 So yesterday I talked mostly about quasi-periodic solutions, so as I promised in the first lecture today I will instead study the diffusive solutions, so the solutions which undergo subal- growth of subal of norms, except that I have to find my notes. So again, let me remind you of the problem, I'll just pick you on T2. And so the problem in itself was that I want to find a solution and the time, which will be a large time, such that the initial datum in HS norms, so here I'm taking S larger than 1 for my result, is very small, and instead the end datum at this time capital T is very large. And so my statement is that for all delta small and K large I can play this game, okay? So this is the question today. So the whole point of the first lecture was to say, as also Professor Posadet pointed out, that instead of studying this system you can study the Birkhoff system, okay? And so let me write the Birkhoff system. I had a compact way of writing it, which I prefer, so I will keep it. So I had the resonant Birkhoff Hamiltonian, which is the sum over the two P plus tuples, which are resonant of the monomial J, and of course I have to define all of this. So this vector J is J1 up to J2P plus 2, H is at 2, and the monomial is UJ1, U bar J2 and so on up to U bar J2P plus 2, as before. And the resonances, so this is a resonance if the sum of the indexes is zero, and the sum of the squares of the indexes is zero, okay? So being in a resonance is three relations, because these are two relations. This is a scalar relation on the two P plus tuples, okay? And then this is the resonant Hamiltonian, meaning that you can find a change of variables such that the NLS equation in these new variables is just the quadratic piece, which gives me the linear dynamics, plus the resonant Hamiltonian plus a term of degree for P plus 2, okay? So my argument, as it was for the quasi-periodic solutions, my argument will be that I want to prove solutions of this type for this Birko-Familtonian here. And since the resonant part is defined by the fact that it Poisson commutes with the term of degree 2, then I just can study H res. And then the point will be that since the solutions I am finding are finite time solutions, then provided I'm sufficiently close to zero, let's say in L1 norm, which is fine, small L1 norm, then I can claim that the solutions, that this piece here is negligible with respect to the solutions, okay? In the case of yesterday, it was more complicated because it was not a finite time solution, so I really had to do a KM theorem. But here the argument is relatively standard, and I think I will not do it, I will concentrate instead on proving the existence of this diffusive solution for the resonant term, okay? But let me remind you that this result for the case of the cubic NLS was proved by the I team, and then there is a result which gives also some estimates on the time of this capital T by Kalashen and Guardia, okay? And in fact, the time that we get, we also get a time estimate for any NLS, but the time estimate is extremely bad, is like some power K to delta to some other power here. This is a large number, so this is an ugly estimate, it's a very long time, okay? But it's very hard to get better estimates on the time, and it becomes also the argument to get better estimates on the time which Kalashen and Guardia did is pretty specific for the NLS, and we are not really able to just push it to our case. And in fact, they got polynomial estimates on the time also, but in that case they required that only the L2 norm of the initial solution was small, and then this simplifies somewhat the problem because you're less close to zero, obviously, in your, it's quite clear that here the problem of diffusion becomes harder and harder the more you are close to zero in your initial data. If you're far away from zero enough, then there were also previous results by Cookson that showed that you had diffusion, but for very large data. So this is a bit of the question, but and at this point I will for some time just go back to my, once I have settled this notation and said these things, I will go back to my notations of the resonances because of course the reason why I got interested in this problem is because you can really use all the machinery that we put up on resonance sets and the genericity conditions to this case, and then this helps you to deal with the non-cubic NLS where you do not have a geometric picture of the resonances. Let me remind you a last thing which I forgot, I hope it fits here, try to make it fit here. A trivial resonance is a list, okay, J such that the J, the odd indexes up to J2P plus 1 coincide with the even indexes, okay, and this occur for any dispersion though and it's all and I have to consider them as special. So I got lost, what was I saying? So how does this connect? Well it's pretty clear what I want so what did I prove last time, let me write it here. I fixed some set S which was composed of N points which I called B1VN, I mean Z2 obviously, and then I said that there exists a polynomial and V1VN and this polynomial was pretty explicit, it was something like this, what is this mean? Square then this is the sum, this is a finite product and these lambdas are in ZN, their sum is 1 and the sum of the moduli is smaller than 2P plus 1 and I'm missing something else, the support is not 1, okay, so I had an explicit polynomial and I wanted to write it down because I'm really going to use this and then if P computed on the VIs is different from zero then I had an invariant subset supported on S, it took me, so you remember, well maybe I'll write it down, this subset is invariant and also the second point which I have to write here and very important is that if you restrict to US then here I just have the trivial