So we should be investing in technologies that actually work instead of in solar panels and in wind and all of this that gives power to the Putin's of the world or even to the Saudi Arabia of the world. I'm not happy about the fact that the Saudis have as much power as they do and they have as much power as they do partially because they stole the oil that was developed by western countries but but then also because we don't develop our own resources. The United States is different because they allowed for fracking so they discovered fracking and then they let it loose you know and one of the reasons the United States is struggling a little bit now is because Biden when he came in said he's launching a war on fossil fuels so nobody's gonna invest in fossil fuels if the president says there's a war on fossil fuels so investment went down and they weren't ready for the crisis.


Finally, we've had our final lecture. So last time I was close to the end of the proof of the main result for SL2. So as I was saying, it was proving that any set in SL2z is at growth, as long as it generates a group. And as long as it's not as large as the entire group. So we were in the middle of this. Let A generate, as I said, the group. For simplicity, we assume that A is its own inverse. That E is that it contains all the inverses of its elements and that it contains its entity. And so the conclusion is that either in a bounded number of steps A grows by a power or in a bounded number of steps, well, not even in a bounded number of steps, right away, really. Let's just put it in the simplest way. In a bounded number of steps, you get something bigger than g to the power of delta. In fact, one can strengthen both of these statements. One can replace A3 and this one can in fact replace this by A cubed being already the entire group. But we will prove this marginally weaker version, which is more than strong enough for all practical purposes. And you can deduce these strengthening from the version I'm giving you. At any rate, we were in the middle of the proof or rather close to the end of the proof. But the case we left for today was the most important one. It's when you have some pivots and some non-pivots in g. It's a typical induction step. So I remind you briefly that a pivot, so it's a reminder. We call psi a pivot if this map going from A to T of k plus minus c, D of k more or less plus minus c. So which map it's given by? Well, it takes A comma t just to A psi T psi minus 1. So it's called a pivot. It's called a pivot if this map is injective. All right, so the induction case is what happens when you have that there exist pivots and non-pivots in g. As I said last time, in order to induction, you don't really need an ordering. You just need generation because A generates g of k. There is going to be some pivots in g or rather some non-pivot in g and some A in A such that psi is not a pivot but A psi is a pivot. In general, whenever you have a set that is non-empty and not the whole group and you have A generating the group, then you are going to have some element of the set that is taken out of the set by the action of that element of A. Very well. So since psi is not a pivot, as we saw last time, this implies that there are many elements of psi T psi inverse in A4. Very well. But at the same time, A psi is a pivot. So this means that phi psi is injective. So as usual, you have that you consider. So we're going to consider what? Phi x. X psi. Yes, phi x. A x, yes, A x, phi A x, exactly. So all right. So it's going to be injective on the elements of the torus when you have A here and when you have the torus A psi T, A psi inverse here. Intersection, we say k if you refer. So here it's going to be injective. So it's image, it's going to be at least one fourth A A4 psi T psi inverse, where the four just comes from the modulus here. And this is at least a constant times A to the 4 3 minus O of delta, simply by this. And so you're done. Because this is, of course, the A psi is going to be, so, yes. And what do we have? Yes, so all of this, all of this map is going to be contained in, did I, so this is correct, right? Oh, the way I defined it, I really mean, I really mean this, excuse me. So, so A psi, this is good. So, simply this way. Excuse me. I wrote out the definition of phi x psi. And, of course, this is at most A to the 7, because all of these elements, this is an intersection, so we can see it from this perspective. And K was just, well, it doesn't matter. K was just 4, I think. So this way, this way I wrote 7. So it's fine now. So we have that A to the 7 is, at least, A to the 4 3 minus O of delta. And we had a, so we have growth. We had a contradiction to the assumption that there was no graph, and we had that. All right. So much for SL2. And the same ideas that I have presented. So it is a contemporary proof, if you wish, an updated proof that is streamlined and that is also generalized as marisily to higher rank groups. So the same ideas apply and give you growth in SLN, in SON, and so on. Now, as I said last time, there is a sticking point still there, and that there is a dependence on N. That really is there for results of this type, but that shouldn't be there in the consequences. Those problems are tied up with the fact that no useful diameter bounds were known for symmetric groups until rather recently. But now they are known. So we have some hope to advance. So let me devote the rest of today's talk to the symmetric group. I will not. So this has been proven three or four times right now, so the proof has become shorter. I cannot yet give you a full proof in one hour or 50 minutes for a cement, but I will try to give you several of the key ideas of the proof. So giving a second proof of this is a difficult exercise for the audience. All right, so what was the previous bound known? So cement is just this group. It's a group with same factorial elements. It's a group of all permutations of N elements. So what was known? So for G, cement, or really this was known for any subgroup. Any subgroup, they're off. We will have that for any A that generates G. So this was known. This was by Cherish from the late 80s. So it was known that the diameter of the group is bounded by, well, basically E to the square root of N log N. So this is, of course, very, very far from being a power of N. So what we showed was not a power of N. We showed something close to that, what's called a quasi-pollinomial bound. So for G, cement of N, or we can also show it for any transitive subgroup, they're off. See, this bound is actually tied for non-transitive subgroups. This is 1,4. It's not 2,019 yet. It looks like 2,019. I don't have the power of prophecy, and even Anna does not take that long. So for any transitive subgroup, so let me explain. So this result is actually tied for some