 So, welcome to this afternoon session. The first speaker will be Janick Seedham, formerly Marseille and now Baltimore. And he will speak about two things that have been developed in the last years. And they are interesting because the connection was not evident between them and with the old staff, which is fractional punkering inequalities and Q-curvatures. Ok, thanks. Do you hear me? No? Do you hear me? So what I would like to talk about, thanks to the organizers for the invitation. This is very nice to be here as the organizer is here. What I would like to talk about is as you know there are many cases where actually you can cook up black metric spaces or some manifolds for which you don't have punkering inequalities. I mean, it's an assumption you have to put on the manifold to have punkering, or on the space to have punkering inequalities, local ones. What I will present is actually a result where for some kind of manifolds where you are considering some type of curvature, there is a strong link between Q-curvature and actually the existence of standard punkering inequalities, local ones, and that you self-improve to non-local punkering inequalities to fractional ones. So let me come to that. Ok, so it's a joint work with my colleague at Hopkins, Yi Wang. So you consider a four manifold and like many years ago Branson found, I mean, many years ago Panitz on this type of manifold introduced this operator, so the bio-applation plus delta of two-third of scalar curvature times the metric. So R is scalar curvature, this is Ritchie, D is differential and delta is co-differential. Panitz introduced this operator because he was studying like some homotopy groups on the sphere, and actually a few years after Branson figured that this operator was linked with a notion of curvature, which is the Q-curvature and which is given by this expression. E is a traceless part of the Ritchie tensor. Ok, so R is the scalar curvature, so this is the trace of the Ritchie tensor. One over N, the trace, one over four, the trace of the Ritchie tensor. So basically Q is the idea of Q-curvature on four manifold, generalizes the notion of Gaussian curvature on two manifolds, on surfaces. So in a sense, if you change conformally the metric, like you consider GU to be exponential 2U times G0, then you change conformally the operator. And actually Q G0, which is Q-curvature in, yes, there is a Z, the co-differential and the differential. Two over three, you are a little bit on the pity side. Each square is a traceless part of the Ritchie tensor, and it's a three also here. I have the feeling to be in calculus class. Ok, so actually if you change conformally the metric, the operator changes conformally also. It's fine. And the Q-curvature changes conformally and satisfies PGU, so I never remember the formula, PG0 U plus 2QG0 equal to 2QGU exponential 4U. So this is the analogous transformation that we have for the Gaussian curvature. You know, when you change conformally the Gaussian curvature on a surface for the Gaussian curvature, you have like minus G0U plus K G0 where K is the Gaussian curvature equal to KGU exponential 2U. Ok, so this is the case of four manifolds. So this is clear, right? On four manifolds you have this operator which is conformally... Exponential U is... 2U is conformal factor, so U is... Yeah, U if you want is conformal factor, but I mean the conformal factor is... So this is clear, exactly. Well, exciting thing, this I don't know, but... Yes, there are simple pleasures. It's a simple pleasure to consider a four manifold and on these four manifolds you have like... Spanish found this operator by means by adopt calculations and you have this change of... This conformal change which is very important geometrically in conformal geometry and you have that on four manifolds you have like the analogous of the Gaussian curvature. I mean un analogous possible of the Gaussian curvature. You see that these two equations are extremely similar, right? So, yeah, this is a fourth order operator. This is a fourth order operator. The bilapresion in itself is not conformally covariant. You have to add the lower order terms, differential lower order terms which can be quite complicated to make it conformally covariant. Well, it happens that this theory can be generalized to higher dimension on manifold of even dimension. Except that in this case the equivalent of the panic operator are not known. The explicit expression are not known. So, for instance, in even dimension, I mean, you would have like equation on the flat metric of this type. Equation of this type, yeah, exponential two times exponential n u, ok? In the flat case, on even dimensions. It's been like, ok, so it seems to be like an ad hoc type of thing. I mean you just cook up some something that you call a curvature. And then why is it relevant geometrically? When it happens that you have the Cherne Gauss-Bonnet formula which tells you that the Euler characteristic of your four manifold is one over four pi squared is integral on the manifold of the veil tensor divided by eight. I will explain you what it is. The veil tensor divided by eight plus q curvature. You know very well what is the Gauss-Bonnet formula for surfaces. In the case of four manifolds you have this formula. So here you have an additional term compared to the Gauss-Bonnet formula which is a veil tensor which is a curvature tensor measuring how far you are from being locally conformally flat. If the veil tensor is zero, this means that you are locally conformally flat. So locally you are like a piece of a sphere. So this formula clearly carries like some geometric and topological information related to this q curvature. So this is a very important formula. And so the question is my goal is to get like on which manifold, which is the role of q curvature on conformally flat manifolds respect to the existence of a Poincare