# Are superstrong stronger than strongly compact cardinals? (or vice versa)

In the last part of Kanamori's excellent "The Higher Infinite" there is a small diagram about the strength and consistency strength of some major large cardinal axioms.

Below supercompact cardinals there are two then-incomparable cardinals. Superstrong cardinals and strongly compact cardinals, both are stronger than Woodin cardinals

Has there been any progress on this problem since then? If yes, is there some reference? If not, can someone give basic outline as to why this problem is difficult? (I am somewhat familiar with the problem of extender models for very large cardinals, so an concise similarities if they exist would suffice).

-
If we can show that superstrog cardinals are stronger than strongly compact cardinals, then it will follow that supercompact cardinals are stronger than strongly compact cardinls which answers an old open question. – Mohammad Golshani Jan 22 '14 at 17:45
Well, that is indeed true. But what about the other way around? Are strongly compact cardinals superstrong? – Asaf Karagila Jan 22 '14 at 17:46

Are strongly compact cardinals superstrong?

The answer is no, not necessarily, and indeed, strongly compact cardinals need not even be strong, nor even a little bit strong. The reason is that it is relatively consistent that the least strongly compact cardinal is the same as the least measurable cardinal, and in this case such a cardinal $\kappa$ will not even be $(\kappa+2)$-strong. This was the first instance of the so-called "identity crises" phenomenon, discovered by Magidor, and similar phenomenon have now been uncovered for many other cardinals (for example, the least weakly compact cardinal can be unfoldable, weakly measurable and nearly $\theta$-supercompact).

That is for outright implication, but meanwhile, for consistency strength, it is a different story. As far as we know, strongly compact and supercompact cardinals have the same consistency strength. Here, the relevant observation is:

Theorem. If $\kappa$ is $2^\kappa$-supercompact, witnessed by $j:V\to M$, then $\kappa$ is superstrong in $M$.

This is the reason that supercompact cardinals are strictly stronger than superstrong cardinals in consistency strength.

Proof: The map $j\upharpoonright V_{\kappa+1}:V_{\kappa+1}\to M_{j(\kappa)+1}$ has size $2^\kappa$ and is therefore inside $M$. Using this part of $j$ to define an extender $E$, the derived extender of $j$, the model $M$ can produce an embedding $j_E:M\to N$, which agrees with $j$ on $V_{\kappa+1}$. In particular, this means $M_{j(\kappa)+1}\subset N$, which means that $E$ witnesses that $\kappa$ is superstrong from the perpsective of $M$. QED

Perhaps one of the inner model theorists will post an answer explaining how much of the inner model theory one can undertake from a strongly compact cardinal, and I suppose that is really what you are asking.

-
The "identity crises" is awesome. As for the consistency, Mohammad pointed out that indeed we don't know if superstrong are as strong as strongly compact (in which case it will solve the problem in the Battle of the Compacts); but do we know anything on the other direction in terms of consistency? – Asaf Karagila Jan 22 '14 at 19:31
I think that from a strongly compact cardinal one can undertake much of the inner model constructions, but I'm not sure exactly how far you get. But I expect one of the inner model experts on MO to tell us before too long. – Joel David Hamkins Jan 23 '14 at 1:21
Thank you, Joel. – Asaf Karagila Jan 23 '14 at 1:28
@AsafKaragila: I find the "identity crises" phenomenon a little unnerving at times, and certainly an eyesore and unwanted complication when thinking about large cardinals. Would you (or really anyone here on MO) mind explaining why they find "identity crises" among large cardinal concepts "awesome"? My naive thinking on this suggests that one of two things follows from this phenomenon: Either certain large cardinal concepts are too robust to be very meaningful or concrete, or the structure of the universe is not only highly complex, but also a little inelegant. Thoughts? – Everett Piper Jan 23 '14 at 3:48
$\kappa^+$-strong compactness already takes us beyond what we can currently do by descriptive inner model theoretic methods. – Andrés Caicedo Jan 23 '14 at 7:54