Farrell–Jones conjecture

One might also try to take for example the family of finite subgroups into account. This family is much easier to handle. Consider the infinite cyclic group ${\displaystyle \mathbb {Z} }$. A model for ${\displaystyle E_{FIN}(\mathbb {Z} )}$ is given by the real line ${\displaystyle \mathbb {R} }$, on which ${\displaystyle \mathbb {Z} }$ acts freely by translations. Using the properties of equivariant K-theory we get

${\displaystyle K_{n}^{\mathbb {Z} }(\mathbb {R} )=K_{n}(S^{1})=K_{n}(pt)\oplus K_{n-1}(pt)=K_{n}(R)\oplus K_{n-1}(R).}$

The Bass-Heller-Swan decomposition gives

${\displaystyle K_{n}^{\mathbb {Z} }(pt)=K_{n}(R[\mathbb {Z} ])\cong K_{n}(R)\oplus K_{n-1}(R)\oplus NK_{n}(R)\oplus NK_{n}(R).}$

Indeed one checks that the assembly map is given by the canonical inclusion.

${\displaystyle K_{n}(R)\oplus K_{n-1}(R)\hookrightarrow K_{n}(R)\oplus K_{n-1}(R)\oplus NK_{n}(R)\oplus NK_{n}(R)}$

So it is an isomorphism if and only if ${\displaystyle NK_{n}(R)=0}$, which is the case if ${\displaystyle R}$ is a regular ring. So in this case one can really use the family of finite subgroups. On the other hand this shows that the isomorphism conjecture for algebraic K-Theory and the family of finite subgroups is not true. One has to extend the conjecture to a larger family of subgroups which contains all the counterexamples. Currently no counterexamples for the Farrell–Jones conjecture are known. If there is a counterexample, one has to enlarge the family of subgroups to a larger family which contains that counterexample.

The transitivity principle is a tool to change the family of subgroups to consider. Given two families ${\displaystyle F\subset F'}$ of subgroups of ${\displaystyle G}$. Suppose every group ${\displaystyle H\in F'}$ satisfies the (fibered) isomorphism conjecture with respect to the family ${\displaystyle F|_{H}:=\{H'\in F|H'\subset H\}}$. Then the group ${\displaystyle G}$ satisfies the fibered isomorphism conjecture with respect to the family ${\displaystyle F}$ if and only if it satisfies the (fibered) isomorphism conjecture with respect to the family ${\displaystyle F'}$.

Given any group homomorphism ${\displaystyle \alpha \colon H\rightarrow G}$ and suppose that G"' satisfies the fibered isomorphism conjecture for a family F of subgroups. Then also H"' satisfies the fibered isomorphism conjecture for the family ${\displaystyle \alpha ^{*}F}$. For example if ${\displaystyle \alpha }$ has finite kernel the family ${\displaystyle \alpha ^{*}VCYC}$ agrees with the family of virtually cyclic subgroups of H.

For suitable ${\displaystyle \alpha }$ one can use the transitivity principle to reduce the family again.

There are also connections from the Farrell–Jones conjecture to the Novikov conjecture. It is known that if one of the following maps

${\displaystyle H_{*}^{G}(E_{VCYC}(G),L_{R}^{\langle -\infty \rangle })\rightarrow H_{*}^{G}(\{\cdot \},L_{R}^{\langle -\infty \rangle })=L_{*}^{\langle -\infty \rangle }(RG)}$

${\displaystyle H_{*}^{G}(E_{FIN}(G),K^{top})\rightarrow H_{*}^{G}(\{\cdot \},K^{top})=K_{n}(C_{r}^{*}(G))}$

is rationally injective, then the Novikov-conjecture holds for ${\displaystyle G}$. See for example,.[6][7]

The Bost conjecture (named for Jean-Benoît Bost) states that the assembly map

${\displaystyle H_{*}^{G}(E_{FIN}(G),K_{l^{1}}^{top})\rightarrow H_{*}^{G}(\{\cdot \},K_{l^{1}}^{top})=K_{*}(l^{1}(G))}$

is an isomorphism. The ring homomorphism ${\displaystyle l^{1}(G)\rightarrow C_{r}(G)}$ induces maps in K-theory ${\displaystyle K_{*}(l^{1}(G))\rightarrow K_{*}(C_{r}(G))}$. Composing the upper assembly map with this homomorphism one gets exactly the assembly map occurring in the Baum–Connes conjecture.

${\displaystyle H_{*}^{G}(E_{FIN}(G),K_{l^{1}}^{top})=H_{*}^{G}(E_{FIN}(G),K^{top})\rightarrow H_{*}^{G}(\{\cdot \},K^{top})=K_{*}(C_{r}(G))}$

The Kaplansky conjecture predicts that for an integral domain ${\displaystyle R}$ and a torsionfree group ${\displaystyle G}$ the only idempotents in ${\displaystyle R[G]}$ are ${\displaystyle 0,1}$. Each such idempotent ${\displaystyle p}$ gives a projective ${\displaystyle R[G]}$ module by taking the image of the right multiplication with ${\displaystyle p}$. Hence there seems to be a connection between the Kaplansky conjecture and the vanishing of ${\displaystyle K_{0}(R[G])}$. There are theorems relating the Kaplansky conjecture to the Farrell Williams–Jones conjecture (compare [8]).