# “Finite part of operator K-theory” I

First of all, I apologize for the hiatus in posting over the past couple of weeks,  Organizing a (non-mathematical) conference has absorbed a big chunk of my time, and then getting back up to speed with routine tasks has absorbed another big chunk.   However…

So I started looking at the recent paper of Shmuel Weinberger and Guoliang Yu,  They are interested in looking at the part of the $$K$$-theory of the maximal C*-algebra of a group $$\Gamma$$ which is generated by the projections

$p_H = \frac{1}{|H|} \sum_{h\in H} h\quad \in {\mathbb C}[G]$

in the complex group algebra of $$G$$, where $$H$$ is a finite cyclic subgroup.   (Question: Why do they restrict attention to finite cyclic subgroups? Wouldn’t any finite subgroup work just as well.)

The claim is that these generate a “large” subgroup of $$K_0(C^*_{max}(G))$$ which is not in the image of the maximal assembly map from $$K_0(BG)$$.  “Large” is expressed in terms of a lower bound for the rank of this abelian group.

The basic strategy, so far as I understand it, can be thought of in terms of a familiar argument for property T groups.  Let $$G$$ be any group.  The maximal group C*-algebra has a homomorphism $$\alpha$$ to $$\mathbb C$$, which just is the regular representation (as a linear map on $${\mathbb C}[G]$$ it sends every group element to 1.   On the other hand, the reduced (and therefore also the maximal) group $$C^*$$ algebras have a different trace $$\tau$$ which sends the identity element to 1 and every other element of $$G$$ to 0 – this is the tracial vector state associated to the unit vector $$\xi_e$$ in the regular representation $$\ell^2(G)$$.  At the level of K-theory we get a diagram

$\begin{array}{ccc} K_0(C^*_{max}(G))&\to^\alpha &{\mathbb Z}\\ \downarrow&&\downarrow\\ K_0(C^*_r(G)&\to^\tau & {\mathbb R}\end{array}$

This diagram need not commute.  In fact, if $$G$$ has property T and we consider at the top left corner the K-theory class of the Kazhdan projection – the projection (whose existence is guaranteed by property T) which maps, under any representation, to the projection onto the G-invariant subspace of that representation – then this class maps to 1 by traversing the diagram via the top right corner and to 0 traversing via the lower left corner.   However, it must commute for any element in the image of the (maximal) assembly map, as follows essentially from Atiyah’s $$L^2$$ index theorem.  Thus, as is well known, we infer that the class of the Kazhdan projection is not in the image of the maximal assembly map.

Weinberger and Yu point out that a similar argument can be applied to the projection $$p_H$$ associated to a finite cyclic subgroup $$H$$ of $$G$$. In fact, the homomorphism $$\alpha$$ takes $$[p_H]$$ to 1, whereas the trace $$\tau$$ takes it to $$|H|^{-1}$$.  This is independent of any property T considerations.  Motivated by this, they conjecture that the rank of the subgroup of $$K_0(C^*_{max}(G)))$$ generated by the $$[p_H]$$ (they call this the “finite part” of this group) is at least equal to the number of distinct orders of cyclic subgroups of $$G$$, and that no non-identity element in the finite part lies in the image of the assembly map.

Next time I hope to talk about their approach to proving this in  certain cases.

Weinberger, Shmuel, and Guoliang Yu. Finite Part of Operator K-theory for Groups Finitely Embeddable into Hilbert Space and the Degree of Non-rigidity of Manifolds. ArXiv e-print, August 21, 2013. http://arxiv.org/abs/1308.4744.

# Maximal Roe algebras, part 2

Let $$X$$ be a bounded geometry metric space.  At the end of the previous post, we observed that if $$\pi \colon {\mathbb C}[X] \to {\mathfrak B}(H)$$ is a Hilbert space representation of the translation algebra of $$X$$, then any unit vector in the range of one of the projections $$\pi(V_{x,x})$$ corresponding to a point of $$X$$ generates a subrepresentation isomorphic to the regular one.  It follows that if $$\pi$$ does not contain a copy of the regular representation, then the projections $$\pi(V_{x,x})$$ must be zero for every $$x\in X$$.

Surprisingly enough, such representations do exist! Continue reading