 Okay, thank you for sticking around and thank you to the organizers for inviting me here It's only the second time in my life that I'm here in Montevideo. I should come more often I hope to come more often if I'm still invited after this talk, we'll see Okay, so I feel a bit of an imposter here. I'm not really a dynamicist What I really do is geometric measure theory fractal geometry and from there I moved to many different areas Include an erotic theory So I'm going to talk about work which is a couple of years old But I think it's the closest to the theme of this conference because Well, there is geometry is fractal geometry. So but still geometry There is dynamics, which is a doubly map and there is a semi-group of the natural numbers. So I guess it can fit into the You didn't specify what kind of geometry what kind of dynamics what kind of group, sorry Okay, so So I'm going to talk about some problems that were posed by a healer Fustenberg in the in the 60s Fustenberg He promoted the following principle. Although he never stated the principle in this way So this is my reinterpretation of his principle He suggested that expansions in basis two and three Principle this is a principle. It's very very big as a principle should be expansions in basis two and three Have no common structure and I'm going to use two and three for the rest of the talk, but You can replace in everything. I will say you can replace two and three by P and Q Which are not powers of the same integer So true can be any P and three can be any Q What should happen is that? Look P over look Q should be rational, but I'm going to use two and three Okay, so As I say, this is very big. So what what could this possibly mean? So let me give an example of a more concrete still a bit big but more concrete Implementation of this principle if you look at Power of two just the natural power of two two to the end for some large N Then the ternary expansion of this number should be pretty random Because this number has a very very special expansion in base two one and all zeros And there should be no common structure with the expansions in base three So it would be very surprising if the ternary expansion of this number Is also special this shouldn't happen So I'm going to write it this way. So this means ternary expansion of a number should Look random to some degree Okay, nothing is known about this For example, it's not even known if for all sufficiently large and the ternary expansion of two to the end Has the digit one for example Even once so this is not known That's much so what I just said just Two to the end having the digit one in the ternary expansion is much weaker than say that it behaves randomly And that's completely open. Okay, so there is no dynamics here. So I'm going to move to So sort of a bit more dynamical version of the principle Still very vague Okay, so post now we have Invariant sets or invariant measures under multiplication by two and three in the circle So let a or maybe mu be invariant Under t2 t2 is just the doubling map Let be or new this means a sets or a measure be invariant under multiplication by three on the circle Then the principle says that a and b and also the measures mu and mu Should have no common structure So being invariant are the doubling map means that if you shift the binary expansion of a point in a you're still in a Being a matter under mu means that the statistics of the Distrubes so if mu is time shooting variant means that the statistics of the binary expansions of the points of points in the interval The new statistics are invariant if they shift the binary expansion So this is related to this principle. This is a bit more dynamical and a little bit more concrete but still very vague because What was the question is what does it mean for two sets or two measures to have no common structure? so Well, let me measure the whole interval the whole circle is invariant under times two and ten three And the big measure is invariant other times two and times three But everything has no structure because it's everything so the whole circle has no structure in the circle and of course There are also Finite sets of rational points or measure supported on finding sets of rational points Which are also invariant under multiplication by two and by three, but these finite sets You should think that these finite sets also have no structure So other than these trivial examples of jointly invariant sets or measures this should happen and Well before you can ask more delicate questions about what does it mean to have no common structure the first thing you have to Believe is that a set on itself or a measure on itself have a lot of common structure because it is the same set so I hope everybody will agree that a and a Have common structure Well, maybe assuming that a is not trivial in the sense that it's neither the whole circle or finite set I'm the same for a measure on itself Okay, if you believe this principle and you also believe that you and yourself have a lot of common structure Then you should believe that there should be no sets which are invariant on the multiplication by two and by three by the way The sets are always closed. Otherwise. They are trivial contra examples to everything. I'm going to say measures, but a probability measures Okay, so if you believe this principle and you also believe this then you should believe that there are no sets Which are jointly invariant under multiplication by two and by three and likewise for measures And as I guess most of you know, this leads us to a very famous theorem and a very famous conjecture in erotic theory So for sets, this is a pioneering result of first number. I think in 1967 there is okay, let's write it this way if a is Compact and times two and times three invariant Then a is finite or everything. So the only jointly invariant sets under times two and three are the trivial ones So this was historically the first result and I would say this was the paper that started I mean this paper started many branches of modern erotic theory and one of the branches that started is the study of