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In chapter 8 of Mitchell's and Steel's FSIT, they prove a central fine structural result, which basically states that if $\mathcal{M}$ is 1-small, $k$-sound, $k$-iterable premouse then the $k+1$-strandard parameter is $k+1$-solid/universal over $\mathcal M$.

The proof is not too complicated but on page 76 they state that if $\rho_{k+1}^{\mathcal M}> lh(E)$ where $E$ is some extender on the $\mathcal M$-sequence then $\mathcal H= \mathcal H_{k+1}^{\mathcal M}(\alpha_s)$ (I omitted the rest of the parameter and $u$ ($u$ is the tuple having the solidity witnesses and the appropriate projectums whenever defined)from the hull so that it is easier to read, this is well written in details page 74) then $\mathcal H$ is an initial segment of $\mathcal M$. I can't see why this is true. Could anyone help me figure out the reason behind this? Thx.

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3 Answers 3

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The assumption is that $\rho=\rho_{k+1}^M>\text{lh}(E)$ for \emph{all} $E$, not just some $E$. But looking at your message of March 6/7, it looks like you are indeed reading it as ``all''. Regarding the question in that message, $H$ agrees with $M$ below $\rho$ because $\rho\leq\alpha_s$, and all ordinals below $\alpha_s$ were put into the uncollapsed hull when forming $H$. So if $\pi:H\to M$ is the uncollapse, then $\text{crit}(\pi)\geq\rho$. (Note you could also just argue that $M=J_\gamma(M|\rho)$ for some ordinal $\gamma$, and $M|\rho$ is also passive. Since $\text{crit}(\pi)\geq\rho$, this gives that $H=J_\beta(M|\rho)$ for some $\beta\leq\gamma$.)

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Although I'm unsure of the details of the question you ask here, I think this is a typical comparison argument for M and H where we have to show that there are no ultrapowers taken on either side. Such a proof is done once somewhere in Steel's Outline of Inner Model Theory and thereafter the argument is taken for granted.

Sorry I couldn't be more helpful, Drake

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Hi Drake. Thank you for the hint, but Steel's article in the handbook doesn't have the fact I quoted above (it has another fact involving comparison also). Indeed this is a typical comparison argument. The argument is very beautiful and uses the copy maps, the Dood-Jensen lemma and iterations of $\mathcal M$ and of $\mathcal H$. I think I have a quick answer to my question above. I am going to post it and then I'll see what the experts will say. –  Carlo Von Schnitzel Mar 6 '13 at 19:41

Let $r=<\alpha_0,...,\alpha_S>$ the $k+1$-standard parameter of $\mathcal M$. If $\mathcal{H} = \mathcal{H}_{k+1}^{\mathcal M}(\alpha_s \cup (\alpha_0,...,\alpha_{s-1},u)$, where $u$ is $u_k(\mathcal M)$ (i.e the tuple having the solidity witnesses and some projecta), was not an initial segment of $\mathcal M$ then this means the iteration used some extender $E$. Now if I can have that $\mathcal H$ and $\mathcal M$ agree below $\rho_{k+1}^{\mathcal M}$, then this will imply that $\rho_{k+1}^{\mathcal M} \leq lh(E)$, which is what we want. So the question is why do $\mathcal H$ and $\mathcal M$ agree below $\rho_{k+1}^{\mathcal M}$?(coherence or $\mathcal M$ is $k$-sound?)

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So now the iteration tree will be simple (it will have exactly one cofinal wellfounded branch at every limit level of the tree) and one can apply Dodd-Jensen. –  Carlo Von Schnitzel Mar 6 '13 at 23:23

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