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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.

Leslie
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1 Answers1

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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!!

Léo
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  • Thank you for your reply! I still don't get both questions but I will first concentrate on A now. How did you get that formula for T? I know that a formula for a Torus is: $$ (c - sqrt(x^2 - y^2))^2 + z^2 = a^2$$ where $c$ is the the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. So in this case: $a=a$ and $c=b$, but you do not have $a^2$ on the right hand side. Also, do you think I have to derive the formula for a Torus in this question, or is something I should have known? – Leslie Feb 18 '14 at 09:40
  • And for B how can I see that te first matrix gives rotation around the y-axis? – Leslie Feb 18 '14 at 15:02
  • By the way in the torus I meant $x^2 + y^2$. Further I want to know why I have to prove that it's periodic? Because for a diffeomorphism it must be smooth, a bijection and it's inverse must be differentiable as wel – Leslie Feb 18 '14 at 21:07
  • Sorry, it's an $a^2$, as well... So it's exactly what you write, right? For understand this formula, i think you have to draw it! – Léo Feb 19 '14 at 15:51
  • First, when $z=+-a$, you find your circle (the meridian), and when $z$ moves, you have to make the other circle (the longitude). – Léo Feb 19 '14 at 15:53
  • My english is not very good, i hope it's understandable! – Léo Feb 19 '14 at 15:54
  • For the matrix in B), it's well-known, the most simple is first to see that it does not touch the y-axe, and then to consider the (2,2) matrix when you forget the 2nd line and row, then it's a plane rotation, whith $\theta$ angle, can you see it? – Léo Feb 19 '14 at 15:57
  • Finally you need to prove it's periodic cause the first map i gave you takes values in $\mathbb{R}^2$, and you need it to take in $\mathbb{S}^1 \times \mathbb{S}^1$. Hope it makes you clear... – Léo Feb 19 '14 at 16:00
  • Thank you Léo! I understand most of it. I don't really see how proving the periodicity of the function shows that my domain is $\mathbb{S}^1 \times \mathbb{S}^1$ further I don't know how to prove the periodicity. Also I want to know if I don't need to prove anything about the inverse because that's in the definition of diffeomorphism. – Leslie Feb 19 '14 at 23:19
  • For proving periodicity, you check that $\forall (m,n) \in \mathbb{Z}^2$, $\phi(\theta +m, \psi +n)=\phi(\theta, \psi)$. Then by basic theory of quotient spaces that you need to learn, $\phi$ gives raise to a smooth map $\tilde{\phi} : \mathbb{R/Z} \times \mathbb{R/Z} \to \mathbb{R}^3$. You can then check that this new map is injective, and it's image is T. Last you need to check it's an immersion, that is its tangent map is injective too. $\mathbb{S}^1 \times \mathbb{S}^1$ is compact, so you have shown that $\tilde{\phi}$ is an embedding! – Léo Feb 19 '14 at 23:51
  • Thank you! Why do I have to check that te tangent map is injective too? And how do I do that? Do I have to calculate $D_{\psi,\theta}\phi$ for that? – Leslie Feb 20 '14 at 00:18
  • Cause you want it to be an embedding! – Léo Feb 20 '14 at 22:14