# Does $V = \textit{Ultimate }L$ imply GCH?

In his Midrasha Mathematicae lectures ("In Search of Ultimate $L$", BSL 23 [2017]: 1–109), Woodin notes that $V = \textit{Ultimate }L$ implies $\textrm{CH}$ (Theorem 7.26, p.103). Is it known whether $V = \textit{Ultimate }L$ implies $\textrm{GCH}$?

## 2 Answers

In his slide Absolutely ordinal definable sets John Steel writes:

At the same time, one hopes that V = ultimate L will yield a detailed ﬁne structure theory for V, removing the incompleteness that large cardinal hypotheses by themselves can never remove. It is known that V = ultimate L implies the CH, and many instances of the GCH. Whether it implies the full GCH is a crucial open problem

During this year's conferene on inner model theory in Münster, Gabriel Goldberg proved that the so-called Ultrapower Axiom implies that $\mathrm{GCH}$ holds above a supercompact cardinal (and since then lowered the bound to a strongly compact cardinal). It seems very likely (it might even be known) that $\mathrm{Ultimate } \ L$ satisfies this requirement. Hence, given enough large cardinals, it will satisfy $\mathrm{GCH}$ at least on a tail end.

For more information, see G. Goldberg. Strong Compactness and the Ultrapower Axiom.