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I am writing a paper for game theorists where I use (countable) amenable groups to do some things. So I am writing up a preliminary section about countable amenable groups whose main purpose is to convince the reader that invariant means are the generalization of the uniform measure.

Well, the first obvious thing to say is that every finite group is amenable and the uniform measure is its unique invariant mean. But this sounds like a merely technical reason. Well, I have at least one deeper reason. Let $G$ be a group, fix $W\subseteq G$ and consider the two-person game where the two players choose $x,y\in G$, respectively, and player 1 wins if $xy\in W$. For generic $W$ we can consider this game as a game without relevant information and therefore, a game theoretical reformulation of the Maximum Entropy Principle, would predict that the players will play casually (whatever it means). Indeed, if $G$ is finite, a Nash equilibrium is given by the uniform measure. The point is that if $G$ is countable amenable, then this game admits Nash equilibria and they are exactly suitable invariant means.

I think this is already a pretty convincing motivation, but I would like to include some more reason.

I am certainly open to any other proposal. In particular, I know that amenable groups and invariant means are used quite a lot in ergodic theory. This field sounds like something able to produce an example of the kind I am looking for. I am not expert in ergodic theory and so I am wondering whether there is some result that can be used as a support of the interpretation of invariant means as the generalization of the uniform measure. I hope there are many and, in case, I would like to know one which is easy to state and to understand by somebody that may have no background in ergodic theory, but may have, potentially, a background in classical probability theory. Summarizing,

Question. Are there ergodic theoretical results that really support the interpretation of invariant means as the generalization of invariant measure for groups?

Thanks in advance,

Valerio

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I can suggest a non-ergodic argument. A finite invariant measure allows you to assign the mean value to every integrable function. An invariant mean assigns a mean value to every bounded function. For example, on Z you get a "limit at infinity" of every bounded sequence. This was the initial motivation of S. Banach, as far as I know. –  Yulia Kuznetsova Jun 5 '12 at 15:01
    
I am not against non-ergodic argument, but in its present form I don't think this is very convincing. Why a mean value defined using an invariant mean is better than a mean value defined used a non-invariant measure? What are properties that definitely force the use of an invariant mean? Concerning the limit of every bounded sequence, I think this is a property of ultrafilters (and not, in general, invariant means), which are one of the least invariant things. Am I misunderstanding anything? –  Valerio Capraro Jun 6 '12 at 19:38
    
It's hard to say. Invariant means are typically non-unique, while uniqueness seems important to my intuitive concept of uniform measure. Also, they tend only to be finitely additive, which rather changes things from the uniform case. (Merely finitely additive probabilities have some rather counterintuitive consequences, mainly due to nonconglomerability.) I have been exploring invariant means as a generalization of uniform measure, but I think how good a generalization they are is up for grabs. –  Alexander Pruss Aug 24 '13 at 19:01
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