 For all L1, L is maybe, no, N, sorry. So it's a product of little n matrices. So this is for all little n, where, which involves the index of the matrices, I just specify which matrix I consider for each term of the product. And if I consider the matrix itself or it's a joint, for all epsilon 1, epsilon n in one star. So if epsilon is one, this means I put the matrix to the power one, that's the matrix. If the epsilon is star, I just mean I consider the cause of transpose on this adjoint. Okay. All this number, all the function, all this in the left hand side, you can put, this is a kind of moment. So I do not have M like Mark, but it's not, it does not depend just on the power I consider here, but it consider here's the data, epsilon 1 and n. For each n, for each family of L's, on each family of epsilon's, we have a number that we do not have. Okay. Why do we care? Because we remind you, I remind you that we want to understand the spectrum of a matrix HN, which is written as a polynomial in several matrices. So I start like this, P of A1, An, and imagine you want to understand the eigenvalue distribution of this guy and start by computing its moments. We want to compute, so we have the empirical eigenvalue distribution of HN, we want to compute its moments. So we apply it to the function X to some power K. What does that mean? It just means that we consider the expectation of one of the sum of this eigenvalues of HN to the power K. Okay. But if we assume that the matrix is Hermitian and we do that, it's equal to the expectation of one of the n, the trace of the matrix HN to the power K. Okay. Just relating the eigenvalue with the trace. That's good. Now, HN is a polynomial in 1, 8L. So it is a HN to the K is also a polynomial in A1, 8L. You can write this guy as the expectation of one of the n, the trace of a polynomial Q in A1, 8L, where Q is P to the power K, obvious. Okay. But this is encoded in the star distribution of A1, 8L. This is what we noted in the right hand side. Phi of the family A apply to Q. Okay. Or forget about this notation. We don't care. This guy is just a linear combination of values like this. Conclusion. If you are able to compute the star distribution of your family of ingredients here. And moreover, if you have a nice description of the star distribution, you can expect to have a nice description of the matrix you're considering here HN. Okay. So typically you have a matrix HN. Let's say it's a sum of your two ingredients. You see that the moment of this empirical again very distribution can always be written as a linear combination of trace of products like this. Okay. So this is okay for the conceptual ingredients of the theory. Let's do a little practice and learn a little bit of techniques. I don't know if I will have the time in 35 minutes to do the whole exercise, but what we want to do is let us compute the limiting star distribution of a family XN. It's one XL of independent. I have already defined the matrices. Okay. So you consider several matrices of this type. You don't necessarily need to have the same distribution for the entries when you change from one matrix to another one. And I will assume that the variance of an under of extraordinary entries is one for every matrix. Okay. So let's do that. And you will occupy a little time. And at the end, we will get a nice formula. I claim. And that will be the opportunity for me to introduce you. And to compare with Gaussian random variable. Okay. So by definition, if we want to compute the limiting star distribution, we must compute some way. The limit when n goes to infinity of the expectation of one other end of the trace of a generic star product in this matrices. So all these matrices are Hermitian. You don't need to put the agent. So we just need to compute the expectation of the product. XL1 up to XLN. This for all N and for all indices L1 up to LN. Okay. With possible repetition. Okay. How do we do that? Let me organize it. Let's call it M of L1 and N. And first, let's describe what we do. First, we do an expansion in terms of entry. Okay. This is a trace of a product. So we can write it explicitly in terms of the entries of the matrices. What does we get? What do we get? So we have this one over N that I can put in front of the expectation. I can put it using linearity in the other trace. On this trace of the product, I can interpret it as a sum of a weekend X. A1, A2 up to AN from one to N. So I have my first matrix. I have my second matrix with A2, A3. And so on up to the last matrix. L1 with this index AN on, because it's a trace. We finish by considering again this first index. Everyone agrees with that. This is the definition of the matrix product and of the trace. Okay. Just for the moment. When we have such an expression, we'll use graph to make computation. I will define it like this in more details. But when I consider a moment like this, I consider, I draw it with a simple cycle. So here I put off length 6 if this little N is 6. And I draw my matrices. XL1, XL2, XL3, and so on. XLN. And I just put the same vertex at the end of age and at the beginning of this one because I have a repetition of index, which appears in this expression. This is very simple. Okay. So now let's compute the large and limit of this guy. First of all, we have matrices. These matrices XL's of the form. XLJ to the square root of N. Okay. And this number. But we have this normalization by square root of N. So let's first, because it's easy factorize this. We have a one N here. So N to the minus one. Each entry for each entry at the square root of N. So we have minus little N, which is a total number of matrices under consideration divided by two. Okay. And now I have the same expression. The sum A1, A2 up to AN of the expectation. So