non-transitive subgroups. Because if you have a two-cycle, a three-cycle, a five-cycle, all these joined, then you will have something of huge diameter, a cyclic group of huge diameter. So if you have a transitive subgroup, you can actually deduce it from this result. But using caveat, using classification. What do you consider that moral or immoral is your choice? But I have to tell you. I am resolved from Babay and Sheresh, 92, using classification. So the classification here, by the way, is also the one that tells us that basically the two cases are the matrix groups. The matrix groups and C-men. There's basically nothing much else to study. All right, so for G in C-men, as I said, G equals to C-men, as I said, and any A-G generating G, we prove that the diameter is what? At most, e to the log n. So as you can see, there's a difference there. And this is the kind of bound that gets called quasi-polynomia. Now again, I have to warn you again. Yes, the proof does use at one step classification, not just to obtain this corollary, but the proof itself. I don't like that. It will be very nice to have a proof without classification. And it will probably give three instead of four. What do you get without classification at all? No, I don't get something in the sense that most of the proof does not use it, but there is one step that uses it. And if I knew how to do that step without classification, I would. But I mean, it could be that you still get something better than that? Well, I suppose one can... Sorry. I imagine, but I haven't tried. And it will be depressing, I think, because everything will work out except for detail. So I think it's better to try to just replace that step so that it works well without classification. Because it's really a trick. It's a way to handle a special case. So some notation. Just to confuse some people and not to confuse others. The nomenclature for actions is different in tracymetric groups. So people write x to the g instead of gx and x to the a or x to the h instead of for the orbit. They also write ax instead of the stabilizer of x and a. So, of course, then... It's actually more intuitive. People in symmetric groups always like to blame the Lambert for the other convention. Now, it's also important there are two kinds of stabilizers. The set-way stabilizer. So that's when you stabilize the set. As I said, that's when you take it to itself, but you might permute its elements. And the point-way stabilizer. So in the case of the... Just so as to wrap one's head around this. In the case of the action of a group of itself on the point-way stabilizer, the set-way stabilizer is a normalizer, and the point-way stabilizer is a centralizer. Very well. So let me talk a bit about random walks, which are exactly what they sound like. You walk on a graph, taking a random decision at each step. But first of all, let me tell you what kind of graphs we will be working with for the most part.  The Schreier graph is not a graph that screams. Rather, you have that the set of vertices is just x, and the set of edges is x, xg, x in x, g in a. So a Schreier graph is a special case of a Schreier graph. It's a Schreier graph for the action of a group in itself. So this possesses several of the nice properties of a Schreier graph. It's regular, connected if the action is transitive, and symmetric if, as is usually the case, a Schreier is our usual convention. Very well. Excuse me. Yes. Is there a conjecture for what the upper bound should be? Enter the hour of one, and we don't really know what hour of one is. Some people believe it's 2 plus epsilon, but yes, enter the 2 plus epsilon will be the most optimistic conjecture. Already enter the log n would be very nice, which would be e to the log n squared. Yeah, 3 would be nice, but not so nice. 2 would be nicer. So a little lemma. So what do we know about random walks? Well, a priori, we know almost nothing, almost nothing. Already, just because a graph is connected, well, for any connected regular graph, you're going to get a very, very, very, very weak expansion type result. In the sense that you will know that lambda 1, the eigenvalue that is not the one that corresponds to constant functions is a bit bigger than zero, and moreover, you can give a lower bound, a non-constant lower bound for lambda 1. So, and that turns out to be useful. So this is folklore. And that tiny gap leads to an upper bound on the mixing time. Regular, meaning every vertex has a greedy connected symmetric graph. The same properties as that. We send vertices. In what? A case such that a random walk with at least k steps ends at any x with probability that it's almost uniform. An infinity, a very strong infinity sense. So after these many steps, yes, a random walk. So you drop somewhere in Paris, nothing about anything. It starts walking very randomly. And after enough time, you will be anywhere in Paris with the same probability without knowing anything about the map of Paris, even at the stage of giving the bound. So just let me say that if any were very large, this bound would be horrible, right? But this is very useful when any's not so large. All right. So what is the idea of the proof? So let me tell you very quickly how do you prove that there is a non-zero gap that you can actually bound if it's not a constant. So it's more comfortable here to work with the adjacency matrix, which is almost the same thing as the laplation only without the subtraction. So for f non-constant and real valued and any r between the max and the min of f and s equal to the set of v's of vertices for which the value of f of r is bigger than r, you're going to have that the sum of a alpha over all vertices in s is a little smaller, really a little smaller. That's the sum of just plain old f of v. This is very easy to prove. It's just that your values over this set of privileged points get slightly contaminated by at least one guy from the outside because the graph is connected. It's just interconnected. And then the only thing we have to do is to apply this to any non-constant eigenfunction. You might say, well, eigenfunctions might be complex valued, of course, but because it's a symmetric graph, the eigenvalues will be real valued and the eigenvectors can be taken to be real valued. Since you can take combinations of eigenvectors with the same eigenvalue, so you worked only with real valued eigenvectors. So you apply this and you have that the adjacency operator makes any non-constant eigenfunction slightly smaller in a set or the sum of a set. That means that the eigenvalue must be a little smaller than one. So that shows the gap. So it's a small, non-constant, let me emphasize, non-constant eigenvalue gap. And this leads, as we know, to abandon the mixing time. So as I said, this is weak, but fortunately what you have for the