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Let $S$ be an infinite set and $f:S\times S\rightarrow\mathbb R$ be bounded.

Question: Are there simple hypotheses on $f$ such that there is a commutative and associative operation $\cdot$ on $S$ such that $f(x,y)=\phi(x\cdot y)$, for some $\phi$?

Example: if $f(x,y)=f(y,x)$, we can construct a commutative $\cdot$ as follows: since $S$ is infinite, the image of $f$ has cardinality at most $|S|$ and then there is an injective mapping $\Psi:Im(f)\rightarrow S$. Define $x\cdot y=\Psi(f(x,y))$. This is a commutative operation which verifies the required property, but, of course, in general it is not associative. And it is not clear to me how to find a simple enough condition that guarantees associativity.

Thanks in advance for any help,

Valerio

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I do not think there is a nice solution. But the problem can be simplified a bit. First the condition that $f$ is bounded is irrelevant since you can always compose with $\arctan$. Second, the condition that $f$ is a function into $\mathbb{R}$ is not important, and in fact the only thing you need from $f$ is its kernel, i.e. the equivalence relation on $S\times S$: $(a,b)\sim (c,d)$ iff $f(a,b)=f(c,d)$. Thus the problem is this: suppose that we are given an equivalence relation $\sim$ on $S\times S$ such that $(a,b)\sim (c,d)\to (b,a)\sim (d,c)$. When will there exists an associative commutative operation $\cdot$ such that $(a,b)\sim (c,d)$ if $a\cdot b=c\cdot d$. One can of course use the structure theory of commutative semigroups (these are more complicated than commutative groups but still manageable) to obtain more information but I am not sure very much can be achieved.

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  • $\begingroup$ Many thanks for the nice reformulation. Could I ask for some reference to find infos about this structure theory of commutative semigroups? $\endgroup$
    – Valerio Capraro
    Commented Jan 10, 2012 at 21:34
  • $\begingroup$ We have one additional piece of information about the equivalence relation, namely that the number of equivalence classes is at most the cardinal of the continuum. Unfortunately, that doesn't seem to make the problem any easier. $\endgroup$
    – Andreas Blass
    Commented Jan 10, 2012 at 22:21
  • $\begingroup$ I do not know books where the theory of commutative semigroups was described. There are standard books: Liapin, Clifford and Preston, Grillet, Higgins, Howie, etc. You can start with an easier case when $\cdot$ is a group operation or even a finite commutative group operation ($S$ is finite). But I do not think that in general one can say more than that $\sim$ comes from a commutative associative operation. If you need specific properties of $\sim$, these may be possible to prove using the structure of commutative semigroups. I do not think a non-tautological characterization is possible. $\endgroup$
    – user6976
    Commented Jan 10, 2012 at 23:21
  • $\begingroup$ My girlfriend, who studies philosophy, told me once that philophers' opinion of Maths is that it is a big tautology (and I don't know how to prove they're wrong). Anyway, also I suspected that one can't say more, but I don't use to trust of me much, above all because I find strange that commutativity can be described so easily but, on the other hand, there is apparently no way to describe associativity. $\endgroup$
    – Valerio Capraro
    Commented Jan 11, 2012 at 8:23
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Here is an approach you might take. Consider the class of , oh, lets call them left-sections of f(x,y): fix x and define g_x(y) = f(x,y). Consider the equivalence classes induced by level sets of g_x. Associativity combined with the shape of the level sets of g_x, g_z, and g_w will determine whether it is feasible to have w = x.z . At the very least you should have some interesting conditions on f which will be necessary for the representation. It may be that Green's relations will prove useful here.

Gerhard "Not an Expert in Semigroups" Paseman, 2012.01.11

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