# Ultrafilter projections and critical points of factor maps

Suppose $$j : V \to M$$ is $$\lambda$$-supercompactness embedding derived from an $$\kappa$$-complete normal ultrafilter $$U$$ on $$P_\kappa(\lambda)$$, $$\lambda$$ regular. Suppose $$\eta$$ is an ordinal such that $$\sup(j[\lambda]) \leq \eta < j(\lambda)$$. Let $$W$$ be the uniform ultrafilter on $$\lambda$$ defined by $$X \in W$$ iff $$\eta \in j(X)$$. Let $$i : V \to N$$ be the ultrapower embedding by $$W$$. There is a factor map $$k : N \to M$$ given by $$k([f]_W) = j(f)(\eta).$$

Question: What is the relationship between $$N$$ and $$M$$? What is the critical point of $$k$$? How closed is $$N$$?

We can ask similar questions about $$\lambda$$-strong embeddings.

You will always have $$N = M$$ and $$k = \text{id}$$. As Joel mentions, this uses Solovay's lemma that $$M = H^M(\text{ran}(j)\cup \{\sup j[\lambda]\})$$. We can use this to show that $$k$$ is surjective, by proving that in fact $$k(\sup i[\lambda]) = \sup j[\lambda]$$. If not, $$k(\sup i[\lambda]) > \sup j[\lambda]$$ is a generator of $$j$$, but $$j$$ has no generators above $$\sup j[\lambda]$$ (since $$\sup j[\lambda]$$ generates everything by Solovay's lemma). Since $$\sup j[\lambda] \in\text{ran}(k)$$ and $$\text{ran}(j)\subseteq \text{ran}(k)$$ and $$M = H^M(\text{ran}(j)\cup\{\sup j[\lambda]\})$$, $$k$$ is surjective, and so $$k = \text{id}$$ and $$N = M$$.

Here are the facts about generators I'm using. Recall that an ordinal $$\nu$$ is a generator of an elementary embedding $$i : W\to N$$ if $$\nu \in N$$ and $$\nu\notin H^N(\text{ran}(i)\cup [\nu]^{<\omega})$$. I need:

1. If $$i : W\to N$$ is an elementary embedding that is discontinuous at a regular cardinal $$\lambda$$ of $$W$$, then $$\sup j[\lambda]$$ is a generator of $$j$$.
2. If $$k : N \to M$$ is a further embedding, then for any generator $$\nu$$ of $$i$$, $$k(\nu)$$ is a generator of $$k\circ i$$.
• Why is $\sup j[\lambda] \in \ran(k)$ if we select $\eta \not= \sup j[\lambda]$? Commented Dec 15, 2023 at 15:33
• Because $k(\sup i[\lambda]) = \sup j[\lambda]$! Commented Dec 15, 2023 at 15:34

If you take $$\eta=\sup j[\lambda]$$, then $$N=M$$ by a theorem of Solovay (the sup function is one-to-one on a measure one set), used in his proof that the SCH holds above any supercompact cardinal. In this case, the measure $$W$$ is isomorphic as an ultrafilter to the original supercompactness measure.

It follows that there are also many other values of $$\eta$$ for which $$N=M$$, since whenever $$\eta$$ codes the ordinal $$\sup j[\lambda]$$ in some canonical way, then $$\sup j[\lambda]$$ will be placed into the seed hull of $$\eta$$, and consequently everything will be there, causing $$N=M$$ again.

For other values of $$\eta$$, however, we do not necessarily have N=M.

The critical point of $$k$$ is the first ordinal missing from the seed hull $$\{j(f)(\eta)\mid f:\lambda\to V\}$$, so of course when it exists it will be at most $$(2^\lambda)^{+V}$$, simply because there only $$2^\lambda$$ many functions to represent such ordinals below $$j(\lambda)$$.