Solovay's discovery of supercompactness in the late 1960's marked the beginning 
of the modern era of large cardinal theory.
In the context of ZFC, supercompactness has both a combinatorial 
characterization in terms of normal ultrafilters and a ``model theoretic'' 
characterization in terms of elementary embeddings \(j : V\to M\) where \(M\) 
is 
an inner model.
In the choiceless context, 
however, the equivalence between 
the usual characterizations 
is no longer 
provable, and instead supercompactness 
splinters into a number of inequivalent but interrelated concepts.

The 
rank-to-rank embeddings \(j : V_{\delta}\to V_{\delta}\) studied here 
exhibit features  reminiscent of
super\-com\-pact\-ness.
In this section we evidence this via a characterization in terms
of normal ultrafilters  in the case
that $\delta=\alpha+2$.
\footnote{
In this section we assume familiarity with ultrapowers as used in set theory;
the reader familiar with supercompactness measures should be fine.}
But since these embeddings force us into the choiceless
realm, we must deal with the subtleties this brings.
A key issue in this regard is that one needs to be more careful regarding
{\L}o\'{s}'s Theorem for ultrapowers, given the role of choice in its usual 
proof. 

\subsection{Ultrapowers and {\L}o\'{s}'s Theorem}\label{subsec:ultrapowers}


In this section we give a quick
review of some standard  
background. We assume familiarity with (ultra)filters, which can be found in 
standard texts. If 
$\mathscr{F}$ is a filter
over a set $X$ (so $X\in\mathscr{F}$)
and $\varphi$ is some property,
 say that $\varphi(x)$ holds for \emph{$\mathscr{F}$-almost all }$x$
(or just \emph{almost all} $x$)
iff $\{x\in X\bigm|\varphi(x)\}\in\mathscr{F}$.


We first recall the definition of \emph{ultrapowers} in our context.
Let $\gamma,\beta\in\OR$ and let $\mathscr{F}$ be any ultrafilter over 
$V_{\gamma}$. Let $\mathscr{U}$ denote the set of all functions 
$f:V_{\gamma}\to V_\beta$. We define a binary relation over $\mathscr{U}$ by
\[ f\approx_{\mathscr{F}}g \iff \{x\in 
V_\alpha\bigm|f(x)=g(x)\}\in\mathscr{F}.\]
Because $\mathscr{F}$ is a filter, it is easy to see that 
$\approx_{\mathscr{F}}$
is an equivalence relation; let $[f]^{V_\beta}_{\mathscr{F}}$ be the 
equivalence class of $f$, where we just write $[f]$ if there is no ambiguity. 
We 
define also the relation
\[ f\in_{\mathscr{F}}g\iff\{x\in V_\alpha\bigm|f(x)\in g(x)\}\in\mathscr{F}.\]
Then $\in_{\mathscr{F}}$ respects $\approx_{\mathscr{F}}$.
The \emph{ultrapower} $\Ult(V_\beta,\mathscr{F})$
of $V_\beta$ by $\mathscr{F}$ is the structure $(U,\in^U)$,
where $U=\{[f]\bigm|f\in\mathscr{U}\}$, and $\in^U$ is the binary relation on 
$U$
induced by $\in_{\mathscr{F}}$. The \emph{ultrapower embedding}
$i^{V_{\beta}}_{\mathscr{F}}:V_\beta\to U$ is defined
$i^{V_{\beta}}_{\mathscr{F}}(x)=[c_x]$ where $c_x\in\mathscr{U}$
is the constant function with constant value $x$.

Now let us say that \emph{$\Sigma_n$-{\L}o\'{s}' Theorem for $U$} holds
iff for all $\Sigma_n$ formulas
$\varphi$ (in the language of set theory) and functions 
$f_1,\ldots,f_n\in\mathscr{U}$, we have
\[\begin{array}{cl} &U\sats\varphi([f_1],\ldots,[f_n])\\
\iff& 
V_\beta\sats\varphi(f_1(x),\ldots,f_n(x))\text{ for almost 
all }x\in V_{\gamma}.\end{array}\]
We just say \emph{{\L}o\'{s}' theorem holds for $U$} if $\Sigma_n$-{\L}o\'{s}' 
theorem holds 
for all $n<\om$.
For atomic formulas $\varphi(u,v)$ (``$u=v$'' and ``$u\in v$'')
the stated equivalence holds by definition.
Assuming $\AC$ it holds for all formulas, as proved by induction on  formula 
complexity. The only step that uses $\AC$ is that for quantifiers:
suppose for example that
\[ \sigma\in\mathscr{F}\text{ and for all }x\in\sigma\text{ we have } 
V_\beta\sats\exists w\ \varphi(w,f(x)).\]
Then we want  
$U\sats\exists w\ \varphi(w,[f])$,
which needs some $w\in\mathscr{U}$ with
$U\sats\varphi([w],[f])$.
So we need $w:V_{\gamma}\to V_\beta$ and 
 by induction, we need some $\sigma'$ such that
\[ \sigma'\in\mathscr{F}\text{ and for all 
}x\in\sigma'\text{, we have }V_\beta\sats\varphi(w(x),f(x)).\]
Using $\AC$, we can in fact take $\sigma'=\sigma$
and $w$ to be an appropriate choice function.
But it is important here that we don't actually require $\sigma'=\sigma$;
so even if $\AC$ fails and there is no choice function with domain $\sigma$,
there might be one with a smaller domain $\sigma'\in\mathscr{F}$.

If {\L}o\'{s}' Theorem holds for $U$ then the ultrapower 
embedding $i:V_\beta\to U$ is  elementary. (However, a key point is that
$U$ need not be wellfounded in general: consider for example
nonprincipal ultrafilters 
over $V_\om$.) If $U$ is wellfounded and extensional,
then by Mostowski's theorem, it is isomorphic
to its (transitive) Mostowski collapse,
and following the usual convention in this case, we then identify
these two. But we will at times need to deal with
ultrapowers without knowing that these properties hold.


