The Weigold--Goldman Conjecture for Compact Lie Groups - Tsachik Gelander

追加: 2026-05-06

[00:00:15]It's a pleasure to have Silander from North who will tell us about the legal government conjecture for compat.

[00:00:23]So it's still it will still be a conjecture after my talk.

[00:00:29]>> We should always audience.

[00:00:32]>> Yeah. Yeah. Yeah. [laughter] There will be Yeah. Good question. So uh so we start with the n uh an integer which I will assume >> [clears throat] >> uh with at least three. We can you can assume that n equals three throughout the [clears throat] and uh soon g will be not infinite but suppose g is a finite or let yeah let g be a finite uh single group [clears throat] um and then uh we look at the space of fitment And okay look at the for fn g if you want it's all the set of all n tles in g it's a finite set and inside I look at the set of all epis from f n to g [clears throat] uh these are all you can think of it as all the n tussle that generates.

[00:01:42]So if G is a finite group then these are finite sets and the group of automorphisms u so you can identify this with G to the n of course uh non-cononically by fixing a generating set for the free group fn.

[00:02:02]So uh the group of automorphisms of fn uh acts on the on both varieties on the nupole and the generate >> and uh my question is about about this.

[00:02:30]So, so uh so how does it acts? So, uh if you have a automorphism h and then it takes uh a homo a subjective map from fn to g this is by definition f of sigma inverse x. Here sigma is a homoph to g and x is a point in g. Uh so that's the action.

[00:03:01]Uh you can associate to this graph the group of n uh as a a very nice generating set called the Nielson transformation a finite generating set for instance if n is three I think there are like 24 neon transformation. So, so you get a a graph uh whose vertices are the three tupil that generate G and it's a regular graph of regularity 24. So, so you get a regular graph and you can ask a question about this graph.

[00:03:40]Um [clears throat] so you would like to know if it's for instance if it's connected or or if it is even an expander uh an expander graph when you change g.

[00:03:54]So by the way every finite simple group is generated by two elements. That's a classical the on the other hand you've never said n equals 2 which you've dismissed which is the only case I've ever thought of.

[00:04:07]>> Yeah. Yeah. I I will say I will say why I'm not talking about any kills too in a moment but uh but uh okay then the question or the viral conjecture conjecture it's a famous conjecture is that if G is finite simple and n is at least three is connected or in my notation then then the action >> it looks like a >> is is is tragic. [laughter] >> Sorry >> your three is horrible.

[00:04:52][laughter] >> My nicest tree. [laughter] So then if uh uh then then this uh this action is is transitive and uh okay at least I heard that uh uh people use this uh what is called product replacement algorithm in computer science to generate a random element say in the symmetric group or in a random permutation uh by taking uh so so to do random work on very large groups it's by multiplying generators is very not effective but here uh uh uh you do random work on the on this on the graph of a tree tool that generate let's say you want to mix 52 cards you you do some nice random work on the three tpples three uh tpples that generate S53 or S which is almost a simple group and then you hope to get a um a random permutation.

[00:06:05]>> So what do you put in the initial thing?

[00:06:08]You put the two generator on the identity or something.

[00:06:12]>> Yeah. Okay. So, so, so, okay, that's that would work [laughter] because okay, so, so first uh I I just the motivation one of the motivation is to get a random element and you can do it by doing random work on the product replacement graph. Uh so this maybe one of the motivation that people were interested in in in this problem >> but the product replacement is the same as the sniff I think.

[00:06:41]>> Yes. Yeah.

[00:06:42]>> Yeah.

[00:06:42]>> So that's a it's exactly the same graph and uh and I heard that people use it and and they knew that this algorithms worked but they didn't know why and uh uh and and and the reason it should work is because this graph is is first it's transitive and secondly it's expanded.

[00:07:06]If it's if that's the case then then then this should work. Today we know we don't know whether it's transitive still but we do know that at least for n= 5 it is expander because we know that uh some polish people prove that the group out fn for n le at least four or five h is kashan property t and this will imply that if this graph is connected then it's expanded Now we still don't know it is connected but we already know that there is a very large connected component and [clears throat] this connected component will be expanded and this is this will be enough for for to generate that the algorithm won't >> you don't it's sufficient that it's one projection to one coordinate so you don't need >> uh uh okay uh usually something right >> that's >> right we don't need transitivity but there is a very large connected component inside so so let let me say uh so this is still an open problem whether this whether this graph is connected that's the vag conjecture it's a famous conjecture let let me say h few words um [clears throat] about uh uh about this uh before going on. So so so a representation row from fn to g is called the redundant uh if one of the generator is is redundant. So if if it's enough if it's enough that um or maybe before I say that I will say why why I don't look at n= 2 always in for n= 2 this this is not true h for n= 2 so if you identify uh a homoph film for fn to g with n tappers.