inequality. So what I will describe first is a case of isoperemetric inequality. And then I will move to Poincare inequality and fractional Poincare inequality. So let me state right away what we proved. And let me explain you the meaning. So what we proved is the following theorem. You take mg to be the following manifold. Rn endowed with the metric exponential 2u. So the conformally flat metric exponential 2u times the flat metric. So mg is this manifold. So let me come to that. So be a complete non-compact even dimensional manifold. Ok? Such that. Hypothesis u is a normal. So I will explain to you what this mean. It means that u of x, so my conformal factor or the log of my conformal factor is of the form 1 over cn, a constant depending only on the dimension, the sum in Rn of log of y divided by x minus y plus qg of y dvg y plus a constant for some constant. So we say that the metric is normal if you have this formula for the conformal factor. I will explain you where it comes from. The second assumption is that if I denote beta plus the total mass of the positive part of the q curvature of this manifold it has to be less than cn, the same constant here which is like an explicit expression which is 2 to the n minus 2, n minus 2 over 2 factorial pi of n over 2 and the negative part beta minus the total negative part has to be finite. Under these assumptions which are basically sharp you have the existence of a fractional Poincaré inequality so for any alpha there exists a constant c which depends only on beta plus, beta minus and the dimension such that for any function and any hidden ball you have this inequality. X is average, I will explain you what it is and omega of x is exponential n u of x and this is the average of f for any set e. Ok? For any alpha between 0 and 1. 0 and 2 sorry with my notation. So what does the CRM say? The CRM say that if you have a conformally flat manifold of this type such that the total curvature basically satisfies this assumption you have to be careful because you have a complete manifold, a complete non-compact so this is not clear to have like a Poincaré inequality in all the possible cases. So basically under the assumption that the total curvature so the sum, the integral of the positive part and the negative part satisfies this assumption plus the fact that the metric is normal then you have like a fractional Poincaré inequality actually we will prove that we have like a Poincaré inequality a p Poincaré inequality e then this p Poincaré inequality for p equal to 2 improves to a, self improves to a fractional one. Here you recognize the Gallardo norm, right? You recognize the Gallardo norm, ok? Ok, omega dx is a way it related to, this is basically to make it geometric if you have g x is of the form exponential to u dx squared dvg, ok? is exponential nu dx the volume element given by the conformal factor so actually here you measure actually the volume element of so not only you have a Poincaré inequality but you have a weighted one, ok? Oh well, alpha equals 2, you have blow up on the constant here Well, if you put it to rescale properly it is by the basis of Bourguin-Mironescu this quantity converges to the standard gradient Well, it's not in Bourguin-Mironescu but this is almost there, right? Ok, so this is basically what I want to prove so I will not prove it completely I will give you like an explanation because actually what I would like to discuss is the way you build functional inequalities on this type of manifold and in particular on complete manifolds let me say a few words about what happens for isoperimetric type inequalities and actually I need a lot of, ok? I need a lot of tools to prove this theorem but let me make a digression because this is important on isoperimetric inequalities on these manifolds because of course if you want I mean this is a simple somehow inequality you would like to prove ok, even if it's a hard one so actually there is a famous inequality due to Fiala and Uber well, Uber which proves the following if you have like a surface which is simply connected and complete I mean on a surface, right? then the volume of any bounded set omega is given by 1 over 2 times 2 pi minus the integral of the positive part of the Gaussian curvature dvg area, ok? so this is like this constant is optimal and it tells you that this type of quantity plays so total Gaussian curvature or at least it's positive part plays a very important role if you want to get isoperimetric inequalities ok? minus I'm not lucky today with my handwriting yeah, yeah, yeah so actually what my colleague did he won so she generalized this inequality in higher dimension consider changing this this total mass of the Gaussian curvature by total mass of q curvature so what she proved is that under like all the assumptions that I was given in my theorem, right? she proved that under the same assumption as the previous theorem she proved that you have indeed any bounded set with smooth boundary you have like an isoperimetric inequality ok? this is not a trivial result in the sense that this is very hard in very general case to obtain isoperimetric think about Cartan adama manifold it's not known the isoperimetric inequality for Cartan adama when all the sectional curvature are negative right? so this is not a priori obvious if you are given any manifold to derive as most as sharp as you can isoperimetric inequalities you know that it depends only on these two quantities so it's not known in general but you know that it depends only on this total mass of the q curvature and the dimension but this is not known I mean she doesn't have like a clean expression as this one for higher dimensional versions it's not really a compactness it's based I will come to that but this is not a compactness this is based on the David Sam theory of strong and A infinity weights that I will develop a little bit ok? so this is