Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider two ultrafilters, $U$ and $V$, on the same cardinal $\kappa$. Let $D(U, V)=\lbrace X\subseteq \kappa: X\in U-V\rbrace$; clearly $D(U, V)$ is a lattice under $\subseteq, \cap, \cup $ since the intersection of two $U$- or $V$-large sets is $U$- or $V$-large, and the union of two $U$- or $V$-small sets is $U$- or $V$-small; by the same reasoning, $D(U, V)$ is a $\lambda$-complete lattice, where $\lambda$ is the minimum of the completeness of $U$ and the completeness of $V$.

My general question is, does this lattice have any interesting properties?

In particular, I'm interested in the following: let $M\models ZFC^-$, let $U\in M$ be a countably complete ultrafilter on some $M$-measurable cardinal $\kappa$, and let $j: M\rightarrow \prod M/U$ be the elementary embedding of $M$ into the ultrapower via $U$. Let $V=\lbrace X\in\wp^M(\kappa): \kappa\in j(X)\rbrace$; then $V\in M$ and $V$ is a normal ultrafilter on $\kappa$. In particular, if $U$ is not normal, then $U\not=V$. Intuitively, however, the difference between $U$ and $V$ is "small" (to be fair, this "intuition" may just be a figment of my not understanding inner model theory); is this somehow reflected by the lattice $D(U, V)$? In general, can anything about the relationship between $U$ and $V$ be read off of the lattice $D(U, V)$?

(Also, is this notion studied somewhere? I've googled around, unsuccessfully.)

EDIT: to clarify, I'm most interested in properties which can be determined from the isomorphism type of the lattice $D(U, V)$ alone.

Thanks in advance; hopefully this isn't too open-ended.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

I've got it!

Theorem. The lattices of the form $D(U,V)$ admit a complete classification by the isomorphism classes of $U$ and $V$ and the question of whether $U\neq V$.

The point is that the lattice isomorphism class of $D(U,V)$, when $U\neq V$, determines and is determined by isomorphism classes of $U$ and $V$ (that is, by their Rudin-Keisler equivalence classes). Meanwhile, when the ultrafilters are the same, $D(U,U)$ is the empty lattice, independently of $U$.

Proof. Notice first that when $U\neq V$ we can recognize whether $U$ is principal from the lattice $D(U,V)$, which will have a least element exactly in this case; similarly, we can recognize whether $V$ is principal from $D(U,V)$, which will have a greatest element exactly in this case.

Next, let me recall from my earlier post how the filters $U$ and $V$ can be reconstructed from $D(U,V)$, using $D(U,V)$ not just as a lattice but specifically as a collection of subsets of $\kappa$. Namely, let $X$ be any element of the lattice, so that $X\in U$ and $X\notin V$. It follows that the complement of $X$ is in $V$, and also any larger set than the complement of $X$ is in $V$. From this, it follows that for $Y\subset X$ we have $Y\in U$ if and only if $Y\in D(U,V)$. So the ultrafilter $U$ and the lattice $D(U,V)$ agree completely on the subsets of $X$. This is enough to reconstruct $U$, since a set is in $U$ if and only if it has $U$-large intersection with $X$. Similarly, we can reconstruct $V$, namely, a set $Y$ is in $V$ if and only if $Y-X\in V$, since $X$ is not in $V$; the complement of $Y-X$ is $X\cup(\kappa-Y)$, and this is not in $V$, but containing $X$ it is in $U$ and hence in $D(U,V)$. In summary, $$Y\in U\ \ \ \iff\ \ \ Y\cap X\in D(U,V)$$ $$Y\in V\ \ \ \iff\ \ \ X\cup(\kappa-Y)\in D(U,V)$$ and this does not depend on the choice of $X\in D(U,V)$.

But let me now explain how one can get access to essentially the same information up to isomorphism, just from knowing $D(U,V)$ as a lattice, and not knowing how these elements sit as subsets of $\kappa$. Assume $U$ is non-principal, and let $x$ be an arbitrary element of $D(U,V)$, viewed now only as a lattice. Let $A$ be the collection of immediate predecessors of $x$ in $D(U,V)$, namely, the set of $a\in D(U,V)$ such that $a\lt x$ and there is nothing between $a$ and $x$. (We know that these $a$ represent the removal of one element of the set representing $x$.) Now, I can define an ultrafilter $U'$ on $A$, by saying that $B\subset A$ is in $U'$ just in case there is a greatest lower bound to $A-B$ in $D(U,V)$. If $x$ is representing the set $X$ in $D(U,V)$, then I claim that $U\upharpoonright X\cong U'\upharpoonright A$, by the map that maps each point in $X$ to the set in $A$ obtained by omitting that point from $X$. Thus, $U'$ is Rudin-Keisler equivalant to $U$, and was constructed purely from viewing $D(U,V)$ as a lattice. Similarly, assume $V$ is non-principal and let $C$ be the set of immediate successors of $x$ in the lattice $D(U,V)$. (These lattice elements correspond exactly to the sets obtained by adding one additional point to the set that $x$ is representing.) Define the ultrafilter $V'$ on $C$ by $D\subset C$ is in $V'$ just in case there is no least upper bound of $D$ in $D(U,V)$. The map that sends the elements of $C$ to the corresponding points of $\kappa$ actually used in $D(U,V)$ is a Rudin-Keisler isomorphism of $V\upharpoonright(\kappa-X)$ with $V'\upharpoonright C$. So again, from $D(U,V)$ viewed purely as a lattice, we are able to extract $V$ up to isomorphism.

Conversely, if $U\cong U'$ and $V\cong V'$, where $U\neq V$ and $U'\neq V'$, then we may find a single function $f$ witnessing the isomorphisms simultaneously (working partly on a $U$-big set and partly on its complement, a $V$-big set), thereby showing that $D(U,V)\cong D(U',V')$ as lattices. So this is a complete classification of $D(U,V)$ up to isomorphism as a lattice. QED

Thus, the properties about $U$ and $V$ that we can determine from the lattice isomorphism class of the lattice $D(U,V)$ are precisely the properties that are determined by the isomorphism classes of $U$ and $V$ themselves, plus the knowledge of whether $U\neq V$.

share|improve this answer
    
To head off a possible confusion for readers, note that the notation in the second paragraph of the answer reverses the roles of $U$ and $V$ in the question. –  Andreas Blass Aug 28 '12 at 15:02
    
This is a really nice answer, but I was most interested in what can be determined just by the isomorphism type of $D(U, V)$ as a lattice (see edit). Is there anything we can say about what lattices are isomorphic to some $D(U, V)$? –  Noah S Aug 28 '12 at 16:58
    
Thanks, Andreas, I have corrected it. And Noah, I added a little something more, but I'll have to think a bit more if you want to characterize the D(U,V) among all lattices. –  Joel David Hamkins Aug 28 '12 at 19:44
    
I updated with an argument providing a classification of all lattices of the form $D(U,V)$. They determine and are determined by the isomorphism classes of $U$ and $V$ and the question of whether $U\neq V$. –  Joel David Hamkins Aug 29 '12 at 1:44
    
Sorry I just noticed this! This is really nice. Thanks! –  Noah S Oct 3 '12 at 22:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.