TITLE: Idempotent measures on the free binary system?
QUESTION [13 upvotes]: 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.

REPLY [2 votes]: 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.