In this section we are only actually interested in the case
that the ordinal $\beta$ above is a successor, so from now on,
we restrict to this case.
In order to analyze ultrapowers and the associated embeddings
defined as above,
we will observe that the coding apparatus from \S\ref{subsec:coding}
allows us to represent functions $f:V_\gamma\to V_\beta$
where $\gamma<\beta$ (such as those forming the ultrapower above), and simple 
properties
thereof, in a simple manner. 
That is, although maybe $f\notin V_\beta$, we  
define the \emph{code} of $f$ as
\[ \widetilde{f}=\{\lceil x,y\rceil\bigm|x\in V_{\gamma}
\text{ and }y\in f(x)\}; \]
note $\widetilde{f}\in V_\beta$ (as $\gamma<\beta$ and $\beta$ is a 
successor).
Unravelling the coding above and the flat pairing function,
it is straightforward to write  a 
$\Sigma_0$ formula $\psi$
such that for all such $\beta,\gamma,f$ we have
\[ \all x\in V_{\gamma}\ \all y\in 
V_{\beta-1}\ [y\in f(x)\Leftrightarrow 
V_\beta\sats\psi(\widetilde{f},x,y)].\]
More generally:
\begin{lem}\label{lem:general_Sigma_0_substitution} There is a recursive 
function $\varphi\mapsto\psi_\varphi$
 such that
for each 
$\Sigma_0$ formula $\varphi$,
$\psi_\varphi$ is a $\Sigma_0$ formula,
and for all successor ordinals $\beta>\om$ and ordinals $\gamma<\beta$
and all finite tuples $\vec{f}=(f_0,\ldots,f_{n-1})$ of functions
 $f_i:V_\gamma\to V_\beta$, and all $x\in V_\gamma$ and $z\in V_\beta$,
 \[
V_\beta\sats\varphi(f_0(x),\ldots,f_{n-1}(x),z)\iff 
V_\beta\sats\psi_\varphi(\widetilde{f_0},\ldots,\widetilde{f_1},x,z).\]
\end{lem}

We leave the straightforward proof to the reader.

\begin{dfn}For a transitive structure $M$ and $k\leq\om$,
 $\mathscr{E}_k(M)$ denotes the set of all $\Sigma_k$-elementary maps $j:M\to 
M$. So $\mathscr{E}_\om(M)=\mathscr{E}(M)$.
\end{dfn}

Now suppose $\beta$ is a successor ordinal and $j\in\mathscr{E}_0(V_\beta)$ and 
$j(V_\alpha)=V_{j(\alpha)}$ for each 
$\alpha<\beta$.
Let $\alpha+1<\beta$ and  $s\in V_{j(\alpha)+1}$.
The ultrafilter $\mathscr{F}$ over $V_{\alpha+1}$ \emph{derived 
from $j$ with seed $s$} is defined as follows: For $\sigma\sub V_{\alpha+1}$, 
set
\[ \sigma\in \mathscr{F}\iff s\in j(\sigma).\]
Note that $\mathscr{F}$ is principal iff $s\in\rg(j)$.

For $f:V_{\alpha+1}\to V_\beta$, we needn't have $f\in V_\beta=\dom(j)$,
but we define
\[ j(f):V_{j(\alpha)+1}\to V_\beta \]
to be the function $g$ such that $\widetilde{g}=j(\widetilde{f})$.


Let $U=\Ult(V_\beta,\mathscr{F})$.
Define the  \emph{natural factor map}
$\pi:U\to V_\beta$
by
\[ \pi([f])=j(f)(s).\]
Then $\pi$ is well-defined.
For if $[f]=[g]$
then there is $\sigma\in\mathscr{F}$ such that
\[ \all x\in\sigma\ [f(x)=g(x)], \]
so by Lemma \ref{lem:general_Sigma_0_substitution},
\[ \all x\in j(\sigma)\ [j(f)(x)=j(g)(x)],\]
and since $\sigma\in\mathscr{F}$, therefore $j(f)(s)=j(g)(s)$.
Similarly, $\pi:U\to\rg(\pi)$ is an isomorphism (with respect $\in^U$ and 
$\in$). In particular, in this case, $U$ is wellfounded.
However, without $\AC$, it is not immediate that $U$ is extensional.
That is, suppose $[f]\neq[g]$. To witness extensionality for $[f],[g]$,
we need some $h:V_{\alpha+1}\to V_\beta$ such that $[h]\in^U[f]$ iff 
$[h]\notin^U[g]$; that is, we need $\sigma\in\mathscr{F}$
such that $h(x)\in f(x)\Delta g(x)$ for all $x\in\sigma$ (where $\Delta$ 
denotes symmetric difference). Because $[f]\neq[g]$,
there is indeed $\sigma\in\mathscr{F}$ such that $f(x)\Delta g(x)\neq\emptyset$
for all $x\in\sigma$, but it is not clear whether there is a corresponding 
choice function (even on some smaller $\sigma'\in\mathscr{F}$).


\subsection{Successor rank-to-rank embeddings as ultrapowers}

In this section we sketch an alternate proof of Theorem 
\ref{tm:cumulative_periodicity}, one which is equivalent
to that presented already, but superficially different,
and maybe more standard for set theory. We will also consider
partial elementarity.


\begin{dfn}
Let $\eta$ be even and $j\in\mathscr{E}_0(V_{\eta+2})$ with 
$j(V_{\eta+1})=V_{\eta+1}$. Then $\mu_j$ denotes
the ultrafilter over $V_{\eta+1}$ derived from $j$ with seed $j\rest 
V_\eta$.\footnote{Note that by our flat pairing convention,
$j\rest V_\eta\in V_{\eta+1}$.}
That is,
\[ \mu_j=\{\sigma\sub V_{\eta+1}:j\rest V_\eta\in j(\sigma)\}.\qedhere\]
\end{dfn}

