 Thank you very much. I'm very happy to be here, so thank you the organizer for inviting me so let me see I hope in this first lecture I mostly plan to say rather easy things and just give an introduction So I'm sorry if I bore the people who already know everything of this, but let me start So I'm going to talk about the non-linear Schrodinger equations. Let me try to not get the signs wrong with any degree because almost everyone is familiar with this notation and The important thing that you have to consider is that I am working on a Taurus. So I'm taking the initial data in some Hs From T2 to C and I will try to give you some reasons of why I'm going to choose T2 This is clearly just a choice Okay, so there's lots of work on these things and I'm taking Sorry, it's larger than one pretty regular initial data. Okay, so there's lots of work known on well-posed net results and But I'm going to look at very specific questions. So I'm going to look at special solutions Here K is some parameter. Actually the only important thing here is design whether it's focusing or de-focusing and I'm going to stick to the de-focusing case but I mean everything but the sign can be scaled out about from away from this. I guess this is trivial And what I'm going to say in fact does not depend whether here I choose design plus or design minus But just to be specific. Okay, so when I say I'm going to look for special solutions means I'm going to take some very Precise problem and you know just say everything about nothing. So these Solutions are absolutely not typical which I'm going to discuss But I think still very interesting and the first thing is since I want to work in a perturbative setting Then I have to decide that my initial data at The hs norm is very small. Okay, this will be my starting point So I am in fact taking an initial datum close to zero But naturally this does not mean that when I evolve The solution will stay with a small hs norm We know that the h1 norms should stay small, but the hs You have bounds on how much it can grow, but This is essentially as much as I can say So for the quasi-period, so what are the special solutions that I want to produce? So I'll have quasi-periodic solutions Which are the ones I call the recurrent So these are global solutions and what is essentially an invariant torus inside the solution space So I want the frequency Which will live in some rn. This is with n frequencies and I want that the coefficients of this rn are rationally independent And then I want to map which From a torus tn into my solution space. So here goes a function capital u of x and phi And this capital u my torus embedding will live in hs of t2 times tn So you see I have an n-dimensional torus And then the relation is that my solution is simply capital u computed at x omega t Okay I'm writing too big because I wanted to put here also the Energy cascade solution. Let's see if they fit Which were the ones I called diffusive These are pretty famous from the paper of the i-team And the point is that I want to fix an x which is larger than 1 A delta which is small a k which is large And then I want a solution And I want that there's it exists a time such that The initial datum in hs norm is small and A kind capital t It is large The hs norm, sorry Okay well in fact These the fact that these kind of solutions exist for The nls at least in the cubic case it was not done by me So these first results were by gangu and shu for the cubic nls on t2 And this the first result was by the i-team, but then I will give some literature But so What did I want to say but the works I am referring to I will have to erase this pretty quickly Are on but if you have to want to consider any piece so not necessarily the cubic nls Then the result above was done by with so by myself With my father Claudio Crocesi This is the quasi periodic solutions And then the second result is a collaboration with Emanuele Haus and Marcel Guardia. I'm not used to writing on the blackboard anymore so Okay, and but these are always for the question of taking p larger than 1 otherwise it was already known since Of the three or four years So why are these solutions interesting? Well They're interesting because Well, first of all, they are interesting per se because they have a very different nature So I have solutions in which the hs norm stays essentially put at all time And these solutions the quasi periodic solutions there If you look at them It's very hard to distinguish them from a solution of the linear equation Except that the linear equation only has periodic solutions, but Apart from that they're very close to linear solutions. And the other hand these solutions there explicitly exhibit behavior that one suspects has to happen meaning that You start very small with the hs norm and you go But the reason why I like these solutions is because they are In some sense they illustrate very nicely the application of paradigm of applying The techniques of finite dimensional dynamical systems to infinite dimensional dynamical systems And they show very well the differences in the two approaches. So I rather like Essentially for this reason here So other things that I probably should justify Well, why do I work on why did I decide to work on Toray? Well, essentially because that's one is what is one is able to do in fact So you want a model of Compact manifold and you want to be pretty to know perfectly well the harmonic analysis Because all the results are very much at least how I construct them are very much based on Fourier series So in fact you could do a little bit better at least for the quasi periodic solutions You could work for instance on spherical varieties or zol manifolds or you could work on Lie groups but still it Quite complicated and the torus gives a very nice example computations are pretty explicit and it's So once this