Lattice of differences between ultrafilters 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.
 A: 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 $U\neq V$. 