[00:09:29][clears throat] Then you you have a you identify this variety with g square.

[00:09:37]Now uh and then you can look at the functions f of xy equals the trace of the commutato.

[00:09:48]It's a trace >> what? Yeah. Okay. Uh let's let's do trace of both commutator.

[00:09:56]>> I mean is G a tuba a matrix group?

[00:09:59]>> Yeah.

[00:10:00]>> Okay.

[00:10:00]>> You can just take the course.

[00:10:02]>> Yeah. So >> you should take >> enough to take one commotator.

[00:10:06]>> No, I think we a conjugacy class in G which if you're matrix group you could take the trace.

[00:10:13]>> Yes I'm looking soon I look only on matrix group. So, so I I allowed to take trace and this is a function uh uh suppose that G suppose that G is a matrix group for instance SL2 FP that's an example of a finite simple book or maybe >> yes >> maybe you do okay then you don't have trace [laughter] >> I don't know you okay so so suppose it's a it's a matrix loop then you can take you look at the trace of the commutator and it's very easy. It's it's a one line or one one minute proof to show that this is an invariant function and usually it will not be constant.

[00:11:01]Uh so so if you have an invariant function which is not constant action is not transitive and why it is invariant because um we know we understand the theorem of Nielson that tells us a very nice generating set for the I will I will mention it in a moment uh and you you show that this function is invariant under all the neilen operation. So what are the neon operation uh ne transformations it's transformation that takes me from one generating tuple to another. So suppose that n equals three now. So I have x y and z and I want to take it to another generating set. One way I I can keep X and multiply Y by X.

[00:12:00]Uh this is example of an transformation.

[00:12:05]Uh so I can I multiply one of them one of the entries by another one from left or from right >> or by the inverse >> or or I can take inverse.

[00:12:16]Yes, that's that's another so so then you count I think there are 24 such for n equals three 24 operation I can multiply each i by each different j from left or right and maybe take in and that these are need some transformation and they generate the group o and you can check check now that this is invariant because we know trace a b is trace ba so we we know so this is invar I added all this operation.

[00:12:50]Um I because of you need to take inverse I also added this uh thing a trace is not invar I think taking any so so um um so that's why that's one reason to look at n equals greater equal three >> when n is large you know something you know >> yeah yeah yeah okay remark so so let let me let me yes for every fixed g if n is large enough then it's true then it's transitive >> okay >> you don't have the same for all >> no no no no very very little is known uh for I think for some family okay I want to leave the finite groups and and move so let me just say what is a redundant or from fn to g is called redundant And >> if I can remove the last uh generator and and it's still epimorphism but I allowed to do it after applying automorphism or after after applying Nelson move. So, so it's it's called redundant if I can write fn as a free as a free product non-triv.

[00:14:33][clears throat] >> Okay. Yeah. another let's let's say it in another way. So let me denote by PN it's a subset of fn uh the set of primitive >> elements.

[00:15:00]So >> you need to define you need to define.

[00:15:03]>> Yeah. Okay. If fn if fn is generated if if f3 is generated by x y and z then x is primitive y is primitive and z. So every generator every free generator is a primitive element. An element is primitive if it belongs to a base or to a n generating set.

[00:15:32]So can you repeat every word which belongs to a basis to a minimal generating set group called >> if if if you look at the group fn and you have a ntle that generated you can show that they cannot satisfy any relation then and so it must be a free base. So so so x is primitive also xy is primitive because I can get it by nilson mir. So, so in in general, if you have a a if you have X and you multiply it by a word of Y and Z from the right and by another word from the left, it's still primitive element.

[00:16:14]So there are many many primitive elements. Uh [clears throat] so a primitive element is a is an element that belongs to a generating set of a minimal size. It's a base for the so the primitive elements are more mysterious than to be primitive in in in a billion group like ZN. It's it's very easy to to to to determine to be primitive in in FN. It's it's a bit more complicated.

[00:16:41]Not that complicated. There is an algorithm to check that Don can tell you many many things about primitive elements. But I'm interested in other things about primitive element.

[00:16:54]So for instance h [clears throat] um I can you can show it's not hard the throw is redundant that's that's that's almost obvious that if and only if row of pn equal g and also it's the same to say that row that pn uh intersect the kernel of four and uh empty >> non empty [clears throat] and android subjective.

[00:17:43]So these are all these are equivalent to epoism being redundant.

[00:17:52]>> Not redundant.

[00:17:53]>> No, this is equivalent to redundant.

[00:17:57]If we redundant means that every element in the image is is every element is an image of primitive and in particular the trivial element is an image of primitive. But if the trivial is an image of primitive then it's also every other. So so so to be a proposition is redundant if and only if the primitive it's objective on the primitive and the vagal conjecture uh is equivalent let me just formulate it again conjecture the famous old conjecture is that every epop is redundant.

[00:18:54]That's it. Everyomorphism should be redundant. It's known that the action that the the set of redundant epomism is invariant of course and people know for a long time that the the action on that set is transitive. You can take any redundant to any other redundant but uh so the val conjecture is equivalent that every representation is redundant.

[00:19:22]So for any okay >> for the here just mean that one of the element one of the three elements is not not needed right >> uh not necessarily so not necessarily I I don't say no >> no you said that people know that redundant is transitive but you mean >> the action on the redundant is like if if you're connected it's clear that you can get to a redundant tpple right >> yes >> and but redundancy is preserved underneath transformation >> it's a there are some old works by I I remember there people like Gilman and Evans people from that prove theorem about finite about varants of the vital conjecture they proved it for family like SL2FP very limited uh results in particular they proved I don't remember who proved that the action on the redundant is transitive.

[00:20:24]>> Uh but this is it's not very hard and >> so G is a finite simple group >> so far. Yeah.

[00:20:32]>> N is greater equal than three.

[00:20:34]>> Yeah.

[00:20:34]>> So so this is basically saying that that for every eporphism we can find a free subgroup with two generators on which is still an eporphism.

[00:20:43]>> No >> it's it's more than [clears throat] that. It's much more than that. Yeah, it's it's a free sub that this is probably easy what you what you said it's probably easy. I I think I think it's easy that forever that you can always find a free subgroup that subject it's a free sub that are generated by a by by a primitive pair by by two element two element that not just not just primitive they belong to a primitive base.

[00:21:10]>> Yeah, you can take a you can find a primitive base and delete one of the >> Yeah. So, so let me let me give another version which Alex what he called the baby vab conjecture which is seems to be even much easier but still open. Uh you can [clears throat] you look at the action of uh or you look at the epigys from fn to g and divide by the automopisms of g. You look at this set up to automous G and then the group that acts it's actually the outer automorphism of Fn and the conjecture is that this action has no fixed point.

[00:22:18]So Val conjecture is more or less equivalent to saying that this action is transitive. So now now I instead of looking at the out of N acting on epism I divided here by the autophysism of G and then the inner automorphism suddenly acts transitively so I here just out automorphisms and it's a much simpler it should be of course if the action is transitive there are no fixed points this is a should be much should be much simpler this is equivalent to say without mention any action and that there is no there is no n inside fn characteristic such that uh it's a finite index uh and simple It's finite >> that equivalent that in open >> characteristic it's more >> under invariant under all >> it's it's characteristic means it's something stronger than being normal normal characteristic yeah so there of course there are normal subgroups such that the question is finite simple every finite simple group is [clears throat] you prevented but but the question is can you find with n characteristic so it's infinite under every auto every autom so that's an open question and this is yeah these are about finite groups it's still wide open there are very very little progress uh for many many year there were no progress maybe I'll just say that uh [clears throat] if you fix g and take n large enough then everything become easy for if you then give me a finite simple >> estimates the estimates like if G is some size how big N has to be >> uh >> so uh the most interesting result that I'm aware of is paper by Neil Avenue and Shelley Girion they know that uh they they estimate n as a function of the rank of g when they consider only find simple group of fle types and and as and they they give estimates on n as a function of the rank. Yeah. So so so the interesting result is they yeah they can find a fixed n depending on the rank >> on the rank not on the size. Yeah, which is much more. Yeah, the size can go to infinity but the rank is smaller.

[00:25:22]>> Yes. So, uh so in particular if you look at the family FS SL DFP or FQN SLDFQ and you fix D then there is N which is good for okay forget about P that was only a motivation for [clears throat] now and now we let G be a comfort group.

[00:25:49]uh compact connected uh lab like so SO3 or so D.

[00:26:06]Okay. So now [clears throat] um so first uh okay so first I consider again oh fn g I can identify it non-cononically with g to the n and uh and then you can ask question about the action of photo on this compact case.

[00:26:38]So first thing you observe is that the measure G is a measure every every league group has has a has a measure which is invariant the left [clears throat] translation if since G is compact it's also invariant at the right and therefore the H measure H is invariant under autom H of this uh thing.

[00:27:14]So I take the the the multip I multiply our measure with itself and time it is outending part.

[00:27:27]So again how do you see that you go back you it's enough to check it for the NSO move.

[00:27:35]So it's enough to check it when I what happened when I multiply one entry by another from the left and by definition of the har measure it's invariant and also because it's a because comp group are only modular it's also invariant under taking the inverse so so so it's so it's invalid and [clears throat] then h there was a conjecture of gold h that turn out to be not difficult.

[00:28:08]uh is that uh if n is at least three uh the action on on g is or so [clears throat] so you look at this space which is like g to the n with the measure and uh I claim Again, you need n >> you don't need n is three. He's got a very beautiful description where n is two on what the measure is on the trace fixed trace of the commutator.

[00:28:48]>> Yes, that's true. Uh >> I'm just saying it because some of us live there.

[00:28:53]>> Yeah. So, so Goldman Goldman worked with also with n equals 2, but it only took G2 to be SU2 I think or product of SU2 and uh and then conjecture that the elodicity holds for any compactly group and uh yeah >> there's a natural measure there coming from a simplectic structure maybe.

[00:29:20]>> Yes, but okay. But again I I I still want this and and this statement is not true when n equals because again this function is is the same function is is invariant and unconstant.

[00:29:39]>> Uh it's true that if you restrict to level sets of this function then uh >> are you assuming that the image is topologically dense?

[00:29:50]So far I don't assume this is true without assuming anything because here is here is a simple clear this >> here is a lema uh uh the set of all let's call it u uh the set of all xy uh in g² set of all pair such that xy is Then I look at all pairs that generate. Uh this is a a full measure in H in G square and uh also a full measure of course dense uh and also open if G is Note if G and and opened unless G is a billion if G is simple or same if G equals if G is frivalization G is equal to the comm so a compactly group is always a product or almost product of a billion like a tom simple group so so if is semi simple the set of pairs that generate is also open. So it's a very nice set actually the complement is is is a risky close the sub variety the pair that don't it's very hard not to generate that's why everything should should be true here it's very hard not to generate if you take two element most surely they will generate and then you can see why uh this theorem is true let let me spend uh two minutes um to tell you >> why because after after that I will not be able to prove anywhere.

[00:32:11]So uh um so >> but one might expect some rigidity properties of the stat sheet.

[00:32:23]>> Yeah, you're going to get to them. Yeah, I'm I'm going to get I will have after that I will have a lot of question and some some ideas but but no results. So this this is a result. So so why so so u so let so here is the proof. So let a be a subset of g cube which is invarian >> I'm going to prove that yeah that the action of3 on gq is already with respect to harmon so uh so a uh I take a invariant set and I want to show that it has it is it is either null or conal it is measured either zero or [clears throat] okay. So um um um so assume assume it's not the case that the measure that's denoted by m the how measure of a uh is not trivial is strictly less than one and strictly bigger than zero I'm going to get a contradiction so then h by fability It means that a is not independent of one there are three coordinates x y and z and and and a a I will say that a is independent of the first coordinate if for every y and z uh the intersection the all the xyz that it's in a has either measure zero or one then I will Okay. Right. So, so, so, so by Fubini everybody understand?

[00:34:27]>> No, but okay. by uh you >> and to be uh there is a coordinate um but there is um a a coordinate a coordinate okay let let me say it that there is x and there is actually there y z inside u.

[00:35:03]So u is the set of generators of dens group such that the measure of all if the how measure such that x y z in a is is is not one and not zero is between one and zero.

[00:35:36]>> So I want to say that by Fubini the set is not independent of the first coordinate or the second it's it's not it could be independent of the first coordinate but then the same idea will work with the second coordinate. So, so, so I I claim that I can find a coordinate. I assume that it's the first coordinate such that the set is not independent of that coordinate. Since U has full measure, I can take the other two coordinate to be U. And then I say now I know that this even this by word of Y and Z, Y and Z generate a dense out.

[00:36:17]So, so by way of why why Z act erodically on the on the first coordinate because a dense the only invariant measure in on G under a dense subgroup is the measure.

[00:36:32]So, so because YZ acts erodically, this set which is invariant and this set the set of all X such that this is in a this set this set is invariant under YZ.

[00:36:50]Okay, maybe I I'm this set this set which I can call it a YZ. Now I fix Y and Z is invariant under the group. Now it's a group Y and Z but Y and Z generate a d subgroup. So this set cannot have a non-trivial measure.

[00:37:16]Okay, maybe it was a bit fast, but but what I want to show is some [clears throat] um it it may have been too fast, but what I want to show is that uh basically the [clears throat] proof holds because generically the vible conjecture is true. So generically h I don't know that every ple is redundant but but this dilemma says that almost every pair is generated then some so almost every fit apple is redundant and this is why this statement is true. So uh okay uh it was a bit fast but the hard measure is what about other measure.

[00:38:17]>> Sorry I'm confused. So you proved now this conjecture for n equals 3.

[00:38:21]>> I pro yeah I proved the conjecture for n three or for any n which is at least three the same proof.

[00:38:28]Okay, wait.

[00:38:30]>> There's a conjecture that's the >> theorem. Prove the theorem.

[00:38:34]>> I think that's a theorem of Goldman, right?

[00:38:35]>> No, that's a theorem of mine. Oh, sorry.

[00:38:37]>> Goldman conjecture it. He wrote a gaffa paper about he proved it for SU.

[00:38:43]>> He proved it for SU2.

[00:38:45]>> Ah, okay.

[00:38:45]>> And he conjectured for and >> that's a that's that's the one that I prove.

[00:38:51]>> Okay. I apologize.

[00:38:52]>> No.

[00:38:54]Yeah. Okay. it was this yeah it wasn't the main theorem in the paper uh I have the paper where I proved to conjecture of morgoulis and soer and and as a byproduct I also prove this deductor of of gman and uh [clears throat] yes anyhow it's this is just for the har uh now um The question is what are classify classify the environment.

[00:39:37]>> He also wrote some papers on the clapping class school.

[00:39:40]>> Yeah. Yeah.

[00:39:42]>> You I thought that at some point he took energy. No, it was okay. Anyway, um >> yeah, >> this is a very clean fast. Yeah, it's it's a it's it's a simple thing I yeah it's yeah uh it's a simple ter but from now on I cannot say much more so so so uh so classified environment so example so if H and G is connected compact Then you can look at the the homorphisms from FN that falls into H if you want if it's like H to the N. And this is a subset of the morphisms that fall into G, right?

[00:40:42]And so you can take so so you see that how on h in the invarant measure right I can take how on hn as a or r on this subvariety and this is an invite measure >> where does the connectivity play >> what >> where does the connectivity >> the connectivity it's example >> I can assume whatever I want >> okay so so then that's an invariant you can if you want you can delete it it's still correct but but I'm afraid of non-connected thing because of library conjunction >> doesn't know anything about the final test >> yeah I I I yes so that such a measure I will call algebraic so so a definition mu on let's say um um let's let's write on gn is algebraic if there is h inside g close compact not necessarily connected H such that the support of M is containing H to the N and you coincide and it coincides with how this is because you are not connected uh H on the connected component on the fibers uh of the map from H to the N to H mod H connected component.

[00:43:01]So that such such a measure uh we call algebraic And the conjecture is that every erodic measure which is autoen every erotic measure is algebraic. That's that's the that's the and and the reason because we don't know viral conjecture and actually v conjecture is the analog of fal conjecture is not it's not true anymore.

[00:43:32]uh when you don't insist uh when you leave the the zone of simple group.

[00:43:45]>> [clears throat] >> Okay. And so the conjecture is that every h every invite measure every erodic measure.

[00:44:05]>> Yes. That'll be and again it is true if n is large enough. If you fix g and take n large enough then it's true. So let let me tell you a much more simp down to earth conjecture.

[00:44:25]>> Can you explain why >> you just said >> sorry >> can you explain why is it?

[00:44:32]Yeah, there is a recent theorem about this by search contact and and and few other people and but uh but okay basically it's it's the the same reason that um okay let me let me say why [clears throat] let me say why >> what was the question again yeah >> why is the But actually is true if we we allow n to be depending on g.

[00:45:07]>> Yeah. All this thing becomes true when n is large and g is fix and but all these things are more interested when we fix n. I don't really care if any I fix it to be seven or three. Uh but I want it to be fixed. Uh let me say uh um again definition f from f n to g is h redundant if um I can write again fn nontrivious as a free product such that f of A is this.

[00:45:58]And here is a stronger conjecture which implies the above. Erh [clears throat] every uh every everyism is redundant.

[00:46:27]This is my conjecture.

[00:46:29]Every conference I give you three element to generate a dense subgroup. I don't say that already two of them but after you apply some neson transformation two of them should generate and it's really hard to imagine that this is not the case right because it's very very easy to generate I can tell you more about this set this set of generator which is open dense a full measure you can also understand exactly its complement it's a very small variety so it's very very hard not to generate a dense subgroup. So if you give me three element that generate a dense subgroup just to to apply transformation to to find let me let me let me tell you equivalent formulation h for every f from fn uh can you name the conjectures and give the implication sorry implies which This two this two are equivalent. This this implies this.