resonances, so this says that the dynamics on Hres is trivial and it's on to Rai and surely I cannot have any behavior of this type and also you have to remember this is a very strong condition so if you, it's a very strong constraint, so if you choose N points in Z2 in a random way then you surely will get them in such a way that this polynomial is not zero, I mean you almost surely will get them in such a way that this polynomial is not zero, so you can ask yourself, okay, can I instead work on the zero set of this polynomial and say something on when the VIs are on the zero set of this polynomial, for instance, could I decide that I want to fix one resonance relation, you see that polynomial there is written as product of irreducibles, which are the single terms, okay, and each single term is one resonance condition for me, okay, it says that the VI satisfy one particular resonance condition, so you can ask yourself, can I make it so if you want to look at the zero set of this polynomial in the space here, so P of V equal to zero, so the VI's I think of them, so let me try to explain, I can think of this as a polynomial inside of C2 to the N, okay, and then I have the zero set which is decomposed in the zero sets of the single resonances, okay, and so I can wonder, can I ask that I am on the zero set of one resonance and not on the zero set of all the other resonances, of course I can because they are separate, irreducible polynomials either they coincide or they are transverse, but unfortunately if I just ask this I'm not really happy because if I just ask this then I cannot impose that this space is invariant, in the cubic case remember that a single resonance relation is a right angle between three points and obviously if I impose that only one right angle holds then this point here cannot be an S and then you have a resonance made out of three points in S and one outside and yesterday I proved to you that if you have a situation of this type then the system US is not invariant, okay, so I really have to require VR and the best thing that I can hope is three independent relations once I have these three angles that are right angles and this is the right angle as well, okay, and this is perfectly reasonable because if you look at one single resonance, one single resonance is three relations so you should get at least a co-dimension three manifold in your condition, so at this point this becomes a reasonable question to ask, my reasonable question is can I choose S so that US is invariant? Let's say I have in the sense there say what I mean here is let's say I have that v1 squared minus v2 squared plus v4 squared v3 squared minus v4 squared is zero same for the vi's but no other resonance occurs, okay, so the only right angles that I can make with three points inside here are the ones drawn in that picture but this is obviously possible because let's see you have to so this is v1 v2 v3 v4 I have to add the point v5 in such a way that no right angles form here well clearly it's quite easy to do that and also if you think about it in the context of saying that you're working on the zero set of that manifold you can say that I have a co-dimension three manifold let me call it M saying that this resonance holds and when I add a new point this is a new variable which has nothing to do with v1 v2 v3 v4 and so obviously the condition that the three points form a right angle is cannot be on this manifold it has to be transferred to this manifold because I have an extra variable so there's nothing more to say okay well then I can become more ambitious and say let's suppose that I want to prescribe a certain number of prefixed resonances can I just choose any set of prefixed resonances and ask that these and only these occur well here I can start with the cubic case and I will show you that you need at least some condition it's not really true that you can do this because if I fix v1 v2 v3 and v4 which form a rectangle and then I need the colored shocks and then say I produce v5 and v6 which form a rectangle with v3 okay I have produced this new rectangle but then obviously I cannot ask that only these two rectangle resonances occurs because I have an extra rectangle joining v1 v2 v5 and v6 it's quite obvious right and in the same way I could say here if I add another couple of points on this circle then again I have more than two rectangles I automatically have three rectangles okay because some right angles are prescribed and this how how so this is perfectly clear in the case p equal 1 if you try to give reasonings in more general cases and it starts to you need to start to be able to do some bookkeeping but what is the point here the point is that clearly if you take two resonance relations so v1 squared I'll just write to the squared ones okay because v3 squared minus v4 squared is equal to zero and then I take v4 squared minus v3 squared plus v5 squared minus v6 squared equal to zero and if I sum them up I still get a resonance okay and naturally since here I have the same two points appearing here and here then I get a resonance which is a rectangle resonance but in general you can it's quite obvious that if you have two resonances so suppose you have that j so if I have a resonance j1 j2p plus 2 now I have to write it down in a better way in this way I never understand I have to give you some notation so in the resonances exactly as we did for the resonance polynomial instead of writing it with the j i's which might be repeating the same value of vi many times I want to write a resonance living on s as some mu i vi equal to zero and some mu i vi squared equal to zero with mu smaller than 2p plus 2 and the sum of the mu i's equal to zero okay this is just rewriting the resonance in a more nice way and then it's obvious