Using AI to study topological 4-manifolds | Slava Krushkal, University of Virginia | IAS/PCMI

from University of California, San Diego in 1996 under currentology. Thanks very much. Uh so uh here's a very brief outline of my talk. Uh I'm going to start with uh topological just a very brief background on topological for manifolds and the round handle problem. So I'm going to write the round handle problem once and then I'm going to just say RHP. uh then I would like to discuss the uh kind of the algebraic combinatorial reform formulation of this. Okay. So this is uh essentially this is all uh geometric topology and if you had one year of I don't know first year of algebraic topology or something then you absolutely equipped to understand the statement and the main problems in the round handle problem. So everybody's Gary said that you know everybody's tired and so on. I don't want to hear any of this. I expect full focus [laughter] anyway. Okay. So this is math and this is really geometric topology and then three is going to be uh sort of AI plus math and the point of this is that uh there is this second part which is the algebraic comatorial formulation and uh so there is really no topology left in here and uh so I'm using AI tools to kind of study that and I view AI as kind of as a research assistant. So I'll just mention how I use it and I think anybody can use it if you have a problem that it's kind of suitable uh you know in this case there's some kind of cominatorial flavor. Okay. So let me explain the origin of the round handle problem. Uh there are these two kind of fundamental or basic tools and for the topology in specifically in topological for manifolds and they're really tools of high dimensional topology that have been very successful uh in dimensions greater than four and they were developed in the 60s uh and I'm not really going to state them. I'm just going to write down the names and then I'm going to write draw a picture which is kind of the geometric uh the key geometric picture that uh is common to both of them. So one is surgery and this is a 4D topological surgery and the other one is that this is a 5D fivedimensional uh esabortism conjecture or theorem depending on the context. So I'll explain when it's a theorem when it's a conjecture. Uh and very roughly surgery is used to construct for manifolds topological for manifolds in a given homotopy type. The escabortism theorem is used to prove that two for manifolds are homeomorphic. If there's a certain type of far dimensional fivedimensional capabortism and algebraically it looks like a product then it actually topologically is a product. So it's a homeomorphic the two boundaries [clears throat] are homeomorphic. Uh okay. So uh both of these were proved by uh Mike Freriedman in the in 1981 for in the simply connected case and uh this has been extended to what's now known as good groups. And I think that Anthony probably mentioned this in his lecture series. Uh so currently there is a restriction uh what's known as sort of good groups. It's a obviously a stupid name but it's just a class of groups for which you can prove them. uh and it's a very small class of groups uh like a billion nil potent and uh the most general statement is groups of subexponential growth in the word metric also closed under extensions direct limits and so on but uh they're all as subset of amunable groups and so it's got a sort of measure zero in the space of all groups and uh so this has question since Freriedman's work in ' 81 whether this actually holds for all groups. So this is the question they work for all groups. And it turns out that uh the key case uh is the free nonabilian groups and if you can prove it for free nonabilian groups so I'm talking about the fundamental group of the for manifold then you can prove it for all groups. So it turns out that so a fact is that uh if uh you can prove them for free fundamental groups then they hold for all groups. So the free groups uh the case of free groups is the key uh case and let me just draw this kind of basic picture that explains why uh the free group case is central and also what the problem is kind of geometrically. So the kind of the basic uh the key geometric picture. So I'm going to draw some uh surfaces and it's actually two complexes in the for manifold and everybody knows here this is a program on surfaces and for manifolds that generically intersect and double points. So, uh, what I'm drawing is a two sphere with a couple of extra self intersections. I'll call this a. This is a one, two sphere. And here's another one. Okay. So, here's B. It's also a two sphere. And let's say there's some kind of a distinguished intersection point between them. And the additional intersection points they're paired up as a plus and a minus. Okay. And moreover, there is this what's known as a witness circle where you connect the double points in one sphere and then return another sphere. And this is a loop. And the assumption here is that uh this loop is nulltopic in the for manifold. And a null homotopy is a map of a disk. So you can represent it as an immersion. So there's a kind of like a null homotopy here and there's another null homotopy here in my picture. Okay. So