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. Dec 22, 2021 at 22:13
• @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 $\mathbb{R}^3$. Dec 22, 2021 at 22:15

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.
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.
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.