 Thank you so thank you very much. It's only my second time in Montevideo, but I'm as glad as being here as the people who are here for the eighth time. I hope so So I thank you for inviting me I hope so I I have nothing new to explain now So I will tell you about the problems I like And so I will tell you about two problems and but I won't give you the solution So the first problem and both are related First problem is a problem about random works on compact groups and about speed of equity distribution So let me first Tell you what I know about random work on compact groups. This is somehow very classical for people in group theory but maybe not everybody handles a Easily group theory, so I will take G which is a compact group Which means a group equipped with a topology which makes it a compact space and which makes the operation of the group continues and the only two examples I will be interested in will be the following so the circle group and And So three the group of orientation preserving isometry is a three-dimensional Euclidean space, and there you can If you are not if you don't like compact groups, you can just think to these two examples and now I Will take new you will be a probability measure a borel probability measure on G and I want to see new you as a low of a random work Which means that I want to define a mark of process a mark of chain on G That is just a mark of chain on the space is just a rule to move randomly Just what it is When I am at the point x This guy I see as a point in the group and I pick G an element in G So which is a translation and I choose it randomly with respect to the loom you I pick it at random and I jump to G X and then I iterate that is I pick a new element spara and I jump and Probability theory when you have a mark of chain you like to define operators which represents the mark of chain Actual functions. So here's the operation if I take just a continuous function of the group I will define an operator So P Phi evaluated at X is just the integral of Phi of G X D mu of D that means if I I start from the point X. I look at all So possible destinations of the random move movement and average the value of my function, okay, and No, if you look at the random work at after n-step its distribution Is given by the nth iterate of the operator this integral over Z to the n It tells you on average Where I am after after n-steps of the random You can ask so saying that the random work equidistributes is saying how this integral behaves for large n Let's call it proposition This is a really classical from the elementary analysis some groups that if for every H Which is a proper closed subgroup of G My probability measure is not concentrated on H Of course if I live on a subgroup we have to study what happens on orbits of the subgroup then For for any continuous function If I look at the Birkhoff sum Birkhoff average The images by the operator of the function this converges to the integral My random work equidistributes in the Birkhoff sense So this converges to int integral which is just the integral towards the of the arm measure That is what is this arm measure? If you don't like if you know it you understand if you don't know it just have to sing to these two examples here. This is The arm measure. This is just a Lebesgue measure Yeah, in this case you can look at a SO3 Just at acting on the sphere Okay, and It acts on the sphere and it acts transitively on the unit and general bundle of the sphere So the arm measure is just the product of the wrong measure of the sphere and the Lebesgue measure on each unit circle above Just what the yes Sorry I Say that you your measure your point your random work on the group It must not leave on a closer group because if it leaves on a closer group, it cannot equidistribute everywhere Okay, but this is what I'm saying is that the only equity the only obstruction to Going everywhere is leaving on a closer group as soon as it is not true. It's okay. For example This proposition tells you that if you take an irrational rotation You have a equidistribution Okay, and Now let me ask make a more if more So you you you will prefer to study these these are you see as soon as you are not a Translation of the group you you already have a lot of guys at every step and you would ask whether they equidistribute or no And the answer is that there is only one obstruction This is just that so let me write it if more over for any H which is now a closed proper normal subgroup of G The measure and for any For any X in G the measure of the left or right coset. This is the same It's smaller than one Then actually you don't have to make a beer cup average for any fire Which is a continuous function on the group If you look at the distribution of the random book after n-step This so this is a uniform convergence. Yeah, and in the same way now I don't have to make a beer cup average. I just go to the integral of fine uniform again because Precisely if your Probability measure is just a dirac mass at a point you cannot equidistribute after n-step You're just a transformation so you cannot equidistribute But what I'm saying and if you are a dirac mass in a quotient of the group the same happens But I'm saying that this is the only obstruction as soon as you are not saying that this probably this measure is one He's exactly saying that my measure is just a dirac mass in some quotient And then he cannot equidistribute But as soon as this is not true, you need you could distribute And now I will prove you serum in three lines I told you this is really harmonic analysis Not example, it's proof What happens is that you will consider the L2 space of the group And when I write L2 of a group, I just mean with respect to harm measure which is and this L2 space It comes with an action of the group on itself with