Okay, first thank you very much for the invitation to speak and I'm really very happy to speak at the conference in honour of the MA. We started about the same time, I mean, it just started a bit earlier and I'm really happy actually that he's part of the geometric group theory community, if he's still part of it. And yeah, I think one of the great things that are still true for academia is to have people like him around, unfortunately now it's quite a big distance, but much less conferences in person, but still it's a great advantage. Okay, what I decided to talk about actually I myself now in recent years work less on groups, but what they choose, I choose this topic which actually emerged from a question that Nathalia Rips asked me exactly a year ago and somehow completely took me out of what I'm doing in this question and this was an attempt to answer what he asked, I'll get to that. So actually I worked in the JSJ starting at the ITS more or less 30 years ago in 1991. I will say something about how it started and for me I looked mainly for applications or for appearances of the JSJ in other algebraic categories, but apparently there are new things that are related to the JSJ or anyway some extensions of the JSJ in groups. I was rather surprised by that. And I should say also that when I was 30 years ago, when I started my career basically, I was very fascinated by the fact that theories from three-dimensional topology not only that they can be transferred to groups, but they are crucial to groups in order to prove all sorts of things. Now I think the JSJ is also fundamentally logic. And by moving to other categories, I thought that it's about time, first of all I found out the JSJ is probably fundamentally not algebraic categories such as people didn't know about this and one of the reasons to go to other categories is to try eventually to find some sort of categorical formulation of the JSJ because I think it's the general. But on the other hand what happened is that whenever I go to another algebraic category there are really lots of new phenomenas and now also in groups. So it's very difficult to make things as a theory categorical if whenever you go to any direction you find some new phenomenas that were not familiar, didn't appear before. I should say also just the closest thing, I plan to write this paper before the conference but I didn't manage to and so I really consider this still as a work in progress. It's not complete and it's not easy so take it with this, I even mentioned at some point what part I didn't write. But I think the work in general is conceptual so I decided to speak about it. It also as you'll see well hopefully if I'm still remembered I mean we had a conversation maybe 25 years ago on things that are related to what I'm going to talk about related at least in terms of the subject if I'll mention that and it's in the appropriate but it's still not completely written so I cannot really regard it as a complete work but anyway it suggests some new directions. So I'll start with the Josh J the composition, it will be a review I think many of you probably know it but it's important also for the continuation so I'll also to put it in context so I'll mention that so there is a now I think well-known theorem of Frédéric Poulenc for I think from 89 maybe 88, I'm not sure about the years that I've written here so it doesn't matter if it's plus minus few years. So what pull up proof is that if the outer automorphism group of hyperbody group is finite is infinite sorry then the group acts in a really free with virtually a billion stabilizers of segments or in fact okay virtually a billion is fine which in this case means either virtually cyclic or finite and then there's a fundamental work by Riebs and by Stina and Fein who prove that okay if you wanted to it's a core well it's a core of their work and the theorem of Poulin that if the outer automorphism group is infinite then a gamma acts on a cyclician tree with virtually a billion edge stabilizers okay so this is a fundamental fact and at the same time I was working on the isomorphism problem for hyperbody groups and during my thesis I managed to prove there's a morphism problem in the torsion free case when the group is rigid so when the outer automorphism group of the hyperbody group is finite or if you want according to these theorems if it doesn't have now it's okay the group is torsion free so it doesn't have cyclic or cyclic splitting or and it's really in the composable so for rigid groups there was a solution and then I tried to generalize it actually while being in the ITS it was my first positive position and I tried all kind of tricks to to get around the the the fact that the outer automorphism group is infinite see it's pretty clear that that when when when the outer automorphism group is infinite the group has many symmetries and this this may be an obstacle for solving the isomorphism problem indeed it is a problem and and I should say also that part of the solution to the to the isomorphism problems in the rigid case was that it's possible to decide if the if the if the group has a splitting or not this this is an output this is in a corollary of this of this of the algorithm for solving the isomorphism problem in there in the rigid case and as I said I tried all kind of tricks to get around the fact that the to get around when the automorphism group is infinite and I couldn't and then it was clear to me that if I want to to to generalize it I must find a theory to that will explain what is the structure of the outer automorphism group and what what are all the splitings of of how to find all the splitings of the group actually it's turkey I'm sorry whatever Bernie musket stated the ITS at the same time and it happened that at the same time he found a proof of the of the jacochelin and Johansen work for for a geometrically finite glanian groups actually a rather difficult proof and I managed to to go to his talk actually together with with l'unique putigailo and that that actually the proof itself has nothing to do with with the JSJ but it's a special case of what the jacochelin and Johansen did actually a rather complicated proof but but but this actually dragged me to the to the direction that I better find some sort of splitting like this for everybody groups and this will explain the structure of the automorphism group of the outer automorphism group and all the splitings of the group so really the lecture of musket somehow without him knowing and never discuss it with him but but this was somehow the trigger okay so so again these are these are some basics about the JSJ but but this will be important for the continuation so I am still going over that so gamma will be a one-ended group and okay I'll discuss it in the torsion free case actually for this method there's no much difference between the torsion free and and case we torsion if you consider automorphisms at least it's readings that come from automorphisms okay so so just think about the torsion free for that matter cycle splitting that divided into three categories okay I'll explain it in a second there are elliptic elliptic splitings in which case one can find a common refinement for the two splitings there's elliptic hyperbolic it's a dual position between two to the compositions but then in the torsion free case there will be a free splitting of the group so if you assume the what the group is one-ended and this doesn't appear and to sort of extend the most important case is the hyperbolic hyperbolic case so this is when splitings are somehow not compatible and it turns out that that in this case okay like in the in the three-dimensional picture there is there is a quadratically hanging subgroup okay in the case of torsion free here we are in the hyperbolic case there will be some some fundamental group of the punctured surface inside unless the group is a surface group itself and and then the two splitings come from intersecting simple closed curves on the on the on this surface and actually these two simple closed curve must fill the surface okay so this is the these are these are the three sort of trichotomy of the compositions that will play a role also in the future but but it's also it will be important for me afterwards to explain why the machine that constructed jfj will fail in the in the place that in the in the situation that I'm going to discuss okay so this is just the picture of what I what I just wrote in words so we have two splitings and we say that they are elliptic elliptic if the extra okay I'm just discussing in the torsion free case later on I'm not going to assume that anymore but but it doesn't really it's not it's not much different so so we say that it's elliptic elliptic if the edge group say z1 here can be conjugated into one of the vertices in the other and okay it's a dual position so z2 is also can be conjugated to the vertus group of the of the first one and in this case it's possible to find common refinement for both so we look at what splitings this vertus group inherit from the from the from the other splitting in the case of hyperbolic hyperbolic okay so a priori we just know that these two splitings are splitings of an abstract group or at the body group later finally presented you but in the beginning it was just for the body group and in this case the way I approached it is by building a machine so each each with each splitting we can associate the cycle group of automorphisms then twist so conjugating one of the vertices this is amalgamated product the hnn has a similar similar group of automorphism so we conjugate one of the vertus group and leave the other one fixed and we look at at the sequence of automorphism that is obtained by each time finite but it's growing and growing to the left we take an automorphism from the first side from the first side group of automorphism then automorphism from the second etc with powers which are growing fast so n1 will be much bigger than m1 m2 will be much bigger than n1 and so on to satisfy certain combinatorial properties that I'm not going to get into and okay using this this machine and and the refinement it's possible to find the composition which will be canonical actually in the case of hyperbolic group it will be invariant under automorphisms there's no problem with that and what happened is that I wrote maybe that's not a common notation but a finite subgroup of the outer automorphism group this is assuming the group is one ended now we'll have a map onto the direct sum okay each each automorphism because it preserve this splitting is up to conjugacy it acts on each of the surfaces okay so it acts in some automorphism and therefore it's mapped into the automorphism in the into the outer automorphism group and we get a map from each automorphism from each automorphism therefore from the whole from this whole group into the mapping class groups of these surfaces and the kernel will be some some free or someone asked something okay I should say actually in my written paper I assumed that this this the short attack sequence please later Gilbert Levit pointed out to me that it's not always the case okay I don't hear if someone asked something but all right I will continue because it's okay okay but this is this is the picture and this will be important also for the for the continuation okay this this short exact sequence sequence yeah is there a question I'm sorry I'm not from the audience okay all right so let's continue okay there was a whole list of continuation actually now it's it's interesting to say somehow sociologically that in the first couple of years practically no one believed that the JFJ actually exists somehow it wasn't accepted at all until what changed the situation was my work with Rips that generalized this work to finally presented groups actually with essentially with the same machine but we just just looking at finally presented groups instead of just hyperbole groups and here here there's no torsion free assumption but we looked only in cyclic splitting and shortly after bodice constructed the JFJ for hyperbole groups again with torsion and not not using automorphism but using the topology of the boundary okay later there were there were other I wrote some I mean there are many papers now the JFJ probably goes in dozens or even hundreds done with the instagive constructed the JFJ for finally presented groups but but over more general groups so say over polycyclic groups and then Koji Fujiwara and Panas Papasoglu constructed actually it's a slightly more general result but