Linear orders in NIP theories
Simon, Pierre
A longstanding open questions asks whether an unstable NIP theory interprets an infinite linear order. I will present a construction giving a type-definable linear (quasi-)order, thus partially answering this question.
 that says that it says something a little bit weaker than saying that in any unstable NIP theory, well, that any unstable NIP theory defines or interprets an infinite linear order, because the order will not be actually definable. But let me just start maybe with a little bit of background. So there's a theorem of Schellach that says that if you have a structure which is NIP and not stable, then some formula has the strict order property. So T has strict order property. So I'll recall what this means. So recall that formula phi has the strict order property if there's a sequence of parameters for y, say, such that the formulas that are defined, or maybe it doesn't matter, are strictly defined sets. If you vary the formula, if you vary the parameters over the sequence, you define sets which are strictly increasing. Having a formula that has a strict order property is equivalent to there being a definable partial order with an infinite chain. Because here, if you have this, you can define an order from that. So I think pre-order and quasi-order means the same thing, no? So pre-quasi-order, by saying that, say, B is less equal to B prime, if the formula defined by B is less, included the formula defined by B prime. So this is, of course, a transitive relation. And the assumption that phi has SOP exactly tells you that this order has an infinite chain. And conversely, you can figure out what to do. So this is telling us we have a partial quasi-order, but if you allow quotients, then you have an interpretable partial order that has an infinite chain. And then there was a question, which I'm not sure to whom it's attributed. I think Schell absolutely raised it. I heard it first from Ryszowski. I don't know. I'm sure other people came up with it. Is can we get a definable, with the same assumption, so if T is nip and stable, does it interpret an infinite linear order? So note that you cannot ask for an infinite definable order because if you just take a structure which has an equivalence relation, all classes of psi 2, and then you have an order on the quotient, so you have an order, but you have two cover of it, then there's no definable order on that structure. But of course, there's an interpretable one. OK, so this is still open. However, what I can prove is the following. So the assumption is still the same. T is nip and stable. Then there is a finite set A of parameter. There's a type over A and some type definable relation such that this defines a strict linear quasi-order on realizations of p, such that r defines a strict linear pre-order or quasi-order on realizations of p. Yeah, but here I have this because it's strict. So there's a choice to make. If you take the one that's not strict, you have v. If you take the one that's strict, you have. Yeah. So some days I mean. We're having a strict type defined one thing. Is this the best possible thing to have a type defined one? Yeah, so that's why I stated like this, so it looks better. Well, we always thought that it's a progress. What do you think it's something? Why is it interpretable? Yeah, exactly. So let me explain what I mean by a pre. So what does it mean, a linear pre-order? It means that it's transitive. So OK, where's the interpretable? So if we have this, so from this, we can define the relation e, which you define as the link that you have neither of the two. You don't have that x is pretty small than y. You don't have that y is pretty small than y. Now this is a v-definable equivalence relation. And the quotient of e by this equivalence relation is linearly ordered by r. So that's what it means to be this linear pre-order. So in the quotient is linearly ordered by r. Oh, yeah. And no, it's more than that. The quotient is infinite. So the quotient is infinite and linearly ordered by r. Is this still? And everything is over r. Over a, sorry. This is over. So is the statement clear? So yeah, for me, it's somewhat easier to think that you have a v-definable equivalence relation. So I'm going to go back to the question. For me, it's somewhat easier to think that you have a v-definable equivalence relation. And then you have a linear order on the quotient. And if you think of what that means, then it means that either the strict order is type-definable or the non-strict order is v-definable as you want. And an immediate corollary is that if t is omega categorical, then the conjecture is true. Because the point is that here a is finite. So if the theory is omega categorical, everything is definable. Right, so that's also a good question. So p can be any, yeah, so p can be a one type. In fact, it can extend any unstable type. That you start with. So what I should say is that, so this looks like a strengthening of this. But it's a theorem of a very different nature. Because this is a non-structure theorem. It tells you you have this thing that says that you have some complexity in it. With this, you should really think of it as a structure theorem. It's giving you a linear order as a very rigid object. It's a very, and so you could hold it. But you have to use those linear orders to actually build some structure theory of models. So build some classification theory of models of an NIP theory, t. So some partly speculative applications. Well, almost all of them are speculative. Only the first one is only partially. So well, the nicest case is when the theory is omega categorical. And so it's natural to start there. And there, we can hope to have a classification of NIP, finitely homogeneous structures, extending the stable case. So the stable case is known by work of a number of people, like Lan, Chirlino-Schofsky, Harrington, well, quasi-finite. Well, there's a very, anyway. So this, when I started doing it, I gave talk about it in Paris, and it's in progress, joined. We found some shows. And I don't really want to talk about it today. It seems that one could hope to really have some coordination. I have no idea how to write this. What? What's the next letter? Hi. I'm very tired. Coordination of models by linear orders with the idea that if you compare with, let's say, the omega stable case, compared with the omega stable case, so a strictly minimal set would be replaced by linear orders. So we do these two things connected? Yeah, yeah, this is the omega stable. This is the omega categorical version of this. But here you can hope for much more than, because here I don't know what that means. Oh, wait. Here you're not assuming omega categorical? No. No, I'm not assuming omega categorical anymore. Well, OK, so in stable, so let's say omega stable, so you coordinateize maybe models by looking at dimensions of regular types. So here those dimensions of regular types, what they could be replaced by are isomorphism types of certain linear orders. So it's no longer going to be a number. It's going to be the isomorphism type of a certain order. And then in the stable case, in very nice cases, you know the model up to isomorphism by knowing those dimensions. Here you could hope for, so for example, what happens in real close fields, in real close fields, let's take maybe divisible order to be in groups. In divisible order to be in groups, if you know, so how does it work? Actually, let's take real close fields. It's a little bit. So what classification could you hope for if, so if you have a real close field, you can look at the valuation given by the convex hull of the reels, and then becomes a valued field. And then the residue field is a subfield of the reels, and then you have the value group. If you know the value group and you know the residue field, so the residue field lives in some bounded thing. So if you know the value group and the residue field, you still don't know the whole field, but there's a maximal one with this given value group and residue field. But actually, if you just fix the value group, then what I said is also true, because the residue field anyway is a subfield of r. So for a fixed value group, there's a maximal model of RCF that has this fixed value group. And this is the idea that one could hope to have something similar, where here the value group is playing the role of my ordered set. So one could hope that you somehow have ordered set, and they don't give you the isomorphism type, but say if the structure is this tall, they will give you at least there's a maximal one that has those. But maximal you mean? Maximum under inclusion. But I haven't really thought about this, so if this looks very vague, it's normal. But this is kind of the idea. When could look at algebraic structures? So this is also very vague. But for example, there's this conjecture that if you have an NIP unstable field, maybe it has a definable valuation, well, if you take such a field, it's going to have a generic type. If the generic type is generically stable, then I don't know what to say. But if the generic type is not generically stable, then some kind of upgraded version of this theorem is going to give you an order such that the generic type is going at minus infinity along that order, which you expect the type to be the generic type at minus infinity along the valuation. So this is going to give you an order such that the generic type goes at minus infinity along that order. And then using the fact that the type is generic, you could hope to somehow create a valuation by looking at stabilizers of convex subsets of the order or something like this. But. So k is a fixed type of field. k is an NIP field. And I'm assuming this generic type is not generically stable. Oh, it has a generic type. It's an F-generic type. F-generic type. I've added two groups. For example, you can have one that does for both. But OK, this is speculative. I haven't really had time. But it's just to give an idea of why this is likely to be useful to give us strong structure theorems, hopefully. But there's some big things that need to be figured out first. Anyway. So now what I want to do is I'd like to give you some idea on how this is proved. And I usually don't like giving proofs. But the reason I'm doing this is because the proof is probably much easier to tell than to read. And I mean, it's the kind of thing that it's not so hard to actually sketch the proof. But it can be very hard to read it. So I think it's worth actually doing it. I'm not going to do the general case. So I'm going to just do a special case that will give you the idea of what's going on. And you'll see there's just one idea. It's not so hard. But we have to play a little bit with indiscernible sequences. And so I need to recall certain things about indiscernible sequences in NIP. OK, so probably you know that if you have, so let's just, from now on, assume that my theory is NIP. And one of the first things, usually, that one learns is that if you have I an indiscernible sequence, say this is I, and you have a formula with some parameter. So the sequence is indiscernible over the empty set. It's not necessarily indiscernible over the parameter B. So when I evaluate that formula along the sequence, then it's going to cut out a finite union of convex sets. So this formula, if x varies along the sequence, I'm going to have finitely many convex sets such that, say, the formula phi of xB here is true. And then here it's false. Then here it's true. And then maybe here I just have one point where it becomes false. And then it's true outside, something like this. But the point is there are only finitely many changes of truth value along the formula. So this is usually one of the first things that one learns about NIP. And it's equivalent to NIP. And this is strengthened by the so-called shrinking of indiscernible. Maybe I'll move here. Which explains what happens when you evaluate not a formula with one variable, but you look at the formula with several variables, each variable having the same size as a tuple of the sequence. And you want to know what happens there. And then the conclusion is essentially the same that there's a finite, that there's an equivalence relation with finitely many convex classes such that the truth value, if you evaluate the formula on a tuple from the sequence, the truth value only depends on, if you look at your tuple, say it's an increasing order on which equivalence relation each element lies. So let's write it down. So i is indiscernible. You have any formula. Then again, you can cut. So maybe some classes have just one element, or finitely many. But there is a convex finite equivalence relation, E on I. So what I mean by convex finite, I mean the classes are convex, and there are finitely many classes. Well, such that the truth value, when I evaluate it on the tuple, so I take a1, a2, a3, a4, a5, and I evaluate my formula