that if mu 1 is a resonance and mu 2 is a resonance then mu 1 plus mu 2 is a resonance in the sense that it satisfies these two equations you just they're linear in the news so you can sum them up in any way what might happen is it might not be true that the modulus of mu 1 plus mu 2 is smaller than 2 p plus 2 and in this case this is still a resonance but it is not a resonance for the nls of degree p okay so go back to the cubic example I do not want this because I have one resonance here one resonance here and the sum of the two resonances is still a resonance for the cubic nls but if I took instead of something like this so let me see I have instead of this situation here I have to I erased this resonance here and then I add another resonance with just one point in common with the first resonance so v5 v6 v7 well then it's still true that the sum of two resonances is a resonance but it will be irrelevant for the cubic nls because if I write this v5 squared plus v6 squared minus v7 squared and I changed equal to zero you can sum them up and you will get a resonance but it's a resonance with the support six it's supported on six modes so it's not interesting for the cubic nls okay so once you have this can I now I have to can I cancel this here because so with all this said I can prove that I can produce a set s with only prescribed resonances occurring so I can choose s such that us is invariant only prescribed occur maybe this one plus some others and no other right angles occur well the answer is yes provided that right here so the answer to that question is here yes if two resonances have at most one point in common okay because if two resonances have at most one point in common when I sum them up they give me a resonance of degree at least six and in the cubic nls it's irrelevant okay so can you play the game the same kind of a game in the nls of higher degree than one well yes but you have to be careful because for instance we have just seen that you cannot take two resonances of a degree four because if you take two resonances of degree four and then you sum them up then you get resonance of degree eight so if you are in the nls of degree eight this is unavoidable this is still a resonance and you have to keep it so you have to be a little bit of careful of what you ask so but you could define maximal resonances so you could say a maximal resonance is a resonance in which exactly two p plus two distinct point appears so this is a maximal resonance for degree four because I have four distinct point I could give a maximal resonance of degree six with six distinct point and then I could ask if I could prescribe these resonances in the nls of degree correspondingly the same and that would be true but otherwise you have to be careful so in general what is the best that you can hope decide that you want to prescribe which means you want some resonances to hold so I prescribe some resonances mu one up to mu capital K I used K and then what do I have to do I have to compute the span so the linear vector field generated by the resonances over Q and intersect it with ZD because obviously I am not interested in a resonance which is not in integer valued but on the other hand I have to consider in principle combinations in Q because still resonance and it's not obvious that this is equal to the span of the mu i's over Z so just be careful okay and then the best I can ask is that only the resonances in this set hold I called the Dn is the number mu is in Zn I erased it mu is exactly in the same space as the number of tangential sites I'm sorry okay and you can require this so you can ask so what is the point you prescribe this certain number of frequencies of resonances maybe you ask I think I'm missing a hypothesis I was writing this down yesterday and I think I miss an hypothesis you need to require that these resonances have at most one point in common to two different resonances cannot have too many points in common okay and then you can prove that you can choose s so that these resonances are satisfied and none of the other resonances are satisfied and more over us is invariant but this is in general not a completely trivial proof okay in the qubit case it's quite obvious you just do it by induction on the points or how however you want but if you want to do this proof in general in any nls so you have to consider many more resonances and you want to try to prove this this is not completely trivial okay and in particular we were interested in the following sub problems we wanted to prescribe a certain number of frequencies of the form of the rectangles of resonances of rectangle form so we wanted to prescribe some resonances which represented rectangles and we wanted to show that we could choose s in such a way that these were the only resonances to hold plus their linear combinations and then still us was invariant and you can do it but it's maybe 10 pages proof so and also if you try to say you decide that you want to work on the quintic nls and you want to prove a statement of this form then it's reasonable the hard thing is always to give a general proof so you don't want to give a proof that holds for some fixed number of degrees you want to give a proof in parole degrees then you have to put up some structure and let me say that in the quintic case well actually in both cases this was proved by myself and Emanuele house and there is a nice explanation I think in the paper on the quintic case which shows why these things are true but you can find the full proof in the paper with Emanuele and Marcel Guardia gives the full proof of this fact so why do I want this well why do I want this is because if I prescribe a number of frequencies and okay accept the linear combinations then I can hope to be able to write