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I am wondering about the following problem: for which (say smooth, complex, connected) algebraic varieties $X$ does the statement any regular map $X\to X$ has a fixed point hold? MathSciNet search does not reveal anything in this topic.

This is true for $\mathbb{P}^n$ (because its cohomology is $\mathbb{Z}$ in even dimensions and $0$ otherwise, and the pullback of an effective cycle is effective, so all summands in the Lefschetz fixed point formula are nonnegative, and the 0-th is positive -- is this a correct argument?). Is it true for varieties with cohomology generated by algebraic cycles (i.e. $h^{p,q}(X)=0$ unless $p=q$ and satisfying Hodge conjecture), for example for Grassmannians, toric varieties, etc.? This is not at all clear that the traces of $f$ on cohomology will be nonnegative.

Probably you have lots of counterexamples. What about positive results?

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    $\begingroup$ Footnote: "MathSciNet search does not reveal anything in this topic" is not surprising, since the search function of MathSciNet has built-in limitations. The actual content of papers cannot be searched, so results depend on what happens to be written in reviews along with titles of papers and subject numbers. No one has yet invented an all-knowing search engine to scan all of the expanding mathematics literature for a particular thought. So humans may still have a minor role to play in answering questions. (For a while anyway.) $\endgroup$
    – Jim Humphreys
    Commented Mar 16, 2010 at 22:17

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By demand I expand a little on my answer. The holomorphic Lefschetz fixed point formula (aka the Woods-Hole formula) considers an endomorphism $f\colon M \to M$ of a smooth and compact complex manifold $M$ (or proper smooth algebraic variety) with only isolated fixed points which are also assumed to be non-degenerate (i.e., the tangent map of $f$ at a fixed point does not have eigenvalue $1$. Then the alternating trace of the action of $f$ on $H^*(M,\mathcal O_M)$ is equal to a sum over the fixed points $p$ of $1/det(1-T_p(f)$. In particular if there are no fixed points the alternating trace is equal to $0$. However, in the case when also $H^i(M,\mathcal O_M)=0$ for $i>0$ then the alternating trace is equal to $1$ so the assumption that there are no fixed points gives a contradiction. Note, that the dimension of $H^i(M,\mathcal O_M)$ is just $h^{0,i}$ so that an assumption that the Hodge numbers vanish off the diagonal gives the required vanishing by a good margin. Furthermore, the vanishing of just $h^{0,i}$ for $i>0$ is much much weaker, it is for instance a birational condition whereas blowing up a smooth curve of genus $>0$ in a variety of dimension at least 3 always give off diagonal Hodge numbers.

As for the question of whether a birational involution of $\mathbb C^n$ always has a fixed point this seems trickier. It is true that the involution can be made to act regularly on a smooth and proper model and hence by the above has a fixed point. It is not clear however that the fixed point will map to a point of $\mathbb C^n$ as $\mathbb C^n$ is not proper.

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    $\begingroup$ So, by Kollar-Miyaoka-Mori Thm, the Fixed Point Property holds for all Fano varieties. Another wide class is that of toric varieties. This gives a broad variety of examples. $\endgroup$
    – Piotr Achinger
    Commented Mar 16, 2010 at 22:10
  • $\begingroup$ The case of Fano varieties is simpler than the KMM theorem, by the Kodaira vanishing theorem $h^{p,0}=0$ for $p<\dim M$ and then Serre duality gives $h^{0,p}=0$ for $p>0$. $\endgroup$
    – Torsten Ekedahl
    Commented Mar 17, 2010 at 9:25
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Here's a positive result: A rationally connected variety over an algebraically closed field is a smooth projective variety such that through any two points, there's a rational curve. One can prove that any automorphism of a rationally connected variety has a fixed point. This is proven using a cohomological argument and I first saw it in Harris' 'Lectures on rationally connected varieties' found here

http://www.mat.uab.es/~kock/RLN/rcv.pdf

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  • $\begingroup$ Are you talking about Homework 1.25? I find this confusing. It says that rationally connected varieties have no higher cohomology and that their Euler characteristic is 1. But the Euler characteristic of CP^n is n+1. $\endgroup$
    – Pete L. Clark
    Commented Mar 16, 2010 at 19:49
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    $\begingroup$ The exercise is no doubt referring to the higher cohomology of the structure sheaf. The holomorphic Lefschetz fixed point formula shows that if its higher cohomology vanishes then any endomorphism has fixed points. This also answers the question as even just the vanishing of $h^{0,p}$ for $p>0$ ensures the existence of a fixed point. $\endgroup$
    – Torsten Ekedahl
    Commented Mar 16, 2010 at 20:14
  • $\begingroup$ Torsten: thank you, your comment really answers my question! Perhaps it would be better to post it as an answer, so I could mark it green... $\endgroup$
    – Piotr Achinger
    Commented Mar 16, 2010 at 20:49
  • $\begingroup$ Torsten: could you elaborate more? And do you have any thought on the applicability of your idea to my related question mathoverflow.net/questions/18347/… ? $\endgroup$
    – AndreA
    Commented Mar 16, 2010 at 21:01
  • $\begingroup$ By the way Harris' exercise 1.25 is essentially Corollary 4.18 in Debarre's book 'Higher dimensional algebraic geometry'. $\endgroup$
    – Frank
    Commented Mar 17, 2010 at 12:21

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