dynamical rigidity So this is a rigidity result. There are extremely many sets which are in bar and are times two and extremely many Like an unbelievably large zoo of sets which are in bar and other times two or times three But the only sets which are invariant under both are the trigger ones So quite amazing So it is a quite non-trivial result But it is it is quite easy if the house of dimension of a zero in fact So the proof goes by considering two cases if a minus a there is metric difference of a with itself What they want to say is everything What they want to say well anyway This is the what they want to say Well, I will just say this if the other dimension of a zero then a minus a also has a mention zero and the result in that case Yeah, it's easy to see. Sorry, let's go again. So if a is not finite And it's in battle times two and ten three then it's easy to see that a minus a is everything But if I has dimension zero then you can see that a minus a also has a mention zero so it cannot be everything So in the dimension zero case this theorem is not difficult. So I just gave you the proof in fact You have to fill in the gaps, but it's just exercises Okay, and I point this because of what happens for measures. So it's still an open problem So the most famous times frequency conjecture still open today is that the same holds for measures Well measures you have to be a little bit more careful because you can take a combination combinations of Lebesgue measure and atomic measures, but this should be the only measures. Let's say Yeah, it should be the only measures which are invariant under times two and times three So let's write it this way if mu is Invariant Under times two and three and Maybe times two are godic something like this Then mu is atomic or lebesgue. Okay. So another famous theorem in a good theory is the rule of Johnson theorem The size of the conjecture is true if mu has positive entropy and Positive entropy is the same as positive dimension. So I prefer to write it instead of dimension to Emphasize the point that I just made So the case which is really well the case which is open and basically nothing is known about for measures It's somehow the case which is easy for sets. So I'm going to talk about other results And let me say that in some sense. So what I'm saying is heuristically. It's not a formal statement But in some sense the results that I'm going to present now Improve dramatically in the case of positive dimension. I would say positive dimension because there are more about sets So in some sense in some sense in the positive entropy case positive dimension positive entropy case We know everything that there is to be known but in the zero dimension case zero entropy case. We know nothing Okay, so Okay, so this gives This gives the answers well answer and We don't know About this principle if you look at the most basic way of interpreting common structure, which is just being equal So if you want to go move beyond equal then We can use some sort of geometry. So these sets are counter sets So the ones that are not trivial. They are counter sets They are uncountable sets if they are not not finite and not the whole circle Then they are uncountable, but they have zero a big measure. In fact, they have dimensionless on one they could have dimension zero, but They are uncountable and they have house or dimensionless on one so they are fractal sets and being invariant under the doubling map or the tripling map is some sort of self similarity or Self-similarity so they are counter sets. So for example the middle third counter set the first fractal everyone encounters in their lives is In variant other times three. So they are these sort of sets. So they are fractals. They are fractal sets. So So in order to look at what common structure could possibly mean to come up with even more concrete Versions of this principle one can use geometry. So I guess measure full measure Substance of the audience is very familiar with house of dimension But I you see I mean very interesting a set of zero measure as well So just in case let me say quickly a few words about house of dimension So how's our dimension is a number which is assigned To all subset of let's say RD and takes values between zero and so measurable no measurable It doesn't matter every set has the house of dimension and it has many good properties. For example, it is country of the stable. It is a So the hazard dimension of a comfortable union is the supreme of the house of dimensions And it gives the right value to sets for which you are ready now the dimension Many fold of dimension cake a hazard dimension cake. So if you've never seen hazard dimension This may be all you need to know and just so our completeness may be let me give the definition because I'm going to use it a little bit later So hazard dimension of a let me give this definition first we try to cover Okay, so So you can take a nap so almost everyone got taken up and then I will see who doesn't know the definition of hazard dimension But you're going to take just one minute. So how's our dimension is based on efficient coverings of the set? So I covered the set but both both of radii are I And I look at the infimum of these sums for there is an exponent s And then I want to find the critical exponent s and that is the house of dimension And maybe I Look now at the infimum of the s for which these infimum is zero and that is the house of dimension They are doing right. I probably didn't do it right something like this. Okay, so now let me say that Two subsets of R. Well, this could be an Rd, but from no one almost everything will be in R So or R2 sometimes so let's say that two subsets of R are resonant So this is definition if The dimension House of dimension of the arithmetic sum So this just means all of the sums a plus b where a is in a and b is in b Just the arithmetic sum is equal to the minimum between the dimension of a plus the dimension of b and One in this case we say that a and b resonate they are resonant and otherwise Well, this number is always an upper bound Okay, so there's going to be a