 just reminds you a bit what was the point in my first lectures and today I will actually give some proofs, not too many, so I hope not to get too technical but still show you a little bit the flavor of all these things but first I have to find my notes. So in the last lecture I talked and then I told you essentially what was the point that well if I start from the NLS equation and with a white chalk so I start from the NLS equation which I think of as a dynamical system and so I write the Hamiltonian in the Fourier variables which is square plus. There with me I will lose this notation very quickly, 2j, 2p plus 2 so you remember it was a non-linear Schrodinger equation of degree 2p plus 1. So the Hamiltonian has a non-linearity of degree 2p plus 2 and then I have the translation variance that is encoded in the Fourier variables by saying that the sums of these Fourier indexes should be zero when you take them with the correct signs. And let me remind you so this is the NLS on T2 by my choice so this J i the Fourier indexes live in Z2 okay and this is nothing but the NLS Hamiltonian. Okay p is an integer okay and then after some discussion we decided that we could pass through to we could do one step of normal form and so instead of make a simplistic change of variables that would send the Hamiltonian in this form here so it's the same piece of degree 2 then here I just have the resonances and here not only I have the conservation of momentum but I have this other condition minus 1 to the i J i mod square equal to zero which represents Poisson commutation with this term here okay and then I have a remainder which is small and remember I told you I'm not going to be very precise on the function spaces but if you want me to I can be and so I require that my sequence is in some hs which is larger than 1 okay and small in this domain. This is a reasonable hypothesis and so still to make everything completely to just remind you other things my point was to find special solutions and today I will concentrate on one of the two types of special solutions that I was proposing to study which is the quasi periodic solutions so these are solutions that are recursive and you start with an initial datum in hs and it's small in hs and this initial datum will stay small in hs for all times they're global solutions okay and so the point was that I can first preliminarily just study this piece here and this is the resonant it's a bulk of normal form Hamiltonian or if you want since this piece here and this piece here Poisson commute you can really just study this Hamiltonian which I will write in red because this is what I will study for the rest of my talk it's the resonant Hamiltonian which is the piece of a degree 2p plus 2 I said I would lose the notation but I will have to i equals 0 to the iji squared equals 0 okay this is the thing that I'm going to study and to make connections with the lessons with the talks of Professor Posadé this if you take p equal to 1 is exactly the Hamiltonian of the resonant system that he was talking about except that naturally since I am not working on t2 times r and working just on t2 I cannot say that I have a limit to solutions of the systems I will have to first study the solutions of the systems and then if I want to find from this system information on the true Hamiltonian I will have to give either have just a finite time result or I will have to do some serious perturbation theory and in fact once I have studied the solutions of these systems I will do a km theorem to prove that from certain types of solutions here I can deduce the existence of corresponding solutions for the full nls it will not but still this system has a very rich dynamics and it's a very interesting and no I think I'll take this down but in order to avoid having two horrible notations I need to give some definitions and here they are so I will a vector j will be a 2p plus 2pl I don't know if the word is English but j3 up to j 2p plus 2 okay and the ji's are in the 2 okay and then let me remind you that I will denote by mj the monomial written there so it's uj1 u bar j2 up to u bar j2p plus 2 so and then I need to define these conditions here so what is a resonance so I will say that j is a resonance and I should say is a resonance of degree to p plus 2 but I will systematically omit this if sum minus 1 to the i ji is equal to 0 and sum minus 1 to the i ji squared is equal to 0 and then I will say that j is a trivial resonance if I take the unordered list of the odd indexes so j1 j3 up to j2p plus 1 and this as an unordered sequence is equal to the even modes j2 j4 up to j2p plus 2 and we saw last time that these are special resonances they always occur no matter what the equation and the point is that if you just look at trivial resonances then the corresponding resonant Hamiltonian is integrable and it preserves the linear modes so all the interesting dynamics is given by the untrivial resonances so here I have a nicer form for my resonant Hamiltonian so I say it's the sum over the resonances of mj it's decidedly more compact notation I think I can send this up well even if I couldn't now it's done supposed to work like that can can somebody read what is written there okay it's good I I can't read nothing but I will point okay so what did I want to prove I want to prove the following fact so I wanted to prove so let me get you to the point where I was now I want to study that resonant Hamiltonian and I want to show that there are many choices of finite sets which I call s which are sets made of v1 vn and en with the vi are in z2 and I want to show that for many choices of these sets I have that these Fourier supports are invariant for the resonant dynamics which is written up in top there okay so there exists what was my statement there exists the polynomial of n complex variables even c2 such that if this polynomial computed at my tangential sites is different from zero then the following holds one so I'll write and then I'll explain such that uj is zero if j is in z2 minus s so this set in which all the Fourier modes are purely concentrated on the set s because I'm placing uj