Zoya Gorodilova. The Popov-Pommerening conjecture for groups of small rank

[00:00:00]My name is and I'm going to tell you about the conjecture for groups of small rock. Uh first of all I want to fix the field. Uh we can see field characteristics here.

[00:00:25]Let the bear and let bey Would you bring his problem uh asks uh then the algebra invariance of action Reaction is generated.

[00:01:31]This is algebra of 30 algebra.

[00:01:44]As I know you have discussed this problem recently. So I would stop on it and move to other question related to this problem. uh if we consider uh a connected group >> as Julian explained us Hbert's problem didn't ask for a fine varieties it asked about linear action on vector spaces and this is some variation of >> yeahation which close to the question which I want to consider we consider connected group and uh closed algebraic success.

[00:02:56]of action. Sub any is generated.

[00:03:16]It's probably question.

[00:03:31]Um, we call uh support ashar uh if it's normalized by some by both of independently conjectured.

[00:04:05]Uh I said if she reluctible and Ash of the book.

[00:04:27]>> Hi.

[00:04:31]Uh then for any >> the answer is yes.

[00:04:45]>> Yeah. uh when uh this algebra is connected to generated but to add some explanation about a regular say uh write down G is SLM.

[00:05:05]What are regular subgroup of SLM? How do they look like?

[00:05:08]>> Uh regular subgroup of uh if we We can say that uh subgroups can be important can be tries to root sets in our existence and uh rebal subgroups will be generated by root subgroups.

[00:06:00]>> But what does that mean on the level of >> seems that the Taurus is regular subgroup but it is not generated by root subs? Yeah.

[00:06:16]But maybe in the picture it's more clear.

[00:06:29]for algebra not for groups.

[00:06:34]So for for at the le algebra level we just have boxes in just the box.

[00:07:06]space.

[00:07:07]Maybe sell one cell.

[00:07:42]So probably you can formulate it in terms of the composition.

[00:07:47]>> Yeah.

[00:08:12]some very Am I right that maybe I too naive but uh when I consider a regular subgroup in JN uh then uh just every element uh is almost uh uh independent on u another ones and uh so uh either this element is non zero or this element is um anything or just zero something like this >> just uh write down uh you you started with un important sub write down the matrix. So if it is uni potent uh then you have units on the diagonal zeros downstairs and some stars upstairs.

[00:09:26]Uh no no put several stars not everywhere somewhere zero somewhere stars.

[00:09:35]Uh okay so draw some pictures some zeros and some stars.

[00:09:41]Certainly it should be a subgroup. So this location of stars should be not arbitrary otherwise it will be not a subgroup. But if it is a subgroup the condition that it is regular means as CJ said that stars are arbitrary numbers and at some locations you have zeros.

[00:09:57]Right? So this is a condition to be.

[00:10:01]Okay. But this are unimportant regular and can you give us example of a reductive regular subgroup in SLM?

[00:10:18]Nucleus please help us.

[00:10:20]>> Taurus >> Taurus SLN minus one. I think >> Taurus is okay. What else?

[00:10:26]>> What is SLN minus one?

[00:10:30]>> How should we subtract one from SLN?

[00:10:35]>> Who said SLN minus one?

[00:10:37]>> Ask me.

[00:10:38]>> What do you mean?

[00:10:39]>> I mean that In the angle in the left upper angle >> can you draw the corresponding pictures centralizers of single on the last plate here somewhere not overlapping subgroup but to draw a separate picture at the bottom.

[00:11:11]>> Do you mean this >> uh but SLn minus one on the diagonal?

[00:11:16]Not units but just big star and zeros.

[00:11:27]Yes. And uh here okay something like this.

[00:11:33]>> So the stars this time this is not the same as in the important case because they are not arbitrary. But the only condition is that this angle of the stars they form either non- degenerate matrix or matrix with determinant one but this is only the condition on the stars model of this they are arbitrary.

[00:11:51]Okay. So this is exactly the condition the the corresponding le algebra is a linear sum of some root subspaces.

[00:12:01]Okay.

[00:12:05]So if we fix the maxly this is all for fixed maximum to for the maximum to of all diagonal matrices then the regular subgroup are given by some combinatorial datas when you fix those root subgroups which generate our algebra but as a linear space. So not as a real algebra but as a linear space. Okay.

[00:12:40]for discussion. We need to fix some mutation.

[00:12:45]uh in our GV fix group and in this group we fix to and maximal And also need to fix this with respect to fix.

