Finding the description of a homeomorphism from $S^1 \times S^1$ onto the surface of doughnut.

Question

The following exercise is given in the book Topology- James Munkres (2nd Edition, ch-9, sec-54). I have some difficulties in solving this one. I try to describe my thoughts here:

Exercise 5. Consider the path $$f(t) = (\cos 2\pi t, \sin 2\pi t) \times (\cos 4\pi t, \sin 4\pi t)$$ in $S^1\times S^1$. Sketch what $f$ looks like when $S^1\times S^1$ is identified with the doughnut surface $D$.

Here, doughnut-shaped surface $D$ in $\Bbb{R}^3$ is obtained by rotating the circle $C_1$ in the $xz$-plane of radius $\frac13$ centered at $(1, 0,0)$ about the $z$-axis.

My Attempt. Here, $f:[0,1]\to S^1\times S^1$. First I try to find out a homeomorphism from $S^1\times S^1$ to $D$. Well, Munkres gave an example of such homeomorphism, $h$, say:

Consider $C_2$ be the circle of radius $1$ in the $xy$-plane centered at origin. Then we can map $C_1 \times C_2$ into $D$ by defining $h(a,b)$ to be the point into which $a$ is carried when one rotates the circle $C_1$ about the $z$-axis until its center hits the point $b$.

Here, $S^1\times S^1$ is homeomorphic with $C_1 \times C_2$.

Therefore $h \circ f:[0, 1]\to D$ is the corresponding path in $D$. So, now everything depends on $h$. I can visualize $h$ as given in its description above. But I need its analytical description also so that I can compute $h(f(t))$.

Let $a=(a_1,0,a_3)\in C_1$ and $b=(b_1, b_2, 0)\in C_2$ so that $(a,b) \in C_1 \times C_2$. I have to find the expression for $h(a,b)=(h_1(a,b), h_2(a,b), h_3(a,b))\in D$.

But I cannot proceed further. Any hints to find out $h_1, h_2, h_3$ or any better way to solve this is appreciated. Thank you.

Answer 1

The circle in plane $xz$ that is rotated to obtain surface $D$ has a parametrization: $$ C_1 = \left\{\left(\Big(1+\frac13\cos\theta\Big),0,\frac13\sin\theta\right): \theta\in[0,2\pi)\right\}$$ As per the description the homeomorphism $h$, you take a point of $C_1$ an rotate it appropriately around the $z$-axis: \begin{align} h\big((\cos\theta,\sin\theta)\times(\cos\phi,\sin\phi)\big) &= \hat{R}_z(\phi)\left(\Big(1+\frac13\cos\theta\Big),0,\frac13\sin\theta\right) = \\ &= \left(\Big(1+\frac13\cos\theta\Big)\cos\phi, \Big(1+\frac13\cos\theta\Big)\sin\phi, \frac13\sin\theta \right)\end{align}

Answer 2

Define $$h : S^1 \times S^1 \to \mathbb{R}^3, h((x,y),(u,v)) = (x(1+\frac{u}{3}),y(1+\frac{u}{3}),\frac{v}{3}) .$$

1) $h$ is injective.

Let $h((x,y),(u,v)) = h((x',y'),(u',v'))$. Then $v = v'$ and therefore $u^2 = 1 -v^2 = 1 - (v')^2 = (u')^2$. But now $$(1+\frac{u}{3})^2 = (x^2+y^2)(1+\frac{u}{3})^2 = x^2(1+\frac{u}{3})^2 + y^2(1+\frac{u}{3})^2 = \\ (x')^2(1+\frac{u'}{3})^2 + (y')^2(1+\frac{u'}{3})^2 = ((x')^2 + (y')^2)(1+\frac{u'}{3})^2 = (1+\frac{u'}{3})^2 .$$ This implies $u = u'$ and we conclude that also $x=x'$, $y=y'$.

2) $h(S^1 \times S^1) = D$.

Let $C_{(x,y)}$ be the circle obtained by rotating $C_1$ around the $z$-axis until its center reaches $(x,y,0) \in C_2$. It is contained in the plane spanned by $(0,0,0)$, $(x,y,0)$ and $(0,0,1)$. Obviously $$C_{(x,y)} = \{ a(x,y,0) + b(0,0,1) \mid (\frac{1}{3})^2 = \lVert a(x,y,0) + b(0,0,1) - (x,y,0) \rVert^2 \\= \lVert ((a-1)x,(a-1)y,b) \rVert^2 = (a-1)^2 + b^2 \} .$$ Substituting $\frac{u}{3} = a-1$ and $\frac{v}{3} = b$, we get $$C_{(x,y)} = \{ (x(1 + \frac{u}{3}),y(1 + \frac{u}{3}),\frac{v}{3}) \mid (u,v) \in S^1 \} = h(\{(x,y)\} \times S^1) .$$ 1) and 2) show that $h$ gives the desired homeomorphism $S^1 \times S^1 \to D$.