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Let $(S,*)$ be the free (non associative) binary system on one generator (so $S$ is just the set of terms in $*$ and $1$). There is an extension of $*$ to the space $P(S)$ of finitely additive probability measures on $S$ defined as follows: $$\mu * \nu (A) = \int \int \mathbf{1}_{*^{-1}(A)} (x,y)\ d \nu (y)\ d \mu (x)$$ Here is the question: Is there an idempotent measure in $P(S)$ (i.e. a $\mu$ such that $\mu * \mu = \mu$)? Has this question been considered in the literature?

Here is the motivation: there is a natural way to identify $(S,*)$ with the positive elements of Thompson's group $F$ (elements of $S$ are "rooted ordered binary trees"). It is not hard to show that an idempotent measure is in fact an invariant measure with respect to the action of $F$. Now, I don't expect someone to produce a positive answer (although I will conjecture the even stronger statement that every compact convex $C \subseteq P(S)$ which is $*$-closed contains an idempotent). I am mostly asking if someone sees how to refute the existence of such a measure or if this question has appeared in the literature.

Some further observations: the map $(\mu,\nu) \mapsto \mu * \nu$ is NOT continuous ($\mu \mapsto \mu * \nu$ is, for each $\nu$, but this is about the extent of continuity). Thus the map $\mu \mapsto \mu * \mu$ is not continuous (if it were, we could apply a fixed point theorem...). If one drops the assumption of freeness, then it is possible to find idempotents if any of the following are true:

  • $S$ is finite (in this case everything is continuous and so fixed point theorems apply).
  • $(S,*)$ is associative (i.e. $S$ is a semigroup) (this is "Ellis's Lemma'').
  • $*$ depends only on one argument (again, fixed point theorems apply).

An auxiliary question is to characterize when a monogenic binary system $(S,*)$ satisfies that $P(S)$ contains an idempotent. It should be noted that for associative binary operations, it is possible to find an idempotent ultrafilter (i.e. taking values in $\{0,1\}$) but that this is impossible for the free (non associative) binary system one one generator.

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This is not an answer but perhaps it gives an idea where to look for it. I think that existence of idempotent measures on $\mathbb N$ is the key point in Furstenberg's proof of van der Waerden's theorem and its generalizations and that one can view van der Waerden's theorem as a reformulation of existence of idempotent measures (I may be wrong here). An analog of van der Waerden's theorem is true for the free binary system (groupoid) with $n$ generators: for every decomposition of $A$ into $k$ parts there exists a polynomial $p(x)$ (i.e. a term with one variable) such that all elements $p(x_i), i=1,...,n,$ belong to the same partition class (it is not difficult and was proved by Bespamyatnyh and myself in 1982 but could have been proved by somebody else before). I do not know if it implies existence of idempotent measure, but it may be worthwhile looking at it.

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@Mark: Isn't this addressed in item 2 in my question (Ellis's Lemma)? Associativity is really key to that argument. It should be noted that this Lemma produces an ultrafilter measure which is idempotent. It is not hard to show that for the free binary system S, there is no idempotent ultrafilter (if one defines the right depth by r(A*B) = 1 + r(B), r(1) = 0, then any idempotent measure must give measure 1/2 to those elements of even and of odd right depth). –  Justin Moore Mar 9 '11 at 14:13
    
@Justin: My point that vdW theorem holds in the non-associative case too. So perhaps there is a way to bi-pass Ellis' lemma. –  Mark Sapir Mar 9 '11 at 14:33
    
I always thought that, van der Waerden's theorem is primarily associated with minimal (left) ideals in the semigroup of ultrafilters, not with idempotent ultrafilters (though those are used in the proof). The statement commonly associated with idempotent ultrafilters is Hindman's theorem. –  Greg Graviton Mar 9 '11 at 15:21
    
@Greg: It is quite possible you are right. As Furstenberg showed Hindman's theorem and vdW theorem are "related". I do not know whether an analog of Hindman's theorem holds in non-associative algebras, never thought about it. –  Mark Sapir Mar 9 '11 at 15:49
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