symmetric group is that even though its representations are in no spatial geometric structure, they do have a very interesting... You do have representations that are spatial in a sense and that they are representations that are much, much smaller than the group. Non-trivial representations that are like that. So we apply this sort of thing to a shire graph. It's useful when n, which is just the number of elements of x, is much smaller than g. So the example would be, in our case, just g acting on the elements one up to n or on k tuples of distinct elements are off. So let A be in Cm, A equals to its inverse, A equals to 3 transitive. So it's an example of how to use this. It's a proposition to the Baba Bil-Sharash. It's by 3 transitive, I mean it acts transitively on 3 tuples. On 3 tuples of distinct elements and you take a g in A whose support is neither the entire thing or zero. The conclusion will be interesting. It will be non-trivial, smaller than the inverse 3. So this will tell you that you can do something interesting when the support of g is not all of n and it's at most the inverse 3. That can also be softened. Then, and we will also use some sort of epsilon, then there is a g' in A to the l where l is not too large. It's n to the 7 log n over epsilon, such that the support of g, oh, by the support of g, what is the support of g? I should have explained. It's just a set of elements moved by g. That's the number of elements moved by g. So this is at most 3 plus 3 plus 3, 1 plus epsilon sub g. By the way, this 3 plus here, it's here because we want it to be there. It's useful. And in fact, I will sketch the proof briefly. I will just keep the detail of how to get this 3 plus. Excuse me. Do you have a g' somewhere? Thank you. This is a prime. But sub g here. So as soon as sub g is, say, 1 fourth, or say if sub g is 1 sixth, this tells you that sub g' will be 1 12. So it's a sketch of a proof. Well, the basic idea is this. So say that you have, it's called s, the soup of g. It's for simplicity. So for sigma and c-man, we can define h to be, what happens when you coordinate g by sigma. Support of h is going to be just s displaced by sigma. So sigma of s. And the support, the simple exercise of gh will be a subset of s intersection of sigma, s section of sigma displaced by g, and s intersection of sigma, and sigma displaced by h. So the support, the size support of gh is going to be at most three times the intersection of s and s-sigma. And now what's the idea? If you could pick sigma just completely at random, then the intersection of s and s-sigma would, on average, be how much? I asked, well I, oh, thank you. So how much for sigma random, how large would the size of s intersection of s-sigma be? In fact, all right, you are s. I now take a random permutation sigma and I end up with s-sigma. You are half of the room. What is the size of s-intersection s-sigma likely to be? People think this is a too personal question. Well, so you're half of the room so s-intersection s-sigma should be of size about n over 4. Right? So for sigma truly random, so uniformly picked in cement, we would have that the size of s-intersection s-sigma is s-n times s. So your size is half because you're half of the room. Now, can we see that, can we see the proof of that statement in detail? Yes, and I will have to give it to you in detail in a few lines because we don't actually have sigma random. So we'll actually have to show you that the argument is based on just a very soft property of sigma that gets stimulated very well by a random walk because of this. All we need is that sigma takes any element to any other one with probability just about the same for any paraphernalia. So we don't really have this but we have a substitute that is actually quite good because this expected value is a sum over all x prime and x of the probability that x be taken to x prime. And so what is a probability for sigma, an outcome of random walk, not of lengths infinity but of reasonable lengths n, big n and d will both be reasonable. They will be bounded by n, little n. So for the outcome of a random walk what is this probability? Well, by this the probability is about 1 over v, 1 over n. This probability will be at least s over n. Very well. By far the problem. That's it. That's almost it. Of course this gives you the result just assuming one transitivity and without the 3 plus here but I will not go into the tricky details. So just let me tell you what an immediate corollary of this is which was a very useful result that was around before us and which we do use to treat the special case. But what you should really take away from this is not just a result itself or a specialist corollary but really the idea that some probabilistic proofs get mimicked very well. So because we really want a random sigma no, we just wanted to show that there exists a sigma for which this is true. Now the probabilistic methods ever since Sardosian company consist in saying well, in order to prove that a good sigma exists we pick a random sigma and we show that it's good with positive probability. So this is the stochastic version of the probabilistic method. You are not allowed to take a random sigma as in any element of sigma in which equal probability sigma is the outcome of a random walk. And in some circumstances it is just as good. This is the moral of the proof. So the corollary, if you have a in cement conclusion, always remember to state the conclusion, then the diameter is bounded by n to the 8 log n very well. Babaik once got upset when I pointed out that he had put 7 in his paper and it's really an 8. It's the same thing.  To G, to G prime you just bracket away. Because every time it gets smaller by actually much more than a constant factor because it decreases more than exponentially. You're done in no time. All right, so this is useful and it's also an example of the probabilistic method. So let me be like every year this is the moral explicitly just with random walks. All right, let me tell you another story now. It has to do again with something that has been a life motive of this mini course, namely how to adapt results valid for subgroups, known for subgroups so that they are valid over sets, general sets. Just sometimes it will be states, it will be h by a and sometimes h by a to the k for some moderate k. Let me call this section large orbits and stabilizer chains. But the set up is really the splitting lemma. So I will actually state the splitting lemma in both it's all the new variant but I will just give you the old variant and then I will actually give the translation of the proof to you as an exercise which you can now do. Following a bit this model user on the walk. So this is called Baba splitting lemma which is again just like large and pink. Baba in fact here was working in the early 