We will again define for all even ordinals $\delta$
a \emph{canonical extension} operation
$k\mapsto k^+$,
with domain $\mathscr{E}_1(V_\delta)$, such that
$k^+:V_{\delta+1}\to V_{\delta+1}$ (but $k^+$ is not claimed
to be elementary in general), and such that $k^+$
is the unique candidate for a $\Sigma_0$-elementary map $\ell:V_{\delta+1}\to 
V_{\delta+1}$ such that $k\sub\ell$ and $\ell(V_\delta)=V_\delta$.
The operation $k\mapsto k^+$, with domain $\mathscr{E}_1(V_\delta)$,
will be definable over $V_{\delta+1}$ without parameters, uniformly in $\delta$
(meaning that there is a formula
$\psi$ such that for all even $\delta$ and
$k\in\mathscr{E}_1(V_\delta)$ and
$x,y\in V_{\delta+1}$, we have
\[ k^+(x)=y\Leftrightarrow V_{\delta+1}\sats\psi(k,x,y),\]
noting  $k\in V_{\delta+1}$ by our flat pairing convention).
The definition of $k\mapsto k^+$ for $k\in\mathscr{E}_1(V_\delta)$,
and proof of its basic properties,
is by induction on $n<\om$, where
$\delta=\lambda+n$ for some limit ordinal $\lambda$.


If $\delta$ is a limit, then $k^+$ is defined as in
Definition \ref{CanonicalExtension}.

Suppose now that $\delta=\eta+2$ where $\eta$ is even.
Let $j\in\mathscr{E}_0(V_{\eta+2})$ with $j(V_{\eta+1})=V_{\eta+1}$; we want to 
define $j^+$ and prove some facts.\footnote{We will end
up seeing that it follows that $j\in\mathscr{E}_1(V_{\eta+1})$.}
Let $\mu=\mu_j$.
Let:
\begin{enumerate}[label=--]
 \item 
$U=\Ult(V_{\eta+2},\mu)$ and 
$i_\mu:V_{\eta+2}\to U$
be the ultrapower map,
\item $\widetilde{U}=\Ult(V_{\eta+3},\mu)$
and $\widetilde{i}_\mu:V_{\eta+3}\to\widetilde{U}$ be the ultrapower map.
\end{enumerate}
We will eventually show that $i_\mu=j$
and $j\sub\widetilde{i}_\mu$, and define $j^+=\widetilde{i}_\mu$.
We don't yet know $U,\widetilde{U}$ are extensional/wellfounded,
so  these ultrapowers are at the ``representation''
level (their elements are  equivalence classes $[f]_\mu$).

Consider the hull
\begin{equation}\label{eqn:H} H=\Hull^{V_{\eta+2}}(\rg(j)\cup\{j\rest 
V_\eta\}),\end{equation}
where
$\Hull^{M}(X)$,
for $X\sub M$, denotes the set of all $x\in 
M$
such that $x$ is definable over $M$ from parameters
in $X$.
The following claim is a typical feature of 
ultrapowers via a measure
derived from an embedding,
although part \ref{item:Ult_iso_Hull} only holds
because $j\rest V_\eta$ encodes enough information,
and for this it is crucial that the canonical
extension $(j\rest V_\eta)^+=j\rest V_{\eta+1}$,
and that this operation is definable over $V_{\eta+2}$ (in fact it is 
definable over 
$V_{\eta+1}$),
a fact we know by induction.

\begin{lem}\label{lem:Ult(V_eta+2,mu)} Recall $U=\Ult(V_{\eta+2},\mu)$
and $H$ is defined in \tu{(}\ref{eqn:H}\tu{)}. We have:
\begin{enumerate}
 \item\label{item:Ult_iso_Hull} $U$
 is extensional and wellfounded; moreover, $U\iso 
H=V_{\eta+2}$.
\item\label{item:i_mu=j} $i_\mu=j$, after we identify $U$ 
with its transitive collapse $V_{\eta+2}$.
\item\label{item:j_Sigma_1-elem} $j:V_{\eta+2}\to V_{\eta+2}$ is 
$\Sigma_1$-elementary.
\end{enumerate}
\end{lem}
\begin{proof}
Part \ref{item:Ult_iso_Hull}: We first show $H=V_{\eta+2}$. As noted above, 
from the parameter $j\rest 
V_\eta$,
$V_{\eta+2}$ can 
compute
$k=(j\rest V_\eta)^+=j\rest V_{\eta+1}$. Now let $x\in 
V_{\eta+2}$.
Then $x\sub V_{\eta+1}$ and
$x=k^{-1}``j(x)$,
and since $j(x)\in\rg(j)$, this suffices.\footnote{Note that
the proof actually shows that 
$V_{\eta+2}=\Hull_{\Sigma_1}^{V_{\eta+2}}(\rg(j)\cup\{j\rest V_\eta\})$,
where $\Hull_{\Sigma_1}^M(X)$ is defined like $\Hull^M(X)$,
except that it only consists of the $y\in M$ such that
for some $\vec{x}\in X^{<\om}$ and $\Sigma_1$ formula $\varphi$,
$y$ is the unique $y'\in M$ such that $M\sats\varphi(\vec{x},y')$.}

Let
$\pi:U\to V_{\eta+2}$ be the factor map
 $\pi([f]^{V_{\eta+2}}_\mu)=j(f)(j\rest V_\eta)$.
By \S\ref{subsec:ultrapowers}, $\pi$ is a well-defined
 $\in$-isomorphism $U\to\rg(\pi)$.
But then $\rg(\pi)=V_{\eta+2}$, because 
given
$x\in V_{\eta+2}$, let
\[ f_x:\mathscr{E}(V_\eta)\to V_{\eta+2} \]
be 
 $f_x(k)=(k^+)^{-1}``x$,
and note
$j(f_x)(j\rest V_\eta)=x$.

Part \ref{item:i_mu=j}:  We have $i_\mu(x)=[c_x]^{V_{\eta+2}}_\mu$.
But note  $\pi\com i_\mu=j$, because
\[ \pi([c_x]^{V_{\eta+2}}_\mu)=j(c_x)(j\rest V_\eta)=c_{j(x)}(j\rest 
V_\eta)=j(x), \]
since $j$ is elementary. But identifying $U$ with $V_{\eta+2}$, 
then $\pi=\id$, so $i_\mu=j$.

