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Suppose that $(S,*)$ is a finite set equipped with a binary operation. Extend the binary operation to the vector space $V$ with basis $S$. The set of probability measures on $S$, viewed as a compact convex subset of $V$ is closed under $*$ and, since $*$ is continuous, there are idempotent measures on $S$.

Must two idempotent measures on $S$ have distinct supports?

I am also interested in the more general question where the assumption of finiteness is dropped and one considers the extension (by convolution) of $*$ to the family of all finitely additive measures on $S$ (in that context, define the support of a measure to be all subsets of $S$ with positive measure).

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  • $\begingroup$ Here "binary operation" on $S$ can be just any map $S\times S\rightarrow S$, or special properties are assumed? $\endgroup$
    – Pietro Majer
    Commented Nov 30, 2012 at 16:24
  • $\begingroup$ Any operation. Nothing is assumed. $\endgroup$
    – Justin Moore
    Commented Nov 30, 2012 at 16:28
  • $\begingroup$ If (S,*) is a group, then the idempotent measures are exactly the uniform measures on the subgroups of S. $\endgroup$
    – Justin Moore
    Commented Nov 30, 2012 at 16:29
  • $\begingroup$ It is false for semigroups. The idempotent measures are obtained as follows. Take a regular J-class, a measure on its set of L-classes, a measure on its set of R-classes and uniform measure on its maximal subgroup. The corresponding product measure on the J-class is idempotent and they all look like this for compact topological semigroups. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 30, 2012 at 16:39
  • $\begingroup$ I meant above uniform measure on a subgp of a max subgp. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 30, 2012 at 16:40

1 Answer 1

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No if I understood. Take the two element left zero semigroup. All measures are idempotent.

Added a left zero semigroup is one satisfying the identity xy=x

Added A finite semigroup $S$ satisfies that distinct idempotent measures have distinct support iff for all idempotents $e,f\in S$ one has $SeS=SfS$ implies $e=f$.

A finite semigroup has a unique idempotent measure with full support iff it is a finite group.

Here is how the proof goes. Suppose $P$ is an idempotent measure on $S$ and assume the support of $P$ is $S$.

Claim 1: $S$ contains no ideal.

Proof. Obviously every state is recurrent for both the right random walk an left random walk on $S$ driven by $P$ because $P$ is idempotent and the support is $S$ (and hence $P$ is stationary for these walks). If $I$ was a proper ideal, then the states in $S\setminus I$ would have to be transient because they fall into $I$ with positive probability. Thus $S$ has no proper ideals.

By Rees's theorem, if $e$ is an idempotent of $S$ then $eSe$ is a group $G$, there are sets $A$ and $B$, and a mapping $P\colon B\times A\to G$ such that $S\cong A\times G\times B$ with multiplication $$(a,g,b)(a',b',b') = (a,gP(b,a')g',b).$$

It is trivial to check that if $Q$ is a measure on $A$, $U$ is uniform measure on $G$ and $R$ is a measure on $B$, then $Q\times U\times R$ is an idempotent measure on $S$. The fact that all idempotent measures $A\times G\times B$ are of this form is an easy calculation using that the only idempotent measure on a finite group with global support is the uniform measure. The key point is to first show that the measure is uniform on subsets of the form $\{a\}\times G\times \{b\}$ using the group result.

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  • $\begingroup$ In general if a finite semigroup has a regular J-class which is not a group then it has distinct idempotent measures with the same support. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 30, 2012 at 16:32
  • $\begingroup$ Of course they all are, unless your product is not the usual one. Suppose x,y are the two elements and p+q=1 with p,q\geq 0. Then $(px+qy)(px+qy) = p^2x+pqx+qpy+q^2y=p(p+q)x+q(p+q)y = px+qy$. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 30, 2012 at 18:21
  • $\begingroup$ Here is a reference for the characterization of idempotent measures on compact semigroups ams.org/journals/proc/1963-014-01/S0002-9939-1963-0169971-2/… There is a generalization by John Pym for compact semitopological semigroups. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 30, 2012 at 18:30
  • $\begingroup$ Yes, sorry you are right. I deleted my comment. $\endgroup$
    – Justin Moore
    Commented Nov 30, 2012 at 18:33
  • $\begingroup$ It is amusing that groups are the only finite semigroups which admit a unique idempotent probability measure with global support. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 30, 2012 at 18:41

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