# 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. – Robert Bryant Mar 16 '15 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. – Henry Segerman Mar 16 '15 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. – Robert Bryant Mar 16 '15 at 23:55
• Yes, thank you Robert! It is a great resource. – Henry Segerman Mar 17 '15 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. – Will Sawin Mar 17 '15 at 18:00