If ${\cal U}$ and ${\cal V}$ are ultrafilters on non-empty sets $A$ and $B$ respectively, then the *tensor product* ${\cal U}\otimes{\cal V}$ is the following ultrafilter on $A\times B$:
$$\big\{X\subseteq A\times B: \{a\in A:\{b\in B: (a,b)\in X\}\in {\cal V}\}\in {\cal U}\big\}.$$
It is a standard exercise to verify that this is an ultrafilter on $A\times B$.


We say two non-principal ultrafilters ${\cal U, V}$ are *Rudin-Keisler equivalent* if there is a bijection $\varphi:\omega\to\omega$ such that ${\cal U} =\varphi(\cal V)$, where $\varphi({\cal V}) = \{\varphi(R): R\in {\cal V}\}$.


We say an ultrafilter ${\cal Z}$ on $\omega$ is *Tensor-representable* if there are non-Keisler-Rudin-equivalent ultrafilters ${\cal U}, {\cal V}$ and a bijection $\psi:\omega^2\to \omega$ such that ${\cal Z} =\psi({\cal U}\otimes {\cal V})$.


What is an example of an ultrafilter ${\cal Z}$ on $\omega$ that is not Tensor-representable?



Non-tensor-representable ultrafilters on $\omega$
Recall that $\mathcal Z$ is a *weak $P$-point* if it is not in the closure of any countable subset of $\omega^\* \setminus \{\mathcal Z\}$. A weak $P$-point is never the tensor product of two non-principal ultrafilters.


To see this, suppose $\mathcal U$ and $\mathcal V$ are two non-principal ultrafilters on $\omega$, and let $\mathcal Z = \mathcal U \times \mathcal V$. I will show that $\mathcal Z$ is not a weak $P$-point in the space of non-principal ultrafilters on $\omega \times \omega$. (This implies that $\psi(\mathcal Z)$ is not a weak $P$-point in $\omega^\*$ for any bijection $\psi: \omega \times \omega \rightarrow \omega$.)


For each $i \in \omega$, let 
$$v\_i = \{\{i\} \times B \,:\, B \in \mathcal V\},$$
and observe that $v\_i$ is a non-principal ultrafilter on $\omega \times \omega$. The set $\{v\_i \,:\, i \in \omega\}$ is a countable, relatively discrete subset of $(\omega \times \omega)^\*$. (It is relatively discrete because for each $i \in \omega$, $U\_i = (\{i\} \times \omega)^\*$ is a (cl)open subset of $(\omega \times \omega)^\*$ such that $U\_i \cap \{v\_i \,:\, i \in \omega\} = \{v\_i\}$.) Clearly $\mathcal Z$ is not equal to any of the $v\_i$. Yet we have $\mathcal Z \in \overline{\{v\_i \,:\, i \in \omega\}}$. Indeed, we see from the definition of $\mathcal Z$ that every neighborhood of $\mathcal Z$ in $(\omega \times \omega)^\*$ contains $\mathcal U$-many of the $v\_i$. (In what might be more familiar notation, this means $\mathcal Z \,=\, \mathcal U\hspace{-.5mm}-\lim\_{i \in \omega} v\_i$.) Thus $\mathcal Z$ is not a weak $P$-point.


Finally, let me mention that Kunen famously proved that $\omega^\*$ contains weak $P$-points. Thus the above observation shows that some ultrafilters are not representable as tensor products.


