# If $j_{1},…,j_{n}:V_{\lambda+1}\rightarrow V_{\lambda+1}$ are elementary embeddings, then does $j_{1}(A)=…=j_{n}(A)=A$ for some linear order $A$?

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.

• What about well-orders? – Monroe Eskew Jan 9 at 9:23
• 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. – Joel David Hamkins Jan 9 at 10:55
• @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 $j(A)=A$. – Joseph Van Name Jan 9 at 22:23
• 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. – Joel David Hamkins Jan 10 at 9:00
• 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. – Joseph Van Name Jan 10 at 21:29