Some kind of idempotence of dense filter

Question

In discussion of following questions question1, question2, question3 became clear (see definitions in question3 ) that for the Frechet filter $\mathcal{N}$ we have $\mathcal{N}\nsim\mathcal{N}\otimes\mathcal{N}$ and for any ultrafilter $\mathcal{U}$ the situation is the same $\mathcal{U}\nsim\mathcal{U}\otimes\mathcal{U}$. But for any bijective mapping $\varphi:\omega\times\omega\to\omega$ there exists the filter $\mathcal{F}$ with $\mathcal{N}\subset\mathcal{F}$ such that $\varphi(\mathcal{F}\otimes\mathcal{F})=\mathcal{F}$ and thus $\mathcal{F}\sim\mathcal{F}\otimes\mathcal{F}$.

So, my following question has chance to receive positive answer. Let us define the density of subset $A\subset\omega$ : $$\rho(A)=\lim_{n\to\infty}\frac{|A\cap n|}{n}$$ if the limit exists. Let $\mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}$. $\mathcal{F_1}$ is the filter and $\mathcal{N}\subset\mathcal{F_1}$. For arbitrary (selective) ultrafilter $\mathcal{U}$ let $\mathcal{F}=\mathcal{F_1}\cap\mathcal{U}$.

Question: is there exists a bijection $\varphi:\omega\times\omega\to\omega$ such that $$ \varphi(\mathcal{F}\otimes\mathcal{F})\subset\mathcal{U} $$

Answer 1

The answer is negative. See proof given by Andreas Blass here. Main idea: filter $\cal{F}\otimes\cal{F}$ must contain $\cal{N}\otimes\cal{N}$ because any free filter contains $\cal{N}$ and thus $\cal{N}\subset\cal{F}$. Then $\varphi^{-1}(\cal{U})$ is selective ultrafilter in $\omega^2$ as bijective image of selective ultrafilter. But selective ultrafilter on $\omega^2$ can not contain $\cal{N}\otimes\cal{N}$.