equal to zero if j does not belong to the set s and I'm calling the complementary s complementary just do you have a note it so this set is invariant and two if I compute the resonant Hamiltonian restricted to this set then I get just the sum over the trivial resonances okay so these are two statements the first statement is saying that for the resonant dynamics I have a finite dimensional invariant manifold and the second statement is telling me essentially that the dynamics on this manifold is integrable and simple okay so how do you prove something and remember last time I told you this was true for generic choices of s and now I can make this genericity more explicit because I can produce a polynomial with variables in c2 and these facts are true provided that this explicitly defined polynomial is different from zero so I just have to tell you what the polynomial is so how does one go on to perform a proof that finite Fourier support is an invariant manifold for Hamiltonian well this is completely trivial you just take your Fourier modes and you divide them in two sets one are the Fourier modes in s and the others are the full all the other ones okay and then I take my resonant Hamiltonian and I decompose it in increasing degrees in the normal variables okay so I will have a piece so this rest is made by a sum over the resonances and the number of j i's in s complementary is zero so all these numbers here are in s and this is mj then I have a term which is linear in the normal variables I'm calling these variables here the normal variables because naturally I'm looking for an invariant subspace supported here and all these directions are normal directions to my invariant subspace okay and so I keep going with my Taylor expansion these are the terms which have one mode in s complementary of degree one and so forth so here I have that the number of j i's can I just is larger equal than two so the number of j i's in s complementary is larger equal than two now and you can do this for any Hamiltonian in any setting is it's just the same kind of reasoning in the same way so what do I have to do I have to compute the Hamiltonian vector field of this Hamiltonian and I have to show that when I compute it on the normal variables equal to zero it has to be tangent to my invariant subspace so to the finite dimensional subspace so I have to make the derivative of h res with respect to the normal variables well naturally this does not depend on the normal variables so it gives zero this term here has degree at least two in the normal variables so when you do the derivative and compute at the normal variables equal to zero it gives zero so my whole point to prove that something is invariant is to show that the linear terms in the normal variables are zero because here it's everything is flat it's extremely simple computation but now here you see it's true that when I write these monomials sometimes there are repetitions so I'm writing the same thing many times but the coefficient is always one so the only way in which this sum can be zero is that this set here is empty okay so what do I have to prove so my point one is equivalent to showing that the j's in the resonance such that the number of ji in s complementary is one this is a set of two p plus two plus this should be empty okay and what about point two well once I have imposed this then the us is invariant and to know this dynamics here I just have to look at this piece here and so what I'm saying is that if I have resonance and all the ji's are let me use the same notation the number of ji's in s complementary is zero so everything is in s then j is a trivial resonance okay if it's a resonance everything is in s then it is trivial so at this point I'm done because to now I have to produce my polynomial and this is extremely simple so let me do it in the case p equal one which has very nice geometric meaning so in p equal one remember so I have that j is a quadruple j2 j3 j4 the resonance sets are made like this you have rectangles j1 j2 j3 j4 okay these are the resonance sets and the trivial resonances are these this is trivial resonances okay this would be j1 equal j2 j3 equal j4 or whatever and these are all together okay so what is my polynomial my polynomial is simple xn you just take the product over all triples i different from l different from k and these are numbers going from one to n and then I just take the scalar product xi minus xl xk minus x so I claim that this is the generality of the polynomial for the cubic case and this is trivial what does it mean that this polynomial is not zero this polynomial is not zero means that if you choose any three points they will not form a right angle okay so if I have say vi vl and vk place like this so this is a right angle then this polynomial is zero and by my ansatz generosity means that the polynomial is zero so I'm choosing my set in such a way that this is never true that they form a rectangle but then it's obvious that my first point is telling me that uh resonance that the resonances with just one point outside of s is empty because what is a resonance with just one point outside of s well it's made like this let's say but this is not possible because then the polynomial is zero or otherwise it could be a trivial resonance but if you see trivial resonant it means that all the points have multiplicity at less two and so I cannot have one point just one point outside of us so it's quite clear so you see it's nothing particularly strange it's just a very explicit polynomial and the first condition is satisfied so why is the second condition satisfied well this is again pretty simple because well this would be vi vl vk and then I'm saying that I have another point v whatever and they form a non-trivial resonance okay this means that there is a rectangle it's all ends here so I guess I must say that probably in this notation it's so evident that you don't really need to say things like this you can try to add points by induction so you might think I get a set s with the end points which satisfies all my conditions and I just want to add one point in such a way that the conditions