Math-1473 · mathematics · Topology

Math-1473

Kirby Problem 4.84

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Statement

Definitions and mathematical statement

Let $X$ be a closed oriented smooth $4$-manifold. Its smooth mapping class group is $\mathcal M(X)=\pi_0\operatorname{Diff}^+(X)$, where $\operatorname{Diff}^+(X)$ has the $C^\infty$ topology and consists of orientation-preserving diffeomorphisms. Let $q:\operatorname{Diff}^+(X)\to\mathcal M(X)$ send a diffeomorphism to its smooth-isotopy class.

A finite subgroup $G\le\mathcal M(X)$ is realized by a smooth finite group action if there is a homomorphism \[ \rho:G\longrightarrow\operatorname{Diff}^+(X) \quad\text{such that}\quad q\circ\rho:G\hookrightarrow\mathcal M(X) \] is the given inclusion. This equality makes $\rho$ injective, so the action is effective. Each group relation must hold exactly for the chosen diffeomorphisms, not just up to isotopy.

Does there exist a closed oriented smooth $X$ for which every finite subgroup of $\mathcal M(X)$ is realized in this sense? The choice of action may vary with the finite subgroup; one section of $q$ on the entire mapping class group is not required.

A finite-order mapping class need not a priori have a representative of that same finite order. The problem asks for simultaneous representatives of all elements of each finite subgroup respecting multiplication. The quantifier on $X$ is existential, while the quantifier on finite subgroups of its mapping class group is universal. There is no simple-connectivity assumption.

Short English statement

Is there a closed oriented smooth four-manifold on which every finite subgroup of the smooth mapping class group is realized by an actual smooth group action?

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