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Let $n>1$. Which smooth manifolds admit a diffeomorphism $f$ of order $n$?

For a closed orientable surface $S_g$ of genus $g$ and $n=2$ the answer is in the affirmative, since $S_g$ can be embedded in $\mathbb{R}^3$ in a symmetric way. A similar argument gives a positive answer for $n=3$: $S_g$ can be embedded in $\mathbb{R}^3$ with rotational symmetry of order 3. The first problem I've encountered in case of $S_g$ is with $g=2$ and $n=4$: all my candidates for $f$ turned out to have order $2$. And I have no idea as to what happens for $g=2$ and $n=5$.

In the case of a general manifold $M$, the existence of a nontrivial global flow $\phi_t$ such that $\phi_0=\phi_1$ would give a positive answer for every $n$, but it seems to be a much stronger condition.

Is a general answer to this question known? If not, maybe there are some simple examples of manifolds without diffeomorphisms of given order?

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Please excuse my ignorance but what is a diffeomorphism order? – Andrey Rekalo Jul 3 '10 at 16:26
The order of a difeomorphism is the number of elements in the cyclic group it generates. – Łukasz Garncarek Jul 3 '10 at 16:32
Thank you, Łukasz. – Andrey Rekalo Jul 3 '10 at 16:32
Dear Andrey Rekalo: The definition given by Łukasz is used for any element of any group. (This has nothing to do with diffeomorphisms.) – Pierre-Yves Gaillard Jul 3 '10 at 18:14
The surface of genus 2 has a diffeomorphism of order 5: Imagine a "carousel" of two parallel closed decagonal 2-disks in 3-space that are connected by 5 equally spaced half-twisted strips between their alternate corresponding edges. This surface has just one boundary component, which we cap off with a 2-disk. It's then easy to check that this is an orientable surface of genus 2, and that rotating the carousel by 2π/5 extends to the 2-disk cap, thereby effecting a diffeomorphism of order 5. – Daniel Asimov Dec 21 '15 at 22:49
up vote 12 down vote accepted

The Nielsen Realisation Problem asks when a (finite) subgroup of the mapping class group (the group of isotopy classes of diffeomorphisms) of a surface can be realised as a group of diffeomorphisms. Kerckhoff proved in the 80s that every finite subgroup of the mapping class group can be realised. (For infinite subgroups, there are various known obstructions, such as the Miller-Morita-Mumford characteristic classes.) Thus, Kerckhoff's theorem implies that a surface admits a diffeomorphism of order n if its mapping class group has an element of order n. Conversely, one can show that any diffeomorphism of a surface must have infinite order if it is isotopic to the identity, every diffeomorphism of order n gives an order n mapping class group element.

If you have a diffeomorphism of finite order on a surface then you can find a complex structure (or a Riemannian metric or a symplectic structure or a conformal structure) for which the diffeomorphism is an automorphism/isometry. This is accomplished by choosing an arbitrary metric and then averaging over all translates of it by powers of the diffeomorphism.

So the point of all this is that in dimension 2 finding an order n diffeomorphism of a genus g surface is the same as finding a complex curve with an order n isometry, or equivalently, a Z/n orbifold point in the moduli space. As Sam and Robin alluded to, there is a bound on the order of n relative to g. Hurwitz's theorem states that the order of the automorphism group of a genus g curve is less than or equal to 84(g−1). There are various other theorems that tell you about what sorts of finite subgroups you can find in mapping class groups.

In higher dimensions, it's harder to give a useful answer. If your manifold has a circle action then you are done because $S^1$ contains Z/n for any n. But there are plenty of manifolds around which do not admit circle actions, such as K3 surfaces. The $\widehat{A}$-genus is an obstruction to admitting a circle action. Some nice things are known about the finite groups of automorphisms of K3 surfaces. In fact, I think they are pretty much completely classified into a finite list.

In general, by averaging over translates of a metric, you can still assume that a given finite order diffeomorphism acts by isometries for some metric. Generally, the isometry group of a compact Riemannian manifold will be a finite dimensional compact Lie group (I think this is a theorem of Yau).

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Actually, you don't need Kerckhoff's theorem for your answer, Nielsen solved the cyclic case: – Ian Agol Jul 4 '10 at 2:47
@Agol: Thanks, yes - I knew that Nielsen had already done the cyclic case. I was trying to simplify the discussion a bit by just pointing to the definitive paper on the topic, but perhaps that was poor scholarship on my part. – Jeffrey Giansiracusa Jul 4 '10 at 19:23
@ Jeffrey: comment abourt (I think this is a theorem of Yau). I don't think so. It is probably that Elie Cartan already know this result. – Entaou Feb 24 '14 at 17:24

One can answer the question "Does $M$ admit a diffeomorphism of order $n$" for any compact 3-manifold using the geometrization theorem, but there is no succinct answer.

