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Let $\Gamma$ be an amenable group. Consider its ultrapower $^*\Gamma$. It is known that $^*\Gamma$ need not be amenable. In fact, there is a stronger notion of uniform amenability for $\Gamma$ (defined using a uniform condition of Folner convergence) and it is known that $^*\Gamma$ is amenable iff $\Gamma$ is uniformly amenable.

My question is: if $\Gamma$ is amenable, is $^*\Gamma$ amenable at least in some weaker sense?

For instance, is there a finitely additive $^*\Gamma$ invariant measure defined on internal subsets of $^*\Gamma$ (as opposed to all subsets as needed for amenability)? One attempt at constructing this would be to take an internal subset $A=(A_n)_{n \in \mathcal{U}}$ and taking the limit (wrt the ultrafilter) of the measures of each individual subset. Similarly, can we also construct an invariant mean for internal bounded functionals using the ultralimit?

However, I am not clear if the right notion should involve a measure that takes values in $^*\mathbb{R}$ or just $\mathbb{R}$. Has this been studied before? Thanks.

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    $\begingroup$ What is a problem in defining the measure of an internal set of the ultraproduct as the ultralimit of measures? $\endgroup$
    – user6976
    Commented Oct 19, 2019 at 22:12
  • $\begingroup$ Isn't the natural setting for such a question arbitrary ultraproducts (rather than ultrapower of a single group?) $\endgroup$
    – YCor
    Commented Oct 20, 2019 at 8:25

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I'm not sure if there's ultimately a good way to say this in a more direct way than just "$^*\Gamma$ is an ultrapower of an amenable group" but you can get pretty much what you want either with $^*\mathbb{R}$ or with $\mathbb{R}$. It's fairly clear how the picture works with $^*\mathbb{R}$ (as in you're just doing non-standard analysis), so lets see how the story works with the standard reals.

A discrete group $\Gamma$ is amenable if and only if there is a left invariant mean $\Lambda$ on $\ell^\infty(\Gamma)$, where a mean is a norm 1 linear functional which is non-negative, i.e. if $f \in \ell^\infty(\Gamma)$ is non-negative, then $\Lambda(f) \geq 0$. These elements together have a fair amount of structure, aside from the left- and right-hand actions of $\Gamma$ on $\ell^\infty (\Gamma)$, $\ell^\infty(\Gamma)$ is also a Banach space and a $C^\ast$-algebra, with its multiplication and involution.

We can bundle all of this information into a single two-sorted 'metric structure' (in the sense of continuous logic) $\mathfrak{G} = (\Gamma, \ell^\infty(\Gamma), \times _ \Gamma , \left\lVert \cdot \right\rVert, +,\times,^\ast, s_\lambda, a_{l}, a_{r}, e, \Lambda)_{\lambda \in \mathbb{C}}$, where $s_\lambda$ is scaling by the complex number $\lambda$, $a_{l,r}:\Gamma \times \ell^\infty(\Gamma) \rightarrow \ell^\infty(\Gamma)$ are the left and right actions, and $e: \Gamma \times \ell^\infty(\Gamma) \rightarrow \mathbb{C}$ is the evaluation map (really we would code this as its real and imaginary parts). (There's an unimportant subtlety I'm glossing over in that we actually need to do something tedious and/or tricky to deal with the fact that $\ell^\infty(\Gamma)$ is unbounded as a metric space, but don't worry about it.) What's important about making this work is that all of the relevant functions are uniformly continuous (on bounded subsets of $\ell^\infty(\Gamma)$) which ensures that they still make sense in the ultrapower. Any continuous first-order properties of this structure will be inherited by the ultrapower. So lets let $\mathfrak{E} = (^\ast \Gamma, A,\dots)$ be the ultrapower of $\mathfrak{G}$. Here are some of the properties that you get easily:

  • As you know $^\ast \Gamma$ is still a discrete group, satisfying any ordinary discrete first-order properties of $\Gamma$.
  • $A$ is a $C^\ast$-algebra, and in particular is the ultrapower of $\ell^\infty(\Gamma)$ in the sense that $C^\ast$-algebraists mean when they talk about ultrapowers. The norm takes values in the standard reals.
  • $a_{l,r}^{\mathfrak{E}}$ are left and right $^\ast \Gamma$ actions on $A$ (preserving all of the $C^\ast$-algebraic structure).
  • $\Lambda^{\mathfrak{E}}$ is a left invariant mean on $A$.
  • $A$ is a sub-$C^\ast$-algebra of $\ell^\infty(^\ast \Gamma)$. This follows from the fact that $\left\lVert f \right\rVert = \sup_{\gamma \in \Gamma} |f(\gamma)| $ is a continuous first-order property of $\mathfrak{G}$ (expressible in terms of the evaluation map $e$).
  • The idempotent elements of $A$ (i.e. $xx = x$) correspond to the internal subsets of $^\ast \Gamma$ as you understand them, the evaluation map $e^{\mathfrak{E}}$ encodes set membership. The fact that $A$ contains all of the indicator functions of finite and co-finite sets is first-order. The mean induces a finitely additive ($\mathbb{R}$-valued) measure on internal subsets.

So at the very least the weak form of amenability that you 'obviously' get is that there is a sub-$C^\ast$-algebra $A$ of $\ell^\infty(^\ast\Gamma)$ which contains $c(^\ast \Gamma)$ and is closed under the induced left- and right-hand actions of $^\ast\Gamma$ and which has a left invariant mean. Another consequence of what we did here is that this weak form of amenability is preserved under passing to ultrapowers.

That weak form of amenability by itself, as far as I know, may be trivial. Someone who actually knows something about amenable groups or $C^\ast$-algebras can probably tell you whether or not every discrete group $\Gamma$ admits such a $C^\ast$-algebra. Getting something stronger would require some work, in that you would need to establish that a property you care about can be expressed in continuous first-order logic.

None of this is terribly surprising, but in my (admittedly biased) opinion this perspective helps organize the structure that is preserved by ultrapowers.

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