of being invariant and being trivial are still satisfied so this is a very easy game to play and you do not need to produce polynomials but maybe this is a formal way of saying things and when things get harder it's useful okay so 20 minutes maybe I'll very quickly give you an idea of what happens in the general case in the general case you just have to be a little bit more formal obviously everything I have to prove that this set is empty okay what does this mean this means that I could suppose from my up there 2p plus 2p I can suppose that the first 2p plus one points are in s and the last one is possibly outside okay but this means that j1 j3 all these points there are in s so they are either v1 or v2 or v3 or one of the v's so I can rephrase the resonance condition instead of writing it as in the second to last line I can write it like this I have a sum over the v i's and i now is going from one to one but I don't know that the coefficient is one so I'm just calling this lambda i and this has to be equal to j2p plus 2 and then the same condition with the same coefficients is for the mod squares okay and this is just rephrasing of that where I'm taking into account multiplicity and you can notice that these lambda i's are a finite number because I know that the sum of the lambda i's must be one so lambda cannot be zero in particular and then the sum of the moduli of the lambda i must be smaller or equal than 2p plus 2p plus 1 okay so this is a finite set but then what does it mean that a resonance occurs well a resonance occurs with a certain lambda if this polynomial which I call p lambda of x1 xn is zero and the polynomial is just I take this I substitute here okay so it's just sum lambda i v i squared minus this is the mod lambda i v i squared okay but you can see that if lambda has does not have support one so if it has at least two different entries then this polynomial is not trivial is zero and I can certainly sorry I wrote v instead of x and so I can certainly ask that this polynomial is not zero computed at the v i's and then I avoid the resonance on the other hand if this lambda has support one so it's just one say v1 then I'm saying that j2p plus 2 is equal to v1 so everything is inside of s I mean you have to go to go I'm going to the proof a little bit fast but you can convince yourself that the polynomial that I'm looking for is the product of these polynomials here over all the possible lambdas and just take you five minutes to check the proof completely so again it's totally explicit and at this point I have my claim so one can go back a little bit and just try to see what does this mean I can erase here because what does it mean that I have an invariance of space for the resonant Hamiltonian and that the dynamic is trivial well the invariance of space the dynamics trivial means that I have produced a number of Fourier supports s on which all the dynamics is given on try because this dynamic here which I will study in a moment is a dynamics which lives only on try it's purely integrable and it's an interesting point which I would like to make but I cannot really pull the argument through to the real nls is that in fact I could play this game also with infinite sets of s in fact I mean what I would get think of the cubic case I would not get a polynomial because I get an infinite product but still if you try to see how many points are such that the us is invariant most of the infinite sequences work because the only thing that you need for an infinite sequence of points s so just place them I cannot place infinite ones but the only thing you need is that you never have right angles between any triple and it's irrelevant the fact that you have product a finite product of polynomial so a polynomial or an infinite set series of products it's exactly the same thing but the problem with this is that in fact I'm not able from these solutions which clearly exist to produce solutions of the true nls so I don't know what to do with this it's true but I don't know what to do with it so let us study the dynamics on the invariance of space so well just with a little bit of computations you see that you get this system here so you have the sum over the alphas in n to the number of tangential frequencies modulus of alpha is smaller or equal than p plus one sorry I have to first write and then explain okay this is the explicit expression of the resonance Hamiltonian computed on a set in which you only have trivial resonances you just have to do the combinatorics this is a multinomial coefficient okay and then well this means that essentially I know everything about this function because I can pass two action angle variables let me do it in a symplectic way these are the standard action angle variables associated to a complex structure and then this Hamiltonian for some reasons in our paper we called it a the only thing that it depends on is the actions because it's integrable and it depends only on the degree it does not have any relation on where s is supported so it's and it's just this expression here plus one alpha squared well I'm not claiming that this is simple it's not really true but you can certainly compute the dynamics of this system and the dynamics of this system is always i dot equal to zero for all i's and then I have that theta i dot is just a derivative of a p of i with respect to ii so it's a constant and this is typically called the ith frequency also if you take the simplest case which would be the cubic and ls then this is really a simple thing to write down so let me write it a1 is just the sum of ii squared plus four i smaller than k hoping that I didn't get anything wrong but it should be this okay so you just compute the Hamiltonian you can compute the frequencies and you know everything but the interesting is that the map that maps i into the set of frequencies omega i so this lives in rn and this lives in rn this map is generically edithiomorphism and well for a1 this is trivial it's a linear map you just compute the determinant it's obvious but in general you can see that the degree of a of p grows with p and so this becomes harder and harder