 I'll just give a brief introduction. Apurvakhar attended the IMO TC in 1996 and 1997 before his BSTAT at ISI Kolkata and PhD in Mathematics at the University of Chicago. After teaching mathematics, including at Yale and Stanford, he returned to join ISBangalore, where he's an assistant professor in mathematics and a Ramanujan fellow. His research interests include matrix analysis and positivity, representation theory, combinatorics, and a bit of probability. And he's going to talk on groups with norms from world games to a polymath project. So what do you have, Apurvakhar? I guess I hope everyone can hear me. There's a mic working. So first of all, as Antur also said, it's a great honor to be back at the Olympiad camp, where more than 20 years ago I was on that side. Unlike Antur, I have no memory of the final day or anything. But it was definitely one of the most, I don't know, landmark or foundation of it. I don't even think of the right adjective, times in my life, because that's really as, and that's what actually goes back to what Prof. Raghunathan was saying, is it brought me in contact with all these incredibly sharp minds, and you learn what is possible. It is you learn if it's possible for you to think faster, to think better. And that for me definitely was the biggest takeaway from the Olympiad camp is I improved my own way or my own level of thinking as a mathematics problem solver. And in a certain sense, the project that I will describe to you brings me back full circle to exactly that philosophy, because it brought me in contact with some of the sharpest minds I've known, especially the last one is very famously very sharp, and the first one actually is very sharp. And again, once you meet these or interact with these people, you really, again, see maybe the next level after the Olympiad camp, the next level of how one can think even faster, even sharper. And that definitely for me was a big learning experience. So I'm very happy that it actually happened just a few months before this Olympiad camp and brought me back to the same kind of school, if you will, where I learned some more as well. Not just about the math itself, of course, there are some very fun and clever ideas here, but also about thinking about math. Anyway, so whatever this, I won't even assume what a group is. I'll try to explain something from the board. Norms everybody here has seen, so I will tell you what they are. And the point is, this is a question that doesn't require any kind of background, any thing. It just requires clever ideas. And the solution is entirely using clever ideas. And I thought it's actually, in some sense, nice enough to present it in the Olympiad camp, where as I said, the technology, the tools, the definitions you need are not so many. The ideas you will see are hopefully interesting. And they are interesting equally to people like Terence Tao, as to people here. So this was a completely, so again in contrast to the previous talk, which was very classical, which goes back to Gauss and Laplace and so on, this is on a problem that can be, as I said, asked without any maps in some sense, almost, and yet nothing was known. Not a single example, not a single counter example. And it got solved in a space of five days. But again, it's also a very modern solution in the sense that the fact that it got solved in five days makes use of the internet. It makes use of blogs. It makes use of, and it crucially made use of a computer, where human intuition sort of seemed to stop working. A computer was the next thing that provided the big. So it's a very 21st century project, and therefore 21st century talk. Without further ado, then so my, yeah, so this is actually based on, this is a full-fledged research paper, and I'm gonna try to give you about 70% of the proof, so of 70% of the paper, if you will. And as I said, it requires pretty much nothing. So joint with Tobias Spitz, he's in Germany. Siddharth Gargill, who is my colleague, and actually was my senior in college in Bangalore. Pace Nielsen in Utah, in USA. Lior Silberman in Vancouver. And Terence Tavern in Los Angeles. So here is the basic problem at hand. So I need to define for you the players in the game, and then, so there's a collection of strings, and there's a function acting on them. So here is the collection of strings. So there's, consider the collection of strings in four letters, so let's just call them alpha, beta, and alpha bar and beta bar, which are essentially one over alpha and one over beta. So you can write down strings like you know, you can write down the long string, like you know. Alpha, beta, alpha, alpha, beta bar, alpha bar. Things like that. There's a collection of strings, a finite line, just that's all. And only two rules, there's one rule, the two rules here are that alpha and alpha bar cancel each other. If I had alpha and that much, then I could say this is the same as alpha, beta, alpha, alpha, beta bar. Yes sir, these just go away. And similarly for beta and beta bar, and it doesn't matter which order they're from, they cancel each other. And if I have, so I'm allowed to also consider the empty string, which has no alphas or betas or these things. So you can call it the empty set, or you can call it one. Okay. And we don't, in particular, we don't assume things like alpha, beta is beta, alpha. So you can't take things past each other. The string is the string. You can't change. A word is a word. You don't say that, you know, you don't say cat is active. So you can't take A and C past each other. So similarly, you can't take alpha. So these are the, the one extra operation permitted is, I can multiply two strings. So if I have one string and another string, I can just put them together. That's all. And so in particular, alpha, beta, alpha, if I have these two strings, that's three letters, two letters, I make it a five-letter string. And then I can, the one rule I'm allowed to use is I can cancel pairs, like a particle and an antiparticle, they cancel, right? So you get alpha, beta, alpha. Okay, so this is the one thing I've permitted. So in math, we call this whole thing the free group on two letters, but forget what we call it. It's just a collection of strings in two letters and they're inverses and they're reciprocal. That's all. Fine. Now here is an example of this string with some additional relations. So suppose I could take the alpha, suppose I could say alpha, beta is beta, alpha. Then what is an example of that? Well, here are, so alpha and beta are going to act as some actions and here are two actions that can be interchanged and which have inverses. Okay, so alpha is one step to the east and beta is one step to the north. And then if I take one step to the east and then one step to the north, I get here. If I took one step to the north and then one step to the east in the reverse order, I would still get there. So alpha, beta equals beta, alpha. I don't assume this in general, but here is an example where I do assume it and then what is alpha bar? What undoes one step to the east? One step to the west. What is beta bar? That's alpha, beta, alpha, beta bar, which the additional relation. Okay, fine. So these are one step, one step, so alpha, beta denote that and then you can consider all strings of this kind and then what this string will denote is the net displacement. You start at the origin, let's say at zero, zero, where do you end up? That is the string, yeah. And of course, if you have south and north or east and west, you can just cancel them because you don't do anything. You come back to the point where you started, so the net displacement doesn't change. Okay, that's an example of what this is and the point about strings is, so here you are allowed to have more tricks, more rules to use to play around with. I'm saying even without those rules, I'm going to ask you a question and we'll see you. So for asking you that question, the problem, here is the next player, the second and last player. Suppose there's a function and I'd call it L because it's supposed to stand for length, so length of that. So I said that this stands for some displacement. That's going to be the length of the displacement. How much distance, like as the crow flies from the origin is there? So I'm sticking to that collection of strings, that free group, but you could ask the same question to model all these additional rules that you can use. Suppose the length satisfies just two properties. One is that it is a sub additive function. So if you have a string, if you have, let's see. So if you have the length of this thing, so the length of this thing, if I can write this as a product of two things, then the length of alpha, beta, alpha, alpha, beta, bar is less than the length of the first part plus the length of the second part. That's one rule. And the second rule is if G and H are equal, so this should be true for every possible pair of strings. And if G and H are equal, then now think of this example, right? So, well, okay, I'll get to the example in a minute, but so if G and H are equal, then L is actually additive. So you get double the length. And that again, so that should be very, here is the problem. Let me just first tell you the problem, fine. So suppose you have some function on the collection of strings, it takes a string and it gives you some number. It can be some real number, negative, positive, zero. It satisfies exactly these two properties. Length of G H is sub-arative and if G equals H, it's additive. Is it true that the length of alpha, beta, alpha bar, beta bar is less equals zero? And I'm gonna draw all these things in this case because I can draw these things in a second. The interesting thing was, so for those of you who know what free groups are and what norms are or metrics are, this is, okay, so if I further assume that this length was non-negative, this should essentially be a metric. And the question is, is it true that there is a metric on the free group which is additive for equal argument? And this was not, no examples are known. And is it that no examples exist? That was not, basically nothing was known about this problem. And it's very simple to ask. You can ask it after, so literally you can understand what this problem really is saying. I'm saying really is saying, not just in these pictures, after say one year of college, map. And, but at the same time, nothing was known. And so the interesting thing then was, as I said, it's a very 21st century talk because it starts on the blog, use the computer and ends in five days. And presumably in older times, it would require collaborations which would take time to form and then slowly exchange letters. And you know, it would probably take more than five days. This happened all over the world. Some people was in, somebody was in Germany, he solved something, then when he slept, somebody was in India, they solved something. When they slept, somebody was in USA and Canada, they solved something and kept going on the clock. As you can see from the, if you look at the blog, there are comments and what time the comments are made, you can see. And you can see that the comments keep happening around the end. So back to the example, I want to write down those two rules for the example and ask this question. So this was the setting, alpha is an east step, beta is a north step. Now this word simply denotes the displacement, as I said. And how do you calculate the distance? How do you calculate the length of the displacement? You use Pythagoras theorem, x coordinate, y coordinate. So more generally, let's say instead of alpha and beta like this, which were unit length, let's say my alpha was some distance a and beta was some distance b. It's still the same things happen. It's still, you can talk about the displacement. It's just that the displacement now becomes like one alpha and one beta goes to the point a comma b, not one comma one, that's all. Nothing has changed. Now given a string s, a string s meaning something like this. This is a string. Simply count the number of times alpha, of course. Call it n of alpha and then count beta and alpha bar and beta bar. Just count the number of times vector and then what is the length? Well, so the length is, what is the coordinate? It is a times the number of east steps minus the number of west steps. If you take steps backwards, then you would have to backtrack. So that's all and then plus b square times this. So and then so and the y coordinate is b times this and then by the same distance Pythagoras theorem you get this formula. Anyway, so why does this satisfy the two properties? So whether or not you had this extra rule, I claim that those two properties are satisfied. Well, l is substitutive because if you have any, so what is a string? A string takes some, start at the origin, you do something and you get somewhere else. That's string one, let's say. A string two starts from the origin and starts from this point and does something else. And the total string is this. And of course, you know that the length of one side of a triangle is less than the sum of the other two lengths. That's called the triangle inequality. That's why this is satisfied. So those two properties are satisfied for this model just because one is called a triangle inequality and the other is saying that if you start at some point and do some action, some string g and then you again do the same action, well then you're just going along the same direction for twice the length. So length is twice. So it's really saying nothing. There's nothing there. In fact, more generally if you have g to the n, meaning you take that same string and do it n times, you just go n times along the same direction for the same thing. So your length becomes n times. And now what was that question? The question was what happens to the length of this element? The question was what happens to alpha, beta, alpha bar, beta bar? Well, but what is alpha, beta, alpha, beta? You could say alpha, beta, backtrack and backtrack. And you come back to zero. And so of course, the length is zero. In particular, it is non positive. I asked was it less equal to zero? In fact, it is equal to zero. So yes, so in this case, clearly the problem is fine. Now, so here is the problem you can state for any group. This is the same question, but instead of saying, yeah, so this is how one says it formally. This is the only time I will use the word group. So you may as well see it once in your life. Well, before, if you see it later, it's very good that you will. You may see it once in the camp, or maybe you already have. Suppose g is any group. The group just means like a set of strings. So this is some set with a way to compose or multiply. For strings, we said take a string, take a string, put them together. We have some way of putting things together, a zero element like the empty string, and an inverse or reciprocal like alpha bar. So for example, oh, by the way, I should ask you. So I'll ask you later. So suppose L is a, so now here is the true problem, which were condition one was not asked earlier. I didn't mention it earlier at all. But now let's say I mention it. Suppose this length is always non negative. And the length, when is the displacement zero? When is the displacement exactly zero? It's zero exactly when you have done nothing. You are at the origin. So the origin is what you call the identity, the thing that's, yeah. So equality if and only if g is that special element, zero element, length of g and length of g inverse are the same. G inverse means you're doing the steps in the opposite direction. So you have this way, you would be on the other side, but they should have the same length. For that, I need that alpha and alpha bar have the same length here, and beta and beta bar here. For instance, and length is sub additive, which we have said, that's the triangle inequality. And there's the extra hypothesis that I had said. And that extra hypothesis is what makes all the difference. When if g is h, then L is positive. Here is the main theorem of the paper that I mentioned. G has to be abelian. G has to be commutative. Meaning, the point is this group. If I, so if I take alpha beta and alpha bar, then I get alpha beta alpha bar. This is not the same as composing in the reverse order, alpha bar first and alpha beta next. Here I just get this. This string and this string are not the same. So this group is not commutative, whatever that means. It means that you can't just take elements and multiply them in any order and expect to get the same answer. But that is forced by that extra hypothesis. So basically the statement says, if you have a group, if you know what these words mean, if you have a group with a norm, meaning this last condition, so you have a group with a metric which is a norm, that group must be abelian and that group must be something called torsion-free. And remarkable, not so hard is the statement that if you have an abelian and torsion-free group, you can have a norm on it. That's whatever that means. Forget these mumbo-jumbos, but yeah. That's the statement. This is actually an if and only if. Anyway, so, but here let me tell you the proof of this statement from the thing. So L satisfies these two properties that we started with, which maybe I should write here anyways because I'll keep referring to them. L of, what is it, G H is less equals L of G plus L of H and L of G squared is equal to twice L of G. So L satisfies these two properties. So if you believe, so if you believe from this, so I might remember my original question. The question was, is it true that L of alpha beta, alpha bar beta bar is less equals zero? If you believe that that's true, meaning G H, G inverse H inverse, right? Alpha bar is just the inverse of alpha and beta bar is the inverse of beta. If you believe this is true from just these two properties, then fine, then that's what I'm saying here. That's all I've said. If the original problem in that free group was true, was had the correct, has a positive answer, this is true, and that's all I'm saying here. But the hypothesis once said that L function is always non-negative. So how can you have this to be non-positive and non-negative, has to be zero? When can it be zero? It can only be zero if this string is the identity. So what do I get? I just got, this is E. From now, what do I do? I multiply on the right by H. So these two cancel. That's what reciprocals do. E times H is just H. That's what identity is true. And now I get G at G inverse is H. Well, multiply by G. Those two cancel and I get G H equals H. So just knowing, if you believe me that this implication is true, if this word to be true, then the group has to be committed. That's sort of very simple. Why is this true? Is this true? And it's so high. And that involves what I mentioned in the title of the talk, word games. So now I wanna try to tell you the word. So back to the original problem at hand. So in some sense, to solve this problem, you can first of all, you can try to work with as many rules on G and H as you want. You can say that G and H commute, in which case we know that this is just zero in some way. You can work with other, you can work with extra rules. Like G square is one, maybe, or X square. You can work with what we call relations in groups. This is the case. The free group is the case where there are no relations. You have no extra rules to play with. There are no extra tricks you can use. Only this much. These are the only two things you have. You can concatenate. You can put two strings together. You can take inverses. Nothing else. And even so, this is like the worst case, the hardest case. If it is true here, you should be able to hope that it's true everywhere else. But is it true here? And why? Very easy to state. And as I said, it still took five days with incredibly sharp people working on it. So first of all, let's see. I claim that this is a group. So that's in bold. Why is it a group? What is the zero element? It's the empty string. Take empty string. Put anything together on left or on right to get what you put together. The empty string exactly is that. What is the inverse of alpha or beta bar that we know? The inverse of alpha was what we called alpha bar. The inverse of beta bar is beta because A is A. Reciprocals are reciprocals. For lack of a better way of saying it. What is the inverse of alpha beta? Is it that? It is not. That's correct. It is the other way. You have to do it in the reverse way. You have to write down it's beta bar and alpha bar. So if you put beta bar here, it first cancels. Then you get alpha and alpha bar. Then they cancel. So you can't just say this. In general. So for example, if you have the inverse of this nice element we were looking at, then that would be first take the inverse of the rightmost. That is beta bar inverse is beta, then alpha, then beta bar alpha. So just usual tricks. Fine. And what we want to understand is if G is, so now if G is this group, and it satisfies these two conditions for all strings G and H, then what can you say about this? Or maybe I'll even, so this was true for all G and H. This was true for all G. The question is maybe I'll make it even simpler. Not for some general G and H. Let's look at the specific string. Alpha beta, alpha bar, beta bar. These G and H are words in these letters. Let me just look at these four letters. Not some G, H, longer. Just the shortest possible system. What is this? Why is this true? And now here are where the word games start. So now I'm going to, so how does one get around to trying to prove this? Well, so here we go. So the first claim is that the length of, as I said, length of G times H is less than length of G plus length of L. Now from that I claim that you can prove this. Yes, how? One word, induction, right? Induction, great. Oh, sorry, I wrote it down, didn't I? Fine. I thought this was supposed to be a quiz, anyway. One last throwback, like, how quickly can you, anyway. The second one is that, so first of all, this one I didn't write down, so good. Suppose I said all of them to be equal, and then I just get the kth power of something, like, you know, perform the same action k times. So clearly you just get this is less equals, length of G plus length of G plus length of G, that's k times length of G. I claim it's equal, always it's equal, why? So some of them are easy to check. For example, length of G square, if K is two, you have two things, this is actually given to us. But from that I claim that length of G fourth is four times length of G, yes. Because you take G square, square, and then, and so by this you can therefore, save an induction that length of, for which K does it work immediately? Powers of two, all powers of two you get the same. How about length of G to the sixth? Six is not a power of two, so again, yes. N implies N minus one, X, oh wow, okay. That is very sharp, yes. So I was gonna do it, yeah, well, that's actually, yeah, just, okay. Saying that is enough, let me actually spell it out, what he said, so take any power of two bigger than that, and I'll do N implies N minus two in this case, but that's twice the application of what he said, right? So you take G to the eighth, that's eight times length of G by power of two, right? G to the eighth, as eight, there are eight of them, so take six of them, what we want, plus one, plus one, that's less than length of G to the sixth, that's length G to the seventh, because I'm using the fact that there are, this is a product of three terms, but I can use this by induction, right? And then now this is, but now length of G to the sixth by the same argument is less than six, right? And now I get back eight, so I start with eight, I get back eight, therefore every inequality in between must be an equality, this is an equality, remove the two, you get the G to the sixth, right? So you can do the same for every power, great. So this is nice, so we have one nice consequence, a specific nice consequence is for equal arguments for any power, integer power, they're equal. Now the third one is interesting, well the title, if you know what conjugation is, that's the third one is saying conjugation preserves length. So let's take any string and then take two elements, left multiply and right multiply that string with one letter or one string each, but those strings should be inverses of each other, okay? So you get some string. Can we prove that the length of S is length of G? That turns out to be slightly harder. So here is how one does, it's called string, the strings are called conjugates of each other. And conjugates, the nice thing is they behave very nicely with powers. So if you take the square of this string, alpha T, alpha bar, alpha T, alpha bar, then you see the alpha and alpha bar cancel. So you get alpha T square, alpha bar. And then you can do by N, if you take one more alpha T, alpha bar, those will cancel, then those will cancel. So in general you just get that S to the N is alpha T to the alpha bar. This is true not just for positive integers, it's true for N equals zero, because you get alpha, alpha bar, which is zero, which is just one. And it's true for negative integers, it's true everywhere. Lemma, L is actually conjugating variable. Take any two strings, G and H, just from these two properties. And the two that we deduced on the previous slide, you can get that the length is Y. Here's the proof, I don't have time, so I'll show you the proof. I'm sure you guys can come up with it, given enough time. So the proof is very simple. Start with the length of G H G inverse multiplied by some number N, positive number N. What is this? This is the length of the Nth power. That's what we did on the previous slide. Length of any string to the N is N times its length. So what is the length of this? It is the length of this to the N. As we just saw what that means, that means G H to the N G inverse. And now I use my property one. So it's less than length of G, so length of G inverse. And then H to the N, as we again know, is less equals N times length of H. That's the same thing applied inductively. This is true for all N. And now we use what the previous speaker, Anthur, told us. The method of divide by N, divide both sides, and the powerful trick is to take N to infinity. It sounds like a very small idea, but very simple thing to do. But here is what it really means. I mean, very simply what it really means. Suppose I know that A, I have some real number A, and that's less than one over N for all N. These are all positive numbers. But from this fact, for all N, I know that this immediately implies, and of course, it can only if A is less equals zero. So that's that idea that you can go from any, this is what calls the Archimedean property of the real numbers. For every real number, there's a rational or one over N smaller than it. But this idea is a very, very powerful one in some sense. And that's what we are going to use. So dividing by N, you get the length, this N goes away, the length of what you want here, is less equals length of H plus this stuff, which doesn't depend on N divided by N. And now when you take N to infinity, that goes away. So you get the less equals here, but I want equality. What do I do now? Apply the same trick, right? If G at G, if this is a conjugate of H, then H is a conjugate of this. I pre-multiply by G inverse and post-multiply by G, which are inverses of each other. So if I call this whole thing H prime, the length of H prime is bigger than the length of G inverse H prime G. So I exactly get the reversal problem, right? So I get equality. So you can start playing these games, and that's why I call them word games. You can start playing these games and deduce more and more facts. So this trick is a powerful method in mathematics and considering this thing, which is called a commutator, whatever it is, is a useful trick for proving that these things commute. Because if you can show that this string is one, is the identity, meaning it has length zero, if you will, then this equal to one, and then we saw how to post-multiply by G and then G inverse of first beta and then alpha and get that, right? So back to ask you. So we have proved all these nice properties. So we have proved, what have we seen? We have seen that length of G to the N is N times length of G, and length of GHG inverse is length of G. We have just proved these extra two properties. And the one by induction, I don't mention, that's it. Ah, thank you very much. Yes, so these are the two we have proved so far. So now, so more generally, can we find what one calls a non-Abelian group? Things don't commute. If they do commute, of course, here is an example of a group where we can prove this very easily. This distance was zero. North, east, west, south, or whatever, was zero. So find this thing or show that no such group exists. And as I said, so this somehow, so this showed up sort of naturally out of some work I was doing in probability theory, but then I actually emailed various people. Somebody at Harvard, somebody at Northwestern, Northeastern, somebody at Maryland, Seattle, Urbana, some people in Bombay, unfortunately nothing was known. And then last December, I was, I had joined ISC by then, I was visiting UCLA, where I met Professor Theron Stahl. And for those of you who don't know who he is, presumably very few, he has a very short biography. So he's the youngest participant in the IMO in history, at the age of 10. That year, he got a bronze. So he's the youngest bronze medalist. Presumably, that's the consequence of the previous one. You can figure it out. The very next year, he got a silver that makes him the youngest silver medalist in IMO history. You can guess what the next slide is. And then, you know, the only other thing he did of importance after that, award he got of importance, IMO and field medalist thing. That's where you all should go. IMO and field medalist. Anyway, so he got, but yeah, he's extremely decorated. He's one of the sharpest minds around, one of the most prolific mathematicians. He's achieved lots of powerful results and gotten most other awards. Anyway, so, yeah, he's incredibly sharp. Let's just put it at that. So, work games are terrible stuff. So this is what we discussed. So suppose, recall, these are the two properties which I wrote down here. Suppose we want to show this thing. So, okay, so, first of all, I can rescale and say that make all of these, whatever these lengths were, right? The length of each of these is some number. If it is negative, fine. If it is positive, divide by some large positive numbers so all of them become less than one. That dividing by positive number doesn't change what we want. That's the point. Zero divided by anything stays here. And positive means the inequality stays there. Now, the strategy then that came out of our discussion was to try to prove that this length is less than C for a smaller and smaller value of the I raised it. But the moment is less than, say, or less than one over n for all n. Then you get that it's less than zero. Okay, so now we come to the work games. So, this is the last time you'll be asked to play games in this camp. Or that can be solved on the board. So from this, from these alone, can you prove this is less equals, oh, I'm sorry, the third property that I mentioned, length of alpha, length of beta, all of those less equals one. From these, can you prove it's less equals four? Yes, that's trivial. That's just the triangle inequality. Okay. Can you prove it's less equals two? Yes, why so? Right, so you break it here. That's right. You do this. Now, the length of this by here is length of beta which is less equals one. And beta bar which is less equals one. And you get two. Perfect. I might have broken it differently, but exactly. So, how about four over three? Let me tell you four over three, because that's, you'll get there. You'll get there, I know because I got there. I'm sure you will. But this is what you'll do. Four over three should remind you of something. This three in the denominator should tell you what to do, cube. So, if you look at the length of alpha, beta, cube. I'm going to call this alpha, beta for convenience. Look at the length of this cube. How many letters does this have? It has four letters. So, how many does the cube have? 12. So, first I write this down. Forget the length. Alpha, beta, alpha bar, beta bar. But I write them three at a time, because I want four. Alpha, beta, alpha bar, beta bar. And then alpha, then sorry, beta, alpha bar, beta bar. There are 12 letters. I write them into groups of three. What is the length of this? This is less than the length of this. The length of this, the this, the this. Each of them is a conjugate. So, I just get the length of beta, length of alpha, length of this, length of this. Each of them is one. At most one, so I get at most four. So, size, the length of this is, at most the length of is less equals four. But this is exactly three times the length that we want. And I get four over three. Word games. The question was, can we keep improving on these word games and come down to here? And that's what he posted on the blog. So, this is where the blog starts. So, here is a curious question posed to me that I don't know the answer to. So, and then he let it have to be the free group on two generators. He described this thing. What is not clear to me is if one can keep, so that's where we ended. And then I left to fly back to India and he posted the blog. If what is not clear is if one can keep arguing like this to continually improve on these bounds and come down all the way to zero, blah, blah, blah, blah. Anyway, this feels like a problem that might be somewhat receptive to a more crowdsourced attack. I'm sure all of you know crowdsourcing. I definitely didn't at my age. It wasn't probably around. So, I'm posting it here in case any readers wish to try to make progress on. And then it started. So, conjugation invariance. This that I erase. L of G H G inverse is elevated. That came within three hours. From that first guy, that's Tobias Fritz from Germany. So, he's this very young and incredibly sharp guy. And then people spent two days trying to find examples. Out of the five days that I said it took to build the world, which you will. The first two days were spent trying to find examples that did work. Maybe, can you introduce additional rules without the rule alpha beta equals beta alpha? That is just always good to work. Without that, can you introduce it? Additional rules and then make it work. Nothing worked out. So, if for those of you who know, I'll say they tried to find solvable groups, nilpotent groups, some kind of perfect groups, maybe nothing worked. So, then they said, okay, fine. Then, then people said, okay, maybe it doesn't work. Let's try to prove it doesn't work. Let's try to bring that bound down to zero. So, four over three was we had already put on the blog. Can you beat four over three? The next bound was five over four. And I'm going to tell you how. More, see, the point of this talk is just to tell you the word game. So, here is the word game. Again, if you want, you can follow it, but this is the word game. Start with a different string. Now, you had alpha beta, alpha bar, beta bar, put an alpha before it, okay? So, anyway, you don't need to follow this. The real idea is on the next slide. This is just some example of how much more involved you have to do. If you write this down, and twice this length, then it means you have five letters, you write 10 letters, so you do something. Alpha, so this is alpha square beta, alpha bar beta bar, alpha square beta, alpha bar beta bar, and then I multiply by alpha bar square, alpha square. So, there were 10 letters already. These are the 10 letters. There are four more letters which are doing nothing. Alpha bar square, alpha square. But then I can do something clever. So, now by sub-additive, I can break off this alpha square and get two. And here I see alpha square beta and beta bar, alpha bar square. If you carefully check, these are inverses of each other because they are in the reverse order also. So, the length of a, so this is a conjugate. The length of this conjugate, you can remove the G and T inverse, and I just get these two. But this is a conjugate, this is a conjugate itself. So, these are conjugates within conjugates. I call them nested conjugates. And so, I get one, two, and two, and so, this is a conjugate, this is a conjugate, these are two, I get four. So, twice length of this is less than four, so this length is less than two. And now I can do the same trick as on that board. So, I take, I want five over four, so I take four times the commutator, I get 16 letters, and I do it cleverly so that what remains here is that extra five letter string. And then you do the math, but then you can just check that this is again a nested, this is a conjugate of four of three conjugates. So, one conjugate gives one, another gives one, another gives one, after breaking this off, which gives two, and you get five. So, you get five over four. Isn't it? You can do various kinds. So, you have a, so the point is the following. One thing is notice the repeated conjugations on the right-hand side, conjugate inside conjugate. The second is that, notice that in each step, you get a finite improvement. I'll say that in the slide right now. So, the blog post one, so now I'm telling you all these times in the Indian standard time of the minions, so 17 December and he posted it on the 16th night in UCLA. Here it was morning. In three hours as I said, literally at the dot of three hours, broke of three hours it came. Then you, there was five over four, so four over three was fairly straightforward. Five over four was not, but it was still doable. You can see why it works on five lines. 19 over 16 became harder and 22 over 23. First of all, this is an important breakthrough because the bound went below one. That was somehow, maybe a psychological barrier for people, will it ever go below one? If it does, then it will go to zero. It did go below one, but you can imagine if it took from here, the next bound was, this is 20 over 16. The next improvement was the 19 over 16. How much harder it would have been to get to here? So it was extremely clever tricks, one after the other, but the point is one has to keep reducing to get to zero and at each stage you are going to a finite number. You have to do this, any finite number of steps with getting to a positive number. It will not get you to zero. In principle, this is an infinite process and there are therefore infinitely many more and more clever tricks and that is going to be hard to do. So what we wanted ideally was something that would recurse, that would sort of apply repeatedly to itself, feed on itself and this is where sort of, maybe let's say intuition broke down or I don't know what to say, but there was a stopping block here for almost 24 hours out of the five days, two were gone and then there was a third day which was almost entirely gone with nothing and then Siddharth Gargill, who was my colleague, he programmed, he had an idea which he asked, but he was verifying their idea manually was sort of very hard. He got a computer to do it for him. So again, this is a very 21st century project, it starts on the blog, it has a computer. The idea was the following. If indeed this was to hold, then if you take powers of this and multiply by alpha, by that sub additive business it should be close to length