explicitly or at least semi explicitly what the resonant nls Hamiltonian is when I have these resonances here and then if I have an explicit expression I can hope again to prove that I have growth of sublifernones for the resonant system by simply so the idea is produce this set where you have prescribed the resonances the set is invariant for the bulk of dynamics so restrict to this set you have a finite dimensional set this finite dimensional set is not integrable because in general because it contains resonances but maybe it's a little bit simpler than just studying the full resonant nls Hamiltonian and maybe there I can apply all the techniques of finite dimensional systems that will allow me to construct this diffusive solutions and that's exactly what one does so the I think it's instructive to show the I teams construction so what is your answer in the case p equal one because in fact what you will need in general p cases it's essentially the same except you have a little bit less control on the resonances so your set s which remember it has to be a set such that us is invariant is written as a union of generation sets okay the cardinality of si which they call n is related to the number of generations by this formula here okay so I have capital N sets each sets containing a lot of points each si is generic in the sense of generic sets which I gave you before so no no right angles okay so this means in particular that if you put all the Fourier support on a single si this is an invariant subset on which I know completely the dynamics okay then I need a further condition which becomes extremely important in the p larger than one case so I'm starting to write small so I have to require that if I take that uj is equal let me write it down for all j in si so let me call this the generation manifold so what I'm requiring is that I can take all the points in one single si and put them in phase put them all equal this is in C and this is in C but I'm asking that all the points are in phase in one single generation so I want that this is invariant and obviously for all i okay this means for instance that if you look at a single generic set the orbits on the single generic set are not quasi periodic orbits but you get periodic orbits because you have a reduction of the degrees of freedom but what I want is that this is invariant for the dynamics of the hs okay and then I have the property of being a generation set which says that I prescribe a certain number of resonances which was I was talking about before so for all j in si there exists a spouse in si and there exists children in the next generation and it has to be true that I have to write the correct order j c1 s and c2 for my rectangle j child j one child the spouse the second child for my rectangle like this and then I write it here for all j in si obviously here I have to require that i is smaller or equal than n minus one because I have capital N generations I do not have children for the last generation okay and then this is the same if I take that i is larger than one then for each j in the eighth generation I have a brother of j I am not becoming completely politically correct but I don't know word sibling but it's s so I needed and also I realized in that in our papers we all use the the Italian notation so we had a figure and so it was f so I have a brother and I have two parents this is in si and then I have two parents and again j one parent the brother and the second parent for my rectangle and the notation this is an ordered list the notation is exactly the same j one parents the brother another parent okay so you see I'm prescribing quite a few frequencies so you can think of this construction as being inductive so you start to fix the first generation I have to fix at least an even number of points I think it's even yes then I choose marriage bounds which means that I have to make these guys here produce children and the children have to be produced in such a way that J&S are opposites on the rectangle so here is a choice of children okay and then I have to take my children I have to make the marriage so I have to make a coupling between choose two points in such a way I did not write it but you have to remember if I want to be able to prescribe resonances I do not want that two points appear in two different resonances so I do not want so these points appear as a couple in one resonance so I cannot make brothers marry which sounds reasonable but apart from the obvious jokes it's it comes from the fact that if I don't ask that then surely I cannot prescribe resonances in the way that I want because I will get some extra points okay so I make my marriage bonds which are circles because remember parents have to be opposite and I draw the children and apart from the fact that the picture becomes very messy if I try to go further from the third generation it's quite clear that this is the mechanism that you have to take the only point that is missing here is the fact that I just drew the points random and instead I have to draw integer points but this is not really a problem because you have to remember that if you have resonance and you rescale it by any rational by any factor well rational integer the important thing is that you get an integer value vector then this is still still a resonance so if I have circles so if I want to have a sufficiently many point integer points on a circle I just have to rescale it or if you want first make this picture on Q and use the fact that Q is dense on these circles and then rescale okay so it's not really a problem the fact that you only have integer points but naturally all this is very nice and what was the thing that I liked a lot but if I do a picture like this unfortunately I do not have the growth of subaliphe norm condition so I have to be a little bit smarter if I want to produce the growth of subaliphe norm condition and so they impose the norm explosion condition which I will write down so why I want I hope I'm not getting the science wrong I think they put n minus 3 but I don't really understand why it's not n minus 1 j to the 2s 3 to the 2s and this has to be large so what I want is I want to choose my points in the generation set in such a way that the hs norm supported on the last generation set is much larger than the hs norm supported on the first generation set and why am I saying this is the hs norm supported on the last generation set because I have the generational equality so remember on the same generation everything is in phase so they're just fix and the only thing you really have to control is this okay well once you have this wonderful construction the point one point is that in the cubic case but I need so I'll write it here obviously I need another condition which I have to write the resonances are the prescribed ones and no others otherwise I mean I would have absolutely no reason of about of what I talked before for the first quarter of an hour so then the point is that I can write the Hamiltonian with all these restrictions I can write the resonant Hamiltonian explicitly and get it wrong because I obviously will it's here well this is the resonant Hamiltonian restricted to the invariant subspace so obviously you have to also divide it by the number of sites in one resonance in one generation otherwise it's not a symplectic change of variables but so restricted to UG it's bjs from 1 to n the square with a possible mistake in front of here but the important thing is that I have this plus 4 the real part bi squared bi plus 1 bar squared this is the Hamiltonian so it's very explicit and it's relatively nice also I would like you to notice that remember the NLS and the resonant Hamiltonian had the mass conservation so we'll have a constant of motion and this is a constant of motion for this dynamics so in fact this is completely irrelevant that's why I wrote it this way so once you get to this nice picture then you start studying this finite dimensional Hamiltonian and what do you get so you get so the first remark does not really depend on the Hamiltonian each generation is generic so for each generation if you put bi equal to 0 for all i different from some fixed point i0 say 1 let me avoid too much notations then you get one periodic orbit and it's unstable okay this is completely generous is just because yes i are a generic sense so it must be true also since the only resonances which occur are the ones that I prescribe if I decide to put the i equal to 0 for all i different from say 1 and 2 and this is invariant in fact I could choose 1 and 3 1 and whatever but the interesting is to put 2 consecutive generations different from 0 and so you have this picture so let's here I have various unstable periodic orbits going like this and unstable means that I know they have a local unstable and stable manifold all of them okay so here I have upcoming ingoing and I'm not going to draw more okay now if you restrict to this set so I am putting all these to 0 so these are not moving and these two exist then you see but this here you have to do by a direct computation on the Hamiltonian you shall see that in fact so in this subspace you have this outgoing and it just reaches the incoming of this one here and the outgoing of this one reaches the incoming of this one okay so they are connected by a heteroclinic in the two-dimensional subspace and this is true for any couple of dimensions because if you look at your equations your equations are completely translation invariance okay I do not see in which generation I am except for the first and the last where maybe where I see some little difference so the point is that once you have this nice picture here you want to do a shadowing them so you say let's suppose that I start very close to this outgoing manifold which means that I'm putting this periodic orbit has some energy on it all the others are very close to 0 and not only I have some energy but I know that I am very close to this manifold on which I have a complete control then well just because these are elliptic and they essentially don't move and also there is very little interaction between say this generation and this generation here so with all these control I can say that essentially the only thing that this initial datum sees is the local dynamics given by these two things here so I start here and I move and I get quite close to the second periodic orbit naturally there is some spreading you start very very close and then you move further away a little bit this is unavoidable but then if I am sufficiently near to this second periodic orbit I can hope that the only thing that this system sees is the linear dynamics close to this periodic orbit so it's a linear hyperbolic dynamics in just two dimensions so one degree of freedom and it's essentially true and so you can see that you start close to this periodic orbit you just go around the periodic orbit sometime and then you go out and you reach this point here this is essential standard linear theory it's quite simple well then keep going and start close enough and you get your picture okay so you can prove I have it written is some really terrible notations but so delta is e to the gamma and I'm sorry I just took them from the paper and didn't have time to think about too much where gamma is very large I guess there is a minus yes okay and then I have a time which is of the order n logarithm of 1 over delta so it's n squared basically n is my number of generations remember and then so I start delta close to be one it's probably not exactly delta it's delta