Let $\theta \,: G \to H$ be a group homomorphism between amenable groups (s.t. $\theta(G)$ is a normal subgroup of $H$, if needed). Is it possible to define amenable means $m_G$ on $L^\infty(G, \mathbb{R})$ and $m_H$ on $L^\infty(H, \mathbb{R})$ in such a way we have a commutative diagram? \begin{equation} \label{} \begin{array}{ccccccccccccccccccccccccccccccc} L^\infty(G, \mathbb R) & \overset{\theta^*}{\longleftarrow} & L^\infty(H, \mathbb {R}) \\ \scriptstyle{m_{G} \Big\downarrow \phantom{m_{G}}} && \scriptstyle{ \phantom{m_{H}} \Big\downarrow m_H} \\ \mathbb R & =& \mathbb R \\ \end{array} \end{equation}

In other words, we should have an equation: $$ m_H(f) = m_G(f \circ \theta) \qquad \forall f \in L^\infty(H, \mathbb R) $$ and $m_H$ and $m_G$ are $H$-invariant and $G$-invariante means respectively [edit: .. and positive and adequately normalized, as in the definition of a mean of an amenable group].

Even an answer in the case $G$ and $H$ are abelian would be interesting!

EDIT consider for example the case of $\theta \,: G \to H$ being surjective [english lessons needed!]. Then we define: $$ m_H(f) := m_G(f \circ \theta) $$ for some amenable mean $m_G$. Hence we get: $$ m_H(h \cdot f) = m_H(f(h \cdot \square)) = m_G(f(h \cdot \theta (\square))) = m_G(f(\theta(g) \cdot \theta (\square))) = m_G(f(\theta (g \square))) = m_G(f(\theta (\square))) = m_H(f) $$

EDIT having thought a little bit more, I think now that the answer is "obviously no!". Suppose that the group homomorphism $\theta \,: G \to H$ is s.t. $[H \,: \theta(G)] = \infty$. Let $f = \delta_{\theta(G)} \,: H \to \mathbb{R}$. Then, by the additivity property and the positivity property of amenable means we must have $m_H(f) = 0$. But: $$ m_H(f) = m_G(f \circ \theta) = m_G(1) = 1 $$

  • $\begingroup$ It seems you answered all your questions. I suggest you write your answer and accept it. $\endgroup$ May 31, 2014 at 17:34
  • $\begingroup$ I agree.. Moreove, I think that surjectivity is a necessary and sufficient hypothesis. I will think about it a little bit and then write it as an answer! $\endgroup$
    – fritz
    May 31, 2014 at 20:27

1 Answer 1


Given a homomorphism $\theta \, : G \to H$, the formula: $$ m_H(f):=m_G (f\circ \theta) \qquad \forall f \in L^\infty(H, \mathbb R) $$ defines an amenable mean on $H$ iff $\theta$ is surjective.

Firstly I prove the sufficiency conditions:

$\\$1) monotonicity: suppose $f \ge 0$. Then $f\circ \theta \ge 0$ too, hence $m_G (f\circ \theta) \ge 0$

$\\$2) $H$-invariance: let $h \in H$, and let $g \in G$ s.t. $\theta(g) = h$. Then: $$ m_H(h \cdot f) = m_H(f (h \cdot \square)) := m_G(f (h \cdot \theta(\square))) = m_G(f (\theta(g) \cdot \theta(\square))) = m_G(f \circ (\theta(g \cdot \square))) = m_G (f \circ \theta) = m_H(f) $$

$\\$3) additivity: $$ m_H(f_1 + f_2) = m_G((f_1 + f_2) \circ \theta) = m_G(f_1 \circ \theta + f_2 \circ \theta) = m_G(f_1 \circ \theta) + m_G(f_2 \circ \theta) = m_H(f_1) + m_H(f_2). $$

$\\$4) ``normalization''. If $f \equiv 1$, then: $$ m_H(f) = m_G(f \circ \theta) = m_G(1) = 1. $$

Now the necessarity condition: suppose that $\theta \, : G \to H$ is not surjective. Then $|H \,: \theta(G)| \ge 2$. It is easy to see that, if $m_H$ is a mean, then $m_H(\delta_{\theta(G)}) = \frac{1}{|H \,: \theta(G)|} \lneq 1$. But we get instead: $$ m_H(\delta_{\theta(G)}) = m_G(\delta_{\theta(G)} \circ \theta) = m_G(1) = 1. $$


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