not a compactness argument but the constant is not known so this is not as good as this fiala uber inequality but at the same time this is like quite remarkable that you can have such a clean inequality in a manifold which can be like so let me make a remark why does it depend on beta minus because you have like in this case you have like I mean this is like you need to have like you need if you don't assume something on the negative part of the q you want a complete manifold so if you don't assume something on the negative part of the q curvature on the whole manifold then you might not have you might not have an isoperemetric inequality in the two-dimensional case it depends only on the positive part of the Gaussian curvature but in higher dimension you have to have the negative part because of different direction so let me make a remark about normality is that if the limit if when x goes to infinity of the scalar curvature is positive then u is normal ok so instead of assuming that this weird condition of normality that I wrote before you can assume like for instance at infinity your your scalar curvature is non-negative which is something giving you some hint about the behavior of course at infinity right of and the normality assumption is a necessary condition there are counter examples to the isoperemetric inequality if you are not normal so the normality assumption I mean what you have to understand is that you are on a complete manifold non-compact so if you want you have necessarily you might have hands and stuff like that you have necessarily to ask something at infinity to get something to get something I mean there is necessary a condition at infinity let me think to seconds the metric is not normal so I think that it doesn't how do you because I'm conformally flat so the high public space is conformally flat but the metric is not normal yeah ok so before going into the proof or how the strategy to prove this fractional let me do a little bit theory of weights because actually this is like this is like the important part somehow the analytical part of the so W is an AP weight is a soup on every ball of the average of W minus P prime over P times P prime is less than a constant for any ball on the manifold ok and P prime is a conjugate of P ok this is an AP weight so A1 weight A1 you are A1 if so you have like some sort of limiting case of that you are A1 if the average is less than the weight or times a constant this is the limiting case this is when you take P go into one in a ok so AP weights are extremely important in harmonic analysis because you can prove that a little wood maximal function from LP weighted into LP weighted if and only if the weight is AP so this is a necessary condition and sufficient condition to be AP to have boundedness on weighted spaces on weighty lobex spaces of the hardy little wood maximal functional so these are very important in functional analysis and they satisfy a lot of properties one of the most important properties that they satisfy is the reversal during quality so any AP weight satisfies a reversal during equality where R is larger than 1 ok in particular by getting its lemas they are a little bit better than LR LR a little bit better than L1 and here you have to take it in average sorry of course ok so you denote the class A infinity as a union when P goes to 1 of AP so you are A infinity if you are one of the AP for P larger than 1 actually you have another characterization of A infinity in terms of the lobex measure but I will not need it so what is important for us is the notion of strong A infinity weight so David and Sam's a few years ago introduced the notion of strong A infinity weight so an example so what is it a strong A infinity weight so you take omega a continuous function which is non negative and you introduce like some sort of distance quasi distance where B x y is a ball of diameter x minus y containing x and y so you can prove that this thing is a quasi distance and if you introduce some sort of intrinsic distance defined by you take any path gamma you take any path gamma in B x y connecting x to y and you introduce like the distance which is the infimum of all the gamas of the integral of a gamma of omega 1 over n dn ds then it's easy to see that omega a infinity implies that there exist a constant such that for all x and y this distance is bounded by this quasi distance and David and Sam's prove that A is strong in A infinity if the reverse holds so delta omega is bounded by c d omega so in other words that these two quasi distances are equivalent to strong A infinity ok so for example any A1 weight any A1 weight is strong A infinity for example mod x to the power alpha which is a canonical A to weight for alpha between minus n and n is strong A infinity for alpha less than 0 since in this case it's a 1 and this is still strong A infinity for alpha positive less than n ok but the weight x1 to the power alpha is not strong A infinity because basically you can choose a path along the x2 axis for instance and you will never have this equivalence between the distances right so this weight is not strong A infinity so the weight of the fractional laplation is not in the kafarelli silvestre extension is not strong A infinity ok this is bad so why do I say that because David and Sam's developed a lot of and try to have like weight inequalities using these weights and what Yi Wang proved is that if u is normal plus the assumption of the manifold that I was mentioning then omega of x equal exponential au nu is a strong infinity ok but the thing is that David and Sam's proved that strong A infinity satisfy sobolefin equality in terms of the weight omega of x where omega is a strong and infinity way but then you just take to get the isoparametric inequality you just need to take the characteristic function well you smooth it out the characteristic function of the set