Let $M$ be a connected closed complex manifold. Assume it has an antiholomorphic involution. Must it have an antiholomorphic involution with a fixed point?



Antiholomorphic involution with a fixed point
No. There exist both non-algebraic and projective counterexamples.


1 *Non-algebraic example.* Take a flat Euclidean torus $T^4=M$ and let $Z$ be its twistor space. It has an antiholomorphic involution without fixed points which is central symmetry in all the fibres. I claim that $Z$ doesn't have an anti-holomorphic involution that has a fixed point.


Suppose by contradiction that such $\sigma$ exists. Let $\tilde Z$ be the universal cover of $Z$. Recall that $\tilde Z$ is the complement to a line in $\mathbb CP^3$. Now, $\sigma$ induces an anti-holomorphic involution $\tilde \sigma$ on $\tilde Z=\mathbb CP^3\setminus \mathbb CP^1$. I claim that $\tilde \sigma$ extends a holomporphic self-map of $\mathbb CP^3$. The point is that $\tilde \sigma $ sends any complex line in $\mathbb CP^3\setminus \mathbb CP^1$ to a complex line in $\mathbb CP^3\setminus \mathbb CP^1$ . One can deduce from this that the map $\tilde \sigma$ is induced by a linear (i.e. degree $1$) map from $\mathbb CP^3$ to itself. It remains to check that the standard real involution on $\mathbb CP^3$ that fixes an $\mathbb RP^3$ doesn't commute with the action of $\mathbb Z^4=\pi\_1(T^4)$ on $\mathbb CP^3$, which is not super hard.


2 *Projective example.* To get a projective example one should take a generic quartic curve in $\mathbb CP^2$ defined by a real equation but without real points. A generic such curve doesn't have a real involution that has a fixed point. Otherwise, by taking a composition of such an involution with the real involution we get a non-trivial holomorphic automorphism (which doesn't exist on a generic quartic)