this picture there are two meanings of this. One is you can imagine that this sits in some for manifold. Another one is that you can just abstractly build this and then take a some kind of canonical fourdimensional thickening. that's sort of untwisted. Some kind of normal bundles are not twisted in some sense. Uh so I'm looking at the four 4D thickening or neighborhood of this. Okay. Okay. And so the way that this comes up in the surgery and the escoortism conjecture is that uh in surgery you're trying to kind of simplify homotopy type and you find it turns out that you know you have these two classes in pi two. You represent them by two spheres. Well generically they intersect. It turns out that they have this extra data and what you'd like to find is a embedded pair of spheres. So here is the question. Can you find an actual embedded pair of two two spheres intersected in a single point uh in those two homotopy classes? So there's a homotopic class of this sphere a homotopic class B and the question is can you represent can you homotope these two spheres that you see so they actually don't have any extra intersections. So in surgery if you do this then you're happy because you can cut this out and just glue in a fourball you simplify your homotopy type. uh if you're trying to prove the 5DS capabortism conjecture then uh this picture is not quite right because this sphere has a self intersection but A and B are supposed to be the belt sphere of the three handles and the attach sorry the belt sphere of the two handles and the attaching sphere of the three handles and then if you get to this picture then you just cancel the handles. So it's actually a common uh picture in both of those uh statements and the question is can you do this and uh the answer is that currently you know it can be done for good groups and it's uh not known for say free groups nonabilian free groups okay so let me say like in one sentence what the current project is uh you know the way that this is proved for good groups is that uh well of course there's a kind of the key theorem of Freriedman about the casting handles to the standard handle but to get to a caston handle uh [clears throat] you know there are some kind of basic moves you can do on these spheres and discs and I'm not going to explain any of that because this is just kind of the background but uh you know for example If you open the Freedman Queen book, it's in chapter one. These are just very basic geometric manipulations. Uh okay. And these manipulations are sufficient for finding some casting handles and then by Freriedman's theorem you can complete the proof and you're done. Okay. Now those methods just don't work if you have a free group. And so what do you do with that? Uh so one way to think about what I'm going to talk about today is to essentially try to study all possible moves. So we have a very concrete sequence of I don't know maybe like four moves they don't work. So AI is helping us to study all possible moves on these surfaces in a slightly different context. And if you can show that none of them work then you're done. Or maybe there are some moves we just don't know about. Maybe it will help us find find those moves. Okay. Anyway, uh so this is the end of my kind of overview. Now let me describe the uh round handle problem. Okay. So I have a link uh in the three sphere. So L is let's say an N component link in S3. And the three sphere is the boundary of the four ball. And for each component of the link I'm going to do the following. uh well first of all actually terminology so I'm going to talk about uh fourdimensional round one handles uh so it's a fourdimensional round one handle is s1 cross d2 cross the interval from minus one to one so you may think of this as a one parameter kind of family of solitator and uh it is attached along S1 cross D2 cross minus one and S1 cross D2 cross + one. So there are two attaching solidi and it's a you know round handle that interpolates between those two. It's a product with the interval. Okay. Okay. And I'm going to illustrate this uh with an AN not but just to show like what the picture is. Uh so let me draw this uh actually my kind of first basic example is going to be the hoff link. So I think of this component this unnot as the one of the two components of the hoff link and I have to specify the two solid to where the round handle is going to be attached. There's going to be a round one handle for each component of the link. And for this component, it's uh actually for each component, it's going to be a pair of solidi where one of them is a thickening of a meridian. So it just is a little thickening of a circle that links uh this component. So this is one solarus and then the other solar runs parallel to the component and in general you know this could be some knot and you pick a parallel with linking number zero. Okay. And I schematically will just draw this arrow here which just is supposed to show that there is a round handle that is attached to the four ball and connects these two solid toi. Okay. So this is one round handle and there is one like that just for for each component. So there's another one over here. Okay. And so there is a so my for manifold is going to