# Idempotent ultrafilters and the Rudin-Keisler ordering

Short version: what can we say about the place of idempotent ultrafilters in the Rudin-Keisler ordering?

Longer version:

If $U$, $V$ are (nonprincipal) ultrafilters on $\omega$, then we write $U\ge_{RK}V$ in case there is some function $f:\omega\rightarrow\omega$ such that $$\forall X\subseteq\omega,\quad f^{-1}(X)\in U\iff X\in V.$$

An ultrafilter $U$ is Ramsey if given any two-coloring of pairs $c: [\omega]^2\rightarrow 2$, there is a homogeneous set for $c$ in $U$. Equivalently, $U$ is Ramsey if whenever $\lbrace C_n\rbrace_{n\in\omega}$ is a partition of $\omega$ with each $C_n\not\in U$, there is some $H\in U$ such that for all $n\in\omega$, $$\vert H\cap C_n\vert=1.$$ An ultrafilter $U$ is idempotent if $U\oplus U=U$, where $$V\oplus W=\lbrace X: \lbrace x: \lbrace y: x+y\in X \rbrace \in W \rbrace \in V\rbrace.$$ Idempotent ultrafilters can be proved to exist in $ZFC$: this amounts to showing that $\oplus$ is left-continuous and associative on the compact space $\beta\mathbb{N}$, and then applying Ellis' theorem that every left-continuous semigroup on a compact space has an idempotent element. By contrast, Ramsey ultrafilters cannot be shown to exist in $ZFC$, although their existence is equiconsistent with $ZFC$ (in particular, if the Continuum Hypothesis - or weaker statements - holds, then there are Ramsey ultrafilters).

Now, an easy argument shows that no Ramsey ultrafilter is idempotent. On the other hand, the Ramsey ultrafilters enjoy a special property with respect to the RK-ordering: they are precisely the RK-minimal ultrafilters. So combining these facts shows that no idempotent ultrafilter can be RK-minimal.

My question is, what else can be said about the idempotent ultrafilters in terms of RK-reducibility? For example, can we have an idempotent ultrafilter U with exactly one RK-class of (necessarily, Ramsey) ultrafilters strictly RK-below U? This seems clearly impossible, but I don't see how to prove it.

Motivation: a few weeks ago, I taught a one-week course on the proof of Hindman's theorem from additive combinatorics using idempotent ultrafilters. On the last day, I talked a bit about other types of ultrafilters, and spent a bit of time defining Ramsey ultrafilters and explaining (not proving) their place in the RK-ordering. One of my students asked whether anything similar could be said about idempotent ultrafilters; besides the obvious, I couldn't come up with anything, so I'm asking here.

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The last section of Andreas's "Ultrafilters: Where Dynamics = Algebra = Combinatorics" has some remarks on the relationship between idempotents, Ramsey ultrafilters, and P-points. –  Todd Eisworth Aug 3 '13 at 15:34
Random Conjecture: Suppose U is idempotent, let $f$ ($g$) be the map sending n to the position of the rightmost (leftmost) 1 in its binary expansion. Then $f(U)$ and $g(U)$ are not RK-equivalent to each other or to $U$. –  Todd Eisworth Aug 3 '13 at 15:36
Todd, see my comment to Andreas's answer -- given any two ultrafilters $p,q$ you'll find an idempotent u with $f(u)=p, g(u)=q$. –  Peter Krautzberger Aug 4 '13 at 1:26

The idempotent ultrafilters closest to being Ramsey are the stable ordered-union ultrafilters. These are officially defined as certain ultrafilters on the set $\mathbb F$ of finite subsets of $\omega$, but they can be transferred to $\omega$ via the "binary expansion" map $\mathbb F\to\omega: s\mapsto\sum_{n\in s}2^n$. The image on $\omega$ of a stable ordered-union ultrafilter is idempotent and has has exactly three non-isomorphic non-principal ultrafilters RK-below it. Two of these are the ones Todd Eisworth mentioned in his comment; the third is the "pairing" of these two, i.e, the image of the idempotent under the map to $\omega^2$ given by $n\mapsto($position of leftmost $1,\,$position of rightmost $1)$. For details, see my paper "Ultrafilters related to Hindman's finite-unions theorem and its extensions" ["Logic and Combinatorics", Contemporary Math 65 (1987) 89-124] also available at http://www.math.lsa.umich.edu/~ablass/uf-hindman.pdf (this is a scanned picture and therefore not searchable; the stable ordered-union stuff starts on page 113).
I'm curious: is it known to what extent (assuming CH, say) the Ramsey ultrafilters below a given ultrafilter $U$ characterize $U$? That is, are there non-RK-equivalent $U, V$ which RK-bound the same Ramsey ultrafilters? Alternatively, is it consistent that there are no such pairs of ultrafilters? (It is certainly consistent that such a pair exists, since all pairs of ultrafilters satisfy this property if there are no Ramsey ultrafilters.) –  Noah S Aug 3 '13 at 18:25
A more relevant question: do idempotent ultrafilters exist RK-above an arbitrary ultrafilter? I suspect the answer is yes, but I'm having trouble proving it: given a function $f: \omega\rightarrow\omega$ and an ultrafilter $U$, the set of $V$ RK-above $U$ via $f$ is compact as a subset of $\beta\mathbb{N}$, but it's not clear to me that it is closed under $\oplus$ (if it were, I could apply Ellis' theorem and be done). Is this known? –  Noah S Aug 3 '13 at 20:39
@NoahS for your first comment: Under CH, the Ramsey ultrafilters below a given ultrafilter U don't characterize U (up to isomorphism). For one thing, there are lots of ultrafilters (including some P-points) with no Ramsey ultrafilters below them. Also, if you fix a Ramsey ultrafilter V, then there are several (undoubtedly $2^{\aleph_1}$, but I haven't checked carefully) isomorphism classes of ultrafilters that are RK-above U and no other Ramsey ultrafilters - in fact above no other ultrafilters at all (except of course themselves and principal ultrafilters). –  Andreas Blass Aug 3 '13 at 21:32
@NoahS for your second comment: I don't know but I would expect that there are ultrafilters on $\omega$ with no idempotent ultrafilters RK-above them. I would be surprised if there exist (say under CH if it helps) any nonprincipal ultrafilter U and any function f such that the set of ultrafilters mapped to U by f is nonempty and closed under addition. –  Andreas Blass Aug 3 '13 at 21:37
There are idempotents RK-above any ultrafilter, e.g., you can extend the inverse filter under the map that maps each $n$ to the minimum (or maximum) of its binary expension. See my dissertation –  Peter Krautzberger Aug 3 '13 at 23:53