we have this product, but with a not capital, but some small letters just mean we discard the normalization. I just put a small letter here and I mean the new normalized matrix. And when we do that, the advantage is that by definition, this quantity does not depend on N. So let's assume today that the big norm matrix is of this form for one variable with distribution is independent of N. So here you have a number. We call it M. No, no, M is already, let's say G. Of L1, LN. Of I. I is an notation for this multi index. Okay. Where I is I1, IN. Right. This is a value which depends on the matrices under consideration and on the entries. But we have big norm matrices, which means that we have independence and identical distribution, which means that, in fact, an observation, that these numbers does not really depend on the actual distribution of these indices. What it depends on is on which indices are equal and which are different. Okay. And this is a strategy we will use today and we will use tomorrow with permutation in both matrices. So let me, in fact, this V of L1. It depends on this L, of course, because this is a choice of the matrices, but it depends on this I, only through something specific on I that we call the kernel of I. What is this kernel of I? Let me define it and illustrate this on this blackboard. The kernel of multi index is a partition. So if you have a multi index I of L1's little n, as we are considering here, the kernel of I is the partition of the integers 1, 2, up to little n, such that two integers from this set, and Q are in the same block of this partition. And this is what I denote P is equivalent to Q with this relation. If and if the corresponding entries of the multi index are the same. This is my definition. An example. If I take the multi index, let's say one, two, okay. The kernel of the multi index. One, two, one, three, two. So this is just a multi index. Okay. So this is the partition. What do I count? I have first, second, third, fourth, and fifth position. First and third position are in the same, I have the same value. So they are in the same block. First and third. The second one is the same as the last one. So my second index to belong to the same block as the last one, which is five and third appears once. So since it appeared in the fourth position, four is a single time. And if I consider this kernel, I get the same kernel. If I change the value, the values of my multi index without modifying the case of multi index. So this kernel is the same as kernel as 17 minus two. I put the same 17 here because it's the same value. Here are five minus two. Okay. Okay. Well, if you take an expectation of products like this, this guy depends only on the kernel. And you can understand this. It's because we have entries which are identically distributed. In this matrix, X one, two has the same distribution as a minus two does not. That's the same distribution. So if you have a moment in like this, and you are the same value at the same moment. This means that because a lot of terms are the same, we can organize the sum depending on which terms are the same. So let's do this. I'm going to use it using the graph approach because it will be relevant. Here I was talking about the partition of the state one, two up to two little n, but now I will be talking about the set of partition of these vertices. They're in bijection. So I don't really care. Okay. So I'm going to remember M or L one and N. I will first consider pie. A partition of this vertex set. V is my vertex set. And it is in bijection with the set. Little n that we consider earlier. Okay. And now I will sum over. So I have my constant that I computed before the one over N from the normalized trace minus N over two from the normalization of the matrices. And when I was writing first is some, I will first, I will write the sum over this routine. This is such that the kernel of a is path. And then I have the same value. There is a question in my back. Yes. When L one and L two are different. Yeah, you can have different distribution here. Then why is the value of V only depending on the terminal because because I am not modifying this. I'm just talking about exchanging the value of this in this matrix. If you have a different matrices, of course the entries of one matrix, you don't want to exchange it with another. We are just permuting entries of the same matrix. Right. Okay. So are we clear with that this V of L one, and pie is the value of the L one, and I for any high, which kernel is. We say they have common values. So that's the dependence in terms of the big object. Okay. What does that mean concretely? It means that you have a sum over indices. A one, a two, a three, a six. Okay. When you're doing the sum, when some indices are equal, you will draw the graph where you identify the two vertices which have the same value. Okay. So for all high partition of this vertex, we'll introduce a new graph. So we call this graph, let's say little t, because in my mind it's a test graph. T is for test. Okay, never mind. We'll introduce t pi, which is a graph that we obtain by identifying vertices in the same block. It's obtained from t. I'm going to use this because it's going to be very chaotic. So I introduced this graph. I will use straight a little bit, and we will use it to make this computation at this level. Okay. Maybe I should not start like this. Okay. For instance, example, if I consider pi is a partition where. I is one is a sangleton. The other guys are sangletons. Yeah. Yeah. So the. I did not finish the computation. No, no, I meant the, then the summand, the V's do not depend on I. Do I see it correctly? So this term does not depend on I. So I should factorize it. So you're just counting the combinatorial