over the same class of groups using complexes of groups and the last two are two asterisk groups more recent one by Scotland's were group somehow at least were group was rather from the onto Peter Scott they were rather bothered by the fact that the JFJ that that we constructed is not exactly like the three-dimensional ones there are some some distinctions so they they managed to do it and they also constructed the composition which is environment and automorphism and the most I don't know more than anyway the most recent I think common reference for that is the work of you know then and the beat um that that that deals a lot with the with the question also of the canonicality of the gesture in case of a world group there's no problem really about canonicality but but the general finally presented groups there is and this is sometimes a problem also for in the applications okay but but what I want to talk about today will not be about general finitely presented groups but um but one should note you know when when I worked on on the JFJ for for at least for ocean free acrobatic groups the JFJ encoded both the automorphisms and the cycling split okay at least for one in the groups but okay but it can be generalized when you go to general finitely presented groups the JFJ in more complicated way encodes still encodes the splitings that's how it's constructed but it doesn't it cannot anymore encode all automorphisms I mean even subgroups of acrobatic groups right like like the example of of no already and you know all kind of example of finitely presented groups say where acrobatic groups fiber over them you know where the fiber group is this finitely presented group um it's impossible that the JFJ will encode this the automorphisms of these groups because if he did then uh mapping tower mapping towers over these groups will not be hyperbolic but um but he doesn't call the splitings and the question is what happened with with automorphisms in uh in other cases so so there are classes like like the body groups like limit groups where sometimes the relatively acrobatic groups where um where the JFJ encode all the automorphisms but in general one cannot expect that and apparently there are some natural questions when um a generalization of the JFJ is supposed to is supposed to do this work okay so so in some sense this work as I said it really started with this question of ribs but but before that okay this is a question that I answered at the time but but okay it didn't at least for me it didn't ring any bell um so Ruchani asked me I think around 2007 it was the time that she worked with Karin Vogman on automorphisms of private and clouding groups I'm not sure about the year but anyway it was around that and she asked if the JFJ encode all the automorphisms of of right-handed outing groups my answer was rather simple that okay it will give some of the automorphism but not all of them this is this is a correct answer but apparently okay if one if one generalize the JFJ then um then it will encode them all but anyway it it left like this there's a very okay I don't know if I managed to speak about right-handed outing groups in this talk but but there's a very nice work by Ruchani and Vogman and part of it with Chris and also there was another group Duncan, Kazachkov and Ramesnikov on automorphisms of of frogs but somehow not from this point of view I mean so one has to go a little further to to find really okay a generous form of JFJ that will encode this these automorphisms and the question that really somehow dragged me into this project was let me left my project and read so for now for a year actually and it was a question of rips he just called me one day during this pandemic and he asked what can one say about the automorphism of cubulated groups I mean he asked that because because he was familiar with with the application of the JFJ for the automorphisms of hyperbole groups and he asked whether whether one can do something like that for for cut zero cube complexes and you know because we know that that lattices in higher-ranked semi-centilit groups don't really have automorphism because of because of super rigidity so and and automorphism say three manifolds all come from the JFJ so so the question is whether all automorphism are somehow low-dimensional as in the case of in the sense of the JFJ so as in the case of hyperbole group and it was clear to me that in order to answer this question please partially it's better to look at cubulated groups although not all of them satisfy that but still okay it's better to look to them as as hierarchically hyperbole spaces over the fundamental groups I look at hierarchically hyperbole groups so HHGs so I don't know if anyone is familiar with it I'm not going to get into definition here because it's technical and and in some sense from my work although I meant to answer this question but but it's not the how to say it's not the main point it's a generalization but but a generalization if you make some additional assumption is not that difficult to to achieve after some preliminary so anyway these spaces were were introduced by Burstock, Hagen and Sisto I wrote the the this is really the year when the paper appeared they produced it you know three years before and what they did is they they took the very important work of major honesty that found hierarchical structure of the of the mapping class group okay starting with the curve complex and using other curve complexes of some subsurfaces again I'm not going to talk about this either and they they axiomatized their work and and talked about spaces that that satisfy these properties and it turns out that many examples satisfy that and therefore it's it's very it's very useful to study certain even certain example using this using the the axiomatization and using some results that they proved for this for these spaces so okay so it holds for the mapping class group but this is the place where they started it holds for example for right angle art in groups it holds for special or visually special cut zero cube cubicle groups and they have many more examples okay this is not anyway it's not the main thing at my top and and really the main thing to for me the main thing to to actually study and to try to understand is what's what is the automorphism group of a group that act on a product of hyperbolic spaces okay the jshc already failed sorry okay sometimes there's noise and I don't know if someone asked something okay so so really is what can one say about the group that acts on on on product like this okay so what is it first so products of hyperbolic spaces we take finitely many so m hyperbolic spaces we can assume that they're all not bounded otherwise the divided ones are not of course not interesting because otherwise we will assume that the the action is is proper properly discontinuous so okay so xj are actually hyperbolic spaces there's a group g that act oh I see that I didn't write here but but I meant cocompactly it doesn't have here it doesn't have to be isometric action because isometric I really mean in the sense you look at if you look at the axiomization of these hhgs but in this very special case of a product somehow this is the the most important case at least under the assumption that I'm going to make and okay what happens in this case yeah that see there are the projection spaces the assumption is that first of all okay the group g it may permute the the projection spaces so we can we can pass to a finite index subgroup if we want the characteristic subgroup well it will be important later to move the characteristic subgroup anyway it's a finite index okay that this finite index subgroup doesn't permute these spaces so it acts as it moves each projection space to itself and these ones I assume that they are isometric this is the assumption in the in the hhg case again well one can relax slightly these assumptions but that's on the point and and the question is what can be said about the structure of the outer automorphism of such a group and what can one say about the dynamics of individual automorphisms as I said okay I thought that if I want to have time I mean I still regard this case as the most important for for hhgs and if I have time I'll say something about how it's how it's generalized but this is anyway this is the in a way this is the main this is the heart of the problem now you may think that I don't okay someone who in in some sense if you look at I don't know pathologies or contrary to group theory then you know product spaces are are rather problematic actually not only in group theory but but but certainly in group theory but it's interesting to know maybe if I if I talk about this just conceptually in logic somehow product spaces are not a problem okay if you if you ask questions first of the question only about elements if you understand the factors you understand the product but but if you ask questions about subgroups about automorphism okay questions that are not first order then products are much more complicated than than than the factors so okay here is here are some problems that they're just good to keep in mind so actually there are two talks in this conference by PRP yesterday and by Martin Breitzen about sub-direct products of of sub-direct products of some direct products of free groups surface groups etc or other groups well also other groups in PRP talk yesterday but anyway Michailo was at least one of the first ones she what she did see this group is a direct sum of free groups is reasonably finite so so the word problem is solvable but what she did is she took a finite represented group that has unsolvable word problem and showed how to translate it using fiber product so that there is a finitely generated subgroup of this direct sum which which doesn't with where the membership problem is unsolvable and since then there are I don't know lots and lots of pathologies that were constructed using fiber products using other things sometimes sometimes problem to to even improve properties of pathologies or counter examples so this is this is one one thing that one should keep in mind subgroups it's impossible to control in in direct products in general even if you know everything about the subgroups of the factors okay another thing that is just good to keep in mind what what I'm going to say will not shed any light on this and these things but still you know if you consider three-dimensional topology in general in some sense this was this was one of the direction in three-dimensional topology before Thurston so any three-dimensional any compact three manifolds let's say close without boundary has a he got splitting so it can be made out of two handled bodies when we identified the boundary and this would be the van Kampen diagram for this for this he got splitting okay this is the fundamental group of the of the he got surface the free group is the fundamental group of the handle body and this is the van Kampen diagram I'm sorry I made a mistake in this notation but anyway it's a fundamental group of this so this handle bodies and the Poincare conjecture for instance ask in this in this formulation this is Stolling's formulation which is equivalent to the Poincare conjecture ask well if we know that this is an epimorphism is it true that there is a simple closed curve in the kernel okay this is equivalent to the to the Poincare conjecture and it's it's important to know that until now there's there's no algebraic approach that gets anywhere near the Poincare conjecture although the Poincare conjecture has okay different proof geometric proof by permanent but that's the only proof that we have to this algebraic statement okay and and one of the reasons that you know at least from my point of view that we that this is hard to approach is that we work with two factors and we want to make any changes in one factor it completely it immediately affects the other factor and controlling both of them somehow coordinating between the factors is a very it's a very major problem and if you think that you know just if you try to apply all sorts of techniques it's better to remember that if you don't look at epimorphism but just sub direct product then all three-dimensional topology are just in this in this in these types of diagrams so you know it's it's it's rather again it's rather pathological situation algebraically okay another another example to to remember is the following so we take a surface and we take we take a pseudonosoph of the surface and we look at the backward and forward iteration of this