 Should I spin this one, learn the marché. Okay, thank you everyone. So for this last lecture, we'll start by giving you references. So here I mentioned two books to learn about the theory of free probability. There is the first one, which is very abstract, and you can pick a lot of materials independently in different chapters on it. I love this book, but there is not a lot of connection with random matrices that you will find in a second book. So Spicer is a contributor of two of these two books, so it is there is really a good continuation. So here you will find a lot of methods, the method of subordination amalgamation, a lot of things that are useful if you really want to compute spectra with your computer, in particular. So the things that you won't find specifically there are there it will be the theory, the combinatorics, if you like it. About what is more specific to what I told you during this lecture is the theory of traffic that I have written a monograph about the subject. So here there is a first introduction it's called traffic distribution and independence. It's the first volume. And then there's a second volume with two other authors, Cébron and Aguist, that you can read to go further into other aspects. But for the content of this lecture today I will talk about freeness over the diagonal. On this connection with traffic independence is made in an article with several authors, which is the team of traffic guys. Benson, I am Cébron and Aguist and Frank Gabriel. So it's a short article where there is not a lot of theory about traffic. We focus on this result of matrices showing numerical applications of how it works. And then the continuation of this aspect of this relationship, we develop it with Jeremy Bigot for the question of finding the spectrum of variance profile matrices and finding the outliers. For covariance matrices or for ambition matrices. So here again, it will be more about the numerics and release mathematical aspects of matrices rather than traffic themselves. Okay. So do you have questions about the last lecture? If not, let me just remember what we did in a second section. I was starting talking about this weird ensembles for which we need more techniques than the ones that free probability offers. So I gave a proof which was a bit technical that shows a convergence of traces for products of permutation invariant matrices. It involves a lot of techniques including graphs. So as you can feel the price you have to pay is investment you have to pay to understand this is quite high. We won't focus really on this, on this aspect we will try to move this complicated combinatorial results into something which is analytic. And especially yesterday with Mark Potter, you discovered that for the classical cases of Gaussian matrices, there exists fixed point equations called the subordination property in the case of the sum, involving the S-transform, S-transform, T-transform, all these voicoles could transform that voicoles could define 40 years ago, and that we use intensively today to compute spectrum. What we will do today after remark and balance profile matrices is to turn this analytical theory to a higher level, which is the level over the diagonal, a theory that also has been constructed by voicoles could 40 years but that has been applied recently for this weird ensembles. So the equation you have met of the subordination property this fixed point equation will meet the same equation but for slightly different objects. And changing a little bit this definition will change the algorithm somehow, it will be more consuming in terms of computational resources, but it will give a good solution computing spectra of this weird ensembles. Okay, so this is the program and let's start. I have to talk about the second kind of weird ensemble I want to consider it will be quite short and just to show you that the tools that we introduce with the trace of test graphs turns out this problem very easily from the moment point of view. It doesn't tells that we will have, obviously a nice solution to an analytic solution to compute the spectra but at the level of combinatorics and moments, this is quite trivial. Let me be more explicit. Assume that we are given a matrix HN that can be written as a Nadamard product a naturalized product of two matrices, but these two matrices does not play a symmetric role. One matrix is like your favorite random matrix you're studying the spectrum, like a GUI matrix or a Bernoulli matrix if you need sparsity. On the other guy, I will call it gamma n, and I'm doing the entry range product. These guys is a variance profile or just a profile function. Let's say your deterministic function. Once you take a shape like this, and you decide that you put zero here on the one here. Okay. So what you're doing is you're taking your matrix and you're erasing your taking your random matrix. And if you do this enterprise product, you're erasing a part of the matrix. We assume that this variance profile is macroscopic. Should I start again? Okay. We assume that this guy is macroscopic example. And it is not a restricted to this example but let's look at this. We fix a function f, a measurable function from zero one square to, let's say, see. And we set gamma n, it will be a n by n matrix with entry a j is this function that we sample at the point i over n j over. Okay. So I draw this picture you can model this by a function, the indicator function of an ensemble on the square. And then you just discretize your matrix and take this value. Of course, because there is this discretization we will assume that this function is at least continuous by path. If you have a measurable function there is a way to invent a relevant profile, where we not taking a point wise as a value because it doesn't make really sense for miserable function but let's look at this. Okay. You can also take a variance profile gamma n, which will be a random matrix. Another example will be to take gamma and random. It will be independent of XN. And for instance, is one with probability alpha