2
$\begingroup$

Background: Let $(Y, \mathcal{B},\mu,T)$ be an ergodic probability system and let $G$ be a compact metrizable group with compact subgroup $H$. Given a measurable map $\rho:Y \to G$. We may define the skew product system $W_{H,\rho}=(Y \times G/H, \mu \times m_{G/H}, T_{\rho})$ where the transformation is given by $$ T_{\rho}(y,u)=(Ty, \rho(y)u)$$ and by $m_{G/H}$ we mean the pushforward onto $G/H$ of the Haar measure on $G$. Note that for $g \in G$ we have a measure preserving map $g:Y \times G/H \to Y \times G/H$ given by $$g.(y,u)=(y,gu)$$ and thus $G$ acts on the Hilbert Space $L^2(W_{H,\rho})$.

Question: If the system $(Y \times G, \mu \times m_G, T_{\rho})$ is ergodic, then is it true that for any eigenfunction $e: Y \times G/H \to \mathbb{C}$ of $W_{H,\rho}=(Y \times G/H, \mu \times m_{G/H}, T_{\rho})$ we have that $e \circ [g_1,g_2]=e$ for $g_1,g_2 \in G$? i.e all eigenfunctions are fixed by the commutator subgroup $[G,G]$?

Motivation: It seems to me that this is suggested in the proof of Lemma 6.1 in Nonconventional ergodic averages and nilmanifolds by Host-Kra (available to public here http://www.math.northwestern.edu/~kra/papers/convnil.pdf)

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, the eigenfunctions of $X = Y\times G/H$ come from $Y\times A$, where $A$ is an abelian quotient of $G$. I'll indicate how to construct $A$, checking the required properties is then routine. Note that we may assume $H$ trivial, since eigenfunctions of $Y\times G/H$ are also eigenfunctions of $Y\times G$.

Note that the $Y$-module spanned by the eigenfunctions is an algebra, and since it is spanned by rank 1 submodules it corresponds to an abelian group extension $Y\times A$ with a minimal cocycle $\alpha$. The trick is now to relate $G$ and $A$. To do this note that we have a measure-theoretic factor map $X \to Y\times (G\times A)$, where the latter system is equipped with the cocycle $\rho\times\alpha$ and the measure on the target system is a pushforward measure.

By the Mackey group construction we may assume that $\rho\times\alpha$ is minimal with values is a subgroup $K\leq G\times A$. Since $\rho$ and $\alpha$ are minimal, $K$ has full projections on both coordinates. Moreover, the projection map $Y\times K \to Y\times G$ is a measure-theoretic isomorphism, so the coordinate projection $K\to G$ is in fact a compact group isomorphism. This gives the desired morphism $G\to K\to A$.

$\endgroup$
2
  • $\begingroup$ Thank you. I have actually found a simple proof not long after posting this question, but had no time to post it here. It is different to yours, but strictly requires the ergodicity of X. I was wondering if your proof requires the ergodicity of X? $\endgroup$ Mar 1, 2015 at 5:15
  • $\begingroup$ I think that all required tools have non-ergodic versions, see arxiv.org/abs/0905.0516 $\endgroup$
    – pavel
    Apr 5, 2015 at 22:04

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.