9
$\begingroup$

If $U$ is a filter on $X$ and $W$ is a filter on $Y$, their product is the filter $U\times W$ on $X\times Y$ generated by rectangles $A\times B$ where $A\in U$ and $B\in W$.

In certain circumstances (e.g., when $U$ is $|W|$-complete), the product of two ultrafilters $U$ and $W$ is again an ultrafilter. In this situation, $U$ and $W$ must commute in the following sense: for any binary relation $R$, we have $\forall^U x\ \forall^W y\ (x\mathrel{R} y)$ if and only if $\forall^W y\ \forall^U x\ (x\mathrel{R} y)$. (Here, for $P$ a unary predicate, we write $\forall^U x\ P(x)$ to mean that $\{x\in X : P(x)\}\in U$.)

The question is whether the converse holds.

Question: If two ultrafilters $U$ and $W$ commute, must $U\times W$ be an ultrafilter?

Conceivably, a positive answer is provable in ZFC or assuming GCH. The question has a vacuous positive answer under the assumption that there are no measurable cardinals since this implies all commuting ultrafilters are principal. Put another way, constructing a counterexample would require the use of large cardinals.

Background: Ultrafilters $U$ and $W$ such that $U\times W$ is ultra were studied by Blass in his thesis. Blass showed that this is equivalent to the statement that $U$ is complete modulo $W$ in the sense that for any sequence $\langle A_i : i\in I\rangle\subseteq U$ defined on a $W$-large set $I$, $\bigcap_{i\in J} A_i\in U$ for some $W$-large set $J$. In particular, despite all appearances, the relation $U$ is complete modulo $W$ is symmetric in $U$ and $W$.

Commuting ultrafilters are related to Kunen's Commuting Ultrapowers Lemma, which says that if $U$ is $|W|$-complete, then $j_U(j_W) = j_W\restriction \text{Ult}(V,U)$ (which is pretty clear) and $j_W(j_U) = j_U\restriction \text{Ult}(V,W)$ (which is nontrivial). Here $j_U : V\to \text{Ult}(V,U)$ denotes the (transitive collapse of the) ultrapower embedding associated to $U$ and $j_U(j_W) = \bigcup_{x\in V} j_U(j_W\restriction x)$. It is a not-so-easy exercise to see that for countably complete ultrafilters $U$ and $W$ commute if and only if $j_U(j_W) = j_W\restriction \text{Ult}(V,U)$. In particular, despite all appearances, the relation $j_U(j_W) = j_W\restriction \text{Ult}(V,U)$ is symmetric in $U$ and $W$.

| cite | improve this question | | | | |
$\endgroup$
3
$\begingroup$

The following may give a hint: Suppose $U$ is a uniform ultrafilter on $\omega$ and $W$ is an ultrafilter on $\kappa$ such that $W$ commutes with $U$, then $W$ is countably complete.
Otherwise, there exist $\langle A_i \in W: i\in \omega\rangle$ decreasing such that $\bigcap_{i\in \omega} A_i =\emptyset$. Let $R\subset \omega\times \kappa$ be such that $i R \gamma$ iff $\gamma\in A_i$.

  • It can't be the case that $\forall^W \gamma \forall^U i \ iR\gamma$. Since fix some such $\gamma\in \kappa$, we have $\gamma\in \bigcap_{i\in \omega} A_i$ which is impossible.
  • By commuting, it can only be that $\neg\forall^U i \forall^W \gamma \ iR\gamma$, hence $\forall^U i \forall^W \gamma \ \neg iR\gamma$. But this is bogus too since fix some such $i\in \omega$, we have $D\in W$ such that for all $\gamma\in D$, $\gamma\not\in A_i$, contradicting with the fact that $A_i\in W$.
| cite | improve this answer | | | | |
$\endgroup$
  • 1
    $\begingroup$ I can show (using ultrapowers) that two commuting ultrafilters $U$ and $W$ can't be $\delta$-decomposable for a common cardinal $\delta$. Maybe this yields a combinatorial proof of that fact. $\endgroup$ – Gabe Goldberg Feb 10 at 3:53
  • 1
    $\begingroup$ you are absolutely right. this is supposed to be a comment but I can't function on a phone app.... $\endgroup$ – Jing Zhang Feb 10 at 5:09
  • 1
    $\begingroup$ Actually I think I can show the converse in the countably complete case assuming GCH: if $U$ and $W$ are countably complete and relatively indecomposable (i.e., for no cardinal $\delta$ are they both $\delta$-decomposable), then $U\times W$ is an ultrafilter $\endgroup$ – Gabe Goldberg Feb 10 at 20:41
  • $\begingroup$ how does that go? $\endgroup$ – Jing Zhang Feb 10 at 23:45
  • $\begingroup$ The proof is a bit long, I'll send you an email. $\endgroup$ – Gabe Goldberg Feb 10 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.