the action of the group by say left translation Here everything acts on the left So I consider the left action my translation of the group on itself, and I know that this space splits as a sum I will call it H. This is a Hilbert space. The name of a Hilbert space is H So it splits as a direct sum and each vi Vi is G invariant and finite dimension This is where complexity comes into play that Every Hilbert unitary representation of a compact group is a direct sum of invariant finite dimensional subspaces So if you don't trust this if you're not familiar with this you can seem to finite groups for a finite group It's clearly true because if you take any vector its orbit spans a finite dimensional subspace Okay, and the same happens for compact groups and what now the invariant The constant of the trivial representation only appears once because you only have once constant function, okay, so You can write it as C plus a sum Vi and each vi Does not contain G invariant vector, but now these are G invariant subspaces, so they are invariant by the action of this operator So I have an operator About an operator P Pi which is the restriction of P to the i it has norm smaller than one and now My assumption here my first assumption first assumption This is a small exercise It's equivalent to saying that it has no not One as an eigenvalue So now it's just a property the elementary property of linear algebra that if you have an operator of norm smaller than one Which does not that mean one as an eigenvalue necessarily Your beer cuff average goes to zero Okay, and the second assumption is equivalent to saying that He has no eigenvalue of modulus one and now it again for the same reason, you know that P to the n if you have An operator which has norm one but which has no eigenvalue of the circle all of the eigenvalues have modulo smaller than one So the power So you see this is a as soon as you know little of Representation station theory of compact groups, which is called a Peter by theory. It's really easy and now the The key question is the question of the speed So knowing the speed is knowing whether so This norm still goes to zero This goes to zero as an In norm that is what I'm proving here. It's what's called a strong convergence when you have an operator in Banner space, you see that it converges strongly if for any vector He what I prove is this okay, but I have a convergence of this form But I didn't prove for the moment that the norm of P n Goes to zero as this is the question So let me give an example. So this would as this is a question of spectral gas That is in other words, let me state it more precisely On each of the vi You know that the eigenvalues are far away from the circle, but you do you have a uniform bound? Okay, that is as you have a sequence an infinite sequence of operator in finite dimension All of them have their eigenvalues far away from the circle But maybe they could get closer and closer to the bound to the circle and in the circle Not the same circle now is when the group g is a circle. This is exactly what may happen So let me stay precisely what happens here In this case, let me describe the vi You have the Fourier coefficient K of x This is I need some care at some point So you have this Fourier coefficients and this decomposition here. It is just the fact that the L2 of the circle Is the sum of K Of the space and by the code function ek Okay, so now each of these functions. So these are the stable subspaces So these are eigenvectors in this case and the abelian case This is a decomposition into like and what I'm saying. I'm saying it for the circle, but it works for any non-discrete abelian And now I have a formula which is just Almost trivial This vector is an eigenvector and what is the eigenvalue? This is the Fourier transform of the major I'm just writing something which is trivial This decomposition here it is in the case of circles. This is just Fourier series and what happens is that for example if mu is Continues to the bag You have Riemann-Lebesgue-Lehmann which tells you that these Fourier coefficients They go to zero So you have the speed here you actually distribute quickly because the eigenvalues they don't go close to the circle They go close to zero. So you are short to a good distribute quickly But when mu is finitely supported so you take a Finitely supported major on the circle. It's really easy to prove that there exists. So this coefficient you should make a probability major Which does not leave on a finite subgroup of the circle These coefficients when k is not zero they are different from one and they have modulo smaller than one But when mu is finitely supported it's really easy to build a sequence kL Such that mu the Fourier transform goes to one actually So in this case You know that there is never a spectral gap. Equidistribution is slow Okay for finitely supported major And the question now Which is still open Is about what happened in this case? The case of SO3 In the case of SO3 So it is a conjecture Sorry What major? Depends on the measure Everything can happen So the conjecture Which maybe can be attributed to Sarnac Is that for any probability major mu On SO3. So I write SO3. I could give you a bunch of Spaces where it is probably true. So where one can hope it to be true, but I prefer to focus on a simple example Such that for any h included in G So this will be G Proper subgroup Mu is not concentrated on G Mu has a spectral gap You see again, it's just I'm just giving you when I give you a measure the question is just Associated to the measure come a sequence of finite dimensional vector spaces equipped with a sequence of Operators with norm smaller than one and with no eigenvalue So more than that monogon value of modules one is there a bound on the way So again values can come close to this okay, so