Non-forking and preservation of NIP types
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Kaplan, Itay
Adler, Casanovas and Pillay proved that if p is a complete stable type over a set B which does not fork over a set A, then the restriction of p to A is also stable. I will address the analogous question, replacing stable with NIP. In addition I will present a new proof for the stable case which uses elementary techniques.
 Joint work with Pedro Andrés Estevan. I think he has another name, but I think I put only three because this is the one I remember. So, okay. So, okay. And I made the slides last night, so I don't remember them very well. So, for me, this is also kind of... So, okay. So, we start with complete theory and we have some type and we want to know... We assume that it doesn't fork over some subset of its domain and we want to know what can we say about its restriction to the subset. And where is the time? Okay. And so, more precisely, suppose P is one of these properties, stable, simple, or NIP, I suppose to be capital letters, is the same true for the restriction. So, let me give all the definitions for those who don't know. So, these definitions work for partial types. So, Pyrovex is stable. If every complete extension of it is definable over the domain in which it is defined, and equivalently, there are no sequences A, I, B, I, such that A, I is realized a type and B, J are something and they witness the order property. Okay. And NIP is the same definition, so partial type is NIP. If there are no sequences A, I, and B, S, S subsets of omega, and some formula such that the A, I is realized the partial type and phi has the independent property with respect to these witnesses. Okay. So, the rules of A, I and B, S may be reversed, meaning that I can ask instead... Here I ask that the A, I realize the type, I can ask instead of the B, S, S, V, I, I, the type and it will give me the same definition. And the same is true for the stable part. For the stable definitions, I can replace the A, I, and B, J. But for simple, this is... Okay. I'll tell you the definition first. Pi is simple. If there are no things, there are no things which witness the property. So, no K less than omega, no tuples indexed by a tree, and some formula such that any branch is consistent with the type, and yes, but every row is gain consistent. Okay. So, unlike the NIP and stable case, it is not true that you can reverse the rules of X and Y in this case, and the triangle free unknown graph is an example where if you reverse the rules, you get one definition but not the other. This is Altium. So, right. Yes. So, the difference, yes. Okay. But I'm not actually going to talk about simple types. I'll just put it here. Okay. Now, let's talk about... So, I started the talk with forking, but somehow everything is easier when you look at co-forking. So, let's say that the type of A over B doesn't co-fork over A if this. Okay. So, this means the type of B over A, the type of B, capital B over A, A doesn't fork over capital A. And so, here's an exercise which I'm going to solve. Don't worry. So, if B is some type which over B, which doesn't co-fork over A, and it is stable on NIP, then so is the restriction to A. And so, I hope it's true. Let's see how we do it. So, yeah. And solution, yes. And it's also it's italic, and that's also not supposed to be like that. So, suppose that you have a partial type. Let's say that we're doing the NIP case. So, what I'm going to say is that the definition of NIP for partial type is equivalent to saying that there is an indecenable sequence, I mean, having IP. Same as saying there is an indecenable sequence over the domain of the partial type, over the realization of the type, and some B such that phi of A, I, B holds if and only if I is even. Okay. So, you can take this as a definition. And then, if you have that B is independent from little A over big A, and the type of A over A has IP, then you can make I, oh, sorry. You can take this indecenable sequence which written as IP for this type, for this type, and by not dividing here, you can make it indecenable over B and you get a contradiction. Okay. So, let's look at the following corollary. If you have a type over set B and M is a model contending B and P doesn't fork over M, then the restriction is stable. Okay. That's a corollary of the previous exercise. So, let me show you the solution. I mean, it's a proof. So, first we can extend P to a global type. Then P is M invariant. You can show for stable types. If you don't fork over something, over a model, you invariant over that model. And actually, also for NIP, I'll get 20 P in a minute, but it's very similar. So, it's M invariant. So, by stability, it means stable types are just types which are definable. Then we know that it's M definable, but then it doesn't co-fork because it's a narrow. Okay. It's a narrow of its restriction to M. Why are you proving the P is stable? What? I'm assuming it. Ah. Corollary. Ah, it's a corollary of the exercise. Right? If it doesn't fork and this one is stable and the big type is stable, then the restriction is also stable if it doesn't co-fork. Oh, it doesn't co-fork. Yeah, but in this case, it doesn't co-fork. Okay. So, again? M invariant, no sufficient. Because if it's M invariant, and you know then it's, since by stability, it follows that it's definable as well. But M invariant doesn't imply that it doesn't conform? No, no, no, but it implies it's definable. Definable implies that it doesn't conform. So, it's in general, right? Yes. Once you have a definable type, it doesn't co-fork. What about the binogilitude? In general? What is it? In the variant? No, no, that's not all. No, I didn't say that. I will use it here. So, it's false and it's true. Okay, so, here's the theorem that, okay, so, I guess, I don't know actually the history, but I guess that this theorem, you can see how this theorem is motivated by this corollary. So, a theorem of Adler as a novice in Pili, which generalized Hasson and Donchos. So, if you have, which I think they only did it in the NIP case. What? No? No. Oh. Who did NIP case? Sorry. These guys. Who did these guys? Hasson and Donchos. I'm sorry. I'm sorry. Okay. So, yeah, so, they proved that if you have a stable type in general without