why t2 okay t2 is a choice or if you want to compromise if When you construct these solutions for quasi periodic solutions the higher the dimension the more complicated the result gets On the other hand for energy cascade solutions if you reduce the dimension things become harder because you have less resonances And in fact if you think p equal one and you take a one dimensional case Then this is an integrable system, which does not exhibit any growth of several of norms. So It's quite clear and also again T2 for this is not a choice It's an necessity because we are absolutely not able to do the result on t1 On the other hand we would be able because it's trivial to have results on higher dimensional try So t2 is a reasonable Compromise but I will try to give some other ideas But then finally well why the nls? Again because one is able to do it on the nls but There are other types of results And especially what I know better because I'm more expert in the field is quasi periodic solutions And there you can play games with other equations like the wave equation or But typically they have to be model equations And if you want to work on a two-dimensional terms you absolutely cannot accept derivatives in the nonlinearity It would be very nice to be able to work with derivatives in the nonlinearity, but as far as I know There is Nobody knows how to do it So once I have said all this let me try to give you the dynamical system point of view So I'm going to be in many ways deliberately naive So the point of view is try to think of this equation as Try to think of this equation as an equation on a sequence space And obviously if you want to think of the equation on the sequence space What you should do is just pass a few re-evaluations Okay And then I am looking as a the problem as a problem on the sequence uj And on the sequence uj. I have the equations I'm just Writing them down for completeness So I said I would choose k equal to to be p plus one and So I have to remember it And then here obviously it does not look particularly nice. So I have so I'll try to write it and then Okay, so I'm what I'm doing is very in a very nice way. I'm just Plugging inside the series P. Ah, sorry. Yes. I always forgot I really don't know what to do if p is not an integer. I would like to try to understand that but I have You're right Okay, so this is just a Fourier series of this expression here and apparently I've done And I've done essentially nothing. I'm just writing it in a more complicated way at this point But I like to stick to this Representation because here I see very nicely. Okay, you see it also in the other notation But what I like is that if I ignore that these are infinite sums and so completely ignore the functional setting of the problem What I have is I have a string of um Oscillators with the linear frequencies j squared and then I have a coupling And I can just try to put plug in all the things that come from finite dimensional dynamical systems on which I knew a little bit about inside this setting And the thing that you have to plug into is clearly the fact that if you if these were a finite dimensional system This would be a finite sum and so it would be clear that everything is well defined Whereas here even this is an infinite sum so even just to say that it's well defined you need to put some structure in it. Okay And then so other notations other things that you have to remember which are obviously Even simpler to write in the full notation But I want to write them here because I like it this way is that this is a Hamiltonian system This is the Hamiltonian minus 1 to di j i is equal to 0 u j 1 r j 2 This is rather boring to write down, but I think that if I start using two compact notations then It will be even heavier And then just To remember you have constants of motions Well, this is all the trivial things that you already know from the nls. There is no need to write it like this This is the mass so the l2 norm And this which is in fact Two-dimensional vector is the momentum which is saying that I have no explicit Spatial dependence here. So I have translation in variance. Okay well, I like to write it here and Uh When I write it in this way. So if I say that this is a Hamiltonian system I should tell you which is the Poisson structure And the Poisson structure is I hope I don't get the sign wrong, but It should be like this possibly minus here because I don't remember Naturally Another objection that you could make is that typically when you write in finite dimensional dynamical systems People usually use real variables and not complex notation. So what does this mean? What is this function of u u bar? What does it mean that this is the complex structure? Well, if you want and again modulus design Just pass your real notation By taking the real and imaginary part of u as a complex as Simplactic variables, and then you will see that this is just Uncoupled harmonica oscillators because this becomes q square plus p squared Which is the usual notation and everything becomes what one is used to Just a question of taste. I like this better because here everything is diagonal Just a question of taste So How do the black parts work so no everything Yes, but if I push this up, I'll never get it down, but probably I don't want to get it down again Yeah, I don't just it's fine Okay, so well Again, yes, I wrote it down because I thought Remember everything that I write needs to be specified. So for instance You could think that you're working. So as I decided that my initial data was in a small ball in hs So you can think that this is a symplectic space and and and the space space is hs Okay, so uj In hs and this is a reasonable Thing which makes everything well posed, but if you just ignore anything and so what happens in finite dimensions Well, so let me I had an order in which I wanted to