factor. I should have finished this computation. Okay. Sorry. So this is because this does depend on this by button on this I, I just have to count the number of guys here. So I have my sum over pi, which is a finite sum because when I fix the size of the moments, this set of vertices is fixed. And now I have some. Power and terms. The one I had before. And what is this term. It is a number of. So it's a number of mutin. This is that are equivalent to some partition. Okay. And if you count this number, let's say we have a disk kernel. We must choose one integer for this group. Another integer for this one. And a third integer different from the two others. For this one. What is the number of choice? It's N. N minus one to take someone different from the previous one. And for this one. And minus two to take an integer different from the two that we have chosen. And in general. This guy. Is. And factorial over N minus the number of. The size of this kernel. That is N times N minus one times N minus two. And that's the number. The size of this kernel. That is N times N minus one times N minus two up to. What we need. Okay. And this guy is equivalent to N to the. Number of blocks. Up to a smaller. This is what I am going to write. Plus. The number of block of blocks of my five partitions. Up to a small. Times. A term. V. Which is independent on. On N. On the poor of this method is that. No. We just have to consider the convergence of each terms like this. Because the sum is finite. We have a finite term which does not depend on N times some poor of N. So we must just check when these guys non zero. Does this guy. Diverse goes to zero or is equal to one. Okay. I should have finished this part before going. Okay. So this is the expansion in terms of the entries where we were group. In some ways moments in the entries that are the same. Okay. Now we go into the next step where we take benefit of this graph representation in order to talk about this limit when N goes to infinity. Okay. So I was starting to introduce this but a bit to earlier. Sorry. You say we must check divergence or not because the V does not depend on N. It's just like we need to take care of that. You. So imagine this exponent is bigger than one. It means that is equal to one or bigger or equal to one. It means that you have N to some power that would be there. Then you should prefer this time to be zero otherwise. There is another term by which compensate, but this is not what happens. What happened is that we will prove that when this guy is non zero, this power is either zero or negative. If it is negative, you discard it, you will not contribute in the limit. So you will must then identify for which five this exponent is zero. There are the terms of order one. And this is what we are going to analyze this. Can we find a nice way to just say, yeah, of course, can it be positive when this is non zero? Yeah, we have one, but we are must explain it a little bit. Okay. Work here. Okay, so how to deal with that. So I will introduce for each pie a graph, which is this one. I will see clearly on the shape of this graph when this power is zero or when it is negative. And I will see that when it is positive, it is actually this term, which is zero. Okay. So this graph, I call it the quotient graph because it's really going to call it like this, but we don't care today. I denoted t exponent five on the idea that if like this, you say that my partition put in the same block to vertices like this. And let's say for the picture that's three and four and also in the same block. Then I will consider the graph where I identify two vertices in the same block, resulting in a single vertex for this guy. So this guy is alone. Let me draw the sandlotons here just to picture. If it is alone, it's going to create a vertex on its own. This group of two vertices will result in a single vertex on this guy in a single vertex. And now to draw my graph, I need my rows, my ages. And I just consider the ages which are induced by starting and finishing at the same group. This vertex represent this group. So this age will come from this one to this one. This is this Excel one. This vertex is between these two groups. So this straight edge start and finish to two different the vertices in this picture, but that has the same vertex at the end. So this creates a loop. We can have loop and we can have multiple ages by this process. And we really want to. And so let me finish that. Excel four, yeah. Excel five. Is it correct? Okay, so this sum is actually a sum over all the quotient graph you can obtain by such a process. And actually this graph, they are encode the traffic moments. Classical moments, they are just a poor, the expectation of a poor. And by knowing values on each graph like this, we're actually computing moments in the, in the sense of traffic. But we will maybe talk about this tomorrow, but not much. This, what we started from is a moment of the product of non commutative variables. Which guy. What we started, we started from a non commutative moments. And we introduce this weird object. And I'm just claiming that they play the role of a moment in the other theory. The reason we made this detour for a reason is because it gives us a convenient proof, but we just because to, we need a book. We don't need this traffic moments. Just one tool. So it's not only a way to express non, let's say usual non commutative moments to count them in a proper way but you say it constructs another alternative. And this proof or with Bernoulli matrices or more general matrices using this and define a theory where this plays a role a month later. And we just see this subject right now. Okay. Hello. No, that we have this subject, we will interpret these quantities where I raise this and