Is this submanifold of [itex]\mathbb{R}^3[/itex] ?

[Karamata]

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

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

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

Then [itex]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)[/itex].

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

 

[jvicens]

Yes, your approach is correct. To prove that this curve is a submanifold of \mathbb{R}^3, we need to show that the differential of the mapping f is non-zero for all t \in \mathbb{R}. As you have correctly shown, if df_t = (0,0,0) for some t \in \mathbb{R}, then t must be either \pi/2 or 3\pi/2. However, at these points, the first and second coordinates of df_t are non-zero, so df_t cannot be equal to (0,0,0). Therefore, the rank of f is always 1, and f(\mathbb{R}) is a submanifold of \mathbb{R}^3.