and zero otherwise. So you just destroying randomly the entries of your matrix. So just let me write something which will be just a little lemon that will end the section that tells us how to understand the moments of HN in terms of XN on gamma n. Remember that we define the trace of a test graph. And if you don't care about all the test graph you can focus on these simple cycles. This is just a way to encode the moment. If you take something more general than moments like add them up more products, you see that you can cook the good test graph to do that. But even for tea which is a cycle this is interesting to have this approach. Let me maybe recall the definition of this guy. This is a morning we need to refresh a little bit of what we had seen before. This is a single matrix, let's say, what does that mean. So it says it's represent an expectation of a normalized trace, but it's a bit more general. If you have an arbitrary graph like this, you have the set of vertices for which for each vertex to choose an index, your test graph on code matrices and the ages. You have a single matrix. So you, you put the same matrix on each edge. Okay. And then given this indices, you make the product of the matrix entries that correspond to each edge, which is represented like this in a short way. This is a sample on this mean fire of the value fire. Okay. Traffic theory tells you that there is a decomposition of this quantity as a finite sum. And this finite sum is organized. The partition, the partitions of the vertex set on there. Another functional called the injective trace denoted with this little nut. Apply on what is called the quotient graph of tea obtained by identifying vertices in the same block of pie. Okay, good. And this guy has actually very good properties when you compute it for classical model of matrices. The limit of this guy is either zero, either something non zero when he has a specific geometric or combinatorial form like being a double tree being a cactus depending on the model. Okay, so it's quite nice and it's also quite nice to see what is traffic independence because the formula is a product when it's zero. Okay. And what about variance profile. It will be quite trivial. In presence of permutation environment matrices. Maybe I should add one. Another definition that we involved that was involved in the proof two days ago. I called it a delta and zero. It's just a different normalization of this two and zero let me write explicitly the definition. It's the expectation of the product, the same product here. So I put five of it as a shortcut. Is it okay online. No one day. No one day. And here there is a fight. And for fight is a given injective map. So let's put a tip I to be consistent we are talking about this guy injective. So if a chain is permutation variant, when you're summing over five injective, the expectation you're not to the, you have an equity which does not depend on fire and so we just see that this guy is just a normalization of this guy right. And this guy is just a product of matrix entries. The expectation of a product of matrix. Okay. Let me change course to gamma n times X n. Assume that the matrices are independent. X n is permutation environment. Assume that the entries are boondi or boondi moments or assume that the quantities I'm going to write makes sense. This content is then. If you want to compute the injective trace of a chain, and you want to do that if you use this formalism as you see because of this formula. It is quite simple. It is what I call the injective density because it's a small quantity on the violence profile time, the injective trace of the original matrix and of course I forget something doesn't make sense. It's not for any test graph, or let's say any question graph if you start from the computation like this. Sorry for some mistake. So you fix a graph you want to compute the injective trace in the matrix of interest, HN. Actually, it is just a product of the injective density of the test graph in the violence profile. The injective trace of this test graph in the permutation environment matrix conclusion is that again, we have a simple formula which just involves a product. Why this is true. Let's look at the definition of this guy. Let me not write everything. We look at the definition of this guy. Let's look at the expectation I can, as before, do it here. But here, HN is the entry-wise product of a gamma N under XN. Okay, so I can just write this product as the product of gamma N of phi of E times XN of phi of E. I can also write a product of this times a product of this. Then I assume that the two matrices are independent. So I can just split my expectation product of XN of phi of E. But XN is permutation invariant. So it does not depend on phi. So this phi and this phi, you can put phi tilde, which is another injective map fixed where I am. Oh, I start from the expression of the trace. I should have a start from the beginning, working on the injective trace of the quotient graph. Sorry, it's injective. And here it's quotient graph. So doing so, you factorize this term, and you rearrange them, and you just get this formula. This is just because you have just you're considering the expectation of products and entries. And of course, you can factorize that where phi tilde is any injective guy, not especially this guy. So you can put this outside. Okay, and you can rearrange, of course, you have the sum on this guy, but Okay, maybe I should. I did not, I define this for permutation invariant model, but I should say that when you have your variance profile, I was confused about this definition is not exactly this. When the matrix is not. So this works for XN. Yes, there is a summation, but I will write something different than the summation, it will be the same. But I will say that it is the same quantity like this, but where phi is a random injective map uniform. So it's just a replacement of the sum, random uniform, but being random uniform you know it's just doing a sum normalizing well by dividing by the number of elements, but