of alpha or it should be less than one. And he did this to a computer assistant search. I will show you on the next slide how he wrote it up and the crazy thing was, so this was the, so 21st, 20th December, 22 or 23 and it took almost 22 hours and then this, so he wrote down this thing. He said here is a computer generated proof where he got to 0.816, starting literally from saying the norm of the length of alpha bar is less than one, therefore the length of this conjugate is less than one. So if you try to write down each step in a proof where you don't jump any steps, it takes you 126 lines to get to the point where you get to 0.816. And of course, I can't imagine doing it manually, maybe somebody can in this audience, I certainly can. He got a computer to generate this and he found, the computer found the path which you can, so this is what computers can actually do when programmed correctly. And so the previous, so this line says that if you take the 17th power of a commutator, these are 68 ST characters and the bars on either side. Then that is less than 13.8596 divided by 17 because of this to the power k thing and you get to 0.81 whatever, 0.814. So he did this and somebody else, Pace Nielsen said this was beautiful, my intuition before something something something. And then he literally said the first 43 lines established something, the next 30 lines of the code he went through it established something, the next 40 lines established something, the last seven lines established. See, actually took it apart and decoded the computer generated proof and found out what it was doing. And then that was the next idea that came. And that was isolated into the following result. So that's the last trick I'm gonna tell you. And I'm almost at the end, yeah, two slides after this. So this is called the internal repetition trick for lack of a better name. So suppose x, y, z, so suppose I have four strings. Just in some four strings. But x is conjugate, conjugate remember is g and g inverse. X is conjugate to something like w times y and x also conjugate to z times w inverse. Yes, so the same w shows up in both of the strings. Then the length of x is at most half the length of y and z. And the remarkable thing here is that it doesn't depend on w. And the proof is strange but I found it very nice to show you. So what you do is you write x, you write two n powers of x. Earlier at some point for conjugation, we wrote n powers of x. Now you write two n powers of x. The first n of them you write as this one. So suppose x is s, w, y, s inverse. Then what is the nth power? It is s, w, y to the n, s inverse. That's what conjugates to do. And the same way if it's conjugate to this one, it is t times this times t inverse. What is the nth power? It is t, z, w inverse to the n times t inverse. So the length of the two nth power is the length of this string. And now don't look at inequality. Look at the picture. This string is written here. s, w, y, w, y, w, y, w, y, s bar. s bar is a single t, z, w, r, z, w, r, z, w, r, t bar. Just write these things down. And now start taking one element at a time out by the subaritivity. So my subaritivity is the length of s plus the rest plus the length of t bar. So I get a s and a t bar come out. Now what remains, I don't know if you can see. If I remove the s and the t bar, I get a conjugate because there's a w here and a w bar here, which is the first one. I remove those because conjugates I can remove. Then what I get is a y and a z. I take those out. So I get s, t bar, y, z, individual term. So s, s, t bar, one, y and one. Then again I get w and a w bar. I remove those. I again get a y and a z and keep going. So ultimately I get n of the y's. All of the n y's I recover. All of the n z's from here I recover. None of the w and w bar, they all cancel, conjugates. And in the between I have s bar and 2. So I get s, s bar, t, t bar, one from here, one from here. And that's it. Divide by what now? 2 n. You get length of x less equals n over 2 n is half. This doesn't depend on n. So by that it goes. And I'll end then with a sample. I still have 10 minutes, maybe 5 minutes, I don't know. Then I'll end with a source. Here is the sample application. Here are some random elements. So x actually is not too random. x is, as I said, the computer was programmed to find, alpha, beta square times alpha. That means alpha, beta, alpha bar, beta bar, alpha, beta, alpha bar, beta bar, alpha. And then choose y, w and z in some clever fashion. Don't try to do this right now. But you can then check that x is constant. So we found somehow cleverly this element and these three. And by that lemma, the length of x is at most, what is x? x was this number. This is a string of nine letters, which is just keep writing them and stop at the ninth point, 1, 2, 3, 2, 3. This. This is my x. The length of x is less than length of y plus length of z. But y is a conjugate of two nested conjugates. That's just 2. z similarly is 2. I get 2. I just did something. I haven't done anything yet. What I'm going to show is the previous bound that was there was 0.816. Using that lemma, I'm going to beat that bound on this board by 8 over 11, which is 0.72 bar. Just using this pattern. Why? 8 over 11, so the trick obviously is to multiply 11 of these copies. If you write down 44 elements, don't write it down. Look at it here. And separate it into four copies, the 11 of them. So now what is each of them? So if you look at the first 11 letters, you get alpha, alpha bar here. You can check. And what is in between is nine successive letters of that kind. But nine successive letters I know how to do. That's two. Whether I take alpha, beta, square, alpha, or beta, alpha bar, square, beta, or alpha bar, beta bar, square, alpha bar, the game is the same. So each of these four is a conjugate of nine left string, which is exactly of the kind you want. So I get 2 plus 2 plus 2 plus 2. I get 8. And so just from that one simple internal repetition trick that I called it, we beat the existing bound on one board. I have not used any of the previous cases. I just used conjugate invariance and the repetition trick. But again, the problem is still, this works very good. It beats the best bound. But still, it's a finite improvement. I need to get down to 0. Getting down to a finite number at each stage doesn't quite help me. So how do I get down to 0? And finally, I'll conclude. It turns out that to show this thing, that trick and the conjugate invariance are enough. It turns out that it's a very, finally, the last bit is a bit of combinatorics involving, again, binomial