Here I use I didn't put the second assumption about normal subgroups because there are no Non trivial normal subgroups in In SO3 Simple So let yeah, you have so let me Describe you so that you understand is a what? The proper subgroups of SO3 are it's really easy So you have a proper subgroups of SO3. It's just these are just the rotation Around the given axis if I fix your given axis you have the rotations around this axis and you have the stabilizer of Regular polyadron these are the maximum these are the Maximals and so I it's difficult to draw because I don't want to take a good so I won't do it Okay, they take a gun. Okay, and so the stabilizers are the maximum proper subgroups of SO3 so what I'm saying is that now you take Let me really state Elementary case of the conjecture which is so First I should say in which cases the conjecture is known to hold so since last 10 years there have been some progress on this conjecture and there is a theorem by Bourguin and Gomblourde, I don't know maybe it's 2005 I'm not sure of the exact maybe 2007 I don't recall exactly And it says that it is true if you is finitely supported or comfortably supported and leaves and leaves on matrices With algebraic entries so you take any finite set of matrices with algebraic coefficients These are three by three matrices. So you take matrices with algebraic coefficients You make a you put weights on them and then you assume that They have not a common axis or they don't fix a common regular polyadron Then you have this operator this average of the action of the element and this has a spectrograph And to my knowledge it is not true in any other case. So for example, let me state an open case take G Which is a gis matrix and H See so these are I'm fixing just to coordinate axes and I'm rotating with angle theta around one and rotating with angle theta around the other and For small theta So I know that there is no common axis and for small theta you are sure not to preserve a regular polyadron for small theta you take For example, I prefer symmetric measure so Nicer to have symmetric measure because this makes self-adjoint operators better And now you can ask you have this operator. Just operator is average of the action of these four elements Does it have a spectral gap or no, it's not So this is the first problem and So now let me go to another field Which is also a group theory with the probability measures and I Will ask questions and we'll see that they are actually related to this one It's not that far away So now I will change the group and I will Take G to be SL2R. This is a group of Unimodular 2 by 2 matrices and again I will take mu which will be a probability measure and Now Again, I don't want to I want to study again the random work on D associated to mu and I don't want to I Don't again there are some not degeneracy assumptions. I want to see here I didn't want to consider matrix matrices with the same axis or preserving a discrete structure So I will assume that for any H Which is included in G but which is a proper Algebraic subgroup mu is not concentrated on H Again, you see this assumption When you see it in this way it seems complicated But concretes very easy to check if I give you an explicit set of matrices You can check because you know the maximum proper subgroups and SL2 it is the same So what is the maximum? What are the maximum proper? algebraic subgroups of SL2R There are three conjugacy classes. The first conjugacy class is the class of SO2 So you are when you ask this you ask that the measure doesn't leave an orthogonal isometrism for certain fixed product You have the group of upper triangular matrices. So you are assuming that The matrices in the support of mu do not have a common Hegel vector, okay, and you just have The union of the set of diagonal and anti-diagonal matrices This is a subgroup and I'm just assuming that I don't have a common pair of lines That is invariant So you see it's easy to check if I give you a Set of matrices to check whether it holds on and Now this group SL2R So it is not compact. So what will happen is that if I take a trajectory of a random walk It will go to infinity under these assumptions. Okay, just And even with the speed There is a Lyapunov exponent, which is positive somewhere, et cetera, but I won't speak of this today I mean there is seen in the properties of equity distribution So I will look at the action of SL2R on a compact space Which is just a projective line So topologically, this is just a circle This is a set of vector lines in R2 And now this is a compact space. So again, I will define an operator With the same formula Now I define it on functions, continuous functions on p1 And Now you can ask what happens With respect to when I iterate this operator on the functions on p1 That is if I start with a point on a boundary, where does it go? And it turns out that it goes to a fixed measure It distributes to a fixed measure So this is a theorem of Fürstenberg I don't know in the 60s somehow There exists nu, which is a probability measure on p1 Such that for any phi when I look at this distribution after n-step p phi, this goes as m goes to infinity Uniformly In particular, phi is nu is a unique p-invariant measure It has a unique p-invariant measure So if for example you have Your measure lives on a guys which contracts in different directions on a kind of IFS You will have a measure on some contour subset of p1 and uh As we got questions related to speed, etc What happens is well known that is a but here the situation is very different And but one knows how that there is a speed in a certain space of holder continuous functions or things like it Okay, and the question I would like to address are questions about the geometry of the measure nu and they are open problems And so for example, so there