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Suppose that $j_{1},\dots,j_{n}:V_{\lambda+1}\rightarrow V_{\lambda+1}$ are elementary embeddings. Then does there necessarily exist a linear ordering $A$ of $V_{\lambda}$ such that $j_{1}(A)=\dots=j_{n}(A)=A$? The result holds if $V_{\lambda}\models(V=HOD)$ which is consistent with the existence of an elementary embedding $j:V_{\lambda+1}\rightarrow V_{\lambda+1}$, but does the result hold in general without any additional hypotheses? I am also interested in the version of this question when $j_{1},\dots,j_{n}$ are only elementary embeddings from $V_{\lambda}$ to $V_{\lambda}$.

The case when $n=1$ is easy since if $j:V_{\lambda+1}\rightarrow V_{\lambda+1}$ is an elementary embedding and if $B$ is a linear ordering on $V_{\mathrm{crit}(j)}$, then $A=\bigcup_{n}j^{n}(B)$ is a linear ordering on $V_{\lambda}$ with $j(A)=A$. I am therefore looking at whether the result holds in general instead of particular instances of the result.

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    $\begingroup$ What about well-orders? $\endgroup$ – Monroe Eskew Jan 9 at 9:23
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    $\begingroup$ Can you answer the question even in the special case where there are only two embeddings, and they agree on the ordinals? I am tending to think we might make a counterexample, by producing embeddings that have fundamental disagreements, so that no linear order can be stretched in the same way by all of them. $\endgroup$ – Joel David Hamkins Jan 9 at 10:55
  • $\begingroup$ @JoelDavidHamkins. Even for your special case, the possible obstruction to the existence of a linear ordering is when we have only two embeddings that agree on the ordinals is probably not a purely algebraic obstruction. My computations suggest that for every finite self-distributive algebra $X$ for which there is a coherent notion of a critical point, there is a linear ordering $\leq$ on $X$ such that $y\leq z$ implies $x*y\leq x*z$ and such linear orderings $\leq$ on $X$ arise from the linear orderings $A$ with $j(A)=A$. $\endgroup$ – Joseph Van Name Jan 9 at 22:23
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    $\begingroup$ My idea had nothing to do with the algebraic considerations. I am imagining making a counterexample by starting with a fixed embedding, and then performing forcing to lift it in two different manners. The resulting embeddings would agree on the ground model and in particular on the ordinals. What I would expect from these lifts is that they stretch the new sets in fundamentally different ways, and that perhaps this might imply that any linear order on $V_{\kappa_0}$ will be stretched differently by them. $\endgroup$ – Joel David Hamkins Jan 10 at 9:00
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    $\begingroup$ Joel. That seems like a good idea since in your scenario, we cannot preserve any well-ordering of order type $\lambda$. In an answer to Monroe Eskew's question, if $j,k$ are the elementary embeddings in Joel's scenario, and $f:V_{\lambda}\rightarrow\lambda$ is the well ordering corresponding to $A$, then for each $x\in V_{\lambda}$, $f(j(x))=j(f(x))=k(f(x))=f(k(x))$ which implies that $j(x)=k(x)$, a contradiction. $\endgroup$ – Joseph Van Name Jan 10 at 21:29

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