[00:13:51]And here G is what? It's just linear algebraic group >> connected to uh to consider this conjecture we can uh impose some reductions on and ash can trans have the prison.

[00:14:51]Uh and if we apply this uh principle to uh our algebra uh we get that to find the algebra of J is when I um it is suffice to consider uh this algebra if algebra or homogeneous space will be finetally generated then we get that uh this algebra will be finetally generated too.

[00:15:37]So write down this implies such an arrow >> that if any sorry clear but uh why the product is finally generated if both multiples.

[00:16:12]So it it's not clear for me this >> is G is reactive and if both algebra >> okay control question if both algebbras are finally generated why is there product is also finally generated you can just uh take off the generators of both parts and consider their product >> like this.

[00:16:44]>> No, bad idea.

[00:16:48]>> Who can say >> take product with one?

[00:16:51]>> Yes.

[00:16:52]>> Take first generators with one and then one the second generator.

[00:16:56]>> Okay.

[00:16:57]>> But here the question is not this G. But okay, if G is the implies generation. So we don't consider we don't need to consider all the fine varieties. We need to consider only one variety fine but this homogeneous faces is always quite projective.

[00:17:17]Yeah, we can switch our consideration only on Gh.

[00:17:27]uh uh also we can uh say that it's uh suffice to consider uh ash important uh if we uh take any uh algebra group and consider it in particle vertical Then motion group will be productive.

[00:18:06]Uh and the algebra of Britain.

[00:18:22]Uh it means that if um This algebra is generated and because of this reductive we get that uh this algebra will be connected generated. So we can say that we assume that u ash h is the importance group in our uh fixed uh max view and it's normized by our fixed uh but some comments firstly this transfer of principle this is not obvious statement this is some claim uh this is I not remember whether it's proved in pop book but maybe it's proof uh so this is some non-trivial statement but >> exactly to prove this isomorphis this between left and right side is this you need some argument this is not immediate but this is true okay so we believe in this second Please write down this kh.

[00:19:39]No like this. Yes, this is equal to the algebra of regular function in the homogeneous space.

[00:19:49]This is more or less definition of the algebra of regular function on the homogeneous space. Again you need to to think about this because this homogeneous space is maybe quasi projective in general maybe nonfine nonp projective uh but difficult analyze the structure of algebraicity. Here you will see that algebra of regular functions here is algebra of >> and H is contained in U because of >> just because this is a unent subgroup and unot subgroup is conjugate subgroup and maximum >> we can consider subgroups up to conjugation.

[00:20:29]>> Okay. So hop critarian is only for quifying homogeneous but you said something about >> no this is for arbitrary this instruction we can conjecture and tell that for any regular importance of group uh algebra of invarant must be finetally generated.

[00:21:03]The other the other reduction we can consider is uh if we suppose that there exists such connected productive sub >> it's not clear for me when we I'm sorry to interrupt but when we conjugate H it can uh became non regular because you conjugate the jurus No, it's unclear for me at all.

[00:21:32]>> No. Or just take this unicotent radical.

[00:21:35]This is normalized by the maximal Taurus.

[00:21:37]>> Yes.

[00:21:38]>> So you you may consider the semiirect product of this Taurus and this H.

[00:21:42]>> Okay.

[00:21:43]>> Uh this is a solvable subgroup connected sol. And by CM this is conjugate to upper triangular matrix and inside of this bar subgroup any unicorn is conjugated.

[00:21:54]>> Okay. Okay. Thank you.

[00:21:55]>> No what I said at the end it's not needed. This was automatically in the universe.

[00:22:03]>> If we consider such that uh if um this algebra is algebraated uh it follows from transport principle.

[00:22:26]If we uh take uh as uh algebra as functions on z also we can say that it's suffice uh to consider J is simple And now I want to some non results about this.

[00:23:24]has been born for uh groups. These uh type less or = 4. U2 and G4 by type. Users uh prove that uh if a uh um important uh critical of the group.

[00:24:33]Uh and then uh for such fail G and H the conjecture is true. Uh for from that nice results I need to >> but let us discuss a bit >> please draw SL4 so prepare such a picture for matrices 4x4 draw uh with such a picture with stars uh unipotent radical of some parabolic subgroup We can say that if you consider the algebra of this probab sub subsets in our groups and then the algebra of robotic group will Well, I think it's easier to just throw this.

[00:26:07]Okay.

[00:26:11]Okay. This you can do different inside.

