Torus, manifolds

I have some trouble with the following questions:

$\mathbb{R}^3$ has standard coördinates $(x, y, z)$. Regard in the plane $x=0$ the circle with centre $(x,y,z) = (0,0,b)$ and radius $a$, $0<a<b$. The area that arise when you turn the circle around the y-axis is called T.

1A. Give the equation of T and prove that it's a manifold of dimension 2.

I thought the following:

$$T= \int_{C} \pi (f(y))^2 dy $$ where C is the circle described above and $f(y)= \sqrt(a-y^2+2zb-b^2)$ But now I don't know how to continue, cause I don't really have any boundaries.

B. Regard now $\mathbb{S}^1 \subseteq \mathbb{R}^2$. Write for the standard 2-Torus $\mathbb{T}^2= \mathbb{S}^1 \times \mathbb{S}^1$, then $\mathbb{T}^2 \subseteq \mathbb{R}^4$ is a two dimensional manifold. Prove that $\mathbb{T}^2$ and $T$ are diffeomorph.

1 Answer

a) Try $T=\lbrace (\sqrt{y^2+z^2} - b)^2 + z^2 = a, (x,y,z) \in \mathbb{R}^3 \rbrace$, i think it gives you the required parametrization of your torus, embedded in the 3 dimensional euclidian space.

It's a 2 manifold because it's the zeros of the submersion $f:\mathbb{R}^3 \to \mathbb{R}, (x,y,z) \mapsto (\sqrt{y^2+z^2} - b)^2 + z^2 - a$ (you can check it easily).

b) For showing it's diffeomorphic to the product of two circles, you need to give first a two dimensional parametrization of T:

$\phi : \mathbb{R}^2 \to \mathbb{R}^3, (\theta, \psi) \mapsto \left( \matrix { \cos 2\pi \theta&0&-\sin 2\pi \theta \\ 0&1&0 \\ \sin 2\pi \theta&0&\cos 2\pi \theta } \right)\left( \matrix { 0 \\ \sqrt{a} \cos 2\pi\psi \\ b+\sqrt{a} \sin 2\pi \psi } \right) $

Note that the vectr of the right defines just your circle C, when the matrix gives you the rotation around the y axis.

Then you have to show that

_this map is smooth

_it's $\mathbb{Z}^2$ periodic, so induces a map $(\mathbb{R/Z})^2 \to \mathbb{R}^3$

_this new map is the diffeomorphism you need!!