# Core Model Induction 101

Mentioning the core model induction to a fellow set theorist is akin to mentioning that you’re a mathematician to the layman — you receive a reaction which is struck by a delightful mix of terror and awe. My humble goal with this blog post is not to offer a “fix-all” solution to this problem, but rather to give a vague (but correct) explanation of what’s actually going on in a core model induction, without getting too bogged down on the details.

I’ll only focus on the (chronologically) first type of core model induction — showing $\textsf{AD}^{L(\mathbb R)}$. This is primarily because I have no idea what happens in the other ones. Now, before we start, why is this an interesting thing to do? First of all, it’s interesting for consistency-strength questions. Woodin has shown that $\textsf{AD}^{L(\mathbb R)}$ is equiconsistent  with the existence of infinitely many Woodins, and showing that your favorite hypothesis entails the consistency of such Woodins can (at this point) only be shown using the core model induction. Unless, of course, you show something a lot stronger from your hypothesis, like the existence of a superstrong — but now you’re being pedantic. Second of all, what’s quite remarkable is that we not only get consistency proofs, but we get something holding in $V$. We get determinacy of a lot of sets. It’s hands-on. That’s useful. Okay, let’s get started.

We call the core model induction an “induction”, which is maths for “showing something holds for all relevant stuff”. Here we’re trying to show that every set of reals in $L(\mathbb R)$ is determined. Another way of phrasing that is to show that every level of $L(\mathbb R)$ satisfies $\textsf{AD}$; i.e. that $J_\alpha(\mathbb R)\vDash\textsf{AD}$ for every ordinal $\alpha$. So we at least sound like we’re doing an induction. Now, what is usually the case in inductive proofs is that we have a clever way of splitting “everything relevant” into chunks that are similar. For induction on the naturals we split up into two chunks: zero and all the successors. We can and will also form some chunks here in the core model induction — note that “everything relevant” in this case is the class of ordinals.

Firstly, a very neat result by Kechris & Woodin (’83) shows that we don’t have have to go through all the ordinals, but it suffices to show that $J_{\alpha+1}(\mathbb R)\vDash\textsf{AD}$ whenever there’s a new scaled set of reals appearing in $J_{\alpha+1}(\mathbb R)$ — call such $\alpha$ critical. The exact definition of a “scaled set” is not important (but if you’re interested then check out my blog post series on scales). The thing is that we have a complete characterisation of all the critical ordinals — we can split up the ordinals into a handful of chunks and we now only have to check that $J_{\alpha+1}(\mathbb R)\vDash\textsf{AD}$ for $\alpha$ a representative for each chunk. I should probably note that the two most important chunks have been given names: they’re called “the inadmissable case” and “the end-of-gap case”.

Okay, we’ve narrowed it down. But if we’re now dealing with a particular chunk, how do we actually show that $J_{\alpha+1}(\mathbb R)\vDash\textsf{AD}$? This is where we do another splitting-into-chunks, aka we do a subinduction. Because all the sets of reals in $J_{\alpha+1}(\mathbb R)$ are precisely the ones definable (with parameters) over $J_\alpha(\mathbb R)$ and we have a definability hierarchy, let’s induct over that! So we have to show that, for all $n<\omega$, every $\bf\Sigma_{n+1}^{J_\alpha(\mathbb R)}$ set of reals is determined. Again we can split up into two chunks here: the zero chunk and the successor chunk.

Let’s focus on the zero case for a while, meaning that we’re trying to show that every $\bf\Sigma_1^{J_\alpha(\mathbb R)}$ set of reals is determined. This is where mice enter the picture. More specifically, Neeman has shown that for a set of reals $A$ to be determined it suffices to find a certain breed of mouse which captures $A$ (again, for details, check out my blog post series on mouse capturing). So what the zero case is about is then finding a mouse which captures a given $\bf\Sigma_1^{J_\alpha(\mathbb R)}$ set.

The successor case is then, naturally, about looking for a mouse which captures a given $\bf\Sigma_{n+2}^{J_\alpha(\mathbb R)}$ set of reals, given that we can find mice capturing $\bf\Sigma_{n+1}^{J_\alpha(\mathbb R)}$ sets. It turns out that it suffices just to stick with the same breed as in the zero case, but where the mouse now has $n+1$ extra Woodin cardinals — I touched upon this in a previous post. The reason why it’s called a “core model induction” and not, say, a “mouse induction”, is what we do at successor stages. To get the extra Woodin we take advantage of a dichotomy, saying that either a mouse with an extra Woodin exists, or the core model exists — the successor stage is then about disproving the existence of such core models.

So, to sum up, a core model induction can be seen as having a zero case and a successor case, where the zero case is split up into the chunks of critical ordinals — two notable chunks are the inadmissable case and the end-of-gap case. This zero case about finding a mouse capturing stuff. The successor case (also sometimes called the projective case) is also about finding a mouse capturing stuff, but here we assume we already got a mouse, and we want to find a new one with an extra Woodin cardinal.