80s back when it wasn't clear even whether the people who were doing classification believed that classification was done. So he was trying to with some success to give much shorter proofs to give much shorter proofs of some consequences of classification and in the process he showed the following. So let's have a two transitive subgroup of cement meaning a subgroup that acts transitively on pairs and say you have some subset of sigma and assume that sigma splits a lot of pairs what do you mean by splitting a lot of pairs? So assume that there are at least a positive portion rho for the pairs which end of pairs just any pairs of distinct elements such that sigma separate so for a positive portion of pairs sigma separates half and b what do you mean by separates there is no g in the stabilizer of h with respect to sigma the point wave stabilizer with alpha g is equal to g so it's what's called sigma separates a and b. So this is a true definition if sigma contains other a or b but it can also happen that sigma separates a and b for example it could happen that this stabilizer is trivial but in this case it separates everything but you could it could also separate something very well and what is the conclusion it's always good to have a conclusion and there exists a sigma h a sigma in h which is not too large just of logarithmic size such that the point wave stabilizer with respect to sigma to ds is equal to the union of course of sigma to the g for all g in s alright so how do you prove that so if you split a positive proportion of full couples then a slightly enhanced version of yourself splits everything so let's point out first that if h is a continuous big h if g prime takes alpha to beta then hg prime h inverse takes alpha h inverse to beta h inverse tautology tautology like all of math but what do we do now so now we consider taking random things so for hh taking at random uniformly h inverse alpha beta to any alpha prime beta prime with equal probability why? simply because all cosets of certain of several parts of the same size it's a basic reason but you could also say that this is intuitively clear so the probability that h inverse takes alpha beta to some alpha prime beta prime such that sigma separates alpha prime and beta prime is how much so we know that sigma separates at least a proportion of full pairs so a probability that h inverse takes alpha beta to some alpha prime beta prime that is separated is at least how much is at least rho very well so but at the same time if this happens if h inverse takes this the consequence would be that no g prime in h sigma h takes alpha to beta meaning sigma h actually separates alpha and beta why? simply because of what I said before alright so now let us do let us choose a lot of random elements independently let s be a set of r random elements of a so by the above the probability that for every h in s there will be some element in h sigma h such that alpha g prime b so the probability that for every h in s sigma to the h will not separate alpha and beta well for every given element of s it is 1 minus rho the probability that this never happens is just 1 minus rho to dr but this would hold this looks like a very strong condition so no wonder the probability is small but this would hold if there were some g in h stabilizer it would go to ds because that g would be good for all for all these purposes if there were a g prime h sigma s with alpha g prime equal to beta so the probability the probability that this happens we have shown is small so the probability that this happens for some pairs is at most n squared is at most n squared minus rho r. But this would happen. So if not true, if these things where you were bounding were not true, then, well, what do you call an element that doesn't take any element to any other element? So what's its identity? So if your group separates everything, then your group is just consistent with the identity. So all you have to do is set r large enough that, so the probability becomes, and of course it's just constant times log n. So in that case, with positive probability, h sigma s is your identity. But if something happens with positive probability, it happens. Again, the probabilistic method. And so there exists an s of size r, of course, such that this is true, which is all we want. So that's it. It's approved by Dito Babay of this using the probabilistic idea. And now your exercise is just to translate this so as to give the same result for sets. And then I will tell you briefly how this result gets used in the overall scheme of the proof. So what was the, I will give you the statement. So if a is in 7, assume as usual that a is in the inverse in a, a generates something to transitive, sigma is some subset, and assume that there exists many pairs such that, well, it's not a subgroup now, it's not a, it's a to the k1, where k1 is 9 and 6 log n separates alpha and beta. Then there exists an s in a to the k2, where k2 is something like that, and 6 log n, such that the point, the point way stabilizer of a a prime sigma s is the identity. And the s is of logarithmic size. So yeah, just translate this using a random walk. Very well. All right, so how do you use this splitting lemma within the proof of growth in cement or of small diameter in cement? Because you see the, actually the, a result of type 8mc, 8mc is bigger than 8 to the 1 plus delta, that's not actually true over cement with delta constant. So the recent idea that goes back in part of Peeber, the concept of stabilizer chain, of building stabilizer chains, comes actually from the theory of algorithms over symmetric groups. The expression comes from, goes from Sims in fact, long before. So all right, so what can you use from a conclusion like that one, that a inverse, the point way stabilizer is the identity. Well just by pigeonhole, you have that the only way that this can happen is for it to be bounded by n to the sigma s. And that implies that sigma is at least log a log n squared. So the assumptions then can happen only once sigma is already fairly large, not before. So the only way that your sigma will split, with the splitting lemma is telling you then, is that if sigma splits a positive proportion of full pairs, sigma must be pretty large. So when sigma is not yet very large, so when sigma is smaller than the constant, so let's find this to be L, log a over log n squared, there are less than the proportion of splits by a to the k1. So we're going to use that. We're going to use that for all sigma that are not very large, you are not separating opposite portion of full pairs. And in particular you have large orbits, because if you don't have large orbits, if every orbit is of size most than over two, then at least half a full pair is separated. Okay, so what do we do then? For any sigma we obtain, so