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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.

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Piotr: The answer depends on the function $f$. Take for instance $f(x)=x$, $x\in {\mathbb Z}$. Maybe you meant to ask for conditions on functions $f$ which ensure this property? –  Misha Jan 11 '13 at 13:26
    
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 $\rho_{g}=\lambda_{g^{-1}}$. –  user2412 Jan 11 '13 at 13:53
    
Piotr: Sorry, I misunderstood the question. –  Misha Jan 11 '13 at 14:00

1 Answer 1

I think, the following is an example. Let $G$ be the infinite dihedral group with involution $s\in G$. Next, let $h: {\mathbb R}\to {\mathbb R}$ be a positive $L^2$-function and $f(n):= \int_{-\infty}^n h(x)dx$ be a function on ${\mathbb Z}$. Now, extend $f$ to $G$ by $f(zs)=f(z)$, $z\in {\mathbb Z}$. Then $f$ determines an $\ell_2$-cocycle $b_g$ but $a_s(x)$ is not in $\ell_2$ (since it does not tend to zero as $|x|\to\infty$.

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Agreed. Any candidates for $W$? –  user2412 Jan 11 '13 at 18:14
    
Piotr: At the moment, the only candidate for $W$ I see is $\ell_\infty(G)$. (I think, if $b_g\in \ell_2(G)$, then $f$ has to be bounded, at least for finitely-generated groups, so $a_g$ is also bounded.) –  Misha Jan 11 '13 at 20:37
    
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. –  user2412 Jan 11 '13 at 20:50
    
Piotr: Right, I forgot that it is $\ell^2$ and not $\ell^1$. –  Misha Jan 11 '13 at 20:57
    
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. –  user2412 Jan 11 '13 at 21:59

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