5
$\begingroup$

For $n$ a natural number, $\alpha$ an ordinal, let $\rho(n,\alpha)$ 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 $$\rho(1,\alpha)=\rho(2,\alpha)=\dots=\rho(x_1,\alpha)\gt$$ $$\rho(x_1+1,\alpha)=\dots=\rho(x_2,\alpha)\gt$$ $$\rho(x_2+1,\alpha)=\dots=\rho(x_3,\alpha)\gt$$ $$\dots=\rho(x_m,\alpha),$$ 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.

$\endgroup$

1 Answer 1

9
$\begingroup$

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 take the Mostowski-Shepherdson collapse of 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.

$\endgroup$
2
  • $\begingroup$ Thanks for this thorough answer! The construction is really elegant. $\endgroup$
    – M Carl
    Nov 19, 2011 at 15:16
  • $\begingroup$ It might be known before but "Patterns of Projecta" By Adam Krawczyk could be one of the references $\endgroup$
    – Jing Zhang
    Aug 25, 2017 at 10:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.