[00:26:13]We want just pictures.

[00:26:15]What are uniform radicals of parabolic?

[00:26:22]>> It will be correspond to this algebra and >> walk upward.

[00:26:48]This is unipotent radical of some parabolic sub.

[00:27:09]So you remove the Good.

[00:27:39]>> Yes. So this is an example.

[00:27:42]>> Okay. This is first example. One more example.

[00:27:46]Let us draw all of them after conjugation. No no produce another picture. We will produce another picture.

[00:27:59]>> There are lots of them.

[00:28:03]>> Okay. But but we will just up to contribution. Okay.

[00:28:20]also an automorphism.

[00:28:23]>> Automorphism, right? Also transpondent transposition >> not with respect to the diagonal.

[00:28:38]No mistake said that right upper should be star.

[00:28:55]So they are encoded by a subset of the set of simple roots. Right?

[00:29:03]So how many simpler roots do you have?

[00:29:10]>> Sorry, how many subsets do you have?

[00:29:19]>> Eight.

[00:29:20]>> Eight. But uh we don't consider symmetric subspace. Okay. So please list all subset of the set of three elements up to such symmetry.

[00:29:34]No, just a combinatoral question. List all subset.

[00:29:40]>> Okay, first one.

[00:29:46]Okay, now simple roots.

[00:29:51]Simple roots.

[00:29:56]Okay.

[00:30:02]Okay.

[00:30:08]And more.

[00:30:20]Yes.

[00:30:22]Oh, this is symmetric to Just alpha 2 is >> okay >> and empty set.

[00:30:31]>> Empty set also.

[00:30:33]>> So which picture correspond? Enumerate them. 1 2 3 4.

[00:30:40]>> Okay. So which correspond to what?

[00:30:52]This Sweet.

[00:31:22]Okay. And give us an example of a regular unipotent subgroup here in the cell for which is not unotical of some parabolic subgroup.

[00:31:44]Okay.

[00:31:49]So, Brkins gave a positive answer in the case of one two three and so on because they are unotic.

[00:32:05]When we consider all important subgroups reactive groups that we can uh say that they corresponds to uh roots and sets uh into in our root system.

[00:32:29]means that um every important subgroup can be parameterized but by some subsets in sets that closed uh under uh it means that uh some of any roots in this set is either remote or is not a root at all.

[00:33:06]And such important groups will be narrated by groups.

[00:33:21]Our set we already considered this as example of sub and for this response this important subgroups um proved that uh if No uh nor this is some company which produce some artificial eating. But Friedick not with one >> P.

[00:34:08]He proved that if the complement to the set fees is linear independent or there is a unique linear dependence then the conjecture is to write it as Rank of this set is ratio of the amount of minus one. Uh for example, according to his results uh in >> in this case the conjecture is true.

[00:34:55]>> Yes.

[00:34:57]So can we assume that this complement is linearly independent?

[00:35:01]>> Uh it's if it's linearly independent or uh there is a unique linear dependence in this set then the conjecture is true.

[00:35:13]>> This condition means this. So if rank is equal to number of elements then they are independent and if minus one then you have only one relations between them and the last result was abouture was obtained by in 2018.

[00:35:45]He put it for some special in his position that uh H might be and there is some uh special uh a standard approach uh to check this is um it tells that if the question loop The following conditions are the algebra is generated um is equal to So this such finite dimension and vector is a stizer of this vector and dimension of The laundry is great.

[00:37:53]Uh and she used this criteria to uh get the results and I want to consider other approach for testing conjecture.

[00:38:13]First of all, uh I want to uh the final deration of this algebra uh is equivalent to fineration.

[00:38:43]of this algebra where minus is the uh opposite maximum group to maximum importance to and minus G on the left and a on the right.

[00:39:13]And we going to consider this algebra.

[00:39:17]Um we adapted to what was that was used by Janka who uh solve the similar problem with the branch.

[00:39:38]We want uh consider the big cell in G.

[00:39:50]Uh >> here h is a is a subgroup of you. uh uh u u H is the group of uh maximum important and we consider opposite two and uh and we have conclusion of algebra. This direct product uh invariance u with respect to the direct product is invariance with respect to group generated by them. Yes.

[00:40:30]And uh so >> no this is indeed a direct product >> or not.

[00:40:35]>> This uh they generate a parabolic group.

[00:40:38]Am I right?

[00:40:39]>> No no no. One of them one of them X on the left and the other one on the right.

[00:40:45]Ah >> so this is not a product of this subgroup inside of group G but it's a direct product acting on both sides.

[00:40:52]Okay, I'm sorry.

[00:41:05]And this algebra can be easily found. Uh it's because of minus x and x.

[00:41:25]We can write and algebra is the right of group and k is amounts in the component. Uh then we can fix uh for uh to each level function that we extend to level function on the um and then level functions and uh have that.

[00:42:38]>> So these are basis characters.

[00:42:40]>> Yes. Uh and our importance group can be represented as the product of um our roots of groups where roots of groups uh taken in uh any order and we can fix uh such order on it.

[00:43:13]Uh that uh first uh carrots form the component.

[00:43:27]and others belongs to then we get that this portion group. Um will be the product of and this just looks and We can fix uh coordinates uh on each of these group and get algebra.

[00:44:35]So what you are doing now? Uh now I find this uh u algebra from variance uh and I want to say that uh this algebra will consist from all functions from this algebra that extends regularly to the whole group.

[00:44:57]Uh we know how this algebra looks. Uh and now we want to uh which functions will be regularly extended to function.

[00:45:15]>> So you explain us some technique how to prove this conjecture in particular cases.

[00:45:21]I want to uh explain a algorithm that let us to uh find uh this algebra explicit.

[00:45:34]>> Yeah. To compute the duration of this algebra.

[00:45:38]>> Okay. But let me understand the logic.

[00:45:40]So you listed some non result. You are finished with this.

[00:45:44]>> Yes.

[00:45:44]>> So this is all non result on this subject.

[00:45:49]Not all those there are some more but >> those which you wanted to mention.

[00:45:54]>> Yes.

[00:45:55]>> Now what is your global aim up to the end of the seminar? You want to prove some positive results.

[00:46:00]>> Uh I want to uh tell about some new approach that let us get some interesting results.

[00:46:08]>> Mhm.

[00:46:09]>> Do you going to report on your own results?

[00:46:12]>> Uh yes >> within this technique.

[00:46:15]>> Yes.

[00:46:16]>> Okay.

[00:46:19]As I said, we want to extend uh to understand which functions from this algebra can be extended regularly to uh our group.

[00:46:33]To do this, we consider the relative position of our I'm sorry. This criterion uh is not important because it's some obvious.

[00:46:57]>> Yes, it's obvious or it's some >> uh >> from left to right. I think it's obvious.

[00:47:03]>> Yes.

[00:47:03]>> So this is the >> from most complicated directionction.

[00:47:18]So left to right can We can support our group uh to the bra cells.

[00:47:55]Uh the big cell corresponds to the case when we take and if we consider Simple reflections.

[00:48:31]These cells uh gives us the open sets of dimension one.

[00:48:45]And it's closure will be devices.

[00:48:59]This will be the only uh cells with one. And if we consider The component of cell group it can be uh written as the union of these prime classes and then it can be represented as the union of these open sets with other components.

[00:49:36]This dimension greater or equal to you can consider a new set. Um set uh from which we exclude uh all the dividers except one fixed.

[00:50:07]Then we have uh conclusions of algebra.

[00:50:34]And I want to propose that uh to check if uh some functions from as big cell can be extended to the whole group.

[00:51:00]Uh It's sufficient to change such um regular every function regular. If it's regular in something boundary of dimension two.

[00:51:35]>> Yeah. Yeah.

[00:51:43]>> So but it's worth mentioning mentioning that uh each subset omega I is obtained by from the big cell by just gluing one divisor.

[00:51:58]Omega, not G. Omega.

[00:52:09]And to uh check what functions uh will be led on every such open set uh we uh need to find the orders of our coordinate functions uh along uh these divisor.

[00:52:33]Uh it's some technical calculation. So I uh just claim that uh if we take uh coordinate on then um uh order of u coordinate on the devices.

[00:53:07]The I uh is equal and uh to the other uh of um cards on important group. Uh we can um consider the um sub group uh generated by and some roots and a single.

[00:54:10]>> So this is a minimal.

[00:54:13]>> Yeah. And As I already mentioned, uh don't take u uh any numbering on the uh root system to get the maximum impossible. And if we uh renumber each Such in such a way that first coordinator will be corresponds to uh the in group to the fixing root. Uh then we provides uh other coordinates on our group and what is um of this uh um coordinates along our fixed devices will be equal - one if I = G and zero otherwise when we consider different coordinates It's uh we can uh get some coordinate changes and if we um write the equation where >> express >> express no we can express uh our first fixed coordinates with using using Our new conjugates is some equation and then our coordinates will be equal.

[00:57:02]The degree of this according to the first coordinate.

[00:57:14]And when we uh know all orders of the coordinate functions, uh we can define a new uh function uh which uh will be uh such which will be multiplied uh to such powers.

[00:57:42]Uh We know possible degree.

[00:57:55]So that uh this uh function has no pole uh along any divisor.

[00:58:05]Uh so it will be regular function uh on hold and uh we can uh write uh the algebra on the big cell using this function.

[00:58:58]We uh want to consider the uh some new algebra.

[00:59:20]It has no at all.

[00:59:25]So actually the problem is to find all functions in in this algebra which are regular on the whole G.

[00:59:32]>> Yes.

[00:59:34]>> So we have this algebra.

[00:59:36]>> Yeah. We want to know we want to understand which functions will be regularly extended to the whole group.

[00:59:45]Uh we want to start from this algebra and oh it's sub algebra in our group in algebra that we are interested and we want to consider >> your version is is it in the other direction >> it's for regular functions >> a regular function okay >> we want It's eventually uh negative degrees of >> powers of uh coordinate of uh to our algebra and check uh which will be regular on and this gives us chain conclusions.

[01:01:11]the last algebra uh will be equal to the algebra which we want to find.

[01:01:23]Uh so we want to calculate uh every this algebra to get the last one.

[01:01:41]To do this, we uh consider uh some new algebraas. Uh if we >> so we start with a z and we want to subsequently compute all these algebas.

[01:02:03]>> Yes. on this increasing chain.

[01:02:07]>> Uhhuh. And if it's final if on the final step it's not finally generated then it will be you know become not finally generated at some step.

[01:02:17]>> Yes.

[01:02:18]And if we at each step we get the finite generated algebra then we uh get that for this uh h actually is two um two calculate these algebraas. We u consider construct the following algebraas and uh take functions on our open sets such that These functions multiplied by some power of first coordinate first coordinate bel and then we again get >> a I Z is just A I right?

[01:03:30]>> Yes.

[01:03:32]>> Uh um it means that we uh sequently add negative degrees of T coordinate on each step and uh uh check which functions will be regular.

[01:03:58]If we >> so this is an infinite series right infinite chain of inclusions >> and in union of this inclusion will be equal to the um >> to the next algebra.

[01:04:28]>> Next algebra.

[01:04:31]>> This is not an algebra.

[01:04:34]>> A I J is it an algebra?

[01:04:37]>> Yes. Because our condition if two elements are sent into algebra after multiplying by something their product is sent to should be multiplied by the square of this element to get it. I think that this is a iig jar elements lie in a i 2 j maybe I I know so the whole a i + one is but a iig j is not I I don't understand >> I don't get what's the problem >> so the problem is that uh if you have two functions with this condition that's multiply by something they belong to ai then their product should be multiplied by the square of this something to belong to ai.

[01:05:35]>> So this >> is not sub algebbras. Yes. Aj this >> other algebra >> subac but their their union is is an algebra.

[01:05:45]>> Yes. Okay.

[01:05:48]something like >> Yeah.

[01:05:58]>> And now we want to sequently reconstruct this to this.

[01:06:10]>> We decided that they are not.

[01:06:14]is this success.

[01:06:26]We want to consider such a deal and This is and if we get that this will be this will be ideal.

[01:07:59]You can get Sorry, but I don't understand that u we decided that aig is not sub algebra but you're right that a i g + one is generated as an algebra by something over a j >> actually it should be it should be an algebra probably we can we just should modify Definition.

[01:09:14]Don't be free.

[01:09:33]seems that you add not f but maybe f divided by something to obtain algebra.

[01:09:48]>> Okay, we have 10 minutes and talk should be useful. So people should learn something from what you say. So what people do to learn now >> um now it's some technical >> okay what is the aim? So what is the final result? Um fin result u uh if we get that uh if this sequence u then uh we get that um for this chain Then we added uh while getting this algebra at each step we added uh the final set of generators. Uh and if at each step our algebra is finally generated in this step. uh we calculate this algebra.

[01:11:04]>> Okay, this is the idea of the proof but does it work? So can does the sequence terminate or so what's the result? Yes.

[01:11:11]Um we deal to prove using this algorithm uh the conjecture uh for the groups of type and uh C3. And from this results we get that um The conjecture is true for all groups uh is rank less or >> this is complimentary to lint result.