white lie for most of the talk So what I'm saying right now is actually not true, but let's imagine that is true So what I'm going to say now is not true, but let's imagine that is true. This is always an upper bound for this Actually, that is true Where I'm going to say in one minute is not true But but don't worry about this either, you know why why it's not true in which case it doesn't matter that I'm lying to you Or you don't know which case believe me that is better than that. I'm lying to you. Okay, so this is always an upper bound for this and Well, you expect to have an equality. There should be a big conspiracy for The equality to fail and I have lied there. So it's not true that this is always well anyway so Generically, there is an equality and there are many ways to make this precise one way is mass and prediction theorem That tells you that if you scale one of the sets by a random real number, then you have equality for example Asterisk because that's not true, but let's imagine that is true So if there is an inequality Something has to be provoking that inequality Basically a and b should look very They should match up the structure of a and b should match up at arbitrary least most scales something like this if there is if there is no Equality so otherwise otherwise means that there is a strict inequality because the right-hand side is always an upper bound So if their dimension of their emetic sum is strictly less than the maximum of the dimensions should be Minimum of the dimensions and one in this case We say that they dissonate We say that a and b are dissonant Okay, and if this happens so In this case if they are dissonant then this really means that a and b do have Common structure at small scale There really has to be a reason for this inequality to occur in general there is equality Inequality is the exception and there should be a reason provoking it Okay, so here is another conjecture of Furstenberg Which also originated from the 60s. So this one is not in print, but Well, you bought Paris was a student of Furstenberg not in the 60s later, but So Furstenberg gave this problem to you while one who was doing his PhD and Mike O'Hman was a master's student of Furstenberg And Furstenberg also gave this problem to him. So I guess yes Yeah, this is generic The dissonant case is the exceptional one. Yeah, so for example Given any two sets a and b for almost all real numbers are a and the scaling of b by r Sorry every yeah, so the definition is wrong But if something is wrong, don't ask a question. Just tell me it's wrong. Of course, it's wrong Okay So this was not the white lie Yeah, but the way so I try to make the talk understandable So if you don't do everyone so if you don't understand, please stop me So it's not a problem to not cover everything. I want you to say this always happens actually so So I want you to understand whatever I am able to say So if you don't understand something for example, because it's strong just ask or let me know, okay Okay, so the conjecture is that if a and b are Times two and times three variant Then they dissonant and again This is in accordance with the general principle because if they resonate they have a lot of common structure And it shouldn't happen by according to this principle So it is one possible way among many others to make this principle concrete Okay, so this was the conjecture and I proved it together with my Kochman. Well together we well Paris removed a special case important special case and then Together with my Kochman will prove the general case So we knew how to prove it in 2008 the preprint came out in 2009 and it was published in 2012 Okay, but here You see that I only drew the arrow going backwards So I said if they are resonant then well, I said it wrong I know I'm saying it right if they are resonant then they have a lot of common structure But this is definitely not an if and only if it's as far as possible. So it's really very far from if and only if So for example most sets Disonate with themselves. So in general the dimension of a palace a is twice the dimension of a or one So the one is here because these are subset of R So the dimension of the sunset cannot be more than one So if the sum of the dimensions is more than one you can you cannot expect to have the sum of the dimensions because it's mounted by one Okay, so for even for a typical set a even for invariance sets So even for times two invariance sets and three invariance sets It can very well happen that the dimension of a plus a is So it is this tries the dimension of a or one So this is definitely not a characterization or not having common structure because it's not even able to distinguish a set on itself So it is a necessary or Condition for not having common structure, but it's very far from sufficient. So this cannot be the end of the story So let me give another definition So I'm not going to say anything else about this Maybe I'm going to mention one word that comes up in the book I'm going to say now CP process. Okay. I said it. No, I'm going to move on Maybe I will say a little bit more about CP process later. So another definition So again, we have two counter sets. I know I'm going to say that they are transversal If no, I'm going to look at the intersections. It's definitely not the sunset. I'm going to look at the intersections So I'm going to say they are transversal if the dimension of the intersection is at most the maximum between zero It cannot be negative on the sum of the dimensions minus one. Okay, again, this is the generic behavior to Sort of random typical whatever you want to say it Contour sets are transversal. So where does this number come from? Where does the number come from? Well, it is what you expect from counting arguments or it is what happened for subspaces So this is saying that the co-dimensions at the co-dimensions at this is what happens for planes. So for from linear algebra And it's also what you expect if you try to You imagine that the dimension comes from covers by both of the same size for example, and you look at some random Well, anyway, so there are