TITLE: Is $V=\textsf{HOD}\not\Rightarrow\textsf{GCH}$ consistent?
QUESTION [5 upvotes]: Whenever $M$ is some fine-structural $L$-like model we can prove the implication $V=M\Rightarrow\textsf{GCH}$. For $L$ this is due to Gödel, and for the modern extender models it follows simply by construction. The most recent direction in the inner model programme searches for fine-structural models of the form $\textsf{HOD}^N$ with $N$ some "nice" model (most recently this includes $N=M_n(x,g)$ for a Turing cone of reals $x$ and $g$ generic for making its least $N$-inaccessible countable). My question is then whether $V=\textsf{HOD}\Rightarrow\textsf{GCH}$ has been shown to be unprovable in general (modulo large cardinals of course)? Is there some pathological $\textsf{HOD}$ which "obviously" doesn't satisfy $\textsf{GCH}$? Also, I'm not even sure if $\textsf{HOD}^{\textsf{HOD}^N}=\textsf{HOD}^N$ holds for these models, as this (apparently) doesn't hold in general.

REPLY [12 votes]: Yes, one can produce a model of $ZFC+V=HOD$ in which the $CH$ fails. This should follow from Consistency results about ordinal definability.
In fact, I think the arguments of my paper   HOD, V and the GCH can be used to produce a model of $ZFC+V=HOD$ in which the $GCH$ fails everywhere.
Also by a result of Apter (see Large cardinals need not be large in HOD for a proof), if there are class many Laver indestructible supercompact cardinals, then $V=HOD$ holds. Note that in such a model $GCH$ must fail at class many regular cardinals.