is a CRM that is a what is this measure and you see it's defined by a In a totally non-explicit form you don't have an So in certain cases you have an integral formula or things like it, but in general The proof of the CRM. It's really a proof at the absurdum You suppose that there are several measures and then you reach a contradiction So you don't have an explicit formula for defining nu So it's rather difficult to handle nevertheless. So it's for example a theorem theorem by givach Which tells you that uh, if mu is compactly supported actually, you can suppose a bit less but let's say It's the low the random work on the group. Yeah, so now when I have to be careful because there are two probability measures So it's difficult. One is mu. This is the distribution of the random work on the group. Okay, and nu Is a limit distribution on the projective line. Okay, I will keep to these notations until the end So if you the distribution is the increment of the random work on the group live in a compact set then nu has positive householder dimension Let me cite a recent advance. So this is uh, I don't know maybe 97 95, I don't know I don't remember. Let me cite a recent result by li 2000 maybe 17 that uh under the same assumption the Fourier transform of nu Goes to zero at infinity so you You equip the circle with the SO2 invariant matrix. So you you you see p1 just at the circle so you can ask for the Fourier coefficient of the measure and this is It looks a little regular that is it satisfies Riemann-Lebesgue-Lehmann Even if it is absolutely Even if it is concentrated on a zero measure Contour set it satisfies Riemann-Lebesgue-Lehmann When has this kind of information And I would like to know I would like to know Whether when can I ensure that nu is absolutely continuous? so of course again If nu is itself is absolutely continuous then nu will be absolutely continuous It's rather easy to prove but in general when does it know I even If mu is finally supported So it's not very clear But at least there are cases where the group span by the Support of the measure is dense enough For the probability measure so the group span by the support of mu Is dense enough in some way for the probability measure nu the distribution the limit distribution To be absolutely continuous with respect to the Lebesgue measure. So there is a theorem Say by Bourguin again, I don't know I didn't prepare this talk very carefully because I I didn't check any bibliography goal refunds Maybe it's uh Let me put a random date Something like it less than 20 Less than five more than 20 years So which tells you that There exists mu Which is a finally supported So again Which does not give mass to any proper algebraic subgroup, etc On sl2 such that nu is Absolutely continuous with respect to Lebesgue measure and the There were previous constructions, but they were random. That is they were modeled And the proof that for almost every parameter Bernoulli convolution are absolutely continuous And the one knew that there were examples But they were not explicit whereas this example is explicit So, uh, at least if you are able to read the paper by Bourguin Which selects particularly small subset of humanity But they are supposed to be explicit And uh, so, uh Uh, so I don't understand this paper I am not able to do it again and But at some point, so we we were reading it you see Benoit and We understood the strategy. So I would like to explain you the strategy And how it is connected to property the spectral gap, etc So what is the strategy to to understand the strategy? I need to talk a little about functional analysis. So we can forget If you didn't like what was before maybe we like it now because uh, I will change topic. So, uh, now I I will do as Andres did this We changed topic and uh, so I will speak about operators in banner spaces and About spectral radius and there is another spectral radius for there are two spectral radii for an operator And once it's called the essential spectral radius, so I would like to talk a little about this So let me start slowly. So I will take E Which is a banner space So in the sequence, this will be a space of functions on the circle Just a two space of the circle for the Lebesgue Major and uh, T Which is a boundary Laniya operator In When I have a banner space and the bounded Laniya operator, I can build its spectral spectrum. So the spectrum of T It is just a set of complex numbers such that lambda minus T Is not invertible So in finite dimensions, this is just the space the set of eigenvalues of T Okay, so, uh, in Any dimension what is known is that it is an unempty Compact subset of So you have somewhere a society to your operator You have a subset. Maybe it is not connected. Maybe it has a related point You have a subset and uh Suspect all ranges of T It is just the uh, uh, supremum Of the moduli of the people living in the spec so you have Zero somewhere and you draw This is supposed to be a circle Now you take this radius. This is the spectral radius we have and these Objects are defined actually in any banner algebra Here's the banner the underlying banner algebra is just the algebra of bounded operators I'm leaving this is just the language of spectrum and spectral radius In the algebra of bounded operators algebra of bounded operators with my banner space The algebra of bounded operators and uh, so in infinite in the infinite dimensional case what happened is that this algebra has an ideal Is not true in the infinite dimension the algebra of matrices is a simple algebra. It has no proper ideal Okay, but here It has an ideal The ideal is the ideal of compact operators So if you don't like compact operators, you will like them in half an hour first And for the moment you can think to e as being a Hilbert space in this case Compact operator is just a norm