4
$\begingroup$

I'm looking for references about the following aspect of cyclic vectors for regular representations.

Let $K$ be a compact Lie group. Let $K$ act on $L^2(K)$ by the left regular representation. Then $L^2(K)$ has a cyclic vector $v$, so that $K\cdot v$ has dense linear span in $L^2(K)$ (for example one can find a proof of this in the 1971 paper of Greenleaf and Moskowitz "Cyclic Vectors for Representations of Locally Compact Groups").

Question: do we have any control over linear dependence of the set $A=\{kv:k\in K\}$? More precisely, is it possible to choose the cyclic vector $v$ so that $A$ is linearly independent; if not, is it possible to choose $v$ so that there exists some measure-zero set $S\subseteq K$ so that $A'=\{kv:k\in K\backslash S\}$ is linearly independent?

PS: I noticed that the paper "Linear Dependency of Translations and Square Integrable Representations" by Linnell, Puls and Roman discusses the problem of finding $v$ so that $A$ becomes linearly dependent; I'm interested in the case when $v$ is cyclic.

$\endgroup$
4
  • $\begingroup$ I’m confused as to what sort of groups you have in mind: won’t it be hard for $A=K/K_v$ to be linearly independent, unless it is countable? $\endgroup$ Sep 2, 2018 at 15:36
  • $\begingroup$ @FrancoisZiegler well the elements of the orbit don't have to be pairwise orthogonal, so I don't immediately see any obstruction to them forming an uncountable lin ind famiy. $\endgroup$
    – Yemon Choi
    Sep 2, 2018 at 20:34
  • $\begingroup$ Are you looking for a single example of a $K$ for which this is possible, or a proof that it is possible for all $K$? $\endgroup$
    – Yemon Choi
    Sep 2, 2018 at 20:36
  • $\begingroup$ @YemonChoi I'm looking for a proof of the general case. As a sidenote, if I'm not mistaken, I think this holds for $K=S^1$. $\endgroup$
    – geometricK
    Sep 3, 2018 at 0:17

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.