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 $\textrm{GCH}$?
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1$\begingroup$ For Woodin's paper: doi.org/10.1017/bsl.2016.34 $\endgroup$– David Roberts ♦Dec 18, 2017 at 6:55
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2 Answers
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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
answered Dec 18, 2017 at 3:40
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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.