any assumption on the theory, if you have any stable type of a set, and now you have some subset, not necessarily a model, and P doesn't fork over the subset, then the restriction is stable. Okay. So, yeah. And the proof, their proof used to generically stable types. And, yeah, I thought about it like why is it so complicated to go for models, because for models it's so easy, like I just did it, and for general set is so difficult. I mean, why? But, okay. So, okay, I thought about it. And what? The stability theory is easy for models. Yeah. Okay. So, yes, I guess that's the reason. So, let's talk about NIP. So, like in stability, like in stable types, if you have a NIP type, yeah, if P is a global NIP type, and it doesn't fork over a set, then it is the scaling variant over that set. Okay. Like in NIP theories. Okay. And so, here's the theorem that we did. So, if P is a global NIP type, and so every Rascals-Torn-Mutomophism fixes the type. So Rascals-Torn-Mutomophism is one which fixes the Rascals-Torn types. So, any Rascals-Torn-Mutomophism fixes this type. Yeah. So, the theorem here is the following, if you have an NIP type which doesn't fork over A, then we'll already show. If it doesn't fork over A, then we know the restriction is NIP. But this has not that, but if you generate a Moli sequence in P over A, then the restriction to A union in the Moli sequence is NIP. Okay. So, to get NIP, you don't necessarily have to have, it's not enough to have A, but it's enough if you add to it a Moli sequence. Okay. But it's not true in general. So, there is an NTP2 theory, and in parenthesis, in fact, about the EIMP minimal. So, it's the simplest NIP, NTP2, sorry. With a NIP type, with a global NIP type, which doesn't fork over a model, in fact, it's a coil over that model. So, it's the best way of non-forking. Okay. Apart from the finability, but this we know you can get. And such that the type is NIP, and in fact, it is distal NDP minimal, but the restriction has IP. So, it seems like it's a very strong negation to any hope that this theorem could hold. Because even of the models, and even if you assume the theory is very nice, and NTP2 is the, by the way, NTP2 is the simplest case you can think of, because for simple theories, the original, I mean, for simple theories, it is true, the result that you can respect to a subset and it doesn't fork. Because if the theory is simple, then forking and co-forking is the same. So, you can just use the previous. Are dependent types stable in simple theory? And also that. Okay. But then you use, okay. But then you use something more stronger than an exercise. Okay. So, I thought, like for the remainder of the talk, I can do, I can do, I can give an idea of this proof, and some applications, and also describe the counter example. Okay. So, let's deduce from this how, deduce from this an easy proof of the stable case, okay, which doesn't use generically stable types. It's a bit cheating because I didn't tell you the proof of the other thing, but I can guarantee it doesn't use generically stable types. So, okay. Yeah. So, first of all, I want to say that the previous theorem, so this one, this one, I can replace NIP by stable, same proof will work. Okay. Okay. So, if you have a stable type that doesn't fork over A, then it's restriction to a moly sequence is stable. Okay. So, let us see how to get the previous theorem of other case numbers in P-line. Okay. So, suppose P is stable but doesn't fork over A. So, suppose towards contradiction that restriction is unstable. Let M be some model, and let such that, yeah, okay, let C0 be a restriction of, a realization of P to that model, a realization of the restriction of P to that model. So, by assumption, there is some any insurmable sequence, J indexed by the integers such that, such that phi of C0 be I if and only if I is greater than 0. Right. That's what instability gives you. That's what the other property gives you. Okay. Now, let I be a model sequence denoted by P over everything we have so far. Okay. So, by stability of the type, it's not very hard to see that this sequence I, the smaller sequence has to be in the same set. Okay. It's not only another sequence, it's in the same set. But then, we get the C0 realized at the time, realized, realized as the type over everything bigger than 0. And also, P is a scarring variant, I said it before, and the NIP type is a scarring variant, in particular any stable type is. So, you get a J is indescribable over, over this, but this is a contradiction. Okay. Because now we have some guy realizing P restricted to A and the model sequence, and a sequence J which is indescribable over this domain, which witnesses instability, which is impossible. Okay. So, okay, so this, this, and, ah, yeah. So now, I want to do. That's the idea to prove. Yes. Yeah. So, in the beginning, we didn't know what to use. But then, somebody came up with the idea of using the blackboard. So, um. Again. Is it your wrong use of blackboard, anyhow, used to have the last slide. So. But then, you're not going to use the blackboard. Yeah. So, which blackboard should I use? This one? I guess I'd use this one. So, um, so we assume that P, um, P is global type, and we assume it's NIP. And P doesn't talk over some set A. Actually, here, the, the proofs for models or sets, I, I, I can't make it simpler for models. So, um, and we have I, maybe it's a bit simpler, but not much. I is a model sequence. A model sequence over A. And we want to show that P restricted to A, I is an. In P, in P. So generated by P. P doesn't talk over A. It doesn't talk over A. It makes sense to generate a model sequence. Um, okay. So. So. Yeah. It means a sky. So what I mean by model sequence by model sequence. I mean, I mean an indescendable sequence, which is generated by P. So just P being NIP, not forking over A. You get the fact that it's invariant over. We know this, right? It's a sky invariant over A. That's going very, very. Not invariant, yeah. But you can, but still I can say a model sequence is just a sequence, which is indescendable and generated by P. So what, what, what the proof gives in the end? So maybe I won't really go into all the details, of course, but so you have some five. So suppose not. Okay. So suppose that's what you want to show. So