[00:47:39]Uh these two are equivalent.

[00:47:42]One side is that side is obvious that side requires some argument. But for every homomorphism which dense image uh is redundant. No. uh uh f of pn dense for every if you have a dense representation in a compact leg group the image of the primitive should be dense I believe but if you prove it you prove this conjecture and and also many other conjectures so that there are list of question of Lubotski Zelmanov of Platano of Potabic list of names that you can connect this uh this result this conjectures and this seems to be it's very hard to imagine that it's not true I think >> one or both of these things you could replace closure by risky closure or >> all this all this conjecture. uh you can uh discuss the retry closure and that when when does this Yani do that usually you want to leave the world of compact group and discuss Dubai group in general and and they're all very interesting and I can talk more about that case than on compact case but uh still I can >> the conjecture on the left which follows from that makes sense that uh rigidity makes sense for n equals 2 as by the way.

[00:49:31]>> Yes.

[00:49:31]>> Then you have the invariant but it's still >> Yeah.

[00:49:34]>> But you what you were pointing out is that if n is greater than equal to three then redundancy is everything.

[00:49:42]>> Yeah.

[00:49:43]>> Which is not equals to.

[00:49:45]>> Yes. But okay another thing let me just say why this is true for large n.

[00:49:51]Suppose you have sufficiently many elements then uh by neielson operation you can bring suffici and then generate a d subgroup by sufficiently many of you can bring by non operation many of them to a small neighborhood of the of the identity and to be in a small neighborhood of the identity you generate a d subgroup if and only if the logarithm then generate the algebra or or if and then you So you can show that if you fix G and N is sufficiently large then any n tupal is redundant and you can use that to prove uh that this conjecture holds if n is is large and but this was done actually I think there are two two two papers by by different groups that actually recently in the last few months published that that prove something like equivalent to the to this implication to to to the fact that for large n and fixed g the thection is the the is true but I don't want to there are details it's not completely obvious I want to say a bit a few more I don't have much time uh I want to say something about baby vot >> is there a group for this simple or >> that what what you call group where this is simple for log but is easy no [laughter] if you take SO3 uh then then it's true for n= 3 so for small enough groups you can do it by hand and also for SL2 oh by the way it's interesting uh you ask about topology. You can ask also about the the standard topology. Is it true that every d representation the image of the primitive is dense?

[00:51:58]Then this you can ask for for SL3 SL2 as well. And there is a theorem of yinsky that showed that it's not true for SL2.

[00:52:08]you can find a dense representation of the free group of on four generators such that the image of the primitive is not dense. Now in the standard house of topology of SL2R or SL2C you do it for SL2C but it also works for SL2R. So there is a representation of of F4 in SL2 it's it's a beautiful geometric construction.

[00:52:36]uh uh so but for the risky the risky topology it should it's still it's it's true for topology you can prove it for SL2R with N= 2 N= 3 uh >> but there is an example of a representation of F4 to SL2C with the image is tense but in the regular topology >> the primitive don't go close to the identity You cannot find primitives close to the identity. Yeah, there is a beautiful example from of Yuminsky geometric. It's by hyperbolic.

[00:53:15]Uh so that's a that's a one remark I I we speak now in the last few minutes about baby vagology.

[00:53:28]I'll be happy to answer. But what is baby vag now?

[00:53:35]compact baby vag again I look at the I look at the epim of pn uh from f3 to g so homorphism with dense image divided by the automopism And I look at F3 acting on this on this and the conjecture is that this has no fixed point.

[00:54:19]In other words, there is no regodic measure which is atomic. There every erodic measure is nonatomic. It's it's equivalent. uh it's much much weaker than saying that every measure is on black.

[00:54:36]So uh so now suppose suppose now uh suppose we have example I want to convince that this must be true uh without proving because I don't know how to prove it.

[00:54:55][clears throat] So first uh remark so so if if row is a fixed point what does it mean? It means that uh means that row of composed with sigma is conjugated to row. So I can >> sigma >> sigma sigma is an any element any automorphism.

[00:55:30]If I apply row after doing automorphism for the free group I should land in something which is >> conjugate >> conjugate by an automorphism.

[00:55:42]The conjugating element is unique.

[00:55:49]It's it may not sit in G but it sit in O G which is still a compa group. Uh so so so it is unique. So this gives a representation from a representation rota from O F3 onto the automorphism group of G.

