Which manifolds admit a diffeomorphism of order $n$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:28:31Z http://mathoverflow.net/feeds/question/30425 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30425/which-manifolds-admit-a-diffeomorphism-of-order-n Which manifolds admit a diffeomorphism of order $n$? Łukasz Garncarek 2010-07-03T16:18:07Z 2010-07-04T20:00:52Z <blockquote> <p>Let $n>1$. Which smooth manifolds admit a diffeomorphism $f$ of order $n$?</p> </blockquote> <p>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 $n=5$.</p> <p>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.</p> <p>Is a general answer to this question known? If not, maybe there are some simple examples of manifolds without diffeomorphisms of given order?</p> http://mathoverflow.net/questions/30425/which-manifolds-admit-a-diffeomorphism-of-order-n/30426#30426 Answer by Robin Chapman for Which manifolds admit a diffeomorphism of order $n$? Robin Chapman 2010-07-03T16:30:01Z 2010-07-03T17:31:25Z <p>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.</p> <p>Because of <a href="http://en.wikipedia.org/wiki/Hurwitz%2527s_automorphisms_theorem" rel="nofollow">Hurwitz's theorem</a> this kind of construction via Riemann surface automorphisms cannot work when $n$ is large compared to $g$.</p> <p><strong>Added</strong></p> <p>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 $4$).</p> http://mathoverflow.net/questions/30425/which-manifolds-admit-a-diffeomorphism-of-order-n/30437#30437 Answer by Sam Nead for Which manifolds admit a diffeomorphism of order $n$? Sam Nead 2010-07-03T18:21:09Z 2010-07-03T18:21:09Z <p>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 $\rm{Diffeo}(S_g)$. </p> <p>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 $\rm{Diffeo}(M)$.</p> http://mathoverflow.net/questions/30425/which-manifolds-admit-a-diffeomorphism-of-order-n/30445#30445 Answer by Jeffrey Giansiracusa for Which manifolds admit a diffeomorphism of order $n$? Jeffrey Giansiracusa 2010-07-03T18:45:02Z 2010-07-03T18:45:02Z <p>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 <strong>every</strong> 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.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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). </p> http://mathoverflow.net/questions/30425/which-manifolds-admit-a-diffeomorphism-of-order-n/30479#30479 Answer by Agol for Which manifolds admit a diffeomorphism of order $n$? Agol 2010-07-04T03:13:24Z 2010-07-04T03:13:24Z <p>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. </p> <p>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. </p> <p>Conversely, given a 3-manifold $M$, one may determine if it admits an $n$-fold symmetry. By the <a href="http://www.ams.org/mathscinet-getitem?mr=806009" rel="nofollow">equivariant sphere theorem</a>, one may find a collection of reducing spheres which are equivariant under the action. Then by the geometrization theorem, one may find <a href="http://www.ams.org/mathscinet-getitem?mr=664520" rel="nofollow">an equivariant decomposition along tori</a> into geometric pieces, such that the action <a href="http://www.ams.org/mathscinet-getitem?mr=856847" rel="nofollow">preserves the homogeneous metric on each piece</a>. In principle, then, one could collate this information to <a href="http:///" rel="nofollow">determine if $M$ admits a symmetry of order $n$</a>. See also <a href="http://front.math.ucdavis.edu/0801.0803" rel="nofollow">Dinkelbach-Leeb</a>. </p> http://mathoverflow.net/questions/30425/which-manifolds-admit-a-diffeomorphism-of-order-n/30546#30546 Answer by Daniel Asimov for Which manifolds admit a diffeomorphism of order $n$? Daniel Asimov 2010-07-04T20:00:52Z 2010-07-04T20:00:52Z <p>For a closed orientable surface M<sub>g</sub> of genus g >= 2, there are only a finite number of possible orders of a diffeomorphism or homeomorphism h:M<sub>g</sub> -> M<sub>g</sub>. One constraint on such orders is that the map induced on first homology </p> <p>&nbsp;&nbsp; h<sub>*</sub>:H<sub>1</sub>(M<sub>2g</sub>) -> H<sub>1</sub>(M<sub>2g</sub>) </p> <p>belongs to GL(Z<sup>2g</sup>), and there is only a finite set of possible orders for elements of this group. (Note that if h<sub>*</sub> is the identity on first homology with g >= 2, then h is homotopic to the identity on M<sub>g</sub> and cannot have order > 1.)</p> <p>For this reason, e.g., on the double torus M<sub>2</sub> there can be no homeomorphism of order greater than 12. For more information see, e.g., <em>Finite groups of matrices whose entries are integers</em>, James Kuzmanovich and Andrey Pavlichenkov, American Mathematical Monthly, Vol. 109, No. 2 (Feb., 2002), pp. 173-186.</p> <p>(In the converse direction, I suspect any isomorphism of the cohomology ring H<sup>*</sup>(M<sub>g</sub>) can be realized by some self-homeomorphism of M<sub>g</sub>, but do not have a reference for this offhand.)</p>