# Finite order automorphisms of complex projective manifolds isotopic to identity

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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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.
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 '11 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 '11 at 16:07