User piotr nowak - MathOverflow

Definition of a uniformly bounded dual of a group

2013-04-07T16:35:57Z

The unitary dual of a group $G$ is the set of equivalence classes of irreducible unitary representations of $G$ with the Fell topology. (This topology is defined using convergence of positive definite functions, or equivalently, diagonal matrix coefficients, associated to the representations).

Question: How to define a "uniformly bounded dual" of $G$ using uniformly bounded representations on a Hilbert space and what topology should it be equipped with?

Such a uniformly bounded dual is mentioned a few times in the literature, but usually vaguely defined to be "analogous to the unitary dual". In particular, there is hardly any explicit statement about what topology is used.

I am interested in this since I would like to understand a series of results of M. Cowling from the early 1980s on a stronger version of Kazhdan's property (T). Cowling proved that for a simple Lie group $\Gamma$, the trivial representation is isolated in such a uniformly bounded dual if and only if the rank of $\Gamma$ is $\ge 2$.

In the case of $Sp(n,1)$ Cowling constructs an explicit family of uniformly bounded representations that approximate the trivial representation. The best description of his construction I have found is in his abstract from an Oberwolfach workshop from 2001 on the geometrization of property (T) (see MFO report, the following abstract is on page 3):

Cocycles for right- and left- regular representations on $\ell_2(G)$

2013-01-11T11:18:37Z

Using the standard notation, let $\lambda_gf(x)=f(g^{-1}x)$ be the left regular representation and $\rho_gf(x)=f(xg)$ be the right regular one, acting on the space $V$ of complex-values functions on $G$.

Let $b:G\to \ell_2(G)$ be a cocycle for $\rho$; i.e., $b_g=\rho_g f-f\in \ell_2(G)$ for some $f\in V$ and every $g\in G$.

Question: Does $a_g=\lambda_g f-f$ belong to $\ell_2(G)$ for every $g\in G$? If not, what is the "smallest" subspace $W$ of $V$, $\ell_2(G)\subseteq W\subseteq V$ for which this happens?

The most interesting case seems to be when $G$ is amenable.

Edit: The metric interpretation is the following. Take all $f\in V$ whose discrete gradient with respect to the left-invariant metric is $\ell_2$-summable. Then $W$ is the space consisting of gradients of these $f$, with respect to the right-invariant metric.

Injectivity for bimodules and Hochschild cohomology

2010-10-21T16:13:46Z

Let $A$ be a Banach algebra and let $X$ be an $A$-bimodule. Is there a notion of (relative) injectivity for $X$ which would imply that $\mathcal{H}^n(A,X)$ vanishes for all $n\ge 1$? Here $\mathcal{H}^n(A,X)$ denotes the continuous Hochschild cohomology of $A$ with coefficients in $X$ ($X$ can be assumed to be dual $A$-bimodule).

I expected there is such a notion, but after reading books by Helemskii, Runde and a few other sources on cohomology of Banach algebras I can't seem to find a general statement of this type, even though versions of projectivity and injectivity are discussed there.

When can closedness of the range of an operator be checked on a positive cone?

2010-07-06T02:32:05Z

Let $T:X\to Y$ be an operator between Banach spaces $X$ and $Y$. Assume that $X$ has a positive cone $C\subset X$, which generates $X$: every element of $X$ can be written as a difference of elements of the positive cone.

Assume now that we can show $$k\Vert x\Vert\le \Vert Tx\Vert$$ for some $k>0$ and every $x\in C$.

Question: Are there any natural, general conditions on $T$, $X$ or both that allow to conclude that the image of $X$ under $T$ is closed in $Y$ (the above inequality holds for all $x\in X$, possibly with a different $k$)?

The motivating example is the following. Let $G=(V,E)$ be an infinite (oriented) graph and take $X=\ell_1(V)$, $Y=\ell_1(E)$. Let $T$ be the discrete gradient, $Tf(x,y)=f(y)-f(x)$. Then it is enough to check the inequality only for positive $f$, then use the triangle inequality.

Update: Given Andreas' response I realized I should probably ask not for general conditions but any sufficient condition that would give the above property, in particular in special cases like the motivating example above..

Answer by Piotr Nowak for An example of a non-amenable exact group without free subgroups.

2010-06-29T02:39:38Z

Owen, I'm a bit late to the party, but I think the answer to your question is ``no", to the best of my knowledge. To phrase it properly, I believe it is not known whether any of the known counterexamples to von Neumann's conjecture is exact.

Jon, one has to be careful with limits hyperbolic groups, for example Gromov's random groups which are not exact are such limits (they are lacunary hyperbolic, in the sense of Olshanskii, Osin and Sapir).

