7
$\begingroup$

Let $H$ be a Hilbert space and $G$ be a finitely generated group. Let $\pi:G\rightarrow GL(H)$ be a representation.

A map $c:G\rightarrow H$ is called cocycle if $c(gh)=π(g)c(h)+c(g)$ for all $g,h$ in the group $G$. A cocycle $c$ is proper if the map $g\mapsto c(g)$ is proper, i.e., for every constant $K$ the number of elements $g$ in the group such that $||c(g)|| \lt K$ is finite.

Q1: Does there always exists a representation (bounded or not) of a group on a Hilbert space which admits a proper cocycle?

the answer to this question is in the comment of Mikael.

Q2: Does there always exists a representation (by bounded operators) of a group on a Hilbert space which admits a proper cocycle?

Here we only assume that $||\pi(g)||<\infty$, but $\pi$ is not necessarily uniformly bounded.

Improve this question
$\endgroup$
34
  • 2
    $\begingroup$ If it means what I think it means, then this question as it currently stands is trivial to answer in the negative. Let the Hilbert space be the complex numbers, and take any group whose cardinality is much larger than that of the complex numbers, acting trivially. Then a cocycle is a homomorphism and it the kernel must be huge for cardinality reasons. $\endgroup$
    – user30035
    Commented Mar 28, 2013 at 21:08
  • 3
    $\begingroup$ My understanding of Kate's question is this: does any f.g. group admit a proper, affine action on a Hilbert space? Am I correct? $\endgroup$
    – Alain Valette
    Commented Mar 28, 2013 at 21:56
  • 1
    $\begingroup$ A natural class of counter-examples to try are groups containing expanders. It is quite likely that a proper affine continuous action would imply existence of a uniform embedding in a Hilbert space (possibly a different one), but I do not see how to prove this at the moment. $\endgroup$
    – Misha
    Commented Mar 28, 2013 at 23:41
  • 4
    $\begingroup$ Alain, I am impressed by your guessing the right question from so few information. Kate, perhaps you should edit your question, adding enough details so that people who are not in the field but interested may have a chance to understand it. $\endgroup$
    – Joël
    Commented Mar 29, 2013 at 1:02
  • 5
    $\begingroup$ If $G$ is finitely generated and coarsely embeddable into a Hilbert space, then the answer is yes. This follows from Bo\.zejko--Fendler's construction (Arch. Math. (Basel) 57 (1991), 290–298). Otherwise, as Misha, I bet on an expander counterexample. $\endgroup$
    – Narutaka OZAWA
    Commented Mar 29, 2013 at 1:33

1 Answer 1

Reset to default
10
$\begingroup$

Q1 and Q2 have a positive answer for all countable groups (conversely a discrete uncountable group cannot bear any proper function). Let $\mu$ be a proper function from $G$ to the positive reals, and view it as a discrete measure on $G$. Assume in addition that $\mu$ grows reasonably, and more precisely satisfies an equality of the form $\mu(gh)\le C_g\mu(h)$ with $C_G\>0$ (e.g. fix a proper subadditive length $|\cdot|$ and define $\mu(g)=|g|+1)$.

Then the action of left action of $G$ on itself induces a well-defined left regular representation $\pi$ of $G$ on $\ell^2(G,\mu)$, which is bounded ($\|\pi(g)\|\le C_g^{1/2}$).

Let $e$ be the unit in $G$ and $\delta_g$ the Dirac function at $g\in G$. Define $b$ as the coboundary $b(g)=\delta_e-g\delta_e=\delta_e-\delta_g$. Then $\|b(g)\|\ge \mu(g)-\|\delta_e\|$, which is proper; thus $b$ is a proper cocycle.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .