Basic Notions Seminar - Polymath 14

Let's hit record. We are recording. Yes. Going to put hide floating meeting controls. Okay. All right, everyone. It's a pleasure to introduce Professor Apurva Chari here to be giving [snorts] this this very nice basic notions seminar. Welcome everyone that is online. Uh the title is Polymath 14 from word games to analysis definition of abelian groups. Apurva, the floor is yours. >> Thank you so much again to Emmanuel and first to ICTP. So, yeah. So, I looked up online and it says that the basic notions seminar you have to explain the basic notion. So, my basic notion is abelian groups, which is probably as basic as, you know, you would start in an in actual math level math degree. So, but the as the other part of the title suggests, I will tell you about something different about this basic notion. The fact that you can define it using analysis. So, this is not something you see in a math degree, if you will. So, this is all based on joint work, which is now, of course, published with five other authors and there were inputs from many many other people. So, if this is as much the story of a theorem as the story itself, the story of how the theorem was proved. Okay, why is this not supposed to be clicked? Okay, so I will start more broadly from some the broader sciences. And since we are in ICTP, of course, there is no the broader sciences are definitely well represented here and studied. And there is a fundamental difference between theoretical or mathematics and the broader sciences. They are very highly collaborative. So, this is one example of one of the most famous modern breakthroughs in science, the genome project of the human genome. There is similarly the particle physics at the small scale the Higgs boson which had 3,000 authors says the same thing with previous year. This is crazy. And part of it is the Atlas collaboration which you can imagine is a ton of people together. This is the paper where they announced it. On the large scale in physics there is gravitational waves which also got on the well. And this is the paper itself. There are 100 and more more than 100 institutions more than 800 authors. And then there is a so okay so there are all these things from you know from the broader sciences. Of course this very theoretical and we don't really do such level such scale projects. And there is one there was one recent thing that started that was an initiative and that's what I will tell you about. So but it started out and still goes on with you know the since we are in Euler's I should have mentioned Euler as well maybe but yes all these people were sort of doing research by themselves. Newton did not know what Leibniz was doing and vice versa and so on. Ramanujan of course in India very famously was doing research alone until he wrote to Hardy. Andrew Wiles worked secretly three decades ago to prove FLT. Two decades ago Perelman worked by himself. One decade ago Yitang Zhang worked by himself. And so these are all these sort of individuals sitting in isolation with pen and paper kind of stories. Now they can sit in isolation with AI. Of course. But so today I will tell you about a modern collaboration mechanism called the Polymath Project. And this is an attempt to sort of involve many more people than just two or three into solving a problem. And I was involved in one of these problems and it helped answer this basic question about groups. So again this is sort of created with a broader audience in mind. And then groups are you know groups what are groups? Groups are really symmetries of course. As as a math audience we understand that a group is something that acts on a set. And so when it acts on a set it means there is a homomorphism from groups into the automorphism group. So into the symmetries of the set. It fixes the set as a whole and it just permutes things inside that entire set, right? Like rotation keeps the circle the same but changes the position of every point in the circle. Right? And there are of course groups show up all over the place. So in the physics there are these Lorentz group symmetries for instance. There are gauge groups in string theory. Uh in my in you know in biology one studies apparently um eigen values and eigen vectors and eigen spaces. And then of course vector spaces are groups under addition. Um and yeah for us groups mean symmetries or permutations. So here is the question. So suppose I have a group with a metric on that which is a metric. So the group is a metric space. The metric is left invariant and right invariant. Okay. When is this a norm? That is the question. So what do I mean by a norm? The norm is exactly what you would expect. The length of a the length of two times a vector should be twice the length of the vector. Norm of twice g or twice let's call it g >> [applause] >> twice the norm of g. Except that this is I would write twice g if my group was abelian and my group need not be abelian. So I write it as length of g squared. So I write it as have a distance satisfying this. Now what is the distance what is a norm or what is a length? Length is the distance from the identity under the group operation. So the identity of the group operation would be distance from e to e squared. And so that's exactly the same thing from there. The distance from e to g power n is n times sorry mod n times the distance. That is what I would mean by a norm. So what are examples of such groups? So one example is literally this board, the plane, and this is the example of the distance is the length. Right? So, this is a So, as we just said on the previous slide, these This is a board R and R 2 or yeah. This is a group under addition, and then there's a norm. And then the norm satisfies this property. So, this is the most standard example. Obviously, more generally, you can talk about a non-linear space or even more general Banach space, whatever. So, uh not just R N, so any Banach space uh any additive subgroup is abelian and torsion-free. So, what does torsion-free mean? Basically, torsion-free means that uh I don't have any elements which are of finite order, except the identity element, okay? I don't have any rotation by 15°, if I have 90°, because four times that is identity. Why can I not have torsion-free? Why can I not have torsion elements? Because if I have suppose G power N is E, then I can take the distance of identity to G power N. On the one hand, this is zero. On the other hand, this is what? It is N times the distance from identity to G. But N is positive, so I can divide by that. So, I just get that the distance is zero. And this means that G equals E. The only element with non-trivial with only torsion element is the identity. Fine. The interesting thing is that the converse is also true. If G is abelian and torsion-free, you can put a norm on it. So, I will [snorts] tell you why. Because this this is if this is one slide, not more than that, and it involves the axiom of choice. Okay, fine. So, what about the question really therefore is what about non-abelian groups? Abelian example is obvious. So, is there a non-abelian example of a group with a norm? And of course, no such examples were known when I thought about this question. So, why did I think about this question? I was doing something in probability theory and with Bala Rajaratnam and there were some stochastic inequalities which were true which we proved in semi groups with a norm with a translation invariant metric not a norm. And then in abelian groups with a norm. And so I was trying to understand if we could prove those for non-abelian groups with a norm. And then I tried to look for examples and as I said no such examples exist. So fine, so what can you do? So I I didn't know what to do. So I emailed lots of people and I asked several experts as well. One of them was at Stanford so I asked her and but anyway nobody could give me any answers. And nobody could tell me any results. So there are no examples for norm and no theorems saying such a group cannot exist or if it exists it has to have these properties or nothing was known. Okay, fine. So fine, so let's go take a step back. Of course I will tell you the answer in 5 minutes. But so let's take a step back. So if you don't insist on this norm condition you just insist on there being a metric then they such groups occur all the time especially in geometric group theory. So there are word metrics in free groups and monoids. There is yeah. So monoids is the same as semi groups of course with identity. So yeah and there are every Lie group of course has a left invariant Riemannian metric. Compact Lie groups have bi-invariant metrics I suppose you can do something. So uh great. So therefore there exist like lots and lots of examples from geometry not of a norm but just of a metric either left invariant or bi-invariant right invariant. So that You mean uh About the norm about the metric? Yeah, it was >> there was no metric on the non There was no norm on the non-abelian group that we knew [clears throat and snorts] about. So, if if So, if your abelianization is torsion-free, then certainly it has a norm. Now, you can ask if it lifts to a norm above, maybe. >> It probably doesn't, but it still could give some information. >> Right. So, in fact so Yeah, okay. Well, I will tell you the answer, but maybe I'll tell you the punchline. Fine. Every norm on a Every So, you can't have a norm on a non-abelian group, but you can have a seminorm where the positivity is not necessary. If norm G is zero, G need not be the identity. It can be others. And the kernel of this the Where is all the G norm G equal to zero? That is precisely the the commutator. >> [clears throat] >> So, every seminorm on a non-abelian group pass factors through its abelianization. So, that's the answer. Yeah. So, there is some information, which is that you lose everything non-abelian in if you want the norm. Yeah. So, anyways, let me define So, So, in the language of geometry, one doesn't call it uh the norm like this. One would call it a length function. Instead of Instead of metric, people use length functions in geometry. So, I'm just telling you different languages. So, length function on either a semigroup, monoid, or a group Oops, sorry. Uh a length function is this this some non-negative valued function. This is basically the triangle inequality. This is positivity, and this is symmetry. So, it's exactly a metric. It's these three metric properties. Uh if there was no inverses because it was a semigroup or a monoid, then the third You can ignore the third condition. But so, how is this a length function? So, basically, what we are saying is What is the length? The length is going to be obviously, the distance from E to G. Now, if you want to say something like So, what So, what we really want to say here is it's not just any distance function. If the distance function has a little extra structure, which is that the length of What do I want to say? So, yeah. So, basically the point is this length function should also be left invariant. >> [clears throat] >> Meaning that the distance from GA to GB is the distance from A to B. Now, how does that help? Because if now I have a triangle inequality So, what is length of GH? This is distance from identity to GH. Which is distance from E to G plus distance from G to GH. And now this is left invariant, so I can cancel and this is exactly length of E plus length of length of G plus length So, so somehow length functions correspond to left invariant metrics and the reverse the reverse way is done. This is how you go backwards as well. If I have a If I take this metric out of the length function, then this is left invariant because the translations go inside and cancel out. Okay, great. So, this is a So, this is in the language of geometry that this left invariant metrics are called length functions. What happens What is called a norm? The norm property is that it's homogeneous. It's said to be homogeneous if this happens. The distance from identity to G power N is Oops. Yeah, A is this is what I mean. Okay. So, this is what is would be called a non-abelian version of a norm because G power N is multiplicative annotation. Great. So, the question then is in the language of geometry, the question is if G is non-abelian, can any length function be homogeneous? Same question. And now there is a result of Milner which says that if G is a connected Lie group and not abelian, then this cannot happen. I will tell you why. If G has nilpotent degree two, this cannot happen either. So, this is a calculation which maybe I can show you. So, what does it mean to have nilpotent degree two? So, nilpotent means you take commutators. You get commutator. Degree two means that the second time you take commutator, you get identity. Which means that the commutator itself, the first time you take was in the center. That's why it commutes with all of G. Right? So, that means that every time I take So, now now look at the right hand side. I take GN and HN, right? So, [clears throat] I'm taking GN and HN. So, this means that basically I'm taking N of these elements plus N of those. There are N squared many uh transactions, right? Every time I create a transaction, I create a copy of the commutator. And that is central. That is in the center. So, I can just push that all the way to wherever. Now, I take one more copy, one more transaction, one more central element to it. So, every time I have N squared transactions, I get N squared of these elements which lie in the center. And so, that's exactly where the left hand side is the right hand side. But now, let's see the length of both sides. Suppose I had a norm, suppose I had a homogeneous length function. This length of GH N squared. Sorry, N squared this side. But what is this? Since it's a length function, since it's a norm, this is N squared length of GH. And on this side, this is the length of GN HN G minus N H minus N. So, whatever this is, it is growing linearly because of triangle inequality. It is N times the length of G to the length of length OF H. ON THE OTHER HAND, IT'S ALSO GROWING QUADRATICALLY. SO, how is something quadratic growing slower than something linear? Only if this length is zero. Otherwise, it cannot grow small. But if this length is then when is the length function zero? The element has to be the identity. When is the common to identity? If they commute. So again non-abelian is really bad. So that's why if G has nilpotent degree two norms cannot exist. So So basically >> [clears throat] >> On the other hand, if you have a semigroup or a monoid, then this was explained to free free monoid. Then this was explained to me by Robert Young of Courant uh that there is something called a Levenshtein metric which is a word metric which always exists. So for certain classes of groups length function homogeneous length function or a norm cannot exist non-abelian groups. On the other hand for every single free monoid such a function always exists. So what is the answer for a generic group? That was still the question. Do Is there Can they produce an example or can they prove that these don't exist? Okay, so without with this preamble here is now the main result. So the following are equivalent and the answer is they don't exist. So G is abelian torsion-free if and only if it has a norm. If and only if in the language of geometry it has a length function that is homogeneous, but I don't even need homogeneous. Just this double condition is enough. So we will prove all of these are equivalent. I'm going to give you almost a complete If and only if G embeds into a Banach space. So as we started out by saying that these are some obvious examples of norm. Take a lattice or take any subgroup in a in a in a real or in a Banach space, clearly the norm scales. It turns out these are the only examples. There are no others. So the natural examples are the only ones and so in some sense this is a slogan for you know also the unity of mathematics. Because you like algebra you like abelian groups, well here is an analysis definition or characterization. If you like geometric group theory, well, you can say that's the same as embedding into a Banach space. So, I mean, all these things are really connected, they are equivalent, and one has to of course mention this phrase of Hermann Weyl here. In these days, the angel of topology and the devil of abstract algebra fight for the soul of every individual discipline of mathematics. So, but sometimes angels and demons, you know, can be one and the same. So, that's that's this that's the thing. That's the main theorem, and the rest of the talk I will try to prove the main theorem. Okay, so clearly, if you embed into a Banach space, then you are abelian and torsion-free, that's obvious. [snorts] And you are a metric space with a norm, also obvious. If you are a metric space with a two-sided invariant metric, then you are left invariant. So, and if you have a norm, then it's two-homogeneous. So, all these implications are trivial. Okay? So, the first thing I will do is tell you why if a group is abelian and torsion-free, you can put a norm on it. You can embed it into a Banach space. Okay? And that is going to use Zorn's Lemma. Next, we're going to show that basically three implies two, meaning if you are two-homogeneous, if your doubling condition happens, then you are then you have the condition for every single n. That's a very cute exercise, but we will do that exercise. And the the finally, the hard part, of course, is if you are, you know, if you have a norm, then you are abelian and torsion-free. Torsion-free is again easy, that we saw already. That you are abelian. So, if you have a norm, why are you abelian? Okay. So, let's start with why is every abelian and torsion-free group embeds into a Banach space. So, exercise. If I take any any real vector space, this has a norm. This is a >> Right. And the answer is basically this. You take any Hamel basis Hamel or whatever called basis. And then you give each of them the norm, let's say the norm equals one. And then you take the element uh alpha beta times v. This is You can take it to the finite sum of course. So, >> [clears throat] >> and that's that's a norm. So, every every real vector space has a norm. Okay, so now let's prove the question. Let's answer the question, right? Does one imply four, meaning the title of the slide. So, suppose I have an abelian and torsion-free group, then it's a subgroup of this. Because it is this it's basically a Q You basically are base changing from like you know, Q to R, but actually you're doing it from Z to R. So, the point is if my abelian group is uh Z, if they if they're integers, I just get the real numbers. But if my abelian group is R itself, then what I get, R tensor R over Z, is an enormous vector space. It's not just one dimension. It is very very big. But it Nevertheless, it is a real vector space. So, then because of torsion-freeness and so on, this is a real vector space. So, I can put a norm on it. Let's put a norm on it. And now restrict it to G tensor Q. So, all we do when we say G tensor Q is we first of all we take a subspace of this with over the rationals. But basically all we are doing is we are just taking G and introducing denominators. We are like one over n, G over n basically. All we are doing here is taking fractions of G. So now if my G was R, then R already is closed under fractions. So then suddenly I reduce from this enormous vector space just down to R itself. Something like that. Okay, great. And now it's very easy. Just do the Cauchy completion. The point is this vector space, we already had a norm and now this this vector space had a norm on it which is scaling invariant. So just now Cauchy complete and that becomes a real vector space again with and the norm extends to the whole thing. And that's therefore a complete R vector space under norm. That's a Banach space. So just the standard constructions you would expect. But you can do this. Fine. So this is why one and four are therefore equivalent. Okay, so now again, why does three imply two? See the title of the slide. This is another cute exercise and now this time I will ask you guys. Okay. Let's just suppose I know that length of G squared, I know that just this doubling property holds. Length of G squared is twice length of G. Can I Can you tell me another number N for which length of G power N is N times the length of G? >> Four. And so you can do that. Length of G squared twice length of G. This implies that the length of G fourth is length of G squared squared. So that's twice length of G squared. And you can keep going. So for every power of two, this is clear. The question now is what about length of G power six? I want to prove this is equal to six times length of G. How do I go about that? So the answer is I don't. I do the following. I take a power of two bigger than six. So I take G to the eight. Yes. This on the one hand I know is length of is eight times length of G. But now because this is a triangle inequality, this is length of G to the six plus one more and one more. I keep them individual or I can I can do here as well. Two of them but whatever. But now because of triangle again, this is less than six times n plus two. G G G G. So I just break it up into lots of G's. And so what I get is this is eight times. So I started with eight and I ended with eight. So everything in the middle is an equality. So this is an equality. And I'm done. So it's a very cute trick. This is I found it somewhere in the literature. I think in the paper by Gajda and Kominek if I remember. And so yeah, so everything is done and then you can do this for any positive integer the same way and for negative integers you can use symmetry and so you're done. Fine. So except the fact that the proof for three implies two is not complete. Why not? Because in three we were assuming that the metric was left invariant and in two we are assuming it's bi-invariant. So I need to prove the right invariance as well. Fine. Except for that the proof the hard part is this. Well, that's no, this is easy part. The hard part is that because we don't know it yet. It is the hard part is like three lines of proof. That's it. But it's it's very clever. We are getting there. Okay, so now finally the main part of it is that I have a group. I have a norm on it which means just I just have this condition, left invariant metric with this. And I want to prove that the group is abelian. So torsion free was already done. We checked that, so that's fine. Okay, so So now here are some other concepts. So this is where I'm telling you about Milnor's theorem while we wait for the abelian proof. So if if G is nilpotent, we saw that G has to be abelian because if G is nilpotent, you can look at the nilpotent degree and then if the nilpotent degree was two, then we proved that the actual nilpotent degree was one. Because anything in the commutator is you know, you can push it back. So GK take the composition series of G and then you get to the subgroup which has importance degree two. But then that reduces to the calculation we did here with the N square transactions. And then actually the importance degree two would have the importance degree one in that context. So that's a problem. Okay, here is now the other part. So as we said in the main theorem which I haven't yet finished proving, if you take any abelian torsion free group it embeds into a Banach space. Can that have a non-abelian counterpart? So first of all what might So this is a very vague question. So I'm I'm going to give you one answer to this, but I mean this is not a theorem. So what is a non-abelian counterpart of a Banach space first of all? Mathematically what would you suggest it is? So in my case I thought of the following. A Banach space is a complete is a geodesically complete space, abelian group of course. And what is a geodesic between two points in a Banach space? It's the straight line. So what are non-abelian geodesically complete groups? They are called Lie groups. And suppose we had translation invariant metric. So suppose I have a group with a translation invariant metric. A theorem of Milner which I mentioned says that any such group any such connected Lie group is actually RN cross K for a compact group K. So now does this have a Can Can this have a non Well, let's restrict the norm. Suppose this was translation invariant. Let's restrict the norm if that metric was a norm to K. Right? So I take an element of K, a non-trivial element of K. And look at the distance from identity to K. So that's just a positive number. I scale. So of course every power of K belongs to the compact subgroup. So I did What what happens to the powers distance to the identity? The length grows unbounded. But this is a compact [snorts] group. So I cannot have an unbounded set of a lot of unbounded sequence of elements. So, there cannot exist a non-trivial element of K in the first place because that would grow. That means K is the identity. K is trivial. But then G is a billion because G is R in under the So, again, this is still a billion. If it is a norm, so a connected Lie group with a bi-invariant metric Yeah, which is a norm has to be R. That was Milnor's theorem. So, again, this doesn't So, there there were reasons to believe that yeah, these such groups don't exist, but that is very far from being a proof. So, now I'm telling you the final bit of the proof. So, this was in 2014 that I first had this question and then I asked lots of people. No answers were found, then I gave up on the question myself and I had other things to do like move back to India. So, then but then I went back to the US that same winter, my first winter, to Terence Tao's office and I asked him the question. And he also couldn't solve it, so we should feel happy that he couldn't He couldn't immediately see why, but so this is the discussion we had. That suppose take you take any group with a norm. If I want to prove it's a billion, he felt we should try to prove a billion. So, but if I want to prove it's a billion, then why I want to check that any two elements commute. So, the key quantity, which I will call this KQ, is the norm of the commutator, which we discussed over there, right? Why is this a key quantity? Because if this is zero for all alpha beta, then the commutator is identity. If that is identity, G is a billion. So, this is why we want to keep we want to show somehow that this length this norm of the the the commutator of any two elements in my group is zero. How do you do that? So, of course, Tao being an analyst, he says, "No, you have to prove that it is less than epsilon for every positive epsilon. You must use the Archimedean property of the real numbers. So, can we show that it is very small? Can we show that it is less than epsilon for every epsilon? So, okay. Some things you can do. So, first of all, let's No. Okay. So, maybe I should say the following, right? So, suppose I have a Let's write this down. So, I have a group with a norm. So, I want to prove to to show that G is a norm. I want to show it's a billion. So, I just basically need to fix two elements. We need for all alpha and beta alpha and beta, it's enough to show that the subgroup they generate is a billion. So, I may as well think of my G as being this. This is my G. And I want to show that this the norm restricts to this, this is a billion, then G is a billion, right? So, on this, I can again, and I have some length function, but I can make sure that the length of alpha, alpha inverse, and beta, beta inverse, all these lengths are less than one. Because I can length function divided by anything is also length function, right? So, this is >> [snorts] >> So, normalize the length function so that they're all one. Then, first of all, here is what it is. The length of the commutator, my key quantity, KQ, is less than four. That is obvious. But now we are going to show later on that the length is actually preserved under conjugation. So, when I said that we haven't yet shown the left invariance implies right invariance if it's a norm, that loose end, that is translates exactly to this condition. We will show that in in a three-line proof, probably in the next slide or something. But now, can anyone Can anyone now tell me why this length this commutator is bounded above by two, not four? >> If you have the property below. >> I'm assuming the line above that. >> Yes, you're assuming that this is conjugation invariant. And you're assuming that all the lengths are less equals one. >> You have this. >> [snorts] >> Yes. So, what you exactly So, what you do is you say alpha, beta, alpha inverse, beta inverse. And this is length of this. Well, I break it into two parts. This length is the same as length of beta, which is at most one. And this length is beta inverse, which is also at most one, and so you get two. Okay, I did it the other way here. So, alpha and then the rest of them, but anyway, I don't need this. Great. So, therefore, this is less equals two. But now I claim that you can do this even better. You can do four over three. So, now we are just going to play these word games. This is Now, how do you show that So, now here is the key fact. How do you show that a number is less than four over three? This is a real number. You show that a number is less than four over three if three times that number is less than four. That is, everybody in class one or two should know this. Maybe they don't need inequalities. Fine. So, how do you show that this is less than four? This is There's no good [snorts] chalk. I'll use the small one. To show that something is less than four, Sorry, I want to show four is bigger than three Q. What is