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Let us assume that there is a non-trivial elementary embedding $j \colon L_\gamma \to L_\gamma$ and $\gamma \geq \omega_1^V$. Can we conclude that $0^{\#}$ exists?

In general, it is known that if there is an elementary embedding $j \colon L_\alpha \to L_\beta$ such that $\mathrm{crit}\, j = \delta$, $(\delta^{+})^L \leq \alpha$ and $\delta \geq \omega_2^V$ then one can conclude that $0^{\#}$ exists, but can we get something better by starting with the assumption that the embedding is from $L_\gamma$ to itself?

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  • $\begingroup$ I think this follows from Theorem 18.27 in Jech. Starting on p. 322 he shows how to get an uncountable set of indiscernibles from this assumption. $\endgroup$ Mar 6, 2023 at 10:18
  • $\begingroup$ If you believe the theorem on $j : L \to L$, then it follows. Since $P(\delta)^L \subseteq L_\gamma$, the derived ultrafilter is an $L$-ultrafilter. $\endgroup$ Mar 6, 2023 at 10:28
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    $\begingroup$ First, I don't see why it is necessary that $P(\delta)^L \subseteq L_\gamma$ in this case. Also, I think that in the proof of $0^{\#}$ from $j \colon L_\alpha \to L_\beta$, Jech is using the assumption that the critical point of $j$ is below $|\alpha|^V$. This is important in order to get that the $L$-ultrafilter produces a well founded ultrapower, but I might be missing something here. $\endgroup$
    – Yair Hayut
    Mar 6, 2023 at 10:35
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    $\begingroup$ Yes, it is an application of covering. I found that in a recent presentation of Usuba, but it seems to be folklore. $\endgroup$
    – Yair Hayut
    Mar 6, 2023 at 14:33
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    $\begingroup$ @AsafKaragila Yes it follows, I think. If $\kappa$ is the critical point of $j$, we have $\kappa > \omega_1$ (or else $0^\#$ exists) and letting $\eta = \sup_{n < \omega} j^n(\kappa)$, $L_\eta\vDash \text{ZFC}$ and $j$ restricts to an elementary embedding from $L_\eta$ to $L_\eta$. $\endgroup$ Mar 9, 2023 at 0:09

1 Answer 1

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No, such embeddings can consistently exist in $L$. In the following, $\omega_1$ and $\omega_2$ will denote these cardinals as computed in $V$.

Let us assume that $0^\sharp$ exists in $V$. There is an increasing sequence $\langle \xi_n\mid n<\omega\rangle\in V$ of order indiscernibles over the structure $(L_{\omega_2};\in, \alpha\mid\alpha<\omega_1)$, for example take the first $\omega$-many Silver indiscernibles $\geq\omega_1$. In $L$, we can define the tree $T$ which attempts to find such a sequence. As $T$ is illfounded in $V$, $T$ is illfounded in $L$, i.e. there is such a sequence $\langle \xi_n\mid n<\omega\rangle$ in $L$. Now let $X=\mathrm{Hull}^{L_{\omega_2}}(\omega_1\cup\{\xi_n\mid n<\omega\})$ and let $\pi\colon X\rightarrow L_\gamma$ be the transitive collapse of $X$. Then $\omega_1<\gamma$. Let us write $\bar \xi_n=\pi(\xi_n)$. We now have

  • $L_\gamma=\mathrm{Hull}^{L_\gamma}(\omega_1\cup\{\bar \xi_n\mid n<\omega\})$ and
  • the $\langle \bar\xi_n\mid n<\omega\rangle$ are order indiscernible over $(L_\gamma;\in,\alpha\mid\alpha<\omega_1)$.

This allows us to build elementary embeddings $j\colon L_\gamma\rightarrow L_\gamma$ similar to how one builds elementary embeddings $j\colon L\rightarrow L$ from Silver indiscernibles. Here is the construction:

Let $i\colon\omega\rightarrow\omega$ be any strictly increasing function. We can define an elementary $j\colon L_\gamma\rightarrow L_\gamma$, $j\in L$, by $j(\tau^{(L_\gamma;\in, \alpha\mid\alpha<\omega_1)}(\bar\xi_0,\dots,\bar\xi_n))=\tau^{(L_\gamma;\in,\alpha\mid\alpha<\omega_1)}(\bar\xi_{i(0)},\dots,\bar\xi_{i(n)})$ for any term $\tau(x_0,\dots, x_n)$ in the language with the $\in$-relation and a constant for every ordinal below $\omega_1$. It is not difficult to check that $j$ is well-defined, elementary and, in case $i\neq\mathrm{id}_{\omega}$, even non-trivial.

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    $\begingroup$ Very nice. So, it seems like one can get this situation just from the existence of an $\omega$-Erdos cardinal (above $\omega_1^V$), and should be strictly weaker than that. Is the consistency strength of this assertion known? $\endgroup$
    – Yair Hayut
    Mar 6, 2023 at 14:25
  • $\begingroup$ Couldn’t you also take the hull of these indiscernibles with ω_2? This would seem to contradict the “folklore” that Yair mentioned. $\endgroup$ Mar 6, 2023 at 19:23
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    $\begingroup$ @MonroeEskew I think the issue is with the $\delta^{+L}\leq \alpha$ condition. $\endgroup$ Mar 6, 2023 at 22:30
  • $\begingroup$ @GabeGoldberg Oh I see. The union of the branch through the tree that L can see actually has cardinality in L the same as whatever lower κ we’re throwing into the hull. $\endgroup$ Mar 7, 2023 at 6:35

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