to prove but you can find a proof in the paper with my father on cmp and you can find another proof with the slightly different notations but it works exactly the same in the paper by weiming bank because this and this well if one has experience with km theory this is a very important fact this is called the fact that your unperturbed system has a twist but the point is that in this way when you move the initial data you move the frequency and so if when you do your small divisor problems which you would have to do in order to do a km theorem if you get at some point that some frequency provides a divergence in your perturbative scheme then you just move the action a little bit and you change the frequency and this is what you're going to use it and in fact you can envision much weaker conditions than this that is very strong but here we have this and so you do not have to think more okay well at this point it seems to me that some questions arise naturally just without even looking at the nls hamiltonian i mean and in fact when i started to get interested in this problem and in particular when i got my father who is an algebraist interested in this problem i never talked about the nls hamiltonian to him i just posed a question of saying okay so by modulating the set s what is the best possible scenario that i can get in the sets of resonances so you see here the best possible scenario if you just look at the resonances restricted all to s or the resonances which have just one point in s is extremely simple this set is empty this set is trivial but let's say that you want to go further and you want to go and look at the sets where the number of ji in s complementary is two what can you say by moving the points vi so this is a question for which has nothing to do with the nls and it can be posed and that's what we asked ourselves so we actually never looked at the dynamics for a long while but only on the for the geometry but since here i really would like to justify why i'm doing things except from the fact that i like the problem what is the interest in studying the second piece so i have at this point proved that my hrs is given by the Hamiltonian which is integrable restricted to us then the linear piece is zero so i don't write it and then i have a quadratic piece i'm just avoiding number of ji's in s complementary this is and then i get all the others okay what is the meaning of this piece here well i have proved that i have an invariant torus so in which the uj's are zero full j in s complementary the ii's are constant and the theta i move as omega i of i times t plus a phase which is irrelevant okay so this piece here has the following meaning so this piece here gives me the linear stability of this solution which is a solution of the resonant Hamiltonian inside the resonant Hamiltonian so if i could say that this this is a quadratic form in the normal variables because it has degree two in the normal variables so if i could say that this is a stable quadratic Hamiltonian i would have linear stability of my solutions in in the resonant system if it's unstable and i can show you unstable directions then it's unstable the point is that when you compute this this is rather complicated because it depends on the angles it's infinite it's an infinite dimensional matrix and i have to find a way to describe it otherwise it's not clear to me what i can say about stability okay and the the simpler is the set of resonances on which i support my sum the simpler is this matrix so it stands to reason that if i try to simplify this set as much as possible i can gain information on the linear stability of the nls Hamiltonian and then let me go even a step further once i have this resonant Hamiltonian i can plug it in above there in my nls Hamiltonian written after one step of Birkhoff normal form so i could add my perturbation of degree four p plus two and then it's not a trivial argument it's in fact the basis of eliesin and coaxing paper on the nls on t2 but you can prove that if you have linear stability of this piece here so the resonant Hamiltonian then when you add the small perturbation p for people of degree four p plus two everything is still stable okay so i really can prove existence and linear stability for the solutions on top of the nl so for the true nls equation so it is a reasonable question to ask oneself it's not just trying to play with numbers but unfortunately so what would i like so the best possible thing that i would like but unfortunately it's not true is that if this set here looks exactly like the set with all the sites on us so the best thing that i could ask is that the only resonances here are trivial i can also ask that resonances are empty so let me i have j in rest the number of ji in s complementary is two so there are always trivial resonances like this you take v1 v1 repeat it as many times as you want and then you have j and j this is a trivial resonance this is a point in s complementary let's say it's useless that i repeat it twice so this is a trivial resonance and it's always there i cannot avoid anything like this okay so trivial resonances are there and there is nothing i can do except that well since you know you can do bookkeeping so i can write explicitly what is the expression of the Hamiltonian restricted to the trivial resonances it's just extremely boring so it's all the alpha in n to the n of alpha is equal to p and then i have something like this i to the alpha okay so the sum of the trivial resonances it's not particularly enlightening but it's explicit and then i am left with the non-trivial now the nice thing is that if this were zero i would be able to do km theory taking this Hamiltonian as an unperturbed system because it would fit exactly in the hypothesis of the km theorem by Cuxing and Eliason who discussed exactly the nls Hamiltonian but they added external parameters and then i would have external parameters which would be these actions here well they're internal but i have parameters and i can check that everything fits in their hypothesis the problem is that unfortunately it's completely true that it's completely false