many ways to understand why this is the natural number and I'm writing a number bound here and not equality because in general this will be bounded sets Or even if they are not bounded they are they are counter sets or they have gaps They have holes so one of them could be contained in a gap of the other for example, and then the intersection will be empty So this can very well happen. So It's not true that generically you have equality here Or at least you have to define generically in a more careful way if you want to have equality generically If you do if you're not careful, you cannot hope to have equality in general, but you do have this inequality Generically in particular here comes another white light for example One example of why this happens generically. So there is something called March Trans into intersection theorem And he says that for any a and b a And a typical image of be are transversal. So for almost all pairs of scale in a translation This country said I'll transfer itself. Okay, so here comes another conjecture of Furstenberg This one does appear in a paper from 1969 or I guess the paper appeared in 1970 So it's a conference proceedings the conference happened in 1969 the paper up here in 1970 Okay, so the conjecture is that if a is times two invariant And be is times three invariant So there is a weak version and a strong version. So the weak version says that a and b are Transversal and the stronger version says that a and any affine image of b are still transversal For all are T because for almost all is true because it's true for any sense without any invariance requirements So the point here is that he's asking these for all RT and in particular the most important that RT is r equals one and t equals zero Of course, R should not be zero. Well, it could be zero because then it's trivial could be zero. Yes Well, it's a principle so again if this doesn't happen That means that a and b should look very much alike at many scales at many small scales This is the way you should think about it. How's our topology has is very bad for hazard dimension It's not the right apology to consider Well, yeah Well, I'm sure you can say something but maybe generic from that point of view generally from that point of view this doesn't happen so Things are weird if you look at dimension. So yeah, so this is a branch I mean, there are many people doing what happens for hazard dimension for generic in that sense and things are weird. It's not So it's maybe more measure theoretically says something like this Or if you take random sets if a and b are generating in some random way Then you definitely have inequality and actually you have equality unless one is contained in a gap of the other or something like this okay, so Heuristically this conjecture and this is a strong version the strong version is stronger than this conjecture and Let me draw a picture to explain why So Let's look at the product set a times b So here we have a for example a could be the middle third count or set so from no one I'm going to just look at one example just for concreteness. So a will be the middle thirds counter set and B will be the middle one quarter counter set So it's the same contraction, but I have two intervals of length one quarter and then I iterate So this is going to be B So a is time string variant B's time times for invariant And you can easily couple times to invariant set out of B as well if you want Okay, and let's look at the product set then what is the arithmetic some arithmetic some is just looking at the product So the product is first generation of the product something like this. There are four rectangles So the sunset is just the projection of the product in this direction So if the sunset is large Then heuristically what would happen is that most fibers of this map So in this case the sum of the dimensions is more than one So the dimension of the middle first country set is look two over look three Which is more than one half and the dimension of B is one half So the dimension is the sum of the dimensions is more than one in this case so according to the previous theorem the projection that the sunset which is a projection in this direction has cause of dimension one By the way, it's an open problem. Whether the some set of these a and these B has positively been measured If you want to torture yourself for a little while then think about this problem Okay, so the dimension the projection has yes Yes, a is a middle third counter set B is a middle one quarter counter set. Is it true that a plus B has positive Lebesgue measure? And if you're if you're very ambitious, you can think whether the sunset has none of the interior Okay, but we know that it has dimension one and Heuristically this means that if you look at the fibers of the map and the fibers of the map are Intersection of intersection of a and translates of B So these are the fibers. So let me pick another color for the fibers So the fibers of this map should be Generically not too large. They should be small Right if all the fibers are very large that means that there will be a drop It has to be compensated in the projection in some way So how sort of dimension doesn't work this way at least not so easily so everything I'm saying is not true But morally it is true that if the projection is large many fibers most fibers should be small and vice versa The second conjecture is saying that all fibers are small not most not almost all but all fibers are small So in this sense heuristically the second conjecture is stronger And actually it's not too hard to prove that it is really stronger. So this conjecture implies the first conjecture. Sorry Was there a question? Okay, so the second the second conjecture is stronger. So this implies The previous conjecture and another way of thinking about this is that this is a better way of characterizing common structure For example a and a in general Could be dissonant. So this is not enough to distinguish a from a but a and a are never Transverse out if the dimension of a is positive. So in dimension zero, this is absolutely nothing if the