limit of finite rank operators. Okay So this is an operator which has a very small image So in the Hilbert case it is just a limit of finite rank operators But what is really nice? You see if you have an operator t And s as finite rank or s is compact, but So what is nice is that it is an ideal And if you think to them as being more or less finite rank operators, it's clear because if s as finite rank T s and s t have finite rank So it is an ideal Okay, it is a closed ideal. So you have a banner algebra You have a ring You have a nice ideal a closed ideal because there is topology here. So there is a quotient So if you like names, I think it is called the calcine algebra That is a quotient of the algebra of bounded operator By the algebra by the ideal of compact operator and this is a proper ideal because now my space has infinite dimension Calcine algebra So again if you take an element t it has a spectrum Sigma e of t it is a spectrum not the spectrum of t But the spectrum of the image of t in this quotient ring Okay, so this is Called the essential spectrum When you take a quotient, it's easier to be invertible here than to be invertible in the large ring Okay, so this is a subset Of the full segment And you have also a new spectral radius the essential spectral risk so an operator Which is Invertible Modulo the compact operators t is invertible in the quotient algebra It is called A Fredholm operator Maybe you know this terminology a succulent you just standard terminology in functional analysis So you can also define if you don't like quotient rings You can define sigma e of t as being the set of complex numbers lambda Such that lambda minus t is not a Fredholm operator and uh What is really fascinating to me, which I still don't understand is that actually You can so you have here you have an inclusion And this inclusion may be proper in some cases it is not in some cases it is and when it is proper it means that Somewhere You have the spectrum The essential spectrum And you have a larger circle so when I'm not saying it's not when the inclusion is proper is that when the essential spectral radius is smaller than the spectral radius And you have eigenvalues which are outside Of the essential disc here. Oh, sorry. My picture is not this is a true spectral radius and here's have the essential spectral radius And it's really easy to describe What are the spectral values which are out of the essential disc? This is a just an abstract phenomenon from functional analysis that if you have lambda Which is in the spectrum But the modulus of lambda is larger than the essential spectral radius Then actually it is an eigenvalue this is Spectral values are difficult to handle in higher dimension in an infinite dimension because they are not eigenvalues But actually these are Okay So what what do you mean precisely that there exists F and G Which are proper which are closed subsets subspaces Of my bina space F is finite dimension And E is just the sum of F plus G And Both places are stable Under my operator The spectrum So now the action of t on F. It is just a finite dimensional operator Okay, the spectrum of t On F. It's just lambda And the spectrum Of t restricted to G. It is just the spectrum of t Deprived So this is why I draw this picture in this way You have this eigenvalues here And in fact you have countably many Also is the space is separable and they will accumulate towards this circle So it's really easy to describe when you have the luck to find an operator for which The essential spectral radius is smaller than the spectral radius. You very well know what happens in this disk Okay, and this is just abstract functional analysis Yes, because you You see this property tells you that lambda is an isolated point on the spectrum Actually, what happens is that The true statement To these tools is lambda Has to be in the same Connected component of the complement of this essential spectrum as infinity. It has to be in the infinite component Just full statement but uh This is a very nice looking statement, but Practically the only way of checking this is checking this So the picture is not this. The picture is that even if you have an eigenvalue here, it will equidistribute to this So now Let me go back To the proof of bourguin's theorem about Absolutely continuous stationary measures So what we did with Yves Benoit was that we read his paper And we understood at some point that there are some formulae And actually this is what he was doing even if he was not stating it explicitly. So This is somehow a very weak talk because I'm explaining you what I understood in a proof of bourguin's It's still kind of an achievement, but So let me state a proposition Assume so I'm I'm back to the notation. I'm back to probability measures on sl2 and stationary measures on p1, etc so I need just to introduce The space on which I want to do functionalities. So I will fix row. This will be the Lebesgue measure On p1. So of course it's cheating you because there are several choices which are possible, but the so2 Which is so2 Invalid so now I'm not cheating anybody. So unique answer. This is just you you parametrize p1 by the angle between lines, okay And uh, and I will E will be called h because it will be a Hilbert space. It is a l2 of p1 for the measure So here is a proposition which is proved in the paper of bourguin that assume That my operator p always the same operator the operator of convolution by the measure has essential spectral radius smaller than one in eight So it is a you see I will assume mu. I will now I will always assume you to be compactly supportive, but you see because My group d my group sl2 it acts on this Hilbert space by composition of the functions But it does not it does not preserve the measure. Okay, so it acts about bounded operators So if you