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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 non-measurable 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 non-measurable cardinals, simply choose a normal measure μ for which the ultrapower jμ has the smallest possible value for jμ(κ). It follows that κ cannot be measurable in the corresponding Mμ, 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 Hδ 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 Hδ.

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.

2 added 341 characters in body

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 non-measurable 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 non-measurable cardinals, simply choose a normal measure μ for which the ultrapower jμ has the smallest possible value for jμ(κ). It follows that κ cannot be measurable in the corresponding Mμ, 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 Hδ 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 Hδ.

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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 non-measurable 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.

A similar argument shows that the supercompactness of κ cannot be characterized by any property in Hδ 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 Hδ.