on that, then to know whether the formula is true or not, I only need to know if the tuple have taken an increasing order, I only need to know in which classes each of them lies. So such that if, say, a1, an, or an i in increasing order, b1, bn, or in i in increasing order, and ai is equivalent to bi for all i, then the formula is true on a1, an. OK, yeah, shouldn't have called b. If and only if it's true on b1, bn. Yes, I'm allowed to take two in the same class. And each of those classes is going to be definable, I mean relatively definable by an instance of that way. So there's a unique, maybe I should say, well, there's a unique finest, yeah, courses one. There's a unique courses one that has this property. So if you do this. So now if you have i an indiscernible sequence and you take, say, a finite tuple b, you can do this for every formula. So every formula is going to give you a finite convex equivalence relation. Now you should think of i, usually indiscernible sequences, especially in stable, they're indexed by omega. That's not what I want to think about. I want to think in a, in a, in a, in a, in a, in a, in a thinking about indiscernible sequences that are indexed by a dense linear order, which is usually very big. But definitely dense. So you should think of all indiscernible sequences as being indexed by a dense order, so that I don't have problems of consecutive elements, things like that. Then if I have a finite tuple b, I can take every formula phi. For every formula phi, I get an equivalence relation. And then I take the intersection of all those equivalence relations. What I get is an equivalence relation with convex classes on i, which now has maybe two to the parameter. The parameter is fixed. i is fixed, b is fixed. Yeah, thank you. Yeah, b is fixed. But phi varies. So now what I get is I get an equivalence relation. So we get convex equivalence relation eb. So it depends on b on i with at most maybe two to the t classes. Such that what? Well, such that the classes, there might be some finite classes. But the infinite classes are mutually indiscernible. And actually also over the finite classes. So maybe older classes, but the finite ones. So the infinite classes are mutually indiscernible. What does it mean? Over b. Yeah, thank you. They're mutually indiscernible over b. What does it mean to be mutually indiscernible? It means if you take two of them, each one is indiscernible over b and the other one. Or well, if you take any number of them, each one is indiscernible over b and all the other ones. OK, so maybe you have to think a moment why this gives you that. And this is really what I prefer to think about. So the way I think about this, you have this sequence. Ah, it's indiscernible, so you don't see any. It's the same everywhere. If you add a finite parameter b, then the sequence is cut into a bounded number of pieces. And those pieces are again mutually indiscernible sequences. OK. So now there's a very nice situation is when you only have finitely many such classes. And in particular, if you only have two such classes. And so there's something that measures that, which is called the DP rank. And I guess I'm only going to define it for finitely many. So in the over categorical case, you have to define it. In the omega categorical case, in the finitely homogeneous case, it will be finite. No, otherwise, no. No. Because they can still be infinitely many formulas with more and more variables. OK. So the DP rank is going to bound the number of classes that you can have. So say that the DP rank of p is at most n. Let's do it only for an integer. If for any indiscernible i and any b realizing p, well, if you do this, you only need to cut into at most n classes so as to get this. So it's not completely true. Well, first I guess it should be n plus 1. And yeah, it's not completely true. OK. What? Yeah. OK. OK. OK. So I'll get back to this. I'll finish this. I want to look a bit more closely at what happens here. So this, I mean, some of the whole idea of the proof rests in understanding the minimal situation. You have an indiscernible sequence, and you have a topol b. The topol b breaks the sequence into many pieces. Let's look at the simplest thing it could do. Well, there are actually two simplest things that it could do. One thing is that i breaks into two infinite pieces over b. So maybe over b, i breaks at some point into two infinite pieces. And imagine that there's no limit point on either side. I mean, there's no end point on either side. And there's some formula with parameter in b that's true here and false here. And that's it. The two pieces are mutually indiscernible over b. OK. So this is the first simplest situation. But there's another one that could happen. It could be that there's just one point that behaves badly. And that if you take off that point, the sequence that remains is indiscernible over b. And the fact that there was those two minimal situations is the main observation behind the definition of distality, actually. So you should think of this situation as being stable like. Because what happens in a stable situation? If the theory is stable, the sequence is totally indiscernible. And you have this stronger property that if you get a finite set, you can just remove a bounded set from the sequence. And whatever is left is indiscernible. So when it happens that you can take a point and you remove it, and the sequence i is indiscernible over b, let's think of this as a stable like situation. And when the sequence is cut in two and there's some formula that's true, below, and false above, then this is like an order like situation. So distal theory, it's not important for this talk, but the distal theory is a theory where this never happens. When you never have this situation, you only have this one. So now if you want to, so why was I in trouble here? Because the number of classes is not the correct number you want to look at. If you have your sequence and you have a parameter b, and let's say it cuts it into finitely many classes. But there could be cuts that are order like cuts, where some formula changes truth value, and then there could be isolated points that you need to take off. And what you want to count are really the number of those isolated points and cuts. This is the relevant number. So you see a point gives you really three classes, but I want to think of it as just two. I want to think of this as one thing that happens in the sequence. And if you count that, then the Dp rank is good. And any b, they are at most n cuts in i over b. And what I mean by a cut is either one of those two situations. So either there's a formula that changes truth value, or there's an element I need to take away. Well, this doesn't happen to consecutive because my sequences are dense. So then the Dp rank is at least, oh, sorry. It's n minus 1. So if it's 1, no, it's good, actually. So the Dp rank would be 2. So it's known, so this is proved by Tai of an Alex, that the Dp rank is additive. Sub-additive, yeah. Thank you. I hope I'll write it correctly. Looks correct. So the Dp rank of a topol, AB of the type of a topol, is bounded by the sum of the Dp rank, we should say rank, the Dp rank of a over a, and the Dp rank of b over a. So something you would expect. It's not so obvious to prove, but it's true. And of course, there's one natural thing to do with the picture that I drew here is to split the Dp rank into two parts, just counting the number of, you could just count how many of those things you can have, and you can count how many of those you can have. And this does work. It does give you two notions of rank, which also sub-additive, each separately. I'm not going to, in this talk, I'm not going to use them because what I'm going to do right now is restrict to the case where the Dp rank is 1. So I was willingly being vague about that. They're cuts given by a formula with one variable and parameters in b plus maybe other parameters from the sequence itself. That accounts for the formulas with more than one variable. So the correct way would be to define it like that. But then, if you'd have to use the sequence, you could get arbitrary many funny things going on. Yeah, yeah, but you're only allowed to cut, yeah, but still, it's going to be bounded. As long as you don't cut on one of the parameters themselves. But let me tell you what the Dp rank 1 case is, and that will be. So from now on, we restrict to the case where the Dp rank of any one type, so the Dp rank of the structure is 1, which means the Dp rank of any one type is 1. And then what does that mean? It means if I take an indiscernible sequence i, and I take now a single ton a, then either the sequence stays indiscernible over a, or there's a cut, and I get two and the two infinite pieces on the left and on the right are mutually indiscernible over a, or there's one point that I need to take off, and when I take it off, the remaining thing is indiscernible over a. OK, and that's a precise statement. This is equivalent to being Dp rank 1. So from now on, I'm going to explain the proof in the Dp rank 1 case, which still shows the idea, but simplifies the number of technical things. And then one has to understand how to restrict to that case. If you could just find the type of Dp rank 1, you'd be good, but that's not always possible, so you have to do more complicated things, but this I won't explain. OK, so the first thing is we want to be in this situation. This is the situation that's order-like. But we can, because if this situation always happens when you take any indiscernible sequence and any single tone, then the theory is stable. So if the theory is unstable, you have this situation happens at least once. If this always happens for every single tone and every indiscernible sequence, then the theory is stable. Because stability is restricted to formula with one variable, and you can test stability by evaluating. So if T is unstable, it is some I and some single tone, A, such that and a formula. So let's write it all out. And there's a formula. OK, let's call it B. There's a formula for your FxB. And here I'm going to assume there's no other parameter in phi. So to answer that, maybe I would need to add things to the base to ensure that. But everything I'm doing is invariant on the adding finitely many constants. So I can certainly get down to this situation. There is I, B, and a formula phi of xB, such that phi holds on the left on an infinite initial segment, and the negation of phi holds on the final segment of the sequence. So I'm in this situation. Just to emphasize that it could be tuples, that the sequence is not B is a single tone, but the sequence might not be. But then I'm going to forget about the bar from now on. So is this OK? You can forget about almost everything and just so that we are done to this situation. We have this indiscernible sequence, and we have one formula that's true, that's not a formula. And we have one formula that's true, then false. And also we have this DP minimal that will play an important role later on. We want to get an order from that. So there are two natural things to do. One natural thing leads to Scheller's theorem that there's a definable partial order. And what is Scheller's argument in this case? It's saying, look at the type. So this is an indiscernible sequence. I can look at the type, let q be the type of an element that fits here in the sequence. So that if you add it here, you get an indiscernible sequence. Let's look at that type q over the sequence. So q over the sequence, which is the type of an extra element in this cut here that fits in here. And you can define naturally a partial order. Well, you have a formula that has the strict order property. Yeah, I didn't need to define q actually at this stage. So if you take a1 and a2 here, it is not possible. OK, let's define p to be the type of b over the sequence. This maybe will be the type more important for me. So it's not possible to have a b prime that realizes p, such that negation of phi of a1 b prime holds and phi of a2 b prime holds. If a1, a2 are two elements such that when you add them both, you get an indiscernible sequence. Why is this not possible? Because look at what the formula phi does. It would be true here. Then it would be false on a1. Then it would be true on a2. And then it would be false again. But that means it changes truth value. That would contradict the p minimality. It changes truth value too much. So this is not possible. So you assume that an a1, a2 preserves this element. Yeah, yeah, I'm assuming that both. So maybe I shouldn't write that. I'm not assuming that not only that they both satisfy q, but that together they fit in the sequence. And then it's inconsistent to have this, which is exactly saying that phi of a1 y and there's a phi of a2 y and there's an inclusion between them, which you probably know what it is. So as a2 goes there, this is bigger. So this should be bigger. And the strict comes from the existence of b. I can find a point that changes truth value between them by indiscernible. So this is how I get the strict inclusion here. And the inclusion is just this statement. The y steps compete, right? Yeah, I think it's correct. OK, so this is the proof of Scherlach's theorem in this easier dp minimal case. Now to get the linear order, we're actually going to look at it a little bit differently. So Scherlach's theorem gives us the partial order on realizations of q on elements that fit in the sequence. I'm going to construct an order in realizations of p. So p is the type of the parameter b. So almost done. There's a little bit more. Just a little bit. Yeah, I know. I approach us for that. OK, so what is the natural way to try to define an order on realizations of p? So if I have b and b prime, if I look at phi of xb, it's true here, false here. Phi of xb prime is true here, false here. That's because they satisfy the same type of the sequence. But now what I can do is I can add another piece of the sequence in the middle. And then phi of xb, let's see what happens. Phi of xb is true here. So let's look at bb prime. So phi of xb is true here. It's false here. And well, it's the same argument I had over there. By dp minimality, there's one point where it changes truth value, where it's from true to false. Be a cut that is sort of indexed by b. Now there's also one point where phi of xb prime changes truth value. A cut, a cut. Sorry, a cut. There's a cut in the sequence where the formula phi of xb prime changes truth value. And now if this happens, I want to think of b as being smaller than b prime. OK, now this doesn't quite work. But let's see what does work, what we can get. So first we can define, well, there's a natural equivalence relation that I won't actually need, but that will turn out to be this equivalence relation. Which is, I can say b and b prime are equivalent. If no matter what I put here, the cut is the same. But you're assuming b is the same type of an i. I'm assuming they have the same type of an i, yeah. P is the type of an i. P is the type of an i. P is the type of an i. It's on the right place. P is the type of b over i. So P knows that the formula is true on the left and false on the right. So there's a natural equivalence relation if the cut, so let's call this the cut defined by b, and this is the cut defined by b prime. So the natural equivalence relation is if cut b equals cut b prime for all, say j, let's call j, whenever I put a piece j in the sequence, the cuts are the same. That's an equivalence relation. Inj, the cut in whatever. Inj, yeah, in this thing. OK, now my strict order, what I want to be a strict order relation, is going to be r of a b prime. And here you have to be careful if you cannot have the cut of b prime less than the cut of b. So if we never have cut of b prime less than cut of b. So no matter how you put j, maybe the cut of b and b prime coincide. That happens in particular if you take j equals is empty. So that's going to happen for sure sometimes. And it could be that b prime can be separated, say, but you cannot have the opposite. So wait, is that what I want to write? No, this is the large one. Yeah, this is the large. Sorry, so this is what is going to be the large relation, the non-strict. And note that this is v definable because it's saying you do not have something. So the complement is there is something. So this is v definable. It's obvious that it's infinite. I mean, that's probably the typical problem. Wait, that's the question. No, no, the fact that the quotient is infinite is obvious. Because it's indiscernible, so whenever I put a j, there is some b that has a cut here, there's some b that has a cut here, there's some b that has a cut here. And those all give different classes. So yeah, so what is obvious? That the quotient by a is infinite. That's obvious. The other thing that's obvious, a little bit less so, but you just have to write 1, 9, is that this is a transitive. And this is just one line. If you write it, maybe I don't have time to write the line. Yeah? Yeah, yeah. What? Yeah, yeah. What? Yeah, yeah. What? What do you want to write? Oh, I have other things to write than the line. Yeah, yeah, what is the question? The formula is always the same phi. Formula is always the same phi. We just vary the. Yes. The formula phi is fixed once and for all. OK, so I claim that this is very easy to check just from the definition. So what is not clear is linearity. So what does linearity mean? It means our two points comparable. So what does it mean that two points are not comparable? It would mean that you have b and b prime. And there is some j such that the cut of b lands below the cut of b prime. But then there's also some other j prime such that it's the opposite, the cut of b prime is below the cut of b. And if you have that, well, you can't compare them. But now, so this is sort of the crux of the proof, is to observe that if this happens too much, we're going to contradict an IP. And what does too much mean? So this might happen. At this stage, it could very well be that this is not yet linear. So if it's linear, we're done. We have all the properties. If it's not linear, it means there are two realizations of p that can be switched by just changing the thing in the middle. So then what we're going to do is we're going to pick one of those realizations, add it to the base, increase the sequence, and then iterate and work here. So you don't need to worry about this. Eventually, if this process keeps going, you will get a sequence like this of points which are all realizations of p. You will get an indiscernible sequence with the property that, well, if you look at phi x bi, it's true up to this cut and then false. And with also the extra property that any two consecutive ones, you can change what's here, put another sequence, so as to switch the two. And now, what we want to do is just show that then we can rearrange this middle piece so that to get any given order on the base. And why can we do it? So this is sort of the main technical thing, is to understand why this works. And the reason is because of this DP rank. So I'll explain it. What do you need to do? What I want to do now is I want to, OK, so maybe I should say the conclusion to the end. So now what I claim is that by changing the sequence in this middle piece, I can arrange to get any permutation of the cuts that I want. Why does this contradict an IP? Because then the formula phi of xy would have IP. Because now if I just take a point here, it's related by phi to everything to the left and to nothing to the right. But everything to the left can be just any subset of the bi's that I want. So if I can do that, I'm done. Now we have to understand why we can do that. And the reason is, so this additivity of DP rank tells us that for the DP rank of, I think I'm only going to need it for 2, the DP rank of any two points is at most 2. Because DP rank of each is 1. So it's actually exactly 2. Because if I take any 2 of, sorry, those are b's. If I take any bi and bj, I do have two cuts that they induce on this sequence. So the main, OK, so now I'm going to do this permutation inductively by switching consecutive points. So I'm going to explain one switch. So say you are at a different situation. You have here some bi. So let's do it just at the start. I want to exchange b3 and b4. What I know by construction is that if I forget about all the other points, I can erase this, put another sequence that exchanges them. But I have no idea a priori what happens over the other points. Maybe the other points now, the sequence changes. Maybe the type has changed. But DP rank will tell me that this cannot actually happen. Because now if I take, so now let's take b3 and here I have b5. So in the new sequence, if I look at what happens over b3, b5, if I look at the new sequence, it's here. I know two places where the top of b3, b5 cuts the sequence, here and here. So there cannot be any other. So what this means is that this whole left piece and this whole right piece have to be mutually indiscernible over b3, b5. And what this means is that this piece that I've added, b3, b5, they don't see anything here. They don't notice that anything has changed. And therefore I can just do the same thing here and replace them and move this here and iterate. And it contradicts an IP. And that's it. Thank you. Thank you. Thank you.
 that says that it says something a little bit weaker than saying  that in any unstable NIP theory, well,  that any unstable NIP theory defines or interprets  an infinite linear order, because the order will not  be actually definable.  But let me just start maybe with a little bit of background.  So there's a theorem of Schellach that says that if you  have a structure which is NIP and not stable, then some  formula has the strict order property.  So T has strict order property.  So I'll recall what this means.  So recall that formula phi has the strict order property if  there's a sequence of parameters for y, say, such that  the formulas that are defined, or maybe it doesn't matter,  are strictly defined sets.  If you vary the formula, if you vary the parameters over  the sequence, you define sets which are strictly increasing.  Having a formula that has a strict order property is  equivalent to there being a definable partial order with  an infinite chain.  Because here, if you have this, you can define an order  from that.  So I think pre-order and quasi-order means the same  thing, no?  So pre-quasi-order, by saying that, say, B is less  equal to B prime, if the formula defined by B is  less, included the formula defined by B prime.  So this is, of course, a transitive relation.  And the assumption that phi has SOP exactly tells you that  this order has an infinite chain.  And conversely, you can figure out what to do.  So this is telling us we have a partial quasi-order, but if  you allow quotients, then you have an interpretable partial  order that has an infinite chain.  And then there was a question, which I'm not sure to whom  it's attributed.  I think Schell absolutely raised it.  I heard it first from Ryszowski.  I don't know.  I'm sure other people came up with it.  Is can we get a definable, with the same assumption, so if  T is nip and stable, does it interpret an infinite linear  order?  So note that you cannot ask for an infinite definable order  because if you just take a structure which  has an equivalence relation, all classes of psi 2, and then  you have an order on the quotient, so you have an  order, but you have two cover of it, then there's no  definable order on that structure.  But of course, there's an interpretable one.  OK, so this is still open.  However, what I can prove is the following.  So the assumption is still the same.  T is nip and stable.  Then there is a finite set A of parameter.  There's a type over A and some type definable relation such  that this defines a strict linear quasi-order on  realizations of p, such that r defines a strict linear  pre-order or quasi-order on realizations of p.  Yeah, but here I have this because it's strict.  So there's a choice to make.  If you take the one that's not strict, you have v.  If you take the one that's strict, you have.  Yeah.  So some days I mean.  We're having a strict type defined one thing.  Is this the best possible thing to have a type defined  one?  Yeah, so that's why I stated like this, so it looks better.  Well, we always thought that it's a progress.  What do you think it's something?  Why is it interpretable?  Yeah, exactly.  So let me explain what I mean by a pre.  So what does it mean, a linear pre-order?  It means that it's transitive.  So OK, where's the interpretable?  So if we have this, so from this, we can define the relation  e, which you define as the link that you have neither of the  two.  You don't have that x is pretty small than y.  You don't have that y is pretty small than y.  Now this is a v-definable equivalence relation.  And the quotient of e by this equivalence relation is  linearly ordered by r.  So that's what it means to be this linear pre-order.  So in the quotient is linearly ordered by r.  Oh, yeah.  And no, it's more than that.  The quotient is infinite.  So the quotient is infinite and linearly ordered by r.  Is this still?  And everything is over r.  Over a, sorry.  This is over.  So is the statement clear?  So yeah, for me, it's somewhat easier to think that you have  a v-definable equivalence relation.  So I'm going to go back to the question.  For me, it's somewhat easier to think that you have a v-definable  equivalence relation.  And then you have a linear order on the quotient.  And if you think of what that means, then it means that  either the strict order is type-definable or the  non-strict order is v-definable as you want.  And an immediate corollary is that if t is omega categorical,  then the conjecture is true.  Because the point is that here a is finite.  So if the theory is omega categorical,  everything is definable.  Right, so that's also a good question.  So p can be any, yeah, so p can be a one type.  In fact, it can extend any unstable type.  That you start with.  So what I should say is that, so this looks like a  strengthening of this.  But it's a theorem of a very different nature.  Because this is a non-structure theorem.  It tells you you have this thing that says that you have  some complexity in it.  With this, you should really think of it as a structure  theorem.  It's giving you a linear order as a very rigid object.  It's a very, and so you could hold it.  But you have to use those linear orders to actually build  some structure theory of models.  So build some classification theory of models of an NIP  theory, t.  So some partly speculative  applications.  Well, almost all of them are speculative.  Only the first one is only partially.  So well, the nicest case is when the theory is omega  categorical.  And so it's natural to start there.  And there, we can hope to have a classification of NIP,  finitely homogeneous structures, extending the  stable case.  So the stable case is known by work of a number of people,  like Lan, Chirlino-Schofsky, Harrington, well, quasi-finite.  Well, there's a very, anyway.  So this, when I started doing it, I gave talk about it in  Paris, and it's in progress, joined.  We found some shows.  And I don't really want to talk about it today.  It seems that one could hope to really have some  coordination.  I have no idea how to write this.  What?  What's the next letter?  Hi.  I'm very tired.  Coordination of models by linear orders with the idea  that if you compare with, let's say, the omega stable case,  compared with the omega stable case, so a strictly minimal  set would be replaced by linear orders.  So we do these two things connected?  Yeah, yeah, this is the omega stable.  This is the omega categorical version of this.  But here you can hope for much more than, because here I  don't know what that means.  Oh, wait.  Here you're not assuming omega categorical?  No.  No, I'm not assuming omega categorical anymore.  Well, OK, so in stable, so let's say omega stable,  so you coordinateize maybe models by looking at  dimensions of regular types.  So here those dimensions of regular types, what they  could be replaced by are isomorphism types of  certain linear orders.  So it's no longer going to be a number.  It's going to be the isomorphism type of a certain order.  And then in the stable case, in very nice cases, you know the  model up to isomorphism by knowing those dimensions.  Here you could hope for, so for example, what happens in  real close fields, in real close fields, let's take  maybe divisible order to be in groups.  In divisible order to be in groups, if you know, so how  does it work?  Actually, let's take real close fields.  It's a little bit.  So what classification could you hope for if, so if you have  a real close field, you can look at the valuation given by  the convex hull of the reels, and then becomes a valued  field.  And then the residue field is a subfield of the reels, and  then you have the value group.  If you know the value group and you know the residue field,  so the residue field lives in some bounded thing.  So if you know the value group and the residue field, you  still don't know the whole field, but there's a maximal  one with this given value group and residue field.  But actually, if you just fix the value group, then what I  said is also true, because the residue field anyway is a  subfield of r.  So for a fixed value group, there's a maximal model of  RCF that has this fixed value group.  And this is the idea that one could hope to have something  similar, where here the value group is playing the role of  my ordered set.  So one could hope that you somehow have ordered set,  and they don't give you the isomorphism type, but say if  the structure is this tall, they will give you at least  there's a maximal one that has those.  But maximal you mean?  Maximum under inclusion.  But I haven't really thought about this, so if this looks  very vague, it's normal.  But this is kind of the idea.  When could look at algebraic structures?  So this is also very vague.  But for example, there's this conjecture that if you have  an NIP unstable field, maybe it has a definable valuation,  well, if you take such a field, it's going to have a  generic type.  If the generic type is generically stable, then I  don't know what to say.  But if the generic type is not generically stable, then some  kind of upgraded version of this theorem is going to give  you an order such that the generic type is going at  minus infinity along that order, which you expect the  type to be the generic type at minus infinity along the  valuation.  So this is going to give you an order such that the generic  type goes at minus infinity along that order.  And then using the fact that the type is generic, you could  hope to somehow create a valuation by looking at  stabilizers of convex subsets of the order or something like  this.  But.  So k is a fixed type of field.  k is an NIP field.  And I'm assuming this generic type is not  generically stable.  Oh, it has a generic type.  It's an F-generic type.  F-generic type.  I've added two groups.  For example, you can have one that does for both.  But OK, this is speculative.  I haven't really had time.  But it's just to give an idea of why this is likely to be  useful to give us strong structure theorems, hopefully.  But there's some big things that need to be  figured out first.  Anyway.  So now what I want to do is I'd like to give you some idea on  how this is proved.  And I usually don't like giving proofs.  But the reason I'm doing this is because the proof is  probably much easier to tell than to read.  And I mean, it's the kind of thing that it's not so hard  to actually sketch the proof.  But it can be very hard to read it.  So I think it's worth actually doing it.  I'm not going to do the general case.  So I'm going to just do a special case that will give you  the idea of what's going on.  And you'll see there's just one idea.  It's not so hard.  But we have to play a little bit with  indiscernible sequences.  And so I need to recall certain things about indiscernible  sequences in NIP.  OK, so probably you know that if you have, so let's just, from  now on, assume that my theory is NIP.  And one of the first things, usually, that one learns is  that if you have I an indiscernible sequence, say  this is I, and you have a formula with some parameter.  So the sequence is indiscernible over the empty  set.  It's not necessarily indiscernible over the  parameter B. So when I evaluate that formula along  the sequence, then it's going to cut out a finite union of  convex sets.  So this formula, if x varies along the sequence, I'm  going to have finitely many convex sets such that, say,  the formula phi of xB here is true.  And then here it's false.  Then here it's true.  And then maybe here I just have one point where it becomes  false.  And then it's true outside, something like this.  But the point is there are only finitely many changes of  truth value along the formula.  So this is usually one of the first things that one learns  about NIP.  And it's equivalent to NIP.  And this is strengthened by the so-called shrinking of  indiscernible.  Maybe I'll move here.  Which explains what happens when you evaluate not a formula  with one variable, but you look at the formula with several  variables, each variable having the same size as a  tuple of the sequence.  And you want to know what happens there.  And then the conclusion is essentially the same that there's  a finite, that there's an equivalence relation with  finitely many convex classes such that the truth value, if  you evaluate the formula on a tuple from the sequence, the  truth value only depends on, if you look at your tuple, say  it's an increasing order on which equivalence relation  each element lies.  So let's write it down.  So i is indiscernible.  You have any formula.  Then again, you can cut.  So maybe some classes have just one element, or  finitely many.  But there is a convex finite equivalence relation, E on I.  So what I mean by convex finite, I mean the classes are  convex, and there are finitely many classes.  Well, such that the truth value, when I evaluate it on the  tuple, so I take a1, a2, a3, a4, a5, and I evaluate my  formula on that, then to know whether the formula is true or  not, I only need to know if the tuple have taken an  increasing order, I only need to know in which classes each  of them lies.  So such that if, say, a1, an, or an i in increasing order, b1,  bn, or in i in increasing order, and ai is equivalent to  bi for all i, then the formula is true on a1, an.  OK, yeah, shouldn't have called b.  If and only if it's true on b1, bn.  Yes, I'm allowed to take two in the same class.  And each of those classes is going to be definable, I  mean relatively definable by an instance of that way.  So there's a unique, maybe I should say, well, there's a  unique finest, yeah, courses one.  There's a unique courses one that has this property.  So if you do this.  So now if you have i an indiscernible sequence and  you take, say, a finite tuple b, you can do this for every  formula.  So every formula is going to give you a finite convex  equivalence relation.  Now you should think of i, usually indiscernible  sequences, especially in stable, they're indexed by  omega.  That's not what I want to think about.  I want to think in a, in a, in a, in a, in a, in a, in a  thinking about indiscernible sequences that are  indexed by a dense linear order, which is usually very big.  But definitely dense.  So you should think of all indiscernible sequences as  being indexed by a dense order, so that I don't have  problems of consecutive elements, things like that.  Then if I have a finite tuple b, I can take every formula phi.  For every formula phi, I get an equivalence relation.  And then I take the intersection of all those  equivalence relations.  What I get is an equivalence relation with convex  classes on i, which now has maybe two to the  parameter.  The parameter is fixed.  i is fixed, b is fixed.  Yeah, thank you.  Yeah, b is fixed.  But phi varies.  So now what I get is I get an equivalence relation.  So we get convex equivalence relation eb.  So it depends on b on i with at most maybe two to the t  classes.  Such that what?  Well, such that the classes, there might be some finite  classes.  But the infinite classes are mutually indiscernible.  And actually also over the finite classes.  So maybe older classes, but the finite ones.  So the infinite classes are mutually indiscernible.  What does it mean?  Over b.  Yeah, thank you.  They're mutually indiscernible over b.  What does it mean to be mutually indiscernible?  It means if you take two of them, each one is indiscernible  over b and the other one.  Or well, if you take any number of them, each one is  indiscernible over b and all the other ones.  OK, so maybe you have to think a moment why this gives you  that.  And this is really what I prefer to think about.  So the way I think about this, you have this sequence.  Ah, it's indiscernible, so you don't see any.  It's the same everywhere.  If you add a finite parameter b, then the sequence is cut  into a bounded number of pieces.  And those pieces are again mutually indiscernible  sequences.  OK.  So now there's a very nice situation is when you only have  finitely many such classes.  And in particular, if you only have two such classes.  And so there's something that measures that, which is called  the DP rank.  And I guess I'm only going to define it for finitely many.  So in the over categorical case, you have to define it.  In the omega categorical case, in the finitely homogeneous  case, it will be finite.  No, otherwise, no.  No.  Because they can still be infinitely many formulas with  more and more variables.  OK.  So the DP rank is going to bound the number of classes  that you can have.  So say that the DP rank of p is at most n.  Let's do it only for an integer.  If for any indiscernible i and any b realizing p, well, if  you do this, you only need to cut into at most n classes so  as to get this.  So it's not completely true.  Well, first I guess it should be n plus 1.  And yeah, it's not completely true.  OK.  What?  Yeah.  OK.  OK.  OK.  So I'll get back to this.  I'll finish this.  I want to look a bit more closely at what happens here.  So this, I mean, some of the whole idea of the proof rests  in understanding the minimal situation.  You have an indiscernible sequence, and you have a  topol b.  The topol b breaks the sequence into many pieces.  Let's look at the simplest thing it could do.  Well, there are actually two simplest things that it could  do.  One thing is that i breaks into two infinite pieces over b.  So maybe over b, i breaks at some point into two infinite  pieces.  And imagine that there's no limit point on either side.  I mean, there's no end point on either side.  And there's some formula with parameter in b that's true here  and false here.  And that's it.  The two pieces are mutually indiscernible over b.  OK.  So this is the first simplest situation.  But there's another one that could happen.  It could be that there's just one point that behaves badly.  And that if you take off that point, the sequence that  remains is indiscernible over b.  And the fact that there was those two minimal situations  is the main observation behind the definition of  distality, actually.  So you should think of this situation as being stable  like.  Because what happens in a stable situation?  If the theory is stable, the sequence is totally indiscernible.  And you have this stronger property that if you get a  finite set, you can just remove a bounded set from the  sequence.  And whatever is left is indiscernible.  So when it happens that you can take a point and you  remove it, and the sequence i is indiscernible over b, let's  think of this as a stable like situation.  And when the sequence is cut in two and there's some formula  that's true, below, and false above, then this is like an  order like situation.  So distal theory, it's not important for this talk, but  the distal theory is a theory where this never happens.  When you never have this situation, you only have  this one.  So now if you want to, so why was I in trouble here?  Because the number of classes is not the correct number you  want to look at.  If you have your sequence and you have a parameter b, and  let's say it cuts it into finitely many classes.  But there could be cuts that are order like cuts, where some  formula changes truth value, and then there could be isolated  points that you need to take off.  And what you want to count are really the number of those  isolated points and cuts.  This is the relevant number.  So you see a point gives you really three classes, but I  want to think of it as just two.  I want to think of this as one thing that happens in the  sequence.  And if you count that, then the Dp rank is good.  And any b, they are at most n cuts in i over b.  And what I mean by a cut is either one of those two  situations.  So either there's a formula that changes truth value, or  there's an element I need to take away.  Well, this doesn't happen to consecutive because my  sequences are dense.  So then the Dp rank is at least, oh, sorry.  It's n minus 1.  So if it's 1, no, it's good, actually.  So the Dp rank would be 2.  So it's known, so this is proved by  Tai of an Alex, that the Dp rank is additive.  Sub-additive, yeah.  Thank you.  I hope I'll write it correctly.  Looks correct.  So the Dp rank of a topol, AB of the type of a topol, is  bounded by the sum of the Dp rank, we should say rank, the  Dp rank of a over a, and the Dp rank of b over a.  So something you would expect.  It's not so obvious to prove, but it's true.  And of course, there's one natural thing to do with the  picture that I drew here is to split the Dp rank into two  parts, just counting the number of, you could just count  how many of those things you can have, and you can count  how many of those you can have.  And this does work.  It does give you two notions of rank, which also sub-additive,  each separately.  I'm not going to, in this talk, I'm not going to use them  because what I'm going to do right now is restrict to the  case where the Dp rank is 1.  So I was willingly being vague about that.  They're cuts given by a formula with one variable and  parameters in b plus maybe other parameters from the  sequence itself.  That accounts for the formulas with more than one variable.  So the correct way would be to define it like that.  But then, if you'd have to use the sequence, you could get  arbitrary many funny things going on.  Yeah, yeah, but you're only allowed to cut, yeah, but  still, it's going to be bounded.  As long as you don't cut on one of the parameters themselves.  But let me tell you what the Dp rank 1 case is, and that  will be.  So from now on, we restrict to the case where the Dp rank of  any one type, so the Dp rank of the structure is 1, which  means the Dp rank of any one type is 1.  And then what does that mean?  It means if I take an indiscernible sequence i,  and I take now a single ton a, then either the sequence  stays indiscernible over a, or there's a cut, and I get two  and the two infinite pieces on the left and on the right are  mutually indiscernible over a, or there's one point that I  need to take off, and when I take it off, the remaining  thing is indiscernible over a.  OK, and that's a precise statement.  This is equivalent to being Dp rank 1.  So from now on, I'm going to explain the proof in the Dp  rank 1 case, which still shows the idea, but simplifies the  number of technical things.  And then one has to understand how to restrict to that case.  If you could just find the type of Dp rank 1, you'd be good,  but that's not always possible, so you have to do more  complicated things, but this I won't explain.  OK, so the first thing is we want to be in this situation.  This is the situation that's order-like.  But we can, because if this situation always happens when  you take any indiscernible sequence and any single tone,  then the theory is stable.  So if the theory is unstable, you have this situation  happens at least once.  If this always happens for every single tone and every  indiscernible sequence, then the theory is stable.  Because stability is restricted to formula with one  variable, and you can test stability by evaluating.  So if T is unstable, it is some I and some single tone, A,  such that and a formula.  So let's write it all out.  And there's a formula.  OK, let's call it B. There's a formula for your FxB.  And here I'm going to assume there's no other parameter  in phi.  So to answer that, maybe I would need to add things to the  base to ensure that.  But everything I'm doing is invariant on the  adding finitely many constants.  So I can certainly get down to this situation.  There is I, B, and a formula phi of xB, such that phi holds  on the left on an infinite initial segment, and the  negation of phi holds on the final segment of the sequence.  So I'm in this situation.  Just to emphasize that it could be tuples, that the  sequence is not B is a single tone, but the sequence might  not be.  But then I'm going to forget about the bar from now on.  So is this OK?  You can forget about almost everything and just so that  we are done to this situation.  We have this indiscernible sequence, and we have one  formula that's true, that's not a formula.  And we have one formula that's true, then false.  And also we have this DP minimal that will play an  important role later on.  We want to get an order from that.  So there are two natural things to do.  One natural thing leads to Scheller's theorem that there's  a definable partial order.  And what is Scheller's argument in this case?  It's saying, look at the type.  So this is an indiscernible sequence.  I can look at the type, let q be the type of an element that  fits here in the sequence.  So that if you add it here, you get an indiscernible sequence.  Let's look at that type q over the sequence.  So q over the sequence, which is the type of an extra  element in this cut here that fits in here.  And you can define naturally a partial order.  Well, you have a formula that has the strict order property.  Yeah, I didn't need to define q actually at this stage.  So if you take a1 and a2 here, it is not possible.  OK, let's define p to be the type of b over the sequence.  This maybe will be the type more important for me.  So it's not possible to have a b prime that realizes p, such  that negation of phi of a1 b prime holds and phi of a2 b  prime holds.  If a1, a2 are two elements such that when you add them  both, you get an indiscernible sequence.  Why is this not possible?  Because look at what the formula phi does.  It would be true here.  Then it would be false on a1.  Then it would be true on a2.  And then it would be false again.  But that means it changes truth value.  That would contradict the p minimality.  It changes truth value too much.  So this is not possible.  So you assume that an a1, a2 preserves this element.  Yeah, yeah, I'm assuming that both.  So maybe I shouldn't write that.  I'm not assuming that not only that they both satisfy q,  but that together they fit in the sequence.  And then it's inconsistent to have this, which is exactly  saying that phi of a1 y and there's a phi of a2 y and  there's an inclusion between them, which you probably  know what it is.  So as a2 goes there, this is bigger.  So this should be bigger.  And the strict comes from the existence of b.  I can find a point that changes truth value between them by  indiscernible.  So this is how I get the strict inclusion here.  And the inclusion is just this statement.  The y steps compete, right?  Yeah, I think it's correct.  OK, so this is the proof of Scherlach's theorem in this  easier dp minimal case.  Now to get the linear order, we're actually going to look  at it a little bit differently.  So Scherlach's theorem gives us the partial order on  realizations of q on elements that fit in the sequence.  I'm going to construct an order in  realizations of p.  So p is the type of the parameter b.  So almost done.  There's a little bit more.  Just a little bit.  Yeah, I know.  I approach us for that.  OK, so what is the natural way to try to define an order on  realizations of p?  So if I have b and b prime, if I look at phi of xb, it's true  here, false here.  Phi of xb prime is true here, false here.  That's because they satisfy the same type of the sequence.  But now what I can do is I can add another piece of the  sequence in the middle.  And then phi of xb, let's see what happens.  Phi of xb is true here.  So let's look at bb prime.  So phi of xb is true here.  It's false here.  And well, it's the same argument I had over there.  By dp minimality, there's one point where it changes  truth value, where it's from true to false.  Be a cut that is sort of indexed by b.  Now there's also one point where phi of xb prime changes  truth value.  A cut, a cut.  Sorry, a cut.  There's a cut in the sequence where the formula phi of xb  prime changes truth value.  And now if this happens, I want to think of b as being  smaller than b prime.  OK, now this doesn't quite work.  But let's see what does work, what we can get.  So first we can define, well, there's a natural equivalence  relation that I won't actually need, but that will turn out  to be this equivalence relation.  Which is, I can say b and b prime are equivalent.  If no matter what I put here, the cut is the same.  But you're assuming b is the same type of an i.  I'm assuming they have the same type of an i, yeah.  P is the type of an i.  P is the type of an i.  P is the type of an i.  It's on the right place.  P is the type of b over i.  So P knows that the formula is true on the left and false on  the right.  So there's a natural equivalence relation if the cut,  so let's call this the cut defined by b, and this is the  cut defined by b prime.  So the natural equivalence relation is if cut b equals  cut b prime for all, say j, let's call j, whenever I put a  piece j in the sequence, the cuts are the same.  That's an equivalence relation.  Inj, the cut in whatever.  Inj, yeah, in this thing.  OK, now my strict order, what I want to be a strict order  relation, is going to be r of a b prime.  And here you have to be careful if you cannot have the cut of  b prime less than the cut of b.  So if we never have cut of b prime less than cut of b.  So no matter how you put j, maybe the cut of b and b prime  coincide.  That happens in particular if you take j equals is empty.  So that's going to happen for sure sometimes.  And it could be that b prime can be separated, say, but you  cannot have the opposite.  So wait, is that what I want to write?  No, this is the large one.  Yeah, this is the large.  Sorry, so this is what is going to be the large relation,  the non-strict.  And note that this is v definable because it's saying  you do not have something.  So the complement is there is something.  So this is v definable.  It's obvious that it's infinite.  I mean, that's probably the typical problem.  Wait, that's the question.  No, no, the fact that the quotient is infinite is  obvious.  Because it's indiscernible, so whenever I put a j, there is  some b that has a cut here, there's some b that has a cut  here, there's some b that has a cut here.  And those all give different classes.  So yeah, so what is obvious?  That the quotient by a is infinite.  That's obvious.  The other thing that's obvious, a little bit less  so, but you just have to write 1, 9, is that this is  a transitive.  And this is just one line.  If you write it, maybe I don't have time to write the line.  Yeah?  Yeah, yeah.  What?  Yeah, yeah.  What?  Yeah, yeah.  What?  What do you want to write?  Oh, I have other things to write than the line.  Yeah, yeah, what is the question?  The formula is always the same phi.  Formula is always the same phi.  We just vary the.  Yes.  The formula phi is fixed once and for all.  OK, so I claim that this is very easy to check just from  the definition.  So what is not clear is linearity.  So what does linearity mean?  It means our two points comparable.  So what does it mean that two points are not comparable?  It would mean that you have b and b prime.  And there is some j such that the cut of b lands below the  cut of b prime.  But then there's also some other j prime such that it's  the opposite, the cut of b prime is below the cut of b.  And if you have that, well, you can't compare them.  But now, so this is sort of the crux of the proof, is to  observe that if this happens too much, we're going to  contradict an IP.  And what does too much mean?  So this might happen.  At this stage, it could very well be that this is not  yet linear.  So if it's linear, we're done.  We have all the properties.  If it's not linear, it means there are two realizations of  p that can be switched by just changing the thing in the  middle.  So then what we're going to do is we're going to pick one of  those realizations, add it to the base, increase the  sequence, and then iterate and work here.  So you don't need to worry about this.  Eventually, if this process keeps going, you will get a  sequence like this of points which are all realizations of  p.  You will get an indiscernible sequence with the  property that, well, if you look at phi x bi, it's true up  to this cut and then false.  And with also the extra property that any two  consecutive ones, you can change what's here, put another  sequence, so as to switch the two.  And now, what we want to do is just show that then we can  rearrange this middle piece so that to get any given order  on the base.  And why can we do it?  So this is sort of the main technical thing, is to  understand why this works.  And the reason is because of this DP rank.  So I'll explain it.  What do you need to do?  What I want to do now is I want to, OK, so maybe I should  say the conclusion to the end.  So now what I claim is that by changing the sequence in this  middle piece, I can arrange to get any permutation of the  cuts that I want.  Why does this contradict an IP?  Because then the formula phi of xy would have IP.  Because now if I just take a point here, it's related by  phi to everything to the left and to nothing to the right.  But everything to the left can be just any subset of the bi's  that I want.  So if I can do that, I'm done.  Now we have to understand why we can do that.  And the reason is, so this additivity of DP rank tells us  that for the DP rank of, I think I'm only going to need it  for 2, the DP rank of any two points is at most 2.  Because DP rank of each is 1.  So it's actually exactly 2.  Because if I take any 2 of, sorry, those are b's.  If I take any bi and bj, I do have two cuts that they  induce on this sequence.  So the main, OK, so now I'm going to do this permutation  inductively by switching consecutive points.  So I'm going to explain one switch.  So say you are at a different situation.  You have here some bi.  So let's do it just at the start.  I want to exchange b3 and b4.  What I know by construction is that if I forget about all the  other points, I can erase this, put another sequence that  exchanges them.  But I have no idea a priori what happens  over the other points.  Maybe the other points now, the sequence changes.  Maybe the type has changed.  But DP rank will tell me that this cannot actually happen.  Because now if I take, so now let's take b3 and here I have  b5.  So in the new sequence, if I look at what happens over b3, b5,  if I look at the new sequence, it's here.  I know two places where the top of b3, b5 cuts the  sequence, here and here.  So there cannot be any other.  So what this means is that this whole left piece and this  whole right piece have to be mutually indiscernible  over b3, b5.  And what this means is that this piece that I've added, b3, b5,  they don't see anything here.  They don't notice that anything has changed.  And therefore I can just do the same thing here and replace  them and move this here and iterate.  And it contradicts an IP.  And that's it.  Thank you.  Thank you.  Thank you.
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that says that it says something a little bit weaker than saying that in any unstable NIP theory, well, that any unstable NIP theory defines or interprets an infinite linear order, because the order will not be actually definable. But let me just start maybe with a little bit of background. So there's a theorem of Schellach that says that if you have a structure which is NIP and not stable, then some formula has the strict order property. So T has strict order property. So I'll recall what this means. So recall that formula phi has the strict order property if there's a sequence of parameters for y, say, such that the formulas that are defined, or maybe it doesn't matter, are strictly defined sets. If you vary the formula, if you vary the parameters over the sequence, you define sets which are strictly increasing. Having a formula that has a strict order property is equivalent to there being a definable partial order with an infinite chain. Because here, if you have this, you can define an order from that. So I think pre-order and quasi-order means the same thing, no? So pre-quasi-order, by saying that, say, B is less equal to B prime, if the formula defined by B is less, included the formula defined by B prime. So this is, of course, a transitive relation. And the assumption that phi has SOP exactly tells you that this order has an infinite chain. And conversely, you can figure out what to do. So this is telling us we have a partial quasi-order, but if you allow quotients, then you have an interpretable partial order that has an infinite chain. And then there was a question, which I'm not sure to whom it's attributed. I think Schell absolutely raised it. I heard it first from Ryszowski. I don't know. I'm sure other people came up with it. Is can we get a definable, with the same assumption, so if T is nip and stable, does it interpret an infinite linear order? So note that you cannot ask for an infinite definable order because if you just take a structure which has an equivalence relation, all classes of psi 2, and then you have an order on the quotient, so you have an order, but you have two cover of it, then there's no definable order on that structure. But of course, there's an interpretable one. OK, so this is still open. However, what I can prove is the following. So the assumption is still the same. T is nip and stable. Then there is a finite set A of parameter. There's a type over A and some type definable relation such that this defines a strict linear quasi-order on realizations of p, such that r defines a strict linear pre-order or quasi-order on realizations of p. Yeah, but here I have this because it's strict. So there's a choice to make. If you take the one that's not strict, you have v. If you take the one that's strict, you have. Yeah. So some days I mean. We're having a strict type defined one thing. Is this the best possible thing to have a type defined one? Yeah, so that's why I stated like this, so it looks better. Well, we always thought that it's a progress. What do you think it's something? Why is it interpretable? Yeah, exactly. So let me explain what I mean by a pre. So what does it mean, a linear pre-order? It means that it's transitive. So OK, where's the interpretable? So if we have this, so from this, we can define the relation e, which you define as the link that you have neither of the two. You don't have that x is pretty small than y. You don't have that y is pretty small than y. Now this is a v-definable equivalence relation. And the quotient of e by this equivalence relation is linearly ordered by r. So that's what it means to be this linear pre-order. So in the quotient is linearly ordered by r. Oh, yeah. And no, it's more than that. The quotient is infinite. So the quotient is infinite and linearly ordered by r. Is this still? And everything is over r. Over a, sorry. This is over. So is the statement clear? So yeah, for me, it's somewhat easier to think that you have a v-definable equivalence relation. So I'm going to go back to the question. For me, it's somewhat easier to think that you have a v-definable equivalence relation. And then you have a linear order on the quotient. And if you think of what that means, then it means that either the strict order is type-definable or the non-strict order is v-definable as you want. And an immediate corollary is that if t is omega categorical, then the conjecture is true. Because the point is that here a is finite. So if the theory is omega categorical, everything is definable. Right, so that's also a good question. So p can be any, yeah, so p can be a one type. In fact, it can extend any unstable type. That you start with. So what I should say is that, so this looks like a strengthening of this. But it's a theorem of a very different nature. Because this is a non-structure theorem. It tells you you have this thing that says that you have some complexity in it. With this, you should really think of it as a structure theorem. It's giving you a linear order as a very rigid object. It's a very, and so you could hold it. But you have to use those linear orders to actually build some structure theory of models. So build some classification theory of models of an NIP theory, t. So some partly speculative applications. Well, almost all of them are speculative. Only the first one is only partially. So well, the nicest case is when the theory is omega categorical. And so it's natural to start there. And there, we can hope to have a classification of NIP, finitely homogeneous structures, extending the stable case. So the stable case is known by work of a number of people, like Lan, Chirlino-Schofsky, Harrington, well, quasi-finite. Well, there's a very, anyway. So this, when I started doing it, I gave talk about it in Paris, and it's in progress, joined. We found some shows. And I don't really want to talk about it today. It seems that one could hope to really have some coordination. I have no idea how to write this. What? What's the next letter? Hi. I'm very tired. Coordination of models by linear orders with the idea that if you compare with, let's say, the omega stable case, compared with the omega stable case, so a strictly minimal set would be replaced by linear orders. So we do these two things connected? Yeah, yeah, this is the omega stable. This is the omega categorical version of this. But here you can hope for much more than, because here I don't know what that means. Oh, wait. Here you're not assuming omega categorical? No. No, I'm not assuming omega categorical anymore. Well, OK, so in stable, so let's say omega stable, so you coordinateize maybe models by looking at dimensions of regular types. So here those dimensions of regular types, what they could be replaced by are isomorphism types of certain linear orders. So it's no longer going to be a number. It's going to be the isomorphism type of a certain order. And then in the stable case, in very nice cases, you know the model up to isomorphism by knowing those dimensions. Here you could hope for, so for example, what happens in real close fields, in real close fields, let's take maybe divisible order to be in groups. In divisible order to be in groups, if you know, so how does it work? Actually, let's take real close fields. It's a little bit. So what classification could you hope for if, so if you have a real close field, you can look at the valuation given by the convex hull of the reels, and then becomes a valued field. And then the residue field is a subfield of the reels, and then you have the value group. If you know the value group and you know the residue field, so the residue field lives in some bounded thing. So if you know the value group and the residue field, you still don't know the whole field, but there's a maximal one with this given value group and residue field. But actually, if you just fix the value group, then what I said is also true, because the residue field anyway is a subfield of r. So for a fixed value group, there's a maximal model of RCF that has this fixed value group. And this is the idea that one could hope to have something similar, where here the value group is playing the role of my ordered set. So one could hope that you somehow have ordered set, and they don't give you the isomorphism type, but say if the structure is this tall, they will give you at least there's a maximal