want to define p and be sure that it is really a bounded operator of this space You have to be careful. You have to make some assumption on the support of the measure You can assume less but you are sure that if it is compactly supported then you get a bounded operator on this space Okay, and uh, but now I have my bounded operator and I assume that it's essential spectral radius is smaller than one then Mu is absolutely continuous to row and with density Inel 2 that is I I will build if I know this I will build my stationary measure as an element of this space and uh The proof is easy Because what happens is that you consider your operator p So what you know is that p Preserves the constant function This is an invariant vector for all the group so, uh, you know That the one belongs to the spectrum of t p sorry But now the spectrum of of an operator is the same as the spectrum of its adjoint operator So you also know that one belongs to the spectrum of the adjoint operator and uh The same holds for the essential spectrum So the essential spectrum of an operator is the same As the essential spectrum Of its adjoint operator Against the elementary functional functionalities So you know that it is smaller than one So now I know that I have Uh, a spectral value which for the adjoint operator I have a spectral value with modulus larger Than the essential spectrum. So I know that it is an eigenvector by the property proposition here. So there exists some function f in eight Which is invariant by the adjoint operator And this is exactly saying in other words saying this Is exactly saying that the measure f rho the measure which is absolutely continuous to rho with the rest with density f Is p invariant. So by uniqueness, I know that my measure lives in h slightly magical because you don't see anything uh, so you have I was a little quick here because you don't know that it is a positive measure But it's really easy to prove that this sign measure is invariant. It's absolute value is also invariant Because the operator is positive Okay, so this is how how he proves it. So now of course There is work still to be done because you have to produce An example of the situation the key proposition which I won't prove The paper barric bourguin Is that there exists? Final case you brought in SL2R such that p has essential spectral essential spectral radius smaller than 1 uh Actually, what we did we see Benoit is that we understood This part this this part of the proof and then we were able to to uh Prove the same statement in SLD But it's actually easier in SLD for D at least three because you use the facts like the anyway, it's easier So this is the most difficult case actually and we are not able to do it. Okay, but Because sorry, so it's difficult. So the point is that so now So what is the strategy of bourguin? So strategy is based on some product estimates and additive combinatorics but uh One part of the proof is to change actually the operator to replace it by another operator I don't know how much time I have left or Ten minutes. Oh I've been much better than usually so, uh You see to define my operator p Uh, I Define the action of an element on a function so You have to be careful when groups act on spaces of functions. You will have to be put inverses at some places. So Later, right is this way. This is how I let the group act on the function on the space. Okay for a function f Which was a function in l2 of P1 which I denoted by h This is how and then I took another h to define my operator P and you see this is not a very good action because It is not an action by unitary automorphisms. Okay, so, uh usually in group theory When you consider this space what you do is that you act So this formula This is a not a good formula the good formula Is to act but to here you add a term Which is the radon which is the derivative here It is a derivative with respect to the so2 invariant metric This is a positive number because I am in sl2 and sl2 preserves orientation And just by the change of variable formula now this operator. This is a unitary operator on the space and uh What happens is that uh in the construction of bourguin The support of this finitely supported measure here is Very close to the identity. It spans a dense subgroup actually, but it is very close to the identity So this derivative is very close to one So if you define a new operator if you define qf to be the operator So you have yes Now I define the operator but by using the twisted action Which is an action by unitary automorphism of this l2 space here You have a much better operator which looks exactly as the operator that I was using in the case of compact groups Because it is another age of isometries It's much a much better operator But what happens is that when the support Of mu goes to the identity The norm the operator norm of p minus q goes to zero because This term here it is a it is a function which which goes to one So if you are able to prove that q has an essential spectral radius Uniformly you have if you have a family what do bourguin proves is that he constructs a sequence. This is Why I told you That's in some way his construction is explicit But still it's difficult to follow the constant but he because he constructs a sequence of measures Which who support converge to zero to the trivial element And what he proved is that the limit As n goes to infinity of the operator's qn of the spectral ready Is smaller than one essential spectral ready. So he knows that the same rolls for p just by Continuity And so Actually in his constructions So we don't have any counter examples. We don't know if you take measures on On sl2, we don't know how to compute this thing in general his construction is of course, it's a very particular example that is his measures These are measures supported on points