[00:56:15]I get a such a thing. Now recently a few years ago it was proven that fn okay let me take it's not true for tree but let's say n h uh h recently it was proven that this group has property t a group with property t has only finitely many representation in any dimension up to fin so it has finitely many uh finitely any maps into compact groups. This is the old theorem of Wang that if you fix a group with property t and you fix a compact group there's only finely many maps from gamma to that group up to conjugacy it's not a hard and therefore this implies that there are only finally many people at most finally many I think there are no I think that even if you take sufficiently complicated automorphism it may have only finitely many fixed point and maybe none and and then you take two of them they will not help but let's suppose that there is h I can claim even more than that h more I claim the following sorry where are you now you prove that there are mostly finite number of >> so so this this is true this I I I I say that this follow from property t >> for bigger than five >> for n it's n= five yeah for any for n for n which is bigger than five.

[00:57:50]>> Why is it true? It's clearly true that you cannot have more than finite one number of point.

[00:57:57]>> Yeah. So that >> why don't you have any?

[00:58:00]>> No, I don't know that that's my conjecture. The conjecture is that there is no any. Now suppose that there is suppose there is a fixed point. Uh I will tell you even more.

[00:58:10]uh [clears throat] claim uh if row is a fixed point.

[00:58:23]So it's a map from F3 to to compactly group with a such that it's [laughter] it's it's conjugate to every if you apply you you can conjugate back. This implies that row of the primitive consist so they all conjugate to each others. uh this plays in some conjugacy class where G has finite order.

[00:59:07]This this is this is relatively elementary. You can prove that if you have a fixed point then every primitive must have a finite order and even more than that they're all conjugate. they will all have the same characteristic polomial and the characteristic polomial will be x to the m minus one. So so so and now it's very hard to imagine. So in particular, let me just say if you have homorphism from F3 and you know that one of the gen one of the primitive goes to an element of infinite order then you can prove that it's not fixed point.

[00:59:54]Uh so but this but now something that I cannot put inside G there are three element generating a dense subgroup you can apply every N transformation of them but whatever you get is always of fixed finite order I don't think it's it is possible um Actually there is a conjecture of Zelmanov which Zelmanov has a conjecture that if you have a representation of the free group that all the primitive element share the same characteristic polomial then the image must be virtually solvable.

[01:00:39]Uh and uh okay time is off I can say more speculation about this the few lemas that you can prove but I don't I cannot prove interesting so I was talking Yeah questions ask. So the picture you took in case is that that the the fraction of the probability that you have a probability one set or measure one full measure set of homorphisms that are redundant. Yeah.

[01:01:28]So it's kind of suggests that one might ask for example if you go back to the finite simple groups I take a sequence of final I fix n and I take a sequence of finite simple groups of order going to infinity whether you can prove some kind of upper bound of the fraction of the do you know that the fraction goes to one that is >> obvious there are result about this that yeah then >> component >> yeah there is this product replacement graph has a very large connected component consist of redundant concentration. There there are estimates about it finite group.

[01:02:02]This was studied. Yeah. Yeah.

[01:02:06]>> So somehow it suggests that the limiting picture you have here is kind of trivial from the point of view of the finite.

[01:02:13]>> Yes. Yeah. Everything I can >> full measure it's kind of already know to be true.

[01:02:20]>> Yes. So this is true. So, so, so everything that I can do is is is trivial but uh what I can conjecture is not trivial. So, so, so, so at least I think so. Uh, so I conjecture you can speak about con but you can also speak about the risky topology of group.

[01:02:43]take SLNC and take a a take [clears throat] a um yeah take a representation of the free suppose that I want to I want to claim that if the image is is risky dense then the image of the primitive is risky dense it's the same as saying that uh it cannot happen that all the primitives are uniro So, so uh so suppose you have a representation >> write down this concept.

[01:03:20]>> Yes. So, so let me let me now uh speak about let me now speak about uh excellency.

[01:03:35]So here is a conjecture [clears throat] conjecture.

[01:03:44]Uh [clears throat] if O from F N again N is at least three to S L3.

[01:04:00]Uh anything.

[01:04:06]>> Maybe you can do any maybe. Okay. And suppose that row of P is unported.

[01:04:16]Uniotent mean that all the values are one.

