2
$\begingroup$

If $G$ is a semigroup and $\beta G$ is the set of ultrafilters on $G$, then $\beta G$ is also a semigroup: given $p,q\in \beta G$, we define $$p*q := \{E\subseteq G : \{g\in G : g^{-1}A\in q\}\in p\}.$$ This is a strange definition at first, but indeed it is an associative binary operation on $\beta G$, and we have an embedding of semigroups $G\hookrightarrow \beta G$ by identifying each element of $G$ with its point-mass.

An ultrafilter $p$ can be identified with its indicator function $\chi_p:\mathcal{P}(G)\rightarrow \{0,1\}$, where $\mathcal{P}(G)$ is the power set of $G$. This is a finitely-additive probability measure on $G$.

Thus let $\mathrm{Pr}^0(G)$ be the space of all finitely-additive probability measures $\mu:\mathcal{P}(G)\rightarrow [0,1]$, so that $\beta G\subseteq \mathrm{Pr}^0(G)$. (Note that we consider every subset of $G$ to be measurable.) Given two probability measures $\mu,\nu\in \mathrm{Pr}^0(G)$, we should be able to convolve them as follows:

  1. Construct the product probability measure $\mu\times \nu$, determined uniquely by the requirement that $(\mu\times\nu)(A\times B)=\mu(A)\nu(B)$ for all $A,B\subseteq G$.

  2. Let the convolution $\mu*\nu$ be the pushforward of $\mu\times \nu$ via the multiplication map $G\times G\rightarrow G$. Thus $$(\mu*\nu)(E) := (\mu\times \nu)\left(\{(x,y)\in G\times G: xy\in E\}\right).$$

Is this correct? This should be an associative binary operation on $\mathrm{Pr}^0(G)$, making it into a semigroup.

Viewing $\beta G$ as a subset of $\mathrm{Pr}^0(G)$ by $p\mapsto \chi_p$, we can ask: is ultrafilter convolution the same as convolution of the corresponding probability measures?

Question: Does $p\mapsto \chi_p$ define a semigroup homomorphism $\beta G \rightarrow \mathrm{Pr}^0(G)$?

$\endgroup$
5
  • $\begingroup$ You have defined $\mu\times\nu$ only on sets of the form $A\times B$. There is no reason to expect this to extend uniquely to all subsets of $G\times G$. $\endgroup$ Mar 6, 2020 at 20:53
  • $\begingroup$ Well, I wouldn't say that there is no reason --- it is indeed true for countably-additive probability measures. Does it fail if we replace "countably-additive" with "finitely-additive"? $\endgroup$
    – Ehsaan
    Mar 6, 2020 at 20:55
  • $\begingroup$ No, even then you can only extend to the $\sigma$-algebra generated by rectangles. $\endgroup$ Mar 6, 2020 at 20:56
  • $\begingroup$ It is reasonable to expect your definition $\mu\times\nu$ to extend to a unique finitely additive probability measure on some algebra of subsets of $G\times G$. But there is no reason that algebra would contain the sets $\{(x,y)\in G\times G:xy\in E\}$ you need to define $\mu*\nu$. $\endgroup$ Mar 6, 2020 at 20:58
  • 1
    $\begingroup$ In the case of commutative $G$, your convolution, if it exists, would also be commutative (just by inspection of the definition). But that's not the case for the product of ultrafilters. $\endgroup$ Mar 6, 2020 at 23:38

0

You must log in to answer this question.

Browse other questions tagged .