if not, then you have some formula phi of X, Y, and some, let's say, A realizing P with six to two AI and some BI, some sequence such that phi of AI, B, sorry, AI, if and only if I even, right? Now, so this formula has IP. However, however, there must be, there is, oh, sorry, there is some sigma, some other formula in P, which implies phi is NIP, right? That's by compactness. So what I mean by that is that there is a, okay, that's like this quickly line is not very precise. This means that whenever you have an indescendable sequence over C, you cannot have some guy realizing this formula, which alternates with respect to phi. This is by compactness. You know that this such a formula exists. So, okay, and further, there is another formula also in P such that sigma is NIP. Okay. So we can go on like this forever, right, but we stop here. And now what the proof gives is that there exists A, I want to say, yeah, sorry, sorry, yeah, there exists some C such that maybe I should call it, there's already a C here, so maybe I should call it C star, but it's the same kind of, same length as this C such that sigma over AIC, oh, yeah, C star, and that's called, okay, maybe, is there an eraser as well? Yeah. Where? Oh, okay. So first of all, the first step in the proof is actually to assume that I is MOLI over AD. Okay? That's the first, you can do it. But if you don't want to do that, let's assume that D is the empty set. What's that symbol? Yeah, it's the same D as the car, the car, is it? Yes, this D. Yeah, yeah, this parameter, yes. It looked like an alpha. Yes, no, it doesn't look like an alpha. Yeah, the other one. This one? No. Ah, this one looks like an alpha. Ah, yeah. Okay, yeah, I agree. Sorry. No, this is fine. There's no alpha at all. Okay, so we assume that I is MOLI over AD. We can assume it without any loss of anything. And then what we'll get is we can find some C star which will incidentally have the same type as C, even the same lascar stone type as C over A, such that sigma of AI C star if and only if I is even. And that will be a contradiction because this sequence, I, is incidental over D, and all of it satisfies high, of course, but the choice of this formula will tell you that this is a contradiction. Okay? Okay, so now, okay, to do this, you have to do something, but it's not very hard. I can promise you. So it's kind of a very local, the proof is very local. You only use the fact that to show that you don't, I mean, if you don't want to show that the whole type is NIP and you only want to show that the restriction to phi is NIP, you need another formula to say it. It would be nice if we could, you know, have this completely local, but it still seems like we need, at least in this proof, a couple of more formulas. Or at least one more formula to be NIP. Okay. By the way, where do you use M being a model? Where is M being a model? Ah, just because it's an, to say that this is, otherwise, why would it be an indescribable sequence? I think that's it. When you generate a type, when you generate a sequence, I think that's the only thing I'm going to use it for. Because you need it to be invariant. It isn't really invariant over the model. It's only the scalar invariant over the set. And also, I would like to mention that this proof here, if the type is, I didn't write it in the slides, but if the type is generically stable, then the result is, so if the type is generically stable, then you can move I. Same proof. Can you move the model? You move the sequence I. Same proof, as for the stable case, gives you this result without the I here for generically stable types. So, generically stable type. If you have, yeah? You go back to that, just this proof here. This one? Click, click. Click, clicking. Yeah? Ah, okay. So, I'm going to click more now. Okay, so, I used the blackboard. So now, let me describe the example. So what the example is, actually, I remember like, okay, anyway, the example is, you look at the theory of trees and you put a random graph structure on the open cones starting at this point. So this is, this sounds like something Pierre would say, but I'll go into a little bit more details. So, let's call this theory the DTRR, if you have a better name. There. So, okay. Here are the axioms. So first axiom is the language. First of all, what is the language? The language is you have a meat tree. So you have less than meat and R. And the theory DTRR is the model completion of the following axioms. So this is going to be a first-class. So first of all, the redact is a meat tree. And this is the part about the random graph. So for any triple of points, x, y, z, so I want to think that x here is like the base, okay? So, and y and z are connected. So you have a random graph associated for each x. That's what you should think of, okay? And this, for instance, this axiom says that it's a graph because you can switch y and z, okay? Now this axiom says that you want that in the tree, x is below y and z, and x equals to the meat, okay? And that's the stuff you put the graph on. That's the axiom. Okay, yeah. And then, next axiom, is that if x, y, z are connected, so if y and z are connected with respect to x, and x is less than z, the meat of z and z prime, then also y and z prime are connected. This means that actually the graph is on the cone, it's not on the points, okay? This relation, the relation x is below z and z prime, z meets z prime, sorry. This is an equivalence relation. Yeah, but vertices are not pointed out, right? Yeah, okay, if you want. But anyway, if you don't, if you don't like the interpretation I gave, you can just look at the axioms and show that. Okay, so yes. Okay, so these are the axioms. Now, yeah, okay. So, let me tell you what the type is. So yeah, I promised you a type. What did I promise? I promised you a type. So actually, maybe I should have said this. The theory is actually omega categorical. It's even better than, okay, so it's in addition to all the other stuff I said. I want to give you a type, which is nip, and it doesn't fork over a model, but its restriction has IP. Okay, so we take some model. So remember, it's a tree with this extra relation. And we take some branch. So branch is just a maximal chain. And we look at this type. This is a type over the branch. It says I'm bigger