one that has those. But maximal you mean? Maximum under inclusion. But I haven't really thought about this, so if this looks very vague, it's normal. But this is kind of the idea. When could look at algebraic structures? So this is also very vague. But for example, there's this conjecture that if you have an NIP unstable field, maybe it has a definable valuation, well, if you take such a field, it's going to have a generic type. If the generic type is generically stable, then I don't know what to say. But if the generic type is not generically stable, then some kind of upgraded version of this theorem is going to give you an order such that the generic type is going at minus infinity along that order, which you expect the type to be the generic type at minus infinity along the valuation. So this is going to give you an order such that the generic type goes at minus infinity along that order. And then using the fact that the type is generic, you could hope to somehow create a valuation by looking at stabilizers of convex subsets of the order or something like this. But. So k is a fixed type of field. k is an NIP field. And I'm assuming this generic type is not generically stable. Oh, it has a generic type. It's an F-generic type. F-generic type. I've added two groups. For example, you can have one that does for both. But OK, this is speculative. I haven't really had time. But it's just to give an idea of why this is likely to be useful to give us strong structure theorems, hopefully. But there's some big things that need to be figured out first. Anyway. So now what I want to do is I'd like to give you some idea on how this is proved. And I usually don't like giving proofs. But the reason I'm doing this is because the proof is probably much easier to tell than to read. And I mean, it's the kind of thing that it's not so hard to actually sketch the proof. But it can be very hard to read it. So I think it's worth actually doing it. I'm not going to do the general case. So I'm going to just do a special case that will give you the idea of what's going on. And you'll see there's just one idea. It's not so hard. But we have to play a little bit with indiscernible sequences. And so I need to recall certain things about indiscernible sequences in NIP. OK, so probably you know that if you have, so let's just, from now on, assume that my theory is NIP. And one of the first things, usually, that one learns is that if you have I an indiscernible sequence, say this is I, and you have a formula with some parameter. So the sequence is indiscernible over the empty set. It's not necessarily indiscernible over the parameter B. So when I evaluate that formula along the sequence, then it's going to cut out a finite union of convex sets. So this formula, if x varies along the sequence, I'm going to have finitely many convex sets such that, say, the formula phi of xB here is true. And then here it's false. Then here it's true. And then maybe here I just have one point where it becomes false. And then it's true outside, something like this. But the point is there are only finitely many changes of truth value along the formula. So this is usually one of the first things that one learns about NIP. And it's equivalent to NIP. And this is strengthened by the so-called shrinking of indiscernible. Maybe I'll move here. Which explains what happens when you evaluate not a formula with one variable, but you look at the formula with several variables, each variable having the same size as a tuple of the sequence. And you want to know what happens there. And then the conclusion is essentially the same that there's a finite, that there's an equivalence relation with finitely many convex classes such that the truth value, if you evaluate the formula on a tuple from the sequence, the truth value only depends on, if you look at your tuple, say it's an increasing order on which equivalence relation each element lies. So let's write it down. So i is indiscernible. You have any formula. Then again, you can cut. So maybe some classes have just one element, or finitely many. But there is a convex finite equivalence relation, E on I. So what I mean by convex finite, I mean the classes are convex, and there are finitely many classes. Well, such that the truth value, when I evaluate it on the tuple, so I take a1, a2, a3, a4, a5, and I evaluate my formula on that, then to know whether the formula is true or not, I only need to know if the tuple have taken an increasing order, I only need to know in which classes each of them lies. So such that if, say, a1, an, or an i in increasing order, b1, bn, or in i in increasing order, and ai is equivalent to bi for all i, then the formula is true on a1, an. OK, yeah, shouldn't have called b. If and only if it's true on b1, bn. Yes, I'm allowed to take two in the same class. And each of those classes is going to be definable, I mean relatively definable by an instance of that way. So there's a unique, maybe I should say, well, there's a unique finest, yeah, courses one. There's a unique courses one that has this property. So if you do this. So now if you have i an indiscernible sequence and you take, say, a finite tuple b, you can do this for every formula. So every formula is going to give you a finite convex equivalence relation. Now you should think of i, usually indiscernible sequences, especially in stable, they're indexed by omega. That's not what I want to think about. I want to think in a, in a, in a, in a, in a, in a, in a thinking about indiscernible sequences that are indexed by a dense linear order, which is usually very big. But definitely dense. So you should think of all indiscernible sequences as being indexed by a dense order, so that I don't have problems of consecutive elements, things like that. Then if I have a finite tuple b, I can take every formula phi. For every formula phi, I get an equivalence relation. And then I take the intersection of all those equivalence relations. What I get is an equivalence relation with convex classes on i, which now has maybe two to the parameter. The parameter is fixed. i is fixed, b is fixed. Yeah, thank you. Yeah, b is fixed. But phi varies. So now what I get is I get an equivalence relation. So we get convex equivalence relation eb. So it depends on b on i with at most maybe two to the t classes. Such that what? Well, such that the classes, there might be some finite classes. But the infinite classes are mutually indiscernible. And actually also over the finite classes. So maybe older classes, but the finite ones. So the infinite classes are mutually indiscernible. What does it mean? Over b. Yeah, thank you. They're mutually indiscernible over b. What does it mean to be mutually indiscernible? It means if you take two of them, each one is indiscernible over b and the other one. Or well, if you take any number of them, each one is indiscernible over b and all the other ones. OK, so maybe you have to think a moment why this gives you that. And this is really what I prefer to think about. So the way I think about this, you have this sequence. Ah, it's indiscernible, so you don't see any. It's the same everywhere. If you add a finite parameter b, then the sequence is cut into a bounded number of pieces. And those pieces are again mutually indiscernible sequences. OK. So now there's a very nice situation is when you only have finitely many such classes. And in particular, if you only have two such classes. And so there's something that measures that, which is called the DP rank. And I guess I'm only going to define it for finitely many. So in the over categorical case, you have to define it. In the omega categorical case, in the finitely homogeneous case, it will be finite. No, otherwise, no. No. Because they can still be infinitely many formulas with more and more variables. OK. So the DP rank is going to bound the number of classes that you can have. So say that the DP rank of p is at most n. Let's do it only for an integer. If for any indiscernible i and any b realizing p, well, if you do this, you only need to cut into at most n classes so as to get this. So it's not completely true. Well, first I guess it should be n plus 1. And yeah, it's not completely true. OK. What? Yeah. OK. OK. OK. So I'll get back to this. I'll finish this. I want to look a bit more closely at what happens here. So this, I mean, some of the whole idea of the proof rests in understanding the minimal situation. You have an indiscernible sequence, and you have a topol b. The topol b breaks the sequence into many pieces. Let's look at the simplest thing it could do. Well, there are actually two simplest things that it could do. One thing is that i breaks into two infinite pieces over b. So maybe over b, i breaks at some point into two infinite pieces. And imagine that there's no limit point on either side. I mean, there's no end point on either side. And there's some formula with parameter in b that's true here and false here. And that's it. The two pieces are mutually indiscernible over b. OK. So this is the first simplest situation. But there's another one that could happen. It could be that there's just one point that behaves badly. And that if you take off that point, the sequence that remains is indiscernible over b. And the fact that there was those two minimal situations is the main observation behind the definition of distality, actually. So you should think of this situation as being stable like. Because what happens in a stable situation? If the theory is stable, the sequence is totally indiscernible. And you have this stronger property that if you get a finite set, you can just remove a bounded set from the sequence. And whatever is left is indiscernible. So when it happens that you can take a point and you remove it, and the sequence i is indiscernible over b, let's think of this as a stable like situation. And when the sequence is cut in two and there's some formula that's true, below, and false above, then this is like an order like situation. So distal theory, it's not important for this talk, but the distal theory is a theory where this never happens. When you never have this situation, you only have this one. So now if you want to, so why was I in trouble here? Because the number of classes is not the correct number you want to look at. If you have your sequence and you have a parameter b, and let's say it cuts it into finitely many classes. But there could be cuts that are order like cuts, where some formula changes truth value, and then there could be isolated points that you need to take off. And what you want to count are really the number of those isolated points and cuts. This is the relevant number. So you see a point gives you really three classes, but I want to think of it as just two. I want to think of this as one thing that happens in the sequence. And if you count that, then the Dp rank is good. And any b, they are at most n cuts in i over b. And what I mean by a cut is either one of those two situations. So either there's a formula that changes truth value, or there's an element I need to take away. Well, this doesn't happen to consecutive because my sequences are dense. So then the Dp rank is at least, oh, sorry. It's n minus 1. So if it's 1, no, it's good, actually. So the Dp rank would be 2. So it's known, so this is proved by Tai of an Alex, that the Dp rank is additive. Sub-additive, yeah. Thank you. I hope I'll write it correctly. Looks correct. So the Dp rank of a topol, AB of the type of a topol, is bounded by the sum of the Dp rank, we should say rank, the Dp rank of a over a, and the Dp rank of b over a. So something you would expect. It's not so obvious to prove, but it's true. And of course, there's one natural thing to do with the picture that I drew here is to split the Dp rank into two parts, just counting the number of, you could just count how many of those things you can have, and you can count how many of those you can have. And this does work. It does give you two notions of rank, which also sub-additive, each separately. I'm not going to, in this talk, I'm not going to use them because what I'm going to do right now is restrict to the case where the Dp rank is 1. So I was willingly being vague about that. They're cuts given by a formula with one variable and parameters in b plus maybe other parameters from the sequence itself. That accounts for the formulas with more than one variable. So the correct way would be to define it like that. But then, if you'd have to use the sequence, you could get arbitrary many funny things going on. Yeah, yeah, but you're only allowed to cut, yeah, but still, it's going to be bounded. As long as you don't cut on one of the parameters themselves. But let me tell you what the Dp rank 1 case is, and that will be. So from now on, we restrict to the case where the Dp rank of any one type, so the Dp rank of the structure is 1, which means the Dp rank of any one type is 1. And then what does that mean? It means