TITLE: The stationary tower and supercompactness
QUESTION [5 upvotes]: I'm currently reading through Larson's book on the stationary tower and a point confused me.
Let $\delta$ be Woodin and let $j:V\to M\subseteq V[G]$ be an elementary embedding associated with the (full) stationary tower $\mathbb P_{<\delta}$. Then $V[G]\models {^{<\delta}}M\subseteq M$. This seems close to $\operatorname{crit}j$ being $\theta$-supercompact for every $\theta<\delta$, in $V[G]$, with the exception that the generic extension might contain more subsets of $\operatorname{crit}j$, so that the measure on $\operatorname{crit}j$ isn't total.
Now say there is a proper class of Woodins, and let $j:V\to V[G]$ be an elementary embedding associated with the stationary tower $\mathbb P_\infty$. Again, is $\operatorname{crit}j$ now 'close' to being supercompact in $V[G]$?
Is this the same thing as $\operatorname{crit}j$ being generically supercompact? Since we can choose $\operatorname{crit}j$ to be any inaccessible, does this then mean that a proper class of Woodins implies a proper class of generically supercompacts?

REPLY [2 votes]: Apparently this has a name. To any large cardinal notion, there is a virtual variant, in which the elementary embeddings in question lie in some generic extension. Reformulating my scenario, I pointed out that a Woodin $\delta$ consistency-wise implies a virtually $\theta$-supercompact for every $\theta<\delta$. Using the countable tower instead, as Yair pointed out, we also get the result that a Woodin consistency-wise implies that $\omega_1$ is virtually almost huge.
But as is pointed out in Gitman's slides, all virtual large cardinals lie consistency-wise below $0^\sharp$, so this observation (that Woodins lie above virtuals) is pretty moot.