6
$\begingroup$

Here is a rather naïve question: to apply the classical holomorphic Lefschetz theorem, you often work with biholomorphisms of a complex compact manifold whose fixed points are non degenerate (i.e. $1$ is not an eigenvalue of the differential) even if work of O'Brian http://www.jstor.org/stable/1998512 allows to get rid of the degeneracy assumption and compute explicitly the multiplicities at the fixed points, provided they remain isolated.

However, I have encountered no interesting examples where some fixed point are isolated and degenerate. For instance, this condition implies (using Bochner's linearization) that the automorphism is not of finite order.

Ideally I would like to construct such an exemple with the manifold $X$ satisfying the additional assumptions $h^0(X, \, T_X)=h^1(X, \, T_X)=0$. A finite product of $\mathbb{P}^2$ blown up at 4 generic points seems a reasonable candidate, but I'm not sure this is the simplest path to start.

$\endgroup$
  • 1
    $\begingroup$ If you want a surface with $h^0(X, \, T_X)=h^1(X, \, T_X)=0$ you can take any infinitesimally rigid surface of general type. For instance, ball quotients and Beauville surfaces do the job. $\endgroup$ – Francesco Polizzi Apr 3 '17 at 6:02
  • 1
    $\begingroup$ Unfortunately, the group of automorphisms of a surface of general type is always finite, in particular there are no automorphisms of infinite order. So my previous comments, although correct, cannot be used in order to find the example you are looking for. $\endgroup$ – Francesco Polizzi Apr 3 '17 at 6:09
3
$\begingroup$

I think the surface constructed in Lemma 4 of https://arxiv.org/abs/1609.06391 should do what you want. No doubt there is a simpler example, so hopefully someone else will chime in.

It's a rational surface $S$, constructed by blowing up $15$ (carefully chosen) points in $\mathbb P^2$. There is a smooth rational curve $C \subset S$ and an automorphism $\phi : S \to S$ which fixes $C$ and restricts to $C$ as the automorphism $z \mapsto z+1$ in suitable coordinates. In particular, there is a unique fixed point $p$ on $C$, given by $\infty$ in these coordinates. This map acts trivially on the tangent space $T_p C \subset T_p S$, which gives the $1$-eigenvector you are looking for.

I think this should have all the properties you want. One should check there is not some curve of fixed points through $p$, but I am pretty sure of it. (Note that $C$ is not a curve of fixed points: although it's mapped to itself, the points on $C$ move around, and only $p$ is fixed.) The automorphism is indeed of positive entropy, hence not finite order, as you note.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thanks! However, you don't have $\mathrm{h}^1(S, \mathrm{T}S)=0$. $\endgroup$ – Julien Grivaux Apr 2 '17 at 22:54
  • 1
    $\begingroup$ (Posting comment as an answer, since I forgot to sign in the first time. Feel free to just delete the original answer, mods -- it missed a condition and isn't really relevant. - "JDL") Oops, you are of course correct! Sorry. I guess you could also look at rigid CY3's, but I think I only know one with an infinite order automorphism, and it doesn't appear to work. $\endgroup$ – user106895 Apr 3 '17 at 2:16
  • 2
    $\begingroup$ @JulienGrivaux, user106895 / JDL to notify moderators of this site about an issue with a post, please click "flag", not "edit". $\endgroup$ – j.c. Apr 3 '17 at 11:33
  • $\begingroup$ @j.c. thanks I did it after your answer, but it didn't change anything... $\endgroup$ – Julien Grivaux Apr 4 '17 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.