 record. All right, everyone. Welcome to this analysis seminar. It's a pleasure to have today Cristina from the University of Barcelona. She's going to be talking to us today about on the number of lower sets with the fixed card event. Cristina, glad to have you here. First of all, thank you very much for organizing this visit for everyone who was involved in this. Thank you very much. Okay, let's start with this definition of integer partition. So what of it? We have a natural number and then its representation as a sum of non-increasing positive integers. It's called an integer partition of n. So let's just consider some examples. Let's say mv and r18 and let us decompose in line over 7 plus 4 plus 3 plus 2 plus 1 and 9. So it is an integer partition of 18 and we can represent it also via the so-called non-diagram. How do we do this? Let me just draw this staircase actually. So the first layer will have seven steps here. The next one is 4, then you have 3, 2, 1, right. So this staircase is actually called the n diagram and this is a representation of this integer partition. It's easy to see that any integer partition can be thrown in this way and vice versa. If you have a n diagram, we can recover the partition. So the important thing is that everywhere here must go smaller and smaller. Okay, so we know a lot of things about this integer partitions. Most of all, we know the generating function but if you stop to think a little bit, you can find this very nature and you can understand why it's the generating function. And we know, so yeah, I didn't say that p2 of n is the number of integer partitions of n. I think I will say something about this too for the next slide. For now, just picture it. And we know the asymptotic. So by having the Ramanujan, we know that p2 of n is for the main part, this is the exponent square of n times some constant. Okay, now we go further and we have these play partitions. So basically it's the same, but now every term in the sum has two indices and you require this monotonicity in two directions. And we will see an example. So how should we understand this? Let's stop here a little bit. So we have mij and let us think that i will stay for the, say, the green axis cube and j will be the red one. And then the value of nij will be the blue axis. Okay. So what do we see? Okay, m11, what is this? So the very high column here is three cubes, right, the very beginning. So we have m11 is three. Then, okay, we fix the green axis and we go along the red one. So what do we have? The next thing is two, right? So m12 is two. Then you have m22, right? One and then m11, yes. What do we have? m13 is one and m14 is one. Okay, then we will go further on this green axis and we have two m11, right? And one and two one is two and two is one and two is three. The next one is just this two and one. So m31 is two and m32 is one and the last one is m41 is one. So we recover this plane partition from this figure here. So you can also know that if you just say, okay, let's be the green axis with the axis of values, then you still have a plane partition, right? But another one where there's still a plane partition, okay? And also that's what is important. What happens if you just choose an axis you want and you slice and you set what you obtain? Well, basically, okay, if you fix one here, the first coordinate, you have a diagonal, right? So you have three and two. Oh, no, it's the wrong thing, right? Three and two and one, one. It is an integer partition, right? So every slice you get is an integer partition itself. And every next slice is smaller than the previous one and it's contained in it, right? So these are things that we have to understand before we do anything. And, okay, we know also the generating function here in this way. And we also know the estimate process. So here in the main part is exponent of m with our two thirds and some constant. So we don't have a D dimensional definition, but we already can make some conjectures about the rules on this value. Okay, let's see the definition. So D dimensional partition is basically when you have D indices for every term and you require this one ethnicity in each of these D directions here. Okay, I can ask me what about generating function? Okay, so far nothing. So there was a conjecture made in McMuffin. So you see here my minus k here. In the previous case, there was minus one. So his idea was that the right thing to put here is like binomial coefficients depending on this D. And so it was staying as a conductor for quite a long time. But then it was proved that it's the wrong thing, even for the case four. And however, there are some numerics that show that probably the order is the right one, but nothing more. So okay, we don't know anything about this about the generating function. And okay, now what's the lower set because my title was lower set. It is actually the visualization of this partition. So in one dimensional case, when the partition was one dimensional, we had this two dimensional picture here. The case of the in partitions, like two dimensional partitions, we had this, okay, yeah, we have this three dimensional picture here. Okay, so now this two in p12 m stands for