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We use controlled topology applied to the action of the infinite dihedral group on a partially compactified plane and deduce two consequences for algebraic K-theory. The first is that the family in the K-theoretic Farrell-Jones conjecture…

K-Theory and Homology · Mathematics 2015-11-30 James F. Davis , Frank Quinn , Holger Reich

We study Farrell Nil-groups associated to a finite order automorphism of a ring $R$. We show that any such Farrell Nil-group is either trivial, or infinitely generated (as an abelian group). Building on this first result, we then show that…

K-Theory and Homology · Mathematics 2016-01-20 Jean-François Lafont , Stratos Prassidis , Kun Wang

We show that for groups acting acylindrically on simplicial trees the $K$- and $L$-theoretic Farrell-Jones Conjecture relative to the family of subgroups consisting of virtually cyclic subgroups and all subconjugates of vertex stabilisers…

Geometric Topology · Mathematics 2017-04-20 Svenja Knopf

For a group G that splits as an amalgamation of A and B over a common subgroup C, there is an associated Waldhausen Nil-group, measuring the "failure" of Mayer-Vietoris for algebraic K-theory. Assume that (1) the amalgamation is…

K-Theory and Homology · Mathematics 2014-02-26 J. -F. Lafont , I. J. Ortiz

We prove that the Waldhausen nilpotent class group of an injective index 2 amalgamated free product is isomorphic to the Farrell-Bass nilpotent class group of a twisted polynomial extension. As an application, we show that the Farrell-Jones…

K-Theory and Homology · Mathematics 2016-05-18 James F. Davis , Qayum Khan , Andrew Ranicki

Let F be a finite group with a Sylow 2-subgroup S that is normal and abelian. Using hyperelementary induction and cartesian squares, we prove that Cappell's unitary nilpotent groups UNil_*(Z[F];Z[F],Z[F]) have an induced isomorphism to the…

Geometric Topology · Mathematics 2008-11-24 Qayum Khan

We show that the relative Farrell-Jones assembly map from the family of finite subgroups to the family of virtually cyclic subgroups for algebraic K-theory is split injective in the setting where the coefficients are additive categories…

K-Theory and Homology · Mathematics 2016-08-31 Wolfgang Lueck , Wolfgang Steimle

In this note, we prove the K- and L-theoretic Farrell-Jones Conjecture with coefficients in an additive category for fundamental groups of graphs of virtually cyclic groups.

K-Theory and Homology · Mathematics 2016-02-23 Xiaolei Wu

We use the controlled algebra approach to study the problem that whether the Farrell-Jones conjecture is closed under passage to over-groups of finite indices. Our study shows that this problem is closely related to a general problem in…

K-Theory and Homology · Mathematics 2015-12-08 Kun Wang

We prove the Farrell-Jones Conjecture for (non-connective) $A$-theory with coefficients and finite wreath products for hyperbolic groups, CAT(0)-groups, cocompact lattices in almost connected Lie groups and fundamental groups of manifolds…

Geometric Topology · Mathematics 2018-10-03 Nils-Edvin Enkelmann , Wolfgang Lück , Malte Pieper , Mark Ullmann , Christoph Winges

We prove the K- and L-theoretic Farrell-Jones Conjecture (with coefficients in additive categories) for virtually solvable groups.

Geometric Topology · Mathematics 2017-05-17 Christian Wegner

For a group G relatively hyperbolic to a family of residually finite groups satisfying the Farrell-Jones conjecture, we reduce the solution of the Farrell-Jones conjecture for G to the case of certain nice cyclic extensions in G.

Group Theory · Mathematics 2013-10-29 Yago Antolín , Giovanni Gandini

Assuming the classical Farrell-Jones conjecture we produce an explicit (commutative) group ring $R$ and a thick subcategory $\mathsf{C}$ of perfect $R$-complexes such that the Waldhausen $K$-theory space $\mathrm{K}(\mathsf{C})$ is…

K-Theory and Homology · Mathematics 2017-10-11 Ilias Amrani

We prove the Farrell-Jones Conjecture for mapping tori of automorphisms of virtually torsion-free hyperbolic groups. The proof uses recently developed geometric methods for establishing the Farrell-Jones Conjecture by…

Geometric Topology · Mathematics 2021-05-28 Mladen Bestvina , Koji Fujiwara , Derrick Wigglesworth

Consider a totally disconnected group G, which is covirtually cyclic, i.e., contains a normal compact open subgroup L such that G/L is infinite cyclic. We establish a Wang sequence, which computes the algebraic K-groups of the Hecke algebra…

K-Theory and Homology · Mathematics 2022-04-19 Arthur Bartels , Wolfgang Lueck

Controlled topology is one of the main tools for proving the isomorphism conjecture concerning the algebraic $K$-theory of group rings. In this article we dive into this machinery in two examples: when the group is infinite cyclic and when…

K-Theory and Homology · Mathematics 2019-08-05 Eugenia Ellis , Emanuel Rodríguez Cirone , Gisela Tartaglia , Santiago Vega

In this paper, we prove the Farrell-Jones Conjecture for the solvable Baumslag-Solitar groups with coefficients in an additive category. We also extend our results to groups of the form, Z[1/p] semidirect product with any virtually cyclic…

Geometric Topology · Mathematics 2014-01-13 Tom Farrell , Xiaolei Wu

In this paper we formulate and lay the foundations for the K-theoretic Farrell-Jones Conjecture for the Hecke algebra of totally disconnected groups. The main result of his paper is the proof that it passes to closed subgroups. Moreover, we…

K-Theory and Homology · Mathematics 2023-06-08 Arthur Bartels , Wolfgang Lueck

We prove the A-theoretic Farrell-Jones Conjecture for virtually solvable groups. As a corollary, we obtain that the conjecture holds for S-arithmetic groups and lattices in almost connected Lie groups.

K-Theory and Homology · Mathematics 2018-09-28 Daniel Kasprowski , Mark Ullmann , Christian Wegner , Christoph Winges

We prove the Farrell-Jones conjecture for free-by-cyclic groups. The proof uses recently developed geometric methods for establishing the Farrell-Jones Conjecture.

Geometric Topology · Mathematics 2021-05-31 Mladen Bestvina , Koji Fujiwara , Derrick Wigglesworth
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