than everything. And notice that this actually gives you a complete type over the model. That's knowing you're bigger than everything, gives you a complete type even in the language with R, because of the cone's business, if you think about it. Okay, so this gives you a complete type over M. And now let's see any realization of that. And pi is a partial type. It just says x is below C. Then, of course, pi is a finite size 5 or in M, because even in B. Okay? And yes. Okay. And now we have to show that pi has IP. And sorry, P has IP. And pi is nip. Okay. Okay, so let's show that this, okay, so this is another place where I kind of plan to use the blackboard. So maybe I'll use this one now. So we have some D realizing P. And let's find some D realizing P. And some BI realizing P. So such that your BI meet BJ is D. So you have B0, B1, B2. And this is D. And this is B. Okay? So this is definitely, this we can definitely do. Okay? Because the type only says bigger than P, than B. Okay. And now for any set, I want to say that the formula says saying that, let's call this point, so yeah, so the formula says that the saying that R of X meet BI, the BI is connected to X with respect to X meet BI for I in S, while the negation for I not in S is consistent with P. Now we have to onboard. So yeah, why is it consistent with P? Because I can find some, let's say I want to find this, to realize this type. So I can find some C, this guy, such that, so it starts a new cone from D. Okay? So C meet BI is OSD in this case. Okay? And C is connected with the graph to the BI's I want. Okay? And not connected to the BI's I don't want. Okay. Good. So, okay, now let's prove that, let's prove, sorry, that the pi is nip. So remember pi says that X is below C. Well, the idea is that once you're below C, then you somehow, you're supposed to look kind of like a linear order. Okay? So, okay, so that's why it should have worked, but then, okay. So, let's say, yeah, I want to show that pi is nip, right? So let's assume we have some A realizing pi. So A is above B and below C. So maybe I should also do this on the board. Okay, so A is this is C and this is B. And we have some I which some sequence witnessing IP with some formula. But the point is it's indescribable over MC. Okay? Because what I want to assume is that pi has IP, which means it's indescribable over the domain of pi. By quantifying relation, and as I know the dense trees are nip, then we can assume that pi looks like that. So R of T1 of XY, T2 of XY, T3 of XY. Well T1 to T2 and T3 are terms in trees. Okay? Okay, so, yes. So I can also assume that I is indescribable over MA, there should be a C there, I think. In the language of trees, I can assume that. Because again, because this theory, just the trees, is nip. Now let's see what the axioms imply. What's the whole thing imply this? The T1 is the meat of T2 and T3. Because why? Because I know that it happens sometimes. Because R holds sometimes. And maybe I should have written the axioms on the board, but one of the axioms is that R of XYZ implies X equals Y meets Z. So if R of these holds sometimes, for some I's in the sequence, because we can assume that the sequence is indescribable over A, then we get this always. Okay. Also we get that A appears in T2 here, but not in T3. Okay, all vice versa. We also get that T3 is not comparable with T2. It also follows from these axioms. Okay. So what we get is that T1 has to be smaller than T2, because A appears here. And the term, maybe I should have said it. If A appears in T2, then this term is below A. It's less or equal than A. So T1 is less or equal than that. So what we get is that, so even if you didn't understand all this, you can still, you have here T1 is the meat of T2 and T3, and you have that, here you have C, here you have T2, here you have T1, and yes, here you have A maybe. And that's T3. Okay, that's what we have. So T1 is the meat of C. So by the drawing you can see, by the drawing you can see this. Okay. So, but then we know that, yes, T1 is smaller than T2 meat C because it's just T2. So by the axioms, we get that, you see, so these two guys are equivalent, model this guy. I told you this is an equivalent solution. So it means that I can replace T2 here by C. Here I place T2 by C. But that's impossible now because this I is MC indescendable. So I can't have this guy alternates. Okay. Yeah, so this is the proof. The example. Yeah, to prove that it's all the nice properties I said, you have to work a little bit harder, but not much harder, surprisingly. And right. Okay. Okay, so let me just end this talk with some questions. So first of all, Artem even asked in his thesis what happens in simple types. So I kind of hope that using this new proof for stable can help solve this question, but so far I couldn't. I didn't think about it too much, but still would be nice to know for simple types the same question, what exactly happens for simple types? What do you expect to happen? I guess I expect this, I guess the same is also stable, I guess. But the problem is that somehow for simple types, you also have inconsistencies. So I mean, you have the three properties, so it's not like in stable. Okay, the same proof cannot work somehow. Is that true in the search and theories? The question? I don't know. Yeah, you can restrict it to, yeah. Good question. Yeah, I don't think we know anything on forking really. Even the models, the model is nothing, even the model. Yeah, I think so, yeah. Even the models. Next question is, okay, that's a little bit, I thought about it during the flight. So in the example, we need one element of the model sequence to get NIP, right? So we had a type, in the example that we had, we had this global type P, and once we realized one element in the model sequence, namely C, then the restriction is NIP. So the question is, maybe the examples, it will be, well, is it always the case that you only need one element from the sequence? I mean, I guess not. But can you find actual examples that you need more than one, you need two or three? Yeah, okay, that's it. Thank you very much. All right. Questions for Ptai? Another question. So your model sequence is, maybe one element of the model sequence. Go ahead. One element.
