# Map of the Klein quartic from $CP^2$ to $R^3$

The Klein quartic $$\mathcal{Q}$$ is cut out of $$\mathbb{CP}^2$$ by the homogeneous equation $$x^3 y + y^3 z + z^3 x = 0.$$ It has 168 orientation preserving automorphisms and includes several copies of the tetrahedral group (with twelve elements).

Is there a nice way to take the points of $$\mathcal{Q}$$ in $$\mathbb{CP}^2$$, map them to $$\mathbb{R}^3$$ (preserving one of the tetrahedral symmetry groups) and so produce an embedded, compact, genus three surface?

There are already a number of models of the Klein quartic in $$\mathbb{R}^3$$. So far we've found the two by Joe Christy and Greg Egan (see this webpage by John Baez) and also a version by Carlo Sequin. As far as we (Saul Schleimer and I) can tell, these are all "topological" models and not obtained by mapping from $$\mathcal{Q} \subset \mathbb{CP}^2$$ in some sensible way.

• If you can get your hands on it, you should have a look at "The Eight-fold Way: The Beauty of Klein's Quartic Curve", edited by Silvio Levy, which has some very beautiful pictures of the Helaman Ferguson sculpture, as well as solid discussions of the mathematics. Mar 16, 2015 at 20:51
• We don't have the physical book, but it is online: (library.msri.org/books/Book35/contents.html). We have looked at Elkies' article somewhat closely, but couldn't work out a suitable map. In trying to figure out how Helaman Ferguson made the sculpture we also looked at his article with Claire Ferguson. In this he says that he: "carved it in a qualitative free form process known as direct carving, paying attention to the combinatorics and topology but not rigid or measured geometry." So the Eight-fold way" is also a topological model. Mar 16, 2015 at 21:05
• Yes, I knew that the Eight-fold way was not an algebraic model, but I thought that you would find it useful or at least interesting if you didn't know it already. Mar 16, 2015 at 23:55
• Yes, thank you Robert! It is a great resource. Mar 17, 2015 at 0:30
• The obvious functions to try are polynomials in $x, \overline{x}, y, \overline{y}, z, \overline{z}$ divided by powers of $\sqrt{ x \overline{x} + y \overline{y} + z \overline{z}}$. You could try the first few polynomials that are invariant by $A_4$ and see if they give you an embedding. Mar 17, 2015 at 18:00