Part \ref{item:j_Sigma_1-elem}: Let $\varphi$ be $\Sigma_0$
and $x,y\in V_{\eta+2}$ with
 $V_{\eta+2}\sats\varphi(j(x),y)$.
We have $y=j(f_y)(j\rest V_\eta)$ where 
$f_y$ is as above.
So
\[ V_{\eta+2}\sats\exists k\in V_{\eta+1}\ [\varphi(j(x),j(f_y)(k))].\]
But since $j$ is $\Sigma_0$-elementary and $j(V_{\eta+1})=V_{\eta+1}$,
therefore
\[ V_{\eta+2}\sats\exists k\in V_{\eta+1}\ [\varphi(x,f_y(k))],\]
hence $V_{\eta+2}\sats\exists z\varphi(x,z)$, as desired.
\end{proof}

Having analyzed $j$ as an ultrapower map,
we now consider extending $j$ to 
$V_{\eta+3}$. Recall $\widetilde{U}=\Ult(V_{\eta+3},\mu)$
and $\widetilde{i}_\mu=i^{V_{\eta+3}}_\mu$.

\begin{dfn}
 Let $R\sub \mathscr{E}(V_\eta)\cross V$ be a relation.
 A \emph{$\mu$-uniformization of $R$}
 is a function $f:\mathscr{E}(V_\eta)\to V$ such that
 for  $\mu$-measure one many $k\in\mathscr{E}(V_\eta)$,
if there is $x$ such that $(k,x)\in R$ then $(k,f(k))\in 
R$.
\end{dfn}

\begin{rem}
 Note that the existence of
 $\mu$-uniformizations is a weak kind of choice principle.
\end{rem}

\begin{lem}\label{lem:when_j_extends_to_Sigma_0-elem_j^+}  We have:
 \begin{enumerate}
\item\label{item:Utilde_wfd} $\widetilde{U}$ is  wellfounded.
\item\label{item:Utilde_extensional_charac} The following are equivalent:
\begin{enumerate}[label=\tu{(}\alph*\tu{)}]
\item\label{item:Utilde_extensional} $\widetilde{U}$ is extensional,
\item\label{item:j_extends_Sigma_0} $j$ extends to a $\Sigma_0$-elementary
$\ell:V_{\eta+3}\to V_{\eta+3}$,
\item\label{item:Sigma_0_unif} for all $R\sub\mathscr{E}(V_\eta)\cross 
V_{\eta+2}$,
there is a $\mu$-uniformization of $R$.
\end{enumerate}
\item \label{item:when_extends_to_Sigma_0} If $\ell:V_{\eta+3}\to V_{\eta+3}$
is a $\Sigma_0$-elementary extension of $j$ then
identifying $\widetilde{U}$ with its transitive collapse,
we have $V_{\eta+2}\psub\widetilde{U}\sub 
V_{\eta+3}$ and
$\ell=\widetilde{i}_\mu$ and $\ell(V_{\eta+2})=V_{\eta+2}$.\footnote{The 
arxiv.org:v1
draft of this paper asserted here  ``$\widetilde{U}\psub V_{\eta+3}$,
and in fact $\mu\notin\widetilde{U}$'', but this was an oversight.
If $\ell$ is \emph{fully} elementary, this holds,
by Theorem \ref{EvenUndefinable}. And
the analogous statement holds with $\eta+2$ replaced by a limit;
see Theorem \ref{tm:no_def_Sigma_1_V_delta_to_V_delta}.
We are not sure in general.}
\end{enumerate}
\end{lem}
\begin{proof}
Part \ref{item:Utilde_wfd}: By Lemma \ref{lem:Ult(V_eta+2,mu)},
the part of the ultrapower formed by functions
with codomain $V_{\eta+2}$ is isomorphic to $V_{\eta+2}$.
It follows that $\widetilde{U}$ is wellfounded.

Part \ref{item:Utilde_extensional_charac}:
Suppose $j\sub\ell\in\mathscr{E}_0(V_{\eta+3})$.  We show
$\ell(V_{\eta+2})=V_{\eta+2}$. Clearly $\ell(V_{\eta+2})\sub V_{\eta+2}$, so
we just need
$V_{\eta+2}\sub\ell(V_{\eta+2})$. Let $x\in V_{\eta+2}$.
Then
$x=j(f_x)(j\rest V_\eta)$.
But
\[ V_{\eta+3}\sats\text{``}f_x(k)\in V_{\eta+2}\text{ for all 
}k\in\mathscr{E}(V_\eta)\text{''},\]
which is a $\Sigma_0$ statement of the parameters
$f_x,V_{\eta+2},\mathscr{E}(V_\eta)$, 
and therefore \[ V_{\eta+3}\sats\text{``}\ell(f_x)(k)\in\ell(V_{\eta+2})\text{
for all }k\in\ell(\mathscr{E}(V_\eta))\text{''},\]
but $j\sub\ell$, and it follows that
$x=\ell(f_x)(j\rest V_\eta)\in\ell(V_{\eta+2})$.

We next show that $\widetilde{i}_\mu=\ell$. For we know
$i_\mu=j$ already, so consider $X\in V_{\eta+3}\cut V_{\eta+2}$,
so $X\sub V_{\eta+2}$. Let $x\in V_{\eta+2}$.
Let
\[ D=\{k\in\mathscr{E}(V_\eta)\bigm|f_x(k)\in X\}.\]
Then $x\in\widetilde{i}_\mu(X)$ iff $D\in\mu$
iff $j\rest V_\eta\in j(D)=\ell(D)$ iff (by $\Sigma_0$-elementarity)
 $\ell(f_x)(j\rest V_\eta)\in\ell(X)$
iff $x=j(f_x)(j\rest V_\eta)\in\ell(X)$.

