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Let $X$ be a smooth projective variety over $\mathbb{C}$. Consider the Kahler structure $(J,g,\omega)$ on $X$ induced by the Fubini-Study metric. Let $Symp(X,\omega)$ be the group of symplectomorphisms of $(X,\omega)$.

QUESTION.

Is there a finite order symplectomorphism $f \in Symp(M,\omega)$ which is not conjugate (in $Symp(X,\omega)$) to an algebraic automorphism ?

I am just looking for one example, preferably such that the dimension of $X$ is as small as possible.

EDIT

How about if we require $f$ to have no fixed points?

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    $\begingroup$ You probably ask whether there exists $X$ with this property? currently it sounds like you're asking whether this holds for all $X$. $\endgroup$
    – YCor
    Commented Oct 15, 2016 at 0:54
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    $\begingroup$ I guess you mean any smooth holomorphic symplectomorphism? Any holomorphic one is algebraic by Chow's theorem. Next, conjugation of a finite order algebraic symplectomorphism by an arbitrary smooth symplectomorphism will probably very often work. Do you rather want a finite order element not conjugate (in the smooth symplectomorphism group) to any algebraic automorphism? $\endgroup$
    – YCor
    Commented Oct 15, 2016 at 0:57
  • $\begingroup$ Thanks for your comments. To be honest, conjugating an automorphism by a general symlectomorphism did not occour to me. I guess this generates a huge (In fact infinite dimensional) amount of examples for my original question - so thanks. I will edit my question. $\endgroup$
    – Nick L
    Commented Oct 15, 2016 at 9:24
  • $\begingroup$ You only considered the last of my comments. $\endgroup$
    – YCor
    Commented Oct 15, 2016 at 17:27
  • $\begingroup$ For the fixed-point-free additional question, one is tempted to take aglearner's example $(C,u)$, pick another elliptic curve $C'$ with a translation $t$ of order 4, and consider the fixed-point-free order 4 symplectomorphism $u\times t$ of $C\times C'$. It sound unlikely that it's conjugated to an automorphism, maybe at least removing some exceptional bad choices for $C'$. $\endgroup$
    – YCor
    Commented Oct 16, 2016 at 23:46

1 Answer 1

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Here is one example - take an elliptic curve without an automorphism of order $4$ with a fixed point. Note at the same time that any torus $T^2$ with an area form has an area preserving automorphism of order $4$ with a fixed point.

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  • $\begingroup$ Isn't adding a torsion point of order 4 on this elliptic curve an automorphism of order 4? $\endgroup$
    – Victor Protsak
    Commented Oct 16, 2016 at 2:56
  • $\begingroup$ @VictorProtsak you're right but it's fine: the translation of order 4 has no fixed point, while the automorphism of order 4 is meant to be a rotation fixing the 0 element, and "Without automorphism of order 4" refers to automorphisms fixing 0. $\endgroup$
    – YCor
    Commented Oct 16, 2016 at 3:44
  • $\begingroup$ @YCor: I know. My point was that under the natural interpretation of the word "automorphism" in the context of this question, every elliptic curve has an automorphism of order 4, so the argument is flawed. $\endgroup$
    – Victor Protsak
    Commented Oct 16, 2016 at 7:36
  • $\begingroup$ Victor, of course you are right, I was thinking of an authomorphism with a fixed point, I corrected the answer accordingly. $\endgroup$
    – aglearner
    Commented Oct 16, 2016 at 9:30
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    $\begingroup$ Allen, unfortunately your example does not work, the situation is a bit more subtle. The problem here is the following: no Kahler form on $F_2$ is symplectomorphic to the symplectic form $w$ on $F_0$ such that the areas of both $\mathbb P^1$ is the same. The point here is that the class of such form $[w]$ on $F_0$ is proportional to $c_1(F_0)$. On the other hand $c_1(F_2)$ vanishes on the exceptional $-2$ curve. But any symplectomorphims send $c_1$ to $c_1$. $\endgroup$
    – aglearner
    Commented Oct 16, 2016 at 15:39

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