# $S^3$ as union of 2 solid tori

I am working through Hatcher's Algebraic Topology and don't get what he means by "standard decomposition of $$S^3$$ into two solid tori",

$$S^3= S^1 \times D^2 \cup D^2 \times S^1$$

Can someone explain (or visualize) what the union looks like? First, I thought it looked like a genus-$$2$$-torus:

but I guess that's not what he means, actually... I have basic knowledge in algebraic topology and I am not familiar with Heegaard Splittings.

To obtain an embedding of $$X$$ in $$S^3-K$$ as a deformation retract we will use the standard decomposition of $$S^3$$ into two solid tori $$S^1\times D^2$$ and $$D^2\times S^1$$, the results of regarding $$S^2$$ as $$\delta D^4=\delta(D^2\times D^2)=\delta D^2\times D^2\cup D^2\times\delta D^2$$. Geometrically, the first solid torus $$S^1\times D^2$$ can be identified with the compact region in $$\Bbb R^3$$ bounded by the standard torus $$S^1\times S^1$$ containing $$K$$, and the second solid torus $$D^2\times S^1$$ is then the closure of the complement of the first solid torus, together with the compactification point at infinity. Notice that meridional circles in $$S^1\times S^1$$ bound disks in the first solid torus, while it is longitudinal circles that bound disks in the second solid torus.

I have read many similar questions and answers, but none I found could help me out, so I started a new one.

• While the answers here are quite nice, the one line version is perhaps useful to hear: "if $X$ is the (closure of the) complement of a solid torus sitting in $\mathbb R^3$ as thickened unknot, then the one-point compactification of $X$ is another solid torus." We are, of course, using here that $S^3$ is the one-point compactification of $\mathbb R^3$. Commented Feb 6, 2021 at 7:52

Here's one way of visualizing it.

In the same way that $$S^2$$ can be mapped to the extended plane $$\mathbb R^2+\infty$$ via the stereoscopic projection, use stereoscopic projection to map $$S^3$$ to $$\mathbb R^3+\infty$$.

Now imagine a solid torus $$T$$ and its complement $$T'$$ in $$\mathbb R^3$$. $$T$$ is evidently $$S^1\times D^2$$. But so is $$T'$$, because it can be decomposed into circles that go through the disk spanned by the smallest circle in $$T$$ (i.e. the circle that travels around the "hole" in the torus) and travel around $$T$$. This decomposition will be preserved topologically when the inverse stereographic projection takes it back to $$S^3$$.

(Lest the verbal description above doesn't make sense, a cross section would look like the following picture from wikipedia)

• Very helpful picture! Commented Feb 6, 2021 at 12:09

if we write the 3 sphere as $$w^2 +x^2 +y^2 +z^2 =1$$ the torus of intersection is $$w^2 +x^2 = \frac{1}{2} \; , \; \; \; y^2+z^2 = \frac{1}{2}$$

One solid torus has $$w^2 + x^2$$ from $$0$$ to $$1/2$$ but $$y^2 + z^2$$ from $$1/2$$ to $$1,$$ so that each fixed value of $$y^2 + z^2$$ gives a disc

You can view the 3-sphere are the points $$(x,y)$$ in $$\mathbb C^2$$ such that $$|x|^2+|y|^2=1$$. Can you see what the set of points in the sphere with $$|x|^2\leq 1/2$$ is? Reversing the inequality gives you the other one.