 Joint work with Pedro Andrés Estevan. I think he has another name, but I think I put only three because this is the one I remember.  So, okay. So, okay. And I made the slides last night, so I don't remember them very well. So, for me, this is also kind of...  So, okay. So, we start with complete theory and we have some type and we want to know...  We assume that it doesn't fork over some subset of its domain and we want to know what can we say about its restriction to the subset.  And where is the time? Okay. And so, more precisely, suppose P is one of these properties, stable, simple, or NIP, I suppose to be capital letters, is the same true for the restriction.  So, let me give all the definitions for those who don't know. So, these definitions work for partial types. So, Pyrovex is stable.  If every complete extension of it is definable over the domain in which it is defined, and equivalently, there are no sequences A, I, B, I, such that A, I is realized a type  and B, J are something and they witness the order property. Okay.  And NIP is the same definition, so partial type is NIP. If there are no sequences A, I, and B, S, S subsets of omega,  and some formula such that the A, I is realized the partial type and phi has the independent property with respect to these witnesses.  Okay. So, the rules of A, I and B, S may be reversed, meaning that I can ask instead... Here I ask that the A, I realize the type, I can ask instead  of the B, S, S, V, I, I, the type and it will give me the same definition. And the same is true for the stable part.  For the stable definitions, I can replace the A, I, and B, J. But for simple, this is... Okay. I'll tell you the definition first.  Pi is simple. If there are no things, there are no things which witness the property. So, no K less than omega, no tuples indexed by a tree,  and some formula such that any branch is consistent with the type, and yes, but every row is gain consistent. Okay.  So, unlike the NIP and stable case, it is not true that you can reverse the rules of X and Y in this case, and the triangle free unknown graph is an example where if you reverse the rules,  you get one definition but not the other. This is Altium. So, right. Yes.  So, the difference, yes. Okay. But I'm not actually going to talk about simple types. I'll just put it here.  Okay. Now, let's talk about... So, I started the talk with forking, but somehow everything is easier when you look at co-forking.  So, let's say that the type of A over B doesn't co-fork over A if this. Okay. So, this means the type of B over A, the type of B, capital B over A, A doesn't fork over capital A.  And so, here's an exercise which I'm going to solve. Don't worry. So, if B is some type which over B, which doesn't co-fork over A, and it is stable on NIP, then so is the restriction to A.  And so, I hope it's true. Let's see how we do it. So, yeah. And solution, yes.  And it's also it's italic, and that's also not supposed to be like that. So, suppose that you have a partial type. Let's say that we're doing the NIP case.  So, what I'm going to say is that the definition of NIP for partial type is equivalent to saying that there is an indecenable sequence, I mean, having IP.  Same as saying there is an indecenable sequence over the domain of the partial type, over the realization of the type, and some B such that phi of A, I, B holds if and only if I is even.  Okay. So, you can take this as a definition. And then, if you have that B is independent from little A over big A, and the type of A over A has IP,  then you can make I, oh, sorry. You can take this indecenable sequence which written as IP for this type, for this type, and by not dividing here, you can make it indecenable over B and you get a contradiction.  Okay. So, let's look at the following corollary. If you have a type over set B and M is a model contending B and P doesn't fork over M, then the restriction is stable.  Okay. That's a corollary of the previous exercise. So, let me show you the solution. I mean, it's a proof.  So, first we can extend P to a global type. Then P is M invariant. You can show for stable types. If you don't fork over something, over a model, you invariant over that model.  And actually, also for NIP, I'll get 20 P in a minute, but it's very similar. So, it's M invariant. So, by stability, it means stable types are just types which are definable.  Then we know that it's M definable, but then it doesn't co-fork because it's a narrow. Okay. It's a narrow of its restriction to M.  Why are you proving the P is stable?  What? I'm assuming it. Ah.  Corollary.  Ah, it's a corollary of the exercise. Right? If it doesn't fork and this one is stable and the big type is stable, then the restriction is also stable if it doesn't co-fork.  Oh, it doesn't co-fork.  Yeah, but in this case, it doesn't co-fork. Okay.  So, again?  M invariant, no sufficient.  Because if it's M invariant, and you know then it's, since by stability, it follows that it's definable as well.  But M invariant doesn't imply that it doesn't conform?  No, no, no, but it implies it's definable. Definable implies that it doesn't conform.  So, it's in general, right?  Yes. Once you have a definable type, it doesn't co-fork.  What about the binogilitude? In general?  What is it?  In the variant?  No, no, that's not all. No, I didn't say that.  I will use it here.  So, it's false and it's true.  Okay, so, here's the theorem that, okay, so, I guess, I don't know actually the history, but I guess that this theorem, you can see how this theorem is motivated by this corollary.  So, a theorem of Adler as a novice in Pili, which generalized Hasson and Donchos.  So, if you have, which I think they only did it in the NIP case.  What?  No?  No.  Oh.  Who did NIP case? Sorry.  These guys.  Who did these guys?  Hasson and Donchos.  I'm sorry.  I'm sorry.  Okay.  So, yeah, so, they proved that if you have a stable type in general without any assumption on the theory, if you have any stable type of a set, and now you have some subset, not necessarily a model, and P doesn't fork over the subset, then the restriction is stable.  Okay.  So, yeah.  And the proof, their proof used to generically stable types.  And, yeah, I thought about it like why is it so complicated to go for models, because for models it's so easy, like I just did it, and for general set is so difficult.  