dimension of a or b is zero Then there is automatics and versatility. This is nothing in the case of dimension zero But if the dimension of a is not zero then certainly a is not transversal to itself. So with this notion of Common structure transversality Well, this is a much finer way of Finding common structure between a and b than looking at the sunset So this was a motivation for the conjecture. Actually first and better In the paper talks about transversality of semi groups. So this is the word where semi groups appear in this talk So you can look at the semi group generated by times two and the semi group generated by times three and four sembergs defines this Traversality between semi groups exactly in this way Saying that invariance sets for one semi group and invariance sets for the other semi group are transversal Then the semi groups are transversal. Okay. Well, the conjecture is true This one. This was a quite a I would say a Daring conjecture. So it was not clear what this should happen actually It's a very strong statement. I would say it's really much more surprising that it's true than the one about some sense The one of the sunset is easy to believe So I think people believed that the conjecture the sunset conjecture was true before anything was proved But they were not so sure about the intersection one, but the intersection one is also true Okay, and so I proved it About two years ago. I know so Men go from the University of Aulu Also proved it. So I'm going to say a couple of words about the proofs So the intersection conjecture Okay, so the proofs are unbelievably different. So it's it's amazing how different they are and so in fact So men at the time was at the University of Aulu in Finland now He's again at the University of Aulu in Finland in between who was in Jerusalem And I have a colleague Ville Osamala the University of Aulu. So I was visiting Ville when I was working on this But I didn't mention that so I thought I could prove the conjecture at that time But he wasn't written down. So I didn't mention anything But he told me that Mengbu thinks that he can prove first numbers conjecture then I had to say me too Okay Yeah, and then may explain the proof to me and it was unbelievably different. That's why there are two papers I mean if we had had the same idea then there would have been one paper, but It wasn't didn't make any sense to write one paper because the proofs are totally different So you see that this is a very concrete sort of one dimensional problem. So it's an easy problem, right? I mean, it's a everything is linear. So the easiest counter sense you could possibly imagine. So it's very concrete very Okay, but Makes proof uses very abstract erotic theory. So he uses dynamics on a space of measures Actually, it's amazing because that he uses the idea that first and bear so in the original paper where he posed the conjecture First and bear proof some partial results in particular Although it's not reading the paper first number put that the conjecture holds if the sum of the dimensions of a and b is at most one half and he did this by introducing dynamical system, which is called the cb process Which is a dynamical system where the face face is a space of measures actually even even bigger, but a Very large face face and the dynamics is basically zoom in So you live in the space of measures and the action of the dynamics is zoom in magnify into some part of the measure and zoom in So that's the dynamics of the cb process So it's a very powerful tool and actually is what we use with my Kochman to prove the sunset conjecture So it's a tool that is all the first number himself developed it than we used it together with my Kochman It's actually what Meng Wu used to prove His intersection conjecture so he He has to construct some cb processes and he does this in the same way So in the same way that you prove that if you have a compact set and an invariant map There is an invariant measure you can construct the cb process The problem when you do this is that you're just averaging So you have no idea of what are the properties of the measure that you obtained So it was known to my Kochman and me and probably even to first and bear That if this dynamical system that you obtain abstractly by averaging if you were very lucky and this dynamical system was which mixing Then everything would be fine Then the method of first and bear Could be used to show that the conjecture holds But it doesn't have to week mixing because this is a complete the abstract measure which is contracted by averaging so Sort of the new completely brilliant idea of Meng Wu was so I'm talking about his proof more than mine, but anyway, so So the brilliant idea was to use tonight's factor theorem. So this dynamical system that he constructs is easy to see how he has positive entropy So there is a factor Bernoulli factor of the same entropy See nice factor theorem classical theorem, but it's all very very abstract. So there is this very large phase space There is this construction of an invariant measure in this very large space space with some properties Which is done by a average and averaging again, and then there is this measure theoretical statement about very generous extremely general theorem about Having but see nice factor theorem. You have a Bernoulli factor and by some magic He's able to sort of pull pull back information. So you have a Bernoulli factor of the same entropy And so the very brilliant insight that Meng had was to realize that having so because the factor has the same entropy the fibers Have zero entropy and even though this factor is only in it's a measure of theoretical factor It doesn't have to be continuous at all is still enough to sort of pull back geometric information And somehow we really a little bit more work. So still use the idea That would work if the original system was Bernoulli Somehow that's his idea. Okay. So what I did was totally different totally different. So my proof is Much more concrete on one dimension as the problem. So I don't have to go to this very general much more aspect setting okay, so