# On the automorphisms of the Klein Quartic

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 :)

• My bad! I'll delete my comment. – rfauffar Jul 26 '13 at 2:01
• Have you tried looking at linear automorphisms? That is, automorphisms of $\mathbb{P}^2$ that descend to the curve? – rfauffar Jul 26 '13 at 2:02
• I do not know how to see the full automorphism group of Klein quartic directly from the equation. I think it might be difficult (or probably unenlightening). We can however try to see it by using uniformization theory. More concretely, the Klein quartic is obtained by taking an smooth punctured Riemann surface (call it $X$) and then filling in the punctures. The punctured Riemann surface is $X$ is just the quotient of the upper half plane $\mathbb{H}$ by a congruent subgroup of $SL_{2}(Z)$. – DBS Jul 26 '13 at 5:56
• Such things are called Modular curves. In your case the Klein Quartic is the curve $X(7)$. some relevant info here en.wikipedia.org/wiki/Modular_curve – DBS Jul 26 '13 at 5:58
• @DBS Thank you, but I am trying to find these automorphisms without recourse to uniformization if possible, because the problem in the book indicates that I can find it directly. – FGerard Jul 29 '13 at 16:33

## 1 Answer

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.

• Thank you but I am still confused. First, is it obvious that an automorphism of a projective curve extends to an automorphism of $P^2$. Second, isn't the transformation you suggested an order seven one? And how do I see that it leaves the Quartic invariant? – FGerard Jul 29 '13 at 16:31
• I have been reading what I can of The Eightfold Way, and that perspective is starting to make an impression on me. One thing that confuses me is when they say that for the tiling of K by 24 regular heptagons, a symmetry of any heptagon extends to the whole surface. Is there an easy way to see that? Thanks! – FGerard Jul 29 '13 at 16:36