<κ定向闭力迫后V中𝒫_κ(λ)超滤的保留与超紧超滤关联问题

Great questions diving into forcing preservation and supercompact cardinals—let’s unpack each one clearly:

First, let’s clarify what "preserving" means here: we’re asking if an ultrafilter $\mathcal{U}$ on $\mathcal{P}_\kappa(\lambda)$ from the ground model $V$ remains an ultrafilter in the generic extension $V[G]$, or if it can be extended to an ultrafilter in $V[G]$.

The breakdown:

• $\mathcal{U}$ does not automatically stay an ultrafilter in $V[G]$. The key reason is that while $<\kappa$-directed closed forcing doesn’t add new sets of size $<\kappa$, it can introduce new subsets of $\mathcal{P}\kappa(\lambda)$ (since $\mathcal{P}\kappa(\lambda)$ itself has size $\lambda^{<\kappa}$, which is typically larger than $\kappa$). For any new subset $A \subseteq \mathcal{P}_\kappa(\lambda)$ in $V[G]$ that wasn’t present in $V$, neither $A$ nor its complement will be in $\mathcal{U}$ (since $\mathcal{U}$ only contains sets from $V$). This means $\mathcal{U}$ can’t satisfy the ultrafilter requirement in $V[G]$.
• That said, if $\mathcal{U}$ is $\kappa$-complete (a standard property for ultrafilters tied to large cardinals like supercompactness), the filter generated by $\mathcal{U}$ in $V[G]$ (i.e., ${A \subseteq \mathcal{P}_\kappa(\lambda)^{V[G]} \mid \exists B \in \mathcal{U}, B \subseteq A}$) is a $\kappa$-complete filter in $V[G]$. Thanks to the $<\kappa$-directed closedness of the forcing, we can always extend this filter to a $\kappa$-complete ultrafilter in $V[G]$.

When $\kappa$ is supercompact and indestructible under $<\kappa$-directed closed forcing, $\kappa$ retains its supercompactness in $V[G]$. Here’s how the associated ultrafilters connect:

• The ultrafilters are not identical. $\mathcal{U}1$ (from $V$) only includes subsets of $\mathcal{P}\kappa(\lambda)$ that exist in the ground model, while $\mathcal{U}_2$ (from $V[G]$) incorporates new subsets added by the forcing.
• They share a tight, natural link: $\mathcal{U}_1$ is exactly the restriction of $\mathcal{U}_2$ to $V$, meaning $\mathcal{U}_1 = \mathcal{U}_2 \cap V$. Here’s the reasoning:
  • Since $<\kappa$-directed closed forcing doesn’t add new sets of size $<\kappa$, $\mathcal{P}\kappa(\lambda)^V = \mathcal{P}\kappa(\lambda)^{V[G]}$. So every subset of $\mathcal{P}_\kappa(\lambda)$ from $V$ is also present in $V[G]$.
  • Let $j: V \to M$ be the elementary embedding that witnesses $\kappa$’s supercompactness via $\mathcal{U}_1$ (so $j(\kappa) > \lambda$ and $M^\lambda \subseteq M$). We can lift this embedding to $j: V[G] \to M[G']$, where $G'$ is a $j(\mathbb{P})$-generic filter over $M$ containing $G$. This lifted embedding proves $\kappa$ is supercompact in $V[G]$, and the associated ultrafilter $\mathcal{U}2$ is defined as ${X \subseteq \mathcal{P}\kappa(\lambda) \mid j"[\lambda] \in j(X)}$.
  • For any $X \in V$, $X \in \mathcal{U}_1$ if and only if $j"[\lambda] \in j(X)$ (by definition of $\mathcal{U}_1$), which is exactly the condition for $X$ to be in $\mathcal{U}_2$. Thus, $\mathcal{U}_1$ is the ground-model "core" of $\mathcal{U}_2$.
• Can $\mathcal{U}_1$ generate $\mathcal{U}_2$? Not in the simple "all supersets of $\mathcal{U}_1$-elements" sense, since $\mathcal{U}2$ includes new subsets of $\mathcal{P}\kappa(\lambda)$ that don’t contain any element of $\mathcal{U}_1$. But $\mathcal{U}_2$ is directly induced by lifting the embedding tied to $\mathcal{U}_1$, so $\mathcal{U}_1$ is the foundation for $\mathcal{U}_2$. Conversely, every $\kappa$-complete normal ultrafilter in $V[G]$ restricts to a $\kappa$-complete normal ultrafilter in $V$, creating a direct correspondence between ground-model and extension ultrafilters.

内容的提问来源于stack exchange，提问作者Amitayu Banerjee