depend on the link. So I start with a link and then I attach if if I have an n component link there will be n one handles that round one handles attached to the forball and this is the four manifold that depends on the link. So this is going to be denoted R subl. It's a fourdimensional manifold which is D4 union N round one handles and uh strictly speaking there's a zero framed one handles. So they anyway there's some kind of data that is a zero framing. So this is the definition of the manifold and the round handle problem is given a link like you know over here L uh is is this link slice in this manifold. So let me write it down over here. So uh this is the the question that depends on a link. Okay. So now uh you can pick your personal favorite flavor topologically or smoothly slice and uh this manifold Okay. And so let me explain the relevance to the surgery and the escaporism conjecture and then state some known results and then kind of start discussing the new stuff. Uh so here is a a statement. So this is a lama or theorem and this is originally due to Mike Freriedman and myself about 10 years ago. And then a better proof was given by Minhung Kim, Mark Powell and Peter Tagner. And this says that if uh surgery and the 5DS escoortism conjectures work topologically for all groups then any link with trivial linking numbers is round handle slice. Sorry this is probably invisible down there. Sorry. Okay, let me write it over here. for all groups would imply that uh the all links with trivial linking numbers All pair wires linking numbers zero are u round candle slice. >> Well, strictly speaking, you can put any category here and the same category here. So smoothly we know that uh surgery and escobortisms conjectures don't work. So that means that for some links with trivial linking numbers they're not around handle slides but topologically this is kind of the key open question. Uh >> sorry >> any any this is the key yes the so the key case that I'm going to talk about in a few minutes is the barman rings you know like a simplest example with a trivial linking numbers and so that's the kind of the main open problem so going back to Freriedman's work in 81 actually in his paper he conjectured that uh surgery serious you know they are both related to what's known as the disk embedding theorem fail for free groups for free groups. So the kind of the reason for having this implication is that you know to try to abstract uh the round handle slice problem for say the barome rings and then you would know that at least one of these is false and so that would be great. Uh I personally have been interested in this for a very long time. So it will be really great to obviously prove this but uh it's kind of interesting. I mean I'm going to say this at the very end but you know I'm talking to these chat bots really a lot. I've been talking for the last few months and that helped me design all these experiments and uh I gave a talk about this right after graduate school. I was interested in this problem and somebody asked me do I think that there's an obstruction or do you think that you know there's a solution to these conjectures and I remember saying something like well four days of the week in a week I think that there's an obstruction and then the other three days I'm trying to solve it uh so because it's just very subtle it kind of balances there and the other day I asked chatbots what they thought and they you know I'm talking to a couple of different ones and both replied that after all these experiments and so on, they think that maybe a little over 50% chance that there's an obstruction and [laughter] so after decades you know it like hasn't moved unfortunately but um I think it kind of narrowed down is narrowed down but uh there's a sharper kind of target now okay uh you can see that I said trivial linking number over here but then I proceeded to for the hoff link. So this is supposed to be a warm-up case. So I'm going to talk about this first. Uh are there any questions about the formulation of the round handle problem? Yes. >> Zero framing. Yes. Okay. So, let me uh state a couple of things. Uh this is not relevant directly to the problem but just to illustrate you know you might wonder what are the some open questions here and that do not involve the bomian rings and so one consequence of uh uh if you just take a knot then okay actually I'm sorry I should have said just from the definition uh pi one of this four manifold are subl is isomeorphic to the free group on n generators and the generators are so there is this four ball you know that's kind of below here and you attach this one handle and what you can imagine is that there's some kind of a base point say in the three sphere and you go to this solurus you go over the one around one handle and come back at this solar and connect back. So every time you go over any one of these round handles, you pick up some generator. So let's say this is the generator A and this is the generator B. And so pi one for the hop link is going to be just the free group on two generators. Okay. Uh so if you have a knot then this is just a single generator. So if n equals 1 then it's a not and then pi 1 is z. Well z is a good group uh surgery