Towers of regular self-covers and linear endomorphisms of tori
Which closed manifolds admit self-covers?
- Tori, with linear maps . Degree is .
- Nilmanifolds, with automorphisms.
- certain fiber bundles with torus or nilmanifold fibers.
These are all examples in dimension 2 (Euler characteristic is multiplicative). Idem in dimension 3 (Tollefson, Mess, Yu-Wang 1999).
In dimension 4, there are other examples. Manifolds with fundamental group a Baumslag-Solitar have been classified by Hambleton-Kreck-Teichner, some of them double cover themselves.
Gromov’s polynomial growth theorem has the following consequence:
Theorem 1 (Gromov) If a closed manifold admits an expanding map (with respect to some Riemannian metric), then is an infrnilmanifold.
1.1. Residual self-covers
We suggest a topological analogue of expansion.
Say a cover is residual if iterates converge to the universal cover, i.e. every loop eventually unwinds.
Question. If has a residual finite self-cover, is infranil ?
1.2. Strongly scale invariant groups
There is a purely group-theoretic analogue. Observe that a self-cover induces an map on fundamental groups whose image has finite index.
Question. Let be a finitely generated group such that has a subgroup
Definition 2 (Nekrashevych-Pete) A finitely generated group is strongly scale invariant if is has a self-homomorphism with finite index image, and such that
They ask wether strongly scalar invariant implies virtually nilpotent.
Previously Benjamini thought that existence of a chain of nested subgroups of finite index, all isomorphic to , with trivial intersection, would suffice, but Nekrashevych-Pete showed this to be false.
One can ask more.
Question. Is a self-cover somehow induced by an endomorphism of a nilpotent group?
This is the case for the Baumslag-Solitar examples . The endomorphims are of the form multiplication by on the first factor.
Definition 3 A finite self-cover is strongly regular if all iterates are regular covers.
Theorem 4 Let be a strongly regular self-cover. Then there exists a free abelian group and an epimorphism
such that induces an isomorphim on ker and a linear endomorphism of of determinant equal to the degree of .
Question. Assume that is nonpositively curved and admits a self-cover. Then is covered by a product with a torus factor?
Corollary 5 If is residual and strongly regular, then is abelian.
Indeed, in this case, for all , so it must be trivial.
I explain the easier result where is merely torsion free nilpotent. Let be the profinite completion of . Then still has finite index image.
Example. . Multiplication by 2 is an example.
Theorem 6 (Reid 2014) Let be a profinite embedding of type . Let be an open embedding. Then is isomorphic to a semi-direct product , where is a contraction group, i.e. it admits an automorphism whose iterates bring everything asymptotically to the origin.
Theorem 7 (Gloeckner-Willis 2006) Any contraction group is a product of a finite product of nilpotent -adic groups and a group of bounded exponent.
Example. equipped with multiplication by .
3.1. Proof of theorem 1
We know that . First show the product is direct. Then that is trivial. This used the abelianization and the description of abelian groups of finite exponents (Prufer).
4. Kahler manifolds and holomorphic maps
Here is a vague question. If is a self-cover has some geometric origin, is a fiber bundle with nilmanifold fibers ? For instance, is an endomorphism of some geometric structure, or is Anosov, or is holomorphic.
Theorem 8 If is Kahler and is strongly regular holomorphic self-cover, then is finitely covered by a product with a complex torus factor and a Kahler factor.