Is it possible to tell if a cardinal number C is supercompact by looking only at the properties of subsets of C and paying no attention to sets whose cardinal numbers are greater than C? This can certainly be done with many smaller infinite cardinal numbers such as measurable cardinal numbers.

No, this is not possible. If κ is supercompact, then in particular it is measurable, and there is a normal measure μ on κ that concentrates on nonmeasurable cardinals. If j:V to M is the corresponding normal ultrapower embedding, then κ is not measurable in M. But M and V have the same power set of κ. Thus, any criterion that looks only at P(κ) cannot distinguish between V and M, but κ is supercompact in V and not in M. (To find a measure μ concentrating on nonmeasurable cardinals, simply choose a normal measure μ for which the ultrapower j_{μ} has the smallest possible value for j_{μ}(κ). It follows that κ cannot be measurable in the corresponding M_{μ}, for if it were, we could make a better μ.) A similar argument shows that the supercompactness of κ cannot be characterized by any property in H_{δ} for any particular δ above κ, since for every λ above δ there will be λ supercompactness ultrapowers j:V to M for which κ is not λ supercompact in M, but V and M have the same H_{δ}. Finally, note that is it not really correct to say that the measurability of a cardinal κ can be characterized by looking only at P(κ), since my argument above shows two models M and V with the same P(κ), but κ is measurable in one of them and not the other. Rather, in order to tell if κ is measurable, you have to look at subsets of P(κ), that is, at P(P(κ)), since this is where the measure lives. 

