 I would like to apologize for my bad English accent because as many French people, I have a very good French accent, even when I speak English. So today I would like to talk about the watermark on the turrets. Here's my plan and I will start with a probability measure on SL to R. So these are the invertible matrices with coefficients in R and with determinant one. And I do a random work on R2 without zero, which is defined like this. I take a point in R2 that is non-zero and I just multiply it by my matrices and this defines a random work. And as the norm is sub-multiplicative, I expect the logarithm of the norm to be sub-additive and so if I normalize by N, by one over N, I expect it to converge to something. But to do so, I have to take care about some things. If I take matrices in that set, you see that what I have is a random work on R and not a random work on SL to R. So I want to prevent this thing from happening. I have to make assumptions on the support of my measure. So first assumption I will ask, I will say that a subgroup of SL to R is strongly irreducible if it doesn't fix any finite union of lines in R2. And I gave you two examples. In the first one, I fixed a line. In the second one, I fixed two lines. And obviously, if you conjugate these examples by any matrix in SL to R, you are still in this example. And you can do the same thing with the three lines and so on. And I will not want that for my measures. My second assumption is a non-compacity one. But actually just asking the subgroup to be non-compact is not enough. So what I will ask is this proximal assumption. And I say that a subgroup of SL to R is proximal if it has an element with an eigenvalue that is bigger than one. Actually, that is smaller also because my matrices are in SL2 so the products of the eigenvalues is equal to one. And it's more general than just asking the non-compacity. You just take these matrices and this group is not proximal and it's not compact. So my assumptions are on the support of the measure but more specifically on the group that it generates. So it is not an assumption about the cardinal of the support. As an example, these two matrices, A and B, generate strongly irreducible and proximal subgroup of SL to R. It is true because each of them has eigenvalues without whose modulus is not equal to one and their eigenvectors are not the same. So in the sequel, I will note J-row of the closure of the subgroup generated by the support of R. And it's still a group and it is strongly irreducible and proximal if and only if, so is the group generated by the support of R. And actually, I could also look at the semi-group or things like that. And the first theorem I have is a result by Fushtenberg and Kesten who say that if I take these assumptions on the support of R and I make a moment assumption, so I want the logarithm of the norm to be integrable and this is really a moment one assumption in the law of large numbers. As I said, the norm is sub-multiplicative so the logarithm of the norm is sub-additive. So I make this moment one assumption and the assumptions on the support of R and then I have the law of large numbers. So one of n times the logarithm of the norm of what I call previously xn conveys to some lambda-run and lambda-run is non-zero. In particular, if I look at the work on R2, it is transient. It means that if I take some point x here, I have g1x somewhere and the wall can come closer to zero, so this is g2g1x. But if I walk long enough, I will escape to infinity. And moreover, if we add moment assumptions on Rho, we have anything we want, the central limit theorem, the law of the Ijatil logarithm and large deviations in equalities. So now I will just look at what's on the torus. So I take matrices with coefficients in z and I ask that since the determinant is equal to one, the inverse also has coefficients in z. So this defines me a nice random goal on the torus. Just putting x to zero equal to x and to go from times n to times n plus one, I just multiply by one of my matrices. So if you look at this, you see the first thing that zero is fixed by this one of work. So if I start at zero, I don't move. And moreover, if I start at any rational point, I will have finite orbit. Since I will stay in, if I start at p over q, I will stay in one over q times zd over zd. And in particular, it's a general thing about random rules on a finite set that if I start at an irrational point, I will equidistribute in its orbit. So the question is, what happens if the starting point is non-rational? And this is a theorem by Bougain-Furman, Linen-Schultz and Moses, who says that always with my two irreducible, strongly irreducible and proximal assumptions, if I ask an exponential moment, then for any non-rational point x, the work equidistributes in the tourist. So for any function, for any continuous function, I have the law of large numbers. And the question that arrives next is what about the CLT and the law of the iterative logarithm? So I will not keep the suspense and it works with the same assumptions. For any gamma, we have the central limit theorem and the law of the iterative logarithm for gamma, a lot of continuous functions, but not for any starting point x. The problem is that if you start, we saw that rational points have a very particular behavior. So you can imagine that if you start close to a rational point, you will equidistribute, but very slowly. And this is my assumption. I have the distance between x and p over q to be bigger than the exponential. And this is exactly saying that I don't want to be too close to rationals. Yes, I just, I ask that this inequality has only finally many solutions. So basically what I'm asking is that the distance between x and the rational is bigger than the exponential. So in other terms, if x is well approximable by a rational p over q, then q has to be very large. This is the assumption. And then I have the central limit theorem and even the law of the iterative logarithm if the variance is non-zero. And in particular, this theorem tells that for almost any starting point, almost every starting point, we have the central limit theorem. But actually it was known before this theorem. This theorem just says that for specific points, this work. So how to prove it? I will not make the proof in, but just the idea is to return to Martin Gales and to use central limit theorems for Martin Gales. So if I have a function that writes f equal to g minus pg, where p is the Markov operator, this means that pg of x is the expectation of g of x1. And if I have a function that writes like this, g minus pg, then I have Martin Gales and I can have the, I can use the central limit theorem for Martin Gales with an increment if g is bounded. So the idea is to solve this equation, but it's called Godin's method. And, okay, so I will not explain to you how we do this. What happens if the variance vanishes? So what we can prove is that we have a formula for the variance. It's given by this, sigma squared of f, is just the integral that I brought on the board. And we can prove that if this variance is zero, then actually the sum of the f of xk is bounded in L infinity. This means that if the variance is zero, you cannot renormalize your sum. It will always converge to zero. Let's be more here in the central limit theorem. I say that one over square root of n is the correct normalization when the variance is not zero. And when the variance is zero, you don't have any good renormalization because your sum is bounded. And so to conclude, the sum works in d dimensions. And I can also add translations by irrational numbers to avoid finite orbits like the point zero if I add the translation. It's okay. And I still have large deviation principle and low of the iterative logarithm. Merci de votre attention. Okay, so thank you. Any questions? So when the variance is equal to zero, your sum remain bounded? Yes, no. I mean, do you have any convergence in law, for example, for this sum or? When the variance is equal to zero, the sum is equal to this g of x zero minus g of xn. So I don't think we can say many things. But actually this is true only if g is continuous. If g is not continuous, I'll just have a result that says that for almost any x, the sum here is bounded. Maybe I have another question. So you had a condition on your starting point, which say more or less that it's not too close to a rational point. And can you give an idea of what the set of points that this was set points that this describes? So it's a very large set of points that has measure one, as an example, actually it has also other dimension two, I guess, and it contains any algebraic number or any number with one algebraic coefficient. It's very, it's a big set, so in the measure sense it's nearly all the point. Yes, but the theorem with almost every point was already known just using more simple things. Just working in L2 to see. Thank you. There's no other question. Let's thank Jean-Baptiste again.