# Is there anything special about the Riemann surface $y^2 = x(x^{10}+11x^5-1)$?

I stumbled upon the fact that the Bolza surface can be obtained as the locus of the equation,

$y^2 = x^5-x$

Its automorphism group has the highest order for genus $2$, namely $48$. I recognized $x^5-x$ as a polynomial invariant of the octahedron. (In fact, the Bolza surface is connected to the octahedron.)

If we use the analogous polynomial invariant of the icosahedron, then does the genus 5 surface,

$y^2 = x(x^{10}+11x^5-1)$

have special properties? How close does the order of its automorphism group get to the bound $84(g-1)$? (For $g = 5$, this would be $336$.)

POSTSCRIPT:

My thanks to Noam Elkies for the highly detailed answer below. The background to this question is an identity I found involving $x^{10}+11x^5-1$. Define,

$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$

and,

$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$

then,

$w^5-10aw^3+45a^2w-a^2 = 0$

for arbitrary r. This in fact is the Brioschi quintic form which the general quintic can be reduced into. Two of the polynomials are easily recognizable as icosahedral invariants, while $r^6+2r^5-5r^4-5r^2-2r+1$ is a polynomial invariant for the octahedron.

Other than in formulas using Ramanujan's continued fractions, I wondered where else those polynomials appear. Since the Bolza surface involved an invariant of the octahedron, it was reasonable to consider if using the corresponding one for the icosahedron would also be special. As Elkies wonderfully showed, it turns out that it is.

• It seems more likely to me that the Jacobian of this curve has extra endomorphisms than that the curve itself has a lot of extra automorphisms. – JSE Feb 29 '12 at 13:42
• The maximum number of automorphisms for a curve of genus 5 is 192. It is easy to show there is no automorphism of order 7, by looking at the map onto the putative quotient by such an automorphism. – roy smith Feb 29 '12 at 21:10
• I ventured to change $11x$ to $11x^5$ since that's what's needed to get the icosahedral polynomial. – Noam D. Elkies Mar 1 '12 at 5:58
• Oh, goodness, how could I have made that typo? Yes, $x^{10}+11x^5-1$ was what I had in mind. Thanks, Prof. Elkies! – Tito Piezas III Mar 1 '12 at 14:28

## 2 Answers

Yes, this Riemann surface, call it $C: y^2 = x^{11}-11x^6-x$, is quite special: not only does it have the maximal number of automorphisms for a hyperelliptic surface of genus $5$, but it is a modular curve in at least two ways, both of which exhibit its full automorphism group.

One is a classical (elliptic) modular curve of level $10$, intermediate between $X(5)$ and $X(10)$, with $[C:X(5)] = 2$ (the hyperelliptic map) and $[X(10):C] = 3$ (a cyclic cover); this modular curve parametrizes elliptic curves $E$ with full level-$5$ structure and odd ${\rm Gal}(E)$, or equivalently full level-$5$ structure and square $j(E)-12^3$. Explicitly, $E$ has Weierstrass equation $Y^2 = X^3 - A(x)X/48 + B(x)/864$ where $A(x) = x^{20} + 228x^{15} + 494x^{10} - 228x^5 + 1$ and $$x^{30} - 522x^{25} - 10005x^{20} - 10005x^{10} + 522x^5 + 1$$ are polynomials with roots at the $20$- and $30$-point orbits of $A_5$. We have $A^3 - B^2 = 12^3 (x^{11}-11x^6-x)^5$, so $j - 12^3 = B^2/(x^{11}-11x^6-x)^5$. The corresponding congruence subgroup $\Gamma$ of ${\rm SL}_2({\bf Z})$ is the index-$2$ subgroup of $\Gamma(5)$ consisting of matrices that reduce mod $2$ to the index-$2$ subgroup of ${\rm SL}_2({\bf Z}/2{\bf Z})$, with $[\Gamma : \Gamma(10)] = 3$. This $\Gamma$ is normal in ${\rm SL}_2({\bf Z})$, and the quotient group is ${\rm Aut}(C)$.

Another modular approach to $C$ is via the $(2,3,10)$ triangle group, call it $G^*$, which appears in class VIII of the nineteen commensurability classes tabulated in

Takeuchi, K.: Commensurability classes of arithmetic triangle groups, J. Fac. Sci. Univ. Tokyo 24 (1977), 201-212.