if I take an indiscernible sequence i, and I take now a single ton a, then either the sequence stays indiscernible over a, or there's a cut, and I get two and the two infinite pieces on the left and on the right are mutually indiscernible over a, or there's one point that I need to take off, and when I take it off, the remaining thing is indiscernible over a. OK, and that's a precise statement. This is equivalent to being Dp rank 1. So from now on, I'm going to explain the proof in the Dp rank 1 case, which still shows the idea, but simplifies the number of technical things. And then one has to understand how to restrict to that case. If you could just find the type of Dp rank 1, you'd be good, but that's not always possible, so you have to do more complicated things, but this I won't explain. OK, so the first thing is we want to be in this situation. This is the situation that's order-like. But we can, because if this situation always happens when you take any indiscernible sequence and any single tone, then the theory is stable. So if the theory is unstable, you have this situation happens at least once. If this always happens for every single tone and every indiscernible sequence, then the theory is stable. Because stability is restricted to formula with one variable, and you can test stability by evaluating. So if T is unstable, it is some I and some single tone, A, such that and a formula. So let's write it all out. And there's a formula. OK, let's call it B. There's a formula for your FxB. And here I'm going to assume there's no other parameter in phi. So to answer that, maybe I would need to add things to the base to ensure that. But everything I'm doing is invariant on the adding finitely many constants. So I can certainly get down to this situation. There is I, B, and a formula phi of xB, such that phi holds on the left on an infinite initial segment, and the negation of phi holds on the final segment of the sequence. So I'm in this situation. Just to emphasize that it could be tuples, that the sequence is not B is a single tone, but the sequence might not be. But then I'm going to forget about the bar from now on. So is this OK? You can forget about almost everything and just so that we are done to this situation. We have this indiscernible sequence, and we have one formula that's true, that's not a formula. And we have one formula that's true, then false. And also we have this DP minimal that will play an important role later on. We want to get an order from that. So there are two natural things to do. One natural thing leads to Scheller's theorem that there's a definable partial order. And what is Scheller's argument in this case? It's saying, look at the type. So this is an indiscernible sequence. I can look at the type, let q be the type of an element that fits here in the sequence. So that if you add it here, you get an indiscernible sequence. Let's look at that type q over the sequence. So q over the sequence, which is the type of an extra element in this cut here that fits in here. And you can define naturally a partial order. Well, you have a formula that has the strict order property. Yeah, I didn't need to define q actually at this stage. So if you take a1 and a2 here, it is not possible. OK, let's define p to be the type of b over the sequence. This maybe will be the type more important for me. So it's not possible to have a b prime that realizes p, such that negation of phi of a1 b prime holds and phi of a2 b prime holds. If a1, a2 are two elements such that when you add them both, you get an indiscernible sequence. Why is this not possible? Because look at what the formula phi does. It would be true here. Then it would be false on a1. Then it would be true on a2. And then it would be false again. But that means it changes truth value. That would contradict the p minimality. It changes truth value too much. So this is not possible. So you assume that an a1, a2 preserves this element. Yeah, yeah, I'm assuming that both. So maybe I shouldn't write that. I'm not assuming that not only that they both satisfy q, but that together they fit in the sequence. And then it's inconsistent to have this, which is exactly saying that phi of a1 y and there's a phi of a2 y and there's an inclusion between them, which you probably know what it is. So as a2 goes there, this is bigger. So this should be bigger. And the strict comes from the existence of b. I can find a point that changes truth value between them by indiscernible. So this is how I get the strict inclusion here. And the inclusion is just this statement. The y steps compete, right? Yeah, I think it's correct. OK, so this is the proof of Scherlach's theorem in this easier dp minimal case. Now to get the linear order, we're actually going to look at it a little bit differently. So Scherlach's theorem gives us the partial order on realizations of q on elements that fit in the sequence. I'm going to construct an order in realizations of p. So p is the type of the parameter b. So almost done. There's a little bit more. Just a little bit. Yeah, I know. I approach us for that. OK, so what is the natural way to try to define an order on realizations of p? So if I have b and b prime, if I look at phi of xb, it's true here, false here. Phi of xb prime is true here, false here. That's because they satisfy the same type of the sequence. But now what I can do is I can add another piece of the sequence in the middle. And then phi of xb, let's see what happens. Phi of xb is true here. So let's look at bb prime. So phi of xb is true here. It's false here. And well, it's the same argument I had over there. By dp minimality, there's one point where it changes truth value, where it's from true to false. Be a cut that is sort of indexed by b. Now there's also one point where phi of xb prime changes truth value. A cut, a cut. Sorry, a cut. There's a cut in the sequence where the formula phi of xb prime changes truth value. And now if this happens, I want to think of b as being smaller than b prime. OK, now this doesn't quite work. But let's see what does work, what we can get. So first we can define, well, there's a natural equivalence relation that I won't actually need, but that will turn out to be this equivalence relation. Which is, I can say b and b prime are equivalent. If no matter what I put here, the cut is the same. But you're assuming b is the same type of an i. I'm assuming they have the same type of an i, yeah. P is the type of an i. P is the type of an i. P is the type of an i. It's on the right place. P is the type of b over i. So P knows that the formula is true on the left and false on the right. So there's a natural equivalence relation if the cut, so let's call this the cut defined by b, and this is the cut defined by b prime. So the natural equivalence relation is if cut b equals cut b prime for all, say j, let's call j, whenever I put a piece j in the sequence, the cuts are the same. That's an equivalence relation. Inj, the cut in whatever. Inj, yeah, in this thing. OK, now my strict order, what I want to be a strict order relation, is going to be r of a b prime. And here you have to be careful if you cannot have the cut of b prime less than the cut of b. So if we never have cut of b prime less than cut of b. So no matter how you put j, maybe the cut of b and b prime coincide. That happens in particular if you take j equals is empty. So that's going to happen for sure sometimes. And it could be that b prime can be separated, say, but you cannot have the opposite. So wait, is that what I want to write? No, this is the large one. Yeah, this is the large. Sorry, so this is what is going to be the large relation, the non-strict. And note that this is v definable because it's saying you do not have something. So the complement is there is something. So this is v definable. It's obvious that it's infinite. I mean, that's probably the typical problem. Wait, that's the question. No, no, the fact that the quotient is infinite is obvious. Because it's indiscernible, so whenever I put a j, there is some b that has a cut here, there's some b that has a cut here, there's some b that has a cut here. And those all give different classes. So yeah, so what is obvious? That the quotient by a is infinite. That's obvious. The other thing that's obvious, a little bit less so, but you just have to write 1, 9, is that this is a transitive. And this is just one line. If you write it, maybe I don't have time to write the line. Yeah? Yeah, yeah. What? Yeah, yeah. What? Yeah, yeah. What? What do you want to write? Oh, I have other things to write than the line. Yeah, yeah, what is the question? The formula is always the same phi. Formula is always the same phi. We just vary the. Yes. The formula phi is fixed once and for all. OK, so I claim that this is very easy to check just from the definition. So what is not clear is linearity. So what does linearity mean? It means our two points comparable. So what does it mean that two points are not comparable? It would mean that you have b and b prime. And there is some j such that the cut of b lands below the cut of b prime. But then there's also some other j prime such that it's the opposite, the cut of b prime is below the cut of b. And if you have that, well, you can't compare them. But now, so this is sort of the crux of the proof, is to observe that if this happens too much, we're going to contradict an IP. And what does too much mean? So this might happen. At this stage, it could very well be that this is not yet linear. So if it's linear, we're done. We have all the properties. If it's not linear, it means there are two realizations of p that can be switched by just changing the thing in the middle. So then what we're going to do is we're going to pick one of those realizations, add it to the base, increase the sequence, and then iterate and work here. So you don't need to worry about this. Eventually, if this process keeps going, you will get a sequence like this of points which are all realizations of p. You will get an indiscernible sequence with the property that, well, if you look at phi x bi, it's true up to this cut and then false. And with also the extra property that any two consecutive ones, you can change what's here, put another sequence, so as to switch the two. And now, what we want to do is just show that then we can rearrange this middle piece so that to get any given order on the base. And why can we do it? So this is sort of the main technical thing, is to understand why this works. And the reason is because of this DP rank. So I'll explain it. What do you need to do? What I want to do now is I want to, OK, so maybe I should say the conclusion to the end. So now what I claim is that by changing the sequence in this middle piece, I can arrange to get any permutation of the cuts that I want. Why does this contradict an IP? Because then the formula phi of xy would have IP. Because now if I just take a point here, it's related by phi to everything to the left and to nothing to the right. But everything to the left can be just any subset of the bi's that I want. So if I can do that, I'm done. Now we have to understand why we can do that. And the reason is, so this additivity of DP rank tells us that for the DP rank of, I think I'm only going to need it for 2, the DP rank of any two points is at most 2. Because DP rank of each is 1. So it's actually exactly 2. Because if I take any 2 of, sorry, those are b's. If I take any bi and bj, I do have two cuts that they induce on this sequence. So the main, OK, so now I'm going to do this permutation inductively by switching consecutive points. So I'm going to explain one switch. So say you are at a different situation. You have here some bi. So let's do it just at the start. I want to exchange b3 and b4. What I know by construction is that if I forget about all the other points, I can erase this, put another sequence that exchanges them. But I have no idea a priori what happens over the other points. Maybe the other points now, the sequence changes. Maybe the type has changed. But DP rank will tell me that this cannot actually happen. Because now if I take, so now let's take b3 and here I have b5. So in the new sequence, if I look at what happens over b3, b5, if I look at the new sequence, it's here. I know two places where the top of b3, b5 cuts the sequence, here and here. So there cannot be any other. So what this means is that this whole left piece and this whole right piece have to be mutually indiscernible over b3, b5. And what this means is that this piece that I've added, b3, b5, they don't see anything here. They don't notice that anything has changed. And therefore I can just do the same thing here and replace them and move this here and iterate. And it contradicts an IP. And that's it. Thank you. Thank you. Thank you.
theory, well, that any unstable NIP theory defines or interprets an infinite linear order, definable relation such that this defines a