this? This is three times the length of alpha, beta, alpha inverse, beta inverse. I must use now the fact that it is a norm. What is three times the length? By definition, this is the length of by the norm property. It's the length of this element Q. This is a string of length four, so I have a string of length 12. Right? Yes. So, I WRITE THIS DOWN. AND SINCE THIS IS AGAIN A BASIC notion seminar, we are going to use the basic notion of associativity. You can write this as product of three different strings of length four, or you can write this as product of four different strings of length three. And now we know what >> [snorts] >> This has length less than one conjugate. This has length less than one, length less than one, length. So, exactly that. And so, we get So, I can say that I've never appreciated or liked the associative law of groups as much as when I did this. Like So, because you can't play with anything more than these very, very basic notions, right? So, this is the notion of associativity. And now we have this is less than four, so therefore we are done. So, three times something less than four. And so, this is less than four times So, the question is, can you come up with these tricks? Can you keep doing better and better tricks to come down to zero? We went from four to two to four over three, but you know, we haven't even gone below one, let alone close to zero. So, this is when we This is the This is the discussion we had in Tao's office, and then then I left that, went back to San Francisco to fly back, and Tao said that uh yeah, can I put it on my blog? Uh so, I said, "Sure, please put it on your blog, and maybe some other people can give some Maybe they'll find a clever proof." So, here is the blog post, and this was posted on 16 December in the US, which is 17 December in India. So, then I I took off after the blog post, and then I landed in India after 15 hours or some direct non-stop flight. And there were so many comments on the blog. Uh this might be a later time, but yeah, there are a bunch of people who have commented. Some of these names actually I think every one of the names here is a collaborator on that paper on that paper. Yeah. So, uh there are four names here. right. I should mention what he said. So, here is a curious question posed to me by Apoorva. I don't know the answer to that. He poses the question, triangle inequality and the linear group. This is the key, the norm condition, right? And uh conjugation invariance, blah blah blah. What is not clear is if one can keep then in in the fold which I have skipped over, he mentions this calculation, 4 over 3 and two of course two and 4 over 3. He He stops here and he says what if is not clear is if one can keep arguing like this to keep improving the lower bound until the slow the upper bound on the norm until the upper bound must vanish. But anyway, this seems like a problem that might be receptive to a crowd source attack. And so I'm posing it here in case readers want to make progress. And so this brings me back to the opening slides where you know mathematics is very individual uh individual profession or research or hobby or whatever you want to call it. But there are some attempts like this where you you know try to get crowds involved. And a lot of people did get involved. They sent lots of suggestions and tips and ideas. So, the first two days so and the point is this got solved in 5 days. But not really 5 days because three out of those 5 days weren't really doing anything. So, the first two days people actually tried to find examples that will will be a countably will have a non-abelian group with a length function. You know, loop groups and winding numbers and algebraic topology and lots of such things. But uh as we now know from the theorem, such examples don't exist. So, at after 2 days I think people mainly gave up. And then they started to think about uh how can we prove the group is abelian? Okay. So, here is also as I said, so the story of the timeline. So, uh the first comment in 3 hours was the loose end that if you have the norm and you are merely left invariant, then you're right invariant as well. That was done in 3 hours, and I was sort of beating my head, you know, why didn't I see it for the last 3 years? But, there are some really clever people. Another one will show up later in the talk. Uh so, here's the lemma. So, this is Tobias Fritz's uh result. If G is a homogeneous length function, then L is conjugation-invariant. And if you translate this condition under the fact that length is distance from identity, then this conjugation-invariance is the same as saying that the distance is right-invariant. So, suddenly it becomes bi-invariant. So, you don't need to assume right-invariance, you get it for free from the length from the norm condition. Okay, great. So, uh Why? So, I have The proof is inside the remaining part of the slide. So, Okay, the proof is still inside the remaining part of the slide. Okay, so here is the proof. So, So, now this is where you use the fact that conjugation powers of conjugates are very nice to calculate. N times the length of this thing is length of this to the N, but that conjugate to the N is this. And now you use the triangle inequality, so you get this. And this is true for all N, right? So, all you do is now divide by N. You get the length of g a g inverse on the left is less than the length of a h plus some error term. And now this is the next basic notion, do you remember, which is called the Archimedean property of real numbers. If you have a real number less than another real number plus anything any as small as possible, then you can get rid of that. You can take limits. So, this is the Archimedean property. Again, I have never appreciated the Archimedean property as much as I did in Tobias's proof. And so, this is of course one side, length of the conjugate less than length of h. Conjugation is an equivalence relation, so you're done. So, great. So, this addresses the loose end, this finishes everything, and we get that it is uh yeah, conjugation-invariant. So, this trick also works. This also works. Good. So, now So then So then people again So with Tobias' result, people got energized. They started to find stuff. As I said, the first 2 days was not really a very productive cuz everybody was trying that way. But then finally people said, "Okay, fine." So this is 2 and 1/2 days later. Some progress had been made. So from 4/3, people improved to 5/4. It's an improvement. 