if you balance this is what you get from the other term, you will get this formula. Okay, which the conclusion is that if you want to understand advanced profile matrix from the point of view of the injective trace, you will have something which is quite immediate to factorize. It doesn't mean that your distribution is easy to compute, because first this method allows you to compute moments, which is just one way to understand the distribution which is quite obscure quite algebraic. And this moment will be, you will have a summation, because you want to compute a moment by you have the sum over the partition of terms that factorize. So from the combinatorics point of view it is easy but it doesn't tells you that you will have a nice spectrum and a nice method to compute. Okay. So in the context of asymptotic traffic independence I want to go back to the techniques of that. It means that if you have a family of independent families of permutation matrices as two days ago. But for each matrix, you put the variance profile. The formula that I show you involving the graph of color right components. It's called, but with the pre factor. We will have we will factorize some terms due to the variance profile. So the phenomenon that the, the structure of three of trees rules the asymptotic of the moments of the injective moments, still be true. But just, you will have a factor, which depends on your variance profile. Again, I'm not going to the details about this but it is important just to see that it doesn't really modify what happened in this topic in this context. The computation we were able to do for pure permutation in round matrices is still valid when we put the variance profile. Just notice that if you have a permutation balance matrix, and you put the variance profile you no longer have a permutation in that matrix. Imagine you your violence profile is like this. Of course, you, you are breaking this. You're breaking the permutation in advance, but you're not breaking the method. Okay. Yes. No, no, no, no, no is sorry for not being very precise about this is a matrix. If, if a matrix is permutation invariant, this delta n zero is just this expectation for a chosen five. If not, you must some over all five and divide by the number of injective map, which is equivalent to take your random uniform injective map. Okay, but what is always true. If you take this more general definition is that the injective trace in a graph, let me put it by is always just the relation between these two quantities is always just a question of normalization. And this is always true. So we computed is n factorial over n minus V by. So this is always true. This is why I call it is injective trace and the injective density because it's just normalization. Is it answering your question. No, no, it's any matrix. If it is not permutation invariant, the definition of these dead ties and zero must be this one. You must average on all your matrix. If it is permutation invariant by just choosing in your matrix if you have a test graph of size three, you will have just to consider a sum matrix of type of size three, you compute your expectation involving this little part of your matrix computing this but permutation invariance tells you that actually this is the same thing as if you were doing this computation in all the matrices. Let me comment a little bit this, how do we compute that. Imagine you have such a test graph. You choose uniformly at random. You choose the matrices. So you choose. Okay, I'm going to be confused. So, a jk. So what does that mean, you choose a j choose a choose a k uniformly at random. It will give you the matrix entries somewhere. So you compute the product of this random variables. Okay, for a given a jk, you're just choosing three entries, but then you average over all the matrix, picking a lot of information on the matrix. Here, this is the same object. The test graph is a test graph and then you can take the question to the test graph you will get a test graph. Just the point is that usually you start computing a moment, then you model in your tea. And the first step is to write as a sum of question graphs. So the question appears as an expansion of a trace, but you can start with a test graph with repetition there. There is no repetition there there will be repetition there, but the nature of the object is the same on the right hand side on the left hand side in general. Of course I start by considering very specific object 30, like the cycles or the double cycles like this, but it was just a choice for the presentation. So this is just a way to pick information on your matrix. Imagine you have your graph you put your graph in your matrix on your average on by putting your graph in your matrix I just mean you come to this. Okay. So this was just to mention that this science profile matrices can be handled with this injective place. So let's stop talking about this injective stuff and combinatorial aspect on the start again with the classical theory of free probability at the level of amalgamation, which is a kind of analog of conditional expectation and introduce the notion of freeness over the day ago. Okay, so I recall you that an uncommitative probability space is the data of a couple. a phi, where a is an algebra phi is a linear map, playing the role of the expectation. You can also consider a star. If you want to do things properly but it will want to be mandatory for the presentation today. So I will introduce what is a conditional expectation in terms of this non primitive language. So this definition of freeness respective to this computational expectation. But for the moment I did not give you the definition of freeness just for fine. So I will start by giving this definition and then adapt it at the level of the conditional expectation, or maybe you will be surprised by the definition of freeness because it's not very similar to the definition of independence so we will take time. So, so you can see there a one a L, which are sub algebras of a classical theory you define the