Now let us deduce that \ref{item:Sigma_0_unif} holds.
So let $R\sub\mathscr{E}(V_\eta)\cross V_{\eta+2}$
and let $D$ be the domain of $R$; that is,
\[ D=\{k\in\mathscr{E}(V_\eta)\bigm|\exists x\ [(k,x)\in R]\}.\]
We may assume $D\in\mu$, so $j\rest V_\eta\in j(D)$.
Now $R\in V_{\eta+3}$ and
\[ V_{\eta+3}\sats\all k\in D\ \exists x\in V_{\eta+2}\ [(k,x)\in R].\]
So by $\Sigma_0$-elementarity and since $\ell(V_{\eta+2})\sub V_{\eta+2}$
(in fact we have equality there),
\[ V_{\eta+3}\sats\all k\in \ell(D)\ \exists x\in V_{\eta+2}\ [(k,x)\in 
\ell(R)],\]
and since $D\in\mu$, therefore we can fix $x\in V_{\eta+2}$
such that $(j\rest V_\eta,x)\in\ell(R)$. We claim 
that $f_x$ is
a $\mu$-uniformization of $R$. For  suppose instead that
\[ C=\{k\in\mathscr{E}(V_\eta)\bigm|(k,f_x(k))\notin R\}\in\mu.\]
 Then $j\rest V_\eta\in j(C)=\ell(C)$,
and by
$\Sigma_0$-elementarity,
 $(j\rest V_\eta,\ell(f_x)(j\rest V_\eta))\notin \ell(R)$,
so $(j\rest V_\eta,x)\notin\ell(R)$, a contradiction.

Now assume \ref{item:Sigma_0_unif} holds ($\mu$-uniformization);
 we will show that $\widetilde{U}$ is extensional and $\Sigma_0$-{\L}o\'{s}' 
theorem 
 holds for $\widetilde{U}$, which implies 
 \[ \widetilde{i}_\mu:V_{\eta+3}\to\widetilde{U}\sub V_{\eta+3} \]
 is $\Sigma_0$-elementary, and therefore in fact
 $\widetilde{i}_\mu:V_{\eta+3}\to V_{\eta+3}$ is $\Sigma_0$-elementary.

 For extensionality, let $f,g:\mathscr{E}(V_\eta)\to V_{\eta+3}$
 be such that $[f]\neq[g]$; that is, 
 \[ D=\{k\in\mathscr{E}(V_\eta)\bigm|f(k)\neq g(k)\}\in\mu.\]
 Then define the relation
 \[ R=\{(k,x)\in\mathscr{E}(V_\eta)\cross V_{\eta+2}\bigm|x\in 
f(k)\Delta g(k)\}.\]
Note that for all $k\in D$,  there is $x$ with $(k,x)\in R$. So we can 
$\mu$-uniformize $R$
with some $h:\mathscr{E}(V_\eta)\to V_{\eta+2}$.
Since $\mu$ is an ultrafilter, either (i) for $\mu$-measure one many
$k$, we have $h(k)\in f(k)\cut g(k)$, or (ii) vice versa.
Suppose (i) holds. Then $[h]\in[f]$ and $[h]\notin[g]$,
verifying extensionality for $[f],[g]$.

It follows now that $\widetilde{U}$ is isomorphic
to some subset of $V_{\eta+3}$ (and we already know
$V_{\eta+2}\sub\widetilde{U}$). Now observe
that  the assumed
$\mu$-uniformization is enough for the usual
proof of $\Sigma_0$-{\L}o\'{s}' theorem to go through
(that is, the usual proof of {\L}o\'{s}' theorem, but just with respect 
to $\Sigma_0$ formulas).
It follows as usual that $\widetilde{i}_\mu$
is $\Sigma_0$-elementary as a map $V_{\eta+3}\to\widetilde{U}$,
and hence as a map $V_{\eta+3}\to V_{\eta+3}$, as desired.

Finally suppose that $\mu$-uniformization as in \ref{item:Sigma_0_unif}
fails; we will show that $\widetilde{U}$ is not extensional.
Let $R\sub\mathscr{E}(V_\eta)\cross V_{\eta+2}$
be a counterexample to $\mu$-uniformization.
We have the constant function $c_\emptyset$.
Define $f:\mathscr{E}(V_\eta)\to V_{\eta+3}$
by
\[ f(k)=\{x\bigm|(k,x)\in R\}.\]
Note that $f(k)\neq\emptyset$ for almost all $k$.
So $[f]\neq[c_\emptyset]$. 
But
 there is no $g$ such that $[g]\in[f]$,
 and therefore $\widetilde{U}$ is non-extensional
 with respect to $[f],[c_\emptyset]$.
 
Part \ref{item:when_extends_to_Sigma_0}: We already
saw these things in the proof of part \ref{item:Utilde_extensional_charac}. 
\end{proof}

\begin{dfn}[Canonical extension via ultrapowers]
Let $\eta$ be even.

For $j\in\mathscr{E}_1(V_{\eta+2})$, we define
 $j^+:V_{\eta+3}\to V_{\eta+3}$
as $j^+=\widetilde{i}_{\mu_j}$, as above.

 For $x\in V_{\eta+2}$, $f_x:\mathscr{E}(V_\eta)\to V_{\eta+2}$
is defined $f_x(k)=(k^+)^{-1}``x$ (really $f_x$ depends on $\eta$, but this 
should be clear in context).
\end{dfn}

\begin{rem}\label{rem:proof_2_cumulative_periodicity}
We now reprove Theorem \ref{tm:cumulative_periodicity},
by induction, using the canonical extension $j^+$ just defined.
The argument is essentially as before, so we just give a sketch.
Let $\lambda$ be a limit and $j:V_{\lambda+2}\to V_{\lambda+2}$
be elementary. Let $\mu=\mu_j$. By Lemma
\ref{lem:Ult(V_eta+2,mu)},
$V_{\lambda+2}=\Ult(V_{\lambda+2},\mu)$
and $j=i^{V_{\lambda+2}}_\mu$ is the ultrapower map.

We claim  $j$ is not definable over $V_{\lambda+2}$ from parameters.
For suppose
$j$ is definable over $V_{\lambda+2}$ from $p\in V_{\lambda+2}$.
Then $p\in\rg(j(f_p))$, since
\[ p=[f_p]^{V_{\lambda+2}}_\mu=j(f_p)(j\rest V_\lambda). \]
  One can now argue as in the proof of Theorem
\ref{EvenUndefinable} to reach a contradiction.

