Assume $\kappa$ is $\lambda$supercompact for some $\lambda$ but not fully supercompact. Are there any known restrictions (or provably nonrestrictions) on the least $\delta$ such that $\kappa$ is not $\delta$supercompact (e.g. $\text{cf}(\delta)\geq\kappa$ or $\text{lim}(\delta)$)?
2 Answers
I'm not sure if this gets to the heart of the question, but here are some observations.
For a successor ordinal $\alpha$, let $\kappa_\alpha$ be the least $\kappa>\alpha$ which is $\kappa^{+\alpha}$supercompact. Then $\kappa_\alpha$ is not $2^{\kappa_\alpha^{+\alpha}}$supercompact. This is because an embedding $j : V \to M$ with critical point $\kappa_\alpha$ witnessing $2^{\kappa_\alpha^{+\alpha}}$supercompactness would have that $M$ has a measure witnessing $\kappa_\alpha$ is $\kappa_\alpha^{+\alpha}$supercompact. (Here, we need to invoke the theorem of Solovay that if $\lambda$ is regular and $\kappa$ is $\lambda$supercompact, then $\lambda^{<\kappa}=\lambda$.) Since $j(\kappa_\alpha) > \kappa_\alpha$, there is $\kappa<\kappa_\alpha$ that is $\kappa^{+\alpha}$supercompact, contradicting the definition of $\kappa_\alpha$.
On the other hand, if $\kappa$ is $\kappa^{+\alpha}$supercompact, where $\alpha<\kappa$ is a limit ordinal, then $\kappa$ is $\kappa^{+\alpha+1}$supercompact, since a $\lambda$supercompactness measure for $\kappa$ witnesses $\lambda^{<\kappa}$supercompactness. See SolovayReinhardtKanamori.
I suppose an interesting specialization of the question is, if $\kappa$ is $\kappa^{+}$supercompact and $2^{\kappa^+} > \kappa^{++}$, is it $\kappa^{++}$supercompact? We can preserve the $\kappa^+$supercompactness and kill the $\kappa^{++}$supercompactness by adding a nonreflecting stationary subset of $\kappa^{++}$, but I don't know how to do it without collapsing $2^{\kappa^+}$ to $\kappa^{++}$ simultaneously.

1$\begingroup$ Do you mean this paper by Solovay, Reinhardt and Kanamori? The theorem that you're referencing is called proposition 3.2 in that paper. $\endgroup$ Apr 23 at 17:03
One can get fairly close bounds by observing that if $\kappa$ is $\lambda$supercompact, then by using a supercompactness measure of least Mitchell rank, there is a $\lambda$supercompactness embedding $j:V\to M$ for which $\kappa$ is not $\lambda$supercompact in $M$. But $\kappa$ will be $\theta$supercompact in $M$ whenever $2^{\theta^{<\kappa}}\leq\lambda^{<\kappa}$. So from the point of view from $M$, we have a good understanding of the exact point of failure of supercompactness of $\kappa$. Indeed, under GCH, these bounds are nearly tight.