 This log M is just nothing else. Okay, do I have, yeah. This is, the last log M is nothing else is basically, it's related with this term, I think. And otherwise, for M larger than one, one has different expressions. This is for M smaller than one. And if M is larger than one, okay, one can write it explicitly. It's not difficult expression, but I just save time not writing it. So similarly, one should consider a contribution from the part B, or I call it rather B plus. Here's B plus, here's B minus. One can show that contribution from B minus, it's immediately clear that it will be subdominant. About B plus, one should be more accurate. And work with this. So let me, as you may guess, here calculations are look at least superficially more complicated just due to the very fact that one should take this piece with this relatively long expression. But nevertheless, this expression, why I said that it's nice? Not because I derived it, maybe it's partly only the reason, but because it's really nice. One can work with it and find when the corresponding exponent is extremized. So you just, in that interval, and you find differentiating that expression a relatively simpler equation. So x plus square root of x squared minus four tau, minus four tau divided by two minus m equal to zero. So one should find just x in terms of m solving this relation, which is really simple. Solution, x is equal to m plus tau divided by m. This is the point where, which is stationary point in the interval from one plus tau to infinity. So one should ensure that really this point belongs to that interval. Again, in fact, it turns out that for m larger than one, now this integral is dominated by settle point, otherwise it's dominated by the boundary. So, and in this way, after taking this value, substituting back to exponential, one finds the value of the integral. And surprisingly, it's really nice functional because when you take everything in substitute back, you expect that it's something cumbersome, but it's not all terms really conspire, and you get surprising, okay, it's not that surprising, but if you look at these horrible expressions, it's surprising. You get that is equal, just everything cancels. You get, with this precision, you get just one. And now you may wonder, you cannot, in this approximation, you don't know really it's one, two, or any constant. There is no exponential terms. There is no terms growing as exponential of n. But as I mentioned, we can resolve this problem even better by resorting to exact evaluation. And then really you find that this is not just all one, which we can conclude from this large deviation calculation, but it's just one. So this is really expression for m larger than one. Now comparing contribution from various intervals for m smaller than one. What happens for m smaller than one? Basically, this term, this contribution dominates. So one gets exponential with exponential, not discussing pre-exponential terms, just exponential of n, one half of m squared minus one, minus log m for m smaller than one. So this is the answer for number of equations standing to infinity, a symptotic answer for number of points of equilibrium. There is exact, so we see that there is type of phase transition here. Namely, asymptotic changes drastically. If you fix parameter m and let n number of equations go to infinity, there is one and only one on average, but this means also, one can show this is minimal possible, but this means that in every realization, equilibrium for may's parameter larger than one. And this is exactly what really linear model predicts. That if you come back and compare with linear model, it's exactly what predicted. There is one stable equilibrium. There in linear model, there was just one, but it's really one in a linear model. There is single equilibrium. Also, we will see later on when we calculate number of stable equilibrium that this is stable equilibrium. So one stable equilibrium for may parameter exceeding threshold value one. When this parameter is smaller than one and fixed and n tends to infinity, we have exponentially many equilibria because it's easy to check that this is positive for m smaller than one. So we really have a picture. This information was inaccessible in linear may model. It's really new bit falling from non-linearity that there is explosion of the number of equilibria. We have exponentially many of them, but we do not yet know what is the nature of this equilibria. So let us continue and apply similar consideration to number of stable equilibria. So now we are interested in the same determinant, but conditioned, I mean, with this indicator function, which requires that its averaging happens or occurs over the matrices, really, which are positive definite. So dn stable, dn stable, it's again the same, and we just rewrite in the same way, of course, this determinational part, but now we have indicator function that max, okay, I will write xm smaller than x where xm is a maximum of a real part of all the i's. And now average. This is the number of stable equilibria. Now it's clear, so recall this, recall this equilibrium density and recall large deviations. So we have one plus tau here, less important for us, but one minus tau is there. So consider various values of x. This depends on x, it's very essential because we need then integrate of x. So first, suppose x is larger than one plus tau. So this