Helicoid

**Bruno J.** · June 7th 2010, 08:42 PM

Prove that the helicoid is homeomorphic to the plane.

**chiph588@** · June 7th 2010, 09:09 PM

Originally Posted by Bruno J.

Prove that the helicoid is homeomorphic to the plane.

We have $\begin{cases} x=r\cos(a\theta)\\y=r\sin(a\theta)\\z=\theta \end{cases}$ .

Can't we just let $a\to0$ (in a continuous manner to make it a homeomorphism)?

**Bruno J.** · June 7th 2010, 09:12 PM

Originally Posted by chiph588@

We have $\begin{cases} x=r\cos(a\theta)\\y=r\sin(a\theta)\\z=\theta \end{cases}$ .

Can't we just let $a\to0$ (in a continuous manner to make it a homeomorphism)?

So what is the image of $(x,y,z)$ under this map? *

*corrected

**chiph588@** · June 7th 2010, 09:22 PM

Originally Posted by Bruno J.

So what is the image of $(x,y,z)$ under this map? *

*corrected

$(x,y,z)\mapsto(r,0,\theta)$ where $r$ and $\theta$ are free variables, so this map is the x-z plane.

**Bruno J.** · June 7th 2010, 09:34 PM

Originally Posted by chiph588@

$(x,y,z)\mapsto(r,0,\theta)$ where $r$ and $\theta$ are free variables, so this map is the x-z plane.

Good!

Here are a few abstract thoughts.

We know the universal covering space of the punctured plane $\mathbb{C}-\{0\}$ is the plane $\mathbb{C}$ , with the projection $\mbox{exp }: \mathbb{C} \to \mathbb{C}-\{0\}$ .

On the other hand, the universal cover of $\mathbb{C}-\{0\}$ is also the Riemann surface of the (multi-valued) inverse function $\mbox{Log}$ , of which the helicoid is the most obvious model. By the universal property of the universal cover, there exists a homeomorphism between the helicoid and the plane.

(Informally : the slit plane $\mathbb{C}-\mathbb{R}_{\geq 0}$ is easily seen to be homeomorphic to a strip without its boundary, (any branch of the logarithm doing the job). Now lift the cut to the helicoid, and map each slice of the helicoid to a strip, identifying the appropriate boundaries of two strips if the corresponding boundaries of the slices are identified on the helicoid. In this way, the helicoid is mapped homeomorphically on the plane.)

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