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Let $(X,\mu)$ be a $G$-space, i.e. a measure space with a Borel quasi-invariant $G$-action. Say that $X$ is amenable (equivalently, that the action is amenable) if there is a $G$-fixed point in every affine space over $X$ with an $\alpha$-twisted action, where $\alpha$ is a corresponding cocycle.

If $X$ is an amenable $G$-space, it follows more or less from definition that for every compact metric space $Y$ there is an $G$-equivariant measurable map $\varphi : X \to M(Y)$, where $M(Y)$ is the collection of probability measures on $Y$.

My question is - is this property equivalent to the above definition of amenable action or strictly weaker?

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What is $G$? a discrete group? countable? what does "Borel" mean in an abstract measure space? –  YCor May 23 '13 at 16:39
    
I meant $G$ to be a locally compact, second countable group. By Borel $G$-action I meant a measurable action with respect to the Borel sets on $G$ and the given $\sigma$-algebra on $X$. Of course, I will be happy to hear an answer under other assumptions (such as $G$ being countable or discrete). –  user33576 May 25 '13 at 14:11
    
What do you mean by an affine space over $X$, and an $\alpha$-twisted action? –  Jesse Peterson May 27 '13 at 22:15
    
@Jesse "unknown (google)" uses the terminology of section 4.3 of Zimmer's Ergodic theory and semi-simple groups, Monographs in Mathematics 81, Birkhäuser (1984), where all the relevant terminology and references can be found. The question asks whether the property proved in Zimmer's proposition 4.3.9 characterizes amenable actions. The definitions are also available in the preliminaries of this article by Zimmer: dx.doi.org/10.1090/S0002-9904-1977-14392-9 (for the purposes of this question one can ignore the ergodicity assumption made there). –  Martin May 28 '13 at 2:17
    
@Martin - you are absolutely right - that is an impressive identification. Thank you for posting a link to the somewhat technical definitions that appear in the question. –  user33576 May 28 '13 at 5:18

1 Answer 1

up vote 5 down vote accepted

Here is a sketch of an argument to show these properties are equivalent which is based on the proof of the same property for amenable groups. To see this I will use the fact that an action $G \curvearrowright (X, \mu)$ is amenable if and only if there is a $G$-equivariant mean from $L^\infty(X, \mu) \overline \otimes L^\infty(G)$ to $L^\infty(X, \mu)$. (A mean in this setting is a positivity preserving map which restricts to the identity on $L^\infty(X, \mu)$.) This is Theorem A (iv) in "Amenable Actions of Groups" by Adams, Elliott, Giordano (1994) (http://www.ams.org/mathscinet-getitem?mr=1250814), which generalizes the case for countable groups established by Proposition 4.1 in "Hyperfinite Factors and Amenable Ergodic Actions" by Zimmer (1977) (http://www.ams.org/mathscinet-getitem?mr=470692).

Theorem: Let $G$ be a locally compact second countable group. Then there exists a $G$-equivariant mean from $L^\infty(X, \mu) \overline \otimes L^\infty(G)$ to $L^\infty(X, \mu)$ if and only if for every continuous action of $G$ on a compact metric space $Y$, there exists a $G$-equivariant map from $X$ to $M(Y)$.

Proof: For the proof I will find it easier to work with function spaces, and so first note that for a compact Hausdorff space $K$, there is a one-to-one correspondence between (equivalence classes of) Borel maps $\pi: X \to M(K)$ and means $\Phi: L^\infty(X, \mu) \otimes C(K) \to L^\infty(X, \mu)$. (Here, $\otimes$ is the $C^*$-tensor product.) This is essentially just the Riesz representation theorem applied to each fiber $K$.

We now show the "only if" direction. If $G \curvearrowright K$ is a continuous action on a compact metric space, by restricting to a closed $G$-invariant subset we may assume that $G \curvearrowright K$ has a dense orbit $K = \overline{G k}$. We then obtain a $G$-equivariant embedding $\rho: C(K) \to C_b(G) \subset L^\infty(G)$ by $\rho(f)(g) = f(gk)$. Restricting our invariant mean to $L^\infty(X, \mu) \otimes \rho(C(K))$ then gives the result.

The converse is a bit more involved, but this is well known for the case of amenable groups. The first step is to show that we have an equivariant mean on the space $UC_b(G)$ of bounded left uniformly continuous functions. In the case when $G$ is countable we are then done. For the general case we then use an approximate identity for convolution to produce an equivariant mean for $L^\infty(G)$.

To produce a mean on the space $UC_b(G)$ note that for any second countable $G$-invariant $C^*$-subalgebra $A \subset UC_b(G)$ we have that the Gelfand spectrum is metrizable, and $G$ acts continuously. Therefore by hypothesis there exists a $G$-equivariant mean $\Phi_A: L^\infty(X, \mu) \otimes A \to L^\infty(X, \mu)$. Since $L^\infty(X, \mu)$ is injective we may then extend $\Phi_A$ to a (perhaps no longer $G$-equivariant) positivity preserving map $\widetilde{\Phi_A} : L^\infty(X, \mu) \otimes UC_b(G) \to L^\infty(X, \mu)$. If we consider the net $\{ \widetilde{\Phi_A} \}$ indexed by the set of all second countable $G$-invariant $C^*$-subalgebras $A$, and ordered by inclusion, then we may take an accumulation point $\Phi$ in the topology of point-wise weak convergence. Since $\Phi$ is $G$-equivariant when restricted to $L^\infty(X, \mu) \otimes A$ for any second countable $C^*$-subalgebra $A$, it follows that $\Phi$ is $G$-equivariant on the whole space.

To produce a mean on $L^\infty(X, \mu) \overline \otimes L^\infty(G)$ we start by taking an approximate identity $\{ \psi_n \} \subset C_c(G)$. Specifically, we want that each $\psi_n \in C_c(G)$ is a non-negative function, $\| \psi_n \|_1 = 1$, ${\rm supp}(\psi_n) \to \{ e \}$, and $\| \psi_n * \delta_g - \delta_g * \psi_n \|_1 \to 0$ for each $g \in G$. If $f \in L^\infty(X, \mu) \overline \otimes L^\infty(G)$, then we have $\psi_n * f \in L^\infty(X, \mu) \otimes UC_b(G)$ for each $n$ (I'm taking convolution pointwise in $X$), and $\| \sigma_g(\psi_n * f) - \psi_n * (\sigma_g(f)) \|_\infty \to 0$ for each $g \in G$, and $f \in L^\infty(X, \mu) \overline \otimes L^\infty(G)$. (We denote by $\sigma_g$ the action of $G$ on $L^\infty(X, \mu) \overline \otimes L^\infty(G)$.) If we set $\Phi_n : L^\infty(X, \mu) \overline \otimes L^\infty(G) \to L^\infty(X, \mu)$, by $\Phi_n( f ) = \Phi( \psi_n * f )$, then it follows that any accumulation point of $\{ \Phi_n \}$ gives us a $G$-equivariant mean. $\blacksquare$

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@Peterson Thank you very much for this detailed answer. This both answers my original question and introduces me to new (for me) concepts. –  user33576 May 29 '13 at 16:43

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