omega you plug it to the sobolefin equality and you get the isoparametric inequality ok like this is one way to prove isoparametric inequality ok so the strategy ah but well let me come to that I'm coming now so the problem is that now I want to get like Poincaré inequalities but what I know now is that I have much stronger information what I know now is that my weight ok I want to get weighted Poincaré inequalities so my weight I know now is strong A infinity by the theorem of Wang right she it's a she yes it's a lady she was a former student of Alice Chang and so now I know that my weight exponential a new is A infinity strong A infinity and actually you have a theorem due to David and sense who proved a point wise Poincaré inequality in the following sense if you are strong A infinity for any X and Y in Rn you have that this point wise Poincaré inequality there are too many use right let me call it V so this is a starting point of our theory if I have a strong infinity way David and sense prove that you have this point wise inequality so you relay it like this distance ok between f of X and f of Y to the gradient of f right this is some sort of generalization of the fundamental theorem of calculus right yes here you have all this weight times the gradient times another weight ok so this is some sort of generalization ok so the first thing that you prove so the strategy of proving for proving a fractional there are many ways to prove a fractional Poincaré inequality but one strategy that I developed a long time ago with Mu and Rus and with Rus is to actually you prove that you have for getting weighted Poincaré inequalities you prove that you have first some sort of weighted standard Poincaré inequality and then you improve it to a non-local one to fractional one by using like harmonic analysis methods so the strategy is first you prove so let me call it star star implies a Poincaré inequality a weighted Poincaré inequality ok for which you have to work a little bit and from the weighted Poincaré inequality by self-improvement you get a fractional Poincaré inequality or I will explain to you how to do the self-improvement this is a CRM this is a CRM of David and Sam's they prove that if you are the weight is one identically one well these these are measures of the border respect to this measure yes yes it looks like not a Poincaré inequality but yes so the strategy is that first this what we prove is that this implies a Poincaré inequality it is much stronger on my manifold with all my assumptions I have like a Poincaré inequality and from this Poincaré inequality I will self-improve it it is going to self-improve automatically to a fractional Poincaré inequality I will explain you why I have time so what we proved first which is not completely obvious but this is not extremely difficult is that you have a Poincaré inequality well I'm going to just state it for p equal to 2 well I state it in general so what we proved first but for a strong a infinity weight omega for any ball euclidean you have like a p Poincaré inequality so something which looks like more of a Poincaré inequality f minus its average to the power p omega of x dx is less than c omega of p over n times the integral over twice the ball of gradient f of u of x to the p omega of x 1 minus p over n dx ok so actually what you can prove is that starting from the David Samson inequality you prove this p Poincaré inequality ok ok so now you know that your manifold that I was mentioning at the beginning rn conformally flat with this exponential u where u is normal with finite total q curvature is supports this manifold supports a p Poincaré inequality weighted p Poincaré inequality of this type it's exactly what you prove ok so now the second step which is the main point is self improving to fractional ones so you take p equal to 2 ok what I'm going to describe I don't have a Wsp Poincaré inequality it's a what I'm going to describe now is a very inversion method it's something related to L2 this is very linear somehow I'm not able to do like p fractional Poincaré inequality so let me take p equal to 2 yeah yeah you can do that also, that's right self improve to give the sobolef but this is different in this situation for the weights for these weights I don't know I think that let me check something because David and Sam's proved already that the sobolef inequality holds and states you what it is well, you don't want no no no but I mean that David and Sam proved that if omega is a stronger infinity way to control a weighted LP star norm by a weighted gradient do the W1P dot norm there is a there is a this one a fractional Poincaré and the sobolef and from the sobolef you get isoperemétrica inequality from local sobolef yes yes yes I don't I don't know for measures how to derive directly a fractional Poincaré inequality for weighted ones you know well you show me afterwards then yeah yeah I don't know I just know when I but do you have an example of a space for which you have a fractional Poincaré without having a Poincaré yeah yeah yeah otherwise okay so the strategy is to improve to self improve so the thing is that you take F so let me let me come to so you take p equal to 2 so the self improvement of Poincaré towards fractional Poincaré so you consider like so here you another way to write this formula is you take like F such that the mean value of F respect to to d omega is 0 and then you have like F squared dvg because of what I wrote before is less than c times the volume of the ball respect to the metric g to the power 2 over n times 2b the gradient of g of f of x dvg now you define the multiplicative operator m of f to be the multiplication by the characteristic function of b and you define a measure d mu let me call it 2 like omega of x characteristic function of the twice of the ball times the Lebesgue measure so then you rewrite your Poincaré inequality as of two