[01:11:59]>> Uh yes she proved >> she >> she >> because it was some scandal with this lint but I will tell you later.

[01:12:10]Uh he he worked for small for some small groups and for a and a for using this.

[01:12:30]So using this technique you consider all regular uni potent subgroup in groups of type B3 and C3 and prove that the algebra of variance is trying to >> uh yes uh >> and how do you do this? You consider just case by case all regular un important subgroups >> uh for some regularly important subgroups uh uh the conjecture is true from previously known results and uh I can say that There are in between street there are 122 uh uh that are closed under addition and have rank three. Uh because we can consider subgroups after conjugate. Uh it left only uh 35 subgroups in E3 and 32.

[01:13:35]And uh uh using the normal results we get that uh we need to apply our algorithm only in six cases in three and for seven >> and this is done by you.

[01:13:54]>> Yes. Uh we implemented this virtual in the safe pass. It's a computer uh algebra system uh which uh allows to calculate the generators of the algebra.

[01:14:14]>> So you can not you can not only prove that this is finally generated but you can in some sense find this generator.

[01:14:21]>> Yes.

[01:14:23]>> And who discovered this approach?

[01:14:27]he used in as branching problem.

[01:14:33]>> So this is a PhD student of Mitashov.

[01:14:37]>> Ah some guy in Moscow state university.

[01:14:39]Yes found this approach.

[01:14:41]>> Yes.

[01:14:42]>> And you learn from him?

[01:14:45]>> Yes.

[01:14:45]>> Does he has some paper on this?

[01:14:48]>> Uh he yes he has >> has submitted a paper.

[01:14:57]And what he did?

[01:14:59]>> He did algebra for F4.

[01:15:09]>> He proves for both conjecture for F4 for F4.

[01:15:14]>> No, no, no. So he used this approach uh to solve another problem which was related to the branching from from the groups of exceptional types to some reductive subgroups.

[01:15:31]>> Okay.

[01:15:31]>> So this is a classical problem from the representation theory.

[01:15:36]>> Okay. And uh his result was related to a case where actually he computed uh this very algebra uh for some concrete case where G is of type F4 and uh and H is a just a maximally important subgroup of B4.

[01:16:03]And who realizes that this can be applied to penarian conjecture?

[01:16:10]>> Sorry.

[01:16:11]>> Uh propose you to study this problem and you did >> but what to publish if it is done by computer. So okay so the algorithm >> computer and go this will be more and more complicated >> if is it the same algorithm that was described by Essen for fighting >> it seems that no it's have the same form but the other is another I think >> it looks somehow >> yeah just somehow similar but I can't say that this is the same object and what is the maybe I don't know uh the most interesting uh open question in this field. So what particular case is interesting now >> before is uh no or not.

[01:17:44]So uh can you uh just uh uh say that okay a uh bigger than four is not yes >> no >> not bigger >> no >> so in type a uh the largest known example is four >> is false >> a4 smaller than four is okay a4 a5 is open before is known before is open for example. So okay and uh are there some uh papers which some condition on h I don't know hes okay there was uh for example hges which are u un important radical of parabolic groups but maybe some other particular cases >> make no if you condition to the >> ah this linear independent or almost linear independent >> and >> but it's rather strange condition >> no it's natural condition because the complement is small >> in some sense >> and the previous examp prove the conjecture for special condition.

[01:19:30]So I uh I guess to hear that there is this group and this subgroup and nobody knows whether the uh homogeneous space have finitely generated algebra.

[01:19:51]Is this uh do you have such a pair with open question that nobody knows or just >> but to do the same calculation for rank four take all possible configuration exclude those which are covered by >> okay okay okay but it's boring I I thought that there is a couple which is some special couple um I don't know okay >> as for me I would propose say considered one dimensional regular subgroup then I think that the answer is yes this should be clear but take two commuting uh unicodent subgroups two commuting group subgroups for them I think the conjecture is open and this is interesting particular >> ah session maybe you're Right? Because each root subgroup is somehow local important derivation and invariants are just general important derivation and maybe this is the same algorithm.

[01:21:06]>> Well maybe that in the there are two parts. First part goes for cables the second part just for localization. the partition is the >> it's that I believe that now I believe that it is the same.

[01:21:24]>> Okay. So you are done.

[01:21:26]>> Yes.

[01:21:27]>> Any more questions?

[01:21:31]If not then our hands