to some little power but it's completely irrelevant and reach delta close to sorry bn b1 equal 1 bj equals 0 and reach delta close to bn equal 1 and all the other bi equals 0 okay so start delta close to the first periodic solution and delta close to the last periodic solution so once you have this construction if I want to have a solution of my resonance system which starts with very small hs norm then I also have to do some very big rescaling because now I use that I can rescale my equations and still get an equation okay and this is a trivial fact and at least I have proved the existence of a solution which does what I wanted for the resonant case okay and the time becomes much worse than this really much worse because I have to do a very large rescaling in order to ensure that the sub-left norm at the beginning is small okay now I have ten fifteen minutes yeah but so at this point so what so what is the basis of all these things so the basis of this condition I explained I think you want this condition because in this way you start here and this orbit has a much smaller sub-left norm than here which is the end orbit so this condition I have to keep for any degree and less I have to produce something like this what I could do is I could try to not ask that the resonances are rectangles so in principle I have many more resonances that I have that I could impose but so I could create a generation set ask that each generation set is each generation is generic which means that I have these unstable periodic orbits and then what is the conclusion of all the thing I can at least try to make this picture provided that I have the heteroclinic connections well the fact that this manifold is invariant is again pretty trivial because I'm just saying that no resonances occur except the ones that I wanted to and the resonance that I want to occur are just between the ith generation and the I plus one generation so this is pretty easy to require so what we thought at the beginning of this problem is okay let's concentrate on this question here can we produce two sets S1 and S2 in such a way that you have two periodic orbits and the periodic orbits are connected well it turned out that it was not so simple so we started well obviously you start simple so instead of trying to produce two sets with a large number of points in each set we started to say okay let's see what happens if I take let's say so I start really really simple I take one rectangle relation and I plug it in the NLS of degree six is it true that this has a tereclinic connections oh well yes it's true so we were pretty happy and essentially once we understood this we said okay then maybe we can do the paper on the NLS of degree five the only problem was that when you go and compute so we started with the generation sets and we arrived here but obviously as I told you you cannot just prescribe rectangle relations if I have two rectangle relations like this and I sum them up then I get a relation of degree six and the relation of degree six is not negligible in the NLS of degree five so what turned out is that we had a more complicated Hamiltonian here but in fact it was not really a problem and you can I mean it it takes a while to prove that you can do it in such a way that the resonances you choose are the only resonances that exist and no other exist but then you can run through this argument in a pretty similar way to the case of the cubic so this seemed nice but we asked ourselves okay but maybe we can try to put a resonance of order six inside just one single resonance made out of which is not a rectangle so the simplest example of a resonance which is not a rectangle is if you choose S to be this set here one minus one two minus two and then you take one minus minus one plus help one minus minus one minus two way I got the wrong sides well it's exactly the same minus two which means plus minus two minus two this is zero and if you play the same games with the squares this is zero so this is a resonance of order six the resonance is J is one minus one one minus one minus two two this I have wrote in Z but you can put it in Z two I explain this in the first lecture okay so what happens if I study the resonant Hamiltonian restricted to this set well it turns out that it is this resonant Hamiltonian is invariant so this set is invariant for the resonant Hamiltonian so I can really study it the Hamiltonian restricted to this set is integrable but non-trivial in the sense that it might have separatrices or heteroclinic connection joining the two periodic orbits which I know exist but the picture is like this so to make a picture I have so I'll make a picture on this cylinder even though it's really incorrect so I have my first periodic orbit I have my second periodic orbit and then instead of having a connection going from here to here I have so say that this is a space in my face space I have some other periodic orbits going like this then I have a separatrix going around this cylinder and I have a stable fixed point on the other side and then I have other periodic orbits going to here which means sorry there which means that in fact these are not unstable and I cannot go from this periodic orbit to this periodic orbit because I have a topological section thank you okay so it just doesn't work I mean I made the picture in a totally incorrect way if you take the subspace restricted to two generation in fact it's a three-sphere so I'm sorry for this picture I made some brutal symplectic reductions and just try to make a picture that was decent okay so well this is a fact and there's nothing we could do about it so we started to think okay so what happens if you plug so if this is a maybe a bad resonance because I'm taking four points and there are the points 1 and