is the value of x. I just put vertical line at value of x larger than one plus tau. Then we know that basically deviations from equilibrium measure are penalized with the rate n square. So typically this constraint is just non-operative. One can make exact computation, show how non-operative is this, but this is completely clear that in this situation this conditioning is empty because with high probability, with probability tending to one, all matrices will satisfy it. Only with exponentially small probability something may happen that this empirical measure will have eigenvalue to the right. So basically in this situation we see that in this situation basically we have the same result is equal to dn equilibrium and equal given by this exponential of n of phi of x. So this is the immediate conclusion. This is first case. Second case, what happens if x is smaller than one plus tau? Now we really consider situation when this line of conditioning is to the left of this and we again know that the probability to fluctuate such that to satisfy this constraint is exponentially suppressed, exponential of minus n square times times really the rate functional. Okay, so again one can make more detailed computations, but basically we conclude that in this situation it's completely clear what we should expect. We should asymptotically expect that a number of this d, this object will be just given by n phi. Again, invoking reasoning of Laplace-Varadan type. It will be n phi x at some measure mu star minus n squared, I will write it k tau of x where k tau of x is a solution of minimization problem. Basically we minimize the large rate, the rate functional, okay, how I call it, I do not remember, let it be I. I of mu, I do not remember what I used, but let it be I or J maybe I used. This large deviation functional which characterized large deviations for elliptic ensemble, but I now need to minimize it of b's in the set bx such that support of the measure is to the left of x. This is complicated problem and in general it's not known how to solve it explicitly. Solution clearly exists, but explicit form of minimizer of this, minimizer, it has minimizer mu star and this is precisely what I should substitute in phi here, but explicit expression is not known. Apart from one single case, tau equal one when we deal with GOE and then Satya Majumdar managed to calculate explicitly the corresponding minimizer. I think he even showed it yesterday, but basically in this only case we can calculate explicitly these two terms. In all other cases we cannot, but fortunately for the modest, our modest goals, namely for the goals of evaluating, of taking these results, substituting back to the integral we need to evaluate and evaluating it in large and limit, this is not needed. This information, these two lines are enough to evaluate this exponential accuracy of that integral and the only thing which we need is that when x tends to one plus tau from the left and from the right phi, this phi just tends to that phi which is just continuity and which is very natural. Although I'm not sure that I know how to prove it, but I cannot, okay, this is assumption of some continuity. I'm sure that eventually it can be verified, but at the moment I don't know how to prove it. If we just make this assumption, this information is absolutely enough to evaluate number of stable equilibria. It's exactly the same, no, not exactly the same, but it's just repeats this type of calculation that I showed for total number of equilibria. It's just that in one of the integrals, intervals, you should use this, but it's very easy because it's clear then that everything will be dominated by minimum of this k and it's known where the minimum of k happens. Along x, of course, it happens exactly at the age and this is just property of large deviation rate. So I will write then explicitly the result of this calculation and we will compare it with that in equilibrium. So n stable, number of stable points behaves as exponential and I will write it sigma stable depending on both m and tau. Here, interestingly, this rate of growth of number of equilibria was independent on tau. It's dependent only on my parameter, but here there is dependence on tau and this sigma is explicitly given by one minus m plus log m plus one minus m squared divided by two tau. This is in the situation when m is smaller than one. If m is larger than one, it's one. Okay, we cannot say that it's one in this situation. We can say that it's only of order of one, but since we know from the calculation, but since we know from exact calculation that the total number of points is one. Clearly in this situation also this can be only one because one can show that one can prove that this type of vector field must have at least one from topological properties must have at least one equilibrium. So clearly this is just one stable equilibrium. Okay, this is the result and using this result, we can draw a phase diagram. In tau m coordinates, we can draw m tau. So tau changes between zero and one, but I still continue slightly this axis above one. So okay, value one, value one. And first we draw this line, m equal one, which clearly separates two regimes and if m is larger than one, we have just one stable point. So this is absolute stability regime of our absolute stability, of our nonlinear model. Now, one can ask, here's the result, but one can notice