pseudonosoph and taking the limit we get actions on two trees which are dual to the stable and unstable filiations on the surface coming from the forward and backward iterations of this of this pseudonosoph so we get an action of the surface on the product of two trees and if you look if you look at the action on the product the action will be properly discontinuous i think it's a reason to throw some but i'm not sure um so so you see what it means that if you look at the two-dimensional object in a sense you don't see the dynamics but you see the dynamics locally but you see the dynamics if you look at the projection if you look at the projection then you'll see the dynamics of the pseudonosoph on each of the of the real trees separately but if you look at the two-dimensional object somehow okay locally we lose the dynamics we have to go further to see what's going on okay so so another example that i want to mention uh which i didn't write here which i'm going to exclude uh by by assuming by by adding an assumption in the minute uh is the is the example of borda and mozes okay mark borda and schacher mozes of a group that act on product of two locally finite trees and which is simple okay and i'm going to make an assumption that in particular we will exclude this example the last one okay so the assumption that i'm going to make is that um you know when you go to when you go to a factor there may be a kernel to the action um so okay in the action of course we assume it's free on the space but um free and properly discontinuous but when we go to the factors okay that it may not be free and it's not properly discontinuous anymore so we assume that the the action is weakly a cylindrical now and this is probably overly overly used terms even tomah has a notion of quickly a cylindrical but somehow tomah notion of quickly a cylindrical implies us okay this is this is a weaker notion than a cylindrical and it's supposed to to to make a classification of the fact that there is a kernel and model of the kernel the action is a cylindrical so what it means okay that if two points that that's how to think about it but if two points are far or far enough okay we fix an epsilon first and if two points are far enough uh there will be boundedly many elements so that if there is an element h that move both of them sorry here i made a mistake uh move both of them more than epsilon that's a mistake in the inequality it's supposed to be bigger than epsilon then okay it's not true that there will be only boundedly many elements but but this element h is belong to boundedly many cosets and cosets are thoughts of elements that acts quasi-trivial in the whole space okay that they move any point not more than that doesn't have to be there i mean uniform constant doesn't have to be the delta of a publicity but but if the space is not aligned it's not easy to see that it's i don't know two deltas of their publicity but but that's what okay that it belongs to to find boundedly many cosets that's what's important and i view that as a classification of of um of the slina bromide and fujibara condition of wwpd but in the h-cylindrical case i don't look only on axis but in general okay so or if you want it's a classification of of just a cylindrical of a cylindrical model okay um so that's that's our assumption on the factors and by the way again i don't know if i managed to get to that but if i go if i talk about it it is then then we also assume somehow there that the stabilizer of each projection space there also acts weakly a cylindrical this is an assumption that okay i have to assume in order for these techniques to work okay so um so okay so we ask again uh what can be said about the automorphism group of a group that act on um um on the product okay we don't care what the group is but only what the automorphism is so we start with the sequence of automorphisms and we look at we look at the action of this of this group eight that stabilize set wise each of the projection space okay and we twist it by the automorphism it's a standard thing and we look at the set sequences that converge should should have written after rescaling for all the for the projection space it at once okay it converges for all of them it's a if the stretching factor goes to infinity so it's a standard now compactness argument that there will always be a subsequence that we that we converge okay it will convert to each to each space so by by passing from subsequence to another n times it will converge to all of them from them okay and the limit gives an action on a real tree and we said um um we said l to be l okay this is also pretty standard now model okay infinity you just we call l a limit group in this setting it just imported me to know for me to know that uh you know there's a various definition for limit groups um you can define it using Gromov-Hausdorf convergence and you can always also define it using more spaces for limit groups over free groups or like a body groups it doesn't matter but here the definition varies so okay so I really could consider here the k infinity in terms of the of the action of the Gromov-Hausdorf convergence okay these are elements that are trivial but we consider automorphism because we consider automorphism there's no there are no kernels for these maps group algebraically but there will be element that act trivially on the on the tree at infinity and therefore this group in general is a is a quotient of H and usually it will be a proper quotient okay so the definition here varies it's important all right okay sorry I'm going much lower than I thought but okay um so I'll explain the principles maybe okay so um so so again I'm just repeating here the the the setup uh X is a is a product space um H okay we pass this we start with G but but we pass to this finite index characteristic subgroup so it's characteristic so the automorphism of the big group act on this as H as well because it's characteristic um so so the group H act on this space and the advantage of H that it preserves set wise each of the factor spaces and we have we have the sequence of automorphisms that uh with which the action using this using this this automorphism and we get that we get a limit action on a real tree okay and for this for this sequence we get a corresponding corresponding limit group which is H modern the kernel now if if the yes okay um now oh I should say maybe I thought it even appeared before but but anyway um this this was very convenient for me uh there is a paper by Daniel Brooks and Michael hard on on um okay they they deal with um with equation in Ethereum families of of a cylindrical everybody groups anyway that they um they do all the technical setup is needed here and uh I can always I can I can refer to them I don't need the analysis of equation Ethereum and so on but but but they are and they all all that is needed in order to to analyze uh this this limit action was done by them um okay so so from the action of L on it's action of H but but this k infinity actually so so it's really a faithful action of L it's possible to write to to read off splitting of the splitting of the group okay it doesn't have to be like this but could be surfaces etc could be splitting it over finite groups free groups etc free splitting etc and each time we do that we get a map to uh from H to a limit group which has a splitting the big problem here okay that doesn't appear in the original construction of the JSJ is that these limit groups are different okay the groups are different the kernels are different you have them uh also correspond to each of these factors and um you know the whole machine that was used to to construct the the JSJ and it doesn't really matter what what machine you use um I mean all all the construction of the JSJ the machine will fail because because it has a kernel there's another um there's another conceptual big problem here and this is you see we look at quotients here and splitting off quotients now the big advantage of the JSJ is that that the automorphism of the group can be read from at one level from the splitting of the group okay it's almost an ideal situation here when you when you look at uh when you look at um at spittings that are obtained from quotients in general there's no there's no chance that something like this will be true even if you consider only free splitting okay all not these spittings but only free splitting it there is no something similar to the JSJ that will that will uh somehow describe all the free spittings of quotients of a given finite represented group actually I have a work on that with Eric Zaligo and reconstructed the McCann-Rosberg diagram that encode all the free spittings of quotients of a finite represented group this was needed afterwards for logic purposes but but there is no chance that it will be in one that okay there could be there is a McCann-Rosberg diagram but it cannot appear in one level and it took me quite a long time to decide whether if one tries to develop a theory here whether it should be in one level or it should be something like McCann-Rosberg diagram and you know McCann-Rosberg diagram in this setup is I'll explain that is much easier to construct but but it's much less powerful okay so what we want is something that will remind will be you know somehow connected to the JSJ in that case of hyperbolic group all right so uh so okay so how to match the different projections and uh you know I knew how to construct some sort of McCann-Rosberg diagram because these are tools that existed from before as I said in Groves and Hall they also constructed they don't construct actually a McCann-Rosberg diagram I think but anyway they they they prepare all the tools to construct maybe I'm wrong maybe they do but but I think not um but they prepare all the tools that are needed for that um as I said it it's it's it's much in a sense it's much weaker but uh but this is something that one could start with and I said that I was really puzzled whether to to look for some sort of JSJ or to look for McCann-Rosberg diagram and at the end I decided you know if I if I cannot beat this this this this McCann-Rosberg diagram but they join it so uh it means that we first construct a McCann-Rosberg diagram and then we use this to construct a JSJ um so so let me yeah I don't know how much I managed to say somehow this is the first time that I talk about it and I somehow assume that I'll be able to say much more um okay so um so how to construct this this McCann-Rosberg diagram um so we we said we start with the with automorphism p I I work now with each projection separately by the way I have an hour or I have less than that um about 55 minutes so you still have 10 minutes or so it's all right all right I'll get whatever I get um so um so so there's a sequence of automorphisms in the beginning we we construct one limit group now this limit group will have will as I said we will have splitting if it diverges it will have a splitting so we can we can associate the modular groups with that now from now on we don't study automorphisms anymore it comes from automorphism what we study quasi-morphisms and um or if you want if it was a real kernel we study homomorphisms this limit group which are in general quotients or proper quotients of the original one and we can construct we can construct that the diagram here but but we are interested only in splitting that are obtained originally from automorphism because this is what we want to study we're not interested in the algebraic structure of these limit groups we are just interested in automorphisms so we cannot use the JSJ of these limit groups although the JSJ exists but we use only splitting that come from these automorphisms and because of that some of this epimorphism will be actually isomorphisms but it's guaranteed that after finitely many steps in fact after bounding the many steps we will move to to a proper quotient and this is this this descending chain conditions was proved by Groves and Huy that after after such such a sequence terminate after finite time and when it terminates we get we get a quasi sequence of quasi-morphism obtained from this automorphism which is non-divergent anymore so there is a bound on that this is important and okay the whole things here are only finitely generated it's not finitely presented in general and finite generate is too big for us in terms of cardinality so we can replace each just resolution by finitely presented object just just for you know to to to think about it just to think that we replace each of these limit group with