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It is a very standard fact that commutative semigroups admit an invariant mean and the proof basically relies on Markov-Kakutani fixed point theorem. Now, it seems to me that the proof of this theorem does not need the associativity (see for instance http://www.springerlink.com/content/g777570166731376/) and then the answer to the following question should be affirmative.

Question: Recall that a magma is a set equipped with a binary operation. Is it true that a commutative countable magma admits an invariant mean?

So, is that trivially true or am I missing some detail?

Thanks in advance,


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mmm.. following the classical proof word by word, it seems that associativity is necessary to prove that the natural action of the magma $M$ on the set $\mathcal M$ of probability measures on $M$ is commutative. Anyway, let me give the explicit example: $M=[-k,k]\subseteq \mathbb Z$ and $$ x\cdot y=\max(\min(x+y,k),-k) $$ Maybe in this case the existence of an invariant mean follows just from the fact that $M$ is finite. –  Valerio Capraro Jun 5 '11 at 13:28

2 Answers 2

Consider the free group $F$ generated by $a,b$ with the new operation: $u*v=$ the shortlex smallest word among $uv$ and $vu$ assumming $a>b>a^{-1}>b^{-1}$. This is a commutative magma. Consider the decomposition $F=F_a\cup F_b\cup F_{a^{-1}}\cup F_{b^{-1}}\cup \{1\}$ where $F_x$ is the set of (reduced) words starting with $x$. Would not this be a paradoxical decomposition of the magma? This would show that the magma does not have an invariant mean.

Update Here is the proof. Suppose that there exists an invariant mean $m$. Note that for every reduced word $au$, $a^{-1}*au=u$ because $a$ is the biggest letter. Hence $a^{-1}F_a=F_b\cup F_{b^{-1}}\cup F_a\cup \{1\}$. Therefore $m(F_b)=m(F_{b^{-1}})=m(\{1\})=0$. Consider the set $F_{b,b'}$ of reduced words starting with $b$ and ending not in $b$. Then $b^{-1}F_{b,b'}$ contains all words starting in $a$ or $a^{-1}$ and ending not in $b$. Therefore the set of all words starting with $a$ or $a^{-1}$ and ending in $b$ has a full measure. Taking a word starting in $b^{-1}$ and ending in $b$ and multiplying it with $b$ on the left produces any word ending in $b$. Therefore the set of such words has measure 0, a contradiction.

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Thank you very much! –  Valerio Capraro Jun 5 '11 at 23:47

For the ones of you that are interested, here is a "finite" counterexample.

(How can I do the braces? I am going to use the round bracket, instead of brace)

Let $M=(-2,-1,0,1,2)$ equipped with the commutative operation $x\cdot y=\max(\min(x+y,2),-2)$. Observe that $1\cdot(-2)=(-1)$, and then the measure of the singleton $(-1)$ should be equal to the measure of the singleton $(-2)$. At the same way, one gets that the only possible invariant mean would be the uniform measure, but this is not invariant since $-2\cdot(-2,-1)=(-2)$.

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I'm very confused, because it seems to me that this same idea shows that any finite 1-generated monoid would be a counterexample unless it's actually a group. Apparently not - what's wrong with my thinking? –  Harry Altman Jun 6 '11 at 0:33
Why is the measure uniform? A finite semigroup is amenable iff it has unique minimal right ideal. But with your proof, the semigroup $0,a$ with zero product is not amenable. –  Mark Sapir Jun 6 '11 at 18:10
Why is not amenable? The measure that gives measure $1$ to the element $0$ and measure $0$ to $a$ is invariant with respect to the zero-product. In my example, I have proved that one can translate the singleton $(−2)$ on the element $(−1)$ and so on, so that all singletons should have the same measure with respect to a possible invariant mean, but this is not the case because at the same time the set $(−2,−1)$ should have the same measure of the singleton $(−2)$. It seems to me this does not contradict your example, because it is not possible to translate $0$ over $a$. Am I missing something? –  Valerio Capraro Jun 6 '11 at 23:15

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