# Cofinality of $j(\kappa)$ for a measurability embedding $j:V\to M$ with critical point $\kappa$

Suppose $\kappa$ is a measurable cardinal and $j:V\to M$ is the ultrapower by a normal measure on $\kappa$. Let's say, for instance, that $2^\kappa=\kappa^{++}$ (note that this assumption has consistency strength greater than just a measurable cardinal). It is easy to see that $j(\kappa)$ has size $\kappa^{++}$ in $V$. So the cofinality of $j(\kappa)$ in $V$ can be either $\kappa^+$ or $\kappa^{++}$ (it cannot be $\kappa$ because $M$ is closed under $\kappa$-sequences). Can each of these possibilities be realized in some model of set theory? More generally, what can be said about the cofinality of $j(\kappa)$ under various GCH assumptions?

As Asaf Karagila notes in the comments below, we can reformulate the question to ask what are the possible cofinalities of the order $\kappa^\kappa/U$ for a normal measure $U$ on $\kappa$. This is equivalent because elements of $j(\kappa)$ are precisely the equivalence classes of functions $f:\kappa\to\kappa$.

• That's a very nice question! You can transform it into just asking what is the cofinality of the order $\kappa^\kappa/U$. This should reduce it to a model theoretic question, methinks. – Asaf Karagila Oct 20 '16 at 10:11
• It's a good thing that I stopped by, then. :) – Asaf Karagila Oct 20 '16 at 10:18
• @AsafKaragila Done! – Victoria Gitman Oct 20 '16 at 12:31
• Suppose $\kappa$ is supercompact and Laver indestructible. Given any regular $\lambda > \kappa$ do a $<\kappa$-support iteraton where at successor steps we add a function $f: \kappa \to \kappa$ where $f$ is eventually above any ground model function. $\kappa$ remains supercompact in the extension and the functions introduced during the iteration should witness the cofinality is equal to $\lambda$ (with respect to any normal measure on $\kappa$). – Mohammad Golshani Oct 20 '16 at 13:49
• @MohammadGolshani: The length of the iteration can be an ordinal. By controlling the cofinality of the length of iteration you will get any desirable cofinality (above $\kappa$) for $j(\kappa)$. For example, if the length of the iteration is $\kappa^{++} + \kappa^{+}$ you will get $2^\kappa = \kappa^{++}$ and $\text{cf } j(\kappa) = \kappa^{+}$. This way you can get full independence between $2^\kappa$ and $\text{cf } j(\kappa)$. – Yair Hayut Oct 20 '16 at 21:12

## 1 Answer

As Mohammad Golshani remarked, it is possible to control the cofinality of $j(\kappa)$ by iterating the forcing that adds a function $f\colon \kappa \to \kappa$ which is eventually larger than any ground model function.

The conditions of the forcing notion are pairs of the form $(s, g)$ where $s{\in} ^{<\kappa}\kappa$ and $g\colon \kappa \to \kappa$, where $(s, g)$ is stronger than $(t, h)$ if $s\supseteq t$, $g \geq h$ everywhere and $\forall \alpha \in \text{dom } s \setminus \text{dom }t$, $s(\alpha) \geq h(\alpha)$.

Assuming $\kappa^{<\kappa} = \kappa$, this forcing is $\kappa$-centred and $\kappa$-directed closed. Let $\mathbb{P}_\alpha$ be the iteration of adding dominating function for $\alpha$ many steps with support ${<}\kappa$. Using the $\kappa$-closure of iteration and standard $\Delta$-system arguments - this iteration is $\kappa^{+}$-c.c. Therefore, it doesn't collapse cardinals.

Let $\langle f_i \mid i < \alpha\rangle$ be the sequence of the generic dominating functions.

Lemma: If $\text{cf }\alpha \geq \kappa^{+}$ then the true cofinality of $\kappa^{\kappa} / J^{bd}$ is $\text{cf }\alpha$. Namely, there is a cofinal, increasing sequence of functions in $\kappa^{\kappa}$ of order type $\text{cf }\alpha$.

Proof: Let $\lambda = \text{cf }\alpha$. Let $\langle \gamma_i \mid i < \lambda\rangle$ be a cofinal sequence. The sequence of functions $g_i = f_{\gamma_i}$ is increasing (modulo bounded error) and by the chain condition of $\mathbb{P}_\alpha$, every function in the generic extension is bounded by one of them. $\square$

Let $\kappa$ be a measurable cardinal. If $\text{tcf } \kappa^{\kappa} / J^{bd} = \lambda$, then for every $\kappa$-complete measure $\mathcal{U}$ on $\kappa$, $\text{cf }j_{\mathcal{U}}(\kappa) = \lambda$. Therefore, in order to construct a model in which $\text{cf }j(\kappa) = \lambda$, $2^\kappa = \mu$ where $\kappa < \lambda \leq \mu$, $\lambda$ regular, $\text{cf } \mu \geq \kappa^{+}$, we may start with indestructible supercompact $\kappa$ such that $2^\kappa = \kappa^{+}$ and force with $\mathbb{P}_{\mu + \lambda}$.

• Is it provable that $\kappa^\kappa/\mathcal J^{bd}$ has a true cofinality in the case of a measurable cardinal? – Asaf Karagila Oct 22 '16 at 18:15
• It's even enough for the non stationary ideal to have true cofinality in this case, when you think about it. – Asaf Karagila Oct 23 '16 at 8:10