Consider a closed 3-manifold which has an order $n$ diffeomorphism. The quotient is an orbifold (possibly a manifold if the cyclic group action is free). Thus, one wants to compute for orbifolds $\mathcal{O}^3$ the homomorphisms $\pi_1(\mathcal{O})\to \mathbb{Z}/n\mathbb{Z}$ such that the fixed point torsion in $\pi_1(\mathcal{O})$ maps non-trivially. This is possible using a homological computation. In principle, one may enumerate orbifolds, and then determine for each orbifold the $n$-fold cyclic covers which are manifolds, to produce all compact 3-manifold which admit an order $n$ diffeomorphism.

Conversely, given a 3-manifold $M$, one may determine if it admits an $n$-fold symmetry. By the equivariant sphere theorem, one may find a collection of reducing spheres which are equivariant under the action. Then by the geometrization theorem, one may find an equivariant decomposition along tori into geometric pieces, such that the action preserves the homogeneous metric on each piece. In principle, then, one could collate this information to determine if $M$ admits a symmetry of order $n$. See also Dinkelbach-Leeb.

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Here's an example with $g=2$ and $n=5$. Consider the hyperelliptic curve defined by the equation $$y^2=x^5-1$$ or to be more precise the corresponding desingularized projective curve. Now this is a Riemann surface of genus two and has an automorphism $(x,y)\mapsto(\zeta x,y)$ where $\zeta$ is a primitive fifth root of unity.

Because of Hurwitz's theorem this kind of construction via Riemann surface automorphisms cannot work when $n$ is large compared to $g$.


Here's another example. Consider the Riemann surface corresponding to the curve $$y^2=x^5-x.$$ It has an order $8$ autmorphism $(x,y)\mapsto(\eta^2 x,\eta y)$ where $\eta$ is a primitive eighth root of unity (and so this genus two surface also has an automorphism of order $4$).

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From a more combinatorial perspective: Let $P_{n}$ be a regular polygon with $n$ sides. Gluing opposite sides of $P$ by translation gives a surface of genus $g$ when $n = 4g$ or $4g+2$. (The most famous examples are when $P$ is a square or hexagon, in which cases this construction gives a two-torus.) It will follow that $S_g$ has diffeomorphisms of order $4g$ and $4g + 2$, respectively. I believe that these generate the largest finite cyclic subgroups of $\rm{Diffeo}(S_g)$.

One dimension higher, Mostow rigidity implies that any finite volume hyperbolic three-manifold $M$ has a finite isometry group. This should bound the size of any finite cyclic subgroup of $\rm{Diffeo}(M)$.

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I'm a bit confused about your last paragraph. Couldn't there be a finite order diffeomorphism which is not an isometry for any hyperbolic metric? – Jason DeVito Jul 3 '10 at 18:31
@Jason - Yes - but every diffeomorphism is isotopic to a isometry. If the order of the diffeo is $k$ then we find that the $k$-th power of that isometry is isotopic to the identity, and so is the identity. – Sam Nead Jul 3 '10 at 18:51
I see - thank you for the clarification! – Jason DeVito Jul 3 '10 at 19:38

For a closed orientable surface Mg of genus g >= 2, there are only a finite number of possible orders of a diffeomorphism or homeomorphism h:Mg -> Mg. One constraint on such orders is that the map induced on first homology

   h*:H1(M2g) -> H1(M2g)

belongs to GL(Z2g), and there is only a finite set of possible orders for elements of this group. (Note that if h* is the identity on first homology with g >= 2, then h is homotopic to the identity on Mg and cannot have order > 1.)

For this reason, e.g., on the double torus M2 there can be no homeomorphism of order greater than 12. For more information see, e.g., Finite groups of matrices whose entries are integers, James Kuzmanovich and Andrey Pavlichenkov, American Mathematical Monthly, Vol. 109, No. 2 (Feb., 2002), pp. 173-186.

(In the converse direction, I suspect any isomorphism of the cohomology ring H*(Mg) can be realized by some self-homeomorphism of Mg, but do not have a reference for this offhand.)

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The converse is false. A 2:1 covering space of a genus $g$ surface gives an involution on genus $2g-1$ whose matrix has eigenvalue $1$ with multiplicity $2g$ and $-1$ with eigenvalue $2(g-1)$. But it's not diagonalizable over $\mathbb Z$. Thus, the diagonal matrix with the same eigenvalues, though conjugate in $Sp_{2g}(\mathbb Q)$ is a conjugacy class in $Sp_{2g}(\mathbb Z)$ that is not in the image of the mapping class group. Also, if you increase the ratio of eigenvalue $1$ to $-1$, no conjugate over $\mathbb Q$ will be realized. – Ben Wieland Jul 5 '10 at 0:59
Wow, that's an interesting example. Thank you. – Daniel Asimov Jul 5 '10 at 19:17

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