# Message #1948

From: Melinda Green <melinda@superliminal.com>

Subject: Re: [MC4D] Visualizing the surface of Klein’s Quartic in 4D

Date: Sat, 03 Dec 2011 18:11:33 -0800

Wow, nice work, Roice! Sometimes we don’t get the answers we were

looking for, but by asking really good questions you almost always end

up in surprisingly interesting places. This journey of yours seems like

a perfect example.

The first thing that struck me when looking at your applet was just how

similar this visualization appears to the flexible polyhedron called the

Steffen Model that I fooled around with here

<http://superliminal.com/geometry/flexible/flexible.htm>. In particular,

the interactive applet

<http://superliminal.com/geometry/flexible/applet/applet.htm> I wrote to

explore potentially flexible polyhedra looks surprisingly like yours.

I suspect that the reason you are not finding pleasing embeddings into

*R*^3 may be due to the way you are identifying vertices. I suspect that

a better way is to try embedding the object into the same flat repeating

3D space as this {4,6} IRP

<http://superliminal.com/geometry/infinite/4_6a.htm> with genus 3. You

may not be able to model KQ with regular planar heptagons but I suspect

that they can at least be modeled as equal polygons in this regularly

repeating 3-space. In fact I suspect that Klein, Séquin

<http://www.cs.berkeley.edu/%7Esequin/PAPERS/Bridges06_PatternsOnTetrus.pdf>

and others have simply missed this possibility because of our natural

bias for the non-repeating 3-space in which we appear to live.

-Melinda

On 12/3/2011 3:56 PM, Roice Nelson wrote:

>

>

> Hi all,

> This isn’t puzzle related, but it is related to higher dimensions. If

> you’d rather just play with an applet, skip to the bottom :)

>

> In the popular writings I’ve seen on Klein’s Quartic, people have said

> it can only be realized as a perfectly regular figure in 4-dimensional

> space. As a member of this group, I found that captivating each time

> I read it, and thought to myself "Let’s get the coordinates and plot

> them! Then we can rotate it around using our 4D controls to get a feel

> for it." So I’ve wanted to look into how to get the 4D KQ coordinates

> for a while. I had trouble finding info about calculating the points,

> and I ultimately only found out what I needed from Klein’s original

> paper <http://library.msri.org/books/Book35/files/klein.pdf>.

>

> It turns out the statements I’ve seen are a bit misleading to

> individuals like me, who are accustomed to thinking of 4-dimensional

> space in terms of Euclidean *R*^4 . The solution of KQ turns out to

> be perfectly regular in the complex projective plane

> <http://en.wikipedia.org/wiki/Complex_projective_plane>, *CP*^2 .

> *CP*^2 is 4-dimensional, but quite a different beast than *R*^4 .

>

> Thinking about the hemi-dodecahedron was a good analogy for me. The

> surface of that object is perfectly regular in the real projective

> plane, *RP*^2 , so it is "regular in 2-dimensions". It is not regular

> in Euclidean *R*^2 though, and furthermore, you can’t even embed the

> hemi-dodec in a regular way in *R*^3 . Try plotting it on a

> hemisphere whose rim has opposite points identified to see this.

> Similarly, KQ can’t be embedded in a regular fashion in *C*^3 space

> (which is 6-dimensional, having two components for each of three

> complex numbers). I don’t know how many Euclidean dimensions would be

> required to plot either the hemi-dodec or KQ in a regular fashion, but

> there is a result giving bounds

> <http://en.wikipedia.org/wiki/Nash_embedding_theorem>. In any case,

> there certainly is no perfectly regular *R*^4 representation of KQ (no

> "isometric embedding" or "isometric immersion").

> Setback but obstinate, I still made an exploration applet which does

> the following steps to plot the 56 triangles of KQ. I haven’t made

> something similar for the dual representation of heptagons.

>

> 1. Calculate the coordinates of KQ vertices in *C*^3 with the help of

> Klein’s paper. There are 168 of these, but the "same point" shows

> up 7 times along a projective ray.

> 2. Identify duplicate points to get a list of representative vertices

> in *CP*^2 (24 of these). This is done by normalizing all 168

> points, and seeing which normalize to the same value. After

> normalization, our 24 points are still living in *C*^3 . (Again,

> as a rough analogy, think of hemi-dodec points on a hemisphere,

> living in 3D.)

> 3. Project the 24 vertices from *C*^3 -> *C*^2 . Now we have 4D

> points we can interpret in *R*^4 . Hallelujah. By the way, these

> 4D points aren’t bound to the 3-sphere (their 4D distance to the

> origin is not necessarily 1).

> 4. Do our typical 4D visualization approach on the 4D points by

> projecting to 3D and 2D, and offering mouse drag controls.

>

> *The result:* a borderline mess, a jumbled looking surface I would not

> want to play a puzzle on. I was disappointed because I had hoped for

> a nice 4D puzzle representation from this study, but at least I

> learned along the way. I think a better option for representing KQ on

> a closed surface is a warped pasting onto the tetrus shape (the shape

> of the sculpture Nan emailed about).

> Here is the applet: Klein Quartic Viewer

> <http://www.gravitation3d.com/magictile/kq_viewer.html>

> Take Care,

> Roice

>

>

>