the dimension of this young diagram. And then in the case of three, it's the dimension of this figure here. So D will stand for this, the dimension of visualization. Okay, but we didn't find this formula yet. Let us do this. So what's the lower sense? It's basically a set, you can define it. So if a sub x belongs to q, then any other point, which have, which had all the coordinates, so we could then better x must belong to ourselves. So it's lower sense. If you have this x belong to q, then a x prime, that is x i less than x i clear, then we must have x prime in the set also. Okay, so how do I understand this? So in this dimensional case, we saw this third case here. And the definition from the beginning was like, I have every layer here less and less and less, but I can define it differently. If I cube here, the set here, give me my set, then all the cubes in this rectangular, not being my set also. So then I define this lower sense, and I will work with them, and not with the partitions. Okay, this will be more convenient. So yeah, you can see that you can do both points, and you also can think of cubes like this. So I mean, it's difficult to understand like d for height values of d, but you can think like you have, you can draw this hyperplace, the coordinate hyperplates, and then every cube, every unit cube you have in your set, you're either touching another one in each direction, or touch the the wall that the hyperplates. Okay, so that's the constant. And yeah, so now d has the right to stay here because it's the dimension of this lower sense. So d dimensional lower sets correspond to d minus one dimensional partitions. Okay, we are interested in this number. Okay, why? Well, first of all, this, the, this structure of lower sets, surprisingly, it appears in a lot of physical settings, like crystal growth, and the microstates in regards to particle inputs, harmonics, oscillators, much special as to this space. But the thing is that this quantity usually appears in physical quantities. But okay, there are some things you can do, you can apply this in mathematics also, in approximation theory at least. So if you have, sometimes it's good to approximate some function by a trigonometric polynomials, and it may sense that the harmonics of your polynomials stay in the lower set. But just by some observation, you can, you can understand this, if you have some harmonic here, you should have all the harmonics, okay. So yeah, what do we know about this video? First of all, we know this theorem by Bracev-Rasevinarov, which was also proved in physical work. So if you fix the dimension when you leave, then you just see that your logarithm of p of n is up to a constant, and to the power one minus one over t. So that's what we saw in this two-dimensional case, three-dimensional case, right? So this is the right thing. However, they didn't care about the constant at all. So these four applications, which is not very good. And for the purposes, so Diet Riemann-Schadt and Siegfried-Consumeric, they provided some explicit values for these constants. So first of all, you can say that this thing here, you always can think of it as constant to the power d, to the power lower d. And if you assume that your n is largely not, say, constant to the power d, then this c1, you can think of it as a constant, actually, because when you write this thing down, you can understand that it's quite at least one, always. So then you are dealing with this lower constant here. So you can think of it as constant, because usually for applications, not always, but usually you're not n being large enough in terms of the k. But the problem with this c2, which is quite huge here, so the main results we'll be talking about today is this one. So once you have n being d to the power d squared, you can just obtain that your logarithm of pd is up to an absolute constant, n to the power 1 minus 1 over d. And if for some reasons you want to release this condition to d to the power d over d, then you can put here additional d squared. Okay, so I will tell some things about ideas on the book. And we will need to introduce a notion of available subset of a lower set. So what is it? Let's draw something here to the mention once again. Please, we will call a cube available if we can take it away without, like, disturbing the lower set structure. So if I take this cube away, the remaining set will be lower set. So this cube is available. This is not available, because if I take it away, I'll get a hole. I can't get a hole. So actually in this set we have just one, two, three cubes available. So this set I will call the maximum available subset of this cube. And this is the definition. And n of q will be the maximum available subset. This is n of q. Now it will be interesting in estimating this point. And the n of q can obtain is that if you have n large enough, like d to the power of d, and if you have a lower set, but cannot add, then you can see that the cardinal is always available subset is at most some absolute constant times n to the power of 1 minus 1 over d. Actually, this is what we would expect. Why? So imagine you have a very strange lower set, like very low here, very low here, I