Next, if $\ell:V_{\lambda+3}\to V_{\lambda+3}$ is elementary
and $j=\ell\rest V_{\lambda+2}$,
then  $\ell=j^+$ by the preceding lemmas.
But $\mu\in V_{\lambda+3}$, and it is straightforward
to see that the ultrapower map
$j^+=\widetilde{i_\mu}=i^{V_{\lambda+3}}_\mu$ is definable over $V_{\lambda+3}$
from the parameter $\mu$, or equivalently, from
$j$. So $\ell$ is definable as desired.

Now suppose $j:V_{\lambda+4}\to V_{\lambda+4}$
is elementary. Let $\mu=\mu_j$
(the measure derived from $j$ with seed
$j\rest V_{\lambda+2}$).
Then since 
$j\rest V_{\lambda+3}=(j\rest V_{\lambda+2})^+$,
the lemmas give that $\Ult(V_{\lambda+4},\mu)=V_{\lambda+4}$
and $j$ is the ultrapower map, so like before, we get that $j$ is not definable 
from parameters. Etc.
\end{rem}

\subsection{Reinhardt ultrafilters}

Let $\lambda$ be even. One can abstract out a notion of filter which corresponds
precisely to elementary embeddings in $\mathscr{E}(V_{\lambda+2})$, and 
also filters which correspond to embeddings in 
$\mathscr{E}_{n+1}(V_{\lambda+2})$,
for each $n<\om$.
The filters below are over $V_{\lambda+1}$,
but one could consider instead filters over $\mathscr{E}(V_\lambda)$,
identifying $j\in\mathscr{E}(V_\lambda)$ with $\rg(j)$,
and  at a small abuse of notation,
we treat the two interchangeably.
Recall that $\trancl(M)$ denotes the transitive collapse of $M$.


\begin{dfn}[Reinhardt ultrafilters]
Let $\lambda$ be even and  $\mu$ be an ultrafilter over 
$V_{\lambda+1}$. We say that $\mu$ is:
 \begin{enumerate}
  \item \emph{rank-J\'onsson} iff
  $\sigma=\{A: A\elem V_\lambda\text{ and 
  }\trancl(A)=V_\lambda\}\in\mu$,
  \item \emph{fine} iff for each $x\in V_\lambda$, we have
 $\tau_x=\{A: x\in A\sub 
V_\lambda\}\in\mu$,
\item \emph{normal} iff for each $\left<\sigma_x\right>_{x\in V_\lambda}\sub 
\mu$,
the diagonal intersection
\[\Delta_{x\in V_\lambda}\sigma_x= \{A: A\sub V_\lambda\text{ and 
}A\in\sigma_x\text{ for each 
}x\in A\}\in \mu, 
\]
\item \emph{pre-Reinhardt} iff rank-J\'onsson, fine and normal,
\item \emph{$\Sigma_1^{\lambda+2}$-Reinhardt} iff pre-Reinhardt and every
$R\sub V_{\lambda+1}\cross V_{\lambda+1}$ can be $\mu$-uniformized,
\item \emph{$\Sigma_{n+2}^{\lambda+2}$-Reinhardt} iff
 pre-Reinhardt
and every
$R\sub V_{\lambda+1}\cross V_{\lambda+2}$
which is $\Pi_{n}$-definable over $V_{\lambda+2}$ from parameters
can be $\mu$-uniformized,
\item \emph{$\Sigma^{\lambda+2}_\om$-Reinhardt} iff
$\Sigma_{n+1}^{\lambda+2}$-Reinhardt for all $n<\om$.\qedhere
\end{enumerate}
\end{dfn}


 Note
 that if $x\in V_{\lambda+i}$, where $i\leq 1$, then 
$f_x:\mathscr{E}(V_\lambda)\to V_{\lambda+i}$.
\begin{lem}\label{lem:pre-Reinhardt}
Let $\mu$ be a pre-Reinhardt ultrafilter over $V_{\lambda+1}$.
Let $U=\Ult(V_{\lambda+1},\mu)$.
Then
$U$ is extensional, wellfounded and isomorphic to 
$V_{\lambda+1}$. Moreover, $[\id]_\mu=i_\mu``V_\lambda$ and
$[f_x]_\mu=x$ for each $x\in V_{\lambda+1}$.
\end{lem}
\begin{proof} We start by considering $V_\lambda$. Write $[f]=[f]_\mu$.
\begin{clm*} $\Ult(V_\lambda,\mu)=V_\lambda$
and $x=[f_x]$ for each $x\in V_\lambda$.\end{clm*}
\begin{proof}
Given $x,y\in V_\lambda$, we have
\[ ([f_x]\in^U[f_y]\Leftrightarrow x\in y)\text{ and }
([f_x]=^U[f_y]\Leftrightarrow x=y).\]
For by rank-J\'onssonness and fineness, for $\mu$-measure one many 
$k\in\mathscr{E}(V_\lambda)$, we have $x,y\in\rg(k)$,
and for all such $k$, note $f_x(k)=k^{-1}(x)$ and 
$f_y(k)=k^{-1}(y)$. This yields the  stated equivalences.
Now let $f:\mathscr{E}(V_\lambda)\to V_\lambda$.
We claim that there is $x\in V_\lambda$ such that $[f]=^U[f_x]$.
For supposing not, then for each $x\in V_\lambda$,
defining
\[ \sigma_x=\{k\in\mathscr{E}(V_\lambda):f(k)\neq f_x(k)\}, \]
we get $\sigma_x\in\mu$. So the diagonal intersection
\[ \sigma=\{k\in\mathscr{E}(V_\lambda):f(k)\neq f_{x}(k)\text{ 
for all }x\in k``V_\lambda\} \in\mu,\]
so $\sigma\neq\emptyset$. Let 
$k\in\sigma$
and $\bar{x}=f(k)\in V_\lambda$. Let $x=k(\bar{x})$. Then 
$f_x(k)=k^{-1}(x)=\bar{x}=f(k)$, a contradiction.
\end{proof}


Now let $x\in V_{\lambda+1}\cut V_\lambda$ and $y\in V_\lambda$.
The $[f_y]\in^U[f_x]$ iff $y\in x$,
because $y\in\rg(k)$ for $\mu$-almost every $k$.
Note also that $[f_x]\notin^U[f_y]$.
It also easily follows that if $x'\in V_{\lambda+1}\cut V_\lambda$ with $x'\neq 
x$ then $[f_x]\neq [f_{x'}]$ (consider $[f_y]$ for some $y\in x\Delta x'$).

