Homeomorphism of toroidal surface with product of two circles using only Cartesian coordinates

Question

Let $T$ be the toroidal surface in $\mathbb{R}^3$ having in Euclidean coordinates $(x, y, z)$ the equation $(R - \sqrt{x^2 + y^2})^2+ z^2 = r^2$, where $0 < r < R$.

Geometrically $T$ is obtained by rotating around the $z$-axis the "small" circle $K$ in the $xz$-plane having equation $(x -R)^2 + z^2 = r^2$ while the center of the rotating $K$ remains on the "big" circle $C$ in the $xy$-plane having equation $x^2+ y^2 = R^2$.

By parametrizing each of $C$ and $K$ by angular coordinates and passing to quotients modulo $2\pi$ one obtains a homemorphism of $C \times K$ with $T$.

But I want to do this strictly in Cartesian coordinates.

To do so, change the names of the coordinates on $K$ from $(x, y, z)$ to $(u, v, w)$ (where $v = 0$ on $K$). Define a map $h :C \times K \rightarrow \mathbb{R}^3$ by $h\bigl( (x, y, 0), \, (u, 0, w)\bigr) = (u \,x/R, u\, y/R, w)$.

Am I correct that $h$ maps into the toral surface $T$ and, in fact, gives the desired homeomorphism of $C \times K$ with $T$?

Since $h$ is continuous and $C \times K$ is compact, of course $h$ will define the desired homeomorphism once it is known to be injective and maps onto $T$. So that’s what my question amounts to. (The formula just looks too suspiciously simple to me!)

Related: Torus as a cartesian product of two circles and Is it true that cartesian product of any two circle 1 define a torus?

Answer 1

The formula is in fact simple. I suggest, however, to regard $C$ and $K$ as cirles in the plane $\mathbb R^2$ which results in $$C = \{(x,y) \mid x^2 + y^2 = R^2\} ,$$ $$K = \{(u,w) \mid (u-R)^2 + w^2 = r^2\} ,$$ $$ h : C \times K \to \mathbb R^3, h((x,y),(u,w)) =\left(\frac{u}{R}x, \frac{u}{R}y,w\right) = \left(\frac{u}{R}(x,y),w\right).$$ Note that $(u,w) \in K$ implies $u > 0$. Indeed, $(u-R)^2 \le (u-R)^2 + w^2 = r^2$ which gives $u \ge \frac{u^2 +R^2 -r^2}{2R} \ge \frac{R^2 -r^2}{2R} > 0$.

The toroidal surface $T$ is defined as $$T = \{(a,b,c) \mid (R- \sqrt{a^2+b^2})^2 + c^2 = r^2 \} = \{(a,b,c) \mid (R - \lVert (a,b) \rVert)^2 + c^2 = r^2 \}$$ where $\lVert - \rVert$ denotes the Euclidean norm on $\mathbb R^2$.

It is now easy to verify that $h((x,y),(u,w)) \in T$. We have $\lVert \frac{u}{R}(x,y) \rVert = \frac{u}{R}\lVert (x,y) \rVert = \frac{u}{R}R = u$, thus $$\left(R - \left\lVert \frac{u}{R}(x,y) \right\rVert \right)^2 + w^2 = (R-u)^2 +w^2 = r^2.$$

Thus we may regard $h$ as a map $h : C \times K \to T$. Instead of checking that it is injective and surjective we shall directly construct the inverse $j : T \to C \times K$. Geometrically it is easy to see what $j$ does. The "$C$-coordinate" of $j(a,b,c)$ is obtained by scaling the vector $(a,b)$ to norm $R$, i.e. $$\frac{R}{\lVert (a,b) \rVert}(a,b)$$ and the "$K$-coordinate" is obtained by $$(\lVert (a,b) \rVert,c)$$ since $\lVert (a,b) \rVert$ is the distance of $(a,b)$ from the origin.

An easy formal check verifies $\frac{R}{\lVert (a,b) \rVert}(a,b) \in C$ and $(\lVert (a,b) \rVert,c) \in K$. Thus $$j : T \to C \times K, j(a,b,c) = \left(\frac{R}{\lVert (a,b) \rVert}(a,b),(\lVert (a,b) \rVert,c)\right)$$ is a well-defined map. We have $$j(h((x,y),(u,w)) = j\left(\frac{u}{R}(x,y),w\right).$$ Since $\lVert \frac{u}{R}(x,y) \rVert = u$, we get $$j\left(\frac{u}{R}(x,y),w\right) = ((x,y),(u,w)).$$ Thus $j \circ h = id$.

Similarly $$h(j(a,b,c)) = h\left(\frac{R}{\lVert (a,b) \rVert}(a,b),(\lVert (a,b) \rVert,c)\right) = \left(\frac{\lVert (a,b) \rVert}{R}\frac{R}{\lVert (a,b) \rVert}(a,b),c)\right) = (a,b,c),$$ i.e. $h \circ j = id$.