say things. So I'm going to check it Yes, so ignore everything and what does this look like? I make a picture here. You have Your Fourier points. So the uj's And if I have no non-linearity nothing moves But if I have a coupling given by the non-linearity you have on all the On all points that are coupled by the first relation that I wrote which is the conservation of momentum You have transmission of energy and so you can spread. So at a linear level And then All these are constants And then the quasi periodic solutions means that I have an initial datum. So that would be nice Well, I just so you have an initial datum that is concentrated on some points wherever They are I knew you had everything So purple Okay, I'll take initial data Essentially concentrated on some points which I'm calling s And s is some subset Of z2 Complexly supported. So it's a finite set and the quasi periodic solutions are solutions which essentially stay put They stay on these sides at all times While the other solutions they start here and then They move they just spread to higher and higher modes And then if you spread the energy to higher and higher modes You will get a higher sub-olefner because increasing the Fourier modes increases the sub-olefner Okay, so this is my picture and again I am justified in doing this because if I think of finite dimensional systems, well This is essentially what you expect In fact, even in finite dimensional systems So one of the the main problem that I will have here is a lack of parameters to modulate Because even in finite dimensional systems, it is well known that these kind of problems Have small devices inside them. And so they are not just The trivial perturbation theory. It's true that You have a linear system plus something which close to zero is small So you can consider as perturbative, but you cannot so try to find these solutions by implicit function theory It just will not work And then what typically one does is you do not look At one equation you look at families of equations So you look at an Hamiltonian made like this where these are parameters And then you have some non-linearity. I'm just not going to write that because it becomes completely relevant Let me just write the degree of the non-linearity Okay, even in finite dimensional systems, you would like to have these parameters here And then what you expect is that you have maximal to write At least for most choices of the parameters, you will have maximal to write which are in living a positive measure set Then you will have a stable and unstable lower dimensional to write So to write which are not of the Lagrangian the smaller dimension So starting from periodic orbits and higher and higher dimension, and then the obviously These are the kind of solutions that I'm looking for in the quasi periodic case And then if you want to see the diffusive solutions Well, what you what you do in finite dimensional system is that you construct co-dimension one Unstable to write okay, and then you prove that if you have sufficiently many of these co-dimension one unstable to write They're stable and unstable manifolds must intersect And this intersection of stable and unstable manifolds gives you the Arnold diffusion So gives you the movement in the action space. Okay, so let me write stable unstable manifold And this will give you So what you will see is that you will start with an initial datum with With some action and then you will get after a long time Same Fourier mode with a very different action So difference of the actions So if you have this picture in mind So the first problem and this is a serious very serious one is that here we have No freedom in the parameters. So what you could do is say well instead of working with this non-linear shredding equation add a potential And try to find solutions for most choices of the potential this surely can be done and in fact There are results in this direction In both cases but Here we really did not want to do that. So we just wanted to see for that equation and that alone and also if you say Add a potential it would be nice to know. So say that the potential is cosine of x and prove that the solutions have the Your behavior and that again would be Not possible to do Once you have removed that problem in which I will discuss in detail after You get all the problems of putting this finite dimensional picture into an infinite dimensional picture So the first point is that we are absolutely not able to do maximal tori Also, I would guess that you do not expect that solutions behave like this I mean you do not expect to have positive measure sets, whatever it means where the solutions are infinite dimensional tori In any case one is essentially unable to provide To construct infinite dimensional tori of any form except by Really cheating a little bit. So adding lots of parameters Or maybe trying to put regularizing Modifying the non-linearity. There are results by Peugeot Burga. There is also Kierchia Perfetti on this But none is really applicable to this model even in one dimension Say you take a one-dimensional cubic and a less plus some higher degree You're not able to do that. So you just have to forget about this and as you forget about this You have to forget about also co-dimension one tori because evidently these are infinite dimensional tori as well They're completely beyond your reach. So what you should do is try to Have a finite dimensional structure embedded in your infinite dimensional setting The problem is that now these kind of things are much less stable So say that if you could prove that you have a finite dimensional invariant subspace Then there you you could work everything with no interest in the infinite dimensional setting, but you can't So this is a question And I think a good example even if I have to wave my hands to tell you this Is when you try to