# Figure $\infty$ is immersion of circle

Where can I find prove of:

Figure $\infty$ is immersion of circle. More thanks for a prove or a function between these manifolds.

• Are you picturing the "$\infty$" figure in a plane? May 7 '13 at 12:30
• @Gustavo Marra :The source says nothing but I suggest in plane. May 7 '13 at 12:38
• Then I believe there's no immersion of a circle into the $\infty$ figure, since you lose injectivity at the self-intersection of "$\infty$". In which book did you see this? May 7 '13 at 12:43
• @HoseynHeydari: This falls under the heading of things Gowers refers to as "Just Do It" proofs. (See gowers.wordpress.com/2008/08/16/just-do-it-proofs) To get you started: what is the definition of an immersion? Can you obtain a figure X as an immersion from the disjoint union of two intervals? OK, now you are done. May 7 '13 at 22:58
• Lemniscate of Bernulli May 7 '13 at 23:51

An immersion of the circle $S^1$ into the plane is tantamount to a periodic function $$t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr),\qquad {\bf z}(t+2\pi)\equiv{\bf z}(t)$$ with ${\bf z}'(t)\ne{\bf 0}$ for all $t$. At the reference given by Yuri Vyatkin we find the following parametric representation of the lemniscate, the "typical" $\infty$-curve: $$x(t)={a\sqrt{2}\cos t\over 1+\sin^2 t},\quad y(t)={a\sqrt{2}\cos t\sin t\over 1+\sin^2 t}\ ,$$ which is obviously smooth and $2\pi$-periodic. To test this map $t\mapsto{\bf z}(t)$ for regularity we compute $$\bigl|{\bf z}'(t)\bigr|^2={2a\sqrt{2}\over 3-\cos(2t)}\ ,$$ and find that this is $>0$ for all $t\in{\mathbb R}$.