a finitely presented object this is not how it's done but never mind and therefore there are only countably many of them if you replace the whole this resolution you call it if you replace it by finitely presented object it is the only finitely there are only countably many of them and this is as I said there is at the end there is a bound on the on the sweating factor we can assume this bound as a positive integer so we can we can we can count all this all this finitely presented approximations of such things and using okay I think by now pretty standard compactness argument it's possible to to be left with only finitely many of them so so it's enough to to use only finitely many of this resolution and what we get is what they call a higher-ranked mechanical spore diagram for products so what does it mean we have m factors okay m factor spaces so each each resolution each element in our diagram will will have finitely many resolution for each factors because we start with sequence of automorphism and we look what happened in each of the factors but you see when we analyze them that's the disadvantage here when we analyze them we analyze them separately we move to quasimorphisms and these quasimorphisms are just appropriate just for the just for the appropriate factor there's no when we continue in this diagram from the first level we lose there's no coordination between the factors anymore so so we have we have this m collection of covers and finitely many of them are enough that means that every automorphism of the group will factor through one of these one of these finitely many m collections okay this is the this is the high-ranked mechanical spore diagram it was obtained from by compactness argument so there's no chance that any like anything like this will be canonical okay here is here is what I call the completion of resolution I'm not going to get into that actually I should say I don't know how much I'll get to that but but one thing that that is appealing for me and actually it appeared before in a in a work with Cody Fujiwara that that tools and techniques that I developed to study the first order the first order theory of free group actually appear here the tools not not the not not anything from logic and this is I think this should happen more and more I mean there are techniques there the solution of taskly problem in generalization that that should have application in group theory and in in in other places and this is an example you see studying automorphisms of groups in the beginning doesn't seem to be related but apparently the tools are related are useful not just really okay so um so the first maybe I'll just describe that I'm assuming that um you know at first that that each each of these resolution is only one level see if it has one level meaning just one map if there were no maps then there are bounds of the stretching factors okay and if there are bounds of the stretching factors and the group acts properly and co-compactly co-compactly it's not matter just act properly then it means that only uh finitely many automorphism will factors or something like this so this is not very interesting so I assume that there there's at most one so every resolution has um has at most two floors of two levels okay so this is this is the first the first thing and in this case okay there's um the group of automorphisms form okay what what I call in the work of doctor problem is a test sequence it's not really a test sequence it's a weak test sequence and what it really means is that this sequence of automorphism just converges to to the splitting that we see here one one is to explain what converge mean because we are going to see it in different scales like in our our entry or something but but anyway from this from the sequence of automorphism we can read off we can read off these splitings all the m-splitings and all factors and now one can one can define a complexity of these resolutions okay complexity of resolution is a is a is a major issue in in studying the first order theory of of of a free group but in there it's far far more complicated here it is it's rather simple we just define say naive uh complexity of of this a billion splitings okay first the the course rank the number of free factors and free elements then the topology these are orbit faults in general they're not not just surfaces um so so the the these orbit faults say the order curve receive the genus order will see graphically in in decreasing order and then then things to do with the with the a billion groups and the edges and we consider only only the resolution this is maximum complexity our order the complexity sorry is well ordered so we can we can look only at maximum complexity and we we look at these these weak decisions these test sequences that converge into maximum complexity resolution now if we take any automorphism any automorphism in in the group and we compose it pre-compose the the sequence with this automorphism we get another test sequence just okay maybe it doesn't factor through this resolution but we have universality of the of this of the diagram that we had before so it will if it doesn't factor through this one at least a subsequence of that will factor through another one okay again because because of universality and using that it's possible to prove okay this is i think this will be my i have more slides but i will i will stop here because of time that that for single for single level level resolution there is exist just one m collection of cover resolution that is preserved by finite index subgroup of the automorphism okay so so there is this you see m m splitings but it's important to know these are not splitings of the original group these are splitings of quotients and usually it will be splitings of proper quotients of the groups the the quotient itself is not canonical but the form of the the composition is is is canonical i didn't study much the the canonicality of this decomposition but but anyway all surfaces that should appear etc and of course it all depends on the finite index subgroup that we chose because if we choose if we choose another finite index then you know we get the splitting which will be in some sense commensurable with this one but it will be different um and and this this the composition will encode okay not all automorphism but the finite set a finite index subgroup of the automorphisms i won't i will not talk about resolutions with higher level now but let me just let me just customize last slide no this is right angle oh not a lot less than okay in general from this uh from this the composition okay i jumped several slides but from this the composition it now it's possible to say all sorts of things on the on the outer automorphism group so for example okay if the group if all splitings are of one ended let's say one ended splitting it's not important if the group is one ended but it's important whether these splitings don't contain um free the compositions or splitings over finite over finite groups then we get also a map from the from the from this finite index subgroup of the outer automorphism group uh onto mapping class group sorry not onto that's important into the mapping class group of these surfaces and outer automorphism group of some virtually virtually a billion um vertex groups as well so it's not the general linear group but outer automorphism group of finitely generated virtually a billion groups and the kernel the kernel has to be has to be finitely generated virtually a billion okay it's a it's a similar picture in the sense to what happened in in in hyperbolic groups or actually it's also similar to what the car what charny and botman did for for right angle awesome group just that there there are no surfaces so um okay now if the group is not one ended then um then you have to work then the outcome cannot be stated uh so so nicely in this well so in such a short way because we will also get the map into into these objects but then there will be a kernel to the map and the kernel to the map you know after we look in the kernel we can we can run off the construction of this hiring jj the composition just for the kernel because you know we we didn't really need to tell that this this machine that it is this we're all automorphism it's okay to take it just from a subgroup and it may be that new surfaces and new a billion groups will emerge from from the free the composition it may be that new of them will emerge so we get another map with another kernel this can happen boundedly many times and at the end what we will get let me see if i have a slide here at the end what we will get is uh you know we'll get factors with only virtually okay i didn't consider with only virtually cycling virtually cycling the composition and maybe some free elements i didn't draw here also edges with free with sorry with finite edge groups and if you like if you consider only stabilizer of the factors if there are no free the composition only stabilizes yes then of all the factors in in all the factors also in the in the in the space okay because the space was a product of certain factors it will be the stabilizer will be finally generated virtually a billion okay i will not talk about the generation to to uh hierarchy at the body group but this is anyway this is not the main the main thing i really think that the main thing is about products i didn't manage also to talk about them about the resolutions with more floors but okay i mean there's a limit what one can do or i didn't i didn't plan it properly okay well stop thank you very much are there questions either online or at the hos yes so you you mentioned the quasi-morphisms occurring somewhere in your in your in your resolutions what could you say more about those quasi-morphisms oh yes okay so let me see when yes because okay you see the okay let me let me just say that the limit group here okay the limit group here is is the group H model model of this kernel now this kernel in general it doesn't act trivially on the it doesn't act trivially on the on the space but but it will but what what it will happen it's this kernel okay not sorry not not this kernel contains more element but but anyway there will be element that move any point just move any point in the space or in the factor just a bounded amount so okay think think about for example think about the limit group here that that which is finally presented so it has finally finally many relations so these relations are not mapped to the identity but they are meant to element that move any point in the space by by this bounded amount and this is a normal support okay that's what quasi-morphisms here means it's actually not very far from from homomorphism just one needs to to get used to work with this thing but but the quasi-morphism we're not interested in all quasi-morphism we are interested only in quasi-morphism that are somehow obtained from automorphisms of the big group through these resolutions okay but again it's not that far from dealing with them using this this mechanical score of diagram machine is not very different in dealing with homomorphisms but again it's important that we are dealing only from those that come from automorphism of the ambient group and not not not just general quasi-morphism at the beginning you mentioned the Mikaerova groups and you have example of interesting Mikaerova groups with a lot of automorphisms okay I didn't I didn't start yeah yeah I didn't study subgroups I didn't study subgroups okay for this it's important Mikaerova group acts on the product of free yeah yeah no I assume that the action is co-compiled oh okay yeah also in the other group sorry on a group also in the antibody group right the JSJ will not give all automorphisms of all subgroups even the JSJ of the subgroup will not we'll often call all automorphisms are there other questions well if not let's thank the speaker again