first of all there is some guy who has long orbit, but even the point way stabilizer of that, because it's just one point, is going to have some long orbit, you assume that, then you can see here the point with stabilizer of alpha 1 and alpha 2, still of size much less than this and so on and so on, alpha L, such that alpha, and then you stop very well. There you stop and this is what's called a stabilizer chain. So it means that using the splitting lemma you show that if we say for a you have any, really any set a that generates something to transitive that is not tiny, you're going to be able to find a point with a large orbit and then you're in the point with stabilizer of that, you're going to find some other element which you can take to be the second one with a large orbit, something like this, and then in the point with stabilizer of those two you're also going to find some sort of large orbit and so on and so on and that's extremely useful. So let me show you some things that you can do with that, very quickly. So yes, so an exercise is that then this implies that a prime L intersects, so a prime is justifying to be fixed, that a prime L intersects at least sigma L, L prime cosets of the point with stabilizer contained in the set with stabilizer. So just choose set to be a set of representatives. And so what are these representing? What are these cosets? What are the cosets of the set with stabilizer in the point with stabilizer look like? Well, you can just restrict them to sigma or rather to the, yeah, you restrict them to sigma and they are acting on sigma and they are giving you these many different elements of sim of sigma. So you end up having, in this way, you can identify these elements to one, two, up to L and what you have is that you have built a prefix, so you have that among the elements that fix one up to L as a set, you have almost all permutations of one up to L. So that's already the beginning of the light in this proof. And now that you have this nice prefix, what can you do with it? Well, cement, sigma, and so in particular set, acts on the point with stabilizer by conjugation. So sigma here is first just alpha one. I say of course, but I'm going to be slightly tricky. I'm not going to go up to alpha L. I'm going to go up to alpha L minus one. All right, so it acts on that by conjugation, but we know that the point with stabilizer has a long orbit. Why? Because it was slightly tricky and I did not put alpha L minus one in sigma. Long orbit, namely, that of alpha L. And well, we can assume, as it turns out, that it acts like cement or all time in it. In fact, you need much less than that. So, but warning, this is a step that uses declassification. In fact, what we need is not really for it to act on cement or all time like that. We just need to act, we need to be able to assume that it acts like a two-transitive sub-problem on that. Or even something a little bit stronger, namely that it's not obvious. We can find an S of constant size. So it is acting on this orbit and it acts, yes, it acts two-transitively on the long orbit or almost the entire long orbit. If you find a way to show this, and it's entirely possible that it could be done without fluctuation, excellent, publish it. It acts on sim L. So on the line above, you want it acts like sim in Sweden? Yeah, no, it's act like sim of the orbit or sim M, whatever it is. Sim or old, yes, very well. All right, so now that you have something, so you just, maybe in clever, you find a set of generators of at least a two-transitive thing on that. So you have on the one hand, yeah, something in the point of a stabilizer that has just a couple of elements that generate something with a long orbit that are two-transitive on that. And on the other hand, you have these long prefix acting on that way conjugation. So what do you get? Surprise, the orbit stabilizer serum comes to the rescue. Because you have this tension in between. So what is going to be your action? So let's act on S, or on the entire group really, but that's on S by conjugation. By orbit stabilizer, either there exists an element of G commuting with every, so an element of that, that is not the identity, commuting with S in S. But if it commutes with every element of S, it commutes with every element of the group generated by S. But the group generated by S is two-transitive. And if you commute with a two-transitive, something that is two-transitive on a set, you must be what on that set? Not you? Yes, in Siemens, if you commute with a two-transitive subgroup and with every element of a two-transitive subgroup, what must you be? So if you have enough elements that, you know, you will always find one that takes you to U, then if you commute with that element, then you're not making a distinction between those two. If you commute with, and if that happens for all pairs, you must guess B, the identity. So G fixes every X in the long orbit. But the long orbit is long, so it means that G has small support. And then you use the result that I told you about by Bill Sherrish. So the alternative is, of course, so either this happens or what happens. Or you're going to have a long orbit. So there exists an S in S. So in particular, because you're doing these two S, there's going to be some element of S. So you're doing these two, to some k-tuples, for a case size of S. So you're going to have some S in S, just like pigeonhole, the one with the largest orbit. You're going to have some S such that this orbit is of size, at least, set to the one of S. So every tuple is taken to the different k-tuple.  But what does this orbit leave in? So this is contained in A to the, well, whatever it is, N, O log N, stabilizing sigma. So this is contained in the point by stabilizer. So you have constructed a lot and a lot of elements, because sigma is of size L factorial, basically, or rho to the L times L factorial. You have constructed a lot and a lot of elements in the point by stabilizer. And now, what can you conclude? You cannot conclude right away that you have growth, but you do have, because maybe you already had a lot of elements in the point by stabilizer. But now, with the elements in the point by stabilizer, where, moreover, you have a long orbit where things do generate something like cement or all 10, you can just recur. You can now work on that point by stabilizer. And there, you will find a long prefix. And there, you will continue working and working. It gets mixed here, but this is a basic idea. So instead of having that, or really working by induction with a set A that grows, you work with a stabilizer chain that grows. So then you have alpha 1 prime, alpha L prime, and so on. In fact, you have to repeat this step log times every time, so this does not get shorter and shorter. But it does work out. So this was a very brief