and escoorism conjectures work for that. So things are you know you can actually study things very carefully and in this case it follows uh from surgery and eskeortism theorem for the fundamental group Z. theorems. Now for Z imply that uh any not is topologically round handle slice. And so this is stated as a theorem in U Kim Powell Titner paper. But essentially the way that this is done is that uh you use surgery to construct the complement of slices. You kind of imagine what the homotopy type of the complement should be. Then you use surgery to actually construct a for manifold with a homotopy type. You glue in the slices and you have some for manifold that should be kind of like this r subl. Well, you don't know it's r subl. It just is homotop simple homotope equivalent to it. But uh then you use the escoortism theorem to prove that it's actually homeomorphic to the manifold. So that means that you actually found slices in the original R subl. Okay. Okay. And smoothly. There's a a paper I believe at the end of last year or maybe early this year. Uh this is Ty Leman, Ellison Miller. Uh say who say okay first examples of knots that are not smoothly sliced. Round handle slice. And my understanding is that they just uh took examples that are uh kind of amunable to some gigard floor techniques. And uh there are some uh kind of highly iterated Taurus knots. I think in the paper they ask like for example how about the teroil like say positive teroil is it round handle slice smoothly and so anyway these are the only examples that are known that are given in that paper Okay. Uh let me so I'd like to start talking about the round handle problem for the top link. And this is again the warm-up case and I'd like to introduce just one piece of terminology which is very helpful here. Uh and you you may ask how do you study these topological slices? Uh you know nobody can really visualize them. They're topological. They're not smooth. But uh one thing that is very helpful to think about is uh if L say is slicing say four ball or any for manifold is topologically slice well let's say in this manifold R subl then you can take your maps of disks you know R subl is a smooth manifold and you have a continuous map well it's a continuous embedding but you can just pair a bit. So it becomes an immersion smooth immersion. Well, you know, there was some distance between the slices to begin with. And if you just take take a small distribution they continue staying disjoint and what you see okay this is just a schematic picture but both of the components uh so they'll bound two disjoint discs and you'll pick up some tiny double points I mean the key point is that you you're taking a small perturbation so it used to be an embedding it became an immersion with very tiny double point loops And uh the generators of the fundamental group of the for manifold has a pretty large size. You know, you have to go over this one hand round handle to connect back to the base point. So, uh the key point here is that these tiny loops you created, they're all trivial in the fundamental group of the for manifold. Okay. So, this this these loops are called the double point loops. This is where you started at the double point and go along one sheet and then come back on the other sheet. And what you see is that uh all double point loops are trivial in the fundamental group of the ambient manifold. And uh this is what's known as a pi one null condition. So I'm going to just write it over here. So these discs are called pi one null and that just means that their emergence smooth emergence but all double point loops are trivial in the fundamental group. So this actually serves several different roles like one is you can study this kind of smoothly like now this is actually a question in a purely in the world of smooth for manifolds uh do the these two components of the hofflink bound by one null discs in this for manifold R subl it is a smooth problem okay that's one but two is uh this notion of pi one null actually is extremely important And back in this context of surgery and esaportism conjectures because uh if you find a py null disc then you can kind of it helps you build a caston handle because you know like once you have a trivial loop that means that there is some kind of null homotopia and that starts looking like a caston handle. So this these are actually extremely important in the subject. Okay. So this is a very basic notion and so here is a claim about the hop link. So uh claim so one is old and uh quite straightforward and this is that uh the cough link is not topologically round handle slice. Okay. And uh when I say old, I mean like really old. And this predates the notion of a round handle problem. And the way you can prove this is just uh you can you know suppose it's topologically sliced. So you can as usual you can look at the complement of slices you get some for manifold with boundary. It turns out that that complement is precisely the complement to some link that should be sliced in the forball. That link in this case should be the white head double of the hoff link. But the white head double of the hoff link is not topologically sliced. So anyway, what I'm I don't want to really write this down, but I'm just saying using kind of elementary methods, you can prove this. And here's a stronger statement. So this is new. Okay, this has some AI contribution here. Uh the components of the huff links do not bound by disjoint py discs. So it's a stronger statement you know that uh if coughlink was topologically round handle slice then you could perturb those discs and you get by one null discs but if you somebody just gives you some arbitrary one null discs you don't know that they came from a topological embedding. So this is a stronger statement and so this is a new result that and I'll try to explain how that's proved and again this is supposed to be a warm up for the main uh kind of the burian case. Okay so uh next I'd like to I promise that this is supposed to be very accessible. So I literally want to just look at this picture and just start manipulating it you know draw some pictures and how would you how would you slice this link in the you know this is a cough link you know it's as basic as it gets there is a linking number so the first question is how do you get rid of the linking number you know uh but let me stop if see there are any questions y any any link with trivial linking numbers. >> So this is a this is this is just a warm-up that we're discussing. Yes, just to you know if you you know to believe again going back to Mike Freriedman's work in 1981 is that there's a there's an abstraction. This is kind of the whole point of this round handle problem. Try to find an obstruction. But before you deal with the barometer rings, how about you know there's more like the basic case of the hover link. So this is what we're discussing. Yes. >> Yes. Well, I mean, I think singularity brings to my mind like some kind of algebraic geometry or something, but this is, you know, this is topological category. I mean, it's I think it's a I don't I'm not aware of a significance of that number. Yes. Now the pi one null disc does not imply slice. The non-existence of pi one null discs implies that there is no slicing because if there was a slice if there were slice discs you could perturb them and you get pi one null discs. So for the negation there is an opposite implication. Well, so this uh so this does not have anything to do with Freriedman's theorem. This is just kind of basic logic that again if if if something was slice then there are pyon null discs. If there are no pi null discs then it's not slice and it's just very basic no freerance theorem. Is that okay? Right. Okay. So let's look at the cough link and uh so uh let me just draw some pictures and invite you to think about this and if you can come up with some kind of independent proof that would be great. I do not know how to prove this without this AI kind of computational component. Uh okay so what so what's going on here? Uh so first I'll discuss this kind of geometric pictures then I'll move on to this cominatorial reformulation that I advertised. Uh okay. So, so let's say uh suppose that you know there is some there are some kind of slices well you can intersect them with these solidi and you you know solidi are threedimensional slices are two dimensional and you would see some kind of links here in this solosurus and in this solurus and what I should have said and let me say it now you can actually assume that the link you see here and here are actually exactly the same because you know your own handle is a product of a solidus cross an interval. As usual, you can look at the middle of the round round handle. It's a solidus cross zero. If the entire interval is from minus one to one, it is the middle of the round handle which is cross zero. You should make you can make those slices transverse to the solar and then after that you can just push stuff out of the round handle. So you can assume that the link you see here is exactly the same as the link you see here. So anything that kind of comes into the solar goes over the one handle round handle and comes out exactly in the same way like out of this solo and the other way. It could actually enter through this solar and come out over there. Okay. Well, [clears throat] so let me call my two components X and Y and And for reasons that will be clear in just a minute, I'm going to call them X sub, E and Y sub E. E is supposed to be the trivial group element in the free group. Okay. Well, again you I'm trying to find some kind of slice discs and there is a linking number. So you have to get rid of this linking number somehow. Well, so I can let me just I'm just kind of explaining some basic manipulations here. uh you can take this component y sub and you know try to make it bound on the disk say in the four ball I'm just I'm drawing this in the three sphere but you can push it in the four ball but while you know let's say in the three sphere there is intersection with this other component well that's you know bad we want them to be disjoint then you can just cut out a little disc out of here and connect it by an annulus and then send it over the one round one handle. So that kind of get gets rid of the linking number between Y sub and X sub. But the disk that I'm trying to build for Y sub is getting kind of complicated. It first there is this disc over