Hi everybody in this video. I want to do a quick walkthrough of the app that you will use in your lab to test out getting app permissions and also getting the location Sorry took so long to get this up. I ran into every technical difficulty including someone running over our power pole That you can imagine. So that's why it's a little late anyway So if you go to your The module for today you will see a lab for Where am I and that is the name of our app that we're going to play around with so in this lab? I give you all the code. I just want you to be able to kind of walk through it Maybe understand it and then later if you so choose apply it in your project I'm going to first grab a copy of the project because I at least want you to play around with this apps and appreciate how it works Okay, so there's a github project It's public access. You can go download a copy of it. So I'm going to grab it and walk through the process of setting it up Whoops, so I'm going to check out a project from version control and paste in the github address that is in the lab right so Android studio has imported it for me. Thank you mr. Studio make this a little bit bigger All right So Gradle is building this right now. So it's going to take a second before this all lines up In the meantime, let's kind of take a look here Go and find the Android Manifest because you're going to need it in a second And is Gradle almost done looks like it Okay Starting to look finished anyway. It's doing its indexing thing. All right in this app There's only there's a couple of activities And we'll we'll go through each of them in a minute Let me just try and run the app though because I want to show you something that you will probably encounter or have encountered already So I tried to run the app and it gave me a build failure with exception And that exception is Google services dot JSON is missing. Hmm. Well, what does that tell you? Well, it tells you Google services dot JSON is missing But it also tells you that this app is using some Google service and probably fire base, right? So in fact this app does use fire base because we are going to be recording location in it and Staving it up into the cloud fire store. So let's go set up our fire base project to deal with this Here I am in the fire base console You can use an existing project if you would like I don't recommend using your like semester project But if you have like the fire base authentication example project still there feel free to use it or you can make a new project It's up to you. I'm going to use this existing project. I have that I use for a bunch of demos And so what you need to do is add Authorize this application to access this fire base project. So we're going to click on add app And then we'll go over here to the Android icon and click on it Okay, register the app Android package name. So this has to be precise So let's go back to our code and the package name it is looking for I'm here in the Android manifest and This is the package name. So this thing whatever this value is. That's what you got to put in that box Okay, so I'm going to grab it. It's edu.uncw. Where am I? Now the rest of this you leave blank Register the app. It's going to think for a second All right, and now it has produced for us the Google services dot json file So even if you are using this with an existing project You still need to do the step and you definitely need to redownload this Google services dot json file If you have an old version of the file on your disk somewhere, you can't just copy it in you got to get this new version Alright, so let me go grab this file. I got to get it off my other screen here Pick it up and I'm going to plop it paste it into my Where am I slash app folder? All right, so now if we go to Run this thing we should have a little bit more luck Okay, I've already got my emulator up and it looks like I've gotten past the Error that I had before so that's a good thing Take just a second to build on my very modern computer In the meantime, let's take just a quick look in the Java here Oh That wasn't so bad So there's only three activities in this app that I want you to kind of poke through and get familiar with The first is the main activity and it's just this screen here. It doesn't actually do much beyond that It just launches new activities Now let's check out the on-demand activity So if you click the button in the emulator that says or on your phone and you can run this on your phone, too No worries click the on-demand button And you'll see an activity that looks like this So I already have some data here your data will be blank and that's okay. That's expected this This activity is going to demonstrate Getting a location like on-demand when the user say presses a button or submits a post or Anything like that, right? It responds to an event. It's a one-time recording like just tell me where the last known location is Okay, so in the code the code is actually there's there's a fair amount to it a Lot of it is setting up this recycler view down here to kind of listen and get the Records as they're saved to the fire store and some of this code is saving to the fire store Which you already know how to do interestingly the fire store has a built-in data type for Latitude and longitude it's called a geo point right and it just wraps up the latitude and longitude inside a little class for you The code to actually get the location You've got to go through a process of requesting permission checking for permission To get the location If the user hasn't granted location permission you have to ask for it Once you have it then you can finally make a call to say hey Android operating system Google libraries tell me the last known location and then that's an asynchronous call and it comes back with a Success listener or a fail listener and then you do something updating your UI or recording to the database I've got it in both the lab page here and A lot of detailed comments in here to kind of help walk you through it I'm not asking you to do anything more than to try and poke at this and understand it because Later in your project you will probably or possibly be you know Adapting this for your code and you'll learn a lot through that process. Anyway, let's record the location Okay, so when I click record location in my app, what do you see? This is the standard kind of Android baked-in dialogue for requesting permissions This dialogue box is Not specified in my code My code makes a call to Android that launches this dialogue box, but I'm not specifying the layout of this thing or anything It's a separate entity, right? It's kind of happening happening at the operating system level If you do something here like click deny Okay, so you get a little toast that says your app won't work until you grant permission But then if you click it again it doesn't prompt you again like for the Settings right to to allow permission and that's kind of by design of Android Android makes it really hard that Once someone denies permission for you to keep pestering them for it. There are ways around that But it's not good practice.  Here's how you can turn it on it for me if you want to let me show you how to do that Whether you're on your phone or in your emulator you want to pull up the settings application If you pull up settings, let's see here Somewhere you'll have an option that looks like apps Okay, and you click on apps and then there's permissions in the permission manager Okay, so your goal is to get to this permission manager in this thing everything is organized by The type of permission so body sensors calendar call logs apps that can get to your camera apps that can access your location That's what we're interested in So if I click on it You can kind of scroll down and here's all the people who are allowed to access it And then here's everybody who's been denied access and here's my app So I can click on this and now change it here. Okay This is the way to do it through the system settings once the user has denied it Of course if they accept the permission whenever it comes up great, you're good to go Let me click record location now. I'm back in my app and here it pulled the location off the phone What's happening behind the scenes is it's pulling this information. It's asking the phone and the Google services What's the last known location of this device? It gets the data back then it uploads a record to the fire store Okay, and then the fire store this recycler view is listening to changes in the fire store and it updates So there's actually a lot of transmission going on here now You'll notice of course as I'm clicking record location here a lot of these locations are the same And that's of course because the emulator isn't moving If you were to do this on your device, you would see different locations as the location fix Just makes a little random noise movements, but you can change the location of the emulator So click the three little dots in the emulator menu right here And you'll get a menu that pops up for location, right? This is the emulator menu one of the options over here is location They have like a Google map integrated here, so let's change this University of North Carolina Wilmington Or none of us are allowed to be right now Let's save this point Okay, and now I'm going to click set location Okay. All right, so the location has been set to here. Let me click record location. Let's see if these things change Nope, there's still still thinking it's there Not what I expected Because as you can see here the the point is three four dot two two dot three nine dot seven seven eight six Look this thing, right? So it's in a different location. Let me go back. Let's go back here Trying to record location. No, it is still convinces there So I'm thinking that maybe something's wrong with my emulator here But if you were to run this on your phone You actually would see maybe let me change this try again. Nope still not updating something going wrong here with the emulator.  I'll try and figure out what's going on later So anyway, this is the the code logic in here will work the problem. I'm having is with the emulator I'm not sure what the problem with the emulator is you've undoubtedly run into problems with the emulator yourself Anyway, maybe you won't even have this bug. Let me go back over Let me go back in my app The other option is to record data continuously and basically what happened. Oh here This is the new location So it was something about the way that the location was being kept in that other activity This is the new location here in the emulator So here you can see There's no record button whenever I launched this activity. It kind of fired off a background process Which again is all managed by Google That's gonna You specify the criteria for when you want the location to update. Whoa. Oh, what happened? Well, when you get a crash like that, what do you do? You got to go to the log cat. Let me get this out of this mode for a second I got to see what happened. All right No debuggable processes. That's cuz it's dead. Let's rerun this and see what happens Investigations, right? Let's just wait and we'll let this run and see what happens Maybe it maybe it was a bug me was a fluke. I doubt it. It should happen again. That's my guess Anyway, there we go. Oh, it's dead, but it's not telling you why it's dead That is very interesting Great, so this is a wonderful demo I have no idea what's going on because this is not throwing an exception. It's just dying inexplicably Let's see if I stop Is it maybe an error with firebase? I think this is probably okay Binding to service failed Uploading is not possible My guess is what's going on here is that firebase Okay, so this is oh here we go. Here's finally an exception user recoverable bad authentication Okay, so I do not have authentication baked into this. Where am I at right now? Long live credential is not available Okay, so there's some little problem going on here with firebase and authentication. I'll get this fixed up for you Actually, I'm not I'm not entirely sure why this is happening So we'll get this fixed up But what we'll see here is that the stuff is updating in the background albeit It's working for a little while and then it's crashing out Sometimes these errors can be due to the fact that you're running firebase on the emulator and firebase doesn't always like that All right, so anyway, it's updating. It's in the background. You can take a look at the code in Continuous look activity kind of see how this works and hopefully you won't run into these issues like I did on the emulator But anyway, have a look at the code try it out Let me know what you run into also try it on your phone see how that works You should get some more exciting data out of that. All right, take care and we'll see you next time


Free your mind. Let's talk about it. Let's talk about it. Nearly three in four young people feel the reaction of their friends when they talk about their mental health issues. The number of depression sufferers has almost doubled since the 80s. One in four people will experience their mental health issue in any given year. Nine in ten people with mental health issues experience discrimination.