Okay, as usual, I don't have time but let me say just I will start talking about the proof and when my time is up Please let me know one. I'll finish It's not funny Okay, maybe I will make it Paul so I will write down some ingredients in the proof and I will ask you Which one do you want me to tell something about? Okay, so there is a little bit of ergodic theory not a match but there is a little bit of ergodic theory that Appears in the form of some properties of Co-cycles so there is a co-cycle. So ergodic theory of co-cycles over uniquely ergodic transformations Then there is a multi fractional analysis In particular there is something called LQ spectrum or a Q dimension of a measure It is a very very classical object in with the fractional analysis that is very important in the proof And then there is lots of additive combinatorics and then So before my paper there was a very significant paper of my cockman on some similar measures So the thing is a bit different on the surface is maybe totally different, but under the surface is quite closely related And somehow the the overall structure of the proof very much follows Oghman's Approach for studying self-similar measures, but all the details in all the steps are totally different somehow the very general structure is a Following my cockman's ideas Okay, so maybe there is not going to be a vote. I'm just going to Start talking about the proof and then Well, my time is up. I will stop. Okay, so So I like projections I like orthogonal projections a lot So if I can reinterpret the problem in terms of projections, maybe I can't do something and if I can't then I can't So in this case even though the problem is about so the sunset you see the sunset was about projections What's exactly about projections the sunset is a projection for me. It's not something. It's a projection Then I was able to do something. Okay, so this is an intersection problem So it doesn't see doesn't seem to be about projections so much So the first step for me and this is already totally different from what men did was to Convert these into a plan of our projections. So let me tell you how to do it So on a you have a measure extremely natural measure the measure mu It's just the counter the big measure on the middle of Thursday's counter set. Let's call it new one. Maybe you want It's a measure of maximal entropy house of measure So it's not natural measure you put weight one half one half and then you keep going in the next intervals And the same here on the set be the middle one quarter set. So this is just an example, but it's basically as difficult as a general theorem. So We only need to understand this example. So on B. You have another measure. Let's call it new two Again, it's a measure of maximal entropy the house of measure The most natural measure you could put on the set Be and then you have the product measure you know, it's a product measure to be here at stairs And then we can project the measure So Okay, something that happens already for the sunset problem I also happens here is that even if you're interested about one prediction in particular You have to study all the projections at once and this is actually an insight that comes on Google Meraida's work Well partly with your cause on the party's conjecture and also Google's work on some of country sets So I have to say that somehow everything started with Google in some sense. So my first result with you at Paris so the main idea was a So Google has a paper has a result which he never wrote that wrote down as usual but He explained to me many many years ago about sums of learning our counter sets and for for him nonlinearity was very essential and What you've all and I realized was basically that you can replace nonlinearity by something else So everything started with Google in some sense. Okay Anyway, and but the idea that you have to look at all the projections at once Even if you only care about one projection comes from Google. Okay, so we have the product measure upstairs mu Maybe let's call me a theta It's just a projection of mu under projection with angle theta So this is angle theta and here we have new theta So it's just a push down of the measure push down of the measure means that the measure of an interval Downstairs is the measure of these street upstairs. So this is me with it Okay, and then let me define the following number for any measure. So the infinity of a measure mu is The largest possible first one exponent of the measure. So it's the supremum So it's a very natural number is the supremum of all s Such that the measure of a ball of radius r is controlled by r to the s And the constant that could depend on s For example, the big measure on r has the infinity dimension 1 Trivially in this case you actually have a quality not just operating quality But this is a very natural condition first one condition that appears everywhere that you're working with fractal measures okay, so Why I'm calling it the infinity is going to become apparent in a minute. So Here is a claim which is easy to prove. Maybe I will even prove it. It is enough to show so to prove the conjecture so It is enough to show That the d infinity dimension of all these projected measures is one Well in this case one because in this case the projection has dimension one So we're in the regime where the sum of the dimensions is more than one if the sum of the dimensions was less than one Then here instead of one we will have the sum of the dimensions So one so here what really goes here is the minimum between the sum of the dimensions and one in this example It is one Okay, so this is very easy. So maybe I will explain it So let me start over with the picture. So it's going to be the same picture of the scatter So what it did here was exactly what I want you to do which is to have a version of the conjecture Involving projections and dimensions. So this is why I like projections on dimensions. Okay, so Here we have this product measure Here we have one of these projections Here is a here is b and what do we want to do we want to show that the intersection of a and b and more