construct the energy cascade solutions because there the construction works like this You can prove that there exist sequences of unstable periodic orbits Okay, which would be very low dimension tori and they are unstable And naturally once you have proved that the periodic orbit is unstable You have a local stable and unstable manifold for these periodic orbits So if you were to be able to say that these low stable and unstable manifolds intersect Then you would have by very simple arguments An energy cascade solution and you could actually do as many as you want. So you could go to infinity, but unfortunately When one tries to give the intersection argument here You really use the co-dimension one because you use that co-dimension one Manifolds have to intersect And instead there you you have to use some much more volatile arguments And that's the reason why here I am not able to go to infinity I'm not able to construct nobody is actually able to construct a solution which goes to infinity Okay, this is because you're really doing computations with low dimensional things which are more fragile And this is I mean this is a fact you cannot modify this So I have said this I have said this At this point, so let's see maybe I can write here a little bit of literature Just to give you Something more of an idea of All the questions so in fact the the first thing that people did is to try to construct the low dimensional tori And they did it for model equations with Adding the parameters by adding a potential or even being even more brutal in their computations And the seal it was not a completely three minutes Not a very simple problem, but there are results essentially starting from the 90s And lots of people that coaxing First the two in combination Craig and Wayne Did a lot of stuff And so This is all if you take on the circle So this is a pretty Well known stuff and now the interesting points in quasi periodic solutions I mean the things that I find most interesting are trying to do the same kind of results At least in dimension one biadic derivatives So here I'd like to mention works by Who worked on fully non-linear equations and also for a fully non-linear nls There is myself with Roberto Feola Okay, so you consider that equation you take it on the circle and you put As many as two derivatives in the non-linearity and prove the existence of stable and unstable periodic objects When you go Uh to higher dimensions. Well, you just have to give up the derivatives, but Again results by Burga and the paper that Is more similar to the things that I am going to try to tell you the papers by Cookson and Eliason Which did that model there but added the potential in order to remove the And the generacy and well as I told you coming to exactly that equation The there's the the result and the cubic nls was gang You and shoe Very tough paper, but extremely nice if you look at want to look at it And then the results that I mentioned with my father and also There are works by long And she also studied the non-linear wave equation. I'd like to mention always without parameters on high dimensional to write and while this result here was very much Given on t2 and it works by their mechanism just on t2 You can do any dimensional to rise. So this result and the result I've write it like this with my father is In any dimension any nls any dimension the interesting thing which Maybe not today, but another another I will try to convince you is that if you take the approach From Our paper you can prove stability or at least linear stability of your solutions Which I feel is interesting because it's nice to know if you have a stable and unstable Solution and also it's nice to be able to work on the linearized equations and this you can do But it's a serious price you have to pay. It's like 70 pages proof even for the cubic nls to constructs to understand which solutions are stable and unstable but You can do it As for the so I can write them here for the energy cascade solutions well The first results were the i-team on the cubic nls and t2 and then kalashin and guardia always cubic nls with time estimates And then there's results. I'm talking about but also I should mention there are also results on um Other systems like for instance, there are the results by patrick gerard by gerard angrelier on the cego equation and the half wave and the the results which are more similar here on the nls equation on t2 times r which are Help honey's vet kof of ishilia and i'm forgetting What's that? Thank you. Everybody knows these results Okay, so at this point I have bored you enough with all the presentation and I can Start making some computations and some proof but No, I still wanted to say something before I I get to the real stuff I wanted to give you an idea Of What these my statements look like at the end when I'm really trying to prove them So I said everything is a question of modulating parameters And quite clearly the only parameters that I can modulate are the initial data So What is the kind of statement then still is in a very vague statement? So let me fix my set s Which I said was a finite subset in z2 And let me call s the points d1 Up to vn Okay, so I'm just calling these points vi and the quasi periodic solutions say for most s and obviously I have to say what this means there exists Many again, I have to say quasi periodic solutions And these quasi periodic solutions stay on s, which was I was telling you Informally when I made that picture there so essentially okay well Who essentially supported on s at all times What I mean is that uniformly in time you can Control With a very small parameter The difference between the whole solution and its projection on this finite subset s here, okay Everything is at all times very much on s in high subolefner. Let's say, okay So well the the two things that are extremely