Hi, I'm Zor. Welcome to Unisor Education. I'd like to solve a couple of problems related to the state of equilibrium. We are talking about statics right now. Now this lecture is part of the course called Physics for Teens. It's presented on Unisor.com. So if you found this lecture on YouTube, for instance, then you will definitely have this lecture only, but if you will go through the website, you will get also well, the whole course basically, but for every lecture you will have notes. They're very detailed like textbook. Then there are certain exams if you wish to take them. It's completely, the site is completely free and results of the exams are not really going anywhere, so you can take as many times as you want. Plus, this website contains the prerequisite course of math for teens. Especially something like vectors and calculus, you really have to be very comfortable with to listen to these lectures. Anyway, let's talk about problems on equilibrium. So the first problem is really very easy. You have a weightless rod and you have two masses and M and M. This is bigger than this one, obviously. Now the rod is weightless, so the weight is concentrated only in these two weights. You can consider them to be basically point objects. Now my problem is I have to put a thread somewhere here, let's say. So the rod is horizontal. Now, obviously, we're talking about the whole thing is happening on earth. So there is a weight here, which is equal to Mg and weight here, which is equal to capital Mg. So what I have to do is find the point where exactly my rod should be hanging from to maintain the equilibrium. Well, let's say the whole length is capital L. So I need to know this length and this length. Well, obviously, this problem is related to equality of the momentum in this particular case. Now to be more rigorous, I have to really have all the forces which are acting on this system to nullify each other and all the moments to nullify each other. Well, as far as the forces, that's basically easy because you have one force and another force and here you have a tension. So the tension should be equal to some of these two forces, but that's not what I'm asking. I'm actually asking where exactly the point should be the hanging point should be located. And as far as the forces, these weights and the tension, they're not related. Wherever the point of hanging is, these forces will still be balancing to each other, so that's irrelevant. The moments are different. So the moment of this guy, so let's say this is x and this is y. Well, obviously x plus y equals to L. I think I have to change my So, let's write it again, x plus y equals L. Okay, that's obvious. Now, what I have to do right now is I have to equate the moments. Well, if this is horizontal, this is 90 degrees and this is 90 degrees. So basically my vector, now you remember that momentum is vector product of radius times the force. So the radius is this way. In this case, and in this case, the radius is opposite, right? Now, the force is mg down. So if I multiply this by this and then this by that, that should be equal, but with opposite signs, and that's why the sum is equal to zero, right? Opposite signs because the forces are directed the same way, but the radius are opposite. So one radius is positive, another is negative. Whatever it is, doesn't matter which one is which, but obviously in one case we will have the positive, in another case we will be having negative. So our purpose is to equate their absolute value, which is actually x times mg, should be equal to y times capital mg. From which, obviously, g is of no interest. We have x divided by y is equal to m divided by m, and considering that, let's say, x is equal to l minus y, so we have l minus y divided by y is equal to m divided by m. So l divided by y minus 1 equals to m divided by m. So this goes to to the right. This is capital M, obviously. So that's m plus m divided by m. So y is equal to m divided by m plus m times l, and x is equal to capital N divided by m plus ml. So this is the answer. Well, obviously, if capital M is greater than lowercase m, then the x is greater than y, and that's how it's supposed to be. The weight is less, so the radius should be greater to equalize the moment with this guy. So that's it for this first problem. Let's go to the next one. Okay, the next one is the following. We have a horizontal surface and an object of weight w. Now the coefficient of static friction is mu. Static is very important in this case. It's static friction. Now we have a rope which makes an angle phi with the horizon. And question is, what is the force F, which is required, the minimum force F, required to start moving. Okay, let's just think about this. Again, we have to equalize the balance. We have to balance all the forces and all the moments. Well, in this case, we don't have any rotation, so it's only to equalize the forces. Now, what forces are acting on this particular object? Well, obviously the weight, definitely. Another is the reaction of the floor. Right? And then we have two components of the F force. Let's call it Fy and Fx. Well, if this is an X and this is a Y, that's why I called it this way. So these are two components. Now force Fx acts horizontally and it's supposed to move the body, move this object forward. Force Fy makes the object lighter, so to speak, not to press on the ground as much. Right? So what we have to do now, we have to equalize the vertical forces and the horizontal forces. Now, what's the horizontal force? Well, this is basically the force of friction and we're talking about the minimum force F, so it's just the beginning whenever you start moving and that's exactly what static friction actually is. This is the resistance to start the move, to start the motion. So basically we have to equalize all these forces. Now, again, let's start from the vertical position, let's say. Well, vertical position is n plus Fy should be equal to w. Now, this is fine and but this is not really very interesting because it's vertical balance. We know that n would be with balance because the force F is not strong enough to start the object flying. Although it might actually be the case, but we are not talking about this. We're talking about much smaller F, so the object doesn't fly anywhere. It's just on the surface and it's supposed to go horizontally. So let's talk about horizontal movement. So what exactly are the forces? So Fx is equal to F times cos 5, right? Now, what is the friction force? Well, friction force is basically the result of weight minus Fy because that's actually the normal reaction of the of the floor times mu, which is the coefficient of static friction. So again, w goes down, Fy goes up. The difference between them is exactly the reaction of the floor, normal perpendicular to the floor and that's what's supposed to be multiplied by the coefficient of friction to get the force which resists the motion and it's supposed to be equal to Fx, which means F times cos 5. Now, Fy is equal to F times sin 5. So what do we have now? Mu times w minus mu times F times sin 5 equals F times cos 5 from which F is equal to, so F goes to this way, so on the top we will have mu w and on the bottom we will have cos sin 5 plus mu sin 5. So that's the answer. This is the force which we have to apply to the rope to start moving this particular object if weight is given and the coefficient of static friction is given. That's it. That's the solution. Number three. Okay, number three also has no rotation. So no moments are involved. We just have to carefully examine the forces which are moving. So let's say we have a platform and on the platform we have a person. This is the person and there is a rope. Which goes through some kind of a pulley and it connects here. I think it's not really a good thing. I'll do it much bigger than that. Let me start from the rope. So we have a pulley, a rope, platform, another rope, another pulley and here we have a person. Okay, now what do we have? We have the weight of the person, W. We have the weight of the platform P and we know that the system is in equilibrium because the person is pulling this particular rope down. Now as he pulls it down, obviously this rope supports the platform from this side and from this side because this is another rope. So what we have to find out is what is the force F. This poor person should pull down the rope to stabilize his position not to fall down and not to go up, basically. So he is in equilibrium. He and the platform are in equilibrium. Let's just think what kind of forces are acting here. Well, obviously there is a tension force here. Well, the tension force here is F. That's the force actually which we have to find out. Now, what also is important? Well, obviously there is a weight and there is a reaction of the floor which actually goes this way. So, now let's consider only the person and all the forces which are acting upon this particular person. Now, since he is in equilibrium, all forces which are acting on the guy must be equal to each other. So what are those forces? Well, since he pulls the rope with the force F, everything is in equilibrium, right? That means that the rope pushes him up or pulls him up, whatever, with the same force F. That's the third law of Newton, right? So, on this person we have the force F which goes up. Reaction of the floor also goes up. His weight goes down. So, basically the weight should be equal to N plus F. To maintain him in balance, that's what's necessary. Now, let's talk about the platform. Now, the platform is also in equilibrium. What acts on the platform? Well, first of all, the platform has its own weight, right? Now, the platform, between the platform and the person, we have this reaction force. Reaction force pushes him up, but at the same time he is pushing the platform down. So, the weight is P and the reaction N is also down. So, these are all down things. Now, what's up? Up is tension of the rope here and tension of the rope here. Well, this is, I don't know what it is, let's put it T. But this tension is equal to force F because that's exactly the effort we are pulling this rope down. So, that's what the tension is, since it's all in equilibrium. Now, finally, this particular pulley is in equilibrium. So, what pulls it down is this force F and the same tension here, right? So, it's 2F, it's unstretchable rope. But the up is the same tension as in this side. So, they are equal. So, these are three conditions to balance our platform and the person and the pulley to balance the whole system. What's necessary to find out is, well, W and P are given and we have to find basically the force F. So, how do we do it? Well, very easily. Let's substitute 2F instead of this T, we will have P plus N equals 3F, right? Now, N from here is equal to W minus F, so we can put P plus W minus F instead of N equals to 3F. So, F goes here, it's 4F, so F is equal to P plus W divided by 4. So, that's the answer. And by the way, you see, divided by 4, we use two pulleys and that allows us to pull with a significantly smaller efforts than the weights which are P and W. Let's say the platform is weightless, right? Then, if the person wants to just pull himself up, it will be just one quarter of his weight, W divided by 4. So, in any case, it's