I mean, why?  But, okay.  So, okay, I thought about it.  And what?  The stability theory is easy for models.  Yeah.  Okay.  So, yes, I guess that's the reason.  So, let's talk about NIP.  So, like in stability, like in stable types, if you have a NIP type, yeah, if P is a global NIP type, and it doesn't fork over a set, then it is the scaling variant over that set.  Okay.  Like in NIP theories.  Okay.  And so, here's the theorem that we did.  So, if P is a global NIP type, and so every Rascals-Torn-Mutomophism fixes the type.  So Rascals-Torn-Mutomophism is one which fixes the Rascals-Torn types.  So, any Rascals-Torn-Mutomophism fixes this type.  Yeah.  So, the theorem here is the following, if you have an NIP type which doesn't fork over A, then we'll already show.  If it doesn't fork over A, then we know the restriction is NIP.  But this has not that, but if you generate a Moli sequence in P over A, then the restriction to A union in the Moli sequence is NIP.  Okay.  So, to get NIP, you don't necessarily have to have, it's not enough to have A, but it's enough if you add to it a Moli sequence.  Okay.  But it's not true in general.  So, there is an NTP2 theory, and in parenthesis, in fact, about the EIMP minimal.  So, it's the simplest NIP, NTP2, sorry.  With a NIP type, with a global NIP type, which doesn't fork over a model, in fact, it's a coil over that model.  So, it's the best way of non-forking.  Okay.  Apart from the finability, but this we know you can get.  And such that the type is NIP, and in fact, it is distal NDP minimal, but the restriction has IP.  So, it seems like it's a very strong negation to any hope that this theorem could hold.  Because even of the models, and even if you assume the theory is very nice, and NTP2 is the, by the way, NTP2 is the simplest case you can think of,  because for simple theories, the original, I mean, for simple theories, it is true, the result that you can respect to a subset and it doesn't fork.  Because if the theory is simple, then forking and co-forking is the same.  So, you can just use the previous.  Are dependent types stable in simple theory?  And also that.  Okay.  But then you use, okay.  But then you use something more stronger than an exercise.  Okay.  So, I thought, like for the remainder of the talk, I can do, I can do, I can give an idea of this proof,  and some applications, and also describe the counter example.  Okay.  So, let's deduce from this how, deduce from this an easy proof of the stable case, okay, which doesn't use generically stable types.  It's a bit cheating because I didn't tell you the proof of the other thing, but I can guarantee it doesn't use generically stable types.  So, okay.  Yeah.  So, first of all, I want to say that the previous theorem, so this one, this one, I can replace NIP by stable, same proof will work.  Okay.  Okay.  So, if you have a stable type that doesn't fork over A, then it's restriction to a moly sequence is stable.  Okay.  So, let us see how to get the previous theorem of other case numbers in P-line.  Okay.  So, suppose P is stable but doesn't fork over A. So, suppose towards contradiction that restriction is unstable.  Let M be some model, and let such that, yeah, okay, let C0 be a restriction of, a realization of P to that model,  a realization of the restriction of P to that model.  So, by assumption, there is some any insurmable sequence, J indexed by the integers such that, such that phi of C0 be I if and only if I is greater than 0.  Right.  That's what instability gives you.  That's what the other property gives you.  Okay.  Now, let I be a model sequence denoted by P over everything we have so far.  Okay.  So, by stability of the type, it's not very hard to see that this sequence I, the smaller sequence has to be in the same set.  Okay.  It's not only another sequence, it's in the same set.  But then, we get the C0 realized at the time, realized, realized as the type over everything bigger than 0.  And also, P is a scarring variant, I said it before, and the NIP type is a scarring variant, in particular any stable type is.  So, you get a J is indescribable over, over this, but this is a contradiction.  Okay.  Because now we have some guy realizing P restricted to A and the model sequence, and a sequence J which is indescribable over this domain, which witnesses instability, which is impossible.  Okay.  So, okay, so this, this, and, ah, yeah.  So now, I want to do.  That's the idea to prove.  Yes.  Yeah.  So, in the beginning, we didn't know what to use.  But then, somebody came up with the idea of using the blackboard.  So, um.  Again.  Is it your wrong use of blackboard, anyhow, used to have the last slide.  So.  But then, you're not going to use the blackboard.  Yeah.  So, which blackboard should I use?  This one?  I guess I'd use this one.  So, um, so we assume that P, um, P is global type, and we assume it's NIP.  And P doesn't talk over some set A. Actually, here, the, the proofs for models or sets,  I, I, I can't make it simpler for models.  So, um, and we have I, maybe it's a bit simpler, but not much.  I is a model sequence.  A model sequence over A. And we want to show that P restricted to A, I is an.  In P, in P. So generated by P. P doesn't talk over A.  It doesn't talk over A. It makes sense to generate a model sequence.  Um, okay.  So.  So.  Yeah.  It means a sky.  So what I mean by model sequence by model sequence.  I mean, I mean an indescendable sequence, which is generated by P.  So just P being NIP, not forking over A.  You get the fact that it's invariant over.  We know this, right?  It's a sky invariant over A.  That's going very, very.  Not invariant, yeah.  But you can, but still I can say a model sequence is just a sequence, which is indescendable  and generated by P.  So what, what, what the proof gives in the end?  So maybe I won't really go into all the details, of course, but so you have some five.  So suppose not.  Okay.  So suppose that's what you want to show.  So if not, then you have some formula phi of X, Y, and some, let's say, A realizing  P with six to two AI and some BI, some sequence such that phi of AI, B, sorry, AI, if and  only if I even, right?  Now, so this formula has IP.  