don't know, something here. Would you expect to have a big available subset? No, right? If you can take away these cubes and put them somehow here, and you'll get much more flexibility here, right? So the intuition is you should uniformize your lower set to get this available subset more, right? And then, well, which sets are the most uniform possible? Okay, this one, right? You can take this set, or this set, or this set, right? So actually, what these sets are? So I have x in my Q, if and only if the sum of its order is less or equal than some prescribed numbers, right? And if you stay thinking and calculating, you will see that these sets have exactly this, I mean, after some absolute constant, they have these available subsets. Okay, now we will not prove the lemma, but I suggest us to prove something weaker. Actually, not weaker because in the previous thing, we had this requirement that n must be large, but now we, without assuming anything on d and n, we will prove that it's less than d times n to the power of one minus one over d, okay? Okay, we will do this by induction on the damage. Once again, we'll start with this two-dimensional case, good one. Okay, I have some Q in two-dimensional case. I claim that we always have, if I have a point x2, x1, x2, then there must also be either this x1 minus x2, it will be large. This or x2 minus x1. Okay, if not, then I have a point here, which is this one. I must have all the points beneath, so I already have points from 0, 1, 2 and a 10, the first one and the second one the same, so I have at least this number of cubes in my set, right? It's not possible. So okay, I have either this or that. This means I can put my set into these two straps in reunion and let me call these parts, which belongs to this strap, the one and the other one will be Q2. So we can think that you can see that these two sets are lower sets, right? Because you just cut apart the weight and that's that's all. So you have Q belongs to the reunion of Q1 and Q2 and the same you can say about the available substance, right? If it's available here, it must be available here, right? So here it is. Here you go. And then the current out of this guy is less than the sum of current out of this here. But what can I say about this guy here? It has the first coordinate being either 1 or either 0 or 1 or 2 of these numbers. And it's not that I cannot take away two cubes which are in the same first coordinate. This is one of these, one of them is beneath and then they're ordered. So there will be a hole. So the maximum value of this guy is square root of L, right? And here the same. So I have that the current out of the hole line is x1 to square root of L. And that's what I wanted. Okay, now I need to proceed my induction somehow. And so let's begin to the picture then. So remember we noticed that we can slice our set so that it became this. Every slice is a lower set itself. And then each next slice is a subset of the groups. Okay, now what I do, I will choose an axis I want. I will slice my set and I will get slices of coordinate as L0, the N1 and so on until some say Nm. Okay. Okay, for each slice I can do something. For example, I can use the induction assumption to get it on case set. What I have, I have there was this, we're proving this. Now what I have with d-1 instead of d is this thing. But also remember that if I have the scale level, then I have table at 1. So this is a slice at here. And it belongs like it lies above this side. So if I take here away q, which also belong here, there will be a hole. Okay, so I can't do that. Also I must have at most this number of cubes. Okay, I have two estimates. This one from the induction and this one just from general reasons. Okay, now I can write the n of q is at most the sum of k from 0 to n, just for the sake of simplicity, let's take that and we'll start with the new. And here I have minimum for the first guy things and this difference. Now I will break this sum into two parts. So the first part will go from 0 to this number like m to the bar 1 over d minus 1. And I will use the first estimate here. And the next sum is just a scope. Okay, I will just use this and I will get m with this. What can I say about this? This is a slice of number m to the bar 1 over d and it's less than m0, it's less than m1 and so on. So at most this can be n divided by this number. So the least in the among all first n to the bar 1 over d slices. Okay, and this, what can I do? The sum of nk is at most n and this power is less than 1, so the maximum is when all they are the same. So I have m to the bar 1 over d things here and I have this way. I will not deal with the full function here because it's boring, but what we get is this thing and it's exactly what we want. So we have 1 here and d minus 1 here and we get what we want. Okay, so we wrote some simple version of the lemma and recorded in our original lemma we had m of d, but an absolute constant here. Okay, so if we denote this tm being the maximal cardinality of available subsets of all sets of cardinality m then when n is large enough we'll have this as and then what can we do? We can estimate the number of lower subset of a set. So if we know the