here, then there's an annulus. It goes over the round one handle and then it comes out over there. Well, you see I created a new linking number. I created the new linking number of this new component that came out and the original Y sub it's actually the same kind of disk goes over the round one handle but you see it's not allowed to intersect itself because it went over the round one handle it means that it picked up a group element so if I were to intersect that new component with Y sub then that intersection will not be pi one null so that's a problem so there's a new linking number well you know but I have this solidus over here. So I can do exactly the same thing kind of try to shrink this component and there's going to be some kind of a annulus and send it over here over this round one handle. It resolve it resolves this linking number that I just created but you know it has to come out on the other side of the round one handle. So here's a new component and it picks up a new group element. Now this is a B because it went over this round one handle. Well, guess what? This new component links both of the original and the new one here. So, you know, there was x sub and maybe I can call this y sub a because it's this y kind of disc, but it went over the round one handle. It picked up the group element a. Then it went over here and it became y subba because it also went over the this other generator. So now there are actually two linking numbers of this new guy with the original x sub and the new y sub a you know the recent y sub a. So you can continue resolving these linking numbers and you create more and more problems and you would think well so what you see is actually there's there are some linear terms and there are quadratic terms. There's a linear term which is a linking number of the original x sub with anything that goes over this round one handle. There is also a linear uh kind of linking number between this original y sub and anything that goes over here. But then there's a quadratic term which is the linking number between anything like over here in this solar and anything in this solar. So the linking number you know you think of that as kind of a linear thing but but here you have linear terms and quadratic terms and you know you're very welcome to try this and I just I don't think there's a single invariant that actually detects the abstraction here uh uh but you know the feeling you get is the more you try the more problems you create and there are so many problems like what do you pick as the abstraction And it's actually a huge problem like you don't know what to pick as an obstruction. There are so many of them that you have to consistently just say well here is my obstruction but any obstruction you try to pick it can be resolved kind of using this piping technique you know method and but you create new problems and anyway uh okay I'm explaining my frustration with trying to solve the problem. [laughter] uh you're welcome to try you know it's very basic you know you can just doodle it uh uh there is a super exponential complexity because you create these linking numbers with anything that existed previously so uh okay so let me uh >> Well, the pi one null condition just came in the fact that so I resolved the first linking number by sending the disk for y subb over this round one handle. It created this new component and I need to cap it off. It has to be a disk, right? But but it starts linking my original component. So I cannot intersect them. So if I did not care about pi one null, I could just >> Yeah. So pi one null is extremely important here. Okay. So, uh I think I'm getting very short on time. So, uh maybe without explaining anything, let me just write down an equation and say that it encodes that problem, that linking number problem. So, here's a claim and I think I don't not have time to really explain where this equation takes place. Uh so the following equation uh encodes the linking problem. So what I'm saying what I mean by that is that if there existed ponal discs disjoint discs bounded by the hoff link then the following equation would have a solution. Uh so strict so this is going to be an infinitely generated a billion group and it's an equation of some commutators in this infinitely generated a billion groups group uh so it's going to be x so let me just write this down and then I'll try to explain it a little Okay. So what what does this mean? Well, uh so I mentioned that you can assume that the two links you see that you know the links are the intersections of these slices with the two solid here they're actually the same link. Uh and what I can uh I can let me draw a one of these solid toi and there is some in you know it could be an incredibly complicated link that passes through here the slices pass through here creating some link in this particular problem before we get to the barome rings I only care about the linking number so what I'm going to do is I'll just draw the meridian to this taurus I'm just going to call this capital x and so there is some kind of stuff here that you know this link You know, some of this could come from the X component, some could be from the Y component. There could be some stuff that