easier to use these system of pulleys to pull something up. And that's exactly how people in practice do, how they use these systems of pulleys. Okay. The last problem also has certain practical implications. Okay, so let's consider you have the wall and the floor and there is a ladder. Well, let's assume that the ladder has the length D and we also have to assume that there is certain static coefficient of friction, which is new. Now, as you understand, if we put this ladder very close to the wall, it will stand very easily. But if you would like to move the base of the ladder out from the wall, the further we put it from the wall, the more chances are that it will start slipping and the friction between the ladder and the floor would not be sufficient to keep the ladder in this position. And at some point, the whole thing might actually start slipping away. So, my question is that given the coefficient of static friction, what would be the angle, the minimum angle, obviously. The maximum is 90 degrees, but the minimum angle will be where the ladder can still actually stand in the equilibrium position without slipping to the right. Okay, so that's the problem. We have to find phi based on mu. Actually, the result will not depend on the D, by the way, but we will come to this later on. Okay, so let's just think about what kind of forces are involved here. Well, there is obviously the friction which is directed this way. Now, there is a reaction of the floor and there is a reaction of the wall. Let's say this is N1, this is N2, and there is a force which is the weight. I put it in the middle. Well, you might say that, look, the weight is not really in the middle. The weight is concentrated basically at every infinitesimally small segment of the ladder. Strictly speaking, you are right. And if you will start doing more precise calculations, we will basically come up with exactly the same result. In cases like this, you can always assume that the weight is concentrated in the center of mass. And obviously, in this particular case, center is in the middle. So the whole body actually behaves as if the weight is concentrated in the center of the mass. So I assume that you agree with me on this particular point, and I hope you do. I put this particular weight here, thinking that the whole weight is here and there is no other forces involved. Now, okay, fine. So let's think about this fact. Now what is the force of the friction? The force of the friction should be equal to the reaction of the floor, which is n, 2 times coefficient of the static friction. I mean, whenever you have this type of contact between the body and the floor or whatever, it's the reaction of this floor which is supposed to be multiplied by the coefficient of the friction to get the force of the friction. So that's all. Basically, that's all horizontal movement, which we are talking about. So if we want this to be an equilibrium, this must be true. At the same time, this is a much more complex problem. There are also other forces which are involved. What are the forces? Well, the forces are which are rotating. You see, when this ladder would shift, the bottom would shift to the right, the ladder would rotate. So we have to really talk about the momentum of rotation. So that's what's very important. So what are the momentum of rotation here? Well, we have two main forces. Well, first of all, we have to choose the center of rotation. We have sufficiently a large number of forces. So we have to really choose the momentum which basically simplifies our calculations. And I suggest we choose as a center of momentum this point because at least two forces already can be completely ignored because their radius is zero. So momentum would be zero. But the momentum of N1 and W is not zero, obviously, relative to this point. So what are the momentum? Well, you always have to represent the force as the sum of two forces. One is along the radius and the other is perpendicular to the radius. That's the easiest way. So if I will do this, so this would be the force which I have to just multiply by the radius. And in this case, again, I will do the perpendicular. That would be my force which I have to multiply by the radius because that's what actual vector product is. So instead of doing sine of whatever, we can always do this. It's kind of easier. So what's the projection? Now if this is phi, then this is phi and this is phi. So this is N1 times sine phi. And the radius is equal to the length of the letter which is D. Now, and that actually turns the whole letter this way. And it should be balanced with another force which is trying to rotate it this way. So if this is phi, this is phi, this is phi. So if this is W, this is, let me see. No, this is phi. No, this is not phi. This is 9 to the degree minus phi. So this is phi. So it's W times cosine. So it should be equal W times cosine phi. And the radius is equal to D divided by 2. So this is the equation of balancing of the momentum. And lo and behold, we are not dependent on the length of the okay. Now, so what do we have here right now? What do we have here right now? Horizontally we have this is the force which goes to the left and this is the force which goes to the right. So that should be equal to N1. Okay, vertically we have N2 here and we have W down. So N2 is supposed to be equal to W. So we have basically balanced, we don't really need this anymore. So we have balanced all the forces. This is horizontal balancing. This is vertical balancing. And this is momentum balancing. Well, that's it. We should really be able to find out phi. Okay, so let's do it this way. So from this N1 is equal to W divided by mu. N1 is equal to N2 which is not multiplied by mu. Sorry. So we will substitute it here. So this is W times mu. And lo and behold, it doesn't depend on the weight of the ladder. And now we have a very simple thing. Mu is equal to cosine phi divided by 2 sine phi. Or tangent phi which is sine over cosine. And 2 goes here so it will be equal to 1 over 2 mu. Right? So phi is equal to arc tangent of 1 over 2 mu. And that's the answer. Well, I was trying to put the problems in increasing difficulty kind of things. But again, there is a common approach. All forces and all momentum should be balanced. And if we are talking about the forces, the best way is to go vertically, separately from the horizontally. So the vertical forces must be equal to each other, balanced to each other. And horizontal forces must balance each other. And obviously momentum should balance each other. So balance means that some of them should be equal to 0 basically. That's what the force which goes in one direction should be equal by magnitude to the force which goes to another direction. Okay, that's it. I suggest you to go to the website unizor.com and read the notes for this particular lecture. I think the first two problems which are relatively easy, I've given just answers. So you can just try to basically solve the same problems just by yourself and check the answer. The third and the fourth problems as a little bit more difficult. I do have solutions, but why don't you just try not to read the solution, solve the problem yourself again, check the answer. And if something is not right, read the solution presented in written format as notes to this lecture. That's it for today. Thank you very much and good luck.


Thanks for checking out this video.  I want to apologize for not getting this video out sooner I usually get it out like as soon as I get the press release for what's coming to shutter for the next month But I ended up being pretty busy for when this one hit me So my apologies for that, but let's get into it. What is coming to shutter in June? There's one film in particular. Well, there are a lot of films that look interesting But there's one film in particular. I'm gonna let you know about when I get to it I think it's the last one on here that I'm very interested in and I'll tell you why Anyway, so the first I'm gonna go through these shutter exclusive and originals And the first one. Oh The disclaimer I usually give on this is that I will always have no spoiler reviews for the shutter exclusive and originals The week they end up coming out on shutter. So they usually come out on Thursdays I will be putting out my no spoiler reviews on Mondays Sometimes Tuesdays depending on when I get the screeners and stuff like that But usually Mondays so you can look for those now for the first one The first one coming out June 3rd is called caveat. Now. This is available in the US Canada UK, Ireland and Australia New Zealand Lone Drifter Isaac accepts a job to look after his landlord's niece Olga for a few days in an isolated house on a Remote island it seems like easy money, but there's a catch He must wear a leather harness and chain that restricts his movements to certain rooms Once Olga's uncle Barrett leaves the two of them alone a game of cat and mouse ensues as Olga displays Increasingly erratic behavior as a trapped Isaac makes a series of horrific discoveries in the house Interesting that is a very interesting up concept. I'm down. I'm very interested in that one Next is the amusement park I've known about this one coming because they kind of when they secured the rights to this they threw it out there Immediately pretty much. This is coming June 8th For US Canada UK Ireland Australia New Zealand I'll just say net from now on. I'm not gonna run through all of them if it is all of them I'll just say all of them got it Recently discovered and restored 46 years after its completion by the George A Romero Foundation and produced by Suzanne de Rocher Romero director George A Romero's the amusement park stars of Martin's Martins out from the film Martin Lincoln Mazzel as an elderly man who finds himself just disoriented and increasingly increasingly isolated as the pains tragedies and humiliation of aging in America are Manifested through roller coasters and chaotic crowds Commissioned by the Lutheran society the film is a is perhaps Romero's wildest and most imaginative movie an allegory about the nightmarish realities of growing older and Is an alluring snapshot of the filmmakers early artistic capacity and Style and would go on to inform his ensuing filmography The lost film was restored in 4k by Indy collect in New York City That's too long of an explanation in my opinion. We just need like a quick synopsis Unfortunately, they tend to go too long on these in my opinion, but regardless. It's a lost George A Romero film I'm very interested to check that one out. So the next one is called super deep and that is coming June 17 I'm gonna turn this sideways might be easier and that is going to be available Everywhere yeah, everyone's getting that I assume all the shutter exclusives and originals for the most part will be everywhere, but So once again June 17th super deep the Cola super deep borehole is the largest Russian secret facility in 1984 at the depth of more than seven miles below the surface Unexplained sounds were recorded Resembling the screams and moans of numerous people since these events the object has been closed a small research team of Scientists and military personnel had gone down below the surface to find out the secret that the world's deepest borehole was hiding What they have