5/4 is 20/16. Somebody improved it to 19/16. About 1 hour later. So this is like Will Sawin in 24 hours posting that unit distance problem improvement or something. But then this was done on the fly. Every Everything was happening on Terence Tao's blog. That was where people were posting all these comments, improvements in the comment section. So you have the whole history still there. And then there were more clever calculations and people tried to get it below 1. Finally, some insane like conjugation, this, that, something insane made it work. The bound went below 22/23. It's like the world record for the 100-m race you know or the marathon or something. Going down. But still I mean you know So now and then there was a 24-hour barrier. So when I said three of the five days were done doing nothing, well, not towards this proof. The first 2 days were gone and this 24-hour barrier was gone. And basically nobody had any idea how to go even crazier or even cleverer. And then this is where the next interesting feature comes up. Everything was happening on the blog and there was a computer involved. So Siddharth Gadgil, who is my colleague, he he had a remarkable idea. He said, "I will do both things at the same time." So that from the beginning, he was spending his time trying to come up with algebraic topology and geometric examples to not work and he programmed his computer to try and make it work. So he did both together. This is a very safe approach, I guess. So, uh and he actually found it or his computer found it. So, here is the So, the right-hand side Okay, well, I can show you the the the conclusion was the for the original conclusion was 0.95, something like this. And Siddharth's proof gave it 0.816. So, okay. Let's look at the right-hand side. This is Siddharth's post. Here is a computer-generated proof on the bound of the length of a commutator for a linear norm the norm. Right? And so, it starts literally with he made the computer write it down into a form that human beings can understand. Statement, next statement, next statement. The previous ones imply the next one. So, that way. And it took 126 lines. So, it started from the norm of alpha inverse of A inverse is less than one. Therefore, the norm of the conjugate is less than one. Norm of B inverse is less than one. So, the norm of this conjugate is less than one. And you keep going. The 125th step is that the norm of ABAB A bar B bar So, A bar means A inverse. And like this is a 68 element string, which is 17 copies of the commutator. So, the length of the commutator to the 17th is less than 12 13 13.8596. Divide by 17, and then the length is less than 13 whatever. 0.815. So, therefore, 0.816. So, this was done on a Saturday morning, apparently. Uh >> [clears throat] >> Saturday morning in America was Friday night. So, Pace Nielsen, and this is the left-hand side now. So, since this is all UCLA time, 20th December, 3 days later, at uh or 4 days later, at 8:00 p.m. So, and Pace Nielsen was in Utah, which is one time zone ahead. He had finished his dinner on Friday night. He was sitting at home. He saw this comment, and then in the next 100 minutes, it took him 100 minutes to read the whole thing, to understand the whole thing, and to improve on it, and then to write all the math here in the comments. So, some people are not only very sharp, they're also very fast. So, he says that my this was beautiful. My intuition was that we have some reason to do something. When I read this computer's proof, I was surprised that this is exactly what the computer did. With some extra ideas thrown in. And then he says that the first 43 lines he summarized established this. The next 30 lines established this. And then here is some improvement from for what we were doing. And then the last few lines established this. And now also he said he can improve on this in the following way. So, all of that in 100 minutes. Fine. So, that was that. So, okay. That is all sort of fun and games, but now let me again tell you the math. So, finally what was the computer doing? People try to understand that. This was 21st December 9:30 a.m. Uh and then I think a few Tao probably didn't answer that day, but the next morning he did and then with discussions we got this lemma out of the whole computer input. Suppose I have four elements in a group with a non with a right and left and homogeneous length function, right? Suppose I have four elements. X is conjugate to WY and to ZW inverse. Then the length of X is less equals this thing. And the key point is it doesn't depend on W. Fine. So, uh let me prove this for you. And then let me show you uh >> [clears throat] >> an application where we will go even more beyond below .816 or whatever it is. The proof is not that bad. So, the idea is the following. Write down X to be a conjugate of WY. Write down also conjugate of this thing. And the claim now is that I write X power 2n. Okay? X power 2n. The first n copies I write in the first way. And the second n copies I write in the second way. And now don't look at this inequality. Look at the picture. Okay? So, the the the length of this big string here this big string is drawn in the picture. There's a S at the start, there's a T inverse at the end, and so on. So, the length of this big string, that is what I want to upper bound. That's less than the length of S, and the length of T inverse, and everything in between. But, everything in between is a conjugate, the outermost arc. The length is conjugate invariant. So, I can kick out like discard those W and W inverse. Right? Then I get a Y. Fine. Then so, what I get is So, maybe I'll write those strings. Maybe it's easier to see. Start with this one. There's a S, there's a W Y W Y W Y S inverse T W Y. No, this is something else. >> W Y S inverse >> W inverse Z, sorry. Thank you. No, Z W inverse. The length of this The single The single length of S plus length of T inverse plus length of the rest. So, I can kick out the rest. So, oh yeah. So, I'll take this out, this out. And so, this is less than this. Now, this is a conjugate, so I can get rid of the W W inverse because it's conjugate invariant. Now again, I get these terms. I get plus length of Y plus length of Z plus length of the rest. Again, of course I have a W and W inverse. So, again I can get rid of these guys. I again get a length of Y, again length of Z, again get rid of the W W inverse, and you keep going. And you can see that all the Ws and all the W inverses just go away. And finally, I get another length of S S inverse and length of T. So, there are N Ys, there are N Zs, and then there are these four terms. And now you know what to do, exactly what to be expected. Divide by 2n and take n to infinity. And then you get length of x because that's twice 2n length of x is less than the average of y and z and some error term that goes away. So, this is what it is. Now, why is this useful? So, this is useful in the following way. Let me write this down here. Let me So, I'll write down the lemma first. Just the statement and then we will apply it. And then interestingly, so x is conjugate to wy and zw inverse is less than the length of x is LESS THAN 2. >> [snorts] >> OKAY, SO HOW DO YOU SPECIALIZE? SO, I SPECIALIZE WITH THIS EXAMPLE. x is alpha beta So, x is alpha beta alpha inverse beta inverse. Again, one more time. There's a nine element string. And then I called them So, y was something weird. y is alpha bar beta inverse alpha beta and then alpha beta Sorry, I should know this. It must be this. Yes, that's it. And the And the w is beta. So, x is conjugate to wy. Let's write down wy. This is beta here. Is this conjugate? Yes, if you see there's two alphas here. If I multiply on the left right by alpha inverse and then by alpha, I get This is a commutator. This is a commutator. So, I get exactly this, but the alpha was moved here. So, it's conjugate. And similarly, you can check that x is conjugate to zw inverse. Fine. So, with these four elements, I claim uh So, first of all, let's write down the length of x. Fine, before I claim anything. It's the length of two copies of this thing plus this. Uh fine, but And what is it? By the lemma, it's less than the length of Y. Well, what is Y? Y is Y is the conjugate of two things. So, the length of Y is conjugate invariant. So, the length of Y is the length of this. By the triangle inequality, 