coefficients and so on. If you want, you can compactly express it. You can explicitly write it down and take calculations. Or you can compactly express it as the previous talk did, don't worry. You can actually use binomial variables or Bernoulli variables to write it much more sharply, much more in small space. And the key idea there is that the sum of n such Bernoulli variables has average spread, not of the order of magnitude of n, but of order of magnitude square root n. The variances of order n, the standard deviations of order square root n, which is what Amser also pointed out. And that is supposed to bound. So you basically would get that n times length of alphabeta, the whole combinator of n times, or four n many letters in the string, is at most something like square root n times the constant. One can check that. That's where you need this bound of the variance of binomial. Now divide by n, take n to infinity. You get the length of this thing. It's less than 1 over n square root of n here. And that's what you need. The order of magnitude is smaller than n. You get it. You get to 0. And that's the proof. That's roughly the proof. This was finally observed by Tau at 3, 5, and in his afternoon, shall we say. So he did the last killing blow. And then that was the proof in roughly 5 days. So to conclude, then what started out is a search for more and more clever tricks and games to reduce this bound on C. Ended up as a 21st century collaboration at breakneck speed. You can see the. I came back from the US, and I was severely jet lagged for the next few days. And I had a Google alert for Gmail alert for every time there's a blog. And I basically couldn't sleep those 5 days because that kept happening very frequently. Lots of people contributing. And the computer that crucially contributed. That's the word. So I'll end with two points. There are more details if you want the full proof with what is a norm and a group, and what are the motivations, and so on. That will be what I talk about at IIT, Bombay, in the colloquium this afternoon. But the second is actually what, again, I will end with something that Professor Agunathan was saying as well, is this is what I do. This is exactly what I do for a living now. And I came to the Olympiad Camp 20-something years ago, not 25, less than that. And I knew I wanted to do math all my life. And even now, you can ask as basic questions as you can write down without any mathematical background. And the sharpest minds in the world are still interested in these questions. And they actually turn out to be. So this actually has some consequences in geometry within what we call geometric group theory and quasi-norms and so on. I won't obviously get there here. But people still love to solve questions with just, can we find cleverer tricks that work? And that turns out to be of interest to people. And I went there to UCLA because I could use a grant to go there. And it turned out to be a very fruitful collaboration. And I was going there for a previous collaboration with Tao. But this came up. And I still, basically, you still get to talk to some of the sharpest minds, work on some of the simplest to state problems, try to apply some of the cleverest tricks, come across some of the most beautiful mathematics. And I really hope, again, as you did, that some of you or maybe many of you will at least pursue and keep up this interest in mathematics. Hopefully, some of you would actually pursue mathematics as a career or maybe computer science. Theoretical computer science is all math, certainly very, very interesting math. When I was in your side, the person who gave a talk at the end, one of them was Suha Chajitpur, who has now recently won the Nevenlina Prize, I believe. He was two years my senior. He was IIT number one. And he went into theoretical computer science and is a professor, not in industry. But again, there are, of course, lots of math jobs for people in industry also. Wall Street, for sure. And data analytics jobs in the Bay Area on the other side of the west coast of the USA. And there are lots of good opportunities, now more than ever, for people doing mathematics. And I sincerely hope that some of you or many of you will at least keep the interest up. You like solving problems. Yes, what problems would like you? Thank you. Thank you, Apurva, for a very exciting talk. Any question? How old are you? This question is, as I said, I came across this question from some probability research about two years ago. But until December, it was not solved. And then in five days, it was solved. This was your question? Yeah. And so that's when he said on the blog, also, he has a question asked to me by me. Asked to him by me. And this lemma is due to whom? That lemma you stated. The internal repetition trick. Right, right. That came up. So after Siddharth put up his proof on the computer and paste Nilsen, decorated it, then I think that's what Tau's bread and butter is. He takes complicated things and makes them as simple as possible. He cleans it up. I think eventually he wrote the lemma down. Terita. That's his specialty. He takes things that look complicated. And in fact, apparently, I think that's what the principal citations for him sort of said, that somehow once he's written the proof, you think, oh, I didn't think of this before. That's his job. I think the lemma, finally, exactly this form was his. But it was isolated by paste Nilsen in some sense. After Siddharth got the idea. Any other question? If there's no more questions, let's thank the speaker again. I would request Professor Praneshacha to present the bouquet to Apurva. We'll take a photograph. Thank you. This one interesting thing came up during the camp via some app message sent by Uday Prasapati to Professor Grover. There was one, Professor A.R. Rao, who was our colleague. He was born in 1909, let me write with you. He passed away in 2011, maybe 12, 1908, sorry. You didn't erase that picture? It was 102 plus. And there was an Iber of Professor A.R. Rao, who was inspired by A.R. Rao, and he was the director of the film, 102, not out. Have you heard of the movie? 102, not out. It had several major actors. The interesting thing we had learned from Professor Grover. Thank you. OK, sorry. I said he was our colleague at the camp. Yeah, he came to the camp when he was 84 and was here until 89. He stopped coming because the lift was not working here at that time. Now the lift is working. Thank you. OK, so that's all that we had for the function and the two talks. Lunch outside for all of us. Please join us for lunch.