For extensionality, let $f:\mathscr{E}(V_\lambda)\to 
V_{\lambda+1}$ and
$x=\{y\in V_\lambda:[f_y]\in^U[f]\}$.
We claim $[f]=^U[f_x]$.
To see this,
for each $y\in V_\lambda$,
let
\[ \sigma_y=\{k\in\mathscr{E}(V_\lambda):f_y(k)\in f_x(k)\Leftrightarrow
f_y(k)\in f(k)\}.\]
Note $\sigma_y\in\mu$. Let $\sigma\in\mu$ be the diagonal intersection, and 
note $f(k)=f_x(k)$ for each $k\in\sigma$. So $[f]=^U[f_x]$, as desired,
and extensionality follows easily.

The fact that $[\id]=i_\mu``V_\lambda$
 is a straightforward consequence of fineness and normality.
 The rest of the lemma now follows easily.
\end{proof}

\begin{rem}\label{rem:choice_princ}
We now characterize the elements of $\mathscr{E}_{n+1}(V_{\lambda+2})$
as the ultrapower maps given by $\Sigma_{n+1}^{\lambda+2}$-Reinhardt 
ultrafilters,
 and hence the elements of $\mathscr{E}(V_{\lambda+2})$ as the ultrapower 
maps via
$\Sigma_\om^{\lambda+2}$-Reinhardt ultrafilters.
Note that because of the $\mu$-uniformization aspect
of Reinhardt ultrafilters, the theorem shows that
weak choice principles follow from the existence
of appropriate elementary embeddings.\end{rem}


\begin{tm}
Let $\lambda$ be even and $n<\om$. Then:
\begin{enumerate}
 \item \label{item:filter_is_Reinhardt}
If $j\in\mathscr{E}_{n+1}(V_{\lambda+2})$ then
 $\mu_j$ is a
$\Sigma_{n+1}^{\lambda+2}$-Reinhardt ultrafilter
and $j=i^{V_{\lambda+2}}_{\mu_j}$.
\item\label{item:ult_map_is_elem} Let $\mu$ be  a 
$\Sigma_{n+1}^{\lambda+2}$-Reinhardt ultrafilter, $U=\Ult(V_{\lambda+2},\mu)$
and $j:V_{\lambda+2}\to U$ be the ultrapower map $j=i^{V_{\lambda+2}}_\mu$.
Then:
\begin{enumerate}[label=\tu{(}\alph*\tu{)}]
\item\label{item:U=V_lambda+2} $U$ is extensional and 
wellfounded, 
$U=V_{\lambda+2}$,
$\mu=\mu_j$,
$[\id]=j``V_\lambda$ and 
$x=[f_x]=j(f_x)(j`` 
V_\lambda)$ for each $x\in V_{\lambda+2}$.
\item\label{item:Sigma_n+1-elem} $j\in\mathscr{E}_{n+1}(V_{\lambda+2})$.
\end{enumerate}
 \end{enumerate}
\end{tm}
\begin{proof}
Part \ref{item:filter_is_Reinhardt}: Let $\mu=\mu_j$. Rank-J\'onssonness and 
fineness
are straightforward. Consider normality, and fix 
$\vec{\sigma}=\left<\sigma_x\right>_{x\in V_\lambda}\sub\mu$, and let 
$\left<\sigma'_x\right>_{x\in V_\lambda}=j(\vec{\sigma})$.
 Let $B=\Delta_{x\in V_\lambda}\sigma_x$.
We must see that
\[ j`` V_\lambda\in j(B)=\Delta_{x\in V_\lambda}\sigma'_x.\]
But if $y\in j``V_\lambda$ then $y=j(x)$ for some $x\in V_\lambda$,
and $\sigma_x\in\mu$, so
$j``V_\lambda\in j(\sigma_x)=\sigma'_{y}$,
as desired.

Now let $U=\Ult(V_{\lambda+2},\mu)$.
By Lemma \ref{lem:Ult(V_eta+2,mu)}, $U=V_{\lambda+2}$ and
$j=i^{V_{\lambda+2}}_{\mu}$. 
Let us verify that $\mu$ is $\Sigma_1^{\lambda+2}$-Reinhardt.
Let $R\sub V_{\lambda+1}\cross V_{\lambda+1}$ and $D\in\mu_j$
such that for all $k\in D$, there is $x\in V_{\lambda+1}$ with $(k,x)\in 
R$. Then by $\Sigma_1$-elementarity and since $j\rest V_\lambda\in j(D)$,
 there is $x\in V_{\lambda+1}$ with $(j\rest V_\lambda,x)\in j(R)$.
Fix such an $x$. We have  $x=j(f_x)(j\rest V_\lambda)$.
So letting $D'$ be the set of all $k\in D$
such that $(k,f_x(k))\in R$, then $D'\in\mu$, so we are done.