1 + 1. And the length of Z for the same reason is also 1 + 1. So, two. So, the length of X is at most two. Now, the claim is that 8 over 11 works. And 8 over 11 is much less. So, let's see. How How do you prove 8 over 11? You know what to do. Multiply by 11. So, 11 times this thing, this is a 44 element string. So, you write down 44 elements. Separate them into groups of four. Okay? And you notice that the in groups of four, the 11 element string, every one of them is a conjugate. So, this is less than the length of the nine element string, 11 element string plus 11 element string plus 11. But, each of them is a conjugate, so the inside nine, inside nine, and nine. And every one of those nine element strings exactly looks like X. So, every one of those is size two, at most two. So, 8 over 11. And so, this is a 0.72. So, suddenly, you know, we have shaved off a tenth of a second from the 100 meter record. But, of course, the problem is this is a finite improvement. And we want to come down to zero. So, if you take If you think about it logarithmically, this is some negative number in log. This is negative in log. 0.72 is even more negative in log, but it's a finite reduction. We want to go down to log of zero, which is minus infinity. So, that's not A finite reduction doesn't help. We need something that will work infinitely many times. So, what is the trick? It turns out that this internal this this trick of using X this this lemma here, you can cleverly use it infinitely many times into one tool, and you can use that tool infinitely many times. So, sort of a quadratic infinity, uh infinity squared many times uh to do the job. And the final step, as it turns out, I have a I can show you the proof later on, but the final step, when you write it down explicitly, can be used either explicitly using binomial combinatorics. So, we have seen some algebra, some geometry, some analysis. You can do some combinatorics, or you can write it in the language of probability theory. So, sort of all kinds of mathematics comes together to prove this. And the key step, finally, is you want to show that something grows faster than something, or something grows slower than something. And so, the the sum of n such IID variables, IID, has average spread, meaning standard deviation of the magnitude square root n. So, when you take IID uh n variables, the variance is n. And the the standard deviation is square root n, and this is going to bound the multiple of the power of this as n goes to infinity. But, what is this? This is This is growing linearly in the length, and this is only square root n. And so, when you divide it out, you get exactly that. So, the upper There's a linearly growing quantity upper bounded by a square root n quantity. So, then, of course, you cannot do that unless your quantity itself was zero. Just like in the commutator case, it was a quadratically growing quantity bounded above by a linearly growing quantity. Right? So, that's that's how one does it. And this was So, this was not This is not exactly this was observed by Tao, but I will show you what he said. And so, in some sense, it finished the proof just less than 5 days later. So, this was the just to remember that show the timeline, this was the 22 of 23, almost 24 hours later, Siddhartha Gadgil had this computer-generated solution. And then, finally, this lemma was abstracted. Also, Tao was awake, I guess, because this was this Oh, no, no, no, this is almost Yeah, yeah, this is only 4 hours later. So, Tao himself was awake and then because I remember he had a part to play in this lemma. So, and then in 4 hours the lemma was made, the lemma was applied, we got to 8 over 11. Within 4 hours more it was 2 over 3. But again, these finite improvements at this point stop mattering. And then there was this idea again by the first guy who suggested this good idea, Tobias Fritz. And he suggested that think of alpha and k. So, think of so, not just the length of alpha power m. No, just not just not saying that this length of alpha power not the length of commutative power k is smaller. Think of this length in two variables and think of it as sort of a discrete heat equation kind of. That was his his words. So, this some kind of discrete heat equation evolution, there is a drift in this equation. When you think of it probabilistically, there is an expectation that is non-zero. So, there is some drift and that drift is what helps us. And the last Yeah, this happened let's see. p.m. a.m. This six This was 6 hours later and then I guess Tao had woken up and then he he's he said that one can finally do this to kill off the problem. One can do some kind of a simple random walk and then move by this these two steps, 1 minus 1 0 or 1 minus 1. That is roughly the drifts in these two steps with half probability of each. And that will hit something with an exponentially high probability by a Chernoff bound. So, he really applied some high power relatively high power tools. Chernoff for me may be high power. Uh but then later on when we did the work, oops, sorry. When we did the when we wrote up the paper, it turned out we just needed Cauchy-Schwarz or Jensen's inequality, which is much more basic than Chernoff's bound. Just a basic thing like weak law of large numbers and Jensen did the job. For the last step. Okay, so the last thing to mention then is that if you can This was saying that the length becomes zero. If you have exact triangle inequality and you have exact norm or doubling condition, the length becomes zero. You can make that quantitative. So, what do I mean? Let G be a group. Let L the this be a quote-unquote almost like almost length function length function up to error. Such that so you have the triangle inequality up to an additive error. So, the point is the error doesn't depend on the whether you take G power N or H power N. This C is the same regardless And you have a homogeneity. You have the norm property up to another additive error. Then for any alpha and beta, there is a global uniform bound or universal bound on the commutator. It is four times C plus five times C prime. So, in our case in the specific question we had, the length was always non-negative. The subadditivity the the triangle inequality was exact that the C was zero and the homogeneity was exact, right? Because the length of G squared is by the first property less than twice length of G. By the second property, it's bigger than twice length of G. So, it is equal. Right? So, that's why C and C prime both being zero give you length of G squared equals twice length of G. So, here of course we have far weaker hypotheses. We don't need length to be zero infinity valued. The length of identity element need not be zero. None of these need to be zero. G need not be the length need not be symmetric. Still, this is the full power of the result we used and this has been applied in some situations in the future to groups with something called the stable commutator length and to some other amenable groups maybe and so on. I don't know much about geometry because I'm not [snorts] a geometer, but there have been some applications. Okay, so that is about it. I will just show you some slides some pictures for the slide. This is the paper when it came out uh in the archive. So, we wrote the we finished the Let's see where was this thing. 