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I'm not sure whether this belongs here or on math stackexchange, but I'll give a try here.

I've heard that an extender is a generalization of an ultrafilter. This is not to say that an ultrafilter is the same sort of object as an extender, since whereas an ultrafilter on an uncountable $\kappa$ is in $V_{\kappa + 2}$, an extender in general is an element of a much higher level $V_\alpha$. But that an extender captures all the information an ultrafilter has about elementary embeddings. Or so I think.

What I am trying to prove is a precise statement of this:

Lemma: given a non-principal $\kappa$-complete ultrafilter on an uncountable $\kappa$, and $j: V \rightarrow M \cong Ult(V,U)$, there is an extender $E$ such that $j_E : V \rightarrow M$ and also $j_E = j$. ($j_E$ is the embedding built from $E$, that is, $j_E$ elementarily maps $V$ into the direct limit of a direct system, one whose elements are ultraproducts of $V$, and whose maps are appropriately defined to commute with the maps from $V$ to these ultraproducts.)

I've tried to take the extender $F$ of a short length such that $F_a = U$. Then all the maps of the direct system are the identity, and the direct limit will be our original $M$, and $j_E = j$. But I'm not convinced this is a genuine extender in that is satisfies one of the (all equivalent?) definitions of extenders.

Another approach is to take $j_U$, and take the embedding $E_{j_U}$ derived from it, and then try to show $j_{E_{j_U}} = j$. This will in fact be an extender, but I can't show whether the target model $M_E$ agrees with $M$.

Any ideas or suggestions?

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please correct "...same sort of object as an extender". A hint to your lemma might be exercise 22.5 in Kanamori's "The higher infinite". – Eran Apr 28 '12 at 6:11
I fixed it, thanks. I'll take a look at the exercise, though I'm not immediately seeing, given the exercise, how a normal ulrafilter over $P_\kappa \kappa$ will give a $(\kappa,\lambda)$-extender... – lesnikow Apr 28 '12 at 8:32
up vote 9 down vote accepted

My preferred account of extenders is the following: an elementary embedding $j:V\to M$ is an extender embedding if every element of $M$ can be expressed in the form $j(f)(\alpha)$, where $f:\kappa\to V$ and $\alpha\lt j(\kappa)$.

(With this extender concept, it is obvious that any ultrapower by a normal measure on $\kappa$ is an extender, since in this case every element of $M$ has the form $j(f)(\kappa)$. More generally, if $j:V\to M$ is the ultrapower by any ultrafilter $U$ on a set $I$, then it is an elementary exercise to see that every element of $M$ has the form $j(f)([id]_U)$, were $id:I\to I$ is the identity map.)

With this extender concept, one never needs all the ordinals $\alpha$ up to $j(\kappa)$, and more generally one has a set $S\subset j(\kappa)$ of generators, such that every element of $M$ has the form $j(f)(s)$, where $s\in S^{\lt\omega}$ and $f:\kappa^{\lt\omega}\to V$.

In this formulation, the extender object itself can be viewed as the set $$E=\{(s,X)\mid X\subset\kappa^{\lt\omega},s\in S^{\lt\omega},s\in j(X)\},$$ which contains all the information necessary to rebuild the embedding. For any fixed seed $s\in S^{\lt\omega}$, one has the induced measure $U_s=\{X\mid s\in j(X)\}$, and the corresponding ultrapower maps $j_s:V\to M_s=V^{\kappa^{|s|}}/U_s$ fit into a directed system of embeddings, for whenever $s\subset t$ then we may build the map $k_{s,t}:M_s\to M_t$ by transforming functions in the obvious way. The extender embedding $j:V\to M$ will be precisely the direct limit of this system of ultrapowers. One way to see this is by using the representation of the direct limit as equivalence classes of threads, and mapping any thread containing $[f]_{U_s}$ to $j(f)(s)$, which will be well-defined, $\in$-preserving and surjective, hence an isomorphism. Alternatively, one can verify the universal property, in that if we have maps $r_s:M_s\to N$, then we can build a factor map $r:M\to N$ via $j(f)(s)\mapsto r_s([f]_{U_s})$, which is well-defined and elementary.

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Hi Joel, thanks for the detailed answer. I'll have to work through this, but this is definitely a good start for me. – lesnikow May 14 '12 at 21:05

I know this question is four years old, but if anyone is stumbling across this question, this might be something worth noting. Maybe it has a way simpler proof, but this should work.

Proposition. A measure is equivalent to an extender with a single generator, in that their ultrapowers are isomorphic.

Proof. Let $U$ be a measure on some $\kappa$ and $E$ a $(\kappa,\lambda)$-extender with a single generator. Then $\text{Ult}(\mathcal M,E)$ is the direct limit of the cross-section ultrapowers $\text{Ult}(\mathcal M,E_a)$, so it suffices to show that if $a\in[\lambda]^{<\omega}$ with $a\not\subseteq\kappa$ and $\xi\in a-\kappa$ then $\text{Ult}(\mathcal M,E_{\{\xi\}})\cong\text{Ult}(\mathcal M,E_a)$. But as $E$ only has a single generator, $\kappa$, we can assume that $a\in[\kappa+1]^{<\omega}$, so that $\xi=\kappa$.

Define the map $\varphi$ as taking $[f]_{E_{\kappa}}$ to $[f^{\{\kappa\},a}]_{E_a}$, which is a well-defined homomorphism as $\kappa\in a$. To show that $\varphi$ is injective, assume $f^{\{\kappa\},a}\sim_{E_a} g^{\{\kappa\},a}$, so that $j(f^{\{\kappa\},a})(a)=j(g^{\{\kappa\},a})(a)$. But by definition of $f^{\{\kappa\},a}$, $j(f^{\{\kappa\},a})(a)=j(f)(\{\kappa\})$ and likewise $j(g^{\{\kappa\},a})(a)=j(g)(\{\kappa\})$, so $f\sim_{E_{\{\kappa\}}}g$.

For surjectivity, let $[f]_{E_a}$ be given and define $\tilde f:[\kappa]^{|a|}\to\mathcal M$ as $\tilde f(u):=f(\{a_0,\dots,a_{|a|-2},u_{|a|-1}\})$ with $u\in[\kappa]^{|a|}$. Then $\tilde f$ is clearly in the image of $\varphi$, so we need to show that $f\sim_{E_a}\tilde f$, i.e. $j(f)(a)=j(\tilde f)(a)$. But since $a_i<\kappa$ for $i<|a|-1$ we have $j(a_i)=a_i$, so $j(\tilde f)(u)=j(f)(\{a_0,\dots,a_{|a|-2},u_{|a|-1}\})$ and thus $j(f)(a)=j(\tilde f)(a)$. QED

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