sketch of the main argument. But what I really want you guys to take home with you is really this action of the set by stabilizer on the point by stabilizer. So once you get the deception, it's really moving around. Otherwise, you'll be yes. Then you are building up a lot of things in the point by stabilizer. That's what makes you win. So I think that's it. And I think I've already stated the main problem. So just to repeat one of them that was already mentioned, people actually believe that the diameter should be genuinely polynomial on n. It might even be n to the two plus epsilon. And some people believe that when you pick a random pair of generators of Cmn or all 10, in fact that should give you or might give you an expander graph with probability arbitrarily close to 1. So that's it. So how complicated is the use of the classification? Is it a direct consequence? It's fairly direct. I mean, we use this. I don't even mess with it directly. There's a paper by Babay and Sheresh that just shows you that the diameter of a group with respect to any set of generators is basically bounded by the diameter of the largest symmetric or alternating factor. So that's the only way in which we use classification. Yes. Is anybody working on trying to do this with SLM as I was talking? So trying to combine the two things? If somebody's trying, they are not telling me. But they might be trying, yes. But if some were in the audience, you would say this is not a mathematical question. Working recorded. Please work on the problem. It's very interesting.


Hey, you see me. Hey, what's up, brother? What's up, bro? How are you doing?  What's going on? Yeah, I have some difference. I have these lights. I have different color lights in my room. So now, anytime I have a red day, I change the color of the light. So this has been pretty green for about two weeks now. So I'm just keeping these same color lights. It's not gonna work, guys. So the whole concept behind all these superstitions is just to get rid of anxiety. Yeah. I'm swaying all that. So as a trader, guys, the worst thing you can have is anything that's giving you stress, right? So whatever it takes, man, riddles, routine, whatever it takes, guys. As ridiculous as it is, it's like you said about, you made a feng shui Instagram live. So that explains a lot of stuff too. It's all about the power of the mind, right? If you feel confident, that's what happens, right? Like where your lucky shirt, right? Where your lucky shirt, something's gonna happen. It turns out you lost. So that's lucky shirt, no longer lucky. I was wearing his lucky pajamas for a year almost. That was pretty crazy, yeah. So let me talk about a couple things, guys. Well, like Val said, we have our meetup coming up. It's going to be August 19th to the 21st in Provo, Utah. So we're going to formally, formally announce all the details pretty soon. But the way it's going to work, guys, is going to be a giant meetup from 11 to three in Utah where we're going to set up a panel, we're going to talk, we'll bring on some members. We'll highlight some stories and kind of talk about the market and what people can do to improve and make money. Because like Val said, the market really isn't hard. People are just not focused on the right things. And then after at night around six or seven o'clock, our broker partner success trader is going to be hosting a mixer where you get to meet the broker behind the scenes. That helps us find shares to short of these companies, guys. So during the day, you'll have access to me, Val, and the entire MIC team. And at night, you'll have access to success trader and their team. So it should be a really, really fun day. And we'll probably have the schedule that we have set up, guys. It's going to be Friday night. It's going to be like a low key networking meet and greet. Saturday is going to be the meetup during the day. Saturday night is going to be the mixer with the broker.  Yep. More information is going to come. You do not have to be a member to attend. Yep. It's free for all, everyone. It's free. No time shares, nothing that we're going to sell you. You guys can show up and just pretty much get the chance to meet us, right? You know, we are not afraid to meet traders. We are not afraid to meet anyone because, you know, we don't scan people. So there's nothing to be fearful of, you know? Yeah. Especially these hard times, guys. It's good to make friends. I mean, this is the opportunity because you know what, man, when the markets are heating up, it's very difficult to take time off. It's very difficult to study and learn new things and watch videos, right, guys? So we always wish like, man, I wish I had more free time.   Yeah. So let's talk about the market, Bao. So you had a really good topic that you were talking about today about the market being tough. And, you know, we're still showing up every single day and we're making money. Both market bear market doesn't matter.  Everyone except us. And that's why we're getting trolled the most. Yep. So let me talk about that right now. Let me put some more. So we always hear people talk about, dang, these markets are hard. I'm not making money. What's going on? Bear market, this bull and giraffe market, that. But to be honest, guys, to me, it's all, I've been trained for a long time, man.   What is a bear market? What is a bull market? When you talk about a bear market, you talk about a macroeconomic cycle. We are day traders.  It doesn't matter to me because we are so short-term. We're a day trader. Within a day, happy to have a bear market. So those are the terms that economists and all these people give you to kind of explain it and also scare you, right? And so the only thing is this guys, the process, whatever your process is, is your process. Okay.  The problem people have is this. It may take you two days, three days, a week in order for your setup to come. And that's because, you know, the markets are slow. And so when you're talking about a bear market, you're talking about a very slow market where stocks have a very hard time of going up because there is no hype. People are scared to buy. People have no money to buy. And so you have to be realistic. You have, back in the, back when the markets were hot during the meme stock, you know, AMC and game stock, they're going crazy. So the people that think that that's a normal market, what's going to happen is when things slow down, they're going to try to force it. They're like, well, we, we gave the analogy in the room today, like trading is like fishing, right?   That's what they call it fishing.  