this, in terms of this new graph. So we see that the quantity we are under interest is the sum of our pie. And to the minus one is minus one. What is little and for this graph or for this graph, it is the number of ages. It is the number of ages of the graph T but it is also the number of ages of the graph T by because they have this quantity is considered. So let me call the EPI for them. The H set of T PI and so cardinal of T PI. I denoted it. It's vertex set. And it's a set. And now what is the last term. It's the number of blocks of the partition by what is it in terms of my graph T PI. It's the number of vertices. I say if I have one block, it constitutes a vertex of my new graph. Plus the number of vertices. This is my term, my term of magnitude, I have a smaller or just coming from the fact that the following factorial. And then I have some weight, which is independent. Okay, it looks good. What can we use here. We have minus one minus the number of ages plus the number of vertices that someone knows some relation between these quantities for graphs. Uh huh. Or maybe you know it or some people can know it for planar graphs, but here we are not talking about planar graphs. There is something which is very similar to there maybe it's different but I have the feeling that it is more or less the same. It is a characterization of the graphs which are trees. So let's consider a generic graph G. This is the E. It's considered a connected graph. Then one minus one minus the number of ages plus the number of vertices is negative or is zero with equality if only if the graph G is a tree. So if you have a tree, what you see that you can remove one vertex and one branch. You still have a tree. Okay. And you conserve this quantity because you remove a vertex and remove an age, which means that this quantity is preserved. And remove everything up to last vertex. And if you have a single vertex and no age, it is true that minus one minus zero plus one is zero. Okay. I know if you have a graph, and you know that it is not a tree. That means that you can remove an age without disconnecting the graph, which means that you have strict inequality. And by recurrence, you, and you can you prove this. So as you can notice, this is not exactly what we have because it's divided by two. So we need another trick and other ID. What we can see is that because the entries are center read my big number trees and trees. So it is another fact that we use together with this. And that is that if an age of discussion graph, if I as multiplicity one, muti piece or multiplicity one such that there you see this is a multiplicity one. Yeah, it is what we called an age of multiplicity to because you have two ages sharing the same extremities. Then the term here will be zero. V of L one, and I apply is zero. I remember that we have an expectation of products of random variables, but because we have ignorant matrices, we have independence of entries, and having an age of multiplicity one, mean exactly that you have one term, which is independent from all others. So the expectation of this product you can factorize the expectation of this random variable, and because you assume that the entries are something like this is your. So let's put these two things together on the finishing time, the proof of the convergence, which is interpretation tomorrow. There is a relevant question on the chat. relevant question. By a specialist, I would say that's all. Hello. Where is it. Comparing this to the bigger, you know, the key difference, possibly independent. The X is more involved. I did not talk about the classical approach of this before using this. Usually, you are usually in there is a lot of people that to deal with this limit. First I assume that the random variable are Gaussian. Why do they assume that they are Gaussian, because starting from this formula. There is a significant benefit of organizing the term with respect to this summer as I explained today, but they use the formula for this. Maybe you see that the formula is specific to Gaussian random variables, and tells you that when you have an expectation of a product of Gaussian random variable. It's the sum of the product of expectation where you do pay pairings in somewhere I'm not going to write this formula. Doing that you get some structure, and you can take benefit of the structure to have the same conclusion. Okay, here we don't assume the variables are Gaussian, because before thanks to this technique we can avoid that. So I think that the question is, what is the difference with this, this approach. My advantage of this approach is that before we are dealing with that first, we really put at the end the treatment of this complicated moment. The only argument we use for this moment is this one which is not very difficult to arguably. And then let's conclude to see what happened. So conclusion we write minus one minus e pi over two plus v pi we decompose is this by introducing what I want. What I call it by bar. This is a number of ages, forgetting the multiplicity of the ages, and assuming that the multiplicity of each age is one. That is, I will replace this double age by one straight age. And if I have an age of mtpcp3, I just in this bar ensemble assume I have a single age. Because I introduced this I intercalate, and to correct my expression, I intercalate this guy here. Plus v pi. Let me just put this part of this. Now we have two terms, which are negative or zero. This guy is either zero is negative when this term is non zero, which is the only case that. Because of this argument, all ages much up here with some multiplicity, at least two for this weight to be non zero. So this guy is positive. It's negative when I have multiplicity three four. Moreover, what does that