According to Takeuchi's table, $G^*$ is the normalizer of the unit-norm group $G_1$ of a maximal order in a quaternion algebra over ${\bf Q}(\sqrt 5)$ ramified over one real place and the prime $(\sqrt 5)$.
Moreover $G_1$ is the $(3,3,5)$ triangle group, contained in $G^*$ with index $2$. Let $G_5$ be the normal subgroup of $G_1$ consisting of units congruent to $1 \bmod (\sqrt 5)$. Then $G^*/G_5 \cong \lbrace \pm 1 \rbrace \times A_5$, and the quotient of the upper half plane $\cal H$ by $G_5$ has genus $5$, so must be our $C$. Moreover, ${\cal H} / G_5$ has no elliptic points, so this identifies the image of the fundamental group $\pi_1(C)$ in ${\rm Aut}{\cal H} = {\rm SL}_2({\bf R})$ with an arithmetic congruence group.

P.S. Roy Smith already noted that if we allow also non-hyperelliptic Riemann surfaces then the maximal number of automorphisms for genus $5$ is not $120$ but $192$. An explicit model for a Riemann surface $S$ with $192$ automorphisms is the intersection of three quadrics $$y^2 = x_0 x_1, \phantom{and} {y'}^2 = x_0^2 - x_1^2, \phantom{and} {y''}^2 = x_0^2 + x_1^2$$ in ${\bf P}^4$. Then $(x_0:x_1:y:y':y'') \mapsto (x_0:x_1)$ gives a normal cover $S \rightarrow {\bf P}^1$ with Galois group $N = ({\bf Z}/2{\bf Z})^3$ acting by arbitrary sign changes on $y,y',y''$, ramified above the vertices of a regular octahedron, with each of $x_0 x_1, x_0^2 - x_1^2, x_0^2 + x_1^2$ vanishing on an opposite pair of vertices. I claim that there is an exact sequence $1 \rightarrow N \rightarrow {\rm Aut}(S) \rightarrow S_4 \rightarrow 1$, so in particular $\#({\rm Aut}(S)) = 2^3 4! = 192$. Indeed let $G$ be the subgroup of ${\rm Aut}(S)$ that stabilizes the span of $\lbrace x_0, x_1 \rbrace$. Then $G$ contains $N$ as the kernel of a homomorphism $G \rightarrow {\rm Aut}({\bf P}^1)$ given by the action on $(x_0:x_1)$. The image is contained in the group $S_4$ of rotations of the octahedron, and indeed equals $S_4$ because any rotation permutes the three opposite pairs of vertices and thus lifts to ${\rm Aut}(S)$. Therefore ${\rm Aut}(S)$ contains a group $G$ of order $2^3 4! = 192$, and by the Hurwitz bound this must be the full group of automorphisms, QED

• Thanks so much, Prof. Elkies! I added a postscript to my original question giving the background why I asked about that particular surface. – Tito Piezas III Mar 1 '12 at 15:36

Your curve is hyperelliptic.

If $X_g$ is a hyperelliptic curve of genus $g$, then $\textrm{Aut}(X_g)$ is a central extension of degree $2$ of one of the groups $$\mathbb{Z}_n, D_n, A_4, S_4, A_5,$$

see this paper.

In the case of Bolza curve the polynomial $x^5-x$ is invariant by the automorphism group of the octahedron, which is $S_4$. In fact, the automorphism group of the Bolza curve is a central extension of $S_4$ by the group of order $2$ generated by the hyperelliptic involution, hence it has order $2 \cdot |S_4|=48$.

Regarding your curve, the polynomial at the right hand side is invariant by the automorphism group of the icosahedron, which is $A_5$. Then the automorphism group is a central extension of $A_5$ by the hyperelliptic involution, hence it has order $2 \cdot |A_5|= 120$.

• The link was broken. I fixed it now. – Francesco Polizzi Feb 29 '12 at 13:22
• Thanks, Francesco. I see in the paper that section 4.3 and 4.4 deals with polynomial invariants for the octahedron, while 4.5 is for the icosahedral ones. – Tito Piezas III Feb 29 '12 at 14:20
• Indeed, looking at Table 1 in the paper it seems that, since $\delta$ must be an integer, the only possibility in your case ($g=5$ and $A_5$-symmetry) is $\delta=(g-5)/30=0$. Then $\textrm{Aut}(G)=\mathbb{Z}_2 \times A_5$. – Francesco Polizzi Feb 29 '12 at 15:21