generally So when I say that the conjecture is true the strong version of the conjecture is true the strong version You cannot prove the weak version without the strong version So to prove that a and b are traversal you have to prove that a and any affine image of me are traversal You cannot do one without the other Okay, so the intersection of a and an affine image of b is exactly the same as looking at a linear fiber of the product of a and b So here I mean so the intersection of the product set with this line is exactly the intersection of a and an affine image of b by definition okay, so I'm assuming that I Was able to prove this and Okay, I want to estimate the hazard dimension of the intersection of a times b and this line from above so I just need to show that I Have good covers. That's enough. So then I cover this intersection By both of some radius are in some optimal way so this is a well an efficient covering of the intersection by both of radius are and then all I have to note is that if I project so So if I take a strip that can contains so has more or less the same size as the radius of these balls So these are both of all of the same radius are well, then the measure downstairs of this segment is going to be How many balls I have times the mass of each ball? But the measure upstairs is extremely nice the measures mu 1 and mu 2 these are these are alphas regular measures So mu 1 of a segment here is are to log 2 over log 3 up to a constant If the segment is entering the set a a and the same for b with exponent one half instead of log 2 over log 3 So the measures mu 1 and mu 2 are extremely irregular extremely uniform extremely nice and so is the product measure So we know what is the measure of each of these balls? We know and they are all equal They are equal to r to the one half plus or two over look three each of these But then we know what is so we have an upper bound Which is what we need to have an upper bound for the measure of this this segment And then we have an upper bound for how many balls we have and believe me that the upper bound is exactly what you Want to have to have the conjecture So that's all So it's trivial. So there is nothing here. It's really trivial. Okay So this is the reduction to a problem about projections and maybe let me try to say something about this co-cycle and What reduction these are erotic theoretic argument achieves? So we want to prove this and in fact is enough to prove this Not for all theta but for an interval of theta once you have it for an in non-trivial interval of theta You can use as a similarity here as a similarity here to move this interval around and cover the whole line So it's enough to prove this for Theta in some interval Okay, unfortunately the infinity dimension is too difficult to handle. So instead of working with the infinity dimensions I work with the VQ with the DQ dimensions. So let me define this So this is the LQ spectrum that I mentioned here So let me define this. So this is a limit In general a limit, but in this case the limit actually exists So this is the limit as n goes to infinity of Snq of nu Okay, no, I need some logarithms. I guess A minus sign Okay limit of minus Snq of I missed the logarithm log of Sn q of nu divided by n And I have to tell you who is Snq of nu This is the sum over Diadic intervals of length 2 to the minus n nu of i to the q And here I have a q minus 1 and the denominator and now it's correct Okay, so I look at this moment sums which are quite natural and Then I look at the power law behavior of this moment sums and this q minus 1 in the denominator It's just a normalization factor that ensures that this is in 0 1 If nu is a measure on r if nu is a measure on rd is between 0 and d So this is a notion of dimension. So if the measure is More or less spread out. It's going to be large if the measure is very concentrated It's going to be small So it's a reasonable notion of dimension of a measure. It's not one dimension It's a family of dimensions parametrized by q And it's very very well known in mutifractor. So it's one of the most basic objects of study in mutifractor analysis Okay, and here is an exercise for you dq of nu tends to the infinity of nu When q tends to infinity So it is enough to show so a further reduction It is enough to show that the q of nu is no the q of mu theta the q of these projections Is one for every q for every theta This is what I really proved So why do I work with this? So at first sight this can look much more complicated than this But this is much better because there is some convexity. So this is closely related to lq norms So this is an lq norm in some sense. So it is an lq norm So there is convexity for q equal infinity. There is no convexity for q finite. There is convexity So this has much better properties even though if you've never seen it before it looks ugly It's wonderful. It's not ugly. Okay. I think I have just enough time to Say Where the erotic theory comes in? Okay, so it turns out that In a previous paper we developed peres and fedia nazaro We realized that these moment sums Are actually a co-cycle if we look at the so not a general measure, but At the projections that we have to look at so we have these projections these projections here mu theta And let's consider the following Sequence of functions of theta fn of theta. So q is fixed q is unfixed large number So it tends to infinity at the very end So q is unfixed large number and this is just the moment sum at level m. So here is the m the q is fixed And then I want to look at mu theta But I'm not going to use theta. Maybe let's call it t. I have to reparameterize. So here theta is an angle I don't want to have an angle. Maybe I want to have something to e to the tangent of Okay, uh, what do I want to do? T something like this Well, I don't know if this is correct, but I have to reparameterize So angles are not good to to get a co-cycle, but it's just a smoother parameterization It doesn't change anything so you can imagine that we have theta here. It doesn't really matter Okay, and then the proposition that it was actually for this