# Changing cofinalities above supercompact cardinals

Question. Suppose $$\kappa$$ is a supercompact cardinal and $$\lambda > \kappa$$ is measurable (or even larger large cardinal if necessary). Is there a set generic extension of the universe in which $$\kappa$$ remains supercompact, $$\lambda$$ is preserved and $$cf(\lambda)=\omega?$$

Remark 1. By Gitik-Shelah indestructibility result, if supercompact cardinal is replaced with strong cardinal, then the answer is yes.

Remark 2. If we require that the forcing preserves $$\lambda^+,$$ then the answer is no, as it is shown by Yair.

Remark 3. In A note on sequences witnessing singularity - following Magidor-Sinapova, Gitik has conjectured the following:

Conjecture. Suppose that

1. $$V ⊆ W$$ models of ZFC with same ordinals,

2. $$κ$$ is a regular cardinal in $$V$$,

3. $$cof(κ) = ω$$ in $$W$$,

4. $$\aleph_1^V=\aleph_1^W,$$

5. $$V, W$$ agree about a final segment of cardinals.

Then there is a subclass $$V′$$ of $$V$$ which is a model of $$ZFC$$, agree with $$V$$ about a final segment of cardinals, and there is a sequence witnessing singularity of $$κ$$ (in $$W$$) which is generic over $$V′$$ for either Namba, Woodin tower or Prikry type forcing.

Assuming this conjecture, it seems quite plausible that the answer to the question might be no in general.

Edition. I realized that the question has connection with recent work of Woodin:

Theorem. Assume $$\kappa$$ is an extendible cardinal. If Woodin's $$HOD$$-conjecture holds, then we can not change the cofinality of some large cardinal $$\lambda > \kappa$$, preserving the supercompactness of $$\kappa,$$ by set forcing without collapsing $$\lambda.$$

• Supercompact can be made immune for $\kappa$-directed closed forcings, unfortunately those are proper, and a proper forcing cannot change cofinality without collapsing cardinals. Mar 17 '16 at 17:46
• I think that if you allow class forcings then Woodin's stationary tower (with height $\delta = On$, assuming "$On$ is Woodin") can get you the desired situation. By forcing below $S^\lambda_\omega$, the critical point of the embedding $j\colon V\to M$ is $\lambda$, $\text{cf }\lambda = \omega$, and therefore $\kappa$ remains supercompact in $M$. Using the closure properties of $M$, the same holds in $V[G]$. Mar 18 '16 at 7:19
• That's quite a conjecture. Mar 26 '16 at 8:30
• Did you realize this by reading the recent review paper about the HOD conjecture? Because I reviewed it recently and felt something similar. But the question is what happens if the HOD conjecture is in fact false, and it is possible to have a dichotomy. Can you still force something like this? May 7 '16 at 12:04
• I recently asked @YairHayut the same question in an email. He showed me an argument of Magidor showing that the strong compactness of $\kappa$ must be destroyed. Jun 13 '18 at 16:50

There is no such forcing that preserves $\lambda^+$. Since $\lambda$ is measurable, $2^{<\lambda} = \lambda$ and therefore $\square_{\lambda,\lambda}$ holds in $V$. Since $\lambda^{+}$ is preserved, the same sequence will witness that there is still a weak square at $\lambda$ in the generic extension. But weak square fails at singular cardinals of small cofinality above a supercompact cardinal, so $\kappa$ cannot be supercompact in the generic extension.
In fact, a theorem of Dzamonja and Shelah shows that if you change the cofinality of an inaccessible cardinal $\lambda$ while preserving $\lambda^+$, there is even a $\square_{\lambda,\omega}$ sequence in the generic extension.