that this piece is empty and at this point i'm not going to make any pictures in the non-cubic case because they're quite hard to draw but in the cubic case this is completely obvious because i take vi i take vj and then i have to form a rectangle with these two points i have these two lines and all the integer points on these two lines form a rectangle so if you take here h you take here k this is the two lines perpendicular yes perpendicular to the line that joins the two tangential sides okay and then clearly this is a right angle and so this is a rectangle and this is unavoidable and if you think about it these two lines they have equations with integer coefficients so they always have infinitely many integer points so it's clearly evident that the set of non-trivial resonances is not empty the best thing that i can do is try to classify them i cannot hope to avoid them also i have another way of forming rectangles because i can form a rectangle by taking vi and vj as the diagonal i don't know the word as the line center of the circle okay so this is h and this is a right angle okay and well these are circles i have very few integer points on them but if you think about it if you want to have many points s then you cannot really avoid having some integer points here okay so we really do have this situation and this is the best you can do so you can ask yourself okay let's see how do i classify things like this well i can ask for instance that a point forms as a low number of resonances as possible so for instance if my set s is made just of these two points here then what is my situation i will have most points here you have z2 inside here so most points in z2 will not be in any resonance relation and then i will have all the integer points which are infinitely many but relatively few on these two lines or on these circles and they will have one resonance okay but you see that this very nice thing fails when you start adding points because okay let me add this useless that i give random numbers this is v1 this is v2 and so maybe i add two points v3 and v4 and maybe i even forget to draw the circles because maybe i can avoid intersections of circles but certainly two lines in r2 generically intersect and so i have this point here let me call this h and this point here belongs to two resonance sets okay and again i cannot even i can move my uh tangential side so my points vi as much as i want but i cannot really avoid this unless i put them all parallel but this is really looking for a one-dimensional solution inside a two-dimensional space so i really do not like this okay so generically these points will intersect so if you just think of this as a problem of intersection of lines well maybe the best thing that you can avoid you can do is you can try to avoid the triple intersection so maybe i would not like to add a point v5 in such a way that i have these two line and intersect here with also with v5 and also with the line coming from v2 this i might hope to avoid because a triple intersection in lines should be non-generic okay but still in fact you have to be careful because i cheated i used five points to make a triple intersection but in fact you have to draw two lines and a circle for every couple of tangential sides so in fact i could try to produce triple intersections which must less points and uh it's not a completely trivial question here it's obvious right i move v5 a little bit and i get out of the triple intersection so we had this example which is rather stupid but i like it because it shows that you should be really careful so here i want to avoid triple intersections between figures made like this okay i have my two points i move my two points and this moves these pictures here so suppose instead of having this picture that you have v1 v2 and instead of this figure i make a simpler figure i take the axis okay well can i ask that if i move the points and then i look at these axis lines i avoid triple intersections no because you have a triangle v1 v2 v3 and then if you take the axis the triple intersection is the center is the incentrum i i don't know the word in english let me make the picture better so you understand you take three points and for every couple of lines you take the axis obviously you cannot avoid the triple intersection okay so in fact these lines are very similar and uh and so you have to be careful because you see i perturbed the problem a little bit i look at the axis instead of this picture and if i try i'm risking of saying totally false things and in fact what you can prove is not that you cannot have triple intersections you have this statement here there exists a polynomial in x1 xn such that if you compute your polynomial at v and it's different from zero then i'll write it and then i'll explain so my triple intersections do occur actually there are multiple intersections but the intersection the multiple intersection points are the vi vj's themselves and there are no multiple intersections except from these special points you see because the incentrum game gets translated and it becomes that you have multiple intersections at these special points but i don't really care of multiple intersections in these special points because the only thing i want to know is the set of resonances where i have two points outside so i'm only really interested in intersections with an outside point okay so this is what you can prove in dimension two okay and what does this mean this means so how do you restate this you restate that that if j so for all points h in z2 minus s so s complementary there exists at most two resonances which are non-trivial obviously otherwise this becomes a stupid statement such that h is the last point of the resonance obviously i have to be careful of multiplicity because i i have lots of groups of permutations which keep the resonance in itself so i have to be a little bit careful of what i do and once you have this you can in fact prove km t but before i go into the km theory i would like to show you what you that you can in fact do but even better than this okay so this is a statement in the spirit of my first statement