vague is this most and this many Well many is easy to say because Here essentially you can parameterize the quasi periodic solutions with the frequency So just let me define a small ball Inside the frequency space in which this is omega in rn Such that squared so I'm saying that the height frequency is very much close to the linear frequency okay And at this point I have a counter set Inside this little ball and this counter set has positive measure And for all frequencies in the counter set There exists a quasi periodic solution of frequency see And this parameterization at this point is bijective because different frequencies will give you different solutions so many in sense which Just this I have a I fix any n and then I fix the frequencies sufficiently close to the expected linear frequencies And I find a solution there What Is uh at this point not defined is what does it mean this most s? But this is a delicate question because I have to give you a definition on how to choose Points in a lattice. So what are good points and bad points? Certainly what is true is that if you give me a box inside z2 So you decide that you want to choose your frequencies inside the box of radius r Then the things that I have to remove are Well, essentially they are a co-dimension one algebraic manifold Okay, but there really you have to remove a very thin a very small number of Frequency omega in a box and anyways, I will make that clear, but I need to give all the definitions So this is what you should be watching for what is this most? Solute s means and then when I have my energy cascade solutions in fact You can play a very simple game if you give me any set s like In the quasi periodic solutions I can produce on this set s an unstable quasi periodic solution actually I will produce an unstable periodic solutions and then I can provide so given s I can give you another set which I can call s2 And I can construct initial data Such that u0 Is on s and at some large time t ut Is on this set s2 Okay, this is a rather simple thing to do you have Set like this you prove that you have on this set an unstable periodic solution And you prove that from this unstable periodic solution You can construct initial data close to this unstable periodic solution which drift The hard thing once you have this is that you want to play this game in such a way that This initial sub all of norm is small and this final sub all of norm is large And this is a much harder question, but just proving that you can drift So that that you can make a change in the actions Provide that this change is not too big. That's quite easy and I will I think I can show that directly So once I have this I will go I'll manage the time better than I thought So once you have all this in order to start doing things I have to take So if you have no external parameters in finite dimensional systems, but I would guess in any case What you try to do to your equation is to apply at least one step of Birkhoff normal form And see if in the Birkhoff coordinates you have a nicer picture And in fact you do So I am trying to decide I think I can erase these things and we have the Construction of the solution So what is the Birkhoff normal form the Birkhoff normal form is a change of variable defined in some ball And here since at this point I'm thinking of the quasi periodic solutions I will take a small ball inside hs And I have my change of variables Which maps this ball into some slightly larger ball And this is an impractic change of variables and when you compute your Hamiltonian in the new variables You will get a nicer picture in this sense Here I have a polynomial of degree 2p plus 2 But which is resonant And I will define what it means and here I have a remainder Which is of higher degree. So this is the degree of this The minimal degree of this as a as a polynomial, okay And here H resonant is just a projection of the non-linearity of the Hamiltonian On to the subspace of analytic functions, which Poisson commutes with the term of degree 2 So in particular what you will have that I will compute Is that the Poisson bracket between This is zero Okay, and this defines This polynomial uniquely I can I will compute it for you explicitly so Well at this point naturally if this were a finite dimensional system everything would be fine because I can say that this Is of degree 4p plus 2 so if you're sufficiently close to zero This is a small perturbation with respect to this system and I can try to look at this system And study its dynamics and deduce The dynamics of the nls over some larger time Interval and then if I want to construct the quasi periodic solutions I can do perturbation theory thinking of this System here as my unperturbed system and of h for p plus 2 as a perturbation Well, naturally in order to say this I have to convince you that this is really small and we are in infinite dimension But in fact, this is not a problem for the nls And you can prove that if you look at the the map From This will be epsilon 0 of hs. So I have you in here To the Hamiltonian vector field Of this small remainder This is an analytic map This I can think of it as inside hs Okay, and so this means that in fact this is really small because The back the Hamiltonian vector field just give is well posed And it gives me a very small correction with respect to the Hamiltonian vector field of this part of the dynamics here This is because I am choosing a rather simple equation. So the the non-linear Schrodinger equation If I had for instance that the linear frequencies these here were not integers So say you have a wave equation with some mass Or if you had derivatives in the non-linearity then the question would be obviously quite more delicate But here it's very simple. You can find it in essentially any book And it's completely trivial question and another remark is