# How many critical points can you have below a Fibonacci term in an algebra of elementary embeddings?

In this question, I asked about how slowly the Fibonacci terms in a algebra of elementary embeddings could grow, but I did not inquire if there was a limit to how quickly the Fibonacci terms could grow.

Define the Fibonacci terms $$t_{1}(x,y)=y,t_{2}(x,y)=x$$ and where $$t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$$ for $$n\geq 1$$. Let $$\mathcal{E}_{\lambda}$$ be the set of all elementary embeddings $$j:V_{\lambda}\rightarrow V_{\lambda}$$. Then $$\mathcal{E}_{\lambda}$$ can be endowed with an operation $$*$$ defined by $$j*k=\bigcup_{\alpha<\lambda}j(k|_{V_{\alpha}})$$. Recall that $$\mathrm{crit}(j)$$ is the least ordinal where $$j(\alpha)>\alpha$$.

Suppose that $$n$$ is a natural number. Then does there exist a natural number $$m$$ such that if $$\lambda$$ is a cardinal and $$X$$ is an $$n$$-generated subalgebra of $$\mathcal{E}_{\lambda}$$ and $$\gamma=\min(\{\mathrm{crit}(t_{2n+1}(j,k))|j,k\in X,\mathrm{crit}(j)\leq\mathrm{crit}(k)\}),$$ then $$|\{\mathrm{crit}(j)|j\in X\}\cap\gamma|\leq m?$$ Are there any good lower or upper bounds on the size of the least such $$m$$?

If $$\gamma<\lambda$$ and $$\gamma$$ is a limit ordinal, then recall that $$j\equiv^{\gamma}k$$ if and only if $$j(x)\cap V_{\gamma}=k(x)\cap V_{\gamma}$$ for each $$x\in V_{\gamma}$$. The motivation for this question comes from the fact that a negative answer shows that the algebras $$\mathcal{E}_{\lambda}/\equiv^{\gamma}$$ generalize to a purely algebraic setting in two very different ways and I still do not know if these two definitions are equivalent.

• Is there any reason why there is a downvote to this question? – Joseph Van Name Mar 23 at 3:11
• I am 100 percent sure that the entities who are downvoting this question are completely ignorant of the subject matter. – Joseph Van Name Mar 24 at 3:41
• It seems like many people on this site are holding a grudge. – Joseph Van Name Mar 24 at 3:50
• The downvoters are downvoting questions like this because they want to see these questions deleted. How sad. – Joseph Van Name yesterday