minus 1 which appear with multiplicity 2 in my resonance relation could I try to make resonance with six distinct point you do compute and I'm not going to even try to make the picture in these variables bit up because I'm absolutely unable to do it but you get rather complicated picture of the first phase portrait and still you have an obstacle you cannot go from one periodic orbit to the second periodic orbit it's explicit computations there's nothing that you can say about it so well then we said okay let's try to understand this problem and we made many simulations and the essentially nothing worked also we have this very depressing fact that if you take one single rectangle resonance so suppose that s is made of these four points here and plug it in the resonant dynamics of the nls of degree I think higher than 9 and 10 that's okay so just plug it in some nls of higher and higher degree and compute the dynamics on this set here instead of getting this wonderful picture where you have the two periodic orbits joined by a heteroclinic you start getting obstructions so it seemed that even if we tried to use rectangles which are the simplest possible resonances we could not go to a degree higher than 10 which seemed to be a catastrophe but what we hadn't realized is that our answers was not correct we are not really asking that the periodic orbit made that the two periodic orbit made of this picture here has a heteroclinic connections what I want is that I have a generation set with the lots of points and each couple of points has a rectangle so this is my first generation and this is my second generation first first first first second second second second and I go on okay what we need is that if we make sufficiently many rectangles then I have this picture here with the number of points in one generation being n very large so this turned out to be the correct approach to take because I'm sorry I because this is the Hamiltonian that you get so so I have my first generation made out of n points my second generation again made out of n points and I have rectangle for each couple of points and no other rectangle no other resonance relations for these points this is possible okay and then I take that that the intergenerational equality so I take that uj is equal to b1 if j is in s1 and uj is equal to b2 if j is in s2 and I show that this is an invariant subspace again this is quite simple and then I try to compute the Hamiltonian and you get this p plus 1 so I'm writing down some rather nasty numbers which are obviously completely irrelevant but since I had them I'll write them down this one so this is the first piece of the Hamiltonian it's a very high degree in n remember that n is a large number but this is a constant of motion because this is the l2 norm for the two-dimensional system and then numbers I hope it fits p minus one and then here I get the Hamiltonian of the cubic case b1 fourth minus b2 fourth plus four real part of b1 b2 this compare with the Hamiltonian that I wrote there with just two generations this piece here is exactly the nls is the cubic nls Hamiltonian restricted to two generations and then I have a correction but this correction is at least and p minus one of degree and p minus two and also I have some control on what it looks like but the important thing is that if you take n large enough the dominant term is this but this is irrelevant because it's a constant of motion so the dominant term is the cubic nls times a high power of a constant of motion well if you write if you draw the face portrait of this interval Hamiltonian you will see that if n is large enough you must necessarily get this picture here and then essentially I'm done because once I have this kind of game here so I prove to you because this is a very low-dimensional system that if this has a heteroclinic connection then this correction should be negligible and once I have this then I am essentially in business because at this point I just have to do my shadowing lemma and shadowing lemmas are much more robust things so I really what what kind of arguments did I use I use the fact that when I move here all the things in far away periodic orbits are essentially fixed and that they have an elliptic behavior in fact this turns out to be true also in the general case so the whole point was to understand this so naturally you can still be curious and ask yourself okay can I maybe take n copies of this relation here and try to see if a similar result works and the answer is it fails if you try to take n copies of a relationship like this or any relation of degree six and try to play the same game as a limiting factor you will get this piece here plus a small correction because this comes from order four as rectangle resonances and naturally this is an integrable Hamiltonian with the b1 and b2 mod square as constants of motion and so nothing moves you obviously cannot have heteroclinic connections and this is why you really have to use rectangles so at this point I'll just make some final comments so maybe I'll cancel ah but I have a whole last blackboard to use which I had completely forgotten about so points the first point is what happens when I change the dimension in this algorithm it seems in my whole construction I essentially only use combinatorics so I could increase the dimension without no particular problem but in fact this is not interesting because naturally if I increase the dimension I can just say I have a two-dimensional solution inside t3 so this is not particularly interesting to point the unfortunate thing which however I have to point out is that I am not able to decrease the dimension because you see well no you don't see because I put it here in principle I could produce resonances of order five even in z1 so even