You're going fishing. It's not called catching, right? It's that simple. They call it catching every day when you sit with no catches. The other days you have 25 fish and you cannot fit it on the board on the boat, right guys? But do you change your, you can. So this is the thing guys. So same thing with trading. Okay. So the bear market for a fish is a drought, right? There's no, nothing to catch. And so he can change his process. That's, that's the whole thing. So I talked about, we talked about this a little bit guys. So for trainings like this, okay? You have a process and your process is revolving around a setups, whatever you define to be an A setup. Okay. And the problem is though, waiting in a bear market, what bear market I mean is like a slower market is that these setups do not happen often. And so you're just sitting there. And so it's like, it's like going to a desert where you have no water and you see illusions in the, you see like lakes everywhere because you want to see that water. You're so thirsty for a trade. And so every, so the C and B setups start to look like A plus setups because you're like, man, I've been sitting here for two days. There's nothing good. This is close enough. And so you're getting desperate. The problem is getting desperate. And when you get desperate to try to offset the lack of opportunity, you oversize, right? Imagine you're the point of using size guys is supposed to be when there's an edge. When you know that the odds are in your favor, that is when you size, you don't size because it's slow. And what a lot of people have done in this market, whether it be, you know, we're in July now, about seven months, what people have been doing is they've been sizing in just because it's slow and not sizing in when there's opportunity. So today for me, we had this stock MTEK, let me show you guys MTEK. We, for me, I identified that this line over here around 170 needed to break. It broke. I shorted it. I got out and now it's going to even lower without me, right? And this is a hundred dollar trade. I didn't get rich off this trade, but you know, these markets were still finding opportunities. And just because slow on that specific trade, I'm not going to use 10 times my size. I'm not going to use 20 times my size. I'm actually using less size because the opportunity is not in my favor, guys. So if you find yourself struggling in this market, are you over sizing? If the answer is yes, size down. If you are trading stocks that you are not normally, like let's say about, you are a bass fisher and the bass season is dried up, you're not going to go into the shark infested waters and try to catch a shark because that's a different type of fish. If there's no bass that day, there's no bass that day.  So that's what a lot of people are doing, guys. They're over sizing. They're over trading. They're getting out of their selected niche and trying other stuff. And it's okay to experiment, but people are experimenting with too much size in this market, guys.  So another word that we always talk about, do not deviate from your process. And that's what you guys are doing. You're deviating from the process because, you know you get, the human nature, you know, you get bored, man. When you get bored, you start to start forcing things. And so that's the only thing. So we always talk about what makes a good day trader, guys. You don't have to be smart to be a good day trader, guys. You have to be disciplined. The best day traders are very disciplined individuals.   It's because of the fact that there's less opportunity. And so you are trading subpar setups. Okay? Imagine if you don't trade those subpar setups, making zero is still better than losing money. So you see a lot of these experts, they, they expect to make the same money that they made last year. That does not work like that. Same thing as a fisherman. You caught 10 fish yesterday. Today you may catch none. But if you force yourself, like Alex says, to start catching other things, a shark might jump out of the water to eat you, right? So, you know, so you have to be very careful, guys. You, you, to me, see the thing with trading is this is not a guaranteed occupation. You catch what you eat, what you catch. It's, it's like, it's like a caveman. If you don't, if you don't have safeway or, or CVS to go buy food, you got to go and hunt the animals, right? And, and today you may not catch any animals. Just because you caught too big, you know, pigs yesterday, he doesn't mean today, you're going to catch the same thing. Those pigs are scarce. So as a trader, guys, you have to be disciplined to be humble to say, so I, I hear a lot of traders say, I'm taking a pay cut this year, pay cut. I didn't know you had a salary job.  Trading's an odd probability game.   They feel entitled. Okay. And they may have one or two good years and they think they're the expert traders and they have not seen the cycle yet.  And the market has been going up for how long now, like 14 straight years. And so they're printing money like crazy. Now, we don't know what's going on guys.  So you cannot be complacent with the traits that you do. You have to understand the markets and, and change and adapt with the market. The algos are changing.  And on top of, on top of changing with the markets who about, I think a lot of people have to understand, adapt their lifestyle as well. A lot of people, they live above their means, right? Let's say you're making $50,000 a year, they're spending $50,000 or $60,000 or even $70,000. A lot of people are going into debt guys. So be able to adjust your life as well. Right. If you went out to dinner five days a week when the market was hot, now you have to adjust to maybe go out to dinner once a week or not even at all. Right. So a lot of people, what they don't do is they're, they don't reduce their expenses as well. They don't live underneath their means. Right. So a lot of people are struggling, right? Forget the regular market. Oil is up like crazy. Groceries are up like crazy. Everything is crazy. Travel is expensive, but a lot of people have not adjusted their lifestyle to say, you know what, the money's not coming in the same way you used to. I'm struggling a little bit right now. Maybe I'm not going to fill up my car with 93 octane.   You got to keep wasting your money. Something that about taught me is you have to save your money, save your money. And a lot of people are still wasting their money. So keep that in mind as well guys. If, if you're struggling a little bit in the market or the real world or whatever it is, it's not time to keep spending and making these crazy credit card purchases and buying yourself an 85 inch TV. It's time to kind of downsize a little bit too, right? That's why I try to live underneath my means so that when hard times like this come, whether it be in the world or whatever it is, I'm not affected because I have not been wasting my money, you know? Well, this is all a part of learning how to run your own business. Trading is running your own business. You're an entrepreneur, right? And so as, as any business, right? And when things are good, you save, you stockpile the money because you know what, man, there may be slopehairs. And so these are the learning experiences that you go through.  