independence of sigma fields, right. And random variables are independent is a sigma fields they generate are independent. This is the same. We define the thing for algebras and different families of non commutative random variable will be free. And the algebras generate are free. So let's talk about this algebra. We say that they are free. And in the classical basic sense of very cool, let's go. And if we have this property. For all integer and for all indices L1 L2 up to LN. So indices that appear that choose one or another algebra. So the first algebra is not the second one. And the second algebra is not the third one but the third one may be the first one. We say that we take an alternating sequence of indices. So this guy, I told you there, they correspond to algebra so they are indices in L. So this algebra in this step, you consider an element in this algebra for all a one in the first algebra a L in the last algebra. But these guys, these n variables are sent to rate. j is zero for all j. Then the product of this variable and because it's non commutative, you must specify another is something. Is there any mistake or. I think it was on in words, we say that an alternating. Alternating, alternating refers to this property products of centred elements. So these guys are my centred elements are sent to my product as expectation zero. So one of the product is the product of the expectation, but the alternating product of centred elements is sent to it. That's it. And if it is the first time you discover what is finesse you should be a bit confused. But remember how you were confused the first time we define in classical independence in little school. It's the same thing. Why this is a good universal rule should be developed in a proper lecture. So if I want to compute a moment with that. It's not obvious maybe I should mention that. Let's say that I take different elements in a one a l and I want to compute the expectation of their product. If I'm not in the situation where it is an alternative product I must work on this formula because they have no no direct rule about that. For an arbitrary guy in the algebra generated by this guy, I can assume that by linearity. It is a product. I can organize my term in such a way I have alternating products. So that we have this property. If you take a monomial, which is a product of element of this, you can just organize a product to regroup the guy in the same algebra. Okay, and then write it like this. If I'm able to compute the expectation of a for any guy of this form by linearity, I'm able to compute five on the algebra generated by this guy. This is knowing the joint law of this element of this algebra. Okay, this whole doesn't tell me directly to this guy because I cannot assume in general that my guy are sent to write. How can I do to compute five of a. I artificially, I start writing the quantity that is involved here by removing for each terms the expectation on this expression is by stop it. Obviously it's not correct because I modify it. So I expand these terms to have a nice expression plus other terms. This guy is actually. So this guy is fire of these guys times the second guys times the other guy, we have this term, this is the one we know. So the other term are the one we obtained by removing all the other expansions, like if I choose first to factorize minus five of a on an arbitrary way the other one is to be another time that I will have to remove. Okay. Then, this guy is zero by fitness, because I'm just made things in such a way as an alternative product of alternative of central elements. And then, by addiction, I do the same trick to compute my terms here. It can be a long process, of course, because it is by induction, and it explains that computing moments in non commutative probability needs some some techniques because there is no direct method. Okay. Okay, so now we can consider what is a conditional expectation. Yes. This is the moment approach. In classical probability, you know you can use the expectation or the characteristic function which has very nice properties. Then, I'm not really mentioned it but there is the cumulant approach, which involves the subject mentioned by Jean, which has a non crossing partitions, and which is related to the Schringer days and formula we have proved for Gaussian matrices in the first Okay, so there is another way to translate what is a fitness and based on these objects, it will require an hour or more to develop this. Okay, so let's skip that and talk about fitness over there. So now we are considering a space which is a little bit richer. We now assume that we have an operator valued probability space that is a triple a B and delta where B is a sub algebra of a and delta is a linear map from a to B. We assume that it is a non commutative conditional expectation in the following sense that should be that should. Okay, let's look at this. The delta of B one a B two is B one delta a B two for all a in a and little bees in B. So what is the classical definition of the conditional expectation, you have the expectation of a product of XY conditionally on X, and you can factorize one of these variable. And you which go outside the filter here it is non commutative. So you do the same thing but on two sides. Okay, it looks just formal, but it works very well. Is it clear. Which example we will consider of non commutative probability space operator valued will take. So for the general for the algebra we take matrices or random matrices but let me write the deterministic setting for B and we will take the diagonal matrices. What I call the again of C on for delta we take the linear map that given a matrix a associate the diagonal matrix of the diagonal element of a. Okay. There are other structure on matrices which are very interesting and useful in practice with today for this weird ensemble this is the one we want to consider. Okay, so just a definition a bit abstract of a structure, but we can adapt the notion of finesse. And we say that sub algebras of this guy a are free over over