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The Bolza curve B double covers the Riemann sphere with branching at the vertices of a regular octahedron. An affine model is given by the locus of $y^2=x^5-x$. How does one show that B does not admit an anticonformal fixedpointfree involution?

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Peter and katz, you are right, I deleted my answer. – Alexandre Eremenko Mar 7 '13 at 22:23

Let $\mathbb C(x,y)$ be the function field of the curve $B$. From $y^2=x^5-x$ we obtain that each element in $\mathbb C(x,y)$ has the form $A(x)+yB(x)$ with rational functions $A$ and $B$. Identify the conformal automorphisms of $B$ with the $\mathbb C$-automorphisms of $\mathbb C(x,y)$.

Thus any conformal automorphism of $B$ has the form $(x,y)\mapsto (R(x)+yU(x),V(x)+yS(x))$ with rational functions $R,S,U,V$. Now it is well known that the hyperelliptic involution $(x,y)\mapsto (x,-y)$ of a hyperelliptic curve commutes with all conformal automorphisms. (The reason for that is that $\mathbb C(x)$ is the unique degree $2$ rational subfield of $\mathbb C(x,y)$.) Writing out what it means that the hyperelliptic involution commutes with conformal automorphisms yields $U=V=0$.

An anti-conformal automorphism is the composition of a map as above and complex conjugation, so it has the form $$\sigma: (x,y)\mapsto (R(\bar x),\bar yS(\bar x))$$ Assuming that $\sigma$ is an involution yields the necessary condition $$R(\bar R(x))=x,$$ where $\bar R$ arises from $R$ by complex conjugation of the coefficients.

In particular, $R$ has degree $1$, so it is a linear fractional map.

As $(R(x),yS(x))$ is on the curve when $(x,y)$ is on the curve, we get $$(x^5-x)S(x)^2=R(x)^5-R(x).$$

Looking at degrees and poles shows that either $R(x)=\gamma(x-\alpha)$ or $R(x)=\gamma/(x-\alpha)$ for some $\alpha$ with $\alpha^5=\alpha$ and $\gamma\in\mathbb C$.

In the first case, degrees tell us that $S(x)$ is a constant $\delta$. Upon replacing $x$ by $x+\alpha$ we get $$((x+\alpha)^5-(x+\alpha))\delta^2=\gamma^5 x^5-\gamma x.$$ Comparing at $x^4$ yields $\alpha=0$. So $R(x)=\gamma x$. But then $R(0)=S(0)=0$, and $(0,0)$ is a fixed point of $\sigma$.

In the second case we have $R(x)=\gamma/(x-\alpha)$. This forces $S(x)=\delta/(x-\alpha)^3$ for some $\delta$. Again, replacing $x$ with $x+\alpha$ and clearing denominators yields $\alpha=0$. We obtain $R(x)=\gamma/x$. From $R(\bar R(x))=x$ we get $\gamma/\bar\gamma=1$, so $\gamma$ is real.

Easy calculations yield $\gamma^4=1$, so $\gamma=\pm1$, and $S(x)=\delta/x^3$ with $\delta^2=-\gamma$. So $$ \sigma: (x,y)\mapsto (-\delta^2/\bar x,\delta\bar y/\bar x^3)$$ with $\delta^2=\pm1$.

We compute $$\sigma^2:(x,y)\mapsto (x,-\frac{\delta}{\bar\delta}y).$$

As $\sigma^2=1$, we get $\delta^2=-1$. But then $(1,0)$ is a fixed point of $\sigma$.

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Thanks, Peter, this is tremendous. I wonder if one could write down something simpler. We have an argument at (see section 8), but it is a bit involved. Will keep thinking. – katz Mar 9 '13 at 20:05

Here is a reasoning which uses (as Peter's one) the fact that on every hyperelliptic curve the hyper-elliptic involution $\sigma$ commutes with every conformal self-map (since hyper-elliptic involution is unique).

Suppose $\sigma_1$ is an orientation reversing fixed point free involution of $B$. Then since $\sigma\sigma_1=\sigma_1\sigma$ the action of $\sigma_1$ on $B$ descends to an orientation reversing involution (call it $\sigma_2$) on $\mathbb CP^1=B/\sigma_1$. Clearly $\sigma_2$ permutes $6$ vertices of the octahedron and it does not fix any of them. So $\sigma_2$ is the central symmetry of (octahedral) $\mathbb CP^1$. Now, take on $\mathbb CP^1$ any big circle $S^1$ (invariant under $\sigma_2$) that does not pass through the vertices. Let $S'^1$ be its preimage on $B$. $S'^1$ is a circle the double covers $S^1$ (because $S^1$ splits $6$ vertices on $\mathbb CP^1$ into two groups of $3$ verities). Since $\sigma_2$ makes the half-turn of $S^1$ it follows that $\sigma_1$ makes a quoter turn of $S'^1$. So it is not an involution of $B$. Contradiction.

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