doesn't go around, some kind of local linking. And uh so this X is kind of measures the components that link that go around the solotaurus. So this is the linking number with the meridian. This is what capital X is. Capital Y is the same thing for the other side. Okay, there's the uh for the other component you know there is some stuff passing through there uh now where this equation takes place is I uh let me just write this down but let me not explain this uh so if you look at this for manifold and delete the slices then uh what actually what I'd like to do is I'd like to take the universal cover of this for manifold so this for manifold has a free fundamental group. You take the universal cover. It's looks like a tree. At each node, you have some forball with like all the stuff drawn on it. And the kind of the vertices of the tree are connected by these round handles. This is the picture. And when you delete the slices, you pick up a meridian. There's a generator of first homology, which is the meridian to each slice. So uh so these capital X and capital Y are supposed to be elements of uh this free aabilian group generated by all possible meridians. I'm just going to call them X subG Y subG where G is a element of the free a billion group on two generators. So all the slices pass through and you can compute the linking number and you know there's a meridian to each of these slices which is a generator of that first homology of the complement. It is a little bit scary because it's infinitely generated. But on the other hand the slices are compact. So only finitely many of them pass through each solar. So these are some kind of finite linear combinations. And then uh to make sense of commutators you have to go to the second term of the lower center series module the third term. So strictly speaking this is in some kind of group modulo the it's in the second term of the lower central serious modulo the third term but I will just completely run out of time if I write everything carefully. Uh so the uh you know the reason why they are commutators is because if you think about the hoff link and the linking number uh you know the really nice thing about the hoff link is that you can draw this Tauros which is a regular boundary of a regular neighborhood of one of the curves. It's actually isotopic to the boundary of the regular neighborhood of the other curve. And then one of these basic kind of curves on the Taurus is a meridian to one of the components and the other one is a meridian to the other component and it's a Taurus so they commute. So everything that I'm drawing here some kind of hop links you know there is this uh the original components X and Y and then there's the soluri and then there are also this small solar. So if you look at all these relations, you find this uh equation. Okay? And so if you look at this equation, there's actually no topology left. It's literally some weird quadratic equation in a infinitely generated ailion group. And the question is do there exist capital X and capital Y that satisfy this equation and uh it really very much encodes that frustration with a trying to find the disk there. Let me just very quickly kind of illustrate this. Uh so suppose you know let's just take like the first try let's say x not y not equal above zero. Let's say you're trying to solve this. Well, that means that like what's left is this is not present. This is not present. This is also not present. What I mean by a x and I'm sorry I didn't explain this. X is this sum of those elements in this free in a infinitely generated group and a you know the free group X on them just by translation. So a* x just means that you take each generator and just translate it by a. So if x and y are both zero, then this is just not there. So what you're left with is just this basic commulator x sub comma y sub. Well, it's not zero. It's like it's one of the generators in this free a group. It's not zero. Well, you want to fix that. You want to make it zero. Well, you know, you could try for example uh take, you know, declare set x1. This is kind of my next correction to be uh y sub. Then if you look at this commutator over here, if I declare x1 as y sub, then this becomes x. This is also x sub x sub comma y sub. And I'm not paying attention to signs here. It cancels this x sub y sub over here. So they cancel but you see but it immediately creates a new problem because the same x is also enters over here with but is shifted by a. So it immediately created this new commutator a times y subb commutator with just y subb. So anyway, if you it precisely mimics those geometric manipulations and what it looks like is kind of a discrete dynamical system where there is some kind of a disturbance here like a seed and it you know you have to kill it. So you introduce a term to kill the seed but it immediately creates a new problem and you try to kill that problem. introduce a new term, it creates several more problems and okay, the goal is to show that it never ends. Okay, so uh okay, let me just uh I'm very short on time, so let me just say the way that this was solved is Actually before I say that let me just write down the corresponding equation for the baromeian rings and just kind of