that in fact, I do not need to take a small ball in hs I could work with a small ball in l1 And then the change of variable would still have all the properties that I'm saying particular this Okay And Having an irrational Would mean that you have no it would be simpler because you have When you look at this commutation who the kernel is much smaller And then you have a much simpler Resident piece, but then well it depends on what you want to do because if you want to do this then It's much yes For the first one it would be significantly simpler. Yes Yes So, okay, so once we are here. I have to just What is the scheme? I I'll say it again because I just got lost and And What my scheme will be that I have to I want to compute this rational piece explicitly I want to say that so for the This result here, which is a finite time result And essentially I'm done because what I want to say is that I can prove The existence of the energy cascade solutions for this system here So we ignore the perturbation And then I want to show that this my energy cascade solution occurs in a time scale in which I can consider this as a perturbation And so just prove a continuation result If you want to do the quasi periodic solution, then this is more subtle because It's not finite time and at some point you will have to consider this as a perturbation But the point is that you will what you can show is that you know sufficiently well the structure of this System here and you really can do a km theorem Thinking as of this as an unperturbed system and just this as the perturbation Use for instance, oh I erased it for instance essentially use the km theorem by coaxing and aliasing And so it was more or less already given So the whole point in everything before taking into account this piece here Is to be able to study the dynamics given by what I would call the Birkoff Hamiltonian So as a first step explicitly compute this piece here and well this So I said I have to compute the projection of The term of degree 2p plus 2 onto the kernel of the adjoint action Of this operator here This is a linear operator acting on Hamiltonians and I want to compute the kernel So the nice thing about complex variables is that this operator is diagonal over monomials So if I just use Excuse me the atrocious notation, but I have to do it. So I'm denoting a monomial u j 1 u bar j 2 u j 2p plus 1 u bar j 2p plus 2 I did not need to okay and j is just a sequence 2p plus 2 plus j 1 j 2 or whatever And so I compute The Poisson bracket in fact of any linear expression of this type against the monomial Then what I get is this lambda j 1 There's an i and possibly there's a sign because I never get signs, right? j 2 plus lambda j 3 and then I go on What do I get I end with a minus lambda j 2p plus 2 times the monomial mj So you see mj is an eigenvector of this adjoint action and this is the eigenvector Okay, so when I have to compute the projection of that expression there and this is extremely simple And I will write my Birkoff Hamiltonian which I call It's Birkoff, but notice if you're used to this notation, it's just the Birkoff normal form after one step of You could do better, but in fact it does not seem to help in any way. So I just Stopped after one step of Birkoff normal form squared and here I have my monomials I'll just write them again j 2 etc to u j 2p plus 2 So this is even so it has a bar, but now I have extra commutation rules. So I have commutation with the momentum Which is the rule that I wrote you see minus 1 to the i j i equal to zero sum over i That's clearly conservation of the rule of Poisson commuting with m there Okay, so I have Zero and then I have my x rule So it's minus 1 to the i my lambda j is just mod j squared okay and now you see that What Is the point the point is quite clearly if this lambda j were rationally independent or say that No combination of 2p plus 2 lambda j can be zero then the only way to have This Poisson structure is that this sequence is trivial and I will define it but Imagine that you have j1 equal to j2 j3 equal to j4 and so on and obviously this term will be zero Okay, so let me give you My definitions because they will be helpful My first definition is a resonance Resonance is a sequence of vectors j1 j2 j2p plus 2 Wait, let me when I put the I don't know how to say the round bracket. I meaning that these are ordered I do not want to mix the components Because at least you see from this formula I cannot cannot exchange an even with an odd with an even number. It will not give me The same monomial. Okay, so a sequence like this such that minus 1 to the i i goes from 1 to 2p plus 2 ji is equal to 0 and the sum minus 1 to the i ji squared is equal to zero okay, so Here I could substitute the sum with the sum over the resonances and sometimes since here I have fixed the number of points I will call this resonance of Order 2p plus 2 so a resonance of order 4 is j1 j2 j3 j4 and so on. Okay And now a trivial resonance Well, a trivial resonant means that if I look at the odd indexes j2 p plus 1 This as an an unordered sequence is equal to the sequence of the events even Modes j2 j4 up to j2 p plus 2 Okay Why do I call this a trivial resonance? well because something of this form is always a resonance and and also Something of this for I mean it's a resonance not because of what this The linear frequencies are no matter what is the linear frequency So take any equation with any linear frequency Laplacian square Whatever you want and the trivial resonance will always appear in the Okay, and also it Essentially does not move anything because if you Have a trivial resonance then you can see that your monomial is in fact the function of the actions Okay, so if you have a trivial resonance uj1 u bar j2, etc u bar j 2p plus 2 I'm starting to hate this Then you can write this as uj1 mod square Uj3 mod