However, however, there must be, there is, oh, sorry, there is some sigma, some other  formula in P, which implies phi is NIP, right?  That's by compactness.  So what I mean by that is that there is a, okay, that's like this quickly line is not  very precise.  This means that whenever you have an indescendable sequence over C, you cannot have some guy realizing  this formula, which alternates with respect to phi.  This is by compactness.  You know that this such a formula exists.  So, okay, and further, there is another formula also in P such that sigma is NIP.  Okay.  So we can go on like this forever, right, but we stop here.  And now what the proof gives is that there exists A, I want to say, yeah, sorry, sorry,  yeah, there exists some C such that maybe I should call it, there's already a C here,  so maybe I should call it C star, but it's the same kind of, same length as this C such  that sigma over AIC, oh, yeah, C star, and that's called, okay, maybe, is there an eraser  as well?  Yeah.  Where?  Oh, okay.  So first of all, the first step in the proof is actually to assume that I is MOLI over  AD.  Okay?  That's the first, you can do it.  But if you don't want to do that, let's assume that D is the empty set.  What's that symbol?  Yeah, it's the same D as the car, the car, is it?  Yes, this D.  Yeah, yeah, this parameter, yes.  It looked like an alpha.  Yes, no, it doesn't look like an alpha.  Yeah, the other one.  This one?  No.  Ah, this one looks like an alpha.  Ah, yeah.  Okay, yeah, I agree.  Sorry.  No, this is fine.  There's no alpha at all.  Okay, so we assume that I is MOLI over AD.  We can assume it without any loss of anything.  And then what we'll get is we can find some C star which will incidentally have the same  type as C, even the same lascar stone type as C over A, such that sigma of AI C star  if and only if I is even.  And that will be a contradiction because this sequence, I, is incidental over D, and all  of it satisfies high, of course, but the choice of this formula will tell you that this is  a contradiction.  Okay?  Okay, so now, okay, to do this, you have to do something, but it's not very hard.  I can promise you.  So it's kind of a very local, the proof is very local.  You only use the fact that to show that you don't, I mean, if you don't want to show that  the whole type is NIP and you only want to show that the restriction to phi is NIP, you  need another formula to say it.  It would be nice if we could, you know, have this completely local, but it still seems  like we need, at least in this proof, a couple of more formulas.  Or at least one more formula to be NIP.  Okay.  By the way, where do you use M being a model?  Where is M being a model?  Ah, just because it's an, to say that this is, otherwise, why would it be an indescribable  sequence?  I think that's it.  When you generate a type, when you generate a sequence, I think that's the only thing  I'm going to use it for.  Because you need it to be invariant.  It isn't really invariant over the model.  It's only the scalar invariant over the set.  And also, I would like to mention that this proof here, if the type is, I didn't write  it in the slides, but if the type is generically stable, then the result is, so if the type  is generically stable, then you can move I.  Same proof.  Can you move the model?  You move the sequence I.  Same proof, as for the stable case, gives you this result without the I here for generically  stable types.  So, generically stable type.  If you have, yeah?  You go back to that, just this proof here.  This one?  Click, click.  Click, clicking.  Yeah?  Ah, okay.  So, I'm going to click more now.  Okay, so, I used the blackboard.  So now, let me describe the example.  So what the example is, actually, I remember like, okay, anyway, the example is, you look  at the theory of trees and you put a random graph structure on the open cones starting  at this point.  So this is, this sounds like something Pierre would say, but I'll go into a little bit more  details.  So, let's call this theory the DTRR, if you have a better name.  There.  So, okay.  Here are the axioms.  So first axiom is the language.  First of all, what is the language?  The language is you have a meat tree.  So you have less than meat and R. And the theory DTRR is the model completion of the following  axioms.  So this is going to be a first-class.  So first of all, the redact is a meat tree.  And this is the part about the random graph.  So for any triple of points, x, y, z, so I want to think that x here is like the base,  okay?  So, and y and z are connected.  So you have a random graph associated for each x.  That's what you should think of, okay?  And this, for instance, this axiom says that it's a graph because you can switch y and  z, okay?  Now this axiom says that you want that in the tree, x is below y and z, and x equals  to the meat, okay?  And that's the stuff you put the graph on.  That's the axiom.  Okay, yeah.  And then, next axiom, is that if x, y, z are connected, so if y and z are connected with  respect to x, and x is less than z, the meat of z and z prime, then also y and z prime  are connected.  This means that actually the graph is on the cone, it's not on the points, okay?  This relation, the relation x is below z and z prime, z meets z prime, sorry.  This is an equivalence relation.  Yeah, but vertices are not pointed out, right?  Yeah, okay, if you want.  But anyway, if you don't, if you don't like the interpretation I gave, you can just look  at the axioms and show that.  Okay, so yes.  Okay, so these are the axioms.  Now, yeah, okay.  So, let me tell you what the type is.  So yeah, I promised you a type.  What did I promise?  I promised you a type.  So actually, maybe I should have said this.  The theory is actually omega categorical.  It's even better than, okay, so it's in addition to all the other stuff I said.  I want to give you a type, which is nip, and it doesn't fork over a model, but its restriction  has IP.  Okay, so we take some model.  So remember, it's a tree with this extra relation.  And we take some branch.  So branch is just a maximal chain.  And we look at this type.  This is a type over the branch.  It says I'm bigger than everything.  And notice that this actually gives you a complete type over the model.  