Fixed Point Property in Algebraic Geometry - MathOverflow

Fixed Point Property in Algebraic Geometry

2010-03-16T18:17:29Z

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?

Answer by Frank for Fixed Point Property in Algebraic Geometry

2010-03-16T19:27:42Z

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

Answer by Torsten Ekedahl for Fixed Point Property in Algebraic Geometry

2010-03-16T21:54:17Z

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.