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 $W$.


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 $A_i\in W$.
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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