like we can stare at both of them together and I'll explain what is happening now. Something is running on my laptop right now. So as we speak something is happening but uh so here is a u just imagine the same problem for the barome rings. Uh so there are three components and I'm going to just draw them maybe like this. And so for the hop link uh there was just a commutator and this is the commutator that's u you know the boundary of a regular neighborhood of one of the components. What you see for the barome rings is uh you can draw this boundary. So you see this Taurus what I'm explaining is that there's going to be a triple commutator. So there's a Taurus here and then if you look at this curve try try to fill it in with a disc. It intersects that component with in two points. This is kind of a very basic thing that people draw for the barome rings. You cut out the two discs. You add the tube and what you see is a triple commutator. There's a base Taurus which is a commutator of two things. And one of those two things itself is a commutator because it has this genus one surface. I'm just explaining that the natural kind of generalization of this equation is going to involve a triple commutator and the equation here is going to be So this is not a billion anymore anymore. This is actually takes place in some group and it's called the Milner group which is was invented by John Miller to study link homotopy. So it's very highly nonabilian uh and uh one way to study this is to use magnus expansion where you can send everything to polomials and study just polomials. And so this is what I did. I uh uh have a subscription to Claude and Chad GPT. I I asked them to write code for me. It was very easy for this. This is literally just this equation over here in this free ailion group. This probably took like a month of very intensive work to really implement the code for you know you you know this is really non-abilian situation. So some kind of non-communive group. Uh I don't know. So let me just share some thoughts about like you know uh how you interact with these chatbots. This is my personal experience. Uh I personally don't code. Uh but you can ask you know claude in my experience is extremely is really fantastic in coding. It is going to immediately not immediately but very quickly write some code for you. You know with probability 100% there is probably going to be something wrong with it. And uh you know I've worked on this problem for years and I did many examples by hand and so like it would return some code and I would try to run some examples that would not match my answer I would ask cloud to like look at that and because it did not match my answer. So after many iterations it converged to a implementation that matched all my answers. You know again uh this is not a proof that the code is correct. This is a proof that it matches my answers. Uh you know I'm not asking any of these chatbots to prove Ethereum for me. I'm just asking them to generate huge number of examples and generate conjectures that I could prove. So this is kind of my main goal in this project over here. But in this case it actually uh there's actually a solution. So the solution here is well I you know kind of unexpected to me you what I suggested was uh sending the free group for the hop link uh to a finite group. This is a cyclic group on of order six and it send a to one and b to two and it's amazing you know it works for zmon 6 it does not work for zod 5 it does not work for zod 7 but this is the kind of thing that AI is actually really good at it just it can just test you know a thousand examples for you a thousand things and it will just find the thing that works this worked and once you send this free group to just a finite cyclic group it actually becomes a finding a search problem. It's actually just a million cases or something and it actually it gave me a Python program. I run on my computer and there's no solution. Okay, in this case there cannot be a finite quotient. uh you can prove that uh just based on theorems that have been proved about this and uh so right now I have this code uh and I'm talking to these chatbots and asking them to kind of test various cases and so they've explored the ball of radius 2. Now in the free group on three generators there are 37 generators there and if you think about non-commutive polomials it's going to be 37 factorial it's actually an abs it's like something like 10 to the 43 uh so you cannot actually compute this but you can explore like some directions that are seem like where the solution could exist and so they eliminated that uh and anyway this is kind of in this balancing situation now where on the one hand they cannot find a solution. This would be really great if they could, but at the same time, the obstructions that they come up with, they're not like some kind of robust invariant. It's just every case is somehow abstracted, but in some kind of weird way that depends on this particular case. So, it is possible that maybe there is some if you increase to like some kind of targeted search in like ball of radius 4, maybe there is a solution. But anyway, I so I told you all I have and anyway, thanks. Sorry. Sorry, I'm over time. >> Dave