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Consider a smooth projective complex surface $S$ with an automorphism $g:S\to S$. A point $p$ is periodic if it has finite orbit under iterates of $g$.

What are some examples of surface automorphisms $g$ with no periodic points?

For example, $S$ can be an abelian variety and $g$ could be translation by some general point $\tau\in S$. Are there others, say on a rational surface, or a K3 surface?

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  • $\begingroup$ Every automorphism of a projective rationally connected variety over any field has a fixed point (you can prove this by reduction to the case of finite fields, where the automorphism is automatically finite order). So you cannot find examples on projective rational surfaces. $\endgroup$ Apr 14, 2015 at 22:49
  • $\begingroup$ By the "Wood's Hole", Atiyah-Bott, holomorphic Lefschetz fixed point formula, an automorphism $g$ of a (projective, complex) K3 surface has a fixed point so long as $g^*$ does not act on $H^0(S,\omega_S)$ as $-1$. However, even if $g^*$ does act as $-1$, then for $h=g\circ g$, then $h^*$ acts as $+1$, and thus $h=g\circ g$ has fixed points. So there will also be no examples on projective, complex K3 surfaces. $\endgroup$ Apr 14, 2015 at 22:58
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    $\begingroup$ Just to point out one more thing: if your automorphism $g$ fixes an ample invertible sheaf, then the action of $g$ on the complete linear system of any tensor power of the invertible sheaf admits a fixed point. Thus, there is a fixed curve inside your surface. Dynamics on curves are easy to understand, and essentially your curve must be arithmetic genus $1$ if it has no periodic orbits. In particular, that rules out Kobayashi hyperbolic surfaces. $\endgroup$ Apr 14, 2015 at 23:32
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    $\begingroup$ @JasonStarr: Kobayashi hyperbolic surfaces are of general type, so the automorphism group is finite. $\endgroup$
    – naf
    Apr 15, 2015 at 5:14
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    $\begingroup$ Another elementary example is the product of an elliptic curve with any other curve. I would guess that the only other examples are fixed point free quotient of ones as above (as well as general abelian surfaces). $\endgroup$
    – naf
    Apr 15, 2015 at 5:59

2 Answers 2

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This is just the collection of my comments above. First of all, for every automorphism $g$ (finite order or not) of every proper, smooth, separably rationally connected variety $X$ over any field $k$, the fixed scheme $X^g$ is nonempty. So there will be no examples with $X$ a rational surface.

Second, by the holomorphic Lefschetz fixed point theorem, if $h^{1,0}(X) = 0$ or $1$ and if $h^{2,0}(X)$ equals $1$, then $g$ or $g\circ g$ has a fixed point. If $h^{1,0}(X)$ equals $0$ and $h^{2,0}(X)$ equals $0$, then $g$ has a fixed point.

Finally, if $g$ preserves an effective divisor class, then $g$ preserves an effective divisor in that class. Thus the effective divisor is genus $1$. This rules out Kobayashi hyperbolic examples.

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Many examples are given in Friedman-Morgan, Page 22.

EDIT The infinite order situation is discussed in the paper by Oguiso (2012).

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    $\begingroup$ Sorry, I don't see how these examples are relevant - $g$ should have infinite order here. $\endgroup$
    – J.C. Ottem
    Apr 14, 2015 at 22:44
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    $\begingroup$ I think the automorphisms in Oguiso's paper are fixed point free but not aperiodic. $\endgroup$
    – J.C. Ottem
    Apr 15, 2015 at 2:33
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    $\begingroup$ I do not see how the paper of Oguiso addresses the question. Even if $g$ has no fixed point, nonetheless, $g\circ g$ must have a fixed point by the holomorphic Lefschetz fixed point theorem. So even if Oguiso calls this automorphism "free", that does not mean it has no periodic orbits. $\endgroup$ Apr 15, 2015 at 2:33
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    $\begingroup$ @J.C.Ottem: We both made the same comment (nearly) simultaneously. $\endgroup$ Apr 15, 2015 at 2:34

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