Which manifolds admit a diffeomorphism of order $n$? - MathOverflow

Which manifolds admit a diffeomorphism of order $n$?

2010-07-03T16:18:07Z

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?

Answer by Robin Chapman for Which manifolds admit a diffeomorphism of order $n$?

2010-07-03T16:30:01Z

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 $g$.

Added

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 $4$).

Answer by Sam Nead for Which manifolds admit a diffeomorphism of order $n$?

2010-07-03T18:21:09Z

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 $\rm{Diffeo}(M)$.

Answer by Jeffrey Giansiracusa for Which manifolds admit a diffeomorphism of order $n$?

2010-07-03T18:45:02Z

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).

Answer by Agol for Which manifolds admit a diffeomorphism of order $n$?

2010-07-04T03:13:24Z

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.

Answer by Daniel Asimov for Which manifolds admit a diffeomorphism of order $n$?

2010-07-04T20:00:52Z

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

h_*:H₁(M_2g) -> H₁(M_2g)

belongs to GL(Z^2g), 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 M_g 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^*(M_g) can be realized by some self-homeomorphism of M_g, but do not have a reference for this offhand.)