On the automorphisms of the Klein Quartic

Question

I am trying to solve a problem from Miranda's book, Algebraic Curves and Riemann Surfaces. On page 84, problem K gives the Klein curve $X$ as a smooth projective plane curve defined by the equation $xy^3+yz^3+zx^3=0$. The problem asks us to show that this Riemann surface, of genus g=3, realizes the Hurwitz bound by finding 168 automorphisms of $X$.

I have found an automorphism subgroup of order 3 (cyclically permuting the variables), and one of order 7 (multiplying the coordinates by appropriate 7th roots of unity), but I just can't figure out how to get a subgroup of order 8, or 4, or 2. Could somebody please give me a hint.

I've spent the last little while reading up on this subject, and most of the discussions involve using a heptagonal tiling etc.. Given where this problem is placed in the book, I can't appeal to that kind of reasoning, so I'm asking for a way to find these automorphisms directly from the defining equation of $X$.

Any help or hint would be appreciated. Even an involution :)

Answer 1

All the automorphisms extend to projective automorphisms of $P^2$. It is natural to look for an involution that is linear in $xyz$-coordinates and represented by a circulant matrix. One matrix that works has rows that are shifts of $[\sin \frac{2 \pi}{7} ,\sin \frac{2 \pi \cdot 4}{7}, \sin \frac{2 \pi \cdot 2}{7}]$.

For more see the MSRI volume The Eightfold Way:The Beauty of Klein's Quartic Curve.

Answer 2

Elkies essay "The Klein Cubic in Number Theory" lists the following symmetry of the Klein cubic: $$ \frac{1}{\sqrt{-7}} \begin{bmatrix} \zeta - \zeta^6 & \zeta^2 - \zeta^5 & \zeta^4 - \zeta^3 \\ \zeta^2 - \zeta^5 & \zeta^4 - \zeta^3 & \zeta - \zeta^6 \\ \zeta^4 - \zeta^3 & \zeta - \zeta^6 & \zeta^2 - \zeta^5 \\ \end{bmatrix}$$ where $\zeta$ is a primitive $7$-th root of unity. Together with the $21$ symmetries you've already found, this generates the full group of $168$.

I'd like to give a conceptual explanation of why this should work, but I don't know if I can; I'll give it some thought.