square and so on up to uj2p plus 1 mod square Okay, so it's obvious that it for some commutes with this whatever is are the linear frequencies But on the other hand something of this type here does not move the actions in any way They're still constants of motion. Okay, so if you only have trivial resonances Then your Birkoff Hamiltonian here is an integrable system and it's integrable in the most trivial way All the linear actions are constants of motion, but the point is that in our case Of the linear Schrodinger equation on t2 and even on the circle if you take p larger than one You can produce non-trivial resonances in a very simple way, which I'm going to show So and which means that it's not true that the actions are constants of motion So let me the the very famous resonances are the resonances for the cubic nls So you can see that j1 j2 j3 j4 Form a non-trivial resonance If they form a rectangle j4 Just plug the rectangle relation and you will see that it is a resonance and also that if it is a resonance of order 4 Then it has to be a rectangle. It's just a simple computation Okay And then well just look at it. It's also why did I put the points like this? I have to remember that I have to be able to exchange j1 and j3 without moving any other points Because if you look at the monomial you can exchange any odd Coefficients one with the other and this is true. I can exchange j1 with j3 But if I want to exchange j1 with j2 I have to exchange j3 with j4 because it's a complex conjugate And then I have to use the fact that the Hamiltonian is real So this is a simple way of remembering how to put place the indexes on the rectangles. Okay, but then once I have a resonance of order 4 It's completely trivial to produce a resonance of higher order from a resonance of order 4 Because you just take j1 j2 j3 j4 then choose any Fourier index j it could be one of these it could be another and just repeat it Until you arrive To your j2 p plus 2 Okay, and this is clearly a resonance because when you substitute in your relations You have that all these cancel out Both from the linear relation here because you have a j and a minus j and from the quadratic relation Okay, so it's very simple to produce higher order resonances But in any case you you do not have only rectangular resonances. You can have other things So for instance, I I never get this so I have to copy it So I can produce a resonance of order 6 already in dimension 1 and this resonance is like this 1 minus 1 1 minus 1 minus 2 and 2 This is from the paper of greber and toman and uh You can check that this is in fact a resonance of order 6 By just computing. So let's try it. I have to do j1 Minus j2. So I get 2 plus j3 minus j4. I get 4 I'm getting lost minus 2 2 minus 2 0 Okay, and then you try it with the squares and it's the same thing just Don't have me. Okay. Well 1 to the square 1 minus here minus 1 is 0 and all these cancel because they're just one the square of the other And so this is obvious But naturally once again once I have a resonance of this type I can produce a resonance of higher order by again adding j j j as many times as I want And also I can make this this is a resonance in which the v Live inside z and not z2 But obviously I can plug this in z2 in many ways the simplest being that I just add 0 here Somewhat stupid way because this is in fact a one-dimensional solution But certainly it works and also I could write other numbers here For instance, I could just plug 1 minus 1 minus 1 minus 1 everywhere and sorry 1 1 all the same number Or plug exactly this thing here or any permutation whatever you can play the game in any way you want And you can also produce more complicated solutions and we had some fun producing resonances at some point But I'm probably not really that amusing but you can do Okay, so Surely you have lots of non trivial resonances Which means that this Hamiltonian does not preserve the linear actions But still I'm I'm not able to use the three blackboards efficiently. I'm I guess a two blackboard person So I'll write the statement of the lemma here and then I will write the proof so Notwithstanding the fact that this is not a system which preserves the linear actions I have this really nice proposition Not only it's nice, but it's also easy to prove Which is even nicer which says the following thing for And now you will finally see what I meant with the most the choices of s for generic S Which I remind you is a set v1 Up to vn, which I wisely just erased but okay For generic s the following holes. So one If you take This finite dimensional subspace so you have your sequence of uj is in hs and Uj is equal to zero for all j Which is in s So I think is in z2 minus s Which is a set which I will call s complementary Okay, so while here I am looking at the solutions which are completely supported on my finite set s So this is a finite dimensional subspace and the interesting thing is that this us Is invariant for Not for the nls dynamics that would be asking really too much, but it's invariant for the birkof Hamiltonian and the other thing is That again for generic choices of s if you compute the birkof Hamiltonian Restricted to us Then this is integrable and All the uj squared Are constant of motion and in fact I can be even more precise The birkof Hamiltonian Restricted to us is just j squared uj squared j in s Plus the sum over the trivial Resonances with j in s I I I did the hash with the notation Ah, yes It's true. You're right Just said I am always afraid that people will forget what this is and so I This point I've become used to writing it in fact What I the notation I like best and since I do have five minutes before proving this thing Is to write this Hamiltonian in this form here. So you have my h Birkof restricted to us And I like to write it like this. This is The piece which is obvious and here