[01:04:21]uh for every primitive then row of fn is important is even unrated to the upper triangular group. This is a conjecture and this implies many things in particular it this in particular it implies there is a conjecture there was a question whether out of n is linear and this was known this beautiful paper by for processi that showed that it's not linear if n is at least three it is linear for n= 2 there is a very very large representation dimension I don't remember 200 or something but for n= 3 and it is not linear but the conjecture is that for any representation of fn the image of fn fn is a subgroup of fn is the inner autom the image of fn must be virtually solvable that's a conjecture this is a stronger conjecture now this conjecture in some cases I can prove if you give me one uni potent if you give me one primitive that goes to a regular uni potent regular >> regular means that the centralizer has minimal damage regular means that the jordan the jordan form is the the second diagonal is full that's regular so in other words the centralizer is minimal dimension any representation of fn row of fn is >> the the [laughter] famous conjecture uh >> if not solvable what you say >> yeah that's this is a famous conjecture and more famous than all the conjecture that I said I was the first to conjecture it I think maybe platonov it's certainly conjecture by Lotski but I think later so conjecture by many people let me write platovski probably Zman know several people actually that if N is at least three and row from O to N to GD C is the representation linear representation then row of fn so if you say what is fn fn I mean by actually inner automation. So FN acts on itself by conjugation and since there is no center this action is faithful. So you can identify FN as a normal subgroup inside ON. So if I have representation of all the fan this image is virtually solar and there are variant of this conjecture and uh and and this is the variant I'm most fond of but I still couldn't prove it.

[01:07:49]uh one thing that I can prove is that if you give me if there is a regular uni potent in the image then it's true then then uh >> so do you know the work of Eskin and his co-workers about character varieties compact character varieties and the kind of rigidity which must say something about your problem.

[01:08:20]>> I know about this work. I I Yeah, I I >> you know what >> I >> I think in some examples it must prove your conjecture. Uh I suspect but >> they have they study not just the action being you know they they talk about the stiffness and then the they identify the measures and the some of these nonlinear actions and the character variety which is closely related.

[01:08:50]Um so they're eventually studying these automorphisms trying to understand the invariant measures stationary measure invar measures I think has used it already for some of these so you >> maybe I I will I will check it familiar >> I mean they're writing up some massive papers I don't know how generally the thing is But he spoke about it here.

[01:09:23]Maybe somebody else knows what they [clears throat] >> Let me just end with I I have I have >> in some sense you would so one of the questions asked is if there's an environment measure that's a single point.

[01:09:42]>> Yes. They >> they also are allergic to finite groups.

[01:09:49]They'll say nothing in >> it's not about sort of I mean so somehow >> I mean you could imagine a situation that they show that the [clears throat] invol needs to be nice but I mean this kind of techniques could tell you when you can have a single point sort of obvious. H I don't know. I'm just asking if you may if you probably talk to them and see what they do.

[01:10:16]>> I Okay. Yeah. Not a good idea. I will >> have maybe similar questions. Yeah.

[01:10:24]Similar question. Is [clears throat] it known? No. Is it known if the possible counter examples for the conjecture are smooth on certain subfolders or is it not? Is it known if the potential counter example was small?

[01:10:46]So I don't understand if suppose you have a a measure that's envirant under the in gant under is it is it there's is it support the manif necessarily manifold and the measure is that >> ah I don't know I I have no idea.

[01:11:13]Yeah, >> that sounds like the best result you can get from me.

[01:11:18]>> Let me let me admit another very very elementary question.

[01:11:30]>> So what I think it's [clears throat] very hard to find coets of subgroups inside a conjugacy class. [laughter] So, so, so, so, so suppose that you have inside a GN, you have MS and you have two element.

[01:11:51]You have G an element and you have a an H sub and suppose that the coset is all contained in in the controversy class.

[01:12:09]So whenever I multiply G by element of H >> sorry >> G is GNC what's G >> GNC itself probably right >> yeah GNC so so yeah so >> we just like G >> the right side GLNC yeah >> suppose that this is contained in in in in all this stuff in all this self-reerential stuff Suppose that you cannot go out of a of a conjugacy class by multiplying by element of age. This is exactly the problem that we're dealing with because if you remember I have X, Y and Z and I can multiply X by whatever I can generate by Y of Z. So I look at a co-et of an element with respect to a group that I hope that it's large and I ask myself what if I cannot get out of the of the conjugacy class. So so I conjecture many things about this. I can prove some things but one of the things that I cannot prove and I still believe is that this implies that H is virtually solvable.

[01:13:33]Or in other words that H does not contain a free group. Or in other words that if H is algebraic it does not contain a representation of SL2.

[01:13:42]You cannot find a representation of SL2C in any dimension and another matrix such that G times the image of SL2 is containing a Gus class.

[01:13:56]And you can use whatever tools from representation theory or from doing dynamic and projective variety by and and try to prove it. Uh I what I can do is very limited. I can tell if anyone is interested in this. I can tell you what I know.

[01:14:15]H I have some ideas but I cannot put it and I have some example but this this is what some of the difficulty lays here. So so if you if you can break this I think uh you can go on and improve the other conjectures.

[01:14:37]So um so yeah [clears throat] no more questions. The fence out here.