B, or free with amalgamation over B but I prefer the shortcut. Even if I have these properties where no, I take my family is a J in the algebra generated by the good algebra I want as before, but also I include B, the operators I play the role of scholars, such that they are centered but not for the complex for the combiner for the conditional expectation. Then the conditional expectation of the product is zero. What is important to understand is that formally, and for, let's write it there because it is where it is important. And we replace the following replacement, we replace the complex numbers by diagonal matrices, and we replace the normalized trace by this diagonal operator. And if you make this translation at this level. So I have written with a file instead of the normalized trace, because the asymptotic finesse is just asymptotic I need the limit. But if you made the replacement you see that it is a is defined finesse with amalgamation. Just by replacing here. We add to the complex numbers. And by replacing the trace by the delta. So the diagonal is zero. Absolutely, which is weird why is it's interesting but it works so give you algorithm at the end so you don't be so picky. Okay. Okay, so it's almost the end. This part was quite technical and I think that if I come back to that to justify to justify that the relation it will be painful for everyone. Just let me sketch a proof that traffic independence implies finesse over the diagonal. And then I will answer your question if you have a moment. I hope that if I have permutation in one matrices I have this relation during this assumption. What we know is a technique to compute the trace of a product that involves this complicated graph of component, component or connected components. Right. Okay, let's try to imagine what happened. Take an expression like this. It is a moment. This is a product that I represent with a cycle like this. And I have several colors let's say we have two families and so talk to colors. If I find this. Okay. So these represent a group of ages in an algebra, a group of ages in the other algebra, and so on. We were considering matrices a one on a two remember. So let's say it is a poor of a one, a poor of a two and so on. So if we want to compute the trace of a poor. Let me just sketch the proof that this fire is zero it's not it's not showing that the delta zero but it will be it will contain all the ingredients. So let's, if we want to compute the fire we compute the total this guy, and we knew that it is a sum of a pie of the injective trace on here. So, up to a small error the indicator function of an event, which is that the quotient graph has a street a tree structure, the graph of color right component is a tree, but it's all pulled by trees, pulled by trees. Okay, let me not be more specific. So if, instead of just taking the trace of a product, I take the trace of products where I remove the diagonal. Let's call the V1 V2 V3 and so on this vertices as a junction between two segments like this, removing the delta. And considering not only a product, but it will be a linear combination, which should be complicated. In general, because as I told you we compute things for moments and so if you have a polynomial you should go for every monomial and have a linear combination should be more complicated. But because you remove the diagonal, it turns out that it actually simplified this formula. This is an equality. You have the same equality but we have a restriction. It won't be the sum for all permutation, but all permutation such that two adjacent vertices are never identified partition. And again, we have not just a monomial but a linear combination, but by some good properties, this actually simplify things instead of complicifying it. Okay, but now if you have a tree, a tree has branches and it has leaves. Okay, this is a very simple. So we should check on the combinatorics, but for the graph structure of tree. It means that if I take a branch of this to get a quotient, which give whose graph of color right component is a tree. We can see that at some point, I must identify one, one, one VI with one VI plus one. Because you just graph of color and component make I'm making an arbitrary graph sub graph like this, and I have a graph structure tree structure. So this condition of being a tree for the level of the graph of correct component is incompatible with the condition we add by removing the data. Okay, this is a sketch of proof. So just try to convince you that five zero. And once you have this technique, you can prove that at some point, and some level, actually the diagonal is zero, not only the trace but any, all the entries. Okay, so this is the end of the first part of the lecture if I'm correct, we have a short break. And just later we'll see a question we forget about these combinatorics and turn this formal notion of freedom servers diagonal into something analytic. Thank you. And you have questions or you can probably answer it right now. Yes. What does that mean. So the question is to specify what does that mean this right. A J. Also, let's say that assumes that AJ is just the algebra generated by a single element. Just to simplify what means being in the algebra generated by a and B, it means that being of the form B zero a B one a B and a B and plus one. B zero B and plus one and for some end on your complex if I because B is non commuting with a. If B is just the algebra of complex numbers, you can just factorize all the B and say it is will be proportional to a board. So you have to have to complete some of these issues. And so you have to to be too unsure that you, you're taking an element of this form or linear combination maybe such that the diagonal of these guys you will. Of course I'm saying diagonal but in the abstract setting it's not good diagonal its just you must find a way to approximate diagonal matrices which is tricky and technical I'm avoiding a lot of or is it more clear other questions okay if not let's have a break thank you other question nine maybe i don't think so