Comment by Piotr Nowak

2013-04-15T12:08:54Z

You will find these proved in appendices G and C in the recent monograph by Bekka, de la Harpe and Valette (see perso.univ-rennes1.fr/bachir.bekka/…).

Comment by Piotr Nowak

2013-04-11T09:08:54Z

@Mikael: I agree that Lafforgue's strong (T) generalizes this in some sense, but he is considering representations which are not necessarily uniformly bounded, but satisfy some growth conditions. Thus the fact you mention for hyperbolic groups should be, at least philosophically, weaker than what Cowling proved for $Sp(n,1)$ (provided the two properties are directly comparable).

Comment by Piotr Nowak

2013-04-11T05:32:14Z

@Mikael: in Cowling's result for $Sp(n,1)$, there is a family of representations $\{\pi_i\}$, such that $\pi_i\to 1$ and for each $i$ we have $\Vert \pi_i\Vert=\sup_{g\in G}\Vert \pi_i(g)\Vert<\infty$, but apparently $\Vert \pi_i\Vert\to \infty$ as $\pi_i\to 1$. This last fact is mentioned in a paper of Julg on the Baum-Connes conjecture in the Comptes Rendus.

Comment by Piotr Nowak

2013-04-10T08:15:47Z

Thanks for the suggestion. I have tried the obvious alternate routes already, moreover, it seemed that MO is the natural place to ask such a question. Several other people also reference or use Cowling's results in their work. For instance, Julg has an approach to the Baum-Connes conjecture with coefficients for $Sp(n,1)$ and mentioned the "u.b. dual" explicitly; Shalom's unpublished proof that $Sp(n,1)$ admits a proper affine action with uniformly bounded linear part is also based on Cowling's result, as far as I know. However, the meaning of the term "isolated" is never defined precisely.

Comment by Piotr Nowak

2013-04-10T07:29:30Z

Understanding how nets converge to the trivial representation is pretty much what I am looking for. More specifically, I would like to know the meaning of the term "isolated" in the context of Cowling's results. One example of his results is that for $Sp(n,1)$ the trivial representation is not isolated within the uniformly bounded representations. Another is that for simple Lie groups of rank at least 2, the trivial representation is isolated within the uniformly bounded ones. Also, of course I meant irreducible in the definition of the unitary dual, I'll edit the question.

Comment by Piotr Nowak

2013-04-08T04:07:48Z

Yemon, I am only considering the Hilbert space.

Comment by Piotr Nowak

2013-04-07T18:26:03Z

Yves, no behavior of norms of the representations is assumed. In fact Cowling has proved that for $Sp(n,1)$ the trivial representation can be approximated by uniformly bounded ones, even though $Sp(n,1)$ has (T). However, the norms of these uniformly bounded representations blow up to infinity as one approaches the trivial one.

Comment by Piotr Nowak

2013-01-11T21:59:02Z

By the way, the metric version is the following: if the gradient of $f$ is in $\ell_2(E)$, where $E$-edges of the Cayley graph, when $G$ has the right-invariant word length metric, then $W$ would denote the space of gradients of all such $f$ with respect to the left-invariant metric.

Comment by Piotr Nowak

2013-01-11T20:50:50Z

Misha, I think $W$ can be much larger than $\ell_{\infty}(G)$ and in general $f$ does not have to be bounded, it just has to have $\ell_2$-summable gradient with respect to the left-invariant metric. More explicitly, consider a function on $\mathbb{Z}$, where $f(n)=f(n-1)+\frac{1}{n}$ for $n$ positive, and $0$ elsewhere.

Comment by Piotr Nowak

2013-01-11T18:14:40Z

Agreed. Any candidates for $W$?

Comment by Piotr Nowak

2013-01-11T13:53:22Z

Misha, I want this to happen for all functions $f$ on $G$, for which $\rho_g f-f$ is a cocycle into $\ell_2(G)$. On $\mathbb{Z}$ the key is commutativity and that $\rho_{g}=\lambda_{g^{-1}}$.

Comment by Piotr Nowak

2010-10-28T17:51:57Z

Indeed, the proof of the "only if" direction in lemma 2.2 is not clear. This is surprising, as far as I know it is considered to be an established fact.

Comment by Piotr Nowak

2010-10-28T14:16:05Z

You're right, although it looks from the proof of lemma 2.2 in Pittet's paper that he actually needs a weaker statement, with "=" replaced by ">" in lemma 2.1.

Comment by Piotr Nowak

2010-10-25T02:06:38Z

This is great, thanks for such a complete answer!

Comment by Piotr Nowak

2010-10-23T14:24:02Z

@Yemon: you should definitely post it as an answer.