number of cubes will take away from this set, the original set was with cardinality n and then we take away at most k cubes and the question is how many lower subsets we can get? So this is an estimate for this number which depends on this t of n which was the cardinality of available subsets. Then we plug in these estimates and we get something like this, but let me give any intuition how do we do this. So I have a lower set once again let me do the mention. It's lower subset just for the sake of whether it would be like this, a and u and it's subset. I want to come from a to b discarding some cubes here, step by step but not one by one. So how can I do this? I will mark all the available cubes so here I have this one, this one, this one, this one and I will see which ones I can take away and which ones must stay to form b at first. Okay so this one must stay, I will not touch it and this one also but these two I can remove. So I remove this one, this one I can choose from 8 primes. Once again I will see the available subsets, this cube, this cube and this two cubes. So these two cubes must stay forever so I will never touch them but they are participating in my available subsets always. So okay I will remove this one and this one and then once again I will see the available subset, this one and this one and I will discard them and finally I will discard this cube and I will come to b. So basically if I know the number of cubes I want to remove and if I know the cardinality of this maximal available subsets only in which step and I also remember that these two cubes they give some value, additional value into this cardinality of available subsets but they are never touched. So if I take all these things into account I can do something here and prove them. Okay basically then the idea is to split your initial lower set into some tricky way into different lower subsets to apply this estimator, this one and the level one after that but I will not enter into details here. So this is about this theory but now I'd like to say change a little bit the sub topic so we're going to talk a little bit about five dimensional lower sets. So remember here we had like m being constant to the power of d and sometimes like d to the power x squared okay forget about this. Now I don't know anything about the relations between n and d so what can I say from nothing I can say this. So this is the theorem by Cohen-Mirat and Navier they say that every in every case you have p of n that's the ripple down to the power 2 to the power dn and in independent estimate which is d to the power n minus 1 n minus 1 look for it. Okay so it turns out that the the second estimate can be refined it was done by that remark shall be taken with 0. So you can just discard this n minus 1 factorial which is that's right here and now just and you put this variable coefficient on the left hand side but now just for the sake of curiosity you have if you assume that you fix the current value of your set and you just write the dimension when you where you leave then d to the power n minus 1 is the right thing this n minus 1 factorial is just constant in this case so but this case is not reasonable but I mean if you assume this you have the right order. So now we will talk about the case when actually d is quite d so at least m divided by some power of m and we have some estimate in this case so first of all this is the very very unnatural case when d is greater than m cubed then you have super sharp estimate so the right hand side you have this variable coefficient from the previous slide and then you have you can estimate from above where the same thing doubled. Okay now there are three estimates the first case when d is much bigger than m squared than when d is of order m squared and then when d is much less than m squared but still with this property here basically you have this sharp order estimate for the lower and yes I would like to prove the first part because it works here. Okay how do you do this just a few minutes ago we were deconstructing these over sets and let us now construct it from the beginning so how do you do it you have n cubes you have to put them somehow so first of all I have no option I have to put the first cube at the origin okay but now you have still n minus 1 cubes and let me do the following I will choose any number j from 1 to pi minus 1 and then I will put j cubes along the x so come here here here okay how many ways I have to do this I have d axis and j cubes to put so it's the the binomial condition d minus 1 plus j choose minus 1 okay now I cannot put any cube more along the x so this is prohibited you know what does it mean any cube now must have at least two non-zero coordinates it means that it must lean on some two previous cubes right okay any two cubes they cannot generate more than one place to put another one and what does it mean I have just j cubes here and I have to choose two to put any one so at most I have this number of possibilities like j choose two okay the next step I already have j plus one cubes and the number of ways to put a new one is at most and so on and so forth I come to this thing so the maximum the maximum number of ways to construct