particular case Proofed in the paper with nazarov And you will pay some myself Is that uh, maybe let's put the log here To get something sub additive instead of some multiplicative Then f m plus k of t is less or equal that f n of t plus f k r to the n of t And what is r? r is translation by angle log 2 over log 3 in the circle So at some point we have to use the log 3 log 2 over log 3 is irrational, right? Because otherwise it would be false for example if we have 2 and 2 if we have the same set here and here It's completely false So it is important that log 2 over log 3 is irrational and it is used exactly here r is an irrational rotation Maybe it's not log 2 over log 3. Maybe it's log 3 over log 2 but something like this So it's an irrational rotation related to answering. This is definitely true I started a couple of minutes late, so I'm going to take two more minutes. Sorry And then I promise I'll stop so I want to mention the ergodic theory partner. I'm really almost there Okay, so this is a uniquely ergodic transformation. It's an irrational rotation. It's uniquely ergodic So this is what really matters. It's uniquely ergodic so One of the most basic results in ergodic theory which is in every textbook is that If you have a uniquely ergodic transformation, the ergodic average is converged uniformly for continuous functions If you have never seen this it's a very good exercise So that you can do In fact, even if function is not continuous if it's continuous almost everywhere. It's enough ergodic average is converged everywhere uniformly If the system is uniquely ergodic But here I have a somebody to co-cycle So we have the sub-adity ergodic theorem. So we know that So we know that fn over n converges almost everywhere by the sub-adity ergodic theorem But the transformation is uniquely ergodic So what is the analog of this elementary result in ergodic theory that ergodic average is converged uniformly When you don't have an ergodic average, but you have a sub-adid to co-cycle Well, there is a one-sided version that that that is the same. It is not true that the It's not true that fn over n converges uniformly If the co-cycle is uniquely ergodic Let's say it converges to some limit l l for limit. But what is true Is that the link soup of fn over n is at most l everywhere And uniformly the uniformly part is not so important for me But so the link soup of fn over n is at most the typical value Everywhere This is the part that uses unique ergodicity So in this way this was stated by and proved by four man But actually there is a proof of the sub-adity ergodic theorem by canelson cut snelson and wise And if you look at their proof, this is there. It is you just have to read the proof It is there Okay, and this performs in the magic So fn it is this so if you look at fn over n What you get in the limit is the lq dimension And what this what this allows is We need to prove that dq of mu theta is one for every theta for every theta Using this it is enough to prove it for almost every theta Because if it's true for almost every theta then it's true for every theta Because the worst case is a typical case You could be worried that the inequalities are going in the wrong way around but there is a minus here. So it is fine so So this is very important because there is a part of the proof that only works for almost every theta It's not true or it's impossible to prove for every theta But in this way it is enough to prove that the typical behavior is a good behavior Okay, I think this is enough. Thank you very much Sorry Yes, so this this I guess I have called it mu theta Yeah, what is the question Sure, okay Okay, so we need to prove that this converges to one when nu is the projections Is any projection of the product measured there? Okay, so if you just if you just look at the numerator without the minus sign And here instead of nu you look at mu theta Well, you can think of this as a sequence of functions of theta Okay, these sequence of functions of theta are actually a co-cycle if you reparameterize theta in the correct way Because it is a co-cycle you can apply the subeditive ergodic theorem That tells you that this converges to some number l almost everywhere This tells you that there is some number l which you have no idea what it is to begin with Such that this so it tells you that So it tells you that this is l for almost every theta Okay, this is what you know from chances of identity of ergodic theorem So it's very far from what we need because we need to know that this l is one and we need to remove the almost everywhere What this does the fact that the system is uniquely ergodic so So you have a one sided inequality which holds everywhere It allows you to remove the almost every if you can prove that l is one But you still have to prove that l is one I haven't said anything about how to prove that l is one. So this takes 25 pages Well, I need to communicate So Basically this measure mu one is a convol has a convolutional structure because You can realize it as the distribution of a random sum where you pick left interval or right interval at every step And the same for mu two. So the same for a mu theta Because you are projecting under so it's a sum distribution of the sum is a convolution So mu theta is a convolution and then there are results in additive combinatorics that tell you that convolutions get smoother unless something happens And then we have to show that this sum this bad something doesn't happen using self similarity Yeah, so there is something called actually my notes for the talk. These are my notes for the talk So I haven't actually gotten to these So I wanted to if I had time I never have time to mention one very important fear I'm called the valox and meredic hours theorem Which is a key result in additive combinatorics and has many applications in particular in many areas in particular geometric group theory and robotic theory And also here So it's really somehow the the magic behind the proof is the valox and meredic hours theorem So sometimes it's all an elaborate application of the valox and meredic hours theorem