I Because what I don't like of this notation is that you have many monomials which are in fact the same monomials So you do not know I mean you have to compute what is the coefficient So if you really want to write the Taylor series Then you get this which I find nicer And I will explain obviously so alpha is So the alpha j's Are defined with j in z2 and this is a sequence indexed by z2. So alpha is in n times z2 Okay And then what this means? u squared to the alpha means the product j in z2 Of u j squared to the alpha j Okay, so this is in fact a monomial and this is a multinomial coefficient. So it's just p plus one factorial by the product and since the sum of the Alpha j is finite. This is a finite number and this is a monomial, okay So what I like in this is that here you really have the honest Taylor series of this expression I'm sorry. I wrote everything in z2, but obviously Sorry Everything is in s. I'm starting to be tired. This is a finite Taylor series. So this is Even better because I was thinking obviously of the representation of the trivial resonances in the full space And not in the finite dimensionals Okay Well, if you want to prove this the simplest thing I have like five minutes Yes, I will prove my statement for the cubic nls where it is trivial and I I actually didn't think I would make it to any proof. I thought I would just state it Oh, no, I'm sorry. I'd never told you what generic means So I have to say that before trying to give any proof. So what is a generic point? So if you have a point x in some space and I think of the complexes cr Then x is generic With respect to some polynomial A polynomial p of x Nothing Here I'm going to see If Sorry, let me write p of y if when you I have a polynomial p of x And then I have a point which I'm calling x0 and what does it mean that x0 is generic If when you take your polynomial and you compute it at x0 this polynomial is not zero. Okay, this is a Very standard definition of generic Obviously the only thing that you have to be careful is that this polynomial should not be identically zero otherwise anything does But this is in fact The heavy part of Everything okay So now Well, when I think of my set s I can surely embed it in some cr because s Is v1 up to vn And each of this is a two-dimensional vector. So s I can think of it as a point inside c2n It's true that these are integer numbers, but certainly they are inside c2n I could embed it inside r2n But I really like giving definition on complexes when I'm thinking about polynomials because then I do not have to Worry about real roots or things like that seems more reasonable and Well, this was done with my father and he doesn't even conceive that things should be real And then okay So what I have to do what I mean when I say generic I mean that I will produce a polynomial for you And this polynomial Will give me the gerenericity and if my point v1 vn is not a zero of this polynomial Then I have my result. So you see it really is most points And also the nice thing and then I stop because I do not have time is that if you take n to be equal to four Let's say and I want to give you know, let's let's do it even simpler n is equal to three I have V2 v3 Okay Then how do I ensure then here I can give you a completely Explicit because my polynomial is a polynomial of v1 v2 v3 And it's just v1 minus v2 against v1 minus sorry v3 minus v2 This is the scalar product And I'm starting to go too fast and I will start again with the proof of this But the point is that you see that four resonances are rectangles And it's quite simple to see that if you take three points And these three points do not form a right angle. So this would be j of v1 This would be v2 and this would be v3 And if they do not form a right angle, which Is exactly requiring that this polynomial is not zero Then you cannot form Resonances with these points Because even if you added a fourth point you could the fourth point you could never make a rectangle You could certainly make choose a fourth point So that this is verified because this is a Parallel outcome, I don't know the word But you cannot make also this because this is a rectangle and this would be the basis of my proof But if I add points and if I had the more complicated resonances Obviously, I will not have such an explicit expression for the polynomial But that's essentially all there is to this proposition here. Okay. Thank you Because I'm thinking of this as a point Okay, so you can always think of a subset of a finite subset as a point inside c1 So Yes, the set is generic if the corresponding point is generic and I have to tell you what the Any triple points which do not form Right angle because if you take these three points and you go and compute this polynomial here So you say this is certainly not generic No, this is generic generic is if the polynomial is not zero So if I take these three points They do not form a right angle and then this polynomial is not zero and then it is generic Okay The three points like this. Yeah, I I went Can I reformulate the proposition by the fact that there exists the polynomial to start from every s So that p of s is not zero then the following exactly there exists a polynomial. Yes But the polynomial will depend naturally on the degree of the ns. Yes, but yes, you're right. I should have written there The degree of the polynomial is always two As in this case. Well, okay. No, if you want just one polynomial the degree will depend on p If you otherwise, no, I was saying it in a wrong way So you can think of your polynomial as one polynomial of degree in the cubic nls. It's n choose three n is the number of point choose three because I have to take all the triples And if I had the higher order nls the degree of the polynomial would grow in fact If you take p going to infinity, you will have less and less generic points