any one was set is to sum this over all j and okay let's do now with this quantity here by a j and then if this is very big you can say that the every next a j is at least twice the previous one that means that I can estimate that the whole thing by twice there's the last guy so it makes sort of sense because if you have too many axes then it was putting cubes along them and not one into another right so that's the very rough thing but if you want to obtain other estimates you will have to do in a more delicate way so you have to create your steps differently so okay first of all you put this one cube at the origin but then I will put just the cubes which has the form zero zero one zero zero so I will choose the axes which will stay in my game and the other axes they will not participate anymore okay so for example I choose this one this one and the next step I will put all the cubes all the form zero zero one zero zero one zero zero or zero zero two zero zero okay so those are the cubes with the sum of coordinates equals two and each step will be just putting the cubes with the sum of x i being k and then you can see that each cube from this set must lean on the cubes with the sum of coordinates equals k minus one okay and this gives you I mean you can carefully estimate this and if you have like sk cubes so you don't have to read this if you have sk cubes on the k step you can estimate that the number of places to put new cubes that most this one so the number of cubes from the previous uh on the previous step squared over two and to choose then sk cubes you have this binomial equation the minus minus squared that's k and now you have to multiply all of them and so p of l will be the sum of these guys here and then the the problem becomes just purely analytical and then you have to analyze these quantities but yeah yeah for this all I wanted to say thank you very much this thing over there won't have questions or comments in the in sub step you mentioned that the cube is available in the sub in the subset uh can it be also available in the cube like in the in the original set uh like when you when you consider nk minus nk minus one yeah you use this fact that you cannot remove from the lower set like I was thinking there might be some cubes in the corner like in available cubes that that are available in both like uh no it's it's possible because they're orders so they are like for example in each of them mentioned this okay so if back here the cube which appears also here it cannot be available yeah that's true but if you just remove the one available cube from this you get the subs like subset yeah if you want to say that if you remove this one yes now that but the the definition of available cube is that you just remove this one and then it's available but if you remove if you have to remove another one in order to remove yours it's yeah what I'm talking about is not that one but the other available cubes uh for example the one on the lower corner this one yeah this is available in both cases right yes and how did you count that that available cubes in nk and nk minus one so I have here n1 yeah and zero it's just four right and three it's oh and one is three so I can just remove the cube which is here but not here and the number of such cubes is and the roman is in one here is just one but all these cubes have to stay because I have something leaning on them but this is just the cubes which appear here but not here and the number of them is that bit of difference okay thank you for the questions yeah you're talking about that part where she used an estimate of a meaning you know yeah one of the things that comes from and you comment a little bit on this generating function stuff for christine yeah beginning so what was the conjecture for the generating function of before so yeah so it has why it's actually doesn't work why it doesn't work I mean it says like what's the best since it doesn't work do you have any kind of new projectors about what the what generating function could be so the first case it was all right yeah it had product here from minus k and here i had minus one so in the second case it had minus k so the conjecture was so this is like two-dimensional things this is three-dimensional and here we have the power which is the binomial coefficient right k zero so here you have k one so the the conjecture was that you have k d minus minus two okay like this but it took a long time to understand that it's not the right thing so there are a lot of numerics and i mean no addition for me at least i'll be from there there's no conjecture just for the case four for the next case which is unknown i don't think so possible generating function yeah that's very interesting so there are people who just they do these numerics and it's not not so easy this gets complicated very fast right the numbers get big very fast yeah and there are tables i mean you can find some tables for quite huge numbers like in six dimensions but i mean that doesn't give you a clue how to think of it it's formal of rtm ramada channel it's very nice okay more questions well if not let's thank you christine again for the nice talk thank you christina thank you