operators I mean you have to introduce a new operator which is L mu 2 of F which is basically the Laplace g of F so in other words this is like the sum from i equal to 1 to n of 1 over omega i omega y dy omega of x 1 minus 2 over n d i f so what does it tells you that in L2 with respect to the measure uh m of f is less than uh 1 half right of L mu 2 times the constant of f right I just use the fact that gradient g of f dvg square is like delta f of square sorry 1 half square dvg right which is a trivial case of the risk transform right so then by spectral theory I know that any power alpha between 1 and 2 between 0 and 2 as operators write this way so now I can if I write it at the level of norms it tells me that uh I want to prove that m alpha f times f in Rn respect to the d mu 2 is less than d times the norm of L alpha of 4 of mu 2 of f L2 of Rn ok I have just rewritten you see what I do I start from this inequality I write it as an operator an inequality on operators I raise it to the power alpha I write to do it by spectral theory I reinterpret it in 12 so now my goal is that this term is a good one this is a norm of f square in the ball of radius b in the ball of b so I have to bound this term by the Gallardo norm and if I do that I'm done the thing that you could think the thing is that there is a tricky part here but you think that you are not far from I mean you have to work but you are in good shape in the sense that this is not true that the Bessel space I mean this is not true let me explain you why this is not a trivial result this is not true that the set of function in L2 such that delta s over 2 of f is in L2 of omega equal the space of f in L2 of omega such that the Gallardo norm is finite this equality between space is not true for omega there is an inclusion but this equality is not true so this space this space is actually included into this one so you can bound so this is something that you can so here my space is more complicated because I have a measure a weight in measure but my bound this thing you can find in Triebel's book well if you dig out you manage to dig out Triebel but basically this is written somewhere so you cannot say that directly is that any power is going to be controlled by what I want in terms of Gallardo norm and if I am on the manifold I don't even know that it's been written somewhere in any case so the first lemma that you want to prove is a spectral lemma you're going to relate so it's a standard thing so I have no time so let me give you the first two lemmas the two crucial lemmas to prove the self-improvement the first crucial lemma is a spectral lemma the spectral lemma tells you the following it tells you that you can bound L alpha mu2 in L2 of Rn with mu2 by a constant times the sum from zero to infinity of T-1-alpha over 2 times the resolvent I mean a resolvent why do you have that you have that because you know very well I mean you have all these functional analytic formulas relating powers of operators of elliptic operators to the resolvent or to the heat kernel by functional calculus so the first thing that you prove is that you bound this term by this term now you need a bound on the resolvent and from the resolvent so the bound that you get on the resolvent is gafne type bounds which are very easy to obtain this is just an energy in equality so the second lemma that you need yeah I write it now lemma is the off diagonal gafne or gafne type estimates so you take two sets E and F on your manifold which are disjoint they are disjoint so what you can prove is that the resolvent in L2 of F plus T times plus the derivative of F is bounded by 8 well universal constant but in this case this is 8 times the universal constant square root of T F L2 into E so you see what I do here mo less right no you don't see you take two disjoint set of your manifold and this inequality this gaf off diagonal inequality tells you that you can control the resolvent in one of the set by the function so you have boundedness but you see that in the other set so this is off diagonal because you are not having the same set here and here they are disjoint and you pay the price of the distance divided by square root of T you pay the price exponential minus the distance no I mean you have this constant coming with the distance between the two ok this is not uniform in the distance if you want I will yeah this is positive but you have a precise bound so actually this type of thing is easy to understand if you call it that U ok identity plus I think it is in the formula this is for T positive yeah yeah yeah yeah L is positive so it comes with a minus here yeah yeah no I agree I agree with you so how do you prove that you take like you call it that U so you want to estimate on U so you have a Pd for U which is U plus T L mu 2 U equal to F and F is given ok and now you multiply that but you do an energy inequality U times Phi squared and Phi is rootably chosen and I don't remember what it is but it has to be cut off so it will be chosen and then you modify Phi I mean this is just an energy you have this inequality for free actually there is nothing hard behind ok so let me finish because I'm running out of time so basically once you have this two lemas you are reaching like what you want because you take your manifold you take a covering of your manifold by perwa's dijon cubes and starting from there you make a Pf minus its average ok and you estimate everything respect to that to that Gaffney estimate because you are on perwa's disjoint cubes with distance to the J something like that and at some point I mean you do the computation and at some point you have to estimate quantities of the type omega of the ball but because your omega is strong a infinity this is comparable to distance to the power n and from there you deduce your fractional perwa inequality so I think I'm going to stop now thank you very much