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 $y^2=x^5x$. How does one show that B does not admit an anticonformal fixedpointfree involution?

Let $\mathbb C(x,y)$ be the function field of the curve $B$. From $y^2=x^5x$ 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 anticonformal 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 $$(x^5x)S(x)^2=R(x)^5R(x).$$ 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 $\sigma$. 


Here is a reasoning which uses (as Peter's one) the fact that on every hyperelliptic curve the hyperelliptic involution $\sigma$ commutes with every conformal selfmap (since hyperelliptic 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 halfturn of $S^1$ it follows that $\sigma_1$ makes a quoter turn of $S'^1$. So it is not an involution of $B$. Contradiction. 