that there is a line in this quadrant which separates values when sigma is positive and sigma is negative. Just solve this equation with respect to tau and you will find that it's vanishes in some point in value of tau depending on m for m smaller than one and there is a line, there is a line, tau of m, this is tau, okay, how to call it? This line, I don't, I call it tau b as a function of m. In fact, it's not quite correct. Here it approaches point one with linear slope, but here slope is infinite. So there is such a line explicit expression for it which separates two regimes. Here, sigma equilibrium, number of, okay, rate of growth or total number of equilibrium is exponentially big, so sigma equilibrium where sigma is basically, is this exponent here is positive, but sigma stable is negative. But what's the meaning of this? It means that on average, we have exponentially few, exponentially few stable points, but what's the meaning of this? Since these points can be one, two, three, four, it means that just there are zero stable points in every realization and with exponentially small, with exponentially small probability only, there will be, say, one stable point. But in every typical realization, no. So we see that really this is regime of absolute instability. Absolute instability. And then here, this is, now when one crosses this line, already appears exponentially many stable equilibria, but of course their number is much smaller than the total number of equilibria, such that the fraction of number of stable equilibrium to the total number is still tending to zero. So this is the picture really. So conclusion is that, okay, last point, you see that there is one special line when this regime does not happen. This is a line of purely gradient flow. In purely gradient flow, immediately after crossing m equal one value, one has exponentially many stable points, but of course fraction is still tending to zero. So I finish here and just I was asked to formulate what are open questions, and I would say that almost all questions are open here because it's really, we know very little, it's just picture starts in emerging, but my time is over, if you ask, I can mention a few problem, but I think that I need to stop here. Thank you. Yes, exactly. This is one, okay, should I comment on this? Okay. One need to evaluate, estimate what I said in one point, variance of this number, just to see how big of fluctuations around this mean. It was done, now it's time to mention, not for this model, for related model, in fact, not in Euclidean space, but on a sphere, it was done for that problem on a sphere and only for gradient type flows, for potential flows. Counting was done, in fact, in very nice paper by Ben Aruz and Alfinger, and counting of the type I presented. But after that paper, there was another nice paper about one year and a half, maybe two years ago, by Eliran Subag, who showed that really, fluctuations are not that big. So he managed to estimate second moment. For general case, I think that eventually, person as clever as Eliran probably will be able to deal also with this problem, but it's open. I mean, I think technically it's quite involved. It's only a purely technical challenge to evaluate, now you won't have this simplification that there is a decoupling between delta functions and terms, and Jacobian terms at different points. So one need to find a handle on evaluating the order of this double integral, which is challenging technical problem. But if one does it, then one sees whether these fluctuations are strong or not. This is one on my list of problem. Yes, thank you for the question. I just did not have time, but maybe I now use just a few minutes to say it. In fact, as always, okay, I reiterate, although this model turns out to be really amenable to quite, okay, to analysis, I think this is nice feature to simulate really this type of system is challenging. But one can hope for universality, and for example, consider really model motivated and used in neural networks with where really these equations are x dot equal to the same. Usually they, in that setting, they take mu equal one, but it does matter just scaling, plus sum of j i j and then functions. Okay, I will write some example of this sigmoid functions, hyperbolic ton of x j, sum over j from one to n. So one sees that, and j i j are just some random coefficients with some covariances. So model of this type, easiest to simulate, and really they were simulated. And there was a very nice work by VineRib and Tubul, by VineRib and Tubul, who really argued that there will be transition in my terminology in m equal one or similar, I mean, parameter, where there will be exponential explosion of points, and they gave some numerics for this. This is basically, I believe, and one can even check that the function that they write as complexity function is exactly the same as this close to the transition point. So maybe one should expect universality although really their model, okay, even if you take j i j as say, simplest case, independent Gaussian, this field is no longer isotropic and translation invariant, and this makes really the tricks which were used to calculate it failing. But I hope that one can give some strong argument of universality, and this is one, but there are also other related models. Okay, so I think we should stop here and let's thank Yann for wonderful lectures.