Some time ago, I looked into arithmetical consequences of the set-theoretic multiverse.

I started with the assumption that there are numbers that are standard, which I will denote $\omega$ and that we have some useful conception of subsets of $\omega$, I found that if two multiverses are compatible in the sense that they are both elements of some larger universe then they must have the same standard system. There is also an interesting converse to this that I discuss in my post.

In detail, if $U$ is a universe then elements $n$ of $\omega^U$ code certain subsets of $\omega$. Namely, $n$ codes the set $x_n$ of all standard $i \in \omega$ such that the $i$-th binary digit of $n$ is $1$. The collection of all sets coded by elements of $\omega^U$ in this way is called the standard system of $U$ and is denoted $\operatorname{SSy}(U)$.

If a universe $U$ is an element of a larger universe $V$ then we must have $\operatorname{SSy}(U) = \operatorname{SSy}(V)$. Indeed, because $V$ can recognize that $\omega^U$ is a model of a reasonable arithmetic theory, $V$ realizes that $\omega^V$ is an initial segment of $\omega^U$ since $V$ believes that $\omega^V$ is *the* standard model of arithmetic. From this, we find that $U$ and $V$ have the same standard system. Indeed, if $p$ is any nonstandard element of $V$ then every $x \in \operatorname{SSy}(V)$ is coded by some $n < 2^p$, and such $n$'s are both in $U$ and $V$.

Note that the assumptions I used are debatable, as discussed by Joel in this post. I never got around to writing the sequel of this post where I weaken the hypotheses somewhat. Perhaps I will find some time now that I am reminded of this...

Also note that compatibility, as defined above, is much stronger than your joint embedding property. However, the underlying reason why I chose to look at compatibility rather than embeddings remains: how would we know that such embeddings exist unless such embeddings exist in some larger universe?