# Extending ground model ultrafilters

Work over a model of $\sf GCH$. Suppose that $\kappa$ is some regular cardinal, and $U$ is a uniform ultrafilter on $\kappa^+$.

Does $U$ have some canonical extension after forcing with $\operatorname{Add}(\kappa,\kappa)$?

(Alternatively, replace $\kappa^+$ by an arbitrary regular $\lambda>\kappa$.)

• I can't imagine any of the extensions deserving the name "canonical". Each of the subsets $A$ of $\kappa$ that you've added produces a subset of $\kappa^+$, by chopping the ordinal $\kappa^+$ into blocks of length $\kappa$ and putting $A$ on each block. An extension of $U$ has to choose between this set and its complement, and I see no way to make that choice canonical. Perhaps a more rigorous argument would be that no extension of $U$ can be invariant under the automorphisms of the forcing, not even the automorphisms that interchange the generic sets with their complements. – Andreas Blass Jul 9 '17 at 15:38
• @Andreas: I feared that would be the answer. Thank you for also addressing the automorphisms issue. If you post this as an answer, I will accept it. – Asaf Karagila Jul 9 '17 at 15:41
• Andreas, are there other possibly "slightly more canonical" uniform ultrafilters perhaps, or do all that we can do is use AC to extend the "co-small subsets filter"? – Asaf Karagila Jul 9 '17 at 19:47

I can't imagine any of the extensions deserving the name "canonical". Each of the subsets $A$ of $\kappa$ that you've added produces a subset of $\kappa^+$, by chopping the ordinal $\kappa^+$ into blocks of length $\kappa$ and putting $A$ on each block. An extension of $U$ has to choose between this set and its complement, and I see no way to make that choice canonical. Perhaps a more rigorous argument would be that no extension of $U$ can be invariant under the automorphisms of the forcing, not even the automorphisms that interchange the generic sets with their complements.