# Compact surface - Tube/Torus

How can we explain, without giving a detailed proof, why the tube $$\sigma (s,\theta )=\gamma (s)+a(n(s)\cos\theta+b(s)\sin\theta )$$ (where $n$ is the principal normal of the curve $\gamma$ and $b$ is its binormal)

around a closed curve in $\mathbb{R}^3$ with no self-intersections is a compact surface diffeomorphic to a torus (provided the tube has sufficiently small radius)?

I don't really have an idea how we could do that...

## 1 Answer

This will be very hand-wavy, so forgive me if it doesn't make sense:

WLOG, suppose $\gamma$ is the unit circle in the $xy-$plane $\gamma(s)=(\cos(s),\sin(s),0), 0\leq s\leq 2\pi$, let $a=1$. Then the we have a surface $\sigma$ that is exactly the product $S^1\times S^1$ (in this case, the normal vectors $n,b$ can be easily calculated explicitly).

• Which surface is the product $S^1\times S^1$ ? – Mary Star Jan 12 '16 at 0:44
• Exactly what you want: en.wikipedia.org/wiki/Torus – charlestoncrabb Jan 12 '16 at 0:46
• Ah ok!! Why should the tube have sufficiently small radius? – Mary Star Jan 12 '16 at 0:48
• So that it is not self-intersecting. The animation at the top of the Wikipedia page is quite illustrative of this – charlestoncrabb Jan 12 '16 at 0:49
• @MaryStar we have a circle in the variable $s$, and a circle in the variable $\theta$. In particular, for a fixed $s$, $\sigma$ is a circle in $\theta$, and for fixed $\theta$, $\sigma$ is a circle in $s$. – charlestoncrabb Jan 12 '16 at 1:32