## Finite order automorphisms of complex projective manifolds isotopic to identity

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Question. Let $V$ be a complex projective manifold of general type (we can even assume that the canonical bundle of $V$ is ample). Suppose $\varphi: V\to V$ is a non-identical automorphism. Can $\varphi$ be isotopic to the identity map (i.e. $\varphi\in Diff_0(V)$)?

I hope the answer is no, and this can be easily proven when $K_V$ is very ample.

More generally what restrictions are known on smooth manifolds that admit self-diffeos of finite order that are isotopic to identity?

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## 1 Answer

The answer to your question is unknown already for surfaces $S$ of general type.

Note that, if $S$ is simply connected, by a result of Quinn (see "Isotopy of 4-manifolds", Journal of Differential Geometry 1986) every automorphism acting trivially on rational cohomology must be topologically isotopic to the identity.

At any rate, it seems that people conjecture that the answer to your question is no for simply connected surfaces of general type. See Catanese's paper "A Superficial Working Guide to Deformations and Moduli" (arXiv:1106.1368), Section 1.4 for further details. In this paper, complex manifolds which do not admit non-trivial automorphisms isotopic to the identity are called rigidified.

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Francesco, huge thaks! This interesting, I am really surprised that this is yet unknown for surfaces of general type. – Dmitri Sep 4 2011 at 12:52
Dear Dmitri, you are welcome. I was really surprised too, when I read Catanese's survey. This seems really a very basic and interesting question. – Francesco Polizzi Sep 4 2011 at 13:05
Francesco, Frank Quinn (J. Diff. Geom. 1986) shows that for a simply connected compact 4-manifold, $\pi_0$ of the group of homeomorphisms is the group of automorphisms of the intersection form. Is this what you are referring to? If so, isn't differentiable isotopy a different story? – Tim Perutz Sep 4 2011 at 15:33
@Tim: let $V$ be a simply-connected compact $4$-manifold and $f \colon V \to V$ an automorphism acting trivially in cohomology. In particular $f$ acts trivially on the intersection form of $H^2$. Therefore by Quinn's result (Theorem 1.1), it follows that $f$ must be in the identity component of $\pi_0 \textrm{Top}(M)$, i.e. $f$ is isotopic to the identity. I'm missing something? – Francesco Polizzi Sep 4 2011 at 16:07

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Francesco, Dmitri's question asked about differentiable isotopy, while Quinn's result is about topological isotopy. In dimension 4, there's a gulf between plain topology and differential topology... – Tim Perutz Sep 4 2011 at 21:18
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