i produced a polynomial which made the first two terms of the degree expansion in the normal variable simple and then i produced a third another polynomial it's not the same polynomial i just wrote with the same letter that makes the term of degree two as simple as possible because for each normal site i have at most two resonances would contain it which means that i can describe the matrix pretty pretty in a new explicit way so with polynomials we are not able to do anything better but in fact the starting point for the results on nls on t2 was not via the principle of constructing a generosity polynomial but it was an idea by gang you and shoe which i find extremely nice and which gives you the really the best possible thing that you can hope so what didn't i use in this picture it's true that if you have two lines generically they intersect this is just the fact but here i never drew the fact that i'm only interested in integer points i don't really care if this intersection is not an integer because it will not appear in my resonance formula it's the resonance our integer vectors so what was the gang you and shoes ideas they tried to move the tangential sites and to see that you could produce tangential sites such that none of the intersection points are integer okay maybe with the circles you can hope to even avoid that the intersection points are real but with the lines intersection points occur and the best thing that you can do is say let's hope it's not an integer the problem of their proof so i don't think you can do it by a polynomial because it's a very different purpose kind of proposition right it's you're not saying that something this is a geometric statement that three points do not intersect while two points are integer one single point is integer this is an arithmetic statement but the point is that they really did it by hand so they took their set and they tried to prove by induction that if you have a set with endpoints which has your intersection property then you can add another point and it still has the intersection property but the proof was very heavy and it could not be pulled through to any degree analyst so it works only for p equal one and also it really works only in dimension one because uh so let let me try to give you an image of what this looks like in t3 you are in z3 so i have an extra direction and instead of having resonant lines i have resonant planes you can imagine that the planes are coming out and this is a section and instead of resonant circles i have resonance spheres but then generically two planes intersect in a line and three planes intersect in a point and so i cannot really hope to avoid any triple intersections i can avoid quadruple intersections and so it becomes more and more complicated you have to be careful but in fact if you just stick to dimension two this works extremely nice so i cannot say there exists a polynomial but i can say that there exists infinitely many s in whatever finite number of points you want such that there exists at most one such that h is equal to the last this is true for any p but just in dimension two and then gengyu and shu's argument was that from this scenario here you can do km theory you cannot really apply any known km theorem directly but you can still perform km theory what i would like to tell you is a slightly different presentation from theirs so i would like to tell you what in the notations that we developed for higher dimensions we could do with a scenario like this because i think it is interesting and so how you catch up to a km theorem once you have this because you see it might seem not so worthwhile to go through all this trouble just to have one resonance instead of two i mean what do you care well the things unfortunately change dramatically so and the point is this so suppose that you're in this setting here then what i'll write here okay and i think i have to take my notes okay so so now just this is a question of tastes but instead of writing my statement for the resonant Hamiltonian i will write it for the full nls Hamiltonian in the variable so i will add my perturbation this is because we're used to thinking about doing well i'm used to thinking about km schemes and so i want to see the perturbation really there i want to see the piece which is the h1 norm really there but this is just possibly a question of tastes so what i say that under the hypothesis s as before oops so the case in just one resonance then there exist symplectic variables and when i say they exist i mean that i know what they are i can write them down it's just that i want to avoid this spare you at least this such that in these variables here is my nls Hamiltonian i have the piece which is the h1 norm which is not touched so i just write it in my variables this is the piece on s and this is the piece outside so this would be my linear dynamics nothing i've said nothing then i have the correction coming from the trivial resonances all restricted to s so my integral piece and then here is what the quadratic form looks like after a simple change of variables under that hypothesis there i have things like this j plus i'll comment later so these are all accounted for these are the terms of degree at least three in the normal variables which i can ignore because they are small in my notation this is the perturbation coming from the Birkhoff normal form and the point is that i can produce symplectic variables in which the Hamiltonian has this form here so i have reduced i have completely diagonalized the normal variables and i have some new corrections of eigenvalues well this means that the system is stable so in fact i am cheating because i have to add that there exist symplectic variables and there exist values of i of the actions so i have symplectic variables and an open set of the actions i such that the nls Hamiltonian is stable at these i's and i have an explicit form for these coefficients here in fact i have the complementary result i also have regions of the actions in which this form is exactly the same but instead of having in standard canonical stable form i have standard canonical unstable