Now suppose $j\in\mathscr{E}_{n+2}(V_{\lambda+2})$
and let $\psi$ be a $\Pi_n$ formula, $p\in V_{\lambda+2}$
and $D\in\mu$, such that for all $k\in D$, there is $x\in V_{\lambda+2}$ with 
$V_{\lambda+2}\sats\psi(p,k,x)$. The assertion
``$\all k\in D\ \exists x\ \psi(p,k,x)$''
is $\Pi_{n+2}$ in parameters $D,p$. So by $\Sigma_{n+2}$-elementarity and since 
$j\rest V_\lambda\in j(D)$,
we can fix $x\in V_{\lambda+2}$ such that
\[ V_{\lambda+2}\sats\psi(j(p),j\rest 
V_\lambda,x).\] Let $D'$ be the set of all $k\in D$
where $V_{\lambda+2}\sats\psi(p,k,f_x(k))$. We claim
 $D'\in\mu$, giving the desired $\mu$-uniformization.
So suppose otherwise. Then $E=\mathscr{E}(V_\lambda)\cut D'\in\mu$,
and $V_{\lambda+2}\sats\all k\in E\ [\neg\psi(p,k,f_x(k))]$.
But then by $\Sigma_{n+1}$-elementarity and since $j\rest V_\lambda\in j(E)$, 
we get $V_{\lambda+2}\sats\neg\psi(j(p),j\rest V_\lambda,x)$,
a contradiction.


Parts \ref{item:ult_map_is_elem}\ref{item:U=V_lambda+2}:
 By Lemma \ref{lem:pre-Reinhardt},
we already know $V_{\lambda+1}=\Ult(V_{\lambda+1},\mu)$
(including extensionality and 
wellfoundedness) and $x=[f_x]$ for all $x\in V_{\lambda+1}$.
Note that it follows that $U=\Ult(V_{\lambda+2},\mu)$ is wellfounded
(though we haven't yet shown extensionality).

Now  $\mu$ is $\Sigma_1^{\lambda+2}$-Reinhardt. Using this, extensionality
is just like in the proof of Lemma \ref{lem:when_j_extends_to_Sigma_0-elem_j^+}.
So we identify $U$ with its Mostowski collapse,
 so $V_{\lambda+1}\sub U\sub V_{\lambda+2}$.
Similarly to extensionality, $\Sigma_0$-{\L}o\'{s}' theorem holds. The 
$\Sigma_1$-elementarity
of $j:V_{\lambda+2}\to U$ follows:
 if $U\sats\exists w\varphi(j(x),w)$ where $\varphi$ is 
$\Sigma_0$, then there is $f$ with $U\sats\varphi(j(x),[f])$, and by
$\Sigma_0$-{\L}o\'{s},
$V_{\lambda+2}\sats\varphi(x,f(k))$ for $\mu$-measure one many $k$,
so $V_{\lambda+2}\sats\exists w\varphi(x,w)$.
And because $[\id]=j``V_\lambda$ by Lemma \ref{lem:pre-Reinhardt},
it is easy to see that $\mu=\mu_j$
(although we haven't shown that $U=V_{\lambda+2}$,
we can still define $\mu_j$ as before).

To see $U=V_{\lambda+2}$,
it suffices to see that $[f_x]=x$ for each $x\in V_{\lambda+2}$,
and for this, given $y\in 
V_{\lambda+1}$, we must see that
 $[f_y]\in^U[f_x]$ iff $y\in x$. 
To see the latter, it suffices to show that $y\in\rg(k^+)$ for $\mu$-measure 
one many $k$,
because for all such $k$, we have $y\in x$ iff
\[ f_k(y)=(k^+)^{-1}(y)\in(k^+)^{-1}``x=f_k(x).\]

Let $D$ be the set of all $k\in \mathscr{E}(V_\lambda)$
such that $y\in\rg(k^+)$. Then since $j$ is $\Sigma_1$-elementary
and $j(V_{\lambda+1})=V_{\lambda+1}$,
$j(D)$ is the set of all $k\in\mathscr{E}(V_\lambda)$
such that $j(y)\in\rg(k^+)$. But 
$j\rest 
V_{\lambda+1}=(j\rest V_\lambda)^+$,
so $j\rest V_\lambda\in j(D)$, so $D$ is $\mu$-measure one,
as desired.

Finally, we already have $[\id]=j``V_\lambda$,
and  $x=[f_x]$ for each $x\in V_{\lambda+2}$.
But then like in the proof of Lemma \ref{lem:Ult(V_eta+2,mu)},
 the factor map $\pi:U\to V_{\lambda+2}$,
defined $\pi([f_x])=j(f_x)(j\rest 
V_\lambda)$, is surjective and in fact the identity,
so $x=j(f_x)(j\rest V_\lambda)$.


Part \ref{item:ult_map_is_elem}\ref{item:Sigma_n+1-elem}:
For $n=0$, this was 
verified above. So suppose $m<\om$ and
$\mu$ is $\Sigma_{m+2}^{\lambda+2}$-Reinhardt;
we show $j$ is $\Sigma_{m+2}$-elementary.
Let $\varphi$ be $\Pi_{m+1}$ and suppose that 
$V_{\lambda+2}\sats\varphi(j(x),y)$. We have $y=[f_y]$.
Let $D$ be the set of all $k\in\mathscr{E}(V_\lambda)$
such that $V_{\lambda+2}\sats\varphi(x,f_y(k))$.
It suffices to see that $D\in\mu$, so suppose 
$E=\mathscr{E}(V_\lambda)\cut D\in\mu$.
Let $\psi$ be a $\Pi_m$ formula
such that
\[ \neg\varphi(u,v)\iff\exists 
w\ \psi(u,v,w).\]
So $V_{\lambda+2}\sats\all k\in E\ \exists 
w\ \psi(x,f_y(k),w)$. Since $\mu$ is $\Sigma_{m+2}^{\lambda+2}$-Reinhardt,
there is $E'\in\mu$ and $g:E'\to V_{\lambda+2}$
such that $V_{\lambda+2}\sats\all k\in E'\ \psi(x,f_y(k),g(k))$.
By induction, $j$ is $\Sigma_{m+1}$-elementary, and as
$y=j(f_y)(j\rest V_\lambda)$
and $j\rest V_\lambda\in j(E')$, we get 
$V_{\lambda+2}\sats\psi(j(x),y,j(g)(j\rest V_\lambda))$,
so $V_{\lambda+2}\sats\neg\varphi(j(x),y))$, contradiction.
\end{proof}
{"config": "arxiv", "file": "2006.01103/reinhardt_filters.tex"}