mean for this. We apply the lemma for the graph obtained from t pi by forgetting the multiplicity of the ages. Let's call it the skeleton graph. Okay, and he tells us that the skeleton graph is a tree. So the graph. So let me write this. Oh, and zero when the skeleton graph is a tree. So the skeleton graph is a tree, and the ages are all multiplicity two. That's the conclusion that we'll explore tomorrow to really talk about friends in the bar the self loops remain self loops. There will not be self loops. If these conditions are satisfied. If this guy zero is this guy zero there is no self loops. What I mean is that when you construct the t pi for t pi there are some t pi which have loops, but none of this guy with a loop will contribute. But the skeleton graph, what you call the scale. It's a loop. No, you don't. It's, it is compatible with this. Okay, so let's, it's time so we must finish the drinks. We don't really should not be late. Let's conclude that we have proved the convergence to the limit. The limit is given by what we call double trees. We have the tree with multiplicity two with multiplicity two and tomorrow, which we see that we interpret this double tree formula to make a link with Gaussian random variable in the non commutative setting. Okay. And if you have questions, we have time to answer it. Thank you very much. Questions. Is there a way to encode the graph in terms of genius. I repeat. I don't think so. I don't think so the theory where you have planar object is a theory of unitary invent matrices. Exactly. I don't think so because this is a story for the unitary invent and symbols on four. Today we consider Vietnam matrices. Yes, for Vietnam matrices, if you write, if you write this approach with the vehicle not trick you will get that. Yes. But if you push you this traffic approach, you will have some operation which are specific to this combinatorial graphs, they are not planar graphs. In general, be about counting genius because it's not relevant. I will say that it is different. It is a different combinator. It is a very similar at the first order. Because at the first order, these double trees is just the way to get a sphere from a, from a cagon. If you go into more detail, or you consider a matrix models, which are not bigger, but this Bernoulli matrix, you will be much outside this planar combinatorial description. If you find a connection between this with this will be very rich but for the moment I'm, I don't think there is something very straightforward to connect this. Can you explain what you mean by the double tree? Yes, but I can also wait tomorrow for having, I can of course, but I just wanted people to be relaxed. And this fact here you just, it's just a fact you, we didn't prove yet. No, I didn't prove yet I tried to convince you. Let's. Yeah. So we have a single view norm at this. Okay. And we consider not this color because it's terrible. You can see there's the expectation of one two times other. To four to the power four and so on. Okay. But here you have different entries of the matrices. Assume that this number one to appears only here. Okay, you know that the entry one two, so one two is first line or second colon is independent from all other entries which has not, which are not the entry to one. Okay. This means that this guy being independent from the other guy you can factorize this expectation. Okay, let me let me put. Okay, this expectation of this. This you're allowing to write this, if this entry does not appear on the right. Okay. And if you do so, you factorize the expectation of a center red random variable. So you get zero, zero times something. Is it more convincing. It's really a fact that should be justified but it's not a big proof on just an observation. This is a little bit more detailed that's combinatorics and graph it's always tricky even if you see that in practice to prove it more, more tricky but I convinced you that the proof by addiction works. Can you tell us of these techniques or other commands question. Yeah. Multi PC team in zero. Absolutely. You don't need that but imagine you are considering more precise statistics on not the trace like this, you will be happy to use this property on that works. This is matrix are the same. For example so it leads to the usual classical transfer matrix formalism yes, the usual what use your classical transfer matrix formalism to solve the. I don't know what is the usual transformation. We get the catalan numbers. No, it's not your question so let's go back to your question. So here you are at the notation of the ML one to. And yes, here you wrote that is the expectation value is this trace of the multiplication of the absolutely. Yes, you're considering the same matrix so I guess you want to consider the same. You want to consider the expectation of the number is trace of matrix to support. Right. Yes. Okay, cool. And so you are mentioning a classical formula that is not classical. So this guy you know that it converge to the case catalan number. Okay, I don't know if everyone know what is the catalan number it's to whenever and and there is another and. That's it. I'm more or less we will check. It's okay. It's the number of double trees with two K ages. We'll see that tomorrow. So you had a question. The expectation value of the whole matrix is the expectation of the it's just the matrix. I'm there. So the problem with it's all. Good. Good. So this is a formula for Eigen values of jury matrix and you assume the matrix is jury. It's not the problem we consider today but yes, if you know if you know how the distribution of Eigen values, you have a formula for the for this we does not use this kind of combinator. This is different way. Time to have a drink. Cool. I took my time. I took my time.