form okay so this new sum also bears an s complement yes sorry yes and so well in fact so there are various comments in order this result we can prove but with lots and lots of effort for the cubic nls in any degree and we can prove for the non-cubic nls in dimension two even without requiring this very strong condition even by just requiring that you're not on the zeros of a polynomial what is the strong difference the strong difference is that here oops well i'll do what i can i know these functions okay these are the zeros of certain completely explicit quadratic polynomials okay so the mu i are solutions i have t square plus some known function of i depending on j times t plus some equal to zero and it's the zeros of these very explicit polynomials when i go to degree higher so if i go to a higher dimensional try so in dimension three or if i give up on this condition and just use polynomials i still can say that these correction functions i can compute them but unfortunately there are zeros of polynomials of higher degree so i i can say that they exist i can describe them but i cannot really write them down and in doing km theorems knowing the structure of these frequencies is extremely important and so really the fact that if i add these conditions i can compute this is not something that you should give up easily but unfortunately it works only in dimension okay so i have 15 minutes so i would like to write just for curiosity for you what are the kinds of condition that one needs in order to perform a km scheme so that you see why i'm saying i really want to know these conditions and then maybe give you a very quick sketch of the proof of the thing that i just cancelled so of the fact that in fact you can avoid that these intersection points are integer and it's in the cubic case it's rather easy proof in the non-cubic it's hard so i'm not even going to attempt so so most km schemes start with a Hamiltonian like this okay the km schemes rely on the fact that you have a number of parameters and the parameters here are the actions and then they rely on the fact that you can impose what are called the melnik of conditions i'm going to just write them for you but the melnik of conditions have in fact a meaning the meaning of the melnik of condition is that if you look at this Hamiltonian here so the piece that you consider your unperturbed Hamiltonian and you compute the adjoint action of this Hamiltonian over the space of Hamiltonians this will give you a matrix and the melnik of conditions are requiring that this matrix is regular semi-simple so it has distinct eigenvalues but typically you just write the melnik of conditions because they are very explicit and that's what I will do so let me remind you that omega of i is the derivative with respect to i i of the function a p of i okay this is just a notation and this is the frequency of my twisted torus so the melnik of conditions say that and then I need another definition sorry square plus so this is another vector so lambda lives in r n and what I want is the lambda scalar product with any integer vectors in z n plus h squared plus mu h of i this is a function of i plus or minus k square plus mu k of i is not identically zero and for what for all l in z n h and k in z 2 so this is infinitely many conditions let's just look at this this are infinitely many analytic functions of i okay in fact there are algebraic functions of i these I just computed out of a polynomial and the mu k's you just have to take me on trust that they are algebraic functions well there are zeros of polynomials in the i variables so it's quite true okay so what I want is that all these combinations do not give zero so this I am taking lambda of i dot l so this is a function of i then I have corrections between two of these and I want this never to be identically zero so if you think about it the simplest possible thing would be if these were zero if these are zero then everything just follows from the fact that this map here is a one to one map from action space to frequency space but even if these did not depend on the actions I would be perfectly happy the point is that you can see that these functions here could cancel with lambda of i so it might be possible that if you just these numbers h square k squared could cancel with these numbers here which are integers and then the functions of i they could cancel and this could be identically zero and you have to avoid this once you avoid this situation then you can plug in a km theorem and you have good hopes of pulling the km theorem through but if you do not have this condition then you have to go through some serious trouble in trying to do perturbation theory on Hamiltonian of this form okay don't you need any kind of estimate from below yes but in fact that that comes automatically from the analyticity and from the fact that this is a diffeomorphism because if l is big this is trivial it's you really play with this and you only need non identically zero but it's a point so well obviously if i can compute these functions i just plug everything in and i check if it's zero or not and in particular since this is a polynomial in the eyes and these are solutions of quadratic expressions if i get square roots and the square roots are non trivial it becomes very easy to show that this is not identically zero but if even if it's degree three as soon as i'm not able to compute i really have to get some nice ideas in order to prove that this cannot be zero even in a very stupid case say like l is equal to zero mod h squared is equal to mod k squared so these cancels and i have to show that i do not have a double occurrence mu h equal to mu k it's not easy to prove it's as it's quite hard okay so and that's why i'm saying i really like this setting i mean you can do it in the cubic NLS we were able to prove for any dimension knowing almost nothing about this bu p except that there were solutions of polynomials of degree the dimension we were able to prove these conditions but you really have to work hard instead here you compute and this is why i like it so how are we with time i really maybe okay those