22nd of December, then over the Christmas break people were busy I guess. But then later on in the new year we wrote the paper. >> [snorts] >> And Gil Kalai has some Polymath blog where he said called it a spontaneous Polymath 14. Because this wasn't Normally the Polymath projects people decide beforehand this is the question we want to solve then you solve it. This just organically grew up because Tao asked it on his blog. And so this was called a spontaneous Polymath 14 success. At that point these were the existing Polymath projects so the last one was classifying homogeneous non-zone groups. The first one was the new proofs and bounds for the density Hales-Jewett theorem. This is some theorem in sort of combinatorics kind of thing which is proved using very high powered ergodic theory and so on. But then the first Polymath project was initiated by Gowers and he said can one find a elementary quote unquote elementary proof of that? That did get found in the first successful project and so ever since that project every time people write a Polymath paper it's called DHJ Polymath by tradition. DHJ stands for density Hales-Jewett the theorem itself. And the first time that got published in the Annals of Mathematics when it got published when it got yeah done. Not all of these problems in the Polymath scheme are successful. For example if you just look at these things proving Roth's conjecture proposed intransitive dice proposed they were never completed. Or improving the bounds for Roth's theorem proposed. Then there were some mini Polymaths where people solved IMO problems which is also another thing people seem to be doing. Now of course you just ask your favorite AI and it will do it. >> [clears throat] >> Anyway at the point when our paper came out these were the papers the papers that came out. The other actual what do you call it famous uh Polymath initiative in between was the part about bounded gaps between primes. When Yitang Zhang came up with 70 million people got together including Tao, and Maynard was part of it, saying, "Can we reduce the 70 million?" They brought it down to 200 something in the project. Now, apparently, there is six or something. Now, you can come down to six. And then there are some things where people literally just wrote down uh So, new equity stations. These are papers. This is the 107 pages is the revised version, focusing solely on these estimates. So, they published that paper, apparently, the same journal. I think this is why Tao suggested that we submit this paper to algebra and number theory. Because I'm There's much more analysis than algebra, but I think Tao had some, you know, positive experience from the previous polynomial project. And this is about groups. So, I guess he suggested that send it to an algebra journal. They fortunately accepted it. Uh and uh yeah, that's about it. So, THANKS AGAIN. >> [applause] >> DOES ANYBODY HAVE ANY COMMENTS OR >> QUESTIONS? >> MAY I go back to the previous question? How when you're passing from the commutative to the non-commutative, do you get the zero level where it's actually commutative? What if you go two levels? So, like when you divide this to find your something, so take all these factors. Suppose you express your problem your product product of that product, so like in non-commutative geometry. >> Okay. >> So, the normal first one divide by everybody and do it again. And now you have a sort of mild level of non- commutativity. You can maybe evaluate that at 1 + 1. >> Right. >> But you can keep going. You know, this is the >> I have never seen Yeah. The You're talking about something like formal deformations, basically? >> Yes. >> Right? So, yeah. No, I haven't seen uh deformation theory connected to this, but as a group as as the statement for groups, it is very I mean, it's very sharp that, you know, if you have anything non-abelian, you will not have a norm. If you have >> can you pretend it's abelian by now taking the first two levels and then fixing and then you go back and then you have a family. >> Right. So, then I haven't seen how this works, no. >> [clears throat] >> Right, right, right. No, I haven't actually I don't think any of us thought about this, but maybe it's worth it. It's worth looking into more carefully. Then you should think about it, absolutely. You should think about it and let me know. I'll probably check here as well just in case there are in case there are what do you call it? Like how do I [snorts] get to that? Escape, I see. In case there are chat questions here. I doubt it, but There are no chapters. It's gone. >> Have there been other successful polymath projects after this one? This is 2016. >> So, interest Yeah, that's fair. All right. So, in fact, I should have I forgot to mention this. So, there was this uh Let's do this. Yeah. So, here. So, the week the week after so, the results submitted January 11th and I think 2 days later Tao announced another project that okay, we have started a new polymath project. He's always very busy. That was about slightly more important stakes than this question, the de Bruijn-Newman constant for the Riemann zeta function. And the So, he had just written a paper with Rogers about that constant being non-negative or something and they were trying to improve the bound the the the Riemann hypothesis is now become that the constant is zero. So, they were trying to bring down that constant bounds like bounds on the length here, but those are slightly harder to do. So, that I don't know if that At some point they stopped it and published their findings. So, they were they So, yeah, they had lots of sort of simulations as well and it's a long paper lots of findings in there. Obviously, they did not bring it down to zero, but so there are partially successful projects. There are, you know, different levels. There was something about sunflower. Yeah, this is okay. But, that's launched. Eldar Shado Sunflower Lemma. So, yeah, I don't know. There are as I said, there are few successful projects. And one was of course the DHJ theorem. One was this uh gaps between primes, which that was being done and that was on the side James Maynard was doing his work and then he published something paralleling their results or between many many primes, right? And then of course he became very famous. And there was our work was successful. And let's see. I Can you even see here? Forget the >> Which one? >> Yeah, which one? So, exactly. A deterministic way to find primes. Launched August 2009. Research results have been published. So, I mean, I think people meet with varying degrees of success, I suppose. And like for something as large as the Riemann hypothesis, you know, maybe what they got on the de Bruijn-Newman constant is should be termed a success. I don't know. >> No, what I mean is I go to this web page right now. >> Oh, where it is right now, I would have to look. But, there there there have been Polymath projects launched afterwards, not just the de Bruijn-Newman constant. Yeah. But, I suspect now with AI coming up, you know, people won't really they will go even more backwards and go even more reclusive and ask AI by themselves. So, this crowdsourcing model is going the other way, you know, because of AI. >> Eventually? >> Maybe. It's my guess. But, we'll see. >> All right. If there are no more questions or comments, let's thank THE BOARD AGAIN. >> [applause] >> I SHOULD PROBABLY DO SOME OF THIS STOP RECORDING. SORRY, I didn't hear you. What? Really? >> [laughter] >> What AI was this? it or Yeah.