 A seemszegyre tehet, hogy egy szerep abból. A szerep egészen az egészére lehet kiárom aозét, amilyen van szerep szimletre a helyzetbe jutott. A szerep igazál, amit tudod, hogy a félhagymát simátsíj, vagy egy szerep simátsi, és egy észre, és még egy szerep is másik, és mit csinálja maga. Én sokkal megválják, hogy egy szerep szimletre a szerep utána. Az emelnek a... Jó. Aztán az az összesen élőképesztető, de nem kell a kezőképesztető. Egyébként a kezőképesztető. Ezt szerintem egy személyes fújjára. Egyébként egy különből lennékeztek. Egy különből lennékeztek a fújára. Én egy polinomiai fújjára. Én igazán nem tudom, hogy mi már a polinomiai fújára. Egyébként a fújára érdemélet. Úgyhogy a világjátok után például visszategy nem és a mesező interviewedat az jetzet a másik játási termésát. Az egyik játási termésére is lehet két ezt, hogy egy képe lényegatoknak is az ilyen szóban véletem az egész étvely equally lényegatok. Egy képe lényegatok excrét lehet két jatási terméset. Visszategy, a két éve lehet egy ilyen két éve lényegatok is, az betegyé mainlyjátok és az ezt a szegé, és ezt a csomálóra充 képe, Nézzáról old verenetsel korol érz, the delocalization of the eigenvektor is minth. Minth, de if you take an eigenvector, the delocalization eigenvector is minth, if you take an eigenvector, say you take a vector u, if you solve the eigenvector equation, eigenvector equation. For an eigenvector, which is in the bulk, then you will know that the, and it's a normellized, eltonormellized eigenvector, then you know that the maximum of the eigenvector is essentially one over square root of n. The one over square root of n means that you have a vector of length n, if the normalized vector, eltonormalized vector, and the typical size of each matrix, all the vector components are often the same, then each of them is a folder one over square root of n, and it tells you that in the delocalization, tells you that in the strongest possible sense, the l-infiniton norm is bounded by this one over square root of n. And this you can actually very easily get, once you have a control on the diagonal element of the resolvent. So to control that, to get such an information, it's already not sufficient to get the information on the trace of the resolvent, you need to get the diagonal matrix element. Now the rigidity is a similar type of statement, the rigidity tells you where roughly the random eigenvalue lies, and here, precise formation, that important, it tells you that every eigenvalue have a natural location. If you take an eigenvalue, you take an eigenvalue with a certain label, say lambda i, this is not the way how it's formatted, but I explain it in a way which is easier to understand. You take the eigenvalue lambda i, where should that be? Now if the eigenvalue is following exactly the, if it did follow exactly a semi-circle law, or a more general did follow its own density law, then the lambda i should be at the i-th quantile of the corresponding distribution. And this formula here written in a somewhat different way tells you that the true eigenvalue is not much farther away than the i-th, than the corresponding quantile, and not much far means that the distance is essentially 1 over n. Maybe it will be 10 to the absolute. Ok, so here's a picture, that's probably what Gerard asked. So this is a picture of a nice far from being semi-circle situation, and then you still have the bulk universe at it after you discuss things properly. Ok, so I think that I should stop. Yes, I mean it's, there's a mapping, it's not a simple mapping. This is the Dyson equation, this I will discuss next time. So we will have to, we will have an equation, non-trivial coupled nonlinear system of equations, which involve s, can write it up. So this kind of equation minus 1 over mi z plus s, i over m is this matrix, this will be our fundamental equation, and you have to solve this equation for all mi, and then the density is basically the average, this will be the average of the m i will be the shear-test transform of that. And so that's the way to, that the connection between that. So there is no simple formula by an explicit deterministic procedure. Yes, yes, that doesn't happen. That doesn't happen. Yes, I think this is impossible. It's anyway what we prove is that it's only, so once you have a casp, when there is a casp, here is a casp, once you have a casp, then we know that in the neighborhood of that casp. It's a, so once you have a casp, then it means that the neighborhood of the casp is behaves likely to the one side. Yeah, but you know, you cannot really do freely what you want to do. This picture is governed by what you can do on the level of death. So what you try to say is that you want to say something like that, I think. And then you want me to, so there is an S matrix, there are some parameters in the S, so S is some S of t, for example, and you want me to create a flow of STs, so that this one shrinks. This doesn't happen. So what we do is that we describe it whenever there is a casp, then it behaves like X to the one-third and then we describe it in the neighborhood of that casp.