seven minutes are perfectly fine no i realized this was very technical so i just didn't want to keep killing you for the other 10 minutes but you gave me permission so how do you go about proving that an intersection point like this is not an integer well as i gave you in the example which i raised of the triple intersections things become easier if i have an intersection point coming from like v1 and v2 and v3 and v4 so these two points and these two points do not look do not have any intersection so it's harder if i have just say v1 v2 and v3 and then i have to make an intersection point it's harder to prove results so let's see if i can say it in a human way the v1 with these points here are your variables and you want to move them in order to avoid some bad behavior the more variables you have the more degrees of freedom you have to move and the simpler is the result to get the less the variables you have the harder okay so i'm just going to go on a hard case because if you have to give an example give an untriggered one so i want i have a point x which isn't an intersection so what does it mean that x is an intersection i will do this example here i had two examples written and so maybe the most instructive is the two lines example so what does it mean that x belongs to this line here it means that x minus v2 against v1 minus v2 this is a scalar product is zero okay what does it mean that x belongs to an extra another condition well x minus i have to choose what to write here if i put here v2 then this is trivial say i write something like this then the solution of this system is v2 itself so this is not an intersection point which is interesting i want to get intersection points outside of my set s if i put here v1 then i would be making the intersection of parallel lines and this is not interesting so this is an interesting case okay this is an intersection between two lines and i want to choose v1 v2 v3 so that x does not belong to z2 okay so let me start by making the ansatz that the v is are inside some box okay it's completely useless that i add other tangential sites i can think that there are three tangential sites and just do it for this okay but then if you do grammar and just use the fact that these are all integer relations you will see that x exists and it is inside some bowl of radius r as well because these are integer lines so they cannot be too parallel if the v is are smaller than r so what do i want to say i want to say that inside of this box here i can remove a few points and avoid that x is integer this is the game that i want to play okay so how do i do this well so this proof was written by my father and i claim no responsibility for the notation so he said but it's nice but maybe it's a little bit formal to for my taste so consider the map that associates to these points the quadruple v1 v2 v3 and x okay so this is a map free and what you want you want to take in general this lives in z2 cube times r two okay and he wants to avoid the preimage instead of z2 to the fourth okay inside here i have z2 to the fourth i want to go back and remove these points so i have to compute how many points are here with these conditions that are all integers okay so i first so let me see which one i have to do so first i fix v1 and v3 in any possible way okay so i get two r to the fourth because there are four parameters okay then i fix x in all the possible ways compatible with the fact that it has to be in this ball so it's integer and it has to be in the ball so i get two r squared points okay now i have left just one parameter and i have to choose v2 but if you look at this equation here v2 is on the circle of diameter x v1 so how many points there are on a circle of diameter of order r well v2 can be fixed in r to the delta ways with delta small okay so these are the points that i have to remove and how many are there they are r to the 6 plus delta how many points do i have in this ball here well wait i made a mess yes obviously this is false fix r you cannot fix r in any possible way i'm sorry here i have an equation for r so i fix one of the components of r and the other one is automatically determined i was starting to go too fast and then i lost myself obviously i have to use the equations otherwise i'm done yes it's you have to need two so fix x fix the first component and the second is given and so now i am in business because now this is r to the five plus delta and my box contains r to the six points so if i take r sufficiently large i certainly have lots of points that are good keep playing these games for all the possible combinations and you will get your result that you have no integer intersection points and you see it's quite easy but it's not a polynomial okay thank you can i just make a minor clarification in your in the statement on the second board there when you say us is invariant instead in very respect with respect to the nls flow no no for the h resonant yeah now that would be too too beautiful to be true no right so and i just like to ask you is that like so when you say there is a polynomial do you mean you have a constructive algorithm for constructing this polynomial or yes okay and maybe just last final question suppose instead that's instead of just a pure power non-narrative got two different non-narratives of different powers we are same uh that we tried to play that game well if if you use this strategy directly you will only see the lowest degree because you're just doing one step of bierch of normal form and then it's just the lowest degree that counts but you could try to plug in other pieces the only problem is that unfortunately we were not able to get any better results by doing it we would have liked to use so what do you mean by better results maybe have okay for instance in higher dimension know better the the corrections or maybe try to have more non-resonance conditions because of course you might like after i've proven linear stability i would might like to go further and so maybe possibly taking into account more resonant pieces would help me to go further in a bierch of normal form but unfortunately we are trying