found poses the greatest threat that humanity has ever faced Now the future of the world is in their hands. So it's a Russian film That sounds very interesting. Love the setup on that's in a quick synopsis. I'm definitely interested in that one Next one is an unquiet grave and that is coming June 24th Everyone is also getting that one a year after losing his wife in a car crash Jamie convinces his sister Ava to return with him to the site of the accident and help him perform a strange ritual But as the night wears on it becomes clear that he has darker intentions an Unquiet grave is an exploration of grief and the harm we cause when we don't take responsibility for our own healing Sounds interesting enough a little vague for how good or not good It could be but we'll see and Then the last of the originals and exclusives is coming June 29th, and that is called vicious fun Great title by the way. This is gonna be available for everyone Joel a caustic 1980s film critic for a national horror magazine finds himself unwittingly trapped in a self-help group for serial killers Love the premise already with no other choice Joel attempts to blend in with his homicidal Surroundings or risk becoming the next victim Love that that sounds great. I really super interested in that one okay They're gonna have featured collection of queer horror that will be available on June 2nd And it's just a bunch of the films that they already have on there or that are coming in that month That'll just be put together as one collection you can go and check out So it'll kind of put all the queer horror in one area So a lot of the films I'm not gonna run through them because a lot of them are already on there But just know June 2nd the queer horror collection will be on there. So coming June 1st Ginger snaps to unleashed unfortunately. This will only be in the United States Ginger sister Brigitte or Brigitte. Sorry Now a werewolf herself must try to find a cure for her bloodlust before the next full moon while hiding out in a rehab clinic From from a relentless werewolf cool, and this is good because the first ginger snaps is already on shutter And then they also have the third ginger snaps. That's coming the same date on the first Which is ginger snaps back the beginning also only in the US actually a lot of these I'm seeing at the moment Or only in the US unfortunately Set in 19th century Canada Brigitte and her sister ginger take refuge in a fort Which later comes under siege by savage werewolves Regardless of how good they are or not I'm gonna check out the the second and third ginger snaps because I've never seen them before So I'm definitely planning on doing those and I'll probably do reviews for them Also coming on the first an American werewolf in London. We all know this one only in the United States unfortunately Two American college students on a walking tour Britain are attacked by a werewolf that none of the locals will admit exists Pretty simple Almost everyone probably watching this this video right now has seen this and if you haven't seen this make sure you see it Also coming the first eaves by you only in the United States again Sorry, what did Eve see and how will it haunt her husband father and womanizer? Louie Battiste is the head of an affluent family But it's the women who rule this gothic world of secrets lies and mystic forces sounds interesting And then the last movie that's coming on the first Available in the US and Canada is burn witch burn a woman who may be a witch defense her husband from forces attempting to harm him Pretty sweet to the point. I don't know if that sells me on it But maybe if you know about this film and you would recommend it Let me know or if you don't recommend it also let me know probably more importantly if you don't recommend it Coming June 2nd islands. This is for the US only characters wander through an erotic maze of love and desire That's it. You have my interest though Also coming June 2nd terror sisters Available just in the US today is a day unlike any before today is the day Cal Thune and their girlfriends imagine their revenge That's really vague. I have no clue what that means like what that is When I was like I was saying that the originals and exclusives like the synopsis were way too long These are way too short at least that one was way too short Also the last one. Oh, no, not the last one coming the second another one coming on the second dare samurai also just in the US Set in a small German village a bloody game of cat and mouse ensues Between a young straight shooting police officer and a cross-dressing villain with a large sword in a predilection for beheading Okay, that sounds interesting. I could I can be interested in that one also coming on the second thirst This is available in the US and Canada Holda holda. I guess is the name is arrested after being accused of murdering her brother After she is let go because of insufficient evidence. She meets Hurtur Sorry, I'm doing my best a thousand year old gay vampire together They have to fight at cult while being investigated by a rogue detective could be interesting Also coming the second rift which is available in the US and Canada Two men in a secluded cabin are haunted by their dead relationship That's it. Also too vague Coming June 7th night of the living dead the 1968 film. This is coming for everyone This is just being re-added because it was already there some time ago, and then it's cycled off So they're adding it back on we all know night of the living dead not gonna go there June 7th reunion coming to the US and Canada a Pregnant woman returns to her recently deceased grandparents old family home to spend time with her estranged mother what begins as a tenuous Reunion solely turns terrifying Sounds interesting. I could be down for that one Also the seventh beyond the door three for the US and Canada an Introverted American student travels to Yugoslavia as part of a school trip to witness an ancient pagan ritual with a deadly secret I'm down pagan rituals. I'm down. I definitely want to check that one out Coming June 14th Monstrous and this is coming for everyone a young won't a young woman goes searching for answers after her friends mysteriously vanishes in Whitehall, New York an Adirondack town known for its bigfoot sightings She quickly learns that an evil more sinister that an evil more sinister than she could ever imagine is hiding in the woods More sinister than bigfoot. I don't know we'll find out also coming the 14th the retreat and this is also coming for everyone a Man finds himself alone and lost after a horrifying encounter with a monster during a backpacking trip in the Adirondack high peaks Adirondacks again Now he must fight for his life and sanity as he battles the evil Native American legend the when ding when to go I know about the when to go and I think there's actually a film called when to go It at least was on shutter. It might still be on shutter I think Larry Fessenden did it and I heard good things. So just throwing that out there Okay, now this the last one coming on the 14th. This is the one. I'm super excited about evil speak Coming to the US and Canada an outcast military cadet taps taps into a way to summon demons and cast spells on his tormentors through his computer now I believe this is the film with Clint Howard in it And if that is the case I have the Blu-ray of ice cream man with Clint Howard in it And there's the Joe Bob Briggs Monster vision version of it and if you watch that he interviews Clint Howard and they talk about evil speak and Joe Bob goes off about how much he loves evil speak. So I'm excited for that Coming June 15th just for the United States is the conspiracy a Documentary about conspiracy theories takes a horrific turn after the filmmakers on a cover an ancient and dangerous secret society Conspiracy theories in general are horror enough. So I don't think I'll be watching that one June 15th coming to the US and Canada house bound. I've seen this film. I recommend it. It is good A young woman is forced to return to her childhood home after being placed under house arrest Where she suspects that something evil may be lurking like I said, I recommend that also coming on the 15th the similars This one is for the US and Canada on the rainy night of October 2nd 1968 eight people waiting in a remote bus station for a bus heading to Mexico City start experiencing a strange phenomenon Maybe although I think someone recently in a comment on one of my videos Recommended the similars so I might have to check that out Coming June 22nd city of the dead now This is for everybody a reporter in a psychic race to close the gates of hell after the suicide of a clergyman Caused them to open allowing the dead to rise from their graves. This is just being re added Lucio Fulci it was on shutter for a while. I guess it's cycled off. It's coming back. If you haven't seen it Definitely watch it. It's great. It's not Rich on story, but it looks awesome and great kills the drill scene. I'll just say that the drill scene June 21st coming to the US and Canada Homewrecker two women befriend each other, but one becomes obsessed with the other That's it. I don't know not enough to get me interested I don't think and then the last film coming on June 22nd and the last film I'm gonna go over period because that's it the antenna and that is available for the US Australia and New Zealand in a dystopian turkey the government begins installing new television antennas at homes throughout the country Mehmet a Superintendent at a crumbling apartment complex must supervise the installation of the new antenna when the broadcast cast is Sorry when the broadcast it transmits begins to menace the residents of the apartment complex Mehmet must seek out the spiteful entity. That's interesting. I Be interested in that one. So that is it everyone Obviously, I'm most interested in the second and third ginger snaps and also evil speak Mostly evil speak if it is the film that I'm thinking it is which I'm pretty sure it is So let me know what you're excited about or you're not that excited. Is this not an exciting month for you for shutter? Honestly, I'm always kind of hoping for months that there isn't much of anything I want to watch because there still are so many films in my queue that I need to watch And I just don't have the time So I want to find that time to kind of get caught up But you know, I probably never will but anyway put some comments down here What are you excited about or are you not excited? Also, what would you really love to see shutter get? Are there any particular films that you're like, man, I really wish shutter would get that I would make me super happy So let's go ahead and do that. Also, do me a quick favor and hit that subscribe button If you like this video or any video or if you find this video helpful, which hopefully someone does Yeah, but just subscribe. I appreciate it. It also encourages me to keep going Motivates me and I'll keep going And then also hit the notification bell button because that we'll know whenever new videos are coming out whether it's one like this In-depth analysis of a film or no spoil review or unboxing or any of that jazz But regardless, I thank you for taking your time to check this out and until next time keep it brutal