with matrices with coefficient in q And he the co the denominators will arise with respect with n but slowly, etc and But in general, you don't know how to compute so So I will just finish by asking questions which are related to the to this proof that is The question is a An optimistic question is there a does there exist epsilon such that for any new for any say I have to be careful as the question is the answer You have to be careful as the answer is purely new. Okay, such that for any gh In sl2 r Which span The risk hidden subgroup that is the risk hidden is just not contained in any proper algebraic group Just explain if you set mu to be One-fourth, so this is some kind It is a version of the question by sarnac about sl2 the question of sarnac for S3 is a kind of version in sl2 If I define q as above by this formula The essential spectral radius of q is smaller than 1 minus epsilon. I don't know if this is true Less optimistic would be same question but for g and h close to identity and even less optimistic G You take it to be this matter You take g to be this matrix So this is exactly the same question as I gave you explicit elements in so3. Okay I somehow I put the same elements here. So this would be to take g this guy and h Would be this guy. This is just These two people are conjugate actually and You take these two people and You look so for t small So here these people it's very easy. This is just g You have the canonical basis and g is diagonalizable and to build h you rotate and And here you see So the you don't have a common eigen basis You don't have a really easy to check that they span the risk hidden subgroup. Okay, and for small t You look to this measure and What is a it's clear That the spectral radius will go to one because the measure is closer and closer to identity. So necessarily the spectral radius Goes to one because the question is how does it go to one? So this it is so you have it is a self-adjoint operator Because I became for I act by isometric and my major is symmetric So it is a self-adjoint operator in your Hilbert space. So you have some spectrum And it feels it it will feel the interval at least it will go to one because the element go to identity But it's not clear how it feels it Do you have more and more eigen values or do you have really? Fat parts of the spectrum I really don't know I have no idea. Okay, and if you know that actually what it does what appears The construction by bourguin is the example where this is what happens The fat part stays far away from one and you have more and more eigen values So, you know that the essential spectral radius stays far away from one in particular You know that the stationary measure is absolutely continuous. Okay, so same question. So either you ask the spectral question or just the Question about stationary measure is it true that the stationary measure of this run of work is absolutely continuous Thank you very much Yes, with fire. Yes, the of course if you mean a densely supported singular measure. Yeah. Yeah, there is an example it is due to Kaiman of each and I don't remember the other guy And the idea is just you you put you take measure of this form and But you put more and more weight and you take two just two guys You put a measure on this form You take a very large g and a very small h And you take this measure And you see what will happen is that as a goes to one Yes, and you have you you use a little rapier you formula To relate dimension with the exponents And you you you have a formula for the liapunov exponents So you know that when a goes to one In the end what you get is a liapunov exponent of just g only And you look at what he tells you with respect to dimension You know that the dimension will be smaller than one Yes, yes Yes, but Is it easier to step but much difficult to much more difficult to prove Yes What what Burga actually does is that he You see actually this technique is very Is very Suppled so so you you can prove this do the same thing in a uh, a sable of spaces So you you you do you you see because The argument is that again when you take measure which are very close to the identity The difference between the unitary operator and the operator For the sable f norm is very small So actually you prove a spectral graph in a sable space with arbitrary high Regularity, so you know that you are absolutely continuous with arbitrary smooth density Yes Because when you live in an algebraic subgroup the situation is very different just that I know how to handle this just These are very degenerate situations If you if you take your own homework as a diagonal group If It would be stupid to forget that we live in the diagonal group. It's just what I mean, okay Because what happens in sl3? So you want me to tell you to tell you about our results. That's very nice from you and What happens in sl3 is that uh In sl3 what you do is that uh to be the you follow the same strategy And the problem is here to build an example of a measure that in sl3 you have s o3 And uh on s o3 you have measures which have a spectral gap There are examples what finally you put in measures with a spectral gap So now you take one and you add a small thing So to make it a risky dance, which s o3 is a is a is a maximal algebraic subgroup of uh Sl3 so as soon as you add anything you are risky dance But still since what you had is very small more or less the spectral radius looks like the spectral radius of this So you have a central spectral radius smaller than one Okay, so but the point in sl2 so that you can do in any semi simple a simple d group except sl2 because in any simple d group the Compacts there are non-abillion compact simple groups But in sl2 you only have s o2 and s o2 is not compact You don't have spectral gaps in s o2 as I explained in the beginning of the lecture