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For $n$ a natural number, $\alpha$ an ordinal, let $p(n,\alpha)$ be the $n$-th projectum of $J_\alpha$, where $J$ is the Jensen hierarchy for $L$.

Call a finite sequence $s:=(x_1,\dots,x_m)$ of integers projectum-representable iff there is an ordinal $\alpha$ such that $$p(1,\alpha)=p(2,\alpha)=\dots=p(x_1,\alpha)\gt$$ $$p(x_1+1,\alpha)=\dots=p(x_2,\alpha)\gt$$ $$p(x_2+1,\alpha)=\dots=p(x_3,\alpha)\gt$$ $$\dots=p(x_m,\alpha),$$ i.e. if the sequence of projecta of $J_\alpha$ consists of a sequence of $x_1$ identical terms, then drops, then has $x_2$ identical terms, then drops again etc.

Which finite sequences are projectum-representable?

By condensation arguments, it can be seen that every representable sequence is already representable with a countable ordinal $\alpha$.

References would also be welcome.

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up vote 7 down vote accepted

The short answer is that all finite sequences are representable.

Suppose you want the general constellation you gave. We work inside $L$. For $X=\langle X,\in \rangle$ a model of V=L (not necessarily transitive) define $S^X_n(Y)$ to be the $\Sigma_n$-Skolem Hull in $X$ of the set $Y$. Let $\tau$ be the least ordinal $> \omega_{m-1}$ with $L_\tau$ a model of $\Sigma_{x_1}$-Separation scheme. (This ensures that $\rho(x_1,\tau)=\tau$.) Set $H_0= L_\tau$. Now define by recursion on $j$ for $1\leq j\leq m-1$: $$H_j = S^{H_{j-1}}_{x_j +1}(\omega_{m-1-j}). $$

Let $L_\alpha \simeq H_m$.

This is admittedly easier to see with an example: say $(x_1,x_2,x_3) = (2,4,6)$ with $m=3$ "drops". Let $\tau$ be the least above $\omega_{m-1} =\omega_2$ with $L_\tau$ a model of $\Sigma_2$-Separation. Then $\tau = \rho(\tau,2)> \omega_2 =\rho(\tau,3) \ldots$. Then $$H_1 = S^{L_\tau}_5(\omega_1).$$ (If we were to transitivise this hull it would be an $L_{\alpha'}$ with $\rho(\alpha',k)=\omega_1$ for $k\geq 5$. One should check, e.g., that $\rho(\alpha',4)\neq \omega_1$: but if this failed then there would be a $\Sigma_4(L_{\alpha'})$ map of $\omega_1$ cofinally into $\alpha'$; but this is a $\Pi_5$ statement, and would then be true in $L_\tau$ which is absurd.) Then take $$H_2 =S^{H_1}_7(\omega).$$

Let $L_\alpha \simeq H_2$. Similar checks reveal that $\rho(\alpha,6)=\omega_1> \rho(\alpha,7) \ldots$ etc. and we have the correct pattern.

I do not know of any references for this.

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Thanks for this thorough answer! The construction is really elegant. – M Carl Nov 19 '11 at 15:16

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