Message #1947

From: Roice Nelson <>
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

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
*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
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
  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

Take Care,