# Is the product of commuting ultrafilters an ultrafilter?

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 $$A_i\in W$$.
• 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. – Gabe Goldberg Feb 10 at 3:53
• 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 – Gabe Goldberg Feb 10 at 20:41