# Message #1947

From: Roice Nelson <roice3@gmail.com>

Subject: Visualizing the surface of Klein’s Quartic in 4D

Date: Sat, 03 Dec 2011 17:56:48 -0600

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.

- 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. - 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.) - 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). - 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