# Is this submanifold of $\mathbb{R}^3$ ?

1. Nov 24, 2013

### Karamata

How to deduce is it $\{\cos(\sqrt{2}t)(2+\cos t), \sin(\sqrt{2}t)(2+\cos t),\sin t \mid t \in \mathbb{R}\}$ submanifold of $\mathbb{R}^3$?

First, this is curve, so I was thinking to find point in which this curve has two intersect, and then some neighborhood of that point isn't homeomorphic to $\mathbb{R}$, and that would be reason why this isn't submanifold, but I failed to prove it on that way.

Maybe it is submanifold, I don't know.

I know that if $f: U \to V, U \subset \mathbb{R}^n, V \subset \mathbb{R}^m, n < m$ is differentiable mapping two manifolds, with rank $n$, then we can prove that $f(U)$ is submanifold of $V$.

In our case, $U=\mathbb{R}, V=\mathbb{R}^3$ and $1<3$. And, let $f(t)=(\cos(\sqrt{2}t)(2+\cos t), \sin(\sqrt{2}t)(2+\cos t),\sin t)$.

Then $df_t=(-\sin t\cos(\sqrt{2}t)+\sqrt{2}\sin(\sqrt{2}t) (2+\cos t),\sqrt{2} \cos(\sqrt{2}t) (2+\cos t)-\sin(\sqrt{2}t) \sin t,\cos t)$.

So if I prove that $df_t \neq (0,0,0)$ for every $t \in \mathbb{R}$ (if we suppose that $df_t = (0,0,0)$ for some $t \in \mathbb{R}$, then $t \in \{\pi/2, 3\pi/2\}$, because of $\cos t = 0$, but then first and second coordinate are different then $0$) then $\operatorname{rank}(f)_t=1$, so I proved that this is submanifold, right?

Last edited: Nov 24, 2013