You expect every day would be good. But then you talk to an experienced trader who's been doing this a long time.   You may have long drawdowns. You know, these are the things that you learn along the way. So you can either learn them on your own or you can have a mentor that's been through there. So you do not have to lose and learn on your own. Most people will never make it out of these cycles because they go broke before they learn.  And then later on, the same thing with life, man, that your parents have been through there. So get your parents advice on where the hell it may be, like dating or business or school or whatever it may be, guys. So same thing. So my whole thing is this, guys, that the arrogance keeps us from asking for help. Like men, I like to ask for directions.   Oh, no, I got it. Right.   Same thing with women, right?  It's true. It's true. That's that's the thing. And a lot of people, they don't know how they made money the last two years, right? Because they got lucky. So these people made money 2020, 2021, all of a sudden 2022, they lost all and more, right? And they don't know why they don't know what the hell's going on. So guys, this is again, you know, we are here to help. We have been here consistently for almost four years now, every single day coming on here teaching. But what we teach is boring. We have no hot girls. We don't have any fancy things to show. And that's why no one really watches these videos. So the small amount of people that watch these videos are changing their lives. And that's why we do it, right? We do it to help anyone that wants to learn how to be helped. We are here to help the serious people only. And, you know, whether that be 100 people, 1,000 people or 100,000 people, it doesn't matter because we're still going to show up. We're still going to do our part. And if you guys are here watching this, you already have an edge. You have a YouTube channel of 1,000 plus free videos of just straight stock market university education. So start using it. The markets are slow. It's a bare market. Why not use the extra time you have to watch some videos? That's it. This is a weird thing with our industry, Alex.  Even LeBron James has a coach. Everybody has a coach. Tom Brady has a coach. What makes you think that trading, you don't need a coach?  Only in trading and it deals with money.  And the weird thing is, no one's successful. Not if people become successful. So what they do is they try to learn on their own, they go broke, and they assume just because they cannot do it, that it is impossible. But the reality is, what is the biggest money generator sector out there? Stocks in real estate.  How come they can make money? It's just because you can do it. So the whole thing is, guys, is the ego guy. Just like I'll give you an example.  I refuse to care.      And that's what it is, guys. So we don't, trading is not for everybody, guys. But my whole thing, Alex is saying, you don't need to make it as hard on yourself. Get help. Ask for help. Okay? If you do not ask for help, we do not humble down. This is not an industry for you because you will go broke.  Yep. It's your chance to come to the meetup, guys. If you guys are struggling traders, if you guys are successful traders, if you guys are business owners, whatever it may be, this is your chance to come and meet hundreds of other like-minded people that want to make money and learn how to improve their lives, right? That's the whole point. You know, when I first started trading, I would kill to have a room of hundreds of successful people helping me out for free. Free. If we wanted, we could charge thousands of dollars for this event. If we wanted, but we don't, right? We don't. So, I mean- Others do actually, right? Maybe we should. I mean, maybe we should, bro.  I think we just charged like 20 bucks, man. Maybe we should, bro. I don't know what that, like it's after a certain point, guys, like, we've done our part, right?  We're going on an island and you guys are fucked. So, you guys might as well enjoy the education.  And like you said, man, there's not going to be any more sales for a while, guys. So, we were done with sales for a while. We didn't enough and right now the markets are sucky, man.  You'd rather be studying and saving your money than forcing trades. And so that's why people are losing, guys. People are losing because they have not seen these market conditions before.  They're just not used to not trading. It's okay to make, like, what would you do, Alex?  I made like 100 bucks, bro. I mean, I made nothing. But to me, I mentioned in the room that I'm not judging myself based on my profits, guys. I judge myself based on my process. Today was a process day for me. And so what? The process didn't spit out money. Some days it spits out money like an ATM. Other days, it spits out 100 bucks, right? And sure, 100 bucks.  So there's nothing to complain about, you know? So, guys, I think we're going to wrap it up here. If you guys want more information, if you guys want to come see us at the Meetup, Utah, August 19th to 21st, we will release the information. If you guys want free education, we have a free webinar going at mindvestingclub.com slash webinar. You do not have to be a member to attend the Meetup. And I hope you guys are getting value from these videos. If you guys just are here looking to learn for free, go to the older videos of MIC. They are still relevant. They are high quality. And you guys could learn for free. And the webinar, tell them the webinar, this is meant for people that do not, are currently not in MIC. Correct. Mindvestingclub.com slash webinars for non-MIC members. It shows a free lesson and a live trade for me. MIC members already have access to this.   Non-fucking-stop trading. Non-stop. This guy is pulling money out of every single thing. Alex. Alex.      Look at this. Yeah, MTEC trade was pretty good. And they're still on, this is only fucking, look, he's still going, bro. And this is my trade, right? This is my trade. Nothing crazy, babe. $100 bucks.  We show everything. We guide you. We mentor you. This is what trading should be. MIC started on the concept of mentorship is the shortcut to stress. And, Bob, I'm very proud to say that we've stuck to our vision from day one and we've helped thousands of people, man. So if you guys want to learn cool, if you guys don't want to learn cool, we're still going to be here every single day. Love you, pal. And we'll see you guys probably tomorrow. All right, thanks for coming on, guys. Thanks, Alex. Love you, man. See you in the back of the room.