That's knowing you're bigger than everything, gives you a complete type even in the language  with R, because of the cone's business, if you think about it.  Okay, so this gives you a complete type over M. And now let's see any realization of that.  And pi is a partial type.  It just says x is below C. Then, of course, pi is a finite size 5 or in M, because even  in B.  Okay?  And yes.  Okay.  And now we have to show that pi has IP.  And sorry, P has IP.  And pi is nip.  Okay.  Okay, so let's show that this, okay, so this is another place where I kind of plan to use  the blackboard.  So maybe I'll use this one now.  So we have some D realizing P. And let's find some D realizing P. And some BI realizing  P. So such that your BI meet BJ is D. So you have B0, B1, B2.  And this is D. And this is B. Okay?  So this is definitely, this we can definitely do.  Okay?  Because the type only says bigger than P, than B. Okay.  And now for any set, I want to say that the formula says saying that, let's call this  point, so yeah, so the formula says that the saying that R of X meet BI, the BI is connected  to X with respect to X meet BI for I in S, while the negation for I not in S is consistent  with P. Now we have to onboard.  So yeah, why is it consistent with P?  Because I can find some, let's say I want to find this, to realize this type.  So I can find some C, this guy, such that, so it starts a new cone from D. Okay?  So C meet BI is OSD in this case.  Okay?  And C is connected with the graph to the BI's I want.  Okay?  And not connected to the BI's I don't want.  Okay.  Good.  So, okay, now let's prove that, let's prove, sorry, that the pi is nip.  So remember pi says that X is below C. Well, the idea is that once you're below C, then  you somehow, you're supposed to look kind of like a linear order.  Okay?  So, okay, so that's why it should have worked, but then, okay.  So, let's say, yeah, I want to show that pi is nip, right?  So let's assume we have some A realizing pi.  So A is above B and below C. So maybe I should also do this on the board.  Okay, so A is this is C and this is B. And we have some I which some sequence witnessing  IP with some formula.  But the point is it's indescribable over MC.  Okay?  Because what I want to assume is that pi has IP, which means it's indescribable over the  domain of pi.  By quantifying relation, and as I know the dense trees are nip, then we can assume that  pi looks like that.  So R of T1 of XY, T2 of XY, T3 of XY.  Well T1 to T2 and T3 are terms in trees.  Okay?  Okay, so, yes.  So I can also assume that I is indescribable over MA, there should be a C there, I think.  In the language of trees, I can assume that.  Because again, because this theory, just the trees, is nip.  Now let's see what the axioms imply.  What's the whole thing imply this?  The T1 is the meat of T2 and T3.  Because why?  Because I know that it happens sometimes.  Because R holds sometimes.  And maybe I should have written the axioms on the board, but one of the axioms is that  R of XYZ implies X equals Y meets Z.  So if R of these holds sometimes, for some I's in the sequence, because we can assume  that the sequence is indescribable over A, then we get this always.  Okay.  Also we get that A appears in T2 here, but not in T3.  Okay, all vice versa.  We also get that T3 is not comparable with T2.  It also follows from these axioms.  Okay.  So what we get is that T1 has to be smaller than T2, because A appears here.  And the term, maybe I should have said it.  If A appears in T2, then this term is below A. It's less or equal than A. So T1 is less  or equal than that.  So what we get is that, so even if you didn't understand all this, you can still, you have  here T1 is the meat of T2 and T3, and you have that, here you have C, here you have  T2, here you have T1, and yes, here you have A maybe.  And that's T3.  Okay, that's what we have.  So T1 is the meat of C. So by the drawing you can see, by the drawing you can see this.  Okay.  So, but then we know that, yes, T1 is smaller than T2 meat C because it's just T2.  So by the axioms, we get that, you see, so these two guys are equivalent, model this  guy.  I told you this is an equivalent solution.  So it means that I can replace T2 here by C. Here I place T2 by C. But that's impossible  now because this I is MC indescendable.  So I can't have this guy alternates.  Okay.  Yeah, so this is the proof.  The example.  Yeah, to prove that it's all the nice properties I said, you have to work a little bit harder,  but not much harder, surprisingly.  And right.  Okay.  Okay, so let me just end this talk with some questions.  So first of all, Artem even asked in his thesis what happens in simple types.  So I kind of hope that using this new proof for stable can help solve this question, but  so far I couldn't.  I didn't think about it too much, but still would be nice to know for simple types the  same question, what exactly happens for simple types?  What do you expect to happen?  I guess I expect this, I guess the same is also stable, I guess.  But the problem is that somehow for simple types, you also have inconsistencies.  So I mean, you have the three properties, so it's not like in stable.  Okay, the same proof cannot work somehow.  Is that true in the search and theories?  The question?  I don't know.  Yeah, you can restrict it to, yeah.  Good question.  Yeah, I don't think we know anything on forking really.  Even the models, the model is nothing, even the model.  Yeah, I think so, yeah.  Even the models.  Next question is, okay, that's a little bit, I thought about it during the flight.  So in the example, we need one element of the model sequence to get NIP, right?  So we had a type, in the example that we had, we had this global type P, and once we realized  one element in the model sequence, namely C, then the restriction is NIP.  So the question is, maybe the examples, it will be, well, is it always the case that  you only need one element from the sequence?  I mean, I guess not.  But can you find actual examples that you need more than one, you need two or three?  Yeah, okay, that's it.  Thank you very much.  All right.  Questions for Ptai?  Another question.  So your model sequence is, maybe one element of the model sequence.  Go ahead.  One element.
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