if I was working on say a quintic nls on a circle because this is a resonance so I do not only have trivial resonances but unfortunately I cannot get a toy model Hamiltonian unless I play with rectangles and clearly rectangles uh on they are degenerate in dimension one so unfortunately is no result for t this is a fact what we can do and what has been done in fact by is to show that there are solutions which exchange a little bit of subolive norms already on t but I cannot one cannot produce the cascade it's a very two-dimensional question also another unfortunate but true thing is that while while I was doing the periodic solutions the quasi-periodic solutions I did not really pointed out but most of the construction I did worked for any dispersive equation so I just need a little bit of dispersion to play with my resonance sets but then I could produce more or less similar theorems as the ones that I gave for the quasi-periodic solutions the hard part is maybe then to do the km theorem but and in fact there is a result by why I'm in one on the wave equation using some kind of really similar kind of fusions obviously the wave equation in dimension higher than one you need dispersion okay but here I would not know how to play this game if I changed the equation but I think it's an interesting problem so say can I do a similar game on the wave equation so I think people tried but I don't think there's been any success in this direction and the problem is that you really are using some explicit properties of the resonance sets of the NLS Hamiltonian so you cannot change what I think that could be interesting is to see if you can do this increasing the dimension because if you increase the dimension resonances should be simpler to understand so it seems to me that is a hopeful point of view so another thing that we're doing and then I close is I think quite interesting but we're not completely sure of success that we have the result is what if we take the cubic NLS and we try to understand the existence of solutions which drift not really close to zero but close to a one-dimensional solution so this is similar to what Zacharhani did for the plane wave solution and it's quite interesting to try to see if you can do it for more complicated but one-dimensional solutions and it seems it might work it's a reasonable question and the last question which I leave you with which I think is totally open but still I think very interesting is can you change the compact domain on which you're working so could I do it on S3 could I do this game on some compactly group and unfortunately the answer is I don't know but I think it's a really good question and there you have quite a lot of control on the resonance sets but you would need to get a little bit serious about the harmonic analysis and try to really understand what these sets looks like so thank you in the cubic case what why is it difficult to know what happens after time so why is it difficult well from here you get no knowledge because at that point P becomes too big okay could I make other techniques to try to know for longer times I really don't know what you could try to see is could I control longer and longer chains but I think that still you you get worse and worse results so the point is I am not really good at these things but the point is that you really cannot go to the limit so you cannot construct a solution which goes to infinity it's simply not within these computations you would be able to do it if you were able to take this picture here and the proof that a picture of this type exists for the true analysis you know that the periodic solutions persist for the true analysis but you don't know you know also these local things but you don't know that they match and then in that case you could do an infinite shadow otherwise so here I understand if you take a small n large enough you somehow reduce to so but why is the analog of the capital M you need them to the number of naturally this is just the two generation case then I have to do this with the capital n generations but the point is that once I am able to do this in fact in n generations I will just get exactly that Hamiltonian there plus a small correction except that it will be more and more generations the size of the s1 does not matter here in this case the size of the original thing does not matter you can take s1 or two people no each si has they have to be always the same number of because you see one of my points was that I needed the small n does not matter in this case the choice of the small n the choice of the small n if I want to do the norm explosion it has to be to and to know exactly at least that's the point is that here every time I tried to perturb this this scheme and the fact that you can construct the generation sets in such a way that to have this and and still maintain all the condition this is the I teams construction and every time I try to perturb it say add a couple of more points it seems that it does not help in any way and actually it hinders your proof so it's quite rigid yes it's quite rigid from the beginning exactly the same yes because you're really very close to zero and so you do not see any effect it's just that maybe it's okay yes because at some point you will have a minus but you see all I discussed was the homogeneous piece of degree 2p plus 2 so I lose the sign I have to be a little bit more careful in all my approximation arguments but since I'm very close to zero it works just that maybe the solution that I have produced has exploded for some reason before I don't know I think not right it's not possible in fact this was pointed to me by the referee in the paper and we corrected it in the paper because we hadn't thought about it but it's absolutely the same thank you